Springer Monographs in Mathematics
Patrice Tauvel Rupert W. T. Yu
Lie Algebras and Algebraic Groups With 44 Figures
123
Patrice Tauvel Rupert W. T. Yu Département de Mathématiques Université de Poitiers Boulevard Marie et Pierre Curie, Téléport 2 – BP 30179 86962 Futuroscope Chasseneuil cedex, France e-mail:
[email protected] [email protected]
Library of Congress Control Number: 2005922400
Mathematics Subject Classification (2000): 17-01, 17-02, 17Bxx, 20Gxx ISSN 1439-7382 ISBN-10 3-540-24170-1 Springer Berlin Heidelberg New York ISBN-13 978-3-540-24170-6 Springer Berlin Heidelberg New York This work is subject to copyright. All rights are reserved, whether the whole or part of the material is concerned, specifically the rights of translation, reprinting, reuse of illustrations, recitation, broadcasting, reproduction on microfilm or in any other way, and storage in data banks. Duplication of this publication or parts thereof is permitted only under the provisions of the German Copyright Law of September 9, 1965, in its current version, and permission for use must always be obtained from Springer. Violations are liable for prosecution under the German Copyright Law. Springer is a part of Springer Science+Business Media springeronline.com © Springer-Verlag Berlin Heidelberg 2005 Printed in Germany The use of general descriptive names, registered names, trademarks, etc. in this publication does not imply, even in the absence of a specific statement, that such names are exempt from the relevant protective laws and regulations and therefore free for general use. Typesetting: by the authors and TechBooks using a Springer LATEX macro package Cover design: Erich Kirchner, Heidelberg Printed on acid-free paper
41/sz - 5 4 3 2 1 0
Preface
The theory of groups and Lie algebras is interesting for many reasons. In the mathematical viewpoint, it employs at the same time algebra, analysis and geometry. On the other hand, it intervenes in other areas of science, in particular in different branches of physics and chemistry. It is an active domain of current research. One of the difficulties that graduate students or mathematicians interested in the theory come across, is the fact that the theory has very much advanced, and consequently, they need to read a vast amount of books and articles before they could tackle interesting problems. One of the goals we wish to achieve with this book is to assemble in a single volume the basis of the algebraic aspects of the theory of groups and Lie algebras. More precisely, we have presented the foundation of the study of finite-dimensional Lie algebras over an algebraically closed field of characteristic zero. Here, the geometrical aspect is fundamental, and consequently, we need to use the notion of algebraic groups. One of the main differences between this book and many other books on the subject is that we give complete proofs for the relationships between algebraic groups and Lie algebras, instead of admitting them. We have also given the proofs of certain results on commutative algebra and algebraic geometry that we needed so as to make this book as selfcontained as possible. We believe that in this way, the book can be useful for both graduate students and mathematicians working in this area. Let us give a brief description of the material treated in this book. As we have stated earlier, our goal is to study Lie algebras over an algebraically closed field of characteristic zero. This allows us to avoid, in considering questions concerning algebraic geometry, the notion of separability, which simplifies considerably our presentation. In fact, under certain conditions of separability, the correspondence between Lie algebras and algebraic groups described in chapter 24 has a very nice generalization when the algebraically closed base field has prime characteristic.
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Chapters 1 to 9 treat basic results on topology, commutative algebra and sheaves of functions that are required in the rest of the book. In chapter 10, we recall some standard results on Jordan decompositions and the theory of abstract groups and group actions. Here, the base field is assumed to be algebraically closed in order to obtain a Jordan decomposition. Chapters 11 to 17 give an introduction to the theory of algebraic geometry which we shall encounter continually in the chapters which follow. We have selected only the notions that we require in this book. The reader should by no means consider these chapters as a thorough introduction to the theory of algebraic geometry. Chapters 18 and 36 are dedicated to root systems which are fundamental to the study of semisimple Lie algebras. We introduce Lie algebras in chapter 19. In this chapter, we prove important results on the structure of Lie algebras such as Engel’s theorem, Lie’s theorem and Cartan’s criterion on solvability. In chapter 20, we define the notions of semisimple and reductive Lie algebras. In addition to characterizing these Lie algebras, we discover in this chapter how the structure of semisimple Lie algebras can be related to root systems. The general theory of algebraic groups is studied in chapters 21 to 28. The relations between Lie algebras and algebraic groups, which are fundamental to us, are established in chapters 23 and 24. Chapter 29 presents applications of these relations to tackle the systematic study of Lie algebras. The reader will observe that the geometrical aspects have an important part in this study. In particular, the orbits of points under the action of an algebraic group plays a central role. Chapter 30 gives a short introduction of the theory of representations of semisimple Lie algebras which we need in order to prove Chevalley’s theorem on invariants in chapter 31. We define in chapter 32 S-triples which are essential to the study of semisimple Lie algebras. Another fundamental notion, treated in chapters 33 to 35, is the notion of nilpotent orbits in semisimple Lie algebras. We introduce symmetric Lie algebras in chapter 37, and semisimple symmetric Lie algebras in chapter 38. In these chapters, we give generalizations of certain results of chapters 32 to 35. In addition to presenting the essential classical results of the theory, some of the results we have included in the final chapters are recent, and some are yet to be published. At the end of each chapter, the reader may find a list of relevant references, and in some cases, remarks concerning the contents of the chapter. There are many approaches to reading this book. We need not read this book linearly. A reader familiar with the theory of commutative algebra may skip chapters 2 to 8, and consider these chapters for references only. Let us also point out that chapters 18, 19 and 20 constitute a short introduction to
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the theory of finite-dimensional Lie algebras and the structure of semisimple Lie algebras. We wish to thank our colleagues A. Bouaziz and H. Sabourin of the university of Poitiers with whom we had many useful discussion during the preparation of this book.
Poitiers, January 2005
Patrice Tauvel Rupert W.T. Yu
Contents
1
Results on topological spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.1 Irreducible sets and spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.2 Dimension . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.3 Noetherian spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.4 Constructible sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.5 Gluing topological spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1 1 4 5 6 8
2
Rings and modules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.1 Ideals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.2 Prime and maximal ideals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.3 Rings of fractions and localization . . . . . . . . . . . . . . . . . . . . . . . . . 2.4 Localizations of modules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.5 Radical of an ideal . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.6 Local rings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.7 Noetherian rings and modules . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.8 Derivations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.9 Module of differentials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
11 11 12 13 17 18 19 21 24 25
3
Integral extensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.1 Integral dependence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.2 Integrally closed domains . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.3 Extensions of prime ideals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
31 31 33 35
4
Factorial rings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.1 Generalities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.2 Unique factorization . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.3 Principal ideal domains and Euclidean domains . . . . . . . . . . . . . 4.4 Polynomials and factorial rings . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.5 Symmetric polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.6 Resultant and discriminant . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
39 39 41 43 45 48 50
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Field extensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.1 Extensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.2 Algebraic and transcendental elements . . . . . . . . . . . . . . . . . . . . . 5.3 Algebraic extensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.4 Transcendence basis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.5 Norm and trace . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.6 Theorem of the primitive element . . . . . . . . . . . . . . . . . . . . . . . . . 5.7 Going Down Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.8 Fields and derivations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.9 Conductor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
55 55 56 56 58 60 62 64 67 70
6
Finitely generated algebras . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.1 Dimension . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.2 Noether’s Normalization Theorem . . . . . . . . . . . . . . . . . . . . . . . . . 6.3 Krull’s Principal Ideal Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.4 Maximal ideals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.5 Zariski topology . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
75 75 76 81 82 84
7
Gradings and filtrations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.1 Graded rings and graded modules . . . . . . . . . . . . . . . . . . . . . . . . . 7.2 Graded submodules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.3 Applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.4 Filtrations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.5 Grading associated to a filtration . . . . . . . . . . . . . . . . . . . . . . . . .
87 87 88 90 91 92
8
Inductive limits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 95 8.1 Generalities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 95 8.2 Inductive systems of maps . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 96 8.3 Inductive systems of magmas, groups and rings . . . . . . . . . . . . . 98 8.4 An example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 100 8.5 Inductive systems of algebras . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 100
9
Sheaves of functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 103 9.1 Sheaves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 103 9.2 Morphisms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 104 9.3 Sheaf associated to a presheaf . . . . . . . . . . . . . . . . . . . . . . . . . . . . 106 9.4 Gluing . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 109 9.5 Ringed space . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 110
10 Jordan decomposition and some basic results on groups . . . 113 10.1 Jordan decomposition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 113 10.2 Generalities on groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 117 10.3 Commutators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 118 10.4 Solvable groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 120 10.5 Nilpotent groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 121
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10.6 Group actions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 122 10.7 Generalities on representations . . . . . . . . . . . . . . . . . . . . . . . . . . . 123 10.8 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 126 11 Algebraic sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 131 11.1 Affine algebraic sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 131 11.2 Zariski topology . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 132 11.3 Regular functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 133 11.4 Morphisms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 134 11.5 Examples of morphisms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 136 11.6 Abstract algebraic sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 138 11.7 Principal open subsets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 140 11.8 Products of algebraic sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 142 12 Prevarieties and varieties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 147 12.1 Structure sheaf . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 147 12.2 Algebraic prevarieties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 149 12.3 Morphisms of prevarieties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 151 12.4 Products of prevarieties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 152 12.5 Algebraic varieties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 155 12.6 Gluing . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 158 12.7 Rational functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 159 12.8 Local rings of a variety . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 162 13 Projective varieties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 167 13.1 Projective spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 167 13.2 Projective spaces and varieties . . . . . . . . . . . . . . . . . . . . . . . . . . . . 168 13.3 Cones and projective varieties . . . . . . . . . . . . . . . . . . . . . . . . . . . . 171 13.4 Complete varieties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 176 13.5 Products . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 178 13.6 Grassmannian variety . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 180 14 Dimension . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 183 14.1 Dimension of varieties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 183 14.2 Dimension and the number of equations . . . . . . . . . . . . . . . . . . . 185 14.3 System of parameters . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 187 14.4 Counterexamples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 190 15 Morphisms and dimension . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 191 15.1 Criterion of affineness . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 191 15.2 Affine morphisms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 193 15.3 Finite morphisms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 194 15.4 Factorization and applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . 197 15.5 Dimension of fibres of a morphism . . . . . . . . . . . . . . . . . . . . . . . . 199 15.6 An example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 203
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16 Tangent spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 205 16.1 A first approach . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 205 16.2 Zariski tangent space . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 207 16.3 Differential of a morphism . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 209 16.4 Some lemmas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 213 16.5 Smooth points . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 215 17 Normal varieties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 219 17.1 Normal varieties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 219 17.2 Normalization . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 221 17.3 Products of normal varieties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 223 17.4 Properties of normal varieties . . . . . . . . . . . . . . . . . . . . . . . . . . . . 225 18 Root systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 233 18.1 Reflections . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 233 18.2 Root systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 235 18.3 Root systems and bilinear forms . . . . . . . . . . . . . . . . . . . . . . . . . . 238 18.4 Passage to the field of real numbers . . . . . . . . . . . . . . . . . . . . . . . 239 18.5 Relations between two roots . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 240 18.6 Examples of root systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 243 18.7 Base of a root system . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 244 18.8 Weyl chambers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 247 18.9 Highest root . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 250 18.10 Closed subsets of roots . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 250 18.11 Weights . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 253 18.12 Graphs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 255 18.13 Dynkin diagrams . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 256 18.14 Classification of root systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 259 19 Lie algebras . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 277 19.1 Generalities on Lie algebras . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 277 19.2 Representations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 279 19.3 Nilpotent Lie algebras . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 282 19.4 Solvable Lie algebras . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 286 19.5 Radical and the largest nilpotent ideal . . . . . . . . . . . . . . . . . . . . . 289 19.6 Nilpotent radical . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 291 19.7 Regular linear forms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 292 19.8 Cartan subalgebras . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 294 20 Semisimple and reductive Lie algebras . . . . . . . . . . . . . . . . . . . . . 299 20.1 Semisimple Lie algebras . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 299 20.2 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 301 20.3 Semisimplicity of representations . . . . . . . . . . . . . . . . . . . . . . . . . . 302 20.4 Semisimple and nilpotent elements . . . . . . . . . . . . . . . . . . . . . . . . 305 20.5 Reductive Lie algebras . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 307
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20.6 Results on the structure of semisimple Lie algebras . . . . . . . . . . 310 20.7 Subalgebras of semisimple Lie algebras . . . . . . . . . . . . . . . . . . . . 313 20.8 Parabolic subalgebras . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 316 21 Algebraic groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 319 21.1 Generalities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 319 21.2 Subgroups and morphisms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 321 21.3 Connectedness . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 322 21.4 Actions of an algebraic group . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 325 21.5 Modules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 326 21.6 Group closure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 327 22 Affine algebraic groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 331 22.1 Translations of functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 331 22.2 Jordan decomposition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 333 22.3 Unipotent groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 335 22.4 Characters and weights . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 338 22.5 Tori and diagonalizable groups . . . . . . . . . . . . . . . . . . . . . . . . . . . 340 22.6 Groups of dimension one . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 345 23 Lie algebra of an algebraic group . . . . . . . . . . . . . . . . . . . . . . . . . . 347 23.1 An associative algebra . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 347 23.2 Lie algebras . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 348 23.3 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 352 23.4 Computing differentials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 354 23.5 Adjoint representation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 359 23.6 Jordan decomposition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 362 24 Correspondence between groups and Lie algebras . . . . . . . . . . 365 24.1 Notations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 365 24.2 An algebraic subgroup . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 365 24.3 Invariants . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 368 24.4 Functorial properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 372 24.5 Algebraic Lie subalgebras . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 375 24.6 A particular case . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 380 24.7 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 383 24.8 Algebraic adjoint group . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 383 25 Homogeneous spaces and quotients . . . . . . . . . . . . . . . . . . . . . . . . 387 25.1 Homogeneous spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 387 25.2 Some remarks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 389 25.3 Geometric quotients . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 391 25.4 Quotient by a subgroup . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 393 25.5 The case of finite groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 397
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26 Solvable groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 401 26.1 Conjugacy classes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 401 26.2 Actions of diagonalizable groups . . . . . . . . . . . . . . . . . . . . . . . . . . 405 26.3 Fixed points . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 406 26.4 Properties of solvable groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 407 26.5 Structure of solvable groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 409 27 Reductive groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 413 27.1 Radical and unipotent radical . . . . . . . . . . . . . . . . . . . . . . . . . . . . 413 27.2 Semisimple and reductive groups . . . . . . . . . . . . . . . . . . . . . . . . . . 415 27.3 Representations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 416 27.4 Finiteness properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 420 27.5 Algebraic quotients . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 422 27.6 Characters . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 424 28 Borel subgroups, parabolic subgroups, Cartan subgroups . . 429 28.1 Borel subgroups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 429 28.2 Theorems of density . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 432 28.3 Centralizers and tori . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 434 28.4 Properties of parabolic subgroups . . . . . . . . . . . . . . . . . . . . . . . . . 435 28.5 Cartan subgroups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 437 29 Cartan subalgebras, Borel subalgebras and parabolic subalgebras . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 441 29.1 Generalities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 441 29.2 Cartan subalgebras . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 443 29.3 Applications to semisimple Lie algebras . . . . . . . . . . . . . . . . . . . . 446 29.4 Borel subalgebras . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 447 29.5 Properties of parabolic subalgebras . . . . . . . . . . . . . . . . . . . . . . . . 450 29.6 More on reductive Lie algebras . . . . . . . . . . . . . . . . . . . . . . . . . . . 453 29.7 Other applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 454 29.8 Maximal subalgebras . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 456 30 Representations of semisimple Lie algebras . . . . . . . . . . . . . . . . 459 30.1 Enveloping algebra . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 459 30.2 Weights and primitive elements . . . . . . . . . . . . . . . . . . . . . . . . . . . 461 30.3 Finite-dimensional modules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 463 30.4 Verma modules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 464 30.5 Results on existence and uniqueness . . . . . . . . . . . . . . . . . . . . . . . 467 30.6 A property of the Weyl group . . . . . . . . . . . . . . . . . . . . . . . . . . . . 469 31 Symmetric invariants . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 471 31.1 Invariants of finite groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 471 31.2 Invariant polynomial functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 475 31.3 A free module . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 478
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32 S-triples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 481 32.1 Jacobson-Morosov Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 481 32.2 Some lemmas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 484 32.3 Conjugation of S-triples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 487 32.4 Characteristic . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 488 32.5 Regular and principal elements . . . . . . . . . . . . . . . . . . . . . . . . . . . 489 33 Polarizations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 493 33.1 Definition of polarizations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 493 33.2 Polarizations in the semisimple case . . . . . . . . . . . . . . . . . . . . . . . 494 33.3 A non-polarizable element . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 497 33.4 Polarizable elements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 499 33.5 Richardson’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 502 34 Results on orbits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 507 34.1 Notations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 507 34.2 Some lemmas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 508 34.3 Generalities on orbits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 509 34.4 Minimal nilpotent orbit . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 511 34.5 Subregular nilpotent orbit . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 513 34.6 Dimension of nilpotent orbits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 517 34.7 Prehomogeneous spaces of parabolic type . . . . . . . . . . . . . . . . . . 518 35 Centralizers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 521 35.1 Distinguished elements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 521 35.2 Distinguished parabolic subalgebras . . . . . . . . . . . . . . . . . . . . . . . 523 35.3 Double centralizers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 525 35.4 Normalizers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 528 35.5 A semisimple Lie subalgebra . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 530 35.6 Centralizers and regular elements . . . . . . . . . . . . . . . . . . . . . . . . . 533 36 σ-root systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 537 36.1 Definition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 537 36.2 Restricted root systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 539 36.3 Restriction of a root . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 544 37 Symmetric Lie algebras . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 549 37.1 Primary subspaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 549 37.2 Definition of symmetric Lie algebras . . . . . . . . . . . . . . . . . . . . . . . 553 37.3 Natural subalgebras . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 554 37.4 Cartan subspaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 555 37.5 The case of reductive Lie algebras . . . . . . . . . . . . . . . . . . . . . . . . . 557 37.6 Linear forms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 559
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38 Semisimple symmetric Lie algebras . . . . . . . . . . . . . . . . . . . . . . . . 561 38.1 Notations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 561 38.2 Iwasawa decomposition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 562 38.3 Coroots . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 565 38.4 Centralizers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 568 38.5 S-triples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 570 38.6 Orbits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 573 38.7 Symmetric invariants . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 579 38.8 Double centralizers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 584 38.9 Normalizers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 588 38.10 Distinguished elements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 589 39 Sheets of Lie algebras . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 593 39.1 Jordan classes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 593 39.2 Topology of Jordan classes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 596 39.3 Sheets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 601 39.4 Dixmier sheets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 603 39.5 Jordan classes in the symmetric case . . . . . . . . . . . . . . . . . . . . . . 605 39.6 Sheets in the symmetric case . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 608 40 Index and linear forms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 611 40.1 Stable linear forms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 611 40.2 Index of a representation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 615 40.3 Some useful inequalities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 616 40.4 Index and semi-direct products . . . . . . . . . . . . . . . . . . . . . . . . . . . 618 40.5 Heisenberg algebras in semisimple Lie algebras . . . . . . . . . . . . . 621 40.6 Index of Lie subalgebras of Borel subalgebras . . . . . . . . . . . . . . . 625 40.7 Seaweed Lie algebras . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 629 40.8 An upper bound for the index . . . . . . . . . . . . . . . . . . . . . . . . . . . . 630 40.9 Cases where the bound is exact . . . . . . . . . . . . . . . . . . . . . . . . . . . 635 40.10 On the index of parabolic subalgebras . . . . . . . . . . . . . . . . . . . . . 638 References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 641 List of notations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 645 Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 647
1 Results on topological spaces
In this chapter, we treat some basic notions of topology such as irreducible and constructible sets, dimension of a topological space, Noetherian space, which are fundamental in algebraic geometry.
1.1 Irreducible sets and spaces 1.1.1 Definition. A topological space X is said to be irreducible if any finite intersection of non-empty open subsets is non-empty. 1.1.2 It follows from the definition that an irreducible topological space is not empty. 1.1.3 Proposition. Let X be a non-empty topological space. Then the following conditions are equivalent: (i) X is irreducible. (ii) X is not the finite union of distinct proper closed subsets. (iii) X is not the union of two proper closed subsets. (iv) Any non-empty open subset of X is dense in X. (v) Any open subset of X is connected. Proof. The implications (i) ⇒ (ii) ⇒ (iii) are clear and (iii) ⇒ (iv) follows from the fact that a subset in X is dense if and only if it meets all nonempty open subsets. Now if U is a non-connected non-empty open subset, then U = U1 ∪ U2 where U1 , U2 are non-empty open subsets and U1 ∩ U2 = ∅. Thus (iv) ⇒ (v). The same argument gives (v) ⇒ (i). Remark. If X is irreducible, then it is connected. The converse is not true. 1.1.4 In the rest of this chapter, X is a topological space. A subset of X is called irreducible if it is non-empty and irreducible as a topological space. From the above definitions, the following result is clear.
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1 Results on topological spaces
Proposition. Let A be a non-empty subset of X. Then the following conditions are equivalent: (i) A is irreducible. (ii) Let F1 , . . . , Fn be closed subsets of X such that A is contained in the union of the Fi ’s, then there exists j ∈ {1, . . . , n} such that A ⊂ Fj . (iii) Let U , V be open subsets of X such that U ∩ A and V ∩ A are nonempty, then U ∩ V ∩ A = ∅. 1.1.5 Proposition. Let A, B be subsets of X. (i) A is irreducible if and only if its closure A is irreducible. (ii) If A is irreducible and A ⊂ B ⊂ A, then B is irreducible. Proof. For any open subset U , we have U ∩ A = ∅ if and only if U ∩ A = ∅. So (i) and (ii) follow. 1.1.6 Proposition. (i) If X is irreducible, then any non-empty open subset of X is also irreducible. (ii) Let (Ui )i∈I be a covering of X by open subsets such that Ui ∩ Uj = ∅ for all i, j ∈ I. If all the Ui ’s are irreducible, then X is irreducible. Proof. (i) Let U , V be non-empty open subsets of X such that V ⊂ U . If X is irreducible, then V is dense in U . Thus U is irreducible. (ii) Let V be a non-empty open subset of X. There exists k ∈ I such that V ∩ Uk = ∅. Since Ui ∩ Uk = ∅ for all i ∈ I and V ∩ Uk is dense in Uk , V ∩ Ui ∩ Uk = ∅. Hence V ∩ Ui = ∅ for all i. It follows that V ∩ Ui is dense in Ui for all i ∈ I, so V is dense in X. 1.1.7 Proposition. Let Y be a topological space and f : X → Y a continuous map. (i) If A ⊂ X is irreducible, then f (A) is irreducible in Y . (ii) Suppose that Y is irreducible, f is an open map and that f −1 (y) is irreducible for all y ∈ Y . Then X is irreducible. Proof. (i) Let U, V be open subsets of Y such that U ∩ f (A) and V ∩ f (A) are non-empty. Then f −1 (U ) and f −1 (V ) are open subsets whose intersection with A is non-empty. It follows that f −1 (U ∩ V ) = f −1 (U ) ∩ f −1 (V ) meets A. Therefore U ∩ V meets f (A) and assertion (i) follows. (ii) Let U, V be non-empty open subsets of X. Since f is open and Y is irreducible, f (U ) meets f (V ) at some point y. Further, f −1 (y) is irreducible, therefore the open subsets U ∩ f −1 (y) and V ∩ f −1 (y) of f −1 (y) have nonempty intersection. Hence U ∩ V = ∅. 1.1.8 Remark. A map f : X → Y is called dominant if f (X) is dense in Y . It follows from 1.1.5 and 1.1.7 that if X is irreducible and f is continuous and dominant, then Y is irreducible.
1.1 Irreducible sets and spaces
3
1.1.9 Definition. A maximal irreducible subset of X will be called an irreducible component of X. 1.1.10 Proposition. (i) All irreducible components of X are closed. (ii) Any irreducible subset of X is contained in an irreducible component and X is the union of its irreducible components. Proof. (i) This is clear from 1.1.5. (ii) Let (Ai )i∈I be a family of irreducible subsets of X totally ordered by inclusion and A = i∈I Ai . Let U, V be open subsets which meet A. Since the family is totally ordered, there exist k ∈ I such that U ∩ Ak = ∅ and V ∩ Ak = ∅. Thus U ∩ V ∩ Ak = ∅ because Ak is irreducible. Now the first assertion follows from Zorn’s lemma. The second assertion follows from the fact that any singleton in X is irreducible. 1.1.11 Corollary. Any connected component of X is a union of irreducible components of X. Proof. Any irreducible subset of X is connected, so it is contained in a connected component of X. Remark. Two distinct irreducible components of X can have a non-empty intersection. 1.1.12 Proposition. Let U ⊂ X be an open subset. The map Z → Z ∩ U is a bijection between the set of irreducible closed subsets of X meeting U and the set of irreducible subsets of U , closed in U . The reciprocal map is V → V . In particular, C → C ∩ U induces a bijection between the set of irreducible components of X meeting U and the set of irreducible components of U . Proof. Let Z be an irreducible closed subset of X meeting U . Since U ∩ Z is open in Z, it is irreducible by 1.1.6 and dense in Z. Therefore Z = Z ∩ U . Conversely, if V is an irreducible closed subset of U , then V is irreducible by 1.1.5 and V = U ∩ V . 1.1.13 Recall that A ⊂ X is nowhere dense in X if the interior of A is empty. Let us suppose that X has only a finite number of distinct irreducible n, denote by Ui = X \ ( j=i Xj ) the set components X1 , . . . , Xn . For 1 i of elements of X which are not in j=i Xj . Lemma. The set Ui is open in X and dense in Xi . The disjoint union of the Ui ’s is dense in X. For i = j, Xi ∩ Xj is nowhere dense in X. Proof. It is obviousthat Ui is open in X and is contained in Xi . If Ui is empty, then Xi ⊂ j=i Xj . Since Xi is irreducible, there exists k = i such that Xi = Xk which is impossible by our hypothesis. Thus Ui is dense in Xi since it is a non-empty open subset. The second assertion is now obvious.
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Let i = j and U an open subset contained in Xi ∩ Xj . So U is contained in the union of the Xk ’s, k = i. Hence U ∩ Ui = ∅. In the same manner, we have U ∩ Uj = ∅. Thus U = ∅ since Ui ∪ Uj is dense in Xi ∪ Xj . 1.1.14 Lemma. Using the hypotheses of 1.1.13, suppose that X is a subset of a topological space Y . Then X1 , . . . , Xn are the distinct irreducible components of X. Proof. By 1.1.5, the Xi ’s are irreducible. Clearly X = X1 ∪ · · · ∪ Xn . If A is an irreducible subset of X, then it is contained in one of the Xi ’s. Finally, if Xi ⊂ Xj , then Xi ⊂ X ∩ Xj = Xj since Xj is closed in X. This contradicts our hypothesis and so the lemma follows.
1.2 Dimension 1.2.1 Definition. Let E be a set. A chain of subsets of length n of E is a strictly increasing sequence of subsets E0 ⊂ E1 ⊂ · · · ⊂ En . 1.2.2 Definition. The dimension of a non-empty topological space X is the supremum of the lengths of chains consisting of irreducible closed subsets. We shall denote it by dim X which is in N ∪ {+∞}. 1.2.3 Proposition. Let us assume that X is non-empty. (i) Let Y be a non-empty subset of X, then dim Y dim X. r (ii) Let X1 , . . . , Xr be closed subsets of X such that X = i=1 Xi , then dim X = max{dim Xi ; 1 i r}. r (iii) Let X1 , . . . , Xr be open subsets of X such that X = i=1 Xi , then dim X = max{dim Xi ; 1 i r}. (iv) Suppose that X is irreducible and dim X is finite. If Y is a closed subset such that dim Y = dim X, then X = Y . Proof. (i) Let Y0 ⊂ Y1 ⊂ · · · ⊂ Yn be a chain of irreducible closed subsets in Y . The closure Yi of Yi in X is irreducible by 1.1.5 and since Yi = Yi ∩ Y , Yi is strictly contained in Yi+1 . Thus dim Y dim X. (ii) Let Z0 ⊂ Z1 ⊂ · · · ⊂ Zn be a chain of irreducible closed subsets in X. Since Zn is irreducible and Zn = (Zn ∩ X1 ) ∪ · · · ∪ (Zn ∩ Xr ), there exists k such that Zn ⊂ Xk . It follows that dim Xk n. But dim Xk dim X by (i), so (ii) follows. (iii) Let Z0 ⊂ Z1 ⊂ · · · ⊂ Zn be a chain of irreducible closed subsets in X. There exists k such that Xk ∩ Z0 = ∅. It follows that Xk ∩ Zi is a non-empty open irreducible subset in Zk for all 1 i r. By 1.1.3 (iv), Xk ∩ Zi = Zi and Xk ∩ Z0 ⊂ · · · ⊂ Xk ∩ Zn is a chain of irreducible closed subsets in Xk . Hence dim X dim Xk and (iii) follows from (i). (iv) Let dim Y = n and Y0 ⊂ Y1 ⊂ · · · ⊂ Yn be a chain of irreducible closed subsets in Y . If Y = X, then Y0 ⊂ Y1 ⊂ · · · ⊂ Yn ⊂ X is a chain of irreducible closed subsets in X. So dim X > dim Y which contradicts our hypothesis.
1.3 Noetherian spaces
5
1.2.4 Definition. Let X be non-empty and finite-dimensional and Y a non-empty subset of X. We define the codimension of Y in X, denoted by codimX Y , to be dim X − dim Y .
1.3 Noetherian spaces 1.3.1 Proposition. The following conditions are equivalent: (i) Any decreasing sequence of closed subsets of X is stationary. (ii) Any ascending sequence of open subsets of X is stationary. (iii) Any non-empty family of closed subsets of X has a minimal element. (iv) Any non-empty family of open subsets of X has a maximal element. If X satisfies the above conditions, then we shall say that X is Noetherian. Proof. This is straightforward. Remark. A Noetherian topological space can be infinite-dimensional. 1.3.2 Proposition. (i) Any subset of a Noetherian space is Noetherian. (ii) Let (Xi )1ip be a finite covering of X such that the Xi ’s are Noetherian. Then X is also Noetherian. Proof. (i) Let Y be a subset of X and (Fn )n be a decreasing sequence of closed subsets of Y . Then (Fn )n is a decreasing sequence of closed subsets of X. Therefore there exists p such that Fp = Fn for all n p. This implies that Fn = Y ∩ Fn = Y ∩ Fp = Fp , and (i) follows. (ii) Let (Fn )n be a decreasing sequence of closed subsets of X. Then for 1 i p, (Xi ∩ Fn )n is a decreasing sequence of closed subsets of Xi . Hence there exists ni such that Xi ∩ Fni = Xi ∩ Fn for n ni . Set q = max(n1 , . . . , np ), then Fn = Fq for n q. 1.3.3 Proposition. The following conditions are equivalent: (i) X is Noetherian. (ii) Any open subset of X is quasi-compact. Proof. (i) ⇒ (ii) By 1.3.2, it suffices to show that X is quasi-compact. Let (Ui )i be an open covering of X. The set of all finite unions of the Ui ’s has a maximal element U since X is Noetherian. It is clear that U = X. (ii) ⇒ (i) Let (Un )n be an ascending sequence of open subsets of X. The union U of the Un ’s is open and therefore it is quasi-compact. Since (Un )n is an open covering of U , it is clear that the sequence (Un )n is stationary. 1.3.4 The following result is called the principle of transfinite induction. Lemma. Let E be a well-ordered set, that is an ordered set such that any non-empty subset has a minimal element. Let F be a subset of E such that: if a ∈ E satisfies {x ∈ E; x < a} ⊂ F , then a ∈ F . Then F = E. Proof. Suppose that F = E. Let b be a minimal element in E \ F . Then {x ∈ E; x < b} ⊂ F . Thus b ∈ F which contradicts our hypothesis on b.
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1.3.5 Proposition. Let X be Noetherian. Then the set of irreducible components of X is finite. Proof. Let E be the set of all closed subsets of X ordered by inclusion, and F be the set of all finite unions of irreducible closed subsets of X. Now let Y ∈ E be such that any proper closed subset of Y is in F . If Y is irreducible, then Y ∈ F . If Y is not irreducible, then Y is the union of two distinct proper closed subsets Y1 , Y2 . But Y1 , Y2 ∈ F , so Y ∈ F . Thus we can apply 1.3.4, and we have E = F . Thus X is the finite union of irreducible closed subsets X1 , . . . , Xn . Further, we can assume that they are non-comparable. Let Z be an irreducible component of X. Then Z = (Z ∩ X1 ) ∪ · · · ∪ (Z ∩ Xn ) which implies that there exists k such that Z = Z ∩ Xk . Thus Z = Xk . This shows that {X1 , . . . , Xn } is the set of irreducible components of X. 1.3.6 Proposition. Let X be non-empty Noetherian, and X1 , . . . , Xn its irreducible components. For 1 i n, set: Xj . Yi = Xi \ (Xi ∩ Xj ) = X \ j=i
j=i
(i) An open subset U is dense in X if and only if U ∩Xi = ∅ for 1 i n. (ii) For 1 i n, Yi is open in X. Further, Y0 = Y1 ∪ · · · ∪ Yn is open dense in X and the irreducible (connected) components of Y0 are Y1 , . . . , Yn . Proof. Clearly, Yi is open in X and it is non-empty. So it is dense in Xi . Thus Y0 is open dense in X. Further, the Yi ’s are pairwise disjoint. So (ii) follows from 1.1.11. Now if U is open dense in X, then U ∩ Xi ⊃ U ∩ Yi = ∅. Conversely, if an open subset V meets each Xi , then V ∩ Xi is dense in Xi and Xi ⊂ V . Hence V is dense.
1.4 Constructible sets 1.4.1 Proposition. Let Y be a subset of X. The following conditions are equivalent: (i) There exists a family (Ui )i∈I of open subsets of X, whose union contains Y , such that Y ∩ Ui is closed in Ui for all i ∈ I. (ii) Any point of Y admits an open neighbourhood U in X such that U ∩Y is closed in U . (iii) Y is open in Y . (iv) Y is the intersection of an open subset and a closed subset of X. If Y satisfies these conditions, then it is called a locally closed subset of X. Proof. Only (ii) ⇒ (iii) needs attention. Let x ∈ Y , and U be an open neighbourhood of x in X such that U ∩ Y is closed in U . Then U ∩ Y = U ∩ Y , which shows that in Y , x is in the interior of Y .
1.4 Constructible sets
7
1.4.2 Let F be a set of subsets of X such that (i) Any open subset of X is in F. (ii) If Z ∈ F, then X \ Z ∈ F. (iii) Any finite union of elements of F is in F. For example, the set of subsets of X satisfies these conditions. The intersection C of all such sets satisfies these conditions. An element of C is called a constructible subset of X. 1.4.3 Proposition. The set C is the set of finite unions of locally closed subsets of X. Proof. Denote by C0 the set of finite unions of locally closed subsets of X. It is clear that C0 ⊂ C. Let C = (U1 ∩ F1 ) ∪ · · · ∪ (Un ∩ Fn ) ∈ C0 where the Ui ’s (resp. Fi ’s) are open (resp. closed) subsets of X. Then it is clear that X \ C is the union of subsets of the form E1 ∩ · · · ∩ En where for each j, there exists i such that Ej is either X \ Ui or X \ Fi . So the proposition follows. 1.4.4 Proposition. Let Y be a subset of X. (i) If C be a constructible subset of X, then C ∩Y is a constructible subset of Y . (ii) Suppose that Y is closed in X. If C is a constructible subset of Y , then C is a constructible subset of X. Proof. Part (i) is obvious and (ii) follows from the fact that if F is closed in Y and U is open in Y , then F is closed in X and U is locally closed in X. 1.4.5 Lemma. Let C be a constructible subset of X whose closure is irreducible. Then C contains an open subset of C. Proof. Let C = (U1 ∩ F1 ) ∪ · · · ∪ (Un ∩ Fn ) ∈ C0 where the Ui ’s (resp. Fi ’s) are open (resp. closed) subsets of X. We can assume further n that Ui ∩ Fi = ∅ for 1 i n. Since C is irreducible and contained in i=1 Fi , there exists j such that C ⊂ Fj . It follows that Uj ∩ Fj = Uj ∩ C and Uj ∩ Fj is a non-empty subset of C, open in C. 1.4.6 Proposition. Let X be Noetherian. Then any constructible subset Y contains an open and dense subset of Y . Proof. Let Y = Y1 ∪ · · · ∪ Yn be the decomposition of Y into irreducible components. Since irreducible components are closed, they are constructible in Y . By 1.1.14, the Yi ’s are the irreducible components of Y . For 1 i n, set: Ti = Yi \ (Yi ∩ Yj ). j=i
We saw in 1.3.6 that the Ti ’s are non-empty and open in Y . By 1.4.5, there exists a subset Ui in Yi which is open and dense in Yi . Let Vi = Ui ∩ Ti . Then
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Vi is open and dense in Yi . But the Vj ’s are pairwise disjoint, so Vi is in fact open in Y . Now it follows from 1.3.6 (i) that V = V1 ∪ · · · ∪ Vn is a subset of Y which is open and dense in Y .
1.5 Gluing topological spaces 1.5.1 Let (Xi )i∈I be a family of sets and X be the sum of the Xi ’s. For i ∈ I, we denote by θi : Xi → X the canonical injection and we shall identify Xi with its image via θi . Suppose that for all i, j ∈ I, there are subsets Xij ⊂ Xi , Xji ⊂ Xj and a map hji : Xij → Xji such that a) For all i ∈ I, Xii = Xi and hii = idXi is the identity map. b) For all i, j, k ∈ I and x ∈ Xij ∩ Xik , we have hji (x) ∈ Xjk and: hki (x) = hkj (hji (x)). By applying b) to the triples (i, j, i) and (j, i, j), we remark that hij is bijective with reciprocal map hji . Let us define a binary relation R on X by xRy if and only if there exist i, j ∈ I such that x ∈ Xij , y ∈ Xji , y = hji (x). The preceding remark implies that R is reflexive and symmetric. Now suppose that x, y, z ∈ X, xRy and yRz. There exist i, j, k ∈ I such that x ∈ Xij , y ∈ Xji ∩ Xjk , z ∈ Xkj , y = hji (x) , z = hkj (y). By b), we have x = hij (y) ∈ Xik and: z = hkj (y) = hkj (hji (x)) = hki (x). Thus R is transitive and it defines an equivalence relation on X. We shall say that the set of equivalence classes X/R is obtained by gluing the Xi ’s along the Xij ’s via the bijections hji . Let us denote by ϕ : X → X/R the canonical surjection. Note that we have the following: (i) Each equivalence class has at most one element in each of the Xi ’s. (ii) For i, j ∈ I, Xij is the set of x ∈ Xi such that there exists a y ∈ Xj with xRy. We deduce from these remarks that for i ∈ I, ϕ ◦ θi induces a bijection from Xi to ϕ ◦ θi (Xi ). 1.5.2 Conserving the hypotheses of 1.5.1, let us suppose further that the following properties are satisfied: c) Each Xi is a topological space with topology Ti . d) For i, j ∈ I, Xij is open in Xi and hji is a homeomorphism from Xij to Xji .
1.5 Gluing topological spaces
9
We can equip X with the topology S, the sum of the topologies Ti ’s (this is the finest topology on X for which the injections θi are continuous). Thus a subset A ⊂ X is open (resp. closed) if and only if each θi−1 (A) is open (resp. closed) in Xi . In particular, each θi (Xi ) is open and closed in X. Let T be the topology on X/R, quotient of the topology S. This is the finest topology for which the maps ϕ ◦ θi are continuous. Therefore a subset B in X/R is open (resp. closed) if and only if B = ϕ(C) where C is an open (resp. closed) subset of X, saturated with respect to R. The topological space X/R is said to be obtained by gluing the Xi ’s along the Xij ’s via the maps hji . 1.5.3 Proposition. Let us suppose that the hypotheses a), b), c), d) of 1.5.1 and 1.5.2 are satisfied. Then each ϕ(Xi ) is open in X/R. Further, the restriction of ϕ to Xi is a homeomorphism from Xi onto ϕ(Xi ). Proof. Set Yi = ϕ(Xi ) , Zi =
θj (Xij ).
j∈I
Then Zi is a subset of X saturated with respect to R, and we have Yi = ϕ(Zi ). Since Zi ∩ Xj = Xij and Zi is open in X by our hypotheses, Yi is open in X/R. Let Ai be an open subset of Xi . Then each Ai ∩ Xij is open in Xij . Consequently, each hji (Ai ∩ Xij ) is open in Xji , and therefore open in Xj . It follows that j∈I θj ◦ hji (Ai ∩ Xij ) is open in X. So ϕ(Ai ) is open in X/R, and hence open in Yi . Now the proposition follows. 1.5.4 Let us conserve the hypotheses and notations of 1.5.1 and 1.5.2. To summarize, we have constructed a topological space Y = X/R, and for each i ∈ I, a homeomorphism ψi from Xi onto an open subset ψi (Xi ) in Y such that Y = i∈I ψi (Xi ), and such that ψi (Xij ) = ψj (Xji ) = ψi (Xi ) ∩ ψj (Xj ) , hji (x) = ψj−1 (ψi (x)) for all i, j ∈ I and x ∈ Xij .
References • [8], [26].
2 Rings and modules
Unless otherwise stated, all algebras over a commutative ring considered in this book are associative. We recall here basic concepts of the theory of commutative rings which will allow us to fix vocabulary and notations for the following chapters. Notions such as local rings, field of fractions and module of differentials will often appear in algebraic geometry. All rings considered in this chapter are commutative. Let A be a ring.
2.1 Ideals 2.1.1 Let us fix some notations. • If B is a ring, we denote by Hom(A, B) the set of ring homomorphisms from A to B. • An element x ∈ A is called a zero divisor if there exists y ∈ A \ {0} such that xy = 0. A ring A is called an integral domain if A = {0} and if A has no zero divisors other than 0. • An element x ∈ A is called nilpotent if there exists n ∈ N∗ such that n x = 0. The set of nilpotent elements of A is an ideal called the nilradical of A. We say that A is reduced if its nilradical is {0}. 2.1.2 Proposition. (Chinese Remainder Theorem) Let a1 , . . . , an be ideals of A such that ai + aj = A for all i = j. Let x1 , . . . , xn ∈ A. Then there exists x ∈ A such that x − xi ∈ ai for 1 i n. Proof. First, suppose that n = 2. Then there exist a1 ∈ a1 , a2 ∈ a2 such that a1 + a2 = 1. Hence x = x2 a1 + x1 a2 satisfies the conclusion of the proposition. Suppose now that n 3. For i 2, there exist ai ∈ a1 and bi ∈ ai such that ai + bi = 1. The product i2 (ai + bi ) is equal to 1 and it is contained in a1 + a2 · · · an . Thus a1 + a2 · · · an = A and from the previous paragraph, there exists y1 ∈ A such that y1 − 1 ∈ a1 and y1 ∈ a2 · · · an . So y1 − 1 ∈ a1 and y1 ∈ ai for all i 2.
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Proceeding in the same manner for an index j 2, we see that there exists yj ∈ A such that yj − 1 ∈ aj and yj ∈ ai for all i = j. We verify easily that x = x1 y1 + · · · + xn yn satisfies the conclusion of the proposition. 2.1.3 Corollary. Let a1 , . . . , an be ideals of A such that ai + aj = A for all i = j. Set b = a1 ∩ · · · ∩ an . (i) We have b = a1 · · · an . (ii) The homomorphism f : A → (A/a1 ) × · · · × (A/an ) induced by the canonical surjections on each factor is surjective and its kernel is b. In particular f induces an isomorphism between A/b and (A/a1 ) × · · · × (A/an ). Proof. (i) From the proof of 2.1.2, a1 + a2 · · · an = A. Therefore by induction, it suffices to consider the case n = 2. Clearly, a1 a2 ⊂ a1 ∩ a2 . Let a1 ∈ a1 , a2 ∈ a2 be such that a1 + a2 = 1. If x ∈ a1 ∩ a2 , then x = xa1 + xa2 ∈ a1 a2 . (ii) It is clear that b is the kernel of f . The surjectivity of f follows from 2.1.2 and the last assertion is now obvious.
2.2 Prime and maximal ideals 2.2.1 An ideal m of A is called maximal if it is maximal in the set of proper ideals of A. This is equivalent to the condition that A/m is a field. Any ideal of A is contained in a maximal ideal and we shall denote by Spm(A) the set of maximal ideals of A. 2.2.2 An ideal p of A is called prime if it satisfies one of the following equivalent conditions: (i) p = A and if x, y ∈ A \ p, then xy ∈ A \ p. (ii) The ring A/p is an integral domain. A maximal ideal is a prime ideal. Let us denote by Spec(A) the set of prime ideals of A. 2.2.3 A chain of prime ideals of length n of A is a sequence p0 ⊂ p1 ⊂ · · · ⊂ p n of pairwise distinct elements of Spec(A). The (Krull) dimension of A, denoted by dim A, is the supremum of the lengths of chains of prime ideals. It is an element of N∪{+∞}. Thus dim A = 0 if and only if prime ideals of A are maximal; this is the case for a commutative field. 2.2.4 Proposition. Let B = A1 ×· · ·×An be a product of rings. Any prime (resp. maximal) ideal of B is of the form A1 ×· · ·×Ai−1 ×ai ×Ai+1 ×· · ·×An , where 1 i n and ai is a prime (resp. maximal) ideal of Ai . Proof. If C and D are rings, then C × D is not an integral domain since (0, 1)(1, 0) = (0, 0). The result now follows easily.
2.3 Rings of fractions and localization
13
2.2.5 Proposition. (i) Let a1 , . . . , an be ideals of A and p a prime ideal of A such that a1 · · · an ⊂ p. Then p contains one of the ai ’s. (ii) Let a be an ideal of A and p1 , . . . , pr be prime ideals of A such that a ⊂ p1 ∪ · · · ∪ pr . Then a is contained in one of the pi ’s. Proof. (i) Suppose that ai ⊂ p for 1 i n. There exists ai ∈ ai \ p. But then a = a1 · · · an ∈ a1 · · · an \ p, which contradicts our hypotheses. (ii) We can assume that pi ⊂ pj for i = j. Let us proceed by induction on r. The result is obvious when r = 1. If a ⊂ p1 ∪ · · · ∪ pr−1 , then we obtain our result by induction. So let us suppose that there exists a ∈ a ∩ pr such that a ∈ p1 ∪ · · · ∪ pr−1 . Suppose that there exists y ∈ ap1 · · · pr−1 \ pr . Then a + y ∈ a and a + y ∈ p1 ∪ · · · ∪ pr−1 . It follows that a + y ∈ pr and so y ∈ pr which gives us a contradiction. So we have ap1 · · · pr−1 ⊂ pr . By our assumption, for i < r, we have pi ⊂ pr . So part (i) implies that a ⊂ pr . 2.2.6 Corollary. Let m ∈ Spm(A) and n ∈ N∗ . Then the only prime ideal containing mn is m. Proof. If p ∈ Spec(A) contains mn , then by 2.2.5 (i), m ⊂ p. Hence m = p.
2.2.7 A prime ideal in A is said to be minimal if it is minimal by inclusion in Spec(A). Proposition. If A = {0}, then the set of minimal prime ideals of A is non-empty. Proof. Since A = {0}, Spec(A) is non-empty. We shall endow Spec(A) with the partial order given by reverse inclusion. Let (pi )i∈I be a totally ordered family of prime ideals. Set r = i∈I pi . Let a, b ∈ A be such that ab ∈ r and a ∈ r. So there exists j ∈ I such that a ∈ pj . Now for all k ∈ I such that pk ⊂ pj , a ∈ pk and since pk is prime, b ∈ pk . Consequently b ∈ r and the result follows from Zorn’s lemma.
2.3 Rings of fractions and localization 2.3.1 Definition. A subset S of A is said to be multiplicative if 1 ∈ S and it is closed under multiplication. 2.3.2 Lemma. Let S be a multiplicative subset of A not containing 0 and a an ideal of A such that a ∩ S = ∅. Then there exists p ∈ Spec(A) such that a ⊂ p and p ∩ S = ∅. Proof. Let A be the set, partially ordered by inclusion, of ideals b of A such that a ⊂ b and b ∩ S = ∅. Since a ∈ A, A is non-empty and it is clearly inductive. By Zorn’s lemma, A has a maximal element p. Suppose that there exist a, b ∈ A \ p such that ab ∈ p. Then the ideals Aa + p and Ab + p contain strictly p. By the maximality of p, there exist
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s ∈ S ∩ (Aa + p) and t ∈ S ∩ (Ab + p). It follows that st ∈ S ∩ p which is a contradiction. We have therefore shown that p is a prime ideal. 2.3.3 Let S be a multiplicative subset of A. We define the binary relation R on S × A as follows: (s, a)R(t, b) if and only if there exists u ∈ S such that u(sb − ta) = 0. This defines an equivalence relation on S × A. We shall endow S × A with the following operations: (s, a) + (t, b) = (st, sb + ta) , (s, a).(t, b) = (st, ab). These operations are compatible with R and we see clearly that they induce a ring structure on (S × A)/R. We call the ring (S × A)/R, denoted by S −1 A, the ring of fractions of A by S. We shall denote the class of (s, a) by a/s or a . s The map i : A → S −1 A defined by a → a/1 is a ring homomorphism. We shall call i the canonical homomorphism. We have the following results: • The canonical homomorphism i is injective if and only if S contains no zero divisor. • We have S −1 A = {0} if and only if 0 ∈ S. This is the case if S contains a nilpotent element. • Let s ∈ S. The element i(s) = s/1 is invertible in S −1 A. Its inverse is 1/s. • If p ∈ Spec(A), then S = A \ p is a multiplicative subset of A. The ring of fractions S −1 A, denoted by Ap , is called the localization of A at p. In particular, Ap = {0}. • Suppose that A = {0} and let S be the set of non zero divisors of A. Then S is a multiplicative subset of A. The ring S −1 A, denoted by Fract(A), is called the full ring of fractions of A. Further, if A is an integral domain, then Fract(A) is a field which is called the quotient field or field of fractions of A. 2.3.4 Theorem. Let A, B be rings, S a multiplicative subset of A, i : A → S −1 A the canonical homomorphism and f ∈ Hom(A, B). Suppose further that any element of f (S) is invertible in B. Then there exists a unique homomorphism g : S −1 A → B such that g ◦ i = f . Proof. Let p : S × A → S −1 A be the map sending (s, a) to a/s and h : S × A → B be the map defined by (s, a) → f (a)f (s)−1 . We verify easily that h(s, a) = h(t, b) if a/s = b/t. It follows therefore that there is a unique map g : S −1 A → B such that g ◦ p = h. A direct computation show that g is a ring homomorphism and g ◦ i = f . The uniqueness of g is immediate. 2.3.5 Corollary. Let S, T be multiplicative subsets of A such that S ⊂ T , iS : A → S −1 A and iT : A → T −1 A the corresponding canonical homomorphisms. There exists a unique homomorphism f : S −1 A → T −1 A such that f ◦ iS = iT . Further, if T contains no zero divisors, then f is injective.
2.3 Rings of fractions and localization
15
Remark. If A is an integral domain and S a multiplicative subset of A which does not contain 0, then S −1 A embeds canonically into the quotient field Fract(A). 2.3.6 Corollary. Suppose that A is a subring of a ring B and f : A → B the canonical injection. Let S be a multiplicative subset of A (hence of B), and i : A → S −1 A, j : B → S −1 B the canonical homomorphisms. There exists a unique homomorphism g : S −1 A → S −1 B such that g ◦ i = j ◦ f . The homomorphism g is injective and we can therefore identify canonically S −1 A as a subring of S −1 B. Proof. The existence and uniqueness of g follows from 2.3.4. Let (s, a) ∈ S ×A be such that g(a/s) = 0. Then 0 = g((a/1).(1/s)) = g ◦ i(a)[g ◦ i(s)]−1 ⇒ g ◦ i(a) = 0 ⇒ f ◦ j(a) = 0 ⇒ j(a) = 0. It follows that there exists t ∈ S such that ta = 0. Hence (s, a)R(t, 0) and a/s = 0. 2.3.7 Let S be a multiplicative subset of A, i : A → S −1 A be the canonical homomorphism and J (A) (resp. J (S −1 A)) the set of ideals of A (resp. S −1 A). Denote by J (A) the set of ideals a of A which verifies the condition: (1)
s ∈ S , a ∈ A , sa ∈ a ⇒ a ∈ a.
For a ∈ J (A), let S −1 a be the set of elements a/s of S −1 A such that a ∈ a, s ∈ S. We verify easily that S −1 a ∈ J (S −1 A) and thus we have a map ψ : J (A) → J (S −1 A) sending a to S −1 a. The proof of the following result is straightforward. Proposition. (i) If a ∈ J (A), then S −1 a is the ideal of S −1 A generated by i(a). Any ideal of S −1 A is of the form S −1 b for some b ∈ J (A). (ii) If n is the nilradical of A, then S −1 n is the nilradical of S −1 A. (iii) For a, b ∈ J (A), we have: S −1 (a + b) = S −1 a + S −1 b, S −1 (ab) = (S −1 a)(S −1 b), S −1 (a ∩ b) = S −1 a ∩ S −1 b. (iv) The map ψ induces a bijection between J (A) and J (S −1 A), whose reciprocal map is given by b → i−1 (b). 2.3.8 Corollary. The map ψ induces a bijection between the set of prime ideals of A not meeting S and the set of prime ideals of S −1 A. Proof. A prime ideal p of A which does not meet S satisfies the condition (1) of 2.3.7. So it suffices to show that S −1 p ∈ Spec S −1 A. Let a, b ∈ A, s, t ∈ S be such that (a/s).(b/t) ∈ S −1 p. There exists u ∈ S such that uab ∈ p. Since u ∈ p, ab ∈ p. So either a ∈ p or b ∈ p. Thus either a/s ∈ S −1 p or b/t ∈ S −1 p.
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2.3.9 Let A be a subring of a ring B and a be an ideal of A. Then an ideal b of B is said to lie above a if b ∩ A = a. Let S be a multiplicative subset of A. Let us conserve the notations i, j, f, g of 2.3.6 and let J (S −1 A), J (S −1 B), J (A), J (B) be as in 2.3.7. Proposition. Let p ∈ Spec A be such that S ∩ p = ∅. (i) The map b → S −1 b induces a surjection from the set E of ideals of B lying above p and the set F of ideals of S −1 B lying above S −1 p, and the map b → j −1 (b) is a bijection between F and E ∩ J (B). (ii) The map q → S −1 q induces a bijection between the set of prime ideals in B lying above p and the set of prime ideals of S −1 B lying above S −1 p. Proof. By 2.3.8, we have S −1 p ∈ Spec S −1 A and i−1 (S −1 p) = p. Let b ∈ J (S −1 B) be such that b ∩ S −1 A = S −1 p. Then we have: p = i−1 (S −1 p) = i−1 (b ∩ S −1 A) = A ∩ j −1 (b ). By 2.3.7, S −1 (j −1 (b )) = b and therefore the image of E via the map b → S −1 b contains F. Let b ∈ E, a ∈ A and s ∈ S. The following conditions are equivalent: (i) g(a/s) ∈ S −1 b. (ii) f (a)/f (s) ∈ S −1 b. (iii) There exists t ∈ S such that f (t)f (a) ∈ b. (iv) There exists t ∈ S such that ta ∈ p. (v) a/s ∈ S −1 p. We deduce that S −1 b ∩ S −1 A = S −1 p. Thus the image of E via the map b → S −1 b is F. The rest of the proposition follows from 2.3.7 and 2.3.8. 2.3.10 Proposition. Let S be a multiplicative subset of A and a ∈ J (A) be such that a ∩ S = ∅. Denote by π : A → A/a the canonical projection. Then π(S) is a multiplicative subset of A/a, and the rings S −1 A/S −1 a and π(S)−1 (A/a) are isomorphic. Proof. It is clear that T = π(S) is a multiplicative subset of B = A/a, and that 0 ∈ T since a ∩ S = ∅. Let iS : A → S −1 A and iT : B → T −1 B be the canonical homomorphisms. If s ∈ S, then iT ◦ π(s) is invertible in T −1 B. By 2.3.4, there exists a unique homomorphism f : S −1 A → T −1 B such that f ◦ iS = iT ◦ π. For a ∈ A, s ∈ S, we have: π(a)/π(s) = (π(a)/π(1)).(π(1)/π(s)) = (iT ◦ π(a))(iT ◦ π(s))−1 = (f ◦ iS (a))(f ◦ iS (s))−1 = f ((a/1).(1/s)) = f (a/s). Thus f is surjective. Now suppose that f (a/s) = 0. The preceding computation implies that π(a)/π(s) = 0, and so there exists t ∈ S such that π(ta) = π(a)π(t) = 0. Thus ta ∈ a and it follows that a/s = (ta)/(ts) ∈ S −1 a.
2.4 Localizations of modules
17
2.3.11 Definition. Let p be a prime ideal of A. We define the height of p, denoted by ht p, to be the dimension of the ring Ap . 2.3.12 Proposition. The height of a prime ideal p is the supremum of chains p0 ⊂ p1 ⊂ · · · ⊂ pr of prime ideals of A such that pr = p. Proof. This is immediate from 2.3.7 (iii) and 2.3.8.
2.4 Localizations of modules 2.4.1 In this section, S will denote a multiplicative subset of A. Let E be an A-module. We define the binary relation S on E × S by: for x, y ∈ E, s, t ∈ S, we have (x, s)S(y, t) if and only if there exists u ∈ S such that u(sy − tx) = 0. It is easy to verify that S defines an equivalence relation on E × S. Let us denote by S −1 E the set of equivalence classes and by x/s the class of (x, s). The proof of the following is similar to the one for rings of fractions. Theorem. Let E be an A-module. (i) Define a structure of S −1 A-module on S −1 E by x/s + y/t = (tx + sy)/st , (a/s)(y/t) = (ay)/(st) where x, y ∈ E, s, t ∈ S and a ∈ A. The map j : E → S −1 E , x → x/1 is a homomorphism of A-modules, called the canonical homomorphism of E to S −1 E. (ii) Let F be a S −1 A-module and f : E → F be a homomorphism of A-modules. Then there exists a unique homomorphism of S −1 A-modules g : S −1 E → F such that f = g ◦ j. 2.4.2 Proposition. Let f : E → F be a homomorphism of A-modules. There exists a unique homomorphism of S −1 A-modules g : S −1 E → S −1 F such that g(x/1) = f (x)/1 for all x ∈ E. Proof. Let jE : E → S −1 E and jF : F → S −1 F be the canonical homomorphisms. By 2.4.1, there exists a S −1 A-module homomorphism g : S −1 E → S −1 F such that jF ◦ f = g ◦ jE . So g(x/1) = f (x)/1 for all x ∈ E. The uniqueness is clear. 2.4.3 Proposition. Let F be a submodule of E. The homomorphism S −1 F → S −1 E induced by the canonical injection F → E in 2.4.2 is injective, and the S −1 A-modules S −1 (E/F ) and S −1 E/S −1 F are isomorphic. Proof. Let x/s be in the kernel of the homomorphism S −1 F → S −1 E. There exists t ∈ S such that tx = 0 which in turn implies that x/s = 0 in S −1 F .
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The map f : E/F → S −1 E/S −1 F defined by sending x+F to x/1+S −1 F is well-defined and we verify easily that it is a homomorphism of A-modules. By 2.4.1, there exists a unique homomorphism of S −1 A-modules g : S −1 (E/F ) → S −1 E/S −1 F , (x + F )/s → x/s + S −1 F. By construction, g is surjective. Let x ∈ E, s ∈ S be such that x/s ∈ S −1 F . Then there exists t ∈ S such that tx ∈ F . So we have t(x + F ) = 0 in E/F and hence (x + F )/s = 0 in S −1 (E/F ). Thus g is injective. 2.4.4 Proposition. Let E be an A-module, jE : E → S −1 E the canonical homomorphism and a an ideal of A. We have (S −1 a)(S −1 E) = a(S −1 E) = (S −1 a)jE (E). Proof. Let a ∈ a, s, t ∈ S and x ∈ E. We have: (a/s).(x/t) = (a/1).(x/st) = (a/st)(x/1). So the proposition follows. 2.4.5 Remarks. 1) Proposition 2.4.3 justifies the notation S −1 a of 2.3.7. 2) Let M be an A-module, then we see easily that S −1 M and S −1 A ⊗A M are isomorphic S −1 A-modules where we consider S −1 A as an A-algebra via the canonical map A → S −1 A.
2.5 Radical of an ideal 2.5.1 Definition. Let a be an ideal of A. We define the radical of a, √ denoted by a, to be the√set of elements a ∈ A for which there exists n ∈ N∗ such that an ∈ a. If a = a, then we say that a is a radical ideal. 2.5.2 Let a be an ideal of A, then by√using the formula of binomial expansion, we obtain easily that its radical a is an ideal of A. Note that the nilradical of A is the radical of the ideal {0}. 2.5.3 Proposition. Let a, b be ideals of A. √ (i) We have a = A if and only if a = A. Further, √ √ √ √ √ √ √ √ √ √ a = a, a + b = a + b, ab = a b = a ∩ b. √ √ ∗ (ii) If a is finitely generated, then ( a)n ⊂ a for √ nsome n ∈ N . ∗ (iii) If p is a prime ideal of A and n ∈ N , then p = p. Proof.√ (i) Let last equality.√From√ab ⊂ a ∩ b ⊂ a, we deduced √ us prove√the √ ∗ that ab ⊂ a ∩ b ⊂ a ∩ b. Let a ∈ a ∩ b, then there √ exist n, p ∈ N n p n+p such that a ∈ a and a ∈ b.√Hence a ∈ ab and so a ∈ ab as required. Aap . There exists m ∈ N∗ such (ii) Let us suppose that a = Aa1 + · · · + √ m that ai ∈ a for all 1 i p. It√follows that ( a)n ⊂ a if n > mp. √ √ √ (iii) We have p ⊂ p and pn = p by (i). Now if a ∈ p, then there ∗ n exists n ∈ N such that a ∈ p. Since p is prime, a ∈ p.
2.6 Local rings
19
2.5.4 Proposition. Let a be√an ideal of A. (i) The nilradical √ of A/a is a/a. (ii) The ring A/ a is reduced. (iii) If A/a is reduced, then a contains the nilradical of A. √ a∈A Proof. (i) It is clear that the nilradical of A/a is contained in a/a. Let √ and n ∈ N∗ be such that an ∈ a. Then (a + a)n ⊂ a and hence a + a ∈ a/a. (ii) This is immediate from (i) and 2.5.3. (iii) If A/a is reduced, then the kernel of the canonical surjection A → A/a contains the nilradical of A. √ 2.5.5 Proposition. Let a be an ideal of A. Then a is the intersection of prime ideals of A containing a. In particular, the nilradical of A is the intersection of prime ideals of A. Proof. By 2.5.4, we can replace A by A/a and assume that a = {0}. Denote by q the intersection of prime ideals of A and by n the nilradical of A. Clearly, n ⊂ q. Now suppose that there exists an element a ∈ q which is not nilpotent. Then S = {an ; n ∈ N} is a multiplicative subset of A which does not contain 0. It follows from 2.3.2 that there exists p ∈ Spec A such that p ∩ S = ∅. So a ∈ p which contradicts the fact that a ∈ q.
2.6 Local rings 2.6.1 Definition. The intersection of maximal ideals of A, denoted by rad A, is an ideal called the Jacobson radical of A. 2.6.2 Proposition. Let a ∈ A. Then a ∈ rad A if and only if 1 − ab is invertible for all b ∈ A. Proof. If a ∈ rad A, then ab ∈ rad A. Therefore 1 − ab is not contained in any maximal ideal. So 1 − ab is invertible. If a ∈ rad A, then there exists a maximal ideal m which does not contain a. So m + Aa = A and there exists b ∈ A, c ∈ m such that c + ab = 1. It follows that 1 − ab = c is not invertible. 2.6.3 Proposition. The following conditions are equivalent: (i) The ring A has a unique maximal ideal. (ii) The set A \ rad A is the set of invertible elements of A. (iii) There exists a proper ideal a of A such that any element of A \ a is invertible. Proof. (i) ⇒ (ii) If m is the unique maximal ideal of A, then m = rad A. So (ii) follows. (ii) ⇒ (iii) Take a = rad A. (iii) ⇒ (i) Suppose that a satisfies the hypotheses of (iii). If a ∈ a, then 1 − ab is invertible for all b ∈ A. So by 2.6.2, a ⊂ rad A. Now let b be an ideal of A containing a. If b = a, then there exists b ∈ b \ a, which is invertible. This implies that b = A. So a = rad A is the unique maximal ideal of A.
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2.6.4 Definition. A ring A satisfying the conditions of 2.6.3 is called a local ring. If A is a local ring with maximal ideal mA , we call A/mA the residual field of A. A homomorphism u : A → B of local rings is said to be local if u(mA ) ⊂ mB . 2.6.5 Remark. Let A be a local ring with maximal ideal m. Then the rings A and Am are canonically isomorphic by 2.3.3 and 2.6.3. 2.6.6 Proposition. Let p ∈ Spec A. Then Ap is a local ring with maximal ideal pAp , and the residual field is canonically isomorphic to the field of fractions of A/p. Proof. Since Ap \pAp consists of invertible elements, the first assertion follows by 2.6.3. Let f : A → A/p be the canonical projection. Then f (A \ p) is the set A/p \ {0}. So the other assertions follow from 2.3.4 and 2.3.10. 2.6.7 Proposition. (Nakayama’s Lemma) Let a be an ideal of A and E be a finitely generated A-module such that aE = E. Then there exists f ∈ 1 + a such that f E = 0. Proof. Denote by Mr (A) the ring of r-by-r matrices with coefficients in A, and by Ir the identity matrix in Mr (A). Let {e1 , . . . , er } be a system of generators of E. Our hypothesis says that we can write for 1 i r: ei = ai1 e1 + · · · + air er where aij ∈ a. Set U = [aij ] ∈ Mr (A) and Λ = Ir − U . Then ⎛ ⎞ ⎛ ⎞ e1 0 ⎜ .. ⎟ ⎜ .. ⎟ Λ⎝ . ⎠ = ⎝ . ⎠. er
0
= (det Λ)Ir . If Λ is the transpose of the matrix of cofactors of Λ, then ΛΛ This implies that (det Λ)ej = 0 for 1 j r. Since det Λ is of the form 1 + f with f ∈ a, our result follows. 2.6.8 Corollary. Let a ⊂ rad A be an ideal of A, E a finitely generated A-module and F a submodule of E. (i) If aE = E, then E = {0}. (ii) If E = aE + F , then E = F . Proof. (i) By 2.6.7, there exists f ∈ a such that (1 + f )E = {0}. But a is contained in rad A, so 1 + f is invertible and hence E = {0}. (ii) Since E/F = a(E/F ), the result follows from (i). 2.6.9 Corollary. Let A be a local ring with maximal ideal m, E a finitely generated A-module. Let e1 , . . . , en ∈ E be such that their images in the A/mmodule E/mE form a system of generators. Then e1 , . . . , en generate the Amodule E.
2.7 Noetherian rings and modules
21
Proof. Let F be the submodule of E generated by e1 , . . . , en . We have E = mE + F . Since m = rad A by 2.6.3, the result follows from 2.6.8. 2.6.10 Let p ∈ Spec A, S = A \ p and E an A-module. We shall denote by Ep the Ap -module S −1 E. Proposition. Let m be a maximal ideal of A and E an A-module. Suppose that there exists an ideal a of A satisfying the following conditions: (i) The only maximal ideal containing a is m. (ii) We have aE = {0}. Then the canonical homomorphism E → Em is bijective. Proof. Condition (i) implies that A/a is local with maximal ideal m/a. Condition (ii) allows us to consider E as an A/a-module. If s ∈ A \ m, then s + a is invertible in A/a. Thus the map µs : E → E defined by x → sx is bijective. If x, y ∈ E satisfy x/1 = y/1, then there exists s ∈ A \ m such that s(x − y) = 0 = µs (x − y). So x = y. Now let y ∈ E and s ∈ A \ m. There exists x ∈ E such that µs (x) = sx = y and so x/1 = y/s. 2.6.11 Corollary. Let m be a maximal ideal of A, E an A-module and k ∈ N be such that mk E = {0}. Then the canonical homomorphism E → Em is bijective. Proof. This is clear if k = 0. For k > 0, it follows from 2.2.6 and 2.6.10. 2.6.12 Proposition. Let m be a maximal ideal of A, E an A-module and k ∈ N. The canonical homomorphism E → Em /mk Em is surjective. Its kernel is mk E and it induces an isomorphism between E/mk E and Em /mk Em . Proof. By 2.6.10, the canonical homomorphism E/mk E → (E/mk E)m is bijective. Further, we can identify canonically (E/mk E)m and Em /(mk E)m by 2.4.3, and (mk E)m = mk Em by 2.4.4. Thus the map E/mk E → Em /mk Em given by x + mk E → x/1 + mk Em is an isomorphism.
2.7 Noetherian rings and modules 2.7.1 Proposition. For any A-module E, the following conditions are equivalent: (i) Any submodule of E is finitely generated. (ii) Any ascending sequence of submodules of E is stationary. (iii) Any non-empty family of submodules of E has a maximal element. Proof. This is straightforward. 2.7.2 Definition. An A-module is called Noetherian if it satisfies the conditions of 2.7.1. The ring A is said to be Noetherian if A itself is a Noetherian A-module.
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2.7.3 following (i) E (ii) F
Proposition. Let E be an A-module and F a submodule of E. The conditions are equivalent: is Noetherian. and E/F are Noetherian.
Proof. Again this is straightforward. 2.7.4 Corollary. (i) Any finite direct sum of Noetherian modules is Noetherian. (ii) Let F, G be submodules of an A-module E such that E = F + G. If F and G are Noetherian, then E is Noetherian. Proof. This is a direct consequence of 2.7.3.
2.7.5 Proposition. Let us suppose that A is Noetherian. For any Amodule E, the following conditions are equivalent: (i) E is Noetherian. (ii) E is finitely generated. Proof. The implication (i) ⇒ (ii) is clear. Suppose that (ii) is verified, and let x ∈ E. The submodule Ax is isomorphic to a quotient of A, so it is Noetherian by 2.7.3. Now by applying 2.7.4 (ii), any finitely generated submodule of E is Noetherian. 2.7.6 Proposition. Let a be an ideal of a Noetherian ring A and S a multiplicative subset of A. (i) The nilradical of A is a nilpotent ideal. (ii) The rings A/a and S −1 A are Noetherian. Proof. Assertion (i) follows from 2.5.3. We deduce from 2.7.3 that A/a is Noetherian because the A/a-submodules of A/a are the A-submodules of A/a. Finally S −1 A is Noetherian by 2.3.7 (iii) and (iv). 2.7.7 Theorem. Let a be a radical ideal of a Noetherian ring A. There exist p1 , . . . , pn ∈ Spec A such that: a = p1 ∩ · · · ∩ pn . If we assume that pi ⊂ pj for i = j, then the preceding decomposition is unique up to a permutation. Proof. Let F be the set of proper radical ideals of A which are not finite intersections of prime ideals. Let us suppose that F is not empty. By 2.7.1, there is a maximal element in F, say r. Since r is not prime, there exist a, b ∈ A \ r such that ab ∈ r, and the ideals a = r + Aa, b = r + Ab contain strictly r. We claim that a and b are proper. If
2.7 Noetherian rings and modules
23
a = A, then 1 = xa + r for some x ∈ A and r ∈ r. But this would imply that b = xab + rb ∈ r, which is impossible. √ It follows that there√exist prime ideals p1 , . . . , pn , q1 , . . . , qm such that a = p1 ∩ · ·√· ∩ pn√and b = q1 ∩ · · · ∩ qn . Set s = a ∩ b. We have r ⊂ s and if u ∈ s, there exists n ∈ N∗ such that un ∈ a ∩ b. This implies in turn that u2n ∈ ab ⊂ r, and so u ∈ r. Hence r = s which is absurd since s ∈ F. Now suppose that p1 , . . . , pn , q ∈ Spec A be such that p1 ∩ · · · ∩ pn ⊂ q. Then p1 · · · pn ⊂ q, and by 2.2.5 (i), there exists i such that pi ⊂ q. Now the uniqueness of the decomposition follows easily. 2.7.8 Corollary. Let A be a Noetherian ring which is not reduced to {0}. (i) The set of minimal prime ideals is finite. (ii) Suppose that A is an integral domain. Then any proper non-zero ideal a of A contains a non-zero finite product of prime ideals. Proof. (i) Let n be the nilradical of A. By 2.7.7, there exist p1 , . . . , pr ∈ Spec A such that n = p1 ∩ · · · ∩ pr . If q is a minimal prime ideal, then n ⊂ q by 2.5.5. So by 2.2.5 (i), there exists i such that q = pi . Thus any minimal prime ideal is one of the pi ’s.√ . . , qs ∈ Spec A are non-zero. (ii) We have a = q1 ∩ · · · ∩ qs where q1 , . √ Since A is Noetherian, by 2.5.3 (ii), for n large, ( a)n ⊂ a. So (q1 · · · qs )n ⊂ a. The result now follows since A is an integral domain. 2.7.9 Theorem. Let A be a Noetherian ring and n ∈ N∗ . The polynomial ring over A in n variables is a Noetherian ring. Proof. By induction on n, it suffices to prove that A[X] is Noetherian. Further, we can suppose that A is not reduced to {0}. Suppose that there exists an ideal a of A which is not finitely generated. Then it is non-zero and let P1 ∈ a \ {0} be such that its degree is minimal among the non-zero elements of a. Since a = A[X]P1 , there exists P2 ∈ a \ A[X]P1 such that its degree is minimal among the elements of a \ A[X]P1 . Repeating this process, we obtain a sequence of elements (Pn )n1 of polynomials in a such that, for any n ∈ N∗ , we have: Pn+1 ∈ a \ (A[X]P1 + · · · + A[X]Pn ), deg Pn+1 deg Q for all Q ∈ a \ (A[X]P1 + · · · + A[X]Pn ). In particular, deg Pn+1 deg Pn for all n ∈ N∗ . Let an be the leading coefficient of Pn and set bn = Aa1 + · · · + Aan . Then (bn )n1 is an ascending sequence of ideals of A. Since A is Noetherian, there exists n ∈ N∗ such that an+1 ∈ bn . So we have an+1 = λ1 a1 + · · · + λn an for some λ1 , . . . , λn ∈ A. Let Q = Pn+1 −
n i=1
λi X deg Pn+1 −deg Pi Pi .
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We have deg Q < deg Pn+1 and Q ∈ a \ (A[X]P1 + · · · + A[X]Pn ). Thus we have obtained a contradiction.
2.8 Derivations 2.8.1 In the rest of this chapter, k will denote a commutative ring and A will denote a commutative k-algebra. Definition. Let M be an A-module. A derivation of A into M is a map δ : A → M such that for all a, b ∈ A: δ(a + b) = δ(a) + δ(b) , δ(ab) = aδ(b) + bδ(a). Denote by Der(A, M ) the set of such derivations. If A = M , we shall write Der A for Der(A, A), and an element of Der A is called a derivation of A. 2.8.2 Proposition. Let δ ∈ Der(A, M ), x ∈ A and n ∈ N∗ . Then: (i) δ(1) = 0. (ii) δ(xn ) = nxn−1 δ(x). (iii) If x is invertible, then δ(x−1 ) = −x−2 δ(x). Proof. Since δ(1) = δ(1.1) = δ(1) + δ(1), part (i) follows. Part (ii) is a simple induction on n and part (iii) follows directly from (i). 2.8.3 Observe that if δ ∈ Der(A, M ), then the set B of elements a ∈ A such that δ(x) = 0, is a subring of A and δ is B-linear. Conversely, if B is a subring of A such that δ is B-linear, then δ(b) = 0 for all b ∈ B. 2.8.4 Definition. A derivation of A over k into M is a derivation δ of A into M such that δ is k-linear. The set of such derivations will be denoted by Derk (A, M ). If A = M , then we write Derk A for Derk (A, A) and an element of Derk A is called a derivation of A over k. 2.8.5 Remarks. 1) The partial derivations of the ring k[T1 , . . . , Tn ] are k-derivations of k[T1 , . . . , Tn ]. 2) Let δ ∈ Derk (A, M ) and a ∈ A. Then the map A → M defined by b → aδ(b) is a k-derivation of A with values in M . This induces a natural structure of A-module on Derk (A, M ). 3) If B is a commutative A-algebra and δ ∈ Der(A, B), then for b ∈ B, the map A → B given by a → bδ(a) is a derivation of A into B. So Der(A, B) admits a natural structure of B-module. 4) Let B be a commutative k-algebra, u : A → B be a homomorphism of k-algebras and N a B-module. We can consider N as an A-module via u. If δ ∈ Derk (B, N ), then we verify easily that δ ◦ u ∈ Derk (A, N ). The map C(u) : Derk (B, N ) → Derk (A, N ), δ → δ◦u is a homomorphism of A-modules. We claim that ker C(u) = DerA (B, N ), that is, the following sequence of Amodules is exact: (2)
0 → DerA (B, N ) → Derk (B, N ) → Derk (A, N ).
2.9 Module of differentials
25
Clearly DerA (B, N ) ⊂ ker C(u) since δ(u(a)) = u(a)δ(1) = 0. Conversely, if δ ∈ ker C(u), then δ(u(a)b) = u(a)δ(b). So ker C(u) = DerA (B, N ). 2.8.6 Proposition. Let A be an integral domain, F the quotient field of A and N a F -vector space (considered also as an A-module). Then any derivation of A over k into N can be extended in a unique way to a derivation of F over k into N . Proof. Let δ ∈ Derk (A, N ) and a, b ∈ A, c, d ∈ A \ {0} be such that ad = bc. Then aδ(d) + dδ(a) = bδ(c) + cδ(b). It follows that: d2 bδ(a) − d2 aδ(b) = bd2 δ(a) − bcdδ(b) = b2 dδ(c) + bcdδ(b) − abdδ(d) − bcdδ(b) = b2 dδ(c) − b2 cδ(d). Define ∆ : F → N as follows, if α = a/b, then: ∆(α) = δ(a)/b − aδ(b)/b2 . The preceding computation implies that ∆ is well-defined and we verify easily that ∆ ∈ Derk (F, N ). Now if ∆ ∈ Derk (F, N ) extends δ to F , then by applying 2.8.2 (iii): ∆ (a/b) = a∆ (1/b) + ∆ (a)/b = ∆ (a)/b − a∆ (b)/b2 = δ(a)/b − aδ(b)/b2 . So ∆ = ∆ and we have uniqueness.
2.9 Module of differentials 2.9.1 Recall that A is a commutative k-algebra. If M, N are A-modules, then we shall write HomA (M, N ) the set of homomorphisms of A-modules from M to N . By a universal derivation for A over k , we mean a pair (Ω, d) verifying the following properties: (i) Ω is an A-module and d ∈ Derk (A, Ω). (ii) For any A-module M and any δ ∈ Derk (A, M ), there exists a unique f ∈ HomA (Ω, M ) such that f ◦ d = δ. Lemma. Let (Ω, d) and (Γ, D) be universal derivations for A over k. (i) There exists a unique λ ∈ HomA (Ω, Γ ) such that D = λ ◦ d. Further, λ is an isomorphism. (ii) Let M be an A-module. If f ∈ HomA (Ω, M ), then f ◦ d ∈ Derk (A, M ) and the map ψ : HomA (Ω, M ) → Derk (A, M ) , f → f ◦ d is an isomorphism of A-modules.
26
2 Rings and modules
Proof. (i) By definition, there exist λ ∈ HomA (Ω, Γ ), µ ∈ HomA (Γ, Ω) such that D = λ ◦ d and d = µ ◦ D. So d = µ ◦ λ ◦ d and D = λ ◦ µ ◦ D. By the uniqueness property, µ ◦ λ = idΩ and λ ◦ µ = idΓ . (ii) That f ◦ d ∈ Derk (A, M ) and ψ is a surjective homomorphism of Amodules are straightforward verifications. The injectivity of ψ follows from the uniqueness property. 2.9.2 The tensor product A ⊗k A has a natural structure of (A, A)bimodule given by a.(x ⊗ y).b = (ax) ⊗ (by) for all a, b, x, y ∈ A. Since A is commutative, we can identify the above structure of (A, A)-bimodule with the structure of A ⊗k A-module induced by the ring structure on A ⊗k A, namely: a.(x ⊗ y).b = (a ⊗ b)(x ⊗ y) Thus A ⊗k A is a left (resp. right) A-module via the ring homomorphism A → A ⊗k A, a → a ⊗ 1 (resp. a → 1 ⊗ a). Denote by m : A ⊗k A → A the k-linear map m(x ⊗ y) = xy. The kernel J of m is an ideal of A ⊗k A and the ring A ⊗k A/J is isomorphic to A. Let δA : A → A ⊗k A be the map given by x → x ⊗ 1 − 1 ⊗ x. Clearly δA (x) ∈ J for any x ∈ A. Lemma. As a left (or right) A-module, J is generated by the image of δA . Proof. Let x1 , . . . , xr , y1 , . . . , yr ∈ A. Then r x i=1 i yi = 0. Thus the lemma follows since r
i=1
xi ⊗ yi = −
r
r
xi δA (yi ) =
i=1
i=1 r
xi ⊗ yi ∈ J implies that δA (xi )yi .
i=1
2.9.3 Set Ωk (A) = J /J 2 . It is clear that Ωk (A) is an A ⊗k A-module. For a, b, x ∈ A, we have: abδA (x) − bδA (x)a = bδA (a)δA (x) ∈ J 2 . In view of 2.9.2, the structures of left and right A-module on Ωk (A) induced by the one on A ⊗k A are the same. We can therefore talk about the A-module ahler) differentials of A over k. Ωk (A), that we call the module of (K¨ If x ∈ A, denote by dA/k (x) (or dx if there is no confusion), the image of δA (x) in Ωk (A). We say that dx is the differential of x. By 2.9.2, the set of dx, x ∈ A, is a system of generators of the A-module Ωk (A). Theorem. (i) We have dA/k ∈ Derk (A, Ωk (A)). (ii) The pair (Ωk (A), dA/k ) is a universal derivation for A over k.
2.9 Module of differentials
27
Proof. (i) Let a, b ∈ A. We have: δA (ab) = aδA (b) + δA (a)b = aδA (b) + bδA (a) + α with α ∈ J 2 . So dA/k ∈ Derk (A, Ωk (A)). (ii) Let M be an A-module and δ ∈ Derk (A, M ). Define ϕ : A ⊗k A → M , a ⊗ b → bδ(a). If a, x ∈ A, then ϕ(δA (x)) = δ(x) − xδ(1) = δ(x), ϕ(aδA (x)) = δ(ax) − xδ(a) = aδ(x). So ϕ|J ∈ HomA (J , M ). On the other hand: ϕ(δA (a)δA (x)) = δ(ax) − aδ(x) − xδ(a) = 0. We deduce that there exists f ∈ HomA (Ωk (A), M ) such that δ = f ◦ dA/k . Uniqueness follows from 2.9.2. 2.9.4 Example. Let A = k[(Ti )i∈I ] be the ring of polynomials over k in the variables Ti , i ∈ I. Denote dA/k by d. We shall show that Ωk (A) is a free A-module with basis (dTi )i∈I . Let P ∈ A. Since d ∈ Derk (A, Ωk (A)), we have: dP =
∂P dTi . i∈I ∂Ti
It follows that (dTi )i∈I is a set of generators of Ωk (A). Now let M be a free A-module with basis (ei )i∈I . Let δ ∈ Derk (A, M ) be defined as follows: for P ∈ A: ∂P ei . δP = i∈I ∂Ti Let f ∈ HomA (Ωk (A), M ) be such that δ = f ◦ d. Then ei = δ(Ti ) = f (dTi ) for all i ∈ I. So the family (dTi )i∈I is linearly independent over A. 2.9.5 Let u : A → B be a homomorphism of commutative k-algebras. 1) By 2.9.1 and 2.9.3, for any A-module M , there exists an isomorphism of A-modules: ϕk,A : HomA (Ωk (A), M ) → Derk (A, M ) , f → f ◦ dA/k . 2) We have dB/k ◦ u ∈ Derk (A, Ωk (B)). So there exists a unique A-module homomorphism Ω(u) : Ωk (A) → Ωk (B) such that Ω(u) ◦ dA/k = dB/k ◦ u. 3) Let iu : Ωk (A) → Ωk (A) ⊗A B be the canonical homomorphism α → α ⊗ 1. Since Ωk (B) is a B-module and Ωk (A) ⊗A B is a B-module via the right action, there exists a B-linear map Ω0 (u) : Ωk (A) ⊗A B → Ωk (B)
28
2 Rings and modules
such that Ω(u) = Ω0 (u) ◦ iu . 4) Note that dB/A is a derivation over A, so it is a derivation over k. It follows that there exists a unique B-linear map Ωu such that dB/A = Ωu ◦dB/k . Let N be a B-module. From the preceding discussion, we have the following commutative diagram of B-modules: Hom(Ωu ,idN )
(3)
HomB (ΩB (A), N ) −−−−−−−−−→ HomB (Ωk (B), N ) ⏐ ⏐ ⏐ϕk,B ⏐ ϕA,B −−−−→
DerA (B, N )
Derk (B, N )
ju
where ju is the canonical injection and Hom(Ωu , idN ) is the map h → h ◦ Ωu . Proposition. The following sequence of B-linear maps is exact: (4)
Ω0 (u)
Ω
u Ωk (A) ⊗A B −−−−→ Ωk (B) −−−− → ΩA (B) −−−−→ 0.
Proof. It suffices to prove that, for any B-module N , the following sequence is exact: h
u u 0 −→ HomB (ΩA (B), N ) −→ HomB (Ωk (B), N ) −→ HomB (Ωk (A) ⊗A B, N )
where hu = Hom(Ωu , idN ) and u = Hom(Ω0 (u), idN ). We have the following commutative diagram: Hom(Ω0 (u),idN )
(5)
HomB (Ωk (B), N ) −−−−−−−−−−→ HomB (Ωk (A) ⊗A B, N ) ⏐ ⏐ ⏐ϕk,A ◦su ϕk,B ⏐ Derk (B, N )
−−−−→
Derk (A, N )
C(u)
where C(u) is as defined in 2.8.5, and su is the canonical isomorphism: su = Hom(iu , idN ) : HomB (Ωk (A) ⊗A B, N ) → HomA (Ωk (A), N ). The vertical arrows of diagram (5) are isomorphisms, and therefore, in view of the diagrams (3) and (5), the result follows from the fact that the sequence (2) in 2.8.5 is exact. 2.9.6 Proposition. Suppose that A is an integral domain. Let B be its field of fractions and u : A → B the canonical injection. Then Ω0 (u) is an isomorphism from Ωk (A) ⊗A B to Ωk (B). Proof. Since the vertical arrows of diagram (5) are isomorphisms, it suffices to prove that the map C(u) : Derk (B, N ) → Derk (A, N ) is bijective for any B-vector space N . So the proposition follows from 2.8.6.
2.9 Module of differentials
29
2.9.7 Corollary. Let K be a commutative field and E = K((Ti )i∈I ) the field of fractions of the ring of polynomials in the variables (Ti )i∈I . Then the family (dTi )i∈I is a basis for the E-vector space ΩK (E). Proof. This is an immediate consequence of 2.9.4 and 2.9.6. 2.9.8 Let kn [T] = k[T1 , . . . , Tn ], P1 , . . . , Pm ∈ kn [T], a the ideal generated by P1 , . . . , Pn and A = kn [T]/a. For 1 i n, let ti be the image of Ti in A. Set t = (t1 , . . . , tn ). Recall that the A-module Ωk (A) is generated by dA/k (ti ), 1 i n. Let (e1 , . . . , en ) be the canonical basis of An and u : An → Ωk (A) the surjective homomorphism given by u(ei ) = dA/k (ti ) for 1 i n. Denote by M the submodule of An generated by ∂Pj ∂Pj (t)e1 + · · · + (t)en ∂T1 ∂Tn where 1 j m. The following lemma extends the result of 2.9.4. Lemma. The kernel of u is M . In other terms, the A-modules Ωk (A) and An /M are isomorphic. Proof. Let α : kn [T] → A and β : An → An /M be the canonical surjections. Since dA/k ∈ Derk (A, Ωk (A)), if Q ∈ kn [T], we have: dA/k (Q(t)) =
∂Q ∂Q (t)dA/k (t1 ) + · · · + (t)dA/k (tn ). ∂T1 ∂Tn
In particular, by taking Q to be one of the Pi ’s, we obtain that M ⊂ ker u. So there exists a unique µ ∈ HomA (An /M, Ωk (A)) such that µ ◦ β = u. Let θ : kn [T] → An be the k-linear map defined by: θ(Q) =
∂Q ∂Q (t)e1 + · · · + (t)en . ∂T1 ∂Tn
We have a ⊂ ker β ◦ θ, and hence there exists a unique ϕ ∈ HomA (A, An /M ) such that ϕ ◦ α = β ◦ θ. It is obvious that ϕ ∈ Derk (A, An /M ). It follows from 2.9.1 and 2.9.5 that there exists a unique ψ ∈ HomA (Ωk (A), An /M ) such that ϕ = ψ ◦dA/k . We obtain that ψ(dA/k (ti )) = ψ ◦dA/k ◦α(Ti ) = β ◦θ(Ti ) = β(ei ) for 1 i n. Since µ◦β(ei ) = dA/k (ti ) and the β(ei )’s generate the A-module An /M , the lemma follows.
References • [8], [10], [11], [52], [56].
3 Integral extensions
In this chapter, we study the notion of integral extensions. In particular, we prove the Going Up Theorem concerning the extension of prime ideals. All rings considered in this chapter are assumed to be commutative. Let A be a subring of a ring B. For x1 , . . . , xn ∈ B, we denote by A[x1 , . . . , xn ] the subring of B generated by A and x1 , . . . , xn .
3.1 Integral dependence 3.1.1 For n ∈ N, denote by Mn (A) the A-algebra of n-by-n matrices with coefficients in A. If M ∈ Mn (A), then we write χM (X) its characteristic polynomial. So χM (X) = det(XIn − M ) where In is the identity matrix. 3.1.2 Definition. An element x ∈ B is said to be integral over A if there exist n ∈ N∗ and a0 , a1 , . . . , an−1 ∈ A such that: xn + an−1 xn−1 + · · · + a1 x + a0 = 0. Such a relation is called an integral equation or equation of integral dependence for x over A. 3.1.3 Let C be a commutative A-algebra, f : A → C the canonical Aalgebra homomorphism. If an element c ∈ C is integral over f (A), then we shall say that c is integral over A. 3.1.4 Theorem. Let A be a subring of B and x ∈ B. Then the following conditions are equivalent: (i) x is integral over A. (ii) A[x] is a finitely generated A-module. (iii) There exists a subring C of B, containing A[x] such that C is a finitely generated A-module. Proof. (i) ⇒ (ii) Let xn +an−1 xn−1 +· · ·+a1 x+a0 = 0 be an integral equation for x over A. Then it is clear that A[x] is generated by 1, x, . . . , xn−1 . (ii) ⇒ (iii) This is obvious.
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3 Integral extensions
(iii) ⇒ (i) Let {y1 , . . . , yn } be a system of generators of the A-module C. For 1 i n, there exist aij ∈ A, 1 j n, such that: xyi = ai1 y1 + · · · + ain yn . Set M = [aij ] ∈ Mn (A) and for k ∈ N∗ , let M k = [aijk ]. Then we have: xk yi = ai1k y1 + · · · + aink yn . Now χM (M ) = 0 and so χM (x)yi = 0 for 1 i n. Thus χM (x)u = 0 for any u ∈ C. By setting u = 1, we obtain an integral equation for x over A. 3.1.5 Corollary. (i) Let x1 , . . . , xn ∈ B. Suppose that x1 is integral over A and that for 2 i n, xi is integral over A[x1 , . . . xi−1 ]. Then A[x1 , . . . , xn ] is a finitely generated A-module. (ii) The set of elements of B integral over A is a subring of B which contains A. Proof. (i) Let us prove this by induction on n. For n = 1, the result follows from 3.1.4. Suppose that n 2, then by induction hypothesis, D = A[x1 , . . . , xn−1 ] is a finitely generated A-module. Let {y1 , . . . , yr } be a system of generators of the A-module D. Now xn is integral over D, so D[xn ] is a finitely generated D-module. Let {z1 , . . . , zs } be a system of generators of the D-module D[xn ]. Then clearly, {yi zj ; 1 i r, 1 j s} is a system of generators of the A-module A[x1 , . . . , xn ]. (ii) Let x, y ∈ B be integral over A. Since xy, x + y ∈ A[x, y], the result follows from (i) and 3.1.4. 3.1.6 Definition. Let A be a subring of B. The set C of elements of B integral over A is called the integral closure of A in B. By 3.1.5, C is a subring of B which contains A. If C = B, we shall say that B is integral over A. If C = A, we shall say that A is integrally closed in B. 3.1.7 Proposition. Let A and D be subrings of a ring B. (i) Suppose that A ⊂ D. If D is integral over A, and B is integral over D, then B is integral over A. (ii) If C is the integral closure of A in B, then C is integrally closed in B. Proof. Let x ∈ B and xn + dn−1 xn−1 + · · · + d1 x + d0 = 0 be an integral equation of x over D. By 3.1.5, the A-module A[d0 , . . . , dn−1 , x] is finitely generated. So x is integral over A. This proves (i), and part (ii) follows. 3.1.8 Proposition. Let b be an ideal of B, a = b∩A and S a multiplicative subset of A. Let us suppose that B is integral over A. Then: (i) B/b is integral over A/a. (ii) S −1 B is integral over S −1 A.
3.2 Integrally closed domains
33
Proof. (i) Let us denote x + b by x, and identify A/a as a subring of B/b. If xn + an−1 xn−1 + · · · + a1 x + a0 = 0 is an integral equation of x over A, then xn + an−1 xn−1 + · · · + a1 x + a0 = 0 is an integral equation of x over A/a. (ii) By 2.3.6, we identify S −1 A as a subring of S −1 B. Let x/s ∈ S −1 B. If n x + an−1 xn−1 + · · · + a1 x + a0 = 0 is an integral equation of x over A, then (x/s)n + (an−1 /s)(x/s)n−1 + · · · + (a1 /sn−1 )(x/s) + a0 /sn = 0 is an integral equation of x/s over S −1 A. 3.1.9 Proposition. Let S be a multiplicative subset of A and C the integral closure of A in B. Then the integral closure of S −1 A in S −1 B is S −1 C. In particular, if A is integrally closed in B, then S −1 A is integrally closed in S −1 B. Proof. By 3.1.8, the integral closure of S −1 A in S −1 B contains S −1 C. Now, if x/s ∈ S −1 B is integral over S −1 A, then there exist a0 , . . . , an−1 ∈ A, t0 , . . . , tn−1 ∈ S such that xn /sn + (an−1 /tn−1 )(xn−1 /sn−1 ) + · · · + (a1 /t1 )(x/s) + a0 /t0 = 0. Set b = t0 · · · tn−1 x and u = (st0 · · · tn−1 )n , then (bn + sn−1 an−1 bn−1 + · · · + s1 a1 b + s0 a0 )/u = 0 where s0 , . . . , sn−1 ∈ S. This implies that there exists v ∈ S such that: v(bn + sn−1 an−1 bn−1 + · · · + s1 a1 b + s0 a0 ) = 0. From this, we deduce that vb = vt0 · · · tn−1 x ∈ C and so x/s ∈ S −1 C.
3.2 Integrally closed domains 3.2.1 Lemma. Let E be an A-module. Then the following conditions are equivalent: (i) The module E is {0}. (ii) For any prime ideal p of A, Ep = {0}. (iii) For any maximal ideal m of A, Em = {0}. Proof. The implications (i) ⇒ (ii) ⇒ (iii) are obvious. Suppose now that (iii) is satisfied. If there exists x ∈ E \ {0}, then a = {a ∈ A; ax = 0} is a proper ideal of A. Let m be a maximal ideal containing a. Since Em = {0}, we have x/1 = 0, or equivalently, there exists s ∈ A \ m such that sx = 0 which implies that s ∈ a ⊂ m. We have therefore a contradiction. 3.2.2 Lemma. Let A be an integral domain. Then Am . A= m∈Spm A
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3 Integral extensions
Proof. For any m ∈ Spm A, Am is a subring of its quotient field K by 2.3.5. The intersection B of the Am ’s is therefore a subring of K containing A. Let m ∈ Spm A, S = A \ m. We have Am ⊂ S −1 B ⊂ Am . So Am = S −1 B. It follows from 2.4.3 that S −1 (B/A) = {0}. So by 3.2.1, B/A = {0}. 3.2.3 Definition. (i) Let A be an integral domain and K its quotient field. By the integral closure of A, we mean the integral closure of A in K. (ii) We say that A is integrally closed domain if A is an integral domain, and it is equal to its integral closure. 3.2.4 Lemma. Let B be an integral domain and (A i )i∈I a family of subrings which are integrally closed domains. Then A = i∈I Ai is an integrally closed domain. Proof. Let K, H, Ki be the field of fractions of A, B, Ai respectively. We have K ⊂ Ki ⊂ H. If x ∈ K is integral over A, then it is integral over Ai for all i ∈ I. Thus x ∈ Ai for all i ∈ I. 3.2.5 Proposition. Let A be an integral domain. The following conditions are equivalent: (i) The ring A is an integrally closed domain. (ii) For any p ∈ Spec A, Ap is an integrally closed domain. (iii) For any m ∈ Spm A, Am is an integrally closed domain. Proof. We have (i) ⇒ (ii) by 3.1.9. The implication (ii) ⇒ (iii) is obvious. Finally, (iii) ⇒ (i) follows from 3.2.2 and 3.2.4. 3.2.6 Lemma. Let A be an integrally closed domain, K its quotient field and P, Q ∈ K[X] monic polynomials such that P Q ∈ A[X]. Then P, Q ∈ A[X]. Proof. Let L be an algebraic closure of K. Then P (X) = (X − α1 ) · · · (X − αm ) , Q(X) = (X − β1 ) · · · (X − βn ), with α1 , . . . , αm , β1 , . . . , βn ∈ L. The αi ’s and the βj ’s are integral over A because they are roots of P Q ∈ A[X]. Since the coefficients of P and Q are polynomials in the αi ’s and the βi ’s with coefficients in A, it follows from 3.1.5 that they are integral over A. Hence P, Q ∈ A[X]. 3.2.7 Lemma. If K is a commutative field, then K[X] is an integrally closed domain. Proof. Let P, Q ∈ K[X] \ {0} be relatively prime and let us suppose that P/Q is integral over K[X]. Let (P/Q)n + Fn−1 (P/Q)n−1 + · · · + F0 = 0 be an integral equation of P/Q over K[X]. Suppose further that n is minimal. So F0 = 0, and we have:
3.3 Extensions of prime ideals
35
P n + Fn−1 P n−1 Q + · · · + F0 Qn = 0. It follows that Q divides P n and so Q is invertible in K[X] because P, Q are relatively prime. Thus Q ∈ K \ {0} which implies that P/Q ∈ K[X]. 3.2.8 Theorem. Let A be an integrally closed domain, then A[X] is also an integrally closed domain. Proof. Denote by K and L, the quotient field of A and A[X] respectively. Let α ∈ L be integral over A[X]. Then it is integral over K[X]. So by 3.2.7, α ∈ K[X]. Let Q = T m + pm−1 T m−1 + · · · + p0 ∈ A[X][T ] be such that Q(α) is an integral equation of α over A[X]. Let s ∈ N be such that deg α < s and set β = α − X s . Then −β is monic and β is a root of Q(T + X s ) = T m + qm−1 T m−1 + · · · + q0 where q0 , . . . , qm−1 ∈ A[X]. Now q0 = Q(X s ) = X sm + pm−1 X s(m−1) + · · · + p0 , and therefore q0 is monic for s large. But β is a root of Q(T + X s ), so: q0 = −β(β m−1 + qm−1 β m−2 + · · · + q1 ) = −βγ, where γ ∈ A[X]. Thus γ is monic in K[X] for s large. Since q0 ∈ A[X], it follows from 3.2.6 that −β ∈ A[X]. Hence α ∈ A[X].
3.3 Extensions of prime ideals 3.3.1 Proposition. Let A be a subring of an integral domain B such that B is integral over A. Then A is a field if and only if B is a field. Proof. Suppose that A is a field. Let b ∈ B\{0} and bn +an−1 bn−1 +· · ·+a0 = 0 an integral equation of b over A. Since B is an integral domain, we can suppose that a0 = 0. Then in Fract(B): n−1 n−2 b−1 = −(a−1 + a−1 + · · · a−1 0 b 0 an−1 b 0 a1 ) ∈ B.
Now let us suppose that B is a field. Let a ∈ A \ {0}, so a−1 ∈ B. Let a + an−1 a−(n−1) + · · · + a0 = 0 be an integral equation of a−1 over A. Then a−1 = −(an−1 + an−2 a + · · · + a0 an−1 ) ∈ A. −n
3.3.2 Corollary. Let A be a subring of a ring B such that B is integral over A and q, q ∈ Spec B. (i) We have q ∈ Spm B if and only if p = q ∩ A ∈ Spm A. (ii) Suppose that q ⊂ q . Then A ∩ q = A ∩ q if and only if q = q . Proof. (i) By 3.1.8, B/q is integral over A/p. So (i) follows from 3.3.1. (ii) Suppose that p = A ∩ q = A ∩ q . Set S = A \ p. Since S ∩ q = S ∩ q = ∅, by 2.3.8, r = S −1 q and r = S −1 q are prime ideals of S −1 B such that r ⊂ r . Since p ⊂ q ⊂ q , we have S −1 p ⊂ r ⊂ r . Hence S −1 p ⊂ r ∩ Ap ⊂ r ∩ Ap . But Ap is a local ring with maximal ideal S −1 p, so S −1 p = r ∩ Ap = r ∩ Ap . It follows from (i) that r, r ∈ Spm S −1 B. Since r ⊂ r , we have equality and we conclude by 2.3.7 that q = q .
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3 Integral extensions
3.3.3 Theorem. (Going Up Theorem) Let A be a subring of a ring B such that B is integral over A. (i) If p ∈ Spec A, then there exists q ∈ Spec B lying above p. (ii) Let p1 , p2 ∈ Spec A and q1 ∈ Spec B lying above p1 . Suppose that p1 ⊂ p2 . Then there exists q2 ∈ Spec B lying above p2 such that q1 ⊂ q2 . Proof. (i) Let S = A \ p, i : A → Ap , j : B → S −1 B the canonical homomorphisms, λ : A → B and µ : Ap → S −1 B the canonical injections. By 2.3.6, we have j ◦ λ = µ ◦ i. By 3.1.8, S −1 B is integral over Ap . Let n ∈ Spm S −1 B, then by 3.3.2, n ∩ Ap ∈ Spm Ap . Since Ap is a local ring with maximal ideal S −1 p, we have S −1 p = n ∩ Ap . So q = j −1 (n) ∈ Spec B and q ∩ A = i−1 (S −1 p) = p. (ii) By 3.1.8, B = B/q1 is integral over A = A/p1 . Denote by p : A → A and q : B → B the canonical surjections, λ : A → B and µ : A → B the canonical injections. We have q ◦ λ = µ ◦ p. Now p(p2 ) ∈ Spec A . It follows from (i) that there exists q2 ∈ Spec B such that A ∩ q2 = p(p2 ). So q2 = q −1 (q2 ) ∈ Spec B satisfies q1 ⊂ q2 and A ∩ q2 = p−1 (A ∩ q2 ) = p2 . 3.3.4 Theorem. Let A be a subring of a ring B such that B is integral over A. Then the following conditions are equivalent: (i) The dimension of A is finite. (ii) The dimension of B is finite. Further, if these conditions are satisfied, then dim A = dim B. Proof. By 3.3.3, any chain of length n of prime ideals of A gives a chain of length n of prime ideals of B. So dim A dim B. Conversely, by using 3.3.2 (ii), we obtain that a chain of length m of prime ideals of B gives a chain of length m of prime ideals of A, so dim B dim A. 3.3.5 Let A be a subring of a ring B. We say that B is a finite A-algebra if B is a finitely generated A-module. By 3.1.4, a finite A-algebra is integral over A. Conversely, by 3.1.5, a finitely generated A-algebra B, integral over A is a finite A-algebra. In general, let C be an A-algebra and f : A → C the ring homomorphism inducing the A-algebra structure on C. We say that C is a finite A-algebra if C is a finite f (A)-algebra. 3.3.6 Lemma. Let K be a subring of A. Suppose that K is a field and A is integral over K. (i) We have Spm A = Spec A. (ii) If A is a finite K-algebra, then Spec A is a finite set. Proof. (i) By 3.3.4, dim A = 0. So Spec A = Spm A. (ii) Since Spec A = Spm A, any prime ideal is a minimal prime ideal. If A is a finite K-algebra, then A is Noetherian by 2.7.5. So the result follows from 2.7.8 (i).
3.3 Extensions of prime ideals
37
3.3.7 Proposition. Let A be a subring of a ring B and p ∈ Spec A. If B is a finite A-algebra, then the set of prime ideals lying above p is finite. Proof. Let S = A \ p. In view of 2.3.9, we can replace A, B, p by S −1 A, S −1 B (which is a finite S −1 A-algebra) and S −1 p. So A is now a local ring with maximal ideal p. Next, replace A and B by A/pA and B/pB so that we can apply 3.3.6 since A is now a field and B a finite A-algebra.
References • [10], [11], [52], [56].
4 Factorial rings
All rings considered in this chapter are commutative. For a ring A, we shall denote by U (A) the group of invertible elements of A.
4.1 Generalities 4.1.1 Let A be a ring. If a ∈ A, then we shall denote the ideal Aa by (a). An ideal of this form is called principal. Let us denote by PA the set of ∗ the set of non-zero principal ideals of A. These principal ideals of A, and PA sets are partially ordered by inclusion. 4.1.2 Let us suppose that A is an integral domain. Let a, b ∈ A. Then we say that a divides b or b is a multiple of a if there exists c ∈ A such that ac = b. We shall write a | b (resp. a b) if a divides (resp. does not divide) b. It is clear that a | b ⇔ (b) ⊂ (a). Note that 0 divides b if and only if b = 0. Further, since A is an integral domain, c is unique when a = 0. We shall call c the quotient of b by a. 4.1.3 Proposition. Let A be an integral domain. The following conditions are equivalent for a, b ∈ A: (i) There exists u ∈ U (A) such that b = au. (ii) a | b and b | a. (iii) (a) = (b). When these conditions are verified, we say that a and b are equivalent, and we shall write a ∼ b. Proof. This is straightforward. 4.1.4 Definition. An element a in an integral domain A is called irreducible if the following conditions are satisfied: (i) a ∈ U (A). (ii) If b, c ∈ A are such that a = bc, then b ∈ U (A) or c ∈ U (A).
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4 Factorial rings
4.1.5 Proposition. Let a be a non-zero element in an integral domain A. If (a) is a prime ideal, then a is an irreducible element of A. Proof. Since (a) is a prime ideal, (a) = A, so a ∈ U (A). Now let b, c ∈ A be such that a = bc. Then b ∈ (a) or c ∈ (a), i.e. a | b or a | c. By 4.1.3, this implies that either c ∈ U (A) or b ∈ U (A). 4.1.6 Definition. Let a = (ai )i∈I be a non-empty family of elements of an integral domain A. Denote by U the family (Aai )i∈I in PA . (i) If U has a least upper bound d in PA , then a generator δ of d is called a greatest common divisor, abbreviated to gcd of a. We shall write: (1)
δ = gcd((ai )i∈I ) =i∈I ai .
(ii) If U has a greatest lower bound d in PA , then a generator µ of m is called a least common multiple, abbreviated to lcm of a. We shall write: (2)
µ = lcm((ai )i∈I ) =i∈I ai .
4.1.7 Remarks. 1) The notations in (1) and (2) are not rigorous since δ (resp. µ) is not unique. However, the ideals d and m are unique. So by 4.1.3, a generator of d (resp. m) is unique up to multiplication by an element in U (A). 2) By convention, when I is empty, we set d = {0}, δ = 0, m = A and µ = 1. 4.1.8 Proposition. Let a = (ai )i∈I , δ, µ be elements of an integral domain A. (i) δ is the gcd of the family a if and only if δ divides each ai , and if any common divisor of the ai ’s is a divisor of δ. (ii) µ is the lcm of the family a if and only if µ is a multiple of each ai , and if any common multiple of the ai ’s is a multiple of δ. Proof. We can assume that I is non-empty. If δ = gcd(a), then (ai ) ⊂ (δ), thus δ divides ai for all i ∈ I. Now if r is a common divisor of the ai ’s, then (ai ) ⊂ (r). It follows from the definition of δ that (δ) ⊂ (r), which implies that r divides δ. Conversely, if δ divides each ai , and if any common divisor r of the ai ’s is a divisor of δ, then (ai ) ⊂ (δ) for all i ∈ I and (δ) ⊂ (r). It follows that δ = gcd(a). This proves (i). The proof of part (ii) is analogue. 4.1.9 Definition. Let a = (ai )i∈I be elements of an integral domain A. (i) We say that the ai ’s are relatively prime if gcd(a) = 1. (ii) We say that the ai ’s are pairwise relatively prime if gcd(ai , aj ) = 1 for all i, j ∈ I, i = j. 4.1.10 Proposition. Let a, b be non-zero elements in an integral domain A. If a is irreducible, then a, b are relatively prime if and only if a does not divide b.
4.2 Unique factorization
41
Proof. If a divides b, then a is a greatest common divisor of a and b. So a, b are not relatively prime since a ∈ U (A). Conversely, if a does not divide b, then any common divisor d of a and b must be invertible since it divides a. It follows that a and b are relatively prime.
4.2 Unique factorization 4.2.1 Definition. A commutative ring A is called factorial (or a unique factorization domain) if it is an integral domain and if it verifies the following conditions: (i) For any a ∈ A \ {0}, there exist m ∈ N, p1 , . . . , pm irreducible elements in A, and u ∈ U (A) such that: (3)
a = up1 · · · pm .
(ii) Let m, n ∈ N, p1 , . . . , pm , q1 , . . . , qn irreducible elements in A and u, v ∈ U (A) be such that up1 · · · pm = vq1 · · · qn . Then m = n and there exists a permutation σ of the set {1, . . . , m} such that qi ∼ pσ(i) for all i. We shall express this condition by saying that the decomposition (3) is essentially unique. 4.2.2 Definition. A subset I of an integral domain A is called a complete system of irreducible elements of A if (i) Any element of I is irreducible. (ii) If q ∈ A is irreducible, then there exists a unique p ∈ I such that p ∼ q. 4.2.3 By 4.2.2, we have another definition of a factorial ring: Definition. Let A be an integral domain and I ⊂ A a complete system of irreducible elements of A. We say that A is factorial if for any a ∈ A non-zero, we can associate a unique pair (ua , ν(a)) where ua ∈ U (A) and ν(a) : I → N, p → νp (a), which is zero almost everywhere such that: (4)
a = ua
pνp (a) .
p∈I
We call (4) the decomposition of a into irreducible elements (relative to I), and we call νp (a), the p-adic valuation of a. 4.2.4 Definition. Let A be an integral domain. (i) We say that A satisfies Euclid’s Lemma if for a, b, p ∈ A with p irreducible, we have p | ab ⇒ p | a or p | b. (ii) We say that A satisfies Gauss’ Lemma if for a, b, c ∈ A, we have:
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4 Factorial rings
a | bc and a b = 1 ⇒ a | c. (iii) Let I be a complete system of irreducible elements of A. We say that A satisfies the condition (D) for I if for any a ∈ A non-zero, there exists a pair (ua , ν(a)) where ua ∈ U (A) and ν(a) : I → N, p → νp (a), such that a decomposes as in (4). Remark. It is clear that if A satisfies condition (D) for I, then it satisfies condition (D) for any other complete system of irreducible elements of A. We shall therefore state condition (D) without specifying I. 4.2.5 Theorem. Let A be an integral domain satisfying condition (D). Then the following conditions are equivalent: (i) A is factorial. (ii) A satisfies Euclid’s Lemma. (iii) An element p ∈ A \ {0} is irreducible if and only if the ideal (p) is prime. (iv) A satisfies Gauss’ Lemma. Proof. Let us fix a complete system I of irreducible elements of A. (i) ⇒ (iv) Let a, b, c ∈ A be such that a | bc and a b = 1. If a, b or c is 0, then clearly, a | c. So let us assume that a, b, c are non-zero. Let d ∈ A be such that ad = bc. Since A is factorial, we have νp (a) + νp (d) = νp (b) + νp (c) for all p ∈ I. But a b = 1, so either νp (a) or νp (b) is zero. It follows in both cases that νp (a) νp (c), and so a | c. (iv) ⇒ (ii) This is clear by 4.1.9. (ii) ⇔ (iii) This is straightforward from Euclid’s Lemma and 4.1.7. (ii) ⇒ (i) Let a ∈ A \ {0} and ωp (a) νp (a) p =v p a=u p∈I
p∈I
be two decompositions of a into irreducible elements. Write a = pνp (a) q = pωp (a) r where q p = 1 = r p. Then by Euclid’s Lemma, we must have νp (a) = ωp (a). Now by induction, we have that νp (a) = ωp (a) for all p ∈ I and u = v. 4.2.6 Proposition. Let A be a factorial ring and I a complete system of irreducible elements of A. Let a, b be non-zero elements of A. Then a divides b if and only if νp (a) νp (b) for all p ∈ I. Proof. This is immediate by 4.2.5. 4.2.7 Theorem. Let A be a factorial ring. Any family a = (ai )i∈I of elements of A has a greatest common divisor and a least common multiple.
4.3 Principal ideal domains and Euclidean domains
43
Proof. We can assume that I = ∅ and ai = 0 for all i ∈ I. Let I be a complete system of irreducible elements of A. Set δ = p∈I pνp where for p ∈ I, νp = inf{νp (ai ), i ∈ I}. Setµ = 0 if there exists q ∈ I such that νq (ai ) is not bounded, otherwise, set µ = p∈I pωp where ωp = sup{νp (ai ), i ∈ I}. By 4.1.7 and 4.2.6, it is clear that δ = gcd(a) and µ = lcm(a). 4.2.8 Proposition. Let a, b be non-zero elements of a factorial ring A. Then (a b)(a b) ∼ ab. Proof. This is an immediate consequence of the proof of 4.2.7. 4.2.9 Theorem. A factorial ring A is an integrally closed domain. Proof. Let x ∈ Fract(A) \ {0} be integral over A, and a0 , · · · , an−1 ∈ A be such that xn + an−1 xn−1 + · · · + a0 = 0. Write x = p/q where p, q ∈ A and p q = 1. Then pn + an−1 qpn−1 + · · · + a0 q n = 0, which implies that p | q n . So q ∈ U (A) and x ∈ A.
4.3 Principal ideal domains and Euclidean domains 4.3.1 Definition. A commutative ring is called a principal ideal domain if it is an integral domain and if all ideals of A are principal. 4.3.2 Remarks. 1) A principal ideal domain is Noetherian. 2) Examples of principal ideal domains are Z and K[X], where K is a commutative field. 4.3.3 Proposition. Let A be a principal ideal domain. Suppose that A is not a field and a an ideal of A. Then the following conditions are equivalent: (i) a is maximal. (ii) a is prime and non-trivial. (iii) There exists an irreducible element p ∈ A such that a = (p). Proof. This is straightforward from the definitions and 4.1.5. 4.3.4 Corollary. (i) Let p be an irreducible element of a principal ideal domain A, then A/(p) is a field. (ii) The ring A[X] is a principal ideal domain if and only if A is a field. Proof. Part (i) is clear from 4.3.3 and the “if” part of (ii) is also clear. Now if A[X] is a principal ideal domain, then clearly A must be an integral domain. Hence U (A[X]) = U (A). It follows that X is an irreducible element of A[X]. By part (i), A[X]/(X) A is a field. 4.3.5 Lemma. A Noetherian ring A satisfies condition (D).
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4 Factorial rings
Proof. Let E be the set of elements a ∈ A\(U (A)∪{0}) which are not products of irreducible elements. Suppose that E = ∅, and let a1 ∈ E. In particular, a1 is not irreducible. So a1 = a2 a2 where a2 , a2 are both non-equivalent to a1 . Further, since a1 ∈ E, one of them must be in E, say a2 . Now, by repeating the argument with a2 , we obtain a strictly increasing sequence (an )n of ideals of A which contradicts the hypothesis that A is Noetherian. 4.3.6 Theorem. A principal ideal domain is factorial. Proof. Let A be a principal ideal domain. By 4.3.2 and 4.3.5, A satisfies condition (D). Now the result follows from 4.3.3 and 4.2.5. 4.3.7 Let A be a principal ideal domain and P the set of ideals of A. If a = (ai )i∈I is a family of elements in A, and δ = gcd(a), µ = lcm(a), then: (ai ) , (µ) = (ai ). (δ) = i∈I
i∈I
We obtain in particular the following well-known result: Theorem. (Bezout’s Theorem) Let a, b be elements of a principal ideal domain A and δ = gcd(a, b). Then there exist x, y ∈ A such that δ = ax + by. 4.3.8 Definition. A commutative integral domain A is called an Euclidean domain if there exists a map ν : A \ {0} → N such that: (i) For all x, y ∈ A \ {0} such that x | y, we have ν(x) ν(y). (ii) If a, b ∈ A \ {0}, there exist q, r ∈ A such that a = bq + r with r = 0 or ν(r) < ν(b). 4.3.9 The principal ideal domains Z and K[X] are examples of Euclidean domains. In fact, we obtain easily (as for Z) that: Theorem. An Euclidean domain is a principal ideal domain. 4.3.10 Theorem. Let A be an integral domain, S a multiplicative set of A and B = S −1 A. (i) If A is factorial, then so is B. (ii) If A is a principal ideal domain, then so is B. (iii) If A is an Euclidean domain, then so is B. Proof. Clearly, the statements are trivial if B is a field. So let us assume that B is not a field. (i) Let I be a complete system of irreducible elements of A, and IS the set of elements of I which do not divide any element of S. Since B is not a field, IS = ∅. Let q ∈ IS . Let a, b ∈ A, s, t ∈ S be such that q = (a/s)(b/t) in B. Thus qst = ab and the factoriality of A implies that q divides a or b. Let us suppose that q | a and a = qc where c ∈ A. Then (c/s)(b/t) = 1 and so b/t ∈ U (B). Thus q is irreducible in B.
4.4 Polynomials and factorial rings
45
Further, it is easy to see that non-zero elements of B are products of elements of IS and elements of U (B). Thus IS is a complete system of irreducibles in B. Finally, to show that B is factorial, we shall show that (ii) of 4.2.5 is satisfied. Let q ∈ IS , b, c ∈ A , s, t ∈ S be such that q/1 divides (b/s)(c/t). We have (qd)/u = (bc)/(st) for some d ∈ A, u ∈ S. Thus qdst = bcu. Further, if qr = u for some r ∈ A, then (q/1)(r/u) = 1 which is absurd since q/1 is irreducible in B. It follows that q divides bc. By Euclid’s Lemma, we have q | b or q | c. It follows that q/1 divides b/s or c/t. Hence B is factorial by 4.2.5. (ii) Let b be an ideal of B and A ∩ b = a. There exists a ∈ A such that a = Aa. So by 2.3.7, b = Ba. (iii) Let ν : A \ {0} → N be the map which makes A an Euclidean domain. By the proof of part (i), B is factorial and IS is a complete system of irreducible elements of A. It follows that for any x ∈ B \ {0}, there exist unique ux ∈ U (B) and ax , a product of irreducibles in IS , such that x = ax ux . Define νS : B \ {0} → N by νS (x) = ν(ax ). This is a well-defined map. On verifies easily that νS (x) = 1 if x ∈ U (B) and νS (x) νS (y) if x | y. Finally, if x, y ∈ B \ {0} and x = ux ax , y = uy ay as above. Then there exist q, r ∈ A such that ax = qay + r with r = 0 or ν(r) < ν(ay ). It follows that x = (qux /uy )y + ux r with ux r = 0 or νS (ux r) = νS (r) ν(r) < ν(ay ) = νS (y).
4.4 Polynomials and factorial rings 4.4.1 Let A be a factorial ring. The content of a polynomial P ∈ A[X]\{0} is defined to be any greatest common multiple of the coefficients of P , and we shall denote it by c(P ). Of course, c(P ) is unique up to a multiple of an element of U (A). A polynomial P is called primitive if c(P ) ∈ U (A). Clearly, P = c(P )P1 where P1 ∈ A[X] is primitive. 4.4.2 Lemma. Let A be a factorial ring and P, Q ∈ A[X] \ {0}. Then c(P Q) ∼ c(P )c(Q). Proof. We can assume that P, Q are primitive. If c(P Q) ∈ U (A), then there exists an irreducible element p of A which divides c(P Q). Let B = A/(p) and consider the algebra morphism ψ : A[X] → B[X] induced from the canonical surjection A → B. Then ψ(P )ψ(Q) = ψ(P Q) = 0. Since B is an integral domain by 4.2.5, we have that either ψ(P ) = 0 or ψ(Q) = 0. This implies that p divides c(P ) or c(Q) which is absurd. 4.4.3 Proposition. Let K be the field of fractions of a factorial ring A and P ∈ A[X]. (i) If P ∈ A, then P is irreducible in A[X] if and only if P is irreducible in A. (ii) If P ∈ A, then P is irreducible in A[X] if and only if P is primitive in A[X] and irreducible in K[X].
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4 Factorial rings
Proof. (i) If P = QR with Q, R ∈ A[X], then deg Q = deg R = 0. So Q, R belong to A. Since U (A) = U (A[X]), the result is clear. (ii) If P is irreducible in A[X], then clearly, it is primitive. Now if P = QR where Q, R ∈ K[X], then there exist primitive polynomials Q , R ∈ A[X] and elements a, b, c, d ∈ A such that Q = ab−1 Q , R = cd−1 Q , a b = 1 = c d. Hence bdP = acQ R and so (bd) = (ac) by 4.4.2. It follows that P = uQ R where u ∈ U (A). Since P is irreducible in A[X], either Q or R is in U (A). This implies that P is irreducible in K[X]. Conversely, suppose that P is primitive in A[X] and irreducible in K[X]. If P = QR with Q, R ∈ A[X] \ {0}, then the irreducibility of P in K[X] implies that either Q or R is in K. Let us suppose that Q ∈ K, then Q ∈ A \ {0}. Now Q | c(P ). Thus Q ∈ U (A) and we conclude that P is irreducible in A[X]. 4.4.4 Corollary. Let the notations be as in 4.4.3. If P ∈ A and P is irreducible in A[X], then P is irreducible in K[X]. 4.4.5 Theorem. Let A be a factorial ring, and X = (Xi )i∈I be a nonempty family of indeterminates over A. The ring A[X] is factorial. Proof. If P ∈ A[X], then there exists a finite subset J of I such that P ∈ A[(Xj )j∈J ]. Thus it suffices to prove the theorem for I finite. Further, by induction, we are reduced to prove that A[X] is factorial. Let P ∈ A[X] and n = deg P 0. We shall first prove by induction on n that P is a product of irreducible elements of A[X]. Since A is factorial, the result is clear for n = 0 by part (i) of 4.4.3. Let us suppose that n 1 and P is not irreducible. By 4.4.1, we can assume further that c(P ) = 1. By part (ii) of 4.4.3, P = QR where 1 deg Q, deg R < n. Applying our induction hypothesis on Q and R, we obtain that P is a product of irreducible elements. We shall show that A[X] satisfies Euclid’s Lemma, which, by 4.2.5, would imply that A[X] is factorial. Let P, Q, R ∈ A[X] \ {0} be such that P is irreducible and P | QR. If P ∈ A, then P | c(QR). Hence P divides c(Q) or c(R) since A is factorial. So P divides Q or R as required. If P ∈ A, then P is irreducible in K[X] by 4.4.4 and c(P ) = 1. Now, K[X] is factorial, so P divides Q or R in K[X]. Let us suppose that P F = Q where F ∈ K[X]. There exist a, b ∈ A with a b = 1 such that a−1 bF = G ∈ A[X] is primitive. It follows that aP G = bQ and so a = bc(Q). Thus b ∈ U (A) and P | Q in A[X]. 4.4.6 Remarks. 1) Let K be a (commutative) field. The ring K[X, Y ] is a factorial ring which is not a principal ideal domain. In fact, Bezout’s identity is not verified here, for X Y = 1 and (X) + (Y ) = K[X, Y ].
4.4 Polynomials and factorial rings
47
2) If I is an infinite set, then A[X] is a factorial ring which is not Noetherian. 4.4.7 Theorem. (Eisenstein’s Criterion) Let A be a factorial ring with field of fractions K. Let P = an X n + an−1 X n−1 + · · · + a0 ∈ A[X] where an and a0 are non-zero. Suppose that there exists an irreducible element p in A such that (i) p | ai , for 0 i n − 1, (ii) an is not divisible by p, (iii) a0 is not divisible by p2 . Then P is irreducible in K[X]. Further, if c(P ) = 1, then P is irreducible in A[X]. Proof. Let us write P = c(P )(an X n + an−1 X n−1 + · · · + a0 ). By (ii), we have p c(P ) = 1, so the polynomial an X n + an−1 X n−1 + · · · + a0 satisfies conditions (i),(ii) and (iii). So by 4.4.3, we are reduced to the case where P is primitive. If P is not irreducible in K[X], then by 4.4.3, we have P = QR with Q = bq X q + bq−1 X q−1 + · · · + b0 , R = cr X r + cr−1 X r−1 + · · · + c0 ∈ A[X], where bq and cr are non-zero, q, r ∈ N∗ and q + r = n. Since a0 = b0 c0 , conditions (i) and (iii) imply that p divides only one of b0 , c0 . Let us suppose that p divides b0 and that c0 is not divisible by p. Let s = min{i such that p bi } > 0. This is well-defined since c(P ) = 1, so c(Q) ∈ U (A) by 4.4.2. We have s q n − 1. But as = b0 cs + b1 cs−1 + · · · + bs c0 where we set cj = 0 if j > r. So condition (i) implies that p | bs which is absurd. 4.4.8 Let a be an ideal of A, B = A/a and ϕ : A → B the canonical surjection. Denote also by ϕ the induced surjection A[X] → B[X]. The image of P ∈ A[X] under ϕ is called the reduction of P modulo a. 4.4.9 Theorem. Let K be the field of fractions of a factorial ring A, p a prime ideal of A, B = A/p, L = Fract(B), ϕ : A[X] → B[X] the reduction map as above. Let P ∈ A[X] be such that deg P = deg ϕ(P ). If ϕ(P ) is irreducible in B[X] or L[X], then P is irreducible in K[X]. Proof. Suppose that P = QR where Q, R ∈ A[X] with: Q = bq X q + bq−1 X q−1 + · · · + b0 , R = cr X r + cr−1 X r−1 + · · · + c0 where q + r = deg P . We have bq cr ∈ p since deg P = deg ϕ(P ). Further p is prime, so bq , cr ∈ p which implies that deg Q = deg ϕ(Q) and deg R = deg ϕ(R). Now ϕ(P ) = ϕ(Q)ϕ(R) is irreducible in B[X] or L[X]. We have therefore that either ϕ(Q) or ϕ(R) has degree 0. Thus either deg Q or deg R is zero. By 4.4.3, P is irreducible in K[X].
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4 Factorial rings
4.5 Symmetric polynomials 4.5.1 In this section, A will be a ring and K a commutative field. We shall fix an integer n > 0 and write A[X] for A[X1 , . . . , Xn ]. For q ∈ N, we denote by Pq the A-submodule of A[X] of homogeneous polynomials of degree q. We shall also denote by Sn the symmetric group of {1, . . . , n}. 4.5.2 Let σ ∈ Sn . For P ∈ A[X], we define the polynomial P σ as follows: P σ (X1 , . . . , Xn ) = P (Xσ(1) , . . . , Xσ(n) ). Clearly, P σ is homogeneous of degree q if and only if P is homogeneous of degree q. Further, the map θσ : A[X] → A[X] , P → P σ is an automorphism of the algebra A[X]. We have θσ ◦ θτ = θστ for σ, τ ∈ Sn . Definition. A polynomial P ∈ A[X] is called symmetric if P σ = P for all σ ∈ S. We denote by A[X]sym the subalgebra of A[X] consisting of symmetric polynomials. 4.5.3 For 1 p n, set Ep =
1i1 <···
Xi1 · · · Xip .
This is a symmetric homogeneous polynomial of degree p. We call this polynomial the elementary symmetric polynomial of degree p in X1 , . . . , Xn . We set E0 = 1 and Ep = 0 if p > n. We see easily that, in A[X1 , · · · , Xn , U, V ], we have (5)
n
n
(U + V Xi ) =
i=1
U n−k V k Ek .
k=0
Thus, in particular, in A[X1 , . . . , Xn , T ]: n
(6)
(1 + T Xi ) =
i=1
(7)
n i=1
(T − Xi ) =
n
Ek T k .
k=0
n
(−1)n−k En−k T k .
k=0
4.5.4 Lemma. Let P ∈ A[X]sym be such that P (X1 , . . . , Xn−1 , 0) = 0. Then P is divisible by En . Proof. We proceed by induction on n. For n = 1, the result is clear. So let n 2.
4.5 Symmetric polynomials
49
Since P (X1 , . . . , Xn−1 , 0) = 0, we have P = P1 Xn + · · · + Pk Xnk where P1 , . . . , Pk ∈ A[X1 , . . . , Xn−1 ]. Further since P is symmetric, the Pj ’s are symmetric in X1 , . . . , Xn−1 . We have P (X1 , . . . , Xn−2 , 0, Xn ) = 0, since P τ = P where τ denotes the transposition (n − 1, n). It follows that Pj (X1 , . . . , Xn−2 , 0) = 0 for all j. Thus the Pj ’s are divisible by X1 · · · Xn−1 . Hence En = X1 · · · Xn divides P as required. 4.5.5 Let ν = (ν1 , . . . , νn ) ∈ Nn and λ ∈ A \ {0}. We define the weight π(λX1ν1 · · · Xnνn ) of the monomial λX1ν1 · · · Xnνn to be: π(λX1ν1 · · · Xnνn ) = ν1 + 2ν2 + · · · + nνn . The weight π(P ) of a non-zero polynomial P is the maximum of the weights associated to the non-zero terms in P . We set π(0) = +∞. 4.5.6 Lemma. Let P be a symmetric polynomial in A[X] of degree d. There exists Q ∈ A[Y1 , . . . , Yn ] such that π(Q) d and P (X1 , . . . , Xn ) = Q(E1 , . . . , En ). Proof. We can suppose that n 2 and d 1. Let e1 , . . . , en−1 be elementary symmetric polynomials in X1 , . . . , Xn−1 . Set B = A[X1 , . . . , Xn−1 ] and for 1 j n − 1, we have: ej = Ej (X1 , . . . , Xn−1 , 0). We shall proceed by induction on n and d. Since P (X1 , . . . , Xn−1 , 0) ∈ B sym , we have by our induction hypothesis that there exists Q1 ∈ B such that: π(Q1 ) d and P (X1 , . . . , Xn−1 , 0) = Q1 (e1 , . . . , en−1 ). Clearly, deg Q1 (E1 , . . . , En−1 ) d and the polynomial P1 (X1 , . . . , Xn ) = P (X1 , . . . , Xn ) − Q1 (E1 , . . . , En−1 ) has degree d, and P1 (X1 , . . . , Xn−1 , 0) = 0. It follows from 4.5.4 that En divides P1 . Let P2 ∈ A[X] be such that P1 = En P2 . Then deg P2 < d and P2 ∈ A[X]sym . By induction hypothesis, there exists Q2 ∈ A[X] such that: π(Q2 ) n − d and P2 (X1 , . . . , Xn ) = Q2 (E1 , . . . , En ) and hence P (X1 , . . . , Xn ) = Q1 (E1 , . . . , En−1 ) + En Q2 (E1 , . . . , En ). Finally, Q1 (Y1 , . . . , Yn−1 ) + Yn Q2 (Y1 , . . . , Yn ) has weight d.
50
4 Factorial rings
4.5.7 Theorem. (i) The algebra A[X]sym is generated by the elementary symmetric polynomials in X1 , . . . , Xn . (ii) The map P → P (E1 , . . . , En ) defines an isomorphism from the polynomial ring A[T1 , . . . , Tn ] to A[X]sym . Proof. Part (i) is a corollary of 4.5.5. The map in part (ii) defines clearly a morphism, and part (i) implies that it is surjective. Let us prove the injectivity of the map by induction on n. The case n = 1 is trivial. Let n 2 and suppose that the map is not injective. Choose P ∈ A[T1 , . . . , Tn ] \ A of minimal degree such that P (E1 , . . . , En ) = 0. Write P = P0 + P1 Tn + · · · + Pk Tnk with Pj ∈ A[T1 , . . . , Tn−1 ]. Since P (e1 , . . . , en−1 , 0) = P0 (e1 , . . . , en−1 ) = 0 where the ei ’s are the elementary symmetric polynomials in X1 , . . . , Xn−1 , we have P0 = 0 by induction. Thus P = Tn Q for some Q ∈ A[T1 , . . . , Tn ]. But deg(Q) < deg(P ) and Q(E1 , . . . , En ) = 0 which contradicts the minimality of deg(P ).
4.6 Resultant and discriminant 4.6.1 Let A be a ring and P, Q ∈ A[X] \ A of degree m and n respectively. Let us write: P = a0 + a1 X + · · · + am X m , Q = b0 + b1 X + · · · + bn X n . We define the resultant of P and Q, denoted by res(P, Q), to be the determinant of the following m + n by m + n matrix: a a1 . . . an−1 an . . . am−1 am 0 ... 0 0 0 . . . an−2 an−1 . . . am−2 am−1 am ... 0 a0 0 0 0 . . . an−3 an−2 . . . am−3 am−2 am−1 . . . ..................................................................... 0 0 ... a0 a1 ... ... ... ... ... am b0 b1 . . . bn−1 bn ... 0 0 ... ... 0 0 . . . bn−2 bn−1 . . . 0 0 ... ... 0 b0 0 ... ... ... ... 0 0 . . . bn−3 bn−2 . . . ..................................................................... 0 0 ... 0 0 ... b0 b1 b2 ... bn In the first n rows, we have the coefficients of P , while in the last m rows, we have the coefficients of Q. Let us suppose that A is an integral domain. For λ, µ ∈ A \ {0}, we have: res(λP, µQ) = λn µm res(P, Q) , res(Q, P ) = (−1)mn res(P, Q). Observe also that res(P, Q) is a homogeneous polynomial in the ai ’s and the bj ’s of degree n (resp. m) in the ai ’s (resp. in the bj ’s). 4.6.2 Lemma. If A is factorial, then the following conditions are equivalent:
4.6 Resultant and discriminant
51
(i) P and Q has a non-constant common factor. (ii) There exist F, G ∈ A[X] \ {0} such that P G = QF , deg(F ) < deg(P ) , deg(G) < deg(Q). Proof. Let R ∈ A[X] be a non constant polynomial dividing P and Q. Then there exist F, G ∈ A[X] such that P = F R and Q = GR. Clearly, F and G satisfies (ii). So we have (i) ⇒ (ii). Conversely, let F, G be as in (ii). Since deg G < deg Q, there exists an irreducible divisor R of Q and n ∈ N∗ such that Rn divides Q, but Rn does not divide G. It follows that R divides P . 4.6.3 Theorem. Let A be factorial, then the following conditions are equivalent: (i) P and Q has a non-constant common factor. (ii) res(P, Q) = 0. Proof. By 4.6.2, (i) is equivalent to the existence of non zero polynomials F = α0 + α1 X + · · · + αm−1 X m−1 and G = β0 + β1 X + · · · + βn−1 X n−1 such that P G = QF . This means that the following homogeneous system in x0 , . . . , xm−1 , y0 , . . . , yn−1 ⎧ a0 y0 = b0 x0 ⎪ ⎪ ⎪ ⎨ a1 y0 + a0 y1 = b1 x0 + b0 x1 (8) .. .. . .. ⎪ . . ⎪ ⎪ ⎩ am yn−1 = bn xm−1 has a non trivial solution in A. This is equivalent to the existence of a non trivial solution in Fract(A). Now the determinant of (8) is ± res(P, Q), so we have proved our result. 4.6.4 Corollary. Let A be a factorial subring of a factorial ring B, and P, Q ∈ A[X]. If P, Q has a non constant common factor in B[X], then P, Q has a non constant common factor in A[X]. 4.6.5 Corollary. Let A be factorial. There exist F, G ∈ A[X] such that: deg(F ) < deg(Q) , deg(G) < deg(P ) , res(P, Q) = P F + QG. Proof. If res(P, Q) = 0, then the assertion follows from 4.6.2 and 4.6.3. So let us suppose that res(P, Q) = 0. Set X = t (1, X, . . . , X m+n−1 ) ∈ Mm+n,1 (A[X]), Y = (P, XP, . . . , X n−1 P, Q, XQ, . . . , X m−1 Q) ∈ Mn+m,1 (A[X]). t
Let M ∈ Mm+n (A) be the matrix in 4.6.1 defining res(P, Q). We have Y = M X. Let M be the transpose of the matrix of cofactors of M . Then,
52
4 Factorial rings
M Y = M M X = res(P, Q)X.
(9)
Let (c0 , . . . , cn−1 , d0 , . . . , dm−1 ) be the first row of M and set : F = c0 + c1 X + · · · + cn−1 X n−1 , G = d0 + d1 X + · · · + dm−1 X m−1 . It follows from (9) that P F + QG = res(P, Q).
4.6.6 Theorem. Let α1 , . . . , αm , β1 , . . . , βn be elements of a factorial ring A, a0 , b0 ∈ A \ {0} and : P = a0 (X − α1 ) · · · (X − αn ) , Q = b0 (X − β1 ) · · · (X − βn ). We have : res(P, Q) = an0 bm 0
n m
(αi − βj ) = an0
i=1 j=1
m
Q(αi ) = (−1)mn bm 0
i=1
n
P (βj ).
j=1
Proof. Let T1 , . . . , Tm , U1 , . . . , Un , V0 , W0 , X be indeterminates over A. Consider the following polynomials in B = A[T1 , . . . , Tm , U1 , . . . , Un ] : f = V0 (X − T1 ) · · · (X − Tm ) = V0 X n + · · · + Vm−1 X + Vm , g = W0 (X − U1 ) · · · (X − Un ) = W0 X n + · · · + Wn−1 X + Wn . With the notations of 4.5.3, we have : Vi = (−1)i V0 Ei (T1 , . . . , Tm ) , Wj = (−1)j W0 Ej (U1 , . . . , Un ). By 4.5.6, we can also view f and g as polynomials over A in the indeterminates V0 , V1 , . . . , Vm , W0 , W1 , . . . , Wn , X. Set : S = V0n W0m
n m
(Ti − Uj ).
i=1 j=1
Clearly, we have : S = V0n
m i=1
g(Ti ) = (−1)mn W0m
n
f (Uj ).
j=1
Let R = res(f, g). Since V0 divides the Vi ’s, and W0 divides the Wj ’s, it follows from the definition of R that R = V0n W0m H, where H ∈ B. On the other hand, by 4.6.3, it is clear that R is divisible by Ti − Uj . Hence, S divides R. Since R and S are both homogeneous of degree m (resp. n) in Vi (resp. Wj ), we deduce that there exists a ∈ A such that R = aS. Finally, since the coefficient of R and S in the monomial V0n W0m is equal to 1, we have R = S. Our theorem follows immediately.
4.6 Resultant and discriminant
53
4.6.7 Definition. Let P ∈ K[X] be a polynomial of degree n 2 whose dominant coefficient is an . We define the discriminant of P , denoted by dis(P ), to be the scalar: dis(P ) = (−1)n(n−1)/2 a−1 n res(P, P ).
4.6.8 Proposition. Let P ∈ K[X] be a polynomial of degree n 2 whose dominant coefficient is an . Suppose further that P is split over K and let α1 , . . . , αn be the roots (not necessarily distinct) of P . Then dis(P ) = an2n−2 (αi − αj )2 . i<j
Proof. This is a direct consequence of 4.6.6 and the definition of the discriminant. 4.6.9 Proposition. Let P ∈ K[X] be a polynomial of degree n 2. (i) P and P are coprime if and only if dis(P ) = 0. (ii) Suppose that the characteristic of K is zero and that P is split over K. Then all the roots of P are simple if and only if dis(P ) = 0. Proof. These are direct consequences of 4.6.3 and 4.6.8.
References • [11], [52], [56].
5 Field extensions
We study certain properties of field extensions in this chapter. In particular, we prove the well-known Going Down Theorem. All fields considered here are assumed to be commutative. Let K be a field.
5.1 Extensions 5.1.1 An extension of K is a pair (L, i) where L is a field and i : K → L is a ring homomorphism. It follows that i is injective and when there is no confusion, we shall identify K with its image i(K). Thus L is an extension of K if K is a subfield of L, and we shall write K ⊂ L. We define the degree of an extension K ⊂ L to be the dimension of L (finite or not) as a vector space over K. We shall denote it by [L : K]. An extension of finite degree is called finite. 5.1.2 Let K ⊂ L be a field extension and S a subset of L. Denote by K[S] (resp. K(S)) the subring (resp. subfield) of L generated by K and S. Clearly, K(S) is the quotient field of K[S]. If S = {x1 , . . . , xn }, we shall write K[x1 , . . . , xn ] and K(x1 , . . . , xn ) for K[S] and K(S). An extension K ⊂ L is finitely generated if there is a finite subset S in L such that L = K(S). If there exists x ∈ L such that L = K(x), then the extension is called cyclic. 5.1.3 The following propositions are straightforward. Proposition. Let K ⊂ L be a field extension, S, T subsets of L and F the set of finite subsets of S. (i) We have K(S ∪ T ) = K(S)(T ) = K(T )(S). (ii) K(S) is the union of K(U ) for U ∈ F. 5.1.4 Proposition. Let K ⊂ L ⊂ M be field extensions. (i) If (ai )i∈I is a K-basis for L and (bj )j∈J is a L-basis for M , then (ai bj )i∈I,j∈J is a K-basis for M . (ii) The extension K ⊂ M is finite if and only if the extensions K ⊂ L and L ⊂ M are finite. Further, if K ⊂ M is finite, then
56
5 Field extensions
[M : K] = [M : L][L : K].
5.2 Algebraic and transcendental elements 5.2.1 Let A be a commutative ring containing K as a subring and S = {x1 , . . . , xn } a finite subset of A. We say that S is algebraically independent over K if P (x1 , . . . , xn ) = 0 for all P ∈ K[X1 , . . . , Xn ] \ {0}. A subset T of A is said to be algebraically independent over K if any finite subset of T is algebraically independent over K. 5.2.2 Let K ⊂ L be a field extension and x ∈ L. We define a K-algebra homomorphism : ϕx : K[X] → L , P → P (x). If ker ϕx = 0, then x is algebraically independent over K, and we say that x is transcendental over K. In particular, ϕx induces an isomorphism between K[X] and K[x] which extends to an isomorphism between K(X) and K(x). Thus the extension K ⊂ K(x) is infinite. If ker ϕx = 0, then x is algebraically dependent over K, and we say that x is algebraic over K. In this case, ker ϕx is a prime ideal and by 4.1.4, there exists a unique irreducible unitary polynomial Px such that ker ϕx = (Px ). We shall call Px , the minimal polynomial of x over K. The homomorphism ϕx then induces an isomorphism between K[X]/(Px ) and K[x] = K(x). Thus deg Px = [K(x) : K] and {1, x, . . . , xdeg Px −1 } is a K-basis of K(x). 5.2.3 Proposition. Let K ⊂ L be a field extension, S a subset of L algebraically independent over K and x ∈ L \ S. The following conditions are equivalent : (i) The element x is algebraic over K(S). (ii) The set S ∪ {x} is algebraically dependent over K. Proof. This is straightforward.
5.3 Algebraic extensions 5.3.1 Definition. An extension K ⊂ L is algebraic if any element of L is algebraic over K. An extension which is not algebraic will be called transcendental. 5.3.2 Clearly, by 5.1.4 and 5.2.2, we have : Proposition. (i) Let K ⊂ L be a field extension. An element x ∈ L is algebraic over K if and only if [K(x) : K] is finite. (ii) A finite extension is algebraic. 5.3.3 Proposition. Let K ⊂ L be a field extension and S a subset of L. Suppose that all the elements of S are algebraic over K. Then : (i) The extension K ⊂ K(S) is algebraic. (ii) If S is finite, then the extension K ⊂ K(S) is finite.
5.3 Algebraic extensions
57
Proof. By 5.1.3 and 5.3.2, it suffices to prove (ii). Part (ii) is clear if S is empty. Let us therefore suppose that S is not empty and proceed by induction on the cardinality of S. Let x ∈ S and set T = S \{s}. We have K(S) = K(T )(x) and [K(S) : K] = [K(T )(x) : K(T )][K(T ) : K] by 5.1.4. By induction, [K(T ) : K] is finite, and x, being algebraic over K, is also algebraic over K(T ). Hence [K(T )(x) : K(T )] is finite and we are done. 5.3.4 Proposition. Let K ⊂ L ⊂ M be field extensions. The following conditions are equivalent : (i) The extension K ⊂ M is algebraic. (ii) The extensions K ⊂ L and L ⊂ M are algebraic. Proof. The implication (i) ⇒ (ii) is trivial. Conversely, suppose that (ii) is satisfied. Let x ∈ M . There exist elements a0 , . . . , an−1 ∈ L such that xn + an−1 xn−1 + · · · + a0 = 0. We have by 5.3.3 that K ⊂ K(a0 , . . . , an−1 ) is finite. On the other hand, x is algebraic over K(a0 , . . . , an−1 ), so K(a0 , . . . , an−1 ) ⊂ K(a0 , . . . , an−1 )(x) is also finite. Hence K ⊂ K(x) is finite and part (i) follows. 5.3.5 Let K ⊂ L be a field extension. By 5.3.3 and 5.3.4, the set of elements of L algebraic over K is a subfield M containing K. We shall call M , the algebraic closure of K in L. If M = K, then we shall say that K is algebraically closed in L. 5.3.6 Lemma. Let K ⊂ L ⊂ M be field extensions and S a subset of M . (i) If K ⊂ L is algebraic, then K(S) ⊂ L(S) is also algebraic. (ii) If S is a finite algebraically independent subset over L, then [L : K] [L(S) : K(S)]. Proof. (i) Let x ∈ L(S), then there exist x1 , . . . , xn ∈ L such that x ∈ K(S)(x1 , . . . , xn ). Since the xi ’s are algebraic over K, x is algebraic over K(S) by 5.3.3. (ii) Let S = {x1 , . . . , xr } and B a K-basis of L. We shall show that B is linearly independent over K(S). Let b1 , . . . , bn be pairwise distinct elements of B and f1 , . . . , fn ∈ K(S) be such that f1 b1 + · · · + fn bn = 0. By multiplying by a suitable element of K(S), we can assume that fi ∈ K[S] for all i. We can therefore write for each i: n fi = λα,i xα , λα,i bi xα = 0 α∈Nr
α∈Nr
i=1
where the λα,i ’s are in K. By the algebraic independence of S over L, we have that n λα,i bi = 0 i=1
for all α ∈ N . Finally, B is a K-basis, so all the λα,i ’s are 0, and the result follows. r
58
5 Field extensions
5.4 Transcendence basis 5.4.1 Definition. Let K ⊂ L be a field extension. A subset S of L is called a transcendence basis of L over K if : (i) S is algebraically independent over K. (ii) The extension K(S) ⊂ L is algebraic. Example. Let L = K(X1 , . . . , Xn ) be the quotient field of the polynomial ring K[X1 , . . . , Xn ]. Then {X1 , . . . , Xn } is a transcendence basis of L over K. 5.4.2 Definition. Let K ⊂ L be a field extension. A subset B of L is called a purely transcendence basis of L over K if B is a transcendence basis of L over K and L = K(B). If L has a purely transcendence basis, then the extension K ⊂ L is called a purely transcendental extension. 5.4.3 Theorem. Let K ⊂ L be a field extension and S, T subsets of L verifying the following conditions : (i) L is algebraic over K(T ). (ii) S is algebraically independent over K. Then there exists a transcendence basis B of L over K such that S ⊂ B ⊂ S ∪ T. Proof. By Zorn’s lemma, there is a maximal (by inclusion) element B among the set of subsets in S ∪ T algebraically independent over K which contain S. Clearly, by 5.2.3, this implies that K(B) ⊂ K(S ∪ T ) is algebraic. On the other hand, by 5.3.3, K(T ) ⊂ K(S ∪ T ) ⊂ L are algebraic extensions. Hence K(B) ⊂ L is algebraic by 5.3.3 and the result now follows. 5.4.4 Corollary. Let K ⊂ L be a field extension. There exists a transcendence basis of L over K. Proof. This follows from 5.4.3 by taking S = ∅ and T = L.
5.4.5 Proposition. Let K ⊂ L be a field extension and S a subset of L. (i) If S is a transcendence basis of L over K, then it is maximal among subsets of L algebraically independent over K. (ii) If L is algebraic over K(S), then any maximal subset of S algebraically independent over K is a transcendence basis of L over K. Proof. These are easy consequences of 5.4.3.
5.4.6 Theorem. Let K ⊂ L be a field extension and B, B two transcendence bases of L over K. Then B and B are equipotent. Proof. It suffices to prove that card B card B . First, suppose that B is finite of cardinal n. If n = 0, then K ⊂ L is algebraic and we have B = ∅. So let n 1 and proceed by induction on n.
5.4 Transcendence basis
59
Let x ∈ B , then applying 5.4.3 with S = {x} and T = B, there exists a proper subset C of B such that C ∪ {x} is a transcendence basis of L over K. Let K1 = K(x) and C = B \ {x}. Clearly, C, C are algebraically independent subsets of L over K1 . It follows that they are transcendence bases of L over K1 . By induction, card C = card C and hence card B card B as required. Let us now consider the case where B is infinite. Since any x in B is algebraic over K(B ), there exists a finite subset Sx of B such that x is algebraic over K(Sx ). Denote by S the union of the Sx for x ∈ B. We have S ⊂ B and since B is infinite, card S card B. But K(B) ⊂ K(S) ⊂ L are algebraic, it follows from 5.4.5 that S = B and hence card B card B as required. 5.4.7 Definition. Let K ⊂ L be a field extension. We define the transcendence degree of L over K, denoted by tr degK L, to be the cardinality of any transcendence basis of L over K. Remark. Let L = K(x1 , . . . , xn ) be a finitely generated extension over K. Then by 5.4.3, tr degK L n. 5.4.8 From the preceding discussion, we obtain easily the following proposition : Proposition. Let K ⊂ L be an extension such that tr degK L = n. (i) Let S ⊂ L be such that K(S) ⊂ L is algebraic. Then card S n, and if card S = n, then S is a transcendence basis of L over K. (ii) Any subset of L algebraically independent over K has at most n elements. If it has n elements, then it is a transcendence basis of L over K. (iii) Suppose that L = K(x1 , . . . , xm ). Then m n, and if m = n, then K ⊂ L is a purely transcendental extension, and x1 , . . . , xm is a purely transcendence basis of L over K. 5.4.9 Theorem. Let K ⊂ L ⊂ M be field extensions, S a transcendence basis of L over K and T a transcendence basis of M over L. Then S ∩ T = ∅ and S ∪ T is a transcendence basis of M over K. Proof. Since T is algebraically independent over L, it is also algebraically independent over K(S). Hence S∩T = ∅ and S∪T is algebraically independent over K. Now, K(S) ⊂ L is algebraic, so L(T ) is algebraic over K(S ∪ T ). This implies in turn that M is algebraic over K(S ∪ T ) since M is algebraic over L(T ). This finishes the proof. 5.4.10 Corollary. Let K ⊂ L ⊂ M be field extensions. Then : tr degK M = tr degK L + tr degL M.
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5 Field extensions
5.4.11 Lemma. Let K ⊂ M be a finitely generated field extension and L the algebraic closure of K in M . Then K ⊂ L is a finite extension. Proof. The extension L ⊂ M , being also finitely generated, has a finite transcendence basis B. Since K ⊂ L is algebraic, K(B) ⊂ L(B) is algebraic by 5.3.6. Now, L(B) ⊂ M is algebraic, so 5.3.4 implies that K(B) ⊂ M is also algebraic. Being also finitely generated, the extension K(B) ⊂ M is finite by 5.3.3. Finally, by 5.3.6, [L : K] [L(B) : K(B)] [M : K(B)], and so the result follows. 5.4.12 Proposition. Let K ⊂ L ⊂ M be field extensions. If K ⊂ M is finitely generated, then so is K ⊂ L. Proof. By 5.4.3, any transcendence basis B of L over K is contained in a transcendence basis of M over K. So the hypothesis implies that B is finite. By 5.4.11, [L : K(B)] is finite since K(B) ⊂ L is algebraic. It follows that if S is a K(B)-basis of L, then S ∪ B is a finite system of generators for the extension K ⊂ L.
5.5 Norm and trace 5.5.1 Let K ⊂ L be a finite extension of degree n. For x ∈ L, the map ux : L → L, y → xy is a homomorphism of vector spaces over K. Set : NormL/K (x) = det ux , TrL/K (x) = tr ux . We call NormL/K (x) (resp. TrL/K (x)) the norm (resp. trace) of x in L relative to K. We have clearly that for λ ∈ K and x, y ∈ L : TrL/K (λx + y) = λ TrL/K (x) + TrL/K (y) , TrL/K (λ) = nλ, NormL/K (λxy) = λn NormL/K (x) NormL/K (y) , NormL/K (λ) = λn . These equalities implies that NormL/K (x) = 0 if and only if x = 0, and if x = 0, then [NormL/K (x)]−1 = NormL/K (x−1 ). 5.5.2 Proposition. Let K ⊂ L be a finite extension of degree n, x ∈ L and X r + ar−1 X r−1 + · · · + a0 the minimal polynomial of x over K. (i) s = n/r is an integer. (ii) We have TrL/K (x) = −sar−1 and NormL/K (x) = [(−1)r a0 ]s . Proof. Part (i) is clear since n/r = [L : K(x)]. Now let (b1 , . . . , bs ) be a K(x)-basis of L, then (xi bj )0i
5.5 Norm and trace
61
so part (ii) follows. 5.5.3 Let K ⊂ L be a finite extension of degree n. The map ϕ : L × L → K , (x, y) → TrL/K (xy) is a symmetric K-bilinear form on L. If B = (a1 , . . . , an ) is a K-basis of L, the determinant of (ϕ(ai , aj ))i,j is called the discriminant with respect to the basis B of L. The discriminant is non-zero exactly when ϕ is non-degenerated. Indeed, this is the case when the characteristic of K is zero since for any x = 0, ϕ(x, x−1 ) = n. 5.5.4 Proposition. Let A be an integral domain, K its quotient field, K ⊂ L a field extension, x ∈ L algebraic over K and P (X) = X r + ar−1 X r−1 + · · · + a0 the minimal polynomial of x over K. Then the following conditions are equivalent : (i) x is integral over A. (ii) a0 , . . . , ar−1 are integral over A. In particular, if x is integral over A and the extension K ⊂ L is finite, then NormL/K (x) and TrL/K (x) are integral over A. Proof. (ii) ⇒ (i) This is clear by 3.1.5 and 3.1.7. (i) ⇒ (ii) Let Ω be an algebraically closed field containing L and x1 , . . . , xr the roots of P in Ω. Since x is integral over A, there exists a polynomial Q(X) = X n + bn−1 X n−1 + · · · + b0 ∈ A[X] such that Q(x) = 0. Then P divides Q in K[X] and x1 , . . . , xr are roots of Q. Thus x1 , . . . , xr are integral over A. Since ±ai is the i-th elementary symmetric polynomial in x1 , . . . , xr , the ai ’s are integral over A by 3.1.5. The last statement follows therefore from 5.5.2. 5.5.5 Theorem. Let A be an integral domain, K its quotient field, K ⊂ L a field extension, B the integral closure of A in L and F the algebraic closure of K in L. (i) The ring B is an integrally closed domain. (ii) We have F = Fract(B). Proof. Clearly, B ⊂ F and so Fract(B) ⊂ F . Conversely let x ∈ F . There exists P (X) = X r + ar−1 X r−1 + · · · + a0 ∈ where bi , ci ∈ A for all i. K[X] such that P (x) = 0. Let us write ai = bi c−1 i Set d = c0 · · · cr−1 ∈ A, we have that ai d ∈ A for all i. It follows that dx is a root of X r + ar−1 dX r−1 + ar−2 d2 X r−2 + · · · + a0 dr ∈ A[X]. This implies that dx ∈ B. Thus F ⊂ Fract(B), which completes the proof of part (ii). By 3.1.7, B is integrally closed in L, hence it is integrally closed in F . So (i) follows from (ii).
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5.5.6 Theorem. Let A be an integrally closed domain, K = Fract(A), K ⊂ L a finite extension of degree n and B the integral closure of A in L. If the characteristic of K is zero, then there exists a K-basis (v1 , . . . , vn ) of L such that B is a submodule of the free A-module Av1 ⊕ · · · ⊕ Avn . Proof. Let (u1 , . . . , un ) be a K-basis of L. By 5.5.5, we can suppose that u1 , . . . , un ∈ B. Since the characteristic of K is zero, the bilinear form ϕ (5.5.3) is nondegenerated. Thus there exists a basis (v1 , . . . , vn ) of L over K verifying ϕ(ui , vj ) = δij for all 1 i, j n. Clearly the sum Av1 ⊕ · · · ⊕ Avn is direct. Let z = a1 v1 + · · · + an vn ∈ B where a1 , . . . , an ∈ K. Since ai = TrL/K (zui ), we have ai ∈ A by 5.5.4. Hence B ⊂ Av1 ⊕ · · · ⊕ Avn as required. 5.5.7 Proposition. Let notations be as in 5.5.6. If A is Noetherian, then so is B. Proof. By 5.5.6, B is a Noetherian A-module. So it is also a Noetherian Bmodule.
5.6 Theorem of the primitive element 5.6.1 Let E, F be field extensions over K, a K-homomorphism from E to F is a K-algebra homomorphism, i.e. a ring homomorphism leaving invariant the elements of K. Clearly, such a homomorphism is injective. A bijective K-homomorphism will be called a K-isomorphism and a bijective K-homomorphism of E onto itself is called a K-automorphism. 5.6.2 Proposition. Let Ω be an algebraically closed field containing K, x, y ∈ Ω algebraic over K. The following conditions are equivalent: (i) x and y have the same minimal polynomial over K. (ii) There exists a K-isomorphism σ : K(x) → K(y) such that σ(x) = y. (iii) There exists a K-homomorphism σ : K(x) → Ω such that σ(x) = y. If these conditions are verified, we shall say that x and y are conjugate over K. Proof. We have (i) ⇒ (ii) by 5.2.2, and (ii) ⇒ (iii) is obvious. Finally if P is the minimal polynomial of x over K. Then P (σ(x)) = 0. Since P is irreducible over K, we have (iii) ⇒ (i). 5.6.3 Lemma. Let Ω be an algebraically closed field containing a field K of characteristic zero, and P ∈ K[T ] \ {0} be an irreducible polynomial of degree n. Then P has exactly n distinct roots in Ω. Proof. We can suppose that n 1 and that P is unitary. If P has a multiple root α ∈ Ω, then P (α) = 0. Since P is irreducible, it is the minimal polynomial of α over K. It follows that P divides P , and therefore P = 0. This is absurd since P ∈ K and the characteristic of K is zero.
5.6 Theorem of the primitive element
63
5.6.4 Let σ : K → L be a ring homomorphism. This induces a ring homomorphism K[X] → L[X] given by : K[X] P = an X n + · · · + a0 → P σ = σ(an )X n + · · · + σ(a0 ) ∈ L[X]. Lemma. Let K ⊂ L be a field extension, x ∈ L algebraic over K, Ω an algebraically closed field and σ a homomorphism from K to Ω. There exists a homomorphism τ : K(x) → Ω extending σ. Proof. Let P be the minimal polynomial of x over K. Then P σ ∈ Ω[X] has non zero degree. It has a root, say y, in Ω. Now if Q(x) = 0, then P divides Q. Thus P σ divides Qσ and therefore σ follows that the map K(x) = K[x] → Ω defined by sending Q (y) = 0. It n n i i a x to i i=0 i=0 σ(ai )y is well-defined. Clearly, this is a homomorphism extending σ. 5.6.5 Theorem. Let K ⊂ L be a finite extension of degree n and σ a homomorphism from K to an algebraically closed field Ω. Then : (i) There exist k extensions of σ to L where 1 k n. (ii) If the characteristic of K is zero, then k = n. Proof. We have L = K(x1 , . . . , xp ) = K[x1 , . . . , xp ] where the xi ’s are algebraic over K. We proceed by induction on p. If p = 1, then the theorem follows from 5.6.3 and the proof of 5.6.4. Suppose that p > 1. Let F = K[x1 , . . . , xp−1 ], then by induction, σ admits r extensions σ1 , . . . , σr to F where 1 r [F : K], and r = [F : K] if the characteristic of K is zero. Again by induction, each σi admits si extensions to L where 1 si [L : F ], and si = [L : F ] if the characteristic of K is zero. The theorem follows immediately. 5.6.6 Corollary. Let K ⊂ Ω be a field extension where Ω is an algebraically closed field, and K ⊂ L a finite extension of degree n. The cardinality k of the set of K-homomorphism from L to Ω verifies 1 k n. Further, if the characteristic of K is zero, then k = n. 5.6.7 Theorem. (Theorem of the Primitive Element) Let K ⊂ L be a finite extension of characteristic zero fields. Then there exists ξ ∈ L such that L = K(ξ). Proof. Clearly, it suffices to prove the theorem in the case where L = K(x, y). Let n = [L : K] and Ω an algebraically closed field containing K. Let {σ1 , . . . , σn } be the set of K-homomorphisms from L to Ω (5.6.6). For 1 i = j n, set: Aij = {a ∈ K ; σi (x) + aσi (y) = σj (x) + aσj (y)}. Clearly, Aij contains at most one element. Since the characteristic of K is zero, it is infinite. So there exists a ∈ K not contained in any of the Aij ’s. It follows that σ1 , . . . , σn induce distinct K-homomorphisms from K(x + ay) to Ω. By 5.6.6, we have [K(x + ay) : K] n. Hence L = K(x + ay).
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Remark. Let K ⊂ L be a finite extension. An element ξ ∈ L is called primitive if L = K(ξ). 5.6.8 Corollary. Let K ⊂ L be a finite extension of characteristic zero fields. Then the set of fields E verifying K ⊂ E ⊂ L is finite. Proof. By 5.6.7, there exists x ∈ L such that L = K(x). Let E be a subfield of L containing K. The minimal polynomial Q of x over E is a unitary divisor in L[X] of the minimal polynomial P of x over K. It suffices therefore to show that E is completely determined by Q because the set of unitary divisors of P in L[X] is finite. Let E0 be the field generated by K and the coefficients of Q. Since E0 ⊂ E, Q is also irreducible in E0 [X]. So Q is the minimal polynomial of x over E0 . But E0 (x) = E(x) = L, so deg Q = [L : E0 ] = [L : E]. Thus E = E0 and E is completely determined by Q. 5.6.9 Theorem. Let Ω, Ω be two algebraic closures of K. There exists a K-isomorphism from Ω to Ω . More precisely, if K ⊂ L ⊂ Ω are field extensions and θ : L → Ω is a K-homomorphism, then there exists a Kisomorphism from Ω to Ω extending θ. Proof. By Zorn’s Lemma, there exists a field E verifying L ⊂ E ⊂ Ω, maximal such that there exists a K-homomorphism σ : E → Ω extending θ. Let x ∈ Ω. Since x is algebraic over K, it is algebraic over E, so by 5.6.5, σ extends to a K-homomorphism ψ : E(x) → Ω . It follows from the maximality of E that x ∈ E. Hence E = Ω. Finally σ(Ω), being isomorphic to Ω, is an algebraically closed subfield of Ω containing K. We deduce that σ(Ω) = Ω .
5.7 Going Down Theorem 5.7.1 Let K ⊂ E ⊂ F be field extensions. Denote by HomK (E, F ) the set of K-homomorphisms from E to F , and Gal(E/K) the group of Kautomorphisms of E. 5.7.2 Lemma. Let K ⊂ E ⊂ Ω be field extensions. Suppose that Ω is algebraically closed and E is algebraic over K. If σ ∈ HomK (E, Ω) verifies σ(E) ⊂ E, then σ(E) = E. Proof. Let x ∈ E, P ∈ K[X] its minimal polynomial over K, and x1 , . . . , xn the (distinct) roots of P contained in E. Since σ is injective and σ(xi ) is again a root of P contained in E for all i, it follows that σ is a permutation on the set {x1 , . . . , xn }. Hence x ∈ σ(E). 5.7.3 Definition. A field extension K ⊂ E is called normal if it is algebraic and if any irreducible polynomial in K[X] admitting a root in E is split over E.
5.7 Going Down Theorem
65
5.7.4 Proposition. Let K ⊂ E ⊂ Ω be field extensions. Suppose that Ω is algebraically closed and E is algebraic over K. The following conditions are equivalent: (i) The extension K ⊂ E is normal. (ii) For all x ∈ E, the conjugates of x in Ω over K are in E. (iii) For all σ ∈ HomK (E, Ω), we have σ(E) ⊂ E. (iv) For all σ ∈ HomK (E, Ω), we have σ(E) = E. Proof. The equivalence (i) ⇔ (ii) is clear. Further we have (iii) ⇔ (iv) by 5.7.2. Now let x ∈ E, since σ(x) and x share the same minimal polynomial, we have (ii) ⇒ (iii). Conversely, if y ∈ Ω is conjugate to x, then by 5.6.2, there exists a K-homomorphism τ : K(x) → K(y) verifying τ (x) = y. We have seen in 5.6.9 that τ can be extended to a K-homomorphism σ : E → Ω. So y = τ (x) = σ(x) ∈ E and (iii) ⇒ (ii). 5.7.5 Proposition. Let K ⊂ E ⊂ Ω be field extensions. Suppose that Ω is algebraically closed and K ⊂ E is a finite extension. Let F be the subfield of Ω generated by the σ(E) where σ runs over the set HomK (E, Ω). Then K ⊂ F is a finite normal extension. Further, if K ⊂ L is a normal extension verifying E ⊂ L ⊂ Ω, then F ⊂ L. Proof. By 5.6.6, the set HomK (E, Ω) is finite. Since K ⊂ σ(E) is a finite extension for any σ, K ⊂ F is a finitely generated algebraic extension. It follows from 5.4.11 that K ⊂ F is a finite extension. Since τ (σ(E)) ⊂ F for all τ ∈ HomK (F, Ω), σ ∈ HomK (E, Ω), normality follows from 5.7.4. The last assertion is a direct consequence of 5.7.4 since any element σ of HomK (E, Ω) can be extended to an element τ ∈ HomK (L, Ω) (5.6.9), so σ(E) = τ (E) ⊂ τ (L) ⊂ L. 5.7.6 Let G be a subgroup of the group of automorphisms of a field E. We shall denote the subfield of G-fixed points by E G . Lemma. Let K be a field of characteristic zero, K ⊂ E a finite normal extension and G = Gal(E/K). Then K = E G . Proof. Clearly, K ⊂ E G . Now suppose that there exists x ∈ E G \K. Then the minimal polynomial of x over K is at least of degree two. Since the characteristic of K is zero and K ⊂ E is normal, there exists y ∈ E, x = y, sharing the same minimal polynomial. By 5.6.2, 5.6.9 and 5.7.4, there exists σ ∈ G such that σ(x) = y which is absurd since x ∈ E G . 5.7.7 Proposition. Let A be an integrally closed domain of characteristic zero, K = Fract A, K ⊂ E a finite normal extension and C the integral closure of A in E. If q, q ∈ Spec C verifies A ∩ q = A ∩ q , then there exists σ ∈ Gal(E/K) such that σ(q) = q .
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Proof. Let Gal(E/K) = {σ1 , . . . , σn } and qi = σi (q). It is clear that σi (C) = C, so qi ∈ Spec C. Suppose that q = qi for all 1 i n. Then by 3.3.2 and 2.2.5, q ⊂ q1 ∪ · · · ∪ qn . Let x ∈ q \ q1 ∪ · · · ∪ qn and y = σ1 (x) · · · σn (x). We have y ∈ E G = K (5.7.5), and since y ∈ C and A is an integrally closed domain, y ∈ A. It follows that y ∈ A ∩ q and by the choice of x, we have y ∈ A ∩ q which contradicts the hypothesis. 5.7.8 Theorem. (Going Down Theorem) Let A be an integrally closed domain of characteristic zero, K = Fract A, and B an integral extension of A whose field of fractions E is a finite extension of K. Let p, p ∈ Spec A be such that p ⊂ p. If q ∈ Spec B verifies p = A ∩ q, then there exists q ∈ Spec B such that p = A ∩ q and q ⊂ q. In particular, ht p = ht q. Proof. Let F be a finite normal extension of K containing E (5.7.5) and denote by C the integral closure of A (so of B also) in F . By 3.3.3, there exist r, r ∈ Spec C verifying r ⊂ r and A∩r = p, A∩r = p . Let s ∈ Spec C be such that B ∩ s = q. Then A ∩ s = p = A ∩ r, so by 5.7.7, there exists σ ∈ Gal(F/K) such that σ(r) = s. Since A ∩ σ(r ) = p , we have that q = B ∩ σ(r ) verifies q ⊂ q and p = A ∩ q . 5.7.9 Let A be an integrally closed domain of characteristic zero, K = Fract A, K ⊂ L a finite extension of degree n, B a subring of L integral over A such that L = Fract B. By 5.6.7, there exists ξ = α/β ∈ L, where α, β ∈ B, such that L = K(ξ). Suppose that β ∈ A and let γ0 , . . . , γm−1 ∈ A be such that γ0 = 0 and β m + γm−1 β m−1 +· · ·+γ0 = 0. It follows that γ0 ξ = α(β m−1 +γm−1 β m−2 +· · ·+γ1 ). Since K(γ0 ξ) = K(ξ), there exists b ∈ B such that L = K(b). Let P (T ) = T n + an−1 T n−1 + · · · + a0 be the minimal polynomial of b over K. By 5.5.4, we have a0 , . . . , an−1 ∈ A. Let Ω be an algebraic closure of L and b1 , . . . , bn the conjugates of b in Ω over K. Recall from 5.6.3 that the bi ’s are the n distinct roots of P in Ω. Set ∆= (bi − bj ) = 0 i<j
whose square is the discriminant d of P . Recall from 4.6.7 that d ∈ A. Lemma. Let us suppose that d is invertible in A. Then: (i) B is a free A-module and {1, b, . . . , bn−1 } is a basis of B over A. (ii) B is the integral closure of A in L. Proof. The set {1, b, . . . , bn−1 } is a basis of L over K. Let x = a0 + a1 b + · · · + an−1 bn−1 ∈ L, a0 , . . . , an−1 ∈ K, be integral over A. Then the conjugates of x in Ω over K are: for 1 i n . xi = a0 + a1 bi + · · · + an−1 bn−1 i
5.8 Fields and derivations
67
This forms a linear system in the ai ’s whose determinant is ∆. It follows that ai = ∆i /∆ where ∆i ∈ L is integral over A since the xi ’s are integral over A. On the other hand, ai = ∆i ∆/d, and d is invertible in A, so the ai ’s are integral over A. Since A is an integrally closed domain, ai ∈ A for all i. Hence x ∈ B and the assertions follow. 5.7.10 Let us conserve the notations and hypotheses of 5.7.9. We suppose further that d is invertible in A. Let m be a maximal ideal of A, k = A/m and for x ∈ A, denote by x its class in k. Let S(T ) = T n + an−1 T n−1 + · · · + a0 ∈ k[T ]. The discriminant of d of S is non-zero because d is invertible in A. Let us suppose that S splits over k and let γ1 , . . . , γn be the distinct roots of S. For 1 i n, set Qi (T ) =
T − γj ∈ k[T ]. j=i γi − γj
Clearly, Qi (γi ) = 1 and Qi (γj ) = 0 if i = j. If R ∈ k[T ] verifies deg R n − 1, then (1)
R(T ) = R(γ1 )Q1 (T ) + · · · + R(γn )Qn (T ).
Let θ denote the class of b in B/mB = B ⊗A k and ei = Qi (θ). Then since {1, θ, . . . , θn−1 } is a basis of B/mB over k by 5.7.9, (1) implies that {e1 , . . . , en } is also a basis of B/mB over k. If i = j, then Qi Qj is divisible by S, so ei ej = 0. On the other hand, Q2i = SM + R where R ∈ k[T ] and deg R n − 1. By (1), we obtain that R = Qi and therefore, ei ei = ei . Thus the ring B/mB is the direct sum of fields kei = B/ni where ni is the inverse image of j=i kej . It follows that n1 , . . . , nn are the n maximal ideals of B lying above m.
5.8 Fields and derivations 5.8.1 We shall use the notations of 2.8 and 2.9. Further let K be a field of characteristic zero. Lemma. Let K ⊂ L be a finite extension. Then any K-derivation of L in a L-vector space N is zero. In particular, ΩK (L) = {0}. Proof. By 5.6.7, there exists x ∈ L algebraic over K such that L = K(x). Let P (T ) be the minimal polynomial of x over K. Let δ ∈ DerK (L, N ). If Q ∈ K[T ], then δ(Q(x)) = Q (x)δ(x). In particular, 0 = δ(P (x)) = P (x)δ(x). By 5.6.3, P (x) = 0. So δ(x) = 0 and the lemma follows since any element of L can be written as Q(x) for some Q ∈ K[T ]. 5.8.2 Lemma. Let K ⊂ E ⊂ F be field extensions such that [F : E] is finite, and N a F -vector space. Any element of DerK (E, N ) extends uniquely to an element of DerK (F, N ).
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Proof. By 5.6.7, there exists x ∈ F such that F = E(x). Let P be the minimal polynomial of x over E and δ ∈ DerK (E, N ). For Q(T ) = n an T n ∈ E[T ], we define Qδ (x) = n xn δ(an ) ∈ N . Suppose that ∆ ∈ DerK (F, N ) extends δ, then for Q ∈ E[T ], we have: (2)
∆(Q(x)) = Qδ (x) + Q (x)∆(x). In particular, since P (x) = 0, u = ∆(x) is the unique solution in N of 0 = P δ (x) + P (x)u.
(3)
So uniqueness follows. Now, let u = −P (x)−1 P δ (x) ∈ N be the unique solution of (3). We claim that the map ∆ : F → N given by (2) and ∆(x) = u is a well-defined map extending δ. This is just an easy consequence of the following equality: (QR)δ (x) + (QR) (x)u = R(x)(Qδ (x) + Q (x)u) + Q(x)(Rδ (x) + R (x)u) for all Q, R ∈ E[T ]. 5.8.3 Let us conserve the notations of 5.8.2. By 2.8.5 and 5.8.2, we obtain an exact sequence of E-vector spaces: 0 → DerE (F, N ) → DerK (F, N ) → DerK (E, N ) → 0. By 2.9.5, this induces an exact sequence of F -vector spaces: α
β
0 → ΩK (E) ⊗E F → ΩK (F ) → ΩE (F ) → 0 where α(dE/K (x) ⊗ 1) = dF/K (x) and β(dF/K (y)) = dF/E (y) for all x ∈ E, y ∈ F . It follows therefore from 5.8.1 that the F -vector spaces ΩK (E) ⊗E F and ΩK (F ) are isomorphic. Now if K ⊂ E is purely transcendental and tr degK (E) = n, then 2.9.7 implies that the dimension of the E-vector space ΩK (E) is n. So, we have obtained: Theorem. Let K ⊂ F be finitely generated. Then the dimension of the F -vector space ΩK (F ) is equal to the transcendence degree of F over K. 5.8.4 Proposition. Let u : A → B be a homomorphism of finitely generated commutative K-algebras which are integral domains. Denote by Ω(u) : ΩK (A) → ΩK (B) the homomorphism of A-modules defined in 2.9.5. Suppose that ΩK (A) (resp. ΩK (B)) is a free A-module (resp. free B-module). Then the following conditions are equivalent: (i) u is injective. (ii) Ω(u) is injective.
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69
Proof. (ii) ⇒ (i) In the notations of 2.9.5, Ω(u) is the unique A-module homomorphism such that the following diagram commutes: A ⏐ ⏐ dA/K
u
−−−−→
B ⏐ ⏐d B/K
Ω(u)
ΩK (A) −−−−→ ΩK (B) Let a ∈ ker u, then dA/K (a) = 0. For any x ∈ A, we have ax ∈ ker u, and therefore 0 = dA/K (ax) = adA/K (x). But the elements dA/K (x), x ∈ A, generate the A-module ΩK (A). So a = 0 because ΩK (A) is a free A-module. (i) ⇒ (ii) Let E (resp. F ) be the field of fractions of A (resp. B). Since u is injective, we may assume that A ⊂ B and E ⊂ F . By 5.8.3, we have dimF ΩK (F ) = tr degK F , dimF ΩE (F ) = tr degE F , dimF ΩK (E) ⊗E F = dimE ΩK (E) = tr degK E. It follows from a dimension count that the following sequence (2.9.5), induced by the canonical injection v : E → F , is exact: 0 → ΩK (E) ⊗E F → ΩK (F ) → ΩE (F ) → 0. We deduce therefore from 2.9.5 that Ω(v) : ΩK (E) → ΩK (F ) is injective. Let ιA : A → E and ιB : B → F denote the canonical injections. By 2.9.5, 2.9.6, the uniqueness of Ω(ιA ) and the hypothesis that ΩK (A) is a free A-module, we obtain that Ω(ιA ) is the composition : ΩK (A) → ΩK (A) ⊗A E → ΩK (E). In particular, Ω(ιA ) is injective. Now we have the following commutative diagram: u
A −−−−→ ⏐ ⏐ ιA
B ⏐ ⏐ιB
v
E −−−−→ F which induces, from the uniqueness of Ω(ιB ◦ u) = Ω(v ◦ ιA ), the following commutative diagram: Ω(u)
ΩK (A) −−−−→ ΩK (B) ⏐ ⏐ ⏐Ω(ι ) ⏐ Ω(ιA ) B Ω(v)
ΩK (E) −−−−→ ΩK (F ) Since Ω(v) and Ω(ιA ) are injective, we deduce that Ω(u) is also injective.
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5.9 Conductor 5.9.1 In this section, A will be a subring of a ring B. The annihilator of the A-module B/A is an ideal of A called the conductor of B in A. It is easy to verify that this is the largest ideal of B contained in A. Proposition. Let S be a multiplicative subset of A, f the conductor of B in A and f1 the conductor of S −1 B in S −1 A. Then S −1 f ⊂ f1 . Furthermore, we have S −1 f = f1 if B is a finite A-algebra. Proof. Since fB ⊂ A, we have (S −1 f)(S −1 B) ⊂ S −1 A. So S −1 f ⊂ f1 . Now let us suppose that {x1 , . . . , xn } is a system of generators of the Amodule B. Let a/s ∈ f1 where a ∈ A, s ∈ S. Then for each i, (a/s)(xi /1) ∈ S −1 A, i.e. there exists ti ∈ S such that ti axi ∈ A. Set t = t1 · · · tn , then clearly taxi ∈ A. Hence ta ∈ f and a/s ∈ S −1 f. 5.9.2 Corollary. Let f be the conductor of B in A, p ∈ Spec A be such that f ⊂ p, and S = A \ p. If B is a finite A-algebra, then Ap = S −1 B. Proof. By 5.9.1, S −1 f is the conductor of S −1 B in Ap . Since f ⊂ p, there exists s ∈ S ∩ f. It follows that 1 = s/s ∈ S −1 f. Hence S −1 f = Ap and Ap = S −1 B by the definition of the conductor. 5.9.3 Proposition. Let A be an integral domain, B be the integral closure of A, f the conductor of B in A and S a multiplicative subset of A not containing 0. (i) A sufficient condition for S −1 A to be an integrally closed domain is that S ∩f = ∅. If B is a finite A-algebra, then this is also a necessary condition. (ii) Suppose that B is a finite A-algebra. Let p ∈ Spec A, then Ap is not an integrally closed domain if and only if f ⊂ p. Proof. By 3.1.9, S −1 B is the integral closure of S −1 A. Since S −1 f = S −1 A if and only if S ∩ f = ∅, part (i) follows from 5.9.1 and 5.9.2. Finally, part (ii) is just a special case of (i). 5.9.4 Proposition. Let B be an integral domain, A integrally closed in B and x ∈ B such that B is integral over A[x]. Suppose that there exists a unitary polynomial P ∈ A[T ] such that P (x) is contained in the conductor of B in A[x]. Then B = A[x]. Proof. Let b ∈ B. Then bP (x) = Q(x) ∈ A[x] for some Q ∈ A[T ]. Since P is unitary, we have Q = F P + R with F, R ∈ A[T ] and deg R < deg P . Let y = b − F (x), thus yP (x) = R(x). If y = 0, then clearly b = F (x) ∈ A[x]. If y = 0, then P (x) = R(x)/y in Fract(B). Since deg R < deg P , we deduce that x is integral over A[1/y]. On the other hand, y is integral over A[x], and hence over A[x, 1/y]. It follows from 3.1.7 that A[x, y, 1/y] is integral over A[1/y]. So y is integral over A[1/y], and there exist a0 , . . . , am−1 ∈ A[1/y] such that
5.9 Conductor
71
y m + am−1 y m−1 + · · · + a0 = 0. Clearly, this implies that y is integral over A. Since A is integrally closed in B, we have y ∈ A, and therefore b = y + F (x) ∈ A[x]. 5.9.5 Lemma. Let B be an integral domain and x ∈ B an element transcendental over Fract A such that B is a finite A[x]-algebra. For any maximal ideal n of B, there exists q ∈ Spec B such that q ⊂ n , q = n , n ∩ A = q ∩ A. Proof. Let A (resp. B ) be the integral closure of A in Fract A (resp. B in Fract B). It is clear that B is integral over A [x] and x is transcendental over Fract A = Fract A. Let n be a prime ideal of B lying above n. By 3.3.2, n ∈ Spm B . Set m = n ∩ A and r = n ∩ A [x]. Note that r ∈ Spm A [x] by 3.3.2, and so m A [x] ⊂ r , m A [x] = r since A [x]/m A [x] (A /m )[T ] is not a field. Since A [x] is an integrally closed domain by 3.2.8 and m A [x] ∈ Spec A [x], we obtain by 5.7.8 that there exists q ∈ Spec B verifying q ⊂ n , q = n and q ∩ A [x] = m A [x]. Hence q ∩ A = m . Now q = q ∩ B is not a maximal ideal of B by 3.3.2, so we have q = n and q ∩ A = n ∩ A. 5.9.6 Lemma. Let B be an integral domain, A integrally closed in B and x ∈ B such that B is a finite A[x]-algebra. Suppose that there exists a maximal ideal n of B which is minimal in the set of q ∈ Spec B such that q∩A = n∩A. (i) For all q ∈ Spec B verifying q ⊂ n, the class of x modulo q is algebraic over Fract(A/(q ∩ A)). (ii) The conductor f of B in A[x] is not contained in n. Proof. (i) Denote by x1 the class of x modulo q. The ring B/q is a finite (A/(q ∩ A))[x1 ]-algebra and n/q ∈ Spm(B/q) is minimal in the set of prime ideals in B/q lying above (n ∩ A)/(q ∩ A). So x1 is algebraic over A/(q ∩ A) by 5.9.5. (ii) Suppose that f ⊂ n. Let q ∈ Spec B be minimal verifying f ⊂ q ⊂ n, and denote p = q ∩ A. Recall from 2.6.6 that since Ap is a local ring with maximal ideal pAp , its residue field K is isomorphic to Fract(A/p). Further the ideal qp in Bp is prime and verifies qp ∩ Ap = pAp (2.3.7). By part (i), the class of x modulo q is algebraic over K. Since Bp is integral over Ap by 3.1.8, all elements of Bq /qp are algebraic over K. It follows therefore that Bq /qp is a field, and so qp ∈ Spm Bq . Further Ap is integrally closed in Bp by 3.1.9, fp is the conductor of Bp in Ap by 5.9.1, and qp is minimal among the prime ideals of Bp containing fp . So by replacing A, B by Ap , Bp , we can assume that q = n. Under this assumption, n/f is a minimal prime ideal of B/f and (n/f)n/f is the minimal prime ideal of (B/f)n/f . So (n/f)n/f is the nilradical of (B/f)n/f .
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Since the class of x modulo n is algebraic over K, there exist y ∈ A \ p ⊂ B \n and P ∈ K[T ] unitary such that yP (x) ∈ n. By the preceding paragraph, the image in (B/f)n/f of the class of yP (x) in B/f is nilpotent. So there exist z ∈ B \ n and k ∈ N∗ such that zy k P (x)k ∈ f, hence P (x)k belongs to the conductor of A[x][zy k B] in A[x]. Now by applying 5.9.4 to A, A[x] and A[x][zy k B], we obtain that zy k B ⊂ A[x]. But this implies that zy k ∈ f ⊂ n which is absurd since y, z ∈ n and n is prime. 5.9.7 Lemma. Let B be an integral domain and A a local ring with maximal ideal m. Let us suppose that: (i) The ring A is integrally closed in B and there exists n ∈ Spm B lying above m. (ii) There exists x ∈ B such that B = A[x] and the class of x modulo n is algebraic over K = A/m. Then B = A. Proof. By (ii), there exists P ∈ A[T ] \ A unitary such that P (x) ∈ mA[x]. Set y = 1 + P (x) and θ, the image of y in C = A[y]/mA[y]. If θ is not invertible in C, then θ is contained in a maximal ideal of C. But A[x]/mA[x] is integral over C, so the image y of y in A[x]/mA[x] is also contained in a maximal ideal by 3.3.2 and 3.3.3. This is absurd since y = 1. Thus we have shown that θ is invertible in C. Next, let X m + αm−1 X m−1 + · · · + α0 be the minimal polynomial of θ over K. Since θ is invertible in C, α0 = 0. For 0 i m − 1, let ai ∈ A be a representative of αi , then there exist n ∈ N and u0 , . . . , un ∈ m such that: y m + am−1 y m−1 + · · · + a0 = un y n + un−1 y n−1 + · · · + u0 . Thus there exists z ∈ A[y] such that yz = a0 − u0 ∈ m. Since A is a local ring, this implies that yz is invertible in A. Hence y is invertible in A[y] ⊂ B, i.e. there exist b1 , . . . , br ∈ A such that br y r + · · · + b1 y = 1. It follows that 1/y is integral over A and (i) implies that 1/y ∈ A. Now if 1/y ∈ m, then 1 ∈ ym ⊂ n which is absurd. So 1/y ∈ A \ m is invertible in A. Thus y ∈ A, which implies that P (x) ∈ A. So the hypotheses on P implies that x is integral over A. Hence x ∈ A and B = A. 5.9.8 Proposition. Let B be an integral domain, A integrally closed in B and x ∈ B such that B is a finite A[x]-algebra. Let m ∈ Spm(A) and n be a maximal ideal of B which is minimal among the prime ideals of B lying above m. Then Am = Bm = Bn . Proof. By 3.1.9, Am is integrally closed in Bm and Bm is a finite Am [x]-algebra. Further, nm is a maximal ideal of Bm which is minimal among the prime ideals of Bm lying above mAm (2.3.7 and 3.3.2). We are therefore reduced to the case where A is a local ring with maximal ideal m and B = Bm .
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73
By 5.9.6, n does not contain the conductor of B in A[x]. It follows from 5.9.2 that B = Br = A[x]r where r = n ∩ A[x]. By 3.3.2, r ∈ Spm A[x], so n = rA[x]r . Our hypothesis on n implies via 3.3.3 that r is minimal among the prime ideals of A[x] lying above m. Now, 5.9.6 and 5.9.7 imply that A = A[x], and so r = m. Hence Am = B = Bm = Bn as required.
References • [9],[19], [22], [26], [52], [56].
6 Finitely generated algebras
We prove three fundamental results in this chapter : Noether’s Normalization Theorem, Krull’s Principal Ideal Theorem, and Hilbert’s Nullstellensatz. We introduce also the Zariski topology on the set of maximal ideals of a ring. All fields and rings considered are commutative. Let us denote by K a field. If A is a ring, then we shall denote by Specm (A) the set of minimal prime ideals of A.
6.1 Dimension 6.1.1 Let A be a finitely generated K-algebra and {x1 , . . . , xn } a system of generators of A. Since A is isomorphic to a quotient of the polynomial ring K[T1 , . . . , Tn ], A is Noetherian by 2.7.9 and 2.7.3. Suppose that A is an integral domain and let F be its quotient field. Then by 5.4.3, we can extract from {x1 , . . . , xn } a transcendence basis of F over K, and hence tr degK F n. In particular, there exists a transcendence basis of F over K contained in A. 6.1.2 Proposition. Let A be a finitely generated K-algebra and p, q be distinct prime ideals of A such that p ⊂ q. Then tr degK Fract(A/q) < tr degK Fract(A/p). Proof. Replace A by A/p, we are reduced to the case where A is an integral domain, p = {0} and q = {0}. Let π : A → A/q be the canonical surjection. Let y1 , . . . , yr ∈ A/q be a transcendence basis of Fract(A/q) over K. For each i, fix xi ∈ A such that π(xi ) = yi . Clearly, the xi ’s are algebraically independent over K. It follows from 6.1.1 that: r = tr degK Fract(A/q) tr degK Fract A. If equality holds, then π induces an isomorphism from K[x1 , . . . , xr ] onto K[y1 , . . . , yr ]. Further for any a ∈ q, a is algebraic over K(x1 , . . . , xr ). So there exist P0 , . . . , Pn ∈ K[X1 , . . . , Xr ] such that P0 (x1 , . . . , xr ) = 0 and
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P0 (x1 , . . . , xr ) + P1 (x1 , . . . , xr )a + · · · + Pn (x1 , . . . , xr )an = 0. Applying π to this equality, we obtain that P0 (y1 , . . . , yr ) = 0 which contra dicts the algebraic independence of the yi ’s. 6.1.3 Corollary. Let A be a finitely generated K-algebra. Then: dim A
max p∈Spec(A)
tr degK Fract(A/p) =
Proof. This is immediate from 6.1.2.
max p∈Specm (A)
tr degK Fract(A/p).
6.1.4 Corollary. We have dim K[X1 , . . . , Xn ] = tr degK K(X1 , . . . , Xn ) = n. Proof. Let A = K[X1 , . . . , Xn ]. Since X1 , . . . , Xn form a transcendence basis of K(X1 , . . . , Xn ), we have dim A tr degK K(X1 , . . . , Xn ) = n. On the other hand, let pi be the ideal of A generated by X1 , . . . , Xi . Clearly A/pi is isomorphic to K[Xi+1 , . . . , Xn ]. So we have a chain of prime ideals {0} ⊂ p1 ⊂ · · · ⊂ pn in A. So dim A n. 6.1.5 Corollary. Let pi , 1 i n be as in the proof of 6.1.4. Then the height of pi is equal to i. Proof. The chain {0} ⊂ p1 ⊂ · · · ⊂ pi implies that ht pi i. Now the result follows from 6.1.2 and 6.1.4.
6.2 Noether’s Normalization Theorem 6.2.1 Theorem. Let n ∈ N∗ , A = K[X1 , . . . , Xn ] and a a proper ideal of A. Then there exist x1 , . . . , xn ∈ A algebraically independent over K verifying the following conditions: (i) The ring A is integral over B = K[x1 , . . . , xn ]. (ii) If a is non trivial, then there exists s ∈ {1, . . . , n} such that the ideal B ∩ a of B is generated by x1 , . . . , xs . Proof. If a = {0}, then it suffices to take xi = Xi for all i. So let us suppose that a = {0}. We shall proceed by induction on n. For n = 1, there exists x ∈ A nonconstant such that a = Ax. Let x1 = x, then (i) and (ii) are satisfied. Let us suppose that n > 1 and x1 ∈ a \ {0}. Then x1 ∈ K and we can write aλ X1i1 · · · Xnin x1 = P (X1 , . . . , Xn ) = λ=(i1 ,...,in )∈Nn
∗
where aλ ∈ K. Let k ∈ N be such that k > i1 + · · · + in for all λ = (i1 , . . . , in ) such that aλ = 0. For λ = (i1 , . . . , in ), write l(λ) = i1 + i2 k + · · · + in k n−1 . Then by the choice of k and the uniqueness of the development of an integer
6.2 Noether’s Normalization Theorem
77
in base k, we have that for λ = (i1 , . . . , in ), λ = (j1 , . . . , jn ) such that aλ = 0, aλ = 0, l(λ) = l(λ ) implies that λ = λ . So there exists a unique µ ∈ Nn verifying aµ = 0 such that l(µ) is maximal. i−1 for 2 i n. Then Set ti = Xi − X1k n−1
0 = P (X1 , t2 + X1k , . . . , tn + X1k ) − x1 l(µ) = aµ X1 + Qj (x1 , t2 , . . . , tn )X1j j
where the Qj ’s are polynomials. It follows that X1 is integral over the algebra i−1 for 2 i n, Xi is integral over C = K[x1 , t2 , . . . , tn ]. Since Xi = ti + X1k C by 3.1.7. Thus A is integral over C. Further, by 5.3.3, the extension Fract(C) ⊂ Fract(A) is algebraic and tr degK Fract(A) = n. Therefore by 5.4.3, x1 , t2 , . . . , tn are algebraically independent over K. In particular, C is isomorphic to A, so it is factorial and an integrally closed domain. Let y ∈ A be such that yx1 ∈ C, then y ∈ Fract(C) and since C is an integrally closed domain and y is integral over C, y ∈ C. Hence C∩Ax1 = Cx1 . Suppose first that a ∩ K[t2 , . . . , tn ] = {0}. Let S = K[t2 , . . . , tn ] \ {0}. The ideal generated by x1 in S −1 C = K(t2 , . . . , tn )[x1 ] is maximal. So S −1 a = S −1 Cx1 . It follows that if y ∈ a ∩ C, then there exists s ∈ S such that sy ∈ Cx1 . Since x1 , t2 , . . . , tn are algebraically independent, Cx1 ∈ Spec(C). Hence y ∈ Cx1 , and a ∩ C = Cx1 . So we obtain the result by setting x2 = t2 , . . . , xn = tn . Suppose now that a ∩ K[t2 , . . . , tn ] = {0}. Let D = K[t2 , . . . , tn ]. By induction, there exist x2 , . . . , xn ∈ D algebraically independent over K, such that D is integral over E = K[x2 , . . . , xn ] and a ∩ E = Ex2 + · · · Exk for some k with 2 k n. By 3.1.7, A is integral over B = K[x1 , x2 , . . . , xn ], and so x1 , x2 , . . . , xn are algebraically independent over K. Since x1 ∈ a, it follows that a ∩ B = Bx1 + a ∩ E = Bx1 + Bx2 + · · · + Bxk . 6.2.2 Theorem. (Noether’s Normalization Theorem) Let A be a finitely generated K-algebra and a1 ⊂ a2 ⊂ · · · ⊂ ap a chain of ideals of A such that ap = A. There exist x1 , . . . , xn ∈ A verifying the following conditions: (i) x1 , . . . , xn are algebraically independent over K, A is integral over B = K[x1 , . . . , xn ] and dim A = n. (ii) There exists a chain of integers 0 h(1) · · · h(p) such that for 1 k p, B ∩ ak is {0} if h(k) = 0 and it is the ideal of B generated by x1 , . . . , xh(k) if h(k) > 0. Proof. First of all, observe that if x1 , . . . , xn ∈ A are algebraically independent over K and A is integral over K[x1 , . . . , xn ], then by 6.1.4 and 3.3.4, we have dim A = n. If A is integral over K, then the conditions are void, i.e. n = 0 and B = K. So let us suppose that A is not integral over K.
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Let us first establish the theorem for A = K[X1 , . . . , Xn ], the ring of polynomials over K. If p = 1, then the theorem is just a consequence of 6.2.1. So let us suppose that p 2. By induction on p, there exist t1 , . . . , tn ∈ A verifying conditions (i) and (ii) with respect to the chain a1 ⊂ · · · ⊂ ap−1 . Set r = h(p − 1), B = K[t1 , . . . , tn ] and C = K[tr+1 , . . . , tn ]. Applying 6.2.1 to C and the ideal C ∩ ap of C, we obtain elements xr+1 , . . . , xn ∈ C algebraically independent over K such that C is integral over D = K[xr+1 , . . . , xn ], and there is an integer h(p) such that D ∩ ap is generated by xr+1 , . . . , xh(p) . Let xi = ti for 1 i r and E = K[x1 , . . . , xn ]. Then by construction, Fract(B) is algebraic over Fract(E). Thus x1 , . . . , xn are algebraically independent over K. Further, by 3.1.7, A is integral over E. So we have (i). For 1 i p, E ∩ ai contains obviously the ideal Ex1 + · · · + Exh(i) of E. If i p − 1, then E ∩ ai = E ∩ (Bx1 + · · · + Bxh(i) ). But by 5.7.8 and 6.1.5, E ∩ ai and Ex1 + · · · + Exh(i) are prime ideals of E of height h(i). So they must be equal. Let z ∈ E ∩ ap . We can write aλ xi11 · · · xirr z= λ=(i1 ,...,ir )
where aλ ∈ D. Clearly aλ xi11 · · · xirr ∈ ap−1 ⊂ ap if λ = λ0 = (0, . . . , 0). It follows that aλ0 ∈ D ∩ ap . This implies that E ∩ ap ⊂ Ex1 + · · · + Exh(p) , and since clearly x1 , . . . , xh(p) ∈ E ∩ ap , we have condition (ii). Now let us suppose that A is a finitely generated K-algebra. Then there exists a surjective ring homomorphism π : A = K[X1 , . . . , Xm ] → A for some m ∈ N. Let a0 = ker π and ai = π −1 (ai ) for 1 i p. Since A is a polynomial ring, there exist x1 , . . . , xm ∈ A verifying conditions (i) and (ii) with respect to the chain a0 ⊂ · · · ⊂ ap . By 3.1.8, A is integral over the K-algebra generated by xj = π(xj ), j > h(0). Thus s > h(0) since A is not integral over K. To prove the theorem, it suffices therefore to show that xh(0)+1 , . . . , xm are algebraically independent over K. Let t = h(0) + 1. If P (xt , . . . , xm ) = 0 for some non zero polynomial over K, then P (xt , . . . , xm ) ∈ a0 ∩ K[x1 , . . . , xm ] = Bx1 + · · · + Bxh(0) where B = K[x1 , . . . , xm ]. But this contradicts the fact that x1 , . . . , xm are algebraically independent. Hence xt , . . . , xm are algebraically independent over K and the theorem follows. 6.2.3 Corollary. Let A be a finitely generated K-algebra. We have dim A =
max p∈Spec(A)
tr degK Fract(A/p) =
max p∈Specm (A)
tr degK Fract(A/p).
In particular, if A is an integral domain, then dim A = tr degK Fract A.
6.2 Noether’s Normalization Theorem
79
Proof. By 6.2.2, A is integral over a polynomial ring B contained in A, and dim A = dim B. By 3.1.8, A/p is integral over B/(B ∩ p) for any p ∈ Spec(A). So tr degK Fract(A/p) = tr degK Fract(B/(B ∩ p)) tr degK Fract B. Thus the result follows from 6.1.3 and 6.1.4. Finally, if A is an integral domain, then {0} is the unique minimal prime ideal of A. 6.2.4 Corollary. Let A be a finitely generated K-algebra such that dim A = 0. If A is an integral domain, then A is a field and the extension K ⊂ A is finite. Proof. If A is an integral domain and dim A = 0, then {0} is a maximal ideal. So A is a field. By 6.2.3, the extension K ⊂ A is algebraic. So it is finite since A is a finitely generated K-algebra. 6.2.5 Corollary. Let A be a finitely generated K-algebra. Then we have dim A[X] = 1 + dim A. Proof. By 6.2.2, A is integral over a polynomial ring B contained in A. So A[X] is integral over B[X]. Hence the result follows from 6.2.2. 6.2.6 Proposition. Let A be a finitely generated K-algebra which is an integral domain. (i) For any p ∈ Spec(A), we have: ht p + dim(A/p) = dim A. (ii) Let p, q ∈ Spec(A) be such that p ⊂ q. Then: ht q = ht p + ht(q/p). Proof. Part (i) is clear for p = {0}. So let us suppose that p = {0}. By 6.2.2, there exist x1 , . . . , xn ∈ A such that B = K[x1 , . . . , xn ] is isomorphic to a polynomial ring, A is integral over B and B ∩ p = Bx1 + · · · + Bxr where 1 r n. By 5.7.8 and 6.1.5, ht p = ht(B ∩ p) = r. On the other hand, B/(B ∩ p) is isomorphic to K[xr+1 , . . . , xn ], so its dimension is n − r. Since A is integral over B, A/p is integral over B/(B ∩ p) (3.1.8). This proves part (i). Finally, since dim A = ht p + dim(A/p) = ht q + dim(A/q), dim(A/p) = ht(q/p) + dim(A/q), part (ii) follows from part (i). 6.2.7 Definition. A chain p0 ⊂ · · · ⊂ pn of prime ideals of length n of a ring A is called maximal if it satisfies the following conditions: (i) p0 ∈ Specm (A) and pn ∈ Spm(A).
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(ii) For any integer i ∈ {0, 1, . . . , n − 1}, the set of q ∈ Spec A verifying pi ⊂ q ⊂ pi+1 is {pi , pi+1 }. 6.2.8 Proposition. Let A be a finitely generated K-algebra which is an integral domain. Then all maximal chains of prime ideals of A has length dim A. Proof. Let {0} = p0 ⊂ · · · ⊂ pn be a maximal chain of prime ideals of A. For 1 i n, the height of pi /pi−1 ∈ Spec(A/pi−1 ) is 1. Hence by 6.2.6, dim(A/pi−1 ) − dim(A/pi ) = 1. On the other hand, dim(A/pn ) = 0 since, A/pn is a field. It follows that dim A = dim(A/p0 ) = n. 6.2.9 Proposition. Let B be a finitely generated K-algebra which is an integral domain, A a finitely generated K-subalgebra of B. There exist a nonzero element t ∈ A and a subalgebra C = A[y1 , . . . , yn ] of B verifying the following conditions: (i) The elements y1 , . . . , yn are algebraically independent over K. (ii) The subalgebra B[1/t] of Fract(B) is integral over C[1/t]. Proof. Let L = Fract(A) and z1 , . . . , zr ∈ B be such that B = A[z1 , . . . , zr ]. By applying 6.2.2 to the L-algebra L[z1 , . . . , zr ], we can find x1 , . . . , xn ∈ B algebraically independent over L such that L[z1 , . . . , zr ] is integral over L[x1 , . . . , xn ]. Let m i −1 Pij (x1 , . . . , xn )zij = 0 zimi + j=0
be an integral equation for zi over L[x1 , . . . , xn ], where Pij ∈ L[T1 , . . . , Tn ]. Let k = max1jmi −1 {deg Pij } and t ∈ A non-zero such that tPij ∈ A[T1 , . . . , Tn ] for all j. Then there exist Qij ∈ A[T1 , . . . , Tn ] such that tk+1 Pij (x1 , . . . , xn ) = Qij (tx1 , . . . , txn ) and so tk+1 zimi +
m i −1 j=0
Qij (tx1 , . . . , txn )zij = 0.
Set yi = txi . Clearly, the yi ’s are algebraically independent over L, hence over K. The above equality implies that zi is integral over A[y1 , . . . , yn ][1/t]. Hence B[1/t] is integral over C[1/t]. 6.2.10 Proposition. Let A be a finitely generated K-algebra which is an integral domain, F = Fract(A), F ⊂ L a finite field extension and B the integral closure of A in L. Suppose that the characteristic of K is zero. Then B is a finitely generated A-module (so B is a finitely generated K-algebra which is an integral domain). Proof. By 6.2.2, A contains a polynomial ring C over which A is integral. Clearly, B is also the integral closure of C in L. Since Fract(C) ⊂ F is a finite extension, we can therefore assume that A is a polynomial ring. The result then follows from 4.2.9 and 5.5.6.
6.3 Krull’s Principal Ideal Theorem
81
6.3 Krull’s Principal Ideal Theorem 6.3.1 Lemma. Let C ⊂ A be integral domains, F = Fract(C) and K = Fract(A). Suppose that C is an integrally closed domain,√A is integral over C and F ⊂ K is a finite extension.√Let x ∈√A be such that Ax ∈ Spec A. Then y = NormK/F (x) ∈ C and C ∩ Ax = Cy. Proof. Let P = X m + cm−1 X m−1 + · · · + c0 be the minimal polynomial of x over F . By 5.5.4, the ci ’s are integral over C. So P ∈ C[X] √ and since ∈ Ax. It follows therefore from 5.5.2 that y ∈ C ∩ Ax. Hence P (x) = 0, c 0 √ √ Cy ⊂ C ∩ Ax. √ n ∈ N∗ and a ∈ A Conversely, if u ∈ C ∩ Ax, √ then there exist n n[K:F ] = NormK/F (un ) = such that u = ax. Hence u ∈ Cy because u NormK/F (a) NormK/F (x) ∈ Cy. 6.3.2 The following theorem is also called Krull’s Hauptidealsatz. Theorem. (Krull’s Principal Ideal Theorem) Let A be a finitely generated K-algebra, x ∈ A non invertible and p ∈ Spec A minimal such that x ∈ p. Then ht p 1 and equality holds if x is not a zero divisor. Proof. Note that such a p exists by applying 2.2.7 to A/Ax. Let us suppose that ht p = r 1. Let q0 ⊂ · · · ⊂ qr = p be a maximal chain of prime ideals. It suffices therefore to show that ht p/q0 = 1 in A/q0 . We are therefore reduced to the case where A is an integral domain and x = 0. By 2.7.8, the set of prime ideals containing x has a finite number of minimal elements, say p, p2 , . . . , ps . Let u ∈ (p2 ∩ · · · ∩ ps ) \ p, S = {un ; u ∈ N} and B = S −1 A. Since B is a finitely generated K-algebra and Fract A = Fract B, we have dim A = dim B and ht p = ht(Bp) by 6.2.3 and 2.3.9. Further, Bp is now the unique prime ideal of B minimal among those containing x. So we can suppose that p is the √ unique minimal element of the set of prime ideals containing x. By 2.7.7, Ax = p. By 6.2.2, A contains a polynomial ring C over which A is integral. Let F = Fract(C), K = Fract(A) and y = NormK/F (x). Then 6.3.1 implies that √ Cy. Now C is √ factorial, so there exists z ∈ C irreducible such that C ∩ p = √ Cy = Cz. So ht p = ht Cy = 1 by 5.7.8. 6.3.3 Corollary. Let A be a finitely generated K-algebra and a a proper ideal of A generated by x1 , . . . , xn . If p is minimal among the prime ideals of A containing a, then ht p n. Proof. Let us proceed by induction on n. We have treated the case n = 1 in 6.3.2. So let us suppose that n 2. By the same argument as in the proof of 6.3.2, we are reduced to the case where A is an integral domain. Now the set E of q ∈ Spec A such that Ax1 + · · · + Axn−1 ⊂ q ⊂ p is not empty. Let q be minimal in E. By our induction hypothesis, ht q n − 1. Further, p/q is minimal among the prime ideals of A/q containing the class of xn modulo q. So the result follows from 6.2.6 and 6.3.2.
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6.3.4 Corollary. Any maximal ideal of A = K[X1 , . . . , Xn ], the polynomial ring in n indeterminates, can be generated by n elements. Proof. Let m ∈ Spm(A). Then by 6.1.4 et 6.2.6, ht m = n. Hence 6.3.3 implies that any system of generators of m has at least n elements. If n = 1, the result is obvious since K[X1 ] is a principal ideal domain. Let us suppose that n 2 and proceed by induction on n. If m ∩ K[X1 ] = {0}, then let S = K[X1 ] \ {0} and S −1 m is a maximal ideal of K(X1 )[X2 , . . . , Xn ]. Therefore ht S −1 m = n − 1. This is not possible since ht S −1 m = ht m = n. Thus m ∩ K[X1 ] is a (non-zero) maximal ideal K[X1 ]x where x ∈ K[X1 ] is irreducible. Clearly, x is also irreducible in A and consequently, Ax is a prime ideal of A. Let y2 , . . . , yn be the images of X2 , . . . , Xn in A/Ax. We have A/Ax = L[y2 , . . . , yn ] where L = K[X1 ]/(K[X1 ]x) and by 6.2.6, dim A/Ax = n − 1. It follows from 6.2.3 that the yi ’s are algebraically independent over L. By the induction hypothesis, the maximal ideal m/Ax of A/Ax is generated by n − 1 elements. Consequently, if we choose representatives of these n − 1 elements in A, then these n − 1 representatives together with x form a system of generators for m.
6.4 Maximal ideals 6.4.1 The following result is a direct consequence of 6.2.4. Proposition. Let A be a finitely generated K-algebra and m ∈ Spm(A). The extension K ⊂ A/m is finite. Moreover, if K is algebraically closed, then the natural map K → A → A/m is an isomorphism of K onto A/m. 6.4.2 Theorem. (Hilbert’s Nullstellensatz) Let a be a proper ideal of a √ finitely generated K-algebra A. Then a is the intersection of maximal ideals of A containing a. Proof. By replacing A by A/a, it suffices to prove that the nilradical n of A is the intersection r of maximal ideals of A. Clearly, n ⊂ r (see also 2.5.5). Let a ∈ r. If a ∈ n, then S = {an ; n ∈ N} does not contain 0. The Kalgebra B = S −1 A is again finitely generated (just add 1/a to a system of generators of A). Let i : A → B be the canonical map (see 2.3.3), m ∈ Spm(B) and i−1 (m) = p ∈ Spec(A) (see 2.3.8). By 6.4.1, K ⊂ B/m is finite. So we deduce, via the canonical injections K → A/p → B/m, that dim A/p = 0. So p ∈ Spm(A) by 6.2.4. But a ∈ m, so a ∈ p which is absurd since a ∈ r. 6.4.3 Proposition. Let A = K[X1 , . . . , Xn ] be the polynomial ring in n indeterminates, and a an ideal of A. Suppose that K is an algebraically closed field. The following conditions are equivalent: n (i) There exist a1 , . . . , an ∈ K such that a = i=1 A(Xi − ai ). (ii) We have a ∈ Spm(A).
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83
Proof. (i) ⇒ (ii) This is clear since A/a is isomorphic to K. (ii) ⇒ (i) Let a ∈ Spm A. Then 6.2.6 implies that K ⊂ A/a is finite and since K is algebraically closed, we have K = A/a. Let ai be the image of Xi in A/a = K. Then m = A(X1 − a1 ) + · · · + A(Xn − an ) ⊂ a. But m ∈ Spm(A), so m = a. 6.4.4 Corollary. Let A, B be finitely generated K-algebras. Suppose that K is algebraically closed. (i) Let m ∈ Spm(A). Then A = K.1 ⊕ m. (ii) Let ρ : A → B be a K-algebra homomorphism. If m ∈ Spm(B), then ρ−1 (m) ∈ Spm(A). Proof. Part (i) is clear since 1 ∈ m and A/m = K by 6.4.1. (ii) Since ρ(1) = 1 and B/m = K by part (i), ρ−1 (m) = A and the induced map A/ρ−1 (m) → B/m is an isomorphism. So ρ−1 (m) ∈ Spm(A). 6.4.5 For any K-algebras A, B, let us denote by Homalg (A, B) the set of K-algebra homomorphisms from A to B. Given a finitely generated K-algebra A, it is clear from 6.4.4 that there is a canonical bijection: Homalg (A, K) → Spm(A) , χ → ker χ whose inverse is given by the projection A = K.1 ⊕ m → K = A/m for m ∈ Spm A. Now if B is another finitely generated K-algebra and ϕ ∈ Homalg (A, B), then the map ϕ∗ : Homalg (B, K) → Homalg (A, K) , χ → χ ◦ ϕ induces a map Spm(ϕ) : Spm(B) → Spm(A) via the canonical bijections above. More precisely, if n ∈ Spm(B), then Spm(ϕ)(n) = ϕ−1 (n). 6.4.6 We can view A as an algebra of functions on Spm(A) as follows: let f ∈ A and m ∈ Spm(A), then we define f (m) to be the class of f in A/m = K. Note that if ϕ ∈ Homalg (A, B), then for n ∈ Spm(B) and f ∈ A, we have: (f ◦ Spm(ϕ))(n) = f (ϕ−1 (n)) = ϕ(f )(n) since the homomorphism A/ϕ−1 (n) → B/n induced by ϕ is just the identity map from K to itself.
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6.5 Zariski topology 6.5.1 Let A be a finitely generated K-algebra. For any subset E of A, we shall denote by V(E) the set of maximal ideals of A which contain E. When E = {f1 , . . . , fn }, we shall also write V(f1 , . . . , fn ) for V(E). 6.5.2 Remark. Suppose that K is algebraically closed. Then considering A as functions as in 6.4.6, V(E) can be identified as the set of common zeros of the elements in E. Note also that if a is an ideal of A generated by E, then V(a) = V(E). 6.5.3 Proposition. Let A be a finitely generated K-algebra, a, b and (ai )i∈I ideals of A.√We have: (i) V(a) = V( a) and if a ⊂ b, then V(b) ⊂ V(a). (ii) V(a) = ∅ if and only if a = A. √ (iii) V(a) if and only if a ⊂ 0. = Spm(A) V(ai ) and V(ab) = V(a ∩ b) = V(a) ∪ V(b). (iv) V( ai ) = i∈I
i∈I
Proof. The first point of (i) and part (iii) are consequences of 6.4.2, while the second point of (i), part (ii) and the first point of (iv) are obvious. If m ∈ Spm(A), then m ⊃ ab if and only if m contains either a or b. So V(ab) = V(a) ∪ V(b). The other equalities follow from part (i) and 2.5.3. 6.5.4 In view of 6.5.3, we define a topology, called Zariski topology on Spm(A), by letting closed subsets to be the V(a)’s where a is an ideal of A. Observe that by part (i) of 6.5.3, we can restrict ourselves to radical ideals. By convention, when we consider Spm(A) as a topological space, we assume that the underlying topology is the Zariski topology. Let a be an ideal of A and B = A/a. Then the map given by m → m/a is a bijection between the set V(a) and Spm B. Clearly, via this identification, the Zariski topology of Spm(B) is the one induced by the Zariski topology of Spm(A). Suppose that K is algebraically closed and ϕ : A → B is the canonical surjection. Then the map Spm(ϕ) of 6.4.5 is the canonical injection V(a) → Spm A. 6.5.5 Let A be a finitely generated K-algebra. For M ⊂ Spm(A), we denote by I(M ) the set of elements a ∈ A such that a ∈ m for all m ∈ M . Clearly, m. I(M ) = m∈M
Proposition. Let a, b be√ideals of A and M ⊂ Spm A. (i) We have I(V(a)) = a. √ √ (ii)√We have √ V(a) ⊂ V(b) (resp. V(a) = V(b)) if and only if b ⊂ a (resp. b = a). (iii) The Zariski closure of M in Spm(A) is V(I(M )).
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85
Proof. Part (i) is proved in 6.4.2 and part (ii) is a direct consequence of 6.5.3 (i) and the fact that the map M → I(M ) reverses inclusions. To prove part (iii), let a be a radical ideal of A such that M ⊂ V(a). Then a = I(V(a)) ⊂ I(M ) by (i). It follows that V(I(M )) ⊂ V(a). On the other hand, it is clear that M ⊂ V(I(M )). Hence we have proved (iii). 6.5.6 Proposition. Let A be a finitely generated K-algebra. (i) Spm(A) is a Noetherian topological space whose points are closed. (ii) Let a be a radical ideal of A. Then a ∈ Spec(A) if and only if V(a) is irreducible. Proof. (i) Since V(m) = {m} for all m ∈ Spm(A), points are closed. Let (Fp )p be a decreasing sequence of closed subsets of Spm(A) where Fp = V(ap ), ap a radical ideal of A. Then by 6.5.5, (ap )p is an increasing sequence of radical ideals of A. Since A is Noetherian (6.1.1), this sequence is stationary. It follows that (Fp )p is also stationary. (ii) Suppose that V(a) is irreducible and let f, g ∈ A be such that f g ∈ a. We have V(a) ⊂ V(f g) = V(f ) ∪ V(g). √ Hence V(a) is contained in either V(f ) or V(g). Now 6.5.5 implies that f ∈ a = a or g ∈ a. So a ∈ Spec(A). Conversely, suppose that a ∈ Spec(A). Let b, c be radical ideals of A such √ that V(a) ⊂ V(b) ∪ V(c). So bc ⊂ a = a by 6.5.3 and 6.5.5. Since a is prime, a contains either b or c. Hence V(a) is contained in either V(b) or V(c). 6.5.7 Corollary. Let a be an ideal of A and p1 , . . . , pn be the minimal elements of the set of prime ideals of A containing a. Then the set of irreducible components of V(a) is the finite set {V(pi ) ; 1 i n}. In particular, the irreducible components of Spm(A) are the closed sets associated to minimal prime ideals of A. So Spm(A) is irreducible if and only √ if 0 ∈ Spec(A). Furthermore, if A is reduced, then Spm(A) is irreducible if and only if A is an integral domain. √ Proof. By 2.7.7 and 2.7.8, we have a = p1 ∩ · · · ∩ pn , and so 6.5.3 implies that V(a) = V(p1 ) ∪ · · · ∪ V(pn ). The result follows from 6.5.6 and 6.5.5. 6.5.8 Proposition. Suppose that K is algebraically closed. Let A, B be finitely generated K-algebras and ϕ ∈ Homalg (B, A). (i) The map Spm(ϕ) : Spm(A) → Spm(B) defined in 6.4.5 is continuous. (ii) If b is an ideal of B, then Spm(ϕ)−1 (V(b)) = V(Aϕ(b)). (iii) For any ideal a of A, the Zariski closure of Spm(ϕ)(V(a)) is the set V(ϕ−1 (a)). Proof. It suffices to prove (ii) and (iii). Let m ∈ Spm(A), then m ∈ V(Aϕ(b)) ⇔ Aϕ(b) ⊂ m ⇔ ϕ(b) ⊂ m ⇔ b ⊂ ϕ−1 (m) = Spm(ϕ)(m) ⇔ Spm(ϕ)(m) ∈ V(b).
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This proves (ii). Let a be an ideal of A. By 6.5.5, the Zariski closure of Spm(ϕ)(V(a)) is V(I(ϕ(V(a)))), the set of maximal ideals of B containing the intersection of the ϕ−1 (m)’s where m ∈ V(a). Now √ ϕ−1 (m) = ϕ−1 m = ϕ−1 ( a) = ϕ−1 (a). m∈V(a)
m∈V(a)
Hence (iii) follows from 6.5.3. 6.5.9 Corollary. Let us keep the notations and hypotheses of 6.5.8. Suppose further that A and B are reduced. Then ϕ is injective if and only if Spm(ϕ) is dominant. Proof. By 6.5.8 (iii), Spm(ϕ) Spm(A) = V(ϕ−1 ({0})). Note that {0} is rad √ ical in the reduced ring A. So the equality ϕ−1 ( a) = ϕ−1 (a) implies that ϕ−1 ({0}) is radical in B. Since B is also reduced, it follows from 6.5.3 and 6.5.5 that V(ϕ−1 ({0})) = Spm(B) if and only if ϕ−1 ({0}) = {0}.
References • [9], [26], [52], [56].
7 Gradings and filtrations
In this chapter, we study the notions of gradings and filtrations on rings and modules. These notions are useful in many areas of mathematics. We shall use them in particular in Chapters 13, 30 and 31.
7.1 Graded rings and graded modules 7.1.1 Definition. Let G be an abelian group. A grading on G is a family (Gn )n∈Z of subgroups of G such that G is the direct sum of the Gn ’s. A group equipped with a grading is called a graded group. 7.1.2 Let G be a graded group. An element x ∈ G is called homogeneous of degree n if x ∈ Gn . If x is non-zero and homogeneous, then x is contained the degree of x. in a unique Gn and we shall denote by n = deg(x) Any y ∈ G can be written uniquely in the form n yn where yn ∈ Gn . We call yn the homogeneous component of degree n of y. When there is no confusion, we shall denote the grading of a graded group G by (Gn )n∈Z . 7.1.3 Definition. Let A be a ring. A grading (An )n∈Z on A is said to be compatible with the ring structure if An Am ⊂ An+m for all m, n ∈ Z. A ring equipped with a compatible grading is called a graded ring. 7.1.4 Definition. Let A be a graded ring and M a left A-module. A grading (Mn )n∈Z on M is said to be compatible with the A-module structure if Am Mn ⊂ Mm+n for all m, n ∈ Z. A left A-module equipped with a compatible grading is called a graded left A-module. 7.1.5 Proposition. Let A be a graded ring. Then A0 is a subring of A. Proof. Clearly, A0 A0 ⊂ A0 . Now let (en )n∈Z be the set of homogeneous components of 1. For any x ∈ Ap , we have x = x.1 = xen = 1.x = en x. n∈Z
n∈Z
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By comparing the homogeneous components of degree p, we obtain that x = xe0 = e0 x. Since any element of A is the sum of its homogeneous components, the equality x = xe0 = e0 x is valid for all x ∈ A. Hence 1 = e0 ∈ A0 . 7.1.6 It follows from 7.1.5 that if M is a graded left A-module, then each Mn is a left A0 -module. We define graded right A-modules similarly. All modules considered in the rest of this chapter are left modules. 7.1.7 Definition. (i) Let A and B be graded rings. A ring homomorphism f : A → B is called graded if f (An ) ⊂ Bn for all n ∈ Z. (ii) Let A be a graded ring and M, N graded A-modules. A homomorphism u : M → N of A-modules is called graded of degree p ∈ Z if u(Mn ) ⊂ Nn+p for all n ∈ Z. A homomorphism u : M → N of A-modules is called graded if it is graded of degree p for some p ∈ Z. Note that if u is non-zero, then p is uniquely determined by u.
7.2 Graded submodules 7.2.1 Proposition. Let M be a graded A-module and N a submodule of M . The following conditions are equivalent: (i) N is a graded submodule, i.e. N = n∈Z (N ∩ Mn ). (ii) The homogeneous components of any element of N are in N . (iii) N is generated by homogeneous elements. Proof. This is straightforward. 7.2.2 Corollary. Let N be a graded submodule of a graded A-module M . (i) If (xi )i∈I is a system of generators of N , then the homogeneous components of the xi ’s form a system of homogeneous generators of N . (ii) If N is finitely generated, then it has a finite system of homogeneous generators. 7.2.3 Let N be a graded submodule of a graded A-module M , then one verifies easily that M/N = n∈Z (N + Mn )/N is a graded (quotient) module. 7.2.4 Let A be a graded ring. A graded submodule of A is called a graded (left) ideal. If a is a bilateral ideal of A, then the preceding paragraph says that the quotient A/a is again a graded ring. 7.2.5 In the rest of this section, A is a commutative graded ring. We call a graded A-module (Mn )n∈Z positively graded if Mn = {0} for all n < 0. 7.2.6 Proposition. Let a be a graded ideal of A. Then a is prime if and only if given any homogeneous elements x, y ∈ A such that xy ∈ a, we have x ∈ a or y ∈ a.
7.2 Graded submodules
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Proof. The condition is obviously necessary. Let us prove that it is sufficient. Let a, b ∈ A be such that ab ∈ a. Write a = ai1 + · · · + air , b = bj1 + · · · + bjs where i1 < · · · < ir , j1 < · · · < js and the aik ’s and bjl ’s are homogeneous of degree ik and jl . Now we can assume that none of the bjl ’s belong to a. Since ab ∈ a and a is graded, the homogeneous component of degree ir js of ab, which is air bjs , belongs to a. It follows that air ∈ a, and so (a − air )b ∈ a. The result follows by induction. √ 7.2.7 Proposition. Let a be a graded ideal of A. Then a is graded. √ Proof. Let x ∈ a. Write x = xi1 + · · · + xir where i1 < · · · < ir and xik is homogeneous of degree ik . There exists n ∈ N∗ such that xn ∈ a. Since xnir is and a is graded, xnir ∈ a. So the homogeneous component of degree nir of xn √ √ √ xir ∈ a and x − xir ∈ a. By induction, xik ∈ a for k = 1, . . . , r. 7.2.8 Theorem. Let A be positively graded. The following are equivalent: (i) A is Noetherian. (ii) A0 is Noetherian and A is a finitely generated A0 -algebra. Proof. (i) ⇒ (ii) Set A+ = n∈N∗ An . This is a graded ideal of A and A0 A/A+ . So A0 is Noetherian. Further, A+ is finitely generated, so by 7.2.2, it is generated by a finite number of homogeneous elements xi1 , . . . , xir where xik ∈ Aik ⊂ A+ . Denote by B the A0 -subalgebra of A generated by the xik ’s. Clearly A0 ⊂ B. Now assume that Ak ⊂ B for k = 0, . . . , n. Let y ∈ An+1 . Then we can find a1 ∈ An+1−i1 , . . ., ar ∈ An+1−ir such that y = a1 xi1 + · · · + ar xir . By our assumption, ak ∈ B for all k = 1, . . . , r. Hence An+1 ⊂ B, and by induction, we have A = B. (ii) ⇒ (i) Since A is a finitely generated A0 -algebra, A is isomorphic to a quotient of a polynomial algebra A0 [X1 , . . . , Xs ]. The result follows by 2.7.3 and 2.7.9. 7.2.9 Let K be a commutative field and A = K[T1 , . . . , Tn ] together with its natural grading. The following result is a graded version of 2.2.5. Proposition. Let p1 , . . . , pr be graded prime ideals of A. If a is a graded ideal of A verifying a ⊂ pi for 1 i r, then there exists a homogeneous element a ∈ a such that a ∈ p1 ∪ · · · ∪ pr . Proof. If a = A, then a = 1 works. So let us assume that a = A. We can further assume that pi ⊂ pj for i = j. We shall proceed by induction on r. The case r = 1 is obvious. So let r 2. For 1 i r, by induction, there exists a homogeneous element ai ∈ a such that ai ∈ j=i pj = Ei . If there exists i such that a ∩ pi ⊂ Ei , then ai ∈ p1 ∪ · · · ∪ pr . If a ∩ pi ⊂ Ei for 1 i r, then we may take ai ∈ a ∩ pi , and by replacing the ai ’s by some suitable powers, we can assume that deg(a1 ) =
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deg(a2 · · · ar ). Then a = a1 + a2 · · · ar is a homogeneous element of a and clearly a ∈ p1 ∪ · · · ∪ pr .
7.3 Applications 7.3.1 In this section, A is a commutative ring. 7.3.2 Theorem. (Artin-Rees Theorem) Let A be Noetherian, a an ideal of A, M a finitely generated A-module et N a submodule of M . There exists q ∈ N∗ such that, for all n q, we have N ∩ (an M ) = an−q (N ∩ aq (M )). Proof. Set B=
n0
an , S =
an M , T =
n0
(N ∩ an M ).
n0
One verifies easily that B is a (positively) graded ring, and S is a (positively) graded B-module. Now B is generated by a as an A-algebra. Since A is Noetherian, B is a finitely generated A-algebra. It follows from 7.2.8 that B is Noetherian. Hence S is finitely generated, and T is a finitely generated graded submodule of S. For q ∈ N, set Fq =
q i=0
(N ∩ ai M ) , Gq = Fq ⊕
ai (N ∩ aq M ) .
i1
Since ai (N ∩ aq M ) ⊂ N ∩ aq+i M , (Gq )q is an increasing sequence of submodules of T . Further, the union of the Gq ’s is T . So there exists q such that T = Gq . Therefore, for n q, we have N ∩ an M = an−q (N ∩ aq M ). 7.3.3 Lemma. Let a be an ideal of A and N a finitely generated A-module. Then N = aN if and only if there exists a ∈ a such that (1 − a)N = 0. Proof. Let x1 , . . . , xs be a system of generators of N . If N = aN , then for 1 i s, there exist aij ∈ a such that xi = ai1 x1 + · · · + ais xs . The determinant of [δij − aij ]1i,js is of the form 1 − a for some a ∈ a. But for 1 i s, (1 − a)xi = 0. Hence (1 − a)N = {0}. The converse is obvious. 7.3.4 Theorem. Let A be Noetherian, a an ideal of A and M a finitely generated A-module. Then x ∈ n0 an M if and only if there exists a ∈ a such that ax = x. In particular, we have n0 an M = {0} if and only if for any a ∈ a \ {0}, x = 0 is the unique solution in M of ax = x. Proof. This is a consequence of 7.3.2 and 7.3.3 with N = n0 an M .
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7.3.5 Corollary. Let A be a Noetherian integral domain and a an ideal of A distinct from A. Then n0 an = {0}. 7.3.6 Corollary. Let A be Noetherian and b an ideal of A. Then b is contained in the Jacobson radical rad A of A if and only if (a + bn ) = a n0
for any ideal a of A. Further, if b ⊂ rad A, then (N + bn M ) = N n0
for any submodule N of a finitely generated A-module M . Proof. Let b ⊂ rad A, M a finitely generated A-module and N a submodule of M . One checks easily that bn (M/N ) = (N + bn M )/N for all n ∈ N. By 7.3.4, if x ∈ n0 bn (M/N ), then there exists b ∈ b such that (1 − b)x = 0. But 1 − b is invertible since b ⊂ rad A. So x = 0 and N is the intersection of the (N + bn M )’s, for n 0. In particular, a is the intersection of the a + bn , for n 0. Conversely, if b ⊂ rad A, then b ⊂ m forsome m ∈ Spm(A). Hence bn ⊂ m, and bn + m = A for all n ∈ N. Thus A = n0 (m + bn ) = m. 7.3.7 Corollary. Let A be Noetherian and a an ideal of A such that a ⊂ rad A. Then, for any finitely generated A-module M : n a M = {0}. n0
7.4 Filtrations 7.4.1 Definition. A filtration on a group G is an increasing sequence (Gn )n∈Z of subgroups of G such that G = n∈Z Gn . 7.4.2 Definition. A filtration on a ring A is a filtration (An )n∈Z of the additive group A which is compatible with the ring structure, i.e.: (1) An Am ⊂ Am+n for all m, n ∈ Z. (2) 1 ∈ A0 . 7.4.3 Definition. Let (An )n∈Z be a filtration on a ring A, M an Amodule. A filtration (Mn )n∈Z is said to be compatible (with the filtration on A) if Am Mn ⊂ Mm+n for all m, n ∈ Z. 7.4.4 Let (Gn )n∈Z be a filtration on G, and H a subgroup of G. Then (H ∩ Gn )n∈Z is called the induced filtration on H. Similarly, if H is normal, then (HGn /H)n∈Z is a filtration on G/H. Let (An )n∈Z be a filtration of a ring A. Then the above procedure provides filtrations on any subring of A and any quotient of A by a bilateral ideal a.
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It is easy to check that the same procedure works for any submodule N of an A-module M with a compatible filtration. 7.4.5 Let (An )n∈Z be a filtration on the ring A and M, N two A-modules with compatible filtrations. A homomorphism u : M → N is said to be compatible if u(Mn ) ⊂ Nn for all n ∈ Z.
7.5 Grading associated to a filtration 7.5.1 Let (Gn )n∈Z be a filtration on an additive group G. For n ∈ Z, let grn (G) = Gn /Gn−1 and grn (G) gr(G) = n∈Z
is a graded commutative group. We call gr(G) the graded group associated to the filtration. 7.5.2 Let (An )n∈Z be a filtration on a ring A, and (Mn )n∈Z a compatible filtration on an A-module M . Let p, q ∈ Z, we define: (1) grp (A) × grq (M ) → grp+q (M ) , (a + Ap−1 , m + Mq−1 ) → am + Mp+q−1 . One verifies easily that this is a well-defined Z-bilinear map. This induces a Z-bilinear map: (2)
gr(A) × gr(M ) → gr(M ).
In particular, when M = A, we obtain a ring structure on gr(A) via this map. We call gr(A) the graded ring associated to A. It follows that gr(M ) is a graded gr(A)-module called the graded module associated to M . 7.5.3 Let (An )n∈Z be a filtration on a ring A, and M , N be A-modules with compatible filtrations. If u : M → N is a compatible homomorphism, then it is easy to see that u induces a homomorphism gr(u) of the associated graded modules given by gr(u)(m + Mp−1 ) = u(m) + Np−1 for m + Mp−1 ∈ grp (M ). Further, if v : N → P is another compatible homomorphism of A-modules with compatible filtrations, then gr(v ◦ u) = gr(v) ◦ gr(u). In the rest of this section, let (An )n∈Z be a filtration on a ring A and let M be an A-module with a compatible filtration (Mn )n∈Z . Proposition. Let N be a submodule of M . Then we have the following exact sequence of graded gr(A)-modules: gr(j)
gr(p)
0 −→ gr(N ) −→ gr(M ) −→ gr(M/N ) −→ 0 where j : N → M is the canonical inclusion, p : M → M/N is the canonical surjection, and we endow N with the induced filtration. Proof. The homomorphisms j and p are clearly compatible and the exactness of the sequence is a straightforward verification.
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7.5.4 Let N = n∈Z Mn . If N = {0}, then we say that the filtration (on M ) is separated. Let x ∈ M , we define the order of x, denoted by ν(x) or νM (x), to be: ν(x) = +∞ if x ∈ N , ν(x) = p if x ∈ Mp , x ∈ Mp−1 . 7.5.5 Proposition. Assume that gr(A) has no zero divisors other than 0, then ν(ab) = ν(a) + ν(b) for all a, b ∈ A. Proof. Since n∈Z An is a bilateral ideal of A, the equality is clear if a or b belongs to n∈Z An . Now if ν(a) = r and ν(b) = s where r, s ∈ Z, then clearly ab ∈ Ar+s . Since gr(A) has no zero divisors other than 0, ab ∈ Ar+s−1 . So ν(ab) = ν(a) + ν(b). 7.5.6 Corollary. If gr(A) has no zero divisors other than 0, then neither has A/ n∈Z An . Proof. By 7.5.4, we have ν(ab) = +∞ for any a, b ∈ A \ n∈Z An . Hence ab ∈ n∈Z An . 7.5.7 Proposition. Let u : M → N be a compatible homomorphism of A-modules with compatible filtrations. If the filtration on M is separated and gr(u) is injective, then u is also injective. Proof. Note that, in general, we have u( n∈Z Mn ) ⊂ n∈Z Nn . Let us assume that the filtration on M is separated and gr(u) is injective. The injectivity of gr(u) implies that νM (x) = νN (u(x)) for any x ∈ M . So if x ∈ ker u, then νM (x) = +∞. Since the filtration on M is separated, u is injective. 7.5.8 If there exists n ∈ Z such that Mn = {0}, then Mp = {0} for all p n, and we say that the filtration on M is discrete. Note that a discrete filtration is separated. Proposition. Let u : M → N be a compatible homomorphism of Amodules with compatible filtrations. If the filtration on N is discrete and gr(u) is surjective, then u is also surjective. Proof. Let us assume that the filtration on N is discrete and gr(u) is surjective. There exists q ∈ Z such that Nq = {0}. This implies that u(Mn ) = Nn for n q. Now let us assume that u(Mn ) = Nn for n p. Since gr(u) is surjective, Np+1 = u(Mp+1 )+Np = u(Mp+1 )+u(Mp ) = u(Mp+1 ). It follows by induction that u(Mn ) = Nn for all n ∈ Z. Hence the surjectivity of u. 7.5.9 In the rest of this section, all the filtrations considered are discrete. Let x ∈ M and n = ν(x). We shall denote the image of x in grn (M ) by σM (x) or simply σ(x), and we shall call σ(x) the principal symbol of x. This induces a map σM : M → gr(M ).
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7.5.10 Proposition. Let (xi )i∈I be a family of elements of M . (i) If (σ(xi ))i∈I is a system of generators of the gr(A)-module gr(M ), then (xi )i∈I generates the A-module M . (ii) If gr(M ) is a free gr(A)-module and (σ(xi ))i∈I is a gr(A)-basis of gr(M ), then M is a free A-module and (xi )i∈I is a A-basis of M . Proof. (i) Let q ∈ Z be such that Aq = {0} and Mq = {0}. Let x ∈ Mn . We shall prove by induction on n that x is contained in the A-submodule generated by (xi )i∈I . This is true for n q. So we may assume that n > q and ν(x) = n. There exist non-zero elements α1 , . . . , αr ∈ gr(A) and indices i1 , . . . , ir ∈ I such that σ(x) = α1 σ(x1 ) + · · · + αr σ(xr ). Set ni = ν(xi ), then we can find a1 , . . . , ar ∈ A such that σ(aj ) = αj and ν(aj ) = n − nj for all 1 j r. It follows that: x−
r
aj xj ∈ Mn−1 .
j=1
By our induction hypothesis, x is contained in the A-submodule generated by (xi )i∈I . (ii) Let N be a free A-module with basis (yi )i∈I . Define a (discrete) filtration on N as follows: for n ∈ Z, Aj yi Nn = j+ni n
where ni = ν(xi ) for i ∈ I. One verifies easily that this is a compatible filtration and the map u : N → M defined by sending yi to xi is a compatible homomorphism of A-modules. The associated map gr(u) is clearly bijective and it follows from 7.5.7 and 7.5.8 that u is an isomorphism. Hence M is free and (xi )i∈I is a basis of M . 7.5.11 Corollary. If gr(M ) is a finitely generated (resp. Noetherian) gr(A)-module, then M is a finitely generated (resp. Noetherian) A-module. Proof. These are direct consequences of 7.2.2, 7.5.3 and 7.5.10. 7.5.12 Corollary. If gr(A) is a left Noetherian ring, then so is A.
References • [10], [52], [56].
8 Inductive limits
8.1 Generalities 8.1.1 Recall that a preorder on a set I is a binary relation which is reflexive and transitive. Let be a preorder on I, then it is filtered or directed if for all i, j ∈ I, there exists k ∈ I such that i k and j k. A subset J ⊂ I is cofinal if for all i ∈ I, there exists j ∈ J such that i j. A set with a filtered preorder is called a filtered set or directed set. 8.1.2 Remarks. 1) A preorder on a set I is filtered if and only if any non-empty finite subset of I has an upper bound. 2) Let X be a topological space and x ∈ X. Then the set of neighbourhoods of x is a filtered set via the preorder U V if and only if V ⊂ U . 8.1.3 In the rest of this chapter, I is a filtered set with preorder . 8.1.4 Definition. An inductive system of sets indexed by I is a family of sets (Ei )i∈I and maps (fji )i,j∈I,ij verifying: for i, j, k ∈ I such that i j k, (i) fji is a map from Ei to Ej . (ii) fii is the identity map on Ei . (iii) fkj ◦ fji = fki . When there is no confusion, we shall denote an inductive system of sets indexed by I simply by (Ei , fji ). 8.1.5 Let (Ei , fji ) be an inductive system of sets indexed by I, and F be the sum of the Ei ’s. If x ∈ F , we define θ(x) to be the unique index i ∈ I such that x ∈ Ei . Define a binary relation R on F as follows: for x, y ∈ F , let i = θ(x) and j = θ(y), then xRy ⇔ there exists k ∈ I such that k i, k j and fki (x) = fkj (y). One verifies easily that R is an equivalence relation and the quotient lim (E −→ i , fji ) = F/R is called the inductive limit of the inductive system. To
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simply notations, we shall write −→ lim Ei for −→ lim(Ei , fji ) when there is no confusion. Note that −→ lim(Ei , fji ) is non-empty if at least one of the Ei ’s is non-empty. For i ∈ I, we write fi for the restriction to Ei of the canonical surjection F → −→ lim Ei . In particular, one verifies easily that for i, j ∈ I, (1)
fi = fj ◦ fji whenever i j.
8.1.6 Examples. 1) Let G be a set and consider the inductive system (Ei , fji ) where Ei = G for all i ∈ I and if i j, then fji is the identity map on G. The maps fi are identical for all i ∈ I, and therefore there is a canonical bijection from G to −→ lim Ei . 2) Let A, B be sets and (Vi )i∈I a family of subsets of A such that Vj ⊂ Vi if i j. Let Ei be the set of maps from Vi to B and define, for i, j ∈ I, i j, fji : Ei → Ej by fji (u) = u|Vj . Then (Ei , fji ) is an inductive system. We call lim Ei the set of germs of functions from Vi to B. This generalizes the case −→ where (Vi )i∈I is the set of neighbourhoods of a point of a topological space A. 8.1.7 Proposition. Let (Ei , fji ) be an inductive system of sets, E = lim −→ Ei and for i ∈ I, fi : Ei → E the canonical map. (i) Let x1 , . . . , xn ∈ E. There exist i ∈ I and y1 , . . . , yn ∈ Ei such that xk = fi (yk ) for 1 k n. (ii) Let i ∈ I and y1 , . . . , yn ∈ Ei be such that fi (y1 ) = · · · = fi (yn ). There exists j ∈ I such that i j and fji (y1 ) = · · · = fji (yn ). Proof. (i) For 1 k n, there exist ik ∈ I and zk ∈ Eik such that xk = fik (zk ). Since I is a directed set, there exists i ∈ I such that i ik for 1 k n. Set yk = fi,ik (zk ). Then (1) implies that xk = fik (zk ) = fi ◦ fi,ik (zk ) = fi (yk ). (ii) For 1 h, k n, there exists αh,k ∈ I such that αh,k i and fαh,k ,i (yh ) = fαh,k ,i (yk ). Let j ∈ I be such that j αh,k for all 1 h, k n. Then fji (yh ) = fji (yk ) since fji = fj,αh,k ◦ fαh,k ,i .
8.2 Inductive systems of maps 8.2.1 Let (Ei , fji ) be an inductive system of sets indexed by I. Set E = lim −→ Ei and for i ∈ I, fi : Ei → E the canonical map. Let G be a set. Proposition. Let (ui )i∈I be a family of maps verifying ui : Ei → G, and (2)
uj ◦ fji = ui if i j. (i) There exists a unique map u : E → G such that
(3)
ui = u ◦ fi for all i ∈ I.
(ii) The map u is surjective if and only if G is the union of the ui (Ei )’s. (iii) The map u is injective if and only if for any i ∈ I and x, y ∈ Ei , ui (x) = ui (y) implies that fji (x) = fji (y) for some j i.
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97
Proof. (i) Let F be the disjoint union of the Ei ’s and f : F → E the canonical surjection. There is a unique map v : F → G such that v|Ei = ui for i ∈ I. Now (2) says that v is compatible with the equivalence relation R. So there is a unique map u : E → G verifying v = u ◦ f . Part (ii) is obvious and part (iii) is just a consequence of 8.1.7. 8.2.2 Remarks. 1) When u is bijective, we shall also call G the inductive limit of (Ei , fji ). 2) If all the fji ’s are injective, then so are the fi ’s. In this case, Ei can be identified with fi (Ei ), and E can be viewed as the union of the Ei ’s. Conversely, let (Fi )i∈I be a family of subsets of a set F such that F = i∈I Fi and Fi ⊂ Fj if i j. For i j, denote by fji : Fi → Fj the canonical injection. Then F is the inductive limit of the inductive system (Fi , fji ), and for i ∈ I, the canonical map fi : Fi → F is just the canonical injection. 8.2.3 Proposition. Let (Ei , fji ) and (Fi , gji ) be inductive systems of sets indexed by I. Let (ui : Ei → Fi )i∈I be a family of maps such that for all i, j ∈ I verifying i j, the following diagram is commutative: u
i Ei −−−− → ⏐ ⏐ fji
Fi ⏐ ⏐gji
Ej −−−−→ Fj uj
Let E = −→ lim Ei , F = −→ lim Fi and for i ∈ I, let fi and gi be the corresponding canonical maps. There exists a unique map u : E → F such that for any i ∈ I, the following diagram is commutative: ui → Fi Ei −−−− ⏐ ⏐ ⏐gi ⏐ fi E −−−−→ F u
Further, u is injective (resp. surjective) if all the ui ’s are injective (resp. surjective). Proof. The family (gi ◦ ui )i∈I satisfies the hypothesis of 8.2.1, so there exists a unique map u : E → F such that u ◦ fi = gi ◦ ui . If the ui ’s are surjective, then gi ◦ ui (Ei ) = u ◦ fi (Ei ) = u( fi (Ei )) = u(E). F = i∈I
i∈I
i∈I
If the ui ’s are injective, then for any i ∈ I, x, y ∈ Ei such that u(fi (x)) = u(fi (y)), we have gi (ui (x)) = gi (ui (y)). Now by 8.1.7, there exists j i such that gji (ui (x)) = gji (ui (y)). Thus uj (fji (x)) = uj (fji (y)), and since uj is injective, fji (x) = fji (y). So u is injective (8.2.1).
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8.2.4 A family (ui : Ei → Fi )i∈I of maps verifying the hypothesis of 8.2.3 shall be called an inductive system of maps from (Ei , fji ) to (Fi , gji ). The map u is called the inductive limit of the family (ui )i∈I and we shall denote it by −→ lim ui when there is no confusion. 8.2.5 Corollary. Let (ui : Ei → Fi )i∈I be an inductive system of maps from (Ei , fji ) to (Fi , gji ), and (vi : Fi → Gi )i∈I an inductive system of maps from (Fi , gji ) to (Gi , hji ). Then (vi ◦ ui )i∈I is an inductive system of maps from (Ei , fji ) to (Gi , hji ) and (4)
lim(vi ◦ ui ) = (lim −→ vi ) ◦ (lim −→ ui ). −→
Proof. This is a straightforward verification. 8.2.6 Let J be a directed subset of I, i.e. J is filtered with respect to the preorder induced from I. Then the subfamily of sets (Ei )i∈J and the subfamily of maps (fji )ij,i,j∈J form an inductive system of sets indexed by J, denoted by (Ei , fji )J . We call (Ei , fji )J the inductive subsystem relative to J. lim(Ei , fji )J . For i ∈ I, denote by fi the Now let E = −→ lim Ei and EJ = −→ canonical map. Then (fi )i∈J is an inductive system of maps from (Ei , fji )J to (Ei , fji ), and g = −→ lim(fi )i∈J : EJ → E. lim(fi )i∈K : EK → Let K ⊂ J be another directed subset of I, and g = −→ lim(fi )i∈K : EK → E, then by 8.2.5, we have EJ , g = −→ (5)
g = g ◦ g .
8.2.7 Proposition. Let J be a cofinal subset of I. The map g = −→ lim(fi )J : EJ → E is bijective. Proof. Let i ∈ J and x, y ∈ Ei be such that fi (x) = fj (y). By 8.1.7, there exists j ∈ J verifying j i and fji (x) = fji (y). By 8.2.1(iii), g is injective. For i ∈ J, let hi : Ei → EJ be the canonical map. If x ∈ E, then there exist i ∈ I and y ∈ Ei such that x = fi (y). If j ∈ J is such that j i, then x = fj ◦ fji (y) = g(hj ◦ fji (y)). It follows that g is surjective. 8.2.8 Remark. The previous statement says that in order to determine the inductive limit, it suffices to determine the inductive limit of the inductive subsystem relative to a cofinal subset of I.
8.3 Inductive systems of magmas, groups and rings 8.3.1 Definition. Let (Ei )i∈I be a family of magmas (resp. semigroups, monoids, groups, rings). An inductive system (Ei , fji ) indexed by I is called an inductive system of magmas (resp. semigroups, monoids, groups, rings) if the fji ’s are homomorphisms of magmas (resp. semigroups, monoids, groups, rings).
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99
8.3.2 Let us denote E = −→ lim Ei and fi : Ei → E the canonical maps. Proposition. There exists a unique structure of magma (resp. semigroups, monoids, groups, rings) on E such that for i ∈ I, fi is a homomorphism of magmas (resp. semigroups, monoids, groups, rings). Further, if the Ei ’s are commutative, then so is E. Proof. Let x, y ∈ E, i ∈ I and xi , yi ∈ Ei be such that x = fi (xi ), y = fi (yi ). Uniqueness follows from the fact that if the fi ’s are homomorphisms of magmas, fi (xi yi ) is the composition of x and y for any law of composition on E. Let us prove its existence. Suppose that aj , bj ∈ Ej verify fj (aj ) = x and fj (bj ) = y. By definition, there exists k ∈ I, k i, k j such that fkj (aj ) = fki (xi ) and fkj (bj ) = fki (yi ). Since the fji ’s are homomorphisms, it follows that: fi (xi yi ) = fk (fki (xi yi )) = fk (fkj (aj bj )) = fj (aj bj ). Hence xy = fi (xi yi ) defines a structure of magma on E and it is clear that with respect to this law of composition, the fi ’s are homomorphisms of magma. The proofs for semigroups, monoids, groups, rings and commutativity are similar. 8.3.3 Proposition. Let (Ei , fji ) be an inductive system of magmas (resp. semigroups, monoids, groups, rings), E = −→ lim Ei and fi : Ei → E the canonical maps. Let F be a magma (resp. semigroup, monoid, group, ring) and ui : Ei → F be homomorphisms verifying ui = uj ◦ fji whenever i j. There exists a unique homomorphism u : E → F such that ui = u ◦ fi . Proof. By 8.2.1, there exists a unique map u : E → F such that ui = u ◦ fi . Therefore, we only have to check that u is a homomorphism. Let x, y ∈ E, i ∈ I and xi , yi ∈ Ei be such that x = fi (xi ), y = fi (yi ). Then u(xy) = u(fi (xi yi )) = ui (xi yi ) = ui (xi )ui (yi ) = u(x)u(y). So the result follows. 8.3.4 Remark. As in 8.2.3, if (Ei , fji ) and (Fi , gji ) are inductive systems of magmas (resp. semigroups, monoids, groups, rings), and (ui : Ei → Fi )i∈I is lim Ei → an inductive system of maps which are homomorphisms, then −→ lim ui : −→ is a homomorphism. lim F i −→ 8.3.5 Proposition. Let (Ai , fji ) be an inductive system of rings and A = lim −→ Ai . (i) If the Ai ’s are non-zero, then A is non-zero. (ii) If the Ai ’s are integral domains, then A is an integral domain. (iii) If the Ai ’s are fields, then A is a field.
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Proof. (i) Let 0i , 1i (resp. 0, 1) be the zero and the identity in Ai (resp. A). There exists i ∈ I such that fi (0i ) = 0 and fi (1i ) = 1. If 0 = 1, then there exists j i such that fji (0i ) = fji (1i ). This implies that 0i = 1i which contradicts the fact that Ai is non-zero. (ii) Let x, y ∈ A be such that xy = 0. There exist i ∈ I, xi , yi ∈ Ai such that fi (xi ) = x, fi (yi ) = y. It follows that fi (xi yi ) = 0i and there exists j i such that fji (xi yi ) = fji (xi )fji (yi ) = 0j . Since Aj is an integral domain, fji (xi ) = 0j or fji (yi ) = 0j . But this implies that fj (fji (xi )) = x = 0 or fj (fji (yi )) = y = 0. So we are done. (iii) Let x ∈ A \ {0}, i ∈ I and xi ∈ Ai with x = fi (xi ). Since the Ai ’s are fields, xi = 0i and fi (x−1 i ) is the inverse of x. So A is a field.
8.4 An example 8.4.1 Let A be a commutative ring, S be the set of multiplicative subsets S of A such that 1 ∈ S, 0 ∈ S. For S ∈ S, let iS : A → S −1 A denote the canonical homomorphism and U (S −1 A) the set of invertible elements in S −1 A. Let S, T, V ∈ S verify S ⊂ T ⊂ V . Note that iT (S) ⊂ U (T −1 A). By 2.3.5, there exists a unique homomorphism fT S : S −1 A → T −1 A such that fT S ◦ iS = iT . Since (fV T ◦ fT S ) ◦ iS = iV = fV S ◦ iS , it follows that: fV T ◦ fT S = fV S . 8.4.2 Let us fix p ∈ Spec(A) and let S1 be the set of S ∈ S such that S ∩ p = ∅. Let S, T ∈ S1 and ST = {st; s ∈ S, t ∈ T }. Since p is prime, S, T ⊂ ST ∈ S1 . We conclude that S1 is a directed set, and (S −1 A, fT S )S∈S1 is an inductive system of commutative rings. Let B = −→ lim S −1 A and fS : S −1 A → B the canonical maps. We claim that B is isomorphic to Ap . Let P = A \ p, then P ∈ S and Ap = P −1 A. From 8.4.1, we obtain that the family (fP S : S −1 A → Ap )S∈S1 satisfies the hypothesis of 8.2.1. It follows that there is a unique ring homomorphism u : B → Ap such that fP S = u ◦ fS for all S ∈ S1 . Now P is the biggest element in S1 and fP P is just the identity map on Ap . So u is surjective. Finally u is injective by 8.2.1(iii). Hence B is isomorphic to Ap . 8.4.3 Let S2 be the set of S ∈ S not containing any zero divisors of A. Then a similar argument shows that S2 is a directed set and the limit of the inductive system (S −1 A, fT S )S∈S2 is isomorphic to the full ring of fractions of A.
8.5 Inductive systems of algebras 8.5.1 Let A be a ring and (Ei , fji ) be an inductive system of abelian groups with E = −→ lim Ei , and fi the canonical maps.
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Let us suppose further that for i ∈ I, Ei is an A-module and whenever i j, fji is a homomorphism of A-modules. Then a simple verification shows that E admits a structure of A-module given by: for any a ∈ A, x ∈ E, i ∈ I, xi ∈ Ei verifying fi (xi ) = x, we set ax = fi (axi ). Note that if (Fi , gji ) is another such inductive system of abelian groups, and (ui : Ei → Fi )i∈I is an inductive system of homomorphisms of A-modules, lim Ei → −→ lim Fi is a homomorphism of Athen we verify readily that −→ lim ui : −→ modules. 8.5.2 Let us conserve the notations and hypotheses of 8.5.1. Suppose further that A is a commutative ring, the Ei ’s are A-algebras and the fji ’s are A-algebra homomorphisms. Then the ring structure and the A-module structure on E make E into an A-algebra and the fi ’s are A-algebra homomorphisms. Again, E is associative (resp. commutative) if the Ei ’s are associative (resp. commutative), and if (Fi , gji ) is another such inductive system and (ui : Ei → Fi )i∈I is an inductive lim ui : system of maps such that ui is a homomorphism of A-algebras, then −→ → lim F is a homomorphism of A-algebras. lim E i i −→ −→
References • [8], [9], [34], [52].
9 Sheaves of functions
The notion of a sheaf is essential in many areas in mathematics, and most particularly in algebraic geometry. Indeed, the theory of sheaves plays an important part in the foundation of modern algebraic geometry. We describe in this chapter some basic facts on sheaves of functions.
9.1 Sheaves 9.1.1 In this chapter, X will be a topological space and E a set. We shall denote the set of open subsets of X by Ω X or Ω, and by ΩxX or Ωx the set of open neighbourhoods in X of an element x ∈ X. Let us fix the partial order on Ω and Ωx given by: U V if and only if V ⊂ U. 9.1.2 Definition. A presheaf of functions on X with values in E is a map F which associates to each U ∈ Ω, a set F(U ) of maps from U to E, such that whenever U, V ∈ Ω verify V ⊂ U , f |V ∈ F(V ) for all f ∈ F(U ). This last condition says that the restriction map, denoted by rVF U or rV U , from F(U ) to F(V ) is well-defined. In the sequel, by a presheaf on X, we shall mean a presheaf of functions on X with values in E. 9.1.3 Remarks. 1) A presheaf on X is an inductive system of sets indexed by Ω. 2) Let F be a presheaf on X. If all the F(U )’s are groups (resp. rings, etc...) and all the rV U ’s are homomorphisms of groups (resp. rings, etc...), we shall say that F is a presheaf of groups (resp. rings, etc...) on X. 3) Let F be a presheaf on X. We shall also use the notation Γ (U, F) for F(U ). The elements of Γ (U, F) are called sections of F over U , and those of Γ (X, F) are called global sections. 9.1.4 Let F be a presheaf on X and x ∈ X. Then (F(U ), rV U )U ∈Ωx is an inductive system of sets indexed by Ωx . Let us denote the inductive limit of
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this system by Fx , and we call Fx the fibre of F at x. The elements of Fx are called the germs of sections of F at x. Further, if F is a presheaf of groups (resp. ring, etc...), then by 8.3.2, Fx is a group (resp. ring, etc...). For U ∈ Ωx , let rU,x : U → Fx be the canonical map. We call rU,x (f ) the germ of f at x. Now let U, V ∈ Ωx , s ∈ Fx , f ∈ F(U ) and g ∈ F(V ). Then s = rU,x (f ) = rV,x (g) if and only if there exists W ∈ Ωx such that W ⊂ U ∩V and rW U (f ) = rW V (g). It follows that f (x) is the same for any f having s as germ at x. We shall call this the value of the germ s at x, and we shall denote it by s(x). 9.1.5 Definition. A presheaf F on X is a sheaf if it satisfies the following condition: if (Ui )i∈I is a family of open subsets of X, U their union and f : U → E a map such that f |Ui ∈ F(Ui ), then f ∈ F(U ). 9.1.6 Examples. 1) Let G be an abelian group and F the presheaf of abelian groups on X defined by F(U ) = Hom(U, G) for all U ∈ Ω, here Hom(U, G) denotes the set of maps from U to G. One verifies easily that F is a sheaf of abelian groups on X. 2) For U ∈ Ω, let F(U ) = C(U, R), the ring of continuous real-valued functions on X. Then F is a sheaf of rings on X. 3) Let X = C and for U ∈ Ω, let F(U ) be the ring of holomorphic functions on U . Then we obtain the sheaf of holomorphic functions on C. 4) Let X = R and for U ∈ Ω, let F(U ) be the ring of bounded continuous real-valued functions on X. It is a presheaf, but not a sheaf. In fact, the identity function is bounded on Un =] − n, n[, but it is not bounded on R.
9.2 Morphisms 9.2.1 Let A, B be sets (resp. groups, rings, etc...), we denote by Hom(A, B) the set of maps (resp. homomorphisms of groups, homomorphisms of rings, etc...) from A to B. 9.2.2 Definition. Let F and G be presheaves on X. A morphism ϕ : F → G is a family of maps (ϕ(U ) : F(U ) → G(U ))U ∈Ω such that whenever U, V ∈ Ω verify V ⊂ U , the following diagram is commutative: ϕ(U )
(1)
F(U ) −−−−→ G(U ) ⏐ ⏐ ⏐G F ⏐ rV rV U U F(V ) −−−−→ G(V ) ϕ(V )
A morphism of sheaves is a morphism of the underlying presheaves. The morphism ϕ is an isomorphism if ϕ(U ) is an isomorphism for all U ∈ Ω. 9.2.3 Remarks. 1) We shall denote the set of morphisms of (pre)sheaves from F to G by Hom(F, G). It is clear that the composition of two morphisms is again a morphism.
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Let us denote by idF the morphism where idF (U ) = idF (U ) for all U ∈ Ω. It follows that given ϕ ∈ Hom(F, G), ϕ is an isomorphism if and only if there exists ψ ∈ Hom(G, F) such that ϕ ◦ ψ = idG and ψ ◦ ϕ = idF . 2) Let x ∈ X and ϕ ∈ Hom(F, G). Recall from 8.2.3 that there is a unique homomorphism ϕx : Fx → Gx such that for all U ∈ Ωx , the following diagram is commutative: ϕ(U )
(2)
F(U ) −−−−→ G(U ) ⏐ ⏐ ⏐rG F ⏐ rU,x U,x Fx
−−−−→ ϕx
Gx
We call ϕx the morphism induced by ϕ. 9.2.4 Theorem. Let ϕ : F → G be a morphism of presheaves on X. The following conditions are equivalent: (i) For all U ∈ Ω, ϕ(U ) is injective. (ii) For all x ∈ X, ϕx is injective. When these conditions are satisfied, we say that ϕ is injective. Proof. The implication (i) ⇒ (ii) was proved in 8.2.3. Let us prove (ii) ⇒ (i). Let U ∈ Ω and f, g ∈ F(U ) be such that ϕ(U )(f ) = ϕ(U )(g). Then by (2) F F (f ) = rU,x (g). So there is an open and the injectivity of ϕx , we have that rU,x neighbourhood Vx ∈ Ωx of x such that Vx ⊂ U and f |Vx = g|Vx . Since the Vx ’s, x ∈ U form a covering of U , we deduce that f = g. 9.2.5 Theorem. Let ϕ : F → G be a morphism of sheaves on X. The following conditions are equivalent: (i) For all U ∈ Ω and g ∈ G(U ), there exists an open covering (Ui )i∈I of U and elements fi ∈ F(Ui ) such that ϕ(Ui )(fi ) = g|Ui for all i ∈ I. (ii) For all x ∈ X, ϕx is surjective. When these conditions are satisfied, we say that ϕ is surjective. Proof. (i) ⇒ (ii) Let x ∈ X, θ ∈ Gx , U ∈ Ω and g ∈ G(U ) be such that G (g). By our hypotheses, there exist an open covering (Ui )i∈I of U and θ = rU,x elements fi ∈ F(Ui ) such that ϕ(Ui )(fi ) = g|Ui for all i ∈ I. Fix k ∈ I such that x ∈ Uk , then G G G G G rU,x (g) = rU ◦ rU (g) = rU ◦ ϕ(Uk )(fk ) = ϕx (rU (fk )). k ,x k ,U k ,x k ,x G (fk ) ∈ Fx , ϕx is surjective. Since rU k ,x (ii) ⇒ (i) Let U ∈ Ω and g ∈ G(U ). For x ∈ U , there exists ρ ∈ Fx such G (g). Let Ux ∈ Ωx and f x ∈ F(Ux ) be such that Ux ⊂ U and that ϕx (ρ) = rU,x F x ρ = rUx ,x (f ). Then: G F G (g) = ϕx ◦ rU (f x ) = rU ◦ ϕ(Ux )(f x ). rU,x x ,x x ,x
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This implies that there exists an open neighbourhood Vx ∈ Ωx such that Vx ⊂ Ux and g|Vx = ϕ(Ux )(f x )|Vx . Hence g|Vx = rVGx ,Ux ◦ ϕ(Ux )(f x ) = ϕ(Vx ) ◦ rVFx ,Ux (f x ). It follows that the family (Vx , rVFx ,Ux (f x ))x∈U satisfies (i).
9.2.6 Remark. Let ϕ : F → G be a morphism of sheaves on X. If ϕ(U ) is surjective for all U ∈ Ω, then by 8.2.3, ϕx is surjective for all x ∈ X. However, the converse is not true. For example, let X = C \ {0} and for U ∈ Ω, let F(U ) be the set of holomorphic functions on U . Consider the morphism of sheaves ϕ : F → F given by g → g . Then ϕ(X) is not surjective since there is no holomorphic function on X such that h (z) = 1/z for all z ∈ X. On the other hand, let U ∈ Ωx and f ∈ F(U ). For any point z ∈ U , fix an open ball Bz of X with centre z, contained in U . Since Bz is simply connected, there exists gz ∈ F(Bz ) such that gz = f |Bz . Hence condition (i) of 9.2.5 is satisfied, and ϕx is surjective. 9.2.7 Theorem. Let ϕ : F → G be a morphism of sheaves on X. Then ϕ is an isomorphism if and only if ϕx is bijective for all x ∈ X. Proof. By 9.2.4 and 9.2.6, we only have to prove that if ϕx is bijective for all x ∈ X, then ϕ(U ) is surjective for all U ∈ Ω. Let U ∈ Ω and g ∈ G(U ). By 9.2.5, for any x ∈ U , there exists Ux ∈ Ωx and f x ∈ F(Ux ) such that ϕ(Ux )(f x ) = g|Ux . It follows that for x, y ∈ U , we have: ϕ(Ux ∩ Uy )(f x |Ux ∩Uy ) = g|Ux ∩Uy = ϕ(Ux ∩ Uy )(f y |Ux ∩Uy ). The injectivity of ϕ implies that f x |Ux ∩Uy = f y |Ux ∩Uy . Since F is a sheaf, there exists f ∈ F(U ) such that f |Ux = f x for all x ∈ U . It follows that ϕ(U )(f ) = g since ϕ(U )(f )|Ux = g|Ux for all x ∈ U . 9.2.8 Lemma. Let F be a sheaf on X, U ∈ Ω and f, g ∈ F(U ) Then f = g if and only if rU,x (f ) = rU,x (g) for all x ∈ U . Proof. It is clear that f = g implies that rU,x (f ) = rU,x (g) for all x ∈ U . Conversely, if rU,x (f ) = rU,x (g) for all x ∈ U , then for any x ∈ U , there exists Vx ∈ Ωx such that Vx ⊂ U and f |Vx = g|Vx . So f = g.
9.3 Sheaf associated to a presheaf 9.3.1 Let F be a presheaf on X and U ∈ Ω. Denote by F + (U ) the set of maps f : U → E verifying: for any x ∈ U , there exist V ∈ Ωx and g ∈ F(V ) such that V ⊂ U and f |V = g. We see immediately that F + is a sheaf on X that we shall call the sheaf associated to the presheaf F. If F is a presheaf of groups (resp. rings, etc...),
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then the above definition implies that F + is a sheaf of groups (resp. rings, etc...). There is a canonical injective morphism of presheaves η F or η : F → F + given by f → f . If F is a sheaf, then F = F + since given U ∈ Ω, f ∈ F + , x ∈ U , there exist Vx ∈ Ωx , Vx ⊂ U and g x ∈ F(Vx ) such that f |Vx = g x . Since the Vx ’s is a covering of U , f ∈ F(U ). 9.3.2 Theorem. Let F be a presheaf on X. (i) For all x ∈ X, the map ηx : Fx → Fx+ is bijective. (ii) If ϕ : F → G is a morphism of presheaves, then there is a unique morphism of sheaves ϕ+ : F + → G + such that ϕ+ ◦ η F = η G ◦ ϕ. (iii) If ϕ : F → G is a morphism of presheaves and G is a sheaf, then there is a unique morphism of sheaves ψ : F + → G such that ψ ◦ η F = ϕ. Proof. (i) Let x ∈ X and fx ∈ Fx+ . There exist U ∈ Ωx , f ∈ F + (U ) such that F+ rU,x (f ) = fx . By definition, there exist V ∈ Ωx , V ⊂ U and g ∈ F(V ) such F (g)) = fx . Hence ηx is surjective. The injectivity of that f |V = g. So ηx (rV,x ηx is a consequence of 9.2.4 and 9.3.1. (ii) Uniqueness follows from (i) and 9.2.8. Now let U ∈ Ω and f ∈ F + (U ). There exist an open covering (Ui )i∈I of U and elements fi ∈ F(Ui ) such that f |Ui = fi . Set gi = ϕ(Ui )(fi ). Since fi |Ui ∩Uj = fj |Ui ∩Uj , we have gi |Ui ∩Uj = gj |Ui ∩Uj . It follows that there exists g ∈ G + (U ) such that g|Ui = gi . It is easy to see that g does not depend on the choice of the covering (Ui )i∈I and of the elements fi . Now set ϕ+ (U )(f ) = g. Then it is clear that ϕ+ is a morphism of sheaves from F + to G + verifying ϕ+ ◦ η F = η G ◦ ϕ. (iii) This is a direct consequence of (ii) since G + = G. 9.3.3 Remarks. 1) Assume that F is not a sheaf. Then ηx is bijective for all x ∈ X (9.3.2), but η is not an isomorphism. Thus 9.2.7 is false for morphisms of presheaves. 2) Let F be the presheaf of 9.1.6 (4), then F + is the sheaf of 9.1.6 (2) with X = R. 3) Let F, G be presheaves on X, ϕ ∈ Hom(F, G). If ϕx is bijective for all + x ∈ X, then so is ϕ+ x (9.3.2). Consequently, ϕ is an isomorphism (9.2.7). 9.3.4 Let F be a presheaf and G a sheaf on X. The results in 9.3.2 induce a map: ξ : Hom(F, G) → Hom(F + , G), ϕ → ϕ+ . Proposition. The map ξ is bijective. Proof. Let ϕ, ψ ∈ Hom(F, G) be such that ξ(ϕ) = ξ(ψ). Then for U ∈ Ω and f ∈ F(U ), we have ϕ+ (U ) ◦ η F (U )(f ) = ψ + (U ) ◦ η F (U )(f ). Hence ϕ(U )(f ) = ψ(U )(f ) and ξ is injective. Now if µ ∈ Hom(F + , G). Let ν(U )(f ) = µ(U ) ◦ η F (U )(f ) for U ∈ Ω and f ∈ F(U ). Then ν ∈ Hom(F, G) and ν + = µ. So ξ is surjective.
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9.3.5 Proposition. Let F be a presheaf, H a sheaf on X and ρ : F → H a morphism of presheaves. Suppose that for any morphism of presheaves ϕ : F → G, there exists a unique morphism of sheaves ψ : H → G such that ϕ = ψ ◦ ρ. Then F + and H are isomorphic. Proof. There exist ψ ∈ Hom(H, F + ) and µ ∈ Hom(F + , H) such that η F = ψ ◦ ρ and ρ = µ ◦ η F . It follows that η F = ψ ◦ µ ◦ η F and ρ = µ ◦ ψ ◦ ρ. By the uniqueness property of our hypothesis, we obtain that ψ ◦ µ = idF + and µ ◦ ψ = idH . 9.3.6 Proposition. Let ϕ : G → F be a morphism of presheaves on X. Suppose that F is a sheaf and that X has a base of open subsets B such that ϕ(U ) is bijective for all U ∈ B. Then F and G + are isomorphic. Proof. For x ∈ X, let Wx = Ωx ∩ B. Then Wx is a cofinal subset of Ωx , it follows from 8.2.7 that: lim G(U ) = Gx . lim F(U ) = Fx , −→ −→
U ∈Wx
U ∈Wx
Our hypothesis and 8.2.3 imply that ϕx is bijective. Hence ϕ+ x is bijective by 9.3.2 and the result follows from 9.2.7. 9.3.7 Let Y ∈ Ω X and F a presheaf on X. The restriction of F to Y , denoted by F |Y is a presheaf on Y endowed with the induced topology. More precisely, F |Y (U ) = F(U ) for U ∈ Ω Y . If F is a sheaf, then F |Y is also a sheaf. Further, if Z ∈ Ω Y , then F |Z = (F |Y )|Z . Let ϕ : F → G be a morphism of presheaves on X. Then we can define the restriction ϕ|Y : F |Y → G|Y in an obvious way. Clearly, ϕ|Y is a morphism of presheaves and it is an isomorphism if ϕ is an isomorphism. 9.3.8 In general, let Y be any subset of X endowed with the induced topology. Let F be a sheaf on X. We define, for V ∈ Ω Y , H(V ) to be the set of maps f : V → E such that there exist U ∈ Ω X and g ∈ F(U ) verifying V = Y ∩ U and f = g|V . It is easy to verify that H is a presheaf on Y . However, it is not a sheaf in general. The sheaf H+ associated to H is then defined as follows: for V ∈ Ω Y , + H (V ) is the set of maps f : V → E such that for all x ∈ V , there exist Ux ∈ ΩxX and g ∈ F(Ux ) verifying f |V ∩Ux = g|V ∩Ux . We shall call H+ the restriction of F to Y , and we shall denote it by F |Y . Again, it is easy to see that if Y is open, then this coincides with the definition in 9.3.7, and that F |Z = (F |Y )|Z if Z ⊂ Y ⊂ X.
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9.4 Gluing 9.4.1 Proposition. Let F be a presheaf, G a sheaf on X, (Xi )i∈I an open covering of X and for i ∈ I, ϕi : F |Xi → G|Xi a morphism of presheaves such that ϕi |Xi ∩Xj = ϕj |Xi ∩Xj for all i, j ∈ I. There exists a unique morphism ϕ : F → G such that ϕ|Xi = ϕi for all i ∈ I. We say that ϕ is obtained by gluing the ϕi ’s. Proof. Let U ∈ Ω, f ∈ F(U ) and for i, j ∈ I, let Ui = U ∩ Xi , fi = f |Ui and fij = f |Ui ∩Uj . If ϕ ∈ Hom(F, G) is such that ϕ|Xi = ϕi for all i, then G ◦ ϕ(U )(f ) = ϕi (Ui )(fi ). rU i ,U
Thus ϕ is unique if it exists. Let us prove the existence of ϕ. By our hypotheses, we have: G ◦ ϕi (Ui )(fi ) = ϕi (Ui ∩ Uj )(fij ) = ϕj (Ui ∩ Uj )(fij ) rU i ∩Uj ,Ui G ◦ ϕj (Uj )(fj ). = rU i ∩Uj ,Uj G Since G is a sheaf, there exists g ∈ G(U ) such that rU (g) = ϕi (Ui )(fi ) for i ,U all i. If we define ϕ(U )(f ) = g, then we verify easily that ϕ : F → G is a morphism such that ϕ|Xi = ϕi for all i.
9.4.2 Proposition. Let (Xi )i∈I be an open covering of X and for i ∈ I, Fi a presheaf on Xi . Suppose that the following conditions are satisfied: (i) There is an isomorphism of sheaves εij : Fi |Xi ∩Xj → Fj |Xi ∩Xj for all i, j ∈ I. (ii) For all i, j, k ∈ I, we have εii = idFi , εjk |Xi ∩Xj ∩Xk ◦ εij |Xi ∩Xj ∩Xk = εik |Xi ∩Xj ∩Xk . Then there exist a sheaf F on X and isomorphisms εi : F |Xi → Fi such that εj |Xi ∩Xj = εij ◦ εi |Xi ∩Xj . Further, the εi ’s and F are unique in the following sense: if F and (εi )i∈I verify the same conclusion, then there is a unique isomorphism ε : F → F such that εi ◦ ε|Xi = εi for all i ∈ I. We say that F is obtained by gluing the sheaves Fi . Proof. Let U ∈ Ω, and for i, j ∈ I, let Ui = U ∩Xi , Uij = U ∩Xi ∩Xj . Set F(U ) to be the set of families (fi ) ∈ i∈I Fi (Ui ) verifying εij (Uij )(fi |Uij ) = fj |Uij for all i, j ∈ I. For V ∈ Ω and V ⊂ U , define rVF U : F → G by rVF U ((fi )) = (rVFii,Ui (fi )) where Vi = Ui ∩V . Then we verify easily that F is a sheaf on X. Finally, define the morphisms εi : F |Xi → Fi by εi (U )((fj )) = fi . Then a straightforward verification proves the proposition.
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9.5 Ringed space 9.5.1 Lemma. Let B be a base of open subsets of X. For U ∈ B, let L(U ) be a set of maps from U to E. Further, suppose that the following conditions are verified: (i) If U, V ∈ B are such that V ⊂ U , then f |V ∈ L(V ) for all f ∈ L(U ). (ii) If (Ui )i∈I is a family of open subsets in B whose union U is in B, and f : U → E is a map verifying f |Ui ∈ L(Ui ) for all i, then f ∈ L(U ). Then there exists a unique sheaf F on X such that F(U ) = L(U ) for all U ∈ B. Proof. Any U ∈ Ω is the union of V ∈ B such that V ⊂ U . Let M(U ) be the set of maps f : U → E verifying f |V ∈ L(V ) for all V ∈ B contained in U . It is then clear that if F exists, then F(U ) = M(U ) for all U ∈ Ω. So we only need to show that M is a sheaf. Clearly, M is a presheaf. Let (Ui )i∈I be a family of open subsets of X whose union is U and the elements fi ∈ M(Ui ) be such that fi |Ui ∩Uj = fj |Ui ∩Uj for all i, j ∈ I. Then there exists a unique map f : U → E such that f |Ui = fi for all i ∈ I. Now if V ∈ B is contained in U , then for any W ∈ B contained in V ∩ Ui , f |W ∈ L(W ). Thus f |V ∩Ui ∈ M(V ∩ Ui ). Since V is the union of the set of W ∈ B such that W ⊂ V ∩ Ui for some i ∈ I, we obtain by (ii) that f |V ∈ L(V ). We have therefore proved that M is a sheaf. 9.5.2 Definition. (i) A Cartan space relative to E is a pair (X, F) where X is a topological space and F a sheaf of functions on X with values in E. (ii) A morphism of Cartan spaces from (X, F) to (Y, G) is a continuous map u : X → Y such that for all V ∈ Ω Y and g ∈ G(V ), the map defined by x → g ◦ u(x) belongs to F(u−1 (V )). 9.5.3 Proposition. Let (X, F) and (Y, G) be Cartan spaces, B and C bases of open subsets for X and Y respectively. The following conditions are equivalent for a map u : X → Y : (i) u is a morphism of Cartan spaces. (ii) For x ∈ X, V ∈ C ∩ Ωu(x) , there exists an open subset U ∈ B ∩ Ωx such that U ⊂ u−1 (V ) and for all g ∈ G(V ), the map (g ◦ u)|U ∈ F(U ). Proof. Obviously, (i) ⇒ (ii). Conversely, suppose that (ii) is verified. Then u is continuous. Let W ∈ Ω Y , g ∈ G(W ) and x ∈ u−1 (W ). There exist an open neighbourhood Ux ∈ B of x and an open neighbourhood Vu(x) ∈ C of u(x) such that: Ux ⊂ u−1 (W ) , u(Ux ) ⊂ Vu(x) ⊂ W and g ◦ u|Ux ∈ F(Ux ). Since the union of the Ux ’s, with x ∈ u−1 (W ), is u−1 (W ), we have that g ◦ u ∈ F(u−1 (W )).
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9.5.4 Let (X, F) be a Cartan space and Y ⊂ X, then we have seen in 9.3.8 that (Y, F |Y ) is again a Cartan space and the canonical injection j : Y → X is a morphism of Cartan spaces. If u : (X, F) → (Z, G) is a morphism of Cartan spaces, then u ◦ j is again a morphism of Cartan spaces. This is just the restriction of u to (Y, F |Y ). Now let u : (Z, G) → (X, F) be a morphism of Cartan spaces such that u(Z) ⊂ Y . Let us denote by ν : Z → Y the induced map of topological spaces. We claim that ν is a morphism of Cartan spaces. Let V ∈ Ω Y , f ∈ F|Y (V ) and U ∈ Ω X be such that V = U ∩Y . Since there is an open covering (Ui )i∈I of V in X and gi ∈ F(Ui ) such that f |V ∩Ui = gi |V ∩Ui , we can suppose (by replacing Ui by U ∩ Ui if necessary) that the unions of the Ui ∩ Y is equal to V . Now gi ◦ u ∈ G(u−1 (Ui )) = G(u−1 (Ui ∩ Y )). On the other hand, if x is an element of u−1 (Ui ) = u−1 (Ui ∩ Y ), then gi ◦ u(x) = f ◦ u(x). Since G is a sheaf, we obtain that f ◦ u ∈ G(u−1 (V )). So we have proved our claim. 9.5.5 Let (X, F) be a Cartan space and (Yi )i∈I an open covering of X. Suppose that for each i ∈ I, we have a morphism of Cartan spaces ui : (Yi , F |Yi ) → (Z, G) and the ui ’s verify ui |Yi ∩Yj = uj |Yi ∩Yj for all i, j ∈ I. then there exists a unique map u : X → Z such that u|Yi = ui for all i ∈ I. Now if Bi is a base of open subsets of Yi , then their union is a base of open subsets of X. It follows from the definitions and 9.5.3 that u is a morphism of Cartan spaces. 9.5.6 As a consequence of the preceding remarks, we have: Proposition. Let (X, F), (Y, G) be Cartan spaces, u : X → Y a map, (Vi )i∈I an open covering of Y and for each i ∈ I, (Uij )j∈J is a family of open subsets of X whose union is u−1 (Vi ). For (i, j) ∈ I × J, let uij : Uij → Vi be the map induced by u. Then u is a morphism of Cartan spaces if and only if for all (i, j) ∈ I × J, the map uij is a morphism of Cartan spaces from (Uij , F |Uij ) to (Vi , G|Vi ). 9.5.7 Let us suppose that E is a commutative ring. A Cartan space (X, F) is a ringed space relative to E if for all U ∈ Ω, F(U ) is a subring of the ring of E-valued functions on U . In particular, F is a sheaf of commutative rings on X. Let (X, F), (Y, G) be ringed spaces and u a morphism from (X, F) to (Y, G) as Cartan spaces. Then for any V ∈ Ω Y , the map G(V ) → F(u−1 (V )), g → g ◦ u, is a ring homomorphism. 9.5.8 Remark. Let us suppose that E is a commutative ring and B a base of open subsets of X. Then a Cartan space (X, F) is a ringed space if and only if F(U ) is a subring of the ring of E-valued functions on U for all U ∈ Ω X (9.5.1).
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References and comments • [5], [26], [34], [37]. For our purposes, we have restricted ourselves to sheaf of functions on a topological space. For a more thorough treatment of the theory of sheaves, the reader may refer to [15], [34] and [37].
10 Jordan decomposition and some basic results on groups
The patient reader will soon be rewarded. After this chapter, we shall begin to study algebraic geometry, Lie algebras and algebraic groups. A large part of this chapter covers basic material on the theory of abstract groups. In this chapter, k will denote an algebraically closed (commutative) field.
10.1 Jordan decomposition 10.1.1 Let V be a finite-dimensional k-vector space. We shall denote by End(V ) (resp. GL(V )) the set of endomorphisms (resp. invertible endomorphisms) of V . An element x ∈ End(V ) is said to be: • semisimple if V has a basis consisting of eigenvectors for x. • nilpotent if xn = 0 for some n ∈ N∗ . • unipotent if x − idV is nilpotent. 10.1.2 Theorem. Let x ∈ End(V ). There exists a unique pair (xs , xn ) of endomorphisms of V such that: (i) xs is semisimple, xn is nilpotent and x = xs + xn . (ii) xs ◦ xn = xn ◦ xs . Furthermore, there exist P, Q ∈ k[T ] such that P (0) = Q(0) = 0 and P (x) = xs , Q(x) = xn . We say that x = xs + xn is the additive Jordan decomposition of x and xs (resp. xn ) is the semisimple (resp. nilpotent) component of x. Proof. Let χx (T ) = (T −λ1 )m1 · · · (T −λr )mr be the characteristic polynomial of x where λ1 , . . . , λr are pairwise distinct eigenvalues of x. Let ai be the ideal of k[T ] generated by (T − λi )mi and Vi = ker(x − λi idV )mi . The Vi ’s are x-stable and V = V1 ⊕ · · · ⊕ Vr . By 2.1.2, there exists P ∈ k[T ] such that P (T ) − λi ∈ ai for all 1 i r and P (T ) ∈ a0 = T k[T ] (this last condition is redundant if 0 ∈ {λ1 , . . . , λr }). Set xs = P (x). Then xs (vi ) = λi vi for vi ∈ Vi . So xs is semisimple. Now let xn = x − xs = Q(x) where Q(T ) = T − P (T ). Then xn is nilpotent and the pair (xs , xn ) verifies conditions (i) and (ii).
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If (ys , yn ) is another such pair, then ys , yn commute with x, and hence they commute with xs , xn . It follows that xs − ys = yn − xn is both semisimple and nilpotent. Thus xs = ys and xn = yn . 10.1.3 Corollary. (i) Let x, y ∈ End(V ) be such that x ◦ y = y ◦ x. Then y commutes with xs and xn . Further, (x + y)s = xs + ys and (x + y)n = xn + yn . (ii) Let x ∈ End(V ) and W a x-stable subspace of V . Then W is xs -stable, xn -stable and x|W = xs |W + xn |W is the Jordan decomposition of x|W . (iii) Let x, W be as in (ii) and z, zs , zn be endomorphisms of V /W induced by x, xs , xn . Then z = zs + zn is the Jordan decomposition of z. 10.1.4 Theorem. Let x ∈ GL(V ). There exists a unique pair (xs , xu ) of elements of GL(V ) such that: (i) xs is semisimple, xu is unipotent and x = xs ◦ xu . (ii) xs ◦ xu = xu ◦ xs . We say that x = xs ◦ xu is the multiplicative Jordan decomposition of x and xu is the unipotent component of x. Proof. Let x = xs + xn be the additive Jordan decomposition of x. Since x ∈ GL(V ) and the eigenvalues of x and xs are identical, xs ∈ GL(V ). Hence −1 x = xs ◦(idV +x−1 s ◦xn ). Let xu = idV +xs ◦xn , then clearly (xs , xu ) satisfies conditions (i) and (ii). Finally, the proof of uniqueness is as in 10.1.2. 10.1.5 We have also analogues of the results of 10.1.3 for the multiplicative Jordan decomposition. 10.1.6 Let E be a k-vector space (not necessarily finite-dimensional). If x ∈ End(E) and v ∈ E, we set Ex (v) to be the subspace generated by xn (v), n ∈ N. This is clearly a x-stable subspace of E. Proposition. Let x ∈ End(E). The following conditions are equivalent: (i) The space E is the union of a family of finite-dimensional x-stable subspaces. (ii) For any v ∈ E, Ex (v) is finite-dimensional. If these conditions are satisfied, we say that x is locally finite. Proof. Clearly E is the union of the Ex (v), v ∈ E, so (ii) ⇒ (i). Now given (i), any element v is contained in some finite-dimensional x-stable subspace V . Hence Ex (v) ⊂ V , and dim Ex (v) is finite. 10.1.7 Remark. Let x be a diagonalisable endomorphism of E and (vi )i∈I a basis of eigenvectors. Let v ∈ E, then there exists a finite subset J ⊂ I such that v belongs to VJ , the span of the vj ’s, j ∈ J. Since VJ is clearly x-stable, it follows that x is locally finite. 10.1.8 Proposition. Let x ∈ End(E). The following conditions are equivalent: (i) x is diagonalisable.
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(ii) x is locally finite, and the restriction of x to any finite-dimensional x-stable subspace is semisimple. If these conditions are satisfied, we say that x is semisimple. Proof. (i) ⇒ (ii) This is clear by 10.1.7 since any finite-dimensional x-stable subspace is contained in some VJ where J is a finite subset of I. (ii) ⇒ (i) Since x is locally finite, V is the union of a family (Vα )α∈A of finite-dimensional x-stable subspaces. The restriction of x to Vα is semisimple and therefore there exists a basis Bα of eigenvectors for Vα . The union B of the Bα ’s, α ∈ A, is a family of eigenvectors spanning V . We can therefore extract a basis, from B, of eigenvectors for x. 10.1.9 Proposition. Let x ∈ End(E) (resp. x ∈ GL(E)). The following conditions are equivalent: (i) E is the union of a family (Vα )α∈A of finite-dimensional x-stable subspaces such that x|Vα is nilpotent (resp. unipotent) for all α ∈ A. (ii) x is locally finite, and the restriction of x to any finite-dimensional x-stable subspace is nilpotent (resp. unipotent). If these conditions are satisfied, we say that x is locally nilpotent (resp. locally unipotent). Proof. Clearly, (ii) ⇒ (i). Conversely, suppose that (i) is verified. Let V be a finite-dimensional x-stable subspace and (v1 , . . . , vn ) be a basis of V . Then x|Ex (vi ) is nilpotent (resp. unipotent) for 1 i n. It follows clearly that x|V is nilpotent (resp. unipotent). 10.1.10 Theorem. Let x be a locally finite endomorphism of E. There exists a unique pair (xs , xn ) of elements of End(E) such that: (i) xs is semisimple, xn is locally nilpotent and x = xs + xn . (ii) xs ◦ xn = xn ◦ xs . Any x-stable subspace is xs -stable and xn -stable. Moreover, if V is a finitedimensional x-stable subspace of E, then x|V = xs |V + xn |V is the Jordan decomposition (10.1.2). We shall call x = xs + xn the additive Jordan decomposition of x. Proof. First of all, the restriction of x to any finite-dimensional x-stable subspace has a Jordan decomposition. Further, given two finite-dimensional xstable subspaces V and W of E, by 10.1.3 (ii) and the uniqueness of the Jordan decomposition, we have (x|V )s |V ∩W = (x|W )s |V ∩W , (x|V )n |V ∩W = (x|W )n |V ∩W . Since x is locally finite, it follows that there exists a unique endomorphism xs (resp. xn ) of E such that xs |V = (x|V )s (resp. xn |V = (x|V )n ) for all finitedimensional x-stable subspace V of E. It is clear by construction that the pair (xs , xn ) satisfies conditions (i) and (ii), and that any x-stable subspace of E is xs -stable and xn -stable.
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10.1.11 We prove in a similar way the following: Theorem. Let x be a locally finite invertible endomorphism of E. There exists a unique pair (xs , xu ) of elements of GL(E) such that: (i) xs is semisimple, xu is locally unipotent and x = xs ◦ xu . (ii) xs ◦ xu = xu ◦ xs . Any x-stable subspace is xs -stable and xu -stable. Moreover, if V is a finitedimensional x-stable subspace of E, then x|V = xs |V ◦ xu |V is the Jordan decomposition (10.1.4). We shall call x = xs ◦ xu the multiplicative Jordan decomposition of x. 10.1.12 Lemma. Let E, F be vector spaces, x ∈ End(E), y ∈ End(F ) and z : E → F a surjective linear map. Suppose further that x is locally finite and z ◦ x = y ◦ z. Then y is locally finite. Furthermore: (i) We have z ◦ xs = ys ◦ z and z ◦ xn = yn ◦ z. (ii) If x, y are automorphisms, then z ◦ xs = ys ◦ z and z ◦ xu = yu ◦ z. Proof. Since z ◦ R(x) = R(y) ◦ z for any R ∈ k[T ], it is clear from the surjectivity of z that y is locally finite and we can assume that E and F are finite-dimensional. Let P, Q ∈ k[T ] be such that xs = P (x) and xn = Q(x). Let (ei )i∈I be a basis of E consisting of eigenvectors of xs . Then (z(ei ))i∈I is a family of eigenvectors of P (y) and it follows from the surjectivity of z that we can extract from this family a basis of F . Hence P (y) is semisimple. Finally Q(y) is clearly nilpotent and y ◦ z = (P (y) + Q(y)) ◦ z, again the surjectivity of z implies that y = P (y) + Q(y). It is now clear that ys = P (y) and yn = Q(y). The proof of part (ii) is similar. 10.1.13 Let E, F be vector spaces, x ∈ End(E), y ∈ End(F ). We denote by x⊗y the endomorphism of E ⊗k F defined as follows: for v ∈ E and w ∈ F : x ⊗ y(v ⊗ w) = x(v) ⊗ y(w). Lemma. Let x, y be locally finite automorphisms of E and F respectively. Then x ⊗ y is a locally finite automorphism of E ⊗k F and (x ⊗ y)s = xs ⊗ ys , (x ⊗ y)u = xu ⊗ yu . Proof. It is clear that x ⊗ y is locally finite and x ⊗ y = (xs ⊗ ys ) ◦ (xu ⊗ yu ) = (xu ⊗ yu ) ◦ (xs ⊗ ys ). Let (ei )i∈I (resp. (fj )j∈J ) be a basis of E (resp. F ) consisting of eigenvectors of xs (resp. ys ). Then clearly (ei ⊗ fj )(i,j)∈I×J is a basis of E ⊗k F consisting of eigenvectors of xs ⊗ ys . So xs ⊗ ys is semisimple. Finally, we have xu ⊗ yu − idE⊗k F = (xu − idE ) ⊗ yu + idE ⊗(yu − idF ). Since (xu − idE ) ⊗ yu commutes with idE ⊗(yu − idF ), it follows that xu ⊗ yu is locally unipotent.
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10.1.14 Proposition. Let x be a locally finite endomorphism of a kalgebra A. (i) If x is a k-algebra automorphism, then so are xs and xu . (ii) If x is a derivation of A, then so are xs and xn . Proof. Let B be the image of the linear map q : A ⊗k A → A given by the multiplication of the k-algebra A. (i) If x is a k-algebra automorphism, then q ◦ (x ⊗ x) = x ◦ q. By 10.1.13, x ⊗ x is locally finite and since q is surjective onto B, we have: q ◦ (xs ⊗ xs ) = q ◦ (x ⊗ x)s = xs ◦ q , q ◦ (xu ⊗ xu ) = q ◦ (x ⊗ x)u = xu ◦ q by 10.1.12 and 10.1.13. Finally, if A has an identity element e, then x(e) = e implies that the line spanned by e is x-stable, hence it is also xs - and xu -stable. Now, xu is unipotent, so xu (e) = e and it follows that xs (e) = e. Hence, xs and xu are k-algebra automorphisms of A. (ii) If x is a derivation, then q ◦ (ι ⊗ x + x ⊗ ι) = x ◦ q where ι denotes the identity map on A. Now we see easily that ι ⊗ x + x ⊗ ι is locally finite and that (ι ⊗ x + x ⊗ ι)s = ι ⊗ xs + xs ⊗ ι and (ι ⊗ x + x ⊗ ι)n = ι ⊗ xn + xn ⊗ ι. It follows from 10.1.12 that: q ◦ (ι ⊗ xs + xs ⊗ ι) = xs ◦ q , q ◦ (ι ⊗ xn + xn ⊗ ι) = xn ◦ q. Consequently, xs and xn are derivations of A.
10.2 Generalities on groups 10.2.1 In general, we shall present group laws multiplicatively. Let G be a group. Denote by µG : G × G → G, (α, β) → αβ the group multiplication and ιG : G → G, α → α−1 , the inverse. When there is no confusion, we shall simply write µ, ι for µG , ιG . Let U , V be subsets of G, then we set U V = {αβ ; α ∈ U , β ∈ V }. and U −1 = {α−1 ; α ∈ U }. Given a subgroup H of G, recall that NG (H) = {α ∈ G; αHα−1 = H} is a subgroup of G called the normalizer of H in G. If K ⊂ NG (H), then we say that K normalizes H. If NG (H) = G, then H is a normal subgroup of G. Similarly CG (H) = {α ∈ G; αβ = βα for all β ∈ H} is a subgroup of G called the centralizer of H in G. If K ⊂ CG (H), then we say that K centralizes H. 10.2.2 Let us denote the group of automorphisms of G by Aut(G). A subgroup H of G is called characteristic if σ(H) = H for all σ ∈ Aut(G). Given α ∈ G, the map iα : G → G given by β → αβα−1 , is an automorphism of G called an inner automorphism of G. The set Int(G) of inner automorphisms of G is a subgroup of Aut(G). Note that a subgroup of G is normal if and only if it is stable under all inner automorphisms of G.
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10.2.3 Let H be a subgroup of G. We define the quotient G/H (resp. H\G) to be the set of equivalence classes, called left (resp. right) cosets, of the equivalence relation Rl (resp. Rr ) defined by αRl β ⇔ α−1 β ∈ H (resp. αRr β ⇔ βα−1 ∈ H). Note that a left (resp. right) coset is of the form αH (resp. Hα) for some α ∈ G. It follows that if H is normal, then G/H = H\G and we have an induced group structure on G/H such that the projection G → G/H is a group homomorphism. In general, we have a bijection between G/H and H\G given by αH → Hα−1 . We define the index of H in G to be the cardinal, denoted by (G : H), of the set G/H (or H\G). Proposition. Let H ⊃ K be subgroups of G. (i) We have card(G) = (G : H) card(H). (ii) We have (G : K) = (G : H)(H : K). 10.2.4 Let H, K be groups and σ : K → Aut(H) a group homomorphism. We define a law on H × K by: (α, β)(α , β ) = (ασ(β)(α ), ββ ) for α, α ∈ H, β, β ∈ K. We check easily that there is a group law on H × K, denoted by H σ K or simply H K, that we call the semi-direct product of K by H with respect to σ. If eH , eK denote the identity elements of H and K respectively, then (eH , eK ) is the identity element of H K and (α, β)−1 = (σ(β −1 )(α−1 ), β −1 ). Proposition. The maps iH : H → H σ K, α → (α, eK ), iK : K → H σ K, β → (eH , β) and pK : H σ K → K, (α, β) → β are group homomorphisms. Further we have the following exact sequence of group homomorphisms i
pK
H H σ K −→ K −→ 1. 1 −→ H −→
10.2.5 Suppose that H, K are subgroups of G such that K normalizes H and K ∩ H = {eG }. Let σ : K → Aut(H), be the group homomorphism given by σ(β)(α) = βαβ −1 . Then we have: Proposition. The map (α, β) → αβ is an isomorphism of H σ K onto the subgroup HK of G.
10.3 Commutators 10.3.1 Let α, β ∈ G. We define the commutator of α and β to be (α, β) = αβα−1 β −1 . Clearly, (α, β)−1 = (β, α) and γ(α, β)γ −1 = (γαγ −1 , γβγ −1 ) for γ ∈ G. If H, K are subgroups of G, we shall denote by (H, K) the subgroup generated by the commutators (α, β) where α ∈ H, β ∈ K. The group (H, K) is normal (resp. characteristic) if H and K are normal (resp. characteristic). In
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particular, the subgroup (G, G) is a characteristic subgroup of G called the derived subgroup or the group of commutators. We shall denote (G, G) also by D(G). It is easy to see that: Proposition. A subgroup H of G contains D(G) if and only if G/H is abelian and H is normal. 10.3.2 Proposition. If the index of the centre Z of G is finite, then D(G) is finite. Proof. Let n = (G : Z) and a1 , . . . , an be representatives of G/Z. Since (ap , aq ) = (ap z, aq z ) for any z, z ∈ Z, the set C of commutators is finite and card(C) n2 . Let c1 , . . . , cr be the (distinct) elements of C. Since (G : Z) = n, (α, β)n ∈ Z for all α, β ∈ G. This implies that (α, β)n = −1 β (α, β)n β = (β −1 αβ, β)n , and hence: (α, β)n+1 = (β −1 αβ, β)n (α, β) = (β −1 αβ, β)n−1 (β −1 αβ, β)(α, β) = (β −1 αβ, β)n−1 (β −1 αβ, β 2 ). Thus (α, β)n+1 can be written as a product of n commutators. −1 On the other hand, ci cj = cj (c−1 j ci cj ) and cj ci cj = ck for some k. Now let x = ci1 · · · cis ∈ D(G), 1 i1 , . . . , is r, be such that s is minimal mr 1 for this property. Then the preceding paragraph says that x = cm 1 · · · cr n+1 where m1 + · · · + mr = s. Since ci is a product of n commutators, it follows from the minimality of s that 0 m1 , . . . , mr n. Thus card(D(G)) 2 (n + 1)r (n + 1)n . 10.3.3 Proposition. Let H, K be subgroups of G such that H normalizes K. If the set S = {(α, β); α ∈ H, β ∈ K} is finite, then so is (H, K). Proof. We can assume that G = HK and so K is normal in G. For β, β1 ∈ H and γ, γ1 ∈ K, we have (γ1 β1 ββ1−1 γ1−1 , γ) = (γ1 , β1 ββ1−1 )(β1 ββ1−1 , γγ1 ). Thus the set T of commutators (αβα−1 , γ), where α ∈ G, β ∈ H and γ ∈ K, is finite and is contained in (H, K). Moreover, T is stable under inner automorphisms. Consequently, (H, K) is normal in G since S ⊂ T . Since an inner automorphism defines a permutation of the set T , this induces a homomorphism from G to the group of permutations of T whose kernel L is exactly CG ((H, K)). This implies that L is normal and (G : L) is finite. Now L ∩ (H, K) is central of finite index in (H, K), so by 10.3.2, D((H, K)) is finite. Since D((H, K)) is normal in G, we can replace G by G/D((H, K)) and assume that (H, K) is abelian. The set U = {(α, β); α ∈ H, β ∈ (H, K)} ⊂ S is finite and generates the subgroup (H, (H, K)). Given α ∈ H, β ∈ (H, K), we have αβα−1 ∈ (H, K) and
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(α, β)2 = (α, β)(αβα−1 )β −1 = (αβα−1 )(α, β)β −1 = (α, β 2 ) ∈ U. Since (H, (H, K)) is abelian and generated by U , it is finite. Further, (H, (H, K)) is normal in G, therefore we can replace G by G/(H, (H, K)) and assume that H centralizes (H, K). Given α ∈ H, β ∈ K, it follows that (α, β)2 = (α, β)α−1 (α, β)α = α−1 (α2 , β)α = (α2 , β) ∈ S. Since S generates (H, K) and (H, K) is abelian, we obtain that (H, K) is finite.
10.4 Solvable groups 10.4.1 Definition. The derived series of G is the sequence (Dn (G))n0 of subgroups of G defined by D0 (G) = G and Dn (G) = (Dn−1 (G), Dn−1 (G)) for n 1. A group G is solvable if there exists an integer n such that Dn (G) = {eG }. 10.4.2 Proposition. (i) Let f : G → H be a homomorphism of groups. We have f (Dn (G)) ⊂ Dn (H) for all n ∈ N. Furthermore, if f is surjective, then f (Dn (G)) = Dn (H). (ii) For any n ∈ N, Dn (G) is a characteristic subgroup of G. Proof. Part (ii) is clear and part (i) is a simple induction. 10.4.3 Definition. A sequence G = G0 ⊃ G1 ⊃ · · · ⊃ Gm of subgroups of G is called a normal chain if Gi+1 is normal in Gi for 0 i m − 1. 10.4.4 Proposition. The following conditions are equivalent for a group G: (i) G is solvable. (ii) There is a sequence G = G0 ⊃ G1 ⊃ · · · ⊃ Gm = {eG } of normal subgroups of G such that Gi /Gi+1 is abelian for 0 i m − 1. (iii) There is a normal chain G = G0 ⊃ G1 ⊃ · · · ⊃ Gm = {eG } of subgroups of G such that Gi /Gi+1 is abelian for 0 i m − 1. Proof. For (i) ⇒ (ii), take Gi = Di (G), while (ii) ⇒ (iii) is trivial. Finally, given (iii), D(Gi ) ⊂ Gi+1 . By induction, we obtain Di (G) ⊂ Gi for all i. 10.4.5 Proposition. Let H be a normal subgroup of G. Then G is solvable if and only if H and G/H are solvable. Proof. Clearly if G is solvable, H and G/H are solvable. Conversely suppose that H and G/H are solvable. Since the canonical surjection π : G → G/H is a homomorphism of groups, by 10.4.2 (i) there exists p ∈ N such that π(Dp (G)) = {eG/H }, i.e. Dp (G) ⊂ H. Now there exists q ∈ N such that Dp+q (G) ⊂ Dq (H) = {eG }. Thus G is solvable.
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10.4.6 Corollary. Let H, K be normal subgroups of G. If H, K are solvable, then HK is also solvable. Proof. Since HK/H is isomorphic to K/(H ∩ K), it is solvable. Now apply 10.4.5 to HK and H.
10.5 Nilpotent groups 10.5.1 Definition. The central descending series of G is the sequence (C n (G))n1 of subgroups of G defined by C 1 (G) = G and for n 1, C n+1 (G) = (G, C n (G)). A group G is nilpotent if there exists an integer n such that C n (G) = {eG }. 10.5.2 Remarks. 1) An abelian group is clearly nilpotent. 2) We verify easily that (C p (G), C q (G)) ⊂ C p+q (G), p, q ∈ N∗ . Thus n n D (G) ⊂ C 2 (G), n ∈ N∗ . So a nilpotent group is solvable. 3) Let G be nilpotent and n be minimal such that C n+1 (G) = {eG }. Then n C (G) is contained in the centre of G. In particular, the centre of G is nontrivial if G is non-trivial. 4) Let f : G → H be a homomorphism of groups. As in 10.4.2, we have by induction that f (C n (G)) ⊂ C n (H) for all n ∈ N∗ , and equality holds if f is surjective. It follows that C n (G) is a characteristic subgroup of G for n ∈ N∗ . Further, the quotient of a nilpotent group is nilpotent. 10.5.3 Proposition. Let H be a subgroup of G contained in the centre of G. Then G is nilpotent if and only if G/H is nilpotent. Proof. We have already remarked (10.5.2) that the quotient of a nilpotent group is nilpotent. Conversely, if G/H is nilpotent, then by 10.5.2 there exists n ∈ N∗ such that C n (G) ⊂ H ⊂ Z(G). Thus C n+1 (G) = {eG }. 10.5.4 Proposition. A group G is nilpotent if and only if there exists a sequence G = G1 ⊃ G2 ⊃ · · · ⊃ Gn+1 = {eG } of subgroups of G such that (G, Gk ) ⊂ Gk+1 for 1 k n. Proof. If G is nilpotent, then take Gk = C k (G). Conversely, we verify easily by induction that C k (G) ⊂ Gk for all k. 10.5.5 Proposition. Let G be a nilpotent group, H a subgroup and K a normal subgroup of G. (i) There exists a normal chain G = H1 ⊃ H2 ⊃ · · · ⊃ Hn+1 = H of subgroups of G such that Hk /Hk+1 is abelian for 1 k n. (ii) There exists a sequence K = K1 ⊃ K2 ⊃ · · · ⊃ Km+1 = {eG } of subgroups of G such that (G, Ki ) ⊂ Ki+1 for 1 i m. Proof. Set Gk = C k (G) and n minimal such that Gn+1 = {eG }. (i) Let Hk = Gk H = HGk since Gk is normal in G. If α ∈ H, β ∈ Gk , γ ∈ Gk+1 , then
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αHk+1 α−1 = (αHα−1 )(αGk+1 α−1 ) ⊂ Hk+1 βαγβ −1 = (β, αγ)αγ ∈ (Gk , G)HGk+1 ⊂ Gk+1 HGk+1 = Hk+1 . Thus Hk+1 is normal in Hk . Finally, Gk /Gk+1 is abelian and the canonical homomorphism Gk /Gk+1 → Hk /Hk+1 is surjective. So Hk /Hk+1 is abelian. (ii) It suffices to take Ki = K ∩ Gi . 10.5.6 Corollary. Let G be a nilpotent group. (i) Let H = G be a subgroup of G, then NG (H) = H and there is a normal subgroup L containing H, L = G and such that G/L is abelian. (ii) Let K = {eG } be a normal subgroup of G. Then K ∩ Z(G) = {eG }. Proof. (i) Let (Hk ) be a sequence of subgroups verifying condition (i) of 10.5.5. If k is maximal such that Hk = H, then Hk ⊂ NG (H) and H = NG (H). If i is minimal such that Hi = G, then L = Hi is a normal subgroup of G containing H and G/L is abelian as required. (ii) Let (Ki ) be a sequence of subgroups verifying 10.5.5 (ii). If j is maximal such that Kj = {eG }, then K ∩ Z(G) ⊃ Kj ∩ Z(G) = Kj = {eG }.
10.6 Group actions 10.6.1 We shall denote the group of permutations of a set E by S(E). If we have a group homomorphism θ from G to S(E), then we say that G acts on E. For simplicity, we shall often write α.x for θ(α)(x) if α ∈ G and x ∈ E. The set G.x = {α.x; α ∈ G} is called the G-orbit of x, and Gx = {α ∈ G; α.x = x} the stabilizer (or isotropy group) of x. We verify easily that Gx is a subgroup of G. 10.6.2 1) Let G acts on E, x, y ∈ E and α ∈ G be such that α.x = y. Then Gy = αGx α−1 . 2) We shall denote by E G the set of fixed points of E. Thus x ∈ E G if and only if Gx = G if and only if G.x = {x}. 3) Let k ∈ N∗ . An action of G on E is called k-transitive if given x1 , . . . , xk , y1 , . . . , yk where xi = xj and yi = yj if i = j, there exists α ∈ G such that α.xi = yi , 1 i k. When k = 1, we say that G acts transitively on E and that E is a G-homogeneous space. 4) Let E , E be subsets of E. We define the transporter TranG (E , E ) of E to E to be the set of α ∈ G such that α(E ) ⊂ E . 5) Let G acts on E and F . A map f : E → F is called G-equivariant if f (α.x) = α.f (x) for all x ∈ E and α ∈ G. 10.6.3 Theorem. Let G acts on E and x ∈ E. There is a bijection between G.x and G/Gx . Thus card(G.x) = (G : Gx ). Proof. The map f : G → G.x defined by α → α.x is surjective and we have f −1 (α.x) = αGx .
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10.6.4 Examples. 1) Left multiplication is an action of G on G called the left regular action. The right regular action is given by α.β = βα−1 . 2) The homomorphism G → Aut(G) induced by inner automorphisms is an action called conjugation. The stabilizer of α is just the centralizer CG (α). Two elements α, β are conjugate in G if they are in the same G-orbit. 3) Let H be a subgroup of G. Then G acts naturally on G/H via left multiplication: α.(βH) = αβH. 10.6.5 Let V be a k-vector space, ϕ : G → GL(V ) a group homomorphism. Then G acts on V and V is called a G-module and (V, ϕ) (or ϕ) a representation of G in V . Let (V, ϕ) be a representation of G and V ∗ be the dual of V . Then there is a natural action of G on V ∗ as follows: let f ∈ V ∗ , v ∈ V and α ∈ G, then (α.f )(v) = f (α−1 .v). We call this the contragredient or dual representation of ϕ. If (W, ψ) is another representation of G, then Homk (V, W ) also has a natural G-module structure as follows: let f ∈ Homk (V, W ), v ∈ V , α ∈ G, then (α.f )(v) = ψ(α)f (ϕ(α−1 )(v)). A homomorphism f ∈ Homk (V, W ) is called a G-module homomorphism (or G-homomorphism) if ψ(α)f (v) = f (ϕ(α)(v)) for all α ∈ G. The set of homomorphisms of G-modules will be denoted by HomG (V, W ). We shall also write EndG (V ) for HomG (V, V ). Clearly, HomG (V, W ) = Homk (V, W )G . We can also put a representation on the tensor product V ⊗k W as follows: let v ∈ V , w ∈ W , α ∈ G, then α.(v ⊗ w) = (ϕ(α)(v)) ⊗ (ψ(α)(w)). We can extend this to the n-th tensor product Tn (V ), and consequently to n n V , the n-th alternating product. S (V ), the n-th symmetric product and
10.7 Generalities on representations 10.7.1 Definition. A G-module E is simple if E = {0} and the only submodules of E are {0} and E. 10.7.2 Proposition. (Schur’s Lemma) (i) If E, F are non-isomorphic simple G-modules, then HomG (E, F ) = {0}. (ii) If E is a simple G-module, then EndG (E) is a field. Furthermore, if E is finite-dimensional, then EndG (E) = k idE . Proof. Since the image and the kernel of a homomorphism of G-modules are G-modules, a homomorphism of simple G-modules is either zero or an isomorphism. So part (i) follows and EndG (E) is a field which contains clearly k idE .
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Now, if E is finite-dimensional, then any u ∈ EndG (E) has an eigenvalue λ (k being algebraically closed). This implies that ker(u − λ idE ) is a non-trivial submodule of E. Hence u = λ idE . 10.7.3 Lemma. Let E be the sum of a family (Ei )i∈I of simple G-modules, and F a submodule of E. There exists J ⊂ I such that E = F ⊕ i∈J Ei . Proof. Let F be the set of subsets K of I such that the sum F + i∈K Ei is direct. Together with the partial order given by inclusion, F is an inductive system. Let J be a maximal element in F. Let V = F ⊕ i∈J Ei . For any k ∈ I, the sum Ek + V is not direct, so Ek ∩ V = {0} and therefore Ek ⊂ V since Ek is simple. Hence E = V . 10.7.4 Proposition. The following conditions are equivalent for a Gmodule E: (i) E is a sum of simple submodules. (ii) E is a direct sum of simple submodules. (iii) Any submodule F of E is a direct summand, i.e. E = F ⊕ F for some submodule F . If these conditions are satisfied, we say that E is semisimple or completely reducible. Proof. We may assume that E = {0}. (i) ⇒ (ii) Suppose that E is the sum of a family (Ei )i∈I of simple submodules. Let F be the set (partially ordered by inclusion) of subsets J of I inductive system If K is such that the sum j∈J Ej is direct. Then F is an a maximal element of F, then clearly we have E = k∈K Ek . (ii) ⇒ (iii) This is obvious by 10.7.3. (iii) ⇒ (i) Let F be a non trivial submodule of E, x ∈ F \ {0} and Vx the submodule of F generated by x. Let F be the set (partially ordered by inclusion) of submodules of Vx which are distinct from Vx . Then it is easy to check that F is an inductive system. If M is a maximal element of F, then Vx /M is clearly a simple module. Let M be a submodule of E such that E = M ⊕ M . Then Vx is the direct sum of M and M ∩ Vx because M ⊂ Vx . Since the submodule M ∩ Vx is isomorphic to Vx /M , it is simple. We have therefore proved that F contains a simple submodule. Now let S be the sum of all simple submodules of E and T a submodule such that E = S ⊕ T . So T has no simple submodule. If T = {0}, then the preceding paragraph says that T contains a simple submodule. So S ∩T = {0} which is absurd. Thus E = S. 10.7.5 Proposition. Let E be the sum of a family (Ei )i∈I of simple Gmodules. (i) Any submodule (resp. quotient) F of E is semisimple. Furthermore, there exists J ⊂ I such that F is isomorphic to i∈J Ei . (ii) Any simple submodule of E is isomorphic to some Ei , i ∈ I.
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Proof. By 10.7.4, there exists a submodule F of E such that E = F ⊕ F . Then by 10.7.3, there exists J ⊂ I such that E = F ⊕ ( i∈J Ei ). Since F is isomorphic to E/F , we have part (i). Part (ii) a just a special case of (i).
10.7.6 Proposition. Any representation of a finite group is semisimple. Proof. Let (E, ρ) be a G-module, F a submodule of E and p ∈ End(E) a projector with respect to F and n = card(G). Set: q=
1 ρ(α)−1 ◦ p ◦ ρ(α). n α∈G
We have q(E) ⊂ F and q(x) = x for x ∈ F . Now for β ∈ G, ρ(β) ◦ q =
1 ρ(αβ −1 )−1 ◦ p ◦ ρ(αβ −1 ) ◦ ρ(β) = q ◦ ρ(β). n α∈G
Thus q is a G-module homomorphism and E = F ⊕ ker q. 10.7.7 Let V be a k-vector space of dimension n.
d Lemma. Let W be a subspace of V of dimension d, y ∈ W \ {0} and d d V such that V = ky ⊕ U . Then V = W ⊕ W where a subspace U ⊂ d−1 W }. W = {x ∈ V ; x ∧ z ∈ U for all z ∈ d−1 d Proof. Let {u1 , . . . , ud } be a basis of W and ϕ ∈ ( V )∗ be such that ϕ(y) = 1 and ϕ|U = 0. Then W = {x ∈ V ; ϕ(x ∧ ui ) = 0 for 1 i d}. It follows that dim W n − d. So it suffices to prove that W ∩ W = {0}. Let x ∈ W \{0} and {x, x2 , . . . , xd } be a basis of W . There exists λ ∈ k\{0} d−1 W , we such that x ∧ (x2 ∧ · · · ∧ xd ) = λy ∈ U . Since x2 ∧ · · · ∧ xd ∈ deduce that x ∈ W . 10.7.8 Lemma. Let G be a subgroup of GL(V ) such that V is a semisimple G-module. Let χ : G → k ∗ be a homomorphism of groups, U = {x ∈ V ; α(x) = χ(α)x for all α ∈ G}, and W the subspace generated by α(x)−χ(α)(x), where α ∈ G and x ∈ V . (i) The subspaces U and W are G-submodules. Furthermore, U (resp. W ) is the unique G-submodule such that V = U ⊕ W . (ii) Let H be a subgroup of GL(V ) containing G as a normal subgroup. If χ extends to a homomorphism of H into k ∗ , then U and W are H-stable. Proof. (i) The subspace U is clearly G-stable. On the other hand, β(α(x) − χ(α)(x)) = (βαβ −1 )(β(x)) − χ(βαβ −1 )(β(x)) ∈ W, so W is also G-stable. Now V is semisimple, there exist submodules U , W such that V = U ⊕ U = W ⊕ W .
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If x ∈ W , then α(x) − χ(α)x ∈ W ∩ W = {0}. Hence W ⊂ U . On the other hand, if x ∈ V , there exist y ∈ U , z ∈ U such that x = y +z. For α ∈ G, α(x) − χ(α)(x) = α(z) − χ(α)(z) ∈ U . Hence W ⊂ U . Since U ∩ W = {0}, part (i) follows. (ii) The proof is analogue to the one for (i). 10.7.9 Let V be a semisimple G-module and W the subspace of V generated by α.x − x where α ∈ G, x ∈ V . By 10.7.8, W is a direct summand of V and V G is the unique submodule such that V = V G ⊕ W . The projection of V onto V G corresponding to this direct sum decomposition, denoted by pV , is called the Reynolds operator associated to V .
10.8 Examples 10.8.1 For n, p ∈ N∗ , we denote by Mn,p (k) the space of n by p matrices with coefficients in k, and we write Mn (k) = Mn,n (k). Let In denote the identity matrix and (Eij )1i,jn the canonical basis of Mn (k). Denote by GLn (k), SLn (k), Dn (k), Tn (k) and Un (k) respectively the group of invertible matrices, the group of invertible matrices of determinant 1, the group of invertible diagonal matrices, the group of invertible upper triangular matrices and the group of upper triangular unipotent matrices. We shall denote by diag(λ1 , . . . , λn ) the matrix λ1 E11 + · · · + λn Enn . 10.8.2 It is clear that any matrix A ∈ GLn (k) can be written uniquely as a product diag(1, . . . , 1, det A)B where B ∈ SLn (k). It follows that GLn (k) is isomorphic to the semidirect product SLk (n)H where H = {diag(1, . . . , 1, λ); λ ∈ k∗ } k∗ . Further, we can verify easily that Tn (k) = Un (k) Dn (k). 10.8.3 Let S1 = {1}, and for n 2, let Sn = {Bij (λ) = In + λEij ; 1 i = j n , λ ∈ k} ⊂ SLn (k). A straightforward computation gives: for i = j, Bij (λ)−1 = Bij (−λ) , Bij (λ)Bij (µ) = Bij (λ + µ). 10.8.4 Proposition. Let λ ∈ k ∗ and A ∈ Mn (k). If A commutes with Bij (λ) for all i = j, then A ∈ kIn . Proof. If A commutes with the Bij (λ)’s, then AEij = Eij A for all i = j. Thus, a1i E1j + · · · + ani Enj = aj1 Ei1 + · · · + ajn Ein . So aii = ajj and aki = 0 = ajl for k = i and l = j.
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10.8.5 Corollary. (i) The centre of GLn (k) is k ∗ In . (ii) The centre of SLn (k) is k ∗ In ∩ SLn (k) = {µIn ; µn = 1}. 10.8.6 Theorem. (i) The group SLn (k) is generated by Sn . (ii) The group Un (k) is generated by Sn ∩ Un (k). Proof. We shall prove (i) by induction on n. The result is obvious for n = 1. Let us suppose that n 2, and let A = [aij ] ∈ SLn (k). If a1j = 0 for all j 2, then a11 = 0. The matrix A = AB12 (1) = [aij ] verifies a12 = 0. So we can assume that there exists j 2 such that a1j = 0. 1 − a11 ) = (aij ) verifies a11 = 1. Hence Now A = AB1p ( a1p 1 0 Bn1 (−an1 ) · · · B21 (−a21 )A B12 (−a12 ) · · · B1n (−a1n ) = 0C where C ∈ SLn−1 (k). By induction, A is a product of elements of Sn (recall that Bij (λ)−1 = Bij (−λ)). (ii) The proof of part (i) applies since given A = [aij ] ∈ Un (k), a11 = 1 and aij = 0 if i > j. 10.8.7 Since k is algebraically closed, k \ {0, 1, −1} is non-empty. We have by straightforward computations the following result. Lemma. Let λ ∈ k. (i) If n 3 and i, j, k ∈ {1, . . . , n} are distinct, then Bij (λ) = (Bik (λ), Bkj (1)). (ii) Let n = 2, µ ∈ k \ {0, −1, 1} and ν = λ(µ2 − 1)−1 . Then, B12 (λ) = (diag(µ, µ−1 ), B12 (ν)) , B21 (λ) = (B21 (ν)−1 , diag(µ−1 , µ)). 10.8.8 Theorem. The derived subgroup of GLn (k) is SLn (k) and SLn (k) is equal to its derived subgroup. Proof. By 10.8.7, any element of Sn is a commutator of elements of SLn (k). Since Sn generates SLn (k) by 10.8.6, and D(SLn (k)) ⊂ D(GLn (k)) ⊂ SLn (k), the result follows. 10.8.9 Since (diag(µ1 , . . . , µn ), Bij (λ)) = Bij (λ−
λµi ) for i = j, we obtain µj
via 10.8.6 (i): Proposition. The derived subgroup of Tn (k) is Un (k). 10.8.10 Let T+ n (k) denote the subspace of strictly upper triangular ma+ r trices. Then Un (k) = In + T+ n (k). For r 1, set Un,r (k) = In + (Tn (k)) .
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Clearly, Un,n (k) = {In }. Further, it is easy to verify that Un,r (k) are normal subgroups of Un (k) and for r, s ∈ N∗ , (Un,r (k), Un,s (k)) ⊂ Un,r+s (k). Hence, by 10.4.6, 10.5.4 and 10.8.2, we have that: Theorem. The group Un (k) (resp. Tn (k)) is nilpotent (resp. solvable). 10.8.11 Lemma. (Burnside’s Theorem) Let E be a finite-dimensional kvector space and A a subalgebra of End(E). If the only A-stable subspaces of E are {0} and E, then A = End(E). Proof. Let n = dim E. The result being obvious for n 1. We can assume that n 2. Note that by our hypothesis, Ax = E for all x ∈ E \ {0}. Let u ∈ A \ {0} be of minimal rank r. We claim that r = 1. If r > 1, there exist x, y ∈ E such that u(x) and u(y) are linearly independent. Since Au(x) = E, there exists v ∈ A such that v ◦ u(x) = y. Since u ◦ v ◦ u(x) and u(x) are linearly independent, u ◦ v ◦ u and u are linearly independent. Let F = u(E) which is u ◦ v-stable. Since k is algebraically closed, u ◦ v has an eigenvalue λ and an eigenvector z = u(t) for some t ∈ E \ ker u. So t ∈ ker(u ◦ v ◦ u − λu) and ker u is strictly contained in ker(u ◦ v ◦ u − λu). We deduce that u ◦ v ◦ u − λu is an element of A of rank strictly less than r. But u ◦ v ◦ u − λu = 0 because u ◦ v ◦ u and u are linearly independent. This contradicts the minimality of r. Thus r = 1 and we have proved our claim. Let u0 ∈ A and u ∈ End(E) be both of rank 1. There exist y0 , y ∈ E \ {0} and ϕ0 , ϕ ∈ E ∗ such that u0 (x) = ϕ0 (x)y0 and u(x) = ϕ(x)y for x ∈ E. Let v be the endomorphism of E defined by v(x) = ϕ(x)y0 . The subspace L = {x ∈ E; ϕ0 (w(x)) = 0 for all w ∈ A} is A-stable and L ⊂ ker ϕ0 = E. It follows from our hypothesis that L = {0} and {ϕ0 ◦ w; w ∈ A} = E ∗ . This implies that there exists w ∈ A such that ϕ = ϕ0 ◦ w, and hence v = u0 ◦ w ∈ A. Since Ay0 = E, there exists s ∈ A such that y = s(y0 ). We obtain that u = s ◦ v ∈ A since v ∈ A. We have therefore proved that the algebra A contains all endomorphisms of E of rank 1. It is now clear that A = End(E). 10.8.12 Definition. Let E be a finite-dimensional k-vector space and G a subgroup of GL(E). We say that G is unipotent if all the elements of G are unipotent. 10.8.13 Theorem. Let E be a finite-dimensional k-vector space and G a unipotent subgroup of GL(E). There exists x ∈ E \ {0} such that α(x) = x for all α ∈ G. Proof. Let us proceed by induction on n = dim E. The case n = 1 is trivial. So let us suppose that n 2. Furthermore, if F is a G-stable subspace of E of dimension < n. We can apply the induction hypothesis on F . We can therefore assume that the only G-stable subspaces of E are {0} and E.
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By Burnside’s Theorem (10.8.11), the subalgebra generated by G is End(E). Now, given u, v ∈ G, tr((idE −u) ◦ v) = tr(v) − tr(u ◦ v) = n − n = 0. It follows that tr((idE −u) ◦ w) = 0 for all w ∈ End(E). Hence u = idE and so G = {idE }, which implies that n = 1. Contradiction. 10.8.14 Let E be a k-vector space of dimension n and for u ∈ End(E), we shall denote the matrix of u with respect to a basis E of E by Mat(u, E). Corollary. Let G be a unipotent subgroup of GL(E). (i) There exists a basis E of E such that Mat(u, E) ∈ Un (k) for all u ∈ G. (ii) The group G is nilpotent. Proof. By applying 10.8.13, part (i) is immediate by induction on n. Now 10.8.10 implies (ii).
References • [5], [8], [10], [23], [40], [52], [56], [78].
11 Algebraic sets
In this chapter, we define algebraic sets, regular functions on them and morphisms between them. These notions will be generalized in the next chapters. From now on, k denotes a commutative algebraically closed field of characteristic zero. Let n ∈ N∗ .
11.1 Affine algebraic sets 11.1.1 We call kn the affine space of dimension n (affine line if n = 1 and affine plane if n = 2). Since k is algebraically closed, k is infinite and therefore an element P ∈ k[T] = k[T1 , . . . , Tn ] can be identified as a k-valued polynomial function on kn sending t = (t1 , . . . tn ) to P (t) = P (t1 , . . . , tn ). 11.1.2 Definition. Let A ⊂ k[T]. Denote by V(A) the set of points t ∈ kn such that P (t) = 0 for all P ∈ A. A subset X of kn is called an affine algebraic set if there exists A ⊂ k[T] such that X = V(A). 11.1.3 Let A ⊂ k[T]. We verify easily that if a is the ideal of k[T] generated by A, then V(A) = V(a). Since k[T] is Noetherian, a is finitely generated. Therefore, there exist P1 , . . . , Pn ∈ k[T] such that V(A) = V({P1 , . . . , Pn }) (denoted simply by V(P1 , . . . , Pn )). A subset of the form V(P ), where P ∈ k[T], is called a hypersurface of kn . 11.1.4 Proposition. Let a, b, (ai )i∈I be ideals of k[T]. Then: n = ∅, and if a ⊂ b, V(b) ⊂ V(a). (i) V(0) = k , V(k[T]) (ii) V( i∈I ai ) = i∈I V(ai ) and V(ab) = V(a ∩ b) = V(a) ∪ V(b). Proof. This is analogue to the proof of 6.5.3. 11.1.5 Let m ∈ Spm k[T]. By 6.4.3, there exists a = (a1 , . . . , an ) ∈ kn such that m = (T1 − a1 )k[T] + · · · + (Tn − an )k[T]. We deduce that V(m) = {a}. It follows from 11.1.4 that:
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Proposition. We have V(a) = ∅ if and only if a = k[T]. 11.1.6 For a subset M ⊂ kn , we denote by I(M ) the set of polynomials P ∈ k[T] such that P (a) = 0 for all a ∈ M . It is clear that I(M ) is a radical ideal of k[T]. We shall call I(M ), the ideal of M . Proposition. (i) We have I(∅) = k[T], and if M ⊂ N ⊂ kn , then I(M ) ⊃ I(N ). (ii) If M is an affine algebraic set, then V(I(M )) = M . (iii) The map sending an affine algebraic set√to its ideal is injective. (iv) If a is an ideal of k[T], then I(V(a)) = a. (v) We have I(kn ) = {0}. Proof. Part (i) is obvious. Clearly M ⊂ V(I(M )). Conversely, if M = V(a) for some ideal a ⊂ k[T], then a ⊂ I(M ). Hence V(I(M )) ⊂ V(a) = M (11.1.4). Now (iii) follows immediately from (ii). Since a ⊂ I(V(a)), part (iv) is clear for a = k[T]. Furthermore, for any m ∈ Spm k[T], V(m) = ∅ (11.1.5). It follows that m ⊂ V(m) = k[T]. Hence I(V(m)) = m. √ Now suppose that a = k[T]. Since I(V(a)) is radical, a ⊂ I(V(a)). Conversely, if m is a maximal√ideal containing a, then I(V(a)) ⊂ I(V(m)) = m intersection of maximal ideals containing by 11.1.4 and part (i). Since a is the √ a by 6.4.2, it follows that I(V(a)) ⊂ a. Finally, part (v) follows from (iv) and 11.1.4. 11.1.7 Corollary. The map a → V(a) induces a bijection between the set of radical ideals of k[T] and the set of affine algebraic subsets of kn .
11.2 Zariski topology 11.2.1 By 11.1.4, we have a topology, called the Zariski topology, on kn where the closed subsets are affine algebraic subsets of kn (or the V(a)’s by 11.1.7). In general, this topology is not Hausdorff. Indeed, closed subsets V(a) of k are finite, thus two non-empty open subsets of k have non-empty intersection. Let a be a radical ideal of k[T] and V = V(a) an affine algebraic subset of kn . The topology induced by the Zariski topology of kn shall also be called the Zariski topology on V . In view of 11.1.4, the closed subsets of V are the V(b)’s where b is an ideal of k[T] containing a. 11.2.2 Proposition. Let M be a subset of kn . The closure of M , with respect to the Zariski topology, is V(I(M )). Proof. Clearly, M ⊂ V(I(M )). Now if a is a radical ideal of k[T] such that M ⊂ V(a), then a = I(V(a)) ⊂ I(M ) by 11.1.6. Thus V(I(M )) ⊂ V(a). 11.2.3 Proposition. (i) Endowed with the Zariski topology, kn is a Noetherian space whose points are closed.
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(ii) Let a be a radical ideal of k[T]. Then V(a) is irreducible if and only if a is prime. Proof. The proof is analogous to the one for 6.5.6. 11.2.4 Corollary. For any ideal a of k[T], the set of irreducible components of V(a) is finite and it is the set of V(pi )’s where the pi ’s, 1 i r, are the minimal prime ideals of k[T] containing a. Proof. These are direct consequences of 1.3.5 and 11.2.3.
11.3 Regular functions 11.3.1 Definition. Let V be an affine algebraic subset of kn . A function from V to k is a regular function if it is the restriction to V of a polynomial function on kn . 11.3.2 For any affine algebraic subset V ⊂ kn , we denote by A(V ) the n algebra of regular functions on V . Note that √ A(k ) = k[T]. Let a = I(V ). Then V = V(a) and a = a (11.1.6). Thus the restrictions to V of P, Q ∈ k[T] are equal if and only if P − Q ∈ a. It follows that the algebras A(V ) and k[T]/a are isomorphic. In particular, A(V ) is finitely generated and reduced. 11.3.3 Proposition. Let V ⊂ kn be an affine algebraic subset. (i) The algebra A(V ) separates the points of V , i.e. if a, b are distinct points of V , then there exists f ∈ A(V ) such that f (a) = f (b). (ii) Elements of A(V ) are continuous maps with respect to the Zariski topology. Proof. (i) Let a = (a1 , . . . , an ), b = (b1 , . . . , bn ) be distinct points in V . Without loss of generality, we can assume that a1 = b1 . The restriction to V of the coordinate function kn → k, (x1 , . . . , xn ) → x1 clearly separates a and b. (ii) For λ ∈ k and ϕ ∈ A(V ), we have ϕ−1 (λ) = V(I(V )) ∩ V(P − λ) where P is an element of k[T] whose restriction to V is ϕ. Thus ϕ−1 (λ) is closed. Since a closed subset of k is either k or a finite set, ϕ is continuous. 11.3.4 Let V ⊂ kn be an affine algebraic subset. Recall from 11.3.2 that A(V ) is a reduced and finitely generated k-algebra. By 6.4.5, the map Homalg (A(V ), k) → Spm(A(V )) , χ → ker χ is a bijection. There is a canonical bijection from Spm(A(V )) to V . Namely, given m ∈ Spm(A(V )), there exists m ∈ Spm(k[T]) such that m ⊃ I(V ) and m /I(V ) = m. By 6.4.3, there exist a1 , . . . , an ∈ k such that
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m = (T1 − a1 )k[T] + · · · + (Tn − an )k[T]. Now m depends only on m and since a = (a1 , . . . , an ) ∈ V(m ), a ∈ V(I(V )) = V . So we have a well-defined map from Spm(A(V )) to V . Conversely, given a = (a1 , . . . , an ) ∈ V , I(V ) ⊂ I({a}) = m where m denotes the maximal ideal of k[T] generated by the Ti − ai , 1 i n. Then a corresponds to the maximal ideal m = m /I(V ) of A(V ). Let us consider the composition of the above bijective maps: θ
ψ
V −→ Spm(A(V )) −→ Homalg (A(V ), k). We verify easily that for a ∈ V , ψ ◦ θ(a) is the homomorphism defined by the evaluation at the point a, namely f → f (a), that we shall denote by χa .
11.4 Morphisms 11.4.1 Definition. Let V ⊂ kn and W ⊂ km be affine algebraic subsets. A map u : V → W is called a morphism (of affine algebraic sets) if for any g ∈ A(W ), g ◦ u ∈ A(V ). 11.4.2 We shall denote by Mor(V, W ) the set of morphisms from V to W . Clearly, the composition of morphisms is again a morphism. An isomorphism of V to W is a bijective morphism whose inverse is a morphism. The function k → k, t → t, is regular on k. It follows that A(V ) is the set of morphisms from V to k. For u ∈ Mor(V, W ), we shall denote by A(u) : A(W ) → A(V ) the map g → g ◦ u. We call A(u) the comorphism of u. If v ∈ Mor(W, Z), then we have A(v ◦ u) = A(u) ◦ A(v). 11.4.3 Proposition. Let V ⊂ kn and W ⊂ km be affine algebraic subsets. For any map u : V → W , denote by ui : V → k, 1 i m, the coordinate maps of u. The following conditions are equivalent: (i) u ∈ Mor(V, W ). (ii) u1 , . . . , um ∈ A(V ). Proof. (i) ⇒ (ii) Let pi : km → k denote the i-th canonical projection. Then ui = pi ◦ u = A(u)(pi ) and since pi |W ∈ A(W ), we have (ii). (ii) ⇒ (i) Let P1 , . . . , Pm ∈ k[T] be such that the restriction of Pi to V is ui . Let g ∈ A(W ) and Q ∈ k[X1 , . . . , Xm ] be such that its restriction to W is g. For any t ∈ V , we have g ◦ u(t) = Q(P1 (t), . . . , Pm (t)). Thus g ◦ u is the restriction of a polynomial function on kn . Hence g ◦ u ∈ A(V ). 11.4.4 Theorem. Let V ⊂ kn , W ⊂ km be affine algebraic subsets. The map u → A(u) is a bijection between Mor(V, W ) and Homalg (A(W ), A(V )). Proof. For ϕ ∈ Homalg (A(W ), A(V )), we define hϕ : Homalg (A(V ), k) → Homalg (A(W ), k) , χ → χ ◦ ϕ.
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135
There exists a unique map m(ϕ) : V → W such that the following diagram is commutative: V ⏐ ⏐
m(ϕ)
−−−−→
W ⏐ ⏐
Homalg (A(V ), k) −−−−→ Homalg (A(W ), k) hϕ
where the vertical maps are the isomorphisms x → χx and y → χy of 11.3.4. For x ∈ V , we have therefore: χm(ϕ)(x) = χx ◦ ϕ.
(1)
It follows that for x ∈ V and g ∈ A(W ), g(m(ϕ)(x)) = χm(ϕ)(x) (g) = χx ◦ ϕ(g) = χx (ϕ(g)) = ϕ(g)(x). We deduce that: g ◦ m(ϕ) = ϕ(g). In particular, g ◦ m(ϕ) ∈ A(V ) which shows that m(ϕ) ∈ Mor(V, W ). Furthermore, we have shown that A(m(ϕ)) = ϕ. Now given u ∈ Mor(V, W ), it is clear that A(u) ∈ Homalg (A(W ), A(V )). Taking ϕ = A(u) in the previous argument, we have: χm(A(u))(x) (g) = (A(u)(g))(x) = g(u(x)) = χu(x) (g). Hence m(A(u)) = u. We have therefore proved that the maps ϕ → m(ϕ) and u → A(u) are mutually inverse. 11.4.5 Corollary. Let u ∈ Mor(V, W ). Then u is an isomorphism if and only if the comorphism A(u) of u is an isomorphism of k-algebras. Proof. If u is an isomorphism and v ∈ Mor(W, V ) is such that u ◦ v = idW , v ◦ u = idV , then A(u) ◦ A(v) = idA(V ) and A(v) ◦ A(u) = idA(W ) . This implies that A(u) is an isomorphism of k-algebras. Conversely, if ϕ ∈ Homalg (A(V ), A(W )) verifies A(u) ◦ ϕ = idA(V ) and ϕ ◦ A(u) = idA(W ) , then m(ϕ) ◦ u = idV and u ◦ m(ϕ) = idW . Hence u is an isomorphism.
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11.5 Examples of morphisms 11.5.1 It follows from 11.4.4 that Mor(kn , km ) is in bijection with the set of k-algebra homomorphisms ϕ : A(km ) → A(kn ). Let A(kn ) = k[T1 , . . . , Tn ] and A(km ) = k[X1 , . . . , Xm ]. Then ϕ is uniquely determined by m polynomials P1 , . . . , Pm ∈ k[T1 , . . . , Tn ] verifying: ϕ(Xi ) = Pi (T1 , . . . , Tn ) for 1 i m. By formula (1) of 11.4.4, the morphism u = m(ϕ) (hence A(u) = ϕ) is defined by: χu(x) = χx ◦ ϕ for x = (x1 , . . . , xn ) ∈ kn . Let u(x) = (y1 , . . . , ym ). Then for 1 i m, yi = χu(x) (Xi ) = χx (ϕ(Xi )) = Pi (x1 , . . . , xn ). Thus (2)
u(x) = (P1 (x1 , . . . , xn ), . . . , Pm (x1 , . . . , xn )). 11.5.2 Let us suppose that n m in 11.5.1, then the map u : kn → km , (x1 , . . . , xn ) → (x1 , . . . , xn , 0, . . . , 0)
is a morphism. Its comorphism A(u) is given by: A(u)(Xi ) = Ti if 1 i n , A(u)(Xi ) = 0 if n + 1 i m. 11.5.3 Let us suppose that m n in 11.5.1, then the map v : kn → km , (x1 , . . . , xn ) → (x1 , . . . , xm ) is a morphism. For 1 i m, we have: A(v)(Xi ) = Ti . 11.5.4 Let V ⊂ kn be an affine algebraic subset and a = I(V ) ⊂ A(kn ). By the definition of A(V ), the canonical injection j : V → kn is a morphism and A(j) is the canonical surjection A(kn ) → A(V ) = A(kn )/a. 11.5.5 Let u ∈ Mor(kn , km ) be given by formula (2) of 11.5.1. Denote by v : kn → kn × km = kn+m the map given by: x = (x1 , . . . , xn ) → (x, u(x)) = (x1 , . . . , xn , P1 (x), . . . , Pm (x)).
11.5 Examples of morphisms
137
The image G of v is called the graph of u. Note that y = (y1 , . . . , yn+m ) ∈ G if and only if y is a zero of the polynomials Tn+i − Pi (T1 , . . . , Tn ) for 1 i m. It follows that G is an affine algebraic set. Further, if we consider v as a map from kn to G, then v ∈ Mor(kn , G) and v is bijective. Now let p : kn+m → kn be the canonical projection (11.5.3) and let j : G → kn+m be the canonical injection. Then by 11.5.3 and 11.5.4, p and j are morphisms (hence p ◦ j is a morphism). But v ◦ (p ◦ j) = idG , (p ◦ j) ◦ v = idkn . Consequently, v is an isomorphism between kn and G. 11.5.6 Again, let u, P1 , . . . , Pm be as in 11.5.1. The ideal I(u(kn )) is the set of polynomials Q ∈ k[X1 , . . . , Xm ] such that for all x ∈ kn , Q(P1 (x), . . . , Pm (x)) = 0. Since k is infinite, I(u(kn )) is the set of polynomials Q ∈ k[X1 , . . . , Xm ] such that Q(P1 (T1 , . . . , Tn ), . . . , Pm (T1 , . . . , Tn )) = 0. The set W = V(I(u(kn ))) is the smallest affine algebraic subset containing u(kn ). Hence W is the Zariski closure of u(kn ). Denote by v : kn → W the morphism induced by u, and j : W → km the canonical injection. We have u = j ◦ v, so A(u) = A(v) ◦ A(j) (by 11.5.4, A(j) is the canonical surjection A(kn ) → A(W )). On the other hand, A(u)(Xi ) = Pi for 1 i m (11.5.1). Let g ∈ A(W ) be such that A(v)(g) = 0. For all x ∈ kn , we have g(v(x)) = 0. But by construction, u(kn ) is dense in W , hence g = 0. We deduced therefore that A(W ) is isomorphic to the subalgebra of A(kn ) generated by P1 , . . . , Pm . Note that in general, u(kn ) = W (i.e. the image of a morphism is not necessarily an affine algebraic set). For example, let u : k2 → k2 , (t1 , t2 ) → (t1 t2 , t2 ). We see easily that the only polynomial Q ∈ k[X1 , X2 ] such that Q(T1 T2 , T2 ) = 0 is zero. Thus the Zariski closure of u(k2 ) is k2 . But clearly, (1, 0) ∈ u(k2 ) and therefore u(k2 ) is not an affine algebraic set. 11.5.7 Let us consider the following morphism u : k → k2 , t → (t2 , t3 ). The image C of u is contained in V(T13 − T22 ). Conversely, given (x1 , x2 ) ∈ V(T13 − T22 ), we have (x1 , x2 ) = u(0) if x1 = 0 and (x1 , x2 ) = u(x2 /x1 ) if x1 = 0. It follows that C = V(T13 −T22 ) and A(C) is
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isomorphic to the k-algebra k[T 2 , T 3 ] (11.5.6). Note however that this algebra, which is an integral domain, is not isomorphic to k[T ], since T ∈ k[T 2 , T 3 ] and T is integral over k[T 2 , T 3 ]. This is an example of a bijective morphism which is not an isomorphism.
11.6 Abstract algebraic sets 11.6.1 Proposition. Any finitely generated reduced k-algebra A is isomorphic to the algebra of regular functions over some affine algebraic set. Proof. Let s1 , . . . , sn be generators for A. The surjection θ : k[T1 , . . . , Tn ] → A, Ti → si , induces an isomorphism between A and k[T1 , . . . , Tn ]/ ker θ. Since A is reduced, ker θ is a radical ideal. Hence I(V(ker θ)) = ker θ and A is isomorphic to A(V(ker θ)). 11.6.2 Remark. Let A be a finitely generated reduced k-algebra and V , W be affine algebraic sets such that A is isomorphic to A(V ) and A(W ). Then by 11.4.5, V and W are isomorphic. 11.6.3 Let V ⊂ kn be an affine algebraic set, a = I(V ) ⊂ A(kn ). To avoid confusion between the notations of 6.5.1 and 11.1.3, for b an ideal of A(V ), we shall denote in this section by V (b), the set of maximal ideals of A(V ) containing b. By 11.3.4, there is a canonical bijection: γV : V → Homalg (A(V ), k) , a → χa . On the other hand, we have another canonical bijection (6.4.5): δV : Homalg (A(V ), k) → Spm(A(V )) , χ → ker χ. Thus εV = δV ◦ γV is a bijection between V and Spm(A(V )). It is clear that if b is an ideal of A(kn ) containing a, then εV induces a bijection between V(b) and V (b/a). It follows that εV is a homeomorphism with respect to the Zariski topology (6.5.4 and 11.2.1). Let W ⊂ km be another affine algebraic set. Then by 6.4.5 and 11.4.4, we have for ϕ ∈ Homalg (A(W ), A(V )), the following commutative diagram: V ⏐ ⏐ γV
m(ϕ)
−−−−→
W ⏐ ⏐γW
hϕ
Homalg (A(V ), k) −−−−→ Homalg (A(W ), k) ⏐ ⏐ ⏐δ ⏐ δV W Spm(A(V ))
Spm(ϕ)
−−−−−→
Spm(A(W ))
11.6 Abstract algebraic sets
139
11.6.4 The preceding discussion allows us to consider the category of affine algebraic sets as a subcategory of a bigger (but equivalent) category, namely (abstract) algebraic sets which are not necessarily subsets of kn . The objects of this new category are the triples (X, A, βX ) where: (i) X is a set. (ii) A is a finitely generated reduced k-algebra. (iii) βX is a bijection from Spm(A) to X. Morphisms of (X, A, βX ) to (Y, B, βY ) are pairs (u, ϕ) where u : X → Y and ϕ ∈ Homalg (B, A) are such that the following diagram is commutative: Spm(ϕ)
Spm(A) −−−−−→ Spm(B) ⏐ ⏐ ⏐β ⏐ βX Y X
u
−−−−→
Y
The composition of morphisms (v, ψ) and (u, ϕ) is (v ◦ u, ϕ ◦ ψ). By abuse of notations, we shall write X and u for (X, A, βX ) and (u, ϕ); we shall also call ϕ the comorphism of u. 11.6.5 To identify the category of affine algebraic sets as a subcategory of the category of algebraic sets, it suffices to identify an affine algebraic set V with the triple (V, A(V ), ε−1 V ) (11.6.3), and a morphism u ∈ Mor(V, W ) with the pair (u, A(u)). By 11.4.4, 11.6.1 and 11.6.3, this category is equivalent to the category of affine algebraic sets. Similarly, we can identify a finitely generated reduced k-algebra A with the triple (Spm(A), A, idSpm(A) ), and a homomorphism ϕ with the pair (Spm(ϕ), ϕ). Again this induces an equivalence between the category of finitely generated reduced k-algebras and the category of algebraic sets. We define regular functions on an algebraic set (X, A, βX ) via transport −1 (x) is a maximal ideal of A. For f ∈ A, we of structure. If x ∈ X, βX −1 set f (x) = f (βX (x)) as defined in 6.4.6. Then regular functions on X are morphisms from X to k, and we can identify them as elements of A. For any morphism u : X → Y , we define A(u) : B → A as in 11.4.2, by g → g ◦ u. 11.6.6 Let (X, A, βX ) be an algebraic set. We define the Zariski topology on X to be the topology induced, via the bijection βX , by the Zariski topology on (Spm(A), A, idSpm(A) ) (6.5.4). In the case where A = A(V ) for some affine algebraic set V , the Zariski topology of Spm(A) is the one obtained from the Zariski topology on V via the canonical bijection from Spm(A(V )) to V (11.6.3). 11.6.7 Let A be a finitely generated reduced k-algebra, a a radical ideal of A and βa the bijection Spm(A/a) → V(a) which is the inverse of the canonical bijection m → m/a. It is clear that (V(a), A/a, βa ) is an algebraic set, that we shall denote by V(a) or Spm(A/a). The Zariski topology on this set is induced by the one on Spm(A). The canonical injection V(a) → Spm(A) is a
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11 Algebraic sets
morphism, whose comorphism is A → A/a. Thus regular functions on V(a) can be viewed as restrictions of regular functions on Spm(A). 11.6.8 Proposition. Let A, B be finitely generated reduced k-algebras and ϕ ∈ Homalg (B, A). The following conditions are equivalent: (i) There exist a closed subset V ⊂ Spm(B) and an isomorphism v : Spm(A) → V such that Spm(ϕ) = j ◦ v where j : V → Spm(B) is the canonical injection. (ii) ϕ is surjective. Proof. (i) ⇒ (ii) Let b be a radical ideal of B such that V = V(b), ψ : B → B/b the canonical surjection and θ ∈ Homalg (B/b, A) such that ϕ = θ ◦ ψ. It follows that j = Spm(ψ), v = Spm(θ) and θ is an isomorphism by 11.4.5 and our hypotheses. Hence ϕ is surjective. (ii) ⇒ (i) Let b = ker ϕ, θ : B/b → A the induced isomorphism and ψ : B → B/b the canonical surjection. The ideal b is radical and by 11.4.5, Spm(θ) : Spm(A) → Spm(B/b) is an isomorphism.
11.7 Principal open subsets 11.7.1 Let A be a finitely generated reduced k-algebra and f ∈ A. We set D(f ) = Spm(A) \ V(Af ). Thus D(f ) is the set of maximal ideals m which do not contain f (or equivalently f (m) = 0). This is an open subset of Spm(A) and an open subset of this form is called a principal open subset of Spm(A). Note that: D(0) = ∅ , D(A) = Spm(A) , D(f g) = D(f ) ∩ D(g). If ϕ ∈ Homalg (B, A) and g ∈ B, then by 6.5.8, we have: Spm(ϕ)−1 (D(g)) = D(ϕ(g)). 11.7.2 Proposition. Principal open subsets of Spm(A) form a base for the Zariski topology on Spm(A). Proof. Let V be a closed subset of Spm(A). If x ∈ V , then there exists f ∈ I(V ) such that f (x) = 0. Thus x ∈ D(f ) and D(f ) ⊂ Spm(A) \ V . 11.7.3 Let A be a finitely generated reduced k-algebra and f ∈ A. Denote by Af the set of maps from D(f ) to k of the form (3)
x →
g(x) f (x)n
where g ∈ A and n ∈ N. It is clear that Af is a finitely generated k-algebra. Proposition. (i) An element g of Af is identically zero on D(f ) if and only if f g = 0. (ii) The algebra Af is reduced. (iii) Let S = {f n ; n ∈ N}. The algebras Af and S −1 A are isomorphic.
11.7 Principal open subsets
141
Proof. Part (i) is obvious. Now if (g(x)/f (x)n )m = 0 for all x ∈ D(f ). Then f g = 0 and g(x)/f (x)n = 0 for all x ∈ D(f ). Hence Af is reduced. Let ε : A → Af be the restriction map g → g|D(f ) . If g ∈ S, then ε(g) is invertible. By 2.3.4, there exists a unique homomorphism θ : S −1 A → Af such that θ ◦ i = ε where i : A → S −1 A is the canonical homomorphism. It is clear that θ is surjective. On the other hand, by part (i), θ is injective. Thus θ is an isomorphism. 11.7.4 We shall prove that a principal open subset D(f ) of Spm(A) can be considered canonically as an algebraic set whose topology is the one induced by the topology of Spm(A). Theorem. There exists a canonical bijection β : Spm(Af ) → D(f ) which is a homeomorphism when D(f ) is endowed with the topology induced from the Zariski topology of Spm(A). Proof. Let ε : A → Af , g → g|D(f ) , and γ = Spm(ε) : Spm(Af ) → Spm(A). Set F = ε(f ). If m ∈ Spm(Af ), then F ∈ m and f ∈ ε−1 (m). This implies that the image of γ is contained in D(f ). For m ∈ D(f ), set I(m) = {ε(g)/F n ; g ∈ m, n ∈ N}. The set I(m) is an ideal of Af . If I(m) = Af , then there exist g ∈ m and n ∈ N such that ε(g)/F n = 1. So g/f n = 1 or equivalently (g − f n )/f n = 0. So (g − f n )f = 0 by 11.7.3. Thus gf = f n+1 , hence f n+1 ∈ m and f ∈ m. Contradiction. So I(m) = Af . There exists n ∈ Spm(Af ) such that n ⊃ I(m). It follows that m ⊂ ε−1 (n), so m = ε−1 (n) and I(m) = n. Thus the image of γ is D(f ). Since I(γ(n)) = n for n ∈ Spm(Af ), γ is injective and therefore it induces a bijection β from Spm(Af ) onto D(f ). As γ is continuous (6.5.8), to obtain the result, it suffices to prove that β maps open sets to open sets. By 11.7.2, we are reduced to prove that γ(D(ε(g)/F n )) = γ(D(ε(g))) is open for all g ∈ A. Let n ∈ Spm(Af ) be such that ε(g) ∈ n. Then g ∈ ε−1 (n) and γ(D(ε(g))) is contained in D(f ) ∩ D(g). Conversely, let m ∈ Spm(A) be such that f ∈ m and g ∈ m. There exists n ∈ Spm(Af ) such that β(n) = m. Thus m = ε−1 (n). Now there exist h ∈ A, θ ∈ m such that 1 = hg + θ. So 1 = ε(h)ε(g) + ε(θ). Since ε(θ) ∈ n, ε(g) ∈ n. Thus γ(D(ε(g))) = D(f ) ∩ D(g), and we have obtained our result. 11.7.5 Remarks. 1) If A is an integral domain and f = 0, then Af can be identified with subring A[1/f ] of Fract(A) (since f g = 0 implies that g = 0, the homomorphism A[1/f ] → Af , g/f n → ε(g)/ε(f )n is injective). 2) Let j : D(f ) → Spm(A) be the canonical injection. The diagram
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11 Algebraic sets γ
Spm(Af ) −−−−→ Spm(A) ⏐ ⏐ ⏐id ⏐ β Spm(A) D(f )
j
−−−−→ Spm(A)
is commutative. Consequently, j is a morphism whose comorphism is ε. 3) Note that regular functions on D(f ) are not necessarily restrictions of regular functions on Spm(A). For example, let A = k[T ], then D(T ) = k \ {0} and x → 1/x is a regular function on D(f ) which is not the restriction of a polynomial. 4) The map δ : Mn (k) → k, M → det M is regular and D(δ) = GLn (k). Thus GLn (k) can be considered as an algebraic set. Left multiplication or right multiplication by A ∈ GLn (k), and the map sending M to M −1 are clearly isomorphisms of the algebraic set GLn (k).
11.8 Products of algebraic sets 11.8.1 Let V ⊂ km and W ⊂ kn be affine algebraic sets. Let us identify A(km ) (resp. A(kn )) as the subring k[T1 , . . . , Tm ] (resp. k[Tm+1 , . . . , Tn+m ]) of k[T1 , . . . , Tn+m ]. Let a = I(V ), b = I(W ) and c the ideal in k[T1 , . . . , Tn+m ] generated by a ∪ b. The affine algebraic set V(c) is defined by the equations: P (x1 , . . . , xm ) = 0 , Q(xm+1 , . . . , xn+m ) = 0 for all P ∈ a and Q ∈ b. Thus √ V(c) = V × W , A(V × W ) = k[T1 , . . . , Tn+m ]/ c. Since we have the canonical isomorphisms A(V ) ⊗k A(W ) (k[T1 , . . . , Tm ]/a) ⊗k (k[Tm+1 , . . . , Tm+n ]/b), k[T1 , . . . , Tn+m ] k[T1 , . . . , Tm ] ⊗k k[Tm+1 , . . . , Tm+n ], the ideal c can be written as: c = a ⊗k k[Tm+1 , . . . , Tm+n ] + k[T1 , . . . , Tm ] ⊗k b. Hence A(V ) ⊗k A(W ) can be identified canonically with k[T1 , . . . , Tm+n ]/c. It follows that there is a canonical surjective homomorphism: ϕ : A(V ) ⊗k A(W ) → A(V × W ) where for f ∈ A(V ), g ∈ A(W ), (x, y) ∈ V × W , ϕ(f ⊗ g)(x, y) = f (x)g(y). Proposition. The homomorphism ϕ is an isomorphism.
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143
Proof. It suffices to prove that ϕ is injective. Let h = f1 ⊗ g1 + · · · + fr ⊗ gr be a non-zero element in ker ϕ. We may assume that h is written with r minimal. It follows that the fi ’s are linearly independent. Now, given any y ∈ W , f1 (x)g1 (y) + · · · + fr (x)gr (y) = 0 for all x ∈ V . Since the fi ’s are linearly independent, gi (y) = 0 for all y ∈ W , i = 1, . . . , r. It follows that the gi ’s are zero, which implies that h = 0. Contradiction. 11.8.2 Corollary. Let A, B be finitely generated reduced k-algebras, then A ⊗k B is a finitely generated reduced k-algebra. Proof. It is clear by 11.6.1 and 11.8.1. 11.8.3 From now on, we shall identify A(V × W ) and A(V ) ⊗k A(W ) via ϕ. Denote by pr1 : V × W → V and pr2 : V × W → W the canonical projections. These are morphisms of affine algebraic sets. For (x, y) ∈ V × W , f ∈ A(V ), we have that f ◦ pr1 (x, y) = f (x). It follows that the comorphisms of pr1 and pr2 are given by: A(pr1 ) : A(V ) → A(V ) ⊗k A(W ) , f → f ⊗ 1, A(pr2 ) : A(W ) → A(V ) ⊗k A(W ) , g → 1 ⊗ g. Proposition. Let U, V, W be affine algebraic sets and v ∈ Mor(U, V ), w ∈ Mor(U, W ). The map u : U → V × W , z → (v(z), w(z)) is the unique morphism such that v = pr1 ◦ u and w = pr2 ◦ u. Proof. Clearly u is the unique map having the required properties. Now let f ∈ A(V ), g ∈ A(W ) and z ∈ U . We have: (f ⊗ g)(u(z)) = (f ⊗ g)(v(z), w(z)) = (f ◦ v(z))(g ◦ w(z)). Since f ◦ v, g ◦ w ∈ A(U ), u is a morphism.
11.8.4 Proposition. Let V, W, V , W be affine algebraic sets. For any v ∈ Mor(V, V ) and w ∈ Mor(W, W ), the map u : V × W → V × W , (x, y) → (v(x), w(y)) is a morphism.
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Proof. Note that u(x, y) = (v ◦ pr1 (x, y), w ◦ pr2 (x, y)). So the result follows from 11.8.3. 11.8.5 Proposition. Let V, W be affine algebraic sets. The canonical projections pr1 , pr2 of V × W onto V and W are open maps. Proof. It suffices to show that the image of a principal open subset is open. Let h = f1 ⊗ g1 + · · · + fn ⊗ gn ∈ A(V × W ), and x0 ∈ U = pr1 (D(h)). Then there exists y0 ∈ W such that (x0 , y0 ) ∈ D(h). Set u = g1 (y0 )f1 + · · · + gn (y0 )fn ∈ A(V ). Then x0 ∈ D(u) and D(u) × {y0 } ⊂ D(h). Hence D(u) ⊂ U and U is an open subset of V . The proof for pr2 is analogue. 11.8.6 Proposition. Let V, W be affine algebraic sets. Then V × W is irreducible if and only if V and W are irreducible. Proof. Since the canonical projections are morphisms, V and W are irreducible if V × W is irreducible. Conversely, suppose that V and W are irreducible. The projections pr1 , pr2 of V × W onto V and W are continuous open maps (6.5.8 and 11.8.5). For x ∈ V , pr−1 1 (x) = {x} × W is clearly irreducible. It follows therefore from 1.1.7 that V × W is irreducible. 11.8.7 Corollary. Let A, B be finitely generated k-algebras. If A and B are integral domains, then A ⊗k B is also an integral domain. 11.8.8 Proposition. Let V, W be affine algebraic sets, V1 (resp. W1 ) a closed subset of V (resp. W ). Then V1 × W1 is a closed subset of V × W . Proof. Let a ⊂ A(V ), b ⊂ A(W ) be ideals such that A(V1 ) = A(V )/a and A(W1 ) = A(W )/b. Then A(V1 × W1 ) = A(V1 ) ⊗k A(W1 ) = A(V × W )/c where c = a ⊗k A(W ) + A(V ) ⊗k b. Hence V1 × W1 = V(c).
11.8.9 Remarks. 1) The Zariski topology on V × W is in general strictly finer than the product of the Zariski topologies on V and W . The open subsets D(f ⊗ g) = D(f ) × D(g), f ∈ A(V ) g ∈ A(W ), form a base with respect to the product topology. However, there exists Zariskiopen subsets of V × W which does not contain any non-empty subsets of this form. For example, if V = W = k, the non-empty subsets D(f ⊗ g) are the complements of finite unions of subsets of the form {x} × W and V × {y}. Let U = k2 \ V(T1 − T2 ). Since k is infinite, none of the preceding subsets contain U. 2) With the notations of example 4 of 11.7.5, the multiplication GLn (k) × GLn (k) → GLn (k) , (M, N ) → M N is a morphism of algebraic sets.
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References and comments • [5], [19], [22], [26], [37], [40], [78]. All the results in this chapter are classic and can be found in the books cited above.
12 Prevarieties and varieties
In this chapter, we define an algebraic variety via the theory of sheaves. Local behaviour of an algebraic variety is considered in the last section.
12.1 Structure sheaf 12.1.1 Let A be a finitely generated reduced k-algebra and X = Spm(A) be endowed with the Zariski topology. For f ∈ A, we shall conserve the notations Af , V(f ) and D(f ) of chapter 11. Theorem. There exists a unique sheaf of functions OX on X, with values in k such that for any f ∈ A, we have: Γ (D(f ), OX ) = Af . We call OX the structure sheaf of the algebraic set X. Proof. By 11.7.2, the D(f )’s, f ∈ A, form a base B for the topology on X. It suffices therefore to verify that B satisfies the hypotheses of 9.5.1. Let f, g ∈ A be such that D(f ) ⊂ D(g). Then V(g) ⊂ V(f ), hence by 11.1.6: Af = I(V(f )) ⊂ I(V(g)) = Ag. Thus there exist n ∈ N∗ and h ∈ A such that f n = gh. If u/g i ∈ Ag , then its restriction to D(f ) is (uhi )/(g i hi ) = ug i /f ni , which belongs to Af . It follows therefore that if D(f ) = D(g), then Af = Ag . Now let f and (fi )i∈I be non zero elements of A such that D(fi ) is a covering of D(f ). Then V(f ) is the intersection of the V(fi )’s, so V(f ) = V(a), where a is the ideal generated by the fi ’s. Since A is Noetherian, we may assume that I = {1, . . . , r}. Let g : D(f ) → k be a function such that g|D(fi ) ∈ Afi for 1 i r. There exist a1 , . . . , ar ∈ A and n ∈ N∗ such that g(x) = ai (x)/fi (x)n for all x ∈ D(fi ).
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If x ∈ D(fi )∩D(fj ) = D(fi fj ), then ai (x)fj (x)n = aj (x)fi (x)n . By 11.7.3, it follows that fi fj (ai fjn − aj fin ) = 0. We have V(f1n , . . . , frn ) = V(f1 , . . . , fr ) ⊂ V(f ). So 11.1.6 implies that there exist m ∈ N∗ and b1 , . . . , br ∈ A such that f m = b1 f1n+1 + · · · + br frn+1 . Hence r r ai fi f m = ai bj fi fjn+1 = fin+1 aj bj fj . j=1
j=1
Set a=
r
aj bj fj .
j=1
For 1 i r, we have therefore ai fi f m = afin+1 which implies that if x ∈ D(fi ), a(x) ai (x) = . f (x)m fi (x)n Thus for x ∈ D(f ), g(x) =
a(x) , f (x)m
and g ∈ Af . 12.1.2 By 9.5.8 and 12.1.1, (X, OX ) is a ringed space. Note that OX determines A since Γ (X, OX ) = A. 12.1.3 Proposition. Let X = Spm(A), Y = Spm(B) where A, B are finitely generated reduced k-algebras. The set of morphisms of the algebraic set X to the algebraic set Y and the set of morphisms of the ringed space (X, OX ) to the ringed space (Y, OY ) are identical. Proof. Let u : (X, OX ) → (Y, OY ) be a morphism of ringed spaces. For any g ∈ B = Γ (Y, OY ), g ◦u ∈ A = Γ (X, OX ). Hence u is a morphism of algebraic sets. Conversely, let u : X → Y be a morphism of algebraic sets and A(u) its comorphism. For g ∈ B, we have u−1 (D(g)) = D(A(u)(g)) (11.7.1). On the other hand, for f = h/g n ∈ Γ (D(g), OY ) where h ∈ B, n ∈ N∗ , we have for x ∈ D(A(u)(g)), (f ◦ u)(x) =
(A(u)(h))(x) h(u(x)) = . g(u(x))n ((A(u)(g))(x))n
Thus f → f ◦ u is a homomorphism (of k-algebras) from Γ (D(g), OY ) = Bg to Γ (u−1 (D(g)), OX ) = AA(u)(g) . Hence by 9.5.3, u is a morphism of ringed spaces. 12.1.4 We can define by transport of structure, the structure sheaf of an (abstract) algebraic set (11.6.4). Proposition 12.1.3 is again valid in this setting.
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Definition. An affine algebraic variety over k, or simply an affine variety is a ringed space (X, OX ) where X is an algebraic set. A morphism of affine varieties is a morphism of the underlying ringed spaces.
12.2 Algebraic prevarieties 12.2.1 Definition. (i) An (algebraic) prevariety over k is a ringed space (X, OX ) relative to k verifying the following conditions: a) X is a Noetherian topological space. b) There is an open covering (Ui )i∈I of X such that each induced ringed space (Ui , OX |Ui ) is an affine algebraic variety over k. (ii) An open subset U of an algebraic prevariety (X, OX ) is called an affine open subset if the induced ringed space (U, OX |U ) is an affine variety. Remarks. 1) An affine variety is a prevariety. 2) When there is no confusion, we shall simply say that X is a prevariety, instead of (X, OX ). 3) A prevariety is quasicompact (1.3.3). Thus there exists a finite covering of X by open affine subsets. 12.2.2 We define the category of algebraic prevarieties over k by letting morphisms of prevarieties to be morphisms of ringed spaces. By 9.5.3, we have the following result: Proposition. Let X, Y be prevarieties and u : X → Y a map. Then u is a morphism of prevarieties if and only if for all open affine subset V of Y and x ∈ u−1 (V ), there exists an open affine subset U of X containing x, and contained in u−1 (V ), such that the restriction of u to U is a morphism of algebraic sets from U to V . 12.2.3 It follows from the previous discussion that: Proposition. Let us conserve the notations of 12.2.2. The map u is an isomorphism of prevarieties if and only if u is bijective and for all x ∈ X, there exists an affine open subset U of X containing x, such that u induces an isomorphism from U to an affine open subset of Y . 12.2.4 Remark. When u satisfies the condition of 12.2.3 without being bijective, we say that u is a local isomorphism from X to Y . 12.2.5 Proposition. Let X be a prevariety. (i) The topological space X admits a base consisting of affine open subsets. (ii) All points of X are closed. (iii) If U is open in X, then the ringed space (U, OX |U ) is a prevariety. Proof. Let (Ui )i∈I be a covering of X by affine open subsets. (i) Each Ui has a base consisting of affine open subsets (11.7.2 and 11.7.4). So X also has a base consisting of affine open subsets. (ii) A subset F of X is closed if and only if F ∩ Ui is closed for all i ∈ I. Thus the assertion follows from 11.2.3.
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(iii) By 1.3.2, U is Noetherian. If B is a base for X consisting of affine open subsets, then those open subsets in B contained in U form a base for the topology of U . The result now follows. 12.2.6 Proposition. Let (X, OX ) be a prevariety and Y a locally closed subset of X. Then the ringed space (Y, OX |Y ) is a prevariety. Proof. By 1.3.2, Y is Noetherian. We may further assume by 9.5.4 and 12.2.5 (iii) that Y is closed in X. Let (Ui )i∈I be a covering of X by affine open subsets. To obtain the result, it suffices to prove that for all i ∈ I, the ringed space (Y ∩ Ui , OX |Y ∩Ui ) is an affine variety. It follows that we may take X = Spm(A) to be an affine variety and Y = V(a) where a is a radical ideal of A. Let us denote ϕ : A → A/a the canonical surjection and j = Spm(ϕ) : Y → X the canonical injection (11.5.4). Let V be an open subset of Y , and G(V ) be the set of maps h : V → k such that there exist an open subset U ⊂ X and g ∈ OX (U ) verifying V = Y ∩ U and g|V = h. The sheaf associated to the presheaf G is OX |Y (9.3.8). Let f ∈ A. We have: D(ϕ(f )) = j −1 (D(f )) = Y ∩ D(f ). It follows from 12.1.1 that OY (D(ϕ(f ))) = (A/a)ϕ(f ) where OY denotes the structure sheaf of Y . Thus if h ∈ OY (D(ϕ(f ))), then there exist g ∈ A and n ∈ N such that h(x) = g(x)/f (x)n , x ∈ Y ∩ D(f ). So the map y → g(y)/f (y)n on D(f ) belongs to OX (D(f )), consequently we obtained that OY (D(ϕ(f ))) ⊂ G(D(ϕ(f ))). Conversely, given h ∈ G(D(ϕ(f ))). There exist an open subset U ⊂ X, and g ∈ OX (U ) such that U ∩ Y = D(ϕ(f )) = Y ∩ D(f ) and g|Y ∩D(f ) = h. Cover U by principal open subsets Ui = D(gi ), gi ∈ A. Then the D(ϕ(gi ))’s cover D(ϕ(f )) and since g|Ui ∈ OX (Ui ) = Agi , ϕ(g)|D(ϕ(gi )) belongs to (A/a)ϕ(gi ) = OY (D(ϕ(gi ))). As OY is a sheaf, we deduce that ϕ(g)|D(ϕ(f )) ∈ OY (D(ϕ(f ))). Hence h ∈ OY (ϕ(f )). The D(ϕ(f ))’s form a base for the topology on Y . It follows therefore (using for example 9.3.6) that OY = OX |Y . 12.2.7 Let (X, OX ) be a prevariety and Y a locally closed subset of X. We call the prevariety (Y, OX |Y ) a sub-prevariety of (X, OX ). From now on, when we considered a locally closed subset of X as a prevariety, it will be as a sub-prevariety of X. 12.2.8 Let u : (X, OX ) → (Y, OY ) be a morphism of prevarieties, Z a locally closed subset of Y containing u(X). Then by 9.5.4, u induces a morphism of prevarieties from (X, OX ) to (Z, OY |Z ). 12.2.9 Let X, Y be prevarieties, Z the sum of the topological spaces X and Y . We identify X and Y as two disjoint open subsets of Z. By 9.5.1, we can define a structure of prevariety on Z by setting OZ (U ) = OX (U ) and
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OZ (V ) = OY (V ) for open subsets U ⊂ X and V ⊂ Y . This prevariety is called the sum of X and Y , and we shall denote it by X Y . If X and Y are affine varieties, then so is X Y , and we have: A(X Y ) = A(X) × A(Y ) , X = V({0} × A(Y )) , Y = V(A(X) × {0}).
12.3 Morphisms of prevarieties 12.3.1 Let (X, OX ) be a prevariety. An element in Γ (X, OX ) is called a regular function on X. Let f be a regular function on X. If U is an affine open subset of X, then the restriction of f to U is a regular function on U . By 12.2.2, we deduce that f is a morphism from X to the affine line k. 12.3.2 Let X, Y be prevarieties and y0 ∈ Y . Then the map u : X → Y sending any x ∈ X to y0 is a morphism (this is clear by 11.4.3 since we may assume that X and Y are affine varieties by 12.2.2). 12.3.3 Let u : (X, OX ) → (Y, OY ) be a morphism of prevarieties. Then we can associated to u the map Γ (u) : Γ (Y, OY ) → Γ (X, OX ) , g → g ◦ u which is a homomorphism of k-algebras. But contrary to 11.4.4, Γ (u) does not in general determine u. However, Proposition. Let X be a prevariety and Y an affine variety. The map u → Γ (u) is a bijection between the set of morphisms of X to Y and the set of k-algebra homomorphisms from Γ (Y, OY ) to Γ (X, OX ). Proof. Let (Ui )i∈I be an covering of X by affine open subsets. 1) Given u : X → Y a morphism of prevarieties and i ∈ I. The map Γ (Y, OY ) → Γ (Ui , OX ) , g → g ◦ u|Ui is a homomorphism of k-algebras. Clearly this is A(ui ) where ui : Ui → Y is the morphism induced by u. If v : X → Y is a morphism verifying g ◦ u = g ◦ v for all g ∈ Γ (Y, OY ), then ui = vi for all i ∈ I (11.4.4). Hence u = v since the Ui ’s cover X. 2) Let ϕ ∈ Homalg (Γ (Y, OY ), Γ (X, OX )) and ϕi the composition ϕ
Γ (Y, OY ) −−−−→ Γ (X, OX ) −−−−→ Γ (Ui , OX ) where the second map is the restriction homomorphism. By 11.4.4, we have ϕi = A(ui ), where ui ∈ Mor(Ui , Y ). To obtain ϕ = A(u), it suffices to show that ui |Ui ∩Uj = uj |Ui ∩Uj . Since affine open subsets form a base for the topology on X, it suffices to show that ui |V = uj |V for any affine open subset V ⊂ Ui ∩ Uj . But these two morphisms have the same comorphism
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Γ (Y, OY ) −−−−→ Γ (X, OX ) −−−−→ Γ (V, OX ) where the second map is the restriction homomorphism. So they are identical by 11.4.4. 12.3.4 Corollary. Let (X, OX ) be a prevariety, U an open subset of X and A = Γ (U, OX ). The following conditions are equivalent: (i) U is an affine open subset. (ii) The algebra A is finitely generated and the morphism u : U → Spm(A), associated to the identity map of A (see 12.3.3), is an isomorphism. Proof. Clearly A is reduced. We have (i) ⇒ (ii) by 11.4.5. Conversely, given (ii), Spm(A) is an affine variety. So by 12.3.3, the morphism u is unique such that Γ (u) = idA . Since u is an isomorphism, (i) follows. 12.3.5 Let us give an example of an open subset which is not affine. Let X = k2 = Spm(k[T1 , T2 ]) and U = X \ {(0, 0)}. We claim that U is not affine. Set U1 = D(T1 ) and U2 = D(T2 ), then U = U1 ∪ U2 . Let f ∈ Γ (U, OX ). Then g = f |U1 ∈ Γ (U1 , OX ), so by 12.1.1, it is of the form P (x1 , x2 ) g(x1 , x2 ) = xm 1 where P ∈ k[T1 , T2 ] and m ∈ N. We may assume that T1 does not divide P . If (x1 , x2 ) ∈ U1 , then xm 1 f (x1 , x2 ) − P (x1 , x2 ) = 0. Now h = T1m f − P defines a regular function on U . Thus h−1 (0) = F is a closed subset of U containing U1 . But U1 is dense in X (so in U ), we deduce that h is identically zero on U . In particular, m = 0 for otherwise P (0, x2 ) = 0 for all x2 = 0 which contradicts the assumption that T1 does not divide P . We have therefore proved that f is the restriction to U of a polynomial, which is unique since k is infinite. It follows that Γ (U, OX ) = Γ (X, OX ). Now let ϕ : Γ (X, OX ) → Γ (U, OX ) be the restriction homomorphism which is the identity map from the preceding discussion. Let u : U → X be the morphism such that Γ (u) = ϕ (12.3.3). If x ∈ U and g ∈ Γ (X, OX ), we have g(x) = g(u(x)). Thus x = u(x) (11.3.3) and u is the canonical injection. So u is not bijective and 12.3.4 says that U is not affine.
12.4 Products of prevarieties 12.4.1 Let X, Y be prevarieties over k, U ⊂ X and V ⊂ Y affine open subsets. We have seen in 11.8 that U × V has a canonical structure of affine variety for which the elements of A(U × V ) are functions of the form (x, y) → f1 (x)g1 (y) + · · · + fn (x)gn (y)
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where n ∈ N∗ , f1 , . . . , fn ∈ A(U ) and g1 , . . . , gn ∈ A(V ). For h ∈ A(U × V ), we set DU,V (h) = {z ∈ U × V ; h(z) = 0 } ⊂ X × Y. The DU,V (h)’s, h ∈ A(U × V ), form a base for the Zariski topology in U × V (11.7.2). Denote by C the set of DU,V (h)’s for U an open affine subset of X, V an open affine subset of Y and h ∈ A(U × V ). Let DU,V (h), DU ,V (h ) ∈ C with h(x, y) = α1 (x)β1 (y) + · · · + αm (x)βm (y) h (x , y ) = γ1 (x )δ1 (y ) + · · · + γn (x )δn (y ) for (x, y) ∈ U × V and (x , y ) ∈ U × V where α1 , . . . , αm ∈ A(U ), β1 , . . . , βm ∈ A(V ), γ1 , . . . , γn ∈ A(U ) and δ1 , . . . , δn ∈ A(V ). Let (x0 , y0 ) ∈ DU,V (h) ∩ DU ,V (h ). There exist affine open subsets U ⊂ U ∩ U and V ⊂ V ∩ V such that (x0 , y0 ) ∈ U × V . Denote by ai , ci (resp. bi , di ) the restrictions to U (resp. V ) of αi , γi (resp. βi , δi ). Then ai , ci ∈ A(U ) and bi , di ∈ A(V ). Thus the maps defined, for (x, y) ∈ U × V , by s(x, y) = a1 (x)b1 (y) + · · · + am (x)bm (y) t(x, y) = c1 (x)d1 (y) + · · · + cn (x)dn (y) belong to A(U × V ). On the other hand: (x0 , y0 ) ∈ DU ,V (st) , DU ,V (st) ⊂ DU,V (h) ∩ DU ,V (h ). It follows that C is a base for a topology T on X × Y . Let us endow X × Y with this topology. By construction, if U ⊂ X and V ⊂ Y are affine open subsets, then T induces on U × V the Zariski topology on U × V . Affine open subsets in the product of affine open subsets U × V form a base for the Zariski topology on U × V . Consequently, the set B of affine open subsets of a product U × V of affine open subsets is a base for the topology T on X × Y . Since X (resp. Y ) is Noetherian, it is the union of a finite number of affine open subsets U1 , . . . , Um (resp. V1 , . . . , Vn ). So X × Y is the finite union of the Noetherian subsets Ui × Vj . Hence X × Y is Noetherian (1.3.2). For W ∈ B, set F(W ) = A(W ). Suppose that F verifies the conditions of 9.5.1. Then there exists a sheaf of functions OX×Y on X × Y such that Γ (W, OX×Y ) = F(W ) for all W ∈ B. The ringed space (X ×Y, OX×Y ) is therefore an algebraic prevariety over k that we shall call the product of the prevarieties X and Y . From now on, when we consider X × Y as a prevariety, it will be with respect to this structure. 12.4.2 Proposition. Let X, Y be prevarieties, X1 ⊂ X and Y1 ⊂ Y . If X1 and Y1 are open (resp. closed, locally closed) in X and Y , then X1 × Y1 is open (resp. closed, locally closed) in X × Y . Furthermore, the subvariety X1 × Y1 of X × Y is the product of the sub-prevarieties X1 and Y1 of X and Y.
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Proof. If X1 and Y1 are open, then the result is clear by the definition of the topology on X × Y . Next, a locally closed subset is open in its closure, so we are reduced to the case where X1 and Y1 are closed. We have to show that if U and V are affine open subsets of X and Y , then (U × V ) ∩ (X1 × Y1 ) = (U ∩ X1 ) × (V ∩ Y1 ) is closed in U × V , and as an affine variety, it is the product of U ∩ X1 and of V ∩ Y1 . But this is clear by 11.8.8. 12.4.3 Proposition. Let X, X , Y, Y , Z be prevarieties. (i) The canonical projections pr1 : X × Y → X and pr2 : X × Y → Y are morphisms. (ii) If u : Z → X and v : Z → Y are morphisms, then the map w : Z → X × Y , z → (u(z), v(z)) is also a morphism. (iii) If u : X → X and v : Y → Y are morphisms, then t : X × Y → X × Y , (x , y ) → (u(x ), v(y )) is also a morphism. (iv) If x0 ∈ X (resp. y0 ∈ Y ), then the map y → (x0 , y) (resp. x → (x, y0 )) is an isomorphism from Y (resp. X) to the closed sub-prevariety {x0 } × Y (resp. X × {y0 }) of X × Y . Proof. (i) Any point (x, y) ∈ X × Y has an open neighbourhood of the form U × V where U, V are affine open subsets of X and Y . The restrictions of pr1 and pr2 are the canonical projections of this product. So the result is a consequence of 11.8.3. (ii) Let z ∈ Z. There exist an affine open neighbourhood U (resp. V ) of u(z) (resp. v(z)) in X (resp. Y ), and an affine open neighbourhood W of z in Z such that u(W ) ⊂ U and v(W ) ⊂ V . So (ii) follows from 11.8.3. (iii) The argument is analogue to the one used in 11.8.4. (iv) The map Y → {x0 } × Y , y → (x0 , y), is clearly bijective. Taking affine open neighbourhoods, we are reduced to the case where X = Spm(A) and Y = Spm(B) are affine varieties. Let m ∈ Spm(A) be such that x = V(m). Since A/m is isomorphic to k, the canonical homomorphism B → (A/m) ⊗k B is an isomorphism. Hence we have (iv). 12.4.4 Remark. Let X, Y, Z be prevarieties. By 12.4.3, the bijections X × Y → Y × X , (x, y) → (y, x), (X × Y ) × Z → X × (Y × Z) , ((x, y), z) → (x, (y, z)) are isomorphisms of prevarieties. We deduce in particular the definition of the product X1 × · · · × Xn for a finite number of prevarieties.
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12.4.5 Proposition. Let X and Y be prevarieties. (i) If X and Y are non-empty, then the topology on X (resp. on Y ) is the quotient topology of the one on X × Y with respect to the relation pr1 (z) = pr1 (z ) (resp. pr2 (z) = pr2 (z )). (ii) If X and Y are irreducible, then X × Y is also irreducible. Proof. (i) Let F ⊂ X be such that F × Y is closed in X × Y . If y ∈ Y , we have (F × Y ) ∩ (X × {y}) = F × {y}. So F × {y} is closed in X × {y}. By 12.4.3 (iv), F is closed in X. So part (i) follows. (ii) Let us write X = U1 ∪ · · · ∪ Um , Y = V1 ∪ · · · ∪ Vn where the Ui ’s and the Vj ’s are non-empty affine open subsets. Then X × Y = Ui × Vj . i,j
Since X, Y are irreducible, (Ui × Vj ) ∩ (Uk × Vl ) = ∅ for 1 i, k m and 1 j, l n. By 1.1.6, it suffices to prove that Ui × Vj is irreducible. So we are reduced to the case where X and Y are affine varieties, and the result follows from 11.8.6.
12.5 Algebraic varieties 12.5.1 The diagonal of a set X, denoted by ∆X , is the subset of X × X defined by ∆X = {(x, x) ; x ∈ X}. Proposition. (i) Let X be an affine variety, then ∆X is closed in X × X. (ii) Let X be a prevariety, then ∆X is locally closed in X × X. Proof. (i) Since X is affine, it is closed in an affine space kn . So X × X is closed in kn × kn (11.8.8). Now, ∆X = (X × X) ∩ ∆kn . So it suffices to show that ∆kn is closed in kn × kn . But this is clear since it is the affine algebraic subset defined by the polynomials xi − xn+i = 0 for 1 i n. (ii) Let z ∈ ∆X , there is an affine open subset U ⊂ X such that U × U is an open neighbourhood of z in X × X. Now (ii) follows from (i) since ∆X ∩ (U × U ) = ∆U . 12.5.2 Corollary. Let X, Y be prevarieties (resp. affine varieties) and u : X → Y a morphism. The graph Gu of u is locally closed (resp. closed) in X × Y . Furthermore, the graph morphism of u X → X × Y , x → (x, u(x)) is an isomorphism onto the sub-prevariety Gu of X × Y .
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Proof. Set v : X × Y → Y × Y , (x, y) → (u(x), y). By 12.4.3 (iii), v is a morphism and Gu = v −1 (∆Y ). So the first part follows from 12.5.1 and the second by observing that pr1 |Gu is the inverse of the graph morphism. 12.5.3 Definition. An algebraic prevariety X over k is called an (algebraic) variety over k if ∆X is closed in X × X. 12.5.4 Proposition. (i) Any sub-prevariety of a variety is a variety. (ii) The product of two varieties is a variety. (iii) Let X be a prevariety, Y a variety and u : X → Y a morphism. Then the graph of u is closed in X × Y . (iv) Let X be a variety, n ∈ N∗ and ∆nX = {(x, . . . , x) ∈ X n ; x ∈ X }. Then ∆nX is closed in X n . Proof. Part (i) follows from the fact that if Y is a sub-prevariety of a variety X, then ∆Y = (Y × Y ) ∩ ∆X . Let X, Y be varieties, then we can identify ∆X×Y ⊂ (X × Y ) × (X × Y ) with ∆X × ∆Y ⊂ (X × X) × (Y × Y ) (12.4.4). So part (ii) follows from 12.4.2. Part (iii) is proved as in 12.5.2, and part (iv) follows from (ii) and (iii) since ∆nX is the graph of the morphism X → X n−1 , x → (x, . . . , x). 12.5.5 Remarks. 1) An affine variety is a variety. 2) A sub-prevariety of a variety X shall be called a subvariety of X. 3) A variety is quasi-affine if it is isomorphic to a subvariety of an affine variety. We have seen in 12.3.5 that a quasi-affine variety is not necessary affine. 12.5.6 Theorem. Let X be a prevariety. The following conditions are equivalent: (i) X is a variety. (ii) For all morphisms u, v : Y → X of prevarieties, the set of y ∈ Y verifying u(y) = v(y) is closed in Y . (iii) For all affine open subsets U, V of X, U ∩ V is an open affine subset and the algebra A(U ∩ V ) is generated by the functions x → f (x)g(x), for x ∈ U ∩ V , where f ∈ A(U ) and g ∈ A(V ). Proof. (i) ⇒ (ii) Let Y , u, v as in (ii), set: w : Y → X × X , y → (u(y), v(y)). Then {y ∈ Y ; u(y) = v(y)} = w−1 (∆X ), and (ii) follows since X is a variety and w is a morphism (12.4.3). (ii) ⇒ (i) This is clear by taking Y = X × X, u = pr1 and v = pr2 . (iii) ⇔ (i) The subsets U ×V , where U, V are affine open subsets of X, form an open covering of X ×X. So ∆X is closed if and only if ∆X ∩(U ×V ) is closed
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in U ×V for all affine open subsets U, V of X. Now the map u : U ∩V → U ×V , x → (x, x), induces an isomorphism of the sub-prevariety U ∩ V onto the subprevariety ∆X ∩ (U ∩ V ) of the affine variety U × V (12.5.2). If we have (i), then the open subset U ∩ V is affine since (U × V ) ∩ ∆X is closed in U ∩ V . Conversely, suppose that the open subset U ∩ V is affine. Denote by j the canonical injection from (U × V ) ∩ ∆X into U × V . Then w
j
U ∩ V −−−−→ (U × V ) ∩ ∆X −−−−→ U × V, where w denotes the isomorphism induced by u. By 11.6.8, (U × V ) ∩ ∆X is a closed subset of U × V if and only if the comorphism A(U ×V ) → A(U ∩V ) of u is surjective. Hence we have obtained the equivalence of (i) and (iii). 12.5.7 Remark. Let (Ui )i∈I be a covering of a prevariety X by affine open subsets. If for all i, j ∈ I, the open subsets Ui and Uj verify condition (iii) of 12.5.6, then the preceding proof says that X is a variety. 12.5.8 Corollary. Let X be a variety, Y a prevariety, Z a subset of Y and u, v be morphisms from Y to X such that u(y) = v(y) for all y ∈ Z. Then u(y) = v(y) for all y ∈ Z. Proof. This is clear by 12.5.6 (ii). 12.5.9 Proposition. Let X be a prevariety such that any two points in X are contained in some affine open subset. Then X is a variety. Proof. Let x, y be two distinct points of X and U an affine open subset of X containing x and y. The set (U × U ) ∩ ∆X is closed in U × U and its complement Ω in U ×U is open in U ×U , so in X ×X as well. Since (x, y) ∈ Ω and Ω ∩ ∆X = ∅, ∆X is closed in X × X. 12.5.10 Proposition. Let u : X → Y be a morphism of prevarieties. Suppose that the following conditions are satisfied: (i) Y is a variety. (ii) There exists an open covering (Vi )i∈I of Y such that each of the open subsets Ui = u−1 (Vi ) is a variety. Then X is a variety. Proof. Since (Ui )i∈I is an open covering of X, it suffices to show that the set (Ui × Uj ) ∩ ∆X is closed in Ui × Uj . Let v : X × X → Y × Y , (x, x ) → (u(x), u(x )). Then (Ui × Uj ) ∩ ∆X = [v −1 ((Vi × Vj ) ∩ ∆Y )] ∩ [(Ui × Ui ) ∩ ∆X ]. By condition (i), (Vi × Vj ) ∩ ∆Y is closed in Vi × Vj . So v −1 ((Vi × Vj ) ∩ ∆Y ) is closed in Ui ∩ Uj . Now condition (ii) implies that (Ui × Ui ) ∩ ∆X is closed in Ui × Ui . Hence (Ui × Uj ) ∩ ∆X is closed in Ui × Uj as required.
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12.5.11 Proposition. Let X, Y be irreducible varieties. A morphism u : X → Y which is a local isomorphism is injective. Thus it is an isomorphism from X onto an open subvariety of Y . Proof. Let x, x ∈ X be distinct points such that u(x) = u(x ) = y. There exists, by our hypothesis, an open subset U (resp. U ) of X containing x (resp. x ) such that u|U (resp. u|U ) is an isomorphism from U (resp. U ) onto an open subset V (resp. V ) of Y . Let g, g be the inverse isomorphisms. Then g, g induce morphisms h, h from the open subset V ∩ V of Y to X. Since X is irreducible, U ∩ U is non-empty, and so u(U ∩ U ) is a non-empty open subset of Y contained in V ∩ V . Hence u(U ∩ U ) is dense in V ∩ V since Y is irreducible. We deduce that h and h are identical on V ∩ V which contradicts the fact that h(y) = h (y).
12.6 Gluing 12.6.1 Let X1 , . . . , Xn be prevarieties. Assume that for 1 i, j n, there exist open subsets Xij of Xi , Xji of Xj , and an isomorphism hji : Xij → Xji of prevarieties, such that a) Xii = Xi and hii = idXi for 1 i n. b) hij and hji are mutually inverse isomorphisms for 1 i, j n. c) For 1 i, j, k n, hji |Xij ∩Xik induces an isomorphism of Xij ∩ Xik onto Xji ∩ Xjk , and hki = hkj ◦ hji are identical on Xij ∩ Xik . 12.6.2 With the above assumption, there exist by 1.5.4 a topological space X, and for 1 i n, a homeomorphism ψi from Xi to an open subset ψi (Xi ) of X such that: 1) X = ψ1 (X1 ) ∪ · · · ∪ ψn (Xn ). 2) ψi (Xij ) = ψj (Xji ) = ψi (Xi ) ∩ ψj (Xj ) for 1 i, j n. 3) hji (x) = ψj−1 (ψi (x)) for 1 i, j n and x ∈ Xij . We shall endow X with a structure of prevariety. The sheaf OXi on Xi transports via ψi into a sheaf Fi on ψi (Xi ). Thus if U is an open subset of ψi (Xi ), Fi (U ) is the set of functions z → g(ψi−1 (z)), with g ∈ Γ (ψi−1 (U ), OXi ). We claim that there exists a unique sheaf OX on X such that OX (U ) = Fi (U ) if 1 i n and U an open subset of ψi (Xi ). For 1 i n, let Bi be a base of open subsets of ψi (Xi ). The union B of the Bi ’s is a base of open subsets of X. In view of 9.5.1, it suffices to prove that Fi and Fj are identical when restricted to ψi (Xi ) ∩ ψj (Xj ). Let W be an open subset of ψi (Xi ) ∩ ψj (Xj ). For x ∈ W , we have ψj−1 (x) = hji (ψi−1 (x)). But the elements of Fj (W ) are the functions g ◦ ψj−1 , with g ∈ Γ (ψj−1 (W ), OXj ). So, by the hypothesis, the functions g ◦ hji are in Γ (ψi−1 (W ), OXi ) (since the hji ’s are isomorphisms of prevarieties). So we have proved our claim (in fact, we have glued the Fi ’s in the sense of 9.4.2). Since X is a finite union of Noetherian open subsets, it is a Noetherian topological space. Further, if we take Bi a base of affine open subsets of
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ψi (Xi ), then B is a base of affine open subsets of X. This implies that X is a prevariety. With respect to this structure of prevariety on X, it is clear that each ψi (Xi ) is an open sub-prevariety of X and each ψi is an isomorphism of prevarieties. 12.6.3 With the above notations, we say that the prevariety X is obtained by gluing the prevarieties Xi along the Xi using the isomorphisms hji . In general, we identify via ψi , the prevarieties Xi and ψi (Xi ). 12.6.4 Using the preceding construction, we give below examples of prevarieties which are not varieties. Let X1 = X2 = k and denote by O1 , O2 the point 0 in X1 and X2 respectively. Let U1 = X1 \ {O1 }, U2 = X2 \ {O2 } and glue X1 and X2 along U1 and U2 via the identity map U1 → U2 . The prevariety X thus obtained is the union X1 ∪ {O2 } = X2 ∪ {O1 } and X1 , X2 are affine open subsets of X. Now the open subset X1 ∩ X2 is just U1 (or U2 ), so it is affine. Moreover, X \ (X1 ∩ X2 ) consists of the two distinct points O1 and O2 . The intersection (X1 × X2 ) ∩ ∆X can be identified with ∆k \ {(0, 0)} where we identify X1 × X2 with k2 . It follows that ∆X is not closed in X × X, and so X is not a variety. Note that if x, y ∈ X \ {O1 } (resp. X \ {O2 }), then x, y are contained in the affine open subset X1 (resp. X2 ). Since X is not a variety, it follows from 12.5.9 that no affine open subset of X contains simultaneously O1 and O2 . The preceding discussion says that any affine open subset of X is contained in X1 or X2 . Hence the intersection of two affine open subsets of X is again an affine open subset. Compare this with condition (iii) of 12.5.6. 12.6.5 Let X1 = X2 = k2 , O1 , O2 the point (0, 0) of X1 and X2 respectively, and X the prevariety obtained by gluing X1 and X2 along X1 \ {O1 } and X2 \ {O2 } via the identity map X1 \ {O1 } → X2 \ {O2 }. Again we may identify X1 and X2 as affine open subsets of X, and identify X1 ∩ X2 with k2 \ {(0, 0)}, which is not affine (12.3.5). So by 12.5.6, X is not a variety.
12.7 Rational functions 12.7.1 In this section, X shall be an irreducible algebraic variety. Lemma. Let U, V be non-empty open subsets of X, u : U → Y and v : V → Y morphisms to a variety Y . If there exists an open subset W in U ∩ V such that u|W = v|W , then u|U ∩V = v|U ∩V . Proof. Since X is irreducible, W is dense in X and so in U ∩ V . The result follows therefore from 12.5.8. 12.7.2 Let U be a non-empty open subset of X and u : U → Y a morphism to a variety Y . By 12.7.1, there exists a largest open subset U0 containing U
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such that u extends to a morphism u0 : U0 → Y . Namely, U0 is the union of all the open subsets containing U where u extends to a morphism to Y . When U0 = U , we say that u is a rational map from X to Y and that U is the domain of definition of the rational map u. So a morphism of X to Y is a rational map whose domain of definition is X, and by 12.7.1, any morphism from a non-empty open subset of X to Y extends uniquely to a rational map from X to Y . A rational map from X to k is called a rational function on X. Such a function is therefore the maximal extension of a regular function on a nonempty open subset of X. There is another way to define rational maps: Let E be the set of pairs (U, u), where U is a non-empty open subset of X and u a morphism from U to Y . Define the following relation on E (U, u) ∼ (V, v) ⇔ u|U ∩V = v|U ∩V . This is an equivalence relation, and a rational map is simply an equivalence class. There is a third way to define rational maps. Order E by setting (U, u) (V, v) ⇔ V ⊂ U and u|V = v. Then we have an inductive system, and the set of rational maps is the set of inductive limits of this system. 12.7.3 Example. Let f, g ∈ Γ (X, OX ) with g = 0. The set U of x ∈ X such that g(x) = 0 is a non-empty open subset. The map x → f (x)/g(x) is regular on U since it is regular on all affine open subsets contained in U (11.7.5 and 12.1.1). It follows that the preceding map extends uniquely to a rational function on X. In general, we do not obtain all the rational functions in this way. 12.7.4 From now on, we shall denote by R(X) the set of rational functions on X. We claim that R(X) has a canonical structure of commutative field. Firstly, the interpretation of rational functions in terms of inductive limit and the results of 8.3 show that R(X) has a canonical structure of commutative ring. Let f ∈ R(X) \ {0}. The set U of points in X where f is non-zero is a non-empty open subset, and the function x → 1/f (x) is regular on U (12.7.3). Denote by 1/f the unique rational function such that its restriction to U is x → 1/f (x). Then it is obvious that 1/f is the inverse of f in the ring R(X). Thus we have proved our claim. 12.7.5 Remark. Let f ∈ R(X) and x ∈ X. If f is not defined at x, it may happen that g = 1/f is defined at x. Then g(x) = 0 for otherwise f = 1/g would be defined in a neighbourhood of x. We shall say that x is a pole of f . For example, the rational function x2 /x1 on k2 is not defined at the points (0, x2 ). We see that such a point (0, x2 ) is a pole if and only if x2 = 0.
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12.7.6 Theorem. Let X be an irreducible variety. For all affine open subset U of X, the ring Γ (U, OX ) is an integral domain and the canonical map Γ (U, OX ) → R(X), sending a regular function on U to the unique rational function extending it, is injective and extends to an isomorphism from the quotient field of Γ (U, OX ) onto R(X). Proof. Since X is irreducible, so is U . Hence Γ (U, OX ) is an integral domain (11.2.3 and 11.3.2). The injectivity of the map Γ (U, OX ) → R(X) is a consequence of 12.7.1. To finish our proof, we only have to show that the extension of this map to the quotient field of Γ (U, OX ) is surjective. Let h ∈ R(X). There exists an affine open subset D(g) ⊂ U , where g ∈ Γ (U, OX ), such that h|D(g) ∈ Γ (D(g), OX ). So h|D(g) is of the form x → f (x)/g(x)m , with f ∈ Γ (U, OX ). If f , g ∈ R(X) are the images of f and g, then h = f / g m . Thus we have obtained our result. 12.7.7 In view of 12.7.6, we can identify R(X) with the quotient field of Γ (U, OX ), for an affine open subset U of X. In particular, if X = Spm(A) is an irreducible affine variety, then R(X) can be identified with the quotient field of A. 12.7.8 Remark. It follows from 12.7.6 that when X is an irreducible variety, R(X) is a finitely generated extension of k. Conversely, let K be a finitely generated extension of k and y1 , . . . , yn a system of generators of K, then K is the quotient field of the finitely generated k-algebra A = k[y1 , . . . , yn ] which is also an integral domain. It follows that there exists an irreducible affine variety X ⊂ kn such that A(X) is isomorphic to A. Thus K is isomorphic to R(X). An irreducible variety Y such that R(Y ) is isomorphic to K is called a model of K. The preceding paragraph says that there is an affine model of K. 12.7.9 Proposition. Let X = Spm(A) be an irreducible variety such that A is a factorial ring, and h ∈ R(X). There exist f, g ∈ A verifying the following conditions: (i) The domain of definition of h is D(g). (ii) For all x ∈ D(g), we have h(x) = f (x)/g(x). Proof. By 12.7.6, there exist f, g ∈ A, g = 0, relatively prime, such that h = f /g (and f, g are unique up to a multiple of an invertible element). Denote by U the domain of definition of h, we have D(g) ⊂ U . If h is defined on a point x ∈ D(g), then h = f1 /g1 where f1 , g1 ∈ A with g1 (x) = 0. So there exists u ∈ A such that f1 = uf and g1 = ug. This is absurd since g(x) = 0. 12.7.10 Remark. We can not drop the hypothesis that A is factorial. In general, the domain of definition of h can contain strictly D(g) for any g ∈ A such that h = f /g with f ∈ A.
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12.7.11 Let X, Y be irreducible varieties, u : X → Y a dominant morphism and V ⊂ Y an affine open subset. There exists an affine open subset U of X such that U ⊂ u−1 (V ). Since u is dominant, u(U ) is dense in V . By 6.5.9, the comorphism ϕ : Γ (V, OY ) → Γ (U, OX ) of the restriction U → V of u is injective. Hence by 12.7.6, ϕ can be extended to an injective homomorphism of the fields θ : R(Y ) → R(X). Let U ⊂ X and V ⊂ Y be affine open subsets such that U ⊂ u−1 (V ). The comorphism ϕ : Γ (V , OY ) → Γ (U , OX ) extends to the same homomorphism θ. In fact, replacing V (resp. U ) by an affine open subset in V ∩ V (resp. U ∩ U ) if necessary, we may assume that V ⊂ V and U ⊂ U . It is then clear that ϕ extends ϕ. When there is no confusion, we shall say that the field homomorphism θ : R(Y ) → R(X) is the comorphism associated to u. 12.7.12 Definition. A dominant morphism u : X → Y of irreducible varieties is called birational if the comorphism θ : R(Y ) → R(X) associated to u is an isomorphism. 12.7.13 Remarks. 1) By 11.5.6, the morphism u : k2 → k2 , (x1 , x2 ) → (x1 x2 , x2 ) is dominant. It is birational since k(T1 , T2 ) = k(T1 T2 , T2 ). However, it is neither surjective (11.5.6) nor injective (since u(x1 , 0) = (0, 0) for all x1 ∈ k). 2) Recall from 11.5.7 the morphism u : k → k2 , t → (t2 , t3 ), whose image C is the algebraic set defined by x31 − x22 = 0. Recall that the morphism v : k → C is bijective, bicontinuous but is not an isomorphism. However, it is a birational morphism since k(T ) = k(T 2 , T 3 ). 3) So a birational morphism can be neither injective nor surjective. If it is bijective and bicontinuous, it does not need to be an isomorphism. Note that a bijective and bicontinuous morphism is not necessarily birational.
12.8 Local rings of a variety 12.8.1 Let (X, OX ) be a variety and x ∈ X. We denote by OX,x , or Ox , the fibre at x of the sheaf OX (9.1.4). If U is an open subset of X containing x and f ∈ Γ (U, OX ), we denote by fx the image of f in OX,x . We defined in 9.1.4 the value of a germ at x. By 8.3, OX,x has a natural structure of commutative ring. 12.8.2 Remarks. 1) If U is an open subset of X containing x, then OX,x = OU,x (8.2.7).
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2) Suppose that X is irreducible, then any regular function on an open neighbourhood of x extends uniquely to a rational function on X defined at x. Moreover, two such functions having the same germ at x extend to the same rational function. Thus OX,x can be identified as a subring of R(X) of rational functions defined at x. In particular, since X is irreducible, the ring OX,x is an integral domain, and if U is an open subset of X, then Γ (U, OX ) = OX,x . x∈U
12.8.3 Proposition. Let (X, OX ) be a variety and x ∈ X. (i) The ring OX,x is local and its maximal ideal mX,x consists of germs of regular functions which are zero at x. The fields OX,x /mX,x and k are isomorphic. (ii) Let U be an affine open subset of X containing x, A = Γ (U, OX ) and n the maximal ideal of A corresponding to x. The rings An and OX,x are isomorphic. In particular, there is a bijection between the set of prime ideal of OX,x and the set of closed irreducible subsets of U containing x. (iii) The ring OX,x is Noetherian. Proof. (i) Let ρ : OX,x → k be the map sending a germ to its value at x. This is a surjective ring homomorphism. Thus mX,x = ker ρ is a maximal ideal and the fields OX,x /mX,x and k are isomorphic. If s ∈ OX,x \ mX,x and (U, f ) a representative of s (9.1.4) where U is an affine open subset of X containing x, then x ∈ D(f ) = {y ∈ U ; f (y) = 0}. Since f is invertible in Γ (D(f ), OU ) = Γ (D(f ), OX ), s is invertible in OX,x . So OX,x is local. (ii) Let ϕ : A → OX,x be the homomorphism sending f ∈ A to its germ at x. If f (x) = 0, that is f ∈ n, then ϕ(f ) is invertible by (i). By 2.3.4, ϕ extends uniquely to a homomorphism ψ : An → OX,x . Let s ∈ OX,x . There exist a principal open subset D(f ) of U containing x and a function g ∈ Γ (D(f ), OX ) such that (D(f ), g) is a representative of s. Since x ∈ D(f ), we have f ∈ A \ n. Moreover, if y ∈ D(f ), then g(y) = h(y)/f (y)n , for some n ∈ N, h ∈ A. Thus ψ(h/f n ) = s, and ψ is surjective. Next, if g ∈ ker ψ, there exist h ∈ A, f ∈ A \ n such that g = h/f . Then ϕ(h) = 0. Thus there exists t ∈ A \ n verifying h|D(t) = 0, and hence ht = 0. So g = 0. We have therefore proved that ψ is an isomorphism, and the second part follows from 2.3.8 and 11.2.3. (iii) This is clear by (ii), 1.3.2 and 2.7.6. 12.8.4 Remark. In general, OX,x is not a finitely generated k-algebra. Take X = k and x = 0, then OX,x is isomorphic to the subring of k(T ) consisting of rational functions of the form P/Q with Q(x) = 0. Since a nonzero polynomial in one variable has a finite number of zeros, it is clear that OX,x is not finitely generated as a k-algebra.
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12.8.5 Proposition. Let (X, OX ) be a variety and x ∈ X Then the following conditions are equivalent: (i) x belongs to a unique irreducible component of X. (ii) The ring OX,x is an integral domain. Proof. Let Y1 , . . . , Yn be the irreducible components of X. (i) ⇒ (ii) Let U = Y2 ∪ · · · ∪ Yn and V = X \ U . By 1.1.13, V ⊂ Y1 is open in X, so it is irreducible. So if x ∈ V , then OX,x = OV,x , and OV,x is an integral domain (12.8.2). (ii) ⇒ (i) Replacing X by an affine open subset containing x, we may assume that X = Spm(A) is affine. Then Yi = V(pi ) where p1 , . . . , pn are the minimal prime ideals of A. Suppose that x ∈ Y1 ∩ Y2 . Again, there is a non-empty affine open subset U1 ⊂ Y1 (resp. U2 ⊂ Y2 ) such that U1 ∩ Yi = ∅ if i 2 (resp. U2 ∩ Yi = ∅ if i = 2). So we can find f1 , f2 ∈ A \ {0} such that D(f1 ) ⊂ U1 and D(f2 ) ⊂ U2 . Thus f1 f2 = 0. Now Y1 and Y2 are irreducible, so U1 and U2 are dense in Y1 and Y2 , which implies that the germs of f1 , f2 at x are non-zero. Hence OX,x is not an integral domain. 12.8.6 Let u : X → Y be a morphism of varieties and x ∈ X. If V is an open neighbourhood of u(x) and g ∈ Γ (V, OY ), then g ◦ u is a regular function in the open subset u−1 (V ). Moreover, if g1 , g2 are equal in a neighbourhood of u(x), then g1 ◦ u and g2 ◦ u are equal in a neighbourhood of x. We obtain therefore a map ux : OY,u(x) → OX,x which is clearly a k-algebra homomorphism. We can also verify easily that ux is a local homomorphism. Let U be an affine open subset of X containing x. Any element of OX,x is of the form fx /hx (notations of 12.8.1), where f, h ∈ A(U ) and h(x) = 0. It follows that ux is completely determined by its values on the germs of functions which belong to A(U ). 12.8.7 Lemma. Let X, Y be varieties, x ∈ X, y ∈ Y and U = Spm(A) (resp. V = Spm(B)) an affine open subset of X (resp. Y ) containing x (resp. y). Suppose further that there is a local homomorphism ϕ : OY,y → OX,x . (i) There exist an affine open neighbourhood U0 ⊂ U of x, an affine open neighbourhood V0 ⊂ V of y and a morphism u0 : U0 → V0 such that u0 (x) = y and ϕ is the local homomorphism (u0 )x . (ii) Let U1 ⊂ U (resp. V1 ⊂ V ) be an affine open neighbourhood of x (resp. y), u1 : U1 → V1 a morphism such that u1 (x) = y and ϕ is the local homomorphism (u1 )x . Then u0 and u1 coincide in a neighbourhood of x. Proof. (i) For b ∈ B, denote by iy (b) the canonical image of b in OY,y . Let {b1 , . . . , bm } be a system of generators of B and α : k[T1 , . . . , Tm ] → B the homomorphism defined by α(Ti ) = bi for 1 i m. Let β = ϕ ◦ iy ◦ α,
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165
a = ker β, {Q1 , . . . , Qr } a system of generators of a, T0 a new indeterminate, and P1 , . . . , Pr homogeneous polynomials of degree h, such that for 1 j r, Qj (T1 /T0 , . . . , Tm /T0 ) = Pj (T0 , T1 , . . . , Tm )/T0h . Each ϕ(iy (bj )) is the germ of a regular function at x which can be written as aj /s where aj , s ∈ A and s(x) = 0. Since the Qj ’s are in a, the regular functions Qj (a1 /s, . . . , am /s) = Pj (s, a1 , . . . , am )/sh are identically zero in a neighbourhood of x. We deduce therefore that there exists t ∈ A verifying t(x) = 0 and Pj (ts, ta1 , . . . , tam ) = 0 in A for 1 j r. Let ρ : k[T1 , . . . , Tm ] → Ats be the homomorphism such that ρ(Ti ) = (tai )/(ts), 1 i m. We have ρ(a) = {0}, so ρ(ker α) = {0}. Thus there exists a homomorphism γ0 : B → Ats such that ρ = γ0 ◦ α; so γ0 (bi ) = (tai )/(ts) for 1 i m. Finally, let u0 = Spm(γ0 ) : U0 = D(st) → V be the associated morphism. If ix : Ats = Γ (D(ts), OX ) → OX,x is the canonical homomorphism, then ix ◦ γ0 = ϕ ◦ iy . It follows that, using the notations of 12.8.3, −1 γ0−1 (I({x})) = γ0−1 (i−1 x (mX,x )) = iy (mY,y ) = I({y}).
Hence ϕ is the local homomorphism (u0 )x . (ii) Replace U0 , U1 (resp. V0 , V1 ) by an affine open subset contained in U0 ∩U1 (resp. V0 ∩V1 ), we may assume that U = U0 = U1 (resp. V = V0 = V1 ). Let γ0 = A(u0 ) and γ1 = A(u1 ). For 1 i m, γ0 (bi ) and γ1 (bi ) are regular functions in a neighbourhood of x having the same germ at x. So there exists s ∈ A verifying s(x) = 0 such that the restriction of γ0 (bi ) and γ1 (bi ) to D(s) are equal. We deduce therefore that u0 |D(s) = u1 |D(s) . 12.8.8 Proposition. Let X, Y be varieties, x ∈ X and y ∈ Y . Suppose further that there is a k-algebra isomorphism ϕ : OY,y → OX,x . (i) There exist an open neighbourhood U0 ⊂ X of x, an open neighbourhood V0 ⊂ Y of y, and an isomorphism u : U0 → V0 such that u(x) = y. (ii) Let U1 ⊂ X (resp. V1 ⊂ Y ) be an open neighbourhood of x (resp. y) and u1 : U1 → V1 an isomorphism. Suppose that the isomorphisms ux and (u1 )x coincide. Then there exists an open neighbourhood U ⊂ U0 ∩ U1 of x such that u|U = u1 |U . Proof. By 12.8.7, it suffices to prove (i). Let ψ be the inverse isomorphism of ϕ. Again by 12.8.7, we can find affine open neighbourhoods U0 of x, V0 of y and a morphism v from V0 to an open subset of X such that ϕ and ψ are the local homomorphisms ux and vy induced by u and v. Replacing U0 by an affine open subset, we may assume that u(U0 ) ⊂ V0 . Then v ◦ u is a morphism from U0 to an open neighbourhood of x such that (v ◦ u)x is the identity map of OX,x . By 12.8.7, (v ◦ u)(z) = z for all z in an open neighbourhood of x. The same arguments show that there exist affine open neighbourhoods U1 of x, V1 of y such that u|U1 is a morphism from U1 to V1 , v|V1 is a morphism from V1 to U1 and (u ◦ v)(z) = z for all z ∈ V1 . Thus our result is proved.
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Remark. We see from 12.8.8 that the local ring OX,x determines a neighbourhood of x in X up to an isomorphism.
References • [5], [19], [22], [26], [37], [40], [78].
13 Projective varieties
We attach to a projective space a natural structure of an irreducible algebraic variety. More generally, we define the notion of projective varieties, and discuss properties of these varieties. In particular, we introduce complete varieties. Certain properties of complete varieties are analogue to those of compact topological spaces.
13.1 Projective spaces 13.1.1 Let n be a non-zero positive integer. We define an equivalence relation R on kn+1 \ {0} by xRy if and only if y = λx for some λ ∈ k \ {0}. The quotient space (kn+1 \ {0})/R is called the projective space of dimension n, and we shall denote it by Pn (k) or Pn . By definition, Pn is the set of 1-dimensional vector subspaces of kn+1 . Let z ∈ Pn , its homogeneous coordinates are the coordinates x0 , x1 , . . . , xn of a point x in kn+1 whose class is z. Homogeneous coordinates are unique up to multiplication by a non-zero scalar. We shall denote by π : kn+1 \ {0} → Pn the canonical surjection. 13.1.2 We shall call a subset C in kn+1 \ {0} a cone if λx ∈ C for all x ∈ C and for all λ ∈ k \ {0}, or equivalently, C = π −1 (π(C)). The union of {0} and a cone C in kn+1 \ {0} shall be called the affine cone over π(C) in kn+1 . Lemma. Let C be a non-empty closed cone in kn+1 \ {0}. Then C ∪ {0} is the closure C of C in kn+1 . Proof. Let D be a 1-dimensional subspace of kn+1 . Then 0 ∈ D \ {0}. Since C is non-empty, C ∪ {0} ⊂ C. Now C is closed, so C = C ∩ (kn+1 \ {0}). Hence C = C ∪ {0}.
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13.1.3 By 13.1.2, a closed cone in kn+1 \{0} is the intersection of kn+1 \{0} and a closed affine cone in kn+1 . 13.1.4 Let C be a closed affine cone in kn+1 and a = I(C) the associated radical ideal in k[T0 , . . . , Tn ]. Let P ∈ a. Since C is a cone, the condition P (x0 , . . . , xn ) = 0 implies that P (λx0 , . . . , λxn ) = 0 for all λ ∈ k. It follows that all the homogeneous components of P are in a, and so a is a graded ideal. Conversely, given a graded ideal a = k[T0 , . . . , Tn ], the set V(a) is an affine cone in kn+1 . The only maximal graded ideal in k[T0 , . . . , Tn ] is the ideal m0 generated by T0 , . . . , Tn . It follows that for any graded ideal a = k[T0 , . . . , Tn ], √ V(a) = {0} ⇔ a = m0 . By 7.2.7, we see that the map a → V(a) induces a bijection between the set of graded radical ideals of k[T0 , . . . , Tn ] which do not contain m0 and the set of closed affine cones of kn+1 which are not reduced to {0}.
13.2 Projective spaces and varieties 13.2.1 For 0 i n, let us denote by Hi0 (resp. Hi1 ) the hyperplane (resp. affine hyperplane) of kn+1 defined by xi = 0 (resp. xi = 1). Let Ui be the set of 1-dimensional subspaces of kn+1 which are not contained in Hi0 . Thus Ui is the set of points z in Pn whose homogeneous coordinates (x0 , . . . , xn ) verify xi = 0. Now any 1-dimensional subspace in Ui intersects Hi1 at a unique point and this induces a bijection ϕi between Ui and Hi1 . Namely, ϕi sends the point z of Ui ⊂ Pn with homogeneous coordinates (x0 , . . . , xn ) to the point x0 xi−1 xi+1 xn ,..., , 1, ,..., . xi xi xi xi It follows clearly that: Pn = U0 ∪ U1 ∪ · · · ∪ Un . 13.2.2 For i, j ∈ {0, 1, . . . , n}, i = j, Ui ∩ Uj is the set of points whose homogeneous coordinates (x0 , . . . , xn ) verify xi = 0 and xj = 0. Such a point corresponds to the points: x0 x0 xi−1 xi+1 xn xj−1 xj+1 xn ,..., , 1, ,..., ,..., , 1, ,..., and xi xi xi xi xj xj xj xj in the hyperplanes Hi1 and Hj1 respectively. 1 1 1 = Hi1 \ Hj0 and Hji = Hj1 \ Hi0 . We define the bijection pji : Hij → Set Hij 1 Hji by: x0 xi−1 1 xi+1 xn (1) (x0 , . . . , xi−1 , 1, xi+1 , . . . , xn ) → ,..., , , ,..., . xj xj xj xj xj
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1 Let z ∈ Ui ∩ Uj ∩ Ul , then ϕi (z) is contained in Hij ∩ Hil1 . We verify readily 1 ∩ Hil1 : that ϕj (z) = pji (ϕi (z)) and ϕl (z) = pli (ϕi (z)). Hence, for any x ∈ Hij
(2)
pli (x) = plj (pji (x)).
Let Hii1 = Hi1 and pii = idHi1 . Then we have shown that Pn can be considered as the set obtained by gluing the sets Hi1 , for 0 i n, along the 1 ’s via the bijections pji (1.5.1). Hij 13.2.3 Using the preceding consideration, we can define a structure of algebraic prevariety on Pn . 1 is an open subvariety of Hi1 The set Hi1 is an affine variety, and Hij 1 1 n+1 (Hij = D(prj ) ∩ Hi where prj : k → k denotes the canonical projection (x0 , . . . , xn ) → xj ). Since regular functions on Hi1 are restrictions of polynomial functions on n+1 , it follows from 11.7.5, 12.1.1 that for i = j, the algebra of regular k 1 is generated by the functions: functions on Hij (x0 , . . . , xn ) →
xl , xj
for 0 l n. We conclude that pji is an isomorphism between the affine 1 1 and Hji . varieties Hij Now equation (2) above is condition c) of 12.6.1. Hence we can endow Pn with a structure of algebraic prevariety over k. From now on, when we consider the projective space as a prevariety, we shall mean the above structure. The main point of the preceding discussion is that the maps ϕi of 13.2.1 allow us to identify Ui with Hi1 , and that we use these maps to define a structure of prevariety on Pn where the Ui ’s are affine open subsets. 13.2.4 For 0 i n, the map ψi : Hi1 → kn (x0 , . . . , xi−1 , 1, xi+1 , . . . , xn ) → (x0 , . . . , xi−1 , xi+1 , . . . , xn ) is an isomorphism of varieties. Let z ∈ Ui , with homogeneous coordinates (x0 , . . . , xn ). The map x0 xi−1 xi+1 xn ψi ◦ ϕi : Ui → kn , z → ,..., , ,..., xi xi xi xi is an isomorphism of Ui onto kn . Thus regular functions on Ui are of the form: x0 xi−1 xi+1 xn z → P ,..., , ,..., xi xi xi xi where P ∈ k[T0 , . . . , Ti−1 , Ti+1 , . . . , Tn ]. Or equivalently, they are of the form: z →
R(x0 , . . . , xn ) xm i
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where m ∈ N and R ∈ k[T0 , . . . , Tn ] is homogeneous of degree m. xl The algebra A(Ui ) is therefore generated by the , where 0 l n. xi The intersection Ui ∩ Uj , i = j, corresponds via the isomorphism ψi ◦ ϕi to the affine open subset D(Tj ) of kn . From this, we deduce that A(Ui ∩ Uj ) is generated by the functions xi 1 xl for 0 l n and = xi xj (xj /xi ) on Ui ∩ Uj . 13.2.5 For 0 i n, denote by θi : D(Ti ) → Hi1 × (k \ {0}), the map x0 xi−1 xi+1 xn (x0 , . . . , xn ) → ,..., , 1, ,..., , xi . xi xi xi xi By 12.1.1 and 12.4.3 (ii), θi is a bijective morphism. Its inverse ((y0 , . . . , yi−1 , 1, yi+1 , . . . , yn ), yi ) → (y0 yi , . . . , yi−1 yi , yi , yi+1 yi , . . . , yn yi ) is clearly a morphism by the definition of regular functions on D(Ti ) and Hi1 × (k \ {0}). Hence θi is an isomorphism. 13.2.6 Proposition. The projective space Pn is an irreducible variety. Proof. That Pn is a variety results from 13.2.4 and 12.5.7. Now Ui is irre ducible and Ui ∩ Uj = ∅ for all i, j. Hence Pn is irreducible (1.1.6). 13.2.7 Proposition. The map π is a morphism of varieties. Proof. Since kn+1 \ {0} = D(T0 ) ∪ D(T1 ) ∪ · · · ∪ D(Tn ), it suffices to prove that the restriction of π to D(Ti ), 0 i n, is a morphism (9.5.5). Now the image of D(Ti ) via π is the affine open subset Ui . Hence by 9.5.4 and 12.2.8, it suffices to show that the map ρi : D(Ti ) → Ui induced by the restriction of π, is a morphism. But we have seen in 13.2.4 that regular functions on Ui are of the form: R(x0 , . . . , xn ) z → xm i where z ∈ Ui with homogeneous coordinates (x0 , . . . , xn ), R ∈ k[T0 , . . . , Tn ] homogeneous of degree m and m ∈ N. So it is clear that if f ∈ A(Ui ), then f ◦ ρi ∈ A(D(Ti )). The result now follows.
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13.3 Cones and projective varieties 13.3.1 Definition. (i) Any algebraic variety isomorphic to a closed subvariety of the projective space Pn is called a projective variety. (ii) An algebraic variety isomorphic to a (not necessarily closed) subvariety of Pn is called a quasi-projective variety. Remark. An affine or quasi-affine variety is a quasi-projective variety. 13.3.2 Proposition. Two points in a projective variety X are contained in an affine open subset. Proof. A closed subvariety of an affine variety is affine, so we may assume that X = Pn , n 1. Let x, y ∈ Pn . There exists a hyperplane in kn+1 which does not contain the subspaces {0}∪π −1 (x) and {0}∪π −1 (y). By applying a linear transformation, we may further assume that this hyperplane is Hi0 . Hence x, y belong to Ui . 13.3.3 Proposition. The map C → π(C) is a bijection between the set of closed cones in kn+1 \ {0} and the set of closed subsets of Pn . Furthermore, for 0 i n, C ∩π −1 (Ui ) is isomorphic to the product (π(C)∩Ui )×(k\{0}). Proof. If F is a closed subset of Pn , then 13.2.7 says that π −1 (F ) is a closed cone in kn+1 \ {0}. Conversely, given a closed cone C in kn+1 \ {0}, we have by identifying Ui and Hi1 (13.2.1): θi (C ∪ D(Ti )) = (π(C) ∩ Ui ) × (k \ {0}) where θi , Ti are as in 13.2.5. Since C ∩ D(Ti ) is closed in D(Ti ) = π −1 (Ui ) and θi is an isomorphism of D(Ti ) onto Ui × (k \ {0}) (13.2.5), π(C) ∩ Ui is closed in Ui by 12.4.5 (i). Hence π(C) is closed in Pn . The last statement follows from 9.5.4 and 12.2.8. 13.3.4 Corollary. (i) The topology on Pn is the quotient topology of the topology on kn+1 \ {0} with respect to the equivalence relation π(x) = π(x ). (ii) The map π is open. Proof. (i) By 13.3.3, closed subsets of Pn are images of closed subsets saturated by the equivalence relation π(x) = π(x ). Hence we have our result. (ii) Let U be an open subset of kn+1 \ {0} and V = λU = π −1 (π(U )). λ∈k\{0}
Since x → λx is an automorphism of the variety kn+1 \ {0}, it follows that V is open in kn+1 \ {0}. So (i) implies that π is open.
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13.3.5 It follows from 13.1.4 and 13.3.3 that any closed subvariety of Pn is defined by a system of equations: (3)
Pi (x0 , . . . , xn ) = 0
where (x0 , . . . , xn ) are homogeneous coordinates of its points and the Pi ’s are homogeneous polynomials in k[T0 , . . . , Tn ]. A closed subvariety defined by a single non-constant homogeneous polynomial (resp. of degree 1) shall be called a hypersurface (resp. hyperplane) of Pn . Thus a hyperplane of Pn is the image π(H \ {0}) where H is a hyperplane in kn+1 . For example, the complement Hi of Ui is the hyperplane π(Hi0 \ {0}). 13.3.6 Let C be a closed cone in kn+1 \ {0} and X = π(C). By 13.1.2 and 13.1.4, C = V(a) \ {0} for some graded radical ideal a of k[T0 , . . . , Tn ]. We shall say that X is defined by the graded ideal a. The intersection of X with Ui , identified as kn as in 13.2.4, is a subvariety V(ai ) of kn where ai is the ideal of k[T0 , . . . , Ti−1 , Ti+1 , . . . , Tn ] generated by the polynomials P (T0 , . . . , Ti−1 , 1, Ti+1 , . . . , Tn ), P ∈ a. Let Q ∈ k[T0 , . . . , Ti−1 , Ti+1 , . . . , Tn ] be a polynomial of degree r. Then T0 Ti−1 Ti+1 Tn r ,..., , ,..., P (T0 , . . . , Tn ) = Ti Q Ti Ti Ti Ti is a homogeneous polynomial in k[T0 , . . . , Tn ]. So we have homogenized Q. Now let Z = V(b) be a closed subvariety of Ui , where b is an ideal of k[T0 , . . . , Ti−1 , Ti+1 , . . . , Tn ]. The closure Z of Z in Pn is equal to π(C) for some closed cone C = V(a)\{0} where a is a graded ideal of k[T0 , . . . , Tn ]. It is easy to see that we can take a to be the graded ideal obtained by homogenizing the elements of b. We can also verify easily that Z = Z ∩ Ui . Conversely, let X be a closed irreducible subvariety of Pn such that X is not contained in the complement Hi of Ui . Then X ∩ Ui is a non-empty open subset of X. So X = X ∩ Ui . 13.3.7 Let A = k[T0 , . . . , Tn ] and Q ∈ A a homogeneous polynomial of degree d > 0. Denote by D+ (Q) the set of points of Pn whose homogeneous coordinates (x0 , . . . , xn ) verify Q(x0 , . . . , xn ) = 0 (this is well-defined since Q is homogeneous). Thus D+ (Q), being the complement of a hypersurface, is open in Pn . Lemma. Any non-empty open subset U of Pn is of the form U = D+ (Q1 ) ∪ · · · ∪ D+ (Qr ) where r ∈ N∗ and Q1 , . . . , Qr ∈ A are homogeneous and non-constant. Proof. The closed subset V = Pn \ U is defined by a graded ideal a of A (13.3.6). By 7.2.2, there exist homogeneous elements Q1 , . . . , Qr ∈ A such that a = AQ1 + · · · + AQr . Clearly, we have U = D+ (Q1 ) ∪ · · · ∪ D+ (Qr ).
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13.3.8 Proposition. Let u be a map from Pn to a prevariety X. The following conditions are equivalent: (i) u is a morphism. (ii) u ◦ π is a morphism from kn+1 \ {0} to X. Proof. (i) ⇒ (ii) This is clear by 13.2.7. (ii) ⇒ (i) Suppose that u ◦ π is a morphism. To show that u is a morphism, it suffices to prove that u|Ui is a morphism for 0 i n. For Q ∈ k[T0 , . . . , Tn ], we set: D (Q) = {x ∈ kn+1 \ {0} ; Q(x) = 0}. In view of 11.7.2, the D (Q)’s form a base for the topology on kn+1 \ {0}. If x ∈ π −1 (Ui ), then our hypothesis says that there exist an affine open subset V of X containing u(π(x)) and Q ∈ k[T0 , . . . , Tn ] verifying: 1) x ∈ D (Q) and u(π(D (Q))) ⊂ V . 2) For all regular function f on V , the function y → f (u(π(y))) is regular on D (Ti Q) = D (Q) ∩ π −1 (Ui ). Let f ∈ Γ (V, OX ) and g the function y → f (u(π(y))) defined on D (Ti Q). Then there exist r ∈ N and a polynomial R ∈ k[T0 , . . . , Tn ] such that for y = (y0 , . . . , yn ) ∈ D (Ti Q), (4)
g(y) =
R(y0 , . . . , yn ) . yir Qr (y0 , . . . , yn )
Let λ ∈ k \ {0}. There exists Sλ ∈ k[T0 , . . . , Tn ] such that y yn 0 ,..., Sλ (y0 , . . . , yn ) = Q λ λ for all (y0 , . . . , yn ) ∈ kn+1 \ {0}. Set Wλ = λD (Ti Q). A point y = (y0 , . . . , yn ) belongs to Wλ if and only if yi = 0 and Sλ (y) = 0. Hence, Wλ = D(Ti Sλ ). Let y hλ : Wλ → k , y → g . λ There exists Rλ ∈ k[T0 , . . . , Tn ] such that if y ∈ Wλ , then: hλ (y) =
Rλ (y0 , . . . , yn ) . yir Sλr (y0 , . . . , yn )
We deduce that hλ is a regular function on Wλ . If x ∈ Wλ ∩ Wµ , then x = λy = µz for some y, z ∈ D (Ti Q). Thus π(z) = π(y), and from the definition of g, we obtain that hλ (x) = hµ (x). It follows that g can be extended to a regular function on the open subset Wλ = π −1 (π(D (Q)) ∩ Ui ), U= λ∈k\{0}
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such that (λy) = (y) for all y ∈ U and λ ∈ k \ {0}. By 13.3.4 (ii), π(D (Q)) is open in Pn , so π(D (Q)) ∩ Ui is also open in Pn . By 13.3.7, there exist homogeneous polynomials Q1 , . . . , Qr ∈ k[T0 , . . . , Tn ] such that π(D (Q)) ∩ Ui = D+ (Q1 ) ∪ · · · ∪ D+ (Qr ). Replacing Q by one of the Qi ’s, we can reduced to the case where Q is homogeneous. Then the function g on D (Ti Q) is invariant by the automorphisms y → λy. This implies that the polynomial R in (4) is homogeneous. From the definition of regular functions on Ui (13.2.4), we conclude that the restriction of u to π(D (Q)) is a morphism. Hence we have our result. 13.3.9 Corollary. A subvariety X of Pn is irreducible if and only if π −1 (X) is an irreducible subvariety of kn+1 \ {0}. Proof. Since π(π −1 (X)) = X, X is irreducible if π −1 (X) is irreducible. Conversely, suppose that X is irreducible. If x ∈ Pn , then π −1 (x), being isomorphic to k \ {0}, is irreducible. It follows from 1.1.7, 13.2.7 and 13.3.4 (ii) that π −1 (X) is irreducible. 13.3.10 Let m, n ∈ N, E = (kn+1 \ {0}) × (km+1 \ {0}) and π : kn+1 \ {0} → Pn , π : km+1 \ {0} → Pm the canonical morphisms. Define p : E → Pn × Pm , (x, x ) → (π(x), π (x )). The map p is a morphism by 12.4.3. We show as in the previous paragraphs that: • The map p is open. • A subset F in Pn × Pm is irreducible if and only if p−1 (F ) is irreducible in E. • Closed subsets of Pn × Pm are the subsets F such that p−1 (F ) is closed in E and invariant by the maps (x, x ) → (λx, λ x ) where λ, λ ∈ k \ {0}. 13.3.11 Let u : Pn → Pm be a morphism of varieties. By 13.3.8, v = u ◦ π is a morphism of kn+1 \ {0} to Pm . Denote by z0 , . . . , zm the homogeneous coordinates of points in Pm , Vj the affine open subset of Pm defined by zj = 0. For 0 i n and x ∈ π −1 (Ui ), the point v(x) belongs to some Vj . The open subset π −1 (Ui ) ∩ v −1 (Vj ) is then a neighbourhood of x, invariant by the automorphisms y → λy, for λ ∈ k \ {0}. It follows from 13.3.4 (i) and 13.3.7 that it contains a neighbourhood Wx of x in kn+1 \ {0} defined by Q(y0 , . . . , yn ) = 0, where Q is a homogeneous polynomial. Let us identify canonically Vj with the set of points (x0 , . . . , xm ) ∈ km+1 \ {0} such that xj = 1 (13.2.1). The coordinates of v(y), for y ∈ Wx , are regular
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functions, invariant by the automorphisms y → λy. Replacing Q by a suitable power, we may write P0 (y) Pj−1 (y) Pj+1 (y) Pm (y) ,..., , 1, ,..., v(y) = Q(y) Q(y) Q(y) Q(y) where the Pl ’s and Q are homogeneous polynomials of the same degree. Set Pj = Q, then if y ∈ Wx , (5)
u(π(y)) = π (w(y))
with (6)
w(y) = (P0 (y), . . . , Pm (y)).
We may further assume that P0 , . . . , Pm are relatively prime. Proceed in the same manner with another point x . Since kn+1 \ {0} is irreducible, Wx ∩ Wx is non-empty and open. It follows that, multiply the Pl ’s by the non-zero scalar if necessary, we may suppose that (5) is verified in Wx with the same polynomials Pl . We deduce therefore that (5) is verified for all y ∈ kn+1 \ {0}. Further, by construction, P0 , . . . , Pn have no common zero in kn+1 \ {0}. Conversely, let P0 , . . . , Pn be homogeneous polynomials of the same degree having no common zero in kn+1 \ {0}. By 13.3.8, formulas (5) and (6) define a morphism from Pn to Pm . 13.3.12 Proposition. The only regular functions on Pn , that is the elements of Γ (Pn , OPn ), are the constant functions. Proof. Such a function is a morphism from Pn to k ⊂ P1 . If z0 , . . . , zn are homogeneous coordinates of z ∈ Pn , then 13.3.11 says that there exist homogeneous polynomials P, Q of the same degree such that f (z) =
P (z0 , . . . , zn ) Q(z0 , . . . , zn )
for any z ∈ Pn and Q(z0 , . . . , zn ) = 0 if (z0 , . . . , zn ) = (0, . . . , 0). Since Q is homogeneous in at least two indeterminates, it is constant. Hence P is also constant. 13.3.13 Let us conserve the notations of 13.3.11, and identify kn with U0 and km with V0 . Let u : kn → km be a morphism. Under what condition can we extend u to a morphism from Pn to Pm ? Note that if such an extension exists, it is unique by 12.5.6 and 13.2.6. Now we have from 11.5.1 that u(x1 , . . . , xn ) = (Q1 (x1 , . . . , xn ), . . . Qm (x1 , . . . , xn )) where Q1 , . . . , Qm are polynomials. Let r = max{deg Qi ; 1 i m }. Then the extension, if it exists, is necessary the morphism:
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(x0 , x1 , . . . , xn ) →
xr0 , xr0 Q1
x1 xn ,..., x0 x0
, . . . , xr0 Qm
x1 xn ,..., x0 x0
.
We deduce that the condition for the extension of u is that the degree r homogeneous components of the Qi ’s are not simultaneously zero on kn \ {0}.
13.4 Complete varieties 13.4.1 Definition. An algebraic variety X is said to be complete if for all algebraic variety Y , the projection pr2 : X × Y → Y is a closed morphism. 13.4.2 The subset {(x, y) ∈ k × k ; xy = 1} ⊂ k × k is closed, but its image under pr2 is k \ {0}, which is not closed in k. So the affine line k is not complete. 13.4.3 Proposition. (i) A closed subvariety of a complete variety is complete. (ii) The product of two complete varieties is complete. (iii) Let X be a complete variety and u a morphism from X to a variety Y . Then u(X) is a closed and complete subvariety in Y . Proof. (i) Let X be a complete variety, Y a closed subvariety of X and Z a variety. We know from 12.4.5 and 12.5.4 that Y × Z is a closed subvariety of X × Z. Thus closed subsets of Y × Z are closed in X × Z. It follows that the restriction to Y × Z of pr2 : X × Z → Z is closed. (ii) Let X, Y be complete varieties and Z a variety. Then X ×Y is a variety (12.5.4). If F is a closed subset of (X × Y ) × Z = X × (Y × Z), its projection F on Y × Z is closed, and the projection F of F on Z is also closed. Since F is the projection of F on Z. We are done. (iii) Let Gu be the graph of u. Since Y is a variety, Gu is closed in X × Y (12.5.4). It follows that u(X), the projection of Gu on Y is closed in Y . Let Z be a variety, F a closed subset in u(X) × Z and F the projection of F on Z. Consider the following projections pr13 : X × Y × Z → X × Z , pr23 : X × Y × Z → Y × Z and set F = pr−1 23 (F ) ∩ (Gu × Z). Since u(X) is closed in Y and F is closed in u(X) × Z, F is closed in Y × Z. This implies that F is closed in Gu × Z. But the restriction of pr13 to Gu × Z is an isomorphism to X × Z (12.5.2). Hence pr13 (F ) is closed in X × Z. On the other hand, F is the projection of pr13 (F ) on Z. So F is closed in Z because X is complete. 13.4.4 Corollary. A complete affine variety is a finite set. Proof. It suffices to prove that if X is a non-empty, irreducible and complete affine variety, then X is a singleton, or equivalently, A(X) = k. Let f ∈ A(X), then by 13.4.3, f (X) is a closed irreducible and complete subvariety of k. Since k is not complete and proper closed subsets of k are finite subsets, f (X) is a singleton, and the result follows.
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177
13.4.5 Theorem. A projective variety is complete. Proof. By 13.4.3 (i), it suffices to prove that Pn is complete. Let Y be a variety, p : Pn × Y → Y the canonical projection and Z a closed subset of Pn × Y . We need to show that p(Z) is closed. a) It suffices to show that for all affine open subset V of Y , V ∩ p(Z) is closed in V , that is p(Z ∩ p−1 (V )) is closed in V . Now Z ∩ p−1 (V ) is closed in Pn × V , so we are reduced to the case where Y = Spm(A) is affine. Set B = A[T0 , . . . , Tn ]. b) Using the notations of 13.2.1, Pn × Y is the union of the affine open subsets Wi = Ui × Y , 0 i n. The algebra Ai = A(Wi ) is given by: T0 T0 Tn Tn ,..., ,..., ⊗k A = A . Ai = k Ti Ti Ti Ti Denote by a the graded ideal of B generated by the homogeneous polynomials Q such that, for 0 i n, Tn T0 ,..., ∈ I(Z ∩ Wi ). Q Ti Ti Let Br (resp. ar ) be the A-module of homogeneous polynomials of degree r in B (resp. a). c) For i ∈ {0, . . . , n} and g ∈ I(Z ∩ Wi ), we shall prove that there exist integers k, r such that Tik g ∈ ar . Since g is a polynomial in Tl /Ti , 0 l n, for a sufficiently large m, Tim g is a homogeneous polynomial of degree m in T0 , . . . , Tn . On the other hand, for any j, (Tim /Tjm )g is zero on Z ∩ Wi ∩ Wj , and (Tim+1 /Tjm+1 )g is zero on (Z ∩ Wj ) \ Wi . Since this holds for any j, we have Tim+1 g ∈ am+1 . d) Let y0 ∈ Y \ p(Z) and m = I({y0 }) ∈ Spm A. To finish our proof, we need to show that there exists f ∈ A verifying y0 ∈ D(f ) and D(f )∩p(Z) = ∅. Now Z ∩ Wi and Ui × {y0 } are disjoint closed subsets of the affine open set Wi , and I(Ui × {y0 }) = mAi , it follows from 11.1.4 and 11.1.6 that: Ai = mAi + I(Z ∩ Wi ). Thus there exist gi ∈ I(Z ∩ Wi ), αij ∈ m and gij ∈ Ai such that for all 0 i n: 1 = gi + αij gij . j
Paragraph c) says that by multiplying the above equality by Tim with m suitably large, we obtain Tim = hi + αij hij j
where hi ∈ am and hij ∈ Bm . This implies that for N suitably large, any monomial of degree N in T0 , . . . , Tn belongs to aN + mBN , that is, BN = aN + mBN .
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Now apply 2.6.7 to the finitely generated A-module BN /aN , there exists f ∈ A \ m such that f BN ⊂ aN . Consequently, f (y0 ) = 0 and f.TiN ∈ aN for all i. Hence f is zero on p(Z) as required. 13.4.6 Remarks. 1) Let X, Y be complete varieties and u : X → Y a morphism. Contrary to the case of compact spaces, u can be a bicontinuous bijection without being an isomorphism. For example, let u : P1 → P2 be the map defined by: (x, y) → (x3 , xy 2 , y 3 ). This is a morphism by 13.3.11. Its image Y is a closed subvariety of P2 (13.4.3 (iii) and 13.4.5). Denote by v : P1 → Y the induced morphism. We check easily that v is bijective. Again by using 13.4.3 and 13.4.5, we see that v is bicontinuous. But the image of the affine open subset U0 of P1 is the algebraic set {(x, y) ∈ k2 ; x3 − y 2 = 0}. So by 11.5.7, v is not an isomorphism. 2) There are complete varieties which are not projective varieties.
13.5 Products 13.5.1 Let z ∈ Pn (resp. z ∈ Pm ) with homogeneous coordinates (x0 , . . . , xn ) (resp. (x0 , . . . , xm )). Denote by s(z, z ) ∈ Pnm+n+m the point with homogeneous coordinates the (n + 1)(m + 1) scalars y(i,j) where y(i,j) = xi xj for 0 i n, 0 j m. Thus we have a map s : Pn × Pm → Pnm+n+m . We shall show that s is a morphism, called the Segre embedding. More precisely, we have the following result: Proposition. The map s is a morphism. Its image S is a closed subvariety of Pnm+n+m and s induces an isomorphism of Pn × Pm onto S. Proof. Let Ui (resp. Uj ) be the affine open subset of Pn (resp. Pm ) defined by xi = 0 (resp. xj = 0) and let Wij be the affine open subset of Pnm+n+m defined by y(i,j) = 0. The restriction of s to Ui × Uj (identified as kn × km ) is clearly a morphism to Wij (identified as knm+n+m ). Hence u is a morphism. Now Pn × Pm is complete (13.4.3 and 13.4.5), so S is a closed subvariety of Pnm+n+m (13.4.3). From the definition of s, it is straightforward to check that: • s−1 (Wij ) = Ui × Uj . • The restriction of s to each Ui × Uj is injective. So u is injective. • Let t : Wij → Ui × Uj be the map defined by sending a point with homogeneous coordinates y(l,h) to the point (z, z ) where z (resp. z ) has homogeneous coordinates the yl,j ’s for 0 l n (resp. the yi,h ’s for 0 h m). Then t is a morphism such that t(s(z, z )) = (z, z ) if (z, z ) ∈ Ui × Uj . Hence we have our result. 13.5.2 Proposition. The product of two projective varieties is projective.
13.5 Products
179
Proof. This is clear from 12.4.2 and 13.5.1. 13.5.3 Let d ∈ N∗ and I the set of α = (α(0), . . . , α(n)) ∈ Nn+1 such that |α| = α(0) + · · · + α(n) = d. n+d The cardinality of I is equal to . d Let J be a set containing I and N its cardinality. By 13.3.13, we can define a morphism v : Pn → PN −1 by sending z ∈ Pn with homogeneous coordinates (x0 , . . . , xn ) to the point of PN −1 with homogeneous coordinates given by the following system: α(0) α(n) if α ∈ I, xα = x0 · · · xn Ph (x0 , . . . , xn ) if h ∈ J \ I, where the Ph ’s are homogeneous polynomials of degree d. Since Pn is complete, the image V of v is a closed subvariety of PN −1 . Proposition. The morphism v induces an isomorphism of Pn onto the closed subvariety V of PN −1 . Proof. For 0 i n, denote by βi = (β(0), . . . , β(n)) ∈ I the elements such that β(i) = d and β(j) = 0 if j = i. Let Wi be the affine open subset of PN −1 consisting of points whose homogeneous coordinate yβi is non-zero. If Ui is the affine open subset of Pn defined by xi = 0, then z ∈ Ui ⇒ v(z) ∈ Wi , z ∈ Ui ⇒ v(z) ∈ Wi . We deduce that v −1 (Wi ) = Ui . Now let w : Wi → Pn be the map sending a point with homogeneous coordinates yh , h ∈ J, to the point of Pn with homogeneous coordinates yγj , 0 j n, where γi = βi and if i = j, γj ∈ I is such that: γj (i) = d − 1 , γj (j) = 1 , γj (h) = 0 if h ∈ {i, j}. The map w is a morphism from Wi to Ui and we verify easily that w(v(z)) = z if z ∈ Ui . Hence the result follows. 13.5.4 Remarks. 1) In the case where J = I, the morphism v is called the Veronese embedding of degree d. 2) Let X be a hypersurface of Pn defined by the homogeneous polynomial P of degree d > 0. Using the notations of 13.5.3, we have aα T α P = α∈L
where L ⊂ I and aα ∈ k \ {0} for α ∈ L. If v is the Veronese embedding of degree d, then v(X) is contained in the hyperplane of PN −1 defined by
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13 Projective varieties
aα Yα = 0.
α∈L
This is one of the reason why the Veronese embedding is useful: it allows us to reduce a problem concerning a hypersurface to a problem concerning a hyperplane. 13.5.5 The following result completes the one in 13.3.7. Proposition. Let Q ∈ k[T0 , . . . , Tn ] be a homogeneous polynomial of degree d > 0. The subset D+ (Q) of Pn is an affine open subset. n+d Proof. Let N = , v : Pn → PN be defined as in 13.5.3 by d v(z) = ((xα )α∈I , Q(x0 , . . . , xn )), and V = v(Pn ) which is closed by 13.5.3. Let DN +1 be the affine open subsets of PN consisting of points such that the last homogeneous coordinate is non-zero. Since V ∩ DN +1 is closed in DN +1 , it is an affine open subset of V . But V ∩ DN +1 = v(D+ (Q)). Hence the result follows since v is an isomorphism of Pn onto V .
13.6 Grassmannian variety
r n 13.6.1 Let Er denote the k-vector space (k ). If (e1 , . . . , en ) is the n canonical basis of k , then the canonical basis of Er is, by definition, the n N= vectors r eH = eii ∧ ei2 ∧ · · · ∧ eir
for any subset H of {1, . . . , n} consisting of r elements i1 < · · · < ir . We have E0 = k, E1 = kn and Er = {0} if r > n. r 13.6.2 Recall that if V is a subspace of k, then V can be identified canonically as a subspace of Er . r-dimensional subspaces of kn . If V ∈ Sn,r , then Let r Sn,r be the setof r V = 1, and so V is a line in Er , or a point in the projective space dim P(Er ) = PN −1 . Thus we have a map ψ : Sn,r → PN −1 . Denote by Gn,r the image of ψ. Note that ψ is injective. Indeed, if V, W ∈ Sn,r and (v1 , . . . , vn ) a basis of kn such that (v1 , . . . , vr ) (resp. (vs , . . . , vs+r−1 )) is a basis of V (resp. W ). Then ψ(V ) = ψ(W ) implies that v1 ∧ · · · ∧ vr and vs ∧ · · · ∧ vs+r−1 are proportional. Hence s = 1 (13.6.1) and V = W . 13.6.3 Proposition. The set Gn,r is closed in PN −1 . Proof. Using the notations of 13.6.1, let WH be the affine open subsets of PN −1 defined by eH = 0. Then the N subsets WH cover Er . It suffices therefore to prove that Gn,r ∩ WH is closed in WH for all H.
13.6 Grassmannian variety
181
Take for example H = {1, . . . , r}, and write W = WH . Let F0 (resp. F0 ) be the subspace generated by e1 , . . . , er (resp. er+1 , . . . , en ) and p the projection on F0 with respect to the direct sum kn = F0 ⊕ F0 . If (v1 , . . . , vr ) is a basis of V ∈ Sn,r , then writing vi = p(vi ) + wi , wi ∈ F0 , we have ψ(V ) ∈ W ⇔ p(v1 ) ∧ · · · ∧ p(vr ) = 0, which is equivalent to the condition that p induces an isomorphism of the vector spaces V and F0 . So if ψ(V ) ∈ W , then V has a unique basis (v1 , . . . , vr ) of the form vi = ei +
(7)
n
aij ej = ei + ui
j=r+1
for 1 i r. Thus v1 ∧ · · · ∧ vr = e1 ∧ · · · ∧ er +
(e1 ∧ · · · ∧ ui ∧ · · · ∧ er ) + w,
1jr
where w is a linear combination of vectors eH such that H contains at least 2 elements in {r + 1, . . . , n}. Moreover, it is clear that the coefficient of such a vector eH is a polynomial PH in the aij ’s, and this polynomial does not depend on V . Finally, we have e1 ∧ · · · ∧ ui ∧ · · · ∧ er = εaij (e1 ∧ · · · ∧ ei−1 ∧ ei+1 ∧ · · · ∧ er ∧ ej ) j>r
where ε ∈ {−1, 1}. Hence the aij ’s determine completely the decomposition of v1 ∧ · · · ∧ vr in the basis (eH )H . Conversely, given r(n − r) scalars aij and define the vi as in (7), the subspace V generated by the vectors vi verifies V ∈ Sn,r and ψ(V ) ∈ W . Identifying W with kN −1 , the preceding arguments show that W ∩ Gn,r is isomorphic to the graph of the morphism kr(n−r) → kN −r(n−r)−1 , (aij ) → (PH (aij )). Hence W ∩ Gn,r is closed in W .
13.6.4 Definition. The closed subvariety Gn,r of PN −1 is called the Grassmannian variety of r-dimensional subspaces of kn . 13.6.5 Remark. Let UH = WH ∩ Gn,r . The proof of 13.6.3 shows that UH is an affine open subset of Gn,r , which is isomorphic to kr(n−r) . 13.6.6 Proposition. The Grassmannian variety Gn,r is irreducible. Proof. Being isomorphic to kr(n−r) , the affine open subsets UH are irreducible. Since the UH ’s cover Gn,r , it suffices to show that UH ∩ UJ = ∅ for H = J. Let
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H ∩ J = {α1 , . . . , αs }, H \ (H ∩ J) = {β1 , . . . , βr−s } , J \ (H ∩ J) = {γ1 , . . . , γr−s }, V = keα1 ⊕ · · · ⊕ keαs ⊕ k(eβ1 + eγ1 ) ⊕ · · · ⊕ k(eβr−s + eγr−s ). Then it is clear from the proof of 13.6.3 that ψ(V ) ∈ UH ∩ UJ .
13.6.7 A flag in kn is a chain of subspaces of kn V1 ⊂ V2 ⊂ · · · ⊂ Vn = kn such that dim Vi = i for 1 i n. Denote by Fn the set of flags of kn . Then Fn can be identified naturally with a subset of G = Gn,1 × Gn,2 × · · · × Gn,n . By 13.5.2, G has a canonical structure of projective variety. Using methods analogue to those in 13.6.3, we can prove that Fn is closed in G. Thus Fn has a structure of projective variety. We shall call Fn , the flag variety of kn .
References and comments • [5], [19], [22], [26], [37], [40], [78]. Being projective, Grassmannian varieties are complete varieties. This fact will prove to be very useful later on.
14 Dimension
The first section of this chapter gives an algebraic interpretation of the dimension of a variety as a topological space. We then use this to establish certain results concerning the dimension of algebraic varieties. All the algebraic varieties considered in this chapter are defined over k.
14.1 Dimension of varieties 14.1.1 Recall that the dimension of a topological space and the dimension of an algebra were defined in chapter 1 and chapter 2 respectively. 14.1.2 Proposition. Let X be an affine variety. (i) We have: dim X = dim A(X). (ii) If X is irreducible, then: dim X = tr degk Fract(A(X)). Proof. (i) The map p → V(p) from the set of prime ideals to the set of irreducible subsets of X is an inclusion-reversing bijection (11.1.4, 11.1.7 and 11.2.3). It follows from the definition of dim X (1.2.2) and the definition of dim A(X) (2.2.3) that dim X = dim A(X). (ii) Recall from 11.2.3 that if X is an affine variety, then the algebra A(X) of regular functions on X is an integral domain if and only if X is irreducible. The result follows therefore from part (i) and 6.2.3. 14.1.3 Theorem. Let X be an irreducible variety and U a non-empty open subset of X. Then dim X = dim U is finite and: dim X = tr degk R(X).
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Proof. Let us first suppose that X is affine. Let U, V be non-empty affine open subsets of X, then U ∩ V = ∅ because X is irreducible. So there exists f ∈ A(X) such that D(f ) = ∅ and D(f ) ⊂ U ∩ V (11.7.2). It follows from 1.2.3 that: dim D(f ) dim U dim X , dim D(f ) dim V dim X. Hence by remark 1 of 11.7.5, Fract(A(X)) = Fract(A(D(f ))), and 14.1.2 (ii) implies that dim D(f ) = dim U = dim V = dim X. Now any non-empty open subset W of X contains an affine open subset U of X. Using the preceding arguments and 1.2.3, we have dim W = dim X. Let us now suppose that X is not affine. The preceding arguments show that all the affine open subsets of X have the same dimension. Since X is the union of a finite number of affine open subsets, dim X = dim U for any affine open subset U of X (1.2.3 (iii)). Finally any non-empty open subset W of X contains a non-empty affine open subset. Hence dim W = dim X (1.2.3 (i)). The equality dim X = tr degk R(X) follows immediately from 12.7.6 and 14.1.2. 14.1.4 Corollary. Let X be a non-empty algebraic variety. Then its dimension is finite and if X1 , . . . , Xn are the irreducible components of X, we have: dim X = max{dim Xi ; 1 i n }. Proof. This follows from 1.2.3 (i), 1.3.5 and 14.1.3. 14.1.5 Examples. 1) A variety of dimension zero is a finite set (since an irreducible variety of dimension zero is clearly a singleton). 2) We have dim kn = n (6.1.4 and 14.1.2). 3) The projective space Pn is irreducible (13.2.6) and contains an open subset isomorphic to kn (13.2.4). Hence dim Pn = n. 4) The Grassmannian variety Gn,r is irreducible (13.6.6) and contains an open subset isomorphic to kr(n−r) (13.6.5). Hence dim Gn,r = r(n − r). 14.1.6 Proposition. Let Y be a non-empty subvariety of a variety X. (i) We have dim Y dim X. (ii) If X is irreducible and Y is closed, then dim Y = dim X if and only if Y = X. (iii) We have dim Y = dim Y . (iv) If X is irreducible, then dim Y = dim X if and only if Y is open. Proof. Parts (i) and (ii) have been proved in 1.2.3. (iii) Let Y1 , . . . , Yn be the irreducible components of Y . Then by 1.1.14, Y1 , . . . , Yn are the irreducible components of Y . Since Yi is closed in Y (1.1.5),
14.2 Dimension and the number of equations
185
it is locally closed in X. So Yi is open in Yi (1.4.1), and dim Yi = dim Yi (14.1.3). The result follows therefore from 14.1.4. (iv) Let us write Y = F ∩ U where U (resp. F ) is an open (resp. closed) subset of X. By 1.2.3, dim Y dim F dim X. It is now clear by part (ii) and 14.1.3 that dim X = dim Y if and only if Y = U is open. 14.1.7 Proposition. Let X, Y be irreducible varieties. Then: dim(X × Y ) = dim X + dim Y. Proof. By 12.4.5, X ×Y is irreducible. Further, if U ⊂ X and V ⊂ Y are affine open subsets, then U × V is an affine open subset of X × Y . We may therefore assume that X = Spm(A) and Y = Spm(B) are affine. Let m = dim X and n = dim Y . By 6.2.2 and 14.1.2, A (resp. B) contains a subalgebra A0 (resp. B0 ) isomorphic to k[T1 , . . . , Tm ] (resp. k[Tm+1 , . . . , Tm+n ]), and A (resp. B) is generated over A0 (resp. B0 ) by a finite number of elements f1 , . . . , fr (resp. g1 , . . . , gs ) algebraic over E0 (resp. F0 ), the quotient field of A0 (resp. B0 ). Since A(X × Y ) = A ⊗k B, it contains the subalgebra A0 ⊗ B0 k[T1 , . . . , Tn+m ] whose quotient field L0 has transcendence degree n + m over k. Now, the quotient field L of A ⊗k B is an extension of L0 generated by f1 , . . . , fr , g1 , . . . , gs . So the degree of transcendence of L over k is n + m. 14.1.8 Let us use the notations π : kn+1 \ {0} → Pn and Ui , 0 i n, of 13.1.1 and 13.2.1. Corollary. If X is a closed and irreducible subvariety of Pn , then dim π −1 (X) = 1 + dim X. Proof. If X ∩Ui = ∅, then by 13.3.3, π −1 (X ∩Ui ) is isomorphic to the product (X ∩ Ui ) × (k \ {0}). Thus the result follows from 14.1.7. 14.1.9 Remarks. 1) Let Y be a closed and irreducible subvariety of an irreducible variety X. The codimension of Y in X, denoted by codimX (Y ) or simply codim(Y ) when there is no confusion, is defined to be dim X − dim Y . Hence codimX (Y ) = 0 if and only if Y = X (14.1.6). 2) Let X1 , . . . , Xn be the irreducible components of a variety X. Then by 14.1.6 and 1.1.14, we have codimXi (Xi ∩ Xj ) 1 whenever i = j.
14.2 Dimension and the number of equations 14.2.1 Definition. A non-empty variety is called of pure dimension or equidimensional if all of its irreducible components has the same dimension. We shall call an irreducible variety of dimension 1 (resp. 2) an algebraic curve (resp. algebraic surface) over k.
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14.2.2 Let X be a variety. As in the affine case, if f ∈ Γ (X, OX ), we denote by V(f ) the closed subvariety of X consisting of elements x ∈ X such that f (x) = 0. If f1 , . . . , fr ∈ Γ (X, OX ), we set: V(f1 , . . . , fr ) = V(f1 ) ∩ · · · ∩ V(fr ). Lemma. We have V(f ) = ∅ if and only if f is invertible in Γ (X, OX ). Proof. If f is invertible, then clearly V(f ) = ∅. Conversely, if V(f ) = ∅, then for any affine open subset U ⊂ X, f |U ∈ Γ (U, OX (U )) verifies V(f |U ) = ∅. Thus f |U is invertible with inverse, say gU . If V is another affine open subset, then gU |U ∩V = gV |U ∩V since it is the inverse of f |U ∩V . So there exists an element g ∈ Γ (X, OX ) such that g|U = gU for any affine open subset U . Hence f g = 1. 14.2.3 The following theorem, also called Krull’s Principal Ideal Theorem (or Hauptidealsatz), is a geometrical translation of 6.3.2. Theorem. Let X be an irreducible variety and f be a non-zero and noninvertible element of Γ (X, OX ). Then V(f ) is a non-empty subvariety of X of pure dimension dim X − 1. Proof. By 14.2.2, V(f ) = ∅. Let Y1 , . . . , Yr be the irreducible components of V(f ). There exists y ∈ Y1 such that y ∈ Y2 ∪ · · · ∪ Yr (1.1.13), and since Y2 ∪ · · · ∪ Yr is closed, there exists an affine open subset U ⊂ X such that y ∈ U and U ∩ V(f ) = U ∩ Y1 is irreducible. By 14.1.3, dim U ∩ Y1 = dim Y1 and dim U = dim X. So we may assume that X = Spm(A) is √ affine with A an integral domain, f ∈ A is non-zero and non-invertible, and Af is a prime ideal. The result follows immediately from 6.2.6 and 6.3.2. 14.2.4 Remark. If X is not irreducible, then V(f ) is not necessary of pure dimension (even if X is of pure dimension). For example, let X = V(T1 T2 ) ⊂ k2 and f the restriction to X of the map (t1 , t2 ) → t1 (t1 + t2 + 1), then we check easily that V(f ) is the disjoint union of a line and a point. 14.2.5 Corollary. Let X be a closed irreducible subvariety of Pn such that dim X 1. Given any homogeneous polynomial P ∈ k[T0 , . . . , Tn ] \ k, the hypersurface V(P ) defined by P (x0 , . . . , xn ) = 0 has a non-empty intersection with X. If V(P ) does not contain X, then X ∩ V(P ) is a subvariety of pure dimension dim X − 1. Proof. Let C = π −1 (X) where π is as in 13.1.1, and C the closure of C in kn+1 . Then dim C 2 (14.1.8) and 0 ∈ C is a zero of P , so the restriction of P to C is non-invertible. Now let C be the zero set of P in C; it is an affine cone and a variety of pure dimension with dim C dim X 1 (14.2.3). It follows that C ∩ (kn+1 \ {0}) = ∅, which implies that π(C ) = V(P ) ∩ X = ∅. The last statement is a direct consequence of 13.3.3.
14.3 System of parameters
187
14.2.6 Proposition. Let X be an irreducible variety and f1 , . . . , fr be elements of Γ (X, OX ). Then for any irreducible component Y of V(f1 , . . . , fr ), we have codimX (Y ) r. Proof. Induction on r. The case r = 1 is just 14.2.3. So let r > 1. There exists an irreducible component Y of V(f1 , . . . , fr−1 ) such that Y ⊂ Y . So: Y ⊂ Y ∩ V(fr ) ⊂ V(f1 , . . . , fr ). It follows that Y is an irreducible component of Y ∩ V(fr ). By induction hypothesis, codimX (Y ) r − 1. Now if fr |Y = 0, then Y = Y . Otherwise, dim Y = dim Y − 1 by 14.2.3. In both cases, we have codimX (Y ) r. 14.2.7 Remark. In 14.2.6, irreducible components of V(f1 , . . . , fr ) do not necessary have the same dimension, and it is possible that they all have codimension strictly less than r. 14.2.8 Proposition. Let X be a closed and irreducible subvariety of Pn and P1 , . . . , Pr ∈ k[T0 , . . . , Tn ] be non-constant homogeneous polynomials. We denote by V(P1 , . . . , Pr ) the set of points of X whose homogeneous coordinates are zeros of P1 , . . . , Pr . (i) Any irreducible component Y of V(P1 , . . . , Pr ) verifies codimX Y r. (ii) If dim X r, then V(P1 , . . . , Pr ) is non-empty. Proof. Part (i) follows from 14.2.6 applied to π −1 (X), while (ii) follows from 14.2.5 and a simple induction. 14.2.9 Corollary. If m < n, then the only morphisms from Pn to Pm are the constant maps. Proof. By 14.2.8 (ii), if P1 , . . . , Pr ∈ k[T0 , . . . , Tn ] are non-constant homogeneous polynomials, with r < n, then V(P1 , . . . , Pr ) = ∅. So the result is a consequence of the description of morphisms given in 13.3.11.
14.3 System of parameters 14.3.1 Proposition. Let X be an irreducible variety and Y be a closed irreducible subvariety of codimension 1 in X. (i) There exist an affine open subset U ⊂ X and f ∈ Γ (U, OX ) \ {0} such that Y ∩ U = ∅ and f |Y ∩U = 0. (ii) For any open subset V ⊂ X and any f ∈ Γ (V, OX ) \ {0} verifying Y ∩ V = ∅ and f |Y ∩V = 0, Y ∩ V is an irreducible component of V(f ). Proof. (i) Being irreducible, Y is non-empty, so there exists an affine open subset U = Spm(A) of X such that Y ∩ U = ∅. Let a be the radical ideal of A such that Y ∩ U = V(a). Then a = {0} by 14.1.6 (iv). Thus any f ∈ a \ {0} works.
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(ii) The irreducibility of Y implies that Y ∩ V is irreducible for any open subset V such that Y ∩ V = ∅. So the open subset Y ∩ V of Y is contained in an irreducible component W of V(f ). By 14.1.6, we have: dim V = dim X , dim(Y ∩ V ) = dim Y = dim X − 1. So we obtain by 14.2.3 that: dim V > dim W dim(Y ∩ V ) = dim V − 1. This implies that dim W = dim(Y ∩ V ), and since W and Y ∩ V are closed irreducible subsets of V , we have W = Y ∩ V (14.1.6). 14.3.2 Corollary. Let X be an irreducible variety, F the set of closed irreducible subvarieties of X which are distinct from X, and F∗ the set of maximal elements of F. If Y ∈ F, then Y ∈ F∗ if and only if codimX (Y ) = 1. Proof. If codimX (Y ) = 1, then Y ∈ F∗ by 14.1.6. Conversely, let Y ∈ F∗ . For any affine open subset U = Spm(A) of X, Y ∩ U is a closed irreducible subset of U , distinct from U . Thus Y ∩ U = V(p) for some p ∈ Spec A. It follows that Y ∩ U ⊂ V(f ) for any f ∈ p \ {0}, and so Y ∩ U is contained in an irreducible component Z of V(f ), and dim Z = dim X − 1. By 1.1.12, Y = Y ∩ U and so Y ⊂ Z. Since Z is closed in U , it is locally closed in X. So Z is open in Z and dim Z = dim Z (14.1.6). So the result follows from the maximality of Y . 14.3.3 Using the notations of 14.3.1, we can ask if for any x ∈ Y , there is an open subset U of X containing x and f ∈ Γ (X, OX ) such that the variety Y ∩ U is exactly V(f ). In general, the answer is no. However: Proposition. Let X be an irreducible affine variety such that A(X) is factorial. If Y is a closed subvariety of X of pure dimension dim X − 1, then there exists f ∈ A(X) such that Y = V(f ). Proof. Let Y1 , . . . , Yr be the irreducible components of Y and pi = I(Yi ) ∈ Spec(A(X)), 1 i r. By 11.2.4, the pi ’s are minimal among the prime ideals of A(X). Hence pi = A(X)fi , where fi is an irreducible element of A(X). Finally, we obtain that Y = V(f ) where f = f1 · · · fr . 14.3.4 Corollary. Any closed subvariety Y of Pn of pure dimension n − 1 is defined by a single equation P (x0 , . . . , xn ) = 0 where P ∈ k[T0 , . . . , Tn ] is a non-constant homogeneous polynomial. In other words, Y is a hypersurface. Proof. Let Y1 , . . . , Yr be the irreducible components of Y and Ci = π −1 (Yi ), C = π −1 (Y ). By 14.1.8, 13.3.3 and 13.3.9, the Ci ’s are closed irreducible cones of dimension n in kn+1 \ {0}. The closure Ci of Ci in kn+1 is again irreducible of dimension n. We see easily that the Ci ’s are the irreducible components of C. Thus C is of pure dimension n in kn+1 . Since it is a closed affine cone, there exists a non-constant homogeneous polynomial P ∈ k[T0 , . . . , Tn ] such that C = V(P ) (14.3.3). Hence our result.
14.3 System of parameters
189
14.3.5 Proposition. Let X = Spm(A) be an irreducible affine variety and Y a closed irreducible subvariety of codimension r 1 in X. For any integer s between 1 and r, there exist f1 , . . . , fs ∈ A such that: (i) Y ⊂ V(f1 , . . . , fs ). (ii) The irreducible components of V(f1 , . . . , fs ) are of codimension s. Consequently, there exist f1 , . . . , fr ∈ A such that Y is an irreducible component of V(f1 , . . . , fr ). We say that the fi ’s form a system of parameters of Y. Proof. Let us proceed by induction on s. Since Y = X, I(Y ) = {0}. If s = 1, any f1 ∈ I(Y ) \ {0} works by 14.2.3. Let us suppose that s > 1. The induction hypothesis says that there exist f1 , . . . , fs−1 ∈ A such that Z = V(f1 , . . . , fs−1 ) is a subvariety of pure dimension containing Y , and its irreducible components Z1 , . . . , Zn have codimension s − 1 in X. Since s − 1 < r, Zi ⊂ Y , so I(Y ) ⊂ I(Zi ). Hence I(Y ) ⊂ I(Z1 ) ∪ · · · ∪ I(Zn ) (2.2.5). Let fs ∈ I(Y ) \ (I(Z1 ) ∪ · · · ∪ I(Zn )). We have Y ⊂ V(f1 , . . . , fs ). If W is an irreducible component of V(f1 , . . . , fs ), then by 14.2.6, we have codimX (W ) s. But W ⊂ V(f1 , . . . , fs−1 ), so W ⊂ Zi for some i and W = Zi since fs ∈ I(Zi ). We conclude by 14.1.6 that codimX (W ) s. Hence codimX (W ) = s. 14.3.6 Corollary. Let A be a finitely generated k-algebra which is an integral domain, p ∈ Spec(A), n = dim A, and r = dim A/p. Then there exists a chain {0} = p0 ⊂ p1 ⊂ · · · ⊂ pn−r = p ⊂ pn−r+1 ⊂ · · · ⊂ pn of prime ideals of A, containing p, of length n. Proof. The existence of the pk ’s, k n − r, is just the definition of dim A/p. Consider A as the algebra of regular functions on an irreducible affine variety X, p corresponds to a closed irreducible subvariety of X of dimension r. So the result follows by applying 14.3.5. 14.3.7 Corollary. Let X be a closed irreducible subvariety of Pn . For any closed irreducible subvariety Y of codimension r in X, there exist r nonconstant homogeneous polynomials P1 , . . . , Pr ∈ k[T0 , . . . , Tn ] such that Y is an irreducible component of X ∩ V(P1 , . . . , Pr ). Proof. We prove, as in 14.3.5, that there exist non-constant homogeneous polynomials P1 , . . . , Ps , 1 s r, such that Y ⊂ V(P1 , . . . , Ps ) and X ∩ V(P1 , . . . , Ps ) is a subvariety of pure dimension. If Z1 , . . . , Zm are the irreducible components of X ∩ V(P1 , . . . , Ps ), then the ideals I(π −1 (Zj )) and I(π −1 (Y )) are graded. We can finish the proof as in 14.3.5 by using 7.2.9.
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14.4 Counterexamples 14.4.1 Let us return to the point mentioned in 12.7.10. Denote by s : P1 × P1 → P3 the Segre embedding (13.5.1). Thus the points z, z ∈ P1 with homogeneous coordinates (x0 , x1 ) and (x0 , x1 ) are sent to the point s(z, z ) of P3 with homogeneous coordinates: (x0 x0 , x0 x1 , x1 x0 , x1 x1 ). It is easy to check that the image C of s is the set of points of P3 whose homogeneous coordinates y0 , y1 , y2 , y3 verify y0 y3 − y1 y2 = 0. By 13.5.1, C is irreducible. Let π : k4 \ {0} → P3 be the canonical surjection (13.1.1), and X the affine cone in k4 defined by x1 x4 − x2 x3 = 0. Then π(X \ {0}) = C. It follows from 13.1.2 and 13.3.9 that X is an irreducible subvariety of k4 . On the open subset D(x2 ) ∩ D(x4 ) = D(x2 x4 ) of X, the functions x1 /x2 and x3 /x4 are identical. They are therefore the restriction of a rational function h on X whose domain of definition U contains D(x2 ) and D(x4 ). Suppose that h ∈ A(X). Then x2 h is regular and is zero on the points (1, 0, γ, 0) of X. But x2 h and x1 are identical on D(x2 x4 ), so the irreducibility of X implies that x2 h and x1 are identical on X. Contradiction. Now suppose that there exists g ∈ A(X) non-invertible such that U = D(g). The complement V(g) of U in X is contained in the plane P ⊂ X of k4 defined by the equations x2 = 0 and x4 = 0. Since P is irreducible, we obtain by 14.1.6 and 14.2.3 that P = V(g). So the restriction of g to the plane P ⊂ X defined by x1 = 0 and x3 = 0 is regular and (0, 0, 0, 0) is the only zero. This is absurd in view of 14.2.3. The preceding argument says that U is contained strictly in D(g) for all functions f, g ∈ A(X) such that h = f /g. As we have seen in 12.7.10, this proves that A(X) is not factorial. 14.4.2 Let us conserve the notations of 14.4.1 and set x = (0, 0, 0, 0). If U is an open subset of X and f ∈ Γ (U, OX ), let VU (f ) = {y ∈ U ; f (y) = 0}. Let us suppose that there exist an open subset U of X containing x, and f ∈ Γ (U, OX ) such that VU (f ) = P ∩ U . Let W = P ∩ U . Then W contains x and is an open subset of P . So dim W = 2. Let g = f |W . Then g ∈ Γ (W, OX |P ) and VW (g) = VU (f ) ∩ P . But VU (f ) ∩ P = {x}, so its dimension is zero. This contradicts 14.2.3. Thus we have proved that there does not exist any open subset U of X containing x and any f ∈ Γ (U, OX ) verifying VU (f ) = P ∩ U (see 14.3.3).
References • [5], [19], [22], [26], [37], [40], [78].
15 Morphisms and dimension
We study properties of morphisms of algebraic varieties in this chapter. The notions of an affine morphism and a finite morphism are introduced. We also consider the relation between dimension and morphism. In particular, the dimension of a fibre of a morphism is considered.
15.1 Criterion of affineness 15.1.1 Let X be a variety and U an open subset of X, we shall denote often Γ (U, OX ) by OX (U ). Furthermore, if U is affine, then we shall also write A(U ) for OX (U ). If f ∈ OX (U ), we set: DU (f ) = {x ∈ U ; f (x) = 0}. 15.1.2 Lemma. Let X be a variety and f ∈ OX (X). Then we have DX (f ) = X if and only if f is invertible in OX (X). Proof. This is just a reformulation of 14.2.2. 15.1.3 Lemma. Let X be a variety, A = OX (X) and f ∈ A. (i) The restriction to DX (f ) of f is invertible in OX (DX (f )). (ii) The identity map A → A induces an isomorphism of Af onto OX (DX (f )). Proof. Let Y = DX (f ) and OY = OX |Y (9.3.7 and 9.3.8). By 12.2.6 and 12.5.4, (Y, OY ) is a variety. If g = f |Y , then DY (g) = Y and so g is invertible in OX (Y ) by 15.1.2. It follows from 2.3.4 that there is a unique ϕ ∈ Homalg (Af , OX (Y )) such that ϕ(h/f n ) = (h|Y )/(f n |Y ). We shall prove that ϕ is an isomorphism. Let (Ui )i∈I be a covering of X by affine open subsets. Since DX (f ) ∩ Ui = DUi (f |Ui ), we deduce that OX (Y ∩ Ui ) = A(Ui )f |Ui (12.1.1).
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Let h/f n ∈ ker ϕ. Then (h|Y )/(f n |Y ) = 0, so (h|Y ∩Ui )/(f n |Y ∩Ui ) = 0. Thus (f |Ui )(h|Ui ) = 0 (11.7.3), which implies that f h = 0. Hence h/f n = 0, and ϕ is injective. Let g ∈ OX (Y ). Then g|Y ∩Ui ∈ OX (Y ∩ Ui ) = A(Ui )f |Ui . So there exist ni ∈ N and hi ∈ A(Ui ) such that (f ni |Y ∩Ui )(g|Y ∩Ui ) = hi |Y ∩Ui . Let m be maximal among the ni ’s and i = (f m−ni |Ui ).hi ∈ A(Ui ). We have ( i −
j )|Y ∩Ui ∩Uj = 0. Since Ui ∩Uj is an affine open subset (12.5.6), (f |Ui ∩Uj ).(( i −
j )|Ui ∩Uj ) = 0 (11.7.3), and so (f i )|Ui ∩Uj = (f j )|Ui ∩Uj . We deduce therefore that there exists ∈ A such that |Ui = (f |Ui ) i for all i. It follows that (f m+1 |Y )g = |Y , and so ϕ is surjective. 15.1.4 Let X, Y be varieties with Y affine. Recall that given a morphism v : X → Y , we have the associated k-algebra homomorphism: Γ (u) : OY (Y ) → OX (X) , g → g ◦ v. By 12.3.3, the map v → Γ (v) is a bijection between Mor(X, Y ) and Homalg (OY (Y ), OX (X)). Suppose further that A = OX (X) is a finitely generated k-algebra, then the identity map A → A = OX (X) induces a morphism u : X → Spm(A). Proposition. The following conditions are equivalent: (i) X is an affine variety. (ii) The k-algebra A is finitely generated and u is an isomorphism. (iii) The k-algebra A is finitely generated and u is a homeomorphism. Proof. We have seen in 12.3.4 that (i) ⇒ (ii), and (ii) ⇒ (iii) is clear. Suppose that we have (iii). Set Y = Spm(A). Since the DY (f )’s, f ∈ A, form a base of open subsets of Y , the subsets DX (f ) = u−1 (DY (f )) form a base of open subsets of X. Moreover, if we denote uf : DX (f ) → DY (f ) the morphism induced by u, it follows from 15.1.3 and the bijectivity of u that Γ (uf ) : OY (DY (f )) = Af → OX (DX (f )) is an isomorphism. Hence u is an isomorphism (12.2.3) and X is affine.
15.1.5 Theorem. Let X be a variety, A = OX (X) and f1 , . . . , fr ∈ A be such that X = DX (f1 ) ∪ · · · ∪ DX (fr ) and each DX (fi ), 1 i r, is an affine variety. The following conditions are equivalent: (i) X is an affine variety. (ii) A = Af1 + · · · + Afr . Proof. (i) ⇒ (ii) If X is affine, the identity map idA induces an isomorphism u : X → Y = Spm(A) (15.1.4). Since u−1 (DY (f )) = DX (f ), we have Spm(A) = DY (f1 ) ∪ · · · ∪ DY (fr ) , V(Af1 + · · · + Afr ) = ∅.
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193
Hence A = Af1 + · · · + Afr as required (6.5.3). (ii) ⇒ (i) Suppose that A = Af1 + · · · + Afr . First, we shall show that A is a finitely generated k-algebra. By our hypotheses and 15.1.3, Afi , being isomorphic to OX (DX (fi )), is finitely generated. Let {si,1 /fini , . . . , si,ri /fini } be a system of generators of Afi where si,k ∈ A. Now there exist 1 , . . . , r ∈ A such that: (1)
1 = 1 f1 + · · · + r fr .
Denote by B the subalgebra of A generated by the fi ’s, the i ’s and the si,k ’s. Let g ∈ A, then g/1 ∈ Afi . So for each i, there exists mi ∈ N such that fimi g = hi ∈ B. Suppose that mi 1. Then by taking a suitable power of (1), there exist t1 , · · · , tr ∈ B such that: 1 = t1 f1m1 + · · · + tr frmr . Hence g = t1 h1 + · · · + tr hr ∈ B, and B = A is finitely generated. Finally, the identity map idA induces an isomorphism ϕi : Afi → OX (DX (fi )) by 15.1.3, which in turn induces an isomorphism (since DX (fi ) is affine) vi : DX (fi ) → Spm(Afi ). Moreover, vi (x) = u(x) for x ∈ DX (fi ) and u−1 (DY (fi )) = DX (fi ). As vi is a homeomorphism, so is u. Hence X is affine (15.1.4).
15.2 Affine morphisms 15.2.1 Definition. Let X, Y be varieties. A morphism u : X → Y is called affine if there is a covering (Vi )i∈I of Y by affine open subsets such that u−1 (Vi ) is an affine open subset of X for all i ∈ I. 15.2.2 Lemma. Let X, Y be affine varieties and u : X → Y a morphism. Then for all affine open subset V in Y , u−1 (V ) is an affine open subset. In particular, u is affine. Proof. Since Y is affine, there exist g1 , . . . , gn ∈ A(Y ) such that V = DY (g1 )∪ · · · ∪ DY (gn ). Let fi = gi ◦ u ∈ A(X) and v : u−1 (V ) → V the morphism induced by u. We have u−1 (DY (gi )) = DX (fi ), DY (gi ) = DV (gi |V ) and so u−1 (DY (gi )) = Du−1 (V ) (hi ) where hi = Γ (v)(gi |V ) (here, the notations are as in 15.1.4). Since V is affine, there exist, by 15.1.5, t1 , . . . , tn ∈ OY (V ) such that 1 = t1 (g1 |V ) + · · · + tn (gn |V ). Hence 1 = Γ (v)(t1 )h1 + · · · + Γ (v)(tn )hn . Since X is affine, so is DX (fi ) = Du−1 (V ) (hi ). It follows from 15.1.5 that u−1 (V ) is affine.
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15.2.3 Proposition. The following conditions are equivalent for a morphism u : X → Y of varieties: (i) u is affine. (ii) For any affine open subset V of Y , u−1 (V ) is an affine open subset. Proof. Clearly, (ii) implies (i). Conversely, let u be affine and (Vi )i∈I a covering of Y by affine open subsets such that u−1 (Vi ) is affine for all i ∈ I. Given any affine open subset V of Y and v : u−1 (V ) → V the morphism induced by u, we have ni DV (gij ), V ∩ Vi = j=1
where gij ∈ OY (V ). We have u−1 (DV (gij )) = Du−1 (V ) (hij ) where hij = Γ (v)(gij ). Applying 15.2.2 to the morphism u−1 (Vi ) → Vi , we see that u−1 (Dv (gij )) is affine. Since V is affine, there exist, by 15.1.5, tij ∈ OY (V ) verifying 1= tij gij . i,j
Hence 1=
Γ (v)(tij )hij .
i,j
Thus u−1 (V ) is affine (15.1.5).
15.3 Finite morphisms 15.3.1 Definition. A morphism u : X → Y of varieties is called finite if there is a covering (Vi )i∈I of Y by affine open subsets such that (i) u−1 (Vi ) is an affine open subset of X for all i ∈ I. (ii) For all i ∈ I, A(u−1 (Vi )) is, via the comorphism A(Vi ) → A(u−1 (Vi )) of the morphism u−1 (Vi ) → Vi induced by u, a finite A(Vi )-algebra. Remarks. 1) In 15.3.1, we do not exclude the possibility that u−1 (Vi ) may be empty. 2) A finite morphism is affine. So the inverse image of an affine open subset under a finite morphism is affine (15.2.3). 15.3.2 Proposition. Let X = Spm(A), Y = Spm(B) be affine varieties, ϕ ∈ Homalg (B, A) and u = Spm(ϕ) : X → Y the morphism associated to ϕ. (i) If A is (via ϕ) a finite B-algebra, then u is finite, and for any open subset V of Y , the morphism u−1 (V ) → V , induced by u, is also finite. (ii) Conversely, if u is finite, then A is (via ϕ) a finite B-algebra. Proof. (i) The first part is clear. As for the second, it suffices to prove the result in the case where V = D(g), g ∈ B. Then u−1 (V ) = D(h) where h = ϕ(g). Let f1 , . . . , fm ∈ A be a system of generators of A viewed as a B-module. If f ∈ A, then there exist t1 , . . . , tm ∈ B such that:
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195
f = ϕ(t1 )f1 + · · · + ϕ(tm )fm . Since regular functions on D(h) are of the form x → f (x)/hn (x), they can be written as: n
n
x → [t1 (u(x))/ (g(u(x))) ]f1 (x) + · · · + [tm (u(x))/ (g(u(x))) ]fm (x). Thus the restriction to D(h) of the fi ’s generates A(D(h)) as a A(D(g))module. (ii) By the second part of (i), we can write Y = D(g1 ) ∪ · · · ∪ D(gm ) with gi ∈ B, and the ring Ahi is a finitely generated Bgi -module where hi = ϕ(gi ). Denote by (fij /hni i )j a finite system of generators of Ahi . Multiplying the fij by some power of hi , we may assume that all the ni ’s are equal to the some integer n. Let f ∈ A, and fi its restriction to D(hi ). We have fi = ϕ(tij )fij /hni ⇒ hni f = ϕ(tij )fij . j
j
Since the set of D(gi ) = D(gin ) covers Y , there exist s1 , . . . , sm ∈ B such that n . But 1 = s1 g1n + · · · + sm gm ϕ(si gin )f = ϕ(si tij )fij , j
so we have f=
ϕ(si tij )fij .
i,j
Hence A is a finite B-algebra. 15.3.3 Corollary. Let u : X → Y be a morphism of varieties. The following conditions are equivalent: (i) u is finite. (ii) For all affine open subset V ⊂ Y , u−1 (V ) is an affine open subset of X and A(u−1 (V )) is a finite A(V )-algebra. Proof. This is clear by 15.2.3 and 15.3.2. 15.3.4 Proposition. Let u : X → Y be a finite morphism of varieties. (i) The map u is closed. (ii) For all y ∈ Y , the set u−1 (y) is finite. Proof. (i) Let (Vi )i∈I be a covering of Y by affine open subsets. If F is a closed subset of X, we have u(F ) ∩ Vi = u(F ∩ u−1 (Vi )). It suffices therefore to show that u(F ) ∩ Vi is closed in Vi for all i. So we are reduced to the case where X = Spm(A) and Y = Spm(B) are affine and u = Spm(ϕ) for some ϕ ∈ Homalg (B, A).
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Let Z = V(a) be a closed subset of X and b = ϕ−1 (a). The points of u(Z) are ϕ−1 (m), with m a maximal ideal of A containing a. Identifying B/b with a subring of A/a via ϕ, A/a is a finitely generated B/b-module. Now by 3.3.2 and 3.3.3, if n ∈ Spm(B/b), then n = m ∩ (B/b) for some m ∈ Spm(A/a). Let m ∈ Spm(A) and n ∈ Spm(B) be such that m = m/a and n = n/b. Then n = ϕ−1 (m). It follows that u(Z) = V(b), and u(Z) is closed in X. (ii) Again, we may assume that X = Spm(A) and Y = Spm(B) are affine and we only need to consider y ∈ u(X). By (i), u(X) = V(b) where b = ker ϕ. Identify B/b as a subring of A as above. According to 3.3.7, there is a finite number of maximal ideal of A lying above a maximal ideal of B/b. So the result follows. 15.3.5 Proposition. (i) The composition of finite morphisms is again finite. (ii) Let Y be a closed subvariety of X. The canonical injection Y → X is finite. (iii) Let u : X → Y be a finite morphism, Z be a closed subvariety of Y containing u(X), j : Z → Y the canonical injection and v : X → Z the morphism such that u = j ◦ v. Then v is finite. (iv) Let ui : Xi → Yi , i = 1, 2, be finite morphisms. Then the morphism v : X1 × X2 → Y1 × Y2 defined by (x1 , x2 ) → (u1 (x1 ), u2 (x2 )), is finite. Proof. (i) This is clear from 15.3.3 and 3.1.7. (ii) Since the intersection of an affine open subset U of X with Y is an affine open subset of Y which is closed in U , we are reduced to the case where X = Spm(A) is affine and Y = V(a) with a a radical ideal of A. The result now follows from the fact that the A-module A/a is generated by 1. (iii) Let z ∈ Z and Vz an affine open neighbourhood of z in Y . Then u−1 (Vz ) = v −1 (Z ∩ Vz ) is an affine open subset and the morphism u−1 (Vz ) → Vz induced by u is finite (15.3.3). Thus we may assume that X = Spm(A), Y = Spm(B) are affine and Z = Spm(B/b) where b is an ideal of B. Since u is finite, A is a finitely generated B-module via A(u) which is just the composition B → B/b → A. Hence A is also a finitely generated B/b-module. (iv) The affine open subsets V ×W , with V (resp. W ) an affine open subset of Y1 (resp. Y2 ), cover Y1 × Y2 . So we may assume that Xi = Spm(Ai ) and Yi = Spm(Bi ), i = 1, 2, are affine. If Ai is a finitely generated Bi -module, then it is clear that A1 ⊗k A2 is a finitely generated B1 ⊗k B2 -module. 15.3.6 Remark. Let Y be a subvariety of X. In general, the canonical injection Y → X is not finite. For example, the injection k \ {0} → k is not finite since its image is not closed (15.3.4). 15.3.7 Example. Let us consider the morphism u : k → k2 , t → (t2 , t3 ), in 11.5.7. Recall that its image is C = V(T13 − T22 ), and A(C) = k[T 2 , T 3 ]. Let v : k → C be the morphism induced by u. Since the A(C)-algebra k[T ]
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197
is finite, v is finite. The canonical injection j : C → k2 is finite by 15.3.5 (ii), and so by 15.3.5 (i), u = j ◦ v is finite. 15.3.8 Corollary. Let u : X → Y be a finite dominant morphism of irreducible varieties. (i) u is surjective and R(X) is a finite algebraic field extension of R(Y ). In particular, dim X = dim Y . (ii) If C ⊂ X is nowhere dense in X, then u(C) is nowhere dense in Y . Proof. (i) Since u(X) is dense in Y and closed (15.3.4), u is surjective. Let V be a non-empty affine open subset of Y . Then A(u−1 (V )) is a finitely generated A(V )-module. Let x1 , . . . , xn be a system of generators, then each xi is integral over A(V ). This implies that the xi ’s are algebraic over R(Y ) = Fract(A(V )) (12.7.6). Since R(X) = Fract(A(u−1 (V ))), the result follows from 14.1.2. (ii) Since u(C) = u(C), we may assume that C is closed. It suffices then to prove that if Z is an irreducible component of C (so Z is closed in X by 1.1.5), then u(Z) is nowhere dense in Y . Since u(Z) is an irreducible closed subset of Y (1.1.7 and 15.3.4), it follows from 15.3.5 that the restriction Z → u(Z) of u is a surjective finite morphism. According to (i) and 14.1.6, dim u(Z) = dim Z < dim X = dim Y. Using 14.1.6 again, we deduce that u(Z) is nowhere dense in Y . 15.3.9 Corollary. Let u : X → Y be a finite morphism of varieties. If Y is complete, then X is complete. Proof. Let Z be a variety, p : X × Z → Z and q : Y × Z → Z the canonical projections. The morphism v : X × Z → Y × Z defined by (x, z) → (u(x), z), satisfies p = q ◦ v. By 15.3.5 (iv), v is finite. So v is a closed map (15.3.4), and since q is closed by hypothesis, so is p. Thus X is a complete variety.
15.4 Factorization and applications 15.4.1 Proposition. let X be an irreducible affine variety of dimension n. There exists a surjective finite morphism u : X → kn . Proof. Let X = Spm(A), where A is a finitely generated k-algebra which is an integral domain. According to 6.2.2 and 14.1.2, there is a subalgebra B of A, isomorphic to k[T1 , . . . , Tn ], such that A is integral over B. So A is a finite Balgebra (3.3.5), and the morphism u : X → Spm(B) = kn which corresponds to the canonical injection B → A, is then a finite morphism (15.3.2). It is dominant by 6.5.9, and so it is surjective (15.3.4). 15.4.2 Proposition. Let u : X → Y be a dominant morphism of irreducible varieties.
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(i) There exist m ∈ N, V ⊂ Y a non-empty affine open subset and a covering (Ui )1ir of u−1 (V ) by affine open subsets such that the restriction ui : Ui → V factors through v
pr
i Ui −→ V × km −→ V,
where pr is the canonical surjection and vi a surjective finite morphism. (ii) The set u(X) contains a non-empty open subset of Y . Proof. We only need to prove (i). Given W = Spm(B) a non-empty affine open subset of Y , u−1 (W ) is a non-empty open subset of X since u is dominant. So u−1 (W ) is the finite union of non-empty affine open subsets U1 , . . . , Ur . Again, since u is dominant, the u(Ui )’s are dense in Y . Let Ui = Spm(Ai ) where Ai is a finitely generated k-algebra which is an integral domain. The morphism B → Ai induced by u is injective (6.5.9). There exist, by 6.2.9, si ∈ B \ {0} and a subalgebra Ci of Ai containing B such that Ci B[T1 , . . . , Tmi ] = B ⊗k k[T1 , . . . , Tmi ] and Ai [1/si ] is a finitely generated Ci [1/si ]-module. The quotient field of Ai [1/si ] is Li = Fract(Ai ) and the quotient field of Ci is isomorphic to K(T1 , . . . , Tmi ) = Li where K = Fract(B). It follows that Li and Li have the same transcendence degree over k (3.3.4 and 6.2.3). This degree is equal to dim Ui = dim X (14.1.2). We deduce therefore that all the mi ’s are equal to some integer m. Let s = s1 · · · sr . Then Ci [1/s] = B[1/s] ⊗k k[T1 , . . . , Tm ]. Take V = Spm(B[1/s]) = DY (s) , Ui = Spm(Ai [1/s]) , 1 i r. Then we are done since the morphism vi is the one associated to the injection Ci [1/s] → Ai [1/s]. 15.4.3 Proposition. Let u : X → Y be a morphism of varieties and C a constructible subset of X. Then u(C) is constructible. Proof. Let X1 , . . . , Xr be the irreducible components of X. Since u(C) = u(C ∩ X1 ) ∪ · · · ∪ u(C ∩ Xr ) and C ∩ Xi is constructible in Xi , we may assume that X is irreducible. We shall proceed by induction on the dimension n of X. The case n = 0 is obvious. Since C is Noetherian, it is the finite union of its irreducible components. They are constructible in X since they are closed in C. Thus we may further assume that C is irreducible. So C is an irreducible closed subvariety of X. Suppose that C = X. Then dim C < X and C is constructible in C. By the induction hypothesis, u(C) is constructible. Suppose that C = X. It suffices to show that u(C) is constructible in u(X). So we may assume that Y is irreducible and u is dominant. Since C is constructible and dense in X, C contains a non-empty open subset U of X (1.4.5), and the restriction of u to U is still dominant. By 15.4.2, u(U ) ⊂ u(C) contains a non-empty open subset V of Y . So we have:
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u(C) = V ∪ u(C ∩ u−1 (Y \ V )). The set u−1 (Y \ V ) is a closed subvariety of X distinct from X since u is dominant and Y \ V is nowhere dense in Y . Moreover, C ∩ u−1 (Y \ V ) is constructible in u−1 (Y \ V ). By the induction hypothesis, u(C ∩ u−1 (Y \ V )) is constructible in Y , and we have finished our proof.
15.5 Dimension of fibres of a morphism 15.5.1 Let u : X → Y be a morphism of varieties and X , Y subvarieties of X and Y such that u(X ) ⊂ Y . We say that X dominates Y if the morphism X → Y induced by u is dominant. 15.5.2 Lemma. Let u : X → Y be a surjective finite morphism of irreducible varieties and W an irreducible closed subvariety of Y . There is at least one irreducible component of u−1 (W ) which dominates W , and any such irreducible component has dimension dim W . Proof. Let Z1 , . . . , Zr be the irreducible components of u−1 (W ). Then u(Zi ) is closed in Y (15.3.4) and since u is surjective, W = u(Z1 ) ∪ · · · ∪ u(Zr ). The irreducibility of W implies that W = u(Zi ) for at least one i. Now if W = u(Zi ), then 15.3.5 implies that the morphism Z → W induced by u is finite, and so dim W = dim Zi (15.3.8). 15.5.3 Theorem. Let u : X → Y be a dominant morphism of irreducible varieties. (i) We have dim X dim Y . (ii) If W is an irreducible closed subvariety of Y and Z an irreducible component of u−1 (W ) which dominates W . Then (2)
dim Z dim W + dim X − dim Y.
(iii) There exists a non-empty open subset V of Y such that for all irreducible closed subvariety W of Y verifying W ∩ V = ∅, there is at least one irreducible component of u−1 (W ) which dominates W , and for such an irreducible component Z, we have (3)
dim Z = dim W + dim X − dim Y.
Proof. Since u is dominant, the comorphism (of fields) R(Y ) → R(X) is injective (12.7.11). Therefore, dim Y = tr degk R(Y ) tr degk R(X) = dim X. By 15.4.2 (i), there exist an integer m ∈ N, a non-empty open subset V ⊂ Y and a covering (Ui )1ir of u−1 (V ) by affine open subsets such that u|Ui = pr ◦vi where vi : Ui → V × km is a surjective finite morphism and pr : V × km → V is the canonical surjection. Since vi is finite and surjective, we have m = dim X − dim Y (15.3.8). Let W be an irreducible subvariety of Y verifying W ∩ V = ∅. Then (W ∩ V ) × km is an irreducible closed subvariety of dimension m + dim W
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in V × km . Applying 15.5.2 to vj , we obtain that there is at least one irreducible component of vi−1 ((V ∩ W ) × km ) which dominates W , and such an irreducible component has dimension m + dim W . Since u−1 (V ∩ W ) is the union of vi−1 ((V ∩ W ) × km ), 1 i r, we have proved that there exists an irreducible component of u−1 (W ) which dominates W . If Z is such an irreducible component, then Z ∩ Ui is dense in Z for some i and Z ∩ Ui dominates (V ∩ W ) × km . Hence dim Z = m + dim W . So we have part (iii). To prove (ii), replace Y by an affine open subset V which meets W , and replace X by u−1 (V ), we may assume that Y is affine. Let s = codimY (W ). According to 14.3.5, there exist f1 , . . . , fs ∈ A(Y ) such that W is an irreducible component of V(f1 , . . . , fs ). Let gi = fi ◦ u ∈ A(X). If Z is an irreducible component of u−1 (W ) dominating W , then Z ⊂ T = V(g1 , . . . , gs ). We claim that Z is an irreducible component of T . Let Z be an irreducible component of T containing Z, then W = u(Z) ⊂ u(Z ) ⊂ V(f1 , . . . , fs ). Since W is an irreducible component of the latter, we have u(Z ) ⊂ W and so Z ⊂ Z ⊂ u−1 (W ). Since Z is an irreducible component of u−1 (W ), we have Z = Z as claimed. From our claim, we deduce that codimX (Z) s = codimY (W ) (14.2.6). Hence we have (2). 15.5.4 Corollary. Let u : X → Y be a dominant morphism of irreducible varieties. (i) If y ∈ u(X), each irreducible component of u−1 (y) has dimension at least dim X − dim Y . (ii) There exists a non-empty open subset of V ⊂ u(X) such that for any y ∈ V , each irreducible component of u−1 (y) has dimension dim X − dim Y . 15.5.5 Corollary. Let u : X → Y be a morphism of varieties. (i) If dim u−1 (y) r for all y ∈ Y , then dim X r + dim Y . (ii) If u is dominant and dim u−1 (y) = r for all y ∈ u(X), then dim X = r + dim Y . Proof. (i) Let Y1 , . . . , Yn be the irreducible components of Y and for 1 i n, (Xij )j be the irreducible components of u−1 (Yi ). Let z ∈ u(Xij ). Applying 15.5.4 to the restriction Xij → u(Xij ), we have dim Xij dim u(Xij ) + dim u−1 (z) r + dim Y. Since X is the union of the Xij ’s, the result follows. (ii) Let us conserve the notations above andlet Yi be an irreducible component of Y of dimension dim Y . Since Yi \ ( j=i Yj ) is open and nonempty in Y (1.3.6), the restriction u−1 (Yi ) → Yi is dominant. Removing certain Xij ’s if necessary, we may assume that all the Xij ’s dominate Yi . If dim Xij < r + dim Y for all j, then applying 15.5.4 (ii) to the restriction Xij → Yi , we can find points y such that dim u−1 (y) < r which contradicts our hypothesis. Hence we have dim X = r + dim Y .
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201
15.5.6 Corollary. Let u : X → Y be a dominant birational morphism of irreducible varieties. There exists a non-empty open subset V of Y such that the restriction u−1 (V ) → V of u is an isomorphism. Proof. By our hypothesis, dim X = dim Y . Replace Y by an affine open subset W and X by u−1 (W ), we are reduced to the case where Y = Spm(B) is affine. Let U = Spm(A) be a non-empty affine open subset of X (so u(U ) is dense in Y ) and W = u(X \ U ). If T is an irreducible component of X \ U , then by 14.1.6, dim T < dim X = dim Y . Thus dim Z < dim Y for any irreducible component Z of W (15.5.3). It follows from 14.1.6 that there exists g ∈ B such that W ∩ DY (g) = ∅, and so u−1 (DY (g)) ⊂ U . If f = g ◦ u ∈ A, then u−1 (DY (g)) = DU (f ). Replacing Y by DY (g) and X by DU (f ), we may suppose that X = Spm(A) and Y = Spm(B) are affine where A, B are integral domains, and the comorphism ϕ : B → A extends to an isomorphism of fields Fract(B) → Fract(A). It follows that if a1 , . . . , am ∈ A are such that A = k[a1 , . . . , am ], then there exist b1 , . . . , bm ∈ B and s ∈ B \ {0} such that ai = bi /s for 1 i m. So A[1/s] and B[1/s] are isomorphic. Thus V = DY (s) works. 15.5.7 Let E be a topological space and f : E → N a map. We say that f is upper semi-continuous if {x ∈ E; f (x) n} is closed in E for all n ∈ N. Theorem. Let u : X → Y be a morphism of irreducible varieties. For x ∈ X, set e(x) to be the maximum dimension of any irreducible component of dim u−1 (u(x)) containing x. Then the map e : X → N is upper semicontinuous. Proof. Replacing Y by u(X), we may assume that u is dominant. Set r = dim X − dim Y 0 (15.5.3), and for n ∈ N, Sn (u) = {x ∈ X ; e(x) n}. We proceed by induction on p = dim Y . The case p = 0 is obvious since Sn (u) is the union of irreducible components of X of dimension at least n. If n r, then Sn (u) = X (15.5.4 (i)). So suppose that n > r. By 15.5.4, there is a non-empty open subset V of Y such that Sn (u) ⊂ X \ u−1 (V ). Let W1 , . . . , Ws be the irreducible components of Y \ V . We have dim Wj < dim Y (14.1.6). For 1 j s, let (Zij )j be the irreducible components of u−1 (Wi ). Denote by vij : Zij → Wi the restriction of u. By the induction hypothesis, the Sn (vij )’s are closed in Zij , and so in X. But if x ∈ X \u−1 (V ), any irreducible components of u−1 (u(x)) containing x is −1 (vij (x)). contained in one of the Zij ’s, so it is an irreducible component of vij Thus Sn (u) is the union of the Sn (vij ), and so Sn (u) is closed in X. 15.5.8 Corollary. Let X, Y be varieties. The canonical projections pr1 : X × Y → X, pr2 : X × Y → Y are open.
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Proof. Let X1 , . . . , Xm (resp. Y1 , . . . , Yn ) be the irreducible components of X (resp. Y ). If U is open in X × Y , then pr1 (U ) is open if pr1 ((Xi × Y ) ∩ U ) = Xi ∩ pr1 (U ) is open in Xi for all i. Moreover, pr1 (U ∩ (Xi × Y )) = pr1 (U ∩ (Xi × Y1 )) ∪ · · · ∪ pr1 (U ∩ (Xi × Yn )). So we may assume that X, Y are irreducible. Let U be an open subset of X × Y and F = X \ pr1 (U ). If x ∈ F and −1 z ∈ pr−1 1 (x), then pr1 (x) = {x} × Y is the unique irreducible component of −1 pr1 (pr1 (z)) ∩ ((X × Y ) \ U ) containing z. Its dimension is dim Y . Now if x ∈ F , U ∩ pr−1 1 (x) is not empty, so the irreducible components of (x) are all of dimension strictly less than dim Y (14.1.6). ((X × Y ) \ U ) ∩ pr−1 1 (F ) is the set of points z of the variety (X × Y ) \ U Hence F × Y = pr−1 1 such that e(z) dim Y (with respect to the restriction of pr1 ). By 15.5.7, F × Y is closed, and so F is closed (12.4.5). 15.5.9 Examples. 1) Let X ⊂ k3 be the cone defined by the equation x1 x3 − (x21 + x22 ) = 0 and u : X → k2 = Y the projection (x1 , x2 , x3 ) → (x1 , x2 ). The morphism u is dominant and birational since k(T1 , T2 , (T12 + T22 )T1−1 ) = k(T1 , T2 ). Clearly V = D(T1 ) ⊂ Y satisfies the conclusion of 15.5.6. Denote by W the line x1 = 0, which is just the complement of V in k2 . The set u−1 (W ) is the line defined by x1 = 0 and x2 = 0 in k3 . Thus u(u−1 (W )) = {(0, 0)}, and no irreducible components of u−1 (W ) dominates W . Moreover dim u−1 (W ) = 1 > dim X − dim Y = 0. Since u−1 (W ) = u−1 (0, 0), we see that, in 15.5.4, we can have dim u−1 (y) > dim X − dim Y . 2) Let us consider the following morphism v : X = k2 → k3 , (t1 , t2 ) → (t21 − 1, t1 (t21 − 1), t2 ). The image Y of v is the cylinder x22 − x21 − x31 = 0. Let u : X → Y be the morphism induced by v. It is a finite morphism. The image W of the morphism w : k → Y , t → (t2 − 1, t(t2 − 1), t), is irreducible of dimension 1. We verify easily that the diagonal ∆ of k and the point (1, −1) are the irreducible components of u−1 (W ). Clearly the irreducible components consisting of the point (1, −1) does not dominate W since dim W = 1. Note that u(k2 \ ∆) = (Y \ W ) ∪ {(0, 0, 1)}. Thus the image of the open set 2 k \ ∆ is not open. So a surjective finite morphism is not necessarily open.
15.6 An example
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15.6 An example 15.6.1 For p, q ∈ N∗ , let Mp,q be the set of p by q matrices with coefficients in k and we write Mp for Mp,p . Endow Mp,q with its natural structure of irreducible affine algebraic variety of dimension pq. Then the map Mr,p × Mp,q × Mq,s → Mr,s , (A, B, C) → ABC is a morphism. 15.6.2 The group GLn (k) is a principal open subset of Mn (11.7.5), so it is an irreducible affine variety of dimension n2 . The map GLn (k) → GLn (k), A → A−1 , is an automorphism of the variety GLn (k). 15.6.3 Let r ∈ {0, 1, . . . , min(p, q)}. Denote by Cr = {A ∈ Mp,q ; rk(A) r} , Cr = {A ∈ Mp,q ; rk(A) = r}. In other words, A ∈ Cr if and only if all the (r + 1)-minors of A are zero. It follows that Cr is closed in Mp,q . Similarly, if A ∈ Cr , then A ∈ Cr if and only if at least one of the r-minors of A is non-zero. Thus Cr is open in Cr . Let Ir ∈ Mr be the identity matrix and Ir 0 ∈ Mp,q . J= 0 0 It is well-known that the image of the morphism w : Mp × Mq → Mp,q , (P, Q) → P JQ, is Cr . So Cr is irreducible in Mp,q . Let u : Mp × Mq → Cr be the morphism induced by w. We have u(GLp (k) × GLq (k)) = Cr . Let v : GLp (k) × GLq (k) → Cr be the morphism induced by u. Let A ∈ Cr and (U, V ) ∈ GLp (k) × GLq (k) be such that A = U JV . If (P, Q) ∈ GLp (k) × GLq (k), then (P, Q) ∈ v −1 (A) ⇔ (U −1 P, QV −1 ) ∈ v −1 (J). We deduce that all the fibres of v are isomorphic. Writing AB XY P = , Q= CD Z T where A, X ∈ Mr , we have P JQ =
AX AY CX CY
.
Hence (P, Q) ∈ v −1 (J) ⇔ A ∈ GLr (k) , X = A−1 , C = 0 , Y = 0.
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It is now clear that the map GLr (k) × GLp−r (k) × GLq−r (k) × Mr,p−r × Mq−r,r → v −1 (J) −1 AD A 0 (A, B, C, D, E) → , 0 B E C is an isomorphism. We deduce from the above discussion that for any A ∈ Cr , v −1 (A) is irreducible and: dim v −1 (A) = r2 + (p − r)2 + (q − r)2 + r(p − r) + r(q − r). Since Cr is open in the irreducible variety Cr , we have dim Cr = dim Cr = r(p + q − r).
References and comments • [5], [19], [22], [26], [37], [40], [78]. Many results of this chapter are due to Chevalley, for example: 15.4.2, 15.4.3, 15.5.7.
16 Tangent spaces
The reader may have come across the notion of a tangent space in differential geometry. We define here the analogous algebraic object, the Zariski tangent space, for an algebraic variety. We shall see later that the Zariski tangent space appears in relating algebraic groups and Lie algebras.
16.1 A first approach 16.1.1 Given P ∈ k[T1 , . . . , Tn ] = k[T] and x ∈ kn , set: Dx (P ) =
∂P ∂P (x)T1 + · · · + (x)Tn . ∂T1 ∂Tn
The map P → Dx (P ) is k-linear, and if P, Q ∈ k[T], then (1)
Dx (P Q) = P (x)Dx (Q) + Q(x)Dx (P ).
16.1.2 Let X be a closed subvariety of kn , a = I(X) the associated radical ideal in k[T] and x = (x1 , . . . , xn ) ∈ X. Denote by nx = I({x}) the maximal ideal of A = A(X) = k[T]/a associated to x. 16.1.3 Let ax be the ideal of k[T] generated by the Dx (P ), P ∈ a. Then by (1), if (Pi )i∈I is a system of generators of a, then (Dx (Pi ))i∈I is a system of generators of ax . We call the tangent space of X at x, denoted by Tanx (X), the vector subspace V(ax ) of kn . If r is the rank of the set of linear forms Dx (P ), P ∈ a, then the dimension of Tanx (X) is n − r. Let {Q1 , . . . , Qr } be a basis of the k-vector space spanned by the Dx (P ), P ∈ a. There exist indices i1 , . . . , in−r such that {Q1 , . . . , Qr , Ti1 , . . . , Tin−r } is a basis of the space of homogeneous polynomials of degree 1. Clearly we can identify k[T]/ax with k[Ti1 , . . . , Tin−r ]. It follows that ax is prime and the algebra of regular functions on Tanx (X) is k[T]/ax .
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16.1.4 Now let L ⊂ kn be a line containing x and v = (v1 , . . . , vn ) ∈ kn be such that L is the set of points of the form x + tv, t ∈ k. Let (Pi )1is be a system of generators of a. Then the points in L ∩ X are obtained by solving the following system of equations in t: (2)
P1 (x + tv) = · · · = Ps (x + tv) = 0. For 1 i n, we have Pi (x + tv) = t
n ∂P i (x)vj + t2 Qi (t) j=1 ∂Tj
where Qi ∈ k[T ]. Thus, t = 0 is a “multiple root” of (2) if and only if, ∂Pi ∂Pi (x)v1 + · · · + (x)vn = 0 ∂T1 ∂Tn for 1 i s. Hence t = 0 is a multiple root of (2) if and only if v ∈ Tanx (X). 16.1.5 Let P, Q ∈ k[T] be such that P − Q ∈ a. By the definition of Tanx (X), the restriction of Dx (P ) − Dx (Q) to Tanx (X) is zero. So we can define a linear map Dx : A → (Tanx (X))∗ , the dual space of Tanx (X), by setting Dx (f ) = Dx (F )|Tanx (X) where f ∈ A and F ∈ k[T] is a representative of f . The map Dx is obviously surjective. Furthermore, since A = k ⊕ nx and Dx (λ) = 0 for any λ ∈ k, Dx induces a linear surjection dx : nx → (Tanx (X))∗ . Proposition. The kernel of dx is n2x . Thus dx induces an isomorphism nx /n2x → (Tanx (X))∗ of k-vector spaces. Proof. Let Nx = k[T](T1 − x1 ) + · · · + k[T](Tn − xn ) be the maximal ideal of k[T] associated to x. Then a ⊂ Nx and nx = Nx /a. It is clear from (1) that n2x ⊂ ker dx . Conversely, let f ∈ ker dx and F ∈ Nx be a representative of f . Since I(Tanx (X)) = ax (16.1.3), there exist F1 , . . . , Fs ∈ a and λ1 , . . . , λs ∈ k such that Dx (F ) = λ1 Dx (F1 ) + · · · + λs Dx (Fs ). Set G = F − λ1 F1 − · · · − λs Fs , then Dx (G) = 0. So ∂G ∂G (x) = · · · = (x) = 0. ∂T1 ∂Tn Since G(x) = 0, it follows from Taylor’s formula that G is contained in the ideal generated by (Ti − xi )(Tj − xj ), 1 i, j n, which is N2x . Finally, G is also a representative of f because F1 , . . . , Fs ∈ a. Hence f ∈ n2x .
16.2 Zariski tangent space
207
16.1.6 It follows from the preceding discussion that we can identify Tanx (X) with (nx /n2x )∗ . We now give another interpretation of Tanx (X). Let ϕ : A → OX,x be the map sending f ∈ A to its germ at x, and mX,x the maximal ideal of OX,x . We have seen in 12.8.3 that ϕ induces a local isomorphism ψ : Anx → OX,x . By 2.6.12, the vector spaces nx /n2x and nx Anx /n2x An are isomorphic. So we have proved: Proposition. The vector spaces Tanx (X) and (mX,x /m2X,x )∗ are isomorphic.
16.2 Zariski tangent space 16.2.1 Using the discussion in 16.1, we can define the tangent space at a point of an arbitrary variety as follows: Definition. Let X be a variety and x ∈ X. We define the Zariski tangent space at x, denoted by Tx (X), the vector space (mX,x /m2X,x )∗ . 16.2.2 Let us denote by ρ : mX,x → mX,x /m2X,x the canonical surjection. If s ∈ OX,x , we denote by s(x) its value at x. Recall that OX,x = k ⊕ mX,x . Let us consider k as an OX,x -module via the homomorphism defined by the evaluation at x, that is s.λ = s(x)λ for all s ∈ OX,x , λ ∈ k. A k-derivation of OX,x in k (2.8.1) is then a k-linear map δ : OX,x → k verifying δ(st) = s(x)δ(t) + t(x)δ(x) where s, t ∈ OX,x . Such a derivation shall be called a point derivation of OX,x , and we denote by Derxk (OX,x , k) the set of these derivations. In the same way, if U is an affine open subset containing x, we can consider k as an A(U )-module by setting f.λ = f (x)λ, for f ∈ A(U ) and λ ∈ k. We shall also call a k-derivation of A(U ) to k, a point derivation of A(U ), that is, a k-linear map ∆ : A(U ) → k verifying for f, g ∈ A(U ), ∆(f g) = f (x)∆(g) + g(x)∆(f ). We shall denote by Derxk (A(U ), k) the set of these derivations. 16.2.3 Let λ ∈ Tx (X) and define a map Lλ : OX,x → k by Lλ |k = 0 , Lλ |mX,x = λ ◦ ρ. If s ∈ OX,x , then s − s(x) ∈ mX,x . We deduce easily that if s, t ∈ OX,x , then (3)
Lλ (st) = s(x)Lλ (t) + t(x)Lλ (s).
Thus Lλ ∈ Derxk (OX,x , k). Conversely, let δ ∈ Derxk (OX,x , k). If s is the germ at x of the constant function 1, then δ(s) = δ(ss) = s(x)δ(s) + s(x)δ(s) = 2δ(s).
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Hence δ(s) = 0. Further, if s, t ∈ OX,x verify s(x) = t(x) = 0, then δ(st) = 0. Thus δ|m2X,x = 0. It follows that there is a linear form λ on mX,x /m2X,x such that δ|mX,x = λ ◦ ρ = Lλ . Hence δ = Lλ . We have therefore identified the tangent space Tx (X) of X at x with the vector space Derxk (OX,x , k). An element of Derxk (OX,x , k) shall be called a tangent vector of X at x. 16.2.4 Let U be an affine open subset of X containing x and let δ ∈ Derxk (OX,x , k). For f ∈ A(U ), set ∆U δ (f ) = δ(fx ), where fx denotes the germ ∈ Derxk (A(U ), k). of f at x. We verify easily that ∆U δ x Conversely, let ∆ ∈ Derk (A(U ), k). If g ∈ A(U ) is such that its germ at x is 0, then by 11.7.3, there exists f ∈ A(U ) verifying f (x) = 0 and f g = 0. So 0 = ∆(f g) = f (x)∆(g) + g(x)∆(f ) = f (x)∆(g). Hence ∆(g) = 0. Now let s ∈ OX,x , there exists f ∈ A(U ) such that f (x) = 0 and s is the germ of some function g/f on D(f ). Let us assume that s is also the germ of a function g1 /f1 with f1 (x) = 0. Then gf1 − g1 f is identically zero in a neighbourhood of x. From the preceding paragraph, we have ∆(gf1 −g1 f ) = 0, which implies that g(x)∆(f1 ) − f (x)∆(g1 ) = g1 (x)∆(f ) − f1 (x)∆(g). A simple computation gives: f12 (x)[f (x)∆(g) − g(x)∆(f )] = f 2 (x)[f1 (x)∆(g1 ) − g1 (x)∆(f1 )]. From this, we deduce that if h = g/f is defined on D(f ) where f, g ∈ A(U ) and f (x) = 0, we can define a linear form δ on OX,x by setting: δ(hx ) = f −2 (x)[f (x)∆(g) − g(x)∆(f )] where hx is the germ of h at x. We now verify easily that δ ∈ Derxk (OX,x , k) and ∆U δ = ∆. We have therefore shown that the tangent space Tx (X) can be identified with Derxk (A(U ), k). In particular, if U is an affine open subset of X containing x, then we can identify Tx (X) with Tx (U ). 16.2.5 We shall give yet another interpretation of Tx (X). Let U be an affine open subset of X containing x. Consider k as an A(U )-module as in 16.2.2, and denote it by kx . Let ΩU = Ωk (A(U )) be the module of differentials of A(U ) over k (see 2.9.3). Recall from 2.9.5 that the map f → f ◦ dA(U )/k is a k-linear isomorphism from HomA(U ) (ΩU , kx ) to Derk (A(U ), kx ) = Derxk (A(U ), k). Let nx = I({x}) be the maximal ideal of A(U ) associated to x, and set
16.3 Differential of a morphism
209
ΩUx = ΩU /nx ΩU . Since A(U ) = k ⊕ nx as vector spaces, we can identify the A(U )-module kx with A(U )/nx via the canonical surjection A(U ) → A(U )/nx . So the vector spaces ΩUx and kx ⊗A(U ) ΩU are isomorphic. On the other hand, it is easy to verify that the map ΩU → kx ⊗A(U ) ΩU , α → 1 ⊗ α, induces a k-linear bijection Ψ : Homk (kx ⊗A(U ) ΩU , k) → HomA(U ) (ΩU , kx ), defined by (Ψ (u))(α) = u(1 ⊗ α) for all u ∈ Homk (kx ⊗A(U ) ΩU , k) and α ∈ ΩU . It follows from the preceding paragraphs that there is a k-linear isomorphism from Derxk (A(U ), k) to Homk (ΩUx , k). So by 16.2.4, we may also identify the tangent space Tx (X) with the dual of ΩUx .
16.3 Differential of a morphism 16.3.1 Let X, Y be varieties, u ∈ Mor(X, Y ) and x ∈ X. We saw in 12.8.6 that u induces a local morphism ux : OY,u(x) → OX,x . Since ux (mY,u(x) ) is contained in mX,x , we have ux (m2Y,u(x) ) ⊂ m2X,x . It follows that ux induces a map ux : mY,u(x) /m2Y,u(x) → mX,x /m2X,x . We shall call the dual map Tx (u) : Tx (X) → Tu(x) (Y ) of ux , the differential of the morphism u at x. It is also sometimes denoted by dux . To avoid confusions with the elements of Ωk (A(U )), we shall not use the notation dux in this chapter. A straightforward verification shows that if v : Y → Z is a morphism, then: Tx (v ◦ u) = Tu(x) (v) ◦ Tx (u). 16.3.2 Let us conserve the notations of 16.3.1. Let y = u(x), and θx : Tx (X) → Derxk (OX,x , k) , θy : Ty (Y ) → Deryk (OY,y , k) be the isomorphisms defined in 16.2.3. Denote by Tx (u) : Derxk (OX,x , k) → Deryk (OY,y , k) the unique linear map such that Tx (u) ◦ θx = θy ◦ Tx (u). If λ ∈ Tx (X) and µ = Tx (u)(λ), then θx (λ)|k = 0 , θx (λ)|mX,x = λ ◦ ρx , θy (µ)|k = 0 , θy (µ)|mY,y = µ ◦ ρy where ρx : mX,x → mX,x /m2X,x and ρy : mY,y → mY,y /m2Y,y denote the canonical surjections. Since µ = λ ◦ ux , we have θy (µ)|mY,y = λ ◦ ux ◦ ρy = λ ◦ ρx ◦ (ux |mY,y ). We deduce therefore that Tx (u) is the map: Derxk (OX,x , k) → Deryk (OY,y , k) , δ → δ ◦ ux .
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16.3.3 Let V be an affine open neighbourhood of y = u(x) in Y and U an affine open neighbourhood of x in X such that u(U ) ⊂ V . Denote by αx : Derxk (OX,x , k) → Derxk (A(U ), k) , αy : Deryk (OY,y , k) → Deryk (A(V ), k) the isomorphisms in 16.2.4. Let Tx (u) : Derxk (A(U ), k) → Deryk (A(V ), k) be the unique linear map such that αy ◦ Tx (u) = Tx (u) ◦ αx . It follows from 16.2.4 and 16.3.2 that if δ ∈ Derxk (OX,x , k) and g ∈ A(V ), (αy ◦ Tx (u)(δ))(g) = (αy ◦ δ ◦ ux )(g) = δ ◦ ux (gy ) = δ((g ◦ u)x ) = αx (δ)(g ◦ u). where gy is the germ of g at y. It follows that if A(u) is the comorphism of u, then Tx (u) : Derxk (A(U ), k) → Deryk (A(V ), k) , ∆ → ∆ ◦ A(u). Remark. When there is no confusion, we shall also denote Tx (u) and by Tx (u).
Tx (u)
16.3.4 Let us take X = km and Y = kn . Then A(X) = k[T1 , . . . , Tm ], A(Y ) = k[S1 , . . . , Sn ], Tanx (X) = km , Tany (Y ) = kn . Let P1 , . . . , Pn ∈ A(X) be such that u(x) = (P1 (x), . . . , Pn (x)) for all x ∈ X. We have the following bijective linear maps: βx : Derxk (A(X), k) → Tanx (X) = km , ∆ → (∆(T1 ), . . . , ∆(Tm )) , βy : Deryk (A(Y ), k) → Tany (Y ) = kn , ∆ → (∆(S1 ), . . . , ∆(Sn )). Let Fx (u) : km → kn be the linear map such that Fx (u) ◦ βx = βy ◦ Tx (u). If Q ∈ A(Y ) and ∆ ∈ Derxk (A(X), k), then 16.3.3 implies that: m ∂(Q ◦ u) (Tx (u)(∆)) (Q) = ∆(Q ◦ u) = (x)∆(Ti ) ∂Ti i=1 m n ∂Q ∂Pj (u(x)) (x) ∆(Ti ) = ∂Ti j i=1 j=1 ∂S n ∂Q m ∂P j (y) (x)∆(Ti ) . = j=1 ∂Sj i=1 ∂Ti
Hence Fx (u) is the linear map k
m
m m ∂P ∂P1 n → k , (v1 , . . . , vm ) → (x)vi , . . . (x)vi . i=1 ∂Ti i=1 ∂Ti n
Let Mx (u) = [aij ] ∈ Mn,m (k) be the matrix of Fx (u) with respect to the canonical bases of km and kn . Then for 1 i m, 1 j n, we have: aij =
∂Pj (x). ∂Ti
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211
16.3.5 Now assume that X is a closed subvariety of Y = kn . Let j : X → Y be the canonical injection, a = I(X) ⊂ A(Y ) = k[T1 , . . . , Tn ] and, for 1 i n, let ti be the image of Ti in A(X) = A(Y )/a. Let v = (v1 , . . . , vn ) ∈ Y . For P ∈ a and v ∈ Tanx (X), we saw in 16.1 that ∂P ∂P (x)v1 + · · · + (x)vn = 0. ∂T1 ∂Tn We can then define an element ∆v ∈ Derxk (A(X), k) by setting for f ∈ A(X) and P any representative of f in A(Y ), ∆v (f ) =
∂P ∂P (x)v1 + · · · + (x)vn . ∂T1 ∂Tn
Conversely, given ∆ ∈ Derxk (A(X), k). We have, for P ∈ A(Y ), ∆(P (t1 , . . . , tn )) =
∂P ∂P (x)∆(t1 ) + · · · + (x)∆(tn ). ∂T1 ∂Tn
In particular, if P ∈ a, then ∆(P (t1 , . . . , tn )) = 0. It follows that we have a bijective linear map γx : Derxk (A(X), k) → Tanx (X) , ∆ → (∆(t1 ), . . . , ∆(tn )). Let Gx (j) : Tanx (X) → Tany (Y ) be the unique linear map such that Gx (j) ◦ γx = βy ◦ Tx (j) where βy is as in 16.3.4. It follows immediately from the preceding discussion that Gx (j) is the canonical injection from Tanx (X) to kn = Tany (Y ). 16.3.6 Now let X and Y be closed subvarieties of km and kn respectively and denote by j1 : X → km and j2 : Y → kn the canonical injections. There exist P1 , . . . , Pn ∈ k[T1 , . . . , Tm ] (not necessarily unique in general) such that u(z) = (P1 (z), . . . , Pn (z)) for all z ∈ X. Let v : km → kn be the morphism defined by z → (P1 (z), . . . , Pn (z)). Then we have j2 ◦ u = v ◦ j1 . From this, we obtain the following commutative diagram: Hx (u)
Tanx (X) ⏐ γx ⏐
−−−−→
Tany (Y ) ⏐γy ⏐
Derxk (A(X), k) ⏐ ⏐ Tx (j1 )
−−−−→
Tx (u)
Deryk (A(Y ), k) ⏐ ⏐T (j ) y 2
Tx (v)
Derxk (k[T1 , . . . , Tm ], k) −−−−→ Deryk (k[S1 , . . . , Sn ], k) where Hx (u) is the unique linear map such that Hx (u) ◦ γx = γy ◦ Tx (u). Consequently, Hx (u) is the linear map
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Tanx (X) → Tany (Y ) , (v1 , . . . , vn ) →
n ∂P n ∂P 1 n (x)vi , . . . , (x)vi . i=1 ∂Ti i=1 ∂Ti
In particular, if : kn → km is the linear map whose matrix [aij ] with re∂Pi (x), then spect to the canonical bases of km and kn is defined by aij = ∂Tj
(Tanx (X)) ⊂ Tany (Y ) and (v) = Hx (u)(v) for all v ∈ Tanx (X). 16.3.7 Proposition. Let X be a subvariety of a variety Y , j : X → Y the canonical injection and x ∈ X. The map Tx (j) : Tx (X) → Tx (Y ) is injective. Proof. The set X is locally closed in Y (12.2.7), so it is the intersection of a closed subset F and an open subset U of Y . Let j1 : X → U , j2 : U → Y be the canonical injections. Since j = j2 ◦ j1 , we have Tx (j) = Tx (j2 ) ◦ Tx (j1 ). By 16.2.4 and 16.3.3, Tx (j2 ) is an isomorphism. So we are reduced to the case where X is closed in Y . We may assume that Y = Spm(A) is affine and X = Spm(A/a) where a is a radical ideal of A. By 16.3.3, if ϕ : A → A/a is the canonical surjection, then Tx (j) is the map Derxk (A/a, k) → Derxk (A, k), ∆ → ∆ ◦ ϕ. It is now clear that Tx (j) is injective. 16.3.8 Let U, V be affine varieties and (x, y) ∈ U × V . Recall from 11.8.1 that the map associating f ⊗ g ∈ A(U ) ⊗k A(V ) to the function (u, v) → f (u)g(v) on U × V is an isomorphism from A(U ) ⊗k A(V ) to A(U × V ). For D ∈ Derxk (A(U ), k) and D ∈ Deryk (A(V ), k), we define a map θD,D : A(U × V ) → k by setting, for f ∈ A(U ), g ∈ A(V ), θD,D (f ⊗ g) = D(f )g(y) + f (x)D (g). (x,y)
We verify easily that θD,D ∈ Derk (A(U × V ), k). So we have a linear map (x,y) θ : Derxk (A(U ), k)×Deryk (A(V ), k) → Derk (A(U ×V ), k), (D, D ) → θD,D . Since θD,D (f ⊗ 1) = D(f ) and θD,D (1 ⊗ g) = D (g), the map θ is injective. (x,y) Now let ∆ ∈ Derk (A(U × V ), k). For f ∈ A(U ) and g ∈ A(V ), we set: D(f ) = ∆(f ⊗ 1) , D (g) = ∆(1 ⊗ g). Then D ∈ Derxk (A(U ), k), D ∈ Deryk (A(V ), k) and ∆ = θD,D . We have therefore proved that θ is an isomorphism. Proposition. Let X, Y be varieties and (x, y) ∈ X × Y . Denote by j : X × {y} → X × Y , j : {x} × Y → X × Y the canonical injections. Then the map T(x,y) (X × {y}) × T(x,y) ({x} × Y ) → T(x,y) (X × Y ) (D, D ) → T(x,y) (j)(D) + T(x,y) (j )(D ) is a bijection. In particular, Tx (X)×Ty (Y ) and T(x,y) (X ×Y ) are isomorphic as k-vector spaces.
16.4 Some lemmas
213
Proof. By 16.2.4, we may assume that X and Y are affine. Let us identify X × {y} with X and {x} × Y with Y . The comorphism of j (resp. j ) is the map sending f ⊗ g to the function u → f (u)g(y) on X (resp. v → f (x)g(v) on Y ). It follows from 16.3.3 that T(x,y) (j)(D) = θD,0 and T(x,y) (j )(D ) = θ0,D . So the result follows.
16.4 Some lemmas 16.4.1 In this section, k is a commutative integral domain and K its quotient field. For λ ∈ k \ {0}, set Sλ = {λn ; n ∈ N} and kλ = Sλ−1 k. Since k is an integral domain, we may identify Sλ−1 k with the subring of K consisting of elements of the form λ−n a with n ∈ N and a ∈ k. For any k-module M , denote by Mλ the kλ -module kλ ⊗k M , which we may identify with Sλ−1 M (2.4.1). Let m, p ∈ N∗ and r ∈ {1, . . . , min(m, p)}. We define the following matrix: Ir 0 Jr = ∈ Mm,p (k). 0 0 16.4.2 Let Λ = [αij ] ∈ Mm,p (k) and (e1 , . . . , ep ) be the canonical basis of k p . Denote by Mk (Λ) or M(Λ), the quotient of k p by the submodule generated by the elements εi = αil e1 + · · · + αip ep , 1 i m. Lemma. (i) Let P ∈ GLm (k), Q ∈ GLp (k). Then M(P Λ) = M(Λ), and the k-modules M(ΛQ) and M(Λ) are isomorphic. (ii) There exist λ ∈ k \ {0}, P ∈ GLm (kλ ) and Q ∈ GLp (kλ ) such that Λ = P Jr Q, where r is the rank of the matrix Λ considered as an element of Mm,p (K). Proof. (i) Let P = [pij ], P Λ = [αij ] and for 1 i m, set εi = αi1 e1 + · · · + αip ep .
Denote by L (resp. L ) the submodule of k p generated by the εi (resp. εi ). Since εi = pi1 ε1 + · · · + pim εm for 1 i m, L ⊂ L. Now the same argument applied to P −1 implies that L ⊂ L . Therefore L = L and M(P Λ) = M(Λ). ] and for 1 i m and 1 j p, set: Next let Q = [qij ], ΛQ = [αij e1 + · · · + αip ep , ej = qj1 e1 + · · · + qjp ep . εi = αi1
So (e1 , . . . , ep ) is a basis of k p and for 1 i m, εi = αi1 e1 + · · · + αip ep . Let u be the automorphism of k p verifying u(ej ) = ej for 1 j p. Then u induces an isomorphism from M(Λ) to M(ΛQ). (ii) It is well-known that there exist P ∈ GLm (K) and Q ∈ GLp (K) such that Λ = P Jr Q. To obtain the result, it suffices to take λ such that the coefficients of P , Q, P −1 , Q−1 belong to kλ .
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16.4.3 Lemma. Let us conserve the notations of 16.4.2. (i) There exists λ ∈ k \ {0} such that M(Λ)λ is a free kλ -module of rank p − r. (ii) We can choose λ such that p − r images of the elements 1 ⊗ ei of kλ ⊗k k p form a basis of the kλ -module M(Λ)λ . Proof. Part (i) is clear by 16.4.2. As for part (ii), for 1 i p, denote by νi the canonical image of 1 ⊗ ei in M(Λ)λ . Let (f1 , . . . , fp−r ) be a basis of the kλ -module M(Λ)λ . Since the νi ’s generate M(Λ)λ , we may assume that there exists M = [βij ] ∈ Mp−r (kλ ) verifying det M = 0 and νj = β1j f1 + · · · + βp−r,j fp−r for 1 j p − r. Modify λ if necessary, we may assume that the coefficients of the matrix M −1 ∈ GLp−r (K) are in kλ . So part (ii) follows. 16.4.4 Let X be an irreducible affine variety and x ∈ X. The ring k = A(X) is an integral domain and recall from 12.7.6 that we can identify its quotient field K with the field of rational functions R(X) on X. Let Λ = [αij ] ∈ Mm,p (k) be as in 16.4.3, r its rank considered as an element of Mm,p (K) and let Λ(x) ∈ Mm,p (k) be the matrix [αij (x)]. Consider k as a k-module, denoted by kx , as in 16.2.2. Set Mk (Λ)x = Mk (Λ)/nx Mk (Λ). Then Mk (Λ)x = kx ⊗k Mk (Λ), and it is clear that we may identify Mk (Λ)x with Mk (Λ(x)) as k-vector spaces. Lemma. Let E = {x ∈ X; dimk Mk (Λ)x = p − r}. (i) E is a non-empty open subset of X. (ii) We have dimK K ⊗k Mk (Λ) = p − r. (iii) If x ∈ E, then there exists f ∈ k = A(X) verifying f (x) = 0, and Mk (Λ)f is a free kf -module of rank p − r. Proof. (i) Let Ls (resp. Ls (x)) be the set of square submatrices of size s in Λ (resp. Λ(x)). Then det Θ = 0 if Θ ∈ Ls with s > r, and there exists Θ ∈ Lr such that det Θ = 0. Similarly, since det Θ ∈ k, we have det A = 0 if A ∈ Ls (x) with s > r, and the set F of x ∈ X for which there exists A ∈ Lr (x) verifying det A = 0, is a non-empty open subset of X. Since we can identify Mk (Λ)x with Mk (Λ(x)), we have E = F. So we have proved part (i). (ii) This is clear because the rank of Λ in Mm,p (K) is r. (iii) Let x ∈ E. There exists A ∈ Lr (x) such that det A = 0. Reindexing if necessary, we may assume that A = [αij (x)]1i,jr and denote by Θ the matrix [αij ]1i,jr . We have Θ ∈ GLr (kf ) where f = det Θ. Let (e1 , . . . , ep ) be the canonical basis of k p and for 1 j p, νj the image of 1 ⊗ ej in Mk (Λ)f . Then for 1 i m, αi1 ν1 + · · · + αip νp = 0, or equivalently
16.5 Smooth points
215
αi1 ν1 + · · · + αir νr = −(αi,r+1 νr+1 + · · · + αip νp ). It follows from our choice of A that ν1 , . . . , νr are kf -linear combinations of νr+1 , . . . , νp . So by (ii), the elements 1 ⊗ νr+1 , . . . , 1 ⊗ νp form a basis of the K-vector space K ⊗kf Mk (Λ)f . Consequently νr+1 , . . . , νp are linearly independent over kf . Thus Mk (Λ)f is a free kf -module of rank p − r.
16.5 Smooth points 16.5.1 Proposition. Let X be a variety and x ∈ X. The dimension of Tx (X) is greater than or equal to the dimension of any irreducible component of X containing x. Proof. By 16.2.4 and 16.3.7, we may assume that X is irreducible and affine. Let X = Spm(A) and nx = I({x}). Since the k-vector spaces nx /n2x and mX,x /m2X,x are isomorphic, it suffices to show that dimk nx /n2x dim X. Let f1 , . . . , fr ∈ nx be such that their images in nx /n2x form a basis of the k-vector space nx /n2x . By 2.6.9, we have nx = Af1 + · · · + Afr . Consider the morphism u : X → kr , y → (f1 (y), . . . , fr (y)). If u(y) = 0, then nx ⊂ ny and so nx = ny , which in turn implies that x = y. For y ∈ X, denote by e(y) the maximum dimension of an irreducible component of u−1 (u(y)) containing y. By 15.5.7, the set of y ∈ X verifying e(y) 1 is closed in X, so the set U of y ∈ X such that e(y) = 0 is open in X. From the previous paragraph, we have x ∈ U . Let v : U → kr denote the restriction of u to U . Then dim v −1 (z) = 0 for all z ∈ v(U ). Thus dim X = dim U dim kr = r (15.5.4 (i)) which proves our result. 16.5.2 Definition. Let X be an irreducible variety of dimension n. A point x ∈ X is called smooth (or simple) if dim Tx (X) = n. The variety X is called smooth (or non-singular) if all its points are smooth. 16.5.3 Remarks. 1) When X is not irreducible, let dimx (X) denote the maximum dimension of irreducible components of X containing x. Then x is a smooth point of X if Tx (X) = dimx X. Similarly, the variety X is smooth if all its points are smooth. 2) We saw in 16.1 that the affine space kn is smooth. Similarly the projective space Pn is also smooth, because any point of Pn is contained in an open affine neighbourhood isomorphic to kn . By using the same argument, we see that the Grassmannian varieties Gn,r are smooth. 3) If x is a smooth point in an irreducible variety X, then we can show that the ring OX,x is factorial, and hence an integrally closed domain. 16.5.4 Proposition. Let X = Spm(A) be an irreducible affine variety, x ∈ X a smooth point, nx the maximal ideal of A associated to x, and f1 , . . . , fn ∈ nx be such that their images in nx /n2x form a basis of the kvector space nx /n2x . Then f1 , . . . , fn are algebraically independent over k, and
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so they form a transcendence basis of the field of rational functions R(X) over k. Proof. Consider the following morphism: u : X → kn , y → (f1 (y), . . . , fn (y)). Suppose that there exists P ∈ k[T1 , . . . , Tn ] \ {0} such that P (f1 , . . . , fn ) = 0, then u(X) ⊂ V(P ) and so dim u(X) n − 1 (14.2.3). Now proceed as in the proof of 16.5.1, we deduce that dim X n − 1 which is absurd. So the fi ’s are algebraically independent over k. The last part follows from 12.7.6 and 5.4.8. 16.5.5 Let X be an irreducible variety, x ∈ X and U an affine open subset of X containing x. Recall that ΩU = Ωk (A(U )) is the module of differentials of A(U ) over k. Theorem. Let X be an irreducible variety of dimension n. (i) Let x be a smooth point of X. There is an affine open neighbourhood U of x such that ΩU is a free A(U )-module of rank n. (ii) The set of smooth points in X is a non-empty open subset of X. Proof. Let U be an affine open subset of X. By 12.7.6, we may identify the quotient field of A(U ) with R(X). Further, the R(X)-vector spaces R(X)⊗A(U ) ΩU and Ωk (R(X)) are isomorphic (2.9.6). Since the characteristic of k is zero, it follows from 5.8.3 and 14.1.2 that dimR(X) Ωk (R(X)) = tr degk R(X) = n. So we may assume that X = Spm(A) is affine with A = k[T1 , . . . , Tp ]/a where a is a radical ideal of k[T1 , . . . , Tp ]. Let {P1 , . . . , Pm } be a system of generators ∂Pi ∈ Mm,p (A(X)) and r the rank of Λ considered as an for a, Λ = ∂Tj ij element of Mm,p (R(X)). Using the notations of 16.4.2, the A(X)-modules ΩX and MA(X) (Λ) are isomorphic (2.9.8). In view of 16.4.4, we obtain that n = p − r. By 16.2.5 and 16.4.4, the k-vector spaces Tx (X) and MA(X) (Λ)x have the same dimension. So part (ii) follows by 16.4.4. Finally, let x ∈ X be a smooth point. By 16.4.4, there exists f ∈ A(X) such that f (x) = 0 and MA(X) (Λ)f is a free A(X)f -module of rank n. If U = D(f ), then U is an affine open subset of X containing f such that A(U ) = A(X)f . So we have part (i). 16.5.6 Remark. If X is equidimensional and if X is the disjoint union of its irreducible components, then 16.5.5 implies that the set of smooth points points of X is a dense open subset of X.
16.5 Smooth points
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16.5.7 Let u : X → Y be a morphism of irreducible affine varieties, x ∈ X and y = u(x). Recall that any A(X)-module can be considered as an A(Y )module via the comorphism A(u) of u. Denote by kx (resp. ky ) the A(X)-module (resp. A(Y )-module) such that f.λ = f (x)λ (resp. g.λ = g(y)λ) for all f ∈ A(X), g ∈ A(Y ) and λ ∈ k. Observe that kx and ky are isomorphic as A(Y )-modules. Denote dA(X)/k and dA(Y )/k by dX and dY . We saw in 2.9.5 that there exists a unique homomorphism θ(u) : ΩY → ΩX of A(Y )-modules such that θ(u) ◦ dY = dX ◦ A(u). Recall from 16.3.3 that Tx (u) can be identified with the map Derk (A(X), kx ) → Derk (A(Y ), ky ) , ∆ → ∆ ◦ A(u). Now, using the following canonical isomorphisms (2.9.5): HomA(X) (ΩX , kx ) → Derk (A(X), kx ), HomA(Y ) (ΩY , ky ) → Derk (A(Y ), ky ), we can identify Tx (u) with the map HomA(X) (ΩX , kx ) → HomA(Y ) (ΩY , ky ) , ϕ → ϕ ◦ θ(u). Finally, with the identifications in 16.2.5, Tx (u) can be identified with the x , k) → Homk (ΩYy , k), ϕ → ψ, where map from Homk (ΩX ψ(λ1 dY (g1 ) + · · · + λs dY (gs )) = ϕ(λ1 dX (g1 ◦ u) + · · · + λs dX (gs ◦ u)) for λ1 , . . . , λs ∈ k and g1 , . . . , gs ∈ A(Y ). Theorem. Let u : X → Y be a morphism of irreducible varieties. (i) Suppose that there exists a smooth point x ∈ X such that y = u(x) is a smooth point in Y and Tx (u) : Tx (X) → Ty (Y ) is surjective. Then u is dominant. (ii) Suppose that u is dominant. There exists a non-empty open subset U of X such that for all x ∈ U , u(x) is a smooth point in Y and Tx (u) is a surjection from Tx (X) onto Tu(x) (Y ). Proof. In view of 16.5.5, we are reduced to the case where X and Y are affine, smooth and irreducible, and ΩX (resp. ΩY ) is a free A(X)-module (resp. A(Y )-module) of rank m = dim X (resp. n = dim Y ). We have the following homomorphism of free A(X)-modules: ϕ(u) : A(X) ⊗A(Y ) ΩY → ΩX , 1 ⊗ α → θ(u)(α). Let us fix bases for these free modules and let Λ = [aij ] ∈ Mm,n (A(X)) be the matrix of ϕ(u) with respect to these bases. For x ∈ X, set Λ(x) = [aij (x)] ∈ Mm,n (k). With the notations of 16.2.5, let x ϕ(u)x : ΩYy = kx ⊗A(X) (A(X) ⊗A(Y ) ΩY ) → kx ⊗A(X) ΩX = ΩX
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be the homomorphism induced by ϕ(u) by scalar extension. In suitable bases, Λ(x) is the matrix of ϕ(u)x . The preceding description shows that ϕ(u)x is the dual map of Tx (u). Thus ϕ(u)x and Tx (u) have the same rank. (i) Suppose that Tx (u) is surjective. Then ϕ(u)x is injective, so the rank of Λ(x) is n. As in the proof of 16.4.4, we deduce that the rank of Λ (over R(X)) is at least n. But the rank of Λ is at most n, so we have equality. It follows that ϕ(u) is injective, which implies in turn that θ(u) is injective. Finally, since ΩX (resp. ΩY ) is a free A(X)-module (resp. A(Y )-module), A(u) is injective (5.8.4). Hence u is dominant (6.5.9). (ii) If u is dominant, then A(u) is injective. By 5.8.4, ϕ(u) is injective. Hence the rank of Λ (over R(X)) is n. There exists therefore a non-empty open subset U of X such that for x ∈ U , the rank of Λ(x) is n. It follows from the discussion above that Tx (u) is surjective for all x ∈ U .
References • [5], [19], [22], [26], [37], [40], [78].
17 Normal varieties
17.1 Normal varieties 17.1.1 Let us begin by a remark which will be useful later on. Let X be a prevariety and U, V affine open subsets of X such that U ∩ V = ∅. Any x ∈ U ∩ V has a fundamental system of neighbourhoods consisting of subsets D(s) such that D(s) ⊂ U ∩ V and s ∈ A(U ). For each of these subsets D(s), there exists t ∈ A(V ) such that D(t) is a neighbourhood of x in D(s). Denote by u : A(V ) → A(U )s = A(D(s)) the comorphism of the canonical injection D(s) → V . We have D(t) = D(u(t)), and u(t) is of the form r/sn for some r ∈ A(U ). Thus D(u(t)) = D(sr) where sr ∈ A(U ). Hence open subsets contained in U ∩ V which are principal in both U and V , form a base for the topology on U ∩ V . 17.1.2 Definition. A point x in a variety X is said to be normal if the local ring OX,x of X at x is an integrally closed domain. The variety X is normal if all its points are normal. 17.1.3 Remarks. 1) If X is irreducible, then it is normal if and only if it admits an open covering consisting of normal subvarieties. 2) If x is a normal point of X, then it is contained in a unique irreducible component of X (12.8.5). We deduce that if X is normal, then irreducible components of X are pairwise disjoint. So we may reduce the study of normal varieties to irreducible normal varieties. 17.1.4 Proposition. Let X = Spm(A) be an irreducible affine variety. Then X is normal if and only if A is an integrally closed domain. Proof. Suppose that X is normal. By 12.8.2, we have OX,x . A= x∈X
It follows from 3.2.4 that A is an integrally closed domain.
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Suppose that A is an integrally closed domain. For p ∈ Spec(A), the ring Ap is an integrally closed domain (3.2.5). So X is normal since OX,x is isomorphic to Anx , where nx = I({x}) is the maximal ideal of A associated to x (12.8.3). 17.1.5 Corollary. The set of normal points of an irreducible variety X contains a non-empty open subset of X. Proof. We may assume that X = Spm(A) is affine. By 6.2.2, there exist t1 , . . . , tn ∈ A algebraically independent over k such that A is integral over C = k[t1 , . . . , tn ]. Let B be the integral closure of A. Then B is integral over C (3.1.7), and by 5.5.6, B is a finitely generated C-module, hence B is a finitely generated A-module. Let Y = Spm(B) and u : Y → X be the morphism induced by the injection A → B. The morphism u is dominant (6.5.9) and birational since Fract(A) = Fract(B). It follows from 15.5.6 that there exists a non-empty open subset U of X such that the restriction u−1 (U ) → U of u is an isomorphism. By 17.1.4, Y is normal, and the result follows. 17.1.6 Example. The affine space kn is normal (4.2.9 and 17.1.4). Since Pn has an open covering by subsets isomorphic to kn , it is normal. In the same way, the Grassmannian varieties Gn,r are also normal. 17.1.7 Proposition. Let X, Y be irreducible varieties and u : X → Y a surjective finite morphism. Assume further that Y is normal. Then for all x ∈ X and for all irreducible subvariety W of Y containing u(x), there exists an irreducible component of u−1 (W ) containing x which dominates W . Proof. Since u is finite, there is an affine open subset V containing u(x) such that U = u−1 (V ) is an affine open neighbourhood of x. We may therefore assume that X = Spm(A), Y = Spm(B) are affine where B ⊂ A are integral domains, B is an integrally closed domain and A is a finitely generated Bmodule. Let p = I(W ), m = I({x}), n = I({u(x)}). We have n = B ∩ m and p ⊂ n. By 5.7.8, there exists q ∈ Spec(A) such that q ⊂ m and p = B ∩ q. So x ∈ V(q) and u(V(q)) = W . Hence there is an irreducible component of u−1 (W ) containing V(q) and it dominates W . 17.1.8 Corollary. Let u : X → Y be a dominant morphism of irreducible varieties. There exists a non-empty open subset V of Y such that: (i) The variety V is normal. (ii) For all y ∈ V , x ∈ u−1 (y) and irreducible subvariety W of Y containing y, there exists an irreducible component of u−1 (W ) containing x which dominates W . Proof. By 15.4.2, there exist a non-empty affine open subset V of Y , m ∈ N, and a covering (Ui )i∈I of u−1 (V ) by affine open subsets such that the
17.2 Normalization v
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pr
restriction vi : Ui → V of u factors through Ui →i V × km → V , where pr denotes the canonical projection. Note that vi is a surjective finite morphism. We may assume that V is normal (17.1.5), and x belongs to one of the Ui ’s. We are reduced to the case where X is one of the Ui ’s, so X is affine. Since A(V × km ) = A(V ) ⊗k k[T1 , . . . , Tm ] = A(V )[T1 , . . . , Tm ], it follows from 3.2.8 that A(V × km ) is an integrally closed domain. So V × km is normal (17.1.4). Now vi is a surjective finite morphism, so the result follows from 17.1.7.
17.2 Normalization 17.2.1 Theorem. Let X be an irreducible variety and K a finite field extension of R(X), the field of rational functions on X. There exist a normal irreducible variety Y such that R(Y ) is isomorphic to K, and a surjective finite morphism u : Y → X such that there is a covering (Ui )i∈I of X by affine open subsets verifying u−1 (Ui ) is affine and A(u−1 (Ui )) is the integral closure of A(Ui ) in R(Y ). Furthermore, if Y1 is an irreducible variety and u1 : Y1 → X is a surjective finite morphism having the same properties, then there is a unique isomorphism v : Y → Y1 such that u = u1 ◦ v. Proof. Let us prove the existence of Y . First, suppose that X = Spm(A) is affine. Let B be the integral closure of A in K, and Y = Spm(B). Then B is a finitely generated A-module (6.2.10). Take u : Y → X to be the morphism induced by the canonical injection A → B. The result follows since K is the quotient field of B (5.5.5). Now let us treat the general case. Let (Ui )i∈I be a covering of X by nonempty affine open subsets. For i ∈ I, let Vi = Spm(Bi ) where Bi is the integral closure of Ai = A(Ui ) in K. Denote by θi : Ai → Bi the canonical injection and ui = Spm(θi ). Let i, j ∈ I. By 12.5.6, Ui ∩ Uj is a non-empty affine open subset, and the algebra Aij = A(Ui ∩ Uj ) is generated by the functions x → f (x)g(x) where x ∈ Ui ∩ Uj , f ∈ Ai and g ∈ Aj . Let Bij be the integral closure of Aij in K, θij : Aij → Bij the canonical injection and uij = Spm(θij ). Let U ⊂ Ui ∩ Uj be a principal open subset of both Ui and Uj (17.1.1). We have U = D(t) = D(si ) = D(sj ) with t ∈ Aij , si ∈ Ai and sj ∈ Aj . Observe that if s ∈ Ai \ {0}, then the ring of regular functions on u−1 i (D(s)) = D(θi (s)) is equal to Bi [1/θi (s)], the integral closure in K of the ring (Ai )s = Ai [1/s] of regular functions on D(s) (3.1.9). The same applies for Aij , Bij , θij . −1 It follows from the above observation that the rings A(u−1 i (U )), A(uj (U )) −1 and A(uij (U )) are all equal to Bij [1/θij (t)], which is the integral closure in K of Aij [1/t]. Since the set of open subsets in Ui ∩ Uj which are principal in both Ui and Uj form a base of the topology on Ui ∩ Uj (17.1.1), it follows −1 from 9.5.1 and 12.1.1 that the open subsets u−1 i (Ui ∩ Uj ) and uj (Ui ∩ Uj ) can be identified canonically.
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Let Y be the prevariety defined by gluing the Vi ’s along the intersections Vi ∩ Vj = Spm(Bij ) via these canonical identifications. Then the map u : Y → X, which is ui on Vi , is a morphism. Since X is a variety and each Vi = u−1 (Ui ) is a variety, Y is also a variety (12.5.10). Furthermore, the Vi ’s are irreducible and the Vi ∩Vj ’s are non-empty, so Y is irreducible (1.1.6). Let us now prove that Y is unique up to an isomorphism. Let Y1 be another such variety. Let (Uj )j∈J be a covering of X by affine open subsets. The open subsets contained in Ui ∩ Uj which are principal in both Ui and Uj form a base of the topology of Ui ∩ Uj (17.1.1). Using the same arguments, we have our result. 17.2.2 The normal variety Y in 17.2.1 is called the normalization of X in K. When K = R(X), we shall simply say that Y is the normalization of X; in this case, the morphism u is birational. 17.2.3 Corollary. Let X be an irreducible complete variety. If Y is the normalization of X in a finite extension of R(X), then Y is complete. Proof. Since the morphism Y → X is finite (17.2.1), the result is a consequence of 15.3.9. 17.2.4 Proposition. Let X, Y be irreducible varieties, u : X → Y a surjective finite morphism and n = [R(X) : R(Y )]. Suppose that Y is normal. (i) For all y ∈ Y , the cardinality of u−1 (y) is at most n. (ii) There exists a non-empty open subset V of Y such that u−1 (V ) is isomorphic to the normalization of V in R(X), and the cardinality of u−1 (y) is n for all y ∈ V . Proof. (i) Since u is finite, u−1 (y) is a finite set for all y ∈ Y (15.3.4). Further, if V is an affine open subset of Y , u−1 (V ) is an affine open subset of X, and A(u−1 (V )) is a finite A(V )-algebra (15.3.3). So we are reduced to the case where X = Spm(A) and Y = Spm(B) are affine. Let y ∈ Y and u−1 (y) = {x1 , . . . , xm }. There exists a ∈ A(X) such that the a(xi ) s, 1 i m, are pairwise distinct. Let P (T ) = T p + αp−1 T p−1 + · · · + α0 the minimal polynomial of a over R(Y ). We have deg(P ) n, and by 5.5.4, αp−1 , . . . , α0 are integral over B. Hence αp−1 , . . . , α0 ∈ B. Let Q(T ) = T p + αp−1 (y)T p−1 + · · · + α0 (y). Then a(x1 ), . . . , a(xm ) are roots of Q. Hence m n. (ii) Again, we may assume that X = Spm(A) and Y = Spm(B) are affine. By 5.6.7, there exists a ∈ R(X) such that R(X) = R(Y )(a). We may assume
17.3 Products of normal varieties
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further that a ∈ A (proceed as in 5.7.9). The degree of the minimal polynomial P of a over R(Y ) is n, and as above, P (T ) = T p + αp−1 T p−1 + · · · + α0 where αp−1 , . . . , α0 ∈ B. Let d ∈ B be the discriminant of P . Then d = 0 since the characteristic of k is zero. Replace Y by D(d) and X by u−1 (D(d)), we may assume that d is invertible in B. In view of 5.7.9, A is the integral closure of B in R(X) and {1, a, . . . , an−1 } is a basis of the B-module A. Further, by 5.7.10, the cardinality of u−1 (y) is n for all y ∈ Y . By 17.1.5, there exists a non-empty open subset U of X which is a normal variety. The set u(X \U ) is closed and nowhere dense in Y (15.3.4 and 15.3.8). So there exists an affine open subset V of Y such that u−1 (V ) is an affine open subset contained in U . Thus the affine variety u−1 (V ) is normal; its field of rational functions is R(X). Finally, A(u−1 (V )) is the integral closure of A(V ) in R(X). Hence we have our result. 17.2.5 Example. Let X = {(x1 , x2 ) ∈ k2 ; x31 − x22 = 0}. We saw in 11.5.7 that X is the image of the morphism u : k → k2 , t → (t2 , t3 ). Thus X is irreducible. However, A(X) = k[T 2 , T 3 ] is not an integrally closed domain (11.5.7). So X is not normal. The set Y = X \ {(0, 0)} is the principal open subset DX (T 2 ) of X, and A(Y ) = k[T, T −1 ]. It follows from 3.2.5 that Y is normal. Thus (0, 0) is the only point in X which is not normal. The morphism v : k → X, t → u(t), is bijective, bicontinuous, birational (12.7.13) and finite (15.3.7). The integral closure of k[T 2 , T 3 ] in k(T ) is k[T ]. Thus k is the normalization of X.
17.3 Products of normal varieties 17.3.1 Lemma. Let X = Spm(A) be an irreducible affine variety and E a finite-dimensional k-vector subspace of A. Denote by F the subspace of A generated by the elements of A which, in R(X), can be represented in the form uv −1 , with u ∈ E and v ∈ E \ {0}. Then F is finite-dimensional. Proof. Let us fix a finite normal extension L of Fract(A) (5.7.5). There exist t1 , . . . , tn ∈ A algebraically independent over k such that A is integral over B = k[t1 , . . . , tn ] (6.2.2). The integral closure B of B in L is also the integral closure of A in L (3.1.7). By 5.5.6, B is a finitely generated B-module, so it is a finitely generated k-algebra. Replace X by Spm(B ), we may assume that A is the integral closure of B in L. Denote by K = Fract(B). The extension K ⊂ L is finite and normal. Let d be its degree, which is also the order of the Galois group Gal(L/K) (5.6.6 and 5.7.4). Let Gal(L/K) = {σ1 , . . . , σd } with σ1 = idL . If u ∈ L, then σ1 (u) · · · σd (u) ∈ K. Replace E by the sum of the σi (E)’s, we may assume that E is Gal(L/K)stable.
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Let E1 be the k-vector space spanned by the products of k elements of E, with k d. Clearly E1 is finite-dimensional, and the same goes for E1 ∩ K. Since any element of E1 is integral over the integrally closed domain B, we deduce that E1 ∩ K ⊂ B. Denote by Br of elements in B[t1 , . . . , tn ] of degree at most r in the ti ’s. Then there exists r ∈ N such that E1 ∩ K ⊂ Br . Now let w ∈ A, u, v ∈ E be such that w = uv −1 . Then: w = uσ2 (v) · · · σd (v)/(σ1 (v) · · · σd (v)). We deduce that w = u (v )−1 where u ∈ A and v ∈ Br \ {0}. There exists a basis (z1 , . . . , zd ) of L over K such that A ⊂ Bz1 ⊕ · · · ⊕Bzd (5.5.6). It follows that there exists s ∈ N such that E1 ⊂ Bs z1 ⊕ · · · ⊕ Bs zd . Let us write u = p1 z 1 + · · · + p d z d , w = q 1 z 1 + · · · + q d z d where p1 , q1 , . . . , pd , qd ∈ B. Then qi v = pi for 1 i d, and we have pi ∈ Bs , for 1 i d, because u = uσ2 (v) · · · σd (v) ∈ E1 . Thus qi ∈ Bs , and hence F is finite-dimensional. 17.3.2 Proposition. Let X, Y be irreducible normal varieties. Then the variety X × Y is normal. Proof. We may assume that X = Spm(A) and Y = Spm(B) are affine. Let f ∈ R(X × Y ) be integral over A ⊗k B and D its domain of definition. Then f can be represented in the form g/h where g = u1 ⊗ v1 + · · · + un ⊗ vn , h = u1 ⊗ w1 + · · · + un ⊗ wn with u1 , . . . , un ∈ A linearly independent and v1 , w1 , . . . , vn , wn ∈ B. For y ∈ Y , define gy , hy ∈ A by: gy = v1 (y)u1 + · · · + vn (y)un , hy = w1 (y)u1 + · · · + wn (y)un . The set V of elements y ∈ Y such that wi (y) is non-zero for some i is a nonempty open subset of Y . If y ∈ V , then hy is a non-zero element of A. So there exists x ∈ X such that (x, y) ∈ DX×Y (h) ⊂ D and fy = gy /hy ∈ R(X) is a regular function on DX (hy ). Denote by F the set of maps from V to k, G the subset of F consisting of maps u such that there exist a non-empty open subset W ⊂ V and v ∈ B verifying u|W = v|W . Since Y is irreducible, it is easy to see that G is a subspace of F. Let y ∈ V . Since f is integral over A⊗k B, we verify easily that fy is integral over A. Since X is normal, there exists a function ty ∈ A which represents fy . On the other hand, fy is the quotient of two elements of the vector subspace E of A generated by the ui ’s. By 17.3.1, there exist s1 , . . . , sm ∈ A linearly independent, such that ty = ρ1 (y)s1 + · · · + ρm (y)sm where ρ1 , . . . , ρm ∈ F. In particular, for all x ∈ X such that (x, y) ∈ D, ty (x) = f (x, y).
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A similar argument shows that there is a non-empty open subset U of X verifying the following conditions: for all x ∈ U , there exist y ∈ Y such that (x, y) ∈ D, and tx ∈ B such that tx (y) = f (x, y) for all (x, y) ∈ D. For x ∈ U , denote by Vx the set of y ∈ V such that (x, y) ∈ D. It is a non-empty open subset of V . Let x ∈ U . For y ∈ Vx , we have ρ1 (y)s1 (x) + · · · + ρm (y)sm (x) = ty (x) = f (x, y) = tx (y). So the restrictions of tx and s1 (x)ρ1 + · · · + sm (x)ρm to Vx are identical. It follows that s1 (x)ρ1 + · · · + sm (x)ρm ∈ G. Hence, if λ is a linear form on F which is zero on G, then s1 (x)λ(ρ1 ) + · · · + sm (x)λ(ρm ) = 0. This being true for all x in the dense open subset U of X, we have: λ(ρ1 )s1 + · · · + λ(ρm )sm = 0. But the si ’s are linearly independent, so λ(ρi ) = 0 for 1 i m. Hence ρi ∈ G, 1 i m. Thus there exist t1 , . . . , tm ∈ B such that f can be represented in the form s1 ⊗ t1 + · · · + sm ⊗ tm ∈ A ⊗k B. So our result follows.
17.4 Properties of normal varieties 17.4.1 Lemma. Let A ⊂ B be finitely generated k-algebras, X = Spm(A), Y = Spm(B), u : Y → X the morphism associated to the canonical injection A → B, n ∈ Spm(B), m = n ∩ A, and y ∈ Y the point associated to n. The following conditions are equivalent: (i) n is minimal among the prime ideals q ∈ Spec(B) such that q∩A = m. (ii) The point y is isolated in the fibre u−1 (u(y)) of the morphism u. Proof. The algebra A/m is a subalgebra of B/n, which is isomorphic to k (6.4.1). So m ∈ √ Spm(A). Thus any proper ideal b of B containing m verifies b ∩ A = m. Let Bm = q1 ∩ · · · ∩ qs where q1 , . . . , qs ∈ Spec(B) with qi ⊂ qj whenever i = j. Then q1 , . . . , qs are exactly the minimal prime ideal lying above m. Also, Z = u−1 (u(y)) = V(q1 ) ∪ · · · ∪ V(qs ) is the decomposition of Z into irreducible components. If part (i) is true, then we may take n = q1 , and Z is the disjoint union of {y} and V(q2 ) ∪ · · · ∪ V(qs ). Hence {y} is open in Z, which proves (ii). Conversely, if y is isolated in Z, then it is an irreducible component of Z. In view of the above discussion, we have (i). 17.4.2 Proposition. Let A ⊂ C ⊂ B be finitely generated k-algebras verifying: they are integral domains, B is integral over C, A is integrally closed in B and Fract(B) is a finite extension of Fract(A). Let m ∈ Spm(A). Suppose that there exists n ∈ Spm(B) minimal in the set of q ∈ Spec(B) verifying q∩A = m. Then the canonical homomorphism Am → Bn is bijective.
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Proof. From our hypotheses, C is a finitely generated A-algebra, so C = A[u1 , . . . , um ], where u1 , . . . , um ∈ C. The case m = 1 is exactly 5.9.8. Let us proceed by induction on m. Suppose that m > 1. Let B be the integral closure of A[u1 , . . . , um−1 ] in B. It is finitely generated (3.3.5) and integrally closed in B. We have C ⊂ B [um ] ⊂ B, so B is integral over B [um ] and Fract(B) is a finite extension of Fract(B ). The ideal n = n ∩ B ∈ Spm(B ) and if q ∈ Spec(B) verifies q ⊂ n and q ∩ B = n , then q ∩ A = n ∩ A = n ∩ A = m, so it follows from our hypothesis on n that q = n and n is minimal among the prime ideals of B lying above n . We may therefore apply 5.9.8 to B ⊂ B [um ] ⊂ B, and we obtain that the homomorphism Bn → Bn is bijective. Denote by u : Spm(B) → Spm(A) , v : Spm(B ) → Spm(A) , w : Spm(B) → Spm(B ) the morphisms induced by the canonical injections. Note that u = v ◦ w. By 17.4.1, the point y corresponding to n is isolated in u−1 (u(y)). It follows that y is also isolated in w−1 (w(y)). Since the homomorphism Bn → Bn is bijective, it follows from 12.8.8 that w induces an isomorphism from a neighbourhood of y in Spm(B) onto a neighbourhood of w(y) in Spm(B ). Hence w(y) is isolated in v −1 (v(w(y))). Applying 17.4.1, we deduce that n is minimal among the prime ideals of B lying above m. By induction on A ⊂ A[u1 , . . . , um−1 ] ⊂ B , the homomorphism Am → Bn is bijective. So we are done. 17.4.3 Let u : X → Y be a morphism of irreducible varieties. Suppose that there exist x ∈ X and an open neighbourhood U of x in X such that u induces an isomorphism from U onto an open subset V of Y . We may assume that U is affine. We see in particular that u is dominant and birational. Further U ∩ u−1 (u(x)) = {x}, and so x is isolated in u−1 (u(x)). In this situation, we have the following more precise statement. Theorem. (Zariski’s Main Theorem) Let u : X → Y be a dominant morphism of irreducible varieties. Suppose that there exists x ∈ X such that x is isolated in the u−1 (u(x)). (i) We have dim X = dim Y , so R(X) is a finite extension of R(Y ). (ii) There is an open neighbourhood V of u(x) in Y and an open neighbourhood U of x in u−1 (V ) such that the restriction of u factorizes through: j
U −→ V −→ V v
where v is finite and surjective, and j is an isomorphism from U onto an open subset of V . (iii) If X is normal, we may suppose that V = Y and V is the normalization of Y in R(X). Proof. Since {x} is an irreducible component of u−1 (u(x)), we have dim X = dim Y by 15.5.3 and 15.5.4. In particular, R(X) is a finite extension of R(Y ).
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To prove the other parts, we may assume that X = Spm(B), Y = Spm(A) are affine, and A ⊂ B. Let A (resp. A ) be the integral closure of A in B (resp. Fract(B)). By 6.2.10, A is a finitely generated A-algebra and since A is Noetherian, A is a finitely generated A-module. Thus A is a finitely generated k-algebra which is also an integral domain. Hence Y = Spm(A ) is irreducible and the morphism v : Y → Y , associated to the canonical injection A → A , is finite (15.3.2). It follows from the injections A → A → B that u factorizes through: v w X −→ Y −→ Y. Since w−1 (w(x)) ⊂ u−1 (u(x)), the point x is isolated in w−1 (w(x)). So by 17.4.1, n = I({x}) is minimal among the prime ideals q of B verifying q∩A = n ∩ A = n = I({w(x)}). Applying 17.4.2 to A ⊂ B ⊂ B, we obtain that the canonical homomorphism An → Bn is bijective. So w induces an isomorphism from a neighbourhood of x in X onto a neighbourhood of w(x) of Y (12.8.8). So we have proved (ii). Let Z be the normalization of Y in R(X). There exist a surjective finite morphism s : Z → Y and a covering (Vi )i∈I of Y by affine open subsets such that s−1 (Vi ) is an affine open subset and A(s−1 (Vi )) is the integral closure of A(Vi ) in R(Z) R(X). Suppose that X is normal. Replacing Y by a Vi containing u(x) and U by an affine open neighbourhood U of x in X such that U ⊂ u−1 (Vi ), we are reduced to the case where Y = Spm(A), X = Spm(B) with B is an integrally closed domain. Then A (as defined above) is isomorphic to A(Z). So (iii) follows. 17.4.4 Corollary. Let u : X → Y be a dominant morphism of irreducible varieties. The following conditions are equivalent: (i) X is normal and for all x ∈ X, x is isolated in u−1 (u(x)). (ii) The extension R(Y ) ⊂ R(X) is finite and u factorizes through: j
X −→ Y −→ Y v
where Y is the normalization of Y in R(X), v is a surjective finite morphism as in 17.2.1, and j is an isomorphism from X onto an open subset of Y . If these conditions are satisfied, there exists an open subset V of Y such that V is normal, and for all y ∈ V , card u−1 (y) = [R(X) : R(Y )]. Proof. The fact that (ii) implies (i) is clear. Furthermore, the last part follows from 17.1.5 and 17.2.4. Let us suppose that (i) is true. Let Y be the normalization of Y in R(X), v : Y → Y be as in 17.2.1 and x ∈ X. By 17.4.3, there exists an isomorphism jU from an affine open neighbourhood U of x in X onto an open subset of Y such that u|U = v ◦ jU . Let U be another affine open neighbourhood of x and jU an isomorphism from U to an open subset of Y such that u |U = v ◦ jU . Then there exists an
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affine open neighbourhood U ⊂ U ∩U of x such that u(U ) is contained in an affine open subset V of Y . Let A = A(V ), B = A(U ) with B an integrally closed domain, and A the integral closure of A in R(X). The restrictions jU |U and jU |U correspond both to the injection A → B. So they are identical. We deduce that there exists a morphism j : X → Y such that j|U = jU for all affine open subset U with the above properties. Thus j is a local isomorphism, and (ii) follows by 12.5.11. 17.4.5 Corollary. Let u : X → Y be a birational morphism of irreducible varieties. Suppose that Y is normal and for all x ∈ X, x is isolated in u−1 (u(x)). Then (i) u is an isomorphism from X onto an open subset of Y . (ii) If X is complete, then u is an isomorphism from X to Y . Proof. Let v : X → X be the normalization of X (17.2.1). The morphism u◦v has the same properties as u. Since Y is normal and R(X ) = R(X) = R(Y ), it follows from 17.4.4 that u ◦ v induces an isomorphism w from X onto an open subset of Y . So w−1 ◦u is the reciprocal isomorphism of v. Hence X = X and (i) follows. If X is complete, then u(X) is closed in Y (13.4.3), so u(X) = Y . 17.4.6 Corollary. Let u : X → Y be a bijective birational morphism of irreducible varieties. If Y is normal, then u is an isomorphism. 17.4.7 Corollary. Let u : X → Y be a morphism of irreducible normal varieties. Suppose that there exists n ∈ N∗ such that card u−1 (y) = n for all y ∈ Y . Then u is a finite morphism and u factorizes through: j
X −→ Y −→ Y v
where v : Y → Y is the normalization of Y in R(X) and j is an isomorphism from X to Y . Proof. Note that our hypothesis implies that u(X) = Y and that for any x ∈ X, x is isolated in u−1 (u(x)) (since it is a finite set). So by 17.4.4, u j
v
factorizes through X → Y → Y , where v is a surjective finite morphism and j is an isomorphism from X onto an open subset of Y . So to obtain the result, we only need to show that j(X) = Y . Suppose that there exists y ∈ Y \ j(X). Then v −1 (v(y )) is a finite set (15.3.4). Since j −1 (v −1 (v(y ))) = u−1 (v(y )), card v −1 (v(y )) n. If card v −1 (v(y )) = n, then y ∈ j(X). So card v −1 (v(y )) n + 1. On the other hand, Y \ j(X) is closed and nowhere dense in Y . By 15.3.4 and 15.3.8, v(Y \ j(X)) is also closed and nowhere dense in Y . Let V = Y \v(Y \j(X)). Then V is open. But v −1 (V ) ⊂ j(X) and card v −1 (y) = n if y ∈ V . It follows from 17.2.4 that card v −1 (v(y )) n. Contradiction.
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17.4.8 Corollary. Let u : X → Y be a bijective morphism of irreducible varieties. If Y is normal, then u is an isomorphism. Proof. Let v : X → X be the normalization of X (17.2.1), and w = u ◦ v. Since v is finite and u is bijective, w−1 (y) is finite for all y ∈ Y (15.3.4). By 17.4.4, R(X) = R(X ) is a finite extension of R(Y ). Let V be a non-empty open subset of Y such that u−1 (V ) is normal (17.1.5), and v1 : v −1 (u−1 (V )) → u−1 (V ) the morphism induced by v. Then v1 is finite and surjective (15.3.3), and in view of 17.2.4, card v1−1 (z) = [R(X ) : R(X)] = 1 for all z ∈ u−1 (V ). Thus card w−1 (y) = 1 for all y ∈ V . The morphism u ◦ v1 : v −1 (u−1 (V )) → V is surjective and finite (17.4.7), so by applying 17.2.4, we obtain that R(X) = R(Y ). Hence u is birational, and the result follows from 17.4.6. 17.4.9 Corollary. Let u : X → Y be a dominant morphism of irreducible varieties. Suppose that Y is normal. If x ∈ X is isolated in u−1 (u(x)), then for all neighbourhood U of x in X, u(U ) is a neighbourhood of u(x) in Y . In particular, if x is isolated in u−1 (u(x)) for all x ∈ X, then u is open. Proof. Let v : X → X be the normalization of X (17.2.1), w = u ◦ v and x ∈ X be isolated in u−1 (u(x)). Then any x ∈ X verifying v(x ) = x is isolated in w−1 (w(x )). If U is a neighbourhood of x in X, then v −1 (U ) is a neighbourhood of x in X . So it suffices to prove that w(v −1 (U )) is a neighbourhood of u(x) in Y . Therefore, we may assume that X is normal. By 17.4.3, there exists an open neighbourhood V of u(x) in Y and an open neighbourhood U of x in X such that the restriction u|U : U → V factorizes j
v
through U → V → V , where V is the normalization of V in R(X) and j is an isomorphism from U onto an open subset of V . So we can further reduce to the case where X = Y is the normalization of Y in a finite extension of R(X). Let W be an open neighbourhood of x in X. Then u(W ) = Y \ T where T = {y ∈ Y ; u−1 (y) ⊂ X \ W }. To obtain the result, it suffices to prove that u(x) ∈ T . Suppose that u(x) ∈ T . Then u(x) is contained in the closure of an irreducible component T0 of T . By 17.1.7, there exists an irreducible component Z of u−1 (T0 ) containing x verifying u(Z) = T0 . If S ⊂ Z is an open subset containing x, then u(S) contains a non-empty open subset of T0 (15.4.2), hence u(S) ∩ T0 = ∅. It follows that S ∩ u−1 (T0 ) = ∅, so x ∈ u−1 (T0 ) ⊂ X \ W which contradicts the fact that W is a neighbourhood of x in X. 17.4.10 Theorem. Let u : X → Y be a dominant morphism of irreducible varieties and r = dim X − dim Y . Suppose that Y is normal. If x ∈ X is such that all the irreducible components of u−1 (u(x)) containing x has dimension r, then the image via u of a neighbourhood of x in X is a neighbourhood of u(x) in Y . Proof. Let U ⊂ X (resp. V ⊂ Y ) be an affine open neighbourhood of x (resp. u(x)) such that U ⊂ u−1 (V ). Since the intersection of U and an irreducible
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component of u−1 (u(x)) containing x is dense in the irreducible component, the restriction U → V satisfies the same hypotheses. So we may assume that X = Spm(B) and Y = Spm(A) are affine, where A ⊂ B are integral domains and A is an integrally closed domain. Also, we only need to prove that u(X) is a neighbourhood of u(x) in Y . Replacing X by a smaller affine open subsets, we may assume further that all the irreducible components of Z = u−1 (u(x)) have dimension r. √ Denote by m = I({u(x)}) ∈ Spm(A). We have Z = V(Bm), and if a = Bm, then A(Z) = B/a. Since all the irreducible components of Z have the same dimension r, it follows from 6.2.2 and 6.2.3 that there exist elements s1 , . . . , sr ∈ B/a algebraically independent over k such that B/a is a finitely generated C-module, where C = k[s1 , . . . , sr ]. For 1 i r, fix ti ∈ B such that its class in B/a is si . Denote by ϕ : A[T1 , . . . , Tr ] → B the A-module homomorphism such that ϕ(Ti ) = ti . Since ϕ the canonical injection A → B factorizes through A → A[T1 , . . . , Tr ] → B, u factorizes through: pr v X −→ Y × kr −→ Y. If W = pr−1 (u(x)) = {u(x)} × kr , then Z = v −1 (W ). Since W = Spm(C), the restriction Z → W of v is finite and surjective. It follows that any point y of u−1 (u(x)) is isolated in u−1 (u(y)). Let S = {x ∈ X; x is isolated in v −1 (v(x ))}. The set S contains Z and is open in X (15.5.7). Since Y × kr is normal (17.1.6 and 17.3.2), the restriction of v to S is open (17.4.9). But pr is also open (15.5.8), we deduce therefore that u(S) is open in Y , and hence u(X) is a neighbourhood of u(x) in Y . 17.4.11 Corollary. Let u : X → Y be a dominant morphism of irreducible varieties. Suppose that Y is normal and for all y ∈ u(X), all the irreducible components of u−1 (y) have dimension dim X − dim Y . Then u is open. 17.4.12 Proposition. Let u : X → Y be a dominant morphism of irreducible varieties. If f is a regular function which is constant on the fibres u−1 (y), y ∈ Y , then f , considered as an element of R(X), belongs to R(Y ). Proof. Let v : X → Y × k be the morphism x → (u(x), f (x)) and p : Y × k → Y the canonical projection. Our hypothesis implies that the restriction of p to v(X) is injective. Let W ⊂ v(X) be such that W is open and dense in v(X) (15.4.2). Denote by w the restriction p to W , then w is injective and p(W ) = Y . Replacing W by a smaller open subset if necessary, we may assume that W is normal (17.1.5). The set u(W ) contains a non-empty open subset V of Y (15.4.2) that we may also assume to be normal. Replacing W by w−1 (W ), we may further assume that w : W → V is bijective. By 17.4.8, w induces an isomorphism from W onto V . The restriction q to W of the canonical projection Y × k → k is a regular function on W , and q ◦ (w|W )−1 is regular on V . For x ∈ u−1 (W ), we have q ◦ (w|W )−1 (u(x)) = f (x). So we are done.
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17.4.13 Proposition. Let x be a normal point of an irreducible variety X and f be a rational function on X whose domain of definition is D. Suppose that x ∈ D, and denote by E the domain of definition of f = 1/f . There exists a subvariety Y of X containing x verifying Y ∩ E = ∅ and such that f (y) = 0 for all y ∈ Y ∩ E. Proof. Recall from 12.8.3 that A = OX,x is Noetherian. By our hypothesis, f ∈ A. Let a = {h ∈ A; f h ∈ A}. We observe that a is an ideal of A contained in the maximal ideal mX,x of A. Let p1 , . . . , pr be the (pairwise distinct) prime ideals minimal among the prime ideals containing a. By 2.5.3 and 2.7.7, for some n ∈ N∗ , we have pn1 pn2 · · · pnr ⊂ a. It follows that pi Ap1 = Ap1 for i 2 since p1 Ap1 is the unique maximal ideal of the local ring Ap1 . Let k ∈ N be minimal such that f \ Ap1 , so pk1 f ⊂ Ap1 . Our hypothesis implies that k > 0. Let g ∈ pk−1 1 p1 g ⊂ Ap1 . Since Ap1 is an integrally closed domain (3.2.5), g is not integral over Ap1 . If p1 g ⊂ p1 Ap1 , then multiplication by g induces a homomorphism of the finitely generated Ap1 -module p1 Ap1 . Proceed as in the proof of 3.1.4, we would obtain that g in integral over Ap1 . Contradiction. So we deduce that p1 Ap1 g ⊂ p1 Ap1 . Thus p1 g contains an invertible element of Ap1 , hence 1/g ∈ p1 Ap1 and (1/g)Ap1 = p1 Ap1 . Let h = f /g k , then h ∈ f pk1 Ap1 ⊂ Ap1 . If h ∈ p1 Ap1 , then f /g k−1 ∈ Ap1 which contradicts the minimality of k. So h is invertible in Ap1 and f = 1/f = h−1 (1/g k ) ∈ p1 Ap1 . Let 1 , . . . , s ∈ A be generators of the ideal p1 and U an affine open neighbourhood of x contained in the domain of definition of the i ’s. Let Y = {y ∈ U ; i (y) = 0 for all 1 i s}. Since p1 ⊂ mX,x , x ∈ Y . But f ∈ p1 Ap1 , so f (z) = 0 for all z ∈ Y ∩ E. 17.4.14 Proposition. Let u : X → Y be a finite morphism of irreducible varieties and x ∈ X a simple point such that u−1 (u(x)) = {x}. Suppose that the map Tx (u) : Tx (X) → Tu(x) (Y ) is injective. Then there exists an open neighbourhood V of y = u(x) in Y such that the restriction of u to u−1 (V ) is an isomorphism onto a closed subvariety of V . Proof. We may assume that X = Spm(B) and Y = Spm(A) are affine where B is a finite A-algebra. Let ϕ be the comorphism of u and m, n the maximal ideals of A and B corresponding to y and x respectively. By our hypothesis, the dual map, m/m2 → n/n2 , of Tx (u) is surjective, and so a system of generators of m gives a system of generators of n/n2 . Thus n = Bϕ(m) + n2 . It follows that n2 ⊂ Bϕ(m) + n3 , so n = Bϕ(m) + n3 . By induction, we obtain that n = Bϕ(m) + nk for all k ∈ N∗ . By 17.4.1, n is the unique prime ideal of B containing Bϕ(m). Thus n/Bϕ(m) is the nilradical of B/Bϕ(m). Since B/Bϕ(m) is Noetherian, it follows from 2.5.2 and 2.5.3 that nk ⊂ Bϕ(m) for some k ∈ N∗ . We deduce therefore that n = Bϕ(m), and so Bϕ(m) + ϕ(A) = B because B = n ⊕ k. The
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A-module B/ϕ(A) is finitely generated and verifies ϕ(m)(B/ϕ(A)) = B/ϕ(A). So by 2.6.7, there exists f ∈ A \ m such that ϕ(f )B = ϕ(A). Hence the homomorphism ϕf : Af → Bϕ(f ) is surjective. By 11.6.8, u is an isomorphism from the open subset D(ϕ(f )) of X onto a closed subvariety of D(f ) in Y .
References and comments • [5], [19], [22], [26], [37], [40], [78]. The result stated and proved in 17.4.11 is very useful for showing that a morphism is an open map.
18 Root systems
In this chapter, we are concerned with the theory of root systems. This simple geometric notion will play an amazingly important role in the structure theory of semisimple Lie algebras. Throughout this chapter, k is a field of characteristic zero and V is a kvector space of dimension l. If u ∈ End(V ), we shall denote by rk(u) the rank of u.
18.1 Reflections 18.1.1 Definition. We say that s ∈ End(V ) is a reflection if s2 = idV and rk(idV −s) = 1. 18.1.2 Let s be a reflection of V and denote by Vs+ (resp. Vs− ) the eigenspace of s with eigenvalue 1 (resp. −1). Then V = Vs+ ⊕ Vs− , dim Vs− = 1 , det s = −1. 18.1.3 Lemma. The following conditions are equivalent for s ∈ End(V ): (i) s is a reflection. (ii) There exist a ∈ V and a∗ ∈ V ∗ such that a∗ (a) = 2 and s(x) = x − a∗ (x)a for all x ∈ V . Proof. The implication (ii) ⇒ (i) is obvious. Let us prove that (i) ⇒ (ii). Since rk(idV −s) = 1, there exist a ∈ V and a∗ ∈ V ∗ such that x − s(x) = a∗ (x)a for all x ∈ V . Finally, s2 = idV implies that a∗ (a) = 2. 18.1.4 If a ∈ V and a∗ ∈ V are such that a∗ (a) = 2, then we shall denote by sa,a∗ the reflection of V defined by: sa,a∗ (x) = x − a∗ (x)a. We can also define a reflection sa∗ ,a of V ∗ by: sa∗ ,a (x∗ ) = x∗ − x∗ (a)a∗ . We have sa∗ ,a = t (sa,a∗ ) = t (sa,a∗ )−1 .
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18.1.5 Proposition. Let s be a reflection of V , u ∈ End(V ) and W ⊂ V a subspace. (i) We have u ◦ s = s ◦ u if and only if Vs+ and Vs− are u-stable. (ii) We have s(W ) ⊂ W if and only if Vs− ⊂ W or W ⊂ Vs+ . Proof. Part (i) is clear. Let us prove part (ii). If W ⊂ Vs+ , then s(W ) ⊂ W . Now if Vs− ⊂ W , then x − s(x) ∈ Vs− ⊂ W for all x ∈ W . Hence s(W ) ⊂ W . Conversely, suppose that s(W ) ⊂ W and W ⊂ Vs+ . Then there exists x ∈ W such that y = x − s(x) = 0. It follows that y ∈ Vs− ∩ W , hence Vs− ⊂ W since dim Vs− = 1. 18.1.6 Proposition. Let R be a finite subset of V which generates V , s, t reflections of V and α ∈ R \ {0} verifying s(α) = t(α) = −α and s(R) = t(R) = R. Then s = t. Proof. Let G ⊂ GL(V ) be the stabilizer of R. Since R generates V , G can be identified with a subgroup of the symmetric group on R, so G is finite. By 18.1.4, we can write s = sα,α∗ and t = sα,β ∗ . A simple induction gives: (s ◦ t)n (x) = x + n (β ∗ (x) − α∗ (x)) α. Taking n to be the order of s ◦ t in G, we see that α∗ = β ∗ . So s = t. 18.1.7 Let B be a bilinear form on V . An endomorphism u of V is an orthogonal transformation (with respect to B) if B(u(x), u(y)) = B(x, y) for all x, y ∈ V . Suppose that B is symmetric or antisymmetric. Then U, W ⊂ V are orthogonal if B(U, W ) = {0}. A vector x ∈ V is isotropic if B(x, x) = 0. A subspace W of V is non-degenerate if W ∩ W ⊥ = {0} where W ⊥ is the set of x ∈ V such that B(x, y) = 0 for all y ∈ W . Proposition. Let B be a non-degenerate symmetric bilinear form on V . (i) A reflection s of V is an orthogonal transformation if and only if Vs+ and Vs− are orthogonal. In this case, Vs+ and Vs− are non-degenerate. (ii) Let H be a non-degenerate hyperplane of V . Then there exists a unique reflection s of V such that s is an orthogonal transformation and s|H = idH . If α ∈ H ⊥ \ {0} , then H ⊥ = kα is non-degenerate and: s(x) = x − 2
B(x, α) α. B(α, α)
We say that s is the reflection orthogonal to the hyperplane H. Proof. (i) Let x ∈ Vs+ , y ∈ Vs− . If s is an orthogonal transformation, then B(x, y) = B(s(x), s(y)) = −B(x, y).
18.2 Root systems
235
Hence Vs+ and Vs− are orthogonal. These eigenspaces are non-degenerate since B is non-degenerate and V = Vs+ ⊕ Vs− . The converse is immediate. (ii) By our assumptions, kα is non-degenerate and V = H ⊕ kα. Clearly, the reflection s described above verifies Vs+ = H and Vs− = kα. So part (i) says that s is an orthogonal transformation. Now let t be another such orthogonal transformation, then Vt+ = H and − Vt = kα by (i). Let α∗ ∈ V ∗ be such that α∗ (α) = 2 and t = sα,α∗ . Then H = ker α∗ and there exists λ ∈ k such that α∗ (x) = λB(x, α) for all x ∈ V . It follows that 2 = α∗ (α) = λB(α, α). So s = t. 18.1.8 Proposition. Let G be a subgroup of GL(V ). Suppose that V is a simple G-module and that G contains a reflection s. (i) The centralizer of G in End(V ) is k idV . (ii) Let B be a G-invariant non-zero bilinear form on V . Then B is nondegenerate, symmetric or antisymmetric. Any bilinear form G-invariant on V is proportional to B. Proof. (i) Let D = (s − idV )(V ). Then dim D = 1. By 18.1.5, if u ∈ End(V ) commutes with s, then u(D) ⊂ D. It follows that there exists λ ∈ k such that (u − λ idV )(D) = 0. Hence ker(u − λ idV ) = V because V is a simple G-module. (ii) Let M = {x ∈ V ; B(x, V ) = {0}} and N = {x ∈ V ; B(V, x) = {0}}. Then M , N are G-submodules of V since B is G-invariant. Hence M = N = {0} and B is non-degenerate. Let b be a G-invariant bilinear form on V . Since B is non-degenerate, there exists u ∈ End(V ) such that b(x, y) = B(u(x), y) for all x, y ∈ V . If α ∈ G and x, y ∈ V , then: B(u◦α(x), y) = b(α(x), y) = b(x, α−1 (y)) = B(u(x), α−1 (y)) = B(α◦u(x), y). So u commutes with G. By (i), u = λ idV for some λ ∈ k. Take, in particular, the bilinear form b(x, y) = B(y, x), then B(y, x) = b(x, y) = λB(x, y) = λb(y, x) = λ2 B(y, x). Hence λ2 = 1 and B is symmetric or antisymmetric.
18.2 Root systems 18.2.1 Definition. A subset R ⊂ V is called a root system in V if the following conditions are satisfied: (R1) R is a finite subset of V which spans V , and 0 ∈ R. (R2) For all α ∈ R, there exists α∨ ∈ V ∗ such that α∨ (α) = 2 and sα,α∨ (R) = R. (R3) For all α ∈ R, α∨ (R) ⊂ Z. Further, if for all α ∈ R, the only elements of R proportional to α are α and −α, we say that R is reduced.
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18.2.2 From now on, let R be a root system in V . The elements of R are called the roots of R, the dimension of V is the rank of R that we shall denote by rk R. Let α ∈ R. By 18.1.6, there exists a unique α∨ satisfying (R2). It follows that (−α)∨ = −α∨ . So we may write sα for sα,α∨ and R∨ = {α∨ ; α ∈ R}. By (R1), the set of u ∈ GL(V ) verifying u(R) ⊂ R is a finite group, denoted by A(R). The subgroup W (R) of A(R) generated by the sα ’s is called the Weyl group of R. Let α ∈ R. We have −α = sα (α) ∈ R by (R2), and so −R = R. Hence − idV ∈ A(R). However, − idV is not necessarily contained in W (R). For α, β ∈ R, set: aα,β = β ∨ (α). From the definition of a root system, we have: ⎧ ⎨ aα,β ∈ Z , aα,α = 2, a−α,β = aα,−β = −aα,β , (1) ⎩ sβ (α) = α − aα,β β. 18.2.3 Let V = V1 ⊕ · · · ⊕ Vn . For 1 i n, let Ri be a root system in Vi . Identify V ∗ with the direct sum of the Vi∗ ’s. If α ∈ Ri , α∨ can be identified with an element of V ∗ which is zero on the Vj for j = i. Thus the union R of the Ri ’s is a root system in V that we shall call the direct sum of the Ri ’s. We have R∨ = R1∨ ∪ · · · ∪ Rn∨ . If α ∈ Ri and j = i, then Vj ⊂ ker α∨ and kα ⊂ Vi . We deduce that sα |Vj = idVj and sα (Vi ) ⊂ Vi by 18.1.5. The group W (R) is the product of the groups W (Ri ). Definition. A root system is irreducible if it is not the direct sum of two non-empty root systems. 18.2.4 Proposition. A root system R is the direct sum of a finite family (Ri )i∈I of irreducible root systems, which is unique up to a permutation of indices. We call the Ri ’s, the irreducible components of R. Proof. If R is non-empty and non-irreducible, then it is the direct sum of two non-empty root systems. By induction on the cardinality of R, we obtain that R is the direct sum of a finite family (Ri )i∈I of irreducible root systems. To prove the uniqueness, it suffices to show that if R is the direct sum of R and R , then Ri ⊂ R or Ri ⊂ R . Let Ri = Ri ∩ R , Ri = Ri ∩ R and V , V , Vi , Vi the vector subspaces spanned by R , R , Ri , Ri . If α ∈ Ri , then sα (R ) ⊂ R , sα (Ri ) ⊂ Ri (18.2.3), so Ri is a root system in Vi . Similarly, Ri is a root system in Vi . It follows easily that Ri is the direct sum of Ri and Ri . But Ri is irreducible, so Ri = ∅ or Ri = ∅. So we are done. 18.2.5 Lemma. Let U be a subspace of V and W the subspace of V spanned by R ∩ U . Then R ∩ U is a root system in W .
18.2 Root systems
237
∨ ∨ Proof. If α ∈ R ∩ U , then αW = α∨ |W . For β, γ ∈ R ∩ U , we have βW (β) = 2 ∨ ∨ (γ) = γ − β (γ)β ∈ U . So sβ,β ∨ (γ) ∈ R ∩ U by (R2). The other and sβ,βW W properties for a root system are clear.
18.2.6 Proposition. Let V = V1 ⊕· · ·⊕Vn and Ri = R∩Vi for 1 i n. The following conditions are equivalent: (i) The subspaces V1 , . . . , Vn are W (R)-stable. (ii) We have R ⊂ V1 ∪ · · · ∪ Vn . (iii) For 1 i n, Ri is a root system in Vi and R is the direct sum of the Ri ’s. Proof. (i) ⇒ (ii) Let α ∈ R. If sα (Vi ) ⊂ Vi , then kα ⊂ Vi or Vi ⊂ ker α∨ (18.1.5). Since V is the direct sum of the Vi ’s, Vj ⊂ ker α∨ for some j. So α ∈ Vj . (ii) ⇒ (iii) This is clear by 18.2.5. (iii) ⇒ (i) This follows from 18.2.3. 18.2.7 Corollary. The root system R is irreducible if and only if V is a simple W (R)-module. Proof. This follows from 10.7.6 and 18.2.6. 18.2.8 Proposition. If R is a root system in V , then R∨ is a root system in V ∗ . Furthermore, α∨∨ = α for all α ∈ R. Proof. Let us first show that R∨ spans V ∗ . We may assume by 18.2.3 and 18.2.4 that R is irreducible. The result is clear if rk R = 1. Suppose that rk R 2. If there exists x ∈ V \ {0} such that α∨ (x) = 0 for all α ∈ R, then the subspace kx is W (R)-stable. But this is absurd by 18.2.7. So R∨ spans V ∗. Now, let α, β ∈ R, γ = sα (β) and θ = β ∨ − β ∨ (α)α∨ . Then θ(γ) = 2 , sγ,θ = sα ◦ sβ ◦ sα . Hence sγ,θ (R) = R, and so θ = γ ∨ = sα∨ ,α (β ∨ ) by 18.1.6. Thus sα∨ ,α (R∨ ) = R∨ . It follows that condition (R2) is satisfied and α∨∨ = α. Finally, we see readily that (R3) is also satisfied. 18.2.9 We call R∨ the dual root system of R. The map R → R∨ , α → α∨ , is a bijection, that we shall call canonical. We observe that aα,β = aβ ∨ ,α∨ for α, β ∈ R. Since sα∨ ,α = t (sα,α∨ )−1 , the map W (R) → W (R∨ ) , w → t w−1 is an isomorphism of groups. So we may identify W (R) and W (R∨ ) via this isomorphism and consider that W (R) acts on V and V ∗ . Let α ∈ R and g ∈ A(R). Set β = g(α), θ = t g −1 (α∨ ). Then
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θ(β) = 2 , sβ,θ = g ◦ sα ◦ g −1 . Hence sβ,θ (R) = R and θ = β ∨ . Hence t g(R∨ ) = R∨ and the map g → t g −1 is an isomorphism from A(R) onto A(R∨ ). So we may consider that A(R) acts on V ∗ . Note that the previous discussion shows that W (R) is a normal subgroup of A(R).
18.3 Root systems and bilinear forms 18.3.1 Proposition. For x, y ∈ V , the map V × V → k defined by: ∨ (x, y) → (x|y) = α (x)α∨ (y) α∈R
is a non-degenerate A(R)-invariant symmetric bilinear form on V . Proof. If g ∈ A(R), we saw in 18.2.9 that t g(R∨ ) = R∨ . Thus t (g(x)|g(y)) = g(α∨ )(x)t g(α∨ )(y) = (x|y). α∈R
So the bilinear form (.|.) is A(R)-invariant. Clearly it is symmetric. For α, β ∈ R, we have α∨ (β) ∈ Z and β ∨ (β) = 2. So (β|β) ∈ N and (β|β) 4. So this bilinear form is non-zero. Now let R1 , . . . , Rn be the irreducible components of R and Vi the subspace spanned by Ri . Then by 18.2.3, for i = j, α∨ (x)α∨ (y) = 0 for all x ∈ Vi , y ∈ Vj and α ∈ R. Thus Vi and Vj are orthogonal with respect to (.|.). We are therefore reduced to the case where R is irreducible and the proposition follows from 18.1.8 and 18.2.7. 18.3.2 Proposition. Let R1 , . . . , Rn be the irreducible components of R, Vi the subspace spanned by Ri and B a W (R)-invariant symmetric bilinear form on V . (i) The restriction of B to Vi is proportional to the one defined in 18.3.1. (ii) The Vi ’s are pairwise orthogonal with respect to B. Proof. (i) By 18.2.7, for 1 i n, Vi is a simple W (Ri )-module. So (i) follows from 18.1.8 and 18.3.1. (ii) Let Ui be the subspace of Vi spanned by w(x) − x, w ∈ W (Ri ) and x ∈ Vi . If w, w ∈ W (Ri ) and x ∈ Vi , then: w (w(x)) − w (x) = [w (w(x)) − w(x)] + [w(x) − x] − [w (x) − x]. So Ui is a W (Ri )-module. Further, sα (α) − α = −2α = 0 if α ∈ Ri . So Ui = {0}. We deduce from 18.2.7 that Ui = Vi . Now let x ∈ Vi , y ∈ Vj with i = j. We saw in 18.2.3 that w|Vj = idVj for all w ∈ W (Ri ). It follows that: B(w(x), y) = B(w(x), w(y)) = B(x, y). So w(x) − x is orthogonal to y. Thus Vi and Vj are orthogonal with respect to B.
18.4 Passage to the field of real numbers
239
18.3.3 Let (.|.) be a non-degenerate W (R)-invariant bilinear form on V . For α, β ∈ R, it follows from 18.1.7 and 18.3.2 that aβ,α = 2
(α|β) . (α|α)
Hence: (α|β) = 0 ⇒
(β|β) aβ,α , aα,β = 0 ⇔ aβ,α = 0 ⇔ (α|β) = 0. = aα,β (α|α)
If x ∈ V , then: sβ ◦ sα (x) − sα ◦ sβ (x) = aα,β α∨ (x)β − aβ,α β ∨ (x)α. Consequently, if α and β are not proportional, then aα,β = 0 ⇔ aβ,α = 0 ⇔ sα ◦ sβ = sβ ◦ sα .
18.4 Passage to the field of real numbers 18.4.1 Let R be a root system in V . Denote by VQ (resp. VQ∗ ) the Q-vector subspace spanned in V (resp. V ∗ ) by R (resp. R∨ ). Let K be an extension of k. We can identify canonically V as a subset of K ⊗k V , and V ∗ as a subset of K ⊗k V ∗ = (K ⊗k V )∗ . Then R is a root system in K ⊗k V and the α∨ ’s are the same. 18.4.2 Proposition. (i) R is a root system in VQ . (ii) The vector space V (resp. V ∗ ) identifies canonically with k⊗Q VQ (resp. k ⊗Q VQ∗ ). The canonical bilinear form on V × V ∗ allows us to identify VQ∗ with (VQ )∗ and VQ with (VQ∗ )∗ . ∨ on VQ . It is Proof. (i) By (R3), α∨ (VQ ) ⊂ Q, so α∨ defines a linear form αQ ∨ now easy to verify that R is a root system in VQ , and αQ is the linear form associated to α in (VQ )∗ . (ii) Let i : k ⊗Q VQ → V be the canonical homomorphism and t i : V ∗ → k ⊗Q (VQ )∗ its dual. The map i is surjective since R spans V . Further, if α ∈ R, ∨ ∨ . By (i) and 18.2.8, the αQ ’s span (VQ )∗ , and so t i is then t i(α∨ ) = 1 ⊗ αQ t surjective. Hence i and i are isomorphisms. They allow us to identify VQ∗ with (VQ )∗ and VQ with (VQ∗ )∗ .
18.4.3 By 18.4.1 and 18.4.2, we can consider R as a root system in VQ and in VR = R ⊗Q VQ , and the corresponding Weyl groups are canonically isomorphic. Let us use the notations of 18.3.1. If x, y ∈ VQ , we have (x|y) ∈ Q. Denote again by (.|.) the canonical extension of this bilinear form to VR . Then for z ∈ VR : ∨ [α (z)]2 0. (z|z) = α∈R
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By 18.3.1, equality holds if and only if z = 0. Thus (.|.) is a W (R)-invariant inner product on VR . Conversely, let us fix a W (R)-invariant inner product on VR . We can consider the length of a root and the angle between two roots. By 18.3.2, this angle does not depend on the choice of the inner product. Similarly, if two roots belong to the same irreducible component, then the ratio of their lengths does not depend on the choice of the inner product.
18.5 Relations between two roots 18.5.1 We shall assume in the rest of this chapter that k = R. Let us fix a root system R in V and a W (R)-invariant inner product (.|.) on V . If α, β ∈ R, denote by θαβ ∈ [0, π[ the angle between α and β, ωαβ the order of sα ◦ sβ and α the length of α. 18.5.2 If α, β ∈ R, then 18.3.3 implies that: aα,β aβ,α = 4 cos2 θα,β 4. Furthermore, aα,β aβ,α ∈ Z. Assuming α β, we see easily that the only possibilities are listed in the following table: θαβ β2 /α2 π undetermined ωαβ 2 π 1 ωαβ 3 2π 1 ωαβ 3 π 2 ωαβ 4 3π 2 ωαβ 4 π 3 ωαβ 6 5π 3 ωαβ 6
aα,β
aβ,α
0
0
1
1
−1
−1
1
2
−1
−2
1
3
−1
−3
2
2
0
1
β=α
−2
−2
π
1
β = −α
1
4
0
4
β = 2α
−1
−4
π
4
β = −2α
=2 =3 =3 =4 =4 =6 =6
18.5 Relations between two roots
241
18.5.3 We deduce from the table above the following proposition. Proposition. Let α and β be roots. ! 1" (i) If β = λα with λ ∈ k, then λ ∈ ± 1, ±2, ± . 2 (ii) If β ∈ kα and α β, then aα,β ∈ {−1, 0, 1}. (iii) If aα,β > 0, that is (α|β) > 0, then α − β ∈ R unless α = β. (iv) If aα,β < 0, that is (α|β) < 0, then α + β ∈ R unless α = −β. (v) If α − β and α + β do not belong to R ∪ {0}, then (α|β) = 0. 1 α ∈ R. 2 2) It may happen that α + β ∈ R and (α|β) = 0. Also, if α − β and α + β do not belong to R ∪ {0}, we say that α and β are strongly orthogonal. Remarks. 1) A root in R is reduced if
18.5.4 Proposition. Let α, β be roots which are not proportional. Set Iα,β = {j ∈ Z; β + jα ∈ R} and Sα,β = {β + jα; j ∈ Iα,β }. (i) The set Iα,β is an interval [−p, q] of Z containing 0. (ii) The set Sα,β is stable under sα and sα (β + qα) = β − pα , sα (β − pα) = β + qα. We say that Sα,β is the α-string through β, and that β − pα is the source, β + qα is the target and p + q the length. (iii) We have aβ,α = p − q. Proof. Let q = max Iα,β , −p = min Iα,β . Since β ∈ R, 0 ∈ Iα,β . Suppose that Iα,β = [−p, q]. Then there exist −p r, s q such that r, s ∈ Iα,β , r < s − 1 , r + i ∈ Iα,β if 1 i s − r − 1. It follows from 18.5.3 that (α|β + sα) 0 and (α|β + rα) 0. But this is absurd since: (α|β + sα) − (α|β + rα) = (s − r)α2 > 0. So Iα,β = [−p, q]. Next, we have sα (β + jα) = β − (aβ,α + j)α. Since sα (R) = R, sα (Sα,β ) = Sα,β . So (ii) follows from the fact that j → −aβ,α − j is an order-reversing bijection from Iα,β to itself. Finally, since sα (β + qα) = β − pα = β − (aβ,α + q)α, part (iii) follows from (ii). 18.5.5 Corollary. Let α, β be roots that are not proportional, S the αstring through β and γ its source. The length of S is −aγ,α and it is equal to 0, 1, 2 or 3. Proof. Note that S = Sα,γ , so the first part follows from 18.5.4 (iii) and the second follows from the table in 18.5.2.
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18.5.6 Proposition. Let α, β be roots which are not proportional and p, q be the integers such that Iα,β = [−p, q]. If α + β ∈ R, then β + α2 p+1 . = β2 q Proof. Let γ = β−pα be the source of the α-string through β. Since α+β ∈ R, the length l = p + q of Sα,γ = Sα,β is in {1, 2, 3} (18.5.5). The results follows since by using 18.5.2 and 18.5.4, the only possibilities are listed in the following table: l p q aβ,α 1 2
β=γ β=γ
0 1 −1 β + α2 = β2 0 2 −2 2 β + α2 = β2
2 3 3
β =γ+α β=γ β =γ+α
1 1 0 β + α2 = 2 β2 0 3 −3 3 β + α2 = β2 1 2 −1 β + α2 = β2
3 β = γ + 2α 2 1
1
β + α2 = 3β2
Let us take for example the case p = 0 and q = 2, then by 18.5.2, aα,β = −1. Now β + α2 = β2 + α2 + 2(β|α). Replacing 2(β|α) by β2 aα,β , we obtain β + α2 = α2 . So we have 2β + α2 = β2 by replacing 2(β|α) by α2 aβ,α in the above equality. 18.5.7 Proposition. Assume that R is irreducible. Let α, β ∈ R, then there exists w ∈ W (R) such that w(β) = α if and only if α = β. Proof. If α = β, then since R is irreducible, there exists w ∈ W (R) such that α and w(β) are not orthogonal. So we may assume that (α|β) = 0. Replacing β by sβ (β) if necessary, we may further assume that (α|β) > 0. By 18.5.2, either α = β or aα,β = aβ,α = 1. In the latter, sα ◦ sβ ◦ sα (β) = sα ◦ sβ (β − α) = sα (−β − α + β) = α. So we are done. The converse is clear since the inner product is W (R)invariant. 18.5.8 Proposition. Assume that R is irreducible and reduced. (i) For α, β ∈ R, we have: # β2 1 1 , ∈ 1, 2, 3, . α2 2 3 (ii) The set {α; α ∈ R} contains at most two elements.
18.6 Examples of root systems
243
Proof. (i) In view of 18.2.7, the subspace spanned by w(β), w ∈ W (R), is V . Thus there exists w ∈ W (R) such that (α|w(β)) = 0. Since w(β) = β, the result follows from 18.5.2. (ii) This is a simple consequence of part (i). 18.5.9 Assume that R is irreducible and reduced. If {α; α ∈ R} = {λ, µ} where λ < µ, then a root of length λ (resp. µ) is called a short (resp. long) root. If all the roots have the same length, then, by convention, they are called long.
18.6 Examples of root systems 18.6.1 In view of the results of the previous sections, we give some examples of root systems of small rank over the real numbers. The reader may verify easily that the examples below are indeed roots systems. • Rank 1. A1 −α
α
Clearly, this is the only irreducible reduced root system of rank 1. • Rank 2. A1 × A1 β −α
α −β
The roots α and β are orthogonal, and this root system is not irreducible. A2 β
β+α
−α
α
−β−α
−β
Here, all the roots have the same length, and the angle between two adjacent roots is π/3. This is an irreducible reduced root system of rank 2.
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B2 β+α
β
−α
β+2α
α
−β−2α
−β−α
−α
√ This is again an irreducible reduced root system of rank 2. We have β = 2α, and the angle between two adjacent roots is π/4. G2 2β+3α
β
β+α
β+2α
−α
−β−3α
β+3α
α
−β−2α
−β−α
−β
−2β−3α
√ In this root system, β = 3α, and the angle between two adjacent roots is π/6. This rank 2 root system is also irreducible and reduced.
18.7 Base of a root system 18.7.1 In this rest of this chapter, all root systems considered are assumed to be reduced. 18.7.2 We define a lexicographic order on V to be an order obtained in the following way: choose a R-basis (e1 , . . . , el ) of V , and write λ1 e1 + · · · + λl el ≺ µ1 e1 + · · · + µl el if for some integer n, we have λ1 = µ1 , . . ., λn = µn , and λn+1 < µn+1 . Clearly, this is a total order on V which is compatible with addition.
18.7 Base of a root system
245
18.7.3 Let ≺ be a lexicographic order on V . Set: R+ = {α ∈ R; α 0} , R− = {α ∈ R; α ≺ 0}. An element of R+ (resp. R− ) is called a positive (resp. negative) root. Denote by B the set of elements of R+ which is not the sum of two positive roots. The set B is called a base of R. 18.7.4 Theorem. Let R be a root system, ≺ a lexicographic order on V , R+ the set of positive roots and B the base of R with respect to this order. Then: (i) Any positive root is the sum of roots in B. (ii) If α, β ∈ B are distinct, then aα,β 0 and α − β ∈ R. (iii) If B = {β1 , . . . , βq }, then (β1 , . . . , βq ) is a basis of V . (iv) If α ∈ R+ \ B, then there exists β ∈ B such that aα,β > 0 and α − β ∈ R+ . (v) If α ∈ R+ , β ∈ B and α = β, then sβ (α) ∈ R+ . In particular, sβ permutes the elements of R+ \ {β}. Proof. (i) Let α1 ≺ · · · ≺ αn be the elements in R+ . It is clear that α1 ∈ B. Suppose that α1 , . . . , αi−1 can all be expressed as a sum of elements of B. If αi ∈ B. then αi = αr + αs where αr , αs ∈ R and r, s < i. We deduce therefore that αi is also a sum of elements of B. So by induction, we have (i). (ii) If α − β ∈ R, then β − α ∈ R. So either α − β or β − α is positive. This is impossible since α, β ∈ B and α = (α − β) + β, β = (β − α) + α. Thus α − β ∈ R, and aα,β 0 by 18.5.3 (iii). (iii) Let λ1 , . . . , λq ∈ R+ be such that, reindexing if necessary, (2)
v = λ1 β1 + · · · + λp βp = λp+1 βp+1 + · · · + λq βq . Then by part (ii), (v|v) = (λ1 β1 + · · · + λp βp |λp+1 βp+1 + · · · + λq βq ) 0.
Hence v = 0. Since βi 0, it follows that λi = 0 for all i. So the elements of B are linearly independent, and part (i) implies that B is a basis of V . (iv) By (i), there exist β1 , . . . , βp ∈ B pairwise distinct, and n1 , . . . , np ∈ N∗ such that α = n1 β1 + · · · + np βp . Since (α|α) > 0, it follows that (βi |α) > 0 for some i. Then 18.5.3 says that α − βi and βi − α are roots. Since βi ∈ B, this forces α − βi to be positive. (v) Let us write α = n1 β1 + · · · + np βp as above. Since α = β, there exists r such that βr = β. It follows that the coefficient of βr in sβ (α) = α − aβ,α β is nr . But by (i) and (iii), the coefficients of a root in the basis B are either all positive or all negative. So sβ (α) ∈ R+ . 18.7.5 Remarks. 1) An element of B is called a simple root of R with respect to B.
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2) In general, let V be a k-vector space and consider R as a root system in VR . The same proof (18.7.4) shows that B is a basis of V , VQ and VR . 18.7.6 Corollary. Set: ρ=
1 α. 2 α∈R+
Then sβ (ρ) = ρ − β and β ∨ (ρ) = 1 for all β ∈ B. Proof. This is a simple consequence of 18.7.4 (v).
18.7.7 Proposition. Let ≺ be a lexicographic order on V and α1 , . . . , αn positive roots such that α = α1 + · · · + αn ∈ R+ . Then there is a permutation σ of {1, 2, . . . , n} such that ασ(1) + · · · + ασ(i) ∈ R+ for 1 i n. Proof. Since (α|α) > 0, there exists j such that (α|αj ) > 0. If α = αj , then n = 1 and the result is obvious. Otherwise by 18.5.3, α − αj ∈ R+ , and the result follows by induction on n. 18.7.8 Proposition. Let ≺ be a lexicographic order on V , B be a base of R and WB the subgroup of W (R) generated by sβ , β ∈ B. (i) We have R = {w(β); w ∈ WB , β ∈ B}. (ii) We have WB = W (R). Proof. Let S = {w(β); w ∈ WB , β ∈ B} ⊂ R. (i) Since sβ (β) = −β, it suffices to show that R+ ⊂ S. Let α1 ≺ · · · ≺ αn be the elements of R+ . Since α1 ∈ B and B ⊂ S, α1 ∈ S. Let us suppose that α1 , . . . , αi−1 ∈ S. If αi ∈ B, then αi ∈ S. Otherwise, by 18.7.4 (iv), there exists β ∈ B such that aα,β > 0. Thus δ = sβ (α) = α − aα,β β ≺ α. Further δ ∈ R+ (18.7.4 (v)). So δ ∈ S and hence α = sβ (δ) ∈ S. Now (i) follows by induction. (ii) It suffices to prove that sα ∈ WB for all α ∈ R. By part (i), there exist w ∈ WB and β ∈ B such that α = w(β). Hence sα = w ◦ sβ ◦ w−1 ∈ WB . 18.7.9 Proposition. (i) Let B be a base of R, B ⊂ B, V the subspace spanned by B and R = R ∩ V . Then B is a base for the root system R . (ii) Let W be a subspace of V , R = R ∩ W and B a base of R . Then there exists a base of R containing B . Proof. (i) We may assume that B = ∅. By 18.2.5, R is a root system in V . = V ∩ R+ . Then it is clear Let B = (β1 , . . . , βl ), B = (β1 , . . . , βp ) and R+ that with respect to the lexicographic order associated to B (resp. B ) on V ) is the set of positive roots. Hence B is the base of (resp. V ), R+ (resp. R+ R with respect to this order. (ii) Let B = (β1 , . . . , βp ) and complete to a basis (β1 , . . . , βl ) of V . The lexicographic order on V associated to this basis defines a base B of R which clearly contains B .
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18.8 Weyl chambers 18.8.1 We shall endow V with the usual topology defined for example by the W (R)-invariant inner product. Let E = (e1 , . . . , el ) be a basis of V and λ1 , . . . , λl ∈ V ∗ such that for any x ∈ V , x = λ1 (x)e1 + · · · + λl (x)el . The set {x ∈ V ; λi (x) > 0, 1 i l} will be called the open simplicial cone associated to the basis E. 18.8.2 Let B = {β1 , . . . , βl } be a base of R and si = sβi for 1 i l. For w ∈ W (R), by 18.7.8 (ii), there exist 1 i1 , . . . , in l such that w = si1 ◦ · · · ◦ sin . Such an expression is called a reduced decomposition if n is minimal, and we say that n is the length of w that we shall denote by (w). By convention, (idV ) = 0. 18.8.3 Lemma. Let B be a base of R. (i) Let β1 , . . . , βp ∈ B not necessary distinct. If s1 ◦ · · · ◦ sp−1 (βp ) ∈ R− , si = sβi , then s1 ◦ · · · ◦ sp = s1 ◦ · · · ◦ sq−1 ◦ sq+1 ◦ · · · ◦ sp−1 for some q ∈ {1, . . . , p − 1}. (ii) Let w ∈ W (R) and β1 , . . . , βp ∈ B such that w = s1 ◦ · · · ◦ sp , si = sβi . If p = (w), then w(βp ) ∈ R− . (iii) Let w ∈ W (R) and n(w) the number of positive roots α such that w(α) ∈ R− . Then n(w) = (w). Proof. (i) Let αp−1 = βp and αi = si+1 ◦ · · · ◦ sp−1 (βp ), 0 i p − 2. We have α0 ∈ R− and αp−1 ∈ R+ . Let q be the smallest integer such that αq ∈ R+ . Then sq (αq ) = αq−1 ∈ R− . So αq = βq by 18.7.4 (v). Since sw(α) = w◦sα ◦w−1 for w ∈ W (R) and α ∈ R, we have: sq = (sq+1 ◦ · · · ◦ sp−1 ) ◦ sp ◦ (sp−1 ◦ · · · ◦ sq+1 ). So part (i) follows. (ii) We have w(βp ) = −s1 ◦ · · · ◦ sp−1 (βp ). If s1 ◦ · · · ◦ sp−1 (βp ) ∈ R− , then p > (w) by part (i). So we have proved (ii). (iii) Proceed by induction on (w). The case (w) = 0 is obvious. Let w = s1 ◦ · · · ◦ sp be a reduced decomposition of w where the notations are as above. By (ii), w(βp ) ∈ R− . So it follows from 18.7.4 (v) that n(w ◦ sp ) = n(w) − 1. But it is clear that (w ◦ sp ) = (w) − 1, so we have by induction that n(w) = (w). 18.8.4 For α ∈ R, define Pα = ker α∨ = {x ∈ V ; (x|α) = 0}. Let P be the union of Pα , α ∈ R. It is a closed subset of V , and the connected components of P = V \ P are open subsets of V , called the Weyl chambers (or chambers) of R. By 18.3.2, they do not depend on the choice of the W (R)-invariant inner product on V . Since the inner product is W (R)-invariant, W (R) acts on P and P , and W (R) permutes the chambers of R.
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18.8.5 Lemma. The group W (R) acts transitively on the set of Weyl chambers of R. Proof. Let C1 , C2 be chambers in R, x1 ∈ C1 and x2 ∈ C2 . Let w ∈ W (R) be such that x1 − w(x2 ) = inf{x1 − w (x2 ); w ∈ W (R)}. Set I = {tx1 + (1 − t)w(x2 ); t ∈ [0, 1]}. Suppose that I ∩ Pα = ∅ for some α ∈ R, then since x1 , w(x2 ) ∈ Pα , there exists t0 ∈]0, 1[ such that (t0 x1 + (1 − t0 )w(x2 )|α) = 0. Thus x1 − sα ◦ w(x2 )2 = x1 − w(x2 )2 − < x1 − w(x2 )2 .
4(1 − t0 ) (w(x2 )|α)2 t0 α2
But this contradicts the assumption on w, and since I is connected, I is contained in a chamber. Hence x1 , w(x2 ) ∈ C1 . Since w permutes the chambers and C1 ∩ w(C2 ) = ∅, we have C1 = w(C2 ). 18.8.6 Proposition. Let B = {β1 , . . . , βl } be a base of R, (β1 , . . . , βl ) the basis of V dual to B with respect to the inner product (.|.), and l # xi βi ; x1 , · · · , xl > 0 . C(B) = {x ∈ V ; (x|βi ) > 0, 1 i l} = i=1
Then C(B) is a Weyl chamber of R. Proof. It is clear from 18.7.4 that C(B) ⊂ P . Since C(B) is convex, it is connected. So C(B) is contained in a chamber, say C, of R. Suppose that there exists x ∈ C \ C(B). Then (x|β) < 0 for some β ∈ B. Set C1 = {x ∈ C; (x|β) > 0} and C2 = {x ∈ C; (x|β) < 0}. Then C is the disjoint union of the open subsets C1 and C2 . Now x ∈ C2 and C(B) ⊂ C1 . This implies that C is not connected. Contradiction. 18.8.7 Theorem. Let R be a root system in V . (i) Any chamber of R is of the form C(B) for some base B of R. In particular, any chamber of R is an open simplicial cone. (ii) The map B → C(B) induces a bijection between the set of bases of R and the set of chambers of R. (iii) The group W (R) acts simply transitively on the set of bases of R and on the set of chambers of R. Proof. (i) If w ∈ W (R) and B a base of R, then clearly w(B) is another base of R. So (i) follows from 18.8.5 and 18.8.6. (ii) The map B → C(B) is surjective by (i). Since C(B) is uniquely determined by R+ βi , 1 i l, (notations of 18.8.6), the map is injective. (iii) By 18.8.5 and (ii), W (R) acts transitively on the set of bases of R. Let B be a base and w ∈ W (R) such that w(B) = B. Then 18.8.3 (ii) forces w to be idV . Hence the action is simply transitive on the set of the bases and, by (ii), on the set of chambers as well.
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18.8.8 Remarks. 1) For a fixed base B of R, the chamber C(B) is called the fundamental Weyl chamber. 2) If C is a Weyl chamber, denote by B = B(C) the corresponding base of R, R+ = R+ (C) and R− = R− (C). For β ∈ B, the hyperplane Pβ is called a wall of the chamber C. 3) Suppose that R is the direct sum of root systems R1 , . . . , Rn . Let Ci be a chamber of Ri and Bi the corresponding base of Ri , then C = C1 × · · · × Cn is a chamber of R corresponding to the base B = B1 ∪ · · · ∪ Bn . By 18.8.7 (iii), all the chambers and bases of R are obtained in this way. 18.8.9 Proposition. Let B = {β1 , . . . , βl } be a base of R and C = C(B). (i) For all x ∈ V , there exists w ∈ W (R) such that w(x) ∈ C. (ii) If x, y ∈ C and w ∈ W (R) verify w(x) = y, then x = y. (iii) Any W (R)-orbit of V intersects C at a unique point. Proof. Let us define the following partial order on V : x y ⇔ y − x ∈ R+ β1 + · · · + R+ βl . (i) Let w be maximal with respect to in the set {w (x); w ∈ W (R)}. If w(x) ∈ C, then there exists β ∈ B such that β ∨ (w(x)) = (β|w(x)) < 0. But sβ ◦ w(x) − w(x) = −β ∨ (w(x))β, so w(x) ≺ sβ ◦ w(x) which contradicts the maximality of w. So w(x) ∈ C. (ii) We shall show by induction on (w) that w is the product of reflections sβ , β ∈ B such that sβ (x) = x. The case (w) = 0 is trivial. So let us suppose that (w) > 0. Then by 18.7.8 (ii) and 18.8.3 (ii), there exists β ∈ B such that w(β) ∈ R− and (w ◦ sβ ) = (w) − 1. Since x, y ∈ C, 0 (β|x) = (β|w−1 (y)) = (w(β)|y) 0. Hence (β|x) = 0. Thus sβ (x) = x and w ◦ sβ (x) = y. The result follows by induction. Thus x = y. (iii) This is clear from (i) and (ii). 18.8.10 Remark. The partial order introduced in the proof of 18.8.9 is called the partial order associated to B (or to C). 18.8.11 Remark. The notions of bases and Weyl chambers can easily be extended to a non-reduced root system R. In particular, R has a base, and W (R) acts simply transitively on the set of bases and the set of Weyl chambers.
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18.9 Highest root 18.9.1 Lemma. Assume that R is irreducible. Let B be a base of R. Then B is not the (disjoint) union of two orthogonal subsets. Proof. Suppose that B is the disjoint union of (non-empty) subsets B1 and B2 such that (β1 |β2 ) = 0 for β1 ∈ B1 , β2 ∈ B2 . By 18.7.8 (ii), the subspace of V spanned by B1 is proper and W (R)-stable. This contradicts 18.2.7. 18.9.2 Theorem. Assume that R is irreducible. Let B = {β1 , . . . , βl } be a base of R, C = C(B) the corresponding chamber. (i) There exists a root α = n1 β1 + · · · + nl βl such that for any root α = p1 β1 + · · · + pl βl , we have ni pi , 1 i l. (ii) We have α ∈ C. (iii) We have α α for all α ∈ R. In particular, α is a long root. α}, then aα, α ∈ {0, 1}. (iv) If α ∈ R+ \ { Proof. Let α = n1 β1 + · · · + nl βl be a maximal element of R with respect to the partial order associated to B. Clearly, α ∈ R+ . Let I = {1, . . . , l} , I1 = {i ∈ I; ni > 0} , I2 = {i ∈ I; ni = 0} = I \ I1 . We claim that I2 = ∅, because otherwise, for each i ∈ I, (α|βi ) 0 (18.7.4), and so by 18.9.1, (βi |βj ) = 0 for some i ∈ I2 and j ∈ I1 . It follows that (α|βi ) < 0 and hence α ≺ α + βi ∈ R+ (18.5.3), which contradicts the maximality of α. So we have our claim. Using 18.5.3 again, we deduce that (α|βi ) 0 for all i ∈ I, and so (α|βi ) > 0 for some i ∈ I. Now, if α is another maximal element of R with respect to , then (α|α ) > 0. So if α = α , then α − α ∈ R (18.5.3), and therefore α α or α α which again contradicts the maximality of α and α . Thus we have proved (i), and part (ii) follows from 18.5.3. Let α ∈ R. To prove (iii), we may assume that α = α and α ∈ C (18.8.9). α| α − α) 0 and Since 0 ≺ α − α, (x| α − α) 0 for all x ∈ C. In particular, ( (α| α − α) 0. Hence ( α| α) (α| α) (α|α). α}, then α α and ( α|α) 0 since α ∈ C. So Finally, if α ∈ R+ \ { (iv) is a consequence of 18.5.2. 18.9.3 Assume that R is irreducible. The root α in 18.9.2 is called the highest root of R with respect to the base B (or chamber C).
18.10 Closed subsets of roots 18.10.1 Definition. Let P be a subset of R. (i) We say that P is symmetric if P = −P . (ii) We say that P is closed if given α, β ∈ P verifying α + β ∈ R, then α + β ∈ P. (iii) We say that P is parabolic if P is closed and R = P ∪ (−P ). 18.10.2 Proposition. Let P ⊂ R be a closed subset such that P ∩(−P ) = ∅. There exists a chamber C of R such that P ⊂ R+ (C).
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251
Proof. 1) Let n ∈ N∗ , α1 , . . . , αn ∈ P and α = α1 + · · · + αn . We claim that α = 0. This is obvious if n = 1. Let us proceed by induction on n. If n 2 and α = 0, then (α1 |α2 +· · ·+αn ) < 0. So there exists some j such that (α1 |αj ) < 0. Since P is closed, it follows from 18.5.3 that α1 + αj ∈ P . Thus α = 0 is the sum of n − 1 elements of P , which contradicts the induction hypothesis. So we have proved our claim. 2) We claim there exists α ∈ P such that (α|β) 0 for all β ∈ P . For otherwise, there exist α1 , α2 ∈ P such that (α1 |α2 ) < 0, then α1 + α2 ∈ P , and there exists α3 ∈ P such that (α1 + α2 |α3 ) < 0, and so α1 + α2 + α3 ∈ P . Repeating this process, we obtain a sequence (αi )i1 of elements of P such that α1 + · · · + αi ∈ P for all i ∈ N∗ . It follows that there exists i < j such that α1 + · · · + αi = α1 + · · · + αj . Hence αi+1 + · · · + αj = 0 which contradicts 1). 3) Finally, let us prove that there is a basis E of V such that the elements of P are positive with respect to the lexicographic order associated to E. The case l = 1 is trivial. Let us proceed by induction on l and assume that l 2. From 2), there exists v1 ∈ V \ {0}, such that (v1 |α) 0 for all α ∈ P . Let H be the hyperplane orthogonal to v1 and V1 the subspace of V spanned by R ∩ H. It follows from 18.2.5 that P ∩ H is a closed subset of R ∩ H. By our induction hypothesis, there exists a basis (v2 , . . . , vl ) of H such that the elements of P ∩H are positive with respect to the associated lexicographic order. It is then clear that the elements of P are positive with respect to the lexicographic order associated to the basis (v1 , . . . , vl ) of V . 18.10.3 Corollary. The following conditions are equivalent for a subset P of R. (i) There exists a chamber C of R such that P = R+ (C). (ii) P is parabolic and P ∩ (−P ) = ∅. If these conditions are verified, the chamber C such that P = R+ (C) is unique. Proof. The equivalence is clear from 18.10.2. If P = R+ (C), then C ∨ is the set of ϕ ∈ V ∗ verifying ϕ(α) > 0 for all α ∈ P . Hence the uniqueness of C. 18.10.4 Lemma. Let C be a chamber of R, B = B(C), R+ = R+ (C), P a closed subset of R containing R+ , Σ = B ∩ (−P ) and Q the set of roots which is the sum of elements of −Σ. Then P = R+ ∪ Q. Proof. Since P is closed, Q ⊂ P (18.7.7). So R+ ∪ Q ⊂ P . Now let α ∈ P \R+ . Then α is the sum of p elements of −B. We shall show by induction on p that α ∈ Q. The case p = 1 is obvious. So let us assume that p 2. By 18.7.7, there exist β ∈ −B, γ ∈ R− such that α = β + γ, and γ is the sum of p−1 elements of −B. Since R+ ⊂ P , −γ, −β ∈ P . So β = (−γ)+α ∈ P and γ = (−β) + α ∈ P , because P is closed. Thus β ∈ −Σ, and γ ∈ Q by our induction hypothesis. Again since P is closed, α = β + γ ∈ Q.
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18.10.5 Proposition. The following conditions are equivalent for a subset P of R. (i) P is parabolic. (ii) P is closed and there exists a chamber C of R such that R+ (C) ⊂ P . (iii) There exists a chamber C of R and a subset Σ of B(C) such that P is union of R+ (C) and the set of roots which can be expressed as a sum of elements of −Σ. Proof. (i) ⇒ (ii) Let C be a chamber such that the cardinal of P ∩ R+ (C) is maximal. Set B = B(C), R+ = R+ (C). Suppose that there exists β ∈ B \ P . So −β ∈ P and −β = sβ (β) ∈ where C = sβ (C) and R+ = R+ (C ). Hence −β ∈ P ∩ R+ . sβ (R+ ) = R+ If α ∈ R+ \ {β}, then sβ (α) ∈ R+ (18.7.4 (v)). So α = sβ ◦ sβ (α) ∈ . Hence P ∩ R+ ⊂ P ∩ R+ . sβ (R+ ) = R+ which contradicts our choice We deduce that {−β} ∪ (P ∩ R+ ) ⊂ P ∩ R+ of C. Hence B ⊂ P and R+ ⊂ P . (ii) ⇒ (iii) This is clear by 18.10.4. (iii) ⇒ (i) We only need to show that P is closed. Let Σ = {β1 , . . . , βp } and B = B(C) = Σ ∪ {βp+1 , . . . , βl }. Let α, β ∈ P be such that α + β ∈ R. Since R+ (C) ⊂ P , we may assume that α + β = −n1 β1 − · · · − nl βl ∈ R− (C), where n1 , . . . , nl ∈ N. Since P = R+ (C) ∪ Q, ni = 0 for p + 1 i l. Hence α + β ∈ Q ⊂ P . 18.10.6 Proposition. Let P be a subset of R, Γ (resp. V ) the subgroup (resp. subspace) of V generated (resp. spanned) by P . Then the following conditions are equivalent: (i) Γ ∩ R = P . (ii) P is closed and symmetric. (iii) P is closed and is a root system in V . If these conditions are verified, then the map α → α∨ |V is a bijection from P onto P ∨ . Proof. (i) ⇒ (ii) This is clear. (ii) ⇒ (iii) With the notations of 18.2.1, P verifies (R1) in V . Let α, β ∈ P . If β = ±α, then sα (β) = −β ∈ P . If β = ±α, then for 0 j aβ,α , β − jα ∈ R. Since P is symmetric and closed, sα (β) = β − aβ,α α ∈ P . Thus sα (P ) ⊂ P , and (R2) is verified. Finally, it is clear that (R3) is verified. So P is a root system V and last part of the proposition follows also. (iii) ⇒ (i) Clearly, P ⊂ Γ ∩ R. Conversely, if α ∈ Γ ∩ R, then α = α1 +· · ·+αn where αi ∈ P , since P = −P . From 0 < (α|α) = (α|α1 +· · ·+αn ), we deduce that (α|αi ) > 0 for some i. If α = αi , then α ∈ P . Otherwise, α − αi ∈ R (18.5.3) and α − αi ∈ Γ . By induction on n, α − αi ∈ P and since P is closed, α = (α − αi ) + αi ∈ P .
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18.11 Weights 18.11.1 Let us use the notations of 18.5.1 and 18.8.1. We define an isomorphism ϕ : V → V ∗ , ϕ(x)(y) = (x|y) for all x, y ∈ V . Denoting (ϕ(x)|ϕ(y)) = (x|y), we obtain a W (R∨ )-invariant inner product on V ∗ . For α ∈ R, we have: 2α∨ ϕ(α) = ∨ ∨ . (α |α ) Thus ϕ(Pα ) = Pα∨ (18.8.4). From this, we deduce that the image of a chamber of R via ϕ is a chamber of R∨ . 18.11.2 Let C be a chamber of R, B = B(C) = {β1 , . . . , βl }, C ∨ = ϕ(C) and B = B(C ∨ ) = {γ1 , . . . , γl }, S(C) = {x ∈ V ; (x|y) 0 for all y ∈ C} and S(C ∨ ) = {λ ∈ V ∗ ; (λ|µ) 0 for all µ ∈ C ∨ }. By 18.8.6, we have: S(C) = R+ β1 + · · · + R+ βl , S(C ∨ ) = R+ γ1 + · · · + R+ γl . It is then clear that S(C ∨ ) = R+ β1∨ + · · · + R+ βl∨ , and we deduce that B = B ∨ = {β1∨ , . . . , βl∨ }. In particular, B ∨ is a base of R∨ . 18.11.3 Let α = n1 β1 + · · · + nl βl ∈ R. Then: l 2βi∨ 2α∨ = ϕ(α) = . n i (α∨ |α∨ ) (βi∨ |βi∨ ) i=1
Hence α∨ =
l
ni βi 2 α−2 βi∨ .
i=1
By 18.11.2, ni βi 2 α−2 ∈ Z for 1 i l. 18.11.4 Denote by Q(R) the sublattice of V spanned by R. By 18.7.4, it is a free abelian subgroup of rank l in V . Any base of R is a basis of Q(R). In the same way, Q(R∨ ) is a sublattice of rank l in V ∗ . The lattice Q(R) is called the root lattice of R and its elements are called radical weights. 18.11.5 Proposition. Let P (R) be the set of x ∈ V such that α∨ (x) ∈ Z for all α ∈ R. We call P (R) the weight lattice of R and its elements are called weights. (i) P (R) is a sublattice of rank l in V . (ii) We have Q(R) ⊂ P (R) ⊂ VQ . (iii) If B ∨ is a base of R∨ , the dual basis of B ∨ in V is a basis of P (R). Proof. If x ∈ V , the following conditions are equivalent: (i) a∗ (x) ∈ Z for all a∗ ∈ Q(R∨ ). (ii) β ∨ (x) ∈ Z for all β ∨ ∈ B ∨ . (iii) The coordinates of x in the dual basis of B ∨ are integers. Consequently, we have parts (i) and (iii). By (R3), we have Q(R) ⊂ P (R).
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Let B = {β1 , . . . , βl } be a base of R, B ∨ = {β1∨ , . . . , βl∨ } as in 18.11.2 and {α1 , . . . , αl } the dual basis of B ∨ . For 1 p, q l, write αp =
l
λpj βj , and δpq =
i=1
l
λpj βq∨ (βj ).
j=1
This is a Cramer’s system with coefficients in Q, so P (R) ⊂ VQ . 18.11.6 Let C be a chamber in R, B = B(C), B ∨ = {β1∨ , . . . , βl∨ } the base of R∨ as in 18.11.2. The dual basis { 1 , . . . , l } of B ∨ is a basis of P (R), and they are called the fundamental weights of R with respect to the base B (or chamber C). We shall conserve this notation for fundamental weights. If x ∈ V , then x ∈ C if and only if β ∨ (x) > 0 for all β ∈ B. Hence C = R∗+
1
+ · · · + R∗+
l
, C = R+
1
+ · · · + R+
l.
18.11.7 Proposition. Let B = (β1 , . . . , βl ) be a base of R, 1 , . . . , l the corresponding fundamental weights and C = C(B). For x ∈ V , the following conditions are equivalent: (i) β ∨ (x) ∈ N for all β ∈ B. (ii) x = n1 1 + · · · + nl l where n1 , . . . , nl ∈ N. (iii) x ∈ C ∩ P (R). (iv) For all w ∈ W (R), x − w(x) ∈ Q(R) and x − w(x) 0 in the partial order associated to B. If x ∈ V verifies these conditions, then x is called a dominant weight of R with respect to B. Proof. The implications (i) ⇒ (ii) ⇒ (iii) are obvious. (iii) ⇒ (iv) Let w ∈ W (R) be such that w(x) is maximal in X = {w (x); w ∈ W (R)} with respect to the partial order associated to B. We saw in the proof of 18.8.9 (i) that w(x) ∈ C. So 18.8.9 (iii) implies that w(x) = x. Thus x is the unique maximal element of X. Since x ∈ P (R), (iv) follows from 18.7.8 (ii). (iv) ⇒ (i) This is clear since x − sβ (x) = β ∨ (x)β for β ∈ B. 18.11.8 Proposition. Let B be a base of R, C = C(B) and the partial order associated to B. If x ∈ V , the following conditions are equivalent: (i) x ∈ C. (ii) sβ (x) x for all β ∈ B. (iii) w(x) x for all w ∈ W (R). Proof. Since x − sβ (x) = β ∨ (x)β, we have (i) ⇔ (ii). Finally, (iii) ⇒ (ii) is obvious and (i) ⇒ (iii) is similar to the implication (iii) ⇒ (iv) of 18.11.7. 18.11.9 Corollary. With the hypotheses of 18.11.8, we have x ∈ C if and only if w(x) ≺ x for all w ∈ W (R) \ {idV }.
18.12 Graphs
255
Proof. The “if” part is clear by 18.11.8 and the definition of C. Conversely, if x ∈ C, then w(x) x for all w ∈ W (R). If w(x) = x and w = idV , then it follows from the proof of 18.8.9 (ii) that w is the product of reflections sβ , β ∈ B, verifying sβ (x) = x. Thus β ∨ (x) = 0 which is absurd since x ∈ C. 18.11.10 Proposition. With the notations of 18.11.6, let ρ be the half sum of the positive roots in R. Then: (i) ρ = 1 + · · · + l ∈ C. (ii) We have β2 = 2(ρ|β) for all β ∈ B. Proof. This is clear since β ∨ (ρ) = 1 = 2(ρ|β)/(β|β) for β ∈ B (18.7.6).
18.11.11 With the notations of 18.2.2 and 18.11.6, if 1 i l, then: βi =
l
αij
j.
j=1
Then A = [αij ] ∈ GLl (k). On the other hand, δij = βi∨ (
j)
=2
(βi | j ) . (βi |βi )
Thus αij = βj∨ (βi ) = aβi ,βj , and the matrix [aβi ,βj ] is invertible.
18.12 Graphs 18.12.1 Definition. A graph is a pair (S, A) where S is a finite set and A is a subset of P(S) (the set of subsets of S) whose elements are subsets of cardinal 2. A normed graph is a pair (Γ, f ) with the following properties: (a) Γ = (S, A) is a graph. (b) Let E be the set of pairs (i, j) such that {i, j} ∈ A. Then f is a map from E to R such that f (i, j)f (j, i) = 1 for all (i, j) ∈ E. 18.12.2 Let Γ = (S, A) be a graph. An element of S is called a vertex and an element of A is called an edge. Two vertices x, y are linked if {x, y} is an edge. A vertex is called a terminal vertex (resp. a ramification point) if it is linked to at most one vertex (resp. at least three vertices). In general, we represent a graph by a picture consisting of points corresponding to the elements of S and two points are linked by a line if and only if the corresponding vertices are linked. 18.12.3 Let Γ = (S, A) and Γ = (S , A ) be graphs. An isomorphism of Γ onto Γ is a bijection from S to S which transports A to A . If S ⊂ S and A ⊂ A, then we say that Γ is a subgraph of Γ . Furthermore, if A = A ∩ P(S ), then we say that Γ is a full subgraph of Γ .
256
18 Root systems
18.12.4 Let Γ = (S, A) be a graph. If a, b ∈ S, then a path from a to b is a sequence C = (x0 , . . . , xn ) of elements of S verifying: (i) x0 = a and xn = b. (ii) For 0 i n − 1, {xi , xi+1 } ∈ A. The integer n is the length of the path C. The path C is injective if the xi ’s are pairwise distinct. Thus a path from a to b of minimal length is injective. We define an equivalence relation ∼ on S as follows: a ∼ b if and only if there is a path from a to b. The equivalence classes are called the connected components. A graph is connected if it has at most one connected component. Thus in a connected graph, there is always a path between two vertices. Let Γ1 , . . . , Γn be the connected components of Γ , considered as full subgraphs of Γ . Then each Γi is connected and for i = j, there is no path between a vertex of Γi and a vertex of Γj . A cycle in Γ is a path C = (x0 , . . . , xn ) of length n 2 verifying x0 = xn and C = (x0 , . . . , xn−1 ) is an injective path. A graph is a forest if Γ has no cycles. A connected forest is called a tree. For n ∈ N∗ , we denote by An the graph with vertices {1, 2, . . . , n} and edges are the subsets {i, j} such that i − j = ±1: 1
2
3
n -1
n
A graph which is isomorphic to An+1 is called a chain of length n 0. 18.12.5 Proposition. Let Γ be a non-empty graph. Then Γ is a chain if and only if Γ is a tree without ramification points. Proof. It is clear that if Γ is chain, then it is a tree without ramification points. Conversely, suppose that Γ is a tree without ramification points. Let C = (x0 , . . . , xn ) be an injective path of maximal length in Γ . Suppose that a is a vertex which is distinct from the xi ’s. Since Γ is connected, there is a path C from a to one of the xi ’s. Replacing a by a vertex in C , we may assume that a and xj are linked. If j = 0 (resp. n), then (a, x0 , . . . , xn ) (resp. (x0 , . . . , xn , a)) is an injective path of length n + 1. But this contradicts our choice of C. If 0 < j < n, then xj is a ramification point which is absurd. Hence {x0 , . . . , xn } is the set of vertices of Γ and since Γ is a tree without ramification points, Γ is necessarily a chain.
18.13 Dynkin diagrams 18.13.1 Definition. Let R, R be root systems in V and V . We say that R and R are isomorphic if there exists a bijective linear map F : V → V such that F (R) = R and for all α, β ∈ R, we have: (F (α)|F (β)) (α|β) = . (F (β)|F (β)) (β|β)
18.13 Dynkin diagrams
257
18.13.2 Observe that the notion of isomorphism in 18.13.1 does not depend on the choice of invariant inner products (18.3.2). Suppose that R and R are isomorphic and F : V → V the corresponding bijective linear map. Then the map w → F ◦ w ◦ F −1 is an isomorphism of the groups W (R) and W (R ). In particular, let R1 , . . . , Rm be the irreducible components of R. Then given λ1 , . . . , λm ∈ R∗ , the direct sum R of the root systems λi Ri is isomorphic to R. 18.13.3 Definition. Let B = (β1 , . . . , βl ) be a base of R. The matrix [aβi ,βj ]1i,jl is called the Cartan matrix of R with respect to B. 18.13.4 It follows from 18.8.7 that, up to a permutation, the Cartan matrix of R does not depend on the choice of the base B. Clearly, isomorphic root systems have the same Cartan matrix, up to a permutation. The converse is given in the following proposition. Proposition. Let R, R be root systems in V, V , B, B bases of R, R . Suppose that there is a bijection f : B → B transforming the Cartan matrix of R to the Cartan matrix of R . Then R and R are isomorphic via an isomorphism F : V → V which extends f . In particular, the Cartan matrix of a root system determines the root system up to an isomorphism. Proof. Let B = (β1 , . . . , βl ), B = (β1 , . . . , βl ) where βi = f (βi ). Since B, B are basis of V and V , it is clear that f extends uniquely to an isomorphism F : V → V . Let α, β ∈ B, α = F (α), β = F (β). Then aβ,α = aβ ,α and: sF (α) (F (β)) = β − aβ ,α α = F (β − aβ,α α) = F (sα (β)). Since B is a basis of V , we deduce that sF (α) ◦ F = F ◦ sα for α ∈ B. Hence 18.7.8 implies that the map w → F ◦ w ◦ F −1 is an isomorphism from W (R) onto W (R ), sending sα to sF (α) if α ∈ B. Let β ∈ R. By 18.7.8, β = w(α) for some α ∈ B and w ∈ W (R). It follows that F (β) = (F ◦ w ◦ F −1 )(F (α)). Hence F (R) = R . On the other hand, since sw(α) = w ◦ sα ◦ w−1 , we have: F ◦sβ = F ◦w◦F −1 ◦F ◦sα ◦w−1 = (F ◦w◦F −1 )◦sF (α) ◦F ◦w−1 = (F ◦w◦F −1 )◦sF (α) ◦(F ◦w◦F −1 )−1 ◦F = sF (w(α)) ◦F = sF (β) ◦F. Finally, for α, β ∈ R, we have: sF (α) F (β) = F (β − aF (β),F (α) α) , F ◦sα (β) = F (β − aβ,α α). Hence aF (β),F (α) = aβ,α .
258
18 Root systems
18.13.5 We associate to R a normed graph Γ (R) = (X, f ), called the Dynkin diagram as follows: Fix a base B = (β1 , . . . , βl ) of R, and for 1 i, j l, let nij = aβi ,βj . The vertices of the graph X is {1, . . . , l} (or {β1 , . . . , βl }) and {i, j} is an edge if and only if (βi |βj ) = 0 (or equivalently nij = 0 or nji = 0). If {i, j} is an edge, we define: βi 2 nij = . f (i, j) = nji βj 2 It is clear that f (i, j)f (j, i) = 1. Let θij be the angle between βi and βj . Then the only possible cases, up to a permutation of i and j, are listed below: π 1) i and j are not linked ; nij = nji = 0, θij = . 2 2π . 2) f (i, j) = f (j, i) = 1 ; nij = nji = −1, βi = βj , θij = 3 √ 1 3π . 3) f (i, j) = 2, f (j, i) = ; nij = −2, nji = −1, βi = 2βj , θij = 2 4 √ 1 5π . 4) f (i, j) = 3, f (j, i) = ; nij = −3, nji = −1, βi = 3βj , θij = 3 6 Thus the Dynkin diagram determines R up to an isomorphism (18.13.4). 18.13.6 In practice, we represent Γ (R) = (X, f ) by a picture with points and lines as follows: the points are the vertices of X, and we draw nij nji lines from i to j. Further, if f (i, j) > 1 and θij = π/2 (cases 3) and 4) above), then we place the symbol > on the lines joining i to j, that is from the long root to the short root: i
j
f(i,j) = 2
i
j
f(i,j) = 3
18.13.7 Proposition. Let R be root system in V and Γ (R) its Dynkin diagram. Then: (i) R is irreducible if and only if Γ (R) is connected. (ii) Γ (R) is a forest, and if R is irreducible, then Γ (R) is a tree. Proof. (i) Suppose that R is the direct sum of two non-empty root systems R1 and R2 . Let B1 , B2 be bases of R1 , R2 . Then B = B1 ∪ B2 is a base of R and Γ (R1 ), Γ (R2 ) are full subgraphs of Γ (R). Further there is no edge joining a vertex of Γ (R1 ) to a vertex of Γ (R2 ). So Γ (R) is not connected. Conversely, suppose that Γ (R) is not connected, then R has a base of the form B = B1 ∪ B2 where B1 , B2 are non-empty orthogonal subsets. Let Vi be the subspace spanned by Bi . If α ∈ B1 , then the orthogonal of α is the direct sum of V2 and a hyperplane in V1 . It follows that sα (Vi ) = Vi for i = 1, 2. The same result applies for α ∈ B2 . So V1 and V2 are W (R)-stable (18.7.8), and R is not irreducible (18.2.7).
18.14 Classification of root systems
259
(ii) Let B be a base of R. If Γ (R) is not a forest, then there exist β1 , . . . , βn ∈ B pairwise distinct such that (β1 , . . . , βn ) is a cycle. Let γi = βi /βi . If {i, j} is an edge, then by 18.13.5, (γi |γj ) −1/2. Hence: $ $2 n $ $ $ γi $ = n + 2 (γi |γj ) n + 2[(γ1 |γ2 ) + · · · + (γn−1 |γn ) + (γn |γ1 )] $ $ i=1
i<j
n − n = 0. This is absurd. So Γ (R) is a forest, and (i) implies that Γ (R) is a tree.
18.13.8 Let B = (β1 , . . . , βl ) be a base of R. If α = n1 β1 + · · · + nl βl ∈ R, we set H(α) to be the set of βi ∈ B such that ni = 0. Proposition. (i) If α ∈ R, then H(α) is a connected subset of B (considered as the set of vertices in the Dynkin diagram of R). (ii) If H is a non-empty connected subset of B, then β∈H β ∈ R. Proof. (i) We may assume that α ∈ R+ and we shall proceed by induction on the cardinality n of H(α). The case n = 1 being trivial, we shall assume that n 2. By 18.7.4, there exists β ∈ B such that α − β ∈ R. Let r be maximal such that γ = α − rβ ∈ R. So r 1 and the β-string through γ is of the form {γ, . . . , γ + qβ}, with q r. Thus aγ,β = −q −r −1 (18.5.4), and so (γ|β) = 0. It follows that there is an edge joining β and an element of H(γ). Since H(α) = H(γ) ∪ {β}, it follows from the induction hypothesis that H(α) is connected. (ii) Let α = β∈H β. We shall prove by induction on the cardinality n of H(α) that α ∈ R. The case n = 1 being trivial, we may assume that n 2. Let (γ0 , γ1 , . . . , γm ) be an injective path of maximal length in H. Since H is a tree (18.13.7) and the length of the path is maximal, γ0 is a terminal vertex in H. Let M = H \ {γ0 }, then M is connected. By the induction hypothesis, α − γ0 ∈ R. Further, γ0 and γ1 are linked, so (α − γ0 |γ0 ) < 0 (18.7.4). Hence α ∈ R (18.5.3).
18.14 Classification of root systems 18.14.1 Theorem. Let R be an irreducible root system in V . Then the Dynkin diagram of R is isomorphic to one of the following : Al
l vertices, l 1
Bl
l vertices, l 2
Cl
l vertices, l 3
Dl
l vertices, l 4
260
18 Root systems
E6
E7
E8
F4 G2 Moreover, these diagrams are pairwise non-isomorphic. Proof. We can check easily that these diagrams are pairwise non-isomorphic. So let us prove the first part. Let B = (β1 , . . . , βl ) be a base of R and Γ be the Dynkin diagram of R. Set γi = βi /βi . Recall from the proof of 18.13.7 that (γi |γj ) −1/2 if {i, j} is an edge, and (γi |γj ) = 0 otherwise. Let f (i, j) be as defined in 18.13.5. 1) Let us suppose that there exist vertices i, j of Γ such that f (i, j) = 3. Since Γ is connected (18.13.7), if Γ is not G2 , then there exists a vertex k distinct from i and j such that √ 3 1 , (γi |γk ) 0 , (γj |γk ) − . (γi |γj ) = − 2 2 It follows that
√ √ √ 1 3 3γi + 2γj + γk 2 3 + 4 + 1 − 2 3 × 2 − 2 × 2 × = 0. 2 2
This is absurd. So Γ G2 . Let us assume from now on that f (i, j) = 3 for all i, j. 2) Let us suppose that there exist vertices i, j, k, l of Γ such that f (i, j) = 2 and f (k, l) = 2. Again, since Γ is connected, renumbering the vertices if necessary, we may assume that Γ contains the following subdiagram : 1
2
3
n 2
n 1
n
1 with n 3. So, (γ1 |γ2 ) = (γn−1 |γn ) = − √ . Hence : 2 $ $2 $ 1 $ $ √ γ1 + γ2 + · · · + γn−1 + √1 γn $ 1 + 1 +(n−2)−4 √1 √1 −2(n−3) 1 = 0. $ 2 2 2 $ 2 2 2 2
18.14 Classification of root systems
261
Again, this is absurd. So we may further assume that Γ has at most one edge {i, j} such that f (i, j) = 2. 3) Let us suppose that Γ has an edge {i, j} such that f (i, j) = 2, and a ramification point. Then Γ contains a subdiagram of the form : n 1
2
3
n 2
n 3
n 1
1 1 with n 4. So (γ1 |γ2 ) = − √ and (γi |γj ) = − for the other edges in this 2 2 subdiagram. Consequently, we have $ $2 $ γ1 $ $ √ + γ2 + · · · + γn−2 + γn−1 + γn $ 1 + (n − 3) + 1 + 1 − 2 √1 √1 $ 2 2 2 $ 2 4 4 2 2 11 11 1 −2 = 0. −2(n − 4) − 2 2 22 22 This is absurd. 4) Suppose that Γ has an edge {i, j} such that f (i, j) = 2. By the previous considerations, Γ is a chain. If Γ is not isomorphic to Bl or Cl or F4 , then it contains a subdiagram of the form : 1
2
3
4
5
It follows that : $ $√ √ √ √ $ 2γ1 + 2 2γ2 + 3γ3 + 2γ4 + γ5 $2 2 + 8 + 9 + 4 + 1 − 2 2.2 2 1 2 √ 1 1 1 −2.2 2.3 √ − 2.3.2 − 2.2 = 0. 2 2 2 Again, this is absurd. So Γ is isomorphic to either Bl , Cl or F4 . We may therefore assume in the rest of the proof that f (i, j) = 1 for all edges {i, j} of Γ . 5) If Γ has no ramification point, then it is isomorphic to Al . 6) Let us suppose that Γ has two distinct ramification points. Then Γ contains a subdiagram of the form : 1
n 1 3
2
4
n 3
n 2 n
We obtain therefore that : $γ γ2 γn $ 1 γn−1 1 $2 $ 1 + γ3 + · · · + γn−2 + + $ 4 + (n − 4) − 2(n − 5) $ + 2 2 2 2 4 2 11 11 11 11 −2 −2 −2 −2 22 22 22 22 = 0.
262
18 Root systems
This is absurd. 7) We are left with the case where Γ has a unique ramification point. In particular, Γ are at least 4 vertices, and it is of the form :
If Γ contains a subdiagram of the form : 1
4
5
6
7
3
2
then 1 γ1 + 2γ2 + 3γ3 + 2γ4 + γ5 + 2γ6 + γ7 2 24 − 2 (2 + 6 + 6 + 6 + 2 + 2) = 0. 2 This is absurd. If Γ contains a subdiagram of the form : 1
2
3
4
5
6
7
8
then we can check easily that γ1 + 2γ2 + 3γ3 + 4γ4 + 3γ5 + 2γ6 + γ7 + 2γ8 2 0, which is absurd. If Γ contains a subdiagram of the form : 1
2
3
4
5
6
7
8
9
then we can check easily that γ1 + 2γ2 + 3γ3 + 4γ4 + 5γ5 + 6γ6 + 4γ7 + 2γ8 + 3γ9 2 0, which is absurd. It is now easy to see that Γ is isomorphic to either Dl , E6 , E7 or E8 .
18.14.2 In the following tables, we give explicit descriptions of irreducible root systems. The irreducible root system R is considered to be a subset of some Rn where n is not necessarily l, the rank of R. Let B = (β1 , . . . , βl ) be a base of R and (ε1 , . . . , εn ) denote the canonical basis of Rn .
18.14 Classification of root systems
263
The vertices of the Dynkin diagram correspond to the elements of B. We add a vertex which corresponds to − α, where α is the highest root of R with respect to B and concerning the edges to this vertex, we use the convention of 18.13.5 and 18.13.6. This new graph is called the extended Dynkin diagram of R. If α = n1 β1 + · · · + nl βl ∈ R, we set: |α| = n1 + · · · + nl . We call |α| the height of α, and we denote it also by ht(α). We fix an order on the roots: let α, β ∈ R+ , then α ≺ β if either |α| < |β| or, |α| = |β| and α is smaller than β in the lexicographic order associated to (βl , βl−1 , . . . , β1 ). For the graphs of type E6 , E7 , E8 , F4 , G2 , we give an explicit list of positive roots in increasing order.
References • [13], [29], [39], [41], [80].
264
18 Root systems
Root system of type Al (l 1) • V is the hyperplane of Rl+1 consisting of points such that the sum of its coordinates is zero. • Roots: εi − εj (1 i, j l + 1, i = j). • Number of roots: n = l(l + 1). • Base: β1 = ε1 − ε2 , β2 = ε2 − ε3 , . . . , βl = εl − εl+1 . • Highest root: α = β1 + · · · + βl . • Cartan matrix: ⎞ ⎛ 2 −1 0 0 · · · 0 0 ⎜ −1 2 −1 0 · · · 0 0 ⎟ ⎟ ⎜ ⎜ 0 −1 2 −1 · · · 0 0 ⎟ ⎟ ⎜ ⎜ 0 0 −1 2 · · · 0 0 ⎟ ⎟ ⎜ ⎜ ··· ··· ··· ··· ··· ··· ···⎟ ⎟ ⎜ ⎝ 0 0 0 0 · · · 2 −1 ⎠ 0 0 0 0 · · · −1 2 • Extended Dynkin diagram (l 2):
1
2
l-1
l
• Order of W (R): (l + 1)!. • Positive roots: εi − εj = βi + · · · + βj−1
(1 i < j l + 1).
• Fundamental weights: i
i (ε1 + · · · + εl+1 ) (1 i l) = (ε1 + · · · + εi ) − l+1 1 [(l − i + 1)β1 + 2(l − i + 1)β2 + · · · + (i − 1)(l − i + 1)βi−1 = l+1 +i(l − i + 1)βi + i(l − i)βi+1 + · · · + iβl ].
18.14 Classification of root systems
Root system of type Bl (l 2) • • • • • •
V = Rl . Roots: ±εi (1 i l), ±εi ± εj (1 i < j l). Number of roots: n = 2l2 . Base: β1 = ε1 − ε2 , β2 = ε2 − ε3 , . . . , βl−1 = εl−1 − εl , βl = εl . Highest root: α = β1 + 2β2 + 2β3 + · · · + 2βl . Cartan matrix: ⎞ ⎛ 2 −1 0 0 · · · 0 0 ⎜ −1 2 −1 0 · · · 0 0 ⎟ ⎟ ⎜ ⎜ 0 −1 2 −1 · · · 0 0 ⎟ ⎟ ⎜ ⎜ 0 0 −1 2 · · · 0 0 ⎟ ⎟ ⎜ ⎜ ··· ··· ··· ··· ··· ··· ···⎟ ⎟ ⎜ ⎝ 0 0 0 0 · · · 2 −2 ⎠ 0 0 0 0 · · · −1 2
• Extended Dynkin diagram: l = 2: 2
1
l3: 1 2
3
l-1
l
• Order of W (R): 2l .l!. • Positive roots: ⎧ (1 i l) ⎨ εi = βi + · · · + βl (1 i < j l) εi − εj = βi + · · · + βj−1 ⎩ εi + εj = βi + · · · + βj−1 + 2(βj + · · · + βl ) (1 i < j l). • Fundamental weights: i
l
= ε1 + · · · + ε i (1 i < l) = β1 + 2β2 + · · · + (i − 1)βi−1 + i[βi + · · · + βl ] 1 1 = (ε1 + · · · + εl ) = (β1 + 2β2 + · · · + lβl ). 2 2
265
266
18 Root systems
Root system of type Cl (l 3) • • • • • •
V = Rl . Roots: ±2εi (1 i l), ±εi ± εj (1 i < j l). Number of roots: n = 2l2 . Base: β1 = ε1 − ε2 , β2 = ε2 − ε3 , . . . , βl−1 = εl−1 − εl , βl = 2εl . Highest root: α = 2β1 + 2β2 + · · · + 2βl−1 + βl . Cartan matrix: ⎞ ⎛ 2 −1 0 0 . . . 0 0 ⎜ −1 2 −1 0 . . . 0 0 ⎟ ⎟ ⎜ ⎜ 0 −1 2 −1 . . . 0 0 ⎟ ⎟ ⎜ ⎜ 0 0 −1 2 . . . 0 0 ⎟ ⎟ ⎜ ⎜ ··· ··· ··· ··· ··· ··· ···⎟ ⎟ ⎜ ⎝ 0 0 0 0 . . . 2 −1 ⎠ 0 0 0 0 . . . −2 2
• Extended Dynkin diagram: 1
2
l-2
l-1
l
• Order of W (R): 2l .l!. • Positive roots: ⎧ (1 i l) ⎨ 2εi = 2(βi + · · · + βl−1 ) + βl (1 i < j l) εi − εj = βi + · · · + βj−1 ⎩ εi + εj = βi + · · · + βj−1 + 2(βj + · · · + βl−1 ) + βl (1 i < j l). • Fundamental weights: i
= ε1 + · · · + ε i
(1 i l)
1 = β1 + 2β2 + (i − 1)βi−1 + i[βi + βi+1 + · · · + βl−1 + βl ]. 2
18.14 Classification of root systems
267
Root system of type Dl (l 4) • • • • • •
V = Rl . Roots: ±εi ± εj (1 i < j l). Number of roots: n = 2l(l − 1). Base: β1 = ε1 − ε2 , β2 = ε2 − ε3 , . . . , βl−1 = εl−1 − εl , βl = εl−1 + εl . Highest root: α = β1 + 2β2 + · · · + 2βl−2 + βl−1 + βl . Cartan matrix: ⎞ ⎛ 2 −1 . . . 0 0 0 0 ⎜ −1 2 . . . 0 0 0 0 ⎟ ⎟ ⎜ ⎜ ··· ··· ··· ··· ··· ··· ···⎟ ⎟ ⎜ ⎜ 0 0 . . . 2 −1 0 0 ⎟ ⎟ ⎜ ⎜ 0 0 . . . −1 2 −1 −1 ⎟ ⎟ ⎜ ⎝ 0 0 . . . 0 −1 2 0 ⎠ 0 0 . . . 0 −1 0 2
• Extended Dynkin diagram (l 4): 1
l- 1 2
3
l-3
l-2
l
• Order of W (R): 2l−1 .l!. • Positive roots:
⎧ (1 i < j l) ⎨ εi − εj = βi + · · · + βj−1 (1 i < l) εi + εl = βi + · · · + βl−2 + βl ⎩ εi + εj = βi + · · · + βj−1 + 2(βj + · · · + βl−2 ) + βl−1 + βl (1 i < j < l). • Fundamental weights: i
l−1
l
= ε1 + · · · + ε i (1 i l − 2) = β1 + 2β2 + · · · + (i − 1)βi−1 1 +i[βi + βi+1 + · · · + βl−2 ] + i[βl−1 + βl ] 2 1 = (ε1 + ε2 + · · · + εl−2 + εl−1 − εl ) 2 1 1 1 = [β1 + 2β2 + · · · + (l − 2)βl−2 ] + lβl−1 + (l − 2)βl 2 4 4 1 = (ε1 + ε2 + · · · + εl−2 + εl−1 + εl ) 2 1 1 1 = [β1 + 2β2 + · · · + (l − 2)βl−2 ] + (l − 2)βl−1 + lβl . 2 4 4
268
18 Root systems
Root system of type E6 • V is the subspace of R8 consisting of points whose coordinates (ξi ) verify ξ6 = ξ7 = −ξ8 . • Roots: 5 1 ±εi ± εj (1 i < j 5), ± (−1)ν(i) εi , ε8 − ε7 − ε6 + 2 i=1 where ν(i) are positive integers such that ν(1) + · · · + ν(5) is even. • Number of roots: n = 72. • Base: 1 (ε1 − ε2 − ε3 − ε4 − ε5 − ε6 − ε7 + ε8 ), 2 β2 = ε1 + ε2 , β3 = ε2 − ε1 , β4 = ε3 − ε2 , β5 = ε4 − ε3 , β6 = ε5 − ε4 . β1 =
• Highest root: α = β1 + 2β2 + 2β3 + 3β4 + 2β5 + β6 . • Cartan matrix: ⎞ ⎛ 2 0 −1 0 0 0 ⎜ 0 2 0 −1 0 0 ⎟ ⎟ ⎜ ⎜ −1 0 2 −1 0 0 ⎟ ⎟ ⎜ ⎜ 0 −1 −1 2 −1 0 ⎟ ⎟ ⎜ ⎝ 0 0 0 −1 2 −1 ⎠ 0 0 0 0 −1 2 • Extended Dynkin diagram: 1
3
4
5
6
2
• Order of W (R): 27 .34 .5. • Positive roots: ⎧ j 5) ⎨ ±εi + εj (1 i < 5 5 1 (−1)ν(i) εi with ν(i) even. ε8 − ε7 − ε6 + ⎩ 2 i=1 i=1 Let us represent the root aβ1 + bβ2 + cβ3 + dβ4 + eβ5 + f β6 by the positive roots are:
a c d e f b
. Then
18.14 Classification of root systems 1 0 0 0 0 1 1 0 1 1 1 1 0 1 1 1
0 0 0 0 0 0 1 1
0 0 0 1 0 0 1 0 1 1 1 0 1 1 2 1
0 0 1 0 1 0 1 0
0 1 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 2 1
0 0 1 1 1 0 1 1 2 1
0 0 1 0 1 1 1 0 0 0 1 1 0 1 2 1 1 2 2 1
0 0 0 0 1 1 1 1 2 1
0 0 0 0 0 1 1 1 0 1 1 0 1 2 2 1 1 2 3 1
1 0 0 0 1 1 1 0 2 1
0 0 0 0 0 0 1 1 1 1 1 1 1 1 2 1 1 2 3 2
0 1 1 0 1 0 1 1
1 1 0 0 0 1 1 0 0 1 2 1 0 1 2 1
0 0 1 0 1 0 2 1
2 1
• Fundamental weights: 1
2
3
4
2 (ε8 − ε7 − ε6 ) 3 1 = [4β1 + 3β2 + 5β3 + 6β4 + 4β5 + 2β6 ] 3 1 = (ε1 + ε2 + ε3 + ε4 + ε5 − ε6 − ε7 + ε8 ) 2 = β1 + 2β2 + 2β3 + 3β4 + 2β5 + β6 =
5 1 (ε8 − ε7 − ε6 ) + (−ε1 + ε2 + ε3 + ε4 + ε5 ) 6 2 1 = [5β1 + 6β2 + 10β3 + 12β4 + 8β5 + 4β6 ] 3 = ε3 + ε4 + ε5 − ε6 − ε7 + ε8 =
= 2β1 + 3β2 + 4β3 + 6β4 + 4β5 + 2β6 5
6
2 (ε8 − ε7 − ε6 ) + ε4 + ε5 3 1 = [4β1 + 6β2 + 8β3 + 12β4 + 10β5 + 5β6 ] 3 1 = (ε8 − ε7 − ε6 ) + ε5 3 1 = [2β1 + 3β2 + 4β3 + 6β4 + 5β5 + 4β6 ]. 3 =
0 0 1 1 0 0 1 0 1 1 1 0 1 2 2 1
269 0 0 1 1 1 1 1 1
270
18 Root systems
Root system of type E7 • V is the hyperplane of R8 orthogonal to ε7 + ε8 . • Roots: 6 1 ±(ε7 − ε8 ), ±εi ± εj (1 i < j 6), ± (−1)ν(i) εi , ε7 − ε8 + 2 i=1 where ν(i) are positive integers such that ν(1) + · · · + ν(6) is odd. • Number of roots: n = 126. • Base: 1 (ε1 − ε2 − ε3 − ε4 − ε5 − ε6 − ε7 + ε8 ), β2 = ε1 + ε2 , 2 β3 = ε2 − ε1 , β4 = ε3 − ε2 , β5 = ε4 − ε3 , β6 = ε5 − ε4 , β7 = ε6 − ε5 . β1 =
• Highest root: α = 2β1 + 2β2 + 3β3 + 4β4 + 3β5 + 2β6 + β7 . • Cartan matrix: ⎞ ⎛ 2 0 −1 0 0 0 0 ⎜ 0 2 0 −1 0 0 0 ⎟ ⎟ ⎜ ⎜ −1 0 2 −1 0 0 0 ⎟ ⎟ ⎜ ⎜ 0 −1 −1 2 −1 0 0 ⎟ ⎟ ⎜ ⎜ 0 0 0 −1 2 −1 0 ⎟ ⎟ ⎜ ⎝ 0 0 0 0 −1 2 −1 ⎠ 0 0 0 0 0 −1 2 • Extended Dynkin diagram: 1
3
4
5
6
7
2
• Order of W (R): 210 .34 .5.7. • Positive roots: ⎧ ε 8 − ε7 ⎪ ⎪ ⎨ ±ε + ε (1 i < j 6) i j 6 6 1 ⎪ ⎪ (−1)ν(i) εi with ν(i) odd. ⎩ ε8 − ε7 + 2 i=1 i=1 Representing the root aβ1 + bβ2 + cβ3 + dβ4 + eβ5 + f β6 + gβ7 by the positive roots are:
a c d e f g , b
18.14 Classification of root systems 1 0 0 0 1 1 0 0 0 1 1 1 0 1 1 1 0 1 1 1 0 1 1 1 1 1 2 1 1 2 3 1 1 2 3 1
0 0 0 0 0 0 0 0 0 1 0 0 1 1 0 1 1 1 2 1 0 2 1 0 2 2 1
0 0 0 1 0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 1 1 2 2 1 1 1 2 1 1 2 2 1 1 2 3 2
0 0 0 0 0 0 1 0 0 1 1 0 1 1 1 1 0 0 1 1 1 2 1 1 2 2 1
0 1 0 0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 1 0 1 1 2 1 0 1 2 1 1 1 2 1 1 2 3 1
0 0 0 0 0 0 1 0 0 1 1 0 1 1 1 1 1 0 2 1 1 2 2 1 3 2 1
0 0 1 0 0 0 1 0 0 0 1 0 0 0 1 0 1 1 2 1 0 1 2 1 1 2 2 1 1 2 3 2 1 2 3 2
0 0 0 1 0 0 1 1 0 1 1 1 1 0 0 2 1 0 2 1 0 2 1 0 3 2 1
0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 2 2 1 1 2 3 1 1 2 4 2
1 0 0 1 1 0 1 1 1 1 0 0 1 1 0 1 1 1 1 1 1 2 1 1 3 2 1
0 0 0 0 0 0 0 0 1 1 1 1 0 1 2 1 0 1 2 1 0 1 2 1 1 1 2 1 1 2 2 1 1 3 4 2
0 1 0 0 1 1 0 0 0 1 0 0 1 1 0 1 1 1 2 1 1 2 2 1 3 2 1
• Fundamental weights: 1
= ε8 − ε 7 = 2β1 + 2β2 + 3β3 + 4β4 + 3β5 + 2β6 + β7
2
=
3
4
5
6
7
1 (ε1 + ε2 + ε3 + ε4 + ε5 + ε6 − 2ε7 + 2ε8 ) 2 1 = [4β1 + 7β2 + 8β3 + 12β4 + 9β5 + 6β6 + 3β7 ] 2 1 = (−ε1 + ε2 + ε3 + ε4 + ε5 + ε6 − 3ε7 + 3ε8 ) 2 = 3β1 + 4β2 + 6β3 + 8β4 + 6β5 + 4β6 + 2β7 = ε3 + ε4 + ε5 + ε6 − 2ε7 + 2ε8 = 4β1 + 6β2 + 8β3 + 12β4 + 9β5 + 6β6 + 3β7 1 (2ε4 + 2ε5 + 2ε6 − 3ε7 + 3ε8 ) 2 1 = [6β1 + 9β2 + 12β3 + 18β4 + 15β5 + 10β6 + 5β7 ] 2 = ε5 + ε 6 − ε 7 + ε 8 = 2β1 + 3β2 + 4β3 + 6β4 + 5β5 + 4β6 + 2β7 =
1 (2ε6 − ε7 + ε8 ) 2 1 = [2β1 + 3β2 + 4β3 + 6β4 + 5β5 + 4β6 + 3β7 ]. 2
=
0 0 0 0 1 1 1 0 1 1 1 0 1 1 1 0 1 1 1 0 1 2 2 1 0 1 2 1 1 2 3 2 2 3 4 2
271 0 0 1 0 0 0 1 0 0 1 1 0 1 1 1 1 1 0 2 2 1 2 1 1 3 2 1
272
18 Root systems
Root system of type E8 • V = R8 . • Roots: ±εi ± εj
(1 i < j),
8 1 (−1)ν(i) εi , 2 i=1
where ν(i) are positive integers such that ν(1) + · · · + ν(8) is even. • Number of roots: n = 240. • Base : 1 (ε1 − ε2 − ε3 − ε4 − ε5 − ε6 − ε7 + ε8 ), β2 = ε1 + ε2 , β3 = ε2 − ε1 , 2 β4 = ε3 − ε2 , β5 = ε4 − ε3 , β6 = ε5 − ε4 , β7 = ε6 − ε5 , β8 = ε7 − ε6 .
β1 =
• Highest root: α = 2β1 + 3β2 + 4β3 + 6β4 + 5β5 + 4β6 + 3β7 + 2β8 . • Cartan matrix: ⎞ ⎛ 2 0 −1 0 0 0 0 0 ⎜ 0 2 0 −1 0 0 0 0 ⎟ ⎟ ⎜ ⎜ −1 0 2 −1 0 0 0 0 ⎟ ⎟ ⎜ ⎜ 0 −1 −1 2 −1 0 0 0 ⎟ ⎟ ⎜ ⎜ 0 0 0 −1 2 −1 0 0 ⎟ ⎟ ⎜ ⎜ 0 0 0 0 −1 2 −1 0 ⎟ ⎟ ⎜ ⎝ 0 0 0 0 0 −1 2 −1 ⎠ 0 0 0 0 0 0 −1 2 • Extended Dynkin diagram: 1
3
5
4
6
7
8
2
• Order of W (R): 214 .35 .52 .7. • Positive roots: ⎧ j 8) ⎨ ±εi + εj (1 i < 7 7 1 ν(i) (−1) εi with ν(i) even. ε8 + ⎩ 2 i=1 i=1 Representing the root aβ1 + bβ2 + cβ3 + dβ4 + eβ5 + f β6 + gβ7 + hβ8 by a c d e f g h , the positives roots are: b 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 1 0 0 0 1 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 1 1 1 0 0 0 0 0 0 0 1 0 0 0 0 0 1 1 0 0 0 0 0 0 1 1 0 0 0 0 0 0 1 0 0 0 0 0 1 1 0 0 0 0 0 0 1 1 0 0 0 0 0 0 1 1 1 1 1 0 0 0 0 0 1 1 0 0 0 0 0 0 1 1 0 0 0 0 0 0 0 1 1
18.14 Classification of root systems 0 1 1 0 0 1 1 1 0 1 2 1 1 1 2 1 0 1 1 0 1 1 1 0 1 1 1 1 1 1 2 1 1 1 2 1 1 1 2 1 1 1 2 1 1 2 3 2 1 2 3 2 2 3 4 2 1 2 4 2 2 3 4 2 2 4 5 3
1 0 0 0 0 0 1 0 1 0 0 0 0 0 1 1 1 0 0 0 1 1 1 0 1 0 0 0 1 1 1 1 1 1 1 1 1 2 2 1 1 1 1 1 0 1 1 1 1 1 1 1 0 1 2 1 1 1 1 1 0 1 2 1 2 1 1 1 0 1 2 1 2 2 1 1 0 1 2 1 2 2 2 1 1 2 3 2 3 2 1 0 1 2 3 2 2 2 2 1 1 2 3 1 3 2 1 0 1 3 4 2 3 3 2 1 2 3 4 2 4 3 2 1 1 3 5 2 4 3 2 1 2 4 6 3
1 1 0 0 0 0 0 0 1 1 0 0 0 1 1 0 1 1 0 0 0 1 1 1 1 1 0 0 0 1 2 1 1 0 0 0 1 1 2 1 1 1 1 1 1 2 2 1 1 1 1 1 1 2 2 1 2 1 1 1 1 2 3 1 2 2 1 1 1 2 3 2 2 2 2 1 1 2 3 2 2 2 1 0 1 2 3 1 2 2 1 1 1 2 3 1 3 2 2 1 1 3 4 2 3 2 1 1 1 2 4 2 3 2 2 1 1 3 4 2 4 3 2 1 2 3 5 2 4 3 2 1 2 4 6 3
1 1 1 0 0 0 0 0 1 1 0 0 0 0 1 0 1 1 0 0 0 0 1 1 1 1 0 0 1 1 1 0 1 1 0 0 0 1 2 1 1 1 0 0 1 1 2 1 2 1 0 0 1 2 2 1 2 1 0 0 1 2 2 1 2 1 0 0 1 2 3 1 2 1 1 0 1 2 3 1 3 2 1 0 1 2 3 2 3 2 1 1 1 2 3 1 3 2 1 0 1 2 4 2 3 2 2 1 1 2 3 2 3 3 2 1 1 2 4 2 4 3 2 1 1 3 5 3 5 3 2 1 2 4 6 3
0 1 1 1 1 1 1 1 1 1 1 0 0 0 0 0 1 1 1 0 0 1 1 0 1 1 1 0 0 1 1 1 2 1 0 0 1 1 1 1 2 1 0 0 1 1 2 1 1 1 1 0 1 1 2 1 2 1 1 0 1 1 2 1 2 1 1 0 1 2 2 1 2 2 1 0 1 2 3 1 2 1 1 1 1 2 3 1 2 2 2 1 1 2 4 2 3 2 1 1 1 2 3 2 3 3 2 1 2 3 4 2 4 3 2 1 2 3 4 2 4 3 2 1 2 3 5 3 5 4 2 1 2 4 6 3
0 0 0 0 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 0 0 0 1 0 1 1 1 0 0 0 1 1 1 1 1 0 0 1 2 1 1 1 1 0 0 1 2 1 2 1 1 0 0 1 2 1 2 2 1 0 1 2 2 1 2 2 1 0 1 2 2 1 2 1 1 1 1 2 2 1 2 2 1 1 1 2 2 1 3 2 1 0 1 2 3 2 3 2 2 1 1 2 3 1 3 2 1 1 1 3 4 2 3 3 2 1 1 3 4 2 4 3 2 1 2 4 5 2 5 4 3 1 2 4 6 3
1 0 0 0 1 0 0 0 1 1 1 1 1 1 1 1 1 1 1 0 2 1 1 0 2 2 1 0 1 1 1 1 2 1 1 1 2 2 1 1 2 2 2 1 3 2 1 1 3 3 2 1 3 2 2 1 4 3 2 1 4 3 2 1 5 4 3 2
• Fundamental weights: 1
2
3
4
5
6
7
8
= 2ε8 = 4β1 + 5β2 + 7β3 + 10β4 + 8β5 + 6β6 + 4β7 + 2β8 1 = (ε1 + ε2 + ε3 + ε4 + ε5 + ε6 + ε7 + 5ε8 ) 2 = 5β1 + 8β2 + 10β3 + 15β4 + 12β5 + 9β6 + 6β7 + 3β8 1 = (−ε1 + ε2 + ε3 + ε4 + ε5 + ε6 + ε7 + 7ε8 ) 2 = 7β1 + 10β2 + 14β3 + 20β4 + 16β5 + 12β6 + 8β7 + 4β8 = ε3 + ε4 + ε5 + ε6 + ε7 + 5ε8 = 10β1 + 15β2 + 20β3 + 30β4 + 24β5 + 18β6 + 12β7 + 6β8 = ε4 + ε5 + ε6 + ε7 + 4ε8 = 8β1 + 12β2 + 16β3 + 24β4 + 20β5 + 15β6 + 10β7 + 5β8 = ε5 + ε6 + ε7 + 3ε8 = 6β1 + 9β2 + 12β3 + 18β4 + 15β5 + 12β6 + 8β7 + 4β8 = ε6 + ε7 + 2ε8 = 4β1 + 6β2 + 8β3 + 12β4 + 10β5 + 8β6 + 6β7 + 3β8 = ε7 + ε 8 = 2β1 + 3β2 + 4β3 + 6β4 + 5β5 + 4β6 + 3β7 + 2β8 .
273
274
18 Root systems
Root system of type F4 • V = R4 . • Roots: ±εi
(1 i 4), ±εi ± εj
(1 i < j 4),
1 (±ε1 ± ε2 ± ε3 ± ε4 ). 2
• Number of roots: n = 48. • Base: β1 = ε2 − ε3 , β2 = ε3 − ε4 , β3 = ε4 , β4 =
1 (ε1 − ε2 − ε3 − ε4 ). 2
• Highest root: α = 2β1 + 3β2 + 4β3 + 2β4 . • Cartan matrix: ⎛ ⎞ 2 −1 0 0 ⎜ −1 2 −2 0 ⎟ ⎜ ⎟ ⎝ 0 −1 2 −1 ⎠ 0 0 −1 2 • Extended Dynkin diagram: 1
3
2
4
• Order of W (R) : 27 .32 . • Positive roots: ⎧ (1 i 4) ⎪ ⎨ εi (1 i < j 4) εi ± ε j 1 ⎪ ⎩ (ε1 ± ε2 ± ε3 ± ε4 ). 2 Denote by
a b c d
the root aβ1 + bβ2 + cβ3 + dβ4 , then the positive roots are:
1 0 0 0
0 1 0 0
0 0 1 0
0 0 0 1
1 1 0 0
0 1 1 0
0 0 1 1
1 1 1 0
0 1 2 0
0 1 1 1
1 1 2 0
1 1 1 1
0 1 2 1
1 2 2 0
1 1 2 1
0 1 2 2
1 2 2 1
1 1 2 2
1 2 3 1
1 2 2 2
1 2 3 2
1 2 4 2
1 3 4 2
2 3 4 2
• Fundamental weights: 1 2 3 4
= ε1 + ε2 = 2β1 + 3β2 + 4β3 + 2β4 = 2ε1 + ε2 + ε3 = 3β1 + 6β2 + 8β3 + 4β4 1 = (3ε1 + ε2 + ε3 + ε3 ) = 2β1 + 4β2 + 6β3 + 3β4 2 = ε1 = β1 + 2β2 + 3β3 + 2β4 .
18.14 Classification of root systems
Root system of type G2 • V is the hyperplane of R3 defined by ξ1 + ξ2 + ξ3 = 0. • Roots: ±(ε1 − ε2 ), ±(ε1 − ε3 ), ±(ε2 − ε3 ), ±(2ε1 − ε2 − ε3 ), ±(2ε2 − ε1 − ε3 ), ±(2ε3 − ε1 − ε2 ). • • • •
Number of roots: n = 12. Base: β1 = ε1 − ε2 , β2 = −2ε1 + ε2 + ε3 . Highest root: α = 3β1 + 2β2 . Cartan matrix: 2 −1 −3 2
• Extended Dynkin diagram: 1
2
• Order of W (R): 12. • Positive roots: β1 , β2 , β1 + β2 , 2β1 + β2 , 3β1 + β2 , 3β1 + 2β2 . • Fundamental weights: 1 2
= −ε2 + ε3 = 2β1 + β2 = −ε1 − ε2 + 2ε3 = 3β1 + 2β2 .
275
19 Lie algebras
The aim of this chapter is to give a detailed account of the general theory of Lie algebras. The representation theory of sl2 (k) in section 19.2 will be used very often in the rest of the book. Results such as Engel’s theorem, Lie’s theorem and Cartan’s Criterion are the backbone of the subject. The notions of regular linear forms and Cartan subalgebras are introduced in the final sections.
19.1 Generalities on Lie algebras 19.1.1 Let g be a k-vector space. A bilinear map g×g → g, (x, y) → [x, y], is a Lie bracket on g if for all x, y, z ∈ g, we have • [x, y] = −[y, x] (antisymmetry), • [x, [y, z]] = [[x, y], z] + [y, [x, z]] (Jacobi identity). Endowed with a Lie bracket, g is called a Lie algebra. Let g be a Lie algebra. A subspace h ⊂ g is a Lie subalgebra (resp. an ideal) if [x, y] ∈ h for all x, y ∈ h. (resp. [x, h] ⊂ h for all x ∈ g). When h is an ideal of g, the quotient g/h inherits a natural structure of Lie algebra given by: [x + h, y + h] = [x, y] + h. Examples. Let us give some examples of Lie algebras. Let A be an associative k-algebra. Then, A has a Lie algebra structure given by: for x, y ∈ A, [x, y] = xy − yx. So if V is a k-vector space of dimension n, then End(V ) (resp. Mn (k)) endowed with the above Lie bracket is a Lie algebra that we shall denote by gl(V ) (resp. gln (k)). The following subsets of gln (k) are clearly Lie subalgebras: • sln (k) (resp. sl(V )): the set of matrices in gln (k) (resp. endomorphisms in gl(V )) whose trace is zero. • tn (k) (resp. nn (k)): the set of upper triangular matrices (resp. strictly upper triangular matrices) in gln (k). • dn (k) the set of diagonal matrices in gln (k).
278
19 Lie algebras
Observe that sl(V ) is an ideal of gl(V ). Let h, h be subsets of g, then we denote by [h, h ] the subspace spanned by the elements [x, y] where x ∈ h, y ∈ h . When h and h are ideals of g, then clearly [h, h ] is also an ideal of g. Let g, g be Lie algebras endowed with Lie brackets [, ] and [, ] , then a map u : g → g is a morphism of Lie algebras if u([x, y]) = [u(x), u(y)] for all x, y ∈ g. So we may talk about isomorphisms and automorphisms. A linear map d : g → g is called a derivation of g if for all x, y ∈ g, d([x, y]) = [d(x), y] + [x, d(y)]. We denote by Der g the set of derivations of g. Note that Der g, endowed with the following Lie bracket, [d, d ] = d ◦ d − d ◦ d, for d, d ∈ Der g, is a Lie algebra. Let x ∈ g. The Jacobi identity implies that the linear map y → [x, y], denoted adg x or ad x, is a derivation of g, called an inner derivation of g. So it induces a linear map g → Der(g) which is called the adjoint representation of g. Convention. In this chapter, we shall only consider finite-dimensional Lie algebras and g will denote a finite-dimensional Lie algebra. Set: ker(ad x)n , g• (x) = (ad x)n (g). g0 (x) = n0
n0
0
We call g (x) the nilspace of ad x. We have: [g0 (x), g0 (x)] ⊂ g0 (x) , [g0 (x), g• (x)] ⊂ g• (x) , g = g0 (x) ⊕ g• (x). Let n = dim g and χx the characteristic polynomial of ad x. Then χx (T ) = an (x)T n + an−1 (x)T n−1 + · · · + a1 (x)T + a0 (x) where ai are polynomial functions on g with an = 1 and a0 = 0 (since [x, x] = 0). If p is the smallest integer such that ap (x) = 0, then by decomposing g into characteristic subspaces of ad x, we obtain that dim g0 (x) = p. Define rk(g) = min{l; al = 0} to be the rank of the Lie algebra g. If x ∈ g verifies dim g0 (x) = rk(g), then we say that x is generic in g. Denote by ggen the set of generic elements of g. It is clear from its definition that ggen is a dense open subset of g. 19.1.2 Lemma. If h is a Lie subalgebra of g, then h ∩ ggen ⊂ hgen . Proof. For x ∈ h, denote by u(x) the endomorphism of g/h induced by adg x. Let χu(x) be the characteristic polynomial of u(x) and χx,h the characteristic polynomial of adh x. Then χx = χu(x) χx,h and the result follows easily.
19.2 Representations
279
19.1.3 Let p, q be subsets of g. The set of elements of q which commute with all the elements of p, denoted by cq (p), is called the centralizer of p in q. The set cg (g), denoted by z(g), is called the centre of g. If g = z(g), then we say that g is abelian (or commutative). If x ∈ g, we shall write qx for cq ({x}). The set of x ∈ q such that [x, p] ⊂ p, denoted by nq (p), is called the normalizer of p in q. If p is a Lie subalgebra, then so is ng (p), and p is an ideal of ng (p). 19.1.4 Let us denote by Aut g the set of automorphisms of g, N the set of elements x ∈ g such that ad x is a nilpotent endomorphism of g. For x ∈ N , define γx ∈ GL(g) as follows: γx =
∞ 1 (ad x)n . n=0 n!
Then γx ∈ Aut g. Such an automorphism is called an elementary automorphism of g. The subgroup Aute g of Aut g generated by elementary automorphisms is called the group of elementary automorphisms of g. Observe that if x ∈ N , α ∈ Aut g, then α(x) ∈ N and ad α(x) = α ◦ (ad x) ◦ α−1 . So γα(x) = α ◦ γx ◦ α−1 , and Aute g is a normal subgroup of Aut g.
19.2 Representations 19.2.1 A representation of g, or a g-module, is a pair (V, σ), where V is k-vector space and σ : g → gl(V ) is a Lie algebra homomorphism. If V is finite-dimensional, then we say that the representation is finite-dimensional. For x ∈ g, v ∈ V , we shall denote σ(x)(v) by x.v. We set V g = {v ∈ V ; x.v = 0 for all x ∈ g}. We define the notions of submodule, quotient module and direct sum in the obvious way. The g-module (V, σ) is said to be simple if V = {0} and the only submodules of V are V and {0}. If V is the direct sum of simple submodules, then we say that V is semisimple, or completely reducible. The analogues of 10.7.2 and 10.7.5 are true for g-modules. Let V be a g-module. A chain 0 = V0 ⊂ V1 ⊂ · · · ⊂ Vn = V of submodules of V is called a Jordan-H¨ older series if the modules Vi /Vi−1 are simple for 1 i n. Let (V, σ) be a representation of g, then we define its dual, or contragredient, representation (V ∗ , π) as follows: for x ∈ g, v ∈ V and f ∈ V ∗ , (π(x)(f ))(v) = −f (σ(x)(v)). The adjoint representation is indeed a representation of g. Its dual representation is called the coadjoint representation of g. If h is a Lie subalgebra of g, then the map h → gl(g/h) induced by x → ad x is called the adjoint representation of h in g/h. A representation (V, σ) is said to be faithful if σ is injective. We shall admit the following result.
280
19 Lie algebras
Theorem. (Ado’s Theorem) Any finite-dimensional Lie algebra g has a faithful finite-dimensional representation. 19.2.2 Let (V, σ) be a finite-dimensional representation of g. We define a symmetric bilinear form βσ on g associated to σ as follows: for x, y ∈ g, βσ (x, y) = tr(σ(x)σ(y)). The symmetric bilinear form associated to the adjoint representation is called the Killing form of g, denoted by Lg . If a is an ideal of g, then the restriction of Lg to a × a is the Killing form of a, and the orthogonal of a with respect to βσ is an ideal of g. Observe that if x, y, z ∈ g, then βσ ([x, y], z) + βσ (y, [x, z]) = 0. If x ∈ g, denote by fx the linear form on g such that fx (y) = Lg (x, y) for all y ∈ g. It follows that the map x → fx is a homomorphism of g-modules, called the Killing homomorphism of g, from the adjoint representation of g to the coadjoint representation of g. 19.2.3 Lemma. Let V be a finite-dimensional vector space and f an endomorphism of V . Then f is nilpotent if and only if tr(f n ) = 0 for all n ∈ N∗ . Proof. This is an elementary result from linear algebra. 19.2.4 In this rest of this section, g is the Lie algebra sl2 (k). Let 01 1 0 00 e= , h= , f= . 00 0 −1 10 Then (e, h, f ) is a basis of g and [e, f ] = h , [h, e] = 2e , [h, f ] = −2f. Let r ∈ N and for 1 i r, let µi = i(r − i + 1) and denote by (kr+1 , σr ) the representation of g given by: ⎞ ⎛ r 0 0 ... 0 ⎜0 r − 2 0 ... 0 ⎟ ⎟ ⎜ ⎟ ⎜ σr (h) = ⎜ 0 0 r − 4 . . . 0 ⎟ , ⎜ .. .. .. . . .. ⎟ ⎝. . . . ⎠ . 0 ⎛
0 µ1 ⎜0 0 ⎜ ⎜ σr (e) = ⎜ ... ... ⎜ ⎝0 0 0 0
0 ... µ2 . . . .. . . . . 0 ... 0 ...
0
0
. . . −r
⎞
⎛
00 ⎟ ⎜1 0 ⎟ ⎜ ⎟ ⎜. . ⎟ , σr (f ) = ⎜ .. .. ⎟ ⎜ ⎝0 0 µr ⎠ 00 0 0 0 .. .
Lemma. The representation σr is simple.
⎞ ... 0 0 0 ... 0 0 0⎟ ⎟ . . .. .. .. ⎟ . . . . .⎟ ⎟ ... 1 0 0⎠ ... 0 1 0
19.2 Representations
281
Proof. Let (ε1 , . . . , εr+1 ) be the canonical basis of kr+1 . Then for any non zero vector v ∈ kr+1 , there exist n ∈ N and λ ∈ k \ {0} such that σr (f )n (v) = λεr+1 . So any non-trivial submodule contains εr+1 , and the result follows since kεi = kσr (e)r−i+1 (εr+1 ). 19.2.5 Let (V, σ) be a finite-dimensional representation of g where dim V = 0. Set H = σ(h), E = σ(e) and F = σ(f ). By a simple induction, we have the following identities for n ∈ N, m ∈ N∗ : [H, E] = 2nE n , [H, F n ] = −2nF n , (1) [F, E m ] = −m(H − (m − 1) idV ) ◦ E m−1 Thus E and F are nilpotent (19.2.3). Let q = min{n ∈ N; E q+1 = 0} and w ∈ V such that v = E q (w) = 0. Set v0 = v and for n ∈ N∗ , vn = F n (v). Then by using (1), we obtain: 0 = [F, E q+1 ](w) ⇒ H(v0 ) = qv0 . Hence, again by (1), we deduce that: (2)
H(vn ) = (q − 2n)vn . Further, a simple induction shows that:
(3)
E(vn ) = n(q − n + 1)vn−1 .
It follows that if vn = 0, it is an eigenvector for H with eigenvalue (q −2n). Let s be such that vs = 0 and vn = 0 for n > s, and let W be the subspace spanned by v0 , . . . , vs . Then W is a g-submodule of V and (v0 , . . . , vs ) is a basis of W by (2). Denote by HW , EW , FW the restrictions of H, E, F to W . Since [EW , FW ] = HW , 0 = tr(HW ) =
s
(q − 2n) = (s + 1)(q − s).
n=0
Hence s = q and the matrices of HW , EW , FW in the basis (v0 , . . . , vq ) are the ones in 19.2.4 with r = q. We deduce that the representations σ|W and σq are equivalent. So we have obtained the following result: Theorem. Let r ∈ N and (V, σ) a simple representation of sl2 (k) of dimension r + 1. Then σ is equivalent to σr . Moreover, the eigenvalues of σ(h) are −r, −r + 2, . . . , r − 2, r, and if v ∈ V \ {0} verifies σ(e)(v) = 0 (resp. σ(f )(v) = 0), then v is a σ(h)-eigenvector of eigenvalue r (resp. −r). 19.2.6 Theorem. Any finite-dimensional representation of g = sl2 (k) is semisimple.
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Proof. Let (V, σ) be a finite-dimensional representation of g. If dim V = 0, there is nothing to prove. So we may assume that dim V 1. By 19.2.5, there is a simple submodule W of dimension q + 1 in W . We shall prove that there is a submodules U of V such that V = V ⊕ W . Then the result would follow by induction on the dimension of V . Denote by η the dual representation of σ. Let w ∈ V and (v0 , . . . , vq ) be the basis of W as in 19.2.5. Let g ∈ V ∗ be such that g(v0 ) = 1. Then (η(E)q (g)) (w) = g((−E)q (w)) = (−1)q . So f = η(E)q (g) = 0 and q is the smallest integer such that η(E)q+1 = 0. By 19.2.5, the subspace M of V ∗ spanned by fn = η(F )n (f ), 0 n q, is a simple submodule of dimension q + 1. Using (3), we verify easily that: fn (vq−n ) = (−1)q−n (q!)2 , fn (vp ) = 0 , if n + p > q.
(4)
Let U be the orthogonal of M in V . Then dim V = dim U + dim W and U is a g-submodule of V . Further using (4) we deduce that U ∩ W = {0}. So V is the direct sum of the submodules U and W . 19.2.7 Corollary. Let (V, σ) be a finite-dimensional representation of sl2 (k). (i) There exist r1 , . . . , rn ∈ N such that σ is equivalent to the direct sum of the representations σri , 1 i n. (ii) We have V = σ(e)(V ) ⊕ ker σ(f ) = σ(f )(V ) ⊕ ker σ(e). (iii) σ(h) is semisimple with integer eigenvalues. (iv) If V is non-trivial, then σ is irreducible if and only if the eigenvalues of σ(h) are simple and they are either all even or all odd. Proof. Part (i) follows from 19.2.6 and 19.2.5. The rest is clear for σr . The general case follows from (i).
19.3 Nilpotent Lie algebras 19.3.1 An ideal of g is said to be characteristic if it is invariant under all the derivations of g. We define by induction the following decreasing chains of characteristic ideals of g: C 1 (g) = g , C 2 (g) = [g, g] , . . . , C i+1 (g) = [g, C i (g)] , . . . D (g) = g , D1 (g) = [g, g] , . . . , Di+1 (g) = [Di (g), Di (g)] , . . . 0
We call (C i (g))i1 the central descending series of g, and (Di (g))i0 the derived series of g. In particular, D(g) = D1 (g) is called the derived ideal of g. If u : g → h is a Lie algebra homomorphism, then u(C i (g)) ⊂ C i (h) and u(Di (g)) ⊂ Di (h) for all i. Further, these inclusions are equalities when u is surjective.
19.3 Nilpotent Lie algebras
283
19.3.2 Proposition. The following conditions are equivalent: (i) There exists an integer i such that C i (g) = {0}. (ii) There exists an integer j such that ad x1 ◦ ad x2 ◦ · · · ◦ ad xj = 0 for all x1 , . . . , xj ∈ g. (iii) There exists a chain g = g1 ⊃ g2 ⊃ · · · ⊃ gn = {0} of ideals of g such that [g, gi ] ⊂ gi+1 for i < n. If these conditions are verified, we say that g is nilpotent. Proof. The equivalence (i) ⇔ (ii) is straightforward by the definition of C i (g). Next, we have (i) ⇒ (iii) by taking gi = C i (g). Finally, to show that (iii) ⇒ (i), it suffices to show that C i (g) ⊂ gi , which is a simple induction. 19.3.3 Proposition. (i) If g is nilpotent, then any subalgebra (resp. quotient) of g is nilpotent. (ii) Let a be a subalgebra of g which is contained in the centre of g. Then g is nilpotent if and only if g/a is nilpotent. Proof. Part (i) is obvious. If g/a is nilpotent, then there exists k ∈ N such that C k (g/a) = {0}, or C k (g) ⊂ a. Since a is in the centre of g, C k+1 (g) = {0}. The converse follows from part (i). 19.3.4 Proposition. Let g be nilpotent. (i) If g = {0}, then z(g) = {0}. (ii) If h is a subalgebra of g distinct from g, then ng (h) = h. (iii) The Killing form of g is zero. Proof. (i) The last non-zero C i (g) is central in g. (ii) Let j = max{i; C i (g) + h = h}. Then C j (g) + h ⊂ ng (h). (iii) By 19.3.2, ad x ◦ ad y is a nilpotent endomorphism for all x, y ∈ g. 19.3.5 Lemma. Let V be a vector space of dimension n > 0. (i) If x ∈ End(V ) is nilpotent, ad x is a nilpotent endomorphism of gl(V ). More precisely, if xp = 0, then (ad x)2p−1 = 0. (ii) If g is a Lie subalgebra of gl(V ) consisting of nilpotent endomorphisms, then V g = {0}. Proof. Part (i) is clear from the fact that if y ∈ gl(V ), then (ad x)n (y) is the sum of terms of the form ±xi yxj with i + j = n. To prove (ii), we proceed by induction on dim g = p. The result is obvious when p 1, so let us assume that p > 1. Let a be a non-zero Lie subalgebra of g of dimension q < p. For x ∈ a, the endomorphism σ(x) induced by adg x on g/a is nilpotent by (i). So by induction, there exists y ∈ g \ a such that [y, a] ⊂ a. Consequently, a is an ideal of a ⊕ ky. By iterating this process, we see that g contains an ideal h of codimension 1. By the induction hypothesis, W = V h = {0}. Now for y ∈ g \ h, z ∈ h and w ∈ W , we have,
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z ◦ y(w) = y ◦ z(w) + [y, z](w) = 0, since [y, z] ∈ h. Thus y(W ) ⊂ W . Since y is nilpotent, y(v) = 0 for some v ∈ W \ {0}. Hence V g = {0}. 19.3.6 Theorem. (Engel’s Theorem) Let (V, σ) be a representation of g of dimension n > 0. Assume that σ(g) is a nilpotent endomorphism for all g ∈ g. Then there is a flag {0} = V0 ⊂ V1 ⊂ · · · ⊂ Vn = V of V such that σ(g)Vi ⊂ Vi−1 for 1 i n. Proof. We may assume that g ⊂ gl(V ). The result is clear when n = 1. So let us proceed by induction on n. By 19.3.5, there exists v ∈ V g which is nonzero. Let ϕ : V → V /kv = W be the canonical surjection. Then by applying the induction hypothesis to W , we have a flag (Wi )0in−1 of W with the required properties. Set V0 = {0} and Vi = ϕ−1 (Wi−1 ) for 1 i n, then the flag (Vi )0in clearly satisfies σ(g)Vi ⊂ Vi−1 for 1 i n. 19.3.7 Corollary. A Lie algebra g is nilpotent if and only if for all x ∈ g, ad x is a nilpotent endomorphism of g. 19.3.8 Let V be a finite-dimensional vector space and x ∈ End(V ). For α ∈ k, set: ker(x − α idV )n . Vα (x) = ker(x − α idV ) , V α (x) = n0
The subspace V α (x) is called the generalized eigenspace of x relative to α, and V 0 (x) is the nilspace of x. Clearly Vα (x) ⊂ V α (x) and α V (x). Vα (x) = {0} ⇔ V α (x) = {0} , V = α∈k
Lemma. Let x, y ∈ End(V ). Then (ad x)n (y) = 0 for some n ∈ N if and only if y(V α (x)) ⊂ V α (x) for all α ∈ k. Proof. Suppose that (ad x)n (y) = 0 for some n ∈ N. Since (ad x)(y) = (ad(x − α idV ))(y), we only need to show that y(V 0 (x)) is contained in V 0 (x). This is trivial when n = 0. Let us proceed by induction on n. Set z = [x, y], then (ad x)n−1 (z) = 0. By the induction hypothesis, z(V 0 (x)) ⊂ V 0 (x). For q ∈ N∗ , we have the following identity: xq y = yxq +
q i=1
xq−i (xy − yx)x−i−1 = yxq +
q
xq−i zxi−1 .
i=1
Since V is finite-dimensional, there exists r ∈ N∗ such that xr (V 0 (x)) = {0}. It follows that x2r−1 y(V 0 (x)) = {0} and therefore y(V 0 (x)) ⊂ V 0 (x).
19.3 Nilpotent Lie algebras
285
Conversely, suppose that y(V α (x)) ⊂ V α (x) for all α ∈ k. For α ∈ k, there exists rα ∈ N∗ such that (x − α idV )rα |V α (x) = 0. This implies that (ad(x − α idV ))2rα −1 (y)|V α = 0 (19.3.5). But (ad x)(y) = (ad(x − α idV ))(y), so (ad x)2rα −1 (y)|V α = 0. Now V is finite-dimensional and V is the sum of the generalized eigenspaces V α (x), so (ad x)n (y) = 0 for some n ∈ N. 19.3.9 Let g be a Lie algebra, P the set of maps from g to k and (V, σ) a finite-dimensional representation of g. For λ ∈ P , set: λ(x) Vλ (g) = Vλ(x) (σ(x)) , V λ (g) = V (σ(x)). x∈g
x∈g
The sum of V λ (g), λ ∈ P , is clearly direct. Proposition. Assume that g is nilpotent. (i) For all λ ∈ P , V λ (g) is a submodule of V . (ii) If V λ (g) = {0}, then Vλ (g) = {0}, and λ is a linear form on g which is zero on [g, g]. (iii) We have: λ V (g). V = λ∈P
If λ ∈ P , there exists a basis of V λ (g) such that for any x ∈ g, the matrix A(x) = [aij (x)] of σ(x)|V λ (g) with respect to this basis is strictly upper triangular, with aii (x) = λ(x) for all i. Proof. (i) Since g is nilpotent, it follows from 19.3.8 (i) that V λ(x) (σ(x)) is a submodule. (ii) If W = V λ (g) = {0}, then by (i), W is a submodule. If x ∈ g, there exists r ∈ N∗ such that (σ(x)|W − λ(x) idW )r = 0. It follows that λ(x) is the unique eigenvalue of σ(x)|W . Hence (dim W )λ(x) = tr σ(x)|W , and therefore λ is a linear form on g which is zero on [g, g]. It is then clear that the map g → End(W ), x → σ(x)|W − λ(x) idW , is a representation of g. The result now follows from Engel’s theorem (19.3.6). (iii) We shall prove part (iii) by induction on dim V = r. The case r = 0 is trivial. So let us assume that r 1. If for all x ∈ g, λ(x) is the unique eigenvalue of σ(x), then V = V λ (g). From the proof of (ii), x → σ(x) − λ(x) idV is a representation of g in V , so by applying Engel’s theorem (19.3.6) to this representation, we have the required basis. Now suppose that there exists x ∈ g such that σ(x) has two distinct eigenvalues. Since V = α∈k V α (x), dim V α (x) < r and the V α (x)’s are submodules of V (19.3.8), the result follows by applying the induction hypothesis.
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19.4 Solvable Lie algebras 19.4.1 Proposition. The following conditions are equivalent: (i) There exists k ∈ N such that Dk (g) = {0}. (ii) There exists a chain g = g0 ⊃ g1 ⊃ · · · ⊃ gn = {0} of ideals of g such that [gi , gi ] ⊂ gi+1 for 0 i n − 1. If these conditions are verified, we say that g is solvable. Proof. The proof is analogue to the one for 19.3.2. 19.4.2 Proposition. (i) A nilpotent Lie algebra is solvable. (ii) Subalgebras and quotients of a solvable Lie algebra are solvable. (iii) Let a be an ideal of g. Then g is solvable if and only if a and g/a are solvable. Proof. Parts (i) and (ii) are straightforward. Now if a and g/a are solvable, then Di (g) ⊂ a for some integer i and Di+j (g) ⊂ Dj (a) = {0} for some integer j. So g is solvable. The converse follows from (ii). 19.4.3 Lemma. Let (V, σ) be a finite-dimensional representation of g, a an ideal of g, λ ∈ a∗ and W = {v ∈ V ; σ(x)(v) = λ(x)(v) for all x ∈ a}. Then W is a submodule of V . Proof. We may assume that g ⊂ gl(V ) and W is non-zero. For v0 ∈ W \ {0} and x ∈ g, set vn = xn v0 , n ∈ N∗ , and p the smallest integer such that v0 , . . . , vp are linearly independent. Denote by U the subspace spanned by v0 , . . . , vp , then it is clear that x(U ) ⊂ U . Now if y ∈ a, then: y(v1 ) = yx(v0 ) = xy(v0 ) + [y, x](v0 ) ∈ λ(y)v1 + kv0 y(v2 ) = yx(v1 ) = xy(v1 ) + [y, x](v1 ) ∈ λ(y)v1 + kx(v0 ) + kv1 + kv0 ⊂ λ(y)v2 + kv1 + kv0 . If 0 i p, then we have by induction that: y(vi ) ∈ λ(y)vi + kvi−1 + · · · + kv0 . It follows that y(U ) ⊂ U and tr(y|U ) = (p + 1)λ(y). Since [x, y] ∈ a, we have that (p + 1)λ([x, y]) = 0. Hence λ([x, y]) = 0. Finally, yx(v0 ) = xy(v0 ) + [y, x](v0 ) = λ(y)x(v0 ) + λ([y, x])(v0 ) = λ(y)x(v0 ). So x(v0 ) ∈ W .
19.4.4 Theorem. (Lie’s Theorem) Let (V, σ) be a finite-dimensional representation of a solvable Lie algebra g. Then there exists a flag {0} = V0 ⊂ V1 ⊂ · · · ⊂ Vn = V of V such that σ(g)(Vi ) ⊂ Vi for 0 i n.
19.4 Solvable Lie algebras
287
Proof. We may assume that g ⊂ gl(V ) and V = {0}. We shall prove by induction on dim g = p that there exists µ ∈ g∗ and v ∈ V \ {0} such that x(v) = µ(x)v for all x ∈ g. We can then conclude as in the proof of 19.3.6. The case p 1 is obvious, so let us assume that p 2. Since D(g) = g, any codimension 1 subspace of g containing D(g) is an ideal of g. Let a be such an ideal. By induction, there exist λ ∈ a∗ and v ∈ V \ {0} such that y(v) = λ(y)v for all y ∈ a. Let W be as in 19.4.3, then W = {0} and g(W ) ⊂ W . Let x ∈ g \ a, then since k is algebraically closed, x|W has an eigenvector w. It follows that z(w) ∈ kw for all z ∈ g. So we are done. 19.4.5 Corollary. (i) Any finite-dimensional simple representation of a solvable Lie algebra is 1-dimensional. (ii) If g is solvable, then there exists a flag {0} ⊂ g0 ⊂ · · · ⊂ gn = g consisting of ideals of g. (iii) If g is solvable, then any element of ad[g, g] is nilpotent. (iv) A Lie algebra g is solvable if and only if D(g) is nilpotent. Proof. Parts (i) and (ii) are consequences of 19.4.4, while part (iii) follows from (ii). Finally, (iv) follows from (iii) and 19.4.2 (iii). 19.4.6 Corollary. Let (V, σ) be a representation of dimension n > 0 of a nilpotent Lie algebra g such that ker σ(x) = {0} for all x ∈ g. Then there exists v ∈ V \ {0} such that σ(x)(v) = 0 for all x ∈ g. Proof. We shall prove the result by induction on n. The case n = 1 is obvious, so we may assume that n > 1. By 19.4.4, there is a basis (v1 , . . . , vn ) of V such that for all x ∈ g, the matrix of σ(x) in this basis is upper triangular with diagonal entries λ1 (x), . . . , λn (x) where λ1 , . . . , λn ∈ g∗ . Since ker σ(x) = {0}, λ1 · · · λn = 0. If λ1 · · · λn−1 = 0, then by applying the induction hypothesis on the restriction of σ to W = kv1 + · · · + kvn−1 , we have our result. So let us assume that λ1 · · · λn−1 = 0. Then λn = 0 since the ring of polynomial functions on g is an integral domain. Let x ∈ g be such that λ1 (x) · · · λn−1 (x) = 0. We have σ(x)(v) = 0 for some v ∈ V \ {0}. By our choice of x, v ∈ W , and σ(g)(v) ∈ W since λn = 0. Define f : g → W by f (y) = σ(y)(v). Then f ([x, y]) = σ(x)(f (y)) for all y ∈ g because σ(x)(v) = 0. It follows that the restriction of σ(x) to f (g) is nilpotent. If f (g) = {0}, then σ(x)(w) = 0 for some w ∈ f (g) \ {0}. But this is absurd since λ1 (x) · · · λn−1 (x) = 0. Thus f (g) = {0} and σ(g)(v) = {0}. 19.4.7 Lemma. Let V be a finite-dimensional vector space and g ⊂ gl(V ) a Lie subalgebra such that tr(xy) = 0 for all x, y ∈ g. Then any element of D(g) is a nilpotent endomorphism of V . Proof. By 19.4.4 and 19.4.5, it suffices to prove that g is solvable. We shall prove this by induction on dim g = r. The case r = 0 is trivial. So let r > 0.
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If D(g) = g, then by the induction hypothesis, D(g) is solvable and therefore g is solvable. So let us assume that D(g) = g and we shall show that this case is not possible. Let h be a maximal (proper) Lie subalgebra of g. Then dim h 1 and h is solvable by induction. Applying 19.4.4 to the adjoint representation of h in g/h, we deduce from the maximality of h that there exist x ∈ g \ h and λ ∈ h∗ verifying g = h ⊕ kx and [y, x] − λ(y)x ∈ h for all y ∈ h. 1) Let (W, σ) be a simple finite-dimensional representation of g and W0 a simple h-submodule of W . Then by 19.4.5, W0 = kw0 for some w0 ∈ W \ {0}. Let µ ∈ h∗ be such that σ(y)(w0 ) = µ(y)w0 for all y ∈ h. Set W−1 = {0} and for n ∈ N, wn = σ(x)n (w0 ) and Wn = kw0 + · · · + kwn . If y ∈ h, then: σ(y)wi+1 = σ(y)σ(x)wi = σ([y, x])wi + σ(x)σ(y)wi = σ([y, x] − λ(y)x)wi + σ(x)(σ(y)wi ) + λ(y)σ(x)wi , and (σ(y) − µ(y) − (i + 1)λ(y))wi+1 is equal to σ([y, x] − λ(y)x)wi + σ(x)(σ(y) − µ(y) − iλ(y))wi . It follows easily by induction that for all i ∈ N and y ∈ h, σ(h)(Wi ) ⊂ Wi and σ(y).wi − (µ(y) + iλ(y))wi ∈ Wi−1 . Since W is a simple g-module, W = Wq where q is the largest integer such that w0 , . . . , wq are linearly independent. The equality in the previous paragraph implies that 1 tr(σ(y)) = (q + 1)µ(y) + q(q + 1)λ(y). 2 Now g = D(g), so tr(σ(y)) = 0 and we have 2µ(y) = −qλ(y), q q q 2 σ(y).wi − (i − )λ(y)wi ∈ Wi−1 , tr(σ(y)2 ) = λ(y)2 . j− 2 2 j=0 2) If {0} = V0 ⊂ · · · ⊂ Vp = V is a Jordan-H¨ older series of the g-module V , and q(j) = dim(Vj+1 /Vj ) − 1, then, using the preceding computation, we obtain that % 2 & q(j) p−1 q(j) 0 = tr(y 2 ) = λ(y)2 . i− 2 j=0 i=0 Since D(g) = g = h, λ = 0 and so q(j) = 0 for all j. Again, the fact that D(g) = g implies that (Vj+1 /Vj )g = Vj+1 /Vj . Hence g consists of nilpotent endomorphisms of V . By 19.3.6, g is nilpotent. But this contradicts the assumption that D(g) = g and dim g > 0. 19.4.8 Theorem. (Cartan’s Criterion) Let (V, σ) be a finite-dimensional representation of g and βσ the bilinear form associated to σ. The following conditions are equivalent: (i) σ(g) is a solvable Lie algebra. (ii) We have βσ (D(g), D(g)) = {0}.
19.5 Radical and the largest nilpotent ideal
289
Proof. We may assume that g ⊂ gl(V ). Then clearly (i) ⇒ (ii) by 19.4.5. Conversely by 19.3.6 and 19.4.7, D2 (g) is nilpotent. So D(g) and g are solvable. Hence (ii) ⇒ (i).
19.5 Radical and the largest nilpotent ideal 19.5.1 Lemma. Let a be an ideal of g and (V, σ) a finite-dimensional representation of g. (i) Assume that σ is simple and all the elements of σ(a) are nilpotent. Then σ(a) = {0}. older series of V . (ii) Let {0} = V0 ⊂ V1 ⊂ · · · ⊂ Vn = V be a Jordan-H¨ Then all the elements of σ(a) are nilpotent if and only if σ(a)(Vi ) ⊂ Vi−1 for 1 i n. Proof. Part (i) follows from 19.3.5 and 19.4.3, while (ii) follows from (i). 19.5.2 Lemma. Let V be a finite-dimensional vector space, g a Lie subalgebra of gl(V ) and a an abelian ideal of g. If V is a simple g-module, then a ∩ D(g) = {0}. Proof. 1) Let S be the associative subalgebra of End(V ) generated by a. Let b ⊂ a be an ideal of g satisfying tr(bs) = 0 for all b ∈ b and s ∈ S. In particular, tr(bn ) = 0 for all b ∈ b and n ∈ N∗ , so b is nilpotent (19.2.3). By 19.5.1, b = {0}. 2) Since a is abelian, we have for x ∈ g, a ∈ a and s ∈ S that tr([x, a]s) = tr(xas − axs) = tr(xas − xsa) = tr(xas − xas) = 0. It follows from point 1) that [g, a] = {0}, and so a ⊂ z(g). Hence xs = sx for all x ∈ g and s ∈ S. 3) Finally, we have tr([x, y]s) = tr(xys − yxs) = tr(xys − xsy) = tr (x(ys − sy)) = 0 for x, y ∈ g and s ∈ S. So point 1) implies that a ∩ D(g) = {0}.
19.5.3 Proposition. There is a unique maximal solvable ideal in g. It is called the radical of g, and is denoted by rad g. Proof. Let a, b be solvable ideals of g. Then a + b is an ideal of g which is solvable (19.4.2) since (a + b)/b and a/a ∩ b are isomorphic. So there is a unique maximal solvable ideal in g. 19.5.4 Proposition. The radical of g is the smallest ideal a of g such that rad(g/a) = {0}.
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Proof. Let r = rad g and p : g → g/r be the canonical surjection. Then o = p−1 (rad(g/r)) is an ideal of g such that o/r is solvable. So o is solvable by 19.4.2 and o = r. Let a be an ideal of g and ϕ : g → g/a the canonical projection. If rad(g/a) = {0}, then ϕ(r) = {0}. Thus a ⊃ r. 19.5.5 Proposition. Let (V, σ) be a finite-dimensional representation of older series of V , βσ the bilinear g, {0} = V0 ⊂ V1 ⊂ · · · ⊂ Vn = V a Jordan-H¨ form on g associated to σ and σi the representation of g on Vi /Vi−1 induced by σ, for 1 i n. (i) There is a unique maximal ideal n among the ideals a of g such that the elements of σ(a) are nilpotent. (ii) We have n = ker σ1 ∩ · · · ∩ ker σn and βσ (n, g) = {0}. We call n the largest nilpotent ideal of σ. Proof. By 19.5.1, for an ideal a of g, σ(x) is nilpotent for all x ∈ a if and only if a ⊂ ker σ1 ∩ · · · ∩ ker σn . So we have (i) and the first part of (ii). Let x ∈ g, y ∈ n, then σ(x)σ(y)(Vi ) ⊂ σ(x)(Vi−1 ) ⊂ Vi−1 , for 1 i n. So σ(x)σ(y) is nilpotent and βσ (n, g) = {0}. 19.5.6 Lemma. Let (V, σ) be a finite-dimensional representation of g. (i) If σ is simple, then σ(rad g ∩ D(g)) = {0}. (ii) If n is the largest nilpotent ideal of σ, then rad g ∩ D(g) ⊂ n. Proof. (i) Let r = rad g and i be the smallest integer such that σ(Di+1 (r)) = {0}. Then the image under σ of a = Di (r) is an abelian ideal of σ(g). By Schur’s Lemma, σ(a) ⊂ k idV , and so σ(D(g) ∩ Di (r)) = {0}. If i > 0, then Di (r) ⊂ D(g) and so σ(a) = {0}. This is impossible by the minimality of i. So i = 0 and σ(r ∩ D(g)) = {0}. Part (ii) follows clearly from (i) and 19.5.5. 19.5.7 Let us conserve the notations of 19.5.5. In general, the fact that σ(x) is nilpotent does not imply that x ∈ n. Nevertheless, we have the following result: Proposition. Let (V, σ) be a finite-dimensional representation of g and n the largest nilpotent ideal of σ. If x ∈ rad g and σ(x) is nilpotent, then x ∈ n. older Proof. Let r = rad g and {0} = V0 ⊂ V1 ⊂ · · · ⊂ Vn = V be a Jordan-H¨ series of the r-module V . By 19.4.5, Wi = Vi /Vi−1 is 1-dimensional, and the representation of r on Wi is given by a linear from λi ∈ r∗ which is zero on D(r). The set of elements x ∈ r such that σ(x) is nilpotent is therefore the ideal a = ker λ1 ∩ · · · ∩ ker λn of r. By 19.5.6, [g, a] ⊂ r ∩ D(g) ⊂ r ∩ n ⊂ a. Thus a is an ideal of g and the result follows from 19.5.5. 19.5.8 Proposition. Let L be the Killing form of g and n the largest nilpotent ideal of the adjoint representation of g. Then n is the largest nilpotent ideal of g, and L(g, n) = {0}.
19.6 Nilpotent radical
291
Proof. Let a be an ideal of g and x ∈ a. Then adg x is nilpotent if and only if ada x is nilpotent. Thus a is nilpotent if and only if adg x is nilpotent for all x ∈ a (19.3.7). Since any nilpotent ideal is contained in the radical of g, the result follows from 19.5.7 and 19.5.5.
19.6 Nilpotent radical 19.6.1 It follows from 19.5.5 that the intersection of the kernels of simple finite-dimensional representations of g is equal to the intersection of the largest nilpotent ideals of finite-dimensional representations of g. This is an ideal of g that we shall call the nilpotent radical of g. In view of 19.5.8, the nilpotent radical of g is contained in the radical of g. 19.6.2 Theorem. Let r = rad g and s the nilpotent radical of g. Then s = r ∩ D(g). In particular, when g is solvable, then s = D(g). Proof. Any λ ∈ g∗ verifying λ(D(g)) = {0} determines a representation of g of dimension 1. So s ⊂ ker λ. Since s ⊂ r, we have s ⊂ r ∩ D(g). The reverse inclusion follows from 19.5.6. 19.6.3 Corollary. Let r = rad g. (i) Let (V, σ) be a finite-dimensional representation of g and β the bilinear form on g associated to σ. Then β(r, D(g)) = {0}. (ii) r is the orthogonal of D(g) with respect to the Killing form L of g. (iii) r is a characteristic ideal of g. (iv) If a is an ideal of g, then rad a = a ∩ r. Proof. (i) Let n be the largest nilpotent ideal of g and s the nilpotent radical of g. By 19.6.2 and 19.6.1, [g, r] ⊂ s ⊂ n. So the result follows from 19.5.5 and 19.2.2. (ii) Let u be the orthogonal of D(g) with respect to L. Then u is an ideal of g containing r (19.2.2). Since the Killing form of u is L|u×u , adg u is solvable (19.4.8). But adg u is isomorphic to u/z(g), so u is solvable, and u = r. (iii) Let x, y ∈ g and d ∈ Der g. Since ad d(x) = [d, ad x], we verify easily that L(d(x), y) + L(x, d(y)) = 0. So the orthogonal of a characteristic ideal with respect to L is characteristic, and the result follows from (ii). (iv) Clearly, a ∩ r ⊂ rad a. On the other hand, by applying (iii) to a, we obtain that rad a is a solvable ideal of g. So rad a ⊂ r. 19.6.4 Proposition. Let r = rad g and n the largest nilpotent ideal of g. (i) We have d(r) ⊂ n for all derivation d of g. (ii) n is a characteristic ideal of g. Proof. It suffices to prove (i). We define a Lie algebra h = kv ⊕ g extending the Lie bracket of g by defining [v, x] = d(x) for all x ∈ g. Then g is an ideal of h and by 19.6.3 (iii), r ⊂ rad h, and so d(r) = [v, r] ⊂ [h, h] ∩ rad h = o. By 19.6.2, o is a nilpotent ideal of h. So d(r) is contained in the nilpotent ideal o ∩ g of g. Hence d(r) ⊂ n (19.5.8).
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19.7 Regular linear forms 19.7.1 Let n = dim g, d ∈ {0, 1, . . . , n} and Gr(g, d) the Grassmannian variety of d-dimensional vector subspaces of g. Recall from 13.6 that given B = (e1 , . . . , en ) a basis of g, the set UB of elements o ∈ Gr(g, d) verifying o ∩ (ked+1 + · · · + ken ) = {0} is open in Gr(g, d). Further, o ∈ UB if and only if o has a unique basis (v1 , . . . , vd ) of the form vi = ei +
n
aij ej .
j=d+1
The map o → (a1,d+1 , . . . , a1,n , . . . , ad,d+1 , . . . , ad,n ) is an isomorphism of UB onto kd(n−d) . 19.7.2 Lemma. (i) The set of d-dimensional Lie subalgebras of g is a closed subset of Gr(g, d). (ii) The set of d-dimensional solvable (resp. nilpotent, abelian) Lie subalgebras of g is a closed subset of Gr(g, d). (iii) The set of (o, f ) ∈ Gr(g, d) × g∗ such that f ([g, o]) = {0} (resp. f ([o, o]) = {0}) is a closed subset of Gr(g, d) × g∗ . Proof. Let us prove (i), and (ii) for solvable Lie subalgebras. The proofs of the rest are similar. We shall use the notations of 19.7.1. An element o ∈ Gr(g, d) is a Lie subalgebra if and only if [vi , vj ] ∈ o for 1 i, j d. These conditions correspond to the vanishing of determinants in the apq ’s. Since the open subsets UB , for B a basis of g, form an affine open covering of Gr(g, d), part (i) follows. Further, a d-dimensional Lie subalgebra o is solvable if and only if Dd (o) = {0}. By expressing this condition in the basis (v1 , . . . , vd ), we see that it is equivalent to the vanishing of certain polynomials in the apq ’s. 19.7.3 For f ∈ g∗ , define Φf to be the alternating bilinear form on g given by: Φf (x, y) = f ([x, y]). For a subset a of g, denote by a(f ) the orthogonal of a with respect to Φf . If a is an ideal of g, then it is easy to check that a(f ) is a Lie subalgebra of g. In particular, the kernel g(f ) of Φf is a Lie subalgebra of g. Since Φf induces a non-degenerate alternating bilinear form on g/g(f ) , dim g − dim g(f ) is an even integer. We define the index of g to be the integer χ(g) = min{dim g(f ) ; f ∈ g∗ }. We say that f ∈ g∗ is regular if dim g(f ) = χ(g), and we denote by g∗reg the set of regular elements of g∗ . 19.7.4 Lemma. Let u be a subspace of g. (i) The set of elements f ∈ g∗ verifying u ∩ g(f ) = {0} is closed in g∗ . (ii) Let f ∈ u∗ and g ∈ g∗ be such that g|u = f and dim g(g) dim g(h) for all h ∈ g∗ verifying h|u = f . Then [g(g) , g(g) ] ⊂ u.
19.7 Regular linear forms
293
Proof. (i) Let E = (e1 , . . . , en ) be a basis of g such that (e1 , . . . , ep ) is a basis of u, and (e∗1 , . . . , e∗n ) the basis of g∗ dual to g. For 1 i, j n, set [ei , ej ] =
n
γijk ek .
k=1
For f = λ1 e∗1 + · · · + λn e∗n , we have: u = µ1 e1 + · · · + µp ep ∈ u ∩ g(f ) ⇔ 0 = f ([u, ej ]) =
p n
µl γljk λk for all j.
l=1 k=1
( ' n So u∩g(f ) = {0} is equivalent to the condition that the matrix k=1 γljk λk , 1 l p and 1 j n, has rank strictly less than p. So we are done. (ii) Fix a subspace w such that g is the direct sum of g(g) and w. Let h ∈ g∗ be such that h|u = 0. If t ∈ k, the assumptions on g imply that g = w ⊕ g(g+th) ⇔ w ∩ g(g+th) = {0}. By (i), this defines an open subset K of k containing 0. Let (e1 , . . . , en ) be a basis of g such that (e1 , · · · , ep ) is a basis of w. For x = λ1 e1 + · · · + λn en ∈ g and t ∈ K, write x = x1 (t) + x2 (t) where x1 (t) = µ1 e1 + · · · + µp ep ∈ w and x2 (t) ∈ g(g+th) . The coefficients µi are obtained via the following equalities: p i=1
µi (g + th)([ei , ej ]) =
n
λk (g + th)([ek , ej ]) , 1 j n.
k=1
' ( The set K of t ∈ K such that the matrix (g + th)([ei , ej ]) , 1 i p, 1 j n, has rank p, is open and it contains 0. It follows that the maps x → x1 (t) and x → x2 (t) are rational on K . Let x, y ∈ g(g) , and x2 (t), y2 (t) their projections in g(g+th) with respect to the direct sum g(g+th) ⊕ w. Then for t ∈ K , we have: (g + th)([x2 (t), y2 (t)]) = 0. Taking the derivatives on both sides at t = 0, we obtain: h([x, y]) + g([x, y2 (0)]) + g([x2 (0), y]) = 0. Hence h([x, y]) = 0. This is true for all h verifying h|u = 0, so we have [x, y] ∈ u. 19.7.5 Proposition. (i) The set g∗reg is open and dense in g∗ . (ii) If f ∈ g∗reg , then the Lie algebra g(f ) is abelian. (iii) For all f ∈ g∗ , g(f ) contains an abelian Lie subalgebra of dimension χ(g).
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Proof. (i) We have f ∈ g∗reg if and only if g(f ) ∩ u = {0} for all subspaces of dimension dim g − χ(g). So the result follows from 19.7.4 (i). (ii) Just take u = {0} in 19.7.4 (ii). (iii) Let χ(g) = d, p : Gr(g, d) × g∗ → g∗ be the canonical surjection and F the set of elements (u, f ) ∈ Gr(g, d) × g∗ such that f ([g, u]) = {0} and [u, u] = {0}. Since p is a closed map (13.4.5 and 13.6.3), it follows from 19.7.2 (iii) that p(F) is closed. But (ii) says that g∗reg ⊂ p(F). So part (i) implies that p(F) = g∗ and we are done. 19.7.6 Fix an integer d 0 and let g∗d = {f ∈ g∗ ; dim g(f ) = d}. The set is locally closed in g∗ , so it is a subvariety of g∗ . Let ϕ : g∗d → Gr(g, d) be the map defined by f → g(f ) .
g∗d
Proposition. The map ϕ is a morphism of varieties. Proof. Let h ∈ g∗d , E = (e1 , . . . , en ) be a basis of g such that (e1 , . . . , ed ) is a basis of g(h) and (e∗1 , . . . , e∗n ) the basis!of g∗ dual to E. Denote " by V the subspace spanned by ed+1 , . . . , en , u = f ∈ g∗d ; V ∩ g(f ) = {0} , and U = {p ∈ Gr(g, d); p ∩ V = {0}}. By 19.7.1 and 19.7.4, u and U are open subsets of g∗d and Gr(g, d). Note that ϕ−1 (U ) = u. Let f = λ1 e∗1 + · · · + λn e∗n ∈ u. For 1 i n and d + 1 j n, set αi,j = f ([ei , ej ]). Since g(f ) ∩ V = {0}, the condition 0 = f ([ei , νd+1 ed+1 + · · · + νn en ]) = νd+1 αi,d+1 + · · · + νn αi,n , 1 i n implies that νd+1 = · · · = νn = 0. So the rank of the matrix [αi,j ] is n − d. Now, for 1 k d, there is a unique (vd+1,k , . . . , vn,k ) ∈ kn−d such that 0 = f ([ei , ek + νd+1,k ed+1 + · · · + νn,k en ]) , 1 i n or equivalently, νd+1,k αi,d+1 + · · · + νn,k αi,n = (λ1 e∗1 + · · · + λn e∗n )([ek , ei ]) , 1 i n. It follows from the formulas for solving a Cramer’s system that there exists an open subset o of u such that the maps f → νj,k , d + 1 j n, 1 k d, are regular on o. So the result follows from the remarks in 19.7.1.
19.8 Cartan subalgebras 19.8.1 For x ∈ g, α ∈ k, we shall denote by gα (x) and gα (x) the subspaces gα (ad x) and gα (ad x) (19.3.8). Let h be a Lie subalgebra of g. Consider g as a h-module via the adjoint representation. If λ ∈ h∗ \ {0} is such that gλ (h) = {0} (19.3.9), then we say that λ is a root of g relative to h. The set of roots, denoted by R(g, h), is called the root system of g relative to h. 19.8.2 For x, y, z ∈ g, n ∈ N and λ, µ ∈ k, we have:
19.8 Cartan subalgebras
(ad x − λ − µ)n ([y, z]) =
295
n n [(ad x − λ)i (y), (ad x − µ)n−i (z)]. i i=0
particular, g0 (x) is a We deduce therefore that [gλ (x), gµ (x)] ⊂ gλ+µ (x). In λ Lie subalgebra of g. Recall also from 19.3.8 that g = λ∈k g (x) and so if y ∈ gλ (x) with λ = 0, then ad y is nilpotent. 19.8.3 Definition. A nilpotent Lie subalgebra h of g verifying h = ng (h) is called a Cartan subalgebra of g. 19.8.4 Lemma. A Cartan subalgebra h of g is a maximal nilpotent Lie subalgebra. Proof. If t is a nilpotent Lie subalgebra containing h, then by applying 19.3.6 to the adjoint representation of h on t/h, we have h = t since h = ng (h). 19.8.5 Lemma. Let x ∈ g and h a Lie subalgebra of g. (i) If g0 (x) ⊂ h, then h = ng (h). (ii) Suppose that there exists an element x ∈ h such that h ⊂ g0 (x) and dim g0 (x) dim g0 (y) for all y ∈ h. Then g0 (x) ⊂ g0 (y) for all y ∈ h. (iii) If x ∈ ggen , then g0 (x) is the unique Cartan subalgebra of g containing x. In particular, g has a Cartan subalgebra. (iv) Let t be a Cartan subalgebra of g. There exists an element x ∈ g such that t = g0 (x). Proof. (i) Since g0 (x) ⊂ h, ad x induces a bijective endomorphism u of ng (h)/h. But x ∈ h implies that [x, ng (h)] ⊂ h, so u = 0 and h = ng (h). (ii) Let t = g0 (x) and y ∈ h. For λ ∈ k, denote by uλ and vλ the endomorphisms of t and g/t induced by ad(x + λy). Let χλ , χλ and χλ be respectively the characteristic polynomial of ad(x + λy), uλ and vλ . Then χλ = χλ χλ , and if n = dim g and r = dim t, then: χλ (T ) = T r +
r−1 i=0
ai (λ)T i , χλ (T ) = T n−r +
n−r−1
bj (λ)T j ,
j=0
where ai and bj are polynomial functions on k. Since t = g0 (x), b0 (0) = 0. So the set E of λ ∈ k such that b0 (λ) = 0 is open dense in k; so k \ E is a finite set. Hence g0 (x + λy) ⊂ t if λ ∈ E, and by our choice of x, this implies that g0 (x + λy) = t if λ ∈ E. It follows that χλ (T ) = T r if λ ∈ E, and thus for all λ ∈ k. This means that t ⊂ g0 (x + λy). So we have our result by replacing y by y − x and by taking λ = 1. (iii) If x ∈ ggen , then the conditions of (ii) are verified for h = g0 (x). So 0 g (x) ⊂ g0 (y) for all y ∈ h. Hence g0 (x) is nilpotent (19.3.7) and g0 (x) is a Cartan subalgebra by (i). Now if t is a Cartan subalgebra containing x, then t ⊂ g0 (x) since t is nilpotent. So t = g0 (x) (19.8.4). (iv) For y ∈ t, denote by σ(y) the endomorphism of g/t induced by ad y. Since t = ng (t), 0 is the only element v ∈ g/t such that σ(y)(v) = 0 for all
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y ∈ t. But t is nilpotent, so by 19.4.6, there exists x ∈ t such that σ(x) is bijective. Consequently g0 (x) = t. 19.8.6 Theorem. Let h be a nilpotent Lie subalgebra of g. If λ ∈ h∗ , let by gλ . us denote gλ (h) simply λ λ µ λ+µ . In particular, g0 is a (i) We have g = λ∈h∗ g , and [g , g ] ⊂ g Lie subalgebra of g containing h. (ii) h is a Cartan subalgebra if and only if h = g0 . (iii) If g0 is nilpotent, then g0 is a Cartan subalgebra of g. (iv) Let λ, µ ∈ h∗ , then L(gλ , gµ ) = {0} if λ = −µ. In particular, L(h, gλ ) = {0} if λ = 0. (v) Let x, y ∈ h. Then: (dim gλ )λ(x)λ(y). L(x, y) = λ∈h∗
Proof. (i) This follows from 19.3.9 and 19.8.2. (ii) Applying 19.3.6 to the adjoint representation of h on g0 /h, we obtain that if g0 = h, then ng (h) = h. So h is not a Cartan subalgebra. Conversely, since [h, ng (h)] ⊂ h and h is nilpotent, h ⊂ ng (h) ⊂ g0 . So if 0 g = h, h is a Cartan subalgebra. (iii) If t = g0 is nilpotent, then t ⊂ g0 (t). Since h ⊂ t, we deduce that 0 g (t) ⊂ t. Hence g0 (t) = t and by (ii), t is a Cartan subalgebra. (iv) Let λ, µ, ν ∈ h∗ , x ∈ gλ and y ∈ gµ . By (i), we have: (ad x◦ ad y)(gν ) ⊂ gλ+µ+ν . If λ + µ = 0, then ad x◦ ad y is nilpotent, hence L(x, y) = 0. Since h ⊂ g0 , we are done. (v) Let x, y ∈ h. There is a basis of gλ such that the matrix of ad x◦ ad y|gλ with respect to this basis is upper triangular with a unique eigenvalue λ(x)λ(y) (19.3.9). The result is now clear. 19.8.7 Corollary. Assume that the Killing form L on g is non-degenerate. Let h be a Cartan subalgebra R = R(g, h). of g and α . g (i) We have g = h ⊕ α∈R (ii) If α ∈ R, then −α ∈ R and L|gα ×g−α is non-degenerate. (iii) The restriction of L to h × h is non-degenerate. (iv) The elements of R span the vector space h∗ . (v) The Lie algebra h is abelian. Proof. Parts (i), (ii) and (iii) are immediate consequences of 19.8.6. Let x ∈ h be such that α(x) = 0 for all α ∈ R. By 19.8.6 (v), L(x, h) = {0}. Now (iii) implies that x = 0. So we have proved (iv). Finally let x, y ∈ h. If α ∈ R, then [x, gα ] ⊂ gα and [y, gα ] ⊂ gα . Since (ad[x, y])|gα = [ad x|gα , ad y|gα ],
19.8 Cartan subalgebras
297
we see that the trace of (ad[x, y])|gα is zero. But by 19.3.9, this is equal to (dim gα )α([x, y]), so α([x, y]) = 0. Thus [x, y] = 0 by (iv).
References and comments • [12], [14], [29], [38], [39], [42], [43], [80]. For a proof of Ado’s theorem (19.2.1), see [12] or [29]. The notion of a Lie algebra extends to any base field k (although there is a slight ambiguity of antisymmetry when the characteristic of k is 2). Some of the results in this chapter can be extended. It is a good exercise for the reader to work out to what extent the results in this chapter remain valid. For example, Lie’s theorem fails in general if k has positive characteristic. Lie algebras over the real numbers are fundamental in the theory of Lie groups. For more details, the reader may refer to [24], [59], [84] and [87].
20 Semisimple and reductive Lie algebras
In this chapter, we study the structure of semisimple and reductive Lie algebras. We introduce certain important objects related to such Lie algebras such as Weyl groups, Borel subalgebras and parabolic subalgebras. Let g be a finite-dimensional Lie algebra over k and L its Killing form.
20.1 Semisimple Lie algebras 20.1.1 Definition. (i) We say that g is simple if dim g > 1 and the only ideals of g are {0} and g. (ii) A Lie algebra g is semisimple if {0} is the only abelian ideal of g. 20.1.2 Theorem. The following conditions are equivalent: (i) g is semisimple. (ii) rad g = {0}. (iii) L is non-degenerate. Proof. (i) ⇒ (ii) Since r = rad g is an ideal, Dr (r) are ideals of g for all r 0. Moreover r is solvable, so if r = {0}, then Dr (r) is non-zero and commutative for some r. (ii) ⇒ (iii) The kernel a of L is an ideal of g and L|a×a is the Killing form of a. So L|a×a = 0 and a is solvable (19.4.8). Hence a ⊂ r = {0}. (iii) ⇒ (i) This follows from 19.5.8. 20.1.3 Corollary. (i) g is semisimple if and only if the Killing homomorphism of g is an isomorphism. (ii) A product of a finite number of semisimple Lie algebras is semisimple. (iii) The Lie algebra g/ rad g is semisimple. 20.1.4 Lemma. Let a be an ideal of g and b the orthogonal of a with respect to L. Suppose that {0} is the only solvable ideal of g contained in a. Then g = a ⊕ b and the Lie algebras g and a × b are isomorphic.
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20 Semisimple and reductive Lie algebras
Proof. Since b is an ideal, so is c = b ∩ a. The Killing form on c is zero, so it is solvable (19.4.8) and we deduce that c = {0}. So the result follows since dim g = dim a + dim b. 20.1.5 Proposition. Assume that g is semisimple. (i) The centre of g is trivial. In particular, the adjoint representation of g is faithful. (ii) We have g = [g, g]. (iii) All derivations of g are inner. (iv) For all finite-dimensional representation (V, σ) of g, σ(g) ⊂ sl(V ). Moreover, if σ is faithful, then the associated bilinear form βσ on g is nondegenerate. Proof. Part (i) is clear and (ii) follows from 19.6.3 and 20.1.2. For (iii), let u = ad g and d = Der g. We have u g, and u is an ideal of d. It follows from 20.1.4 that d = u × v where v is the orthogonal of u in d with respect to the Killing form. Since [d, ad x] = ad d(x) for all d ∈ d and x ∈ g, it is now clear that v = {0}. The first part of (iv) is a direct consequence of (ii). Let a be the orthogonal of g with respect to βσ . By Cartan’s Criterion, σ(a) is solvable and so a = {0} if σ is faithful. 20.1.6 Proposition. Let a be an ideal of a semisimple Lie algebra g and b the orthogonal of a in g with respect to the Killing form L. (i) a and g/a are semisimple. (ii) We have g = a ⊕ b and the Lie algebras g and a × b are isomorphic. Proof. These are direct consequences of 19.6.3 (iv), 20.1.2 and 20.1.4. 20.1.7 Theorem. The following conditions are equivalent: (i) g is semisimple. (ii) g is a direct product of simple Lie algebras. Proof. A simple Lie algebra is semisimple since its radical is {0}. So by 20.1.6, the implication (i) ⇒ (ii) is a simple induction on dim g. Conversely, if g is a direct product of simple Lie algebras, then its Killing form is clearly non-degenerate (since simple Lie algebras are semisimple). So g is semisimple by 20.1.2. 20.1.8 Proposition. Suppose that g = a1 × · · · × an where the ai ’s are simple Lie algebras. Then any ideal of g is a direct product of certain ai ’s. In particular, the ai ’s are the minimal ideals of g. We call the ai ’s the minimal components of g.
20.2 Examples
301
Proof. Let a be an ideal of g. By permuting the indices, we may assume that a ∩ ai = {0} if and only if 1 i p for some integer p. Then by the simplicity of the ai ’s, a ∩ ai = ai for 1 i p and [a, ai ] ⊂ a ∩ ai = {0} for i > p. It follows that a ⊃ a1 × · · · × ap and [a, ap+1 × · · · × an ] = {0}. Thus a ∩ (ap+1 × · · · × an ) = {0} since it is in the centre of ap+1 × · · · × an which is semisimple. As a ⊃ a1 × · · · × ap , we have a = a1 × · · · × ap .
20.2 Examples 20.2.1 Lemma. Let g be a Lie subalgebra of gl(V ) where V is a finitedimensional vector space. Assume that z(g) = {0} and that V is a simple g-module. Then g is semisimple. Proof. By 19.5.6, an abelian ideal a in g verifies [a, g] ⊂ a ∩ D(g) = {0}. So a ⊂ z(g) = {0}. 20.2.2 Theorem. Let V be a finite-dimensional vector space and a the set of scalar endomorphisms of V . (i) We have z(gl(V )) = a and z(sl(V )) = {0}. (ii) gl(V ) a × sl(V ). (iii) The Lie algebra sl(V ) is semisimple. Proof. Part (i) is just 10.8.4 and (ii) is clear. Part (ii) implies that any sl(V )submodule of V is a gl(V )-submodule. So (iii) follows from (i) and 20.2.1. 20.2.3 Theorem. Let V be a n-dimensional vector space, β a nondegenerate symmetric or alternating bilinear form on V and g the Lie subalgebra of gl(V ) consisting of elements f ∈ gl(V ) such that β(f (v), w) + β(v, f (w)) = 0 for all v, w ∈ V . Then g is semisimple except when β is symmetric and n = 2. Proof. For a subspace W of V and we denote by W ⊥ its orthogonal with respect to β. If n 1, then g = {0} which is semisimple. If n = 2 and β is symmetric, then by choosing a suitable basis of V , we see that dim g = 1. Thus g is not semisimple. If n = 2 and β is alternating, then by choosing a suitable basis of V , we see that g sl2 (k) which is semisimple by 20.2.2. Suppose that n 3. Let W be a g-submodule of V distinct from V and {0} and u, v ∈ V , w ∈ W be such that β(u, w) = 0, β(v, w) = 0. Define f ∈ End(V ) as follows: for x ∈ V ,
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f (x) = β(x, u)v − β(v, x)u. We verify easily that f ∈ g and since f (w) = −β(v, w)u, we have u ∈ f (W ) ⊂ W . We conclude that (kw)⊥ ⊂ W and therefore dim W n − 1. But V = W , so W = (kw)⊥ . This is true for all w ∈ W \ {0}. It follows that W = W ⊥ and 2(n − 1) = dim W + dim W ⊥ = n. But this is impossible for n 3. We have therefore proved that V is a simple g-module. In view of 20.2.1, we are left to show that z(g) = {0}. By Schur’s Lemma, z(g) ⊂ k idV and it is clear from the hypotheses on β and g that g∩k idV = {0}. So we are done.
20.3 Semisimplicity of representations 20.3.1 Lemma. If dim g = 1, then g has a 2-dimensional representation which is not semisimple. Proof. Let g = kx and (u, v) a basis of a 2-dimensional vector space. Define σ : g → gl(V ) by setting σ(λx)(u) = λv and σ(λx)(v) = 0. Then it is clear that (V, σ) is a representation of g in which kv is the unique submodule of dimension 1. In particular, V is not semisimple. 20.3.2 Lemma. Assume that g is semisimple. Let (V, σ) be a simple finitedimensional faithful representation of g and β the bilinear form on g associated to σ. (i) β is non-degenerate. (ii) Suppose that dim g = n = 0. Let (x1 , . . . , xn ) be a basis of g and (y1 , . . . , yn ) the basis of g dual to (x1 , . . . , xn ) with respect to β. Then: c = σ(x1 )σ(y1 ) + · · · + σ(xn )σ(yn ) =
n idV . dim V
Proof. Since σ is faithful, we may assume that σ(g) = g ⊂ gl(V ). (i) If a is the kernel of β, then a is a solvable ideal (19.4.8). So a = {0}. (ii) For z ∈ g and 1 i n, we have: [z, xi ] =
n j=1
λij xj , [z, yi ] =
n
µij yj .
j=1
So λij = β([z, xi ], yj ) = −β(xi , [z, yj ]) = −µji . It follows easily that [z, c] = 0. Since tr(c) = β(x1 , y1 ) + · · · + β(xn , yn ) = n, part (ii) follows from Schur’s Lemma. 20.3.3 Theorem. (Whitehead’s First Lemma) Let g be a semisimple Lie algebra, (V, σ) a finite-dimensional representation of g and f : g → V a linear map. The following conditions are equivalent: (i) f ([x, y]) = σ(x)f (y) − σ(y)f (x) for all x, y ∈ g. (ii) There exists v ∈ V such that f (x) = σ(x)(v) for all x ∈ g.
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303
Proof. The implication (ii) ⇒ (i) is straightforward. Let us prove (i) ⇒ (ii). Suppose that (i) is verified and we may assume that g = {0}. First, suppose that σ is simple. Let ker σ = a. By 20.1.7 and 20.1.8, g a × b where a, b are ideals which are semisimple. So a = [a, a] and by our hypothesis on f , f (a) = {0}. Let (x1 , . . . , xn ) be a basis of b and (y1 , . . . , yn ) the basis of b dual to (x1 , . . . , xn ) with respect to the bilinear form on b associated to the restriction n idV . Set of σ to b. By 20.3.2, c = σ(x1 )σ(y1 ) + · · · + σ(x1 )σ(y1 ) = dim V v = c−1 (σ(x1 )f (y1 ) + · · · + σ(xn )f (yn )). Let z ∈ g. We have by the proof of 20.3.2 that for 1 i n, [z, xi ] =
n
λij xj , [z, yi ] =
j=1
n
−λji yj .
j=1
Since f satisfies (i), we obtain that c ◦ f (z) = = =
n i=1 n i=1 n
σ(xi )σ(yi )f (z) = σ(xi )f ([yi , z]) +
n
σ(xi )f ([yi , z]) +
n
i=1 n
i=1 n
i=1
i=1
σ([xi , z])f (yi ) +
σ(xi )σ(z)f (yi )
σ(z)σ(xi )f (yi )
σ(z)σ(xi )f (yi ) = c ◦ σ(z)(v).
i=1
So σ(z)(v) = f (z) for all z ∈ g as required. In the general case, we proceed by induction on dim V . From the previous paragraph, we may assume that V contains a submodule W distinct from V and {0}. Let π : V → V /W be the canonical surjection. Then by the induction hypothesis, there exists v ∈ V such that (π ◦ f )(x) = π ◦ σ(x)(v) for all x ∈ g. Let θ(x) = f (x) − σ(x)(v). Then θ(x) ∈ W and θ ∈ Hom(g, W ) satisfies (i). By the induction hypothesis, there exists w ∈ W such that θ(x) = σ(x)(w) for all x ∈ g. So f (x) = σ(x)(v + w) for all x ∈ g. 20.3.4 Theorem. (Weyl’s Theorem) Let g be a Lie algebra. The following conditions are equivalent: (i) All finite-dimensional representations of g is semisimple. (ii) g is semisimple. Proof. We may assume that g = {0}. (i) ⇒ (ii) If (i) is verified, then the adjoint representation of g is semisimple. If a is an abelian ideal of g, then g = a⊕b where b is an ideal of g. So a ⊂ z(g). If z(g) = {0}, then g has an ideal of codimension 1. But 20.3.1 would imply that g has a non-semisimple representation of dimension 2. Contradiction. (ii) ⇒ (i) Let (V, σ) be a finite-dimensional representation of g, U a nontrivial submodule of V , π : V → V /U the canonical surjection and (V /U, τ )
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the representation of g induced by σ. Denote by M the set of linear maps from V /U to V and set N = {φ ∈ M ; φ(V /U ) ⊂ U }. The map λ : g → End(M ), defined by setting for x ∈ g and φ ∈ M : λ(x)(φ) = σ(x) ◦ φ − φ ◦ τ (x). then defines a representation of g in which N is a submodule. Let us fix φ0 ∈ M such that π ◦ φ0 = idV /U . The linear map f : g → M , x → λ(x)(φ0 ) verifies f (g) ⊂ N since for any w ∈ V /U , π (f (x)(w)) = π◦σ(x)◦φ0 (w) − π◦φ0 ◦τ (x)(w) = τ (x)◦π◦φ0 (w) − τ (x)(w) = τ (x)(w) − τ (x)(w) = 0. So by 20.3.3, condition (i) of 20.3.3 is valid for the representation (N, µ) induced by λ. Thus there exists ψ0 ∈ N such that f (x) = µ(x)(ψ0 ) for all x ∈ g. Set θ0 = φ0 − ψ0 , then λ(x)(θ0 ) = 0 for all x ∈ g. Hence π ◦ θ0 = idV /U and θ0 ◦ τ (x) = σ(x) ◦ θ0 for all x ∈ g. It follows that V is the direct sum of the submodules U and θ0 (V /U ). So σ is semisimple. 20.3.5 Theorem. (Levi-Malcev Theorem) Let g be a Lie algebra and r = rad g. There exists a Lie subalgebra s of g such that g = s ⊕ r. Moreover, s is semisimple and it is called a Levi subalgebra of g. Proof. Observe that such a Lie algebra is isomorphic to g/r, so it is semisimple by 20.1.3. We shall prove by induction on dim r that such a Lie subalgebra exists. The case r = {0} is obvious, so we may assume that r = {0}. 1) Suppose that there is an ideal a of g contained in r such that a = {0} and a = r. Then rad g/a = r/a and by our induction hypothesis, there is a Lie subalgebra b of g such that b ∩ r = a , g = b + r. Since b/a g/r is semisimple, rad b ⊂ a and so rad b = a. By induction, there is a Lie subalgebra s of g such that b = s ⊕ a. It follows from the definition of b that g = s ⊕ r. 2) We are left with the case where r is commutative. In this case, z(g) = {0} or r since [g, r] is an ideal of g contained in r. When z(g) = r, the adjoint representation of g is semisimple since ad g g/r is semisimple. Thus there is an ideal s of g such that g = s ⊕ r. When z(g) = {0}, g is isomorphic to the Lie subalgebra ad g in End(g). Let (End(g), σ) be the representation of g defined by σ(x)(u) = [ad x, u] for x ∈ g and u ∈ End(g), and M = {u ∈ End(g); u(g) ⊂ r and u|r = λu idr for some λu ∈ k}. Denote by N the kernel of the linear form u → λu on M . Since r is abelian, R = adg r ⊂ N . Also, we have σ(g)(M ) ⊂ N , and σ(g)(R) ⊂ R. A straightforward computation shows that for x ∈ r and u ∈ M , we have σ(x)(u) = −λu ad x. So σ(r)(M ) ⊂ R. It follows that the representation
20.4 Semisimple and nilpotent elements
305
(M/R, θ) of g induced by σ is semisimple since θ(r) = {0} (20.3.4). Moreover, θ(g)(M/R) ⊂ N/R, so there exists v ∈ M such that λv = −1 and for all x ∈ g, σ(x)(v) ∈ R. Since z(g) = {0}, we have a well-defined linear map α : g → r, given by ad α(x) = σ(x)(v). We have α|r = idr because σ(x)(u) = −λu ad x for x ∈ r and u ∈ M . It follows that g = s ⊕ r where s = ker α = {x ∈ g; [ad x, v] = 0} is clearly a Lie subalgebra of g. 20.3.6 Corollary. Let r = rad g and n the nilpotent radical of g. Then n = r ∩ D(g) = [g, r]. Proof. Let s be a Levi subalgebra of g. Then we have D(g) = [s, s] + [s, r] + [r, r] = s + [s, r] + [r, r] = s ⊕ [g, r]. So n = r ∩ D(g) = [g, r] (19.6.2).
20.4 Semisimple and nilpotent elements 20.4.1 Lemma. Let V be a finite-dimensional vector space and x an endomorphism of V . (i) If x is nilpotent (resp. semisimple), then so is ad x. (ii) Let x = xs + xn be the Jordan decomposition of x. Then ad xs + ad xn is the Jordan decomposition of ad x. Proof. Part (ii) is clear from (i). In view of 19.3.5, we only need to prove (i) when x is semisimple. Let (v1 , . . . , vn ) be a basis of eigenvectors of V for x, then it is easy to check that the endomorphisms zij ∈ End(V ) defined by zij (vi ) = vj , zij (vk ) = 0 if k = i, form a basis of eigenvectors of End(V ) for ad x. 20.4.2 Proposition. Let V be a finite-dimensional vector space and g a semisimple Lie subalgebra of gl(V ). Then g contains the semisimple and nilpotent components of its elements. Proof. Let V be the set of g-submodules of V , and for W ∈ V, we define gW = {x ∈ gl(V ); x(W ) ⊂ W, tr(x|W ) = 0}. Since g = [g, g], g ⊂ gW . Let t = ngl(V ) (g) ∩ gW . W ∈V
Let x ∈ t, X = adgl(V ) x, S = adgl(V ) xs and N = adgl(V ) xn . Then the Jordan decomposition of X is S +N (20.4.1). So xs , xn ∈ t because S, N (resp. xs , xn ) are polynomials in X (resp. x) with no constant term. We verify easily that t is a Lie subalgebra of gl(V ) and g is an ideal of t. So t = g × a for some ideal a of t (20.1.4). Moreover [a, g] ⊂ a ∩ g = {0}. Take W to be a minimal element of V \ {0}. Then W is simple. For a ∈ a, tr(a|W ) = 0, so Schur’s Lemma says that a|W = 0. Since V is the sum of minimal elements of V \ {0}, we deduce that a = {0}. Hence t = g and xs , xn ∈ g.
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20.4.3 Corollary. With the hypotheses of 20.4.2, an element x ∈ g is semisimple (resp. nilpotent) if and only if adg x is semisimple (resp. nilpotent). Proof. Since (g, ad) is a g-submodule of (g, adgl(V ) ), 20.4.1 says that if x is semisimple (resp. nilpotent), then so is adg x. Conversely, if adg x is semisimple (resp. nilpotent), then since the adjoint representation is faithful, it follows by 20.4.2 that x = xs (resp. x = xn ). 20.4.4 Theorem. Let g be semisimple and x ∈ g. Then the following conditions are equivalent: (i) adg x is semisimple (resp. nilpotent). (ii) There is a finite-dimensional faithful representation (V, σ) of g such that σ(x) is semisimple (resp. nilpotent). (iii) For all finite-dimensional representation (V, σ) of g, σ(x) is semisimple (resp. nilpotent). Proof. We have (iii) ⇒ (ii) since the adjoint representation is faithful. Also (ii) ⇒ (i) follows from 20.4.3. Let us prove (i) ⇒ (iii). Let (V, σ) be a finite-dimensional representation of g, h = g/ ker σ and (V, π) the representation of h obtained from σ. Let y be the image of x in h. If adg x is semisimple (resp. nilpotent), then so is adh y. Thus π(y) = σ(x) is semisimple (resp. nilpotent) by 20.4.3. 20.4.5 Let g be semisimple. An element x ∈ g is called semisimple (resp. nilpotent) if it satisfies the conditions of 20.4.4. In view of 20.4.2, an element x ∈ g can be written uniquely in the form s + n where s is semisimple, n is nilpotent and [s, n] = 0. We say that s (resp. n) is the semisimple component (resp. nilpotent component) of x. By 20.4.2, we have the following result: Corollary. Let h be a semisimple Lie subalgebra of a semisimple Lie algebra g, x ∈ h and s, n the semisimple and nilpotent components of x in g. Then s, n ∈ h. 20.4.6 Proposition. Let g be semisimple. Then g is generated, as a Lie algebra, by its nilpotent elements. Proof. Let h be a Cartan subalgebra of g and R the root system of g relative to h. By 19.8.6 and 19.8.7, h is abelian and [h, gα ] ⊂ gα for all α ∈ R. Since g = h ⊕ ( α∈R gα ), we have: g = [g, g] ⊂
α∈R
gα +
[gα , gβ ].
α,β∈R
But all the elements of gα are nilpotent (19.8.2). So the result follows.
20.5 Reductive Lie algebras
307
20.5 Reductive Lie algebras 20.5.1 Definition. (i) A Lie algebra g is said to be reductive if its adjoint representation is semisimple. (ii) A Lie subalgebra h of g is reductive in g if the adjoint representation of h in g is semisimple. 20.5.2 Remark. If h is reductive in g, then h is reductive since h is a h-submodule of g. 20.5.3 Lemma. An abelian Lie algebra h has a finite-dimensional representation such that the associated bilinear form is non-degenerate. Proof. Let n = dim h, then the representation on kn defined by identifying h with the set of diagonal matrices works. 20.5.4 Proposition. Let r = rad g and n the nilpotent radical of g. The following conditions are equivalent: (i) g is reductive. (ii) g = s × a where s is semisimple and a is abelian. (iii) g has a finite-dimensional representation such that the associated bilinear form is non-degenerate. (iv) g has a faithful semisimple representation of finite dimension. (v) n = {0}. (vi) r is the centre of g. Proof. (i) ⇒ (ii) If g is reductive, then g is the direct sum of its minimal (non-trivial) ideals a1 , . . . , an . It follows that ai is simple or dim ai = 1 for all i. So we have (ii). (ii) ⇒ (iii) By 20.5.3, a has a finite-dimensional representation (V, σ) such that the associated bilinear form on a is non-degenerate. Also the bilinear form on s associated to the adjoint representation (s, ads ) of s is non-degenerate. So the bilinear form on g associated to the representation (s ⊕ V, ads ⊕σ) of g is non-degenerate. (iii) ⇒ (iv) In the notations of 19.5.5 (ii), take the direct sum of the σi ’s. (iv) ⇒ (v) This is clear by the definition of n and 19.5.1. (v) ⇒ (vi) Since z(g) ⊂ r and n = [g, r] (20.3.6), the result is clear. (vi) ⇒ (i) If r = z(g), then ad g is a semisimple Lie subalgebra of gl(g) (20.3.5). So the adjoint representation is semisimple (20.3.4). 20.5.5 Corollary. A Lie algebra g is reductive if and only if D(g) is semisimple. In particular, if g is reductive, then g D(g) × z(g). Proof. If D(g) is semisimple, then g = D(g) × a for some ideal a of g (20.1.4). Since [a, g] ⊂ a ∩ D(g) = {0}, a = z(g). So the result follows from 20.5.4.
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Remark. By 20.2.2, gl(V ) is a reductive Lie algebra. 20.5.6 Let (V, σ) and (W, π) be representations of g. We define the representation (V ⊗k W, σ ⊗ π) of g by setting for x ∈ g: (σ ⊗ π)(x) = σ(x) ⊗ idW + idV ⊗π(x). Corollary. If σ and π are semisimple of finite dimension, then σ ⊗ π is semisimple. Proof. We may assume that σ and π are simple and since the nilpotent radical n of g is contained in ker σ ∩ ker π (19.5.1), replacing g by g/n, we may assume further that n = {0}, so g is reductive (20.5.4). Let g = s×a be as in 20.5.4 (ii). By Schur’s Lemma, σ(a) ⊂ k idV and π(a) ⊂ k idW . So σ⊗π(a) ⊂ k idV ⊗ idW . Since σ ⊗ π|s is semisimple (20.3.4), the result follows. 20.5.7 Proposition. Assume that g is reductive. Let x ∈ D(g), y ∈ z(g) and z = x + y. The following conditions are equivalent: (i) x is semisimple in D(g). (ii) There exists a faithful semisimple finite-dimensional representation (V, σ) of g such that σ(z) is semisimple. (iii) For all semisimple finite-dimensional representation (V, σ) of g, σ(z) is semisimple. Proof. By Schur’s Lemma, σ(y) is a scalar multiple of idV for any simple finite-dimensional representation (V, σ) of g. In particular, σ(y) is semisimple. Since y ∈ z(g), we may assume that y = 0. Thus we have (i) ⇒ (iii) by 20.4.4, (iii) ⇒ (ii) by 20.5.4 (iv), and finally (ii) ⇒ (i) by 20.4.4. 20.5.8 Proposition. With the notations of 20.5.7, the following conditions are equivalent: (i) x is nilpotent in D(g) and y = 0. (ii) There exists a faithful semisimple finite-dimensional representation (V, σ) of g such that σ(z) is nilpotent. (iii) For all semisimple finite-dimensional representation (V, σ) of g, σ(z) is nilpotent. Proof. We have (i) ⇒ (iii) by 20.4.4 and (iii) ⇒ (ii) by 20.5.4 (iv). So let us prove (ii) ⇒ (i). Let (V, σ) be as in (ii). Let s, n be the semisimple and nilpotent components of x in D(g). By 20.4.4 and 20.5.7, σ(n) is nilpotent and σ(s + y) is semisimple. Furthermore, they commute and their sum σ(z) is nilpotent. So σ(s + y) = 0. Since σ is faithful and D(g) ∩ z(g) = {0}, we deduce that y = s = 0, and hence x = n is nilpotent.
20.5 Reductive Lie algebras
309
20.5.9 Assume that g is reductive. An element z ∈ g is called semisimple (resp. nilpotent) if it verifies the conditions of 20.5.7 (resp. 20.5.8). We define as in 20.4.5 the semisimple and nilpotent components of z. We shall denote by Sg (resp. Ng ) the set of semisimple (resp. nilpotent) elements of g. 20.5.10 Theorem. Let r = rad g, (V, σ) a finite-dimensional representation of g, g = σ(g) and r = σ(r). The following conditions are equivalent: (i) σ is semisimple. (ii) g is reductive and the elements of its centre are semisimple. (iii) The elements of r are semisimple. (iv) The restriction of σ to r is semisimple. Proof. (i) ⇒ (ii) This is clear from 20.5.4 (iv) and 20.5.7. (ii) ⇒ (iii) This is obvious. (iii) ⇒ (iv) If (iii) is verified, then by 20.3.6, [g , r ] = {0}. So r is abelian, and since its elements are semisimple, (iv) follows. (iv) ⇒ (i) Using the notations of 19.3.9, we have V = λ∈r∗ Vλ (r) by 19.4.5 (ii). Since [g , r ] = {0} (20.3.6), each Vλ (r) is a g-module. So (i) follows from 20.3.4. 20.5.11 Lemma. We have V = V g ⊕ σ(g)(V ) for all semisimple representation (V, σ) of g. Proof. Clearly, we may assume that V is simple. Since σ(g)(V ) is a submodule of V , the result follows. 20.5.12 Lemma. Let t be a Lie subalgebra of a semisimple Lie algebra g verifying: a) L|t×t is non-degenerate. b) If x ∈ t, then its semisimple and nilpotent components are in t. Then t is reductive in g. Proof. Condition a) and 20.5.4 (iii) imply that t is reductive. Let x ∈ z(t) and s, n the semisimple and nilpotent components of x. Then condition b) and 10.1.2 imply that s, n ∈ z(t). It follows that for all y ∈ t, (ad n) ◦ (ad y) is nilpotent (since [y, n] = 0) and hence L(n, y) = 0. We deduce from condition a) that n = 0 and so all the elements of z(t) are semisimple. The result now follows from 20.5.10 (ii). 20.5.13 Proposition. Assume that g is semisimple. Let t be a Lie subalgebra of g which is reductive in g, and c = cg (t). Then: (i) c verifies the conditions of 20.5.12, and so it is reductive in g. (ii) g = c ⊕ [t, g] and [t, g] is the orthogonal of c with respect to L. Proof. By 10.1.2 and 20.4.5, c contains the semisimple and nilpotent components of its elements. For (x, y, z) ∈ t × c × g, we have:
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20 Semisimple and reductive Lie algebras
L([x, z], y) = L(z, [y, x]) = L(z, 0) = 0. Thus c and [t, g] are orthogonal. So by 20.5.11, we have (ii). Finally since L is non-degenerate, it is non-degenerate on c × c. So (i) follows. 20.5.14 Proposition. Assume that g is semisimple. Let (s, n) ∈ Sg × Ng be such that [s, n] = 0. Then n ∈ [gs , gs ]. Proof. By 20.5.13, gs is reductive in g. So the result follows by applying 20.5.8 to the adjoint representation of gs in g. 20.5.15 Definition. Let g be semisimple. A Lie subalgebra of g is called a torus if all its elements are semisimple. 20.5.16 Lemma. Assume that g is semisimple. Let t be a torus of g. (i) t is abelian and reductive in g. (ii) If t is a maximal torus, then cg (t) = t. Proof. (i) Since rad t consists of semisimple elements, t is reductive (20.5.10) and so D(t) is semisimple. Thus D(t) = {0} (20.4.6), and t is abelian. (ii) By 20.5.13, c = cg (t) is reductive in g, so c is the direct product of its centre z and D(c). Since t is maximal, z = t and D(c) contains no non-zero semisimple elements. Hence D(c) = {0}.
20.6 Results on the structure of semisimple Lie algebras 20.6.1 In this section, g is a semisimple Lie algebra and h is a Cartan subalgebra of g. We set R = R(g, h) and gλ = gλ (h). If x, y ∈ g, we shall write x, y for L(x, y). Let λ ∈ h∗ . By 19.8.7 (iii), there exists a unique hλ ∈ h such that λ(h) = hλ , h for all h ∈ h. The map h∗ → h, λ → hλ , is a linear isomorphism. For λ, µ ∈ h∗ , define λ, µ = hλ , hµ . Thus we obtain a non-degenerate symmetric bilinear form on h∗ . Note that we have: λ, µ = hλ , hµ = λ(hµ ) = µ(hλ ). It follows also from 19.8.7 that hα , α ∈ R, span h∗ . 20.6.2 Proposition. Let t be a Lie subalgebra of g. Then t is a Cartan subalgebra of g if and only if t is maximal torus of g. Proof. Suppose that t is a Cartan subalgebra. Let h ∈ t and h = s + n its Jordan decomposition. By 10.1.2, s, n ∈ ng (t) = t. So 19.8.7 (iii) and 20.5.12 imply that t is reductive in g. Hence t is abelian. By 20.5.10 (ii), t is a torus of g and it is maximal since ng (t) = t. Conversely, suppose that t is a maximal torus in g. By 20.5.16 (i), t is abelian and there is a subspace m of g such that g = t ⊕ m and [t, m] ⊂ m. It follows that ng (t) = cg (t) = t by 20.5.16 (ii).
20.6 Results on the structure of semisimple Lie algebras
311
20.6.3 Corollary. (i) Any generic element of g is semisimple, and any semisimple element is contained in a Cartan subalgebra of g. So Sg is the union of Cartan subalgebras of g. (ii) If h ∈ h, λ ∈ h∗ and x ∈ gλ , then [h, x] = λ(h)x. Proof. It is immediate from 19.8.5 and 20.6.2. 20.6.4 Proposition. Let α ∈ R. (i) If x ∈ gα , y ∈ g−α and h ∈ h, then: h, [x, y] = α(h)x, y = h, hα x, y , [x, y] = x, yhα . (ii) The vector space [gα , g−α ] is khα . (iii) We have α(hα ) = 0. Proof. (i) By 20.6.3, [h, x] = α(h)x, so h, [x, y] = [h, x], y = α(h)x, y = h, hα x, y = h, x, yhα . Since [x, y] ∈ g0 = h and L|h×h is non-degenerate, we have [x, y] = x, yhα . (ii) This follows from (i) since gα , g−α = {0} (19.8.7). (iii) Let (x, y) ∈ gα × g−α be such that x, y = 1, so [x, y] = hα . If α(hα ) = 0, then a = khα + kx + ky is a nilpotent Lie algebra. Applying 19.4.4 to the adjoint representation of a in g, we see that if z ∈ [a, a], then adg z is nilpotent. Thus, hα is nilpotent. By 20.6.2, we obtain that hα = 0. Contradiction. 20.6.5 In view of 20.6.4 (iii), for α ∈ R, we may define: Hα = Then: α(Hα ) = 2 , Hα , Hα =
2hα . hα , hα
4 , Hα = Hα , Hα hα . hα , hα 2
Let Xα ∈ gα \ {0}. There exists X−α ∈ g−α verifying Xα , X−α = 0. By 20.6.4, [Xα , X−α ] = Xα , X−α hα . Choose X−α so that [Xα , X−α ] = Hα . Then we obtain: [Hα , Xα ] = 2Xα , [Hα , X−α ] = −2X−α , [Xα , X−α ] = Hα . So the Lie algebra sα = kHα + kXα + kX−α is isomorphic to sl2 (k). 20.6.6 Proposition. (i) For α, β ∈ R, we have dim gα = 1 and β(Hα ) is an integer. (ii) If x, y ∈ h, then x, y = α∈R α(x)α(y).
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Proof. (i) Suppose that dim gα > 1. If y ∈ g−α \ {0}, there exists Xα ∈ gα \ {0} such that Xα , y = 0. Let X−α and sα be as in 20.6.5. By 20.6.4, [Xα , y] = Xα , yhα = 0. So it follows from 19.2.5 and 19.2.7 that y is a linear combination of eigenvectors of ad Hα whose eigenvalues are in N. But [Hα , y] = −α(Hα )y = −2y. Contradiction. Hence dim gα = 1. By 20.6.3, [Hα , x] = β(Hα )x for α, β ∈ R, x ∈ gβ . So β(Hα ) ∈ Z (19.2.7 and 20.6.5). (ii) This is clear from part (i) and 19.8.6. 20.6.7 For α, β ∈ R, set aβα = β(Hα ). The integers aβα are called the Cartan integers of (g, h). We have aαα = 2 and: aβα = hβ , Hα = 2
β, α Hβ , Hα hβ , hα =2 =2 . hα , hα α, α Hα , Hα
By 20.6.6, Hα , Hβ ∈ Z. It follows that the ratio between hα and Hα is rational, and that hα , hβ ∈ Q. 20.6.8 Theorem. Let α, β ∈ R. (i) There exist p, q ∈ N such that [−p, q] is the set of integers t such that β + tα ∈ R ∪ {0}. Moreover, we have aβα = p − q. (ii) We have β − aβα α ∈ R. (iii) If β − α ∈ R ∪ {0}, then aβα 0, p = 0 and q = −aβα . (iv) The only elements of R proportional to α are α and −α. (v) If α + β ∈ R, then [gα , gβ ] = gα+β . Proof. (i) Set u = t∈Z gβ+tα , s = gα + g−α + kHα sl2 (k). Since [s, u] ⊂ u, u is a s-module. The set of eigenvalues for Hα in u is {(β +tα)(Hα ) = aβα +2t, t ∈ Z and β + tα ∈ R ∪ {0}}. But by 19.2.5, 19.2.6 and 19.2.7, it is also of the form {−r, −r + 2, . . . , r − 2, r} for some r ∈ N. Hence the set of t ∈ Z such that β + tα ∈ R ∪ {0} is [−p, q] for some p, q ∈ Z. Since β ∈ R, 0 ∈ [−p, q], so p, q ∈ N. Finally, p − q = aβα since r = 2q + aβα and −r = −2p + aβα . (ii) By (i), −p q − p = −aβα q, so β − aβα α ∈ R ∪ {0}. If β = aβα α, then by 20.6.7, we have: aβα = 2
α, α β, α = 2aβα = 2aβα . α, α α, α
So aβα = 0, which implies that β = 0. Contradiction. (iii) This is clear from (i). (iv) Let λ ∈ k be such that γ = λα ∈ R. Then γ(Hα ) = 2λ ∈ Z. Exchanging the roles of γ and α, we deduce that 2λ−1 ∈ Z. Thus λ belongs to {−2, −1, −1/2, 1/2, 1, 2}. Since R = −R, we need only to prove that if α ∈ R, then 2α ∈ R. Suppose that 2α ∈ R and y ∈ g2α \ {0}. Then from the preceding discussion, 3α ∈ R and so [Xα , y] = 0. Since [X−α , y] ∈ gα = kXα , we have
20.7 Subalgebras of semisimple Lie algebras
313
[Xα , [X−α , y]] = 0 and 4y = [Hα , y] = [[Xα , X−α ], y] = 0. Thus y = 0. Contradiction. (v) If α + β ∈ R, then q 1. So [Xα , gβ ] = 0 (19.2.5, 19.2.6 and 19.2.7). Since dim gα+β = 1 (20.6.6), we are done. 20.6.9 For α ∈ R, let sα ∈ End(h∗ ) be defined by: sα (λ) = λ − λ(Hα )α = λ − 2
λ, α α. α, α
Thus: s2α = idh∗ , sα (λ), sα (µ) = λ, µ. If β ∈ R, then sα (β) = β − aβα α ∈ R. It follows that sα (R) = R. The subgroup W (g, h) of GL(h∗ ) generated by the sα ’s, α ∈ R, is called the Weyl group relative to (g, h). Since w(R) = R for all w ∈ W (g, h) and R spans h∗ (19.8.7), W (g, h) is a finite group. 20.6.10 From the discussion in the preceding paragraphs, we deduce that R is a reduced root system in h∗ (18.2.1). Moreover, the set of Hα ’s form a dual root system of R in h. Let QHα , h∗Q = Qα , hR = R ⊗Q hQ , h∗R = R ⊗Q h∗Q hQ = α∈R
α∈R
Then we may identify h with k⊗Q hQ , and h∗ with k⊗Q h∗Q . The Weyl chambers are defined in hR and h∗R . The group W (g, h) acts on h∗ and h∗Q . It acts also on h et hQ (18.2.9), so on hR and h∗R . The form ., . takes on rational values on hQ and h∗Q , and defines a nondegenerate positive-definite form on them. It can be extended to an inner product on hR and h∗R . Finally these forms are all W (g, h)-invariant.
20.7 Subalgebras of semisimple Lie algebras 20.7.1 In this section, g is a semisimple Lie algebra, h is a Cartan subalgebra of g. We denote R(g, h) by R and W (g, h) by W . For α ∈ R, we denote hα = kHα = [gα , g−α ]. For a subset P of R, we set: α gP = g , hP = hα . α∈P
α∈P
If P, Q ⊂ R, then it is easy to check that: (1)
[h, gP ] = gP , [gP , gQ ] = hP ∩(−Q) + g(P +Q)∩R . We deduce immediately the following result from these equalities.
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Lemma. Let h be a subspace of h and P ⊂ R. The following conditions are equivalent: (i) h + gP is a Lie algebra of g. (ii) P is closed subset of R and hP ∩(−P ) ⊂ h . 20.7.2 Proposition. The following conditions are equivalent for a Lie subalgebra a of g: (i) [h, a] ⊂ a. (ii) There exist a closed subset P ⊂ R and a subspace h of h such that hP ∩(−P ) ⊂ h and a = h + gP . Proof. The implication (ii) ⇒ (i) is straightforward. Suppose that (i) is verified. Let P = {α ∈ R; gα ⊂ a} and h = a ∩ h. By 19.3.9, we have a = h + α∈P gα . So (ii) follows from 20.7.1. 20.7.3 Lemma. Let a be a Lie subalgebra of g such that [h, a] ⊂ a, h the subspace of h and P ⊂ R such that a = h + gP . Let h be a subspace of h and Q ⊂ R verifying h ⊂ h and Q ⊂ P . Then the following conditions are equivalent: (i) h + gQ is an ideal of a. (ii) We have (P + Q) ∩ R ⊂ Q and hP ∩(−Q) ⊂ h ⊂ α∈P \Q ker α. Proof. We have [h + gP , h + gQ ] = hP ∩(−Q) + [h , gQ ] + [h , gP ] + g(P +Q)∩R . Thus h + gQ is an ideal of a if and only if hP ∩(−Q) ⊂ h , [h , gP ] ⊂ gQ , g(P +Q)∩R ⊂ gQ . So the result is clear. 20.7.4 Proposition. Let a be a Lie subalgebra of g such that [h, a] ⊂ a, h the subspace of h and P ⊂ R such that a = h + gP . Set: Q = {α ∈ P ; −α ∈ / P } , t = {x ∈ h ; α(x) = 0, α ∈ P ∩ (−P )}. (i) The radical of a is t + gQ . (ii) gQ is a nilpotent ideal of a. Proof. Let α ∈ P , β ∈ Q be such that α + β ∈ R. Since P is closed, α + β ∈ P . However, α + β ∈ −P , for otherwise −β = −(α + β) + α ∈ P , this contradicts the fact that β ∈ Q. So α + β ∈ Q, and (P + Q) ∩ R ⊂ Q. Since P ∩ (−Q) = ∅ (so Q ∩ (−Q) = ∅), it follows from 18.10.2 and 20.7.3 that gQ is a nilpotent ideal of a. We have, from the definition of t, that [t, gP ] ⊂ gQ . So t + gQ is an ideal of a and it is solvable since gQ is nilpotent. Let r = rad a. Then by 19.6.3, [h, r] ⊂ r, and hence r = h + gS where h ⊂ h and S ⊂ P . To prove (i), it suffices to prove that S ⊂ Q and h ⊂ t. If α ∈ S is such that −α ∈ P , then hα = [gα , g−α ] ⊂ r. This implies in turn that g−α = [hα , g−α ] ⊂ r. So r contains a Lie subalgebra isomorphic to sl2 (k). This contradicts the fact that r is solvable. Hence S ⊂ Q. Next, if x ∈ h and α ∈ P ∩ (−P ), then [x, gα ] ⊂ gα ∩ r = {0}. So α(x) = 0 and h ⊂ t.
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315
20.7.5 Proposition. Let a be a Lie subalgebra of g such that [h, a] ⊂ a, h the subspace of h and P ⊂ R such that a = h + gP . (i) All the elements of a are nilpotent if and only if h = {0}. If this is the case, then P ∩ (−P ) = ∅ and a is nilpotent. (ii) a is solvable if and only if P ∩ (−P ) = ∅. If this is the case, then [a, a] = gS where S = ((P + P ) ∩ R) ∪ {α ∈ P ; α(h ) = {0}}. (iii) a is reductive in g if and only if P = −P . (iv) a is semisimple if and only if P = −P and h = hP . If this is the case, hP is a Cartan subalgebra of a and the root system of (a, hP ) is {α|hP ; α ∈ P }. Proof. (i) If all the elements of a are nilpotent, then a is nilpotent (19.3.7) and h = {0} (20.6.2). Conversely, if h = {0}, then P ∩ (−P ) = ∅ (20.7.1). It follows from 18.10.2 that all the elements of gP are nilpotent. (ii) If P ∩(−P ) = ∅, then gP is a Lie subalgebra (20.7.1) which is nilpotent by (i). Since [a, a] = [h , gP ] + g(P +P )∩R ⊂ gP , we see that a is solvable and we have the description [a, a] = gS as required. Conversely, if α ∈ P ∩ (−P ), then a contains the Lie subalgebra hα + gα + g−α which is isomorphic to sl2 (k). So a is not solvable. (iii) In the notations of 20.7.4, since rad a = t + gQ , we have from 20.7.4 (i) that a is reductive in g ⇔ adg x is semisimple for all x ∈ t + gQ ⇔ Q = ∅ ⇔ P = −P. (iv) If a is semisimple, P = −P by (iii). Since a = [a, a] = hP + gP , we have h = hP . Conversely, if P = −P and h = hP , then a is reductive by (iii) and its radical is {0} by 20.7.4. So a is semisimple. Finally, suppose that a is semisimple. Since hP is ad h-stable, so is n = na (hP ). It follows that n = hP + gS where S ⊂ P . For α ∈ S, we have gα = [hα , gα ] ⊂ [hP , gα ] ⊂ hP which is absurd. So S = ∅ and hP = n is a Cartan subalgebra of a. The remaining statements now follow. 20.7.6 Proposition. Let a be a Lie subalgebra of g containing h. (i) We have a = ng (a). (ii) a contains the semisimple and nilpotent components of its elements. Proof. (i) Write a = h + gP with P ⊂ R. Then ng (a) = h + gQ where Q ⊃ P . Since gQ = [h, gQ ] ⊂ a, we have Q = P . (ii) Let x ∈ a, and s, n its semisimple and nilpotent components. Then [s, a] ⊂ a and [n, a] ⊂ a (10.1.2). Hence s, n ∈ a by (i). 20.7.7 Proposition. Let P be a closed subset of R and a = h + gP . (i) a is solvable if and only if P ∩ (−P ) = ∅. If this is the case, then [a, a] = gP . (ii) a is reductive if and only if P = −P .
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Proof. Part (i) follows from 20.7.5 (ii) and if P = −P , then a is reductive by 20.7.5 (iii). Finally, if a is reductive, then [a, a] = h + gP , where h ⊂ h, is semisimple. So P = −P by 20.7.5 (iv).
20.8 Parabolic subalgebras 20.8.1 Let us conserve the notations of 20.7.1. Let B be a base of R and denote by R+ (resp. R− ) the corresponding set of positive (resp. negative) roots. Set: α α n+ = g , n− = g , b+ = h ⊕ n+ , b− = h ⊕ n− . α∈R+
α∈R−
By 20.7.5, n+ and n− (resp. b+ and b− ) are nilpotent (resp. solvable) Lie subalgebras of g. The decomposition: (2)
g = n− ⊕ h ⊕ n+
is called the triangular decomposition of g relative to B. The Lie subalgebra b+ is called the Borel subalgebra of g relative to h and B. In general, a Lie subalgebra b of g is called a Borel subalgebra of (g, h) if there exists a base B of R such that b is the Borel subalgebra of g relative to h and B . 20.8.2 Proposition. (i) The Lie algebra n+ (resp. n− ) is generated by gα , α ∈ B (resp. α ∈ −B). (ii) The Lie algebras b+ and b− are maximal solvable Lie subalgebras of g; moreover, ng (b± ) = b± and z(b± ) = {0}. Proof. (i) Let α ∈ R+ . By 18.7.7, there exist α1 , . . . , αn ∈ B such that α = α1 + · · · + αn and α1 + · · · + αi ∈ R for 1 i n. So the result follows from 20.6.8 (v). (ii) We have ng (b± ) = b± by 20.7.6. If t is a Lie subalgebra of g containing b+ , then it contains h. So by 20.7.2, t = h + gP for some closed subset P of R containing R+ . If P contains a negative root α, then t contains the Lie subalgebra gα + g−α + hα sl2 (k), so t is not solvable. Hence b+ is a maximal solvable Lie subalgebra. The same argument applies for b− . Finally, if X = H + α∈R+ λα Xα ∈ z(b+ ), then [h, X] = {0} implies that X = H. On the other hand [X, n+ ] = {0}, so X = 0. Again the same argument applies for b− . 20.8.3 Proposition. Let l = dim h. (i) For (h, x) ∈ h × n+ , the characteristic polynomial of adg (h + x) is: Xl X − α(h) . α∈R
(ii) The largest nilpotent ideal, the nilpotent radical and the set of nilpotent elements of b+ are all equal to n+ . Furthermore, we have:
20.8 Parabolic subalgebras
[n+ , n+ ] =
317
gα .
α∈R+ \B ∗ α Proof. (i) Let us fix a lexicographic order on hβR defining B. For y ∈ g , we have: [h + x, y] = α(h)y + z with z ∈ β>α g . So in a suitable basis, the matrix of ad(h + x) is triangular with diagonal entries 0 (l times) and α(h), α ∈ R. (ii) This is clear by (i), 19.6.2, 18.7.7 and 20.6.8 (v).
20.8.4 Proposition. Let b = h + gP be a Lie subalgebra of g containing h. Then the following conditions are equivalent: (i) b is a Borel subalgebra of (g, h). (ii) There exists a chamber C of R such that P = R+ (C). (iii) R is the disjoint union of −P and P . Proof. The subset P is closed by 20.7.1. So the equivalence (ii) ⇔ (iii) follows from 18.10.3. If b is a Borel subalgebra of (g, h), then P = R+ , and so (i) ⇒ (iii) ⇒ (ii). Finally if (ii) is verified, then it is clear from the definition of a Borel subalgebra that b is a Borel subalgebra of (g, h). Hence (ii) ⇒ (i). 20.8.5 Proposition. Let p = h + gP be a Lie subalgebra of g containing h. Then the following conditions are equivalent: (i) p contains a Borel subalgebra of (g, h). (ii) There exists a chamber C of R such that P ⊃ R+ (C). (iii) P is parabolic. If these conditions are verified, we say that p is a parabolic subalgebra of (g, h). Proof. We have (i) ⇔ (ii) by 20.8.4 and (ii) ⇔ (iii) by 18.10.5. 20.8.6 Theorem. Let p = h + gP be a parabolic subalgebra of (g, h), Q = P \ (−P ) and s = h + gP ∩(−P ) . Then (i) p = s ⊕ gQ . (ii) s is reductive in g. (iii) gQ is the largest nilpotent ideal and the nilpotent radical of p. It is also the orthogonal of p with respect to the Killing form of g. (iv) z(p) = {0}. (v) p = ng (p). (vi) rad p = t ⊕ gQ , where t = {x ∈ h; α(x) = 0 for all α ∈ P ∩ (−P )}. Proof. Part (i) is clear and (ii), (v), (vi) follow from 20.7.5 (iii), 20.7.6 and 20.7.4 respectively. (iii) By 20.7.4 (ii), gQ is a nilpotent ideal of p. If n is the largest nilpotent ideal of p, then gQ ⊂ n ⊂ h + gQ (20.7.4). If x ∈ h ∩ n, then adp x is nilpotent (19.5.8), so α(x) = 0 for all α ∈ P , and x = 0 since P is parabolic. Thus n = gQ . Since [h, gQ ] = gQ , the nilpotent radical of p contains gQ (20.3.6),
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and so it is equal to gQ . Finally, part (i) and 19.8.6 imply that gQ is the orthogonal of p with respect to the Killing form of g. (iv) Let z = x + y ∈ z(p) where x ∈ h, y ∈ α∈P gα . Since [z, h] = {0}, y = 0. Now [z, gα ] = {0} for all α ∈ P . But P is parabolic, so z = 0. 20.8.7 Let h ∈ h \ {0}, e, f ∈ g verifying the relations of 19.2.4. If p ∈ Z, denote by gp the ad h-eigenspace of eigenvalue p. By 19.2.5, 19.2.6 and 19.2.7, we have: g = p∈Z gp . Set: p= g p , s = g0 , n = gp . p0
p>0
Then h ⊂ s and p = h ⊕ gP where P = {α ∈ R; α(h) 0}. It is easy to check that P contains a base of R. So p is a parabolic subalgebra of (g, h) whose nilpotent radical is n and s = h ⊕ gP ∩(−P ) is reductive in g. 20.8.8 Let B be a base of R and b the Borel subalgebra of g relative to B. For J ⊂ B, denote by Q(J) the set of roots which can be written as a sum of elements of −J. Set: P (J) = R+ ∪ Q(J) , pJ = h ⊕ gP (J) . It follows from 18.10.5 that pJ is a parabolic subalgebra of g containing b, and any parabolic subalgebra of g containing b is obtained in this way. It is clear that if J, J ⊂ B, then pJ ⊂ pJ if and only if J ⊂ J , and pJ = pJ if J = J . Thus the parabolic subalgebras of g containing b, distinct from g, and maximal with these properties are of the form pK , where K = B \ {β} for some β ∈ B.
References • [14], [29], [38], [39], [42], [43], [80].
21 Algebraic groups
In this chapter, we present some general properties of algebraic groups. All the varieties considered in this chapter are algebraic varieties over k.
21.1 Generalities 21.1.1 Let G be a group. Denote by µG : G × G → G, (α, β) → αβ, the group multiplication and ιG : G → G, α → α−1 , the inverse. As in 10.2.1, we set, for U, V ⊂ G, U V = {αβ; α ∈ U, β ∈ V } and U −1 = {α−1 ; α ∈ U }. Definition. Let G be a group endowed with a structure of algebraic variety. If the group multiplication µG and the inverse ιG are morphisms of algebraic varieties, then we say that G is an algebraic group. Furthermore, if G is an affine algebraic variety, we say that G is an affine algebraic group. Remark. Note that the topology on G × G here is the Zariski topology, and not the usual product topology. So an algebraic group is not a topological group. 21.1.2 Let G, G be algebraic groups. If α ∈ G, then translations β → αβ, β → βα, and conjugation, β → αβα−1 , are isomorphisms of the algebraic variety G. A map G → G is a morphism of algebraic groups if it is a homomorphism of groups and a morphism of algebraic varieties. We define in an obvious way the notion of isomorphisms and automorphisms of algebraic groups. 21.1.3 Example. 1) The additive group Ga = k and the multiplicative group Gm = k \ {0} are algebraic groups. 2) For n ∈ N∗ , GLn (k) is an algebraic group (11.7.5 and 11.8.9). 3) Any closed subgroup of an algebraic group is an algebraic group. Thus SLn (k) and the subgroup of GLn (k) consisting of diagonal (resp. upper triangular, lower triangular) matrices are algebraic groups. 4) Let n ∈ N∗ . Denote by Jn = [aij ] ∈ GLn (k) the matrix where aij = 1 if i + j = n + 1 and aij = 0 otherwise. Set:
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21 Algebraic groups
# 0 Jn 0 Jn A= . −Jn 0 −Jn 0 ⎧ ⎛ ⎞ ⎞⎫ ⎛ 1 0 0 ⎬ 1 0 0 ⎨ O2n+1 (k) = A ∈ GL2n+1 (k) ; t A ⎝ 0 0 Jn ⎠ A = ⎝ 0 0 Jn ⎠ . ⎭ ⎩ 0 Jn 0 0 Jn 0 # 0 Jn 0 Jn O2n (k) = A ∈ GL2n (k) ; t A A= . Jn 0 Jn 0
Sp2n (k) =
A ∈ GL2n (k) ; t A
Since these subgroups are closed in the corresponding GLp (k), they are algebraic groups. 5) Let A be a k-algebra of finite dimension (not necessarily associative). Then the group of automorphisms of A, denoted by Aut A, is an algebraic group. 6) If G, G are algebraic groups, then their direct product G × G endowed with the Zariski product topology is an algebraic group. 21.1.4 Lemma. Let G be an algebraic group. Then there is a unique irreducible component of G which contains the identity element e. Proof. Let X1 , X2 be two irreducible components of G which contain e, and X = X1 X2 = µ(X1 × X2 ). Since the Xi ’s are irreducible, so is X1 × X2 , and hence X is irreducible. Thus X ⊂ X1 or X ⊂ X2 . On the other hand e ∈ Xi , i = 1, 2, it follows that Xi ⊂ X, i = 1, 2. Hence X = X1 = X2 and we are done. 21.1.5 We shall denote by G◦ the unique irreducible component of G containing e, and we shall call G◦ the identity component of G. 21.1.6 Theorem. Let G be an algebraic group. (i) The identity component G◦ is a closed normal subgroup of G of finite index. The cosets of G modulo G◦ are at the same time the irreducible components and the connected components of G. In particular, G is a pure variety. (ii) If H is a closed subgroup of finite index in G, then H contains G◦ . Proof. Since µG (G◦ × G◦ ) = G◦ G◦ and ιG (G◦ ) = (G◦ )−1 are irreducible subsets of G containing e, they are contained in G◦ (21.1.4). So G◦ is a subgroup of G. Similarly, for α ∈ G, αG◦ α−1 is an irreducible subset of G◦ containing e, so αG◦ α−1 ⊂ G◦ and G◦ is normal. Since translations are isomorphisms of the algebraic variety G (21.1.2), the coset αG◦ is the unique irreducible component of G containing α. So part (i) follows from the fact that G is the disjoint union of the cosets of G modulo G◦ . (ii) Let H be such a subgroup. Let H1 , . . . , Hn be the left cosets of G modulo H having non-empty intersection with G◦ . Then G◦ is the disjoint union of the closed subsets H1 ∩ G◦ , . . . , Hn ∩ G◦ . Since G0 is irreducible, we obtain that n = 1 and G◦ ⊂ H.
21.2 Subgroups and morphisms
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21.1.7 Remarks. 1) By 21.1.6, an algebraic group is irreducible if and only if it is connected. We shall use the term connected algebraic groups instead of irreducible algebraic group. 2) Let G be a connected algebraic group. If α, β ∈ G, we have β = (βα−1 )α. Since the map γ → (βα−1 )γ is an automorphism of the variety G, we see that the point β is smooth if and only if α is smooth. Since the set of smooth points is non-empty (16.5.5), we deduce that G is a smooth variety. We obtain in the same way that G is a normal variety (17.1.5). 21.1.8 In the rest of this chapter, G will denote an algebraic group.
21.2 Subgroups and morphisms 21.2.1 Lemma. Let U, V be dense open subsets of G. Then G = U V . Proof. Let α ∈ G. By 21.1.2, αV −1 is a dense open subset of G. So it follows that αV −1 ∩ U = ∅ and α ∈ U V . 21.2.2 Proposition. Let H be a subgroup of G. (i) H is a subgroup of G. (ii) If H contains a non-empty open subset of H, then H = H. In particular, if H is constructible, then H is closed. Proof. (i) Let α ∈ H. Since H = αH ⊂ αH and αH is closed by 21.1.2, H ⊂ αH. Hence α−1 H ⊂ H and HH ⊂ H. If β ∈ H, then Hβ ⊂ H. It follows −1 that Hβ = Hβ ⊂ H. Hence H H ⊂ H. Similarly, by 21.1.2, H = H −1 = H. So H is a subgroup of G. (ii) Let U be an open subset of H contained in H. If α ∈ H and β ∈ U , then α ∈ αβ −1 U ⊂ H and αβ −1 U is an open subset of H. Thus H is open in H. It follows that H is a constructible subset of G and therefore H contains a dense open subset of H (1.4.6). In view of 21.2.1, we have H = HH = H. 21.2.3 Corollary. If H, K are closed subgroups of G such that K normalizes H, then HK is a closed subgroup of G. Proof. By our hypothesis, HK is a subgroup of G. On the other hand, it is equal to µG (H × K). Since µ is a morphism of varieties, HK is constructible (15.4.3). So the result follows from 21.2.2 (ii). 21.2.4 Proposition. Let u : G → H be a morphism of algebraic groups. (i) ker u and im u are closed subgroups of G and H respectively. (ii) u(G◦ ) = u(G)◦ . (iii) dim G = dim(ker u) + dim(im u).
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Proof. (i) Since u is a homomorphism of groups, ker u and im u are subgroups of G and H respectively. Since u is morphism of varieties, ker u is closed and im u is constructible (15.4.3). So the result follows from 21.2.2. (ii) By (i), u(G◦ ) is a closed connected subgroup of im u. Since G◦ has finite index in G (21.1.6), u(G◦ ) has finite index in im u. Thus, again by 21.1.6, we obtain that u(G◦ ) = u(G)◦ . (iii) Since the fibres of the morphism G → u(G) induced by u are the cosets of G modulo ker u, they all have dimension dim(ker u). So (iii) follows from 15.5.5 (ii). 21.2.5 Let A ∈ Mn (k) be a nilpotent matrix and p ∈ N be such that Ap+1 = 0. For t ∈ k, we set: ϕA (t) = In +
t2 t tp A + A2 + · · · + Ap . 1! 2! p!
Then this defines a morphism of algebraic groups ϕA : Ga → GLn (k). It follows from the previous results that ϕA (Ga ) is a closed connected subgroup of GLn (k). 21.2.6 Proposition. Let u : G → H be a morphism of algebraic groups. Then the following conditions are equivalent: (i) u is an isomorphism of groups. (ii) u is an isomorphism of algebraic groups. Proof. The implication (ii) ⇒ (i) is obvious. Suppose that u is an isomorphism of groups. By 21.2.4 (ii), u(G◦ ) = H ◦ . Since G◦ and H ◦ are normal varieties, the result is just a consequence of 17.4.8 and 21.1.6.
21.3 Connectedness 21.3.1 Theorem. Let (Xi )i∈I be a family of irreducible varieties and for i ∈ I, ui : Xi → G a morphism. Assume that e ∈ Yi = ui (Xi ) for all i ∈ I. Denote by H the subgroup of G generated by the Yi ’s and K the smallest closed subgroup of G containing the Yi ’s. (i) The closed subgroup K is connected and is equal to H. (ii) There exist n ∈ N, i1 , . . . , in ∈ I and ε1 , . . . , εn ∈ {−1, 1} such that: H = Yiε11 · · · Yiεnn . Proof. If i ∈ I, the map Xi → G, α → ui (α)−1 , is again a morphism. So replacing I if necessary, we may assume that for any i ∈ I, Yi−1 = Yj for some j ∈ I. Observe that we have clearly K = H. For m ∈ N and λ = (i1 , . . . , im ) ∈ I m , set Yλ = Yi1 · · · Yim . Since Yλ is the image of the morphism Xi1 × · · · × Xim → G, (α1 , . . . , αm ) → α1 · · · αm , it follows that Yλ and Yλ are irreducible.
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Let µ = (j1 , . . . , jn ) ∈ I n , n ∈ N. Set (λ, µ) = (i1 , . . . , im , j1 , . . . , jn ) ∈ . We claim that Yλ Yµ ⊂ Y(λ,µ) . Let us prove our claim. For α ∈ Yλ , the I translation morphism β → αβ maps Yµ into Y(λ,µ) . It follows that Yλ Yµ ⊂ Y(λ,µ) . Now for β ∈ Yµ , we have Yλ β = Yλ β ⊂ Y(λ,µ) . Hence Yλ Yµ ⊂ Y(λ,µ) as claimed. Let F = {Yλ ; n ∈ N, λ ∈ I n }. Since the topological space G is Noetherian, F has a maximal element, say Yµ . Let n ∈ N and λ ∈ I n . Since e ∈ Yλ , it follows from the claim above that: m+n
Yµ ⊂ Yµ Yλ ⊂ Y(λ,µ) . Since Yµ is maximal, we have Yµ = Yµ Yλ = Y(µ,λ) . Taking λ = µ, we deduce that Yµ is stable under multiplication. Similarly, −1 taking λ such that Yλ = Yµ−1 , we see that Yµ ⊂ Yµ . So Yµ is a subgroup of G which is closed and connected by definition. Clearly K = Yµ . Furthermore, being the image of a morphism Yµ is a constructible subset of G (15.4.3) and therefore by 1.4.6 and 21.2.1, Yµ = Yµ Yµ = Y(µ,µ) . Hence H ⊂ H = K = Yµ = Y(µ,µ) ⊂ H. 21.3.2 Let (Gi )i∈I be a family of subgroups of G, and ji : Gi → G the canonical injections. In order to apply 21.3.1 in this situation, the Gi ’s need to have a structure of algebraic variety and the ji ’s need to be morphisms. Corollary. Let (Gi )i∈I be a family of closed connected subgroups of G and H the subgroup of G generated by the Gi ’s. Then H is a closed connected subgroup of G. 21.3.3 Corollary. With the notations of 10.8.1, the groups SLn (k), Dn (k), Tn (k) and Un (k) are closed connected subgroups of GLn (k). Proof. These are clearly closed subgroups of GLn (k). The group Dn (k) is connected because it is isomorphic to (Gm )n . The connectedness of SLn (k) and Un (k) follows from 10.8.6 and 21.3.2. Finally, Tn (k) is connected because it is isomorphic to Dn (k) × Un (k). 21.3.4 Let g be a finite-dimensional Lie algebra over k. By 21.2.5 and 21.3.2, we have: Corollary. The group Aute g is a connected algebraic group which is a closed normal subgroup of Aut g. 21.3.5 Let H, K be closed subgroups of G. It is not true in general that the subgroup (H, K) is closed in G. However, we have the following result: Proposition. Let H, K be closed subgroups of G.
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(i) If either H or K is connected, then (H, K) is a closed connected subgroup of G. (ii) Assume that H normalizes K. Then the (H, K) is a closed subgroup of G. In particular, (G, G) is a closed subgroup of G. Proof. (i) Assume that H is connected. For β ∈ K, the map uβ : H → G, α → αβα−1 β −1 , is a morphism of varieties. Since (H, K) is the subgroup of G generated by uβ (H), β ∈ K, the result follows from 21.3.1. (ii) We may assume that G = HK (21.2.3). Then K and (H, K) are normal subgroups of G. By (i), (H ◦ , K) and (H, K ◦ ) are closed connected subgroups of G. Let L be the subgroup of G generated by α(H ◦ , K)α−1 and β(H, K ◦ )β −1 , α, β ∈ G. Since (H, K) is normal, we have L ⊂ (H, K). Moreover, L is closed (21.3.2). So we are done if L has finite index in (H, K). Let θ : G → G/L be the canonical surjection. Then θ(H ◦ ) centralizes θ(K), and θ(K ◦ ) centralizes θ(H). Since H ◦ (resp. K ◦ ) has finite index in H (resp. K), it follows that S = {(γ, δ); γ ∈ θ(H), δ ∈ θ(K)} is a finite set. So (H, K)/L is finite by 10.3.3. 21.3.6 Corollary. If G is connected, then Dn (G) and C n+1 (G) are closed connected subgroups of G for all n ∈ N. 21.3.7 Proposition. Let α ∈ G and H a closed subgroup of G. (i) The centralizers CG (α) of α in G and CG (H) of H is G are closed subgroups of G. (ii) The set {β ∈ G; βHβ −1 ⊂ H} is the closed subgroup NG (H) of G. Proof. (i) Let u, v denote the translations β → αβ and β → βα respectively. Then CG (α) = {β ∈ G; u(β) = v(β)}. So it is closed by 12.5.8. Since CG (H) is the intersection of CG (β), β ∈ H, it is also closed. (ii) For β ∈ G, the map iβ : G → G, α → βαβ −1 , is an automorphism of the algebraic group G. So Kβ = iβ (H) is a closed subgroup of G of dimension dim H and iβ (H ◦ ) = (Kβ )◦ . On the other hand, (H : H ◦ ) = (Kβ : (Kβ )◦ ). If βHβ −1 ⊂ H, then Kβ ⊂ H and Kβ◦ is a closed connected subgroup of H containing e of dimension dim H ◦ . So (Kβ )◦ = H ◦ . It follows that H = Kβ , and so NG (H) = {β ∈ G; βHβ −1 ⊂ H}. Let β ∈ H. The map uβ : G → G, α → αβα−1 , is a morphism because it is the composition of G → G × G , α → (αβ, α−1 ) and G × G → G , (α, γ) → αγ. −1 So u−1 ⊂ H} is the β (H) is closed in G. Since NG (H) = {β ∈ G; βHβ −1 intersection of uβ (H), β ∈ H, it is closed in G.
21.4 Actions of an algebraic group
325
21.4 Actions of an algebraic group 21.4.1 Definition. Let G acts on a variety X. We say that G acts rationally on X, or X is a G-variety if the map G × X → X , (α, x) → α.x is a morphism of algebraic varieties. So, if X is a G-variety, the map x → α.x is an automorphism of X for all α ∈ G. 21.4.2 Proposition. Let X be a G-variety. (i) If Y and Z are subsets of X with Z closed in X, then TranG (Y, Z) is a closed subset of G. (ii) For all x ∈ X, Gx is a closed subgroup of G. (iii) If α ∈ G, then the set of fixed points of α is closed in X. Thus X G is closed in X. (iv) If G is connected, then G stabilizes each irreducible component of X. Proof. (i) If x ∈ X, the map ux : G → X, α → α.x is a morphism. So u−1 x (Z) is closed in G. Since TranG (Y, Z) is the intersection of u−1 y (Z), y ∈ Y , it is closed in G. (ii) This is clear by (i) since Gx = TranG ({x}, {x}). (iii) The diagonal ∆ of X in X × X is closed (12.5.3). For α ∈ G, the map u : X → X × X, x → (x, α.x), is a morphism. The set of fixed points of α is u−1 (∆), so it is closed in X. (iv) Let G be connected, Y an irreducible component of X and H the stabilizer of Y in H. By (i), H is closed in G and since the elements of G permutes the irreducible components of X, we obtain that (G : H) is finite. Hence G = G◦ ⊂ H by 21.1.6. 21.4.3 Proposition. Let X be a G-variety, x ∈ X and Y = G.x. (i) The set Y is locally closed in X, so it is a subvariety of X. (ii) If G is connected, then Y is irreducible and smooth. (iii) We have dim(G.x) = dim G − dim Gx . Proof. It is clear that Y is stable under the action of G. (i) The map ux : G → X, α → α.x is a morphism and Y = im ux . So Y is constructible (15.4.3) and it contains a dense open subset of Y (1.4.6). Since G acts transitively on Y and Y is stable under the action of G, we deduce that Y is open in Y . (ii) Since Y = im ux , Y is irreducible when G is connected. It is smooth since G acts transitively on Y . (iii) Let vx : G → Y be the morphism induced by ux . Then the fibres of vx are left cosets of G modulo Gx . So the results follows from 15.5.5 (ii). 21.4.4 Proposition. Let X and Y be G-varieties. Assume that X is a G-homogeneous space.
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(i) The G◦ -orbits of X are open and closed in X. They all have the same dimension and they are the pairwise disjoint irreducible components of X. In particular, X is a pure variety. (ii) For n ∈ N, {y ∈ Y ; dim(G.y) n} is a closed subset of Y . Proof. (i) Let x ∈ X. For α ∈ G, the right coset G◦ α defines a G◦ -orbit, G◦ α.x of X. Since G acts transitively on X, and (G : G◦ ) is finite, there exist elements α1 , . . . , αn ∈ G such that X = (G◦ α1 .x) ∪ · · · ∪ (G◦ αn .x) and G◦ αi .x ∩ G◦ αj .x = ∅ if i = j. Let Xi = G◦ αi .x. So the Xi ’s are exactly the G◦ -orbits in X. Suppose for example that dim X1 dim Xi for 2 i n and Y = X1 \ X1 is non-empty. Since X1 is irreducible (21.4.3) and Y is closed in X1 , dim Y < dim X1 = dim X1 . But this is absurd because Y , being G◦ -stable, is the union of certain Xi ’s with i 2. So X1 is closed in X. Now Xi = G◦ αi .x = αi G◦ .x = (αi α1−1 ).X1 , thus Xi is closed in X for all i. The result is now clear. (ii) By (i), G.y is a pure variety whose irreducible components are G◦ orbits. So we may assume that G is connected. By 21.4.2 (iv), we may further assume that X is irreducible. Let u : G × X → X × X be the morphism sending (α, x) to (x, α.x). Then u−1 (u(α, x)) = (αGx , x) for all α ∈ G and x ∈ X. By 21.1.4 and 21.4.2 (ii), (α(Gx )◦ , x) is the unique irreducible component of (αGx , x) containing (α, x). Its dimension is dim Gx . So the result is a consequence of 21.4.3 (iii) and 15.5.7. 21.4.5 Proposition. Let X be a G-variety, x ∈ X and Y = G.x. (i) The set Y is the union of Y and orbits of dimension strictly less than dim Y . (ii) Orbits of minimal dimension are closed. In particular, X contains a closed orbit. Proof. Let Y1 , . . . , Yn be the irreducible components of Y . They are G◦ -orbits by 21.4.4. By 1.1.14, Y1 , . . . , Yn are the irreducible components of Y . Since Y \ Y ⊂ (Y1 \ Y1 ) ∪ · · · ∪ (Yn \ Yn ), we have dim(Y \ Y ) < dim Y . So we have (i) since Y \ Y is G-stable. Finally (ii) follows easily from (i).
21.5 Modules 21.5.1 Let V be a k-vector space of dimension n. By identifying V with kn via a basis B of V , we can transport the Zariski topology of kn on V . It is clear that the topology obtained on V does not depend on the choice of B. Hence we have a notion of Zariski topology on V , and in a similar way, on End(V ) and GL(V ). 21.5.2 Let (V, ρ) be a finite-dimensional G-module. We say that V is a rational G-module if the map ρ : G → GL(V ) is a morphism of algebraic groups. Lemma. The following conditions are equivalent:
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327
(i) V is a rational G-module. (ii) The map θ : G×V → V , (α, x) → ρ(α)(x), is a morphism of varieties. (iii) For all x ∈ V , the map θx : G → V , α → ρ(α)(x), is a morphism of varieties. Proof. Since θ is the composition of G × V → End(V ) × V , (α, x) → (ρ(α), x), End(V ) × V → V , (u, x) → u(x), where the second map is a morphism of varieties, we have (i) ⇒ (ii). The implication (ii) ⇒ (iii) is obvious. Let B = (x1 , . . . , xn ) be a basis of V and (ϕ1 , . . . , ϕn ) the basis of V ∗ dual to B. The map V → k, v → ϕi (v), is a morphism of varieties for all i. So if (iii) is verified, then the map G → Mn (k) sending α ∈ G to the matrix ρ(α) with respect to B is a morphism of varieties. Hence (iii) ⇒ (i). 21.5.3 Remark. Let V be a finite-dimensional rational G-module, then the map θ in 21.5.2 (ii) makes V into a G-variety. So we may apply the results of 21.4 to rational G-modules. 21.5.4 Definition. Let V be a G-module. We say that V is locally finite if for all x ∈ V , there exists a finite-dimensional G-submodule Vx of V containing x. If V is a locally finite G-module, then V is a rational G-module if for all x ∈ V , we can choose Vx to be a rational G-module. 21.5.5 Proposition. Let (V, ρ) be a rational G-module and W a submodule of V . Then W is a rational G-module. Proof. For x ∈ W , let Vx be a rational G-submodule of finite dimension containing x. By 21.5.2, for y ∈ Vx , the map G → V , α → ρ(α)(y), is a morphism of varieties. If y ∈ W ∩Vx , then the map G → W ∩Vx , α → ρ(α)(y), is again a morphism of varieties. It follows from 21.5.2 that W ∩ Vx is a finitedimensional rational G-submodule containing x.
21.6 Group closure 21.6.1 Let H be a subgroup of G. In this section, we study the properties of the map H → H. Recall from 21.2.2 that H is a subgroup of G. Lemma. (i) If H is commutative, then so is H. (ii) If H is normal in G, then so is H. Proof. (i) For β ∈ G, denote by uβ and vβ the morphisms from G to itself defined by uβ (α) = αβ and vβ (α) = βα. Then if β ∈ H, then uβ and vβ are identical on H, so they are identical on H (12.5.8). This implies in turn that for β ∈ H, uβ and vβ are identical on H, hence on H. Thus H is commutative. (ii) For β ∈ G, the morphism of varieties ϕβ : G → G, α → βαβ −1 , verifies ϕβ (H) ⊂ H, and so ϕβ (H) ⊂ H.
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21.6.2 Let P ⊂ G. The intersection of all the closed subgroups of G containing P , denoted by A(P ), is called the group closure of P in G. Clearly, A(P ) is the smallest closed subgroup of G containing P . If H is a subgroup of G, then A(H) = H. So A(P ) can also be identified as the closure of the subgroup of G generated by P . 21.6.3 Lemma. Let G, H be algebraic groups, P, Q ⊂ G, R ⊂ H and u : G → H a morphism of algebraic groups. (i) If P is a dense subset of Q, then A(P ) = A(Q). (ii) If P normalizes (resp. centralizes) Q, then A(P ) normalizes (resp. centralizes) A(Q). (iii) We have A(P × R) = A(P ) × A(R). (iv) We have u(A(P )) = A(u(P )). Proof. (i) Since Q ⊂ P ⊂ A(P ), A(Q) ⊂ A(P ). The reverse inclusion is obvious. (ii) If x ∈ G normalizes Q, then it normalizes the subgroup generated by Q, and hence A(Q) (same proof as in 21.6.1 (ii)). Thus P ⊂ NG (A(Q)). Since NG (A(Q)) is closed in G (21.3.7), A(P ) ⊂ NG (A(Q)). (iii) Since A(P ) × A(R) is a closed subgroup of G × H containing P × R, it contains A(P × R). Conversely, A(P × R) contains P × {eH } and {eG } × Q, so it contains A(P ) × {eH } and {eG } × A(Q). So A(P × R) ⊃ A(P ) × A(R). (iv) Since the subgroup u(A(P )) is closed in H (21.2.4) and it contains u(P ), we have A(u(P )) ⊂ u(A(P )). Moreover, P ⊂ u−1 (A(u(P ))) ⇒ A(P ) ⊂ u−1 (A(u(P ))) ⇒ u(A(P )) ⊂ A(u(P )). So we have proved our result. 21.6.4 Proposition. Let H, K be subgroups of G. Then the groups (H, K) and (H, K) have the same closure in G. Proof. The map u : G × G → G , (α, β) → αβα−1 β −1 , is a morphism of varieties. Since H × K is dense in H × K, u(H × K) is dense in u(H ×K). So the result follows from 21.6.3 (i) since A(u(H ×K)) = (H, K) and A(u(H × K)) = (H, K). 21.6.5 Proposition. Let H, K be subgroups of G. If H normalizes K, then H normalizes K and (H, K) = (H, K). Proof. By 21.6.3 (ii), H ⊂ NG (K), so HK is a closed subgroup of G (21.2.3). We may therefore assume that G = HK. Then K is normal in G and (H, K) is closed in G (21.3.5). So the result follows from 21.6.4.
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21.6.6 Corollary. Let H, K be subgroups of G. (i) For all n ∈ N, we have: Dn (H) = Dn (H) , C n+1 (H) = C n+1 (H). (ii) Assume that K is a normal subgroup of H. If H/K is abelian (resp. nilpotent, solvable), then so is H/K. Proof. By 21.6.5, (i) is a simple induction on n. If H/K is solvable, then Dn (H) ⊂ K for some n ∈ N. So (i) implies that Dn (H) = Dn (H) ⊂ K. Thus H/K is solvable. The proofs of the other cases are similar. 21.6.7 Corollary. The following conditions are equivalent: (i) G is solvable. (ii) There is a sequence G = G0 ⊃ G1 ⊃ · · · ⊃ Gn+1 = {e} of closed subgroups of G such that (Gi , Gi ) ⊂ Gi+1 for 0 i n. Proof. In view of 21.3.6 and 21.6.6 (i), we have (i) ⇒ (ii) by taking Gi = Di (G). Conversely, given (ii), a simple induction shows that Di (G) ⊂ Gi . So we have (i). 21.6.8 Corollary. The following conditions are equivalent: (i) G is nilpotent. (ii) There is a sequence G = G0 ⊃ G1 ⊃ · · · ⊃ Gn+1 = {e} of closed subgroups of G such that (G, Gi ) ⊂ Gi+1 for 0 i n. Proof. This is analogue to the proof of 21.6.7.
References • [5], [23], [38], [40], [78].
22 Affine algebraic groups
In the rest of this book, all the algebraic groups that we are going to come across are affine. We shall therefore assume from now on that by an algebraic group, we mean an affine algebraic group. In fact, we shall see that under this assumption, an algebraic group can be identified with a closed subgroup of a general linear group. By convention, G will always denote an (affine) algebraic group.
22.1 Translations of functions 22.1.1 Let G be an algebraic group, X an affine G-variety and π : G × X → X , (α, x) → α.x the morphism defining the action of G on X. The comorphism of π A(π) : A(X) → A(G × X) = A(G) ⊗k A(X) sends f ∈ A(X) to the function F on G × X defined by F (α, x) = f (α.x). From this, we obtain that if α ∈ G and f ∈ A(X), then the function τα f on X defined by (τα f )(x) = f (α−1 .x) belongs to A(X). We verify easily that this defines a group homomorphism τ : G → GL(A(X)), α → τα . We shall endow A(X) with this G-module structure. Proposition. Let E be a finite-dimensional subspace of A(X). (i) There exists a finite-dimensional G-submodule of A(X) containing E. (ii) The subspace E is a G-submodule of A(X) if and only if A(π)(E) is contained in A(G) ⊗k E. Proof. (i) We may assume that dim E = 1. So E is spanned by an element f ∈ A(X). Let
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A(π)(f ) = u1 ⊗ v1 + · · · + un ⊗ vn where u1 , . . . , un ∈ A(G) and v1 , . . . , vn ∈ A(X). For α ∈ G and x ∈ X, (τα f )(x) = A(π)(f )(α−1 , x) = u1 (α−1 )v1 (x) + · · · + un (α−1 )vn (x). Thus the G-submodule generated by f is contained in the subspace spanned by v1 , . . . , vn . (ii) Let (v1 , . . . , vn ) be a basis of E. Suppose that A(π)(E) ⊂ A(G) ⊗k E. If f ∈ E, then there exist u1 , . . . , un ∈ A(G), such that A(π)(f ) = u1 ⊗ v1 + · · · + un ⊗ vn . Then for α ∈ G, we have as seen in the proof of (i) that τα f = u1 (α−1 )v1 + · · · + un (α−1 )vn . So E is a G-submodule of A(X). Suppose that E is a G-submodule of A(X). For f ∈ E, there exist w1 , . . . , wm ∈ A(X), u1 , . . . , un+m ∈ A(G) such that the elements v1 , . . . , vn , w1 , . . . , wm are linearly independent and A(π)(f ) = u1 ⊗ v1 + · · · + un ⊗ vn +
m
un+i ⊗ wi .
i=1
It follows that for α ∈ G, τα f = u1 (α−1 )v1 + · · · + un (α−1 )vn +
m
un+1 (α−1 )wi .
i=1
Since E is a G-submodule, un+i (α−1 ) = 0 for all α ∈ G and 1 i m. So un+i = 0 for 1 i m and we have A(π)(f ) ∈ A(G) ⊗k E as required. 22.1.2 Proposition. Let G be an algebraic group and X an affine Gvariety. Then the G-module A(X) is rational. Proof. Let f ∈ A(X). By 22.1.1, f is contained in a finite-dimensional Gsubmodule E of A(X). Let (v1 , . . . , vn ) be a basis of E. Then as we have seen in the proof of 22.1.1, there exist u1 , . . . , un ∈ A(G), such that τα f = u1 (α−1 )v1 + · · · + un (α−1 )vn for all α ∈ G. Since the maps G → k, α → ui (α−1 ), are morphisms of varieties, the map G → E, α → τα f , is also a morphism of varieties. So E is a rational G-module (21.5.2). 22.1.3 Proposition. Let G be an algebraic group and X an affine Gvariety. Then there exist a finite-dimensional rational G-module V and a Gequivariant morphism u : X → V such that u is an isomorphism from X onto a G-stable closed subvariety of V .
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Proof. Let v1 , . . . , vn be generators of the algebra A(X). By 22.1.2, there is a finite-dimensional rational G-submodule W of A(X) containing all the vi ’s. The dual W ∗ of W is also a rational G-module. Denote by ϕ : X → W ∗ the map defined as follows: ϕ(x)(w) = w(x) for all w ∈ W , x ∈ X. Then it is clear that ϕ is a G-equivariant morphism. Since the vi ’s generate A(X), it is easy to see that A(ϕ) is surjective. So we can conclude by 11.6.8. 22.1.4 Let G be an affine algebraic group. The group G acts on itself by left and right translations. We obtain from the preceding discussion two G-module structures λ, ρ : G → GL(A(G)) on A(G) defined as follows: for α, β ∈ G, u ∈ A(G), (λα u)(β) = u(α−1 β) , (ρα u)(β) = u(βα). We call λα (resp. ρα ) the left translation (resp. right translation) of functions. 22.1.5 Theorem. Let G be an affine algebraic group. There exists n ∈ N such that G is isomorphic (as an algebraic group) to a closed subgroup of GLn (k). Proof. Applying 22.1.3 to the representation of G on itself by left translations, we obtain that there exist a rational G-module (V, ρ) and a G-equivariant morphism u : G → V such that u is an isomorphism from G onto a Gstable closed subvariety of V . Let α ∈ G be such that ρ(α) = idV . Since u is G-equivariant, we have u(αβ) = ρ(α)u(β) = u(β). So α = e because u is injective. We conclude that ρ is injective and therefore G is isomorphic as an algebraic group to the closed subgroup ρ(G) of GL(V ) (21.2.6). 22.1.6 Lemma. Let G be an affine algebraic group, H a closed subgroup, and a = I(H) the defining ideal of H in A(G). Then H = {α ∈ G; λα (a) ⊂ a} = {α ∈ G; ρα (a) ⊂ a}. Proof. Let α, β ∈ H, u ∈ a. Since α−1 β ∈ H, we have u(α−1 β) = 0, so λα u, ρβ u ∈ a. Conversely, if λα (a) ⊂ a, then u(α−1 ) = (λα u)(e) = 0 for all u ∈ a. So −1 α ∈ H and hence α ∈ H. The proof for ρα is analogue.
22.2 Jordan decomposition 22.2.1 Let us consider A(G) as a G-module by right translation. Then A(G) is rational (22.1.2) and by 10.1.11, ρα decomposes into (ρα )s (ρα )u with (ρα )s semisimple, (ρα )u locally unipotent and (ρα )s (ρα )u = (ρα )u (ρα )s . Theorem. If α ∈ G, there exists a unique pair (αs , αu ) of elements of G such that: (ρα )s = ραs , (ρα )u = ραu , α = αs αu = αu αs . The decomposition α = αs αu is called the Jordan decomposition of α and αs (resp. αu ) is called the semisimple (resp. unipotent) component of α. If α = αs (resp. α = αu ), then we say that α is semisimple (resp. unipotent).
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22 Affine algebraic groups
Proof. Since ρα is an automorphism of the k-algebra A(G), so are (ρα )s and (ρα )u (10.1.14). Thus the maps f → ((ρα )s f ) (e) , f → ((ρα )u f ) (e) from A(G) → k are k-algebra homomorphisms. By 11.3.4, there exist αs , αu ∈ G such that ((ρα )s f ) (e) = f (αs ) and ((ρα )u f ) (e) = f (αu ) for all f ∈ A(G). Let β ∈ G. Since ρα and λβ commute, (ρα )s and λβ commute also. It follows that ((ρα )s f ) (β) = λβ −1 ((ρα )s f ) (e) = (ρα )s (λβ −1 f ) (e) = (λβ −1 f )(αs ) = f (βαs ). Therefore (ρα )s = ραs and in the same way, we have (ρα )u = ραu . Finally, the map G → GL(A(G)), α → ρα , is injective, so α = αs αu = αu αs since ρα = (ρα )s (ρα )u = (ρα )u (ρα )s . 22.2.2 Remark. As in 22.2.1, we have (λα )s = λαs and (λα )u = λαu . 22.2.3 Proposition. Let G and H be algebraic groups. (i) If θ : G → H is a morphism of algebraic groups, then θ(αs ) = θ(α)s and θ(αu ) = θ(α)u for all α ∈ G. (ii) Suppose that G is a closed subgroup of GL(V ) where V is a finitedimensional k-vector space. Let α ∈ G. Then its semisimple and unipotent components defined in 22.2.1 coincide with those defined in 10.1.4. Proof. (i) Since K = im θ is a closed subgroup of H (21.2.4) and θ factorizes through G → K = im θ → H, it suffices to prove the result in the following two cases: 1) G is a closed subgroup of H and θ is the canonical injection. Since A(G) = A(H)/I(G) and I(G) is stable under ρα , α ∈ G (22.1.6), the result follows from 10.1.3 and 10.1.5. 2) θ is surjective. Then A(θ) is injective and A(θ) ◦ ρθ(α) = ρα ◦ A(θ). Thus A(H) can be identified as a subspace of A(G) stable under ρα , α ∈ G. So we may conclude as in the previous case. (ii) In view of (i), it suffices to prove the result in the case where G = GL(V ) and by fixing a basis, we may assume that G = GLn (k). Then A(G) = k[T11 , T12 , . . . , Tnn ]D where D is the determinant (11.7.5). Writing α = [αij ], then for 1 i n, ρα T1i = α1i T11 + · · · + αni T1n . Thus the subspace spanned by T1i , 1 i n, is stable under ρα . Again we may conclude by 10.1.3 and 10.1.5. 22.2.4 Corollary. Let α ∈ G and θ an isomorphism of algebraic groups from G onto a closed subgroup of GL(V ) where V is a finite-dimensional k-vector space. The following conditions are equivalent:
22.3 Unipotent groups
335
(i) α is semisimple (resp. unipotent). (ii) θ(α) is a semisimple (resp. unipotent) automorphism of V . 22.2.5 Proposition. The set of unipotent elements of G is closed in G. Proof. By 22.1.5 and 22.2.3 (i), it suffices to prove the result for G = GL(V ) where V is a finite-dimensional k-vector space. Since α ∈ GL(V ) is unipotent if and only if (α − idV )dim V = 0, the result follows. Remark. In general, the set of semisimple elements of G is neither open nor closed in G. 22.2.6 Proposition. Let G be commutative and denote by Gs (resp. Gu ) the set of semisimple (resp. unipotent) elements of G. (i) Gs and Gu are closed subgroups of G. (ii) The map π : Gs × Gu → G, (α, β) → αβ, is an isomorphism of algebraic groups. (iii) If G is connected, then Gs and Gu are also connected. Proof. We may assume that G is a closed subgroup of GLn (k) (22.1.5). So we may apply 22.2.4 with V = kn . (i) Since G is commutative, we have (αβ)s = αs βs and (αβ)u = αu βu . So Gs and Gu are subgroups of G. We saw in 22.2.5 that Gu is closed. There exists α ∈ GLn (k) such that αGs α−1 ⊂ Dn (k) because Gs is commutative. It follows that Gs = G ∩ (α−1 Dn (k)α) and it is closed in G since Dn (k) is closed in GLn (k). (ii) The uniqueness of the Jordan decomposition implies that π is a bijective morphism of algebraic groups. So the result follows from 21.2.6. (iii) This follows from (ii) and 11.8.6.
22.3 Unipotent groups 22.3.1 Definition. Let G be an algebraic group. (i) We say that G is unipotent if all the elements of G are unipotent. (ii) A morphism λ : Ga → G of algebraic groups is called an additive one-parameter subgroup of G. 22.3.2 Proposition. The following conditions are equivalent: (i) G is unipotent. (ii) G is isomorphic, as an algebraic group, to a closed subgroup of Un (k). Proof. (i) ⇒ (ii) Since G is isomorphic to a closed subgroup H of GLn (k) (22.1.5) and the elements of H are unipotent automorphisms by 22.2.4, the result follows from 10.8.14. (ii) ⇒ (i) This is clear by 22.2.3 and 22.2.4.
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22.3.3 Let V be a finite-dimensional k-vector space. If n is a nilpotent endomorphism of V and u an unipotent automorphism of V , we may define: exp(n) = en =
∞ 1 ∞ (−1)k+1 nk , log(u) = (u − idV )k . k! k k=0 k=1
Then en is an unipotent automorphism of V and log(u) is a nilpotent endomorphism of V . Furthermore, it is well-known that: exp(log(u)) = u and log(exp(n)) = n. Now let ϕn : Ga → GL(V ) , t → etn . We saw in 21.2.5 that ϕn is a morphism of algebraic groups. It follows that ϕn (Ga ) is a connected closed subgroup of GL(V ). Suppose that n = 0. For t ∈ k, we have log(etn ) = tn. So ϕn induces a bijection between Ga and ϕn (Ga ) whose inverse map is log |ϕn (Ga ) . It follows from the definition of log that ϕn is an isomorphism of algebraic groups from Ga onto ϕn (Ga ). 22.3.4 Proposition. Let G be a closed subgroup of GL(V ), u an unipotent element of G and n = log(u). Then ϕn (Ga ) ⊂ G. Consequently, ϕn (Ga ) is the smallest closed subgroup of G (and so of GL(V )) containing u. Proof. Let k ∈ Z. Since exp(log(uk )) = uk = exp(k log(u)), we deduce that ϕn (Z) ⊂ G. Since G is closed and Z is dense in Ga , we have by the continuity of ϕn that ϕn (Ga ) = ϕn (Z) ⊂ ϕn (Z) ⊂ G. So the result follows. 22.3.5 Proposition. Let V be a finite-dimensional k-vector space and λ : Ga → GL(V ) a group homomorphism. The following conditions are equivalent: (i) λ is an additive one-parameter subgroup of GL(V ). (ii) There exists n ∈ End(V ) nilpotent such that λ(t) = etn for all t ∈ k. Proof. The implication (ii) ⇒ (i) is clear. So let us suppose that (i) is verified. By 22.2.4 and 22.3.3, Ga is unipotent. So λ(Ga ) is a connected closed unipotent subgroup of GL(V ). Let n = log(λ(1)). Then ϕn (Ga ) ⊂ λ(Ga ) (22.3.4) and for k ∈ Z, λ(k) = λ(1)k = ekn . Since Z is dense in the irreducible variety Ga , we obtain that λ = ϕn (12.5.8). 22.3.6 Theorem. Let G be a unipotent algebraic group. (i) G is connected. (ii) If V is a non-zero rational G-module, then V G = {0}. (iii) Let X be an affine G-variety. Then the G-orbits in X are closed.
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Proof. (i) We may assume that G ⊂ GL(V ) for some finite-dimensional kvector space V . Let N be the set of log(u), u ∈ G. By 22.3.4, G is generated by ϕn (Ga ), n ∈ N . Thus the result follows from 21.2.5 and 21.3.1. (ii) We may assume that V is finite-dimensional and the result was proved in 10.8.13. (iii) Suppose that there exists x ∈ X such that Z = G.x is not closed in X. Set Y = Z \ Z. Then Y is a non-empty closed subset of Z (21.4.3). Let a = I(Y ). Then a = {0} and by (ii), there exists f ∈ a \ {0} such that α.f = f for all α ∈ G. Hence f (α.x) = f (x) for all α ∈ G and f is constant on Z, and so on Z. But f is zero on Y , and so f = 0. Contradiction. 22.3.7 Let X be an irreducible affine G-variety. We can extend the Gaction on A(X) (22.1.1) to Fract(A(X)) as follows: for α ∈ G, u, v ∈ A(X) with v = 0, α(uv −1 ) = (α.u)(α.v)−1 . It is clear that Fract(A(X))G is a subfield of Fract(A(X)). Recall also from 12.7.7 that we may identify Fract(A(X)) with R(X). Corollary. Let G be a unipotent algebraic group and X an irreducible affine G-variety. Then R(X)G = Fract A(X)G . Proof. Clearly R(X)G ⊃ Fract A(X)G . Let u be an element of R(X)G . Denote by V the set of s ∈ A(X) such that us ∈ A(X). Then V is a Gsubmodule of A(X). Since u ∈ Fract A(X), V = {0}. So by 22.3.6, there exists v ∈ V G \ {0} ⊂ A(X)G \ {0}. It follows that w = uv ∈ A(X)G and u = wv −1 ∈ Fract(A(X)G ). 22.3.8 Lemma. Let K be a commutative field and G a group of automorphisms α of the polynomial ring K[T ] such that α.K ⊂ K. (i) We have K(T )G = Fract(K[T ]G ). (ii) There exists p ∈ K[T ]G such that K(T )G = K G (p). Proof. Let α ∈ G and u = α.T . If deg(u) = 0, then since α.K ⊂ K, we have α.K[T ] ⊂ K which is absurd. Similarly, if deg(u) 2, then deg(α.v) 2 for all v ∈ K[T ] \ K, and so α.K[T ] = K[T ], which is again absurd. We deduce that deg(u) = 1 and hence deg(α.v) = deg(v) for all v ∈ K[T ]. (i) Let f ∈ K(T )G \ Fract K[T ]G . Replacing f by f −1 if necessary, we may assume that f = uv −1 where u, v ∈ K[T ] are relatively prime and deg(u) deg(v) > 0. If α ∈ G, then u(α.v) − v(α.u) = 0. Since u and v are relatively prime, we deduce that there exists χ(α) ∈ K \ {0} such that α.v = χ(α)v and α.u = χ(α)u. Now there exist q, r ∈ K[T ] such that u = qv + r and deg(r) < deg(v). Then χ(α)u = χ(α)(α.q)v + α.r. It follows that α.q = q and α.r = χ(α)r. Moreover f = q + rv −1 and α.(rv −1 ) = rv −1 . So we obtain (i) by induction on deg(u) + deg(v).
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(ii) If K[T ]G ⊂ K, then by (i), we have K(T )G = K G = K G (1). So let us suppose that K[T ]G \ K = ∅. Take p ∈ K[T ]G \ K of minimal degree. For u ∈ K[T ]G \ K, there exist q, r ∈ K[T ] unique such that u = qp + r and deg(r) < deg(p). It follows from our choice of p that r ∈ K G . By induction on deg(u), we obtain that u ∈ K G [p]. Now the result follows from (i). 22.3.9 The following result applies in particular for unipotent groups. Proposition. Let n ∈ N and G a subgroup of Tn (k). Consider the natural action of G on kn , which induces an action of G on k(T1 , . . . , Tn ). Then k(T1 , . . . , Tn )G is a purely transcendental extension of k. Proof. Let K = k(T1 , . . . , Tn−1 ) and T = Tn , then G is a group of automorphisms of K[T ] leaving K stable. So the result follows from 22.3.8 and induction on n.
22.4 Characters and weights 22.4.1 Lemma. Let Γ be a group and X the set of homomorphisms from Γ to k\{0}. Then X is a family of linearly independent elements of the vector space of maps from Γ to k. Proof. Suppose that there exist χ1 , . . . , χn ∈ X pairwise distinct linearly dependent with n minimal with this property. Then χi = 0, n > 1 and there exist λ1 , . . . , λn−1 ∈ k \ {0} such that: χn = λ1 χ1 + · · · + λn−1 χn−1 . Since χ1 = χn , χ1 (β) = χn (β) for some β ∈ Γ . It follows that if α ∈ Γ , then χn (α)χn (β) = λ1 χ1 (α)χ1 (β) + · · · + λn−1 χn−1 (α)χn−1 (β) = (λ1 χ1 (α) + · · · + λn−1 χn−1 (α))χn (β) So λ1 (χ1 (β) − χn (β))χ1 + · · · + λn−1 (χn−1 (β) − χn (β))χn−1 = 0. But this is absurd by our choice of n. 22.4.2 Definition. Let G be an algebraic group. (i) A morphism χ : G → Gm of algebraic groups is called a (multiplicative) character of G. (ii) A morphism λ : Gm → G is called a multiplicative one-parameter subgroup of G. 22.4.3 Denote by X ∗ (G) the set of characters of G. It is a multiplicative subgroup of the set of maps from G to k and by 22.4.1, the elements of X ∗ (G) are linearly independent in the k-vector space of maps from G to k. If χ ∈ X ∗ (G) and α, β ∈ G, then it is clear that χ(αβα−1 β −1 ) = 1, and hence χ|D(G) = 1. In particular, X ∗ (SLn (k)) = {1} (10.8.8).
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Let H be a closed normal subgroup of G. If χ ∈ X ∗ (H) and α ∈ G, then the map χα : H → Gm , β → χ(αβα−1 ), is a character of H. Thus the map (α, χ) → χα defines an action of G on X ∗ (H). 22.4.4 Denote by X∗ (G) the set of multiplicative one-parameter subgroups of G. If G is abelian, then X∗ (G) is a multiplicative subgroup of the set of maps from Gm to G. Let λ ∈ X∗ (G) and α ∈ G. Then the map λα : Gm → G, t → αλ(t)α−1 , is again a multiplicative one-parameter subgroup of G. Thus the map (α, λ) → λα defines an action of G on X∗ (G). 22.4.5 Let (V, π) be a rational G-module. For χ ∈ X ∗ (G), denote by Vχ = {v ∈ V ; π(α)(v) = χ(α)v for all α ∈ G}. Then Vχ is a subspace of V . We say that χ is a weight of G in V if Vχ = {0}. If χ is a weight of G in V , then Vχ is called the weight space of weight χ of G in V , and a non-zero vector of Vχ is called a weight vector of weight χ of G in V . A non-zero vector of V is called a weight vector if it is a weight vector for some χ. Let v ∈ V \ {0}. Then it is easy to see that v is a weight vector if and only if the subspace spanned by π(G)(v) is 1-dimensional. Let H be a closed normal subgroup of G. We saw in 22.4.3 that G acts on X ∗ (H). Let χ ∈ X ∗ (H) and denote by Vχ,H the weight space of weight χ of H in V . Then we see that π(α)(Vχ,H ) ⊂ Vχα ,H . 22.4.6 Lemma. Let (V, π) be a rational G-module. Then the sum Vχ χ∈X ∗ (G)
is direct. In particular, if V is finite-dimensional, the set of weights of G in V is finite. Proof. If the sum is not direct, then there exist an integer n 2, pairwise distinct elements χ1 , · · · , χn of X ∗ (G), and vi ∈ Vχi \ {0}, 1 i n, such that v1 + · · · + vn = 0. We may assume that n is minimal with this property. Since χ1 = χ2 , there exists α ∈ G such that χ1 (α) = χ2 (α). Then 0 = π(α)(v1 + · · · + vn ) = χ1 (α)v1 + · · · + χn (α)vn , and hence, (χ2 (α) − χ1 (α))v2 + · · · + (χn (α) − χ1 (α))vn = 0. But this contradicts the minimality of n.
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22.5 Tori and diagonalizable groups 22.5.1 Definition. Let G be an algebraic group. (i) We say that G is a torus if it is isomorphic, as an algebraic group, to Dn (k) for some n ∈ N. (ii) We say that G is diagonalizable if it is isomorphic, as an algebraic group, to a closed subgroup of Dn (k) for some n ∈ N. Remarks. 1) A torus is a connected diagonalizable algebraic group. 2) It is clear that a diagonalizable algebraic group is commutative, and that all the elements of a diagonalizable group are semisimple. 22.5.2 Let n ∈ N∗ and G = Dn (k). We have: A(G) = k[T1 , T1−1 , . . . , Tn , Tn−1 ] =
(a1 ,...,an )∈Zn
kT1a1 · · · Tnan .
Moreover, we verify easily that invertible elements of A(G) are the elements λT1a1 · · · Tnan with λ ∈ k \ {0} and (a1 , . . . , an ) ∈ Zn . For (a1 , . . . , an ) ∈ Zn , set χ(a1 ,...,an ) = T1a1 · · · Tnan ∈ A(G). Then we verify easily that χ(a1 ,...,an ) ∈ X ∗ (G). Conversely, given χ ∈ X ∗ (G), then χ is invertible in A(G) and χ(1) = 1, so it is of the form χ(a1 ,...,an ) for some (a1 , . . . , an ) ∈ Zn . It follows that the map: Zn → X ∗ (G) , (a1 , . . . , an ) → χ(a1 ,...,an ) , is an isomorphism of groups, and the elements of X ∗ (G) form a basis of A(G). Similarly, for (b1 , . . . , bn ) ∈ Zn , the map λ(b1 ,...,bn ) : Gm → G , t → diag(tb1 , . . . , tbn ), defines an element of X∗ (G), and so the map Zn → X∗ (G) , (b1 , . . . , bn ) → λ(b1 ,...,bn ) , is an isomorphism of groups. Proposition. Let n ∈ N∗ and G = Dn (k). (i) The groups X ∗ (G) and X∗ (G) are free of rank n and the elements of X ∗ (G) form a basis of A(G). (ii) If χ ∈ X ∗ (G) and λ ∈ X∗ (G), then there exists an integer χ, λ such that χ(λ(t)) = t χ,λ for all t ∈ Gm . Furthermore, the pairing X ∗ (G) × X∗ (G) → Z , (χ, λ) → χ, λ, is bilinear and non-degenerate. (iii) If (V, π) is a rational G-module, then V = Vχ . χ∈X ∗ (G)
In particular, V is the direct sum of G-submodules of dimension 1.
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Proof. Part (i) is just a summary of the preceding discussion, and (ii) follows from: χ(a1 ,...,an ) (λ(b1 ,...,bn ) (t)) = ta1 b1 +···+an bn . Finally for (iii), we may assume that V is finite-dimensional. Let B be a basis of V . For α ∈ G, denote by [πij (α)] the matrix of π(α) with respect to B. Since πij ∈ A(G), part (i) says that they are linear combinations of characters of G. So we have: χ(α)uχ , π(α) = χ∈X ∗ (G)
where uχ ∈ End(V ). If α, β ∈ G, then: χ(α)χ(β)uχ = π(α)◦π(β) = π(αβ) = χ∈X ∗ (G)
χ,χ ∈X ∗ (G)
χ(α)χ (β)uχ ◦uχ .
Applying 22.4.1 to the characters of G × G, we deduce that uχ ◦uχ = uχ and uχ ◦uχ = 0 if χ = χ . On the other hand: uχ = π(e) = idV . χ∈X ∗ (G)
So if Mχ = uχ (V ), then: V =
χ∈X ∗ (G)
Mχ .
Finally, if v ∈ Mχ , then uχ (v) = v and: π(α)(v) = χ (α)uχ ◦uχ (v) = χ(α)u2χ (v) = χ(α)v. χ ∈X ∗ (G)
So Mχ = Vχ and we are done. 22.5.3 Theorem. The following conditions are equivalent for an algebraic group G. (i) G is diagonalizable. (ii) X ∗ (G) is a finitely generated abelian group whose elements form a basis of A(G). (iii) Any rational G-module is the direct sum of G-submodules of dimension 1. Proof. (i) ⇒ (ii) Suppose that G is closed subgroup of H = Dn (k). Then A(G) is a quotient of A(H). If χ ∈ X ∗ (H), then its image in A(G) belongs to X ∗ (G). Hence X ∗ (G) is a basis of A(G) by 22.5.2 and 22.4.1. Let (ε1 , . . . , εn ) be the canonical basis of the Z-module Zn and φi the image of χεi in A(G). Then φa1 1 · · · φann , a1 , . . . , an ∈ Z, span A(G), and they are characters of G. It follows that any φ ∈ X ∗ (G) is a linear combination of the φa1 1 · · · φann ’s. By 22.4.1, φ is one of the φa1 1 · · · φann ’s. Thus X ∗ (G) is finitely generated.
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(ii) ⇒ (iii) The proof is the same as the one for 22.5.2. (iii) ⇒ (i) We may assume that G is a closed subgroup of GL(V ) where V is a finite-dimensional k-vector space (22.1.5). Then V is the direct sum of G-submodules of dimension 1, and the result is clear. 22.5.4 Corollary. Let G be diagonalizable and H a closed subgroup of G. (i) H is diagonalizable. (ii) There exist χ1 , . . . , χn ∈ X ∗ (G) such that H = (ker χ1 ) ∩ · · · ∩ (ker χn ). (iii) If θ ∈ X ∗ (H), then there exists χ ∈ X ∗ (G) such that χ|H = θ. Proof. (i) This is obvious. ∗ ∗ (ii) Let XH (G) = {χ ∈ X ∗ (G); H ⊂ ker χ}. If χ ∈ XH (G), then χ − 1 belongs to I(H). Let f ∈ I(H) \ {0}. By 22.5.3, there exist λ1 , . . . , λn ∈ k \ {0} and χ1 , . . . , χn ∈ X ∗ (G) pairwise distinct such that f = λ1 χ1 + · · · + λn χn . Reindexing if necessary, we may assume that there exist m0 = 0 < m1 < · · · < mr = n such that χi |H = θj for mj−1 + 1 i mj , where θ1 , . . . , θr are pairwise distinct elements of X ∗ (H). Applying 22.4.1 to the θj ’s, we obtain that λmj−1 +1 + · · · + λmj = 0 for 1 j r. Set: µj =
m j i=mj−1 +1
λi (χ−1 mj χi − 1).
∗ Then f = χm1 µ1 + · · · + χmr µr , and we have χ−1 mj χi ∈ XH (G) if mj−1 + 1 ∗ (G). Since A(G) i mj . Thus the ideal I(H) is generated by χ − 1, χ ∈ XH ∗ is Noetherian, there exist χ1 , . . . , χr ∈ XH (G) such that
I(H) = A(G)(χ1 − 1) + · · · + A(G)(χr − 1). So the result follows. (iii) By 22.5.3, there exist λ1 , . . . , λn ∈ k \ {0} and χ1 , . . . , χn ∈ X ∗ (G) such that θ = λ1 (χ1 |H ) + · · · + λn (χn |H ). In view of 22.4.1, χi |H = θ for all i, and the sum λ1 + · · · + λn = 1. 22.5.5 Lemma. (i) The only closed connected subgroups of Gm are {1} and Gm . (ii) If G is a connected algebraic group, then X ∗ (G) is torsion free. Proof. (i) Let H be a closed connected subgroup of Gm . Since dim H dim Gm = 1 (14.1.6), we have dim H = 0 or 1. But H is connected, so H is either {1} or Gm . (ii) Let χ ∈ X ∗ (G). Then χ(G) is a closed connected subgroup of Gm (21.2.4). By (i), χ(G) = {1} or Gm . In the first case, χ = 1. In the second case, χn = 1 for all n.
22.5 Tori and diagonalizable groups
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22.5.6 The group algebra k[Zn ] of the abelian group Zn is isomorphic to k[T1 , T1−1 , . . . , Tn , Tn−1 ]. It follows from 22.5.3 that if G is diagonalizable, then the group algebra k[X ∗ (G)] is isomorphic to A(G). 22.5.7 Theorem. Let n ∈ N∗ and G an affine algebraic group. The following conditions are equivalent: (i) As an algebraic group, G is isomorphic to Dn (k). (ii) G is a torus of dimension n. (iii) G is a connected diagonalizable algebraic group of dimension n. (iv) G is diagonalizable and X ∗ (G) is isomorphic to Zn . Proof. Clearly, we have (i) ⇒ (ii) ⇒ (iii). (iii) ⇒ (iv) By 22.5.3 and 22.5.5, X ∗ (G) is a finitely generated torsion free abelian group. So it is of the form Zr for some r ∈ N. In view of 22.5.6, A(G) is isomorphic to k[T1 , T1−1 , . . . , Tr , Tr−1 ]. Thus r = n by 6.1.4, 6.2.3 and 14.1.2. (iv) ⇒ (i) Let χ1 , . . . , χn be a basis of the free abelian group X ∗ (G). The map u : G → Dn (k) , α → diag χ1 (α), . . . , χn (α) is a morphism of algebraic groups. Since that A(G) = k[T1 , T1−1 , . . . , Tn , Tn−1 ] by 22.5.6, A(u) is bijective. Thus u is an isomorphism of algebraic groups (11.4.5). 22.5.8 Lemma. Let A be a commutative ring and f : E → F a surjective homomorphism of A-modules. Suppose that F is a free A-module. Then there is a submodule E of E such that (i) E = E ⊕ ker f . (ii) f induces an isomorphism from E onto F . Proof. Let (yi )i∈I be a basis of F . For i ∈ I, fix xi ∈ E such that f (xi ) = yi . Then (xi )i∈I is a linearly independent subset. So the submodule E = i∈I Axi satisfies the required properties. 22.5.9 Theorem. Let G be diagonalizable. Then G◦ is a torus, and there exists a finite subgroup H (not unique in general) of G such that G is, isomorphic as an algebraic group, to the product H × G◦ . Proof. That G◦ is a torus follows from 22.5.4 and 22.5.7. Let π : G → Dn (k) be a morphism of algebraic groups inducing an isomorphism from G to a closed subgroup of Dn (k). By 22.5.4 (iii), restriction to H induces a surjective homomorphism from Zn X ∗ (Dn (k)) to X ∗ (G◦ ). By 22.5.5, X ∗ (G◦ ) is a free Z-module. Applying 22.5.8 and using the fact that Z is a principal ideal domain (so any submodule of a free module is free), we obtain a basis χ1 , . . . , χn of X ∗ (Dn (k)) such that χ1 , . . . , χr generate the subgroup consisting of characters χ verifying χ|G◦ = 1. The automorphism α → diag(χ1 (α), . . . , χn (α))
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of Dn (k) then sends π(G◦ ) to the set of matrices diag(µ1 , . . . , µn ) verifying µ1 = · · · = µr = 1. Thus Dn (k) = Dr (k) × π(G◦ ). So as abstract groups, π(G) = H × π(G◦ ) where H = π(G) ∩ Dr (k). Since H is isomorphic to G/G◦ , it is finite (21.1.6). Finally π induces an isomorphism of algebraic groups from G onto H × π(G◦ ) since the restriction of π to each irreducible component of G is an isomorphism. 22.5.10 The following result is called the rigidity of diagonalizable groups. Proposition. Let G, H be diagonalizable algebraic groups, X an irreducible affine variety and u : X × G → H a morphism of varieties. Assume that for all x ∈ V , the map vx : G → H , α → u(x, α), is a morphism of algebraic groups. Then u(x, α) does not depend on x. Proof. Let (θ, x) ∈ X ∗ (H) × X. By our hypotheses, the map α → θ(u(x, α)) is a character of G and the map (x, α) → θ(u(x, α)) is a regular function on X × G. By 22.5.3, we have: fθ,χ (x)χ(α), θ u(x, α) = χ∈X ∗ (G)
where fθ,χ ∈ A(X) for all χ ∈ X ∗ (G). It follows from 22.4.1 that given x ∈ X, there exists χθ,x ∈ X ∗ (G) such that θ(u(x, α)) = χθ,x (α) for all α ∈ G. So fθ,χθ,x (x) = 1 and fθ,χ (x) = 0 if χ ∈ X ∗ (G) \ {χθ,x }. Thus fθ,χ (x) ∈ {0, 1}. For ε ∈ {0, 1}, set Xε = {x ∈ X; fθ,χ (x) = ε}. Since X is the disjoint union of the closed subsets X0 and X1 , we deduce that χθ,x depends only on θ. But 11.3.3 and 22.5.3 say that the elements of X ∗ (H) separate the points of H, so the result follows. 22.5.11 Corollary. Let ◦ H be a closed ◦ diagonalizable subgroup of G. (i) We have NG (H) = CG (H) . (ii) The index of CG (H) in NG (H) is finite. Proof. The subgroups NG (H) and CG (H) are closed in G by 21.3.7. Applying 22.5.10 to the following map ◦ u : NG (H) × H → H , (β, α) → βαβ −1 , ◦ we obtain that u(β, α) = u(e, α) = α for all α ∈ H and β ∈ NG (H) . So we have proved (i). Finally, (ii) follows from (i) and 21.1.6 (i).
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22.6 Groups of dimension one 22.6.1 Lemma. If G is connected and dim G = 1, then G is commutative. Proof. Let us fix β ∈ G and denote by u : G → G the morphism α → αβα−1 . Since u(G) is irreducible (1.1.5 and 1.1.7) and dim G = 1, either u(G) = {β} or u(G) = G. Suppose that u(G) = G. Then u(G) contains a non-empty open subset of G (15.4.3 and 1.4.6). Thus G \ u(G) is a finite set (14.1.3, 14.1.5 and 14.1.6). We may assume that G is a closed subgroup of GLn (k) (22.1.5). For α ∈ G, denote by χα its characteristic polynomial. There exist f0 , . . . , fn−1 ∈ A(G) such that χα (T ) = T n + fn−1 (α)T n−1 + · · · + f0 (α). Since χβ = χαβα−1 , the morphism G → kn , α → (f0 (α), . . . , fn−1 (α)), is constant on u(G) and hence on G. Thus χα (T ) = χe (T ) = (T − 1)n for all α ∈ G, and G is unipotent. Now u(G)β −1 ⊂ D(G) and so G \ D(G) is finite. It follows that D(G) = G, so D(G) = G (21.3.6). But this contradicts the fact that G is unipotent (10.8.10 and 22.3.2). So αβα−1 = {β} for all α, β ∈ G and G is commutative. 22.6.2 Theorem. Let G be a connected algebraic group of dimension 1. (i) If G is not unipotent, then G is isomorphic to Gm as algebraic groups. (ii) If G is unipotent, then G is isomorphic to Ga as algebraic groups. Proof. We may assume that G is a closed subgroup of GLn (k). By 22.6.1, G is commutative. So we have G = Gs × Gu . Since dim G = 1 and G is connected, we have either G = Gs or G = Gu . If G = Gs , then since G is commutative, G is diagonalizable (22.2.4). Therefore G is isomorphic to Gm by 22.5.7. If G = Gu , then we may identify G with a closed subgroup of Un (k) (22.3.2). Let α ∈ G \ {e}, then the image of the additive one-parameter subgroup ϕlog(α) (Ga ) is a closed subgroup of G isomorphic to Ga (22.3.4). But dim G = 1, so G is isomorphic to Ga .
References and comments • [5], [23], [38], [40], [78]. Examples of non-affine algebraic groups are abelian varieties. The reader may refer to [53] and [58] for a detailed account of the subject.
23 Lie algebra of an algebraic group
This chapter explains how to associate a Lie algebra to an algebraic group, and studies some basic properties of this association.
23.1 An associative algebra 23.1.1 In this chapter, G will denote an algebraic group with identity element eG or e, µG or µ the group multiplication and ιG or ι the inverse. For α ∈ G, define χα : A(G) → k, χα (f ) = f (α) to be the evaluation at α. It is clear that χα is an element of the dual A(G)∗ of A(G). 23.1.2 If x, y ∈ A(G)∗ , then x ⊗ y ∈ (A(G) ⊗k A(G))∗ is the element defined as follows: (x ⊗ y)(f ⊗ g) = x(f )y(g) for all f, g ∈ A(G). Denote by x · y = (x ⊗ y) ◦ A(µ). So x · y ∈ A(G)∗ . Lemma. Endowed with the operation (x, y) → x · y, A(G)∗ is an associative k-algebra whose identity element is χe . The map G → A(G)∗ , α → χα , induces an injective homomorphism of G into the group of invertible elements of A(G)∗ . Proof. Let f ∈ A(G) and A(µ)(f ) = s1 ⊗t1 +· · ·+sn ⊗tn where s1 , t1 , . . . , sn , tn ∈ A(G). For α ∈ G, we have: f (α) = f (eα) =
n
si (e)ti (α) = f (αe) =
i=1
n
si (α)ti (e).
i=1
Thus f = s1 (e)t1 + · · · + sn (e)tn = t1 (e)s1 + · · · + tn (e)sn . If x ∈ A(G)∗ , then: (χe · x)(f ) = (χe ⊗ x)◦A(µ)(f ) =
n
n si (e)x(ti ) = x si (e)ti = x(f ).
i=1
i=1
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It follows that χe · x = x. Similarly, we have x · χe = x. Consequently, we have χα · χβ = χαβ for all α, β ∈ G and the map α → χα is injective. The associativity of the group multiplication implies that: idA(G) ⊗A(µ) ◦A(µ) = A(µ) ⊗ idA(G) ◦A(µ). So for all x, y, z ∈ A(G)∗ : (x ⊗ y ⊗ z)◦I ⊗ A(µ)◦A(µ) = x ⊗ (y ⊗ z)◦A(µ) ◦A(µ) = x · (y · z), (x ⊗ y ⊗ z)◦ A(µ) ⊗ I ◦A(µ) = (x ⊗ y)◦A(µ) ⊗ z ◦A(µ) = (x · y) · z, where I = idA(G) . We have therefore proved the result. 23.1.3 Lemma. Let u : G → H be a morphism of algebraic groups and A (u) the map dual to A(u). Then A (u) ∈ Homalg (A(G)∗ , A(H)∗ ), and A (u)(χα ) = χu(α) for all α ∈ G. Proof. Since u is a morphism of algebraic groups, we have: A(µG )◦A(u) = A(u) ⊗ A(u) ◦A(µH ). So for x, y ∈ A(G)∗ : (x · y)◦A(u) = (x ⊗ y)◦A(µ G )◦A(u) = (x ⊗ y)◦ A(u) ⊗ A(u) ◦A(µH ) = x◦A(u) · y◦A(u) . Thus A (u) is a homomorphism of algebras and the second part follows immediately.
23.2 Lie algebras 23.2.1 For α ∈ G, we shall denote the left and right translation of functions by α by λα and ρα as in 22.1.4. Since αG◦ is the unique irreducible component of G containing α, we have dim G = dim G◦ . It follows from 16.3.7 and the smoothness of points in G that Te (G◦ ) can be identified with Te (G), which in turn, can be identified with Derek (A(G), k) (16.2.4). Let Lie(G) be the set of derivations X of A(G) verifying: X◦λα = λα ◦X for all α ∈ G. Such a derivation is said to be left invariant and Lie(G) is a Lie subalgebra of Derk (A(G)). Let X ∈ Lie(G). For f ∈ A(G), set: θ(X)(f ) = X(f )(e) = χe X(f ) . Clearly θ(X) ∈ Derek (A(G), k) and so we have a linear map:
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θ: Lie(G) → Derek (A(G), k) = Te (G) , X → θ(X). We shall prove that θ is bijective. 23.2.2 Let f ∈ A(G) and x ∈ Derek (A(G), k). Define a function f ∗ x on G as follows: for α ∈ G, set (f ∗ x)(α) = x(λα−1 f ). This is called the right convolution of f by x. Let A(µ)(f ) = s1 ⊗ t1 + · · · + sn ⊗ tn where s1 , t1 , . . . , sn , tn ∈ A(G). For α, β ∈ G, we have A(µ)(f )(α, β) = f (αβ) = s1 (α)t1 (β) + · · · + sn (α)tn (β). So λα−1 f = s1 (α)t1 + · · · + sn (α)tn . Since (f ∗ x)(α) = x(λα−1 f ), it follows that: f ∗ x = x(t1 )s1 + · · · + x(tn )sn . Hence f ∗ x ∈ A(G). So right convolution by x is an endomorphism of A(G). We claim that it is a derivation of A(G). Let f, g ∈ A(G), x ∈ Derek (A(G), k) and α ∈ G. Then: (f g ∗ x)(α) = x λα−1 f λα−1 g = x(λα−1 f )(λα−1 g)(e) + x(λα−1 g)(λα−1 f )(e) = x(λα−1 f )g(α) + x(λα−1 g)f (α) = (f ∗ x)(α)g(α) + (g ∗ x)(α)f (α). Hence (f g) ∗ x = (f ∗ x)g + (g ∗ x)f and we have proved our claim. For β ∈ G. we have: λβ (f ∗ x) (α) = (f ∗ x)(β −1 α) = x(λ α−1 β f ) = x λα−1 (λβ f ) = (λβ f ) ∗ x (α). So λβ (f ∗ x) = (λβ f ) ∗ x and the derivation f → f ∗ x is left invariant. It follows that x → ∗x defines a map η : Derek (A(G), k) → Lie(G). We claim that η is the inverse of θ. If X ∈ Lie(G), f ∈ A(G) and α ∈ G, then: (η◦θ(X)(f ))(α) = (f ∗ θ(X))(α) =θ(X)(λα−1f ) = X(λα−1 f )(e) = λα−1 (Xf ) (e) = X(f )(α). So η◦θ(X) = X. Similarly, if x ∈ Derek (A(G), k), then θ◦η(x) (f ) = (f ∗ x)(e) = x(λe−1 f ) = x(f ). Hence θ◦η(x) = x. So our claim follows and we have obtained the following result.
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Proposition. The linear map θ : Lie(G) → Derek (A(G), k) = Te (G) is an isomorphism. In particular, Lie(G) is a Lie algebra over k of dimension dim G. 23.2.3 We can transport the Lie algebra structure on Lie(G) to Te (G) via the isomorphism θ, and the Lie algebra thus obtained, denoted by L(G), shall be called the Lie algebra of G. This Lie algebra structure can be recovered from the structure of associative algebra on A(G)∗ defined in 23.1. Let x, y ∈ L(G) = Te (G), f ∈ A(G) and A(µ)(f ) = s1 ⊗ t1 + · · · + sn ⊗ tn where s1 , t1 , . . . , sn , tn ∈ A(G). Then: (f ∗ x) ∗ y (e) = x(t1 )(s1 ∗ y)(e) + · · · + x(tn )(sn ∗ y)(e) = y(s1 )x(t1 ) + · · · + y(sn )x(tn ) = (y · x)(f ). Hence θ [η(x), η(y)] = x · y − y · x. and the Lie bracket on L(G) is given by: [x, y] = x · y − y · x. 23.2.4 Let u : G → H be a morphism of algebraic groups and denote by du : L(G) → L(H) the differential due (or Te (u)) of u at e (16.3.1). Proposition. The map du is a Lie algebra homomorphism. Proof. Let x, y, z ∈ L(G). Since du(z) = z◦A(u) (16.3.3) and we saw in the proof of 23.1.3 that: (x · y − y · x)◦A(u) = x◦A(u) · y◦A(u) − y◦A(u) · x◦A(u) , the result follows from 23.2.3. 23.2.5 Let H be a closed subgroup of G. By 16.3.7 and 23.2.4, we may identify L(H) with a Lie subalgebra of L(G). More precisely: Lemma. Let a = I(H). Then: L(H) = {x ∈ L(G); a ∗ x ⊂ a} = {x ∈ L(G); x(a) = {0}}. Proof. Let j : H → G be the canonical injection, then A(j) is just the restriction of an element of A(G) to H and if x ∈ L(H), then dj(x) = x◦A(j). Hence dj(x)(f ) = 0 for all f ∈ a. Let f ∈ a, x ∈ L(H) and α ∈ H. Since λα−1 f ∈ a (22.1.6), we have f ∗ dj(x) (α) = dj(x)(λα−1 f ) = 0. Thus f ∗ dj(x) ∈ a Conversely, let x ∈ L(G) be such that a ∗ x ⊂ a. If f ∈ a, then (f ∗ x)(e) = 0 = x(f ). We deduce that x induces an element y of Derek (A(G)/a, k) = Derek (A(H), k), because A(H) = A(G)/a. Clearly, x = dj(y). Finally if f ∈ a and x ∈ L(G) verify f ∗ x ∈ a, then x(f ) = (f ∗ x)(e) = 0. Conversely, if x(g) = 0 for all g ∈ a, then, since λα−1 g ∈ a for α ∈ H, we have (g ∗ x)(α) = x(λα−1 g) = 0, and so g ∗ x ∈ a.
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23.2.6 A derivation X ∈ Derk (A(G), k) is said to be right invariant if X ◦ ρα = ρα ◦ X for all α ∈ G. We shall denote the set of right invariant derivation of A(G) by Lie (G). It is a Lie subalgebra of Derk (A(G)). As in the case of left invariant derivations, we have a linear map θ : Lie (G) → Te (G) defined as follows: for X ∈ Lie (G) and f ∈ A(G), set θ (X)(f ) = X(f )(e) = χe (X(f )). Also for f ∈ A(G) and x ∈ Te (G), we have the notion of left convolution x ∗ f of f by x defined as follows: for α ∈ G, (x ∗ f )(α) = x(ρα f ). Using the same arguments as in 23.2.2, right convolution defines a map η which is the inverse of θ . Thus we have, via θ , another Lie algebra structure on Te (G) that we shall denote by L (G). Then we see that the Lie bracket on L (G) is given by: [x, y] = y · x − x · y. So L (G) is the opposite Lie algebra of L(G). The analogue of 23.2.5 for L (H) (with left convolution) remains valid. 23.2.7 Let α ∈ G, x ∈ L(G), and f, s1 , t1 , . . . , sn , tn ∈ A(G) be such that A(µ)(f ) = s1 ⊗ t1 + · · · + sn ⊗ tn . Since (χα · x)(f ) = s1 (α)x(t1 ) + · · · + sn (α)x(tn ), it follows from 23.2.2 that: (χα · x)(f ) = (f ∗ x)(α). Similarly, we have: (x · χα )(f ) = (x ∗ f )(α). Let us evaluate χα · x · χα−1 . For 1 i n, let A(µ)(ti ) = ui1 ⊗ vi1 + · · · + uip ⊗vip where uij , vij ∈ A(G). Then ρα−1 ti = vi1 (α−1 )ui1 +· · ·+vip (α−1 )uip . Thus x(ρα−1 ti ) = x(ui1 )vi1 (α−1 ) + · · · + x(uip )vip (α−1 ) and we deduce that: ' ( (χα · x · χα−1 )(f ) = χα ⊗ (x ⊗ χα−1 )◦A(µ) ◦A(µ)(f ) p n n si (α) x(uij )vij (α−1 ) = si (α)x(ρα−1 ti ). = i=1
j=1
i=1
On the other hand, let iα : G → G, β → αβα−1 . Then: (f ◦iα )(β) = f (αβα−1 ) = s1 (α)t1 (βα−1 ) + · · · + sn (α)tn (βα−1 ). Hence f ◦iα = s1 (α)(ρα−1 t1 ) + · · · + sn (α)(ρα−1 tn ). So we have: (χα · x · χα−1 )(f ) = x(f ◦iα ).
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23.3 Examples 23.3.1 Let G = Ga and A(G) = k[T ]. The Lie algebras Lie(G) and L(G) are 1-dimensional, so they are abelian. Let X ∈ Derk (A(G)) be the derivation defined by X(P ) = P (T ) for all P ∈ k[T ]. If α ∈ G, then λα ◦ X = X ◦ λα . So X spans Lie(G). It follows that L(G) is spanned by x, where x(P ) = P (0) for P ∈ k[T ]. 23.3.2 Let G = Gm and A(G) = k[T, T −1 ]. Again Lie(G) and L(G) are 1-dimensional. The unique derivation X of A(G) verifying X(T ) = T is left invariant, and therefore it spans Lie(G). It follows that L(G) is spanned by x ∈ Derek (A(G), k) where x(P ) = P (1) for P ∈ A(G). 23.3.3 Let G = GLn (k) and denote by I the identity matrix. Since GLn (k) is an affine open subset of Mn (k), TI (GLn (k)) may be identified with TI (Mn (k)) (16.2.4), which is Mn (k). Let A(G) = k[T11 , T12 , . . . , Tnn , D−1 ] where D is the polynomial in Tij such that D(M ) = det M for all M ∈ Mn (k). It follows that an element x ∈ L(G) is completely determined by its values on Tij . Denote by ϕ : L(G) → gln (k) the map given by x → [x(Tij )]. It is clear that ϕ is linear and injective. Since L(G) and gln (k) are both of dimension n2 , ϕ is bijective. Let x, y ∈ L(G), X = [xij ] = ϕ(x) and Y = [yij ] = ϕ(y). The formulas for matrix multiplication imply that A(µ)(Tij ) = Ti1 ⊗ T1j + · · · + Tin ⊗ Tnj for 1 i, j n. Thus (x · y)(Tij ) = xi1 y1j + · · · + xin ynj . It follows from 23.2.3 that [x, y] = ϕ−1 ([X, Y ]). So we have proved that ϕ is a Lie algebra isomorphism and we may identify L(G) with gln (k) via ϕ. Now let V be a finite-dimensional k-vector space, H = GL(V ) and h = L(H). Then the same applies for H. In fact, the map End(V )∗ → A(H), ψ → ψ|H , is injective since H is dense in End(V ). So any x ∈ h identifies with a linear form on End(V )∗ . It follows that given x ∈ h, there is a unique X ∈ End(V ) such that x(g) = g(X) for all g ∈ End(V )∗ . This defines a linear map ϕ : h → gl(V ) = End(V ) which is injective since A(H) is generated by End(V )∗ and 1/ det. They have the same dimension, and by using the same arguments as above, it is clear that ϕ is a Lie algebra isomorphism. So we may identify h with gl(V ) via ϕ. 23.3.4 Let H be a closed subgroup of G = GLn (k) and a (resp. b) the ideal of H in A(G) (resp. A(Mn (k)) = k[T11 , T12 , . . . , Tnn ]). Then clearly a is the ideal of A(G) generated by b.
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Take H = Tn (k) the set of upper triangular matrices in G. Then a is generated by Tij , i > j. If x ∈ L(G) and X = ϕ(x) = [xij ] as in 23.3.3, then by 23.2.2, we have, for 1 i, j n, Tij ∗ x = x1j Ti1 + x2j Ti2 + · · · + xnj Tin . Thus if i > j, then Tij ∗ x ∈ a if and only if xkj = 0 for k i. By 23.2.5, we see that L(Tn (k)) identifies with tn (k), the set of upper triangular matrices in Mn (k). The same arguments show that the Lie algebra of Dn (k) is dn (k), and the Lie algebra of Un (k) is nn (k) (in the notations of 10.8.1 and 19.1.1). 23.3.5 Let G = GLn (k) and A = [aij ] ∈ Mn (k) a nilpotent matrix. Recall from 22.3.3 that the image H of the map u : Ga → G, t → etA , is a connected closed subgroup of G and u induces an isomorphism from Ga onto H. We saw in 23.3.1 that L(Ga ) is spanned by x, where x(P ) = P (0) for P ∈ k[T ]. We deduce that L(H) is spanned by du(x). We have du(x)(Tij ) = x(Tij ◦ u) = aij , so du(x) = ϕ−1 (A) where ϕ is as in 23.3.3. Identifying L(G) with gln (k), we see that L(H) = kA. 23.3.6 Let us use the notations of 23.3.3. We may consider the algebra A = A(Mn (k)) = k[T11 , T12 , . . . , Tnn ] as a subalgebra of A(G). If X ∈ Mn (k) and f ∈ A(G), we denote f ∗ ϕ−1 (X) by f ∗ X (there is no confusion since ϕ is a Lie algebra isomorphism). The map f → f ∗ X is a derivation of A(G) and if X = [xij ], then Tij ∗ X = x1j Ti1 + · · · + xnj Tin . We deduce that f ∗ X ∈ A if f ∈ A. In the same way, if f ∈ A and α ∈ G, then ρα f and λα f belong to A. Proposition. Let H be a closed subgroup of G = GLn (k) and b the ideal of H in A. Then: H = {α ∈ G; ρα (b) = b} = {α ∈ G; λα (b) = b}, L(H) = {X ∈ gln (k); b ∗ X ⊂ b}. Proof. Let S = {Dn ; n ∈ N} and a the ideal of H in A(G). Then a = S −1 b (2.3.7). So a = A(G)b, b = A ∩ a. Let α ∈ G and X ∈ gln (k). For f ∈ A(G) and g ∈ A, we have ρα (f g) = (ρα f )(ρα g) , (f g) ∗ X = (f ∗ X)g + f (g ∗ X). It follows that if ρα (b) ⊂ b and b ∗ X ⊂ b, then ρα (a) ⊂ a and a ∗ X ⊂ a. So 22.1.6 and 23.2.5 imply that α ∈ H and X ∈ L(H). Conversely, given α ∈ H and X ∈ L(H). Since a ∗ X ⊂ a (23.2.5) and A ∗ X ⊂ A, we have b ∗ X ⊂ b. Similarly, we have ρα (a) ⊂ a and ρα−1 (a) ⊂ a, so ρα (a) = a. But ρα (A) = A, we deduce therefore that ρα (b) = b.
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23.4 Computing differentials 23.4.1 Let G be an algebraic group and u : G → X a morphism of varieties. When there is no confusion, we shall denote due by du. Recall that we may identify L(GLn (k)) with gln (k) via the map x → [x(Tij )] (23.3.3). 23.4.2 Lemma. Let u : G → GLn (k) be a morphism of algebraic groups and uij ∈ A(G), 1 i, j n, be such that u(α) = [uij (α)] for all α ∈ G. If x ∈ L(G), then du(x) = [x(uij )]. Proof. This is clear since du(x)(Tij ) = x(Tij ◦ u) = x(uij ).
23.4.3 Let V be a finite-dimensional subspace of A(G) such that for all α ∈ G, ρα (V ) ⊂ V , and ρ : G → GL(V ) the morphism of algebraic groups given by α → ρα |V . Proposition. For x ∈ L(G) and f ∈ V , we have: dρ(x)(f ) = f ∗ x. Proof. Let B = (f1 , . . . , fn ) be a basis of V . If α ∈ G and 1 j n, then ρα fj = ρ1j (α)f1 +ρ2j (α)f2 +· · ·+ρnj (α)fn , where ρij ∈ A(G) for 1 i, j n. By 23.4.2, if x ∈ L(G), then dρ(x)(fj ) = x(ρ1j )f1 + · · · + x(ρnj )fn . On the other hand, we have: fj (βα) = (λβ −1 fj )(α) = (ρα fj )(β) = ρ1j (α)f1 (β) + · · · + ρnj (α)fn (β) for α, β ∈ G. Hence: (fj ∗ x)(β) = x(λβ −1 fj ) = x(ρ1j )f1 (β) + · · · + x(ρnj )fn (β) = dρ(x)(fj )(β), and we have proved the required result. 23.4.4 Corollary. Let V be a finite-dimensional subspace of A(G) verifying ρα (V ) ⊂ V for all α ∈ G. Then V ∗ x ⊂ V for all x ∈ L(G). 23.4.5 In relation with 23.4.3, we give a result concerning the determinant. Let n ∈ N∗ , Sn the symmetric group of {1, . . . , n} and if σ ∈ Sn , denote by ε(σ) the signature of σ. If α ∈ G = GLn (k), denote by D(α) the determinant of α. As an element of A(G), we have: D= ε(σ)T1σ(1) · · · Tnσ(n) . σ∈Sn
For α, β ∈ G and x = [x(Tij )] ∈ L(G) = gln (k), we have λα−1 D = D(α)D because (λα−1 D)(β) = D(αβ) = D(α)D(β). Since x ∈ L(G) = Derek (A(G), k), we see that x(λα−1 D) is equal to D(α)
σ∈Sn
ε(σ)
n i=1
δ1σ(1) · · · δi−1,σ(i−1) x(Tiσ(i) )δi+1,σ(i+1) · · · δn,σ(n) ,
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355
where δij denotes the symbol of Kronecker. Thus (D ∗ x)(α) = D(α) tr(x). Consequently: D ∗ x = tr(x)D. Now H = SLn (k) = {X ∈ Mn (k); D(X) = 1} is a connected closed subgroup of G of dimension n2 − 1 (21.3.3 and 14.2.3). Denote by b the ideal of H in A(Mn (k)). Then D − 1 ∈ b. If x ∈ L(H), then (D − 1) ∗ x = tr(x)D ∈ b (23.3.6). Since b = A(Mn (k)), it follows that tr(x) = 0. Thus L(H) ⊂ sln (k). Since dim L(H) = dim sln (k) = n2 − 1, we deduce that L(H) = sln (k). 23.4.6 Let G, H be algebraic groups, x ∈ L(G) and y ∈ L(H). For f ∈ A(G) and g ∈ A(H), set: θx,y (f ⊗ g) = x(f )g(eH ) + f (eG )y(g). We saw in 16.3.8 that the map θ : L(G) × L(H) → L(G × H), (x, y) → θx,y , is a linear bijection. We claim that it is a Lie algebra isomorphism. Let x, x ∈ L(G), y, y ∈ L(H), z = θx,y , z = θx ,y . Then it is easy to check that: (f ⊗ g) ∗ z = (f ∗ x) ⊗ g + f ⊗ (g ∗ y). It follows that (f ⊗ g) ∗ z ∗ z is equal to: (f ∗ x) ∗ x ⊗ g + (f ∗ x) ⊗ (g ∗ y ) + (f ∗ x ) ⊗ (g ∗ y) + f ⊗ (g ∗ y) ∗ y . This implies that (f ⊗ g) ∗ z ∗ z − (f ⊗ g) ∗ z ∗ z is equal to: (f ∗ x) ∗ x − (f ∗ x ) ∗ x ⊗ g + f ⊗ (g ∗ y) ∗ y − (g ∗ y ) ∗ y . Hence we have proved our claim (23.2.3). We shall identify the Lie algebras L(G) × L(H) and L(G × H) via θ. 23.4.7 Proposition. Let x, y ∈ L(G), then: dµ(x, y) = x + y , dι(x) = −x. Proof. Let f ∈ A(G) and A(µ)(f ) = s1 ⊗t1 +· · ·+sn ⊗tn where si , ti ∈ A(G) pour 1 i n. Set z = θx,y (23.4.6). Then: dµ(z)(f ) = z
n i=1
n x(si )ti (e) + si (e)y(ti ) . si ⊗ ti = i=1
On the other hand, we saw in the proof of 23.1.2 that: f = s1 (e)t1 + · · · + sn (e)tn = t1 (e)s1 + · · · + tn (e)sn . So it is clear that dµ(z)(f ) = x(f ) + y(f ). Next, let π : G → G × G denote themorphism α → α, ι(α) . Then µ◦π(α) = e, so dµ◦dπ = 0. But dπ(x) = x, dι(x) , so we obtain from the formula for dµ that x + dι(x) = 0.
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23.4.8 Corollary. Let f, g ∈ A(G) be such that g(α) = f (α−1 ) for all α ∈ G. If x ∈ L(G), then x(g) = −x(f ). Proof. Note that g = f ◦ ι and f, g are morphisms from G to Ga . So dg = df ◦dι = −df (23.4.7). But df (x)(h) = x(h◦f ) and dg(x)(h) = x(h◦g) for x ∈ L(G) and h ∈ k[T ] = A(Ga ). So the result follows by taking h = T . 23.4.9 Corollary. Let (V, π) be a finite-dimensional rational representation of G, (V ∗ , σ) its contragredient representation (which is clearly rational), x ∈ L(G), f ∈ V ∗ , and v ∈ V . Then: (dσ(x)f (v) = −f dπ(x)v . Proof. Let (v1 , . . . , vn ) be a basis of V and (ϕ1 , . . . , ϕn ) its dual basis. We have n n πij (α)vi , σ(α)ϕj = σij (α)ϕi , π(α)vj = i=1
i=1
where πij , σij ∈ A(G). Then we check easily that σij (α) = πji (α−1 ). It follows from 23.4.2 and 23.4.8 that for x ∈ L(G), we have: dσ(x)(ϕj ) = −x(πj1 )ϕ1 − · · · − x(πjn )ϕn . Hence dσ(x)(ϕj )(vi ) = −x(πji ) = −ϕj dπ(x)vi . 23.4.10 Proposition. Let u : G → G be a morphism of varieties such that u(e) = e, and set v : G → G, α → u(α)α−1 . Then dv = du − idL(G) . Proof. Let θ : G → G × G be the morphism α → (u(α), α−1 ). Then v = µ ◦ θ. If x ∈ L(G), then by 23.4.7, we have dv(x) = dµ(du(x), −x) = du(x) − x. 23.4.11 Remark. Let V be a finite-dimensional subspace of A(G) verifying λα (V ) ⊂ V for all α ∈ G. Denote by λ : G → GL(V ) the morphism of algebraic groups α → λα |V . Using the same computations as in 23.4.3 and the result obtained in 23.4.8, we obtain that for x ∈ L(G) and f ∈ V , dλ(x)(f ) = −x ∗ f.
23.4.12 Let π : G → GL(U ) and τ : H → GL(V ) be finite-dimensional rational representations of the algebraic groups G and H, and S = U ⊕ V , T = U ⊗k V . We define the rational representations (S, π ⊕ τ ) and (T, π ⊗ τ ) of G × H as follows: for α ∈ G, β ∈ H, u ∈ U and v ∈ V , (π ⊕ τ )(α, β)(u, v) = π(α)(u), τ (β)(v) , (π ⊗ τ )(α, β)(u ⊗ v) = π(α)(u) ⊗ τ (β)(v).
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Proposition. Let x ∈ L(G), y ∈ L(H), u ∈ U and v ∈ V . Then: d(π ⊕ τ )(x, y)(u, v) dτ (y)(v) = dπ(x)(u), , d(π ⊗ τ )(x, y)(u ⊗ v) = dπ(x)(u) ⊗ v + u ⊗ dτ (y)(v) . Proof. Let (e1 , . . . , em ) and (f1 , . . . , fn ) be bases of U and V . Then π(α)ej =
m
πij (α)ei , τ (β)fl =
i=1
n
τkl (β)fk ,
k=1
where α ∈ G, β ∈ H, πij ∈ A(G) and τkl ∈ A(H). Let I = {1, . . . , m} × {1, . . . , n} and for (j, l) ∈ I, set g(j,l) = ej ⊗ fl . The g(j,l) ’s, for (j, l) ∈ I, form a basis of U ⊗k V . We have (π ⊗ τ )(α, β)g(j,l) =
(i,k)∈I
(j,l)
(j,l)
θ(i,k) (α, β)g(i,k) , (j,l)
where θ(i,k) ∈ A(G × H) = A(G) ⊗k A(H). So θ(i,k) = πij ⊗ τkl . Let z = (x, y) ∈ L(G × H). By 23.4.6, we have: (j,l)
z(θ(i,k) ) = πij (eG )y(τkl ) + x(πij )τkl (eH ). Hence we obtain the second equality (23.4.2). The proof for the first equality is analogue. 23.4.13 Let us conserve the notations of 23.4.12 with H = G. We shall also denote by (π ⊗ σ, U ⊗k V ) the rational representation of G defined by: (π ⊗ τ )(α)(u ⊗ v) = π(α)u ⊗ τ (α)v . It follows from the preceding discussion that for x ∈ L(G): d(π ⊗ τ )(x) = dπ(x) ⊗ idV + idU ⊗dτ (x). 23.4.14 Let (V, π) be a finite-dimensional rational representation of G. n (π) of G. We define, as in 10.6.5, the representations Tn (π), Sn (π) and They are also rational and we have for x ∈ L(G), v1 , . . . , vn ∈ V : d Tn (π)(x)(v1 ⊗ · · · ⊗ vn ) =
n
v1 ⊗ · · · ⊗ vi−1 ⊗ dπ(x)vi ⊗ vi+1 ⊗ · · · ⊗ vn ,
i=1
d Sn (π)(x)(v1 · · · vn ) = d
n
(π)(x)(v1 ∧ · · · ∧ vn ) =
n
n
v1 · · · vi−1 dπ(x)vi vi+1 · · · vn ,
i=1
v1 ∧ · · · ∧ vi−1 ∧ dπ(x)vi ∧ vi+1 ∧ · · · ∧ vn .
i=1
23.4.15 Let A be a finite-dimensional k-algebra (not necessarily associative or with unit). A k-linear map δ : A → A is called a derivation of A if
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for all x, y ∈ A, δ(x.y) = δ(x).y + x.δ(y). Let Der A (resp. Aut A) denote the set of derivations (resp automorphisms) of A. By 21.1.3, Aut A is an algebraic group. Proposition. Let π : G → Aut A be a rational representation of G. Then for x ∈ L(G), dπ(x) ∈ Der A. Proof. Let (e1 , . . . , en ) be a basis of A. There exist γijk ∈ k and πij ∈ A(G) such that: n n γijk ek , π(α)ej = πij (α)ei . ei .ej = i=1
k=1
Since π(α)(ei .ej ) = π(α)ei . π(α)ej ), we have:
n
γijk πlk (α) =
n
γrsl πri (α)πsj (α).
r,s=1
k=1
Denote by δuv the symbol of Kronecker, then: n
γijk x(πlk ) =
k=1
=
γrsl δri x(πsj ) + x(πri )δsj
n r,s=1 n
n
s=1
r=1
γisl x(πsj ) +
γrjl x(πri ).
On the other hand, 23.4.2 implies that: dπ(x)(ei .ej ) =
n n l=1
γijk x(πlk ) el .
k=1
Similarly, dπ(x)ei .ej + ei . dπ(x)ej is equal to: n n l=1
n n γrjl x(πri ) el + γisl x(πsj ) el .
r=1
l=1
s=1
So the result follows. 23.4.16 Proposition. Let g be a semisimple Lie algebra and G the group of elementary automorphisms of g. Then L(G) = g. Proof. Since G ⊂ Aut g, L(G) ⊂ ad g (23.4.15 and 20.1.5) and L(G) is a Lie subalgebra of ad g (16.3.7 and 23.2.4). Finally, if ad x is nilpotent, then ad x ∈ L(G) by 23.3.5. So the result is a consequence of 20.4.6. 23.4.17 Proposition. Let (U, π) and (V, τ ) be finite-dimensional rational representations of an algebraic group G.Assume that there exists a linear map θ:U → V such that θ π(α)u = τ (α) θ(u) pour (α, u) ∈ G × U . Then we have θ dπ(x)u = dτ (x) θ(u) for x ∈ L(G) and u ∈ U .
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359
Proof. Let U and V be bases of U and V . Denote by A(α) = [πij (α)] (resp. B(α) = [τkl (α)]) the matrix of π(α) (resp. τ (α)) with respect to the basis U (resp. V). If x ∈ L(G), 23.4.2 implies that M (x) = [x(πij )] (resp. N (x) = [x(τkl )]) is the matrix of dπ(x) (resp. dτ (x)) with respect to the basis U (resp. V). Let Λ be the matrix of θ with respect to the bases U and V. Then our hypothesis says that ΛA(α) = B(α)Λ. From this, we deduce immediately that ΛM (x) = N (x)Λ. This proves the result.
23.5 Adjoint representation 23.5.1 In the rest of this chapter, G is an algebraic group and g its Lie algebra. 23.5.2 For α ∈ G, denote by iα the inner automorphism of G defined by β → αβα−1 . Recall that the set Int G of inner automorphisms of G is a subgroup of Aut G. Let AdG (α) or Ad(α) be the differential of iα . By 16.3.1 and 23.2.4, it is an automorphism of the Lie algebra g. Since iαβ = iα ◦ iβ for α, β ∈ G, we deduce that Ad(αβ) = Ad(α) ◦ Ad(β). Thus the map Ad : G → GL(g) , α → Ad(α) defines a representation of G, called the adjoint representation of G. By 23.2.7, if α ∈ G and x ∈ g, then: Ad(α)(x) = χα · x · χα−1 . Let u : G → H a morphism of algebraic groups. If α ∈ G, then u◦iα = iu(α) ◦u. It follows that: du◦ AdG (α) = AdH u(α) ◦du. Let α ∈ G, x ∈ g and y = Ad(α)(x). In the notations of 23.2.2, let X, Y ∈ Lie(G) be defined by X = η(x), Y = η(y). It is clear that ρα ◦X◦ρα−1 is a derivation of A(G). Moreover, since left and right translations commute, ρα ◦X◦ρα−1 ∈ Lie(G). Let f ∈ A(G) and β ∈ G. Then: (ρα ◦X◦ρα−1 )(f )(β) = X(ρ α−1 f )(βα) = x λ(βα)−1 (ρα−1 f ) = x (λβ −1 f )◦iα = y(λβ −1 f ) = Y (f )(β). Hence Y = ρα ◦X◦ρα−1 . Let f1 , . . . , fr ∈ A(G), be linearly independent elements which generate A(G) as an algebra. Complete this family to a basis (fn )n∈N of the k-vector space A(G). Since the G-module A(G) is rational (for the representations λ and ρ), we have an,k (α)fn fk ◦iα = n∈N
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where, for 1 k r, (an,k )n is a family of regular functions on G verifying for α ∈ G, an,k (α) = 0 only if n belongs to a finite subset (which depends on k and α) of N. Let x ∈ g, y = Ad(α)(x) and xn = x(fn ). For n r + 1, xn is a polynomial in x1 , . . . , xr . Moreover: ank (α)xn . y(fk ) = x(fk ◦iα ) = n∈N
An element z ∈ g is completely determined by z(fk ), 1 k r. It follows that if x ∈ g, then the map α → Ad(α)(x) is a morphism from G to g. By 21.5.2, the adjoint representation of G in g is rational. 23.5.3 Lemma. Let α ∈ G and u : G → G be the morphism of varieties β → αβα−1 β −1 . Then du = Ad(α) − idg . Proof. Apply 23.4.10 with u = iα . 23.5.4 Lemma. Assume that G is a closed subgroup of GLn (k), so we may identify g with a Lie subalgebra of gln (k) (23.2.5 and 23.3.3). If α ∈ G and x ∈ g, then Ad(α)(x) = αxα−1 . Proof. Let α = [αij ], β = [βij ] ∈ G, α−1 = [αij ] and γ = αβα−1 = [γij ]. If x ∈ g, we may identify x with the matrix [x(Tij )]. Now
n n Ad(α)(x)(Tij ) = x(Tij ◦iα ) = x αik αlj Tkl = αik x(Tkl )αlj . k,l=1
k,l=1
On the other hand, we have γij =
n
αik βkl αlj .
k,l=1
So Ad(α)(x) = αxα−1 . 23.5.5 Let us determine the differential of Ad. Let α, β ∈ G, x, y ∈ g, f ∈ A(G) and A(µ)(f ) = s1 ⊗ t1 + · · · + sn ⊗ tn where s1 , t1 , . . . , sn , tn ∈ A(G). We obtain ρα f = t1 (α)s1 + · · · + tn (α)sn as in 23.2.2. Hence (1)
(y ∗ f )(α) = y(ρα f ) = y(s1 )t1 (α) + · · · + y(sn )tn (α). It follows that:
(2)
x(y ∗ f ) = y(s1 )x(t1 ) + · · · + y(sn )x(tn ) = (y.x)(f ).
Denote Ad by u. If x ∈ g, we saw in 23.3.3 that du(x) may be identified with an element (x) of End(g) and for all ψ ∈ End(g)∗ ⊂ A GL(g) , we have: ψ (x) = du(x)(ψ).
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361
Let f ∈ A GL(g) and y ∈ g. Define ϑf,y ∈ End(g)∗ as follows: for ϕ ∈ End(g), ϑf,y (ϕ) = ϕ(y) (f ). Then for x, y ∈ g and f ∈ A GL(g) , we have: du(x)(y) (f ) = (x)(y) (f ) = ϑf,y (x) = du(x)(ϑf,y ) = x(ϑf,y ◦u). On the other hand, if α ∈ G, then: (ϑf,y ◦ u)(α) = u(α)(y) (f ) = diα (y) (f ) = y(f ◦iα ). But if A(µ)(f ) = s1 ⊗ t1 + · · · + sn ⊗ tn , we saw in 23.2.7 that: f ◦iα = s1 (α)(ρα−1 t1 ) + · · · + sn (α)(ρα−1 tn ). It follows from 23.2.6 that: −1 y(f ◦iα ) = s1 (α)(y ∗ t1 )(α−1 ) + · · · + sn (α)(y ∗ tn )(α ) = s1 (y ∗ t1 )◦ι + · · · + sn (y ∗ tn )◦ι (α). We deduce therefore that: ϑf,y ◦u = s1 (y ∗ t1 )◦ι + · · · + sn (y ∗ tn )◦ι. In view of 23.4.7, we obtain that: n n x(ϑf,y ◦u) = x(si )(y ∗ ti )(e) − si (e)x(y ∗ ti ). i=1
i=1
It follows from (1), (2) and the definition of y ∗ ti that: n n x(ϑf,y ◦u) = x(si )y(ti ) − si (e)(y · x)ti . i=1
i=1
Finally 23.1.2 and 23.2.3 imply that: du(x)(y)(f ) = x(ϑf,y ◦u) = (x · y)(f ) − (χe · y · x)(f ) = (x · y)(f ) − (y · x)(f ) = [x, y](f ). Thus we have proved the following result: Theorem. The differential of the adjoint representation AdG is adg . 23.5.6 Corollary. If H is a closed normal subgroup of G and h is the Lie algebra of H. Then h is an ideal of g. Proof. Let a be the ideal of H in A(G), α ∈ G, f ∈ a and x ∈ h. If H is normal in G, then f ◦ iα ∈ a, so x(f ◦ iα ) = 0 (23.2.5). Thus Ad(α)(x)(f ) = 0, and this implies that Ad(α)(x) ∈ h (23.2.5). Let (x1 , . . . , xn ) be a basis of g such that (x1 , . . . , xp ) is a basis of h. If y ∈ g, then 23.4.2 implies that for 1 j n, we have πij ∈ A(G) such that: n n πij (α)xi , (d Ad)(y)(xj ) = y(πij )xi . Ad(α)(xj ) = i=1
i=1
So Ad(α)(h) ⊂ h implies that πij = 0 if 1 j p and p + 1 i n. Thus (d Ad)(x)(h) ⊂ h and the result follows from 23.5.5.
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23.6 Jordan decomposition 23.6.1 Let (V, π) be a representation of g. We say that the representation π is locally finite if any v ∈ V is contained in a finite-dimensional g-submodule of V . 23.6.2 Let us consider A(G) as a rational G-module via ρ. If x ∈ g, denote by η(x) the derivation of A(G) defined by η(x)(f ) = f ∗ x for f ∈ A(G) (see 23.2.2). The map g × A(G) → A(G) , (x, f ) → f ∗ x defines a representation of g which is locally finite (23.4.4). Theorem. Let x ∈ g. There exists a unique pair (xs , xn ) of elements of g such that: x = xs + xn , [xs , xn ] = 0 , η(x)s = η(xs ) , η(x)n = η(xn ). The element xs (resp. xn ) is called the semisimple component (resp. nilpotent component) of x, and x = xs + xn is called the Jordan decomposition of x. Proof. Let X = η(x). If α ∈ G, then the automorphism λα of A(G) commutes with X. By 10.1.2, λα commutes also with Xs and Xn . On the other hand, Xs and Xn are derivations of A(G) (10.1.14). So Xs , Xn ∈ Lie(G) and we are done since η : g = L(G) → Lie(G) is a Lie algebra isomorphism (23.2.2). 23.6.3 Proposition. Let u : G → H be a morphism of algebraic groups. For all x ∈ g, we have: (du(x))s = du(xs ) , (du(x))n = du(xn ). Proof. Set X = ηG (x) and Y = ηH du(x) . Let f ∈ A(H), then we have du(x)(f ) = x(f ◦u). So if α ∈ G, then: X(f ◦u)(α) = x λα−1 (f ◦u) = x (λu(α)−1 f )◦u = du(x)(λu(α)−1 f ) = Y (f ) u(α) . It follows that X◦A(u) = A(u)◦Y . Let E = A(u) A(H) . Then X(E) ⊂ E. Since A(u) induces a surjection from A(H) onto E, 10.1.12 implies that Xs ◦A(u) = A(u)◦Ys and Xn ◦A(u) = A(u)◦Yn . Thus: du(xs )(f ) = xs (f ◦u) = Xs (f ◦u)(eG ) = Ys (f ) u(eG ) = Ys (f )(eH ) = (du(x))s (f ). It follows that du(xs ) = (du(x))s . We obtain in a similar way du(xn ) = (du(x))n . 23.6.4 Proposition. Let V be a finite-dimensional k-vector space and G a closed subgroup of GL(V ). If x ∈ g, then the Jordan decomposition of x defined in 23.6.2 coincide with the one defined in 10.1.2.
23.6 Jordan decomposition
363
Proof. By 23.6.3, we may assume that G = GL(V ) and by fixing a basis of V , we are reduced to the case where G = GLn (k). We have A(G) = k[T11 , T12 , . . . , Tnn ]D where D is the determinant. Let x = [xij ] ∈ g. Then as in 23.3.4: Tij ∗ x = x1j Ti1 + x2j Ti2 + · · · + xnj Tin . Thus Wi = kTi1 + · · · + kTin is a subspace of A(G) such that η(x)(Wi ) ⊂ Wi . Moreover, the action of η(x) on Wi may be identified as the matrix operation of x on kn . We deduce therefore that η(x)|Wi = η(xs )|Wi + η(xn )|Wi is the Jordan decomposition of η(x)|Wi . It follows that if W is the sum of the Wi ’s, for 1 i n, then η(x)|W = η(xs )|W + η(xn )|W is the Jordan decomposition of η(x)|W . By 10.1.10, we obtain that η(xs )|W = η(x)s |W and η(xn )|W = η(x)n |W . But derivations of A(G) which are identical on W are equal. So we are done.
References • [5], [23], [38], [40], [78].
24 Correspondence between groups and Lie algebras
We describe more precisely the correspondence between algebraic groups and their Lie algebras in this chapter. We give functorial properties between them and how certain properties are preserved. We also explain how to attach to a Lie algebra an algebraic group (namely, its algebraic adjoint group).
24.1 Notations 24.1.1 Let G and H be algebraic groups. We say that G and H are isomorphic if they are isomorphic as algebraic groups. A closed subgroup of G will also be called an algebraic subgroup of G. Recall that L(G) denote the Lie algebra of G and Lie(G) the Lie algebra of left invariant derivations of A(G) (23.2.1). We shall denote by µG or µ (resp. ιG or ι) the group multiplication (resp. inverse) of G, and for α ∈ G, iα denotes the inner automorphism of G defined by iα (β) = αβα−1 . 24.1.2 Let g be the Lie algebra of G. Recall that AdG or Ad denotes the adjoint representation of G. If a is a subspace of g, we denote by CG (a) (resp. NG (a)) the set of elements α ∈ G verifying Ad(α)(x) = x (resp. Ad(α)(x) ∈ a) for all x ∈ a. If x ∈ g, we shall also denote CG (kx) by CG (x).
24.2 An algebraic subgroup 24.2.1 In this section, g denote the Lie algebra of an algebraic group G. For α ∈ G, let χα be the linear form on A(G) defined by χα (f ) = f (α). Recall from 23.1.2 that A(G)∗ has a structure of associative algebra whose unit is χe , where e denotes the identity element of G. If x ∈ A(G)∗ and n ∈ N∗ , we shall denote by xn the product in A(G)∗ of n factors of x. By convention, x0 = χe . 24.2.2 Let us fix x ∈ g. Set jx to be the subspace of A(G) consisting of elements f verifying xn (f ) = 0 for all n ∈ N.
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Denote by ηG (x) or η(x), the left invariant derivation of A(G) defined by η(x)(f ) = f ∗ x for f ∈ A(G) (see 23.2.2). Thus x(f ) = η(x)(f ) (e). Let f ∈ A(G), then A(µ)(f ) = s1 ⊗ t1 + · · · + sp ⊗ tp where si , ti ∈ A(G). We saw in 23.2.2 that η(x)(f ) = x(t1 )s1 + · · · + x(tp )sp . 1) If n ∈ N, then: η n+1 (x)(f ) = η n (x) η(x)(f ) = η n (x) x(t1 )s1 + · · · + x(tp )sp = η n (x)(s1 )x(t1 ) + · · · + η n (x)(sp )x(tp ). We have x(f ) = η(x)(f ) (e). Now suppose that xn (g) = η n (x)(g) (e) for all g ∈ A(G), then: n ) = xn (s1 )x(t xn+1 (f ) = (xn ⊗ x)◦A(µ)(f n1 ) + · · · + x (sp )x(tp ) n = η (x)(s1 ) (e)x(t1 ) + · · · + η (x)(sp ) (e)x(tp ).
It follows that xn+1 (f ) = η n+1 (x)(f ) (e), and so by induction, we have xn (f ) = η n (x)(f ) (e) for all n ∈ N (the case n = 0 is clear) and f ∈ A(G). 2) Let n ∈ N and f ∈ jx . Then: xn η(x)(f ) = η n (x) η(x)(f ) (e) = η n+1 (x)(f ) (e) = 0. Thus η(x)(jx ) ⊂ jx . 3) Let f ∈ jx and g ∈ A(G). Then: χe (f g) = f (e)g(e) = 0 , x(f g) = x(f )g(e) + f (e)x(g) = 0. Set h = η(x)(f ) and = η(x)(g). Suppose that xn (uv) = 0 for all u ∈ jx and v ∈ A(G). Then: xn+1 (f g) = η n (x) η(x)(f g) (e) = η n (x) hg + f (e) = xn (hg) + xn (f ). Hence xn+1 (f g) = 0 since h ∈ jx . We have therefore proved that jx is an ideal of A(G). 4) Let f ∈ A(G) and g = f ◦ι. For α, β ∈ G, we have: f (αβ) = A(µ)(f )(α, β) = s1 (α)t1 (β) + · · · + sp (α)tp (β). It follows that: A(µ)(g)(α, β) = g(αβ) = f (β −1 α−1 ) = A(µ)(f )(β −1 , α−1 ) = s1 (β −1 )t1 (α−1 ) + · · · + sp (β −1 )tp (α−1 ). Hence: A(µ)(g) = (t1 ◦ι) ⊗ (s1 ◦ι) + · · · + (tp ◦ι) ⊗ (sp ◦ι). Now χe (g) = f (e) = χe (f ) and x(g) = −x(f ) (23.4.8). Suppose that we have shown that xn (u◦ι) = (−1)n xn (u) for all u ∈ A(G). Then:
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xn+1 (g) = (xn ⊗ x)◦A(µ)(g) = xn (t1 ◦ι)x(s1 ◦ι) + · · · + xn (tp ◦ι)x(sp ◦ι) = (−1)n+1 [xn (t1 )x(s1 ) + · · · + xn (tp )x(sp )] = (−1)n+1 (x ⊗ xn )A(µ)(f ) = (−1)n+1 xn+1 (f ). Thus xn (g) = (−1)n xn (f ) for all n ∈ N. So if f ∈ jx , then f ◦ι ∈ jx . 5) Let f, g ∈ A(G) \ jx . Denote by p and q the smallest integers such that xp (f ) and xq (g) are non-zero. Since η(x) is a derivation of A(G), we have: xp+q (f g) = η p+q (x)(f g) (e) =
=
m+n=p+q
(p + q)! m η (x)(f ) η n (x)(g) (e) m+n=p+q m!n! (p + q)! m (p + q)! p x (f )xn (g) = x (f )xq (g) = 0. m!n! p!q!
Thus f g ∈ jx , and we have shown that jx is a prime ideal of A(G). 6) Let (x) = V(jx ) be the set of α ∈ G verifying f (α) = 0 for all f ∈ jx . By definition, e ∈ (x). We shall show that if α ∈ (x) and f ∈ jx , then λα−1 f ∈ jx where λα is the left translation of functions (22.1.4). Since the derivation η(x) is left invariant (23.2.2), we have η n (x)(λα−1 f ) = n λα−1 η (x)(f ) . From this, we deduce that: xn (λα−1 f ) = η n (x)(λα−1 f ) (e) = λα−1 η n (x)(f ) (e) = η n (x)(f ) (α).
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It follows from point 2 that η(x)(jx ) ⊂ jx . Hence our result. 7) Points 2, 4 and 6 prove that (x) is an algebraic subgroup of G. Moreover, by 10.1.6 and point 5, the defining ideal of (x) in A(G) is: I (x) = jx = jx .
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It follows that (x) is connected. 8) By 11.8.1 and the preceding discussions, the ideal Jx of (x) × (x) in the algebra A(G×G) = A(G)⊗k A(G) is equal to Jx = jx ⊗k A(G)+A(G)⊗k jx . Let f ∈ jx . If α, β ∈ (x), then A(µ)(f )(α, β) = f (αβ) = 0. Thus A(µ)(jx ) ⊂ Jx . For p, q ∈ N, denote by Tp,q,x the kernel of the linear form xp ⊗ xq on A(G) ⊗k A(G), and let Tx be the intersection of the Tp,q,x ’s, p, q ∈ N. We have xp+q (f ) = (xp ⊗xq )◦A(µ)(f ), so f ∈ jx implies that A(µ)f ∈ Tx . It follows that Tx ⊂ Jx . Since we have clearly Jx ⊂ Tx , we obtained that Tx = Jx .
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24.2.3 Theorem. Let G be an algebraic group and x ∈ L(G). (i) The group (x) is a connected and commutative algebraic subgroup of G such that x ∈ L( (x)). (ii) Any closed subgroup H of G verifying x ∈ L(H) contains (x). Proof. Points 2 and 7 of 24.2.2 imply that I (x) = jx and η(x)(jx ) ⊂ jx . So by 23.2.5, x ∈ L (x) . Let us prove that (x) is abelian.
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Denote by ζ : A(G) ⊗k A(G) → A(G) ⊗k A(G) the linear map defined by ζ(f ⊗ g) = g ⊗ f . If p, q ∈ N and f ∈ A(G), we have: (xp ⊗ xq )◦ζ◦A(µ)(f ) = (xq ⊗xp )◦A(µ)(f ) = xq+p (f ) = (xp ⊗ xq )◦A(µ)(f ). Thus (xp ⊗ xq )◦ ζ◦A(µ) − A(µ) = 0. By point 8 of 24.2.2, we deduce that ζ◦A(µ) − A(µ) A(G) ⊂ Ix . If α, β ∈ (x) and f ∈ A(G), then:
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f (αβ) − f (βα) = A(µ)(f )(α, β) − A(µ)(f )(β, α) p p si (α)ti (β) − si (β)ti (α) = i=1 i=1 = A(µ) − ζ◦A(µ) (f )(α, β) = 0. So we have αβ = βα. (ii) Let H be a closed subgroup of G verifying x ∈ L(H), and a = I(H) the ideal of H. We have η(x)(a) ⊂ a (23.2.5), so η n (x)(a) ⊂ a and xn (a) = {0}, if n ∈ N by point 1 of 24.2.2. Thus a ⊂ jx , and (x) ⊂ H.
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24.2.4 Remark. Let V be a finite-dimensional vector space, G a closed subgroup of GL(V ) and x ∈ L(G). If x is a nilpotent endomorphism, then by 22.3.4 and 23.3.5, the group (x) is the set of etx , t ∈ k.
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24.2.5 Corollary. Let h be a Lie subalgebra of g, and (h) the intersection of all algebraic subgroups of G whose Lie algebra contains h. Then (h) is connected and h ⊂ L( (h)).
a a Proof. It is clear that a(h) is an algebraic subgroup of G, and since L(a(h)) = L(a(h)◦ ), a(h) is connected. By 24.2.3, we have a(x) ⊂ a(h) for all x ∈ h. So it follows from 16.3.7, 23.2.4 and 24.2.3 that x ∈ L(a(x)) ⊂ L(a(h)). Hence h ⊂ L(a(h)). 24.2.6 Remark. Let H be a closed connected subgroup of G, h its Lie algebra. Since (h) ⊂ H, we have L( (h)) ⊂ h. It follows from 24.2.5 that h = L( (h)). Thus dim (h) = dim H = dim h. Since H is connected, we deduce that (h) = H.
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24.3 Invariants 24.3.1 Let g be the Lie algebra of an algebraic group G, (V, π) a rational representation of G of dimension n, and B a basis of V . If u ∈ End(V ), denote of u with respect to B. We by [ϕij (u)] the matrix define πij ∈ A(G) by πij (α) = ϕij π(α) , and if x ∈ g, set dij (x) = ϕij dπ(x) , 1 i, j n. Let x1 , . . . , xr ∈ g. Then: ϕij dπ(x1 )◦ · · · ◦dπ(xr ) = di,k1 (x1 )dk1 ,k2 (x2 ) · · · dkr−1 ,j (xr ). 1k1 ,...,kr−1 n
∗ Moreover, for x ∈ g and ϕ ∈ End(V ) , we have by 23.3.3 that:
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ϕ dπ(x) = dπ(x)(ϕ) = x(ϕ◦π). Thus: ϕij dπ(x1 )◦ · · · ◦dπ(xr ) =
1k1 ,...,kr−1 n
x1 (πi,k1 )x2 (πk1 ,k2 ) · · · xr (πkr−1 ,j ).
Hence: ϕij dπ(x1 )◦ · · · ◦dπ(xr ) = (x1 · · · xr )(πij ) = (x1 · · · xr )(ϕij ◦π). So by linearity, we obtain the following result:
∗ Lemma. Let r ∈ N∗ , x1 , . . . , xr ∈ g and ϕ ∈ End(V ) . Then: ϕ dπ(x1 )◦ · · · ◦dπ(xr ) = (x1 · · · xr )(ϕ◦π).
24.3.2 Theorem. Let (V, π) be a finite-dimensional rational representation of an algebraic group G, E the subspace of End(V ) spanned by π(α)−idV , α ∈ G, and F the subspace of End(V ) spanned by the products dπ(x1 ) ◦ · · · ◦ dπ(xr ) where r ∈ N∗ and x1 , . . . , xr ∈ g = L(G). (i) For x ∈ g, we have dπ(x) ∈ E. (ii) Assume that G is connected. If α ∈ G, then π(α) − idV ∈ F . Proof. (i) Let ϕ ∈ End(V )∗ be such that ϕ|E = 0. Then α → ϕ ◦ π(α) is a constant function on G and so x(ϕ ◦ π) = 0 for all x ∈ g. So by 23.3.3, we have 0 = dπ(x)(ϕ) = ϕ(dπ(x)). Hence dπ(x) ∈ E. (ii) Let x ∈ g, n ∈ N∗ and ϕ ∈ End(V )∗ verifying ϕ|F = 0. We have xn (ϕ ◦ π) = 0 by 24.3.1. So ϕ ◦ π − ϕ(idV ) ∈ jx (notations of 24.2.2). It follows that if α ∈ (x), then ϕ(π(α)−idV ) = 0. Hence π(α)−idV ∈ F for α ∈ (x). If α, β ∈ G, then: π(αβ) − idV = π(α) − idV ◦ π(β) − idV + π(α) − idV + π(β) − idV .
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So for n ∈ N∗ , x1 , . . . , xn ∈ g and αi ∈ (xi ), 1 i n, we obtain that π(α1 · · · αn ) − idV ∈ F . Let H be the subgroup of G generated by (x), x ∈ g. It algebraic subgroup of G (21.3.2 and 24.2.3). If α ∈ H, is a connected ) which are zero on F , then ϕ π(α) − idV = 0 for all linear forms on End(V so π(α) − idV ∈ F . If x ∈ g, then x ∈ L (x) ⊂ L(H) (24.2.3). We have therefore g = L(H), hence dim G = dim H. Since G is connected, we have G = H, and the result follows.
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24.3.3 Corollary. Let G be a connected algebraic group, g its Lie algebra and V a finite-dimensional rational G-module. (i) We have V G = V g . (ii) A subspace W of V is a G-submodule if and only if W is a gsubmodule.
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24.3.4 Let (E, π) be a rational (not necessarily finite-dimensional) Gmodule (see 21.5.4) and V, W two finite-dimensional G-submodules of E. Denote by σ : G → GL(V ), τ : G → GL(W ) and θ : G → GL(V + W ) the corresponding subrepresentations of G. If α ∈ G, then θ(α)|V = σ(α) and θ(α)|W = τ (α). We deduce easily that for x ∈ g, dθ(x)|V = dσ(x) and dθ(x)|W = dτ (x). In particular, dσ(x)|V ∩W = dτ (x)|V ∩W and so there exists a unique (x) ∈ End(E) such that (x)|U = d(π|U )(x) for all finite-dimensional G-stable subspace U of E. We shall also denote (x) by dπ(x). It is clear that the map g → gl(E), x → dπ(x), is a Lie algebra homomorphism. Thus E is a g-module. We deduce immediately from 24.3.3 the following result: Proposition. Let G be a connected algebraic group, g its Lie algebra and E a rational G-module. (i) We have E G = E g . (ii) A subspace F of E is a G-submodule if and only if F is a g-submodule. 24.3.5 Proposition. Let H, K be algebraic subgroups of G whose Lie algebras are h and k respectively. (i) If H ⊂ K, then h ⊂ k. (ii) Assume that H and K and connected. If h ⊂ k, then H ⊂ K. (iii) We have L(H ∩ K) = h ∩ k. Proof. (i) This is a consequence of 16.3.7 and 23.2.4. (ii) If h ⊂ k, then (h) ⊂ (k). So H ⊂ K by 24.2.6. (iii) We have L(H ∩K) ⊂ h∩k by (i). On the other hand, (24.2.5). So h ∩ k ⊂ L( (h ∩ k)) ⊂ L(H ∩ K).
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24.3.6 Let (V, π) be a finite-dimensional rational representation of G, W a subspace of V , v ∈ V and: NG (W ) = {α ∈ G; π(α)(W ) = W } , ng (W ) = {x ∈ g; dπ(x)(W ) ⊂ W }, CG (v) = {α ∈ G; π(α)(v) = v} , cg (v) = {x ∈ g; dπ(x)(v) = 0}, CG (v) , cg (W ) = cg (v). CG (W ) = v∈W
v∈W
If H is an algebraic subgroup of G, set: NH (W ) = H ∩ NG (W ) , CH (v) = H ∩ CG (v). These are clearly algebraic subgroups of G (21.4.2). Proposition. With the above notations, we have: (i) LNG (W) = ng (W ). (ii) LCG (v) = cg (v). (iii) LCG (W ) = cg (W ). (iv) L NH (W ) = h ∩ ng (W ) , L CH (v) = h ∩ cg (v).
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Proof. (i) Let us fix a basis (v1 , . . . , vn ) of V such that (v1 , . . . , vp ) is a basis of W . For α ∈ G, denote by [πij (α)] the matrix of π(α) with respect to this basis. Then α ∈ NG (W ) if and only if πij (α) = 0 for 1 i p and p + 1 j n. So 23.2.5 and 23.4.2 imply that L(NG (W )) ⊂ ng (W ). For f ∈ V ∗ and v ∈ V , define ϑf,v ∈ End(V )∗ by ϑf,v (ϕ) = f ϕ(v) if ϕ ∈ End(V ). If n ∈ N∗ and x ∈ g, then by 24.3.1: n f dπ(x) (v) = ϑf,v (dπ(x))n = xn (ϑf,v ◦π). n So x ∈ ng (W ) if and only if f dπ(x) (v) = 0 for all n ∈ N∗ , v ∈ W and f ∈ W ⊥ , the orthogonal of W in V ∗ . Let x ∈ ng (W ). We have therefore ∈ jx (notations of 24.2.2). It follows that if α ∈ (x), ϑf,v ◦π − ϑf,v ◦π(e) then f π(α)(v) = ϑf,v ◦π(α) = ϑf,v ◦π(e) = 0 for all v ∈ W and f ∈ W ⊥ . (x) ⊂ NG (W ). By 24.2.3 and 24.3.5, we have x ∈ L (x) ⊂ Consequently, L NG (W ) . So we have proved (i). (ii) G (v)) ⊂ cg (v), and x ∈ cg (v) if and only if As innpart (i), we have L(C = 0 for all n ∈ N∗ and f ∈ V ∗ . So we can finish our proof as ϑf,v dπ(x) in (i). p p (iii) Since CG (W ) = i=1 CG (vi ) and cg (W ) = i=1 cg (vi ), the result follows from part (ii) and 24.3.5 (iii). (iv) This follows from (i), (ii) and 24.3.5 (iii).
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24.3.7 Let (V, σ) and (W, τ ) be finite-dimensional rational representations of G. We have already seen that there is a natural structure of rational representation π of G on Hom(V, W ) given by: for f ∈ Hom(V, W ) and v ∈ V , π(α)(f ) (v) = τ (α) f σ(α−1 )(v) . Similarly, let (V, σ ) and (W, τ ) be representations of a Lie algebra g, then there is a natural structure of representation π of g on Hom(V, W ) given by: for x ∈ g, f ∈ Hom(V, W ) and v ∈ V , π (x)(f ) (v) = τ (x) f (v) − f σ (x)(v) . Suppose that g = L(G) and σ = dσ, τ = dτ . Then we deduce easily from 23.4 that π = dπ. Now let A be a finite-dimensional k-algebra (not necessarily associative or with unit). The algebra structure on A is defined by ν ∈ End(A ⊗k A, A) such that ν(a ⊗ b) = ab for all a, b ∈ A. Denote by τ (resp. τ ) the identity representation of G = GL(A) (resp. g = gl(A)) and let σ = τ ⊗ τ (see 23.4.13), σ = dσ. Construct π and π as above with V = A ⊗k A and W = A. Then we see easily that for α ∈ G (resp. x ∈ g), α ∈ Aut A (resp. x ∈ Der A) if and only if π(α)(ν) = ν (resp. π (x)(ν) = 0). Since π = dπ, it follows from 24.3.6 (ii) that: Proposition. The Lie algebra of Aut A is Der A.
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24.4 Functorial properties 24.4.1 Theorem. Let u : G → H be a morphism of algebraic groups and K the kernel of u. Then L(K) is the kernel of du. Proof. Let a = I(K) be the defining ideal of K in A(G) and b the set √ of elements f ◦ u − f ◦ u(eG ) for f ∈ A(H). Clearly K = V(b) and so a = b. Let x ∈ L(K) and f ∈ A(H). Then by 23.2.5, we have: 0 = x f ◦u − f ◦u(eG ) = x(f ◦u) = du(x)(f ). So x belongs to the kernel of du. Let x ∈ L(G) and f ∈ A(H). Then for α ∈ G, η(x)(f ◦u)(α) = λα−1 η(x)(f◦u) (eG ) = η(x) λα−1 (f ◦u) (eG ) = x λα−1 (f ◦u) = x (λu(α)−1 f )◦u = du(x)(λu(α)−1 f ).
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So if x ∈ ker(du), then b ⊂ jx = I( (x)) (notations as in 24.2.2). Since jx is prime, we have a ⊂ I( (x)), hence (x) ⊂ K. So by 24.2.3 and 24.3.5, x ∈ L(K).
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24.4.2 Theorem. Let G be an algebraic group, Z(G) its centre, g its Lie algebra and z(g) the centre of g. (i) We have Z(G) ⊂ ker(AdG ) and L ker(AdG ) =z(g). (ii) If G is connected, then Z(G) = ker(AdG ) and L Z(G) = z(g). Proof. (i) Let α ∈ Z(G), then iα = idG . So for x ∈ g and f ∈ A(G), we have: Ad(α)(x)(f ) = diα (x)(f ) = x(f ◦iα ) = x(f ). Hence Ad(α) = idg and α ∈ ker(AdG ). The second part of (i) follows easily from 24.4.1 and 23.5.5. (ii) By 22.1.5, G has a faithful finite-dimensional rational representation (V, π). So 24.4.1 implies that the representation (V, dπ) of g is also faithful. If α ∈ G, then we have (23.5.2): dπ◦ AdG (α) = AdGL(V ) π(α) ◦dπ. So if α ∈ ker(AdG ) and x ∈ g, then (23.5.4) dπ(x) = π(α)◦dπ(x)◦π(α)−1 , which means that π(α) belongs to the centralizer of dπ(g) in End(V ). If G is connected, then 24.3.2 says that the subalgebra (with unit) of End(V ) generated by dπ(g) is equal to the subalgebra (with unit) in End(V ) generated by π(G). So if α ∈ ker AdG , then π(αβ) = π(βα) for all β ∈ G. Since π is injective, α ∈ Z(G). So by (i), we have Z(G) = ker(AdG ) and L Z(G) = z(g).
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24.4.3 Remark. If G is not connected, then it may happen that Z(G) = ker(AdG ) and L Z(G) = z(g). Let us give an example. Using the notations of example 4 of 21.1.3, take G = O2 (k). We verify easily that G has two connected components G◦ and G1 given by: # # a 0 0 b ◦ ; b ∈ k \ {0} . G = ; a ∈ k \ {0} , G1 = 0 a−1 b−1 0 Let I2 denote the identity matrix of M2 (k), then: Z(G) = {−I2 , I2 } , L Z(G) = {0}. Since G◦ = SL2 (k) ∩ D2 (k), it follows from 23.3.4, 23.4.5 and 24.3.5 (iii) that g = L(G) is the set of matrices diag(u, −u) with u ∈ k. By 23.5.4, we have ker(AdG ) = G◦ . So Z(G) = ker(AdG ) and L Z(G) = z(g). 24.4.4 Theorem. Let G be an algebraic group. (i) If G is commutative, then L(G) is commutative. (ii) If G is connected and L(G) is commutative, then G is commutative. Proof. (i) If G is commutative, then G = ker(AdG ). So L(G) is commutative (24.4.2 (i)). (ii) By 22.1.5, G has a faithful finite-dimensional rational representation (V, π). If g = L(G) is commutative, then the subalgebra (with unit) of End(V ) generated by dπ(g) is also commutative. If G is connected, then 24.3.2 implies that π(G) is commutative. Since π is injective, we conclude that G is commutative. 24.4.5 Remark. If G is not connected, then the example in 24.4.3 shows that although L(G) is commutative, G is not commutative. 24.4.6 Lemma. Let α ∈ G, H an algebraic subgroup of G and h its Lie algebra. We have: Ad(α)(h) = L(αHα−1 ). Proof. Let a = I(H) (resp. b = I(αHα−1 )) be the ideal of H (resp. αHα−1 ) in A(G). It is clear that b = {f ◦iα−1 ; f ∈ a}. Let x ∈ L(G), y = Ad(α)(x) and f ∈ A(G). Since y(f ) = x(f ◦iα ), we have by 23.2.5 that: y ∈ L(αHα−1 ) ⇔ y(b) = {0} ⇔ x(a) = {0} ⇔ x ∈ h. So we have obtained the result. 24.4.7 Theorem. Let G be an algebraic group, H an algebraic subgroup of G, g = L(G) and h = L(H). (i) If H is normal in G, then h is an ideal of g. (ii) Assume that G and H are connected. If h is an ideal of g, then H is normal in G.
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Proof. We have already seen (i) in 23.5.6. Let us prove (ii). We have ad(g)(h) ⊂ h, so Ad(G)(h) = h (24.3.3). It follows from 24.4.6 that for all α ∈ G, L(αHα−1 ) = h, hence αHα−1 = H (24.3.5). 24.4.8 Theorem. Let G be an algebraic group, H an algebraic subgroup of G, g = L(G) and h = L(H). (i) We have L NG (H) ⊂ ng (h) and L CG (H) ⊂ cg (h). (ii) If H is connected, then: N G (H) = NG (h) , CG (H) = CG (h), L NG (H) = L NG (h) = ng (h) , L CG (H) = L CG (h) = cg (h). Moreover, Proof. The groups NG (H) and CG (H) are closed in G (21.3.7). (H), h is an ideal of L N (H) (24.4.7 (i)), so since H is normal in N G G L NG (H) ⊂ ng (h). Let (V, π) be a finite-dimensional faithful rational representation of G (22.1.5), A (resp. B) the subalgebra with unit of End(V ) generated by π(H) (resp. dπ(h)). By 24.3.2 (i), we have B ⊂ A. It follows that CG (H) ⊂ CG (h). Therefore, by 24.3.5 (i) and 24.3.6 (iii), we have L CG (H) ⊂ cg (h). If H is connected, then A = B (24.3.2). Thus NG (H) = NG (h) et CG (H) = CG (h) and the rest follows from 24.3.6. 24.4.9 Let H, K be subgroups of G. Recall that (H, K) denotes the subgroup of G generated by αβα−1 β −1 , α ∈ H, β ∈ K. Recall also that the derived subgroup (G, G) of G is closed (21.3.5) and it is connected if G is connected (21.3.6). Proposition. Let H, K be closed subgroups of G, L the closure of (H, K) in G (it is a subgroup by 21.2.2), and g, h, k, l the Lie algebras of G, H, K, L respectively. Let α ∈ H, β ∈ K, x ∈ h, and y ∈ k. Then [x, y] , Ad(α)(y) − y , Ad(β)(x) − x are elements of l. Proof. For α ∈ H, denote by uα : K → L the map β → αβα−1 β −1 . If y ∈ k, then duα (y) = Ad(α)(y) − y (see 23.4.10). Thus Ad(α)(y) − y ∈ l. Similarly, we have Ad(β)(x) − x ∈ l. It follows that for x ∈ h, we have a map vx : K → l given by vx (β) = Ad(β)(x) − x for β ∈ K. If y ∈ k, then dvx (y) = [y, x]. Hence [x, y] ∈ l. 24.4.10 Corollary. Let G be an algebraic group and g its Lie algebra. Then [g, g] ⊂ L (G, G) . 24.4.11 Remark. We shall see later that [g, g] = L (G, G) if G is connected. This is however not true if G is not connected. In the example of 24.4.3, [g, g] = {0}, (G, G) = G◦ and so L (G, G) = g.
24.5 Algebraic Lie subalgebras
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24.5 Algebraic Lie subalgebras 24.5.1 In this section, G will denote an algebraic group, g its Lie algebra. Recall that if x ∈ g (resp. h is a subspace of g), then (x) (resp. (h)) is the intersection of algebraic subgroups of G whose Lie algebra contains x (resp. h). Moreover, (x) and (h) are connected and (x) is commutative (24.2.3 and 24.2.5). Let m be a subset of g and H an algebraic subgroup of G such that m ⊂ L(H). Then (x) ⊂ H for all x ∈ m, and so H contains the subgroup K generated by the (x)’s, x ∈ m. Since K is a connected algebraic subgroup of G (21.3.2) whose Lie algebra contains m (24.3.5), we deduce that K is the smallest algebraic subgroup of G whose Lie algebra contains m. We shall denote K by (m).
a
a
a
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a
a a
a
Definition. A Lie subalgebra h of g is called algebraic if h = L(H) for some algebraic subgroup H of G. 24.5.2 Proposition. Let (hi )i∈I be a family of algebraic Lie subalgebras of g, and Hi an algebraic subgroup of G such that L(Hi ) = hi . Set: h= hi , H = Hi . i∈I
i∈I
Then h is an algebraic Lie subalgebra of g, and h = L(H). Proof. The group H is closed in G and verifies L(H) ⊂ h by 24.3.5. Moreover,
a(h) ⊂ Hi, so a(h) ⊂ H. Thus h ⊂ La(h) ⊂ L(H).
24.5.3 Proposition. Let u : G → H be a morphism of algebraic groups, and h = L(H). (i) The Lie algebra of u(G) is du(g). (ii) If k is an algebraic Lie subalgebra of g, then du(k) is an algebraic Lie subalgebra of h. (iii) If l is an algebraic Lie subalgebra of h, then (du)−1 (l) is an algebraic Lie subalgebra of g. Proof. (i) By 21.2.4, u(G) is a closed subgroup of H, and we may assume that G is connected. The morphism v : G → u(G) induced by u is dominant. Let α, β ∈ G. Denote by w : G → G (resp. w : u(G) → u(G)) the automorphism of varieties defined by γ → (αβ −1 )γ (resp. δ → v(αβ −1 )δ). We have w(β) = α, w (v(β)) = v(α) and v ◦ w = w ◦ v. So dwβ (resp. dwv(β) ) is an isomorphism from Tβ (G) onto Tα (G) (resp. from Tv(β) (u(G)) onto Tv(α) (u(G))) and we have dvα ◦ dwβ = dwv(β) ◦ dvβ . Thus 16.5.7 (ii) implies that du induces a surjection from g onto L(u(G)). (ii) We may assume that k = g. So the result follows from (i). (iii) Applying (i) and 24.5.2, we may assume that u and du are surjective. Let L = (l) and K = u−1 (L). Since ker u ⊂ K, ker du = L(ker u) ⊂ L(K) (24.3.5 and 24.4.1). Applying (i) to the surjective morphism K → L induced
a
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24 Correspondence between groups and Lie algebras
by u, we obtain that du(L(K)) = l. So we have (du)−1 (l) = L(K) + ker du = L(K). 24.5.4 Let m be a non-empty subset of g. Denote by a(m) the intersection of all the algebraic Lie subalgebras of g containing m. By 24.5.2, a(m) is the smallest algebraic Lie subalgebra of g containing m. Clearly, we have a(m) = L( (m)). If m is a Lie subalgebra of g, then we call a(m) the algebraic hull of m in g.
a
Proposition. Let u : G → H be a morphism of algebraic groups and m a non-empty of subset g. Then: (i) du a(m) = a du(m) . du(m) . (ii) u (m) =
a
a
Proof. (i) By 24.5.3, du(a(m)) is an algebraic Lie subalgebra of L(H) containing du(m), so it contains a(du(m)). Conversely, again by 24.5.3, (du)−1 (a(du(m))) is an algebraic Lie subalgebra of g containing m, so it contains a(m). It follows that du(a(m)) ⊂ a(du(m)). (ii) We have L( (du(m))) = a(du(m)) and L(u( (m))) = du(a(m)) (24.5.3). Since the groups u( (m)) and (du(m)) are connected, the result follows from (i) and 24.3.5.
a
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a
24.5.5 Theorem. Let (V, π) be a finite-dimensional rational representation of G, and E, F vector subspaces of V such that E ⊂ F . Then H = {α ∈ G; (π(α) − idV )(F ) ⊂ E} is an algebraic subgroup of G and h = {x ∈ g; dπ(x)(F ) ⊂ E} is a Lie subalgebra of g verifying h = L(H). Proof. By definition, H ⊂ NG (F ) and h ⊂ ng (F ). So we may assume that V = F , G = NG (F ) and g = ng (F ) (24.3.6). Let B = (v1 , . . . , vn ) be a basis of V such that (v1 , . . . , vp ) is a basis of E. For α ∈ G and x ∈ g, denote by [πij (α)] (resp. [xij ]) the matrix of π(α) (resp. dπ(x)) with respect to B. Then x ∈ h (resp. α ∈ H) if and only if xij = 0 (resp. πij (α) = δij where δij denotes the Kronecker symbol) for p + 1 i n and 1 j n. We deduce therefore that H is an algebraic subgroup of G, h is a Lie subalgebra of g, and we have L(H) ⊂ h by ∗23.4.2. For f ∈ V ∗ and v ∈ V , define ϑf,v ∈ End(V ) as in the proof of 24.3.6, and denote by E ⊥ the orthogonal of E in V ∗ . We have: x ∈ h ⇔ xn (ϑf,v ◦π) = 0 for all n ∈ N∗ , (f, v) ∈ E ⊥ × V, α ∈ H ⇔ ϑf,v ◦π(α) = ϑf,v ◦π(e) for all (f, v) ∈ E ⊥ × V. Proceeding as in the proof of 24.3.6, we obtain that h ⊂ L(H).
24.5.6 Corollary. Let m, n be subspaces of g such that m ⊂ n. Then H = {α ∈ G; (Ad(α) − idg )(n) ⊂ m} is an algebraic subgroup of G, and
24.5 Algebraic Lie subalgebras
377
h = {x ∈ g; ad(x)(n) ⊂ m} is an algebraic Lie subalgebra of g verifying h = L(H). 24.5.7 Theorem. Let h be a Lie subalgebra of g. (i) Any ideal of h is an ideal of a(h). (ii) We have [a(h), a(h)] = [h, h]. Proof. (i) Let a be an ideal of h. By 24.5.6, the set of x ∈ g verifying [x, a] ⊂ a is an algebraic Lie subalgebra k of g. Since h ⊂ k, we have a(h) ⊂ k. Thus a is an ideal of a(h). (ii) As in (i), the set of x ∈ g verifying [x, h] ⊂ [h, h] contains a(h). So the set of y ∈ g such that [y, a(h)] ⊂ [h, h] is an algebraic Lie subalgebra g containing h, so it contains a(h). We deduce therefore that [a(h), a(h)] ⊂ [h, h]. Since h ⊂ a(h), we have [a(h), a(h)] ⊃ [h, h]. 24.5.8 Let (Xi )i∈I be a family of irreducible subvarieties of G. Suppose that the following conditions are satisfied: (i) e ∈ Xi for all i ∈ I. (ii) If i ∈ I, there exists j ∈ I such that Xi−1 = {α−1 ; α ∈ Xi } = Xj . Let H be the subgroup of G generated by the Xi ’s. By 21.3.1, H is a connected algebraic subgroup of G. Let h be its Lie algebra. Proposition. The vector space h is spanned by the vector subspaces Ad(α)(Te (βi−1 Xi )) of g, for α ∈ H, i ∈ I and βi ∈ Xi . Proof. 1) By 21.3.1, there exist i1 , . . . , in ∈ I such that the morphism u : Xi1 × · · · × Xin → H , (β1 , . . . , βn ) → β1 · · · βn is surjective. To simplify notations let us write Xk for Xik if 1 k n. Let X = X1 × · · · × X n . There exists a = (α1 , . . . , αn ) ∈ X such that the map dua : Ta (X) → Tu(a) (H) is surjective (16.5.7). For 1 i n, set: γi = α1 · · · αi , Yi = αi−1 Xi , Zi = γi Yi γi−1 . Let Y = Y1 × · · · × Yn and Z = Z1 × · · · × Zn . Define the following maps: s : Y → X , (β1 , . . . , βn ) → (α1 β1 , . . . , αn βn ), t : Y → Z , (β1 , . . . , βn ) → (γ1 β1 γ1−1 , . . . , γn βn γn−1 ), v : Z → H , (β1 , . . . , βn ) → β1 · · · βn , −1 . θ : H → H , β → β u(a) We verify easily that θ◦u◦s = v◦t. Moreover if ε = (e, . . . , e) ∈ Y , then s(ε) = a. The maps s and θ are isomorphisms of varieties and dua is surjective. It follows that the differential
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24 Correspondence between groups and Lie algebras
d(v◦t)ε : Tε (Y ) → Te (H) = h of v◦t = θ◦u◦s at the point ε is surjective. 2) Let x = (x1 , . . . , xn ) ∈ Tε (Y ), so xi ∈ Te (αi−1 Xi ) for 1 i n. By 23.4.7, we have: d(v◦t)ε (x) =
n k=1
d(iγk )e (xk ) =
n
Ad(γk )(xk ).
k=1
Since γk ∈ H and d(v◦t)ε is surjective, we have proved that h is contained in the sum of the vector subspaces Ad(α)(Te (βi−1 Xi )), for α ∈ H, i ∈ I and βi ∈ Xi . Now, each Ad(α)(Te (βi−1 Xi )) is clearly contained in h, so the result follows. 24.5.9 Theorem. Let (Hi )i∈I be a family of connected algebraic subgroups of G, H the connected algebraic subgroup of G generated by the Hi ’s (see 21.3.2) and h, hi the Lie algebras of H, Hi , i ∈ I respectively. (i) The vector space h is the sum of the vector spaces Ad(α)(hi ), α ∈ H, i ∈ I. (ii) As a Lie algebra, h is generated by hi , i ∈ I. Proof. (i) Since Hi is a group, we have βi−1 Hi = Hi for all βi ∈ Hi and Te (βi−1 Hi ) = hi . So the result follows from 24.5.8. (ii) Let k be the Lie subalgebra of g generated by hi , i ∈ I. Since Hi ⊂ H, hi ⊂ h, and as h is a Lie algebra, we deduce that k ⊂ h. To show that h ⊂ k, it suffices, by (i), to establish that Ad(α)(k) ⊂ k for α ∈ H, that is, H ⊂ NG (k). Since NG (k) is a group, we are reduced to prove that Hi ⊂ NG (k) for all i ∈ I. The Hi ’s are connected, so this is equivalent to proving that hi ⊂ ng (k) (24.4.8). But this is obvious since hi ⊂ k. 24.5.10 Corollary. The following conditions are equivalent for a Lie subalgebra h of g: (i) h is an algebraic Lie subalgebra of g. (ii) For all x ∈ h, a(x) ⊂ h. (iii) The vector space h is a sum of algebraic Lie subalgebras of g. (iv) As a Lie algebra, h is generated by algebraic Lie subalgebras of g. Proof. The implications (i) ⇒ (ii) ⇒ (iii) ⇒ (iv) are clear, while (iv) ⇒ (i) follows from 24.5.9. 24.5.11 Let H, K be connected normal algebraic subgroups of G. By 21.3.5, L = (H, K) is a connected algebraic subgroup of G. Let h, k, l be the Lie algebras of H, K, L. Theorem. We have l = [h, k].
24.5 Algebraic Lie subalgebras
379
Proof. 1) We have [h, k] ⊂ l by 24.4.9. 2) For α, β ∈ G, let cα (β) = αβα−1 β −1 . Since (αβα−1 β −1 )−1 = βαβ −1 α−1 = α−1 (αβα−1 )α(αβα−1 )−1 , we deduce that cα (K)−1 = cα−1 (K). By 21.3.1, there exist n ∈ N∗ and α1 , . . . , αn ∈ H such that the map u : K n → L , (γ1 , . . . , γn ) → cα1 (γ1 ) · · · cαn (γn ) is surjective. So by 16.5.7, there exists b = (β1 , . . . , βn ) ∈ K n such that dub : Tb (K n ) → Tu(b) (L) is surjective. 3) Let s : K n → K n and θ : L → L be defined as follows: −1 . s(γ1 , . . . , γn ) = (β1 γ1 , . . . , βn γn ) , θ(γ) = γ u(b) These are isomorphisms of varieties. For 1 k n and γ ∈ K, let: −1 −1 Ck (γ) = cα1 (β1 ) · · · cαk−1 (βk−1 )cαk (βk γ) cαk (βk ) · · · cα1 (β1 ) . Finally, define t : K n → K n and v : K n → H by: t(γ1 , . . . , γn ) = C1 (γ1 ), . . . , Cn (γn ) , v(γ1 , . . . , γn ) = γ1 · · · γn . We check easily that: v◦t = θ◦u◦s. Denote by ε = (e, . . . , e) ∈ K n . Since θ and s are isomorphisms of varieties, the differential d(v◦t)ε : Tε (K n ) → Te (L) = l of v◦t at the point ε is surjective. Let δk = cα1 (β1 ) · · · cαk (βk ), 1 k n. Then we obtain easily that: Ck (γ) = iδk iβk αk (γ)iβk (γ −1 ) . n If x = (x1 , . . . , xn ) ∈ Te (K) = Tε (K n ), then 23.4.7 implies that: d(v◦t)ε (x) =
n
Ad(δk ) Ad(βk ) Ad(αk )(xk ) − xk .
k=1
Since [k, [h, k]] ⊂ [h, k] and K is connected, 24.5.5 implies that Ad(γ)([h, k]) ⊂ [h, k] for all γ ∈ K. As δk , βk ∈ K, to obtain the result, it suffices therefore to prove that if x ∈ k and α ∈ H, then Ad(α)(x) − x ∈ [h, k]. But this follows again from 24.5.5 and the fact that H is connected. 24.5.12 Theorem. Let h be a Lie subalgebra of g. Then [h, h] is an algebraic Lie subalgebra of g. Furthermore, if H is a connected subgroup of G such that h = L(H), then [h, h] is the Lie algebra of D(H).
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Proof. Since [h, h] = [a(h), a(h)] (24.5.7), we may assume that h is algebraic. Then H = (h) and the result follows from 24.5.11.
a
24.5.13 Corollary. (i) If G is solvable (resp. nilpotent), then g is solvable (resp. nilpotent). (ii) Assume that G is connected. If g is solvable (resp. nilpotent), then G is solvable (resp. nilpotent). Proof. We have L(G) = L(G◦ ) and if G is solvable (resp. nilpotent), then so is G◦ . So we may assume that G is connected. Then 24.5.11 implies that L Dn (G) = Dn (g) and L(C n (G) = C n (g) (notations of 19.3.1). So the result follows.
24.6 A particular case 24.6.1 In this section, G denotes an algebraic group and g its Lie algebra. If x ∈ g, then an element of a(x) is called a replica of x. and nilpotent components. 24.6.2 Let x ∈ g and xs , xn its semisimple 1) We have xs , xn ∈ a(x) = L (x) . Thus a(xs ) ⊂ a(x), a(xn ) ⊂ a(x) (24.5.10), and (xs ) ⊂ (x), (xn ) ⊂ (x) (24.2.3). Note that the Lie algebra a(x) is commutative (24.4.4). 2) Let H ⊂ (x) be the subgroup of G generated by (xs ) and (xn ). Then H is a connected algebraic subgroup of G (21.3.1). Moreover, xs ∈ L (xs ) ⊂ L(H) and xn ∈ L( (xn ) ⊂ L(H). Hence x ∈ L(H), and (x) ⊂ H (24.2.3). It follows that H = (x). Since (x) is commutative, (x) = (xs ) (xn ). Applying 24.5.9 we obtain that a(x) = a(xs ) + a(xn ). 3) Let us reduce to the case where G is a closed subgroup of GL(V ) with V a finite-dimensional vector space (22.1.5). Suppose that x is semisimple. Using a basis of diagonalization for x, we may assume that G ⊂ GLn (k) and x ∈ dn (k). Since dn (k) = L Dn (k) , (x) ⊂ Dn (k) and a(x) ⊂ dn (k). Thus the elements of (x) (resp. a(x)) are semisimple. Suppose that x is nilpotent. We may assume that x ∈ nn (k). Again we see that (x) ⊂ Un (k) and a(x) ⊂ nn (k). Thus the elements of (x) are unipotent, and those of a(x) are nilpotent. In the general case, the preceding discussion shows that a(xs ) (resp. a(xn )) is the set of semisimple (resp. nilpotent) elements of a(x). Similarly, (xs ) (resp. (xn )) is the set of semisimple (resp. unipotent) elements of (x). As in 22.2.6, we obtain:
a a
a a
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Proposition. Let us conserve the above notations. (i) We have a(x) = a(xs ) ⊕ a(xn ). (ii) The map (xs ) × (xn ) → (x), (α, β) → αβ, is an isomorphism of algebraic groups.
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24.6.3 Suppose that G is a closed subgroup of GL(V ). To determine (x) (resp. a(x)), it suffices to know (xs ) and (xn ) (resp. a(xs ) and a(xn )).
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24.6 A particular case
381
a
For (xn ) and a(xn ), we may use 23.3.5 and 24.2.4. Let us consider xs . For simplicity, let us suppose that x semi-simple. We may assume, via a basis of diagonalization, that G ⊂ GLn (k) and x = diag(s1 , . . . , sn ) where s1 , . . . , sn ∈ k. Denote by: Lx = {(a1 , . . . , an ) ∈ Zn ; a1 s1 + · · · + an sn = 0}.
a
We saw in 24.6.2 that (x) ⊂ Dn (k) and a(x) ⊂ dn (k). Moreover, by 22.5.4 and 22.5.7, (x) is a torus and it is the intersection of the kernel of certain characters of Dn (k). Let H = Dn (k), h = dn (k), A(H) = k[T1 , T1−1 , . . . , Tn , Tn−1 ] and we identify L(H) with h via the map y → [y(Ti )] (see 23.3.3 and 23.3.4). Thus if P ∈ A(H), then: n ∂P x(P ) = si (e). ∂Ti i=1
a
Let of H. By 22.5.2, there exist integers a1 , . . . , an such that χ be a character χ diag(t1 , . . . , tn ) = ta1 1 · · · tann for all t1 , . . . , tn ∈ k \ {0}. If Q ∈ k[T, T −1 ] ∈ A(Gm ), then: n n ∂(Q◦χ) si (e) = si ai Q (1). dχ(x)(Q) = x(Q◦χ) = ∂Ti i=1 i=1 Hence (23.3.2):
d dχ(x) = (a1 s1 + · · · + an sn ) . dT T =1
a
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By 24.4.1 and the definition of (x), we deduce that (x) ⊂ ker χ if and only if (a1 , . . . , an ) ∈ Lx . Since (x) is the intersection of the kernel of certain characters of H, we have proved that (x) is the set of matrices diag(t1 , . . . , tn ) such that ta1 1 · · · tann = 1 for all (a1 , . . . , an ) ∈ Lx . On the other hand, Lx is a subgroup of Zn , so it is free and finitely generated. Applying 24.3.5 (iii) and 24.4.1, we see that a(x) is the set of matrices diag(u1 , . . . , un ) such that a1 u1 + · · · + an un = 0 for all (a1 , . . . , an ) ∈ Lx . To summarize, we have obtained the following result:
a
a
Theorem. Let x = diag(s1 , . . . , sn ) ∈ dn (k) and
a
Lx = {(a1 , . . . , an ) ∈ Zn ; a1 s1 + · · · + an sn = 0}.
The group (x) is the set of matrices diag(t1 , . . . , tn ) ∈ Dn (k) verifying ta1 1 · · · tann = 1 for all (a1 , . . . , an ) ∈ Lx . The Lie algebra a(x) is the set of matrices diag(u1 , . . . , un ) ∈ dn (k) such that a1 u1 + · · · + an un = 0 for all (a1 , . . . , an ) ∈ Lx . 24.6.4 Proposition. Let V be a finite-dimensional vector space and let x ∈ gl(V ). (i) Any element of (x) can be written as P (x) for some P ∈ k[T ]. (ii) Any replica of x can be written as Q(x) for some Q ∈ k[T ] without constant term.
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24 Correspondence between groups and Lie algebras
a
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Proof. Since (x) = (xs ) (xn ), a(x) = a(xs ) + a(xn ) (24.6.2) and xs , xn are polynomials without constant term in x, it suffices to consider on the one hand the case where x is nilpotent, and on the other hand, the case where x is semisimple. If x is nilpotent, then (x) = {etx ; t ∈ k} and a(x) = kx (23.3.5 and 24.2.4). So the result is clear. Suppose that x is semisimple. Denote by s1 , . . . , sp the pairwise distinct eigenvalues of x and V1 , . . . , Vp the corresponding eigenspaces. For i ∈ {1, . . . , p} and v ∈ Vi , set:
a
Gv = {α ∈ GL(V ); α(kv) ⊂ kv} , gv = {y ∈ gl(V ); y(kv) ⊂ kv}. By 24.5.6, Gv is an algebraic subgroup of GL(V ) whose Lie algebra is gv . Since x ∈ gv , we deduce easily that for all α ∈ (x) and y ∈ a(x), V1 , . . . , Vp are eigenspaces of α and y. So let α ∈ (x) and t1 , . . . , tp ∈ k \ {0} be such that α|Vi = ti idVi for 1 i p. There exists P ∈ k[T ] such that ti = P (si ) for 1 i p. Thus α = P (x), and we have proved (i). We see similarly that if y ∈ a(x), then there exists Q ∈ k[T ] such that y = Q(x). If x is invertible, by using the minimal polynomial of x, we see that idV is a polynomial without constant term in x, so we may assume that Q has no constant term. If x is not invertible, then we have, for example, s1 = 0. The set of elements y ∈ gl(V ) verifying y|V1 = 0 is an algebraic Lie subalgebra of gl(V ) containing x (24.5.6), hence y|V1 = 0 for all y ∈ a(x). But y = Q(x), so Q has no constant term.
a
a
24.6.5 Proposition. Let V be a finite-dimensional vector space and x ∈ gl(V ). Then x is nilpotent if and only if tr(xy) = 0 for all replica y of x. Proof. If x is nilpotent, then so is xp for all p ∈ N∗ , so tr(xp ) = 0. So 24.6.4 implies that tr(xy) = 0 for all replica y of x. Conversely, suppose that tr(xy) = 0 for all replica y of x. Since xy = yx, xn y = yxn and xn y is nilpotent. Consequently, tr(xs y) = 0 for all replica y of x. Since a(xs ) ⊂ a(x), it suffices to prove that if x is semisimple and tr(xy) = 0 for all y ∈ a(x), then x = 0. Let (v1 , . . . , vr ) be a basis of V consisting of eigenvectors of x with eigenvalues s1 , . . . , sr , and E the Q-vector subspace of k spanned by the si ’s. For a Q-linear form ϕ on E, define y ∈ gl(V ) to be the element such that y(vi ) = ϕ(si )vi for 1 i r. It is clear from 24.6.3 that y ∈ a(x). So: tr(xy) = s1 ϕ(s1 ) + · · · + sr ϕ(sr ) = 0. We deduce from this that: 2 2 ϕ(s1 ) + · · · + ϕ(sr ) = 0. Thus ϕ(s1 ) = · · · = ϕ(sr ) = 0. Since the si ’s span E, we have E = {0}.
24.8 Algebraic adjoint group
383
24.7 Examples 24.7.1 In this section, G denotes an algebraic group whose Lie algebra is g. We shall give examples of algebraic Lie subalgebras of g. 24.7.2 Let h be a Lie subalgebra of g. • If h = [h, h], then h is an algebraic Lie subalgebra of g (24.5.12). In particular, if h is semisimple, h is an algebraic Lie subalgebra of g. • Suppose that h = ng (a), where a is a Lie subalgebra of g. Then by 24.5.6, h is algebraic in g. In particular, if h is its own normalizer in g, then it is an algebraic Lie subalgebra of g. • Similarly, if h = cg (a), where a is a subspace of g, then h is an algebraic Lie subalgebra of g. 24.7.3 Lemma. Let h be an ideal of g. Then [g, a(h)] ⊂ h. In particular, a(h) is an ideal of g. Proof. Let k = {x ∈ g; [x, g] ⊂ h}. Then 24.5.6 says that k is algebraic in g. If h is an ideal of g, then h ⊂ k. Hence a(h) ⊂ k. 24.7.4 Proposition. Let h be a Lie subalgebra of g and r the radical of h. The following conditions are equivalent: (i) h is an algebraic Lie subalgebra of g. (ii) r is an algebraic Lie subalgebra of g. Proof. (i) ⇒ (ii) Suppose that h = a(h). To obtain the result, we may assume that g = h. So r is the largest solvable ideal of g. By 24.7.3, a(r) is an ideal of g and it is solvable since [a(r), a(r)] = [r, r] (24.5.7). Hence r = a(r). (ii) ⇒ (i) By 20.3.5, h = s ⊕ r where s is a semisimple Lie subalgebra of g. Since s is algebraic (24.7.2), if r is algebraic, then so is h (24.5.10). 24.7.5 Remark. Let r be the radical of h and s a semisimple Lie subalgebra of g verifying h = s ⊕ r. Then we obtain easily that a(h) = s ⊕ a(r). Furthermore, a(r) is the radical of a(h). 24.7.6 Let h be a Lie subalgebra of g such that x = xn for all x ∈ h. We claim that h is algebraic in g. We may assume that G = GL(V ). By 23.6.4, any x ∈ h is a nilpotent endomorphism. So by 23.3.5, kx is algebraic in g for all x ∈ h. Hence 24.5.10 implies that h is an algebraic Lie subalgebra of g.
24.8 Algebraic adjoint group 24.8.1 Definition. We define the algebraic adjoint group of a Lie algebra g to be the smallest algebraic subgroup of GL(g) whose Lie algebra contains adg g.
a
24.8.2 For any Lie subalgebra h of gl(g), denote by (h) the smallest algebraic subgroup of GL(g) whose Lie algebra contains h, and a(h) the smallest algebraic Lie subalgebra of gl(g) containing h.
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24 Correspondence between groups and Lie algebras
a
The algebraic adjoint group of g is therefore G = (ad g) and we have L(G) = a(ad g). If ad g is an algebraic Lie subalgebra of gl(g), then we say that G is the adjoint group of g, and that g is ad-algebraic. Let Aut g be the group of automorphisms of g. We have already seen that Aut g is an algebraic subgroup of GL(g) whose Lie algebra is Der g (24.3.7). So G ⊂ Aut g. If α ∈ Aut g, then α normalizes ad g, and therefore normalizes G. Thus G is a normal subgroup of Aut g. Applying 22.3.4 and 23.3.5, we see also that G contains the group of elementary automorphisms of g. We have also from 23.4.16 that G = Aute g if g is semisimple. 24.8.3 Let g be a Lie algebra and G its algebraic adjoint group. • Let x ∈ g. If α ∈ (ad x), there exists P ∈ k[T ] such that α = P (ad x) (24.6.4). It follows that any ideal of g is G-stable. Conversely, if a subspace of g is G-stable, then it is also ad g-stable (24.3.3). Thus G-stable subspaces of g are exactly the ideals of g. • The derivation ad x of g extends to a derivation of the symmetric algebra S(g) of g, denoted by adS(g) x. An ideal of S(g) is said to be g-invariant if it is stable under adS(g) x for all x ∈ g. In view of 23.4.14 and 24.3.4, an ideal of S(g) is g-invariant if and only if it is G-stable. By 24.3.4, we also have:
a
S(g)G = S(g)g . 24.8.4 Let us look at an example. Let α ∈ k \ {0} and g the Lie algebra with basis (x, y, z), and Lie bracket defined as follows: [x, y] = y , [x, z] = αz , [y, z] = 0, all other brackets can be deduced from these. Let X = ad x, Y = ad y, Z = ad z. To determine the algebraic adjoint group G of g and L(G), it suffices to determine (X), (Y ), (Z), and a(X), a(Y ), a(Z) (24.5.9). We shall identify an element of End(g) with its matrix with respect to the basis (x, y, z). By 23.3.5 and 24.2.4, we have:
a
a
a(Y ) = {etY ; t ∈ k} , a(Z) = {etZ ; t ∈ k} , a(Y ) = kY
a
, a(Z) = kZ.
The eigenvalues of X are 0, 1, α. In the notations of 24.6.3, LX is the set of elements (a1 , a2 , a3 ) ∈ Z3 such that a2 +a3 α = 0. Let us distinguish two cases: r 1) α = ∈ Q where r, s ∈ Z \ {0} are coprime. Then: s LX = {(n, pr, −ps); n, p ∈ Z}.
a
Thus a(X) = kX. Therefore g is ad-algebraic. The group (X) is the set of −ps = 1 for all n, p ∈ Z. So it is the set of matrices diag(t1 , t2 , t3 ) with tn1 tpr 2 t3 s r matrices diag(1, t , t ) where t ∈ k \ {0}. We deduce easily that:
24.8 Algebraic adjoint group
385
⎫ ⎧⎛ ⎞ ⎬ ⎨ 1 0 0 G = ⎝ u ts 0 ⎠ ; u, v ∈ k , t ∈ k \ {0} . ⎭ ⎩ v 0 tr 2) α ∈ Q. Then LX = Z × {0} × {0}. Thus a(X) is 2-dimensional and it is spanned, as a vector space, by X and T = diag(0, 0, 1). We have a(ad g) = kX ⊕ kT ⊕ kY ⊕ kZ, so a(ad g) is of dimension 4. The group (X) is the set of matrices diag(1, t, t ) with tt = 0. We obtain that: ⎧⎛ ⎫ ⎞ ⎨ 100 ⎬ G = ⎝ u t 0 ⎠ ; u, v ∈ k , t, t ∈ k\{0} . ⎩ ⎭ v 0 t
a
In particular, g is not ad-algebraic. 24.8.5 Proposition. Let g be a Lie algebra, G its algebraic adjoint group, h a Lie subalgebra of g, K = (adg h) and k = a(adg h). (i) For x ∈ h, α ∈ K and X ∈ k, we have α(x) ∈ h and X(x) ∈ h. (ii) Let L = {α|h ; α ∈ K} and l = {X|h ; X ∈ k}. Then L is the algebraic adjoint group of h, and l is the smallest algebraic Lie subalgebra of gl(h) containing adh h.
a
a
Proof. (i) Since K is generated by the (adg y), y ∈ h, the fact that K(h) ⊂ h follows from 24.6.4 (i). In the same way, X(h) ⊂ h follows from 24.6.4 (ii). (ii) Let u : K → GL(h) be the map α → α|h . Then is a rational (h, u) du(adg h) by representation of K. We have L = u(K) = u (adg h) = 24.5.4, hence u(K) = (adh h). Thus L is the algebraic adjoint group of h. The second part of (ii) is clear since l = L(u(K)) by 24.5.3.
a
a
a
References and comments • [5], [23], [38], [40], [78]. The presentation of 24.2 given here is inspired by [38]. Most of the notions (like replica) of sections 24.5, 24.6, 24.7 and 24.8 are taken from [23]. The fact that the base field is of characteristic zero is essential in establishing the results of this chapter. When the characteristic of the base field is positive, the correspondence between algebraic groups and their Lie algebras becomes less “nice”.
25 Homogeneous spaces and quotients
We consider in this chapter actions of an algebraic group on an algebraic variety. The notions of homogeneous space and geometric quotient are studied, and we show that the quotient of an algebraic group by a closed subgroup has a natural structure of algebraic variety. In this chapter, G is an algebraic group.
25.1 Homogeneous spaces 25.1.1 Proposition. Let X be an irreducible G-variety. If X is a Ghomogeneous space, then X is a smooth normal variety. Proof. This follows from 16.5.5 and 17.1.5. 25.1.2 Theorem. Let X, Y be G-varieties, u : X → Y a G-equivariant morphism and r = dim X − dim Y . Suppose further that X and Y are Ghomogeneous spaces. (i) The map u is open. For all y ∈ Y , each irreducible component of u−1 (y) has dimension r. (ii) Let W be a closed irreducible subvariety of Y and Z an irreducible component of u−1 (W ). Then Z dominates W and r = dim Z − dim W . (iii) Let Z be a variety and v : X × Z → Y × Z be the morphism defined by (x, z) → (u(x), z). Then v is open. (iv) The map u is an isomorphism if and only if u is bijective. Proof. In view of 21.4.4, we may reduce easily to the case where G is connected and X, Y are irreducible. Note that since X and Y are G-homogeneous spaces and u is G-equivariant, u is surjective. (i) By 15.5.4, there exists a non-empty open subset V of Y such that for all y ∈ V , all the irreducible components of u−1 (y) have dimension r. Let y ∈ Y . Since G acts transitively on Y , we have α.y ∈ V for some α ∈ G. It follows easily from the G-equivariance of u that if Z1 , . . . , Zs are the irreducible components of u−1 (y), then α.Z1 , . . . , α.Zs are the irreducible components of
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u−1 (α.y). Now the map x → α.x is an isomorphism of the variety X, so dim Zi = r for all 1 i s. We deduce from 17.4.11 and 25.1.1 that u is open. (ii) By 17.1.8 and 15.5.3, there exists a non-empty open subset V of Y such that V is normal and for all y ∈ V , x ∈ u−1 (y) and irreducible subvariety Y of Y containing y, there is at least one irreducible component of u−1 (Y ) containing x which dominates Y and for such an irreducible component X , we have r = dim X − dim Y . Let Z be an irreducible component of u−1 (W ), then there exists z ∈ Z which is contained in no other irreducible components of u−1 (W ). Moreover, there exists α ∈ G such that u(α.z) = α.u(z) ∈ V . It is clear that α.Z is an irreducible component of u−1 (α.W ) containing α.z. Applying the preceding result with x = α.z, y = u(α.z), and Y = α.W , we obtain that α.Z dominates α.W and dim(α.Z) − dim(α.W ) = r. So (ii) follows. (iii) We may assume that Z is affine. Moreover, if (iii) is true for Z, then it is also true for any closed subvariety of Z. So we may further assume that Z = kn . By (i), all the fibres of v have dimension r. So the result follows from 17.4.11 since Y × kn is normal (17.3.2 and 25.1.1). (iv) Since Y is normal (25.1.1), the result follows from 17.4.8. 25.1.3 Let X be a G-variety and x ∈ X. The stabilizer Gx of x in G is an algebraic subgroup of G (21.4.2) and the orbit G.x of x is a locally closed subset of X (21.4.3). Further, we have by 21.4.3: dim(G.x) = dim G − dim Gx . Recall (21.4.4) that G.x is pure and its irreducible components are exactly the G◦ -orbits (thus pairwise disjoint) in G.x. Let C be an irreducible component of G.x. Then it is a G◦ -homogeneous space, so it is normal and smooth (25.1.1). Let π : G → G.x, α → α.x, be the orbit morphism. If we consider the G-action on G by left translation, then π is G-equivariant. By 25.1.2, π is open. Note also that if α ∈ G, then π −1 (α.x) = αGx . Proposition. The map dπe : L(G) → Tx (G.x) is surjective and its kernel is L(Gx ). Proof. Let Y be an irreducible component of G.x containing x. We have Tx (G.x) = Tx (Y ) since irreducible components of G.x are pairwise disjoint, and Y is the image of the restriction of π to G◦ . So we may consider π to be a surjective morphism from G◦ onto Y . By 16.5.7, there exists a non-empty open subset U of G◦ such that dπα : Tα (G◦ ) → Tα.x (Y ) is surjective for all α ∈ U . Since G◦ and Y are both G◦ -homogeneous spaces, we deduce that dπe is surjective. Thus dim ker(dπe ) = dim G − dim(G.x) = dim L(Gx ). But it is clear that L(Gx ) ⊂ ker(dπe ). So we have equality.
25.2 Some remarks
389
25.2 Some remarks 25.2.1 Let u : (X, OX ) → (Y, OY ) be a morphism of varieties, U ⊂ X and V ⊂ Y open subsets such that u(U ) ⊂ V . If g ∈ OY (V ), we define U U uU V (g) : U → k by uV (g)(x) = g ◦ u(x) for all x ∈ U . Clearly uV (g) ∈ OX (U ). 25.2.2 In this section, X denotes a G-variety. If U is an open subset of X, we denote by OX (U )G the set of f ∈ OX (U ) such that f (x) = f (y) if x and y belong to the same G-orbit. Lemma. Let U be an open subset of X, f ∈ OX (U )G and V = α∈G α.U . There exists a unique element g ∈ OX (V )G such that g|U = f . Proof. The uniqueness is obvious. So let us prove the existence of g. For α ∈ G, the map uα : α.U → U , x → α−1 .x is an isomorphism of varieties. So gα : α.U → k, x → f (α−1 .x), is an element of OX (α.U ). For α, β ∈ G and x ∈ α.U ∩ β.U , there exist y, z ∈ U such that x = α.y = β.z. It follows that y, z ∈ G.x, so gα (x) = f (y) = f (z) = gβ (y). By the properties of a sheaf, there exists g ∈ OX (V ) such that g|α.U = gα for all α ∈ G. 25.2.3 Assume that U is an open G-stable subset of X. If α ∈ G, the map α.f : U → U , x → f (α−1 .x), is an element of OX (U ). It is then easy to check that G × OX (U ) → OX (U ) , (α, f ) → α.f is an action of G on OX (U ). The set of fixed points under this action is clearly OX (U )G . 25.2.4 Lemma. Assume that X is irreducible. Let U, V be non-empty open subsets of X such that U ⊂ V , and f ∈ OX (U ), g ∈ OX (V ) verifying g|U = f . Then the following conditions are equivalent: (i) g ∈ OX (V )G . (ii) f ∈ OX (U )G . Proof. (i) ⇒ (ii) This is obvious. (ii) ⇒ (i) Suppose that there exist α ∈ G and x ∈ V such that α.x ∈ V and g(x) = g(α.x). Then x ∈ α−1 .V . Set U = U ∩(α−1 .U ) and V = V ∩(α−1 .V ). Let θ : X → X ×X be the morphism y → (y, α.y). The map h = g⊗1−1⊗g belongs to OX×X (V ×V ), and the restriction of h◦θ to V belongs to OX (V ). We have h ◦ θ(x) = 0 and h ◦ θ|U = 0. But this is absurd since U is dense in V . Hence g ∈ OX (V )G . 25.2.5 Let us assume that X is irreducible. Let f ∈ OX (U ) and h ∈ R(X) the rational function defined by f . Denote by V the domain of definition of h and g ∈ OX (V ) the function induced by h. If f ∈ OX (U )G , then 25.2.4 says that g ∈ OX (V )G . Hence the open subset V is G-stable (25.2.2).
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Conversely, given h ∈ R(X), V its domain of definition, and g ∈ OX (V ) the function induced by h. By 25.2.2, if g ∈ OX (V )G , then V is G-stable and g|U ∈ OX (U )G for all open subsets U in V . Let us denote by R(X)G the set of rational functions h verifying the following condition: if V is the domain of definition of h, and g ∈ OX (V ) the function induced by h, then g ∈ OX (V )G . We can interpret R(X)G in another way. Namely, let α ∈ G, U an open subset of X and uU α : α.U → U the isomorphism of varieties defined by x → α−1 .x. If f ∈ OX (U ), then f ◦ uU α ∈ OX (α.U ), and if V is an open subset )| of U , then (f |V ) ◦ uVα = (f ◦ uU α α.V . Denote by uα : R(X) → R(X) the inductive limit of the family (uU α )U . Thus we obtain an action of G on R(X). Then R(X)G is the set of invariants of G under this action. 25.2.6 Assume that X is irreducible. We deduce from the preceding discussion that for a non-empty affine open subset U of X, we have Fract(OX (U )G ) ⊂ R(X)G . 25.2.7 Let x ∈ X, Z = G.x, H = Gx and π : G → Z be as in 25.1.3. Endow G with the structure of H-variety via right translation. Let V be a non-empty open subset of Z and U = π −1 (V ). For α ∈ U and β ∈ H, we have αβ ∈ U , so U is H-stable, and OG (U )H is the set of functions h ∈ OG (U ) verifying f (αβ) = f (α) for all (α, β) ∈ U × H. Let g ∈ OZ (V ) and f = πVU (g) ∈ OG (U ). Then for (α, β) ∈ U × H, we have: f (αβ) = g((αβ).x) = g(α.x) = f (α). Thus f ∈ OG (U )H and the map OZ (V ) → OG (U )H , g → πVU (g), is injective. We shall show that it is bijective. Let C1 , . . . , Cn be the irreducible components of Z. By 21.4.4 and 25.1.1, they are open and closed, smooth, normal, pairwise disjoint G◦ -orbits in Z. Let Vi = V ∩ Ci and Ui = π −1 (Vi ), 1 i n. We may assume that Vi = ∅ for 1 i r and Vi = ∅ if i > r. Clearly, the open subsets Ui are pairwise disjoint and H-stable. So to obtain the result, it suffices to consider the case where V is irreducible. Assume that V is irreducible. Then V = Ci = G◦ .y for some i and y ∈ Z. Moreover, V is a normal variety by the preceding discussion. Let f ∈ OG (U )H . Set ∆ ⊂ U × k = {(α, f (α)); α ∈ U }. It is closed in U × k (12.5.4). Let u : G × k → Z × k , (α, λ) → (π(α), λ). By 25.1.2 (iii), u is an open morphism. Let ∆ = u(∆) ⊂ V × k. Let (π(β), f (β)) ∈ ∆ where β ∈ U , and (α, λ) ∈ (U × k) \ ∆. If (π(β), f (β)) = u(α, λ), then π(β) = π(α) and λ = f (β). So α ∈ βH and f (α) = f (β) = λ, hence (α, λ) ∈ ∆ which is absurd. It follows that u−1 (∆ ) = ∆ and ∆ is a closed subvariety of V × k. Let G◦ β1 , . . . , G◦ βr be the irreducible components of G having non-empty intersection with U . Set Ui = G◦ βi ∩ U . Then we have π(Ui ) = V because π(G◦ βi ) = G◦ .y.
25.3 Geometric quotients
391
Let γ ∈ U \U1 , then there exists γ1 ∈ U1 such that γ.x = γ1 .x. So γ ∈ γ1 H, and hence π(γ) = π(γ1 ) and f (γ) = f (γ1 ). It follows that ∆ is the image of the morphism U1 → V × k, β → (π(β), f (β)). Since U1 is open in G◦ β1 , it is irreducible. We obtain therefore that ∆ is also irreducible. Let p : ∆ → V (resp. q : ∆ → k) be the restriction to ∆ of the canonical projection Z × k → Z (resp. Z × k → k). Then p is bijective, and so by 17.4.8, p is an isomorphism. Thus q ◦ p−1 ∈ OZ (V ) and if α ∈ U , then q ◦ p−1 (π(α)) = f (α). Hence f = πVU (q ◦ p−1 ) and we are done.
25.3 Geometric quotients 25.3.1 Definition. Let X be a G-variety. A geometric quotient of X by G is a pair (Y, π) where Y is a variety and π : X → Y is a morphism verifying the following conditions: (i) π is open, constant on G-orbits and defines a bijection from the set of G-orbits onto Y . (ii) If V is an open subset of Y and U = π −1 (V ), πVU induces a bijection from OY (V ) onto OX (U )G . 25.3.2 Let G and X be as in 25.3.1. If a geometric quotient (Y, π) of X by G exists, then the fibres of π are the G-orbits, so they are closed. We can further precise condition (i) of 25.3.1. Lemma. Let X be a G-variety and θ : X → Y a surjective morphism of varieties whose fibres are the G-orbits in X. Then the following conditions are equivalent: (i) The map θ is open. (ii) For all G-stable open subset U of X, θ(U ) is an open subset of Y . (iii) For all G-stable closed subset F of X, θ(F ) is a closed subset of Y . In particular, if θ is a closed map, then it is also open. Proof. Clearly, (i) ⇒ (ii). Conversely, if U is an open subset of X, then U = α∈G α.U is a G-stable open subset of X such that θ(U ) = θ(U ). So (ii) ⇒ (i). Let F be a G-stable closed subset of X, then U = X \ F is a G-stable open subset of X. Thus θ(F ) = Y \ θ(U ). So (ii) ⇒ (iii). Conversely, if U is a G-stable open subset of X, then F = X \ U is a G-stable closed subset of X and θ(U ) = Y \ θ(F ). Hence (iii) ⇒ (ii). 25.3.3 In the rest of this section, X is a G-variety. Proposition. Assume that a geometric quotient (Y, π) of X by G exists. Let Z be a variety and u : X → Z a morphism which is constant on G-orbits in X. Then there exists a unique morphism v : Y → Z such that u = v ◦ π. Proof. The hypothesis implies that there is a unique map v : Y → Z such that u = v ◦ π. Let W be an open subset of Z. Then u−1 (W ) = π −1 (v −1 (W ))
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is open in X. Since π is surjective and open, v −1 (W ) = π(u−1 (W )) is open in Y . Thus v is continuous. Now let V = v −1 (W ), U = u−1 (W ) and f ∈ OZ (W ). To obtain the result, we need to prove that f ◦ v ∈ OY (V ). But f ◦ v ◦ π = f ◦ u ∈ OX (U ) and since π is constant on G-orbits, f ◦ v ◦ π ∈ OX (U )G . As πVU induces a bijection from OY (V ) onto OX (U )G , f ◦ v ∈ OY (V ). 25.3.4 Corollary. If (Y, π) and (Y , π ) are geometric quotients of X by G, then there exists a unique isomorphism v : Y → Y such that π = v ◦ π. 25.3.5 In view of 25.3.4, we can talk about, if it exists, the geometric quotient of X by G. The existence of a geometric quotient of X by G requires very strong conditions on the G-orbits as we shall see in the following result. Proposition. Assume that X is irreducible. Let θ : X → Y be a surjective morphism of varieties whose fibres are the G-orbits in X. (i) The G-orbits in X have dimension r = dim X − dim Y . (ii) If Y is normal, then (Y, θ) is the geometric quotient of X by G. Proof. (i) By 15.5.4, dim G.x r for all x ∈ X and there is a non-empty open subset V of Y such that dim G.x = r for all x ∈ θ−1 (V ). Let T = {(α, x, α.x); α ∈ G, x ∈ X} ⊂ G × X × X be the graph of the morphism (α, x) → α.x. It is closed in G × X × X (12.5.4). Let ∆ be the diagonal of X, Z = T ∩ (G × ∆) and p : Z → ∆ the canonical projection. If x ∈ X, then p−1 (x, x) = Gx × {(x, x)}, so all the irreducible components of p−1 (x, x) are of the same dimension. Let C be an irreducible component of Z containing {e} × ∆ and q = p|C . Then q is surjective and if x ∈ X, then q −1 (x, x) is the union of irreducible components of p−1 (x, x). Applying 15.5.4, we obtain that dim Gx dim C − dim ∆ for all x ∈ X and dim Gx = dim C − dim ∆ if x belongs to a non-empty open subset U of X. It follows that for x ∈ X, r dim G.x = dim G − dim Gx dim G − dim C + dim ∆ = s. Since dim G.x = r = s for x ∈ U ∩ θ−1 (V ), we obtain that dim G.x = r for all x ∈ X. (ii) Assume that Y is normal. Since G-orbits are pure varieties (21.4.4), it follows from 17.4.11 and part (i) that θ is open. Let V be a non-empty open subset of Y and U = θ−1 (V ). The open subset U is G-stable and if g ∈ OY (V ), then θVU (g) ∈ OX (U )G . The map g → θVU (g) is therefore an injection from OY (V ) into OX (U )G . We need to prove that it is surjective. Let f ∈ OX (U )G and ∆ = {(x, f (x)); x ∈ U } ⊂ U × k. By 12.5.4, ∆ is closed in U × k. Let u : X × k → Y × k be the morphism (x, λ) → (θ(x), λ).
25.4 Quotient by a subgroup
393
For x ∈ X and λ ∈ k, u−1 (θ(x), λ) = G.x × {λ}. So all the irreducible components of the fibres of u have dimension r. Since Y ×k is normal (17.3.2), it follows from 17.4.11 that u is open. Since X is irreducible, so is ∆ and hence ∆ = u(∆) is also irreducible. Proceeding as in 25.2.7, we obtain that f ∈ θVU (OY (V )). 25.3.6 Proposition. Assume that X is irreducible and that (Y, π) is a geometric quotient of X by G. Then the comorphism ϕ of π is an isomorphism from R(Y ) onto R(X)G . Proof. Since π is dominant, the comorphism ϕ : R(Y ) → R(X) is welldefined. Clearly, ϕ(R(Y )) ⊂ R(X)G . Conversely, let h ∈ R(X)G . Its domain of definition U is G-stable and V = π(U ) is an open subset such that π −1 (V ) = U . Since ϕ induces a bijection between OY (V ) and OX (U )G , we have h ∈ ϕ(R(Y )).
25.4 Quotient by a subgroup 25.4.1 Let V be a finite-dimensional k-vector space, W a subspace d of V dimension d > 0, G = GL(V ) and g = gl(V ). The exterior power d W . Denote by π (resp. dπ) the representation of G contains the line L = d V given in 10.6.5 (resp. 23.4.14). (resp. g) on Lemma. Let α ∈ G and x ∈ g. (i) We have α(W ) = W if and only if π(α)(L) = L. (ii) We have x(W ) ⊂ W if and only if dπ(x)(L) ⊂ L. Proof. If α(W ) = W (resp. x(W ) ⊂ W ), then it is clear that π(α)(L) = L (resp. dπ(x)(L) ⊂ L). (i) Let (e1 , . . . , en ) be a basis of V such that (e1 , . . . , ed ) is a basis of W and (el+1 , . . . , el+d ) a basis of α(W ). Clearly L = ke1 ∧ · · · ∧ ed and π(α)(L) = kel+1 ∧ · · · ∧ el+d . Since the elements ei1 ∧ · · · ∧ eid , 1 i1 < · · · < id n, d form a basis of V , it is clear that if π(α)(L) = L, then l = 0 and hence α(W ) = W . n aij ej and v = e1 ∧ · · · ∧ vd . Then: (ii) For 1 i d, let x(ei ) = i=1
dπ(x)(v) = =
d
e1 ∧ j=1 n d
· · · ∧ ej−1 ∧ x(ej ) ∧ ej+1 ∧ · · · ∧ ed
aij e1 ∧ · · · ∧ ej−1 ∧ ei ∧ ej+1 ∧ · · · ∧ ed .
i=1 j=1
If x(W ) ⊂ W , then aij = 0 for some d < i n, 1 j d, and we obtain clearly that dπ(x)(L) ⊂ L.
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25 Homogeneous spaces and quotients
25.4.2 Let g be the Lie algebra of G, H a closed subgroup of G and h the Lie algebra of H. Let us consider the action of G on A(G) by right translation and conserve the notation ρα of 22.1.4. We shall also used the right convolution defined in 23.2.2. Lemma. There exists a finite-dimensional subspace V of A(G), stable by right translations, and a subspace W of V such that: H = {α ∈ G; ρα (W ) = W } , h = {x ∈ g; W ∗ x ⊂ W }. Proof. Let (u1 , . . . , ur ) be a system of generators of the defining ideal a = I(H) in A(G). There exists a finite-dimensional G-submodule V of A(G) containing u1 , . . . , ur (22.1.1). Set W = V ∩ a. Let α ∈ H. Then ρα (W ) = W (22.1.6). Conversely, if α ∈ G verifies ρα (W ) = W , then ρα (ui ) ∈ a for 1 i r, so ρα (a) ⊂ a. Hence α ∈ H by 22.1.6. The second part is proved in the same manner by using 23.2.5 and 23.4.4. 25.4.3 Theorem. Let g be the Lie algebra of G, H a closed subgroup of G and h its Lie algebra. There exists a rational finite-dimensional G-module V and a line L in V such that: H = {α ∈ G; π(α)(L) = L} , h = {x ∈ g; dπ(x)(L) ⊂ L}. Proof. This is immediate by 25.4.1 and 25.4.2. 25.4.4 Corollary. Let us conserve the notations of 25.4.3. There exists a quasi-projective G-variety X which is a G-homogeneous space, and an element x ∈ X such that: (i) H is the stabilizer of x in G. (ii) Let θ : G → X be the morphism θ(α) = α.x for α ∈ G. The fibres of θ are the left cosets αH, α ∈ G. Proof. Let V and L be as in 25.4.3, P(V ) the projective space associated to V , ϕ : V \ {0} → P(V ) the canonical surjection and x = ϕ(L). Set α.ϕ(v) = ϕ π(α)(v) for α ∈ G and v ∈ V \{0}. This defines a rational action of G on P(V ). Denote by X the G-orbit of x in P(V ). By 21.4.3, X is a quasi-projective variety. So the corollary follows from 25.4.3. 25.4.5 Theorem. Let g be the Lie algebra of G, H a closed normal subgroup of G and h its Lie algebra. Then there exists a finite-dimensional rational G-module (W, ϕ) such that H = ker ϕ and h = ker dϕ. Proof. Let V, π, L be as in 25.4.3 and X ∗ (H) the group of characters of H. For χ ∈ X ∗ (H), denote by Vχ the weight space of weight χ of H in V . Recall from 22.4.5 that since H is normal, π(G) permutes the Vχ ’s. Further L ⊂ Vχ0
25.4 Quotient by a subgroup
395
for some χ0 ∈ X ∗ (H), so we may assume that V is the direct sum of the weight spaces Vχ (22.4.6). Let E be the subspace of gl(V ) consisting of endomorphisms such that
(Vχ ) ⊂ Vχ for all χ. Observe that E is isomorphic to the direct sum of the gl(Vχ )’s. Then for α ∈ G, ∈ E and χ ∈ X ∗ (H), we have π(α) ◦ ◦ π(α−1 )(Vχ ) = π(α) ◦ (Vχα ) ⊂ π(α)(Vχα ) ⊂ Vχ where χα is as defined in 22.4.3. So π(α) ◦ ◦ π(α−1 ) ∈ E and this defines a −1 map ϕ : G → GL(E) where ϕ(α)( ) = π(α) ◦ ◦ π(α ). It follows from 23.5.4 that ϕ(α) = AdGL(V ) π(α) . So ϕ is a rational representation of G. Since π(α), α ∈ H, acts on each Vχ by scalar multiplication, we obtain that ϕ(α)( ) = for all ∈ E, and so α ∈ ker ϕ. Conversely, if α ∈ ker ϕ, then π(α) ◦ = ◦ π(α) for all ∈ E. It follows that π(α)(Vχ ) ⊂ Vχ for all χ ∈ X ∗ (H) and π(α)|Vχ is central in gl(Vχ ). Hence π(α) acts on Vχ by scalar multiplication. In particular π(α)(L) ⊂ L and so α ∈ H (25.4.3). The fact that h = ker dϕ now follows from 24.4.1. 25.4.6 Let X be a G-variety, x ∈ X, H = Gx and θ : G → G.x the morphism α → α.x. If α ∈ G, then θ−1 (α.x) = αH. Now H acts on G in the following way: for α ∈ G and β ∈ H, β.α = αβ −1 . Then the fibres of π are the H-orbits of G. Proposition. With the above notations, (G.x, θ) is the geometric quotient of G by H. Proof. By 25.1.2, θ is open. So the result is just a reformulation of 25.3.5. 25.4.7 Let H be a closed subgroup of G and π : G → G/H the canonical surjection. By 25.4.4 and 25.4.6, there exists a geometric quotient (X, θ) of G by H (action of H on G as in 25.4.6) and x ∈ X, verifying: • X is a smooth quasi-projective G-variety. • Gx = H. • If α ∈ G, then θ(α) = α.x. Define ϕ : G/H → X by ϕ(αH) = α.x. This defines a bijection from G/H onto X. Transporting the topology of X to G/H, (G/H, π) is then a geometric quotient of G by H. Moreover, by 25.3.4, the topology thus obtained on G/H (which makes it into a quasi-projective variety) is the unique topology for which (G/H, π) is a geometric quotient of G by H. Unless otherwise stated, we shall always endow G/H with this structure. 25.4.8 Proposition. Let H, K be closed subgroups of G such that H ⊂ K. Then the canonical map G/H → G/K is a morphism of algebraic varieties. Proof. This is clear by 25.3.3 and 25.4.7.
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25 Homogeneous spaces and quotients
25.4.9 For i = 1, 2, let Gi be an algebraic group, Hi a closed subgroup of Gi and πi : Gi → Gi /Hi the canonical surjection. Let G = G1 × G2 , H = H1 × H2 and u : G/H → (G1 /H1 ) × (G2 /H2 ) , (α1 , α2 )H → (π1 (α1 ), π2 (α2 )). Proposition. (i) The map u is an isomorphism of varieties. (ii) Let v : G1 → G2 be a morphism of algebraic groups such that v(αH1 ) ⊂ v(α)H2 for all α ∈ G. Then the map w : G1 /H1 → G2 /H2 such that w ◦ π1 = π2 ◦ v is a morphism of varieties. Proof. (i) The group G acts naturally on G1 /H1 × G2 /H2 and the stabilizer (resp. G-orbit) of (H1 , H2 ) is H (resp. (G1 /H1 ) × (G2 /H2 )). Let θ : G → (G1 /H1 ) × (G2/H2 ) be the morphism (α1 , α2 ) → π1 (α1 ), π2 (α2 ) . Then in view of 25.4.6, (G1 /H1 ) × (G2 /H2 ), θ is a geometric quotient of G by H. So (i) follows from 25.3.4. (ii) Since π2 ◦ v : G1 → G2 /H2 is a morphism, the result follows clearly by 25.3.3. 25.4.10 Proposition. Let H be a closed subgroup of G. The map u : G × (G/H) → G/H , (α, βH) → αβH is a morphism of varieties. Proof. Applying 25.4.9 with G1 = G2 = G, H1 = {e} and H2 = H, we obtain that the varieties (G × G)/({e} × H) and G × (G/H) are isomorphic. Let π : G → G/H be the canonical surjection. Then the morphism v = π ◦ µG : G × G → G/H is constant on the ({e} × H)-orbits in G × G (via the right translation action). So the result follows from 25.3.3. 25.4.11 Theorem. Let H be a closed normal subgroup of G and π : G → G/H the canonical surjection. (i) The group structure of G/H together with its structure of variety define a structure of affine algebraic group on G/H. The map π is a morphism of algebraic groups. (ii) The map dπ induces an isomorphism from L(G)/L(H) onto L(G/H). Proof. By 25.4.5, there is a finite-dimensional rational G-module (V, ϕ) such that H = ker ϕ and L(H) = ker dϕ. Let K = ϕ(G), then K is a closed subgroup of GL(V ), so it is affine. Endow GL(V ) with the following G-variety structure: for α ∈ G and β ∈ GL(V ), α.β = ϕ(α) ◦ β. Then K is the G-orbit of idV and GidV = H. Let π : G → G. idV = K be the morphism α → ϕ(α) ◦ idV . By 25.4.6, (K, π) is a geometric quotient of G by H. Thus G/H is an affine variety. Moreover, by construction, the group structure on G/H is the one from the algebraic group K. So (i) follows. By 24.4.7, L(H) is an ideal of L(G), so (ii) follows from 25.1.3.
25.5 The case of finite groups
397
25.4.12 Let H, K be closed normal subgroups of G such that H ⊂ K. Denote by π : G → G/H the canonical surjection. Proposition. (i) The set π(K) is a closed normal subgroup of G/H, isomorphic to the algebraic group K/H. (ii) The algebraic groups G/K and (G/H)/π(K) are isomorphic. Proof. (i) Clearly, π(K) is a normal subgroup of G/H. It is closed by 21.2.4 and 25.4.11 (or by 25.3.2). Let θ : K → π(K) be the morphism induced by π. By 25.3.3, there is a unique morphism of varieties u : K/H → π(K) such that θ = u ◦ π where π : K → K/H is the canonical surjection. It is clear that u is a bijective group homomorphism. Hence by 21.2.6, u is an isomorphism of algebraic groups. (ii) The kernel of the composition G → G/H → (G/H)/π(K) of canonical surjections is K. By 25.3.3, this defines a morphism of varieties v : G/K → (G/H)/π(K). The morphism v is a bijective group homomorphism, so the result follows again from 21.2.6. 25.4.13 Let H be a closed subgroup of G. In general, if H is not normal, G/H is not affine. However, we have the following result: Proposition. Let G be a unipotent algebraic group and H a closed subgroup of G. Then the variety G/H is affine. Proof. Let V, L, π be as in 25.4.3. Since H is unipotent, LH = {0} (22.3.6). It follows that LH = L and for x ∈ L \ {0}, the map θ : G → G.x, α → α.x, induces an isomorphism from G/H to G.x (25.4.6). Since G.x is affine (22.3.6), we are done.
25.5 The case of finite groups 25.5.1 Proposition. Let G be a finite group consisting of automorphisms of an affine variety X. Then the algebra of invariants A(X)G is finitely generated. Moreover, if X is irreducible, then R(X)G = Fract(A(X)G ). Proof. Let f1 , . . . , fn be a system of generators for A(X) and α1 , . . . , αs the elements of G with α1 = eG . For 1 i n, set: Pi (T ) = (T − α1 .fi ) · · · (T − αs .fi ). Then G permutes the roots of Pi , so the coefficients of Pi are in A(X)G . Let B be the subalgebra of A(X)G generated by the coefficients of the Pi ’s. Since Pi (fi ) = 0, fi is integral over B, and so A(X) is integral over B (3.1.5). Moreover, A(X) is a finite B-algebra (3.3.5). Since B is noetherian, A(X)G is a finitely generated B-module, so the algebra A(X)G is finitely generated. Suppose that X is irreducible. Let f = uv −1 ∈ R(X)G where u, v ∈ A(X) and: v1 = (α1 .v) · · · (αs .v) , u1 = u(α2 .v) · · · (αs .v). Then v1 ∈ A(X)G and f = u1 v1−1 . Hence u1 ∈ A(X)G and so we have f ∈ Fract(A(X)G ).
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25 Homogeneous spaces and quotients
25.5.2 Let G be a finite group consisting of automorphisms of an irreducible affine variety X. By 25.5.1, there is an affine variety Y such that Y = Spm(A(X)G ). Let u : X → Y be the morphism defined by the canonical injection A(X)G → A(X). Proposition. The pair (Y, u) is the geometric quotient of X by G. Proof. We saw in the proof of 25.5.1 that A(X) is integral over A(X)G . If n ∈ Spm(A(X)G ), then there exists m ∈ Spm(A(X)) such that n = m ∩ A(X)G (3.3.2 and 3.3.3). Thus u is surjective. Given x ∈ X, we have G.x ⊂ u−1 (u(x)). Suppose that there exists y ∈ −1 u (u(x)) such that y ∈ G.x. Since the finite sets G.x and G.y are disjoint and closed, it follows from 11.3.3 that there exists f ∈ A(X) verifying f |G.x = 1 and f |G.y = 0. Set g = (α1 .f ) · · · (αs .f ) where α1 , . . . , αs are the elements of G. Then g ∈ A(X)G and g|G.x = 1, g|G.y = 0. This is absurd since u(x) = u(y). Thus the fibres of u are the G-orbits. By 15.3.2, u is a finite morphism, and therefore u is closed (15.3.4). Since the fibres of u are the G-orbits, 25.3.2 implies that u is open. Let V be an open subset of Y and U = u−1 (V ). Then U is G-stable and G G it is clear that uU V (OY (V )) ⊂ OX (U ) . Conversely, let f ∈ OX (U ) . Since G G G X is irreducible, we have OX (U ) ⊂ R(X) = Fract(A(X) ) (12.8.2 and 25.5.1). It follows that there exist g, h ∈ A(X)G = A(Y ) such that f = g/h and h|U = 0. Hence g/h ∈ OY (V ), and uU V is bijective. So we have proved that (Y, u) is the geometric quotient of X by G. 25.5.3 In the case of a linear action, we have the following more precise statement: Proposition. Let V be a finite-dimensional vector space, G a finite subgroup of GL(V ) and N the order of G. Then the algebra A(V )G is generated by homogeneous invariants of degree at most N . Proof. Let p : A(V ) → A(V ) denote the map defined by: p(u) =
1 α.u. N α∈G
We have up(v) = p(uv) if u ∈ A(V )G and v ∈ A(V ). Moreover, the image of p is A(V )G and deg(p(u)) deg(u) for all u ∈ A(V ). Let B be the subalgebra of A(V )G generated by invariants of degree at most N , and AN the subspace of A(V ) consisting of polynomial functions of degree at most N −1. We claim that A(V ) = BAN . Since A(V ) is generated by powers of linear forms, to prove our claim, it suffices to prove that n ∈ BAN for all ∈ V ∗ and n ∈ N. This is clear if n < N . Now, (T − α. ) = T N + aN −1 T N −1 + · · · + a0 α∈G
25.5 The case of finite groups
399
where a0 , . . . , aN −1 ∈ B, so N ∈ B + B + · · · + B N −1 . By induction, we obtain that n ∈ BAN if n N . Let f ∈ A(V )G . It follows from our claim that there exist a1 , . . . , ar ∈ B and f1 , . . . , fr ∈ AN such that f = a1 f1 + · · · + ar fr . So: f = p(f ) = a1 p(f1 ) + · · · + ar p(fr ). Since deg(p(fi )) deg(fi ) N − 1, we have p(fi ) ∈ B, and so f ∈ B as required.
References and comments • [5], [21], [38], [40], [78]. More general results concerning quotients of an algebraic variety can be found in [30]. In general, for a G-variety X, the geometric quotient of X by G does not exist. In [72], Rosenlicht proved that there exists a dense open G-stable subvariety Y of X such that the geometric quotient of Y by G exists. Other proofs of this result can be found in [35], [67], [75]. In particular, we deduce from this that tr degk R(X)G = dim X − ρ where ρ is the maximal dimension of G-orbits in X.
26 Solvable groups
Structure of diagonalizable groups and solvable groups are studied in this chapter. We prove in particular two important results : Lie-Kolchin Theorem and Borel’s Fixed Point Theorem. In this chapter, G denotes an algebraic group whose Lie algebra is g.
26.1 Conjugacy classes 26.1.1 Let H be a closed subgroup of G, h its Lie algebra and x ∈ g. We set: CH (x) = {α ∈ H; (Ad α)(x) = x} , ch (x) = {y ∈ h; [x, y] = 0}. The set CH (x) (resp. ch (x)) is a closed subgroup of H (resp. Lie subalgebra of h). We have CH (x) = H ∩ CG (x) and ch (x) = h ∩ cg (x). Moreover, by 23.5.5, 24.3.5 and 24.3.6, L CH (x) = ch (x). For γ ∈ G, set: CH (γ) = {α ∈ H; αγ = γα} , ch (γ) = {x ∈ h; (Ad γ)(x) = x}. Again, CH (γ) (resp. ch (γ)) is a closed subgroup of H (resp. Lie subalgebra of h). More generally, if L is a subset G, we set CH (L) (resp. ch (L)) to be the intersection of the CH (γ)’s, with γ ∈ L. (resp. ch (γ)’s, γ ∈ L). It follows that CH (L) (resp. ch (L)) is a closed subgroup of H (resp. Lie subalgebra of h). Lemma. (i) If γ ∈ G, then L CH (γ) = ch (γ). (ii) If L is a subset of G, then L CH (L) = ch (L). Proof. By 24.5.2, it suffices to prove (i). Since CH (γ) = H ∩ CG (γ) and ch (γ) = h ∩ cg (γ), it follows from 24.3.5 (iii) that it suffices to prove the result for H = G. Let u, v, w : G → G be the morphisms defined as follows: u(α) = αγα−1 , v(α) = αγα−1 γ −1 , w(α) = αγ, α ∈ G. Then u = w ◦ v and due = dwe ◦ dve = dwe ◦(Ad(γ)−idg ) (23.5.3). Since CG (γ) is the stabilizer of γ under the action
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26 Solvable groups
of G given by conjugation, 25.1.3 implies that ker(due ) = L(CG (γ)). So the result follows because dwe is bijective. 26.1.2 For γ ∈ G and x ∈ g, set: ClH (γ) = {αγα−1 ; α ∈ H} , clH (x) = {(Ad α)(x) ; α ∈ H}. The set ClH (γ) is the H-orbit of γ under the action of H on G by conjugation, so it is locally closed in G (21.4.3). In the same way, clH (x) is locally closed in g. Lemma. If (Ad α)(x) − x ∈ h for all α ∈ H, then [x, h] ⊂ h. Proof. Apply 24.5.5 to the subspaces E = h and F = kx + h of g. 26.1.3 Let us conserve the above notations and let π : H → ClH (γ) be dπe induces a surjection from h onto the morphism α → αγα−1. By 25.1.3, Tγ (ClH (γ)) with kernel L CH (γ) = ch (γ). Suppose further that γ is semisimple and γHγ −1 ⊂ H. Then (Ad γ)(h) ⊂ h and the restriction of Ad γ to h is semisimple. So h = ch (γ) ⊕ m where m is the sum of eigenspaces of (Ad γ)|h whose eigenvalues are not equal to 1. We see, via the automorphism α → αγ −1 of G, that Te (ClH (γ)γ −1 ) = m. 26.1.4 Theorem. Let H be a closed subgroup of G and h its Lie algebra. (i) Let γ be a semisimple element of G verifying γHγ −1 ⊂ H. Then ClH (γ) is closed in G. (ii) Let x be a semisimple element of g verifying (Ad α)(x) − x ∈ h for all α ∈ H. Then clh (γ) is closed in g. Proof. We may assume that G = GL(V ) for some finite-dimensional vector space V . (i) For α ∈ G, let µα be its minimal polynomial. For α ∈ NG (H), we have (Ad α)(h) ⊂ h and we shall denote by χα the characteristic polynomial of (Ad α)|h . Let W be the set of elements α ∈ NG (H) verifying: µγ (α) = 0 , χα = χγ . Then γ ∈ W , and if α ∈ W , then βαβ −1 ∈ W for all β ∈ H. Moreover, if α ∈ W , then µα divides µγ . It follows that the elements of W are semisimple. Let α ∈ W . Then by 21.4.3 and 26.1.1, we have: dim ClH (α) = dim H − dim CH (α) = dim h − dim ch (α). Since χα = χγ and (Ad α)|h is semisimple, dim ch (α) = dim ch (γ). Thus the H-orbits of the variety W (under conjugation) are all of the same dimension. By 21.4.5, these orbits are closed in W . Since W is clearly closed in NG (H) and NG (H) is closed in G, the result follows.
26.1 Conjugacy classes
403
(ii) Let ng (H) = {y ∈ g; (Ad α)(y) − y ∈ h for all α ∈ H}. We have x ∈ ng (H) and we saw in 26.1.2 that if y ∈ ng (H), then [y, h] ⊂ h. Let us denote by χy the characteristic polynomial of (ad y)|h . Let µx be the minimal polynomial of x as an element of End(V ) and set: w = {y ∈ ng (H); µx (y) = 0 , χy = χx }. Then w contains x, and it is a closed subset of g verifying (Ad α)(w) ⊂ w for all α ∈ H. Using the same argument as in (i), we obtain that the H-orbits of w are all of the same dimension, and so they are closed. 26.1.5 Theorem. Let U be a closed unipotent subgroup of G, γ a semisimple element of G verifying γU γ −1 ⊂ U , and M the set of elements αγα−1 γ −1 , with α ∈ U . (i) The group CU (γ) is closed and connected. The set M is a closed irreducible subset of U . (ii) The map u : M × CU (γ) → U , (α, β) → αβ, is an isomorphism of varieties. (iii) The morphism v : G → G, α → αγα−1 γ −1 , induces an automorphism of the variety M . Proof. Since γU γ −1 ⊂ U , we have M ⊂ U . Moreover, U is connected (22.3.6). 1) Clearly, CU (γ) is closed in U and it is connected (22.3.6). The fact that M is closed follows from 26.1.4 and M is irreducible because M = v(U ). So we have proved (i). 2) Let α ∈ U , β = αγα−1 γ −1 . If β ∈ CU (γ), then γβ = βγ = αγα−1 . Since β is unipotent and γ is semisimple, βγ = γβ is the Jordan decomposition of αγα−1 . But this element is semisimple. So β = e and M ∩ CU (γ) = {e}. 3) Let α, β ∈ U . Then: v(αβ) = αβγβ −1 α−1 γ −1 = α(βγβ −1 γ −1 )α−1 (αγα−1 γ −1 ) = αv(β)α−1 v(α). So if β ∈ CU (γ), then v(β) = e and: v(αβ) = v(α) = v(α)v(β). 4) Using the equalities in point 3, we obtain that if α is central in U , then: v(βα) = v(αβ) = v(β)v(α). 5) Let us prove (ii) in the case where U is commutative. By point 4, the map v induces a homomorphism from the group U to itself whose image is M and kernel is CU (γ). So dim U = dim M + dim CU (γ) (21.2.4) and u is a homomorphism of algebraic groups. The connectedness of U , CU (γ) and M imply therefore that u is surjective. Let (α, β) ∈ M × CU (γ) be such that u(α, β) = αβ = e. Then α = β −1 ∈ M ∩ CU (γ) = {e} by point 2. It follows that α = β = e. The homomorphism u
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26 Solvable groups
is therefore injective, and we have proved that u is an isomorphism of algebraic groups (21.2.6). 6) Let us prove (ii) in the general case. We proceed by induction on the dimension of U . The case dim U = 0 is obvious, so let us assume that dim U > 0. Since U is unipotent, it is nilpotent (22.3.2 and 10.8.14). Its centre Z is a non-trivial closed unipotent subgroup, so it is also connected. If Z = U , then the result follows from point 5. Otherwise, Z = U . Note that γ and U are contained in the normalizer NG (Z) of Z in G. Let π : NG (Z) → NG (Z)/Z be the canonical surjection, G = NG (Z)/Z, U = U/Z and γ = π(γ). Set: M = {α γ (α )−1 (γ )−1 ; α ∈ U } = π(M ) , N = {αγα−1 γ −1 ; α ∈ Z}. Applying the induction hypothesis to (G , U , γ ) and (G, Z, γ), we obtain that the morphisms u : M × CU (γ ) → U and u0 : N × CZ (γ) → Z are isomorphisms of varieties. • Let us show that u is injective. Let α1 , α2 ∈ CU (γ) and β1 , β2 ∈ M be such that β1 α1 = β2 α2 . Let α = α1 α2−1 ∈ CU (γ), then β1 α = β2 . So π(β1 )π(α) = π(β2 ). Since π(α) ∈ CU (γ ) and π(β1 ), π(β2 ) ∈ M , we obtain that π(α) = e, or equivalently α ∈ Z. Let us write β1 = v(δ1 ), β2 = v(δ2 ) where δ1 , δ2 ∈ U . The fact that α ∈ Z ∩ CU (γ) implies that δ2 γδ2−1 = β2 γ = β1 αγ = (δ1 γδ1−1 )α = α(δ1 γδ1−1 ). The elements δ1 γδ1−1 et δ2 γδ2−1 are semisimple and α is unipotent. We deduce therefore that α = e, and so α1 = α2 and β1 = β2 . • Next, we show that v(Z) = Z ∩ M . If α ∈ Z, then γα−1 γ −1 ∈ Z, so v(Z) ⊂ Z ∩ M . Conversely, since u0 is surjective, any α ∈ Z is of the form δβ, where β ∈ CZ (γ) and δ ∈ N = v(Z). Furthermore, if α ∈ M , then α = δ ∈ v(Z) because u is injective. • We claim that CU (γ ) = π CU (γ) . Let α ∈ U be such that π(α)γ = γ π(α). Then v(α) ∈ M ∩ ker π = Z ∩ M = v(Z). Thus v(α) = v(β) for some β ∈ Z. By applying point 4 to β and β −1 α, we have β −1 α ∈ CU (γ). As π(β −1 α) = π(α), we have therefore proved our claim. u is surjective. Since u is surjective, π(M ) = M and • Letus show that π CU (γ) = CU (γ ), we have U = M CU (γ)Z = M ZCU (γ). The surjectivity of u0 implies then that U = M v(Z)CZ (γ)CU (γ). Finally, as M v(Z) = M by point 4, we have U = M CU (γ). • So u is bijective and it follows from 17.4.8 that isomorphism. u is an 7) Since u is surjective, we have M = v(U ) = v M CU (γ) = v(M ) by point 3. On the other hand, if α, β ∈ M verify v(α) = v(β), then β −1 α ∈ CU (γ) and u(α, e) = u(β, β −1 α). It follows from the injectivity of u that α = β. Thus v induces a bijection from M onto itself. But M = ClU (γ)γ −1 , so 25.1.1 implies that M is normal. Therefore (iii) follows from 17.4.8.
26.2 Actions of diagonalizable groups
405
26.2 Actions of diagonalizable groups 26.2.1 In this section, T denotes a diagonalizable algebraic group. Let us suppose that T acts rationally on G and that if α ∈ T , the morphism θα : G → G , β → α.β is an automorphism of the group G. Denote (dθα )e by πα . The map πα is an automorphism of the vector space g, and α → πα defines a rational representation of T on g. Let GT (resp. gT ) be the set of elements β ∈ G (resp. x ∈ g) verifying θα (β) = β (resp. πα (x) = x) for all α ∈ T . Recall that X ∗ (T ) denotes the set of characters of T . If χ ∈ X ∗ (T ), denote by gχ the set of x ∈ g such that πα (x) = χ(α)x for all α ∈ T . Denote by 1T ∈ X ∗ (T ) the trivial character of T , that is 1T (α) = 1 for all α ∈ T , and we set: ! " Φ(T, G) = χ ∈ X ∗ (T ) \ {1T }; gχ = {0} . Then we have: g = gT ⊕
gχ .
χ∈Φ(T,G)
Let χ ∈ X ∗ (T ) and T χ the kernel of χ. Denote by CG (T χ ) the set of β ∈ G verifying θα (β) = β for all α ∈ T χ . By identifying G and T as subgroups of the semidirect product G θ T , we may always assume that G and T are closed subgroups of an algebraic group. Proposition. (i) We have L(GT ) = gT . (ii) Assume that G is connected and G = GT . Then G is generated (as a group) by the subgroups CG (T χ ), χ ∈ X ∗ (T ). Proof. (i) Identify G and T as closed subgroups of an algebraic group, then apply 26.1.1 (ii). (ii) The condition G = GT implies that Φ(G, T ) = ∅. Denote by G the subgroup generated by the CG (T χ ), χ ∈ Φ(G, T ). Applying (i) with T = T χ , χ we obtain that L(CG (T χ )) = gT . So gT + gχ ⊂ L(G ) if χ ∈ Φ(T, G). Since g is the sum of the gT + gχ , χ ∈ Φ(T, G), it follows that g ⊂ L(G ). So G = G◦ ⊂ G and the result follows. 26.2.2 Let us suppose that T acts rationally on the algebraic groups G and H via group automorphisms of G and H respectively, as in 26.2.1, and denote them by θαG and θαH for α ∈ T . Proposition. Let σ : G → H be a T -equivariant surjective morphism of algebraic groups. Then σ induces a surjective morphism from (GT )◦ onto (H T )◦ . Proof. Since σ◦θαG = θαH ◦σ for all α ∈ T , we have dσ◦dθαG = dθαH ◦dσ for all α ∈ T . The surjectivity of u implies that du is surjective. Thus du induces a surjection from L(G)T onto L(H)T . By 26.2.1 (i), this surjection is the differential of the restriction GT → H T . So the result follows.
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26.2.3 Proposition. Let T be a torus of G and K a closed subgroup of G such that T ⊂ NG (K), and k = L(K). Then there exists α ∈ T such that: CK (α) = CK (T ) , ck (α) = ck (T ). Proof. We may assume that G ⊂ GL(V ) where V is a finite-dimensional vector space. Denote by χ1 , . . . , χr the pairwise distinct weights of T in V , and let Vi = Vχi , 1 i r. We have V = V1 ⊕ · · · ⊕ Vr (22.5.2). If M = GL(V1 ) × · · · × GL(Vr ) ⊂ GL(V ) and m = L(M ), then it is clear that CK (T ) = M ∩ K and ck (T ) = m ∩ k. is a non-trivial character of T , its kernel If 1 i, j r and i = j, χi χ−1 j Tij is a subgroup of T distinct from T . Since the field k is infinite, we deduce that there exists α ∈ T which is not contained in any of the Tij ’s, i = j. Therefore CK (α) = M ∩ K and ck (α) = m ∩ k.
26.3 Fixed points 26.3.1 Lemma. Let V be a k-vector space of dimension n > 0 and G a connected solvable subgroup of GL(V ). Then there exists v ∈ V \ {0} such that Gv ⊂ kv. Proof. In view of 21.6.6, we may assume that G is an algebraic subgroup of GL(V ). We shall proceed by induction on dim G. If dim G = 1, then G is commutative (22.6.1), and the result is then clear because G = Gs or G = Gu (22.2.6). So let us suppose that dim G 2. Let H = (G, G) be the group of commutators of G, then it is a connected solvable algebraic subgroup of G (21.3.6) verifying dim H < dim G. So by induction and 22.4.5, there exists χ ∈ X ∗ (H) such that Vχ = 0 where Vχ denotes the weight space of weight χ of H in V . Since H is normal in G, we saw in 22.4.5 that the (direct) sum W of the Vχ ’s, χ ∈ X ∗ (H), is G-stable. So we may assume that V = W . Note that the set of weights of H in V is finite (22.4.6). Let us fix a basis of V consisting of weight vectors of H, and identify GL(V ) with GLn (k) via this basis. Then the elements of H are diagonal matrices. For β ∈ H, the map G → H, α → αβα−1 , is a morphism of varieties whose image is connected since G is connected. But the image is a finite set since its elements are diagonal matrices having the same eigenvalues as for the diagonal matrix β. We deduce therefore that each Vχ is G-stable. So we may further assume that V = Vχ where χ ∈ X ∗ (H). Now H = (G, G), so det β = 1 for all β ∈ H. But for β ∈ H, we have β = χ(β) idV . Hence the connected subgroup H is finite. So H = {e} and G is commutative. But the result is clear when G is commutative. So we are done. 26.3.2 Theorem. (Lie-Kolchin Theorem) Let V be a finite-dimensional kvector space and G a connected solvable subgroup of GL(V ). Then there exists a basis of V with respect to which the elements of G are upper triangular.
26.4 Properties of solvable groups
407
Proof. Let us proceed by induction on dim V . The case dim V 1 is obvious. So let us assume that dim V 2. By 26.3.1, there exists a weight vector v1 of G in V . For α ∈ G, denote by α the automorphism of W = V /kv1 induced by α, and v → v , the canonical surjection V → W . Then clearly the set G of the α , α ∈ G, is a connected solvable subgroup of GL(W ). So by the induction hypothesis, there exists v2 , . . . , vn ∈ V such that (v 2 , . . . , v, n ) is a basis of W with respect to which the elements of G are upper triangular. It follows that (v1 , . . . , vn ) is a basis of V with respect to which the elements of G are upper triangular. 26.3.3 Corollary. Let G be a connected solvable algebraic group, then its derived subgroup D(G) = (G, G) is a closed connected unipotent subgroup. Proof. In view of 21.3.6, we only need to show that D(G) is unipotent. We may assume that G is a closed subgroup of some GLn (k) (22.1.5). By 26.3.2, there exists α ∈ GLn (k) such that αGα−1 ⊂ Tn (k) (notations of 10.8.1). Hence αD(G)α−1 ⊂ Un (k) (10.8.9), and D(G) is unipotent. 26.3.4 Theorem. (Borel’s Fixed Point Theorem) Let G be a connected solvable algebraic group and X a non-empty complete G-variety. Then the set X G is not empty. Proof. We shall prove the result by induction on d = dim G. If d = 0, then G = {e} and X = X G . So let us assume that d > 0 and let H = D(G). Since dim H < d, the set Y = X H is a non-empty closed subvariety of X (21.4.2). So Y is complete (13.4.3). Now H is normal in G, so if x ∈ Y , α ∈ G and β ∈ H, then: β.(α.x) = α. (α−1 βα).x = α.x, so Y is G-stable. There exists by 21.4.5 x ∈ Y such that the orbit G.x is closed, and therefore complete. The group Gx contains H, so Gx is normal in G. By 25.4.11 and 25.4.7, the variety G/Gx is affine and it is isomorphic to G.x. But G.x is irreducible since G is connected, it follows therefore from 13.4.4 that G.x = {x}.
26.4 Properties of solvable groups 26.4.1 Let Gs (resp. Gu ) denote the set of semisimple (resp. unipotent) elements of G. Recall from 22.2.5 that Gu is a closed subset of G. 26.4.2 Proposition. Let G be connected and solvable. (i) The set Gu is a closed connected nilpotent subgroup of G. It contains D(G) and so it is normal in G. (ii) The group G/Gu is a torus. Proof. (i) By 22.1.5 and 26.3.2, we may assume that G ⊂ Tn (k) for some n ∈ N∗ . Since the set of unipotent elements of Tn (k) is Un (k), we have
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Gu = G ∩ Un (k) by 22.2.3. So part (i) follows from 10.8.9, 10.8.10, 22.2.5 and 22.3.6 (i). (ii) Since D(G) ⊂ Gu , G/Gu is abelian. It is connected because it is the image of G under the canonical surjection G → G/Gu . Let β ∈ G/Gu and α ∈ G such that π(α) = β. Denote by αs and αu the semisimple and unipotent components of α. It follows that β = π(αs ). Hence β is semisimple (22.2.3) and the result follows from 22.5.7. 26.4.3 Proposition. Let G be connected and solvable. (i) There exists a chain G = Gp ⊃ Gp−1 ⊃ · · · ⊃ G1 ⊃ G0 = closed connected normal subgroups of G such that dim(Gi /Gi−1 ) = 1 1 i p. (ii) There exists a chain G = Gn ⊃ Gn−1 ⊃ · · · ⊃ G1 ⊃ G0 = closed connected normal subgroups of G such that dim(Gi /Gi−1 ) = 1 1 i n.
Gu of for all {e} of for all
Proof. (i) Let π : G → G/Gu = T be the canonical surjection. By 26.4.2, T is isomorphic to Dp (k) (Gm )p for some p. It is therefore clear that there is a chain T = Tp ⊃ Tp−1 ⊃ · · · ⊃ T1 ⊃ T0 = {e} of closed connected subgroups of T such that Ti /Ti−1 Gm for 1 i p. The subgroups Hi = π −1 (Ti ) of G are closed and normal in G, and so are the subgroups Gi = Hi◦ . Moreover, dim(Hi /Hi−1 ) = dim(Ti /Ti−1 ) = 1 (25.4.12) and (Hi−1 ∩ Gi )/Gi−1 is a finite set. So dim(Gi /Gi−1 ) 1 and the result follows. (ii) By 26.3.2, we may assume that G ⊂ Tr (k) and Gu ⊂ Ur (k) for some r ∈ N∗ . Let J be the set of pairs (i, j) where i, j ∈ {1, . . . , r} and i < j, and endow J with the following total order: (i, j) < (k, l) if j < l or if j = l and i > k. Let λ1 < · · · < λs be the elements of J in ascending order, and for (i, j) ∈ J, let U(i,j) be the set of matrices [xkl ] ∈ Ur (k) verifying xkl = 0 for (k, l) (i, j). Then we verify easily that Ur (k) = Uλ0 ⊃ Uλ1 ⊃ · · · ⊃ Uλs = {Ir } is a chain of closed normal subgroups of Tr (k). Moreover, we have Uλi / Uλi+1 Ga for 0 i < s. Let Gi = (G ∩ Uλi )◦ , then Gi is a closed normal subgroup of G contained in Gu and we have dim(Gi /Gi+1 ) 1. So the result follows from (i). 26.4.4 Proposition. Let G be connected and solvable. Then the following conditions are equivalent: (i) The set Gs is a subgroup of G. (ii) The group G is nilpotent. If these conditions are verified, then Gs is a connected subgroup of G contained in the centre of G, and the map θ : Gs × Gu → G, (α, β) → αβ, is an isomorphism of algebraic groups. Proof. Let us denote by π : G → G/Gu the canonical surjection. (i) ⇒ (ii) Suppose that Gs is a subgroup of G. Since π|Gs is injective, it follows from 26.4.2 (ii) that Gs is abelian.
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409
We may assume that G ⊂ GLn (k) and since Gs is abelian, we may assume further that Gs ⊂ Dn (k). So Gs = G ∩ Dn (k) is closed in G. Thus the restriction of π to Gs is a bijective morphism of algebraic groups, hence an isomorphism (21.2.6). In particular, Gs is connected. Since αβα−1 ∈ Gs for all β ∈ Gs and α ∈ G, we have G = NG (Gs ). So by 22.5.11: G = NG (Gs ) = NG (Gs )◦ = CG (Gs )◦ . Thus Gs is contained in the centre of G. Since G/Gs is unipotent, it is nilpotent (10.8.14), and so G is nilpotent (10.5.3). Note that since Gs is contained in the centre of G, the map θ is a bijective morphism of algebraic groups, and hence it is an isomorphism (21.2.6). (ii) ⇒ (i) Suppose that G is nilpotent. Fix α ∈ Gs and define the following morphisms of varieties: u : G → G , β → αβα−1 and v : G → G , β → αβα−1 β −1 . Since G is nilpotent, there exists n ∈ N∗ such that v n (β) = e for all β ∈ G. So (dv)n = 0, and dv is a nilpotent endomorphism of g = L(G). As dv = Ad(α) − idg and α ∈ Gs , it is also a semisimple endomorphism of g (22.2.3). Hence Ad(α) = idg . It follows from 24.4.2 and the connectedness of G that α is central in G. It is now clear that Gs is a subgroup of G. 26.4.5 Proposition. Let G be connected and solvable and g its Lie algebra. (i) L(Gu ) is the set of nilpotent elements of g. (ii) If G is nilpotent, then L(Gs ) is the set of semisimple elements of g. Proof. (i) We may assume that G ⊂ Tn (k) and Gu ⊂ Un (k). Then L(Gu ) ⊂ L(Un (k)) = nn (k), so all the elements of L(Gu ) are nilpotent. Let π : G → G/Gu be the canonical surjection. If x is a nilpotent element of g, then since G/Gu is a torus, 23.6.3 implies that dπ(x) = 0, or x ∈ ker dπ. So x ∈ L(ker π) = L(Gu ) by 24.4.1. (ii) Assume that G is nilpotent. Then π induces an isomorphism from Gs onto the torus G/Gu (26.4.2), so all the elements of L(Gs ) are semisimple. By 26.4.4, Gs is contained in the centre of G, so L(Gs ) is in the centre of g (24.4.2). Again by 26.4.4, G and Gs × Gu are isomorphic, so g is isomorphic to the direct product L(Gs ) × L(Gu ). The result follows therefore from (i) and the uniqueness of the Jordan decomposition in g.
26.5 Structure of solvable groups 26.5.1 In this section, G denotes a connected solvable algebraic group whose Lie algebra is g. We saw in 26.4.2 that Gu is normal in G, so its Lie algebra n is an ideal of g (24.4.7) and n is the set of nilpotent elements of g (26.4.5). Moreover, since D(G) ⊂ Gu (26.4.2), we have [g, g] ⊂ n (24.3.5 and 24.5.12).
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Proposition. There exists a Lie subalgebra t verifying the following conditions: (i) t is an algebraic Lie subalgebra of g. (ii) t is abelian and all its elements are semisimple. (iii) As a vector space, g = t ⊕ n. Proof. We may assume that G ⊂ GL(V ) and g ⊂ gl(V ) for some finitedimensional vector space V . The elements of n are then nilpotent endomorphisms of V (23.6.4). Let A be the set of abelian Lie subalgebras of g whose elements are semisimple endomorphisms of V . Then A = ∅ since {0} ∈ A. Let t be an element of A of maximal dimension. By 23.6.4, the representation t → adg t of t in g is semisimple, and t + n is stable under this representation. So there exists a subspace m such that g = m ⊕ (t + n) and [t, m] ⊂ m. Since [g, g] ⊂ n, we deduce that [t, m] = {0}. Let x ∈ m and xs , xn its semisimple and nilpotent components. By 23.6.4, xs , xn ∈ g and since [x, t] = {0}, we have [xs , t] = {0} (10.1.3 and 23.6.4). It follows that kxs + t is an abelian subalgebra whose elements are semisimple. Our choice of t implies that xs = 0. So all the elements of m are nilpotent. Hence m ⊂ n, which in turn implies that m = {0}. Thus g = t + n and it is clear that t ∩ n = {0}. So we are left to prove that t is algebraic. Let x ∈ t and y a replica of x. Then by 24.6.4 (ii), y is semisimple and [y, t] = {0}. Again by our choice of t, we deduce that y ∈ t. It follows from 24.5.10 that t is an algebraic Lie subalgebra of g. 26.5.2 Let us conserve the hypotheses and notations of 26.5.1 and let T be a connected algebraic subgroup of G verifying L(T ) = t. Proposition. (i) The group T is a torus of dimension dim(G/Gu ). (ii) The map u : T × Gu → G, (α, β) → αβ is an isomorphism of varieties. Proof. (i) We may again assume that G ⊂ GL(V ) and g ⊂ gl(V ). Note that T = (t) (see 24.5.1). If x ∈ t, then any element of (x) can written as P (x) for some P ∈ k[X] (24.6.4) and T is generated by the (x)’s, x ∈ t. It follows that T is abelian and its elements are semisimple. Since T is connected, it is a torus (22.5.7) of dimension dim(G/Gu ) (26.5.1 (iii)). (ii) Since T is a torus, T ∩ Gu = {e}. Let T Gu be the set of products αβ, α ∈ T and β ∈ Gu . Then T Gu is a subgroup of G because Gu is normal in G. It is also closed in G by 21.2.3 and connected since it is the image of the morphism u. Let us consider the following actions of T × Gu on T × Gu and G:
a
a
a
(α, β).(α , β ) = (αα , β β −1 ) , (α, β).γ = αγβ −1 . where α, α ∈ T , β, β ∈ Gu and γ ∈ G. Then u is injective and equivariant with respect to these actions. Moreover, the stabilizer of e in T × Gu is {e}
26.5 Structure of solvable groups
411
because T ∩ Gu = {e}. As the (T × Gu )-orbit of e is T Gu , it follows from 21.4.3 that: dim(T Gu ) = dim(T × Gu ) = dim T + dim Gu = dim G. Hence the connectedness of G implies that G = T Gu . So u is a bijective morphism of (T × Gu )-homogeneous spaces, and 25.1.2 says that u is an isomorphism of varieties. 26.5.3 Set:
C ∞ (G) =
C n (G).
n1
We have C ∞ (G) ⊂ D(G) ⊂ Gu , so C ∞ (G) is a normal unipotent subgroup of G. If T is a torus of G, then T ∩ Gu = {e}, so dim T dim(G/Gu ). In particular, if dim T = dim(G/Gu ), then T is a maximal torus. Now 26.5.2 says that such a torus exists and if T is such a torus, then the map T × Gu → G is an isomorphism of varieties. 26.5.4 Lemma. Let T ⊂ G be a torus of dimension dim(G/Gu ). If α is an element of Gs , then there exists β ∈ C ∞ (G) such that βαβ −1 ∈ T . Proof. We have already noted that G = T Gu . Let us proceed by induction on dim G. Suppose that G is nilpotent. Then T = Gs by 26.4.4 and so the result is obvious. Suppose that G is not nilpotent, then C ∞ (G) = {e}. The unipotent group ∞ C (G) is connected and nilpotent (10.8.10 and 21.3.6). The last non-trivial term of the central descending series of C ∞ (G) is connected (21.3.6), so it is contained in the identity component N of the centre of C ∞ (G). It follows that N = {e} and N is a normal subgroup of G. In particular, N ⊂ Gu and N ∩ T = {e}. Let G = G/N and π : G → G the canonical surjection. It is clear that the map π(T ) × (G )u → G , (γ, δ) → γδ, is an isomorphism of varieties. Applying the induction hypothesis on G , we obtain that there exists γ ∈ C ∞ (G ) = π(C ∞ (G)) such that γπ(α)γ −1 ∈ π(T ). Let δ ∈ C ∞ (G) be such that π(δ) = γ. Then by replacing α by δαδ −1 , we may assume that α ∈ T N = N T . Let β ∈ N and γ ∈ T be such that α = βγ. By 26.1.5, there exists δ ∈ N and ε ∈ CN (γ) such that β = (δγδ −1 γ −1 )ε. Then α = (δγδ −1 )ε = ε(δγδ −1 ) since ε commutes with γ and N is abelian. But δγδ −1 is semisimple and ε is unipotent, so α = (δγδ −1 )ε is the Jordan decomposition of α. Hence ε = e and δ −1 γδ = γ ∈ T . 26.5.5 Theorem. Let S, T be maximal tori of a connected solvable algebraic group G. (i) There exists α ∈ C ∞ (G) such that S = αT α−1 . In particular, dim S = dim T = dim(G/Gu ). (ii) The map T × Gu → G, (α, β) → αβ, is an isomorphism of varieties.
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Proof. In view of 26.5.2 and 26.5.3, it suffices to show that there exists an element α ∈ C ∞ (G) such that S = αT α−1 where T is a maximal torus of dimension dim(G/Gu ). By 26.2.3, there exists γ ∈ S such that CG (S) = CG (γ). By 26.5.4, we may assume that γ ∈ T . Then T ⊂ CG (γ)◦ = CG (S)◦ . Let C = CG (S)◦ . Any torus of C is a torus of G, and so by 26.5.2 (i) and 26.5.3, T is a torus of C of dimension dim(C/Cu ). It follows by 26.5.4 that if β ∈ S, then δβδ −1 ∈ T for some δ ∈ C. Since S is central in C, we deduce that S ⊂ T . But S is a maximal torus, so S = T . 26.5.6 Proposition. Let G be a connected solvable algebraic group. (i) If T is a maximal torus of G, then any semisimple element of G is conjugate to a unique element of T . (ii) Let S be a (not necessarily closed) subgroup of G consisting of semisimple elements. Then S is contained in a torus. Moreover, the group CG (S) is connected and is equal to NG (S). Proof. (i) Recall that the canonical surjection π : G → G/Gu induces an isomorphism from T onto G/Gu . Let α ∈ Gs . By 26.5.4, α is conjugate to an element of T . Now if β ∈ G, then π(βαβ −1 ) = π(α) since G/Gu is abelian. So the result follows. (ii) The restriction of the canonical surjection π to S is injective, so S is abelian. If α ∈ NG (S) and β ∈ S, then π(βαβ −1 ) = π(α) since G/Gu is abelian. So βαβ −1 = α from the injectivity of π|S . Hence CG (S) = NG (S). Now, S is a closed diagonalizable subgroup of G verifying CG (S) = CG (S) (21.6.3). So we may assume that S is closed in G. Let T be a maximal torus in G. If S is central in G, then CG (S) = G is connected and since any element of S is conjugate to an element of T (26.5.4), we have S ⊂ T . Suppose that S ⊂ Z(G) and s ∈ S \Z(G). Again by 26.5.4, we may assume that s ∈ T . Then T ⊂ CG (s) = T (CG (s))u , which proves that H = CG (s) is connected (22.3.6). Since H = G, we can argue by induction on dim G, that S is conjugated in H to a subgroup of T and that CH (S) = CG (S) is connected. So we have proved (ii).
References and comments • [5], [40], [78], [79]. In this chapter, we have followed closely the presentation of solvable groups given in [5].
27 Reductive groups
Semisimple groups and reductive groups are very useful objects in the study of algebraic groups. In this chapter, we consider these groups and their representations. We prove Hilbert-Nagata Theorem on invariants of rational reductive group actions, and we introduce the notion of algebraic quotient.
27.1 Radical and unipotent radical 27.1.1 In this section, H is an algebraic group and G is a closed normal subgroup of H. Denote by R1 (G) the set of normal solvable subgroups of G and R2 (G) (resp. R(G)) the set of elements of R1 (G) which are closed (resp. closed and connected) in G. Let R1 (G), R2 (G) and R(G) be the subgroups of G generated by R1 (G), R2 (G) and R(G) respectively. These subgroups are normal in G and the subgroup R(G) is called the radical of G. Lemma. (i) We have R1 (G) = R2 (G). The group R1 (G) is the largest normal solvable subgroup of G. It is a closed normal subgroup of H. (ii) The radical R(G) of G is a connected solvable algebraic subgroup of G. It is the largest connected normal solvable subgroup of G. It is normal in H and we have R(G) = R(G◦ ) = R1 (G)◦ . Proof. (i) If H ∈ R1 (G), then H ∈ R2 (G) (21.6.1 and 21.6.6). So it is clear that R1 (G) = R2 (G). If R1 (G) = {e}, then it is solvable. Otherwise, there exists K ∈ R2 (G) such that R1 (G) ⊃ K = {e}. Since the group R1 (G)/K is contained in R1 (G/K), we obtain by induction on dim G that R1 (G)/K is solvable. Hence R1 (G) is solvable (10.4.5). Again by using 21.6.1 and 21.6.6, we see that R1 (G) is closed. We deduce therefore that R1 (G) is the largest normal solvable subgroup of G. If α ∈ H, then αR1 (G)α−1 is again a normal solvable subgroup of G. So it is contained in R1 (G). It follows that R1 (G) is normal in H.
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(ii) Since R(G) ⊂ R1 (G), it is solvable. Also, it is clear that R(G) is normal in G, and it is closed and connected by 21.3.2. So R(G) ⊂ R1 (G)◦ . Now R1 (G) is normal in G, so R1 (G)◦ is also normal in G. Moreover R1 (G)◦ is solvable, so R(G) = R1 (G)◦ . We deduce therefore that R(G) is the largest connected normal solvable subgroup of G. By (i), R1 (G) is normal in H, so R(G) = R1 (G)◦ is also normal in H. Finally, R(G) is connected, so it is contained in G◦ . Therefore R(G) ∈ R(G◦ ). Hence R(G) ⊂ R(G◦ ). Conversely, since G◦ is normal in G, R(G◦ ) is a connected normal solvable algebraic subgroup of G, so R(G◦ ) ⊂ R(G). 27.1.2 Let us conserve the notations of 27.1.1. Let U(G) be the set of normal unipotent algebraic subgroups of G and denote by Ru (G) the subgroup of G generated by U(G). We call Ru (G) the unipotent radical of G. If K ∈ U(G), then K ∈ R(G), so Ru (G) ⊂ R(G). Moreover, by 21.3.2, Ru (G) is a closed connected subgroup of G. Lemma. The group Ru (G) is the set R(G)u of unipotent elements of R(G). It is the largest normal unipotent subgroup of G. It is normal in H and we have Ru (G) = Ru (G◦ ). Proof. If K ∈ U(G), then K ∈ R(G) and K ⊂ R(G)u . So Ru (G) ⊂ R(G)u . Since R(G) is normal in H and R(G)u is a closed normal subgroup of R(G) (26.4.2), it is clear that R(G)u is a normal algebraic subgroup of H. In particular, R(G)u ⊂ Ru (G), and hence R(G)u = Ru (G). Finally, R(G) = R(G◦ ) by 27.1.1, so Ru (G) = R(G)u = R(G◦ )u = Ru (G◦ ). 27.1.3 Proposition. Let g be the Lie algebra of G, r the radical of g and t the nilpotent radical of g. (i) The Lie algebra of R(G) is r. (ii) We have t ⊂ L Ru (G) . Proof. In view of 27.1.1 and 27.1.2, we may assume that G is connected. (i) By 24.4.7 and 24.5.13, L R(G) is a solvable ideal of g, so L R(G) ⊂ r. On the other hand, r is an algebraic Lie subalgebra of g (24.7.4). Let H be the smallest closed subgroup of G such that L(H) = r. Then by 24.4.7 and 24.5.13, H is a connected normal solvable algebraic of G, so H ⊂ R(G). subgroup Hence r = L(H) ⊂ L R(G) . Thus r = L R(G) . (ii) Let x ∈ t. Then ρ(x) is nilpotent for all finite-dimensional representation ρ of g. It follows that η(x) is locally nilpotent where η is the representation of g in A(G) defined in 23.6.2. Thus x is a nilpotent element of r. Since is the set of nilpotent elements of r (26.4.5 and 27.1.2), we have L Ru (G) x ∈ L Ru (G) . 27.1.4 Remarks. 1) In general, we have t = L R (G) . For example, if u G = Ga , then Ru (G) = G, L Ru (G) = g and t = {0}. 2) The group (G, R(G)) is connected (21.3.5) and its Lie algebra is equal to [g, r] = t (20.3.6 and 24.5.11). So 27.1.3 implies that (G, R(G)) ⊂ Ru (G).
27.2 Semisimple and reductive groups
415
27.2 Semisimple and reductive groups 27.2.1 Definition. An algebraic group G is said to be semisimple if R(G) = {e}. It is said to be reductive if Ru (G) = {e}. 27.2.2 Proposition. Let G be an algebraic group and g its Lie algebra. (i) The group G is semisimple if and only if g is semisimple. (ii) The group G is reductive if and only if g is reductive and all the elements of the centre of g are semisimple. Proof. Let r be of g, and t the nilpotent radical of g. the radical (i) Since L R(G) = r and R(G) is connected, we have R(G) = {e} if and only if r = {0}. So the result follows from the Levi-Malcev Theorem. (ii) Suppose that G is reductive. Then by 27.1.3, t = {0}. So g is reductive and r is the centre of g. Thus R(G) is the centre of G◦ (24.4.2). But Ru (G) = R(G)u (27.1.2), we deduce therefore from 26.4.5 that all the elements of r are semisimple. Conversely, suppose that g is reductive and all the elements of r = z(g) are semisimple. Since Ru (G) = R(G)u is connected and L Ru (G) is the set of nilpotent elements of r (26.4.5), Ru (G) = {e}. Thus G is reductive. 27.2.3 Remarks and examples. 1) Let G is a connected algebraic group. Recall from 21.3.6 and 24.5.12 that (G, G) is a connected algebraic subgroup of G whose Lie algebra is [g, g]. So if G is semisimple, then G = (G, G) (20.1.5). Similarly, if G is reductive, then (G, G) is semisimple (20.5.5). 2) The product of two semisimple (resp. reductive) algebraic groups is semisimple (resp. reductive). 3) Recall from 20.2.2 that sln (k) is semisimple. So the group SLn (k) is semisimple (23.4.5 and 27.2.2). 4) The Lie algebra g can be reductive while G is not reductive. This is the case for the additive group G = Ga (27.1.4). 27.2.4 Proposition. Let G be a connected algebraic group. Then the following conditions are equivalent: (i) G is semisimple. (ii) The only connected normal abelian algebraic subgroup of G is {e}. Proof. The implication (i) ⇒ (ii) is clear since such a subgroup is contained in R(G) = {e}. Conversely if R(G) = {e}, then the last non-trivial term of the derived series of R(G) is a connected normal abelian algebraic subgroup of G (21.3.6). 27.2.5 Proposition. Let G be a connected reductive algebraic group. (i) Its radical R(G) is a torus, and it is equal to the identity component of Z(G). (ii) The group (G, G) ∩ R(G) is finite.
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Proof. (i) Since R(G)u = Ru (G) = {e}, we have R(G) = R(G)s . As R(G) is connected, it is a torus (22.5.7). By 27.1.3, L R(G) isthe radical of g, which is equal to z(g) (20.5.4 and 27.2.2). Since L Z(G) = z(g) (24.4.2) and R(G) is connected, we have R(G) = Z(G)◦ (24.3.5). (ii) Let H = (G, G) ∩ R(G). Then H ◦ is a connected normal abelian algebraic subgroup of (G, G). By 27.2.4, H ◦ = {e}. So H is finite.
27.3 Representations 27.3.1 Let us start by a remark. Let G be an algebraic group, E a rational G-module and F a simple submodule of E. If x ∈ F \ {0}, then the submodule generated by x is clearly F . It follows that F is finite-dimensional. 27.3.2 Theorem. Let G be an algebraic group, (V, ρ) a finite-dimensional rational representation of G and σ the restriction of ρ to G◦ . Then the following conditions are equivalent: (i) ρ is semisimple. (ii) σ is semisimple. Proof. (i) ⇒ (ii) By 10.7.4, we may assume that ρ is simple. So V is finitedimensional and it contains a simple G◦ -submodule W . Now G◦ is normal in G, so ρ(α)(W ) is a simple G◦ -module for all α ∈ G. Set: S= ρ(α)(W ). α∈G
Then S is a non-zero G-submodule of V . Hence S = V and therefore σ is semisimple. (ii) ⇒ (i) Let W be a d-dimensional G-submodule of V . We shall show that V = W ⊕ W for some G-submodule W of V , and the result would follow. Let g be the Lie algebra of G. By 27.2.2, dσ(g) = g is reductive, and the elements of its centre are semisimple. So if (E, θ) is a finite-dimensional rational representation of σ(G), then p the representation p dθ(g ) is semisimple. V ) (resp. (σp , V )) the representation For an integer p, denote by (ρp , of G (resp. G◦ ) associated to ρ (resp. σ) as defined in 23.4.14. Then by the above discussion and 20.5.10, dσp is semisimple, and so 24.3.3 implies that σp is semisimple. d d W be a non-zero element. Then W = ky and for α ∈ G, Let y ∈ we have ρd (α)(y) = θ(α)y where θ(α) = det(ρ(α)|W ) is a character of G. Let d V ; σd (α)(x) = θ(α)(x) for all α ∈ G◦ } and T be the subspace of S = {x ∈ d d V spanned by σd (α)(x) − θ(α)x, α ∈ G◦ and x ∈ V. d V = S⊕T Applying 10.7.8 to θ, G◦ and G, we obtain the decomposition d V into G-submodules. of We claim that the restriction λ of ρd to S is a semisimple representation of G. If α ∈ G, set λ1 (α) = θ(α)−1 λ(α). Then (λ1 , S) is a representation of G
27.3 Representations
417
whose kernel contains G◦ . Thus λ1 (α) depends only on the class of α modulo G◦ . This induces a representation (λ2 , S) of the finite group G/G◦ which is semisimple by 10.7.6. It follows that λ1 , and hence λ, is semisimple. We have therefore proved our claim. Since ky is a G-submodule of S, it followsthat S = ky ⊕ U for some d V = ky ⊕ U for some GG-submodule U . This implies, in turn, that submodule U . It follows from 10.7.7 that V = W ⊕ W where W is the set of d−1 W. x ∈ V such that x ∧ z ∈ U for all z ∈ To finish the proof, it suffices therefore to prove that W is a G-submodule. d−1 W . Since ρ(α)(W ) = W , we have Let x ∈ W , α ∈ G and u ∈ d−1 d−1 d−1 ρd−1 (α)( W) = W . So u = ρd−1 (α)(u ) for some u ∈ W. We deduce therefore that ρ(α)(x) ∧ u = ρ(α)(x ∧ u ) ∈ U since U is a G submodule. Hence W is a G-submodule as required. 27.3.3 Theorem. The following conditions are equivalent for an algebraic group G: (i) G is reductive. (ii) Any finite-dimensional rational representation of G is semisimple. (iii) Any rational representation of G is semisimple. Proof. (i) ⇒ (ii) Suppose that G is reductive. Then g = L(G) is reductive and all the elements of r are semisimple (27.2.2). Let (V, ρ) be a finite-dimensional rational G-module. By 20.5.10, the representation (V, dρ) of g is semisimple. Therefore, 24.3.3 implies that the restriction of ρ to G◦ is semisimple. So ρ is semisimple by 27.3.2. (ii) ⇒ (i) We may assume that G ⊂ GL(V ) for some finite-dimensional vector space V . Suppose that G is not reductive, then H = Ru (G) = {idV } and the group H acts non-trivially on V . So V H = V and V H = {0} (22.3.6). Now since H is normal in G, V H is a G-submodule of V . If V is a semisimple G-module, then V = V H ⊕ W for some non-zero G-submodule W . But again by 22.3.6, W H = {0} which implies that V H ∩ W = {0}. Contradiction. (ii) ⇒ (iii) If V is a rational G-module, then it is the sum of finitedimensional rational G-submodules. So if (ii) is verified, then V is the sum of simple G-modules. (iii) ⇒ (ii) This is obvious. 27.3.4 Let G be an algebraic group and let us consider the action ρ of G on A(G) via right translations (22.1.4). Assume that G is reductive. The subspace A(G)G is the set of constant functions and we may identify it with k. By 10.7.9, the subspace V of A(G) spanned by ρα f − f , α ∈ G and f ∈ A(G), is the unique submodule of A(G) such that A(G) = k ⊕ V . Let us denote by IG or I the Reynolds operator associated to A(G). Note that I ◦ρα = I for all α ∈ G. It is clear that λα (k) = k if α ∈ G, where λα denotes the left translation of functions (22.1.4). Furthermore, we see immediately that λα (V ) = V . Thus I ◦λα = I.
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27.3.5 In the rest of this section, for an algebraic group G, we denote by G the set of isomorphism classes of finite-dimensional simple rational G-modules. - then Mω denotes a representative of ω, and M ∗ denotes the dual If ω ∈ G, ω G-module (see 10.6.5). If E is a rational G-module, then we set E(ω) to be the sum of submodules of Eω which are isomorphic to Mω . The G-submodule E(ω) is called the isotypic component of type ω of E. If E and F are finite-dimensional rational G-modules, we shall endow Hom(E, F ) and E ⊗k F with the G-module structures as defined in 10.6.5. 27.3.6 Theorem. Let G be a reductive algebraic group and E, F rational G-modules. - then any simple submodule of E(ω) is isomorphic to Mω . (i) If ω ∈ G, (ii) We have: E(ω) . E= ω∈G
- then f (E(ω) ) ⊂ F(ω) . (iii) If f ∈ HomG (E, F ) and ω ∈ G, - the map (iv) For all ω ∈ G, θω : HomG (Mω , E) ⊗k Mω → E(ω) , f ⊗ v → f (v) , is an isomorphism of G-modules. Proof. (i) This is clear by 10.7.5 (ii). (ii) Since E is a semisimple G-module (27.3.3), it is equal to the sum of the - and E the sum of the E(ω) ’s where ω ∈ G - \ {θ}. E(ω) ’s (27.3.1). Let θ ∈ G, If E ∩ E(θ) is non-zero, then it contains a simple submodule E . But part (i) and 10.7.5 imply that E is isomorphic to Mθ and Mω for some ω = θ. This is absurd (10.7.2). So the sum of the E(ω) ’s is direct. (iii) In view of 10.7.2, this is obvious. (iv) By (iii), θω is well-defined, and a simple verification shows that θω is a homomorphism of G-modules. It is also clear that θω is surjective. Let (Vi )i∈I be a family of submodules of E verifying Vi Mω for all i ∈ I, and: Vi . E(ω) = i∈I
Denote by (pi )i∈I the family of projectors associated to this direct sum decomposition. If f ∈ HomG (Mω , E), then the image of f is contained in E(ω) and f = i∈I pi ◦ f . It follows easily that: HomG (Mω , E) = HomG (Mω , Vi ). i∈I
Let us fix ui ∈ HomG (Mω , Vi ) \ {0}. By 10.7.2, ui is bijective and all the elements of HomG (Mω , Vi ) are proportional to ui . Let f1 , . . . , fn ∈ HomG (Mω , E) and x1 , . . . , xn ∈ Mω be such that
27.3 Representations
419
θω (f1 ⊗ x1 + · · · + fn ⊗ xn ) = 0. For all i ∈ I, we have pi f1 (x1 ) + · · · + fn (xn ) = 0. So we may assume that f1 , . . . , fn ∈ HomG (Mω , Wi ), fk = λk ui , with λi ∈ k for 1 k n. Thus λ1 x1 + · · · + λn xn = 0 since ui is injective. We deduce therefore that f1 ⊗ x1 + · · · + fn ⊗ xn = 0. Hence θω is injective. 27.3.7 Remarks. Let us conserve the hypotheses of 27.3.6. 1) In the proof of 27.3.6 (iv), the cardinality of the set I is called the multiplicity of Mω in E, and we shall denote it by multω (E). The module HomG (Mω , E) has dimension multω (E), and we have the following isomorphism of G-modules: multω (E)Mω . E ω∈G
2) If L is a submodule of E, then L(ω) = L ∩ E(ω) for all ω ∈ G. 27.3.8 Proposition. Let G be a reductive algebraic group acting rationally on a commutative k-algebra A by automorphisms, and let B = AG . (i) If a is a G-stable ideal of A, then the G-modules B/(B∩a) and (A/a)G are isomorphic. (ii) Let (ai )i∈I be a family of G-stable ideals of A. Then: ai = (B ∩ ai ). B∩ i∈I
i∈I
- But a(ω) = Proof. (i) By 27.3.6, a is the direct sum of the a(ω) ’s, for ω ∈ G. a ∩ A(ω) by 27.3.7. We deduce therefore that: A/a = (A(ω) /a(ω) ). ω∈G
It follows that (A/a)(ω) = A(ω) /a(ω) , and we obtain the result by taking ω to be the isomorphism class of the trivial G-module. (ii) We have: ai = (ai )(ω) = (ai )(ω) . i∈I
So
i∈I
ai
(ω)
=
i∈I ω∈G -
i∈I (ai )(ω)
- i∈I ω∈G
- and we conclude as in (i). for all ω ∈ G,
27.3.9 Let E, F be G-modules and consider the action of G × G on E ⊗k F defined in 23.4.12, and the action of G × G on A(G) defined as follows: for α, β, γ ∈ G and f ∈ A(G), (α, β).f (γ) = f (α−1 γβ). Theorem. Let G be a reductive algebraic group and
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φ:
ω∈G
Mω∗ ⊗k Mω → A(G)
the map sending f ⊗ x ∈ Mω∗ ⊗k Mω to the function α → f (α.x). (i) φ is an isomorphism of (G × G)-modules. (ii) We have multω A(G) = dim Mω for all ω ∈ G. Proof. Let x ∈ Mω , f ∈ Mω∗ , and α, β, γ ∈ G. Then: (α, β).φ(f ⊗ x) (γ) = φ(f ⊗ x)(α−1 γβ) = f (α−1 γβ.x) = (α.f )(γβ.x) = φ (α.f ) ⊗ (β.x) (γ). So φ is a homomorphism of (G × G)-modules. 1) Let us fix f ∈ Mω \ {0}. Then (e, α).(f ⊗ x) = f ⊗ (α.x) for x ∈ Mω . Thus f ⊗k Mω is a simple G-module isomorphic to Mω . It follows immediately that since ρα φ(f ⊗ x) = φ f ⊗ (α.x) , φ induces a homomorphism of the simple G-module f ⊗k Mω into the Gmodule A(G). It is clear that φ(f ⊗k Mω ) is non-zero, so it is isomorphic to Mω and contained in A(G)(ω) . Hence A(G)(ω) = {0} for all ω ∈ G. 2) Now, let V be a simple submodule of A(G) which is isomorphic to Mω and x → ux a G-module isomorphism from Mω onto V . Define g ∈ Mω∗ by g(x) = ux (e), x ∈ Mω . Then for α ∈ G and x ∈ Mω , we have: φ(g ⊗ x)(α) = g(α.x) = uα.x (e) = (α.ux )(e) = ux (α). We deduce that φ(g ⊗k Mω ) = V . Since φ(Mω∗ ⊗k Mω ) ⊂ A(G)(ω) by point 1, it follows that φ(Mω∗ ⊗k Mω ) = A(G)(ω) . So φ is surjective by 27.3.6 (ii). - and M = Mω . If V is a submodule of M ∗ , its 3) Finally, let ω ∈ G orthogonal in M is a submodule of M . So M ∗ is a simple G-module. We claim that M ∗ ⊗k M is a simple (G × G)-module. Denote by σ and σ the representations of G on M and M ∗ , and θ the associated representation of G × G on M ∗ ⊗k M . By 10.8.11, the subalgebra of End(M ) (resp. End(M ∗ )) generated by σ(G) (resp. σ (G)) is equal to End(M ) (resp. End(M ∗ )). Since End(M ∗ ⊗k M ) is isomorphic to End(M ∗ )⊗k End(M ), we deduce that the subalgebra of End(M ∗ ⊗k M ) generated by θ(G × G) is equal to End(M ∗ ⊗k M ). So we have proved our claim. Since A(G)(ω) = φ(Mω∗ ⊗k Mω ) = {0}, our claim and point 2 imply that φ is bijective. So we have proved (i), and (ii) follows immediately.
27.4 Finiteness properties 27.4.1 In this section, G denotes a reductive algebraic group. Let A be a commutative k-algebra on which G acts rationally by automorphisms. Denote by p the Reynolds operator associated to A. Lemma. For u ∈ AG and v ∈ A, we have p(uv) = up(v). Consequently, the unique complementary submodule ker(p) of AG in A, is an AG -module.
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421
Proof. There exist, by 10.7.9, α1 , . . . , αn ∈ G and w1 , . . . , wn ∈ A such that: v = p(v) +
n
(αi .wi − wi ).
i=1
Since αi .(uwi ) = u(αi .wi ), we have: uv = up(v) +
n
αi .(uwi ) − uwi .
i=1
On the other hand, up(v) ∈ AG , so the result follows from 10.7.9. 27.4.2 Let V be a finite-dimensional rational G-module and A the algebra of polynomial functions on V . Endow A with the G-module structure defined in 22.1.1. Proposition. The algebra AG is finitely generated. Proof. For n ∈ N, let An be the subspace of A consisting of homogeneous polynomial functions of degree n. Then An is G-stable and so: G AG = An . n∈N
Let I be the ideal of A generated by AG n , n 1. Since A is Noetherian, I is finitely generated. Let f1 , . . . , fr ∈ AG be a system of generators of I that we may assume to be homogeneous. Now let f ∈ AG be homogeneous of degree non-zero, then: f = f1 g1 + · · · + fr gr where g1 , . . . , gr are homogeneous. Let p denote the Reynolds operator of A. It follows from 27.4.1 that: f = p(f ) = f1 p(g1 ) + · · · + fr p(gr ). But hi = p(gi ) ∈ AG for 1 i r, so if hi ∈ k[f1 , . . . , fr ] for 1 i r, then f ∈ k[f1 , . . . , fr ]. Since deg(hi ) < deg f , we obtain by induction (on the degree of f ) that AG = k[f1 , . . . , fr ]. 27.4.3 Theorem. (Hilbert-Nagata Theorem) Let G be a reductive algebraic group and A a finitely generated commutative k-algebra on which G acts rationally by automorphisms. (i) The algebra B = AG is finitely generated, and is a direct summand of the B-module A. (ii) If b is an ideal of B, then B ∩ (Ab) = b. Proof. The fact that B is a direct summand of the B-module A follows from 27.4.1. Let us denote by p the Reynolds operator associated to A.
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Let b be an ideal of B. Then b ⊂ B∩(Ab). Conversely, if f ∈ B∩(Ab), then there exist f1 , . . . , fn ∈ b and g1 , . . . , gn ∈ A such that f = g1 f1 + · · · + gn fn . By 27.4.1, we have: f = p(f ) = f1 p(g1 ) + · · · + fn p(gn ). Thus f ∈ b. Now let x1 , . . . , xn be a system of generators of A. There exists a finitedimensional submodule W of A containing x1 , . . . , xn . Denote by V the dual of W . The injection V ∗ = W → A extends to a surjective G-module homomorphism u from A(V ) onto A. Since u◦pA(V ) = p◦u, u induces a surjective homomorphism A(V )G → AG . Hence the result follows from 27.4.2. 27.4.4 Let us conserve the notations and hypotheses of 27.4.3. Let V be a rational G-module. If u ∈ HomG (V, A) and f ∈ AG , then clearly the map x → f u(x) belongs to HomG (V, A). Thus HomG (V, A) has a natural structure of AG -module. Proposition. Let V be a finite-dimensional rational G-module. Then HomG (V, A) is a finitely generated AG -module. Proof. We have: G G G A ⊗ A(V ) = A ⊗ A(V )n A ⊗ A(V )n = n0 n0 G A ⊗ A(V )n = AG ⊕ (A ⊗ V ∗ )G ⊕ n2 G A ⊗ A(V )n . = AG ⊕ HomG (V, A) ⊕ n2
G By 27.4.3, A ⊗ A(V ) is a finitely generated commutative AG -algebra. In G particular, the ideal n1 A ⊗ A(V )n of this AG -algebra is finitely generated. Since HomG (V, A) = (A ⊗ A(V )1 )G , the result follows. 27.4.5 Corollary. Let us conserve the notations and hypotheses of 27.4.3. - the subspace A(ω) is a finitely generated AG -module. For all ω ∈ G, Proof. This is clear by 27.3.6 (iv) and 27.4.4.
27.5 Algebraic quotients 27.5.1 In this section, G is a reductive algebraic group and X an affine G-variety. Recall from 22.1.1 that G acts naturally on A(X). By 27.4.3, the reduced algebra A(X)G is finitely generated. So 11.6.1 implies that A(X)G is the algebra of regular functions of an affine variety that we shall denote by X//G, instead of Spm(A(X)G ). Let π : X → X//G be the morphism induced by the injection A(X)G → A(X). The pair (X//G, π), or X//G, is called the algebraic quotient of X by G.
27.5 Algebraic quotients
423
Proposition. (i) The map π is surjective and constant on G-orbits. (ii) Let Y be an affine G-variety and u : X → Y be a morphism which is constant on the G-orbits of X. Then there exists a unique morphism v : X//G → Y such that u = v ◦ π. (iii) If X is normal, then X//G is also normal. Proof. (i) For α ∈ G, x ∈ X et f ∈ A(X)G = A(X//G), we have: f (x) = A(π)(f )(x) = f π(x) = f (α.x) = f π(α.x) . So π(x) = π(α.x). Let ξ ∈ X//G and a its ideal in A(X)G . Then aA(X) ∩ A(X)G = a by 27.4.3. It follows that aA(X) = A(X). Let b be a maximal ideal of A(X) containing aA(X) and x ∈ X be the point defined by b. Then we have ξ = π(x). (ii) The assumptions imply that A(u) A(Y ) ⊂ A(X)G . Thus A(u) = A(π)◦θ where θ is an algebra homomorphism from A(Y ) to A(X)G . The morphism v = Spm(θ) verifies u = v◦π. Clearly, v is unique with this property. (iii) If f ∈ Fract A(X)G is integral over A(X)G , then f is integral over A(X), so f belongs to A(X) since X is normal. Hence f ∈ A(X)G . 27.5.2 Proposition. Let us conserve the hypotheses and notations of 27.5.1. (i) If Y is a closed G-stable subset of X, then π(Y ) is a closed subset of X//G. Furthermore, π(Y ), π|Y is the algebraic quotient of Y by G. (ii) Let (Yi )i∈I be a family of closed G-stable subsets of X. Then: Yi = π(Yi ). π i∈I
i∈I
Proof. (i) Let j : Y → X be the canonical injection and (Y //G, ω) the algebraic quotient of Y by G. The map A(j) is the surjection f → f |Y . Let us denote by pA(X) , pA(Y ) the corresponding Reynolds operators. Since A(j)◦pA(X) = pA(Y ) ◦A(j), A(j) induces a surjection θ : A(X)G → A(Y )G . The map A(ω) is the canonical injection A(Y )G → A(Y ) and we have A(ω)◦θ = A(j)◦A(π), so Spm(θ)◦ω = π◦j. By 11.6.8, Spm(θ) = i◦v where v is an isomorphism of Y //G onto a closed subvariety Z of X//G, and i : Z → X//G is the canonical injection. So (i) follows. (ii) If a is an ideal of A(X) (resp. A(X)G ), denoteby V(a) (resp. V (a)) the variety of zeros of a in X (resp. X//G). We have i∈I Yi = V i∈I ai . So (i) and 27.3.8 imply that: π Yi = π Yi = V A(X)G ∩ ai = V (A(X)G ∩ ai ) i∈I i∈I i∈I i∈I = V (A(X)G ∩ ai ) = π(Yi ) = π(Yi ). i∈I
i∈I
i∈I
424
27 Reductive groups
27.5.3 Proposition. Let us conserve the hypotheses and notations of 27.5.1. If x ∈ X, then π −1 (π(x)) contains a unique closed G-orbit O and we have: π −1 π(x) = {y ∈ X; O ⊂ G.y}. Proof. The set Y = π −1 π(x) is closed and G-stable, so it contains a closed orbit by 21.4.5. Suppose that O1 and O2 are two distinct closed orbits contained in Y . Then 27.5.2 (ii) implies that π(O1 ) ∩ π(O2 ) is empty, which is absurd. So Y contains a unique closed G-orbit O. If y ∈ Y , then the closed G-stable subset G.y of Y contains a closed orbit (21.4.5), so O ⊂ G.y. Conversely, as π is constant on the closures of orbits, if O ⊂ G.y, then π(x) = π(y). 27.5.4 Remark. In view of 27.5.3, we can interpret X//G as the set of closed G-orbits of X.
27.6 Characters 27.6.1 Definition. Let G be an algebraic group. (i) A function f ∈ A(G) is said to be central if f (αβα−1 ) = f (β) for all α, β ∈ G. We shall denote by C(G) the subalgebra of A(G) consisting of central functions. (ii) Let (V, π) be a finite-dimensional rational G-module. The character of V is defined to be the map χV : G → k, α → tr π(α). 27.6.2 If V, W are finite-dimensional rational G-modules, then: χV ⊕W = χV + χW , χV ⊗W = χV χW . Moreover, it is clear that χV ∈ C(G) and that χV = χW if V and W are - then we denote χM simply by χω . isomorphic G-modules. If ω ∈ G, ω Proposition. Let G be a reductive algebraic group. The space C(G) is spanned by the χω ’s, with ω ∈ G. - and set M = Mω . Let (e1 , . . . , en ) be a basis of M , Proof. Fix ω ∈ G ∗ ∗ (e1 , . . . , en ) the dual basis and φ as defined in 27.3.9. If α ∈ G, then: χω (α) =
n i=1
∗
e∗i (α.ei ) =
n
φ(e∗i ⊗ ei )(α).
i=1
Thus χω ∈ φ(M ⊗k M ). By 27.3.9, to obtain the result, it suffices to prove that any central function contained in φ(M ∗ ⊗k M ) is proportional to χω . Let θ : M ∗ ⊗k M → Endk (M ) be the map sending f ⊗ x to the endomorphism y → f (y)x. It is well-known that θ is a linear isomorphism and we verify easily that θ is a G-module homomorphism. Moreover, for α, β ∈ G, f ∈ M ∗ and x ∈ M , we have: φ α.(f ⊗ x) (β) = f (α−1 βαx) = φ(f ⊗ x)(α−1 βα).
27.6 Characters
425
We deduce therefore that if ξ ∈ M ∗ ⊗k M , then the following conditions are equivalent: (i) φ(ξ) ∈ C(G). (ii) α.ξ = ξ for all α ∈ G. (iii) α.θ(ξ) = θ(ξ) for all α ∈ G. (iv) α.θ(ξ)(α−1 .x) = θ(ξ)(x) for all (α, x) ∈ G × M . (v) θ(ξ) ∈ EndG (M ) = k idM (10.7.2). But, we have: idM = θ(e∗1 ⊗ e1 + · · · + e∗n ⊗ en ). Since θ is bijective, the result follows from equivalence of conditions (i) and (v). 27.6.3 Let G be a reductive algebraic group, I the Reynolds operator associated to A(G) and (V, σ) a finite-dimensional rational G-module. Let (e1 , . . . , en ) be a basis of V and (e∗1 , . . . , e∗n ) the dual basis. Set α.ej =
n
σij (α)ei
i=1
for α ∈ G and 1 j n. With φ in the notations of 27.3.9, we have σij = φ(e∗i ⊗ ej ). If α, β ∈ G, then: n n σlj (αβ)el = σij (β) σli (α)el i=1 l=1 l=1 n n = σli (α)σij (β) el .
(αβ).ej =
n
i=1
l=1
Hence: (1)
ρβ σlj =
n
σij (β)σli , λα−1 σlj =
i=1
n
σli (α)σij .
i=1
Since I ◦ρβ = I ◦λα−1 = I (27.3.4), we have therefore (2)
I(σlj ) =
n
σij (β) I(σli ) =
i=1
n
σli (α) I(σij )
i=1
for all α, β ∈ G. We deduce that: (3)
I(σlj ) =
n
I(σli ) I(σij ).
i=1
Define q ∈ End(V ) as follows: for 1 j n: q(ej ) =
n i=1
I(σij )ei .
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27 Reductive groups
If x ∈ V and f ∈ V ∗ , then: f q(x) = I φ(f ⊗ x) .
(4)
Lemma. The map q is the Reynolds operator pV associated to V . Proof. The equalities in (2) imply that q◦σ(α) = σ(α)◦q = q for all α ∈ G and G ∗ (3) implies that q 2 = q. Finally if x ∈ V , for f ∈ V , we have φ(f ⊗x)(α) = f (α.x) = f (x), so I φ(f ⊗ x) = f (x). By (4), we obtain that f q(x) = f (x) for all f ∈ V ∗ , hence q(x) = x. Now the result follows from 10.7.9. 27.6.4 Let us define an involution f → f ∗ in A(G) and a bilinear form (f, g) → f, g on A(G) as follows: f ∗ (α) = f (α−1 ) , f, g = I(f g ∗ ). If h ∈ A(G) and α ∈ G, then (ρα h−h)∗ = λα h∗ −h∗ . By 10.7.9 and 27.3.4, we deduce that I(f ) = I(f ∗ ) for f ∈ A(G). The form . , . is therefore symmetric. Let us endow A(G) with this bilinear form in the rest of this section. 27.6.5 Let (V, σ) and (W, τ ) be non-isomorphic simple rational G-modules of dimension m and n, E (resp. F) a basis of V (resp. W ). For α ∈ G, denote by S(α) = [σij (α)] and T (α) = [τij (α)] the matrices of σ(α) and τ (α) with respect to the basis E et F. Lemma. (i) For 1 i, j m and 1 k, l n, we have σij , τkl = 0. (ii) For 1 i, j, k, l m, we have mσij , σkl = δil δjk , where δ denotes the Kronecker symbol. Proof. (i) For A ∈ Mm,n (k), set A(α) = S(α)AT (α−1 ) = [θij (α)] and let B = [I(θij )] ∈ Mm,n (k). If α, β ∈ G, then A(αβ) = S(α)A(β)T (α−1 ). It follows as in 27.6.3 that S(α)B = BT (α). So 10.7.2 implies that B = 0. But if A = [aij ], then: θij (α) = σik (α)akl τlj∗ (α). k,l
Hence: I(θij ) =
akl σik , τlj .
k,l
So (i) follows by taking A to be the matrices of the canonical basis of Mm,n (k). (ii) If σ = τ , then 10.7.2 implies that B = µA Im , with µA ∈ k. So I(θij ) = 0 if i = j, and the same argument as in (i) proves the result for i = j. On the other hand, we have tr A(α) = tr(A) for all α ∈ G, so tr(B) = tr(A). It follows that µA = m−1 tr(A). So (ii) follows as in (i) by taking A to be the matrices of the canonical basis of Mm (k).
27.6 Characters
427
27.6.6 Theorem. Let G be a reductive algebraic group. (i) The decomposition A(G) = (Mω∗ ⊗ Mω ) ω∈G
in 27.3.9 is orthogonal. Moreover, the restriction of . , . to each Mω∗ ⊗ Mω is non-degenerate. - form an orthonormal basis of C(G). (ii) The characters χω , ω ∈ G, - then: (iii) If V is a finite-dimensional rational G-module and ω ∈ G, multω (V ) = χω , χV . (iv) Any finite-dimensional rational G-module is determined up to an isomorphism by its character. Proof. Let V and W be two simple rational G-modules of finite dimension, (e1 , . . . , em ) (resp. (ε1 , . . . , εn )) a basis of V (resp. W ), and (e∗1 , . . . , e∗m ) (resp. (ε∗1 , . . . , ε∗n )) the dual basis. Let φ be as in 27.3.9. Suppose that V and W are not isomorphic. Then φ(e∗i ⊗ej ), φ(ε∗k ⊗εl ) = 0 by 27.6.3 and 27.6.5. Hence we have (i). - where ω = θ. We saw in the proof of 27.6.2 that χω ∈ Let ω, θ ∈ G, φ(Mω∗ ⊗ Mω ). So (i) implies that χω , χθ = 0. Moreover: χω , χω = φ(e∗i ⊗ ei ), φ(e∗j ⊗ ej ). i,j
So 27.6.5 says that χω , χω = 1, and (ii) follows from 27.6.2. Part (iii) follows from (ii) and 27.3.7, and (iv) is an immediate consequence of (iii).
References • [5], [38], [40], [51], [77], [78].
28 Borel subgroups, parabolic subgroups, Cartan subgroups
In this chapter, our interest turns to certain fundamental notions of the theory of algebraic groups : Borel subgroups, parabolic subgroups, Cartan subgroups. One of the main result we prove in this chapter is that Borel subgroups (resp. Cartan subgroups) of an algebraic group are conjugate. Let G be a connected algebraic group and g its Lie algebra. Two elements x, y ∈ g are conjugate if there exists α ∈ G such that y = Ad(α)(x).
28.1 Borel subgroups 28.1.1 Definition. A maximal connected solvable subgroup of G is called a Borel subgroup of G. 28.1.2 Remarks. Let H be a subgroup of G. If H is solvable and connected, then so is H. Thus a Borel subgroup is a maximal connected solvable algebraic subgroup of G. In particular, for reasons of dimension, G has a Borel subgroup. If H is unipotent, then H is also unipotent (10.8.14), so it is nilpotent and connected (10.8.10 and 22.3.6). Thus H is contained in a Borel subgroup of G. In the same way, a torus of G is contained in a Borel subgroup of G. 28.1.3 Theorem. (i) If B is a Borel subgroup, then G/B is projective. (ii) All the Borel subgroups of G are conjugate. Proof. Let P be a Borel subgroup of G of maximal dimension. By 25.4.3, there exists a finite-dimensional rational representation (V, π) of G and a line L of V such that: P = {α ∈ G; π(α)(L) = L} , p = L(P ) = {x ∈ g; dπ(x)(L) ⊂ L}. Applying 26.3.2 to the action of P on V /L, we obtain that P fixes a flag H0 : L = V1 ⊂ V2 ⊂ · · · ⊂ Vn = V of V . Since G acts rationally on the flag variety F of V , our choice of L implies that P is the stabilizer of H0 in G.
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28 Borel subgroups, parabolic subgroups, Cartan subgroups
Now let H ∈ F and H the stabilizer of H in G. Then π(H) is solvable. On the other hand, ker π ⊂ P , so ker π is solvable. It follows that H is solvable. Hence dim H dim P . We deduce therefore that the G-orbit of H0 is of minimal dimension. So it is closed (21.4.5) and projective (13.6.7). Since the morphism G/P → G.H0 , αP → α.H0 , is G-equivariant and bijective, it follows from 25.1.2 that it is an isomorphism. Thus G/P is projective. Finally, let B be a Borel subgroup of G. Then B acts on G/P by (β, αP ) → βαP . By 13.4.5 and 26.3.4, there exists α ∈ G such that BαP = αP . So α−1 Bα ⊂ P , or B ⊂ αP α−1 . It follows from the definition of a Borel subgroup that α−1 Bα = P . 28.1.4 Corollary. Let P be a closed subgroup of G. Then the following conditions are equivalent: (i) The variety G/P is projective. (ii) The variety G/P is complete. (iii) The group P contains a Borel subgroup of G. If these conditions are verified, then P is called a parabolic subgroup of G. Proof. (i) ⇒ (ii) This is clear by 13.4.5. (ii) ⇒ (iii) Let B be a Borel subgroup of G. Then by 26.3.4, the action of B on G/P , given by (β, αP ) → βαP , has a fixed point, say αP . So BαP = αP and P contains the Borel subgroup α−1 Bα of G. (iii) ⇒ (i) Let B be a Borel subgroup contained in P . Since G/B is projective (28.1.3) and the canonical morphism G/B → G/P is surjective, 13.4.3 implies that G/P is complete. But G/P is quasi-projective (25.4.7), so it is a projective variety. 28.1.5 Remark. Let H be a closed connected subgroup of G. By 28.1.4, the following conditions are equivalent: (i) H is a Borel subgroup of G (ii) H is solvable and G/H is projective. 28.1.6 Corollary. Let B be a Borel subgroup of G. (i) Let u : G → G be an automorphism of the algebraic group G. If u(α) = α for all α ∈ B, then u = idG . (ii) We have Z(G)◦ ⊂ Z(B) ⊂ CG (B) = Z(G). Proof. (i) Let π : G → G/B be the canonical surjection and v : G → G the morphism α → u(α)α−1 . If u(α) = α for all α ∈ B, then v is constant on αB for all α ∈ G. By 25.3.3 and 25.4.7, there exists a morphism w : G/B → G such that w◦π = v. It follows therefore from 13.4.3, 13.4.5 and 28.1.3 that v(G) is closed in G. So v(G) is affine and complete. Moreover, v(G) is connected. Thus v(G) = v(B) = {e} by 13.4.4, and u = idG . (ii) The subgroup Z(G)◦ is connected and solvable. So Z(G)◦ is contained in a Borel subgroup P of G. By 28.1.3, there exists α ∈ G such that αP α−1 = B. So Z(G)◦ ⊂ Z(B) since αZ(G)◦ α−1 = Z(G)◦ . Clearly, Z(G) ∪ Z(B) ⊂
28.1 Borel subgroups
431
CG (B). Finally, let α ∈ CG (B) and iα the inner automorphism of G defined by α. Then iα |B = idB , so iα ∈ idG by part (i). Hence α ∈ Z(G). Remark. We shall see in 28.2.3 that Z(B) = Z(G). 28.1.7 Corollary. (i) The set of maximal tori of G is equal to the set of maximal tori of the Borel subgroups of G. Two maximal tori of G are conjugate. (ii) The set of maximal unipotent subgroups of G is equal to the set of maximal unipotent subgroups of the Borel subgroups of G. Two maximal unipotent subgroups of G are conjugate. Proof. (i) Let T be a maximal torus. We observed in 28.1.2 that T is contained in a Borel subgroup B and T is a maximal torus in B. By 26.5.5, any maximal torus of B is conjugate to T . So the result follows from the fact that Borel subgroups are conjugate (28.1.3). (ii) Again, we saw in 28.1.2 that any maximal unipotent subgroup U is contained in a Borel subgroup B and U is a maximal unipotent subgroup in B. So U = Bu by 26.4.2. The result follows again from 28.1.3. 28.1.8 Corollary. Let B be a Borel subgroup of G. (i) If B = Bs , then G is a torus. (ii) If {e} is the only torus of B, then G is unipotent. Proof. (i) By 26.5.5, B = T Bu where T is a maximal torus of B. If B = Bs , then B = T , and T = Z(B) ⊂ Z(G) by 28.1.6. So T is normal in G, and the variety G/T is irreducible, affine and complete (25.4.11 and 28.1.3). So it is reduced to a single point (13.4.4). (ii) If B = Bu , then H = Z(B)◦ is not trivial unless G = {e}, and H ⊂ Z(G) (28.1.6). The group B/H is a unipotent Borel subgroup of G/H. By induction on dim G, we deduce that G/H = B/H. Hence G = B is unipotent. 28.1.9 Corollary. The following conditions are equivalent: (i) G is nilpotent. (ii) G has a nilpotent Borel subgroup. (iii) G has a unique maximal torus. (iv) There exists a maximal torus of G contained in Z(G). Proof. (i) ⇒ (ii) This is obvious. (ii) ⇒ (iii) Let B be a nilpotent Borel subgroup. By 26.4.4, B = T × Bu where T = Bs is a maximal torus of G. By 28.1.6, T ⊂ Z(B) ⊂ Z(G). Since any maximal torus of G is conjugate to T (28.1.7), T is the unique maximal torus of G. (iii) ⇒ (iv) If G has a unique maximal torus T , then T is normal in G (28.1.7). For α ∈ G, the map T → T , β → αβα−1 , is a morphism of the algebraic group T . So by 22.5.10, we have T ⊂ Z(G).
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28 Borel subgroups, parabolic subgroups, Cartan subgroups
(iv) ⇒ (i) Let T be a maximal torus of G contained in Z(G). By 26.5.5 and 28.1.7, we have B = T × Gu and T = Bs . So B is nilpotent (26.4.4). The group B/T is then a nilpotent Borel subgroup of G/T . By induction on dim G, we deduce that G/T is nilpotent, and since T ⊂ Z(G), G is nilpotent (10.5.3). 28.1.10 Corollary. (i) If T is a maximal torus of G, then C = CG (T )◦ is nilpotent and C = NG (C)◦ . (ii) If dim G 2, then G is solvable. Proof. (i) By 28.1.7 and 28.1.9, T is the unique maximal torus of C and C is nilpotent. Moreover if α ∈ NG (C), then αT α−1 is a maximal torus of C, so αT α−1 = T . Thus T is normal in NG (C), and so T commutes with NG (C)◦ by 22.5.10. Hence NG (C)◦ ⊂ C and the result follows. (ii) Let B be a Borel subgroup of G. Then B = T Bu where T is a maximal torus of G. If B = G, then dim B 1, so B = T or B = Bu . But 28.1.8 would imply that G = B. Contradiction. 28.1.11 Theorem. Let B be the set of Borel subgroups of G. Then: ◦ ◦ R(G) = B , Ru (G) = Bu B∈B
B∈B
Proof. Let H (resp. K) be the identity component of the intersection of B (resp. Bu ), B ∈ B. Since R(G) is connected, normal and solvable, it is contained in H (28.1.3). On the other hand, H is normal (28.1.3), connected and solvable. So H ⊂ R(G). The equality K = Ru (G) is proved in the same way.
28.2 Theorems of density 28.2.1 Proposition. Let H be a closed subgroup of G and X= αHα−1 . α∈H
(i) X contains a dense open subset of X. (ii) If the variety G/H is complete, then X is closed in G. (iii) Suppose that there exists γ ∈ G which is contained only in a finite number of conjugates of H. Then H = G. Proof. Denote by pr1 : (G/H) × G → G/H and pr2 : (G/H) × G → G the canonical projections, π : G → G/H the canonical surjection and u, v the morphisms: u : G × G → G × G , (α, β) → (α, αβα−1 ), v : G × G → (G × G)/(H × {e}) = (G/H) × G , (α, β) → (π(α), β).
28.2 Theorems of density
433
Finally, let M = {(α, β) ∈ G × G; β ∈ αHα−1 }, N = v(M ). (i) Since X = pr2 ◦v◦u(G × H), it is a constructible subset of G (15.4.3). So the result follows from 1.4.6. (ii) We have M = u(G × H) and u is an isomorphism of varieties. Thus M is closed in G × G. It follows from 25.3.2 and 25.4.7 that N is closed in (G/H) × G. So if G/H is complete, then X = pr2 (N ) is closed. (iii) Let p = pr1 |N . If α ∈ G, then p−1 π(α) = {π(α)} × (αHα−1 ). We obtain therefore from 15.5.5 that dim N = dim(G/H) + dim H = dim G. Consider the action of G on G/H given by α.π(β) = π(αβ) and denote q = pr2 |N . If α ∈ G, then q −1 (α) = Eα × {α} where Eα = {π(β); β −1 αβ ∈ H} = {π(β); α.π(β) = π(β)}. The set q −1 (γ) is finite by assumption. Let us consider the morphism N → X = pr2 (N ) induced by q. Then 15.5.4 (i) says that dim X = dim N . Thus dim X = dim G and, since G is connected, we have X = G by 1.2.3. 28.2.2 Theorem. Let B be a Borel subgroup of G, T a maximal torus in G and C = CG (T )◦ . Then: G= αBα−1 , Gu = αBu α−1 , Gs = αT α−1 . α∈G
Furthermore, the set
α∈G
α∈G
αCα−1 contains a dense open subset of G.
α∈G
Proof. By 28.1.10, C is nilpotent. So 26.4.4 implies that C = T ×Cu . Fix α ∈ T such that CG (α) = CG (T ) (26.2.3) and let β ∈ G be such that βαβ −1 ∈ C. Then βαβ −1 ∈ T and CG (T ) ⊂ CG (βαβ −1 ). But dim CG (βαβ −1 ) = dim C, so we deduce that CG (βαβ −1 ) = CG (βαβ −1 )◦ = C. Thus βCβ −1 = C, that is β ∈ NG (C). Since C = NG (C)◦ (28.1.10), we have proved that the set of conjugates of α which are contained in C is finite. We deduce therefore from 28.2.1 that the union of the conjugates of C contains a dense open subset of G. The group C is connected and nilpotent, so up to conjugation, we may assume that C ⊂ B (28.1.3). So the union D of the set of Borel subgroups contains a dense open subset of G. But G/B is projective (28.1.3), so D is closed in G by 28.2.1. Hence D = G. The other equalities follow now from 26.5.6 and 28.1.7. 28.2.3 Corollary. (i) We have Z(B) = Z(G) for any Borel subgroup B of the group G. (ii) The set Z(G)s is the intersection of all the maximal tori of G. Proof. (i) Let α ∈ Z(G). By 28.2.2, a conjugate of α is in B, so α ∈ B and Z(G) ⊂ Z(B). The result follows therefore from 28.1.6. (ii) Let α ∈ Z(G)s . As in the proof of (i), we have α ∈ B. So by 26.5.6, α is contained in a maximal torus of B. Hence α belongs to the intersection H of
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28 Borel subgroups, parabolic subgroups, Cartan subgroups
all the maximal tori of G (28.1.7). Conversely, H is a diagonalizable subgroup of G and clearly H is normal in G. So 22.5.10 implies that H is central in G. Hence H = Z(G)s .
28.3 Centralizers and tori 28.3.1 Recall that for α ∈ G, αs (resp. αu ) denotes the semisimple (resp. unipotent) component of α. Proposition. (i) Let H be a connected solvable subgroup of G. For any α ∈ CG (H), there exists a Borel subgroup of G containing H and α. (ii) Let S be a torus of G and α ∈ CG (S). There exists a torus of G containing S and αs . The group CG (S) is connected. (iii) If α ∈ G, then α ∈ CG (αs )◦ . Proof. (i) The group H is solvable, connected and α ∈ CG (H). We may therefore assume that H is closed. Let B be a Borel subgroup containing α (28.2.2). Consider the action of G on G/B given by: (β, γB) → (βγ)B. The variety X of fixed points of α is non-empty since it contains eB, and it is H-stable. It follows from 26.3.4 that there exists β ∈ G such that βB ∈ X and γ.βB = βB for all γ ∈ H. So βBβ −1 is a Borel subgroup of G containing H and α. (ii) By (i), S and α are contained in a Borel subgroup B of G. We have αs ∈ B and the first part follows from 26.5.6. Again by 26.5.6, CB (S) is connected, so α ∈ CG (S)◦ . Hence CG (S) = CG (S)◦ . (iii) Let B be a Borel subgroup containing α. Then it contains also αs , and α ∈ CB (αs ) = CB (αs )◦ (26.5.6). Hence α ∈ CG (αs )◦ . 28.3.2 Proposition. Let S be a torus of G and C = CG (S). (i) If B is a Borel subgroup of G containing S, then C ∩ B is a Borel subgroup of C. (ii) Any Borel subgroup of C is of the form C ∩ B, where B is a Borel subgroup of G containing S. Similar results hold for maximal tori and maximal unipotent subgroups. Proof. If D is a Borel subgroup of C, then S ⊂ Z(C) ⊂ Z(D) (28.2.3). Since all the Borel subgroups of C are conjugate (28.1.3), it suffices to prove (i). The group C ∩ B = CB (S) is solvable and connected (26.5.6), so the result would follows from 28.1.4 if the variety C/(C ∩ B) is complete. Observe that SBu is a connected algebraic subgroup of G (21.2.3). Consider the action of C on G/B given by (α, βB) → (αβ)B. Then C ∩ B is the stabilizer of eB = eB in C. So the orbit map C/(C ∩ B) → CeB is a bijective C-equivariant morphism, hence, it is an isomorphism by 25.1.2. It suffices therefore to show that CeB is closed in G/B. Let π : G → G/B be the canonical surjection. Then X = CB = π −1 (CeB ) is B-stable (on the right), it follows that X is also B-stable. Since π is open (25.4.7), π sends B-stable
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closed subsets to closed subsets of G/B (25.3.2). So we are reduced to show that X is closed in G. If α ∈ X, then α−1 Sα ⊂ B, and this is still true for α ∈ X by continuity. Let us consider the morphism X × S → B/Bu , (α, β) → α−1 βαBu . It verifies the hypotheses of 22.5.10. It follows that α−1 βα ∈ βBu for α ∈ X and β ∈ S. Thus α−1 Sα is a maximal torus of the group SBu . So 28.1.7 implies that there exists γ ∈ SBu such that α−1 Sα = γ −1 Sγ. Hence αγ −1 ∈ NG (S), and so α ∈ NG (S)B. Let Y = NG (S)B. We obtain therefore that CB = X ⊂ X ⊂ Y . Since Y = π −1 (NG (S)eB ) is the inverse image of a locally closed subset (21.4.3), it is locally closed, and hence it is a subvariety of G. Let us consider the action of C × B on Y given by (α, β) · γ = αγβ −1 . The orbits under this action are of the form CγB, γ ∈ NG (S). By 22.5.11 and 28.3.1, C = NG (S)◦ and so C is normal in NG (S) (21.1.6). It follows that for α ∈ NG (S), αCB = CαB, and for α, β ∈ NG (S), the morphism αCB → βCB, γ → βα−1 γ, is an isomorphism of irreducible subvarieties of G. Consequently, since C has finite index in NG (S) (21.1.6), it follows from the preceding discussion that Y has a finite number of C × B-orbits all of the same dimension. In particular, these orbits are closed (21.4.5). Since X is also the closure of X in Y , we have X = X (21.4.5). The proofs for the cases of maximal tori and maximal unipotent subgroups are analogue. 28.3.3 Corollary. Let T be a maximal torus of G. If B is a Borel subgroup of G containing T , then B contains CG (T ). Proof. The group CG (T ) is connected (28.3.1) and nilpotent (28.1.9). So the result follows from 28.3.2.
28.4 Properties of parabolic subgroups 28.4.1 Lemma. Let u : G → H be a surjective morphism of algebraic groups and B = T Bu a Borel subgroup of G where T is a maximal torus of G. We have u(B) = u(T )u(Bu ) and u(B) is a Borel subgroup of H. Moreover, u(T ) is a maximal torus of H and u(Bu ) = u(B)u . In particular, any Borel subgroup of H is the image of a Borel subgroup of G. Proof. The composition G → H → H/u(B) of morphisms induces a surjective morphism G/B → H/u(B). So H/u(B) is a complete variety (28.1.3 and 13.4.3). Thus u(B) is a parabolic subgroup of H (28.1.4). Since u(B) is solvable and connected, it is a Borel subgroup of H. Finally, it is clear that u(B) = u(T )u(Bu ), and we have u(Bu ) = u(B)u by 22.2.3. So u(T ) is a maximal torus of u(B), and hence of H.
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28.4.2 Theorem. Let G be a connected algebraic group. (i) If B is a Borel subgroup of G, then NG (B) = B. (ii) For any parabolic subgroup P of G, we have NG (P ) = P = P ◦ . Proof. (i) Let N = NG (B) and proceed by induction on dim G. The result is clear if dim G 2 by 28.1.10. Let α ∈ N and T a maximal torus of B. Since αT α−1 is a maximal torus of B, there exists β ∈ B such that αT α−1 = βT β −1 (28.1.7). So αβ −1 ∈ NG (T ) and by replacing α by αβ −1 , we may assume that α ∈ NG (T ). Consider the morphism u : T → T , γ → αγα−1 γ −1 . Since T is commutative and α normalizes T , we see that u is an endomorphism of the group T . We have two cases: 1) u is not surjective. Then S = (ker u)◦ is a non-trivial torus of T and α ∈ CG (S). Moreover, α normalizes CG (S) ∩ B which is a Borel subgroup of CG (S) (28.3.2). If CG (S) = G, then α ∈ B by the induction hypothesis. Otherwise if CG (S) = G, then S ⊂ Z(G) = Z(B). So by passing to the quotient G/S, we obtain the result by 28.4.1 and the induction hypothesis. 2) u is surjective. By 25.4.3, there exists a finite-dimensional rational Gmodule V and v ∈ V \ {0} such that N = {α ∈ G; α.v ∈ kv}. If β ∈ Bu , then β.v = v by 22.3.6. If β ∈ T , then the surjectivity of u implies that β is the commutator of two elements of N , so β.v = v. It follows that β.v = v for all β ∈ B. The map γ → γ.v induces therefore a morphism G/B → V . By 13.4.3, 13.4.4 and 13.4.5, this morphism is constant. So γ.v = v for all γ ∈ G. Hence G = N , and therefore G = N = B by 28.2.2. (ii) Let P be a parabolic subgroup of G and B a Borel subgroup of G contained in P . In particular, B ⊂ P ◦ . Let α ∈ NG (P ). Then αBα−1 is a Borel subgroup of P ◦ , so there exists β ∈ P ◦ such that αBα−1 = βBβ −1 . Now part (i) says that β −1 α ∈ NG (B) = B. So α ∈ βB ⊂ P ◦ . We have therefore proved that NG (P ) = P ◦ = P . 28.4.3 Corollary. Let B be a Borel subgroup of G. (i) We have NG (Bu ) = B. (ii) Any parabolic subgroup of G is conjugate to a unique parabolic subgroup containing B. (iii) If H is a solvable subgroup (not necessarily connected or closed) of G containing B, then H = B. Proof. (i) Let U = Bu and N = NG (U ). Since B normalizes U , it is a Borel subgroup of N ◦ . By 28.2.2, any unipotent element of N ◦ is conjugate in N ◦ to an element of U . As U is normal in N ◦ , it follows that U = Nu . We deduce therefore that the group N ◦ /U is a torus, and so, it is solvable. Thus N ◦ is solvable and B = N ◦ . Now N normalizes N ◦ , so N ⊂ NG (B) = B.
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(ii) Let P, Q be parabolic subgroups containing B verifying Q = αP α−1 with α ∈ G. Then B and αBα−1 are Borel subgroups of Q. So there exists β ∈ Q such that αBα−1 = βBβ −1 (28.1.3). Thus β −1 α ∈ NG (B) = B, and so α ∈ βB ⊂ Q. Hence P = Q. (iii) It suffices to show that H = B. But H is a parabolic subgroup. So H is connected 28.4.2, and since it is solvable, we have H = B (28.1.5). Remark. Condition (iii) of 28.4.3 does not say that a maximal solvable subgroup of G is a Borel subgroup of G.
28.5 Cartan subgroups 28.5.1 Definition. A subgroup of G is called a Cartan subgroup of G if it is equal to the centralizer of a maximal torus of G. 28.5.2 Cartan subgroups of G are connected and nilpotent (28.1.10 and 28.3.1). Two Cartan subgroups are conjugate (28.1.7) and their common dimension is called the rank of G, denoted by rk(G). By 28.2.2, if C is a Cartan subgroup of G, then the union of the conjugates of C contains a dense open subset of G. By 28.1.7 and 28.3.2, the map T → CG (T ) is a bijection from the set of maximal tori of G onto the set of Cartan subgroups of G. We have also, for a maximal torus T , CG (T ) = T × CG (T )u . 28.5.3 Lemma. (i) Let H be a connected nilpotent algebraic group and K a closed subgroup of H distinct from H. Then dim K < dim NH (K). (ii) If C is a Cartan subgroup of G, then it is a maximal connected nilpotent subgroup of G. Proof. (i) Since H = K, H = {e}. The last non-trivial term in the central descending series of H is connected (21.3.6) and central in H. So Z = Z(H)◦ = {e}. If Z ⊂ K, then the connected algebraic subgroup ZK contains strictly K and normalizes K. So dim NH (K) dim ZK > dim K. Suppose that Z ⊂ K. Let π : H → H/Z be the canonical surjection. We have: NH/Z (K/Z) = π(NH (K)) , dim NH (K) = dim Z + dim NH/Z (K/Z). So the result follows by induction on dim H. (ii) Suppose that H is a connected nilpotent subgroup of G containing strictly C. We may assume that H is closed. By (i), C = NH (C)◦ . Let α ∈ NH (C) and write C = CG (T ) where T is a maximal torus of G. Then αT α−1 is a maximal torus of C, and so αT α−1 = T since T is central in C. Thus NH (C)◦ ⊂ NH (T )◦ = CH (T ) = C (28.1.10 and 28.3.1). We have therefore obtained a contradiction.
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28.5.4 Let α ∈ G. By 28.2.2, αs belongs to a maximal torus of G. We deduce therefore that dim CG (αs ) rk(G). We say that α is generic in G if dim CG (αs ) = rk(G). Denote by Ggen the set of generic elements of G. Lemma. Let T be a maximal torus of G and α ∈ T . Then the following conditions are equivalent: (i) α ∈ Ggen . (ii) CG (α)◦ = CG (T ). (iii) χ(α) = 1 for all χ ∈ Φ(T, G) (in the notations of 26.2.1). Proof. The equivalence (i) ⇔ (ii) is clear since CG (T ) is connected (28.3.1), while (ii) ⇔ (iii) follows from 26.1.1 and 26.2.1. 28.5.5 Proposition. Let α be a semisimple element of G. The following conditions are equivalent: (i) α ∈ Ggen . (ii) CG (α)◦ is a Cartan subgroup of G. (iii) CG (α)◦ is a nilpotent subgroup of G. (iv) α belongs to a unique maximal torus of G. (v) α belongs to a finite number of maximal torus of G. Proof. The equivalence (i) ⇔ (ii) (resp. (ii) ⇔ (iii)) follows from 28.5.4 (resp. 26.4.4 and 28.5.3). Let T be a maximal torus of G containing α (28.2.2). Then T ⊂ CG (α)◦ . If CG (α)◦ is nilpotent, then T is its unique maximal torus (26.4.4). So (iii) ⇒ (iv) ⇒ (v). To finish the proof, we shall show that (v) ⇒ (ii). Let H = CG (α)◦ and T a maximal torus of G containing α. Since maximal tori of H are conjugate in H, the connected variety H/NH (T ) is finite, and hence it is reduced to a point. Thus T is normal in H. It follows from 22.5.10 that T is central in H. So H ⊂ CG (T ). But CG (T ) ⊂ CG (α)◦ = H, hence H = CG (T ). 28.5.6 Lemma. Let H be a connected algebraic group which contains a unipotent element α belonging to a unique Cartan subgroup C of H. Then H is nilpotent. Proof. Let T be a maximal torus such that CG (T ) = C, and B a Borel subgroup of H containing T . By 28.3.3, C ⊂ B. Let Bu = Nn ⊃ Nn−1 ⊃ · · · ⊃ N0 = {e} be the central descending series of Bu . We have N0 ⊂ C. Suppose that Ni ⊂ C. Since α ∈ Bu , we have α−1 βαβ −1 ∈ Ni if β ∈ Ni+1 , so βαβ −1 ∈ αNi ⊂ C, and α ∈ β −1 Cβ. Since β −1 Cβ is a Cartan subgroup of H containing α, our hypothesis on α implies that β −1 Cβ = C. We deduce therefore that Ni+1 ⊂ NH (C). Since Ni+1 is connected (21.3.6), we have Ni+1 ⊂ NH (C)◦ = C. We have therefore proved that Bu ⊂ C. Now T ⊂ C and B = T Bu , so we have B = C. Consequently B is nilpotent (28.1.10 and 28.3.1), and so H is nilpotent (28.1.9).
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28.5.7 Theorem. Let G be a connected algebraic group. (i) An element α ∈ G is generic if and only if it belongs to a unique Cartan subgroup of G. (ii) The set Ggen contains a dense open subset of G. Proof. (i) If α is generic, then C = CG (αs )◦ is the unique Cartan subgroup containing αs (28.5.5), and α ∈ C by 28.3.1. Now let D be a Cartan subgroup containing α. Then αs ∈ Ds and D = Ds × Du . Thus D ⊂ CG (Ds )◦ ⊂ CG (αs )◦ , and hence D = C. Conversely, suppose that α belongs to a unique Cartan subgroup C. Since αs ∈ Cs ⊂ Z(C), we deduce that H = CG (αs )◦ contains C. So any Cartan subgroup of H contains αs ∈ Z(H). Since α is contained in a unique Cartan subgroup, it follows that αu = αs−1 α is contained in a unique Cartan subgroup of H. Hence H is nilpotent by 28.5.6, and so H is a Cartan subgroup of G (28.5.5) which implies that αs , and hence α, is generic. (ii) Observe that by 28.5.4 (iii), the set Ggen is non-empty. Let C = CG (T ) = T × Cu be a Cartan subgroup of G and T0 the set of α ∈ T verifying χ(α) = 1 for all χ ∈ Φ(T, G) (notations of 26.2.1). The set C0 = T0 × Cu is dense and open in C and by 28.5.4, C0 = C ∩ Ggen . Since any generic element belong to a Cartan subgroup, we deduce that Ggen is the image of the morphism u : G × C0 → G , (α, γ) → αγα−1 . It follows from the inclusion C0 ⊂ u(G × C0 ) = Ggen that C = C0 ⊂ Ggen . As Ggen is stable by conjugation, it contains the union of the conjugates of C, which contains a dense open subset of G (28.5.2). 28.5.8 Let H be a nilpotent algebraic group. By 26.4.4, T = (H ◦ )s is the unique maximal torus of H ◦ and H ◦ = T × (H ◦ )u . So T ⊂ Z(H ◦ ) and T is normal in H. Lemma. We have T ⊂ Z(H). Proof. Let t = L(T ) and for α ∈ H, let uα and vα be the morphisms of T into itself defined by: uα (γ) = αγα−1 , vα (γ) = αγα−1 γ −1 . Then uα (e) = vα (e) = e, so duα and dvα are endomorphisms of t. Since H/H ◦ is finite, uα and duα are automorphisms of finite order. So duα is a semisimple automorphism of t. Similarly, since H is nilpotent, dvα is nilpotent. But dvα = duα − idt , so uα = idT . Hence T ⊂ Z(H). 28.5.9 Theorem. Let G be a connected algebraic group and C a subgroup (not necessarily closed) of G. The following conditions are equivalent: (i) C is a Cartan subgroup of G. (ii) C is a maximal nilpotent subgroup of G, and any subgroup of C of finite index in C, has finite index in NG (C). (iii) C is a closed connected nilpotent subgroup of G and C = NG (C)◦ .
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Proof. (i) ⇒ (ii) Let C = CG (T ) be a Cartan subgroup of G where T is a maximal torus of G, and H a nilpotent subgroup containing C. We may assume that H is closed. Then T is a maximal torus of H ◦ . By 28.5.8, we have T ⊂ Z(H), and so H ⊂ CG (T ) = C. Now C is connected, so if K is a subgroup of C of finite index in C, then K = C. Thus NG (K) ⊂ NG (C). Since NG (C)◦ = C by 28.1.10, C has finite index in NG (C). The inclusions K ⊂ C ⊂ NG (C) and 10.2.3 show that K has finite index in NG (C). (i) ⇒ (iii) This follows from 28.1.10, 28.3.1 and 28.5.3. (iii) ⇒ (i) We have C = S × Cu where S = Cs (26.4.4). Let B be a Borel subgroup containing C, T a maximal torus of B containing S, and D the centralizer of S in B. The group D is connected (28.3.1) and B = T Bu , D = T Du . Since S ⊂ Z(D), the group SDu is connected, nilpotent and it contains C. Now C = NG (C)◦ , it follows that SDu = C (28.5.3). The group D is connected and solvable. So (D, D) ⊂ Du ⊂ C (26.4.2), and hence C is normal in D. It follows from the equality C = NG (C)◦ that D = C, so T ⊂ C, and S = T . Thus C = CB (T ) is a Cartan subgroup of B, and hence of G. (ii) ⇒ (i) If C verifies (ii), then C is closed since C is nilpotent. Next, C ◦ has finite index in C and therefore C ◦ has finite index in NG (C). Hence C ◦ = NG (C)◦ . It follows from (iii) ⇒ (i) that C ◦ is a Cartan subgroup of G, and (i) ⇒ (ii) shows that C ◦ is maximal nilpotent. So C = C ◦ .
References • [5], [38], [40], [78].
29 Cartan subalgebras, Borel subalgebras and parabolic subalgebras
In this chapter, we apply the results in Chapter 28 on Cartan subgroups, Borel subgroups and parabolic subgroups to their Lie algebras. We generalize the definitions of Borel subalgebras and parabolic subalgebras to arbitrary Lie algebras, and we establish the relations between the group objects and Lie algebra objects. In particular, we prove analogues of results on conjugacy of such subalgebras.
29.1 Generalities 29.1.1 In this chapter, g is a finite-dimensional k-Lie algebra and G its algebraic adjoint group (24.8.1). By 24.5.7 and 24.5.12, ad[g, g] = [ad g, ad g] is an algebraic Lie subalgebra of gl(g). Let us denote by G1 the connected algebraic subgroup of G such that L(G1 ) = ad[g, g]. 29.1.2 As in 19.8.1, we let gλ (x) to be the generalized eigenspace of ad x relative to λ ∈ k, and denote by N(g) the set of x ∈ g verifying x ∈ gλ (y) for some y ∈ g and λ ∈ k \ {0}. We saw in 19.8.2 that if x ∈ N(g), then ad x is nilpotent. Let Auts g be the subgroup of Aute g generated by ead x , x ∈ N(g). If h is a Lie subalgebra of g, then N(h) ⊂ N(g). So if we denote by Auts (g, h) the group generated by eadg x , x ∈ N(h), then we may identify {γ|h ; γ ∈ Auts (g, h)} with Auts h. The group Auts g is a connected algebraic subgroup of Aute g (21.2.5 and 21.3.2). Lemma. Let f : g → h be a surjective morphism of Lie algebras and β ∈ Auts h. Then there exists α ∈ Auts g such that β ◦ f = f ◦ α. Proof. Let y ∈ h and x ∈ g be such that y = f (x). Then f (gλ (x)) = hλ (y), so f (N(g)) = N(h). To obtain the result, we may assume that β = ead y where y ∈ N(h). Then y = f (x) for some x ∈ N(g). It is then easy to see that α = ead x verifies the conclusion of the lemma.
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29.1.3 Assume that g is semisimple. Let h be a Cartan subalgebra of g, R = R(g, h) the root system of g relative to h, and W = W (g, h) its Weyl group. Recall from 18.2.9 that W acts on h. Lemma. Let w ∈ W be considered as an automorphism of h. There exists θ ∈ Auts g such that θ|h = w. Proof. We may assume that w = sα , where α ∈ R. Let Xα ∈ gα , X−α ∈ g−α be such that [Xα , X−α ] = Hα . Let: θ = ead Xα ◦e− ad X−α ◦ead Xα . Then θ ∈ Auts g. To obtain the result, it suffices to prove that θ(h) = h − α(h)Hα for all h ∈ h. This is clear if α(h) = 0. On the other hand, we have: θ(Hα ) = ead Xα ◦e− ad X−α (Hα −2Xα ) = −Hα +2Xα −2Xα = −Hα = sα (Hα ). So we are done.
29.1.4 Proposition. Let H be an algebraic group, h its Lie algebra and (V, π) a finite-dimensional rational H-module. Define θ to be the morphism of varieties: θ : H × V → V , (β, w) → π(β)(w). (i) If (α, v) ∈ H × V and (x, w) ∈ Tα (H) × V , then: dθ(α,v) (x, w) = dπα (x)(v) + π(α)(w). (ii) For all (α, v) ∈ H × V , the linear maps dθ(α,v) and dθ(eH ,v) have the same rank. Proof. (i) Define the morphisms: p : H → H × V , β → (β, v) and q : V → H × V , w → (α, w). Then θ ◦ p(β) = π(β)(v) and θ ◦ q(w) = π(α)(w) and so: d(θ ◦ p)α (x) = dπα (x)(v) , d(θ ◦ q)v (w) = π(α)(w). On the other hand, we have dpα (x) = (x, 0) and dqv (w) = (0, w). Thus: dπα (x)(v) = d(θ◦p)α (x) = dθ(α,v) ◦dpα (x) = dθ(α,v) (x, 0), π(α)(w) = d(θ◦q)v (w) = dθ(α,v) ◦dqv (w) = dθ(α,v) (0, w). Since dθ(α,v) (x, w) = dθ(α,v) (x, 0) + dθ(α,v) (0, w), the result follows. (ii) Fix α ∈ H and consider the following isomorphisms of varieties: σ : H × V → H × V , (β, w) → (αβ, w) , ρ : V → V, w → π(α)(w). Then θ ◦ σ = ρ ◦ θ. If v ∈ V , then d(θ ◦ σ)(eH ,v) = d(ρ ◦ θ)(eH ,v) , that is, dθ(α,v) ◦ dσ(eH ,v) = π(α) ◦ dθ(eH ,v) . So the result follows.
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29.2 Cartan subalgebras 29.2.1 We shall use the notations of chapter 19. Lemma. Let G1 be as in 29.1.1, h a Lie subalgebra of g and x ∈ h be such that ad x induces an automorphism of g/h. The image of the morphism θ : G1 × h → g , (β, y) → β(y) contains a dense open subset of g. Proof. The hypothesis on x implies that g = h + [x, g]. In particular, we have g = h + [g, g] and so g = h + [x, g] = h + [x, h + [g, g]] = h + [x, [g, g]]. Now 29.1.4 says that dθ(e,x) (u, y) = u(x) + y for u ∈ L(G1 ) and y ∈ h. So the image t of dθ(e,x) is h + L(G1 )(x). But L(G1 ) = ad[g, g], and so t = g. It follows from 16.5.7 that θ is dominant, and so the result follows from 15.4.2. 29.2.2 Lemma. Any Cartan subalgebra h of g contains a generic element of g. Proof. For x ∈ h, let π(x) denote the endomorphism of g/h induced by ad x. Since h = ng (h), 0 is the only element v of g/h verifying π(x)(v) = 0 for all x ∈ h. As h is nilpotent, it follows from 19.4.6 that there exists x0 ∈ h such that π(x0 ) is bijective. By applying 29.2.1 and using the fact that the set of generic elements ggen of g contains a dense open subset of g (19.1.1), we deduce that G1 (h) ∩ ggen = ∅. Hence we obtain h ∩ ggen = ∅ because G1 (ggen ) = ggen . 29.2.3 Theorem. (i) A Lie subalgebra of g is a Cartan subalgebra of g if and only if h = g0 (x) for some x ∈ ggen . (ii) Let h and t be Cartan subalgebras of g. There exists α ∈ Auts g such that t = α(h). In particular, all the Cartan subalgebras of g have dimension the rank of g. Proof. (i) This is clear by 19.8.5 (iii) and 29.2.2. (ii) By (i), h = g0 (x) for some x ∈ ggen . Let λ1 , . . . , λn be the pairwise distinct non-zero eigenvalues of ad x. Then: g = g0 (x) ⊕ gλ1 (x) ⊕ · · · ⊕ gλn (x). If yi ∈ gλi , then yi ∈ N(g), and so ad yi belongs to the Lie algebra of Auts g. Consider the morphism θ : (Auts g) × h → g , (β, z) → β(z). If z ∈ h et yi ∈ gλi , then 29.1.4 implies that: dθ(e,x) (0, y) = y , dθ(e,x) (ad yi , 0) = [yi , x].
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Since ad x induces a bijective endomorphism of gλi , the image of dθ(e,x) is g. The same argument as in 29.2.1 shows that (Auts g)(h) contains a dense open subset of g. So the same is true for (Auts g)(k). We deduce therefore from 26.2.2 that h ∩ ggen ∩ (Auts g)(k) = ∅. So the result follows easily by 19.8.5 (iii). 29.2.4 Proposition. Let f : g → g be a surjective morphism of Lie algebras. (i) The image under f of a Cartan subalgebra h of g is a Cartan subalgebra of g . (ii) Let h be a Cartan subalgebra of g . Then any Cartan subalgebra of −1 f (h ) is a Cartan subalgebra of g. (iii) Any Cartan subalgebra of g is the image under f of a Cartan subalgebra of g. Proof. (i) Let a = ker f and π : g → g/a g be the canonical surjection. If π(x) ∈ ng/a (π(h)), then x ∈ ng (h + a) = h + a (19.8.5 (i) and 29.2.3 (i)). Thus ng/a (π(h)) = π(h), and ng (f (h)) = f (h). Since f (h) is nilpotent, it is a Cartan subalgebra of g . (ii) Let h be a Cartan subalgebra of t = f −1 (h ). By (i), f (h) is a Cartan subalgebra of h , so f (h) = h (19.8.4). If x ∈ ng (h), then f (x) ∈ ng (h ) = h . Thus x ∈ t, and therefore ng (h) = nt (h) = h. As h is nilpotent, it is a Cartan subalgebra of g. (iii) This is direct consequence of (i) and (ii). 29.2.5 Proposition. Let L be a connected algebraic group and l its Lie algebra. A Lie subalgebra of l is a Cartan subalgebra of l if and only if it is the Lie algebra of a Cartan subgroup of L. Proof. Let H be a Cartan subgroup of L and h its Lie algebra. Then H is a connected nilpotent closed subgroup verifying H = NL (H)◦ (28.5.9). So 24.4.8 and 24.5.13 imply that h is a Cartan subalgebra of l. Conversely, let h be a Cartan subalgebra of l. Since h = nl (h), h is an algebraic Lie algebra of l (24.7.2). Let H be a connected algebraic subgroup of L such that L(H) = h. Then H is nilpotent (24.5.13) and H = NL (H)◦ by 24.3.5 since L(NL (H)◦ ) = L(NL (H)) = nl (h) = h (24.4.8). So H is a Cartan subgroup of L by 28.5.9. 29.2.6 Proposition. Let l be the Lie algebra of a connected algebraic group L. (i) An element x ∈ l is semisimple if and only if x belongs to the Lie algebra of a torus of L. (ii) If x ∈ l is semisimple, then x belongs to a Cartan subalgebra of l. Proof. (i) If T is a torus of L, then all the elements of L(T ) are semisimple (26.4.5). Conversely, suppose that x ∈ l is semisimple. Let H = CL (x). Then
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cl (x) = L(H) (24.3.6), so x ∈ L(H). Let T be a maximal torus of H and C = CH (T )◦ . Then L(C) = ch (T ) (26.1.1), so x ∈ L(C). Now C is nilpotent (28.1.10) and so C = T × Cu (26.4.4). It follows therefore from 26.4.5 that x ∈ L(T ). (ii) Let T be a maximal torus of L such that x ∈ L(T ). Then C = CL (T ) is a Cartan subgroup of L containing T . Hence 29.2.5 implies that L(C) is a Cartan subalgebra of l containing x. 29.2.7 Proposition. Let m be the Lie algebra of a connected algebraic group M , a a Lie subalgebra of m and s = a(a). (i) We have rk(s) = rk(a) + dim(s/a). (ii) If k is a Cartan subalgebra of s, then a ∩ k is a Cartan subalgebra of a. (iii) If h is a Cartan subalgebra of a, then k = a(h) is a Cartan subalgebra of s such that h = k ∩ a. Proof. 1) Let k be a Cartan subalgebra of s and h = a ∩ k. Since k is algebraic (29.2.5), we have a(h) ⊂ k. Let x ∈ k ∩ sgen (29.2.2). We have s = k ⊕ l where l is a subspace of s which is (ad x)-stable, and ad x induces an automorphism of l. Thus l = [x, l] ⊂ [s, s], and 24.5.7 implies that s = k + [a, a]. We deduce therefore that a = h + [a, a]. Moreover, a(h) and [a, a] are algebraic, so a(h) + [a, a] is an algebraic Lie subalgebra of m containing a (24.5.10). It follows that s = a(h) + [a, a]. Consequently, if x ∈ k, then x = y + z where y ∈ a(h) and z ∈ [a, a]. Hence z = x − y ∈ k, and so z ∈ k ∩ a = h which in turn implies that x ∈ a(h). We have therefore proved that a(h) = k. 2) Denote by S the connected algebraic subgroup of M such that L(S) = s. Let N = {α ∈ S; Ad(α)(h) = h} and n = L(N ). Then n = ns (h) (24.3.6). But if α ∈ N and p is an algebraic Lie subalgebra of s containing h. Then Ad(α)(p) also contains h. It follows from the definition of a(h) = k that Ad(α)(k) = k. We deduce therefore that n ⊂ ns (k) = k (24.3.5). Hence n ∩ a ⊂ h, and so h = na (h). As h is nilpotent, it is a Cartan subalgebra of a. We have therefore proved (ii). 3) Now s = k + a, so dim(s/a) = dim(k/h). So (i) follows. 4) Finally, let h be a Cartan subalgebra of a, x ∈ h ∩ agen , and u the endomorphism of s/a induced by ads x. Since a is an ideal of s containing [s, s] = [a, a] (24.5.7), we have u = 0. Thus dim s0 (x) = dim a0 (x) + dim(s/a), and by (i), we deduce that x ∈ sgen . It follows that k = s0 (x) is a Cartan subalgebra of s (19.8.5). Since h = a0 (x), we have h = k ∩ a. So part (iii) follows from point 1.
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29.3 Applications to semisimple Lie algebras 29.3.1 Assume that g is semisimple, and denote by L its Killing form. Since L is non-degenerate, the Killing homomorphism x → fx defined in 19.2.2 is an isomorphism. In the notations of 19.7.3, we have for x, y, z ∈ g: Φfx (y, z) = fx ([y, z]) = L(x, [y, z]). We deduce immediately that: g(fx ) = gx . Thus we say that x ∈ g is regular if fx ∈ g∗reg . Let us denote by greg the set of regular elements of g; it is a dense open subset of g. 29.3.2 Proposition. If g is semisimple, then its rank is equal to its index. Proof. Since ggen and greg are both dense open subsets of g, there exists x ∈ g which is regular and generic. So the result is clear since gx = g0 (x) (19.8.7 (v)). 29.3.3 Theorem. Let g be a semisimple Lie algebra, h1 and h2 two Cartan subalgebras of g. Then the root systems R1 = R(g, h1 ) and R2 = R(g, h2 ) are isomorphic. Proof. By 29.2.3, there exists θ ∈ Aute g such that θ(h1 ) = h2 . Let us define τ : h∗1 → h∗2 by λ → λ◦θ−1 . Let x ∈ g and α ∈ R1 be such that [h, x] = α(h)x for all h ∈ h1 . If t ∈ h2 , then: [t, θ(x)] = θ [θ−1 (t), x] = α θ−1 (t) θ(x) = τ (α)(t) θ(x). Thus τ (R1 ) = R2 , and τ induces an isomorphism from (h∗1 )R onto (h∗2 )R . If x, y ∈ h1 , then: L(x, y) = α(x)α(y) = τ (α) θ(x) τ (α) θ(y) α∈R α∈R 1 1 β θ(x) β θ(y) = L θ(x), θ(y) . = β∈R2
For i = 1, 2, λ ∈ h∗i , let hiλ be the unique element of hi verifying L(h, hiλ ) = λ(h) for all h ∈ hi . It follows that θ(h1λ ) = h2τ (λ) . Hence in the notations of 20.6.1, λ, µ = L(h1λ , h1µ ) = L(h2τ (λ) , h2τ (µ) ) = τ (λ), τ (µ) for all λ, µ ∈ h∗1 . So 18.13.1 implies that R1 and R2 are isomorphic.
29.3.4 Let h be a Cartan subalgebra of a semisimple Lie algebra g. Then 29.3.3 says that up to an isomorphism, R = R(g, h) does not depend on the
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choice of h. We shall say that R is the root system of g and similarly, for all the objects relative to R such as Weyl group, weights and so on. 29.3.5 Proposition. Assume that g is semisimple and denote by R its root system. (i) If g = g1 × g2 and R1 , R2 are the root systems of g1 and g2 , then R is the direct sum of R1 and R2 . (ii) If R is the direct sum of root systems R1 and R2 , then g = g1 × g2 where α α kHα + g , g2 = kHα + g , g1 = α∈R1
α∈R1
α∈R2
α∈R2
where Hα is as defined in 20.6.5. Proof. (i) The ideal g1 and g2 are orthogonal with respect to L. So if hi is a Cartan subalgebra of gi , then h = h1 × h2 is a Cartan subalgebra of g. We have h∗ = h∗1 × h∗2 , and by taking Ri = R(gi , hi ) and R = R(g, h), it is clear that R = R1 ∪ R2 and that R1 and R2 are orthogonal. (ii) If α, β ∈ Ri verify α + β ∈ R, then α + β ∈ Ri . So gi is a subalgebra of g. Let α ∈ R1 and β ∈ R2 . Then α + β ∈ R, so [gα , gβ ] = {0}. Moreover, (α|β) = 0, so β(Hα ) = 0 and [Hα , gβ ] = {0}. Hence [g1 , g2 ] = {0}. 29.3.6 Corollary. Let g be a semisimple Lie algebra and R its root system. Then g is simple if and only if R is irreducible. 29.3.7 We shall admit the following result: Theorem. (i) Two semisimple Lie algebras are isomorphic if and only if their root systems are isomorphic. (ii) Each of the diagrams Al , Bl , Cl , Dl , E6 , E7 , E8 , F4 , G2 of 18.14 corresponds to an isomorphism class of simple Lie algebras. 29.3.8 Let g be a simple Lie algebra. We shall say, for example, that g is of type Al if its root system is of type Al .
29.4 Borel subalgebras 29.4.1 Definition. (i) A maximal solvable Lie subalgebra of a Lie algebra g is called a Borel subalgebra of g. (ii) A Lie subalgebra of a Lie algebra g is a parabolic subalgebra of g if it contains a Borel subalgebra of g. Remark. By 20.8.2, these definitions are compatible with the ones for semisimple Lie algebras in 20.8.1 and 20.8.5. 29.4.2 Lemma. Let b be a Borel subalgebra of a Lie algebra g. Then ng (b) = b. Proof. If x ∈ ng (b), then b + kx is a solvable Lie subalgebra of h. So x ∈ b.
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29.4.3 Proposition. Let h be the Lie algebra of a connected algebraic group H. (i) A Borel subalgebra of h is algebraic in h. (ii) A Lie subalgebra of h is a Borel subalgebra if and only if it is the Lie algebra of a Borel subgroup of H. In particular, all the Borel subalgebras of h are conjugate. (iii) If p is a parabolic subalgebra of h, then it is algebraic in h, it contains a Cartan subalgebra of h, and p = nh (p). (iv) A Lie subalgebra of h is a parabolic subalgebra if and only if it is the Lie algebra of a parabolic subgroup of H. Proof. (i) This follows from 24.7.2 and 29.4.2. (ii) Let b be a Borel subalgebra of h. By (i), there exists a connected algebraic subgroup B of H such that L(B) = b. So B is solvable by 24.5.13. If R is a connected solvable algebraic subgroup of H containing B, then L(R) is a solvable Lie algebra containing b. It follows that R = B. Hence B is a Borel subgroup of H. Conversely, let B be a Borel subgroup of H and b = L(B). Then b is solvable. If r is a solvable Lie subalgebra of h containing b, then (r) is a connected solvable algebraic subgroup of H containing B. Thus r = b and b is a Borel subalgebra of h. (iii) The group (p) contains a Borel subgroup of H (24.3.5), and so a Cartan subgroup of H (28.1.7 and 28.3.3). By 29.2.5, p contains a Cartan subalgebra of h. The fact that p = nh (p) follows from 19.8.5 and 29.2.2, hence p is algebraic (24.7.2). (iv) This is an immediate consequence of (iii).
a
a
29.4.4 Proposition. Let m be the Lie algebra of a connected algebraic group M , a a Lie subalgebra of m and s = a(a). (i) Let k be a Borel subalgebra of s. Then b = k ∩ a is a Borel subalgebra of a such that k = a(b). Moreover, we have: dim(k/b) = dim(s/a). (ii) If b is a Borel subalgebra of a, then a(b) is a Borel subalgebra of s such that a(b) ∩ a = b. Proof. (i) Since k is an algebraic Lie subalgebra of s, a(b) ⊂ k and k contains a Cartan subalgebra l of s (29.4.3). By 29.2.7 and its proof, we have a(h) = l and s = l + [a, a] where h = l ∩ a. So an element x ∈ k is of the form y + z where y ∈ l and z ∈ [a, a]. So z = x − y ∈ k ∩ a = b. Hence x ∈ a(b) and we obtain that a(b) = k. Let b be a solvable Lie subalgebra of a containing b. Then a(b ) is a solvable Lie subalgebra of s containing a(b) = k. Thus a(b ) = k and b ⊂ k ∩ a = b. We have therefore proved that b is a Borel subalgebra of a.
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Observe that the Lie subalgebras l and h above are Cartan subalgebras of s and a (29.2.7). We deduce therefore that rk(a) = rk(b) and rk(s) = rk(k). Since k = a(b), the equality of dimensions follows immediately from 29.2.7 (i). (ii) The Lie algebra k = a(b) is solvable (24.5.7). If k is a Borel subalgebra of s containing k, then b = a ∩ k . Hence k = a(b) = k by (i). 29.4.5 Corollary. (i) Any Borel subalgebra of g contains a Cartan subalgebra of g. (ii) All the Borel subalgebras of g have the same dimension. Proof. By 19.2.1, we may assume that g is a subalgebra of gl(V ) = L(GL(V )) for some finite-dimensional vector space V . Then the results are consequences of 29.2.7, 29.4.3 and 29.4.4. 29.4.6 Assume that g is semisimple. Let b be a Borel subalgebra of g. Then 29.4.5 says that b contains a Cartan subalgebra h of g. Let R = R(g, h). Then 20.7.1 and 20.7.2 imply that there is a closed subset P of R such that b = h + gP (notations of 20.7.1). Since b is solvable, P ∩ (−P ) = ∅ (20.7.5). By 18.10.2, there exists a chamber C of R such that P ⊂ R+ (C). Furthermore, P = R+ (C) since b is maximal solvable. We have therefore proved that a Lie subalgebra of g is a Borel subalgebra if and only if it is a Borel subalgebra of (g, h) (20.8.1) for some Cartan subalgebra h of g. 29.4.7 Theorem. Let b and b be Borel subalgebras of g. Then there exists θ ∈ Auts g such that b = θ(b). Proof. Let us proceed by induction on dim g. The result is clear if g is solvable. So let us assume that g is not solvable, and let r = rad g. Suppose that r = {0}. Let a = g/r and f : g → a be the canonical surjection. The Lie algebra b + r is solvable, so r ⊂ b. Similarly r ⊂ b . Since f (b) and f (b ) are Borel subalgebras of a, by the induction hypothesis, there exists σ ∈ Auts a such that f (b ) = σ(f (b)). As r ⊂ b ∩ b , it follows from 29.1.2 that there exists θ ∈ Auts g such that b = θ(b). We may therefore assume that g is semisimple. Then by 29.4.6, there exists a Cartan subalgebra h (resp. h ) of g such that b (resp. b ) is a Borel subalgebra of (g, h) (resp. (g, h )). Let B (resp. B ) be a base of R(g, h) (resp. R(g, h )) which defines b (resp. b ). By 29.2.3 (ii), we may assume that h = h . Now 18.8.7 says that there exists w ∈ W (g, h) such that w(B) = B . So the result follows from 29.1.3. 29.4.8 Corollary. Assume that g is semisimple. Let L denote its Killing form. (i) If n is a Lie subalgebra of g whose elements are nilpotent, then n is contained in the nilpotent radical of a Borel subalgebra of g. (ii) If x ∈ g, then x is contained in a Borel subalgebra of g. In particular, there exists a Borel subalgebra of g verifying the following conditions:
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x ∈ b , 2 dim b = dim g + rk(g) , L(x, [b, b]) = {0}. Proof. (i) By 19.3.7, n is nilpotent, so it is contained in a Borel subalgebra b of g. So by 20.8.3 and 29.4.6, n is contained in the nilpotent radical of b. (ii) The first part is clear and the second follows from 29.4.6. 29.4.9 Proposition. Let b and b be Borel subalgebras of a semisimple Lie algebra g. Then b ∩ b contains a Cartan subalgebra of g. Proof. Let n = [b, b], n = [b , b ], p = b ∩ b , s a complementary subspace of b + b in g, and h a Cartan subalgebra of b (and so of g by 29.4.5). Also let l, n, p be the dimensions of h, n and p. We have dim b = dim b = l + n, and: dim s⊥ = dim(b + b ) = 2(l + n) − p, where s⊥ is the orthogonal of s with respect to the Killing form of g. Since dim(s⊥ ∩ p) dim s⊥ + dim p − dim g, we have: (1)
dim(s⊥ ∩ p) 2(l + n) − p + p − (l + 2n) = l.
By 20.8.3 and 20.8.6, we have n = b⊥ , n = (b )⊥ and p ∩ n ⊂ n . It follows that p ∩ n ⊂ n ∩ n = b⊥ ∩ (b )⊥ . Consequently, s⊥ ∩ p ∩ n = {0}. We deduce therefore from (1) that b = n ⊕ (s⊥ ∩ p) and b = n ⊕ (s⊥ ∩ p). Let z ∈ h ∩ ggen (29.2.2). There exists y ∈ n such that x = y + z ∈ s⊥ ∩ p. By 20.8.3 (i), adg x and adg z have the same characteristic polynomial, so x ∈ ggen . It follows therefore from 19.1.2 that x is generic simultaneously in g, b and b . As g, b and b have the same rank (29.4.5), we deduce that g0 (x) = b0 (x) = (b )0 (x) is a Cartan subalgebra of g, b and b . 29.4.10 Corollary. Assume that g is semisimple. The group Aute g acts transitively on the set of pairs (h, b) where b is a Borel subalgebra of g and h is a Cartan subalgebra of g contained in b. Proof. Given (h1 , b1 ) and (h2 , b2 ) two such pairs. By 29.4.9, there exists a Cartan subalgebra h of g contained in b1 ∩ b2 . By 29.2.3, we may therefore assume that h = h1 = h2 . Let R = R(g, h) and Bi a base of R associated to bi , i = 1, 2. Then there exists w ∈ W (g, h) such that w(B1 ) = B2 (18.8.7). By 29.1.3, there exists θ ∈ Aute g such that θ|h = w. Thus θ(h) = h and θ(b1 ) = b2 .
29.5 Properties of parabolic subalgebras 29.5.1 Definition. A special automorphism of a Lie algebra a is an automorphism of the form ead x where x belongs to the nilpotent radical of a. 29.5.2 The following result is a refinement of 20.3.5. Theorem. Let a be a Lie algebra and s, t be Levi subalgebras of a. Then there exists a special automorphism of a which sends s to t.
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Proof. Let r = rad a and n the nilpotent radical of a. By 20.3.6, we have n = [a, r] = [a, a] ∩ r. 1) If [a, r] = n = {0}, then s = [a, a] = t (20.5.4), and the result is clear. 2) Suppose that [a, r] = {0} and the only ideals of a contained in r are r and {0}. Then [a, r] = r, [r, r] = {0}, and the centre of a is {0}. For x ∈ t, let ϕ(x) be the unique element of r such that x + ϕ(x) ∈ s. If x, y ∈ t, then: [x + ϕ(x), y + ϕ(y)] = [x, y] + [x, ϕ(y)] + [ϕ(x), y] ∈ s. Hence: ϕ([x, y]) = (ad x)(ϕ(y)) − (ad y)(ϕ(x)). By 20.3.3, there exists r ∈ r such that ϕ(x) = −[x, r] for all x ∈ t. Since r is commutative, we obtain, for x ∈ t, that: x + ϕ(x) = x + [r, x] = ead r (x). Hence our result since we have r = n here. 3) Let us consider the general case. We proceed by induction on the dimension n of r. We may suppose that n > 0 and n = [a, r] = {0}. The centre c of n is non-trivial (19.3.4). Let m be a minimal non-zero ideal of a contained in c. If m = r, then the result follows from point 2. So let us assume that m = r. Let a = a/m, r = r/m = rad(a ) and f : a → a the canonical surjection. Since the Lie algebras f (s) and f (t) are Levi subalgebras of a , it follows from the induction hypothesis that there exists a ∈ [a , r ] such that f (t) = ead a (f (s)). Fix a ∈ [a, r] such that a = f (a). Then: s1 = ead a (s) ⊂ m + t = h. Since s1 and t are Levi subalgebras of h, it follows again by the induction hypothesis that ead b (s1 ) = t for some b ∈ m. The element b belongs to the centre of [a, r], so: t = ead b ead a (s) = ead(a+b) (s). Hence the result since a + b ∈ [a, r].
29.5.3 Corollary. Let s be a Levi subalgebra of a and h a semisimple subalgebra of a. (i) There is a special automorphism of a which sends h into a subalgebra of s. (ii) h is contained in a Levi subalgebra of a. (iii) The Levi subalgebras of a are the maximal semisimple subalgebras of the Lie algebra a.
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Proof. It suffices to prove (i). Let b = h + rad a. It is clear that h is a Levi subalgebra of b and b ∩ s is a complementary Lie subalgebra of rad a in b. So b ∩ s is a Levi subalgebra of b. It follows from 29.5.2 that ead a (h) = b ∩ s for some a ∈ [b, rad a]. 29.5.4 Proposition. Let g be a semisimple Lie algebra, p a parabolic subalgebra of g, H the set of Cartan subalgebras of p, S the set of commutative Lie subalgebras of p consisting of semisimple elements and S0 the set of maximal elements of S. Then H = S0 . Proof. Since Cartan subalgebras of p are exactly the Cartan subalgebras of g contained in p (29.4.3 and 29.2.3), it follows from 20.6.2 and 20.5.16 that H ⊂ S0 . Now let t ∈ S0 and c = cp (t), and h a Cartan subalgebra of c. Then t ⊂ h. Since t consists of semisimple elements which commute pairwise, we have a decomposition q = np (h) = m ⊕ h of (adp t)-stable subspaces. So: [t, m] ⊂ [t, q] ⊂ [h, q] ⊂ h. Thus [t, m] ⊂ h∩m = {0}, and m ⊂ c. Consequently, q = nc (h), hence q = h. It follows that h is a Cartan subalgebra of p, hence of g. Thus h is a commutative Lie subalgebra of p consisting of semisimple elements. So t = h and S0 ⊂ H. 29.5.5 Corollary. Let g be a semisimple Lie algebra, p a parabolic subalgebra of g and t a Lie subalgebra of p consisting of semisimple elements. Then t is contained in a Cartan subalgebra of p. Proof. This follows from 29.5.4 since t is commutative (20.5.16).
29.5.6 Definition. Let g be a semisimple Lie algebra, p a parabolic subalgebra of g and n the nilpotent radical of p. A Lie subalgebra l of p is called a Levi factor of p if it is reductive in g and p = l ⊕ n. 29.5.7 Theorem. Let g be a semisimple Lie algebra and p a parabolic subalgebra of g. (i) p has a Levi factor. (ii) If l and s are Levi factors of p, then there is an elementary automorphism of p which sends l into s. (iii) Let c be the centre of l. Then c = cg (l) and l = cg (c). Proof. (i) This follows from 20.8.6. (ii) Let r = rad p and n the nilpotent radical of p. We have l = l ⊕ c, s = s ⊕ d, where l = [l, l] and s = [s, s] are semisimple, and c = z(l) (resp. d = z(s)) consists of semisimple elements (20.5.5 and 20.5.10). Take l to be the Levi factor of p defined in 20.8.6, we see that if h and k are Cartan subalgebras of l and s , then h ⊕ c and k ⊕ d are Cartan subalgebras of p. By 29.2.3, there exists α ∈ Aute p such that α(h ⊕ c) = α(k ⊕ d). But:
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α(r) = r , (h ⊕ c) ∩ r = c , (k ⊕ d) ∩ r = d. We deduce therefore that α(c) = d. So we may assume that d = c. Let z = cp (c). Then l and c are Levi subalgebras of z. So by 29.5.2, there exists an element x in the nilpotent radical of z such that s = ead x (l ). Since [x, c] = {0}, we also have ead x (c) = c. Hence ead x (l) = s. (iii) This follows easily from (ii) and 20.8.6. Remark. Note that Levi factors and Levi subalgebras are not the same.
29.6 More on reductive Lie algebras 29.6.1 Let (V, σ) and (W, τ ) be finite-dimensional g-modules and (V ∗ , σ ∗ ) the contragredient representation of σ (19.2.1). We define the the representation π of g on Hom(V, W ) as follows: π(x)(f ) = τ (x) ◦ f − f ◦ σ(x). ∗
The map V ⊗k W → Hom(V, W ) sending ϕ ⊗ y, where ϕ ∈ V ∗ and y ∈ W , to the map v → ϕ(v)y, is a vector space isomorphism. We verify easily that it is a homomorphism of g-modules if we endow V ∗ ⊗k W with the representation σ ∗ ⊗ τ (20.5.6). It follows therefore from 20.5.6 that we have the following result: Lemma. If σ and τ are semisimple, then so is π. 29.6.2 Proposition. Let (V, σ) be a g-module, h a reductive Lie subalgebra of g and W the sum of simple h-submodules in V . Then σ(g)(W ) ⊂ W . Proof. Let S be a simple h-submodule of V and consider g as a h-module via the adjoint representation. Then 20.5.6 implies that g ⊗k S is a semisimple h-module. Let θ : g⊗k S → V be the linear map defined by θ(x⊗v) = σ(x)(v). Then we verify easily that θ is homomorphism of h-modules. It follows that θ(g ⊗k S) is a semisimple h-submodule of V , and so it is contained in W . Hence σ(g)(S) ⊂ W and the result follows. 29.6.3 Corollary. Let h be a reductive Lie subalgebra of g and (V, σ) a finite-dimensional semisimple representation of g. The restriction of g to h is semisimple. Proof. We may assume that σ is simple. Let W be as in 29.6.2 and S a nonzero h-submodule of V of minimal dimension. Then S is a simple h-module, so S ⊂ W and W = {0}. It follows again by 29.6.2 and the simplicity of V that W = V . 29.6.4 Proposition. Let V be a finite-dimensional vector space and g a Lie subalgebra of gl(V ). The following conditions are equivalent: (i) The natural representation of g on V is semisimple. (ii) g is reductive and all the elements of the centre of g are semisimple endomorphisms of V . (iii) g is reductive in gl(V ).
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Proof. We have (i) ⇒ (iii) by 29.6.1, (iii) ⇒ (i) by 29.6.3, while the equivalence (i) ⇔ (ii) was proved in 20.5.10.
29.7 Other applications 29.7.1 Let a, b be Lie algebras and y → δy a homomorphism from b to the Lie algebra of derivations of a. Then we verify easily that for x, x ∈ a and y, y ∈ b, [(y, x), (y , x )] = [y, y ], [x, x ] + δy (x ) − δy (x) defines a Lie bracket on b × a. We obtain therefore a Lie algebra g that we call the semi-direct product of b by a (with respect to the homomorphism y → δy ). The underlying vector space of g is b × a in which we may identify canonically a and b as subspaces, and a (resp. b) is an ideal (resp. Lie subalgebra) of g. 29.7.2 Lemma. Let a be a commutative Lie subalgebra of g and c = cg (a). Suppose that adg x is semisimple for all x ∈ a. Then Cartan subalgebras of c are exactly the Cartan subalgebras of g which contains a. Proof. Let h be a Cartan subalgebra of c. Then a ⊂ z(c) ⊂ h. Let n = ng (h). We have [a, n] ⊂ [h, n] ⊂ h. Since adg x is semisimple for all x ∈ a and a is commutative, we have the decomposition n = h ⊕ m of a-stable subspaces. It follows that [a, m] ⊂ h ∩ m = {0}, so m ⊂ c. Hence n = nc (h) and n = h. Thus h is a Cartan subalgebra of g. Conversely, let h be a Cartan subalgebra of g which contains a. Then h = g0 (h) ⊂ g0 (a) = c. Since h = ng (h), we have h = nc (h), and h is a Cartan subalgebra of c. 29.7.3 In the rest of this section, V is a finite-dimensional k-vector space, g is a Lie subalgebra of gl(V ), r = rad g, u the set of nilpotent endomorphisms of V in r and n the largest nilpotent ideal of the natural representation of g in V . By 19.3.5, 19.3.7 and 19.5.7, we have n = u. Also [g, g] ∩ r ⊂ u by 19.6.2. Finally, if h is a Lie subalgebra of gl(V ), we denote by u(h) the set of nilpotent endomorphisms of V in rad h. Thus u = u(g) = u(r). 29.7.4 Lemma. Let g be nilpotent and algebraic in gl(V ) and t the set of semisimple endomorphisms of V in g. Then t ⊂ z(g) and g is the direct product of t and u. Proof. This follows easily from 24.5.13, 26.4.4 and 26.4.5. 29.7.5 Proposition. Let g be algebraic in gl(V ) and denote by S the set of commutative Lie subalgebras of g consisting of semisimple elements, S0 the set of maximal elements of S and H the set of Cartan subalgebras of g. (i) For h ∈ H, the set ϕ(h) of semisimple elements of h belongs to S0 . (ii) If t ∈ S0 , then ψ(t) = cg (t) ∈ H. (iii) The maps ϕ : H → S0 , ψ : S0 → H are bijections and they are mutually inverse. (iv) The group Aute g acts transitively on S0 .
29.7 Other applications
455
Proof. (i) Since h = ng (h), h is algebraic in g. By 29.7.4, ϕ(h) ∈ S and h = ϕ(h) × u. It follows that h ⊂ cg (ϕ(h)) ⊂ g0 (h) = h (19.8.6), because the elements of u are nilpotent. Thus h = cg (ϕ(h)). Let t ∈ S be such that ϕ(h) ⊂ t. Then t ⊂ cg (ϕ(h)) ⊂ h. So t = ϕ(h), and we obtain that ϕ(h) ∈ S0 . (ii) Let t ∈ S0 and h a Cartan subalgebra of ψ(t). By 29.7.2, we have h ∈ H and t ⊂ h. So ϕ(h) ∈ S and t ⊂ ϕ(h). Hence t = ϕ(h), and h = ψ(ϕ(h)) = ψ(t). Thus ψ(t) ∈ H and ϕ(ψ(t)) = t. (iii) This is clear from the proof of (ii). (iv) This follows from (iii) and 29.2.3. 29.7.6 Corollary. Let us conserve the hypotheses of 29.7.5 and suppose further that g is solvable. If t ∈ S0 , then g is the semi-direct product of t by u. Proof. Since t ∩ u = {0}, it suffices to prove that g = t + u. Let h = ψ(t) ∈ H. Since g/[g, g] is commutative, by applying 29.2.4 to the canonical surjection g → g/[g, g], we obtain that g = h + [g, g]. But g is solvable, so [g, g] ⊂ u. So g = h + u. Using again the fact that g is solvable, it is clear that u(h) ⊂ u. Since h = t + u(h) by 29.7.4, we have g = t + u as required. 29.7.7 Let g be algebraic in gl(V ). Then for x ∈ g, its semisimple component s and nilpotent component n are also in g (23.6.2 and 23.6.4). Moreover, 10.1.3 and 20.4.1 say that adg s and adg n are the semisimple and nilpotent components of adg x. Let h be a Lie subalgebra of gl(V ) which contains g. If adg x is nilpotent, then adg x = adg n, and adh n extends adg x. Since n is nilpotent and n ∈ h, it follows again from 10.1.3 and 20.4.1 that adh n is nilpotent. Thus we have obtained the following result: Lemma. Any element of Aute g extends to an element of Aute h. 29.7.8 Proposition. Let g be algebraic in gl(V ). (i) There exists a Lie subalgebra m of g, reductive in gl(V ), such that g is the semi-direct product of m by u. (ii) If m and m are two Lie subalgebras verifying the conditions of (i), then there exists θ ∈ Aute g such that m = θ(m). Moreover, m and m are algebraic Lie subalgebras of g. Proof. (i) We have r = t⊕u by 29.7.6, where t is a commutative Lie subalgebra consisting of semisimple elements. Also, by 20.5.11, we have g = z⊕[t, g] where z = cg (t). Since [t, g] ⊂ r, we have obtained that g = z + r. Let s be a Levi subalgebra of z (20.3.5) and rz the radical of z. Then z = s ⊕ rz , and g = s + rz + r. As rz + r is clearly a solvable ideal of g, we deduce that rz + r = r. Thus g = s ⊕ r, and s is a Levi subalgebra of g. Now let m = s ⊕ t. Then g = m ⊕ u, and since the centre of m is t, 29.6.4 implies that m is reductive in gl(V ).
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(ii) Let m be another such Lie subalgebra of g. We have m = s ⊕ t where s = [m , m ] is semisimple and t = z(m ) consists of semisimple elements. Then r = t ⊕ u = t ⊕ u. So by 29.7.5, there exists σ ∈ Aute r such that t = σ(t) and σ(u) = u. So 29.7.7 says that σ extends to an element of Aute g. Thus we may assume that t = t . So s ⊂ cg (t) and since dim s = dim s , s is a Levi subalgebra of cg (t). By 29.5.2, there exists an elementary automorphism θ of cg (t) such that θ(s) = s . Since t is central in cg (t), we have θ(t) = t, and so θ(m) = m . Finally, cg (t) is an algebraic Lie subalgebra of g (24.7.2), so we may extend θ to an element of Aute g (29.7.7). To show that m = s ⊕ t is algebraic, it suffices to show that t is algebraic (24.7.4). Since r is algebraic (24.7.4), a(t) is a commutative Lie subalgebra of r consisting of semisimple elements (24.5.7, 24.5.10 and 24.6.4). Hence a(t)∩u = {0}, and so t = a(t) is algebraic.
29.7.9 As in 29.5.6, a Lie subalgebra m of g verifying condition (i) of 29.7.8 is called a Levi factor of g. 29.7.10 Corollary. Let g be algebraic in gl(V ). There exist a Borel sub algebra b of g and θ ∈ Aute g such that u = u b ∩ θ(b) . Proof. Let m be a Levi factor of g, s = [m, m] and t = z(m). Let h be a Cartan subalgebra of s and R+ a positive root system associated to the root system of (s, h). Set: −α α g , b = t ⊕ u ⊕ g . b=t⊕u⊕ α∈R+
α∈R+
Then it is clear that b and b are Borel subalgebras of g, and b ∩ b = t ⊕ u. So u = u(b ∩ b ), and the result follows from 29.4.7.
29.8 Maximal subalgebras 29.8.1 Theorem. Let g be a semisimple Lie algebra, n a Lie subalgebra of g consisting of nilpotent elements and p = ng (n). If n is the set of nilpotent elements of rad p, then p is a parabolic subalgebra of g. Proof. Since the adjoint representation is faithful, we may assume that g is an algebraic Lie subalgebra of gl(V ) for some finite-dimensional vector space V (24.7.2), and we shall use the notations of 29.7.3. By 20.4.4, x ∈ g is semisimple (resp. nilpotent) if and only if it is a semisimple (resp. nilpotent) endomorphism of V . So n = u(p). Note also that p is algebraic in g (24.7.2). Let b be a Borel subalgebra of g whose nilpotent radical contains n (29.4.8), and s = b∩p. By 29.7.10, there exist a Borel subalgebra b1 of p and θ ∈ Aute p such that n = u b1 ∩ θ(b1 ) . It is clear that n is stable under elementary automorphisms of p, and as s is solvable, we may assume that s ⊂ b1 (29.4.7). Further, θ extends to an element σ ∈ Aute g (29.7.7). Let b = σ(b).
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457
Since θ(n) = n, we deduce that n ⊂ p ∩ b ∩ b . As p ∩ b ∩ b is solvable, n ⊂ u(p ∩ b ∩ b ). On the other hand, b ∩ p = s ⊂ b1 , so p ∩ b ∩ b ⊂ b1 ∩ θ(b1 ). It follows that u(p ∩ b ∩ b ) ⊂ n. Thus we have proved that n = u(p∩b∩b ) and hence n ⊂ u(b∩b ). Suppose that n = u(b ∩ b ), then by applying Lie’s Theorem (19.4.4) to the adjoint representation of n in u(b∩b ), we obtain an element x ∈ u(b∩b )\u(p∩b∩b ) such that [x, n] ⊂ n. This is absurd since p = ng (n) and n = u(p ∩ b ∩ b ). Consequently, we have n = u(b ∩ b ). Since b ∩ b normalizes u(b ∩ b ) = n, we deduce that b ∩ b ⊂ p. It follows from 29.4.9 that b ∩ b , and hence p, contains a Cartan subalgebra h of g. Let R = R(g, h), B the base of R defined by b and R+ the set of corresponding positive roots. Then there exists a closed subset P of R such that p = h + gP (in the notations of 20.7.1). Let α ∈ B and Q = R+ \ {α}. Then 18.7.4 and 18.8.7 say that Q ∪ {−α} is a system of positive roots, and b = h ⊕ gQ∪{−α} is a Borel subalgebra of g. Let S ⊂ R be the closed subset such that b = h ⊕ gS . Clearly, we have u(b) = gR+ , u(b ) = gS , n = u(b ∩ b ) = gR+ ∩S and u(b ∩ b ) = gQ . If α ∈ R+ ∩ S, then α ∈ S, and we have −α ∈ S and R+ ∩ S ∩ Q = R+ ∩ S. Thus u(b ∩ b ∩ b ) = u(b ∩ b ) = n. It follows that [g−α , n] ⊂ n because S and Q are closed subsets not containing α. Thus −α ∈ P . As ±α ∈ R+ ∩ S, 20.7.4 implies that α ∈ P . We have therefore proved that B ⊂ P . So p is a parabolic subalgebra of g containing b. 29.8.2 Theorem. Let g be a semisimple Lie algebra and a a Lie subalgebra of g. If a is maximal in the set of Lie subalgebras of g distinct from g, then a is either parabolic or reductive in g. Proof. We may assume that g is an algebraic Lie subalgebra of gl(V ) for some finite-dimensional vector space V . Recall that a(a) is the smallest algebraic Lie subalgebra of g which contains a. If a(a) = g, then by 24.5.7, a is an ideal of g, so it is semisimple (20.1.6), and therefore reductive in g (20.3.4). If a(a) = g, then a(a) = a. Let m be a Levi factor of a. Then a = m ⊕ u(a) (29.7.8). If u(a) = {0}, then a = m is reductive in g. Otherwise, the normalizer p of u(a) in g contains a and p = g because g is semisimple. So p = a, and 29.8.1 says that a is parabolic in g. 29.8.3 Theorem. Let g be a semisimple Lie algebra and n a Lie subalgebra of g consisting of nilpotent elements. There exists a parabolic subalgebra p of g verifying the following conditions: (i) n is contained in the nilpotent radical of p. (ii) The normalizer of n in g is contained in p. (iii) p is stable under any automorphism of g which leaves n invariant.
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Proof. We may assume that g is an algebraic Lie subalgebra of gl(V ) for some finite-dimensional vector space V . Let n0 = n and p1 = ng (n). Define by induction the two sequences (ni )i0 and (pi )i1 as follows: pi = ng (ni−1 ) , ni = u(pi ). They are bothincreasing. So there exists an integer j such that pj = pj+1 , that is pj = ng u(pj ) . It follows from 29.8.1 that pj is a parabolic subalgebra of g. We have therefore n ⊂ nj = u(pj ) and p1 ⊂ p. Finally, an automorphism of g, which leaves n invariant, also leaves invariant the sequences (ni )i0 and (pi )i1 . 29.8.4 Corollary. Let g be a semisimple Lie algebra, n a Lie subalgebra of g consisting of nilpotent elements and s a semisimple Lie subalgebra of g which normalizes n. Then there exists a parabolic subalgebra p of g verifying the following conditions: (i) n is contained in the nilpotent radical of p. (ii) There is a Levi factor of p which contains s. Proof. By 29.8.3, there is a parabolic subalgebra p of g such that s ⊂ p and n is contained in the nilpotent radical of p. Let p = l⊕m where m is the nilpotent radical of p and l a Levi factor of p (29.5.7). The Lie algebra t = [l, l] is a Levi subalgebra of p. So by 29.5.3, there exists θ ∈ Aute p such that θ(s) ⊂ t. Thus θ−1 (l) is a Levi factor of p containing s.
References and comments • [5], [14], [29], [32], [38], [39], [40], [42], [43], [78], [80]. For a proof of 29.3.7, see [14], [59].
30 Representations of semisimple Lie algebras
We define the enveloping algebra of a Lie algebra in this chapter. This infinitedimensional associative algebra is a very useful tool to studying representations of the Lie algebra. We obtain in the case of semisimple Lie algebras a complete description, up to isomorphisms, of finite-dimensional irreducible representations.
30.1 Enveloping algebra 30.1.1 Let g be a Lie algebra. Denote by T(g) the tensor algebra of g and I the bilateral ideal of T(g) generated by the elements x ⊗ y − y ⊗ x − [x, y] for x, y ∈ g. The quotient T(g)/I, denote by U(g), is called the universal enveloping algebra of g. We shall state without proof in this section some properties of U(g). Theorem. (Poincar´e-Birkhoff-Witt Theorem) The map g → U(g), x → x + I, is injective. We shall identify g as a subspace of U(g) via this map. If (x1 , . . . , xn ) is a basis of g, then the elements xν11 · · · xνnn , with ν1 , . . . , νn ∈ N, form a basis of U(g). 30.1.2 We shall call the injection of g in U(g) the canonical injection. It is clear that this injection is a Lie algebra homomorphism if we endow U(g) with the Lie bracket given by the commutator (19.1.1). 30.1.3 Proposition. Any representation of g extends uniquely to a representation of the associative algebra U(g). In particular, this induces a bijection from the set of representations of g onto the set of representations of U(g). 30.1.4 Let ϕ : g → h be a Lie algebra homomorphism. Then ϕ extends uniquely to a homomorphism U(ϕ) : U(g) → U(h) of associative algebras with unit.
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Proposition. Let ϕ : g → h and ψ : h → k be homomorphisms of Lie algebras. (i) We have U(ψ ◦ ϕ) = U(ψ) ◦ U(ϕ). (ii) ϕ is injective (resp. surjective) if and only if U(ϕ) is injective (resp. surjective). 30.1.5 Let h be a Lie subalgebra of g and j : h → g the canonical injection. Then 30.1.4 says that U(j) : U(h) → U(g) is injective. We may therefore identify U(h) as a subalgebra of U(g) via U(j). Proposition. The algebra U(g) is a free (left or right) U(h)-module. If (x1 , . . . , xp ) is a basis of a complementary subspace of h in g, then the elements ν xν11 · · · xpp , with ν1 , . . . , νp ∈ N, form a basis of U(g) as a (left or right) U(h)module. 30.1.6 The adjoint representation of g extends naturally to a representation on Tn (g) as follows: x.(x1 ⊗ · · · ⊗ xn ) =
n
x1 ⊗ · · · ⊗ xi−1 ⊗ [x, xi ] ⊗ xi+1 ⊗ · · · ⊗ xn .
i=1
So g acts on T(g), and it is easy to check that the bilateral ideal I is stable under this action. It follows that U(g) is a g-module. Moreover, the natural grading of T(g) induces a filtration Un (g) n0 of g-submodules of U(g). We can describe this filtration explicitly: if n ∈ N, then Un (g) is the subspace of U(g) spanned by the products x1 · · · xp , where x1 , . . . , xp ∈ g and p n. In particular, U0 (g) = k. We shall call this filtration the canonical filtration of U(g). It follows from 30.1.1 that if (x1 , . . . , xr ) is a basis of g, then the elements xν11 · · · xνr r , with ν1 , . . . , νr ∈ N and ν1 + · · · + νr n, form a basis of Un (g). 30.1.7 Let U−1 (g) = {0}. For n 0, let Gn = Un (g)/ Un−1 (g) and n G= G n0
is the graded algebra associated to the canonical filtration of U(g). It follows from the explicit description of Un (g) in 30.1.6 that we may identify G0 with k, G1 with g and Gn with the space spanned by the product of exactly n elements of G1 = g. Moreover, it is not difficult to see from the defining relations of I that G is commutative. Thus the canonical injection g → G extends uniquely to a homomorphism, that we shall call canonical, of associative algebras S(g) → G, where S(g) denotes the symmetric algebra of g. Theorem. The canonical homomorphism S(g) → G is an isomorphism.
30.2 Weights and primitive elements
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30.2 Weights and primitive elements 30.2.1 In the rest of this chapter, g is a semisimple Lie algebra, h is a Cartan subalgebra of g, R = R(g, h) and W = W (g, h) (see 19.8.1 and 20.6.9). Let us fix a base B of R, and denote by R+ (resp. R− ) the corresponding set of positive (resp. negative) roots of R. Let P, P++ , Q be respectively the weight lattice of R, the set of dominant weights of R relative to B and the root lattice of R (18.12). Denote by Q+ ⊂ Q the set of linear combination with coefficients in N of the elements of B. We define the following partial order on h∗ : λ µ ⇔ µ − λ ∈ Q+ . Let L be the Killing form on g and we shall use the notations hλ , Hα , Xα in 20.6.1 and 20.6.5. Finally, we let: α α g , n− = g , b+ = h + n+ , b− = h + n− . n+ = α∈R+
α∈R−
30.2.2 Let (V, σ) be a representation of g. For simplicity, we shall denote σ(x)(v) by x.v. If µ ∈ h∗ , let Vµ be the set of v ∈ V verifying x.v = µ(x)v for all x ∈ h. Then Vµ is a subspace of V and the sum of the Vµ ’s is direct (19.3.9). A non-zero element of V is called a weight vector of weight µ. The dimension of Vµ , denote by mV (µ), is called the multiplicity of µ in V (or σ). If Vµ = {0}, then µ is said to be a weight of V (or σ), and we denote by P(V ) the set of weights of V . Lemma. Let α ∈ R and µ ∈ h∗ . Then σ(gα )(Vµ ) ⊂ Vµ+α . Proof. Let x ∈ h, y ∈ gα and v ∈ Vµ . Then: x.(y.v) = y.(x.v) + [x, y].v = µ(x)y.v + α(x)y.v = (µ(x) + α(x))y.v. So y.v ∈ Vµ+α . 30.2.3 Lemma. Let (V, σ) be a representation of g and v ∈ V . The following conditions are equivalent: (i) σ(b+ )(v) ⊂ kv. (ii) σ(h)(v) ⊂ kv and σ(n+ )(v) = {0}. (iii) σ(h)(v) ⊂ kv and σ(gα )(v) = {0} for all α ∈ B. If these conditions are verified and v = 0, then v is called a primitive vector of V . Proof. Suppose that (i) is verified, then v ∈ Vµ for some µ ∈ h∗ . So if α ∈ R+ , then σ(gα )(v) ∈ Vµ ∩ Vµ+α = {0} (30.2.2). Hence we have (ii). The implication (ii) ⇒ (iii) is clear and (iii) ⇒ (i) follows from the fact that n+ is generated by the gα , α ∈ B (20.8.2).
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30.2.4 Remark. It follows from the proof of 30.2.3 that if v is a primitive vector of V , then v ∈ Vµ for some µ ∈ h∗ . 30.2.5 Proposition. Let (V, σ) be a g-module, v a primitive vector of V of weight λ and U the n− -submodule of V generated by v. Suppose further that V is generated by v as a g-module. (i) We have U = V . Vµ is finite-dimensional and dim Vλ = 1. (ii) For all µ ∈ h∗ , (iii) We have V = µ∈h∗ Vµ . (iv) For all µ ∈ P(V ), we have λ µ. (v) The centralizer of σ(g) in End(V ) is k idV . Proof. For α ∈ R, let Yα = σ(Xα ). The space U is spanned by the vectors u(α1 , . . . , αn ) = Yα1 ◦ · · · ◦Yαn (v), with n ∈ N, α1 , . . . , αn ∈ R− . If x ∈ g, then σ(x)(u(α1 , . . . , αn )) is equal to: n j=1
Yα1 ◦ · · · ◦Yαj−1 ◦[σ(x), Yαj ]◦Yαj+1 ◦ · · · ◦Yαn (v) + Yα1 ◦ · · · ◦Yαn ◦σ(x)(v).
Hence for h ∈ h: (1)
σ(h)(u(α1 , . . . , αn )) = α1 (h) + · · · + αn (h) + λ(h) u(α1 , . . . , αn ). Similarly, if x ∈ n+ , then:
(2) σ(x)(u(α1 , . . . , αn )) =
n j=1
Yα1 ◦ · · · ◦Yαj−1 ◦[σ(x), Yαj ]◦Yαj+1 ◦ · · · ◦Yαn (v).
The equality (1) shows that U is σ(h)-stable. On the other hand, in (2), we have: kYα . [σ(x), Yαi ] ∈ σ(h) + α>αi
A simple induction then proves that σ(x)(u(α1 , . . . , αn )) ∈ U . Thus U is a g-submodule of V , so U = V by the hypothesis on v. In view of (1), we have also obtained (iii) and (iv). Let α1 , . . . , αr be pairwise distinct elements of R+ . By (1), if µ ∈ h∗ , dim Vµ is at most the cardinality of the set of elements (p1 , . . . , pr ) ∈ Nr verifying p1 α1 + · · · + pr αr = λ − µ. So (ii) follows. Finally suppose that c ∈ End(V ) commutes with σ(g). Then for h ∈ h, we have: σ(h)◦c(v) = c◦σ(h)(v) = λ(h)c(v). Thus c(v) ∈ Vλ . By (ii), there exists t ∈ k such that c(v) = tv. So we obtain that c(u(α1 , . . . , αn )) = tu(α1 , . . . , αn ). Hence c = t idV by (i). 30.2.6 Lemma. Let (V, σ) be a simple representation of g and λ ∈ P(V ). The following conditions are equivalent:
30.3 Finite-dimensional modules
463
(i) For all µ ∈ P(V ), we have µ λ. (ii) λ is the largest weight in V . (iii) If α ∈ B, then α + λ ∈ P(V ). (iv) There exists a primitive vector of V of weight λ. If these conditions are verified, then we say that λ is the highest weight of V. Proof. The implications (i) ⇒ (ii) ⇒ (iii) are clear, while (iii) ⇒ (iv) follows from 30.2.2 and 30.2.3. Finally, if v is a primitive vector of V of weight λ, then since V is simple, V is generated by v, and (i) follows from 30.2.5. 30.2.7 Consequently, we have obtained the following result: Proposition. Let (V, σ) be a simple representation of g such that P(V ) has a maximal element λ. (i) For all µ ∈ h∗ , Vµ is finite-dimensional and dim Vλ = 1. Primitive vectors of V are the non-zero vectors of Vλ . We have µ λ for all µ ∈ P(V ), or equivalently, λ is the highest weight of V . (ii) The h-module V is semisimple and V = µ∈h∗ Vµ . (iii) The centralizer of σ(g) in End(V ) is k idV .
30.3 Finite-dimensional modules 30.3.1 Theorem. Let (V, σ) be a finite-dimensional representation of g. (i) We have V = µ∈h∗ Vµ . (ii) Any weight of V belongs to the weight lattice P . (iii) If µ ∈ h∗ and w ∈ W , then dim Vµ = dim Vw(µ) . In particular, P(V ) is W -stable. (iv) Suppose that V is simple. Then the centralizer of σ(g) in End(V ) is k idV . The g-module V has a highest weight λ and we have λ ∈ P++ , dim Vλ = 1. Proof. (i) Let α ∈ B. By 19.2.7 and 20.6.5, σ(Hα ) is diagonalizable with integer eigenvalues. Since h is commutative, the elements σ(h), h ∈ h, can be diagonalized simultaneously. So V µ (h) = Vµ (notations of 19.3.9). Hence we have (i). (ii) We have already noted that the eigenvalues of σ(Hα ) are integers. So µ(Hα ) ∈ Z for all α ∈ B. Hence µ ∈ P (18.11.5). (iii) Let α ∈ B, n = µ(Hα ) and v ∈ Vµ \ {0}. If n 0, the vector σ(X−α )n (v) is non-zero (19.2) and it belongs to Vµ−nα (30.2.2). Thus σ(X−α )n |Vµ is injective, hence mV (µ − nα) mV (µ). If n 0, then σ(Xα )−n |V (µ) is injective, so mV (µ − nα) mV (µ). But µ − nα = µ − µ(Hα )α = sα (µ). Consequently, if µ ∈ P(V ), then sα (µ) ∈ P(V ) and mV (sα (µ)) mV (µ). By 18.7.8, for w ∈ W , we have w(µ) ∈ P(V ) and mV (w(µ)) mV (µ). By exchanging the roles of µ et w(µ), we obtain (iii).
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(iv) If V is simple, then (i) implies that P(V ) is a finite non-empty set. So by 30.2.7, V has a highest weight λ and dim Vλ = 1. Let v ∈ Vλ \ {0} and α ∈ B. Then Xα .v = 0, so λ(Hα ) 0 (19.2). Hence λ ∈ P++ by (ii) and 18.11.7. 30.3.2 Let (. | .) be a W -invariant inner product on h∗R and . the norm associated to it. Proposition. Let (V, σ) be a finite-dimensional simple representation of g and λ its highest weight. Then for all µ ∈ P(V ), we have µ λ. Proof. Let µ ∈ P(V ). There exists w ∈ W such that ν = w(µ) ∈ P++ (18.8.9). By 30.3.1 (iii), we have ν ∈ P(V ), so λ − ν ∈ Q+ , λ + ν ∈ P++ . Thus: 0 (λ + ν|λ − ν) = λ2 − ν2 . Since µ = ν, the result follows. 30.3.3 Remark. Recall from 18.8.7 (iii) that there exists a unique element w0 ∈ W such that w0 (B) = −B. It follows that if λ is the highest weight of the simple g-module V , then w0 (λ) is the lowest weight of V . By 30.3.1, mV (w0 (λ)) = 1. 30.3.4 Proposition. Let (V, σ) be a finite-dimensional representation of g and w0 ∈ W the unique element such that w0 (B) = −B. Endow V ∗ with the contragredient representation θ of σ. (i) If µ ∈ h∗ , the orthogonal of Vµ in V ∗ is ν=−µ Vν∗ , so we may identify ∗ as the dual of Vµ . V−µ (ii) Assume that V is simple with highest weight λ. Then V ∗ is simple with highest weight −w0 (λ). Proof. (i) Since V = ν∈h∗ Vν (30.3.1), we may identify V ∗ with the direct sum ν∈h∗ (Vν )∗ . If h ∈ h, then θ(h) = −t σ(h), so θ(h)|(Vν )∗ = −ν(h) id(Vν )∗ . So the result follows. (ii) Assume that V is simple. If M ∗ is a g-submodule of V ∗ , then its orthogonal M in V is a g-submodule. So V ∗ is simple. By (i), the highest weight of V ∗ is −ξ, where ξ is the lowest weight of V . By 30.3.3, we have ξ = w0 (λ).
30.4 Verma modules 30.4.1 Recall from 30.1.3 that there is a canonical bijection between the set of representations of a Lie algebra a and the set of representations of its universal enveloping algebra. By abuse of notation, if σ is a representation of a, then we shall also denote by σ the representation of U(a). The representation θ of U(a) on itself by left multiplication is called the left regular representation of U(a).
30.4 Verma modules
465
Let k be a Lie subalgebra of g and (x1 , . . . , xn ) a basis of a complementary subspace of k in g. For ν = (ν1 , . . . , νn ) ∈ Nn , we set xν = xν11 · · · xνnn ∈ U(g). Recall from 30.1.5 that the elements xν , ν ∈ Nn , form a basis of the right U(k)-module U(g). Let (M, π) be a representation of k, so M is a left U(k)-module. By considering U(g) as a left U(g)-module via the left regular representation, and as a right U(k)-module, we obtain a left U(g)-module: V = U(g) ⊗U(k) M. Then V = ν∈Nn kxν ⊗ M . Let (V, σ) denote the corresponding U(g)-module. If m ∈ M , then we may identify m with 1⊗m, and we have xν ⊗m = σ(xν )(m). Thus σ(xν )|M induces an isomorphism from M onto σ(xν )(M ). Therefore, we have: V = σ(xν )(M ). ν∈Nn
Let u ∈ U(g). Then there exist elements aµν of U(k) such that: µ uxν = x aµν . µ∈Nn
So if m ∈ M , then:
µ µ x aµν (1 ⊗ m) = x ⊗ π(aµν )(m) σ(u)(xν ⊗ m) = σ µ µ µ = σ(x ) π(aµν )(m) . µ
30.4.2 Let λ ∈ h∗ . We define a representation (τλ , kλ ) of b+ as follows: as a vector space kλ = k, and if h ∈ h and n ∈ n+ , τλ (h + n)(1) = λ(h)1. As in 30.4.1, we can define a left U(g)-module: M (λ) = U(g) ⊗U(b+ ) kλ . We call M (λ) the Verma module associated to g, h, B (or b+ ) and λ. 30.4.3 Remark. If we consider U(n− ) as a left U(n− )-module via its left regular representation, then 30.4.1 implies that the map U(n− ) → M (λ), u → u ⊗ 1, is an isomorphism of U(n− )-modules. by B(µ) the number of families (nα )α∈R+ of 30.4.4 If µ ∈ h∗ , denote elements of N verifying µ = α∈R+ nα α. We have B(µ) > 0 if and only if µ ∈ Q+ . Proposition. Let λ ∈ h∗ and α1 , . . . , αn the pairwise distinct elements of R+ .
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(i) The module M (λ) has highest weight λ, and 1 ⊗ 1 is a primitive vector of V of weight λ. In particular, M (λ)λ = 1 ⊗ k and M (λ) = U(n− )M (λ)λ . (ii) The weights of M (λ) are the elements of h∗ of the form λ − ν, with ν ∈ Q+ . (iii) If µ ∈ h∗ , then dim M (λ)µ = B(λ − µ). (iv) For µ ∈ h∗ , let S(µ) be the set of elements (ν1 , . . . , νn ) ∈ Nn such that λ − ν1 α1 − · · · − νn αn = µ. Then: ν1 νn M (λ)µ = X−α · · · X−α ⊗ k. 1 n (ν1 ,...,νn )∈S(µ)
ν1 νn · · · X−α . For h ∈ h, we have: Proof. If (ν1 , . . . , νn ) ∈ Nn , denote X ν = X−α 1 n
h.(X ν ⊗ 1) = [h, X ν ] ⊗ 1 + X ν h ⊗ 1 = (−ν1 α1 − · · · − νn αn )(h)X ν ⊗ 1 + X ν ⊗ λ(h) = (λ − ν1 α1 − · · · − νn αn )(h)(X ν ⊗ 1). Hence (iv). The other parts follows immediately. 30.4.5 We shall denote the element 1 ⊗ 1 of M (λ) by vλ , and we shall call vλ the canonical generator of M (λ). 30.4.6 The module M (λ) is universal in the following sense: Proposition. Let λ ∈ h∗ and (V, σ) a representation of g which has a primitive vector v of weight λ. (i) There exists a unique g-module homomorphism ϕ : M (λ) → V such that ϕ(vλ ) = v. (ii) Suppose further that V is generated by v as a g-module. Then ϕ is surjective. A necessary and sufficient condition for ϕ to be bijective is that σ(u) is injective for all u ∈ U(n− ) \ {0}. Proof. (i) Since vλ generates M (λ), it is clear that if ϕ exists, it is unique. Now let ψ ∈ Hom(kλ , V ) be such that ψ(1) = v. Then if h ∈ h and n ∈ n+ , we have: ψ ◦ τλ (h + n)(1) = λ(h)v = σ(h + n)v. So ψ is a homomorphism of b+ -modules. Now by the universal property of the tensor product, ψ extends uniquely to a g-module homomorphism ϕ : M (λ) → V verifying ϕ(vλ ) = v. (ii) Suppose that V is generated by v as a g-module. Then clearly ϕ is surjective. If ϕ is bijective, then 30.4.3 implies that σ(u) is injective for all u ∈ U(n− ) \ {0}. If ϕ is not bijective, then there exists u ∈ U(n− ) such that ϕ(u ⊗ 1) = 0. Hence: σ(u)(v) = σ(u)ϕ(vλ ) = ϕ(u.vλ ) = ϕ(u ⊗ 1) = 0. Thus σ(u) is not injective.
30.5 Results on existence and uniqueness
467
30.4.7 Proposition. Let λ ∈ h∗ and: M (λ)+ = M (λ)µ . µ=λ
(i) Any g-submodule of M (λ) distinct from M (λ) is contained in M (λ)+ . (ii) There exists a largest g-submodule K(λ) of M (λ) distinct from M (λ). The quotient M (λ)/K(λ) is a simple g-module. Proof. Let E be a g-submodule of M (λ). By 30.2.5 and 30.4.4, we have: (E ∩ M (λ)µ ). E= µ∈h∗
Since M (λ)λ = kvλ and M (λ) is generated by vλ , it follows that if E = M (λ), then E ∩ M (λ)λ = {0}. Thus E ⊂ M (λ)+ . Now let K(λ) be the sum of all the g-submodules of M (λ) which are distinct from M (λ). Then we have just proved that K(λ) ⊂ M (λ)+ . Hence (ii) follows easily. 30.4.8 In the notations of 30.4.7, we shall denote: L(λ) = M (λ)/K(λ), and vλ the image of vλ in L(λ). It is clear that vλ is a primitive vector of L(λ) of weight λ, and it generates the g-module L(λ). We shall call vλ the canonical generator of L(λ). 30.4.9 Proposition. Let λ ∈ h∗ and V a simple g-module with highest weight λ. Then V is isomorphic to L(λ). Proof. This follows from 30.4.6 and 30.4.7.
30.5 Results on existence and uniqueness 30.5.1 Lemma. Let β ∈ B, sβ = kHβ + kXβ + kX−β and (V, σ) a representation of g. (i) The sum Vβ of sβ -submodules of finite dimension of V is a gsubmodule. (ii) Suppose that there exist m ∈ N and v ∈ V verifying σ(Xβ )(v) = 0 and σ(Hβ )(v) = mv. For j 0, let vj = σ(X−β )j (v). Then σ(Xβ )(vm+1 ) = 0. Moreover, if vm+1 = 0, then kv0 + · · · + kvm is a sβ -submodule of V . Proof. (i) This follows from 19.2.6 and 29.6.2. (ii) Set v−1 = 0. We verify easily that σ(Hβ )(vj ) = (m − 2j)vj for l 0. Suppose that σ(Xβ )(vj ) = j(m − j + 1)vj−1 (This is true if j = 0). Then: σ(Xβ )(vj+1 ) = σ(X−β )σ(Xβ )(vj ) + σ(Hβ )(vj ) = j(m − j + 1)σ(X−β )(vj−1 ) + (m − 2j)vj = (j + 1)(m − j)vj . In particular, σ(Xβ )(vm+1 ) = 0. The last part follows immediately.
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30.5.2 Theorem. Let g be a semisimple Lie algebra and (V, σ) a simple g-module with highest weight λ. The following conditions are equivalent: (i) V is finite-dimensional. (ii) λ ∈ P++ . Proof. We have already seen that (i) implies (ii) in 30.3.1. So let us assume that λ ∈ P++ and let v ∈ V be a primitive vector of weight λ. Let β ∈ B, sβ = kHβ + kXβ + kX−β , m = λ(Hβ ) ∈ N. For j 0, let vj = σ(X−β )j (v). By 30.5.1 (ii), σ(Xβ )(vm+1 ) = 0. Let α ∈ B \ {β}. Since [Xα , X−β ] = 0 (18.7.4), we have: σ(Xα )(vm+1 ) = σ(Xα )σ(X−β )m+1 (v) = σ(X−β )m+1 σ(Xα )(v) = 0. If vm+1 = 0, then it is a primitive vector of V of weight distinct from λ (because σ(Hβ )(vm+1 ) = −(m − 2)vm+1 ). This is absurd by 30.2.7, hence vm+1 = 0, and kv0 + · · · + kvm is a sβ -submodule of V . The sum of all the finite-dimensional simple sβ -submodules of V is therefore non-zero. and by 30.5.1 (i), it is a g-submodule, so it is equal to V . It follows that there is family -modules such that V = i∈I Mi . (Mi )i∈I of finite-dimensional simple sβ Let u ∈ Vµ \ {0} and write u = i ui , where ui ∈ Mi for all i. Then σ(Hβ )(ui ) = µ(Hβ )ui for all i. It follows from 19.2.7 that n = µ(Hβ ) ∈ Z because the ui ’s are not all equal to zero. This being true for all β ∈ B, we deduce that P(V ) ⊂ P . If n 0 and ui = 0, then σ(X−β )n (ui ) = 0 (19.2), so σ(X−β )n (u) = 0. If sβ (µ) = µ − nβ ∈ P(V ). Since n 0, then σ(Xβ )n (u) = 0. So in both cases, the sβ ’s generate W (18.7.8), we have W P(V ) ⊂ P(V ). Now any W -orbit of an element of P meets P++ (18.8.9 and 18.11.7). Moreover, by 30.2.7, if µ ∈ P(V ) ∩ P++ , then λ − µ ∈ Q+ and λ + µ ∈ P++ , so λ µ as in the proof of 30.3.2. It follows that P(V ) is contained in the intersection of the discrete set P and the compact set {µ ∈ h∗R ; µ λ}. So P(V ) is finite. Finally, for µ ∈ P(V ), Vµ is finite-dimensional (30.2.7), so V is finitedimensional. 30.5.3 If V is a g-module, denote by [V ] its isomorphism class. Theorem. The map λ → [L(λ)] is a bijection from P++ to the set of isomorphism classes of finite-dimensional simple g-modules. Proof. This follows from 30.3.1, 30.4.9 and 30.5.2. 30.5.4 Let B = (β1 , . . . , βl ) and ( 1 , . . . , l ) the set of fundamental weights of R with respect to B (18.11.6). By 30.5.3, for 1 i l, there is, up to an isomorphism, a unique finite-dimensional simple representation σi of g with highest weight i . The representations σ1 , . . . , σl are called the fundamental simple representations of g.
30.6 A property of the Weyl group
469
30.6 A property of the Weyl group 30.6.1 Denote by Aute (g, h) (resp. Aut0e (g, h)) the subgroup of Aute g consisting of elements θ ∈ Aute g verifying θ(h) = h (resp. θ(h) = h for all h ∈ h). It is clear that Aut0e (g, h) is a normal subgroup of Aute (g, h). The group Aute g acts on g∗ as follows: θ(µ) = µ◦θ−1 for θ ∈ Aute g and µ ∈ g∗ . 30.6.2 Let (V, σ) be a finite-dimensional representation of g. If x ∈ g is nilpotent, then σ(x) is a nilpotent endomorphism of V (20.4.4). We may therefore define eσ(x) ∈ End(V ). Lemma. Let x, y ∈ g be such that x is nilpotent. Then: σ ead x (y) = eσ(x) ◦ σ(y) ◦ e−σ(x) . Proof. We obtain easily that σ(ead x (y)) = ead σ(x) (σ(y)). So it suffices to prove that if f, g ∈ End(V ) with f nilpotent, then ead f (g) = ef ◦g◦e−f . Define the endomorphisms L, R of End(V ) as follows: for g ∈ End(V ): L(g) = f ◦g , R(g) = −g◦f. We have L◦R = R◦L, ad f = L + R, and L, R are nilpotent. We deduce therefore that: ead f (g) = eL+R (g) = eL eR (g) = Hence the result.
Li Rj (g) = ef ◦g◦e−f . i! j! i,j0
30.6.3 Let (V, σ) be a finite-dimensional representation of g. If θ ∈ Aute g, then 30.6.2 says that there exists Sθ ∈ GL(V ) such that for all x ∈ g: σ θ(x) = Sθ ◦ σ(x) ◦ Sθ−1 . Suppose that θ ∈ Aute (g, h). Let µ ∈ h∗ , v ∈ Vµ and h ∈ h. Then: σ(h) Sθ (v) = (µ◦θ−1 )(h)Sθ (v) = θ(µ)(h)Sθ (v). It follows that Sθ (Vµ ) = Vθ(µ) , so P(V ) is stable under θ. Applying this to the adjoint representation of g, we see that R and Q are stable under θ. 30.6.4 Lemma. Let θ ∈ Aute (g, h) be such that θ(B) = B. Then θ(β) = β for all β ∈ B and θ ∈ Aut0e (g, h).
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Proof. We saw in 30.6.3 that θ(Q) = Q, and since θ(B) = B, we have θ(Q+ ) = Q+ . Let λ ∈ P++ . There exists a finite-dimensional representation (V, σ) of g with highest weight λ (30.5.3). By 30.6.3, θ(λ) is a weight of V , so θ(λ) λ (30.2.7). In the same way, θ−1 (λ) λ. From this, we deduce that θ(λ) = λ since θ(Q+ ) = Q+ . Using, for example, the fundamental weights of R, we see that θ|h∗ = idh∗ , and so θ|h = idh . 30.6.5 Theorem. Let us conserve the hypotheses and notations of 30.2.1 and 30.6.1. We define a homomorphism of groups ϕ : Aute (g, h) → GL(h∗ ) by setting ϕ(θ)(µ) = µ ◦ θ−1 for θ ∈ Aute (g, h) and µ ∈ h∗ . Then ϕ induces an isomorphism of groups from Aute (g, h)/ Aut0e (g, h) onto W . Proof. We have ker ϕ = Aut0e (g, h), so ϕ induces an injective homomorphism ϕ : Aute (g, h)/ Aut0e (g, h) → GL(h∗ ). It follow easily from 29.1.3 that W ⊂ im ϕ. Let θ ∈ Aute (g, h). Then ϕ(θ)(R) = R (30.6.3). Also, it is clear that B = ϕ(θ)(B) is a base of R. Let w ∈ W be such that w(B ) = B (18.8.7) and θ ∈ Aute (g, h) verifying ϕ(θ ) = w. Then ϕ(θ ◦θ)(B) = B. So we have θ ◦θ ∈ Aut0e (g, h) (30.6.4). Hence im ϕ = W , and we are done.
References and comments • [12], [14], [29], [38], [39], [42], [43], [80]. We have only given a few properties of enveloping algebras in this chapter. A complete treatment of these algebras, in particular for the properties admitted in section 30.1, can be found in [29].
31 Symmetric invariants
In this chapter, we prove two fundamental theorems of Chevalley and Kostant on invariants.
31.1 Invariants of finite groups 31.1.1 Lemma. Let l ∈ N and P, Q ∈ A = k[T1 , . . . , Tl ]. Suppose that P is homogeneous of degree 1 and Q(x) = 0 for all x ∈ kl such that P (x) = 0. Then there exists R ∈ A such that Q = P R. Proof. We may assume that P is of the form T1 +a2 T2 +· · ·+al Tl . Then there exist U ∈ A and V ∈ k[T2 , . . . , Tl ] such that Q = P U + V . If V = 0, then since k is infinite, there exist α2 , . . . , αl ∈ k such that V (α2 , . . . , αl ) = 0. Let α1 ∈ k be such that P (α1 , . . . , αl ) = 0 (recall that k is algebraically closed). Then Q(α1 , . . . , αl ) = V (α2 , . . . , αl ) = 0. Contradiction. 31.1.2 In this section, V is a k-vector space of dimension l, S = S(V ∗ ) the symmetric algebra of V ∗ and F = Fract(S). The algebra S is isomorphic to k[T1 , . . . , Tl ], and for n ∈ N, we denote by S n the set of homogeneous elements of degree n in S. Let G be a finite subgroup of GL(V ) of order m. Then G acts on S as follows: for α ∈ G, P ∈ S and x ∈ V , (α.P )(x) = P (α−1 (x)). This action extends to an action of G on F by setting α.(u−1 v) = (α.u)−1 (α.v) for α ∈ G, u, v ∈ S and u = 0. Let us denote by R = S G and K = F G the set of fixed points under these actions. Observe that R is a graded subalgebra of S and K is a subfield of F . Denote by R+ the set of elements of R without constant term and I = SR+ . Since S is Noetherian, there exist homogeneous elements u1 , . . . , ur of R+ such that I = Su1 + · · · + Sur . Since the characteristic of k is zero, we may define for u ∈ S:
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u• =
1 α.u. m α∈G
Then u• ∈ R. Also if u ∈ S and v ∈ R, then (uv)• = u• v. 31.1.3 Lemma. (i) The algebra S is integral over R. (ii) The field K is the field of fractions of R. (iii) The extension K ⊂ F is finite of degree m, and the transcendence degree of K over k is equal to l. Proof. For x ∈ F , let Px (T ) =
(T − α.x).
α∈G
(i) If u ∈ S, then Pu is G-invariant, so its coefficients are in R. Since Pu (u) = 0, we deduce that u is integral over R. (ii) This follows from 25.5.1. (iii) By 5.4.10, it suffices to prove that [F : K] = m. Let x ∈ F . As in (i), Px (x) = 0 and Px ∈ K[T ], so x is algebraic over K, and [K(x) : K] m. Let x ∈ F be such that [K(x) : K] is maximal. For any y ∈ F , there exists, by 5.6.7, z ∈ F such that K(x, y) = K(z). It follows from the choice of x that F = K(x) and [F : K] m. Let Q ∈ K[T ] be an irreducible polynomial which has a root x ∈ F . Then Q divides Px , and so Q splits over F . Thus the extension K ⊂ F is normal. Let H = Gal(F/K) and Ω an algebraic closure of K. We have K = F H (5.7.6). It follows from 5.6.6 and the normality of the extension K ⊂ F that: [F : K] m = card G card H card HomK (F, Ω) = [F : K]. So we have proved the lemma.
31.1.4 Lemma. Let u1 , . . . , ur be homogeneous elements of R+ which generate the ideal I = SR+ . Then the k-algebra R is generated by u1 , . . . , ur . Proof. We shall show by induction on n that if u ∈ R is homogeneous of degree n, then it is a polynomial in the ui ’s. This is obvious if n = 0. If n > 0, then u ∈ I, so u = s1 u1 + · · · + sr ur with s1 , . . . , sr ∈ S. Since u and the ui ’s are homogeneous, we may also assume that the si ’s are homogeneous of degree deg u − deg ui < n. Now as observed earlier, we have: u = u• = s•1 u1 + · · · + s•r ur . Since s•i ∈ R, the induction hypothesis implies that u is a polynomial in the ui ’s. 31.1.5 Lemma. Let us suppose that G is generated by reflections. Let u1 , . . . , ur ∈ R be such that u1 ∈ Ru2 + · · · + Rur , and v1 . . . , vr ∈ S homogeneous such that: u1 v1 + · · · + ur vr = 0.
(1) Then v1 ∈ I.
31.1 Invariants of finite groups
473
Proof. Let J = Su2 + · · · + Sur . If u1 ∈ J, then u1 = w2 u2 + · · · + wr ur with w2 , . . . , wr ∈ S. But then, we would have u1 = u•1 = w2• u2 + · · · + wr• ur ∈ Ru2 + · · · + Rur which contradicts the hypothesis. Thus u1 ∈ J. We shall prove v1 ∈ I by induction on deg v1 . If deg v1 0, then v1 = 0 because u1 ∈ J. So let us suppose that deg v1 > 0. Let s ∈ G be a reflection with respect to the hyperplane H and u ∈ V ∗ verifying ker u = H. Since s.vi − vi is zero on H, it follows from 31.1.1 that there exists wi ∈ S such that svi − vi = uwi . As vi and s.vi are homogeneous of the same degree, wi is homogeneous of degree < deg vi . On the other hand, we have: u1 (s.v1 ) + · · · + ur (s.vr ) = 0.
(2)
Equations 1 and 2 imply therefore that u(u1 w1 +· · ·+ur wr ) = 0. It follows by the induction hypothesis that w1 ∈ I, so s.v1 ≡ v1 (mod I). Since R+ is stable under the action of G, so is I. Thus G acts on the quotient S/I. Let v1 be the image of v1 in S/I, then we have just seen that s.v1 = v1 . But G is generated by reflections, so α.v1 = v1 for all α ∈ G. Hence v1• ≡ v1 (mod I). As v1• ∈ R+ ⊂ I, we have v1 ∈ I. 31.1.6 Theorem. (Chevalley’s Theorem) Let V be a vector space of dimension l, and G a finite subgroup of GL(V ) generated by reflections. Then the k-algebra of G-invariants of S(V ∗ ) is generated by l algebraically independent homogeneous elements. Proof. By fixing a basis of V ∗ , we shall identify S = S(V ∗ ) with k[T1 , . . . , Tl ]. Let u1 , . . . , ur ∈ R+ be homogeneous elements which generate the ideal I, and let r be minimal with this property. By 31.1.4, R is generated by the ui ’s, so it suffices to prove that the ui ’s are algebraically independent over k, and 31.1.3 would imply that r = l. Let us suppose that P (u1 , . . . , ur ) = 0 for some P ∈ k[Y1 , . . . , Yr ] \ {0}. Let λY1m1 · · · Yrmr be a monomial which appears in P . Then the degree of m1 r is d = m1 d1 + · · · + mr dr where di = deg ui . We may therefore λu1 · · · um r assume that P is the sum of monomials λY1m1 · · · Yrmr such that the sum m1 d1 + · · · + mr dr are all equal to d. For 1 k l, we have: r i=1
Pi
∂ui ∂P = 0, where Pi = (u1 , . . . , ur ). ∂Tk ∂Yi
Note that Pi is homogeneous of degree d − di in R and that the ∂ui /∂Tk ’s are homogeneous elements of S. Without loss of generality, we may assume that (P1 , . . . , Pm ) is a minimal set of generators of the ideal of R generated by the Pi ’s. So for m < i r, there exist Qij ∈ R, 1 j m, such that Pi = Qi1 P1 + · · · + Qim Pm .
474
31 Symmetric invariants
We may further assume that the Qij ’s are homogeneous of degree dj − di . For 1 k l, we have: ∂u m r ∂uj i = 0. Pi + Qji ∂Tk j=m+1 ∂Tk i=1 Applying 31.1.5, we see that there exist t1k , . . . , trk ∈ S such that: (3)
r r ∂u1 ∂uj + Qj1 = tik ui . ∂Tk j=m+1 ∂Tk i=1
Multiplying both sides of equation (3) by Tk , and sum over k, we obtain by Euler’s formula that (4)
d1 u 1 +
r
dj Qj1 uj =
j=m+1
r
vi u i ,
i=1
where v1 , . . . , vs ∈ S. But v1 = T1 t11 + · · · + Tl t1l has no constant term, and the left hand side of (4) is homogeneous of degree d1 . So by comparing the homogeneous components of degree d1 in equation (4), we see that d1 u1 , and hence u1 , belongs to the ideal of S generated by u2 , . . . , ur . This contradicts our choice of r. 31.1.7 Proposition. Let us conserve the hypotheses and notations of 31.1.6. Let u1 , . . . , ul and v1 , . . . , vl be two sets of algebraically independent homogeneous generators of R. Then there exists a permutation σ of {1, . . . , l} such that deg ui = deg vσ(i) for 1 i l. Proof. Each ui is a polynomial in the vi ’s and vice versa. So we have for 1 i, j l: l ∂u ∂v i k = δij . ∂v ∂u k j k=1 This proves that the matrices [∂ui /∂vj ] and [∂vi /∂uj ] are invertible, so their determinant are non-zero. It follows from the definition of the determinant that there is a permutation σ of {1, . . . , l} such that l ∂ui = 0. ∂v σ(i) i=1
Up to a reindexation, we may assume that σ is the identity permutation. So if we write ui = Pi (v1 , . . . , vl ), then vi appears in Pi . Removing if necessary the terms which are redundant in Pi , we may assume that for each monomial v1m1 · · · vlml in Pi , we have: deg ui = m1 deg v1 + · · · + ml deg vl . Hence deg ui deg vi . Now by exchanging the roles of ui and vi , we obtain that deg ui = deg vi .
31.2 Invariant polynomial functions
475
31.1.8 Proposition. With the hypotheses of 31.1.6, S is a free R-module of rank m = card G. Proof. Let θ : S → S/I denote the canonical surjection. 1) Let (vλ )λ∈Λ be a set of homogeneous elements of S such that the θ(vλ )’s span the k-vector space S/I. Let T be the R-submodule of S generated by the vλ ’s. It is a graded submodule of S. We claim that S = T . We shall prove our claim by showing, by induction, that S n = T n for all n ∈ N where S n and T n denote the homogeneous component of degree n of S and T . Since S 0 ∩ I = {0}, one of the vλ ’s is of degree 0, so S 0 = T 0 . Let u ∈ S n , n > 0. Then there exist αλ ∈ k, w1 , . . . , wr ∈ R+ and t1 , . . . , tr ∈ S homogeneous of degree < n such that u=
αλ vλ +
λ∈Λ
r
ti wi .
i=1
By the induction hypothesis, ti ∈ T for all i. Hence u ∈ T n . So we have proved our claim. 2) Let v1 , . . . , vr ∈ S be homogeneous elements such that the θ(vi )’s are linearly independent. We shall show by induction on r that the vi ’s are independent over R. The result is obvious if r = 1. So let us suppose that r > 1 and that u1 v1 + · · · + ur vr = 0 with ui ∈ R homogeneous. Since v1 ∈ I, there exist homogeneous elements w2 , . . . , wr ∈ R such that u1 = w2 u2 + · · · + wr ur (31.1.5). Thus u2 (v2 + w2 v1 ) + · · · + ur (wr + wr v1 ) = 0. But the elements vi +wi v1 , 2 i r, are homogeneous and their image under θ are linearly independent. By the induction hypothesis, u2 = · · · = ur = 0. Hence u1 = 0. 3) We deduce from points 1 and 2 that S is a free R-module. Moreover, 31.1.3 says that K = Fract(R) and [F : K] = m. It is now clear that the rank of S over R is m.
31.2 Invariant polynomial functions 31.2.1 In this section, V is a vector space of dimension l. For n ∈ N, we denote by Sn (V ) be the set of homogeneous elements of degree n in the symmetric algebra S(V ) of V . The algebra S(V ∗ ) can be identified as the algebra of polynomial functions on V with values in k. 31.2.2 Let a be a Lie algebra and (V, σ) a representation of a. This induces a representation θ of a in S(V ) given by: for x ∈ a and u = u1 · · · un ∈ Sn (V ): x.u = θ(x)u =
n
u1 · · · ui−1 (σ(x)ui )ui+1 · · · un .
i=1
Thus for x ∈ a and u, v ∈ S(V ), we have:
476
31 Symmetric invariants
(5)
θ(x)(uv) = (θ(x)u)v + u(θ(x)v).
Recall that S(V )a (resp. Sn (V )a ) is the set of elements u ∈ S(V ) (resp. S (V )) verifying θ(x)(u) = 0 for all x ∈ a. By (5), S(V )a is a graded subalgebra of S(V ). In this section, we are interested in the case where V = a the adjoint representation or V = a∗ the coadjoint representation. An element of S(a∗ )a is called an invariant polynomial function on a. n
31.2.3 Lemma. Let a be a Lie algebra, (V, σ) a finite-dimensional amodule and n ∈ N. The map f : a → k, x → tr(σ(x)n ), is an invariant polynomial function on a. Proof. It is clear that f ∈ S(a∗ ). Now if x, y ∈ a, then: −(x.f )(y) =
n
tr σ(y)i−1 σ([x, y])σ(y)n−i
i=1 n
n tr σ(y)i−1 σ(x)σ(y)n−i+1 − tr σ(y)i σ(x)σ(y)n−i i=1 i=1 = tr σ(x)σ(y)n − σ(y)n σ(x) = 0.
=
So the result follows. 31.2.4 In the rest of this section, g is a semisimple Lie algebra and h a Cartan subalgebra of g. Proposition. For f ∈ S(g∗ ), we have f ∈ S(g∗ )g if and only if f ◦ α = f for all α ∈ Aute g. Proof. This follows from 23.4.14, 23.4.16 and 24.3.4. 31.2.5 We shall use the notations of 30.2.1. The Weyl group W = W (g, h) can be considered as a group of automorphisms of h (18.2.9). Let S(h∗ )W (resp. Sn (h∗ )W ) be the set of f ∈ S(h∗ ) (resp. f ∈ Sn (h∗ )) verifying f ◦ w = f for all w ∈ W . Lemma. If n ∈ N, the vector space Sn (h∗ )W is spanned by polynomial functions on h of the form x → tr(ρ(x)n ) where ρ is a finite-dimensional representation of g. Proof. If f ∈ S(h∗ ), we set θ(f ) ∈ S(h∗ )W to be: θ(f ) = f ◦w. w∈W
By 18.11.5, P is a sublattice of h∗ spanning h∗ . It follows that the λn ’s, for λ ∈ P , span the vector space Sn (h∗ ). Moreover, by 18.8.9 and 18.11.7, {w(λ) ; w ∈ W } ∩ P++ = ∅ for all λ ∈ P . We deduce therefore that θ(λn ), with λ ∈ P++ , span the vector space Sn (h∗ )W .
31.2 Invariant polynomial functions
477
For λ ∈ P++ , let Eλ = {µ ∈ P++ ; µ < λ} (with respect to the partial order defined in 30.2.1). Recall from the proof of 30.5.2 that Eλ is a finite set. Let (V, σ) be a finite-dimensional representation of g with highest weight λ. Define g : h → k, x → tr(σ(x)n ). ∗ W Let P(V ) be thenset of weights of V . By 30.3.1, g ∈ S(h ) . Hence g = µ∈P(V )∩P++ aµ θ(µ ) with aµ ∈ k are non-zero for all µ ∈ P(V ) ∩ P++ . Thus θ(λn ) is a linear combination of g and the θ(µn )’s, with µ ∈ Eλ (30.2.7). So the result follows by induction on the cardinality of Eλ . 31.2.6 Theorem. Let g be a semisimple Lie algebra, h a Cartan subalgebra of g, W = W (g, h) and i : S(g∗ ) → S(h∗ ) the homomorphism of restriction. (i) The map i induces an isomorphism from S(g∗ )g onto S(h∗ )W . (ii) If n ∈ N, the vector space Sn (g∗ )g is spanned by functions of the form x → tr(σ(x)n ) where σ is a finite-dimensional representation of g. Proof. Let w ∈ W . By 29.1.3, there exists α ∈ Aute g such that α|h = w. Thus 31.2.4 implies that: f (x) = f ◦ α(x) = f ◦ w(x) for f ∈ S(g∗ )g and x ∈ h. Hence i(S(g∗ )g ) ⊂ S(h∗ )W . Now let f ∈ S(g∗ )g be such that i(f ) = 0. If x ∈ g is semisimple, then α(x) ∈ h for some α ∈ Aute g (29.2.3 and 29.2.6). So f (x) = f ◦ α(x) = 0 and hence by 29.2.2 and the fact that ggen is dense in g, we have f = 0. So i is injective on S(g∗ )g . For n ∈ N, let Tn be the linear span of the functions x → tr(σ(x)n ), where σ is a finite-dimensional representation of g. By 31.2.3 and 31.2.5, we have: Tn ⊂ Sn (g∗ )g , Sn (h∗ )W ⊂ i(Tn ). So the theorem follows.
31.2.7 Let ϕ : g → g∗ be the Killing isomorphism of g. Then ϕ extends to an isomorphism, also denoted by ϕ, from S(g) onto S(g∗ ), that we shall call the Killing isomorphism from S(g) onto S(g∗ ). Denote by ψ : S(h) → S(h∗ ) the isomorphism induced by restriction to h of the Killing form L of g. Since L|h×h is W -invariant, ψ commutes with the actions of W on S(h) and S(h∗ ). Let g = h ⊕ n+ ⊕ n− be as in 30.2.1. Lemma. Let J be the ideal of S(g) generated by n+ ∪ n− . (i) We have S(g) = S(h) ⊕ J. Let j : S(g) → S(h) be the homomorphism of algebras defined by the projection with respect to this decomposition. (ii) Let i : S(g∗ ) → S(h∗ ) be the restriction map. Then i ◦ ϕ = ψ ◦ j.
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31 Symmetric invariants
Proof. Part (i) is obvious. Let us prove part (ii). Let x ∈ g and h ∈ h. We have L(h, n+ + n− ) = {0}, so: i◦ϕ(x)(h) = ϕ(x)(h) = L(x, h) = L(j(x), h) = ψ◦j(x)(h). Thus (ii) follows since i, j, ϕ, ψ are homomorphisms of algebras. 31.2.8 Theorem. Let g be a semisimple Lie algebra of rank l and h a Cartan subalgebra of g. Let us conserve the notations of 31.2.7. (i) The homomorphism j induces an isomorphism from the algebra S(g)g onto the algebra S(h)W of W -invariants in S(h). (ii) There exist l algebraically independent elements of S(g)g which generate S(g)g . Proof. Part (i) follows from 31.2.6 and 31.2.7, while (ii) follows from (i) and 31.1.6.
31.3 A free module 31.3.1 Let E be a finite-dimensional k-vector space and A(E) = S(E ∗ ). For n ∈ N, let An (E) be the set of homogeneous elements of degree n of A(E). For n < 0, set Fn (E) = {0}, and for n 0: Fn (E) = A0 (E) + A1 (E) + · · · + An (E). Thus (Fn (E))n defines a filtration on A(E). 31.3.2 Let V be a vector space of dimension l and U a subspace of V . Denote by i : A(V ) → A(U ) the restriction map. Since (V /U )∗ identifies with linear forms on V which are zero on U , we have an injective homomorphism of algebras s : A(V /U ) → A(V ). We define a filtration on A(V ) as follows: for n ∈ Z, Gn (V ) = s A(V /U ) .Fn (V ) = s A(V /U ) .Fn (U ). Let grG (V ) be the graded algebra associated to this filtration. If u ∈ A(V ), denote by σG (u) its principal symbol. Define a grading (Hn )n∈Z on the algebra A(V /U ) ⊗k A(U ) as follows: Hn = A(V /U ) ⊗k An (U ). Lemma. (i) The graded algebras grG (V ) and A(V /U ) ⊗k A(U ) are isomorphic. (ii) If f is a homogeneous element of A(V ) such that i(f ) is non-zero. Then σG (f ) = i(f ) ∈ A(U ) ⊂ A(V /U ) ⊗k A(U ). Proof. Let E be a basis of V which contains a basis of U . Then A(V ) is generated by the dual basis of E. The proof of the lemma is now straightforward.
31.3 A free module
479
31.3.3 Proposition. Let A be a graded subalgebra of A(V ), A = i(A) and R = A(V /U ) ⊗k A. Suppose that the following conditions are verified: (i) The restriction of i to A is injective. (ii) The A -module A(U ) is free. Then A(V ) is a free R-module. In particular, it is a free A-module. Let (vj )j∈J be a basis of the A -module A(U ) consisting of homogeneous elements. For j ∈ J, let uj ∈ A(V ) be homogeneous such that i(uj ) = vj . then (uj )j∈J is a basis of the R-module A(V ). Proof. Using the notations of 31.3.2, let us consider the filtration induced on A by the Gn (V )’s. Since i|A is injective, the subalgebra σG (A) of grG (V ) can be identified with A ⊂ A(U ), and we have σG (uj ) = i(uj ) = vj for all j ∈ J (31.3.2). Now (vj )j∈J is also a basis of the A(V /U ) ⊗k A -module A(V /U ) ⊗k A(U ). Since A(V /U ) ⊗k A(U ) is isomorphic to grG (V ) (31.3.2), we deduce by 7.5.10 that the R-module A(V ) is free, and (uj )j∈J is a basis of this module. 31.3.4 Theorem. Let g be a semisimple Lie algebra, h a Cartan subalgebra of g and W = W (g, h). (i) The algebra S(g∗ ) is a free S((g/h)∗ )⊗k S(g∗ )g -module of rank card W . (ii) The algebra S(g∗ ) is a free S(g∗ )g -module, and the algebra S(g) is a free S(g)g -module. Proof. Part (i) is a consequence of 31.1.8, 31.2.8 and 31.3.3, while (ii) follows from (i) and by using the Killing isomorphism.
References and comments • [4], [29], [41], [48], [49], [80]. Theorem 31.2.8 is due to Chevalley. Theorem 31.3.4 is due to Kostant, and the proof given here is taken from [4].
32 S-triples
In the first section, we prove the Jacobson-Morosov Theorem on Lie subalgebras of a semisimple Lie algebra which are isomorphic to sl2 (k). We then show how such Lie subalgebras can be used to obtain information on nilpotent elements in a semisimple Lie algebra. The results in this chapter will be used to study certain orbits in semisimple Lie algebras. Let g be a finite-dimensional k-Lie algebra and L its Killing form. If a is a subset of g, a⊥ denotes the orthogonal of a with respect to L.
32.1 Jacobson-Morosov Theorem 32.1.1 Definition. A triple (x, t, y) of elements of g is called a S-triple if (x, t, y) = (0, 0, 0) and [t, x] = 2x , [t, y] = −2y , [x, y] = t. If (x, t, y) is a S-triple of g, then t is called its simple element, x its positive element and y its negative element. An element of g is simple if it is the simple element of a S-triple. 32.1.2 Lemma. Let u, v be endomorphisms of a finite-dimensional vector space V . Suppose that [u, [u, v]] = 0. Then: (i) [u, v] is nilpotent. (ii) If u is nilpotent, then so is u ◦ v. Proof. (i) Let w = [u, v] and p ∈ N. Since [u, w] = 0, we have: [u, v◦wp ] = [u, v]◦wp = wp+1 . Hence tr(wp+1 ) = 0 for all p ∈ N. So w is nilpotent (19.2.3). (ii) Let p ∈ N∗ . Again since [u, w] = 0, we have: [v, up ] =
p−1 i=0
ui ◦[v, u]◦up−i−1 = p[v, u]◦up−1 .
482
32 S-triples
Let λ ∈ k and x ∈ V \{0} be such that u◦v(x) = λx. If r ∈ N∗ is the smallest integer such that ur (x) = 0, then: λur−1 (x) = ur ◦v(x) = v◦ur (x) − [v, ur ](x) = −r[v, u]◦ur−1 (x). As [v, u] is nilpotent and ur−1 (x) = 0, we have λ = 0. Thus 0 is the only eigenvalue of u◦v, so u◦v is nilpotent. 32.1.3 Lemma. Let t, x ∈ g be such that [t, x] = 2x and t ∈ [x, g]. Then there exists y ∈ g such that [t, y] = −2y and [x, y] = t. Proof. The Lie algebra h = kt + kx is solvable and x ∈ [h, h]. So it follows from 19.4.4 that ad x is nilpotent. Let z ∈ g be such that [x, z] = t, and set: I = idg , X = ad x , T = ad t , Z = ad z. Recall from 19.2.5 that [T, X p ] = 2pX p , [Z, X p ] = −p(T − (p − 1)I)X p−1 , so n = ker X and mp = im X p are T -stable. Furthermore, if u ∈ mp−1 , then: −p(T − (p − 1)I)(u) ∈ ZX(u) + mp . Moreover if u ∈ n, then X(u) = 0 and we obtain for p 1, (T − (p − 1)I)(n ∩ mp−1 ) ⊂ n ∩ mp . As X is nilpotent, there exists k ∈ N such that mk+1 = {0}. So (T − kI) · · · (T − I)T (n) = {0}. This shows that the eigenvalues of T |n are greater than or equal to zero, and so (T +2I)|n is invertible. But [x, [t, z]+2z] = −2[x, z]+2t = 0. Thus [t, z]+2z ∈ n. So there exists v ∈ n such that [t, z] + 2z = (T + 2I)(v) = [t, v] + 2v. Let y = z − v, then we have [x, y] = t, [t, y] = −2y. 32.1.4 Lemma. Let g be a semisimple Lie algebra and x a nilpotent element of g. Then x ∈ [x, [x, g]]. Proof. Let n = ker(ad x)2 , m = (ad x)2 (g). If y ∈ n, then 0 = ad[x, [x, y]] = [ad x, [ad x, ad y]], so by 32.1.2, (ad x) ◦ (ad y) is nilpotent. Hence L(x, y) = 0. Thus x ∈ n⊥ . Now for z ∈ g, we have: 0 = L([x, [x, y]], z) = −L([x, y], [x, z]) = L(y, [x, [x, z]]). So m ⊂ n⊥ and since L is non-degenerate, we have m = n⊥ , and x ∈ m.
32.1 Jacobson-Morosov Theorem
483
32.1.5 Theorem. (Jacobson-Morosov Theorem) Let g be a semisimple Lie algebra and x a non-zero nilpotent element of g. Then there exist t, y ∈ g such that (x, t, y) is a S-triple of g. Proof. By 32.1.4, there exists z ∈ g such that [x, [x, z]] = x. Let t = −2[x, z]. Then [t, x] = 2x and t ∈ [x, g]. So the result follows from 32.1.3. 32.1.6 Corollary. If g is semisimple, then Auts g = Aute g. Proof. Let x be a non-zero nilpotent element of g. By 32.1.5, there exists t ∈ g such that [t, x] = 2x. So the result follows. 32.1.7 Proposition. Let g be a semisimple Lie algebra and h, x ∈ g. Suppose that h is semisimple and x is a non-zero nilpotent element such that [h, x] = λx, with λ ∈ k. Then there is a S-triple (x, t, y) such that: [h, t] = 0 , [h, y] = −λy. Proof. For µ ∈ k, denote by gµ the (ad h)-eigenspace of eigenvalue µ. By 32.1.5, there is a S-triple (x, t, y). Let us write: tν , y = yν , t= ν∈k
ν∈k
with tν , yν ∈ gν . Since [tν , x] ∈ gλ+ν , we deduce from [t, x] = 2x that [t0 , x] = 2x and [tν , x] = 0 if ν = 0. Moreover: [x, yν ]. [x, y] = t = ν∈k
So t0 = [x, y−λ ]. Thus [t0 , x] = 2x and t0 ∈ [x, g]. By 32.1.3, there is a S-triple (x, t0 , z). If we write z = ν∈k zν where zν ∈ gν , then it follows from the relations of a S-triple that [t0 , z−λ ] = −2z−λ and [x, z−λ ] = t0 . Hence (x, t0 , z−λ ) is a S-triple with the required properties. 32.1.8 Proposition. Let g be a semisimple Lie algebra, λ ∈ k and x, h, t ∈ g \ {0} verifying: [h, x] = 2x , [h, t] = 0 , [t, x] = λx. Let us suppose that t is semisimple and h ∈ (gt ∩gx )⊥ . Then there exists y ∈ g such that (x, h, y) is a S-triple. Proof. Let gν be the (ad t)-eigenspace of eigenvalue ν. Since [t, x] = λx, gx is (ad t)-stable. Hence we have: x (g ∩ gν ). gx = ν∈k
Note that g = g0 and L(gµ , gν ) = {0} if µ + ν = 0. So h ∈ (gx )⊥ if h ∈ (gt ∩ gx )⊥ . Since (gx )⊥ = [x, g], the result follows from 32.1.3. t
484
32 S-triples
32.1.9 Proposition. Let t be a simple element of a semisimple Lie algebra g. Let q ∈ Z, then the dimension of the (ad t)-eigenspace of eigenvalue 2q + 1 is even. Proof. For q ∈ Z, denote by gq the (ad t)-eigenspace of eigenvalue q. Let p ∈ N and (x, t, y) be a S-triple of g. We define an alternating bilinear form on g2p+1 as follows: Bp (u, v) = L u, (ad y)2p+1 (v) where u, v ∈ g2p+1 . Since (ad y)2p+1 (g2p+1 ) = g−2p−1 and the restriction of L to g2p+1 × g−2p−1 is non-degenerate, Bp is non-degenerate. It follows that dim g2p+1 is even, and since dim g2p+1 = dim g−2p−1 , the result follows.
32.2 Some lemmas 32.2.1 Lemma. Let h1 , h2 be Cartan subalgebras of a semisimple Lie algebra g. There exists θ ∈ Aute g such that h1 = θ(h2 ) and θ(x) = x for all x ∈ h1 ∩ h 2 . Proof. The algebra c = cg (h1 ∩ h2 ) is reductive in g (20.5.13 and 20.5.16). So c = [c, c] ⊕ z(c). Let k1 , k2 be Cartan subalgebras of [c, c] such that hi = ki ⊕ z(c) for i = 1, 2. By 29.2.3, there exists α ∈ Auts [c, c] such that α(k1 ) = k2 . Let x1 , . . . , xn be nilpotent elements of [c, c] such that α = exp(ad[c,c] x1 ) · · · exp(ad[c,c] xn ). It follows from 20.5.8 that adg xi is a nilpotent endomorphism of g for 1 i n. Thus θ = exp(adg x1 ) · · · exp(adg xn ) satisfies the required properties. 32.2.2 Lemma. Let s = sl2 (k) and e, f, h be as in 19.2.4. Let x be a nonzero nilpotent element of s. Then there exists α ∈ Aute s such that α(x) = e. Proof. Let x = ae + bh + cf with a, b, c ∈ k. Since x is a nilpotent endomorphism of k2 (20.4.4), we have b2 + ac = 0. For λ ∈ k \ {0}, let −1 θλ = eλ ad e ◦ e−λ ad f ◦ eλ ad e . Then: (1)
θλ (e) = −λ−2 f , θλ (f ) = −λ2 e , θλ (h) = −h.
If a = 0, then b = 0, c = 0 and θλ (x) = −cλ2 e. So we are reduced to the case where a = 0. Since k is algebraically closed, the formulas (1) imply that we may assume a = 1. Then eb ad f (x) = e + (c + b2 )f = e. 32.2.3 Lemma. If (x, t, y) and (x, t, z) are S-triples in g, then y = z. Proof. We have [x, y − z] = 0, [t, y − z] = −2(y − z), and the restriction of ad x to ker(ad t + 2 idg ) is injective (19.2), so y = z. 32.2.4 Lemma. Let x ∈ g, p = ker(ad x), q = [x, g] and n = p ∩ q. (i) We have [p, q] ⊂ q and n is a Lie subalgebra of g. (ii) If x is a positive element of a S-triple (x, t, y), then n is (ad t)-stable, and for all u ∈ n, ad u is nilpotent.
32.2 Some lemmas
485
Proof. (i) Let u ∈ p, v ∈ q and w ∈ g be such that v = [x, w]. Then: [u, v] = [u, [x, w]] = [[u, x], w] + [x, [u, w]] = [x, [u, w]] ∈ q. As p is a Lie subalgebra of g, [p ∩ q, p ∩ q] ⊂ p ∩ q. (ii) Since [t, x] = 2x, p and q are (ad t)-stable, so n is also (ad t)-stable. For t ∈ Z, let gp = ker(ad t − p idg ). Then: gp , [gp , gq ] ⊂ gp+q . g= p∈Z
We deduce therefore that if u ∈ r = result follows since n ⊂ r by 19.2.
p>0
gp , then ad u is nilpotent. So the
32.2.5 Lemma. Let n be a Lie subalgebra of g such that adg x is nilpotent for all x ∈ n. Let t ∈ g be such that [t, n] = n. Then exp(adg n)(t) = t + n. Proof. We have clearly exp(ad n)(t) ⊂ t + n. Let v ∈ n. Since n is nilpotent, it suffices to prove that for all p 1: t + v ∈ ead n (t) + C p (n). This is clear if p = 1. Suppose that there exist x ∈ n and y ∈ C p (n) such that t + v = ead x (t) + y. Since [t, n] = n, adn t is bijective, and as the subspaces C p (n) are (ad t)-stable, the restriction of ad t to C p (n) is also bijective. So there exists z ∈ C p (n) such that y = [z, t], and so: ead(x+z) (t) − ead x (t) ∈ [z, t] + C p+1 (n). Hence: ead(x+z) (t) ∈ t + v − y + [z, t] + C p+1 (n) = t + v + C p+1 (n). We have therefore proved the lemma. 32.2.6 Lemma. Let (x, t, y) and (x, t , y ) be S-triples of g. Denote by n = ker(ad x) ∩ im(ad x). Then there exists z ∈ n such that: ead z (x) = x , ead z (t) = t , ead z (y) = y . Proof. By 32.2.4, ad z is nilpotent for any z ∈ n, and we have [t, n] = n by 19.2. Also, [x, t − t ] = 0 and [x, y − y ] = t − t , so t ∈ t + n. Thus 32.2.5 says that ead z (t) = t for some z ∈ n. Since z ∈ n, we have ead z (x) = x. Finally by 32.2.3, we obtain that ead z (y) = y . 32.2.7 Lemma. Let (x, t, y) be a S-triple of g. For p ∈ Z, set gp = ker(ad t − p idg ), and for λ ∈ k \ {0}, let τλ be the automorphism of g verifying τλ |gp = λp idgp . Then τλ ∈ Aute g.
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32 S-triples
Proof. Let s = kx + kt + ky and V ⊂ g a simple s-module of dimension r + 1. By 19.2.5 and 19.2.7, there is a basis (u0 , . . . , ur ) of V such that [t, ui ] = (r − 2i)ui , [x, ui ] = i(r − i + 1)ui−1 , [y, ui ] = ui+1 , where by convention, we set u−1 = ur+1 = 0. The endomorphisms ad x, ad y of g are nilpotent (19.2.5). So if λ ∈ k \ {0}, we may define θλ = −1 eλ ad x ◦e−λ ad y ◦eλ ad x ∈ Aute g. By the relations (1) of 32.2.2, there exist νi ∈ k \ {0} such that θλ (ui ) = νi ur−i for 0 i r. Since ui = (ad y)i (u0 ), we obtain (by using again (1)): θλ (ui ) = (−1)i λ2i (ad x)i (θλ (u0 )) = (−1)i λ2i ν0 (ad x)i (ur ) = (−1)i λ2i ν0 r(r − 1) · · · (r − i + 1)i!ur−i . Let us compute ν0 . We have eλ ad x (u0 ) = u0 and: −1
e−λ
ad y
(u0 ) = u0 +
(−1) (−1)r u1 + · · · + ur . λ r!λr
Since θλ (u0 ) is proportional to ur , we have: θλ (u0 ) =
(−1)r ur . r!λr
Thus we have obtained that: θλ (ui ) =
(−1)r−i i! 2i−r λ ur−i . (r − i)!
It follows that θλ ◦θ13 (ui ) = λr−2i ui = τλ (ui ). So by 19.2, we are done. 32.2.8 Let g be semisimple, G = Aute g (it is the adjoint group of g by 21.3.4 and 23.4.16), t a semisimple element of g, Gt the stabilizer of t in G and: g0 = ker(ad t) , g2 = ker(ad t − 2 idg ). Then by 24.3.6 (ii), L(Gt ) = L(G◦t ) = g0 . It is also clear that G◦t (g2 ) ⊂ g2 . Lemma. Assume that there exists x ∈ g2 such that [x, g0 ] = g2 . Then Gt (x) contains a dense open subset of g2 . Proof. The morphism u : G◦t → g2 , α → α(x), is dominant by 16.5.7. So the result follows from 15.4.2. 32.2.9 Lemma. Let g be semisimple, (x, t, y) a S-triple of g and Gt , g0 , g2 = {u ∈ g2 ; [u, g0 ] = g2 } is a dense open subset of g2 be as in 32.2.8. Then g2 . Moreover, it is a Gt -orbit. Proof. The set g2 is non-empty since it contains x (19.2). The condition u ∈ g2 is equivalent to the condition dim[u, g0 ] < dim g2 . But it is equivalent g2 \ to the vanishing of certain polynomials. So g2 is a dense open subset of g2 . The fact that g2 is a Gt -orbit follows easily from 32.2.8.
32.3 Conjugation of S-triples
487
32.3 Conjugation of S-triples 32.3.1 In this section, g is a semisimple Lie algebra and G = Aute g. If (x, t, y) and (x , t , y ) are S-triples of g, then we say that they are Gconjugate if there exists θ ∈ G such that x = θ(x), t = θ(t) and y = θ(y). Theorem. Let (x, t, y) and (x , t , y ) be S-triples of g, t = kx + kt + ky and t = kx + kt + ky . Then the following conditions are equivalent: (i) There exists θ ∈ G such that x = θ(x). (ii) There exists θ ∈ G such that t = θ(t). (iii) The S-triples (x, t, y) and (x , t , y ) are G-conjugate. (iv) There exists θ ∈ G such that t = θ(t). Proof. The implications (iii) ⇒ (i), (iii) ⇒ (ii) and (iii) ⇒ (iv) are clear, while (i) ⇒ (iii) follows from 32.2.6. (ii) ⇒ (i) If (ii) is verified, we may assume that t = t . Then [x, g0 ] = [x , g0 ] = g2 , and 32.2.9 implies that x = θ(x) for some θ ∈ Gt . (iv) ⇒ (i) If (iv) is verified, we may assume that t = t . Since t is isomorphic to sl2 (k) and x, x are nilpotent, it follows from 32.2.2 that x = θ(x) for some θ ∈ Aute t. But if u ∈ t is nilpotent, then adg u is also nilpotent (20.4.4). So θ extends to an element of G. 32.3.2 Corollary. (i) The map sending a S-triple of g to its positive element induces a bijection from the set of G-conjugacy classes of S-triples onto the set of G-orbits of non-zero nilpotent elements of g. (ii) The map sending a S-triple of g to its simple element induces a bijection from the set of G-conjugacy classes of S-triples onto the set of G-orbits of non-zero simple elements of g. Proof. This is clear by 32.1.5 and 32.3.1. 32.3.3 Proposition. Let (x, t, y) be a S-triple of g. (i) Let n = ker(ad x) ∩ im(ad x). Then t + n is the set of elements t of g such that there exists a S-triple of the form (x, t , y ). The map n → g, u → t + u, induces a bijection from n onto the set of S-triples of g whose positive element is x. (ii) Let g2 be defined as in 32.2.9. There exists a S-triple of the form g2 . The map (x , t, y ) → x is a bijection from the (x , t, y ) if and only if x ∈ set of S-triples of g whose simple element is t onto g2 . Proof. Part (ii) follows immediately from 32.2.9. Let us prove (i). If u ∈ n, then t + u ∈ im(ad x) and [t + u, x] = 2x. So by 32.1.3, there is a S-triple of the form (x, t + u, y ). Conversely, if (x, t , y ) is a S-triple, then [t − t , x] = 0 and [x, y − y ] = t − t , so t − t ∈ n. 32.3.4 Remark. In the above notations, we have (32.2.5 and 32.2.9): t + n = {exp(ad z)(t); z ∈ n} , g2 = Gt (x).
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32.4 Characteristic 32.4.1 In this section, g is assumed to be semisimple. Let G = Aute g. If x ∈ g, denote by Ox the G-orbit of x. Let us also fix a Cartan subalgebra h of g and a base B of the root system R = R(g, h). 32.4.2 Let t ∈ h. If t is a simple element of g, then its eigenvalues are integers. It follows that α(t) ∈ Z for all α ∈ R. Lemma. Let t ∈ h be a simple element of g such that β(t) ∈ N for all β ∈ B. Then β(t) ∈ {0, 1, 2} for all β ∈ B. Proof. Let gp = ker(ad t − p idg ), p ∈ Z, S = {α ∈ B; α(t) = 2} and for α ∈ R, Xα ∈ gα \ {0}. Denote by b the Borel subalgebra associated to h and B, and n the nilpotent radical of b. Recall from 18.7.4 that [n, X−β ] ⊂ b for all β ∈ B. Let (x, t, y) be a S-triple of g. Since x ∈ g2 , we have x ∈ n and we can write x = α∈S λα Xα where λα ∈ k. If β ∈ B, then: λα [Xα , X−β ] ∈ b ∩ g2−β(t) . [x, X−β ] = α∈S
If λα = 0 and [Xα , X−β ] = 0 for some α ∈ S, then β(t) 2. Otherwise, we have [x, X−β ] = 0. But (ad x)|gγ is injective for all γ ∈ R verifying γ(t) < 0. Hence β(t) = 0. 32.4.3 Proposition. If t is a simple element of g, then there exists θ ∈ G such that θ(t) ∈ h and β(θ(t)) ∈ {0, 1, 2} for all β ∈ B. Proof. By 29.2.3 and 29.2.6, we may assume that t ∈ h and so t ∈ hQ because t is simple. There exists w ∈ W (g, h) such that β(w(t)) 0 for all β ∈ B (18.8.7). So the result follows from 32.4.2 and 29.1.3. 32.4.4 Corollary. Let l be the rank of g. (i) The number of G-orbits of simple elements of g is at most 3l . (ii) The number of G-orbits of nilpotent elements of g is at most 3l . (iii) The number of G-conjugacy classes of Lie subalgebras of g which are isomorphic to sl2 (k) is at most 3l . Proof. Part (i) follows from 32.4.3 and the others from 32.3.1. 32.4.5 Lemma. Let t be a simple element of g. There is a unique t1 ∈ Ot such that t1 ∈ h and β(t1 ) 0 for all β ∈ B. Proof. The existence of t1 follows from 32.4.3. Now suppose that t1 , t2 verify the required properties. Let σ1 , σ2 ∈ G be such that σ1 (t) = t1 , σ2 (t) = t2 , and let hi = σi−1 (h) for i = 1, 2. We have t ∈ h1 ∩ h2 . By 32.2.1, there exists σ ∈ G such that σ(h1 ) = h2 and σ(t) = t. So θ = σ2 ◦ σ ◦ σ1 verifies θ(t1 ) = t2 and θ(h) = h. Thus θ ∈ Aute (g, h) (in the notations of 30.6.1). By 30.6.5, there exists w ∈ W (g, h) such that θ|h = w|h (18.2.9). In particular, w(t1 ) = t2 , hence t1 = t2 (18.8.9).
32.5 Regular and principal elements
489
32.4.6 Let B = (β1 , . . . , βl ) and x a non-zero nilpotent element of g. Then 32.1.5 says that there is a S-triple of the form (x, t, y). Let t1 be the unique element of Ot such that t1 ∈ h and βi (t1 ) 0 for all 1 i l (32.4.5). Let ni = βi (t1 ), 1 i l. We have ni ∈ {0, 1, 2} by 32.4.2. By 32.2.6, we see that the sequence (n1 , . . . , nl ) depends only on x (or Ox ). We shall denote this sequence by C(x), and we call C(x) the characteristic of x. Note that if x is a nilpotent element of g such that C(x ) = C(x), then x ∈ Ox (32.3.1). By convention, we set C(0) = (0, . . . , 0). The map Ox → C(x) defines an injection of the set of G-orbits of nilpotent elements of g into the set of sequence of l elements of {0, 1, 2}. This map is not bijective in general. We often represent a nilpotent orbit of g by attaching to each vertex βi of the Dynkin diagram of g, the number ni .
32.5 Regular and principal elements 32.5.1 Let us conserve the hypotheses and notations of 32.4.1. Let V be a finite-dimensional g-module. Recall that if V = V1 ⊕ · · · ⊕ Vn = W1 ⊕ · · · ⊕ Wp are two decompositions of V into simple g-submodules, then n = p and it is equal to the length of V . Recall from 29.3.1 and 29.3.2 that an element x ∈ g is regular if and only if dim gx = rk g. 32.5.2 Definition. (i) A simple element t of g is called principal if it is regular and the eigenvalues of ad t are even integers. (ii) A S-triple (x, t, y) is called principal if the length of g as a module over kx + kt + ky is equal to the rank of g. 32.5.3 Proposition. Let (x, t, y) be a S-triple of g. The following conditions are equivalent: (i) x is regular. (ii) t is principal. (iii) (x, t, y) is principal. Proof. Let s = kx + kt + ky, gp = ker(ad t − idg ), p ∈ Z, and a the sum of the subspaces g2p for p ∈ Z. The subspace a is a s-submodule of g. Denote by lg and la the lengths of the s-modules g and a. From the results of 19.2, 19.7, 29.3.1 and 29.3.2, we have: (2)
dim gx = lg la = dim gt rk(g).
So it follows from the definitions that all three conditions are equivalent to the condition that the inequalities in (2) are equalities.
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32 S-triples
32.5.4 Let B = (β1 , . . . , βl ) and R+ the set of positive roots with respect to B. Let us use the notations of 20.6.5 and for α ∈ R, let Xα ∈ gα be elements such that [Xα , X−α ] = Hα . For 1 i l, denote Hi = Hβi . Let t0 = Hα . α∈R+
Lemma. (i) We have βi (t0 ) = 2 for 1 i l, and there exist non-zero positive integers n1 , . . . , nl such that: t0 = n1 H1 + · · · + nl Hl . (ii) Let (ai )1il and (bi )1il be families of scalars such that ai bi = ni for 1 i l, and: x = a1 Xβ1 + · · · + al Xβl , y = b1 X−β1 + · · · + bl X−βl . Then (x, t0 , y) is a S-triple of g. Proof. (i) By 18.11.5 and 20.6.10, (Hi )1il is a base of the root system (Hα )α∈R . Hence the existence of the ni ’s, and βi (t0 ) = 2 by 18.7.6. (ii) It follows from 18.7.4 (ii) that: [x, y] =
l i,j=1
ai bj [Xβi , X−βj ] =
l i=1
ai bi [Xβi , X−βi ] =
l
ni Hi = t0 .
i=1
Hence (x, t0 , y) is a S-triple of g. 32.5.5 Proposition. Suppose that g = {0} and let us conserve the notations of 32.5.4. (i) The element t0 is principal. (ii) The elements u of g contained in a S-triple (u, t0 , v) are the elements of the form β∈B xβ where xβ ∈ gβ \ {0} for β ∈ B. (iii) The Lie algebra g contains regular nilpotent elements. Proof. (i) By 32.5.4, β(t0 ) = 2 for β ∈ B. It follows that the eigenvalues of 0 0 ad t0 are even, and the centralizer of t β in g is h. So t is principal. (ii) Let u = β∈B xβ where xβ ∈ g \ {0} for β ∈ B. Then it is contained in a S-triple (u, t0 , v) by 32.5.4. Conversely, let (u, t0 , v) be a S-triple. Since [t0 , u] = 2u and [t0 , v] = −2v, we deduce that: −β β g , v∈ g . u∈ β∈B
Let u=
β∈B
xβ , v =
β∈B
β∈B
x−β , t0 =
β∈B
nβ Hβ ,
32.5 Regular and principal elements
491
with xβ ∈ gβ , x−β ∈ g−β and nβ ∈ N∗ for all β ∈ B (32.5.4). It follows from 18.7.4 (ii) that: nβ Hβ = t0 = [u, v] = [xα , x−β ] = [xβ , x−β ]. β∈B
α,β∈B
β∈B
0 for all β ∈ B. Hence [xβ , x−β ] = 0, and so xβ = (iii) This is clear by (i), (ii) and 32.5.3. 32.5.6 Theorem. Let g be a semisimple Lie algebra, G its adjoint group. (i) The set of simple principal elements of g is a G-orbit. (ii) The set of regular nilpotent elements of g is a G-orbit. Proof. (i) Let t be a simple principal element of g. Up to a conjugation by an element of G, we may assume that t ∈ h and β(t) ∈ {0, 1, 2} for all β ∈ B (32.4.3). Since t is regular, gt = h, so β(t) = 0 if β ∈ B. Also β(t) ∈ 2Z (t being principal). Thus, β(t) = 2 for all β ∈ B, so t = t0 . (ii) Let x, x be regular nilpotent elements and (x, t, y), (x , t , y ) S-triples of g. By 32.5.3, t and t are principal. So the result follows from part (i) and 32.3.1. 32.5.7 Remark. If g = {0}, then 32.5.6 implies that the characteristic of a regular nilpotent element of g is (2, 2, . . . , 2) with rk(g) factors.
References and comments • [14], [20], [24], [32], [48], [49], [80]. Many results on S-triples presented in this chapter are due to Kostant ([48] and [49]). We can find explicit tables for the characteristic of nilpotent elements of simple Lie algebras in [20], [24] and [32].
33 Polarizations
In this chapter, we introduce certain subalgebras of a Lie algebra which generalize Cartan subalgebras. One of the main results proved in this chapter is Richardson’s Theorem (33.5.6). Throughout this chapter, g will denote a finite-dimensional semisimple Lie algebra over k and L its Killing form. If a ⊂ g, a⊥ denotes the orthogonal of a with respect to L.
33.1 Definition of polarizations 33.1.1 Let a be a Lie algebra. Recall from 19.7.3 that if f ∈ a∗ , Φf denotes the alternating bilinear form on a defined by: Φf (x, y) = f ([x, y]). If b is an ideal of a, then b(f ) denotes the orthogonal of b with respect to Φf . 33.1.2 Let a be ad-algebraic (24.8.2) and A its adjoint group. Then A acts naturally on a and a∗ which correspond to the adjoint and coadjoint representations of a (23.4.9 and 23.5.5). Let f ∈ a∗ and A.f its orbit. By 24.3.6, ada a(f ) is the Lie algebra of the stabilizer of f in A. We deduce from 21.4.3 that: dim A.f = dim(ada a) − dim(ada a(f ) ). Since z(a) is in the kernel of the adjoint representation, it is clear that z(a) is contained in a(f ) . So: dim A.f = dim a − dim a(f ) . In particular, dim A.f is even (19.7.3). 33.1.3 Definition. Let f ∈ a∗ . (i) A Lie subalgebra b of a is said to be subordinate to f if it is a totally isotropic subspace with respect to Φf , that is f ([b, b]) = {0}.
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33 Polarizations
(ii) A polarization of a at f is a Lie subalgebra which is a maximal totally isotropic subspace with respect to Φf of a. In particular, it contains a(f ) , and a subordinate Lie subalgebra b for f is a polarization for f if and only if: 2 dim b = dim a + dim a(f ) . 33.1.4 Recall from 19.7.3 that a∗reg denotes the set of regular linear forms. We shall state without proof the following result which will not used in the sequel. Theorem. (Duflo’s Theorem) Let b be a solvable ideal of a Lie algebra a and f ∈ a∗ verifying dim a(f ) dim a(g) for all g ∈ a∗ such that g|b = f |b . Then there exists a solvable polarization of a at f . In particular, if f ∈ a∗reg , then there is a solvable polarization of a at f .
33.2 Polarizations in the semisimple case 33.2.1 Recall from 29.3.1 that when g is semisimple, the Killing homomorphism x → fx of g is an isomorphism, and g(fx ) = gx . We shall say that a Lie subalgebra a of g is subordinate to x ∈ g if L(x, [a, a]) = {0}. Similarly, a polarization of g at x is a Lie subalgebra p of g verifying: fx ([p, p]) = L(x, [p, p]) = {0} , 2 dim p = dim g + dim gx . We shall denote by Pol(x, g) or Pol(x) the set of polarizations of g at x, and by Polr (x, g) or Polr (x) the set of solvable polarizations of g at x. An element x ∈ g is polarizable if Pol(x) = ∅. Clearly, if p is subordinate to x, then p + gx is a totally isotropic subspace with respect to Φfx . So by definition, if p ∈ Pol(x), then p ⊃ gx . By 29.3.1, 29.4.5 and 29.4.8, we have the following result: Proposition. If x ∈ g, then Polr (x) = ∅ if and only if x ∈ greg . Moreover, if x ∈ greg , then there is a Borel subalgebra of g which is a polarization of g at x. 33.2.2 Lemma. Let n ∈ N∗ and (µ1 , . . . , µn ), (ν1 , . . . , νn ) ∈ kn be such that the µi ’s are pairwise distinct and if µi = 0, then νi = 0. There exists P ∈ k[T ] without constant term such that P (µi ) = νi for 1 i n. Proof. Replacing (µ1 , . . . , µn ), (ν1 , . . . , νn ) by (µ1 , . . . , µn , 0), (ν1 , . . . , νn , 0) if necessary, we may assume that one of the µi ’s is equal to 0. Then: P (T ) =
n ' T −µ ( j νi i=1 j=i µi − µj
is a polynomial which verifies P (µi ) = νi and P (0) = 0. 33.2.3 Lemma. Let V be a finite-dimensional vector space, X, Y subspaces of gl(V ) such that Y ⊂ X, and T = {t ∈ gl(V ); [t, X] ⊂ Y }. If z ∈ T verifies tr(zt) = 0 for all t ∈ T , then z is nilpotent.
33.2 Polarizations in the semisimple case
495
Proof. Let d = dim V and z = s + n the Jordan decomposition of x with s semisimple and n nilpotent. Let (e1 , . . . , ed ) be a basis of V such that s(ei ) = λi ei for 1 i d, and denote (gij )1i,jd the basis of gl(V ) such that gij (ek ) = δjk ei for 1 i, j, k d. Let U be the Q-subspace of k spanned by the λi ’s and f a Q-linear form on U . We shall prove that f = 0, for it would imply that U = {0} and therefore z = n is nilpotent. Let t ∈ gl(V ) be defined by t(ei ) = f (λi )ei , 1 i d. If 1 i, j d, then: (ad s)(gij ) = (λi − λj )gij , (ad t)(gij ) = (f (λi ) − f (λj ))gij . Note also that since f is Q-linear, we have: λi − λj = 0 ⇒ f (λi ) − f (λj ) = 0 λi − λj = λh − λl ⇒ f (λi ) − f (λj ) = f (λh ) − f (λl ). So by 33.2.2, there exists P ∈ k[T ] without constant term such that ad t = P (ad s). But ad s is a polynomial in ad z without constant term. Since (ad z)(X) ⊂ Y and Y ⊂ X, we have t ∈ T . As n is also a polynomial in z, we deduce that [t, n] = 0, so nt is nilpotent. It follows that: 0 = tr(zt) = tr(st) = λ1 f (λ1 ) + · · · + λd f (λd ). Hence:
2 2 0 = f tr(zt) = f (λ1 ) + · · · + f (λd ) .
Since f (λi ) ∈ Q and the λi ’s span U , we have f = 0. 33.2.4 Lemma. Let x ∈ g and p a Lie subalgebra of g. The following conditions are equivalent: (i) p ∈ Pol(x). (ii) [x, p] is the orthogonal of p with respect to L. If these conditions are verified, then [x, p] is an ideal of p. Proof. Let q = p⊥ . Then dim q = dim g − dim p. (i) ⇒ (ii) Suppose that p ∈ Pol(x). Then: {0} = L(x, [p, p]) = L([x, p], p) , dim q = dim p − dim gx . So [x, p] ⊂ q. Since p ∈ Pol(x), we have gx ⊂ p. Thus: dim[x, p] = dim p − dim gx = dim q. Hence [x, p] = q. (ii) ⇒ (i) Suppose that [x, p] = q. Since L([x, p], gx ) = L(p, [x, gx ]) = {0}, we deduce that gx ⊂ q⊥ = p. So: dim q = dim g − dim p = dim[x, p] = dim p − dim gx .
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33 Polarizations
Hence 2 dim p = dim g + dim gx . Moreover: {0} = L(q, p) = L([x, p], p) = L(x, [p, p]). So p ∈ Pol(x). Finally if conditions (i) and (ii) are verified, then: L([[x, p], p], p) = L([x, p], [p, p]) ⊂ L([x, p], p) = {0}. Thus [[x, p], p] ⊂ p⊥ = [x, p], and [x, p] is an ideal of p.
33.2.5 Lemma. Let x be a polarizable element of g and p ∈ Pol(x). Then all the elements of [x, p] are nilpotent. Proof. For x ∈ g, let σ(x) = adg x, and T the set of elements t ∈ gl(g) verifying [t, σ(p)] ⊂ σ([x, p]). Let z ∈ p and u ∈ T . Then by 33.2.4: tr uσ([x, z]) = tr [u, σ(x)]σ(z) ∈ tr σ([x, p])σ(z) ⊂ L([x, p], p) = {0}. So the result follows by 33.2.3. 33.2.6 Theorem. Let x be a polarizable element of g and p ∈ Pol(x). Then: (i) p is a parabolic subalgebra of g whose nilpotent radical is [x, p]. (ii) If x is nilpotent, then L(x, p) = {0}. Proof. (i) By 33.2.4 and 33.2.5, [x, p] is an ideal of p consisting of nilpotent elements. So there exists a Borel subalgebra b of g such that [x, p] ⊂ n where n denotes the nilpotent radical of b (29.4.8). Since n is the orthogonal of b with respect to L, we have b ⊂ [x, p]⊥ = p (33.2.4), and so p is a parabolic subalgebra. The second part of (i) follows from 33.2.4 and 20.8.6. (ii) In view of (i) and the results of 20.8, there exists a Cartan subalgebra h of g, a positive root system R+ of R(g, h), and R1 , R2 ⊂ R+ such that: α R + = R 1 ∪ R 2 , R1 ∩ R 2 = ∅ , p = h ⊕ g α∈S
where S = R+ ∪ (−R1 ). Let us write x = h + α∈S Xα with h ∈ h and Xα ∈ gα . Since L(x, [p, p]) = {0}, we have Xα = 0 for α ∈ ±R1 . If x is nilpotent, then h = 0 (20.8.3). Thus x belongs to the nilpotent radical of p, hence L(x, p) = {0} (20.8.6). 33.2.7 Corollary. Let x ∈ greg . Then Polr (x) = Pol(x) is the set of Borel subalgebras of g containing x. 33.2.8 Proposition. (i) Any semisimple element of g is polarizable. (ii) If p is a parabolic subalgebra of g, then there is a semisimple element x ∈ g such that p ∈ Pol(x).
33.3 A non-polarizable element
497
Proof. (i) If x is semisimple, then gx contains a Cartan subalgebra h of g (20.6.3). Let R = R(g, h) and R = {α ∈ R; α(x) = 0}. Then α g . gx = h ⊕ α∈R
Since R = R ∩ V where V = {λ ∈ h∗ ; λ(x) = 0}, it is a root system in V (18.2.5). Let B be a base of R , B a base of R containing B (18.7.9) and R+ the corresponding positive root system. Set S = R+ ∪ R and: α g . p=h⊕ α∈S
We obtain therefore that 2 dim p = dim g + dim gx , and p is a parabolic subalgebra of g. Moreover, in the notations of 20.7.1, we have: α g . [p, p] = hR ⊕ α∈S
We can now verify easily that L(x, [p, p]) = {0}, and hence p ∈ Pol(x). (ii) There is a Cartan subalgebra h of g and a base B of R(g, h) such that p = h ⊕ gP (J) where J ⊂ B (notations of 20.8.8). Let x ∈ h be such that α(x) = 0 if α ∈ J, and α(x) = 1 if α ∈ B \ J. Then it is clear that p ∈ Pol(x).
33.3 A non-polarizable element 33.3.1 In this section, g is assumed to be simple of rank l. Let h be a Cartan subalgebra of g, R = R(g, h), B a base of R and R+ , R− the corresponding system of positive and negative roots. Let us conserve the notation Hα of 20.6.5. Let θ be the highest root of R with respect to B and H = Hθ . Fix E ∈ gθ and F ∈ g−θ such that [E, F ] = H. Then (E, H, F ) is a S-triple. Set: g(i) where for i ∈ Z , g(i) = {x ∈ g; [H, x] = ix}. p= i∈N
By 20.8.7, p is a parabolic subalgebra of g and it is a polarization of g at H. In view of the results of 19.2, we have: dim gE = dim g(0) + dim g(1). If α ∈ R+ , then α(H) ∈ {0, 1, 2} (18.9.2). Thus g(i) = {0} if i 3. Let α ∈ R+ \ {θ} be such that α(H) > 0. Then α(H) = 1 (18.9.2). Hence: g(2) = kE , p = g0 + g(1) + kE. Since [H, E] = 2E, we deduce that p = gE ⊕ kH. 33.3.2 Let
498
33 Polarizations
π = {α ∈ B; α(H) > 0} , δ =
α.
α∈B
Since g is simple, we have δ ∈ R+ (18.13.8). We have, in the notation of 20.8.8, p = pJ where J = B \ π. Lemma. Suppose that l 2. (i) If α ∈ B, then α(H) ∈ {0, 1}. (ii) We have 0 < card(π) 2 and card(π) = 2 if and only if δ = θ, which is equivalent to the condition that g is of type Al . Proof. (i) This is clear since g(2) = kE, l = 1 and θ ∈ B. (ii) Since g(2) = {0}, π = ∅. Now if card(π) 3, then δ(H) 3 which is absurd. Suppose that π = {α, β}, with α = β. Then α(H) = β(H) = 1 by (i), so δ(H) = 2 and δ = θ. The converse is clear by (i). Finally by using the classification of irreducible root systems in 18.14, we deduce that δ = θ if and only if g is of type Al . 33.3.3 Lemma. With the above notations. (i) If E is polarizable, then p ⊂ q for all q ∈ Pol(E). (ii) If l 2 and card(π) = 1, then E is not polarizable. Proof. (i) For x, y ∈ g, let Φ(x, y) = L(E, [x, y]). We have L(E, H) = 0 and L(E, q) = {0} (33.2.6). So it follows easily that q + kH is a totally isotropic subspace with respect to Φ. But q ∈ Pol(E), so H ∈ q, and as gE ⊂ q, we deduce from 33.3.1 that p ⊂ q. (ii) Since p = pB\π , if card(π) = 1, then p is a maximal (proper) parabolic subalgebra of g. If E is polarizable, then (i) implies that Pol(E) = {p}. But p ∈ Pol(H), so dim g(0) = dim gH = dim gE = dim g(0) + dim g(1). Thus dim g(1) = 0 which is absurd because card(π) dim g(1) by 33.3.2. 33.3.4 The preceding result shows that if g is not of type Al , then it contains a non-polarizable nilpotent element. 33.3.5 When card(π) = 2, or equivalently, g is of type Al with l 2, it is easy to check that the elements of π are exactly the two terminal vertices α1 and αl . The parabolic subalgebras containing pB\π are exactly pB\{α1 } and pB\{αl } . They are both subordinate subalgebras of g at E, and they are both of codimension l in g. By 33.3.3 (i), they are the only polarizations of g at E.
33.4 Polarizable elements
499
33.4 Polarizable elements 33.4.1 Lemma. Let p be a parabolic subalgebra of g, n its nilpotent radical, h a semisimple element of p and s = [gh , gh ]. Then p ∩ s is a parabolic subalgebra of s with nilpotent radical n ∩ s. Proof. Let us fix a Cartan subalgebra h of p which contains h (29.5.5). Let R = R(g, h) and R = {α ∈ R; α(h) = 0}. Then in the notation of 20.7.1, α α gh = h ⊕ g , s = hR ⊕ g . α∈R
α∈R
There is a positive root system R+ of R and a partition R+ = R1 ∪ R2 such that α α g , n= g p=h⊕ α∈S R+
α∈R2
where S = R+ ∪ (−R1 ). Let = R ∩ R+ , Ri = R ∩ Ri , i = 1, 2, and S = R ∩ S. Then S = R+ ∪ (−R1 ) is a parabolic subset of R , and: α α p ∩ s = hR ⊕ g , n∩s= g . α∈S
Hence the result follows.
α∈R2
33.4.2 Let x ∈ g, and h, e be the semisimple and nilpotent components of x. The Lie algebra gh is reductive in g (20.5.13), and gh = g ⊕ c where c = z(gh ) and g = [gh , gh ]. Also, we have e ∈ g (20.5.14). Theorem. With the above notations, the following conditions are equivalent: (i) x is polarizable in g. (ii) e is polarizable in g . Proof. (i) ⇒ (ii) Let p ∈ Pol(x). Then p is a parabolic subalgebra of g with nilpotent radical n = [x, p] (33.2.6). Since h ∈ c ⊂ gh ∩ ge = gx ⊂ p, we have p ∩ gh = c + p ∩ g . So by 33.4.1, p∩g is a parabolic subalgebra of g with nilpotent radical n∩g , and e ∈ p∩g since e ∈ gx . Let ϕ = (ad x)|p and ψ = (ad h)|p . Then ψ is semisimple and ϕ◦ψ = ψ ◦ϕ. Since p = ker ψ ⊕ ψ(p), we have: ϕ(p) = ϕ(ker ψ) ⊕ ϕ(ψ(p)) = ϕ(ker ψ) ⊕ ψ(ϕ(p)). So ϕ(ker ψ) = ϕ(p) ∩ ker ψ, that is [x, p ∩ gh ] = gh ∩ [x, p] = gh ∩ n. Thus we obtain: [e, p ∩ g ] = [e, c + (p ∩ g )] = [e, p ∩ gh ] = [x, p ∩ gh ] = gh ∩ n. Since [e, p ∩ g ] ⊂ g , we deduce therefore that:
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33 Polarizations
[e, p ∩ g ] = gh ∩ n = g ∩ n. It follows from 33.2.4 that p ∩ g is a polarization of g at e. (ii) ⇒ (i) Let p ∈ Pol(e, g ). Then p is a parabolic subalgebra of g with nilpotent radical n = [e, p ], and ge ∩ g ⊂ p . Let q ∈ Pol(h, g) (33.2.8) with nilpotent radical m. We have: q = gh ⊕ m , [h, q] = m , g = gh ⊕ [h, g] , gh = g ⊕ c, so we may define:
p = p ⊕ c ⊕ m , n = n ⊕ m.
We shall prove that p is a polarization of g at x. 1) We have p + c ⊂ gh ⊂ q, so [p + c, m] ⊂ m and p is a Lie subalgebra of g. 2) Let h be a Cartan subalgebra of p . Then h = h ⊕ c is a Cartan subalgebra of gh , and hence of g. Let R = R(g, h) and R = {α ∈ R; α(h) = 0}. In the notations of 20.7.1, we have:
p = h ⊕ (g )P , q = h ⊕ gR ∪S , with P ∪ (−P ) = R , R ∪ S ∪ (−S) = R. We deduce therefore that p = h ⊕ gT where T ∪ (−T ) = R. Thus p is a parabolic subalgebra of g and it is easy to see that n is the nilpotent radical of p. 3) We obtain 2 dim p = 2(dim p + dim c + dim m) = dim g + dim(ge ∩ g ) + 2 dim c + dim g − dim gh = dim g + dim gx , because gh = g ⊕ c and gx = gh ∩ ge = (g ∩ ge ) ⊕ c. 4) Since e ∈ gh and gh ⊂ q, we have: [x, p] = [x, c + p + m] = [x, p ] + [h + e, m] ⊂ [x, p ] + m = [e, p ] + m = n + m = n. Hence: L(x, [p, p]) = L([x, p], p) ⊂ L(n, p) = {0}. We have therefore obtained our result. 33.4.3 Theorem 33.4.2 shows that in the search of polarizable elements of g, it is important to determine polarizable nilpotent elements. Let us give some sufficient conditions for a nilpotent element to be polarizable. Let (e, h, f ) be a S-triple of g and set: g(i) where for i ∈ Z , g(i) = {x ∈ g; [h, x] = ix}. p= i∈N
33.4 Polarizable elements
501
Then: g(0) = gh , L(e, g(i)) = {0} if i = −2, 2 dim p = dim g + dim g(0) , dim ge = dim g(0) + dim g(1). Lemma. (i) If g(2i + 1) = {0} for all i ∈ Z, then e is polarizable. (ii) If g is simple of type Al , then e is polarizable. Proof. (i) If g(2i + 1) = {0} for all i ∈ Z, then dim ge = dim g(0), and so p ∈ Pol(e). (ii) Let g be simple of type Al , h a Cartan subalgebra of g containing h, R = R(g, h) and B a base of R such that β(h) ∈ {0, 1, 2} for all β ∈ B (32.4.6). Let β1 , . . . , βl be the pairwise distinct elements of B such that βi +· · ·+βj ∈ R for 1 i j l (see 18.14), and (h1 , . . . , hl ) the basis of h dual to (β1 , . . . , βl ). We have h = λ1 h1 + · · · + λl hl with λ1 , . . . , λl ∈ {0, 1, 2}. Let r1 < r2 < · · · < rm be the indices i such that λi = 2. Set r0 = 0, rm+1 = l + 1 and for 0 j l, Ej = {k ∈ N; rj < k < rj+1 }. Thus λk ∈ {0, 1} for all k ∈ Ej . For 0 j m, let Ej1 = {s ∈ Ej ; λs = 1}. If s1 < s2 < · · · < spj are the elements of Ej1 in increasing order, we define: Hj+ =
i even
hsi , Hj− =
i odd
hsi .
Now let: + − − R+ j = {α ∈ R ; α(h) = α(Hj ) = 1} , Rj = {α ∈ R; α(h) = α(Hj ) = 1}, m m −α −α R+ = R+ g , R− = R− g . j , s+ = j , s− = j=0
α∈R+
j=0
α∈R−
Since g is of type Al , any positive root is of the form βi + · · · + βj with 1 i j l. It follows that if (βi + · · · + βj )(h) = 1, then there exist a unique k and a unique s ∈ Ek1 such that i, j ∈ Ek and i s j. We deduce easily from this that: g(−1) = s+ ⊕ s− , [s+ , s+ ] = [s− , s− ] = {0}, [g(0), s+ ] ⊂ s+ , [g(0), s− ] ⊂ s− . From this, we deduce that p ⊕ s+ and p ⊕ s− are Lie subalgebras of g such that L(e, p ⊕ s+ ) = L(e, p ⊕ s− ) = {0}. Thus: 2 dim s+ dim g(−1) , 2 dim s− dim g(−1). Consequently, 2 dim s+ = 2 dim s− = dim g(−1), and p ⊕ s+ , p ⊕ s− are polarizations of g at e. 33.4.4 Remark. Let g be simple of type Al . Then the proof of 33.4.3 says that: 1) dim g(−1) is even (we have a more general result in 32.1.9).
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33 Polarizations
2) If 1 appears in the characteristic (32.4.6) of e, then there exists at least two distinct polarizations of g at e. 33.4.5 Lemma. Let g = g1 × · · · × gn and x = x1 + · · · + xn with xi ∈ gi , for 1 i n. The following conditions are equivalent: (i) For 1 i n, xi is polarizable in g. (ii) x is polarizable in gi . If these conditions are verified, then any polarization of g at x is of the form p1 × · · · × pn where pi ∈ Pol(xi , gi ). Proof. Note that the restriction of L to gi × gi is the Killing form of gi . Moreover, it is clear that gx = gx1 1 × · · · × gxnn . (i) ⇒ (ii) If pi ∈ Pol(xi , gi ), then clearly p1 × · · · × pn ∈ Pol(x). (ii) ⇒ (i) Let p ∈ Pol(x). By 33.2.6, p is a parabolic subalgebra of g. It follows from 20.8.8 that p = p1 × · · · × pn where pi is a parabolic subalgebra of gi . Moreover, L(x, [p, p]) = {0}, so L(xi , [pi , pi ]) = {0} and pi is a subordinate Lie subalgebra of gi at xi , and 2 dim pi dim gi + dim gxi i . But: 2 dim p = 2
n
dim pi = dim g + dim gx =
i=1
n i=1
(dim gi + dim gxi i ).
Hence 2 dim pi = dim gi + dim gxi i , and pi ∈ Pol(xi , gi ). 33.4.6 Let g1 , . . . , gr be the minimal components of the semisimple Lie algebra g. We say that g is of type A if there exist l1 , . . . , lr ∈ N∗ such that gi is simple of type Ali for 1 i r. Proposition. Let g be of type A, then all the elements of g are polarizable. Proof. Let x ∈ g. If x is nilpotent, then it is polarizable by 33.4.3 and 33.4.5. Otherwise, let h, e be the semisimple and nilpotent components of x. We have h = 0 and e ∈ g = [gh , gh ]. Since g is also of type A, the result follows from 33.4.2.
33.5 Richardson’s Theorem 33.5.1 Let G = Aute g and Ng the set of nilpotent elements of g. One of the goals of this section is to prove that if p is a parabolic subalgebra of g, then p ∈ Pol(x) for some x ∈ Ng . This is much harder to prove than part (ii) of 33.2.8. 33.5.2 Lemma. Let x ∈ g. (i) We have x ∈ [g, x] if and only if x ∈ Ng . (ii) If x ∈ Ng , then dim G.(kx) = 1 + dim G(x). Proof. (i) We may assume that x = 0. Then the result follows from 32.1.5 and 19.8.2.
33.5 Richardson’s Theorem
503
(ii) Let k∗ = k \ {0}. The map G × k∗ × g → g, (α, λ, y) → λα(y), defines an action of the algebraic group H = G × k∗ on g. The H-orbit of x is G(k∗ x) and we have G(k∗ x) = G(kx). Denote by π : H → g, the orbit map (α, λ) → λα(x). By 25.1.3, dπ(e,1) is a surjection from L(H) = (ad g)×k onto Te G(k∗ x) . But π is the composition of H → End(g) , (α, λ) → λα , and End(g) → g , u → u(x). We deduce therefore that for µ ∈ k and y ∈ g: dπ(e,1) (ad y, µ) = µx + [y, x]. The image of dπ(e,1) is therefore kx + [g, x]. So the result follows from (i). 33.5.3 Proposition. Let z ∈ g. There exists x ∈ G(kz) such that: x ∈ Ng and dim G(x) = dim G(z). Proof. We may assume that z is not nilpotent. For x ∈ g, let χx denote the characteristic polynomial of ad x. We have χx (T ) = T n + fn−1 (x)T n−1 + · · · + f1 (x)T + f0 (x), where n = dim g and f0 , . . . , fn−1 are G-invariant polynomial functions on g. Since z is not nilpotent, there exist r ∈ {0, 1, . . . , n − 1} and ar , . . . , an−1 ∈ k such that ar = 0 and: χz (T ) = T n + an−1 T n−1 + · · · + ar T r . If λ ∈ k and α ∈ G, then: χα(λz) (T ) = T n + an−1 λT n−1 + · · · + ar λn−r T r . We deduce therefore that if x ∈ G(kz), then: f0 (x) = · · · = fr−1 (x) = 0, (1) [fk (x)]n−r = an−r [fr (x)]n−k , r k n − 1. an−k r k The relations in (1) are clearly verified if x ∈ G(kz). But x is nilpotent if and only if χx (T ) = T n . So: G(kz) ∩ Ng = {x ∈ G(kz) ; fr (x) = 0}. Since z ∈ / G(kz) ∩ Ng and 0 ∈ G(kz) ∩ Ng , it follows from 14.2.3 and 33.5.2 that any irreducible component Z of G(kz) ∩ Ng verifies: dim Z = dim G(kz) − 1 = dim G(z). The set Z is closed and G-stable. We have Z = G(x1 ) ∪ · · · ∪ G(xs ), with x1 , . . . , xs ∈ G(kz) ∩ Ng , because the nilpotent orbits are irreducible and the number of nilpotent orbits is finite (32.4.4). As Z is irreducible, Z = G(xi ) for some i and dim G(xi ) = dim G(z).
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33 Polarizations
33.5.4 Let p be a parabolic subalgebra of g. By 29.4.3 (iv), adg p is the Lie algebra of a parabolic subgroup P of G. In view of 28.4.2, 29.4.3 (iii) and 24.3.6, P is the identity component of the normalizer of p in G. Let m be the nilpotent radical of p, d = dim p and Gr(g, d) the Grassmannian variety of d-dimensional subspaces of g. Then G acts naturally on Gr(g, d). Let P be the G-orbit of p. Then by 21.4.3, we obtain that: dim P = dim G − dim P = dim g − dim p = dim m. Proposition. Let z ∈ g be a polarizable element and p ∈ Pol(z). There exists x ∈ G(kz) verifying the following conditions: (i) x is nilpotent and dim G(x) = dim G(z). (ii) x is polarizable and p ∈ Pol(x). Proof. We may assume that z is not nilpotent. Let d = dim p, P the G-orbit of p in Gr(g, d) and p : Gr(g, d) × g → g the canonical projection on g. Recall from 21.4.3 and 21.4.5 that P is open in P and P \ P is a union of G-orbits of dimension strictly less than dim P. Since p is a Lie subalgebra, the elements of P are Lie subalgebras of g (19.7.2). Denote by L the set of elements (q, y) ∈ P × G(kz) verifying: L(y, [q, q]) = {0}. Then L is a closed subset of P × G(kz) containing (p, z), and p(L) contains G(kz). As Gr(g, d) is a complete variety (13.6.1 and 13.4.5), p(L) contains G.(kz). Thus if y ∈ G.(kz), then there exists a d-dimensional Lie subalgebra q of g such that q ∈ P and L(y, [q, q]) = {0}. By 33.5.3, there exists x ∈ G.(kz) verifying (i). Let q ∈ P be such that L(x, [q, q]) = {0}. Then: dim q = dim p =
1 1 (dim g + dim gz ) = (dim g + dim gx ). 2 2
So q ∈ Pol(x). Let n be the nilpotent radical of q. We have [x, q] = n by 33.2.6 and gx ⊂ q. Hence: dim n = dim q − dim gx = dim p − dim gz = dim m. Thus the G-orbits of p and q in Gr(g, d) have the same dimension. It follows that q ∈ P, and so there exists α ∈ G such that q = α(p). The element α−1 (x) verifies therefore the required conditions. 33.5.5 Corollary. If z ∈ g is polarizable, then there exists a polarizable nilpotent element x ∈ g such that dim G(x) = dim G(z). 33.5.6 Theorem. (Richardson’s Theorem) Let g be a semisimple Lie algebra, G its adjoint group, p a parabolic subalgebra of g with nilpotent radical m, and P a parabolic subgroup of G such that L(P ) = adg p.
33.5 Richardson’s Theorem
505
(i) There exists a unique nilpotent G-orbit O such that O ∩ m is a dense open subset of m. (ii) The set O ∩ m is a P -orbit. (iii) If x ∈ O ∩ m, then [x, p] = m and p ∈ Pol(x). Proof. By 32.4.4, there exist nilpotent orbits O1 , . . . , Or such that m = (O1 ∩ m) ∪ · · · ∪ (Or ∩ m). Hence: m = O1 ∩ m ∪ · · · ∪ Or ∩ m. Since m is irreducible, there is a nilpotent orbit O such that m = O ∩ m. So m ⊂ O, and as an orbit is open in its closure, O ∩ m is a dense open subset of m. It is now clear that O is the unique G-orbit U such that U ∩ m is a dense open subset of m. Let s = inf{dim gx ; x ∈ m} , M = {x ∈ m ; dim gx = s}. The set M is a non-empty open subset of m. So it has a non-empty intersection with O, hence dim O = dim g − s, and O ∩ m ⊂ M. Conversely, if y ∈ M, then dim Oy = dim O, y ∈ O, so y ∈ O (21.4.5). Thus M = O ∩ m. There is a Cartan subalgebra h of g such that h ⊂ p. Let R = R(g, h), B a base of R and R+ the corresponding set of positive roots. Since p is parabolic, for a suitable choice of B, there is a partition R+ = R1 ∪ R2 of R+ such that: α gα , m = g . p=h⊕ α∈R+ ∪(−R1 )
α∈R2
If h = {h ∈ h; α(h) = 0 for all α ∈ R1 }, then the radical r of p is r = h ⊕ m and gα l=h⊕ α∈R1 ∪(−R1 )
is a Levi factor of p. Let z ∈ h be such that α(z) = 0 if α ∈ R1 and α(z) = 0 if α ∈ R2 . We have gz = l, and since z ∈ r, p is subordinate to z (19.6.3). We deduce therefore that p ∈ Pol(z). It follows from 33.5.4 that there exists x ∈ Ng such that dim gx = dim gz and p ∈ Pol(x). But L(x, p) = {0} (33.2.6), so x ∈ m (20.8.6). Now, as p ∈ Pol(x), we have gx ⊂ p. So dim gx = dim gz = dim l implies that: dim P (x) = dim[x, p] = dim p − dim gx = dim p − dim l = dim m. Since (adg p)(m) ⊂ m, we have P (m) = m. So the orbit P (x) is a dense open subset of m, and hence P (x) ∩ O = ∅. Consequently, s = dim l and dim O = 2 dim m.
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33 Polarizations
If y ∈ O ∩ m, the same arguments show that p ∈ Pol(y), and again we obtain that P (y) is a dense open subset of m. It follows that P (y) = P (x). Thus P (x) = O ∩ m. 33.5.7 Corollary. Let p be a parabolic subalgebra of g and m its nilpotent radical. (i) There exists x ∈ m such that p ∈ Pol(x). (ii) The set G(m) is irreducible and its dimension is 2 dim m. 33.5.8 Remarks. 1) The proof of 33.5.6 shows that O is the unique orbit U verifying dim U = 2 dim m and U ∩ m = ∅. 2) An element x verifying the conditions of 33.5.6 is called a Richardson element of p, and the corresponding orbit O is called the Richardson orbit associated to p. 3) Let x ∈ O ∩ m and: −α g . m− = α∈R2
Then: [x, g] = [x, p] + [x, m− ] = m + [x, m− ]. Since dim[x, g] = 2 dim m, it follows that the restriction of ad x to m− is injective and m ∩ [x, m− ] = {0} because [x, m− ] ⊂ l.
References and comments • [27], [28], [29], [31], [70], [80]. Polarizations are very useful for the determination of primitive ideals of the enveloping algebra. In the case of a solvable Lie algebra, they provide a way to determine all primitive ideals of the enveloping algebra via the orbit method of Kirillov-Dixmier ([29]). Duflo’s Theorem stated in 33.1.4 is proved in [29] and [31]. Richardson’s Theorem (33.5.6) is fundamental in the study of semisimple Lie algebras. The result formulated in this chapter is in the Lie algebra setting. For other formulations, see [70].
34 Results on orbits
This chapter is devoted to the description of properties of adjoint orbits in a semisimple Lie algebra.
34.1 Notations 34.1.1 In this chapter, g is a semisimple Lie algebra of rank l, L is the Killing form of g and G is the adjoint group of g. If a is a subspace of g, then a⊥ denotes the orthogonal of a with respect to L. Denote by Sg and Ng respectively the set of semisimple and nilpotent elements of g. Recall also that ggen and greg denote the set of generic and regular elements of g. Let us fix a Cartan subalgebra h of g. Let R = R(g, h), W = W (g, h), B a base of R and R+ (resp. R− ) the corresponding set of positive (resp. negative) roots. Let α g , b = h ⊕ n. n= α∈R+
A S-triple (e, h, f ) is called standard (with respect to h and B) if h ∈ h and β(h) ∈ {0, 1, 2} for all β ∈ B. By 32.3.2 and 32.4.3, we obtain that if u is a non-zero nilpotent element of g, then there exist θ ∈ G and t, v ∈ g such that (θ(u), t, v) is a standard S-triple. 34.1.2 We shall denote the G-orbit of an element x ∈ g by Ox , and we say that Ox is nilpotent (resp. semisimple, polarizable) if x is nilpotent (resp. semisimple, polarizable). Note that dim Ox is even since dim Ox = dim g−dim gx (29.3.1 and 19.7.3). Also, by 25.1.3, the tangent space of Ox at x is [x, g].
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34 Results on orbits
34.2 Some lemmas 34.2.1 Lemma. Let x, y ∈ h be such that y ∈ W (x). Then there exists f ∈ S(h∗ )W such that f (x) = f (y). Proof. If w ∈ W , then there exists gw ∈ S(h∗ ) such that gw (y) = 1 and gw w(x) = 0. Set: g =1− gw , f = g◦w. w∈W
w∈W
Then g(y) = 0 and g(z) = 1 if z ∈ W (x). Hence f (y) = 0 and f (x) = 1. Finally, it is clear that f ∈ S(h∗ )W . 34.2.2 Lemma. Let x, y ∈ Sg be such that f (x) = f (y) for all f ∈ S(g∗ )g . Then there exists θ ∈ G such that y = θ(x). Proof. By 29.2.3 and 31.2.4, we may assume that x, y ∈ h. Then f (x) = f (y) for all f ∈ S(h∗ )W (31.2.6). So the result follows by 34.2.1 and 29.1.3. 34.2.3 Lemma. Let x ∈ g and s, n its semisimple and nilpotent components. Then f (x) = f (s) for all f ∈ S(g∗ )g . Proof. Let (V, σ) be a finite-dimensional representation of g. Then p p p σ(x) = σ(s)k σ(n)p−k . k k=0 for p ∈ N∗ . But σ(s)k σ(n)p−k is nilpotent for 0 k < p. Thus tr[σ(x)p ] = tr[σ(s)p ], and the result follows by 31.2.6. 34.2.4 Lemma. Let x, y ∈ g. The following conditions are equivalent: (i) The semisimple components of x and y are G-conjugate. (ii) We have f (x) = f (y) for all f ∈ S(g∗ )g . Proof. This is clear by 34.2.2 and 34.2.3. 34.2.5 Lemma. Let us conserve the notations x, s, n of 34.2.3. (i) If n = 0, then dim gx < dim gs . (ii) If λ ∈ k \ {0}, then there exists θλ ∈ G such that θλ (x) = s + λ2 n. (iii) We have s ∈ Ox . (iv) If n = 0, then x ∈ Os . Proof. (i) We have gx = gs ∩ gn , gs is reductive in g and n ∈ a = [gs , gs ]. Thus gx = z(gs ) ⊕ an . If n = 0, then an = a. Hence gx = gs . (ii) We may assume that n = 0. Let a be as in the proof of (i). Then by 32.1.5, there exist h, m ∈ a such that (n, h, m) is a S-triple. For λ ∈ k \ {0}, let:
34.3 Generalities on orbits
509
γλ = exp(ad λ−1 n) exp(− ad λm) exp(ad λ−1 n) and θλ = γ1 ◦γλ . Then θλ (n) = λ2 n. Since m, n ∈ a, we have θλ (s) = s. Hence θλ (x) = s + λ2 n. (iii) By (ii), we have s + µn ⊂ Ox if µ ∈ k \ {0}, so s + kn ⊂ Ox . (iv) This follows from (i) and 34.1.2. 34.2.6 Lemma. Let x, y ∈ g be such that y ∈ Ox . Then: (i) dim gx dim gy , or equivalently, dim[y, g] dim[x, g]. (ii) dim(ad y)2 (g) dim(ad x)2 (g). Proof. (i) Since dim Ox = dim g − dim gx , this is clear by 21.4.5. (ii) For z ∈ g, denote by M (z) the matrix of (ad z)2 with respect to some basis of g and ρ(z) = dim(ad z)2 (g) its rank. Then (ii) follows from the fact that the map z → ρ(z) is lower semi-continuous, and constant on Ox .
34.3 Generalities on orbits 34.3.1 Proposition. If g = {0}, then it contains an infinite number of semisimple G-orbits. Proof. For x ∈ g, let V (x) be the set of eigenvalues of ad x. If u, v ∈ g verify Ou = Ov , then V (u) = V (v). So if x is a non-zero semisimple element of g, then there is a strictly increasing sequence of integers (np )p0 such that V (np x) = V (nq x) if p = q. 34.3.2 Proposition. The following conditions are equivalent for x ∈ g: (i) x is semisimple. (ii) Ox is closed. Proof. (i) ⇒ (ii) Let y ∈ Ox and y = ys + yn its Jordan decomposition. Then 34.2.5 (iii) implies that ys ∈ Oy ⊂ Ox . Now 34.2.4 says that if f ∈ S(g∗ )g , then f is constant on Ox , and hence on Ox . We deduce therefore that f (x) = f (ys ). Thus 34.2.4 implies that ys ∈ Ox , and so y ∈ Ox = Oys . It follows from 34.2.5 (iv) that y = ys . Hence y ∈ Ox , and Ox is closed. (ii) ⇒ (i) This is clear by 34.2.5 (iii). 34.3.3 Let us make a few remarks on nilpotent elements. 1) Let x ∈ Ng and λ ∈ k \ {0}. Then by 34.2.5 (ii), λx ∈ Ox . 2) If (e, h, f ) is a S-triple in g, then Oe = Of by point 1 and the computation in the proof of 32.2.7. 3) It follows from the first remark that Ng is a cone. It is closed in g since: Ng = {x ∈ g; (ad x)dim g = 0}. Let J+ be the set of elements of S(g∗ )g having no constant term. By 19.2.3 and 20.4.4, x ∈ Ng is equivalent to the condition tr(σ(x)n ) = 0 for all n ∈ N∗
510
34 Results on orbits
and all finite-dimensional representation σ of g. In view of 31.2.6, we have that Ng = V(S(g∗ )J+ ) (notations of 11.1.3). 4) By 29.4.7 and 29.4.8, any nilpotent element is conjugate to an element of n, thus Ng = G.n. 5) The set of regular nilpotent elements of g is a G-orbit (32.5.6) whose characteristic is (2, . . . , 2). We shall denote this orbit by Oreg , and it is called the regular nilpotent orbit of g. 34.3.4 Proposition. We have Ng = Oreg . In particular, Ng is irreducible. Proof. Since n ∩ greg is a dense open subset of n, we have n ⊂ Oreg . So the result follows from remark 4 of 34.3.3. 34.3.5 Proposition. The following conditions are equivalent for x ∈ g: (i) x is nilpotent. (ii) 0 ∈ Ox . Proof. The implication (i) ⇒ (ii) follows from 34.2.5 (ii). If (ii) is verified, then f (x) = f (0) for all f ∈ S(g∗ )g . In particular, tr(ad x)n = 0 for all n ∈ N∗ (31.2.3). So x is nilpotent (19.2.3). 34.3.6 Proposition. Let x ∈ g. Then: (i) Ox is a union of a finite number of G-orbits. (ii) Ox contains a unique closed G-orbit. It is the G-orbit of the semisimple component of x. Proof. (i) By 21.4.5, Ox is a union of G-orbits. If x is semisimple, then Ox = Ox is a single G-orbit (34.3.2). If x is nilpotent, then Ox ⊂ Ng (34.3.4) is a union of a finite number of nilpotent G-orbits (32.4.4). So let us assume that x = s+n where s, n are the semisimple and nilpotent components of x which are both non-zero. Let F = {y ∈ g; f (x) = f (y) for all f ∈ S(g∗ )g }. It is a closed subset of g which contains Ox (31.2.4), so Ox ⊂ F . Let y ∈ g, and u, v its semisimple and nilpotent components. By 34.2.2 and 34.2.3, y ∈ F if and only if u ∈ Os . We deduce therefore that if y ∈ F , then there exists θ ∈ G such that y = θ(s + w) where w is nilpotent and [s, w] = 0. Moreover, w belongs to the semisimple Lie algebra s = [gs , gs ]. Let v1 , . . . , vr be representatives of the set of nilpotent (Aute s)-orbits in s. Then there exist i ∈ {1, 2, . . . , r} and γ ∈ Aute s such that y = θ(s + γ(vi )) = θ ◦ γ(s + vi ). Hence F contains at most 3m G-orbits where m is the rank of s (32.4.4). (ii) The result is clear if x is semisimple (34.3.2), and if x is nilpotent, {0} is the unique closed orbit in Ox (34.3.5). So let us assume that x = s+n where s, n are the semisimple and nilpotent components of x which are both non-zero. By 34.2.5 and 34.3.2, Os is a closed orbit in Ox . Moreover, we saw in the proof of (i) that the orbits contained in Ox are of the form Oy where y = s + v with v nilpotent and [v, s] = 0. By 34.3.2, such an orbit is closed if and only if v = 0. So we are done.
34.4 Minimal nilpotent orbit
511
34.4 Minimal nilpotent orbit 34.4.1 Lemma. Let (e, h, f ) be a S-triple of g and s = ke + kh + kf . The following conditions are equivalent: (i) The image of (ad e)2 is ke. (ii) The s-module g/s is the sum of simple modules of dimension 1 or 2. (iii) The only eigenvalues of ad h distinct from −1, 0 and 1 are −2 and 2, and they are both of multiplicity 1. Proof. By 19.2.6, we have a decomposition g = s ⊕ a1 ⊕ · · · ⊕ ap of g into simple s-submodules. So the lemma follows from the classification of simple finite-dimensional representations of sl2 (k) in 19.2. 34.4.2 In the rest of this section, g is assumed to be simple. Denote by θ the highest root of R and H the unique element h of [gθ , g−θ ] such that θ(h) = 2. Fix also E ∈ gθ and F ∈ g−θ verifying [E, F ] = H. Lemma. Let (e, h, f ) be a standard S-triple of g. The following conditions are equivalent: (i) (e, h, f ) verifies the conditions of 34.4.1. (ii) h = H. If these conditions are verified, then e ∈ gθ and f ∈ g−θ . Proof. Since (E, H, F ) is a standard S-triple verifying the conditions of 34.4.1, we have (ii) ⇒ (i) by 32.3.1 (ii). Conversely, suppose that (i) is verified. If l = 1, then the result is clear. Suppose that l 2. As (e, h, f ) is standard, we have α(h) ∈ {0, 1, 2} for all α ∈ B. So α(h) θ(h) for all α ∈ R+ and we deduce from condition (iii) of 34.4.1 that θ(h) = 2 and α(h) ∈ {0, 1} if α ∈ R+ \ {θ}. Hence e ∈ gθ , f ∈ g−θ and since [e, f ] = h ∈ [gθ , g−θ ], we have h = H. 34.4.3 Lemma. (i) The S-triples verifying the conditions of 34.4.2 are G-conjugate. (ii) The set of non-zero nilpotent elements of g verifying the conditions of 34.4.1 is the G-orbit OE . We have OE = {0} ∪ OE . We shall denote OE also by Omin , and Omin is called the (non-zero) minimal nilpotent orbit of g. Proof. (i) Any S-triple is G-conjugate to a standard S-triple (32.4.3). So the result follows by 34.4.2 and 32.3.1. (ii) The first part is clear by (i) and 32.3.1. Now if y ∈ OE , then dim im(ad y)2 1 (34.2.6). So it follows that OE ⊂ {0} ∪ OE . The reverse inclusion is clear by 34.3.5. 34.4.4 Let us define n(g) = min{dim Ox ; x ∈ g \ {0}} , E = {x ∈ g ; dim Ox = n(g)}, n1 (g) = min{dim Ox ; x ∈ Ng \ {0}} , n2 (g) = min{dim Ox ; x ∈ Sg \ {0}}, and l(g) the minimal codimension of parabolic subalgebras of g. Lemma. (i) An element of E is either nilpotent or semisimple. (ii) We have n(g) = n1 (g) 2l(g) = n2 (g).
512
34 Results on orbits
Proof. (i) This is clear by 34.2.5 (i). (ii) By 33.2.8, we have 2l(g) = n2 (g). Let p be a parabolic subalgebra of g and x a non-zero element of its nilpotent radical. Then L(x, p) = {0} (20.8.6), so p is a Lie subalgebra of g subordinate to x. Thus 2 dim p dim g + dim gx . Hence n1 (g) 2l(g). Finally (i) implies that n(g) = n1 (g). 34.4.5 Let n ∈ Ng \ {0} and (n, s, m) a S-triple of g. Up to a conjugation by G, we may assume that it is standard, that is, n ∈ n, s ∈ h and for all α ∈ B, α(s) ∈ {0, 1, 2} (32.4.3). Then θ(s) α(s) for all α ∈ R+ . For p ∈ Z, let gs (p) = ker(ad s − p idg ) and ps = gs (p) , g2s = gs (p). p0
p2
Then gθ ⊂ g2s and ps ∈ Pol(s), so ps is a parabolic subalgebra of g (33.2.6). Recall that in view of 19.2, we have dim gn = dim gs (0) + dim gs (1), and the map ps → g2s , x → [x, n], is surjective. 34.4.6 Proposition. The orbit Omin is the unique nilpotent G-orbit of dimension n(g). For any n ∈ Ng \ {0}, we have Omin ⊂ On . Proof. Let n ∈ Ng \ {0}. Up to a conjugation by G, we may assume that we are in the situation of 34.4.5. Let Ps be the parabolic subgroup of G whose Lie algebra is adg ps (29.4.3). Then by 24.3.3 (ii), we have Ps (n) ⊂ g2s . We deduce therefore a morphism Ps → g2s , σ → σ(n), whose differential is the surjective map adg ps → g2s , x → [x, n]. Hence g2s ⊂ Ps (n) ⊂ On (16.5.7). Since E ∈ g2s , we deduce that Omin = OE ⊂ On . Finally, as n(g) = n1 (g) (34.4.4), the result follows. 34.4.7 Let us consider the case where (n, s, m) = (E, H, F ) and let us use the notations π, δ of 33.3.2. Suppose that card(π) = 1 and l 2. We have n(g) 2l(g) (34.4.4). If n(g) = 2l(g), then any parabolic subalgebra of codimension l(g) in g which contains b belongs to Pol(E). But this is absurd by 33.3.3. Hence n(g) < 2l(g). We deduce from this and from the results of 33.3 the following result: Proposition. If g is a simple Lie algebra which is not of type Al , then there is a unique non-zero G-orbit of minimal dimension. This orbit is nilpotent and it is contained in the closure of any non-zero nilpotent orbit of g.
34.5 Subregular nilpotent orbit
513
34.5 Subregular nilpotent orbit 34.5.1 In this section, g is assumed to be simple of dimension n. Definition. (i) An element x ∈ g is subregular if dim gx = l + 2, or equivalently, dim Ox = n − l − 2. (ii) A G-orbit Ox of g is subregular if x is subregular. 34.5.2 Let α ∈ B. Define h ∈ h by α(h) = 0 and β(h) = 1 if β ∈ B \ {α}. Then it is clear that h is subregular. Thus, subregular semisimple orbits exist. The goal of this section is to prove that there is a unique subregular nilpotent orbit in g. 34.5.3 For α ∈ B, we set: pα = b + kg−α , nα =
gβ .
β∈R+ \{α}
Thus pα is a parabolic subalgebra of g whose nilpotent radical is nα . Similarly, let α, β be distinct elements of B, S = (Zα + Zβ) ∩ R and: γ g , nαβ = gγ . pαβ = b + γ∈R+ \S
γ∈S
Then pαβ is a parabolic subalgebra of g whose nilpotent radical is nαβ . Let x be a Richardson element of pα (33.5.8). We have x ∈ nα , [x, pα ] = nα , and pα ∈ Pol(x) (33.5.6). Thus: dim gx = 2 dim pα − dim g = l + 2. It follows that x is subregular, so there is a subregular nilpotent orbit in g. 34.5.4 Let (x, h, y) be a standard S-triple such that x is subregular. If p ∈ Z, let gh (p) = ker(ad h − p idg ). For α ∈ R, let Xα be a non-zero element of gα . We have: dim gh l , dim gx = dim gh + dim gh (1) = l + 2. So dim gh (1) ∈ {0, 2} since dim g − dim gu is even for all u ∈ g (34.1.2). If dim gh (1) = 0, then dim gx = dim gh , and gh (i) = {0} if i is odd. As dim gh = l + 2, gh = h + kXα + kX−α for some α ∈ B. Thus gh (p) pα = p0
and pα is a polarization of g at h. If dim gh (1) = 2, then h is regular, so α(h) ∈ {1, 2} for all α ∈ B. So there exist α, β ∈ B distinct such that α(h) = β(h) = 1 and γ(h) = 0 for γ ∈ B \ {α, β}. We have x ∈ nα , and pα ∈ Pol(x). If α + β ∈ R, then x ∈ nαβ and pαβ is a Lie subalgebra of g subordinate to x. We deduce then that:
514
34 Results on orbits
2 + dim b dim pαβ
1 (dim g + dim gx ) = 1 + dim b. 2
Contradiction. So α + β ∈ R. 34.5.5 It follows from 34.5.4 that any subregular nilpotent element x is polarizable. Since a polarization is a parabolic subalgebra of g (33.2.6), by a dimension argument, we obtain that if p is a polarization of g at x, then p is G-conjugate to pα for some α ∈ B. Note that x is a Richardson element in p. 34.5.6 Let Γ denote the Dynkin diagram of R and we shall identify B with the set of vertices of Γ . Lemma. Let α, β ∈ B be linked in Γ (see 18.12.2). Then pα and pβ have a common Richardson element. Proof. Let S = (Zα + Zβ) ∩ R. Then S is a root system of rank 2 in the subspace of h∗ spanned by α and β. Let T = R+ \ S and : γ a= g . γ∈T
For λ ∈ k, eλ ad Xα and eλ ad X−α leave a stable. 1) Suppose that S is of type G2 . Then by the classification given in 18.14, we have R = S. So B = {α, β}, and we may assume that α is a short root. Let x = Xα+β + X2α+β , and h ∈ h be defined by α(h) = 0, β(h) = 2. Then: [h, x] = 2x , gh = h + kXα + kX−α . Let u ∈ h, λ, µ ∈ k be such that t = u + λXα + µX−α verifies [t, x] = 0. Since the coefficient of X3α+β (resp. Xβ ) in [t, x] is zero, we obtain λ = µ = 0. Hence u = 0. Thus gh ∩ gx = {0}. By 32.1.8, there is a S-triple of the form (x, h, y). As gh (1) = {0}, we have dim gx = dim gh = l + 2, and x is subregular. Note that x ∈ nα ∩ nβ , and: [x, pα ] = nα , [x, pβ ] = nβ . We have therefore proved that x is a Richardson element for both pα and pβ . In the rest of the proof, x is a Richardson element of pα . 2) Suppose that S is of type A2 , and x ∈ nβ . Then x = aXβ + bXα+β + u, with u ∈ a, a, b ∈ k and a = 0. By replacing x by ead Xα (x), we may assume that b = 0. Then, for a suitable λ ∈ k, y = eλ ad X−α (x) = bXα+β + v,
34.5 Subregular nilpotent orbit
515
where v ∈ a. Thus y ∈ nα ∩ nβ . Since y is a Richardson element for pα and dim pα = dim pβ , y is also a Richardson element for pβ . / nβ . Then 3) Suppose that S is of type B2 , α is long and x ∈ x = aXβ + bXα+β + cXα+2β + u, with u ∈ a, a, b, c ∈ k and a = 0. Again we may assume that b = 0, and for a suitable λ ∈ k, y = eλ ad X−α (x) = bXα+β + cXα+2β + v, where v ∈ a. So y is the element with the required properties. / nβ . Then 4) Suppose that S is of type B2 , α is short and x ∈ x = aXβ + bXα+β + cX2α+β + u, where u ∈ a, a, b, c ∈ k and a = 0. Again we may assume b = 0. Let λ ∈ k. Then eλ ad X−α (aXβ + bXα+β + cX2α+β ) is of the form (a + bλµ + λ2 cν)Xβ + dXα+β + cX2α+β , where d ∈ k and µ, ν ∈ k \ {0} do not depend on λ. So there is a λ such that y = eλ ad X−α (x) ∈ nβ . Thus y is a Richardson element for pα and pβ . 34.5.7 Proposition. (i) There is a unique subregular nilpotent orbit O in g.
(ii) We have O = Ng , the set of non-regular nilpotent elements of g.
Proof. (i) We saw in 34.5.3 that there are subregular nilpotent elements. Let x, y be subregular nilpotent elements. Up to a conjugation by G, we may assume that there exist α, β ∈ B such that pα and pβ are polarizations of g at x and at y (34.5.5). The Dynkin diagram of g is connected and Richardson elements in a parabolic subalgebra are conjugate (33.5.6), so by repeated applications of 34.5.6, we obtain that Ox = Oy . (ii) Let (x, h, y) a standard S-triple with x non-regular. Then there is an α ∈ B such that α(h) = 2 (32.5.7). We have x ∈ nα . If z ∈ nα , then L(z, pα ) = {0} (20.8.6). So pα is subordinate to z, and hence z is not regular. So x ∈ nα ⊂ Ng . By (i), there is a subregular element in nα . So it follows that the set of subregular elements of g in nα is the set of elements z ∈ nα verifying gz gt for all t ∈ nα , which is a non-empty open subset of nα . We deduce again from (i) that nα ⊂ O. In particular x ∈ nα ⊂ O, and hence Ng ⊂ O. Finally, if z ∈ O, then dim Oz dim O, so z is not a regular element and z ∈ Ng . 34.5.8 We list below, for the different types of simple Lie algebra, the characteristic of the unique subregular nilpotent orbit.
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34 Results on orbits
• Type A2r+1 , r 0: 2
2
2
0
2
2
2
• Type A2r , r 1: 2
1
2
2
1
2
2
2
• Type Bl , l 2: 2
2
2
2
2
0
2
2
2
0
2
2
2
2
0
• Type Cl , l 3: 2
• Type Dl , l 4: 2 2
2
• Type E6 :
2
2
0
2
2
2
• Type E7 : 2
2
2
0
2
2
2
• Type E8 : 2
2
0
2
2
2
2
2
34.6 Dimension of nilpotent orbits
• Type F4 :
• Type G2 :
2
2
0
2
0
517
2
34.6 Dimension of nilpotent orbits 34.6.1 In this section, g is assumed to be simple of dimension n. Let I be the set of integers p ∈ N such that n − l − 2p is the dimension of a nilpotent orbit. Note that, in general, I is not an interval of N. We shall determine an interval J inside I. 34.6.2 Lemma. Let m ∈ N and P ⊂ B. (i) If card P = 2m, then there is a subset Q of P of cardinal m such that α + β ∈ R for all α, β ∈ Q. (ii) If card P = 2m + 1, then there is a subset Q of P of cardinal m + 1 such that α + β ∈ R for all α, β ∈ Q. Proof. Let Γ be the Dynkin diagram of R. The results are clear if m = 0 or if m = 1 since Γ has no cycles. Let us proceed by induction on m. Let Λ be the full subgraph of Γ whose vertices are the elements in P . Then Λ has a terminal vertex. Let α ∈ P be a terminal vertex of Λ. If β ∈ P is linked to α, then set P = P \ {α, β}. If α is not linked to any vertex of Λ, then let β ∈ P \ {α} and set P = P \ {α, β}. By the induction hypothesis, there is a subset Q of P verifying the conclusion of the lemma for P . We obtain therefore the result by letting Q = Q ∪ {α}. 34.6.3 Proposition. Let r be the largest integer such that 2r l + 1. For 0 p r, there is a polarizable nilpotent orbit of dimension n − l − 2p. Proof. By 34.6.2, there exist α1 , . . . , αr ∈ B such that αi + αj ∈ R for 1 i, j r. For 0 p r, set: pp = b + g−α1 + · · · + g−αp . Then pp is a parabolic subalgebra of g. Let xp be a Richardson element of pp (33.5.8), then pp ∈ Pol(xp ) (33.5.6), and its orbit is a polarizable nilpotent orbit of dimension n − l − 2p.
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34 Results on orbits
34.7 Prehomogeneous spaces of parabolic type 34.7.1 In this section, g is a semisimple Lie algebra and we assume that there is a decomposition of g into subspaces gi such that [gi , gj ] ⊂ gi+j for i, j ∈ Z. g= i∈Z
Let ∂ ∈ End(g) be defined by ∂(x) = ix if x ∈ gi . Then ∂ is a semisimple derivation of g. By 20.1.5, there is a unique semisimple element t0 of g such that ∂ = ad t0 . Thus gi is the (ad t0 )-eigenspace with eigenvalue i. Also, we have L(gi , gj ) = {0} if i + j = 0 and L|gi ×g−i is non-degenerate. The Lie subalgebra g0 = gt0 is a reductive in g (20.5.13) and it contains a Cartan subalgebra of g (20.6.3). Denote by K the identity component of the stabilizer of t0 in G, and if x ∈ g, we denote by Kx the stabilizer of x in K. 34.7.2 Definition. Let i ∈ Z. (i) A S-triple (e, h, f ) of g is called a Si -triple if e ∈ gi and f ∈ g−i (so h ∈ g0 ). (ii) We say that h ∈ g is i-simple if there is a Si -triple of the form (e, h, f ). 34.7.3 Remark. Let e be a non-zero nilpotent element of gi . Then 32.1.7 implies that there is a Si -triple of the form (e, h, f ). 34.7.4 Let h ∈ Sg ∩ g0 , and for λ ∈ k, let gλ = ker(ad h − λ idg ). Let us assume that there exist i ∈ Z and e ∈ gi such that e ∈ g2 , [e, g0 ] = g2 . Lemma. The set Kh (e) contains a dense open subset of gi ∩ g2 . Proof. First of all, note that from the definition of K, we have α(gn ) = gn for all α ∈ K and n ∈ Z. Similarly, H = Kh◦ leaves gλ stable for all λ ∈ k. We may therefore define the morphism ϕ : H → gi ∩ g2 , α → α(e). By 24.3.6, L(K) = ad g0 and L(H) = ad(g0 ∩ g0 ) (24.3.5). So the image of the differential of ϕ at the identity of H is [e, g0 ∩ g0 ]. So by 15.4.2 and 16.5.7, it suffices to prove that [e, g0 ∩ g0 ] = gi ∩ g2 . Let x ∈ gi ∩ g2 . Then by our assumption on e, we have x = [e, y] for some y ∈ g0 . Since [t0 , h] = 0, we can write y= yj j
with yj ∈ gj ∩ g0 for all j ∈ Z. So we deduce from x = [e, y] that x = [e, y0 ]. Thus we have [e, g0 ∩ g0 ] = gi ∩ g2 as required.
34.7 Prehomogeneous spaces of parabolic type
519
34.7.5 Two S-triples (e, h, f ) and (e , h , f ) are K-conjugate if there exists α ∈ K such that e = α(e), h = α(h) and f = α(f ). Proposition. Let i ∈ Z \ {0} and (e, h, f ), (e , h , f ) two Si -triples. The following conditions are equivalent: (i) There exists α ∈ K such that α(e) = e . (ii) There exists α ∈ K such that α(h) = h . (iii) The S-triples (e, h, f ) and (e , h , f ) are K-conjugate. Proof. (i) ⇒ (ii) Since K(gi ) = gi for all i ∈ Z, we may assume that e = e . Let n = ker(ad e) ∩ im(ad e) ∩ g0 . By 32.2.4, n is a Lie subalgebra of g whose elements are nilpotent. Since [e, h − h ] = 0 and [e, f − f ] = h − h , we have h − h ∈ n. But [h, n] = n (19.2.7). So there exists z ∈ n such that ead z (h) = h (32.2.5). Since z ∈ g0 , we have ead z ∈ K. (ii) ⇒ (iii) Again, we may assume that h = h . Since e and e satisfy the hypotheses of 34.7.4, we have Kh (e) ∩ Kh (e ) = ∅. Thus if α ∈ Kh verifies α(e) = e , then α(h) = h and α(f ) = f (32.2.3). (iii) ⇒ (i) This is obvious. 34.7.6 Lemma. Let h be a Cartan subalgebra of g, c a subspace of h and θ an automorphism of g verifying θ−1 (c) ⊂ h. Then there exists α ∈ G such that α ◦ θ(h) = h and α(x) = x for all x ∈ c. Proof. Since θ(h) is a Cartan subalgebra of g containing c, there exists, by 32.2.1, α ∈ G such that α ◦ θ(h) = h and α(x) = x for all x ∈ h ∩ θ(h). 34.7.7 Lemma. Let h be a Cartan subalgebra of g containing t0 and h a semisimple element of g0 . There exists θ ∈ Aute ([g0 , g0 ]) such that θ(h) ∈ h. Proof. We have g0 = [g0 , g0 ] ⊕ z(g) and z(g) ⊂ h (20.5.5). Thus h = h1 + h2 with h1 ∈ [g0 , g0 ] and h2 ∈ z(g). Since h = h ∩ [g0 , g0 ] is a Cartan subalgebra of [g0 , g0 ], it follows from 29.2.3 and 29.2.6 that there exists θ ∈ Aute ([g0 , g0 ]) such that h1 = θ(h1 ) ∈ h . Hence θ(h) = h1 + h2 ∈ h. 34.7.8 Let h be a Cartan subalgebra of g containing t0 , H the set of α ∈ G verifying α(h) = h and W0 the set of w ∈ W such that w(t0 ) = t0 (see 18.2.9). Denote by ϕ the natural map from H to W as defined in 30.6.5. Thus W0 = ϕ(Ht0 ), where Ht0 is the stabilizer of t0 in H. Proposition. If i ∈ Z \ {0}, then there exists a natural injection from the set of K-orbits of i-simple elements into the set of double cosets W0 \W/W0 . Proof. Let h ∈ g be i-simple. By 34.7.7, K(h) ∩ h = ∅, so we may assume that h ∈ h. Let (e, h, f ) be a Si -triple and σ = e− ad e ◦ead f ◦e− ad e ∈ G. Then: σ(h) = −h , σ(t0 ) = t0 − ih. Let c = kt0 ⊕ kh. By 34.7.6, there exists ρ ∈ G such that ρ(h) = h, ρ(t0 ) = t0 and ρ◦σ(h) = h. Set θ = ρ◦σ, then:
520
34 Results on orbits
θ ∈ H , θ(h) = −h , θ(t0 ) = t0 − ih.
(1)
Let h, h ∈ h be i-simple elements verifying h ∈ K(h). In view of 34.7.6, we may assume that h = α(h) with α ∈ Ht0 . Let θh ∈ H (resp. θh ∈ H) be the element of G associated to h (resp. h ) verifying the relations in (1). Then: (θh )−1 ◦α−1 ◦θh (t0 ) = (θh )−1 ◦α−1 (t0 − ih ) = (θh )−1 (t0 − ih) = t0 . Thus
ϕ[(θh )−1 ◦α−1 ◦θh ] = [ϕ(θh )]−1 [ϕ(α)]−1 ϕ(θh ) ∈ W0 .
Hence: W0 ϕ(θh )W0 = W0 ϕ(θh )W0 . So this induces a map from the set of K-orbits of i-simple elements to set of double cosets W0 \W/W0 given by K(h) → W0 ϕ(θh )W0 . Let us show that this map is injective. If h, h are i-simple elements such that W0 ϕ(θh )W0 = W0 ϕ(θh )W0 , then there exist α1 , α2 ∈ Ht0 such that: ϕ(α1 ◦θh ◦α2 ) = ϕ(θh ). Hence:
t0 − iα1 (h) = α1 ◦θh ◦α2 (t0 ) = θh (t0 ) = t0 − ih .
Thus α1 (h) = h (since i = 0), and so K(h) = K(h ). 34.7.9 Corollary. Let i ∈ Z \ {0}. The number of K-orbits in gi is finite, and gi contains an open K-orbit. Proof. If i = 0, then all the elements of gi are nilpotent. So 34.7.3, 34.7.5 and 34.7.8 imply that the number of K-orbits in gi is finite. If O1 , . . . , Os are the K-orbits in gi , then gi = O1 ∪ · · · ∪ Os . Since gi is irreducible, the result follows from 21.4.3.
References and comments • [24], [29], [44], [48], [49], [57], [79], [80], [85], [86]. The proofs given in 34.5 on subregular orbits are inspired by the methods of [79]. The reader can find in [44] results relating the minimal nilpotent orbit to primitive ideals of the enveloping algebra of g.
35 Centralizers
We study the centralizer of an element in a semisimple Lie algebra in this chapter. We characterize in particular certain nilpotent elements which are called distinguished. Let g be a semisimple Lie algebra. We shall use the notations and conventions of 34.1.
35.1 Distinguished elements 35.1.1 Definition. (i) An element x ∈ g is called distinguished (in g) if gx contains only nilpotent elements. (ii) Let x ∈ Ng . We say that x is even if its characteristic contains only the integers 0 and 2. 35.1.2 Lemma. If x ∈ g is nilpotent, then all the elements of z(gx ) are nilpotent. Proof. Let u ∈ z(gx ) and u = y + z be its Jordan decomposition, with y semisimple and z nilpotent. Then y, z ∈ z(gx ). If y = 0, then there exist λ ∈ k and v ∈ g, both non-zero, such that [y, v] = λv. Thus: [y, [x, v]] = [[y, x], v] + [x, [y, v]] = [x, [y, v]] = λ[x, v]. Let a = {w ∈ g ; [y, w] = λw}, then we have shown that [x, a] ⊂ a. Since x is nilpotent, there exists w ∈ a \ {0} such that [x, w] = 0, or equivalently, w ∈ gx . So [y, w] = λw = 0 and y ∈ z(gx ). Contradiction. 35.1.3 Proposition. The following conditions are equivalent for x ∈ g: (i) x is distinguished in g. (ii) 0 is the only semisimple element of g which commutes with x. (iii) [x, g] contains a parabolic subalgebra of g. (iv) gx ⊂ [x, g]. (v) x is nilpotent and gx is a nilpotent Lie algebra.
522
35 Centralizers
Proof. (i) ⇔ (ii) This is clear since gx contains the semisimple and nilpotent components of its elements. (i) ⇒ (iii) Let x be distinguished in g, then 29.4.8 implies that gx is contained in the nilpotent radical n of a Borel subalgebra q of g. Taking the orthogonals with respect to the Killing form on g, we obtain that q ⊂ [g, x]. (iii) ⇒ (iv) Let p be a parabolic subalgebra of g, and m its nilpotent radical. If p ⊂ [g, x], then, as above, we have gx ⊂ m, so gx ⊂ [g, x] as required. (iv) ⇒ (i) If y ∈ g verifies [x, [x, y]] = 0, then [x, y] is nilpotent (32.1.2). Hence we have (i). (i) ⇒ (v) This is obvious. (v) ⇒ (ii) Let us suppose that gx contains a non-zero semisimple element h. Then ad h leaves gx stable, and since gx is nilpotent, we have clearly h ∈ z(gx ). But this contradicts 35.1.2. 35.1.4 Remarks. 1) It follows from 35.1.3 (v) and 19.7.5 that a regular nilpotent element of g is distinguished. 2) Let x ∈ Ng \ {0}. Assume that x is not distinguished in g, and h ∈ gx is a non-zero semisimple element. Then x ∈ s = [gh , gh ], and s is semisimple. If x is not distinguished in s, then we repeat the above procedure by replacing g by s. By repeating this process, we see that there exists a semisimple Lie subalgebra t of g containing x such that x is distinguished in t. 35.1.5 Let (x, h, y) be a S-triple of g. For p ∈ Z, we set gp to be the eigenspace of ad h associated to the eigenvalue p, and: gp , m = gp . l = g0 , p = p0
p>0
Then p is a parabolic subalgebra of g with nilpotent radical m, and l is a Levi factor of p. Recall from 19.2 that the map g0 → g2 , z → [x, z], is surjective. Proposition. The following conditions are equivalent: (i) x is distinguished in g. (ii) ad x induces a bijection from g0 onto g2 . (iii) dim g0 = dim g2 . Proof. Let s = kx + kh + ky. By the theory of representations of s described in 19.2, we have: x (g ∩ gp ) = (l ∩ gx ) ⊕ mx . gx = p0
Since [gi , gj ] ⊂ gi+j if i, j ∈ Z, the elements of mx are nilpotent. Moreover, l ∩ gx = cg (s), so it is reductive in g (20.5.13). Hence all the elements of gx are nilpotent if and only if l ∩ gx = {0}. The proposition is now clear. 35.1.6 Theorem. (Bala-Carter Theorem) Any distinguished element in a semisimple Lie algebra g is even.
35.2 Distinguished parabolic subalgebras
523
Proof. Let us use the notations in 35.1.5 and assume that x is distinguished in g. By 33.5.6, there exists u ∈ m such that [p, u] = m. Let us write ui , u= i>0 ∗
with ui ∈ gi for i ∈ N . From .
gi ,
/ ui = gi ,
i>0
i0
i>0
we deduce that: [g0 , u1 + u2 ] + [g1 , u1 ] = g1 ⊕ g2 . Suppose that g1 = {0}. Since u1 ∈ g1 , we have dim[g1 , u1 ] < dim g1 . We obtain therefore from 35.1.5 that: dim[g0 , u1 + u2 ] + dim[g1 , u1 ] < dim g0 + dim g1 = dim g2 + dim g1 . Contradiction. Thus g1 = {0}, and hence gi = {0} if i is odd. So x is even. 35.1.7 Corollary. Any distinguished element in g is polarizable. Proof. In the notations of 35.1.5, if x is even, then dim gh = dim gx . So 2 dim p = dim g + dim gx , and p ∈ Pol(x) since x ∈ m.
35.2 Distinguished parabolic subalgebras 35.2.1 Let J be a subset of B, denote by RJ the set of roots which are linear combinations of elements of J. and: α g , pJ = b + lJ , mJ = gα . lJ = h ⊕ α∈R+ \RJ
α∈RJ
Then pJ is a parabolic subalgebra of g with nilpotent radical mJ , and lJ is a Levi factor of pJ . 35.2.2 For β ∈ B, set:
ηJ (β) =
2 if β ∈ B \ J, 0 if β ∈ J.
We define a map ηJ : R → 2Z by setting for α ∈ R: nβ β ⇒ ηJ (α) = nβ ηJ (β). α= β∈B
For i ∈ Z, let:
β∈B
⎧ ⎪ ⎨
kgα if i = 0, ηJ (α)=i gJ (i) = kgα si i = 0. ⎪ ⎩h + ηJ (α)=0
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35 Centralizers
We have [gJ (i), gJ (j)] ⊂ gJ (i + j) for all i, j ∈ Z. Moreover, lJ = gJ (0), and g is the direct sum of the gJ (i) for i ∈ Z. Recall that if a is a Lie algebra and k ∈ N∗ , then C k (a) denotes the k-th term of the central descending series of a. The following result is a generalization of 20.8.3. Proposition. For k ∈ N∗ , we have: C k (mJ ) = gJ (2p). pk
In particular, dim gJ (2) = dim(mJ /[mJ , mJ ]). Proof. For k ∈ N∗ , let
Gk =
gJ (2p).
pk
Clearly mJ = C 1 (mJ ) = G1 and C k (mJ ) ⊂ Gk . To establish the reverse inclusion, we proceed by induction on k. Let α ∈ R+ be such that ηJ (α) 2k. By 18.7.7, α = β1 + · · · + βr , with β1 , . . . , βr ∈ B and β1 + · · · + βi ∈ R for 1 i r. Let s = max{i; βi ∈ B \ J}. We have s 2 since ηJ (α) 2k. Set γ = β1 + · · · + βs−1 . Then γ ∈ R and α = γ + βs + · · · + βr . It follows that we may choose non-zero vectors Xδ in gδ , for δ ∈ R, such that Xα = [[· · · [[Xγ , Xβs ], Xβs+1 ] · · ·], Xβr ]. From this, we deduce that Xα ∈ [[· · · [[Gk , G1 ], gJ (0)] · · ·], gJ (0)] ⊂ [Gk , G1 ]. By the induction hypothesis, we obtain that Gk+1 ⊂ C k+1 (mJ ). 35.2.3 Definition. A parabolic subalgebra p of g is called distinguished if dim(p/m) = dim(m/[m, m]), where m is the nilpotent radical of p. 35.2.4 Proposition. (i) We have dim lJ dim(mJ /[mJ , mJ ]). (ii) The parabolic subalgebra pJ of g is distinguished if and only if we have dim gJ (0) = dim gJ (2). Proof. By 35.2.2, it suffices to prove (i). Let x be a Richardson element of pJ (33.5.6). We have [pJ , x] = mJ . The adjoint representation of pJ on mJ induces a representation σ of x) = mJ /[mJ , mJ ] where pJ /mJ on mJ /[mJ , mJ ]. It follows that σ(pJ /mJ )( x denotes the image of x in mJ /[mJ , mJ ]. So we have the result. 35.2.5 We shall now give the relation between distinguished elements and distinguished parabolic subalgebras. Proposition. Let x be a distinguished element in g, (x, h, y) a S-triple and p the parabolic subalgebra of g associated to this S-triple as defined in 35.1.5. (i) p is a distinguished parabolic subalgebra of g. (ii) x belongs to the Richardson orbit of p.
35.3 Double centralizers
525
Proof. Using the notations of 35.1.5, we have: [p, x] = gi . i2
Since x is even (35.1.6), we have [p, x] = m and p ∈ Pol(x). We deduce therefore that dim Ox = 2 dim m and x is a Richardson element of p (33.5.8). Now there exists a Cartan subalgebra h such that h ∈ h ⊂ p. For a suitable choice of base B of the root system (g, h), we have p = pJ , where J ⊂ B. It is then obvious that gJ (i) = gi for all i ∈ Z. So (i) follows from 35.1.5 (iii) and 35.2.4 (ii). 35.2.6 Proposition. Suppose that pJ is a distinguished parabolic subalgebra of g. If x is a Richardson element of pJ , then it is distinguished in g and gx = pxJ = mxJ . Proof. We have [pJ , x] = mJ , pJ ∈ Pol(x), and so gx ⊂ pJ , and gx = pxJ . Let y ∈ gJ (−2) \ {0}. If [x, y] ∈ mJ , then y ∈ [pJ , x]⊥ = m⊥ J = pJ which is absurd. So [x, y] ∈ mJ . It follows easily from (35.2.4) that [pJ + gJ (−2), x] = pJ . Hence by 35.1.3 (iii), x is a distinguished element in g. Finally since pJ ⊂ [g, x], we have gx ⊂ mJ by taking orthogonals with respect to the Killing form. We have therefore gx = pxJ = mxJ . 35.2.7 Corollary. The following conditions are equivalent for x ∈ g: (i) x is distinguished in g. (ii) We have gx ⊂ Ox . Proof. (i) ⇒ (ii) Let (x, h, y) be a S-triple containing x and p the associated parabolic subalgebra as defined in 35.1.5. Denote by m the nilpotent radical of p. By 35.2.5 and 35.2.6, we have gx ⊂ m and m ⊂ Ox . (ii) ⇒ (i) Suppose that (ii) is verified, and let f be a polynomial function which vanishes on Ox . If t ∈ k and y ∈ gx , then f (x+ty) = 0, so Dx (f )(y) = 0. Thus gx ⊂ Tx (Ox ) (see 16.1). Now Tx (Ox ) = Tx (Ox ) = [x, g] (16.2.4, 21.4.3 and 34.1.2). So we may conclude by 35.1.3 (iv).
35.3 Double centralizers 35.3.1 Proposition. Let x ∈ g. Then: (i) [x, g] = [z(gx ), g] = (gx )⊥ . (ii) z(gx ) = [gx , g]⊥ . Proof. If y ∈ g, then: L([gx , g], y) = {0} ⇔ L(g, [y, gx ]) = {0} ⇔ [y, gx ] = {0} ⇔ y ∈ z(gx ). Hence z(gx ) = [gx , g]⊥ . Similarly: L([x, g], y) = {0} ⇔ L([x, y], g) = {0} ⇔ y ∈ gx , L([z(gx ), g], y) = {0} ⇔ L([z(gx ), y], g) = {0} ⇔ y ∈ gx . So we are done.
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35 Centralizers
35.3.2 Corollary. The following conditions are equivalent for x, y ∈ g: (i) y ∈ z(gx ). (ii) gx ⊂ gy . (iii) [g, gx ] ⊂ [g, gy ]. (iv) [g, y] ⊂ [g, x]. (v) z(gy ) ⊂ z(gx ). (vi) [z(gy ), g] ⊂ [z(gx ), g]. Proof. The equivalences (ii) ⇔ (iv), (iv) ⇔ (vi) and (iii) ⇔ (v) follow from 35.3.1, and it is clear that (ii) ⇒ (iii). (i) ⇒ (ii) If [y, gx ] = {0}, then gx ⊂ gy . (v) ⇒ (i) If z(gy ) ⊂ z(gx ), then y ∈ z(gx ) since y ∈ z(gy ). 35.3.3 Let x ∈ g. If y ∈ g verifies gy = gx , then y ∈ z(gx ) (35.3.2). Set: z(gx )• = {y ∈ g ; gx = gy }. Proposition. The set z(gx )• is a non-empty open subset of z(gx ), and: z(gx )• = {y ∈ g ; z(gy ) = z(gx )} = {y ∈ z(gx ) ; rk(ad y) = rk(ad x)}. Proof. Let y ∈ z(gx ). By 35.3.2, we have rk(ad y) rk(ad x) and: rk(ad y) = rk(ad x) ⇔ z(gy ) = z(gx ) ⇔ gy = gx . The proposition is now clear. 35.3.4 Proposition. Let x ∈ g and O its G-orbit. (i) If x is semisimple, then the set O ∩ z(gx ) is finite. (ii) If x is nilpotent, then O ∩ z(gx ) = z(gx )• . Proof. (i) If x is semi-simple, then there exist a Cartan subalgebra h of g, a base B of the root system of (g, h) and a subset J of B such that z(gx ) is the set of h ∈ h such that α(h) = 0 for all α ∈ J. By 34.2.4, 31.2.6 and 34.2.1, the set O ∩ h is finite. (ii) Let us suppose that x is nilpotent. Since the number of nilpotent orbits is finite, there is a nilpotent orbit U such that U ∩ z(gx ) = U is an open dense subset of z(gx ) (35.1.2). By 35.3.3, we have U ∩ z(gx ) = ∅, dim U = dim O, and U ⊂ z(gx )• . Conversely, if y ∈ z(gx )• , then y ∈ U and dim Oy = dim U, so y ∈ U (21.4.5). Thus z(gx )• = U and U = O. 35.3.5 Proposition. Let x ∈ g. Then: x (i) ng (gx ) = {y∈ g ; [x, y] ∈ z(g )}. x x (ii) ng (g ) = ng z(g ) .
35.3 Double centralizers
527
Proof. (i) Let y ∈ g be such that [x, y] ∈ z(gx ). If u ∈ gx , then [[y, x], u] = 0. So: 0 = [[y, u], x] + [y, [x, u]] = [[y, u], x]. Thus [y, u] ∈ gx , and y ∈ ng (gx ). Conversely, let y ∈ n(gx ). If u ∈ gx and v ∈ z(gx ), then: 0 = [y, [u, v]] = [[y, u], v] + [u, [y, v]] = [u, [y, v]]. We deduce therefore that [y, v] ∈ cg (gx ) = z(gx ). In particular, [y, x] ∈ z(gx ). (ii) From the preceding argument, we obtain that ng (gx ) ⊂ ng (z(gx )). Conversely, if y ∈ ng (z(gx )), then [y, x] ∈ z(gx ), and so y ∈ ng (gx ) by (i). 35.3.6 Corollary. If x is semisimple, then ng (gx ) = ng (z(gx )) = gx . Proof. Here we have g = [x, g] ⊕ gx . So the result is clear (35.3.5 (i)). 35.3.7 We shall study in detail the structure of z(gx ) and n(gx ) when x is nilpotent. Let (e, h, f ) be a S-triple of g and s = ke + kh + kf . If t is a semisimple element of g, a an (ad t)-stable subspace of g and λ ∈ k, we set: at,λ = {x ∈ a ; [t, x] = λx}. When t = h, we shall denote ah,λ by aλ . 35.3.8 It is clear that ge and z(ge ) are (ad h)-stable. Lemma. We have z(ge )n = {0} if n 1. Proof. By the theory of representations of s (see 19.2), we have (ge )n = {0} if n < 0. So z(ge )n = {0} if n < 0. Since (ge )0 = cg (s), we have z(ge )0 ⊂ cg (s). If y ∈ z(ge )0 and z ∈cg (s), that y ∈ z cg (s) . But then [z, y] = 0 because z ∈ ge . We deduce therefore cg (s) is reductive in g, so the elements of z cg (s) are semisimple. In view of 35.1.2, we have y = 0. Hence z(ge )0 = {0}. Suppose that z(ge )1 = {0}. Then the sum r of 2-dimensional simple ssubmodules of g is non-zero. We see easily that L|r×r is non-degenerate. Let u ∈ z(ge )1 \ {0}. We have [f, u] ∈ r−1 \ {0}. So there exists v ∈ r1 such that L(v, [f, u]) = 1. Thus L([u, v], f ) = 1. But it is clear that v ∈ ge . So we have a contradiction. Hence z(ge )1 = {0}. 35.3.9 Proposition. Let e ∈ Ng \ {0} and (e, h, f ) a S-triple. Then: z(ge ) = z(ge )2n . n1
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Proof. If e is distinguished, then the result follows from 35.1.6 and 35.3.8. Otherwise, there is a non-zero semisimple element x ∈ ge . Let a = [gx , gx ], then: gx = a ⊕ z(gx ). The Lie algebra a is semisimple and all the elements of z(gx ) are semisimple. Now z(ge ) ⊂ gx , so by 35.1.2, we obtain that z(ge ) ⊂ a. Hence: z(ge ) ⊂ z(ae ) ⊂ ae ⊂ ge . If e is not distinguished in a, then we repeat the same argument by replacing g by a. So we are reduced to the following situation: there exists a semisimple subalgebra t of g containing e such that: (i) z(ge ) ⊂ z(te ). (ii) e is distinguished in t. Let (e, h1 , f1 ) be a S-triple of t. The eigenvalues of ad h1 |t are even (35.1.6), so the eigenvalues of ad h1 |z(ge ) are also even. Now there exists θ ∈ G such that θ(e) = e and θ(h) = h1 (32.3.1). Clearly, θ(ge ) = ge and θ(z(ge )) = z(ge ). Consequently, the eigenvalues of ad h|z(ge ) are even. Since z(ge )0 = {0} (35.3.8), we have obtained the result. 35.3.10 Corollary. Let e ∈ Ng \ {0}, λ ∈ k and t ∈ Sg be such that [t, e] = λe. Then z(ge ) is (ad t)-stable and: z(ge )t,nλ . z(ge ) = n1
Proof. It is obvious that ge and z(ge ) are (ad t)-stable. We may assume that [t, h] = 0 and [t, f ] = −λf (32.1.7). So 2t − λh ∈ ge . By 35.3.9, the eigenvalues of ad h|z(ge ) are of the form 2n, with n ∈ N∗ . Since t and h are semisimple, [t, h] = 0, and ad(2t − λh)|z(ge ) = 0, the eigenvalues of ad t|z(ge ) are of the form nλ where n ∈ N∗ .
35.4 Normalizers 35.4.1 Let us conserve the hypotheses and notations of 35.3.7 and set: d = [f, z(ge )]. It is clear that d is (ad h)-stable, and by 35.3.9, we have: dn and ad h|z(ge ) is injective. d= n0
Hence: [e, d] = [e, [f, z(ge )]] = [h, z(ge )] = z(ge ). It follows that dim d = dim z(ge ), d ∩ ge = {0} and d ⊂ ng (ge ) (35.3.5). In particular, the sum d + ge is direct.
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529
Proposition. We have: (i) [ng (ge ), e] = z(ge ). (ii) ng (ge ) = ge ⊕ d, so dim ng (ge ) = dim ge + dim z(ge ). Furthermore: n(ge )n . ng (ge ) = n0
Proof. (i) We have z(ge ) = [e, d] ⊂ [e, ng (ge )] ⊂ z(ge ) by 35.3.5 (i). So part (i) follows. (ii) Since ge ⊂ ng (ge ), part (i) implies that: dim ng (ge ) = dim z(ge ) + dim ng (ge ) ∩ ge = dim z(ge ) + dim ge . So ng (ge ) = ge ⊕ d since dim d = dim z(ge ) and d ∩ ge = {0}. The last part follows from 35.3.9. 35.4.2 We have cg (s) = ge ∩gh = gf ∩gh = ge ∩gf , or cg (s) = (ge )0 = (gf )0 . So we have by 35.3.9 and 35.4.1 that: ng (ge )0 = (ge )0 ⊕ d0 = cg (s) ⊕ d0 = cg (s) ⊕ [f, z(ge )2 ]. Note also that ng (ge )0 is a Lie subalgebra of g. Proposition. The vector space d0 is the orthogonal of cg (s) in ng (ge )0 with respect to L, and it is an abelian Lie algebra. We have [d0 , cg (s)] = {0}, and so ng (ge )0 is the product of the Lie algebras d0 and cg (s). Proof. 1) We deduce from cg (s) ⊂ gf that: L(d, cg (s)) = L([f, z(ge )], cg (s)) = L(z(ge ), [f, cg (s)]) = {0}. Thus d0 is contained in the orthogonal of cg (s). Since L|cg (s)×cg (s) is nondegenerate (20.5.13), the first part follows. 2) Let u ∈ z(ge ), v ∈ cg (s) = ge ∩ gf . We have [u, v] = [f, v] = 0, and so: [[f, u], v] = [[f, v], u] + [f, [u, v]] = 0. It follows that [d, cg (s)] = {0}. In particular, [d0 , cg (s)] = {0}. 3) If u, v ∈ d, w ∈ cg (s), then point 2 implies that: L([u, v], w) = L(u, [v, w]) = 0. Since ng (ge )0 is a Lie subalgebra of g, it follows from point 1 that [u, v] ∈ d0 . Thus d0 is a Lie subalgebra of g. 4) Let w1 , w2 ∈ d0 . There exist z1 , z2 ∈ z(ge )2 such that: w1 = [f, z1 ] , w2 = [f, z2 ]. For i = 1, 2, we have:
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35 Centralizers
[e, wi ] = [e, [f, zi ]] = [[e, f ], zi ] + [f, [e, zi ]] = [[e, f ], zi ] = [h, zi ] = 2zi . So: [e, [w1 , w2 ]] = [[e, w1 ], w2 ] + [w1 , [e, w2 ]] = 2[z1 , [f, z2 ]] + 2[[f, z1 ], z2 ] = 2[f, [z1 , z2 ]] = 0. Thus [w1 , w2 ] ∈ ge , and point 3 implies that [w1 , w2 ] ∈ d0 ∩ ge . As the sum d + ge is direct, we have proved that d0 is commutative.
35.5 A semisimple Lie subalgebra 35.5.1 Lemma. Let a be a reductive Lie subalgebra of g, c = cg (a), c⊥ the orthogonal of c with respect to the Killing form of g and t the Lie subalgebra of g generated by c⊥ . Then t is an ideal of g. Proof. We have [c⊥ , t] ⊂ t. Also we verify easily that [c, c⊥ ] ⊂ c⊥ , so [c, t] ⊂ t. Since g = c ⊕ c⊥ by 20.5.13, t is an ideal of g. 35.5.2 Let us conserve the hypotheses of 35.3.7, and set d = dim z(ge )2 . Let θ = ead e ◦e− ad f ◦ead e ∈ G. We saw in 32.2.2 that: θ(e) = −f , θ(h) = −h , θ(f ) = −e. Hence for n ∈ Z:
θ z(ge )n = z(gf )−n .
Let v be the s-submodule of g generated by z(ge )2 . By the definition of θ, we see easily that v is θ-stable, so it contains z(gf )−2 . It contains also d0 = [f, z(ge )2 ], which is of dimension d, because ad f |gi is injective if i > 0. Let v0 ∈ z(ge )2 \ {0}. For j 0, set vj = (ad f )j (v0 ). By 30.5.1, v3 = 0, and since [f, v0 ] = 0, kv0 + kv1 + kv2 is a 3-dimensional simple s-submodule of g (19.2). Thus v = v−2 ⊕ v0 ⊕ v2 where: v2 = z(ge )2 , v0 = d0 , v−2 = z(gf )−2 . Proposition. The subspace v is a Lie subalgebra of g. It is isomorphic to the product of d copies of sl2 (k), and d0 is a Cartan subalgebra of v. Proof. First, let us prove that v is a Lie subalgebra of g and d0 is a Cartan subalgebra of v. We have [z(ge )2 , z(ge )2 ] = [z(gf )−2 , z(gf )−2 ] = {0}, and by 35.4.2, d0 is an abelian Lie subalgebra of g. Further, 35.4.1 and 35.3.5 (ii) say that [d0 , z(ge )2 ] ⊂ z(ge )2 . Taking r = 2 in the computations of 32.2.7, we obtain that θ(x) = −x for all x ∈ d0 . It follows that [d0 , z(gf )−2 ] ⊂ z(gf )−2 . Since d0 ∩ cg (s) = {0}, we have by the theory of representations of sl2 (k) (see 19.2) that: (1)
[e, d0 ] = z(ge )2 , [e, z(gf )−2 ] = d0 .
35.5 A semisimple Lie subalgebra
531
We deduce therefore that: [z(ge )2 , z(gf )−2 ] = [[e, d0 ], z(gf )−2 ] = [e, [d0 , z(gf )−2 ]] + [d0 , [e, z(gf )−2 ]] ⊂ [e, z(gf )−2 ] + [d0 , d0 ] ⊂ d0 . So we have proved that v is a Lie subalgebra of g, and d0 is a Cartan subalgebra of g since it is abelian (35.4.2) and d0 = nv (d0 ) because h ∈ d0 , [h, z(ge )2 ] = z(ge )2 , [h, z(gf )−2 ] = z(gf )−2 . Next, let us prove that v is semisimple. Applying θ to the first equality in (1), we obtain that [f, d0 ] = z(gf )2 . It follows therefore again by (1) that [v, v] = v. Thus the radical r of v is exactly the nilpotent radical of v, and the elements of r are all nilpotent in g. Let us write v = t⊕r where t is a Levi subalgebra of v containing s (29.5.3). So by 29.8.4, there exists a parabolic subalgebra p of g such that its nilpotent radical contains r, and it has a Levi factor l which contains t. If r is non-zero, then since v is the sum of simple s-submodules of dimension 3, r is also the sum of simple s-submodules of dimension 3. Thus r ∩ z(ge )2 contains a non-zero element w. But if c denotes the centre of l, then c = cg (l) and l = cg (c) (29.5.7). Thus there exists x ∈ g such that [x, t] = {0}, hence x ∈ ge , and [x, w] = 0. This is absurd since w ∈ z(ge ). So r = {0} and v is semisimple. Finally, as d0 is a Cartan subalgebra of v and d = dim d0 , a base of a root system of v has d elements. Since dim v = 3d, it is obvious that v is a product of d copies of sl2 (k). 35.5.3 Theorem. Let e be a non-zero nilpotent element of a semisimple Lie algebra g and d = dim z(ge )2 be as defined in 35.5.2. Then d is the number of simple ideals of g on which the projection of e is non-zero. Proof. We shall often use the results on the representations of semisimple Lie algebras (see 19.2 and chapter 30). Also, it is easy to see that to prove the theorem, it suffices to verify that if g is simple, then d = 1. We shall use the notations of 35.5.2. Let a1 , . . . , ad be the simple components of v; they are isomorphic to sl2 (k) (35.5.2). For 1 i d, let pi : v → ai be the projector associated to the decomposition v = a1 ⊕ · · · ⊕ ad ; the pi ’s are Lie algebra homomorphisms. Set: ei = pi (e) , hi = pi (h) , fi = pi (f ). As the Lie algebra s is simple, pi (s) is either simple or is equal to {0}. In the latter, we would have [s, ai ] = {0}, hence ai ⊂ v0 = d0 which is absurd since d0 is abelian (35.5.2). So pi (s) is simple. Note that [hi , ei ] = 2ei , so [h, ei ] = 2ei . Thus ei ∈ z(ge )2 . Similarly, fi ∈ f z(g )−2 , and hi ∈ d0 . Let us consider the bases of the root systems of (s, kh) and (v, d0 ) associated to the Borel subalgebras kh + ke and d0 + z(ge )2 .
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35 Centralizers
Let V be a simple v-submodule of g and u a primitive vector of V (u is unique up to multiplication by a non-zero scalar). For 1 i d, we have [ei , u] = 0 and [hi , u] = ki u, with ki ∈ N. Set Fi = ad fi |V . We verify easily that the set of vectors F1r1 ◦ · · · ◦Fdrd (u), with 0 ri ki for all i, is a basis of V . We deduce immediately that ku is the set of v ∈ V such that [ei , v] = 0 for all i, and dim V = (k1 + 1) · · · (kd + 1). Now consider V as a s-module, and let w ∈ V be a primitive vector for s. Then [e, w] = 0, that is w ∈ ge . Since ei ∈ z(ge ), we have [ei , w] = 0. Thus w is a scalar multiple of u. It follows that V is a simple s-module. Now [h, u] = (k1 + · · · + kd )u, so the dimension of V is k1 + · · · + kd + 1. Hence: k1 + · · · + kd + 1 = (k1 + 1) · · · (kd + 1). Consequently, if dim V > 1, all except one of the ki ’s are equal to zero, and if ki = 0, then V is a simple ai -module such that [aj , V ] = {0} for j = i. The v-module g being semisimple, the discussion above implies that (2)
g = cg (v) ⊕ r1 ⊕ · · · ⊕ rd
where, for 1 i d, ri is a ai -module, which is the sum of simple ai -modules of dimensions not equal to 1 (so cri (ai ) = {0}), and [aj , ri ] = {0} for j = i. We have ri = {0} since ai = pi (s) ⊂ ri . Set: aj , mi = rj . bi = j=i
j=i
Thus cg (ai ) = cg (v) ⊕ mi . Let i, j ∈ {1, . . . , d} be distinct. Since [ai , ri ] = ri and [ai , rj ] = {0}, we have: L(ri , rj ) = L([ai , ri ], rj ) = L(ri , [ai , rj ]) = {0}, L ri , cg (v) = L ri , [ai , cg (v)] = {0}. The factors in the decomposition (2) are therefore pairwise orthogonal. As ri = [ai , g] by (2), we obtain that: [ri , rj ] ⊂ [[ai , [aj , g]], g] + [ai , [g, [aj , g]]] ⊂ [[ai , rj ], g] + [ai , [g, [aj , g]]] ⊂ ri . Similarly, [ri , rj ] ⊂ rj . Hence [ri , rj ] = {0}. We deduce therefore that: L [ri , cg (v) + ri ], mi = L cg (v) + ri , [ri , rj ] = {0}. j=i
Thus [ri , cg (v) + ri ] ⊂ m⊥ i = cg (v) + ri . The Lie algebra bi is reductive in g, and: cg (bi ) = cg (v) ⊕ ri , cg (bi )⊥ = mi . Let t be the Lie subalgebra of g generated by cg (bi )⊥ . Then the inclusion ri ⊂ cg (v) + ri implies that:
35.6 Centralizers and regular elements
533
mi ⊂ t ⊂ r⊥ i = cg (v) ⊕ mi = cg (ai ). If d > 1, then t = {0} and t = g. By 35.5.1, t is an ideal of g. But this is impossible if g is simple. So we are done.
35.6 Centralizers and regular elements 35.6.1 Lemma. Let τ be an automorphism of finite order of a non-zero semisimple Lie algebra a. Then there exists x ∈ a \ {0} such that τ (x) = x. Proof. Let r > 1 be the order of τ . We may assume that a does not contain a non-zero semisimple τ -stable Lie subalgebra other than a itself. Denote S = {λ ∈ k; λr = 1} and for λ ∈ S, let a(λ) be the eigenspace of τ associated to the eigenvalue λ. Then a is the direct sum of the vector subspaces a(λ), for λ ∈ P = {λ ∈ S; a(λ) = {0}}. We verify easily that [a(λ), a(µ)] ⊂ a(λµ), and the restriction of the Killing form L of a on a(λ) × a(µ) is zero if λµ = 1, while L|a(λ)×a(λ−1 ) is nondegenerate. Let us suppose that a(1) = {0}. Then [a(λ), a(λ−1 )] = {0} for all λ ∈ S. 1) Let λ ∈ P . It is clear that a(λ) contains the semisimple and nilpotent components of its elements. Let x be a nilpotent element of a(λ). Since [a(λ),a(λ−1 )] = {0}, (ad x)◦(ad y) is nilpotent for all y ∈ a(λ−1 ), so L x, a(λ−1 ) = {0}. Since L|a(λ)×a(λ−1 ) is non-degenerate, x = 0. Consequently any element x ∈ a(λ) is semisimple, and so ax is a τ -stable reductive Lie subalgebra and it has the same rank as a. Our assumption on a implies that the τ -stable semisimple Lie algebra is equal to {0}. Thus ax is a Cartan subalgebra of a, and so, x is regular. We have therefore shown that any non-zero element of a(λ) is generic in a. 2) Let λ ∈ P and: a(λm ). u(λ) = m∈Z
Let x ∈ a(λ) \ {0}. If m ∈ Z, then there exists p ∈ N such that λm+p = 1. Thus ad x|u(λ) is nilpotent, so it is zero because x is generic. Thus u(λ) is contained in the Cartan subalgebra ax of a. 3) Let λ be a primitive r-th root of unity. We have: a=
r−1 i=0
a(λi ) =
r−1
a(λi ).
i=1
If λ ∈ P , then point 2 implies that a is abelian, which is absurd. So λ ∈ P . Let p be the smallest positive integer such that a(λp ) = {0}. By using the earlier remarks on the Killing form, we have: a(1) = · · · = a(λp−1 ) = a(λr−p+1 ) = · · · = a(λr ) = {0} , a(λr−p ) = {0}. Let us fix a non-zero element x in a(λr−p ) = a(λ−p ), and q an integer such that p q r − 1. If q is a multiple of p, then ad x|a(λq ) = 0 by point 2. Otherwise,
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35 Centralizers
q = np+s, with n, s ∈ N∗ and 1 s p−1. Then q +n(r −p) = s+nr, which proves that (ad x)n |a(λq ) = 0. We have therefore shown that ad x is nilpotent. It follows from point 1 that x = 0. Contradiction. 35.6.2 Lemma. Let a be a non-solvable Lie algebra and τ an automorphism of a of finite order. There exists x ∈ a \ {0} such that τ (x) = x. Proof. By 35.6.1, we may assume that a is not semisimple. Let r be the radical of a. It is τ -stable and there exists a τ -stable subspace f of a such that a = f⊕r. Let τ be the automorphism of a/r induced by τ . By 35.6.1, there exists x ∈ f such that τ (x + r) = x + r. This implies that τ (x) ∈ x + r, and hence τ (x) = x since f is τ -stable. 35.6.3 Let us use the notations of 35.3.7 and let us assume that e is a distinguished element in g. So g2i+1 = {0} for all i ∈ Z (35.1.6). Let m be the largest integer such that g2m = {0}; it is also the largest integer such that g−2m = {0}; Lemma. Let y ∈ g−2m and x = e + y. (i) The element x is nilpotent if and only if y = 0. (ii) We have gx ∩ g0 = {0}. Proof. (i) Note that x is non-zero. If x is nilpotent, then there exists t ∈ g such that [t, x] = 2x (32.1.5). Let us write t = tn with tn ∈ gn for all n ∈ Z. Then [t0 , e] = 2e and [t0 , y] = 2y. So t0 − h ∈ ge ∩ g0 . Hence t0 = h (35.1.5). Consequently 2y = −2my, and so y = 0. The converse is clear. (ii) If z ∈ gx ∩ g0 , then [z, e] = [z, y] = 0. Therefore z ∈ ge ∩ g0 = {0} by 35.1.5. 35.6.4 Let us conserve the hypotheses of 35.6.3. Let λ be a primitive (2m + 2)-th root of unity. Denote by τ the automorphism of g such that τ (x) = λ2n x for x ∈ g2n , −m n m. We have τ ∈ Aute g (32.2.7), and the eigenspaces of τ are a0 , . . . , am where ai = g2i + g2i−2m−2 for 0 i m. In particular, g0 is the space of fixed points of τ . Let y ∈ g−2m and x = e + y. Then τ (x) = λ2 x, so gx is τ -stable. Lemma. The Lie algebra gx is solvable. Proof. By 35.6.3 (ii), 0 is the only fixed point of τ in gx . So the result follows from 35.6.2. 35.6.5 Lemma. Let us conserve the hypotheses of 35.6.3. The following conditions are equivalent: (i) gx is a Cartan subalgebra of g.
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535
(ii) x is a generic element of g. (iii) x is semisimple. There exists y ∈ g−2m such that x verifies these conditions if and only if a1 contains a generic element of g. Proof. The implications (i) ⇒ (ii) ⇒ (iii) are obvious. So let x be semisimple. Then gx is a reductive subalgebra whose rank is the rank of g. As gx is solvable by 35.6.4, it is a Cartan subalgebra of g. Finally, if x verifies these conditions, it is a generic element of g contained in a1 . Conversely, let u be the set of generic elements of g in a1 and let us suppose that u = ∅. Then u is a dense open subset of a1 . Let Gh be the stabilizer of h in G, then by 32.2.9, Gh (e) + g−2m is a dense open subset of a1 . So there exist α ∈ Gh and z ∈ g−2m such that α(e) + z is generic. Hence e + α−1 (z) is generic and α−1 (z) ∈ g−2m . 35.6.6 Lemma. Let us conserve the hypotheses of 35.6.3. Suppose that dim g2 − dim g4 = 1. For all y ∈ g−2m \ {0}, x = e + y is a semisimple element of g. Proof. Let xs and xn be the semisimple and nilpotent component of x. It is clear that xs , xn ∈ a1 . Let us write xn = u + v with u ∈ g2 and v ∈ g−2m . Since [xn , x] = 0, we have [e, u] = 0. Since the map ad e : g2 → g4 is surjective with kernel ke, we have u ∈ ke. If u = 0, then [xn , x] = 0 implies that v ∈ ge ∩ g−2m = {0}. Thus xn = 0 and x is semisimple. If u = λe with λ = 0, then v = 0 (35.6.3 (i)). The condition [xn , x] = 0 implies in this case that y ∈ ge ∩ g−2m = {0}. 35.6.7 Lemma. Let us assume that g is simple and suppose that e is a non-zero nilpotent element in g such that ge is a commutative Lie algebra. Then e is regular. Proof. By 35.1.3 (v), e is distinguished. For i ∈ N, we have: dim g2i − dim g2(i+1) = dim(g2i ∩ ge ) = dim z(ge )2i . Since g is simple, it follows from 35.5.3 that: dim g2 − dim g4 = 1. Let m be as in 35.6.3. Then 35.6.5 and 35.6.6 imply that there exists y ∈ g−2m such that x = e + y is a generic element of g. 1) Let us first show that for all u ∈ ge , there is a unique v ∈ g verifying: v∈ gi , u + v ∈ g x . i<0
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35 Centralizers
If v1 , v2 verify these conditions, then v1 − v2 is a nilpotent element of the Cartan subalgebra gx of g. So v1 = v2 . Now consider the following equation in v: 0 = [u + v, e + y] = [u, y] + [v, e]. Note that [y, v] = 0 by the hypotheses on y and v. Since ge is abelian, [u, y] ∈ (35.3.1). So there exists v ∈ g such that [v, e] + [u, y] = 0. We (ge )⊥ = [g, e] have [u, y] ∈ i0 gi and e ∈ g2 . So we may assume that v ∈ i<0 gi . 2) It follows from point 1 that the map ge → gx , u → u + v is linear and injective, so dim ge dim gx . Since x is regular, so is e. 35.6.8 Let g be a reductive Lie algebra of rank r, s = [g, g] and c = z(g). If x ∈ s and y ∈ c, then gx+y = c + sx . This implies that dim gx+y r, and that dim gx+y = r if and only if x is regular in s. If this is the case, we shall say that x + y is a regular element of g. Theorem. Let g be a reductive Lie algebra and x ∈ g. The following conditions are equivalent: (i) x is a regular element of g. (ii) The Lie algebra gx is abelian. Proof. (i) ⇒ (ii) This follows from 19.7.5 and 29.3.1. (ii) ⇒ (i) We shall prove this by induction on dim g. The case g = {0} is obvious. We may assume that g is semisimple. If x is semisimple, then gx is a reductive subalgebra whose rank is equal to the rank of g. So if gx is abelian, then it is a Cartan subalgebra of g, and x is regular. If x is nilpotent, then the result follows easily from 35.6.7. Now suppose that x is neither semisimple nor nilpotent. Let xs and xn be the semisimple and nilpotent component of x. Then gxs is a reductive subalgebra whose rank is equal to the rank of g and it contains xn . Now (gxs )xn = gx is abelian. By the induction hypothesis, dim(gxs )xn is equal to the rank of gxs , and so it is equal to the rank of g. Consequently, the element x is regular.
References and comments • [2], [3], [17], [18], [20], [46], [63], [76], [80]. The proof of the Bala-Carter Theorem (35.1.6) given here is due to Jantzen, and it is taken from [46]. A great number of results presented in sections 35.3 to 35.6 are proved in [17] and [18]. Certain proofs given here are due to Panyushev ([63]), for example 35.5.3. The proof of 35.6.8 is also due to Panyushev ([63]), and the proof of 35.6.6 is due to Springer ([76]).
36 σ-root systems
In this chapter, we study properties of certain root systems related to involutive automorphisms. We shall use these properties in Chapters 37 and 38. Let k be a commutative field of characteristic zero, V be a k-vector space of dimension l with l = 0 and R be a reduced root system in V . If v ∈ V and f ∈ V ∗ , we denote f (v) by v, f . We shall use the notations of chapter 18.
36.1 Definition 36.1.1 Let us fix an involutive automorphism σ of V such that σ = ± idV . Set: V = {x ∈ V ; σ(x) = x} , V = {x ∈ V ; σ(x) = −x}, R = R ∩ V = {α ∈ R; σ(α) = α} , R = {α ∈ R; σ(α) = α}.
By 18.2.5, R is a root system in the subspace of V spanned by R . Moreover, if B is a base of R , then there exists a base B of R containing B , and R is the set of elements of R which are linear combinations of elements of B (18.7.9). ∗ ∗ Let us identify V (resp. V ) with the set of elements of V ∗ which ∗ ∗ vanishes on V (resp. V ). Thus V ∗ = V ⊕ V . If θ is the transpose of σ, then: ∗
∗
V = {ϕ ∈ V ∗ ; θ(ϕ) = ϕ} , V = {ϕ ∈ V ∗ ; θ(ϕ) = −ϕ}. For x ∈ V and ϕ ∈ V ∗ , we shall write x = x + x , ϕ = ϕ + ϕ , ∗
∗
with x ∈ V , x ∈ V , ϕ ∈ V , ϕ ∈ V . 36.1.2 Definition. (i) We say that R is a σ-root system if σ(R) = R. (ii) A σ-root system R is called normal if for all α ∈ R, we have:
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36 σ-root systems
α + σ(α) ∈ R.
36.1.3 In the rest of this section, we assume that R is a σ-root system in V. Let α, β ∈ R, x ∈ V . Then: sα σ(x) = σ(x) − σ(x), α∨ α = σ x − x, θ(α∨ )σ(α) , ∨ σ(α), θ(α∨ ) = α, α∨ = 2, β, θ(α∨ ) = σ(β), α ∈ Z, ∨ ∨ β − β, θ(α )σ(α) = σ σ(β) − σ(β), α α ∈ R. Thus:
σ◦sα ◦σ = sσ(α) , σ(α)∨ = θ(α∨ ).
36.1.4 Let B be a base of R and R+ (resp. R− ) the corresponding set of positive (resp. negative) roots. Set: = R ∩ R+ , R− = R ∩ R− . R+
Lemma. In the notations above, we can choose B so that the following conditions are verified: ) = R− . (i) σ(R+ . (ii) If α ∈ R+ and β ∈ R verify α − β ∈ V , then β ∈ R+ (iii) (R+ + R+ ) ∩ R ⊂ R+ . Proof. Denote by VQ the Q-vector subspace of V spanned by R; we know that V = VQ ⊗Q k, and we have σ(VQ ) = VQ (because R is a σ-root system). Set VQ = V ∩ VQ et VQ = V ∩ VQ . 1) We claim that V = VQ ⊗Q k et V = VQ ⊗Q k. Let (α1 , . . . , αl ) be a base of R. If x ∈ V , then there exist scalars λ1 , . . . , λl such that x = λ1 α1 + · · · + λl αl . Thus x=
l λ l λ i i αi + σ(αi ) + αi − σ(αi ) . 2 2 i=1 i=1
So the claim follows easily. 2) Let (e1 , . . . , ep ) be a basis of VQ and (ep+1 , . . . , el ) a basis of VQ . Endow VQ with the lexicographic order defined by the basis (e1 , . . . , el ) of VQ . Let R+ (resp. R− ) be the corresponding set of positive (resp. negative) roots, and = V ∩ R+ , R− = V ∩ R− . R+ , with λr = 0. So we have λr > 0 and r p. Let α = λr er + · · · + λl el ∈ R+ It follows that: σ(α) = −λr er − · · · − λp ep + λp+1 ep+1 + · · · + λl el . We deduce from λr > 0 that σ(α) ∈ R− . So we have (i). If β = µs es + · · · + µl el ∈ R, with s l and µs = 0, then α − β ∈ V implies that s = r and µs = λr . Thus β ∈ R+ , and we have (ii). Finally, (iii) follows from our choice of basis.
36.2 Restricted root systems
539
36.1.5 Let us conserve the notations of 36.1.4. Let x ∈ VQ . There exist λ1 , . . . , λl ∈ Q such that x = λ1 α1 + · · · + λl αl . So: x =
l λ l λ x + σ(x) x − σ(x) i i = αi + σ(αi ) , x = = αi − σ(αi ) . 2 2 i=1 2 i=1 2
Let S = {α ; α ∈ R }. Then it follows that VQ is the Q-vector subspace of V spanned by S.
36.2 Restricted root systems 36.2.1 Theorem. Let R be a normal σ-root system in V and S = {α ; α ∈ R }. Then S is a (not necessarily reduced) root system in V . Proof. It is clear that S is finite, does not contain 0 and spans V . Let λ, µ ∈ S. Fix α, β ∈ R verifying α = λ, β = µ. Let us consider the following cases. 1) We have σ(α) = −α, or α = λ. Then σ(α)∨ = −α∨ , so α∨ = α∨ ∗ belongs to V . For x ∈ V , set:
s(x) = x − x, α∨ λ.
Then the fact that λ, α∨ = α, α∨ = 2, implies that s is a reflection; in particular, s(µ) = 0. Moreover:
µ, α∨ = β + µ, α∨ = β, α∨ ∈ Z, sα (β) = β + s(µ) ⇔ s(µ) = [sα (β)] . 2) We have σ(α) = −α and α, σ(α)∨ = 0. We have:
2λ = α − σ(α) , 2α∨ = α∨ − σ(α)∨ . For x ∈ V , set:
s(x) = x − x, 2α∨ λ.
Since α, σ(α)∨ = σ(α), α∨ = 0, we obtain that:
λ, 2α∨ =
1 1 α, α∨ + σ(α), σ(α)∨ = 2. 2 2
Thus s is a reflection, so s(µ) = 0. Moreover:
µ, 2α∨ = β, α∨ − σ(α)∨ − β , α∨ − σ(α)∨ = β, α∨ − β, σ(α)∨ ∈ Z. Set γ = sσ(α) ◦sα (β). As α, σ(α)∨ = σ(α), α∨ = 0, we obtain that:
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36 σ-root systems
γ = β − β, α∨ α − β, σ(α)∨ σ(α). So γ = ν + s(µ) where ν ∈ V . Hence s(µ) = γ . 3) We have σ(α) = −α and σ(α), α∨ = 0. Since α + σ(α) ∈ R, we have σ(α), α∨ > 0 (18.5.3). Moreover: σ(α), α∨ = α, θ(α∨ ) = α, σ(α)∨ . So, by 18.5.2:
either α, σ(α)∨ = σ(α), α∨ = 1 or α, σ(α)∨ = σ(α), α∨ = 2 and α = σ(α)
Since α = σ(α) (because α ∈ R ), it follows that: α, σ(α)∨ = σ(α), α∨ = 1. For x ∈ V , set:
s(x) = x − x, 4α∨ λ.
We have:
λ, 4α∨ = α − σ(α), α∨ − σ(α)∨ = α, α∨ + σ(α), σ(α)∨ − α, σ(α)∨ − σ(α), α∨ = 2 + 2 − 1 − 1 = 2. Thus s is a reflection, so s(µ) = 0. Moreover:
µ, 4α∨ = 2β, α∨ − σ(α)∨ − 2β , α∨ − σ(α)∨ = 2β, α∨ − σ(α)∨ ∈ Z. Set γ = sα ◦sσ(α) ◦sα (β). As α, σ(α)∨ = σ(α), α∨ , we obtain that: γ = β − β, α∨ − σ(α)∨ α − σ(α) = β − β, 2α∨ (2λ) = β + s(µ). Hence s(µ) = γ . 4) It follows finally from the preceding discussion that S is a root system in V . 36.2.2 Let us assume again that R is a normal σ-root system in V . Denote by W the Weyl group of R and set: Wσ = {w ∈ W ; w◦σ = σ◦w}. Let W be the subgroup of W generated by the sα ’s, α ∈ R . The group W can be identified as the Weyl group of R , and we verify easily that W is a normal subgroup of Wσ . Lemma. Let λ ∈ S. There exists w ∈ Wσ such that sλ = w|V .
36.2 Restricted root systems
541
Proof. Let us fix α ∈ R such that α = λ. We shall consider the same cases as in 36.2.1. 1) We have σ(α) = −α. We saw in 36.2.1 that sλ = sα |V , and it follows easily that sα ∈ Wσ . 2) We have σ(α) = −α and α, σ(α)∨ = σ(α), α∨ = 0. By 36.2.1, we have λ∨ = 2α∨ . Let w = sσ(α) ◦sα . Since α, σ(α)∨ = 0, it follows that for x ∈ V : ∨ w(x) = x − x, α∨ α − σ(x),α σ(α) ∨ = x − x, α α − σ(α) = x − x, α∨ (2α ) = x − x, α∨ (2α ) = sλ (x).
We have therefore obtained that w(V ) ⊂ V and w|V = sλ . On the other hand, if x ∈ V , then: w(x) = x − x, α∨ α − x, σ(α)∨ σ(α) = x − x, α∨ (2α ) ∈ V . So we have w ∈ Wσ . 3) We have σ(α) = −α and σ(α), α∨ = 0. Again we saw in the proof of 36.2.1 that we have α, σ(α)∨ = σ(α), α∨ = 1 and λ∨ = 4α∨ . Let w = sα ◦sσ(α) ◦sα . If x ∈ V , then:
w(x) = x − x, 4α∨ α . This shows that w stabilizes V and V , and so w ∈ Wσ and w|V = sλ . 36.2.3 Let us conserve the notations of 36.2.2. Let l be the rank of R and (αl−l +1 , . . . , αl ) a base of R . In the construction of 36.1.4, take ei = αi , for l − l + 1 i l. We see easily that αl−l +1 , . . . , αl are simple roots with respect to the lexicographic order on VQ defined by the basis (e1 , . . . , el ). By 36.1.4, there exists a base B = (α1 , . . . , αl ) of R verifying the following conditions: ). (i) (αl−l +1 , . . . , αl ) is a base of R (so α1 , . . . , αl−l ∈ R+ (ii) σ(R+ ) = R− . Lemma. There exists a permutation of order 2, i → i• , of {1, . . . , l − l } such that, for 1 i l − l , we have σ(αi ) = −αi• +
l j=l−l +1
where −nij ∈ N for l − l + 1 j l. Proof. Let i ∈ {1, . . . , l − l }. We have σ(αi ) =
l j=1
nij αj ,
nij αj ,
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36 σ-root systems
with −nij ∈ N. Moreover, since σ(αi ) ∈ R− , there exists i• ∈ {1, . . . , l − l } such that nii• < 0. By applying σ, we obtain:
αi = So: αi =
l j=l−l +1
l−l
nij
j=1
nij αj +
l−l
nij
l
j=1
l−l
njk αk .
k=1
l−l l−l njk αk = nij njk αk .
k=1
k=1
j=1
If i, k ∈ {1, . . . , l − l } are distinct, then: 1=
l−l
nij nji ,
j=1
l−l
nij njk = 0.
j=1
Since nij ∈ −N and nii• = 0, we have ni• k = 0 for k = i. But: 1=
l−l
ni• k nki• .
k=1
We deduce therefore that nii• = ni• i = −1. As ni• i = 0, we have shown that nik = 0 for k = i• . So the result follows immediately. 36.2.4 Let us conserve the notations of 36.2.3. From the preceding discussions, we may assume that there exist integers l1 , l2 , l such that l − l = l1 + 2l2 , l = l1 + l2 and:
⎧ if 1 i l1 ⎨i i• = i + l2 if l1 + 1 i l1 + l2 ⎩ i − l2 if l1 + l2 + 1 i l1 + 2l2
If x ∈ V , then 2x = x − σ(x). Consequently: ⎧ α if 1 i l1 ⎪ ⎪ i ⎪ ⎨1 (α + αi+l2 ) if l1 + 1 i l1 + l2 (1) αi = 2 i ⎪ ⎪ ⎪ ⎩ 1 (α + α i i−l2 ) if l1 + l2 + 1 i l1 = 2l2 . 2 From this, we deduce that (α1 , . . . , αl ) is a basis of V . Hence: dim V = l = l1 + l2 . On the other hand, it is clear that any element of S is a linear combination (with integer coefficients of the same sign) of α1 , . . . , αl . So (α1 , . . . , αl ) is a base of S (18.8.11).
36.2 Restricted root systems
543
Lemma. Let w ∈ Wσ . Then w ∈ W if and only if w|V = idV , and for w to be an element of W , it suffices that: {w(α1 ), . . . , w(αl )} = {α1 , . . . , αl }. ∗
Proof. If α ∈ V , then α∨ ∈ V , so sα |V = idV . It follows that if w ∈ W , then w|V = idV . Suppose that w|V = idV . Since w ∈ Wσ , we have w(R ) = R , and it is clear that {w(αl−l +1 ), . . . , w(αl )} is a base of R . It follows from 18.8.7 that there exists w ∈ W such that: {w(αl−l +1 ), . . . , w(αl )} = {w (αl−l +1 ), . . . , w(αl )}. Replace w by w
−1
◦w, we are reduced to the case where:
{w(αl−l +1 ), . . . , w(αl )} = {αl−l +1 , . . . , αl }. Let i ∈ {1, . . . , l − l }. We have αi − σ(αi ) ∈ V , and so: w(αi ) − σ(w(αi )) = w(αi − σ(αi )) = αi − σ(αi ). As w(R ) = R , we have w(R ) = R . Also, σ(αi ) ∈ R− (36.2.3). We deduce therefore from the above equality and 36.2.3 (i) that w(αi ) ∈ R+ . Consequently, the positive roots with respect to the base {w(α1 ), . . . , w(αl )} are the positive roots with respect to the base {α1 , . . . , αl }. It follows from 18.8.7 that w = idV . Now suppose that {w(α1 ), . . . , w(αl )} = {α1 , . . . , αl }. Replace w by −1 w ◦w, with w ∈ W , we may assume that {αl−l +1 , . . . , αl } is invariant by w. Let i ∈ {1, . . . , l−l }. In view of our assumption, there exists k ∈ {1, . . . , l− l } such that w(αi ) = αk . So w(αi ) = αk − αk + w(αi ). We deduce therefore from w(V ) ⊂ V that αk − w(αi ) = αk − w(αi ) ∈ V . By 36.1.4 (ii), w(αi ) ∈ . So we can conclude as in the previous case. R+
36.2.5 Theorem. Let R be a normal σ-root system in V , S be defined as in 36.2.1 and WS the Weyl group of S. The map w → w|V induces a surjective homomorphism Wσ → WS , with kernel W . Proof. Let us use the notations of 36.2.3 and 36.2.4. Let w ∈ Wσ . We have w(R ) = R and w(R ) = R , so w(S) = S. The set {w(α1 ), . . . , w(αl )} is a base of S (18.8.11). So there exists w ∈ WS such that {w(α1 ), . . . , w(αl )} = {w (α1 ), . . . , w (αl )}. Let w1 ∈ Wσ be such that w = w1 |V (36.2.2). We have w1−1 ◦w ∈ Wσ , and w1−1 ◦w stabilizes {α1 , . . . , αl }. By 36.2.4, we have w = w1 ◦w2 , with w2 ∈ W . So w|V = w1 |V = w . The map w → w|V induces therefore a surjective map from W onto WS (36.2.2), with kernel W (36.2.4).
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36 σ-root systems
36.2.6 We shall use the notation VQ of 36.1.4 and 36.1.5. Proposition. Let λ, µ ∈ VQ . The following conditions are equivalent: (i) µ ∈ WS (λ). (ii) µ ∈ W (λ). Thus if x ∈ VQ , then WS (x) = W (x) ∩ VQ (x). Proof. (i) ⇒ (ii) This is clear by 36.2.5. (ii) ⇒ (i) In the notations of 36.2.3 and 36.2.4, there exists a base {α1 , . . . , αl } of R verifying the following conditions: a) {αl−l +1 , . . . , αl } is a base of R . . b) α1 , . . . , αl−l ∈ R+ c) {α1 , . . . , αl } is a base of S. d) The relations (1) of 36.2.4 are verified. Set: D = {x ∈ VQ ; x, αi∨ 0, 1 i l}, DS = {x ∈ VQ ; x, (αi )∨ 0, 1 i l }. . Thus: For l1 + 1 i l1 + l2 , we have αi = αi+l 2
DS = {x ∈ VQ ; x, (αi )∨ 0, 1 i l − l }.
On the other hand, if α ∈ R , then (α )∨ ∈ {α∨ , 2α∨ , 4α∨ } (see the proof of 36.2.1). Hence DS ⊂ DS . Let w ∈ W be such that µ = w(λ). By 18.8.11 (where we take only the intersections of the chambers with VQ ), there exist w1 , w2 ∈ WS and ν ∈ DS such that w2 (µ) ∈ DS and λ = w1 (ν). Using 36.2.5, we see that there exist w1 , w2 ∈ Wσ verifying w1 = w1 |V , w2 = w2 |V . So w2 (µ) = w2 ◦w◦w1 (ν). Since w2 (µ), ν ∈ DS ⊂ D , it follows from 18.8.9 that w2 (µ) = ν. Hence −1 −1 µ = w2 ◦w1 (λ), and so µ ∈ WS (λ) (36.2.5).
36.3 Restriction of a root 36.3.1 Let R be a normal σ-root system in V , and let us conserve the notations of the preceding sections. If λ ∈ S, denote by S(λ) the set of α ∈ R such that α = λ. Set m(λ) = card S(λ). We say that λ ∈ S is special if λ ∈ R. So λ ∈ S is special if and only if there exists α ∈ S(λ) verifying σ(α) = −α. 36.3.2 Lemma. The following conditions are equivalent for λ ∈ S: (i) λ is special. (ii) m(λ) is odd. Proof. If α = α + α ∈ R, then σ(α) = α − α . Thus: α ∈ S(λ) ⇔ −σ(α) ∈ S(λ). It follows that if λ is special, then S(λ) \ {λ} is the disjoint union of sets of the form {α, −σ(α)}. So m(λ) is odd. The converse is analogue.
36.3 Restriction of a root
545
36.3.3 Lemma. Let λ ∈ S. Then: (i) If m(λ) is odd, then 2λ ∈ S. (ii) If m(λ) is even and 2λ ∈ S, then m(2λ) is odd. Proof. (i) Suppose that m(λ) is odd and 2λ ∈ S. Then λ ∈ R (36.3.2). Since ∗ σ(λ) = −λ, we have (−λ)∨ = −λ∨ = σ(λ)∨ = θ(λ∨ ), so λ∨ ∈ V . Let α ∈ R verify α = 2λ. Then: α, λ∨ = α , λ∨ = 4. This implies, by 18.5.2, that α = 2λ. This is absurd since R is reduced. (ii) Suppose that m(λ) and m(2λ) are even. Let us fix α ∈ S(λ) and β ∈ S(2λ). It follows from 36.3.2 that α = −σ(α) and β = −σ(β). • Since R is normal, we have α + σ(α) ∈ R, hence σ(α), α∨ 0 (18.5.3). On the other hand, α − σ(α) = 2λ ∈ R, because m(2λ) is even (36.3.2). We deduce therefore that σ(α), α∨ 0. Thus σ(α), α∨ = 0. Hence:
2 = α + λ, α∨ + α∨ = α , α∨ + λ, α∨ 0 = α − λ, α∨ + α∨ = α , α∨ − λ, α∨ Consequently:
α , α∨ = λ, α∨ = 1.
• Similarly, β + σ(β) ∈ R, where σ(β), β ∨ 0. If σ(β), β ∨ > 0, then β − σ(β) = 4λ ∈ R. So 4λ ∈ S which is absurd because λ ∈ S (18.5.3). We deduce therefore that σ(β), β ∨ = 0. As in the preceding point, we obtain that: β , β ∨ = 2λ, β ∨ = 1. • Suppose that α, β ∨ < 0. Then α + β ∈ R, so 3λ ∈ S which is absurd. Hence α, β ∨ 0. Similarly, we have: α, β ∨ 0 , β, α∨ 0 , σ(α), β ∨ 0 , σ(β), α∨ 0. Now:
1 + α , β ∨ 0 2 1 σ(α), β ∨ = α , β ∨ − λ, β ∨ = − + α , β ∨ 0 2 1 Since α, β ∨ and σ(α), β ∨ are integers, we have α , β ∨ = ± . Also, we 2 have: ∨ ∨ ∨ ∨ β, α = β , α + 2λ, α = β , α + 2 0 σ(β), α∨ = β , α∨ − 2λ, α∨ = β , α∨ − 2 0.
α, β ∨ = α , β ∨ + λ, β ∨ =
1 If α , β ∨ = − , then α, β ∨ = 0, so β, α∨ = 0. We deduce that 2 β , α∨ = −2, and so σ(β), α∨ = −4. By 18.5.2, we obtain that σ(β) = 2α. This is absurd since R is reduced. 1 If α , β ∨ = , then we obtain β = 2α. Contradiction. 2 • We have therefore proved that m(2λ) is odd.
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36 σ-root systems
36.3.4 Lemma. Let λ ∈ S. If m(λ) is odd, then σ(α), α∨ = 0 for all α ∈ S(λ) \ {λ}. Proof. If σ(α), α∨ > 0, then α − σ(α) ∈ R, so 2λ ∈ S. But this contradicts 36.3.3 (i). If σ(α), α∨ < 0, then since α = −σ(α) (because α = λ), we have α + σ(α) ∈ R. This is absurd because R is normal. 36.3.5 Lemma. Let λ ∈ S be such that m(λ) is even. (i) We have 2λ ∈ S if and only if there exists α ∈ S(λ) σ(α), α∨ > 0. If these conditions are verified, σ(β), β ∨ > β ∈ S(λ). (ii) We have 2λ ∈ S if and only if there exists α ∈ S(λ) σ(α), α∨ = 0. If these conditions are verified, σ(β), β ∨ = β ∈ S(λ).
such that 0 for all such that 0 for all
Proof. (i) If σ(α), α > 0, then α − σ(α) = 2λ ∈ R, so 2λ ∈ S. Conversely, suppose that 2λ ∈ S. Since m(λ) is even, we have λ ∈ R (36.3.2), so σ(α) = −α for α ∈ S(λ). The system R being normal, α + σ(α) ∈ R, hence α, σ(α)∨ 0. On the other hand, m(2λ) is odd (36.3.3), hence 2λ = α − σ(α) ∈ R (36.3.2). So the σ(α)-string of α is of the form α − qσ(α), . . . , α − σ(α), α, with q 1. But −q = −α, σ(α)∨ (18.5.4). So α, σ(α)∨ > 0. (ii) Suppose that σ(α), α∨ = 0. By (i), 2λ ∈ S. Conversely, if 2λ ∈ S, then (i) implies that σ(α), α∨ 0. If σ(α), α∨ < 0, then α + σ(α) ∈ R, unless σ(α) = −α. But if σ(α) = −α, then α ∈ S, so m(λ) is odd (36.3.2). Contradiction. Since the system is normal, the result follows. 36.3.6 Lemma. Let λ ∈ S and α, β ∈ S(λ) be such that α = β and α = −σ(β). (i) If 2λ ∈ S, then α, β ∨ > 0. (ii) If 2λ ∈ S, then α, β ∨ 0. Furthermore, α, β ∨ > 0 if and only if α, σ(β)∨ = 0. Proof. (i) Suppose that 2λ ∈ S. a) If λ = α, then α, β ∨ = λ, β ∨ . Moreover, m(λ) is odd (36.3.2) and ∨ σ(β), β = 0 (36.3.5). We deduce therefore that β , β ∨ = λ, β ∨ . So:
α, β ∨ = λ, β ∨ =
1 1 (β , β ∨ + λ, β ∨ ) = β, β ∨ = 1. 2 2
b) Suppose that α = λ and β = λ. By 36.3.4 and 36.3.5:
σ(α), α∨ = 0 ⇒ α , α∨ = λ, α∨ σ(β), β ∨ = 0 ⇒ β , β ∨ = λ, β ∨
36.3 Restriction of a root
Since
547
2 = α, α∨ = α , α∨ + λ, α∨ = β, β ∨ = β , β ∨ + λ, β ∨ ,
we obtain that:
α, β ∨ = α , β ∨ + λ, β ∨ = α , β ∨ + 1 σ(α), β ∨ = α , β ∨ − λ, β ∨ = α , β ∨ − 1. Hence:
α, β ∨ − σ(α), β ∨ = 2.
Exchanging the roles of α and β, we have also: β, α∨ − σ(β), α∨ = 2. Now, since 2λ ∈ S, we have α + β ∈ R, so α, β ∨ 0. Similarly, from σ(α) − β ∈ R, we deduce that σ(α), β ∨ 0. If α, β ∨ = 0, then β, α∨ = 0 and σ(β), α∨ = −2. By 18.5.2, we have either α = −σ(β), which is not possible by our hypothesis, or α, σ(β)∨ = −1, so σ(α), β ∨ = −1. But this is absurd since σ(α), β ∨ = −2. So we have proved that α, β ∨ > 0. (ii) Suppose that 2λ ∈ S. Then m(λ) is even, and by 36.3.5: σ(α), α∨ > 0 , σ(β), β ∨ > 0. Also, we have 2λ ∈ R (36.3.2 and 36.3.3). Since 2λ + α ∈ R (otherwise 3λ ∈ S), 2λ, α∨ 0. Moreover: 0 2λ, α∨ = α, α∨ − σ(α), α∨ = 2 − σ(α), α∨ < 2. So 2λ, α∨ ∈ {0, 1}. If 2λ, α∨ = 0, then σ(α), α∨ = 2. By 18.5.2, either σ(α) = α, which is absurd, or α, σ(α)∨ = 1, that is σ(α), α∨ = 1. Contradiction. Thus 2λ, α∨ = 1. Similarly 2λ, β ∨ = 1. It follows that: β, α∨ − σ(β), α∨ = 1 = α, β ∨ − σ(α), β ∨ . Since we have:
2λ, α∨ = 1 = 2 − σ(α), α∨ = 2 − σ(β), β ∨ , σ(α), α∨ = σ(β), β ∨ = 1.
Suppose that β, α∨ < 0. Then α + β ∈ R. Moreover 2α + β ∈ R because 3λ ∈ S. So the α-string of β is of the form β − qα, . . . , β, β + α.
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We have 1 − q = −β, α∨ > 0 (18.5.4), hence q = 0. Thus β − α ∈ R and β, α∨ = −1. We deduce therefore that σ(β), α∨ = −2. As α = −σ(β), we have α, σ(β)∨ = −1 (18.5.2), hence σ(α), β ∨ = −1. Since 1 = α, β ∨ − σ(α), β ∨ , we deduce that α, β ∨ = 0, and so β, α∨ = 0. Contradiction. We have therefore proved that β, α∨ 0. Replacing β by −σ(β), we obtain that σ(β), α∨ 0. As β, α∨ and σ(β), α∨ are integers and β, α∨ − σ(β), α∨ = 1 = α, β ∨ − σ(α), β ∨ , it follows that: β, α∨ > 0 ⇔ β, α∨ = 1 ⇔ σ(β), α∨ = 0. So we have obtained (ii).
References and comments • [1], [29], [87]. Most of the results presented in this chapter were first obtained in [1].
37 Symmetric Lie algebras
37.1 Primary subspaces 37.1.1 In this section, we shall generalize certain notions in chapter 19. Let g be a Lie algebra and (V, σ) a finite-dimensional representation of g. When V = g, we shall assume that σ is the adjoint representation of g. Let A be a subset of g and denote by P the set of maps from A to the field k. For λ ∈ P , set Vλ (A) (resp. V λ (A)) to be the subspace of V consisting of vectors v such that for all x ∈ A, we have σ(x)(v) = λ(x)v (resp. (σ(x) − λ(x) idV )n (v) = 0 for some integer n). We shall call Vλ (A) the eigenspace of V relative to λ (and σ), and V λ (A) the primary subspace of V relative to λ (and σ). If V λ (A) is non-zero, we say that λ is a weight of A in V , and any non-zero vector in V λ (A) shall be called a weight vector. The set of weights of A in V will be denoted by ∆A (V ), and the subspace V 0 (A), corresponding to the map x → 0 for all x ∈ A, is called the nilspace of V (with respect to σ). When A is reduced to a singleton x, we can identify P with k. In this case, we shall denote Vλ (A) and V λ (A) simply by Vλ(x) (x) and V λ(x) (x). It follows in the general case that: λ(x) Vλ(x) (x) , V λ (A) = V (x). Vλ (A) = x∈A
x∈A
Let u ∈ End(V ), then by considering the abelian Lie algebra ku and the representation induced by the inclusion ku ⊂ End(V ), we may define for α ∈ k the subspaces Vα (u) and V α (u) as above. Thus we recover the definitions of 19.3.8. 37.1.2 Proposition. The sums Vλ (A) and λ∈P
are direct, and the set ∆A (V ) is finite.
λ∈P
V λ (A)
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37 Symmetric Lie algebras
Proof. Since V is finite-dimensional, it is clear that ∆A (V ) is finite. Also, it is clear that we only need to prove that the second sum is direct. 1) If A is empty, then there is nothing to prove. 2) If A = {x}, then this is just a consequence of the decomposition of V into generalized eigenspaces of σ(x). 3) Let us assume that card A 2 and suppose that there exists a nonempty subset I of ∆A (V ) minimal with the property that we can find a non-zero vector vλ in V λ (A) verifying λ∈I vλ = 0. Clearly card I 2. Let µ, ν ∈ I be distinct. Then there exists x ∈ A such that µ(x) = ν(x). Let Q = {λ(x); λ ∈ I} and J = {λ ∈ I; λ(x) = µ(x)}. Since µ(x) = ν(x), card Q 2. Moreover, for α ∈ Q, we verify easily that vλ ∈ V α (x) if and only if λ(x) = α. We saw in point 2 that the sum α∈Q V α (x) is direct. This implies for α = µ(x) that λ∈J vλ = 0. Since ν ∈ J, this contradicts the minimality of the set I. 37.1.3 Proposition. The following conditions are equivalent: (i) A ⊂ g0 (A). (ii) For all λ ∈ P , V λ (A) is σ(A)-stable, and we have: V λ (A). V = λ∈∆A (V )
Proof. (i) ⇒ (ii) The subspace V λ (A) is the intersection of the V λ(x) (x), for x ∈ A. By the hypothesis and 19.3.8, each V λ (A) is σ(A)-stable. We shall prove by induction on dim V that V is the sum of the V λ (A)’s. If σ(x) has a unique eigenvalue λ(x) for all x ∈ A, then clearly V = V λ (A). Otherwise, there exists x ∈ A such that σ(x) has at least two distinct eigenvalues. Now V is the direct sum of the subspaces V α (x), α ∈ k, and dim V α (x) < dim V for all α. Since each V α (x) is σ(A)-stable (19.3.8), the result follows by the induction hypothesis. (ii) ⇒ (i) This follows from 19.3.8. 37.1.4 Proposition. (i) Let λ, µ ∈ P . Then: σ gλ (A) V µ (A) ⊂ V λ+µ (A) , [gλ (A), gµ (A)] ⊂ gλ+µ (A). In particular, g0 (A) is a Lie subalgebra of g. (ii) Suppose that A is a subspace of g and A ⊂ g0 (A). Then all the elements of ∆A (V ) are linear forms on A. Proof. (i) Let x ∈ A, α = λ(x) and β = µ(x). Set S = ad x − α idg and T = σ(x) − β idV . If y ∈ gλ (A) and v ∈ V µ (A), then we obtain by induction on n that: n n n i n−i σ(y)(v) = (v) . σ(x) − (α + β) idV σ S (y) T i=0 i
37.1 Primary subspaces
551
Hence the first inclusion. The second inclusion is just a special case of the first. (ii) Let λ ∈ ∆A (V ). By the hypotheses and 19.3.8, V λ (A) is σ(A)-stable. So we are reduced to the case where V = V λ (A). If x ∈ A, n then the characteristic polynomial χx of σ(x) verifies χx (X) = X − λ(x) , where n = dim V . On the other hand, χx (X) = X n + a1 (x)X n−1 + · · · + an (x), where ai is a polynomial function on A, of degree i, 0 i n. Thus nλ(x) = −a1 (x) for all x ∈ A. Since the characteristic of k is zero, we are done. 37.1.5 Proposition. Let A be a subset of g verifying A ⊂ g0 (A), and a the subspace of g spanned by A. Then there exists x ∈ a such that σ(x) induces an automorphism of V /V 0 (A). Proof. Let λ ∈ ∆A (V ) \ {0}. If we set dλ (y) = det(σ(y)|V λ (A) ), then dλ is a non-zero polynomial function on a. Since k is infinite, the product d of the dλ ’s, for λ ∈ ∆A (V ) \ {0}, is non-zero on a. So any x ∈ a verifying d(x) = 0 works. 37.1.6 Set:
V • (A) =
i σ(x) (V ) . x∈A
i1
Proposition. Suppose that A ⊂ g0 (A). Then: (i) V 0 (A) and V • (A) are σ(A)-stable. (ii) V = V 0 (A) ⊕ V • (A). (iii) Any σ(A)-stable subspace W of V verifying W 0 (A) = {0}, is contained in V • (A). (iv) We have: σ(x)(V • (A)) = V • (A). x∈A •
Moreover, V (A) is the only σ(A)-stable subspace of V which is also a complementary subspace of V 0 (A) in V . Proof. Each V λ (A) is σ(A)-stable and V is the sum of the V λ (A)’s, for λ ∈ P (37.1.3). If x ∈ A, the characteristic polynomial of σ(x)|V λ (A) is (X − λ(x))n(λ) , where n(λ) = dim V λ (A). Thus the intersection i1 (σ(x)i )(V λ (A)) is zero if λ(x) = 0, and is equal to V λ (A) if λ(x) = 0. It follows that: V λ (A). V • (A) = λ∈P \{0}
Hence (i), (ii) and (iv). Let W be a subspace of V verifying the conditions of (iii). Then:
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37 Symmetric Lie algebras
W =
W λ (A) , W λ (A) = W ∩ V λ (A).
λ∈P
Since W 0 (A) = {0}, we have (iii). Finally, let U be a σ(A)-stable subspace of V verifying U ∩ V 0 (A) = {0}. Then U 0 (A) = {0}, and so U ⊂ V • (A) by (iii). The last assertion is now clear. 37.1.7 In the notations of 37.1.6, the decomposition V = V 0 (A) ⊕ V • (A) is called the Fitting decomposition of V (with respect to σ). When A = {x}, we shall denote V 0 (A) and V • (A) simply by V 0 (x) and V • (x). As in 37.1.1, if u ∈ End(V ), we shall also use the notations V 0 (u) and • V (u). Proposition. Let u ∈ End(V ), s and n the semisimple and nilpotent components of u. For all λ ∈ k, we have V λ (u) = V λ (s) = Vλ (s). Proof. Let g = ku. Then u ∈ g0 (u) and V is the sum of the V λ (u), λ ∈ k. Since s is a polynomial without constant term in u, V λ (u) is s-stable. But s|V λ (u) = λ idV λ (u) , so the result follows. 37.1.8 A bilinear form b on V is called A-invariant if b(σ(x)v, w) + b(v, σ(x)w) = 0 for all v, w ∈ V and x ∈ A. Proposition. Let b be an A-invariant bilinear form on V and λ, µ ∈ P . If λ + µ = 0, then V λ (A) and V µ (A) are orthogonal with respect to b. Proof. Let x ∈ A, f = σ(x) − λ(x) idV , g = σ(x) − µ idV , v ∈ V λ (A) and w ∈ V µ (A). Since b is A-invariant, we have: − λ(x) + µ(x) b(v, w) = b f (v), w + b v, g(w) . By induction on n, we obtain that: n i b f (v), g n−i (w) . i=0 i
n n (−1)n λ(x) + µ(x) b(v, w) =
The result follows immediately.
37.2 Definition of symmetric Lie algebras
553
37.2 Definition of symmetric Lie algebras 37.2.1 Definition. A symmetric Lie algebra is a triple (g; k, p) where g is a Lie algebra and k, p are subspaces of g verifying: g = k ⊕ p , [k, k] ⊂ k , [k, p] ⊂ p , [p, p] ⊂ k. The direct sum decomposition g = k ⊕ p will be called the symmetric decomposition of g. 37.2.2 Let (g; k, p) be a symmetric Lie algebra. By definition, k is a Lie subalgebra of g. Let θ ∈ End(g) be defined by θ(x) = x if x ∈ k, and θ(x) = −x if x ∈ p. It is easy to check that θ is an involution of g, and k, p are its eigenspaces with eigenvalues 1 and −1. It is obvious that we have the following converse: Proposition. Let θ be an involutive automorphism of a Lie algebra g, and: k = {x ∈ g; θ(x) = x} , p = {x ∈ g; θ(x) = −x}. Then (g; k, p) be a symmetric Lie algebra. 37.2.3 In other words, 37.2.2 says that a symmetric Lie algebra is a pair (g, θ) where g is a Lie algebra and θ an involutive automorphism of g. The involution θ determines k and p, and vice versa. We shall also use the notation (g, θ) for a symmetric Lie algebra. Let L be the Killing form on g. Since, for x ∈ k and y ∈ p, we have: (ad x) ◦ (ad y)(k) ⊂ p and (ad x) ◦ (ad y)(p) ⊂ k, it follows that L(x, y) = {0}. Thus k and p are orthogonal with respect to L. In particular, if L is non-degenerate, then k is the orthogonal of p, and vice versa. A Lie subalgebra of a Lie algebra g will be called symmetrizing if it is the set of fixed points of an involutive automorphism of g. 37.2.4 Examples. 1) Let n be an integer greater than or equal to 2, and g = sln (k). For x ∈ g, set θ(x) = −t x where t x is the transpose of the matrix x. Then θ is an involutive automorphism of g, and k (resp. p) is the set of skew-symmetric (resp. symmetric) matrices in g. 2) Let s be a Lie algebra, g = s × s and θ the involutive automorphism of g defined by θ(x, y) = (y, x). The symmetric Lie algebra (g, θ) is called the diagonal defined by s. Here k = {(x, x) ; x ∈ s} and p = {(x, −x) ; x ∈ s}. 3) A Lie algebra g is said to be graded if g = p∈Z gp is Z-graded as a vector space, and for all p, q ∈ Z, we have [gp , gq ] ⊂ gp+q . Let g be such a Lie algebra. Take k (resp. p) to be the direct sum of the gp ’s for p even (resp. odd). Then (g; k, p) is a symmetric Lie algebra.
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37 Symmetric Lie algebras
37.3 Natural subalgebras 37.3.1 Let (g; k, p) or (g, θ) be a symmetric Lie algebra. For A ⊂ g, using the notations of 37.1, let: k0 (A) = k ∩ g0 (A) , k• (A) = k ∩ g• (A), p0 (A) = p ∩ g0 (A) , p• (A) = p ∩ g• (A). If A = {x}, then we shall denote p0 (A) by p0 (x). 37.3.2 Proposition. Let A ⊂ k or A ⊂ p be such that A ⊂ g0 (A). Then: g0 (A) = k0 (A) ⊕ p0 (A) , g• (A) = k• (A) ⊕ p• (A), k = k0 (A) ⊕ k• (A) , p = p0 (A) ⊕ p• (A), g = k0 (A) ⊕ k• (A) ⊕ p0 (A) ⊕ p• (A). Proof. Since g = g0 (A) ⊕ g• (A) (37.1.6), we only need to prove the equalities on the first line. Let y = k + p, with k ∈ k, p ∈ p. Now y ∈ g0 (A) if and only if for all x ∈ A, there exists n ∈ N such that: 0 = (ad x)2n (y) = (ad x)2n (k) + (ad x)2n (p). As (ad x)2n (k) ⊂ k and (ad x)2n (p) ⊂ p (because A ⊂ k or A ⊂ p), we deduce that g0 (A) = k0 (A) ⊕ p0 (A). Similarly: g• (A) = (ad x)2i (g) . x∈A
i0
So g• (A) = k• (A) ⊕ p• (A) since (ad x)2i (g) = (ad x)2i (k) ⊕ (ad x)2i (p). 37.3.3 Definition. Let (g; k, p) be a symmetric Lie algebra. (i) A subspace q of p is called a p-subalgebra of g (or of p) if (ad x)2 (q) ⊂ q for all x ∈ q. (ii) A p-subalgebra q of g is said to be natural if there exists x0 ∈ q such that (ad x0 )2 induces an automorphism of the vector space p/q. 37.3.4 Theorem. Let (g; k, p) be a symmetric Lie algebra, A ⊂ p verifying A ⊂ g0 (A), and a the subspace of p spanned by A. (i) The subspace p0 (A) is a natural p-subalgebra of g. (ii) There exists x ∈ a such that (ad x)2 induces an automorphism of the vector space p/p0 (A). Proof. By 37.1.4, g0 (A) is a Lie subalgebra of g. If x ∈ p0 (A), then (ad x)2 (p) ⊂ p and (ad x)2 (g0 (A)) ⊂ g0 (A). So p0 (A) is a p-subalgebra of g. By 37.1.5, there exists x ∈ a ⊂ p0 (A) such that ad x induces an automorphism of g/g0 (A) = (k/k0 (A)) ⊕ (p/p0 (A)). So (ad x)2 induces an automorphism of p/p0 (A). So we have part (ii) and p0 (A) is a natural p-subalgebra of g.
37.4 Cartan subspaces
555
37.4 Cartan subspaces 37.4.1 Definition. Let (g; k, p) be a symmetric Lie algebra. An element x ∈ p is called p-generic in g if for all y ∈ p : dim p0 (x) dim p0 (y). 37.4.2 For x ∈ p, denote by χx the characteristic polynomial of (ad x)2 |p : χx (X) = X r + ar−1 (x)X r−1 + · · · + as (x)X s where dim p = r and ar−1 , . . . , as are polynomial functions on p, with as = 0. The subspace p0 (x) is the generalized eigenspace of (ad x)2 associated to the eigenvalue 0. So we have dim p0 (x) = s if x is p-generic, and the set of p-generic elements in g is a non-empty Zariski-open subset of p. 37.4.3 Definition. Let (g; k, p) be a symmetric Lie algebra. A subset a of p is called a Cartan subspace of g if a = p0 (a). Remark. It is clear that a Cartan subspace of g is indeed a subspace of the vector space p. 37.4.4 Theorem. Let (g; k, p) be a symmetric Lie algebra and a a subset of p. The following conditions are equivalent: (i) a is a Cartan subspace of g. (ii) a is a minimal natural p-subalgebra of g. (iii) a ⊂ g0 (a) and there exists x ∈ p such that a = p0 (x). (iv) a is a minimal natural p-subalgebra of g verifying a ⊂ g0 (a). Proof. (i) ⇒ (ii) Let a be a Cartan subspace of g. By 37.3.4, a is a natural p-subalgebra of g. Let q ⊂ a be a natural p-subalgebra of g and x ∈ q be such that (ad x)2 induces an automorphism of p/q. Then (ad x)2 induces an automorphism of a/q. Since (ad x)n (a) = {0} for some large n, we deduce that q = a. (ii) ⇒ (iii) For x ∈ a, let δ(x) be the determinant of the endomorphism of p/a induced by (ad x)2 . If (ii) is verified, then there exists x0 ∈ a such that δ(x0 ) = 0. It follows that p0 (x0 ) ⊂ a. But p0 (x0 ) is a natural p-subalgebra by 37.3.4. Thus a = p0 (x0 ). Let n = dim a. For x ∈ a, let [aij (x)] be the matrix of (ad x)2n |a with respect to a basis of a. If a ⊂ g0 (a), then there exists x ∈ a such that (ad x)2n |a = 0. So we can find integers p, q such that apq (x) = 0. As k is infinite, the polynomial function x → apq (x)δ(x) is non-zero on a. Therefore if y ∈ a verifies apq (y)δ(y) = 0, then p0 (y) ⊂ a and (ad y)2n |a = 0, so (ad y)2 |a is not nilpotent. Thus p0 (y) = a. This is absurd since p0 (y) is natural (37.3.4). (iii) ⇒ (iv) By 37.3.4, a is a natural p-subalgebra of g, and we may conclude as in the proof of (i) ⇒ (ii). (iv) ⇒ (i) Since a ⊂ g0 (a), we have a ⊂ p0 (a). If a is a natural p-subalgebra of g, then a = p0 (a) as in the proof of (i) ⇒ (ii).
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37 Symmetric Lie algebras
37.4.5 Corollary. (i) Any symmetric Lie algebra has a Cartan subspace. (ii) If x ∈ g is p-generic, then p0 (x) is a Cartan subspace of g. Proof. (i) Since g has natural p-subalgebras, (i) follows from 37.4.4. (ii) Let x be a p-generic element of g. If p0 (x) is not a Cartan subspace, then there exists a natural p-subalgebra q which is contained strictly in p0 (x). Let x0 ∈ q be such that (ad x0 )2 induces an automorphism of p/q. Then p0 (x0 ) ⊂ q p0 (x). This is absurd since x is p-generic. 37.4.6 Proposition. Let q be a p-subalgebra of g. The following conditions are equivalent: (i) q is natural. (ii) q contains a natural p-subalgebra of g. (iii) q contains a Cartan subspace of g. (iv) There exists x ∈ p such that p0 (x) ⊂ q. Proof. (i) ⇒ (ii) This is clear. (ii) ⇒ (iii) If (ii) is verified, then q contains a minimal natural p-subalgebra of g. So q contains a Cartan subspace (37.4.4). (iii) ⇒ (iv) This follows clearly from the equivalence (i) ⇔ (iii) of 37.4.4. (iv) ⇒ (i) If p0 (x) ⊂ q, then as (ad x)2 induces an automorphism of p/p0 (x) (37.3.4), it induces an automorphism of p/q. 37.4.7 Theorem. Let A be a subset of p verifying the following conditions: a) [x, y] = 0 for all x, y ∈ A. b) adg x is semisimple for all x ∈ A. Then A is contained in a Cartan subspace a of g. In particular, if x ∈ p and adg x is semisimple, then x belongs to a Cartan subspace of g. Proof. Condition a) implies that A ⊂ g0 (A), so p0 (A) is a natural p-subalgebra of g (37.3.4). Now condition b) implies that p0 (A) is the centralizer of A in p. Let a be a Cartan subspace of g contained in p0 (A) (37.4.6), then [a, x] = {0} for all x ∈ A, hence A ⊂ p0 (a) = a. 37.4.8 Proposition. Let a be a Cartan subspace of g and h the Lie subalgebra of g generated by a. Then h = a ⊕ [a, a], and h is nilpotent. Proof. We have a = p0 (a) ⊂ g0 (a), and g0 (a) is a Lie subalgebra of g. It follows that: [a, [a, a]] ⊂ g0 (a) ∩ p = a , [[a, a], [a, a]] ⊂ [a, a]. Thus h = a ⊕ [a, a]. Let L be the Killing form of h. Since (h; [a, a], a) is a symmetric Lie algebra, a and [a, a] are orthogonal with respect to L (37.2.3). As (ad x)2 |h is nilpotent for x ∈ a, L(x, x) = 0. It follows that L(a, a) = {0}, and hence L(h, a) = {0}. This implies that a is contained in the radical of h. Since ad x|h is nilpotent for all x ∈ a, it follows from 19.5.7 that a is contained in the largest nilpotent ideal of h. But h = a ⊕ [a, a], so h is nilpotent.
37.5 The case of reductive Lie algebras
557
37.4.9 Corollary. Let a be a Cartan subspace of g and (V, σ) a finitedimensional representation of g. There exists a basis E of V such that with respect to E, the matrix of the σ(x)’s, x ∈ a, are upper triangular. Proof. This is clear by 19.4.4 and 37.4.8. 37.4.10 Let K be the smallest algebraic subgroup of Aut g such that its Lie algebra contains adg k. Then the set {α|k ; α ∈ K} is the algebraic adjoint group of k (24.8.5), and we have α(k) = k, α(p) = p for all α ∈ K. Theorem. Let (g; k, p) be a symmetric Lie algebra and K as above. If a1 , a2 are Cartan subspaces of g, then there exists α ∈ K such that a2 = α(a1 ). Proof. For i = 1, 2, let Ui be the set of x ∈ ai such that (ad x)2 induces an automorphism of p/ai . Then Ui is a non-empty Zariski-open subset of ai , and if x ∈ Ui , then ai = p0 (x). Let ϕi : K × Ui → p , (α, x) → α(x). If x ∈ Ui , denote by fi,x the differential of ϕi at the point (e, x). Since Ui is open, if x ∈ Ui , the image of fi,x contains ai and [k, x] (see 29.1.4), so it contains (ad x)2 (p). As p = ai + (ad x)2 (p), the image of fi,x contains also p, and the interior of the image of ϕi is non-empty (15.4.2 and 16.5.7). From this, we deduce that there exist x0 ∈ p, αi ∈ K and xi ∈ Ui such that x0 = α1 (x1 ) = α2 (x2 ). If α = α2−1 α1 , then x2 = α(x1 ). So we have a2 = α(a1 ) because ai = p0 (xi ). 37.4.11 Corollary. Let (g; k, p) be a symmetric Lie algebra. Then: (i) All the Cartan subspaces of g have the same dimension. (ii) Any Cartan subspace of g is of the form p0 (x) where x is a p-generic element of g. Proof. (i) This is clear by 37.4.10. (ii) If x is p-generic, then p0 (x) is a Cartan subspace (37.4.5). Now if a is a Cartan subspace of g, then there exists y ∈ p such that a = p0 (y) (37.4.4). By (i), y is p-generic.
37.5 The case of reductive Lie algebras 37.5.1 Proposition. Let (g; k, p) or (g, θ) be a semisimple symmetric Lie algebra, that is, g is semisimple. (i) Let x ∈ g and s, n its semisimple and nilpotent components. If x ∈ k (resp. x ∈ p), then s, n ∈ k (resp. s, n ∈ p). (ii) The Lie algebra k is reductive in g and k = ng (k). Proof. (i) This is obvious since θ(x) = θ(s)+θ(n) is the Jordan decomposition of θ(x).
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37 Symmetric Lie algebras
(ii) Let L be the Killing form on g. By 37.2.3, the restriction of L to k and p are non-degenerate. So k is reductive in g by (i) and 20.5.12. Now the normalizer of k in g is of the form k ⊕ u where u is a subspace of p. We have [k, u] ⊂ k ∩ p = {0}, and as k is reductive in g, there exists a subspace o of p complementary to u such that [k, o] ⊂ o. From: [g, k + v] = [k + u + v, k + v] ⊂ k + [k, u + v] ⊂ k + v, we deduce that k ⊕ o is an ideal of g. But [u, u] ⊂ k, so g/(k ⊕ u) is abelian. Hence g = k ⊕ o (because g is semisimple), and u = {0}. 37.5.2 Theorem. Let (g; k, p) be a semisimple symmetric Lie algebra, L its Killing form, a a Cartan subspace of g and m the centralizer of a in k. (i) a is an abelian Lie subalgebra of g consisting of semisimple elements. (ii) The centralizer of a in g is m ⊕ a. (iii) The restrictions of L to a × a and m × m are non-degenerate. (iv) The Lie algebras a and m are reductive in g. (v) a is contained in the centre of g0 (a) and (g0 (a); m, a) is a reductive symmetric Lie algebra, that is, g0 (a) is reductive. (vi) Let l be a Cartan subalgebra of m. Then l ⊕ a is a Cartan subalgebra of g. Proof. By 37.1.3 and 37.1.4, we have: λ g= g (a). λ∈a∗
Also L(gλ (a), gµ (a)) = {0} if λ + µ = 0 (37.1.8). It follows that L|g0 (a)×g0 (a) is non-degenerate, and so g0 (a) is reductive (20.5.4 (iii)). Let L0 be the Killing form of g0 (a). If x ∈ a, (ad x)2 |g0 (a) is nilpotent, hence L0 (x, x) = 0. Thus L0 |a×a is both symmetric and alternating, so it is identical zero. As L0 (k0 (a), a) = {0} (37.2.3) and g0 (a) = k0 (a) ⊕ a, we see that a is contained in the kernel of L0 . Since g0 (a) is reductive, a is contained in the centre of g0 (a). In particular, a is abelian. It follows that the centralizer of a in g (resp. k, p) is g0 (a) (resp. m, a). So we have g0 (a) = m ⊕ a, and part (v) is clear. Recall from above that L|g0 (a)×g0 (a) is non-degenerate. Since L(m, a) = {0}, the restrictions of L to a × a and m × m are therefore non-degenerate. So by 20.5.4 (iii), a and m are reductive. Let l be as in (vi) and h = l ⊕ a. The Lie algebras l and a are nilpotent and [l, a] = {0}. So h is nilpotent. Now g0 (h) ⊂ g0 (a) = a ⊕ m. Thus if x ∈ g0 (h), we can write x = y + z with y ∈ a and z ∈ m. But z ∈ m0 (l), so z ∈ l because l is a Cartan subalgebra of m. It follows that h is a Cartan subalgebra of g. As a ⊂ h, the elements of a are semi-simple. So a is a torus of g, hence it is reductive in g (20.5.16). By 20.5.13, m ⊕ a is also reductive in g. We obtain therefore by 20.5.12 that m is reductive in g.
37.6 Linear forms
559
37.5.3 Theorem. Let (g; k, p) be a reductive symmetric Lie algebra, a a Cartan subspace of g and m the centralizer of a in k. (i) a is an abelian Lie subalgebra of g consisting of semisimple elements. (ii) The centralizer of a in g is m ⊕ a. (iii) The Lie algebras a and m are reductive in g. (iv) a is contained in the centre of g0 (a) and (g0 (a); m, a) is a reductive symmetric Lie algebra. (v) Let l be a Cartan subalgebra of m. Then l ⊕ a is a Cartan subalgebra of g. Proof. All the subspaces u considered here verify: u = (u ∩ [g, g]) ⊕ (u ∩ z(g)). So the assertions are direct consequences of 37.5.2.
37.5.4 Corollary. The following conditions are equivalent for a subset a of p: (i) a is a Cartan subspace of g. (ii) a is maximal among the set of subspaces of p consisting of semisimple elements which commute pairwise. (iii) a is a maximal subset of pairwise commuting semisimple elements in p. Proof. This is obvious by 37.4.7 and 37.5.3. 37.5.5 Remarks. 1) Let g = kx ⊕ ky be the Lie algebra defined by [x, y] = y. Set k = kx and p = ky, then (g; k, p) is a symmetric Lie algebra. Clearly a = p is a Cartan subspace of g and ck (p) = {0}. But a is not a Cartan subalgebra of g. 2) Let (Eij )1i,j3 be the canonical basis of gl3 (k). Set: g0 = kE11 + kE22 + kE33 , g1 = kE12 + kE23 , g2 = kE13 , gi = {0} if i ∈ Z \ {0, 1, 2}. The direct sum g of the gi ’s, i ∈ Z, is a graded Lie algebra. As we have seen in example 3 of 37.2.4, (g; g0 ⊕ g2 , g1 ) is a symmetric Lie algebra. It is easy to see that p = g1 is a Cartan subspace of g which is not abelian.
37.6 Linear forms 37.6.1 Let us identify k∗ (resp. p∗ ) with the set of linear forms on g which is zero on p (resp. k). Then g∗ = k∗ ⊕ p∗ . Let us consider the coadjoint representation of g on g∗ and denote the action by (x, f ) → x.f . If x ∈ k, y ∈ p, f ∈ k∗ and g ∈ p∗ , then: x.f ∈ k∗ , y.f ∈ p∗ , x.g ∈ p∗ , y.g ∈ k∗ .
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37 Symmetric Lie algebras
Using the notations of 19.7.3, we set, for f ∈ g∗ : k(f ) = k ∩ g(f ) , p(f ) = p ∩ g(f ) . If f ∈ k∗ ∪ p∗ , then we obtain easily that g(f ) = k(f ) + p(f ) . Proposition. If f ∈ p∗ , then: dim k − dim k(f ) = dim p − dim p(f ) . Proof. The bilinear form Φf of 19.7.3 induces a non-degenerate alternating f on g/g(f ) = (k/k(f ) ) × (p/p(f ) ). Since k/k(f ) et p/p(f ) are bilinear form Φ f , the result is clear. totally isotropic with respect to Φ 37.6.2 In view of 37.6.1, we have the following result: Proposition. The following conditions are equivalent for f ∈ p∗ : (i) dim k(f ) dim k(g) for all g ∈ p∗ . (ii) dim p(f ) dim p(g) for all g ∈ p∗ . (iii) dim g(f ) dim g(g) for all g ∈ p∗ . 37.6.3 We say that f ∈ p∗ is p-regular if it verifies the conditions of 37.6.2. Let us denote by p∗reg the set of p-regular elements of p∗ . It is an open dense subset of p∗ . 37.6.4 Proposition. Let f ∈ p∗reg . Then: (i) [k(f ) , p(f ) ] = {0}. (ii) k(f ) is an ideal of g(f ) , and [p(f ) , p(f ) ] is an abelian ideal of g(f ) . Proof. Since f ∈ p∗reg , we have dim g(f ) dim g(g) for all g ∈ p∗ , or equivalently, for all g ∈ g∗ such that f |k = g|k = 0. It follows from 19.7.4 that [g(f ) , g(f ) ] ⊂ k. Thus [p(f ) , k(f ) ] = {0}, and k(f ) is an ideal of g(f ) . Now, if x ∈ k(f ) , y, z ∈ p(f ) , then: [x, [y, z]] = [[x, y], z] + [y, [x, z]] = 0. So [p
(f )
,p
(f )
] is an abelian ideal of g(f ) .
37.6.5 Let g = kx ⊕ ky ⊕ kz be the Lie algebra defined as follows: [x, y] = z, [x, z] = [y, z] = 0. Then (g; k = kz, p = kx ⊕ ky) is a symmetric Lie algebra. For any f ∈ p∗ , we have g(f ) = g, p(f ) = p. Thus f is p-regular, but p(f ) is not abelian. Note that: kx = k , px = kx , dim k − dim kx = 0 , dim p − dim px = 1. So the result of 37.6.1 does not extend to the adjoint representation.
References and comments • [29], [50], [54], [55]. The notion of natural p-subalgebra comes from [55].
38 Semisimple symmetric Lie algebras
In this chapter, we examine to what extent the results for in Chapters 31, 32, 33, 34 and 35 for semisimple Lie algebras have analogues for semisimple symmetric Lie algebras.
38.1 Notations 38.1.1 In this chapter, (g; k, p) or (g, θ) (see 37.2.1 and 37.2.3) will be a semisimple symmetric Lie algebra whose Killing form will be denoted by L. We shall denote by a a Cartan subspace of g, m the centralizer of a in k and l a Cartan subalgebra of m. We saw in 37.5.2 that h = l ⊕ a is a Cartan subalgebra of g. 38.1.2 Let us denote by R the root system of (g, h) and W its Weyl group. Then θ(h) = h, and if σ is the transpose of θ|h , then σ(R) = R. For α ∈ R, gα denotes the generalized eigenspace of g relative to α. Set: R = {α ∈ R; α|a = 0} , R = {α ∈ R; α|a = 0}. We have:
R = {α ∈ R; σ(α) = α}.
38.1.3 Let B be a base of R and R+ the corresponding set of positive roots. Set: = R ∩ R+ , R− = R ∩ R− . R+ From now on, we shall assume that B is chosen in such a way so that the conditions of 36.1.4 are verified.
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38 Semisimple symmetric Lie algebras
38.2 Iwasawa decomposition 38.2.1 Proposition. (i) The set {α|l ; α ∈ R } is the root system of (m, l). (ii) If α ∈ R , then mα|l = gα , and: m=l⊕ gα . α∈R
Proof. Let α ∈ R and x ∈ gα . Then [x, a] = {0}, so x = y + z where y ∈ m and z ∈ a (37.5.2). On the other hand, if l ∈ l, then: α(l)(y + z) = [l, y + z] = [l, y] ∈ m. Thus z = 0. So x ∈ m and α|l is a root of (m, l) and gα = mα|l . Conversely, let β be a root of (m, l) and x ∈ mβ . Denote by α the linear form on h extending β which is zero on a. For a ∈ a and l ∈ l, we have: [a + l, x] = [l, x] = β(l)x = α(a + l)x. Thus α ∈ R and the result follows. 38.2.2 Lemma. If α ∈ R, then α + σ(α) ∈ R. Proof. Let x ∈ gα \ {0}. We have θ(x) ∈ gσ(α) \ {0}. If β = α + σ(α) ∈ R, then [x, θ(x)] ∈ gβ \ {0}. But β ∈ R , so gβ ⊂ m ⊂ k (38.2.1). Hence: [x, θ(x)] = θ([θ(x), x]) = [θ(x), x] = −[x, θ(x)]. Thus [x, θ(x)] = 0. Contradiction. 38.2.3 Let S be the set of restrictions to a of the elements of R . By 36.2.1 and 38.2.2, S is a (not necessarily reduced) root system in a∗ . We call S the root system of (g, a) or the restricted root system of h to a. We shall denote the Weyl group of S by WS . 38.2.4 Lemma. The following conditions are equivalent for λ ∈ a∗ \ {0}: (i) λ ∈ S. (ii) There exists x ∈ g \ {0} such that [a, x] = λ(a)x for all a ∈ a. (iii) There exists p ∈ p \ {0} such that [a, [a, p]] = λ(a)2 p for all a ∈ a. (iv) There exists k ∈ k \ {0} such that [a, [a, k]] = λ(a)2 k for all a ∈ a. Proof. (i) ⇒ (ii) This is obvious. (ii) ⇒ (i) Let r be the set of x ∈ g verifying [a, x] = λ(a)x for all a ∈ a. For (a, x, m) ∈ a × r × m, we have: λ(a)[m, x] = [m, [a, x]] = [a, [m, x]]. Thus r is ad m-stable, so it is ad h-stable. So there exist µ ∈ h∗ and x ∈ r \ {0} such that [h, x] = µ(h)x for all h ∈ h. As µ|a = λ, we have λ ∈ S.
38.2 Iwasawa decomposition
563
(ii) ⇒ (iii) Let us write x = k + p, with (k, p) ∈ k × p. If a ∈ a, then [a, k] ⊂ p and [a, p] ⊂ k. If [a, x] = λ(a)x for all a ∈ a, then: [a, k] = λ(a)p , [a, p] = λ(a)k. This proves that k et p are non-zero and: [a, [a, k]] = λ2 (a)k , [a, [a, p]] = λ2 (a)p. So we have also proved (ii) ⇒ (iv). (iii) ⇒ (i) Let p ∈ p \ {0} be such that [a, [a, p]] = λ2 (a)p for all a ∈ a. Let us write p = i xi , where the xi ’s are linear independent elements of g verifying [a, xi ] = µi (a)xi for all a ∈ a and the µi ’s are pairwise distinct linear forms on a. Then: λ2 (a) xi = µ2i (a)xi . i 2
i
µ2i (a),
which implies that the µi ’s have the same kernel as λ. Thus So λ (a) = there exists εi ∈ {−1, 1} such that µi = εi λ. In view of the implication (ii) ⇒ (i), we deduce that λ ∈ S or −λ ∈ S, hence λ ∈ S. (iv) ⇒ (i) The proof is analogue to the proof of (iii) ⇒ (i). 38.2.5 From now on, if λ ∈ a∗ , we shall denote by gλa the set of x ∈ g such that [a, x] = λ(a)x for all a ∈ a. Thus: gλa . g=h⊕ λ∈S
If λ ∈ S, then gλa is the sum of the gα ’s for α ∈ R verifying α|a = λ. w(λ)
38.2.6 Lemma. Let λ ∈ S and w ∈ WS . Then dim gλa = dim ga
.
Proof. Let A (resp. B) be the set of roots of R whose restriction to a is λ (resp. w(λ)). By 36.2.5, there exists w ∈ W such that w (l∗ ) = l∗ , w (a∗ ) = a∗ and w |a∗ = w. Thus w (A) = B. Since: ν ν w(λ) g , ga = g gλa = ν∈A
ν∈B
and dim g = 1 for all ν ∈ R, the result follows. ν
38.2.7 Let S+ be a system of positive roots of S and: λ n= ga . λ∈S+
Proposition. (i) The subspace n of g is a nilpotent Lie subalgebra and: (1)
g = k ⊕ a ⊕ n. This is called the Iwasawa decomposition of g defined by k, a and S+ . (ii) The orthogonal of a with respect to L is k ⊕ n.
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38 Semisimple symmetric Lie algebras
Proof. (i) We have: (2)
g=m⊕a⊕
λ∈S+
. gλa ⊕ g−λ a λ∈S+
⊂ k + n for all To establish g = k + a + n, it suffices to prove that g−λ a −λ λ , then θ(x) ∈ θ(g ) = g . It follows that: λ ∈ S+ . Let x ∈ g−λ a a a x = (x + θ(x)) − θ(x) ∈ k + n. Now if (x, y, z) ∈ k × a × n verifies x + y + z = 0, then: 0 = θ(x + y + z) = x − y + θ(z), and 2y + (z − θ(z)) = 0. So (2) implies that y = z = 0, hence x = 0. Thus g = k ⊕ a ⊕ n. Finally, it is clear that n is a nilpotent Lie subalgebra. (ii) We have L(a, k) = {0} (37.2.3) and L(a, n) = {0} (19.8.6). So (ii) follows from (i). 38.2.8 Let B be a base of R verifying the conditions of 36.1.4. The re to a form a system of positive roots S+ of strictions of the elements of R+ S. The corresponding decomposition g = k ⊕ a ⊕ n will be called the Iwasawa decomposition of g defined by k, a, h, B. Proposition. The subspace q = m ⊕ a ⊕ n is a Lie subalgebra of g and n is an ideal in q. and β ∈ R . If α + β ∈ R, then Proof. We have [a, n] ⊂ n. Let α ∈ R+ α β α+β ⊂ n. We deduce from 38.2.1 that α + β ∈ R+ (36.1.4), so [g , g ] ⊂ g [m, n] ⊂ n. So the result follows.
38.2.9 In the notations of example 2 of 37.2.4 with s semisimple, let t be a Cartan subalgebra of s and a = {(x, −x); x ∈ t}. Then m = {(x, x); x ∈ t}, and h = a ⊕ m t × t. We shall identify h∗ with t∗ × t∗ . Let T = R(s, t), R = R(g, h). Then R = (T × {0}) × ({0} × T ) and R = ∅. If C is a base of T , then B = (C ×{0})×({0}×−C) is a base of R verifying the conditions of 36.1.4. Moreover, if T+ , T− , R+ , R− are the corresponding sets of positive and negative roots, then R+ = (T+ × {0}) × ({0} × T− ). Let s = t ⊕ y+ ⊕ y− be the triangular decomposition of s defined by C, n+ = y+ × y− and n− = y− × y+ . Then g = h ⊕ n+ ⊕ n− is the triangular decomposition of g defined by B and g = k⊕a⊕n+ is the Iwasawa decomposition of g defined by k, a, h, B. 38.2.10 Now let us consider example 1 of 37.2.4. Let a be the set of diagonal matrices of g. Then a is a Cartan subalgebra of g which is also a Cartan subspace of g. Here, we have R = ∅. Let (Eij )1i,jn be the canonical basis of gln (k) and for 1 i n, let εi be the linear form on a given by λ1 E11 + · · · + λn Enn → λi . Set βi = εi − εi+1 , 1 i n − 1. Then B = {β1 , . . . , βn−1 } is a base of R. The Lie subalgebra n associated to B is therefore the set of strictly upper triangular matrices of g.
38.3 Coroots
565
38.3 Coroots 38.3.1 We define a non-degenerate symmetric bilinear form Lθ on g by setting for x, y ∈ g: Lθ (x, y) = L(x, θ(y)). If x ∈ gλa and y ∈ gµa , then L(x, y) = 0 when λ+µ = 0. Thus the decomposition g=m⊕a⊕ gλa λ∈S
is Lθ -orthogonal. Note that the restrictions of Lθ to a × a and to gλa × gλa for λ ∈ S, are non-degenerate. 38.3.2 Since L|a×a is non-degenerate, if φ ∈ a∗ , there exists a unique vφ ∈ a such that for all a ∈ a: L(vφ , a) = φ(a). Lemma. Let λ ∈ S. (i) If e ∈ gλa , then [e, θ(e)] = L(e, θ(e))vλ = Lθ (e, e)vλ . (ii) We have L(vλ , vλ ) = λ(vλ ) = 0. Proof. As θ(e) ∈ θ(gλa ) = g−λ a , we have [e, θ(e)] ∈ m ⊕ a. Also: θ([e, θ(e)]) = [θ(e), e] = −[e, θ(e)]. Hence [e, θ(e)] ∈ a. So for a ∈ a: L([e, θ(e)], a) = L([a, e], θ(e)) = λ(a)L(e, θ(e)) = λ(a)Lθ (e, e). The result follows since L|a×a is non-degenerate. (ii) There exists e ∈ gλa such that Lθ (e, e) = L(e, θ(e)) = 0 (because Lθ |gλa ×gλa is non-degenerate). We deduce therefore from (i) that there exists (x, y) ∈ gλa × g−λ such that [x, y] = vλ . a Suppose that λ(vλ ) = 0. Then [vλ , x] = [vλ , y] = 0. Thus u = kvλ ⊕kx⊕ky is a nilpotent Lie algebra. If z ∈ [u, u], then the eigenvalues of adg z are equal to zero by Lie’s Theorem. Since vλ ∈ [u, u], we have α(vλ ) = 0 for all α ∈ R. Contradiction. 38.3.3 We set, for λ ∈ S, tλ =
2vλ . λ(vλ )
Thus λ(tλ ) = 2. It follows from the proof of 38.3.2 that there exist eλ ∈ gλa and fλ ∈ g−λ such that (eλ , tλ , fλ ) is a S-triple of g. a Lemma. Let λ, µ ∈ S. (i) We have µ(tλ ) ∈ Z. (ii) The set of n ∈ Z such that µ + nλ ∈ S ∪ {0} is an interval [−n , n ] with n , n ∈ N. Moreover, n − n = µ(tλ ). (iii) We have µ − µ(tλ )λ ∈ S.
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38 Semisimple symmetric Lie algebras
Proof. Let:
gλ = ktλ ⊕ keλ ⊕ kfλ sl(2, k) , b =
gµ+nλ . a
n∈Z
The Lie algebra gλ acts on b via the adjoint representation, that we shall denote by ρ. The eigenvalues of ρ(tλ ) are of the form µ(tλ ) + 2n. They are consecutive even (or odd) integers from −r to r. So µ(tλ ) ∈ Z, and the set of n ∈ Z such that µ + nλ ∈ S ∪ {0} is an interval [−n , n ]. Since 0 ∈ [−n , n ], n , n ∈ N. Furthermore: µ(tλ ) + 2n = r , µ(tλ ) + 2n = −r. We deduce therefore that µ(tλ ) = n − n . As −n n − n n , we have −n µ(tλ ) n ; hence µ − µ(tλ )λ ∈ S ∪ {0}. If µ(tλ )λ = µ, then: µ(tλ ) = µ(tλ )λ(tλ ) = 2µ(tλ ). It follows that µ(tλ ) = 0, hence µ = µ(tλ )λ = 0 which is absurd. 38.3.4 Let λ ∈ S and sλ be the reflection of a∗ associated to λ. By 38.3.3, we have, for all φ ∈ a∗ , sλ (φ) = φ − φ(tλ )λ. 38.3.5 Let us use the notations hα and Hα of 20.6.1 and 20.6.5 for the pair (g, h). For α ∈ R, we have θ(Hα ) = Hσ(α) . In particular, if σ(α) = α, then Hα ∈ l, and if σ(α) = −α, then θ(Hα ) = H−α = −Hα , so Hα ∈ a. If h ∈ h, we can write h = h + h , with h ∈ l and h ∈ a. Let us identify ∗ l (resp. a∗ ) with the set of elements of h∗ which are zero on a (resp. l). If µ ∈ h∗ , then we may write µ = µ + µ , with µ ∈ l∗ and µ ∈ a∗ . Let λ ∈ S and α ∈ R be such that α = λ. We shall express tλ in terms of Hα . There are three cases. 1) If σ(α) = −α, then λ = α and Hα ∈ a. So it is clear that hα = vλ , and we have tλ = Hα . 2) If σ(α) = −α and α, σ(α)∨ = 0, then: 0 = α(Hσ(α) ) = α(θ(Hα )) = σ(α)(Hα ). In other words: 0 = L(θ(hα ), hα ) = L(hα , hα ) − L(hα , hα ). It follows that: L(hα , hα ) = L(hα , hα ) + L(hα , hα ) = 2L(hα , hα ).
38.3 Coroots
567
If a ∈ a, then: λ(a) = L(vλ , a) = α(a) = L(hα + hα , a) = L(hα , a). We deduce therefore that: 2vλ = 2hα = hα − θ(hα ). Hence: Hα =
2hα hα 2hα = = . L(hα , hα ) 2L(vλ , vλ ) L(vλ , vλ )
Now: tλ =
2hα 2vλ = . L(vλ , vλ ) L(vλ , vλ )
Consequently, we obtain that: tλ = Hα − θ(Hα ) = Hα − Hσ(α) . 3) If σ(α) = −α and α, σ(α)∨ = 0, then we saw in the proof of 36.2.1 that: α, σ(α)∨ = σ(α), α∨ = 1, that is: α(Hσ(α) ) = σ(α)(Hα ) = α(θ(Hα )) = 1. This implies that: L(hα , 2hα − 2hα ) = L(hα , 2θ(hα )) = L(hα , hα ). Hence: 2L(hα , hα ) − 2L(hα , hα ) = 2L(hα , θ(hα )) = L(hα + hα , hα + hα ) = L(hα , hα ) + L(hα , hα ). Consequently: L(hα , hα ) = 3L(hα , hα ) , L(hα , hα ) = 4L(hα , hα ). As in case 2, we have 2vλ = 2hα = hα − θ(hα ). Hence: Hα =
hα 2hα = . L(hα , hα ) 2L(vλ , vλ )
It follows that: Hα − θ(Hα ) = We have therefore obtained that:
vλ 2hα = . 2 L(vλ , vλ ) L(vλ , vλ )
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38 Semisimple symmetric Lie algebras
tλ = 2Hα − 2θ(Hα ) = 2Hα − 2Hσ(α) . 38.3.6 Let λ ∈ S. Since Lθ |gλa ×gλa is non-degenerate, there exists e ∈ gλa such that L(e, e) = 0. As k is algebraically closed, there exists r ∈ k such that: 2 . r2 = λ(vλ )Lθ (e, e) Set: xλ = re , yλ = θ(xλ ). We verify easily that: [tλ , xλ ] = 2xλ , [tλ , yλ ] = −2yλ , [xλ , yλ ] = tλ .
(3)
Similarly, we obtain: ⎧. / 1 ⎪ ⎪ = −xλ + yλ (x + y ), t λ λ λ ⎪ ⎪ ⎪ ⎨.2 / 1 (4) (xλ + yλ ), xλ − yλ = −tλ ⎪ ⎪ / .2 ⎪ ⎪ ⎪ ⎩ 1 (xλ + yλ ), z = 0 if z ∈ ker λ ⊂ a 2
38.4 Centralizers 38.4.1 We shall denote by S (resp. N ) the set of semisimple (resp. nilpotent) elements of g in p, and G the set of p-generic elements of g (37.4.1). By 37.4.5 and 37.5.2, we have G ⊂ S and G is a dense open subset of p (37.4.2). Let G = Aute g. Then L(G) = adg g and G is the algebraic adjoint group of g (23.4.16 and 24.8.2). We saw in 37.5.1 that k = ng (k). Since the adjoint representation of g is injective, it follows from 24.7.2 that there exists a connected algebraic subgroup K of G such that L(K) = adg k. As [k, k] ⊂ k and [k, p] ⊂ p, we have K(k) = k and K(p) = p (24.3.4). Moreover, by 24.3.6, K is the identity component of the normalizer of k in G. Finally, it follows from 24.8.5 that we may identify {α|k ; α ∈ K} with the algebraic adjoint group of k. 38.4.2 For x ∈ g, we set: k x = k ∩ gx , p x = p ∩ g x . If x ∈ k ∪ p, then clearly, gx = kx ⊕ px . Proposition. For all x ∈ p, we have: dim kx − dim px = dim k − dim p. In particular, dim g − dim gx = 2(dim k − dim kx ) = 2(dim p − dim px ).
38.4 Centralizers
569
Proof. Since g is semisimple, the Killing homomorphism of g is an isomor phism of g onto g∗ . The result is therefore a reformulation of 37.6.1. 38.4.3 Corollary. The following conditions are equivalent for x ∈ p: (i) dim gx dim gy for all y ∈ p. (ii) dim kx dim ky for all y ∈ p. (iii) dim px dim py for all y ∈ p. 38.4.4 An element x ∈ p is said to be p-regular if x satisfies the conditions of 38.4.3. We shall denote by R the set of p-regular elements of p. It is a dense open subset of p. So R ∩ G = ∅, and G = R ∩ S. Corollary. The following conditions are equivalent for x ∈ p: (i) x ∈ R. (ii) dim G(x) dim G(y) for all y ∈ p. (iii) dim K(x) dim K(y) for all y ∈ p. Proof. Let Gx (resp. Kx ) be the stabilizer of x in G (resp. K). Then L(Gx ) = adg gx and L(Kx ) = adg kx (24.3.6). So the result follows from 21.4.3 and 38.4.3. 38.4.5 Let a be a Cartan subspace of g and m = ck (a). Recall from 37.5.2 that cg (a) = m ⊕ a. Proposition. We have: dim a − dim m = dim p − dim k. Proof. Since a is an abelian Lie subalgebra of g consisting of semisimple elements (37.5.2), there exist pairwise distinct linear forms λ1 , . . . , λn on a such that: g = a ⊕ m ⊕ gλa 1 ⊕ · · · ⊕ gλa n . If x ∈ a \ (ker λ1 ∪ · · · ∪ ker λn ), then kx = m and px = a. So by 38.4.2, we are done. 38.4.6 Corollary. The following conditions are equivalent for x ∈ p: (i) x ∈ R. (ii) dim kx = dim m. (iii) dim px = dim a. (iv) dim gx = dim a + dim m. Proof. The equivalence (i) ⇔ (iii) follows from the fact that G ∩ R = ∅, while (ii) ⇔ (iii) is clear by 38.4.2 and 38.4.5. Finally (iv) ⇔ (i) is obvious by the previous equivalences and 38.4.3.
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38 Semisimple symmetric Lie algebras
38.4.7 Proposition. Let (g; k, p) be a semisimple symmetric Lie algebra and r the dimension of Cartan subspaces of g. (i) If x ∈ p, then px contains an abelian Lie subalgebra of dimension r. (ii) If x ∈ R, then [kx , px ] = [px , px ] = {0}, so px ⊂ z(gx ), and we may identify gx with the direct product of the Lie algebras kx and px . (iii) If x ∈ R ∩ N , then any element of px is nilpotent. (iv) Any element of R is polarizable. Proof. For n ∈ N, we shall denote by Gr(p, n) the Grassmannian variety of n-dimensional subspaces of p, and let p : Gr(p, r) × p → p be the canonical projection. (i) The set L of (q, x) ∈ Gr(p, r) × p verifying [q, q] = [q, x] = {0} is a closed subset of Gr(p, r) × p. As Gr(p, r) is a complete variety, it follows that p(L) is closed in p. But by 37.4.11 and 37.5.4, p(L) contains the dense subset G of p. So p(L) = p, and we have proved (i). (ii) This is obvious since for x ∈ R, we have [px , px ] = {0} by (i), while x x [k , p ] = {0} by 37.6.4. (iii) This is clear by (ii) and 35.1.2. (iv) For x ∈ R, 38.4.6 says that: dim g + dim gx = dim g + dim m + dim a. Thus dim g + dim m + dim a is an even integer, say 2s. The set M of (q, x) ∈ Gr(p, s) × p such that q is a Lie subalgebra of g verifying L(x, [q, q]) = {0}, is a closed subset of Gr(p, s). The elements of G, being semisimple, are polarizable (33.2.8), so G ⊂ p(M). Since G is dense in p and p(M) is closed in p, we have p(M) = p. The result is now clear. 38.4.8 In general, if x ∈ p is polarizable, there is no polarization q of g in x which verifies q = (q ∩ k) ⊕ (q ∩ p). For example, let (Eij )1i,j2 be the canonical bases of gl2 (k), g = sl2 (k) , k = k(E12 − E21 ) , p = k(E11 − E22 ) ⊕ k(E12 + E21 ). The only polarizations of g in x = E11 − E22 is kx ⊕ E12 and kx ⊕ E21 . Neither of them is the sum of their intersections with k and p.
38.5 S-triples 38.5.1 Let us denote by U the set of Lie subalgebras of g which are isomorphic to sl2 (k). Definition. (i) An element s ∈ U is called normal if s ⊂ k and θ(s) = s. (ii) A S-triple (e, h, f ) is called normal if h ∈ k and e, f ∈ p. (iii) Let H be a subgroup of Aut g. Two S-triples (e, h, f ) and (e , h , f, ) are H-conjugate if there exists α ∈ H such that e = α(e), h = α(h) and f = α(f ).
38.5 S-triples
571
(iv) An element s ∈ U is said to be principal if it is normal and it contains a nilpotent p-regular element. (v) A normal S-triple (e, h, f ) is called principal if e is p-regular. 38.5.2 Remarks. Let s ∈ U. Then it is easy to prove the following results: 1) If s is normal, then: s = (s ∩ k) ⊕ (s ∩ p) , dim(s ∩ k) = 1 , dim(s ∩ p) = 2. 2) Since k is algebraically closed, s is normal if and only if there exists a normal S-triple (e, h, f ) such that s = ke + kh + kf . 38.5.3 Proposition. Let e be a non-zero nilpotent element of p. Then there exists a normal S-triple (e, h, f ) containing e. Proof. By 32.1.5, there exists a S-triple (e, h, f ) containing e. Let us write h = h1 + h2 , f = f1 + f2 , with h1 , f1 ∈ k and h2 , f2 ∈ p. Then: 2e = [h, e] = [h1 , e] + [h2 , e] , h1 + h2 = h = [e, f ] = [e, f1 ] + [e, f2 ]. Hence [h1 , e] = 2e, [e, f2 ] = h1 . It follows from 32.1.3 that there exists f3 ∈ g such that (e, h1 , f3 ) is a S-triple. Let f3 = f4 + f5 , with f4 ∈ k, f5 ∈ p. Now −2f3 = [h1 , f3 ] = [h1 , f4 ] + [h1 , f5 ] , h1 = [e, f3 ] = [e, f4 ] + [e, f5 ], so [h1 , f5 ] = −2f5 , [e, f5 ] = h1 . Thus (e, h1 , f5 ) is a normal S-triple. 38.5.4 Lemma. Let (e, h, f ) and (e, h , f ) be normal S-triples having the same positive element. Then there exists a nilpotent element x ∈ ke such that: e = ead x (e) , h = ead x (h) , f = ead x (f ). Proof. The subspace ke is ad h-stable. Let te+ be the subspace of te spanned by the elements x ∈ te verifying [h, x] = λx, with λ ∈ N∗ . Then te+ = te ∩ [e, g] and by 32.2.4, te+ is a Lie subalgebra of g whose elements are nilpotent. As [h, te+ ] = te+ , 32.2.5 implies that exp(te+ )(h) = h + te+ . Now [h − h, e] = 0 and h − h = [e, f − f ], so h − h ∈ ke+ . Consequently, there exists x ∈ te+ such that h = ead x (h). Hence f = ead x (f ) (32.2.3). 38.5.5 Let (e, h, f ) be a normal S-triple. For n ∈ Z, set: gn = {x ∈ g; [h, x] = nx} , kn = k ∩ gn , pn = p ∩ gn . We have gn = kn ⊕ pn . The following result is analogue to the result of 32.1.9. Proposition. Let n be an integer. (i) We have dim kn = dim k−n and dim pn = dim p−n . (ii) If n is odd, then dim tn = dim pn .
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38 Semisimple symmetric Lie algebras
Proof. Recall that L|gn ×g−n is non-degenerate. Since k and p are orthogonal with respect to L, we deduce that the restrictions of L to kn ×k−n and pn ×p−n are non-degenerate. This proves (i). Let j ∈ N and ϕj : g−2j−1 → g2j+1 , x → (ad e)j (x). The map ϕj is bijective, and since 2j + 1 is odd, we obtain: ϕj (k−2j−1 ) = p2j+1 , ϕj (p−2j−1 ) = k2j+1 . So (ii) follows from (i). 38.5.6 Lemma. Let (e, h, f ) be a S-triple of g and s be the element of U spanned by e, h, f . Set hc = e + f . (i) There exists α ∈ G such that hc = α(h). (ii) There exists a S-triple (x, hc , y), G-conjugate to (e, h, f ), such that x, hc , y span s. (iii) Assume that (e, h, f ) is normal. Then there exists a S-triple (ec , hc , fc ) verifying fc = θ(ec ) and ec , hc , fc span s. Furthermore, ec is uniquely determined up to a scalar multiplication by ±1. Proof. Let α ∈ G be defined by: 1 α = exp(ad f )◦ exp − ad e . 2 Then: 1 α(h) = e + f , α(e) = e − h − f , α(f ) = − (e + h − f ). 4 From this, we deduce immediately (i) and (ii). Set: 1 1 ec = (e − h − f ) , fc = − (e + h − f ). 2 2 Then (ec , hc , fc ) is a S-triple. By part (i) and 32.3.1, it is G-conjugate to (e, h, f ). Moreover, if (e, h, f ) is normal, then fc = θ(ec ). Let us assume that (e, h, f ) is normal, and let (e , hc , f ) be another Striple verifying the conclusion of (iii). From θ(hc ) = −hc , we deduce that [hc , θ(e )] = −2θ(e ). So θ(e ) = λfc , with λ ∈ k \ {0}. Therefore e = θ2 (e ) = λec . But: hc = [e , f ] = [λec , θ(e )] = λ[ec , θ(λec )] = λ2 [ec , fc ] = λ2 hc . So λ = ±1.
38.5.7 Lemma. Let s ∈ U and (e, h, f ) a S-triple spanning s. (i) Let x be a non-zero nilpotent element of s. Then there exists α ∈ G such that α(x) = e. (ii) Assume that s is principal. Any element x ∈ s ∩ N \ {0} is p-regular.
38.6 Orbits
573
Proof. (i) There exists α ∈ Aute s such that α(x) = e (32.2.2). If z ∈ s is such that ads z is nilpotent, then adg z is also nilpotent (20.4.4). Thus α extends to an element of G. (ii) Let x ∈ s ∩ N \ {0}, then (i) implies that x = α(e) for some α ∈ G. So dim ge = dim gx and (ii) follows from 38.4.6. 38.5.8 Proposition. Let (e, h, f ) be a normal S-triple and s the element of U spanned by e, h, f . (i) Assume that s is principal. Then hc = e+f ∈ G, and if x ∈ s∩S \{0}, then x ∈ G. Moreover, the dimension of any irreducible s-submodule of g is odd. (ii) Conversely, let us assume that the following conditions are verified: a) s ∩ G = ∅. b) The dimension of any irreducible s-submodule of g is odd. Then s is principal in U. Proof. (i) By the theory of representations of sl2 (k), we have dim gh dim ge with equality if and only if the dimensions of all the irreducible s-submodules of g are odd. On the other hand, dim gh = dim ge+f (38.5.6). As s is normal, we have e + f ∈ p. Finally, e ∈ R (38.5.7). Hence: dim ge dim gh = dim ge+f dim ge . This shows that hc ∈ G, and the dimension of any irreducible s-submodule of g is odd. Let x = ae + bh + cf be a non-zero semisimple element of s sl2 (k). As an endomorphism of k2 , x is semisimple (20.4.4). So b2 + ac = 0. We deduce therefore that there exist λ, µ ∈ k such that: cλ2 + 2bλ − a = 0 , µ2 (cλ2 + 2bλ − a) + 2µ(b + cλ) + c = 0. A simple computation shows that exp(µ ad f )◦ exp(λ ad e)(x) = νh, where ν ∈ k \ {0}. So dim gx = dim gh , and x ∈ G. (ii) Condition b) implies that dim ge = dim gh , while condition a) implies that x ∈ G for x ∈ s ∩ S non-zero. Hence e ∈ N ∩ R, and s is normal.
38.6 Orbits 38.6.1 Throughout this section, S, N , G, G and K are as in 38.4.1. Theorem. (i) Let O be the G-orbit of an element of p. Any irreducible component Z of O ∩ p is a K-orbit and verifies: 2 dim Z = dim O. (ii) The number of K-orbits of the elements of N is finite.
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38 Semisimple symmetric Lie algebras
Proof. By 32.4.4, it suffices to prove (i). Let Z be an irreducible component of O ∩ p. Since K.Z is an irreducible subset of p ∩ O, Z is K-stable. Let x ∈ Z and K = K(x). Then we have by 38.4.2: dim O = dim g − dim gx = 2(dim k − dim kx ) = 2 dim K. It follows that all the K-orbits in Z have the same dimension. Now: Tx (O) = [x, g] = [x, k] + [x, p] , Tx (K) = [x, k]. Thus Tx (O) ∩ p = [x, k]. On the other hand: Tx (Z) ⊂ Tx (O) ∩ p = [x, k] = Tx (K) ⊂ Tx (Z). We deduce therefore that Tx (Z) = Tx (K). Let y be a smooth point in Z. It follows from the previous paragraph that dim Z = dim K(y), and so K(y) is dense in Z (14.1.6). Since all the K-orbits in Z have the same dimension, we conclude by 21.4.5 that Z = K(y), and the result follows from the equality of dimensions above. 38.6.2 Let (e, h, f ) be a normal S-triple. For n ∈ Z, set: gn = {x ∈ g; [h, x] = nx} , kn = k ∩ gn , pn = p ∩ gn . Then gn = kn ⊕ pn . Let ξ ∈ k \ {0}. The automorphism τξ of g, given by τξ (x) = ξ n x for x ∈ gn , is an element of G (32.2.7). Let H = {τξ ; ξ ∈ k \ {0}}; it is an irreducible subgroup of G. Let s1 , . . . , sm be the eigenvalues of adk h and L the smallest algebraic subgroup of G whose Lie algebra contains k adg h. If ξ ∈ k \ {0} and (a1 , . . . , am ) ∈ Zm verifying a1 s1 + · · · + am sm = 0, then: (ξ s1 )a1 · · · (ξ sm )am = ξ a1 s1 +···+am sm = 1. It follows from 24.6.3 that τξ ∈ L. Since τξ (k) = k, by 38.4.1, we have H ⊂ K (because H is irreducible). 38.6.3 Let us conserve the notations of 38.6.2 and denote by Kh the stabilizer of h in K. If x ∈ p2 and α ∈ Kh , then 2α(x) = α([h, x]) = [α(h), α(x)] = [h, α(x)]. Thus α(x) ∈ g2 , and so α(x) ∈ p2 because α(p) = p. Now [e, g0 ] = g2 , so [e, k0 ] = p2 . Lemma. The set Kh (e) contains a dense open subset of p2 . Proof. The proof is analogue to the proof of 32.2.8. 38.6.4 Lemma. Two normal S-triples (e, h, f ), (e , h, f ) having the same simple element are K-conjugate.
38.6 Orbits
575
Proof. By 38.6.3, we have Kh (e) ∩ Kh (e ) = ∅. So there exists α ∈ K such that α(e) = e and α(h) = h. So by 32.2.3, α(f ) = f . 38.6.5 Theorem. Let (g; k, p) be a semisimple symmetric Lie algebra, and K be as in 38.4.1. (i) Let (e, h, f ) and (e , h , f ) be normal S-triples of g. The following conditions are equivalent: a) The S-triples (e, h, f ) and (e , h , f ) are K-conjugate. b) There exists α ∈ K such that α(e) = e . c) There exists α ∈ K such that α(h) = h . (ii) The map (e, h, f ) → e induces a bijection between the set of Kconjugacy classes of normal S-triples of g and the set of K-orbits of non-zero nilpotent elements of p. Proof. This follows immediately from 38.5.3, 38.5.4 and 38.6.4. 38.6.6 Remarks. 1) If (e, h, f ) is a normal S-triple, then e and f are not in general K-conjugate. 2) Let (e, h, f ) and (e , h , f ) be normal S-triples and s, s the corresponding elements in U. If there exists α ∈ K such that α(s) = s , then the two S-triples are not necessarily K-conjugate. So there is no natural bijection between the set of K-orbits of non-zero nilpotent elements of p and the set of normal elements of U. 38.6.7 Lemma. Let x ∈ p and s, n its semisimple and nilpotent components. (i) If ξ ∈ k \ {0}, then s + ξ 2 n ∈ K(x). (ii) We have s ∈ K(x). Proof. Using 38.5.3 and the fact that τξ ∈ K (38.6.2), the proof is the same as in 34.2.5. 38.6.8 Proposition. (i) We have x ∈ N satisfied, then λx ∈ K(x) (ii) We have x ∈ S if
Let x ∈ p. if and only if 0 ∈ K(x). If these conditions are for all λ ∈ k \ {0}. and only if K(x) is closed in p.
Proof. Since K(x) ⊂ G(x), part (i) follows from 34.3.5 and 38.6.7. If x ∈ S, then G(x) is closed by 34.3.2, and 38.6.1 implies that K(x) is closed. If x is not semisimple, then 38.6.7 says that K(x) is not closed. 38.6.9 Theorem. Let x ∈ p. (i) There exists z ∈ N such that: z ∈ K(kx) , dim K(z) = dim K(x).
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(ii) If x is polarizable, then we may choose z to be polarizable. (iii) Assume that x is polarizable and there exists a polarization r of g in x such that r = (r ∩ k) ⊕ (r ∩ p). Then we may choose z such that r is a polarization of g in z. Proof. (i) The proof is the same as in 33.5.3 by replacing G by K. (ii) Assume that x is polarizable. Let d be the dimension of a polarization of g in x and p : Gr(g, d) × p → p the canonical surjection. The set: F = {(q, y) ∈ Gr(g, d) × p; [q, q] ⊂ q , L(y, [q, q]) = {0}} is closed in Gr(g, d) × p and the closed subset p(F) contains K(kx). It follows that K(kx) ⊂ p(F). Let z ∈ K(kx) ∩ N be such that dim K(z) = dim K(x). Then dim G(z) = dim G(x) by 38.6.1, so dim gz = dim gx . But there exists a d-dimensional Lie subalgebra q of g such that L(z, [q, q]) = {0}. Hence q is a polarization of g in z. (iii) Let r be a parabolic subalgebra of g such that r = (r ∩ k) ⊕ (r ∩ p). Set r = dim r, k = dim(r ∩ k), p = dim(r ∩ p), and let q : Gr(g, r) × p → p be the canonical surjection. Let H be the K-orbit of r in Gr(g, r). The normalizer of r in k is k ∩ r. So: dim H = dim k − k. The K-orbit H is open in its closure H and H \ H is the union of K-orbits of dimension strictly smaller than dim H (21.4.3 and 21.4.5). For r ∈ H, we have: r = (k ∩ r ) ⊕ (p ∩ r ) , dim(k ∩ r ) = k , dim(p ∩ r ) = p. Let q ∈ H. Then q is a Lie subalgebra of g (19.7.2) and: q = (k ∩ q) ⊕ (p ∩ q) , dim(k ∩ q) = k , dim(p ∩ q) = p. Now, let r be a polarization of g in x. The set L of (q, y) ∈ H×K(kx) verifying L(y, [q, q]) = {0} is a closed subset of Gr(g, r) × p, and the closed subset q(L) contains K(kx), hence K(kx). Let z ∈ K(kx) ∩ N as in part (i). There exists q ∈ H such that L(z, [q, q]) = {0}. We have dim K(z) = dim K(x), so dim kz = dim kx , hence dim gz = dim gx . It follows that q is a polarization of g in x, so it is a parabolic subalgebra of g (33.2.6). Since dim(q ∩ k) = dim(r ∩ k), the discussion above says that q and r are in the same K-orbit of Gr(g, r). Thus there exists α ∈ K such that q = α(r). Taking z = α−1 (z), we have that z ∈ K(kx) ∩ N , dim K(z ) = dim K(x) and r is a polarization of g in z . 38.6.10 Remarks. 1) In the notation x = E11 − E22 of 38.4.8, we can verify easily that if q is a polarization of g in x, then there does not exist a nilpotent element z ∈ p such that q is a polarization of g in z. 2) Let g be a simple Lie algebra of type B2 , with basis, in the notations 18.6.1 and 20.6.5:
38.6 Orbits
577
Hα , Hβ , Xα , Xβ , Xα+β , X2α+β , X−α , X−β , X−α−β , X−2α−β . Denote by k (resp. p) the subspace of g spanned by Hα , Hβ , X±β , X±(2α+β) (resp. X±α , X±(α+β) ). Then (g; k, p) is a symmetric Lie algebra. Set: b = kHα + kHβ + kXα + kXβ + kXα+β + kX2α+β . The Lie algebra b is a Borel subalgebra of g verifying b = (b ∩ k) ⊕ (b ∩ p). We verify easily that kXα + kXα+β does not contain any regular element of g. So if x ∈ p, then b is not a polarization of g in x (33.2.1). 38.6.11 The following result is a version of Richardson’s Theorem for semisimple symmetric Lie algebras. Theorem. Let q be a parabolic subalgebra of g, m its radical nilpotent. Assume that the following conditions are verified: a) q = (q ∩ k) ⊕ (q ∩ p). b) There exists z ∈ p such that q is a polarization of g in z. Let us denote by K0 the identity component of the stabilizer of q in K. (i) There exists a unique nilpotent K-orbit K in p such that m ∩ p ∩ K is a dense open subset of m ∩ p. (ii) The set m ∩ p ∩ K is a K0 -orbit. (iii) If x ∈ m ∩ p ∩ K, then [x, q ∩ k] = m ∩ p , [x, q ∩ p] = m ∩ k, and q is a polarization of g in x. Proof. Part (i) is clear because the number of nilpotent K-orbits of p is finite. Set: s = min{dim kx ; x ∈ m ∩ p} , M = {x ∈ m ∩ p; dim kx = s}. The set M is a non-empty open subset of m ∩ p, so K ∩ M = ∅. Hence dim K = dim k − s, and K ∩ m ∩ p ⊂ M. Conversely, if u ∈ M, then dim K(u) = dim K and u ∈ K. So u ∈ K. Thus M = m ∩ p ∩ K. By 38.6.9 (iii), there exists y ∈ p ∩ N such that q is a polarization of g in y. As y is nilpotent, we have L(y, q) = {0} (33.2.6), hence y ∈ m ∩ p. Now, for u ∈ M, we have L(u, q) = {0}. Hence (38.4.2): 2 dim q dim g + dim gu = 2 dim p + 2s. Since q is a polarization of g in y, we have: 2 dim q = dim g + dim gy = 2 dim p + 2 dim ky . Thus s = dim ky . This proves that if u ∈ m ∩ p ∩ K, then q is a polarization of g in u. So [u, q] = m (33.2.6). Part (iii) now follows.
578
38 Semisimple symmetric Lie algebras
Let u ∈ m ∩ p ∩ K. Since q is a polarization of g in u, we have gu ⊂ q. From [u, q ∩ k] = m ∩ p, we deduce that dim(m ∩ p) = dim(q ∩ k) − dim ku . The Lie algebra of K0 is adq (k ∩ q). So we have: dim K0 (u) = dim(k ∩ q) − dim(k ∩ q ∩ ku ) = dim(k ∩ q) − dim ku = dim(m ∩ p). Thus K0 (u) is a dense open subset of m ∩ p contained in m ∩ p ∩ K. This being true for all u ∈ m ∩ p ∩ K, we have (ii). 38.6.12 Let q be a parabolic subalgebra of g verifying q = (q ∩ k) ⊕ (q ∩ p), and m its nilpotent radical. It is natural to ask whether there exists x ∈ p ∩ m such that [x, k ∩ q] = p ∩ m. This would give a more interesting version of Richardson’s Theorem than 38.6.11. We shall give an example to show that the answer to the above question is in general negative. Let g be a simple Lie algebra of type D4 and B = {α1 , α2 , α3 , α4 } a base of the root system of g. In the notations of 18.14.2 with βi = αi and 20.6.5, let k be the subspace of g spanned by: H1 , H2 , H3 , H4 , X±α1 , X±α3 , X±α4 , X±(α1 +2α2 +α3 +α4 ) . Let p be the subspace of g spanned by: X±α2 , X±(α1 +α2 ) , X±(α2 +α3 ) , X±(α2 +α4 ) , X±(α1 +α2 +α3 ) , X±(α1 +α2 +α4 ) , X±(α2 +α3 +α4 ) , X±(α1 +α2 +α3 +α4 ) . The triple (g; k, p) is a symmetric Lie algebra. Let q be the Borel subalgebra of g with respect to B. Then: q = (k ∩ q) ⊕ (k ∩ p) , dim(k ∩ q) = dim(p ∩ q) = 8. The subspace p ∩ q = p ∩ m of q is spanned by: Xα2 , Xα1 +α2 , Xα2 +α3 , Xα2 +α4 , Xα1 +α2 +α3 , Xα1 +α2 +α4 , Xα2 +α3 +α4 , Xα1 +α2 +α3 +α4 . If x ∈ p ∩ m, then [x, Xα1 +2α2 +α3 +α4 ] = 0. Hence: dim([x, k ∩ q]) 7 < dim(p ∩ m). Thus [x, k ∩ q] = p ∩ m.
38.7 Symmetric invariants
579
38.7 Symmetric invariants 38.7.1 In this section, we shall use the notations a, m, l, h, R, S, W, WS of 38.1.1 and 38.2.3. Lemma. The set K(a) is dense in p. Proof. Since G is dense in p, the result follows from 37.4.10 and 37.4.11. 38.7.2 Let us consider the action of WS on a given by: w(a), λ = a, w−1 (λ) for (a, λ, w) ∈ a × a∗ × WS . Thus w(a) = t w−1 (a). In particular, if λ ∈ S, then: sλ (a) = a − λ(a)tλ , where tλ is as in 38.3.3. Let uλ = ker λ ⊂ a, then: sλ (tλ ) = −tλ , sλ |uλ = iduλ . Lemma. Let w ∈ WS . There exists α ∈ K such that α(a) = w(a) for all a ∈ a. Proof. It suffices to consider the case where w = sλ , with λ ∈ S. In the notations of 38.3.6, set: h = xλ + yλ , e =
1 1 (xλ − yλ − tλ ) , f = − (xλ − yλ + tλ ). 2 2
We see easily that (e, h, f ) is a S-triple of g. Moreover, tλ ∈ a and θ(xλ ) = yλ (see 38.3.6). It follows that (e, h, f ) is normal. Let τ = τξ ∈ K be defined with respect to (e, h, f ) as in 38.6.2, with ξ 2 = −1. Then τ (e) = −e, τ (f ) = −f , tλ = −(e + f ), so τ (tλ ) = −tλ . Now if z ∈ uλ , then [h, z] = 0, because [z, xλ ] = [z, yλ ] = 0. Hence τ (z) = z. So we have proved that τ |a = sλ . 38.7.3 The Lie algebra k acts on p∗ via the coadjoint representation. This action extends to a representation of k in S(p∗ ) via derivations of S(p∗ ). Denote by S(p∗ )k the algebra of k-invariants, that is, the set of f ∈ S(p∗ ) such that x.f = 0 for all x ∈ k. Since the group K acts on p∗ , it acts on S(p∗ ) and by 24.3.4, we have S(p∗ )k = S(p∗ )K . In the same way, WS acts on a∗ , and so on S(a∗ ). Denote by S(a∗ )WS the set of WS -invariants. 38.7.4 Lemma. Let x, y ∈ a be such that y ∈ WS (x). Then there exists f ∈ S(a∗ )WS such that f (x) = f (y). Proof. This is the same as in 34.2.1.
580
38 Semisimple symmetric Lie algebras
38.7.5 Lemma. Let x ∈ p and s, n its semisimple and nilpotent components. Then f (x) = f (s) for all f ∈ S(p∗ )k . Proof. Since S(p∗ )k = S(p∗ )K , if f ∈ S(p∗ )k and α ∈ K, then f (x) = f (α(x)). So if y ∈ K(x), then f (y) = f (x). The result is therefore a consequence of 38.6.7. 38.7.6 Let j : S(p∗ ) → S(a∗ ) be the restriction map. Lemma. (i) The restriction of j to S(p∗ )k is injective. (ii) We have j(S(p∗ )k ) ⊂ S(a∗ )WS . Proof. Let f ∈ S(p∗ )k = S(p∗ )K be such that f |a = 0. If a ∈ a and α ∈ K, then f (α(a)) = f (a) = 0. Since K(a) is dense in p (38.7.1), we have f = 0. Finally part (ii) follows from 38.7.2. 38.7.7 Let us denote by π : S(p∗ )k → S(a∗ )WS the injective map induced by j. Let r = dim p. For x ∈ p, set: Tx = (ad x)2 |p . The characteristic polynomial χx of Tx is of the form: χx (λ) = det(λ idp −Tx ) = λr + pr−1 (x)λr−1 + · · · + ps (x)λs , where pr−1 , . . . , ps ∈ S(p∗ ). It is clear that pr−1 , . . . , ps ∈ S(p∗ )K . For x ∈ a and λ ∈ S, we have χx (λ2 (x)) = 0. Thus λ2 , and hence λ is integral over im π. Since S spans a∗ , we have proved the following result: Lemma. The ring S(a∗ ) is integral over im π. 38.7.8 Lemma. Let f, g ∈ S(p∗ )k \ {0} and p ∈ S(p∗ ) be such that f = gp. Then p ∈ S(p∗ )k . Proof. For x ∈ k, we have: 0 = x.f = (x.g)p + g(x.p) = g(x.p). So the result is clear. 38.7.9 Lemma. The ring S(p∗ )k is integrally closed. Proof. Let Q be the field of fractions of S(p∗ )k . If x ∈ Q is integral over S(p∗ )k , then it is integral over S(p∗ ). Since S(p∗ ) is factorial, we have x ∈ S(p∗ ). Hence x ∈ S(p∗ )k by 38.7.8.
38.7 Symmetric invariants
581
38.7.10 Since π is injective, we have therefore obtained the following result: Lemma. The ring im π is integrally closed. 38.7.11 Let us denote by aQ the Q-vector subspace of a spanned by the tλ ’s, λ ∈ S (38.3.3). Lemma. Let x ∈ aQ and y ∈ a be such that f (x) = f (y) for all f ∈ im π. Then y ∈ aQ and y ∈ WS (x). Proof. The restriction maps S(g∗ ) → S(p∗ ) → S(a∗ ) induce the homomorphisms of algebras S(g∗ )g → S(p∗ )k → S(a∗ )WS . By assumption, g(x) = g(y) for all g ∈ S(g∗ )g . It follows therefore from 31.2.6 and 34.2.1 that y ∈ W (x). In particular, y ∈ aQ . Thus by 36.2.6, we have y ∈ WS (x). 38.7.12 Theorem. Let (g; k, p) be a semisimple symmetric Lie algebra, a a Cartan subspace of g, S the root system of (g, a), WS the Weyl group of (g, a). The restriction map S(p∗ ) → S(a∗ ) induces an isomorphism of algebras from S(p∗ )k onto S(a∗ )WS . Proof. We shall use the notations of 38.7.7. Let E (resp. F) be the field of fractions of S(a∗ ) (resp. im π) and set qi = π(pi ), s i < r. We define a polynomial F with coefficients in im π by: F (X) = X 2r + qr−1 X 2r−2 + · · · + qs X 2s . In a suitable algebraic closure of F, the roots of F are 0 and the λ’s, where λ ∈ S. Since S spans a∗ , E is the splitting field of F over F. So the extension F ⊂ E is normal and finite. Let H = Gal(E/F). Let σ ∈ H. Then σ fixes each element of F and permutes the non-zero roots of F . Thus S(a∗ ) is σ-invariant. If f ∈ S(a∗ ), let us denote by f σ its image under σ. Let x ∈ aQ and σ ∈ H. We define a homomorphism of k-algebras: θ : S(a∗ ) → k , f → f σ (x). So there exists y ∈ a such that θ(f ) = f σ (x) = f (y) for all f ∈ S(a∗ ). In particular, if g ∈ im π, then: g(y) = g σ (x) = g(x). It follows from 38.7.11 that y ∈ aQ and y ∈ WS (x). If h ∈ S(a∗ )WS , then: hσ (x) = h(y) = h(x).
582
38 Semisimple symmetric Lie algebras
This being true for all x ∈ aQ , we have obtained that hσ = h, and so h ∈ F. Taking into account the results of 38.7.7 and 38.7.10, we have the following results: a) S(a∗ )WS ⊂ F. b) im π is integrally closed. c) The ring S(a∗ )WS is integral over im π. Consequently S(a∗ )WS ⊂ im π, and the result follows from 38.7.6. 38.7.13 Let us denote by I the set of elements of S(p∗ )k without constant term. Proposition. Let x ∈ p. Then x ∈ N if and only if f (x) = 0 for all f ∈ S(p∗ )I. Proof. Let J+ be as in 34.3.3. The restriction map S(g∗ ) → S(p∗ ) sends J+ into I. If f (x) = 0 for all f ∈ S(p∗ )I, then x ∈ N (34.3.3). Conversely, let x ∈ N and f ∈ I. Then for α ∈ K, we have f (α(x)) = f (x). Thus f is constant on K(x), and hence on K(x). Since 0 ∈ K(x) (38.6.8), we have f (x) = f (0) = 0. 38.7.14 Lemma. If x, y ∈ S verify f (x) = f (y) for all f ∈ S(p∗ )k , then y ∈ K(x). Proof. We may assume that x and y belong to a Cartan subspace of g (37.4.10). By 38.7.12, we have f (x) = f (y) for all f ∈ S(a∗ )WS . So y ∈ WS (x) (38.7.4), and hence by 38.7.2, y ∈ K(x). 38.7.15 Lemma. The following conditions are equivalent for x, y ∈ p: (i) The semisimple components of x and y belong to the same K-orbit in p.
(ii) We have f (x) = f (y) for all f ∈ S(p∗ )k .
Proof. This is a direct consequence of 38.7.5 and 38.7.14. 38.7.16 Proposition. Let x ∈ p and s, n its semisimple and nilpotent components. (i) If n is non-zero, then dim px < dim ps . (ii) K(x) is the union of a finite number of K-orbits. (iii) K(s) is the unique closed K-orbit contained in K(x). Proof. (i) This follows from 34.2.5 and 38.4.2. (ii) The set K(x) is a union of K-orbits in p. If x is nilpotent or semisimple, then the result follows from 38.6.1 or 38.6.8. Suppose now that s and n are non-zero. The set F of y ∈ p verifying f (x) = f (y) for all f ∈ S(p∗ )k = S(p∗ )K is a closed subset of p containing K(x).
38.7 Symmetric invariants
583
By 38.7.15, y ∈ p belongs to F if and only if its semisimple component u belongs to K(s). From this, we deduce that if y ∈ F , then there exists α ∈ K such that y = α(s + w) for some w nilpotent verifying [s, w] = 0. In particular, w belongs to the semisimple Lie algebra s = [gs , gs ]. The triple (s; s∩k, s∩p) is a semisimple symmetric Lie algebra and w ∈ s∩p. Let K be the smallest algebraic subgroup of G whose Lie algebra contains adg s ∩ k. Clearly K ⊂ K and γ(s) = s for all γ ∈ K . By 38.6.1, the number of nilpotent K -orbits in s ∩ p is finite. Let w1 , . . . , wn be representatives of these nilpotent orbits. Then there exist γ ∈ K and an index i such that y = α(s + γ(wi )) = α ◦ γ(s + wi ). Thus y ∈ K(s + wi ), and the result follows. (iii) Any closed K-orbit in K(x) is of the form K(y) with y semisimple (38.6.8). So f (x) = f (s) = f (y) for all f ∈ S(p∗ )k . Hence we have y ∈ K(s) by 38.7.15. 38.7.17 Let r = dim a. By 38.7.12 and 31.1.6, there exist homogeneous and algebraically independent elements u1 , . . . , ur in S(p∗ )k which generate the algebra S(p∗ )k . Thus: S(p∗ )I =
r
S(p∗ )ui .
i=1
Theorem. The set N of nilpotent elements in p is a closed subset of codimension r in p. Let N1 , . . . , Nk be its irreducible components. For 1 i k, we have that: (i) Ni is a K-invariant subset of codimension r in p. In particular, it is the union of a finite number of K-orbits. (ii) R ∩ Ni is non-empty and there exists ei ∈ R ∩ Ni such that: K(ei ) = R ∩ Ni . The orbit K(ei ) is a dense open subset of Ni . The set R∩N =
k
(R ∩ Ni )
i=1
is a dense open subset of N . Proof. It is clear that N is closed in p and that each Ni is the union of a finite number of nilpotent K-orbits (38.6.1). It follows that Ni = K(ei ) for some ei ∈ Ni , and K(ei ) is the unique dense open K-orbit in Ni . Now by 38.4.2, 38.4.3 and 38.4.6, we have: codimp K(ei ) = dim p − dim K(ei ) = dim p − dim k + dim kei = dim pei dim a = r. Moreover, since N is the closed subset defined by u1 , . . . , ur (38.7.13), it follows that codimp Ni r (14.2.6). Thus:
584
38 Semisimple symmetric Lie algebras
codimp Ni = codimp K(ei ) = r. We deduce therefore that ei ∈ R ∩ Ni . The other parts are now clear. 38.7.18 Let us consider example 1 of 37.2.4 with n = 2. Set: 1 i 1 1 −i 1 01 e= , h=i , f= , −1 0 1 i 2 1 −i 2 where i2 = −1. Then (e, h, f ) is a normal S-triple of g. We see readily that N is the set of matrices of the form: a b , b −a where a, b ∈ k verify a2 + b2 = 0. From this, we deduce the following results: a) We have N = (ke) ∪ (kf ), so N is not irreducible. b) The K-orbits of e and f are distinct.
38.8 Double centralizers 38.8.1 We shall conserve the notations of 38.4.1. Recall that if a and b are subspaces of g, then: na (b) = {x ∈ a; [x, b] ⊂ b} , ca (b) = {x ∈ a; [x, b] = {0}}. The proofs of the following two lemmas are analogue to the ones for 35.3.1 and 35.3.2. Lemma. Let x ∈ p. Then: (i) [x, k] + k = [cp (px ), k] + k = (px )⊥ . (ii) cp (px ) = ([px , k] + k)⊥ . (iii) [x, p] + p = [cp (kx ), p] + p = (kx )⊥ . (iv) ck (px ) = ([px , p] + p)⊥ . 38.8.2 Lemma. The following conditions are equivalent for x, y ∈ p: (i) y ∈ cp (px ). (ii) px ⊂ py . (iii) [px , k] ⊂ [py , k]. (iv) [y, k] ⊂ [x, k]. (v) cp (py ) ⊂ cp (px ). (vi) [cp (py ), k] ⊂ [cp (px ), k]. 38.8.3 Proposition. Let x ∈ p. (i) The element x is semisimple if and only if cp (px ) is an abelian Lie subalgebra consisting of semisimple elements. If this is the case, then cp (px ) is contained in z(gx ), and so cp (px ) = p ∩ z(gx ).
38.8 Double centralizers
585
(ii) The element x is nilpotent if and only if cp (px ) is an abelian Lie subalgebra consisting of nilpotent elements. (iii) Let s, n be the semisimple and nilpotent components of x. Let us write [gs , gs ] = k1 ⊕ p1 where k1 ⊂ k and p1 ⊂ p. Then: cp (px ) = cp (ps ) ⊕ cp1 (pn1 ). Proof. Note that x ∈ px , so cp (px ) ⊂ px and the Jacobi identity says that cp (px ) is an abelian ideal of the Lie algebra gx . (i) If x is semisimple, then gx is reductive in g. So cp (px ) ⊂ z(gx ), and the result is clear since x ∈ cp (px ). (ii) Let x be nilpotent, u ∈ cp (px ) and y, z the semisimple and nilpotent components of u. We have y, z ∈ cp (px ). If y = 0, then there exist λ ∈ k \ {0} and v ∈ g \ {0} such that [y, v] = λv. Then: [y, [x, v]] = [[y, x], v] + [x, [y, v]] = [x, [y, v]] = λ[x, v]. Thus a = {w ∈ g; [y, w] = λw} is ad x-stable. Since x is nilpotent, there exists w ∈ a \ {0} such that [x, w] = 0, hence w ∈ gx . Let us write w = w1 + w2 with w1 ∈ kx et w2 ∈ px . As y ∈ p and w ∈ a, we obtain that: [y, w1 ] = λw2 , [y, w2 ] = λw1 . But [y, w2 ] = 0 since w2 ∈ px . So w1 = w = 0. Contradiction. (iii) We have px ⊂ ps and px ⊂ pn , hence cp (ps ) ⊂ cp (px ) and cp (pn ) ⊂ cp (px ). By parts (i) and (ii), we see that cp (ps )⊕cp (pn ) ⊂ cp (px ). On the other hand: gs = ks ⊕ ps = z(gs ) ⊕ [gs , gs ] = z(gs ) ⊕ ⊕k1 ⊕ p1 . We know also that n ∈ [gs , gs ], hence n ∈ p1 . Finally, since gx = gs ∩ gn , we deduce that px = ps ∩ pn = pn1 ⊕ (z(gs ) ∩ p). So by (i), we obtain that cp (px ) = (z(gs ) ∩ p) ⊕ cp1 (pn1 ) = cp (ps ) ⊕ cp1 (pn1 ). 38.8.4 Let x, y ∈ p. If py = px , then cp (py ) = cp (px ) (38.8.2). In particular, y ∈ cp (px ). Set: cp (px )• = {y ∈ p; py = px } ⊂ cp (px ). Lemma. The set cp (px )• is a non-empty open subset of cp (px ) and: cp (px )• = {y ∈ p; cp (py ) = cp (px )} = {y ∈ cp (px ); rk(ad y) = rk(ad x)}. Proof. The first equality follows from 38.8.2. Let y ∈ cp (px ). Again by 38.8.2, we have that dim px dim py , hence dim gx dim gy (38.4.2) and rk(ad x) rk(ad y). Similarly, rk(ad x) = rk(ad y) if and only if dim px = dim py . But since y ∈ cp (px ), we have px ⊂ py . Thus rk(ad x) = rk(ad y) if and only if px = py .
586
38 Semisimple symmetric Lie algebras
38.8.5 Proposition. Let x ∈ p and K its K-orbit. (i) If x is nilpotent, then K ∩ cp (px ) = cp (px )• . (ii) If x is semisimple, then K ∩ cp (px ) is a finite set. Proof. (i) The number of nilpotent K-orbits in p being finite, it follows from 38.8.3 (ii) that there exists such an orbit U such that U = U ∩ cp (px ) is a dense open subset of cp (px ). Now 38.8.4 implies that U ∩ K = ∅. Hence dim U = dim K, and U ⊂ cp (px )• . Conversely, if y ∈ cp (px )• , then y ∈ U and dim K(y) = dim U. It follows that y ∈ U. Thus cp (px )• = U and U = K. (ii) If x is semisimple, then 35.3.4 says that the set z(gx ) ∩ G(x) is finite, and the result follows since cp (px ) ⊂ z(gx ) (38.8.3 (i)). 38.8.6 Let (e, h, f ) be a normal S-triple in g and s = ke + kh + kf . Denote by gs the centralizer of s in g and ks = k ∩ gs , ps = p ∩ gs . In the notations of 35.3.7, we have gs = (ge )0 , ks = (ke )0 and ps = (pe )0 . 38.8.7 Proposition. In the above notations, we have: cp (pe )0 = ck (pe )1 = {0}. Proof. 1) Let x ∈ cp (pe )0 . Then x ∈ ps and [x, (pe )0 ] = {0} = [x, ps ]. If y ∈ ks and z ∈ pe , then: [[y, x], z] = [[y, z], x] + [y, [x, z]]. We have [x, z] = 0 and [y, z] ∈ pe , so [[y, z], x] = 0, hence [y, x] ∈ cp (pe ). Thus cp (pe )0 is an ideal of gs which is reductive in g. It follows from 38.8.3 that any element of cp (pe )0 is nilpotent. Hence cp (pe )0 ⊂ a = [gs , gs ]. Thus cp (pe )0 is a nilpotent ideal of the semisimple Lie algebra a. Consequently, cp (pe )0 = {0}. 2) Let r be the sum of 2-dimensional simple s-submodules of g. The restriction of L to r × r is non-degenerate and r = r1 ⊕ r−1 . Moreover, it is clear that L(r1 , r1 ) = L(r−1 , r−1 ) = {0}. If x ∈ ck (pe )1 \ {0}, then x ∈ r1 ∩ k \ {0}. Since L(k, p) = {0}, there exists y ∈ r−1 ∩ k such that L(x, y) = 0. We may write y = [f, z] where z ∈ r1 ∩ p. In particular, z ∈ pe . So: L([z, x], f ) = L(x, [f, z]) = L(x, y) = 0. This is absurd because [z, x] = 0. 38.8.8 Examples and remarks. 1) Let g = sl3 (k) and (Eij )1i,j3 the canonical basis of gl3 (k). Set: k = k(E11 − E22 ) + k(E22 − E33 ) + kE13 + kE31 , p = kE12 + kE21 + kE23 + kE32 .
38.8 Double centralizers
587
Then (g; k, p) is a symmetric Lie algebra. Let (e, h, f ) be the normal S-triple defined by e = E12 , h = E11 − E22 , f = E21 . Then: pe = cp (pe ) = kE12 + kE32 , cp (pe )2 = kE12 , cp (pe )1 = kE32 . In particular, cp (pe )1 = {0}. We verify easily that the dimension of a Cartan subspace of g is 1. So this example shows also that px can be abelian without the condition that x ∈ R (compare with 35.6.8). 2) Let g = sl2n (k). For AB X= ∈ sl2n (k), CD with A, B, C, D ∈ gln (k), set: θ(X) =
A −B −C D
.
This defines an involutive automorphism of the Lie algebra g. Let k be the set of fixed points of θ and p the set of X ∈ g such that θ(X) = −X. Then (g; k, p) is a symmetric Lie algebra. More precisely, we have: 1 0 A 0 k= ; A, D ∈ gln (k), tr(A + D) = 0 , 0 D 1 0 0 B p= ; B, C ∈ gln (k) . C 0 Suppose that n = 3. We define a normal S-triple (e, h, f ) by: e = E15 + E26 + E63 , h = E11 + 2E22 − 2E33 − E55 , f = E51 + 2E62 + 2E36 . Then ke is spanned by: E13 , E21 + E65 , E23 , E45 , E11 − 2E44 + E55 , E22 + E33 − 3E44 + E66 , and pe is spanned by: E14 , E15 , E16 + E53 , E24 , E25 , E26 + E63 , E43 . It follows that: ck (pe ) = kE13 + kE23 , cp (pe ) = ke. Thus ck (pe )3 = kE13 , ck (pe )4 = kE23 . In particular, ad h|ck (pe ) has even and odd eigenvalues (compare with 35.3.9). 3) Let us consider example 2 of 38.6.10 with: e = Xα+β , h = Hα + 2Hβ , f = X−α−β . Then pe = ke, ck (pe ) = kHα + kXβ + kX2α+β . So ck (pe )0 = kHα , ck (pe )2 = kXβ + kX2α+β .
588
38 Semisimple symmetric Lie algebras
38.9 Normalizers 38.9.1 Proposition. Let x ∈ p. Then: (i) nk (px ) = {y ∈ k; [x, y] ∈ cp (px )}. (ii) nk (px ) = nk (cp (px )). (iii) If x is semisimple, then nk (px ) = kx . Proof. (i) Let y ∈ k, u ∈ px and suppose that [x, y] ∈ cp (px ). Then [[y, x], u] = 0, so: 0 = [[y, u], x] + [y, [x, u]] = [[y, u], x]. Thus y ∈ nk (px ). Conversely, if y ∈ nk (px ), then for u ∈ px and v ∈ cp (px ), we have: 0 = [[y, u], v] = [[y, v], u] + [y, [u, v]] = [[y, v], u]. Thus [y, v] ∈ cp (px ). In particular, [y, x] ∈ cp (px ). (ii) We have just seen that nk (px ) ⊂ nk (cp (px )). Conversely if y ∈ n(cp (px )), then [y, x] ∈ cp (px ). Hence y ∈ nk (px ) by (i). (iii) If x is semisimple, then g = gx ⊕ [x, g]. We deduce therefore that k = kx ⊕ [x, p] and p = px ⊕ [x, k]. Let u ∈ kx , v ∈ [x, p] and y = u + v. We have [y, x] = [v, x] ∈ [x, k]. So [y, x] ∈ cp (px ) ⊂ px if and only if [v, x] = 0, that is v = 0. So the result follows from (i). 38.9.2 Let (e, h, f ) be a normal S-triple of g. Since cp (pe )0 = {0} by 38.8.7, we have: [e, [f, cp (pe )]] = [h, cp (pe )] = cp (pe ). This shows that dim[f, cp (pe )] = dim cp (pe ), and ke ∩ [f, cp (pe )] = {0}. So we deduce that the sum ke + [f, cp (pe )] is direct. Moreover, it follows from 38.9.1 (i) that: [f, cp (pe )] ⊂ nk (pe ). Proposition. Let (e, h, f ) be a normal S-triple of g. Then: (i) [nk (pe ), e] = cp (pe ). (ii) nk (pe ) = ke ⊕ [f, cp (pe )]. In particular: nk (pe ) = nk (pe )n . n−1
Proof. (i) Since [f, cp (pe )] ⊂ nk (pe ) and [e, [f, cp (pe )]] = cp (pe ), the result is a consequence of 38.9.1 (i). (ii) Since ke ⊂ nk (pe ), it follows from (i) that: dim nk (pe ) = dim ke + dim cp (pe ) = dim ke + dim[f, cp (pe )]. So we have proved the first part. The second part is then a consequence of 38.8.7.
38.10 Distinguished elements
589
38.9.3 In view of 38.9.2, we have: ng (pe ) = ke ⊕ [f, cp (pe )] ⊕ cp (pe ) , [e, ng (pe )] = cp (pe ).
38.10 Distinguished elements 38.10.1 Definition. An element x ∈ p is k-distinguished (resp. pdistinguished) if all the elements of kx (resp. px ) are nilpotent. 38.10.2 In the rest of this section, (e, h, f ) is a normal S-triple of g, and s = ke + kh + kf . We shall denote by gs (resp. ks , ps ) the centralizer of s in g (resp. k, p). We have: g s = ge ∩ g f = ge ∩ g h = gf ∩ g h . Since the S-triplet is normal, we have analogous identities for k and p. 38.10.3 Let x ∈ p. It is clear that px contains the semisimple and nilpotent components of its elements. Thus x is p-distinguished if and only if 0 is the only semisimple element of px . Note that, by 38.4.7 (iii), a p-regular element is p-distinguished. Similarly, x is k-distinguished if and only if 0 is the only semisimple element of kx . Proposition. Let x ∈ p. Then x is p-distinguished if and only if it is nilpotent and the Lie subalgebra of g generated by px is nilpotent. Proof. We can check easily that b = px ⊕ [px , px ] is the Lie subalgebra generated by px . If x is nilpotent and b is nilpotent, then any semisimple element y ∈ px is central in b. Hence y ∈ cp (px ). By 38.8.3, y = 0. Thus x is p-distinguished. Conversely, suppose that x is p-distinguished. Let L denote the Killing form of b. Since (b; [px , px ], px ) is a symmetric Lie subalgebra of (g, θ), [px , px ] and px are orthogonal with respect to L . Let y ∈ px . Then ad y|b is nilpotent, and therefore L (y, y) = 0. Hence L (px , px ) = {0}. It follows that L (px , b) = {0}, and px is contained in the radical of b. We deduce then from 19.5.7 that px is contained in the largest nilpotent ideal of b. We conclude therefore from the definition of b that it is nilpotent. 38.10.4 Proposition. In the notations of 38.10.2, the following conditions are equivalent: (i) e is p-distinguished. (ii) f is p-distinguished. (iii) The restriction of L to pe × pe is identically zero. (iv) The restriction of L to pf × pf is identically zero. (v) The restriction of L to ps × ps is identically zero. (vi) The vector space ps is {0}.
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38 Semisimple symmetric Lie algebras
Proof. (v) ⇔ (vi) The restriction of L to gs × gs is non-degenerate (20.5.13) and gs = ks ⊕ps . Since k and p are orthogonal with respect to L, the equivalence of the two conditions is clear. (iii) ⇔ (v) The subspace ps is ad h-stable, and the eigenvalues of ad h|pe are in N. Since L(gi , gj ) = {0} if i + j = 0 and ps = pe ∩ g0 , we see that L|pe ×pe is identically zero if and only if L|ps ×ps is identically zero. (i) ⇒ (vi) Suppose that e is p-distinguished and ps = {0}. Then from the equivalence of (iii), (v) and (vi), we deduce that L|pe ×pe is not identically zero. Since 2L(x, y) = L(x + y, x + y) − L(x, x) − L(y, y), for x, y ∈ pe , it follows that there exists x ∈ pe such that L(x, x) = tr(ad x)2 = 0. This is absurd since x is nilpotent. (vi) ⇒ (i) Let x ∈ pe be semisimple and non-zero. Then gx = kx ⊕ px is reductive in g. It follows from 32.1.7 that there is a S-triple (e, h , f ) such that h , f ∈ gx . Let us write h = h1 + h2 , f = f1 + f2 with h1 , f1 ∈ kx and h2 , f2 ∈ px . Then: 2e = [h1 , e] + [h2 , e] , h1 + h2 = [e, f1 ] + [e, f2 ]. Thus [h1 , e] = 2e, [e, f2 ] = h1 . By 32.1.3, there exists f3 ∈ g such that (e, h1 , f3 ) is a S-triple. As in 38.5.3, we may assume that f3 ∈ p. Let t = ke + kh1 + kf3 . Then x ∈ pt . But the S-triples (e, h, f ) and (e, h1 , f3 ) are K-conjugate (38.6.5). Consequently ps = {0}. The proofs of the equivalences (vi) ⇔ (ii) and (iv) ⇔ (v) are similar. 38.10.5 Corollary. The following conditions are equivalent: (i) e is p-distinguished. (ii) dim p0 = dim k2 . (iii) pe ⊂ [e, k]. (iv) [e, k] + k contains a parabolic subalgebra of g. Proof. (i) ⇔ (ii) The map ad e induces a surjection from g0 onto g2 , so it induces a surjective map ϕ from p0 onto k2 . The kernel of ϕ is ps . The equivalence is therefore clear by 38.10.4. (i) ⇒ (iii) If ps = {0}, then by the theory of representations of sl2 (k), we have pe ⊂ [e, g] = [e, k] + [e, p]. Hence pe ⊂ [e, k]. (iii) ⇒ (i) Suppose that (iii) is verified, and x ∈ pe . There exists y ∈ k such that x = [e, y]. We have [e, [e, y]] = 0, so x = [e, y] is nilpotent (32.1.2). (i) ⇒ (iv) Set gn , m = gn . q= n0
n1
Then q is a parabolic subalgebra of g and m is its nilpotent radical. If e is p-distinguished, then ps = {0}, so pe ⊂ m. Taking the orthogonals with respect to L, we obtain by 38.8.1 that q ⊂ [e, k] + k.
38.10 Distinguished elements
591
(iv) ⇒ (i) Let q be a parabolic subalgebra of g contained in [e, k] + k and m its nilpotent radical. By taking the orthogonals with respect to L, we have pe ⊂ m . So the elements of pe are nilpotent.
38.10.6 We have also the following analogue of 35.2.7: Proposition. The following conditions are equivalent: (i) e is p-distinguished. (ii) pe ⊂ K(e). Proof. (i) ⇒ (ii) Recall that N denotes the set of nilpotent elements in p. Let N2 = {(x, y) ∈ N × N ; [x, y] = 0} and π : N2 → N , (x, y) → x, the canonical projection. Note that K acts diagonally on N2 and π is K-equivariant. 1) By 38.6.1, the number of K-orbits of the elements of N is finite. Let O1 , . . . , Or be the distinct nilpotent K-orbits of p, and for 1 i r, fix xi ∈ Oi . For x ∈ N and α ∈ K, we have π −1 (α.x) = {α.x} × (pα.x ∩ N ) = α {x} × (px ∩ N ) = α π −1 (x) . So dim π −1 (x) = dim π −1 (α.x) and π −1 (Oi ) = K. {xi } × (pxi ∩ N ) . Consequently, dim π −1 (Oi ) = dim Oi + dim(pxi ∩ N ) dim Oi + dim pxi = dim p. Since pxi ∩ N is closed in pxi , we deduce further that dim π −1 (Oi ) = dim p if and only if xi is p-distinguished. In particular, we have dim N2 = dim p. 2) The map ab · (x, y) → (ax + by, cx + dy), cd defines an action of GL2 (k) on N2 . Since GL2 (k) is irreducible, if C is an irreducible component of N2 , then GL2 (k) · C is an irreducible subset of N2 containing C. So GL2 (k) · C = C. 3) If e is p-distinguished and O = K(e), then π −1 (O) = K. {e} × pe is irreducible of dimension dim p. It follows from point 1 that π −1 (O) is an irreducible component of N2 . Now, by point 2, we have: 01 01 · ({e} × pe ) ⊂ · π −1 (O) = π −1 (O). pe × {e} = 10 10 Hence by applying π, we obtain that: pe ⊂ π π −1 (O) ⊂ O. (ii) ⇒ (i) If pe ⊂ K(e), then 0 ∈ K(e) and e is nilpotent (38.6.8). Hence p ⊂ K(e) ⊂ N . So e is p-distinguished. e
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38 Semisimple symmetric Lie algebras
38.10.7 As in 38.10.4 and 38.10.5, we have the following result: Proposition. The following conditions are equivalent: (i) e is k-distinguished. (ii) f is k-distinguished. (iii) The restriction of L to ke × ke is identically zero. (iv) The restriction of L to kf × kf is identically zero. (v) The restriction of L to ks × ks is identically zero. (vi) The vector space ks is {0}. (vii) ke ⊂ [e, p]. (viii) [e, p] + p contains a parabolic subalgebra of g. 38.10.8 Let us consider the example given in the second remark of 38.8.8 with n = 2. The elements: e = E14 + E42 , h = 2E11 − 2E22 , f = 2E41 + 2E24 . form a normal S-triple, and: ke = kE12 + k(E11 + E22 − 3E33 + E44 ). So ke is an abelian Lie algebra which contains non-zero semisimple elements Thus in contrast 38.10.3, the following conditions are not equivalent: (i) e is k-distinguished. (ii) e is nilpotent and the Lie algebra ke is nilpotent.
References and comments • [29], [50], [54], [55]. Many results presented in sections 38.4 and 38.5 are taken from [50]. The proof of 38.7.12 is due to Lepowsky ([54]). Semisimple symmetric Lie algebras appear essentially in the theory of symmetric spaces. The reader may refer to [36] for a detailed account of the theory of symmetric spaces. The notion of p-distinguished elements was also considered in [64], [65], [66]. In particular, a classification of p-distinguished elements was given in [66].
39 Sheets of Lie algebras
In this chapter, we study the geometry of orbits via the notions of Jordan classes and sheets. We treat both the semisimple and semisimple symmetric cases. Throughout this chapter, g is a semisimple Lie algebra with Killing form L and adjoint group G. We shall denote by Sg (resp. Ng ) the set of semisimple (resp. nilpotent) elements of g.
39.1 Jordan classes 39.1.1 Proposition. Let x ∈ g, and s, n its semisimple and nilpotent components. (i) x ∈ Ng (resp. x ∈ Sg ) if and only if z(gx ) is an abelian Lie subalgebra contained in Ng (resp. in Sg ). (ii) If s = [gs , gs ], then we have: z(gx ) = z(gs ) ⊕ z(sn ). Moreover, z(gs ) (resp. z(sn )) is the set of semisimple (resp. nilpotent) elements of z(gx ). Proof. (i) If x ∈ Sg , then gx is reductive in g (20.5.13). So the elements of z(gx ) are semisimple. The converse is clear. The case where x is nilpotent follows from 35.1.2. (ii) We have n ∈ s (20.5.14), gx = gs ∩ gn and gs = s ⊕ z(gs ). Thus x g = z(gs ) ⊕ sn . Hence z(gx ) = z(gs ) ⊕ z(sn ). The last statement follows from part (i). 39.1.2 For x ∈ g, denote by z(gx )• the set of elements y ∈ g verifying g = gx . By 35.3.3, z(gx )• is a dense open subset of z(gx ), and: y
z(gx )• = {y ∈ g; z(gx ) = z(gy )} = {y ∈ z(gx ); rk(adg y) = rk(adg x)}.
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39 Sheets of Lie algebras
More generally, let s be a semisimple Lie subalgebra of g and x ∈ s. Then we denote by z(sx )• the set of elements y ∈ s verifying sy = sx . Lemma. Let x ∈ g, s, n its semisimple and nilpotent components and s = [gs , gs ]. We have: z(gx )• = {y + z; y ∈ z(gs )• , z ∈ z(sn )• }. Proof. Let y ∈ z(gs ), z ∈ z(sn ) and u = y + z. By 39.1.1, y and z are respectively the semisimple and nilpotent components of u. Suppose that y ∈ z(gs )• and z ∈ z(sn )• . Then we have gy = gs , sz = sn . Thus gu = z(gs ) + sn = gx , and u ∈ z(gx )• . Let t = [gy , gy ]. By the first paragraph of the proof and 39.1.1, z(gy ) (resp. z z(t )) is the set of semisimple (resp. nilpotent) elements of z(gu ). If u ∈ z(gx )• , then z(gy ) = z(gs ) and z(tz ) = z(sn ). It follows from 35.3.2 that gy = gs , hence s = t. This proves that y ∈ z(gs )• and z ∈ z(sn )• . 39.1.3 Let x, y ∈ g. Denote respectively by xs , ys (resp. xn , yn ) the semisimple (resp. nilpotent) components of x and y. G We say that x and y are G-Jordan equivalent, in which case we write x ∼ y, if there exists α ∈ G such that: gys = gα(xs ) = α(gxs ) , yn = α(xn ). This defines an equivalence relation on g. The equivalence class of x, that we shall denote by JG (x), is called the G-Jordan class or the Jordan class of x in g. A Jordan class is clearly a G-stable set. 39.1.4 Examples. 1) If x ∈ Ng , then JG (x) = G(x). 2) Let ggen be the set of generic elements of g. Then x ∈ ggen if and only if gx is a Cartan subalgebra of g. By 29.2.2 and 29.2.3, ggen is a G-Jordan class in g. 39.1.5 Proposition. Let x ∈ g. (i) If s, n are the semisimple and nilpotent components of x, then: JG (x) = G z(gs )• + n . (ii) The set JG (x) is irreducible in g. Proof. (i) Let z ∈ z(gs )• and y = z + n. By 39.1.1, z and n are the semisimple and nilpotent components of y, and we have gz = gs as in the proof of 39.1.2. Hence y ∈ JG (x), and so G(y) ⊂ JG (x). Conversely, let y ∈ JG (x), t and m its semisimple and nilpotent components, and α ∈ G verifying gt = gα(s) , m = α(n). We have t ∈ z(gα(s) )• , so t ∈ α z(gs )• . We have therefore obtained (i). (ii) Let us consider the ϕ : G × g → g, (α, z) → α(z + n). By morphism (i), we have JG (x) = ϕ z(gs )• . Since z(gs )• is irreducible, part (ii) follows.
39.1 Jordan classes
595
39.1.6 Proposition. Let x, y ∈ g. The following conditions are equivalent: G (i) x ∼ y. (ii) There exists α ∈ G such that gy = α(gx). (iii) There exists α ∈ G such that z(gy ) = α z(gx ) . Proof. The equivalence (ii) ⇔ (iii) is immediate by 35.3.2. (i) ⇒ (ii) We shall use the notations of 39.1.3. If there exists α ∈ G such that gys = gα(xs ) and yn = α(xn ), then: gy = gys ∩ gyn = α(gxs ) ∩ α(gxn ) = α(gxs ∩ gxn ) = α(gx ). G
(iii) ⇒ (i) It suffices to prove that if z(gy ) = z(gx ), then x ∼ y. If x is nilpotent, then it follows from 35.3.4 and 39.1.4 that y ∈ G(x) = JG (x). Let us consider the general case. Let s, n be the semisimple and nilpotent components of x, s = [gs , gs ], and S the smallest algebraic subgroup of G The set {α|s ; α ∈ S} whose Lie algebra is adg s (24.7.2). is the adjoint group of s (24.8.5). We have S z(gs ) = z(gs ), so S z(gs )• = z(gs )• . In view of 39.1.2, 39.1.4, 39.1.5, and the case where x is nilpotent, we obtain that: y ∈ z(gs )• + z(sn )• ⊂ z(g s )• + S(n) = S z(gs )• + n ⊂ G z(gs )• + n) = JG (x). Hence the result.
39.1.7 Corollary. (i) We have JG (x) = G z(gx )• . (ii) The set JG (x) is locally closed in g, so it is a subvariety of g. Proof. Part (i) is clear by 39.1.6. Let us prove (ii). Let d = dim gx and u = {y ∈ g; dim gy = d}. By 19.7.6 and 29.3.1, the map ϕ : u → Gr(g, d), y → gy is a morphism of varieties. Now 39.1.6 says that JG (x) = ϕ−1 (Ω) where Ω is the G-orbit of gx in Gr(g, d). Since Ω is locally closed in Gr(g, d) (21.4.3), so is JG (x). 39.1.8 Let us fix a Cartan subalgebra h of g, and denote by R the root system of g relative to h. 1) Let P be a subset of R which is the intersection of R with a vector subspace of h∗ . Denote by hP the set of elements x of h such that α(x) = 0 for all α ∈ P, and h•P the set of elements x of hP verifying α(x) = 0 for all α ∈ R \ P. The set of the subsets hP of h thus constructed is finite. Let x ∈ h and R the set of elements of R which are zero at x. Then R is the intersection of R with a vector subspace of h∗ . We have z(gx ) = hR and z(gx )• = h•R .
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39 Sheets of Lie algebras
2) Let sx = [gx , gx ] and Sx the smallest algebraic subgroup of G whose Lie algebra is adg sx . We saw in 39.1.6 that the sets z(gx ) and z(gx )• are stable under the action of Sx . By 32.4.4, there exist n1 , . . . , nr ∈ sx such that Sx (ni ), 1 i r, are the distinct nilpotent Sx -orbits of Sx in sx . Let n ∈ gx ∩ Ng . We have n ∈ sx , so n = θ(ni ), for some θ ∈ Sx and 1 i r. So: z(gx )• + n = z(gx )• + θ(ni ) = θ z(gx )• + ni . By 39.1.5, we obtain that JG (x + ni ) = JG (x + n). 3) Finally let y ∈ g, and t, u its semisimple and nilpotent components. There exists θ ∈ G such that θ(t) ∈ h. Consequently, JG (y) = JG (x + v), with x ∈ h and v ∈ gx nilpotent. It follows therefore from points 1, 2 and 3 that: Proposition. The set of Jordan classes in g is finite.
39.2 Topology of Jordan classes 39.2.1 Lemma. Let H be an algebraic group, K a parabolic subgroup of H, X an H-variety and Y a K-stable closed subset of X. Then H(Y ) is a closed subset of X. Proof. Endow H/K with the structure of variety as defined in 25.4.7, and denote by π : H → H/K the canonical surjection. The group H acts rationally on H and H/K by left multiplication (25.4.10). Thus the H-varieties H and H/K are H-homogeneous and π is an H-equivariant morphism. It follows from 25.1.2 that the morphism u : H × X → (H/K) × X , (α, x) → (π(α), x) is open. Let S = {(α, x) ∈ H × X; α−1 (x) ∈ Y }. Since S is the inverse image of Y of the morphism H × X → X, (α, x) → α−1 (x), it is closed in H × X. Set S = u(S). Let us consider the action of K on H × X given by β(α, x) = (αβ −1 , x), (α, β, x) ∈ H × K × X. The fibres of u are the K-orbits in H × X. It follows that S is K-stable because Y is K-stable. Hence by 25.3.2, S is a closed subset of (H/K) × X. Let p : H × X → X and q : (H/K) × X → X denote the canonical surjections. Then clearly, we have q(S ) = p(S) = H(Y ). Now, H/K is complete (28.1.4), so q(S ) is closed in X, and we have obtained the result. 39.2.2 Let x ∈ g, and s, n its semisimple and nilpotent components. Let us fix a Cartan subalgebra h of g contained in gs , and let R be the root system of g relative to h. Set R = {α ∈ R; α(s) = 0}. Then in the notations of 39.1.8, we have:
39.2 Topology of Jordan classes
gs = h ⊕
α∈R
597
gα , z(gs ) = hR , z(gs )• = h•R .
The set R is a root system in the subspace that it spans. Let B be a base of R , B a base of R containing B (18.7.9), R+ the set of positive roots with respect to B and R = R ∪ R+ . Then α g p1 = h ⊕ α∈R
is a parabolic subalgebra of g having gs as a Levi factor, and whose nilpotent radical is: gα . n1 = α∈R+ \R
Let P1 be a parabolic subgroup of G such that L(P1 ) = adg p1 (29.4.3), and N1 the subgroup of G generated by exp(adg u), u ∈ n1 . By 21.3.2, N1 is a closed connected subgroup of G whose Lie algebra is adg n1 (23.3.5 and 24.5.9). Thus N1 is a normal subgroup of P1 (24.4.7). On the other hand, by 24.7.2, adg gs is an algebraic Lie subalgebra of adg p1 . So there is a closed connected subgroup M of P1 such that L(M ) = adg gs . Since N1 is normal in P1 , it follows from 24.5.9 that M N1 = N1 M is a closed connected subgroup of P1 whose Lie algebra is adg gs +adg n1 = adg p1 . We have therefore obtained that P1 = N1 M = M N1 . Now let u ∈ z(gs )• and y = u + n. Then u and n are the semisimple and nilpotent components of y, so gy = gu ∩gn = gs ∩gn ⊂ gs . Hence gy ∩n1 = {0}. We deduce therefore that [y, n1 ] = n1 , and by 32.2.5, exp(adg n1 )(y) = y + n1 . Thus y + n1 ⊂ N1 (y). On the other hand, it is clear from the definition of N1 that N1 (y) ⊂ y + n1 , so we have N1 (y) = y + n1 . It follows that: N1 z(gs )• + n = z(gs )• + n + n1 , N1 z(gs )• + n = z(gs ) + n + n1 . On the other hand: M z(gs ) + n + n1 = z(gs ) + M (n) + n1 , and, Consequently:
M z(gs ) + n + n1 = z(gs ) + M (n) + n1 . P1 z(gs )• + n = z(gs ) + M (n) + n1 .
In view of 39.1.5 and 39.2.1, we have obtained the following result: Proposition. We have: JG (x) = G z(gs ) + M (n) + n1 .
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39 Sheets of Lie algebras
39.2.3 Remark. Let x ∈ g. Since JG (x) = G z(g)• (39.1.7), it follows from 39.1.2 that dim G(y) = dim G(x) for all y ∈ JG (x). Hence by 21.4.4, dim G(y) dim G(x) if y ∈ JG (x). It may happen that dim G(y) = dim G(x) and y ∈ JG (x) \ JG (x). For example, if g = {0} and x ∈ ggen , then JG (x) = ggen and JG (x) = g. Now, for any y ∈ Ng ∩ greg , we have dim G(y) = dim G(x). 39.2.4 Proposition. The following conditions are equivalent for a subset J of g: (i) There exists s ∈ Sg such that J = JG (s). (ii) There exists a parabolic subalgebra p of g such that J = G(r), where r is the radical of p. Proof. (i) ⇒ (ii) Using 39.2.2 with n = 0, we have JG (s) = G z(gs ) + n1 , and the result follows from the fact that z(gs ) + n1 is the radical of p1 . (ii) ⇒ (i) Let l be a Levi factor of p, k the centre of l and n the nilpotent radical of p. Then p = k ⊕ [l, l] ⊕ n, and r = k ⊕ n is the radical of p. There exists s ∈ k such that gs = l and z(gs ) = k (see for example the proof of 33.2.8 (ii)). By 39.2.2, we have G(r) = JG (s). 39.2.5 Theorem. Let J be a Jordan class in g. There exist a parabolic subalgebra p of g and a solvable ideal w of p such that J = G(w). Proof. Let x ∈ g be such that JG (x) = J, and s, n its semisimple and nilpotent components. We have already treated the case n = 0 in 39.2.4. So we may assume that n = 0. Set s = [gs , gs ], k = z(gs ). Then n ∈ s. The Lie algebras adg s and adg k are algebraic Lie subalgebras of adg g. Let us denote by S and K the closed connected subgroups of G such that L(S) = adg s and L(K) = adg k. 1) Let us use the notations p1 , n1 , P1 , N1 , M of 39.2.2. As in 39.2.2, we see that M = KS = SK, and K is central in M . Let (n, h, m) be a S-triple in s containing n as positive element. For i ∈ Z, let si be the set of elements y ∈ s verifying [h, y] = iy. If p2 = si , p2 = k ⊕ p2 , n2 = si , i0
i2
then p2 is a parabolic subalgebra of s, and n2 is an ideal of p2 contained in the nilpotent radical of p2 . Set P2 , P2 and N2 to be the connected algebraic subgroups of G whose Lie algebras are respectively adg p2 , adg p2 and adg n2 . We have P2 = KP2 = P2 K, and the set P2 (n) = P2 (n) is open and dense in n2 (see the proof of 34.4.6). 2) Set: p = k + p2 + n1 , n = n1 + n2 , w = k + n. We have [k + p2 , n1 ] ⊂ [gs , n1 ] ⊂ n1 . From this, we deduce that p is a Lie subalgebra of g which is clearly parabolic. Moreover:
39.2 Topology of Jordan classes
599
[p, w] ⊂ [p, k + n1 ] + [p, n2 ] ⊂ n1 + [n1 + p2 , n2 ] ⊂ n1 + n2 ⊂ w. So w is an ideal of p which is clearly solvable. If P is the parabolic subgroup of G whose Lie algebra is adg p, then P (w) = w. It follows from 39.2.1 that G(w) is a closed subset of g. Hence 39.1.5 and 39.2.2 imply that JG (x) = G(k + n). Thus JG (x) ⊂ G(w) = G(w). 3) Now P = P2 N1 = N1 P2 , and N1 is a normal subgroup of P . Let α ∈ P2 and u ∈ z(gs )• . We saw in 39.2.2 that N1 (u + n) = u + n + n1 . But p2 ⊂ cg (k) = gu , so α(u) = u. Consequently: N1 u + α(n) = N1 α(u + n) = α N1 (u + n) = α(u + n + n1 ) = u + α(n) + n1 . Thus: P (u + n) = N1 P2 (u + n) = N1 u + P2 (n) = u + P2 (n) + n1 . Hence:
P z(gs )• + n = z(gs )• + P2 (n) + n1 .
Recall from point 1 that P2 (n) = n2 . So: P z(gs )• + n = k + P2 (n) + n1 = k + n2 + n1 = w. We have therefore JG (x) = G(w) as required. 39.2.6 Corollary. Let J be a Jordan class in g. There exists a unique nilpotent G-orbit O in g such that J ∩ Ng = O. Proof. Let us conserve the notations of 39.2.5 so that J = G(w). By the construction of w, and by 20.8.3 (i), n is the set of nilpotent elements of w. This implies that J ∩ Ng = G(n). Thus J ∩ Ng is the image of the morphism G × n → g, (α, x) → α(x). It is therefore an irreducible subset of g. Since J is G-stable, it follows from 32.4.4 that J ∩ Ng = O1 ∪ · · · ∪ Or = O1 ∪ · · · ∪ Or , where O1 , . . . , Or are nilpotent G-orbits. Hence J ∩ Ng = Oj for some j. Finally, uniqueness is a consequence of the fact that any G-orbit is open in its closure. 39.2.7 Proposition. Let x ∈ g, s, n its semisimple and nilpotent components, and J = JG (x). (i) J is a union of Jordan classes in g. of all its ele(ii) J contains the semisimple and nilpotent components ments, and the set of semisimple elements of J is G z(gs ) .
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39 Sheets of Lie algebras
Proof. We shall use the notations of 39.2.5. Let u ∈ w be a semisimple element which commutes with the elements of k. Let us write u = t + n1 + n2 , with t ∈ k, n1 ∈ n1 and n2 ∈ n2 . Since [u, k] = [n1 , k], n1 = 0. So t and n2 are the semisimple and nilpotent components of u. Hence u = t. Thus k is maximal among the set of commutative Lie subalgebras of w consisting of semisimple elements. Let y ∈ J = G(w). To prove (i), it suffices to show that JG (y) ⊂ J. Since Jordan classes are G-stable (39.1.3), so we may assume that y ∈ w. Let t, m be the semisimple and nilpotent components of y. Then t, m ∈ w because, by construction, adg w is the Lie algebra of an algebraic subgroup of G. It follows from the preceding discussion, 29.7.5 (iv) and 29.7.7 that we may further assume that t ∈ k. So gt ⊃ gs , and hence z(gt ) ⊂ z(gs ) = k. Consequently, z(gt )• + m ⊂ k + n1 + n2 = w. Hence JG (y) ⊂ G(w) = J. Finally, part (ii) follows easily from the above statements. 39.2.8 Lemma. Let x ∈ g, and s, n its semisimple and nilpotent components. Then [x, g] + z(gx ) = [x, g] + z(gs ), and the sum [x, g] + z(gs ) is direct. Proof. Let s = [gs , gs ]. We saw in 39.1.1 that z(gx ) = z(gs ) + z(sn ). But by 35.3.9, z(sn ) ⊂ [n, s] = [s + n, s] = [x, s] ⊂ [x, g]. So we have the required equality. of L to z(gs) is nonThe Lie algebra gs is reductive sin g, so therestriction s degenerate (20.5.13). Since L z(g ), [s + n, g] = L [z(g ), s + n], g = {0}, [x, g] ∩ z(gs ) = {0}. 39.2.9 Proposition. Let x ∈ g, s, n its semisimple and nilpotent components and N the normalizer of gx in G. Then: dim JG (x) = dim G(x) + dim z(gs ) = dim[x, g] + dim z(gs ) = dim g − dim N + dim z(gx ). Proof. 1) The map ϕ: G × z(gx )• → JG (x), (α, y) → α(y) is a dominant x • morphism (39.1.7). So there exists a smooth point (α, y) of G × z(g ) such that α(y) is a smooth point of JG (x), and the tangent space Tα(y) JG (x) is the image of dϕ(α,y) (16.5.7). For β ∈ G, the maps (γ, z) → (βγ, z) and γ(z) → βγ(z) are isomorphisms of varieties. We may therefore assume that α is the identity element e of G. As in 29.1.4, the image of dϕ(e,y) is [y, g] + z(gy ). If t is the semisimple component of y, then dim JG (x) = dim[y, g] + dim z(gt ) (39.2.8). Hence we have the first G two equalities because y ∼ x. x 2) Set d = dim g , and let O be the G-orbit of gx in Gr(g, d). As we have seen in 39.1.7, the map ψ : JG (x) → O, y → gy is a surjective morphism of varieties. If y ∈ JG (x), then ψ −1 (gy ) = z(gy )• . So the fibres of ψ have dimension dim z(gx )• = dim z(gx ). But dim O = dim G − dim N (21.4.3). So the last equality follows by 15.5.4.
39.3 Sheets
601
39.3 Sheets 39.3.1 Recall that a cone C (of g) is a subset of g such that λx ∈ C for all x ∈ C, λ ∈ k \ {0}. Let n ∈ N and U a subset of g. We shall denote by U (n) the set of elements x ∈ U verifying dim G(x) = n. If p is the largest integer such that U (p) = ∅, then we shall write U • for U (p) (this generalizes the notation of 39.1.2). Definition. An irreducible component of g(n) , n ∈ N, is called a sheet of g. 39.3.2 Remarks. Let N be a sheet of g. 1) By 21.4.4, N is a locally closed subset of g. 2) Since G.N is an irreducible subset in the corresponding g(n) , N is G-stable. Let x ∈ N and F = (G(k \ {0})).x. Since N is a G-stable cone, F ⊂ N. On the other hand, the set of y ∈ F = G(kx) verifying dim G(y) = dim G(x) is irreducible, it is therefore contained in N. Consequently, N contains a nilpotent G-orbit (33.5.3). 3) The set greg = g• is a sheet of g, called the regular sheet of g. 39.3.3 Proposition. Let N be a sheet of g. There exists a unique Jordan class J in g such that J ⊂ N and J = N. Moreover N = (J)• . Proof. Let n ∈ N. If x, y belong to the same Jordan class, then the G-orbits of x and y have the same dimension (39.1.6 (ii)). We deduce therefore from 39.1.8 that there exist Jordan classes J1 , . . . , Jr in g such that g(n) = J1 ∪ · · · ∪ Jr . Let us denote by N1 , . . . , Ns the irreducible components of g(n) . For 1 i s, we have Ni ⊂ J1 ∪ · · · ∪ Jr . Since Ni is irreducible, Ni ⊂ Jt , for some t. Hence Ni ⊂ Jt . Conversely, since Jt is irreducible, Jt ⊂ Nj for some j. So Ni ⊂ Jt ⊂ Nj . Thus i = j (1.1.14). Uniqueness follows from 39.2.7 (i) and 39.1.8. Finally, let J be a Jordan class verifying J ⊂ N and J = N. For x ∈ J and y ∈ J, we have dim G(y) dim G(x) (21.4.4). Consequently, N ⊂ (J)• . The reverse inclusion follows from the fact that (J)• is an open subset of J, so it is irreducible. Remark. With the above notations, we have N = J in general. This is the case if g = {0}, and N = greg , J = ggen . 39.3.4 Corollary. Let N be a sheet of g. (i) There exists a parabolic subalgebra p of g and a solvable ideal w of p verifying: • N = G(w) , N = G(w• ) = G(w) . (ii) N is a union of Jordan classes in g.
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39 Sheets of Lie algebras
Proof. (i) This is clear by 39.2.5 and 39.3.3. (ii) Let J be the unique Jordan class verifying the conditions of 39.3.3 for N. By 39.2.7, J is a union of Jordan classes. Since N = (J)• (39.3.3) and the elements of a Jordan class belong to a single g(n) (39.1.6 (ii)), the result is clear. 39.3.5 Proposition. Let N be a sheet of g. Then N contains a unique nilpotent G-orbit. Proof. Let d be the dimension of the G-orbits contained in N and U ⊂ N a nilpotent G-orbit. Such an orbit exists by 39.3.2. Denote by O the unique nilpotent G-orbit such that N ∩ Ng = O (39.2.6 and 39.3.3). Then dim O = dim O d (21.4.4), and U ⊂ O. So U = O, and hence U = O (21.4.5). 39.3.6 Let N be a sheet of g. Denote by J = JG (x) the unique Jordan class verifying N = (J)• (39.3.3). Let s, n be the semisimple and nilpotent components of x, and we shall use the notations of 39.2.5. We saw in 39.2.6 that G(n) = J ∩ Ng . By 39.3.4 and 39.3.5, we deduce • that G(n• ) = G(n) is the unique nilpotent G-orbit contained in N. 39.3.7 Proposition. Let y ∈ w and z ∈ w• ⊂ N. Then: (i) dim G(z) = dim g + dim n − dim p. (ii) P (z) = z + n. (iii) w ∩ G(z) is a finite union of P -orbits. (iv) N ∩ (y + n) = ∅. Proof. (i) We may assume that z = x because x, z belong to the same sheet, so G(x) and G(z) have the same dimension. We have gx = gs ∩ gn = k ⊕ sn . Also: dim sn = dim s0 + dim s1 = dim p2 − dim n2 . Since p = k ⊕ p2 ⊕ n1 and n = n1 ⊕ n2 , we obtain that dim gx = dim(p/n). (ii) We verify easily that [p, w] ⊂ n. Consequently, P (u) ⊂ u + n for u ∈ w (24.5.5). It follows from part (i) that: dim(p/n) = dim gz dim pz = dim p − dim P (z) dim p − dim(z + n) = dim(p/n). Thus dim P (z) = dim n, and hence P (z) = z + n. (iii) Let h be a Cartan subalgebra of g containing k and W the Weyl group relative to (g, h). Recall that w = k ⊕ n. Also, we saw in 39.2.6 that n is the set of nilpotent elements of w, and in 39.2.7, we saw that for any semisimple element u of w, there exists α ∈ Aute w ⊂ P such that α(u) ∈ k. Let r ∈ w, and t, v its semisimple and nilpotent components. We have v ∈ n, and there exists α ∈ P such that α(t) ∈ k. But P (t) ∈ t + n by the proof of (ii). From this, we deduced that r + n = τr + n, where τr ∈ k is P -conjugate to the semisimple component of r.
39.4 Dixmier sheets
603
If r, z ∈ w are G-conjugate, then τr , τz ∈ k are G-conjugate. It follows from 31.2.6, 34.2.1 and 34.2.4 that τr and τz belong to the same W -orbit in h. We deduce that G(z) meets only a finite number of classes in w modulo n. Let us denote these classes by z1 + n, . . . , zs + n, with z1 , . . . , zs ∈ w ∩ G(z). Then: P (z1 ) ∪ · · · ∪ P (zs ) ⊂ w ∩ G(z) ⊂ (z1 + n) ∪ · · · ∪ (zs + n). For 1 i s, zi ∈ G(z), so zi ∈ w• , and P (zi ) = zi + n by (ii). For u ∈ P (zi ) \ P (zi ), we have dim P (u) < dim P (zi ) = dim P (z) = dim n (see (ii) / G(z). Consequently: and 21.4.5), hence u ∈ / w• , and u ∈ w ∩ G(z) = P (z1 ) ∪ · · · ∪ P (zs ). (iv) Since G(n• ) ⊂ N, the result is clear if y ∈ n (39.3.6). Let us suppose that y ∈ n. We have ky + n ⊂ w ⊂ G(w) = N. Since N is locally closed and n• ⊂ N, we deduce that N ∩ (ky + n) is a non-empty open subset of ky + n. There exists therefore t ∈ k \ {0} such that (ty + n) ∩ N = ∅. But N is a cone, so we have also (y + n) ∩ N = ∅.
39.4 Dixmier sheets 39.4.1 Lemma. Let X, Y be irreducible smooth varieties such that Y is normal, and ϕ : X → Y a dominant morphism. Assume that for all x ∈ X, the differential of ϕ at the point x is surjective. (i) Let x ∈ X and Z an irreducible component of ϕ−1 (ϕ(x)). Then: dim Z = dim X − dim Y. (ii) The map ϕ is open. Proof. By 17.4.11, it suffices to prove (i). Our assumptions imply that dim X dim Y and dim Z dim X − dim Y for all irreducible component Z of ϕ−1 (ϕ(x)) (15.5.4). Let z ∈ Z. Then: dim ker dϕz = dim Tz (X) − dim Tϕ(z) (Y ) = dim X − dim Y. Since the restriction ψ of ϕ to Z is constant, dψz = 0. So Tz (Z) ⊂ ker dϕz , and dim Z dim Tz (Z) dim X − dim Y . Hence (i). 39.4.2 Lemma. Let x ∈ Sg and y ∈ gx . The following conditions are equivalent: (i) gy ⊂ gx . (ii) g = [y, g] + gx . Proof. (i) ⇒ (ii) Since x ∈ Sg , g = gx + [x, g]. So if gy ⊂ gx , then [x, g] ⊂ [y, g] (35.3.1), and hence (i) ⇒ (ii). (ii) ⇒ (i) For λ ∈ k, set gλ to be the eigenspace of ad x corresponding to the eigenvalue λ. We have:
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39 Sheets of Lie algebras
g=
λ∈k
gλ , [x, g] =
gλ .
λ∈k\{0}
Let y ∈ gx be such that g = [y, g] + gx . Taking the orthogonals with respect to L, we obtain that gy ∩ [x, g] = {0}. But since y ∈ gx , gy is (ad x)-stable, so gy is the sum of its intersections with the gλ ’s. Thus gy ⊂ g0 = gx . 39.4.3 Lemma. Let s ∈ Sg and ux the set of y ∈ gx verifying gy ⊂ gx . Then ux is an open subset of gx . Proof. Let us consider the map ϕ : G × gx → g, (α, y) → α(y). For y ∈ gx , the image of dϕ(e,y) is [y, g] + gx . By 39.4.2, y ∈ ux if and only if the rank of dϕ(e,y) is equal to dim g. So the result follows. 39.4.4 Lemma. Let x ∈ Sg and vx the set of y ∈ g such that gy is G-conjugate to a subspace of gx . Then vx is an open subset of g. Proof. Let ux be as in 39.4.3. It is open in gx , so it is a smooth variety. Let us consider the map ϕ : G × ux → g, (α, y) → α(y). For (α, y) ∈ G × ux , the image of dϕ(α,y) is α([y, g] + gx ). So 39.4.2 implies that dϕ(α,y) is surjective, hence ϕ is dominant (16.5.7). The result follows therefore from 39.4.1 because vx is the image of ϕ. • 39.4.5 Theorem. Let x ∈ Sg . Then G(z(gx )) is a sheet of g. It is the unique sheet of g containing x and its dimension is dim G(x) + dim z(gx ). Proof. Let N be a sheet containing x, and vx be as in 39.4.4. Then N ∩ vx y x is contained in the set of y ∈ g such that g is G-conjugate to g , that is in G z(gx )• (39.1.6 and 39.1.7). By 39.4.4, N ∩ vx is a non-empty open subset of N. Hence: N = N ∩ vx ⊂ G(z(gx )• ) ⊂ G(z(gx )). • Let U = G(z(gx )) . Then N ⊂ U. Since U is open in G(z(gx )), it meets G(z(gx )). Now if y ∈ G(z(gx )), then dim gy dim gx . We deduce therefore that U is contained in the g(n) which contains N. So N = U because U is irreducible. Finally, 39.2.9 says that its dimension is dim G(x) + dim z(gx ). 39.4.6 Definition. A sheet of g which contains a semisimple element is called a Dixmier sheet of g. Remarks. 1) By 39.4.5, a semisimple element of g is contained in a unique (Dixmier) sheet of g. 2) Let h be a Cartan subalgebra of g and R the root system of g relative to h. In the notations of 39.1.8, • 39.4.5 says that Dixmier sheets of g are subsets of g of the form G(hP ) , where P is the intersection of R with a vector subspace of h∗ . 39.4.7 Theorem. Dixmier sheets of g are precisely the subsets of g of the form G(r)• , where r is the radical of a parabolic subalgebra of g.
39.5 Jordan classes in the symmetric case
605
Proof. Let p be a parabolic subalgebra of g, r its radical, l a Levi factor of • x p, and•c the centre of l. If x ∈ c , then z(g ) = c. It follows from 39.4.5 that G(c) is a Dixmier sheet of g. But G(r) = JG (x) = G(c) (39.2.2, 39.2.4 and 39.1.7). So G(r)• is a Dixmier sheet of g. The proof of the converse is analogue by using again 39.4.5. 39.4.8 Theorem. Let g be a semisimple Lie algebra. The set of polarizable elements of g is the union of the Dixmier sheets of g. Proof. Let us fix a Cartan subalgebra h of g and denote by R the root system of g relative to h. • 1) Let D = G(hP ) be a Dixmier sheet of g (see Remark 2 of 39.4.6), B a base of P, B a base of R containing B , R+ the set of positive roots with respect to B and gα . p=h⊕ α∈P∪R+
Let q = dim p, Gr(g, q) the Grassmannian variety of q-dimensional subspaces of g, and p : Gr(g, q) × g → g the canonical surjection. Denote by F the set of elements of Gr(g, q) of the form θ(p), with θ ∈ G, and: L = {(q, x) ∈ F × D; L(x, [q, q]) = {0}}. The set L is closed in F × D and we check easily that the closed subset p(L) contains G(hP ). Thus D ⊂ p(L). Let x ∈ D. Then there exists q ∈ F such that L(x, [q, q]) = {0}. Since F consists of Lie subalgebras of g (19.7.2), we see that q is a polarization of g at x. 2) Let x be a polarizable element of g and p a polarization of g at x. We saw in 33.2.6 that p is a parabolic subalgebra of g whose nilpotent radical is n = [x, p]. It follows that x belong to the radical r of p. Up to a conjugation by an element of G, we may assume that h ⊂ p. For a suitable choice of positive roots R+ , there exists a subset P of R which is the intersection of R with a vector subspace of h∗ such that: gα , n = gα . p=h⊕ α∈P∪R+
α∈R+ \P
As in 33.2.8, if h ∈ h•P , then p is a polarization of g at h. Consequently, h x dim g =• dim g . Since exp(ad n)(h) = h + n (32.2.5), it is clear that x ∈ G(hP ) .
39.5 Jordan classes in the symmetric case 39.5.1 In the rest of this chapter, (g; k, p) or (g, θ) is a semisimple symmetric Lie algebra (see Chapter 38). We shall denote by K the connected algebraic subgroup of G such that L(K) = adg k. Finally, set Sp = Sg ∩ p, and Np = Ng ∩ p.
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39 Sheets of Lie algebras
Let x ∈ p. By 38.8.3, cp (px ) is an abelian ideal of gx . If x ∈ Np , then cp (px ) contains only nilpotent elements. If x ∈ Sp , then cp (px ) contains only semisimple elements and cp (px ) = p ∩ z(gx ). Recall also that cp (px )• is the set of elements y ∈ p verifying py = px . It is an open subset of cp (px ), and we have (38.8.4): cp (px )• = {y ∈ cp (px ); rk(ad y) = rk(ad x)}. By using 38.8.3, the proof of the following result is analogue to the one for 39.1.2. Lemma. Let x ∈ p, s, n its semisimple and nilpotent components and [gs , gs ] = k1 ⊕ p1 , with k1 ⊂ k et p1 ⊂ p. Then: cp (px )• = {y + z; y ∈ cp (ps )• , z ∈ cp1 (pn1 )• }. 39.5.2 Let x, y ∈ p. We have x = xs + xn and y = ys + yn , where xs , ys are semisimple, xn , yn nilpotent, and [xs , xn ] = [ys , yn ] = 0. We say that x and y are K-Jordan equivalent if there exists α ∈ K such that: pys = pα(xs ) = α(pxs ) , yn = α(xn ). This defines an equivalence relation on p. The class of x, denoted by JK (x), K is called the K-Jordan class of x in p. We shall write x ∼ y if x and y are K-Jordan equivalent. The proof of the following result is analogue to the one for 39.1.5. Proposition. Let x ∈ p and s, n its semisimple and nilpotent components. The set JK (x) is an irreducible subset of p and: JK (x) = K cp (ps )• + n . 39.5.3 Remark. Let x ∈ p and s, n its semisimple and nilpotent components. If t ∈ cp (ps )• , then pt = ps , and rk(ad s) = rk(ad t) (39.5.1). Since cp (ps ) ⊂ z(gs ) (39.5.1), cp (ps )• ⊂ z(gs )• (notation of 39.1.2). It follows from K
39.1.5 and 39.5.2 that JK (x) ⊂ JG (x). Thus, for x, y ∈ p, the condition x ∼ y G implies that x ∼ y. G However, it may happen that x ∼ y, but x and y are not K-Jordan equivalent. This is the case for the elements e and f of 38.7.18. 39.5.4 Proposition. Let x, y ∈ p. The following conditions are equivalent: K (i) x ∼ y. (ii) There exists α ∈ K such that py = α(px ). (iii) There exists α ∈ K such that cp (py ) = α cp (px ) .
39.5 Jordan classes in the symmetric case
607
Proof. The equivalence (ii) ⇔ (iii) is straightforward by 38.8.2, and the implication (i) ⇒ (ii) is proved as in 39.1.6 by replacing G by K and g by p. Let us suppose that there exists α ∈ K such that py = α(px ) = pα(x) . If x is nilpotent, then 38.8.4 and 38.8.5 imply that y ∈ K α(x) = K(x), K
so x ∼ y. Let us consider the general case. Let s, n be the semisimple and nilpotent components of α(x). We have y ∈ cp (pα(x) ). Let us write [gs , gs ] = k1 ⊕ p1 , subgroup of with k1 ⊂ k and p1 ⊂ p, and denote by K1 the smallest algebraic K with Lie algebra adg k1 . Then K1 z(gs ) = z(gs ), so K1 cp (ps ) = cp (ps ) by 38.8.3 (i). Hence K1 cp (ps )• = cp (ps )• . It follows from the preceding paragraph and 39.5.1 that: s • s • y ∈ cp (ps )• + cp1 (pn1 )• ⊂ cp (p ) s+•K1 (n) = K1 cp (p) + n ⊂ K cp (p ) + n) = JK α(x) = JK (x). K
Thus x ∼ y. 39.5.5 Corollary. Let x ∈ p. (i) We have: JK (x) = K cp (px )• . (ii) The set JK (x) is locally closed in p, so it is a subvariety of p. Proof. (i) This is clear by 39.5.4. (ii) Let d ∈ N and u the set of y ∈ p such that dim py = d. The set u is a locally closed subset of p, so it is a subvariety of p. As in 19.7.6 and identifying p with p∗ via the Killing form, the map ϕ : u → Gr(p, d), y → py , is a morphism of varieties. In view of 39.5.4, we may finish the proof as in 39.1.7 (ii). 39.5.6 Let us fix a Cartan subspace a of p, and denote by S the root system of (g, a). If P is a subset of S which is the intersection of S with a vector subspace of a∗ , denote by aP the set of x ∈ a such that α(x) = 0 for all α ∈ P. Let a•P be the set of x ∈ aP verifying α(x) = 0 for all α ∈ S \ P. The set of the subsets aP and a•P of a thus constructed is finite. Let x ∈ a and P the set of elements of S which are zero at x. We obtain easily that cp (px ) = aP . Let us write [gx , gx ] = kx ⊕ px , with kx ⊂ k and px ⊂ p. Let Kx be the smallest algebraic of K with Lie subgroup algebra adg kx . We have Kx z(gx ) = z(gx ), Kx cp (px ) = cp (px ) and Kx cp (px )• = cp (px )• . By 38.6.1, there exist n1 , . . . , nr ∈ px such that Kx (n1 ), . . . , Kx (nr ) are the nilpotent Kx -orbits in px . Let n ∈ p be a nilpotent element which commutes with x. Then n ∈ px , so n = α(ni ), for some 1 i r and α ∈ Kx . Hence: cp (px )• + n = cp (px )• + α(ni ) = α cp (px )• + ni .
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Consequently, JK (x + n) = JK (x + ni ). Finally, let y ∈ p and s, n its semisimple and nilpotent components. There exists α ∈ K such that α(s) ∈ a (37.4.10 and 37.5.4). We deduce therefore that JK (y) = JK (x + u), where x ∈ a and u ∈ p is a nilpotent element which commutes with x. So we have obtained the following result: Proposition. The number of K-Jordan classes in p is finite. 39.5.7 Lemma. Let x ∈ p, and s, n its semisimple and nilpotent components. We have [x, k] + cp (px ) = [x, k] + cp (ps ), and the sum [x, k] + cp (ps ) is direct. Proof. Since cp (ps ) ⊂ z(gs ) (39.5.1), 39.2.8 implies that the sum [x, k] + cp (ps ) is direct. Let us write s = [gs , gs ] = k1 + p1 , with k1 ⊂ k and p1 ⊂ p. By 38.8.3, we have cp (px ) = cp (ps ) + cp1 (pn1 ). Denote by S ⊂ K the connected algebraic subgroup with Lie algebra adg s, and let O be the S-orbit of n. It follows from 38.8.5 that cp1 (pn1 ) ⊂ O. Hence: cp1 (pn1 ) ⊂ Tn cp1 (pn1 ) ⊂ Tn (O) = Tn (O) = [n, s] = [s + n, s] = [x, s] ⊂ [x, g]. Since cp1 (pn1 ) ⊂ p, cp1 (pn1 ) ⊂ [x, k]. Hence the result. 39.5.8 In view of 39.5.7, the proof of the following result is analogue to the one for 39.2.9. Proposition. Let x ∈ p, s, n its semisimple and nilpotent components and N the normalizer of px in K. Then: dim JK (x) = dim K(x) + dim cp (ps ) = dim[x, k] + dim cp (ps ) = dim k − dim N + dim cp (px ).
39.6 Sheets in the symmetric case 39.6.1 We shall conserve the hypotheses and notations of 39.5.1. For n ∈ N, we denote by p(n) the set of elements x ∈ p such that dim K(x) = n. It is a locally closed subset of p. We shall call an irreducible component of p(n) a sheet of p. As in 39.3.2, we observe that a sheet of p is a K-stable cone. The set R of p-regular elements of p (see 38.4.4) is a sheet of p that we shall call the regular sheet of p. 39.6.2 Proposition. If Q is a sheet of p, then Q contains a nilpotent K-orbit. Proof. We may proceed as in remark 2 of 39.3.2 by using 38.6.9 (i) instead of 33.5.3.
39.6 Sheets in the symmetric case
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39.6.3 Remark. Contrary to the Lie algebra case (39.3.5), a sheet of p can contain several nilpotent K-orbits. For example, the regular sheet of example 38.7.18 contains two distinct nilpotent K-orbits. 39.6.4 Lemma. Let x ∈ Sp and y ∈ px . The following conditions are equivalent: (i) py ⊂ px . (ii) p = [y, k] + px . Proof. Since x ∈ Sp , we have g = [x, g] ⊕ gx . Hence: k = [x, p] ⊕ kx , p = [x, k] ⊕ px , [x, k] = [x, [x, p]]. If py ⊂ px , then [x, k] ⊂ [y, k] (38.8.2). So (i) ⇒ (ii). Let us prove (ii) ⇒ (i). Consider the restriction u of (ad x)2 to p. It is a semisimple endomorphism of p. For λ ∈ k, denote by pλ the eigenspace of u corresponding to the eigenvalue λ. Then: pλ , u(p) = pλ . p= λ∈k
λ∈k\{0}
Let y ∈ px be such that p = [y, k] + px . Then by 38.8.1, we obtain: k = p⊥ = ([y, k] + px )⊥ = (py + k) ∩ ([x, k] + k). From this, we deduce that py ∩ [x, k] = {0}, that is py ∩ u(p) = {0}. But since y ∈ px , py is u-stable, so it is the sum of the py ∩ pλ for λ ∈ k. Hence py ⊂ p 0 = p x . 39.6.5 Lemma. Let x ∈ Sp and ux the set of y ∈ px verifying py ⊂ px . Then ux is an open subset of px . Proof. Let ϕ : K × px → p be the map (α, y) → α(y). The image of dϕ(e,y) is [y, k] + px . By 39.6.4, y ∈ ux if and only if the rank of dϕ(e,y) is dim p. So the result follows. 39.6.6 Lemma. Let x ∈ Sp and vx the set of y ∈ px such that py is K-conjugate to a subspace of px . Then vx is an open subset of p. Proof. In view of 39.6.5, the proof is analogue to the one for 39.4.4. 39.6.7 Theorem. Let x ∈ Sp . Then (K(cp (px )))• is a sheet of p. It is the unique sheet of p containing x, and its dimension is dim cp (px ) + dim K(x). Proof. In view of the preceding results, the proof is analogue to the one for 39.4.5. 39.6.8 Proposition. Let Q1 , . . . , Qr be the sheets of p which contain a semisimple element. Then any element Q1 ∪ · · · ∪ Qr is polarizable.
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Proof. If x ∈ Sp , then cp (px ) ⊂ z(gx ) (39.5.1). Hence: (K(cp (px )))• ⊂ (G(z(gx )))• . Since (G(z(gx )))• is a sheet of G in g containing x (39.4.5), the result is a consequence of 39.4.8. 39.6.9 Let us consider example 1 of 38.8.8. By 33.4.6, any element of g is polarizable. Let x ∈ p be a non-zero semisimple element. Then it is contained in a Cartan subspace a of p (37.4.7). By 37.4.11, a = ps where s is p-generic. But dim a = 1 (38.8.8 and 37.4.11). So x is p-generic. We have therefore shown that any non-zero semisimple element of p is p-generic, and so, the regular sheet R is the unique sheet of p which contain a non-zero semisimple element of p. Let y = E12 ∈ p. Then py = kE12 + kE32 , so y ∈ R but y is polarizable. Thus, contrary to the Lie algebra case (39.4.8), the union of the sheets of p which contain a semisimple element is not the set of polarizable elements of p.
References and comments • [6], [7], [16], [28], [81] The notion of sheets was first introduced in [28] for determining polarizable elements. The definition of a sheet was generalized in [6] and [7]. For the presentation of Jordan classes in this chapter, we have followed [16]. The fact that the set of polarizable elements is the union of Dixmier sheets (39.4.8) was used in [71] to prove the irreducibility of the commuting variety {(x, y) ∈ g × g; [x, y] = 0} of a reductive Lie algebra g. Generalization of the problem of the irreducibility of the commuting variety to the symmetric case was studied in [60], [73], [61], [74]. In particular, the notion of sheets and p-distinguished elements were used in [74].
40 Index and linear forms
In this chapter, we study certain properties related to the coadjoint representation of a Lie algebra. In particular, we determine the index of certain classes of Lie algebras. Throughout this chapter, g will denote a finite-dimensional Lie algebra.
40.1 Stable linear forms 40.1.1 Let us consider the coadjoint representation of g on g∗ and let us conserve the notations of section 19.7. For a subspace a of g, we denote by a⊥ the orthogonal of a in g∗ . Recall that for f ∈ g∗ , we denote by Φf the alternating bilinear form on g defined by Φf (x, y) = f ([x, y]) for x, y ∈ g, and a(f ) denote the orthogonal of a with respect to Φf . In particular, g(f ) is the kernel of Φf , and we have: (f ) ⊥ g = g.f , [g, g(f ) ]⊥ = {g ∈ g∗ ; g(f ) ⊂ g(g) }. In the rest of this section, K will be an algebraic subgroup of Aut g with Lie algebra k. The group K acts rationally on g and g∗ . For α ∈ K, f ∈ g∗ and x ∈ g, we have (α.f )(x) = f (α−1 (x)). The differential of this action defines an action of k on g∗ given by (X.f )(x) = f (−X(x)) for X ∈ k. We set: kf = {X ∈ k; X.f = 0} , kf = {x ∈ g; (k.f )(x) = {0}}. Thus we have (kf )⊥ = k.f . If K contains the algebraic adjoint group of g, then adg g ⊂ k, and so kf ⊂ g(f ) . The bilinear form (X, x) → f (X(x)) on k × g induces a non-degenerate bilinear form on (k/kf ) × (g/kf ). It follows that: dim k − dim kf = dim g − dim kf .
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Definition. An element f ∈ g∗ is said to be K-stable if there exists a neighbourhood V of f in g∗ such that for all g ∈ V , g(f ) and g(g) are Kconjugate. If K is the algebraic adjoint group of G (resp. if K = Aut g), a K-stable linear form is called stable (resp. weakly stable). 40.1.2 Lemma. Let f ∈ g∗ . Suppose that K is connected and that the morphism ψ : K × [g, g(f ) ]⊥ −→ g∗ , (α, h) → α(h) is dominant. Then: (i) f ∈ g∗reg . (ii) Let W = [g, g(f ) ]⊥ ∩ g∗reg . The restriction of ψ to K × W is open. Proof. (i) The image of ψ contains an open dense subset of g∗ by 15.4.2, so it contains a regular element g of g∗ because g∗reg is a non-empty open subset of g∗ . If α ∈ K and h ∈ [g, g(f ) ]⊥ verify g = α.h, then α(g(f ) ) ⊂ α(g(h) ) = g(g) . Hence f ∈ g∗reg . (ii) Let ϕ : K × W → g∗ denote the restriction of ψ to K × W . By (i), W is a non-empty open subset of [g, g(f ) ]⊥ . So ϕ is dominant. Let N be the normalizer of g(f ) in K. If α, β ∈ K and g, h ∈ W verify β.h = α.g, or equivalently h = (β −1 α).g, then since g(h) = g(g) = g(f ) ,we have β −1 α ∈ N , or β ∈ αN . It follows that the fibre S = ϕ−1 ϕ(α, g) is equal to {(αβ, β −1 .g); β ∈ N }. Now S is also the graph of the morphism αN → g∗ , γ → (γ −1 α).g, so it is isomorphic to N (12.5.4). As N is a pure variety (21.1.6), it follows from 15.5.4 and 17.4.11 that ϕ is an open map. 40.1.3 Lemma. Let K ◦ be the identity component of K and f ∈ g∗ . The following conditions are equivalent: (i) f is K ◦ -stable. (ii) f is K-stable. If these conditions are verified, then f ∈ g∗reg . Proof. (i) ⇒ (ii) This is clear. (ii) ⇒ (i) Let us fix an open neighbourhood U of f such that g(h) and g(f ) are K-conjugate for all h ∈ U . Since U ∩ g∗reg = ∅, f ∈ g∗reg , and therefore W = [g, g(f ) ]⊥ ∩ g∗reg is a non-empty open subset of [g, g(f ) ]⊥ ; W is in fact the set of h ∈ g∗ verifying g(h) = g(f ) . −1 If g ∈ U , then g(g) = α(g(f ) ) for some α ∈ K. So g(α .g) = g(f ) , and g ∈ α(W ). Thus the image of the morphism ψ : K × W → g∗ , (α, h) → α.h, contains U . Now let K0 = K ◦ , K1 = α1 K ◦ , . . . ,Kr = αr K ◦ be the (distinct) irreducible components of K, where α1 , . . . , αr ∈ K. Set Ti = ψ(Ki × W ). Since U ⊂ T0 ∪ · · · ∪ Tr , we obtain that g∗ = T0 ∪ · · · ∪ Tr . It follows from the irreducibility of g∗ that g∗ = Ti for some i. But Ti = αi (T0 ) = αi T0 , so g∗ = T0 , and the restriction ϕ of ψ to K ◦ × W is dominant. Now 40.1.2 implies that ϕ
40.1 Stable linear forms
613
is an open map, so T0 is an open neighbourhood of f in g∗ with the property that for any g ∈ T0 , g(g) and g(f ) are K ◦ -conjugate. 40.1.4 Theorem. Let g be a Lie algebra, f ∈ g∗ , K an algebraic subgroup of Aut g and k the Lie algebra of K. Then the following conditions are equivalent: (i) [g, g(f ) ] ∩ kf = {0}. (ii) The linear form f is K-stable. Proof. By 40.1.3, we may assume that K is connected. Let e denote the identity element of K. (i) ⇒ (ii) Let us consider the morphism ψ : K × [g, g(f ) ]⊥ → g∗ given by (α, h) → α.h. By 29.1.4, we have, for X ∈ k and h ∈ [g, g(f ) ]⊥ : dψ(e,f ) (X, h) = h + X.f. The image of dψ(e,f ) is therefore [g, g(f ) ]⊥ +k.f , which is g∗ by our hypothesis. Thus ψ is dominant (16.5.7). Let W = [g, g(f ) ]⊥ ∩ g∗reg . Then 40.1.2 says that ψ(K × W ) is an open subset of g∗ containing f , and for any g ∈ ψ(K × W ), g(g) and g(f ) are K-conjugate. (ii) ⇒ (i) We have f ∈ g∗reg (40.1.3). So W = [g, g(f ) ]⊥ ∩ g∗reg is a nonempty open subset of [g, g(f ) ]⊥ ; recall that W is the set of h ∈ g∗ such that g(h) = g(f ) . The image of the morphism ϕ : K × W → g∗ , (α, h) → α.h is therefore the set of h ∈ g∗ such that g(h) is K-conjugate to g(f ) . For g ∈ W and (X, h) ∈ k × [g, g(f ) ]⊥ , we have: dϕ(e,g) (X, h) = h + X.g. Thus the image Tg of dϕ(e,g) is [g, g(f ) ]⊥ + k.g. Let us denote by n the normalizer of g(f ) in k. As: (X.g)([g, g(f ) ]) = g [X(g), g(f ) ] + g [g, X(g(f ) )] = g [g, X(g(f ) )] , and g(g) = g(f ) , we deduce that X.g ∈ [g, g(f ) ]⊥ if and only if X ∈ n. Hence: dim(k.g ∩ [g, g(f ) ]⊥ ) = dim n.g. On the other hand, if X ∈ kg , then X ∈ n because X.g = 0. So: dim n.g = dim n − dim kg = dim n − dim k + dim k.g. We deduce therefore that: dim Tg = dim k − dim n + dim([g, g(f ) ]⊥ ). In particular, dim Tg does not depend on g. It follows from 29.1.4 (ii) that the dimension of the image T(α,g) of dϕ(α,g) does not depend on α or on g. Now, our hypothesis implies that ϕ is dominant, so there exists (α, g) ∈ K × W such that T(α,g) = g∗ (16.5.7). Hence T(e,f ) = g∗ , and [g, g(f ) ]⊥ + k.f = g∗ . Consequently, [g, g(f ) ] ∩ kf = {0}.
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40.1.5 Corollary. Let f ∈ g∗ . (i) If g is ad-algebraic, then f is stable if and only if [g, g(f ) ] ∩ g(f ) = {0}. (ii) If [g, g(f ) ] ∩ g(f ) = {0}, then f is K-stable for any algebraic subgroup K of Aut g whose Lie algebra contains adg g, in particular, this is the case if K contains the algebraic adjoint group of g. 40.1.6 The following proposition gives a sufficient condition for applying part (ii) of 40.1.5: Proposition. Let g be a Lie subalgebra of a semisimple Lie algebra and f ∈ g∗ . If g(f ) is a commutative Lie algebra consisting of semisimple elements, then [g, g(f ) ] ∩ g(f ) = {0}. Proof. If g(f ) is a commutative Lie algebra consisting of semisimple elements, then there exists a vector subspace r of g such that g = g(f ) ⊕r and [g(f ) , r] ⊂ r. So [g, g(f ) ] ⊂ r, and the result follows. Remark. More generally, if a is a commutative Lie subalgebra of g consisting of semisimple elements, then [g, a] ∩ a = {0}. 40.1.7 We shall now give an example of a Lie algebra which does not admit any weakly stable linear form. Let s = sl2 (k) and V be a simple s-module of dimension 5. The Lie bracket of the semi-direct product s V is given by: for x1 , x2 ∈ s and v1 , v2 ∈ V , [(x1 , v1 ), (x2 , v2 )] = ([x1 , x2 ], x1 .v2 − x2 .v1 ). The endomorphism t of s V , (x, v) → v, is a derivation of the Lie algebra s V . We may therefore define the semi-direct product g = kt (s V ) where the Lie bracket is given by: for λ1 , λ2 ∈ k and z1 , z2 ∈ s V , [(λ1 t, z1 ), (λ2 t, z2 )] = (0, [z1 , z2 ] + λ1 t(z2 ) − λ2 t(z1 )). It is easy to check that kt V and V are respectively the radical and the nilpotent radical of g. We shall show that g does not admit any weakly stable linear form. The Lie algebra of Aut g is d = Der g. We claim that d = adg g. Let us proof our claim. Let δ ∈ d. By 20.3.3, there exists w ∈ g such that δ(x) = [x, w] for all x ∈ s. Replacing δ by δ − adg w, we may assume that δ|s = 0. Since the ideals kt V and V are stable under δ, we have, for v ∈ V : δ(v) = δ([t, v]) = [δ(t), v] + [t, δ(v)] = [δ(t), v] + δ(v). So [δ(t), V ] = {0}, and hence δ(t) ∈ V . Now, if x ∈ s, then: 0 = δ([x, t]) = [δ(x), t] + [x, δ(t)] = [x, δ(t)]. Since V is simple and δ(t) ∈ V , we deduce that δ(t) = 0.
40.2 Index of a representation
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For x ∈ s and v ∈ V , we have δ([x, v]) = [x, δ(v)]. So δ|V is an endomorphism of V which commutes with the action of s. It follows by Schur’s lemma that δ = λ adg t for some λ ∈ k. So we have proved our claim. Let f ∈ g∗ . The subspace g(f ) is the kernel of the linear map uf : g → g∗ , x → x.f . Since [V, V ] = {0}, the restriction of uf to V induce a linear map from V to (kt ⊕ s)∗ . But dim V = 5 and dim(kt ⊕ s) = 4, so g(f ) ∩ V = {0}, and g(f ) ∩ V ⊂ [g, g(f ) ] because [t, v] = v for all v ∈ V . Finally, our claim implies that g(f ) = df , so condition (i) of 40.1.4 is not satisfied, and f is not weakly stable.
40.2 Index of a representation 40.2.1 Let V be a finite-dimensional g-module. We set: gv = {x ∈ g; x.v = 0} , g.v = {x.v; x ∈ g}. We say that v is g-regular or simply regular if: dim gv = min{dim gw ; w ∈ V }. Since dim gv + dim g.v = dim g, v is regular if and only if: dim g.v = max{dim g.w; w ∈ V }. The set of regular elements of V is a non-empty Zariski open subset of V . 40.2.2 In the notations of 40.2.1, we consider the contragredient representation of V on V ∗ . Definition. The integer dim V − max{dim g.f ; f ∈ V ∗ } is called the index of the g-module V , and we shall denote it by χ(g, V ). 40.2.3 Remarks. 1) The index of the adjoint representation of g is just the index χ(g) of g as defined in 19.7.3. 2) Let G be a connected algebraic group, g its Lie algebra, and V a rational G-module. Then V has a natural structure of g-module, and we have χ(g, V ) = dim V − max{dim G.f ; f ∈ V ∗ } = min{dim Gf ; f ∈ V ∗ }. 40.2.4 Let f ∈ V ∗ . We define a linear map Lf : g → V ∗ as follows: for x ∈ g and v ∈ V , Lf (x)(v) = f (x.v). The kernel (resp. image) of Lf is gf (resp. g.f ). Thus: χ(g, V ) = dim V − max{rk(Lf ); f ∈ V ∗ }.
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40.2.5 Let f ∈ V ∗ . We define a bilinear form Ψf on g × V as follows: for x ∈ g and v ∈ V , Ψf (x, v) = Lf (x)(v) = f (x.v). Let us fix bases X = (x1 , . . . , xn ) of g and V = (v1 , . . . , vm ) of V . Let M (X , V) be the matrix [xi .vj ]1in,1jm , considered as a matrix with coefficients in the field of fractions of the symmetric algebra S(V ) of V . Then Mf (X , V) = of Ψf with respect to the bases X and V. [f (xi .vj )] ∈ Mn,m(k) is the matrix Thus rk(Lf ) = rk Mf (X , V) . Consequently, we have: χ(g, V ) = dim V − rk M (X , V) .
40.3 Some useful inequalities 40.3.1 Let V be a finite-dimensional g-module and v ∈ V . For x ∈ gv and y ∈ g, we have: x.(y.v) = [x, y]v + y.(x.v) = [x, y].v. So g.v is a gv -module. Let us denote by V the quotient gv -module V /g.v. Proposition. (Vinberg’s Lemma) In the above notations, we have: max{dim g.u; u ∈ V } dim g.v + max{dim gv .ξ; ξ ∈ V }. Proof. 1) Let (x1 , . . . , xn ) be a basis of g such that (x1 , . . . , xp ) is a basis of gv . For t ∈ k, the dimension of Wt = g.(v + tu) + gv .u is the rank of the family of vectors (x1 .(v + tu), . . . , xn .(v + tu), x1 .u, . . . , xp .u). So there exists an open subset O of k containing 0 such that for all t ∈ O, we have: dim W0 dim Wt . 2) Let O• = O \ {0}. For t ∈ k \ {0}, we have: gv .u = tgv .u = gv (v + tu) ⊂ g.(v + tu). Thus for t ∈ O• , we have:
dim W0 dim (g.(v + tu) .
3) Let π : V → V be the canonical surjection. Since π(gv .u) = gv .π(u), we have, for ξ = π(u), that: dim(gv .ξ) = dim(gv .u) − dim (g.v) ∩ (gv .u) . Hence dim W0 = dim g.v + dim(gv .ξ). 4) It follows from point 2 that if t ∈ O• , then: dim g.v + dim(gv .ξ) dim (g.(v + tu) . So the result follows.
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40.3.2 Corollary. Let g be a Lie algebra and f ∈ g∗ . Then: χ(g) χ(g(f ) ). Proof. Let R : g∗ → (g(f ) )∗ be the restriction map. The kernel of R is g.f . So R induces an isomorphism : g∗ /g.f −→ (g(f ) )∗ . R Let π : g∗ → g∗ /g.f denote the canonical surjection. Then for any λ ∈ g∗ , we π(λ) . have R(λ) = R For x ∈ g(f ) , we have x.π(λ) = π(x.λ), and for y ∈ g(f ) , we have: R(x.λ)(y) = (x.λ)(y) = λ([y, x]) = R(λ)([y, x]) = x.R(λ) (y). Thus
x.π(λ) = R π(x.λ) = R(x.λ) = x.R(λ) = x.R π(λ) . R
is an isomorphism of g(f ) -modules, and the action of g(f ) This shows that R ∗ on g /g.f can be identified with the coadjoint representation of g(f ) . So the result follows from 40.3.1. 40.3.3 Remark. If f is regular in g∗ , then 40.3.2 implies that: dim g(f ) = χ(g) χ(g(f ) ) dim g(f ) . Thus χ(g(f ) ) = dim g(f ) , which proves that g(f ) is a commutative Lie algebra. So we recover 19.7.5 (ii). 40.3.4 Theorem. Let g be a Lie algebra and a an ideal of g. Then: χ(g) + χ(a) dim(g/a) + 2χ(g, a). Proof. Let f ∈ g∗ , and h ∈ a∗ the restriction of f to a. Then it is clear that a(h) + g(f ) ⊂ a(f ) . It follows therefore that dim(a(h) + g(f ) ) dim a(f ) . On the other hand, we have also: dim a(f ) = dim g − dim a + dim a ∩ g(f ) = dim(g/a) + dim a ∩ g(f ) . Now, a(h) ∩ g(f ) = a ∩ g(f ) is the orthogonal in a of the subspace g.h of a∗ . Consequently, dim(a(h) ∩ g(f ) ) = dim a − dim g.h. Hence: dim a(h) + dim g(f ) dim(g/a) + 2 dim a − 2 dim g.h. From this, we deduce that: χ(g) + χ(a) dim(g/a) + 2(dim a − dim g.h). This inequality holds for all h ∈ a∗ , so the result follows.
618
40 Index and linear forms
40.4 Index and semi-direct products 40.4.1 Let q be a Lie algebra, a an ideal of q, f ∈ q∗ , and f0 the restriction of f to a. Since a is an ideal, the Lie subalgebra h = a(f ) depends only on f0 . Let us denote by h the restriction of f to h. As q(f ) ⊂ a(f ) , we have: h(h) = h ∩ h(f ) = a(f ) ∩ (a + q(f ) ) = (a ∩ a(f ) ) + q(f ) . Consequently: h(h) = a(f0 ) + q(f ) .
(1)
Suppose that a is commutative. Then a = a(f0 ) and a ⊂ h(h) . Moreover, a ∩ q(f ) is the orthogonal of q.f0 in a. It follows therefore from (1) that: dim q(f ) = dim h(h) − dim(q.f0 ).
(2)
40.4.2 Let V be a finite-dimensional g-module. Endow V × g with the Lie algebra structure given by: for x1 , x2 ∈ g and v1 , v2 ∈ V , [(v1 , x1 ), (v2 , x2 )] = (x1 .v2 − x2 .v1 , [x1 , x2 ]). Let q = V × g and we may identify q∗ with V ∗ × g∗ . Let f ∈ q∗ , f0 = f |V and f1 = f |g . Then in the notations of 40.4.1, h = gf0 × V and q.f0 = g.f0 . Let = ( 0 , 1 ) ∈ V ∗ × g∗ = q∗ be such that 0 = f0 . Set h = |h and g = |gf0 = 1 |gf0 . For (v, x), (w, y) ∈ h, we have:
[(v, x), (w, y)] = 1 ([x, y]) = g([x, y]) It follows that: h(h) = V × (gf0 )(g) . In view of (2), we obtain that: dim q() = dim(gf0 )(g) + dim V − dim(g.f0 ). From this, we deduce immediately the following result: Proposition. With the hypotheses and notations above, let f = (f0 , f1 ) be a regular element of q∗ . (i) We have: χ(q) = dim q(f ) = χ(gf0 ) + dim V − dim(g.f0 ). (ii) Let us assume further that f0 is g-regular in V ∗ . Then: χ(q) = χ(gf0 ) + χ(g, V ).
40.4 Index and semi-direct products
619
40.4.3 Let G be a connected algebraic group, g its Lie algebra and V a finite-dimensional rational G-module. Then V has a natural structure of g-module, and we construct the Lie algebra q as in 40.4.2. Proposition. Assume that χ(g, V ) = 0. Then for any g-regular element f0 in V ∗ , we have: χ(q) = χ(gf0 ). Proof. The assumption χ(g, V ) = 0 implies that the G-orbit of any g-regular element is an open subset of V ∗ . Thus g-regular elements are G-conjugate. So the result follows by 40.4.2. 40.4.4 Let us now illustrate how the above considerations give estimations on the index of certain Lie algebras. Let us suppose that there exist a Lie subalgebra a of g and an ideal m of g such that g = a ⊕ m. For λ ∈ k, we define a Lie algebra gλ whose underlying vector space is g, and whose Lie bracket [., .]λ is given as follows: [x, y]λ = [x, y] if (x, y) ∈ (a × a) ∪ (a × m) ∪ (m × a) , [x, y]λ = λ[x, y] if (x, y) ∈ m × m. Thus g1 = g, while g0 verifies the hypotheses of 40.4.2. For λ = 0, the linear map fλ : g → gλ defined by fλ (x) = x if x ∈ a and fλ (x) = λ−1 x if x ∈ m, defines an isomorphism of Lie algebras. Thus χ(gλ ) = χ(g) for all λ = 0. Let X = (e1 , . . . , en ) be a basis of g such that (e1 , . . . , ep ) is a basis of a and (ep+1 , . . . , en ) is a basis of m. In the notations of 40.2.5, we denote by Mλ the matrix M (X , X ) relative to the Lie algebra gλ . We have: A B A B , M = M1 = λ −t B C −t B λC where A ∈ Mp S(g) and C ∈ Mn−p S(g) . Since | rk(M0 ) − rk(M1 )| rk(M0 − M1 ) = rk C, we obtain that: |χ(g0 ) − χ(g)| dim m − χ(m). As there is an open subset O of k containing 0 verifying rk(M0 ) rk(Mλ ) for all λ ∈ O, we deduce that rk(M0 ) rk(M1 ). Hence χ(g) χ(g0 ), and we have proved: Proposition. In the above notations, we have: χ(g0 ) + χ(m) − dim m χ(g) χ(g0 ).
620
40 Index and linear forms
40.4.5 We shall now give another application of 40.4.2. Suppose that g is of the form: g = g0 ⊕ g1 ⊕ g2 , where [gi , gj ] ⊂ gi+j where we set gk = {0} if k 3. The subspaces g0 and h = g0 ⊕ g2 are Lie subalgebras of g, while g2 is an abelian ideal of g. For i = 0, 1, 2, we can identify (gi )∗ with the subspace g∗−i of g∗ consisting of elements f ∈ g∗ which are identically zero on gj for all j = i. If x ∈ gi and f ∈ g∗−j , then x.f ∈ g∗i−j . Proposition. Suppose that there exists a g0 -regular element f2 in g∗−2 verifying g(f2 ) ∩ g1 = {0}. (i) There exists a g0 -regular element g ∈ g∗−2 such that: χ(g) = χ(h) = χ (g0 )g + χ(g0 , g2 ). (ii) Let us suppose that g0 is the Lie algebra of an algebraic group G0 , g2 is a rational G0 -module and that the g0 -action on g2 is the one given by the differential of the G0 -action on g2 . If χ(g0 , g2 ) = 0, then χ(g) = χ(h) = χ (g0 )g for any g0 -regular element g ∈ g∗−2 . Proof. (i) We need only to prove χ(g) = χ(h) since the second equality follows from 40.4.2. Let g = g0 + g1 + g2 ∈ g∗reg where gi ∈ g∗−i for i = 0, 1, 2. Then in view of 19.7.4 (i) and our hypothesis, we may assume that g2 is g0 -regular and g(g2 ) ∩ g1 = {0}. We have therefore: g1 .g∗0 = {0} , g1 .g∗−1 ⊂ g∗0 , g1 .g2 = g∗−1 . Thus there exists x ∈ g1 such that h = ead x .g verifies h = h0 + g2 where h0 ∈ g∗0 . So h ∈ g∗reg , and h(h|h ) ⊂ g(h) . Hence χ(h) χ(g). Conversely, let S be the set of g0 -regular elements f2 ∈ g∗−2 verifying (f2 ) g ∩ g1 = {0}. Our hypothesis implies that S is a non-empty open subset of g∗−2 . So there exists ∈ h∗reg such that = 0 + 2 with 0 ∈ g∗0 and
2 ∈ S. Denote by ∈ g∗ the linear form such that |h = and |g1 = 0. Since g1 . 2 = g∗−1 , we deduce that g( ) ⊂ h, and hence g( ) ⊂ h() . Therefore χ(g) χ(h). (ii) In view of (i), the proof is the same as the one for 40.4.3.
40.5 Heisenberg algebras in semisimple Lie algebras
621
40.5 Heisenberg algebras in semisimple Lie algebras 40.5.1 Let g be a semisimple Lie algebra, G its adjoint group, L its Killing form, h a Cartan subalgebra of g and R the root system of g relative to h. Let us fix a base Π of R, and denote by R+ (resp. R− ) the corresponding set of positive (resp. negative) roots. Set: −α α g , n− = g , b = h ⊕ n , b− = h ⊕ n− . n= α∈R+
α∈R+
Let α, β ∈ R. Denote by Hα the unique element of [gα , g−α ] such that α(Hα ) = 2. Recall that β(Hα ) ∈ Z. For λ ∈ h∗ , we shall write λ, α∨ for λ(Hα ). So in the notations of 18.3.3 and 20.6.7, we have β, α∨ = aβα . If P is a subset of R, we set: kHα . hP = α∈P
For any subset S of Π, denote by ZS (resp. NS) the subgroup (resp. subsemigroup) of h∗ generated by S. We set: S = R+ ∩ RS = R ∩ NS. RS = R ∩ ZS , R+
The set RS is clearly closed and symmetric, so by 18.10.6, it is a root system in the vector subspace that it spans in h∗ . Moreover, S is a base of RS and S is the corresponding set of positive roots. When S is a connected subset R+ of Π, the root system RS is irreducible, and we shall denote the highest root of RS by εS . S \ {εS }, we have Let us suppose that S is connected. For any root α ∈ R+ ∨ S ∨ α, εS ∈ {0, 1} (18.9.2). If T = {α ∈ R ; α, εS = 0}, then T is a root system in the subspace that it spans in h∗ , and S ∩ T is a base of T . S . Then α + εS ∈ R. It follows by 18.5.4 that α − εS ∈ R. Let α ∈ T ∩ R+ Hence α and εS are strongly orthogonal (see 18.5.3). 40.5.2 Let us suppose that g is simple, or equivalently, R is irreducible. Denote by P the set of roots that are not orthogonal to εΠ , and P+ = P ∩R+ . Set: α g , m= gα , m− = g−α , g = hR\P ⊕ α∈R\P α∈R+ \P+ α∈R+ \P+ α p= g , p0 = gα . α∈P+
α∈P+ \{εΠ }
Proposition. (i) The Lie algebra g is semisimple and we have the direct sum decomposition g = hR\P ⊕ m ⊕ m− . (ii) We have: n = m ⊕ p , [m, p0 ] ⊂ p0 , [n, gεΠ ] = {0}. In particular, p is an ideal of n.
622
40 Index and linear forms
(iii) If α ∈ P+ \ {εΠ }, then there exists a unique α ∈ P+ \ {εΠ } such that α + α = εΠ . For any α ∈ P+ \ {εΠ }, we may choose Xα ∈ gα \ {0} so that for α, β ∈ P+ \ {εΠ }: ±XεΠ if β = α , [Xα , Xβ ] = 0 if β = α . Proof. (i) This is obvious. (ii) The two equalities are clear. Let α ∈ P+ \ {εΠ } and β ∈ R+ \ P+ . ∨ ∨ Then α, ε∨ Π = 1 and β, εΠ = 0. So α + β, εΠ = 1, and hence [m, p0 ] ⊂ p0 . ∨ (iii) Let α ∈ P+ \ {εΠ }. Since α, εΠ = 1, the εΠ -string through α is {α − εΠ , α}. Set α = εΠ − α. Then α ∈ R+ (18.9.2) and: ∨ ∨ α , ε∨ Π = εΠ , εΠ − α, εΠ = 2 − 1 = 1.
So α ∈ P+ \ {εΠ }. It is obvious that α is unique. Finally, let α, β ∈ P+ \ {εΠ } be such that α + β = εΠ . Then α + β ∈ R because α + β, ε∨ Π = 2. So the last part follows. 40.5.3 Remark. A Heisenberg algebra is a Lie algebra admitting a basis (x1 , y1 , . . . , xn , yn , z), where n ∈ N, such that: for 1 i = j n, [xi , yi ] = z , [xi , yj ] = [xi , z] = [yj , z] = 0. So 40.5.2 says that p is a Heisenberg algebra. 40.5.4 Let S be a subset of Π. We define a set K(S) by induction on the the cardinality of S as follows: a) K(∅) = ∅. b) If S1 , . . . , Sr are the connected components of S, then: K(S) = K(S1 ) ∪ · · · ∪ K(Sr ). c) If S is connected, then: K(S) = {S} ∪ K({α ∈ S; α, ε∨ S = 0}). An element of K(S) is therefore a subset of S. The following results are immediate consequences of the definition. Lemma. (i) Each M ∈ K(S) is a connected subset of Π. (ii) If M, M ∈ K(S), then either M ⊂ M , or M ⊂ M , or M and M are disjoint subsets of S verifying α + β ∈ R for all α ∈ RM and β ∈ RM . (iii) If M, M ∈ K(S) and M = M , then εM and εM are strongly orthogonal. 40.5.5 Using the classification of irreducible root systems in 18.14, we give, in the table below, the cardinality k of K(Π) for the different types of simple Lie algebras. Here, for r ∈ Q, [r] denotes the largest integer r.
40.5 Heisenberg algebras in semisimple Lie algebras
A , 1 B , 2 C , 3 D , 4 E6
+1 k
2 4 2 2
623
E7
E8
F4
G2
7
8
4
2
40.5.6 Using the numbering of simple roots in 18.14, we describe below the elements of the set {εM ; M ∈ K(Π)} for the different types of simple Lie algebras. the set of odd positive For a strictly positive integer n, we denote by Nodd n integers less than or equal to n. • Type A , 1: #
+1 βi + · · · + βi+(−2i+1) ; 1 i . 2 • Type B , 2: odd {βi + 2βi+1 + · · · + 2β ; i ∈ Nodd −1 } ∪ {βi ; i ∈ N }.
• Type C , 3: {2βi + · · · + 2β−1 + β ; 1 i − 1} ∪ {β }. • Type D , 4 and even: odd {βi + 2βi+1 + · · · + 2β−2 + β−1 + β ; i ∈ Nodd −3 } ∪ {βi ; i ∈ N } ∪ {β }.
• Type D , 5 and odd: {βi + 2βi+1 + · · · + 2β−2 + β−1 + β ; i ∈ Nodd −3 } ∪ {βi ; i ∈ Nodd } ∪ {β + β + β }. −2 −1 −2 • Type E6 : (in the notations of chapter 18) 1 2 3 2 1 1 1 1 1 1 0 1 1 1 0 0 0 1 0 0 , , , . 2 0 0 0
• Type E7 : (in the notations of chapter 18) 2 3 4 3 2 1 , 0 1 21 2 2 1 , 0 1 21 1 0 0 , 0 0 00 0 0 1 2 0 0 0 0 0 0 , 0 1 00 0 0 0 , 0 0 00 1 0 0 . 1
,
• Type E8 : (in the notations of chapter 18) 2 4 6 5 4 3 2 3 0 0 0 0 0 1 0 0
, ,
2 3 4 3 2 1 0 2 0 0 0 0 0 0 0 1
, ,
0 1 2 2 2 1 0 1 0 1 0 0 0 0 0 0
, ,
0 1 2 1 0 0 0 , 1 0 0 0 1 0 0 0 . 0
• Type F4 : (in the notations of chapter 18) 2 3 4 2
,
0 1 2 2
,
0 1 2 0
,
0 1 0 0.
624
40 Index and linear forms
• Type G2 : 3β1 + 2β2 , β1 . 40.5.7 In the rest of this section, we shall denote K(Π) by K. For M ∈ K, set: M M \ {εM }. Γ M = {α ∈ RM ; α, ε∨ M > 0} , Γ0 = Γ
Lemma. Let M, M be distinct elements of K, α, β ∈ Γ M and γ ∈ Γ M . M M \ {δ ∈ R+ ; δ, ε∨ (i) We have Γ M = R+ M = 0}. M (ii) R+ is the disjoint union of Γ , for M ∈ K. (iii) If α + β ∈ R, then α + β = εM . (iv) If α + γ ∈ R, then either we have M ⊂ M and α + γ ∈ Γ0M or M M ⊂ M and α + γ ∈ Γ0 . Proof. Part (i) is clear by 18.9.2. Part (ii) is obvious and (iii) follows from 40.5.2. Let us prove (iv). By 40.5.4, we have either M ⊂ M or M ⊂ M . Let us suppose, for ∨ example, that M ⊂ M . Then γ, ε∨ M = 0. Hence α + γ, εM > 0, so α + γ ∈ M M Γ . Finally, if α + γ = εM , then γ = εM − α. So α ∈ Γ0 , and ∨ ∨ 1 = α, ε∨ M = α + γ, εM = εM , εM = 2,
which is absurd. So α + γ ∈ Γ0M . 40.5.8 For M ∈ K, set: aM =
α∈Γ M
gα , nM =
gα .
M α∈R+
By 40.5.2, aM is a Heisenberg algebra. The following result is an immediate consequence of 40.5.4: Lemma. Let M ∈ K. M is the disjoint union of Γ M , for M ∈ K verifying M ⊂ (i) The set R+ M.
(ii) The subspace nM is the direct sum of aM , for M ∈ K verifying M ⊂ M.
40.5.9 In the rest of this section, we shall fix a vector Xα ∈ gα \ {0} for each α ∈ R. If m is an ideal of b contained in n, we set: L(m) = {M ∈ K; m ∩ nM = {0}}. Lemma. (i) Let M ∈ K and M ∈ L(m) be such that M ⊂ M . Then M ⊂ L(m). (ii) We have: L(m) = {M ∈ K; XεM ∈ m}.
(iii) Suppose that m is the nilpotent radical of a parabolic subalgebra of g containing b. Let M ∈ L(m) and α ∈ Γ0M . Then we have either Xα ∈ m or XεM −α ∈ m.
40.6 Index of Lie subalgebras of Borel subalgebras
625
Proof. (i) This is clear because if M ⊂ M , then nM ⊂ nM . (ii) Since XεM ∈ nM , we have {M ∈ K; XεM ∈ m} ⊂ L(m). Conversely, let M ∈ L(m). The subspace m ∩ nM is an (ad h)-stable Lie subalgebra of g. It M such that gα ⊂ m. But, in view of follows by 20.7.2 that there exists α ∈ R+ 18.7.7 and 18.9.2 (i), XεM belongs to the nM -module generated by Xα . Since m is an ideal of b, we obtain that XεM ∈ m. So we have proved (ii). (iii) Let M ∈ L(m) and α ∈ Γ0M . We have εM − α ∈ Γ0M (40.5.2). Now, if Xα ∈ m, then X−α ∈ p. Since XεM ∈ m by (ii), we deduce that [X−α , XεM ] ∈ m. So we are done. 40.5.10 Let m be an ideal of b contained in n. We set: kHεM , d(m) = aM , c(m) = l(m) ⊕ d(m). l(m) = M ∈L(m)
M ∈L(m)
Note that the εM ’s, being pairwise strongly orthogonal, form a linearly independent family. In particular, we have dim l(m) = card L(m). Lemma. The subspaces d(m) and c(m) are ideals of b containing m. Proof. Let α ∈ R+ , M ∈ L(m) and β ∈ Γ M . We have [HεM , Xα ] = α(HεM )Xα = α, ε∨ M Xα . It follows from the definitions that [b, l(m)] ⊂ d(m). We have α ∈ Γ M for some M ∈ K. If [Xα , Xβ ] = 0, then 40.5.7 implies that either M ⊂ M or M ⊂ M . If M ⊂ M , then M ∈ L(m) (40.5.9), and α + β ∈ Γ M (40.5.7). So [Xα , Xβ ] ∈ d(m). If M ⊂ M , then α + β ∈ Γ M and again [Xα , Xβ ] ∈ d(m). Hence we have proved that [b, d(m)] ⊂ d(m).
40.6 Index of Lie subalgebras of Borel subalgebras 40.6.1 In this section, we shall conserve the notations and assumptions of section 40.5, and we shall denote by K the set K(Π). Let a be a Lie subalgebra of g. If Y ∈ g, then we define a linear form ϕYa on a by setting for X ∈ a: ϕYa (X) = L(X, Y ). If c is a Lie subalgebra of g such that [c, a] ⊂ a, then c acts on a∗ via the coadjoint representation, and if Z ∈ c, then: [Z,Y ]
Z.ϕYa = ϕa Thus:
.
Y
gY ∩ a ⊂ a(ϕa ) . 40.6.2 For α ∈ R, we fix a non-zero element Xα of gα , and we set: U= X−εK . K∈K
626
40 Index and linear forms
Lemma. Let X ∈ b. The following conditions are equivalent: (i) X ∈ h and εK (X) = 0 for all K ∈ K. (ii) [X, U ] ∈ n. Proof. (i) ⇒ (ii) If (i) is verified, then [X, U ] = 0. (ii) ⇒ (i) There exist H ∈ h, aα ∈ k for α ∈ R+ , such that X=H+ aα Xα . α∈R+
So [X, U ] ∈
K∈K
aεK λK HεK + n + n−
where λK ∈ k \ {0} for K ∈ K. As the εK ’s are pairwise strongly orthogonal, the HεK ’s are linearly independent. It follows that aεK = 0 for all K ∈ K. Let K ∈ K, α ∈ Γ0K and β = εK − α. Then [Xα , X−εK ] [X, U ] = λaα X−β + aα K ∈K\{K} aγ [Xγ , X−εK ] − εK (H)X−εK , + K ∈K
K ∈K,γ∈R+ \{α}
where λ ∈ k \ {0}. Observe that since β ∈ Γ0K , β, ε∨ K = 1, so β = εK for any K ∈ K. So if [X, U ] ∈ n, then either aα = 0 or γ − εK = −β for some K ∈ K and γ ∈ R+ \ {α}. Suppose that aα = 0, then we would have β + γ = εK for some K ∈ K and γ ∈ R+ \ {α}. By 40.5.7, we deduce that K = K and γ = α which is absurd. We have therefore proved that aα = 0 for all α ∈ R+ . Hence X ∈ h, and since εK (X)X−εK , [X, U ] = − K∈K
we deduce that εK (X) = 0 for all K ∈ K.
40.6.3 Proposition. Let U be as in 40.6.2, m an ideal of b contained in n and f = ϕU m. (i) We have b.f = m∗ , so f is a b-regular element of m∗ . In other words, χ(b, m) = 0. (ii) Let B be the Borel subgroup of G whose Lie algebra is adg b. Then the B-orbit of f is open and dense in m∗ . (iii) Any b-regular element of m∗ is a regular element of m∗ . Proof. (i) It is clear that we may assume m = n. Then it suffices to show that N = {X ∈ n; f ([b, X]) = {0}} = {0}. But N is the set of elements X ∈ n verifying L([U, X], b) = {0}. Since the orthogonal of b with respect to L is n,
40.6 Index of Lie subalgebras of Borel subalgebras
627
it is the set of elements X ∈ n such that [X, U ] ∈ n. So the result follows from 40.6.2. (ii) This is clear by (i), 16.5.7 and 21.4.3. (iii) Since m(α.f ) = α(m(f ) ) for α ∈ B, this is an immediate consequence of (i), (ii) and 19.7.5. 40.6.4 Proposition. Let m be an ideal of b contained in n, and L(m), d(m), l(m) and c(m) be as in 40.5.9 and 40.5.10. (i) We have χ d(m) = card L(m) . In particular, χ(n) = card(K). (ii) Suppose that m is the nilpotent radical of a parabolic subalgebra p of g containing b. Then: χ(m) = dim c(m) − dim m. Proof. Let K1 , . . . , Kr be the elements of L(m). For 1 i r, let pi be the integer such that 2pi + 1 = dim aKi . Let αi,1 , . . . , αi,pi ∈ Γ0Ki be such that {αi,1 , εKi − αi,1 , . . . , αi,pi , εKi − αi,pi } = Γ0Ki . Choose Xi,j ∈ gαi,j and Yi,j ∈ gεKi −αi,j such that [Xi,j , Yi,j ] = XεKi (This is possible by 40.5.2). Let Bi = (Xi,1 , Yi,1 , . . . , Xi,pi , Yi,pi ). Then Bi ∪ {XεKi } is a basis of aKi . (i) Let f = ϕU d(m) , and denote by Φf the alternating bilinear form (X, Y ) → f ([X, Y ]) on d(m) × d(m). By 40.5.10 and 40.6.3, f is a regular element of d(m)∗ . Let 1 i < j r. If Ki ∩ Kj = ∅, then [aKi , aKj ] = {0} (40.5.4). Otherwise, we have either Ki ⊂ Kj or Kj ⊂ Ki (40.5.4). Let us suppose for example that Kj ⊂ Ki . Let α ∈ Γ Ki and β ∈ Γ Kj be such that α + β ∈ R. Then by 40.5.7, α + β ∈ Γ0Ki . So f ([Xα , Xβ ]) = 0. Now let B = (XεK1 , . . . , XεKr ) and B = B1 ∪ · · · ∪ Br . Then B = B ∪ B is a basis of d(m). We deduce from the arguments above that the matrix of Φf with respect to the basis B is of the form: ⎞ ⎛ 0r,r 0r,2 · · · · · · 0r,2 ⎜ 02,r J 02,2 · · · 02,2 ⎟ ⎜ .. . ⎟ ⎜ . 02,2 J . . . .. ⎟ ⎜ . ⎟ .. . . . . ⎝ .. . . 0 ⎠ . 2,2
02,r 02,2 · · · 02,2 J
where 0r,s denotes the zero r by s matrix, and J = Hence
0 1 . −1 0
rk(Φf ) = dim d(m) − card L(m) .
We have therefore proved the first part of (i). If m = n, then d(m) = n and L(m) = K. So the second part follows. (ii) Let g = ϕU m and Φg be the bilinear form (X, Y ) → g([X, Y ]) on m. For 1 i r, mKi = m ∩ aKi is h-stable because m is an ideal of b. So there exists a subset Bi of Bi such that Bi ∪ {X−εKi } is a basis of mKi .
628
40 Index and linear forms
Let E1 (resp. E2 ) be the set of indices k such that {Xi,k , Yi,k } ∈ Bi (resp. {Xk , Yk } ∈ Bi ). By 40.5.9 (iii), we have card(Bi ∩ {Xi,k , Yi,k }) = 1 if k ∈ E1 . Since XεKi ∈ m (40.5.9 (ii)), it follows that: dim aKi = 1 + 2 card E1 + 2 card E2 , dim mKi = 1 + card E1 + 2 card E2 . On the other hand, by considering the matrix of Φg |mKi ×mKi with respect to Bi ∪ {X−εKi }, we verify easily that: ri = rk(Φg |mKi ×mKi ) = 2 card E2 . Consequently, we deduce that: 1 + dim aKi − dim mKi = dim mKi − ri . In view of the proof of part (i), we obtain that: card L(m) + dim d(m) − dim m = dim m − rk(Φg ). So we are done because g is a regular element of m∗ (40.6.3).
40.6.5 Let m be an ideal of b contained in n, and c(m) be as in 40.5.10. By 40.5.10, c(m) is an ideal of b. Set t(m) = h + c(m) = h ⊕ d(m). Proposition. We have χ c(m) = 0 and χ t(m) = rk(g) − card L(m) . Moreover, c(m)∗ and t(m)∗ contain a stable element. Proof. Let B be the basis of d(m) as in the proof of 40.6.4 (i), and C = (HεK1 , . . . , HεKr ) where K1 , . . . , Kr are the elements of L(m). Then D = C ∪B is a basis of c(m). Let f = ϕU c(m) . Then the matrix of Φf with respect to D is of the form: ⎞ ⎛ 0r,r A 0r,2 · · · · · · 0r,2 ⎜ −t A 0r,r 0r,2 · · · · · · 0r,2 ⎟ ⎟ ⎜ ⎜ 02,r 02,r J 02,2 · · · 02,2 ⎟ ⎟ ⎜ . . . .. M =⎜ . .. 0 . .. ⎟ 2,2 J ⎟ ⎜ .. .. .. . . . . ⎝ .. . . 0 ⎠ . . 2,2
02,r 02,r 02,2 · · · 02,2 J
where 0r,s and J are as in the proof of 40.6.4, and A ∈ Mr (k) is a diagonal matrix of rank r. Hence rk(Φf ) = dim c(m), and so χ c(m) = 0. Finally, let g = ϕU t(m) . Let z = {H ∈ h; εK (H) = 0} for all K ∈ L(m). Then dim z = rk(g) − card L(m) = s. Let E be a basis of z. Then E ∪ D is a basis of t(m), and the matrix of Φg with respect to this basis is of the form: 0s,s 0s,m 0m,s M
40.7 Seaweed Lie algebras
629
where m = dim c(m). It follows that t(m)(g) = z. Since z is a commutative Lie algebra consisting of semisimple elements, g is stable by 40.1.6 and 40.1.5. So we are done since a stable linear form is regular (40.1.3). 40.6.6 Proposition. Let f = ϕU b be as in 40.6.2. (i) f is a stable element of b∗ . In particular, it is regular in b∗ . (ii) We have χ(b) = rk(g) − card(K). In particular, χ(b) + χ(n) = rk(g). Proof. By applying 40.6.5 with m = n, we have (i) and the first part of (ii). The second part of (ii) follows from 40.6.4.
40.7 Seaweed Lie algebras 40.7.1 In the rest of this chapter, g will be a semisimple Lie algebra, and G its adjoint group. If P is a subset of R, set: α g . gP = α∈P
Definition. (i) Two parabolic subalgebras p and p are said to be weakly opposite if p + p = g. (ii) A seaweed Lie algebra of g is a Lie subalgebra q of g of the form q = p ∩ p where p and p are weakly opposite parabolic subalgebras of g. S T 40.7.2 Let S, T be subsets of Π. We check easily that R+ ∪ R− is a closed subset of R. Set: S T p = gR+ ⊕ b− , p = gR− ⊕ b.
Then p and p are weakly opposite parabolic subalgebras of g, and gS,T = p ∩ p = h ⊕ gR+ ⊕ gR− S
T
is a seaweed Lie algebra of g. Conversely, let p and p be parabolic subalgebras verifying b ⊂ p and b− ⊂ p . Then p + p = g, and there exist subsets S and T of Π such that S T p ∩ p = h ⊕ gR+ ⊕ gR− = gS,T . We shall say that q is a standard seaweed Lie algebra of g (relative to h S T and Π) if there exist S, T ⊂ Π such that q = h ⊕ gR+ ⊕ gR− = gS,T . 40.7.3 Proposition. Let q be a Lie subalgebra of g. The following conditions are equivalent: (i) q is a seaweed Lie algebra of g. (ii) q is G-conjugate to a standard seaweed Lie algebra of g.
630
40 Index and linear forms
Proof. It is clear that (ii) implies (i). Let us suppose that (i) is verified. Let p and p be weakly opposite parabolic subalgebras of g such that q = p ∩ p . By 29.4.9, q contains a Cartan subalgebra t of g. Denote by R the root system of g relative to t. Then there exist parabolic subsets P, Q of R such that p = t ⊕ gP and p = t ⊕ gQ . Since q is a seaweed Lie algebra, we have: P ∪ (−P ) = Q ∪ (−Q) = P ∪ Q = R . So we have also (−P ) ∪ (−Q) = R . Let T = P ∩ (−Q). It is a closed subset of R . We shall show that T is in fact parabolic, that is T ∪ (−T ) = R . Let α ∈ R \ T . We have three cases: a) If α ∈ P and α ∈ −Q, then α ∈ −P and α ∈ Q. b) If α ∈ P and α ∈ −Q, then α ∈ −P and α ∈ Q since (−P )∪(−Q) = R . c) If α ∈ P and α ∈ −Q, then α ∈ −P and α ∈ Q because P ∪ Q = R . So in all the cases, α ∈ −T . Thus T is a parabolic subset of R , and the implication (i) ⇒ (ii) follows immediately.
40.8 An upper bound for the index 40.8.1 The notations we introduce in this section will be used in the rest of this chapter. Let us conserve the assumptions of 40.5. Denote by the rank of g. We shall fix for each α ∈ R, a non-zero element Xα of gα , and we shall also fix a total order > on R+ . 40.8.2 Let (H1 , . . . , H ) be a basis of h, {Hi ; 1 i } ∪ {Xα ; α ∈ R}, is then a basis of g, and we shall denote by {Hi∗ ; 1 i } ∪ {Xα∗ ; α ∈ R} the corresponding dual basis. 40.8.3 Let S, T be subsets of Π, S = K(S), T = K(T ) and ES (resp. ET , ES,T ) the subspace of h∗ spanned by εM , M ∈ S (resp. T , S ∪ T ). For M ∈ S (resp. N ∈ T ), we denote by aM (resp. bN ) a non-zero element of k. Set: ∗ aM Xε∗M + bN X−ε ∈ q∗ , f= N M ∈S
N ∈T
where q = gS,T is the standard seaweed Lie algebra of g associated to S and T as defined in 40.7. X We have also f = ϕq S,T , where XS,T = aM X−εM + bN XεN , M ∈S
where aM =
N ∈T
aM bN , bN = . L(XεM , X−εM ) L(XεN , X−εN )
Let mS (resp. mT ) be the cardinality of S (resp. T ) and m = mS + mT . We define
40.8 An upper bound for the index
631
Ωf = (aM1 , . . . , aMmS , bN1 , . . . , bNmT ) ∈ (k \ {0})m , where εM1 < · · · < εMmS and εN1 < · · · < εNmT . We shall say that Ωf defines f . 40.8.4 For M ∈ S and N ∈ T , we define the following sets: • H1 (M ) is the set of pairs (α, εM − α) with α ∈ Γ0M and α < εM − α. • H2 (N ) is the set of pairs (−β, −εN + β) with β ∈ Γ0N and β < εN − β. • I1 (M, N ) is the set of pairs (−β, εM ) with β ∈ Γ0N , and εM − β = −εN for some N ∈ T . • I2 (M, N ) is the set of pairs (α, −εN ) with α ∈ Γ0M , and α − εN = εM for some M ∈ S. • J1 (M, N ) is the set of pairs (α, −β) with α ∈ Γ0M and β ∈ Γ0N , such that α − β = εM for some M ∈ S. • J2 (M, N ) is the set of pairs (α, −β) with α ∈ Γ0M and β ∈ Γ0N , such that α − β = −εN for some N ∈ T . Let us write: H1 (M ) , H2 = H2 (N ) , H = H1 ∪ H2 , M ∈S N ∈T I1 (M, N ) , I2 = I2 (M, N ) , I1 = (M,N )∈S×T (M,N )∈S×T J = J1 (M, N ) ∪ J2 (M, N ) , H1 =
(M,N )∈S×T
(M,N )∈S×T
Z = H ∪ I1 ∪ I2 ∪ J , r = card H , s = dim ES,T . If z = (α, β) ∈ Z, then we shall denote by z = {α, β} the underlying set of z. 40.8.5 Lemma. Let M ∈ S and N ∈ T . Then: (i) If (−β, εM ) ∈ I1 (M, N ) verifies εM − β = −εN , then N N εM ∈ Γ0N . (ii) If (α, −εN ) ∈ I2 (M, N ) verifies α − εN = εM , then M M εN ∈ Γ0M . (iii) If (α, −β) ∈ J1 (M, N ) verifies α − β = εM , then M M β ∈ Γ0M . (iv) If (α, −β) ∈ J2 (M, N ) verifies α − β = −εN , then N N α ∈ Γ0N .
and and and and
Proof. This is clear by lemmas 40.5.4 and 40.5.7. 40.8.6 For M ∈ S and N ∈ T , we set: α −α a•M = g , b•N = g , aM = kXεM + a•M , bN = kX−εN + b•N , M α∈Γ0 α∈Γ0N • • kXεM + kX−εN , v = aM + bN . u= M ∈S
N ∈T
M ∈S
N ∈T
632
40 Index and linear forms
Recall from 40.5.8 that aM and bN are Heisenberg algebras whose centres are kXεM and kX−εN respectively. The dimension of a•M (resp. b•N ) is an even integer 2nM (resp. 2nN ), and we have: nM + nN . r= M ∈S
∗
∗
N ∈T
∗
We shall consider h , u and v as vector subspaces of q∗ via the direct sum decomposition q = h ⊕ u ⊕ v. 40.8.7 Using the notations of 40.8.2, for z ∈ (α, β) ∈ Z, we set: vz = Xα∗ ∧ Xβ∗ ∈
2
q∗ .
Let M, M ∈ S, N, N ∈ T , α ∈ Γ0M and β ∈ Γ0N . Recall from lemmas 40.5.4 and 40.5.7 that: • εM − εN ∈ {εM , −εN }. • α − εN = −εN . • εM − β = εM . It follows that if we identify the alternating bilinear form Φf on q with an 2 ∗ element of q , then Φf = Ψf + Θf , ∗ where Θf ∈ ES,T u and λ z vz Ψf = z∈Z
with λz ∈ k for all z ∈ Z. If z = (α, β) ∈ Z, then Θf (Xα , Xβ ) = 0. Moreover, we have either [Xα , Xβ ] = µz XεM for some M ∈ S and µz ∈ k \ {0}, or [Xα , Xβ ] = µz X−εN for some N ∈ T and non zero scalar µz . Consequently, µz aM if [Xα , Xβ ] = µXεM , λz = Φf (Xα , Xβ ) = f ([Xα , Xβ ]) = µz bN if [Xα , Xβ ] = µX−εN . Thus λz is non-zero. 40.8.8 Lemma. In the above notations, we have: (i) q(f ) contains a commutative Lie subalgebra of g consisting of semisimple elements of dimension: dim h − dim ES,T + card(S ∩ T ). (ii) We have ∧s Θf = 0 and ∧s+1 Θf = 0. (iii) There exists a non-empty open subset U of (k\{0})m such that ∧r Ψf = 0 whenever Ωf ∈ U . (iv) If Ωf ∈ U , then ∧r+s Φf = 0.
40.8 An upper bound for the index
633
Proof. (i) Let t be the orthogonal of ES,T in h. Then: dim t = dim h − dim ES,T . Let XS,T be as in 40.8.3. If H ∈ t, then [H, XS,T ] = 0, and so H.f = 0. Next let M ∈ S ∩ T , and YM = aM XεM + bM X−εM . Set e = aM XεM , f = bM X−εM and h = [e, f ]. Then 38.5.6 says that YM is G-conjugate to a non-zero multiple of h. Thus YM is semisimple. Now by the definition of XS,T and part (iii) of lemma 40.5.4, we obtain that YM .f = 0. In view of the previous paragraphs, we have proved (i). (ii) Let (α1 , . . . , αs ) be a basis of ES,T such that αi = εMi if 1 i p, and αi = −εNi if p + 1 i + 1 s, where the Mi ’s and the Ni ’s are elements of S and T respectively. Complete this basis to a basis B = (α1 , . . . , α ) of h∗ , and let B = (h1 , . . . , h ) be the basis of h dual to B . In the same manner, we complete the system (XεM1 , . . . , XεMp , X−εNp+1 , . . . , X−εNs ) of linearly independent vectors to a basis C of u. Then D = B ∪ C is a basis of h + u. Observe that [hk , u] = {0} if s + 1 k . Since the εM ’s (resp. εN ’s) are pairwise strongly orthogonal, it follows from 40.8.7 that in the basis D, the matrix of the restriction of Φf , which is also the restriction of Θf , to h + u is of the form: ⎛ ⎞ 0s,s 0s,−s D A ⎜ 0−s,s 0−s,−s 0−s,s 0−s,m−s ⎟ ⎟ , M =⎜ ⎝ −t D 0s,−s 0s,s 0s,m−s ⎠ −t A 0m−s,−s 0m−s,s 0m−s,m−s where A ∈ Ms,m−s (k), (m is as defined in 40.8.3), and D ∈ Ms,s (k) is diagonal with entries: (aM1 , . . . , aMp , bNp+1 , . . . , bNs ). Thus the rank of M is 2s, and therefore ∧s Θf = 0 and ∧s+1 Θf = 0. (iii) Let z1 , . . . , zn be the elements of Z such that z1 , . . . , zr are the elements of H. For simplicity, let us write λi vi and µi , instead of λzi vzi and µzi . Observe that for 1 i, j n, vi ∧vj = vj ∧vi and vi ∧vi = 0. Consequently: (3) ∧r Ψf = r! λi1 · · · λir vi1 ∧ · · · ∧ vir . 1i1 <···
So the coefficient λ1 · · · λr of v1 ∧ · · · ∧ vr in the sum of (3) is: nN nM aM µz bN µz . M ∈S
z∈H1 (M )
N ∈T
z∈H2 (N )
Now suppose that vi1 ∧ · · · ∧ vir = λv1 ∧ · · · ∧ vr with λ ∈ k \ {0}, 1 i1 < · · · < ir n and (i1 , . . . , ir ) = (1, . . . , r). Then we have:
634
40 Index and linear forms
E = z 1 ∪ · · · ∪ z r = z, , i1 ∪ · · · ∪ z ir . It follows that if zit ∈ H, then zit ∈ J and there exist M ∈ S and N ∈ T N , such that Γ0M ∩ z, it = ∅ and (−Γ0 ) ∩ z it = ∅. Let zj1 , . . . , zjk be the elements zit which do not belong to H. Let M0 ∈ S be maximal by inclusion among the elements M of S verifying: Γ0M ∩ (z, j1 ∪ · · · ∪ z, jk ) = ∅. There exists t ∈ {1, . . . , k} such that if z = zjt , then Γ0M0 ∩ z = ∅. Let α ∈ Γ0M0 ∩ z . Then x ∈ J and (α, εM0 − α) = zil for 1 l r. It follows therefore from 40.8.7 and lemma 40.8.5 that either λz = µz aM for some M ∈ S, M = M0 , or λz = µz bN for some N ∈ T . In both cases, we deduce that the coefficient λi1 · · · λir in the sum of (3) is of the form: mM mN aM bN µi1 · · · µir M ∈S
N ∈T
where mM0 < nM0 . It is now clear that there exists a non-empty open subset U of (k \ {0})m verifying ∧r Ψf = 0 if Ωf ∈ U . (iv) We have: r+s r+s r+s (∧k Ψf ) ∧ (∧r+s−k Θf ). ∧ Φf = k k=0 j j Since ∧j Θf ∈ ( ES,T ) ( u∗ ), to show that ∧r+s Φf = 0, it suffices to prove that (∧r Ψf ) ∧ (∧s Θf ) = 0. We saw in the proof of (iii) that if Ωf ∈ U , then ∧r Ψf = λv1 ∧ · · · ∧ vr + w, where λ ∈ k \ {0} and w is a linear combination of terms of the form vzi1 ∧ r s , · · · ∧ vzir with z, i1 ∪ · · · ∪ z ir = E. It is therefore clear that (∧ Ψf ) ∧ (∧ Θf ) = 0 if Ωf ∈ U . Remarks. 1) Observe that the preceding proof shows that if Ωf ∈ U , then the restriction of Ωf to v × v is non-degenerate. 2) If S or T is empty, then 40.6.6 says that ∧r+s Φf = 0 for all Ωf ∈ (k \ {0})m . However, this is not true in general. 40.8.9 Theorem. In the notations above, we have: χ(gS,T ) rk(g) + card S + card T − 2 dim ES,T . Proof. Recall from 40.8.6 that gS,T = h ⊕ u ⊕ v. So: dim g = h + card S + card T + 2r.
40.9 Cases where the bound is exact
635
If Ωf ∈ U , then the fact that ∧r+s Φf = 0 implies that rk(Φf ) 2(r + s). Thus: (f ) dim gS,T dim g − 2(r + s). Hence: (f )
dim gS,T dim h + card S + card T − 2s. So we are done.
40.8.10 Remark. Since card S = dim ES , card T = dim ET and ES,T = ES + ET , the preceding inequality can be rewritten as follows: χ(gS,T ) rk(g) + dim ES + dim ET − 2 dim(ES + ET ). 40.8.11 Corollary. Let q be a seaweed Lie algebra of g. Then: (i) χ(q) rk(g). (ii) χ(q) = rk(g) if and only if q is a Levi factor of g. Proof. By proposition 40.7.3, we may assume that q is standard. Let S, T be S T subsets of Π such that q = h ⊕ gR+ ⊕ gR− . (i) This is a direct consequence of 40.8.10. (ii) If q is a Levi factor of g, then q is reductive, and it is clear by 29.3.2 that χ(q) = rk(g). Conversely, suppose that χ(q) = rk(g). Then 40.8.10 says that ES = ET . Let N be a connected component of T . We have εN ∈ ET = ES ⊂ kS. Since εN is a linear combination of all the roots of N with strictly positive integral coefficients and Π is a basis of h∗ , it follows that N ⊂ S, and hence T ⊂ S. By exchanging the roles of S and T , we obtain that S ⊂ T . Thus S = T , and q is a Levi factor of g. 40.8.12 Corollary. If the elements εM , M ∈ S ∪ T , form a basis of h∗ , then χ(gS,T ) = 0. Proof. By 40.8.10, χ(gS,T ) 0. So the result follows.
40.9 Cases where the bound is exact 40.9.1 Lemma. Let V be a finite-dimensional vector space, V a hyperplane of V , Φ an alternating bilinear form on V and Φ its restriction to V . Denote by N and N the kernel of Φ and Φ . (i) If N ⊂ V , then N is a hyperplane of N . (ii) If N ⊂ V , then N = N ∩ V and N is a hyperplane of N . Proof. (i) If N ⊂ V , then N ⊂ N , and if x ∈ V \ V , then: N = {v ∈ N ; Φ(v, x) = 0}.
636
40 Index and linear forms
It follows that dim(N /N ) 1. Being the rank of Φ and Φ , the integers dim V − dim N and dim V − dim N are even. So dim(N /N ) = 1. (ii) Suppose that N ⊂ V , and x ∈ N \ V . If v ∈ N , then Φ(x, v) = 0 and Φ(v, V ) = {0}. So v ∈ N , and hence N ⊂ N ∩ V . Clearly, N ∩ V ⊂ N . Thus N = N ∩ V , and N is a hyperplane of N . 40.9.2 In the notations of 40.8, we set: dS,T = rk(g) + card K(S) + card K(T ) − 2 dim ES,T = rk(g) + dim ES + dim ET − 2 dim(ES + ET ). By 40.8.9, we have χ(gS,T ) dS,T . We shall show that equality holds in certain cases. 40.9.3 Proposition. If dS,T ∈ {0, 1}, then χ(gS,T ) = dS,T . Proof. The case dS,T = 0 is clear by 40.8.9. So let us suppose that dS,T = 1. We have dim gS,T − dS,T = 2(r + s). Since dim gS,T − χ(gS,T ) is an even integer (it is the rank of an alternating bilinear form on gS,T ), we deduce that dS,T and χ(gS,T ) are of the same parity. So χ(gS,T ) = 1. 40.9.4 Proposition. Let S and T be subsets of Π. Suppose that the set {εM ; M ∈ K(S) ∪ K(T )} consists of linear independent elements. Then, if Ωf ∈ U , f is a stable element of g∗S,T and χ(gS,T ) = dS,T . Proof. Our hypothesis implies that dim ES,T = card(K(S) ∪ K(T )). So we deduce from the definition of dS,T that: dS,T = rk(g) − dim ES,T + card(K(S) ∩ K(T )). (f )
It follows from 40.8.8 and 40.8.9 that if Ωf ∈ U , then dim gS,T = dS,T and (f )
gS,T is a commutative Lie algebra consisting of semisimple elements. So the result follows by 40.1.6 and 40.1.5. 40.9.5 Proposition. If card(S) = 1 or card(T ) = 1, then χ(gS,T ) = dS,T . Proof. Without loss of generality, we may assume that card(T ) = 1 and T = {α}. For simplicity, we shall denote gS,T by q. In view of 40.9.4, we may further assume that α ∈ S, {α} ∈ K(S), and α ∈ ES . Set: S a = h ⊕ gR+ . Let U=
M ∈S
V aM X−εM , V = Xα + U , f = ϕU a , g = ϕq ,
40.9 Cases where the bound is exact
637
where the coefficients aM are chosen so that f is a stable element of a∗ (this is possible by 40.9.4). The restriction of Φg to a is therefore Φf . The proof of 40.8.8 (i) shows that: ker εM . a(f ) = M ∈S
As α ∈ ES , we deduce that [Xα , a(f ) ] = {0}, so a(f ) ⊂ q(g) . By lemma 40.9.1, we obtain that: q(g) = kX ⊕ a(f ) where X = X−α + Y for some Y ∈ a. Since a(f ) ⊂ h, we have [a, a(f ) ] ∩ a(f ) = {0}, and therefore [X, a(f ) ] = {0}. Thus q(g) is a commutative Lie algebra. Let A be the smallest algebraic subgroup of the adjoint group G of g whose Lie algebra is adg a. Then the set of restrictions of the elements of A to a is the adjoint group of a. Since the set of stable elements of a∗ is a non-empty A-invariant open subset of a∗ , we deduce that there exists h ∈ q∗ regular such that λ = h|a is stable. If θ ∈ A, then θ(h)|a = θ(λ). Consequently, we may assume that: a(λ) = a(f ) = ker εM ⊂ h. M ∈S
Now h([a, a(f ) ]) = {0}, and X−α commutes with a(f ) = a(λ) . It follows therefore that h([X−α , a(λ) ]) = {0}. Hence h([q, a(λ) ]) = {0} and a(λ) ⊂ q(h) . It follows by 40.9.1 that dim q(h) = 1 + dim a(λ) , hence χ(q) = 1 + χ(a). Since ES,T = ES,∅ , card K(T ) = 1 and χ(a) = dS,∅ , we deduce that χ(q) = dS,T . 40.9.6 A Lie algebra verifying the hypothesis of 40.9.5 does not possess necessarily a stable linear form. Let us give an example of such a Lie algebra. Suppose that g is of type D4 , and using the notations of 18.14, let S = {β1 , β2 , β3 , β4 }, T = {β2 }. Then q = kX−β2 + b. Let B (resp. Q) be the smallest algebraic subgroup of G whose Lie algebra is adg b (resp. adg q). Then the set of restrictions of the elements of B (resp. Q) to b (resp. q) is the adjoint group of b (resp. q). Set: U = X−β1 + X−β3 + X−β4 + X−β1 −2β2 −β3 −β4 , g = ϕU b. By 40.5.6, 40.6.2 and 40.6.6, g is a stable element of b∗ . Moreover, χ(b) = 0 (40.6.6). It follows that the B-orbit of g is the open set b∗reg of b∗ . Now suppose that the open subset E of stable elements of q∗ is non-empty. Then there exists f ∈ E such that f |b ∈ b∗reg . If θ ∈ B, then θ(f ) ∈ E and θ(f )|b = θ(f |b ). So we may assume that f is of the form f = ϕVq where V = λXβ2 + U where λ ∈ k. If λ = 0, then we can check easily that q(f ) = kX, where
638
40 Index and linear forms
X = X−β2 + µ1 Xβ1 + µ2 Xβ3 + µ3 Xβ4 +µ4 Xβ1 +β2 +β3 + µ5 Xβ1 +β2 +β4 + µ6 Xβ2 +β3 +β4 + µ7 Xβ1 +2β2 +β3 +β4 , and µ1 , . . . , µ7 are non-zero elements of k. If λ = 0, then again, we can check easily that q(f ) = kX, where X = X−β2 + µ1 Xβ1 +β2 +β3 + µ2 Xβ1 +β2 +β4 + µ3 Xβ2 +β3 +β4 , and µ1 , µ2 , µ3 are non-zero elements of k. Let H ∈ h be the element such that β1 (H) = β3 (H) = β4 (H) = −β2 (H) = 1. Then in both cases, [H, X] = X. So [q, q(f ) ]∩q(f ) = {0}. But this contradicts 40.1.5. Thus, q∗ does not possess any stable linear form.
40.10 On the index of parabolic subalgebras 40.10.1 Lemma. Let S be a subset of Π and n = card K(S). Then there exist subsets S1 , . . . , Sn of Π verifying the following conditions: (i) S1 ⊂ S2 ⊂ · · · ⊂ Sn = S. (ii) K(S1 ) ⊂ K(S2 ) ⊂ · · · ⊂ K(Sn ). (iii) card K(Si ) = i for 1 i n. Proof. Let M be a connected component of S and M the set of elements α ∈ M such that α, ε∨ M = 0. Set Sn−1 = S \ (M \ M ). Then it is clear by 40.5.4 that card K(Sn−1 ) = n − 1 and K(Sn−1 ) ⊂ K(S). So the result follows. 40.10.2 Theorem. Let g be a semisimple Lie algebra of rank . For any integer i ∈ {0, 1, . . . , }, there exists a parabolic subalgebra p of g such that χ(p) = i. Proof. It is clear that we may assume g to be simple. Let n = card K(Π). Using the numbering of root systems in 18.14, we set Π = {β1 , . . . , β }. There exist subsets S0 , S1 , . . . , Sn of Π such that S0 = ∅, Sn = Π and S1 , . . . , Sn verify the conditions (i), (ii) and (iii) of lemma 40.10.1. By 40.9.4, for 0 i n, we have: χ(gΠ,Si ) = + i − n. Recall from table 40.5.5 that if g is of type Bk , Ck , D2k , E7 , E8 , F4 or G2 , then n = . So the theorem follows in these cases. 1) Let us suppose that g is of type D2k+1 . Then card K(Π) = 2k and for 1 i , gΠ,Si−1 is a parabolic subalgebra of g of index i. As β2k ∈ EΠ (40.5.6), by 40.9.4 (iii) or 40.9.5, we have χ(gΠ,{β2k } ) = 0. 2) Let us suppose that g is of type E6 . Then card K(Π) = 4 and again by using the Sj ’s, we obtain parabolic subalgebras of g of index i for 2 i 6. By 40.5.6, {β1 , β5 }∪{εM ; M ∈ K(M )} is basis of h∗ . Again by using 40.9.4 (iii), we obtain that:
40.10 On the index of parabolic subalgebras
639
χ(gΠ,{β1 } ) = 1 , χ(gΠ,{β1 ,β5 } ) = 0. 3) Finally, let us suppose that g is of type A . Let ρ be the largest integer ( + 1)/2. By 40.5.5, ρ = card K(Π). So by using the Sj ’s, we obtain parabolic subalgebras of g of index i for − ρ i . We are therefore left to construct parabolic subalgebras of g of index 0, 1, . . . , − ρ − 1. Set T0 = ∅, and define Tk , for 1 k − ρ = , by: Tk−1 ∪ {βk } if k is odd, Tk = Tk−1 ∪ {β+1−k } if k is even. By 40.5.6, the εM ’s, M ∈ K(Π) ∪ K(Tk ), are linearly independent. So 40.9.4 (iii) implies that χ(gΠ,Tk ) = − k for 0 k . So we are done because gΠ,Tk is a parabolic subalgebra of g.
References and comments • [25], [33], [43], [45], [47], [63], [62], [68], [82], [83]. The notion of stable linear form was introduced in [47], and was studied in an algebraic point of view in [82]. The results 40.3.4 and 40.4.5 are taken from [63], and 40.4.2 is taken from [68]. The proof of 40.3.4 given here is due to Ra¨ıs. The construction given in 40.5.4 can be found in [45] and [43], and it is often called the Kostant’s cascade construction. The results of 40.6.3 was established in [45], and 40.6.5 in [82] The notion of seaweed Lie algebra was first introduced in [25] for type A, and in [62] in the general case. If g is of type A, the index of a seaweed Lie algebra was computed in [25] via a combinatorial formula. For types A, B, C, D, inductive formulas relating the index of seaweed Lie algebras were established in [62]. The results of sections 40.8 to 40.10 are taken from [83]. Note also that 40.10.2 was announced without proof in [33]. The index of a certain class of Lie algebras related to semisimple Lie algebras is computed in [69]. Let g be an ad-algebraic Lie algebra. Then by Rosenlicht’s Theorem stated in the comments of Chapter 25, we have the equality χ(g) = tr degk R(g∗ ).
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List of notations
dim X, 4 codimX Y , 5 Spm(A), 12 Spec(A), 12 dim A, 12 S −1 A, 14 Ap , 14 Fract(A), 14 ht p, 17 S −1 √ E, 17 a, 18 rad A, 19 Ep , 21 Der(A, M ), Der A, 24 Derk (A, M ), Derk A, 24 Ωk (A), 26 a ∼ b, 39 c(P ), 45 res(P, Q), 50 dis P , 53 [L : K], 55 tr degK L, 59 Gal(E/K), 64 Specm (A), 75 V(E), 84 I(M ), 84 lim(Ei , fji ), −→ lim Ei , 96 −→ lim u i , 98 −→ rVF U , 103 Γ (U, F ), 103 Fx , 104 Int(G), 117 k, 131
V(A), 131 I(M ), 132 A(V ), 133 Mor(V, W ), 134 A(u), 134 D(f ), 140 X Y , 151 ∆X , 155 R(X), 160 Pn (k), 167 Gn,r , 181 Fn , 182 codimX (Y ), 185 Tanx (X), 205 Tx (X), 207 Derxk (OX,x , k), 207 Tx (u), 209 rk(u), 233 sa∗ ,a , 233 rk R, 236 sα , 236 A(R), 236 W (R), 236 (w), 247 C(B), 248 R+ (C),R− (C), 249 , 254 |α|, ht(α), 263 gl(V ), gln (k), 277 sl(V ), sln (k), 277 tn (k), nn (k), 277 dn (k), 277 [h, h ], 278
646
List of notations
Der g, 278 adg x, ad x, 278 rk(g), 278 ggen , 278 cq (p), 279 z(g), 279 qx , 279 nq (p), 279 Aut g, 279 Aute g, 279 Lg , 280 C i (g), Di (g), D(g), 282 rad g, 289 Gr(g, d), 292 Φf , 292 a(f ) , 292 χ(g), 292 g∗reg , 292 R(g, h), 294 Sg , Ng , 309 W (g, h), 313 Ga , Gm , 319 Sp2n (k), 320 O2n+1 (k), O2n (k), 320 Aut A, 320 G◦ , 320 A(P ), 328 X ∗ (G), 338 X∗ (G), 339 Lie(G), 348 f ∗ x, 349 L(G), 350 (h), 368 a(m), 376 CH (x), ch (x), 401 R(G), 413 Ru (G), 414 multω (E), 419
a
X//G, 422 C(G), 424 rk(G), 437 Ggen , 438 N(g), 441 Auts g, 441 greg , 446 U(g), 459 P++ , 461 mV (µ), 461 P(V ), 461 M (λ), 465 vλ , 466 L(λ), 467 Aute (g, h), Aut0e (g, h), 469 C(x), 489 Pol(x, g), Pol(x), 494 Polr (x, g), Polr (x), 494 Oreg , 510 Omin , 511 Vλ (A), V λ (A), 549 V • (A), 551 (g; k, p), 553 (g, θ), 553 p∗reg , 560 WS , 562 gλa , 563 Lθ , 565 S, N , 568 R, 569 z(gx )• , 593 G
x ∼ y, 594 JG (x), 594 U (n) , 601 JK (x), 606 K
x ∼ y, 606 χ(g, V ), 615 gS,T , 629
Index
A-invariant bilinear form, 552 G-Jordan class, 594 G-Jordan equivalent, 594 G-equivariant, 122 G-homogeneous space, 122 G-variety, 325 K-Jordan class, 606 K-Jordan equivalent, 606 K-stable, 612 α-string, 241 g-regular, 615 k-distinguished, 589 p-distinguished, 589 p-generic, 555 p-regular, 560, 569 p-subalgebra, 554 natural, 554 σ-root system, 537 normal, 537 i-simple, 518 p-adic valuation, 41 abelian Lie algebra, 279 acts rationally, 325 ad-algebraic, 384 adjoint group, 384 algebraic, 383 adjoint representation, 278, 359 Ado’s Theorem, 280 affine algebraic group, 319 affine algebraic variety, 149 affine cone, 167 affine line, 131
affine morphism, 193 affine open subset, 149 affine plane, 131 affine space, 131 affine variety, 149 algebraic, 56 algebraic adjoint group, 383 algebraic closure, 57 algebraic curve, 185 algebraic group, 319 automorphism of, 319 connected, 321 diagonalizable, 340 isomorphism of, 319 morphism of, 319 reductive, 415 semisimple, 415 algebraic hull, 376 algebraic Lie subalgebra, 375 algebraic quotient, 422 algebraic set abstract, 139 affine, 131 algebraic subgroup, 365 algebraic surface, 185 algebraic variety, 156 algebraically closed, 57 algebraically independent, 56 Artin-Rees Theorem, 90 associated partial order, 249 Bala-Carter Theorem, 522 base of a root system, 245
648
Index
Bezout’s Theorem, 44 birational morphism, 162 Borel subalgebra, 316, 447 Borel subgroup, 429 Borel’s Fixed Point Theorem, 407 Burnside’s Theorem, 128 canonical generator, 466, 467 canonical homomorphism, 14, 17 Cartan integers, 312 Cartan matrix, 257 Cartan space, 110 morphism of, 110 Cartan subalgebra, 295 Cartan subgroup, 437 Cartan subspace, 555 Cartan’s Criterion, 288 central descending series, 121, 282 central function, 424 centralizer, 117, 279 centre of a Lie algebra, 279 chain, 4, 12, 256 maximal, 79 character, 424 character (multiplicative), 338 characteristic, 489 characteristic ideal, 282 characteristic subgroup, 117 Chevalley’s Theorem, 473 Chinese Remainder Theorem, 11 closed subset of roots, 250 coadjoint representation, 279 codimension, 5, 185 cofinal, 95 commutator, 118 comorphism, 134, 139, 162 compatible grading, 87 complete system, 41 complete variety, 176 completely reducible, 124, 279 conductor, 70 cone, 167 affine, 167 conjugate, 62, 429 conjugation, 123 connected algebraic group, 321 connected components of a graph, 256 constructible, 7 content, 45
contragredient representation, 123, 279 coset, 118 cycle, 256 degree, 55, 87 derivation, 24 universal, 25 derivation of a Lie algebra, 278 derivation over a field, 24 derived ideal, 282 derived series, 120, 282 derived subgroup, 119 diagonal, 155 diagonal symmetric Lie algebra, 553 diagonalizable, 340 differential, 26, 209 dimension Krull, 12 pure, 185 dimension of a topological space, 4 directed set, 95 discriminant, 53, 61 distinguished k-, 589 p-, 589 distinguished element, 521 distinguished parabolic subalgebra, 524 divides, 39 Dixmier sheet, 604 dominant, 2 dominant weight, 254 dominates, 199 dual representation, 123, 279 dual root system, 237 Duflo’s Theorem, 494 Dynkin diagram, 258 extended, 263 edge, 255 eigenspace, 549 Eisenstein’s Criterion, 47 elementary automorphism, 279 group of, 279 elementary symmetric polynomial, 48 Engel’s Theorem, 284 equation of integral dependence, 31 equidimensional, 185 equivalent, 39 essentially unique, 41
Index Euclid’s Lemma, 41 Euclidean domain, 44 even element, 521 extension, 55 algebraic, 56 cyclic, 55 finite, 55 finitely generated, 55 normal, 64 purely transcendental, 58 transcendental, 56 factorial, 41 faithful representation, 279 fibre, 104 field of fractions, 14 filtered set, 95 filtration, 91 discrete, 93 graded ring associated to a, 92 induced, 91 separated, 93 finite A-algebra, 36 finite morphism, 194 Fitting decomposition, 552 flag, 182 flag variety, 182 forest, 256 full ring of fractions, 14 fundamental simple representation, 468 fundamental weight, 254 fundamental Weyl chamber, 249 Gauss’ Lemma, 41 generalized eigenspace, 284 generic, 278, 438 geometric quotient, 391 germs, 104 value of, 104 germs of functions, 96 gluing, 8, 9 gluing morphisms of presheaves, 109 gluing prevarieties, 159 gluing sheaves, 109 Going Down Theorem, 66 Going Up Theorem, 36 graded positively, 88 graded A-module, 87
graded group, 87 graded ideal, 88 graded Lie algebra, 553 graded quotient module, 88 graded ring, 87 graded ring homomorphism, 88 graded submodule, 88 grading, 87 compatible, 87 graph, 137, 255 connected, 256 normed, 255 Grassmannian variety, 181 greatest common divisor, 40 group nilpotent, 121 solvable, 120 unipotent, 128 group closure, 328 group of commutators, 119 height, 17, 263 Heisenberg algebra, 622 highest root, 250 highest weight, 463 Hilbert’s Nullstellensatz, 82 Hilbert-Nagata Theorem, 421 homogeneous component, 87 homogeneous coordinates, 167 homogeneous of degree n, 87 homomorphism of G-modules, 123 hyperplane, 172 hypersurface, 131, 172 ideal maximal, 12 minimal prime, 13 prime, 12 principal, 39 ideal lying above, 16 ideal of a Lie algebra, 277 ideal of a subset, 132 identity component, 320 index, 118, 292, 615 inductive limit, 95 inductive limit of maps, 98 inductive system, 95 inductive system of maps, 98 inner automorphism, 117
649
650
Index
inner derivation, 278 integral, 31, 32 integral closure, 32, 34 integral domain, 11 integral equation, 31 integrally closed, 32 integrally closed domain, 34 invariant polynomial function, 476 irreducible, 1 irreducible component, 3 irreducible components of a root system, 236 irreducible element, 39 irreducible root system, 236 isomorphism of affine algebraic sets, 134 isotropic, 234 isotropy group, 122 isotypic component, 418 Iwasawa decomposition, 563, 564 Jacobi identity, 277 Jacobson radical, 19 Jacobson-Morosov Theorem, 483 Jordan class, 594 Jordan decomposition, 333, 362 additive, 113, 115 multiplicative, 114, 116 Jordan-H¨ older series, 279 Killing form, 280 Killing homomorphism, 280 Killing isomorphism, 477 Krull dimension, 12 Krull’s Hauptidealsatz, 81, 186 Krull’s Principal Ideal Theorem, 81, 186 largest nilpotent ideal, 290 least common multiple, 40 left convolution, 351 left invariant derivation, 348 left regular representation, 464 left translation, 333 length, 247, 489 length of a chain, 4 length of a path, 256 length of an α-string, 241 Levi factor, 452, 456 Levi subalgebra, 304 Levi-Malcev Theorem, 304
lexicographic order, 244 Lie algebra, 277 abelian, 279 graded, 553 morphism of, 278 nilpotent, 283 radical of a, 289 reductive, 307 semisimple, 299 simple, 299 solvable, 286 symmetric, 553 Lie algebra of an algebraic group, 350 Lie bracket, 277 Lie subalgebra, 277 Lie’s Theorem, 286 Lie-Kolchin Theorem, 406 linked, 255 local homomorphism, 20 local isomorphism, 149 local ring, 20 localization, 14 locally closed, 6 locally finite, 114, 327, 362 locally nilpotent, 115 locally unipotent, 115 long root, 243 maximal ideal, 12 minimal component, 300 minimal nilpotent orbit, 511 minimal polynomial, 56 model, 161 module, 123 g-, 279 semisimple, 124, 279 simple, 123, 279 module of differentials, 26 morphism affine, 193 birational, 162 differential of a, 209 finite, 194 graph of a, 137 morphism of affine algebraic sets, 134 multiple, 39 multiplicative subset, 13 multiplicity, 419, 461
Index Nakayama’s Lemma, 20 natural p-subalgebra, 554 negative element, 481 negative root, 245 nilpotent, 11 locally, 115 nilpotent component, 113, 306, 362 nilpotent element, 306, 309 nilpotent endomorphism, 113 nilpotent group, 121 nilpotent Lie algebra, 283 nilpotent orbit, 507 nilpotent radical, 291 nilradical, 11 nilspace, 278, 284, 549 Noether’s Normalization Theorem, 77 Noetherian, 5 Noetherian module, 21 Noetherian ring, 21 non-degenerate subspace, 234 non-singular variety, 215 norm, 60 normal chain, 120 normal point, 219 normal S-triple, 570 normal subgroup, 117 normal variety, 219 normalization, 222 normalizer, 117, 279 normed graph, 255 nowhere dense, 3 one-parameter subgroup (additive), 335 one-parameter subgroup (multiplicative), 338 open simplicial cone, 247 opposite Lie algebra, 351 orbit, 122 order, 93 orthogonal, 234 pairwise relatively prime, 40 parabolic subalgebra, 317, 447 distinguished, 524 parabolic subgroup, 430 parabolic subset of roots, 250 path, 256 injective, 256 Poincar´e-Birkhoff-Witt Theorem, 459
point derivation, 207 polarizable, 494 polarizable orbit, 507 polarization, 494 pole, 160 polynomial content of a, 45 elementary symmetric, 48 primitive, 45 symmetric, 48 weight of a, 49 positive element, 481 positive root, 245 preorder, 95 directed, 95 filtered, 95 presheaf, 103 injective morphism of, 105 isomorphism of, 104 morphism of, 104 restriction, 108 prevariety algebraic, 149 isomorphism of, 149 product, 153 sub-, 150 sum of, 151 primary subspace, 549 prime ideal, 12 minimal, 13 primitive element, 64 primitive vector, 461 principal element, 489 principal ideal, 39 principal ideal domain, 43 principal open set, 140 principal S-triple, 489, 571 principal symbol, 93 principle of transfinite induction, 5 projective space, 167 projective variety, 171 quasi-affine variety, 156 quasi-projective variety, 171 quotient, 39, 277 quotient field, 14 radical, 289 Jacobson, 19
651
652
Index
radical ideal, 18 radical of an algebraic group, 413 radical of an ideal, 18 radical weight, 253 ramification point, 255 rank of a Lie algebra, 278 rank of a root system, 236 rank of an algebraic group, 437 rational G-module, 326, 327 rational function, 160 rational map, 160 reduced decomposition, 247 reduced ring, 11 reduced root, 241 reduced root system, 235 reduction modulo an ideal, 47 reductive algebraic group, 415 reductive Lie algebra, 307 reductive Lie subalgebra, 307 reductive symmetric Lie algebra, 558 reflection, 233 reflection orthogonal to a hyperplane, 234 regular, 446, 615 p-, 569 regular element, 536 regular function, 133, 151 regular linear form, 292 regular nilpotent orbit, 510 regular sheet, 601, 608 relatively prime, 40 pairwise, 40 replica, 380 representation, 123, 279 coadjoint, 279 contragredient, 123, 279 dual, 123, 279 faithful, 279 residual field, 20 restricted root system, 562 resultant, 50 Reynolds operator, 126 Richardson element, 506 Richardson orbit, 506 Richardson’s Theorem, 504 right convolution, 349 right invariant derivation, 351 right translation, 333 rigidity of diagonalizable groups, 344
ring of fractions, 14 ringed space, 111 root, 236, 294 highest, 250 long, 243 negative, 245 positive, 245 reduced, 241 short, 243 simple, 245 special, 544 root lattice, 253 root system, 235, 294 σ-, 537 direct sum of, 236 dual, 237 irreducible, 236 isomorphic, 256 normal σ-, 537 reduced, 235 root system of (g, a), 562 Si -triple, 518 S-triple, 481 normal, 570 principal, 571 standard, 507 Schur’s Lemma, 123 seaweed Lie algebra, 629 standard, 629 section, 103 global, 103 Segre embedding, 178 semi-direct product, 118, 454 semisimple algebraic group, 415 semisimple component, 113, 306, 333, 362 semisimple element, 306, 309, 333 semisimple endomorphism, 113, 115 semisimple Lie algebra, 299 semisimple module, 124, 279 semisimple orbit, 507 semisimple symmetric Lie algebra, 557 sheaf, 104 isomorphism of, 104 morphism of, 104 restriction, 108 surjective morphism of, 105 sheaf associated to a presheaf, 106
Index sheet, 601, 608 short root, 243 simple element, 481 simple Lie algebra, 299 simple module, 123, 279 simple point, 215 simple root, 245 smooth point, 215 smooth variety, 215 solvable group, 120 solvable Lie algebra, 286 source of an α-string, 241 special automorphism, 450 special root, 544 stabilizer, 122 stable linear form, 612 weakly, 612 strongly orthogonal roots, 241 structure sheaf, 147 subgraph, 255 full, 255 subgroup characteristic, 117 derived, 119 normal, 117 subordinate, 493, 494 subregular element, 513 subregular orbit, 513 subvariety, 156 symmetric decomposition, 553 symmetric Lie algebra, 553 symmetric polynomial, 48 symmetric subset of roots, 250 symmetrizing, 553 system of parameters, 189 tangent space, 205, 207 tangent vector, 208 target of an α-string, 241 terminal vertex, 255 Theorem of the Primitive Element, 63 torus, 340 torus subalgebra, 310 trace, 60 transcendence basis, 58 purely, 58 transcendence degree, 59 transcendental, 56 transitive, 122 k-, 122 transporter, 122
653
tree, 256 triangular decomposition, 316 unipotent locally, 115 unipotent component, 114, 333 unipotent element, 333 unipotent endomorphism, 113 unipotent group, 128, 335 unipotent radical, 414 unique factorization domain, 41 universal derivation, 25 universal enveloping algebra, 459 upper semi-continuous, 201 variety algebraic, 156 complete, 176 non-singular, 215 normal, 219 projective, 171 quasi-affine, 156 quasi-projective, 171 smooth, 215 variety of pure dimension, 185 Verma module, 465 Veronese embedding, 179 vertex, 255 Vinberg’s Lemma, 616 wall, 249 weakly opposite parabolic subalgebras, 629 weakly stable linear form, 612 weight, 253, 339, 461, 549 dominant, 254 fundamental, 254 radical, 253 weight lattice, 253 weight space, 339 weight vector, 339, 461, 549 Weyl chamber, 247 fundamental, 249 Weyl group, 236, 313 Weyl’s Theorem, 303 Whitehead’s First Lemma, 302 Zariski tangent space, 207 Zariski topology, 84, 132, 139 Zariski’s Main Theorem, 226 zero divisor, 11
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