Smooth Manifolds

Rajnikant Sinha Smooth Manifolds Smooth Manifolds Rajnikant Sinha Smooth Manifolds 123 Rajnikant Sinha Departme...

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Rajnikant Sinha

Smooth Manifolds

Smooth Manifolds

Rajnikant Sinha

Smooth Manifolds

123

Rajnikant Sinha Department of Mathematics Magadh University Bodh Gaya, Bihar India

ISBN 978-81-322-2103-6 DOI 10.1007/978-81-322-2104-3

ISBN 978-81-322-2104-3

(eBook)

Library of Congress Control Number: 2014948765 Springer New Delhi Heidelberg New York Dordrecht London © Springer India 2014 This work is subject to copyright. All rights are reserved by the Publisher, whether the whole or part of the material is concerned, specifically the rights of translation, reprinting, reuse of illustrations, recitation, broadcasting, reproduction on microfilms or in any other physical way, and transmission or information storage and retrieval, electronic adaptation, computer software, or by similar or dissimilar methodology now known or hereafter developed. Exempted from this legal reservation are brief excerpts in connection with reviews or scholarly analysis or material supplied specifically for the purpose of being entered and executed on a computer system, for exclusive use by the purchaser of the work. Duplication of this publication or parts thereof is permitted only under the provisions of the Copyright Law of the Publisher’s location, in its current version, and permission for use must always be obtained from Springer. Permissions for use may be obtained through RightsLink at the Copyright Clearance Center. Violations are liable to prosecution under the respective Copyright Law. The use of general descriptive names, registered names, trademarks, service marks, 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. While the advice and information in this book are believed to be true and accurate at the date of publication, neither the authors nor the editors nor the publisher can accept any legal responsibility for any errors or omissions that may be made. The publisher makes no warranty, express or implied, with respect to the material contained herein. Printed on acid-free paper Springer is part of Springer Science+Business Media (www.springer.com)

Preface

The theory of smooth manifolds is a frequently discussed area in modern mathematics among the broad class of audience. A number of popular books on this subject are available for general readers with little prerequisite in mathematics. On the other hand, numerous textbooks and monographs for advanced graduate students and research fellows are available as well. The intermediate books, allowing senior undergraduate students to enter the exciting ﬁeld of smooth manifolds in a pedagogical way, are very few. It is argued that new concepts involved in this ﬁeld are too complicated to allow a simple introduction. Thus, the only way to master this theory is to be a timeconsuming effort to accumulate the intelligible parts of advanced textbooks into a comprehensible collection of notes. No doubt, this approach has the advantage of making young students, with perseverance to go through such a learning process, a very well-trained future research worker. But unfortunately, this severely limits the number of students who ever really master the subject. This book modestly attempts to bridge the gap between a university curriculum and the more advanced books on Smooth Manifolds. Actually, the manuscript has evolved from my habit of writing elaborate proofs of different theorems over the past one decade. This book intentionally supplies to the readers a high level of detail in arguments and derivations. More lengthy proofs of various theorems are, in general, outlined in such a way that they can be digested by an interested senior undergraduate student with little risk of ever getting lost. This book has been written in such a style that it invites the young students to ﬁll in the gaps everywhere during its study. It is another matter that the author is there for the needy students with complete solutions. That is why, supplying a separate list of problems will only deviate us from the main aim of this book. In my experience, some common problems for a young university student, trying to master mathematics, are the phrases in literatures like “it is clear that …”, “it is easy to see that …”, “it is straightforward”, etc. If a student cannot supply the proof of such a “clearly,” which most likely is the case, the common reaction under the time pressure of the studies is to accept the statement as true. And from this point

v

vi

Preface

on, throughout the rest part of the course, the deeper understanding of the subject is lost. I have beneﬁted from a number of advanced textbooks on the subject, and some unpublished works of my own and my distinguished friends. Some of these sources are mentioned in the bibliography. However, in an introductory book such as this, it is not possible to mention all literatures and all the people who have contributed to the understanding of this exciting ﬁeld. It is hoped that our readers (active and passive both) will ﬁnd that I have substantially fulﬁlled the objective of bridging the gap between university curriculum and more advanced texts, and that they will enjoy the reading this book as much as I did while writing it. Rajnikant Sinha

Contents

1

Differentiable Manifolds . . . . . . . . . 1.1 Topological Manifolds . . . . . . . 1.2 Smooth Manifolds . . . . . . . . . . 1.3 Examples of Smooth Manifolds .

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1 1 5 14

2

Tangent Spaces . . . . . . . . . . . . . . . . . . . . . . . . 2.1 Smooth Functions . . . . . . . . . . . . . . . . . . . 2.2 Algebra of Smooth Functions . . . . . . . . . . . 2.3 Smooth Germs on Smooth Manifolds . . . . . . 2.4 Derivations . . . . . . . . . . . . . . . . . . . . . . . . 2.5 Cotangent Spaces. . . . . . . . . . . . . . . . . . . . 2.6 Tangent Space as a Dual Space . . . . . . . . . . 2.7 Contravariant Vectors and Covariant Vectors. 2.8 Tangent Maps . . . . . . . . . . . . . . . . . . . . . .

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31 31 37 54 59 72 91 108 117

3

Multivariable Differential Calculus . . . . . . . 3.1 Linear Transformations . . . . . . . . . . . . . 3.2 Differentiation . . . . . . . . . . . . . . . . . . . 3.3 Inverse Function Theorem. . . . . . . . . . . 3.4 Implicit Function Theorem . . . . . . . . . . 3.5 Constant Rank Theorem (Easy Version) . 3.6 Smooth Bump Functions. . . . . . . . . . . . 3.7 The Constant Rank Theorem in Rn . . . . 3.8 Taylor’s Theorem . . . . . . . . . . . . . . . .

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125 125 141 163 177 191 199 212 226

4

Topological Properties of Smooth Manifolds 4.1 Constant Rank Theorem . . . . . . . . . . . . 4.2 Lie Groups . . . . . . . . . . . . . . . . . . . . . 4.3 Locally Path Connected Spaces . . . . . . . 4.4 Smooth Manifold as Paracompact Space .

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231 231 246 265 273

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4.5 4.6 4.7

Partitions of Unity Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . Topological Manifolds With Boundary . . . . . . . . . . . . . . . . . . . Smooth Covering Maps . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

288 297 304

5

Immersions, Submersions, and Embeddings 5.1 Pointwise Pushforward . . . . . . . . . . . . . 5.2 Open Submanifolds . . . . . . . . . . . . . . . 5.3 Tangent Bundles . . . . . . . . . . . . . . . . . 5.4 Smooth Immersion. . . . . . . . . . . . . . . . 5.5 Inverse Function Theorem for Manifolds 5.6 Shrinking Lemma . . . . . . . . . . . . . . . . 5.7 Global Rank Theorem . . . . . . . . . . . . . 5.8 Properly Embedded . . . . . . . . . . . . . . . 5.9 Regular Level Sets. . . . . . . . . . . . . . . . 5.10 Smooth Submanifolds. . . . . . . . . . . . . . 5.11 Tangent Space to a Submanifold . . . . . .

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307 307 314 330 343 350 364 382 392 406 416 421

6

Sard’s Theorem . . . . . . . . . . . . . 6.1 Measure Zero in Manifolds . . 6.2 Taylor’s Inequality. . . . . . . . 6.3 Sard’s Theorem on Rn . . . . . 6.4 Sard’s Theorem on Manifolds

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431 431 448 453 475

7

Whitney Embedding Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.1 Compact Whitney Embedding in RN . . . . . . . . . . . . . . . . . . . . 7.2 Compact Whitney Embedding in R2mþ1 . . . . . . . . . . . . . . . . . .

477 477 480

Bibliography . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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About the Author

Rajnikant Sinha is former professor of mathematics at Magadh University, Bodh Gaya, India. A passionate mathematician by heart, Prof. Sinha has published several interesting researches in international journals and contributed a book Solutions to Weatherburn’s Elementary Vector Analysis. His research areas are topological vector spaces, differential geometry, and manifolds.

ix

Chapter 1

Differentiable Manifolds

The pace of this chapter is rather slow for the simple reason that the deﬁnition of smooth manifold is itself forbidding for many unfamiliar readers. Although many young university students feel comfortable with the early parts of multivariable calculus, if one feels uncomfortable, then he should go ﬁrst through some parts of Chap. 3. How veriﬁcations are done in various examples of smooth manifolds is a crucial thing to learn for a novice, so this aspect has been dealt here with some more detail.

1.1 Topological Manifolds Deﬁnition Let M be a Hausdorff topological space. Let m be a positive integer. If, for every x in M; there exists an open neighborhood U of x such that U is homeomorphic to some open subset of Euclidean space Rm , then we say that M is an m-dimensional topological manifold. Deﬁnition Let M be an m-dimensional topological manifold. Let x be an element of M: So there exists an open neighborhood U of x such that U is homeomorphic to some open subset of Euclidean space Rm : Hence, there exists a homeomorphism uU : U ! uU ðU Þ such that uU ðUÞ is an open subset of Rm : Here, the ordered pair ðU; uU Þ is called a coordinate chart of M: ðU; uU Þ is also simply denoted by ðU; uÞ: Deﬁnition Let M be an m-dimensional topological manifold. Let ðU; uU Þ be a coordinate chart of M: So uU : U ! uU ðU Þ ð Rm Þ:

© Springer India 2014 R. Sinha, Smooth Manifolds, DOI 10.1007/978-81-322-2104-3_1

1

2

1 Differentiable Manifolds

Hence, for every x in U; uU ðxÞ is in Rm : So there exist real numbers u1 ; . . .; um such that uU ð xÞ ¼ u1 ; . . .; um : Here, we say that ui ði ¼ 1; . . .; mÞ are the local coordinates of the point x. In short, we write ðuU ð xÞÞi ui : Lemma 1.1 Let M be an m-dimensional topological manifold. Let ðU; uU Þ; ðV; uV Þ be coordinate charts of M such that U \ V is nonempty. Then, 1. 2. 3. 4. 5. 6. 7.

domðuV ðuU Þ1 Þ ¼ uU ðU \ VÞ; ranðuV ðuU Þ1 Þ ¼ uV ðU \ VÞ; uV ðuU Þ1 is a function, uV ðuU Þ1 is 1–1, uU ðU \ VÞ; uV ðU \ VÞ are open subsets of Rm ; ðuV ðuU Þ1 Þ1 ¼ uU ðuV Þ1 ; ðuV ðuU Þ1 Þ : uU ðU \ VÞ ! uV ðU \ VÞ, and ðuU ðuV Þ1 Þ : uV ðU \ VÞ ! uU ðU \ VÞ are homeomorphisms.

Proof of 1 Let us take any x in domðuV ðuU Þ1 Þ: We want to show that x is in uU ðU \ VÞ: Since x is in domðuV ðuU Þ1 Þ, by the deﬁnition of domain, there exists y such that ðx; yÞ is in uV ðuU Þ1 : Since ðx; yÞ is in uV ðuU Þ1 , by the deﬁnition of composition, there exists z such that ðx; zÞ is in ðuU Þ1 ; and ðz; yÞ is in uV : Since ðx; zÞ is in ðuU Þ1 ; ðz; xÞ is in uU : Since ðz; xÞ is in uU ; and uU : U ! uU ðUÞ; uU ðzÞ ¼ x; and z is in U: Since ðz; yÞ is in uV ; and uV : V ! uV ðVÞ; uV ðzÞ ¼ y; and z is in V: Here, z is in U; and z is in V; so z is in U \ V: Since z is in U \ V; and x ¼ uU ðzÞ; x is in uU ðU \ VÞ: Thus, (see Fig. 1.1), dom uV ðuU Þ1 uU ðU \ V Þ: Next, let x be in uU ðU \ VÞ: We want to prove that x is in domðuV ðuU Þ1 Þ: Since x is in uU ðU \ VÞ; there exists y in U \ V such that uU ðyÞ ¼ x: Since y is in U \ V; y is in U; and y is in V: Since y is in U; and uU : U ! uU ðUÞ; there exists z such that ðy; zÞ is in uU : Since y is in V; and uV : V ! uV ðVÞ; there exists w such that ðy; wÞ is in uV : Since ðy; zÞ is in uU ; ðz; yÞ is in ðuU Þ1 : Since ðz; yÞ is in ðuU Þ1 ; and

1.1 Topological Manifolds

3

Fig. 1.1 m-dimensional topological manifold

ðy; wÞ is in uV ; ðz; wÞ is in uV ðuU Þ1 ; and hence, z is in domðuV ðuU Þ1 Þ: Since ðy; zÞ is in uU ; and uU is a function, uU ðyÞ ¼ z: Since uU ðyÞ ¼ z; and uU ðyÞ ¼ x; x ¼ z: Since x ¼ z; and z is in domðuV ðuU Þ1 Þ; x is in domðuV ðuU Þ1 Þ: Thus, uU ðU \ V Þ dom uV ðuU Þ1 : h Proof of 2 On using 1, we have domðuU ðuV Þ1 Þ ¼ uV ðV \ UÞ: So 1 LHS ¼ ran uV ðuU Þ1 ¼ dom uV ðuU Þ1 1 ¼ dom ðuU Þ1 ðuV Þ1 ¼ dom uU ðuV Þ1 ¼ uV ðV \ U Þ ¼ uV ðU \ V Þ ¼ RHS:

h

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1 Differentiable Manifolds

Proof of 3 Since ðU; uU Þ is a coordinate chart of M, uU is a 1–1 function, and hence, ðuU Þ1 is a function. Since ðuU Þ1 is a function, and uV is a function, their h composite uV ðuU Þ1 is a function. Proof of 4 Since ðU; uU Þ is a coordinate chart of M; uU is a 1–1 function, and hence, ðuU Þ1 is a 1–1 function. Since ðV; uV Þ is a coordinate chart of M; uV is a h 1–1 function. Since ðuU Þ1 is 1–1, and uV is 1–1, uV ðuU Þ1 is 1–1. Proof of 5 Since ðU; uU Þ is a coordinate chart of M; uU : U ! uU ðUÞ is a homeomorphism from open subset U of M onto open subset uU ðUÞ of Rm : Similarly, V is an open subset of M: Since U; V are open subsets of M; U \ V is an open subset of M: Since U; U \ V are open subset of M; and U \ V U; U \ V is open in U: Since U \ V is open in U; and uU : U ! uU ðUÞ is a homeomorphism, uU ðU \ VÞ is open in uU ðUÞ: Since uU ðU \ VÞ is open in uU ðUÞ; and uU ðUÞ is an open subset of Rm ; uU ðU \ VÞ is an open subset of Rm : Similarly, uV ðU \ VÞ is an open subset of Rm : h Proof of 6

LHS ¼ uV ðuU Þ

1

1

¼

ðuU Þ

1

1

ðuV Þ1 ¼ uU ðuV Þ1 ¼ RHS: h

Proof of 7 Since ðuV ðuU Þ1 Þ1 ¼ uU ðuV Þ1 ; it sufﬁces to prove that

uV ðuU Þ1

: uU ðU \ V Þ ! uV ðU \ V Þ;

and

uU ðuV Þ1

: uV ðU \ V Þ ! uU ðU \ V Þ

are continuous. Since ðU; uU Þ is a coordinate chart of M; uU is a homeomorphism, and hence, ðuU Þ1 is continuous. Since ðV; uV Þ is a coordinate chart of M; uV is a homeomorphism, and hence, uV is continuous. Since ðuU Þ1 ; uV are continuous, their composite uV ðuU Þ1 is continuous. Similarly, uU ðuV Þ1 is continuous. Hence, uV ðuU Þ1 is a homeomorphism. h Similarly, uU ðuV Þ1 is a homeomorphism.

1.2 Smooth Manifolds

5

1.2 Smooth Manifolds Deﬁnition Let M be an m-dimensional topological manifold. Let ðU; uU Þ; ðV; uV Þ be coordinate charts of M; and let U \ V be nonempty. So

uV ðuU Þ1

: uU ðU \ V Þ ! uV ðU \ V Þ

is a homeomorphism from uU ðU \ VÞ onto uV ðU \ VÞ; where uU ðU \ VÞ and uV ðU \ VÞ are open subsets of Rm : For every ðx1 ; . . .; xm Þ in uU ðU \ VÞ; put uV ðuU Þ1 x1 ; . . .; xm f 1 x1 ; . . .; xm ; . . .; f m x1 ; . . .; xm ¼ y1 ; . . .; ym ; where ðy1 ; . . .; ym Þ is in uV ðU \ VÞ; and

uU ðuV Þ1

y1 ; . . .; ym g1 y1 ; . . .; ym ; . . .; gm y1 ; . . .; ym ¼ x1 ; . . .; xm :

Since uV ðuU Þ1 is continuous, each f i : uU ðU \ VÞ ! R is continuous. Similarly, each gi : uV ðU \ VÞ ! R is continuous. For ﬁxed i; and for every j ¼ 1; . . .; m; if ðDj f i Þðx1 ; . . .; xm Þ exists at every point 1 ðx ; . . .; xm Þ of uU ðU \ VÞ; then we get functions Dj f i : uU ðU \ VÞ ! R: If every Dj f i : uU ðU \ VÞ ! R is continuous, then we say that f i is C 1 : Similarly, by gi is C 1 ; we shall mean that each Dj gi : uV ðU \ VÞ ! R ðj ¼ 1; . . .; mÞ is continuous. By ðU; uU Þ and ðV; uV Þ are C1 -compatible, we mean that either ðU \ V is an empty setÞ; or ðf i ; gi are C1 for every i ¼ 1; . . .; m; whenever U \ V is a nonempty setÞ: For ﬁxed i; and for every j; k ¼ 1; . . .; m; if Dk ðDj f i Þ ð Dkj f i Þ exists at every point of uU ðU \ VÞ; then we get functions Dk ðDj f i Þ : uU ðU \ VÞ ! R: If each Dk ðDj f i Þ : uU ðU \ VÞ ! R is continuous, then we say that f i is C 2 : Similarly, by gi is C 2 ; we shall mean that each Dk ðDj gi Þ : uV ðU \ VÞ ! R ðj; k ¼ 1; . . .; mÞ is continuous. By ðU; uU Þ and ðV; uV Þ are C 2 -compatible, we mean that

6

1 Differentiable Manifolds

either ðU \ V is an empty setÞ; or ðf i ; gi are C2 for every i ¼ 1; . . .; m; whenever U \ V is a nonempty setÞ: Similar deﬁnitions for ðU; uU Þ and ðV; uV Þ are C 3 -compatible can be supplied, etc. It is clear that if ðU; uU Þ and ðV; uV Þ are C 3 -compatible, then ðU; uU Þ and ðV; uV Þ are C 2 -compatible, etc. Deﬁnition Let M be an m-dimensional topological manifold. Let r be a positive integer. Let A fðU; uU Þ; ðV; uV Þ; ðW; uW Þ; . . .g be a nonempty collection of coordinate charts of M: If 1. fU; V; W; . . .g is a cover of M; that is, [fU : ðU; uU Þ is in Ag ¼ M; 2. all pairs of members in A are C r -compatible, e ; u Þ is a coordinate chart of M, but not a 3. A is maximal (in the sense that if ð U e U e ; u Þ and ðU; uU Þ member of A; then there exists ðU; uU Þ in A; such that ð U e U r are not C -compatible), then we say that A is a Cr -differentiable structure on M; and the ordered pair ðM; AÞ is called a Cr -differentiable manifold. Here, members of A are called admissible coordinate charts of M: Theorem 1.2 Let M be an m-dimensional topological manifold. Let r be a positive integer. Let A fðU; uU Þ; ðV; uV Þ; ðW; uW Þ; . . .g be a nonempty collection of coordinate charts of M satisfying 1. fU; V; W; . . .g is a cover of M; that is, [fU : ðU; uU Þ is in Ag ¼ M; 2. all pairs of members in A are Cr -compatible. Then, there exists a unique C r -differentiable structure B on M which contains A: (This theorem suggests that in constructing a differentiable manifold, it is enough to ﬁnd a collection A of coordinate charts for which conditions 1 and 2 are satisﬁed.) Proof Existence Let B be the collection of all coordinate charts ðU; uU Þ of M such that ðU; uU Þ is Cr -compatible with every member of A: First of all, we shall try to show that B contains A: For this purpose, let us take any ðU; uU Þ in A: By the condition 2, ðU; uU Þ is C r -compatible with all members of A; so by the deﬁnition of B; ðU; uU Þ is in B: Hence, A is a subset of B: Further since A is nonempty, and A is a subset of B; B is nonempty.

1.2 Smooth Manifolds

7

Now we want to prove that B is a Cr -differentiable structure on M; that is, 1. [fU : ðU; uU Þ is in Bg ¼ M; 2. all pairs of members of B are C r -compatible. e ; u Þ is a coordinate chart of M, but not a 3. B is maximal (in the sense that, if ð U e U e ; u Þ and ðU; uU Þ are member of B; then there exists ðU; uU Þ in B such that ð U e U r not C -compatible). For 1: Since A is contained in B; [fU : ðU; uU Þis in Ag [fU : ðU; uU Þ is in Bg M: Now, by the given condition 1, [fU : ðU; uU Þ is in Bg ¼ M: For 2: Let us take any ðU; uU Þ in B; and ðV; uV Þ in B: We have to prove that ðU; uU Þ and ðV; uV Þ are C r -compatible. Here, two cases arise: either U \ V is an empty set, or U \ V is a nonempty set.h Case I: when U \ V is an empty set. In this case, by the deﬁnition of C r compatible, ðU; uU Þ and ðV; uV Þ are C r -compatible. Case II: when U \ V is a nonempty set. For every ðx1 ; . . .; xm Þ in uU ðU \ VÞ; put

uV ðuU Þ1

x1 ; . . .; xm f 1 x1 ; . . .; xm ; . . .; f m x1 ; . . .; xm ¼ y1 ; . . .; ym ;

where ðy1 ; . . .; ym Þ is in uV ðU \ VÞ; and

uU ðuV Þ1

y1 ; . . .; ym g1 y1 ; . . .; ym ; . . .; gm y1 ; . . .; ym ¼ x1 ; . . .; xm :

We have to show that f i ; gi are C r for every i ¼ 1; . . .; m: Since U \ V is nonempty, there exists an element x in U \ V ð MÞ: Since x is in M, by condition 1, there exists ðW; uW Þ in A such that x is in W: Since ðU; uU Þ is in B; and ðW; uW Þ is in A, by the deﬁnition of B; ðU; uU Þ and ðW; uW Þ are Cr -compatible. Since x is in W; and x is in U \ Vð UÞ; x is in W \ U; and hence, U \ W is a nonempty set. Since ðU; uU Þ and ðW; uW Þ are Cr -compatible, and U \ W is nonempty, by the deﬁnition of C r -compatible, f1i ; gi1 are C r for every i ¼ 1; . . .; m, where for every ðx1 ; . . .; xm Þ in uU ðU \ WÞ; uW ðuU Þ1 x1 ; . . .; xm f11 x1 ; . . .; xm ; . . .; f1m x1 ; . . .; xm ¼ y1 ; . . .; ym 2 uW ðU \ W Þ; and 1 1 m ¼ x ; . . .; xm : uU ðuW Þ1 y1 ; . . .; ym g11 y1 ; . . .; ym ; . . .; gm 1 y ; . . .; y

8

1 Differentiable Manifolds

Similarly, f2i ; gi2 are Cr for every i ¼ 1; . . .; m, where for every ðx1 ; . . .; xm Þ in uV ðV \ WÞ; uW ðuV Þ1 x1 ; . . .; xm f21 x1 ; . . .; xm ; . . .; f2m x1 ; . . .; xm ¼ y1 ; . . .; ym ; 2 uW ðV \ W Þ and 1 1 m ¼ x ; . . .; xm : uV ðuW Þ1 y1 ; . . .; ym g12 y1 ; . . .; ym ; . . .; gm 2 y ; . . .; y Now for every ðx1 ; . . .; xm Þ in uU ðU \ V \ WÞ

1 1 f x ; . . .; xm ; . . .; f m x1 ; . . .; xm ¼ uV ðuU Þ1 x1 ; . . .; xm x1 ; . . .; xm ¼ uV ðuW Þ1 uW ðuU Þ1 uW ðuU Þ1 x1 ; . . .; xm ¼ uV ðuW Þ1 ¼ uV ðuW Þ1 f11 x1 ; . . .; xm ; . . .; f1m x1 ; . . .; xm 1 1 m m 1 m : ¼ g12 f11 x1 ; . . .; xm ; . . .; f1m x1 ; . . .; xm ; . . .; gm 2 f1 x ; . . .; x ; . . .; f1 x ; . . .; x

Hence,

f 1 x1 ; . . .; xm ¼ g12 f11 x1 ; . . .; xm ; . . .; f1m x1 ; . . .; xm :

Since g12 ; and each f1k ðk ¼ 1; . . .; mÞ are Cr ; og12 f11 ðx1 ; . . .; xm Þ; . . .; f1m ðx1 ; . . .; xm Þ ox1 ! n 1 1 1 X og2 f1 ðx ; . . .; xm Þ; . . .; f1m ðx1 ; . . .; xm Þ of1k ðx1 ; . . .; xm Þ ¼ ox1 of1k ðx1 ; . . .; xm Þ k¼1

exists and is continuous. It follows that og12 f11 ðx1 ; . . .; xm Þ; . . .; f1m ðx1 ; . . .; xm Þ of 1 ðx1 ; . . .; xm Þ ¼ ox1 ox1 Þ exists and is continuous. Similarly, of ðxox;...;x exists and is continuous, etc. 2 1 1 1 2 So f is C : Similarly, f is C ; etc. Hence, f 1 is C r : Similarly, f 2 is C r ; etc. So each f i is C r : Similarly, each gi is C r : Thus, we have shown that, in all cases, ðU; uU Þ and ðV; uV Þ are C r -compatible. For 3: We claim that B is maximal. If not, otherwise, let B be not maximal. We have to arrive at a contradiction. Since B is not maximal, there exists a 1

1

m

1.2 Smooth Manifolds

9

coordinate chart ðU; uU Þ of M such that ðU; uU Þ is not a member of B; and ðU; uU Þ is Cr -compatible with every member of B: Since ðU; uU Þ is C r -compatible with every member of B; and A is contained in B; ðU; uU Þ is C r -compatible with every member of A; and hence, by the deﬁnition of B; ðU; uU Þ is in B; a contradiction. Thus, we have shown that B is a C r -differentiable structure on M; which contains A: Uniqueness If not, otherwise, let B1 and B2 be two distinct C r -differentiable structures on M; which contains A: We have to arrive at a contradiction. Since B1 and B2 are two distinct C r -differentiable structures on M; either (there exists a coordinate chart ðU; uU Þ of M such that ðU; uU Þ is in B1 ; and ðU; uU Þ is not in B2 Þ or (there exists a coordinate chart ðU; uU Þ of M such that ðU; uU Þ is in B2 ; and ðU; uU Þ is not in B1 Þ: Case I: when there exists a coordinate chart ðU; uU Þ of M such that ðU; uU Þ is in B1 ; and ðU; uU Þ is not in B2 : First of all, we shall show that B1 is contained in B; where B is the differential structure constructed above. For this purpose, let us take any coordinate chart ðV; uV Þ of M in B1 : We have to show that ðV; uV Þ is in B: Since ðV; uV Þ is in B1 ; and B1 is a C r -differentiable structure on M which contains A; ðV; uV Þ is C r -compatible with every member of A; and hence, by the deﬁnition of B; ðV; uV Þ is in B: Thus, we have shown that B1 is contained in B: Similarly, B2 is contained in B: Since ðU; uU Þ is not in B2 ; and B2 is Cr differentiable structure on M; there exists ðV; uV Þ in B2 ; such that ðV; uV Þ and ðU; uU Þ are not C r -compatible. Since ðV; uV Þ is in B2 ; and B2 is contained in B; ðV; uV Þ is in B: Since ðU; uU Þ is in B1 ; and B1 is contained in B; ðU; uU Þ is in B: Since ðU; uU Þ; ðV; uV Þ are in B; and B is a C r -differentiable structure on M; ðV; uV Þ and ðU; uU Þ are Cr -compatible, a contradiction. Case II: when there exists a coordinate chart ðU; uU Þ of M such that ðU; uU Þ is in B2 ; and ðU; uU Þ is not in B1 . This case is similar to the case I. h Theorem 1.3 Let M be an m-dimensional topological manifold. Let A fðU; uU Þ; ðV; uV Þ; ðW; uW Þ; . . .g be a nonempty collection of coordinate charts of M satisfying 1. fU; V; W; . . .g is a cover of M; that is, [fU : ðU; uU Þ is in Ag ¼ M; 2. all pairs of members in A are C2 -compatible. Then, there exists a unique C 1 -differentiable structure B on M which contains A: Proof First of all, we shall show that 20 : all pair of members in A are C 1 -compatible. For this purpose, let us take any ðU; uU Þ and ðV; uV Þ in A: Here, two cases arises: either U \ V is an empty set, or U \ V is a nonempty set. h Case I: when U \ V is an empty set. In this case, by the deﬁnition of C 1 -compatible, ðU; uU Þ and ðV; uV Þ are C1 -compatible.

10

1 Differentiable Manifolds

Case II: when U \ V is a nonempty set. For every ðx1 ; . . .; xm Þ in uU ðU \ VÞ; put uV ðuU Þ1 x1 ; . . .; xm f 1 x1 ; . . .; xm ; . . .; f m x1 ; . . .; xm ¼ y1 ; . . .; ym ; where ðy1 ; . . .; ym Þ is in uV ðU \ VÞ; and

uU ðuV Þ1

y1 ; . . .; ym g1 y1 ; . . .; ym ; . . .; gm y1 ; . . .; ym ¼ x1 ; . . .; xm :

It is enough to show that f i ; gi are C1 for every i ¼ 1; . . .; m: Since ðU; uU Þ and ðV; uV Þ in A, by condition 2, ðU; uU Þ is C 2 -compatible with ðV; uV Þ: Since ðU; uU Þ is C 2 -compatible with ðV; uV Þ; and U \ V is nonempty, by the deﬁnition of C 2 -compatible, f i ; gi are C2 for every i ¼ 1; . . .; m. Since, for every i ¼ 1; . . .; m; f i ; gi are C 2 ; we have f i ; gi are C1 : Now, on using conditions 1, 20 , and Theorem 1.2, there exists a unique C1-differentiable structure B on M which contains A: h Note 1.4 As above, we can prove the following result: Let M be an m-dimensional topological manifold. Let r; s be positive integers satisfying 0\s r: Let A fðU; uU Þ; ðV; uV Þ; ðW; uW Þ; . . .g be a nonempty collection of coordinate charts of M satisfying 1. fU; V; W; . . .g is a cover of M; that is, [fU : ðU; uU Þ is in Ag ¼ M; 2. all pairs of members in A are C r -compatible. Then, there exists a unique C s -differentiable structure B on M which contains A: Deﬁnition Let M be an m-dimensional topological manifold. Let A be a nonempty collection of coordinate charts of M: If 1. [fU : ðU; uU Þ is in Ag ¼ M; 2. all pairs of members in A are C 1 -compatible (in the sense that every pair of members in A is C r -compatible for every positive integer rÞ, then we say that A is an atlas on M: Deﬁnition Let M be an m-dimensional topological manifold. Let A fðU; uU Þ; ðV; uV Þ; ðW; uW Þ; . . .g be a nonempty collection of coordinate charts of M: If

1.2 Smooth Manifolds

11

1. A is an atlas on M; e ; u Þ is a coordinate chart of M, but not a 2. A is maximal (in the sense that if ð U e U e ; u Þ and ðU; uU Þ member of A; then there exists ðU; uU Þ in A; such that ð U e U are not C1 -compatible), then we say that A is a C1 -differentiable structure on M; and the pair ðM; AÞ is called a C 1 -differentiable manifold. C1 -differentiable structure is also called smooth structure, and C 1 -differentiable manifold is also called smooth manifold. Here, members of A are called admissible coordinate charts of M: Note 1.5 We shall try to prove: if A is an atlas on an m-dimensional topological manifold M, then there exists a unique C1 -differentiable structure B on M which contains A: Existence Let B be the collection of all coordinate charts ðU; uU Þ of M such that ðU; uU Þ is C 1 -compatible with every member of A: First of all, we shall try to show that B contains A: For this purpose, let us take any ðU; uU Þ in A: Since A is an atlas, and ðU; uU Þ is in A; ðU; uU Þ is C 1 compatible with all members of A, and hence, by the deﬁnition of B; ðU; uU Þ is in B: Hence, A is a subset of B: Further since A is nonempty, and A is a subset of B; B is nonempty. Now we want to prove that B is a C 1 -differentiable structure on M; that is, 1. [fU : ðU; uU Þ is in Bg ¼ M; 2. all pairs of members of B are C 1 -compatible. e ; u Þ a coordinate chart of M, but not a 3. B is maximal (in the sense that if ð U e U e ; u Þ and ðU; uU Þ are member of B; then there exists ðU; uU Þ in B such that ð U e U 1 not C -compatible. For 1: Since A is contained in B; [fU : ðU; uU Þ is in Ag [fU : ðU; uU Þ is in Bg M: Since A is an atlas, [fU : ðU; uU Þ is in Ag ¼ M: Thus, [fU : ðU; uU Þ is in Bg ¼ M: For 2: Let us take any ðU; uU Þ in B; and ðV; uV Þ in B: We have to prove that ðU; uU Þ and ðV; uV Þ are C1 -compatible. Here, two cases arise: either U \ V is an empty set, or U \ V is a nonempty set. Case I: when U \ V is an empty set. In this case, by the deﬁnition of C r compatible, ðU; uU Þ and ðV; uV Þ are Cr -compatible for every positive integer r, and hence, ðU; uU Þ and ðV; uV Þ are C1 -compatible. Case II: when U \ V is a nonempty set. For every ðx1 ; . . .; xm Þ in uU ðU \ VÞ; put

uV ðuU Þ1

x1 ; . . .; xm f 1 x1 ; . . .; xm ; . . .; f m x1 ; . . .; xm ¼ y1 ; . . .; ym ;

where ðy1 ; . . .; ym Þ is in uV ðU \ VÞ; and

12

1 Differentiable Manifolds

uU ðuV Þ1

y1 ; . . .; ym g1 y1 ; . . .; ym ; . . .; gm y1 ; . . .; ym ¼ x1 ; . . .; xm :

We have to show that f i ; gi are C 1 for every i ¼ 1; . . .; m: Since U \ V is nonempty, there exists an element x in U \ V ð MÞ: Since x is in M; and A is an atlas, there exists ðW; uW Þ in A such that x is in W: Since ðU; uU Þ is in B; and ðW; uW Þ is in A, by the deﬁnition of B; ðU; uU Þ and ðW; uW Þ are C 1 -compatible. Since x is in W; and x is in U \ Vð UÞ; x is in W \ U; and hence, U \ W is a nonempty set. Since ðU; uU Þ and ðW; uW Þ are C1 -compatible, and U \ W is nonempty, by the deﬁnition of C1 -compatible, f1i ; gi1 are C 1 for every i ¼ 1; . . .; m, where for every ðx1 ; . . .; xm Þ in uU ðU \ WÞ;

uW ðuU Þ1

x1 ; . . .; xm f11 x1 ; . . .; xm ; . . .; f1m x1 ; . . .; xm ¼ y1 ; . . .; ym 2 uW ðU \ W Þ;

and

uU ðuW Þ1

1 1 m y1 ; . . .; ym g11 y1 ; . . .; ym ; . . .; gm ¼ x ; . . .; xm : 1 y ; . . .; y

Similarly, f2i ; gi2 are C 1 for every i ¼ 1; . . .; m, where for every ðx1 ; . . .; xm Þ in uV ðV \ WÞ;

uW ðuV Þ1

x1 ; . . .; xm f21 x1 ; . . .; xm ; . . .; f2m x1 ; . . .; xm ¼ y1 ; . . .; ym 2 uW ðV \ W Þ;

and 1 1 m ¼ x ; . . .; xm : uV ðuW Þ1 y1 ; . . .; ym g12 y1 ; . . .; ym ; . . .; gm 2 y ; . . .; y Now for every ðx1 ; . . .; xm Þ in uU ðU \ V \ WÞ 1 1 f x ; . . .; xm ; . . .; f m x1 ; . . .; xm ¼ uV ðuU Þ1 x1 ; . . .; xm ¼ uV ðuW Þ1 uW ðuU Þ1 x1 ; . . .; xm ¼ uV ðuW Þ1 uW ðuU Þ1 x1 ; . . .; xm ¼ uV ðuW Þ1 f11 x1 ; . . .; xm ; . . .; f1m x1 ; . . .; xm 1 1 m m 1 m : ¼ g12 f11 x1 ; . . .; xm ; . . .; f1m x1 ; . . .; xm ; . . .; gm 2 f1 x ; . . .; x ; . . .; f1 x ; . . .; x

1.2 Smooth Manifolds

Hence,

13

f 1 x1 ; . . .; xm ¼ g12 f11 x1 ; . . .; xm ; . . .; f1m x1 ; . . .; xm :

Since g12 ; and each f1k ðk ¼ 1; . . .; mÞ are C1 ; og12 f11 ðx1 ; . . .; xm Þ; . . .; f1m ðx1 ; . . .; xm Þ ox1 ! n 1 1 1 m X og2 f1 ðx ; . . .; x Þ; . . .; f1m ðx1 ; . . .; xm Þ of1k ðx1 ; . . .; xm Þ ¼ ox1 of1k ðx1 ; . . .; xm Þ k¼1 exists and is continuous. It follows that og12 f11 ðx1 ; . . .; xm Þ; . . .; f1m ðx1 ; . . .; xm Þ of 1 ðx1 ; . . .; xm Þ ¼ ox1 ox1 Þ exists and is continuous. Similarly, of ðxox;...;x exists and is continuous, etc. 2 So f 1 is C 1 : Similarly, f 1 is C2 ; etc. Hence, f 1 is C1 : Similarly, f 2 is C 1 ; etc. So each f i is C 1 : Similarly, each gi is C 1 : Thus, we have shown that, in all cases, ðU; uU Þ and ðV; uV Þ are C1 -compatible. For 3: We claim that B is maximal. If not, otherwise, let B be not maximal. We have to arrive at a contradiction. Since B is not maximal, there exists a coordinate chart ðU; uU Þ of M such that ðU; uU Þ is not a member of B; and ðU; uU Þ is C 1 compatible with every member of B: Since ðU; uU Þ is C 1 -compatible with every member of B; and A is contained in B; ðU; uU Þ is C1 -compatible with every member of A; and hence, by the deﬁnition of B; ðU; uU Þ is in B; a contradiction. Thus, we have shown that B is a C 1 -differentiable structure on M; which contains A: Uniqueness If not, otherwise, let B1 and B2 be two distinct C 1 -differentiable structures on M; which contains A: We have to arrive at a contradiction. Since B1 and B2 are two distinct C 1 -differentiable structures on M; either (there exists a coordinate chart ðU; uU Þ of M such that ðU; uU Þ is in B1 ; and ðU; uU Þ is not in B2 Þ or (there exists a coordinate chart ðU; uU Þ of M such that ðU; uU Þ is in B2 ; and ðU; uU Þ is not in B1 Þ: 1

1

m

Case I: when there exists a coordinate chart ðU; uU Þ of M such that ðU; uU Þ is in B1 ; and ðU; uU Þ is not in B2 . First of all, we shall show that B1 is contained in B; where B is the differential structure constructed above. For this purpose, let us take any coordinate chart ðV; uV Þ of M in B1 : We have to show that ðV; uV Þ is in B: Since ðV; uV Þ is in B1 ; and B1 is a C1 -differentiable structure on M which contains A; ðV; uV Þ is

14

1 Differentiable Manifolds

C 1 -compatible with every member of A; and hence, by the deﬁnition of B; ðV; uV Þ is in B: Thus, we have shown that B1 is contained in B: Similarly, B2 is contained in B: Since ðU; uU Þ is not in B2 ; and B2 is C 1 differentiable structure on M; there exists ðV; uV Þ in B2 such that ðV; uV Þ and ðU; uU Þ are not C 1 -compatible. Since ðV; uV Þ is in B2 ; and B2 is contained in B; ðV; uV Þ is in B: Since ðU; uU Þ is in B1 ; and B1 is contained in B; ðU; uU Þ is in B: Since ðU; uU Þ; ðV; uV Þ are in B; and B is a C 1 -differentiable structure on M; ðV; uV Þ and ðU; uU Þ are C1 -compatible, a contradiction. Case II: when there exists a coordinate chart ðU; uU Þ of M such that ðU; uU Þ is in B2 ; and ðU; uU Þ is not in B1 . This case is similar to the case I. Thus, in all cases, we get a contradiction. h From now on, we shall assume without further mention that in our differentiable manifold ðM; AÞ; M is a second countable topological space.

1.3 Examples of Smooth Manifolds Example 1.6 (i) Let G be a nonempty open subset of Rm : Let us take G for M: Since Rm is a Hausdorff second countable topological space, G with the induced topology is a Hausdorff second countable topological space. For every x in G; G is an open neighborhood of x: Since the identity mapping IdG : G ! G given by IdG ð xÞ ¼ x for every x in G is a homeomorphism from G onto G; G is an m-dimensional topological manifold. For A; let us take the singleton set fðG; IdG Þg: It is easy to observe that A satisﬁes the conditions 1 and 2 of Theorem 1.2 for every positive integer r: Hence, there exists a unique C1 -differentiable structure on G which contains fðG; IdG Þg: This C1 -differentiable structure on G is called the standard differentiable structure of G: Thus, every nonempty open subset of Rm is an example of a smooth manifold. (ii) Let M be an m-dimensional smooth manifold, whose topology is O; and differential structure is A: Let ðU; uU Þ 2 A: Since ðU; uU Þ 2 A; U is an open subset of M: Let O1 be the induced topology over U; that is, O1 ¼ fG : G U; and G 2 Og: Put A1 fðU; uU Þg. Since M is an m-dimensional smooth manifold, M is a Hausdorff space. Since M is a Hausdorff space, and U is a subspace of M; U with the induced topology is a Hausdorff space. Since M is an m-dimensional smooth manifold, M is a second countable space. Since M is a second countable space, and U is a subspace of M; U with the induced

1.3 Examples of Smooth Manifolds

15

topology is a second countable space. It is easy to observe that A satisﬁes the conditions 1 and 2 of Theorem 1.2 for every positive integer r: Hence, there exists a unique C 1 -differentiable structure on U which contains fðU; uU Þg: Thus for every admissible coordinate chart ðU; uU Þ of an m-dimensional smooth manifold M; U is an m-dimensional smooth manifold. Example 1.7 For m ¼ 1; 2; 3; . . .; by the m-dimensional unit sphere Sm ; we mean the set

x ; . . .x 1

mþ1

: x ; . . .x 1

mþ1

2R

mþ1

qﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ 2 2 1 mþ1 and ðx Þ þ þ ðx Þ ¼ 1 :

Since Rmþ1 is a Hausdorff second countable topological space, and Sm is a subset of Rmþ1 ; Sm with the induced topology is a Hausdorff second countable topological space. Here,

qﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ x1 ; x2 : x1 ; x2 2 R2 and ðx1 Þ2 þðx2 Þ2 ¼ 1 ;

qﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ 1 2 3 1 2 3 2 2 2 2 3 1 2 3 S x ; x ; x : x ; x ; x 2 R and ðx Þ þðx Þ þðx Þ ¼ 1 ; etc.

S1

Now we shall try to prove that the unit circle S1 is a 1-dimensional topological manifold. For this purpose, let us take any ða1 ; a2 Þ in S1 : We have to ﬁnd an open neighborhood G of ða1 ; a2 Þ, and a homeomorphism u : G ! uðGÞ such that uðGÞ is an open subset of R1 : Since ð0; 0Þ does not lie in S1 ; and ða1 ; a2 Þ is in S1 ; both of a1 ; a2 cannot be 0. So two cases arise: either a1 6¼ 0 or a2 6¼ 0: Case I: when a1 6¼ 0 Now two subcases arise: either 0\a1 or a1 \0: Subcase I: when 0\a1 . Let us take

x1 ; x2 : x1 ; x2 2 S1 and 0\x1

for G1 : Since ða1 ; a2 Þ 2 S1 and 0\a1 ; ða1 ; a2 Þ 2 G1 : Since

x1 ; x2 : x1 ; x2 2 S1 and 0\x1 ¼ x1 ; x2 : x1 ; x2 2 R2 and 0\x1 \ S1 ;

and fðx1 ; x2 Þ : ðx1 ; x2 Þ 2 R2 and 0\x1 gð¼ ð0; 1Þ ð1; 1ÞÞ is open in R2 ; fðx1 ; x2 Þ : ðx1 ; x2 Þ 2 S1 and 0\x1 gð¼ G1 Þ is open in S1 relative to the induced topology. Thus, G1 is open in S1 : Since ða1 ; a2 Þ 2 G1 ; and G1 is open in S1 ; G1 is an open neighborhood of ða1 ; a2 Þ: Since

16

1 Differentiable Manifolds

G1 ¼ x1 ; x2 : x1 ; x2 2 S1 and 0\x1

qﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ ¼ x1 ; x2 : ðx1 Þ2 þðx2 Þ2 ¼ 1 and 0\x1 ; it is clear that fx2 : ðx1 ; x2 Þ 2 G1 g is equal to the open interval ð1; 1Þ: Let us deﬁne a function p1 : G1 ! x2 : x1 ; x2 2 G as follows: for every ðx1 ; x2 Þ in G1 ; p1 x 1 ; x 2 ¼ x 2 : Clearly, p1 is a 1–1 function from G1 onto the open interval ð1; 1Þ: Further, since projection map is a continuous and open map, p1 : G1 ! ð1; 1Þ is a homeomorphism. Thus, we have shown that there exist an open neighborhood G1 of ða1 ; a2 Þ and a homeomorphism p1 : G1 ! ð1; 1Þ; where ð1; 1Þ is an open subset of R1 : Subcase II: when a1 \0. Proceeding as in subcase I, there exist an open neighborhood G2 x1 ; x2 : x1 ; x2 2 S1 and x1 \0 of ða1 ; a2 Þ and a homeomorphism p2 : G2 ! ð1; 1Þ deﬁned as: for every ðx1 ; x2 Þ in G2 ; p2 x 1 ; x 2 ¼ x 2 : Case II: when a2 6¼ 0 Now two subcases arise: either 0\a2 or a2 \0: Subcase I: when 0\a2 . Proceeding as above, there exist an open neighborhood G3

1 2 1 2 x ; x : x ; x 2 S1 and 0\x2

of ða1 ; a2 Þ and a homeomorphism p3 : G3 ! ð1; 1Þ

1.3 Examples of Smooth Manifolds

17

deﬁned as: for every ðx1 ; x2 Þ in G3 ; p3 x 1 ; x 2 ¼ x 1 : Subcase II: when a2 \0. Proceeding as above, there exist an open neighborhood G4

1 2 1 2 x ; x : x ; x 2 S1 and x2 \0

of ða1 ; a2 Þ and a homeomorphism p4 : G4 ! ð1; 1Þ deﬁned as: for every ðx1 ; x2 Þ in G4 ; p4 x 1 ; x 2 ¼ x 1 : Thus, we see that in all cases, there exist an open neighborhood G of ða1 ; a2 Þ, and a homeomorphism u : G ! uðGÞ; where uðGÞ is an open subset of R1 : Hence, by the deﬁnition of topological manifold, S1 is a 1-dimensional topological manifold. Here, we get four coordinate charts ðG1 ; p1 Þ; ðG2 ; p2 Þ; ðG3 ; p3 Þ; ðG4 ; p4 Þ: Let us take fðG1 ; p1 Þ; ðG2 ; p2 Þ; ðG3 ; p3 Þ; ðG4 ; p4 Þg for A: We want to show that A has the following two properties: 1. [fU : ðU; uU Þ is in Ag ¼ S1 ; 2. all pairs of members in A are C 1 -compatible. For 1: Here, [fU : ðU; uU Þ; is in Ag ¼ G1 [ G2 [ G3 [ G4 ¼ x1 ; x2 : x1 ; x2 2 S1 and 0\x1 [ x1 ; x2 : x1 ; x2 2 S1 and x1 \0 [ x1 ; x2 : x1 ; x2 2 S1 and 0\x2 [ x1 ; x2 : x1 ; x2 2 S1 and x2 \0 ¼ S1 :

For 2: Let us take ðG1 ; p1 Þ; ðG2 ; p2 Þ as a pair of members of A: Since G1 \ G2 ¼ ð

x1 ; x2 : x1 ; x2 2 S1 and 0\x1 \ x1 ; x2 : x1 ; x2 2 S1 and x1 \0 Þ

is an empty set, by the deﬁnition of C r -compatible, ðG1 ; p1 Þ; ðG2 ; p2 Þ are C r -compatible for every positive integer r: Hence, ðG1 ; p1 Þ; ðG2 ; p2 Þ are C 1 -compatible.

18

1 Differentiable Manifolds

Next, let us take ðG1 ; p1 Þ; ðG3 ; p3 Þ as a pair of members of A: Here, x1 ; x2 : ¼ x1 ; x2 :

G1 \ G3 ¼

1 2 x ; x 2 S1 and 0\x1 \ x1 ; x2 : x1 ; x2 2 S1 and 0\x2 1 2 x ; x 2 S1 ; 0\x1 and 0\x2 :

We want to prove that

p1 p1 : p3 ðG1 \ G3 Þ ! p1 ðG1 \ G3 Þ 3

is C 1 : By Lemma 1.1, p1 ðp1 3 Þ is a homeomorphism from open subset p3 ðG1 \ G3 Þ of R1 onto open subset p1 ðG1 \ G3 Þ of R1 : Now let us take any p3 ðxÞ in p3 ðG1 \ G3 Þ; where x is in G1 \ G3 : Since x is in G1 \ G3 ; and G1 \ G3 ¼ fðx1 ; x2 Þ : ðx1 ; x2 Þ 2 S1 ; 0\x1 and 0\x2 g; there exist real numbers x1 ; x2 such that x ¼ ðx1 ; x2 Þ 2 S1 ; 0\x1 and 0\x2 . Since 1 2 x ; x 2 S1 ¼ and 0\x2 ; x2 ¼

x1 ; x2

:

qﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ x1 ; x2 2 R2 and ðx1 Þ2 þðx2 Þ2 ¼ 1 ;

qﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ 1 ðx1 Þ2 : Since 0\x1 ; 0\x2 ; and ðx1 Þ2 þ ðx2 Þ2 ¼ 1; it follows

that 0\x1 \1; and 0\x2 \1: Hence, t p3 ðxÞ ¼ p3 ðx1 ; x2 Þ ¼ x1 2 ð0; 1Þ: Now it is clear that p3 ðG1 \ G3 Þ is equal to the open interval ð0; 1Þ: Further,

ðtÞ ¼ p1 p1 p1 p1 ðp3 ð xÞÞ ¼ p1 p1 3 3 3 ðp3 ð xÞÞ qﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ ¼ p1 ð xÞ ¼ p1 x1 ; x2 ¼ x2 ¼ 1 ðx1 Þ2 qﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ qﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ ¼ 1 ðp3 ðx1 ; x2 ÞÞ2 ¼ 1 ðp3 ð xÞÞ2 ¼ 1 t2

or,

qﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ p1 p1 ð t Þ ¼ 1 ðt Þ2 3

for every t in p3 ðG1 \ G3 Þð¼ ð0; 1ÞÞ: Therefore,

or,

0 1 t ðtÞ ¼ pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ ð2tÞ ¼ pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ p1 p1 3 2 1 t2 2 1t

0 t p1 p1 ðtÞ ¼ pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ ðt 2 ð0; 1ÞÞ: 3 1 t2

1.3 Examples of Smooth Manifolds

19

It follows that t 7! ðp1 ðp1 3 ÞÞðtÞ is a differentiable function on the open subset 1 p3 ðG1 \ G3 Þ of R : Also, since 0 t p1 p1 ðtÞ ¼ pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ : 3 1 t2 t is a continuous function, and t 7! pﬃﬃﬃﬃﬃﬃﬃ 1t2 : p3 ðG1 \ G3 Þ ! p1 ðG1 \ G3 Þ p1 p1 3

is C 1 : Similarly, we can show that

is C 2 ; etc. Thus,

p1 p1 : p3 ðG1 \ G3 Þ ! p1 ðG1 \ G3 Þ 3

p1 p1 : p3 ðG1 \ G3 Þ ! p1 ðG1 \ G3 Þ 3

is C r for every positive integer r: Hence,

: p3 ðG1 \ G3 Þ ! p1 ðG1 \ G3 Þ p1 p1 3

is C 1 : Thus, we have shown that the pair ðG1 ; p1 Þ; ðG3 ; p3 Þ are C1 -compatible. Similarly, we can show that all other pairs of members in A are C1 -compatible. This proves 2. Hence, by Theorem 1.2, there exists a unique C 1 -differentiable structure B on S1 which contains A: Thus, the ordered pair ðS1 ; BÞ is an example of a smooth manifold of dimension 1. Similarly, we can construct a C 1 -differentiable structure on S2 ; etc. In short, the unit sphere Sm is a smooth manifold of dimension m. Example 1.8 Let us deﬁne a relation over R2 fð0; 0Þg as follows: For every ðx1 ; x2 Þ; ðy1 ; y2 Þ in R2 fð0; 0Þg; by 1 2 1 2 x ;x y ;y we shall mean there exists a real number t such that 1 2 x ; x ¼ t y1 ; y2 : We shall try to show that is an equivalence relation over R2 fð0; 0Þg; that is, 1. ðx1 ; x2 Þ ðx1 ; x2 Þ for every ðx1 ; x2 Þ in R2 fð0; 0Þg; 2. if ðx1 ; x2 Þ ðy1 ; y2 Þ; then ðy1 ; y2 Þ ðx1 ; x2 Þ; 3. if ðx1 ; x2 Þ ðy1 ; y2 Þ; and ðy1 ; y2 Þ ðz1 ; z2 Þ then ðx1 ; x2 Þ ðz1 ; z2 Þ:

20

1 Differentiable Manifolds

For 1: Since ðx1 ; x2 Þ ¼ 1ðx1 ; x2 Þ; ðx1 ; x2 Þ ðx1 ; x2 Þ: For 2: Since ðx1 ; x2 Þ ðy1 ; y2 Þ; there exists a real number t such that ðx1 ; x2 Þ ¼ tðy1 ; y2 Þ: Since ðx1 ; x2 Þ; ðy1 ; y2 Þ are in R2 fð0; 0Þg; ðx1 ; x2 Þ and ðy1 ; y2 Þ are nonzero. Since ðx1 ; x2 Þ and ðy1 ; y2 Þ are nonzero, and ðx1 ; x2 Þ ¼ tðy1 ; y2 Þ; t is nonzero, and hence, ðy1 ; y2 Þ ¼ 1t ðx1 ; x2 Þ: Since ðy1 ; y2 Þ ¼ 1t ðx1 ; x2 Þ; ðy1 ; y2 Þ ðx1 ; x2 Þ: For 3: Since ðx1 ; x2 Þ ðy1 ; y2 Þ; there exists a real number t such that ðx1 ; x2 Þ ¼ tðy1 ; y2 Þ: Since ðy1 ; y2 Þ ðz1 ; z2 Þ; there exists a real number s such that ðy1 ; y2 Þ ¼ sðz1 ; z2 Þ: Since ðx1 ; x2 Þ ¼ tðy1 ; y2 Þ; and ðy1 ; y2 Þ ¼ sðz1 ; z2 Þ; ðx1 ; x2 Þ ¼ tðsðz1 ; z2 ÞÞ ¼ ðstÞðz1 ; z2 Þ; and hence, ðx1 ; x2 Þ ðz1 ; z2 Þ: Thus, we have shown that is an equivalence relation over R2 fð0; 0Þg: If ðx1 ; x2 Þ 6¼ ð0; 0Þ; then the equivalence class of ðx1 ; x2 Þ is given by

x1 ; x2

y1 ; y2 : y1 ; y2 x1 ; x2 1 2 1 2 1 2 ¼ y ; y : y ; y ¼ t x ; x for some real t

or,

x1 ; x2

¼ t x1 ; x2 : t 6¼ 0 :

We shall denote the quotient space

R2 fð0; 0Þg =

of all equivalence classes by P1 : Thus, P1 or, P1 ¼

1 2 1 2 x ;x : x ; x 6¼ ð0; 0Þ

1 2 t x ; x : t 6¼ 0 : x1 ; x2 6¼ ð0; 0Þ :

The topology of R2 fð0; 0Þg is the induced topology of the standard topology of R2 ; and the topology of the quotient space ðR2 fð0; 0ÞgÞ= ð¼ P1 Þ is the quotient topology. Thus, a subset G of P1 is open means the set [ x1 ; x2 : x1 ; x2 2 G is open in R2 fð0; 0Þg; that is, [ t x1 ; x2 : t 6¼ 0 : x1 ; x2 2 G

1.3 Examples of Smooth Manifolds

21

is open in R2 fð0; 0Þg; that is, either [ t x1 ; x2 : t 6¼ 0 : x1 ; x2 2 G or

[ t x1 ; x2 : t 6¼ 0 : x1 ; x2 2 G [fð0; 0Þg

is open in R2 : Since R2 is a Hausdorff space, its subspace R2 fð0; 0Þg is a Hausdorff space, and hence, its quotient space ðR2 fð0; 0ÞgÞ= ð¼ P1 Þ is a Hausdorff space. Since R2 is a second countable space, its subspace R2 fð0; 0Þg is a second countable space, and hence, its quotient space ðR2 fð0; 0ÞgÞ= ð¼ P1 Þ is a second countable space. Put 1 2 U1 x ;x : x1 6¼ 0 : We want to show that U1 is open in P2 : For this purpose, we must show that [ t x1 ; x2 : t 6¼ 0 : x1 ; x2 2 U1 is open in R2 fð0; 0Þg: Since (see Fig. 1.2), [ t x1 ; x2 : t 6¼ 0 : x1 ; x2 2 U1 ¼ [ t x1 ; x2 : t 6¼ 0 : x1 6¼ 0 ¼ R2 0; x2 : x2 is any real ; and R2 fð0; x2 Þ : x2 is any realg is open in R2 ; [fftðx1 ; x2 Þ : t 6¼ 0g : ½ðx1 ; x2 Þ 2 U1 g is open in R2 fð0; 0Þg: Thus, we have shown that U1 is open in P1 :

O

Fig. 1.2 Hausdorff space

O

22

1 Differentiable Manifolds

Now let us deﬁne a function u1 : U1 ! R as follows: for every ½ðx1 ; x2 Þ in U1 u1

x1 ; x2

x2 : x1

We shall try to show that 1. 2. 3. 4. 5.

u1 is well-deﬁned, u1 is 1–1, u1 is continuous, u1 ðU1 Þ is open in R; u1 1 : u1 ðU1 Þ ! U1 is continuous.

For 1: Let ½ðx1 ; x2 Þ ¼ ½ðy1 ; y2 Þ ; where ½ðx1 ; x2 Þ ; ½ðy1 ; y2 Þ are in U1 : Since ½ðx1 ; x2 Þ ; ½ðy1 ; y2 Þ are in U1 ; ðx1 ; x2 Þ; ðy1 ; y2 Þ are nonzero members of R2 ; and 2 2 x1 ; y1 are nonzero. Since x1 ; y1 are nonzero, xx1 ; yy1 are real numbers. We must prove that x2 y2 ¼ : x1 y1 Since ½ðx1 ; x2 Þ ¼ ½ðy1 ; y2 Þ ; ðx1 ; x2 Þ ðy1 ; y2 Þ: Further since ðx1 ; x2 Þ; ðy1 ; y2 Þ are nonzero members of R2 ; there exists a nonzero real number t such that

2

x x1

x1 ; x2 ¼ t y1 ; y2 ¼ ty1 ; ty2 :

So x1 ¼ ty1 ; and x2 ¼ ty2 : Since x1 ¼ ty1 ; x2 ¼ ty2 ; and x1 ; y1 ; t are nonzero, 2 2 ¼ tyty1 ¼ yy1 : This proves 1. For 2: Let u1 ð½ðx1 ; x2 Þ Þ ¼ u1 ð½ðy1 ; y2 Þ Þ: We have to prove that ½ðx1 ; x2 Þ ¼ ½ðy1 ; y2 Þ ; that is, there exists a real number t such that ðx1 ; x2 Þ ¼ tðy1 ; y2 Þ: Here, y2 x2 ¼ u 1 y1 ; y2 ¼ 1: ¼ u1 x1 ; x2 1 x y Case I: when x2 6¼ 0 2 2 2 2 Since xx1 ¼ yy1 ; and x2 6¼ 0; y2 6¼ 0; and hence, x1 ¼ xy2 y1 : So ðx1 ; x2 Þ ¼ xy2 ðy1 ; y2 Þ:

1.3 Examples of Smooth Manifolds

23

Case II: when x2 ¼ 0 2 2 1 1 Since xx1 ¼ yy1 ; and x2 ¼ 0; y2 ¼ 0: So ðx1 ; x2 Þ ¼ ðx1 ; 0Þ ¼ xy1 ðy1 ; 0Þ ¼ xy1 ðy1 ; y2 Þ:

Thus, we see that in all cases, there exists a real number t such that ðx1 ; x2 Þ ¼ tðy1 ; y2 Þ: This proves 2. For 3. Let us note that the equivalence class ½ðx1 ; x2 Þ of ðx1 ; x2 Þ is simply the straight line joining the origin O and ðx1 ; x2 Þ; but origin O deleted. Further, the slope ð¼ tan hÞ of this line is the u1 -image of ½ðx1 ; x2 Þ : Since tan h is a continuous function over ð p2 ; p2Þ; it is clear that u1 : U1 ! R

is a continuous function. For 4. It is clear that u1 ðU1 Þ ¼ ð1; þ1Þ ¼ R ; and R is an open set, so u1 ðU1 Þ is an open set of real numbers. For 5. In 4, we have seen that u1 ðU1 Þ ¼ R; so it remains to prove that u1 1 : R ! U1 is continuous. Since, for every m in R; ½ð1; mÞ is in U1 ; and u1 ð½ð1; mÞ Þ ¼ m1 ¼ m; u1 1 ðmÞ ¼ ½ð1; mÞ : Since m 7! ð1; mÞ is continuous, and ðx1 ; x2 Þ 7! ½ðx1 ; x2 Þ is continuous, their composite m 7! ½ð1; mÞ is continuous, that is, u1 1 is a continuous function. This proves 5. Similarly, we can show that 1 2 x ;x U2 : x2 6¼ 0 is an open subset of P1 ; and the function u2 : U2 ! R deﬁned as follows: for every ½ðx1 ; x2 Þ in U2 u2

x1 ; x2

x1 ; x2

is a homeomorphism from U2 onto R: Thus, the open subset U2 of P2 is homeomorphic to R:

24

1 Differentiable Manifolds

Since P1 is a Hausdorff space, and fU1 ; U2 g is an open cover of P1 , P1 is a 1-dimensional topological manifold. Here, we get two coordinate charts ðU1 ; u1 Þ; ðU2 ; u2 Þ: Let us take fðU1 ; u1 Þ; ðU2 ; u2 Þg: for A: We want to show that A has the following two properties: 1. U1 [ U2 ¼ P1 ; 2. all pairs of members in A are C r -compatible, for every r ¼ 1; 2; 3; . . .: For 1: It is clear from the construction of U1 and U2 : For 2: Let us take ðU1 ; u1 Þ; ðU2 ; u2 Þ as a pair of members in A: Here, U1 \ U2 ¼

x1 ; x2

: x1 6¼ 0; and x2 6¼ 0 :

Further u2 ðu1 Þ1 : u1 ðU1 \ U2 Þ ! u2 ðU1 \ U2 Þ is a continuous function. Also, for every m in u1 ðU1 \ U2 Þ; there exists ½ðx1 ; x2 Þ

in U1 \ U2 such that m ¼ u1 x1 ; x2 : 2

Since ½ðx1 ; x2 Þ is in U1 \ U2 ; x1 6¼ 0 and x2 6¼ 0: Now m ¼ u1 ð½ðx1 ; x2 Þ Þ ¼ xx1 , and hence, u2 ðu1 Þ1 ðmÞ ¼ u2 ðu1 Þ1 ðmÞ ¼ u2 ðu1 Þ1 u1 x1 ; x2 x1 1 ¼ 2¼ : ¼ u 2 x1 ; x2 m x Now since u1 ðU1 \ U2 Þ ¼ R f0g ¼ u2 ðU1 \ U2 Þ; ðu2 ðu1 Þ1 Þ : m 7! m1 is Cr for every r ¼ 1; 2; 3; . . .: This proves 2. Hence, by Theorem 1.2, there exists a unique C1 -differentiable structure B on P1 which contains A: Thus, the ordered pair ðP1 ; BÞ is an example of a smooth manifold of dimension 1. Similarly, we can construct a C 1 -differentiable structure on P2 ; etc. In short, we say that the projective space Pm is a smooth manifold of dimension m. Example 1.9 Let M be a 2-dimensional smooth manifold with differentiable structure A; and let N be a 3-dimensional smooth manifold with differentiable structure B:

1.3 Examples of Smooth Manifolds

25

Now we want to show that the Cartesian product M N; with the product topology, is a ð2 þ 3Þ-dimensional topological manifold. For this purpose, we must prove that 1. M N; with the product topology, is a Hausdorff topological space, 2. for every ðx; yÞ in M N; there exists an open neighborhood of ðx; yÞ in M N which is homeomorphic to some open subset of the Euclidean space R2þ3 (see Fig 1.3). For 1: Since M is a smooth manifold, the topology of M is Hausdorff. Similarly the topology of N is Hausdorff. Since the topologies of M and N are Hausdorff, their product topology is Hausdorff. This proves 1. For 2: Let us take any ðx; yÞ in M N; where x is in M; and y is in N: Since x is in M; and M is a 2-dimensional smooth manifold with differentiable structure A; there exists a coordinate chart ðU; uU Þ 2 A such that x is in U: Similarly, there exists a coordinate chart ðV; wV Þ 2 B such that y is in V: Since ðU; uU Þ 2 A; and A is a differentiable structure on M; uU is a homeomorphism from the open subset U of M onto the open subset uU ðUÞ of R2 : Similarly, wV is a homeomorphism from the open subset V of N onto the open subset wV ðVÞ of R3 : Since x is in U; and U is open in M; U is an open neighborhood of x: Similarly, V is an open neighborhood of y: Since U is an open neighborhood of x; and V is an open neighborhood of y; the Cartesian product U V is an open neighborhood of ðx; yÞ in M N: Since uU ðUÞ is open in R2 ; and wV ðVÞ is open in R3 ; their Cartesian product ðuU ðUÞÞ ðwV ðVÞÞ is open in R2 R3 ð¼ R2þ3 Þ: Now let us deﬁne a function ðuU wV Þ : U V ! ððuU ðU ÞÞ ðwV ðV ÞÞÞ as follows: for every ðx; yÞ in U V; ðuU wV Þðx; yÞ ðuU ð xÞ; wV ð yÞÞ: It remains to prove that ðuU wV Þ is a homeomorphism, that is,

Fig. 1.3 Hausdorff topological space

26

1 Differentiable Manifolds

1: ðuU wV Þ is 1 1; 2: ðuU wV Þ is onto; 3: ðuU wV Þ is continuous, 4: ðuU wV Þ1 is continuous: For 1: Let ðuU wV Þðx; yÞ ¼ ðuU wV Þðz; wÞ; where x; z are in U; and y; w are in V: We have to show that ðx; yÞ ¼ ðz; wÞ; that is, x ¼ z; and y ¼ w: Since ðuU ð xÞ; wV ð yÞÞ ¼ ðuU wV Þðx; yÞ ¼ ðuU wV Þðz; wÞ ¼ ðuU ðzÞ; wV ðwÞÞ; uU ð xÞ ¼ uU ðzÞ; and wV ð yÞ ¼ wV ðwÞ: Since uU is a homeomorphism, uU is 1–1. Since uU is 1–1, and uU ðxÞ ¼ uU ðzÞ; x ¼ z: Similarly y ¼ w: This proves 1. For 2: We have to prove that ðuU wV Þ : U V ! ððuU ðU ÞÞ ðwV ðV ÞÞÞ is onto. For this purpose, let us take any ðs; tÞ in ðuU ðUÞÞ ðwV ðVÞÞ: Now we have to ﬁnd a member of U V; whose ðuU wV Þ-image is ðs; tÞ: Since ðs; tÞ is in ðuU ðUÞÞ ðwV ðVÞÞ; s is in uU ðUÞ; and t is in wV ðVÞ: Since s is in uU ðUÞ; there exists an element x in U such that s ¼ uU ðxÞ. Similarly, there exists an element y in V such that t ¼ wV ðyÞ: Since x is in U; and y is in V; ðx; yÞ is in U V: Further ðuU wV Þðx; yÞ ¼ ðuU ð xÞ; wV ð yÞÞ ¼ ðs; tÞ: This proves 2. For 3: Since uU is a homeomorphism, uU is a continuous function, and hence, x 7! uU ðxÞ is continuous. Since, in the product topology, ðx; yÞ 7! x is continuous, and x 7! uU ðxÞ is continuous, their composite ðx; yÞ 7! uU ðxÞ is continuous. Similarly, ðx; yÞ 7! wV ðyÞ is continuous. Since ðx; yÞ 7! uU ðxÞ is continuous, and ðx; yÞ 7! wV ðyÞ is continuous, with respect to product topology, ðx; yÞ 7! ðuU ðxÞ; wV ðyÞÞð¼ ðuU wV Þðx; yÞÞ is continuous, and hence, ðuU wV Þ : U V ! ððuU ðU ÞÞ ðwV ðV ÞÞÞ is a continuous function. This proves 3. For 4: We have to prove that ðuU wV Þ1 : ððuU ðUÞÞ ðwV ðVÞÞÞ ! U V is continuous, that is, ðuU wV Þ : U V ! ððuU ðU ÞÞ ðwV ðV ÞÞÞ

1.3 Examples of Smooth Manifolds

27

is an open mapping. For this purpose, let us take any open subset G H of U V; where G is open in U; and H is open in V: We have to prove that ðuU wV ÞðG HÞ is open in ðuU ðUÞÞ ðwV ðVÞÞ. By the deﬁnition of ðuU wV Þ; ðuU wV ÞðG H Þ ¼ ðuU ðGÞ wV ðH ÞÞ: Since uU : U ! ðuU ðUÞÞ is a homeomorphism, and G is open in U; uU ðGÞ is an open subset of uU ðUÞ. Similarly, wV ðHÞ is an open subset of wV ðVÞ: Since uU ðGÞ is an open subset of uU ðUÞ; and wV ðHÞ is an open subset of wV ðVÞ; their Cartesian product uU ðGÞ wV ðHÞð¼ ðuU wV ÞðG HÞÞ is an open subset of the product space uU ðUÞ wV ðVÞ: This proves 4. Thus, ðuU wV Þ : U V ! ððuU ðU ÞÞ ðwV ðV ÞÞÞ is a homeomorphism. This proves 2. Hence, M N; with the product topology, is a ð2 þ 3Þ-dimensional topological manifold. Since M is a 2-dimensional smooth manifold, M is a second countable space. Similarly, N is a second countable space. Since M; N are second countable spaces, their product M N is a second countable space. Here, for every ðU; uU Þ 2 A; and for every ðV; wV Þ 2 B, we get the coordinate chart ðU V; ðuU wV ÞÞ of M N: Let us denote the collection fðU V; ðuU wV ÞÞ : ðU; uU Þ 2 A; ðV; wV Þ 2 Bg of coordinate charts of M N by C: We want to show that C has the following two properties: 1: [fU V : ðU V; ðuU wV ÞÞ 2 Cg ¼ M N; 2: all pairs of members in C are Cr -compatible, for every r ¼ 1; 2; 3; . . .: For 1: This is clear from the above discussion. ^ V; ^ ðu ^ w ^ ÞÞ as a pair of For 2: Let us take any ðU V; ðuU wV ÞÞ; ðU U V members in C: Here, ^ : ^ V \V ^ V ^ ¼ U \U ðU V Þ \ U Now we want to prove that the mapping

^ V ^ uU^ wV^ ðuU wV Þ1 : ðuU wV Þ ðU V Þ \ U ^ V ^ ! uU^ wV^ ðU V Þ \ U

is C 1 ; that is,

28

1 Differentiable Manifolds

^ ^ V \V uU^ wV^ ðuU wV Þ1 : ðuU wV Þ U \ U ^ ^ V \V ! uU^ wV^ U \ U

is C 1 : Here, ðU V; ðuU wV ÞÞ is in C; so ðU; uU Þ 2 A and ðV; wV Þ 2 B: ^ w ^ Þ 2 B: Since ðU; uU Þ 2 A; ðU; ^ u ^ Þ 2 A; and ^ u ^ Þ 2 A and ðV; Similarly, ðU; U V U r ^ u ^ Þ are C -compatible for every A is a differentiable structure, ðU; uU Þ and ðU; U ^ u ^ Þ are Cr -compatible for every positive positive integer r: Since ðU; uU Þ and ðU; U ^ u ^ Þ are C 1 -compatible. Since ðU; uU Þ and ðU; ^ u ^ Þ are integer r; ðU; uU Þ and ðU; U U 1 1 2 1 2 1 C -compatible, f ; f ; g ; g are C ; where, for every ðx1 ; x2 Þ in ^ R2 Þ; uU ðU \ UÞð

uU^ ðuU Þ1

^ ; x1 ; x2 f 1 x1 ; x2 ; f 2 x1 ; x2 ¼ y1 ; y2 2 uU^ U \ U

and 1 1 2 1 1 2 2 1 2 1 2 uU uU^ y ;y g y ;y ;g y ;y ¼ x ;x : Similarly, h1 ; h2 ; h3 ; k 1 ; k2 ; k3 ^ R3 Þ; wV ðV \ VÞð

are

C1 ;

where

for

every

ðx1 ; x2 ; x3 Þ

in

wV^ ðwV Þ1 x1 ; x2 ; x3 h1 x1 ; x2 ; x3 ; h2 x1 ; x2 ; x3 ; h3 x1 ; x2 ; x3 ^ ; ¼ y1 ; y2 ; y3 2 wV^ V \ V and

1 1 2 3 1 1 2 3 2 1 2 3 3 1 2 3 wV wV^ y ;y ;y k y ;y ;y ;k y ;y ;y ;k y ;y ;y ¼ x1 ; x2 ; x3 :

^ ^ ðV \ VÞÞð R2þ3 Þ: So Let us take any ðx1 ; x2 ; x3 ; x4 ; x5 Þ in ðuU wV ÞððU \ UÞ ^ such that ^ and x in V \ V there exist x in U \ U ðuU ð xÞ; wV ðx ÞÞ ¼ ðuU wV Þðx; x Þ ¼ x1 ; x2 ; x3 ; x4 ; x5 : Hence, ðuU wV Þ1 x1 ; x2 ; x3 ; x4 ; x5 ¼ ðx; x Þ; or

1.3 Examples of Smooth Manifolds

29

u U ð xÞ ¼ x1 ; x2 ; and wV ðx Þ ¼ x3 ; x4 ; x5 : Now uU^ wV^ ðuU wV Þ1 x1 ; x2 ; x3 ; x4 ; x5 ¼ uU^ wV^ ðuU wV Þ1 x1 ; x2 ; x3 ; x4 ; x5 ¼ uU^ wV^ ðx; x Þ ¼ uU^ ðuU Þ1 uU ð xÞ; wV^ ðwV Þ1 wV ðx Þ ¼ uU^ ðuU Þ1 uU ð xÞ; wV^ ðwV Þ1 wV ðx Þ ¼ uU^ ðuU Þ1 ðuU ð xÞÞ; wV^ ðwV Þ1 ðwV ðx ÞÞ ¼ uU^ ðuU Þ1 x1 ; x2 ; wV^ ðwV Þ1 x3 ; x4 ; x5 ¼ uU^ ðuU Þ1 x1 ; x2 ; wV^ ðwV Þ1 x3 ; x4 ; x5 ¼ f 1 x 1 ; x 2 ; f 2 x 1 ; x 2 ; h1 x 3 ; x 4 ; x 5 ; h2 x 3 ; x 4 ; x 5 ; h3 x 3 ; x 4 ; x 5 ¼ f 1 x 1 ; x 2 ; f 2 x 1 ; x 2 ; h1 x 3 ; x 4 ; x 5 ; h2 x 3 ; x 4 ; x 5 ; h3 x 3 ; x 4 ; x 5 : ^ ðV \ VÞÞð ^ R2þ3 Þ; Thus, for every ðx1 ; x2 ; x3 ; x4 ; x5 Þ in ðuU wV ÞððU \ UÞ uU^ wV^ ðuU wV Þ1 x1 ; x2 ; x3 ; x4 ; x5 ¼ f 1 x 1 ; x 2 ; f 2 x 1 ; x 2 ; h1 x 3 ; x 4 ; x 5 ; h2 x 3 ; x 4 ; x 5 ; h3 x 3 ; x 4 ; x 5 : Now we want to prove that F 1 : ðx1 ; x2 ; x3 ; x4 ; x5 Þ 7! f 1 ðx1 ; x2 Þ is C1 : Here, F ¼ f 1 a; where a : R5 ! R2 is deﬁned by aðx1 ; x2 ; x3 ; x4 ; x5 Þ ðx1 ; x2 Þ: So for x ¼ ðx1 ; x2 ; x3 ; x4 ; x5 Þ; 1

1 DF ð xÞ ¼ D f 1 a ð xÞ ¼ Df 1 ðað xÞÞ ððDaÞð xÞÞ or,

h 1 of DF 1 ð xÞ ¼ ox1 ðað xÞÞ h 1 of ¼ ox1 ðað xÞÞ

1 i 1 0 0 0 of ð a ð x Þ Þ ox2 0 1 00 1 i of ð a ð x Þ Þ 0 0 0 : 2 ox

0 0

30

1 Differentiable Manifolds

Hence,

DF 1 ¼ ðD1 f 1 Þ a

ðD2 f 1 Þ a

0

0

0 :

Since f 1 is C 1 ; ðD1 f 1 Þ is continuous. Since ðD1 f 1 Þ is continuous, and a is continuous, their composite ðD1 f 1 Þ a is continuous. Similarly, ðD2 f 1 Þ a is continuous. Since ðD1 f 1 Þ a; ðD2 f 1 Þ a are continuous, and

DF 1 ¼ ðD1 f 1 Þ a

ðD2 f 1 Þ a

0 0

0 ;

F 1 is C 1 : Thus, we have shown that ðx1 ; x2 ; x3 ; x4 ; x5 Þ 7! f 1 ðx1 ; x2 Þ is C 1 : Similarly, the functions ðx1 ; x2 ; x3 ; x4 ; x5 Þ 7! f 2 ðx1 ; x2 Þ; ðx1 ; x2 ; x3 ; x4 ; x5 Þ 7! 1 3 4 5 h ðx ; x ; x Þ;

x1 ; x2 ; x3 ; x4 ; x5 ! 7 h2 x3 ; x4 ; x5 ; x1 ; x2 ; x3 ; x4 ; x5 7! h3 x3 ; x4 ; x5

are C1 : Since ðx1 ; x2 ; x3 ; x4 ; x5 Þ 7! f 1 ðx1 ; x2 Þ; ðx1 ; x2 ; x3 ; x4 ; x5 Þ 7! f 2 ðx1 ; x2 Þ; 1 2 3 4 5 1 2 3 4 5 1 3 4 5 2 3 4 5 x1 ; x2 ; x3 ; x4 ; x5 7! h3 x3 ; x4 ; x5 ; x ; x ; x ; x ; x 7! h x ; x ; x ; x ; x ; x ; x ; x 7! h x ; x ; x are C1 ; and uU^ wV^ ðuU wV Þ1 x1 ; x2 ; x3 ; x4 ; x5 ¼ f 1 x 1 ; x 2 ; f 2 x 1 ; x 2 ; h1 x 3 ; x 4 ; x 5 ; h2 x 3 ; x 4 ; x 5 ; h3 x 3 ; x 4 ; x 5 ;

^ ^ V \V uU^ wV^ ðuU wV Þ1 : ðuU wV Þ U \ U ^ ^ V \V ! uU^ wV^ U \ U

is C 1 : Similarly, we can prove that ðuU^ wV^ Þ ðuU wV Þ1 is C 2 ; etc. Hence, the ^ V; ^ ðu ^ w ^ ÞÞ of members in C are Cr -compatible, pair ðU V; ðuU wV ÞÞ; ðU U V for every r ¼ 1; 2; 3; . . .: This proves 2. Hence, by Theorem 1.2, there exists a unique C 1 -differentiable structure D on M N which contains C: Thus, the ordered pair ðM N; DÞ is an example of a smooth manifold of dimension ð2 þ 3Þ. Similarly, if M is an m-dimensional smooth manifold, and N is an n-dimensional smooth manifold, then the Cartesian product M N; with the product topology, becomes an ðm þ nÞ-dimensional smooth manifold. In short, we say that the product manifold M N is an ðm þ nÞdimensional smooth manifold.

Chapter 2

Tangent Spaces

Above dimension three, Euclidean space loses its quality of being visible. So, in order to investigate higher-dimensional smooth manifolds, we need to generalize the concept of tangent line in the case of curves in two-dimensional Euclidean space, and tangent plane in the case of surfaces in three-dimensional Euclidean space, using abstract tools of algebra, and analysis. This generalized concept has been named the tangent space. Among many other concepts that facilitate the enquiry into the properties of higher-dimensional smooth manifolds as we shall see later, the notion of tangent space is foremost. It is possible to introduce tangent space in various ways. Because each method has its own merit, we shall introduce it in multiple ways in this chapter for deeper understanding. Remember that in the process of generalization, unlike curves and surfaces, smooth manifolds are devoid of any ambient space. It was in this sense a challenging task for early mathematicians.

2.1 Smooth Functions Deﬁnition Let M be an m-dimensional smooth manifold. Let f : M ! R be any function. Let p be an element of M. If, for every admissible coordinate chart ðU; uU Þ of M satisfying p 2 U, ðf ðuU Þ1 Þ : uU ðUÞ ! R is C 1 at the point uU ðpÞ in Rm , then we say that f is C 1 at p in M. Observe that, if f is C 1 at p in M, then f is continuous at p. Reason: Since M is an m-dimensional smooth manifold, and p is an element of M, there exists an admissible coordinate chart ðU; uU Þ of M satisfying p 2 U. Since f is C 1 at p in M, ðf ðuU Þ1 Þ : uU ðUÞ ! R is C1 at the point uU ðpÞ in Rm : Since ðf ðuU Þ1 Þ : uU ðUÞ ! R is C 1 at the point uU ðpÞ in Rm , ðf ðuU Þ1 Þ : uU ðUÞ ! R is continuous at the point uU ðpÞ. Since ðU; uU Þ is an admissible coordinate chart of M, uU : U ! uU ðUÞ is continuous. Since uU : U ! uU ðUÞ is continuous, and p 2 U, uU : U ! uU ðUÞ is continuous at p. Since uU : U ! uU ðUÞ is continuous at p, and ðf ðuU Þ1 Þ : uU ðUÞ ! R is continuous at the point uU ðpÞ, © Springer India 2014 R. Sinha, Smooth Manifolds, DOI 10.1007/978-81-322-2104-3_2

31

32

2 Tangent Spaces

their composite function ðf ðuU Þ1 Þ uU ð¼ f ððuU Þ1 uU Þ ¼ f Þ is continuous at p. Thus, f is continuous at p. Theorem 2.1 Let M be an m-dimensional smooth manifold. Let f : M ! R be any function. Let p be an element of M. If there exists an admissible coordinate chart ðU; uU Þ of M satisfying p 2 U such that ðf ðuU Þ1 Þ : uU ðUÞ ! R is C1 at the point uU ðpÞ in Rm , then f is C1 at p in M. Proof Let us take any admissible coordinate chart ðV; uV Þ of M satisfying p 2 V. We have to show that ðf ðuV Þ1 Þ : uV ðVÞ ! R is C1 at the point uV ðpÞ in Rm . Since

f ðuV Þ1 ¼ f ðuU Þ1 uU ðuV Þ1 ¼ f ðuU Þ1 uU ðuV Þ1 ¼ f ðuU Þ1 uU ðuV Þ1 ;

ðuU ðuV Þ1 Þ is C 1 at the point uV ðpÞ in Rm , and ðf ðuU Þ1 Þ is C1 at the point uU ðpÞ in Rm , their composite ðf ðuU Þ1 Þ ðuU ðuV Þ1 Þð¼ ðf ðuV Þ1 ÞÞ is C 1 at the point uV ðpÞ in Rm : Thus, we have shown that ðf ðuV Þ1 Þ : uV ðVÞ ! R is C 1 at the point uV ðpÞ in Rm : h Deﬁnition Let M be an m-dimensional smooth manifold. Let f : M ! R be any function. By f is C 1 on M (or f is smooth on M), we mean that f is C 1 at every point p in M. The set of all smooth functions f : M ! R on M is denoted by C 1 ðMÞ. Observe that, if f is C 1 on M, then f is a continuous function. Deﬁnition Let M be an m-dimensional smooth manifold, and N be an n-dimensional smooth manifold. Let f : M ! N be any continuous function. Let p be an element of M. If for every admissible coordinate chart ðU; uU Þ of M satisfying p 2 U, and for every admissible coordinate chart ðV; wV Þ of N satisfying f ðpÞ 2 V, each of the n component functions of the mapping wV f ðuU Þ1 : uU U \ f 1 ðVÞ ! wV ðVÞ is C 1 at the point uU ðpÞ in Rm (in short, the mapping wV f ðuU Þ1 : uU U \ f 1 ðV Þ ! wV ðV Þ is C 1 at the point uU ðpÞ in Rm ); then, we say that f is C 1 at p in M. As above we observe that if f : M ! N is C 1 at p in M, then f is continuous at p. Theorem 2.2 Let M be an m-dimensional smooth manifold, and N be an n-dimensional smooth manifold. Let f : M ! N be any continuous function. Let p

2.1 Smooth Functions

33

be an element of M. If there exist an admissible coordinate chart ðU; uU Þ of M satisfying p 2 U, and an admissible coordinate chart ðV; wV Þ of N satisfying f ðpÞ 2 V such that each of the n component functions of the mapping wV f ðuU Þ1 : uU U \ f 1 ðV Þ ! wV ðV Þ is C 1 at the point uU ðpÞ in Rm ; then, f is C1 at p in M. b, b ; u Þ of M satisfying p 2 U Proof Let us take any admissible coordinate chart ð U b U b . We have to b ; w Þ of N satisfying f ðpÞ 2 V and any admissible coordinate chart ð V b V show that

1 b \ f 1 V b b ! wb V wb f u b : ub U V U U V is C 1 at the point u b ðpÞ in Rm . Since U 1 1 wb f u b f ðuU Þ1 uU ub ¼ wb V U V U 1 f ðuU Þ1 uU u b ¼ wb V U 1 1 1 ðwV Þ wV f ðuU Þ uU u b ¼ wb V U 1 ðwV f Þ ðuU Þ1 uU u b ; ¼ wb ðwV Þ1 V

U

1

ðuU ðuU Þ1 Þ is C 1 at the point u b ðpÞ in Rm , wV ðf ðuU Þ Þ is C 1 at the point U

uU ðpÞ in Rm , and ðwb ðwV Þ1 Þ is C1 at the point wV ðf ðpÞÞ in Rn , their composite V

ðwb ðwV Þ1 Þ ðððwV f Þ ðuU Þ1 Þ ðuU ðu b Þ1 ÞÞ ð¼ wb ðf ðu b Þ1 ÞÞ is C 1 V U V U

at the point u b ðpÞ in Rm . U b \ f 1 ð V b ÞÞ ! w ð V b Þ is Thus, we have shown that wb ðf ðu b Þ1 Þ : u b ð U b V U U V m 1 C at the point u b ðpÞ in R . h U

Lemma 2.3 Let M be an m-dimensional smooth manifold, N be an n-dimensional smooth manifold, and K be a k-dimensional smooth manifold. Let p be an element of M. Let f : M ! N be C 1 at p in M, and g : N ! K be C 1 at f(p) in N. Then, g f is C1 at p in M. Proof We have to prove that g f is C1 at p in M. By Theorem 2.2, we must ﬁnd an admissible coordinate chart ðU; uU Þ of M satisfying p 2 U; and an admissible coordinate chart ðW; vw Þ of K satisfying ðg f ÞðpÞ 2 W such that the mapping

34

2 Tangent Spaces

vw ðg f Þ ðuU Þ1 : uU U \ ðg f Þ1 ðW Þ ! vw ðW Þ is C 1 at the point uU ðpÞ in Rm . Since p is in M, and M is an m-dimensional smooth manifold, there exists an admissible coordinate chart ðU; uU Þ of M satisfying p 2 U. Since f : M ! N, and p is in M, f(p) is in N. Since f(p) is in N, and N is an ndimensional smooth manifold, there exists an admissible coordinate chart ðV; wV Þ of N satisfying f ðpÞ 2 V. Since f : M ! N is C 1 at p in M, ðU; uU Þ is an admissible coordinate chart of M satisfying p 2 U, and ðV; wV Þ is an admissible coordinate chart of N satisfying f ðpÞ 2 V, the mapping wV f ðuU Þ1 : uU U \ f 1 ðV Þ ! wV ðV Þ is C 1 at the point uU ðpÞ in Rm . Similarly, there exists an admissible coordinate chart ðW; vw Þ of K satisfying gðf ðpÞÞ 2 W such that the mapping vw g ðwV Þ1 : wV V \ g1 ðW Þ ! vw ðW Þ is C 1 at the point wV ðf ðpÞÞ in Rn . Since wV ðf ðuU Þ1 Þ is C 1 at the point uU ðpÞ in Rm , and vw ðg ðwV Þ1 Þ is C 1 at the point wV ðf ðpÞÞ in Rn , their composite ðvw ðg ðwV Þ1 ÞÞ ðwV ðf ðuU Þ1 ÞÞð¼ vw ððg f Þ ðuU Þ1 ÞÞ is C 1 at the point uU ðpÞ in Rm : h Deﬁnition Let M be an m-dimensional smooth manifold, and N be an n-dimensional smooth manifold. Let p be an element of M, and let G be an open neighborhood of p in M. Let f : G ! N be any continuous function. If for every admissible coordinate chart ðU; uU Þ of M satisfying p 2 U, and for every admissible coordinate chart ðV; wV Þ of N satisfying f ðpÞ 2 V, each of the n component functions of the mapping wV f ðuU Þ1 : uU U \ G \ f 1 ðV Þ ! wV ðV Þ is C 1 at the point uU ðpÞ in Rm (in short, the mapping wV f ðuU Þ1 : uU U \ G \ f 1 ðV Þ ! wV ðV Þ is C 1 at the point uU ðpÞ in Rm ); then, we say that f is C 1 at p in M. Deﬁnition Let M be an m-dimensional smooth manifold, and N be an n-dimensional smooth manifold. Let G be a nonempty open subset of M. Let f : G ! N be any continuous function. By f is a smooth map from G to N, we mean that f is C 1 at every point p in G.

2.1 Smooth Functions

35

Deﬁnition Let M and N be m-dimensional smooth manifolds. Let f : M ! N be a 1–1, onto mapping. If f is a smooth map from M to N, and f 1 is a smooth map from N to M, then we say that f is a diffeomorphism from M onto N. We observe that if f is a diffeomorphism from M onto N, then f is a homeomorphism from M onto N. Deﬁnition Let M be an m-dimensional smooth manifold, and N be an m-dimensional smooth manifold. If there exists a function f : M ! N such that f is a diffeomorphism from M onto N, then we say that the manifolds M and N are isomorphic (or, diffeomorphic). Deﬁnition Let M be an m-dimensional smooth manifold. Let a and b be any real numbers such that a < b. Let c be a continuous function from the open interval ða; bÞ to M. Recall that the open interval ða; bÞ is a 1-dimensional smooth manifold. Hence, it is meaningful to say that c is a smooth map from ða; bÞ to M. If c is a smooth map from ða; bÞ to M, then we say that c is a parametrized curve in the manifold M. Deﬁnition Let M be an m-dimensional smooth manifold. Let p 2 M. Let c be a parametrized curve in the manifold M. If there exists a real number e [ 0 such that c is deﬁned on the open interval ðe; eÞ, and cð0Þ ¼ p, then we say that c is a parametrized curve in M through p. The set of all parametrized curves in M through p is denoted by Cp ðMÞ (or, simply Cp Þ. Example 2.4 In the Example 1.9, let us deﬁne p1 : M N ! M as follows: for every ðx; yÞ in M N, p1 ðx; yÞ ¼ x: We want to prove that p1 is a smooth function from the product manifold M N onto M. For this purpose, let us take any ðp; qÞ in M N. Now, there exists a coordinate chart ðU V; ðuU wV ÞÞ of the product manifold M N such that ðU; uU Þ 2 A; ðV; wV Þ 2 B, and ðp; qÞ is in U V. Hence, ðU; uU Þ is a coordinate chart of M such that pð¼ p1 ðp; qÞÞ is in U. We want to show that uU p1 ðuU wV Þ1 : ðuU wV ÞðU V Þ ! uU ðU Þ is C 1 at the point ðuU wV Þðp; qÞð¼ ðuU ðpÞ; wV ðqÞÞÞ in R2þ3 . Let us take any ðx1 ; x2 ; x3 ; x4 ; x5 Þ in ðuU wV ÞðU VÞ. So, there exist u in U, and v in V such that ðuU ðuÞ; wV ðvÞÞ ¼ ðuU wV Þðu; vÞ ¼ x1 ; x2 ; x3 ; x4 ; x5 : Hence,

ðuU wV Þ1 x1 ; x2 ; x3 ; x4 ; x5 ¼ ðu; vÞ;

36

2 Tangent Spaces

or uU ðuÞ ¼ x1 ; x2 ; wV ðvÞ ¼ x3 ; x4 ; x5 : So

uU p1 ðuU wV Þ1

x1 ; x2 ; x3 ; x4 ; x5 ¼ ðuU p1 Þ ðuU wV Þ1 x1 ; x2 ; x3 ; x4 ; x5 ¼ ðuU p1 Þðu; vÞ ¼ uU ðp1 ðu; vÞÞ ¼ uU ðuÞ ¼ x1 ; x2 :

Hence, ðuU p1 ðuU wV Þ1 Þ : ðx1 ; x2 ; x3 ; x4 ; x5 Þ 7! ðx1 ; x2 Þ, which is trivially C 1 . So, by Theorem 2.2, the natural projection p1 of the product manifold M N is a smooth map. Similarly, the natural projection p2 of the product manifold M N is a smooth map. Deﬁnition Let M be an m-dimensional smooth manifold. Let p be an element of M. By Cp1 ðMÞ (or, simply Cp1 Þ, we mean the collection of all real-valued functions f whose dom f is an open neighborhood of p in M, and for every admissible coordinate chart ðU; uU Þ of M satisfying p 2 U,

f ðuU Þ1 : uU ððdom f Þ \ U Þ ! R

is C 1 at the point uU ðpÞ in Rm . Observe that if f is in Cp1 ðMÞ, then f is continuous on some open neighborhood of p. Deﬁnition Let M be an m-dimensional smooth manifold. Let f : M ! Rk be any function. By f is smooth, we mean for every p in M there exists an admissible coordinate chart ðU; uU Þ of M satisfying p 2 U such that the composite function f ðuU Þ1 : uU ðUÞ ! Rk is smooth in the sense of ordinary calculus. Remember, uU ðUÞ is an open subset of Rm containing uU ðpÞ. Note 2.5 Let G be a nonempty open subset of Rm . We know that G is an mdimensional smooth manifold with the standard differentiable structure containing fðG; IdG Þg. Let f : G ! Rk be any smooth function in the sense just deﬁned. We want to show that f : G ! Rk is a smooth function in the sense of ordinary calculus. Since G is nonempty, there exists a in G. Since a is in G, and f : G ! Rk be any smooth function in the sense just deﬁned, there exists an admissible coordinate chart ðU; uU Þ of G satisfying a 2 U such that the composite function f ðuU Þ1 : uU ðUÞ ! Rk is smooth in the sense of ordinary calculus. Since ðU; uU Þ is in fðG; IdG Þg, U ¼ G, and uU ¼ IdG . Since U ¼ G, and uU ¼ IdG , f ðuU Þ1 ¼

2.1 Smooth Functions

37

f ðIdG Þ1 ¼ f IdG ¼ f , and uU ðUÞ ¼ IdG ðUÞ ¼ U ¼ G. Since uU ðUÞ ¼ G; and f ðuU Þ1 ¼ f , and f ðuU Þ1 : uU ðUÞ ! Rk is smooth in the sense of ordinary calculus, f : G ! Rk is smooth in the sense of ordinary calculus. Conversely, let f : G ! Rk be smooth in the sense of ordinary calculus. We want to show that f : G ! Rk is a smooth function in the sense just deﬁned. Let us take any a in G. We have to ﬁnd an admissible coordinate chart ðU; uU Þ of G satisfying a 2 U such that the composite function f ðuU Þ1 : uU ðUÞ ! Rk is smooth in the sense of ordinary calculus. Let us take ðG; IdG Þ for ðU; uU Þ. Hence, uU ðUÞ ¼ IdG ðUÞ ¼ U ¼ G; and f ðuU Þ1 ¼ f ðIdG Þ1 ¼ f IdG ¼ f . Since uU ðUÞ ¼ G; f ðuU Þ1 ¼ f , and f : G ! Rk is smooth in the sense of ordinary calculus, f ðuU Þ1 : uU ðUÞ ! Rk is smooth in the sense of ordinary calculus. Thus, we have shown that f : G ! Rk is smooth in the sense of ordinary calculus if and only if f : G ! Rk is a smooth function in the sense just deﬁned. Note 2.6 Let M be an m-dimensional smooth manifold. Let f : M ! Rk be any smooth function. Similar to the proof of Theorem 2.1, it can be shown that for every p in M, and for every admissible coordinate chart ðU; uU Þ of M satisfying p 2 U; the composite function f ðuU Þ1 : uU ðUÞ ! Rk is smooth in the sense of ordinary calculus.

2.2 Algebra of Smooth Functions Note 2.7 Let M be an m-dimensional smooth manifold. By C 1 ðMÞ, we mean the collection of all smooth functions f : M ! R. For every f, g in C 1 ðMÞ, we deﬁne ðf þ gÞ : M ! R as follows: for every x in M, ðf þ gÞð xÞ f ð xÞ þ gð xÞ: For every f in C1 ðMÞ, and for every real t, we deﬁne ðtf Þ : M ! R as follows: for every x in M, ðtf Þð xÞ tðf ð xÞÞ: It is clear that the set C 1 ðMÞ, together with vector addition, and scalar multiplication deﬁned as above, constitutes a real linear space. In short, C 1 ðMÞ is a real linear space. For every f, g in C 1 ðMÞ, we deﬁne ðf gÞ : M ! R as follows: for every x in M, ðf gÞð xÞ ðf ð xÞÞðgð xÞÞ:

38

2 Tangent Spaces

It is easy to see that 1. 2. 3. 4.

for for for for

every every every every

f ; g; h in C1 ðMÞ, ðf gÞ h ¼ f ðg hÞ, f ; g in C1 ðMÞ, f g ¼ g f , f ; g; h in C1 ðMÞ, ðf þ gÞ h ¼ f h þ g h, real t, and for every f ; g in C1 ðMÞ, tðf gÞ ¼ ðtf Þ g.

In short, C 1 ðMÞ is an algebra. Note 2.8 Let M be an m-dimensional smooth manifold. Let p 2 M. Let ðU; uU Þ be an admissible coordinate chart of M satisfying p 2 U. Let c be in Cp ðMÞ. Since c is in Cp ðMÞ, there exists a real number d [ 0 such that c is a smooth map from ðd; dÞ to M, and cð0Þ ¼ p. Since c is a smooth map from ðd; dÞ to M, and 0 is in ðd; dÞ, c is continuous at 0. Since c is continuous at 0, cð0Þ ¼ p, and U is an open neighborhood of p, c1 ðUÞ is an open neighborhood of 0, and hence, ðd; dÞ \ c1 ðUÞ is an open neighborhood of 0. Also since domðuU Þ ¼ U, and domðcÞ ¼ ðd; dÞ, domðuU cÞ ¼ ðd; dÞ \ c1 ðUÞ. Since ðuU cÞ : ððd; dÞ \ c1 ðUÞÞ ! Rm , ðd; dÞ \ c1 ðUÞ is an open neighborhood of 0, uU is a smooth, and c is a smooth map, ðuU cÞ0 ð0Þ exists and is a member of Rm . Deﬁnition Let M be an m-dimensional smooth manifold. Let p 2 M. Let c; c1 be in Cp ðMÞ. By c c1 , we shall mean that for every admissible coordinate chart ðU; uU Þ of M satisfying p 2 U, ðuU cÞ0 ð0Þ ¼ ðuU c1 Þ0 ð0Þ. Lemma 2.9 Let M be an m-dimensional smooth manifold. Let p 2 M. Let c; c1 be in Cp ðMÞ. If there exists an admissible coordinate chart ðU; uU Þ of M satisfying p 2 U, and ðuU cÞ0 ð0Þ ¼ ðuU c1 Þ0 ð0Þ, then c c1 . Proof Let us take any admissible coordinate chart ðV; wV Þ of M satisfying p 2 V: We have to prove that ðwV cÞ0 ð0Þ ¼ ðwV c1 Þ0 ð0Þ. 0 LHS ¼ ðwV cÞ0 ð0Þ ¼ wV ðuU Þ1 uU c ð0Þ 0 ¼ wV ðuU Þ1 ðuU cÞ ð0Þ 0 ¼ wV ðuU Þ1 ððuU cÞð0ÞÞ ðuU cÞ0 ð0Þ 0 ¼ wV ðuU Þ1 ððuU cÞð0ÞÞ ðuU c1 Þ0 ð0Þ 0 ¼ wV ðuU Þ1 ðuU ðcð0ÞÞÞ ðuU c1 Þ0 ð0Þ 0 ¼ wV ðuU Þ1 ðuU ð pÞÞ ðuU c1 Þ0 ð0Þ 0 ¼ wV ðuU Þ1 ðuU ðc1 ð0ÞÞÞ ðuU c1 Þ0 ð0Þ 0 ¼ wV ðuU Þ1 ððuU c1 Þð0ÞÞ ðuU c1 Þ0 ð0Þ ;

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39

and 0 RHS ¼ ðwV c1 Þ0 ð0Þ ¼ wV ðuU Þ1 uU c1 ð0Þ 0 ¼ wV ðuU Þ1 ðuU c1 Þ ð0Þ 0 ¼ wV ðuU Þ1 ððuU c1 Þð0ÞÞ ðuU c1 Þ0 ð0Þ : h

Hence, LHS = RHS.

Lemma 2.10 Let M be an m-dimensional smooth manifold. Let p 2 M. Then, is an equivalence relation over Cp ðM Þ. Proof Here, we must prove 1. for every c in Cp ðM Þ, c c, 2. if c c1 , then c1 c, 3. if c c1 and c1 c2 , then c c2 . For 1: Since p 2 M, and M is an m-dimensional smooth manifold, there exists an admissible coordinate chart ðU; uU Þ of M satisfying p 2 U: Since ðuU cÞ0 ð0Þ ¼ ðuU cÞ0 ð0Þ, by Lemma 2.9, c c: For 2: Let c c1 : We have to prove that c1 c: For this purpose, let us take any admissible coordinate chart ðU; uU Þ of M satisfying p 2 U. We must show that ðuU c1 Þ0 ð0Þ ¼ ðuU cÞ0 ð0Þ. Since c c1 , ðuU cÞ0 ð0Þ ¼ ðuU c1 Þ0 ð0Þ, and hence, ðuU c1 Þ0 ð0Þ ¼ ðuU cÞ0 ð0Þ. For 3: Let c c1 and c1 c2 . We have to prove that c c2 . For this purpose, let us take any admissible coordinate chart ðU; uU Þ of M satisfying p 2 U. We must show that ðuU cÞ0 ð0Þ ¼ ðuU c2 Þ0 ð0Þ. Since c c1 , ðuU cÞ0 ð0Þ ¼ ðuU c1 Þ0 ð0Þ. Since c1 c2 , ðuU c1 Þ0 ð0Þ ¼ ðuU c2 Þ0 ð0Þ. Since ðuU cÞ0 ð0Þ ¼ ðuU c1 Þ0 ð0Þ; and ðuU c1 Þ0 ð0Þ ¼ ðuU c2 Þ0 ð0Þ, ðuU cÞ0 ð0Þ ¼ ðuU c2 Þ0 ð0Þ. h Note 2.11 Let M be an m-dimensional smooth manifold. Let p 2 M. By the Lemma 2.10, the quotient set Cp ðMÞ= is the collection of all equivalence classes ½½c, where c is in Cp ðMÞ. Thus, Cp ðM Þ= ½½c : c 2 Cp ðM Þ where ½½c c1 2 Cp ðM Þ : c1 c : Intuitively, the Lemma 2.10 says that in Cp ðMÞ we will not distinguish between c1 and c whenever c1 c.

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Deﬁnition Let M be an m-dimensional smooth manifold. Let p 2 M. Let ðU; uU Þ be any admissible coordinate chart of M satisfying p 2 U. Let c; c1 be in Cp ðMÞ. If ½½c ¼ c1 , then c c1 , and hence, ðuU cÞ0 ð0Þ ¼ ðuU c1 Þ0 ð0Þ. This shows that ½½c 7! ðuU cÞ0 ð0Þ is a well-deﬁned function from Cp ðMÞ= to Rm . We shall denote this function by uU : Thus, uU : ðCp ðMÞ=Þ ! Rm is the function deﬁned as follows: for every ½½c in Cp ðMÞ= where c is in Cp ðMÞ, uU ð½½cÞ ðuU cÞ0 ð0Þ: Lemma 2.12 Let M be an m-dimensional smooth manifold. Let p 2 M. Let ðU; uU Þ be any admissible coordinate chart of M satisfying p 2 U. Then, the function uU : Cp ðM Þ= ! Rm is 1–1 and onto. Proof 1–1: Let uU ð½½cÞ ¼ uU ð½½c1 Þ, where c; c1 are in Cp ðMÞ. We have to prove that ½½c ¼ ½½c1 , that is, c c1 . Since ðuU cÞ0 ð0Þ ¼ uU ð½½cÞ ¼ uU ð½½c1 Þ ¼ ðuU c1 Þ0 ð0Þ, ðuU cÞ0 ð0Þ ¼ ðuU c1 Þ0 ð0Þ, and hence, by Lemma 2.9, c c1 . Onto: Let us take any v in Rm . We have to ﬁnd c in Cp ðMÞ such that U cÞð0Þ ¼ v; that is, uU ð½½cÞ ¼ v; that is, ðuU cÞ0 ð0Þ ¼ v; that is, limt!0 ðuU cÞðtÞðu t

U ðcð0ÞÞ U ðpÞ limt!0 uU ðcðtÞÞu ¼ v; that is, limt!0 uU ðcðtÞÞu ¼ v. Let us deﬁne a function t t m c1 : ð1; 1Þ ! R as follows: For every t in the open interval ð1; 1Þ,

c1 ðtÞ tv þ uU ð pÞ: Put c ðuU Þ1 c1 : We shall try to prove that 1. 2. 3. 4.

cð0Þ ¼ p; 0 is an interior point of the domain of c, c is a smooth map from ðd; dÞ to M for some d [ 0, U ðpÞ limt!0 uU ðcðtÞÞu ¼ v. t

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41

For 1: Here, LHS ¼ cð0Þ ¼ ðuU Þ1 c1 ð0Þ ¼ ðuU Þ1 ðc1 ð0ÞÞ ¼ ðuU Þ1 ð0v þ uU ð pÞÞ ¼ ðuU Þ1 ðuU ð pÞÞ ¼ p ¼ RHS: This proves 1. For 2: By the deﬁnition of c1 , the function c1 : ð1; 1Þ ! Rm is continuous. Since c1 : ð1; 1Þ ! Rm is continuous, and 0 is in the open interval ð1; 1Þ, c1 is continuous at 0. Since ðU; uU Þ is a coordinate chart of M, ðuU Þ1 : uU ðUÞ ! U is a 1–1, onto, continuous function, and uU ðUÞ is an open subset of Rm . Since p 2 U, uU ðpÞ 2 uU ðUÞ. Since c1 ð0Þ ¼ 0v þ uU ðpÞ ¼ uU ðpÞ 2 uU ðUÞ, and ðuU Þ1 : uU ðUÞ ! U is continuous, ðuU Þ1 : uU ðUÞ ! U is continuous at the point c1 ð0Þ. Since c1 ð0Þ 2 uU ðUÞ, and uU ðUÞ is an open subset of Rm , uU ðUÞ is an open neighborhood of c1 ð0Þ. Since c1 : ð1; 1Þ ! Rm is continuous, uU ðUÞ is an open neighborhood of c1 ð0Þ, there exists d [ 0 such that d\1 and, for every t in ðd; dÞ, we have c1 ðtÞ 2 uU ðUÞ, and hence, cðtÞ ¼ ððuU Þ1 c1 ÞðtÞ ¼ ðuU Þ1 ðc1 ðtÞÞ 2 U. Since cðtÞ 2 U for every t in ðd; dÞ, it follows that ðd; dÞ is a subset of the domðcÞ. Hence, 0 is an interior point of the domain of the function c. This proves 2. For 3: In 2, we have seen that c is deﬁned over ðd; dÞ. Now, it remains to be proved that c is a smooth map from ðd; dÞ to M; that is, c is C 1 at every point t in ðd; dÞ. Here, c is deﬁned over ðd; dÞ, so cðtÞ is meaningful for every t in ðd; dÞ. Since cðtÞ is meaningful for every t in ðd; dÞ, and cðtÞ ¼ ðuU Þ1 c1 ðtÞ ¼ ðuU Þ1 ðc1 ðtÞÞ ¼ ðuU Þ1 ðtv þ uU ð pÞÞ; ðuU Þ1 ðtv þ uU ðpÞÞ is meaningful for every t in ðd; dÞ. Since ðU; uU Þ is a coordinate chart of M, ðuU Þ1 : uU ðUÞ ! U. Since ðuU Þ1 : uU ðUÞ ! U, and ðuU Þ1 ðtv þ uU ðpÞÞ is meaningful for every t in ðd; dÞ, ðuU Þ1 ðtv þ uU ðpÞÞ is in U for every t in ðd; dÞ. Since ðuU Þ1 ðtv þ uU ðpÞÞ is in U for every t in ðd; dÞ, and cðtÞ ¼ ðuU Þ1 ðtv þ uU ðpÞÞ, cðtÞ is in U for every t in ðd; dÞ: Thus, c maps ðd; dÞ to U: Now, it remains to be proved that c is a smooth map from ðd; dÞ to M; that is, c is C 1 at every point t in ðd; dÞ: For this purpose, let us ﬁx any t in ðd; dÞ: Since c1 is C1 at t; c ¼ ðuU Þ1 c1 ; c maps ðd; dÞ to U; and t is in ðd; dÞ; uU c is C 1 at t: Since uU c is C1 at t; and ðU; uU Þ is an admissible coordinate chart of M satisfying cðtÞ 2 U, by Theorem 2.2, c is C1 at t in ðd; dÞ: This proves 3. From 1, 2, and 3, we ﬁnd that c is a parametrized curve in M through p; that is, c is in Cp ðMÞ.

42

2 Tangent Spaces uU ðcðtÞÞ uU ðpÞ u ðððuU Þ1 c1 ÞðtÞÞ uU ðpÞ c ðtÞ uU ðpÞ ¼ lim U ¼ lim 1 t!0 t!0 t t t ðtv þ uU ðpÞÞ uU ðpÞ ¼ lim ¼ v ¼ RHS: t!0 t

4. LHS ¼ lim t!0

This proves that uU : ðCp ðMÞ=Þ ! Rm is onto.

h

Note 2.13 Let M be an m-dimensional smooth manifold. Let p 2 M: Let ðU; uU Þ be any admissible coordinate chart of M satisfying p 2 U: Since uU : ðCp ðMÞ=Þ ! Rm is 1–1 onto, ðuU Þ1 : Rm ! ðCp ðMÞ=Þ exists, and is 1–1 onto. This fact allows us to deﬁne a binary operation + over Cp ðMÞ= as follows: For every ½½c; ½½c1 in Cp ðMÞ= where c; c1 are in Cp ðMÞ; by ½½c þ ½½c1 , we mean ðuU Þ1 ðuU ð½½cÞ þ uU ð½½c1 ÞÞ: Next, for every ½½c in Cp ðMÞ= where c is in Cp ðMÞ; and for every real t; by t½½c, we mean ðuU Þ1 ðtðuU ðcÞÞÞ: Hence, uU ð½½c þ ½½c1 Þ ¼ uU ð½½cÞ þ uU ð½½c1 Þ; and uU ðt½½cÞ ¼ tðuU ð½½cÞÞ: Now, we want to verify that Cp ðMÞ= is a real linear space: 1. þ is associative: Take any ½½c; ½½c1 ; ½½c2 in Cp ðMÞ= where c; c1 ; c2 are in Cp ðMÞ: We have to show that ð½½c þ ½½c1 Þ þ ½½c2 ¼ ½½c þ ð½½c1 þ ½½c2 Þ: LHS ¼ ð½½c þ ½½c1 Þ þ ½½c2 ¼ ðuU Þ1 ðuU ð½½cÞ þ uU ð½½c1 ÞÞ þ ½½c2 ¼ ðuU Þ1 uU ðuU Þ1 ðuU ð½½cÞ þ uU ð½½c1 ÞÞ þ uU ð½½c2 Þ ¼ ðuU Þ1 ððuU ð½½cÞ þ uU ð½½c1 ÞÞ þ uU ð½½c2 ÞÞ ¼ ðuU Þ1 ðuU ð½½cÞ þ ðuU ð½½c1 Þ þ uU ð½½c2 ÞÞÞ; RHS ¼ ½½c þ ð½½c1 þ ½½c2 Þ ¼ ½½c þ ðuU Þ1 ðuU ð½½c1 Þ þ uU ð½½c2 ÞÞ ¼ ðuU Þ1 uU ð½½cÞ þ uU ðuU Þ1 ðuU ð½½c1 Þ þ uU ð½½c2 ÞÞ ¼ ðuU Þ1 ðuU ð½½cÞ þ ðuU ð½½c1 Þ þ uU ð½½c2 ÞÞÞ: Hence, LHS ¼ RHS: 2. Existence of zero element: Here uU : ðCp ðMÞ= Þ ! Rm is 1–1 onto, and 0 2 Rm ; so ðuU Þ1 ð0Þ is in ðCp ðMÞ= Þ: We shall try to prove that ðuU Þ1 ð0Þ

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43

serves the purpose of zero element. For this purpose, let us take any ½½c in Cp ðMÞ= where c is in Cp ðMÞ: We have to show that ðuU Þ1 ð0Þ þ ½½c ¼ ½½c: LHS ¼ ðuU Þ1 ð0Þ þ ½½c ¼ ðuU Þ1 uU ðuU Þ1 ð0Þ þ uU ð½½cÞ ¼ ðuU Þ1 ð0 þ uU ð½½cÞÞ ¼ ðuU Þ1 ðuU ð½½cÞÞ ¼ ½½c ¼ RHS: 3. Existence of negative element: Let us take any ½½c in Cp ðMÞ= where c is in Cp ðMÞ: Since ½½c is in Cp ðMÞ=, and uU : ðCp ðMÞ= Þ ! Rm ; uU ð½½cÞ is in Rm ; and hence, ðuU ð½½cÞÞ is in Rm : So ðuU Þ1 ððuU ð½½cÞÞÞ is in ðCp ðMÞ=Þ: We have to show that ðuU Þ1 ððuU ð½½cÞÞÞ þ ½½c ¼ ðuU Þ1 ð0Þ: LHS ¼ ðuU Þ1 ððuU ð½½cÞÞÞ þ ½½c ¼ ðuU Þ1 uU ðuU Þ1 ððuU ð½½cÞÞÞ þ uU ð½½cÞ ¼ ðuU Þ1 ðððuU ð½½cÞÞÞ þ uU ð½½cÞÞ ¼ ðuU Þ1 ð0Þ ¼ RHS:

4. þ is commutative: Take any ½½c; ½½c1 in Cp ðMÞ= where c; c1 are in Cp ðMÞ: We have to show that ½½c þ ½½c1 ¼ ½½c1 þ ½½c: LHS ¼ ½½c þ ½½c1 ¼ ðuU Þ1 ðuU ð½½cÞ þ uU ð½½c1 ÞÞ ¼ ðuU Þ1 ðuU ð½½c1 Þ þ uU ð½½cÞÞ ¼ ½½c1 þ ½½c ¼ RHS: 5. Take any ½½c; ½½c1 in Cp ðMÞ= where c; c1 are in Cp ðMÞ: Take any real number t: We have to prove that tð½½c þ ½½c1 Þ ¼ t½½c þ t½½c1 : LHS ¼ tð½½c þ ½½c1 Þ ¼ ðuU Þ1 ðtðuU ð½½c þ ½½c1 ÞÞÞ ¼ ðuU Þ1 t uU ðuU Þ1 ðuU ð½½cÞ þ uU ð½½c1 ÞÞ ¼ ðuU Þ1 ðtðuU ð½½cÞ þ uU ð½½c1 ÞÞÞ;

RHS ¼ t½½c þ t½½c1 ¼ ðuU Þ1 ðtðuU ð½½cÞÞÞ þ ðuU Þ1 t uU ½½c1 :

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So it remains to be proved that ðuU Þ1 ðtðuU ð½½cÞ þ uU ð½½c1 ÞÞÞ ¼ ðuU Þ1 ðtðuU ð½½cÞÞÞ þ ðuU Þ1 ðtðuU ð½½c1 ÞÞÞ that is, tðuU ð½½cÞ þ uU ð½½c1 ÞÞ ¼ uU ðuU Þ1 ðtðuU ð½½cÞÞÞ þ ðuU Þ1 ðtðuU ð½½c1 ÞÞÞ : RHS ¼ uU ðuU Þ1 ðtðuU ð½½cÞÞÞ þ ðuU Þ1 ðtðuU ð½½c1 ÞÞÞ ¼ uU ðuU Þ1 ðtðuU ð½½cÞÞÞ þ uU ðuU Þ1 ðtðuU ð½½c1 ÞÞÞ ¼ tðuU ð½½cÞÞ þ tðuU ð½½c1 ÞÞ ¼ tðuU ð½½cÞ þ uU ð½½c1 ÞÞ ¼ LHS: 6. Take any ½½c in Cp ðMÞ= where c is in Cp ðMÞ: Take any real numbers s; t: We have to prove that ðs þ tÞ½½c ¼ s½½c þ t½½c: LHS ¼ ðs þ tÞ½½c ¼ ðuU Þ1 ððs þ tÞðuU ð½½cÞÞÞ ¼ ðuU Þ1 ðsðuU ð½½cÞÞ þ tðuU ð½½cÞÞÞ; RHS ¼ s½½c þ t½½c ¼ ðuU Þ1 ðuU ðs½½cÞ þ uU ðt½½cÞÞ: So it remains to be proved that ðuU Þ1 ðsðuU ð½½cÞÞ þ tðuU ð½½cÞÞÞ ¼ ðuU Þ1 ðuU ðs½½cÞ þ uU ðt½½cÞÞ; that is, sðuU ð½½cÞÞ þ tðuU ð½½cÞÞ ¼ uU ðs½½cÞ þ uU ðt½½cÞ: RHS ¼ uU ðs½½cÞ þ uU ðt½½cÞ ¼ sðuU ð½½cÞÞ þ tðuU ð½½cÞÞ ¼ LHS: 7. Take any ½½c in Cp ðMÞ= where c is in Cp ðMÞ: Take any real numbers s; t: We have to prove that ðstÞ½½c ¼ sðt½½cÞ:

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45

LHS ¼ ðstÞ½½c ¼ ðuU Þ1 ððstÞðuU ð½½cÞÞÞ; RHS ¼ sðt½½cÞ ¼ ðuU Þ1 ðsðuU ðt½½cÞÞÞ ¼ ðuU Þ1 ðsðtðuU ð½½cÞÞÞÞ ¼ ðuU Þ1 ððstÞðuU ð½½cÞÞÞ: Hence, LHS ¼ RHS: 8. Take any ½½c in Cp ðMÞ= where c is in Cp ðMÞ: We have to prove that 1½½c ¼ ½½c: LHS ¼ 1½½c ¼ ðuU Þ1 ð1ðuU ð½½cÞÞÞ ¼ ðuU Þ1 ðuU ð½½cÞÞ ¼ ½½c ¼ RHS: h Thus, we have shown that Cp ðMÞ= is a real linear space. Since Cp ðMÞ= is a real linear space, uU : ðCp ðMÞ= Þ ! Rm is 1–1, onto, uU ð½½c þ ½½c1 Þ ¼ uU ð½½cÞ þ uU ð½½c1 Þ; and uU ðt½½cÞ ¼ tðuU ð½½cÞÞ; so uU is an isomorphism between real linear space Cp ðMÞ= and the real linear space Rm : Further since the dimension of Rm is m; Cp ðMÞ= is a real linear space of dimension m: Conclusion: Let M be an m-dimensional smooth manifold. Let p be an element of M: Then, Cp ðMÞ= is isomorphic to Rm for some vector addition, and scalar multiplication over Cp ðMÞ=: Deﬁnition Let M be an m-dimensional smooth manifold. Let p be an element of M: The set Cp ðMÞ= is denoted by Tp M or Tp ðMÞ (see Fig. 2.1). Note 2.14 Let M be a 2-dimensional smooth manifold. Let p 2 M: Let ðU; uU Þ be any admissible coordinate chart of M satisfying p 2 U: Let us deﬁne functions u1 : U ! R; u2 : U ! R as follows: for every x in U;

Fig. 2.1 Dimensional smooth manifold

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uU ð xÞ u1 ð xÞ; u2 ð xÞ : (In short, we say that u1 ; u2 are the component functions of uU :) We shall try to prove that u1 is in Cp1 ðMÞ: By the deﬁnition of Cp1 ðMÞ; it remains to be proved that for every admissible coordinate chart ðV; uV Þ of M satisfying p 2 V, u1 ðuV Þ1 : uV ðU \ V Þ ! R is C 1 at the point uV ðpÞ in R2 : Since M is a 2-dimensional smooth manifold, and ðU; uU Þ; ðV; uV Þ are admissible coordinate charts of M satisfying p 2 U \ V, by the deﬁnition of smooth manifold, uU ðuV Þ1 : uV ðU \ V Þ ! R is C 1 at the point uV ðpÞ in R2 : For every ðy1 ; y2 Þ in uV ðU \ VÞ; we have ðuV Þ1 ðy1 ; y2 Þ 2 U \ V U; and hence, uU ðuV Þ1 y1 ; y2 ¼ uU ðuV Þ1 y1 ; y2 ¼ u1 ðuV Þ1 y1 ; y2 ; u2 ðuV Þ1 y1 ; y2 ¼ u1 ðuV Þ1 y1 ; y2 ; u2 ðuV Þ1 y1 ; y2 So ðu1 ðuV Þ1 Þ; ðu2 ðuV Þ1 Þ are the component functions of ðuU ðuV Þ1 Þ: Next, since ðuU ðuV Þ1 Þ is C 1 at the point uV ðpÞ in R2 ; its component functions ðu1 ðuV Þ1 Þ; ðu2 ðuV Þ1 Þ are C 1 at the point uV ðpÞ in R2 : This proves that u1 is in Cp1 ðMÞ: Similarly, u2 is in Cp1 ðMÞ: Theorem 2.15 Let M be an m-dimensional smooth manifold. Let p be an element of M: Let be a relation over Cp1 ðMÞ deﬁned as follows: for every f and g in 1 Cp ðMÞ; by f g; we shall mean that there exists an open neighborhood H of p such that f ðxÞ ¼ gðxÞ for every x in H: Then, is an equivalence relation over Cp1 ðMÞ: Proof Here, we must prove 1. for every f in Cp1 ðMÞ; f f ; 2. if f g, then g f ; 3. if f g; and g h, then f h:

2.2 Algebra of Smooth Functions

47

For 1. Let us take any f in Cp1 ðMÞ: So, by the deﬁnition of Cp1 ðMÞ; dom f is an open neighborhood of p in M: Since dom f is an open neighborhood of p in M; and f ðxÞ ¼ f ðxÞ for every x in dom f , by the deﬁnition of ; f f : This proves 1. For 2. Let f g where f and g are in Cp1 ðMÞ: So, by the deﬁnition of ; there exists an open neighborhood H of p such that f ðxÞ ¼ gðxÞ for every x in H: Hence, gðxÞ ¼ f ðxÞ for every x in H; where H is an open neighborhood of p: Therefore, by the deﬁnition of , g f , this proves 2. For 3. Let f g; and g h; where f ; g; h are in Cp1 ðMÞ: Since f g, by the deﬁnition of ; there exists an open neighborhood H of p such that f ðxÞ ¼ gðxÞ for every x in H: Similarly, there exists an open neighborhood K of p such that gðxÞ ¼ hðxÞ for every x in K: Since H; K are open neighborhoods of p; H \ K is an open neighborhood of p: Also we have f ðxÞ ¼ hðxÞ for every H \ K: Hence, by the deﬁnition of ; f h: This proves 3. Hence, is an equivalence relation over Cp1 ðMÞ:

h

Deﬁnition Let M be an m-dimensional smooth manifold. Let p be an element of M: From Theorem 2.15, the relation is an equivalence relation over Cp1 ðMÞ: The quotient set Cp1 ðMÞ= ; of all equivalence classes is denoted by F p ðMÞ (or, simply F p Þ: Thus, n o F p ðM Þ ½ f : f 2 Cp1 ðM Þ where

n o ½ f g : g 2 Cp1 ðM Þ and g f :

Intuitively, Theorem 2.15 says that in Cp1 ðMÞ we will not distinguish between f and g whenever f g: Deﬁnition Let M be an m-dimensional smooth manifold. Let p be an element of M: Let f ; g be in Cp1 ðMÞ: By the sum f þ g, we mean the function whose domain is ðdom f Þ \ ðdom gÞ; and for every x in ðdom f Þ \ ðdom gÞ; ðf þ gÞð xÞ f ð xÞ þ gð xÞ: By the product f g, we mean the function whose domain is ðdom f Þ \ðdom gÞ; and for every x in ðdom f Þ \ðdom gÞ; ðf gÞð xÞ f ð xÞ gð xÞ: For any real t; by the scalar multiple tf , we mean the function whose domain is dom f ; and for every x in dom f ;

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2 Tangent Spaces

ðtf Þð xÞ tðf ð xÞÞ: Theorem 2.16 Let M be an m-dimensional smooth manifold. Let p be an element of M: If f ; g are in Cp1 ðMÞ, then f þ g is in Cp1 ðMÞ: Proof Here, let f ; g be in Cp1 ðMÞ: We must prove: 1. domðf þ gÞ is an open neighborhood of p in M; 2. for every admissible coordinate chart ðU; uU Þ of M satisfying p 2 U, ðf þ gÞ ðuU Þ1 : uU ððdomðf þ gÞÞ \ U Þ ! R is C 1 at the point uU ðpÞ in Rm : For 1. Since f ; g are in Cp1 ðMÞ; dom f and dom g are open neighborhoods of p in M; and hence, their intersection ðdom f Þ \ðdom gÞ is an open neighborhood of p in M: Since ðdom f Þ \ðdom gÞ is an open neighborhood of p in M; and domain of f þ g is ðdom f Þ \ðdom gÞ; domðf þ gÞ is an open neighborhood of p in M: This proves 1. For 2. Let us take any admissible coordinate chart ðU; uU Þ of M satisfying p 2 U: Since f is in Cp1 ðMÞ, by the deﬁnition of Cp1 ðMÞ;

f ðuU Þ1 : uU ððdom f Þ \ U Þ ! R

is C 1 at the point uU ðpÞ in Rm : Similarly,

g ðuU Þ1 : uU ððdom gÞ \ U Þ ! R

is C 1 at the point uU ðpÞ in Rm : Further, since

f ðuU Þ1 : uU ððdom f Þ \ U Þ ! R

and

g ðuU Þ1 : uU ððdom gÞ \ U Þ ! R;

their sum is the function

f ðuU Þ1 þ g ðuU Þ1 : ðuU ððdom f Þ \ U ÞÞ \ ðuU ððdom gÞ \ U ÞÞ ! R;

2.2 Algebra of Smooth Functions

49

and is C 1 at the point uU ðpÞ in Rm : Now, since dom

f ðuU Þ1 þ g ðuU Þ1 ¼ ðuU ððdom f Þ \ U ÞÞ \ ðuU ððdom gÞ \ U ÞÞ ¼ uU ðððdom f Þ \ U Þ \ ððdom gÞ \ U ÞÞ ¼ uU ðððdom f Þ \ ðdom gÞÞ \ U Þ ¼ uU ððdomðf þ gÞÞ \ U Þ ¼ dom ðf þ gÞ ðuU Þ1

and, for every x in dom ðf þ gÞ ðuU Þ1 ; f ðuU Þ1 þ g ðuU Þ1 ð xÞ ¼ f ðuU Þ1 ð xÞ þ g ðuU Þ1 ð xÞ ¼ f ðuU Þ1 ð xÞ þ g ðuU Þ1 ð xÞ ¼ ðf þ gÞ ðuU Þ1 ð xÞ ¼ ðf þ gÞ ðuU Þ1 ð xÞ

so,

f ðuU Þ1 þ g ðuU Þ1 ¼ ðf þ gÞ ðuU Þ1 :

Since ðf ðuU Þ1 Þ þ ðg ðuU Þ1 Þ ¼ ðf þ gÞ ðuU Þ1 ; and ðf ðuU Þ1 Þ þ ðg ðuU Þ1 Þ is C 1 at the point uU ðpÞ in Rm ; ðf þ gÞ uU Þ1 is C1 at the point uU ðpÞ in Rm : This proves 2. Hence, f þ g is in Cp1 ðMÞ: h Theorem 2.17 Let M be an m-dimensional smooth manifold. Let p be an element of M: If f ; g are in Cp1 ðMÞ, then f g is in Cp1 ðMÞ: Proof Its proof is quite similar to the proof of Theorem 2.16.

h

Theorem 2.18 Let M be an m-dimensional smooth manifold. Let p be an element of M: If t is a real, and f is in Cp1 ðMÞ, then tf is in Cp1 ðMÞ: Proof Its proof is similar to the proof of Theorem 2.16.

h

Theorem 2.19 Let M be an m-dimensional smooth manifold. Let p be an element of M: If f ; g; h; k are in Cp1 ðMÞ; f g; and h k, then ðf þ hÞ ðg þ kÞ: Proof Let us take any f ; g; h; k in Cp1 ðMÞ; and let f g; h k: Since f ; g are in Cp1 ðMÞ, by Theorem 2.16, f þ h is in Cp1 ðMÞ: Similarly g þ k is in Cp1 ðMÞ: We have to prove that ðf þ hÞ ðg þ kÞ: So, by the deﬁnition of ; we must ﬁnd an

50

2 Tangent Spaces

open neighborhood H of p such that ðf þ hÞðxÞ ¼ ðg þ kÞðxÞ for every x in H: Since f g; there exists an open neighborhood H1 of p such that f ðxÞ ¼ gðxÞ for every x in H1 : Since h k; there exists an open neighborhood H2 of p such that hðxÞ ¼ kðxÞ for every x in H2 : Since H1 ; H2 are open neighborhoods of p; H1 \ H2 is an open neighborhoods of p: Also, for every x in H1 \ H2 ; ðf þ hÞðxÞ ¼ f ðxÞþ hðxÞ ¼ gðxÞ þ kðxÞ ¼ ðg þ kÞðxÞ: h Theorem 2.20 Let M be an m-dimensional smooth manifold. Let p be an element of M: If f ; g; h; k are in Cp1 ðMÞ; f g; and h k, then ðf hÞ ðg kÞ: Proof Its proof is quite similar to the proof of Theorem 2.19.

h

Theorem 2.21 Let M be an m-dimensional smooth manifold. Let p be an element of M: If f ; g are in Cp1 ðMÞ; f g; and t is a real, then ðtf Þ ðtgÞ: Proof Let us take any f ; g in Cp1 ðMÞ; and let f g: Since f ; g are in Cp1 ðMÞ, by Theorem 2.18, tf is in Cp1 ðMÞ: Similarly, tg is in Cp1 ðMÞ: We have to prove that ðtf Þ ðtgÞ: So, by the deﬁnition of ; we must ﬁnd an open neighborhood H of p such that ðtf ÞðxÞ ¼ ðtgÞðxÞ for every x in H: Since f g; there exists an open neighborhood H of p such that f ðxÞ ¼ gðxÞ for every x in H: So, for every x in H; ðtf ÞðxÞ ¼ tðf ðxÞÞ ¼ tðgðxÞÞ ¼ ðtgÞðxÞ: h Note 2.22 Theorems 2.19, 2.20, and 2.21 give guarantee that the following deﬁnitions are legitimate. Deﬁnition For every f ; g in Cp1 ðMÞ; and for every real t; ½f þ ½g ½f þ g; t½f ½tf ; ½f ½g ½f g: Theorem 2.23 Let M be an m-dimensional smooth manifold. Let p be an element of M: The quotient set Cp1 ðMÞ= ; together with vector addition, and scalar multiplication is deﬁned as follows: for every f ; g in Cp1 ðMÞ; and for every real t; ½f þ ½g ½f þ g;

t½f ½tf ;

constitute a real linear space. In short, F p ðMÞ is a real linear space. Let us deﬁne multiplication operation over Cp1 ðMÞ= as follows: for every f ; g in Cp1 ðMÞ; ½f ½g ½f g:

2.2 Algebra of Smooth Functions

51

Then, 1. 2. 3. 4.

for for for for

every every every every

f ; g; h in Cp1 ðMÞ; ð½f ½gÞ½h ¼ ½f ð½g½hÞ; f ; g in Cp1 ðMÞ; ½f ½g ¼ ½g½f ; f ; g; h in Cp1 ðMÞ; ð½f þ ½gÞ½h ¼ ½f ½h þ ½g½h; real t; and for every f ; g in Cp1 ðMÞ; tð½f ½gÞ ¼ ðt½f Þ½g:

In short, F p ðMÞ is an algebra. Proof The conditions for linear space remain to be veriﬁed. 1. + is associative: Let us take any f ; g; h in Cp1 ðMÞ: We have to prove that ð½ f þ ½gÞ þ ½h ¼ ½ f þ ð½g þ ½hÞ: LHS ¼ ð½ f þ ½gÞ þ ½h ¼ ½f þ g þ ½h ¼ ½ðf þ gÞ þ h ¼ ½f þ ðg þ hÞ ¼ ½ f þ ½g þ h ¼ ½ f þ ð½g þ ½hÞ ¼ RHS: 2. Existence of zero element: Let us deﬁne the constant function 0 : M ! R as follows: For every x in M; 0ðxÞ 0: We want to prove that 0 is in Cp1 ðMÞ: For this purpose, we must prove 1. The domain of the function 0 is an open neighborhood of p in M; 2. for every admissible coordinate chart ðU; uU Þ of M satisfying p 2 U,

0 ðuU Þ1 : uU ððdom 0Þ \ U Þ ! R

is C 1 at the point uU ðpÞ in Rm : For 1: Here, the domain of the function 0 is M which is an open neighborhood of p in M: This proves 1. For 2: Let us take any admissible coordinate chart ðU; uU Þ of M satisfying p 2 U. Now, by the deﬁnition of the function 0; ð0 ðuU Þ1 Þ is the constant function 0 deﬁned on the open subset uU ðUÞ of Rm ; which is known to be C 1 at the point uU ðpÞ in Rm : This proves 2. Thus, we have shown that 0 is in Cp1 ðMÞ: Now, it remains to be showed that ½0 þ ½f ¼ ½f ¼ ½f þ ½0 for every f in Cp1 ðMÞ: Here, LHS ¼ ½0 þ ½ f ¼ ½0 þ f ¼ ½ f ¼ ½f þ 0 ¼ ½ f þ ½0 ¼ RHS: This proves 2.

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2 Tangent Spaces

3. Existence of negative element: Let us take any f in Cp1 ðMÞ: So, by the deﬁnition of Cp1 ðMÞ; f is a real-valued function f whose dom f is an open neighborhood of p in M; and for every admissible coordinate chart ðU; uU Þ of M satisfying p 2 U,

f ðuU Þ1 : uU ððdom f Þ \ U Þ ! R

is C 1 at the point uU ðpÞ in Rm : Now, let us deﬁne a function ðf Þ : ðdom f Þ ! R as follows: For every x in dom f ; ðf ÞðxÞ ðf ðxÞÞ: We want to prove that ðf Þ is in Cp1 ðMÞ: For this purpose, we must prove 1. The domain of the function ðf Þ is an open neighborhood of p in M; 2. For every admissible coordinate chart ðU; uU Þ of M satisfying p 2 U,

ðf Þ ðuU Þ1 : uU ððdomðf ÞÞ \ U Þ ! R

is C 1 at the point uU ðpÞ in Rm : For 1: Since f is in Cp1 ðMÞ, by the deﬁnition of Cp1 ðMÞ; dom f is an open neighborhood of p in M: Since dom f is an open neighborhood of p in M; and, by the deﬁnition of ðf Þ; domðf Þ ¼ dom f ; domðf Þ is an open neighborhood of p in M: This proves 1. For 2: Let us take an admissible coordinate chart ðU; uU Þ of M satisfying p 2 U. Now, by the deﬁnition of the function ðf Þ; ðf Þ ðuU Þ1 ¼ ðf ðuU Þ1 Þ: Since ðf ðuU Þ1 Þ is C 1 at the point uU ðpÞ in Rm ; ðf ðuU Þ1 Þ is C1 at the point uU ðpÞ in Rm : Since ðf ðuU Þ1 Þ is C1 at the point uU ðpÞ in Rm ; and ðf Þ ðuU Þ1 ¼ ðf ðuU Þ1 Þ; ðf Þ ðuU Þ1 is C 1 at the point uU ðpÞ in Rm : This proves 2. Thus, we have shown that ðf Þ is in Cp1 ðMÞ: Now, it remains to be showed that ½ðf Þ þ ½f ¼ ½0 ¼ ½f þ ½ðf Þ. Here, LHS ¼ ½ðf Þ þ ½ f ¼ ½ðf Þ þ f ¼ ½0 ¼ ½f þ ðf Þ ¼ ½ f þ ½ðf Þ ¼ RHS:

This proves 3. 4. + is commutative: Let us take any f ; g in Cp1 ðMÞ: We have to prove that ½ f þ ½g ¼ ½g þ ½ f :

2.2 Algebra of Smooth Functions

53

Here LHS ¼ ½ f þ ½g ¼ ½f þ g ¼ ½g þ f ¼ ½g þ ½ f ¼ RHS: This proves 4. 5. For every real s; t; and f in Cp1 ðMÞ; ðs þ tÞ½f ¼ s½f þ t½f : Here LHS ¼ ðs þ tÞ½ f ¼ ½ðs þ tÞf ¼ ½ðsf Þ þ ðtf Þ ¼ ½ðsf Þ þ ½ðtf Þ ¼ s½ f þ t½ f ¼ RHS: 6. For every real s; t; and f in Cp1 ðMÞ; ðstÞ½f ¼ sðt½f Þ : Here, LHS ¼ ðstÞ½ f ¼ ½ðstÞf ¼ ½sðtf Þ ¼ s½ðtf Þ ¼ sðt½ f Þ ¼ RHS: 7. For every real t; and for every f ; g in Cp1 ðMÞ; tð½f þ ½gÞ ¼ t½f þ t½g : Here, LHS ¼ tð½ f þ ½gÞ ¼ t½ðf þ gÞ ¼ ½tðf þ gÞ ¼ ½ðtf Þ þ ðtgÞ ¼ ½ðtf Þ þ ½ðtgÞ ¼ t½ f þ t½g ¼ RHS: 8. For every f in Cp1 ðMÞ; 1½f ¼ ½f : Here, LHS ¼ 1½ f ¼ ½1f ¼ ½ f ¼ RHS: Thus, we have shown that quotient set Cp1 ðMÞ= is a real linear space. Now, we want to prove that F p ðMÞ is an algebra. 1. Multiplication is associative: LHS ¼ ð½ f ½gÞ½h ¼ ½f g þ ½h ¼ ½ðf gÞ h ¼ ½f ðg hÞ ¼ ½ f ½g h ¼ ½ f ð½g½hÞ ¼ RHS: 2. Multiplication is commutative: LHS ¼ ½ f ½g ¼ ½f g ¼ ½g f ¼ ½g½ f ¼ RHS: 3. Multiplication distributes over +: LHS ¼ ð½ f þ ½gÞ½h ¼ ½f þ g½h ¼ ½ðf þ gÞ h ¼ ½ðf hÞ þ ðg hÞ ¼ ½f h þ ½g h ¼ ½ f ½h þ ½g½h ¼ RHS: 4. LHS ¼ tð½f ½gÞ ¼ t½f g ¼ ½tðf gÞ ¼ ½ðtf Þ g ¼ ½tf ½g ¼ ðt½f Þ½g ¼ RHS:

h

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2 Tangent Spaces

2.3 Smooth Germs on Smooth Manifolds Deﬁnition Let M be an m-dimensional smooth manifold. Let p be an element of M: Under the vector addition, and scalar multiplication as deﬁned in Theorem 2.23, F p ðMÞ is a real linear space. The members of F p ðMÞ are called C 1 -germs at p on M. Thus, if f is in Cp1 ðMÞ, then the equivalence class ½f is in F p ðMÞ; and hence, ½f is a C1 -germ at p on M: Theorem 2.24 Let M be an m-dimensional smooth manifold. Let p be an element of M: Let c be in Cp ðMÞ: Let f be in Cp1 ðMÞ: Then, 1. 0 is an interior point of the domain of the real-valued function f c; cÞð0Þ 2. limt!0 ðf cÞð0þtÞðf exists, t 3. if f g, then lim t!0

ðf cÞð0 þ tÞ ðf cÞð0Þ ðg cÞð0 þ tÞ ðg cÞð0Þ ¼ lim : t!0 t t

Proof 1: Since c is in Cp ðMÞ, by the deﬁnition of Cp ðMÞ; there exists a real number d [ 0 such that c : ðd; dÞ ! M, cð0Þ ¼ p; and c is a smooth map from ðd; dÞ to M: Since cis a smooth map from ðd; dÞ to M; c is a continuous map. Since f is in Cp1 ðMÞ; f : ðdom f Þ ! R and dom f is an open neighborhood of p in M: Since cð0Þ ¼ p; and p is in dom f ; 0 is an element of the domain of f c: Since f is in Cp1 ðMÞ; f is continuous on some open neighborhood U of p. Since c : ðd; dÞ ! M is a continuous map, cð0Þ ¼ p; and U is an open neighborhood of p; there exists e [ 0 such that e\d and, for every t in ðe; eÞ, we have cðtÞ 2 U ð ðdom f ÞÞ; and hence, ðf cÞðtÞ ¼ f ðcðtÞÞ 2 R: It follows that ðe; eÞ is a subset of domðf cÞ: Hence, 0 is an interior point of the domain of the function f c: This proves 1. 2: Since 0 is an interior point of the domain of the real-valued function f c; it is meaningful to write ð f c Þ ð 0 þ t Þ ð f c Þ ð 0Þ ; t!0 t

lim

cð0Þ provided it exists. Now we shall try to show that limt!0 ðf cÞð0þtÞðf exists. t

2.3 Smooth Germs on Smooth Manifolds

55

Since p 2 M; and M is an m-dimensional smooth manifold, there exists an admissible coordinate chart ðU; uU Þ of M satisfying p 2 U: Since f is in Cp1 ðMÞ; and ðU; uU Þ is an admissible coordinate chart of M satisfying p 2 U; f ðuU Þ1 : uU ððdom f Þ \ U Þ ! R is C1 at the point uU ðpÞ in Rm : Observe that the domain of ðf ðuU Þ1 Þ; that is, uU ððdom f Þ \ UÞ is an open neighborhood of uU ðpÞ ð¼ uU ðcð0ÞÞ ¼ ðuU cÞð0ÞÞ in Rm : Since c is a smooth map from ðd; dÞ to M; and 0 2 ðd; dÞ; c is C 1 at 0 in R: Since c is C1 at 0 in R; cð0Þ ¼ p; and ðU; uU Þ is an admissible coordinate chart of M satisfying p 2 U; ðuU cÞ : ðd; dÞ ! Rm is C1 at the point 0 in R: Since ðuU cÞ : ðd; dÞ ! Rm is C 1 at the point 0;

f ðuU Þ1 : uU ððdom f Þ \ U Þ ! R

is C 1 at the point ðuU cÞð0Þ in Rm ; and the domain of ðf ðuU Þ1 Þ is an open neighborhood of ðuU cÞð0Þ in Rm ; the composite function ðf ðuU Þ1 Þ ðuU cÞð¼ f cÞ is C 1 at the point 0 in R: Hence,

dðf cÞðtÞ

dt t¼0 that is, ð f c Þ ð 0 þ t Þ ð f c Þ ð 0Þ t!0 t

lim exists. This proves 2. 3: Let f g: We have to prove that lim t!0

ðf cÞð0 þ tÞ ðf cÞð0Þ ðg cÞð0 þ tÞ ðg cÞð0Þ ¼ lim : t!0 t t

Since f g, there exists an open neighborhood H of pð¼ cð0ÞÞ such that f ðxÞ ¼ gðxÞ for every x in H: Since f ðxÞ ¼ gðxÞ for every x in H; and cð0Þ is in H; ðf cÞð0Þ ¼ f ðcð0ÞÞ ¼ gðcð0ÞÞ ¼ ðg cÞð0Þ:

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Since c : ðd; dÞ ! M is a continuous map, and H is an open neighborhood of cð0Þ; there exists e1 [ 0 such that e1 \d; and for every t in the open interval ðe1 ; e1 Þ; cðtÞ is in H: Since 0 is an interior point of the domain of the function f c; there exists e2 [ 0 such that e2 \d; and the open interval ðe2 ; e2 Þ is contained in the domain of the function f c: Similarly, there exists e3 [ 0 such that e3 \d; and the open interval ðe3 ; e3 Þ is contained in the domain of the function g c: Put e minfe1 ; e2 ; e3 g: Clearly, e [ 0: If t is in the open interval ðe; eÞ; then t is in the open interval ðe2 ; e2 Þ; and hence, ðf cÞðtÞ is meaningful. Similarly, if t is in the open interval ðe; eÞ; then ðg cÞðtÞ is meaningful. If t is in the open interval ðe; eÞ; then t is in the open interval ðe1 ; e1 Þ; and hence, cðtÞ is in H: Since, for every t in ðe; eÞ; cðtÞ is in H; and since f ðxÞ ¼ gðxÞ for every x in H; for every t in ðe; eÞ; ðf cÞðtÞ ¼ f ðcðtÞÞ ¼ gðcðtÞÞ ¼ ðg cÞðtÞ: Hence, ð f c Þ ð 0 þ t Þ ð f c Þ ð 0Þ ¼ t ðf cÞðtÞ ðf cÞð0Þ lim ¼ lim t!0 t t ! 0; lim t!0

ðf cÞðtÞ ðf cÞð0Þ t

t in ðe; eÞ ¼

lim t ! 0;

ð g c Þ ð t Þ ð g c Þ ð 0Þ t

t in ðe; eÞ ¼

lim t ! 0;

ð g c Þ ð 0 þ t Þ ð g c Þ ð 0Þ t

t in ðe; eÞ ðg cÞð0 þ tÞ ðg cÞð0Þ : ¼ lim t!0 t This proves 3.

h

Note 2.25 Let M be an m-dimensional smooth manifold. Let p be an element of M: Let c be in Cp ðMÞ: Let ½f be in F p ðMÞ where f is in Cp1 ðMÞ: Let ½g be in F p ðMÞ where g is in Cp1 ðMÞ: If ½f ¼ ½g, then f g; and hence,

2.3 Smooth Germs on Smooth Manifolds

lim t!0

57

ðf cÞð0 þ tÞ ðf cÞð0Þ ðg cÞð0 þ tÞ ðg cÞð0Þ ¼ lim : t!0 t t

So the following deﬁnition is well deﬁned. Deﬁnition Let M be an m-dimensional smooth manifold. Let p be an element of M: Let c be in Cp ðMÞ: Let ½f be in F p ðMÞ where f is in Cp1 ðMÞ: Then, ð f c Þ ð 0 þ t Þ ð f c Þ ð 0Þ t!0 t

lim

exists, and is denoted by c; ½f : Thus, c; ½ f lim t!0

ðf cÞð0 þ tÞ ðf cÞð0Þ t

or,

dðf cÞðtÞ

c; ½ f

: dt t¼0 Theorem 2.26 Let M be an m-dimensional smooth manifold. Let p be an element of M: Let c be in Cp ðMÞ: Let ½f be in F p ðMÞ where f is in Cp1 ðMÞ: Let ½g be in F p ðMÞ where g is in Cp1 ðMÞ: Then, 1. c; ½ f þ ½g ¼ c; ½ f þ c; ½g ; 2. c; a½f ¼ a c; ½f for every real a: In short, we say that ; is linear in the second variable. Proof 1. Here ððf þ gÞ cÞð0 þ tÞ ððf þ gÞ cÞð0Þ t ðf þ gÞðcðtÞÞ ðf þ gÞðcð0ÞÞ ðf ðcðtÞÞ þ gðcðtÞÞÞ ðf ðcð0ÞÞ þ gðcð0ÞÞÞ ¼ lim ¼ lim t!0 t!0 t t ðf ðcðtÞÞ f ðcð0ÞÞÞ þ ðgðcðtÞÞ gðcð0ÞÞÞ ¼ lim t!0 t ðf ðcðtÞÞ f ðcð0ÞÞÞ ðgðcðtÞÞ gðcð0ÞÞÞ þ ¼ lim t!0 t t ðf ðcðtÞÞ f ðcð0ÞÞÞ ðgðcðtÞÞ gðcð0ÞÞÞ þ lim ¼ lim t!0 t!0 t t ðf ðcðt þ 0ÞÞ f ðcð0ÞÞÞ ðgðcðt þ 0ÞÞ gðcð0ÞÞÞ þ lim ¼ lim t!0 t!0 t t ¼ c; ½ f þ c; ½g ¼ RHS:

LHS ¼ c; ½ f þ ½g ¼ c; ½f þ g ¼ lim t!0

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2. Here ððaf Þ cÞð0 þ tÞ ððaf Þ cÞð0Þ t ðaf ÞðcðtÞÞ ðaf Þðcð0ÞÞ að f ð c ð t Þ Þ Þ að f ð c ð 0Þ Þ Þ ¼ lim ¼ lim t!0 t!0 t t f ðcðtÞÞ f ðcð0ÞÞ f ð c ð t Þ Þ f ð c ð 0Þ Þ ¼ lim a ¼ a lim t!0 t!0 t t f ðcðt þ 0ÞÞ f ðcð0ÞÞ ¼ a lim ¼ a c; ½ f ¼ RHS: h t!0 t h

LHS ¼ c; a½ f ¼ c; ½af ¼ lim t!0

Note 2.27 Let M be an m-dimensional smooth manifold. Let p 2 M: Let ½½c; ½½c1 be in Tp Mð¼ Cp ðMÞ=Þ where c; c1 are in Cp ðMÞ: Let ½f ; ½f1 be in F p ðMÞ where f ; f1 are in Cp1 ðMÞ: Let ½½c ¼ ½½c1 ; and ½f ¼ ½f1 : We shall try to show that

dðf cÞðtÞ

dðf1 c1 ÞðtÞ

¼

: dt dt t¼0 t¼0 Since p 2 M; and M is an m-dimensional smooth manifold, there exists an admissible coordinate chart ðU; uU Þ of M satisfying p 2 U: Now since ½½c ¼ ½½c1 ; c c1 ; and hence, ðuU cÞ0 ð0Þ ¼ ðuU c1 Þ0 ð0Þ: Since ½f ¼ ½f1 ; f g; and hence, there exists an open neighborhood H of pð¼ cð0ÞÞ such that f ðxÞ ¼ gðxÞ for every x in H:

d f ðuU Þ1 uU c ðtÞ

¼

dt

t¼0 t¼0

1

ðuU cÞ ðtÞ

d f ðuU Þ

¼

dt

t¼0 0 1 ¼ f ðuU Þ ððuU cÞð0ÞÞ ðuU cÞ0 ð0Þ 0 ¼ f ðuU Þ1 ððuU cÞð0ÞÞ ðuU c1 Þ0 ð0Þ 0 ¼ f ðuU Þ1 ððuU ðcð0ÞÞÞÞ ðuU c1 Þ0 ð0Þ 0 ¼ f ðuU Þ1 ððuU ð pÞÞÞ ðuU c1 Þ0 ð0Þ 0 ¼ f ðuU Þ1 ððuU ðc1 ð0ÞÞÞÞ ðuU c1 Þ0 ð0Þ 0 ¼ f ðuU Þ1 ððuU c1 Þð0ÞÞ ðuU c1 Þ0 ð0Þ

d f ðuU Þ1 ðuU c1 Þ ðtÞ

¼ dðf c1 ÞðtÞ ¼ RHS: ¼

dt dt

t¼0

dðf cÞðtÞ

LHS ¼

dt

t¼0

2.3 Smooth Germs on Smooth Manifolds

59

Hence, the following deﬁnition is legitimate: Deﬁnition Let M be an m-dimensional smooth manifold. Let p 2 M: Let ½½c be in Tp M where c is in Cp ðMÞ: Let ½f be in F p ðMÞ where f is in Cp1 ðMÞ: By ½½c½f , we dðf cÞðtÞ shall mean jt¼0 ð¼ c; ½f Þ: dt Lemma 2.28 Let M be an m-dimensional smooth manifold. Let p 2 M: Let ½½c be in Tp M where c is in Cp ðMÞ: Then 1. ½½cð½f þ ½f1 Þ ¼ ½½c½f þ ½½c½f1 for every ½f ; ½f1 in F p ðMÞ where f ; f1 are in Cp1 ðMÞ; 2. ½½cðt½f Þ ¼ tð½½c½f Þ for every ½f in F p ðMÞ where f is in Cp1 ðMÞ; and for every real t; 3. ½½cð½f ½f1 Þ ¼ ð½½c½f Þðf1 ðpÞÞ þ ðf ðpÞÞð½½c½f1 Þ for every ½f ; ½f1 in F p ðMÞ where f ; f1 are in Cp1 ðMÞ: Proof 1: LHS ¼ ½½cð½f þ ½f1 Þ ¼ ½½cð½f þ f1 Þ ¼ c; ½f þ f1 ¼ c þ ½f1 ¼ c; ½f þ c; ½f1 ¼ c½f þ c½f1 ¼ RHS: 2: LHS ¼ ½½cðt½f Þ ¼ ½½cð½tf Þ ¼ c; t½f ¼ t c; ½f ¼ tð½½c½f Þ ¼ RHS: dððf f1 Þ cÞðtÞ dððf f1 ÞðcðtÞÞÞ jt¼0 ¼ jt¼0 3: LHS ¼ ½½cð½f ½f1 Þ ¼ ½½cð½f f1 Þ ¼ dt dt dððf ðcðtÞÞ f1 ðcðtÞÞÞÞ dððððf cÞðtÞÞ ððf1 cÞðtÞÞÞÞ ¼ jt¼0 ¼ jt¼0 dt dt dððf cÞðtÞÞ dððf1 cÞðtÞÞ h ¼ jt¼0 ððf1 cÞð0ÞÞ þ ððf cÞð0ÞÞ jt¼0 dt dt ¼ ð½½c½f Þððf1 cÞð0ÞÞ þ ððf cÞð0ÞÞð½½c½f Þ ¼ ð½½c½f Þððf1 ðcð0ÞÞÞÞ þ ðf ðcð0ÞÞÞð½½c½f1 Þ h ¼ ð½½c½f Þððf1 ðpÞÞÞ þ ðf ðpÞÞð½½c½f1 Þ ¼ RHS:

2.4 Derivations Deﬁnition Let M be an m-dimensional smooth manifold. Let p be an element of M: Let D : F p ðMÞ ! R be any function. If 1. Dð½f þ ½f1 Þ ¼ D½f þ D½f1 for every ½f ; ½f1 in F p ðMÞ where f ; f1 are in Cp1 ðMÞ; 2. Dðt½f Þ ¼ tðD½f Þ for every ½f in F p ðMÞ where f is in Cp1 ðMÞ; and for every real t; 3. Dð½f ½f1 Þ ¼ ðD½f Þðf1 ðpÞÞ þ ðf ðpÞÞðD½f1 Þ for every ½f ; ½f1 in F p ðMÞ where f ; f1 are in Cp1 ðMÞ; then we say that D is a derivation at p. Here, the collection of all derivations at p is denoted by Dp ðMÞ:

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Example 2.29 Let M be an m-dimensional smooth manifold. Let p 2 M: Let ½½c be in Tp M where c is in Cp ðMÞ: By the Lemma 2.28, the mapping ½f 7! ½½c½f is an example of a derivation at p: We shall denote this derivation by D½ c : Thus, D½ c ð½ f Þ ½½c½ f for every ½f in F p ðMÞ where f is in Cp1 ðMÞ: Also Dc : ½½c 2 Tp M Dp ðM Þ: Note 2.30 Let M be an m-dimensional smooth manifold. Let p be an element of M: For every D; D1 in Dp ðMÞ; we deﬁne D þ D1 : F p ðMÞ ! R as follows: For every ½f in F p ðMÞ where f is in Cp1 ðMÞ; ðD þ D1 Þð½f Þ D½f þ D1 ½f : We shall try to show that D þ D1 is in Dp ðMÞ; that is, 1. ðD þ D1 Þð½f þ ½f1 Þ ¼ ðD þ D1 Þ½f þ ðD þ D1 Þ½f1 for every ½f ; ½f1 in F p ðMÞ where f ; f1 are in Cp1 ðMÞ; 2. ðD þ D1 Þðt½f Þ ¼ tððD þ D1 Þ½f Þ for every ½f in F p ðMÞ where f is in Cp1 ðMÞ; and for every real t; 3. ðD þ D1 Þð½f ½f1 Þ ¼ ððD þ D1 Þ½f Þðf1 ðpÞÞ þ ðf ðpÞÞððD þ D1 Þ½f1 Þ for every ½f ; ½f1 in F p ðMÞ where f ; f1 are in Cp1 ðMÞ: For 1: LHS ¼ ðD þ D1 Þð½f þ ½f1 Þ ¼ Dð½f þ ½f1 Þ þ D1 ð½f þ ½f1 Þ ¼ ðD½f þ D½f1 Þ þ ðD1 ½f þ D1 ½f1 Þ ¼ ðD½f þ D1 ½f Þ þ ðD½f1 þ D1 ½f1 Þ ¼ ðD þ D1 Þ½f þ ðD þ D1 Þ½f1 ¼ RHS: For 2: LHS ¼ ðD þ D1 Þðt½f Þ ¼ Dðt½f Þ þ D1 ðt½f Þ ¼ tðD½f Þ þ tðD1 ½f Þ ¼ tðD½f þ D1 ½f Þ ¼ tððD þ D1 Þ½f Þ ¼ RHS: For 3: LHS ¼ ðD þ D1 Þð½f ½f1 Þ ¼ Dð½f ½f1 Þ þ D1 ð½f ½f1 Þ ¼ ððD½f Þðf1 ðpÞÞ þ ðf ðpÞÞðD½f1 ÞÞ þ ððD1 ½f Þðf1 ðpÞÞ þ ðf ðpÞÞðD1 ½f1 ÞÞ ¼ ððD½f Þ þ D1 ½f Þðf1 ðpÞÞ þ ðf ðpÞÞðD½f1 þ D1 ½f1 Þ ¼ ððD þ D1 Þ½f Þðf1 ðpÞÞ þ ðf ðpÞÞððD þ D1 Þ½f1 Þ ¼ RHS: This proves that D + D1 is in Dp ðMÞ. For every D in Dp ðMÞ; and for every real t; we deﬁne tD : F p ðMÞ ! R as follows: For every ½f in F p ðMÞ where f is in Cp1 ðMÞ; ðtDÞð½f Þ tðD½f Þ: We shall try to show that tD is in Dp ðMÞ; that is, 1. ðtDÞð½f þ ½f1 Þ ¼ ðtDÞ½f þ ðtDÞ½f1 for every ½f ; ½f1 in F p ðMÞ where f ; f1 are in Cp1 ðMÞ; 2. ðtDÞðs½f Þ ¼ sððtDÞ½f Þ for every ½f in F p ðMÞ where f is in Cp1 ðMÞ; and for every real s;

2.4 Derivations

61

3. ðtDÞð½f ½f1 Þ ¼ ððtDÞ½f Þðf1 ðpÞÞ þ ðf ðpÞÞððtDÞ½f1 Þ for every ½f ; ½f1 in F p ðMÞ where f ; f1 are in Cp1 ðMÞ: For 1: LHS ¼ ðtDÞð½f þ ½f1 Þ ¼ tðDð½f þ ½f1 ÞÞ ¼ tðD½f þ D½f1 Þ ¼ tðD½f Þ þ tðD½f1 Þ ¼ ðtDÞ½f þ ðtDÞ½f1 ¼ RHS: For 2: LHS ¼ ðtDÞðs½f Þ ¼ tðDðs½f ÞÞ ¼ tðsðD½f ÞÞ ¼ sðtðD½f ÞÞ ¼ sððtDÞ½f Þ ¼ RHS: For 3: LHS ¼ ðtDÞð½f ½f1 Þ ¼ tðDð½f ½f1 ÞÞ ¼ tððD½f Þðf1 ðpÞÞ þ ðf ðpÞÞðD½f1 ÞÞ ¼ tðD½f Þðf1 ðpÞÞ þ ðf ðpÞÞtðD½f1 Þ ¼ ððtDÞ½f Þðf1 ðpÞÞ þ ðf ðpÞÞððtDÞ½f1 Þ ¼ RHS: This proves that tD is in Dp ðMÞ: Now, we shall try to verify that Dp ðMÞ is a real linear space. 1. + is associative: Let us take any D; D1 ; D2 in Dp ðMÞ: We have to prove that ðD þ D1 Þ þ D2 ¼ D þ ðD1 þ D2 Þ; that is, for every ½f in F p ðMÞ where f is in Cp1 ðMÞ; ððD þ D1 Þ þ D2 Þ½f ¼ ðD þ ðD1 þ D2 ÞÞ½f : LHS ¼ ððD þ D1 Þ þ D2 Þ½ f ¼ ðD þ D1 Þ½ f þ D2 ½ f ¼ ðD½ f þ D1 ½ f Þ þ D2 ½ f D½ f þ ðD1 ½ f þ D2 ½ f Þ ¼ D½ f þ ðD1 þ D2 Þ½ f ¼ ðD þ ðD1 þ D2 ÞÞ½ f ¼ RHS:

2. Existence of zero element: Let us deﬁne the constant function 0 : F p ðMÞ ! R as follows: for every ½f in F p ðMÞ; 0½f 0: We want to prove that 0 is in Dp ðMÞ: For this purpose, we must prove 1. 2. 3.

0ð½f þ ½f1 Þ ¼ 0½f þ 0½f1 for every ½f ; ½f1 in F p ðMÞ where f ; f1 are in Cp1 ðMÞ; 0ðt½f Þ ¼ tð0½f Þ for every ½f in F p ðMÞ where f is in Cp1 ðMÞ; and for every real t; 0ð½f ½f1 Þ ¼ ð0½f Þðf1 ðpÞÞ þ ðf ðpÞÞð0½f1 Þ; for every ½f ; ½f1 in F p ðMÞ where f ; f1 are in Cp1 ðMÞ:

For 1: LHS ¼ 0ð½f þ ½f1 Þ ¼ 0 ¼ 0 þ 0 ¼ 0½f þ 0½f1 ¼ RHS: For 2: LHS ¼ 0ðt½f Þ ¼ 0 ¼ t 0 ¼ tð0½f Þ ¼ RHS: For 3: LHS ¼ 0ð½f ½f1 Þ ¼ 0ð½f f1 Þ ¼ 0 ¼ 0ðf1 ðpÞÞ þ ðf ðpÞÞ0 ¼ ð0½f Þðf1 ðpÞÞ þ ðf ðpÞÞð0½f1 Þ ¼ RHS: Thus, we have shown that 0 is in Dp ðMÞ: Now, it remains to be showed that 0 þ D ¼ D for every D in Dp ðMÞ; that is, for every ½f in F p ðMÞ where f is in Cp1 ðMÞ; ð0 þ DÞ½f ¼ D½f : LHS ¼ ð0 þ DÞ½ f ¼ 0½ f þ D½ f ¼ 0 þ D½ f ¼ D½ f ¼ RHS: Hence, 0 serves the purpose of zero element in Dp ðMÞ: 3. Existence of negative element: Let us take any D in Dp ðMÞ: Now, let us deﬁne ðDÞ : F p ðMÞ ! R as follows: For every ½f in F p ðMÞ; ðDÞ½f ðD½f Þ: We want to prove that ðDÞ is in Dp ðMÞ: For this purpose, we must prove

62

2 Tangent Spaces

1. ðDÞð½f þ ½f1 Þ ¼ ðDÞ½f þ ðDÞ½f1 for every ½f ; ½f1 in F p ðMÞ where f ; f1 are in Cp1 ðMÞ; 2. ðDÞðt½f Þ ¼ tððDÞ½f Þ for every ½f in F p ðMÞ where f is in Cp1 ðMÞ; and for every real t; 3. ðDÞð½f ½f1 Þ ¼ ððDÞ½f Þðf1 ðpÞÞ þ ðf ðpÞÞððDÞ½f1 Þ; for every ½f ½f1 in F p ðMÞ where f ; f1 are in Cp1 ðMÞ For 1: LHS ¼ ðDÞð½f þ ½f1 Þ ¼ ðDð½f þ ½f1 ÞÞ ¼ ðD½f þ D½f1 Þ ¼ ðD½f Þ þ ððD½f1 ÞÞ ¼ ðDÞ½f þ ðDÞ½f1 ¼ RHS: This proves 1. For 2: LHS ¼ ðDÞðt½f Þ ¼ ðDðt½f ÞÞ ¼ ðtðD½f ÞÞ ¼ tððD½f ÞÞ ¼ tððDÞ½f Þ ¼ RHS: For 3: LHS ¼ ðDÞð½f ½f1 Þ ¼ ðDð½f ½f1 ÞÞ ¼ ððD½f Þðf1 ðpÞÞ þ ðf ðpÞÞðD½f1 ÞÞ ¼ ððD½f ÞÞðf1 ðpÞÞ þ ðf ðpÞÞððD½f1 ÞÞ ¼ ððDÞ½f Þðf1 ðpÞÞ þ ðf ðpÞÞððDÞ½f1 Þ ¼ RHS:

Thus, we have shown that ðDÞ is in Dp ðMÞ: Now, it remains to be showed that ðDÞ þ D ¼ 0; that is, for every ½f in F p ðMÞ where f is in Cp1 ðMÞ; ððDÞ þ DÞ½f ¼ 0½f : LHS ¼ ððDÞ þ DÞ½ f ¼ ðDÞ½ f þ D½ f ¼ ðD½ f Þ þ D½ f ¼ 0 ¼ 0½ f ¼ RHS:

Hence, ðDÞ serves the purpose of negative element of D in Dp ðMÞ: 4. + is commutative: Let us take any D; D1 in Dp ðMÞ: We have to prove that D þ D1 ¼ D1 þ D; that is, for every ½f in F p ðMÞ where f is in Cp1 ðMÞ; ðD þ D1 Þ½f ¼ ðD1 þ DÞ½f : LHS ¼ ðD þ D1 Þ½ f ¼ D½ f þ D1 ½ f ¼ D1 ½ f þ D½ f ¼ ðD1 þ DÞ½ f ¼ RHS: This proves 4. 5. For every real s; t; and D in Dp ðMÞ; we have to prove that ðs þ tÞD ¼ sD þ tD; that is, for every ½f in F p ðMÞ where f is in Cp1 ðMÞ; ððs þ tÞDÞ½f ¼ ðsD þ tDÞ½f : Here LHS ¼ ððs þ tÞDÞ½ f ¼ ðs þ tÞðD½ f Þ ¼ sðD½ f Þ þ tðD½ f Þ ¼ ðsDÞ½ f þ ðtDÞ½ f ¼ ðsD þ tDÞ½ f ¼ RHS: 6. For every real s; t; and D in Dp ðMÞ; we have to prove that ðstÞD ¼ sðtDÞ; that is, for every ½f in F p ðMÞ where f is in Cp1 ðMÞ; ððstÞDÞ½f ¼ ðsðtDÞÞ½f : Here, LHS ¼ ððstÞDÞ½ f ¼ ðstÞðD½ f Þ ¼ sðtðD½ f ÞÞ ¼ sððtDÞ½ f Þ ¼ ðsðtDÞÞ½ f ¼ RHS:

2.4 Derivations

63

7. For every real t; and for every D; D1 in Dp ðMÞ; we have to prove that tðD þ D1 Þ ¼ tD þ tD1 ; that is, for every ½f in F p ðMÞ where f is in Cp1 ðMÞ; ðtðD þ D1 ÞÞ½f ¼ ðtD þ tD1 Þ½f : Here LHS ¼ ðtðD þ D1 ÞÞ½ f ¼ tððD þ D1 Þ½ f Þ ¼ tðD½ f þ D1 ½ f Þ ¼ tðD½ f Þ þ tðD1 ½ f Þ ¼ ðtDÞ½ f þ ðtD1 Þ½ f ¼ ðtD þ tD1 Þ½ f ¼ RHS:

8. For every D in Dp ðMÞ; we have to show that 1D ¼ D; that is, for every ½f in F p ðMÞ where f is in Cp1 ðMÞ; ð1DÞ½f ¼ D½f : Here, LHS ¼ ð1DÞ½ f ¼ 1ðD½ f Þ ¼ D½ f ¼ RHS: Thus, we have shown that Dp ðMÞ is a real linear space. Lemma 2.31 Let M be an m-dimensional smooth manifold. Let p be an element of M: Let D : F p ðMÞ ! R be a derivation at p. Let ½f be in F p ðMÞ where f is in Cp1 ðMÞ: If f is a constant function, then D½f ¼ 0: Proof Since f is a constant function, there exists a real number c such that f ðxÞ ¼ c for every x in the domain of f : Since Dð1Þ ¼ Dð1 1Þ ¼ ðDð1ÞÞð1ð pÞÞ þ ð1ð pÞÞðDð1ÞÞ ¼ ðDð1ÞÞ1 þ 1ðDð1ÞÞ ¼ 2ðDð1ÞÞ; Dð1Þ ¼ 0: Now, LHS ¼ D½ f ¼ DðcÞ ¼ Dðc1Þ ¼ cðDð1ÞÞ ¼ c 0 ¼ 0 ¼ RHS:

h h

Note 2.32 Let M be a 2-dimensional smooth manifold. Let p 2 M: Let ðU; uU Þ be any admissible coordinate chart of M satisfying p 2 U: Let u1 ; u2 be the component functions of uU : So u1 : U ! R; u2 : U ! R; and for every x in U; uU ð xÞ ¼ u1 ð xÞ; u2 ð xÞ : Since u1 is in Cp1 ðMÞ; ½u1 is in F p ðMÞ: Similarly, ½u2 is in F p ðMÞ:

64

2 Tangent Spaces

Further, let us observe that D1 u1 ðuU Þ1 ðuU ð pÞÞ ¼ D1 u1 ðuU Þ1 u1 ð pÞ; u2 ð pÞ u1 ðuU Þ1 ðu1 ð pÞ þ t; u2 ð pÞÞ u1 ðuU Þ1 ðu1 ð pÞ; u2 ð pÞÞ ¼ lim t!0 t 1 1 1 2 u ðuU Þ ððu ð pÞ; u ð pÞÞ þ ðt; 0ÞÞ u1 ðuU Þ1 ðu1 ð pÞ; u2 ð pÞÞ ¼ lim t!0 t 1 1 u ðuU Þ ðuU ð pÞ þ ðt; 0ÞÞ u1 ðuU Þ1 ðuU ð pÞÞ ¼ lim t!0 t 1 1 u ðuU Þ ðuU ð pÞ þ ðt; 0ÞÞ u1 ð pÞ ¼ lim t!0 t 1 1 u ðuU Þ ðuU ð pÞ þ ðt; 0ÞÞ u1 ð pÞ ¼ lim t!0 t u1 ð qÞ u1 ð pÞ ¼ lim t!0 t where q ðuU Þ1 ðuU ðpÞ þ ðt; 0ÞÞ; that is,

u1 ðqÞ; u2 ðqÞ ¼ uU ðqÞ ¼ uU ð pÞ þ ðt; 0Þ ¼ u1 ð pÞ; u2 ð pÞ þ ðt; 0Þ ¼ u1 ð pÞ þ t; u2 ð pÞ :

Hence, u1 ðqÞ ¼ u1 ðpÞ þ t: This shows that

u1 ð qÞ u1 ð pÞ ð u1 ð pÞ þ t Þ u1 ð pÞ ¼ lim ¼ 1: D1 u1 ðuU Þ1 ðuU ð pÞÞ ¼ lim t!0 t!0 t t Thus,

D1 u1 ðuU Þ1 ðuU ð pÞÞ ¼ 1:

2.4 Derivations

65

Next D2 u1 ðuU Þ1 ðuU ð pÞÞ ¼ D2 u1 ðuU Þ1 u1 ð pÞ; u2 ð pÞ u1 ðuU Þ1 ðu1 ð pÞ; u2 ð pÞ þ tÞ u1 ðuU Þ1 ðu1 ð pÞ; u2 ð pÞÞ ¼ lim t!0 t 1 1 1 2 u ðuU Þ ððu ð pÞ; u ð pÞÞ þ ð0; tÞÞ u1 ðuU Þ1 ðu1 ð pÞ; u2 ð pÞÞ ¼ lim t!0 t 1 1 u ðuU Þ ðuU ð pÞ þ ð0; tÞÞ u1 ðuU Þ1 ðuU ð pÞÞ ¼ lim t!0 t 1 1 u ðuU Þ ðuU ð pÞ þ ð0; tÞÞ u1 ð pÞ ¼ lim t!0 t 1 1 u ðuU Þ ðuU ð pÞ þ ð0; tÞÞ u1 ð pÞ ¼ lim t!0 t u1 ð qÞ u1 ð pÞ ¼ lim t!0 t where q ðuU Þ1 ðuU ðpÞ þ ð0; tÞÞ; that is,

u1 ðqÞ; u2 ðqÞ ¼ uU ðqÞ ¼ uU ð pÞ þ ð0; tÞ ¼ u1 ð pÞ; u2 ð pÞ þ ð0; tÞ ¼ u1 ð pÞ; u2 ð pÞ þ t :

Hence, u1 ðqÞ ¼ u1 ðpÞ: This shows that u1 ð pÞ u1 ð pÞ ¼ 0: D2 u1 ðuU Þ1 ðuU ð pÞÞ ¼ lim t!0 t Thus,

D2 u1 ðuU Þ1 ðuU ð pÞÞ ¼ 0:

Similarly, ðD2 ðu2 ðuU Þ1 ÞÞðuU ðpÞÞ ¼ 1; and ðD1 ðu2 ðuU Þ1 ÞÞðuU ðpÞÞ ¼ 0: Deﬁnition Let M be a 2-dimensional smooth manifold. Let p be an element of M: Let ðU; uU Þ be any admissible coordinate chart of M satisfying p 2 U: Let u1 ; u2 be the component functions of uU : The function ouo1 jp : F p ðMÞ ! R is deﬁned as follows: For every ½f in F p ðMÞ;

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2 Tangent Spaces

! o

½ f D1 f ðuU Þ1 ðuU ð pÞÞ:

1 ou p The function

o ou2 jp

: F p ðMÞ ! R is deﬁned as follows: For every ½f in F p ðMÞ;

! o

1 f ð u Þ ðuU ð pÞÞ: ½ f D 2 U

ou2 p

From the Note 2.32,

!

o

j 1 u ¼ 0 oui p

Note 2.33 We shall prove that

o ou1 jp

if if

i¼j i 6¼ j:

is in Dp ðMÞ: For this purpose, we must prove

1. ðouo1 jp Þð½f þ ½f1 Þ ¼ ðouo1 jp Þ½f þ ðouo1 jp Þ½f1 for every ½f ; ½f1 in F p ðMÞ where f ; f1 are in Cp1 ðMÞ; 2. ðouo1 jp Þðt½f Þ ¼ tððouo1 jp Þ½f Þ for every ½f in F p ðMÞ where f is in Cp1 ðMÞ; and for every real t; 3. ðouo1 jp Þð½f ½f1 Þ ¼ ððouo1 jp Þ½f Þðf1 ðpÞÞ þ ðf ðpÞÞððouo1 jp Þ½f1 Þ; for every ½f ; ½f1 in F p ðMÞ where f ; f1 are in Cp1 ðMÞ

!

! o

o

ð ½ Þ ¼ ð½f þ f1 Þ ¼ D1 ðf þ f1 Þ ðuU Þ1 ðuU ð pÞÞ ½ f þ f 1

1 1 ou p ou p 1 ¼ D1 f ðuU Þ ðuU ð pÞÞ ¼ D1 f ðuU Þ1 ðuU ð pÞÞ þ f1 ðuU Þ1

!

! o

o

1 ½f þ ðuU ð pÞÞ ¼ ½f1 ¼ RHS: þ D1 f1 ðuU Þ ou1 p ou1 p

For 1: LHS ¼

For 2:

! o

ðuU ð pÞÞ ðt½ f Þ ¼ D1 ðtf Þ ðuU Þ1 ðuU ð pÞÞ ¼ D1 t f ðuU Þ1

1 ou p

! ! o

¼ t D1 f ðuU Þ1 ðuU ð pÞÞ ¼ t ½ f ¼ RHS: 1 ou p

LHS ¼

! o

! o

1 For 3: LHS ¼ ou1

ð½ f ½f1 Þ ¼ ou1

ð½f f1 Þ ¼ D1 ðf f1 Þ ðuU Þ p p

ðuU ð pÞÞ

f ðuU Þ1 f1 ðuU Þ1 ðuU ð pÞÞ ¼ D1 1 ¼ D1 f ðuU Þ ðuU ð pÞÞ f1 ðuU Þ1 ðuU ð pÞÞ D1 f1 ðuU Þ1 ðuU ð pÞÞ þ f ðuU Þ1 ðuU ð pÞÞ ¼ D1 f ðuU Þ1 ðuU ð pÞÞ ðf1 ð pÞÞ þ ðf ð pÞÞ D1 f1 ðuU Þ1 ðuU ð pÞÞ

! !

! ! o

o

¼ ½ f f ð ð p Þ Þ þ ð f ð p Þ Þ ½f1 ¼ RHS: 1 ou1 p ou1 p

2.4 Derivations

67

This proves that

o ou1 jp

is in Dp ðMÞ: Similarly,

o ou2 jp

is in Dp ðMÞ:

Note 2.34 We shall prove that ouo1 jp ; ouo2 jp are linearly independent. For this purpose, let t1 ðouo1 jp Þ þ t2 ðouo2 jp Þ ¼ 0: We have to show that t1 ¼ t2 ¼ 0: Since t1 ðouo1 jp Þ þ t2 ðouo2 jp Þ ¼ 0; so 0 ¼ 0 u1 ¼

t1

!

!! 1 o

o

þ t2 u ¼ t1 ou1 p ou2 p

!

! o

1 þ t2 u ou1 p

!

! o

1 u ou2 p

¼ t1 ð1Þ þ t2 ð0Þ ¼ t1 :

Thus, t1 ¼ 0: Similarly, t2 ¼ 0: Thus, we have shown that ouo1 jp ; ouo2 jp are linearly independent. Lemma 2.35 Let M be a 2-dimensional smooth manifold. Let p be an element of M: Let ðU; uU Þ be any admissible coordinate chart of M satisfying p 2 U: Let us deﬁne functions u1 : U ! R; u2 : U ! R as follows: For every x in U; uU ð xÞ u1 ð xÞ; u2 ð xÞ : If uU ðpÞ ¼ ð0; 0Þ, then fouo1 jp ; ouo2 jp g is a basis of the real linear space Dp ðMÞ: Proof From the above discussion, it remains to be proved that fouo1 jp ; ouo2 jp g generates the whole space Dp ðMÞ: For this purpose, let us take any D in Dp ðMÞ: We have to ﬁnd real numbers t1 ; t2 such that D ¼ t1 ðouo1 jp Þ þ t2 ðouo2 jp Þ; that is, for every ½f in F p ðMÞ where f is in Cp1 ðMÞ; we have

!

!! o

o

D ½ f ¼ t1 þ t2 ½ f : ou1 p ou2 p Let us take any ½f in F p ðMÞ where f is in Cp1 ðMÞ: It sufﬁces to prove that

!

!! 1 o

2 o

þ D u

D½ f ¼ D u ½ f ; ou1 p ou2 p that is, D½ f ¼ D u1

! ! o

½ f þ D u2

1 ou p

! ! o

½f : ou2 p

Let us note that for every uU ðqÞð¼ ðu1 ðqÞ; u2 ðqÞÞÞ in uU ðUÞ; where q is in U;

68

2 Tangent Spaces f ðqÞ f ð pÞ ¼ f ðqÞ f ðuU Þ1 ð0; 0Þ ¼ f ðuU Þ1 u1 ðqÞ; u2 ðqÞ f ðuU Þ1 ð0; 0Þ ¼ f ðuU Þ1 u1 ðqÞ; u2 ðqÞ f ðuU Þ1 ð0; 0Þ

t¼1 ¼ f ðuU Þ1 u1 ðqÞ t; u2 ðqÞ t

t¼0

Z1 d f ðuU Þ1 u1 ðqÞ t; u2 ðqÞ t dt ¼ dt 0

d 1 u ð qÞ t u1 ðqÞ t; u2 ðqÞ t D1 f ðuU Þ1 dt 0 d 2 u ðqÞ t u1 ðqÞ t; u2 ðqÞ t dt þ D2 f ðuU Þ1 dt Z 1 1 u ð qÞ u1 ðqÞ t; u2 ðqÞ t D1 f ðuU Þ1 ¼

¼

Z 1

0

þ

2 u ðqÞ dt u1 ðqÞ t; u2 ðqÞ t D2 f ðuU Þ1

¼ u1 ðqÞ

Z 1

u1 ðqÞ t; u2 ðqÞ t dt D1 f ðuU Þ1

0

þ u2 ðqÞ

Z1

u1 ðqÞ t; u2 ðqÞ t dt D2 f ðuU Þ1

0

¼ u1 ðqÞ

Z 1

t u1 ðqÞ; u2 ðqÞ dt D1 f ðuU Þ1

0

þ u ðqÞ 2

Z1

t u1 ðqÞ; u2 ðqÞ dt D2 f ðuU Þ1

0

¼ u1 ðqÞ

Z 1

D1 f ðuU Þ1 ðtðuU ðqÞÞÞ dt

0

þ u2 ðqÞ

Z1 0

¼ u ðqÞ 1

Z1 0

D2 f ðuU Þ1 ðtðuU ðqÞÞÞ dt

! !

o

½ f dt ou1 ðuU Þ1 ðtðuU ðqÞÞÞ

! !

o

½ f dt þ u ðqÞ ou2 ðuU Þ1 ðtðuU ðqÞÞÞ 0 ¼ u1 ðqÞ ðg1 ðqÞÞ þ u2 ðqÞ ðg2 ðqÞÞ;

2

Z1

2.4 Derivations

69

where Z1 g1 ð qÞ 0

! !

o

½ f dt; ou1 ðuU Þ1 ðtðuU ðqÞÞÞ

and Z1 g2 ðqÞ 0

! !

o

½ f dt ou2 ðuU Þ1 ðtðuU ðqÞÞÞ

for every q in U: Hence, f ð qÞ ¼ u1 ð qÞ ð g1 ð qÞ Þ þ u2 ð qÞ ð g2 ð qÞ Þ þ f ð pÞ ¼ u1 g1 ð qÞ þ u2 g2 ð qÞ þ f ð pÞ ¼ u1 g1 þ u2 g2 þ f ð pÞ ð qÞ for every q in U: So f ¼ u1 g1 þ u2 g2 þ f ð pÞ: Next D½ f ¼ D f ð pÞ þ u1 g1 þ u2 g2 ¼ D½f ð pÞ þ D u1 g1 þ D u2 g2 ¼ 0 þ D u1 g1 þ D u2 g2 ¼ D u1 g1 þ D u2 g2 ¼ D u1 ðg1 ð pÞÞ þ u1 ð pÞ ðD½g1 Þ þ D u2 ðg2 ð pÞÞ þ u2 ð pÞ ðD½g2 Þ 1 2 ¼ D u ðg1 ð pÞÞ þ ð0ÞðD½g1 Þ þ D u ðg2 ð pÞÞ þ ð0ÞðD½g2 Þ ¼ D u1 ðg1 ð pÞÞ þ D u2 ðg2 ð pÞÞ 0 1 ! ! 1

Z

1 o

½ f dtA ¼ D u @ ou1 ðuU Þ1 ðtðuU ð pÞÞÞ 0 0 1 ! ! 1

Z

2 o

½ f dtA þ D u @ ou2 ðuU Þ1 ðtðuU ð pÞÞÞ 0 0 1 ! ! 1

Z

1 o

¼ D u @ ½ f dtA ou1 ðuU Þ1 ðtð0;0ÞÞ 0 0 1 ! ! 1

Z

2 o

½ f dtA þ D u @ ou2 1 0

ðuU Þ ðtð0;0ÞÞ

70

2 Tangent Spaces

0 1 Z 1 ¼ D u @

o

ou1

0

0 1 Z 2 @ þ D u 0

¼ D u1 @ 0

¼ D u1 @

0

Z1 0

¼ D u1

ðuU Þ1 ð0;0Þ

! 1 ½ f dtA

! ! 1

o

½ f dtA ou2 ðuU Þ1 ð0;0Þ 1 0 1

! ! Z 2 o

Aþ D u @ ½ f dt ou1

p

0

1 0

! ! Z1

o

1dtA þ D u2 @ ½f ou1

p

¼ D u1

!

0

! ! o

½ f 1 þ D u2 ou1 p

! ! o

½ f þ D u2

1 ou p

1

! ! o

½ f dtA ou2 p

1

! ! Z1

o

1dtA ½f ou2

p

! ! o

½f 1 ou2 p

! ! o

½ f ¼ RHS: ou2 p

0

h h

Lemma 2.36 Let M be a 2-dimensional smooth manifold. Let p be an element of M: Then, Dp ðMÞ is a 2-dimensional real linear space. Proof Since M is a 2-dimensional smooth manifold, and p is in M; there exists an admissible coordinate chart ðU; uU Þ of M satisfying p 2 U: Here, we can ﬁnd admissible coordinate chart ðU; wU Þ of M satisfying wU ðxÞ ¼ uU ðxÞ uU ðpÞ for every x in U: Hence, wU ðpÞ ¼ 0: Let us deﬁne functions u1 : U ! R;

u2 : U ! R

as follows: For every x in U;

u1 ð xÞ; u2 ð xÞ wU ð xÞ:

Then, by Lemma 2.35, fouo1 jp ; ouo2 jp g is a basis of the real linear space Dp ðMÞ: Hence, the dimension of Dp ðMÞ is 2: h As above, we can prove the following Lemma 2.37 Let M be an m-dimensional smooth manifold. Let p be an element of M: Then, Dp ðMÞ is an m-dimensional real linear space. Also, if ðU; uU Þ is any admissible coordinate chart of M satisfying p 2 U; then fouo1 jp ; . . .; ouom jp g is a basis of Dp ðMÞ; where u1 ; . . .; um are the component functions of uU :

2.4 Derivations

71

Theorem 2.38 Let M be an m-dimensional smooth manifold. Let p be an element of M: Put Hp ðM Þ ½ f : ½ f 2 F p ðM Þ; and c; ½ f ¼ 0

for every c in Cp ðM Þ :

Then, Hp ðMÞ is a linear subspace of the real linear space F p ðMÞ: In short, we say that Hp is a linear subspace of F p : Proof Let us try to prove that Hp ðMÞ is nonempty. Let us recall that ½0 2 F p ðMÞ; where 0 denotes the constant function zero deﬁned on M: Since, for every c in Cp ðMÞ, 0ðcðt þ 0ÞÞ 0ðcð0ÞÞ 00 ¼ lim ¼ lim 0 ¼ 0; t!0 t!0 t!0 t t

c; ½0 ¼ lim

by the deﬁnition of Hp ðMÞ; ½0 2 Hp ðMÞ: Hence, Hp ðMÞ is nonempty. Now, it remains to be proved that 1. If ½f 2 Hp ðMÞ; ½g 2 Hp ðMÞ; then ½f þ ½g 2 Hp ðMÞ; 2. For every real a; and for every ½f 2 Hp ðMÞ; a½f 2 Hp ðMÞ: For 1: Let ½f 2 Hp ðMÞ; ½g 2 Hp ðMÞ: Since ½f 2 Hp ðMÞ; ½g 2 Hp ðMÞ; and Hp ðMÞ is contained in F p ðMÞ; ½f 2 F p ðMÞ; ½g 2 F p ðMÞ: Since ½f 2 F p ðMÞ; ½g 2 F p ðMÞ; and by Theorem 2.23, F p ðMÞ is a real linear space, ½f þ ½g 2 F p ðMÞ: Now, let us take any c in Cp ðMÞ. Since ½f 2 Hp ðMÞ; by the deﬁnition of Hp ðMÞ; c; ½f ¼ 0: Similarly, c; ½g ¼ 0: Now, by Theorem 2.26, c; ½ f þ ½g ¼ c; ½ f þ c; ½g ¼ 0 þ 0 ¼ 0: So, by the deﬁnition of Hp ðMÞ; ½f þ ½g 2 Hp ðMÞ: This proves 1. For 2: Let a be any real number, and let ½f 2 Hp ðMÞ: Since ½f 2 Hp ðMÞ; and Hp ðMÞ is contained in F p ðMÞ; ½f 2 F p ðMÞ: Since a is a real number, ½f 2 F p ðMÞ;and by Theorem 2.23, F p ðMÞ is a real linear space, a½f 2 F p ðMÞ: Now, let us take any c in Cp ðMÞ. Since ½f 2 Hp ðMÞ; by the deﬁnition of Hp ðMÞ; c; ½f ¼ 0: Now, by Theorem 2.26, c; a½ f ¼ a c; ½ f ¼ að0Þ ¼ 0: So, by the deﬁnition of Hp ðMÞ; a½f 2 Hp ðMÞ: This proves 2. Hence, Hp ðMÞ is a linear subspace of the real linear space F p ðMÞ: h

72

2 Tangent Spaces

2.5 Cotangent Spaces Deﬁnition Let M be an m-dimensional smooth manifold. Let p be an element of M: The set

½ f : ½ f 2 F p ðM Þ; and c; ½ f ¼ 0 for every c in Cp ðM Þ

is denoted by Hp ðMÞ (or, simply Hp Þ: From Theorem 2.38, Hp is a linear subspace F of the real linear space F p : So it is meaningful to write the quotient space Hpp : We know that Fp ½ f þ Hp : ½ f 2 F p : Hp Here, vector addition, and scalar multiplication over For every ½f ; ½g 2 F p ; and for every real t;

Fp Hp

are deﬁned as follows:

½ f þ Hp þ ½g þ Hp ð½ f þ ½gÞ þ Hp ¼ ½f þ g þ Hp ; t ½ f þ Hp ðt½ f Þ þ Hp ¼ ½tf þ Hp : F

The quotient space Hpp is denoted by Tp ðMÞ (or, simply Tp ) and is called the F cotangent space of M at p: Since Hpp is a real linear space, the cotangent space Tp of

M at p is a real linear space. Intuitively, in Tp , we will not distinguish between ½f and ½g whenever ½f ½g is in Hp :

Deﬁnition Let M be an m-dimensional smooth manifold. Let p be an element of M: Let ½f 2 F p : ½f þ Hp is denoted by ½~f ðor; ½f Þ or ðdf Þp ; and is called the cotangent vector on M at p determined by the term ½f : Thus, we can write Tp ¼ ½~f : ½ f 2 F p or, n o Tp ¼ ðdf Þp : ½ f 2 F p : From the above discussion, we get the following formulae:

or,

½ f þ½g ¼ ½f þ g t½ f ¼ ½tf

2.5 Cotangent Spaces

73

(

ðdf Þp þðdgÞp ¼ ðd ðf þ gÞÞp t ðdf Þp ¼ ðd ðtf ÞÞp :

Theorem 2.39 Let M be an m-dimensional smooth manifold. Let p be an element of M: Let ½f be in F p ðMÞ where f is in Cp1 ðMÞ: Let ðU; uU Þ be an admissible coordinate chart of M satisfying p 2 U: Then, ½f 2 Hp ðMÞ if and only if

D1 f ðuU Þ1 ðuU ð pÞÞ ¼ D2 f ðuU Þ1 ðuU ð pÞÞ ¼ ¼ Dm f ðuU Þ1 ðuU ð pÞÞ ¼ 0:

In other words, ½f 2 Hp ðMÞ if and only if ðouo1 jp Þ½f ¼ ðouo2 jp Þ½f ¼ ¼ ¼ 0:

ðouom jp Þ½f

Proof (if part): Let

D1 f ðuU Þ1 ðuU ð pÞÞ ¼ ¼ Dm f ðuU Þ1 ðuU ð pÞÞ ¼ 0:

We have to show that ½f 2 Hp ðMÞ; that is, c; ½f ¼ 0 for every c in Cp ðMÞ. dðf cÞðtÞ jt¼0 ; we For this purpose, let us take any c in Cp ðMÞ. Since c; ½f ¼ dt must prove that

dðf cÞðtÞ

¼ 0: dt t¼0 Since c is in Cp ðMÞ, by the deﬁnition of Cp ðMÞ; c is a parametrized curve in M through p; and hence, there exists a real number d [ 0 such that c is deﬁned on the open interval ðd; dÞ; cð0Þ ¼ p; and c is a smooth map from ðd; dÞ to M: As in the proof of Theorem 2.24, we can show that 0 is an interior point of the domain of the mapping uU c: Since 0 is an interior point of the domain of the mapping uU c; there exists e [ 0 such that e\d; and each of the m component functions of the mapping ðuU cÞ : ðe; eÞ ! Rm is C 1 at 0 in ðe; eÞ; and hence, each of the m functions F1 : ðe; eÞ ! R; . . .; Fm : ðe; eÞ ! R is C 1 at 0 in ðe; eÞ where ðF1 ðtÞ; . . .; Fm ðtÞÞ ðuU cÞðtÞ for every t in ðe; eÞ: Since f is in Cp1 ðMÞ, by the deﬁnition of Cp1 ðMÞ; dom f is an open neighborhood of p in M; and

74

2 Tangent Spaces

f ðuU Þ1 : uU ððdom f Þ \ U Þ ! R

is C 1 at the point uU ðpÞ ð¼ uU ðcð0ÞÞ ¼ ðuU cÞð0ÞÞ in Rm ; and hence, the partial derivatives ðD1 ðf ðuU Þ1 ÞÞððuU cÞð0ÞÞ; . . .; ðDm ðf ðuU Þ1 ÞÞððuU cÞð0ÞÞ exist. Now,

d f ðuU Þ1 ðuU cÞ ðtÞ

dðf cÞðtÞ

LHS ¼

¼ dt dt

t¼0

¼

h D1 f ðuU Þ1 ððuU cÞð0ÞÞ

2

3 dF 1 ðtÞ 6 7 i6 dt 7 6 7 ... Dm f ðuU Þ1 ððuU cÞð0ÞÞ 6 ... 7 6 7 4 dF ðtÞ 5 t¼0

m

2

¼

h D1 f ðuU Þ1 ðuU ð pÞÞ

dt

3

dF1 ðtÞ 6 7 i6 dt 7 6 7 ... Dm f ðuU Þ1 ðuU ð pÞÞ 6 .. 7 6. 7 4 dF ðtÞ 5

t¼0

m

dt

2

3 dF1 ðtÞ 6 dt 7 6 7 6 7 ¼ ½ 0 . . . 0 6 ... 7 6 7 4 dF ðtÞ 5

t¼0

dF1 ðtÞ dFm ðtÞ ¼ 0 ¼ 0 ¼ RHS: þ ... þ 0 dt dt t¼0

m

dt

t¼0

h Proof (only if part): Let ½f 2 Hp ðMÞ: We have to show that D1 f ðuU Þ1 ðuU ð pÞÞ ¼ ¼ Dm f ðuU Þ1 ðuU ð pÞÞ ¼ 0: If not, otherwise, let ðD1 ðf ðuU Þ1 ÞÞðuU ðpÞÞ (for simplicity) be nonzero. We have to arrive at a contradiction. Let us deﬁne a function c1 : ð1; 1Þ ! Rm as follows: For every t in the open interval ð1; 1Þ; c1 ðtÞ ðt; 0; . . .; 0Þ þ uU ð pÞ: Put c ðuU Þ1 c1 : We shall try to prove that 1. cð0Þ ¼ p; 2. 0 is an interior point of the domain of c; 3. c is a smooth map from ðd; dÞ to M for some d [ 0:

2.5 Cotangent Spaces

75

For 1: Here, LHS ¼ cð0Þ ¼ ðuU Þ1 c1 ð0Þ ¼ ðuU Þ1 ðc1 ð0ÞÞ ¼ ðuU Þ1 ðð0; 0; . . .; 0Þ þ uU ð pÞÞ ¼ ðuU Þ1 ðuU ð pÞÞ ¼ p ¼ RHS:

This proves 1. For 2: By the deﬁnition of c1 ; the function c1 : ð1; 1Þ ! Rm is continuous. Since c1 : ð1; 1Þ ! Rm is continuous, and 0 is in the open interval ð1; 1Þ; c1 is continuous at 0. Since ðU; uU Þ is a coordinate chart of M; ðuU Þ1 : uU ðUÞ ! U is a 1–1; onto, continuous function, and uU ðUÞ is an open subset of Rm : Since p 2 U; uU ðpÞ 2 uU ðUÞ: Since c1 ð0Þ ¼ ð0; 0; . . .; 0Þ þ uU ðpÞ ¼ uU ðpÞ 2 uU ðUÞ; and ðuU Þ1 : uU ðUÞ ! U is continuous, ðuU Þ1 : uU ðUÞ ! U is continuous at the point c1 ð0Þ: Since c1 ð0Þ 2 uU ðUÞ; and uU ðUÞ is an open subset of Rm ; uU ðUÞ is an open neighborhood of c1 ð0Þ: Since c1 : ð1; 1Þ ! Rm is continuous, uU ðUÞ is an open neighborhood of c1 ð0Þ; there exists d [ 0 such that d\1 and, for every t in ðd; dÞ we have c1 ðtÞ 2 uU ðUÞ; and hence, cðtÞ ¼ ððuU Þ1 c1 ÞðtÞ ¼ ðuU Þ1 ðc1 ðtÞÞ 2 U: Since cðtÞ 2 U for every t in ðd; dÞ; it follows that ðd; dÞ is a subset of the domðcÞ: Hence, 0 is an interior point of the domain of the function c: This proves 2. For 3: In 2, we have seen that c is deﬁned over ðd; dÞ: Now, it remains to be proved that c is a smooth map from ðd; dÞ to M; that is, c is C 1 at every point t in ðd; dÞ: Here, c is deﬁned over ðd; dÞ; so cðtÞ is meaningful for every t in ðd; dÞ: Since cðtÞ is meaningful for every t in ðd; dÞ; and cðtÞ ¼ ðuU Þ1 c1 ðtÞ ¼ ðuU Þ1 ðc1 ðtÞÞ ¼ ðuU Þ1 ððt; 0; . . .; 0Þ þ uU ð pÞÞ; ðuU Þ1 ððt; 0; . . .; 0Þ þ uU ðpÞÞ is meaningful for every t in ðd; dÞ: Since ðU; uU Þ is a coordinate chart of M; ðuU Þ1 : uU ðUÞ ! U: Since ðuU Þ1 : uU ðUÞ ! U; and ðuU Þ1 ððt; 0; . . .; 0Þ þ uU ðpÞÞ is meaningful for every t in ðd; dÞ; ðuU Þ1 ððt; 0; . . .; 0Þ þ uU ðpÞÞ is in U for every t in ðd; dÞ: Since ðuU Þ1 ððt; 0; . . .; 0Þ þ uU ðpÞÞ is in U for every t in ðd; dÞ; and cðtÞ ¼ ðuU Þ1 ððt; 0; . . .; 0Þ þ uU ðpÞÞ; cðtÞ is in U for every t in ðd; dÞ: Thus, c maps ðd; dÞ to U: Now, it remains to be proved that c is a smooth map from ðd; dÞ to M; that is, c is C 1 at every point t in ðd; dÞ: For this purpose, let us ﬁx any t in ðd; dÞ: From the deﬁnition of c1 ; c1 is C 1 at t: Since t is in ðd; dÞ; and c maps ðd; dÞ to U; cðtÞ is in U: Since c1 is C 1 at t; c ¼ ðuU Þ1 c1 ; c maps ðd; dÞ to U; and t is in ðd; dÞ; uU c is C1 at t: Since uU c is C1 at t; and ðU; uU Þ is an admissible

76

2 Tangent Spaces

coordinate chart of M satisfying cðtÞ 2 U, by Theorem 2.2, c is C 1 at t in ðd; dÞ: This proves 3. From 1, 2, and 3, we ﬁnd that c is a parametrized curve in M through p; that is, c is in Cp ðMÞ: Since ½f 2 Hp ðMÞ; and c is in Cp ðMÞ, by the deﬁnition of Hp ðMÞ; c; ½f ¼ 0: Put uU ð pÞ ða1 ; . . .; am Þ: Next, let us deﬁne m functions F1 : ðd; dÞ ! R; . . .; Fm : ðd; dÞ ! R such that for every t in ðd; dÞ; ðF1 ðtÞ; . . .; Fm ðtÞÞ ðuU cÞðtÞ ¼ c1 ðtÞ ¼ ðt; 0; . . .; 0Þ þ uU ð pÞ ¼ ðt; 0; . . .; 0Þ þ ða1 ; . . .; am Þ ¼ ðt þ a1 ; a2 ; . . .; am Þ: Hence,

d dðf cÞðtÞ

0 ¼ c; ½ f ¼ ¼

dt t¼0

¼

h

f ðuU Þ1 ðuU cÞ ðtÞ

dt

D1 f ðuU Þ1 ððuU cÞð0ÞÞ

t¼0

2

3 dðt þ a1 Þ 6 7 dt 6 7 6 7 7 i6 dða2 Þ 6 dt 7 1 Dm f ð u U Þ ððu U c Þð0 ÞÞ 6 7 6 . 7 6 . 7 . 6 7 4 dða Þ 5 m

2

3 1 h i6 0 7 6 7 ¼ D1 f ðuU Þ1 ðuU ð pÞÞ Dm f ðuU Þ1 ðuU ð pÞÞ 6 .. 7 4 . 5 0 t¼0 1 1 ¼ D 1 f ðu U Þ ðuU ð pÞÞ 1 þ 0 þ þ 0 ¼ D1 f ðuU Þ ðuU ð pÞÞ 6¼ 0;

dt

t¼0

h

which is a contradiction.

Theorem 2.40 Let M be a m-dimensional smooth manifold. Let p be an element of M: Let f 1 ; f 2 2 Cp1 ðMÞ: Let G be an open neighborhood of ðf 1 ðpÞ; f 2 ðpÞÞ in R2 : Let F : G ! R be a smooth function. Then, there exists f in Cp1 ðMÞ such that for every x in dom f ; f ð xÞ ¼ F f 1 ð xÞ; f 2 ð xÞ :

2.5 Cotangent Spaces

77

Proof For this purpose, by the deﬁnition of Cp1 ðMÞ; we must ﬁnd a function f such that 1. dom f is an open neighborhood of p in M; 2. f ðxÞ ¼ Fðf 1 ðxÞ; f 2 ðxÞÞ for every x in dom f ; and 3. for every admissible coordinate chart ðU; uU Þ of M satisfying p 2 U,

f ðuU Þ1 : uU ððdom f Þ \ U Þ ! R

is C 1 at the point uU ðpÞ in Rm : Since f 1 2 Cp1 ðMÞ, by the deﬁnition of Cp1 ðMÞ; f 1 is a real-valued function whose domðf 1 Þ is an open neighborhood of p in M: Similarly, f 2 is a real-valued function whose domðf 2 Þ is an open neighborhood of p in M: Since domðf 1 Þ is an open neighborhood of p in M; and domðf 2 Þ is an open neighborhood of p in M; their intersection domðf 1 Þ \ domðf 2 Þ is an open neighborhood of p in M: Put V1 dom f 1 \ dom f 2 : Let us deﬁne a function g : V1 ! R2 as follows: For every x in V1 gð xÞ f 1 ð xÞ; f 2 ð xÞ : Since f 1 2 Cp1 ðMÞ; f 1 is continuous at p: Similarly, f 2 is continuous at p: Since f ; f 2 are continuous at p; and gðxÞ ¼ ðf 1 ðxÞ; f 2 ðxÞÞ for every x in V1 ; g : V1 ! R2 is continuous at p. Since g : V1 ! R2 is continuous at p, and G is an open neighborhood of ðf 1 ðpÞ; f 2 ðpÞÞð¼ gðpÞÞ in R2 ; there exists an open neighborhood Vð V1 ¼ domðf 1 Þ \ domðf 2 ÞÞ of p such that gðVÞ is contained in G. Since gðVÞ is contained in G; and F : G ! R, for every x in V; FðgðxÞÞð¼ Fðf 1 ðxÞ; f 2 ðxÞÞÞ is a real number. Now, let us deﬁne a function 1

f :V !R as follows: For every x in V; f ð xÞ F f 1 ð xÞ; f 2 ð xÞ : Clearly, the conditions 1 and 2 are satisﬁed.

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2 Tangent Spaces

For 3, let us take an admissible coordinate chart ðU; uU Þ of M satisfying p 2 U. We have to show that the function f ðuU Þ1 : uU ðV \ U Þ ! R is C1 at the point uU ðpÞ in Rm : Here, for every uU ðxÞ in uU ðV \ UÞ; where x is in V \ U; we have f ðuU Þ1 ðuU ð xÞÞ ¼ f ðuU Þ1 uU ð xÞ ¼ f ðuU Þ1 uU ð xÞ ¼ f ð xÞ ¼ F f 1 ð xÞ; f 2 ð xÞ ¼ F f 1 ðuU Þ1 uU ð xÞ ; f 2 ðuU Þ1 uU ð xÞ ¼F f 1 ðuU Þ1 uU ð xÞ ; f 2 ðuU Þ1 uU ð xÞ ¼F f 1 ðuU Þ1 uU ð xÞ; f 2 ðuU Þ1 uU ð xÞ ¼ F f 1 ðuU Þ1 ðuU ð xÞÞ; f 2 ðuU Þ1 ðuU ð xÞÞ : Since f 1 2 Cp1 ðMÞ; and ðU; uU Þ is an admissible coordinate chart ðU; uU Þ of M satisfying p 2 U, by the deﬁnition of Cp1 ðMÞ;

f 1 ðuU Þ1 : uU dom f 1 \ U ! R

is C1 at the point uU ðpÞ in Rm : Since ðdom f Þ \ U ¼ V \ U domðf 1 Þ \ domðf 2 Þ \ U domðf 1 Þ \ U; uU ððdom f Þ \ UÞ uU ððdomðf 1 ÞÞ \ UÞ: Since uU ððdom f Þ \ UÞ uU ððdomðf 1 ÞÞ \ UÞ; and ðf 1 ðuU Þ1 Þ : uU ðdomðf 1 Þ \ UÞ ! R is C1 at the point uU ðpÞ in Rm ; the restriction of f 1 ðuU Þ1 to uU ððdom f Þ \ UÞ is C1 at the point uU ðpÞ in Rm : Therefore, for every x in ðdom f Þ \ U; uU ð xÞ 7! f 1 ðuU Þ1 ðuU ð xÞÞ is C 1 at the point uU ð pÞ in Rm : Similarly, for every x in ðdom f Þ \ U; uU ð xÞ 7! f 2 ðuU Þ1 ðuU ð xÞÞ is C 1 at the point uU ð pÞ in Rm : It follows that, for every x in ðdom f Þ \ U; uU ð xÞ 7! f 1 ðuU Þ1 ðuU ð xÞÞ; f 2 ðuU Þ1 ðuU ð xÞÞÞ is C1 at the point uU ð pÞ in Rm : Since, for every x in ðdom f Þ \ U; uU ðxÞ 7! ððf 1 ðuU Þ1 ÞðuU ðxÞÞ; ðf 2 ðuU Þ1 ÞðuU ðxÞÞÞ is C 1 at the point uU ðpÞ in Rm ; G is an open neighborhood of ðf 1 ðpÞ; f 2 ðpÞÞ in R2 ; and F : G ! R is a smooth function, for every x in ðdom f Þ \ Uð¼ V \ UÞ; the function uU ðxÞ 7! Fððf 1 ðuU Þ1 ÞðuU ðxÞÞ; ðf 2 ðuU Þ1 ÞðuU ðxÞÞÞð¼ ðf ðuU Þ1 ÞðuU ðxÞÞÞ is C 1 at the point uU ðpÞ in Rm ; and hence,

2.5 Cotangent Spaces

79

f ðuU Þ1 : uU ðV \ U Þ ! R is C 1 at the point uU ðpÞ in Rm : This proves 3.

h

Theorem 2.41 Let M be an m-dimensional smooth manifold. Let p be an element of M: Let f 1 ; f 2 2 Cp1 ðMÞ: Let G be an open neighborhood of ðf 1 ðpÞ; f 2 ðpÞÞ in R2 : Let F : G ! R be a smooth function. Let f ; g be in Cp1 ðMÞ such that for every x in dom f ; f ð xÞ ¼ F f 1 ð xÞ; f 2 ð xÞ and, for every x in dom g; gð xÞ ¼ F f 1 ð xÞ; f 2 ð xÞ : Then, f g: Proof By Theorem 2.40, the existence of f ; g are guaranteed. Since f is in Cp1 ðMÞ; dom f is an open neighborhood of p in M: Similarly, dom g is an open neighborhood of p in M: Since dom f is an open neighborhood of p in M; and dom g is an open neighborhood of p in M; their intersection ðdom f Þ \ ðdom gÞ is an open neighborhood of p in M: Next, let us take any x in ðdom f Þ \ðdom gÞ: It remains to be showed that f ðxÞ ¼ gðxÞ: Since x is in ðdom f Þ \ðdom gÞ; x is in dom f : Since x is indom f ; f ðxÞ ¼ Fðf 1 ðxÞ; f 2 ðxÞÞ: Similarly, gðxÞ ¼ Fðf 1 ðxÞ; f 2 ðxÞÞ: Since f ðxÞ ¼ h Fðf 1 ðxÞ; f 2 ðxÞÞ and gðxÞ ¼ Fðf 1 ðxÞ; f 2 ðxÞÞ; f ðxÞ ¼ gðxÞ: Theorem 2.42 Let M be an m-dimensional smooth manifold. Let p be an element of M: Let f 1 ; f 2 2 Cp1 ðMÞ: Let G be an open neighborhood of ðf 1 ðpÞ; f 2 ðpÞÞ in R2 : Let F : G ! R be a smooth function. Then, there exists a unique ½f in F p ðMÞ; where f is in Cp1 ðMÞ; such that for every x in dom f ; f ð xÞ ¼ F f 1 ð xÞ; f 2 ð xÞ : Also ½ f ¼ ðD1 F Þ f 1 ð pÞ; f 2 ð pÞ f 1 þ ðD2 F Þ f 1 ð pÞ; f 2 ð pÞ f 2 : Proof Existence: By Theorem 2.40, there exists f in Cp1 ðMÞ such that for every x in dom f ; f ð xÞ ¼ F f 1 ð xÞ; f 2 ð xÞ :

80

2 Tangent Spaces

Further, since f is in Cp1 ðMÞ; the C1 -germ ½f is in F p ðMÞ: This proves the existence part of the theorem. Uniqueness: For this purpose, let ½f ; ½g be in F p ðMÞ; where f ; g are in Cp1 ðMÞ; such that for every x in dom f ; f ð xÞ ¼ F f 1 ð xÞ; f 2 ð xÞ and, for every x in dom g;

gð xÞ ¼ F f 1 ð xÞ; f 2 ð xÞ :

We have to prove that ½f ¼ ½g: By Theorem 2.41, f g: Since f ; g are in Cp1 ðMÞ; f g; and is an equivalence relation over Cp1 ðMÞ; the equivalence class ½f determined by f , and the equivalence class ½g determined by g are equal, that is, ½f ¼ ½g: This proves the uniqueness part of the theorem. h Since ½f 2 F p ðMÞ; the cotangent vector ½f ðthat is; ½f þ Hp Þ on M at p F determined by the germ ½f is a member of the quotient space Hpp ðthat is; the cotangent space Tp ðMÞ of M at pÞ: Next, since f 1 2 Cp1 ðMÞ; the equivalence class ½f 1 determined by f 1 is a member of F p ðMÞ; and hence, the cotangent vector ½f 1 on M at p determined by F the term ½f 1 is a member of the quotient space Hpp : Similarly, the cotangent vector

½f 2 onM at p determined by the term ½f 2 is a member of the quotient space Since ½f 1 ; ½f 2 are

F in Hpp 1

; and

Fp Hp

2

:

is a real linear space, the linear combination

ððD1 FÞðf ðpÞ; f ðpÞÞÞ½f þ ððD2 FÞðf 1 ðpÞ; f 2 ðpÞÞÞ½f 2 of ½f 1 ; ½f 2 is in Finally, we have to show that 1

Fp Hp

½ f ¼ ðD1 F Þ f 1 ð pÞ; f 2 ð pÞ f 1 þ ðD2 F Þ f 1 ð pÞ; f 2 ð pÞ f 2 : Since ðD1 F Þ f 1 ð pÞ; f 2 ð pÞ f 1 þ ðD2 F Þ f 1 ð pÞ; f 2 ð pÞ f 2 ¼ ðD1 F Þ f 1 ð pÞ; f 2 ð pÞ f 1 þ ðD2 F Þ f 1 ð pÞ; f 2 ð pÞ f 2 ¼ ðD1 F Þ f 1 ð pÞ; f 2 ð pÞ f 1 þ ðD2 F Þ f 1 ð pÞ; f 2 ð pÞ f 2 we have to show that ½ f ¼ ðD1 F Þ f 1 ð pÞ; f 2 ð pÞ f 1 þ ðD2 F Þ f 1 ð pÞ; f 2 ð pÞ f 2 ;

Fp Hp

:

2.5 Cotangent Spaces

81

that is, f þ Hp ¼

ðD1 F Þ f 1 ð pÞ; f 2 ð pÞ f 1 þ ðD2 F Þ f 1 ð pÞ; f 2 ð pÞ f 2 þ Hp ;

that is,

ðD1 F Þ f 1 ð pÞ; f 2 ð pÞ f 1 þ ðD2 F Þ f 1 ð pÞ; f 2 ð pÞ f 2 f 2 Hp ;

that is, for every c in Cp ðMÞ, c;

ðD1 F Þ f 1 ð pÞ; f 2 ð pÞ f 1 þ ðD2 F Þ f 1 ð pÞ; f 2 ð pÞ f 2 f ¼ 0;

that is, for every c in Cp ðMÞ, c; ðD1 F Þ f 1 ð pÞ; f 2 ð pÞ f 1 þ ðD2 F Þ f 1 ð pÞ; f 2 ð pÞ f 2 ½ f ¼ 0; that is, for every c in Cp ðMÞ,

ðD1 F Þ f 1 ð pÞ; f 2 ð pÞ c; f 1 þ ðD2 F Þ f 1 ð pÞ; f 2 ð pÞ c; f 2

c; ½ f ¼ 0; that is, for every c in Cp ðMÞ,

ðD1 F Þ f 1 ð pÞ; f 2 ð pÞ c; f 1 þ ðD2 F Þ f 1 ð pÞ; f 2 ð pÞ c; f 2 ¼ c; ½ f :

For this purpose, let us take any c in Cp ðMÞ. Since c is in Cp ðMÞ; there exists a real number d [ 0 such that c is deﬁned on the open interval ðd; dÞ; and cð0Þ ¼ p: We have to prove that

ðD1 F Þ f 1 ð pÞ; f 2 ð pÞ c; f 1 þ ðD2 F Þ f 1 ð pÞ; f 2 ð pÞ c; f 2 ¼ c; ½ f :

LHS ¼ ðD1 F Þ f 1 ð pÞ; f 2 ð pÞ c; f 1 þ ðD2 F Þ f 1 ð pÞ; f 2 ð pÞ c; f 2

1 dðf 1 cÞðtÞ

1 dðf 2 cÞðtÞ

2 2

¼ ðD1 F Þ f ð pÞ; f ð pÞ

þ ðD2 F Þ f ð pÞ; f ð pÞ

dt dt t¼0

t¼0 1 2

dðf ðcðtÞÞÞ

þ ðD2 F Þ f 1 ð pÞ; f 2 ð pÞ dðf ðcðtÞÞÞ : ¼ ðD1 F Þ f 1 ð pÞ; f 2 ð pÞ

dt dt t¼0 t¼0

dðf cÞðtÞ

dðf ðcðtÞÞÞ

dðF ðf 1 ðcðtÞÞ; f 2 ðcðtÞÞÞÞ

RHS ¼ c; ½ f ¼

¼

¼

dt dt dt t¼0 t¼0 t¼0

dðF ððf 1 cÞðtÞ; ðf 2 cÞðtÞÞÞ

¼

dt t¼0

82

2 Tangent Spaces

dðf 1 cÞðtÞ

¼ ðD1 F Þ f c ð0Þ; f c ð0Þ

dt t¼0

1 2 dðf 2 cÞðtÞ

þ ðD2 F Þ f c ð0Þ; f c ð0Þ

dt t¼0

1

1 d ð f ð c ð t Þ Þ Þ

¼ ðD1 F Þ f ðcð0ÞÞ; f 2 ðcð0ÞÞ

dt t¼0

1 dðf 2 ðcðtÞÞÞ

2

þ ðD2 F Þ f ðcð0ÞÞ; f ðcð0ÞÞ

dt t¼0

1

1 1 dðf 2 ðcðtÞÞÞ

d ð f ð c ð t Þ Þ Þ 2 2

: ¼ ðD1 F Þ f ð pÞ; f ð pÞ

þ ðD2 F Þ f ð pÞ; f ð pÞ

dt dt t¼0 t¼0

1

2

Hence, LHS = RHS.

h

Note 2.43 Since ½f is also denoted by ðdf Þp ; the formula in Theorem 2.17 can be written as: ðdf Þp ¼ ðD1 F Þ f 1 ð pÞ; f 2 ð pÞ df 1 p þ ðD2 F Þ f 1 ð pÞ; f 2 ð pÞ df 2 p : oF 1 2 Since ðD1 FÞðf 1 ðpÞ; f 2 ðpÞÞ is classically written as ox 1 ðf ðpÞ; f ðpÞÞ; etc., we can write: 1 2 1 2 oF 1 oF 1 2 2 f ð pÞ; f ð pÞ f ð pÞ; f ð pÞ dF f ; f p ¼ ðdf Þp ¼ df p þ df p : ox1 ox1

Note 2.44 Since the multiplication operation of real numbers is a smooth function from R2 to R, in Theorem 2.42, we can take multiplication operation in place of F: oF 1 2 2 oF 1 2 1 So ox 1 ðf ðpÞ; f ðpÞÞ becomes f ðpÞ; and ox2 ðf ðpÞ; f ðpÞÞ becomes f ðpÞ: Hence, we get 1 2 1 2 oF 1 oF 1 2 2 d f f p ¼ ðdf Þp ¼ f ð pÞ; f ð pÞ f ð pÞ; f ð pÞ df p þ df p ox1 ox2 2 1 1 2 ¼ f ð pÞ df p þ f ð pÞ df p or,

1 2 2 1 1 2 d f f p ¼ f ð pÞ df p þ f ð pÞ df p :

Similarly, for every f 1 ; f 2 ; f 3 in Cp1 ðMÞ; we have 1 2 3 2 3 1 3 1 2 1 2 3 d f f f p ¼ f ð pÞ f ð pÞ df p þ f ð pÞ f ð pÞ df p þ f ð pÞ f ð pÞ df p ;

etc. If we recollect our previous results, we get the following theorem.

2.5 Cotangent Spaces

83

Theorem 2.45 Let M be an m-dimensional smooth manifold. Let p be an element of M: Let f ; g 2 Cp1 ðMÞ: Then, 1. ðdðf þ gÞÞp ¼ ðdf Þp þ ðdgÞp ; 2. ðdðtf ÞÞp ¼ tððdf Þp Þ; for every real t; 3. ðdðf gÞÞp ¼ ðgðpÞÞðdf Þp þ ðf ðpÞÞðdgÞp : h

Proof Proofs have already been supplied.

Theorem 2.46 Let M be an m-dimensional smooth manifold. Let p be an element of M: Let f 1 ; f 2 ; f 3 2 Cp1 ðMÞ: Let G be an open neighborhood of ðf 1 ðpÞ; f 2 ðpÞ; f 3 ðpÞÞ in R3 : Let F : G ! R be a smooth function. Then, there exists a unique ½f in F p ðMÞ; where f is in Cp1 ðMÞ; such that for every x in dom f ; f ð xÞ ¼ F f 1 ð xÞ; f 2 ð xÞ; f 3 ð xÞ : Also ½ f ¼ ðD1 F Þ f 1 ð pÞ; f 2 ð pÞ; f 3 ð pÞ f 1 þ ðD2 F Þ f 1 ð pÞ; f 2 ð pÞ; f 3 ð pÞ f 2 þ ðD3 F Þ f 1 ð pÞ; f 2 ð pÞ; f 3 ð pÞ f 3 : Proof Its proof is quite similar to the proof of Theorem 2.42, etc.

h

Theorem 2.47 Let M be a 2-dimensional smooth manifold. Let p be an element of M: Let ðU; uU Þ be an admissible coordinate chart of M satisfying p 2 U: Let u1 ; u2 be the component functions of uU : Then, 1. u1 ; u2 are in Cp1 ðMÞ; 2. ½u1 ; ½u2 are linearly independent in the real linear space Tp ðMÞ; 3. for every ½f in Tp ðMÞ; where f is in Cp1 ðMÞ; ½ f ¼

D1 f ðuU Þ1 u1 ð pÞ; u2 ð pÞ u1 þ D2 f ðuU Þ1 u1 ð pÞ; u2 ð pÞ u2 :

In other words, ðdf Þp ¼

! ! 1 o

½ f du p þ ou1 p

! ! 2 o

½ f du p : ou2 p

4. f½u1 ; ½u2 g is a basis of the real linear space Tp ðMÞ; 5. dimðTp ðMÞÞ ¼ 2:

84

2 Tangent Spaces

Fig. 2.2 Cotangent space

Proof 1. This fact has been shown earlier. 2. In 1, we have seen that u1 is in Cp1 ðMÞ: Since u1 is in Cp1 ðMÞ, the C 1 -germ ½f is in F p ðMÞ; and hence, ½u1 ð¼ ½u1 þ Hp Þ is in the cotangent space Tp of M at p: Similarly, ½u2 ð¼ ½u2 þ Hp Þ is in the cotangent space Tp of M at p (see Fig. 2.2). We have to show that ½u1 ; ½u2 are linearly independent. For this purpose, let a1 ½u1 þ a2 ½u2 ¼ 0; where a1 ; a2 are real numbers. We have to show that a1 ¼ Since 0 ¼ a1 ½u1 þ a2 ½u2 ¼ ½a1 u1 þ ½a2 u2 ¼ 0; and a2 ¼ 0. ½a1 u1 þ a2 u2 ; ½a1 u1 þ a2 u2 is in Hp ; and hence, by the deﬁnition of Hp ; c; ½a1 u1 þ a2 u2 ¼ 0 for every c in Cp ðMÞ. Therefore, for every c in Cp ðMÞ, 0 ¼ c; a1 u1 þ a2 u2 ¼ c; a1 u1 þ a2 u2 ¼ c; a1 u1 þ a2 u2

1 2 dðu1 cÞðtÞ

dðu2 cÞðtÞ

¼ a1 c; u þ a2 c; u ¼ a1

þa2

; dt dt t¼0 t¼0 that is,

dðu1 cÞðtÞ

dðu2 cÞðtÞ

a1

þa2

¼ 0. . .ð Þ dt dt t¼0 t¼0 for every c in Cp ðMÞ. Let us deﬁne a function c1 : R ! R 2 as follows: For every t in R; c1 ðtÞ u1 ð pÞ þ t; u2 ð pÞ :

2.5 Cotangent Spaces

85

Clearly, c1 is a smooth function. Here, c1 is continuous, c1 ð0Þ ¼ ðu1 ðpÞ þ 0; u2 ðpÞÞ ¼ ðu1 ðpÞ; u2 ðpÞÞ ¼ uU ðpÞ; and uU ðUÞ is an open neighborhood of uU ðpÞ; so there exists e [ 0 such that for every t in the open interval ðe; eÞ; we have c1 ðtÞ is in uU ðUÞ; and hence, ðuU Þ1 ðc1 ðtÞÞ is in U: Now, let us deﬁne a function c : ðe; eÞ ! M as follows: For every t in ðe; eÞ; cðtÞ ðuU Þ1 ðc1 ðtÞÞ ¼ ðuU Þ1 c1 ðtÞ: We shall try to see that c is in Cp ðMÞ. By the deﬁnition of Cp ðMÞ; we must prove that 1. cð0Þ ¼ p; 2. c is a parametrized curve in the manifold M; that is, c : ðe; eÞ ! M is a smooth map from ðe; eÞ to M; that is, c is C 1 at every point p of ðe; eÞ: LHS ¼ cð0Þ ¼ ðuU Þ1 ðc1 ð0ÞÞ ¼ ðuU Þ1 ðu1 ðpÞ þ 0; u2 ðpÞÞ ¼ ðuU Þ1 ðu1 ðpÞ; u2 ðpÞÞ ¼ ðuU Þ1 ðuU ðpÞÞ ¼ p ¼ RHS: For 2: For this purpose, let us take any t0 in ðe; eÞ: We have to show that c : ðe; eÞ ! M is C1 at t0 : We want to apply Theorem 2.2. Since t0 is in ðe; eÞ, by the deﬁnition of c; cðt0 Þð¼ ðuU Þ1 ðc1 ðt0 ÞÞÞ is in U; and hence, ðU; uU Þ is an admissible coordinate chart of M satisfying cðt0 Þ 2 U: Since uU c ¼ uU ððuU Þ1 c1 Þ ¼ ðuU ðuU Þ1 Þ c1 ¼ c1 ; and c1 is a smooth function, uU c is a smooth function. Since c : ðe; eÞ ! M; t0 is in ðe; eÞ; ðU; uU Þ is an admissible coordinate chart of M satisfying cðt0 Þ 2 U; and uU c is a smooth function so, by Theorem 2.2, c is C 1 at t0 in ðe; eÞ: This proves 2. Thus, we have shown that c is in Cp ðMÞ. Hence, from ð Þ; For 1: Here,

d u1 ðuU Þ1 c1 ðtÞ

d u2 ðuU Þ1 c1 ðtÞ

þa2

0 ¼ a1

dt dt

t¼0 t¼0

d u1 ðuU Þ1 c1 ðtÞ

d u2 ðuU Þ1 c1 ðtÞ

¼ a1

þa2

dt dt

t¼0 t¼0 2 3 h i d ðu1 ð pÞ þ tÞ 1 1 dt 1 1 4 5 ¼ a 1 D 1 u ðuU Þ ðc1 ð0ÞÞ D 2 u ðuU Þ ðc1 ð0ÞÞ d ðu2 ð pÞÞ dt t¼0

86

2 Tangent Spaces 2 3 i d ðu1 ð pÞ þ tÞ 1 dt 5 þ a2 D1 u2 ðuU Þ ðc1 ð0ÞÞ D2 u2 ðuU Þ ðc1 ð0ÞÞ 4 d 2 dt ðu ð pÞÞ t¼0 h i 1 1 1 ¼ a1 D1 u1 ðuU Þ ðc1 ð0ÞÞ D2 u1 ðuU Þ ðc1 ð0ÞÞ 0 h i 1 1 1 2 2 ðc1 ð0ÞÞ D 2 u ð uU Þ ðc1 ð0ÞÞ þ a2 D1 u ðuU Þ 0 ¼ a1 D1 u1 ðuU Þ1 ðc1 ð0ÞÞ þ a2 D1 u2 ðuU Þ1 ðc1 ð0ÞÞ ¼ a1 D1 u1 ðuU Þ1 ðuU ð pÞÞ þ a2 D1 u2 ðuU Þ1 ðuU ð pÞÞ ¼ a1 ð1Þ þ a2 ð0Þ ¼ a1 : h

1

Thus, a1 ¼ 0: Similarly, a2 ¼ 0: Thus, we have shown that ½u1 ; ½u2 are linearly independent. This proves 2. 3. Let us take any ½f ; where f is in Cp1 ðMÞ: We have to ﬁnd reals a1 ; a2 such that ½f ¼ a1 ½u1 þ a2 ½u2 : Since f is in so, by the deﬁnition of Cp1 ðMÞ; f is a real-valued function whose dom f is an open neighborhood of p in M; and

Cp1 ðMÞ

f ðuU Þ1 : uU ððdom f Þ \ U Þ ! R

is C1 at the point uU ðpÞ in R2 : Here, uU ððdom f Þ \ UÞ is an open neighborhood of uU ðpÞð¼ ðu1 ðpÞ; u2 ðpÞÞÞ: Since ðf ðuU Þ1 Þ : uU ððdom f Þ \ UÞ ! R is C1 at the point uU ðpÞ in R2 , for some open neighborhood Gð uU ððdom f Þ \ UÞ R2 Þ of uU ðpÞ; ðf ðuU Þ1 Þ is smooth on G: For every x in ðdom f Þ \ U; we have f ð xÞ ¼ f ðuU Þ1 ðuU ð xÞÞ ¼ f ðuU Þ1 u1 ð xÞ; u2 ð xÞ : Now, we want to apply Theorem 2.42. Since, for every x in ðdom f Þ \ U; f ð xÞ ¼ f ðuU Þ1 u1 ð xÞ; u2 ð xÞ ;

u1 ; u2 2 Cp1 ðM Þ;

G is an open neighborhood of ðu1 ðpÞ; u2 ðpÞÞð¼ uU ðpÞÞ in R2 ; and ðf ðuU Þ1 Þ : G ! R is a smooth function so, by Theorem 2.42, ½ f ¼

u1 ð pÞ; u2 ð pÞ u1 þ D2 f ðuU Þ1 u1 ð pÞ; u2 ð pÞ u2 : D1 f ðuU Þ1

This proves 3. 4. From 2,3, we ﬁnd that f½u1 ; ½u2 g is a basis of the real linear space Tp ðMÞ: This proves 4.

2.5 Cotangent Spaces

87

5. From 4, f½u1 ; ½u2 g is a basis of the real linear space Tp ðMÞ; and the number of elements in the basis is 2; so the dimension of Tp ðMÞ is 2: This proves 5. h As above, we can prove the following theorem. Theorem 2.48 Let M be an m-dimensional smooth manifold. Let p be an element of M: Let ðU; uU Þ be an admissible coordinate chart of M satisfying p 2 U: Let us deﬁne m functions u1 : U ! R; . . .; um : U ! R as follows: For every x in U; uU ð xÞ u1 ð xÞ; . . .; um ð xÞ : Then, 1. u1 ; . . .; um are in Cp1 ðMÞ: 2. ½u1 ; . . .; ½um are linearly independent in the real linear space Tp ðMÞ: 3. For every ½f in Tp ðMÞ; where f is in Cp1 ðMÞ; ½ f ¼

D1 f ðuU Þ1 u1 ð pÞ; . . .; um ð pÞ u1 þ þ Dm f ðuU Þ1 u1 ð pÞ; . . .; um ð pÞ ½um :

In other words, ðdf Þp ¼

! ! 1 o

½ f du p þ þ

ou1 p

! ! o

m ð Þ : ½ f du p

oum p

4. f½u1 ; . . .; ½um g is a basis of the real linear space Tp ðMÞ: 5. dimðTp ðMÞÞ ¼ m: Proof Its proof is quite similar to the proof of Theorem 2.47.

h

Deﬁnition Let M be an m-dimensional smooth manifold. Let p be an element of M: Let ðU; uU Þ be an admissible coordinate chart of M satisfying p 2 U: The basis f½u1 ; . . .; ½um g of the cotangent space Tp ðMÞ of M at p; as deﬁned in Theorem 2.48, is called the natural basis of Tp ðMÞ determined by ðU; uU Þ: Note 2.49 Let M be an m-dimensional smooth manifold. Let p be an element of M: Let c be in Cp ðMÞ. Let ½f ¼ ½g ; where f ; g are in Cp1 ðMÞ: We shall show that c; ½f ¼ c; ½g : Since ½f ¼ ½g ; ½f g ¼ ½f ½g 2 Hp ; and hence,

88

2 Tangent Spaces

0 ¼ c; ½ f ½g ¼ c; ½ f c; ½g or, c; ½ f ¼ c; ½g : This shows that the following deﬁnition is unambiguous. Deﬁnition Let M be an m-dimensional smooth manifold. Let p be an element of M: Let c be in Cp ðMÞ. Let ½f be in Tp ðMÞ; where f is in Cp1 ðMÞ: By c; ½f ; we mean c; ½f : Theorem 2.50 Let M be an m-dimensional smooth manifold. Let p be an element of M: Let us deﬁne a relation over Cp ðMÞ as follows: for every c; c0 in Cp ðMÞ; by c c0 ; we shall mean that for every ½f in Tp ðMÞ; c; ½f ¼ c0 ; ½f : Then, is an equivalence relation over Cp ðMÞ: Proof Here, we must prove: 1. c c for every c in Cp ðMÞ; 2. if c c0 then c0 c; 3. if c c0 and c0 c00 , then c c00 : For 1: Let us take any c in Cp ðMÞ: We have to prove that c c: For this purpose, let us take any ½f in Tp ðMÞ; where f is in Cp1 ðMÞ: Since c; ½f ¼ c; ½f , by the deﬁnition of ; c c: This proves 1. For 2: Let c c0 . We have to prove that c0 c; that is, c0 ; ½f ¼ c; ½f for every ½f in Tp ðMÞ; where f is in Cp1 ðMÞ: For this purpose, let us take any ½f in Tp ðMÞ; where f is in Cp1 ðMÞ: Since c c0 , by the deﬁnition of ; c; ½f ¼ c0 ; ½f ; and hence, c0 ; ½f ¼ c; ½f : This proves 2. For 3: Let c c0 , and c0 c00 . We have to prove that c c00 ; that is, c; ½f ¼ c00 ; ½f for every ½f in Tp ðMÞ; where f is in Cp1 ðMÞ: For this purpose, let us take any ½f in Tp ðMÞ; where f is in Cp1 ðMÞ: Since c c0 , by the deﬁnition of ; c; ½f ¼ c0 ; ½f : Since c0 c00 , by the deﬁnition of ; c0 ; ½f ¼ c00 ; ½f : Since c; ½f ¼ c0 ; ½f and c0 ; ½f ¼ c00 ; ½f ; c; ½f ¼ c00 ; ½f : This proves 3. Thus, we have shown that is an equivalence relation over Cp ðMÞ:

h

Deﬁnition Let M be an m-dimensional smooth manifold. Let p be an element of M: We have seen that the relation ; as deﬁned in Theorem 2.50, is an equivalence relation over Cp ðMÞ: For any c in Cp ðMÞ; the equivalence class of c is denoted by ½c: Thus, for any c in Cp ðMÞ; ½c c0 : c0 2 Cp ðM Þ and c0 c :

2.5 Cotangent Spaces

89

We shall denote the quotient set Cp ðMÞ= by C p ðMÞ (or, simply, C p ). Thus, C p ðM Þ ½c : c 2 Cp ðM Þ : Intuitively, in C p ðMÞ, we will not distinguish between c0 and c whenever c0 c: Note 2.51 Let M be an m-dimensional smooth manifold. Let p be an element of M: Let ½f be in Tp ðMÞ; where f is in Cp1 ðMÞ: Let ½c ¼ ½c0 ; where c; c0 are in Cp ðMÞ: We shall show that c; ½f ¼ c0 ; ½f : Since ½c ¼ ½c0 ; c c0 ; and hence, by Theorem 2.22, c; ½ f ¼ c0 ; ½ f : This shows that the following deﬁnition is unambiguous. Deﬁnition Let M be an m-dimensional smooth manifold. Let p be an element of M: Let ½f be in Tp ðMÞ; where f is in Cp1 ðMÞ: Let c be in Cp ðMÞ. By h½c; ½f i; we mean c; ½f : Thus,

dðf cÞðtÞ

h½c; ½ f i ¼ c; ½ f ¼ c; ½ f ¼

dt t¼0 f ðcð0 þ tÞÞ f ðcð0ÞÞ ; ¼ lim t!0 t which is a real number. Theorem 2.52 Let M be a 2-dimensional smooth manifold. Let p be an element of M: Let ðU; uU Þ be an admissible coordinate chart of M satisfying p 2 U: Let c be in Cp ðMÞ. Let us deﬁne 2 functions u1 : U ! R;

u2 : U ! R

as follows: For every x in U; uU ð xÞ u1 ð xÞ; u2 ð xÞ : (In short, u1 ; u2 are the component functions of uU :) Then, for every ½f in Tp ðMÞ; where f is in Cp1 ðMÞ; h½c; ½ f i ¼

d 1 u c ðtÞ

D1 f ðuU Þ1 u1 ð pÞ; u2 ð pÞ dt t¼0

d

1 1 2 2 u c ðtÞ

: þ D2 f ðuU Þ u ð pÞ; u ð pÞ dt t¼0

90

2 Tangent Spaces

In other words,

h½c; ½ f i ¼

! !

o

d 1

þ u c ð t Þ ½ f

1 ou p dt t¼0

! !

o

d 2

; u c ð t Þ ½ f

2 ou p dt t¼0

or

! ! 1 o

d

h½c; ½ f i ¼ c ð t Þ u ½ f ou1 p dt t¼0

! ! 2 o

d

þ c ð t Þ : u ½ f ou2 p dt t¼0

Proof Since c is in Cp ðMÞ, by the deﬁnition of Cp ðMÞ, there exists a real number d [ 0 such that c is deﬁned on the open interval ðd; dÞ; cð0Þ ¼ p; and c is a smooth map from ðd; dÞ to M: Also, c is continuous at 0: Since c is continuous at 0; U is an open neighborhood of pð¼ cð0ÞÞ, there exists e [ 0 such that e\d; and for every t in the open interval ðe; eÞ we have cðtÞ 2 U: So, for every t in ðe; eÞ; ðuU cÞðtÞ ¼ uU ðcðtÞÞ ¼ u1 ðcðtÞÞ; u2 ðcðtÞÞ ¼ u1 c ðtÞ; u2 c ðtÞ : Thus, in the open neighborhood ðe; eÞ of 0, the component functions of uU c are u1 c; and u2 c: Also, ðuU cÞð0Þ ¼ uU ðcð0ÞÞ ¼ uU ð pÞ ¼ u1 ð pÞ; u2 ð pÞ : Now

dðf cÞðtÞ

LHS ¼ h½c; ½ f i ¼ c; ½ f ¼ c; ½ f ¼

dt t¼0 1 d f ðuU Þ uU c ðtÞ ¼ jt¼0 dt

d f ðuU Þ1 ðuU cÞ ðtÞ

¼

dt

t¼0 h i 1 ¼ D1 f ðuU Þ ð ð u U c Þ ð 0Þ Þ D2 f ðuU Þ1 ðuU cÞð0Þ 2 3

d ðu1 cÞðtÞ

5

4 dt d ðu2 cÞðtÞ

dt t¼0

2.5 Cotangent Spaces

91

i D1 f ðuU Þ1 ðu1 ð pÞ; u2 ð pÞÞ D2 f ðuU Þ1 ðu1 ð pÞ; u2 ð pÞÞ 2 3

d ðu1 cÞðtÞ

5

4 dt d ðu2 cÞðtÞ

dt t¼0

d 1 1 1 2 u c ðtÞ

¼ D1 f ðuU Þ u ð pÞ; u ð pÞ dt t¼0

d

u2 c ðtÞ

¼ RHS: þ D2 f ðuU Þ1 u1 ð pÞ; u2 ð pÞ dt t¼0 ¼

h

h

2.6 Tangent Space as a Dual Space Note 2.53 Here, we will digress slightly from the current topic. Let V be any real linear space. Let V be the collection of all linear functionals of V: We know that V is also a real linear space under pointwise vector addition, and pointwise scalar multiplication operations. V is called the dual space of V: Let n be the dimension of V: Let fe1 ; . . .; en g be a basis of V: Let us deﬁne a function e 1 : V ! R as follows: For every real t1 ; . . .; tn e 1 t1 e1 þ þ tn en ¼ t1 : Clearly, e 1 is in V : Similarly, we deﬁne function e 2 : V ! R as follows: For every real t1 ; . . .; tn e 2 t1 e1 þ þ tn en ¼ t2 ; etc. We shall try to show that fe 1 ; . . .; e n g constitutes a basis of V : For this purpose, we must prove 1. e 1 ; . . .; e n are linearly independent, that is, t1 e 1 þ þ tn e 1 ¼ 0 implies t1 ¼ 0; . . .; tn ¼ 0: 2. fe 1 ; . . .; e n g generates V ; that is, for every f in V ; there exist reals t1 ; . . .; tn such that t1 e 1 þ þ tn e n ¼ f : For 1: Let t1 e 1 þ þ tn e 1 ¼ 0: So, 0 ¼ 0 e1 ¼ t1 e 1 þ þ tn e 1 e1 ¼ t1 e 1 e1 þ þ tn e n e1 ¼ t1 1 þ t2 0 þ þ tn 0 ¼ t1 : Hence, t1 ¼ 0: Similarly, t2 ¼ 0; . . .; tn ¼ 0:

92

2 Tangent Spaces

For 2: Let us take any f in V : We shall show that ðf ðe1 ÞÞe 1 þ þ ðf ðen ÞÞe n ¼ f . For this purpose, it is enough to prove that 1 f e e1 þ þ ðf ðen ÞÞe n ðe1 Þ ¼ f e1 ; . . .; f e1 e 1 þ þ ðf ðen ÞÞe n ðen Þ ¼ f ðen Þ:

Here, 1 f e e1 þ þ ðf ðen ÞÞe n ðe1 Þ ¼ f e1 e 1 ðe1 Þ þ þ ðf ðen ÞÞe n ðe1 Þ ¼ f e1 1 þ f e2 0 þ þ ðf ðen ÞÞ 0 ¼ f e1 ¼ RHS;

LHS ¼

etc. Thus, we have shown that fe 1 ; . . .; e n g constitutes a basis of V : The basis fe 1 ; . . .; e n g of the dual space V is called the dual basis of the basis fe1 ; . . .; en g: Observe that 1. e j ðei Þ ¼ dij : 2. dimðV Þ ¼ n ¼ dimðVÞ: Now, let us consider the mapping m : V ! V deﬁned as follows: For every real t1 ; . . .; tn m t1 e1 þ þ tn en ¼ t1 e 1 þ þ tn e n : We shall show that m is an isomorphism from V onto V : For this purpose, we must prove 1. m is 1–1 2. m is onto. 3. mðsx þ tyÞ ¼ sðmðxÞÞ þ tðmðyÞÞ; for every real s; t; and for every x; y in V: For 1: Let mðxÞ ¼ mðyÞ: We have to prove that x ¼ y: Let x ¼ s1 e1 þ þ sn en ; and y ¼ t1 e1 þ þ tn en . Now, we have s1 e 1 þ þ sn e n ¼ m s1 e1 þ þ sn en ¼ mð xÞ ¼ mð yÞ ¼ m t1 e1 þ þ tn en ¼ t1 e 1 þ þ tn e n ; and hence, s1 ¼ t1 ; . . .; sn ¼ tn : Since s1 ¼ t1 ; . . .; sn ¼ tn ; x ¼ s1 e1 þ þ sn en ¼ t1 e1 þ þ tn en ¼ y: This proves 1. For 2: For this purpose, let us take any f in V : We have to ﬁnd an element in V whose m-image is f : Since f is in V ; and fe 1 ; . . .; e n g constitutes a basis of V ; there exist reals t1 ; . . .; tn such that

2.6 Tangent Space as a Dual Space

93

t1 e 1 þ þ tn e n ¼ f : Now m t1 e1 þ þ tn en ¼ t1 e 1 þ þ tn e n ¼ f ; where t1 e1 þ þ tn en is in V: This proves 2. For 3: Let us take any x s1 e1 þ þ sn en ; and y t1 e1 þ þ tn en : Let us take any reals s; t: Now LHS ¼ mðsx þ tyÞ ¼ m s s1 e1 þ þ sn en þ t t1 e1 þ þ tn en ¼ m ðss1 þ tt1 Þe1 þ þ ðssn þ ttn Þen ¼ ðss1 þ tt1 Þe 1 þ þ ðssn þ ttn Þe n ¼ s s1 e 1 þ þ sn e n þ t t1 e 1 þ þ tn e n ¼ s m s1 e1 þ þ sn en 1 þ t m t1 e þ þ tn en ¼ sðmðxÞÞ þ tðmð yÞÞ ¼ RHS:

Thus, we have shown that m is an isomorphism from V onto V : Hence, V and V are isomorphic real linear spaces. Deﬁnition Let M be an m-dimensional smooth manifold. Let p be an element of M: We will denote the dual space of the real linear space Tp ðMÞ by T p ðMÞ: Thus, T p ðMÞ is the collection of all linear functionals deﬁned on Tp ðMÞ: It is known that T p ðMÞ is a real linear space. Now, let ðU; uU Þ be an admissible coordinate chart of M satisfying p 2 U: Let us deﬁne m functions u1 : U ! R; . . .; um : U ! R as follows: For every x in U; uU ð xÞ u1 ð xÞ; . . .; um ð xÞ : By Theorem 2.48, f½u1 ; . . .; ½um g is a basis of the real linear space Tp ðMÞ: Let us deﬁne the function ½u1 : Tp ðMÞ ! R as follows: For every real t1 ; . . .; tm ; ½u1 t1 u1 þ þ tm ½um t1 : It is easy to see that ½u1 is in T p ðMÞ: As above, we deﬁne ½u2 : Tp ðMÞ ! R as follows: For every real t1 ; . . .; tm ;

94

2 Tangent Spaces

½u2 t1 u1 þ þ tm ½um t2 ; etc. We know that f½u1 ; . . .; ½um g constitutes a basis of the dual space T p ðMÞ of Tp ðMÞ. f½u1 ; . . .; ½um g is called the dual basis of the basis f½u1 ; . . .; ½um g Clearly, ½ ui u j ¼ dij ; where dij denotes the Kronecker delta symbol. In other words, ½ ui

du j

p

¼ dij :

Also, dim T p ðM Þ ¼ dim Tp ðM Þ ¼ m: Theorem 2.54 Let M be a 2-dimensional smooth manifold. Let p be an element of M: Let c be in Cp ðMÞ. Let us deﬁne a function ½c : Tp ðM Þ ! R; as follows: For every ½f in Tp ðMÞ; where f is in Cp1 ðMÞ; ½c ð½ f Þ h½c; ½ f i: Then, ½c is a linear functional on the real linear space Tp ðMÞ; that is, f½c : c 2 Cp ðMÞg is a subset of the dual space T p ðMÞ of Tp ðMÞ: Proof Here, we must prove: For every real a; b; and for every ½f ; ½g in Tp ðMÞ; where f ; g are in Cp1 ðMÞ; ½c ða½ f þb½g Þ ¼ að½c ð½ f ÞÞ þ bð½c ð½g ÞÞ: LHS ¼ ½c ða½ f þb½g Þ ¼ ½c ð½af þ½bg Þ ¼ ½c ð½af þ bg Þ ¼ h½c; ½af þ bg i ¼ c; ½af þ bg ¼ c; ½af þ bg ¼ c; ½af þ ½bg ¼ c; a½ f þ b½g ¼ a c; ½ f þ b c; ½g ¼ a c; ½ f þ b c; ½g ¼ ha½c; ½ f i þ bh½c; ½g i ¼ að½c ð½ f ÞÞ þ bð½c ð½g ÞÞ ¼ RHS: h

h Lemma 2.55 Let M be a 2-dimensional smooth manifold. Let p be an element of M: Let ðU; uU Þ be an admissible coordinate chart ðU; uU Þ of M such that p is in U:

2.6 Tangent Space as a Dual Space

95

Let u1 ; u2 be the component functions of uU : Let T be a linear functional of Tp ðMÞ: Let c1 : R ! R 2 be the function deﬁned as follows: For every t in R; c1 ð t Þ

1 2 t; T u t þ uU ð pÞ: T u

Then 1. ðuU Þ1 c1 is in Cp ðMÞ 2. for every f in Cp1 ðMÞ; h i T ð½ f Þ ¼ h ðuU Þ1 c1 ; ½ f i:

Proof 1: Clearly, c1 is a smooth function. Here, c1 is continuous, c1 ð0Þ ¼ ððTð½u1 ÞÞ 0; ðTð½u2 ÞÞ 0Þ þ uU ðpÞ ¼ uU ðpÞ; and uU ðUÞ is an open neighborhood of uU ðpÞ; so there exists e [ 0 such that for every t in the open interval ðe; eÞ; we have c1 ðtÞ in uU ðUÞ; and hence, ðuU Þ1 ðc1 ðtÞÞ is in U: Now, let us deﬁne a function c : ðe; eÞ ! M as follows: For every t in ðe; eÞ; cðtÞ ðuU Þ1 ðc1 ðtÞÞ ¼ ðuU Þ1 c1 ðtÞ: So, c ¼ ðuU Þ1 c1 : We shall try to see that c is in Cp ðMÞ: By the deﬁnition of Cp ðMÞ; we must prove that 1. cð0Þ ¼ p; 2. c is a parametrized curve in the manifold M; that is, c : ðe; eÞ ! M is a smooth map from ðe; eÞ to M; that is, c is C 1 at every point of ðe; eÞ:

96

2 Tangent Spaces

For 1: Here LHS ¼ cð0Þ ¼ ðuU Þ1 ðc1 ð0ÞÞ ¼ ðuU Þ1 ðuU ðpÞÞ ¼ p ¼ RHS: This proves 1. For 2: For this purpose, let us take any t0 in ðe; eÞ: We have to show that c : ðe; eÞ ! M is C 1 at t0 : We want to apply Theorem 2.2. Since t0 is in ðe; eÞ, by the deﬁnition of c; cðt0 Þð¼ ðuU Þ1 ðc1 ðt0 ÞÞÞ is in U; and hence, ðU; uU Þ is an admissible coordinate chart of M satisfying cðt0 Þ 2 U: Since uU c ¼ uU ððuU Þ1 c1 Þ ¼ ðuU ðuU Þ1 Þ c1 ¼ c1 ; and c1 is a smooth function, uU c is a smooth function. Since c : ðe; eÞ ! M; t0 is in ðe; eÞ; ðU; uU Þ is an admissible coordinate chart of M satisfying cðt0 Þ 2 U; and uU c is a smooth function, by Theorem 2.2, c is C 1 at t0 in ðe; eÞ: This proves 2. Thus, we have shown that c is in Cp ðMÞ. This proves 1. 2: Since ½f is in Tp ðMÞ so, by Theorem 2.47, ½ f ¼

u1 ð pÞ; u2 ð pÞ u1 þ D2 f ðuU Þ1 u1 ð pÞ; u2 ð pÞ u2 ; D1 f ðuU Þ1

and hence, 1 1 2 1 LHS ¼ T f ¼ T D1 f uU u p ;u p u 1 1 2 2 þ D2 f uU u p ;u p u 1 1 2 1 ¼ D1 f uU u p ;u p T u 1 1 2 2 u p ;u p T u : þ D2 f uU Next, by Theorem 2.52, Dh i E ðuU Þ1 c1 ; ½ f ¼ h½c; ½ f i

d 1 1 1 2 u c ðtÞ

¼ D1 f ðuU Þ u ð pÞ; u ð pÞ dt t¼0

d

u2 c ðtÞ

: þ D2 f ðuU Þ1 u1 ð pÞ; u2 ð pÞ dt t¼0

RHS ¼

Now, it remains to be proved that

d u1 ðuU Þ1 c1 ðtÞ

dt

; ¼ T u1 t¼0

2.6 Tangent Space as a Dual Space

and

97

d u2 ðuU Þ1 c1 ðtÞ

dt

Here,

d u1 ðuU Þ1 c1 ðtÞ

LHS ¼

dt

: ¼ T u2 t¼0

d u1 ðuU Þ1 c1 ðtÞ

¼

dt

2 3 i d ððT ð½u1 ÞÞt þ u1 ð pÞÞ dt 4 5 D 2 u ð uU Þ ðc1 ð0ÞÞ d ððT ð½u2 ÞÞt þ u2 ð pÞÞ dt t¼0 h i T ð½u1 Þ 1 1 1 1 ¼ D1 u ðuU Þ ðc1 ð0ÞÞ D 2 u ð uU Þ ðc1 ð0ÞÞ T ð½u2 Þ 1 1 þ D2 u1 ðuU Þ ðuU ð pÞÞ T u1 ðuU ð pÞÞ T u2 ¼ D1 u1 ðuU Þ 1 þ ð0Þ T u2 ¼ T u ¼ RHS: ¼ ð1Þ T u1 t¼0

h ¼ D1 u1 ðuU Þ1 ðc1 ð0ÞÞ

Similarly,

t¼0

1

1

d u2 ðuU Þ1 c1 ðtÞ

dt

: ¼ T u2 t¼0

h h

Theorem 2.56 Let M be a 2-dimensional smooth manifold. Let p be an element of M: Let T be a linear functional on Tp ðMÞ: Then, there exists c in Cp ðMÞ such that for every ½f in Tp ðMÞ; T ð½ f Þ ¼ h½c; ½ f i: Proof Since M is a 2-dimensional smooth manifold, and p is in M; there exists an admissible coordinate chart ðU; uU Þ of M such that p is in U: Let us deﬁne 2 functions u1 : U ! R;

u2 : U ! R

as follows: For every x in U;

uU ð xÞ u1 ð xÞ; u2 ð xÞ :

Let c1 : R ! R 2

98

2 Tangent Spaces

be the function deﬁned as follows: for every t in R; c1 ð t Þ

1 2 T u t; T u t þ uU ð pÞ:

Then, by Lemma 2.55, ðuU Þ1 c1 is in Cp ðMÞ and for every ½f in Tp ðMÞ; h i T ð½ f Þ ¼ h ðuU Þ1 c1 ; ½ f i:

h h

Theorem 2.57 Let M be a 2-dimensional smooth manifold. Let p be an element of M: For every c in Cp ðMÞ we deﬁne a function ½c : Tp ðM Þ ! R; as follows: For every ½f in Tp ðMÞ; where f is in Cp1 ðMÞ; ½c ð½ f Þ h½c; ½ f i: Then, the mapping ½c 7! ½c is a 1–1 mapping from f½c : c 2 Cp ðMÞg onto T p ðMÞ: Proof Here, we must prove 1. ½c 7! ½c is a well-deﬁned mapping from f½c : c 2 Cp ðMÞg to T p ðMÞ; 2. if ½c ¼ ½c0 , where c; c0 2 Cp ðMÞ, then ½c ¼ ½c0 ; 3. for every T in T p ðMÞ; there exists c in Cp ðMÞ such that ½c ¼ T: For 1: Let ½c ¼ ½c0 ; that is, c c0 : We have to show that ½c ¼ ½c0 ; that is, for every ½f in Tp ðMÞ; ½c ð½f Þ ¼ ½c0 ð½f Þ: For this purpose, let us take any ½f in Tp ðMÞ: Since c c0 ; and ½f is in Tp ðMÞ, by the deﬁnition of ; c; ½f ¼ c0 ; ½f : Now LHS ¼ ½c ð½ f Þ ¼ h½c; ½ f i ¼ c; ½ f ¼ c0 ; ½ f ¼ h½c0 ; ½ f i ¼ ½c0 ð½ f Þ ¼ RHS: So ½c 7! ½c is a well-deﬁned mapping. Next, by Theorem 2.54, ½c is in T p ðMÞ for every c in Cp ðMÞ. This proves 1. For 2: Let ½c ¼ ½c0 , where c; c0 2 Cp ðMÞ: We have to prove that ½c ¼ ½c0 ; that is, c c0 ; that is, for every ½f in Tp ðMÞ; c; ½f ¼ c0 ; ½f : For this purpose, let us take any ½f in Tp ðMÞ: We have to show that c; ½ f ¼ c0 ; ½ f :

2.6 Tangent Space as a Dual Space

99

Since ½c ¼ ½c0 , ½c : Tp ðMÞ ! R; ½c0 : Tp ðMÞ ! R; and ½f is in Tp ðMÞ;

LHS ¼ c; ½ f ¼ h½c; ½ f i ¼ ½c ð½ f Þ ¼ ½c0 ð½ f Þ ¼ h½c0 ; ½ f i ¼ c0 ; ½ f ¼ RHS: For 3: Let us take any T in T p ðMÞ: Since T is in T p ðMÞ; and T p ðMÞ is dual space of the real linear space Tp ðMÞ; T is a linear functional on Tp ðMÞ; and hence, by Theorem 2.56, there exists c in Cp ðMÞ such that for every ½f in Tp ðMÞ; T ð½ f Þ ¼ h½c; ½ f i: Since, for every ½f in Tp ðMÞ; we have Tð½f Þ ¼ h½c; ½f i ¼ ½c ð½f Þ; T ¼ ½c : This proves 3. h Similarly, we can prove the following theorem. Theorem 2.58 Let M be an m-dimensional smooth manifold. Let p be an element of M. For every c in Cp ðMÞ, we deﬁne a function ½c : Tp ðM Þ ! R; as follows: For every ½f in Tp ðMÞ, where f is in Cp1 ðMÞ, ½c ð½ f Þ h½c; ½ f i: Then, the mapping ½c 7! ½c is a 1–1 mapping from f½c : c 2 Cp ðMÞg onto T p ðMÞ: Proof Its proof is quite similar to the proof of Theorem 2.57.

h

Deﬁnition Let M be an m-dimensional smooth manifold. Let p be an element of M. Let g be the mapping ½c 7! ½c from f½c : c 2 Cp ðMÞg to T p ðMÞ; as deﬁned in Theorem 2.58. By Theorem 2.58, g is a 1–1 mapping from f½c : c 2 Cp ðMÞg onto T p ðMÞ; and hence, g1 is a 1–1 mapping from T p ðMÞ onto f½c : c 2 Cp ðMÞg: Deﬁne vector addition, and scalar multiplication over the set f½c : c 2 Cp ðMÞg as follows: For every c; c0 in Cp ðMÞ; and for every real t, ½c þ ½c0 ¼ g1 ðgð½cÞ þ gð½c0 ÞÞ;

100

2 Tangent Spaces

and t½c ¼ g1 ðtðgð½cÞÞÞ: Since T p ðMÞ is a real linear space, and g is a 1–1 mapping from f½c : c 2 Cp ðMÞg onto T p ðMÞ; f½c : c 2 Cp ðMÞg is also a real linear space. Clearly, g : ½c 7! ½c is an isomorphism from real linear space f½c : c 2 Cp ðMÞg onto real linear space T p ðMÞ: Thus, we have the following formulae: gð½c þ ½c0 Þ ¼ gð½cÞ þ gð½c0 Þ and gðt½cÞ ¼ tðgð½cÞÞ: In other words,

ð½c þ ½c0 Þ ¼ ð½cÞ þð½c0 Þ

and ðt½cÞ ¼ tð½cÞ ; because g is the mapping ½c 7! ½c from f½c : c 2 Cp ðMÞg to T p ðMÞ, as deﬁned in Theorem 2.58. Since g : ½c 7! ½c is an isomorphism from real linear space f½c : c 2 Cp ðMÞg onto real linear space T p ðMÞ, g1 : ½c 7! ½c is also an isomorphism. Also, dim ½c : c 2 Cp ðM Þ ¼ dim T p ðM Þ :

so

Further, since dimðT p ðMÞÞ ¼ dimðTp ðMÞÞ ¼ ðthe dimension of the manifoldÞ, dim ½c : c 2 Cp ðM Þ ¼ ðthe dimension of the manifoldÞ: It follows that the real linear space f½c : c 2 Cp ðMÞg is isomorphic to Rm :

Theorem 2.59 Let M be a 2-dimensional smooth manifold. Let p be an element of M. Let ðU; uU Þ be an admissible coordinate chart of M satisfying p 2 U: Let c : ðd; dÞ ! M; and c0 : ðd0 ; d0 Þ ! M be in Cp ðMÞ; where d [ 0; and d0 [ 0: Then, ½c ¼ ½c0 if and only if

dðuU cÞðtÞ

dðuU c0 ÞðtÞ

¼

: dt dt t¼0 t¼0

2.6 Tangent Space as a Dual Space

101

In other words, c c0 if and only if c c0 ; that is, ½c ¼ c: And hence, ½c : c 2 Cp ðM Þ ¼ Tp M: Proof of “only if” part: Let ½c ¼ ½c0 . Since ½c ¼ ½c0 , c c0 : Let us deﬁne 2 functions u1 : U ! R;

u2 : U ! R

as follows: For every x in U, uU ð xÞ u1 ð xÞ; u2 ð xÞ : By Theorem 2.48, ½u1 ; ½u2 are in Tp ðMÞ: Since c c0 ; and ½u1 ; ½u2 are in Tp ðMÞ, by the deﬁnition of , c; ½u1 ¼ c0 ; ½u1 ; and c; ½u2 ¼ c0 ; ½u2 : Now

dðuU cÞðtÞ

dðuU ðcðtÞÞÞ

dðu1 ðcðtÞÞ; u2 ðcðtÞÞÞ

LHS ¼

¼

¼

dt dt dt t¼0 t¼0 t¼0 1

2 1 2

dðu ðcðtÞÞÞ dðu ðcðtÞÞÞ

dððu cÞðtÞÞ dððu cÞðtÞÞ

; ; ¼ ¼

dt dt dt dt t¼0 t¼0

1 2 dððu1 cÞðtÞÞ

dððu2 cÞðtÞÞ

¼

; c; u

¼ c; u

;

dt dt t¼0 t¼0

0 1 2 dððu1 c0 ÞðtÞÞ

dððu2 c0 ÞðtÞÞ

0 ¼ c ; u

; c ; u

¼

;

dt dt t¼0 t¼0

1 0 2 0 1 0 2 0 dððu c ÞðtÞÞ dððu c ÞðtÞÞ

dðu ðc ðtÞÞÞ dðu ðc ðtÞÞÞ

; ; ¼

¼

dt dt dt dt t¼0 t¼0

dðu1 ðc0 ðtÞÞ; u2 ðc0 ðtÞÞÞ

dðuU ðc0 ðtÞÞÞ

dðuU c0 ÞðtÞ

¼

¼

¼

¼ RHS: dt dt dt t¼0 t¼0 t¼0 Proof of “if” part Let

dðuU cÞðtÞ

dðuU c0 ÞðtÞ

¼

: dt dt t¼0 t¼0 We have to show that ½c ¼ ½c0 , that is, c c0 ; that is, for every ½f in Tp ðMÞ; c; ½f ¼ c0 ; ½f : For this purpose, let us take any ½f in Tp ðMÞ: We have to show that c; ½f ¼ c0 ; ½f : Now, since

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2 Tangent Spaces

so,

dððu1 cÞðtÞÞ

dððu2 cÞðtÞÞ

;

dt dt t¼0 t¼0

1 2

dððu cÞðtÞÞ dððu cÞðtÞÞ

; ¼

dt dt t¼0

1

2

dðu ðcðtÞÞÞ dðu ðcðtÞÞÞ

dðu1 ðcðtÞÞ; u2 ðcðtÞÞÞ

; ¼ ¼

dt dt dt t¼0 t¼0

dðuU ðcðtÞÞÞ

¼

dt

t¼0

dðuU cÞðtÞ

dðuU c0 ÞðtÞ

¼ ¼

dt dt t¼0

t¼0

0 1 0

dðuU ðc ðtÞÞÞ

dðu ðc ðtÞÞ; u2 ðc0 ðtÞÞÞ

¼ ¼

dt dt t¼0 t¼0 1 0

2 0

dðu ðc ðtÞÞÞ dðu ðc ðtÞÞÞ

; ¼

dt dt t¼0

1 0 2 0 dððu c ÞðtÞÞ dððu c ÞðtÞÞ

; ¼

dt dt t¼0

dððu1 c0 ÞðtÞÞ

dððu2 c0 ÞðtÞÞ

¼ ;

dt dt t¼0 t¼0 2 0 1 0

; c ; u

; ¼ c ; u

c; u1 ; c; u2 ¼

c; u1 ¼ c0 ; u1

and c; u2 ¼ c0 ; u2 :

Since ½f in Tp ðMÞ, by Theorem 2.48, ½ f ¼

u1 ð pÞ; u2 ð pÞ u1 þ D2 f ðuU Þ1 u1 ð pÞ; u2 ð pÞ u2 : D1 f ðuU Þ1

Now, LHS ¼ c; ½ f ¼ c; D1 f ðuU Þ1 u1 ð pÞ; u2 ð pÞ u1 þ D2 f ðuU Þ1 u1 ð pÞ; u2 ð pÞ u2

¼ c; D1 f ðuU Þ1 u1 ð pÞ; u2 ð pÞ u1 þ D2 f ðuU Þ1 u1 ð pÞ; u2 ð pÞ u2 1 2 1 1 1 2 1 2 ¼ c; D 1 f ðuU Þ u ð pÞ; u ð pÞ u þ D2 f ðuU Þ u ð pÞ; u ð pÞ u 1 1 1 1 2 u ð pÞ; u ð pÞ c; u þ D2 f ðuU Þ u1 ð pÞ; u2 ð pÞ c; u2 ¼ D 1 f ðuU Þ u1 ð pÞ; u2 ð pÞ c0 ; u1 þ D2 f ðuU Þ1 u1 ð pÞ; u2 ð pÞ c0 ; u2 ¼ D1 f ðuU Þ1 D1 f ðuU Þ1 u1 ð pÞ; u2 ð pÞ u1 þ D2 f ðuU Þ1 u1 ð pÞ; u2 ð pÞ u2 ¼ c0 ; D1 f ðuU Þ1

u1 ð pÞ; u2 ð pÞ u1 þ D2 f ðuU Þ1 u1 ð pÞ; u2 ð pÞ u2 ¼ c0 ; 1 2 1 1 0 1 2 1 2 þ D 2 f ðuU Þ

¼ c ; D1 f ðuU Þ u ð pÞ; u ð pÞ u u ð pÞ; u ð pÞ u ¼ c0 ; ½ f ¼ RHS:

h

2.6 Tangent Space as a Dual Space

103

Note 2.60 Let M be an m-dimensional smooth manifold. Let p be an element of M. Since f½c : c 2 Cp ðMÞg ¼ Tp M; and f½c : c 2 Cp ðMÞg is a real linear space, Tp M is a real linear space. Since g : ½c 7! ½c is an isomorphism from real linear space f½c : c 2 Cp ðMÞg onto real linear space T p ðMÞ, f½c : c 2 Cp ðMÞg and T p ðMÞ are isomorphic, and hence, Tp M and T p ðMÞ are isomorphic. Deﬁnition Let M be an m-dimensional smooth manifold. Let p be an element of M: The real linear space Tp M is called the tangent space of M at p; and the elements of Tp M are called tangent vectors at p: Theorem 2.61 Let M be a 2-dimensional smooth manifold. Let p be an element of M. Then, the mapping ð½c; ½f Þ 7! h½c; ½f i from the real linear space Tp ðMÞ Tp ðMÞ to R is bilinear, that is, linear in each variable. Proof Let ðU; uU Þ be an admissible coordinate chart of M satisfying p 2 U: We have to show that 1. For every ½c in f½c : c 2 Cp ðMÞg; for every ½f ; ½g in Tp ðMÞ, and for every real s; t, h½c; s½ f þt½g i ¼ sh½c; ½ f i þ th½c; ½g i: 2. For every c; c0 in Cp ðMÞ; for every ½f in Tp ðMÞ; h½c þ ½c0 ; ½ f i ¼ h½c; ½ f i þ h½c0 ; ½ f i: 3. For every ½f in Tp ðMÞ; for every c in Cp ðMÞ, and for every real s, hs½c; ½ f i ¼ sh½c; ½ f i: For 1: LHS ¼ h½c; s½f þ t½g i ¼ h½c; ½sf þ ½tg i ¼ h½c; ½sf þ tg i ¼ c; ½sf þ tg ¼ s c; ½ f þ t c; ½g ¼ s c; ½ f þ t c; ½g ¼ sh½c; ½ f i þ th½c; ½g i ¼ RHS: For 2: Let us take any c; c0 in Cp ðMÞ: Let us take any ½f in Tp ðMÞ: We have to show that h½c þ ½c0 ; ½f i ¼ h½c; ½f i þ h½c0 ; ½f i: LHS ¼ h½c þ ½c0 ; ½f i ¼ h½c00 ; ½f i where ½c00 ½c þ ½c0 for some c00 in Cp ðMÞ: Since ½c00 ½c þ ½c0 ; ð½c00 Þ ¼ ð½c þ ½c0 Þ ¼ ð½cÞ þ ð½c0 Þ ; and hence, h½c00 ; ½ f i ¼ ð½c00 Þ ð½ f Þ ¼ ð½cÞ þð½c0 Þ ð½ f Þ ¼ ðð½cÞ Þð½ f Þ þ ð½c0 Þ ð½ f Þ ¼ h½c; ½ f i þ h½c0 ; ½ f i:

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2 Tangent Spaces

Thus, we have shown that h½c þ ½c0 ; ½f i ¼ h½c; ½f i þ ½c0 ; ½f : This proves 2. For 3: Let us take any ½f in Tp ðMÞ: Let us take any c in Cp ðMÞ; and any real s: We have to show that hs½c; ½f i ¼ sh½c; ½f i: LHS ¼ hs½c; ½f i ¼ h½c0 ; ½f i where ½c0 s½c for some c0 in Cp ðMÞ: Since ½c0 ¼ s½c; ð½c0 Þ ¼ ðs½cÞ ¼ sðð½cÞ Þ; and hence,

h½c0 ; ½ f i ¼ ð½c0 Þ ð½ f Þ ¼ ðsðð½cÞ ÞÞð½ f Þ ¼ sððð½cÞ Þð½ f ÞÞ ¼ sh½c; ½ f i: Thus, we have shown that hs½c; ½f i ¼ sh½c; ½f i. This proves 3.

h

Theorem 2.62 Let M be an m-dimensional smooth manifold. Let p be an element of M. Then, the mapping ð½c; ½f Þ 7! h½c; ½f i from the real linear space Tp ðMÞ Tp ðMÞ to R is bilinear, that is, linear in each variable. Proof Its proof is quite similar to that of Theorem 2.61, etc.

h

Deﬁnition Let M be a 2-dimensional smooth manifold. Let p be an element of M. Let ðU; uU Þ be an admissible coordinate chart of M such that p is in U. Let u1 ; u2 be the component functions of uU . Let us deﬁne the function U 1 : Tp ðMÞ ! R as follows: For every real t1 ; t2 ; t1 : U 1 t1 u1 þt2 u2 Clearly, U 1 is a linear functional on Tp ðMÞ; and hence, U 1 is in T p ðMÞ. Let us deﬁne the function U 2 : Tp ðMÞ ! R as follows: For every real t1 ; t2 , t2 : U 2 t1 u1 þt2 u2 Clearly, U 2 is a linear functional on Tp ðMÞ; and hence, U 2 is in T p ðMÞ: Since g : ½c 7! ½c is an isomorphism from real linear space f½c : c 2 Cp ðMÞg onto real linear space T p ðMÞ; and f½c : c 2 Cp ðMÞg ¼ Tp ðMÞ; g : ½c 7! ½c is an isomorphism from tangent space Tp ðMÞ onto real linear space T p ðMÞ, and hence, there exists a unique ½k1 in Tp ðMÞ such that ½k1 ¼ U 1 : Similarly, there exists a unique ½k2 in Tp ðMÞ such that ½k2 ¼ U 2 : We know that fU 1 ; U 2 g is a basis of the dual space T p ðMÞ of Tp ðMÞ: Further, since ½k1 ¼ U 1 ; and ½k2 ¼ U 2 ; f½k1 ; ½k2 g is a basis of T p ðMÞ: Also, h½ki ; u j i ¼ ½ki u j ¼ dij : Since f½k1 ; ½k2 g is a basis of T p ðMÞ; and g : ½c 7! ½c is an isomorphism from tangent space Tp ðMÞ onto T p ðMÞ, f½k1 ; ½k2 g is a basis of Tp ðMÞ. Here, we say that f½k1 ; ½k2 g is the natural basis of the tangent space Tp ðMÞ under local coordinate system ðui Þ: Similar deﬁnitions can be supplied for a 3dimensional smooth manifold, etc.

2.6 Tangent Space as a Dual Space

105

In continuation with the above discussion for 2-dimensional smooth manifold M, let us note that for every ½f in Tp ðMÞ; h½k1 ; ðdf Þp i ¼ h½k1 ; ½ f i ¼ h½k1 ; D1 f ðuU Þ1 u1 ð pÞ; u2 ð pÞ u1 þ D2 f ðuU Þ1 u1 ð pÞ; u2 ð pÞ u2 i ¼ D1 f ðuU Þ1 u1 ð pÞ; u2 ð pÞ h½k1 ; u1 i þ D2 f ðuU Þ1 u1 ð pÞ; u2 ð pÞ h½k1 ; u2 i ¼ D1 f ðuU Þ1 u1 ð pÞ; u2 ð pÞ ð1Þ þ D2 f ðuU Þ1 u1 ð pÞ; u2 ð pÞ ð0Þ ¼ D1 f ðuU Þ1 u1 ð pÞ; u2 ð pÞ : Thus, for every ½f in Tp ðMÞ;

h½k1 ; ðdf Þp i ¼ D1 f ðuU Þ1 u1 ð pÞ; u2 ð pÞ ¼

! o

½ f : ou1 p

So h½k1 ; ðdf Þp i ¼

! o

½ f : ou1 p

Similarly, we have h½k2 ; ðdf Þp i ¼

! o

½ f : ou2 p

So, for every i ¼ 1; 2; and for every ½f in F p ðMÞ;

h½ki ; ½ f i ¼

! o

½ f : oui p

Hence, when there is no confusion, we denote ½ki by ouo i jp : Further, since f½k1 ; ½k2 g is the natural basis of the tangent space Tp ðMÞ under local coordinate system ðui Þ; fouo1 jp ; ouo2 jp g is the natural basis of the tangent space Tp ðMÞ under local coordinate system ðui Þ. Thus, for every ½f in F p ðMÞ;

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2 Tangent Spaces

o

h i

; ðdf Þp i ¼ ou p

! o

½ f : oui p

Also, we have

o h i

; du j p i ¼ dij : ou p Theorem 2.63 Let M be a 2-dimensional smooth manifold. Let p be an element of M. Let ðU; uU Þ be an admissible coordinate chart of M satisfying p 2 U. Let c be in Cp ðMÞ. Let u1 ; u2 be the component functions of uU . Then,

d 1 d 2 u c ðtÞ

; u c ðtÞ

dt dt t¼0 t¼0

are the components of the tangent vector ½c at p, with respect to the natural basis (

) o

o

; ou1 p ou2 p

of the tangent space Tp ðMÞ under local coordinate system ðui Þ. Proof Let us take any ½f in Tp ðMÞ; where f is in Cp1 ðMÞ: Since

d 1 u c ðtÞ

u1 ð pÞ; u2 ð pÞ D1 f ðuU Þ1 dt t¼0

d 2 1 1 2 u c ðtÞ

u ð pÞ; u ð pÞ þ D 2 f ðuU Þ dt t¼0

2 d 1 d u c ðtÞ

þh½k2 ; ðdf Þp i u c ðtÞ

¼ h½k1 ; ðdf Þp i dt dt t¼0 t¼0

2 d 1 d u c ðtÞ

u c ðtÞ

¼ h½k1 ; ½ f i þ h½k2 ; ½ f i dt dt t¼0 t¼0

d 1 d 2

u c ðt Þ

u c ðt Þ

¼ ð½k1 ð½ f ÞÞ þ ð½k2 ð½ f ÞÞ dt dt t¼0 t¼0

d 1 d 2

u c ðt Þ

u c ðtÞ

¼ ½k1 ð½ f Þ þ ½k2 ð½ f Þ dt dt t¼0 t¼0

d 1 d 2

u c ðt Þ

u c ðtÞ

½k1 ð½ f Þ þ ½k2 ð½ f Þ ¼ dt dt t¼0 t¼0

d 1 d 2 u c ðtÞ

u c ðtÞ

¼ ½k1 þ ½k2 ð½ f Þ dt dt t¼0 t¼0

d 1 d 2 u c ðtÞ

u c ðtÞ

½k1 þ ½k2 ð½ f Þ; ¼ dt dt t¼0 t¼0

½c ð½ f Þ ¼ h½c; ½ f i ¼

2.6 Tangent Space as a Dual Space

107

so ½c ¼

d 1 d 2 u c ðtÞ

u c ðtÞ

½ k1 þ ½ k2 : dt dt t¼0 t¼0

Further, since

d 1 d 2 u c ðtÞ

u c ðtÞ

gð½cÞ ¼ ½c ¼ ½ k1 þ ½k2 dt dt t¼0 t¼0

d 1 d 2 u c ðtÞ

u c ðtÞ

¼g ½ k1 þ ½ k2 ; dt dt t¼0 t¼0 and g is a 1–1 mapping from Tp ðMÞ onto T p ðMÞ;

d 1 d 2 u c ðtÞ

u c ðtÞ

½ k1 þ ½ k2 dt dt t¼0 t¼0

!

!

d 1 o

d 2 o

u c ðt Þ

u c ðtÞ

¼ þ : dt ou1 p dt ou2 p t¼0 t¼0

½ c ¼

Further, since f½k1 ; ½k2 g is a basis of the tangent space Tp ðMÞ, that is, (

!

!) o

o

; ou1 p ou2 p is a basis of the tangent space Tp ðMÞ,

d 1 d 2

u c ðt Þ ; u c ðtÞ

dt dt t¼0 t¼0

are the components of the tangent vector ½c at p, with respect to the natural basis.h Note 2.64 As above, for m-dimensional smooth manifold M, we get the formula: ½ c ¼

!

!

d 1 o

d m o

u c ðt Þ

ð u cÞ ð t Þ

þ þ : dt ou1 p dt oum p t¼0 t¼0

Further, since f½k1 ; . . .; ½km g is a basis of the tangent space Tp ðMÞ; that is, (

!

!) o

o

; . . .; ou1

oum

p

p

108

2 Tangent Spaces

is a basis of the tangent space Tp ðMÞ,

d 1 d m

u c ðtÞ ; . . .; ðu cÞðtÞ

dt dt t¼0 t¼0 are the components of the tangent vector ½c at p, with respect to the natural basis. Note 2.65 When there is no confusion, we generally avoid writing the lower index p of the tangent vectors, cotangent vectors, etc. Theorem 2.66 Let M be an m-dimensional smooth manifold. Let p be an element of M. Let ðdf Þp ; ðdgÞp be in Tp ; where f ; g are in Cp1 . Let X be in Tp M. Let s; t be any real numbers. Then, 1. hX; ðdðsf þ tgÞÞp i ¼ s hX; ðdf Þp i þ t hX; ðdgÞp i; 2. hX; ðdðf gÞÞp i ¼ f ðpÞ hX; ðdgÞp i þ gðpÞ hX; ðdf Þp i: Proof 1: By Theorem 2.45, hX; ðdðsf þ tgÞÞp i ¼ hX; s ððdf Þp Þ þ t ððdgÞp Þi: Further, since the mapping ð½c; ½f Þ 7! h½c; ½f i from the real linear space Tp Tp to R is linear in the second variable, hX; s ððdf Þp Þ þ t ððdgÞp Þi ¼ s hX; ðdf Þp þ t X; ðdgÞp i; and hence,

hX; ðd ðsf þ tgÞÞp i ¼ s hX; ðdf Þp þt X; ðdgÞp i: This proves 1. 2: By Theorem 2.45, hX; ðdðf gÞÞp i ¼ hX; ðgðpÞÞðdf Þp þ ðf ðpÞÞðdgÞp i: Further, since the mapping ð½c; ½f Þ 7! h½c; ½f i from the real linear space Tp Tp to R is linear in the second variable, hX; ðgðpÞÞðdf Þp þ ðf ðpÞÞðdgÞp i ¼ ðgðpÞÞ hX; ðdf Þp i þ ðf ðpÞÞ hX; ðdgÞp i; and hence, hX; ðd ðf gÞÞp i ¼ ðgð pÞÞ hX; ðdf Þp i þ ðf ð pÞÞ hX; ðdgÞp i: This proves 2.

h

2.7 Contravariant Vectors and Covariant Vectors Deﬁnition Let M be an m-dimensional smooth manifold. Let p be an element of M. Let ðdf Þp be in Tp , where f is in Cp1 . Let X be in Tp M: By Xf, we mean hX; ðdf Þp i; and is called the directional derivative of the function f along the tangent vector X.

2.7 Contravariant Vectors and Covariant Vectors

109

Now, the Theorem 2.66 can be restated as follows: Theorem 2.67 Let M be an m-dimensional smooth manifold. Let p be an element of M. Let X be in Tp M: Then, for every f ; g in Cp1 ; and for every real s; t; 1. Xðsf þ tgÞ ¼ s ðXf Þ þ t ðXgÞ; 2. Xðf gÞ ¼ f ðpÞ Xðf Þ þ gðpÞ XðgÞ: Theorem 2.68 Let M be a 2-dimensional smooth manifold. Let p be an element of M. Let ½c be in Tp M, where c is in c 2 Cp ðMÞ. Let ðU; uU Þ; ðU 0 ; u0U Þ be two admissible coordinate charts of M satisfying p 2 U, and p 2 U 0 . Let u1 ; u2 be the component functions of uU . Let u01 ; u02 be the component functions of u0U . Then,

01 01

d 01 ou d 1 ou d 2

u c ðt Þ ¼ u c ðt Þ þ u c ðtÞ

; 1 2 dt dt dt ou ou t¼0 t¼0 t¼0

where ou01 01 1 D u ð u Þ u1 ð pÞ; u2 ð pÞ ¼ 1 U ou1

! o

01 u ; ou1 p

ou01 01 1 D u ð u Þ u1 ð pÞ; u2 ð pÞ ¼ 2 U ou2

! o

01 u : ou2 p

and

Proof Here,

d 1 d 2

u c ðt Þ ; u c ðtÞ

dt dt t¼0 t¼0

are the components of the tangent vector ½c at p, with respect to the natural basis (

) o

o

; ou1 p ou2 p

of the tangent space Tp M under local coordinate system ðui Þ. Also,

d 01 d 02 u c ðtÞ

; u c ðtÞ

dt dt t¼0 t¼0

110

2 Tangent Spaces

are the components of the tangent vector ½c at p, with respect to the natural basis (

) o

o

; ou01 p ou02 p

of the tangent space Tp M under local coordinate system ðu0i Þ. Since c is in Cp ðMÞ, cð0Þ ¼ p. We have to prove that

01 01

d 01 ou d 1 d 2

þ ou

: u c ðtÞ

¼ u u c ð t Þ c ð t Þ

1 2 dt dt dt ou ou t¼0 t¼0 t¼0

1 d 01 d 0 u c ðtÞ

¼ uU c ðtÞ

dt dt t¼0 t¼0 1

d 1 0 uU ðu U Þ uU c ðtÞ

¼ dt t¼0 1

d 1 0 ðuU cÞ uU ðu U Þ ðtÞ

¼ dt t¼0 1

d 1 u0U ðuU Þ ¼ ðuU cÞ ðtÞ

dt t¼0 1 1 1 1 0 0 ¼ D 1 uU ðu U Þ ððu U c Þð0 ÞÞ D 2 u U ðuU Þ ððuU cÞð0ÞÞ 2 3 d ðu cÞ1 ðtÞ

U 6 dt 7 t¼0 5 4 d ðu cÞ2 ðtÞ

U dt t¼0 1 1 1 ¼ D1 u0U ðuU Þ ðuU ðcð0ÞÞÞ D2 u0U ðuU Þ1 ðuU ðcð0ÞÞÞ 2 3 d ðu cÞ1 ðtÞ

U 6 dt 7 t¼0 5 4 d ðu cÞ2 ðtÞ

U dt t¼0 1 1 1 0 ðu U ð p ÞÞ D2 u0U ðuU Þ1 ðu U ð p ÞÞ ¼ D 1 uU ðu U Þ 2 3 d ðu cÞ1 ðtÞ

U 6 dt 7 t¼0 5 4 d ðu cÞ2 ðtÞ

U dt t¼0 1 1 1 0 ¼ D 1 uU ðu U Þ ðu U ð p ÞÞ D2 u0U ðuU Þ1 ðu U ð p ÞÞ 2 3

1 d c ðt Þ

ð Þ u U 6 dt 7 t¼0 5 4

2 d c ð t Þ ð Þ u

U dt

LHS ¼

t¼0

2.7 Contravariant Vectors and Covariant Vectors

111

1 1 D1 u0U ðuU Þ1 ðuU ð pÞÞ D2 u0U ðuU Þ1 ðu U ð p ÞÞ 2 3

d ðu U Þ1 ðc ðt ÞÞ

dt 6 7 t¼0 5 4

2 d ð c ð t Þ Þ ð Þ u

U dt t¼0 1 1 1 0 ¼ D 1 u U ðu U Þ ðuU ð pÞÞ D2 u0U ðuU Þ1 ðu U ð p ÞÞ 2 3 d ððu1 ÞðcðtÞÞÞ

dt 6 7 t¼0 5 4 d ððu2 ÞðcðtÞÞÞ

dt t¼0 1 1 1 0 ðuU ð pÞÞ D2 u0U ðuU Þ1 ðu U ð p ÞÞ ¼ D 1 u U ðu U Þ 2 3 d ðu1 cÞðtÞ

6 dt 7 t¼0 5 4 d ðu2 cÞðtÞ

dt t¼0 1 1 1 ¼ D1 u0U ðuU Þ ðu1 ð pÞ; u2 ð pÞÞ D2 u0U ðuU Þ1 ðu1 ð pÞ; u2 ð pÞÞ 2 3 d ðu1 cÞðtÞ

6 dt 7 t¼0 5 4 d ðu2 cÞðtÞ

dt t¼0

1

d 1 1 0 u c ðtÞ

u1 ð pÞ; u2 ð pÞ ¼ D 1 u U ðu U Þ dt t¼0

1

d 1 u2 c ðtÞ

: u1 ð pÞ; u2 ð pÞ þ D2 u0U ðuU Þ dt

¼

t¼0

01

ou d 1 d 2

þ ou

u u RHS ¼ c ð t Þ c ð t Þ

1 2 dt dt ou ou t¼0 t¼0

d 1 1 01 1 2 u c ðtÞ

¼ D1 u ðuU Þ u ð pÞ; u ð pÞ dt t¼0

d

1 01 1 2 2 u c ðtÞ

þ D2 u ðuU Þ u ð pÞ; u ð pÞ dt t¼0

1

d 1 1 0 1 2 u c ðtÞ

¼ D1 uU ðuU Þ u ð pÞ; u ð pÞ dt t¼0

1

d 1 u2 c ðtÞ

: þ D2 u0U ðuU Þ u1 ð pÞ; u2 ð pÞ dt t¼0

01

Hence, LHS ¼ RHS.

h

112

2 Tangent Spaces

Note 2.69 Similarly, under the same conditions as in theorem 34, we get

02 02

d 02 ou d 1 ou d 2

u c ðt Þ ¼ u c ðt Þ þ u c ðtÞ

: 1 2 dt dt dt ou ou t¼0 t¼0 t¼0 In short, we write

0i 0i

d 0i ou d 1 d 2

þ ou

; u c ðtÞ

¼ u u c ð t Þ c ð t Þ

1 2 dt dt dt ou ou t¼0 t¼0 t¼0

or

2 0i X

d 0i ou d j

; u c ðtÞ

¼ u c ð t Þ

j dt dt ou t¼0 t¼0 j¼1

and we say that tangent vectors are contravariant vectors. Similar results can be obtained for 3-dimensional smooth manifold, etc. Theorem 2.70 Let M be a 2-dimensional smooth manifold. Let p be an element of M. Let ðdf Þp be in Tp , where f is in Cp1 . Let ðU; uU Þ; ðU 0 ; u0U Þ be two admissible coordinate charts of M satisfying p 2 U; p 2 U 0 . Let u1 ; u2 be the component functions of uU : Let u01 ; u02 be the component functions of u0U . Then, of ¼ ou01

2 ou1 of ou of þ ; ou01 ou1 ou01 ou2

where of 1 D1 f u0U u01 ð pÞ; u02 ð pÞ ¼ ou01 of 1 D f ð u Þ u1 ð pÞ; u2 ð pÞ ¼ 1 U 1 ou of 1 D f ð u Þ u1 ð pÞ; u2 ð pÞ ¼ 2 U ou2

! o

½ f ; ou01 p

! o

½ f ; ou1 p

! o

½ f : ou1 p

Proof By Theorem 2.48,

of ou1

p

of u1 ð pÞ; u2 ð pÞ ; u1 ð pÞ; u2 ð pÞ D1 f ðuU Þ1 D2 f ðuU Þ1 2 ou p

2.7 Contravariant Vectors and Covariant Vectors

113

are the components of the cotangent vector ðdf Þp at p, with respect to the natural basis fðdu1 Þp ; ðdu2 Þp g of the cotangent space Tp under local coordinate system ðui Þ. Also,

of ou01

p

1 01 of 1 u01 ð pÞ; u02 ð pÞ ; u ð pÞ; u02 ð pÞ D1 f u0U D2 f u0U ou02 p

are the components of the cotangent vector ðdf Þp at p, with respect to the natural basis fðdu01 Þp ; ðdu02 Þp g of the cotangent space Tp under local coordinate system ðu0i Þ. We have to prove that of ¼ ou01

2 ou1 of ou of þ : ou01 ou1 ou01 ou2

of 1 ¼ D1 f u0U u01 ð pÞ; u02 ð pÞ 01 ou 1 ¼ D1 f ðuU Þ1 uU u0U u01 ð pÞ; u02 ð pÞ 1 01 ¼ D1 f ðuU Þ1 uU u0U u ð pÞ; u02 ð pÞ

LHS ¼

¼ 1st column of the matrix product h

1 01 D1 f ðuU Þ1 uU u0U u ð pÞ; u02 ð pÞ 1 01 i uU u0U u ð pÞ; u02 ð pÞ D2 f ðuU Þ1 12 2 3 0 1 1 0 1 1 01 02 01 02 u u D u u ð ð p Þ; u ð p Þ Þ D u u ð ð p Þ; u ð p Þ Þ 1 2 U U 7 6 U U 7 6 7 6 2 0 1 2 5 4 01 02 0 1 01 02 ðu ð pÞ; u ð pÞÞ D2 u U u U ðu ð pÞ; u ð pÞÞ D 1 uU uU 1 01 ¼ D1 f ðuU Þ1 uU u0U u ð pÞ; u02 ð pÞ 1 1 01 u ð pÞ; u02 ð pÞ D1 uU u0U 1 01 þ D2 f ðuU Þ1 uU u0U u ð pÞ; u02 ð pÞ 1 2 01 u ð pÞ; u02 ð pÞ D1 uU u0U 0 1 0 1 1 uU u0U uU ð p Þ uU ð pÞ D1 uU u0U ¼ D1 f ðuU Þ1 0 1 0 1 2 uU ð pÞ D1 uU u0U þ D2 f ðuU Þ1 uU u0U uU ð pÞ 1 1 1 0 uU u0U u0U ð pÞ D1 uU u0U uU ð pÞ ¼ D1 f ðuU Þ1 1 1 2 0 uU ð pÞ þ D2 f ðuU Þ1 uU u0U u0U ð pÞ D1 uU u0U

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2 Tangent Spaces

1 1 0 1 uU u0U u0U ð pÞ D1 uU u0U D1 f ðuU Þ1 uU ð pÞ 0 1 2 1 uU u0U u0U ð pÞ D1 uU u0U u U ð pÞ þ D2 f ðuU Þ1 0 1 1 ¼ D1 f ðuU Þ1 ðuU ð pÞÞ uU ð pÞ D1 uU u0U 0 1 2 D1 uU u0U þ D2 f ðuU Þ1 ðuU ð pÞÞ u U ð pÞ 0 1 1 ¼ D1 uU u0U D1 f ðuU Þ1 ðuU ð pÞÞ u U ð pÞ 0 1 2 D2 f ðuU Þ1 ðuU ð pÞÞ : þ D1 uU u0U u U ð pÞ ¼

2 ou1 of ou of RHS ¼ þ 01 1 01 ou ou ou ou2 1 01 of ¼ D1 u1 u0U u ð pÞ; u02 ð pÞ ou1 of 1 þ D1 u2 u0U u01 ð pÞ; u02 ð pÞ 2 ou 01 0 1 1 of ¼ D1 uU uU u ð pÞ; u02 ð pÞ ou1 1 2 01 of þ D1 uU u0U u ð pÞ; u02 ð pÞ ou2 01 1 1 ¼ D1 uU u0U u ð pÞ; u02 ð pÞ D1 f ðuU Þ1 u1 ð pÞ; u2 ð pÞ 0 1 2 01 02 þ D1 uU uU u ð pÞ; u ð pÞ D2 f ðuU Þ1 u1 ð pÞ; u2 ð pÞ 0 1 1 0 ¼ D1 uU uU uU ð pÞ D1 f ðuU Þ1 ðuU ð pÞÞ 0 1 2 0 þ D1 uU uU u U ð pÞ D2 f ðuU Þ1 ðuU ð pÞÞ : So, LHS = RHS.

h

2.7 Contravariant Vectors and Covariant Vectors

115

Note 2.71 Similarly, under the same conditions as in Theorem 2.70, we get of ¼ ou02

2 ou1 of ou of þ : ou02 ou1 ou02 ou2

In short, we write 2 X of ou j of ¼ ; ou0i ou0i ou j j¼1 and we say that cotangent vectors are covariant vectors. Similar results can be obtained for 3-dimensional smooth manifold, etc. Note 2.72 Let M be an m-dimensional smooth manifold, and N be an n-dimensional smooth manifold. Let p be an element of M, and let F : M ! N be any 1 smooth map. Then, for every f in CFðpÞ ðNÞ, f F is in Cp1 ðMÞ. 1 Reason: Since f is in CFðpÞ ðNÞ, f is a real-valued function such that dom f is an open neighborhood of FðpÞ in N. Since F : M ! N is a smooth map, and p is in M, F : M ! N is continuous at p. Since F : M ! N is continuous at p, and dom f is an open neighborhood of FðpÞ in N, F 1 ðdom f Þ is an open neighborhood of p in M. Since domðf FÞ ¼ F 1 ðdom f Þ, and F 1 ðdom f Þ is an open neighborhood of p in M, domðf FÞ is an open neighborhood of p in M. Now, it remains to be proved that f F is C1 at p. Since F : M ! N is a smooth map, and p is in M, F is C 1 at p. Since f is in 1 CFðpÞ ðNÞ, f is a real-valued function such that dom f is an open neighborhood of FðpÞ in N, and f is C1 at FðpÞ in N. Since the F-image of domðf FÞ is contained in dom f, F is C 1 at p, and f is C 1 at FðpÞ, the composite function f F is C1 at p. Hence, f F is in Cp1 ðMÞ.

Theorem 2.73 Let M be any m-dimensional smooth manifold, and let N be any ndimensional smooth manifold. Let F : M ! N be any smooth map, and let p be in M. Let F : TF ð pÞ ! Tp be the function deﬁned as follows: For every ðdf ÞFðpÞ in TFðpÞ , where f is in 1 CFðpÞ ðNÞ,

F ðdf ÞF ð pÞ ¼ ðd ðf F ÞÞp :

116

2 Tangent Spaces

Then, F is linear. 1 Proof For every f in CFðpÞ ðNÞ, from the above note, f F is in Cp1 ðMÞ, and hence, 1 ðNÞ, F ððdf ÞFðpÞ Þ ð ðdðf FÞÞp Þ is ðdðf FÞÞp is in Tp . Since, for every f in CFðpÞ 1 ðNÞ, and any in Tp , F : TFðpÞ ! Tp is a function. Next, let us take any f ; g in CFðpÞ real s; t. We have to show that

F ðd ðsf þ tgÞÞF ð pÞ ¼ s F ðdf ÞF ð pÞ þ t F ðdgÞF ð pÞ : On using Theorem 2.45, we have LHS ¼ F ðd ðsf þ tgÞÞF ð pÞ ¼ ðd ððsf þ tgÞ F ÞÞp ¼ ðd ðððsf Þ F þ ðtgÞ F ÞÞÞp ¼ ðd ððsðf F Þ þ tðg F ÞÞÞÞp ¼ s ðd ðf F ÞÞp þ t ðd ðg F ÞÞp h ¼ s F ðdf ÞF ð pÞ þ t F ðdgÞF ð pÞ ¼ RHS: h Note 2.74 Let M be any m-dimensional smooth manifold, and N be any n-dimensional smooth manifold. Let p be in M. Let F : M ! N be any smooth map. Fix any c in Tp ðMÞ, where c is in Cp ðMÞ. As in Theorem 2.73, F : TFðpÞ ! Tp . So for every 1 ðdf ÞFðpÞ in TFðpÞ where f is in CFðpÞ ðNÞ, F ððdf ÞFðpÞ Þ is in Tp . Since F ððdf ÞFðpÞ Þ is in Tp and, ½c is in Tp ðMÞ, h½c; F ððdf ÞFðpÞ Þi is a real number. We shall try to show that to R is linear, the mapping ðdf ÞFðpÞ 7! h½c; F ððdf ÞFðpÞ Þi from real linear space TFðpÞ 1 that is, for every f ; g in CFðpÞ ðNÞ, and for every real s; t; h½c; F s ðdf ÞF ð pÞ þ t ðdgÞF ð pÞ i ¼ sh½c; F ðdf ÞF ð pÞ i þ th½c; F ðdgÞF ð pÞ i:

By Theorem 2.73, LHS ¼ h½c; F s ðdf ÞF ð pÞ þ t ðdgÞF ð pÞ i ¼ h½c; s F ðdf ÞF ð pÞ þ t F ðdgÞF ð pÞ i: Further, by Theorem 2.62, h½c; s F ðdf ÞF ð pÞ þ t F ðdgÞF ð pÞ i ¼ sh½c; F ðdf ÞF ð pÞ i þ th½c; F ðdgÞF ð pÞ i ¼ RHS

2.7 Contravariant Vectors and Covariant Vectors

117

so, LHS = RHS. So, ðdf ÞFðpÞ 7! h½c; F ððdf ÞFðpÞ Þi is a linear functional on the real linear space TFðpÞ , and hence, the mapping ðdf ÞFðpÞ 7! h½c; F ððdf ÞFðpÞ Þi is a member of T FðpÞ . Let us denote the mapping ðdf ÞFðpÞ 7! h½c; F ððdf ÞFðpÞ Þi by b ð½cÞ. Thus, for every ½c in Tp ðMÞ, F b ð½cÞ is in T FðpÞ . Since F b ð½cÞ is in T FðpÞ , F b ð½cÞ; that is, for there exists a unique F ð½cÞ in TFðpÞ ðNÞ such that ðF ð½cÞÞ ¼ F 1 every ðdf ÞFðpÞ in TFðpÞ where f is in CFðpÞ ðNÞ, ðF ð½cÞÞ ðdf ÞF ð pÞ ¼ h½c; F ðdf ÞF ð pÞ i: Since for every ½c in Tp ðMÞ, F ð½cÞ is a unique member of TFðpÞ ðNÞ, F : Tp ðMÞ ! TFðpÞ ðNÞ:

2.8 Tangent Maps Deﬁnition Let M be any m-dimensional smooth manifold, and N be any ndimensional smooth manifold. Let p be in M. Let F : M ! N be any smooth map. Let p be in M. The mapping F : Tp ðMÞ ! TFðpÞ ðNÞ; as deﬁned in the Note 2.74, is called the tangent map induced by F. 1 We have seen that, for every ðdf ÞFðpÞ in TFðpÞ where f is in CFðpÞ ðNÞ, and for every ½c in Tp ðMÞ where c is in Cp ðMÞ,

ðF ð½cÞÞ ðdf ÞF ð pÞ ¼ h½c; F ðdf ÞF ð pÞ i: Theorem 2.75 Let M be any m-dimensional smooth manifold, and let N be any ndimensional smooth manifold. Let F : M ! N be any smooth map, and let p be in M. Then, the tangent map F : Tp ðMÞ ! TFðpÞ ðNÞ induced by F is linear. Proof Let us take any ½c1 ; ½c2 in Tp ðMÞ, and any real s; t. We have to show that F ðs½c1 þ t½c2 Þ ¼ sðF ð½c1 ÞÞ þ tðF ð½c2 ÞÞ: We ﬁrst try to prove that, for every ðdf ÞFðpÞ in TFðpÞ ;

ðF ðs½c1 þ t½c2 ÞÞ ðdf ÞF ð pÞ ¼ ðsðF ð½c1 ÞÞ þ tðF ð½c2 ÞÞÞ ðdf ÞF ð pÞ :

118

2 Tangent Spaces

LHS ¼ ðF ðs½c1 þ t½c2 ÞÞ ðdf ÞF ð pÞ ¼ hs½c1 þ t½c2 ; F ðdf ÞF ð pÞ i ¼ sh½c1 ; F ðdf ÞF ð pÞ i þ th½c2 ; F ðdf ÞF ð pÞ i ¼ s ðF ð½c1 ÞÞ ðdf ÞF ð pÞ þ t ðF ð½c2 ÞÞ ðdf ÞF ð pÞ ¼ ðsððF ð½c1 ÞÞ ÞÞ ðdf ÞF ð pÞ þ ðtððF ð½c2 ÞÞ ÞÞ ðdf ÞF ð pÞ ¼ ðsððF ð½c1 ÞÞ Þ þ tððF ð½c2 ÞÞ ÞÞ ðdf ÞF ð pÞ ¼ ðsðF ð½c1 ÞÞ þ tðF ð½c2 ÞÞÞ ðdf ÞF ð pÞ ¼ RHS: , Hence, LHS = RHS. Thus, we have shown that, for every ðdf ÞFðpÞ in TFðpÞ

ðF ðs½c1 þ t½c2 ÞÞ ðdf ÞF ð pÞ ¼ ðsðF ð½c1 ÞÞ þ tðF ð½c2 ÞÞÞ ðdf ÞF ð pÞ : Hence, ðF ðs½c1 þ t½c2 ÞÞ ¼ ðsðF ð½c1 ÞÞ þ tðF ð½c2 ÞÞÞ : Now, by the deﬁnition of η, gðF ðs½c1 þ t½c2 ÞÞ ¼ ðF ðs½c1 þ t½c2 ÞÞ ¼ ðsðF ð½c1 ÞÞ þ tðF ð½c2 ÞÞÞ ¼ gðsðF ð½c1 ÞÞ þ tðF ð½c2 ÞÞÞ so, gðF ðs½c1 þ t½c2 ÞÞ ¼ gðsðF ð½c1 ÞÞ þ tðF ð½c2 ÞÞÞ: Further, since η is 1–1, F ðs½c1 þ t½c2 Þ ¼ sðF ð½c1 ÞÞ þ tðF ð½c2 ÞÞ:

h h

Note 2.76 Let M be any 2-dimensional smooth manifold, and N be any 3dimensional smooth manifold. Let p be in M. Let F : M ! N be any smooth map. Let ðU; uU Þ be an admissible coordinate chart of M satisfying p 2 U, and let ðV; wV Þ be an admissible coordinate chart of N satisfying FðpÞ 2 V. Let u1 ; u2 be the component functions of uU , and let v1 ; v2 ; v3 be the component functions of wV . Since ðV; wV Þ is an admissible coordinate chart of N satisfying FðpÞ 2 V, and v1 ; v2 ; v3 are the component functions of wV , fðdv1 ÞFðpÞ ; ðdv2 ÞFðpÞ ; ðdv3 ÞFðpÞ g is the ðNÞ determined by ðV; wV Þ. Since ðU; uU Þ natural basis of the cotangent space TFðpÞ is an admissible coordinate chart of M satisfying p 2 U, and u1 ; u2 are the component functions of uU ; fðdu1 Þp ; ðdu2 Þp g is the natural basis of the cotangent space

2.8 Tangent Maps

119

Tp ðMÞ determined by ðU; uU Þ. Since F : TFðpÞ ! Tp , and ðdv1 ÞFðpÞ ; ðdv2 ÞFðpÞ ;

, F ððdv1 ÞFðpÞ Þ; F ððdv2 ÞFðpÞ Þ; F ððdv3 ÞFðpÞ Þ are in Tp . Also, by ðdv3 ÞFðpÞ are in TFðpÞ the deﬁnition of F ,

F

dv1 F ð pÞ ¼ d v1 F p :

Next, by Theorem 2.48, 1 d v F p ¼ D1 v1 F ðuU Þ1 u1 ð pÞ; u2 ð pÞ d u1 p þ D2 v1 F ðuU Þ1 u1 ð pÞ; u2 ð pÞ d u2 p ¼ D1 v1 F ðuU Þ1 u1 ð pÞ; u2 ð pÞ d u1 p þ D2 v1 F ðuU Þ1 u1 ð pÞ; u2 ð pÞ d u2 p 1 1 ¼ D1 wV F ðuU Þ1 u1 ð pÞ; u2 ð pÞ d u p 1 2 þ D2 wV F ðuU Þ1 u1 ð pÞ; u2 ð pÞ d u p 2 1 X i 1 1 2 d u p: ¼ Di wV F ðuU Þ u ð pÞ; u ð pÞ i¼1

Thus, F

dv1

F ð pÞ

¼

2 1 X i d u p: Di wV F ðuU Þ1 u1 ð pÞ; u2 ð pÞ i¼1

Similarly, F

dv

2

F ð pÞ

2 2 X i 1 1 2 Di wV F ðuU Þ ¼ u ð pÞ; u ð pÞ d u p; i¼1

and F

dv3

F ð pÞ

¼

2 3 X i Di wV F ðuU Þ1 u1 ð pÞ; u2 ð pÞ d u p: i¼1

120

2 Tangent Spaces

In short, we write 2 X a Di wV F ðuU Þ1 F ðdva ÞF ð pÞ ¼ u1 ð pÞ; u2 ð pÞ d ui p i¼1

or 2 X oF a i a F ðdv ÞF ð pÞ ¼ d u p; oui p i¼1

where a a oF 1 D w F ð u Þ u1 ð pÞ; u2 ð pÞ : i U V i ou p If there is no confusion, we simply write F ðdva Þ ¼

2 X oF a i¼1

oui

dui :

If M is an m-dimensional smooth manifold, and N is an n-dimensional smooth manifold, then as above, we get the formula: For every a ¼ 1; . . .; n; F ðdva Þ ¼

m X oF a i¼1

oui

dui :

Note 2.77 Let M be any m-dimensional smooth manifold, and N be any n-dimensional smooth manifold. Let p be in M. Let F : M ! N be any smooth map. Let ðU; uU Þ be an admissible coordinate chart of M satisfying p 2 U, and let ðV; wV Þ be an admissible coordinate chart of N satisfying FðpÞ 2 V. Let u1 ; . . .; um be the component functions of uU , and let v1 ; . . .; vn be the component functions of wV . For i ¼ 1; . . .; m, and for a ¼ 1; . . .; n, ouo i jp is in Tp ðMÞ, and ðdva ÞFðpÞ is in TFðpÞ . So, by the deﬁnition of F ,

F

*

+

!!

o

o

ðdva ÞF ð pÞ ¼

; F ðdva ÞF ð pÞ oui p oui

p *

+

m o X oF a j ¼ d u p

; oui j¼1 ou j p p

2.8 Tangent Maps

121

m X oF a o

j ¼ ; d u pi j ou p oui p j¼1 a m X oF a oF j d ¼ : ¼ i j ou p oui p j¼1 Thus, F

!! oF a o

a ðdv ÞF ð pÞ ¼ : oui p oui p

We want to prove that

F

!! X n o

oF b o ¼ j ; oui p ovb F ð pÞ oui p b¼1

1 , where f is in CFðpÞ ðNÞ, that is, for every ðdf ÞFðpÞ in TFðpÞ

F

!! o

ð df Þ F ð pÞ ¼ oui p

! n X oF b o j ð df Þ F ð pÞ : ovb F ð pÞ oui p b¼1

Here,

LHS ¼

F

!! o

o

; F ðdf Þ ð df Þ ¼ F ð pÞ F ð pÞ oui p oui p

!

n a o

X 1 1 n ¼ i ;F Da f ðwV Þ v ð pÞ; . . .; v ð pÞ ðdv ÞF ð pÞ p a¼1 X a o

n 1 1 n ; Da f ðwV Þ v ð pÞ; . . .; v ð pÞ F ðdv ÞF ð pÞ ¼ oui p a¼1

n X o a 1 1 n

¼ Da f ðwV Þ ; F ðdv ÞF ð pÞ v ð pÞ; . . .; v ð pÞ oui p a¼1 ! !! n X o

a F Da f ðwV Þ1 ð dv Þ ¼ v1 ð pÞ; . . .; vn ð pÞ F ð pÞ oui p a¼1 n X oF a Da f ðwV Þ1 ; ¼ v1 ð pÞ; . . .; vn ð pÞ oui p a¼1

122

2 Tangent Spaces

and RHS ¼

! !! o

ð df Þ F ð pÞ ovb F ð pÞ ! !

o

ðdf ÞF ð pÞ

b ov F ð pÞ

! o

ovb F ð pÞ

n X oF b

oui n X oF b ¼ oui p b¼1 n X oF b ¼ oui p b¼1 b¼1

p

n X v1 ð pÞ; . . .; vn ð pÞ ðdva ÞF ð pÞ Da f ðwV Þ1

!!

a¼1

¼

n X oF b b¼1

oui

o

ovb

¼

n X b¼1

¼ ¼

a¼1

p

!

F ð pÞ

n X oF b b¼1

¼

n X

oui b

oF oui

ðdv ÞF ð pÞ

p

p

a¼1

oui

!!

!

o

a

v ð pÞ; . . .; v ð pÞ Da f ðwV Þ ; ðdv ÞF ð pÞ ovb F ð pÞ a¼1 ! n X a 1 1 n Da f ðwV Þ v ð pÞ; . . .; v ð pÞ db n X

1

1

n

a¼1

n X oF b b¼1 n X

a

Da f ðwV Þ1 v1 ð pÞ; . . .; vn ð pÞ

Db f ðwV Þ1 v1 ð pÞ; . . .; vn ð pÞ

p

oF Da f ðwV Þ1 v1 ð pÞ; . . .; vn ð pÞ : i ou p a

Hence, LHS = RHS. Thus, we have shown that F

!! n X o

oF b ¼ oui p oui p b¼1

! ! o

: ovb F ð pÞ

Since, by the deﬁnition of η, g F

!! o

¼ oui p

!! X

! ! n o

oF b o

¼ F oui p ovb F ð pÞ oui p b¼1 ! !

!! n n X X oF b o

oF b o

g ¼ ¼g ovb F ð pÞ oui p oui p ovb F ð pÞ b¼1 b¼1

2.8 Tangent Maps

so, g F

123

!!

!! n X o

oF b o

¼g : oui p oui p ovb F ð pÞ b¼1

Further, since η is 1–1,

!

!

! n n X X o

oF b o

oF a o

¼ ¼ F oui p oui p ovb F ð pÞ oui p ova F ð pÞ a¼1 b¼1 In short, we write F

o oui

¼

n X oF a o : oui ova a¼1

Note 2.78 Let M and N be any 3-dimensional smooth manifolds. Let p be in M. Let F : M ! N be any smooth map. Let ðU; uU Þ be an admissible coordinate chart of M satisfying p 2 U, and let ðV; wV Þ be an admissible coordinate chart of N satisfying FðpÞ 2 V. Let u1 ; u2 ; u3 be the component functions of uU , and let v1 ; v2 ; v3 be the component functions of wV . Here, (

) o

o

o

; ; ou1 p ou2 p ou3 p is the natural basis of the tangent space Tp ðMÞ under local coordinate system ðui Þ, and (

) o

o

o

; ; ov1 F ð pÞ ov2 F ð pÞ ov3 F ð pÞ

is the natural basis of the tangent space TFðpÞ ðNÞ under local coordinate system ðvi Þ. Also, by Theorem 2.75, the tangent map F is a linear transformation from linear space Tp ðMÞ to linear space TFðpÞ ðNÞ. From the above note,

!

! 3 X o

oF j o

F ¼ : oui p oui p ov j F ð pÞ j¼1 So the matrix representation of F* under the natural bases (

) o

o

o

; ; ou1 p ou2 p ou3 p

( and

) o

o

o

; ; ov1 F ð pÞ ov2 F ð pÞ ov3 F ð pÞ

124

2 Tangent Spaces

is 2

oF 1 1 6 ou p 6 oF 2 6 ou1 6 4 3 p oF ou1 p

oF 1 ou2 2 p oF ou2 3 p oF ou2 p

3

oF 1 ou3 2 p oF ou3 3 p oF ou3 p

7 7 7 7 5

;

33

1 j 1 2 2 where ðoF oui Þp stands for ðDi ððwV ðF ðuU Þ ÞÞ ÞÞðu ðpÞ; u ðpÞ; u ðpÞÞ: Further, the linear transformation F : Tp ðMÞ ! TFðpÞ ðNÞ is an isomorphism if and only if j

2

oF 1 1 6 ou p 6 oF 2 det6 6 ou1 p 4 3 oF ou1 p

oF 1 ou2 2 p oF ou2 3 p oF ou2 p

3

oF 1 ou3 2 p oF ou3 3 p oF ou3 p

7 7 7 6¼ 0: 7 5

Note 2.79 Let E be an open subset of R3 . Let F : E ! R3 . Let p be in E. By Example 1.6, E is a smooth manifold with the standard differentiable structure of E. Here, differentiable structure on E contains fðE; IdE Þg. Similarly, R3 is a smooth manifold with the standard differentiable structure of R3 . Here, differentiable structure on R3 contains fðR3 ; IdR3 Þg. Now, let F : E ! R3 be a smooth function. From the Note 2.78, the tangent map F induced by F is an isomorphism if and only if 2

oF 1 1 6 ou p 6 oF 2 det6 6 ou1 p 4 3 oF ou1 p

oF 1 ou2 2 p oF ou2 3 p oF ou2 p

3

oF 1 ou3 2 p oF ou3 3 p oF ou3 p

7 7 7 6¼ 0; 7 5

where j j oF 1 3 ¼ D i F ð i Þ ðiE ð pÞÞ ¼ Di F j ðiE ð pÞÞ ¼ Di F j ð pÞ: i E R i ou p

Thus, F* is an isomorphism if and only if 2

ðD1 F 1 Þð pÞ ðD2 F 1 Þð pÞ det4 ðD1 F 2 Þð pÞ ðD2 F 2 Þð pÞ ðD1 F 3 Þð pÞ ðD2 F 3 Þð pÞ

3 ðD3 F 1 Þð pÞ ðD3 F 2 Þð pÞ 5 6¼ 0: ðD3 F 3 Þð pÞ

Chapter 3

Multivariable Differential Calculus

In the very deﬁnition of smooth manifold, there is a statement like “…for every point p in the topological space M there exist an open neighborhood U of p, an open subset V of Euclidean space Rn ; and a homeomorphism u from U onto V …”. This phrase clearly indicates that homeomorphism u can act as a messenger in carrying out the well-developed multivariable differential calculus of Rn into the realm of unornamented topological space M. So the role played by multivariable differential calculus in the development of the theory of smooth manifolds is paramount. Many theorems of multivariable differential calculus are migrated into manifolds like their generalizations. Although most students are familiar with multivariable differential calculus from their early courses, but the exact form of theorems we need here are generally missing from their collection of theorems. So this is somewhat long chapter on multivariable differential calculus that is pertinent here for a good understanding of smooth manifolds.

3.1 Linear Transformations Deﬁnition The set of all linear transformations from Rn to Rm is denoted by LðRn ; Rm Þ: By LðRn Þ, we mean LðRn ; Rn Þ: Thus, LðRn Þ denotes the set of all linear transformations from Rn to Rn . We know that LðRn ; Rm Þ is a real linear space under pointwise vector addition and pointwise scalar multiplication. Deﬁnition Let A be in LðRn ; Rm Þ and B be in LðRm ; Rk Þ: By the product BA, we mean the mapping BA : Rn ! Rk deﬁned as follows: For every x in Rn ; ðBAÞðxÞ BðAðxÞÞ:

© Springer India 2014 R. Sinha, Smooth Manifolds, DOI 10.1007/978-81-322-2104-3_3

125

126

3 Multivariable Differential Calculus

It is easy to see that BA is a linear transformation from Rn to Rk ; and hence, BA is in LðRn ; Rk Þ: Note 3.1 Let A be in LðR3 ; R2 Þ: Let (t1, t2, t3) be any member of R3 satisfying qﬃﬃﬃﬃﬃﬃﬃﬃﬃ qﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ |(t1, t2, t3)| ≤ 1. Since jt1 j ¼ ðt1 Þ2 ðt1 Þ2 þ ðt2 Þ2 þ ðt3 Þ2 ¼ jðt1 ; t2 ; t3 Þj 1; so jt1 j 1: Similarly, jt2 j 1; and jt3 j 1: Further, since A is a linear transformation from R3 to R2 ; jAððt1 ; t2 ; t3 ÞÞj ¼ jAðt1 ð1; 0; 0Þ þ t2 ð0; 1; 0Þ þ t3 ð0; 0; 1ÞÞj ¼ jt1 ðAðð1; 0; 0ÞÞÞ þ t2 ðAðð0; 1; 0ÞÞÞ þ t3 ðAðð0; 0; 1ÞÞÞj jt1 ðAðð1; 0; 0ÞÞÞj þ jt2 ðAðð0; 1; 0ÞÞÞj þ jt3 ðAðð0; 0; 1ÞÞÞj ¼ jt1 jjAðð1; 0; 0ÞÞj þ jt2 jjAðð0; 1; 0ÞÞj þ jt3 jjAðð0; 0; 1ÞÞj 1 jAðð1; 0; 0ÞÞj þ 1 jAðð0; 1; 0ÞÞj þ 1 jAðð0; 0; 1ÞÞj ¼ jAðð1; 0; 0ÞÞj þ jAðð0; 1; 0ÞÞj þ jAðð0; 0; 1ÞÞj: Thus, we have seen that for every (t1, t2, t3) in R3 satisfying jðt1 ; t2 ; t3 Þj 1; we have jAððt1 ; t2 ; t3 ÞÞj ðjAðð1; 0; 0ÞÞj þ jAðð0; 1; 0ÞÞj þ jAðð0; 0; 1ÞÞjÞ: So, ðjAðð1; 0; 0ÞÞj þ jAðð0; 1; 0ÞÞj þ jAðð0; 0; 1ÞÞjÞ is an upper bound of the set

jAððt1 ; t2 ; t3 ÞÞj : ðt1 ; t2 ; t3 Þ is in R3 ; and jðt1 ; t2 ; t3 Þj 1 :

Thus, fjAððt1 ; t2 ; t3 ÞÞj : ðt1 ; t2 ; t3 Þ is in R3 ; and jðt1 ; t2 ; t3 Þj 1g is a nonempty bounded above set of real numbers, and hence, sup jAððt1 ; t2 ; t3 ÞÞj : ðt1 ; t2 ; t3 Þ is in R3 ; and jðt1 ; t2 ; t3 Þj 1 exists. Further, we have seen that sup jAððt1 ; t2 ; t3 ÞÞj : ðt1 ; t2 ; t3 Þ is in R3 ; and jðt1 ; t2 ; t3 Þj 1 ðjAðð1; 0; 0ÞÞj þ jAðð0; 1; 0ÞÞj þ jAðð0; 0; 1ÞÞjÞ: Similarly, if A is in LðRn ; Rm Þ; then supfjAðxÞj : x is in Rn ; and jxj 1g exists, and supfjAðxÞj : x is in Rn ; and j xj 1g ðjAðð1; . . .; 0ÞÞj þ þ jAðð0; . . .; 1ÞÞjÞ:

3.1 Linear Transformations

127

Deﬁnition Let A be in LðRn ; Rm Þ: By kAk, we mean the real number supfjAðxÞj : x is in Rn ; and j xj 1g: kAk is called the norm of A. From the above discussion, we ﬁnd that for every A in LðRn ; Rm Þ; k Ak ðjAðð1; . . .; 0ÞÞj þ þ jAðð0; . . .; 1ÞÞjÞ: Theorem 3.2 1. If A is in LðRn ; Rm Þ; then for every x in Rn ; jAðxÞj k Akj xj: 2. If A is in LðRn ; Rm Þ; then the mapping A : Rn ! Rm is uniformly continuous. 3. If A, B are in LðRn ; Rm Þ; then kA þ Bk k Ak þ kBk: 4. If A is in LðRn ; Rm Þ; and t is a real number, then ktAk ¼ jtjk Ak: 5. If, for every A, B in LðRn ; Rm Þ; dðA; BÞ kA Bk; then LðRn ; Rm Þ is a metric space with the metric d. 6. The mapping A 7! kAk from LðRn ; Rm Þ to R is continuous. 7. If A is in LðRn ; Rm Þ; and B is in LðRm ; Rk Þ; then kBAk kBkk Ak: Proof 1: Let A be in LðRn ; Rm Þ: Let x be in Rn : Case I: when x = 0

128

3 Multivariable Differential Calculus

jAðxÞj ¼ jAð0Þj ¼ j0j ¼ 0 0 ¼ k Ak0 ¼ k Akj0j ¼ k Akj xj: 1 Case II: when x ≠ 0. Since x ≠ 0, jxj is a nonzero real number, and hence, jxj 1 1 1 is a real number. Also, jjxj xj ¼ jjxj jjxj ¼ jxj jxj ¼ 1 1: It follows that 1 jAðjxj xÞj is a member of the set fjAðyÞj : y is in Rn ; and jyj 1g; and hence, 1 xÞj kAk. Now, since from the deﬁnition of kAk , jAðjxj

1 1 1 1 x k Ak; jAðxÞj ¼ jAðxÞj ¼ ðAðxÞÞ ¼ A j xj j xj j xj j xj jAðxÞj k Akj xj: Thus, we see that in all cases, jAðxÞj kAkjxj: This proves 1. 2: Let us take ɛ > 0. e Case I: when kAk 6¼ 0. For any x, y in Rn satisfying jx yj\ kAk ; we have

jAðxÞ AðyÞj ¼ jAðx yÞj k Akjx yj\k Ak

e k Ak

¼ e:

Case II: when kAk ¼ 0. For any x, y in Rn satisfying jx yj\1; we have jAðxÞ AðyÞj ¼ jAðx yÞj k Akjx yj ¼ 0 jx yj ¼ 0\e: Thus, we see that in all cases, for every ɛ > 0, there exists δ > 0 such that for every x, y in Rn satisfying jx yj\d; we have jAðxÞ AðyÞj\e: So, by the deﬁnition of uniform continuity, A is uniformly continuous. This proves 2. 3: Let A, B be in LðRn ; Rm Þ: Let us take any x in Rn satisfying jxj 1: Now, jðA þ BÞðxÞj ¼ jAðxÞ þ BðxÞj jAðxÞj þ jBðxÞj k Akj xj þ jBðxÞj k Akj xj þ kBkj xj ¼ ðk Ak þ kBkÞj xj ðk Ak þ kBkÞ 1 ¼ k Ak þ kBk: So, kAk þ kBk is an upper bound of the set fjðA þ BÞðxÞj : x is in Rn ; and jxj 1g; and hence, kA þ Bk ¼ supfjðA þ BÞðxÞj : x is in Rn ; and j xj 1g k Ak þ kBk: This proves 3.

3.1 Linear Transformations

129

4: Let A be in LðRn ; Rm Þ; and let t be a real number. Let us take any x in Rn satisfying jxj 1: Now, jðtAÞðxÞj ¼ jtðAðxÞÞj ¼ jtjjAðxÞj jtjðk Akj xjÞ ¼ ðjtjk AkÞj xj ðjtjk AkÞ 1 ¼ jtjk Ak: So, jtjkAk is an upper bound of the set fjðtAÞðxÞj : x is in Rn ; and jxj 1g; and hence, ktAk ¼ supfjðtAÞðxÞj : x is in Rn ; and j xj 1g jtjk Ak: So, for every real t, ktAk jtjk Ak ðÞ:

Case I: when t 6¼ 0. Since t 6¼ 0;

1 t

is a real number, and hence, by ðÞ;

1 1 1 ktAk ¼ ktAk: k Ak ¼ ðtAÞ t t jt j Thus, jtjkAk ktAk: Hence, by (*), ktAk ¼ jtjk Ak: Case II: when t = 0. Here, ktAk ¼ k0Ak ¼ k0k ¼ supfj0ðxÞj : x is in Rn ; and j xj 1g ¼ supfj0jg ¼ supf0g ¼ 0 ¼ 0 k Ak ¼ j0jk Ak ¼ jtjk Ak: Thus, we see that in all cases, ktAk ¼ jtjk Ak: This proves 4. 5: Here, we must prove i. For every A in LðRn ; Rm Þ; kA Ak ¼ 0:

130

3 Multivariable Differential Calculus

ii. For every A, B in LðRn ; Rm Þ; if kA Bk ¼ 0, then A = B. iii. For every A, B in LðRn ; Rm Þ; kA Bk ¼ kB Ak: iv. for every A, B, C in LðRn ; Rm Þ; kA Bk kA C k þ kC Bk: For i: kA Ak ¼ k0k ¼ supfj0ðxÞj : x is in Rn ; and jxj 1g ¼ supfj0jg ¼ supf0g ¼ 0: For ii: Let kA Bk ¼ 0: Let us take any x in Rn : We have to prove that AðxÞ ¼ BðxÞ: Since 0 jðA BÞðxÞj kA Bkjxj ¼ 0 jxj ¼ 0; |(A − B) (x)| = 0, and hence, A(x) − B(x) = (A − B)(x) = 0. Thus, A(x) = B(x). For iii: By 4, LHS ¼ kA Bk ¼ kð1ÞðB AÞk ¼ j1jkðB AÞk ¼ 1 kB Ak ¼ kB Ak ¼ RHS: For iv: By 3, kA Bk ¼ kðA CÞ þ ðC BÞk kA C k þ kC Bk: This completes the proof of iv. 6: Let us take any A in LðRn ; Rm Þ: We have to prove that the mapping A 7! kAk is continuous. For this purpose, let us take any ɛ > 0. Let us take any B in LðRn ; Rm Þ satisfying kB Ak\e: We have to prove that jkBk kAkj\e: Since kBk ¼ kðB AÞ þ Ak kB Ak þ kAk; kBk kAk kB Ak: Similarly, ðkBk kAkÞ ¼ kAk kBk kA Bk ¼ kB Ak: Hence, jkBk k Akj ¼ maxfkBk k Ak; ðkBk k AkÞg kB Ak\e: This proves 6. 7: Let us take any A in LðRn ; Rm Þ and B in LðRm ; Rk Þ: Let us take any x in Rn satisfying jxj 1: Since jðBAÞðxÞj ¼ jBðAðxÞÞj kBkjAðxÞj kBk ðk Akj xjÞ ¼ ðkBkk AkÞj xj kBkk Ak 1 ¼ kBkk Ak; kBAk ¼ supfjðBAÞðxÞj : x is in Rn ; and j xj 1g kBkk Ak: This proves 7.

h

3.1 Linear Transformations

131

Theorem 3.3 Let X be the set of all invertible (i.e., 1–1 and onto) members of LðRn Þ: 1. If A is in X; then kA1 k is nonzero. 2. If A is in X, then the open sphere Sð1=kA1 kÞ ðAÞ is contained in X; that is, if A is in X, B is in LðRn Þ; and kB Ak\ kA11 k, then B is in X. 3. X is an open subset of LðRn Þ: 4. The mapping A 7! A1 from X to X is continuous. Proof 1: Let us take any A in X: We claim that kA1 k is nonzero. If not, otherwise, let kA1 k ¼ 0: We have to arrive at a contradiction. Since A is in X, by the deﬁnition of X; A−1 exists and A−1 is in X: Also, AA−1 = I, where I denotes the identity transformation from Rn to Rn : For every x in Rn ; 0 jA1 ðxÞj kA1 kjxj ¼ 0 jxj ¼ 0; so jA1 ðxÞj ¼ 0; and hence, A−1(x) = 0. Therefore, ð1; 0; . . .; 0Þ ¼ Iðð1; 0; . . .; 0ÞÞ ¼ ðAA1 Þðð1; 0; . . .; 0ÞÞ

¼ A A1 ðð1; 0; . . .; 0ÞÞ ¼ Aðð0; 0; . . .; 0ÞÞ ¼ ð0; 0; . . .; 0Þ; and hence, 1 = 0, a contradiction. So, our claim is true, that is, kA1 k is nonzero. This proves 1. 2: Let A be in X; and let B be in LðRn Þ: Since A is in X, by 1, kA1 k is nonzero, and hence, kA11 k is a real number. Also, by the deﬁnition of norm kk, kA11 k is a positive real number. Let 1 kB Ak\ 1 : kA k We have to show that B is in X; that is, B : Rn ! Rn is 1–1 and onto. 1–1ness: Let B(x) = B(y). We have to prove that x = y, that is, x y ¼ 0. If not, otherwise, let x y 6¼ 0. We have to arrive at a contradiction. Since x y 6¼ 0, 0\jx yj: Since B(x) = B(y), and B is a linear transformation, 0 ¼ BðxÞ BðyÞ ¼ Bðx yÞ. Hence,

jx yj ¼ A1 A ðx yÞ ¼ A1 ðAðx yÞÞ A1 jAðx yÞj ¼ A1 jAðx yÞ 0j ¼ A1 jAðx yÞ Bðx yÞj ¼ A1 jðA BÞðx yÞj A1 ðkA Bkjx yjÞ

¼ A1 kA Bk jx yj 1

¼ A1 kB Ak jx yj\ A1 1 jx yj ¼ jx yj: kA k

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3 Multivariable Differential Calculus

Thus, jx yj\jx yj; which gives a contradiction. This proves that B is 1–1. Ontoness: Let us take any y in Rn : Since B : Rn ! Rn is 1–1, B ((1, 0, …, 0)), …, B((0, …, 0, 1)) are n distinct elements of Rn : Now, we shall show that B((1, 0, …, 0)), …, B((0, …, 0, 1)) are linearly independent. For this purpose, let t1(B((1, 0, …, 0))) + ⋯ + tn(B((0, …, 0, 1))) = 0. We have to show that t1 = 0, …, tn = 0. Here, Bðð0; . . .; 0ÞÞ ¼ ð0; . . .; 0Þ ¼ 0 ¼ t1 ðBðð1; 0; . . .; 0ÞÞÞ þ þ tn ðBðð0; . . .; 0; 1ÞÞÞ ¼ Bðt1 ð1; 0; . . .; 0ÞÞ þ þ Bðtn ð0; . . .; 0; 1ÞÞ ¼ Bððt1 ; 0; . . .; 0ÞÞ þ þ Bðð0; . . .; 0; tn ÞÞ ¼ Bððt1 ; 0; . . .; 0Þ þ þ ð0; . . .; 0; tn ÞÞ ¼ Bððt1 ; . . .; tn ÞÞ; and B is 1–1, so (0,…, 0) = (t1,…, tn), and hence, t1 = 0,…, tn = 0. Thus, B ((1, 0,…, 0)),…, B((0,…, 0, 1)) constitute a basis of the n-dimensional real linear space Rn : Since B((1, 0,…, 0)),…, B((0, …, 0, 1)) constitute a basis of space Rn ; and y is in Rn ; there exist real numbers s1, …, sn such that y ¼ s1 ðBðð1; 0; . . .; 0ÞÞÞ þ þ sn ðBðð0; . . .; 0; 1ÞÞÞ ¼ Bðs1 ð1; 0; . . .; 0ÞÞ þ þ Bðsn ð0; . . .; 0; 1ÞÞ ¼ Bððs1 ; . . .; sn ÞÞ: Thus, B((s1, …, sn)) = y, where (s1, …, sn) is in Rn : This proves that B is onto. This proves 2. 3: Let us take any A in X. Since A is in X, by 1, kA1 k is nonzero. Now by 2, the open sphere fB : B 2 LðRn Þ; and kB Ak\ kA11 kg is contained in X. Hence, by the deﬁnition of open set, X is an open subset of LðRn Þ: This proves 3. 4: Let us take any A in X. We have to prove that the mapping B 7! B1 is continuous at A Since A is in X, by 1, kA1 k is nonzero. If B is in LðRn Þ satisfying kB Ak\ kA11 k ; then by 2, B is in X, and hence, B−1 exists. Now, it is enough to prove that limB!A kB1 A1 k exists, and its value is 0. Here,

1

B A1 ¼ A1 A B1 A1 BB1 ¼ A1 AB1 A1 BB1

¼ A1 AB1 BB1 ¼ A1 ðA BÞB1 A1 kA BkB1

¼ A1 kA Bk sup B1 ðxÞ : x is in Rn ; and j xj 1

or, 1

B A1 A1 kA Bk 1 sup B ðxÞ : x is in Rn ; and j xj 1 ðÞ:

3.1 Linear Transformations

133

Observe that for any x in Rn satisfying jxj 1, we have 1 1 1 1 1

1 1 B ðxÞ ¼ A A B ðxÞ ¼ A A B ðxÞ A A B ðxÞ 1 1 1

¼ A ðA BÞ B ðxÞ þ x A ðA BÞ B1 ðxÞ þ j xj

A1 kA BkB1 ðxÞ þ j xj A1 kB AkB1 ðxÞ þ 1 so, 1

B ðxÞ 1 A1 kB Ak A1 : Further, since kB Ak\ kA11 k, 1 B ðxÞ

1 A 1 kA1 kkB Ak

:

Hence,

sup B1 ðxÞ : x is in Rn

and

1 A : j xj 1 1 kA1 kkB Ak

Now, from (*), 1 A 1 kB AkA1 ¼ A 1 kA1 kkB Ak 1 kA1 kkB Ak 1 1 ¼ A 1 þ : 1 kA1 kkB Ak

1

B A1 A1 kB Ak

Thus, if kB Ak\ kA11 k, then 1 1 1 0 B A A 1 þ

1 : 1 kA1 kkB Ak

By Theorem 3.2, the mapping B 7! kBk from X to R is continuous, so limB!A kB Ak ¼ 0: Also, 1 1 1 1 lim A 1 þ 1 þ ¼ A B!A 1 kA1 kkB Ak 1 kA1 k 0 1 ¼ A ð1 þ 1Þ ¼ 0:

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3 Multivariable Differential Calculus

Since, for every B satisfying kB Ak\ kA11 k, we have 0 B1 A1 A1 1 þ

1 ; 1 kA1 kkB Ak

and lim A1 1 þ

B!A

1 1 1 kA kkB Ak

¼0

so, limB!A kB1 A1 k exists, and its value is 0. This shows that the mapping h B 7! B1 from X to X is continuous at A. This proves 4. Deﬁnition Let {e1, e2, e3} be any basis of real linear space R3 ; and let {f1, f2} be any basis of real linear space R2 : Let A be in LðR3 ; R2 Þ: Let Aðe1 Þ a11 f1 þ a21 f2 ; Aðe2 Þ a12 f1 þ a22 f2 ; Aðe3 Þ a13 f1 þ a23 f2 : The matrix

a11 a21

a12 a22

a13 a23

23

is denoted by [A]. Similar deﬁnition can be supplied for A in LðRn ; Rm Þ: Theorem 3.4 Let M be the collection of all 2 × 3 matrices with real entries. Let {e1, e2, e3} be a basis of real linear space R3 ; and let {f1, f2} be a basis of real linear space R2 : Then, the mapping A 7! ½A from LðR3 ; R2 Þ to M is 1–1 and onto. Proof 1–1ness: Let [A] = [B], where

a ½ A 11 a21

a12 a22

a13 a23

and 23

b ½B 11 b21

b12 b22

b13 b23

: 23

We have to prove that the mappings A : R3 ! R2 and B : R3 ! R2 are equal, that is, for every t1e1 + t2e2 + t3e3 in R3 ; Aðt1 e1 þ t2 e2 þ t3 e3 Þ ¼ Bðt1 e1 þ t2 e2 þ t3 e3 Þ: For this purpose, let us take any t1e1 + t2e2 + t3e3 in R3 : Since [A] = [B], aij = bij for every i = 1, 2, and for every j = 1, 2, 3. Further,

3.1 Linear Transformations

a ½ A ¼ 11 a21

135

a12 a22

a13 a23

and 23

b ½B ¼ 11 b21

b12 b22

b13 b23

23

so A(e1) = a11f1 + a21f2 = b11f1 + b21f2 = B(e1). Similarly, A(e2) = B(e2) and A (e3) = B(e3). So, LHS ¼ Aðt1 e1 þ t2 e2 þ t3 e3 Þ ¼ t1 ðAðe1 ÞÞ þ t2 ðAðe2 ÞÞ þ t3 ðAðe3 ÞÞ ¼ t1 ðBðe1 ÞÞ þ t2 ðBðe2 ÞÞt3 ðBðe3 ÞÞ ¼ Bðt1 e1 þ t2 e2 þ t3 e3 Þ ¼ RHS: This proves that A 7! ½A is 1–1. Ontoness: Let us take any 2 × 3 matrix

a11 a21

a12 a22

a13 a23

: 23

Let us deﬁne a function A : R3 ! R2 as follows: For every t1e1 + t2e2 + t3e3 in R ; 3

Aðt1 e1 þ t2 e2 þ t3 e3 Þ t1 ða11 f1 þ a21 f2 Þ þ t2 ða12 f1 þ a22 f2 Þ þ t3 ða13 f1 þ a23 f2 Þ: We shall show that A is in LðR3 ; R2 Þ: For this purpose, let us take any x ≡ s1e1 + s2e2 + s3e3, y ≡ t1e1 + t2e2 + t3e3, and any real number t. We have to prove that 1. Aðx þ yÞ ¼ AðxÞ þ AðyÞ; 2. AðtxÞ ¼ tðAðxÞÞ: For 1: LHS ¼ Aðx þ yÞ ¼ Aððs1 e1 þ s2 e2 þ s3 e3 Þ þ ðt1 e1 þ t2 e2 þ t3 e3 ÞÞ ¼ Aððs1 þ t1 Þe1 þ ðs2 þ t2 Þe2 þ ðs3 þ t3 Þe3 Þ ¼ ðs1 þ t1 Þða11 f1 þ a21 f2 Þ þ ðs2 þ t2 Þða12 f1 þ a22 f2 Þ þ ðs3 þ t3 Þða13 f1 þ a23 f2 Þ: and RHS ¼ AðxÞ þ AðyÞ ¼ Aðs1 e1 þ s2 e2 þ s3 e3 Þ þ Aðt1 e1 þ t2 e2 þ t3 e3 Þ ¼ s1 ða11 f1 þ a21 f2 Þ þ s2 ða12 f1 þ a22 f2 Þ þ s3 ða13 f1 þ a23 f2 Þ þ t1 ða11 f1 þ a21 f2 Þ þ t2 ða12 f1 þ a22 f2 Þ þ t3 ða13 f1 þ a23 f2 Þ ¼ ðs1 þ t1 Þða11 f1 þ a21 f2 Þ þ ðs2 þ t2 Þða12 f1 þ a22 f2 Þ þ ðs3 þ t3 Þða13 f1 þ a23 f2 Þ:

So, LHS = RHS. This proves 1.

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3 Multivariable Differential Calculus

For 2: LHS ¼ AðtxÞ ¼ Aðtðs1 e1 þ s2 e2 þ s3 e3 ÞÞ ¼ Aððts1 Þe1 þ ðts2 Þe2 þ ðts3 Þe3 Þ ¼ ðts1 Þða11 f1 þ a21 f2 Þ þ ðts2 Þða12 f1 þ a22 f2 Þ þ ðts3 Þða13 f1 þ a23 f2 Þ ¼ ðts1 a11 þ ts2 a12 þ ts3 a13 Þf1 þ ðts1 a21 þ ts2 a22 þ ts3 a23 Þf2 ¼ tðs1 a11 þ s2 a12 þ s3 a13 Þf1 þ tðs1 a21 þ s2 a22 þ s3 a23 Þf2 ¼ ts1 ða11 f1 þ a21 f2 Þ þ ts2 ða12 f1 þ a22 f2 Þ þ ts3 ða13 f1 þ a23 f2 Þ ¼ tðs1 ða11 f1 þ a21 f2 Þ þ s2 ða12 f1 þ a22 f2 Þ þ s3 ða13 f1 þ a23 f2 ÞÞ: and RHS ¼ tðAðxÞÞ ¼ tðAðs1 e1 þ s2 e2 þ s3 e3 ÞÞ ¼ tðs1 ða11 f1 þ a21 f2 Þ þ s2 ða12 f1 þ a22 f2 Þ þ s3 ða13 f1 þ a23 f2 ÞÞ: So, LHS = RHS. This proves 2. Finally, it remains to prove that

a ½ A ¼ 11 a21

a12 a22

a13 : a23

Since Aðe1 Þ ¼ Að1e1 þ 0e2 þ 0e3 Þ ¼ 1ða11 f1 þ a21 f2 Þ þ 0ða12 f1 þ a22 f2 Þ þ 0ða13 f1 þ a23 f2 Þ; Aðe1 Þ ¼ a11 f1 þ a21 f2 : Similarly, Aðe2 Þ ¼ a12 f1 þ a22 f2 ; Aðe3 Þ ¼ a13 f1 þ a23 f2 : Hence,

½ A ¼

a11 a21

a12 a22

a13 : a23

h h

Note 3.5 The result similar to Theorem 3.4 can be proved as above for LðR ; R Þ: n

m

Theorem 3.6 Let {e1, e2, e3, e4} be any basis of real linear space R4 ; let {f1, f2, f3} be any basis of real linear space R3 ; and let {g1, g2} be any basis of real linear space R2 : Let A be in LðR4 ; R3 Þ; and let B be in LðR3 ; R2 Þ: Then,

3.1 Linear Transformations

137

½BA ¼ ½B½ A: Proof Here, A is in LðR4 ; R3 Þ and B is in LðR3 ; R2 Þ; so their product BA (i.e., the composite B ∘ A) is in LðR4 ; R2 Þ; and hence, LHSð¼ ½BAÞ is a 2 × 4 matrix. Since A is in LðR4 ; R3 Þ; [A] is a 3 × 4 matrix. Since B is in LðR3 ; R2 Þ; [B] is a 2 × 3 matrix. Since [B] is a 2 × 3 matrix, and [A] is a 3 × 4 matrix, RHSð¼ ½B½AÞ is a 2 × 4 matrix. So, matrices in LHS and RHS have the same order. Let 2

a11 ½ A ¼ 4 a21 a31

a12 a22 a32

a13 a23 a33

3 a14 a24 5 a34 34

and ½B ¼

b11 b21

b12 b22

b13 b23

: 23

Hence, ðBAÞðe1 Þ ¼ BðAðe1 ÞÞ ¼ Bða11 f1 þ a21 f2 þ a31 f3 Þ ¼ a11 ðBðf1 ÞÞ þ a21 ðBðf2 ÞÞ þ a31 ðBðf3 ÞÞ ¼ a11 ðb11 g1 þ b21 g2 Þ þ a21 ðb12 g1 þ b22 g2 Þ þ a31 ðb13 g1 þ b23 g2 Þ ¼ ða11 b11 þ a21 b12 þ a31 b13 Þg1 þ ða11 b21 þ a21 b22 þ a31 b23 Þg2 :

So the ﬁrst column of [BA] is a11 b11 þ a21 b12 þ a31 b13 a11 b21 þ a21 b22 þ a31 b23 Now, ½B½ A ¼

b11

b12

b13

b21

b22

b23

2

23

a11

6 4 a21 a31

b11 a11 þ b12 a21 þ b13 a31 ¼ b21 a11 þ b22 a21 þ b23 a31 a11 b11 þ a21 b12 þ a31 b13 ¼ a11 b21 þ a21 b22 þ a31 b23

a12

a13

a22

a23

a32

a33

a14

3

7 a24 5 a34 34 24 24

Hence, ﬁrst column of [BA] is the same as the ﬁrst column of [B][A]. Similarly, second column of [BA] is the same as the second column of [B][A], third column of [BA] is the same as the third column of [B][A], and fourth column of [BA] is the same as the fourth column of [B][A]. This shows that [BA] = [B][A]. h

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3 Multivariable Differential Calculus

Note 3.7 The result similar to Theorem 3.6 can be proved as above for A in LðRn ; Rm Þ and B in LðRm ; Rp Þ: Note 3.8 Let A be in LðR3 ; R2 Þ: Let ½ A ¼

a11 a21

a12 a22

a13 a23

23

relative to the basis {(1, 0, 0), (0, 1, 0), (0, 0, 1)} of R3 and the basis {(1, 0), (0, 1)} of R2 : Let (t1, t2, t3) be any member of R3 satisfying jðt1 ; t2 ; t3 Þj 1: Since A is a linear transformation from R3 to R2 ; Aððt1 ; t2 ; t3 ÞÞ ¼ Aðt1 ð1; 0; 0Þ þ t2 ð0; 1; 0Þ þ t3 ð0; 0; 1ÞÞ ¼ t1 ðAðð1; 0; 0ÞÞÞ þ t2 ðAðð0; 1; 0ÞÞÞ þ t3 ðAðð0; 0; 1ÞÞÞ ¼ t1 ða11 ; a21 Þ þ t2 ða12 ; a22 Þ þ t3 ða13 ; a23 Þ ¼ ðt1 a11 þ t2 a12 þ t3 a13 ; t1 a21 þ t2 a22 þ t3 a23 Þ ¼ ððt1 ; t2 ; t3 Þ ða11 ; a12 ; a13 Þ; ðt1 ; t2 ; t3 Þ ða21 ; a22 ; a23 ÞÞ; and hence, jAððt1 ; t2 ; t3 ÞÞj ¼ jððt1 ; t2 ; t3 Þ ða11 ; a12 ; a13 Þ; ðt1 ; t2 ; t3 Þ ða21 ; a22 ; a23 ÞÞj qﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ ¼ jðt1 ; t2 ; t3 Þ ða11 ; a12 ; a13 Þj2 þjðt1 ; t2 ; t3 Þ ða21 ; a22 ; a23 Þj2 qﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ ðjðt1 ; t2 ; t3 Þjjða11 ; a12 ; a13 ÞjÞ2 þðjðt1 ; t2 ; t3 Þjjða21 ; a22 ; a23 ÞjÞ2 qﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ ð1jða11 ; a12 ; a13 ÞjÞ2 þð1jða21 ; a22 ; a23 ÞjÞ2 qﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ ¼ jða11 ; a12 ; a13 Þj2 þjða21 ; a22 ; a23 Þj2 qﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ ¼ ða11 Þ2 þða12 Þ2 þða13 Þ2 þða21 Þ2 þða22 Þ2 þða23 Þ2 : This shows that k Ak ¼ supfjAððt1 ; t2 ; t3 ÞÞj : ðt1 ; t2 ; t3 Þ is in Rn ; and jðt1 ; t2 ; t3 Þj 1g vﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ ! qﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ u 3 2

X uX 2 2 2 2 2 2 2 t ða11 Þ þða12 Þ þða13 Þ þða21 Þ þða22 Þ þða23 Þ ¼ aij : j¼1

Conclusion: If A is in LðR3 ; R2 Þ and ½ A ¼

a11 a21

a12 a22

a13 a23

23

i¼1

3.1 Linear Transformations

139

relative to the basis {(1, 0, 0), (0, 1, 0), (0, 0, 1)} of R3 and the basis {(1, 0), (0, 1)} of R2 ; then vﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ ! u 3 2

uX X 2 aij : k Ak t j¼1

i¼1

The result similar to this conclusion can be proved as above for Rn in place of R3 and Rm in place of R2 : Theorem 3.9 Let X be a metric space. Let s be the 1–1, onto function A 7! ½A of Theorem 3.4, relative to the basis {(1, 0, 0), (0, 1, 0), (0, 0, 1)} of R3 and the basis {(1, 0), (0, 1)} of R2 : If, for i = 1, 2, and j = 1, 2, 3, each aij : X ! R is continuous, then the mapping x 7! s1

a11 ðxÞ a12 ðxÞ a21 ðxÞ a22 ðxÞ

a13 ðxÞ a23 ðxÞ

23

from X to LðR3 ; R2 Þ is also continuous. Proof Let us observe that for every x, y in X, 1 a11 ðxÞ a12 ðxÞ s a21 ðxÞ a22 ðxÞ ¼ Ax Ay ;

a13 ðxÞ a23 ðxÞ

s

1

a11 ðyÞ a12 ðyÞ a21 ðyÞ a22 ðyÞ

a13 ðyÞ a23 ðyÞ

where

Ay s

a12 ðxÞ a13 ðxÞ 2 L R3 ; R2 ; a22 ðxÞ a23 ðxÞ

a12 ðyÞ a13 ðyÞ 2 L R3 ; R2 : a22 ðyÞ a23 ðyÞ

a11 ðxÞ a21 ðxÞ a11 ðyÞ 1

Ax s1

a21 ðyÞ

Since Ax ¼ s1

a11 ðxÞ a12 ðxÞ a21 ðxÞ a22 ðxÞ

½Ax ¼ sðAx Þ ¼

a13 ðxÞ a23 ðxÞ

a11 ðxÞ a12 ðxÞ a21 ðxÞ a22 ðxÞ

;

a13 ðxÞ ; a23 ðxÞ

140

3 Multivariable Differential Calculus

and hence, Ax ðð1; 0; 0ÞÞ ¼ ða11 ðxÞÞð1; 0Þ þ ða21 ðxÞÞð0; 1Þ ¼ ða11 ðxÞ; a21 ðxÞÞ; Ax ðð0; 1; 0ÞÞ ¼ ða12 ðxÞÞð1; 0Þ þ ða22 ðxÞÞð0; 1Þ ¼ ða12 ðxÞ; a22 ðxÞÞ; Ax ðð0; 0; 1ÞÞ ¼ ða13 ðxÞÞð1; 0Þ þ ða23 ðxÞÞð0; 1Þ ¼ ða13 ðxÞ; a23 ðxÞÞ: Similarly, Ay ðð1; 0; 0ÞÞ ¼ ða11 ðyÞ; a21 ðyÞÞ; Ay ðð0; 1; 0ÞÞ ¼ ða12 ðyÞ; a22 ðyÞÞ; Ay ðð0; 0; 1ÞÞ ¼ ða13 ðyÞ; a23 ðyÞÞ: Since Ax is in LðR3 ; R2 Þ; and Ay is in LðR3 ; R2 Þ; Ax − Ay is in LðR3 ; R2 Þ; and hence, by Note 3.1,

Ax Ay Ax Ay ðð1; 0; 0ÞÞ þ Ax Ay ðð0; 1; 0ÞÞ

þ Ax Ay ðð0; 0; 1ÞÞ

¼ Ax ðð1; 0; 0ÞÞ Ay ðð1; 0; 0ÞÞ þ Ax ðð0; 1; 0ÞÞ Ay ðð0; 1; 0ÞÞ

þ Ax ðð0; 0; 1ÞÞ Ay ðð0; 0; 1ÞÞ ¼ jðða11 ðxÞ; a21 ðxÞÞ ða11 ðyÞ; a21 ðyÞÞÞj þ jðða12 ðxÞ; a22 ðxÞÞ ða12 ðyÞ; a22 ðyÞÞÞj þ jðða13 ðxÞ; a23 ðxÞÞ ða13 ðyÞ; a23 ðyÞÞÞj ¼ jða11 ðxÞ a11 ðyÞ; a21 ðxÞ a21 ðyÞÞj þ jða12 ðxÞ a12 ðyÞ; a22 ðxÞ a22 ðyÞÞj þ jða13 ðxÞ a13 ðyÞ; a23 ðxÞ a23 ðyÞÞj ðja11 ðxÞ a11 ðyÞj þ ja21 ðxÞ a21 ðyÞjÞ þ ðja12 ðxÞ a12 ðyÞj þ ja22 ðxÞ a22 ðyÞjÞ þ ðja13 ðxÞ a13 ðyÞj þ ja23 ðxÞ a23 ðyÞjÞ: Thus, for every x, y in X, 1 a11 ðxÞ a12 ðxÞ s a21 ðxÞ a22 ðxÞ

a13 ðxÞ a23 ðxÞ

s1

a11 ðyÞ a12 ðyÞ a21 ðyÞ a22 ðyÞ

ðja11 ðxÞ a11 ðyÞj þ ja21 ðxÞ a21 ðyÞjÞ þ ðja12 ðxÞ a12 ðyÞj þ ja22 ðxÞ a22 ðyÞjÞ þ ðja13 ðxÞ a13 ðyÞj þ ja23 ðxÞ a23 ðyÞjÞ ðÞ: Now, let us ﬁx any x in X. It is enough to prove that the function

a23 ðyÞ a13 ðyÞ

3.1 Linear Transformations

x 7! s

141

1

a11 ðxÞ a12 ðxÞ a21 ðxÞ a22 ðxÞ

a13 ðxÞ a23 ðxÞ

23

from X to LðR3 ; R2 Þ is continuous at x. For this purpose, let us take any ɛ > 0. Since each aij : X ! R is continuous at x, there exists δ > 0 such that for every y in the open neighborhood Bδ(x) of x, we have jaij ðxÞ aij ðyÞj\ 6e ðfor every i ¼ 1; 2; and j ¼ 1; 2; 3Þ: Hence, for every y in the open neighborhood Bδ(x) of x, we have from (*), 1 a11 ðxÞ a12 ðxÞ a13 ðxÞ a11 ðyÞ a12 ðyÞ a13 ðyÞ 1 s s a21 ðxÞ a22 ðxÞ a23 ðxÞ a21 ðyÞ a22 ðyÞ a23 ðyÞ ðja11 ðxÞ a11 ðyÞj þ ja21 ðxÞ a21 ðyÞjÞþ_ ðja12 ðxÞ a12 ðyÞj þ ja22 ðxÞ a22 ðyÞjÞ e e e e e e þ ðja13 ðxÞ a13 ðyÞj þ ja23 ðxÞ a23 ðyÞjÞ \ þ þ þ þ þ ¼ e: 6 6 6 6 6 6

h h

Note 3.10 The result similar to Theorem 3.9 can be proved as above for R in place of R3 and Rm in place of R2 : n

3.2 Differentiation Theorem 3.11 Let E be an open subset of R3 : Let f : E ! R2 : Let f1 : E ! R; f2 : E ! R be the component functions of f, that is, f(h) = (f1(h), f2(h)) for every h in E. Let x be in E. If there exist A and B in LðR3 ; R2 Þ such that

1 lim ððf1 ðx þ hÞ; f2 ðx þ hÞÞ ðf1 ðxÞ; f2 ðxÞÞ AðhÞÞ h!0 jhj

¼ 0;

and lim

h!0

1 ððf1 ðx þ hÞ; f2 ðx þ hÞÞ ðf1 ðxÞ; f2 ðxÞÞ BðhÞÞ j hj

¼ 0;

then A = B. Proof We have to prove that A = B, that is, for every h in R3 ; AðhÞ BðhÞ ¼ 0: For this purpose, let us take any h* ≡ (h1, h2, h3) in R3 : It is sufﬁcient to prove that |(A − B)(h*)| = 0. Case I: when h ¼ 0

142

3 Multivariable Differential Calculus

LHS ¼ jðA BÞðh Þj ¼ jðA BÞð0Þj ¼ j0j ¼ 0 ¼ RHS: Case II: when h* ≠ 0 1 ððf1 ðx þ hÞ; f2 ðx þ hÞÞ ðf1 ðxÞ; f2 ðxÞÞ AðhÞÞÞ ¼ 0; t 7! th is a Since limh!0 ðjhj continuous function, and h* ≠ 0, and hence,

1 lim ððf1 ðx þ th Þ; f2 ðx þ th ÞÞ ðf1 ðxÞ; f2 ðxÞÞ Aðth ÞÞ t!0 jth j

¼ 0:

Similarly,

1 lim ððf1 ðx þ th Þ; f2 ðx þ th ÞÞ ðf1 ðxÞ; f2 ðxÞÞ Bðth ÞÞ t!0 jth j

¼ 0:

Hence,

1 ððf1 ðx þ th Þ; f2 ðx þ th ÞÞ ðf1 ðxÞ; f2 ðxÞÞ Aðth ÞÞ jth j 1 ððf1 ðx þ th Þ; f2 ðx þ th ÞÞ ðf1 ðxÞ; f2 ðxÞÞ Bðth ÞÞ jth j

lim t!0

exists, and its value is 0 ð¼ 0 0Þ: Further, since jh j [ 0; 1 0 ¼ lim ððf1 ðx þ th Þ; f2 ðx þ th ÞÞ ðf1 ðxÞ; f2 ðxÞÞ Aðth ÞÞ t!0 jth j 1 ððf1 ðx þ th Þ; f2 ðx þ th ÞÞ ðf1 ðxÞ; f2 ðxÞÞ Bðth ÞÞ jth j 1 ¼ lim ðBðth Þ Aðth ÞÞ t!0 jth j 1 1 ¼ lim tðBðh Þ Aðh ÞÞ ðtðBðh ÞÞ tðAðh ÞÞÞ ¼ lim t!0 jth j t!0 jth j 1 1 ¼ lim tððB AÞðh ÞÞ ¼ lim tððB AÞðh ÞÞ : t!0 jth j t!0 jt jjh j Since limt!0 ðjtjjh1 j tððB AÞðh ÞÞÞ ¼ 0; 1 1 1 t ð B A Þ h ð ð Þ Þ 0 ¼ lim jðB AÞðh Þj ¼ jðB AÞðh Þj: ¼ lim t!0 jtjjh j t!0 jh j jh j Since jh1 j jðB AÞðh Þj ¼ 0, LHS ¼ jðA BÞðh Þj ¼ jðB AÞðh Þj ¼ 0 ¼ RHS: Thus, we have shown that in all cases, LHS = RHS. h

3.2 Differentiation

143

Note 3.12 The result similar to Theorem 3.11 can be proved as above for Rn in place of R3 and Rm in place of R2 : According to Theorem 3.11, if there exists A in LðR3 ; R2 Þ such that 1 lim ððf1 ðx þ hÞ; f2 ðx þ hÞÞ ðf1 ðxÞ; f2 ðxÞÞ AðhÞÞ ¼ 0; h!0 jhj then such an A is unique. Deﬁnition Let E be an open subset of R3 : Let f : E ! R2 : Let f1 : E ! R; f2 : E ! R be the component functions of f, that is, f ðhÞ ¼ ðf1 ðhÞ; f2 ðhÞÞ for every h in E. Let x be in E. If there exists A in LðR3 ; R2 Þ such that 1 lim ððf1 ðx þ hÞ; f2 ðx þ hÞÞ ðf1 ðxÞ; f2 ðxÞÞ AðhÞÞ ¼ 0; h!0 jhj then we say that f is differentiable at x, and we write f 0 ðxÞ ¼ A: Here, f 0 ðxÞ is called the differential of f at x, or the total derivative of f at x. It is also denoted by Df(x). 1 ðf ðx þ hÞ f ðxÞ AðhÞÞÞ ¼ 0: In short, f 0 ðxÞ ¼ A means limh!0 ðjhj If f is differentiable at every point of E, then we say that f is differentiable on E. Similar deﬁnition can be supplied for Rn in place of R3 and Rm in place of R2 : Theorem 3.13 Let E be an open subset of Rn : Let f : E ! Rm : Let x be in E. Let A be in LðRn ; Rm Þ: f 0 ðxÞ ¼ A if and only if there exists a function r : fh : x þ h 2 Eg ! Rm such that (i) rð0Þ ¼ 0; (ii) f ðx þ hÞ ¼ f ðxÞ þ AðhÞ þ jhjðrðhÞÞ for every h in fh : x þ h 2 Eg; (iii) r is continuous at 0. Proof of “if” part Suppose that there exists a function r : fh : x þ h 2 Eg ! Rm such that (i) rð0Þ ¼ 0; (ii) f(x + h) = f(x) + A(h) + |h|(r(h)) for every h in fh : x þ h 2 Eg; (iii) r is continuous at 0. We have to show that f 0 ðxÞ ¼ A, that is, 1 ðf ðx þ hÞ f ðxÞ AðhÞÞ ¼ 0: lim h!0 jhj

144

3 Multivariable Differential Calculus

Since r is continuous at 0,

1 ðf ðx þ hÞ f ðxÞ AðhÞÞ RHS ¼ 0 ¼ rð0Þ ¼ lim rðhÞ ¼ lim h!0 h!0 jhj

¼ LHS:

Proof of “only if” part Let f 0 ðxÞ ¼ A, that is,

1 ðf ðx þ hÞ f ðxÞ AðhÞÞ lim h!0 jhj

¼ 0 ðÞ:

Let us deﬁne r : fh : x þ h 2 Eg ! Rm as follows: For every h in fh : x þ h 2 Eg;

rðhÞ

1 jhj ðf ðx

0;

þ hÞ f ðxÞ AðhÞÞ; if h 6¼ 0 : if h ¼ 0

Clearly, r(0) = 0. Also, it is clear that for every h in fh : x þ h 2 Eg; f ðx þ hÞ ¼ f ðxÞ þ AðhÞ þ jhjðrðhÞÞ. By (*),

1 lim rðhÞ ¼ lim ðf ðx þ hÞ f ðxÞ AðhÞÞ ¼ 0 ¼ rð0Þ; h!0 h!0 jhj h

and hence, r is continuous at 0.

Theorem 3.14 Let E be an open subset of R3 : Let f : E ! R2 : Let f1 : E ! R; f2 : E ! R be the component functions of f, that is, f(h) = (f1(h), f2(h)) for every h in E. Let x be in E. Let A 2 LðR3 ; R2 Þ; where ½ A ¼

a11 a21

a12 a22

a13 a23

23

relative to the basis {(1, 0, 0), (0, 1, 0), (0, 0, 1)} of R3 and the basis {(1, 0), (0, 1)} of R2 : f 0 ðxÞ ¼ A if and only if

1 lim ðf1 ðx þ hÞ f1 ðxÞ ðh1 ; h2 ; h3 Þ ða11 ; a12 ; a13 ÞÞ ¼ 0; ðh1 ;h2 ;h3 Þh!0 jhj 1 lim ðf2 ðx þ hÞ f2 ðxÞ ðh1 ; h2 ; h3 Þ ða21 ; a22 ; a23 ÞÞ ¼ 0: ðh1 ;h2 ;h3 Þh!0 jhj

3.2 Differentiation

145

Proof f 0 ðxÞ ¼ A means

1 ðf ðx þ hÞ f ðxÞ AðhÞÞ h!0 jhj 1 ¼ lim ððf1 ðx þ hÞ; f2 ðx þ hÞÞ ðf1 ðxÞ; f2 ðxÞÞ AðhÞÞ : ðh1 ;h2 ;h3 Þh!0 jhj

ð0; 0Þ ¼ 0 ¼ lim

Since AðhÞ ¼ Aððh1 ; h2 ; h3 ÞÞ ¼ h1 ðAðð1; 0; 0ÞÞÞ þ h2 ðAðð0; 1; 0ÞÞÞ þ h3 ðAðð0; 0; 1ÞÞÞ ¼ h1 ða11 ; a21 Þ þ h2 ða12 ; a22 Þ þ h3 ða13 ; a23 Þ ¼ ðh1 a11 þ h2 a12 þ h3 a13 ; h1 a21 þ h2 a22 þ h3 a23 Þ ¼ ððh1 ; h2 ; h3 Þ ða11 ; a12 ; a13 Þ; ðh1 ; h2 ; h3 Þ ða21 ; a22 ; a23 ÞÞ; f 0 ðxÞ ¼ A if and only if ð0; 0Þ ¼

lim

ðh1 ;h2 ;h3 Þh!0

1 ððf1 ðx þ hÞ; f2 ðx þ hÞÞ ðf1 ðxÞ; f2 ðxÞÞÞ j hj

ððh1 ; h2 ; h3 Þ ða11 ; a12 ; a13 Þ; ðh1 ; h2 ; h3 Þ ða21 ; a22 ; a23 ÞÞ 1 ¼ lim ðf1 ðx þ hÞ f1 ðxÞ ðh1 ; h2 ; h3 Þ ða11 ; a12 ; a13 ÞÞ; ðh1 ;h2 ;h3 Þh!0 jhj 1 ðf2 ðx þ hÞ f2 ðxÞ ðh1 ; h2 ; h3 Þ ða21 ; a22 ; a23 ÞÞ j hj 1 ¼ lim ðf1 ðx þ hÞ f1 ðxÞ ðh1 ; h2 ; h3 Þ ða11 ; a12 ; a13 ÞÞ; ðh1 ;h2 ;h3 Þh!0 jhj 1 lim ðf2 ðx þ hÞ f2 ðxÞ ðh1 ; h2 ; h3 Þ ða21 ; a22 ; a23 ÞÞ : ðh1 ;h2 ;h3 Þh!0 jhj Thus, f 0 ðxÞ ¼ A if and only if lim

1

lim

1

ðh1 ;h2 ;h3 Þh!0 jhj ðh1 ;h2 ;h3 Þh!0 jhj

ðf1 ðx þ hÞ f1 ðxÞ ðh1 ; h2 ; h3 Þ ða11 ; a12 ; a13 ÞÞ ¼ 0; ðf2 ðx þ hÞ f2 ðxÞ ðh1 ; h2 ; h3 Þ ða21 ; a22 ; a23 ÞÞ ¼ 0:

h h

Note 3.15 The result similar to Theorem 3.14 can be proved as above for R in place of R3 and Rm in place of R2 : n

146

3 Multivariable Differential Calculus

Theorem 3.16 Let E be an open subset of Rn : Let f : E ! Rm : Let x be in E. If f is differentiable at x, then f is continuous at x. Proof Let f be differentiable at x. We have to prove that f is continuous at x, that is, limh!0 ðf ðx þ hÞ f ðxÞÞ ¼ 0: For this purpose, let us take any ɛ > 0. Since f is differentiable at x, there exists a linear transformation A from Rn to Rm 1 such that 0 ¼ limh!0 ðjhj ðf ðx þ hÞ f ðxÞ AðhÞÞÞ; and hence, there exists δ > 0 such that for every h in fh : x þ h 2 Eg satisfying 0\jhj\d, we have e 1 1 jf ðx þ hÞ f ðxÞ AðhÞj ¼ ðf ðx þ hÞ f ðxÞ AðhÞÞ 0\ : 2 j hj j hj Case I: when kAk 6¼ 0 e Now, if h is in fh : x þ h 2 Eg satisfying 0\jhj\minf1; 2kAk ; dg, then 1 ðf ðx þ hÞ f ðxÞ AðhÞÞ þ AðhÞ jðf ðx þ hÞ f ðxÞÞ 0j ¼ jhj jhj 1 jhj ðf ðx þ hÞ f ðxÞ AðhÞÞ þ jAðhÞj j hj 1 jhj ðf ðx þ hÞ f ðxÞ AðhÞÞ þ k Akjhj j hj 1 1 ðf ðx þ hÞ f ðxÞ AðhÞÞ þ k Akjhj jhj e e \1 þ k Ak ¼ e: 2 2k Ak

So, f is continuous at x. Case II: when kAk ¼ 0 Here, kAk ¼ 0, so A = 0. Now, if h is in fh : x þ h 2 Eg satisfying 0\jhj\minf1; dg, then 1 ðf ðx þ hÞ f ðxÞ 0ðhÞÞ jðf ðx þ hÞ f ðxÞÞ 0j ¼ jhj jhj 1 ¼ j hj jðf ðx þ hÞ f ðxÞ 0ðhÞÞj j hj 1 e 1 jðf ðx þ hÞ f ðxÞ AðhÞÞj \1 \e: 2 j hj So, f is continuous at x. Thus, we see that in all cases, f is continuous at x.

h

3.2 Differentiation

147

Note 3.17 If A 2 LðRn ; Rm Þ, then A is differentiable on Rn ; and for every x in Rn ; A0 ðxÞ ¼ A: Reason: Let us take any x in Rn : Since limh!0

1 ðAðx þ hÞ AðxÞ AðhÞÞ j hj 1 ¼ limh!0 ððAðxÞ þ AðhÞÞ AðxÞ AðhÞÞ j hj ¼ limh!0 ð0Þ ¼ 0; A0 ðxÞ ¼ A:

Further, since A is differentiable on Rn , by Theorem 3.16, A is continuous on Rn : Theorem 3.18 Let E be an open subset of Rn : Let x be in E. Let f : E ! Rm be differentiable at x. Let G be an open subset of Rm ; and G contains the range of f. Let g : G ! Rp be differentiable at f ðxÞ: Then, 1. the composite function ðg f Þ : E ! Rp is differentiable at x, 2. ðg f Þ0 ðxÞ ¼ ðg0 ðf ðxÞÞÞðf 0 ðxÞÞ: Proof Here, f : E ! Rm is differentiable at x, so by Theorem 3.13, there exists a function r : fh : x þ h 2 Eg ! Rm such that (i) rð0Þ ¼ 0; (ii) f ðx þ hÞ ¼ f ðxÞ þ ðf 0 ðxÞÞðhÞ þ jhjðrðhÞÞ for every h in fh : x þ h 2 Eg; (iii) r is continuous at 0. Here, g : G ! Rp is differentiable at f ðxÞ, so by Theorem 3.17, there exists a function s : fk : f ðxÞ þ k 2 Gg ! Rp such that (i′) sð0Þ ¼ 0; (ii′) gðf ðxÞ þ kÞ ¼ gðf ðxÞÞ þ ðg0 ðf ðxÞÞÞðkÞ þ jkjðsðkÞÞ for every k in fk : f ðxÞþ k 2 Gg; (iii′) s is continuous at 0. We have to show that the composite function ðg f Þ : E ! Rp is differentiable at x, and ðg f Þ0 ðxÞ ¼ ðg0 ðf ðxÞÞÞðf 0 ðxÞÞ: By Theorem 3.13, it is enough to ﬁnd a function t : fh : x þ h 2 Eg ! Rp such that (I) tð0Þ ¼ 0; (II) ðg f Þðx þ hÞ ¼ ðg f ÞðxÞ þ ððg0 ðf ðxÞÞÞðf 0 ðxÞÞÞðhÞ þ jhjðtðhÞÞ for every h in fh : x þ h 2 Eg; (III) t is continuous at 0.

148

3 Multivariable Differential Calculus

Here, for every h in fh : x þ h 2 E g; ðg f Þðx þ hÞ ¼ gðf ðx þ hÞÞ ¼ gðf ðxÞ þ ðf 0 ðxÞÞðhÞ þ jhjðrðhÞÞÞ ¼ gðf ðxÞ þ ððf 0 ðxÞÞðhÞ þ jhjðrðhÞÞÞÞ ¼ gðf ðxÞÞ þ ðg0 ðf ðxÞÞÞððf 0 ðxÞÞðhÞ þ jhjðrðhÞÞÞ þ jðf 0 ðxÞÞðhÞ þ jhjðrðhÞÞjðsððf 0 ðxÞÞðhÞ þ jhjðrðhÞÞÞÞ ¼ gðf ðxÞÞ þ ðg0 ðf ðxÞÞÞððf 0 ðxÞÞðhÞÞ þ ðg0 ðf ðxÞÞÞðjhjðrðhÞÞÞ þ jðf 0 ðxÞÞðhÞ þ jhjðrðhÞÞjðsððf 0 ðxÞÞðhÞ þ jhjðrðhÞÞÞÞ ¼ ðg f ÞðxÞ þ ððg0 ðf ðxÞÞÞ ðf 0 ðxÞÞÞðhÞ þ jhjððg0 ðf ðxÞÞÞðrðhÞÞÞ þ jðf 0 ðxÞÞðhÞ þ jhjðrðhÞÞjðsððf 0 ðxÞÞðhÞ þ jhjðrðhÞÞÞÞ ¼ ðg f ÞðxÞ þ ððg0 ðf ðxÞÞÞðf 0 ðxÞÞÞðhÞ 1 þ jhj ðg0 ðf ðxÞÞÞðrðhÞÞ þ jðf 0 ðxÞÞðhÞ þ jhjðrðhÞÞjðsððf 0 ðxÞÞðhÞ þ jhjðrðhÞÞÞÞ : j hj

So, for every h in fh : x þ h 2 Eg, ðg f Þðx þ hÞ ¼ ðg f ÞðxÞ þ ððg0 ðf ðxÞÞÞðf 0 ðxÞÞÞðhÞ þ jhjððg0 ðf ðxÞÞÞðrðhÞÞ 1 þ jðf 0 ðxÞÞðhÞ þ jhjðrðhÞÞjðsððf 0 ðxÞÞðhÞ þ jhjðrðhÞÞÞÞÞ: jhj Now, let us deﬁne the function t : fh : x þ h 2 Eg ! Rp as follows: For every h in fh : x þ h 2 Eg;

tðhÞ

ðg0 ðf ðxÞÞÞðrðhÞÞ þ j1hj jðf 0 ðxÞÞðhÞ þ jhjðrðhÞÞjðsððf 0 ðxÞÞðhÞ þ jhjðrðhÞÞÞÞ; 0;

if h 6¼ 0 : if h ¼ 0

It sufﬁces to prove that t is continuous at 0, that is, lim tðhÞ ¼ 0:

h!0

For this purpose, let us take any e [ 0: Since ðg0 ðf ðxÞÞÞðrðhÞÞ ¼ ððg0 ðf ðxÞÞÞ rÞðhÞ; the linear transformation g0 ðf ðxÞÞ is continuous, and r is continuous at 0, the mapping h 7! ðg0 ðf ðxÞÞÞðrðhÞÞ is continuous at 0. Hence, limh!0 ðg0 ðf ðxÞÞÞðrðhÞÞ ¼ ðg0 ðf ðxÞÞÞðrð0ÞÞ ¼ ðg0 ðf ðxÞÞÞð0Þ ¼ 0: It follows that there exists d1 [ 0 such that for every h in fh : x þ h 2 Eg satisfying 0\jhj\d1 , we have e jðg0 ðf ðxÞÞÞðrðhÞÞj\ : 2 Since, for every nonzero h in fh : x þ h 2 Eg, we have

3.2 Differentiation

149

1 1 ðjðf 0 ðxÞÞðhÞj þ jjhjðrðhÞÞjÞ jðf 0 ðxÞÞðhÞ þ jhjðrðhÞÞj j hj j hj 1 1 ¼ ðjðf 0 ðxÞÞðhÞj þ jhjjrðhÞjÞ ðkf 0 ðxÞkjhj þ jhjjrðhÞjÞ ¼ kf 0 ðxÞk þ jrðhÞj: j hj j hj Since limh!0 rðhÞ ¼ 0; there exists d2 [ 0 such that for every h in fh : x þ h 2 Eg satisfying 0\jhj\d2 , we have jrðhÞj\1: So, for every nonzero h in fh : x þ h 2 Eg satisfying 0\jhj\d2 ; we have 1 jðf 0 ðxÞÞðhÞ þ jhjðrðhÞÞj jjf 0 ðxÞjj þ 1: j hj Since limh!0 rðhÞ ¼ 0; and limk!0 sðkÞ ¼ 0; limh!0 sððf 0 ðxÞÞðhÞ þ jhjðrðhÞÞÞ ¼ sððf 0 ðxÞÞð0Þ þ j0jðrð0ÞÞÞ ¼ sð0 þ 0Þ ¼ sð0Þ ¼ 0; and hence, there exists d3 [ 0 such that for every h in fh : x þ h 2 Eg satisfying 0\jhj\d3 , we have e jsððf 0 ðxÞÞðhÞ þ jhjðrðhÞÞÞj\ 2ðkf 0 ðxÞkþ1Þ : Hence, for every h in fh : x þ h 2 Eg satisfying 0\jhj\minfd1 ; d2 ; d3 g, we have 1 jðf 0 ðxÞÞðhÞ þ jhjðrðhÞÞjðsððf 0 ðxÞÞðhÞ þ jhjðrðhÞÞÞÞ jhj 1 ¼ jðf 0 ðxÞÞðhÞ þ jhjðrðhÞÞjjsððf 0 ðxÞÞðhÞ þ jhjðrðhÞÞÞj jhj e e ðkf 0 ðxÞk þ 1Þjsððf 0 ðxÞÞðhÞ þ jhjðrðhÞÞÞj\ðkf 0 ðxÞk þ 1Þ ¼ : 2ðkf 0 ðxÞk þ 1Þ 2 Further, 1 jtðhÞ 0j ¼ jtðhÞj ¼ ðg0 ðf ðxÞÞÞðrðhÞÞ þ jðf 0 ðxÞÞðhÞ þ jhjðrðhÞÞj j hj ðsððf 0 ðxÞÞðhÞ þ jhjðrðhÞÞÞÞj jðg0 ðf ðxÞÞÞðrðhÞÞj 1 e e þ jðf 0 ðxÞÞðhÞ þ jhjðrðhÞÞjðsððf 0 ðxÞÞðhÞ þ jhjðrðhÞÞÞÞ\ þ ¼ e: 2 2 j hj

This proves that limh!0 tðhÞ ¼ 0:

h

Note 3.19 Theorem 3.18 is known as the chain rule of derivative. Theorem 3.20 Let E be an open subset of R3 : Let f : E ! R2 : Let f1 : E ! R; f2 : E ! R be the component functions of f, that is, f ðhÞ ¼ ðf1 ðhÞ; f2 ðhÞÞ for every h in E. Let x ðx1 ; x2 ; x3 Þ be in E. Let f be differentiable at x.

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3 Multivariable Differential Calculus

Then, all the partial derivatives ðD1 f1 ÞðxÞ; ðD2 f1 ÞðxÞ; ðD3 f1 ÞðxÞ; ðD1 f2 ÞðxÞ; ðD2 f2 Þ ðxÞ; ðD3 f2 ÞðxÞ exist. Also, ðf 0 ðxÞÞðð1; 0; 0ÞÞ ¼ ððD1 f1 ÞðxÞ; ðD1 f2 ÞðxÞÞ; ðf 0 ðxÞÞðð0; 1; 0ÞÞ ¼ ððD2 f1 ÞðxÞ; ðD2 f2 ÞðxÞÞ; ðf 0 ðxÞÞðð0; 0; 1ÞÞ ¼ ððD3 f1 ÞðxÞ; ðD3 f2 ÞðxÞÞ; and for every h in R3 ; ðf 0 ðxÞÞðhÞ ¼ ðððrf1 ÞðxÞÞ h; ððrf2 ÞðxÞÞ hÞ; where ðrf1 ÞðxÞ ððD1 f1 ÞðxÞ; ðD2 f1 ÞðxÞ; ðD3 f1 ÞðxÞÞ; ðrf2 ÞðxÞ ððD1 f2 ÞðxÞ; ðD2 f2 ÞðxÞ; ðD3 f2 ÞðxÞÞ: (Here, ðrf1 ÞðxÞ is called the gradient of f1 ; etc.) Proof Since f is differentiable at x, f 0 ðxÞ is in LðR3 ; R2 Þ: Let

a ½f ðxÞ ¼ 11 a21 0

a12 a22

a13 a23

23

relative to the basis fð1; 0; 0Þ; ð0; 1; 0Þ; ð0; 0; 1Þg of R3 and the basis fð1; 0Þ; ð0; 1Þg of R2 : By Theorem 3.14, lim

ðh1 ;h2 ;h3 Þh!0

1 ðf1 ðx þ hÞ f1 ðxÞ ðh1 ; h2 ; h3 Þ ða11 ; a12 ; a13 ÞÞ jhj

¼ 0:

Hence, lim t!0

1 ðf1 ððx1 ; x2 ; x3 Þ þ ð0; t; 0ÞÞ f1 ððx1 ; x2 ; x3 ÞÞ ð0; t; 0Þ ða11 ; a12 ; a13 ÞÞ jð0; t; 0Þj

or,

lim t!0

or,

1 ðf1 ððx1 ; x2 þ t; x3 ÞÞ f1 ððx1 ; x2 ; x3 ÞÞ ta12 Þ ¼ 0 jtj

1 ðf1 ððx1 ; x2 þ t; x3 ÞÞ f1 ððx1 ; x2 ; x3 ÞÞ ta12 Þ ¼ 0 t!0 t

lim

¼0

3.2 Differentiation

or,

151

1 ðf1 ððx1 ; x2 þ t; x3 ÞÞ f1 ððx1 ; x2 ; x3 ÞÞÞ a12 ¼ 0 t!0 t

lim or,

ðD2 f1 ÞðxÞ ¼ lim t!0

f1 ððx1 ; x2 þ t; x3 ÞÞ f1 ððx1 ; x2 ; x3 ÞÞ ¼ a12 : t

It follows that ðD2 f1 ÞðxÞ exists, and its value is a12 : Similarly, all ðDj fi ÞðxÞ exist, and ðDj fi ÞðxÞ ¼ aij . Hence,

ðD1 f1 ÞðxÞ ½f ðxÞ ¼ ðD1 f2 ÞðxÞ 0

ðD2 f1 ÞðxÞ ðD3 f1 ÞðxÞ ðD2 f2 ÞðxÞ ðD3 f2 ÞðxÞ

23

relative to the basis fð1; 0; 0Þ; ð0; 1; 0Þ; ð0; 0; 1Þg of R3 and the basis fð1; 0Þ; ð0; 1Þg of R2 : Also, ðf 0 ðxÞÞðð1; 0; 0ÞÞ ¼ ððD1 f1 ÞðxÞ; ðD1 f2 ÞðxÞÞ; ðf 0 ðxÞÞðð0; 1; 0ÞÞ ¼ ððD2 f1 ÞðxÞ; ðD2 f2 ÞðxÞÞ; ðf 0 ðxÞÞðð0; 0; 1ÞÞ ¼ ððD3 f1 ÞðxÞ; ðD3 f2 ÞðxÞÞ: Also, for every h in R3 ; LHS ¼ ðf 0 ðxÞÞðhÞ ¼ ðf 0 ðxÞÞððh1 ; h2 ; h3 ÞÞ ¼ ðf 0 ðxÞÞðh1 ð1; 0; 0Þ þ h2 ð0; 1; 0Þ þ h3 ð0; 0; 1ÞÞ ¼ h1 ððD1 f1 ÞðxÞ; ðD1 f2 ÞðxÞÞ þ h2 ððD2 f1 ÞðxÞ; ðD2 f2 ÞðxÞÞ þ h3 ððD3 f1 ÞðxÞ; ðD3 f2 ÞðxÞÞ ¼ ðh1 ððD1 f1 ÞðxÞÞ þ h2 ððD2 f1 ÞðxÞÞ þ h3 ððD3 f1 ÞðxÞÞ; h1 ððD1 f2 ÞðxÞÞ þ h2 ððD2 f2 ÞðxÞÞ þ h3 ððD3 f2 ÞðxÞÞÞ ¼ ðððD1 f1 ÞðxÞ; ðD2 f1 ÞðxÞ; ðD3 f1 ÞðxÞÞ ðh1 ; h2 ; h3 Þ; ððD1 f2 ÞðxÞ; ðD2 f2 ÞðxÞ; ðD3 f2 ÞðxÞÞ ðh1 ; h2 ; h3 ÞÞ ¼ ðððrf1 ÞðxÞÞ ðh1 ; h2 ; h3 Þ; ððrf2 ÞðxÞÞ ðh1 ; h2 ; h3 ÞÞ ¼ ðððrf1 ÞðxÞÞ h; ððrf2 ÞðxÞÞ hÞ ¼ RHS:

h Note 3.21 The result similar to Theorem 3.20 can be proved as above for R in place of R3 and Rm in place of R2 : n

Theorem 3.22 Let E be an open subset of R3 : Let f : E ! R2 : Let f1 : E ! R; f2 : E ! R be the component functions of f, that is, f ðhÞ ¼ ðf1 ðhÞ; f2 ðhÞÞ for every h in E. Let x ðx1 ; x2 ; x3 Þ be in E. Let f be differentiable at x. Let u ðu1 ; u2 ; u3 Þ be a unit vector in R3 : Then,

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3 Multivariable Differential Calculus

ðDu f1 ÞðxÞ ¼ u ððrf1 ÞðxÞÞ; ðDu f2 ÞðxÞ ¼ u ððrf2 ÞðxÞÞ; where f1 ððx1 þ tu1 ; x2 þ tu2 ; x3 þ tu3 ÞÞ f1 ððx1 ; x2 ; x3 ÞÞ ; t f2 ððx1 þ tu1 ; x2 þ tu2 ; x3 þ tu3 ÞÞ f2 ððx1 ; x2 ; x3 ÞÞ : ðDu f2 ÞðxÞ lim t!0 t ðDu f1 ÞðxÞ lim t!0

(Here, ðDu f1 ÞðxÞ is called the directional derivative of f1 in the direction of u, etc.) Proof Since f is differentiable at x; f 0 ðxÞ is in LðR3 ; R2 Þ: Let

a ½f ðxÞ ¼ 11 a21 0

a12 a22

a13 a23

23

relative to the basis fð1; 0; 0Þ; ð0; 1; 0Þ; ð0; 0; 1Þg of R3 and the basis fð1; 0Þ; ð0; 1Þg of R2 : By Theorem 3.14, lim

ðh1 ;h2 ;h3 Þh!0

1 ðf1 ðx þ hÞ f1 ðxÞ ðh1 ; h2 ; h3 Þ ða11 ; a12 ; a13 ÞÞ jhj

¼ 0:

Hence, lim t!0

1 ðf1 ððx1 ; x2 ; x3 Þ þ tðu1 ; u2 ; u3 ÞÞ f1 ððx1 ; x2 ; x3 ÞÞ j t ð u1 ; u2 ; u3 Þ j ðtðu1 ; u2 ; u3 ÞÞ ða11 ; a12 ; a13 ÞÞ ¼ 0

or, lim t!0

1 ðf1 ððx1 þ tu1 ; x2 þ tu2 ; x3 þ tu3 ÞÞ f1 ððx1 ; x2 ; x3 ÞÞ jtjjðu1 ; u2 ; u3 Þj tððu1 ; u2 ; u3 Þ ða11 ; a12 ; a13 ÞÞÞ ¼ 0

or, lim t!0

1 ðf1 ððx1 þ tu1 ; x2 þ tu2 ; x3 þ tu3 ÞÞ f1 ððx1 ; x2 ; x3 ÞÞ jt j 1 tððu1 ; u2 ; u3 Þ ða11 ; a12 ; a13 ÞÞÞ ¼ 0

3.2 Differentiation

153

or, 1 lim ðf1 ððx1 þ tu1 ; x2 þ tu2 ; x3 þ tu3 ÞÞ f1 ððx1 ; x2 ; x3 ÞÞ t!0 t tððu1 ; u2 ; u3 Þ ða11 ; a12 ; a13 ÞÞÞ ¼ 0 or,

f1 ððx1 þ tu1 ; x2 þ tu2 ; x3 þ tu3 ÞÞ f1 ððx1 ; x2 ; x3 ÞÞ t!0 t ððu1 ; u2 ; u3 Þ ða11 ; a12 ; a13 ÞÞ ¼ 0

lim

or, f1 ððx1 þ tu1 ; x2 þ tu2 ; x3 þ tu3 ÞÞ f1 ððx1 ; x2 ; x3 ÞÞ t ¼ ðu1 ; u2 ; u3 Þ ððD1 f1 ÞðxÞ; ðD2 f1 ÞðxÞ; ðD3 f1 ÞðxÞÞ ¼ ðu1 ; u2 ; u3 Þ ððrf1 ÞðxÞÞ ¼ u ððrf1 ÞðxÞÞ

ðDu f1 ÞðxÞ ¼ lim t!0

Thus, ðDu f1 ÞðxÞ ¼ u ððrf1 ÞðxÞÞ: Similarly, ðDu f2 ÞðxÞ ¼ u ððrf2 ÞðxÞÞ:

h h

Note 3.23 The result similar to Theorem 3.22 can be proved as above for Rn in place of R3 and Rm in place of R2 : Theorem 3.24 Let E be an open subset of Rn : Let f : E ! R: Let x ðx1 ; . . .; xn Þ be in E Let f be differentiable at x. Then, 0 1 maxfðDu f ÞðxÞ : u 2 Rn and juj ¼ 1g ¼ @D

f AðxÞ:

1 ððrf ÞðxÞÞ jðrf ÞðxÞj

(In short, if f is differentiable at x, then the maximum value of the directional derivative of f at x is obtained in the direction of the gradient of f at x.) Proof Let u ðu1 ; . . .; un Þ be any unit vector in Rn : We have to prove that 0 1 ðDu f ÞðxÞ @D

1 ððrf ÞðxÞÞ jðrf ÞðxÞj

f AðxÞ:

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3 Multivariable Differential Calculus

By Theorem 3.22, ðDu f ÞðxÞ ¼ u ððrf ÞðxÞÞ: Hence, ðDu f ÞðxÞ jðDu f ÞðxÞj ¼ ju ððrf ÞðxÞÞj jujjðrf ÞðxÞj 1 ¼ 1 jðrf ÞðxÞj ¼ jðrf ÞðxÞj ¼ jðrf ÞðxÞj2 jðrf ÞðxÞj 1 ðððrf ÞðxÞÞ ððrf ÞðxÞÞÞ ¼ jðrf ÞðxÞj 1 ððrf ÞðxÞÞ ððrf ÞðxÞÞ ¼ jðrf ÞðxÞj 0 1 ¼ @D

f AðxÞ:

1 ððrf ÞðxÞÞ jðrf ÞðxÞj

Hence, 0

1

ðDu f ÞðxÞ @D

1 ððrf ÞðxÞÞ jðrf ÞðxÞj

f AðxÞ: h

Theorem 3.25 Let a\b: Let c : ða; bÞ ! R2 be any function differentiable on the open interval ða; bÞ: Let G be an open subset of R2 ; and G contains the range of c: Let f : G ! R be differentiable on G Then, for every t in the open interval ða; bÞ; d cðtÞ : ðf cÞ ðtÞ ¼ ððrf ÞðcðtÞÞÞ dt 0

Proof Let c1 : ða; bÞ ! R; c2 : ða; bÞ ! R be the component functions of c; that is, cðtÞ ¼ ðc1 ðtÞ; c2 ðtÞÞ for every t in ða; bÞ: Let us take any t in the open interval ða; bÞ. By the chain rule of derivative, we have ðf cÞ0 ðtÞ ¼ ðf 0 ðcðtÞÞÞðc0 ðtÞÞ: Hence, ðf cÞ0 ðtÞ ¼ ½ðf 0 ðcðtÞÞÞðc0 ðtÞÞ ¼ ½ðf 0 ðcðtÞÞÞ½ðc0 ðtÞÞ 3 2d c1 ðtÞ 7 6 ¼ ½ ðD1 f ÞðcðtÞÞ ðD2 f ÞðcðtÞÞ 12 4 dt 5 d c ðtÞ dt 2 21 d d c ðtÞ þ ððD2 f ÞðcðtÞÞÞ c ðtÞ : ¼ ððD1 f ÞðcðtÞÞÞ dt 1 dt 2

3.2 Differentiation

155

So, d d c ðtÞ þ ððD2 f ÞðcðtÞÞÞ c ðtÞ LHS ¼ ðf cÞ ðtÞ ¼ ððD1 f ÞðcðtÞÞÞ dt 1 dt 2 d d ¼ ððD1 f ÞðcðtÞÞ; ðD2 f ÞðcðtÞÞÞ c ðtÞ; c2 ðtÞ dt 1 dt d ¼ ððD1 f ÞðcðtÞÞ; ðD2 f ÞðcðtÞÞÞ ðc1 ðtÞ; c2 ðtÞÞ dt d cðtÞ ¼ ððD1 f ÞðcðtÞÞ; ðD2 f ÞðcðtÞÞÞ dt d ¼ ððrf ÞðcðtÞÞÞ cðtÞ ¼ RHS: dt 0

h

Note 3.26 The result similar to Theorem 3.25 can be proved as above for R in place of R2 : m

Note 3.27 Let a be in Rn : The function t 7! ta from R to Rn is denoted by a, provided that there is no confusion. Here, kak ¼ supfjtaj : t 2 R and jtj 1g ¼ supfjtjjaj : t 2 R and jtj 1g jaj ¼ j1aj supfjtaj : t 2 R and jtj 1g ¼ kak: This shows that kak ¼ jaj: Theorem 3.28 Let E be an open convex subset of R3 : Let f : E ! R2 be differentiable on E. Let the set fkf 0 ðxÞk : x 2 Eg be bounded above. Then, for every a, b in E, jf ðbÞ f ðaÞj jb ajðsupfkf 0 ðxÞk : x 2 E gÞ Proof First of all, we shall prove the following lemma.

h

Lemma 3.29 Let a\b: Let f : ½a; b ! Rn be any continuous function. If f is differentiable on ða; bÞ; then there exists a real number t in ða; bÞ such that jf ðbÞ f ðaÞj ðb aÞjf 0 ðtÞj: Proof of Lemma Case I: when f ðaÞ ¼ f ðbÞ. The lemma is trivial in this case. Case II: when f ðaÞ 6¼ f ðbÞ. Since f ðaÞ 6¼ f ðbÞ; 0\jf ðbÞ f ðaÞj; and hence, n 1 1 jf ðbÞf ðaÞj ðf ðbÞ f ðaÞÞ is in R : Put c jf ðbÞf ðaÞj ðf ðbÞ f ðaÞÞ: Clearly, jcj ¼ 1: Let us deﬁne a function

156

3 Multivariable Differential Calculus

g : ½a; b ! R as follows: For every x in ½a; b; gðxÞ c ðf ðxÞÞ: Now, we want to apply Lagrange’s mean value theorem on g. Since f : ½a; b ! Rn is continuous, g : ½a; b ! R is continuous. Since f : ½a; b ! Rn is differentiable on ða; bÞ; g : ½a; b ! R is differentiable on ða; bÞ; and for every x in ða; bÞ; g0 ðxÞ ¼ c ðf 0 ðxÞÞ: Hence, by the Lagrange’s mean value theorem, there exists a real number t in ða; bÞ such that jf ðbÞ f ðaÞj ¼

1 ðf ðbÞ f ðaÞÞ ðf ðbÞ f ðaÞÞ ¼ c ðf ðbÞ f ðaÞÞ jf ðbÞ f ðaÞj

¼ c ðf ðbÞÞ c ðf ðaÞÞ ¼ gðbÞ gðaÞ ¼ ðb aÞg0 ðtÞ ¼ ðb aÞðc ðf 0 ðtÞÞÞ ¼ c ððb aÞðf 0 ðtÞÞÞ jc ððb aÞðf 0 ðtÞÞÞj jcjjðb aÞðf 0 ðtÞÞj ¼ 1jðb aÞðf 0 ðtÞÞj ¼ ðb aÞjf 0 ðtÞj:

Thus, there exists a real number t in ða; bÞ such that jf ðbÞ f ðaÞj ðb aÞjf 0 ðtÞj:

h

Proof of the main theorem Let us take any a ða1 ; a2 ; a3 Þ; b ðb1 ; b2 ; b3 Þ in E. Let us take a nonzero element c ðc1 ; c2 Þ in R2 : Let f1 ; f2 be the component functions of f. Let us deﬁne a function c : ½0; 1 ! R3 as follows: For every t in the closed interval ½0; 1; cðtÞ ð1 tÞa þ tb: Clearly, c is continuous on [0,1]. Also, it is clear that c is differentiable on the open interval (0,1), and for every t in (0,1), c0 ðtÞ ¼ b a: Since E is convex, a and b are in E, the range of c is contained in E. Since c : ½0; 1 ! R3 ; the range of c is contained in E, and f : E ! R2 ; the composite function f c is deﬁned on ½0; 1:

3.2 Differentiation

157

We want to apply Lemma 3.29 on f c: Since f : E ! R2 is differentiable on E, f is continuous on E. Since c is continuous on ½0; 1; the range of c is contained in E, and f is continuous on E, f c is continuous on [0, 1]. Since c is differentiable on the open interval ð0; 1Þ; the range of c is contained in E, and f : E ! R2 is differentiable on E, by Theorem 3.18, the composite function ðf cÞ : ½0; 1 ! R2 is differentiable on the open interval ð0; 1Þ; and for every x in ð0; 1Þ; ðf cÞ0 ðxÞ ¼ ðf 0 ðcðxÞÞÞðc0 ðxÞÞ: Hence, by Lemma 3.29 and Theorem 3.2, there exists a real number t in ð0; 1Þ such that jf ðbÞ f ðaÞj ¼ jf ðcð1ÞÞ f ðcð0ÞÞj ¼ jðf cÞð1Þ ðf cÞð0Þj ð1 0Þðf cÞ0 ðtÞ ¼ ðf cÞ0 ðtÞ ¼ ðf cÞ0 ðtÞ ¼ kðf 0 ðcðtÞÞÞðc0 ðtÞÞk kf 0 ðcðtÞÞkkc0 ðtÞk ¼ kf 0 ðcðtÞÞkkb ak ¼ kf 0 ðcðtÞÞkjb aj ¼ jb ajkf 0 ðcðtÞÞk jb ajðsupfkf 0 ðxÞk : x 2 E gÞ: Thus, jf ðbÞ f ðaÞj jb ajðsupfkf 0 ðxÞk : x 2 E gÞ:

h

Note 3.30 The result similar to Theorem 3.28 can be proved as above for R in place of R3 and Rm in place of R2 : n

Theorem 3.31 Let E be an open subset of R3 : Let f : E ! R2 : Let f1 : E ! R; f2 : E ! R be the component functions of f, that is, f ðxÞ ¼ ðf1 ðxÞ; f2 ðxÞÞ for every x in E. If f : E ! R2 is differentiable on E, and the mapping x 7! f 0 ðxÞ from Eð R3 Þ to the metric space LðR3 ; R2 Þ is continuous, then f is C1 function (i.e., for i = 1, 2, and j = 1, 2, 3, each function Dj fi : E ! R exists and is continuous). Proof Let us take any x in E. Since x is in E, and f : E ! R2 is differentiable on E, f is differentiable at x. Since f is differentiable at x, by Theorem 3.20, all the partial derivatives ðDj fi ÞðxÞ exist. This shows that each Dj fi : E ! R is a function for i = 1, 2, and j = 1, 2, 3. Now, we shall try to prove that D1 f2 : E ! R is continuous. For this purpose, let us ﬁx any x in E. Let us take any e [ 0: Since the mapping x 7! f 0 ðxÞ from E ð R3 Þ to the metric space LðR3 ; R2 Þ is continuous, there exists d [ 0 such that for every y in E satisfying jy xj\d, we have kf 0 ðyÞ f 0 ðxÞk\e: Let us take any y in E satisfying jy xj\d: It is enough to prove that jðD1 f2 ÞðyÞ ðD1 f2 ÞðxÞj\e: Since

158

3 Multivariable Differential Calculus

½f 0 ðyÞ f 0 ðxÞ ¼ ½f 0 ðyÞ ½f 0 ðxÞ ðD1 f1 ÞðyÞ ðD2 f1 ÞðyÞ ðD3 f1 ÞðyÞ ¼ ðD1 f2 ÞðyÞ ðD2 f2 ÞðyÞ ðD3 f1 ÞðyÞ 23 ðD1 f1 ÞðxÞ ðD2 f1 ÞðxÞ ðD3 f1 ÞðxÞ ðD1 f2 ÞðxÞ ðD2 f2 ÞðxÞ ðD3 f1 ÞðxÞ 23 ðD1 f1 ÞðyÞ ðD1 f1 ÞðxÞ ðD2 f1 ÞðyÞ ðD2 f1 ÞðxÞ ¼ ðD1 f2 ÞðyÞ ðD1 f2 ÞðxÞ ðD2 f2 ÞðyÞ ðD2 f2 ÞðxÞ

ðD3 f1 ÞðyÞ ðD3 f1 ÞðxÞ ðD3 f1 ÞðyÞ ðD3 f1 ÞðxÞ

; 23

so jðD1 f2 ÞðyÞ ðD1 f2 ÞðxÞj jððD1 f1 ÞðyÞ ðD1 f1 ÞðxÞ; ðD1 f2 ÞðyÞ ðD1 f2 ÞðxÞÞj ¼ jðf 0 ðyÞ f 0 ðxÞÞðð1; 0; 0ÞÞj sup jðf 0 ðyÞ f 0 ðxÞÞðtÞj : t 2 R3 ; jtj 1 ¼ kf 0 ðyÞ f 0 ðxÞk\e: Hence, jðD1 f2 ÞðyÞ ðD1 f2 ÞðxÞj\e: This proves that D1 f2 : E ! R is continuous. Similarly, all other partial derivatives are continuous. Hence, f is a C1 function. h Theorem 3.32 Let E be an open subset of R3 : Let f : E ! R2 : Let f1 : E ! R; f2 : E ! R be the component functions of f, that is, f ðxÞ ¼ ðf1 ðxÞ; f2 ðxÞÞ for every x in E. If f is a C1 function (i.e., for i ¼ 1; 2; and j ¼ 1; 2; 3; each function Dj fi : E ! R exists and is continuous), then f : E ! R2 is differentiable on E: Proof Let us take any a ða1 ; a2 ; a3 Þ in E. We have to prove that f : E ! R2 is differentiable at a. By Theorem 3.14, it is enough to prove that (i)

1 ðf1 ðða1 ; a2 ; a3 Þ jðh1 ; h2 ; h3 Þj

lim

ðh1 ;h2 ;h3 Þh!0

þ ðh1 ; h2 ; h3 ÞÞ f1 ðða1 ; a2 ; a3 ÞÞ ðh1 ; h2 ; h3 Þ ððD1 f1 Þðða1 ; a2 ; a3 ÞÞ; ðD2 f1 Þðða1 ; a2 ; a3 ÞÞ; ðD3 f1 Þðða1 ; a2 ; a3 ÞÞÞÞ

! ¼ 0;

and

(ii)

lim

ðh1 ;h2 ;h3 Þh!0

1 ðf2 ðða1 ; a2 ; a3 Þ jðh1 ; h2 ; h3 Þj

þ ðh1 ; h2 ; h3 ÞÞ f2 ðða1 ; a2 ; a3 ÞÞ ðh1 ; h2 ; h3 Þ

ððD1 f2 Þðða1 ; a2 ; a3 ÞÞ; ðD2 f2 Þðða1 ; a2 ; a3 ÞÞ; ðD3 f2 Þðða1 ; a2 ; a3 ÞÞÞÞ ¼ 0:

3.2 Differentiation

159

For (i): We have to prove that 0¼

lim

ðh1 ;h2 ;h3 Þh!0

1 ðf1 ðða1 ; a2 ; a3 Þ þ ðh1 ; h2 ; h3 ÞÞ f1 ðða1 ; a2 ; a3 ÞÞ jðh1 ; h2 ; h3 Þj

ðh1 ; h2 ; h3 Þ ððD1 f1 Þðða1 ; a2 ; a3 ÞÞ; ðD2 f1 Þðða1 ; a2 ; a3 ÞÞ; ðD3 f1 Þðða1 ; a2 ; a3 ÞÞÞÞ ;

that is, 0¼

lim

ðh1 ;h2 ;h3 Þh!0

1 ðf1 ðða1 þ h1 ; a2 þ h2 ; a3 þ h3 ÞÞ f1 ðða1 ; a2 ; a3 ÞÞ jðh1 ; h2 ; h3 Þj

ðh1 ððD1 f1 Þðða1 ; a2 ; a3 ÞÞÞ þ h2 ððD2 f1 Þðða1 ; a2 ; a3 ÞÞÞ þ h3 ððD3 f1 Þðða1 ; a2 ; a3 ÞÞÞÞÞ

For this purpose, let us take any e [ 0: Since D1 f1 : E ! R; D2 f1 : E ! R; D3 f1 : E ! R are continuous at ða1 ; a2 ; a3 Þ; there exists d [ 0 such that for every ðx1 ; x2 ; x3 Þ in E satisfying jðx1 ; x2 ; x3 Þ ða1 ; a2 ; a3 Þj\d ; we have

Dj f1 ððx1 ; x2 ; x3 ÞÞ Dj f1 ðða1 ; a2 ; a3 ÞÞ\ e ðj ¼ 1; 2; 3Þ ðÞ 3 Since ða1 ; a2 ; a3 Þ is in E, and E is an open subset of R3 ; we can ﬁnd d [ 0 such and the Cartesian product K ½a1 d; a1 þ d that 0\d\d ½a2 d; a2 þ d ½a3 d; a3 þ d E: Let us ﬁx any h ðh1 ; h2 ; h3 Þ such that jðh1 ; h2 ; h3 Þj\d: Since jh1 j jðh1 ; h2 ; h3 Þj\d; d\h1 \d; or a1 d\a1 þ h1 \a1 þ d; or a1 þ h1 2 ½a1 d; a1 þ d: Similarly, a2 þ h2 2 ½a2 d; a2 þ d; and a3 þ h3 2 ½a3 d; a3 þ d: Since a1 ; a1 þ h1 2 ½a1 d; a1 þ d; and for every t in ½a1 d; a1 þ d; ðt; a2 þ h2 ; a3 þ h3 Þ 2 ½a1 d; a1 þ d ½a2 d; a2 þ d ½a3 d; a3 þ d E; we can deﬁne a function g1 : ½a1 d; a1 þ d ! R as follows: For every t in ½a1 d; a1 þ d; g1 ðtÞ ¼ f1 ððt; a2 þ h2 ; a3 þ h3 ÞÞ: Now, we want to apply Lagrange’s mean value theorem on g1. For ﬁxed t0 in ½a1 d; a1 þ d; we have

160

lim t!0

3 Multivariable Differential Calculus

g1 ðt0 þ tÞ g1 ðt0 Þ f1 ðt0 þ t; a2 þ h2 ; a3 þ h3 Þ f1 ððt0 ; a2 þ h2 ; a3 þ h3 ÞÞ ¼ lim t!0 t t ¼ ðD1 f1 Þððt0 ; a2 þ h2 ; a3 þ h3 ÞÞ:

Since 1 ðt0 Þ limt!0 g1 ðt0 þtÞg ¼ ðD1 f1 Þððt0 ; a2 þ h2 ; a3 þ h3 ÞÞ; ðt0 ; a2 þ h2 ; a3 þ h3 Þ is in E, t and D1 f1 : E ! R exists, g1 is differentiable at t0. Further, g0 ðt0 Þ ¼ ðD1 f1 Þððt0 ; a2 þ h2 ; a3 þ h3 ÞÞ: Since g1 is differentiable at t0, g1 is continuous at t0. Thus, we see that g1 : ½a1 d; a1 þ d ! R is continuous and differentiable on ½a1 d; a1 þ d: Hence, by the Lagrange’s mean value theorem, there exists a real number c1 lying between a1 þ h1 and a1 such that f1 ðða1 þ h1 ; a2 þ h2 ; a3 þ h3 ÞÞ f1 ðða1 ; a2 þ h2 ; a3 þ h3 ÞÞ ¼ g1 ð a1 þ h1 Þ g1 ð a1 Þ ¼ ðða1 þ h1 Þ a1 Þg0 ðc1 Þ ¼ h1 ðg0 ðc1 ÞÞ ¼ h1 ððD1 f1 Þððc1 ; a2 þ h2 ; a3 þ h3 ÞÞÞ: Thus, we see that there exists a real number c1 lying between a1 þ h1 and a1 such that f1 ðða1 þ h1 ; a2 þ h2 ; a3 þ h3 ÞÞ ¼ f1 ðða1 ; a2 þ h2 ; a3 þ h3 ÞÞ þ h1 ððD1 f1 Þððc1 ; a2 þ h2 ; a3 þ h3 ÞÞÞ: Similarly, there exists a real number c2 lying between a2 þ h2 and a2 such that f1 ðða1 ; a2 þ h2 ; a3 þ h3 ÞÞ ¼ f1 ðða1 ; a2 ; a3 þ h3 ÞÞ þ h2 ððD2 f1 Þðða1 ; c2 ; a3 þ h3 ÞÞÞ: Again, there exists a real number c3 lying between a3 þ h3 and a3 such that f1 ðða1 ; a2 ; a3 þ h3 ÞÞ ¼ f1 ðða1 ; a2 ; a3 ÞÞ þ h3 ððD3 f1 Þðða1 ; a2 ; c3 ÞÞÞ: On adding the last three equations, we get f1 ðða1 þ h1 ; a2 þ h2 ; a3 þ h3 ÞÞ f1 ðða1 ; a2 ; a3 ÞÞ ¼ h1 ð ð D 1 f 1 Þ ð ð c 1 ; a2 þ h2 ; a3 þ h3 Þ Þ Þ þ h2 ððD2 f1 Þðða1 ; c2 ; a3 þ h3 ÞÞÞ þ h3 ððD3 f1 Þðða1 ; a2 ; c3 ÞÞÞ or,

3.2 Differentiation

161

f1 qðða1 þ h1 ; a2 þ h2 ; a3 þ h3 ÞÞ f1 ðða1 ; a2 ; a3 ÞÞðh1 ððD1 f1 Þðða1 ; a2 ; a3 ÞÞÞ þ h2 ððD2 f1 Þðða1 ; a2 ; a3 ÞÞÞ þ h3 ððD3 f1 Þðða1 ; a2 ; a3 ÞÞÞÞ ¼ ðh1 ððD1 f1 Þððc1 ; a2 þ h2 ; a3 þ h3 ÞÞÞ þ h2 ððD2 f1 Þðða1 ; c2 ; a3 þ h3 ÞÞÞ þ h3 ððD3 f1 Þðða1 ; a2 ; c3 ÞÞÞÞ ðh1 ððD1 f1 Þðða1 ; a2 ; a3 ÞÞÞ þ h2 ððD2 f1 Þðða1 ; a2 ; a3 ÞÞÞ þ h3 ððD3 f1 Þðða1 ; a2 ; a3 ÞÞÞÞ ¼ h1 ððD1 f1 Þððc1 ; a2 þ h2 ; a3 þ h3 ÞÞ ðD1 f1 Þðða1 ; a2 ; a3 ÞÞÞ þ h2 ððD2 f1 Þðða1 ; c2 ; a3 þ h3 ÞÞ ðD2 f1 Þðða1 ; a2 ; a3 ÞÞÞ þ h3 ððD3 f1 Þðða1 ; a2 ; c3 ÞÞ ðD3 f1 Þðða1 ; a2 ; a3 ÞÞÞ: So, 1 ¼ ðf1 ðða1 þ h1 ; a2 þ h2 ; a3 þ h3 ÞÞ f1 ðða1 ; a2 ; a3 ÞÞ jðh1 ; h2 ; h3 Þj

ðh1 ððD1 f1 Þðða1 ; a2 ; a3 ÞÞÞ þ h2 ððD2 f1 Þðða1 ; a2 ; a3 ÞÞÞ þ h3 ððD3 f1 Þðða1 ; a2 ; a3 ÞÞÞÞÞ 1 ðh1 ððD1 f1 Þððc1 ; a2 þ h2 ; a3 þ h3 ÞÞ ðD1 f1 Þðða1 ; a2 ; a3 ÞÞÞ ¼ jðh1 ; h2 ; h3 Þj þ h2 ððD2 f1 Þðða1 ; c2 ; a3 þ h3 ÞÞ ðD2 f1 Þðða1 ; a2 ; a3 ÞÞÞ þ h3 ððD3 f1 Þðða1 ; a2 ; c3 ÞÞ ðD3 f1 Þðða1 ; a2 ; a3 ÞÞÞÞ jh1 j jðD1 f1 Þððc1 ; a2 þ h2 ; a3 þ h3 ÞÞ ðD1 f1 Þðða1 ; a2 ; a3 ÞÞj jðh1 ; h2 ; h3 Þj jh2 j þ jðD2 f1 Þðða1 ; c2 ; a3 þ h3 ÞÞ ðD2 f1 Þðða1 ; a2 ; a3 ÞÞj jðh1 ; h2 ; h3 Þj jh3 j þ jðD3 f1 Þðða1 ; a2 ; c3 ÞÞ ðD3 f1 Þðða1 ; a2 ; a3 ÞÞj jðh1 ; h2 ; h3 Þj ¼ jðD1 f1 Þððc1 ; a2 þ h2 ; a3 þ h3 ÞÞ ðD1 f1 Þðða1 ; a2 ; a3 ÞÞj þ jðD2 f1 Þðða1 ; c2 ; a3 þ h3 ÞÞ ðD2 f1 Þðða1 ; a2 ; a3 ÞÞj

þ jðD3 f1 Þðða1 ; a2 ; c3 ÞÞ ðD3 f1 Þðða1 ; a2 ; a3 ÞÞj:

Since c1 lies between a1 þ h1 and a1, ðc1 a1 Þ2 ðh1 Þ2 , and hence, jðc1 ; a2 þ h2 ; a3 þ h3 Þ ða1 ; a2 ; a3 Þj ¼ jðc1 a1 ; h2 ; h3 Þj qﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ qﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ ¼ ðc1 a1 Þ2 þðh2 Þ2 þðh3 Þ2 ðh1 Þ2 þðh2 Þ2 þðh3 Þ2 ¼ jðh1 ; h2 ; h3 Þ ð0; 0; 0Þj\d\d : Since jðc1 ; a2 þ h2 ; a3 þ h3 Þ ða1 ; a2 ; a3 Þj\d , by ðÞ;

162

3 Multivariable Differential Calculus

e jðD1 f1 Þððc1 ; a2 þ h2 ; a3 þ h3 ÞÞ ðD1 f1 Þðða1 ; a2 ; a3 ÞÞj\ : 3 Similarly, e jðD2 f1 Þðða1 ; c2 ; a3 þ h3 ÞÞ ðD2 f1 Þðða1 ; a2 ; a3 ÞÞj\ ; 3 and e jðD3 f1 Þðða1 ; a2 ; c3 ÞÞ ðD3 f1 Þðða1 ; a2 ; a3 ÞÞj\ : 3 Hence, 1 jðh ; h ; h Þj ðf1 ðða1 þ h1 ; a2 þ h2 ; a3 þ h3 ÞÞ f1 ðða1 ; a2 ; a3 ÞÞ 1 2 3

ðh1 ððD1 f1 Þðða1 ; a2 ; a3 ÞÞÞ þ h2 ððD2 f1 Þðða1 ; a2 ; a3 ÞÞÞ þ h3 ððD3 f1 Þðða1 ; a2 ; a3 ÞÞÞÞÞ

jðD1 f1 Þððc1 ; a2 þ h2 ; a3 þ h3 ÞÞ ðD1 f1 Þðða1 ; a2 ; a3 ÞÞj þ jðD2 f1 Þðða1 ; c2 ; a3 þ h3 ÞÞ ðD2 f1 Þðða1 ; a2 ; a3 ÞÞj þ jðD3 f1 Þðða1 ; a2 ; c3 ÞÞ ðD3 f1 Þðða1 ; a2 ; a3 ÞÞj e e e \ þ þ ¼ e: 3 3 3

This proves (i). Similarly, (ii) is proved.

h

Theorem 3.33 Let E be an open subset of R3 : Let f : E ! R2 : Let f1 : E ! R; f2 : E ! R be the component functions of f, that is, f ðxÞ ¼ ðf1 ðxÞ; f2 ðxÞÞ for every x in E. f is a C1 function (i.e., for i ¼ 1; 2; and j ¼ 1; 2; 3; each function Dj fi : E ! R exists and is continuous), if and only if f : E ! R2 is differentiable on E and f 0 : E ! LðR3 ; R2 Þ is continuous. Proof By using Theorems 3.31 and 3.32, we ﬁnd that the only part that remains to be proved is the following: if f is a C1 function, then the mapping f 0 : E ! LðR3 ; R2 Þ is continuous. For this purpose, let us take any x in E. We have to prove that f 0 is continuous at x. Let us take any e [ 0: Since, for i ¼ 1; 2; and j ¼ 1; 2; 3; each function Dj fi : E ! R is continuous at x, there exists a real number d [ 0 such that for every y in E satisfying jy xj\d; we have

e Dj fi ðyÞ Dj fi ðxÞ\ pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ ðfor every i ¼ 1; 2; and j ¼ 1; 2; 3Þ: 23 Now, let us take any y in E satisfying jy xj\d: It is enough to prove that

3.2 Differentiation

163

kf 0 ðyÞ f 0 ðxÞk\e: Here, ½f 0 ðxÞ ¼

ðD1 f1 ÞðxÞ ðD2 f1 ÞðxÞ ðD3 f1 ÞðxÞ ðD1 f2 ÞðxÞ ðD2 f2 ÞðxÞ ðD3 f3 ÞðxÞ

; 23

and ½f 0 ðyÞ ¼

ðD1 f1 ÞðyÞ ðD2 f1 ÞðyÞ ðD3 f1 ÞðyÞ ðD1 f2 ÞðyÞ ðD2 f2 ÞðyÞ ðD3 f3 ÞðyÞ

; 23

so ½f 0 ðyÞ f 0 ðxÞ ¼

ðD1 f1 ÞðyÞ ðD1 f1 ÞðxÞ ðD1 f2 ÞðyÞ ðD1 f2 ÞðxÞ

ðD2 f1 ÞðyÞ ðD2 f1 ÞðxÞ ðD2 f2 ÞðyÞ ðD2 f2 ÞðxÞ

ðD3 f1 ÞðyÞ ðD3 f1 ÞðxÞ ðD3 f3 ÞðyÞ ðD3 f3 ÞðxÞ

; 23

and hence, vﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ !ﬃ v u 3 u 3 2 ! 2 2 uX X uX X

2 e t t pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ Dj fi ðyÞ Dj fi ðxÞ \ ¼ e: kf ðyÞ f ðxÞk 23 j¼1 i¼1 j¼1 i¼1 0

0

h Note 3.34 The result similar to Theorem 3.33 can be proved as above for Rn in place of R3 and Rm in place of R2 :

3.3 Inverse Function Theorem Theorem 3.35 Let X be a complete metric space, with metric d. Let f : X ! X be any function. Let c be a real number such that 0\c\1; and for every x, y in X, d ðf ðxÞ; f ðyÞÞ cðd ðx; yÞÞ: Then, there exists a unique x in X such that f ðxÞ ¼ x: Proof Existence: Let us take any a in X. Since a is in X, and f : X ! X; f ðaÞ is in X. Similarly, f ðf ðaÞÞ is in X, f ðf ðf ðaÞÞÞ is in X, etc. Thus, we get a sequence fa; f ðaÞ; f ðf ðaÞÞ; f ðf ðf ðaÞÞÞ; . . .g in X. Put a0 a; a1 f ðaÞ; a2 f ðf ðaÞÞ; a3 f ðf ðf ðaÞÞÞ; etc. Here, fa0 ; a1 ; a2 ; a3 ; . . .g is a sequence in X. Also, f ðan Þ ¼ anþ1 : Let us observe that by using the given condition,

164

3 Multivariable Differential Calculus

d ðf ðaÞ; f ðf ðaÞÞÞ cðd ða; f ðaÞÞÞ: Also, d ðf ðf ðaÞÞ; f ðf ðf ðaÞÞÞÞ cðd ðf ðaÞ; f ðf ðaÞÞÞÞ cðcðd ða; f ðaÞÞÞÞ ¼ c2 ðd ða; f ðaÞÞÞ; so d ðf ðf ðaÞÞ; f ðf ðf ðaÞÞÞÞ c2 ðd ða; f ðaÞÞÞ: Similarly, d ðf ðf ðf ðaÞÞÞ; f ðf ðf ðf ðaÞÞÞÞÞ c3 ðd ða; f ðaÞÞÞ; etc. Now, d ðf ðf ðaÞÞ; f ðf ðf ðf ðf ðf ðaÞÞÞÞÞÞÞ cðd ðf ðaÞ; f ðf ðf ðf ðf ðaÞÞÞÞÞÞÞ cðcðd ða; f ðf ðf ðf ðaÞÞÞÞÞÞÞ ¼ c2 ðd ða; f ðf ðf ðf ðaÞÞÞÞÞÞ c2 ðd ða; f ðaÞÞ þ d ðf ðaÞ; f ðf ðaÞÞÞ þ d ðf ðf ðaÞÞ; f ðf ðf ðaÞÞÞÞ þd ðf ðf ðf ðaÞÞÞ; f ðf ðf ðf ðaÞÞÞÞÞÞ

c2 d ða; f ðaÞÞ þ cðd ða; f ðaÞÞÞ þ c2 ðd ða; f ðaÞÞÞ þ c3 ðd ða; f ðaÞÞÞ

¼ c2 1 þ c þ c2 þ c3 ðd ða; f ðaÞÞÞ c2 1 þ c þ c2 þ c3 þ c4 þ ðd ða; f ðaÞÞÞ 1 ðd ða; f ðaÞÞÞ: ¼ c2 1c Thus, we have seen that d ðf ðf ðaÞÞ; f ðf ðf ðf ðf ðf ðaÞÞÞÞÞÞÞ c2

1 ðd ða; f ðaÞÞÞ: 1c

Similarly, d ðf ðf ðf ðaÞÞÞ; f ðf ðf ðf ðf ðf ðf ðaÞÞÞÞÞÞÞÞ c3

1 ðd ða; f ðaÞÞÞ; 1c

1 1 etc. Thus, dða2 ; a6 Þ ð1c Þðdða; f ðaÞÞÞc2 ; dða3 ; a7 Þ ð1c Þðdða; f ðaÞÞÞc3 ; etc. We shall try to prove that the sequence fa0 ; a1 ; a2 ; a3 ; . . .g is Cauchy. For this purpose, let us take e [ 0: Since 0\c\1; limn!1 cn ¼ 0; and hence, 1 Þðdða; f ðaÞÞÞcn ¼ 0: Therefore, there exists a positive integer N such limn!1 ð1c

3.3 Inverse Function Theorem

165

1 that for every n N; we have jð1c Þðdða; f ðaÞÞÞcn 0j\e: Hence, m n N implies

1 1 n n d ð an ; am Þ ðd ða; f ðaÞÞÞc ¼ ðd ða; f ðaÞÞÞc 0\e: 1c 1c

This proves that fa0 ; a1 ; a2 ; a3 ; . . .g is Cauchy in X. Since fa0 ; a1 ; a2 ; a3 ; . . .g is Cauchy in X, and X is a complete metric space, there exists an element b in X such that lim an ¼ b:

n!1

Now, we shall try to show that the function f : X ! X is continuous at b. For this purpose, let us take any e [ 0: Now, let us take any y in X satisfying dðy; bÞ\e. We have to show that dðf ðyÞ; f ðbÞÞ\e. Here, by using the given conditions, we have d ðf ðyÞ; f ðbÞÞ cðd ðy; bÞÞ\cðeÞ\1e ¼ e: This proves that the function f : X ! X is continuous at b. Since f : X ! X is continuous at b, and limn!1 an ¼ b; limn!1 f ðan Þ ¼ f ðbÞ: Hence, f ðbÞ ¼ limn!1 f ðan Þ ¼ limn!1 anþ1 ¼ limn!1 an ¼ b: Thus, f ðbÞ ¼ b: This proves the existence part of the theorem. Uniqueness: If not, otherwise, suppose that there exist x, y in X such that x 6¼ y; f ðxÞ ¼ x; and f ðyÞ ¼ y: We have to arrive at a contradiction. From the given condition, we have 0 6¼ d ðx; yÞ ¼ d ðf ðxÞ; f ðyÞÞ cðd ðx; yÞÞ: This implies that 1 c; which contradicts the supposition.

h

Theorem 3.36 Let E be an open subset of R3 : Let f : E ! R3 : Let a be in E. If (i) f is a C1 function, (ii) f 0 ðaÞ is invertible, then there exists a connected open neighborhood U of a such that 1. 2. 3. 4. 5.

U is contained in E, f is 1–1 on U (i.e., if x and y are in U, and f ðxÞ ¼ f ðyÞ; then x = y), f(U) is connected and open in R3 ; the 1–1 function f−1 from f(U) onto U is a C1 function, if f is a C2 function, then f−1 from f(U) onto U is a C2 function, etc.

Proof Let f1 : E ! R; f2 : E ! R; f3 : E ! R be the component functions of f, that is, f ðxÞ ¼ ðf1 ðxÞ; f2 ðxÞ; f3 ðxÞÞ for every x in E. Let a ða1 ; a2 ; a3 Þ: Let X be

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3 Multivariable Differential Calculus

the set of all invertible (i.e., 1–1 and onto) members of LðR3 Þ: Since f 0 ðaÞ is invertible, f 0 ðaÞ is in X Since f 0 ðaÞ is in X, by Theorem 3.3, the open sphere S

1

kðf 0 ðaÞÞ1 k

ðf 0 ðaÞÞ;

with center f 0 ðaÞ and radius 1 ; 0 ðf ðaÞÞ1 is contained in X Since f is a C1 function, by Theorem 3.33, f 0 : E ! LðR3 Þ is continuous. Since f 0 : E ! LðR3 Þ is continuous at a, and S

1

2

kðf 0 ðaÞÞ1 k

ðf 0 ðaÞÞ

is an open neighborhood of f 0 ðaÞ in X, there exists a real number r > 0 such that the open sphere Sr ðaÞ is contained in E, and for every x in Sr ðaÞ; we have f 0 ðxÞ in S

1

2

kðf 0 ðaÞÞ1 k

ðf 0 ðaÞÞ:

For every x in Sr ðaÞ; we have f 0 ðxÞ in S

1

2

kðf 0 ðaÞÞ1 k

ðf 0 ðaÞÞ;

so 1 kf 0 ðxÞ f 0 ðaÞk\ 0 2 ðf ðaÞÞ1 or, 1 1 1 I ðf 0 ðaÞÞ ðf 0 ðxÞÞ ¼ ðf 0 ðaÞÞ ðf 0 ðxÞÞ I ¼ ðf 0 ðaÞÞ ðf 0 ðxÞÞ I 1 1 ¼ ðf 0 ðaÞÞ ðf 0 ðxÞÞ ðf 0 ðaÞÞ ðf 0 ðaÞÞ 1 1 ¼ ðf 0 ðaÞÞ ðf 0 ðxÞ f 0 ðaÞÞ ¼ ðf 0 ðaÞÞ ðf 0 ðxÞ f 0 ðaÞÞ 1 1 ðf 0 ðaÞÞ kf 0 ðxÞ f 0 ðaÞk\ : 2

Thus, we see that for every x in Sr ðaÞ; we have

3.3 Inverse Function Theorem

167

1 1 I ðf 0 ðaÞÞ ðf 0 ðxÞÞ 2

ðÞ

For every k ðk1 ; k2 ; k3 Þ in R3 ; let us deﬁne a function uk : Sr ðaÞ ! R3 as follows: For every x ðx1 ; x2 ; x3 Þ in Sr ðaÞ; 1 uk ðxÞ ¼ x ðf 0 ðaÞÞ ðf ðxÞÞ þ k: Now, we shall try to show that for every x in Sr ðaÞ; 1

u0k ðxÞ ¼ I ðf 0 ðaÞÞ ðf 0 ðxÞÞ: Let 2 a11 h i 1 ðf 0 ðaÞÞ 4 a21 a31

a12 a22 a32

3 a13 a23 5 a33 33

relative to the basis fð1; 0; 0Þ; ð0; 1; 0Þ; ð0; 0; 1Þg of R3 : Since 1 uk ðxÞ ¼ x ðf 0 ðaÞÞ ðf ðxÞÞ þ k 1 ððf1 ðxÞ; f2 ðxÞ; f3 ðxÞÞÞ þ ðk1 ; k2 ; k3 Þ ¼ ðx1 ; x2 ; x3 Þ ðf 0 ðaÞÞ ¼ ðx1 ; x2 ; x3 Þ ððf1 ðxÞÞða11 ; a21 ; a31 Þ þ ðf2 ðxÞÞða12 ; a22 ; a32 Þ þ ðf3 ðxÞÞða13 ; a23 ; a33 ÞÞ þ ðk1 ; k2 ; k3 Þ ¼ ðx1 ða11 ðf1 ðxÞÞ þ a12 ðf2 ðxÞÞ þ a13 ðf3 ðxÞÞÞ þ k1 ; x2 ða21 ðf1 ðxÞÞ þ a22 ðf2 ðxÞÞ þ a23 ðf3 ðxÞÞÞ þ k2 ; x3 ða31 ðf1 ðxÞÞ þ a32 ðf2 ðxÞÞ þ a33 ðf3 ðxÞÞÞ þ k3 Þ or, uk ðxÞ ¼ ðx1 ða11 ðf1 ðxÞÞ þ a12 ðf2 ðxÞÞ þ a13 ðf3 ðxÞÞÞ þ k1 ; x2 ða21 ðf1 ðxÞÞ þ a22 ðf2 ðxÞÞ þ a23 ðf3 ðxÞÞÞ þ k2 ; x3 ða31 ðf1 ðxÞÞ þ a32 ðf2 ðxÞÞ þ a33 ðf3 ðxÞÞÞ þ k3 Þ

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3 Multivariable Differential Calculus

so,

u0k ðxÞ ðð1; 0; 0ÞÞ ¼ ðD1 ðx1 ða11 ðf1 ðxÞÞ þ a12 ðf2 ðxÞÞ þ a13 ðf3 ðxÞÞÞ þ k1 Þ; D1 ðx2 ða21 ðf1 ðxÞÞ þ a22 ðf2 ðxÞÞ þ a23 ðf3 ðxÞÞÞ þ k2 Þ; D1 ðx3 ða31 ðf1 ðxÞÞ þ a32 ðf2 ðxÞÞ þ a33 ðf3 ðxÞÞÞ þ k3 ÞÞ ¼ ð1 ða11 ððD1 f1 ÞðxÞÞ þ a12 ððD1 f2 ÞðxÞÞ þ a13 ððD1 f3 ÞðxÞÞÞ; ða21 ððD1 f1 ÞðxÞÞ þ a22 ððD1 f2 ÞðxÞÞ þ a23 ððD1 f3 ÞðxÞÞÞ; ða31 ððD1 f1 ÞðxÞÞ þ a32 ððD1 f2 ÞðxÞÞ þ a33 ððD1 f3 ÞðxÞÞÞÞ:

Also,

1 1 I ðf 0 ðaÞÞ ðf 0 ðxÞÞ ðð1; 0; 0ÞÞ ¼ I ðð1; 0; 0ÞÞ ðf 0 ðaÞÞ ððf 0 ðxÞÞðð1; 0; 0ÞÞÞ 1 ððD1 f1 ÞðxÞ; ðD1 f2 ÞðxÞ; ðD1 f3 ÞðxÞÞ ¼ ð1; 0; 0Þ ðf 0 ðaÞÞ 1 ¼ ð1; 0; 0Þ ððD1 f1 ÞðxÞÞ ðf 0 ðaÞÞ ðð1; 0; 0ÞÞ 1 1 þ ððD1 f2 ÞðxÞÞ ðf 0 ðaÞÞ ðð0; 1; 0ÞÞ þ ððD1 f3 ÞðxÞÞ ðf 0 ðaÞÞ ðð0; 0; 1ÞÞ ¼ ð1; 0; 0Þ ðððD1 f1 ÞðxÞÞða11 ; a21 ; a31 Þ þ ððD1 f2 ÞðxÞÞða12 ; a22 ; a32 Þ þ ððD1 f3 ÞðxÞÞða13 ; a23 ; a33 ÞÞ

¼ ð1 ða11 ððD1 f1 ÞðxÞÞ þ a12 ððD1 f2 ÞðxÞÞ þ a13 ððD1 f3 ÞðxÞÞÞ; ða21 ððD1 f1 ÞðxÞÞ þ a22 ððD1 f2 ÞðxÞÞ þ a23 ððD1 f3 ÞðxÞÞÞ; ða31 ððD1 f1 ÞðxÞÞ þ a32 ððD1 f2 ÞðxÞÞ þ a33 ððD1 f3 ÞðxÞÞÞÞ:

This shows that

1 ðuk 0 ðxÞÞðð1; 0; 0ÞÞ ¼ I ðf 0 ðaÞÞ ðf 0 ðxÞÞ ðð1; 0; 0ÞÞ:

Similarly,

1 ðuk 0 ðxÞÞðð0; 1; 0ÞÞ ¼ I ðf 0 ðaÞÞ ðf 0 ðxÞÞ ðð0; 1; 0ÞÞ; 1 ðuk 0 ðxÞÞðð0; 0; 1ÞÞ ¼ I ðf 0 ðaÞÞ ðf 0 ðxÞÞ ðð0; 0; 1ÞÞ:

Hence, for every y in R3 ; we have 1 ðuk 0 ðxÞÞðyÞ ¼ I ðf 0 ðaÞÞ ðf 0 ðxÞÞ ðyÞ: This shows that for every x in Sr ðaÞ; 1

u0k ðxÞ ¼ I ðf 0 ðaÞÞ ðf 0 ðxÞÞ; and hence, by ðÞ;

3.3 Inverse Function Theorem

169

kuk 0 ðxÞk

1 : 2

Therefore, 12 is an upper bound of the set fkuk 0 ðxÞk : x 2 Sr ðaÞg; and hence, fkuk 0 ðxÞk : x 2 Sr ðaÞg is bounded above. Now, we want to apply Theorem 3.28. Since Sr ðaÞ is an open convex subset of 3 R ; uk : Sr ðaÞ ! R3 is differentiable on Sr ðaÞ; and the set fkuk 0 ðxÞk : x 2 Sr ðaÞg is bounded above, for every x, y in Sr ðaÞ;

1 juk ðyÞ uk ðxÞj jy xj sup u0k ðxÞ : x 2 Sr ðaÞ jy xj : 2 Hence, for every k in R3 ; and for every x, y in Sr ðaÞ; we have 1 juk ðyÞ uk ðxÞj jy xj 2

ðÞ:

Next, for every x, y in Sr ðaÞ; we have 1 ðf ðyÞ f ðxÞÞ jy xj ðf 0 ðaÞÞ 1 1 ¼ jy xj ðf 0 ðaÞÞ ðf ðyÞÞ ðf 0 ðaÞÞ ðf ðxÞÞ 1 1 ðy xÞ ðf 0 ðaÞÞ ðf ðyÞÞ ðf 0 ðaÞÞ ðf ðxÞÞ 1 1 ¼ y ðf 0 ðaÞÞ ðf ðyÞÞ þ k x ðf 0 ðaÞÞ ðf ðxÞÞ þ k 1 ¼ juk ðyÞ uk ðxÞj jy xj: 2 Hence, for every x, y in Sr ðaÞ; we have 1 1 jy xj ðf 0 ðaÞÞ ðf ðyÞ f ðxÞÞ 2

ð Þ

For 1: We have seen that the open neighborhood Sr ðaÞ of a is contained in E. We will take Sr ðaÞ for U. Since Sr ðaÞ is connected, U is connected. For 2: We have to show that f is 1–1 on Sr ðaÞ: If not, otherwise, let f be not 1–1 on Sr ðaÞ: We have to arrive at a contradiction. Since f is not 1–1 on Sr ðaÞ; there exist b and c in Sr ðaÞ such that b 6¼ c; and f ðbÞ ¼ f ðcÞ: So, from ð Þ; 1 1 1 0\ jc bj ðf 0 ðaÞÞ ðf ðcÞ f ðbÞÞ ¼ ðf 0 ðaÞÞ ðf ðbÞ f ðbÞÞ 2 1 ¼ ðf 0 ðaÞÞ ð0Þ ¼ j0j ¼ 0; which is a contradiction. Hence, f is 1–1 on Sr ðaÞ:

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3 Multivariable Differential Calculus

For 3: We have to show that f ðSr ðaÞÞ is an open subset of R3 and is connected. Since Sr ðaÞ is connected, and f is continuous, f ðSr ðaÞÞ is connected. Now, we want to show that f ðSr ðaÞÞ is an open subset of R3 : For this purpose, let us take any f ðbÞ in f ðSr ðaÞÞ; where b is in Sr ðaÞ: We have to ﬁnd a real number e [ 0 such that the sphere S

e

2

kðf 0 ðaÞÞ1 k

ðf ðbÞÞ

is contained in f ðSr ðaÞÞ; that is, we have to ﬁnd a real number e [ 0 such that if e jy f ðbÞj\ 0 2 ðf ðaÞÞ1 then there exists x in R3 such that jx aj\r; and f ðx Þ ¼ y: Since b is in Sr ðaÞ; we can ﬁnd a real number e [ 0 such that Se ½b fx : x 2 R3 and jx bj eg Sr ðaÞ: Next, let us take any y in R3 such that e : jy f ðbÞj\ 0 2 ðf ðaÞÞ1 Now, we will try to show that uððf 0 ðaÞÞ1 ÞðyÞ : Se ½b ! Se ½b: Observe that 1 1 ðf ðbÞÞ þ ðf 0 ðaÞÞ ðyÞ b uððf 0 ðaÞÞ1 ÞðyÞ ðbÞ b ¼ b ðf 0 ðaÞÞ 1 1 1 ¼ ðf 0 ðaÞÞ ðyÞ ðf 0 ðaÞÞ ðf ðbÞÞ ¼ ðf 0 ðaÞÞ ðy f ðbÞÞ e e 1 1 ¼ : ðf 0 ðaÞÞ jy f ðbÞj ðf 0 ðaÞÞ 1 0 2 2 ðf ðaÞÞ

Thus, e uððf 0 ðaÞÞ1 ÞðyÞ ðbÞ b : 2 Next, let us take any z in Se ½b: We have to prove that uððf 0 ðaÞÞ1 ÞðyÞ ðzÞ is in Se ½b; that is, uððf 0 ðaÞÞ1 ÞðyÞ ðzÞ b e:

3.3 Inverse Function Theorem

171

Since z is in Se ½b; jz bj e; and hence, by ðÞ; uððf 0 ðaÞÞ1 ÞðyÞ ðzÞ b uððf 0 ðaÞÞ1 ÞðyÞ ðzÞ uððf 0 ðaÞÞ1 ÞðyÞ ðbÞ þ uððf 0 ðaÞÞ1 ÞðyÞ ðbÞ b 1 jz bj þ uððf 0 ðaÞÞ1 ÞðyÞ ðbÞ b 2 1 1 e e þ uððf 0 ðaÞÞ1 ÞðyÞ ðbÞ b e þ ¼ e: 2 2 2 This proves that uððf 0 ðaÞÞ1 ÞðyÞ : Se ½b ! Se ½b: Since fx : x 2 R3 and jx bj eg is a closed subset of the complete metric space R3 ; Se ½b is a complete metric space. Also, from ðÞ; for every s, t in Se ½b; we have 1 uððf 0 ðaÞÞ1 ÞðyÞ ðsÞ uððf 0 ðaÞÞ1 ÞðyÞ ðtÞ js tj: 2 Hence, by Theorem 3.35, there exists a unique x in Se ½bð Sr ðaÞÞ such that uððf 0 ðaÞÞ1 ÞðyÞ ðx Þ ¼ x : Hence, jx aj\r; and 1 1 x ¼ uððf 0 ðaÞÞ1 ÞðyÞ ðx Þ ¼ x ðf 0 ðaÞÞ ðf ðx ÞÞ þ ðf 0 ðaÞÞ ðyÞ: Therefore, 1 1 1 ðf 0 ðaÞÞ ðf ðx ÞÞ þ ðf 0 ðaÞÞ ð0Þ ¼ 0 ¼ ðf 0 ðaÞÞ ðyÞ 1 0 ðy f ðx ÞÞ: ¼ ðf ðaÞÞ Now, since ðf 0 ðaÞÞ1 is invertible, 0 ¼ y f ðx Þ; or f ðx Þ ¼ y: Thus, we have shown that f ðSr ðaÞÞ is an open subset of R3 : For 4: We ﬁrst prove that the 1–1 function f−1 from f ðSr ðaÞÞ onto Sr ðaÞ is differentiable at every point of f ðSr ðaÞÞ: For this purpose, let us take any f(x) in f ðSr ðaÞÞ; where x is in Sr ðaÞ: Since f : E ! R3 is a C1 function, by Theorem 3.33, f is differentiable at every point of E. Since f is differentiable at every point of E, and x is in Sr ðaÞ ð EÞ; f 0 ðxÞ exists. Observe that if t is in Sr ðaÞ; then f 0 ðtÞ 2 S

1

2

kðf 0 ðaÞÞ1 k

ðf 0 ðaÞÞ S

1

kðf 0 ðaÞÞ1 k

ðf 0 ðaÞÞ X;

and hence, ðf 0 ðtÞÞ1 exists. Thus, if t is in Sr ðaÞ; then

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3 Multivariable Differential Calculus 1

f 0 ðtÞ; ðf 0 ðtÞÞ 2 X:

ð0 Þ

Since x is in Sr ðaÞ; from ð0 Þ; f 0 ðxÞ; ðf 0 ðxÞÞ1 exist. Now, we will try to prove that f−1 is differentiable at f(x) and ðf 1 Þ0 ðf ðxÞÞ ¼ ðf 0 ðxÞÞ1 ; that is, lim

k!0

1 1 1 f ðf ðxÞ þ k Þ f 1 ðf ðxÞÞ ðf 0 ðxÞÞ ðkÞ jk j

¼ 0;

that is, lim y ! f ðxÞ y 2 f ðSr ðaÞÞ

1 1 1 1 0 f ðyÞ f ðf ðxÞÞ ðf ðxÞÞ ¼ 0; ðy f ðxÞÞ jy f ðxÞj

that is, 1 1 1 lim ðy f ðxÞÞ ¼ 0; f ðyÞ f 1 ðf ðxÞÞ ðf 0 ðxÞÞ jy f ðxÞj y ! f ðxÞ y 2 f ðSr ðaÞÞ that is, 1 1 1 lim ðy f ðxÞÞ ¼ 0: f ðyÞ x ðf 0 ðxÞÞ y f ðxÞ j j y ! f ðxÞ y 2 f ðSr ðaÞÞ For this purpose, let us take a sequence ff ðan Þg in f ðSr ðaÞÞ such that each an is in Sr ðaÞ; each f ðan Þ is different from f ðxÞ; and limn!1 f ðan Þ ¼ f ðxÞ: We have to prove that 1 1 1 ðf ðan Þ f ðxÞÞ ¼ 0; f ðf ðan ÞÞ x ðf 0 ðxÞÞ n!1 jf ðan Þ f ðxÞj lim

that is, 1 1 ðf ðan Þ f ðxÞÞ ¼ 0: an x ðf 0 ðxÞÞ n!1 jf ðan Þ f ðxÞj lim

Since x and an are in Sr ðaÞ, from ð Þ; 1 1 1 jan xj ðf 0 ðaÞÞ ðf ðan Þ f ðxÞÞ ðf 0 ðaÞÞ jf ðan Þ f ðxÞj; 2

3.3 Inverse Function Theorem

173

and hence, 1 0 jan xj 2ðf 0 ðaÞÞ jf ðan Þ f ðxÞj ðn ¼ 1; 2; 3; Þ: This, together with limn!1 f ðan Þ ¼ f ðxÞ; implies that limn!1 jan xj ¼ 0: Thus, lim an ¼ x:

n!1

Since each f ðan Þ is different from f ðxÞ; x 6¼ an : Since f 0 ðxÞ exists, each an is in Sr ðaÞ; x is in Sr ðaÞ; each an is different from x, and limn!1 an ¼ x; lim

n!1 jan

1 jf ðan Þ f ðxÞ ððf 0 ðxÞÞðan xÞÞj ¼ 0: xj

Since each an 6¼ x; jan xj 6¼ 0: Also, each jf ðan Þ f ðxÞj 6¼ 0: It follows that for every positive integer n, 1 1 1 2ðf 0 ðaÞÞ : jf ðan Þ f ðxÞj j an x j Now, for every positive integer n, we have 1 1 ðf ðan Þ f ðxÞÞ an x ðf 0 ðxÞÞ jf ðan Þ f ðxÞj 1 1 1 2ðf 0 ðaÞÞ ðf ðan Þ f ðxÞÞ an x ðf 0 ðxÞÞ jan xj 1 1 1 1 2ðf 0 ðaÞÞ ðf 0 ðxÞÞ ðan xÞ ðf 0 ðxÞÞ ðf ðan Þ f ðxÞÞ ðf 0 ðxÞÞ jan xj 1 0 1 1 1 2ðf 0 ðaÞÞ ððf 0 ðxÞÞðan xÞÞ ðf 0 ðxÞÞ ðf ðan Þ f ðxÞÞ ðf ðxÞÞ jan xj 1 0 1 1 2ðf 0 ðaÞÞ ðððf 0 ðxÞÞðan xÞÞ ðf ðan Þ f ðxÞÞÞ ðf ðxÞÞ jan xj 1 0 1 1 2ðf 0 ðaÞÞ ðf ðxÞÞ jððf 0 ðxÞÞðan xÞÞ ðf ðan Þ f ðxÞÞj jan xj 1 0 1 0 1 ¼ 2ðf ðaÞÞ ðf ðxÞÞ jf ðan Þ f ðxÞ ððf 0 ðxÞÞðan xÞÞj : jan xj

Since 1 1 ðf ðan Þ f ðxÞÞ an x ðf 0 ðxÞÞ jf ðan Þ f ðxÞj 1 1 1 2ðf 0 ðaÞÞ ðf 0 ðxÞÞ jf ðan Þ f ðxÞ ððf 0 ðxÞÞðan xÞÞj ; jan xj

0

174

3 Multivariable Differential Calculus

and lim

n!1 jan

1 jf ðan Þ f ðxÞ ððf 0 ðxÞÞðan xÞÞj ¼ 0; xj

so, 1 1 ðf ðan Þ f ðxÞÞ ¼ 0: an x ðf 0 ðxÞÞ n!1 jf ðan Þ f ðxÞj lim

This proves that f−1 is differentiable at f(x) and 1 0 1 f ðf ðxÞÞ ¼ ðf 0 ðxÞÞ ð00 Þ for every x in Sr ðaÞ: Hence, f−1 is differentiable on the open subset f ðSr ðaÞÞ of R3 : Since f−1 is differentiable on the open subset f ðSr ðaÞÞ of R3 ; f−1 is continuous on the open subset f ðSr ðaÞÞ of R3 : Now, we want to prove that the mapping ðf 1 Þ0 : f ðSr ðaÞÞ ! LðR3 Þ is continuous. Let us take any y in f ðSr ðaÞÞ: Since y is in f ðSr ðaÞÞ; there exists x in Sr ðaÞ; such that f ðxÞ ¼ y: Since f ðxÞ ¼ y 2 f ðSr ðaÞÞ; f 1 ðyÞ ¼ x: Since x is in Sr ðaÞ, by ð00 Þ;

0 0

1 1 f 1 ðyÞ ¼ f 1 ðf ðxÞÞ ¼ ðf 0 ðxÞÞ ¼ f 0 f 1 ðyÞ :

Hence, for every y in f ðSr ðaÞÞ;

1 1 0 f ðyÞ ¼ f 0 f 1 ðyÞ : We have seen that y 7! f 1 ðyÞ is differentiable on f ðSr ðaÞÞ; so y 7! f 1 ðyÞ is a continuous function from f ðSr ðaÞÞ to Sr ðaÞ: Since f is a C1 function on E, by Theorem 3.33, x 7! f 0 ðxÞ is continuous on E. Since y 7! f 1 ðyÞ is a continuous function from f ðSr ðaÞÞ to Sr ðaÞ; and x 7! f 0 ðxÞ is continuous on Eð Sr ðaÞÞ; by ð0 Þ; their composite function y 7! f 0 ðf 1 ðyÞÞ is continuous from f ðSr ðaÞÞ to X: Since y 7! f 0 ðf 1 ðyÞÞ is continuous from f ðSr ðaÞÞ to X, and by Theorem 3.3, the mapping A 7! A1 from X to X is continuous, their composite function y 7! ðf 0 ðf 1 ðyÞÞÞ1 is continuous from f ðSr ðaÞÞ to X. Since y 7! ðf 0 ðf 1 ðyÞÞÞ1 is continuous from f ðSr ðaÞÞ to X, and ðf 0 ðf 1 ðyÞÞÞ1 ¼ ðf 1 Þ0 ðyÞ; the function y 7! ðf 1 Þ0 ðyÞ is continuous from f ðSr ðaÞÞ to X, and hence, by Theorem 3.33, f−1 is a C1 function on the open subset f ðSr ðaÞÞ of R3 : For 5: Let f be a C2 function. Let g1 : f ðSr ðaÞÞ ! R; g2 : f ðSr ðaÞÞ ! R; g3 : f ðSr ðaÞÞ ! R be the component functions of f−1 that is, f 1 ðxÞ ¼ ðg1 ðxÞ; g2 ðxÞ; g3 ðxÞÞ for every x in f ðSr ðaÞÞ: We have to prove that f−1 from f ðSr ðaÞÞ onto Sr ðaÞ is a C2 function, that is, each of the nine second-order partial derivatives Djk ðgi Þ : f ðSr ðaÞÞ ! R exists and is continuous. Let us take any y in f ðSr ðaÞÞ: We will try to show that ðD12 ðg3 ÞÞðyÞ exists. Since y is in f ðSr ðaÞÞ;

3.3 Inverse Function Theorem

175

1

0 f 1 ðyÞ ¼ f 0 f 1 ðyÞ ;

or,

0

f 1 ðyÞ f 0 f 1 ðyÞ ¼ I;

and hence, by Theorem 3.6, 3 1 0 0 h 0

i h 1 0 i 0 1 7 6 ¼ ½I ¼ f 1 ðyÞ f 0 f 1 ðyÞ ðyÞ f f ðyÞ ¼ f 40 1 05 0 0 1 33 3 2 ðD1 ðg1 ÞÞðyÞ ðD2 ðg1 ÞÞðyÞ ðD3 ðg1 ÞÞðyÞ 7 6 ¼ 4 ðD1 ðg2 ÞÞðyÞ ðD2 ðg2 ÞÞðyÞ ðD3 ðg2 ÞÞðyÞ 5 ðD1 ðg3 ÞÞðyÞ ðD2 ðg3 ÞÞðyÞ ðD3 ðg3 ÞÞðyÞ 33 3 2 ðD1 ðf1 ÞÞðf 1 ðyÞÞ ðD2 ðf1 ÞÞðf 1 ðyÞÞ ðD3 ðf1 ÞÞðf 1 ðyÞÞ 7 6 4 ðD1 ðf2 ÞÞðf 1 ðyÞÞ ðD2 ðf2 ÞÞðf 1 ðyÞÞ ðD3 ðf2 ÞÞðf 1 ðyÞÞ 5 2

ð D 1 ðf 3 ÞÞ ðf

1

ðyÞÞ ðD2 ðf3 ÞÞðf

1

ðyÞÞ ðD3 ðf3 ÞÞðf

1

ðyÞÞ

33

This shows that 2

3 ðD1 ðg1 ÞÞðyÞ ðD2 ðg1 ÞÞðyÞ ðD3 ðg1 ÞÞðyÞ 4 ðD1 ðg2 ÞÞðyÞ ðD2 ðg2 ÞÞðyÞ ðD3 ðg2 ÞÞðyÞ 5 ðD1 ðg3 ÞÞðyÞ ðD2 ðg3 ÞÞðyÞ ðD3 ðg3 ÞÞðyÞ 33 2 31 ðD1 ðf1 ÞÞðf 1 ðyÞÞ ðD2 ðf1 ÞÞðf 1 ðyÞÞ ðD3 ðf1 ÞÞðf 1 ðyÞÞ ¼ 4 ðD1 ðf2 ÞÞðf 1 ðyÞÞ ðD2 ðf2 ÞÞðf 1 ðyÞÞ ðD3 ðf2 ÞÞðf 1 ðyÞÞ 5 : ðD1 ðf3 ÞÞðf 1 ðyÞÞ ðD2 ðf3 ÞÞðf 1 ðyÞÞ ðD3 ðf3 ÞÞðf 1 ðyÞÞ Hence, 1 ðD2 ðg3 ÞÞðyÞ ¼ 1 1 ð ð f Þ Þ ð f ðyÞ Þ ð D ð f Þ Þ ðyÞÞ ðD3 ðf1 ÞÞðf 1 ðyÞÞ D 2 1 ðf 1 1 ðD1 ðf2 ÞÞðf 1 ðyÞÞ ðD2 ðf2 ÞÞðf 1 ðyÞÞ ðD3 ðf2 ÞÞðf 1 ðyÞÞ ðD1 ðf3 ÞÞðf 1 ðyÞÞ ðD2 ðf3 ÞÞðf 1 ðyÞÞ ðD3 ðf3 ÞÞðf 1 ðyÞÞ ðD1 ðf1 ÞÞðf 1 ðyÞÞ ðD2 ðf1 ÞÞðf 1 ðyÞÞ ðD1 ðf3 ÞÞðf 1 ðyÞÞ ðD2 ðf3 ÞÞðf 1 ðyÞÞ ðD1 ðf1 ÞÞðf 1 ðyÞÞ ðD2 ðf1 ÞÞðf 1 ðyÞÞ ðD ðf ÞÞðf 1 ðyÞÞ ðD ðf ÞÞðf 1 ðyÞÞ 1 3 2 3 : ¼ 1 1 1 ðD1 ðf1 ÞÞðf ðyÞÞ ðD2 ðf1 ÞÞðf ðyÞÞ ðD3 ðf1 ÞÞðf ðyÞÞ 1 1 1 ðD1 ðf2 ÞÞðf ðyÞÞ ðD2 ðf2 ÞÞðf ðyÞÞ ðD3 ðf2 ÞÞðf ðyÞÞ ðD1 ðf3 ÞÞðf 1 ðyÞÞ ðD2 ðf3 ÞÞðf 1 ðyÞÞ ðD3 ðf3 ÞÞðf 1 ðyÞÞ

:

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3 Multivariable Differential Calculus

Now, since f is a C2 function, each of the nine second-order partial derivatives Djk ðfi Þ : Sr ðaÞ ! R of f1 exists, and for every y in f ðSr ðaÞÞ; we have 1 1 1 ðD1 ðf1 ÞÞðf ðyÞÞ ðD2 ðf1 ÞÞðf ðyÞÞ C B C B ðD1 ðf3 ÞÞðf 1 ðyÞÞ ðD2 ðf3 ÞÞðf 1 ðyÞÞ C B num C ðD12 ðg3 ÞÞðyÞ ¼ D1 B B ðD ðf ÞÞðf 1 ðyÞÞ ðD ðf ÞÞðf 1 ðyÞÞ ðD ðf ÞÞðf 1 ðyÞÞ C ¼ denom ; 2 1 3 1 C B 1 1 @ ðD ðf ÞÞðf 1 ðyÞÞ ðD ðf ÞÞðf 1 ðyÞÞ ðD ðf ÞÞðf 1 ðyÞÞ A 2 2 3 2 1 2 ðD1 ðf3 ÞÞðf 1 ðyÞÞ ðD2 ðf3 ÞÞðf 1 ðyÞÞ ðD3 ðf3 ÞÞðf 1 ðyÞÞ 0

where ðD1 ðf1 ÞÞðf 1 ðyÞÞ ðD2 ðf1 ÞÞðf 1 ðyÞÞ D1 ðD1 ðf3 ÞÞðf 1 ðyÞÞ ðD2 ðf3 ÞÞðf 1 ðyÞÞ ðD1 ðf1 ÞÞðf 1 ðyÞÞ ðD2 ðf1 ÞÞðf 1 ðyÞÞ ðD3 ðf1 ÞÞðf 1 ðyÞÞ ðD1 ðf2 ÞÞðf 1 ðyÞÞ ðD2 ðf2 ÞÞðf 1 ðyÞÞ ðD3 ðf2 ÞÞðf 1 ðyÞÞ ðD1 ðf3 ÞÞðf 1 ðyÞÞ ðD2 ðf3 ÞÞðf 1 ðyÞÞ ðD3 ðf3 ÞÞðf 1 ðyÞÞ ðD1 ðf1 ÞÞðf 1 ðyÞÞ ðD2 ðf1 ÞÞðf 1 ðyÞÞ ðD1 ðf3 ÞÞðf 1 ðyÞÞ ðD2 ðf3 ÞÞðf 1 ðyÞÞ 1 0 ðD1 ðf1 ÞÞðf 1 ðyÞÞ ðD2 ðf1 ÞÞðf 1 ðyÞÞ ðD3 ðf1 ÞÞðf 1 ðyÞÞ C B @D1 ðD1 ðf2 ÞÞðf 1 ðyÞÞ ðD2 ðf2 ÞÞðf 1 ðyÞÞ ðD3 ðf2 ÞÞðf 1 ðyÞÞ A 1 1 1 ðD1 ðf3 ÞÞðf ðyÞÞ ðD2 ðf3 ÞÞðf ðyÞÞ ðD3 ðf3 ÞÞðf ðyÞÞ ðD11 ðf1 ÞÞðf 1 ðyÞÞ ðD2 ðf1 ÞÞðf 1 ðyÞÞ ðD1 ðf1 ÞÞðf 1 ðyÞÞ ðD12 ðf1 ÞÞðf 1 ðyÞÞ þ ¼ ðD11 ðf3 ÞÞðf 1 ðyÞÞ ðD2 ðf3 ÞÞðf 1 ðyÞÞ ðD1 ðf3 ÞÞðf 1 ðyÞÞ ðD12 ðf3 ÞÞðf 1 ðyÞÞ ðD1 ðf1 ÞÞðf 1 ðyÞÞ ðD2 ðf1 ÞÞðf 1 ðyÞÞ ðD3 ðf1 ÞÞðf 1 ðyÞÞ ðD1 ðf2 ÞÞðf 1 ðyÞÞ ðD2 ðf2 ÞÞðf 1 ðyÞÞ ðD3 ðf2 ÞÞðf 1 ðyÞÞ ðD1 ðf3 ÞÞðf 1 ðyÞÞ ðD2 ðf3 ÞÞðf 1 ðyÞÞ ðD3 ðf3 ÞÞðf 1 ðyÞÞ ðD1 ðf1 ÞÞðf 1 ðyÞÞ ðD2 ðf1 ÞÞðf 1 ðyÞÞ ðD1 ðf3 ÞÞðf 1 ðyÞÞ ðD2 ðf3 ÞÞðf 1 ðyÞÞ 0 ðD11 ðf1 ÞÞðf 1 ðyÞÞ ðD2 ðf1 ÞÞðf 1 ðyÞÞ ðD3 ðf1 ÞÞðf 1 ðyÞÞ B @ ðD11 ðf2 ÞÞðf 1 ðyÞÞ ðD2 ðf2 ÞÞðf 1 ðyÞÞ ðD3 ðf2 ÞÞðf 1 ðyÞÞ ðD11 ðf3 ÞÞðf 1 ðyÞÞ ðD2 ðf3 ÞÞðf 1 ðyÞÞ ðD3 ðf3 ÞÞðf 1 ðyÞÞ ðD1 ðf1 ÞÞðf 1 ðyÞÞ ðD12 ðf1 ÞÞðf 1 ðyÞÞ ðD3 ðf1 ÞÞðf 1 ðyÞÞ þ ðD1 ðf2 ÞÞðf 1 ðyÞÞ ðD12 ðf2 ÞÞðf 1 ðyÞÞ ðD3 ðf2 ÞÞðf 1 ðyÞÞ ðD1 ðf3 ÞÞðf 1 ðyÞÞ ðD12 ðf3 ÞÞðf 1 ðyÞÞ ðD3 ðf3 ÞÞðf 1 ðyÞÞ 1 ðD1 ðf1 ÞÞðf 1 ðyÞÞ ðD2 ðf1 ÞÞðf 1 ðyÞÞ ðD13 ðf1 ÞÞðf 1 ðyÞÞ C ðD1 ðf2 ÞÞðf 1 ðyÞÞ ðD2 ðf2 ÞÞðf 1 ðyÞÞ ðD13 ðf2 ÞÞðf 1 ðyÞÞ A; ðD1 ðf3 ÞÞðf 1 ðyÞÞ ðD2 ðf3 ÞÞðf 1 ðyÞÞ ðD13 ðf3 ÞÞðf 1 ðyÞÞ

num

3.3 Inverse Function Theorem

177

and ðD1 ðf1 ÞÞðf 1 ðyÞÞ ðD2 ðf1 ÞÞðf 1 ðyÞÞ ðD3 ðf1 ÞÞðf 1 ðyÞÞ 2 denom ðD1 ðf2 ÞÞðf 1 ðyÞÞ ðD2 ðf2 ÞÞðf 1 ðyÞÞ ðD3 ðf2 ÞÞðf 1 ðyÞÞ : ðD1 ðf3 ÞÞðf 1 ðyÞÞ ðD2 ðf3 ÞÞðf 1 ðyÞÞ ðD3 ðf3 ÞÞðf 1 ðyÞÞ It follows that ðD12 ðg3 ÞÞðyÞ exists. So D12 ðg3 Þ : f ðSr ðaÞÞ ! R: Next, since each of the nine second-order partial derivatives Djk ðfi Þ : Sr ðaÞ ! R of fi is continuous, and f−1is continuous on f ðSr ðaÞÞ; by the above formula for num and denom, we ﬁnd that num and denom are continuous functions on f ðSr ðaÞÞ; and num Þ is continuous on f ðS ðaÞÞ: hence, D12 ðg3 Þð¼ denom r Similarly, all the other Djk ðgi Þ : f ðSr ðaÞÞ ! R are continuous. This proves that f−1 from f(U) onto U is a C2 function, etc. h Note 3.37 The result similar to Theorem 3.36 can be proved as above for Rn in place of R3 : This theorem is known as the inverse function theorem. Theorem 3.38 Let E be an open subset of Rn : Let f : E ! Rn be a function. If 1. f is a C1 function, 2. f 0 ðaÞ is invertible for every a in E, then f is an open mapping (i.e., f-image of every open subset of E in an open set). Proof Let us take any open subset G of E. We have to prove that f(G) is an open set. For this purpose, let us take any f(a) in f(G) where a is in G. Since a 2 G E, by the given assumption, f 0 ðaÞ is invertible. Now, by the inverse function theorem, there exists an open neighborhood U of a such that U is contained in G and f(U) is open in R3 : Since a 2 U G; f ðaÞ 2 f ðUÞ f ðGÞ: Since f ðaÞ 2 f ðUÞ; and f(U) is open in R3 ; f(U) is an open neighborhood of f(a). Since f(U) is an open neighborhood of f(a) and f ðUÞ f ðGÞ; f(a) is an interior point of f(G), and hence, f(G) is an open set. h

3.4 Implicit Function Theorem Deﬁnition Let a ða1 ; a2 ; a3 Þ 2 R3 ; b ðb1 ; b2 Þ 2 R2 : By the ordered pair ða; bÞ, we mean ða1 ; a2 ; a3 ; b1 ; b2 Þ: Clearly, if a 2 R3 ; b 2 R2 , then ða; bÞ 2 R3þ2 : Let A 2 LðR3þ2 ; R3 Þ: By Ax , we mean the function Ax : R3 ! R3 deﬁned as follows: For every ða1 ; a2 ; a3 Þ in R3 ;

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3 Multivariable Differential Calculus

Ax ðða1 ; a2 ; a3 ÞÞ Aða1 ; a2 ; a3 ; 0; 0Þ: We shall show that Ax 2 LðR3 ; R3 Þ: We must prove (i) Ax ða þ bÞ ¼ Ax ðaÞ þ Ax ðbÞ for every a, b in R3 ; (ii) Ax ðtaÞ ¼ tðAx ðaÞÞ for every a in R3 and for every real t. For (i): Let us take any a ða1 ; a2 ; a3 Þ; b ðb1 ; b2 ; b3 Þ in R3 : LHS ¼ Ax ða þ bÞ ¼ Ax ðða1 ; a2 ; a3 Þ þ ðb1 ; b2 ; b3 ÞÞ ¼ Ax ðða1 þ b1 ; a2 þ b2 ; a3 þ b3 ÞÞ ¼ Aðða1 þ b1 ; a2 þ b2 ; a3 þ b3 ; 0; 0ÞÞ ¼ Aðða1 ; a2 ; a3 ; 0; 0Þ þ ðb1 ; b2 ; b3 ; 0; 0ÞÞ ¼ Aðða1 ; a2 ; a3 ; 0; 0ÞÞ þ Aððb1 ; b2 ; b3 ; 0; 0ÞÞ ¼ Ax ðða1 ; a2 ; a3 ÞÞ þ Ax ððb1 ; b2 ; b3 ÞÞ ¼ Ax ðaÞ þ Ax ðbÞ ¼ RHS: For (ii): Let us take any a ða1 ; a2 ; a3 Þ in R3 and any real t. LHS ¼ Ax ðtaÞ ¼ Ax ðtða1 ; a2 ; a3 ÞÞ ¼ Ax ððta1 ; ta2 ; ta3 ÞÞ ¼ Aððta1 ; ta2 ; ta3 ; 0; 0ÞÞ ¼ Aðtða1 ; a2 ; a3 ; 0; 0ÞÞ ¼ tðAðða1 ; a2 ; a3 ; 0; 0ÞÞÞ ¼ tðAx ðða1 ; a2 ; a3 ÞÞÞ ¼ tðAx ðaÞÞ ¼ RHS: Thus, we have shown that if A 2 LðR3þ2 ; R3 Þ, then Ax 2 LðR3 ; R3 Þ: Again, by Ay, we mean the function Ay : R2 ! R3 deﬁned as follows: For every ða1 ; a2 Þ in R2 ; Ay ðaÞ Að0; 0; 0; a1 ; a2 Þ: We shall show that Ay 2 LðR2 ; R3 Þ: We must prove (i) Ay ða þ bÞ ¼ Ay ðaÞ þ Ay ðbÞ for every a, b in R2 ; (ii) Ay ðtaÞ ¼ tðAy ðaÞÞ for every a in R2 and for every real t.

3.4 Implicit Function Theorem

179

For (i): Let us take any a ða1 ; a2 Þ; b ðb1 ; b2 Þ in R2 : LHS ¼ Ay ða þ bÞ ¼ Ay ðða1 ; a2 Þ þ ðb1 ; b2 ÞÞ ¼ Ay ðða1 þ b1 ; a2 þ b2 ÞÞ ¼ Aðð0; 0; 0; a1 þ b1 ; a2 þ b2 ÞÞ ¼ Aðð0; 0; 0; a1 ; a2 Þ þ ð0; 0; 0; b1 ; b2 ÞÞ ¼ Aðð0; 0; 0; a1 ; a2 ÞÞ þ Aðð0; 0; 0; b1 ; b2 ÞÞ ¼ Ay ðða1 ; a2 ÞÞ þ Ay ððb1 ; b2 ÞÞ ¼ Ay ðaÞ þ Ay ðbÞ ¼ RHS: For (ii): Let us take any a ða1 ; a2 Þ in R2 and any real t. LHS ¼ Ay ðtaÞ ¼ Ay ðtða1 ; a2 ÞÞ ¼ Ay ððta1 ; ta2 ÞÞ ¼ Aðð0; 0; 0; ta1 ; ta2 ÞÞ ¼ Aðtð0; 0; 0; a1 ; a2 ÞÞ

¼ tðAðð0; 0; 0; a1 ; a2 ÞÞÞ ¼ t Ay ðða1 ; a2 ÞÞ

¼ t Ay ðaÞ ¼ RHS: Thus, we have shown that if A 2 LðR3þ2 ; R3 Þ, then Ay 2 LðR2 ; R3 Þ: Now, we will try to prove that for every a in R3 ; and for every b in R2 ; Aðða; bÞÞ ¼ Ax ðaÞ þ Ay ðbÞ ðÞ: For this purpose, let a ða1 ; a2 ; a3 Þ 2 R3 and b ðb1 ; b2 Þ 2 R2 : LHS ¼ Aðða; bÞÞ ¼ Aðða1 ; a2 ; a3 ; b1 ; b2 ÞÞ ¼ Aðða1 ; a2 ; a3 ; 0; 0Þ þ ð0; 0; 0; b1 ; b2 ÞÞ ¼ Aðða1 ; a2 ; a3 ; 0; 0ÞÞ þ Aðð0; 0; 0; b1 ; b2 ÞÞ ¼ Ax ðða1 ; a2 ; a3 ÞÞ þ Ay ððb1 ; b2 ÞÞ ¼ Ax ðaÞ þ Ay ðbÞ ¼ RHS: Note 3.39 Similar notations and results as above can be supplied for Rn in place of R3 and Rm in place of R2 : Note 3.40 Let A 2 LðR3þ2 ; R3 Þ; and let 2

a11 ½ A ¼ 4 a21 a31

a12 a22 a32

a13 a23 a33

a14 a24 a34

3 a15 a25 5 : a35 35

So, Ax ðð1; 0; 0ÞÞ ¼ Aðð1; 0; 0; 0; 0ÞÞ ¼ ða11 ; a21 ; a31 Þ; Ax ðð0; 1; 0ÞÞ ¼ Aðð0; 1; 0; 0; 0ÞÞ ¼ ða12 ; a22 ; a32 Þ; Ax ðð0; 0; 1ÞÞ ¼ Aðð0; 0; 1; 0; 0ÞÞ ¼ ða13 ; a23 ; a33 Þ:

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3 Multivariable Differential Calculus

Hence, 2

a11 ½Ax ¼ 4 a21 a31

a12 a22 a32

3 a13 a23 5 : a33 33

Next, Ay ðð1; 0ÞÞ ¼ Aðð0; 0; 0; 1; 0ÞÞ ¼ ða14 ; a24 ; a34 Þ; Ay ðð0; 1ÞÞ ¼ Aðð0; 0; 0; 0; 1ÞÞ ¼ ða15 ; a25 ; a35 Þ: Hence, 2 a14 Ay ¼ 4 a24 a34

3 a15 a25 5 : a35 32

Conclusion: If A 2 LðR3þ2 ; R3 Þ; and 2

a11 ½ A ¼ 4 a21 a31

a12 a22 a32

a13 a23 a33

a14 a24 a34

3 a15 a25 5 ; a35 35

then 2

a11 ½Ax ¼ 4 a21 a31

a12 a22 a32

3 2 a14 a13 a23 5 and Ay ¼ 4 a24 a33 33 a34

3 a15 a25 5 : a35 32

Also, if Ax is invertible, then with the usual meaning of symbols, 2 a11 h i 1 ðAx Þ ¼ 4 a21 a31

a12 a22 a32

31 2 A11 a13 1 a23 5 ¼ 4 A21 D a33 A31

A12 A22 A32

3T 2 A11 A13 1 A23 5 ¼ 4 A12 D A33 A13

and hence,

1 ðAx Þ1 ðð1; 0; 0ÞÞ ¼ ðA11 ; A12 ; A13 Þ; D 1 ðAx Þ1 ðð0; 1; 0ÞÞ ¼ ðA21 ; A22 ; A23 Þ; D 1 ðAx Þ1 ðð0; 0; 1ÞÞ ¼ ðA31 ; A32 ; A33 Þ: D

A21 A22 A23

3 A31 A32 5; A33

3.4 Implicit Function Theorem

Conclusion: If A 2 LðR3þ2 ; R3 Þ; Ax 2 a11 a12 ½ A ¼ 4 a21 a22 a31 a32

181

is invertible, and 3 a13 a14 a15 a23 a24 a25 5 ; a33 a34 a35 35

then

1 ðAx Þ1 ðð1; 0; 0ÞÞ ¼ ðA11 ; A12 ; A13 Þ; D 1 ðAx Þ1 ðð0; 1; 0ÞÞ ¼ ðA21 ; A22 ; A23 Þ; D 1 ðAx Þ1 ðð0; 0; 1ÞÞ ¼ ðA31 ; A32 ; A33 Þ: D Theorem 3.41 Let A 2 LðR3þ2 ; R3 Þ: If Ax is invertible, then for every k in R2 ; there exists a unique h in R3 ; such that Aðh; kÞ ¼ 0: The value of h is ðððAx Þ1 ÞðAy ðkÞÞÞ: Proof Existence: Let us take any k in R2 : Since k is in R2 ; and Ay : R2 ! R3 ; Ay ðkÞ is in R3 : Since Ay ðkÞ is in R3 ; Ax : R3 ! R3 ; and Ax is invertible, ðððAx Þ1 ÞðAy ðkÞÞÞ is in R3 . Also, from the formula ðÞ,

A ðAx Þ1 Ay ðkÞ ; k ¼ Ax ðAx Þ1 Ay ðkÞ þ Ay ðkÞ

¼ Ax ðAx Þ1 Ay ðkÞ þ Ay ðkÞ

¼ Ax ðAx Þ1 Ay ðkÞ þ Ay ðkÞ

¼ Ay ðkÞ þ Ay ðkÞ ¼ 0: Uniqueness: If not, otherwise, let there exist k in R2 ; h in R3 ; and h1 in R3 such that h 6¼ h1 ; Aððh; kÞÞ ¼ 0; and Aððh1 ; kÞÞ ¼ 0: We have to arrive at a contradiction. Since 0 ¼ Aððh; kÞÞ ¼ Ax ðhÞ þ Ay ðkÞ; and Ax is invertible,

0 ¼ ðAx Þ1 ð0Þ ¼ ðAx Þ1 Ax ðhÞ þ Ay ðkÞ

¼ ðAx Þ1 ðAx ðhÞÞ þ ðAx Þ1 Ay ðkÞ

¼ ðAx Þ1 Ax ðhÞ þ ðAx Þ1 Ay ðkÞ ¼ IðhÞ þ ðAx Þ1 Ay ðkÞ

¼ h þ ðAx Þ1 Ay ðkÞ :

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3 Multivariable Differential Calculus

It follows that h¼

ðAx Þ1 Ay ðkÞ :

Similarly,

h1 ¼ ðAx Þ1 Ay ðkÞ ; which contradicts the assumption h 6¼ h1 : Further, we have shown that for a given h k, the value of h is ðððAx Þ1 ÞðAy ðkÞÞÞ: Note 3.42 The result similar to Theorem 3.41 can be proved as above for Rn in place of R3 and Rm in place of R2 : Theorem 3.43 Let E be an open subset of R3þ2 : Let f : E ! R3 : Let a 2 R3 ; b 2 R2 ; and ða; bÞ 2 E: If (i) f is a C 1 function, (ii) ðf 0 ðða; bÞÞÞx is invertible, (iii) f ðða; bÞÞ ¼ 0; then there exist an open neighborhood Uð EÞ of ða; bÞ in R3þ2 ; an open neighborhood W of b in R2 ; and a function g : W ! R3 such that 1. 2. 3. 4. 5.

for every y in W, ðgðyÞ; yÞ 2 U; for every y in W, f ððgðyÞ; yÞÞ ¼ 0; gðbÞ ¼ a; g is a C 1 function, for every y in W, g0 ðy Þ ¼ ððf 0 ððgðy Þ; y ÞÞÞx Þ1 ðf 0 ððgðy Þ; y ÞÞÞy

6. g0 ðbÞ ¼ ððf 0 ðða; bÞÞÞx Þ1 ðf 0 ðða; bÞÞÞy 7. if f is a C 2 function, then g is a C 2 function, etc.

Proof Let f1 : E ! R; f2 : E ! R; f3 : E ! R, be the component functions of f, that is, f ððx; yÞÞ ¼ ðf1 ððx; yÞÞ; f2 ððx; yÞÞ; f3 ððx; yÞÞÞ for every ðx; yÞ in Eð R3þ2 Þ: Let us deﬁne a function F : E ! R3þ2 as follows: For every ðx; yÞ in E, where x ðx1 ; x2 ; x3 Þ is in R3 and y ðy1 ; y2 Þ is in R2 ;

3.4 Implicit Function Theorem

183

F ððx; yÞÞ ðf ððx; yÞÞ; yÞ ¼ ððf1 ððx; yÞÞ; f2 ððx; yÞÞ; f3 ððx; yÞÞÞ; ðy1 ; y2 ÞÞ ¼ ðf1 ððx; yÞÞ; f2 ððx; yÞÞ; f3 ððx; yÞÞ; y1 ; y2 Þ: Clearly, the component functions of F are ðx; yÞ 7! f1 ððx; yÞÞ; ðx; yÞ 7! y2

ðx; yÞ 7! f2 ððx; yÞÞ;

ðx; yÞ 7! f3 ððx; yÞÞ;

ðx; yÞ 7! y1 ;

from E to R: We want to apply the inverse function theorem on F. For this purpose, we ﬁrst prove (a) F is a C1 function, (b) F 0 ðða; bÞÞ is invertible. For (a): Since f is a C 1 function, for i ¼ 1; 2; 3 and j ¼ 1; 2; 3; 4; 5ð¼ 3 þ 2Þ, each function Dj fi : E ! R exists and is continuous. Also, ðx; yÞ 7! y1 and ðx; yÞ 7! y2 are smooth functions. This shows that F is a C1 function. For (b): Here, det½F 0 ðða; bÞÞ 2 ðD1 f1 Þðða; bÞÞ 6 ðD1 f2 Þðða; bÞÞ 6 6 ¼ det6 ðD1 f3 Þðða; bÞÞ 6 40 0

ðD2 f1 Þðða; bÞÞ ðD2 f2 Þðða; bÞÞ ðD2 f3 Þðða; bÞÞ

ðD3 f1 Þðða; bÞÞ ðD3 f2 Þðða; bÞÞ ðD3 f3 Þðða; bÞÞ

ðD4 f1 Þðða; bÞÞ ðD5 f1 Þðða; bÞÞ ðD4 f2 Þðða; bÞÞ ðD5 f2 Þðða; bÞÞ ðD4 f3 Þðða; bÞÞ ðD5 f3 Þðða; bÞÞ

0

0

1

0

0

0

0

1

ðD1 f1 Þðða; bÞÞ 6 ¼ det4 ðD1 f2 Þðða; bÞÞ

ðD2 f1 Þðða; bÞÞ ðD2 f2 Þðða; bÞÞ

3 ðD3 f1 Þðða; bÞÞ 7 ðD3 f2 Þðða; bÞÞ 5:

ðD1 f3 Þðða; bÞÞ

ðD2 f3 Þðða; bÞÞ

ðD3 f3 Þðða; bÞÞ

2

Since ðf 0 ðða; bÞÞÞ 2 LðR3þ2 ; R3 Þ; ðf 0 ðða; bÞÞÞx : R3 ! R3 : Also, ðf 0 ðða; bÞÞÞx ðð1; 0; 0ÞÞ ¼ ðf 0 ðða; bÞÞÞðð1; 0; 0; 0; 0ÞÞ ¼ ððD1 f1 Þðða; bÞÞ; ðD1 f2 Þðða; bÞÞ; ðD1 f3 Þðða; bÞÞÞ; ðf 0 ðða; bÞÞÞx ðð0; 1; 0ÞÞ ¼ ðf 0 ðða; bÞÞÞðð0; 1; 0; 0; 0ÞÞ ¼ ððD2 f1 Þðða; bÞÞ; ðD2 f2 Þðða; bÞÞ; ðD2 f3 Þðða; bÞÞÞ; and ðf 0 ðða; bÞÞÞx ðð0; 0; 1ÞÞ ¼ ðf 0 ðða; bÞÞÞðð0; 0; 1; 0; 0ÞÞ ¼ ððD3 f1 Þðða; bÞÞ; ðD3 f2 Þðða; bÞÞ; ðD3 f3 Þðða; bÞÞÞ;

3 7 7 7 7 7 5 55

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3 Multivariable Differential Calculus

so,

ðf 0 ðða; bÞÞÞx

2

3 ðD1 f1 Þðða; bÞÞ ðD2 f1 Þðða; bÞÞ ðD3 f1 Þðða; bÞÞ ¼ 4 ðD1 f2 Þðða; bÞÞ ðD2 f2 Þðða; bÞÞ ðD3 f2 Þðða; bÞÞ 5: ðD1 f3 Þðða; bÞÞ ðD2 f3 Þðða; bÞÞ ðD3 f3 Þðða; bÞÞ

Next, since ðf 0 ðða; bÞÞÞx is invertible, 2 ðD1 f1 Þðða; bÞÞ 0 0 6¼ det ðf ðða; bÞÞÞx ¼ det4 ðD1 f2 Þðða; bÞÞ ðD1 f3 Þðða; bÞÞ ¼ det½F 0 ðða; bÞÞ:

ðD2 f1 Þðða; bÞÞ ðD2 f2 Þðða; bÞÞ ðD2 f3 Þðða; bÞÞ

Since det½F 0 ðða; bÞÞ 6¼ 0; so F 0 ðða; bÞÞ is invertible. 2 ðD1 f1 Þððx; yÞÞ ðD2 f1 Þððx; yÞÞ ðx; yÞ 7! det4 ðD1 f2 Þððx; yÞÞ ðD2 f2 Þððx; yÞÞ ðD1 f3 Þððx; yÞÞ ðD2 f3 Þððx; yÞÞ

3 ðD3 f1 Þðða; bÞÞ ðD3 f2 Þðða; bÞÞ 5 ðD3 f3 Þðða; bÞÞ

Since f is a C1 function, 3 ðD3 f1 Þððx; yÞÞ ðD3 f2 Þððx; yÞÞ 5 ðD3 f3 Þððx; yÞÞ

is continuous. Further, since det½ðf 0 ðða; bÞÞÞx is nonzero, there exists a neighborhood U of ða; bÞ such that det½ðf 0 ððx; yÞÞÞx is nonzero for every ðx; yÞ in U : It follows that ðf 0 ððx; yÞÞÞx is invertible for every ðx; yÞ in U : Now, by the inverse function theorem and Theorem 3.38, there exists an open neighborhood U of ða; bÞ such that (1) (2) (3) (4) (5)

U is contained in E, F is 1–1 on U, FðUÞ is open in R3þ2 ; the 1–1 function F 1 from FðUÞ onto U is a C 1 function, U is contained in U :

Put

W y : y 2 R2 and for some x in R3 ðx ; yÞ 2 U; f ððx ; yÞÞ ¼ 0 :

Since ða; bÞ 2 U; and f ðða; bÞÞ ¼ 0; b 2 W: Let us observe that W ¼ y : y 2 R2 and for some x in R3 ; ðx ; yÞ 2 U; f ððx ; yÞÞ ¼ 0 ¼ y : y 2 R2 and ð0; yÞ ¼ ðf ðx ; y Þ; y Þ for some ðx ; y Þ in U ¼ y : y 2 R2 and ð0; yÞ ¼ F ððx ; y ÞÞ for some ðx ; y Þ in U ¼ y : y 2 R2 and ð0; yÞ 2 FðUÞ : Thus,

W ¼ y : y 2 R2 and ð0; yÞ 2 FðUÞ :

Now, we will show that W is an open subset of R2 : For this purpose, let us take any y0 2 W: Since y0 2 W, ð0; y0 Þ 2 FðUÞ: Since ð0; y0 Þ 2 FðUÞ; and FðUÞ is

3.4 Implicit Function Theorem

185

open in R3þ2 ; there exist open neighborhood V1 of 0 in R3 and open neighborhood V2 of y0 in R2 such that V1 V2 FðUÞ. Since 0 2 V1 ; and V1 V2 FðUÞ; f0g V2 V1 V2 FðUÞ; and hence, ð0; yÞ 2 FðUÞ for every y in V2 : It follows that V2 W: Since V2 W; and V2 is an open neighborhood of y0 ; y0 is an interior point of W. Hence, W is an open subset of R2 : Since W is an open subset of R2 ; and b 2 W; W is an open neighborhood of b in R2 : Now, we will try to prove that if y 2 W; then there exists a unique x in R such that ðx ; yÞ 2 U; f ððx ; yÞÞ ¼ 0: For this purpose, let us take any y 2 W: The existence of x is clear from the deﬁnition of W. Uniqueness: If not, otherwise, let there exist x ; x in R3 such that ðx ; yÞ 2 U; f ððx ; yÞÞ ¼ 0; ðx ; yÞ 2 U; f ððx ; yÞÞ ¼ 0; x 6¼ x : We have to arrive at a contradiction. Here, x 6¼ x ; so ðx ; yÞ 6¼ ðx ; yÞ: Since ðx ; yÞ 6¼ ðx ; yÞ; ðx ; yÞ 2 U; ðx ; yÞ 2 U; and F is 1–1 on U, ð0; yÞ ¼ ðf ðx ; yÞ; yÞ ¼ Fððx ; yÞÞ 6¼ Fððx ; yÞÞ ¼ ðf ððx ; yÞÞ; yÞ ¼ ð0; yÞ: Thus, ð0; yÞ 6¼ ð0; yÞ; which is a contradiction. Thus, we have shown that if y 2 W, then there exists a unique x in R3 such that ðx ; yÞ 2 U; f ððx ; yÞÞ ¼ 0: Let us denote x by gðyÞ: Thus, g : W ! R3 such that for every y in W, ðgðyÞ; yÞ 2 U; f ððgðyÞ; yÞÞ ¼ 0: This proves 1, 2. Further, since b 2 W; ðgðbÞ; bÞ 2 U; and f ðða; bÞÞ ¼ 0 ¼ f ððgðbÞ; bÞÞ: Since f ðða; bÞÞ ¼ f ððgðbÞ; bÞÞ; Fðða; bÞÞ ¼ ðf ðða; bÞÞ; bÞ ¼ ðf ððgðbÞ; bÞÞ; bÞ ¼ FððgðbÞ; bÞÞ: Since Fðða; bÞÞ ¼ FððgðbÞ; bÞÞ; ða; bÞ 2 U; ðgðbÞ; bÞ 2 U; and F is 1–1 on U, ða; bÞ ¼ ðgðbÞ; bÞ; and hence, gðbÞ ¼ a: This proves 3. 4: Let g1 : W ! R; g2 : W ! R; g3 : W ! R be the component functions of g, that is, gððy1 ; y2 ÞÞ ¼ ðg1 ððy1 ; y2 ÞÞ; g2 ððy1 ; y2 ÞÞ; g3 ððy1 ; y2 ÞÞÞ for every ðy1 ; y2 Þ in Wð R2 Þ: Let h1 : FðUÞ ! R; h2 : FðUÞ ! R; h3 : FðUÞ ! R; h4 : FðUÞ ! R; h5 : FðUÞ !R be the component functions of F 1 ; that is, 1

F ððx; yÞÞ ¼ ðh1 ððx; yÞÞ; h2 ððx; yÞÞ; h3 ððx; yÞÞ; h4 ððx; yÞÞ; h5 ððx; yÞÞÞ for every ðx; yÞ in FðUÞð R3þ2 Þ: Since the function F 1 from FðUÞ onto U is C 1 , each of the 25 ﬁrst-order partial derivatives Dj hi : FðUÞ ! R exists and is continuous. We have to prove that g is a

186

3 Multivariable Differential Calculus

C 1 function, that is, each of the six ﬁrst-order partial derivatives Dj gi : W ! R exists and is continuous. Here, we will try to prove that D2 g1 : W ! R exists. For this purpose, let us ﬁx any y ðy1 ; y2 Þ in W. We will try to prove that g ððy ;y þtÞÞg ððy ;y ÞÞ

ðD2 g1 Þððy1 ; y2 ÞÞ exists, that is, limt!0 1 1 2 t 1 1 2 exists. Since y is in W, ðgðy Þ; y Þ 2 U; 0 ¼ f ððgðy Þ; y ÞÞ; and hence, ð0; y Þ ¼ ðf ððgðy Þ; y ÞÞ; y Þ ¼ F ððgðy Þ; y ÞÞ 2 FðUÞ: Since ð0; y Þ 2 FðUÞ; and D5 h1 : FðUÞ ! R exists, ðD5 h1 Þðð0; y ÞÞ exists. Since F 1 is a 1–1 function from FðUÞ onto U, and for every ðx; yÞ in U, where x ðx1 ; x2 ; x3 Þ is in R3 and y ðy1 ; y2 Þ is in R2 ; Fððx; yÞÞ ¼ ðf ððx; yÞÞ; yÞ;

ðx; yÞ ¼ F 1 ððf ððx; yÞÞ; yÞÞ: Further, since for every y ðy1 ; y2 Þ in W, we have ðgðyÞ; yÞ 2 U; 0 ¼ f ððgðyÞ; yÞÞ; ðg1 ððy1 ; y2 ÞÞ; g2 ððy1 ; y2 ÞÞ; g3 ððy1 ; y2 ÞÞ; y1 ; y2 Þ ¼ ððg1 ððy1 ; y2 ÞÞ; g2 ððy1 ; y2 ÞÞ; g3 ððy1 ; y2 ÞÞÞ; ðy1 ; y2 ÞÞ

¼ ðgðyÞ; yÞ ¼ F 1 ððf ððgðyÞ; yÞÞ; yÞÞ ¼ F 1 ðð0; yÞÞ ¼ ðh1 ðð0; yÞÞ; h2 ðð0; yÞÞ; h3 ðð0; yÞÞ; h4 ðð0; yÞÞ; h5 ðð0; yÞÞÞ: Hence, for every y ðy1 ; y2 Þ in W, g1 ððy1 ; y2 ÞÞ ¼ h1 ðð0; yÞÞ ¼ h1 ðð0; 0; 0; y1 ; y2 ÞÞ: Now,

g1 y1 ; y2 þ t g1 y1 ; y2 h1 0; 0; 0; y1 ; y2 þ t h1 0; 0; 0; y1 ; y2 ¼ lim lim t!0 t!0 t

t

¼ ðD5 h1 Þðð0; y ÞÞ: ¼ ðD5 h1 Þ ð0; 0; 0Þ; y1 ; y2 This proves that D2 g1 : W ! R exists. We have seen that for every y in W,

g1 y1 ; y2 þ t g1 y1 ; y2 ¼ ðD5 h1 Þðð0; y ÞÞ: ðD2 g1 Þððy ÞÞ ¼ lim t!0 t Further, since y 7! ð0; y Þ and ðx; yÞ 7! ðD5 h1 Þððx; yÞÞ are continuous functions, their composite function y 7! ðD5 h1 Þðð0; y ÞÞð¼ ðD2 g1 Þððy ÞÞÞ is continuous. Thus, we have shown that D2 g1 : W ! R exists and is continuous. Similarly, all other Dj gi : W ! R exist and are continuous.

3.4 Implicit Function Theorem

187

This proves 4. 5: Let us ﬁx any y ðy1 ; y2 Þ in W. Since y ¼ ðy1 ; y2 Þ is in W, ðgðy Þ; y Þ 2 U; f ððgðy Þ; y ÞÞ ¼ 0: Let us deﬁne a function hy : W ! R3þ2 as follows: For every y ðy1 ; y2 Þ in W, hy ðyÞ ðgðyÞ; yÞ ¼ ðg1 ððy1 ; y2 ÞÞ; g2 ððy1 ; y2 ÞÞ; g3 ððy1 ; y2 ÞÞ; y1 ; y2 Þ: If y is in W, then ðgðyÞ; yÞ 2 U; f ððgðyÞ; yÞÞ ¼ 0: Hence, hy : W ! Uð EÞ: Here, component functions of hy are y 7! g1 ðyÞ; y 7! g2 ðyÞ; y 7! g3 ðyÞ; y 7! y1 ; y 7! y2 : Since g is a C 1 function, y 7! g1 ðyÞ; y 7! g2 ðyÞ; y 7! g3 ðyÞ are continuously differentiable functions. Further, y 7! y1 ; y 7! y2 are smooth functions. Thus, we see that all component functions of hy are continuously differentiable functions, and hence, hy is a C 1 function on W. Since hy is a C 1 function on W, and y is in W, hy is differentiable at y : Since f is a C 1 function on E, and hy ðy Þ ¼ ðgðy Þ; y Þ 2 U E; f is differentiable at hy ðy Þ: For every y in W, we have ðgðyÞ; yÞ 2 U; and

f hy ðyÞ ¼ f ððgðyÞ; yÞÞ ¼ 0: Let us take any y in W. Since hy is a C 1 function on W, and y is in W, hy is differentiable at y. Since y is in W, hy : W ! Uð EÞ; hy ðyÞ is in E. Since hy ðyÞ is in E, and f is a C 1 function, f is differentiable at hy ðyÞ: Now, since y is in W, f is differentiable at hy ðyÞ; and hy is differentiable at y, by the chain rule of derivative, ðf 0 ððgðyÞ; yÞÞÞ

0

0 hy ðyÞ ¼ f 0 hy ðyÞ hy ðyÞ ¼ 0:

Hence, 0 ð0; 0; 0Þ ¼ 0ðð1; 0ÞÞ ¼ ðf 0 ððgðyÞ; yÞÞÞ hy ðyÞ ðð1; 0ÞÞ 0 ¼ ðf 0 ððgðyÞ; yÞÞÞ hy ðyÞ ðð1; 0ÞÞ ¼ ðf 0 ððgðyÞ; yÞÞÞðððD1 g1 ÞðyÞ; ðD1 g2 ÞðyÞ; ðD1 g3 ÞðyÞ; 1; 0ÞÞ ¼ ððD1 g1 ÞðyÞÞððf 0 ððgðyÞ; yÞÞÞðð1; 0; 0; 0; 0ÞÞÞ þ ððD1 g2 ÞðyÞÞððf 0 ððgðyÞ; yÞÞÞðð0; 1; 0; 0; 0ÞÞÞ þ ððD1 g3 ÞðyÞÞððf 0 ððgðyÞ; yÞÞÞðð0; 0; 1; 0; 0ÞÞÞ þ 1ðf 0 ððgðyÞ; yÞÞÞðð0; 0; 0; 1; 0ÞÞ þ 0 ¼ ððD1 g1 ÞðyÞÞððD1 f1 ÞððgðyÞ; yÞÞ; ðD1 f2 ÞððgðyÞ; yÞÞ; ðD1 f3 ÞððgðyÞ; yÞÞÞ þ ððD1 g2 ÞðyÞÞððD2 f1 ÞððgðyÞ; yÞÞ; ðD2 f2 ÞððgðyÞ; yÞÞ; ðD2 f3 ÞððgðyÞ; yÞÞÞ þ ððD1 g3 ÞðyÞÞððD3 f1 ÞððgðyÞ; yÞÞ; ðD3 f2 ÞððgðyÞ; yÞÞ; ðD3 f3 ÞððgðyÞ; yÞÞÞ þ ððD4 f1 ÞððgðyÞ; yÞÞ; ðD4 f2 ÞððgðyÞ; yÞÞ; ðD4 f3 ÞððgðyÞ; yÞÞÞ

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3 Multivariable Differential Calculus

or, ð0; 0; 0Þ ¼ ððD1 g1 ÞðyÞÞððD1 f1 ÞððgðyÞ; yÞÞ; ðD1 f2 ÞððgðyÞ; yÞÞ; ðD1 f3 ÞððgðyÞ; yÞÞÞ þ ððD1 g2 ÞðyÞÞððD2 f1 ÞððgðyÞ; yÞÞ; ðD2 f2 ÞððgðyÞ; yÞÞ; ðD2 f3 ÞððgðyÞ; yÞÞÞ þ ððD1 g3 ÞðyÞÞððD3 f1 ÞððgðyÞ; yÞÞ; ðD3 f2 ÞððgðyÞ; yÞÞ; ðD3 f3 ÞððgðyÞ; yÞÞÞ þ ððD4 f1 ÞððgðyÞ; yÞÞ; ðD4 f2 ÞððgðyÞ; yÞÞ; ðD4 f3 ÞððgðyÞ; yÞÞÞ: So, for every i ¼ 1; 2; 3; and for every y in W, 0 ¼ ððD1 g1 ÞðyÞÞððD1 fi ÞððgðyÞ; yÞÞÞ þ ððD1 g2 ÞðyÞÞððD2 fi ÞððgðyÞ; yÞÞÞ þ ððD1 g3 ÞðyÞÞððD3 fi ÞððgðyÞ; yÞÞÞ þ ðD4 fi ÞððgðyÞ; yÞÞ: Now, since y is in W, for every i ¼ 1; 2; 3; 0 ¼ ððD1 g1 Þðy ÞÞððD1 fi Þððgðy Þ; y ÞÞÞ þ ððD1 g2 Þðy ÞÞððD2 fi Þððgðy Þ; y ÞÞÞ þ ððD1 g3 Þðy ÞÞððD3 fi Þððgðy Þ; y ÞÞÞ þ ðD4 fi Þððgðy Þ; y ÞÞ ðÞ: Next, we will prove that

1 g0 ðy Þ ¼ ðf 0 ððgðy Þ; y ÞÞÞx ðf 0 ððgðy Þ; y ÞÞÞy that is,

ðf 0 ððgðy Þ; y ÞÞÞx ðg0 ðy ÞÞ ¼ ðf 0 ððgðy Þ; y ÞÞÞy

that is, (i) ðððf 0 ððgðy Þ; y ÞÞÞx Þðg0 ðy ÞÞÞðð1; 0ÞÞ ¼ ððf 0 ððgðy Þ; y ÞÞÞy Þðð1; 0ÞÞ; and (ii) ðððf 0 ððgðy Þ; y ÞÞÞx Þðg0 ðy ÞÞÞðð0; 1ÞÞ ¼ ððf 0 ððgðy Þ; y ÞÞÞy Þðð0; 1ÞÞ: For (i):

0

ðf ððgðy Þ; y ÞÞÞx ðg0 ðy ÞÞ ðð1; 0ÞÞ ¼ ðf 0 ððgðy Þ; y ÞÞÞx ððg0 ðy ÞÞðð1; 0ÞÞÞ

¼ ðf 0 ððgðy Þ; y ÞÞÞx ðððD1 g1 Þðy Þ; ðD1 g2 Þðy Þ; ðD1 g3 Þðy ÞÞÞ ¼ ðf 0 ððgðy Þ; y ÞÞÞðððD1 g1 Þðy Þ; ðD1 g2 Þðy Þ; ðD1 g3 Þðy Þ; 0; 0ÞÞ ¼ ððD1 g1 Þðy ÞÞððD1 f1 Þððgðy Þ; y ÞÞ; ðD1 f2 Þððgðy Þ; y ÞÞ; ðD1 f3 Þððgðy Þ; y ÞÞÞ þ ððD1 g2 Þðy ÞÞððD2 f1 Þððgðy Þ; y ÞÞ; ðD2 f2 Þððgðy Þ; y ÞÞ; ðD2 f3 Þððgðy Þ; y ÞÞÞ

LHS ¼

þ ððD1 g3 Þðy ÞÞððD3 f1 Þððgðy Þ; y ÞÞ; ðD3 f2 Þððgðy Þ; y ÞÞ; ðD3 f3 Þððgðy Þ; y ÞÞÞ ¼ ððD4 f1 Þððgðy Þ; y ÞÞ; ðD4 f2 Þððgðy Þ; y ÞÞ; ðD4 f3 Þððgðy Þ; y ÞÞÞ;

3.4 Implicit Function Theorem

189

by ðÞ: Now, RHS ¼ ðf 0 ððgðy Þ; y ÞÞÞy ðð1; 0ÞÞ ¼ ðf 0 ððgðy Þ; y ÞÞÞy ðð1; 0ÞÞ ¼ ððf 0 ððgðy Þ; y ÞÞÞðð0; 0; 01; 0ÞÞÞ ¼ ððD4 f1 Þððgðy Þ; y ÞÞ; ðD4 f2 Þððgðy Þ; y ÞÞ; ðD4 f3 Þððgðy Þ; y ÞÞÞ ¼ ððD4 f1 Þððgðy Þ; y ÞÞ; ðD4 f2 Þððgðy Þ; y ÞÞ; ðD4 f3 Þððgðy Þ; y ÞÞÞ:

This proves (i). Similarly, (ii) holds. This proves 5. 6: Since b 2 W; and gðbÞ ¼ a; by 5,

1

1 g0 ðbÞ ¼ ðf 0 ððgðbÞ; bÞÞÞx ðf 0 ððgðbÞ; bÞÞÞy ¼ ðf 0 ðða; bÞÞÞx ðf 0 ðða; bÞÞÞy : This proves 6. 7: Let f be a C2 function. We have to prove that g : W ! R3 is a C 2 function, that is, each of the 12 second-order partial derivatives Djk gi : W ! R exists and is continuous. We will try to prove that D12 g3 : W ! R exists and is continuous. Let us take any y in W. By 5,

1 g0 ðyÞ ¼ ðf 0 ððgðyÞ; yÞÞÞx ðf 0 ððgðyÞ; yÞÞÞy so, ððD1 g1 ÞðyÞ; ðD1 g2 ÞðyÞ; ðD1 g3 ÞðyÞÞ ¼ ðg0 ðyÞÞðð1; 0ÞÞ

1 ¼ ðf 0 ððgðyÞ; yÞÞÞx ðf 0 ððgðyÞ; yÞÞÞy ðð1; 0ÞÞ

1 0 ðf ððgðyÞ; yÞÞÞy ðð1; 0ÞÞ ¼ ðf 0 ððgðyÞ; yÞÞÞx

1 0 ¼ ðf 0 ððgðyÞ; yÞÞÞx ððf ððgðyÞ; yÞÞÞðð0; 0; 0; 1; 0ÞÞÞ

1 ððD4 f1 ÞððgðyÞ; yÞÞ; ðD4 f2 ÞððgðyÞ; yÞÞ; ðD4 f3 ÞððgðyÞ; yÞÞÞ ¼ ðf 0 ððgðyÞ; yÞÞÞx

1 ð1; 0; 0Þ ðf 0 ððgðyÞ; yÞÞÞx ¼ ððD4 f1 ÞððgðyÞ; yÞÞÞ

1 ð0; 1; 0Þ ððD4 f2 ÞððgðyÞ; yÞÞÞ ðf 0 ððgðyÞ; yÞÞÞx

1 ððD4 f3 ÞððgðyÞ; yÞÞÞ ðf 0 ððgðyÞ; yÞÞÞx ð0; 0; 1Þ 1 1 ðA11 ; A12 ; A13 Þ ððD4 f2 ÞððgðyÞ; yÞÞÞ ðA21 ; A22 ; A23 Þ D D 1 ððD4 f3 ÞððgðyÞ; yÞÞÞ ðA31 ; A32 ; A33 Þ; D

¼ ððD4 f1 ÞððgðyÞ; yÞÞÞ

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where ðD1 f1 ÞððgðyÞ; yÞÞ D ðD1 f2 ÞððgðyÞ; yÞÞ ðD1 f3 ÞððgðyÞ; yÞÞ

ðD2 f1 ÞððgðyÞ; yÞÞ ðD2 f2 ÞððgðyÞ; yÞÞ ðD2 f3 ÞððgðyÞ; yÞÞ

ðD3 f1 ÞððgðyÞ; yÞÞ ðD3 f2 ÞððgðyÞ; yÞÞ ; ðD3 f3 ÞððgðyÞ; yÞÞ

and Aij denotes the cofactor of ðDj fi ÞððgðyÞ; yÞÞ in the above determinant. Hence, ðD1 g3 ÞðyÞ ¼

1 ðððD4 f1 ÞððgðyÞ; yÞÞÞA13 þ ððD4 f2 ÞððgðyÞ; yÞÞÞA23 þ ððD4 f3 ÞððgðyÞ; yÞÞÞA33 Þ: D

Now, since f is a C 2 function, and g is a C 1 function, the second-order partial derivative ðD12 g3 ÞðyÞ exists and is continuous. Similarly, all other ðDjk gi ÞðyÞ exist and are continuous. This proves that g is a C 2 function. h Note 3.44 The result similar to Theorem 3.43 can be proved as above for Rn in place of R3 and Rm in place of R2 : This theorem is known as the implicit function theorem. As above, we can prove the following theorem: Theorem 3.45 Let E be an open subset of R3þ2 : Let f : E ! R2 : Let a 2 R3 ; b 2 R2 ; and ða; bÞ 2 E: If (i) f is a C 1 function, (ii) ðf 0 ðða; bÞÞÞy is invertible, (iii) f ðða; bÞÞ ¼ 0; then there exist an open neighborhood Uð EÞ of ða; bÞ in R3þ2 ; an open neighborhood W of a in R3 ; and a function g : W ! R2 such that 1. 2. 3. 4. 5.

for every x in W, ðx; gðxÞÞ 2 U; for every x in W, f ðx; gðxÞÞ ¼ 0; gðaÞ ¼ b; g is a C 1 function, for every x in W, g0 ðx Þ ¼ ððf 0 ððx ; gðx ÞÞÞÞy Þ1 ðf 0 ððx ; gðx ÞÞÞÞx

6. g0 ðaÞ ¼ ððf 0 ðða; bÞÞÞy Þ1 ðf 0 ðða; bÞÞÞx 7. if f is a C 2 function, then g is a C 2 function, etc.

3.5 Constant Rank Theorem (Easy Version)

191

3.5 Constant Rank Theorem (Easy Version) Deﬁnition Let i1 : Rm ! Rm Rn be the function deﬁned as follows: For every x in Rm i1 ðxÞ ðx; 0Þ: Here, i1 is called the ﬁrst injection of factor into product. Similarly, by the second injection i2 of factor into product, we mean the function i2 : Rn ! Rm Rn deﬁned as follows: For every y in Rn i2 ðyÞ ð0; yÞ: Let p1 : Rm Rn ! Rm be the function deﬁned as follows: For every x in Rm and for every y in Rn p1 ðx; yÞ x: Here, p1 is called the ﬁrst projection from product onto factor. Similarly, by the second projection p2 from product onto factor, we mean the function p2 : Rm Rn ! Rn deﬁned as follows: For every x in Rm and for every y in Rn p2 ðx; yÞ y: Theorem 3.46 Let U be an open neighborhood of ða; bÞð ða1 ; a2 ; a3 ; b1 ; b2 ÞÞ in R3 R2 : Let f : U ! R2 be a smooth function. Let f ða1 ; a2 ; a3 ; b1 ; b2 Þ ¼ ð0; 0Þ: If ðf 0 ða1 ; a2 ; a3 ; b1 ; b2 ÞÞ i2 : R2 ! R2 is a linear isomorphism, then 1. there exists a local diffeomorphism u at ða1 ; a2 ; a3 ; b1 ; b2 Þ in R3 R2 such that f u ¼ p2 : (i.e., there exist an open neighborhood V of ða1 ; a2 ; a3 Þ in R3 ; an open neighborhood W of ðb1 ; b2 Þ in R2 ; an open neighborhood U1 of ða1 ; a2 ; a3 ; b1 ; b2 Þ in R3 R2 ; and a function u : U1 ! V W such that V W U; u is a diffeomorphism, and for every ðx1 ; x2 ; x3 ; y1 ; y2 Þ in U1 ; f ðuðx1 ; x2 ; x3 ; y1 ; y2 ÞÞ ¼ ðy1 ; y2 Þ:) 2. there exist an open neighborhood W1 of ða1 ; a2 ; a3 Þ in R3 ; an open neighborhood W2 of ðb1 ; b2 Þ in R2 , and a smooth function ðg1 ; g2 Þ g : W1 ! R2 such that (a) for every ðx1 ; x2 ; x3 Þ in W1 ; ðx1 ; x2 ; x3 ; g1 ðx1 ; x2 ; x3 Þ; g2 ðx1 ; x2 ; x3 ÞÞ 2 U; (b) for every ðx1 ; x2 ; x3 Þ in W1 ; f ðx1 ; x2 ; x3 ; g1 ðx1 ; x2 ; x3 Þ; g2 ðx1 ; x2 ; x3 ÞÞ ¼ ð0; 0Þ;

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(c) g is unique, (i.e., if ðh1 ; h2 Þ h : W1 ! R2 is a smooth function such that for every ðx1 ; x2 ; x3 Þ in W1 ; ðx1 ; x2 ; x3 ; h1 ðx1 ; x2 ; x3 Þ; h2 ðx1 ; x2 ; x3 ÞÞ 2 U; and f ðx1 ; x2 ; x3 ; h1 ðx1 ; x2 ; x3 Þ; h2 ðx1 ; x2 ; x3 ÞÞ ¼ ð0; 0Þ; then g ¼ h:) (d) if a : W1 ! f 1 ð0; 0Þ is the function deﬁned such that for every ðx1 ; x2 ; x3 Þ in W1 aðx1 ; x2 ; x3 Þ ðx1 ; x2 ; x3 ; g1 ðx1 ; x2 ; x3 Þ; g2 ðx1 ; x2 ; x3 ÞÞ; then (i) (ii) (iii) (iv)

aðW1 Þ is an open neighborhood of ða; bÞ in f 1 ð0; 0Þ; a is a homeomorphism from W1 onto aðW1 Þ; a ¼ u s1 and a1 ¼ p1 on their common domain, a; a1 are smooth functions.

Proof Let f ðf1 ; f2 Þ: Since ðf 0 ða1 ; a2 ; a3 ; b1 ; b2 ÞÞ i2 : R2 ! R2 is an isomorphism, and for every ðy1 ; y2 Þ in R2 ; ððf 0 ða1 ; a2 ; a3 ; b1 ; b2 ÞÞ i2 Þðy1 ; y2 Þ ¼ ðf 0 ða1 ; a2 ; a3 ; b1 ; b2 ÞÞði2 ðy1 ; y2 ÞÞ ¼ ðf 0 ða1 ; a2 ; a3 ; b1 ; b2 ÞÞð0; 0; 0; y1 ; y2 Þ 2 3 0 6 7 6 0 7 ðD1 f1 Þða; bÞðD2 f1 Þða; bÞðD3 f1 Þða; bÞðD4 f1 Þða; bÞðD5 f1 Þða; bÞ 6 7 60 7 ¼ 7 ðD1 f2 Þða; bÞðD2 f2 Þða; bÞðD3 f2 Þða; bÞðD4 f2 Þða; bÞðD5 f2 Þða; bÞ 6 6 7 4 y1 5 y2 ððD4 f1 Þða; bÞÞy1 þ ððD5 f1 Þða; bÞÞy2 ðD4 f1 Þða; bÞðD5 f1 Þða; bÞ y1 ¼ ¼ ; ððD4 f2 Þða; bÞÞy1 þ ððD5 f2 Þða; bÞÞy2 ðD4 f2 Þða; bÞðD5 f2 Þða; bÞ y2 it follows that

ðD4 f1 Þða; bÞ ðD5 f1 Þða; bÞ 6 0: ¼ det ðD4 f2 Þða; bÞ ðD5 f2 Þða; bÞ Let us deﬁne a function w : U ! R3 R2 as follows: For every ðx; yÞ in U where x ðx1 ; x2 ; x3 Þ is in R3 and y ðy1 ; y2 Þ is in R2 ; wðx1 ; x2 ; x3 ; y1 ; y2 Þ ¼ wðx; yÞ ðx; f ðx; yÞÞ ¼ ðx1 ; x2 ; x3 ; f1 ðx1 ; x2 ; x3 ; y1 ; y2 Þ; f2 ðx1 ; x2 ; x3 ; y1 ; y2 ÞÞ:

3.5 Constant Rank Theorem (Easy Version)

193

Now, since f : U ! R2 is a smooth function, w : U ! R3 R2 is also a smooth function. Further, since 2

1000

0

3

7 6 0 7 60 1 0 0 7 6 7 6 detðw ða1 ; a2 ; a3 ; b1 ; b2 ÞÞ ¼ det6 0 0 1 0 0 7 7 6 4 0 0 0 ðD4 f1 Þða; bÞ ðD5 f1 Þða; bÞ 5 0

0 0 0 ðD4 f2 Þða; bÞ ðD5 f2 Þða; bÞ ðD4 f1 Þða; bÞ ðD5 f1 Þða; bÞ ¼ det ðD4 f2 Þða; bÞ ðD5 f2 Þða; bÞ

is nonzero, w0 ða1 ; a2 ; a3 ; b1 ; b2 Þ is invertible. Hence, by the inverse function theorem, there exist an open neighborhood V of ða1 ; a2 ; a3 Þ in R3 ; an open neighborhood W of ðb1 ; b2 Þ in R2 ; and an open subset U1 of R3 R2 such that V W U and w acts as a diffeomorphism from V W onto U1 : Hence, w1 : U1 ! V W is a diffeomorphism. Put u w1 : Clearly, u : U1 ! V W is a diffeomorphism. Now, for every x in V, and for every y in W, we have ðp2 wÞðx; yÞ ¼ p2 ðwðx; yÞÞ ¼ p2 ððx; f ðx; yÞÞÞ ¼ f ðx; yÞ: So, p2 w ¼ f ; and hence, f u ¼ ðp2 wÞ u ¼ p2 ðw uÞ ¼ p2 ðw w1 Þ ¼ p2 : This proves 1. Since ðf 0 ða1 ; a2 ; a3 ; b1 ; b2 ÞÞ i2 : R2 ! R2 is an isomorphism, h i ðD4 f1 Þða; bÞ ðD5 f1 Þða; bÞ 0 det ðf ða1 ; a2 ; a3 ; b1 ; b2 ÞÞy ¼ det 6¼ 0; ðD4 f2 Þða; bÞ ðD5 f2 Þða; bÞ and hence, ðf 0 ða1 ; a2 ; a3 ; b1 ; b2 ÞÞy is invertible. Now, by the implicit function theorem, there exist an open neighborhood W1 of ða1 ; a2 ; a3 Þ in R3 ; an open neighborhood W2 of ðb1 ; b2 Þ in R2 , and a smooth function ðg1 ; g2 Þ g : W1 ! R2 such that W1 W2 U and for every ðx1 ; x2 ; x3 Þ in W1 ; ðx1 ; x2 ; x3 ; g1 ðx1 ; x2 ; x3 Þ; g2 ðx1 ; x2 ; x3 ÞÞ 2 W1 W2 ; and f ðx1 ; x2 ; x3 ; g1 ðx1 ; x2 ; x3 Þ; g2 ðx1 ; x2 ; x3 ÞÞ ¼ ð0; 0Þ: This proves 2(a) and 2(b). Next, let h : W1 ! R2 be a smooth function such that for every x ðx1 ; x2 ; x3 Þ in W1 ; ðx1 ; x2 ; x3 ; h1 ðx1 ; x2 ; x3 Þ; h2 ðx1 ; x2 ; x3 ÞÞ 2 U; and f ðx1 ; x2 ; x3 ; h1 ðx1 ; x2 ; x3 Þ;

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h2 ðx1 ; x2 ; x3 ÞÞ ¼ ð0; 0Þ; where h ðh1 ; h2 Þ: We have to prove that g ¼ h; that is, for every ðx1 ; x2 ; x3 Þ in W1 ; gðx1 ; x2 ; x3 Þ ¼ hðx1 ; x2 ; x3 Þ: Since ðf1 ðx1 ; x2 ; x3 ; g1 ðx1 ; x2 ; x3 Þ; g2 ðx1 ; x2 ; x3 ÞÞ; f2 ðx1 ; x2 ; x3 ; g1 ðx1 ; x2 ; x3 Þ; g2 ðx1 ; x2 ; x3 ÞÞÞ ¼ f ðx1 ; x2 ; x3 ; g1 ðx1 ; x2 ; x3 Þ; g2 ðx1 ; x2 ; x3 ÞÞ ¼ ð0; 0Þ so, wðx1 ; x2 ; x3 ; g1 ðx1 ; x2 ; x3 Þ; g2 ðx1 ; x2 ; x3 ÞÞ ¼ ðx1 ; x2 ; x3 ; f1 ðx1 ; x2 ; x3 ; g1 ðx1 ; x2 ; x3 Þ; g2 ðx1 ; x2 ; x3 ÞÞ; f2 ðx1 ; x2 ; x3 ; g1 ðx1 ; x2 ; x3 Þ; g2 ðx1 ; x2 ; x3 ÞÞÞ ¼ ðx1 ; x2 ; x3 ; 0; 0Þ; and hence, ðx1 ; x2 ; x3 ; g1 ðx1 ; x2 ; x3 Þ; g2 ðx1 ; x2 ; x3 ÞÞ ¼ w1 ðx1 ; x2 ; x3 ; 0; 0Þ ¼ uðx1 ; x2 ; x3 ; 0; 0Þ for every ðx1 ; x2 ; x3 Þ in W1 : Similarly, ðx1 ; x2 ; x3 ; h1 ðx1 ; x2 ; x3 Þ; h2 ðx1 ; x2 ; x3 ÞÞ ¼ uðx1 ; x2 ; x3 ; 0; 0Þ for every ðx1 ; x2 ; x3 Þ in W1 : It follows that for every x ðx1 ; x2 ; x3 Þ in W1 , ðx; gðxÞÞ ¼ ðx1 ; x2 ; x3 ; g1 ðx1 ; x2 ; x3 Þ; g2 ðx1 ; x2 ; x3 ÞÞ ¼ ðx1 ; x2 ; x3 ; h1 ðx1 ; x2 ; x3 Þ; h2 ðx1 ; x2 ; x3 ÞÞ ¼ ðx; hðxÞÞ: Hence, g ¼ h: This proves 2(c). (d) (i): For every ðx1 ; x2 ; x3 Þ in W1 ; f ðx1 ; x2 ; x3 ; g1 ðx1 ; x2 ; x3 Þ; g2 ðx1 ; x2 ; x3 ÞÞ ¼ ð0; 0Þ; so aðx1 ; x2 ; x3 Þ ¼ ðx1 ; x2 ; x3 ; g1 ðx1 ; x2 ; x3 Þ; g2 ðx1 ; x2 ; x3 ÞÞ 2 f 1 ð0; 0Þ: Hence, a : W1 ! f 1 ð0; 0Þ is a function. Since ða1 ; a2 ; a3 Þ is in W1 ; aða1 ; a2 ; a3 Þ ¼ ða1 ; a2 ; a3 ; g1 ða1 ; a2 ; a3 Þ; g2 ða1 ; a2 ; a3 ÞÞ ¼ uða1 ; a2 ; a3 ; 0; 0Þ:

3.5 Constant Rank Theorem (Easy Version)

195

Further, since wða1 ; a2 ; a3 ; b1 ; b2 Þ ¼ ða1 ; a2 ; a3 ; f1 ða1 ; a2 ; a3 ; b1 ; b2 Þ; f2 ða1 ; a2 ; a3 ; b1 ; b2 ÞÞ ¼ ða1 ; a2 ; a3 ; 0; 0Þ so ða; bÞ ¼ ða1 ; a2 ; a3 ; b1 ; b2 Þ ¼ uða1 ; a2 ; a3 ; 0; 0Þ ¼ aða1 ; a2 ; a3 Þ 2 aðW1 Þ: Now, we want to show that aðW1 Þ is an open set. Let us observe that

aðW1 Þ ¼ u W1 R2 \ f 1 ð0; 0Þ: (Reason: For this purpose, let us take any aðx1 ; x2 ; x3 Þ 2 LHS; where ðx1 ; x2 ; x3 Þ is in W1 : Here, aðx1 ; x2 ; x3 Þ ¼ ðx1 ; x2 ; x3 ; g1 ðx1 ; x2 ; x3 Þ; g2 ðx1 ; x2 ; x3 ÞÞ ¼ uðx1 ; x2 ; x3 ; 0; 0Þ

2 u W1 R 2 : Also, since f ðaðx1 ; x2 ; x3 ÞÞ ¼ f ðx1 ; x2 ; x3 ; g1 ðx1 ; x2 ; x3 Þ; g2 ðx1 ; x2 ; x3 ÞÞ ¼ ð0; 0Þ; aðx1 ; x2 ; x3 Þ 2 f 1 ð0; 0Þ: It follows that aðx1 ; x2 ; x3 Þ 2 uðW1 R2 Þ \ f 1 ð0; 0Þ ¼ RHS: Hence, LHS RHS: Next, let us take any ðx1 ; x2 ; x3 ; y1 ; y2 Þ 2 RHS ¼ uðW1 R2 Þ \ f 1 ð0; 0Þ: We have to prove that ðx1 ; x2 ; x3 ; y1 ; y2 Þ 2 aðW1 Þ. Since ðx1 ; x2 ; x3 ; y1 ; y2 Þ 2 uðW1 R2 Þ \ f 1 ð0; 0Þ; ðx1 ; x2 ; x3 ; y1 ; y2 Þ 2 U; ðx1 ; x2 ; x3 ; y1 ; y2 Þ 2 V W; ðf1 ðx1 ; x2 ; x3 ; y1 ; y2 Þ; f2 ðx1 ; x2 ; x3 ; y1 ; y2 ÞÞ ¼ f ðx1 ; x2 ; x3 ; y1 ; y2 Þ ¼ ð0; 0Þ; and there exists ðw1 ; w2 ; w3 ; t1 ; t2 Þ 2 W1 R2 such that uðw1 ; w2 ; w3 ; t1 ; t2 Þ ¼ ðx1 ; x2 ; x3 ; y1 ; y2 Þ: Thus, ðw1 ; w2 ; w3 ; t1 ; t2 Þ ¼ wðx1 ; x2 ; x3 ; y1 ; y2 Þ ¼ ðx1 ; x2 ; x3 ; f1 ðx1 ; x2 ; x3 ; y1 ; y2 Þ; f2 ðx1 ; x2 ; x3 ; y1 ; y2 ÞÞ ¼ ðx1 ; x2 ; x3 ; 0; 0Þ: Hence, ðx1 ; x2 ; x3 Þ ¼ ðw1 ; w2 ; w3 Þ 2 W1 :Also ðx1 ; x2 ; x3 ; y1 ; y2 Þ ¼ uðw1 ; w2 ; w3 ; t1 ; t2 Þ ¼ uðx1 ; x2 ; x3 ; 0; 0Þ ¼ ðx1 ; x2 ; x3 ; g1 ðx1 ; x2 ; x3 Þ; g2 ðx1 ; x2 ; x3 ÞÞ ¼ aðx1 ; x2 ; x3 Þ 2 aðW1 Þ: This proves that aðW1 Þ ¼ uðW1 R2 Þ \ f 1 ð0; 0Þ:)

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Since W1 is an open subset of R3 ; W1 R2 is an open subset of R3 R2 : Further, since u : U1 ! V W is a diffeomorphism, and W1 R2 is an open subset of R3 R2 ; uðW1 R2 Þ is an open subset of R3 R2 ; and hence, aðW1 Þð¼ uðW1 R2 Þ \ f 1 ð0; 0ÞÞ is open in f 1 ð0; 0Þ: This proves (d)(i). (d)(ii): Here, we must prove that (A) a : W1 ! aðW1 Þ is 1–1, (B) a is continuous, (C) a1 : aðW1 Þ ! W1 is continuous. For (A): Let aðx1 ; x2 ; x3 Þ ¼ aðy1 ; y2 ; y3 Þ: We have to prove that ðx1 ; x2 ; x3 Þ ¼ ðy1 ; y2 ; y3 Þ: Since ðx1 ; x2 ; x3 ; g1 ðx1 ; x2 ; x3 Þ; g2 ðx1 ; x2 ; x3 ÞÞ ¼ aðx1 ; x2 ; x3 Þ ¼ aðy1 ; y2 ; y3 Þ ¼ ðy1 ; y2 ; y3 ; g1 ðy1 ; y2 ; y3 Þ; g2 ðy1 ; y2 ; y3 ÞÞ; ðx1 ; x2 ; x3 Þ ¼ ðy1 ; y2 ; y3 Þ: For (B): Since g : W1 ! R2 is a smooth function, and a : ðx1 ; x2 ; x3 Þ 7! ðx1 ; x2 ; x3 ; g1 ðx1 ; x2 ; x3 Þ; g2 ðx1 ; x2 ; x3 ÞÞ; each of the component functions of a is continuous, and hence, a is continuous. For (C): Since for every x ðx1 ; x2 ; x3 Þ 2 W1 ; a1 ðx1 ; x2 ; x3 ; g1 ðx1 ; x2 ; x3 Þ; g2 ðx1 ; x2 ; x3 ÞÞ ¼ ðx1 ; x2 ; x3 Þ ¼ p1 ðx1 ; x2 ; x3 ; g1 ðx1 ; x2 ; x3 Þ; g2 ðx1 ; x2 ; x3 ÞÞ;

and p1 is continuous, a1 is continuous. This proves (d)(ii). (d)(iii) Since for every x ðx1 ; x2 ; x3 Þ 2 W1 ; aðx1 ; x2 ; x3 Þ ¼ uðx1 ; x2 ; x3 ; 0; 0Þ ¼ uðs1 ðx1 ; x2 ; x3 ÞÞ ¼ ðu s1 Þðx1 ; x2 ; x3 Þ; a ¼ u s1 : Next, since a1 ðx1 ; x2 ; x3 ; g1 ðx1 ; x2 ; x3 Þ; g2 ðx1 ; x2 ; x3 ÞÞ ¼ ðx1 ; x2 ; x3 Þ ¼ p1 ðx1 ; x2 ; x3 ; g1 ðx1 ; x2 ; x3 Þ; g2 ðx1 ; x2 ; x3 ÞÞ;

a1 ¼ p1 on their common domain. This proves (d)(iii). (iv) Since u and s1 are smooth functions, their composite u s1 ð¼ aÞ is smooth, h and hence, a is smooth. Since p1 ð¼ a1 Þ is smooth, a1 is smooth. Theorem 3.47 Let U be an open neighborhood of ða; bÞ in Rm Rp : Let f : U ! Rp be a smooth function. Let f ða; bÞ ¼ 0: If ðf 0 ða; bÞÞ i2 : Rp ! Rp is a linear isomorphism, then

3.5 Constant Rank Theorem (Easy Version)

197

1. there exists a local diffeomorphism u at ða; bÞ in Rm Rp such that f u ¼ p2 : (i.e., there exist an open neighborhood V of a in Rm ; an open neighborhood W of b in Rp ; an open neighborhood U1 of ða; bÞ in Rm Rp ; and a function u : U1 ! V W such that V W U; u is a diffeomorphism, and for every ðx; yÞ in U1 ; f ðuðx; yÞÞ ¼ y:) 2. there exist an open neighborhood W1 of a in Rm ; an open neighborhood W2 of b in Rp , and a smooth function g : W1 ! Rp such that (a) for every x in W1 ; ðx; gðxÞÞ 2 U; (b) for every x in W1 ; f ðx; gðxÞÞ ¼ 0; (c) g is unique, (i.e., if h : W1 ! Rp is a smooth function such that for every x in W1 ; ðx; hðxÞÞ 2 U; and f ðx; hðxÞÞ ¼ 0; then g ¼ h:) (d) if a : W1 ! f 1 ð0Þ is the function deﬁned such that for every x in W1 aðxÞ ðx; gðxÞÞ; then (i) (ii) (iii) (iv)

aðW1 Þ is an open neighborhood of ða; bÞ in f 1 ð0Þ; a is a homeomorphism from W1 onto aðW1 Þ; a ¼ u s1 and a1 ¼ p1 on their common domain, a; a1 are smooth functions.

Proof Its proof is quite similar to the proof of Theorem 3.46.

h

Theorem 3.48 Let f : Rm Rp ! Rp be a smooth function. Let a be in Rm and b be in Rp : Let f ða; bÞ ¼ 0: If for every ðx; yÞ in f 1 ð0Þ; the rank of linear transformation f 0 ðx; yÞ is p, then f 1 ð0Þ is a smooth manifold of dimension m. Proof Since f : Rm Rp ! Rp is a smooth function, f is a continuous function. Since f : Rm Rp ! Rp is a continuous function, and {0} is a closed subset of Rp ; f 1 ð0Þ is closed subset of Rm Rp : Since Rm Rp is homeomorphic to Rmþp ; and Rmþp is Hausdorff and second countable space, Rm Rp is Hausdorff and second countable. Since Rm Rp is Hausdorff and second countable, and f 1 ð0Þ is a subspace of Rm Rp ; f 1 ð0Þ is Hausdorff and second countable. Let us take any ða; bÞ in f 1 ð0Þ: Since ða; bÞ is in f 1 ð0Þ; f ða; bÞ ¼ 0: By assumption, the rank of linear transformation f 0 ða; bÞ : Rm Rp ! Rp is p, and hence, ðf 0 ða; bÞÞ i2 : Rp ! Rp is a linear isomorphism. Now, by Theorem 3.47, there exist an open neighborhood W1 of a in Rm and a function a : W1 ! f 1 ð0Þ such that (i) aðW1 Þ is an open neighborhood of ða; bÞ in f 1 ð0Þ; (ii) a is a homeomorphism from W1 onto aðW1 Þ; (iii) a; a1 are smooth functions.

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Hence, aðW1 Þ is an open neighborhood of ða; bÞ in f 1 ð0Þ; and a1 : aðW1 Þ ! W1 is a diffeomorphism. This shows that f 1 ð0Þ is a smooth manifold of dimension m. h Deﬁnition Let N be a smooth manifold of dimension m þ p: Let M N: If for every x in M, there exists an admissible coordinate chart ðU; uU Þ in N satisfying x 2 U such that uU ðU \ M Þ ¼ ðuU ðUÞÞ \ ðRm f0gÞ then we say that M is a smooth m-submanifold of N. Theorem 3.49 Let N be a smooth manifold of dimension m þ p: Let M N: If M is a smooth m-submanifold of N, then M is a smooth manifold of dimension m. Proof Let us take any a in M. Since a is in M, and M is a smooth m-submanifold of N, there exists an admissible coordinate chart ðU; uU Þ in N satisfying a 2 U; and uU ðU \ M Þ ¼ ðuU ðUÞÞ \ ðRm f0gÞ: Since a is in M, and a 2 U; a is in U \ M. Since U is open in N, and M N; U \ M is open in M. Since a is in U \ M; and U \ M is open in M, U \ M is an open neighborhood of a in M. Since ðU; uU Þ is an admissible coordinate chart in the m þ p-dimensional smooth manifold N, uU ðUÞ is open in Rmþp ðﬃ ðRm Rp Þ ðRm f0gÞÞ; and hence, uU ðU \ MÞð¼ ðuU ðUÞÞ\ ðRm f0gÞÞ is open in Rm ðﬃ ðRm f0gÞÞ: Since ðU; uU Þ is an admissible coordinate chart in N, uU is 1–1, and hence, the restriction uU jU\M of uU to U \ M; is 1–1. Thus, the restriction uU jU\M of uU to U \ M is a 1–1 mapping from the open neighborhood U \ M of a in M onto the open set uU ðU \ MÞ of Rm : Since ðU; uU Þ is an admissible coordinate chart in N, uU is continuous, and hence, the restriction uU jU\M of uU to U \ M is continuous. Since ðU; uU Þ is an admissible coordinate chart in N, ðuU Þ1 : uU ðUÞ ! U is continuous, and hence, the restriction ððuU Þ1 ÞjuU ðU\MÞ of ðuU Þ1 to uU ðU \ MÞð¼ ðuU ðUÞÞ \ ðRm f0gÞÞ is continuous. Further, since ðuU jU\M Þ1 ¼ ððuU Þ1 ÞjuU ðU\MÞ ; it follows that the inverse

function ðuU jU\M Þ1 is continuous. Thus, the ordered pair ðU \ M; uU jU\M Þ is a coordinate chart of M satisfying a 2 M: Next, let ðU \ M; uU jU\M Þ and ðV \ M; wV jV\M Þ be two coordinate charts of M, where ðU; uU Þ and ðV; wV Þ are admissible coordinate charts in N, and ðU \ MÞ \ ðV \ MÞ is nonempty. We have to prove that ðwV jV\M Þ ðuU jU\M Þ1 is smooth. Since ðU; uU Þand ðV; wV Þ are admissible coordinate charts in N, wV ðuU Þ1 is smooth, and hence, ðwV jV\M Þ ðuU jU\M Þ1 ð¼ ðwV ðuU Þ1 Þ jðU\MÞ\ðV\MÞ Þ is smooth. Hence, M is a smooth manifold of dimension m. h

3.6 Smooth Bump Functions

199

3.6 Smooth Bump Functions Theorem 3.50 Let e [ 0: Let a be in R3 : Let Be ðaÞ fx : x 2 R3 ; jx aj\eg: Then, there exists a function h : R3 ! ½0; 1 such that 1. 2. 3. 4. 5.

h is smooth, h is onto, for every x in Be ðaÞ; hðxÞ [ 0; for every x in fx : x 2 R3 ; jx aj\ 2e g; hðxÞ ¼ 1; for every x 62 Be ðaÞ; hðxÞ ¼ 0:

Proof Let us deﬁne a function f : R ! R as follows: For every x in R;

1 ex if 0\x; f ðxÞ 0 if x 0: Thus, f ðxÞ is positive if and only if x is positive, and f ðxÞ is never negative. We shall ﬁrst try to prove by induction (a): For every x [ 0; and for every nonnegative 1 integer k, the kth derivative f ðkÞ ðxÞ is of the form ðp2k ð1xÞÞex for some polynomial 1 p2k ðyÞ of degree 2k in y. Let us denote the statement, f ðkÞ ðxÞ is of the form p2k ð1xÞex for some polynomial p2k ðyÞ of degree 2k in y, by PðkÞ: We must prove 1. P(1) holds, 2. if P(k) holds, then Pðk þ 1Þ holds. For 1: Pð1Þ means f 0 ðxÞ is of the form p2 ð1xÞex for some polynomial p2 ðyÞ of degree 2 in y. From the given deﬁnition of f, 1 1 1 1 f 0 ðxÞ ¼ 2 ex ¼ p2 e x; x x 1

where p2 ðyÞ y2 : Here, p2 ðyÞ is a polynomial of degree 2 in y. This proves 1. 1 For 2: Let PðkÞ be true, that is, f ðkÞ ðxÞ is of the form p2k ð1xÞex for some polynomial p2k ðyÞ of degree 2k in y. We have to prove that Pðk þ 1Þ is true, that is, f ðkþ1Þ ðxÞ is 1 of the form p2kþ2 ð1xÞex for some polynomial p2kþ2 ðyÞ of degree 2k þ 2 in y. Here, 0 d d 1 1 e x p2k f ðkþ1Þ ðxÞ ¼ f ðkÞ ðxÞ ¼ f ðkÞ ðxÞ ¼ dx dx x 1 1 1 1 1x 1 x þ p e ¼ p02k e 2k x x2 x x2 1 1 1 1 p02k e x 2 ¼ p2k x x x 1 1 e x; ¼ q2kþ2 x

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3 Multivariable Differential Calculus

where q2kþ2 ðyÞ ðp2k ðyÞ p02k ðyÞÞy2 ; which is a polynomial of degree 2k þ 2 in y. This proves 2. Hence, by the principle of mathematical induction, PðkÞ is true for every positive integer k. This proves (a). Now, we shall try to prove (b): f is C1 on R and that 1 f ðkÞ ð0Þ ¼ 0 for all k 0: Here, for x\0; f 0 ðxÞ ¼ 0; and for 0\x; f 0 ðxÞ ¼ ddx ex ¼

1 1x x2 e :

Next,

lim e!0 e[0

f ð0 eÞ f ð0Þ f ðeÞ 0 f ðeÞ 0 ¼ lim ¼ lim ¼ lim ¼ 0; e e e e!0 e!0 e!0 e e[0 e[0 e[0

and

lim e!0 e[0

f ð0 þ eÞ f ð0Þ f ðeÞ 0 ¼ lim e e e!0 e[0 f ðeÞ e e ¼ lim ¼ lim e!0 e e!0 e e[0 e[0

1

¼ lim

t

t!1 et

¼ lim

1

t!1 et

¼ 0:

Hence, lim

h!0

f ð0 þ hÞ f ð0Þ ¼ 0; h

that is, f 0 ð0Þ ¼ 0: Thus, f 0 ðxÞ ¼

0

1 1x x2 e

if if

x 0; 0\x:

Here, for x\0; f 00 ðxÞ ¼ 0; and for 0\x; f 00 ðxÞ ¼

d 0 d 1 1 2 1 1 1 1 2 1 1 x x þ x f ðxÞ ¼ e e e þ ¼ ¼ e x: dx dx x2 x3 x2 x2 x3 x4

3.6 Smooth Bump Functions

201

Now, lim e!0 e[0

f 0 ð0 eÞ f 0 ð0Þ f 0 ðeÞ 0 f 0 ðeÞ 0 ¼ lim ¼ lim ¼ lim e e e!0 e!0 e e!0 e e[0 e[0 e[0 ¼ 0;

and lim e!0 e[0

f 0 ð0 þ eÞ f 0 ð0Þ f 0 ðeÞ 0 f 0 ðeÞ ¼ lim ¼ lim ¼ lim e e e!0 e!0 e e!0 e[0 e[0 e[0 ¼ lim

t!1

1 1e e2 e

e

1 t3 3t2 ¼ lim 1 ¼ lim t ¼ lim 3 t!1 t!1 e et e ! 0 e ee e[0

3 2t 321 ¼ lim ¼ 0: t!1 et et

Hence, f 0 ð0 þ hÞ f 0 ð0Þ ¼ 0; h!0 h lim

that is, f 00 ð0Þ ¼ 0: Thus, f 00 ðxÞ ¼

02 x3

1 þ x14 ex

if if

x 0; 0\x:

Here, for x\0 f 000 ðxÞ ¼ 0 and for 0\x; d d 2 1 1 6 4 1 1 1 2 1 x x þ x f 000 ðxÞ ¼ f 00 ðxÞ ¼ þ þ e þ e ¼ e dx dx x3 x4 x4 x5 x2 x3 x4 6 6 1 1 ¼ 4 þ 5 þ 6 e x : x x x Now, lim e!0 e[0

f 00 ð0 eÞ f 00 ð0Þ f 00 ðeÞ 0 f 00 ðeÞ 0 ¼ lim ¼ lim ¼ lim e e e e!0 e!0 e!0 e e[0 e[0 e[0 ¼ 0;

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3 Multivariable Differential Calculus

and f 00 ð0 þ eÞ f 00 ð0Þ f 00 ðeÞ 0 f 00 ðeÞ ¼ lim ¼ lim ¼ lim lim e e e!0 e!0 e e!0 e!0 e[0 e[0 e[0 e[0 ¼ lim e!0 e[0

2 e3

1 þ e14 e e e

2 1 1 2t4 þ t5 5! þ ¼ lim t ¼ 0: 1 ¼ lim 4 5 t!1 e e e ee t!1 et

Hence, f 00 ð0 þ hÞ f 00 ð0Þ ¼ 0; h!0 h lim

that is, f 000 ð0Þ ¼ 0: Thus, 000

f ðxÞ ¼

0 6 x4

1 þ 6 þ x16 ex x5

if x 0; if 0\x:

Similarly, f ð4Þ ð0Þ ¼ 0; f 5 ð0Þ ¼ 0; etc. This completes the proof of (b). Thus, f : R ! R is a smooth function. Now, let us deﬁne a function g : R3 ! R as follows: For every x ðx1 ; x2 ; x3 Þ in R3 ; f e2 ðx1 Þ2 þðx2 Þ2 þðx3 Þ2 : gðxÞ f e2 ðx1 Þ2 þðx2 Þ2 þðx3 Þ2 þ f ðx1 Þ2 þðx2 Þ2 þðx3 Þ2 14 e2 Clearly, in the expression for gðxÞ, the denominator 1 2 2 2 2 2 2 þ f ð x1 Þ þ ð x2 Þ þ ð x3 Þ e 2 f e ð x1 Þ þ ð x2 Þ þ ð x3 Þ 4

2

is nonzero, and hence, g : R3 ! R is a well-deﬁned function. Since 0 f ; it is clear that 0 g 1: Further, since f is C1 on R; g is C 1 on R: Next, let us observe that if x ðx1 ; x2 ; x3 Þ satisﬁes e2 ðx1 Þ2 þðx2 Þ2 þðx3 Þ2 0; then f e2 ðx1 Þ2 þðx2 Þ2 þðx3 Þ2 ¼ 0;

3.6 Smooth Bump Functions

203

and hence, 1 f e2 ðx1 Þ2 þðx2 Þ2 þðx3 Þ2 A is 0: gðxÞ@¼ f e2 ðx1 Þ2 þðx2 Þ2 þðx3 Þ2 þ f ðx1 Þ2 þðx2 Þ2 þðx3 Þ2 14 e2 0

For every x ðx1 ; x2 ; x3 Þ satisfying

1 ðx1 Þ2 þðx2 Þ2 þðx3 Þ2 e2 0; 4

f

1 ðx1 Þ2 þðx2 Þ2 þðx3 Þ2 e2 ¼ 0; 4

and hence, f e2 ðx1 Þ2 þðx2 Þ2 þðx3 Þ2 gð x Þ @ ¼ f e2 ðx1 Þ2 þðx2 Þ2 þðx3 Þ2 þ f ðx1 Þ2 þðx2 Þ2 þðx3 Þ2 14 e2 1 f e 2 ð x1 Þ 2 þ ð x2 Þ 2 þ ð x3 Þ 2 A is 1: ¼ f e2 ðx1 Þ2 þðx2 Þ2 þðx3 Þ2 þ 0 0

Finally, for every x ðx1 ; x2 ; x3 Þ satisfying ðx1 Þ2 þ ðx2 Þ2 þ ðx3 Þ2 \e2 ; we have e ððx1 Þ2 þ ðx2 Þ2 þ ðx3 Þ2 Þ is positive, and hence, f ðe2 ððx1 Þ2 þ ðx2 Þ2 þ ðx3 Þ2 ÞÞ is positive. Thus, for every x ðx1 ; x2 ; x3 Þ satisfying ðx1 Þ2 þ ðx2 Þ2 þ ðx3 Þ2 \e2 ; 2

1 f e2 ðx1 Þ2 þðx2 Þ2 þðx3 Þ2 A gðxÞ@¼ f e2 ðx1 Þ2 þðx2 Þ2 þðx3 Þ2 þ f ðx1 Þ2 þðx2 Þ2 þðx3 Þ2 14 e2 0

is positive. Since g : R3 ! ½0; 1; and g is C 1 on R; g is continuous. Since g : R3 ! ½0; 1 is continuous, g assumes 0, and g assumes 1, g assumes all values of ½0; 1: Hence, g : R3 ! ½0; 1 is onto. Thus, we have seen that g is a smooth function from R3 onto ½0; 1 such that for every x in R3 ; 8 if j xj 2e ; <1 if e j xj; gðxÞ ¼ 0 : positive if j xj\e: Next, let us deﬁne a function h : R3 ! ½0; 1 as follows: For every x in R3 ; hðxÞ gðx aÞ:

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3 Multivariable Differential Calculus

1. Since g is a smooth function, and x 7! ðx aÞ is a smooth function, their composite h : x 7! gðx aÞ is also a smooth function. 2. Since x 7! ðx aÞ is a function from R3 onto R3 ; and g from R3 onto ½0; 1; their composite h : x 7! gðx aÞ is onto. 3. If x is in Be ðaÞ, then jx aj\e; and hence, hðxÞð¼ gðx aÞÞ is positive. 4. For every x in fx : x 2 R3 ; jx aj\ 2e g, we have jx aj\ 2e ; and therefore, hðxÞ ¼ gðx aÞ ¼ 1: 5. For every x 62 Be ðaÞ; we have e jx aj; and hence, hðxÞ ¼ gðx aÞ ¼ 0: h Note 3.51 We shall try to prove that there exists a smooth function h : R ! R such that 1. hðtÞ ¼ 1 if t 1; 2. 0\hðtÞ\1 if 1\t\2; 3. hðtÞ ¼ 0 if 2 t: In the proof of Theorem 3.50, we have seen that the function f : R ! R is deﬁned as follows: For every t in R;

1 e t if 0\t f ðtÞ 0 if t 0; is smooth. Now, let us deﬁne the function h : R ! R as follows: For every t in R; hðtÞ

f ð2 t Þ : f ð2 tÞ þ f ðt 1Þ

If t 1; then 1 t; and hence, 1 2 t: If t 1; then t 1 0; and hence, f ðt 1Þ ¼ 0: Thus, for t 1; e2t 1

hðtÞ ¼

e2t þ 0 1

¼ 1:

If 2 t; then 2 t 0; and hence, f ð2 tÞ ¼ 0: If 2 t; then 1 t 1: Thus, for 2 t; hðtÞ ¼

0 0 þ et1 1

¼ 0:

If 1\t\2; then 0\t 1; and hence, 0\et1 ¼ f ðt 1Þ: 1

3.6 Smooth Bump Functions

205

If 1\t\2; then 0\2 t; and hence, 0\e2t ¼ f ð2 tÞ: It follows that for 1\t\2; 1

e2t 1

hðtÞ ¼

e2t þ et1 1

1

2 ð0; 1Þ:

It remains to show that h is smooth. Since f is smooth, and t 7! 2 t is smooth, their composite t 7! f ð2 tÞ is smooth. Similarly, t 7! f ðt 1Þ is smooth. This shows that t 7!

f ð2 t Þ ð¼ hðtÞÞ f ð2 tÞ þ f ðt 1Þ h

is smooth, and hence, h is smooth. Here, we observe that 1. hðtÞ ¼ 1 if and only if t 1; 2. hðtÞ ¼ 0 if and only if 2 t; 3. 0\hðtÞ\1 if and only if 1\t\2:

Note 3.52 We shall try to prove that there exists a function H : R3 ! ½0; 1 such that 1. H is smooth, 2. HðxÞ ¼ 1; if x is in the closed ball B1 ½0ð¼ fx : x 2 R3 ; jxj 1gÞ; 3. the closure ðH 1 ðð0; 1ÞÞ of H 1 ðð0; 1Þ is equal to the closed ball B2 ½0ð¼ fx : x 2 R3 ; jxj 2gÞ: Let us deﬁne the function H : R3 ! ½0; 1 as follows: For every x ðx1 ; x2 ; x3 Þ in R3 ; qﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ HðxÞ h ðx1 Þ2 þðx2 Þ2 þðx3 Þ2 ; where h denotes the function as deﬁned in Note 3.51. For 1: Since x ðx1 ; x2 ; x3 Þ 7! ð

qﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ ðx1 Þ2 þ ðx2 Þ2 þ ðx3 Þ2 Þ is a smooth function

over R3 fð0; 0; 0Þg; and h : R ! ½0; 1 is a smooth function, their comqﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ posite x 7! hð ðx1 Þ2 þ ðx2 Þ2 þ ðx3 Þ2 Þð¼ HðxÞÞ is a smooth function from

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3 Multivariable Differential Calculus

R3 fð0; 0; 0Þg to ½0; 1: It remains to show that ðx1 ; x2 ; x3 Þ 7! h qﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ ð ðx1 Þ2 þ ðx2 Þ2 þ ðx3 Þ2 Þ is smooth at ð0; 0; 0Þ: Here, qﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ qﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ 2 2 2 h ð0 þ tÞ þð0Þ þ ð0Þ h ð0Þ2 þ ð0Þ2 þ ð0Þ2 lim t!0 t[0

t

¼ lim t!0 t[0

hðtÞ hð0Þ 11 ¼ lim ¼ lim 0 ¼ 0; t t!0 t t!0 t[0 t[0

and qﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ qﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ 2 2 2 h ð0 tÞ þð0Þ þ ð0Þ h ð0Þ2 þ ð0Þ2 þ ð0Þ2 lim t!0 t[0

t

¼ lim t!0 t[0

hðtÞ hð0Þ 11 ¼ lim ¼ lim 0 ¼ 0: t t ! 0 t t!0 t[0 t[0

Since h

qﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ qﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ ð0 þ tÞ2 þð0Þ2 þ ð0Þ2 h ð0Þ2 þ ð0Þ2 þ ð0Þ2

lim t!0 t[0

¼0

t

qﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ qﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ 2 2 2 h ð0 tÞ þð0Þ þ ð0Þ h ð0Þ2 þ ð0Þ2 þ ð0Þ2

¼ lim t!0 t[0

t

;

ðD1 HÞð0; 0; 0Þ exists, and its value is 0. Also, for ðx1 ; x2 ; x3 Þ 6¼ ð0; 0; 0Þ; qﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ 1 qﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ ð2x1 þ 0 þ 0Þ ðx1 Þ2 þðx2 Þ2 þðx3 Þ2 h0 2 2 ðx1 Þ þðx2 Þ2 þðx3 Þ2 qﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ x1 ðx1 Þ2 þðx2 Þ2 þðx3 Þ2 : ¼ qﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ h0 2 2 2 ðx1 Þ þðx2 Þ þðx3 Þ

ðD1 H Þðx1 ; x2 ; x3 Þ ¼

3.6 Smooth Bump Functions

207

Thus, 8 qﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ < pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ x1 0 ðx1 Þ2 þðx2 Þ2 þðx3 Þ2 h ðD1 H Þðx1 ; x2 ; x3 Þ ¼ ðx1 Þ2 þðx2 Þ2 þðx3 Þ2 : 0

if

ðx1 ; x2 ; x3 Þ 6¼ ð0; 0; 0Þ

if

ðx1 ; x2 ; x3 Þ ¼ ð0; 0; 0Þ:

Now, we want to show that D1 H is continuous at ð0; 0; 0Þ: We recall that the smooth function h : R ! R is deﬁned as follows: For every t in R; f ð2 t Þ : Also f ðxÞ f ð2 tÞ þ f ðt 1Þ 0 if x 0; ¼ 1 1x e if 0\x: 2 x

hðtÞ ¼

e x 0

1

if if

0\x; and f 0 ðxÞ x 0;

So h0 ðtÞ ¼

ðf 0 ð2 tÞÞð1Þðf ð2 tÞ þ f ðt 1ÞÞ ðf ð2 tÞÞðf 0 ð2 tÞ þ f 0 ðt 1ÞÞ ð f ð 2 t Þ þ f ð t 1Þ Þ 2

and hence, h0 ð0Þ ¼ ¼ ¼

ðf 0 ð2 0ÞÞð1Þðf ð2 0Þ þ f ð0 1ÞÞ ðf ð2 0ÞÞðf 0 ð2 0Þ þ f 0 ð0 1ÞÞ ðf ð2 0Þ þ f ð0 1ÞÞ2 ðf ð2ÞÞð1Þðf ð2Þ þ f ð1ÞÞ ðf ð2ÞÞðf 0 ð2Þ þ f 0 ð1ÞÞ 0

ðf ð2Þ þ f ð1ÞÞ2 ðf ð2ÞÞð1Þðf ð2Þ þ 0Þ ðf ð2ÞÞðf 0 ð2Þ þ 0Þ 0

ðf ð2Þ þ 0Þ2

¼

0

1 2 e2

¼ 0:

Thus, h0 ð0Þ ¼ 0: Since for ðx1 ; x2 ; x3 Þ 6¼ ð0; 0; 0Þ; we have x1 0 qﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ 1; 2 2 2 ðx1 Þ þðx2 Þ þðx3 Þ and h0 ð0Þ ¼ 0; lim

ðx1 ;x2 ;x3 Þ!ð0;0;0Þ

¼

ðD1 H Þðx1 ; x2 ; x3 Þ

x1 qﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ h0 ðx1 ;x2 ;x3 Þ!ð0;0;0Þ 2 ðx1 Þ þðx2 Þ2 þðx3 Þ2 lim

qﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ ðx1 Þ2 þðx2 Þ2 þðx3 Þ2 ¼ 0 ¼ ðD1 H Þð0; 0; 0Þ:

;

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3 Multivariable Differential Calculus

This shows that D1 H is continuous at ð0; 0; 0Þ: Similarly, D2 H is continuous at ð0; 0; 0Þ; and D3 H is continuous at ð0; 0; 0Þ: Similarly, higher partial derivatives of H exist at ð0; 0; 0Þ: This shows that H is smooth. For 2: Let us take any x ðx1 ; x2 ; x3 Þ satisfying jxj 1: So qﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ HðxÞ ¼ hð ðx1 Þ2 þ ðx2 Þ2 þ ðx3 Þ2 Þ ¼ hðjxjÞ ¼ 1: For 3: Let us take any a ða1 ; a2 ; a3 Þ in ðH 1 ðð0; 1ÞÞ : We have to show that qﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ qﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ ða1 Þ2 þ ða2 Þ2 þ ða3 Þ2 2: If not, otherwise, let 2\ ða1 Þ2 þ ða2 Þ2 þ ða3 Þ2 : qﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ We have to arrive at a contradiction. Since 2\ ða1 Þ2 þ ða2 Þ2 þ ða3 Þ2 ¼ jaj; there exists a real number e [ 0 such that the open ball Be ðaÞ contains no point of B2 ½0: Since Be ðaÞ is an open neighborhood of a, and a is in ðH 1 ðð0; 1ÞÞ ; there exists b in H 1 ðð0; 1Þ such that b is in Be ðaÞ: Since b is in H 1 ðð0; 1Þ; hðjbjÞ ¼ HðbÞ 2 ð0; 1: Since hðjbjÞ 2 ð0; 1; and hðtÞ ¼ 0 if and only if 2 t; jbj\2; and hence, b is in B2 ½0: Since b is in B2 ½0; and Be ðaÞ contains no point of B2 ½0; b is not in Be ðaÞ; a contradiction. Thus, we have shown that ðH 1 ðð0; 1ÞÞ B2 ½0: It remains to prove that B2 ½0 ðH 1 ðð0; 1ÞÞ : Let us take any a in B2 ½0: We have to prove that a 2 ðH 1 ðð0; 1ÞÞ : Since a is in B2 ½0; jaj 2: Case I: when jaj\2. Since jaj\2; and hðtÞ ¼ 0 if and only if 2 t; 0\hðjajÞ ¼ HðaÞ 1; and hence, a 2 H 1 ðð0; 1Þ ðH 1 ðð0; 1ÞÞ : Hence, a 2 ðH 1 ðð0; 1ÞÞ : Case II: when jaj ¼ 2. For every positive integer n [ 1; 1 lim 1 a ¼ a; n!1 2n and

1 1 a ¼ 1 1 jaj ¼ 1 1 2 ¼ 2 1 2 ð1; 2Þ: 2n 2n 2n n

Since

1 1 a 2 ð1; 2Þ; 2n H

and hence,

1 1 1 a ¼h 1 a 2 ð0; 1Þ ð0; 1; 2n 2n 1 1 a 2 H 1 ðð0; 1Þ: 2n

3.6 Smooth Bump Functions

209

Since, for every positive integer, n [ 1; 1 1 a 2 H 1 ðð0; 1Þ; 2n and

1 a ¼ a; so a 2 H 1 ðð0; 1Þ : lim 1 n!1 2n

Thus, we see that in all cases, a 2 ðH 1 ðð0; 1ÞÞ :

h

As above, we can show that there exists a function H : Rn ! ½0; 1 such that 1. H is smooth, 2. HðxÞ ¼ 1; if x is in the closed ball B1 ½0ð¼ fx : x 2 Rn ; jxj 1gÞ; 3. the closure ðH 1 ðð0; 1ÞÞ of H 1 ðð0; 1Þ is equal to the closed ball B2 ½0ð¼ fx : x 2 Rn ; jxj 2gÞ: Deﬁnition Let h : R ! R be a smooth function. If 1. hðtÞ ¼ 1 if t 1; 2. 0\hðtÞ\1 if 1\t\2; 3. hðtÞ ¼ 0 if 2 t; then we say that h is a cutoff function. By Note 3.51, there exists a cutoff function. Deﬁnition Let X be a topological space. Let f : X 7! R be any function. The closure ðf 1 ðR f0gÞÞ of f 1 ðR f0gÞ is called the support of f and is denoted by supp f : By f is compactly supported, we mean that supp f is a compact set. By the note 3.52, there exists a function H : Rn ! ½0; 1 such that 1. H is smooth, 2. HðxÞ ¼ 1; if x is in the closed ball B1 ½0; 3. supp H ¼ B2 ½0: Deﬁnition Let X be a topological space. Let f : X 7! R be any function. Let S be any nonempty subset of X. If supp f is contained in S, then we say that f is supported in S. Deﬁnition Let H : Rn ! ½0; 1 be a function. Let F be a closed subset of Rn ; and let G be an open subset of Rn : If 1. H is smooth, 2. HðxÞ ¼ 1; if x is in F, 3. H is supported in G,

210

3 Multivariable Differential Calculus

Fig. 3.1 F and K are disjoint

then we say that H is a smooth bump function corresponding to closed set F and open set G. By Note 3.52, there exists a smooth bump function corresponding to closed set B1 ½0 and open set B3 ð0Þ: Theorem 3.53 Let F be a nonempty closed subset of R3 ; and let K be a nonempty compact subset of R3 : If F and K are disjoint, then there exists a function r : R3 ! ½0; 1 such that 1. 2. 3. 4.

r is smooth, for every x in F; rðxÞ ¼ 0; for every x in K; rðxÞ ¼ 1; r is onto.

Proof Since F and K are disjoint, K F c where F c denotes the complement of F. Since F is a closed subset of R3 ; F c is an open set. Let us take any x in K. Since x 2 K F c ; x 2 F c : Since x 2 F c ; and F c is an open set, there exists a real number dx [ 0 such that x 2 B12dx ðxÞ F c : It follows that the collection fB12dx ðxÞ : x 2 Kg is an open cover of K. Further, since K is compact, there exist ﬁnite many x1 ; x2 ; . . .; xn in K such that K is contained in B12dx ðx1 Þ [ B12dx ðx2 Þ [ [ B12dxn ðxn Þ: 1 2 Since (see Fig. 3.1) B12dx ðx1 Þ F c ; B12dx ðx2 Þ F c ; . . .; B12dxn ðxn Þ F c ; 1

2

ðB12dx ðx1 Þ [ B12dx ðx2 Þ [ [ B12dxn ðxn ÞÞ F c : Now, for every i ¼ 1; 2; . . .; n; by 1 2 Theorem 3.50, there exist functions hi : R3 ! ½0; 1 such that 1. hi is smooth, 2. hi is onto, 3. for every x in B12dx ðxi Þ; hi ðxÞ [ 0; i

4. for every x in fx : x 2 R3 ; jx xi j\ 12 dxi g; hi ðxÞ ¼ 1; 5. for every x 2 6 B12dx ðxi Þ; hi ðxÞ ¼ 0: i

3.6 Smooth Bump Functions

211

Now, let us deﬁne a function r : R3 ! ½0; 1 as follows: For every x in R3 ; rðxÞ 1 ðð1 h1 ðxÞÞð1 h2 ðxÞÞ ð1 hn ðxÞÞÞ: Since 0 h1 1, for every x in R3 ; 0 1 h1 ðxÞ 1: Similarly, for every x in R ; 0 1 h2 ðxÞ 1; . . .; 0 1 hn ðxÞ 1: Hence, for every x in R3 ; 0 ð1 h1 ðxÞÞð1 h2 ðxÞÞ ð1 hn ðxÞÞ 1: It follows that for every x in R3 ; 0 1 ðð1 h1 ðxÞÞð1 h2 ðxÞÞ ð1 hn ðxÞÞÞ 1: Thus, we have shown that r : R3 ! ½0; 1 is a well-deﬁned function. 3

1. Since each hi is smooth, r : x 7! 1 ðð1 h1 ðxÞÞð1 h2 ðxÞÞ ð1 hn ðxÞÞÞ is smooth. 2. Let us take any x in F. Since x is in F, and ðB12dx ðx1 Þ [ B12dx ðx2 Þ [ [ B12dxn ðxn ÞÞ F c ; x 62 ðB12dx ðx1 Þ [ B12dx ðx2 Þ [ [ 1 2 1 2 B12dxn ðxn ÞÞ; and hence, for every i ¼ 1; 2; . . .; n; we have x 62 B12dx ðxi Þ: Thus, i rðxÞ ¼ 1 ðð1 0Þð1 0Þ ð1 0ÞÞ ¼ 0: 3. Let us take any x in K. Since x is in K, and K is contained in B12dx ðx1 Þ [ 1 B12dx ðx2 Þ [ [ B12dxn ðxn Þ; there exists j 2 f1; 2; . . .; ng such that x 2 B12dx ðxj Þ: 2

j

For simplicity, let us take 1 for j. Now, since x 2 B12dx ðx1 Þ; h1 ðxÞ ¼ 1; and 1 hence, rðxÞ ¼ 1 ðð1 1Þð1 h2 ðxÞÞ ð1 hn ðxÞÞÞ ¼ 1: 4. Since F is nonempty, there exists a in F. Hence, rðaÞ ¼ 0: Thus, the function r assumes 0. Similarly, r assumes 1. Since r : R3 ! ½0; 1; and r is smooth, r is continuous. Since r : R3 ! ½0; 1 is continuous, r assumes 0, and r assumes 1, r assumes all values of ½0; 1: Hence, r : R3 ! ½0; 1 is onto. h

Note 3.54 The result similar to Theorem 3.53 can be proved as above for Rn in place of R3 : Theorem 3.55 Let a be in R3 : Let U be an open neighborhood of a in R3 : Let f : U ! R be any function. If f is smooth, then there exist an open neighborhood V of a and a function F : R3 ! R such that 1. 2. 3. 4.

V is contained in U, for every x in V, FðxÞ ¼ f ðxÞ; for every x 62 U; FðxÞ ¼ 0; F is smooth.

Proof Since U is an open neighborhood of a, there exists a real number e [ 0 such that Be ðaÞ U: Since Be=3 ½að fx : x 2 R3 ; jx aj 3e gÞ is a closed and bounded subset of R3 ; Be=3 ½a is compact. Since Be=3 ½a Be=2 ðaÞ; Be=3 ½a and R3 Be=2 ðaÞ

212

3 Multivariable Differential Calculus

are disjoint sets. Since Be=2 ðaÞ is an open set, R3 Be=2 ðaÞ is a closed set. Now, by Theorem 3.53, there exists a function r : R3 ! ½0; 1 such that 1. 2. 3. 4.

r is smooth, for every x in R3 Be=2 ðaÞ; rðxÞ ¼ 0; for every x in Be=3 ½a; rðxÞ ¼ 1; r is onto.

Let us take Be=3 ðaÞ for V. Let us deﬁne a function F : R3 ! R as follows: For every x in R3 ;

FðxÞ

f ðxÞrðxÞ if 0 if

x 2 U; x 62 U:

1. Since V ¼ Be=3 ðaÞ Be ðaÞ U; V is contained in U. 2. For every x in V, we have x 2 V ¼ Be=3 ðaÞ Be=3 ½a Be ðaÞ U; and hence, FðxÞ ¼ f ðxÞrðxÞ ¼ f ðxÞ1 ¼ f ðxÞ: 3. Let us take any x 62 U: Hence, by the deﬁnition of F, we have FðxÞ ¼ 0: 4. Since f and r both are smooth on the open set U, their product x 7! f ðxÞrðxÞ is smooth on U. Further, for x 2 U, we have FðxÞ ¼ f ðxÞrðxÞ; so F is smooth over U. Since for every x in U Be=2 ðaÞ; we have FðxÞ ¼ f ðxÞrðxÞ ¼ f ðxÞ0 ¼ 0: Hence, by 3, for every x in R3 Be=2 ðaÞ; FðxÞ ¼ 0: This shows that F is smooth over the open set R3 Be=2 ½a: Since F is smooth over the open set U, and F is smooth over the open set R3 Be=2 ½a; F is smooth over the open set U [ ðR3 Be=2 ½aÞð¼ R3 Þ: Thus, F : R3 ! R is a smooth function. h Note 3.56 The result similar to Theorem 3.55 can be proved as above for Rn in place of R3 :

3.7 The Constant Rank Theorem in Rn Theorem 3.57 Let A0 be an open neighborhood of ð0; 0; 0Þ in R3 : Let B0 be an open subset of R4 : Let F : A0 ! B0 be a function. If (i) F is a C2 function, (ii) for every x in A0 ; the linear transformation F 0 ðxÞ has rank 2, (iii) Fð0; 0; 0Þ ¼ ð0; 0; 0; 0Þ; then there exist an open neighborhood A of ð0; 0; 0Þ in R3 ; an open neighborhood B of ð0; 0; 0; 0Þ in R4 ; an open subset U of R3 ; an open subset V of R4 ; a function G : A ! U; and a function H : B ! V such that

3.7 The Constant Rank Theorem in Rn

213

1. A is contained in A0 ; and B is contained in B0 ; 2. G is a C 2 diffeomorphism (i.e., G is 1–1 onto, G is a C 2 function, and G1 is a C 2 function), 3. H is a C2 diffeomorphism, 4. FðAÞ B; 5. For every x ðx1 ; x2 ; x3 Þ in U, ðH F G1 Þðx1 ; x2 ; x3 Þ ¼ ðx1 ; x2 ; 0; 0Þ: Proof Let F ðF1 ; F2 ; F3 ; F4 Þ: Since ð0; 0; 0Þ is in A0 ; the linear transformation F 0 ð0; 0; 0Þ has rank 2. Hence, 2

ðD1 F1 Þð0; 0; 0Þ 6 ðD1 F2 Þð0; 0; 0Þ 6 rank4 ðD1 F3 Þð0; 0; 0Þ ðD1 F4 Þð0; 0; 0Þ

ðD2 F1 Þð0; 0; 0Þ ðD2 F2 Þð0; 0; 0Þ ðD2 F3 Þð0; 0; 0Þ ðD2 F4 Þð0; 0; 0Þ

3 ðD3 F1 Þð0; 0; 0Þ ðD3 F2 Þð0; 0; 0Þ 7 7 ¼ 2: ðD3 F3 Þð0; 0; 0Þ 5 ðD3 F4 Þð0; 0; 0Þ

This shows that there exists a nonzero 2 2 minor determinant of 2

ðD1 F1 Þð0; 0; 0Þ 6 ðD1 F2 Þð0; 0; 0Þ 6 4 ðD1 F3 Þð0; 0; 0Þ ðD1 F4 Þð0; 0; 0Þ

ðD2 F1 Þð0; 0; 0Þ ðD2 F2 Þð0; 0; 0Þ ðD2 F3 Þð0; 0; 0Þ ðD2 F4 Þð0; 0; 0Þ

3 ðD3 F1 Þð0; 0; 0Þ ðD3 F2 Þð0; 0; 0Þ 7 7: ðD3 F3 Þð0; 0; 0Þ 5 ðD3 F4 Þð0; 0; 0Þ

For simplicity, let

ðD1 F1 Þð0; 0; 0Þ det ðD1 F2 Þð0; 0; 0Þ

ðD2 F1 Þð0; 0; 0Þ 6 0: ¼ ðD2 F2 Þð0; 0; 0Þ

Now, let us deﬁne a function G : A0 ! R3 as follows: For every ðx1 ; x2 ; x3 Þ in A0 ; Gðx1 ; x2 ; x3 Þ ¼ ðF1 ðx1 ; x2 ; x3 Þ; F2 ðx1 ; x2 ; x3 Þ; x3 Þ: Since ðF1 ; F2 ; F3 ; F4 Þ F : A0 ! B0 R4 is a C 2 function, each Fi : A0 ! R is a C 2 function, and hence, each component function of G is C2 : This shows that G is a C2 function. Also, for every x ðx1 ; x2 ; x3 Þ in A0 ; 2

3 ðD1 F1 Þðx1 ; x2 ; x3 Þ ðD2 F1 Þðx1 ; x2 ; x3 Þ ðD3 F1 Þðx1 ; x2 ; x3 Þ G0 ðxÞ ¼ 4 ðD1 F2 Þðx1 ; x2 ; x3 Þ ðD2 F2 Þðx1 ; x2 ; x3 Þ ðD3 F2 Þðx1 ; x2 ; x3 Þ 5: 0 0 1

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3 Multivariable Differential Calculus

Hence, 2

ðD1 F1 Þð0; 0; 0Þ ðD2 F1 Þð0; 0; 0Þ ðD3 F1 Þð0; 0; 0Þ

6 detðG0 ð0; 0; 0ÞÞ ¼ det4 ðD1 F2 Þð0; 0; 0Þ 0 ðD1 F1 Þð0; 0; 0Þ ¼ det ðD1 F2 Þð0; 0; 0Þ

3

7 ðD2 F2 Þð0; 0; 0Þ ðD3 F2 Þð0; 0; 0Þ 5 0 1 ðD2 F1 Þð0; 0; 0Þ 6¼ 0: ðD2 F2 Þð0; 0; 0Þ

Since detðG0 ð0; 0; 0ÞÞ is nonzero, G0 ð0; 0; 0Þ is invertible. Now, we can apply the inverse function theorem on G : A0 ! R3 : There exists an open neighborhood A1 of ð0; 0; 0Þ such that 1. 2. 3. 4.

A1 is contained in A0 ; G is 1–1 on A1 (i.e., if x and y are in A1 ; and GðxÞ ¼ GðyÞ; then x ¼ y), GðA1 Þ is open in R3 ; the 1–1 function G1 from GðA1 Þ onto A1 is a C 2 function.

Now, since ðF1 ð0; 0; 0Þ; F2 ð0; 0; 0Þ; F3 ð0; 0; 0Þ; F4 ð0; 0; 0ÞÞ ¼ Fð0; 0; 0Þ ¼ ð0; 0; 0; 0Þ; each Fi ð0; 0; 0Þ ¼ 0: Next, since ð0; 0; 0Þ is in A0 ; and G : A0 ! R3 ; Gð0; 0; 0Þ ¼ ðF1 ð0; 0; 0Þ; F2 ð0; 0; 0Þ; 0Þ ¼ ð0; 0; 0Þ: Thus, Gð0; 0; 0Þ ¼ ð0; 0; 0Þ: Since G is 1–1 on A1 ; Gð0; 0; 0Þ ¼ ð0; 0; 0Þ; and ð0; 0; 0Þ is in A1 ; G1 ð0; 0; 0Þ ¼ ð0; 0; 0Þ: Now, ðF G1 Þð0; 0; 0Þ ¼ FðG1 ð0; 0; 0ÞÞ ¼ Fð0; 0; 0Þ ¼ ð0; 0; 0; 0Þ: Thus, ðF G1 Þð0; 0; 0Þ ¼ ð0; 0; 0; 0Þ: Since A1 is contained in A0 ; and F : A0 ! B0 ; ðF G1 ÞðGðA1 ÞÞ ¼ FðA1 Þ B0 : Thus, F G1 : GðA1 Þ ! B0 . For every ðx1 ; x2 ; x3 Þ in GðA1 Þ, there exists ðy1 ; y2 ; y3 Þ in A1 such that ðx1 ; x2 ; x3 Þ ¼ Gðy1 ; y2 ; y3 Þ ¼ ðF1 ðy1 ; y2 ; y3 Þ; F2 ðy1 ; y2 ; y3 Þ; y3 Þ. Further, since G is 1–1 on A1 ;

F G1 ðx1 ; x2 ; x3 Þ ¼ F G1 ðx1 ; x2 ; x3 Þ ¼ F ðy1 ; y2 ; y3 Þ ¼ ðF1 ðy1 ; y2 ; y3 Þ; F2 ðy1 ; y2 ; y3 Þ; F3 ðy1 ; y2 ; y3 Þ; F4 ðy1 ; y2 ; y3 ÞÞ ¼ ðx1 ; x2 ; F3 ðy1 ; y2 ; y3 Þ; F4 ðy1 ; y2 ; y3 ÞÞ

¼ x1 ; x2 ; F3 G1 ðx1 ; x2 ; x3 Þ ; F4 G1 ðx1 ; x2 ; x3 Þ

¼ x1 ; x2 ; F3 G1 ðx1 ; x2 ; x3 Þ; F4 G1 ðx1 ; x2 ; x3 Þ : Thus, for every ðx1 ; x2 ; x3 Þ in GðA1 Þ;

F G1 ðx1 ; x2 ; x3 Þ ¼ x1 ; x2 ; F3 G1 ðx1 ; x2 ; x3 Þ; F4 G1 ðx1 ; x2 ; x3 Þ :

3.7 The Constant Rank Theorem in Rn

215

Hence, for every x ðx1 ; x2 ; x3 Þ in GðA1 Þ; 2

1 60

1 0 6 F G ðxÞ ¼ 4 ðD1 ðF3 G1 ÞÞðx1 ; x2 ; x3 Þ ðD1 ðF4 G1 ÞÞðx1 ; x2 ; x3 Þ

0 1 ðD2 ðF3 G1 ÞÞðx1 ; x2 ; x3 Þ ðD2 ðF4 G1 ÞÞðx1 ; x2 ; x3 Þ

3 0 7 0 7: ðD3 ðF3 G1 ÞÞðx1 ; x2 ; x3 Þ 5 1 ðD3 ðF4 G ÞÞðx1 ; x2 ; x3 Þ

Also, for every x ðx1 ; x2 ; x3 Þ in GðA1 Þ; 2

3 100

4 0 1 0 5 ¼ I 0 ðxÞ ¼ G G1 0 ðxÞ ¼ G0 G1 ðxÞ G1 0 ðxÞ ; 001

so 2

3 100

0

0 6¼ 1 ¼ det4 0 1 0 5 ¼ det G0 G1 ðxÞ G1 ðxÞ 001 0 1

1 0 ¼ det G G ðxÞ det G ðxÞ : It follows that detððG1 Þ0 ðxÞÞ 6¼ 0; and hence, ðG1 Þ0 ðxÞ is invertible for every x in GðA1 Þ: Since for every x ðx1 ; x2 ; x3 Þ in GðA1 Þ; ðF G1 Þ0 ðxÞ ¼ ðF 0 ðG1 ðxÞÞÞ ððG1 Þ0 ðxÞÞ; and ðG1 Þ0 ðxÞ is invertible, 3 0 0 1 7 60 1 0 7 6 rank6 7 1 1 1 4 ðD1 ðF3 G ÞÞðx1 ; x2 ; x3 Þ ðD2 ðF3 G ÞÞðx1 ; x2 ; x3 Þ ðD3 ðF3 G ÞÞðx1 ; x2 ; x3 Þ 5 ðD1 ðF4 G1 ÞÞðx1 ; x2 ; x3 Þ ðD2 ðF4 G1 ÞÞðx1 ; x2 ; x3 Þ ðD3 ðF4 G1 ÞÞðx1 ; x2 ; x3 Þ

0

0 ¼ rank F G1 ðxÞ ¼ rank F 0 G1 ðxÞ G1 ðxÞ ¼ rank F 0 G1 ðxÞ ¼ 2: 2

Thus, for every x ðx1 ; x2 ; x3 Þ in GðA1 Þ;

D3 F3 G1 ðx1 ; x2 ; x3 Þ 3 2 0 0 1 7 6 1 0 ¼ det4 0 5 ¼ 0; 1 1 1 ðD1 ðF3 G ÞÞðx1 ; x2 ; x3 Þ ðD2 ðF3 G ÞÞðx1 ; x2 ; x3 Þ ðD3 ðF3 G ÞÞðx1 ; x2 ; x3 Þ

and

D3 F4 G1 ðx1 ; x2 ; x3 Þ 3 2 1 0 0 7 6 ¼ det4 0 1 0 5 ¼ 0: 1 1 1 ðD1 ðF4 G ÞÞðx1 ; x2 ; x3 Þ ðD2 ðF4 G ÞÞðx1 ; x2 ; x3 Þ ðD3 ðF4 G ÞÞðx1 ; x2 ; x3 Þ

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3 Multivariable Differential Calculus

It follows that for every x ðx1 ; x2 ; x3 Þ in GðA1 Þ; F3 G1 and F4 G1 are functions of x1 ; x2 only. Since Gð0; 0; 0Þ ¼ ð0; 0; 0Þ; A1 is an open neighborhood of ð0; 0; 0Þ; and GðA1 Þ is open in R3 ; GðA1 Þ is an open neighborhood of ð0; 0; 0Þ; and hence, GðA1 Þ R is an open neighborhood of ð0; 0; 0; 0Þ in R4 : Let us deﬁne a function T : GðA1 Þ R ! R4 as follows: For every ðy1 ; y2 ; y3 ; y4 Þ in GðA1 Þ R;

T ðy1 ; y2 ; y3 ; y4 Þ y1 ; y2 ; y3 þ F3 G1 ðy1 ; y2 ; y3 Þ; y4 þ F4 G1 ðy1 ; y2 ; y3 Þ : Here, each component function of T is C2 ; so T is a C 2 function, and hence, T is continuous. Here,

F1 G1 ð0; 0; 0Þ; F2 G1 ð0; 0; 0Þ; F3 G1 ð0; 0; 0Þ; F4 G1 ð0; 0; 0Þ

¼ F1 G1 ð0; 0; 0Þ ; F2 G1 ð0; 0; 0Þ ; F3 G1 ð0; 0; 0Þ ; F4 G1 ð0; 0; 0Þ

¼ F G1 ð0; 0; 0Þ ¼ F G1 ð0; 0; 0Þ ¼ ð0; 0; 0; 0Þ;

so

T ð0; 0; 0; 0Þ ¼ 0; 0; 0 þ F3 G1 ð0; 0; 0Þ; 0 þ F4 G1 ð0; 0; 0Þ

¼ 0; 0; F3 G1 ð0; 0; 0Þ; F4 G1 ð0; 0; 0Þ ¼ ð0; 0; 0; 0Þ: Since T : GðA1 Þ R ! R4 is continuous, Tð0; 0; 0; 0Þ ¼ ð0; 0; 0; 0Þ; and B0 be an open neighborhood of ð0; 0; 0; 0Þ in R4 , there exists an open neighborhood V1 of ð0; 0; 0; 0Þ such that V1 is contained in GðA1 Þ R; and TðV1 Þ is contained in B0 : Now, for every y ðy1 ; y2 ; y3 ; y4 Þ in V1 ; T 0 ðy1 ; y2 ; y3 ; y4 Þ 2 1 60 6 ¼6 4 D1 ðy3 þ ðF3 G1 Þðy1 ; y2 ; y3 ÞÞ D1 ðy4 þ ðF4 G1 Þðy1 ; y2 ; y3 ÞÞ 0 0 D3 ðy3 þ ðF3 G1 Þðy1 ; y2 ; y3 ÞÞ D3 ðy4 þ ðF4 G1 Þðy1 ; y2 ; y3 ÞÞ

0 1 D2 ðy3 þ ðF3 G1 Þðy1 ; y2 ; y3 ÞÞ D2 ðy4 þ ðF4 G1 Þðy1 ; y2 ; y3 ÞÞ 3 0 7 0 7 7 1 D4 ðy3 þ ðF3 G Þðy1 ; y2 ; y3 ÞÞ 5 D4 ðy4 þ ðF4 G1 Þðy1 ; y2 ; y3 ÞÞ

3.7 The Constant Rank Theorem in Rn

2

217

1 60 6 ¼6 4 0 þ D1 ðF3 G1 Þðy1 ; y2 ; y3 Þ 0 þ D1 ðF3 G1 Þðy1 ; y2 ; y3 Þ 0

0

1 0 þ D2 ðF3 G1 Þðy1 ; y2 ; y3 Þ 0 þ D2 ðF4 G1 Þðy1 ; y2 ; y3 Þ 3 0 7 0 0 7 7 1 5 1 þ D3 ðF3 G Þðy1 ; y2 ; y3 Þ 0 0 þ D3 ðF4 G1 Þðy1 ; y2 ; y3 Þ 1 þ D4 ðF4 G1 Þðy1 ; y2 ; y3 Þ 2 1 0 60 1 6 ¼6 4 D1 ðF3 G1 Þðy1 ; y2 ; y3 Þ D2 ðF3 G1 Þðy1 ; y2 ; y3 Þ D1 ðF3 G1 Þðy1 ; y2 ; y3 Þ D2 ðF4 G1 Þðy1 ; y2 ; y3 Þ 3 0 0 7 0 0 7 7 1 5 1 þ D3 ðF3 G Þðy1 ; y2 ; y3 Þ 0 D3 ðF4 G1 Þðy1 ; y2 ; y3 Þ 1 þ D4 ðF4 G1 Þðy1 ; y2 ; y3 Þ 1 0 0 0 60 0 1 0 6 ¼6 4 D1 ðF3 G1 Þðy1 ; y2 ; y3 Þ D2 ðF3 G1 Þðy1 ; y2 ; y3 Þ 1 þ 0 0

3

2

D1 ðF3 G1 Þðy1 ; y2 ; y3 Þ 1 60 6 ¼6 4 D1 ðF3 G1 Þðy1 ; y2 ; y3 Þ 2

D2 ðF4 G1 Þðy1 ; y2 ; y3 Þ 0 1 D2 ðF3 G1 Þðy1 ; y2 ; y3 Þ

7 7 7 5

1þ0 0 3 0 0 0 07 7 7 1 05

D1 ðF3 G1 Þðy1 ; y2 ; y3 Þ D2 ðF4 G1 Þðy1 ; y2 ; y3 Þ 0 1

Hence, for every y ðy1 ; y2 ; y3 ; y4 Þ in V1 ; 2

1 60 0 detðT ðy1 ; y2 ; y3 ; y4 ÞÞ ¼ det6 4 D1 ðF3 G1 Þðy1 ; y2 ; y3 Þ D1 ðF3 G1 Þðy1 ; y2 ; y3 Þ

0 1 D2 ðF3 G1 Þðy1 ; y2 ; y3 Þ D2 ðF4 G1 Þðy1 ; y2 ; y3 Þ

0 0 1 0

3 0 07 7 ¼ 1 6¼ 0: 05 1

Since for every y ðy1 ; y2 ; y3 ; y4 Þ in V1 ; detðT 0 ðy1 ; y2 ; y3 ; y4 ÞÞ is nonzero, T ðy1 ; y2 ; y3 ; y4 Þ is invertible, and hence, T 0 ð0; 0; 0; 0Þ is invertible. Now, we can apply the inverse function theorem on T : V1 ! R4 : There exists an open neighborhood V of ð0; 0; 0; 0Þ such that 0

1. 2. 3. 4.

V is contained in V1 ; T is 1–1 on V (i.e., if x and y are in V, and TðxÞ ¼ TðyÞ; then x ¼ y), TðVÞ is open in R4 ; the 1–1 function T 1 from TðVÞ onto V is a C 2 function.

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3 Multivariable Differential Calculus

Here, ð0; 0; 0; 0Þ is in V, and Tð0; 0; 0; 0Þ ¼ ð0; 0; 0; 0Þ; so ð0; 0; 0; 0Þ is in TðVÞ: Further, since TðVÞ is open in R4 ; TðVÞ is an open neighborhood of ð0; 0; 0; 0Þ in R4 : Since G1 is a C 2 function from GðA1 Þ onto A1 ; A1 is contained in A0 ; and F : A0 ! B0 is a C2 function, their composite function F G1 GðA1 Þ ! B0 is C 2 : Since GðA1 Þ is an open neighborhood of ð0; 0; 0Þ; ð0; 0; 0Þ is in GðA1 Þ: Since ð0; 0; 0Þ is in GðA1 Þ; and F G1 : GðA1 Þ ! B0 is C 2 ; F G1 : GðA1 Þ ! B0 is continuous at ð0; 0; 0Þ: Since F G1 : GðA1 Þ ! B0 is continuous at ð0; 0; 0Þ; ðF G1 Þð0; 0; 0Þ ¼ ð0; 0; 0; 0Þ; and TðVÞ is an open neighborhood of Tð0; 0; 0; 0Þð¼ ð0; 0; 0; 0ÞÞ; there exists an open neighborhood U of ð0; 0; 0Þ such that U is contained in GðA1 Þ and ðF G1 ÞðUÞ is contained in TðVÞ: Now, let us take the 1–1 function T 1 from TðVÞ onto V for H, the open set TðVÞ for B, and G1 ðUÞ for A. It remains to prove the following: (i) (ii) (iii) (iv) (v) (vi) (vii) (viii) (ix) (x) (xi) (xii) (xiii) (xiv) (xv) (xvi)

A is an open neighborhood of ð0; 0; 0Þ; B is an open neighborhood of ð0; 0; 0; 0Þ; U is an open subset of R3 ; V is an open subset of R4 ; G : A ! U is 1–1, G : A ! U is onto, G : A ! U is C 2 ; G1 : U ! A is C 2 ; H : B ! V is 1–1, H : B ! V is onto, H : B ! V is C 2 ; H 1 : V ! B is C2 ; A is contained in A0 ; B is contained in B0 ; FðAÞ B For every x ðx1 ; x2 ; x3 Þ in U, ðH F G1 Þðx1 ; x2 ; x3 Þ ¼ ðx1 ; x2 ; 0; 0Þ:

For (i): Since G : A1 ! GðA1 Þ is a C 2 function, and ð0; 0; 0Þ is in A1 ; G : A1 ! GðA1 Þ is continuous at ð0; 0; 0Þ: Since G is 1–1 on A1 ; G : A1 ! GðA1 Þ is continuous at ð0; 0; 0Þ; and U is an open neighborhood of Gð0; 0; 0Þð¼ ð0; 0; 0ÞÞ such that U is contained in the open set GðA1 Þ; G1 ðUÞ is an open neighborhood of ð0; 0; 0Þ and is contained in A1 : Thus, A is an open neighborhood of ð0; 0; 0Þ: For (ii): Since TðVÞ is an open neighborhood of ð0; 0; 0; 0Þ in R4 ; and B stands for TðVÞ; B is an open neighborhood of ð0; 0; 0; 0Þ in R4 : For (iii): This has been shown above. For (iv): This has been shown above. For (v): Since U is contained in GðA1 Þ; and G is 1–1 on A1 ; G1 ðUÞ is contained in A1 : Since G is 1–1 on A1 ; G1 ðUÞ is contained in A1 ; and A stands for G1 ðUÞ; G : A ! U is 1–1. For (vi): Since G : A ! U is 1–1, GðAÞ ¼ GðG1 ðUÞÞ ¼ U: So G : A ! U is onto.

3.7 The Constant Rank Theorem in Rn

219

For (vii): Since G is C2 on A0 ; and A ¼ G1 ðUÞ A1 A0 , G is C2 on A. For (viii): Since G1 ðUÞ A1 ; and G is 1–1 on A1 ; U GðA1 Þ: Since U

GðA1 Þ; and the function G1 from GðA1 Þ onto A1 is a C 2 function, G1 : U ! G1 ðUÞ is C2 ; that is, G1 : U ! A is C2 . For (ix): Since T is 1–1 on V, T 1 : TðVÞ ! V is a 1–1 function. Since H stands for the function T 1 and B stands for TðVÞ; H : B ! V is 1–1. For (x): Since T is 1–1 on V, T 1 : TðVÞ ! V is onto, and hence, H : B ! V is onto. For (xi): Since the function T 1 from TðVÞ onto V is a C 2 function, H : B ! V is C 2 . For (xii): Since T : GðA1 Þ R ! R4 is a C2 function, and V is an open subset of the open set GðA1 Þ R; T : V ! TðVÞ is a C 2 function, and hence, H 1 : V ! B is C 2 . For (xiii): Since U is contained in GðA1 Þ; and G is 1–1 on A1 ; G1 ðUÞ is contained in A1 ð A0 Þ and hence, A is contained in A0 . For (xiv): Since V is contained in V1 ; and TðV1 Þ is contained in B0 ; TðVÞ is contained in B0 ; and hence, B is contained in B0 . For (xv): Here, FðAÞ ¼ FðG1 ðUÞÞ ¼ ðF G1 ÞðUÞ TðVÞ ¼ B: Thus, FðAÞ B. For (xvi): Let us take any x ðx1 ; x2 ; x3 Þ in U. Since ðx1 ; x2 ; x3 Þ is in U, ðx1 ; x2 ; ðF3 G1 Þðx1 ; x2 ; x3 Þ; ðF4 G1 Þðx1 ; x2 ; x3 ÞÞ ¼ ðF G1 Þðx1 ; x2 ; x3 Þ 2 ðF G1 ÞðUÞ TðVÞ. Next, since T : V ! TðVÞ is 1–1 onto,

H F G1 ðx1 ; x2 ; x3 Þ ¼ H F G1 ðx1 ; x2 ; x3 Þ

¼ H x1 ; x2 ; F3 G1 ðx1 ; x2 ; x3 Þ; F4 G1 ðx1 ; x2 ; x3 Þ

¼ T 1 x1 ; x2 ; F3 G1 ðx1 ; x2 ; x3 Þ; F4 G1 ðx1 ; x2 ; x3 Þ :

It remains to prove that

T 1 x1 ; x2 ; F3 G1 ðx1 ; x2 ; x3 Þ; F4 G1 ðx1 ; x2 ; x3 Þ ¼ ðx1 ; x2 ; 0; 0Þ: Let T 1 ðx1 ; x2 ; ðF3 G1 Þðx1 ; x2 ; x3 Þ; ðF4 G1 Þðx1 ; x2 ; x3 ÞÞ ¼ ðy1 ; y2 ; y3 ; y4 Þ 2 V V1 GðA1 Þ R: It follows that ðy1 ; y2 ; y3 Þ 2 GðA1 Þ: Since ðy1 ; y2 ; y3 Þ 2 GðA1 Þ; and ðx1 ; x2 ; x3 Þ 2 U GðA1 Þ;

x1 ; x2 ; F3 G1 ðx1 ; x2 ; x3 Þ; F4 G1 ðx1 ; x2 ; x3 Þ ¼ T ðy1 ; y2 ; y3 ; y4 Þ

¼ y1 ; y2 ; y3 þ F3 G1 ðy1 ; y2 ; y3 Þ; y4 þ F4 G1 ðy1 ; y2 ; y3 Þ :

220

3 Multivariable Differential Calculus

Hence,

y1 ¼ x1 ; y2 ¼ x2 ; F3 G1 ðx1 ; x2 ; x3 Þ ¼ y3 þ F3 G1 ðy1 ; y2 ; y3 Þ; and

F4 G1 ðx1 ; x2 ; x3 Þ ¼ y4 þ F4 G1 ðy1 ; y2 ; y3 Þ:

It follows that

F3 G1 ðx1 ; x2 ; x3 Þ ¼ y3 þ F3 G1 ðx1 ; x2 ; y3 Þ;

F4 G1 ðx1 ; x2 ; x3 Þ ¼ y4 þ F4 G1 ðx1 ; x2 ; y3 Þ:

and

Since ðy1 ; y2 ; y3 Þ 2 GðA1 Þ; y1 ¼ x1 ; y2 ¼ x2 ; ðx1 ; x2 ; y3 Þ 2 GðA1 Þ: Since for every ðn; g; 1Þ in GðA1 Þ; F3 G1 is a function of n; g only, and ðF3 G1 Þðx1 ; x2 ; y3 Þ ¼ ðF3 G1 Þðx1 ; x2 ; x3 Þ: ðx1 ; x2 ; y3 Þ; ðx1 ; x2 ; x3 Þ 2 GðA1 Þ; 1 Since ðF3 G Þðx1 ; x2 ; y3 Þ ¼ ðF3 G1 Þðx1 ; x2 ; x3 Þ; and ðF3 G1 Þðx1 ; x2 ; x3 Þ ¼ y3 þ ðF3 G1 Þðx1 ; x2 ; y3 Þ; y3 ¼ 0: Since for every ðn; g; 1Þ in GðA1 Þ; F4 G1 is a function of n; g only, and ðx1 ; x2 ; y3 Þ; ðx1 ; x2 ; x3 Þ 2 GðA1 Þ; ðF4 G1 Þðx1 ; x2 ; y3 Þ ¼ ðF4 G1 Þðx1 ; x2 ; x3 Þ: Now, since ðF4 G1 Þðx1 ; x2 ; y3 Þ ¼ ðF4 G1 Þðx1 ; x2 ; x3 Þ; and ðF4 G1 Þðx1 ; x2 ; x3 Þ ¼ y4 þ ðF4 G1 Þðx1 ; x2 ; y3 Þ; y4 ¼ 0:Since y3 ¼ 0; y4 ¼ 0; y1 ¼ x1 ; y2 ¼ x2 ; and T 1 ðx1 ; x2 ; ðF3 G1 Þðx1 ; x2 ; x3 Þ; ðF4 G1 Þðx1 ; x2 ; x3 ÞÞ ¼ ðy1 ; y2 ; y3 ; y4 Þ;

T 1 x1 ; x2 ; F3 G1 ðx1 ; x2 ; x3 Þ; F4 G1 ðx1 ; x2 ; x3 Þ ¼ ðx1 ; x2 ; 0; 0Þ:

h

Theorem 3.58 Let A0 be an open neighborhood of ð0; 0; 0Þ in R3 : Let B0 be an open neighborhood of b ðb1 ; b2 ; b3 ; b4 Þ in R4 : Let F : A0 ! B0 be a function. If (i) F is a C2 function, (ii) for every x in A0 ; the linear transformation F 0 ðxÞ has rank 2, (iii) Fð0; 0; 0Þ ¼ b; then there exist an open neighborhood A of ð0; 0; 0Þ in R3 ; an open neighborhood B of b in R4 ; an open subset U of R3 ; an open subset V of R4 ; a function G : A ! U; and a function H : B ! V such that 1. A is contained in A0 ; and B is contained in B0 ; 2. G is a C 2 diffeomorphism (i.e., G is 1–1 onto, G is a C 2 function, and G1 is a C 2 function), 3. H is a C2 diffeomorphism,

3.7 The Constant Rank Theorem in Rn

221

4. FðAÞ B; 5. For every x ðx1 ; x2 ; x3 Þ in U, ðH F G1 Þðx1 ; x2 ; x3 Þ ¼ ðx1 ; x2 ; 0; 0Þ: ^ 0 A0 ; and B ^0 ! B ^ 0 B0 b: Let us deﬁne a function F ^:A ^ 0 as Proof Put A ^ 0 (since x is in A ^ 0 ð¼ A0 Þ; x is in A0 ; and hence, FðxÞ is follows: For every x in A deﬁned.) ^ FðxÞ FðxÞ b ¼ ðTb F ÞðxÞ; ^ ¼ Tb F: Hence, Tb F ^ ¼ F: where Tb : y 7! ðy bÞ is a translation. Thus, F ^ 0; 0Þ ¼ Fð0; 0; 0Þ b ¼ b b ¼ ð0; 0; 0; 0Þ: Thus, Fð0; ^ 0; 0Þ ¼ ð0; 0; 0; 0Þ: Also, Fð0; ^ 0 ð¼ A0 Þ is an open neighborhood Since A0 is an open neighborhood of ð0; 0; 0Þ; A ^ 0 ð¼ B0 bÞ is an open neighborof 0. Since B0 is an open neighborhood of b, B hood of 0. Since F is a C 2 function, and Tb is a C 2 function, their composite ^0; F ^ is a C 2 function. Thus, F ^ is a C 2 function. For every x in A ^ 0 ðxÞ ¼ Tb Fð¼ FÞ 0 0 0 0 0 ðTb FÞ ðxÞ ¼ ððTb Þ ðFðxÞÞÞ ðF ðxÞÞ ¼ I4 ðF ðxÞÞ ¼ F ðxÞ: Thus, for every ^0; F ^ 0 ; then x is in A0 ; and hence, F ^ 0 ðxÞ ¼ F 0 ðxÞ: If x is in A ^ 0 ðxÞð¼ F 0 ðxÞÞ has x in A rank 2. Thus, we see that all the conditions of Theorem 3.57 are satisﬁed for ^0 ! B ^ of ^ ^ 0 . Hence, by Theorem 3.57, there exist an open neighborhood A F:A 3 4 ^ of ð0; 0; 0; 0Þ in R ; an open subset U ^ of ð0; 0; 0Þ in R ; an open neighborhood B 3 4 ^ ^ ^ ^ ^ ^ ^ R ; an open subset V of R ; a function G : A ! U, and a function H : B ! V such that ^ is contained in A ^ 0 ; and B ^ is contained in B ^0; A 2 ^ ^ ^ is a C2 function, and G ^ 1 is a G is a C diffeomorphism (i.e., G is 1–1 onto, G 2 C function), ^ is a C2 diffeomorphism, 3. H ^ B; ^ ^ 4. FðAÞ ^ 1 Þðx1 ; x2 ; x3 Þ ¼ ðx1 ; x2 ; 0; 0Þ: ^ ðH ^ F ^ G 5. For every x ðx1 ; x2 ; x3 Þ in U; 1. 2.

^ BB ^ þ b; U U; ^ and V V: ^ Let us deﬁne a function G : A ! U Put A A; ^ as follows: For every x in A, GðxÞ GðxÞ: Let us deﬁne a function H : B ! V as ^ and ^ bÞ ¼ ðH ^ Tb ÞðyÞ: Hence, G ¼ G; follows: For every y in B, HðyÞ Hðy 1 ^ ^ H ¼ H Tb : Further, H ¼ H ðTb Þ ¼ H Tb : It remains to prove the following: (i) (ii) (iii) (iv) (v) (vi)

A is an open neighborhood of ð0; 0; 0Þ; B is an open neighborhood of b, U is an open subset of R3 ; V is an open subset of R4 ; G : A ! U is 1–1, G : A ! U is onto,

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(vii) (viii) (ix) (x) (xi) (xii) (xiii) (xiv) (xv) (xvi)

G : A ! U is C 2 ; G1 : U ! A is C 2 ; H : B ! V is 1–1, H : B ! V is onto, H : B ! V is C 2 ; H 1 : V ! B is C2 ; A is contained in A0 ; B is contained in B0 ; FðAÞ B; For every x ðx1 ; x2 ; x3 Þ in U, ðH F G1 Þðx1 ; x2 ; x3 Þ ¼ ðx1 ; x2 ; 0; 0Þ:

^ is an open neighborhood of ð0; 0; 0Þ; Að¼ AÞ ^ is an open For (i): Since A neighborhood of ð0; 0; 0Þ. ^ is an open neighborhood of 0, Bð¼ B ^ þ bÞ is an open neighFor (ii): Since B borhood of b. ^ is an open subset of R3 . ^ is an open subset of R3 ; Uð¼ UÞ For (iii): Since U 4 ^ ^ For (iv): Since V is an open subset of R ; Vð¼ VÞ is an open subset of R4 . ^!U ^ and U ¼ U; ^ :A ^ ¼ G; A ¼ A; ^ is 1–1, G ^ G : A ! U is For (v): Since G 1–1. ^!U ^ and U ¼ U; ^ :A ^ A ¼ A; ^ is onto, G ¼ G; ^ G : A ! U is For (vi): Since G onto. ^ is C2 ; and G ¼ G; ^ G is C2 . For (vii): Since G ^ 1 is C 2 ; and G ¼ G; ^ G1 is C 2 . For (viii): Since G ^þb!B ^ is 1–1, ðH ¼ÞH ^ Tb : ^ ^ ^ For (ix): Since H : B ! V is 1–1, and Tb : B ^ þ bð¼ BÞ ! Vð¼ ^ B VÞ is 1–1. Hence, H : B ! V is 1–1. ^ ^ ^ is onto, and Tb : B ^þb!B ^ is onto, ðH ¼ÞH ^ Tb : For (x): Since H : B ! V ^ þ bð¼ BÞ ! Vð¼ ^ B VÞ is onto. Hence, H : B ! V is onto. ^ :B ^!V ^ is C2, and Tb : B ^þb!B ^ isC 2 ; ðH ¼ÞH ^ Tb : For (xi): Since H 2 2 ^ ^ B þ bð¼ BÞ ! Vð¼ VÞ is C . Hence, H : B ! V is C . ^ Tb Þ1 ¼ ðTb Þ1 H ^ 1 ¼ Tb H ^ 1 : Since H ^ 1 : For (xii): Here H 1 ¼ ðH ^ ^ is C 2 ; and the translation Tb : B ^!B ^ þ bð¼ BÞ is C 2 ; their Vð¼ VÞ ! B 1 1 2 ^ : V ! B is C ; and hence, H 1 : V ! B is C2. composite ðH ¼ÞTb H ^ is contained in A ^0; A ^ ¼ A; and A ^ 0 ¼ A0 ; A is contained in A0. For (xiii): Since A ^ ^ 0 ¼ B0 b; B − b is ^ ^ For (xiv): Since B is contained in B0 ; B ¼ B b; and B contained in B0 b; and hence, B is contained in B0. For (xv): Let us take any x in A. We have to prove that F(x) is in B. Here, x is in ^ so x is in A: ^ Since x is in A; ^ and Fð ^ B ^ AÞ ^ ¼ B b; FðxÞ b ¼ A, and A ¼ A, ^ FðxÞ 2 B b; and hence, FðxÞ is in B.

3.7 The Constant Rank Theorem in Rn

223

For (xvi): Let us take any x ðx1 ; x2 ; x3 Þ in U. We have to show that ðH F G1 Þðx1 ; x2 ; x3 Þ ¼ ðx1 ; x2 ; 0; 0Þ:

^ G1 ðx1 ; x2 ; x3 Þ LHS ¼ H F G1 ðx1 ; x2 ; x3 Þ ¼ H Tb F

^ 1 ðx1 ; x2 ; x3 Þ ^ G1 ðx1 ; x2 ; x3 Þ ¼ H ^ F ^ G ¼ ðH Tb Þ F ¼ ðx1 ; x2 ; 0; 0Þ ¼ RHS:

h

Theorem 3.59 Let A0 be an open neighborhood of a ða1 ; a2 ; a3 Þ in R3 : Let B0 be an open neighborhood of b ðb1 ; b2 ; b3 ; b4 Þ in R4 : Let F : A0 ! B0 be a function. If (i) F is a C2 function, (ii) for every x in A0, the linear transformation F 0 ðxÞ has rank 2, (iii) FðaÞ ¼ b; then there exist an open neighborhood A of a in R3 ; an open neighborhood B of b in R4 ; an open subset U of R3 ; an open subset V of R4 ; a function G : A ! U; and a function H : B ! V such that 1. A is contained in A0, and B is contained in B0, 2. G is a C2 diffeomorphism (i.e., G is 1–1 onto, G is a C2 function, and G−1 is a C2 function), 3. H is a C2 diffeomorphism, 4. FðAÞ B; 5. For every x ðx1 ; x2 ; x3 Þ in U, ðH F G1 Þðx1 ; x2 ; x3 Þ ¼ ðx1 ; x2 ; 0; 0Þ: ^ 0 A0 a; and B ^0 ! B ^ 0 B0 : Let us deﬁne a function F ^:A ^ 0 as Proof Put A ^ 0 (since x is in A ^ 0 ¼ A0 a; x þ a is in A0, and hence, follows: For every x in A Fðx þ aÞ is deﬁned.) ^ FðxÞ Fðx þ aÞ ¼ ðF Ta ÞðxÞ; ^ ¼ F Ta : Hence, F ^ Ta ¼ F: where Ta : x 7! ðx þ aÞ is a translation. Thus, F ^ ^ Also, Fð0; 0; 0Þ ¼ FðaÞ ¼ b: Thus, Fð0; 0; 0Þ ¼ b: Since A0 is an open neighbor^ 0 ð¼ A0 aÞ is an open neighborhood of 0. Since B0 is an open hood of a, A ^ 0 ð¼ B0 Þ is an open neighborhood of b. Since F is a C2 neighborhood of b, B ^ is a C2 function. function, and Ta is a C2 function, their composite F Ta ð¼ FÞ 2 ^0; ^ is a C function. For every x in A Thus, F

^ 0 ðxÞ ¼ ðF Ta Þ0 ðxÞ ¼ ðF 0 ðTa ðxÞÞÞ ðTa Þ0 ðxÞ ¼ ðF 0 ðx þ aÞÞ I3 ¼ F 0 ðx þ aÞ: F ^0, F ^ 0 ; then x þ a is in A0, and ^ 0 ðxÞ ¼ F 0 ðxÞ: If x is in A Thus, for every x in A 0 0 ^ hence, F ðxÞð¼ F ðx þ aÞÞ has rank 2. Thus, we see that all the conditions of

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3 Multivariable Differential Calculus

^0 ! B ^ :A ^ 0 : Hence, by Theorem 3.58, there exist an Theorem 3.58 are satisﬁed for F 3 ^ ^ of b in R4 ; an open neighborhood A of (0, 0, 0) in R ; an open neighborhood B 3 4 ^ ! U; ^ :A ^ of R ; an open subset V ^ of R ; a function G ^ and a function open subset U ^ ^ ^ H : B ! V such that ^ is contained in A ^ 0 ; and B ^ is contained in B ^0; 1. A 2 ^ is a C diffeomorphism (i.e., G ^ is 1–1 onto, G ^ is a C2 function, and G ^ 1 is a 2. G 2 C function), ^ is a C2 diffeomorphism, 3. H ^ B; ^ AÞ ^ 4. Fð ^ 1 Þðx1 ; x2 ; x3 Þ ¼ ðx1 ; x2 ; 0; 0Þ: ^ ðH ^ F ^ G 5. For every x ðx1 ; x2 ; x3 Þ in U; ^ þ a; B B; ^ U U; ^ and V V: ^ Let us deﬁne a function G : A ! U Put A A ^ ^ Ta ÞðxÞ: Let us deﬁne a as follows: For every x in A, GðxÞ Gðx aÞ ¼ ðG ^ function H : B ! V as follows: For every y in B, HðyÞ HðyÞ: Hence, G ¼ ^ Ta ; and H ¼ H: ^ ¼ G Ta : ^ Further, G G It remains to prove the following: (i) (ii) (iii) (iv) (v) (vi) (vii) (viii) (ix) (x) (xi) (xii) (xiii) (xiv) (xv) (xvi)

A is an open neighborhood of a, B is an open neighborhood of b, U is an open subset of R3 ; V is an open subset of R4 ; G : A ! U is 1–1, G : A ! U is onto, G : A ! U is C 2 ; G1 : U ! A is C 2 ; H : B ! V is 1–1, H : B ! V is onto, H : B ! V is C 2 ; H 1 : V ! B is C2 ; A is contained in A0 ; B is contained in B0 ; FðAÞ B; For every x ðx1 ; x2 ; x3 Þ in U, ðH F G1 Þðx1 ; x2 ; x3 Þ ¼ ðx1 ; x2 ; 0; 0Þ:

^ is an open neighborhood of ð0; 0; 0Þ; Að¼ A ^ þ aÞ is an open For (i): Since A neighborhood of a. ^ is an open neighborhood of b, Bð¼ BÞ ^ is an open neighborhood For (ii): Since B of b. ^ is an open subset of R3 . ^ is an open subset of R3 ; Uð¼ UÞ For (iii): Since U 4 ^ is an open subset of R4 . ^ is an open subset of R ; Vð¼ VÞ For (iv): Since V ^!U ^þa!A ^ is 1–1, G ^ :A ^ Ta : ^ is 1–1, and Ta : A For (v): Since G ^ ^ Ta : A ! U is 1–1. ^ is 1–1, and hence, G Aþa!U

3.7 The Constant Rank Theorem in Rn

225

^!U ^þa!A ^ is onto, G ^ :A ^ Ta : ^ is onto, and Ta : A For (vi): Since G ^ ^ ^ A þ a ! U is onto, and hence, G Ta : A ! U is onto. ^!U ^þa!A ^ is C2 ; G ^ :A ^ Ta : ^ is C 2 ; and Ta : A For (vii): Since G 2 2 ^ ^ ^ A þ a ! U isC , and hence, G Ta : A ! U is C . ^ 1 : Since ^ 1 ¼ ðG Ta Þ1 ¼ Ta G1 ; G1 ¼ Ta G For (viii): Since G ^ C2 ; and Ta : A ^!A ^ þ a is C 2 ; their composite G1 ¼ Ta ^ 1 : U ^ ! Ais G 1 2 ^ þ a is C ; and hence, G1 : U ! A is C 2 . ^ :U ^ !A G ^ :B ^!V ^ is 1–1, H ¼ H; ^ B ¼ B; ^ and V ¼ V; ^ H : B ! V is For (ix): Since H 1–1. ^ :B ^!V ^ is onto, H ¼ H; ^ B ¼ B; ^ and V ¼ V; ^ H : B ! V is For (x): Since H onto. ^ B ¼ B; ^ and V ¼ V; ^ H : B ! V is C 2 . ^ :B ^!V ^ is C 2 ; H ¼ H; For (xi): Since H ^ !B ^ is C 2 ; H ¼ H; ^ B ¼ B; ^ and V ¼ V; ^ H 1 : V ! B ^ 1 : V For (xii): Since H 2 is C . ^ A aÞ is contained in A ^ 0 ð¼ A0 aÞ; A is contained in A0. For (xiii): Since Að¼ ^ ^ For (xiv): Since Bð¼ BÞ is contained in B0 ð¼ B0 Þ; B is contained in B0. For (xv): Let us take any x in A. We have to prove that FðxÞ 2 B: Since x is in ^ þ a; x a 2 A: ^ Now, since Fð ^ B; ^ AÞ ^ FðxÞ ¼ ðF ^ Ta ÞðxÞ ¼ Fðx ^ aÞ A¼A ^ 2 B. ^ ðx1 ; x2 ; x3 Þ is For (xvi): Let us take any x ðx1 ; x2 ; x3 Þ in U. Now, since U ¼ U; ^ and hence, in U;

^ Ta 1 ðx1 ; x2 ; x3 Þ ^ F ^ Ta G LHS ¼ H F G1 ðx1 ; x2 ; x3 Þ ¼ H

^ 1 ðx1 ; x2 ; x3 Þ ¼ H ^ 1 ðx1 ; x2 ; x3 Þ ^ F ^ G ^ F ^ Ta Ta G ¼ H ¼ ðx1 ; x2 ; 0; 0Þ ¼ RHS: h Theorem 3.60 Let A0 be an open neighborhood of a ða1 ; a2 ; . . .; an Þ in Rn : Let B0 be an open neighborhood of b ðb1 ; b2 ; . . .; bm Þ in Rm : Let F : A0 ! B0 be a function. If (i) F is a Cr function, (ii) For every x in A0 ; the linear transformation F 0 ðxÞ has rank k; (iii) FðaÞ ¼ b; then there exist an open neighborhood A of a in Rn ; an open neighborhood B of b in Rm ; an open subset U of Rn ; an open subset V of Rm ; a function G : A ! U; and a function H : B ! V such that 1. A is contained in A0 ; and B is contained in B0 ; 2. G is a C r diffeomorphism (i.e., G is 1–1 onto, G is a Ck function, and G−1 is a C k function),

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3 Multivariable Differential Calculus

3. H is a Ck diffeomorphism, 4. FðAÞ B; 5. For every x ðx1 ; x2 ; . . .; xn Þ in U, ðH F G1 Þðx1 ; x2 ; . . .; xn Þ ¼ ðx1 ; x2 ; . . . xk ; 0; . . .; 0Þ: |ﬄﬄﬄ{zﬄﬄﬄ} mk

Proof Its proof is quite similar to the proof of Theorem 3.59.

h

Note 3.61 Theorem 3.60 is known as the constant rank theorem.

3.8 Taylor’s Theorem Deﬁnition Let p 2 S Rn : If for every x in S, the line segment ½p; x fð1 kÞp þ kx : 0 k 1g S, then we say that S is star-shaped with respect to point p. Theorem 3.62 Let p 2 U R2 : Let U be star-shaped with respect to point p ðp1 ; p2 Þ; and let U be an open set. Let f : U ! R be C1 on U. Then, there exist C 1 functions g1 ðxÞ; g2 ðxÞ on U such that 1. For every x ðx1 ; x2 Þ in U, f ðxÞ ¼ f ðpÞ þ ðg1 ðxÞÞðx1 p1 Þ þ ðg2 ðxÞÞðx2 p2 Þ; of 2. g1 ðpÞ ¼ oxof1 ðpÞ and g2 ðpÞ ¼ ox 2 ðpÞ: Proof Let us take any x in U and any t satisfying 0 t 1: Now, since U is starshaped with respect to point p, ð1 tÞp þ tx is in S, and hence, f ðð1 tÞp þ txÞ is a real number. Here,

d d f ðð1 tÞp þ txÞ ¼ f p1 þ t x1 p1 ; p2 þ t x2 p2 dt dt 1

of d 1 1 p ¼ ð ð 1 t Þp þ tx Þ þ t x p ox1 dt 2

of d 2 2 p þ ð ð 1 t Þp þ tx Þ þ t x p ox2 dt 1

of ¼ ðð1 tÞp þ txÞ x p1 1 ox

of þ ðð1 tÞp þ txÞ x2 p2 ; 2 ox

3.8 Taylor’s Theorem

227

hence, 2

1 of of 2 Z ð ð 1 t Þp þ tx Þ dt þ x p ð ð 1 t Þp þ tx Þ dt ox1 ox2 0 0 1

2

Z1 of of 1 2 ¼ ðð1 tÞp þ txÞ x p þ ðð1 tÞp þ txÞ x p dt ox1 ox2 0

x 1 p1

Z1

t¼1 ¼ f ðxÞ f ðpÞ: ¼ f ðð1 tÞp þ txÞjt¼0

Now,

f ðxÞ ¼ f ðpÞ þ x1 p1 ðg1 ðxÞÞ þ x2 p2 ðg2 ðxÞÞ; where g1 ðxÞ

Z1

0

of Z1 of ðð1 tÞp þ txÞ dt; and g2 ðxÞ ðð1 tÞp þ txÞ dt: ox1 ox2 0

of It remains to prove that g1 ; g2 are C1 on U, and gi ðpÞ ¼ ox i ðpÞði ¼ 1; 2Þ: Here,

og1 o Z 1 of ¼ ðð1 tÞp þ txÞ dt ox1 ox1 0 ox1 Z o of 1 ¼ 0 ðð1 tÞp þ txÞ dt ox1 ox1 1

2 2

Z o of 1 1 1 2 p þ t x p ;p þ t x p ¼ 0 dt ox1 ox1 1

2 2

o of 1 oð p1 þ t ð x 1 p1 Þ Þ Z 1 2 ; p p þ t x p þ t x p ¼ 10 ox1 ox1 ox1 1 1

2 2

o of 1 oð p þ t ð x 1 p1 Þ Þ 1 2 ; p þ p þ t x p þ t x p dt ox2 ox1 ox2

Z o of 1 p þ t x 1 p1 ; p2 þ t x 2 p2 t ¼ 10 1 1 ox ox

o of 1 þ p þ t x 1 p1 ; p2 þ t x 2 p2 0 dt 2 1 ox ox !!

Z o2 f 1 p þ t x 1 p1 ; p2 þ t x 2 p2 t dt ¼ 10 2 1 oð x Þ ! Z o2 f 1 ðð1 tÞp þ txÞ dt: ¼ 0t oð x 1 Þ 2

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3 Multivariable Differential Calculus

Thus, ! og1 Z1 o2 f ¼ t ðð1 tÞp þ txÞ dt: ox1 0 oðx1 Þ2 Similarly,

og1 o2 g ; ; etc. can be obtained. This proves that g1 is C 1 on U. ox2 oðx1 Þ2

Similarly, g2 is C1 on U. Now, Z1 of Z1 of of Z1 ð ð 1 t Þp þ tp Þ dt ¼ ðpÞ dt ¼ ðpÞ dt g1 ðpÞ ¼ 1 1 1 ox ox ox 0 0 0 of of ¼ ðpÞ 1 ¼ 1 ðpÞ: 1 ox ox of of Thus, g1 ðpÞ ¼ ox 1 ðpÞ: Similarly, g2 ðpÞ ¼ ox2 ðpÞ:

h

Note 3.63 The result similar to above is also valid for R ; R ; . . .: 3

4

Theorem 3.64 Prove that if f : R2 ! R is C 1 ; then there exist C1 functions g11 ; g12 ; g22 on R2 such that

of of ð0; 0Þ x þ ð0; 0Þ y þ x2 g11 ðx; yÞ þ xyg12 ðx; yÞ f ðx; yÞ ¼ f ð0; 0Þ þ ox oy þ y2 g22 ðx; yÞ: Proof Here, f : R2 ! R is C 1 , by Theorem 3.62, there exist C1 functions g1 ðx; yÞ; g2 ðx; yÞ on R2 such that for every ðx; yÞ in R2 ; f ðx; yÞ ¼ f ð0; 0Þ þ ðg1 ðx; yÞÞðx 0Þ þ ðg2 ðx; yÞÞðy 0Þ ¼ f ð0; 0Þ þ xg1 ðx; yÞ þ yg2 ðx; yÞ and g1 ð0; 0Þ of 1 function, there exist C 1 ¼ of ox ð0; 0Þ; g2 ð0; 0Þ ¼ oy ð0; 0Þ: Since g1 ðx; yÞ is a C functions g11 ðx; yÞ; g12 ðx; yÞ on R2 such that for everyðx; yÞ in R2 ; g1 ðx; yÞ ¼ of g1 ð0; 0Þ þ ðg11 ðx; yÞÞðx 0Þ þ ðg12 ðx; yÞÞðy 0Þ ¼ ox ð0; 0Þ þ xg11 ðx; yÞ þ yg12 ðx; yÞ og1 og1 and g11 ð0; 0Þ ¼ ð ox Þð0; 0Þ; g12 ð0; 0Þ ¼ oy ð0; 0Þ: Similarly, there exist C 1 functions g21 ðx; yÞ; g22 ðx; yÞ on R2 such that for every ðx; yÞ in R2 ; g2 ðx; yÞ ¼ g2 ð0; 0Þ þ ðg21 ðx; yÞÞðx 0Þþ ðg22 ðx; yÞÞðy 0Þ ¼ of oy ð0; 0Þ þ xg21 ðx; yÞ þ yg22 ðx; yÞ og2 2 and g21 ð0; 0Þ ¼ ðog ox Þð0; 0Þ; g22 ð0; 0Þ ¼ oy ð0; 0Þ: Hence,

3.8 Taylor’s Theorem

229

of ð0; 0Þ þ xg11 ðx; yÞ þ yg12 ðx; yÞ f ðx; yÞ ¼ f ð0; 0Þ þ x ox of þy ð0; 0Þ þ xg21 ðx; yÞ þ yg22 ðx; yÞ oy of of ¼ f ð0; 0Þ þ ð0; 0Þ x þ ð0; 0Þ y ox oy

þ ðg11 ðx; yÞÞx2 þ ðg12 ðx; yÞ þ g21 ðx; yÞÞxy þ ðg22 ðx; yÞÞy2 : Since g12 ðx; yÞ; g21 ðx; yÞ are C1 functions, g12 ðx; yÞ þ g21 ðx; yÞ is also a C 1 h function.

Chapter 4

Topological Properties of Smooth Manifolds

In real analysis, its deep theorems (like Lagrange’s mean value theorem, fundamental theorem of integral calculus, etc.) are quite difﬁcult to prove without the help of topological theorems (like Weierstrass theorem, Heine–Borel theorem, etc.) for real line. Similar is the situation in complex analysis. Here, in the study of smooth manifolds, in order to prove its deep rooted theorems, we will need its topological properties vigorously. So, for smoothening later work, it is better to collect all the results, to be needed in future, relating to topological properties of smooth manifolds at one place. This chapter aims at this end and is largely self-contained. The pace of this chapter is moderate so that the topic is well assimilated.

4.1 Constant Rank Theorem Theorem 4.1 Let E be an open subset of R3 . Let f : E ! R3 : Let a be in E. If (i) f is a smooth function, (ii) the determinant of the Jacobian matrix of f at a is nonzero, that is, 2

ðD1 f 1 ÞðaÞ 4 det ðD1 f 2 ÞðaÞ ðD1 f 3 ÞðaÞ

ðD2 f 1 ÞðaÞ ðD2 f 2 ÞðaÞ ðD2 f 3 ÞðaÞ

3 ðD3 f 1 ÞðaÞ ðD3 f 2 ÞðaÞ 5 6¼ 0; ðD3 f 3 ÞðaÞ 33

where f 1 ; f 2 ; f 3 are the component functions of f, then there exists a connected open neighborhood U of a such that 1. U is contained in E, 2. f ðUÞ is connected and open in R3 ; 3. f has a smooth inverse on f ðUÞ:

© Springer India 2014 R. Sinha, Smooth Manifolds, DOI 10.1007/978-81-322-2104-3_4

231

232

4 Topological Properties of Smooth Manifolds

Proof We want to apply Theorem 3.36. Since f is a smooth function, f is a C 1 function. Since 2 3 ðD1 f 1 ÞðaÞ ðD2 f 1 ÞðaÞ ðD3 f 1 ÞðaÞ det4 ðD1 f 2 ÞðaÞ ðD2 f 2 ÞðaÞ ðD3 f 2 ÞðaÞ 5 6¼ 0; ðD1 f 3 ÞðaÞ ðD2 f 3 ÞðaÞ ðD3 f 3 ÞðaÞ 33 2

ðD1 f 1 ÞðaÞ ðD2 f 1 ÞðaÞ 4 ðD1 f 2 ÞðaÞ ðD2 f 2 ÞðaÞ ðD1 f 3 ÞðaÞ ðD2 f 3 ÞðaÞ

31 ðD3 f 1 ÞðaÞ ðD3 f 2 ÞðaÞ 5 ðD3 f 3 ÞðaÞ

exists, and hence, f 0 ðaÞ is invertible. Hence, by the Theorem 3.36, there exists a connected open neighborhood U of a such that 1. 2. 3. 4. 5.

U is contained in E, f is 1–1 on U (that is, if x, y are in U, and f ðxÞ ¼ f ðyÞ; then x = y), f ðUÞ is connected and open in R3 ; the 1–1 function f 1 from f ðUÞ onto U is a C 1 function, if f is a C2 function, then f 1 from f ðUÞ onto U is a C 2 function, etc.

By the conclusions 1, 2, 3, and 4, we ﬁnd the f 1 is a C 1 function on f ðUÞ: Now, it remains to prove that f 1 is smooth on f ðUÞ; that is, f 1 is a Cn function for every n ¼ 2; 3; 4; . . .: Since f is a smooth function, f is a C 2 function, and hence by conclusion 5, f 1 is a C 2 function. Again, since f is a smooth function, f is a C 3 function, and hence by conclusion 5, f 1 is a C3 function, etc. Hence, h f 1 is smooth on f ðUÞ: Note 4.2 The result similar to Theorem 4.1 can be proved as above for Rn in place of R3 : Theorem 4.3 Let E be an open subset of R3 : Let f : E ! R3 : Let a be in E. If (i) f is a smooth function, (ii) the tangent map f of f at a is an isomorphism, then there exists an open neighborhood U of a, and an open neighborhood V of f ðaÞ such that the restriction of f to U is a diffeomorphism from U onto V. Proof By the Note 2.77, the tangent map f of f at a is an isomorphism if and only if 2 3 ðD1 f 1 ÞðaÞ ðD2 f 1 ÞðaÞ ðD3 f 1 ÞðaÞ det4 ðD1 f 2 ÞðaÞ ðD2 f 2 ÞðaÞ ðD3 f 2 ÞðaÞ 5 6¼ 0: ðD1 f 3 ÞðaÞ ðD2 f 3 ÞðaÞ ðD3 f 3 ÞðaÞ

4.1 Constant Rank Theorem

233

This shows that we can apply Theorem 4.1. By Theorem 4.1, there exists an open neighborhood U of a such that 1. U is contained in E, 2. f ðUÞ is open in R3 ; 3. f has a smooth inverse on f ðUÞ: This shows that f ðUÞ is an open neighborhood of f ðaÞ such that the restriction of f to U is a diffeomorphism from U onto f ðUÞ: h Note 4.4 The result similar to Theorem 4.3 can be proved as above for Rn in place of R3 : Theorem 4.5 Let M and N be 3-dimensional smooth manifolds. Let f : M ! N: Let p be in M. If (i) f is a smooth map, (ii) the tangent map f : Tp ðMÞ ! Tf ðpÞ ðNÞ is an isomorphism, then there exists an open neighborhood U of p in M such that 1. V f ðUÞ is an open neighborhood of f ðpÞ in N, 2. the restriction of f to U is a diffeomorphism from U onto V. Proof Here, p is in M, and f : M ! N; so f ðpÞ is in N. Since f ðpÞ is in N, and N is a 3-dimensional smooth manifold, there exists an admissible coordinate chart ðV; wV Þ in N satisfying f ðpÞ 2 V: Since f : M ! N is a smooth map, f : M ! N is a continuous map. Since f : M ! N is a continuous map, and V is an open neighborhood of FðpÞ; there exists an open neighborhood U1 of p in M such that f ðU1 Þ is contained in V. Here, p is in M, and M is a 3-dimensional smooth manifold, so there exists an admissible coordinate chart ðU; uU Þ in M satisfying p 2 U: Since U and U1 are open neighborhoods of p in M, their intersection U \ U1 is an open neighborhood of p in M. Since uU is a homeomorphism from open set U onto open set uU ðUÞ; and U \ U1 is an open set, uU ðU \ U1 Þ is an open set. Here f : M ! N is a smooth map, ðuU Þ1 is a smooth map, and wV is a smooth map, so their composite wV ðf ðuU Þ1 Þ is a smooth function on the open neighborhood uU ðU \ U1 Þ of uU ðpÞ in R3 : We want to apply Theorem 4.1 on the function wV ðf ðuU Þ1 Þ: Since the tangent map f : Tp ðMÞ ! Tf ðpÞ ðNÞ is an isomorphism so, by the Note 2.77,

234

4 Topological Properties of Smooth Manifolds

2

1 1 1 1 u w f ð u Þ ð ð p Þ Þ D w f ð u Þ ðuU ð pÞÞ D 1 2 U U U V V 6 6 6 2 2 6 det6 D1 wV f ðuU Þ1 ðuU ð pÞÞ D2 wV f ðuU Þ1 ðuU ð pÞÞ 6 6 4 3 3 ðuU ð pÞÞ D2 wV f ðuU Þ1 ðuU ð pÞÞ D1 wV f ðuU Þ1 3 1 D3 wV f ðuU Þ1 ðu U ð p ÞÞ 7 7 7 2 7 1 ðuU ð pÞÞ 7 6¼ 0; D3 wV f ðuU Þ 7 7 3 5 1 D3 wV f ðuU Þ ðu U ð p ÞÞ

and hence, the determinant of the Jacobian matrix of wV ðf ðuU Þ1 Þ at uU ðpÞ is nonzero. Hence, by Theorem 4.1, there exists an open neighborhood W of uU ðpÞ such that 1. W is contained in uU ðU \ U1 Þ; 2. ðwV ðf ðuU Þ1 ÞÞðWÞ is open in R3 ; 3. wV ðf ðuU Þ1 Þ has a smooth inverse on ðwV ðf ðuU Þ1 ÞÞðWÞ: Since W is contained in uU ðU \ U1 Þ; and uU is 1–1, ðuU Þ1 ðWÞ is contained in U \ U1 : Since uU ðpÞ is in W, p is in ðuU Þ1 ðWÞ: Since uU is a homeomorphism, and W is an open subset of open set uU ðUÞ; ðuU Þ1 ðWÞ is open. Thus, ðuU Þ1 ðWÞ is an open neighborhood of p in M. For 1: We have to prove that f ððuU Þ1 ðWÞÞ is open in N. Since ðwV ðf ðuU Þ1 ÞÞðWÞ is open, and wV is a homeomorphism, ðwV Þ1 ððwV ðf ðuU Þ1 ÞÞ ðWÞÞð¼ f ððuU Þ1 ðWÞÞÞ is open. For 2: Since wV ðf ðuU Þ1 Þ is 1–1, ðwV Þ1 is 1–1 and uU is 1–1, their composite ðwV Þ1 ðwV ðf ðuU Þ1 ÞÞ uU ð¼ f Þ is 1–1. Since f is 1–1, the function f 1 exists. It remains to prove that f 1 is smooth. Since ðwV ðf ðuU Þ1 ÞÞ1 ð¼ uU f 1 ðwV Þ1 Þ is smooth, ðuU Þ1 is smooth, and wV is smooth, their composite function ðuU Þ1 ðuU f 1 ðwV Þ1 Þ wV ð¼ f 1 Þ is smooth. h Note 4.6 The result similar to Theorem 4.5 can be proved as above for Rn in place of R3 : Theorem 4.7 Let M be an m-dimensional differentiable manifold, N be an ndimensional smooth manifold, and F : N ! M be a smooth function. Let p be in N. Let ðU; uU Þ be an admissible coordinate chart in N satisfying p 2 U; ðV; wV Þ be an b;u Þ admissible coordinate chart in M satisfying FðpÞ 2 V; and FðUÞ V: Let ð U b U

4.1 Constant Rank Theorem

235

b ; ðV b ; w Þ be an admissible be an admissible coordinate chart in N satisfying p 2 U b V b ; and Fð U bÞ V b : Then, the rank of coordinate chart in M satisfying FðpÞ 2 V ðwV F ðuU Þ1 Þ0 ðuU ðpÞÞ and the rank of ðwb F ðu b Þ1 Þ0 ðu b ðpÞÞ are equal. V

U

U

Proof Here,

1 0 1 0 wb F u b u b ð pÞ ¼ wb F ðuU Þ1 uU u b u b ð pÞ V U U V U U 1 0 u b ð pÞ ¼ wb ðwV Þ1 wV F ðuU Þ1 uU u b V U U 1 0 ¼ wb ðwV Þ1 wV F ðuU Þ1 uU u b u b ð pÞ V U U 0 1 ¼ wb ðwV Þ1 wV F ðuU Þ1 uU u b u b ð pÞ V U U 0 1 1 0 wV F ðuU Þ1 uU u b u b ð pÞ uU u b u b ð pÞ U U U U 0 1 ¼ wb ðwV Þ1 wV F ðuU Þ1 uU u b u b ð pÞ V U U 0 1 0 wV F ðuU Þ1 ðuU ð pÞÞ uU u b u b ð pÞ U U 0 0 1 1 ¼ w b ðwV Þ ððwV F Þð pÞÞ wV F ðuU Þ ðuU ð pÞÞ V 1 0 u b ð pÞ : uU u b U

U

b Þ ! w ðV \ V b Þ is a smooth function, and Since wb ðwV Þ1 : wV ðV \ V b V V 1 1 1 b b Þ is a smooth funcðwb ðwV Þ Þ ¼ wV ðwb Þ : wb ðV \ V Þ ! wV ðV \ V V V V b Þ ! w ðV \ V b Þ is a diffeomorphism, and hence tion, wb ðwV Þ1 : wV ðV \ V b V V 1 0 ðwb ðwV Þ Þ ððwV FÞðpÞÞ is invertible. It follows that the rank of composite V

0 0 wb ðwV Þ1 ððwV F Þð pÞÞ wV F ðuU Þ1 ðuU ð pÞÞ V

and the rank of ððwV F ðuU Þ1 Þ0 ðuU ðpÞÞÞ are equal. Similarly the rank of 0 0 wb ðwV Þ1 ððwV F Þð pÞÞ wV F ðuU Þ1 ðuU ð pÞÞ V 1 0 uU u b u b ð pÞ U

U

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4 Topological Properties of Smooth Manifolds

is equal to the rank of 0 0 wb ðwV Þ1 ððwV F Þð pÞÞ wV F ðuU Þ1 ðuU ð pÞÞ : V

Hence, rank of 0 0 wb ðwV Þ1 ððwV F Þð pÞÞ wV F ðuU Þ1 ðuU ð pÞÞ V

is equal to rank of ððwV F ðuU Þ1 Þ0 ðuU ðpÞÞÞ:

h

In light of the above theorem, the following deﬁnition is well deﬁned: Deﬁnition Let M be an m-dimensional smooth manifold, N be an n-dimensional smooth manifold, and F : N ! M be a smooth function. Let p be in N. Since F : N ! M is a smooth function, there exist an admissible coordinate chart ðU; uU Þ in N satisfying p 2 U; and an admissible coordinate chart ðV; wV Þ in M satisfying FðpÞ 2 V; such that FðUÞ V; and wV ðF ðuU Þ1 Þ : uU ðUÞ ! wV ðVÞ is C 1 at the point uU ðpÞ in R3 : Hence, the linear transformation ðwV F ðuU Þ1 Þ0 ðuU ðpÞÞ from Rn to Rm exists. The rank of ðwV F ðuU Þ1 Þ0 ðuU ðpÞÞ is called the rank of F at p. Since rankððwV F ðuU Þ1 Þ0 ðuU ðpÞÞÞ minfm; ng; the rank of F at p is less than or equal to minfm; ng: Deﬁnition Let M be an m-dimensional differentiable manifold, N be an n-dimensional smooth manifold, and F : N ! M be a smooth function. Let k be any nonnegative integer. By F has rank k, we mean that k is the rank of F at every point of N. Theorem 4.8 Let N be a 4-dimensional smooth manifold, M be a 3-dimensional smooth manifold, F : M ! N be a smooth map, and 2 be the rank of F. Let p be in M. Then, there exist admissible coordinate chart ðU ; uU Þ in M satisfying p 2 U ; and admissible coordinate chart ðV ; wV Þ in N satisfying FðpÞ 2 V ; such that FðU Þ V ; and for every ðx1 ; x2 ; x3 Þ in uU ðU Þ; wV F ðuU Þ1 ðx1 ; x2 ; x3 Þ ¼ ðx1 ; x2 ; 0; 0Þ: Proof Since p is in M, and F : M ! N is a smooth map, there exist an admissible coordinate chart ðU; uU Þ in M satisfying p 2 U; and an admissible coordinate chart ðV; wV Þ in N satisfying FðpÞ 2 V such that FðUÞ V; and each of the 4 component functions of the mapping wV F ðuU Þ1 : uU ðU Þ ! wV ðV Þ is C 1 at the point uU ðpÞ in R3 :

4.1 Constant Rank Theorem

237

Now, we want to apply Theorem 3.59. Here, uU ðUÞ acts for A0 ; uU ðpÞ acts for a, wV ðVÞ acts for B0, wV ðFðpÞÞ acts for b, wV ðF ðuU Þ1 Þ acts for F. Now, we must verify the following conditions of the theorem: uU ðUÞ is an open neighborhood of uU ðpÞ in R3 ; wV ðVÞ is an open neighborhood of wV ðFðpÞÞ in R4 ; wV ðF ðuU Þ1 Þ : uU ðUÞ ! wV ðVÞ is a C 1 function, for every x in uU ðUÞ; the linear transformation ðwV ðF ðuU Þ1 ÞÞ0 ðxÞ has rank 2, 5. ðwV ðF ðuU Þ1 ÞÞðuU ðpÞÞ ¼ wV ðFðpÞÞ:

1. 2. 3. 4.

For 1: Since ðU; uU Þ is an admissible coordinate chart in M satisfying p 2 U; uU ðUÞ is an open neighborhood of uU ðpÞ in R3 : For 2: Since ðV; wV Þ is an admissible coordinate chart in N satisfying FðpÞ 2 V; wV ðVÞ is an open neighborhood of wV ðFðpÞÞ in R4 : For 3: Since F : M ! N is a smooth function, ðU; uU Þ is an admissible coordinate chart in M satisfying p 2 U; and ðV; wV Þ an admissible coordinate chart in N satisfying FðpÞ 2 V; wV ðF ðuU Þ1 Þ : uU ðUÞ ! wV ðVÞ is a C 1 function. For 4: Let us take any uU ðqÞ in uU ðUÞ where q is in U. We have to prove that the rank of ðwV F ðuU Þ1 Þ0 ðuU ðqÞÞ is 2. Since 2 is the rank of F : M ! N; and p is in N, the rank of F at p is 2, and hence, the rank of ðwV F ðuU Þ1 Þ0 ðuU ðpÞÞ is 2. For 5: Here, LHS ¼ ðwV ðF ðuU Þ1 ÞÞðuU ðpÞÞ ¼ ðwV FÞðpÞ ¼ wV ðFðpÞÞ ¼ RHS: Hence, by Theorem 3.59, there exists an open neighborhood A of uU ðpÞ in b of R3 ; an R3 ; an open neighborhood B of ðwV FÞðpÞ in R4 ; an open subset U 4 b :A!U b of R ; a function G b ; and a function H b :B!V b such that open subset V 1. A is contained in uU ðUÞ; and B is contained in wV ðVÞ; b is a C 1 diffeomorphism (that is, G b is 1–1 onto, G b is a C 1 function, and 2. G b 1 is a C1 function), G b is a C 1 diffeomorphism, 3. H 4. ðwV ðF ðuU Þ1 ÞÞðAÞ B; b 1 Þ b ; ðH b ðwV ðF ðuU Þ1 ÞÞ G 5. For every x ðx1 ; x2 ; x3 Þ in U ðx1 ; x2 ; x3 Þ ¼ ðx1 ; x2 ; 0; 0Þ: b uU for uU ; H b wV for wV ; ðuU Þ1 ðAÞ for U ; ðwV Þ1 ðBÞ Now, let us take G for V : It remains to prove: (i) ðuU Þ1 ðAÞ is open in M, b uU ÞððuU Þ1 ðAÞÞ is open in R3 ; (ii) ð G b uU ÞððuU Þ1 ðAÞÞ is a homeomorphism, b uU : ðuU Þ1 ðAÞ ! ð G (iii) G

238

4 Topological Properties of Smooth Manifolds

b uU Þ is an admissible coordinate chart in M satisfying (iv) ððuU Þ1 ðAÞ; G 1 p 2 ðuU Þ ðAÞ; b wV Þ is an admissible coordinate chart in N satisfying (v) ððwV Þ1 ðBÞ; H 1 FðpÞ 2 ðwV Þ ðBÞ; b uU ÞððuU Þ1 ðAÞÞ; ðð H b b wV Þ F ð G (vi) for every ðx1 ; x2 ; x3 Þ in ð G 1 uU Þ Þðx1 ; x2 ; x3 Þ ¼ ðx1 ; x2 ; 0; 0Þ; (vii) FððuU Þ1 ðAÞÞ ðwV Þ1 ðBÞ: For (i): Since A is an open neighborhood of uU ðpÞ in R3 ; A is contained in uU ðUÞ; and ðU; uU Þ is an admissible coordinate chart in M satisfying p 2 U; ðuU Þ1 ðAÞ is open in M. b :A!U b :A!U b is a diffeomorphism, G b is a homeomorFor (ii): Since G 1 b b phism, and hence, ð G uU ÞððuU Þ ðAÞÞð¼ GðAÞÞ is open in R3 . b :A!U b :A!U b is a homeob is a C 1 diffeomorphism, G For (iii): Since G 1 b :A!U b is morphism. Since uU : ðuU Þ ðAÞ ! A is a homeomorphism, and G 1 1 b uU : ðuU Þ ðAÞ ! ð G b uU ÞððuU Þ ðAÞÞð¼ GðAÞ b bÞ a homeomorphism, G ¼U is a homeomorphism. For (iv): Let ðW; vW Þ be any admissible coordinate chart in M satisfying W \ððuU Þ1 ðAÞÞ is nonempty. Since A is an open neighborhood of uU ðpÞ; uU ðpÞ 2 A; and hence p 2 ðuU Þ1 ðAÞ: b uU Þ1 ; and ð G b uU Þ ðvW Þ1 are C1 : Now, we have to prove that vW ð G Since ðW; vW Þ; ðU; uU Þ are admissible coordinate charts in M, vW ðuU Þ1 is b is a C1 diffeomorphism, G b 1 is C 1 , and hence, vW C1 : Now, since G 1 1 1 1 b uU Þ ð¼ ðvW ðuU Þ Þ G b Þ is C : Similarly, ð G b uU Þ ðvW Þ1 is C 1 : ðG For (v): Its proof is similar to that of (iv). b uU ÞððuU Þ1 ðAÞÞ: We have to prove For (vi): Let us take any ðx1 ; x2 ; x3 Þ in ð G that 1 b uU b wV F G H ðx1 ; x2 ; x3 Þ ¼ ðx1 ; x2 ; 0; 0Þ: b uU ÞððuU Þ1 ðAÞÞ ¼ GðAÞ b b; ¼U Since ðx1 ; x2 ; x3 Þ 2 ð G 1 b uU b wV F G H ð x1 ; x2 ; x3 Þ b 1 ðx1 ; x2 ; x3 Þ b wV F ðuU Þ1 G ¼ H

LHS ¼

¼ ðx1 ; x2 ; 0; 0Þ ¼ RHS:

4.1 Constant Rank Theorem

239

For (vii): Since wV ððF ðuU Þ1 ÞðAÞÞ ¼ ðwV ðF ðuU Þ1 ÞÞðAÞ B; and wV is 1–1, FððuU Þ1 ðAÞÞ ðwV Þ1 ðBÞ: h Note 4.9 As in Theorem 4.8, we can prove the following result: Let N be an n-dimensional smooth manifold, M be an m-dimensional smooth manifold, F : M ! N be a smooth map, and r be the rank of F. Then, for every p in M, there exist admissible coordinate chart ðU; uÞ in M satisfying p 2 U; and admissible coordinate chart ðV; wÞ in N satisfying FðpÞ 2 V such that FðUÞ V; and for every ðx1 ; . . .; xr ; xrþ1 ; . . .; xm Þ in uðUÞ; 0 1 w F u1 ðx1 ; . . .; xr ; xrþ1 ; . . .; xm Þ ¼ @x1 ; . . .; xr ; 0; . . .; 0 A: |ﬄﬄﬄ{zﬄﬄﬄ} nr

This theorem is also known as the constant rank theorem. Lemma 4.10 Let M be an m-dimensional smooth manifold. Let ðU; uU Þ be an admissible coordinate chart of M. Let V be a nonempty open subset of U. Then, ðV; wV Þ is an admissible coordinate chart of M, where wV denotes the restriction of uU on V. Proof Here, we must prove: 1. 2. 3. 4. 5. 6.

V is open, wV ðVÞ is open, wV : V ! wV ðVÞ is 1–1 onto, wV is continuous, ðwV Þ1 is continuous, if ðW; vW Þ is an admissible wV ðW \ VÞ ! vW ðW \ VÞ is a 7. if ðW; vW Þ is an admissible vW ðW \ VÞ ! wV ðW \ VÞ is a

coordinate chart of M, then vW ðwV Þ1 : smooth function, coordinate chart of M, then wV ðvW Þ1 : smooth function.

For (1): It is given. For (2): Since ðU; uU Þ is a coordinate chart of M, uU ðUÞ is open in R4 ; and uU : U ! uU ðUÞ is an open mapping. Since uU : U ! uU ðUÞ is an open mapping, and V is a nonempty open subset of U, uU ðVÞ is an open subset of the open set uU ðUÞ; and hence, uU ðVÞ is an open set. Since uU ðVÞ is an open set, and wV is the restriction of uU on V, wV ðVÞ is an open set. For (3): Since ðU; uU Þ is a coordinate chart of M, uU : U ! uU ðUÞ is 1–1. Since uU : U ! uU ðUÞ is 1–1, and wV is the restriction of φU on V, wV : V ! wV ðVÞ is 1–1. Further, clearly wV : V ! wV ðVÞ is onto. For (4): Since ðU; uU Þ is a coordinate chart of M, uU : U ! uU ðUÞ is continuous, and hence, its restriction wV : V ! wV ðVÞ is continuous.

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4 Topological Properties of Smooth Manifolds

For (5): Since V is contained in U, uU ðVÞð¼ wV ðVÞÞ is contained in uU ðUÞ: Since ðU; uU Þ is a coordinate chart of M, ðuU Þ1 : uU ðUÞ ! U is continuous, and hence, its restriction ðuU Þ1 jwV ðVÞ ð¼ ðwV Þ1 Þ is continuous. For (6): Let us take any admissible coordinate chart ðW; vW Þ of M. We shall try to prove that vW ðwV Þ1 : wV ðW \ VÞ ! vW ðW \ VÞ is a smooth function. Since ðW; vW Þ; and ðU; uU Þ are admissible coordinate charts of M, vW ðuU Þ1 : uU ðW \ UÞ ! vW ðW \ UÞ is a smooth function. Since V is contained in U, W \ U; is contained in W \ U; and hence, uU ðW \ VÞ is contained in uU ðW \ UÞ: Since wV is the restriction of uU : U ! uU ðUÞ on V, and W \ V is contained in V, wV ðW \ VÞ ¼ uU ðW \ VÞ: Since wV ðW \ VÞ ¼ uU ðW \ VÞ; and uU ðW \ VÞ is contained in uU ðW \ UÞ; wV ðW \ VÞ is contained in uU ðW \ UÞ: Since wV ðW \ VÞ is contained in uU ðW \ UÞ; and vW ðuU Þ1 : uU ðW \ UÞ ! vW ðW \ UÞ is a smooth function, the restriction of vW ðuU Þ1 to wV ðW \ VÞ is a smooth function. If wV ðxÞ 2 wV ðW \ VÞ; where x 2 W \ Vð VÞ; then, ðvW ðuU Þ1 ÞðwV ðxÞÞ ¼ ðvW ðuU Þ1 ÞðuU ðxÞÞ ¼ vW ðxÞ ¼ ðvW ðwV Þ1 ÞðwV ðxÞÞ: It follows that the restriction of vW ðuU Þ1 to wV ðW \ VÞ; and vW ðwV Þ1 : wV ðW \ VÞ ! vW ðW \ VÞ are the same functions. Since the restriction of vW ðuU Þ1 to wV ðW \ VÞ; and vW ðwV Þ1 : wV ðW \ VÞ ! vW ðW \ VÞ are the same functions, and the restriction of vW ðuU Þ1 to wV ðW \ VÞ is a smooth function, vW ðwV Þ1 : wV ðW \ VÞ ! vW ðW \ VÞ is a smooth function. For (7): Let us take any admissible coordinate chart ðW; vW Þ of M. We shall try to prove that wV ðvW Þ1 : vW ðW \ VÞ ! wV ðW \ VÞ is a smooth function. Since ðW; vW Þ; and ðU; uU Þ are admissible coordinate charts of M, uU ðvW Þ1 : vW ðW \ UÞ ! uU ðW \ UÞ is a smooth function. Since V is contained in U, W \ V is contained in W \ U; and hence, vW ðW \ VÞ is contained in vW ðW \ UÞ: Since vW ðW \ VÞ is contained in vW ðW \ UÞ; and uU ðvW Þ1 : vW ðW \ UÞ ! uU ðW \ UÞ is a smooth function, the restriction of uU ðvW Þ1 to vW ðW \ VÞ is a smooth function. If vW ðxÞ 2 vW ðW \ VÞ; where x 2 W \ Vð VÞ; then uU ðxÞ ¼ wV ðxÞ; and hence, ðuU ðvW Þ1 ÞðvW ðxÞÞ ¼ uU ðxÞ ¼ wV ðxÞ ¼ ðwV ðvW Þ1 ÞðvW ðxÞÞ: It follows that restriction of uU ðvW Þ1 to vW ðW \ VÞ; and wV ðvW Þ1 : vW ðW \ VÞ ! wV ðW \ VÞ are the same the functions. Since the restriction of uU ðvW Þ1 to vW ðW \ VÞ; and the function wV ðvW Þ1 : vW ðW \ VÞ ! wV ðW \ VÞ are the same, and the restriction of uU ðvW Þ1 to vW ðW \ VÞ is a smooth function, h wV ðvW Þ1 : vW ðW \ VÞ ! wV ðW \ VÞ is a smooth function. Lemma 4.11 Let M be an m-dimensional smooth manifold. Let ðU; uU Þ be an admissible coordinate chart of M. Let G be a nonempty open subset of the open set

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uU ðUÞð Rm Þ: Then ððuU Þ1 ðGÞ; wðuU Þ1 ðGÞ Þ is an admissible coordinate chart of M, where wðuU Þ1 ðGÞ denotes the restriction of uU on ðuU Þ1 ðGÞ:

Proof Since ðU; uU Þ is an admissible coordinate chart of M, and G is a nonempty open subset of the open set uU ðUÞ; ðuU Þ1 ðGÞ is a nonempty open subset of U. So, by Lemma 4.10, ððuU Þ1 ðGÞ; wðuU Þ1 ðGÞ Þ is an admissible coordinate chart of M,

where wðuU Þ1 ðGÞ denotes the restriction of uU on ðuU Þ1 ðGÞ:

h

Theorem 4.12 Let N be a 4-dimensional smooth manifold, M be a 3-dimensional smooth manifold, F : M ! N be a smooth function, and 2 be the rank of F. Let p be in M. Then, there exist admissible coordinate chart ðU; uU Þ in M satisfying p 2 U; and admissible coordinate chart ðV; wV Þ in N satisfying FðpÞ 2 V such that uU ðpÞ ¼ ð0; 0; 0Þ; wV ðFðpÞÞ ¼ ð0; 0; 0; 0Þ and for every ðx1 ; x2 ; x3 Þ in uU ðUÞ;

wV F ðuU Þ1 ðx1 ; x2 ; x3 Þ ¼ ðx1 ; x2 ; 0; 0Þ:

Further, we may assume uU ðUÞ ¼ ðe; eÞ ðe; eÞ ðe; eÞ, and wV ðVÞ ¼ ðe; eÞ ðe; eÞ ðe; eÞ ðe; eÞ for some e [ 0: In short, uU ðUÞ ¼ Ce3 ð0Þ; and wV ðVÞ ¼ Ce4 ð0Þ for some e [ 0: Proof By Theorem 4.8, there exist admissible coordinate chart ðU ; uU Þ in M satisfying p 2 U ; and admissible coordinate chart ðV ; wV Þ in N satisfying FðpÞ 2 V such that for every ðx1 ; x2 ; x3 Þ in uU ðU Þ; wV F ðuU Þ1 ðx1 ; x2 ; x3 Þ ¼ ðx1 ; x2 ; 0; 0Þ: Put U U ; uU uU uU ðpÞ; V V ; wV wV ððuU ðpÞÞ1 ; ðuU ðpÞÞ2 ; 0; 0Þ: It remains to prove: (i) (ii) (iii) (iv) (v)

ðU; uU Þ is an admissible coordinate chart in N satisfying p 2 U; ðV; wV Þ is an admissible coordinate chart in M satisfying FðpÞ 2 V; uU ðpÞ ¼ ð0; 0; 0Þ; for every ðx1 ; x2 ; x3 Þ in uU ðUÞ; ðwV F ðuU Þ1 Þðx1 ; x2 ; x3 Þ ¼ ðx1 ; x2 ; 0; 0Þ; wV ðFðpÞÞ ¼ ð0; 0; 0; 0Þ:

For (i): Since ðU ; uU Þ is an admissible coordinate chart in N satisfying p 2 U , U ð¼ UÞ is an open neighborhood of p, and hence, U is an open neighborhood of p. Since ðU ; uU Þ is an admissible coordinate chart in N, uU ðU Þ is open in R3 : Now, since U ¼ U ; uU ðUÞ is open in R3 ; and hence, ðuU þ uU ðpÞÞ ðUÞð¼ uU ðUÞ þ uU ðpÞÞ is open in R3 : Since uU ðUÞ þ uU ðpÞ is open in R3 ; uU ðUÞ is open in R3 : Since ðU ; uU Þ is an admissible coordinate chart in M,

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uU : U ! uU ðU Þ is a homeomorphism. Since uU : U ! uU ðU Þ is a homeomorphism, and the translation TuU ð pÞ : uU ðU Þ ! ðuU ðU Þ uU ð pÞÞ is a homeomorphism, their composite ðuU ¼ÞTuU ð pÞ uU : U ð¼ U Þ ! ðuU ðU Þ uU ð pÞÞð¼ ðuU uU ð pÞÞðU Þ¼ uU ðU ÞÞ

is a homeomorphism. Thus, we have shown that ðU; uU Þ is a coordinate chart in M satisfying p 2 U: Next, let ðW; vW Þ be an admissible coordinate chart in M such that U \ W is nonempty. Now, we have to prove that vW ðuU Þ1 ; and uU ðvW Þ1 are C1 : Here, 1 vW ðuU Þ1 ¼ vW ðuU uU ð pÞÞ1 ¼ vW TuU ð pÞ uU ¼ vW ðuU Þ1 TuU ð pÞ ¼ vW ðuU Þ1 TuU ð pÞ ; so vW ðuU Þ1 ¼ ðvW ðuU Þ1 Þ TuU ðpÞ : Since ðW; vW Þ; and ðU ; uU Þ are

admissible coordinate charts in M, vW ðuU Þ1 is C 1 . Since vW ðuU Þ1 is C1 ; and translation TuU ðpÞ is C1 , their composite ðvW ðuU Þ1 Þ TuU ðpÞ ð¼ vW ðuU Þ1 Þ is C1 , and hence, vW ðuU Þ1 is C1 . Similarly, uU ðvW Þ1 is C1 :

For (ii): Its proof is similar to that of (i). For (iii): uU ðpÞ ¼ ðuU uU ðpÞÞðpÞ ¼ uU ðpÞ uU ðpÞ ¼ ð0; 0; 0Þ: Thus uU ðpÞ ¼ ð0; 0; 0Þ: For (iv): Let us take any ðx1 ; x2 ; x3 Þ in uU ðUÞ: We have to prove that ðwV F ðuU Þ1 Þðx1 ; x2 ; x3 Þ ¼ ðx1 ; x2 ; 0; 0Þ: Here, ðuU Þ1 ¼ ðuU uU ðpÞÞ1 ¼ ðTuU ðpÞ uU Þ1 ¼ ðuU Þ1 TuU ðpÞ : Since uU ¼ uU uU ðpÞ; uU ðUÞ ¼ uU ðUÞ uU ðpÞ ¼ uU ðU Þ uU ðpÞ: Now, since ðx1 ; x2 ; x3 Þ is in uU ðUÞ; ðx1 þ ðuU ðpÞÞ1 ; x2 þ ðuU ðpÞÞ2 ; x3 þ ðuU ðpÞÞ3 Þ is in uU ðU Þ; and hence,

wV F ðuU Þ1 x1 þ ðuU ð pÞÞ1 ; x2 þ ðuU ð pÞÞ2 ; x3 þ ðuU ð pÞÞ3 ¼ x1 þ ðuU ð pÞÞ1 ; x2 þ ðuU ð pÞÞ2 ; 0; 0 :

4.1 Constant Rank Theorem

243

Now, LHS ¼ wV F ðuU Þ1 ðx1 ; x2 ; x3 Þ ¼ wV F ðuU Þ1 ðx 1 ; x 2 ; x 3 Þ uU ð pÞ ¼ wV ðuU ð pÞÞ1 ; ðuU ð pÞÞ2 ; 0; 0 F ðuU Þ1 TuU ð pÞ ðx1 ; x2 ; x3 Þ ¼ TððuU ð pÞÞ1 ;ðuU ð pÞÞ2 ;0; 0Þ wV F ðuU Þ1 TuU ð pÞ ðx1 ; x2 ; x3 Þ ¼ TððuU ð pÞÞ1 ;ðuU ð pÞÞ2 ;0; 0Þ wV F ðuU Þ1 TuU ð pÞ ðx1 ; x2 ; x3 Þ ¼ TððuU ð pÞÞ1 ;ðuU ð pÞÞ2 ;0; 0Þ wV F ðuU Þ1 x1 þ ðuU ð pÞÞ1 ; x2 þ ðuU ð pÞÞ2 ; x3 þ ðuU ð pÞÞ3 ¼ TððuU ð pÞÞ1 ;ðuU ð pÞÞ2 ;0; 0Þ x1 þ ðuU ð pÞÞ1 ; x2 þ ðuU ð pÞÞ2 ; 0; 0 ¼ ðx1 ; x2 ; 0; 0Þ ¼ RHS:

For (v): Since p 2 U; uU ðpÞ 2 uU ðUÞ; and ððuU ðpÞÞ1 ; ðuU ðpÞÞ2 ; ðuU ðpÞÞ2 Þ ¼ uU ðpÞ ¼ ð0; 0; 0Þ so ð0; 0; 0; 0Þ ¼ ðuU ð pÞÞ1 ; ðuU ð pÞÞ2 ; 0; 0 ¼ wV F ðuU Þ1 ðuU ð pÞÞ ¼ ðwV F Þð pÞ ¼ wV ðF ð pÞÞ: This completes the proof of the main part. Since wV ðVÞ is an open neighborhood of wV ðFðpÞÞð¼ ð0; 0; 0; 0ÞÞ in R4 ; and uU ðUÞ is an open neighborhood of uU ðpÞð¼ ð0; 0; 0ÞÞ; there exists e [ 0 such that ð0; 0; 0; 0Þ 2 ðe; eÞ ðe; eÞ ðe; eÞ ðe; eÞð¼ Ce4 ð0ÞÞ wV ðVÞ, and ð0; 0; 0Þ 2 ðe; eÞ ðe; eÞ ðe; eÞð¼ Ce3 ð0ÞÞ uU ðUÞ: Now, by Lemma 4.11, ððwV Þ1 ðCe4 ð0ÞÞ; wV Þ is an admissible coordinate chart in N satisfying FðpÞ 2 ðwV Þ1 ðCe4 ð0ÞÞ; and ððuU Þ1 ðCe3 ð0ÞÞ; uU Þ is an admissible coordinate chart in M satisfying p 2 ðuU Þ1 ðCe3 ð0ÞÞ: Since uU ðpÞ ¼ ð0; 0; 0Þ 2 Ce3 ð0Þ; p 2 ðuU Þ1 ðCe3 ð0ÞÞ: Next, since wV ðFðpÞÞ ¼ ð0; 0; 0; 0Þ 2 Ce4 ð0Þ; so FðpÞ 2 ðwV Þ1 ðCe4 ð0ÞÞ. Also since uU is 1–1, uU ððuU Þ1 ðCe3 ð0ÞÞÞ ¼ Ce3 ð0Þ: Similarly, wV ððwV Þ1 ðCe4 ð0ÞÞÞ ¼ Ce4 ð0Þ: Now, it remains to prove: for every ðx1 ; x2 ; x3 Þ in Ce3 ð0Þ; ðwV F ðuU Þ1 Þ ðx1 ; x2 ; x3 Þ ¼ ðx1 ; x2 ; 0; 0Þ. For this purpose, let us take any ðx1 ; x2 ; x3 Þ in Ce3 ð0Þ: Since ðx1 ; x2 ; x3 Þ is in Ce3 ð0Þ; and Ce3 ð0Þ uU ðUÞ; ðx1 ; x2 ; x3 Þ is in uU ðUÞ; and hence, ðwV F ðuU Þ1 Þðx1 ; x2 ; x3 Þ ¼ ðx1 ; x2 ; 0; 0Þ: h Theorem 4.13 Let M be an m-dimensional smooth manifold, N be an n-dimensional smooth manifold, F : M ! N be a smooth function, and k be the rank of F. Let p be in M. Then, there exist admissible coordinate chart ðU; uU Þ in M satisfying

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p 2 U; and admissible coordinate chart ðV; wV Þ in N satisfying FðpÞ 2 V such that uU ðpÞ ¼ 0; wV ðFðpÞÞ ¼ 0; and for every, ðx1 ; x2 ; . . .; xn Þ in uU ðUÞ; 0 1 wV F ðuU Þ1 ðx1 ; x2 ; . . .; xm Þ ¼ @x1 ; x2 ; . . .; xk ; 0; . . .; 0A: |ﬄﬄﬄ{zﬄﬄﬄ} nk

Further, we may assume uU ðUÞ ¼ Cem ð0Þ; and wV ðVÞ ¼ Cen ð0Þ for some e [ 0: Proof Its proof is quite similar to the proof of Theorem 4.12.

h

Lemma 4.14 Let M be an m-dimensional smooth manifold, N be an n-dimensional smooth manifold, and F : N ! M be a 1–1 smooth function. Let O1 be the topology of N. Let A1 be the differential structure on N. Put O fFðGÞ : G 2 O1 g and A fðFðUÞ; uU F 1 Þ : ðU; uU Þ 2 A1 g: Then, I. O is a topology over FðNÞ; II. A determines a unique differential structure on FðNÞ, and III. F is a diffeomorphism from N onto FðNÞ: Proof of I (i) Since / 2 O1 ; Fð/Þð¼ /Þ is in O; and hence, / 2 O: Next, since N 2 O1 ; so FðNÞis in O: (ii) Let FðGÞ 2 O; where G 2 O1 : Let FðHÞ 2 O; where H 2 O1 : We have to prove that FðGÞ \ FðHÞ 2 O: Since G 2 O1 ; H 2 O1 ; and O1 is a topology, G \ H 2 O1 ; and hence, FðG \ HÞ 2 O: Since F : N ! M is 1–1, FðG \ HÞ ¼ FðGÞ \ FðHÞ: Since FðG \ HÞ ¼ FðGÞ \ FðHÞ; and FðG \ HÞ 2 O; FðGÞ \ FðHÞ 2 O: (iii) Let FðGi Þ 2 O; where Gi 2 O1 for every index i in index set I. We have to prove that [i2I FðGi Þ 2 O: Since for every index i, Gi 2 O1 ; and O1 is a topology, [i2I Gi 2 O1 ; and hence, Fð[i2I Gi Þð¼ [i2I FðGi ÞÞ 2 O: Thus, h [i2I FðGi Þ 2 O: Hence, O is a topology over FðNÞ: Proof of II It is clear that F is a homeomorphism from topological space ðN; O1 Þ onto topological space ðFðNÞ; OÞ: Since N is an n-dimensional smooth manifold, and O1 is the topology of N, ðN; O1 Þ is a Hausdorff topological space. Since ðN; O1 Þ is a Hausdorff topological space, and ðN; O1 Þ is homeomorphic onto topological space ðFðNÞ; OÞ; ðFðNÞ; OÞ is a Hausdorff topological space. Since N is an n-dimensional smooth manifold, and O1 is the topology of N, ðN; O1 Þ is a second countable space. Since ðN; O1 Þ is a second countable space, and ðN; O1 Þ is homeomorphic onto topological space ðFðNÞ; OÞ; ðFðNÞ; OÞ is a second countable space. We want to prove that ðFðNÞ; OÞ is an n-dimensional topological manifold. For this purpose, let us take any FðxÞ in FðNÞ; where x is in N. Since x is in N, and A1 is a differential structure on N, there exists ðU; uU Þ 2 A1 such that x 2 U: It follows

4.1 Constant Rank Theorem

245

that FðxÞ 2 FðUÞ: Since ðU; uU Þ 2 A1 ; ðFðUÞ; uU F 1 Þ 2 A: It remains to prove: (a) FðUÞ 2 O; (b) ðuU F 1 ÞðFðUÞÞ is an open subset of Rn ; (c) uU F 1 is a homeomorphism from FðUÞ (relative to topology O) onto ðuU F 1 ÞðFðUÞÞ. For (a): Here, ðU; uU Þ 2 A1 ; and A1 is a differential structure on N, U 2 O1 ; and hence, FðUÞ 2 O: For (b): Here, U 2 O1 ; and O1 is the topology of N, U is contained in N. Since U is contained in N, F is 1–1, ðuU F 1 ÞðFðUÞÞ ¼ uU ðUÞ: Since ðU; uU Þ 2 A1 ; and A1 is a differential structure on N, uU ðUÞð¼ ðuU F 1 ÞðFðUÞÞÞ is open in Rn : Hence, ðuU F 1 ÞðFðUÞÞ is an open subset of Rn : For (c): Since ðU; uU Þ 2 A1 ; uU is 1–1. Since uU is 1–1, and F is 1–1, the composite uU F 1 is 1–1. Thus, uU F 1 : FðUÞ ! ðuU F 1 ÞðFðUÞÞ is a 1–1 onto function. Since ðU; uU Þ 2 A1 ; uU is a homeomorphism from U onto uU ðUÞ: Since F is a homeomorphism from topological space ðN; O1 Þ onto topological space ðFðNÞ; OÞ; F 1 is a homeomorphism from topological space ðFðNÞ; OÞ onto topological space ðN; O1 Þ, and hence, the restriction of F 1 on FðUÞ (relative to topology OÞ is a homeomorphism from FðUÞ onto U. It follows that uU F 1 is a homeomorphism from FðUÞ (relative to topology O) onto uU ðUÞð¼ ðuU F 1 ÞðFðUÞÞÞ: Thus, we have shown that ðFðNÞ; OÞ is an n-dimensional topological manifold. Now we want to prove: All pairs of members in A are C 1 -compatible. For this purpose, let us take any ðFðUÞ; uU F 1 Þ 2 A, where ðU; uU Þ is in A1 ; and ðFðVÞ; wV F 1 Þ 2 A where ðV; wV Þ is in A1 satisfying FðUÞ \ FðVÞ is nonempty. Since F is 1–1, FðUÞ \ FðVÞ ¼ FðU \ VÞ: Since FðUÞ \ FðVÞ ¼ FðU \ VÞ; and FðUÞ \ FðVÞ is nonempty, FðU \ VÞis nonempty, and hence, U \ V is nonempty. Since U \ V is nonempty, A1 is a differential structure on N, ðU; uU Þ is in A1 ; and ðV; wV Þ is in A1 ; ðU; uU Þ and ðV; wV Þ are C1 -compatible, that is, ðwV ðuU Þ1 Þ : uU ðU \ VÞ ! uV ðU \ VÞ is a smooth function. We have to show that ðFðUÞ; uU F 1 Þ and ðFðVÞ; wV F 1 Þ are C1 -compatible, that is, ððwV F 1 Þ ðuU F 1 Þ1 Þ : ðuU F 1 ÞðFðUÞ \ FðVÞÞ ! ðwV F 1 Þ ðFðUÞ \ FðVÞÞ is a smooth function, that is, ðwV F 1 Þ ðF ðuU Þ1 Þ : ðuU F 1 ÞðFðU \ VÞÞ ! ðwV F 1 ÞðFðU \ VÞÞ is a smooth function, that is, ðwV ðuU Þ1 Þ : uU ðU \ VÞ ! wV ðU \ VÞ is a smooth function. This has already been shown. Thus, we have shown that A determines a unique differentiable structure on FðNÞ that contains A: In short, FðNÞ is an n-dimensional smooth manifold. h Proof of III Now, we want to prove that F is a diffeomorphism from N onto FðNÞ: For this purpose, let us take any p in N. We have to prove that F is C1 at p. Now let us take any admissible coordinate chart ðU; uU Þ of N satisfying p 2 U; an

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admissible coordinate chart ðFðVÞ; wV F 1 Þ of FðNÞ satisfying FðpÞ 2 FðVÞ where ðV; wV Þ is in A1 : Since F is 1–1, and FðpÞ 2 FðVÞ; p 2 V: Since p 2 U; and p 2 V; p 2 U \ V; and hence, U \ V is nonempty. We shall try to prove that ðwV F 1 Þ F ðuU Þ1 is C 1 at uU ðpÞ. Since ðU; uU Þ is in A1 ; ðV; wV Þ is in A1 ; and U \ V is nonempty, wV ðuU Þ1 ð¼ ðwV F 1 Þ F ðuU Þ1 Þ is C 1 at uU ðpÞ: Thus ðwV F 1 Þ F ðuU Þ1 is C 1 at uU ðpÞ: Thus F is C1 : Similarly, F 1 is C 1 : This proves that F is a diffeomorphism from N onto FðNÞ: h

4.2 Lie Groups Deﬁnition Let G be an n-dimensional smooth manifold that is also a group. Let m : G G ! G be the mapping deﬁned as follows: for every g, h in G, mðg; hÞ gh: Let i : G ! G be the mapping deﬁned as follows: for every g in G, iðgÞ g1 : If m is a smooth map from 2n-dimensional product manifold G × G to ndimensional smooth manifold G, and i is a smooth map from n-dimensional smooth manifold G to itself, then we say that G is a Lie group. Here, m is the binary operation of the group G and is smooth. Further, i is the (algebraic) inverse mapping of the group G, which is smooth. Since smooth mappings are continuous mappings, and m, i are smooth mappings, m, i are continuous mappings. Since the binary operation m of group G is continuous, and the inverse mapping i of the group G is continuous, G is a topological group. Thus, every Lie group is also a topological group. Note 4.15 Let G be an n-dimensional smooth manifold, and M be an m-dimensional smooth manifold. Let a be in G. Let us deﬁne a mapping F : M ! G M as follows: for every h in M, F ðhÞ ða; hÞ: We shall try to show that F is a smooth mapping. For this purpose, let us ﬁx any h0 in M. We have to prove that F is smooth at h0 in M. For this purpose, let us take any admissible coordinate chart ðV; wV Þ of M satisfying h0 2 V; and any admissible coordinate chart ðU; uU Þ of G satisfying a 2 U: Here, ðU; uU Þ is an admissible coordinate chart of G satisfying a 2 U; ðV; wV Þ is an admissible coordinate chart of M satisfying h0 2 V; and Fðh0 Þ ¼ ða; h0 Þ 2 U V; ðU V; ðuU wV ÞÞ is an admissible coordinate chart of G M satisfying Fðh0 Þ 2 U V: We have to prove that each of the 2 component functions of the mapping

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247

ðuU wV Þ F ðwV Þ1 : wV V \ F 1 ðU V Þ ! ðuU wV ÞðU V Þ is C1 at the point wV ðh0 Þ in Rm ; that is, ðr; sÞ 7! p1 ðððuU wV Þ ðF ðwV Þ1 ÞÞ ðr; sÞÞ; and ðr; sÞ 7! p2 ðððuU wV Þ ðF ðwV Þ1 ÞÞðr; sÞÞ are C 1 at the point wV ðh0 Þin Rm : For every ðr; sÞ in wv ðV \ F 1 ðU VÞÞ, there exists y in V such that wV ðyÞ ¼ ðr; sÞ: So y ¼ ðwV Þ1 ðr; sÞ: Now, since

ðuU wV Þ F ðwV Þ1 ðr; sÞ ¼ ðuU wV Þ F ðwV Þ1 ðr; sÞ ¼ ðuU wV ÞðF ð yÞÞ ¼ ðuU wV Þða; yÞ ¼ ðuU ðaÞ; wV ð yÞÞ;

ðr; sÞ 7! p1 ðððuU wV Þ ðF ðwV Þ1 ÞÞðr; sÞÞ ¼ p1 ðuU ðaÞ; wV ðyÞÞ ¼ uU ðaÞ; which is

a constant function. Hence, ðr; sÞ 7! p1 ðððuU wV Þ ðF ðwV Þ1 ÞÞðr; sÞÞ is a smooth function. Next, since ðr; sÞ 7! p2 ðððuU wV Þ ðF ðwV Þ1 ÞÞðr; sÞÞ ¼ p2 ðuU ðaÞ; wV ðyÞÞ ¼ wV ðyÞ ¼ ðr; sÞ is the identity function, ðr; sÞ 7! p2 ðððuU wV Þ ðF ðwV Þ1 ÞÞðr; sÞÞ is a smooth function. Thus, we have shown that F is a smooth mapping, that is, h 7! ða; hÞ is a smooth mapping. Similarly, h 7! ðh; aÞ is a smooth mapping. Note 4.16 Let G; H; K be smooth manifolds. Let f : H ! K be a smooth mapping. Let F : G H ! G K be the mapping deﬁned as follows: for every ðx; yÞ in G H; F ðx; yÞ ðx; f ð yÞÞ:

We shall try to show that F is a smooth mapping. For simplicity, let each of G,H, K have dimension 2. Let us ﬁx any ðx0 ; y0 Þ in G H: We have to prove that F is smooth at ðx0 ; y0 Þ: For this purpose, let us take any admissible coordinate chart ðU V; ðuU wV ÞÞ of the product manifold G H; where ðU; uU Þ is an admissible coordinate chart of G satisfying x0 2 U; and ðV; wV Þ is an admissible coordinate chart of H satisfying ~ ðuU wV~ ÞÞ of y0 2 V: Next, let us take any admissible coordinate chart ðU V; ~ ~ G K; where ðV; wV~ Þ is an admissible coordinate chart of K satisfying f ðy0 Þ 2 V: ~ ~ ~ Here, since x0 2 U; and f ðy0 Þ 2 V; Fðx0 ; y0 Þ ¼ ðx0 ; f ðy0 ÞÞ 2 U V: Thus, ðU V; ~ ðuU wV~ ÞÞ is an admissible coordinate chart of G K satisfying Fðx0 ; y0 Þ 2 U V: Now, we have to prove that each of the 4 component functions of the mapping

~ uU wV~ F ðuU wV Þ1 : ðuU wV Þ ðU V Þ \ F 1 U V ~ ! uU wV~ U V

248

4 Topological Properties of Smooth Manifolds

is C 1 at the point ðuU wV Þðx0 ; y0 Þ in R4 ; that is, ðr; s; t; uÞ 7! p1 ðððuU wV~ Þ ðF ðuU wV Þ1 ÞÞðr; s; t; uÞÞ; ðr; s; t; uÞ 7! p2 ðððuU wV~ Þ ðF ðuU wV Þ1 ÞÞ ðr; s; t; uÞÞ; ðr; s; t; uÞ 7! p3 ðððuU wV~ Þ ðF ðuU wV Þ1 ÞÞðr; s; t; uÞÞ; and ðr; s; t; uÞ 7! p4 ðððuU wV~ Þ ðF ðuU wV Þ1 ÞÞðr; s; t; uÞÞ are C1 at the point ðuU wV Þðx0 ; y0 Þ in R4 : ~ there exist x in U, For every ðr; s; t; uÞ in ðuU wV ÞððU VÞ \ F 1 ðU VÞÞ; and y in V such that ðuU wV Þðx; yÞ ¼ ðr; s; t; uÞ: So ðr; s; t; uÞ ¼ ðuU wV Þðx; yÞ ¼ ðuU ðxÞ; wV ðyÞÞ; and hence, uU ðxÞ ¼ ðr; sÞ; and wV ðyÞ ¼ ðt; uÞ: Now, since uU wV~ F ðuU wV Þ1 ðr; s; t; uÞ ¼ uU wV~ F ðuU wV Þ1 ððuU wV Þðx; yÞÞ ¼ uU wV~ ðF ðx; yÞÞ ¼ uU wV~ ðx; f ð yÞÞ ¼ uU ð xÞ; wV~ ðf ð yÞÞ ¼ uU ð xÞ; wV~ f ð yÞ ¼ r; s; wV~ f ð yÞ ¼ r; s; wV~ f ðwV Þ1 ðt; uÞ ¼ r; s; wV~ f ðwV Þ1 ðt; uÞ ; so

uU wV~ F ðuU wV Þ1 ðr; s; t; uÞ ¼ p1 r; s; wV~ f ðwV Þ1 ðt; uÞ ¼ r;

ðr; s; t; uÞ 7! p1

which is a smooth function. Hence, ðr; sÞ 7! p1 ðððuU wV~ Þ ðF ðuU wV Þ1 ÞÞ ðr; s; t; uÞÞ is a smooth function. Next, since uU wV~ F ðuU wV Þ1 ðr; s; t; uÞ ðr; s; t; uÞ 7! p2 p2 r; s; wV~ f ðwV Þ1 ðt; uÞ ¼ s; which is a smooth function, ðr; s; t; uÞ 7! p2 ðððuU wV~ Þ ðF ðuU wV Þ1 ÞÞ ðr; s; t; uÞÞ is a smooth function. Next, uU wV~ F ðuU wV Þ1 ðr; s; t; uÞ ðr; s; t; uÞ 7! p3 ¼ p3 r; s; wV~ f ðwV Þ1 ðt; uÞ ¼ p1 wV~ f ðwV Þ1 ðt; uÞ :

4.2 Lie Groups

249

Now, since f : H ! K is a smooth mapping, ðV; wV Þ is an admissible coordinate ~ wV~ Þ is an admissible coordinate chart of chart of H satisfying y0 2 V; and ðV; ~ K satisfying f ðy0 Þ 2 V; ~ ! wV~ V ~ wV~ f ðwV Þ1 : wV V \ f 1 V is C 1 at the point wV ðy0 Þ in R2 ; that is, ðt; uÞ 7! p1 ðwV~ ðf ðwV Þ1 Þðt; uÞÞ; and ðt; uÞ 7! p2 ðwV~ ðf ðwV Þ1 Þðr; sÞÞ are C1 at the point wV ðy0 Þ in R2 : Since ðr; s; t; uÞ 7! ðt; uÞ is smooth, and ðt; uÞ 7! p1 ðwV~ ðf ðwV Þ1 Þðt; uÞÞ is smooth, their composite ðr; s; t; uÞ 7! p1 ðwV~ ðf ðwV Þ1 Þðt; uÞÞ is smooth, and hence, uU wV~ F ðuU wV Þ1 ðr; s; t; uÞ ðr; s; t; uÞ 7! p3 is a smooth map. Similarly, ðr; s; t; uÞ 7! p4 ðððuU wV~ Þ ðF ðuU wV Þ1 ÞÞ ðr; s; t; uÞÞ is a smooth function. Thus, we have shown that F is a smooth mapping. In short, if y 7! f ðyÞ is smooth map, then ðx; yÞ 7! ðx; f ðyÞÞ is smooth. Similarly, if x 7! f ðxÞ is a smooth map, then ðx; yÞ 7! ðf ðxÞ; yÞ is smooth. Lemma 4.17 Let G be a Lie group. Let a be in G. Then, the mapping F : y 7! ay is a smooth mapping from smooth manifold G to itself. The mapping F is denoted by La and is called the left translation by a. Thus, for every y 2 G; La ðyÞ ¼ ay: Observe that La : G ! G is a 1–1, onto mapping, and ðLa Þ1 ¼ La1 : Now, since La ; La1 ð¼ ðLa Þ1 Þ are smooth, La is a diffeomorphism. Proof Since G is a Lie group, G is a smooth manifold. Hence y 7! ða; yÞ is a smooth map. Since G is a Lie group, ðx; yÞ 7! xy is a smooth map. Since y 7! ða; yÞ is a smooth map, and ðx; yÞ 7! xy is a smooth map, their composite map y 7! ay is a smooth map. h Note 4.18 Let G be a Lie group. Let a be in G. Then, as above, it can be shown that the mapping F : x 7! xa is a smooth mapping from smooth manifold G to itself. The mapping F is denoted by Ra and is called the right translation by a. Thus, for every x 2 G; Ra ðyÞ ¼ ya: Observe that Ra : G ! G is a 1–1, onto mapping, and ðRa Þ1 ¼ Ra1 : Now, since Ra ; Ra1 ð¼ ðRa Þ1 Þ are smooth, Ra is a diffeomorphism. Lemma 4.19 Let G be an n-dimensional smooth manifold that is also a group. If ðg; hÞ 7! gh1 is a smooth mapping from 2n-dimensional product manifold G × G to n-dimensional smooth manifold G, then G is a Lie group. Proof For simplicity, let n = 2. Next, let e be the identity element of the group. Let us deﬁne a mapping m : G G ! G as follows: for every g; h in G,

250

4 Topological Properties of Smooth Manifolds

mðg; hÞ gh: Let us deﬁne a mapping i : G ! G as follows: for every g in G, iðgÞ g1 : We have to show that (i) i is smooth, (ii) m is smooth. For (i): Since h 7! ðe; hÞ is a smooth mapping, and ðg; hÞ 7! gh1 is a smooth mapping, their composite h 7! eh1 ¼ h1 ¼ iðhÞ is a smooth mapping, and hence, i is a smooth mapping. For (ii): Since h 7! h1 is a smooth mapping, ðg; hÞ 7! ðg; h1 Þ is a smooth mapping. Since ðg; hÞ 7! ðg; h1 Þ is a smooth mapping, and ðg; hÞ 7! gh1 is a smooth mapping, their composite ðg; hÞ 7! gðh1 Þ1 ¼ gh ¼ mðghÞ is a smooth mapping, and hence, m is a smooth mapping. h Note 4.20 Let M be an m-dimensional smooth manifold, whose topology is O; and differential structure is A: Let G be a nonempty open subset of M. Let O1 be the induced topology over G, that is, O1 ¼ fG1 : G1 G; and G1 2 Og: Put AG fðU; uU Þ : ðU; uU Þ 2 A; and U Gg: Since M is an m-dimensional smooth manifold, M is a Hausdorff space. Since M is a Hausdorff space, and G is a subspace of M, G with the induced topology is a Hausdorff space. Since M is an m-dimensional smooth manifold, M is a second countable space. Since M is a second countable space, and G is a subspace of M, G with the induced topology is a second countable space. Now, we shall try to show that AG is an atlas on G that is, 1. [fU : ðU; uU Þ is in AG g ¼ G; 2. all pairs of members in AG are C 1 -compatible. For (1): By the deﬁnition of AG ; it is clear that [fU : ðU; uU Þ is in AG g G: So, it remains to prove that G [fU : ðU; uU Þ is in AG g: For this purpose, let us take any p in G. Since p is in G, and G is a subset of M, p is in M. Since p is in M, and M is an m-dimensional smooth manifold, there exists ðU; uU Þ in A such that p is in U. Since ðU; uU Þ is in A; U is an open subset of M. Since U is open, and G is open, U \ G is open. Thus, U \ G is an open neighborhood of p.

4.2 Lie Groups

251

We shall try to show that ðU \ G; uU jðU\GÞ Þ is in AG ; that is, ðU \ G; uU jðU\GÞ Þ 2 A; and U \ G G. Since U \ G G; it remains to show that ðU \ G; uU jðU\GÞ Þ 2 A: Since ðU; uU Þ is in A; ðU; uU Þ is an admissible coordinate chart of M. Now, since U \ G is a nonempty open subset of U, by the Lemma 4.10, ðU \ G; uU jðU\GÞ Þ is an admissible coordinate chart of M, and hence, ðU \ G; uU jðU\GÞ Þ 2 A: For (2): Let us take any ðU; uU Þ; ðV; wV Þ 2 AG : We have to show that ðU; uU Þ andðV; wV Þ are C 1 -compatible. Since ðU; uU Þ 2 AG , by the deﬁnition of AG ; ðU; uU Þ 2 A: Similarly ðV; wV Þ 2 A: Since ðU; uU Þ 2 A; ðV; wV Þ 2 A; and A is a differential structure, ðU; uU Þ and ðV; wV Þ are C 1 -compatible. Thus, we have shown that AG is an atlas on G. Hence, AG determines a unique smooth structure on G. Thus, the open subset G of smooth manifold M becomes an mdimensional smooth manifold. Here, G is called an open submanifold of M. Example 4.21 Let Mð2 3; RÞ be the collection of all 2 × 3 matrices with real entries, that is, Mð2 3; RÞ

abc : a; b; c; d; e; f 2 R : def

We deﬁne addition of matrices as follows:

a d

b e

a c þ 1 f d1

b1 e1

c1 f1

a þ a1 d þ d1

b þ b1 e þ e1

c þ c1 : f þ f1

Clearly, Mð2 3; RÞ is a commutative group under addition as binary operation. We deﬁne scalar multiplication of matrix as follows:

a t d

c ta f td

b e

tb te

tc : tf

Clearly, Mð2 3; RÞ becomes a real linear space of dimension ð2 3Þ: Here, one of the basis of Mð2 3; RÞ is

1 0

0 0

0 1 0 ; 0 0 0

0 0 ; 0 0

0 0

0 0 1 ; 1 0 0

0 0 ; 0 0

0 1

0 0 0 ; 0 0 0

0 1

:

Clearly, the real linear space R23 is isomorphic to the real linear space Mð2 3; RÞ: Since R23 is isomorphic to Mð2 3; RÞ; we will not distinguish between

252

4 Topological Properties of Smooth Manifolds

a d

b e

c and ða; b; c; d; e; f Þ: f

Now since R23 is a ð2 3Þ-dimensional smooth manifold, Mð2 3; RÞ is also a ð2 3Þ-dimensional smooth manifold. Similarly, Mðm n; RÞ is a mn-dimensional smooth manifold. Clearly, a Mð2 3; CÞ d

b e

c : a; b; c; d; e; f are complex numbers f

is a 2 ð2 3Þ-dimensional smooth manifold. Similarly, Mðm n; CÞ is a 2mndimensional smooth manifold. The n2 -dimensional smooth manifold Mðn n; RÞ is denoted by Mðn; RÞ: The 2n2 -dimensional smooth manifold Mðn n; CÞ is denoted by Mðn; CÞ: Let GLð2; RÞ be the collection of all invertible 2 2 matrices with real entries, that is,

x1 x 2 x1 x2 GLð2; RÞ : 2 Mð2; RÞ andx1 x4 x2 x3 6¼ 0 : x3 x4 x3 x4 x x It is clear that ½ 1 2 7! ðx1 ; x2 ; x3 ; x4 Þ is in 1–1 correspondence from x3 x4 GLð2; RÞ onto fðx1 ; x2 ; x3 ; x4 Þ : ðx1 ; x2 ; x3 ; x4 Þ 2 R4 and x1 x4 x2 x3 6¼ 0g: Let us deﬁne a function f : R4 ! R as follows: for every ðx1 ; x2 ; x3 ; x4 Þ in R4 f ð x1 ; x2 ; x3 ; x4 Þ x1 x4 x2 x3 : We shall try to show that f is a continuous function. Since ðx1 ; x2 ; x3 ; x4 Þ 7! x1 ; and ðx1 ; x2 ; x3 ; x4 Þ 7! x4 are continuous functions, ðx1 ; x2 ; x3 ; x4 Þ 7! ðx1 ; x4 Þ is continuous. Since ðx1 ; x2 ; x3 ; x4 Þ 7! ðx1 ; x4 Þ is continuous, and the multiplication operation ðx1 ; x4 Þ 7! x1 x4 is continuous, their composite function ðx1 ; x2 ; x3 ; x4 Þ 7! x1 x4 is continuous. Similarly, ðx1 ; x2 ; x3 ; x4 Þ 7! x2 x3 is continuous. Since ðx1 ; x2 ; x3 ; x4 Þ 7! x1 x4 is continuous, and ðx1 ; x2 ; x3 ; x4 Þ 7! x2 x3 is continuous, ðx1 ; x2 ; x3 ; x4 Þ 7! ðx1 x4 ; x2 x3 Þ is continuous. Since ðx1 ; x2 ; x3 ; x4 Þ 7! ðx1 x4 ; x2 x3 Þ is continuous, and the difference operation ðy; zÞ 7! y z is continuous, their composite function ðx1 ; x2 ; x3 ; x4 Þ 7! x1 x4 x2 x3 ¼ f ðx1 ; x2 ; x3 ; x4 Þ is continuous, and hence, f is a continuous function. Since f0g is a closed subset of R; R f0g is an open subset of R: Since R f0g is an open subset of R; and f : R4 ! R is a continuous function, f 1 ðR f0gÞ ð¼ fðx1 ; x2 ; x3 ; x4 Þ : ðx1 ; x2 ; x3 ; x4 Þ 2 R4 and x1 x4 x2 x3 6¼ 0gÞ is an open subset of R4 ; and hence, fðx1 ; x2 ; x3 ; x4 Þ : ðx1 ; x2 ; x3 ; x4 Þ 2 R4 and x1 x4 x2 x3 6¼ 0g is an open subset of R4 : Since fðx1 ; x2 ; x3 ; x4 Þ : ðx1 ; x2 ; x3 ; x4 Þ 2 R4 and x1 x4 x2 x3 6¼ 0g is an x x open subset of R4 ; and ½ 1 2 7! ðx1 ; x2 ; x3 ; x4 Þ is in 1–1 correspondence from x3 x4 GLð2; RÞ onto fðx1 ; x2 ; x3 ; x4 Þ : ðx1 ; x2 ; x3 ; x4 Þ 2 R4 and x1 x4 x2 x3 6¼ 0g; GLð2; RÞ

4.2 Lie Groups

253

is an open subset of Mð2; RÞð¼ R4 Þ: Since GLð2; RÞ is an open subset of R4 ; and R4 is a 4-dimensional smooth manifold, GLð2; RÞ is a 4-dimensional smooth manifold. Clearly, fðGLð2; RÞ; IdGLð2;RÞ Þg is an atlas of GLð2; RÞ: This shows that ðGLð2; RÞ GLð2; RÞ; ðIdGLð2;RÞ IdGLð2;RÞ ÞÞ is an admissible coordinate chart of the product manifold GLð2; RÞ GLð2; RÞ: Since, for every

e a b ; g c d

f h

in GLð2; RÞ GLð2; RÞ;

we have

IdGLð2;RÞ IdGLð2;RÞ

e f ; g h c d

a b e f ¼ IdGLð2;RÞ ; IdGLð2;RÞ c d g h

e a b e f a b ; ¼ IdGLð2;RÞGLð2;RÞ ; ¼ g c d g h c d a

b

f h

;

IdGLð2;RÞ IdGLð2;RÞ ¼ IdGLð2;RÞGLð2;RÞ ; and hence, ðGLð2; RÞ GLð2; RÞ; IdGLð2;RÞGLð2;RÞ Þ is an admissible coordinate chart of the product manifold GLð2; RÞ GLð2; RÞ: We deﬁne multiplication of matrices as follows:

a c

b d

e g

ae þ bg f ce þ dg h

af þ bh : cf þ dh

We know that GLð2; RÞ becomes a group under multiplication as binary operation. Thus, GLð2; RÞ is a 4-dimensional smooth manifold that is also a group. We want to show that GLð2; RÞ is a Lie group. Let us deﬁne m : GLð2; RÞ GLð2; RÞ ! GLð2; RÞ as follows: for every

a b e ; c d g

f h

in GLð2; RÞ GLð2; RÞ;

we have

af þ bh m : cf þ dh

a b Let us deﬁne i : GLð2; RÞ ! GLð2; RÞ as follows: for every in c d GLð2; RÞ;

a c

b e ; g d

f h

a c

b d

e g

ae þ bg f ¼ ce þ dg h

254

4 Topological Properties of Smooth Manifolds

1 d a b i c c d ad bc

2

d b 6 ad bc ¼ 4 c a ad bc

3 b ad bc 7 5: a ad bc

We have to show: (i) m is a smooth map from (4 + 4)-dimensional product manifold GLð2; RÞ GLð2; RÞ to 4-dimensional smooth manifold GLð2; RÞ, (ii) i is a smooth map from 4-dimensional smooth manifold GLð2; RÞ to 4dimensional smooth manifold GLð2; RÞ. For (i): For this purpose, let us ﬁx any

e f a b in GLð2; RÞ GLð2; RÞ: ; g h c d

We have to prove that m is smooth at

a c

e b ; g d

f h

:

For this purpose, let us take the admissible coordinate chart ðGLð2; RÞ GLð2; RÞ; IdGLð2;RÞGLð2;RÞ Þ of the product manifold GLð2; RÞ GLð2; RÞ: Here,

e f a b 2 GLð2; RÞ GLð2; RÞ: ; g h c d

Next, let us take the admissible coordinate chart ðGLð2; RÞ; IdGLð2;RÞ Þ of GLð2; RÞ: Clearly, a m c

e b ; g d

f h

ae þ bg ¼ ce þ dg

af þ bh 2 GLð2; RÞ: cf þ dh

Now, we have to prove that each of the 4 component functions of the mapping 1 IdGLð2;RÞ m IdGLð2;RÞGLð2;RÞ : IdGLð2;RÞGLð2;RÞ ðGLð2; RÞ GLð2; RÞÞ \ m1 ðGLð2; RÞÞ ! IdGLð2;RÞ ðGLð2; RÞÞ is C 1 at the point IdGLð2;RÞGLð2;RÞ

a c

e b ; g d

f h

¼

a c

e b ; g d

f h

in R8 :

4.2 Lie Groups

255

Let us observe that IdGLð2;RÞ ðm ðIdGLð2;RÞGLð2;RÞ Þ1 Þ ¼ m; ðIdGLð2;RÞGLð2;RÞ Þ ððGLð2; RÞ GLð2; RÞÞ \ m1 ðGLð2; RÞÞÞ ¼ ðGLð2; RÞ GLð2; RÞÞ \ m1 ðGL ð2; RÞÞ ¼ ðGLð2; RÞ GLð2; RÞÞ \ðGLð2; RÞ GLð2; RÞÞ ¼ ðGLð2; RÞ GLð2; RÞÞ; and IdGLð2;RÞ ðGLð2; RÞÞ ¼ GLð2; RÞ; so we have to prove that each of the 4 component functions of the mapping m : GLð2; RÞ GLð2; RÞ ! GLð2; RÞ is C 1 at the point

e b ; g d

a c

f h

;

that is,

x1

x3 x1 x3

x2

y1

y2

x4 y3

x2 y1

y4

y2

x4

y4

y3

7! x1 y1 þ x2 y3 ; 7! x3 y1 þ x4 y3 ;

x1

x3 x1 x3

x2

y1

y2

x4 y3

x2 y1

y4

y2

x4

y4

y3

7! x1 y2 þ x2 y4 ; 7! x3 y2 þ x4 y4

are smooth, that is, ðx1 ; x2 ; x3 ; x4 ; y1 ; y2 ; y3 ; y4 Þ 7! x1 y1 þ x2 y3 ; ðx1 ; x2 ; x3 ; x4 ; y1 ; y2 ; y3 ; y4 Þ 7! x1 y2 þ x2 y4 ; ðx1 ; x2 ; x3 ; x4 ; y1 ; y2 ; y3 ; y4 Þ 7! x3 y1 þ x4 y3 ; ðx1 ; x2 ; x3 ; x4 ; y1 ; y2 ; y3 ; y4 Þ 7! x3 y2 þ x4 y4 are smooth. Clearly, all these four functions are smooth. Thus, we have shown that m is a smooth map. For (ii): For this purpose, let us ﬁx any

a c

b d

in GLð2; RÞ:

We have to prove that i is smooth at

a b : c d

For this purpose, let us take the admissible coordinate chart ðGLð2; RÞ; IdGLð2;RÞ Þ of the smooth manifold GLð2; RÞ: Here,

a c

b d

2 GLð2; RÞ:

256

4 Topological Properties of Smooth Manifolds

Next, let us take the admissible coordinate chart ðGLð2; RÞ; IdGLð2;RÞ Þ of the smooth manifold GLð2; RÞ: Clearly, 3 2 d b

1 d b a b 6 bc ad bc 7 ¼ 4 adc ¼ i 5 2 GLð2; RÞ: a c d ad bc c a ad bc ad bc Now, we have to prove that each of the 4 component functions of the mapping 1 IdGLð2;RÞ i IdGLð2;RÞ : IdGLð2;RÞ ðGLð2; RÞÞ \ i1 ðGLð2; RÞÞ ! IdGLð2;RÞ ðGLð2; RÞÞ is C 1 at the point IdGLð2;RÞ

a c

b d

a ¼ c

b d

in R4 :

Let us observe that IdGLð2;RÞ ði ðIdGLð2;RÞ Þ1 Þ ¼ i; ðIdGLð2;RÞ ÞððGLð2; RÞÞ \ i ðGLð2; RÞÞÞ ¼ ðGLð2; RÞÞ \ i1 ðGLð2; RÞÞ ¼ ðGLð2; RÞÞ \ ðGLð2; RÞÞ ¼ GLð2; RÞ; and IdGLð2;RÞ ðGLð2; RÞÞ ¼ GLð2; RÞ; so we have to prove that each of the 4 component functions of the mapping 1

i : GLð2; RÞ ! GLð2; RÞ is C 1 at the point

that is,

x1

x3 x1 x3

x2

x4

x2 x4

a b ; c d

7!

x4 ; x1 x4 x2 x3

7!

x3 ; x1 x4 x2 x3

x1

x3 x1 x3

x2

x4

x2 x4

7!

x2 ; x1 x4 x2 x3

7!

x1 x1 x4 x2 x3

are smooth, that is, x4 ; x1 x4 x2 x3 x3 ðx1 ; x2 ; x3 ; x4 Þ 7! ; x1 x4 x2 x3 ðx1 ; x2 ; x3 ; x4 Þ 7!

x2 ; x1 x4 x2 x3 x2 ðx1 ; x2 ; x3 ; x4 Þ 7! x1 x4 x2 x3 ðx1 ; x2 ; x3 ; x4 Þ 7!

are smooth. Clearly, all these four functions are smooth. Thus, we have shown that i is a smooth map. Hence, GLð2; RÞ is a 4-dimensional Lie group. Similarly,

4.2 Lie Groups

257

GLðn; RÞ is a n2 -dimensional Lie group. GLðn; RÞ is known as the general linear group. As above, it can be shown that GLðn; CÞ is a 2n2 -dimensional Lie group. GLðn; CÞ is known as the complex general linear group. Example 4.22 Let V be a real linear space. Let GLðVÞ be the collection of all invertible linear transformations from V onto V. We shall try to show that GLðVÞ is a group under composition of mappings as binary operation. (1) Closure law: Let us take any T1 ; T2 in GLðVÞ: We have to show that T1 T2 is in GLðVÞ; that is, (a) T1 T2 is 1–1 from V to V, (b) T1 T2 maps V onto V, (c) for every x; y in V, and for every s; t in R; ðT1 T2 Þðsx þ tyÞ ¼ sððT1 T2 ÞðxÞÞ þ tððT1 T2 ÞðyÞÞ: For (a): Since T1 is in GLðVÞ; T1 is invertible, and hence, T1 is 1–1. Similarly, T2 is 1–1. Since T1 is 1–1, and T2 is 1–1, their composite T1 T2 is 1–1. For (b): Since T1 is in GLðVÞ; T1 is invertible, and hence, T1 maps V onto V. Similarly, T2 maps V onto V. Since T1 maps V onto V, and T2 maps V onto V, their composite T1 T2 maps V onto V. For (c): LHS ¼ ðT1 T2 Þðsx þ tyÞ ¼ T1 ðT2 ðsx þ tyÞÞ ¼ T1 ðsðT2 ðxÞÞ þ tðT2 ðyÞÞÞ ¼ sðT1 ðT2 ðxÞÞÞ þ tðT1 ðT2 ðyÞÞÞ ¼ sððT1 T2 ÞðxÞÞ þ tððT1 T2 ÞðyÞÞ ¼ RHS: Thus, we have shown that T1 T2 is in GLðVÞ: (2) is associative: Let T1 ; T2 ; T3 be in GLðVÞ: We have to show that ðT1 T2 Þ T3 ¼ T1 ðT2 T3 Þ: Since the composition of mappings is associative, ðT1 T2 Þ T3 ¼ T1 ðT2 T3 Þ: (3) Existence of identity element: Let us deﬁne the mapping IdV : V ! V as follows: for every x in V, IdV ðxÞ ¼ x: It is clear that IdV is an invertible linear transformation from V onto V, and hence, IdV is in GLðVÞ: Also, for any T in GLðVÞ; it is clear that IdV T ¼ T IdV ¼ T: Hence, IdV serves the purpose of identity element in GLðVÞ: (4) Existence of inverse element: Let us take any T in GLðVÞ: Since T is in GLðVÞ; T is invertible, and hence, T 1 exists. Now, we shall try to show that T 1 is in GLðVÞ; that is, T 1 : V 7! V is linear. For this purpose, let us take any x; y in V, and any real s; t: We have to show that T 1 ðsx þ tyÞ ¼ sðT 1 ðxÞÞ þ tðT 1 ðyÞÞ: Since x is in V, and T maps V onto V, there exists x1 in V such that Tðx1 Þ ¼ x: Hence, T 1 ðxÞ ¼ x1 : Similarly Tðy1 Þ ¼ y; and T 1 ðyÞ ¼ y1 : Now, LHS ¼ T 1 ðsx þ tyÞ ¼ T 1 ðsðT ðx1 ÞÞ þ tðT ðy1 ÞÞÞ ¼ T 1 ðT ðsx1 þ ty1 ÞÞ ¼ sx1 þ ty1 ¼ s T 1 ð xÞ þ t T 1 ð yÞ ¼ RHS:

258

4 Topological Properties of Smooth Manifolds

Thus, T 1 is in GLðVÞ: It is clear that T 1 T ¼ T T 1 ¼ IdV : Hence, T 1 serves the purpose of the inverse element of T in GLðVÞ: Thus, we have shown that GLðVÞ is a group. Next, let V be ﬁnite-dimensional real linear space. Let n be the dimension of V. For simplicity, we shall take 2 for n. Let fe1 ; e2 g be a basis of the real linear space V. Let us take any T in GLðVÞ: So, T : V 7! V: Let Tðe1 Þ a11 e1 þ a21 e2 ; and Tðe2 Þ a12 e1 þ a22 e2 : Since T is in GLðVÞ; T is invertible, and hence, there exists unique ðt1 ; t2 Þ such that ða11 t1 þ a12 t2 Þe1 þ ða21 t1 þ a22 t2 Þe2 ¼ t1 ða11 e1 þ a21 e2 Þ þ t2 ða12 e1 þ a22 e2 Þ ¼ t1 ðT ðe1 ÞÞ þ t2 ðT ðe2 ÞÞ ¼ T ðt1 e1 þ t2 e2 Þ ¼ e1 ¼ 1e1 þ 0e2 : Thus, the system of equations

a11 t1 þ a12 t2 ¼ 1 a21 t1 þ a22 t2 ¼ 0

has a unique solution for ðt1 ; t2 Þ; and hence, a11 a22 a12 a21 6¼ 0: Since a11 a22 a12 a21 6¼ 0;

a11 a21

a12 a22

is in GLð2; RÞ:

Now, we can deﬁne a mapping g : GLðVÞ ! GLð2; RÞ as follows:

a gðT Þ 11 a21

a12 : a22

We shall try to show that g is an isomorphism from group GLðVÞ onto group GLð2; RÞ: For this purpose, we must show: (i) g is 1–1, (ii) g maps GLðVÞ onto GLð2; RÞ; (iii) for every S; T in GLðVÞ; gðS TÞ ¼ ðgðSÞÞðgðTÞÞ: For (i): Let gðSÞ ¼ gðTÞ; where Sðe1 Þ ¼ a11 e1 þ a21 e2 ; Sðe2 Þ ¼ a12 e1 þ a22 e2 ; Tðe1 Þ ¼ b11 e1 þ b21 e2 ; and Tðe2 Þ ¼ b12 e1 þ b22 e2 : We have to show that S ¼ T; that is, Sðe1 Þ ¼ Tðe1 Þ; and Sðe2 Þ ¼ Tðe2 Þ: Since

a11 a21

a12 a22

b ¼ gðSÞ ¼ gðT Þ ¼ 11 b21

b12 ; b22

a11 ¼ b11 ; and a21 ¼ b21 ; and hence, Sðe1 Þ ¼ a11 e1 þ a21 e2 ¼ b11 e1 þ b21 e2 ¼ Tðe1 Þ: Thus, Sðe1 Þ ¼ Tðe1 Þ: Similarly, Sðe2 Þ ¼ Tðe2 Þ:

4.2 Lie Groups

259

For (ii): We have to show that g : GLðVÞ ! GLð2; RÞ is onto. For this purpose, let us take any

a11 a21

a12 a22

in GLð2; RÞ:

We have to construct invertible linear transformation T from V to V such that

a gðT Þ ¼ 11 a21

a12 : a22

Let us deﬁne T : V ! V as follows: for every t1 ; t2 2 R; T ðt1 e1 þ t2 e2 Þ ¼ ða11 t1 þ a12 t2 Þe1 þ ða21 t1 þ a22 t2 Þe2 : It is clear that T is linear. Also Tðe1 Þ ¼ Tð1e1 þ 0e2 Þ ¼ ða11 1 þ a12 0Þe1 þ ða21 1 þ a22 0Þe2 ¼ a11 e1 þ a21 e 2 : Thus, Tðe

1 Þ ¼ a11 e1 þ a21 e2 : Similarly, Tðe2 Þ ¼ a11 a12 ¼ a11 a22 a12 a21 6¼ 0; T is invertible. a12 e1 þ a22 e2 : Now, since det a21 a22 Since T is an invertible linear transformation from V onto V, T is in GLðVÞ: Since Tðe1 Þ ¼ a11 e1 þ a21 e2 ; and Tðe2 Þ ¼ a12 e1 þ a22 e2 ;

a gðT Þ ¼ 11 a21

a12 : a22

For (iii): Let gð SÞ ¼

a11 a21

a12 ; a22

where Sðe1 Þ ¼ a11 e1 þ a21 e2 ; and Sðe2 Þ ¼ a12 e1 þ a22 e2 : Let

b gðT Þ ¼ 11 b21

b12 ; b22

where Tðe1 Þ ¼ b11 e1 þ b21 e2 ; and Tðe2 Þ ¼ b12 e1 þ b22 e2 : So, ðS T Þðe1 Þ ¼ SðT ðe1 ÞÞ ¼ Sðb11 e1 þ b21 e2 Þ ¼ b11 ðSðe1 ÞÞ þ b21 ðSðe2 ÞÞ ¼ b11 ða11 e1 þ a21 e2 Þ þ b21 ða12 e1 þ a22 e2 Þ ¼ ða11 b11 þ a12 b21 Þe1 þ ða21 b11 þ a22 b21 Þe2 :

260

4 Topological Properties of Smooth Manifolds

Thus, ðS T Þðe1 Þ ¼ ða11 b11 þ a12 b21 Þe1 þ ða21 b11 þ a22 b21 Þe2 : Next, ðS T Þðe2 Þ ¼ SðT ðe2 ÞÞ ¼ Sðb12 e1 þ b22 e2 Þ ¼ b12 ðSðe1 ÞÞ þ b22 ðSðe2 ÞÞ ¼ b12 ða11 e1 þ a21 e2 Þ þ b22 ða12 e1 þ a22 e2 Þ ¼ ða11 b12 þ a12 b22 Þe1 þ ða21 b12 þ a22 b22 Þe2 : Thus, ðS T Þðe2 Þ ¼ ða11 b12 þ a12 b22 Þe1 þ ða21 b12 þ a22 b22 Þe2 : It follows that

a11 b11 þ a12 b21

a11 b12 þ a12 b22

a21 b11 þ a22 b21 ¼ ðgðSÞÞðgðT ÞÞ ¼ RHS:

a21 b12 þ a22 b22

LHS ¼ gðS T Þ ¼

¼

a11

a12

a21

a22

b11

b12

b21

b22

Thus, we have shown that g is an isomorphism between group GLðVÞ onto group GLð2; RÞ: Hence, group GLðVÞ is isomorphic onto group GLð2; RÞ: Since GLðVÞ and GLð2; RÞ are isomorphic groups, and GLð2; RÞ is a Lie group, we can bestow a topology on GLðVÞ so that GLðVÞ is also a Lie group. It also shows that the smooth structure on GLðVÞ is independent of the choice of basis of V. Thus, if V is a real linear space of dimension n, then GLðVÞ is isomorphic to the Lie group GLðn; RÞ: Similarly, if V is a complex linear space of dimension n, then GLðVÞ is isomorphic to the Lie group GLðn; CÞ: Example 4.23 Observe that ðR; þÞ is a group, and R is a 1-dimensional smooth manifold. We want to show that R is a Lie group. By Lemma 4.19, it is enough to show that the mapping f : ðx; yÞ 7! x y is a smooth mapping from 2-dimensional product manifold R R to 1-dimensional smooth manifold R: For this purpose, let us ﬁx any ða; bÞ in R R: We have to prove that f is smooth at ða; bÞ. For this purpose, let us take the admissible coordinate chart ðR R; IdRR Þ of the product manifold R R: Here, ða; bÞ 2 R R: Next, let us take the admissible coordinate chart ðR; IdR Þ of R. Clearly f ða; bÞ ¼ a b 2 R: Now, we have to prove that the function IdR f ðIdRR Þ1 : ðIdRR Þ ðR RÞ \ f 1 ðRÞ ! IdR ðRÞ is C 1 at the point IdRR ða; bÞð¼ ða; bÞÞ in R2 : Let us observe that IdR ðf ðIdRR Þ1 Þ ¼ f ; ðIdRR ÞððR RÞ \ f 1 ðRÞÞ ¼ ðR RÞ \ f 1 ðRÞ ¼ ðR RÞ \ ðR RÞ ¼ R R; and IdR ðRÞ ¼ R; so we have to prove that the function

4.2 Lie Groups

261

f :RR!R is C1 at the point ða; bÞ: But this is known to be true. Thus, we have shown that ðx; yÞ 7! x y is a smooth map. This proves that R is a Lie group under addition. Similarly, R2 is a Lie group under addition, R3 is a Lie group under addition, etc. Since each ordered pair ðx; yÞ of real numbers x; y can be identiﬁed with the pﬃﬃﬃﬃﬃﬃﬃ complex number x þ 1y; and R2 is a Lie group under addition, clearly C is a Lie group under addition. Example 4.24 By R we shall mean the set of all nonzero real numbers. We know that R is a group under multiplication as binary operation. Since the elements of R can be identiﬁed, in a natural way, with the elements of GLð1; RÞ; and GLð1; RÞ; is a 1-dimensional Lie group, R is also a 1-dimensional Lie group under multiplication. Next, since R is a 1-dimensional Lie group, R is a smooth manifold. Since R is a smooth manifold, and fx : x 2 R and 0\xg is an open subset of R ; fx : x 2 R and 0\xg is a smooth manifold. We know that fx : x 2 R and 0\xg is a subgroup of the multiplicative group R : Thus, fx : x 2 R and 0\xg is a 1dimensional smooth manifold that is also a group. Since the function ðx; yÞ 7! xy is smooth so, by Lemma 4.19, fx : x 2 R and 0\xg is a 1-dimensional Lie group under multiplication. Example 4.25 By C we shall mean the set of all nonzero complex numbers. We know that C is a group under multiplication as binary operation. Since the elements of R can be identiﬁed, in a natural way, with the elements of GLð1; CÞ; and GLð1; CÞ is a 2-dimensional Lie group, C is also a 2-dimensional Lie group under multiplication. Example 4.26 We know that the unit circle S1 ð¼ fðx; yÞ : ðx; yÞ 2 R2 and pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ x2 þ y2 ¼ 1gÞ is a 1-dimensional smooth manifold. Observe that we can identify pﬃﬃﬃﬃﬃﬃﬃ ðx; yÞ with the complex number x þ 1y: Thus, we can identify the unit circle S1 with fz : z 2 C and jzj ¼ 1g: Further, we know that fz : z 2 C and jzj ¼ 1g is a multiplicative group. Thus, unit circle S1 is a 1-dimensional smooth manifold that is also a group. Since the function ðz; wÞ 7! wz is smooth, by Lemma 4.19, the circle S1 is a 1-dimensional Lie group. The Lie group S1 is called the circle group. Example 4.27 Let G1 and G2 be Lie groups. Since G1 and G2 are Lie groups, G1 and G2 are groups. We know that the Cartesian product G1 G2 is also a group under the binary operation given by ðg1 ; h1 Þðg2 ; h2 Þ ðg1 g2 ; h1 h2 Þ: Since G1 and G2 are Lie groups, G1 and G2 are smooth manifolds, and hence, their Cartesian product G1 G2 is a smooth manifold (called the product manifold of G1 and G2 ). Thus, G1 G2 is a 1-dimensional smooth manifold that is also a group.

262

4 Topological Properties of Smooth Manifolds

We want to show that G1 G2 is a Lie group. By Lemma 4.19, it is enough to show that the mapping ððx; yÞ; ðz; wÞÞ 7! ðxz1 ; yw1 Þ is smooth. Since the projection map ððx; yÞ; ðz; wÞÞ 7! ðx; yÞ is smooth, and ðx; yÞ 7! x is smooth, their composite ððx; yÞ; ðz; wÞÞ 7! x is smooth. Similarly, ððx; yÞ; ðz; wÞÞ 7! y is smooth. Since ððx; yÞ; ðz; wÞÞ 7! x and ððx; yÞ; ðz; wÞÞ 7! z are smooth, ððx; yÞ; ðz; wÞÞ 7! ðx; zÞ is smooth. Since G1 is a Lie group, and x; z are in G1 ; ðx; zÞ 7! xz1 is smooth. Since ððx; yÞ; ðz; wÞÞ 7! ðx; zÞ is smooth, and ðx; zÞ 7! xz1 is smooth, their composite ððx; yÞ; ðz; wÞÞ 7! xz1 is smooth. Similarly, ððx; yÞ; ðz; wÞÞ 7! yw1 is smooth. Since ððx; yÞ; ðz; wÞÞ 7! xz1 and ððx; yÞ; ðz; wÞÞ 7! yw1 are smooth, ððx; yÞ; ðz; wÞÞ 7! ðxz1 ; yw1 Þ is smooth. Thus, we have shown that G1 G2 is a Lie group. Similarly, if G1 ; G2 ; G3 are Lie groups, then G1 G2 G3 is a Lie group, etc. Here, G1 G2 G3 is called the direct product of Lie groups G1 ; G2 ; G3 ; etc. The direct product of circle group S1 with itself, that is, S1 S1 is an abelian Lie group under pointwise multiplication. S1 S1 is called the 2-torus and is denoted by T2 : Similarly, by 3-torus T3 , we mean the abelian Lie group S1 S1 S1 ; etc. Example 4.28 Let G be a ﬁnite group. Let us bestow discrete topology over G. Observe that discrete topology over G is Hausdorff and second countable. Here, G is called a 0-dimensional Lie group or discrete group. Deﬁnition Let G and H be Lie groups. Let F : G ! H be a mapping. If F is a smooth mapping, and F is a group homomorphism, then we say that F is a Lie group homomorphism from G to H. Deﬁnition Let G and H be Lie groups. Let F : G ! H be a mapping. If F is a group isomorphism, and F is a diffeomorphism, then we say that F is a Lie group isomorphism from G onto H. Deﬁnition Let G and H be Lie groups. If there exists a mapping F : G ! H such that F is a Lie group isomorphism from G onto H, then we say that G and H are isomorphic Lie groups. Example 4.29 We have seen that the circle S1 ; and the set C of all nonzero complex numbers are Lie groups. Let us deﬁne a function I : S1 ! C as follows: for every z in S1 ; IðzÞ ¼ z: Clearly, the inclusion map I is a Lie group homomorphism from S1 to C : Example 4.30 We have seen that R is a Lie group under addition, and R is a Lie group under multiplication. Consider the function exp : R ! R deﬁned as follows: for every x in R; expðxÞ ex : Since x 7! ex is a smooth function and is a group homomorphism from R to R ; exp is a Lie group homomorphism from R to R : Observe that the range of exp is fx : x 2 R and 0\xg: We have seen that fx : x 2 R and 0\xg is a Lie group under multiplication. Here, the inverse function of exp is ln : fx : x 2 R and 0\xg ! R: Since ln is a smooth function and is 1–1 onto, exp is a Lie group isomorphism from R to fx : x 2 R and 0\xg:

4.2 Lie Groups

263

Example 4.31 We have seen that C is a Lie group under addition, and C is a Lie group under multiplication. Consider the function exp : C ! C deﬁned as follows: for every z in C; pﬃﬃﬃﬃﬃﬃﬃ expðzÞ ez ¼ eReðzÞ cosðImðzÞÞ þ 1 sinðImðzÞÞ : Since ðx; yÞ ¼ z 7! ez ¼ ðex cos y; ex sin yÞ is a smooth function and is a group homomorphism from C to C ; exp is a Lie group homomorphism from Cto C : Observe that the range of exp is C ; exp is a function from C onto C : Since exp is not 1–1, exp is not a Lie group isomorphism from C to C : Example 4.32 We have seen that R is a Lie group under addition, and the circle S1 is a Lie group under multiplication. Consider the function e : R ! S1 deﬁned as follows: for every t in R; pﬃﬃﬃﬃﬃﬃﬃ eðtÞ cosð2ptÞ þ 1 sinð2ptÞ: pﬃﬃﬃﬃﬃﬃﬃ Since t 7! ðcosð2ptÞ þ 1 sinð2ptÞÞ ¼ ðcosð2ptÞ; sinð2ptÞÞ is a smooth function and is a group homomorphism from R onto S1 ; e is a Lie group homomorphism from R onto S1 : Here, the kernel kerðeÞ of homomorphism e is given by kerðeÞ ¼ e1 ð1Þ ¼ pﬃﬃﬃﬃﬃﬃﬃ ft : t 2 R and eðtÞ ¼ 1g ¼ ft : t 2 R and cosð2ptÞ þ 1 sinð2ptÞ ¼ 1g ¼ f0;

1; 2; 3; . . .g ¼ Z. We have seen that R2 is a Lie group under addition and the 2-torus T2 ð¼ S1 S1 Þ is a Lie group under pointwise multiplication. Consider the function e2 : R2 ! S1 S1 deﬁned as follows: for every s; t in R; pﬃﬃﬃﬃﬃﬃﬃ pﬃﬃﬃﬃﬃﬃﬃ e2 ðs; tÞ cosð2psÞ þ 1 sinð2psÞ; cosð2ptÞ þ 1 sinð2ptÞ : Since pﬃﬃﬃﬃﬃﬃﬃ pﬃﬃﬃﬃﬃﬃﬃ ðs; tÞ 7! cosð2psÞ þ 1 sinð2psÞ; cosð2ptÞ þ 1 sinð2ptÞ ¼ ððcosð2psÞ; sinð2psÞÞ; ðcosð2ptÞ; sinð2ptÞÞÞ ¼ ðcosð2psÞ; sinð2psÞ; cosð2ptÞ; sinð2ptÞÞ is a smooth function and is a group homomorphism from R2 onto S1 S1 ; e2 is a Lie group homomorphism from R2 onto 2-torus T2 ð¼ S1 S1 Þ: Here, the kernel kerðe2 Þ of homomorphism e2 is given by kerðe2 Þ ¼ ðe2 Þ1 ð1; 1Þ ¼

pﬃﬃﬃﬃﬃﬃﬃ fðs; tÞ : ðs; tÞ 2 R2 and e2 ðs; tÞ ¼ ð1; 1Þg ¼ fðs; tÞ : ðs; tÞ 2 R2 ; cosð2psÞ þ 1 sinð2psÞ ¼ pﬃﬃﬃﬃﬃﬃﬃ 1; and cosð2ptÞ þ 1 sinð2ptÞ ¼ 1g¼ f0; 1; 2; 3; . . .g f0; 1; 2; 3; . . .g ¼ Z2 :

264

4 Topological Properties of Smooth Manifolds

Similar discussion can be given for e3 ; e4 ; etc. Example 4.33 We have seen that

x1 x2 xx GLð2; RÞ ¼ : 1 2 2 Mð2; RÞ and x1 x4 x2 x3 6¼ 0 x3 x4 x3 x4 is a Lie group under matrix multiplication, and R is a Lie group under ordinary multiplication. Consider the function det : GLð2; RÞ 7! R deﬁned as follows: For every

x1 x3

x2 x4

x1 x3

x2 x4

2 Mð2; RÞ;

we have det

x1 x4 x2 x3 :

Since ð x1 ; x2 ; x3 ; x4 Þ ¼

x1 x3

x2 7 ð x1 x4 x2 x3 Þ ! x4

is a smooth function, and by a property of determinant det is a group homomorphism from GLð2; RÞ onto R ; det is a Lie group homomorphism from GLð2; RÞ onto R : Example 4.34 Let us take any n-dimensional Lie group G. Let us ﬁx any a in G. For simplicity, let us take 2 for n. Consider the mapping Ca : G 7! G deﬁned as follows: for every x in G, Ca ðxÞ axa1 : We want to show that Ca is a Lie group homomorphism from G onto G; that is, 1. Ca is a group homomorphism, 2. Ca is onto, 3. Ca is smooth. For 1: Since for every x; y in G; Ca ðxyÞ aðxyÞa1 ¼ ðaxa1 Þðaya1 Þ ¼ ðCa ðxÞÞðCa ðyÞÞ; Ca is a group homomorphism. For 2: For this purpose, let us take any y in G. Since a1 ya is in G, and Ca ða1 yaÞ ¼ aða1 yaÞa1 ¼ y; it follows that Ca is onto. For 3: Since G is a Lie group, and a is in group G, the map x 7! ax is smooth. Similarly x 7! xa1 ; is smooth. Since x 7! ax is smooth, and x 7! xa1 ; their composite x 7! axa1 ð¼ Ca ðxÞÞ is smooth, and hence Ca is smooth. Thus we have shown that Ca is a Lie group homomorphism from G onto G:

4.3 Locally Path Connected Spaces

265

4.3 Locally Path Connected Spaces Note 4.35 Let X be a topological space. Let S be a connected subset of X. We shall try to show that the closure S– of S is also connected. We claim that S– is connected. If not, otherwise, let S– be not connected. We have to arrive at a contradiction. Since S– is not connected, there exist open subsets G1 ; G2 of X such that G1 \ðS Þ is nonempty, G2 \ðS Þ is nonempty, G1 \ G2 \ðS Þ is empty, and S G1 [ G2 : Since G1 \ðS Þ is nonempty, there exists a 2 G1 \ðS Þ; and hence a 2 G1 : Since a 2 G1 and G1 is open, G1 is an open neighborhood of a. Since a 2 G1 \ðS Þ; a 2 S : Since a 2 S and G1 is an open neighborhood of a, G1 \ S is nonempty. Similarly, G2 \ S is nonempty. Since S S ; and G1 \ G2 \ðS Þ is empty, G1 \ G2 \ S is empty. Since S S G1 [ G2 ; S G1 [ G2 : Since G1 ; G2 are open subsets of X, G1 \ S is nonempty, G2 \ S is nonempty, G1 \ G2 \ S is empty, and S G1 [ G2 ; S is not connected, a contradiction. Note 4.36 Let X be a topological space. Let S be a nonempty subset of X. For every x; y in S, by x y we shall mean that there exists a connected subset C of X such that C S; x 2 C; and y 2 C: We shall show that * is an equivalence relation over S, that is, 1. for every x in S, x x; 2. if x y then y x; 3. if x y; and y z then x z: For 1: Let us take any x in S. Since the singleton set {x} is connected subset of X, and x 2 fxg S, by the deﬁnition of *, x x: For 2: Let x y: Now, by the deﬁnition of *, there exists a connected subset C of X such that C S; x 2 C; and y 2 C: Thus, C is a connected subset of X such that C S; y 2 C; and x 2 C; and hence, by the deﬁnition of *, y x: For 3: Let x y; and y z: Since x y; so by the deﬁnition of *, there exists a connected subset C of X such that C S; x 2 C; and y 2 C: Similarly, there exists a connected subset C1 of X such that C1 S; y 2 C1 ; and z 2 C1 : Since C is a connected subset of X, C1 is a connected subset of X, and y 2 C \ C1 ; C [ C1 is a connected subset of X. Also x 2 C C [ C1 ; and z 2 C1 C [ C1 : Since C S; and C1 S; so C [ C1 S: It follows, by the deﬁnition of *, that x z: Thus, we have shown that * is an equivalence relation over S. Hence, S is partitioned into equivalence classes. Here, each equivalence class is called a component of S. Now, we shall try to show that each component of S is a connected subset of X. Let us take any component [a] of S, where a is in S, and ½a fx : x 2 S; and x ag: We claim that [a] is connected. If not, otherwise, let [a] be not connected, that is, there exist open subsets G1 ; G2 of X such that G1 \½a is nonempty, G2 \½a is nonempty, G1 \ G2 \½a is empty, and ½a G1 [ G2 : We have to arrive at a contradiction. Since G1 \½a is nonempty, there exists b in G1 \½a : Similarly, there exists c in G2 \½a : Here, b and c are in [a], and [a] is an

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equivalence class, so b c; and hence, by the deﬁnition of *, there exists a connected subset C of X such that C S; b 2 C; and c 2 C: Thus, b is in G1 \ C; and hence, G1 \ C is nonempty. Similarly, G2 \ C is nonempty. Clearly, C is contained in [a]. (Reason: If not, otherwise, there exists t 2 C such that t 62 ½a : Since C is a connected subset of X, C S; b 2 C; and t 2 C; b t: Since b is in G1 \½a ; b a. Since b t; and b a; t a; and hence t 2 ½a ; a contradiction.) Since C is contained in [a], and G1 \ G2 \½a is empty, G1 \ G2 \ C is empty. Next since C is contained in [a] and ½a G1 [ G2 ; C G1 [ G2 : Since G1 ; G2 are open subsets of X, G1 \ C is nonempty, G2 \ C is nonempty, G1 \ G2 \ C is empty, and C G1 [ G2 ; C is not connected, a contradiction. Thus, we have shown that [a] is a connected subset of S. Clearly, [a] is a maximal connected subset of S. (Reason: If not, otherwise, let [a] be not a maximal connected subset of S. So, there exists a connected subset C of S such that [a] is a proper subset of C. So, there exists t in C such that t is not in [a]. Since a 2 ½a C S; t is in C, and C is a connected subset of X, t a; and hence, t is in [a], a contradiction.) Finally, we want to show that every component of X is a closed set. Let us take any component C of X. Since C is a component of X, C is connected, and hence, C is connected. Since C is a component of X, C is connected, and C C X, C ¼ C ; and hence, C is closed. Thus, we have seen that (i) every nonempty subset S topological space X is partitioned into components, and (ii) every component of S is connected. (iii) if C is a component of S, C1 is connected, and C C1 S then C ¼ C1 : (iv) every component of X is a closed set. Note 4.37 Let X be a topological space. Let S be a nonempty subset of X. Let a, b be elements of S. Let f be a function from closed interval [0,1] to S. If (i) f is continuous, (ii) f ð0Þ ¼ a and f ð1Þ ¼ b, then, we say that f is a path in S from a to b. Let X be a topological space. Let S be a nonempty subset of X. If for every a, b in S, there exists a path f in S from a to b, then we say that S is a path connected subset of X. By X is path connected we mean: X is a path connected subset of X. In other words, a topological space X is path connected means for every x, y in X, there exists a continuous function f : ½0; 1 7! X such that f ð0Þ ¼ x and f ð1Þ ¼ y: Let X be a topological space. Let S be a nonempty subset of X. For every x, y in S, by x y, we shall mean that the following statement is true:

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there exists a path f in S from x to y: We shall try to show that * is an equivalence relation over S. By the deﬁnition of equivalence relation, we must prove: (1) x x for every x in S, (2) if x y, then y x; (3) if x y and y z, then x z: For (1): Let us take any x in S. By the deﬁnition of *, we must ﬁnd a path f in S from x to x. For this purpose, consider the constant function f : ½0; 1 ! S deﬁned by f ðt Þ x for every t in [0,1]. Since every constant function is a continuous function, and f is a constant function, f is a continuous function. Further, by the deﬁnition of f, f ð0Þ ¼ x and f ð1Þ ¼ x. Hence, by the deﬁnition of path, f is a path in S from x to x. This proves (1). For (2): Let x y. So, by the deﬁnition of *, there exists a path f in S from x to y. Since f is a path in S from x to y so, by the deﬁnition of path, f is a function from the closed interval [0,1] to S such that (i) f is continuous, (ii) f ð0Þ ¼ x and f ð1Þ ¼ y. We have to prove: y x. So, by the deﬁnition of *, we must ﬁnd a path in S from y to x, and hence, by the deﬁnition of path, we must ﬁnd a function g : ½0; 1 7! S such that (10 ) g is continuous, (20 ) gð0Þ ¼ y and gð1Þ ¼ x: Let us deﬁne g : ½0; 1 ! X as follows: gðtÞ f ð1 tÞ for every t in [0, 1]. Clearly, the range of g is equal to the range of f. Since the range of g is equal to the range of f, and f : ½0; 1 ! S; g : ½0; 1 ! S: For ð10 Þ: Since t 7! 1 t is a polynomial function, this is continuous. Since t 7! 1 t is a continuous function, and f is a continuous function, their composite function t 7! f ð1 tÞ ¼ gðtÞ is continuous, and hence, g is continuous. This proves (10 ). For ð20 Þ : Here, by the deﬁnition of g, gð0Þ ¼ f ð1 0Þ ¼ f ð1Þ ¼ y; and gð1Þ ¼ f ð1 1Þ ¼ f ð0Þ ¼ x: This proves 20 : Thus we have shown: If x y, then y x: This proves ð2Þ. For (3): Given that x y and y z. We have to prove: x z; that is, by the deﬁnition of *, we must ﬁnd a path h in S from x to z, that is, we must ﬁnd a function h : ½0; 1 ! S such that

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(10 ) h is continuous, (20 ) hð0Þ ¼ x and hð1Þ ¼ z. Since x y, by the deﬁnition of *, there exists a path in S from x to y, and hence, there exists a function f : ½0; 1 ! S such that (100 ) f is continuous, (200 ) f ð0Þ ¼ x and f ð1Þ ¼ y. Since y z, by the deﬁnition of *, there exists a path in S from y to z, and hence, there exists a function g : ½0; 1 7! S such that (1000 ) g is continuous, (2000 ) gð0Þ ¼ y and gð1Þ ¼ z. Let us deﬁne a function h : ½0; 1 7! X as follows: f ð2tÞ if 0 t\ 12 hð t Þ gð2t 1Þ if 12 t 1: Clearly, the range of h is the union of the range of f, and the range of g. Further, since f : ½0; 1 ! S; and g : ½0; 1 ! S; the range of h is contained S, and hence h : ½0; 1 7! S: For ð10 Þ: we have to prove that h is continuous over [0, 1], that is, h is continuous at all points of [0, 1]. By the deﬁnition of h over 0 t\ 12, h is the composite function of two continuous functions, namely t 7! 2t and f so, h is continuous at all points of ft : 0 t\ 12g: Similarly, h is continuous at all points of ft : 12 \t 1g: So, it remains to prove that h is continuous at 12 : For this purpose, we should calculate the left-hand limit of h at 12, the right-hand limit of h at 12 and hð12Þ. If all the three values are equal, then we can say that h is continuous at 12. In short, we have to prove: 1 lim hðtÞ ¼ lim þ hðtÞ ¼ h : 1 2 t!ð2Þ t!ð12Þ Here, limt!ð12Þ hðtÞ ¼ limt!12 f ð2tÞ; by the deﬁnition of h. Since t 7! f ð2tÞ is a

continuous function over ½0; 12 , limt!12 f ð2tÞ ¼ f ð2 12Þ ¼ f ð1Þ ¼ y: Hence, limt!ð12Þ hðtÞ ¼ y: Similarly, limt!ð1Þþ hðtÞ ¼ limt!12 gð2t 1Þ ¼ gð2ð12Þ 1Þ ¼ 2 gð0Þ ¼ y:Hence limt!ð12Þ hðtÞ ¼ limt!ð1Þþ hðtÞ ¼ y. Further, by the deﬁnition of 2

h, hð12Þ ¼ gð2ð12Þ 1Þ ¼ gð0Þ ¼ y: Hence, limt!ð12Þ hðtÞ ¼ limt!ð1Þþ hðtÞ ¼ hð12Þ. 2

Thus, we have shown that h is continuous over [0,1]. This proves ð10 Þ For ð20 Þ: Here hð0Þ ¼ f ð2ð0ÞÞ ¼ f ð0Þ ¼ x, and hð1Þ ¼ gð2ð1Þ 1Þ ¼ gð1Þ ¼ z. This proves ð20 Þ: Thus, we have shown: If x y and y z, then x z: This proves (3). Hence, * is an equivalence relation over S. Hence, S is partitioned into equivalence classes. Here, each equivalence class is called a path component of S.

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Note 4.38 Let X be a topological space. Let S be a nonempty subset of X. Now, we shall try to prove that each path component of S is a path connected subset of X. For this purpose, let us take any path component [a] of S, where a 2 S; ½a fx : x 2 S; and x ag; and x a stands for: There exists a path in S from x to a. We have to show that [a] is a path connected subset of S. For this purpose, let us take any x,y in [a]. We have to ﬁnd a path in [a] from x to y. Since [a] is a path component of S, [a] is an equivalence class. Since [a] is an equivalence class, and x is in [a], x a; and hence, there exists a function f : ½0; 1 ! S such that (10 ) f : ½0; 1 ! S is continuous, (20 ) f ð0Þ ¼ x and f ð1Þ ¼ a. Clearly, the range of f is contained in [a]. (Reason: Let us take any b f ðt0 Þ 2 ran f ; where t0 2 ð0; 1Þ: We have to prove that b 2 ½a ; that is, a b: We must ﬁnd a function F : ½0; 1 ! S such that (i) F is continuous, (ii) Fð0Þ ¼ a and Fð1Þ ¼ b. Let us deﬁne F : ½0; 1 ! S as follows: for every t in [0,1], FðtÞ f ð1 þ ðt0 1ÞtÞ: Clearly, the range of [0,1] under t 7! ð1 þ ðt0 1ÞtÞ is ½t0 ; 1 ; and hence, the ran F f ð½t0 ; 1 Þ S: Since ran F f ð½t0 ; 1 Þ S; F : ½0; 1 ! S. Since F is the composite of two continuous functions t 7! ð1 þ ðt0 1ÞtÞ; and f, F is continuous. Clearly, Fð0Þ ¼ f ð1 þ ðt0 1Þ0Þ ¼ f ð1Þ ¼ a; and Fð1Þ ¼ f ð1 þ ðt0 1Þ1Þ ¼ f ðt0 Þ ¼ b.) Since the range of f is contained in [a], and f : ½0; 1 ! S; f : ½0; 1 ! ½a : Now, since f : ½0; 1 ! S; is continuous, f : ½0; 1 ! ½a : is continuous. Further, f ð0Þ ¼ x and f ð1Þ ¼ a. Thus we get a path in [a] from x to a. Similarly, we get a path in [a] from a to y. Since there exist a path in [a] from x to a, and a path in [a] from a to y, there exists a path in [a] from x to y. Let X be a topological space. Let S be a nonempty subset of X. Now, we shall try to prove that if C is a path component of S, C1 is path connected, and C C1 S, then C ¼ C1 : If C is a component of S, C1 is connected, and C C1 S, then C ¼ C1 : Let us take a path component [a] of S, and a path connected set C satisfying ½a C S: We claim that ½a ¼ C1 : If not, otherwise, let [a] be a proper subset of C. So, there exists t in C such that t is not in [a]. Since a 2 ½a C; a is in C. Since a is in C, t is in C, and C is a path connected subset of X, there exists a path f in C from t to a. Since f is a path in C, and C S; f is a path in S, and hence, t is in [a], a contradiction. Thus, we have seen that (i) every nonempty subset S of a topological space X is partitioned into path components, (ii) every path component of S is path connected, (iii) if C is a path component of S, C1 is path connected, and C C1 S then C ¼ C1 : Note 4.39 Let X be a topological space. Let S be a nonempty subset of X. We shall try to prove: Each component of S is a union of path components of S. Let us take any component fy : y 2 S and y ag of S, where a is in S, and y a stands for:

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There exists a connected subset A of X such that A S; and y; a are in A. It is enough to show that fy : y 2 S and y 0 ag fy : y 2 S and y ag where y 0 a stands for: There exists a path in S from y to a. For this purpose, let us take any y in LHS. We have to prove that y is in RHS, that is, y a: By the deﬁnition of *, we should ﬁnd a connected subset A of X such that A S; and y, a are in A. Since y is in LHS, y 0 a. Since y 0 a, by the deﬁnition of 0 , there exists a path f in S from y to a. Since f is a path in S from y to a, f is a function from closed interval [0,1] to S satisfying (I) f is continuous, (II) f ð0Þ ¼ y and f ð1Þ ¼ a. Since f : ½0; 1 ! S is continuous, and [0,1] is connected, so the f-image set f ð½0; 1 Þ of [0, 1] is a connected subset of X. Since 0 is in [0, 1], f ð0Þ is in f ð½0; 1 Þ. Similarly, f ð1Þ is in f ð½0; 1 Þ: Further, f ð0Þ ¼ y and f ð1Þ ¼ a: Thus, f ð½0; 1 Þ is a connected subset of X such that f ð½0; 1 Þ S; and y, a are in f ð½0; 1 Þ: Note 4.40 Let X be a topological space. We shall try to show: If X is path connected, then X is connected. If not, otherwise, let X be not connected. We have to arrive at a contradiction. Since X is not connected, there exist open subsets G1 ; G2 of X such that G1 is nonempty, G2 is nonempty, G1 \ G2 is empty, and X ¼ G1 [ G2 : Since G1 is nonempty, there exists a in G1 : Similarly, there exists b in G2 Now, since X is path connected, there exists a continuous function f : ½0; 1 ! X such that f ð0Þ ¼ a; and f ð1Þ ¼ b: Since f : ½0; 1 ! X is continuous, and [0, 1] is connected, the image set f ð½0; 1 Þ is connected. Here, a is in G1 ; and a ¼ f ð0Þ 2 f ð½0; 1 Þ; so a 2 G1 \ f ð½0; 1 Þ; and hence, G1 \ f ð½0; 1 Þ is nonempty. Similarly, G2 \ f ð½0; 1 Þ is nonempty. Next, since G1 \ G2 is empty, G1 \ G2 \ f ð½0; 1 Þ is empty. Since G1, G2 are open subsets of X such that G1 \ f ð½0; 1 Þ is nonempty, G2 \ f ð½0; 1 Þ is nonempty, G1 \ G2 \ f ð½0; 1 Þ is empty, and f ð½0; 1 Þ X ¼ G1 [ G2 ; f ð½0; 1 Þ is not connected, a contradiction. Deﬁnition Let X be a topological space. If for every x in X, and for every open neighborhood U of x, there exists an open neighborhood V of x such that (i) x 2 V U; (ii) V is a connected subset of X, then we say that X is a locally connected space. Deﬁnition Let X be a topological space. If for every x in X, and for every open neighborhood U of x, there exists an open neighborhood V of x such that (i) x 2 V U; (ii) V is a path connected subset of X, then we say that X is a locally path connected space.

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Note 4.41 Let X be a topological space. We shall try to prove: If X is locally connected space, and G is a nonempty open subset of X, then each component of G is open. Let X be a locally connected space, and let G be an open subset of X. We have to show that each component of G is open in X. By the deﬁnition of component, every component of G is of the form ½a fx : x 2 G and x ag for some a 2 G, where x a stands for: There exists a connected subset A of X such that A G; x 2 A; and a 2 A: We have to show that [a] is an open subset of X, that is, every point of [a] is an interior point of [a]. For this purpose, let us take any x 2 ½a . We must ﬁnd an open neighborhood V of x such that V ½a . Since x 2 ½a and ½a G, x 2 G. Since x 2 G and G is open in X, G is an open neighborhood of x. Since X is a locally connected space, and G is an open neighborhood of x, there exists an open neighborhood V of x such that (I) x 2 V G; (II) V is a connected subset of X. We claim that V ½a : If not, otherwise, let V 6 ½a : So, there exists b in V such that b 62 ½a . Since [a] is an equivalence class of a, and b 62 ½a , b ¿ a: Here, b 2 V G so b 2 G: Similarly, x 2 G. Further, since V is a connected subset of X, V G; and x, b are in V, by the deﬁnition of *, x b: Since x b; and * is an equivalence relation, ½x ¼ ½b : Since x 2 ½a ; and * is an equivalence relation, ½x ¼ ½a : Since ½x ¼ ½b ; and ½x ¼ ½a , ½a ¼ ½b . Since ½a ¼ ½b , a b. This is a contradiction. So, our claim is true, that is, V ½a : Clearly, if X is locally connected space, then each component of X is open. Note 4.42 Let X be a topological space. We shall try to prove: If X is locally path connected space, and G is a nonempty open subset of X, then each path component of G is open. Let X be a locally path connected space, and let G be an open subset of X. We have to show that each path component of G is open in X. By the deﬁnition of path component, every path component of G is of the form ½a fx : x 2 G and x ag for some a 2 G, where x a stands for: There exists a path in G from x to a. We have to show that [a] is an open subset of X, that is, every point of [a] is an interior point of [a]. For this purpose, let us take any x 2 ½a . We must ﬁnd an open neighborhood V of x such that V ½a . Since x 2 ½a ; and ½a G, x 2 G. Since x 2 G and G is open in X, G is an open neighborhood of x. Since X is a locally path connected space, and G is an open neighborhood of x, there exists an open neighborhood V of x such that (I) x 2 V G; (II) V is a path connected subset of X.

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It remains to prove: V ½a : If not, otherwise, let V 6 ½a : We have to arrive at a contradiction. Since V 6 ½a , there exists b in V such that b 62 ½a . Since [a] is an equivalence class of a, and b 62 ½a , b ¿ a: Here, b 2 V G so b 2 G: Similarly, x 2 G. Further, since V is a path connected subset of X and x, b are in V, there exists a path in V from x to b. Since there exists a path in V from x to b, and V G, by the deﬁnition of *, x b: Since x b and * is an equivalence relation, ½x ¼ ½b : Since x 2 ½a ; and * is an equivalence relation, ½x ¼ ½a : Since ½x ¼ ½b ; and ½x ¼ ½a , ½a ¼ ½b . Since ½a ¼ ½b , a b. Thus, we get a contradiction. Clearly, if X is a locally path connected space, then each path component of X is open. Note 4.43 Let X be a locally path connected space. We shall try to prove that the collection of all path components of X, and the collection of all components of X are equal. For any a in X, let fy : y 2 X and y 0 ag be a path component of X, where y 0 a stands for: There exists a path in X from y to a. It is enough to prove that fy : y 2 X and y 0 ag ¼ fy : y 2 X and y ag where y a stands for: There exists a connected subset A of X such that A G; y 2 A; and a 2 A: We know that fy : y 2 X and y 0 ag fy : y 2 X and y ag: So, it remains to prove that fy : y 2 X and y ag fy : y 2 X and y 0 ag

. . . ð Þ

If not, otherwise, let there exists y 2 fy : y 2 X and y ag such that y 62 fy : y 2 X and y 0 ag. We have to arrive at a contradiction. Here, y a is true and y 0 a is false. Since y a is true, by the deﬁnition of *, there exists a connected subset A of X such that y ; a are in A. Since fy : y 2 X and y 0 y g is the path component of X containing y ; and X is a locally path connected space, P fy : y 2 X and y 0 y g is an open set containing y : Since P contains y ; and y is in A, y 2 A \ P, and therefore, A \ P is nonempty. Since X is partitioned into path components, complement of each path component in X is a union of path components of X. Since X is a locally path connected space, each path component is open. Since union of open sets is open, complement of each path component in X is union of path components, and each path component is open, complement of each path component in X is open. Since fy : y 2 X and y 0 y g is a path component, and complement of each path component in X is open, the complement P0 of P is open. Thus, we have obtained two open sets P and P0 such that A \ P is nonempty. Now, we shall try to prove that A \ ðP0 Þ in nonempty. Since y 62 fy : y 2 X and y 0 ag; and 0 is an equivalence relation over X, fy : y 2 X and y 0 y gð¼ PÞ and fy : y 2 X and y 0 ag are disjoint sets, and therefore, fy : y 2 X and y 0 ag is contained in P0 : Since 0 is reflexive over X,

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a is in fy : y 2 X and y 0 ag. Since a 2 fy : y 2 X and y 0 ag P0 ; a is in P0 . Since a is in P0 and a is in A, a is in A \ðP0 Þ, and therefore, A \ðP0 Þ is nonempty. Since P and P0 are two open sets such that (I) P \ A and P0 \ A are nonempty sets, (II) P \ P0 \ A is empty, (III) A P [ðP0 Þ; by the deﬁnition of connected set, A is not a connected set. This is a contradiction. This proves (*). h

4.4 Smooth Manifold as Paracompact Space Deﬁnition Let n be a ﬁxed positive integer. Let M be a topological space. If the following conditions are satisﬁed: 1. M is a Hausdorff space, 2. M is a second countable space, 3. for every x in M, there exists an open neighborhood U of x in M, and an open e of Euclidean space Rn such that U and U e are homeomorphic, then we subset U say that M is a topological manifold of dimension n. In other words, M is a topological manifold of dimension n means M is a topological space that satisﬁes the following conditions: 10 : If x and y are distinct points of M, then there exist an open neighborhood U of x and an open neighborhood V of y such that U and V have no common element, 20 . There exists a countable collection C of open subsets of M such that for every neighborhood U of x, there exists a member G of C such that x 2 G U; 30 . For every x in M, there exist an open neighborhood U of x in M, an open e of Euclidean space Rn , and a 1–1 mapping φ from U onto U e such that subset U 1 φ and its inverse u , both are continuous functions. Here, ðU; uÞ is referred as a coordinate chart of M. Note 4.44 Let M be a topological manifold of dimension n. We shall try to show that M is locally path connected. For this purpose, let us take an element x in M and an open neighborhood U of x in M. We have to ﬁnd an open set V of x such that (I) x 2 V U; (II) V is a path connected subset of M. Since M is a topological manifold, and x is in M so, there exist an open e 1 of Euclidean space Rn such that neighborhood U1 of x in M and an open subset U e 1 are homeomorphic. U1 and U

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Since U and U1 are open neighborhoods of x, their intersection U \ U1 is also an e 1 are homeomorphic so, there exists a open neighborhood of x. Here that U1 and U e 1–1 mapping φ from U1 onto U 1 such that φ and its inverse u1 , both are continuous onto e 1 11; functions. Since U \ U1 is an open subset of U1, and u1 : U ! U1 is cone 1 Since uðU \ U1 Þ is an open subset of U e 1; tinuous, uðU \ U1 Þ is an open subset of U n n e and U 1 is an open subset of R , uðU \ U1 Þ is an open subset of R : Since U \ U1 is an open neighborhood of x, x is in U \ U1 ; and hence, uðxÞ is in uðU \ U1 Þ: Since uðxÞ is in uðU \ U1 Þ and uðU \ U1 Þ is an open subset of Rn , uðU \ U1 Þis an open neighborhood of uðxÞ in Rn . So, there exists e [ 0 such that the open ball B fy 2 Rn : jy uð xÞj\eg e 1 ; B is contained in U e 1 : We is contained in uðU \ U1 Þ: Since B uðU \ U1 Þ U n e know that B is open in R . Since φ is a continuous function from U1 onto U 1 ; and e 1 , u1 ðBÞ is an open subset of U1. Let us take u1 ðBÞ for B is an open subset of U V in (I) and (II). So, we must prove: (I) x 2 u1 ðBÞ U; (II) u1 ðBÞ is a path connected subset of M. For (I): Since juðxÞ uðxÞj ¼ 0\e, uðxÞ 2 fy 2 Rn : jy uðxÞj\eg ¼ B; and hence, x 2 u1 ðBÞ: Since U \ U1 is contained in U1, uðU \ U1 Þ uðUÞ: Since B uðU \ U1 Þ uðUÞ, B uðUÞ and hence u1 ðBÞ U: This proves (I). For (II): We have to prove that u1 ðBÞ is a path connected subset of M. For this purpose, let us take any a and b in u1 ðBÞ. We have to ﬁnd a path f in M from a to b, that is, we have to ﬁnd a continuous function f from closed interval [0, 1] to M such that f ð0Þ ¼ a and f ð1Þ ¼ b: Consider the function f : ½0; 1 7! M deﬁned by f ðtÞ u1 ðð1 tÞuðaÞ þ tuðbÞÞ for every t in [0, 1]. Here, t 7! u1 ðð1 tÞuðaÞ þ tuðbÞÞ is a composite function of two functions t 7! ð1 tÞuðaÞ þ tuðbÞ and u1 . Since a is in u1 ðBÞ, uðaÞ 2 B, and hence, by the deﬁnition of B; juðaÞ uðxÞj\e: Similarly, juðbÞ uðxÞj\e. Since juðaÞ uðxÞj\e; and juðbÞ uðxÞj\e; for every t satisfying 0 t 1, we have jðð1 tÞuðaÞ þ tuðbÞÞuð xÞj ¼ jð1 tÞðuðaÞ uð xÞÞ þ tðuðbÞ uð xÞÞj ð1 tÞjuðaÞ uð xÞj þ tjuðbÞ uð xÞj \ð1 tÞe þ te ¼ e: It follows, by the deﬁnition of B, that t 7! ð1 tÞuðaÞ þ tuðbÞ is a function from e 1 Þ; and [0,1] to B. Since t 7! ð1 tÞuðaÞ þ tuðbÞ is a function from [0,1] to Bð U e 1 7! U1 M; their composite t 7! u1 ðð1 tÞuðaÞ þ tuðbÞÞ maps [0, 1] to u1 : U M. Since t 7! ð1 tÞuðaÞ þ tuðbÞ and u1 both are continuous, their composite

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function t 7! u1 ðð1 tÞuðaÞ þ tuðbÞÞ is continuous. Further, since f ðtÞ ¼ u1 ðð1 tÞuðaÞ þ tuðbÞÞ and , t 7! u1 ðð1 tÞuðaÞ þ tuðbÞÞ is continuous, f : ½0; 1 7! M is continuous. Also f ð0Þ ¼ u1 ðð1 0ÞuðaÞ þ 0uðbÞÞ ¼ u1 ðuðaÞÞ ¼ a; and f ð1Þ ¼ u1 ðð1 1ÞuðaÞ þ 1uðbÞÞ ¼ u1 ðuðbÞÞ ¼ b. Thus f is a path in M from a to b. This proves (II). Note 4.45 Let M be a topological manifold of dimension n. We shall try to prove: (i) M is connected if and only if it is path connected. (ii) The collection of all path components of M is equal to the collection of all components of M. (iii) M has countable many components, each of which is an open subset of M, and a connected topological manifold. For (i): Let M be path connected. We have to prove that M is connected. Since M is path connected, M is a path component. Since M is a topological manifold, M is locally path connected. Since M is locally path connected, and M is a path component, M is a component. Since M is a component, M is connected. Converse: Let M be connected. We have to prove that M is path connected. Since M is connected, M is a component. Since M is a topological manifold, M is locally path connected. Since M is locally path connected, and M is a component, M is a path component. Since M is a path component, M is path connected. This completes the proof of (i). For (ii): Since M is a topological manifold, M is locally path connected. Since M is locally path connected, the collection of all path components of M is equal to the collection of all components of M. This proves (ii). For (iii): We have to prove that M has countable many components. If not, otherwise, let M has uncountable many components. We have to arrive at a contradiction. Let C be a component of M. Since M is a topological manifold, M is locally path connected. Since M is locally path connected, and C is a component of M, C is a path component. Since M is locally path connected, and C is a path component, C is open in M. Since C is open in M, and a is in C, C is an open neighborhood of a in M. Since M is a topological manifold, M is a second countable space. Since M is a second countable space, there exists a countable collection C of open subsets of M such that for every neighborhood U of x, there exists a member G of C such that x 2 G U: Now, since C is an open neighborhood of a, there exists a member G of C such that x 2 G C: Thus, we see that corresponding to each component C, there exists a member G of C such that G is contained in C. Since M is partitioned into components, each component contains a member of C, and the collection of all components is uncountable, the collection C is uncountable. This is a contradiction. This proves that M has countable many components.

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Now, let us take a component C. We have to prove that C is open and connected. Since M is a topological manifold, M is locally path connected. Since M is locally path connected and C is a component of M, C is a path component. Since M is locally path connected, and C is a path component, C is open in M. Since C is a component, C is connected. Hence, C is open and connected. Lastly, we have to show that C is a topological manifold of dimension n, where topology on C is the induced topology of M. Since C is an open subset of topological space M, under induced topology of M, open subsets of C are exactly those subsets of C which are open in M. By the deﬁnition of topological manifold, we must prove: 1. C is a Hausdorff space, 2. C is a second countable space, e1 3. for every x in C, there exist a neighborhood U1 of x in C, and an open subset U n e 1 are homeomorphic: of Euclidean space R such that U1 and U For 1: Since M is a topological manifold, M is a Hausdorff space. Since M is a Hausdorff space, and C is a subspace of M, C is a Hausdorff space. This proves 1. For 2: Since M is a topological manifold, M is a second countable space. Since M is a second countable space, and C is a subspace of M, C is a second countable. This proves 2. For 3: Let us ﬁx any a in C. We have to ﬁnd an open neighborhood U1 of a in C, e 1 of Euclidean space Rn such that U1 and U e 1 are and an open subset U homeomorphic. Since a is in C, and C is contained in M so, a is in M. Since a is in M, and M is a topological manifold, there exists an open neighborhood U of a in M, an open e of Euclidean space Rn , and a 1–1 mapping φ from U onto U e such that φ subset U 1 and its inverse u , both are continuous functions. Since a is in C, and C is open, C is an open neighborhood of a. Since C is an open neighborhood of a, and U is an open neighborhood of a, C \ U is an open neighborhood of a, and C \ U is cone is 1–1, so its restriction ujC U to C \ U tained in U. Here, function φ from U to U \ e is continuous, its restriction ujC U to is also 1–1. Since function φ from U to U \ onto e 11; C \ U is also continuous. Since the mapping u1 : U ! U is continuous, and e . Since uðC \ UÞ is open in U e; C \ U is an open subset of U, uðC \ UÞ is open in U n n e is open in R , uðC \ UÞ is open in R Since u : U ! U e is a homeomorand U e; phism, C \ U is an open subset of U, and uðC \ UÞ is an open subset of U ujC \ U : C \ U ! uðC \ UÞ is a homeomorphism. This proves 3.

Note 4.46 Let X be a second countable space. We shall try to show that every open cover of X has a countable subcover. Let G be an open cover of X. We have to ﬁnd a countable subcover of G. Since X is a second countable space, there exists a basis B of X such that B is countable. Put

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B1 fU : U 2 B; and there exists G 2 G such that U Gg: Since B1 B; and B is countable, B1 is countable. Now, we shall try to show that B1 is a basis of X. For this purpose, let us take any open neighborhood G of x in X. We have to ﬁnd a member U of B1 such that x 2 U G: Since x is in X, and G is a cover of X, there exists G1 in G such that x is in G1. Since G1 is in G; and G is an open cover of X, G1 is open. Since G1 is open, and x is in G1, G1 is an open neighborhood of x. Since G1 is an open neighborhood of x, and G is an open neighborhood of x, their intersection G1 \ G is an open neighborhood of x. Since G1 \ G is an open neighborhood of x, and B is a basis of X, there exists U in B such that x 2 U G1 \ G G: Since U G1 \ G G1 ; U G1 Since U is in B; G1 2 G; and U G1 , by the deﬁnition of B1 ; U is in B1 : Thus, we have shown that B1 is a basis of X. By the deﬁnition of B1 ; for each U in B1 ; the set fG : G 2 G; and U Gg is nonempty, and hence, we can choose a unique GU in G such that U GU : Put G1 fGU : U 2 B1 g: We shall try to show that G1 is a countable subcover of G: Since B1 is countable, by the deﬁnition of G1 ; G1 is countable. Clearly, G1 G: It remains to prove that G1 is a cover of X. For this purpose, let us take any x in X. We have to ﬁnd a member of G1 that contains x. Since X is an open neighborhood of x, and B1 is a basis of X, there exists U in B1 such that x 2 U X: Since U is in B1 , by the deﬁnition of G1 ; GU 2 G1 ; and hence U GU : Since x 2 U; and U GU ; x 2 GU ; where GU 2 G1 : Thus, G1 is a cover of X. Thus, we have shown that G1 is a countable subcover of G: Lemma 4.47 Let M be a topological manifold of dimension n. Then, there exists a countable collection fðU1 ; u1 Þ; ðU2 ; u2 Þ; ðU3 ; u3 Þ; . . .g of coordinate charts of M such that 1. fU1 ; U2 ; U3 ; . . .g is a basis of M, 2. each ui ðUi Þ is an open ball in Rn ; 3. each closure Ui of Ui is a compact subset of M. Proof Let C be the collection of all coordinate charts ðU; uÞ of M. Since M is a topological manifold, fU : ðU; uÞ 2 Cg is an open cover of M. Since M is a topological manifold, M is second countable. Since M is a second countable space, and fU : ðU; uÞ 2 Cg is an open cover of M, there exists a countable subcover fU1 ; U2 ; U3 ; . . .g of fU : ðU; uÞ 2 Cg: Here, U1 2 fU1 ; U2 ; U3 ; . . .g fU : ðU; uÞ 2 Cg; U1 2 fU : ðU; uÞ 2 Cg; and hence, there exists a function u1 e 1 of Rn such such that ðU1 ; u1 Þ 2 C: It follows that there exists an open subset U e 1 is a homeomorphism. that u1 : U1 7! U Let B be the collection of all open balls Br ðxÞ with center x and radius r such that r is a positive rational, and each coordinate of x is a rational number. We know that

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B is a basis of Rn : Since the set of all rational numbers is countable, the collection B is countable. Let B1 be the collection of all Br ðxÞ 2 B such that the closure ðBr ðxÞÞ e 1 : We shall try to show that B1 is a basis of U e 1: of Br ðxÞ is contained in U ~ ~ is For this purpose, let us take any open neighborhood V of y such that V n e 1 : Since V ~ is an open neighborhood of y, and B is a basis of R ; there contained in U ~ where ry is a positive exists an open ball Bry ðxy Þ in B such that y 2 Bry ðxy Þ V; rational, and each coordinate of xy is a rational number. Since y 2 Bry ðxy Þ; jy xy j\ry : There exists a positive rational number sy such that jy xy j\sy \ry : ~U e 1 : Since Hence, y 2 Bsy ðxy Þ ðBsy ðxy ÞÞ fz : jz xy j sy g Bry ðxy Þ V sy is a positive rational, and each coordinate of xy is a rational number, Bsy ðxy Þ is in e 1 ; Bsy ðxy Þ is in B1 . Also y 2 Bsy ðxy Þ B: Since Bsy ðxy Þ is in B; and ðBsy ðxy ÞÞ U ~ and Bsy ðxy Þ is an open neighborhood of y, B1 is a basis of U e 1 : Thus, we have V; e 1: shown that B1 is a basis of U Since B is countable, and B1 is contained in B; B1 is countable. Thus, B1 is a e 1 : Let us take any Br ðxÞ in B1 : Since Br ðxÞ is in B1 ; and B1 is a countable basis of U e e 1 : Since Br ðxÞ is an open basis of U 1 ; Br ðxÞ is an open subset of the open set U e 1 ; and u1 : U1 ! U e 1 is a homeomorphism, u1 e subset of U 1 ðBr ðxÞÞ is open in U 1 : 1 e e Since u1 ðBr ðxÞÞ is open in U 1 ; and u1 : U1 ! U 1 is a homeomorphism, the restriction u1 ju1 ðBr ðxÞÞ : u1 1 ðBr ðxÞÞ ! Br ðxÞ is a homomorphism. Thus, the 1

ordered pair ðu1 1 ðBr ðxÞÞ; u1 ju1 ðBr ðxÞÞ Þ is a coordinate chart of U 1 : 1

Since B1 is countable, the collection fðu1 1 ðBr ðxÞÞ; u1 ju1 ðBr ðxÞÞ Þ : Br ðxÞ is in B 1 g 1

of coordinate charts is countable. Thus, fðu1 ðBr ðxÞÞ Þ : Br ðxÞ is in B 1 g 1 ðBr ðxÞÞ; u1 ju1 1 is a countable collection of coordinate charts of M. Now, we shall try to prove that 1. fu1 1 ðBr ðxÞÞ : Br ðxÞ is in B1 g is a basis of U1 ; n 2. for each Br ðxÞ in B1 ; ðu1 ju1 ðBr ðxÞÞ Þðu1 1 ðBr ðxÞÞÞ is an open ball in R ; 1

1 3. for each Br ðxÞ in B1 ; the closure ðu1 1 ðBr ðxÞÞÞ of u1 ðBr ðxÞÞ is a compact subset of U1 :

e 1 is a homeomorphism, B1 is a basis of U e 1; For 1: Since u1 : U1 7! U ðB ðxÞÞ : B ðxÞ is in B g is a basis of U : fu1 r r 1 1 1 For 2: Since ðu1 ju1 ðBr ðxÞÞ Þðu1 1 ðBr ðxÞÞÞ ¼ Br ðxÞ; and Br ðxÞ is an open ball in 1

n Rn ; ðu1 ju1 ðBr ðxÞÞ Þðu1 1 ðBr ðxÞÞÞ is an open ball in R : 1 e 1 ; u1 e For 3: Since, for each Br ðxÞ in B1 ; ðBr ðxÞÞ U 1 : U 1 ! U1 is continu ous, and, by Heine–Borel theorem, ðBr ðxÞÞ is compact, u1 1 ððBr ðxÞÞ Þ 1 1 ð¼ ðu1 ðBr ðxÞÞÞ Þ is compact, and hence, ðu1 ðBr ðxÞÞÞ is a compact subset of U1 : Here, U2 2 fU1 ; U2 ; U3 ; . . .g fU : ðU; uÞ 2 Cg; so U2 2 fU : ðU; uÞ 2 Cg; and hence, there exists a function u2 such that ðU2 ; u2 Þ 2 C; e 2 of Rn ; such that u2 : U2 7! U e 2 is a and hence, there exists an open subset U

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homeomorphism. Let B2 be the collection of all Br ðxÞ 2 B such that the closure e2 ðBr ðxÞÞ of Br ðxÞ is contained in U As above, fðu1 ðBr ðxÞÞ Þ : Br ðxÞis in B 2 g is a countable collection 2 ðBr ðxÞÞ; u2 ju1 2 of coordinate charts of M such that (a) fu1 2 ðBr ðxÞÞ : Br ðxÞis in B2 g is a basis of U2 ; n (b) for each Br ðxÞin B2 ; ðu2 ju1 ðBr ðxÞÞ Þðu1 2 ðBr ðxÞÞÞ is an open ball in R ; 2

1 (c) for each Br ðxÞin B2 ; the closure ðu1 2 ðBr ðxÞÞÞ of u2 ðBr ðxÞÞ is a compact subset of U2 ; etc.

Since the countable union of countable sets is countable, the collection : Br ðxÞis in Bi g of coordinates charts is countable. It remains to prove that

1 Þ [1 i¼1 fðui ðBr ðxÞÞ; ui ju1 i ðBr ðxÞÞ

1 I. [1 i¼1 fui ðBr ðxÞÞ : Br ðxÞis in Bi g is a basis of M, II. for each i ¼ 1; 2; 3; . . .; and for each Br ðxÞ in Bi ; ðui ju1 ðBr ðxÞÞ Þðu1 i ðBr ðxÞÞÞ is i n an open ball in R ; III. for each i ¼ 1; 2; 3; . . .; and for each Br ðxÞ in Bi ; the closure ðu1 i ðBr ðxÞÞÞ of 1 ui ðBr ðxÞÞ is a compact subset of M, etc.

For I: Let G be any open neighborhood of a in M. Since fU1 ; U2 ; U3 ; . . .g is a cover of M, there exists a positive integer i such that a 2 Ui : Since Ui \ G is an open neighborhood of a, Ui \ G is contained in Ui ; and fu1 i ðBr ðxÞÞ : Br ðxÞ is in Bi g is a basis of Ui ; there exists Br ðxÞ in Bi such that a 2 u1 i ðBr ðxÞÞ Ui \ G G: 1 This shows that [1 fu ðB ðxÞÞ : B ðxÞ is in B g is a basis of M. r r i i¼1 i For II: This is clear. For III: This is clear. h Note 4.48 Let M be an m-dimensional topological manifold. Let p 2 M: Let U be an open neighborhood of p. We shall try to prove: There exists an open neighborhood V of p such that V U: Since M is an m-dimensional topological manifold so, by Lemma 4.47, there exists a countable collection fðU1 ; u1 Þ; ðU2 ; u2 Þ; ðU3 ; u3 Þ; . . .g of coordinate charts of M such that 1. fU1 ; U2 ; U3 ; . . .g is a basis of M, 2. each ui ðUi Þ is an open ball in Rm ; 3. each closure Ui of Ui is a compact subset of M. Since U is an open neighborhood of p, and fU1 ; U2 ; U3 ; . . .g is a basis of M, there exists a positive integer k such that p 2 Uk U: Here uk ðUk Þ is an open ball in Rm ; and uk ðpÞ 2 uk ðUk Þ; there exists a real number r [ 0 such that the closed ball Br ½uk ðpÞ uk ðUk Þ; and hence, u1 k ðBr ½uk ðpÞ Þ Uk : Since ðUk ; uk Þ is a

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coordinate chart of M, uk : Uk ! uk ðUk Þ is a homeomorphism. Since uk : Uk ! 1 uk ðUk Þ is a homeomorphism, u1 k : uk ðUk Þ ! Uk is continuous. Since uk : uk ðUk Þ ! Uk is continuous, and Br ½uk ðpÞ is compact, u1 k ðBr ½uk ðpÞ Þ is compact. Since M is topological manifold, the topology of M is Hausdorff. Since the topology 1 of M is Hausdorff, and u1 k ðBr ½uk ðpÞ Þ is compact, uk ðBr ½uk ðpÞ Þ is closed in M. 1 Clearly, uk ðBr ðuk ðpÞÞÞ is an open neighborhood of p, and u1 k ðBr ðuk ðpÞÞÞ 1 u1 ðB ½u ðpÞ Þ: Let us take u ðB ðu ðpÞÞÞ for V. Thus, V is an open neighborr k r k k k 1 1 hood of p, and V uk ðBr ½uk ðpÞ Þ: Since V uk ðBr ½uk ðpÞ Þ; and 1 u1 k ðBr ½uk ðpÞ Þ is closed in M, V uk ðBr ½uk ðpÞ Þ Uk U: Thus, V U: Lemma 4.49 Let M be an n-dimensional smooth manifold. Then, there exists a countable collection fðU1 ; u1 Þ; ðU2 ; u2 Þ; ðU3 ; u3 Þ; . . .g of admissible coordinate charts of M such that 1. fU1 ; U2 ; U3 ; . . .g is a basis of M. 2. each ui ðUi Þ is an open ball in Rn ; 3. each closure Ui of Ui is a compact subset of M. Proof Let C be the collection of all admissible coordinate charts ðU; uÞ of M. Since M is a smooth manifold, fU : ðU; uÞ 2 Cg is an open cover of M. Since M is a smooth manifold, M is second countable. Since M is a second countable space, and fU : ðU; uÞ 2 Cg is an open cover of M, there exists a countable subcover fU1 ; U2 ; U3 ; . . .g of fU : ðU; uÞ 2 Cg: Here, U1 2 fU1 ; U2 ; U3 ; . . .g fU : ðU; uÞ 2 Cg; U1 2 fU : ðU; uÞ 2 Cg; and e 1 of Rn such that hence, there exists a function u1 ; and an open subset U e 1 , and ðU1 ; u1 Þ 2 C: Let B be the collection of all open balls Br ðxÞ u1 : U1 ! U with center x and radius r such that r is a positive rational, and each coordinate of x is a rational number. We know that B is a basis of Rn : Since the set of all rational numbers is countable, the collection B is countable. Let B1 be the collection of all e 1: Br ðxÞ 2 B such that the closure ðBr ðxÞÞ of Br ðxÞ is contained in U e We shall try to show that B1 is a basis of U 1 : For this purpose, let us take any ~ of y such that V ~ is contained in U e 1 : Since V ~ is an open open neighborhood V neighborhood of y, and B is a basis of Rn ; there exists an open ball Bry ðxy Þ in B such ~ where ry is a positive rational, and each coordinate of xy is a that y 2 Bry ðxy Þ V; rational number. Since y 2 Bry ðxy Þ; jy xy j\ry : There exists a positive rational number sy such that jy xy j\sy \ry : Hence, y 2 Bsy ðxy Þ ðBsy ðxy ÞÞ ~U e 1 : Since sy is a positive rational, and each fy : jy xy j sy g Bry ðxy Þ V coordinate of xy is a rational number, Bsy ðxy Þ is in B: Since Bsy ðxy Þ is in B; and e 1 ; Bsy ðxy Þ is in B1 . Also y 2 Bsy ðxy Þ V; ~ and Bsy ðxy Þ is an open ðBsy ðxy ÞÞ U e neighborhood of y, B1 is a basis of U 1 : e 1 : Since B is countable, and B1 is Thus, we have shown that B1 is a basis of U e 1 : Let us take any contained in B; B1 is countable. Thus, B1 is a countable basis of U

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e 1 ; Br ðxÞ is an open subset of Br ðxÞ in B1 : Since Br ðxÞ is in B1 ; and B1 is a basis of U e e 1 ; and u1 : U1 ! U e 1 is a the open set U 1 : Since Br ðxÞ is an open subset of U 1 1 homeomorphism, u1 ðBr ðxÞÞ is open in U1 : Since u1 ðBr ðxÞÞ is open in U1 ; and e 1 is a homeomorphism, the restriction u1 ju1 ðB ðxÞÞ : u1 u1 : U1 ! U 1 ðBr ðxÞÞ ! r 1

Br ðxÞ is a homomorphism. Thus, the ordered pair ðu1 ðBr ðxÞÞ Þ is a 1 ðBr ðxÞÞ; u1 ju1 1 coordinate chart of U1 : Now, we want to show that ðu1 ðBr ðxÞÞ Þ 2 C: For this purpose, let 1 ðBr ðxÞÞ; u1 ju1 1 us take any ðW; wÞ 2 C: Now, we must prove that ðBr ð xÞÞ \ W u1 ju1 ðBr ðxÞÞ w1 : w u1 1 1 u1 ! u1 ju1 ðBr ðxÞÞ 1 ð B r ð xÞ Þ \ W ; 1

and 1 u1 : u1 ju1 ðBr ðxÞÞ w u1 ju1 ðBr ðxÞÞ 1 ðBr ð xÞÞ \ W 1 1 ! w u1 1 ðBr ð xÞÞ \ W are C 1 : Let us take any ðy1 ; . . .; yn Þ 2 wððu1 1 ðBr ðxÞÞÞ \ WÞ: So, there exists p in ðu1 ðB ðxÞÞÞ W such that wðpÞ ¼ ðy ; . . .; yn Þ; and hence w1 ðy1 ; . . .; yn Þ ¼ p; \ r 1 1 and p 2 u1 1 ðBr ðxÞÞ: Now,

u1 ju1 ðBr ðxÞÞ w1 ðy1 ; . . .; yn Þ 1 ¼ u1 ju1 ðBr ðxÞÞ w1 ðy1 ; . . .; yn Þ ¼ u1 ju1 ðBr ðxÞÞ ð pÞ 1 1 1 1 ¼ u1 ð pÞ ¼ u1 w ðy1 ; . . .; yn Þ ¼ u1 w ðy1 ; . . .; yn Þ 1 ðy1 ; ; yn Þ; ¼ u1 w wððu1 ðBr ðxÞÞÞ \ W Þ 1

so ðu1 ju1 ðBr ðxÞÞ Þ w1 ¼ ðu1 w1 Þjwððu1 ðBr ðxÞÞÞ \ WÞ : Since ðU1 ; u1 Þ 2 C; ðW; wÞ 2 1

1

C; and C is the collection of all admissible coordinate charts of M, ðu1 w1 Þ : wðU1 \ WÞ ! u1 ðU1 \ WÞ is C 1 : Since u1 1 ðBr ðxÞÞ is an open subset of U1 ; 1 wððu1 ðB ðxÞÞÞ \ WÞ is an open subset of wðU r 1 \ WÞ: Since wððu1 ðBr ðxÞÞÞ \ WÞ 1 is an open subset of wðU1 \ WÞ; and ðu1 w1 Þ : wðU1 \ WÞ ! u1 ðU1 \ WÞ is C1 ; ðu1 w1 Þjwððu1 ðBr ðxÞÞÞ \ WÞ ð¼ ðu1 ju1 ðBr ðxÞÞ Þ w1 Þ is C1 ; and hence, ðu1 ju1 ðBr ðxÞÞ Þ 1

1

1

w1 is C 1 : Thus, we have shown that ððu1 ju1 ðBr ðxÞÞ Þ w1 Þ : wððu1 1 ðBr ðxÞÞÞ 1

1 1 \WÞ ! ðu1 ju1 ðBr ðxÞÞ Þððu1 1 ðBr ðxÞÞÞ \ WÞis C : Similarly, ðw ðu1 ju1 ðBr ðxÞÞ Þ Þ : 1

1

1 1 1 ðu1 ju1 ðBr ðxÞÞ Þððu1 1 ðBr ðxÞÞÞ \ WÞ ! wððu1 ðBr ðxÞÞÞ \ WÞ is C . Thus, ðu1 ðBr 1 ðxÞÞ; u1 ju1 ðBr ðxÞÞ Þ 2 C: 1

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4 Topological Properties of Smooth Manifolds

Since B1 is countable, the collection fðu1 1 ðBr ðxÞÞ; u1 ju1 ðBr ðxÞÞ Þ : Br ðxÞ is in B 1 g 1

of coordinate charts is countable. Thus, fðu1 1 ðBr ðxÞÞ; u1 ju1 ðBr ðxÞÞ Þ : Br ðxÞ is in B 1 g 1 is a countable collection of admissible coordinate charts of M. Now, we shall try to prove that 1. fu1 1 ðBr ðxÞÞ : Br ðxÞ is in B1 g is a basis of U1 ; n 2. for each Br ðxÞ in B1 ; ðu1 ju1 ðBr ðxÞÞ Þðu1 1 ðBr ðxÞÞÞ is an open ball in R ; 1

1 3. for each Br ðxÞ in B1 ; the closure ðu1 1 ðBr ðxÞÞÞ of u1 ðBr ðxÞÞ is a compact subset of U1 :

e 1 is a homeomorphism, B1 is a basis of U e 1; For 1: Since u1 : U1 ! U ðB ðxÞÞ : B ðxÞ is in B g is a basis of U : fu1 r r 1 1 1 For 2: Since ðu1 ju1 ðBr ðxÞÞ Þðu1 1 ðBr ðxÞÞÞ ¼ Br ðxÞ; and Br ðxÞ is an open ball in 1

n Rn ; ðu1 ju1 ðBr ðxÞÞ Þðu1 1 ðBr ðxÞÞÞis an open ball in R : 1 e 1 ; u1 e For 3: Since, for each Br ðxÞ in B1 ; ðBr ðxÞÞ U 1 : U 1 ! U1 is continuous, 1 and, by Heine–Borel theorem, ðBr ðxÞÞ is compact, u1 1 ððBr ðxÞÞ Þð¼ ðu1 1 ðBr ðxÞÞÞ Þ is compact, and hence, ðu1 ðBr ðxÞÞÞ is a compact subset of U1 :

Here, U2 2 fU1 ; U2 ; U3 ; . . .g fU : ðU; uÞ 2 Cg; U2 2 fU : ðU; uÞ 2 Cg; and hence, there exists a function u2 such that ðU2 ; u2 Þ 2 C; and hence, there exists an e 2 of Rn such that u2 : U2 ! U e 2 is a homeomorphism. Let B2 be the open subset U e 2: collection of all Br ðxÞ 2 B such that the closure ðBr ðxÞÞ of Br ðxÞ is contained in U 1 As above, fðu2 ðBr ðxÞÞ; u2 ju1 ðBr ðxÞÞ Þ : Br ðxÞ is in B2 g is a countable collection 2 of admissible coordinate charts of M satisfying (a) fu1 2 ðBr ðxÞÞ : Br ðxÞ is in B2 g is a basis of U2 ; n (b) for each Br ðxÞin B2 ; ðu2 ju1 ðBr ðxÞÞ Þðu1 2 ðBr ðxÞÞÞ is an open ball in R ; 2

1 (c) for each Br ðxÞin B2 ; the closure ðu1 2 ðBr ðxÞÞÞ of u2 ðBr ðxÞÞ is a compact subset of U2 ; etc.

Since the countable union of countable sets is countable, the collection 1 Þ : Br ðxÞ is in Bi g of admissible coordinates charts is [1 i¼1 fðui ðBr ðxÞÞ; ui ju1 i ðBr ðxÞÞ countable. It remains to prove that 1 I. [1 i¼1 fui ðBr ðxÞÞ : Br ðxÞ is in B i g is a basis of M, II. for each i ¼ 1; 2; 3; . . .; and for each Br ðxÞ in Bi ; ðui ju1 ðBr ðxÞÞ Þðu1 i ðBr ðxÞÞÞ is i an open ball in Rn ; III. for each i ¼ 1; 2; 3; . . .; and for each Br ðxÞ in Bi ; the closure ðu1 i ðBr ðxÞÞÞ of u1 i ðBr ðxÞÞ is a compact subset of M, etc.

4.4 Smooth Manifold as Paracompact Space

283

For I: Let G be any open neighborhood of a in M. Since fU1 ; U2 ; U3 ; . . .g is a cover of M, there exists a positive integer i such that a 2 Ui : Since Ui \ G is an open neighborhood of a, Ui \ G is contained in Ui ; and fu1 i ðBr ðxÞÞ : Br ðxÞ is in Bi g is a basis of Ui ; there exists Br ðxÞ is Bi such that a 2 u1 i ðBr ðxÞÞ Ui \ G G: 1 fu ðB ðxÞÞ : B ðxÞ is in B g is a basis of M. This shows that [1 r r i i¼1 i For II: This is clear. For III: This is clear. h Deﬁnition Let X be a topological space. Let U be a collection of subsets of X. If for every x in X, there exists an open neighborhood G of x such that fU : U 2 U and U \ G is nonemptyg is a ﬁnite set, then we say that U is locally ﬁnite. In other words, U is locally ﬁnite means for every x in X, there exists an open neighborhood G of x such that all members of U except ﬁnite many are “outside” G. In short, if for every x in X, there exists an open neighborhood G of x such that G intersects only ﬁnite many members of U; then we say that U is locally ﬁnite. Clearly, every subcollection of a locally ﬁnite collection is locally ﬁnite. Note 4.50 Let X be a topological space. Let U be an open cover of X. Let each member of U intersects only ﬁnite many members of U: We shall try to show that U is locally ﬁnite. For this purpose, let us take any x in X. Since x is in X, and U is a cover of X, there exists G in U such that x is in G. Since G is in U; and U is an open cover of X, G is open. Since G is open, and x is in G, G is an open neighborhood of x. Since G is in U; and each member of U intersects only ﬁnite many members of U; G intersects only ﬁnite many members of U: This proves that U is locally ﬁnite. Lemma 4.51 Let M be a topological manifold. There exists a countable collection fG1 ; G2 ; G3 ; . . .g of open sets such that 1. fG1 ; G2 ; G3 ; . . .g is a cover of M, 2. fG1 ; G2 ; G3 ; . . .g is locally ﬁnite, 3. for each n ¼ 1; 2; 3; . . .; the closure G n of Gn is compact. Proof By Lemma 4.47, there exists a countable collection fðU1 ; u1 Þ; ðU2 ; u2 Þ; ðU3 ; u3 Þ; . . .g of coordinate charts of M such that (i) fU1 ; U2 ; U3 ; . . .g is a basis of M, (ii) each closure Un of Un is a compact subset of M. Put B1 U1 : By (i), clearly fU1 ; U2 ; U3 ; . . .g is an open cover of M. Since B1 ¼ U1 ; B 1 ¼ ¼ U ; and U is compact, B is compact. Since fU ; U ; U ; . . .g is U1 : Since B 1 2 3 1 1 1 1 an open cover of M, and B1 is compact, there exists a positive integer m1 [ 1 such that B 1 U1 [ U2 [ [ Um1 :

284

4 Topological Properties of Smooth Manifolds

Put B2 U1 [ U2 [ [ Um1 : Since B2 ¼ U1 [ U2 [ [ Um1 ; B 2 ¼ ðU1 [ U2 [ [ Um1 Þ ¼ ðU1 Þ [ðU2 Þ [ [ðUm1 Þ: Since each of U1 ; U2 ; . . .; Um1 is compact, the ﬁnite union ðU1 Þ [ðU2 Þ [ [ðUm1 Þð¼ B 2 Þ is compact, and hence, is compact. Since fU ; U ; U ; . . .g is an open cover of M, and B B 1 2 3 2 2 is compact, there exists a positive integer m2 [ m1 [ 1 such that B2 U1 [ U2 [ [ Um1 [ [Um2 : Put B3 U1 [ U2 [ [ Um2 : Since B3 ¼ U1 [ U2 [ [ Um2 ; B 3 ¼ ðU1 [ U2 [ [ Um2 Þ ¼ ðU1 Þ [ðU2 Þ [ [ðUm2 Þ: Since each of U1 ; U2 ; . . .; Um2 is compact, the ﬁnite union ðU1 Þ [ðU2 Þ [ [ðUm2 Þð¼ B 3 Þ is compact, and hence, B3 is compact. Since fU1 ; U2 ; U3 ; . . .g is an open cover of M, and B 3 is compact, there exists a positive integer m3 [ m2 [ m1 [ 1 such that B 3 U1 [ U2 [ [ Um1 [ [ Um2 [ [ Um3 : Put B4 U1 [ U2 [ [ Um3 ; etc. Clearly (a) (b) (c) (d) (e)

for every positive integer n, Un Bn ; fB1 ; B2 ; B3 ; . . .g is an open cover of M, B1 B2 B3 ; each closure B n of Bn is a compact subset of M, for every positive integer n, Bn B n Bnþ1 :

Put G1 B1 ; G2 B2 ; G3 B3 ðB 1 Þ; G4 B4 ðB2 Þ; G5 B5 ðB3 Þ; etc. Clearly fG1 ; G2 ; G3 ; . . .g is a countable collection of open sets (see Fig. 4.1). It remains to prove that

1. fG1 ; G2 ; G3 ; . . .g is a cover of M, 2. fG1 ; G2 ; G3 ; . . .g is locally ﬁnite, 3. for each n ¼ 1; 2; 3; . . .; the closure G n of Gn is compact.

Fig. 4.1 A countable collection of open sets

4.4 Smooth Manifold as Paracompact Space

285

For 1: This is clear. For 2: Here, fG1 ; G2 ; G3 ; . . .g is an open cover of M, and each member of fG1 ; G2 ; G3 ; . . .g intersects only ﬁnite many members of fG1 ; G2 ; G3 ; . . .g; fG1 ; G2 ; G3 ; . . .g is locally ﬁnite. For 3: Since G1 ¼ B1 ; G 1 ¼ B1 : Since B1 is compact, G1 is compact. Since G2 ¼ B2 ; G2 ¼ B2 : Since B2 is compact, G2 is compact. Since G3 B 3; B : Since the closed set G is contained in the compact set B ; G G 3 3 3 3 3 is compact. Since G4 B ; G B : Since the closed set G is contained in the 4 4 4 4 ; G is compact, etc. This proves 3. h compact set B 4 4 Deﬁnition Let X be a topological space. Let A; B be open covers of X. By A is a reﬁnement of B we mean: For every A in A; there exists B in B such that A B: Lemma 4.52 Let M be an m-dimensional smooth manifold. Let X be any open cover of M. There exists a countable collection fðU1 ; u1 Þ; ðU2 ; u2 Þ; ðU3 ; u3 Þ; . . .g of admissible coordinate charts of M such that 1. 2. 3. 4. 5.

fU1 ; U2 ; U3 ; . . .g covers M, fU1 ; U2 ; U3 ; . . .g is locally ﬁnite, fU1 ; U2 ; U3 ; . . .g is a reﬁnement of X ; each un ðUn Þ is equal to the open ball B3 ð0Þ; 1 1 fu1 1 ðB1 ð0ÞÞ; u2 ðB1 ð0ÞÞ; u3 ðB1 ð0ÞÞ; . . .g covers M.

Proof Since M is a smooth manifold, M is a topological manifold, and hence, by Lemma 4.51, there exists a countable collection fG1 ; G2 ; G3 ; . . .g of open sets such that a. fG1 ; G2 ; G3 ; . . .g is a cover of M, b. fG1 ; G2 ; G3 ; . . .g is locally ﬁnite, c. for each n ¼ 1; 2; 3; . . .; the closure G n of Gn is compact. Let us ﬁx any p in M. Since fG1 ; G2 ; G3 ; . . .g is locally ﬁnite, and p is in M, there exists an open neighborhood Wp of p such that

Gk : Gk \ Wp is nonempty

is a ﬁnite set. Since p 2 Wp ; fGk : p 2 Gk g ¼ fGk : p 2 Gk \ Wp g fGk : Gk \ Wp is nonemptyg: Since fGk : p 2 Gk g fGk : Gk \ Wp is nonemptyg; and fGk : Gk \ Wp is nonemptyg is ﬁnite, fGk : p 2 Gk g is ﬁnite. Since fG1 ; G2 ; G3 ; . . .g is a cover of M, and p 2 M; fGk : p 2 Gk g is nonempty. Since fGk : p 2 Gk g is a nonempty ﬁnite collection of open neighborhoods of p, \p2Gk Gk is an open neighborhood of p.

286

4 Topological Properties of Smooth Manifolds

Since X is an open cover of M, and p is in M, there exists Xp in X such that p 2 Xp : Since X is an open cover of M, and Xp is in X ; Xp is open. Since Xp is open, and p 2 Xp ; Xp is an open neighborhood of p. Since p is in M, and M is a smooth manifold, there exists an admissible coordinate chart ðVp ; up Þ of M such that Vp is an open neighborhood of p. Since \p2Gk Gk is an open neighborhood of p, Wp is an open neighborhood of p, Xp is an open neighborhood of p, and Vp is an open neighborhood of p, their intersection ð\p2Gk Gk Þ \ Wp \ Xp \ Vp is an open neighborhood of p. Since ðVp ; up Þ is an admissible coordinate chart of M, ð\p2Gk Gk Þ \ Wp \ Xp \ Vp Vp ; and ð\p2Gk Gk Þ \ Wp \ Xp \ Vp is an open neighborhood of p, the pair ðð\p2Gk Gk Þ \ Wp \ Xp \ Vp ; up jð\p2G Gk Þ \ Wp \ Xp \ Vp Þ is an admissible coordinate k chart of M: Since ðð\p2Gk Gk Þ \ Wp \ Xp \ Vp ; up jð\p2G Gk Þ \ Wp \ Xp \ Vp Þ is an k

admissible coordinate chart of M, and p 2 ð\p2Gk Gk Þ \ Wp \ Xp \ Vp ; there exists an admissible coordinate chart ðUp ; wp Þ such that Up ð\p2Gk Gk Þ \ Wp \ Xp \ Vp ; Up is an open neighborhood of p, wp ðUp Þ ¼ B3 ð0Þ; and wp ðpÞ ¼ 0: Thus, wp : Up ! B3 ð0Þ is a homeomorphism. Since Up ð\p2Gk Gk Þ \ Wp \ Xp \ Vp Wp ; and fGk : Gk \ Wp is nonemptyg is a ﬁnite set, fGk : Gk \ Up is nonemptyg is a ﬁnite set. It is clear that if q 2 Gk0 then Uq Gk0 : (Reason: Let q 2 Gk0 : Since q 2 Gk0 ; Uq ð\q2Gk Gk Þ \ Wq \ Xq \ Vq ð\q2Gk Gk Þ Gk0 :) Since ðUp ; wp Þ is a coordinate chart, and wp ðUp Þ ¼ B3 ð0Þ; wp : Up ! B3 ð0Þ is continuous. Since wp : Up ! B3 ð0Þ is continuous, and B1 ð0Þ is an open subset of

B3 ð0Þ; ðwp Þ1 ðB1 ð0ÞÞ is open in Up : Since ðwp Þ1 ðB1 ð0ÞÞ is open in Up ; and Up is

open in M; ðwp Þ1 ðB1 ð0ÞÞ is open in M. Since wp ðpÞ ¼ 0 2 B1 ð0Þ; p 2

ðwp Þ1 ðB1 ð0ÞÞ: Thus, ðwp Þ1 ðB1 ð0ÞÞ is an open neighborhood of p: Since wp :

Up ! B3 ð0Þ; and B1 ð0Þ B3 ð0Þ; ðwp Þ1 ðB1 ð0ÞÞ Up :

Since for every p in M; ðwp Þ1 ðB1 ð0ÞÞ is an open neighborhood of p; the

collection 1

fðwp Þ1 ðB1 ð0ÞÞ : p 2 G 1g

fðwp Þ ðB1 ð0ÞÞ : p 2

G 1g

is

an

is an open cover of

open

G 1;

and

cover G 1

of

G 1:

Since

is compact, there exists

1 1 ﬁnite many p11 ; p12 ; . . .; p1k1 2 G 1 such that G1 ðwp11 Þ ðB1 ð0ÞÞ [ðwp12 Þ

ðB1 ð0ÞÞ [ [ðwp1k Þ1 ðB1 ð0ÞÞ Up11 [ Up12 [ [ Up1k1 ; where ðUp11 ; wp11 Þ; 1 ðUp12 ; wp12 Þ; . . .; ðUp1k1 ; wp1k Þ are admissible coordinate charts of M: 1

4.4 Smooth Manifold as Paracompact Space

287

Similarly, there exists ﬁnite many p21 ; p22 ; . . .; p2k2 2 G 2 such that G2 Up21 [ Up22 [ [ Up2k2 ; etc. Clearly, ðUp11 ; wp11 Þ; ðUp12 ; wp12 Þ; . . .; ðUp1k1 ; wp1k1 Þ, ðUp21 ; wp21 Þ; ðUp22 ; wp22 Þ; . . .; ðUp2k2 ; wp2k Þ; ðUp31 ; wp31 Þ; ðUp32 ; wp32 Þ; . . .; ðUp3k3 ; wp3k Þ; . . . 2 3 constitute a countable collection of admissible coordinate charts of M: 1 1 1 For 1: Since G1 G 1 ðwp11 Þ ðB1 ð0ÞÞ [ðwp12 Þ ðB1 ð0ÞÞ [ [ðwp1k Þ 1

1 1 ðB1 ð0ÞÞ Up11 [ Up12 [ [ Up1k1 ; G2 G 2 ðwp21 Þ ðB1 ð0ÞÞ [ðwp22 Þ ðB1 ð0ÞÞ

1 [ [ðwp2k2 Þ ðB1 ð0ÞÞ Up21 [ Up22 [ [ Up2k2 ; . . .; and fG1 ; G2 ; G3 ; g is a cover of M; fUp11 ; Up12 ; . . .; Up1k1 ; Up21 ; Up22 ; . . .; Up2k2 ; Up31 ; Up32 ; . . .; Up3k3 ; . . .g is an open cover of M: For 2: We have to prove that fUp11 ; Up12 ; . . .; Up1k1 ; Up21 ; Up22 ; . . .; Up2k2 ; Up31 ; Up32 ; . . .; Up3k3 ; . . .g is locally ﬁnite.

Since fG1 ; G2 ; G3 ; . . .g is a cover of M; and p11 2 M; there exists a positive integer n11 such that p11 2 Gn11 : Since p11 2 Gn11 ; Up11 Gn11 : Similarly, there exists a positive integer n12 such that Up12 Gn12 ; etc. Since fUp11 ; Up12 ; . . .; Up1k1 ; Up21 ; Up22 ; . . .; Up2k2 ; Up31 ; Up32 ; . . .; Up3k3 ; . . .g is a cover of M; and each Upij is contained in Gnij ; fGn11 ; Gn12 ; . . .; Gn1k1 ; Gn21 ; Gn22 ; . . .; Gn2k2 ; Gn31 ; Gn32 ; . . .; Gn3k3 ; . . .g is a cover of M: Since fG1 ; G2 ; G3 ; . . .g is locally ﬁnite, and fGn11 ; Gn12 ; . . .; Gn1k1 ; Gn21 ; Gn22 ; . . .; Gn2k2 ; Gn31 ; Gn32 ; . . .; Gn3k3 ; . . .g fG1 ; G2 ; G3 ; . . .g; fGn11 ; Gn12 ; . . .; Gn1k1 ; Gn21 ; Gn22 ; . . .; Gn2k2 ; Gn31 ; Gn32 ; . . .; Gn3k3 ; . . .g is locally ﬁnite. Since fGn11 ; Gn12 ; . . .; Gn1k1 ; Gn21 ; Gn22 ; . . .; Gn2k2 ; Gn31 ; Gn32 ; . . .; Gn3k3 ; . . .g is locally ﬁnite, and each Upij is contained in Gnij ; fUp11 ; Up12 ; . . .; Up1k1 ; Up21 ; Up22 ; . . .; Up2k2 ; Up31 ; Up32 ; . . .; Up3k3 ; . . .g is locally ﬁnite. For 3: We have to show that the open cover fUp11 ; Up12 ; . . .; Up1k1 ; Up21 ; Up22 ; . . .; Up2k2 ; Up31 ; Up32 ; . . .; Up3k3 ; . . .g is a reﬁnement of the open cover X : Here Up11 ð\p11 2Gk Gk Þ \ Wp11 \ Xp11 \ Vp11 Xp11 2 X : Thus, Up11 Xp11 2 X : Similarly, Up12 Xp12 2 X ; etc. For 4: This is clear. For 5: We have to show that n

1 1 1 1 1 w1 p11 ðB1 ð0ÞÞ; wp12 ðB1 ð0ÞÞ; . . .; wp1k ðB1 ð0ÞÞ; wp21 ðB1 ð0ÞÞ; wp22 ðB1 ð0ÞÞ; . . .; wp2k ðB1 ð0ÞÞ; . . . 1

o

2

covers M. Since

1 1 1 G1 G 1 ðwp11 Þ ðB1 ð0ÞÞ [ðwp12 Þ ðB1 ð0ÞÞ [ [ðwp1k Þ ðB1 ð0ÞÞ; 1

1 1 1 G2 G 2 ðwp21 Þ ðB1 ð0ÞÞ [ðwp22 Þ ðB1 ð0ÞÞ [ [ðwp2k Þ ðB1 ð0ÞÞ; . . .; 2

and

288

4 Topological Properties of Smooth Manifolds

1 1 fG1 ; G2 ; G3 ; g is a cover of M; fw1 p11 ðB1 ð0ÞÞ; wp12 ðB1 ð0ÞÞ; ; wp1k ðB1 1 1 ð0ÞÞ; w1 p21 ðB1 ð0ÞÞ; wp22 ðB1 ð0ÞÞ; ; wp2k ðB1 ð0ÞÞ; g is a cover of M: 2

1

h

Deﬁnition Let X be a topological space. If for every open cover A of X; there exists an open cover C of M such that C is locally ﬁnite, and C is a reﬁnement of A; then we say that X is a paracompact space. Note 4.53 By the Lemma 4.52, we ﬁnd that every smooth manifold is a paracompact space.

4.5 Partitions of Unity Theorem Note 4.54 (Pasting Lemma) Let X and Y be topological spaces. Let A and B be subsets of X such that the union of A and B is X: Let f : A ! Y be a continuous function, where the topology of A is the induced topology of X: Let g : B ! Y be a continuous function, where the topology of B is the induced topology of X: Let f ðxÞ ¼ gðxÞ for every x in A \ B. Let F : X ! Y be a function deﬁned as follows: F ð xÞ

f ð xÞ if x is in A gð xÞ if x is in B:

Then, if A and B are open subsets of X, then F is continuous. Proof Since f : A ! Y is a function, g : B ! Y is a function, for every x in A \ B; f ðxÞ ¼ gðxÞ; and F ð xÞ

f ð xÞ if x is in A gð xÞ if x is in B;

F is well defined on A [ Bð¼ X Þ: Let us take any x in X. We have to prove that F is continuous at x. Since x is in X, and X ¼ A [ B; x 2 A or x 2 B: Case I: When x 2 A. Since f : A ! Y is a continuous function, and x 2 A; f is continuous at x: Since f is continuous at x; and FjA ¼ f ; FjA is continuous at x: Since x 2 A; and A is open, x is an interior point of A: Since x is an interior point of A; and FjA is continuous at x; F is continuous at x: Case II: When x 2 B. As in case I, F is continuous at x: h Lemma 4.55 Let X be a topological space. Let U be a collection of subsets of X: If U is locally ﬁnite, then ð[fU : U 2 UgÞ ¼ [fU : U 2 Ug: Proof Let U be locally ﬁnite. For any U 2 U; U ð[fU : U 2 UgÞ; so U ð[fU : U 2 UgÞ : This shows that [fU : U 2 Ug ð[fU : U 2 UgÞ : It remains to show that ð[fU : U 2 UgÞ [fU : U 2 Ug: For this purpose, let us

4.5 Partitions of Unity Theorem

289

take any a in ð[fU : U 2 UgÞ : We claim that a is in [fU : U 2 Ug: If not, otherwise, let for every U 2 U; a 62 U : We have to arrive at a contradiction. h Now, since U is locally ﬁnite, there exists an open neighborhood Ga of a such that all but ﬁnite many U 2 U are outside Ga : For every U 2 U, we have a 62 U ; there exists an open neighborhood HU of a such that HU contains no point of U ; and hence, HU contains no point of U: Since all but ﬁnite many U 2 U are outside Ga ; the ﬁnite intersection ð\Ga \ U6¼; HU Þ \ Ga is an open neighborhood of a: Since for every U 2 U; HU contains no point of U; and all but ﬁnite many U 2 U are outside Ga ; ð\Ga \ U6¼; HU Þ \ Ga contains no point of [fU : U 2 Ug: It follows that a is not h in ð[fU : U 2 UgÞ ; a contradiction. Deﬁnition Let X be a topological space. Let f : X ! Rk be any function. The closure fx : f ðxÞ 6¼ 0g of fx : f ðxÞ ¼ 6 0g is denoted by supp f ; and is called the support of f : Lemma 4.56 Let M be a 3-dimensional smooth manifold. Let fXi gi2I be an open cover of M: Then, there exists a family fwi gi2I of functions wi : M ! ½0; 1 such that 1. 2. 3. 4.

for each i 2 I; wi is smooth, for each i 2 I; supp wi is contained in Xi ; fsupp wi gi2I is locally ﬁnite, P i2I wi ¼ 1:

Proof By the Lemma 4.52, there exists a countable collection fðU1 ; u1 Þ; ðU2 ; u2 Þ; ðU3 ; u3 Þ; . . .g of admissible coordinate charts of M such that i. ii. iii. iv. v.

fU1 ; U2 ; U3 ; . . .g covers M; fU1 ; U2 ; U3 ; . . .g is locally ﬁnite, fU1 ; U2 ; U3 ; . . .g is a reﬁnement of fXi gi2I ; each un ðUn Þ is equal to the open ball B3 ð0Þ; 1 1 fu1 1 ðB1 ð0ÞÞ; u2 ðB1 ð0ÞÞ; u3 ðB1 ð0ÞÞ; . . .g covers M.

By the Note 3.52, there exists a function H : R3 ! ½0; 1 such that I. H is smooth, II. HðxÞ ¼ 1; if x is in the closed ball B1 ½0 ; III. supp H ¼ B2 ½0 : Let us ﬁx any positive integer n. Observe that 1 Un ¼ u1 n ðB3 ð0ÞÞ un ðB2 ½0 Þ: 1 If x 2 Un u1 n ðB2 ½0 Þ; then x 62 un ðB2 ½0 Þ; and hence, un ðxÞ 62 B2 ½0 ¼ 1 1 supp H ¼ ðH ðR f0gÞÞ H ðR f0gÞ: It follows that if x 2 Un

290

4 Topological Properties of Smooth Manifolds

1 u1 n ðB2 ½0 Þ; then un ðxÞ 62 H ðR f0gÞ; and hence, Hðun ðxÞÞ ¼ 0: This shows that the function fn : M ! ½0; 1 deﬁned as follows: For every x in M;

fn ð xÞ

H ðun ð xÞÞ 0

if x 2 Un ; if x 2 6 u1 n ðB2 ½0 Þ;

1 is well deﬁned. Clearly, fn1 ðR f0gÞ u1 n ðB2 ½0 Þ; and hence, supp fn ¼ ðfn ðR 1 1 f0gÞÞ ðu1 n ðB2 ½0 ÞÞ ¼ un ðB2 ½0 Þ un ðB3 ð0ÞÞ ¼ Un : Thus, supp fn Un : Since ðUn ; un Þ is an admissible coordinate chart of 3-dimensional smooth manifold M, un ðUn Þ is open in R3 ; and un : Un ! un ðUn Þð¼ B3 ð0ÞÞ is a homeomorphism. Since un : Un ! un ðUn Þ; and H : R3 ! ½0; 1 ; H un : Un ! ½0; 1 : Clearly, H un : Un ! ½0; 1 is smooth. (Reason: Take any p in Un : We have to prove that H un is smooth at p: For this purpose, let us take any admissible coordinate chart ðV; wÞ of M such that p 2 V: We have to prove that ðH un Þ w1 is smooth. Since ðUn ; un Þ; ðV; wÞ are admissible coordinate charts of M; and Un \ V is nonempty, un w1 is smooth. Since un w1 is smooth, and H is smooth, their composite H ðun w1 Þð¼ ðH un Þ w1 Þ is smooth, and hence, ðH un Þ w1 is smooth.) Since

f n ð xÞ ¼

ðH un Þð xÞ if x 2 Un ; 0 if x 2 M ðun 1 ðB2 ½0 ÞÞ

is a well-deﬁned function, Un and M ðu1 n ðB2 ½0 ÞÞ are open sets, the restriction ð¼ 0Þ is smooth, fn is fn jUn ð¼ H un Þ is smooth, and the restriction fn jMðu1 n ðB2 ½0 ÞÞ smooth. Let us ﬁx any a in M: Since fU1 ; U2 ; U3 ; . . .g is a locally ﬁnite cover of M, and a is in M; there exists an open neighborhood Ga of a; and a positive integer N such that 1 k N implies Uk is outside Ga : For k N; Uk ¼ u1 k ðB3 ð0ÞÞ uk ðB2 ½0 Þ; and Uk is outside Ga ; u1 k ðB2 ½0 Þ is outside Ga for every k N: Since for every k N; Ga is outside u1 ðB 2 ½0 Þ; by the deﬁnition of fk ; for every x in the open neighborhood Ga of k a; and k N; we have fk ðxÞ ¼ 0: Now, since a 2 Ga ; fN ðaÞ ¼ 0 ¼ fNþ1 ðaÞ ¼ fNþ2 ðaÞ ¼ : It follows that f1 ðaÞ þ f2 ðaÞ þ f3 ðaÞ þ is a real number. 1 1 Since fu1 1 ðB1 ð0ÞÞ; u2 ðB1 ð0ÞÞ; u3 ðB1 ð0ÞÞ; . . .g covers M; and a is in M; there 1 exists a positive integer N0 such that a 2 u1 N0 ðB1 ð0ÞÞ uN0 ðB3 ð0ÞÞ ¼ UN0 ; and hence uN0 ðaÞ 2 B1 ð0Þ B1 ½0 : Since uN0 ðaÞ 2 B1 ½0 ; a 2 UN0 ; and HðxÞ ¼ 1 for every x in the closed ball B1 ½0 ; fN0 ðaÞ ¼ HðuN0 ðaÞÞ ¼ 1: Since f1 : M ! ½0; 1 ; f2 : M ! ½0; 1 ; f3 : M ! ½0; 1 ; . . .; and fN0 ðaÞ ¼ 1; f1 ðaÞ þ f2 ðaÞ þ f3 ðaÞ þ fN0 ðaÞ ¼ 1: Further, since 1 f1 ðaÞ þ f2 ðaÞ þ f3 ðaÞ þ ; and f1 ðaÞ þ f2 ðaÞ þ f3 ðaÞ þ is a real number,

4.5 Partitions of Unity Theorem

291

fn ðaÞ 2 ½0; 1 : f1 ðaÞ þ f2 ðaÞ þ f3 ðaÞ þ Thus, the function gn : M ! ½0; 1 deﬁned as follows: For every x in M; gn ð x Þ

f n ð xÞ f 1 ð xÞ þ f 2 ð xÞ þ f 3 ð xÞ þ

is well deﬁned. Further, we have shown that for every a in M; there exists an open neighborhood Ga of a; and a positive integer N such that for every x in the open neighborhood Ga ; gn ð x Þ ¼

f n ð xÞ : f1 ð xÞ þ f2 ð xÞ þ þ fN ð xÞ

Further, since f1 ; f2 ; f3 ; . . . are smooth, x 7!

f n ð xÞ ð ¼ gn ð x Þ Þ f 1 ð xÞ þ f 2 ð xÞ þ þ f N ð xÞ

is smooth in Ga : This shows that gn is smooth. Clearly, supp gn ¼ supp fn Un : Since for every x in the open neighborhood Ga of a; we have fN ðxÞ ¼ 0 ¼ fNþ1 ðxÞ ¼ fNþ2 ðxÞ ¼ ; f N ð xÞ 0 ¼ ¼ 0; f1 ð xÞ þ f2 ð xÞ þ þ fN ð xÞ f1 ð xÞ þ f2 ð xÞ þ þ fN ð xÞ fNþ1 ð xÞ 0 gNþ1 ð xÞ ¼ ¼ ¼ 0; f1 ð xÞ þ f2 ð xÞ þ þ fN ð xÞ f1 ð xÞ þ f2 ð xÞ þ þ fN ð xÞ gNþ1 ðxÞ ¼ 0; . . .: gN ð x Þ ¼

Hence, ðg1 þ g2 þ g3 þ ÞjGa ¼ ðg1 þ g2 þ þ gN ÞjGa : Clearly, for every x in M, g1 ð x Þ þ g2 ð x Þ þ g3 ð x Þ þ 1 1 f 1 ð xÞ þ f2 ð xÞ ¼ f1 ð xÞ þ f2 ð xÞ þ f3 ð xÞ þ f1 ð xÞ þ f2 ð xÞ þ f3 ð xÞ þ 1 f 3 ð xÞ þ þ f 1 ð xÞ þ f 2 ð xÞ þ f 3 ð xÞ þ 1 ¼ ðf1 ð xÞ þ f2 ð xÞ þ f3 ð xÞ þ Þ ¼ 1: f1 ð xÞ þ f2 ð xÞ þ f3 ð xÞ þ In short, g1 þ g2 þ g3 þ ¼ 1:

292

4 Topological Properties of Smooth Manifolds

Since fU1 ; U2 ; U3 ; . . .g is a reﬁnement of fXi gi2I ; for every positive integer m, there exists an index iðmÞ 2 I such that Um XiðmÞ : It follows that [i 2I fm : iðmÞ ¼ i g ¼ f1; 2; 3; . . .g: Let us ﬁx any i in I: Since fx : 0\f1 ðxÞg fx : 0\f1 ðxÞg ¼ supp f1 U1 ; fx : 0\f1 ðxÞg U1 : Similarly, fx : 0\f2 ðxÞg U2 ; fx : 0\f3 ðxÞg U3 ; . . .. Further, since fU1 ; U2 ; U3 ; . . .g is locally ﬁnite, ffx : 0\f1 ðxÞg; fx : 0\f2 ðxÞg; fx : 0\f3 ðxÞg; . . .g is locally ﬁnite, and hence, its subcollection ffx : 0\fm ðxÞg : iðmÞ ¼ i g is locally ﬁnite. It follows that ð[iðmÞ¼i fx : 0\fm ðxÞgÞ ¼ [iðmÞ¼i

ðfx : 0\fm ðxÞg Þ ¼ [iðmÞ¼i ðsupp fm Þ [iðmÞ¼i Um [iðmÞ¼i XiðmÞ ¼ [iðmÞ¼i Xi ¼ Xi :

Thus ð[iðmÞ¼i fx : 0\fm ðxÞgÞ Xi : P Since for every x in M; iðmÞ¼i gm ðxÞ g1 ðxÞ þ g2 ðxÞ þ g3 ðxÞ þ ¼ 1; it follows that the function wi : M ! ½0; 1 deﬁned as follows: ( P gm if fm : iðmÞ ¼ i g is nonempty; wi ¼ iðmÞ¼i 0 if fm : iðmÞ ¼ i g is the empty set; is well deﬁned. For 1: Case I: When Pfm : iðmÞ ¼ i g is nonempty. Let us take any a in M: We have to prove that iðmÞ¼i gm ð¼ wi Þ is smooth at a: We have seen that there integers m1 ; m2 ; . . .; mk such exist an open neighborhood Ga of a; and positiveP that i ¼ iðm1 Þ ¼ iðm2 Þ ¼ ¼ iðmk Þ; and ð iðmÞ¼i gm ÞjGa ¼ ðgm1 þ gm2 þ þ gmk ÞjGa : Since each gn is smooth, ðgm1 þ gm2 þ þ gmk ÞjGa is smooth P at a: Since ðgm1 þ gm2 þ þ gmk ÞjGa is smooth at a; and ð iðmÞ¼i gm ÞjGa ¼ P ðgm1 þ gm2 þ þ gmk ÞjGa ; ð iðmÞ¼i gm ÞjGa is smooth at a: This shows that P wi ð¼ iðmÞ¼i gm Þ is smooth. Case 2: When fm : iðmÞ ¼ i gis the empty set. Since the constant function 0 is smooth, and wi ¼ 0; wi is smooth. So, in all cases, wi is smooth. For 2: Case I: When fm : iðmÞ ¼ i g is nonempty. We have to prove that supp wi Xi : Here

0 supp wi ¼ supp @ 8 <

X

1

00

gm A ¼ @ @

iðmÞ¼i

X

gm iðmÞ¼i 9 8

11

1

A ð R f0gÞ A

9 = < = X ¼ x : 0\@ gm Að xÞ ¼ x : 0\ ð gm ð x Þ Þ : ; : ; iðmÞ¼i iðmÞ¼i 9 8 = < X 1 f m ð xÞ ¼ x : 0\ ; : f 1 ð xÞ þ f 2 ð xÞ þ f 3 ð xÞ þ iðmÞ¼i 0

X

1

4.5 Partitions of Unity Theorem

293

8 9 < = X 1 ¼ x : 0\ ðfm ð xÞÞ : ; f1 ð xÞ þ f2 ð xÞ þ f3 ð xÞ þ iðmÞ¼i 8 9 < = X ¼ x : 0\ fm ð xÞ ¼ [iðmÞ¼i fx : 0\fm ð xÞg Xi : : ; iðmÞ¼i Hence, supp wi Xi : Case II: When fm : iðmÞ ¼ i g is the empty set. Here, supp wi ¼ supp ð0Þ ¼ ; Xi ; and hence, supp wi Xi :Hence, in all cases, supp wi Xi : For 3: We have seen that if fm : iðmÞ ¼ i g is nonempty; then supp wi ¼ ð[iðmÞ¼i fx : 0\fm ðxÞgÞ ¼ [iðmÞ¼i ðfx : 0\fm ðxÞg Þ ¼ [iðmÞ¼i ðsupp fm Þ [iðmÞ¼i Um : Thus, if fm : iðmÞ ¼ i g is nonempty; then supp wi [iðmÞ¼i Um : Also, if fm : iðmÞ ¼ i g is the empty set; then supp wi ¼ supp 0 ¼ ;: Now, it sufﬁces to show that f[iðmÞ¼i Um : i 2 Ig is locally ﬁnite. Since ffm : iðmÞ ¼ i g : i 2 Ig is a partition of f1; 2; 3; . . .g; and fU1 ; U2 ; U3 ; . . .g is locally ﬁnite, f[iðmÞ¼i Um : i 2 Ig is locally ﬁnite. For 4: Here LHS ¼

X

wi ¼

i2I

¼

X

X i 2I

¼

X i 2I fm:iðmÞ¼i gis non empty

X i 2I fm:iðmÞ¼i gis empty set

0 @

X

X

wi þ

i 2I fm:iðmÞ¼i gis non empty

wi þ

i 2I fm:iðmÞ¼i gis non empty

X

wi ¼

0¼

wi

i 2I fm:iðmÞ¼i gis empty set

X

wi

i 2I fm:iðmÞ¼i gis non empty

1 gm A ¼ g1 þ g2 þ g3 þ ¼ 1 ¼ RHS:

h

iðmÞ¼i

h

Note 4.57 Clearly, the above lemma is also valid for m-dimensional smooth manifold M. Deﬁnition Let M be a topological space. Let fXi gi2I be an open cover of M: Let fwi gi2I be a family of functions wi : M ! ½0; 1 satisfying the following conditions: 1. 2. 3. 4.

for each i 2 I; wi is continuous, for each i 2 I; supp wi is contained in Xi ; fsupp wi gi2I is locally ﬁnite, P wi ¼ 1: i2I

294

4 Topological Properties of Smooth Manifolds

Then, we say that fwi gi2I is a partition of unity subordinate to fXi gi2I : Since fsupp wi gi2I is locally P ﬁnite, for P every x in M; there are only ﬁnite many nonzero wi ðxÞ’s in the sum i2I wi ðxÞ; i2I wi ¼ 1 is meaningful. Further, if M is a smooth manifold, and each wi is smooth, then we say that fwi gi2I is a smooth partition of unity subordinate to fXi gi2I : Now, the Lemma 4.56 can be abbreviated as: “If M is an n-dimensional smooth manifold, and fXi gi2I is an open cover of M; then there exists a smooth partition of unity subordinate to fXi gi2I :” This result is known as the partitions of unity theorem. Deﬁnition Let M be a topological space. Let A be a closed subset of M: Let U be an open subset of M such that A U: Let w : M ! ½0; 1 be a continuous function. If 1. for every x in A, wðxÞ ¼ 1; 2. supp w U; then we say that w is a bump function for A supported in U: Lemma 4.58 Let M be a smooth manifold. Let A be a closed subset of M: Let U be an open subset of M such that A U: Then, there exists a smooth function w : M ! ½0; 1 such that w is a bump function for A supported in U: Proof Clearly, fM A; Ug is an open cover of M: So, by the partitions of unity theorem, there exist functions w1 : M ! ½0; 1 ; w : M ! ½0; 1 such that 1. w1 ; w are smooth, 2. supp w1 is contained in M A; and supp w is contained in U; 3. w1 þ w ¼ 1: It remains to prove: For every x in A; wðxÞ ¼ 1; that is, for every x in A; 1 w1 ðxÞ ¼ ð1 w1 ÞðxÞ ¼ 1; that is, for every x in A; w1 ðxÞ ¼ 0: Since fx : 0 6¼ w1 ðxÞg supp w1 M A; A fx : 0 ¼ w1 ðxÞg ¼ fx : 0 ¼ 1 wðxÞg ¼ fx : wðxÞ ¼ 1g: h Lemma 4.59 Let M be an m-dimensional smooth manifold. Let A be a nonempty closed subset of M: Let U be an open subset of M such that A U: Let f : A ! Rk be a smooth function (that is, for every p in A; there exists an open neighborhood Wp of p; and a smooth function Fp : Wp ! Rk such that f jWp \ A ¼ Fp jWp \ A Þ: Then, there exists a smooth function ~f : M ! Rk such that 1. ~fA ¼ f ; 2. supp ~f U: Proof Let us ﬁx any p in A: Since f : A ! Rk is a smooth function, and p is in A, there exists an open neighborhood Vp of p, and a smooth function Fp : Vp ! Rk such that f jVp \ A ¼ Fp jVp \ A : Since p is in A, and A U; p is in U. Since p is in U,

4.5 Partitions of Unity Theorem

295

and U is open, U is an open neighborhood of p. Since U is an open neighborhood of p, and Vp is an open neighborhood of p, Vp \ U is an open neighborhood of p. Clearly, the collection fVp \ U : p 2 Ag [fM Ag is an open cover of smooth manifold M: By the partitions of unity theorem, then there exist a collection fwp : p 2 Ag [fwg of smooth functions wp : M ! ½0; 1 ; and w : M ! ½0; 1 such that (a) for each p 2 A; supp wp is contained in Vp \ U; supp w is contained in M A; (b) fsupp wp : p 2 Ag [fsupp wg is locally ﬁnite, P (c) p2A wp ¼ 1 w: Since for every p 2 A; wp : M ! ½0; 1 is smooth, and Vp \ U is an open neighborhood of p; wp jVp \ U is smooth. Since Fp : Vp ! Rk is smooth, Vp is an open neighborhood of p; Vp \ U is an open neighborhood of p; and Vp \ U Vp ; Fp jVp \ U is smooth. Since wp jVp \ U is smooth, Fp jVp \ U is smooth, and Vp \ U is an open neighborhood of p; ðwp jVp \ U ÞðFp jVp \ U Þ is smooth on Vp \ U: For every p in A, supp wp Vp \ U; so M ðVp \ UÞ M ðsupp wp Þ: If t 2 ðVp \ UÞ \ ðM ðsupp wp ÞÞ then t 62 ðsupp wp Þ; and hence ðwp jVp \ U ÞðtÞ ¼ 0: Thus, for every t 2 ðVp \ UÞ \ðM ðsupp wp ÞÞ; we have w p V

p

\U

Fp Vp \ U ðtÞ ¼ wp V \ U ðtÞ Fp Vp \ U ðtÞ p ¼ 0 Fp ðtÞ ¼ 0: Vp \ U

It follows that we can deﬁne a function vp : M ! Rk as follows: For every x in M; vp ð x Þ

8 < wp :0

Vp

F ð xÞ p Vp \ U \U

if x 2 Vp \ U if x 2 M supp wp :

Since M ðsupp wp Þ fx : vp ðxÞ ¼ 0g; fx : vp ðxÞ 6¼ 0g supp wp ; and hence, supp vp ¼ fx : vp ðxÞ 6¼ 0g ðsupp wp Þ ¼ supp wp : Thus, for every p 2 A; supp vp supp wp : Since fsupp wp : p 2 Ag [fsupp wg is locally ﬁnite, its subcollection fsupp wp : p 2 Ag is also locally ﬁnite. Since fsupp wp : p 2 Ag is locally ﬁnite, and for every p 2 A; supp vp supp wp ; fsupp vp : p 2 Ag is locally ﬁnite. Further, since Vp \ U is open, M ðsupp wp Þ is open, ðwp jVp \ U ÞðFp jVp \ U Þ is smooth on Vp \ U; and 0 is smooth on M ðsupp wp Þ; by the deﬁnition of vp ; vp : M ! Rk is smooth. Let us take any x in M: Since fsupp wp : p 2 Ag is also locally ﬁnite, and x is in M; there exists an open neighborhood Gx of x such that all but ﬁnite many

296

4 Topological Properties of Smooth Manifolds

supp wp ’s are outside Gx : Since all but ﬁnite many supp wp ’s are outside Gx ; there P exists p1 ; p2 ; . . .; pn 2 A such that ð p2A vp ÞjGx ¼ ðvp1 þ vp2 þ þ vpn ÞjGx : Since vp1 ; vp2 ; . . .; vpn are smooth, and Gx is an open neighborhood of x; ðvp1 þ vp2 þ þ vpn ÞjGx is smooth on Gx : Since ðvp1 þ vp2 þ þ vpn ÞjGx is P P smooth on Gx ; and ð p2A vp ÞjGx ¼ ðvp1 þ vp2 þ þ vpn ÞjGx ; ð p2A vp ÞjGx is P smooth on Gx : It follows that y 7! ð p2A vp ÞðyÞ is smooth. Now, we can deﬁne a function ~f : M ! Rk as follows: ~f

X

vp :

p2A

Clearly, ~f is smooth. For 1: Let us ﬁx any x 2 A: We have to show that ~f ðxÞ ¼ f ðxÞ: Since fy : wðyÞ 6¼ 0g supp w M A; x 2 A fy : wðyÞ ¼ 0g; and hence wðxÞ ¼ 0: Since supp wp Vp \ U; M ðVp \ UÞ M ðsupp wp Þ: If x 62 Vp \ U; then x 2 M ðVp \ UÞ M ðsupp wp Þ; and hence vp ðxÞ ¼ 0: Also if x 62 Vp \ U then x 62 supp wp ¼ fy : wp ðyÞ 6¼ 0g fy : wp ðyÞ 6¼ 0g; and hence wp ðxÞ ¼ 0: Thus, if x 62 Vp \ U; then vp ðxÞ ¼ 0 ¼ wp ðxÞ: Now, ~f ð xÞ ¼

X

! vp ðxÞ ¼

X

X

vp ðxÞ ¼

vp ðxÞ þ

X

fp:p2A;x2Vp \ U g fp:p2A;x62Vp \ U g X X X ¼ vp ð xÞ þ 0¼ vp ð xÞ fp:p2A;x2Vp \ U g fp:p2A;x62Vp \ U g fp:p2A;x2Vp \ U g X X ¼ Fp ð xÞ ¼ wp wp p2A

fp:p2Ag

fp:p2A;x2Vp \ U g X

Vp \ U

Vp \ U

vp ð xÞ

Vp \ U

ð xÞ

Fp Vp \ U ð xÞ

fp:p2A;x2Vp \ U g X ¼ wp V \ U ð xÞ ðf ðxÞÞ p p fp:p2A;x2Vp \ U g fp:p2A;x2Vp \ U g 0 1 0 1 X X B C B C wp V \ U ð xÞAðf ðxÞÞ ¼ @ wp ð xÞAðf ð xÞÞ ¼@ p fp:p2A;x2Vp \ U g fp:p2A;x2Vp \ U g 0 1 X X B C wp ð x Þ þ 0Aðf ðxÞÞ ¼@ fp:p2A;x2Vp \ U g fp:p2A;x62Vp \ U g 0 1 0 1 X X X B C @ ¼@ wp ðxÞAðf ð xÞÞ wp ð x Þ þ wp ðxÞAðf ð xÞÞ ¼ f p:p2A g p:p2A;x2V p:p2A;x6 2 V U U \ \ f g f g p p 1 1 00 X ¼ @@ wp AðxÞAðf ð xÞÞ ¼ ðð1 wÞðxÞÞðf ðxÞÞ ¼ ð1 wð xÞÞðf ðxÞÞ ¼ ð1 0Þðf ðxÞÞ ¼ f ðxÞ: fp:p2Ag

wp V

ð xÞ Fp ðxÞ ¼ \U

4.5 Partitions of Unity Theorem

297

Thus, ~f ðxÞ ¼ f ðxÞ: So, for any x 2 A we have ~f ðxÞ ¼ f ðxÞ: Thus, ~f jA ¼ f : This proves 1. For 2: Let us take any y 2 M such that ~f ðyÞ 6¼ 0: Here, 0 6¼ ~f ðyÞ ¼

X

! vp ð yÞ ¼

X

X

vp ðyÞ ¼

vp ð yÞ þ

X

fp:p2A;y2Vp \ U g fp:p2A;y62Vp \ U g X X X ¼ vp ð yÞ þ 0¼ vp ðyÞ fp:p2A;y2Vp \ U g fp:p2A;y62Vp \ U g fp:p2A;y2Vp \ U g X X Fp ð yÞ ¼ w w ¼ p2A

fp:p2Ag

p Vp \ U

fp:p2A;y2Vp \ U g X ¼ wp ðyÞ Fp ð xÞ : fp:p2A;y2Vp \ U g

Vp \ U

fp:p2A;y2Vp \ U g

vp ð yÞ

p Vp \ U

ð yÞ Fp Vp \ U ðyÞ

P Since fp:p2A;y2Vp \ Ug ðwp ðyÞÞðFp ðxÞÞ 6¼ 0; there exists p 2 A such that wp ðyÞ 6¼ 0: It follows that fy : ~f ðyÞ 6¼ 0g [p2A fy : wp ðyÞ 6¼ 0g; and hence, supp ~f ¼ fy : ~f ðyÞ 6¼ 0g ð[p2A fy : wp ðyÞ 6¼ 0gÞ : Since fsupp wp : p 2 Ag is locally ﬁnite, ð[p2A fy : wp ðyÞ 6¼ 0gÞ ¼ [p2A ðfy : wp ðyÞ 6¼ 0g Þ: It follows that supp ~f [p2A ðfy : wp ðyÞ 6¼ 0g Þ ¼ [p2A ðsupp wp Þ [p2A ðVp \ UÞ ¼ ð[p2A Vp Þ \ U U: supp ~f U: This proves 2.

Thus, h

4.6 Topological Manifolds With Boundary Deﬁnition By H3 we mean the set fðx1 ; x2 ; x3 Þ : ðx1 ; x2 ; x3 Þ 2 R3 and 0 x3 g; and is called the closed 3-dimensional upper half-space. By Int H3 we mean fðx1 ; x2 ; x3 Þ : ðx1 ; x2 ; x3 Þ 2 R3 and 0\x3 g: By oH3 we mean fðx1 ; x2 ; 0Þ : ðx1 ; x2 ; 0Þ 2 R3 g: Observe that the collection of all open subsets of H3 which do not intersect oH3 is contained in the collection of all open subsets of R3 : Let M be a topological space, whose topology is Hausdorff, and second countable. If for every p in M, there exist an open neighborhood U of p; “a subset G of H3 which is open in H3 or a subset G of R3 which is open in R3 ,” and a homeomorphism u from U onto G; then we say that M is a 3-dimensional topological manifold with boundary. Here, the ordered pair ðU; uÞ is called a chart of M: If uðUÞð¼ GÞ is open in R3 ; then we say that ðU; uÞ is an interior chart of M: If uðUÞ ð¼ GÞ is a subset of H3 ; uðUÞ is open in H3 ; and uðUÞ \ oH3 6¼ ;; then we say that ðU; uÞ is a boundary chart of M:

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Let p 2 M: By p is an interior point of M; we mean: There exists an interior chart ðU; uÞ of M such that p 2 U: By p is a boundary point of M; we mean: There exists a boundary chart ðU; uÞ of M such that p 2 U; and uðpÞ 2 oH3 : By Int M we mean the set of all interior points of M; and is called the interior of M: By oM we mean the set of all boundary points of M; and is called the boundary of M: Similar deﬁnitions can be supplied for 4-dimensional topological manifolds with boundary, etc. Note 4.60 Let M be a 3-dimensional topological manifold with boundary. We want to prove: 1. M ¼ ðInt MÞ [ðoMÞ 2. ðInt MÞ \ðoMÞ ¼ ;: For 1: Clearly, ðInt MÞ M; and ðoMÞ M: So, ðInt MÞ [ðoMÞ M: It remains to prove that M ðInt MÞ [ðoMÞ: If not, otherwise, let there exists p 2 M such that p 62 Int M; and p 62 oM: We have to arrive at a contradiction. Since p 2 M; and M is a 3-dimensional topological manifold with boundary, there exist an open neighborhood U of p; “a subset G of H3 which is open in H3 or a subset G of R3 which is open in R3 ,” and a homeomorphism u from U onto G: Thus, ðU; uÞ is a chart for M: Since p 62 Int M; p is not an interior point of M: Since p is not an interior point of M; ðU; uÞ is a chart for M; and p 2 U; ðU; uÞ is not an interior chart of M: Since ðU; uÞ is not an interior chart of M; uðUÞð¼ GÞ is not open in R3 ; and hence, G is not open in R3 : Since G is not open in R3 ; and “G is a subset of H3 which is open in H3 or G is a subset of R3 which is open in R3 ,” G is a subset of H3 which is open in H3 : Since p 2 U; uðpÞ 2 uðUÞð¼ GÞ: Since G is not open in R3 ; G is a subset of H3 ; and G is open in H3 ; G \ oH3 6¼ ;: (Reason: If not, otherwise, let G \ oH3 ¼ ;: Since G H3 ¼ Int H3 [ oH3 ; and G \ oH3 ¼ ;; G Int H3 : Since G is open in H3 ; there exists an open set G1 in R3 such that G1 \ H3 ¼ G: Since G ¼ G1 \ H3 G1 ; and G Int H3 ; G G1 \ Int H3 : Since G1 \ Int H3 Int H3 H3 ; and G1 \ Int H3 G1 ; G1 \ Int H3 G1 \ H3 ¼ G: Since G1 \ Int H3 G; and G G1 \ Int H3 ; G ¼ G1 \ Int H3 : Since G1 ; Int H3 are open in R3 ; their intersection G1 \ Int H3 ð¼ GÞ is open in R3 ; and hence, G is open in R3 ; a contradiction.) Since uðUÞð¼ GÞ is a subset of H3 which is open in H3 ; and uðUÞ \ oH3 6¼ ;; ðU; uÞ is a boundary chart for M: Since p 62 oM; p is not a boundary point of M: Since p is not a boundary point of M; ðU; uÞ is a chart for M satisfying p 2 U; and ðU; uÞis a boundary chart for M; uðpÞ 62 oH3 : Clearly, oH3 ð¼ fðx1 ; x2 ; 0Þ : ðx1 ; x2 ; 0Þ 2 R3 gÞ is closed in H3 : It follows that G oH3 is open in H3 : Since

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uðpÞ 62 oH3 ; uðpÞ 2 G; and G oH3 is open in H3 ; there exists an open neighborhood HuðpÞ of uðpÞ in H3 such that HuðpÞ G oH3 G; and hence HuðpÞ \ oH3 ¼ ;: Since HuðpÞ \ oH3 ¼ ;; and HuðpÞ is open in H3 ; HuðpÞ is open in R3 : Since u is a homeomorphism from U onto G; G is open in H3 ; HuðpÞ G; and HuðpÞ is an open neighborhood of uðpÞ in H3 ; u1 ðHuðpÞ Þ is an open neighborhood of p in M; and uju1 ðHuðpÞ Þ is a homeomorphism from u1 ðHuðpÞ Þ onto HuðpÞ : Thus,

the ordered pair ðu1 ðHuðpÞ Þ; uju1 ðHuðpÞ Þ Þ is a chart for M: Now, since ðuju1 ðHuðpÞ Þ Þ

ðu1 ðHuðpÞ ÞÞ ¼ HuðpÞ ; and HuðpÞ is open in R3 ; the chart ðu1 ðHuðpÞ Þ; uju1 ðHuðpÞ Þ Þ is an interior chart for M: Since ðu1 ðHuðpÞ Þ; uju1 ðHuðpÞ Þ Þ is an interior chart for M;

and p 2 u1 ðHuðpÞ Þ; p is an interior point of M; and hence p 2 Int M; a contradiction. For 2: Its proof is postponed right now. This result is known as the topological invariance of the boundary. Similar results can be supplied for 4-dimensional topological manifolds with boundary, etc. Note 4.61 Let M be a topological n-manifold with boundary. We want to show that Int M is an open subset of M. Here, Int M ¼ fp : p 2 M; there exists a chartðU; uÞ of M such that p 2 U; and uðU Þ is open in Rn g:

For this purpose, let us take any p 2 Int M: We have to ﬁnd an open neighborhood Up such that if q 2 Up then q 2 Int M: Since p 2 Int M; p 2 M; there exists a chart ðU; uÞ of M such that p 2 U; and uðUÞ is open in Rn : Since ðU; uÞ is a chart, U is open in M: Since U is open in M; and p 2 U; U is an open neighborhood of p: Next, let us take any q 2 U: Here, ðU; uÞ is a chart of M such that q 2 U; and uðUÞ is open in Rn : Hence, by deﬁnition of the Int M; q 2 Int M: This proves that Int M is an open subset of M: Note 4.62 Clearly, a topological n-manifold M “in the original sense” is a topological n-manifold M with boundary. If M is a topological n-manifold M “in the original sense,” then Int M ¼ M: Hence, by the topological invariance of the boundary, if M is a topological nmanifold “in the original sense,” then oM ¼ ;: Conversely, let M be a topological n-manifold M with boundary such that oM ¼ ;: Hence, M ¼ Int M [ oM ¼ Int M [ ; ¼ Int M: Since Int M ¼ M; M is a topological n-manifold “in the original sense.”

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Thus, we have shown that a topological n-manifold M with boundary is a topological n-manifold if and only if oM ¼ ;: Sometimes, a manifold in the original sense is referred as manifold without boundary. Note 4.63 Let M be a topological n-manifold with boundary. We want to show that Int M is a topological n-manifold without boundary. Here, Int M is contained in M: The topology of Int M is assumed to be the subspace topology of M: Since M is a topological n-manifold with boundary, the topology of M is Hausdorff, and second countable, and hence, the subspace topology over Int M is Hausdorff, and second countable. Next, let us take any p 2 Int M: Since p 2 Int M; p is an interior point of M; and hence, there exists an interior chart ðU; uÞ of M such that p 2 U: Since ðU; uÞ is an internal chart of M, ðU; uÞ is a chart of M; and uðUÞ is open in Rn : Since ðU; uÞ is a chart of M, U is open in M; and u : U ! uðUÞ is a homeomorphism. Since U is open in M; and Int M is open in M; U \ Int M is open in the topological space Int M: Since p 2 U; and p 2 Int M; p 2 U \ Int M: Thus, U \ Int M is an open neighborhood of p in the topological space Int M: Since U is open in M; and Int M is open in M; U \ Int M is open in M: Since U \ Int M; U are open in M; and U \ Int M U; U \ Int M is open in U: Since u : U ! uðUÞ is a homeomorphism, and U \ Int M is open in U; uðU \ Int MÞ is open in uðUÞ: Since uðU \ Int MÞ is open in uðUÞ; and uðUÞ is open in Rn ; uðU \ Int MÞ is open in Rn : Since U \ Int M is open in U; uðU \ Int MÞ is open in uðUÞ; and u : U ! uðUÞ is a homeomorphism, ujU \ IntM is a homeomorphism from U \ Int M onto uðU \ Int MÞ: Since U \ Int M is an open neighborhood of p in the topological space Int M; uðU \ Int MÞ is open in Rn ; and ujU \ IntM is a homeomorphism from U \ Int M onto uðU \ Int MÞ; the ordered pair ðU \ Int M; ujU \ IntM Þ is an interior chart of the topological space Int M: Also p 2 U \ Int M: This shows that Int M is a topological n-manifold without boundary. Note 4.64 Let M be a topological 3-manifold with boundary. We want to show that 1. oM is a closed subset of M, 2. oM is a topological 2-manifold without boundary. For 1: Since Int M is open in M; M Int M is closed in M: By the topological invariance of the boundary, oM ¼ M Int M: Since oM ¼ M Int M; and M Int M is closed in M; oM is a closed subset of M: For 2: Here, oM is contained in M: The topology of oM is assumed to be the subspace topology of M: Since M is a topological 3-manifold with boundary, the topology of M is Hausdorff, and second countable, and hence, the subspace topology over oM is Hausdorff, and second countable.

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Next, let us ﬁx any p 2 oM: Since p 2 oM; p is a boundary point of M; and hence, there exists a boundary chart ðU; uÞ of M such that p 2 U; and uðpÞ 2 oH3 ¼ fðx1 ; x2 ; 0Þ : ðx1 ; x2 ; 0Þ 2 R3 g: Since ðU; uÞ is a boundary chart of M, U is open in M; uðUÞ is open in H3 ; and u : U ! uðUÞ is a homeomorphism. Since p 2 U; uðpÞ 2 uðUÞ: Since uðpÞ 2 uðUÞ; and uðUÞ is open in H3 ; uðUÞ is an open neighborhood of uðpÞ in H3 : Since uðUÞ is an open neighborhood of uðpÞ in H3 ; and uðpÞ 2 oH3 H3 ; uðUÞ \ oH3 is an open neighborhood of uðpÞ in oH3 : Since u : U ! uðUÞ is a homeomorphism, and uðUÞ \ oH3 uðUÞ; the restriction uju1 ðuðUÞ \ oH3 Þ is a homeomorphism from u1 ðuðUÞ \ oH3 Þ onto uðUÞ \ oH3 : Since uðpÞ 2 uðUÞ; and uðpÞ 2 oH3 ; uðpÞ 2 uðUÞ \ oH3 ; and hence p 2 u1 ðuðUÞ \ oH3 Þ: Now, we shall try to prove: (a) u1 ðuðUÞ \ oH3 Þ oM; (b) u1 ðuðUÞ \ oH3 Þ is open in oM; (c) uðUÞ \ oH3 is open in oH3 : For a: Let us take any a 2 u1 ðuðUÞ \ oH3 Þ: Since a 2 u1 ðuðUÞ \ oH3 Þ; and u : U ! uðUÞ; a 2 U: Since a 2 u1 ðuðUÞ \ oH3 Þ; uðaÞ 2 uðUÞ \ oH3 ; and hence, uðaÞ 2 oH3 : Since ðU; uÞ is a boundary chart of M; a 2 U; and uðaÞ 2 oH3 ; a is a boundary point of M; and hence a 2 oM: For b: First of all, we want to prove that u1 ðuðUÞ \ oH3 Þ ¼ U \ oM: For this purpose, let us take any a in u1 ðuðUÞ \ oH3 Þ: Since a 2 u1 ðuðUÞ \ oH3 Þ; and u : U ! uðUÞ; a 2 U: Hence, u1 ðuðUÞ \ oH3 Þ U: Since u1 ðuðUÞ \ oH3 Þ U; and from (a), u1 ðuðUÞ \ oH3 Þ oM; u1 ðuðUÞ \ oH3 Þ U \ oM: Next, let us take any b in U \ oM: Since b is inU \ oM; b 2 U; and hence, uðbÞ 2 uðUÞ: Clearly, uðbÞ 2 oH3 : (Reason: If not, otherwise, let uðbÞ 62 oH3 : We have to arrive at a contradiction. Since b is in U \ oM; b is in U: Since b is in U; and u : U ! uðUÞ; uðbÞ 2 uðUÞ: Since uðbÞ 2 uðUÞ; and uðUÞ is open in H3 ; uðbÞ 2 H3 ¼ Int H3 [ oH3 :Since uðbÞ 2 Int H3 [ oH3 ;and uðbÞ 62 oH3 ; uðbÞ 2 Int H3 ¼ fðx1 ; x2 ; x3 Þ : ðx1 ; x2 ; x3 Þ 2 R3 and 0\x3 g: Since uðbÞ 2 fðx1 ; x2 ; x3 Þ : ðx1 ; x2 ; x3 Þ 2 R3 and 0\x3 g; uðbÞ 2 uðUÞ; and uðUÞ is open in H3 ; there exists a real number r [ 0 such that the open ball Br ðuðbÞÞ is contained in uðUÞ \ Int H3 : It follows that Br ðuðbÞÞ is open in uðUÞ: Since Br ðuðbÞÞ is open in uðUÞ; and u : U ! uðUÞ is a homeomorphism, u1 ðBr ðuðbÞÞÞ is open in U; and the restriction uju1 ðBr ðuðbÞÞÞ is a homeomorphism from u1 ðBr ðuðbÞÞÞ onto Br ðuðbÞÞ: Since uðbÞ 2 Br ðuðbÞÞ; b 2 u1 ðBr ðuðbÞÞÞ: Since u1 ðBr ðuðbÞÞÞ is open in U; and U is open in M; u1 ðBr ðuðbÞÞÞ is open in M: Since u1 ðBr ðuðbÞÞÞ is open in M; and b 2 u1 ðBr ðuðbÞÞÞ; u1 ðBr ðuðbÞÞÞ is an open neighborhood of b in M: It follows that the ordered pair

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ðu1 ðBr ðuðbÞÞÞ; uju1 ðBr ðuðbÞÞÞ Þ is a chart of M: Also since b 2 u1 ðBr ðuðbÞÞÞ;

ðuju1 ðBr ðuðbÞÞÞ Þðu1 ðBr ðuðbÞÞÞÞ ¼ Br ðuðbÞÞ; and Br ðuðbÞÞ is open in R3 ; ðu1 ðBr ðuðbÞÞÞ; uju1 ðBr ðuðbÞÞÞ Þ is an interior chart of M; and hence b 2 Int M: Since b 2 Int M, by the topological invariance of the boundary, b 62 oM; and hence, b 62 U \ oM; a contradiction.) Since uðbÞ 2 oH3 ; and uðbÞ 2 uðUÞ; uðbÞ 2 uðUÞ \ oH3 ; and hence, b 2 1 u ðuðUÞ \ oH3 Þ: Thus, U \ oM u1 ðuðUÞ \ oH3 Þ: Thus, we have shown that u1 ðuðUÞ \ oH3 Þ ¼ U \ oM: Since U is open in M; and oM M; U \ oMð¼ u1 ðuðUÞ \ oH3 ÞÞ is open in oM; and hence, u1 ðuðUÞ \ oH3 Þ is open in oM: This proves (b). For c: Since uðUÞ is open in H3 ; and oH3 H3 ; uðUÞ \ oH3 is open in oH3 : This proves (c).

Since the mapping p3 : oH3 ! R2 deﬁned by p3 ðx1 ; x2 ; 0Þ ðx1 ; x2 Þ is a homeomorphism, and uju1 ðuðUÞ \ oH3 Þ is a homeomorphism from u1 ðuðUÞ \ oH3 Þ onto uðUÞ \ oH3 ; their composite p3 ðuju1 ðuðUÞ \ oH3 Þ Þ is a homeomorphism from

u1 ðuðUÞ \ oH3 Þ onto p3 ðuðUÞ \ oH3 Þ: Since uðUÞ \ oH3 is open in oH3 ; and p3 : oH3 ! R2 is a homeomorphism, p3 ðuðUÞ \ oH3 Þ is open in R2 : Here, p3 ðuju1 ðuðUÞ \ oH3 Þ Þ is a homeomorphism from u1 ðuðUÞ \ oH3 Þ onto p3 ðuðUÞ

\oH3 Þ; u1 ðuðUÞ \ oH3 Þ is open in oM; p 2 u1 ðuðUÞ \ oH3 Þ; and p3 ðuðUÞ \ oH3 Þ is open in R2 ; the ordered pair ðu1 ðuðUÞ \ oH3 Þ;Þp3 ðuju1 ðuðUÞ \ oH3 Þ Þ is

an internal chart of oM; where p 2 u1 ðuðUÞ \ oH3 Þ: Hence, oM is a topological 2-manifold without boundary. This completes the proof of 2. Similar results can be supplied for 4-dimensional topological manifolds with boundary, etc.

Deﬁnition Let A be a nonempty subset of Rn : Let f : A ! Rm : By f is smooth we mean: For every x in A; there exist an open neighborhood Vx of x in Rn ; and a function fx : Vx ! Rm such that fx jA \ Vx ¼ f jA \ Vx ; and fx is smooth. Deﬁnition Let G be a nonempty subset of Hn : Let G be open in Hn : Let f : G ! Rm : By f is smooth we mean: For every x in G; there exist an open neighborhood Vx of x in Rn ; and a function fx : Vx ! Rm such that fx jG \ Vx ¼ f jG \ Vx ; and fx is smooth in the usual sense. Note 4.65 Let G be a nonempty subset of Hn : Let G be open in Hn : Let f : G ! Rm : Let f be smooth. We ﬁrst show that G \ðInt Hn Þ is open in Rn : Since G is open in Hn ; there exists an open subset U of Rn such that G ¼ U \ Hn : Now, G \ðInt Hn Þ ¼ ðU \ Hn Þ \ðInt Hn Þ ¼ U \ðHn \ Int Hn Þ ¼ U \ Int Hn : Since U; Int Hn are open in Rn ; U \ Int Hn ð¼ G \ðInt Hn ÞÞ is open in Rn ; and hence, G \ðInt Hn Þ is open in Rn :

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Since G \ðInt Hn Þð GÞ is open in Rn ; and f : G ! Rm ; it is meaningful to say “f jG \ðIntHn Þ is smooth in the usual sense.” We want to show that f jG \ðIntHn Þ is smooth in the usual sense. For this purpose, let us take any a in G \ðInt Hn Þ: We have to show that f jG \ðIntHn Þ is smooth at a in the usual sense. Since a is in G \ðInt Hn Þ; a is in G: Since a is in G; and f : G ! Rm is smooth, there exist an open neighborhood Va of a in Rn ; and a function fa : Va ! Rm such that fa jG \ Va ¼ f jG \ Va ; and fa is smooth in the usual sense. Since a is in G \ðInt Hn Þ; and G \ðInt Hn Þ is open in Rn ; G \ðInt Hn Þ is an open neighborhood of a in Rn : Since G \ðInt Hn Þ is an open neighborhood of a in Rn ; and Va is an open neighborhood of a in Rn ; G \ðInt Hn Þ \ Va is an open neighborhood of a in Rn ; and is contained in the open set Va : Now, since fa : Va ! Rm is smooth in the usual sense, the restriction fa jG \ðIntHn Þ \ Va is smooth at a in the usual sense. Since fa jG \ Va ¼ f jG \ Va ; G \ðInt Hn Þ \ Va G \ Va ; fa jG \ðIntHn Þ \ Va ¼ f jG \ðIntHn Þ \ Va : Since fa : Va ! Rm is smooth in the usual sense, G \ðInt Hn Þ \ Va is an open neighborhood of a in Rn ; and is contained in the open set Va ; fa jG \ðIntHn Þ \ Va is smooth at a in the usual sense. Since fa jG \ðIntHn Þ \ Va is smooth at a in the usual sense, and fa jG \ðIntHn Þ \ Va ¼ f jG \ðIntHn Þ \ Va ; f jG \ðIntHn Þ \ Va is smooth at a in the usual sense. Since f jG \ðIntHn Þ \ Va is smooth at a in the usual sense, G \ðInt Hn Þ \ Va is an open neighborhood of a in Rn ; G \ðInt Hn Þ is an open neighborhood of a in Rn ; and G \ðInt Hn Þ \ Va is contained in G \ðInt Hn Þ; f jG \ðIntHn Þ is smooth at a in the usual sense. Thus, we have shown that if G is open in Hn ; and f : G ! Rm is smooth, then f jG \ðIntHn Þ is smooth in the usual sense. Note 4.66 Let G be a nonempty subset of H3 : Let G be open in H3 : Let f : G ! Rm be smooth. We want to prove that f : G ! Rm is continuous. For this purpose, let us take any a in G: Now, let us take any open neighborhood Vf ðaÞ of f ðaÞ in Rm : We have to ﬁnd an open neighborhood Ua of a in R3 such that 1. Ua \ H3 G; 2. f ðUa \ H3 Þ Vf ðaÞ : Since G is open in H3 ; there exists an open subset W1 of R3 such that G ¼ W1 \ H3 : Since a is in Gð¼ W1 \ H3 Þ; a is in W1 : Since a is in W1 ; and W1 is open in R3 ; W1 is an open neighborhood of a in R3 : Since f : G ! Rm is smooth, and a is in G; there exist an open neighborhood Va of a in Rn ; and a function fa : Va ! Rm such that fa jG \ Va ¼ f jG \ Va ; and fa is smooth in the usual sense. Since Va is an open neighborhood of a in Rn ; and fa : Va ! Rm is smooth in the usual sense, fa : Va ! Rm is continuous. Since fa : Va ! Rm is continuous, and a is in Va ; fa : Va ! Rm is continuous at a:

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Since Va is an open neighborhood of a in R3 ; and W1 is an open neighborhood of a in R3 ; Va \ W1 is an open neighborhood of a in R3 : Since a 2 G; and a 2 Va ; a 2 G \ Va : Since a 2 G \ Va ; and fa jG \ Va ¼ f jG \ Va ; fa ðaÞ ¼ f ðaÞ: Since Vf ðaÞ is an open neighborhood of f ðaÞ; and fa ðaÞ ¼ f ðaÞ; Vf ðaÞ is an open neighborhood of fa ðaÞ: Since fa : Va ! Rm is continuous at a; Va is an open neighborhood of a in R3 ; Va \ W1 is an open neighborhood of a in R3 ; and Vf ðaÞ is an open neighborhood of fa ðaÞ in Rm ; there exists an open neighborhood Ua of a in R3 ; such that Ua Va \ W1 ; and fa ðUa Þ Vf ðaÞ : For 1: Take any x in Ua \ H3 : Since x is in Ua \ H3 ; x is in Ua Va \ W1 ; and hence, x is in W1 : Since x is in Ua \ H3 ; x is in H3 : Since x is in H3 ; and x is in W1 ; x is in W1 \ H3 ð¼ GÞ; and hence, x is in G: Thus, Ua \ H3 G: For 2: Since fa jG \ Va ¼ f jG \ Va ; and G ¼ W1 \ H3 ; fa jW1 \ H3 \ Va ¼ f jW1 \ H3 \ Va : Since Ua Va \ W1 ; Ua \ H3 Va \ W1 \ H3 ¼ W1 \ H3 \ Va : Since Ua \ H3 W1 \ H3 \ Va ; and fa jW1 \ H3 \ Va ¼ f jW1 \ H3 \ Va ; f ðUa \ H3 Þ ¼ fa ðUa \ H3 Þ fa ðUa Þ Vf ðaÞ : Thus f ðUa \ H3 Þ Vf ðaÞ :

4.7 Smooth Covering Maps e and M be connected smooth manifolds. Let p : M e ! M be a Deﬁnition Let M smooth map. If e onto M, 1. p maps M 2. for every p in M; there exists a connected open neighborhood U of p such that for each component C of the open set p1 ðUÞ; the restriction pjC : C ! U is a diffeomorphism, then we say that p is a smooth covering map, M is a base of the e is a covering manifold of M: covering, and M e is a locally path connected, and Observe that, in the condition (2), since M 1 e p ðUÞ is an open subset of M each component of p1 ðUÞ is open. Lemma 4.67 Every smooth covering map is a topological covering map. e ! M be a smooth covering map, where M e andM are connected Proof Let p : M smooth manifolds. Since M is a smooth manifold, M is a topological manifold, and e ! M is a smooth covering map, hence, M is locally path connected. Since p : M e e p : M ! M is a continuous map. Since M is a connected topological space, M is a e ! M is a continuous map, it is locally path connected topological space, and p : M meaningful to talk about “p is a topological covering map,” that is,

4.7 Smooth Covering Maps

305

e onto M; 1. p maps M 2. for every p in M; there exists a connected open neighborhood U of p such that for each component C of the open set p1 ðUÞ; the restriction pjC : C ! U is a homeomorphism, e ! M is a smooth covering map, p maps M e onto M: For 1: Since p : M e For 2: Let us take any p in M: Since p : M ! M is a smooth covering map, there exists a connected open neighborhood U of p such that for each component C of the open set p1 ðUÞ; the restriction pjC : C ! U is a diffeomorphism, and hence, pjC : C ! U is a homeomorphism. h e and M be connected smooth manifolds. Let p : M e ! M be a Lemma 4.68 Let M e smooth covering map. Then, p is a local diffeomorphism, that is, for every p in M; there exists an open neighborhood C of p such that pðCÞ is open in M, and the restriction pjC : C ! pðCÞ is a diffeomorphism. e : We have to ﬁnd an open neighborhood C of Proof Let us take an element p in M p such that pðCÞ is open in M, and the restriction pjC : C ! pðCÞ is a diffeomorphism. e ! M; and p is in M; e pðpÞ is in M: Since pðpÞ is in M; and Since p : M e p : M ! M is a smooth covering map, there exists a connected open neighborhood U of pðpÞ such that for each component C of the open set p1 ðUÞ; the restriction pjC : C ! U is a diffeomorphism. Since pðpÞ is in U; p 2 p1 ðUÞ: Since p 2 p1 ðUÞ; and p1 ðUÞ is partitioned into components, there exists a component C of p1 ðUÞ such that p 2 C: Since e ! M is a smooth map, p : M e ! M is a continuous map. Since p : M e !M p:M 1 is a continuous map, and U is an open neighborhood of pðpÞ; p ðUÞ is an open e is a smooth manifold, M e is a topological manifold, and neighborhood of p: Since M e e hence M is locally path connected. Since M is locally path connected, and p1 ðUÞ e and C is a component of p1 ðUÞ; C is open. Since C is is an open subset of M e open, and p 2 C; C is an open neighborhood of p in M. Since pjC : C ! U is a diffeomorphism, pðCÞ ¼ ðpjC ÞðCÞ ¼ U: Since pðCÞ ¼ U; and U is open, pðCÞ is open in M. Since pjC : C ! U is a diffeomorphism, p is a local diffeomorphism. h

Chapter 5

Immersions, Submersions, and Embeddings

The counterpart of “homeomorphism in topological spaces” and “linear isomorphism in real linear spaces” are the concepts of immersion, submersion, and embedding in smooth manifolds. Because a smooth manifold associates with a topological space together with a collection of linear spaces (i.e., tangent spaces at various points of manifold), there emerges different types of “homomorphisms” among smooth manifolds. Their deﬁnitions and various theorems on relationship between immersion, submersion, and embedding constitute a beautiful area of studies. In this chapter, we shall prove some of the important theorems on immersion, submersion, embedding, and its related notions. Deﬁnitely, the pace of this chapter is a bit slower for obvious reason.

5.1 Pointwise Pushforward Deﬁnition Let a 2 R3 : By R3a we mean the Cartesian product fag R3 : Members of R3a are called geometric tangent vectors in R3 at a, and R3a is called the geometric tangent space to R3 at a. A geometric tangent vector ða; vÞ in R3 at a is denoted by va or vja : We deﬁne vector addition and scalar multiplication over R3a as follows: va þ wa ðv þ wÞa ;

tðva Þ ðtvÞa :

Clearly, R3a becomes a real linear space. Here, the zero vector is ð0; 0; 0Þa ; and the negative vector of va is ðvÞa : It is easy to see that the mapping v 7! va is an isomorphism from R3 onto R3a ; and hence, R3 and R3a are isomorphic. Note 5.1 Since members of a smooth manifold M may not be added, the above deﬁnition of tangent space cannot be generalized. Observe that C1 ðR3 Þ is a real linear space under pointwise vector addition and scalar multiplication. Also, corresponding to each geometric tangent vector va in R3 at a, we can deﬁne a function Dv ja : C1 ðR3 Þ ! R as follows: For every f 2 C1 ðR3 Þ; © Springer India 2014 R. Sinha, Smooth Manifolds, DOI 10.1007/978-81-322-2104-3_5

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Dv ja ðf Þ ðDv Þðf ðaÞÞ ¼

d dt

f ða þ tvÞ f ðaÞ : ðf ða þ tvÞÞ ¼ lim t!0 t t¼0

We know that Dv ja is linear and satisﬁes the Leibnitz rule:

Dv ja ðf gÞ ¼ Dv ja ðf Þ ðgðaÞÞ þ ðf ðaÞÞ Dv ja ðgÞ :

Also,

Dðv1 ;v2 ;v3 Þ a ðf Þ ¼ v1 Dð1;0;0Þ a ðf Þ þ v2 Dð0;1;0Þ a ðf Þ þ v3 Dð0;0;1Þ a ðf Þ f ða þ tð1; 0; 0ÞÞ f ðaÞ f ða þ tð0; 1; 0ÞÞ f ðaÞ þ v2 lim t!0 t t f ð a þ t ð 0; 0; 1 Þ Þ f ðaÞ þ v3 lim t!0 t 1 2 ¼ v ððD1 f ÞðaÞÞ þ v ððD2 f ÞðaÞÞ of 2 of 3 of ðaÞ þ v ðaÞ þ v ðaÞ þ v3 ððD3 f ÞðaÞÞ ¼ v1 ox1 ox2 ox3 3 X of of vi ðaÞ ¼ vi ðaÞ : ¼ i i ox ox i¼1 ¼ v1 lim t!0

Thus,

i of Dðv1 ;v2 ;v3 Þ a ðf Þ ¼ v ðaÞ : oxi

It follows that of of Dð1;0;0Þ a ðf Þ ¼ 1 ðaÞ; Dð0;1;0Þ a ðf Þ ¼ 2 ðaÞ; ox ox of and Dð0;0;1Þ a ðf Þ ¼ 3 ðaÞ: ox Deﬁnition Let a 2 R3 : Let w : C 1 ðR3 Þ ! R be a linear function. If w satisﬁes the Leibnitz rule: wðf gÞ ¼ ðwðf ÞÞðgðaÞÞ þ ðf ðaÞÞðwðgÞÞ for every f, g in C 1 ðR3 Þ; then we say that w is a derivation at a. Let us denote the collection of all derivations at a by Ta R3 : Clearly, Ta R3 is a real linear space under pointwise vector addition and scalar multiplication. It is clear that if va 2 R3a ; then the mapping Dv ja : C1 ðR3 Þ ! R; deﬁned as above, is a derivation at a. Thus, for every va 2 R3a ; Dv ja 2 Ta R3 :

5.1 Pointwise Pushforward

309

Note 5.2 Let a 2 R3 : Let w 2 Ta R3 : Let f ; g 2 C 1 ðR3 Þ: We shall try to show: 1. If f is a constant function, then wðf Þ ¼ 0; 2. If f ðaÞ ¼ gðaÞ ¼ 0, then wðf gÞ ¼ 0: For 1: Here, f is a constant function, so there exists a real number c such that f ðxÞ ¼ c for every x in R3 : Let us deﬁne a function f1 : R3 ! R as follows: For every x in R3 ; f1 ðxÞ ¼ 1: Clearly, f1 2 C 1 ðR3 Þ; and f ¼ cf1 : Now, wðf1 Þ ¼ w ðf1 f1 Þ ¼ ðwðf1 ÞÞ ðf1 ðaÞÞ þ ðf1 ðaÞÞ ðwðf1 ÞÞ ¼ ðwðf1 ÞÞ 1 þ 1ðwðf1 ÞÞ ¼ 2ðwðf1 ÞÞ; and hence, wðf1 Þ ¼ 0: Next, wðf Þ ¼ wðcf1 Þ ¼ cðwðf1 ÞÞ ¼ c 0 ¼ 0: This proves 1. For 2: LHS ¼ wðf gÞ ¼ ðwðf ÞÞðgðaÞÞ þ ðf ðaÞÞ ðwðgÞÞ ¼ ðwðf ÞÞ 0 þ 0 ðwðgÞÞ ¼ 0 ¼ RHS: Note 5.3 Let us ﬁx any a ða1 ; a2 ; a3 Þ 2 R3 : Let g : R3a ! Ta R3 be the mapping deﬁned as follows: For every va 2 R3a ; gðva Þ Dv ja : Then, g is a linear isomorphism from real linear space R3a onto real linear space Ta R3 : Reason: Let va ; wa 2 R3a ; where v ðv1 ; v2 ; v3 Þ; and w ðw1 ; w2 ; w3 Þ: Let s; t 2 R: We have to prove that gðsðva Þ þ tðwa ÞÞ ¼ sðgðva ÞÞ þ tðgðwa ÞÞ; that is, for every f in C 1 ðR3 Þ; ðgðsðva Þ þ tðwa ÞÞÞ ðf Þ ¼ ðsðgðva ÞÞ þ tðgðwa ÞÞðf Þ. LHS ¼ ðgðsðva Þ þ tðwa ÞÞÞðf Þ ¼ g ðsvÞa þðtwÞa ðf Þ ¼ g ðsv þ twÞa ðf Þ ¼ DðsvþtwÞ a ðf Þ of ðaÞ ¼ Dðsv1 þtw1 ;sv2 þtw2 ;sv3 þtw3 Þ a ðf Þ ¼ svi þ twi oxi of of ¼ s vi ðaÞ þ t wi ðaÞ ¼ s Dv ja ðf Þ þ t Dw ja ðf Þ i i ox ox ¼ sððgðva ÞÞðf ÞÞ þ tððgðwa ÞÞðf ÞÞ ¼ ðsðgðva ÞÞ þ tðgðwa ÞÞÞðf Þ ¼ RHS: This proves that g : R3a ! Ta R3 is linear. Now, we want to prove that g is 1–1. For this purpose, let gððv1 ; v2 ; v3 Þa Þ ¼ gððw1 ; w2 ; w3 Þa Þ: We have to prove that ðv1 ; v2 ; v3 Þ ¼ ðw1 ; w2 ; w3 Þ: Let us deﬁne a function p1 : R3 ! R; as follows: For every ðx1 ; x2 ; x3 Þ in R3 ; p1 ðx1 ; x2 ; x3 Þ x1 : Clearly, p1 2 C 1 ðR3 Þ: Since Dðv1 ;v2 ;v3 Þ a ¼ g v1 ; v2 ; v3 a ¼ g w1 ; w2 ; w3 a ¼ Dðw1 ;w2 ;w3 Þ a ; and p1 2 C 1 ðR3 Þ;

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of 2 of 3 of ðaÞ þ v ðaÞ þ v ðaÞ ox1 ox2 ox3 of ¼ Dðv1 ;v2 ;v3 Þ a ðp1 Þ ¼ Dðw1 ;w2 ;w3 Þ a ðp1 Þ ¼ w1 ðaÞ ox1 of of ðaÞ þ w3 ðaÞ þ w2 ox2 ox3

v1 ¼ v1 ð1Þ þ v2 ð0Þ þ v3 ð0Þ ¼ v1

¼ w1 ð1Þ þ w2 ð0Þ þ w3 ð0Þ ¼ w1 ;

and hence, v1 ¼ w1 : Similarly, v2 ¼ w2 ; v3 ¼ w3 : Hence, ðv1 ; v2 ; v3 Þ ¼ ðw1 ; w2 ; w3 Þ: Now, it remains to be proved that g : R3a ! Ta R3 is onto. For this purpose, let us take any w 2 Ta R3 : We have to ﬁnd an element va 2 R3a such that gðva Þ ¼ w; that is, Dv ja ¼ w; that is, for every f 2 C 1 ðR3 Þ; ðDv ja Þðf Þ ¼ wðf Þ: For every i ¼ 1; 2; 3, let us deﬁne a function pi : R3 ! R; as follows: For every ðx1 ; x2 ; x3 Þ in R3 ; pi ðx1 ; x2 ; x3 Þ xi : Clearly, each pi 2 C1 ðR3 Þ: Since w 2 Ta R3 ; w : C1 ðR3 Þ ! R: Now, since pi 2 C 1 ðR3 Þ; wðpi Þ is a real number. It follows that ðwðp1 Þ; wðp2 Þ; wðp3 ÞÞ 2 R3 : Let us 3 1 take ðwðp1 Þ; wðp2 Þ; wðp3 ÞÞ for v: It remains to be proved1that 3for every f 2 C ðR Þ; ðDðwðp1 Þ;wðp2 Þ;wðp3 ÞÞ a Þðf Þ ¼ wðf Þ; that is, for every f 2 C ðR Þ; ðwðpi ÞÞð

of ðaÞÞ ¼ wðf Þ: oxi

Since f 2 C 1 ðR3 Þ; f : R3 ! R; and is smooth. Now, by Theorem 3.62, there exist C 1 functions g1 ðxÞ; g2 ðxÞ; g3 ðxÞ on R3 such that 1. for every x ðx1 ; x2 ; x3 Þ in R3 ; f ðxÞ ¼ f ðaÞ þ ðg1 ðxÞÞðx1 a1 Þ þ ðg2 ðxÞÞ ðx2 a2 Þ þ ðg3 ðxÞÞðx3 a3 Þ; that is, f ¼ f ðaÞ þ g1 ðp1 a1 Þ þ g2 ðp2 a1 Þ þg3 ðp3 a1 Þ; of 2. g1 ðaÞ ¼ oxof1 ðaÞ; g2 ðaÞ ¼ oxof2 ðaÞ; g3 ðaÞ ¼ ox 3 ðaÞ: RHS ¼ wðf Þ ¼ w f ðaÞ þ g1 p1 a1 þ g2 p2 a2 þ g3 p3 a3 ¼ wðf ðaÞÞ þ ðwðg1 ÞÞ p1 a1 ðaÞ þ w p1 a1 ðg1 ðaÞÞ þ ðwðg2 ÞÞ p2 a2 ðaÞ þ w p2 a2 ðg2 ðaÞÞ þ ðwðg3 ÞÞ p3 a3 ðaÞ þ w p3 a3 ðg1 ðaÞÞ ¼ 0 þ ðwðg1 ÞÞ a1 a1 þ ðwðp1 Þ 0Þðg1 ðaÞÞ þ ðwðg2 ÞÞ a2 a2 þ ðwðp2 Þ 0Þðg2 ðaÞÞ þ ðwðg3 ÞÞ a3 a3 þ ðwðp3 Þ 0Þðg3 ðaÞÞ ¼ ðwðp1 ÞÞðg1 ðaÞÞ þ ðwðp2 ÞÞðg2 ðaÞÞ þ ðwðp3 ÞÞðg3 ðaÞÞ of ¼ ðwðpi ÞÞðgi ðaÞÞ ¼ ðwðpi ÞÞ ðaÞ oxi ¼ LHS:

5.1 Pointwise Pushforward

311

Thus, we have shown that g is a linear isomorphism from real linear space R3a onto real linear space Ta R3 : Hence, R3a is isomorphic to Ta R3 : Since R3a is a linear space of dimension 3, and R3a is isomorphic to Ta R3 ; Ta R3 is of dimension 3. Further, since fð1; 0; 0Þa ; ð0; 1; 0Þa ; ð0; 0; 1Þa g is a basis of R3a ; and g is a linear isomorphism from R3a onto Ta R3 ; fgðð1; 0; 0Þa Þ; gðð0; 1; 0Þa Þ; gðð0; 0; 1Þa Þg is a basis of Ta R3 ; that is, n o Dð1;0;0Þ a ; Dð0;1;0Þ a ; Dð0;0;1Þ a is a basis of Ta R3 ; that is,

o o o ; ; ox1 a ox2 a ox3 a

is a basis of Ta R3 : Similarly,

o o o ; ; . . .; n ox1 a ox2 a ox a

is a basis of Ta Rn : Note 5.4 Since R3a is the geometric tangent space to R3 at a, and R3a is isomorphic to Ta R3 ; Ta R3 can be thought of as the geometric tangent space to R3 at a. Similarly, for every a in Rk ; Rka is isomorphic to Ta Rk . The interesting thing is that Ta R3 ; but not R3a ; can be generalized on a smooth manifold. Deﬁnition Let M be an m-dimensional smooth manifold. Let p 2 M: Let v : C 1 ðMÞ ! R be a linear function. If v satisﬁes the Leibnitz rule: vðf gÞ ¼ ðvðf ÞÞðgðpÞÞ þ ðf ðpÞÞðvðgÞÞ; for every f, g in C 1 ðMÞ; then we say that v is a derivation at p. The collection of all derivations at p is denoted by Tp M: Clearly, Tp M is a real linear space under pointwise vector addition and scalar multiplication. Here, the real linear space Tp M is called the tangent space to M at p and the members of Tp M are called tangent vectors at p. Note 5.5 Let M be an m-dimensional smooth manifold. Let p 2 M: Let v 2 Tp M: Let f ; g 2 C1 ðMÞ: As above, it is easy to show: 1. If f is a constant function, then vðf Þ ¼ 0: 2. If f ðpÞ ¼ gðpÞ ¼ 0, then vðf gÞ ¼ 0: Note 5.6 Let M be an m-dimensional smooth manifold and N be an n-dimensional smooth manifold. Let F : M ! N be a smooth map. Let v 2 Tp M: Let f 2 C1 ðNÞ:

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Since f 2 C 1 ðNÞ; f : N ! R is a smooth map. Since F : M ! N is a smooth map, and f : N ! R is a smooth map, their composite f F : M ! R is a smooth map, and hence, f F 2 C1 ðMÞ: Since v 2 Tp M; v : C 1 ðMÞ ! R: Since v : C 1 ðMÞ ! R; and f F 2 C1 ðMÞ; vðf FÞ is a real number. Let us deﬁne the mapping ððdFp ÞðvÞÞ : C 1 ðNÞ ! R as follows: For every f 2 C 1 ðNÞ;

dFp ðvÞ ðf Þ vðf F Þ:

We shall try to show that ððdFp ÞðvÞÞ 2 TFðpÞ N; that is, ðdFp ÞðvÞ is a derivation at FðpÞ: Observe that ðdFp ÞðvÞ : C1 ðNÞ ! R is linear. (Reason: Let f ; g 2 C 1 ðNÞ; and let s, t be any real. We must prove: dFp ðvÞ ðsf þ tgÞ ¼ s dFp ðvÞ ð f Þ þ t dFp ðvÞ ðgÞ : LHS ¼ ððdFp ÞðvÞÞðsf þ tgÞ ¼ vððsf þ tgÞ FÞ ¼ vðsðf FÞ þ tðg FÞÞ ¼ sðvðf FÞÞþ

tðvðg FÞÞ ¼ sðððdFp ÞðvÞÞðf ÞÞ þ tðððdFp ÞðvÞÞðgÞÞ ¼ RHS:Þ Now, we shall try to prove that ðdFp ÞðvÞ : C 1 ðNÞ ! R satisﬁes the Leibnitz rule:

dFp ðvÞ ðf gÞ ¼ dFp ðvÞ ðf Þ ðgðFðpÞÞÞ þ ðf ðFðpÞÞÞ dFp ðvÞ ðgÞ

for every f ; g in C 1 ðNÞ: LHS ¼

dFp ðvÞ ðf gÞ ¼ vððf gÞ F Þ ¼ vððf F Þ ðg F ÞÞ

¼ ðvðf F ÞÞððg F ÞðpÞÞ þ ððf F ÞðpÞÞðvðg F ÞÞ ¼ dFp ðvÞ ðf Þ ððg F ÞðpÞÞ þ ððf F ÞðpÞÞ dFp ðvÞ ðgÞ ¼ dFp ðvÞ ðf Þ ðgðFðpÞÞÞ þ ðf ðFðpÞÞÞ dFp ðvÞ ðgÞ ¼ RHS: Thus, we have shown that ðdFp ÞðvÞ 2 TFðpÞ N: Deﬁnition Let M be an m-dimensional smooth manifold, and N be an n-dimensional smooth manifold. Let F : M ! N be a smooth map. If v 2 Tp M; then ðdFp ÞðvÞ 2 TFðpÞ N: Thus, dFp : Tp M ! TFðpÞ N: Here, dFp is called the differential of F at p (or tangent map, total derivative, or derivative of F, or pointwise pushforward of F). dFp is also denoted by F 0 ðpÞ; DF; F (see Fig. 5.1). Note 5.7 Let M be an m-dimensional smooth manifold, and N be an n-dimensional smooth manifold. Let F : M ! N be a smooth map. We shall try to show that dFp : Tp M ! TFðpÞ N is a linear mapping from tangent space Tp M to tangent space TFðpÞ N: Let us take any v1 ; v2 in Tp M; and any real s; t: We have to show that

dFp ðsv1 þ tv2 Þ ¼ s dFp ðv1 Þ þ t dFp ðv2 Þ ;

that is, for every f in C1 ðNÞ;

5.1 Pointwise Pushforward

313

Fig. 5.1 Tangent map

dFp ðsv1 þ tv2 Þ ðf Þ ¼ s dFp ðv1 Þ þ t dFp ðv2 Þ ðf Þ: LHS ¼ dFp ðsv1 þ tv2 Þ ðf Þ ¼ ðsv1 þ tv2 Þðf F Þ ¼ s ð v1 ð f F Þ Þ þ t ð v2 ð f F Þ Þ ¼ s dFp ðv1 Þ ðf Þ þ t dFp ðv2 Þ ðf Þ ¼ s dFp ðv1 Þ ðf Þ þ t dFp ðv2 Þ ðf Þ ¼ s dFp ðv1 Þ þ t dFp ðv2 Þ ðf Þ ¼ RHS:

Note 5.8 Let M; N; P be smooth manifolds. Let F : M ! N; and G : N ! P be smooth maps. Let p 2 M: It follows that the composite map G F : M ! P is a smooth map. Here, dFp : Tp M ! TFðpÞ N; dGFðpÞ : TFðpÞ N ! TGðFðpÞÞ P; and dðG FÞp : Tp M ! TðGFÞðpÞ P: Since dFp : Tp M ! TFðpÞ N; and dGFðpÞ : TFðpÞ N ! TGðFðpÞÞ P; ðdGFðpÞ Þ ðdFp Þ : Tp M ! TGðFðpÞÞ P; that is, ðdGFðpÞ Þ ðdFp Þ : Tp M ! TðGFÞðpÞ P: Thus, dðG FÞp : Tp M ! TðGFÞðpÞ P; and ðdGFðpÞ Þ ðdFp Þ : Tp M ! TðGFÞðpÞ P: We shall try to prove: dðG FÞp ¼ ðdGFðpÞ Þ ðdFp Þ; that is, for every v in Tp M; ðdðG FÞp ÞðvÞ ¼ ððdGFðpÞ Þ ðdFp ÞÞðvÞ: Further, since ðdðG FÞp ÞðvÞ 2 TðGFÞðpÞ P; ðd ðG FÞp ÞðvÞ : C1 ðPÞ ! R; and hence, we must prove: For every f in C 1 ðPÞ; dðG F Þp ðvÞ ðf Þ ¼ dGFðpÞ dFp ðvÞ ðf Þ: LHS ¼ dðG F Þp ðvÞ ðf Þ ¼ vðf ðG F ÞÞ ¼ vððf GÞ F Þ ¼ dFp ðvÞ ðf GÞ ¼ dGFðpÞ dFp ðvÞ ðf Þ ¼ dGFðpÞ dFp ðvÞ ðf Þ ¼ RHS: Thus, we have shown that dðG FÞp ¼ ðdGFðpÞ Þ ðdFp Þ:

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Note 5.9 Let M be a smooth manifold. Let p 2 M: Here, IdM : M ! M is given by IdM ðxÞ ¼ x for every x in M: Clearly, IdM is a smooth function. Also, dðIdM Þp : Tp M ! Tp M; and IdTp M : Tp M ! Tp M: We shall try to prove: dðIdM Þp ¼ IdTp M ; that is, for every v in Tp M; ðdðIdM Þp ÞðvÞ ¼ ðIdTp M ÞðvÞ; that is, for every v in Tp M; ðdðIdM Þp ÞðvÞ ¼ v: Further, since v is in Tp M; v : C 1 ðMÞ ! R; and hence, we must prove: For every f in C 1 ðMÞ; ððdðIdM Þp ÞðvÞÞðf Þ ¼ vðf Þ: LHS ¼

dðIdM Þp ðvÞ ðf Þ ¼ vðf ðIdM ÞÞ ¼ vðf Þ ¼ RHS:

Thus, we have shown that dðIdM Þp ¼ IdTp M :

5.2 Open Submanifolds Note 5.10 Let M; N be smooth manifolds. Let F : M ! N be a diffeomorphism. Let p 2 M: Since F : M ! N is a diffeomorphism, F : M ! N is a smooth map. We shall try to show: dFp : Tp M ! TFðpÞ N is an isomorphism. Since dFp is linear, it sufﬁces to show that dFp is 1–1 and onto. dFp is 1–1: Let ðdFp ÞðvÞ ¼ ðdFp ÞðwÞ; where v; w are in Tp M: We have to prove that v ¼ w: Since dFp : Tp M ! TFðpÞ N; and v is in Tp M; ðdFp ÞðvÞ is in TFðpÞ N; and hence, ðdFp ÞðvÞ : C1 ðNÞ ! R: Since v is in Tp M; v : C 1 ðMÞ ! R: Similarly, w : C 1 ðMÞ ! R: Now, we have to prove that v ¼ w; that is, for every f in C 1 ðMÞ; vðf Þ ¼ wðf Þ: For this purpose, let us take any f in C1 ðMÞ: Since f is in C1 ðMÞ; f : M ! R is smooth. Since F : M ! N is a diffeomorphism, F 1 : N ! M is smooth. Since F 1 : N ! M is smooth, and f : M ! R is smooth, their composite f ðF 1 Þ : N ! R is smooth, and hence, f ðF 1 Þ 2 C1 ðNÞ: Now, since ðdFp ÞðvÞ ¼ ðdFp ÞðwÞ; for every g in C 1 ðNÞ; vðg FÞ ¼ ððdFp ÞðvÞÞðgÞ ¼ ððdFp ÞðwÞÞðgÞ ¼ wðg FÞ: Since for every g in C1 ðNÞ; vðg FÞ ¼ wðg FÞ; and f ðF 1 Þ 2 C 1 ðNÞ; LHS ¼ vðf Þ ¼ vðf IdM Þ ¼ vðf ðF 1 FÞÞ ¼ vððf ðF 1 ÞÞ FÞ ¼ wððf ðF 1 ÞÞ FÞ ¼ wðf ðF 1 FÞÞ ¼ wðf IdM Þ ¼ wðf Þ ¼ RHS: Thus, we have shown that if F : M ! N is a diffeomorphism, then dFp : Tp M ! TFðpÞ N is an isomorphism. Note 5.11 Let M; N be smooth manifolds. Let F : M ! N be a diffeomorphism. Since F : M ! N is a diffeomorphism, F 1 : N ! M is a smooth map, and hence, dðF 1 ÞFðpÞ : TFðpÞ N ! Tp M is an isomorphism. Since dFp : Tp M ! TFðpÞ N is an isomorphism, ðdFp Þ1 : TFðpÞ N ! Tp M: Thus

5.2 Open Submanifolds

d F 1 FðpÞ : TFðpÞ N ! Tp M;

315

and

dFp

1

: TFðpÞ N ! Tp M:

Now, we want to prove that dðF 1 ÞFðpÞ ¼ ðdFp Þ1 ; that is, ðdðF 1 ÞFðpÞ Þ ðdFp Þ ¼ IdTp M : LHS ¼ d F 1 FðpÞ dFp ¼ d F 1 F p ¼ dðIdM Þp ¼ IdTp M ¼ RHS: Thus, we have shown that if F : M ! N is a diffeomorphism, then dðF 1 ÞFðpÞ

¼ ðdFp Þ1 :

Theorem 5.12 Let M be a smooth manifold. Let p 2 M; v 2 Tp M: Let f ; g 2 C 1 ðMÞ: If there exists an open neighborhood U of p such that f jU ¼ gjU ; then vðf Þ ¼ vðgÞ: In short, we say that tangent vectors act “locally.” Proof Let U be an open neighborhood of p such that f jU ¼ gjU : We want to show that vðf Þ ¼ vðgÞ: Let us put h f g: Since f ; g 2 C1 ðMÞ; h ¼ f g 2 C 1 ðMÞ; and hence, h 2 C1 ðMÞ: Since f jU ¼ gjU ; hjU ¼ 0: Clearly, p 62 supp h: (Reason: If not, otherwise, let p 2 supp h: We have to arrive at a contradiction. Since p 2 supp h ¼ fx : hðxÞ 6¼ 0g ; and U is an open neighborhood of p; there exists t 2 M such that hðtÞ 6¼ 0; and t 2 U: Since t 2 U; and hjU ¼ 0; hðtÞ ¼ 0; a contradiction.) Since p 62 supp h; hðpÞ ¼ 0: Here, M is a smooth manifold, so M is Hausdorff. Since M is Hausdorff, and p 2 M; fpg is closed, and hence, M fpg is open: Since p 62 supp h; supp h M fpg: Here, supp h M fpg; supp h is a closed subset of M; M fpg is an open subset of M, so, by Lemma 4.58, there exists a smooth function w : M ! ½0; 1 such that w is a bump function for supp h supported in M fpg; and hence, 1. for every x in supp h; wðxÞ ¼ 1; 2. supp w M fpg: Since supp w M fpg; p 62 supp w; and hence, wðpÞ ¼ 0: Since w : M ! ½0; 1 is smooth, w 2 C 1 ðMÞ: Since w 2 C 1 ðMÞ; and h 2 C 1 ðMÞ; their product ðw hÞ 2 C 1 ðMÞ: Since v 2 Tp M; v : C 1 ðMÞ ! R: Now, vðw hÞ ¼ ðvðwÞÞ ðhðpÞÞ þ ðwðpÞÞ ðvðhÞÞ ¼ ðvðwÞÞð0Þ þ ð0ÞðvðhÞÞ ¼ 0: Thus, vðw hÞ ¼ 0: Clearly, w h ¼ h: (Reason: If x 2 supp h; then from condition 1, ðw hÞðxÞ ¼ ðwðxÞÞðhðxÞÞ ¼ 1ðhðxÞÞ ¼ hðxÞ: If x 62 supp h; then hðxÞ ¼ 0; and hence, ðw hÞðxÞ ¼ ðwðxÞÞðhðxÞÞ ¼ ðwðxÞÞ0 ¼ 0 ¼ hðxÞ: Thus, for all x in M; ðw hÞðxÞ ¼ hðxÞÞ. Now, 0 ¼ vðw hÞ ¼ vðhÞ ¼ vðf gÞ ¼ vðf Þ vðgÞ; so vðf Þ ¼ vðgÞ: h Theorem 5.13 Let M be an m-dimensional smooth manifold. Let p 2 M: Let U be an open neighborhood of p. We know that U is also a smooth manifold (called open submanifold U of M). Let i : U ! M be the mapping deﬁned as follows: For every x in U, iðxÞ x: (Here, we say that i is the inclusion map of U.) Then, the

316

5 Immersions, Submersions, and Embeddings

differential dip : Tp U ! TiðpÞ Mð¼ Tp MÞ is an isomorphism, that is, dip : Tp U ! Tp M is 1–1 and onto. Proof Since M is a smooth manifold, M is a topological manifold. Since M is a topological manifold, and U is an open neighborhood of p, there exists an open neighborhood V of p such that V U: We know that dip is linear. dip is 1–1: For this purpose, let ðdip ÞðvÞ ¼ 0 where v 2 Tp U: We have to prove that v ¼ 0: Since v 2 Tp U; v : C 1 ðUÞ ! R: We have to show that v ¼ 0; that is, for every f 2 C1 ðUÞ; vðf Þ ¼ 0: For this purpose, let us take any f 2 C 1 ðUÞ: Since f 2 C 1 ðUÞ; f : U ! R is smooth. Since f : U ! R is smooth, and V U, f jV is smooth. Here, V U; V is closed, U is open, and f jV is smooth, so, by Lemma 4.59, there exists a smooth function ~f : M ! Rm such that ~f V ¼ f jV : Since ~f V ¼ f jV ; and V V ; ~f V ¼ f jV : Since ~f : M ! Rm is smooth, ~f U is smooth on U; and hence, ~f U 2 C 1 ðUÞ: Since V V U; ð~f U Þ ¼ ~f V ¼ f jV ; V and hence, ð~f U Þ ¼ f jV : Since U is a smooth manifold, p 2 U; v 2 Tp U; ~f U ; f 2 V C 1 ðUÞ; V is an open neighborhood of p such that V U; and ð~f U Þ ¼ f jV , by V Theorem 5.12, vð~f U Þ ¼ vðf Þ: Now, LHS ¼ vðf Þ ¼ v ~f U ¼ v ~f i ¼ dip ðvÞ ~f ¼ 0 ~f ¼ 0 ¼ RHS: This proves that dip is 1–1. dip : Tp U ! Tp M is onto: For this purpose, let us take any w 2 Tp M: We have to ﬁnd v 2 Tp U such that ðdip ÞðvÞ ¼ w: Since w 2 Tp M; w : C 1 ðMÞ ! R: It follows that we must ﬁnd v 2 Tp U such that for every f 2 C 1 ðMÞ; ððdip ÞðvÞÞðf Þ ¼ wðf Þ; that is, for every f 2 C 1 ðMÞ; vðf iÞ ¼ wðf Þ: Now, let us take any g 2 C1 ðUÞ: Since g 2 C1 ðUÞ; g : U ! R is smooth. Since M is a topological manifold, and U is an open neighborhood of p; there exists an open neighborhood V of p such that V U: Since g : U ! R is smooth, and V U, gjV is smooth. Here, V U; V is closed, U is open, and gjV is smooth, by Lemma 4.59, there exists a smooth function g : M ! Rm such that g jV ¼ gjV : Next, let g : M ! Rm be another smooth function such that g jV ¼ gjV : Since g jV ¼ gjV ¼ g jV ; and V V ; g jV ¼ g jV : Since M is a smooth manifold, p 2 M; w 2 Tp M; g ; g 2 C1 ðMÞ; V is an open neighborhood of p; and g jV ¼ g jV , by Theorem 5.12, wðg Þ ¼ wðg Þ: Let us deﬁne a function v : C 1 ðUÞ ! R as follows: For every g 2 C1 ðUÞ; vðgÞ wðg Þ where g 2 C 1 ðMÞ; and g jV ¼ gjV : From the above discussion, we ﬁnd that v : C 1 ðUÞ ! R is well deﬁned. Now, we want to prove that v is linear. For this purpose, let us take any g1 ; g2 2 C1 ðUÞ; and s; t any real numbers. We have to prove that vðsg1 þ tg2 Þ ¼ sðvðg 1 ÞÞ þ tðvðg2 ÞÞ: Let g1 2 C1 ðMÞ such that g1 V ¼ g1 jV ; and g2 2 C 1 ðMÞ such that g2 V ¼ g2 jV : It follows that sðg1 Þ þ tðg2 Þ 2 C 1 ðMÞ: Since g1 V ¼ g1 jV ; and

5.2 Open Submanifolds

317

g2 V ¼ g2 jV ; ðsðg1 Þ þ tðg2 ÞÞV ¼ ðsg1 þ tg2 ÞjV : Since ðsg1 þ tg2 ÞjV ¼ ðsðg1 Þ þ

tðg2 ÞÞjV ; and sðg1 Þ þ tðg2 Þ 2 C1 ðMÞ;

LHS ¼ vðsg1 þ tg2 Þ ¼ w s g1 þ t g2 ¼ s w g1 þ t w g2

¼ sðvðg1 ÞÞ þ tðvðg2 ÞÞ ¼ RHS: Now, we want to prove the Leibnitz rule: For every f ; g 2 C 1 ðUÞ; vðf gÞ ¼ ðvðf ÞÞðgðpÞÞ þ ðf ðpÞÞðvðgÞÞ: Let f 2 C 1 ðMÞ such that f jV ¼ f jV ; and g 2 C 1 ðMÞ such that g jV ¼ gjV : It follows that ðf Þ ðg Þ 2 C 1 ðMÞ: Since f jV ¼ f jV ; and g jV ¼ gjV ; ððf Þ ðg ÞÞjV ¼ ðf gÞjV : Since f jV ¼ f jV ; and p 2 V V ; f ðpÞ ¼ f ðpÞ. Similarly, g ðpÞ ¼ gðpÞ: Now, LHS ¼ vðf gÞ ¼ wððf Þ ðg ÞÞ ¼ ðwðf ÞÞðg ðpÞÞ þ ðf ðpÞÞðwðg ÞÞ ¼ ðvðf ÞÞðg ðpÞÞ þ ðf ðpÞÞðvðgÞÞ ¼ ðvðf ÞÞðgðpÞÞ þ ðf ðpÞÞðvðgÞÞ ¼ RHS: Thus, we have shown that v : C 1 ðUÞ ! R is a derivation at p; and hence, v 2 Tp U. Next, let us take any f 2 C 1 ðMÞ: It remains to be proved that vðf iÞ ¼ wðf Þ: Since i : U ! M is smooth, and f : M ! R is smooth, f i : U ! R is smooth, and hence, f i 2 C 1 ðUÞ: Now, since V U; f jV ¼ ðf iÞjV : Since f 2 C 1 ðMÞ; f i 2 C 1 ðUÞ; and f jV ¼ ðf iÞjV , by the deﬁnition of v; vðf iÞ ¼ wðf Þ: h Theorem 5.14 Let M be an m-dimensional smooth manifold. Let p 2 M: Then, the real linear space Tp M is of dimension m. Here, M is an m-dimensional smooth manifold, and p 2 M; so there exists an admissible coordinate chart ðU; uÞ of M such that p 2 U: Since ðU; uÞ is an admissible coordinate chart of M, there exists an open neighborhood G of uðpÞ in Rm such that u : U ! G is a diffeomorphism. Since u : U ! G is a diffeomorphism, F : M ! N is an isomorphism, and hence, Tp U is isomorphic onto TuðpÞ G: Since M is an m-dimensional smooth manifold, p 2 M; and U is an open neighborhood of p, by Theorem 5.13, the differential dip : Tp U ! TiðpÞ Mð¼ Tp MÞ is an isomorphism, where i denotes the inclusion map of U. Since dip : Tp U ! Tp M is an isomorphism, Tp U is isomorphic onto Tp M: Since Rm is an m-dimensional smooth manifold, uðpÞ 2 Rm ; and G is an open neighborhood of uðpÞ, by Theorem 5.13, the differential d^iuðpÞ : TuðpÞ G ! T^iðuðpÞÞ Rm ð¼ TuðpÞ Rm Þ is an isomorphism, where ^i denotes the inclusion map of G. Since d^iuðpÞ : TuðpÞ G ! TuðpÞ Rm is an isomorphism, TuðpÞ G is isomorphic onto TuðpÞ Rm : m We have seen that Rm uðpÞ is isomorphic to TuðpÞ R : Since Tp U is isomorphic onto TuðpÞ G; Tp U is linear isomorphic onto Tp M; TuðpÞ G is isomorphic onto TuðpÞ Rm ; and

318

5 Immersions, Submersions, and Embeddings

m m Rm uðpÞ is isomorphic to TuðpÞ R ; Tp M is isomorphic to RuðpÞ ; and hence, dim Tp M ¼ m m dim Rm uðpÞ : Since dim Tp M ¼ dim RuðpÞ ; and dim RuðpÞ ¼ m; dim Tp M ¼ m:

Note 5.15 Let M be a two-dimensional smooth manifold with differentiable structure A; and let N be a three-dimensional smooth manifold with differentiable structure B: Hence, the product manifold M N is a ð2 þ 3Þ-dimensional smooth manifold. Let ðp; qÞ 2 M N: Since ðp; qÞ 2 M N; p 2 M; and q 2 N: Since p 2 M; and M is a twodimensional smooth manifold, the tangent space Tp M is a two-dimensional real linear space. Similarly, Tq N is a three-dimensional real linear space, and the tangent space Tðp;qÞ ðM NÞ is a ð2 þ 3Þ-dimensional real linear space. Since Tp M is a twodimensional real linear space, and Tq N is a three-dimensional real linear space, their direct product Tp M Tp N is a ð2 þ 3Þ-dimensional real linear space. Let p1 : M N ! M be the mapping deﬁned as follows: For every ðx; yÞ 2 M N; p1 ðx; yÞ x: Let p2 : M N ! N be the mapping deﬁned as follows: For every ðx; yÞ 2 M N; p2 ðx; yÞ y: We know that p1 : M N ! M is a smooth mapping. Since p1 : M N ! M is a smooth mapping, and ðp; qÞ 2 M N; its differential dðp1 Þðp;qÞ : Tðp;qÞ ðM NÞ ! Tp1 ðp;qÞ Mð¼ Tp MÞ is a linear map, that is, dðp1 Þðp;qÞ : Tðp;qÞ ðM NÞ ! Tp M is a linear map. Similarly, dðp2 Þðp;qÞ : Tðp;qÞ ðM NÞ ! Tq N is a linear map. Let us deﬁne a mapping a : Tðp;qÞ ðM NÞ ! Tp M Tq N as follows: For every v in Tðp;qÞ ðM NÞ; aðvÞ ¼

dðp1 Þðp;qÞ ðvÞ; dðp2 Þðp;qÞ ðvÞ :

Since v 7! ðdðp1 Þðp;qÞ ÞðvÞ is linear, and v 7! ðdðp2 Þðp;qÞ ÞðvÞ is linear, v 7! ððdðp1 Þðp;qÞ ÞðvÞ; ðdðp2 Þðp;qÞ ÞðvÞÞ is linear, and hence, a : Tðp;qÞ ðM NÞ ! Tp M Tp N is linear. Similarly, we can show that let M be an m-dimensional smooth manifold, N be an n-dimensional smooth manifold, and ðp; qÞ 2 M N: Deﬁne a : Tðp;qÞ ðM NÞ ! Tp M Tq N as follows: For every v in Tðp;qÞ ðM NÞ; aðvÞ ððdðp1 Þðp;qÞ ÞðvÞ; ðdðp2 Þðp;qÞ ÞðvÞÞ: Then, α is linear, etc. Note 5.16 Let M be an m-dimensional smooth manifold. Let p 2 M: Let U be an open neighborhood of p. We know that U is also a smooth manifold (called open submanifold U of M). Let i : U ! M be the mapping deﬁned as follows: For every x in U, iðxÞ x: We know that the differential dip : Tp U ! Tp M is an isomorphism. So, from now on, we shall not distinguish between a derivation v in Tp U; and the derivation ðdip ÞðvÞ in Tp M: Also, we shall not distinguish between Tp U and Tp M: Here, let v be in Tp U: So v : C1 ðUÞ ! R: Since ðdip ÞðvÞ is in Tp M; ðdip ÞðvÞ : 1 C ðMÞ ! R: Observe that the action of “derivation at a point” on a function depends only on the values of the function in any open neighborhood of the point.

5.2 Open Submanifolds

319

Also, we shall not distinguish between a derivation v in Tp M; and the derivation ðdip Þ1 ðvÞ in Tp U: Note 5.17 Let M be an m-dimensional smooth manifold. Let ðU; uÞ be an admissible coordinate chart of M. Let us ﬁx any p 2 U: Since ðU; uÞ is an admissible coordinate chart of M satisfying p 2 U; there exists an open neighborhood G of uðpÞ in Rm such that u : U ! G is a diffeomorphism. Since u : U ! G is a diffeomorphism, dup : Tp U ! TuðpÞ G is an isomorphism. Thus, for every v 2 Tp U; and for every f 2 C1 ðGÞ;

dup ðvÞ ðf Þ ¼ vðf uÞ:

Since M is an m-dimensional smooth manifold, p 2 M; and U is an open neighborhood of p, by Theorem 5.13, the differential dip : Tp U ! TiðpÞ Mð¼ Tp MÞ is an isomorphism, where i denotes the inclusion map of U. Since dip : Tp U ! Tp M is an isomorphism, ðdip Þ1 : Tp M ! Tp U is an isomorphism. Thus, for every v 2 Tp U; and for every f 2 C 1 ðMÞ;

dip ðvÞ ðf Þ ¼ vðf iÞ ¼ v f jU :

Since Rm is an m-dimensional smooth manifold, uðpÞ 2 Rm ; and G is an open neighborhood of uðpÞ, by Theorem 5.13, the differential d^iuðpÞ : TuðpÞ G ! T^iðuðpÞÞ Rm ð¼ TuðpÞ Rm Þ is an isomorphism, where ^i denotes the inclusion map of G. Thus, for every v 2 TuðpÞ G; and for every f 2 C 1 ðRm Þ; d^iuðpÞ ðvÞ ðf Þ ¼ vðf ^iÞ ¼ v f jG : Since ðdip Þ1 : Tp M ! Tp U is an isomorphism, dup : Tp U ! TuðpÞ G is an isomorphism, and d^iuðpÞ : TuðpÞ G ! TuðpÞ Rm is an isomorphism, their composite ðd^iuðpÞ Þ ðdup Þ ðdip Þ1 : Tp M ! TuðpÞ Rm is an isomorphism. It follows that ðdip Þ ðdup Þ1 ðd^iuðpÞ Þ1 : TuðpÞ Rm ! Tp M is an isomorphism. Further, since u : U !

G is a diffeomorphism, ðdup Þ1 ¼ dðu1 ÞuðpÞ ; and hence, ðdip Þ dðu1 ÞuðpÞ ðd^iuðpÞ Þ1 : TuðpÞ Rm ! Tp M is an isomorphism. Since ðdip Þ dðu1 ÞuðpÞ ðd^iuðpÞ Þ1 : TuðpÞ Rm ! Tp M is an isomorphism, and (

is a basis of TuðpÞ Rm ;

) o o o ; ; . . .; m ox1 uðpÞ ox2 uðpÞ ox uðpÞ

320

(

5 Immersions, Submersions, and Embeddings

! 1 o ; dip d u1 uðpÞ d^iuðpÞ 1 ox uðpÞ ! !) 1 o o 1 ; . . .; dip d u uðpÞ d^iuðpÞ ox2 uðpÞ oxm uðpÞ

1 dip d u1 uðpÞ d^iuðpÞ

is a basis of Tp M: Here, we can write:

1

dip d u ¼ ¼

uðpÞ

d^iuðpÞ

1

1

dip d u

dip d u

uðpÞ

uðpÞ

! o ox1 uðpÞ

1

d^iuðpÞ

1

!! o ox1 uðpÞ

! !! 1 o o d u uðpÞ ¼ dip ox1 uðpÞ ox1 uðpÞ !

o : ox1 uðpÞ

¼ d u1 uðpÞ Thus,

1 dip d u1 uðpÞ d^iuðpÞ

! ! 1 o o ¼ d u uðpÞ : ox1 uðpÞ ox1 uðpÞ

Similarly,

1

dip d u

uðpÞ

d^iuðpÞ

1

! ! 1 o o ¼ d u uðpÞ ; etc; ox2 uðpÞ ox2 uðpÞ

Thus, ( d u1 uðpÞ

! ! !) 1 o o o 1 ; d u uðpÞ ; . . .; d u uðpÞ ox1 uðpÞ ox2 uðpÞ oxm uðpÞ

is a basis of Tp M: For i ¼ 1; 2; . . .; m; the tangent vector d u1 uðpÞ

! o oxi uðpÞ

5.2 Open Submanifolds

at p is denoted by

o oxi p .

321

Thus, (

) o o o ; ; . . .; m ox1 p ox2 p ox p

is a basis of Tp M: Conclusion: Let M be an m-dimensional smooth manifold. Let ðU; uÞ be an admissible coordinate chart of M. Then, for every p 2 U; (

is a basis of Tp M; where

o oxi p

) o o o ; ; . . .; m ox1 p ox2 p ox p stands for

d u1 uðpÞ

! o : oxi uðpÞ

Deﬁnition Let M be an m-dimensional smooth manifold. Let ðU; uÞ be an admissible coordinate chart of M. For every p 2 M; the ordered basis ! o o o ; ; . . .; m ; ox1 p ox2 p ox p as deﬁned in the Note 5.17, is called the coordinate basis of Tp M: If v 2 Tp M, then there exists a unique m-tuple ðv1 ; v2 ; . . .; vm Þ of real numbers such that v¼v

1

! ! ! o o o 2 m þv þ þ v : ox1 p ox2 p oxm p

Here, we say that ðv1 ; v2 ; . . .; vm Þ are the components of v with respect to coordinate basis. Note 5.18 Let U be an open subset of Rn ; V be an open subset of Rm ; and p 2 U: Let F : U ! V be smooth, where F ðF 1 ; F 2 ; . . .; F m Þ: Since U is a nonempty open subset of the n-dimensional smooth manifold Rn ; U is also an n-dimensional smooth manifold. Similarly, V is an m-dimensional smooth manifold. It follows that dFp : Tp U ! TFðpÞ V is a linear map from real linear space Tp U to real linear space TFðpÞ V: Here,

322

5 Immersions, Submersions, and Embeddings

! o o o ; ; . . .; n ox1 p ox2 p ox p is the coordinate basis of Tp Rn ; and we do not distinguish between Tp U and Tp Rn ; so ! o o o ; ; . . .; n ox1 p ox2 p ox p is the coordinate basis of Tp U: Similarly, ! o o o ; ; . . .; m oy1 FðpÞ oy2 FðpÞ oy FðpÞ is the coordinate basis of TFðpÞ V: Now, since dFp : Tp U ! TFðpÞ V; for every f 2 C 1 ðVÞ; dFp

!! o ðf Þ ¼ ox1 p

! o ðf F Þ ox1 p

¼ 1st column of ½ðD1 f ÞðFðpÞÞ 3 2 ðD1 F 1 ÞðpÞ ðDn F 1 ÞðpÞ 7 6. .. . 7 . .. ðDm f ÞðFðpÞÞ 6 5 4 .. m m ðD1 F ÞðpÞ ðDn F ÞðpÞ ¼ ððD1 f ÞðFðpÞÞÞ D1 F 1 ðpÞ þ þ ððDm f ÞðFðpÞÞÞððD1 F m ÞðpÞÞ i m X of oF ¼ ðFðpÞÞ ðpÞ i oy ox1 i¼1 m X oF i of ¼ ðpÞ ðFðpÞÞ oyi ox1 i¼1 ! !! i m X oF o ðpÞ ¼ f oyi FðpÞ ox1 i¼1 !! ! i m X oF o ¼ ðpÞ f oyi FðpÞ ox1 i¼1 ! ! m X oF i o ¼ ðpÞ f oyi FðpÞ ox1 i¼1

5.2 Open Submanifolds

323

and hence,

dFp

! X ! m o oF i o ðpÞ ¼ : ox1 p oyi FðpÞ ox1 i¼1

Similarly,

dFp

! ! m X o oF i o ðpÞ ¼ ; etc: ox2 p oyi FðpÞ ox2 i¼1

Since for every j ¼ 1; 2; . . .; n; dFp

! ! m X o oF i o ðpÞ ¼ ; ox j p oyi FðpÞ ox j i¼1

and ! o o o ; ; . . .; m oy1 FðpÞ oy2 FðpÞ oy FðpÞ is the coordinate basis of TFðpÞ V; the matrix representation of linear map dFp is the following m n matrix: 2

oF 1 ox1 6 oF 2 6 1 6 ox

ðpÞ

oF m ox1

4

ðpÞ

ðpÞ

oF 1 ox2 oF 2 ox2

ðpÞ

oF m ox2

ðpÞ

ðpÞ

oF 1 oxn oF 2 oxn

ðpÞ

3

7 ðpÞ 7 7: 5 m oF ðpÞ oxn

Let us recall that this matrix is the matrix representation of the total derivative DFðpÞ : Rn ! Rm : Thus, we ﬁnd that in this case, differential dFp : Tp Rn ! TFðpÞ Rm corresponds to the total derivative DFðpÞ : Rn ! Rm ; tangent space Tp Rn corresponds Rn ; and tangent space TFðpÞ Rm corresponds Rm : Note 5.19 Let M be an m-dimensional smooth manifold, and N be an n-dimensional smooth manifold. Let p 2 M: Let F : M ! N be a smooth mapping. Let ðU; uÞ be an admissible coordinate chart of M such that p 2 U: Let ðV; wÞ be an admissible coordinate chart of N such that FðpÞ 2 V: Since F : M ! N is a smooth ^ 1; F ^ 2 ; . . .; mapping, ðw F u1 Þ : uðU \ F 1 ðVÞÞ ! wðVÞ is smooth. Put ðF n 1 1 ^ ÞF ^ w F u : Thus, F ^ : uðU \ F ðVÞÞ ! wðVÞ is smooth, that is, each F 1 pi ðw F u Þ is smooth, where pi denotes the ith projection function from Rn to R:

324

5 Immersions, Submersions, and Embeddings

Since F : M ! N is a smooth, F : M ! N is continuous. Since F : M ! N is continuous, and V is open in N, F 1 ðVÞ is open in M. Since F 1 ðVÞ is open in M, and U is open in M, U \ F 1 ðVÞ is open in M. Clearly, p 2 U \ F 1 ðVÞ: Thus, U \ F 1 ðVÞ is an open neighborhood of p. Since U \ F 1 ðVÞ U; U \ F 1 ðVÞ is open in M, and U is open in M, U \ F 1 ðVÞ is open in U. Since U \ F 1 ðVÞ is an open neighborhood of p in U, and u is a homeomorphism from U onto an open subset of Rm ; uðU \ F 1 ðVÞÞ is an open neighborhood of uðpÞ in Rm : Clearly, ^ : uðU \ F 1 ðVÞÞ ! wðVÞ is smooth, as in the wðVÞ is open in Rn : Now, since F n ^uðpÞ : TuðpÞ Rm ! T ^ Note 5.18, the matrix representation of dF ^ FðuðpÞÞ R ð¼ TðFuÞðpÞ n n R ¼ TðwFÞðpÞ R Þ is 2 oF^ 1 ox1 6 ^2 6 oF1 6 ox

4 ^n oF ox1

ðuðpÞÞ

ðuðpÞÞ

^1 oF ox2 ^2 oF ox2

ðuðpÞÞ

^n oF ox2

ðuðpÞÞ

ðuðpÞÞ

ðuðpÞÞ

^1 oF oxm ^2 oF oxm

ðuðpÞÞ

3

7 ðuðpÞÞ 7 7: 5 ^n ooxFm ðuðpÞÞ

^ w F u1 ; and u; w are 1–1, w1 F ^ ¼ F u1 : Here, F u1 : Since F m 1 ^ uðUÞ ! N; so dðF u ÞuðpÞ : TuðpÞ R ! TFðpÞ N: Similarly dðw1 FÞ uðpÞ : m TuðpÞ R ! TFðpÞ N: Here ! o o o ; ; . . .; m ox1 p ox2 p ox p is the coordinate basis of Tp M; where ! o 1 o d u uðpÞ : oxi p oxi uðpÞ Also, ! o o o ; ; . . .; n oy1 FðpÞ oy2 FðpÞ oy FðpÞ is the coordinate basis of TFðpÞ N; where ! o o 1 : d w wðFðpÞÞ oy j FðpÞ oy j wðFðpÞÞ Now, we want to obtain the matrix representation of dFp :

5.2 Open Submanifolds

325

Since F : M ! N is a smooth mapping, dFp : Tp M ! TFðpÞ N: Next, for i ¼ 1; 2; . . .; m;

dFp

! !! 1 o o d u uðpÞ ¼ dFp oxi p oxi uðpÞ ! 1 o ¼ dFp d u uðpÞ oxi uðpÞ ¼

dFu1 ðuðpÞÞ d u1 uðpÞ

! o oxi uðpÞ ! o 1 ^ ¼ d w F uðpÞ oxi uðpÞ

! o oxi uðpÞ

¼ d F u1 uðpÞ

! o oxi uðpÞ !! o 1 ^ dFuðpÞ ¼ d w F^ ðuðpÞÞ oxi uðpÞ ¼ d w1 F ðuðpÞÞ ðith column of 31 2 ^1 ^1 ^1 oF oF oF ox1 ðuðpÞÞ ox2 ðuðpÞÞ oxm ðuðpÞÞ 7C 6 ^2 6 oF1 ðuðpÞÞ oF^22 ðuðpÞÞ oF^m2 ðuðpÞÞ 7C 7C 6 ox ox ox 7C 6 5A 4 ^n ^n ^n oF oF oF ð uðpÞ Þ ð uðpÞ Þ ð uðpÞ Þ m 1 2 ox ox ox !! n ^j X oF o 1 ¼ d w F^ ðuðpÞÞ ð uðpÞ Þ oy j F^ ðuðpÞÞ oxi j¼1 !! n ^j X oF o 1 ¼ d w ðFu ðuðpÞÞ ^ ÞðpÞ oy j ðFu oxi ^ ÞðpÞ j¼1 !! n X ^j oF o 1 ¼ d w ðwF ÞðpÞ ðuðpÞÞ oy j ðwF ÞðpÞ oxi j¼1 !! n o X ^j oF 1 ¼ ð uðpÞ Þ d w ðwF ÞðpÞ oy j ðwF ÞðpÞ oxi j¼1 !! n ^j o X 1 oF ¼ ðuðpÞÞ d w wðFðpÞÞ oy j wðFðpÞÞ oxi j¼1 ! n ^j X oF o ¼ ð uðpÞ Þ : oy j FðpÞ oxi j¼1 ^uðpÞ ¼ d w1 F^ ðuðpÞÞ dF

326

5 Immersions, Submersions, and Embeddings

Thus,

dFp

! ! n ^j X o oF o ðuðpÞÞ ¼ : oxi p oy j FðpÞ oxi j¼1

Since for every i ¼ 1; 2; . . .; m; ! ! n ^j X o oF o dFp ðuðpÞÞ ¼ ; oxi p oy j FðpÞ oxi j¼1 and

! o o o ; ; . . .; n oy1 FðpÞ oy2 FðpÞ oy FðpÞ

is the coordinate basis of TFðpÞ N; the matrix representation of linear map dFp is the following n m matrix: 2

^1 oF ox1

ðuðpÞÞ

6 6 oF^ 2 6 ox1 ðuðpÞÞ 6 4 ^n oF ox1 ðuðpÞÞ 2

oðp1 ðwFu1 ÞÞ ðuðpÞÞ ox1 6

^1 oF ox2

ðuðpÞÞ

^2 oF ox2

ðuðpÞÞ

^n oF ox2

ðuðpÞÞ

^1 oF oxm

ðuðpÞÞ

3

7 7 ^2 ooxFm ðuðpÞÞ 7 7 5 ^n ooxFm ðuðpÞÞ

oðp1 ðwFu1 ÞÞ ðuðpÞÞ ox2

6 oðp2 ðwFu1 ÞÞ 6 o p wFu1 ÞÞ ¼ 6 ð 2 ð ox1 ðuðpÞÞ ðuðpÞÞ ox2 6 4 oðpn ðwFu1 ÞÞ oðpn ðwFu1 ÞÞ ð uðpÞ Þ ðuðpÞÞ 1 ox ox2

3

oðp1 ðwFu1 ÞÞ ðuðpÞÞ oxm 7

7 oðp2 ðwFu1 ÞÞ 7 7: ð uðpÞ Þ oxm 7 5 oðpn ðwFu1 ÞÞ ðuðpÞÞ oxm

It follows that the matrix representation of dFp is the same as the matrix representation of dðw F u1 ÞuðpÞ : Note 5.20 Let M be an m-dimensional smooth manifold. Let p 2 M: Let ðU; uÞ be an admissible coordinate chart of M such that p 2 U: Let ðV; wÞ be an admissible coordinate chart of M such that p 2 V: Let uðpÞ ðx1 ; x2 ; . . .; xm Þ; and wðpÞ ~x1 ; ~x2 ; . . .; ~xm : Let ! o o o ; ; . . .; m ox1 p ox2 p ox p

5.2 Open Submanifolds

327

be the coordinate basis of Tp M corresponding to ðU; uÞ; where ! o 1 o d u uðpÞ : oxi p oxi uðpÞ Let ! o o o ; ; . . .; m o~x1 p o~x2 p o~x p be the coordinate basis of Tp M corresponding to ðV; wÞ; where ! o 1 o d w wðpÞ : o~xi p o~xi wðpÞ Let v 2 Tp M: Let ðv1 ; v2 ; . . .; vm Þ be the components of v with respect to coordinate basis ! o o o ; ; . . .; m ox1 p ox2 p ox p of Tp M: Let ð~v1 ; ~v2 ; . . .; ~vm Þ be the components of v with respect to coordinate basis ! o o o ; ; . . .; m o~x1 p o~x2 p o~x p of Tp M: Thus, 0

0 1 1 ! o o A 1@ 2@ o A m ~v þ ~v þ þ ~v ¼v m 1 2 ox p ox p ox p ! ! ! o o o þ v2 þ þ vm : ¼ v1 ox1 p ox2 p oxm p Since ðU; uÞ and ðV; wÞ are admissible coordinate charts of M such that p 2 U \ V; w u1 : uðU \ VÞ ! wðU \ VÞ is smooth. Here, with the hope that there will be no confusion, for every i ¼ 1; 2; . . .; m; the ith component of the “transition map” w u1 is generally denoted by ~xi : Thus, ~xi has two meanings: the ith component of wðpÞ; and the ith component function of w u1 : The situation indicates which meaning is to be attached with. Here, for every x 2 uðU \ VÞ; ðw u1 ÞðxÞ ¼ ð~x1 ðxÞ; ~x2 ðxÞ; . . .; ~xm ðxÞÞ:

328

5 Immersions, Submersions, and Embeddings

Since uðU \ VÞ is an open neighborhood of uðpÞ in Rm ; wðU \ VÞ is an open neighborhood of wðpÞ in Rm ; and w u1 : uðU \ VÞ ! wðU \ VÞ is a diffeomorphism, the differential dðw u1 ÞuðpÞ : TuðpÞ uðU \ VÞ ! TwðpÞ wðU \ VÞ is an isomorphism, and hence, dðw u1 ÞuðpÞ : TuðpÞ Rm ! TwðpÞ Rm is an isomorphism. Here, ! o o o ; ; . . .; m ox1 uðpÞ ox2 uðpÞ ox uðpÞ is a basis of TuðpÞ Rm : Hence, for j ¼ 1; 2; . . .; m; ! ! o 1 o o 1 1 d u uðpÞ ¼ d w wu uðpÞ ox j p ox j uðpÞ ox j uðpÞ ! o ¼ d w1 ðwu1 ÞðuðpÞÞ d w u1 uðpÞ ox j uðpÞ ! o 1 1 ¼ d w wðpÞ d w u uðpÞ ox j uðpÞ !! o 1 1 ¼ d w wðpÞ d w u uðpÞ ox j uðpÞ ¼ d w1 wðpÞ 0 31 2 1 1 1 oðwu1 Þ oðwu1 Þ oðwu1 Þ ð uðpÞ Þ ð uðpÞ Þ ð uðpÞ Þ oxm ox2 B 7C 6 ox1 B 7C 6 2 2 2 oðwu1 Þ oðwu1 Þ B 7C 6 oðwu1 Þ Bjth column of 6 ox1 ðuðpÞÞ ox2 ðuðpÞÞ oxm ðuðpÞÞ 7C B 7C 6 B 7C 6 @ 5A 4 m m m oðwu1 Þ oðwu1 Þ oðwu1 Þ ðuðpÞÞ ox2 ðuðpÞÞ oxm ðuðpÞÞ ox1 ! !! i m X oðw u1 Þ o 1 ¼ d w wðpÞ ðuðpÞÞ o~xi ðwu1 ÞðuðpÞÞ ox j i¼1 ! !! i m X oðw u1 Þ o 1 ðuðpÞÞ ¼ d w wðpÞ o~xi wðpÞ ox j i¼1 ! ! ! i m X ox o 1 ¼ d w wðpÞ ðuðpÞÞ o~xi wðpÞ ox j i¼1 !! m X o o~xi 1 ¼ ð uðpÞ Þ d w wðpÞ o~xi wðpÞ ox j i¼1 ! m i X o~x o ¼ ðuðpÞÞ : o~xi p ox j i¼1

5.2 Open Submanifolds

329

Thus, for j ¼ 1; 2; . . .; m; ! m i o X o~x o ¼ ðuðpÞÞ : ox j p i¼1 ox j o~xi p Now, m X i¼1

! ! ! ! o o o o 1 2 m ~v ¼ ~v þ ~v þ þ ~v ¼v o~xi p o~x1 p o~x2 p o~xm p ! ! ! o o o 1 2 m þv þ þ v ¼v ox1 p ox2 p oxm p ! !! m m m i X X X o o~x o j j ¼ v v ðuðpÞÞ ¼ ox j p o~xi p ox j j¼1 j¼1 i¼1 !!! m m i X X o~x o j ¼ ðuðpÞÞ v o~xi p ox j j¼1 i¼1 !!! m m i X X o~x o j ¼ ð uðpÞ Þ v o~xi p ox j i¼1 j¼1 !! ! m m i X X o~x o j ¼ ðuðpÞÞ v o~xi p ox j i¼1 j¼1 ! ! m m i X X o~x o j ¼ ðuðpÞÞ v ; o~xi p ox j i¼1 j¼1 i

so m X i¼1

! X ! ! m m i X o o~x o j ~v ðuðpÞÞ v ¼ ; o~xi p o~xi p ox j i¼1 j¼1 i

and hence ~vi ¼

m i X o~x j¼1

ox

ð uðpÞ Þ v j: j

In short, ~v ¼ i

o~xi ðuðpÞÞ v j : ox j

330

5 Immersions, Submersions, and Embeddings

5.3 Tangent Bundles Lemma 5.21 Let M be a nonempty set. Let fUi gi2I be a nonempty family of subsets of M. Let fui gi2I be a family of maps ui from Ui to Rm : If 1. 2. 3. 4. 5.

for each i 2 I; ui is a 1–1 mapping from Ui onto an open subset ui ðUi Þ of Rm ; for each i; j 2 I; ui ðUi \ Uj Þ; and uj ðUi \ Uj Þ are open in Rm ; if Ui \ Uj 6¼ ;; then the map uj u1 i : ui ðUi \ Uj Þ ! uj ðUi \ Uj Þ is smooth, there exists a countable subcollection of fUi gi2I that covers M; if p; q are distinct points of M; then either ðthere exists i 2 I such that p; q 2 Ui Þ; or ðthere exist i; j 2 I such that p 2 Ui ; q 2 Uj ; and Ui \ Uj ¼ ;Þ; then there exists a unique smooth differential structure A over M such that for each i 2 I; ðUi ; ui Þ 2 A:

Proof First of all, we shall try to show that the collection fu1 k ðGÞ : k 2 I; and G is open in Rm g is closed with respect to ﬁnite intersection. 1 1 For this purpose, let us take any u1 i ðG1 Þ; uj ðG2 Þ 2 fuk ðGÞ : k 2 I; m m and G is open in R g; where i; j 2 I; G1 ; G2 are open in R : 1 1 1 Case I: when i ¼ j. Since i ¼ j; u1 i ðG1 Þ \ uj ðG2 Þ ¼ ui ðG1 Þ \ ui ðG2 Þ ¼ m 1 ui ðG1 \ G2 Þ: Since G1 ; G2 are open in R ; G1 \ G2 is open in Rm : Now, 1 since i 2 I; and G1 \ G2 is open in Rm ; u1 i ðG1 \ G2 Þ 2 fuk ðGÞ : k 2 I; m 1 1 and G is open in R g; and hence, ui ðG1 Þ \ uj ðG2 Þ 2 fu1 k ðGÞ : k 2 I; and G is open in Rm g: Case II: when i 6¼ j. 1 Subcase I: when u1 i ðG1 Þ \ uj ðG2 Þ ¼ ;. Since fUi gi2I is a nonempty family of subsets of M, I is nonempty. Since I is nonempty, there exists i0 2 I: Since i0 2 I; 1 1 and the empty set ; is open in Rm ; u1 i ðG1 Þ \ uj ðG2 Þ ¼ ; ¼ ui0 ð;Þ 2 m 1 1 fu1 k ðGÞ : k 2 I; and G is open in R g; and hence ui ðG1 Þ \ uj ðG2 Þ 2 m 1 fuk ðGÞ : k 2 I; and G is open in R g: 1 Subcase II: when u1 i ðG1 Þ \ uj ðG2 Þ 6¼ ;. 1 1 1 Since u1 i ðG1 Þ \ uj ðG2 Þ 6¼ ;; there exists x 2 ui ðG1 Þ \ uj ðG2 Þ; and hence, ui ðxÞ 2 G1 ; uj ðxÞ 2 G2 ; x 2 Ui ; and x 2 Uj : Since x 2 Ui ; and x 2 Uj ; Ui \ Uj 6¼ ;;

1 and hence, by condition 3, ðuj u1 i Þ ðG2 Þ is open in ui ðUi \ Uj Þ: Since

1 ðuj u1 i Þ ðG2 Þ is open in ui ðUi \ Uj Þ and, by condition 2, ui ðUi \ Uj Þ is open in

1 m 1 1 Rm ; ðuj u1 i Þ ðG2 Þð¼ ðui uj ÞðG2 ÞÞ is open in R ; and hence, ðui uj Þ ðG2 Þ m m is open in Rm : Since ðui u1 j ÞðG2 Þ is open in R ; and G1 is open in R ; G1 \ m m 1 ðui u1 j ÞðG2 Þ is open in R : Since G1 \ ðui uj ÞðG2 Þ is open in R ; m 1 1 1 u1 i ðG1 \ ðui uj ÞðG2 ÞÞ 2 fuk ðGÞ : k 2 I; and G is open in R g: Since fuk 1 1 1 ðGÞ : k 2 I; and G is open in Rm g3 u1 i ðG1 \ ðui uj ÞðG2 ÞÞ ¼ ui ðG1 Þ \ ui 1 1 1 1 1 1 ððui uj ÞðG2 ÞÞ ¼ ui ðG1 Þ \ uj ðG2 Þ; ui ðG1 Þ \ uj ðG2 Þ 2 fuk ðGÞ : k 2 I; and G is open in Rm g:

5.3 Tangent Bundles

331

1 1 Thus, we see that, in all cases, u1 i ðG1 Þ \ uj ðG2 Þ 2 fuk ðGÞ : k 2 I; m and G is open in Rm g: Hence, fu1 k ðGÞ : k 2 I; and G is open in R g is closed with respect to ﬁnite intersection. Let O be the collection of arbitrary unions of the m members of fu1 i ðGÞ : i 2 I; and G is open in R g: We shall show that O is a topology over M:

(a) Since fUi gi2I is a nonempty collection of subsets of M; I is nonempty. Since I is nonempty, there exists i0 2 I. Since i0 2 I; and the empty set ; is open in m 1 Rm ; ; ¼ u1 i0 ð;Þ 2 fui ðGÞ : i 2 I; and G is open in R g O; and hence, ; 2 O: m m (b) Since for each i 2 I; ui maps from Ui to Rm ; u1 i ðR Þ ¼ Ui : Now, since R is m m 1 1 open in R ; fUi : i 2 Ig ¼ fui ðR Þ : i 2 Ig fui ðGÞ : i 2 I; and G is open in Rm g; and hence, [i2I Ui 2 O: By condition 4, there exists a countable subcollection of fUi gi2I that covers M, so fUi gi2I also covers M, and hence, [i2I Ui ¼ M: Since M ¼ [i2I Ui 2 O; M 2 O: (c) Clearly, O is closed with respect to arbitrary union. m 1 (d) Let [fu1 i ðGi1 Þ : i 2 I; and Gi1 is open in R g 2 O; and [fui ðGi2 Þ : i 2 I; m and Gi2 is open in R g 2 O: We have to prove that ð[fu1 i ðGi1 Þ : i 2 I; m 1 and Gi1 is open in R gÞ \ ð[fui ðGi2 Þ : i 2 I; and Gi2 is open in Rm gÞ 2 O: m 1 Here, ð[fu1 i ðGi1 Þ : i 2 I; and Gi1 is open in R gÞ \ ð[fui ðGi2 Þ : i 2 I; and Gi2 m 1 1 is open in R gÞ ¼ [fuj ðGj1 Þ \ uk ðGk2 Þ : ðj; kÞ 2 I I; Gj1 ; Gk2 are open in Rm g: m Since fu1 i ðGÞ : i 2 I; and G is open in R g is closed with respect to ﬁnite 1 1 intersection, fuj ðGj1 Þ \ uk ðGk2 Þ : ðj; kÞ 2 I I; Gj1 ; Gk2 are open in Rm g fu1 i 1 ðGÞ : i 2 I; and G is open in Rm g; and hence, [fu1 ðG Þ \ u ðG Þ : ðj; kÞ 2 j1 k2 j k I I; Gj1 ; Gk2 are open in Rm g 2 O: It follows that ð[fu1 ðG Þ : i 2 I; i1 i m m 1 and Gi1 is open in R gÞ \ ð[fui ðGi2 Þ : i 2 I; and Gi2 is open in R gÞ 2 O: Thus, O is a topology over M: m It is clear that fu1 i ðGÞ : i 2 I; and G is open in R g is a basis of the topological space ðM; OÞ. Now, we want to prove that for each i 2 I; Ui 2 O: For this purpose, let us ﬁx any i0 2 I: We have to prove that Ui0 2 O: Clearly, fu1 i ðGÞ : m 1 i 2 I; and G is open in Rm g O; and Ui0 ¼ u1 ðR Þ 2 fu ðGÞ : i 2 I; and G i0 i m is open in R g; so Ui0 2 O: Now, we want to prove that for each i 2 I; ui : Ui ! ui ðUi Þ is continuous. For this purpose, let us take any open set G in ui ðUi Þ: We have to show that u1 i ðGÞ is open in Ui : Since G is open in ui ðUi Þ; and ui ðUi Þ is open in Rm ; G is open in Rm ; 1 and hence, by the deﬁnition of O; for every i 2 I; u1 i ðGÞ 2 O: Since ui ðGÞ 2 O; 1 u1 i ðGÞ \ Ui is open in Ui : Since ui : Ui ! ui ðUi Þ; ui ðGÞ Ui ; and hence, 1 1 1 1 ui ðGÞ \ Ui ¼ ui ðGÞ: Since ui ðGÞ \ Ui ¼ ui ðGÞ; and u1 i ðGÞ \ Ui is open in Ui ; u1 ðGÞ is open in U : i i Thus, we have shown that for each i 2 I; ui : Ui ! ui ðUi Þ is continuous. Now, we want to show that ui is an open mapping. For this purpose, let us take any

332

5 Immersions, Submersions, and Embeddings

open set G in Ui : We have to show that ui ðGÞ is open in ui ðUi Þ: Since m fu1 j ðGÞ : j 2 I; and G is open in R g is a basis of the topological space ðM; OÞ; m fu1 j ðGÞ \ Ui : j 2 I; and G is open in R g is a basis of Ui : Now, it sufﬁces to prove that ui ðu1 j ðGÞ \ Ui Þ is open in ui ðUi Þ for every j 2 I; and for every G which is open in Rm : 1 Case I: when u1 j ðGÞ \ Ui ¼ ;. Here, ui ðuj ðGÞ \ Ui Þ ¼ ui ð;Þ ¼ ;; which is open in ui ðUi Þ: 1 1 Case II: when u1 j ðGÞ \ Ui 6¼ ;: Since uj ðGÞ Uj ; and uj ðGÞ \ Ui 6¼ ;; Uj \ 1 Ui 6¼ ;: Here, ui is 1–1, so ui ðu1 j ðGÞ \ Ui Þ ¼ ui ðuj ðGÞÞ \ ui ðUi Þ ¼ ðui 1 u1 j ÞðGÞ \ ui ðUi Þ: Since Uj \ Ui 6¼ ;; by the condition 3, ui uj is open. Since m 1 1 G is open in R ; and ui uj is open, ðui uj ÞðGÞ is open in Rm ; and hence, 1 ðui u1 j ÞðGÞ \ ui ðUi Þ is open in ui ðUi Þ: Since ðui uj ÞðGÞ \ ui ðUi Þ ð¼ ui 1 1 ðuj ðGÞ \ Ui ÞÞ is open in ui ðUi Þ; ui ðuj ðGÞ \Ui Þ is open in ui ðUi Þ:

Thus, in all cases, ui ðu1 j ðGÞ \ Ui Þ is open in ui ðUi Þ: This shows that ui is an open mapping. Thus, we have shown that ui : Ui ! ui ðUi Þ is a homeomorphism. It follows that for each i 2 I; the ordered pair ðUi ; ui Þ is a coordinate chart of M. Now, we want to show that the topology O of M is Hausdorff. For this purpose, let us take any two distinct points p, q in M. We have to ﬁnd an open neighborhood G of p, and an open neighborhood G1 of q such that G; G1 are disjoint. Since p, q are distinct points of M, by condition 5, either ðthere exists i 2 I such that p; q 2 Ui Þ or ðthere exist i; j 2 I such that p 2 Ui ; q 2 Uj ; and Ui \ Uj ¼ ;Þ: Case I: when there exist i; j 2 I such that p 2 Ui ; q 2 Uj ; and Ui \ Uj ¼ ;. Since i 2 I; Ui 2 O: Since Ui 2 O; and p 2 Ui ; Ui is an open neighborhood of p in M: Similarly, Uj is an open neighborhood of q in M: Also, Ui \ Uj ¼ ;: Thus, we have found an open neighborhood Ui of p; and an open neighborhood Uj of q such that Ui ; Uj are disjoint. Case II: when there exists i 2 I such that p; q 2 Ui . Since ui : Ui ! ui ðUi Þ is continuous, ui is 1–1, and p; q are distinct elements of Ui ; ui ðpÞ; ui ðqÞ are distinct points of ui ðUi Þ: Since, by condition 1, ui ðUi Þ is open in Rm ; and ui ðpÞ; ui ðqÞ are distinct points of ui ðUi Þ; there exist an open neighborhood G1 of ui ðpÞ in Rm and an open neighborhood G2 of ui ðqÞ in Rm such that G1 ui ðUi Þ; G2 ui ðUi Þ; and G1 ; G2 are disjoint. Since ui : Ui ! ui ðUi Þ is continuous, and G1 is an open neighborhood of ui ðpÞ in Rm ; u1 i ðG1 Þ is an open neighborhood of p in Ui : Since u1 ðG Þ is an open neighborhood of p in Ui ; and 1 i Ui 2 O; u1 ðG Þ is an open neighborhood of p in M: Similarly, u1 1 i i ðG2 Þ is an 1 1 1 open neighborhood of q in M: Also, ui ðG1 Þ \ ui ðG2 Þ ¼ ui ðG1 \ G2 Þ ¼ 1 u1 i ð;Þ ¼ ;: Thus, we have found an open neighborhood ui ðG1 Þ of p, and an 1 1 1 open neighborhood ui ðG2 Þ of q such thatui ðG1 Þ; ui ðG2 Þ are disjoint. Thus, we see that, in all cases, there exist disjoint open neighborhoods of p and q. Hence, the topology O over M is Hausdorff.

5.3 Tangent Bundles

333

Now, we want to show that the topology O of M is second countable. Observe that for every j 2 I; the subspace topology of Uj is second countable. (Reason: For every j 2 I; uj ðUj Þ is a subset of Rm : Since each uj ðUj Þ is a subset of Rm ; and the topology of Rm is second countable, the subspace topology of uj ðUj Þ is second countable. Since the topology of uj ðUj Þ is second countable, and uj : Uj ! uj ðUj Þ is a homeomorphism, the topology of Uj is second countable.) By condition 4, there exists a countable subcollection fUi1 ; Ui2 ; Ui3 ; . . .g of fUi gi2I such that fUi1 ; Ui2 ; Ui3 ; . . .g covers M: Since fUi1 ; Ui2 ; Ui3 ; . . .g is a subcollection of fUi gi2I ; and for each i 2 I; the topology of Ui is second countable, each Uik ðk ¼ 1; 2; 3; . . .Þ is a second countable space, and hence, there exists a countable basis fVk1 ; Vk2 ; Vk3 ; . . .g of Uik : Since each Vkl is open in Uik ; and Uik 2 O; each Vkl 2 O: Observe that fVkl : k; l are positive integersg is a countable collection of open subsets of M. We shall try to show that fVkl : k; l are positive integersg is a basis of M. For this purpose, let us take any open neighborhood G of p in M. Since p 2 M; and fUi1 ; Ui2 ; Ui3 ; . . .g covers M, there exists a positive integer k such that p 2 Uik : Since p 2 Uik ; p 2 G; p 2 Uik \ G: Since G are open in M, Uik \ G is open in Uik : Thus, Uik \ G is an open neighborhood of p in Uik : Since Uik \ G is an open neighborhood of p in Uik ; and fVk1 ; Vk2 ; Vk3 ; . . .g is a basis of Uik ; there exists a positive integer l such that p 2 Vkl Uik \ G G: This shows that fVkl : k; l are positive integersg is a countable basis of M. Hence, the topology of M is second countable. Now, we want to show that M is an m-dimensional topological manifold. For this purpose, let us take any p 2 M: Since p 2 M; and fUi gi2I is a cover of M; there exists i0 2 I such that p 2 Ui0 : Here, ðUi0 ; ui0 Þ is a coordinate chart of M, and p 2 Ui0 : It follows that M is an m-dimensional topological manifold. Next, let B be the collection of all coordinate charts ðUi ; ui Þ; where i 2 I: We shall try to prove that B is an atlas on M: For this purpose, let us take any coordinate charts ðUi ; ui Þ; ðUj ; uj Þ, where Ui \ Uj 6¼ ;: We have to prove that uj u1 i : ui ðUi \ Uj Þ ! uj ðUi \ Uj Þ is smooth. Since Ui \ Uj 6¼ ;; by the condition 3, uj u1 i : ui ðUi \ Uj Þ ! uj ðUi \ Uj Þ is smooth. This proves that B is an atlas on M. Since B is an atlas on an m-dimensional topological manifold M, by Note 1.5, there exists a unique smooth differential structure A on M which contains B: h Deﬁnition Let‘fXi gi2I be any family of sets. The set fði; xÞ : i 2 I and x 2 Xi g is denoted by i2I Xi ; and is called the disjoint union of fXi gi2I . Thus, ‘ X ¼ [ ðfig Xi Þ: i i2I i2I Note 5.22 Let M be an m-dimensional smooth manifold. Let A be the smooth structure of M. Let ðU; uÞ 2 A: For every p 2 U; let ! o o o ; ; . . .; m ox1 p ox2 p ox p

334

5 Immersions, Submersions, and Embeddings

be the coordinate basis of Tp M corresponding to ðU; uÞ; where ! o 1 o d u uðpÞ : oxi p oxi uðpÞ ‘ ‘ Here, r2U Tr M r2M Tr M; and uðUÞ Rm Rm Rm ¼ R2m : Let us deﬁne ‘ ~ : r2U Tr M ! uðUÞ Rm as follows: For every a function u p; v

~ p; v u

1

! !! a o o m Tr M; þ þ v 2 1 m ox p ox p r2U

! !! o o m þ þ v ðp1 uÞðpÞ; . . .; ðpm uÞðpÞ; v1 ; . . .; vm : 1 m ox p ox p

1

~ is well deﬁned. First of all, observe that u (Reason:Let ðp; v1 ðoxo1 p Þ þ þ vm ðoxom p ÞÞ ¼ ðq; w1 ðoxo1 q Þ þ þ wm ðoxom q ÞÞ where ‘ Tr M; ðp; v1 ð o1 Þ þ þ vm ð om ÞÞ 2 ox

ox p

p

r2U

and q; w1

! !! a o o m Tr M: þ þ w 2 ox1 q oxm q r2U

We have to show that

ðp1 uÞðpÞ; . . .; ðpm uÞðpÞ; v1 ; . . .vm ¼ ðp1 uÞðqÞ; . . .; ðpm uÞðqÞ; w1 ; . . .; wm :

Since p; v

! !! o o m þ þ v ¼ ox1 p oxm p

1

1

q; w

! !! o o m þ þ w ; ox1 q oxm q

p ¼ q and v

1

! ! ! ! o o o o m 1 m þ þ v ¼w þ þ w : ox1 p oxm p ox1 q oxm q

5.3 Tangent Bundles

335

Since p ¼ q; for every i ¼ 1; . . .; m; ðpi uÞðpÞ ¼ ðpi uÞðqÞ: Since p ¼ q; and 1

v

so v1

Since v

! ! ! ! o o o o m 1 m þ þ v ¼w þ þ w ; ox1 p oxm p ox1 q oxm q ! ! ! ! o o o o m 1 m þ þ v ¼ w þ þ w : ox1 p oxm p ox1 p oxm p

! ! ! ! o o o o m 1 m þ þ v ¼w þ þ w ; ox1 p oxm p ox1 p oxm p

1

and

! o o o ; ; . . .; m ox1 p ox2 p ox p

is a basis of Tp M; for every i ¼ 1; . . .; m; vi ¼ wi : Since for every i ¼ 1; . . .; m; vi ¼ wi ; and ðpi uÞðpÞ ¼ ðpi uÞðqÞ; ððp1 uÞðpÞ; . . .; ðpm uÞðpÞ; v1 ; . . .vm Þ ‘ ¼ ððp1 uÞðqÞ; . . .; ðpm uÞðqÞ; w1 ; . . .; wm Þ:) Thus, f r2U Tr MgðU;uÞ2A is a ‘ ugðU;uÞ2A is a family of functions from family of subsets of r2M Tr M; and f~ ‘ m 2m r2U Tr M onto uðUÞ R ð R Þ: Now, we shall verify all the conditions 1–5 of Lemma 5.21: ‘ ~ : r2U Tr M ! uðUÞ Rm is 1–1: For this purpose, let 1. u ~ p; v u

1

! !! ! !! o o o o m 1 m ~ q; w þ þ v ¼u þ þ w : ox1 p oxm p ox1 q oxm q

We have to show that 1

p; v

! !! o o m þ þ v ¼ ox1 p oxm p

Here,

ðp1 uÞðpÞ; . . .; ðpm uÞðpÞ; v ; . . .; v 1

m

! !! o o m þ þ w : ox1 q oxm q

1

q; w

! !! o o m ~ p; v ¼u þ þ v ox1 p oxm p ! !! o o m ~ q; w1 ¼u þ þ w ox1 q oxm q ¼ ðp1 uÞðqÞ; . . .; ðpm uÞðqÞ; w1 ; . . .; wm ; 1

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5 Immersions, Submersions, and Embeddings

and hence, for every i ¼ 1; . . .; m; ðpi uÞðpÞ ¼ ðpi uÞðqÞ; and vi ¼ wi : Since for every i ¼ 1; . . .; m; ðpi uÞðpÞ ¼ ðpi uÞðqÞ; uðpÞ ¼ uðqÞ: Since uðpÞ ¼ uðqÞ; and u is 1–1, p ¼ q: Since for every i ¼ 1; . . .; m; vi ¼ wi ; v

1

Since v

1

so v1

! ! ! ! o o o o m 1 m þ þ v ¼w þ þ w : ox1 p oxm p ox1 p oxm p ! ! ! ! o o o o m 1 m þ þ v ¼w þ þ w ; and p ¼ q; ox1 p oxm p ox1 p oxm p ! ! ! ! o o o o m 1 m þ þ v ¼w þ þ w : ox1 p oxm p ox1 q oxm q

It follows that p; v1

! !! o o m þ þ v ¼ ox1 p oxm p

q; w1

! !! o o m þ þ w : ox1 q oxm q

‘ ~ : r2U Tr M ! uðUÞ Rm is 1–1. Now, we want to Thus, we have shown that u ‘ ~ : r2U Tr M ! uðUÞ Rm is onto. For this purpose, let us take any show that u ððp1 uÞðpÞ; . . .; ðpm uÞðpÞ; v1 ; . . .; vm Þ 2 uðUÞ Rm ; where p 2 U: Here, p 2 U; and ! ! o o 1 m v þ þ v 2 Tp M; ox1 p oxm p so v

1

! ! a o o m Tr M: þ þ v 2 1 m ox p ox p r2U

Also, ~ v1 u

! !! o o þ þ vm ¼ ðp1 uÞðpÞ; . . .; ðpm uÞðpÞ; v1 ; . . .; vm : 1 m ox p ox p

‘ ~ : r2U Tr M ! uðUÞ Rm is a 1–1 onto mapping. This shows that u Now, we want to show that uðUÞ Rm is open in R2m : Since ðU; uÞ 2 A; and A is a smooth structure of M; uðUÞ is open in Rm ; and hence, uðUÞ Rm is open in Rm Rm ð¼ R2m Þ:

5.3 Tangent Bundles

337

2. Let us take any ðU; uÞ 2 A; and ðV; wÞ 2 A: We have to prove that ‘ ‘ ‘ ~ ðð r2U Tr MÞ \ ð r2V Tr MÞÞ is open in R2m : Here, u ~ ðð r2U Tr MÞ\ u ‘ ‘ ‘ ~ ð r2U \ V Tr MÞ: Here, u ~ : r2U Tr M ! uðUÞ Rm is ð r2V Tr MÞÞ ¼ u deﬁned as follows: For every ! !! a o o m p; v Tr M; þ þ v 2 ox1 p oxm p r2U ! !! o o m ~ p; v1 u þ þ v ðp1 uÞðpÞ; . . .; ðpm uÞðpÞ; v1 ; . . .; vm ox1 p oxm p 1

‘ ‘ ~ ð r2U \ V Tr MÞ ¼ uðU \ VÞ Rm ; and hence, u ~ ðð r2U Tr MÞ It follows that u ‘ \ð r2V Tr MÞÞ ¼ uðU \ VÞ Rm : Since ðU; uÞ 2 A; and ðV; wÞ 2 A; uðU \ VÞ is open in Rm ; and hence, uðU \ VÞ Rm is open in R2m : This shows that ‘ ‘ ~ ‘ ~ ðð r2U Tr MÞ \ð r2V Tr MÞÞ is open in R2m : Similarly, wðð u r2U Tr MÞ\ ‘ 2m ð r2V Tr MÞÞ is open in R : ‘ ‘ 3. Let us take any ðU; uÞ 2 A; and ðV; wÞ 2 A: Let ð r2U Tr MÞ \ ð r2V Tr MÞ 6¼ ;: We have to prove that ~ u ~ w

1

a

~ :u

! Tr M

\

r2U

a

!! ~ !w

Tr M

r2V

a

! \

Tr M

r2U

r2V

is smooth, that is, ~ u ~ w

1

~ :u

a r2U\V

! Tr M

~ !w

a

! Tr M

r2U\V

is smooth. For every p 2 U \ V; let ! o o o ; ; . . .; m ox1 p ox2 p ox p be the coordinate basis of Tp M corresponding to ðU; uÞ; where ! o 1 o d u uðpÞ ; oxi p oxi uðpÞ

a

!! Tr M

338

5 Immersions, Submersions, and Embeddings

! o o o ; ; . . .; m o~x1 p o~x2 p o~x p

and let

be the coordinate basis of Tp M corresponding to ðV; wÞ; where ! o 1 o d w wðpÞ : o~xi p o~xi wðpÞ Also,

i ! o o~x o ¼ ðuðpÞÞ : ox j p o~xi p ox j

~ð Here, for every ðx1 ; . . .; xm ; v1 ; . . .; vm Þ 2 u

~ u ~ 1 w

‘ r2U\V

Tr MÞ; we have

1 1 ~ u ~ x ; . . .; xm ; v1 ; . . .; vm x1 ; . . .; xm ; v1 ; . . .; vm ¼ w ! 1 1 o 1 m ~ u x ; . . .; x ; v ¼w ox1 u1 ðx1 ;...;xm Þ !! o þ þ vm oxm u1 ðx1 ;...;xm Þ !! ~ u1 x1 ; . . .; xm ; vi o ¼w oxi u1 ðx1 ;...;xm Þ !!! j i o~x 1 o 1 1 m m ~ x ; . . .; x ¼ w u x ; . . .; x ; v o~x j u1 ðx1 ;...;xm Þ oxi !! j x 1 o 1 1 m i o~ m ~ x ; . . .; x ¼ w u x ; . . .; x ; v o~x j u1 ðx1 ;...;xm Þ oxi 1 1 1 1 m ¼ ðp1 wÞ u x ; ; x ; . . .; ðpm wÞ u x ; . . .; xm ; 1 m o~x 1 x 1 m i o~ m x ; . . .; x x ; . . .; x vi ; . . .; v oxi oxi 1 1 m ¼ p1 w u x ; . . .; x ; . . .; pm w u1 x1 ; . . .; xm ; 1 m o~x 1 o~x 1 x ; . . .; xm ; . . .; vi x ; . . .; xm vi oxi oxi ¼ p1 w u1 x1 ; . . .; xm ; . . .; pm w u1 x1 ; . . .; xm ; ! ! 1 1 m oðw u1 Þ 1 m i oð w u Þ 1 m x ; . . .; x x ; . . .; x vi ; . . .; v oxi oxi 1 m 1 x ; . . .; xm ; ¼ w u1 x1 ; . . .; xm ; . . .; w u1 ! ! 1 1 m oðw u1 Þ 1 m i oð w u Þ 1 m x ; . . .; x x ; . . .; x ; . . .; v : vi oxi oxi

5.3 Tangent Bundles

339

Thus,

~ u ~ 1 w

v

i

1 m 1 x ; . . .; xm ; x1 ; . . .; xm ; v1 ; . . .; vm ¼ w u1 x1 ; . . .; xm ; . . .; w u1

! 1 1 m oðw u1 Þ 1 m i oð w u Þ 1 m x ; . . .; x x ; . . .; x Þ: ; . . .; v oxi oxi

Here, w u1 is smooth, so each of ðx1 ; . . .; xm ; v1 ; . . .; vm Þ 7! ðw u1 Þi ðx1 ; . . .; xm Þ is smooth, and each of

x ; . . .; x ; v ; . . .; v 1

m

1

m

7! v

i

j oðw u1 Þ 1 x ; . . .; xm i ox

!

~ u ~ 1 is smooth. is smooth, and hence, w 4. Since fMg is an open cover of M, by the Lemma 4.52, there exists a countable collection fðU1 ; u1 Þ; ðU2 ; u2 Þ; ðU3 ; u3 Þ; g of admissible coordinate charts of M such that fU1 ; U2 ; U3 ; g covers M. Here, each ðUk ; u‘ k Þ is an admissible coordinate charts of M, so each ðUk ; uk Þ 2 A, and hence, f r2Uk Tr Mgk2N is a ‘ countable subcollection of f r2U Tr MgðU;uÞ2A : ‘ ‘ T M: Since Now, we want to prove that f r2Uk Tr Mgk2N covers r2M ‘ r 1 1 fU1 ; U2 ; U3 ; . . .g covers M, M ¼ [k¼1 Uk ; and hence, [k¼1 ð q2Uk Tq MÞ ¼ ‘ ‘ ‘ ‘ ‘ Tq M ¼ q2M Tq M ¼ r2M Tr M: It follows that f q2U1 Tq M; q2U2 q2[1 k¼1 ‘ Uk ‘ Tq M; q2U3 Tq M; . . .g covers r2M Tr M: ‘ 5. Let ðp; vÞ 2 TM; ðq; wÞ 2 r2M Tr M; and ðp; vÞ 6¼ ðq; wÞ: Since ðp; vÞ 2 ‘ r2M Tr M; v 2 Tp M: Similarly, w 2 Tq M: Case I: when p ¼ q. Since p ¼ q; and ðp; vÞ 6¼ ðq; wÞ; v 6¼ w: Since p ¼ q; and w 2 Tq M; w 2 Tp M: Since p 2 M; and M is an m-dimensional smooth ‘ manifold, there exists ðV; wÞ 2 A such that p 2 V: Since ðV; wÞ 2 A; r2V Tr M 2 ‘ ‘ f r2U Tr MgðU;uÞ2A : Since p 2 V; and v 2 Tp M; ðp; vÞ 2 r2V Tr M: Since p 2 ‘ V; and p ¼ q; q 2 V: Since q 2 V; and w 2 Tq M; ðq; wÞ 2 r2V Tr M: Thus, there ‘ ‘ ‘ exists r2V Tr M 2 f r2U Tr MgðU;uÞ2A such that ðp; vÞ; ðq; wÞ 2 r2V Tr M: Case II: when p 6¼ q. Since M is an m-dimensional smooth manifold, its topology is Hausdorff. Since the topology of M is Hausdorff, p 6¼ q; and p; q 2 M; there exist an open neighborhood G1 of p and an open neighborhood G2 of q such that G1 \ G2 ¼ ;: Since p 2 M; and M is a smooth manifold, there exists ðV; wÞ 2 A such that p 2 V: Since ðV; wÞ 2 A; and p 2 V; V is an open neighborhood of p. Since V is an open neighborhood of p, and G1 is an open neighborhood of p, G1 \ V is an open neighborhood of p. Since ðV; wÞ 2 A; G1 \ V is an open subset of V, and A is a smooth structure on M, ðG1 \ V; wjG1 \V Þ 2 A: Similarly,

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5 Immersions, Submersions, and Embeddings

there exists ðW; vÞ 2 A such that ðG2 \ W; vjG2 \W Þ 2 A: Since ðG1 \ V; ‘ ‘ ‘ wjG1 \V Þ 2 A; r2G1 \V Tr M 2 f r2U Tr MgðU;uÞ2A : Similarly, r2G2 \W Tr M 2 ‘ f r2U Tr MgðU;uÞ2A : Here, G1 \ V is an open neighborhood of p, so p 2 G1 \ V: ‘ Since p 2 G1 \ V; and v 2 Tp M; ðp; vÞ 2 r2G1 \V Tr M: Similarly, ðq; wÞ 2 ‘ ‘ ‘ ‘ r2G2 \W Tr M: Next, ð r2G1 \V Tr MÞ \ ð r2G2 \W Tr MÞ ¼ r2ðG1 \VÞ\ðG2 \WÞ ‘ ‘ ‘ Tr M ¼ r2ðG1 \G2 Þ\ðV\WÞ Tr M ¼ r2;\ðV\WÞ Tr M ¼ r2; Tr M ¼ ;: Hence, the condition 5 is satisﬁed. Now, by Lemma 5.21, there exists a unique ‘smooth differential structure ‘ B over ‘ ~ T M such that for each ðU; uÞ 2 A; ð T M; u Þ 2 B: Thus, r2M r r2U r r2M Tr M becomes a 2m-dimensional smooth manifold. This 2m-dimensional smooth mani‘ fold r2M Tr M is called the tangent bundle of M. It is denoted by TM. Here, ‘ ~ Þ is corresponding to each admissible coordinate chart ðU; uÞ of M, ð r2U Tr M; u an admissible coordinate chart of TM. Theorem 5.23 Let M be an m-dimensional smooth manifold. Let ðM; uÞ be an admissible coordinate chart of M. Then, TM is diffeomorphic onto M Rm : Proof ‘ Since ðM; uÞ is an admissible coordinate chart of M, by the ‘ Note 5.22, ~ Þ is an admissible coordinate chart of TM, where u ~ : r2M Tr M ! ð r2M Tr M; u ‘ ~ Þ is an admissible coordinate uðMÞ Rm : Now, since r2M Tr M ¼ TM; ðTM; u ~ : TM ! uðMÞ Rm : Since ðM; uÞ is an admissible coorchart of TM, where u dinate chart of M, φ is a diffeomorphism from M onto φ(M). Since φ is a diffeo~ Þ is an morphism from M onto φ(M), M is diffeomorphic onto φ(M). Since ðTM; u ~ : TM ! uðMÞ Rm ; u ~ : TM ! admissible coordinate chart of TM; where u uðMÞ Rm is a diffeomorphism, and hence, TM is diffeomorphic onto uðMÞ Rm : Since TM is diffeomorphic onto uðMÞ Rm ; and M is diffeomorphic onto uðMÞ; TM is diffeomorphic onto M Rm : h Deﬁnition Let M be an m-dimensional smooth manifold, N be an n-dimensional smooth manifold, and F : M ! N be smooth. ‘ The mapping dF : TM ! TN is deﬁned as follows: For every ðp; vÞ 2 TMð¼ r2M Tr MÞ; where p 2 M; and v 2 Tp M; ðdF Þðp; vÞ FðpÞ; dFp ðvÞ : Here, dF is called the global differential of F (or, global tangent map of F). Theorem 5.24 Let M be an m-dimensional smooth manifold, N be an n-dimensional smooth manifold, and F : M ! N be smooth. Then, dF : TM ! TN is smooth. ‘ Proof Let us take any ðp; vÞ 2 TMð¼ r2M Tr MÞ where p 2 M; and ðp; vÞ 2 Tp M: Let us take any admissible coordinate chart ðU; uÞ of M such that p 2 U: For every q 2 U; let

5.3 Tangent Bundles

341

! o o o ; ; . . .; m ox1 q ox2 q ox q be the coordinate basis of Tq M corresponding to ðU; uÞ; where ! o 1 o : d u uðqÞ oxi q oxi uðqÞ ‘ ‘ Here, r2U Tr Mð r2M Tr M ¼ TMÞ is an open neighborhood of ðp; vÞ in TM; and uðUÞ Rm ð Rm Rm ¼‘ R2m Þ is open in R2m : ~ : r2U Tr M ! uðUÞ Rm as follows: For every Let us deﬁne a function u ! !! a o o m q; v Tr M; þ þ v 2 ox1 q oxm q r2U ! !! o o 1 m ~ q; v u þ þ v ðp1 uÞðqÞ; . . .; ðpm uÞðqÞ; v1 ; . . .; vm ; ox1 q oxm q 1

where pi : Rm ! R‘ denotes the ith projection function of Rm : ~ Þ is an admissible coordinate chart of TM. Here, We know that ð r2U Tr M; u F : M ! N; and p 2 M; so FðpÞ 2 N: Let us take any admissible coordinate chart ðV; wÞ of N such that FðpÞ 2 V: For every r 2 V; let o o o ; ; . . .; n oy1 r oy2 r oy r

be the coordinate basis of TrN corresponding to ðV; wÞ; where ! o 1 o d w wðrÞ : oyi r oyi wðrÞ ‘ ‘ Here, r2V Tr Nð r2N Tr N ¼ TNÞ is an open neighborhood of ðdFÞðp; vÞ ð¼ ðFðpÞ; ðdFp ÞðvÞÞÞ; and wðVÞ Rn ð Rn Rn ¼ R2n Þ is open in R2n : Let us ~ : ‘ Tr N ! wðVÞ Rn as follows: For every deﬁne a function w r2V a o o n r; w1 Tr N; þ þ w 2 1 oy r oyn r r2V o o 1 n ~ w r; w þ þ w p01 w ðrÞ; . . .; p0n w ðrÞ; w1 ; . . .; wn ; oy1 r oyn r

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5 Immersions, Submersions, and Embeddings

where p0i : Rn ! R denotes the ith projection function of Rn : We know that ‘ ~ is an admissible coordinate chart of TN: We have to prove that ð r2V Tr N; wÞ ‘ 1 ~ ‘ Tr NÞÞ is smooth. Here, for every ~ Fu ~ ðð r2U Tr MÞÞ ! wðð ~ :u w r2V ‘ ~ ð r2U\V Tr MÞ; we have ðx1 ; . . .; xm ; v1 ; . . .; vm Þ 2 u 1 1 ~ Fu ~ u ~ 1 x1 ; . . .; xm ; v1 ; . . .; vm ¼ w ~ x ; . . .; xm ; v1 ; . . .; vm w ! !! o m ~ F u1 x1 ; . . .; xm ; v1 o ¼ w þ þ v ox1 u1 ðx1 ;...;xm Þ oxm u1 ðx1 ;...;xm Þ !! ~ F u1 x1 ; . . .; xm ; vi o ¼ w oxi u1 ðx1 ;...;xm Þ !!! ~ F u1 x1 ; . . .; xm ; vi o ¼w oxi u1 ðx1 ;...;xm Þ !!! ~ F u1 x1 ; . . .; xm ; dFp vi o ¼w oxi u1 ðx1 ;...;xm Þ !!! o ~ F u1 x1 ; . . .; xm ; vi dFp ¼w oxi u1 ðx1 ;...;xm Þ 1 0 00 !11 o p0j ðw F u1 Þ 1 1 i o m 1 1 m ~ @ A @ @ AA u u x ; . . .; x ¼ w F u x ; . . .; x ; v oy j F ðu1 ðx1 ;...;xm ÞÞ oxi 1 0 00 !11 o p0j ðw F u1 Þ 1 1 i o m 1 m A ~ @ @ @ AA ¼ w F u x ; . . .; x ; v x ; . . .; x oy j 1 1 m oxi F ðu ðx ;...;x ÞÞ

0 0 11 !1 o p0j ðw F u1 Þ 1 1 o m i@ 1 m AA ~ @ @ A ¼ w F u x ; . . .; x ; v x ; . . .; x oy j F ðu1 ðx1 ;...;xm ÞÞ oxi ¼ p01 w F u1 x1 ; . . .; xm ; . . .; p0n w F u1 x1 ; . . .; xm ; 0 o p1 ðw F u1 Þ 1 o p0n ðw F u1 Þ 1 vi x ; . . .; xm ; . . .; vi x ; . . .; xm i i ox ox ¼ p01 w F u1 x1 ; . . .; xm ; . . .; p0n w F u1 x1 ; . . .; xm ; 0 0 o p1 w F u1 1 o pn w F u1 1 vi : x ; . . .; xm ; . . .; vi x ; . . .; xm i i ox ox 0

Thus,

~ Fu ~ 1 x1 ; . . .; xm ; v1 ; . . .; vm w ¼ p01 w F u1 x1 ; . . .; xm ; . . .; p0n w F u1 x1 ; . . .; xm ; 0 0 1 o p1 w F u1 1 m i o pn w F u 1 m vi x ; . . .; x x ; . . .; x ; . . .; v oxi oxi

Since F : M ! N is smooth, ðU; uÞ is an admissible coordinate chart of M such that p 2 U; ðV; wÞ is an admissible coordinate chart of N such that FðpÞ 2 V; w F u1 is smooth. Since w F u1 is smooth, each function

5.3 Tangent Bundles

343

ðx1 ; . . .; xm ; v1 ; . . .; vm Þ ! ðp0j w F u1 Þðx1 ; . . .; xm Þ is smooth, and each function 0 1 o p0j w F u1 ðx1 ; . . .; xm ; v1 ; . . .; vm Þ 7! vi @ x1 ; . . .; xm A oxi ~ Fu ~ 1 is smooth. is smooth, and hence, w

h

5.4 Smooth Immersion Note 5.25 Let M; N; P be smooth manifolds. Let F : M ! N; and G : N ! P be smooth maps. Since F : M ! N is smooth, by Theorem 5.24, dF : TM ! TN is smooth. Similarly, dG : TN ! TP is smooth. Since dF : TM ! TN is smooth, and dG : TN ! TP is smooth, ðdGÞ ðdFÞ : TM ! TP is smooth. Since F : M ! N is smooth, and G : N ! P is smooth, G F : M ! P is smooth, and hence, by Theorem 5.24, dðG FÞ : TM ! TP is smooth. We shall ‘ try to show that dðG FÞ ¼ ðdGÞ ðdFÞ; that is, for every ðp; vÞ 2 TMð¼ r2M Tr MÞ; where p 2 M; and v 2 Tp M; ðdðG FÞÞðp; vÞ ¼ ððdGÞ ðdFÞÞðp; vÞ: LHS ¼ ðdðG F ÞÞðp; vÞ ¼ ðG F ÞðpÞ; dðG F Þp ðvÞ ¼ GðFðpÞÞ; dðG F Þp ðvÞ ¼ GðFðpÞÞ; dGFðpÞ dFp ðvÞ ¼ GðFðpÞÞ; dGFðpÞ dFp ðvÞ ; RHS ¼ ððdGÞ ðdF ÞÞðp; vÞ ¼ ðdGÞððdF Þðp; vÞÞ ¼ ðdGÞ FðpÞ; dFp ðvÞ ¼ GðFðpÞÞ; dGFðpÞ dFp ðvÞ :

Hence, LHS ¼ RHS: Thus, we have shown that dðG FÞ ¼ ðdGÞ ðdFÞ: Note 5.26 Let M be a smooth manifold. Here, IdM : M ! M is given by IdM ðxÞ ¼ x for every x in M. Clearly, IdM is a smooth function. Hence, by Theorem 5.24, dðIdM Þ : TM ! TM is smooth. Here, IdTM : TM ! TM is a smooth function. We ‘ shall try to show that dðIdM Þ ¼ IdTM ; that is, for every ðp; vÞ 2 TMð¼ r2U Tr MÞ; where p 2 M; and v 2 Tp M; ðdðIdM ÞÞðp; vÞ ¼ ðIdTM Þðp; vÞ: LHS ¼ ðdðIdM ÞÞðp; vÞ ¼ IdM ðpÞ; dðIdM Þp ðvÞ ¼ p; dðIdM Þp ðvÞ ¼ p; dðIdM Þp ðvÞ ¼ p; IdTp M ðvÞ ¼ ðp; vÞ ¼ ðIdTM Þðp; vÞ ¼ RHS: Thus, we have shown that dðIdM Þ ¼ IdTM : Note 5.27 Let M, N be smooth manifolds. Let F : M ! N be a diffeomorphism. Since F : M ! N is a diffeomorphism, F : M ! N is smooth, and hence, dF :

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5 Immersions, Submersions, and Embeddings

TM ! TN is smooth. We want to prove that dF : TM ! TN is a diffeomorphism, that is, dF is 1–1 onto, and ðdFÞ1 is smooth. dF : TM ! TN is 1–1: For this purpose, let ðdFÞðp; vÞ ¼ ðdFÞðq; wÞ; where p; q 2 M; and v; w 2 Tp M: We have to show that ðp; vÞ ¼ ðq; wÞ; that is, p = q, and v = w. Here, ðFðpÞ; ðdFp ÞðvÞÞ ¼ ðdFÞðp; vÞ ¼ ðdFÞðq; wÞ ¼ ðFðqÞ; ðdFq ÞðwÞÞ; so FðpÞ ¼ FðqÞ; and ðdFp ÞðvÞ ¼ ðdFq ÞðwÞ: Since F : M ! N is a diffeomorphism, F is 1–1. Since F is 1–1, and FðpÞ ¼ FðqÞ; p = q. Since p = q, and ðdFp ÞðvÞ ¼ ðdFq ÞðwÞ; ðdFp ÞðvÞ ¼ ðdFp ÞðwÞ: Since F : M ! N is a diffeomorphism, and p 2 M; dFp : Tp M ! TFðpÞ N is an isomorphism, and hence, dFp is 1–1. Since dFp is 1–1, and ðdFp ÞðvÞ ¼ ðdFp ÞðwÞ; v = w. Thus, dF : TM ! TN is 1–1. dF : TM ! TN is onto: For this purpose, let us take any ðq; wÞ 2 TN; where q 2 N; and w 2 Tq N: Since F : M ! N is a diffeomorphism, F : M ! N is onto. Since F : M ! N is onto, and q 2 N; there exists p 2 M such that FðpÞ ¼ q: Since FðpÞ ¼ q; and w 2 Tq N; w 2 TFðpÞ N: Since F : M ! N is a diffeomorphism, and p 2 M; dFp : Tp M ! TFðpÞ N is an isomorphism, and hence, Tp M ! TFðpÞ N is onto. Since dFp : Tp M ! TFðpÞ N is onto, and w 2 TFðpÞ N; there exists v 2 Tp M such that ðdFp ÞðvÞ ¼ w: Since p 2 M; and v 2 Tp M; ðp; vÞ 2 TM: Also, ðdFÞðp; vÞ ¼ ðFðpÞ; ðdFp ÞðvÞÞ ¼ ðq; ðdFp ÞðvÞÞ ¼ ðq; wÞ: This shows that dF : TM ! TN is onto. ðdFÞ1 : TN ! TM is smooth: Since F : M ! N is a diffeomorphism, F 1 : N ! M is smooth, and hence, dðF 1 Þ : TN ! TMis smooth. Now, we shall try to show that ðdFÞ1 ¼ dðF 1 Þ; that is, for every ðq; wÞ 2 TN; where q 2 N; and w 2 Tq N; ðdFÞ1 ðq; wÞ ¼ ðdðF 1 ÞÞðq; wÞ; that is, for every ðq; wÞ 2 TN; where q 2 N; and w 2 Tq N; ðq; wÞ ¼ ðdFÞððdðF 1 ÞÞðq; wÞÞ: For this purpose, let us take any ðq; wÞ 2 TN; where q 2 N; and w 2 Tq N: Now, since dF : TM ! TN is onto, there exists ðp; vÞ 2 TM such that ðFðpÞ; ðdFp ÞðvÞÞ ¼ ðdFÞðp; vÞ ¼ ðq; wÞ: It follows that FðpÞ ¼ q; and ðdFp ÞðvÞ ¼ w: Since F : M ! N is a diffeomorphism, and p 2 M; dFp : Tp M ! TFðpÞ N is an isomorphism, and hence, dFp : Tp M ! TFðpÞ N is 1–1 onto. Since dFp : Tp M ! TFðpÞ N is 1–1 onto, and ðdFp ÞðvÞ ¼ w; v ¼ ðdFp Þ1 ðwÞ: Since F : M ! N is a diffeomorphism, F : M ! N is a 1–1 onto. Since F : M ! N is a 1–1 onto, and FðpÞ ¼ q; p ¼ F 1 ðqÞ: Now, RHS ¼ ðdF Þ d F 1 ðq; wÞ ¼ ðdF Þ F 1 ðqÞ; d F 1 q ðwÞ ¼ ðdF Þ F 1 ðqÞ; d F 1 q ðwÞ ¼ ðdF Þ F 1 ðqÞ; d F 1 FðpÞ ðwÞ 1 1 ¼ ðdF Þ F 1 ðqÞ; dFp ðwÞ ¼ ðdF Þ p; dFp ðwÞ ¼ ðdF Þðp; vÞ ¼ FðpÞ; dFp ðvÞ ¼ q; dFp ðvÞ ¼ ðq; wÞ ¼ LHS:

5.4 Smooth Immersion

345

Thus, ðdFÞ1 ¼ dðF 1 Þ: Since ðdFÞ1 ¼ dðF 1 Þ; and dðF 1 Þ : TN ! TM is smooth, ðdFÞ1 : TN ! TMis smooth. This proves that if F : M ! N is a diffeomorphism, then dF : TM ! TN is a diffeomorphism. Note 5.28 Let M; N be smooth manifolds. Let F : M ! N be a diffeomorphism. From the Note 5.27, dF : TM ! TN is a diffeomorphism. Since dF : TM ! TN is a diffeomorphism, dF : TM ! TN is 1–1 onto, and hence, ðdFÞ1 : TN ! TM: Since F : M ! N is a diffeomorphism, F 1 : N ! M is a diffeomorphism, and hence, by the Note 5.27, dðF 1 Þ : TN ! TM is a diffeomorphism. Thus, ðdFÞ1 : TN ! TM; and dðF 1 Þ : TN ! TM: Now, we shall try to show that ðdFÞ1 ¼ dðF 1 Þ; that is, for every ðq; wÞ 2 TN; where q 2 N; and w 2 Tq N; ðdFÞ1 ðq; wÞ ¼ ðdðF 1 ÞÞðq; wÞ; that is, for every ðq; wÞ 2 TN; where q 2 N; and w 2 Tq N; ðq; wÞ ¼ ðdFÞððdðF 1 ÞÞðq; wÞÞ: For this purpose, let us take any ðq; wÞ 2 TN; where q 2 N; and w 2 Tq N: Now, since dF : TM ! TN is onto, there exists ðp; vÞ 2 TM such that ðFðpÞ; ðdFp ÞðvÞÞ ¼ ðdFÞðp; vÞ ¼ ðq; wÞ: It follows that FðpÞ ¼ q; and ðdFp ÞðvÞ ¼ w: Since F : M ! N is a diffeomorphism, and p 2 M; dFp : Tp M ! TFðpÞ N is an isomorphism, and hence, dFp : Tp M ! TFðpÞ N is 1–1 onto. Since dFp : Tp M ! TFðpÞ N is 1–1 onto, and ðdFp ÞðvÞ ¼ w; v ¼ ðdFp Þ1 ðwÞ: Since F : M ! N is a diffeomorphism, F : M ! N is a 1–1 onto. Since F : M ! N is a 1–1 onto, and FðpÞ ¼ q; p ¼ F 1 ðqÞ: Now, RHS ¼ ðdF Þ d F 1 ðq; wÞ ¼ ðdF Þ F 1 ðqÞ; d F 1 q ðwÞ ¼ ðdF Þ F 1 ðqÞ; d F 1 q ðwÞ ¼ ðdF Þ F 1 ðqÞ; d F 1 FðpÞ ðwÞ 1 1 ¼ ðdF Þ F 1 ðqÞ; dFp ðwÞ ¼ ðdF Þ p; dFp ðwÞ ¼ ðdF Þðp; vÞ ¼ FðpÞ; dFp ðvÞ ¼ q; dFp ðvÞ ¼ ðq; wÞ ¼ LHS: Thus, we have shown that if F : M ! N is a diffeomorphism, then ðdFÞ1 ¼ dðF 1 Þ: Deﬁnition Let t0 be a real number. Let e be any positive real number. Let M be an m-dimensional smooth manifold. Let c : ðt0 e; t0 þ eÞ ! M be a smooth map. Observe that the open interval ðt0 e; t0 þ eÞ is a one-dimensional smooth manifold. Since c : ðt0 e; t0 þ eÞ ! M is a smooth map, and t0 2 ðt0 e; t0 þ eÞ; dct0 : Tt0 ðt0 e; t0 þ eÞ ! Tcðt0 Þ M: The standard coordinate basis vector of Tt0 R is d : denoted by dt t0

346

5 Immersions, Submersions, and Embeddings

d is the standard coordinate Since Tt0 ðt0 e; t0 þ eÞ is the same as Tt0 R; and dt t0 d is the standard coordinate basis vector of T ðt e; t þ eÞ: basis vector of Tt0 R; dt t0 0 0 t0 d is in T ðt e; t þ eÞ; and dc : T ðt e; t þ eÞ ! T M; Since dt t0 0 0 t0 0 0 cðt0 Þ t0 t0 d Þ 2 T M: Here, ðdc Þð d Þ is denoted by c0 ðt Þ; and is called the ðdct0 Þðdt 0 cðt0 Þ t0 dt t0

t0

velocity of c at t0 : Since

! d c0 ðt0 Þ ¼ dct0 2 Tcðt0 Þ M; so c0 ðt0 Þ 2 Tcðt0 Þ M: dtt0

Since c0 ðt0 Þ 2 Tcðt0 Þ M; c0 ðt0 Þ is a function from C 1 ðMÞ to R: Hence, for every f 2 C 1 ðMÞ; we have 0

ðf cÞ ðt0 Þ ¼

! d ð f cÞ ¼ dtt0

!! d dct0 ðf Þ ¼ ðc0 ðt0 ÞÞðf Þ: dtt0

Thus, for every f 2 C 1 ðMÞ; we have ðc0 ðt0 ÞÞðf Þ ¼ ðf cÞ0 ðt0 Þ: Here, cðt0 Þ 2 M; and M is an m-dimensional smooth manifold, so there exists an admissible coordinate chart ðU; uÞ of M such that cðt0 Þ 2 U: Let ! o o ; . . .; m ox1 cðt0 Þ ox cðt0 Þ be the coordinate basis of Tcðt0 Þ M corresponding to ðU; uÞ; where ! o o 1 d u uðcðt0 ÞÞ : oxi cðt0 Þ oxi uðcðt0 ÞÞ So the matrix representation of linear map dct0 is the following m 1 matrix: 3 dðp1 ðucÞÞ ðt Þ 0 dt 7 6 6 dðp2 ðucÞÞ ðt Þ 7 0 7 6 dt 7; 6 7 6 .. 5 4. dðpm ðucÞÞ ðt Þ 0 dt 2

5.4 Smooth Immersion

347

where pi : Rm ! R denotes the ith projection function of Rm : It follows that ! ! d dðp1 ðu cÞÞ o c ðt0 Þ ¼ dct0 ðt 0 Þ ¼ dtt0 dt ox1 cðt0 Þ ! dðpm ðu cÞÞ o þ þ ðt 0 Þ dt oxm cðt0 Þ ! dðpi ðu cÞÞ o ¼ ðt 0 Þ : dt oxi cðt0 Þ

0

Thus, ! dðpi ðu cÞÞ o c ðt0 Þ ¼ ðt 0 Þ : dt oxi cðt0 Þ 0

In the case of Euclidean space Rm for M, the above formula takes the form: ! d ð pi c Þ o c ðt0 Þ ¼ ðt0 Þ ; dt oxi cðt0 Þ 0

which is essentially the same as in the classical calculus. Theorem 5.29 Let M be an m-dimensional smooth manifold. Let p 2 M: Let v 2 Tp M: Then, there exist a positive real e and a smooth curve c : ðe; eÞ ! M such that cð0Þ ¼ p; and c0 ð0Þ ¼ v: Proof Since p 2 M; and M is an m-dimensional smooth manifold, there exists an admissible coordinate chart ðU; uÞ of M such that p 2 U; and uðpÞ ¼ 0: For every q 2 U; let ! o o ; . . .; m ox1 q ox q be the coordinate basis of Tq M corresponding to ðU; uÞ; where ! o 1 o d u uðqÞ : oxi q oxi uðqÞ Here, ! o o ; . . .; m ox1 p ox p

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5 Immersions, Submersions, and Embeddings

is the coordinate basis of Tp M; and v 2 Tp M; there exist real numbers v1 ; . . .; vm such that v¼v

1

! ! ! o o o m i þ þ v ¼v : ox1 p oxm p oxi p

Since ðU; uÞ is an admissible coordinate chart of M, u is a mapping from U onto uðUÞ; where uðUÞ is an open neighborhood of uðpÞð¼ 0Þ in Rm : Since t 7! ðtv1 ; . . .; tvm Þ is a continuous map from R to Rm ; and uðUÞ is an open neighborhood of uðpÞð¼ ð0; . . .; 0Þ ¼ ð0v1 ; . . .; 0vm ÞÞ in Rm ; there exists e [ 0 such that for every t 2 ð0 e; 0 þ eÞ, we have ðtv1 ; . . .; tvm Þ 2 uðUÞ; and hence, for every t 2 ðe; eÞ, we have u1 ðtv1 ; . . .; tvm Þ 2 U: Now, let us deﬁne c : ðe; eÞ ! Uð MÞ as follows: For every t 2 ðe; eÞ; cðtÞ u1 ðtv1 ; . . .; tvm Þ: Here, cð0Þ ¼ u1 ð0v1 ; . . .; 0vm Þ ¼ u1 ð0Þ ¼ u1 ðuðpÞÞ ¼ p: Thus, cð0Þ ¼ p: We shall try to show that c is smooth. Since ðU; uÞ is an admissible coordinate chart of M; u is a diffeomorphism from U onto uðUÞ; and hence, u1 is smooth. Since u1 is smooth, and t 7! ðtv1 ; . . .; tvm Þ is smooth, their composite t 7! u1 ðtv1 ; . . .; tvm Þð¼ cðtÞÞ is smooth, and hence, c is smooth. It follows that for every t 2 ðe; eÞ we have ðpi ðu cÞÞðtÞ ¼ pi ðuðcðtÞÞÞ ¼ pi u u1 tv1 ; . . .; tvm ¼ pi tv1 ; . . .; tvm ¼ tvi ¼ vi t; so d ð pi ð u c Þ Þ ð0Þ ¼ vi : dt Now, we shall try to show that c0 ð0Þ ¼ v: Here, ! ! ! d ð pi ð u c Þ Þ o o o i i ð0Þ c ð0Þ ¼ ¼v ¼v ¼ v: dt oxi cð0Þ oxi cð0Þ oxi p 0

Hence, c0 ð0Þ ¼ v:

h

Note 5.30 Let M be an m-dimensional smooth manifold, N be an n-dimensional smooth manifold, and F : M ! N be a smooth map. Let c : ðt0 e; t0 þ eÞ ! M be a smooth map. Since c : ðt0 e; t0 þ eÞ ! M is a smooth map, and F : M ! N is a smooth map, their composite F c : ðt0 e; t0 þ eÞ ! N is a smooth map, and hence, ðF cÞ0 ðt0 Þ 2 TðFcÞðt0 Þ N ¼ TFðcðt0 ÞÞ N: Since c : ðt0 e; t0 þ eÞ ! M is a smooth map, c0 ðt0 Þ 2 Tcðt0 Þ M: Since c : ðt0 e; t0 þ eÞ ! M; cðt0 Þ 2 M: Since cðt0 Þ 2 M; and F : M ! N is smooth, dFcðt0 Þ : Tcðt0 Þ M ! TFðcðt0 ÞÞ N. Since dFcðt0 Þ :

5.4 Smooth Immersion

349

Tcðt0 Þ M ! TFðcðt0 ÞÞ N; and c0 ðt0 Þ 2 Tcðt0 Þ M; ðdFcðt0 Þ Þðc0 ðt0 ÞÞ 2 TFðcðt0 ÞÞ N: Thus, ðdFcðt0 Þ Þðc0 ðt0 ÞÞ and ðF cÞ0 ðt0 Þ are elements of TFðcðt0 ÞÞ N: We shall try to show that ðdFcðt0 Þ Þðc0 ðt0 ÞÞ ¼ ðF cÞ0 ðt0 Þ: ! ! d d ¼ dFcðt0 Þ dct0 RHS ¼ ðF cÞ ðt0 Þ ¼ dðF cÞt0 dtt0 dtt0 !! d ¼ dFcðt0 Þ ¼ dFcðt0 Þ ðc0 ðt0 ÞÞ ¼ LHS: dct0 dt t0 0

Thus, we have shown that ðdFcðt0 Þ Þðc0 ðt0 ÞÞ ¼ ðF cÞ0 ðt0 Þ: Note 5.31 Let M be an m-dimensional smooth manifold, N be an n-dimensional smooth manifold, and F : M ! N be a smooth map. Let p 2 M: Let v 2 Tp M: Let c : ðe; eÞ ! M be a smooth curve on M; such that cð0Þ ¼ p; and c0 ð0Þ ¼ v: (The existence of such a c has been established in Theorem 5.29.) Since F : M ! N is a smooth map, dFp : Tp M ! TFðpÞ N: Since dFp : Tp M ! TFðpÞ N; and v 2 Tp M; ðdFp ÞðvÞ 2 TFðpÞ N: Here, c : ðe; eÞ ! M is smooth, and F : M ! N is smooth, so their composite F c : ðe; eÞ ! N is smooth, and hence, ðF cÞ0 ð0Þ 2 TðFcÞð0Þ N ¼ TFðcð0ÞÞ N ¼ TFðpÞ N: Thus, ðdFp ÞðvÞ and ðF cÞ0 ð0Þ are members of TFðpÞ N: We shall try to show that ðdFp ÞðvÞ ¼ ðF cÞ0 ð0Þ: RHS ¼ ðF cÞ0 ð0Þ ¼ dFcð0Þ ðc0 ð0ÞÞ ¼ dFp ðc0 ð0ÞÞ ¼ dFp ðvÞ ¼ LHS: This proves that ðdFp ÞðvÞ ¼ ðF cÞ0 ð0Þ: ðc0 ð0ÞÞ ¼ ðF cÞ0 ð0Þ:

In

short,

we

write

ðdFcð0Þ Þ

Deﬁnition Let M be an m-dimensional smooth manifold, N be an n-dimensional smooth manifold, and F : M ! N be a smooth map. Let p 2 M: It follows that dFp : Tp M ! TFðpÞ N: We know that dFp is a linear map from real linear space Tp M to real linear space TFðpÞ N: Here, ðdFp ÞðTp MÞ is a linear subspace of TFðpÞ N: We know that dim Tp M ¼ m; and dim TFðpÞ N ¼ n: Since ðdFp ÞðTp MÞ is a linear subspace of TFðpÞ N; dimððdFp ÞðTp MÞÞ dim TFðpÞ N ¼ n: Since dFp is a linear map from real linear space Tp M to real linear space TFðpÞ N; dimððdFp Þ ðTp MÞÞ dim Tp M ¼ m: Since dimððdFp ÞðTp MÞÞ m; and dimððdFp ÞðTp MÞÞ

n; dimððdFp ÞðTp MÞÞ minfm; ng: Here, dimððdFp ÞðTp MÞÞ is called the rank of F at p. It follows that the rank of F at p is r if and only if the matrix representation of linear map dFp has rank r. Clearly, the rank of F at p is less than or equal to minfm; ng: Observe that the linear map dFp : Tp M ! TFðpÞ N is onto if and only if dimððdFp ÞðTp MÞÞ ¼ dim TFðpÞ Nð¼ nÞ: Thus, the linear map dFp : Tp M ! TFðpÞ N is onto if and only if the rank of F at p is n.

350

5 Immersions, Submersions, and Embeddings

Observe that the linear map dFp : Tp M ! TFðpÞ N is 1–1 if and only if dimððdFp ÞðTp MÞÞ ¼ dim Tp Mð¼ mÞ: Thus, the linear map dFp : Tp M ! TFðpÞ N is 1–1 if and only if the rank of F at p is m. Deﬁnition Let M be an m-dimensional smooth manifold, N be an n-dimensional smooth manifold, and F : M ! N be a smooth map. Let k be a positive integer. If for every p 2 M; the rank of F at p is k, then we write rank F ¼ k; and say that F has constant rank k. Let M be an m-dimensional smooth manifold, N be an ndimensional smooth manifold, and F : M ! N be a smooth map. Here, the linear map dFp : Tp M ! TFðpÞ N is onto for every p 2 M if and only if the rank of F at p is n for every p 2 M: Thus, the linear map dFp : Tp M ! TFðpÞ N is onto for every p 2 M if and only if rank F ¼ n: Note 5.32 Let M be an m-dimensional smooth manifold, N be an n-dimensional smooth manifold, and F : M ! N be a smooth map. Here, the linear map dFp : Tp M ! TFðpÞ N is 1–1 for every p 2 M if and only if the rank of F at p is m for every p 2 M: Thus, the linear map dFp : Tp M ! TFðpÞ N is 1–1 for every p 2 M if and only if rank F ¼ m: Deﬁnition Let M be an m-dimensional smooth manifold, N be an n-dimensional smooth manifold, and F : M ! N be a smooth map. If for every p 2 M; the linear map dFp : Tp M ! TFðpÞ N is onto, then we say that F is a smooth submersion. Thus, F is a smooth submersion if and only if rank F ¼ n: Deﬁnition Let M be an m-dimensional smooth manifold, N be an n-dimensional smooth manifold, and F : M ! N be a smooth map. If for every p 2 M; the linear map dFp : Tp M ! TFðpÞ N is 1–1, then we say that F is a smooth immersion. Thus, F is a smooth immersion if and only if rank F ¼ m:

5.5 Inverse Function Theorem for Manifolds Note 5.33 Let Mð2 3; RÞ be the collection of all 2 3 matrices with real entries, that is, Mð2 3; RÞ

a d

c : a; b; c; d; e; f 2 R : f

b e

We deﬁne addition of matrices as follows:

a d

b e

a c þ 1 f d1

b1 e1

c1 f1

a þ a1 b þ b1 c þ c 1 : d þ d1 e þ e1 f þ f1

Clearly, Mð2 3; RÞ is a commutative group under addition as binary operation.

5.5 Inverse Function Theorem for Manifolds

351

We deﬁne scalar multiplication of matrix as follows: t

a d

ta c td f

b e

tc : tf

tb te

Clearly, Mð2 3; RÞ becomes a real linear space of dimension ð2 3Þ: Here, one of the basis of Mð2 3; RÞ is

1 0

0 0

0 1 0 ; 0 0 0

0 0 ; 0 0

0 0 1 ; 1 0 0

0 0

0 0 ; 0 0

0 1

0 0 0 ; 0 0 0

0 1

:

Clearly, the real linear space R23 is isomorphic to the real linear space Mð2 3; RÞ: Since R23 is isomorphic to Mð2 3; RÞ; we will not distinguish between

a d

b e

c f

and

ða; b; c; d; e; f Þ:

Now, since R23 is a ð2 3Þ-dimensional smooth manifold, Mð2 3; RÞ is also a ð2 3Þ-dimensional smooth manifold. Let M2 ð2 3; RÞ be the collection of all

a d

c 2 Mð2 3; RÞ f

b e

such that

a rank d

b e

c ¼ 2: f

If

a d

b e

c 2 M2 ð2 3; RÞ; f

then

a rank d

b e

c ¼ 2; f

that is, if

a d

b e

c 2 M2 ð2 3; RÞ f

352

5 Immersions, Submersions, and Embeddings

then

det

b e

a c 6 0 or det ¼ d f

a c 6 0 or det ¼ d f

b 6 0 : ¼ e

We want to show that M2 ð2 3; RÞ is an open subset of Mð2 3; RÞ: For this purpose, let us ﬁx any

a d

b e

c 2 M2 ð2 3; RÞ: f

It follows that

b det e

c 6 0 ¼ f

or

c 6 0 ¼ f

a det d

or

a det d

b 6 0: ¼ e

Case I: when det

b e

c 6 0 ¼ f

Here,

b bf ce ¼ det e

c 6 0: ¼ f

Since there exists e [ 0, the open interval

b det e

c b e; det f e

c þe f

does not contain 0. Let us deﬁne a function f : R6 ! R as follows: For every ðx1 ; x2 ; x3 ; x4 ; x5 ; x6 Þ in R6

x f ðx1 ; x2 ; x3 ; x4 ; x5 ; x6 Þ x2 x6 x5 x3 ¼ det 2 x5

x3 : x6

Clearly, f is a continuous function. Also,

b c f ða; b; c; d; e; f Þ ¼ det : e f Since f is a continuous at ða; b; c; d; e; f Þ; there exists an open neighborhood U of ða; b; c; d; e; f Þ in R6 such that for every ðx1 ; x2 ; x3 ; x4 ; x5 ; x6 Þ 2 U;

5.5 Inverse Function Theorem for Manifolds

x2 det x5

x3 x6

353

¼ f ðx1 ; x2 ; x3 ; x4 ; x5 ; x6 Þ 2 ðf ða; b; c; d; e; f Þ e; f ða; b; c; d; e; f Þ þ eÞ b c b c þ e 63 0: e; det ¼ det e f e f

Thus, for every ðx1 ; x2 ; x3 ; x4 ; x5 ; x6 Þ 2 U; we have x2 x3 det 6¼ 0; x5 x6 and hence,

x rank 1 x4

x2 x5

x3 x6

¼2

for every ðx1 ; x2 ; x3 ; x4 ; x5 ; x6 Þ 2 U: Since U is an open neighborhood of ða; b; c; d; e; f Þ in R6 ;

x1 x2 x3 : ð x1 ; x2 ; x3 ; x4 ; x5 ; x6 Þ 2 U x4 x5 x6 is an open neighborhood of

a d

c in Mð2 3; RÞ: f

b e

Since

rank

x1 x4

x2 x5

x3 x6

¼2

for every ðx1 ; x2 ; x3 ; x4 ; x5 ; x6 Þ 2 U;

x1 x4

x2 x5

x3 x6

: ð x1 ; x2 ; x3 ; x4 ; x5 ; x6 Þ 2 U

M2 ð2 3; RÞ:

Since

x1 x4

x2 x5

x3 x6

: ð x1 ; x2 ; x3 ; x4 ; x5 ; x6 Þ 2 U

is an open neighborhood of

a d

b e

c in Mð2 3; RÞ; f

354

and

5 Immersions, Submersions, and Embeddings

x1 x4

x2 x5

x3 x6

: ð x1 ; x2 ; x3 ; x4 ; x5 ; x6 Þ 2 U

a d

is an interior point of M2 ð2 3; RÞ: Case II: when a det d This case is similar to the case I. Case III: when

a det d

b e

c f

M2 ð2 3; RÞ;

c 6 0 ¼ f

b 6 0 ¼ e

This case is similar to the case I. Hence, in all cases,

a d

b e

c f

is an interior point of M2 ð2 3; RÞ: This proves that M2 ð2 3; RÞ is an open subset of Mð2 3; RÞ: Since M2 ð2 3; RÞ is an open subset of Mð2 3; RÞ; and Mð2 3; RÞ is a ð2 3Þ-dimensional smooth manifold, M2 ð2 3; RÞ is also a ð2 3Þ-dimensional smooth manifold. Similarly, M2 ð3 2; RÞ is a ð3 2Þ-dimensional smooth manifold. Its generalization to higher dimensions can also be given. Such smooth manifolds are known as the manifolds of matrices of full rank. Theorem 5.34 Let M be an m-dimensional smooth manifold, N be an n-dimensional smooth manifold, and F : M ! N be a smooth map. Let p 2 M: If dFp : Tp M ! TFðpÞ N is onto, then there exists an open neighborhood W of p such that for every q 2 W; the linear map dFq : Tq M ! TFðqÞ N is onto. Proof Here, p 2 M; and M is an m-dimensional smooth manifold, so there exists an admissible coordinate chart ðU; uÞ of M such that p 2 U: For every q 2 U; let ! o o o ; ; . . .; m ox1 q ox2 q ox q be the coordinate basis of Tq M corresponding to ðU; uÞ; where

5.5 Inverse Function Theorem for Manifolds

355

! o 1 o d u uðqÞ : oxi q oxi uðqÞ Here, p 2 M; and F : M ! N; so FðpÞ 2 N: Since FðpÞ 2 N; and N is an n-dimensional smooth manifold, there exists an admissible coordinate chart ðV; wÞ of N such that FðpÞ 2 V: For every r 2 V; let o o o ; ; . . .; oy1 oy2 oyn r

r

r

be the coordinate basis of Tr N corresponding to ðV; wÞ; where ! o 1 o d w wðrÞ : oyi r oyi wðrÞ Here, the matrix representation of linear map dFp is the following n m matrix: 2

oðp1 ðw F u1 ÞÞ oðp1 ðw F u1 ÞÞ ðuðpÞÞ ðuðpÞÞ 6 1 ox ox2 6 6 oðp ðw F u1 ÞÞ 1 oðp2 ðw F u ÞÞ 6 2 ðuðpÞÞ ðuðpÞÞ 6 6 ox1 ox2 6... ... 6 4 oðpn ðw F u1 ÞÞ oðpn ðw F u1 ÞÞ ðuðpÞÞ ðuðpÞÞ 2 ox1 3 ox 1 oðp1 ðw F u ÞÞ ... ðuðpÞÞ 7 oxm 7 7 1 oðp2 ðw F u ÞÞ 7 ðuðpÞÞ 7: ... m 7 ox 7 ... ... 7 5 oðpn ðw F u1 ÞÞ ð uðpÞ Þ ... m ox

Since the linear map dFp : Tp M ! TFðpÞ N is onto, and dim TFðpÞ N ¼ n; the rank of the matrix representation of linear map dFp is n; that is, 2

oðp1 ðw F u1 ÞÞ oðp1 ðw F u1 ÞÞ ðuðpÞÞ ðuðpÞÞ 6 1 ox ox2 6 6 1 1 oðp2 ðw F u ÞÞ 6 oðp2 ðw F u ÞÞ ðuðpÞÞ ðuðpÞÞ rank6 6 ox1 ox2 6... ... 6 4 oðpn ðw F u1 ÞÞ oðpn ðw F u1 ÞÞ ðuðpÞÞ ðuðpÞÞ 2 ox1 3 ox 1 oðp1 ðw F u ÞÞ ... ðuðpÞÞ 7 oxm 7 7 oðp2 ðw F u1 ÞÞ 7 ... ð uðpÞ Þ 7 7 oxm 7 ... ... 7 1 5 oðpn ðw F u ÞÞ ... ð uðpÞ Þ m ox

356

5 Immersions, Submersions, and Embeddings

is n. Since the rank of F at p is less than or equal to minfm; ng; and the rank of F at p is n, n minfm; ng m; and hence, n m: Here, n m; and 2

oðp1 ðw F u1 ÞÞ oðp1 ðw F u1 ÞÞ ð uðpÞ Þ ðuðpÞÞ 6 ox1 ox2 6 6 oðp ðw F u1 ÞÞ 1 oðp2 ðw F u ÞÞ 6 2 ðuðpÞÞ ðuðpÞÞ rank6 6 ox1 ox2 6... . . . 6 4 oðpn ðw F u1 ÞÞ oðpn ðw F u1 ÞÞ ðuðpÞÞ ðuðpÞÞ 1 ox ox2 3 oðp1 ðw F u1 ÞÞ ... ðuðpÞÞ 7 oxm 7 7 1 oðp2 ðw F u ÞÞ 7 ðuðpÞÞ 7 ... m 7 ox 7 ... ... 7 5 oðpn ðw F u1 ÞÞ ... ðuðpÞÞ oxm

is n, so

2

oðp1 ðw F u1 ÞÞ oðp1 ðw F u1 ÞÞ ðuðpÞÞ ðuðpÞÞ 6 1 ox ox2 6 6 oðp ðw F u1 ÞÞ oðp2 ðw F u1ÞÞ 6 2 ðuðpÞÞ ðuðpÞÞ 6 6 ox1 ox2 6... ... 6 4 oðpn ðw F u1 ÞÞ oðpn ðw F u1 ÞÞ ðuðpÞÞ ðuðpÞÞ 1 2 ox 3 ox 1 oðp1 ðw F u ÞÞ ... ðuðpÞÞ 7 oxm 7 7 oðp2 ðw F u1 ÞÞ 7 ... ð uðpÞ Þ 7 7 oxm 7 ... ... 7 1 5 oðpn ðw F u ÞÞ ... ð uðpÞ Þ oxm

2 Mn ðn m; RÞ Mðn m; RÞ: Since Mn ðn m; RÞ is an open subset of Mðn m; RÞ: It follows that Mn ðn m; RÞ is an open neighborhood of 2

oðp1 ðw F u1 ÞÞ oðp1 ðw F u1 ÞÞ ðuðpÞÞ ðuðpÞÞ 6 1 ox ox2 6 6 oðp ðw F u1 ÞÞ oðp2 ðw F u1 ÞÞ 6 2 ðuðpÞÞ ðuðpÞÞ 6 6 ox1 ox2 6... . . . 6 4 oðpn ðw F u1 ÞÞ oðpn ðw F u1 ÞÞ ðuðpÞÞ ðuðpÞÞ 1 2 ox 3 ox 1 oðp1 ðw F u ÞÞ ... ðuðpÞÞ 7 oxm 7 7 1 oðp2 ðw F u ÞÞ 7 ðuðpÞÞ 7: ... m 7 ox 7 ... ... 7 5 oðpn ðw F u1 ÞÞ ... ð uðpÞ Þ m ox

5.5 Inverse Function Theorem for Manifolds

357

Since F : M ! N is a smooth map, ðU; uÞ is an admissible coordinate chart of M such that p 2 U; and ðV; wÞ is an admissible coordinate chart of N such that FðpÞ 2 V; w F u1 is smooth, and hence, each pi ðw F u1 Þ is smooth. Since each pi ðw F u1 Þ is smooth, and u is smooth, each q 7!

oðpi ðw F u1 ÞÞ ðuðqÞÞ ox j

is continuous, and hence, 2

oðp1 ðw F u1 ÞÞ oðp1 ðw F u1 ÞÞ ðuðqÞÞ ðuðqÞÞ 6 1 ox ox2 6 6 oðp2 ðw F u1 ÞÞ 6 oðp2 ðw F u1 ÞÞ ð uðqÞ Þ ðuðqÞÞ q 7! 6 6 ox1 ox2 6... ... 6 4 oðpn ðw F u1 ÞÞ oðpn ðw F u1 ÞÞ ðuðqÞÞ ðuðqÞÞ ox1 ox2 3 oðp1 ðw F u1 ÞÞ ... ðuðqÞÞ 7 oxm 7 7 oðp2 ðw F u1 ÞÞ 7 ðuðqÞÞ 7 ... m 7 ox 7 ... ... 7 1 5 o ð pn ð w F u Þ Þ ... ð uðqÞ Þ oxm is continuous. It follows that there exists an open neighborhood W of p in M such that for every q 2 W, 2

oðp1 ðw F u1 ÞÞ oðp1 ðw F u1 ÞÞ ðuðqÞÞ ðuðqÞÞ 6 1 ox ox2 6 6 oðp2 ðw F u1 ÞÞ 6 oðp2 ðw F u1 ÞÞ ð uðqÞ Þ ðuðqÞÞ 6 6 ox1 ox2 6... ... 6 4 oðpn ðw F u1 ÞÞ oðpn ðw F u1 ÞÞ ðuðqÞÞ ðuðqÞÞ ox1 ox2 3 oðp1 ðw F u1 ÞÞ ... ðuðqÞÞ 7 oxm 7 7 oðp2 ðw F u1 ÞÞ 7 ðuðqÞÞ 7 2 Mn ðn m; RÞ; ... m 7 ox 7 ... ... 7 1 5 oðpn ðw F u ÞÞ ... ð uðqÞ Þ oxm and hence, for every q 2 W, we have

358

5 Immersions, Submersions, and Embeddings

2

oðp1 ðw F u1 ÞÞ oðp1 ðw F u1 ÞÞ ðuðqÞÞ ðuðqÞÞ 6 1 ox ox2 6 6 oðp ðw F u1 ÞÞ 1 oðp2 ðw F u ÞÞ 6 2 ðuðqÞÞ ðuðqÞÞ rank6 6 ox1 ox2 6... ... 6 4 oðpn ðw F u1 ÞÞ oðpn ðw F u1 ÞÞ ðuðqÞÞ ðuðqÞÞ ox1 ox2 3 oðp1 ðw F u1 ÞÞ ... ðuðqÞÞ 7 oxm 7 7 oðp2 ðw F u1 ÞÞ 7 ð uðqÞ Þ ... 7 ¼ n: 7 oxm 7 ... ... 7 5 oðpn ðw F u1 ÞÞ ... ðuðqÞÞ m ox

It follows that for every q 2 W; the linear map dFq : Tq M ! TFðqÞ N is onto.

h

Theorem 5.35 Let M be an m-dimensional smooth manifold, N be an n-dimensional smooth manifold, and F : M ! N be a smooth map. Let p 2 M: If dFp : Tp M ! TFðpÞ N is 1–1, then there exists an open neighborhood W of p such that for every q 2 W; the linear map dFq : Tq M ! TFðqÞ N is 1–1. Proof Here, p 2 M; and M is an m-dimensional smooth manifold, so there exists an admissible coordinate chart ðU; uÞ of M such that p 2 U: For every q 2 U; ! o o o ; ; . . .; m ox1 q ox2 q ox q is the coordinate basis of Tq M corresponding to ðU; uÞ; where ! o 1 o d u uðqÞ : oxi q oxi uðqÞ Here, p 2 M; and F : M ! N; so FðpÞ 2 N: Since FðpÞ 2 N; and N is an ndimensional smooth manifold, there exists an admissible coordinate chart ðV; wÞ of N such that FðpÞ 2 V: For every r 2 V; o o o ; ; . . .; oy1 r oy2 r oyn r is the coordinate basis of Tr N corresponding to ðV; wÞ; where ! o 1 o d w wðrÞ : oyi r oyi wðrÞ Here, the matrix representation of linear map dFp is the following n m matrix:

5.5 Inverse Function Theorem for Manifolds

359

2

oðp1 ðw F u1 ÞÞ oðp1 ðw F u1 ÞÞ ðuðpÞÞ ðuðpÞÞ 6 1 ox ox2 6 6 oðp ðw F u1 ÞÞ oðp2 ðw F u1 ÞÞ 6 2 ðuðpÞÞ ðuðpÞÞ 6 6 ox1 ox2 6... . . . 6 4 oðpn ðw F u1 ÞÞ oðpn ðw F u1 ÞÞ ðuðpÞÞ ðuðpÞÞ 1 ox ox2 3 oðp1 ðw F u1 ÞÞ ... ðuðpÞÞ 7 oxm 7 7 1 oðp2 ðw F u ÞÞ 7 ð uðpÞ Þ ... 7: m 7 ox 7 ... ... 7 1 5 oðpn ðw F u ÞÞ ... ðuðpÞÞ m ox

Since the linear map dFp : Tp M ! TFðpÞ N is 1–1, and dim Tp M ¼ m; the rank of the matrix representation of linear map dFp is m; that is, 2

oðp1 ðw F u1 ÞÞ oðp1 ðw F u1 ÞÞ ðuðpÞÞ ðuðpÞÞ 6 1 ox ox2 6 6 oðp2 ðw F u1 ÞÞ 6 oðp2 ðw F u1 ÞÞ ðuðpÞÞ ðuðpÞÞ rank6 6 ox1 ox2 6... . . . 6 4 oðpn ðw F u1 ÞÞ oðpn ðw F u1 ÞÞ ðuðpÞÞ ðuðpÞÞ 1 ox ox2 3 oðp1 ðw F u1 ÞÞ ... ðuðpÞÞ 7 oxm 7 7 1 oðp2 ðw F u ÞÞ 7 ð uðpÞ Þ ... 7 m 7 ox 7 ... ... 7 1 5 oðpn ðw F u ÞÞ ðuðpÞÞ ... m ox

is m. Since the rank of F at p is less than or equal to minfm; ng; and the rank of F at p is m, m minfm; ng n; and hence, m n: Here, m n; and 2

oðp1 ðw F u1 ÞÞ oðp1 ðw F u1 ÞÞ ðuðpÞÞ ðuðpÞÞ 6 1 ox ox2 6 6 oðp ðw F u1 ÞÞ oðp2 ðw F u1 ÞÞ 6 2 ðuðpÞÞ ðuðpÞÞ rank6 6 ox1 ox2 6... . . . 6 4 oðpn ðw F u1 ÞÞ oðpn ðw F u1 ÞÞ ðuðpÞÞ ðuðpÞÞ 2 ox1 3 ox 1 oðp1 ðw F u ÞÞ ... ðuðpÞÞ 7 oxm 7 7 1 oðp2 ðw F u ÞÞ 7 ðuðpÞÞ 7 ... m 7 ox 7 ... ... 7 5 oðpn ðw F u1 ÞÞ ... ð uðpÞ Þ m ox

360

5 Immersions, Submersions, and Embeddings

is m, so 2

oðp1 ðw F u1 ÞÞ oðp1 ðw F u1 ÞÞ ðuðpÞÞ ðuðpÞÞ 6 1 ox ox2 6 6 oðp ðw F u1 ÞÞ oðp2 ðw F u1 ÞÞ 6 2 ð uðpÞ Þ ðuðpÞÞ 6 6 ox1 ox2 6... . . . 6 4 oðpn ðw F u1 ÞÞ oðpn ðw F u1 ÞÞ ð uðpÞ Þ ðuðpÞÞ ox1 ox2 3 oðp1 ðw F u1 ÞÞ ... ðuðpÞÞ 7 oxm 7 7 oðp2 ðw F u1 ÞÞ 7 ð uðpÞ Þ ... 7 7 oxm 7 ... ... 7 1 5 oðpn ðw F u ÞÞ ... ð uðpÞ Þ oxm 2 Mm ðn m; RÞ Mðn m; RÞ: Since Mm ðn m; RÞ is an open subset of Mðn m; RÞ; it follows that Mm ðn m; RÞ is an open neighborhood of 2

oðp1 ðw F u1 ÞÞ oðp1 ðw F u1 ÞÞ ðuðpÞÞ ðuðpÞÞ 6 1 ox ox2 6 6 oðp ðw F u1 ÞÞ oðp2 ðw F u1 ÞÞ 6 2 ð uðpÞ Þ ðuðpÞÞ 6 6 ox1 ox2 6... . . . 6 4 oðpn ðw F u1 ÞÞ oðpn ðw F u1 ÞÞ ð uðpÞ Þ ðuðpÞÞ ox1 ox2 3 oðp1 ðw F u1 ÞÞ ... ðuðpÞÞ 7 oxm 7 7 oðp2 ðw F u1 ÞÞ 7 ð uðpÞ Þ ... 7: 7 oxm 7 ... ... 7 1 5 oðpn ðw F u ÞÞ ... ð uðpÞ Þ oxm Since F : M ! N is a smooth map, ðU; uÞ is an admissible coordinate chart of M such that p 2 U; and ðV; wÞ is an admissible coordinate chart of N such that FðpÞ 2 V; w F u1 is smooth, and hence, each pi ðw F u1 Þ is smooth. Since each pi ðw F u1 Þ is smooth, and u is smooth, each

5.5 Inverse Function Theorem for Manifolds

q 7!

361

oðpi ðw F u1 ÞÞ ðuðqÞÞ ox j

is continuous, and hence, 2

oðp1 ðw F u1 ÞÞ oðp1 ðw F u1 ÞÞ ðuðqÞÞ ðuðqÞÞ 6 1 ox ox2 6 6 oðp ðw F u1 ÞÞ oðp2 ðw F u1 ÞÞ 6 2 ð uðqÞ Þ ðuðqÞÞ q 7! 6 6 ox1 ox2 6... . . . 6 4 oðpn ðw F u1 ÞÞ oðpn ðw F u1 ÞÞ ðuðqÞÞ ðuðqÞÞ 1 ox ox2 3 oðp1 ðw F u1 ÞÞ ... ðuðqÞÞ 7 oxm 7 7 oðp2 ðw F u1 ÞÞ 7 ð uðqÞ Þ ... 7 7 oxm 7 ... ... 7 1 5 o ð pn ð w F u Þ Þ ... ð uðqÞ Þ oxm is continuous. It follows that there exists an open neighborhood W of p in M such that for every q 2 W; 2

oðp1 ðw F u1 ÞÞ oðp1 ðw F u1 ÞÞ ðuðqÞÞ ðuðqÞÞ 6 1 ox2 6 6 oðp ðw F u1 ÞÞ oðp2 ðw F u1 ÞÞ 6 2 ð uðqÞ Þ ðuðqÞÞ 6 6 ox1 ox2 6... . . . 6 4 oðpn ðw F u1 ÞÞ oðpn ðw F u1 ÞÞ ðuðqÞÞ ðuðqÞÞ 1 ox ox2 3 oðp1 ðw F u1 ÞÞ ... ðuðqÞÞ 7 oxm 7 7 oðp2 ðw F u1 ÞÞ 7 ð uðqÞ Þ ... 7 2 Mm ðn; RÞ; 7 oxm 7 ... ... 7 1 5 oðpn ðw F u ÞÞ ... ð uðqÞ Þ oxm and hence, for every q 2 W, we have

362

5 Immersions, Submersions, and Embeddings

2

oðp1 ðw F u1 ÞÞ oðp1 ðw F u1 ÞÞ ð uðqÞ Þ ðuðqÞÞ 6 ox1 ox2 6 6 oðp ðw F u1 ÞÞ oðp2 ðw F u1 ÞÞ 6 2 ðuðqÞÞ ðuðqÞÞ rank6 1 6 ox ox2 6... ... 6 4 oðpn ðw F u1 ÞÞ oðpn ðw F u1 ÞÞ ðuðqÞÞ ðuðqÞÞ ox1 ox2 3 oðp1 ðw F u1 ÞÞ ... ðuðqÞÞ 7 oxm 7 7 1 oðp2 ðw F u ÞÞ 7 ð uðqÞ Þ ... 7 ¼ m: 7 oxm 7 ... ... 7 1 5 oðpn ðw F u ÞÞ ... ð uðqÞ Þ oxm It follows that for every q 2 W; the linear map dFq : Tq M ! TFðqÞ N is 1–1.

h

Theorem 5.36 Let M be an m-dimensional smooth manifold, N be an m-dimensional smooth manifold, and F : M ! N be a smooth map. Let p 2 M: If dFp : Tp M ! TFðpÞ N is invertible (i.e., 1–1 and onto); then, there exists a connected open neighborhood U0 of p, and a connected open neighborhood V0 of FðpÞ such that FjU0 : U0 ! V0 is a diffeomorphism. Proof Here, p 2 M; and M is an m-dimensional smooth manifold, so there exists an admissible coordinate chart ðU; uÞ of M such that p 2 U: For every q 2 U; let ! o o o ; ; . . .; m ox1 q ox2 q ox q be the coordinate basis of Tq M corresponding to ðU; uÞ; where ! o 1 o d u uðqÞ : oxi q oxi uðqÞ Here, p 2 M; and F : M ! N; so FðpÞ 2 N: Since FðpÞ 2 N; and N is an m-dimensional smooth manifold, there exists an admissible coordinate chart ðV; wÞ of N such that FðpÞ 2 V; and wðFðpÞÞ ¼ 0: For every r 2 V; let o o o ; ; . . .; m oy1 r oy2 r oy r be the coordinate basis of Tr N corresponding to ðV; wÞ; where

5.5 Inverse Function Theorem for Manifolds

363

! o 1 o d w wðrÞ : oyi r oyi wðrÞ Here, the matrix representation of linear map dFp is the following m m matrix: 2

oðp1 ðw F u1 ÞÞ oðp1 ðw F u1 ÞÞ ðuðpÞÞ ðuðpÞÞ 6 1 ox ox2 6 6 oðp ðw F u1 ÞÞ oðp2 ðw F u1 ÞÞ 6 2 ð uðpÞ Þ ðuðpÞÞ 6 6 ox1 ox2 6... ... 6 4 oðpm ðw F u1 ÞÞ oðpm ðw F u1 ÞÞ ðuðpÞÞ ðuðpÞÞ ox1 ox2 3 oðp1 ðw F u1 ÞÞ ... ðuðpÞÞ 7 oxm 7 7 oðp2 ðw F u1 ÞÞ 7 ðuðpÞÞ 7: ... m 7 ox 7 ... ... 7 1 5 oðpm ðw F u ÞÞ ... ð uðpÞ Þ oxm Since the linear map dFp : Tp M ! TFðpÞ N is 1–1 onto, and dim Tp M ¼ m; 2

oðp1 ðw F u1 ÞÞ oðp1 ðw F u1 ÞÞ ð uðpÞ Þ ðuðpÞÞ 6 ox1 ox2 6 6 oðp ðw F u1 ÞÞ oðp2 ðw F u1 ÞÞ 6 2 ðuðpÞÞ ðuðpÞÞ det6 1 6 ox ox2 6... ... 6 4 oðpn ðw F u1 ÞÞ oðpn ðw F u1 ÞÞ ðuðpÞÞ ðuðpÞÞ ox1 ox2 3 oðp1 ðw F u1 ÞÞ ... ðuðpÞÞ 7 oxm 7 7 oðp2 ðw F u1 ÞÞ 7 ð uðpÞ Þ ... 7 6¼ 0: 7 oxm 7 ... ... 7 1 5 oðpn ðw F u ÞÞ ... ðuðpÞÞ m ox Since F : M ! N is a smooth map, ðU; uÞ is an admissible coordinate chart of M such that p 2 U; and ðV; wÞ is an admissible coordinate chart of N such that FðpÞ 2 V; w F u1 : uðU \ F 1 ðvÞÞ ! wðVÞ is smooth. Also, ðw F u1 ÞðuðpÞÞ ¼ wðFðpÞÞ ¼ 0: Clearly, uðU \ F 1 ðVÞÞ is an open neighborhood of

364

5 Immersions, Submersions, and Embeddings

uðpÞ in Rm : Further, wðVÞ is an open neighborhood of 0 in Rm : Now, by Theorem ^ of uðpÞ such that 4.1, there exists a connected open neighborhood U ^ is contained in uðU \ F 1 ðVÞÞ; 1. U ^ is connected and is an open neighborhood of 0, 2. ðw F u1 ÞðUÞ 1 ^ 3. ðw F u Þ has a smooth inverse on ðw F u1 ÞðUÞ: ^ and V0 ðF u1 ÞðUÞ: ^ We want to prove that U0 is conPut U0 u1 ðUÞ; ^ is connected, U ^ is contained nected and is an open neighborhood of p in M. Since U 1 1 1 ^ in uðU \ F ðVÞÞ; and u is continuous, u ðUÞð¼ U0 Þ is connected, and hence, ^ is open, U ^ contained in uðU \ F 1 ðVÞÞð uðUÞÞ; and U0 is connected. Since U ^ is open in uðUÞ; and hence, u1 ðUÞ ^ is open in U. Since u1 ðUÞ ^ is uðUÞ is open, U 1 ^ open in U, and U is open in M, u ðUÞð¼ U0 Þ is open in M, and hence, U0 is open ^ ^ p 2 u1 ðUÞð¼ U0 Þ; and hence, p 2 U0 : in M. Since uðpÞ 2 U; 1 Since u ; F are continuous, F u1 is continuous. Since F u1 is continu^ ^ is connected, and U ^ contained in uðU \ F 1 ðVÞÞ; ðF u1 ÞðUÞð¼ V0 Þ is ous, U 1 connected, and hence, V0 is connected. Since ðw F u Þ has a smooth inverse ^ u F 1 w1 is continuous. Since u F 1 w1 is conon ðw F u1 ÞðUÞ; tinuous, and w is continuous, their composite u F 1 is continuous. Since u F 1 ^ is open, ðu F 1 Þ1 ðUÞð¼ ^ ^ ¼ V0 Þ is open, and is continuous, and U ðF u1 ÞðUÞ 1 ^ V0 ¼ ðF u ÞðUÞ3ðF ^ u1 ÞðuðpÞÞ ¼ FðpÞ: hence, V0 is open. Since uðpÞ 2 U; Thus, V0 is a connected open neighborhood of FðpÞ: Since the inverse of ðw F u1 Þ exists, w F u1 is 1–1. Also, u; w1 are 1–1. Since w F u1 ; u; w1 are 1–1, their composite F is 1–1. It remains to be proved that ðFjU0 Þ1 : V0 ! U0 is smooth, that is, ðF 1 V0 Þ is smooth. Since w F u1 has a smooth inverse, u F 1 w1 is smooth, and hence, F 1 is smooth on V0 :

h

Note 5.37 The Theorem 5.36 is known as the inverse function theorem for manifolds. Deﬁnition Let M and N be smooth manifolds. Let F : M ! N be any function. If for every p in M, there exists an open neighborhood U of p such that F(U) is open in N, and the restriction FjU : U ! FðUÞ is a diffeomorphism, then we say that F is a local diffeomorphism.

5.6 Shrinking Lemma Note 5.38 Let M and N be smooth manifolds. Let F : M ! N be a local diffeomorphism. We shall try to show that F is an open map. For this purpose, let us take any nonempty open subset U of M.

5.6 Shrinking Lemma

365

We have to show that F(U) is open in N, that is, each member of F(U) is an interior point of F(U), that is, for every a in U, there exists an open neighborhood V of F(a) such that V FðUÞ. For this purpose, let us take any a in U. Since F : M ! N is a local diffeomorphism, and a is in M, there exists an open neighborhood U1 of a such that FðU1 Þ is open in N, and the restriction FjU1 : U1 ! FðU1 Þ is a diffeomorphism. Since U; U1 are open neighborhoods of a, U \ U1 is an open neighborhood of a, and hence, U \ U1 is open in U1 . Since FjU1 : U1 ! FðU1 Þ is a diffeomorphism, FjU1 : U1 ! FðU1 Þ is a homeomorphism. Since FjU1 : U1 ! FðU1 Þ is a homeomorphism, and U \ U1 is open in U1 , FjU1 ðU \ U1 Þð¼ FðU \ U1 ÞÞ is open in the open set FðU1 Þ, and hence, FðU \ U1 Þ is open in the open set FðU1 Þ. Since FðU \ U1 Þ is open in the open set FðU1 Þ, FðU \ U1 Þ is open in N. Since a is in U \ U1 , F(a) is in FðU \ U1 Þ. Thus, FðU \ U1 Þ is an open neighborhood of F(a). Also, FðU \ U1 Þ FðUÞ. Note 5.39 Let M, N, P be smooth manifolds. Let F : M ! N; and G : N ! P be local diffeomorphisms. We shall try to show that their composite G F : M ! P is a local diffeomorphism. Let us take any a 2 M. We have to ﬁnd an open neighborhood U of a such that ðG FÞðUÞ is open in P, and the restriction ðG FÞjU : U ! ðG FÞðUÞ is a diffeomorphism. Since F : M ! N is a local diffeomorphism, and a 2 M, there exists an open neighborhood U1 of a such that FðU1 Þ is open in N, and the restriction FjU1 : U1 ! FðU1 Þ is a diffeomorphism. Since FðaÞ 2 N, and G : N ! P is a local diffeomorphism, there exists an open neighborhood V of F(a) such that G(V) is open in P, and the restriction GjV : V ! GðVÞ is a diffeomorphism. Since FðU1 Þ is an open neighborhood of F(a), and V is an open neighborhood of F(a), FðU1 Þ \ V is an open neighborhood of F(a). Here, FðU1 Þ \ V is an open neighborhood of F(a) in FðU1 Þ, and FjU1 : U1 ! FðU1 Þ is a diffeomorphism, ðFjU1 Þ1 ðFðU1 Þ \ VÞ is an open neighborhood of a, and FjðFj

U1 Þ

1

ðFðU1 Þ\VÞ :

ðFjU1 Þ1 ðFðU1 Þ \ VÞ ! FðU1 Þ \

V is a diffeomorphism. Since FðU1 Þ \ V is an open neighborhood of F(a) in N, and GjV : V ! GðVÞ is a diffeomorphism, GðFðU1 Þ \ VÞ is an open neighborhood of GðFðaÞÞð¼ ðG FÞðaÞÞ, and GjðFðU1 Þ\VÞ : FðU1 Þ \ V ! GðFðU1 Þ \ VÞ is a diffeomorphism. Since FjðFj

U1 Þ

1

ðFðU1 Þ\VÞ :

ðFjU1 Þ1 ðFðU1 Þ \ VÞ ! FðU1 Þ \ V is a dif-

feomorphism, and GjðFðU1 Þ\VÞ : FðU1 Þ \ V ! GðFðU1 Þ \ VÞ is a diffeomorphism, their

composite

GðFðU1 Þ \ VÞ

is

ðGjðFðU1 Þ\VÞ Þ ðFjðFj

U1 Þ

a

diffeomorphism.

ðFðU1 Þ \ VÞÞ ¼ ðG FÞjðFj

1 U1 Þ ðFðU1 Þ\VÞ

1

ðFðU1 Þ\VÞ Þ

Clearly,

: ðFjU1 Þ1 ðFðU1 Þ \ VÞ ! ðGjðFðU1 Þ\VÞ Þ ðFjðFj

. Thus, ðG FÞjðFj

U1 Þ

1 U1 Þ ðFðU1 Þ\VÞ

: ðFjU1 Þ

1

1

ðFðU1 Þ \ VÞ ! GðFðU1 Þ \ VÞ is a diffeomorphism, ðFjU1 Þ1 ðFðU1 Þ \ VÞ is an open neighborhood of a, and

366

5 Immersions, Submersions, and Embeddings

1 1 ðG F Þj Fj 1 ðF ðU Þ\V Þ ðF ðU1 Þ \ V Þ ¼ ðG F Þ FjU1 ðF ðU 1 Þ \ V Þ FjU1 ð U1 Þ 1 1 ¼ G F FjU1 ðF ðU 1 Þ \ V Þ ¼ G ðF ðU 1 Þ \ V Þ

is an open neighborhood of ðG FÞðaÞ. It follows that G F : M ! P is a local diffeomorphism. Note 5.40 Let M, N be smooth manifolds. Let F : M ! N be a local diffeomorphism. We shall try to show that F : M ! N is a local homeomorphism, that is, F : M ! N is continuous, and for every a 2 M, there exists an open neighborhood U of a such that F(U) is an open neighborhood of F(a) such that the restriction FjU : U ! FðUÞ is a homeomorphism. First of all, we shall try to show that F : M ! N is continuous. For this purpose, let us take any a 2 M. Next, let us take any open neighborhood V of F(a). We have to ﬁnd an open neighborhood U of a such that FðUÞ V: Since a 2 M, and F : M ! N is a local homeomorphism, there exists an open neighborhood U1 of a such that FðU1 Þ is open in N, and the restriction FjU1 : U1 ! FðU1 Þ is a diffeomorphism. Here, FðU1 Þ is an open neighborhood of F(a), and V is an open neighborhood of F(a), so FðU1 Þ \ V is an open neighborhood of F (a), and hence, FðU1 Þ \ V is open in FðU1 Þ. Since FðU1 Þ \ V is open in FðU1 Þ, and FjU1 : U1 ! FðU1 Þ is a diffeomorphism, ðFjU1 Þ1 ðFðU1 Þ \ VÞ is open in U1 , and hence, ðFjU1 Þ1 ðFðU1 Þ \ VÞ is open in M. Since a 2 U1 , ðFjU1 ÞðaÞ ¼

FðaÞ 2 FðU1 Þ \ V, a 2 ðFjU1 Þ1 ðFðU1 Þ \ VÞ, and hence, ðFjU1 Þ1 ðFðU1 Þ \ VÞ is an open neighborhood of a. It remains to be proved that FððFjU1 Þ1 ðFðU1 Þ \ VÞÞ V. Here, FjU1 : U1 ! FðU1 Þ is 1–1, so F

1 FjU1

ðF ðU1 Þ \ V Þ

¼ F ðU1 Þ \ V V:

Thus, we have shown that F : M ! N is continuous. Now, we shall try to show that for every a 2 M, there exists an open neighborhood U of a, such that F(U) is an open neighborhood of F(a) and the restriction FjU : U ! FðUÞ is a homeomorphism. For this purpose, let us ﬁx any a 2 M. Since a 2 M, and F : M ! N is a local diffeomorphism, there exists an open neighborhood U of a such that F(U) is open in N, and the restriction FjU : U ! FðUÞ is a diffeomorphism. Since FjU : U ! FðUÞ is a diffeomorphism, FjU : U ! FðUÞ is a homeomorphism.

5.6 Shrinking Lemma

367

Note 5.41 Let M, N be smooth manifolds. Let M1 be a nonempty open subset of M. Let F : M ! N be a local diffeomorphism. We shall try to prove that FjM1 : M1 ! N is a local diffeomorphism. For this purpose, let us take any a 2 M1 . We have to ﬁnd an open neighborhood U of a in M1 such that ðFjM1 ÞðUÞ is open in N, and the restriction ðFjM1 ÞU : U ! ðFjM1 ÞðUÞ is a diffeomorphism. Since a 2 M1 M, a 2 M. Since a 2 M, and F : M ! N is a local diffeomorphism, there exists an open neighborhood U1 of a such that FðU1 Þ is open in N, and the restriction FjU1 : U1 ! FðU1 Þ is a diffeomorphism. Since M1 is an open neighborhood of a, M1 \ U1 is an open neighborhood of a in U1 . Since M1 \ U1 is an open neighborhood of a in U1 , and FjU1 : U1 ! FðU1 Þ is a diffeomorphism, ðFjU1 ÞðM1 \ U1 Þ is open in FðU1 Þ, and hence, ðFjU1 ÞðM1 \ U1 Þ is open in N. Clearly, ðFjM1 ÞðM1 \ U1 Þ is open in N. (Reason: Here, ðFjU1 ÞðM1 \ U1 Þ is open in N, and ðFjM1 ÞðM1 \ U1 Þ ¼ ðFjM1 \U1 ÞðM1 \ U1 Þ ¼ ðFjU1 ÞðM1 \ U1 Þ, so ðFjM1 ÞðM1 \ U1 Þ is open in N.) Now, it remains to be showed that ðFjM1 ÞðM \U Þ : ðM1 \ U1 Þ ! ðFjM1 Þ ðM1 \ U1 Þ 1

1

is a diffeomorphism, that is, FjðM1 \U1 Þ : ðM1 \ U1 Þ ! ðFjM1 ÞðM1 \ U1 Þ is a diffeomorphism. Since FjU1 : U1 ! FðU1 Þ is a diffeomorphism, its restriction FjðM1 \U1 Þ : ðM1 \ U1 Þ ! ðFjM1 ÞðM1 \ U1 Þ is a diffeomorphism.

Note 5.42 Let M be an m-dimensional smooth manifold and N be an n-dimensional smooth manifold. Let F : M ! N be a diffeomorphism. We shall try to show that F : M ! N is a local diffeomorphism. For this purpose, let us take any a 2 M. We have to ﬁnd an open neighborhood U of a in M such that F(U) is open in N, and the restriction FjU : U ! FðUÞ is a diffeomorphism. Since a 2 M, and M is a smooth manifold, there exists an admissible coordinate chart ðU; uÞ of M such that a 2 U. Since F : M ! N is a diffeomorphism, F : M ! N is 1–1. Since F is 1–1, so FjU : U ! FðUÞ is 1–1 onto. Since F : M ! N is a diffeomorphism, F : M ! N is a homeomorphism. Since F : M ! N is a homeomorphism, and U is an open neighborhood of a, F(U) is an open neighborhood of F(a). Now, it remains to be proved that: 1. FjU : U ! FðUÞ is smooth and 2. ðFjU Þ1 : FðUÞ ! U is smooth. For 1: Since F : M ! N is a diffeomorphism, F : M ! N is smooth. Since F : M ! N is smooth, U is an open neighborhood of a, and F(U) is an open neighborhood of F(a), the restriction FjU : U ! FðUÞ is smooth. For 2: Observe that ðFjU Þ1 ¼ ðF 1 ÞFðUÞ .

368

5 Immersions, Submersions, and Embeddings

(Reason: Let ðy; xÞ 2 LHS. So ðy; xÞ 2 ðFjU Þ1 . Since ðy; xÞ 2 ðFjU Þ1 , ðx; yÞ 2 FjU . Since ðx; yÞ 2 FjU , x 2 U, and ðx; yÞ 2 F. Since ðx; yÞ 2 F, ðy; xÞ 2 F 1 . Since ðx; yÞ 2 F, FðxÞ ¼ y. Since x 2 U, y ¼ FðxÞ 2 FðUÞ. Since ðy; xÞ 2 F 1 , and y 2 FðUÞ, ðy; xÞ 2 ðF 1 ÞFðUÞ ¼ RHS. Thus, LHS RHS. Next, let ðy; xÞ 2 RHS. So ðy; xÞ 2 ðF 1 Þ . Since ðy; xÞ 2 ðF 1 Þ , y 2 FðUÞ, and FðUÞ

FðUÞ

ðy; xÞ 2 F 1 . Since y 2 FðUÞ, there exists x1 2 U such that Fðx1 Þ ¼ y. Since ðy; xÞ 2 F 1 , ðx; yÞ 2 F, and hence, FðxÞ ¼ y. Since FðxÞ ¼ y, and Fðx1 Þ ¼ y, FðxÞ ¼ Fðx1 Þ. Since FðxÞ ¼ Fðx1 Þ and F is 1–1, x ¼ x1 . Since x ¼ x1 , and x1 2 U, x 2 U. Since x 2 U, and ðx; yÞ 2 F, ðx; yÞ 2 FjU , and hence, ðy; xÞ 2 ðFjU Þ1 ¼ LHS. Thus, RHS LHS. Hence, RHS LHS:Þ Since F : M ! N is a diffeomorphism, F 1 : N ! Mis smooth, and hence, ðF 1 ÞFðUÞ : FðUÞ ! U is smooth. Since ðF 1 ÞFðUÞ : FðUÞ ! U is smooth, and , ðFj Þ1 : FðUÞ ! U is smooth. ðFj Þ1 ¼ ðF 1 Þ U

FðUÞ

U

Note 5.43 Let M be an m-dimensional smooth manifold and N be an n-dimensional smooth manifold. Let F : M ! N be 1–1 onto. Let F : M ! N be a local diffeomorphism. We shall try to show that F : M ! Nis a diffeomorphism. Since F : M ! N is a local diffeomorphism, by Note 5.40, F : M ! N is a local homeomorphism, and hence, F : M ! N is continuous. Since F : M ! N is a local diffeomorphism, by Note 5.38, F : M ! N is an open map. Since F : M ! N is 1–1 onto, F : M ! N is continuous, and F : M ! N is an open map, F : M ! N is a homeomorphism. Now, we want to show that 1. F : M ! N is smooth, and 2. F 1 : N ! M is smooth. For 1: Let us ﬁx any a 2 M. We have to ﬁnd an admissible coordinate chart ðU; uÞ of M satisfying a 2 U, and an admissible coordinate chart ðV; wÞ of N satisfying FðaÞ 2 V such that w F u1 : uðU \ F 1 ðVÞÞ ! wðVÞ is smooth. Since a 2 M, and F : M ! N is a local diffeomorphism, there exists an open neighborhood U1 of a such that FðU1 Þ is an open neighborhood of F(a) in N, and the restriction FjU1 : U1 ! FðU1 Þ is a diffeomorphism. Since a 2 M, and M is an m-dimensional smooth manifold, there exists an admissible coordinate chart ðU2 ; u2 Þ of M such that a 2 U2 . Since a 2 U2 , and F : M ! N, FðaÞ 2 N. Since FðaÞ 2 N, and N is a smooth manifold, there exists an admissible coordinate chart ðV2 ; w2 Þ of N such that FðaÞ 2 V2 . Since U1 is an open neighborhood of a, and U2 is an open neighborhood of a, U1 \ U2 is an open neighborhood of a. Since ðU2 ; u2 Þ is an admissible coordinate chart of M satisfying a 2 U2 , and U1 \ U2 is an open neighborhood of a, ðU1 \ U2 ; u2 jU1 \U2 Þ is an admissible coordinate chart of M satisfying a 2 U1 \ U2 . Since FjU1 : U1 ! FðU1 Þ is a diffeomorphism, the restriction FjU1 \U2 : U1 \ U2 ! FðU1 \ U2 Þ is a

5.6 Shrinking Lemma

369

diffeomorphism. Since F : M ! N is a homeomorphism, and U1 \ U2 is an open neighborhood of a, FðU1 \ U2 Þ is an open neighborhood of F(a). Since ðV2 ; w2 Þ is an admissible coordinate chart of N such that V2 is an open neighborhood of F(a), and FðU1 \ U2 Þ is an open neighborhood of F(a), V2 \ FðU1 \ U2 Þ is an open neighborhood of F(a). Since ðV2 ; w2 Þ is an admissible coordinate chart ðV2 ; w2 Þ of N, and V2 \ FðU1 \ U2 Þ is an open neighborhood of F (a), ðV2 \ FðU1 \ U2 Þ; w2 jV2 \FðU1 \U2 Þ Þ is an admissible coordinate chart of N such that FðaÞ 2 V2 \ FðU1 \ U2 Þ. Now, it remains to be showed that ðw2 jV2 \FðU1 \U2 Þ Þ F ðu2 jU1 \U2 Þ1 is smooth. Since FjU1 : U1 ! FðU1 Þ is a diffeomorphism, FjU1 : U1 ! FðU1 Þ is smooth. Since FjU1 : U1 ! FðU1 Þ, ðU1 \ U2 ; u2 jU1 \U2 Þ is an admissible coordinate chart of M satisfying a 2 U1 \ U2 , and ðV2 \ FðU1 \ U2 Þ; w2 jV2 \FðU1 \U2 Þ Þ is an admissible coordinate chart of N such that FðaÞ 2 V2 \ FðU1 \ U2 Þ, ðw2 jV2 \FðU1 \U2 Þ Þ

FjU1 ðu2 jU1 \U2 Þ1 ð¼ ðw2 jV2 \FðU1 \U2 Þ Þ F ðu2 jU1 \U2 Þ1 Þ is smooth, and hence,

ðw2 jV2 \FðU1 \U2 Þ Þ F ðu2 jU1 \U2 Þ1 is smooth. Thus, we have shown that F : M ! N is smooth.

For 2: Since F : M ! N is 1–1 onto, F 1 : N ! M is 1–1 onto. Now, since F : M ! N is a local diffeomorphism, F 1 : N ! M is a local diffeomorphism. Since F 1 : N ! M is 1–1 onto and local diffeomorphism, as in 1, F 1 : N ! M is a diffeomorphism. Theorem 5.44 Let M be an m-dimensional smooth manifold and N be an n-dimensional smooth manifold. Let F : M ! N be a local diffeomorphism. Then, F is a smooth immersion and smooth submersion. Proof We have to prove that rank F ¼ m, and rank F ¼ n, that is, for every p 2 M, the rank of F at p is m and m = n. For this purpose, let us take any p 2 M. Since p 2 M, and F : M ! N is a local diffeomorphism, there exists an open neighborhood U of p in M such that F(U) is open in N and the restriction FjU : U ! FðUÞ is a diffeomorphism. Since U is a nonempty open subset of M, and M is an m-dimensional smooth manifold, U is an m-dimensional smooth manifold, and hence, dim Tp U ¼ dim U ¼ m. Since F(U) is open in N, and N is an n-dimensional smooth manifold, F(U) is an n-dimensional smooth manifold, and hence dim TFðpÞ ðFðUÞÞ ¼ dimðFðUÞÞ ¼ n. Since FjU : U ! FðUÞ is a diffeomorphism, and p 2 U, the linear map dFp : Tp U ! TFðpÞ ðFðUÞÞ is an isomorphism. Since dFp : Tp U ! TFðpÞ ðFðUÞÞ is an isomorphism, dFp : Tp U ! TFðpÞ ðFðUÞÞ is 1–1, and hence, rank of F at p is dim Tp Uð¼ mÞ. Thus, for every p 2 M, the rank of F at p is m. Since dFp : Tp U ! TFðpÞ ðFðUÞÞ is an isomorphism, m ¼ dim Tp U ¼ dim TFðpÞ ðFðUÞÞ ¼ n. h Theorem 5.45 Let M be an m-dimensional smooth manifold and N be an ndimensional smooth manifold. Let F : M ! N be a smooth immersion and smooth submersion map. Then, F : M ! N is a local diffeomorphism.

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5 Immersions, Submersions, and Embeddings

Proof For this purpose, let us take any p 2 M. We have to ﬁnd an open neighborhood U of p in M such that F(U) is open in N, and the restriction FjU : U ! FðUÞ is a diffeomorphism. Since p 2 M, and F : M ! N is a smooth immersion, the linear map dFp : Tp M ! TFðpÞ N is 1–1. Since p 2 M, and F : M ! Nis a smooth submersion, the linear map dFp : Tp M ! TFðpÞ N is onto. Since the linear map dFp : Tp M ! TFðpÞ N is 1–1 onto, m ¼ dim M ¼ dim Tp M ¼ dimðTFðpÞ NÞ ¼ n. Since m = n, and N is an n-dimensional smooth manifold, N is an m-dimensional smooth manifold. Since M, N are m-dimensional smooth manifolds, F : M ! N is a smooth map, and dFp : Tp M ! TFðpÞ N is 1–1 onto, by Theorem 5.36, there exists an open neighborhood U of p, and an open neighborhood V of F(p) such that FjU : U ! V is a diffeomorphism. Since FjU : U ! V is a diffeomorphism, FjU : U ! V is 1–1 onto, and hence, FðUÞ ¼ ðFjU ÞðUÞ ¼ V. Since FðUÞ ¼ V, and V is open, F(U) is open. Since FjU : U ! V is a diffeomorphism, and FðUÞ ¼ V, FjU : U ! FðUÞ is a diffeomorphism. Thus, U is an open neighborhood of p, F(U) is h open in N, and FjU : U ! FðUÞ is a diffeomorphism. Theorem 5.46 Let M, N be m-dimensional smooth manifolds. Let F : M ! N be a smooth immersion. Then, F : M ! N is a local diffeomorphism. Proof We ﬁrst try to show that F : M ! N is a smooth submersion. For this purpose, let us take any p 2 M. We have to show that the linear map dFp : Tp M ! TFðpÞ N is onto. Since p 2 M, and F : M ! N is a smooth immersion, the linear map dFp : Tp M ! TFðpÞ N is 1–1. Since dim Tp M ¼ dim M ¼ m ¼ dim N ¼ dimðTFðpÞ NÞ, dim Tp M ¼ dimðTFðpÞ NÞ. Since dim Tp M ¼ dimðTFðpÞ NÞ, and the linear map dFp : Tp M ! TFðpÞ N is 1–1, dFp : Tp M ! TFðpÞ N is onto. Since F : M ! N is a smooth immersion and smooth submersion, by Theorem 5.45, F : M ! N is a local diffeomorphism. h Theorem 5.47 Let M, N be m-dimensional smooth manifolds. Let F : M ! N be a smooth submersion. Then, F : M ! N is a local diffeomorphism. Proof We ﬁrst try to show that F : M ! N is a smooth immersion. For this purpose, let us take any p 2 M. We have to show that the linear map dFp : Tp M ! TFðpÞ N is 1–1. Since p 2 M, and F : M ! N is a smooth submersion, the linear map dFp : Tp M ! TFðpÞ N is onto. Since dim Tp M ¼ dim M ¼ m ¼ dim N ¼ dimðTFðpÞ NÞ, dim Tp M ¼ dimðTFðpÞ NÞ. Since dim Tp M ¼ dimðTFðpÞ NÞ, and the linear map dFp : Tp M ! TFðpÞ N is onto, dFp : Tp M ! TFðpÞ N is 1–1. Since F : M ! N is a smooth submersion and smooth immersion, by Theorem 5.45, F : M ! N is a local diffeomorphism. h Note 5.48 Before going ahead, let us recall the constant rank Theorem 4.9: Let M be an m-dimensional smooth manifold, N be an n-dimensional smooth manifold, F : M ! N be a smooth map, and r be the rank of F. Then, for every p in

5.6 Shrinking Lemma

371

M, there exist admissible coordinate chart ðU; uÞ in M satisfying p 2 U and admissible coordinate chart ðV; wÞ in N satisfying FðpÞ 2 V, such that FðUÞ V, and for every ðx1 ; . . .; xr ; xrþ1 ; . . .; xm Þ in uðUÞ; 0 1 1 w F u ðx1 ; . . .; xr ; xrþ1 ; . . .; xm Þ ¼ @x1 ; . . .; xr ; 0; . . .; 0 A: |ﬄﬄﬄ{zﬄﬄﬄ} nr

Theorem 5.49 Let M be a m-dimensional smooth manifold, N be an n-dimensional smooth manifold, and F : M ! N be smooth submersion (i.e., F is of constant rank n). Then, for every p in M, there exist admissible coordinate chart ðU; uÞ in M satisfying p 2 U and admissible coordinate chart ðV; wÞ in N satisfying FðpÞ 2 V such that FðUÞ V, and for every ðx1 ; . . .; xn ; xnþ1 ; . . .; xm Þ in uðUÞ; w F u1 ðx1 ; . . .; xm Þ ¼ ðx1 ; . . .; xn Þ: h

Proof Its proof is clear.

Theorem 5.50 Let M be a m-dimensional smooth manifold, N be an n-dimensional smooth manifold, and F : M ! N be smooth immersion (i.e., F is of constant rank m). Then, for every p in M, there exist admissible coordinate chart ðU; uÞ in M satisfying p 2 U and admissible coordinate chart ðV; wÞ in N satisfying FðpÞ 2 V such that FðUÞ V, and for every ðx1 ; . . .; xm Þ in uðUÞ, 0 1 w F u1 ðx1 ; . . .; xm Þ ¼ @x1 ; . . .; xm ; 0; . . .; 0 A: h |ﬄﬄﬄ{zﬄﬄﬄ} nm

h

Proof Its proof is clear.

Theorem 5.51 Let M be an m-dimensional smooth manifold, N be an n-dimensional smooth manifold, F : M ! N be smooth, and M be connected. Then, the following statements are equivalent: 1. For every p 2 M, there exist an admissible coordinate chart ðU; uÞ in M satisfying p 2 U and an admissible coordinate chart ðV; wÞ in N satisfying FðpÞ 2 V such that w F u1 behaves linearly, i.e., there exists a positive integer r such that for every ðx1 ; . . .; xr ; xrþ1 ; . . .; xm Þ in uðUÞ, 0 1 w F u1 ðx1 ; . . .; xr ; xrþ1 ; . . .; xm Þ ¼ @x1 ; . . .; xr ; 0; . . .; 0 A: |ﬄﬄﬄ{zﬄﬄﬄ} nr

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5 Immersions, Submersions, and Embeddings

2. F has constant rank. Proof 1 ) 2: Let us ﬁx any p 2 M. By the condition 1, there exist an admissible coordinate chart ðU; uÞ in M satisfying p 2 U, an admissible coordinate chart ðV; wÞ in N satisfying FðpÞ 2 V, and a positive integer r such that for every ðx1 ; . . .; xr ; xrþ1 ; . . .; xm Þ in uðUÞ, 0 1 w F u1 ðx1 ; . . .; xr ; xrþ1 ; . . .; xm Þ ¼ @x1 ; . . .; xr ; 0; . . .; 0 A: |ﬄﬄﬄ{zﬄﬄﬄ} nr

For every q 2 U,

! o o o ; ; . . .; m ox1 q ox2 q ox q

is the coordinate basis of Tq M corresponding to ðU; uÞ, where ! o 1 o d u uðqÞ : oxi q oxi uðqÞ For every r 2 V,

o o o ; ; . . .; n oy1 r oy2 r oy r

is the coordinate basis of Tr N corresponding to ðV; wÞ, where ! o 1 o : d w wðrÞ oyi r oyi wðrÞ We know that for every q 2 U, the matrix representation of linear map dFq is the following n m matrix: 2

oðp1 ðw F u1 ÞÞ oðp1 ðw F u1 ÞÞ ðuðqÞÞ ðuðqÞÞ 6 1 ox ox2 6 6 oðp ðw F u1 ÞÞ 1 oðp2 ðw F u ÞÞ 6 2 ðuðqÞÞ ðuðqÞÞ 6 6 ox1 ox2 6... . . . 6 4 oðpn ðw F u1 ÞÞ oðpn ðw F u1 ÞÞ ðuðqÞÞ ðuðqÞÞ 1 2 ox 3 ox 1 oðp1 ðw F u ÞÞ ... ðuðqÞÞ 7 oxm 7 7 1 oðp2 ðw F u ÞÞ 7 ... ð uðqÞ Þ 7: 7 oxm 7 ... ... 7 5 oðpn ðw F u1 ÞÞ ... ðuðqÞÞ oxm

5.6 Shrinking Lemma

373

It follows that for every q 2 U, the rank of the matrix representation of the linear map dFq is r. Since for every q 2 U, the rank of F at q is equal to the rank of the matrix representation of the linear map dFq , and the rank of the matrix representation of the linear map dFq is r, the rank of F at q is r. Thus, we have shown that, corresponding to each p 2 M, there exist a positive integer r and an open neighborhood Up;r of p such that for every q 2 Up;r , the rank of F at q is r. Observe that f[p2M Up;r : r ¼ 1; 2; . . .; minfm; ngg is a ﬁnite partition of M into open sets. Since f[p2M Up;r : r ¼ 1; 2; . . .; minfm; ngg is a ﬁnite partition of M into open sets, and M is connected, f[p2M Up;r : r ¼ 1; 2; . . .; minfm; ngg is a singleton set. It follows that for every q 2 M, the rank of F at q is a constant. Thus, F has constant rank. 2 ) 1: Let F has constant rank r. By the constant rank theorem, for every p in M, there exist an admissible coordinate chart ðU; uÞ in M satisfying p 2 U and an admissible coordinate chart ðV; wÞ in N satisfying FðpÞ 2 V such that FðUÞ V, and for every ðx1 ; . . .; xr ; xrþ1 ; . . .; xm Þ in uðUÞ, 0 1 w F u1 ðx1 ; . . .; xr ; xrþ1 ; . . .; xm Þ ¼ @x1 ; . . .; xr ; 0; . . .; 0 A: |ﬄﬄﬄ{zﬄﬄﬄ} nr

h h

Deﬁnition Let X be a topological space. If for every x 2 X, there exist an open neighborhood Ux of x and a compact set C such that Ux C, then we say that X is a locally compact space. Clearly, every compact space is locally compact. Note 5.52 Let M be an n-dimensional smooth manifold. We shall try to show that M is locally compact space. For this purpose, let us take any p 2 M. By Lemma 4.49 of Chap. 4, there exists a countable collection fðU1 ; u1 Þ; ðU2 ; u2 Þ; ðU3 ; u3 Þ; . . .g of admissible coordinate charts of M such that 1. fU1 ; U2 ; U3 ; . . .g is a basis of M, 2. each ui ðUi Þ is an open ball in Rn , 3. each closure Ui of Ui is a compact subset of M. Since fU1 ; U2 ; U3 ; . . .g is a basis of M, fU1 ; U2 ; U3 ; . . .g is a cover of M, and hence, there exists a positive integer k such that p 2 Uk . By 3, Uk is a compact subset of M. Further, Uk Uk . Hence, M is locally compact. Note 5.53 Let X be a Hausdorff topological space. Then, we shall try to show that the following statements are equivalent: 1. X is locally compact. 2. For every x 2 X, there exists an open neighborhood Ux of x such that Ux is compact. 3. There exists a basis B of X such that for every U 2 B; U is compact. 3 ) 2: Let B be a basis of X such that for every U 2 B; U is compact. We have to prove 2. For this purpose, let us take any x 2 X. Since x 2 X, and B is a

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5 Immersions, Submersions, and Embeddings

basis of X, there exists an open neighborhood Ux of x such that Ux 2 B, and Ux is compact. 2 ) 1: Let us take any x 2 X. We have to ﬁnd an open neighborhood Ux of x and a compact set C such that Ux C. Since x 2 X, so by 2, there exists an open neighborhood Ux of x such that Ux is compact. Also, Ux Ux . 1 ) 3: Let B be the collection of all open sets U for which U is compact. We shall try to show that B is a basis. For this purpose, let us take any open neighborhood G of x. Since x 2 X, and X is locally compact, there exist an open neighborhood Ux of x and a compact set C such that Ux C. Since Ux and G are open neighborhoods of x, Ux \ G is an open neighborhood of x. Clearly, Ux \ G C. Since C is compact, and X is Hausdorff, C is closed, and hence, C ¼ C. Since Ux \ G C, ðUx \ GÞ C ¼ C. Since ðUx \ GÞ C, C is compact, and ðUx \ GÞ is closed, ðUx \ GÞ is compact. Since Ux \ G is open, and ðUx \ GÞ is compact, Ux \ G is in B. Also, Ux \ G is an open neighborhood of x. This shows that B is a basis. If V 2 B, then, by the deﬁnition of B; V is compact. Note 5.54 Let X be a locally compact Hausdorff space. Let G be a nonempty open subset of X. Then, we shall try to show that G is locally compact Hausdorff. Since the topology of X is Hausdorff, and G has subspace topology of X, the topology of G is Hausdorff. Now, we want to show that G is locally compact. For this purpose, let us take any x in G. We have to ﬁnd an open neighborhood Vx of x in G, and a compact subset C1 of G such that Vx C1 . Since x is in X, and X is locally compact, there exist an open neighborhood Ux of x in X and a compact subset C of X such that Ux C. Since Ux is an open neighborhood of x in X, Ux \ G is an open neighborhood of x in G. Now, we want to show that ðUx \ GÞ \ G is compact in G. Since C is compact, and X is Hausdorff, C is closed in X. Since C is closed in X, and Ux \ G C, ðUx \ GÞ C. Since ðUx \ GÞ C, C is compact, and ðUx \ GÞ is closed, ðUx \ GÞ is compact in X. Since ðUx \ GÞ is compact in X, and G is open in X, ðUx \ GÞ \ G is compact in G. It remains to be showed that Ux \ G ðUx \ GÞ \ G. Since Ux \ G ðUx \ GÞ , and Ux \ G G, Ux \ G ðUx \ GÞ \ G. Note 5.55 Let X be a locally compact Hausdorff space. Let F be a nonempty closed subset of X. Then, we shall try to show that F is locally compact Hausdorff space. Since the topology of X is Hausdorff, and F has subspace topology of X, the topology of F is Hausdorff. Now, we want to show that F is locally compact. For this purpose, let us take any x in F. We have to ﬁnd an open neighborhood Vx of x in F, and a compact subset C1 of F such that Vx C1 . Since x is in X, and X is locally compact, there exist an open neighborhood Ux of x in X and a compact subset C of X such that Ux C. Since Ux is an open neighborhood of x in X, Ux \ F is an open neighborhood of x in F.

5.6 Shrinking Lemma

375

Now, we want to show that ðUx \ FÞ \ F is compact in F. Since Ux C, Ux \ F C \ F C. Since C is compact, and X is Hausdorff, C is closed in X. Since C is closed in X, and Ux \ F C, ðUx \ FÞ C. Since ðUx \ FÞ and F are closed in X, ðUx \ FÞ \ F is closed in X. Since ðUx \ FÞ \ F ðUx \ FÞ C, C is compact, and ðUx \ FÞ \ F is closed, ðUx \ FÞ \ F is compact in X. Since ðUx \ FÞ \ F is compact in X, and ðUx \ FÞ \ F F, ðUx \ FÞ \ F is compact in F. It remains to be showed that Ux \ F ðUx \ FÞ \ F. Since Ux \ F ðUx \ FÞ , and Ux \ F F, Ux \ F ðUx \ FÞ \ F. Note 5.56 Let X be a locally compact Hausdorff space. Let x 2 X. Let G be an open neighborhood of x. We shall try to show that there exists an open neighborhood Vx of x such that Vx G, and Vx is compact. Since X is a locally compact Hausdorff space, and G is a nonempty open subset of X, by Note 5.54, G is a locally compact Hausdorff space. Since G is a locally compact Hausdorff space, and x 2 G, there exist an open neighborhood Vx of x in G and a compact subset C of G such that Vx C. Since Vx is open in G, and G is open in X, Vx is open in X. Since Vx is open in X, and x 2 Vx , Vx is an open neighborhood of x in G. Since C is a compact subset of G, and G X, C is a compact subset of X. Since C is a compact subset of X, and X is a Hausdorff space, C is a closed subset of X. Since C is a closed subset of X, and Vx C, Vx C G. Since Vx C; Vx is closed, and C is a compact subset of X, Vx is a compact subset of X. This result is known as the shrinking lemma. Note 5.57 Let X be a compact space. Let fFn g be a sequence of closed subsets of X. T1Let each Fn be nonempty. Let F1 F2 TF13 . We shall try to show that n¼1 Fn is nonempty. If not, otherwise, let n¼1 Fn ¼ ;. We have to arrive at a contradiction. T T1 c c Here, X ¼ ;c ¼ ð 1 since each Fn is closed, n¼1 Fn Þ ¼ n¼1 ððFn Þ Þ. Further, S c c c c ððF each ðFn Þ is open. Since each ðFn Þ is open, and 1 n Þ Þ ¼ X, fðFn Þ g is an n¼1 c open cover of X. Since fðFn Þ g is an open cover of X, and X is compact, there exist S positive integers n1 \n2 \ \nk such that kr¼1 ððFnr Þc Þ ¼ X. Since F1 F2 F3 , ðF1 Þc ðF2 Þc ðF3 Þc . Now, since n1 \n2 S \ \nk , ðFn1 Þc ðFn2 Þc ðFnk Þc , and hence, X ¼ kr¼1 ððFnr Þc Þ ¼ ðFnk Þc . Since ðFnk Þc ¼ X, Fnk is empty, which contradicts the assumption that each Fn is nonempty. Theorem 5.58 Let X be a locally compact Hausdorff space.TLet fUn g be a sequence of subsets of X. Let each Un be open and dense. Then, 1 n¼1 Un is dense. T1 T1 Proof If not, otherwise, let n¼1 Un be not dense, that is, ð n¼1 Un Þ is a proper T subset of X. We have to arrive at a contradiction. Since ð 1 n¼1 Un Þ is a proper T1 subset of X, there exists a 2 X such that a 62 ð n¼1 Un Þ . It follows that there exists T an open neighborhood Va of a such that Va \ ð 1 n¼1 Un Þ ¼ ;.

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5 Immersions, Submersions, and Embeddings

Since Va is an open neighborhood of a, and U1 is open, Va \ U1 is open. Since Va is an open neighborhood of a, and U1 is dense, Va \ U1 is nonempty. Since Va \ U1 is a nonempty open subset of X, and X is a locally compact Hausdorff space, there exists a nonempty open set W1 such that W1 Va \ U1 , and W1 is compact. Here, W1 W1 Va \ U1 . Since W1 is a nonempty open subset of X, and U2 is open and dense, W1 \ U2 is a nonempty open subset of X. Since W1 \ U2 is a nonempty open subset of X, and X is a locally compact Hausdorff space, there exists a nonempty open set W2 such that W2 W1 \ U2 , and W2 is compact. Clearly, W2 W1 W1 . Also, W1 ; W2 are nonempty closed sets. Here, W2 W2 W1 \ U2 ðVa \ U1 Þ \ U2 ¼ Va \ ðU1 \ U2 Þ. Since W2 is a nonempty open subset of X, and U3 is open and dense, W2 \ U3 is a nonempty open subset of X. Since W2 \ U3 is a nonempty open subset of X, and X is a locally compact Hausdorff space, there exists a nonempty open set W3 such that W3 W2 \ U3 , and W3 is compact. Clearly, W3 W2 W2 . Also, W1 ; W2 ; W3 are nonempty closed sets. Here, W3 W3 W2 \ U3 ðVa \ ðU1 \ U2 ÞÞ \ U3 ¼ Va \ ðU1 \ U2 \ U3 Þ, etc. Thus, we get a decreasing sequence fWn g of nonempty closed subsets of a T compact set W1 . Hence, by Note 5.57, 1 W V \ n¼1 ðWn Þ is nonempty. Since T1 1 a U1 ; W2 Va \ ðU1 \ U2 Þ; W3 Va \ ðU1 \ U2 \ U3 Þ, etc., ; n¼1 ðWn Þ T T1 h Va \ ð 1 n¼1 Un Þ ¼ ;. It follows that n¼1 ðWn Þ is empty, a contradiction. Deﬁnition Let X be a topological space. Let A be any subset of X. If A contains no nonempty open set, then we say that A is nowhere dense. Theorem 5.59 Let X be a locally compact Hausdorff space. Let fAn g be a S sequence of nowhere dense subsets of X. Then, 1 A has no interior point. n n¼1 Proof For every n, An is nowhere dense, so ; ¼ ððAn Þ Þo ¼ ððððAn Þ Þc Þ Þc , and hence,TðððAn Þ Þc Þ ¼ X.S It follows that ððAn Þ Þc is an open dense set.SSo, by Theorem 1 c c 5.58, n¼1 ððAn Þ Þ ð¼ ð 1 ðA Þ Þc Þ is dense, and hence, X ¼ ðð 1 n¼1 ðAn Þ Þ Þ . S1 S1 n¼1 n c c o h It follows that ; ¼ ððð n¼1 ðAn Þ Þ Þ Þ ¼ ð n¼1 ðAn Þ Þ . This theorem is known as Baire category theorem. Theorem 5.60 Let M be an m-dimensional smooth manifold, N be an n-dimensional smooth manifold, and F : M ! N be smooth. Let F has constant rank r. Let F : M ! N be onto. Then, F is a smooth submersion. Proof We have to prove that rank F ¼ n. If not, otherwise, let r rank F\n. We have to arrive at a contradiction. Here, r rank F\n, so n − r is a positive integer. Let us take any p 2 M. By the constant rank theorem, there exist an admissible coordinate chart ðUp ; up Þ in M satisfying p 2 Up and an admissible coordinate chart ðVFðpÞ ; wFðpÞ Þ in

5.6 Shrinking Lemma

377

N satisfying FðpÞ 2 VFðpÞ , such that FðUp Þ VFðpÞ , and for every ðx1 ; . . .; xr ; xrþ1 ; . . .; xm Þ in up ðUp Þ, 1 0 C B ðx1 ; . . .; xr ; xrþ1 ; . . .; xm Þ ¼ @x1 ; . . .; xr ; 0; . . .; 0A wFðpÞ F u1 p |ﬄﬄﬄ{zﬄﬄﬄ} ð1 Þnr 2 wFðpÞ VFðpÞ : By Lemma 4.47, there exists an open neighborhood Wp of p such that Wp ðWp Þ Up , and ðWp Þ is compact. We shall try to show that FððWp Þ Þ is a nowhere dense subset of N. Since F : M ! N is smooth, F : M ! N is continuous. Since F : M ! N is continuous, and ðWp Þ is compact, FððWp Þ Þ is a compact subset of N. Since FððWp Þ Þ is a compact subset of N, and the topology of N is Hausdorff, FððWp Þ Þ is a closed subset of N. Since FððWp Þ Þ is a closed subset of N, ðFððWp Þ ÞÞ ¼ FððWp Þ Þ. It remains to be showed that FððWp Þ Þ has no interior point. Let us take any q 2 ðWp Þ . Since q 2 ðWp Þ ð Up Þ, q 2 Up , and hence, up ðqÞ 2 up ðUp Þ. Put up ðqÞ ðy1 ; . . .; yr ; yrþ1 ; . . .; ym Þ. So wFðpÞ ðFðqÞÞ ¼ wFðpÞ F u1 up ðqÞ p

0

1

B C ¼ wFðpÞ F u1 ðy1 ; . . .; yr ; yrþ1 ; . . .; ym Þ@y1 ; . . .; yr ; 0; . . .; 0A p |ﬄﬄﬄ{zﬄﬄﬄ} ð1 Þnr

2 |ﬄﬄﬄﬄﬄﬄﬄﬄ{zﬄﬄﬄﬄﬄﬄﬄﬄ} R R f0g f0g : |ﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄ{zﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄ} r

ð1 Þnr

Since q 2 Up , FðqÞ 2 FðUp Þ VFðpÞ . Since FðqÞ 2 VFðpÞ , and 1

0

C B FðqÞ 2 w1 R R f0g f0g A ; FðpÞ @|ﬄﬄﬄﬄﬄﬄﬄﬄ{zﬄﬄﬄﬄﬄﬄﬄﬄ} |ﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄ{zﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄ} r

ð1 Þnr

1

0

C B FðqÞ 2 VFðpÞ \ w1 R R f0g f0g A : FðpÞ @|ﬄﬄﬄﬄﬄﬄﬄﬄ{zﬄﬄﬄﬄﬄﬄﬄﬄ} |ﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄ{zﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄ} r

ð1 Þnr

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5 Immersions, Submersions, and Embeddings

It follows that 1

0

C B R R f0g f0g A : F Wp VFðpÞ \ w1 FðpÞ @|ﬄﬄﬄﬄﬄﬄﬄﬄ{zﬄﬄﬄﬄﬄﬄﬄﬄ} |ﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄ{zﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄ} r

ð1 Þnr

Clearly, 1

0

C B VFðpÞ \ w1 R R f0g f0g A FðpÞ @|ﬄﬄﬄﬄﬄﬄﬄﬄ{zﬄﬄﬄﬄﬄﬄﬄﬄ} |ﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄ{zﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄ} r

ð1 Þnr

has no interior point. (Reason: Here, 0

11

0

B CC B wFðpÞ @VFðpÞ \ w1 R f0g f0gAA FðpÞ @R |ﬄﬄﬄﬄﬄﬄﬄﬄ{zﬄﬄﬄﬄﬄﬄﬄﬄ} |ﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄ{zﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄ} r

0

ð1 Þnr

0

11

B CC B ¼ wFðpÞ VFðpÞ \ wFðpÞ @w1 R f0g f0g AA FðpÞ @R |ﬄﬄﬄﬄﬄﬄﬄﬄ{zﬄﬄﬄﬄﬄﬄﬄﬄ} |ﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄ{zﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄ} r

0

ð1 Þnr

1

1

0

C B C B wFðpÞ VFðpÞ \ @R R f0g f0g A; R f0g f0g A @R |ﬄﬄﬄﬄﬄﬄﬄﬄ{zﬄﬄﬄﬄﬄﬄﬄﬄ} |ﬄﬄﬄﬄﬄﬄﬄﬄ{zﬄﬄﬄﬄﬄﬄﬄﬄ} |ﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄ{zﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄ} |ﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄ{zﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄ} r

r

ð1 Þnr

ð1 Þnr

and 0

1

B C R R f0g f0g A @|ﬄﬄﬄﬄﬄﬄﬄﬄ{zﬄﬄﬄﬄﬄﬄﬄﬄ} |ﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄ{zﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄ} r

ð1 Þnr

have no interior point, so 0

11

0

B CC B wFðpÞ @VFðpÞ \ w1 R R f0g f0g A A FðpÞ @|ﬄﬄﬄﬄﬄﬄﬄﬄ{zﬄﬄﬄﬄﬄﬄﬄﬄ} |ﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄ{zﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄ} r

ð1 Þnr

has no interior point. Since ðVFðpÞ ; wFðpÞ Þ is an admissible coordinate chart of N, wFðpÞ is a homeomorphism from VFðpÞ onto wFðpÞ ðVFðpÞ Þ. Since wFðpÞ is a homeomorphism from VFðpÞ onto

5.6 Shrinking Lemma

379

0

11

0

B CC B wFðpÞ VFðpÞ ; @VFðpÞ \ w1 R R f0g f0g AA VFðpÞ ; FðpÞ @|ﬄﬄﬄﬄﬄﬄﬄﬄ{zﬄﬄﬄﬄﬄﬄﬄﬄ} |ﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄ{zﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄ} r

and

0

ð1 Þnr

11

0

B CC B wFðpÞ @VFðpÞ \ w1 R R f0g f0g A A FðpÞ @|ﬄﬄﬄﬄﬄﬄﬄﬄ{zﬄﬄﬄﬄﬄﬄﬄﬄ} |ﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄ{zﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄ} r

has no interior point,

ð1 Þnr

1

0

C B VFðpÞ \ w1 R R f0g f0g A FðpÞ @|ﬄﬄﬄﬄﬄﬄﬄﬄ{zﬄﬄﬄﬄﬄﬄﬄﬄ} |ﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄ{zﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄ} r

ð1 Þnr

has no interior point in VFðpÞ . Since 0

1

C B R R f0g f0g A VFðpÞ \ w1 FðpÞ @|ﬄﬄﬄﬄﬄﬄﬄﬄ{zﬄﬄﬄﬄﬄﬄﬄﬄ} |ﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄ{zﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄ} r

ð1 Þnr

has no interior point in VFðpÞ , and VFðpÞ is open in N, 0

1

C B R R f0g f0g A VFðpÞ \ w1 FðpÞ @|ﬄﬄﬄﬄﬄﬄﬄﬄ{zﬄﬄﬄﬄﬄﬄﬄﬄ} |ﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄ{zﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄ} r

ð1 Þnr

has no interior point in N.) Since 0

1

B C VFðpÞ \ w1 R R f0g f0g A FðpÞ @|ﬄﬄﬄﬄﬄﬄﬄﬄ{zﬄﬄﬄﬄﬄﬄﬄﬄ} |ﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄ{zﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄ} r

has no interior point in N, and

ð1 Þnr

0

1

B C F Wp R R f0g f0g A ; VFðpÞ \ w1 FðpÞ @|ﬄﬄﬄﬄﬄﬄﬄﬄ{zﬄﬄﬄﬄﬄﬄﬄﬄ} |ﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄ{zﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄ} r

ð1 Þnr

FððWp Þ Þ has no interior point. Thus, FððWp Þ Þ is a nowhere dense subset of N. Here, fWp : p 2 Mg is an open cover of M. Since M is a smooth manifold, the topology of M is second countable. Since M is a second countable space, and

380

5 Immersions, Submersions, and Embeddings

fWp : p 2 Mg is an open cover of M, there exist p1 ; p2 ; p3 ; . . . in M such that fWpn : n 2 Ng is an open cover of M. Since fWpn : n 2 Ng is a cover of M, and F : M ! N is onto, fFðWpn Þ : n 2 Ng is a cover of N. Since Wpn ðWpn Þ , FðWpn Þ FððWpn Þ Þ. Since, for every n 2 N; FðWpn Þ FððWpn Þ Þ, and S fFððWpn Þ Þ : n 2 Ng is a cover of N, N ¼ 1 n¼1 FððWpn Þ Þ. Since N is a smooth manifold, N is a Hausdorff locally compact space. Since N is a Hausdorff locally compact space, and for every n 2 N; FððWp Þ Þ is a nowhere dense subset of N, by S Baire category theorem, 1 n¼1 FððWpn Þ Þð¼ NÞ has no interior point, and hence, N has no interior point. This is a contradiction. h Theorem 5.61 Let M be an m-dimensional smooth manifold, N be an n-dimensional smooth manifold, and F : M ! N be smooth. Let F has constant rank r. Let F : M ! N be 1–1. Then, F is a smooth immersion. Proof We have to prove that rank F ¼ m. If not, otherwise, let r rank F\m. We have to arrive at a contradiction. Here, r rank F\m, so m − r is a positive integer. Let us take any p 2 M. By the constant rank theorem, there exist an admissible coordinate chart ðU; uÞ in M satisfying p 2 U and an admissible coordinate chart ðV; wÞ in N satisfying FðpÞ 2 V such that FðUÞ V, and for every ðx1 ; . . .; xr ; xrþ1 ; . . .; xm Þ in uðUÞ,

0 1 0 1 B C B C w F u1 @x1 ; . . .; xr ; xrþ1 ; . . .; xm A ¼ @x1 ; . . .; xr ; 0; . . .; 0A 2 wðVÞ: |ﬄﬄﬄ{zﬄﬄﬄ} |ﬄﬄﬄﬄﬄﬄﬄ{zﬄﬄﬄﬄﬄﬄﬄ} ð0 Þnr

ð1 Þmr

Since F : M ! N is 1–1, u1 is 1–1, and w is 1–1, their composite w F u1 from uðUÞ to V is 1–1. Put 1

0

C B uðpÞ @a1 ; . . .; ar ; arþ1 ; . . .; am A: |ﬄﬄﬄﬄﬄﬄﬄﬄ{zﬄﬄﬄﬄﬄﬄﬄﬄ} ð1 Þmr

Since 1

0

C B @a1 ; . . .; ar ; arþ1 ; . . .; am A ¼ uðpÞ 2 uðUÞ; |ﬄﬄﬄﬄﬄﬄﬄﬄ{zﬄﬄﬄﬄﬄﬄﬄﬄ} ð1 Þmr

and uðUÞ is an open subset of Rm , there exists e [ 0 such that ða1 e; a1 þ eÞ ðar e; ar þ eÞ ðarþ1 e; arþ1 þ eÞ ðam e; am þ eÞ uðUÞ:

5.6 Shrinking Lemma

381

Since 1 0

0

1

C B C B Ba1 ; . . .; ar ; arþ1 ; . . .; am e C; Ba1 ; . . .; ar ; arþ1 ; . . .; am þ e C A A @ @ |ﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄ{zﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄ2} |ﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄ{zﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄ2} ð1 Þmr

ð1 Þmr

2 ða1 e; a1 þ eÞ ðar e; ar þ eÞ ðarþ1 e; arþ1 þ eÞ ðam e; am þ eÞ uðUÞ; 1

0

0 1 C B eC @ w F u1 B . . .; 0A @a1 ; . . .; ar ; arþ1 ; . . .; am 2A ¼ a1 ; . . .; ar ; 0; |ﬄﬄﬄ{zﬄﬄﬄ} |ﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄ{zﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄ} nr ð1 Þmr

1

0

B eC C ¼ w F u1 B @a1 ; . . .; ar ; arþ1 ; . . .; am þ 2 A: |ﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄ{zﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄ} ð1 Þmr

Since 0

1

B eC C w F u1 B @a1 ; . . .; ar ; arþ1 ; . . .; am 2A |ﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄ{zﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄ} ð1 Þmr

0

1

B eC C ¼ w F u1 B @a1 ; . . .; ar ; arþ1 ; . . .; am þ 2A |ﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄ{zﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄ} ð1 Þmr

and w F u1 is 1 1; 1

0

1

0

C B C B Ba1 ; . . .; ar ; arþ1 ; . . .; am e C ¼ Ba1 ; . . .; ar ; arþ1 ; . . .; am þ e C; A A @ @ |ﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄ{zﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄ2} |ﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄ{zﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄ2} ð1 Þmr

which is a contradiction.

ð1 Þmr

h

382

5 Immersions, Submersions, and Embeddings

5.7 Global Rank Theorem Theorem 5.62 Let M be an m-dimensional smooth manifold, N be an n-dimensional smooth manifold, and F : M ! N be smooth. Let F has constant rank. Let F : M ! N be 1–1 and onto. Then, F is a diffeomorphism. Proof Since F : M ! N is onto, by Theorem 5.60, F is a smooth submersion. Since F : M ! N is 1–1, by Theorem 5.61, F is a smooth immersion. Since F : M ! N is submersion, and immersion, by Theorem 5.45, F : M ! N is a local diffeomorphism. Since F : M ! N is a local diffeomorphism, 1–1, and onto, by Note 5.43, F : M ! N is a diffeomorphism. h Note 5.63 Theorems 5.60, 5.61, and 5.62 together are known as the global rank theorem. Deﬁnition Let X, Y be topological spaces. Let F : X ! Y be a mapping. If F is 1–1 and continuous map, and F is a homeomorphism from X onto F(X), where F(X) has the subspace topology of Y, then we say that F is a topological embedding. Note 5.64 Let X, Y be topological spaces. Let F : X ! Y be a mapping. Let F be 1–1 and continuous. Let F be an open mapping, that is, for every open subset U of X, F(U) is open in Y. We shall try to show that F : X ! Y is a topological embedding. Here, let us take any open subset U of X. It remains to be proved that F(U) is open in F(X). Since U is open in X, and F is an open mapping, F(U) is open in Y. Since F(U) is open in Y, FðUÞ \ FðXÞ is open in F(X). Since FðUÞ FðXÞ, FðUÞ \ FðXÞ ¼ FðUÞ. Since FðUÞ \ FðXÞ ¼ FðUÞ, and FðUÞ \ FðXÞ is open in F (X), F(U) is open in F(X). Note 5.65 Let X, Y be topological spaces. Let F : X ! Y be a mapping. Let F be 1–1 and continuous. Let F be a closed mapping, that is, for every closed subset A of X, F(A) is closed in Y. We shall try to show that F : X ! Y is a topological embedding. Here, let us take any open subset U of X. It remains to be proved that F(U) is open in F(X). Since U is open in X, the complement U c of U is closed in X. Since U c is closed in X, and F is a closed mapping, FðU c Þ is closed in Y. Since FðU c Þ is closed in Y, and FðU c Þ FðXÞ, FðU c Þ \ FðXÞð¼ FðU c ÞÞ is closed in F(X), and hence, FðU c Þ is closed in F(X). Since FðU c Þ is closed in F(X), FðXÞ FðU c Þ is open in F(X). Since F : X ! Y is 1–1, FðXÞ FðU c Þ ¼ FðX U c Þ ¼ FðUÞ. Since FðXÞ FðU c Þ ¼ FðUÞ, and FðXÞ FðU c Þ is open in F(X), F(U) is open in F(X). Note 5.66 Let X be a compact space and Y be a Hausdorff space. Let F : X ! Y be a continuous mapping and F be 1–1. We shall try to show that F : X ! Y is a topological embedding. First of all, we shall try to show that F : X ! Y is a closed map. For this purpose, let us take any closed subset A of X. We have to show that F(A) is closed in Y. Since A is closed in the compact set X, A is compact. Since A is compact, and

5.7 Global Rank Theorem

383

F : X ! Y is continuous, F(A) is compact. Since F(A) is a compact subset of Y, and Y is Hausdorff, F(A) is a closed subset of Y. Thus, F : X ! Y is a closed map. Since F : X ! Y is 1–1, continuous, and closed, by the above note, F : X ! Y is a topological embedding. This result is known as closed map lemma. Deﬁnition Let X, Y be topological spaces. Let F : X ! Y be a mapping. If for every compact subset C of Y, F 1 ðCÞ is compact in X, then we say that F : X ! Y is a proper map. Note 5.67 Let X be a Hausdorff topological space and Y be a Hausdorff locally compact space. Let F : X ! Y be a continuous mapping. Let F be a proper map. We shall try to show that F is a closed mapping. For this purpose, let us take any closed subset A of X. We have to show that F(A) is closed in Y. If not, otherwise, let F(A) be not closed. We have to arrive at a contradiction. Since F(A) is not closed, there exists b 2 Y such that b is a limit point of F(A), and b 62 FðAÞ. Since b 2 Y, and Y is a Hausdorff locally compact space, there exists an open neighborhood Ub of b such that Ub is compact. Clearly, b is a limiting point of FðAÞ \ ðUb Þ: (Reason: Let us take any open neighborhood Vb of b. It follows that Vb \ Ub is an open neighborhood of b. Since Vb \ Ub is an open neighborhood of b, and b is a limit point of F(A), there exists a point c 2 Vb \ Ub , and c 2 FðAÞ. Here, c 2 Vb \ Ub , so c 2 Ub ð ðUb ÞÞ, and hence, c 2 ðUb Þ. Since c 2 ðUb Þ, and c 2 FðAÞ, c 2 FðAÞ \ ðUb Þ. Also, since c 2 Vb \ Ub , c 2 Vb . Since c 2 FðAÞ, and b 62 FðAÞ, c 6¼ b. This shows that b is a limiting point of FðAÞ \ ðUb Þ.) Since F : X ! Y is a continuous mapping, F is a proper map, and Ub is a compact subset of Y, F 1 ðUb Þ is a compact subset of X. Since F 1 ðUb Þ is a compact subset of X, and the topology of X is Hausdorff, F 1 ðUb Þ is a closed subset of X. Since F 1 ðUb Þ is a closed subset of X, and A is a closed subset of X, F 1 ðUb Þ \ A is a closed subset of X. Since F 1 ðUb Þ \ A is a closed subset of compact set F 1 ðUb Þ, F 1 ðUb Þ \ A is compact. Since F 1 ðUb Þ \ A is a compact subset of X, and F : X ! Y is a continuous mapping, FðF 1 ðUb Þ \ AÞ is compact. Clearly, FðF 1 ðUb Þ \ AÞ ¼ FðAÞ \ ðUb Þ: (Reason: LHS ¼ FðF 1 ðUb Þ \ AÞ FðF 1 ðUb ÞÞ \ FðAÞ ðUb Þ \ FðAÞ ¼ FðAÞ\ ðUb Þ ¼ RHS. Next, let us take any y 2 FðAÞ \ ðUb Þ. We have to show that y 2 FðF 1 ðUb Þ \ AÞ. Since y 2 FðAÞ \ ðUb Þ, y 2 ðUb Þ, and there exists x 2 A such that FðxÞ ¼ y. Since FðxÞ ¼ y 2 ðUb Þ, x 2 F 1 ðUb Þ. Since x 2 F 1 ðUb Þ, and x 2 A, x 2 F 1 ðUb Þ \ A. Therefore, y ¼ FðxÞ 2 FðF 1 ðUb Þ \ AÞ.) Since FðF 1 ðUb Þ \ AÞ ¼ FðAÞ \ ðUb Þ, and FðF 1 ðUb Þ \ AÞ is compact, FðAÞ \ ðUb Þ is compact. Since FðAÞ \ ðUb Þ is a compact subset of Y, and the topology of Y is Hausdorff, FðAÞ \ ðUb Þ is closed. Since FðAÞ \ ðUb Þ is closed, and b is a limiting point of FðAÞ \ ðUb Þ, b 2 FðAÞ \ ðUb Þð FðAÞÞ, and hence, b 2 FðAÞ. This is a contradiction.

384

5 Immersions, Submersions, and Embeddings

Deﬁnition Let M be an m-dimensional smooth manifold, N be an n-dimensional smooth manifold, and F : M ! N be smooth. If F is a smooth immersion, and F : M ! N is a topological embedding, then we say that F is a smooth embedding of M into N. Note 5.68 Let M be an m-dimensional smooth manifold, N be an n-dimensional smooth manifold, P be a p-dimensional smooth manifold, F : M ! N be a smooth embedding, and G : N ! P be a smooth embedding. We shall try to show that the composite map G F : M ! P is a smooth embedding, that is, 1. G F : M ! P is a smooth immersion, and 2. G F : M ! P is a topological embedding. For 1: We have to show that G F : M ! P is a smooth immersion. For this purpose, let us take any a 2 M: We have to show that the linear map dðG FÞa : Ta M ! TðGFÞðaÞ P is 1–1. Since F : M ! N is a smooth embedding, F : M ! N is a smooth immersion. Since F : M ! N is a smooth immersion, and a 2 M; the linear map dFa : Ta M ! TFðaÞ N is 1–1. Since G : N ! P is a smooth embedding, G : N ! P is a smooth immersion. Since G : N ! P is a smooth immersion, and FðaÞ 2 N; the linear map dGFðaÞ : TFðaÞ N ! TGðFðaÞÞ P is 1–1. Since dGFðaÞ : TFðaÞ N ! TGðFðaÞÞ P is 1–1, and dFa : Ta M ! TFðaÞ N is 1–1, their composite ðdGFðaÞ Þ ðdFa Þð¼ dðG FÞa Þ is 1–1, and hence, dðG FÞa is 1–1. For 2: Here, we have to prove that G F : M ! P is a topological embedding, that is, (a) G F : M ! P is 1–1. (b) G F : M ! P is continuous. (c) G F is a homeomorphism from M onto ðG FÞðMÞ; where the topology of ðG FÞðMÞ is the subspace topology of P. For a: Since F : M ! N is a smooth embedding, F : M ! N is a topological embedding, and hence, F : M ! N is 1–1. Similarly, G : N ! P is 1–1. Since F : M ! N is 1–1, and G : N ! P is 1–1, their composite G F : M ! P is 1–1. For b: Since F : M ! N is a smooth embedding, F : M ! N is a topological embedding, and hence, F : M ! N is continuous. Similarly, G : N ! P is continuous. Since F : M ! N is continuous, and G : N ! P is continuous, their composite G F : M ! P is continuous. For c: Since F : M ! N is a smooth embedding, F : M ! N is a topological embedding, and hence, F is a homeomorphism from M onto F(M), where F (M) has the subspace topology of N. Since G : N ! P is a smooth embedding, G : N ! P is a topological embedding, and hence, G is a homeomorphism from N onto G(N), where G(N) has the subspace topology of P. Since G is a homeomorphism from N onto G(N), the restriction G|F(M) is a homeomorphism from F(M) onto G(F(M)), where the topology of F(M) is the subspace topology

5.7 Global Rank Theorem

385

of N, and the topology of GðFðMÞÞ is the subspace topology of G(N). Since the topology of GðFðMÞÞ is the subspace topology of GðNÞ; and the topology of G (N) is the subspace topology of P, the topology of GðFðMÞÞ as the subspace topology of G(N) and the topology of G(F(M)) as the subspace topology of P are the same. It follows that G|F(M) is a homeomorphism from F(M) onto G(F (M)), where the topology of F(M) is the subspace topology of N, and the topology of G(F(M)) is the subspace topology of P. Since F is a homeomorphism from M onto F(M), where F(M) has the subspace topology of N, and GjFðMÞ is a homeomorphism from F(M) onto G(F(M)), where the topology of F (M) is the subspace topology of N, and the topology of G(F(M)) is the subspace topology of P, the composite ðGjFðMÞ Þ F is a homeomorphism from M onto GðFðMÞÞð¼ ðG FÞðMÞÞ; where the topology of ðG FÞðMÞ is the subspace topology of P. Further, since ðGjFðMÞ Þ F ¼ G F; G F is a homeomorphism from M onto (G F)(M), where the topology of (G F)(M) is the subspace topology of P. Note 5.69 Let M be an m-dimensional smooth manifold, N be an n-dimensional smooth manifold, and F : M ! N be a 1–1 smooth immersion. Let F be an open map. We shall try to show that F : M ! N is a smooth embedding, that is, F is a smooth immersion, and F : M ! N is a topological embedding. Since F : M ! N is a smooth immersion, it sufﬁces to show that F : M ! N is a topological embedding. Since F : M ! N is a smooth map, so F : M ! N is continuous. Since F is 1–1, continuous, and open, by Note 5.64, F : M ! N is a topological embedding. Note 5.70 Let M be an m-dimensional smooth manifold, N be an n-dimensional smooth manifold, and F : M ! N be a 1–1 smooth immersion. Let F be a closed map. We shall try to show that F : M ! N is a smooth embedding, that is, F is a smooth immersion, and F : M ! N is a topological embedding. Since F : M ! N is a smooth immersion, it sufﬁces to show that F : M ! N is a topological embedding. Since F : M ! N is a smooth map, F : M ! N is continuous. Since F is 1–1, continuous, and closed, by Note 5.65, F : M ! N is a topological embedding. Note 5.71 Let M be an m-dimensional smooth manifold, N be an n-dimensional smooth manifold, and F : M ! N be a 1–1 smooth immersion. Let F be a proper map. We shall try to show that F : M ! N is a smooth embedding. Since M is an m-dimensional smooth manifold, M is a Hausdorff locally compact space. Similarly, N is a Hausdorff locally compact space. Since F : M ! N is a smooth map, F : M ! N is continuous. Further, since F is a proper map, by Note 5.67, F is a closed mapping. Since F : M ! N is a 1–1 smooth immersion, and F is a closed map, by Note 5.70, F : M ! N is a smooth embedding.

386

5 Immersions, Submersions, and Embeddings

Note 5.72 Let M be an m-dimensional smooth manifold, N be an n-dimensional smooth manifold, and F : M ! N be a 1–1 smooth immersion. Let M be compact. We shall try to show that F : M ! N is a smooth embedding. We ﬁrst try to show that F : M ! N is proper map. For this purpose, let us take any compact subset C of N. We have to show that F 1 ðCÞ is compact in M. Since C is compact in N, and the topology of N is Hausdorff, C is closed. Since F : M ! N is smooth, F : M ! N is continuous. Since F : M ! N is continuous, and C is a closed subset of N, F 1 ðCÞ is closed in M. Since F 1 ðCÞ is closed in M, and M is compact, F 1 ðCÞ is compact in M. This shows that F : M ! N is proper map. Since F : M ! N is a 1–1 smooth immersion, and F : M ! N is proper map, by Note 5.71, F : M ! N is a smooth embedding. Note 5.73 Let M be an m-dimensional smooth manifold, N be an n-dimensional smooth manifold, F : M ! N be a 1–1 smooth immersion, and m ¼ n: We shall try to show that F : M ! N is a smooth embedding. By Theorem 5.46, F : M ! N is a local diffeomorphism, and hence, by Note 5.40, F is an open map. Since F : M ! N is a 1–1 smooth immersion, and F is an open map, by Note 5.69, F : M ! N is a smooth embedding. Theorem 5.74 Let M be an m-dimensional smooth manifold, N be an n-dimensional smooth manifold, and F : M ! N be smooth. Suppose that for every p 2 M, there exists an open neighborhood U of p such that the restriction FjU : U ! N is a smooth embedding. Then, F : M ! N is a smooth immersion. Proof We have to show that F : M ! N is a smooth immersion, that is, rank F ¼ m; that is, for every p 2 M; the rank of F at p is m. For this purpose, let us take any p 2 M: By the given condition, there exists an open neighborhood U of p such that the restriction FjU : U ! N is a smooth embedding. Since FjU : U ! N is a smooth embedding, FjU : U ! N is a smooth immersion. Since FjU : U ! N is a smooth immersion, and p 2 U, the linear map dðFjU Þp : Tp U ! TðFjU ÞðpÞ N is 1–1, that is, dðFjU Þp : Tp U ! TFðpÞ N is 1–1. Since U is an open neighborhood of p in M, by Note 5.16, Tp U and Tp M are essentially the same. Since Tp U and Tp M are essentially the same, and dðFjU Þp : Tp U ! TFðpÞ N is 1–1, dFp : Tp M ! TFðpÞ N is 1–1. Since dFp : Tp M !FðpÞ N is 1–1, ðrank of F at pÞ ¼ dimððdFp ÞðTp MÞÞ ¼ dimðTp MÞ ¼ dim M ¼ m: h Theorem 5.75 Let M be an m-dimensional smooth manifold, N be an n-dimensional smooth manifold, and F : M ! N be a smooth immersion. Let p 2 M: Then, there exists an open neighborhood U of p such that the restriction FjU : U ! N is a smooth embedding.

5.7 Global Rank Theorem

387

Proof By Theorem 5.50, there exist an admissible coordinate chart ðU1 ; uÞ in M satisfying p 2 U1 ; an admissible coordinate chart ðV; wÞ in N satisfying FðpÞ 2 V such that FðU1 Þ V; and for every ðx1 ; . . .; xm Þ in uðU1 Þ;

wFu

1

0

1

ðx1 ; . . .; xm Þ ¼ @x1 ; . . .; xm ; 0; . . .; 0 A: |ﬄﬄﬄ{zﬄﬄﬄ} nm

Here, ðU1 ; uÞ is an admissible coordinate chart ðU1 ; uÞ in M satisfying p 2 U1 ; so U1 is an open neighborhood of p in M. Since for every ðx1 ; . . .; xm Þ in uðU1 Þ;

wFu

1

0

1

ðx1 ; . . .; xm Þ ¼ @x1 ; . . .; xm ; 0; . . .; 0 A; |ﬄﬄﬄ{zﬄﬄﬄ} nm

and FðU1 Þ V; ðw F u1 Þ : uðU1 Þ ! wðVÞ and is 1–1. Since ðw F u1 Þ; u; w1 are 1–1, their composite w1 ðw F u1 Þ uð¼ FjU1 Þ is a 1–1 mapping from U1 to Vð NÞ: It follows that FjU1 : U1 ! N is 1–1. Since U1 is an open neighborhood of p in M, and M is an m-dimensional smooth manifold, there exists an open neighborhood U of p such that U U U1 ; and U− is compact. Here, U is a nonempty open subset of M and M is an m-dimensional smooth manifold, U is also an m-dimensional smooth manifold. Now, since F : M ! N is a smooth mapping, FjU : U ! N is a smooth map. Now, it remains to be proved that FjU : U ! N is a smooth embedding, that is, 1. FjU : U ! N is a smooth immersion and 2. FjU : U ! N is a topological embedding. For 1: Clearly, FjU : U ! N is a smooth immersion, that is, for every q 2 U; the rank of F|U at q is m. (Reason: For this purpose, let us take any q 2 U: Since F : M ! N is a smooth immersion, and q 2 M, the linear map dFq : Tq M ! TFðqÞ N is 1–1. Since U is an open neighborhood of q in M, Tq U and Tq M are essentially the same. Since Tq U and Tq M are essentially the same, and dFq : Tq M ! TFðqÞ N is 1–1, dðFjU Þq : Tq U ! TFðqÞ N is 1–1. Since the linear map dðFjU Þq : Tq U ! TFðqÞ N is 1–1, ðrank of FjU at qÞ ¼ dimððdðFjU Þq ÞðTq UÞÞ ¼ dimðTq UÞ ¼ dimðTq MÞ ¼ dim M ¼ m: Hence, rank of FjU at q is m.) For 2: First of all, we shall try to show that FjU : U ! N is a topological embedding. Since FjU1 : U1 ! N is 1–1, and U U1 , FjU : U ! N is 1–1. Since F : M ! N is smooth, F : M ! N is continuous, and hence, the restriction FjU : U ! N is continuous. Further, since U is compact, and N is a Hausdorff space, by closed map Lemma 5.66, FjU : U ! N is a topological embedding. Since FjU : U ! N is a topological embedding, and U U ; the

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restriction ðFjU ÞU : U ! N is a topological embedding. Since ðFjU ÞU : U ! N is a topological embedding, and ðFjU ÞU ¼ FjU ; FjU : U ! N is a topological embedding. h Theorem 5.76 Let M be an m-dimensional smooth manifold, N be an n-dimensional smooth manifold, and p : M ! N be a smooth submersion. Let p 2 M: Then, there exist an open neighborhood V of pðpÞ in N and a smooth map r : V ! M such that p r ¼ IdV ; and rðpðpÞÞ ¼ p: Proof Since p : M ! N is a smooth submersion, and p 2 M, by Theorem 5.49, there exist admissible coordinate chart ðU; uÞ in M satisfying p 2 U; admissible coordinate chart ðV; wÞ in N satisfying pðpÞ 2 V such that pðUÞ V; and for every ðx1 ; . . .; xn ; xnþ1 ; . . .; xm Þ in uðUÞ; w p u1 ðx1 ; . . .; xn ; xnþ1 ; . . .; xm Þ ¼ ðx1 ; . . .; xn Þ: Here, p 2 U; and ðU; uÞ is an admissible coordinate chart in M, so uðpÞ is an element of the open subset uðUÞ of Rm : Put uðpÞ ða1 ; . . .; an ; anþ1 ; . . .; am Þ: Here, ða1 ; . . .; an ; anþ1 ; . . .; am Þ is an element of the open subset uðUÞ of Rm ; so there exists e [ 0 such that ða1 e; a1 þ eÞ ðan e; an þ eÞ ðanþ1 e; anþ1 þ eÞ ðam e; am þ eÞ uðUÞ: For every ðx1 ; . . .; xn ; xnþ1 ; . . .; xm Þ 2 ða1 e; a1 þ eÞ ðan e; an þ eÞ ðanþ1 e; anþ1 þ eÞ ðam e; am þ eÞ; we have ðw p u1 Þðx1 ; . . .; xn ; xnþ1 ; . . .; xm Þ ¼ ðx1 ; . . .; xn Þ 2 wðVÞ: It follows that w p u1 ðða1 e; a1 þ eÞ ðan e; an þ eÞ ðanþ1 e; anþ1 þ eÞ ðam e; am þ eÞÞÞÞ ¼ w p u1 ðða1 e; a1 þ eÞ ðan e; an þ eÞ ðanþ1 e; anþ1 þ eÞ ðam e; am þ eÞÞ ¼ ða1 e; a1 þ eÞ ðan e; an þ eÞ 3 ða1 ; . . .; an Þ: Hence,

ða1 ; . . .; an Þ ¼ w p u1 ða1 ; . . .; an ; anþ1 ; . . .; am Þ ¼ w p u1 ðuðpÞÞ ¼ wðpðpÞÞ

is an interior point of w p u1 ðða1 e; a1 þ eÞ ðan e; an þ eÞ ðanþ1 e; anþ1 þ eÞ ðam e; am þ eÞÞÞÞ:

5.7 Global Rank Theorem

389

Since wðpðpÞÞ is an interior point of w p u1 ðða1 e; a1 þ eÞ ðan e; an þ eÞ ðanþ1 e; anþ1 þ eÞ ðam e; am þ eÞÞÞÞ; w p u1 ðða1 e; a1 þ eÞ ðan e; an þ eÞ ðanþ1 e; anþ1 þ eÞ ðam e; am þ eÞÞÞÞ is an open neighborhood of wðpðpÞÞ: Since w p u1 ðða1 e; a1 þ eÞ ðan e; an þ eÞ ðanþ1 e; anþ1 þ eÞ ðam e; am þ eÞÞÞÞ is an open neighborhood of wðpðpÞÞ; and w is a homeomorphism from V onto wðVÞ, p u1 ðða1 e; a1 þ eÞ ðan e; an þ eÞ ðanþ1 e; anþ1 þ eÞ ðam e; am þ eÞÞÞ is an open neighborhood of pðpÞ. Put V p u1 ðða1 e; a1 þ eÞ ðan e; an þ eÞ ðanþ1 e; anþ1 þ eÞ ðam e; am þ eÞÞÞ: Here, V is an open neighborhood of pðpÞ: Let us deﬁne r : V ! M as follows: For every q in V ; rðqÞ u1 ðp1 ðwðqÞÞ; . . .; pn ðwðqÞÞ; anþ1 ; . . .; am Þ: Since w; u1 ; and all projection maps pi are smooth, r is smooth. Here, we want to prove that p r ¼ IdV ; that is;for every q 2 V ; pðrðqÞÞ ¼ q. LHS ¼ pðrðqÞÞ ¼ p u1 ðp1 ðwðqÞÞ; . . .; pn ðwðqÞÞ; anþ1 ; . . .; am Þ ¼ p u1 ðp1 ðwðqÞÞ; . . .; pn ðwðqÞÞ; anþ1 ; . . .; am Þ ¼ w1 w p u1 ðp1 ðwðqÞÞ; . . .; pn ðwðqÞÞ; anþ1 ; . . .; am Þ ¼ w1 ðp1 ðwðqÞÞ; . . .; pn ðwðqÞÞÞ ¼ w1 ðwðqÞÞ ¼ q ¼ RHS:

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5 Immersions, Submersions, and Embeddings

It remains to be proved that rðpðpÞÞ ¼ p: LHS ¼ rðpðpÞÞ ¼ u1 p1 w p u1 ðuðpÞÞ ; . . .; pn w p u1 ðuðpÞÞ ; anþ1 ; . . .; am ¼ u1 p1 w p u1 ða1 ; . . .; an ; anþ1 ; . . .; am Þ ; . . .; pn w p u1 ða1 ; . . .; an ; anþ1 ; . . .; am ÞÞ; anþ1 ; . . .; am Þ ¼ u1 ðp1 ða1 ; . . .; an Þ; . . .; pn ða1 ; . . .; an Þ; anþ1 ; . . .; am Þ ¼ u1 ða1 ; . . .; an ; anþ1 ; . . .; am Þ ¼ u1 ðuðpÞÞ ¼ p ¼ RHS:

h

Theorem 5.77 Let M be an m-dimensional smooth manifold, N be an n-dimensional smooth manifold, and p : M ! N be a smooth map. If for every p 2 M; there exist an open neighborhood V of pðpÞ in N and a smooth map r : V ! M such that p r ¼ IdV ; and rðpðpÞÞ ¼ p; then p : M ! N is a smooth submersion. Proof We have to prove that p : M ! N is a smooth submersion, that is, for every p 2 M; the linear map dpp : Tp M ! TpðpÞ N is onto. For this purpose, let us take any p 2 M: By the given condition, there exist an open neighborhood V of pðpÞ in N a smooth map r : V ! M such that p r ¼ IdV ; and rðpðpÞÞ ¼ p: Since V is an open neighborhood of pðpÞ in N, TpðpÞ V and TpðpÞ N are essentially the same. Now, since p; r are smooth maps, IdTpðpÞ N ¼ IdTpðpÞ V ¼ dðIdV ÞpðpÞ ¼ dðp rÞpðpÞ ¼ dprðpðpÞÞ drpðpÞ ¼ dpp drpðpÞ : Since

is onto.

dpp drpðpÞ ¼ IdTpðpÞ N ;

so dpp : Tp M ! TpðpÞ N h

Note 5.78 If we combine the Theorems 5.76, and 5.77, then we get the result: Let M be an m-dimensional smooth manifold, N be an n-dimensional smooth manifold, and p : M ! N be a smooth map. p : M ! N is a smooth submersion if and only if for every p 2 M, there exist an open neighborhood V of π(p) in N and a smooth map r : V ! M such that p r ¼ IdV and rðpðpÞÞ ¼ p. Theorem 5.79 Let M be an m-dimensional smooth manifold, N be an n-dimensional smooth manifold, and p : M ! N be a smooth submersion. Then, p : M ! N is an open map. Proof Let G be a nonempty open subset of M. We have to show that π(G) is open in N. For this purpose, let us take any pðpÞ 2 pðGÞ; where p 2 G: Since p 2 M; and p : M ! N is a smooth submersion, by Theorem 5.76, there exist an open

5.7 Global Rank Theorem

391

neighborhood V of π(p) in N and a smooth map r : V ! M such that p r ¼ IdV ; and rðpðpÞÞ ¼ p: Since r : V ! M is smooth, r : V ! M is continuous. Since r : V ! M is continuous, and G is an open neighborhood of pð¼ rðpðpÞÞÞ in M, r1 ðGÞ is an open neighborhood of pðpÞ in V. Since r1 ðGÞ is an open neighborhood of π(p) in V, and V is open, r1 ðGÞ is an open neighborhood of π(p) in N. Now, it sufﬁces to show that r1 ðGÞ pðGÞ: For this purpose, let us take any q 2 r1 ðGÞ: Since q 2 r1 ðGÞ V; q ∊ V. Since q ∊ V, q ¼ IdV ðqÞ ¼ ðp rÞðqÞ ¼ pðrðqÞÞ: Since q 2 r1 ðGÞ; rðqÞ 2 G: Since rðqÞ 2 G; pðrðqÞÞ 2 pðGÞ: Since pðrðqÞÞ 2 pðGÞ; and pðrðqÞÞ ¼ q; q 2 pðGÞ: Thus, we have shown h that r1 ðGÞ pðGÞ: Deﬁnition Let X be a topological space and Y be a nonempty set. Let F : X ! Y be any onto map. Let O be the collection of all subsets G of Y such that F 1 ðGÞ is open in X. We shall try to show that O is a topology over Y. 1. Since F 1 ð;Þ ¼ ;; and ; is open in X, ; 2 O: Since F 1 ðYÞ ¼ X; and X is open in X, Y 2 O: 2. Let G1 ; G2 2 O: We have to show that G1 \ G2 2 O; that is, F 1 ðG1 \ G2 Þ is open in X. Since G1 2 O; F 1 ðG1 Þ is open in X. Similarly, F 1 ðG2 Þ is open in X. It follows that F 1 ðG1 Þ \ F 1 ðG2 Þð¼ F 1 ðG1 \ G2 ÞÞ is open in X, and hence, F 1 ðG1 \ G2 Þ is open in X. 3. Let Gi 2 O for every i 2 I: We have to show that [i2I Gi 2 O; that is, F 1 ð[i2I Gi Þ is open in X. Since, for every i 2 I; Gi 2 O; F 1 ðGi Þ is open in X. It follows that [i2I ðF 1 ðGi ÞÞð¼ F 1 ð[i2I Gi ÞÞ is open in X, and hence, F 1 ð[i2I Gi Þ is open in X. Thus, we have shown that O is a topology over Y. The topology O is called the quotient topology on Y determined by F. Deﬁnition Let X; Y be topological spaces. Let F : X ! Y be any continuous, onto mapping. If the quotient topology on Y determined by F is the topology of Y, then we say that F is a quotient map. Theorem 5.80 Let M be an m-dimensional smooth manifold, N be an n-dimensional smooth manifold, and p : M ! N be a smooth submersion. Let p : M ! N be onto. Then, p : M ! N is a quotient map. Proof For this purpose, let G be any nonempty subset of N such that p1 ðGÞ is open in M. It sufﬁces to show that G is open in N. Since p : M ! N is a smooth submersion, by Theorem 5.79, p : M ! N is an open map. Since p : M ! N is an onto map, and G is a subset of N, pðp1 ðGÞÞ ¼ G: Since p : M ! N is an open map, and π-1(G) is open in M, pðp1 ðGÞÞð¼ GÞ is open in N, and hence, G is open in N. h

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5 Immersions, Submersions, and Embeddings

Theorem 5.81 Let M be an m-dimensional smooth manifold, N be an n-dimensional smooth manifold, and P be a p-dimensional smooth manifold. Let p : M ! N be a smooth submersion. Let p : M ! N be onto. Let F : N ! P be any map such that F p : M ! P is smooth. Then, F : N ! P is smooth. Proof We have to prove that F : N ! P is smooth, that is, for every q in N, there exists an open neighborhood Vq of q in N such that FjVq : Vq ! P is smooth. For this purpose, let us take any q in N. Since q is in N, and π : M ! N is onto, there exists p 2 M such that pðpÞ ¼ q: Since p : M ! N is a smooth submersion, and p 2 M; by Theorem 5.76, there exist an open neighborhood Vq of pðpÞ in N and a smooth map r : Vq ! M such that p r ¼ IdVq ; and rðpðpÞÞ ¼ p: Since r : Vq ! M is smooth, and F p : M ! P is smooth, their composite ðF pÞ r : Vq ! P is smooth. Here, ðF pÞ r ¼ F ðp rÞ ¼ F IdVq ¼ FjVq : Since FjVq ¼ ðF pÞ r; and ðF pÞ r is smooth, FjVq is smooth. h

5.8 Properly Embedded Deﬁnition Let X; Y be any nonempty sets. Let F : X ! Y be any mapping. Let b 2 Y: The set F 1 ðbÞ is called the ﬁber of F over b. Clearly, X is partitioned into ﬁbers of F. Theorem 5.82 Let M be an m-dimensional smooth manifold, N be an n-dimensional smooth manifold, and P be a p-dimensional smooth manifold. Let p : M ! N be a smooth submersion. Let π : M ! N be onto. Let F : M ! P be any smooth map. Let F be constant on ﬁbers of π, that is, if pðxÞ ¼ pðyÞ, then FðxÞ ¼ FðyÞ: ^ : N ! P such that F ^ p ¼ F: Then, there exists a unique smooth map F Proof Existence: We ﬁrst try to show that F p1 is a function. For this purpose, let ðx; yÞ 2 F p1 ; and ðx; zÞ 2 F p1 : We have to show that y ¼ z: Since ðx; yÞ 2 F p1 ; there exists u such that ðx; uÞ 2 p1 ; and ðu; yÞ 2 F: Since ðx; uÞ 2 p1 ; ðu; xÞ 2 p: Since ðx; zÞ 2 F p1 ; there exists v such that ðx; vÞ 2 p1 ; and ðv; zÞ 2 F: Since ðx; vÞ 2 p1 ; ðv; xÞ 2 p: Since ðv; xÞ 2 p; pðvÞ ¼ x: Similarly, pðuÞ ¼ x: It follows that pðuÞ ¼ pðvÞ: Since pðuÞ ¼ pðvÞ; and F is constant on ﬁbers of π, FðuÞ ¼ FðvÞ: Since ðv; zÞ 2 F; FðvÞ ¼ z: Since ðu; yÞ 2 F; FðuÞ ¼ y: Since FðuÞ ¼ y; FðvÞ ¼ z; and FðuÞ ¼ FðvÞ; y ¼ z: This proves that F p1 is a function. Since π : M ! N is onto, and F : M ! P; dom ðF p1 Þ ¼ N; and ran ðF p1 Þ P: Since F p1 is a function, dom ðF p1 Þ ¼ N; and ran ðF p1 Þ P; we can write F p1 : N ! P: ^ ¼ F p1 ; and F p1 : N ! P; F ^ : N ! P: Now, ^ F p1 : Since F Put F 1 1 1 ^ F p ¼ ðF p Þ p ¼ F ðp pÞ: Clearly, F ðp pÞ ¼ F:

5.8 Properly Embedded

393

(Reason: Let ðx; yÞ 2 LHS ¼ F ðp1 pÞ: So there exist u, v such that ðx; uÞ 2 p; ðu; vÞ 2 p1 ; and ðv; yÞ 2 F: Since ðu; vÞ 2 p1 ; ðv; uÞ 2 p: It follows that pðvÞ ¼ u; pðxÞ ¼ u; and hence, pðxÞ ¼ pðvÞ: Since pðxÞ ¼ pðvÞ; and F is constant on ﬁbers of π, FðxÞ ¼ FðvÞ: Since ðv; yÞ 2 F; FðvÞ ¼ y: Since FðvÞ ¼ y; and FðxÞ ¼ FðvÞ; FðxÞ ¼ y; and hence, ðx; yÞ 2 F ¼ RHS: Hence, LHS RHS: Next, let ðx; yÞ 2 RHS ¼ F: Since ðx; yÞ 2 F; and F : M ! P; x 2 M; and y 2 P: Since x 2 M; and p : M ! N; there exists u 2 N such that ðx; uÞ 2 p: Since ðx; uÞ 2 p; ðu; xÞ 2 p1 : Since ðx; uÞ 2 p; ðu; xÞ 2 p1 ; and ðx; yÞ 2 F; ðx; yÞ 2 F ðp1 pÞ ¼ LHS: Hence, RHS LHS: Since LHS RHS; and RHS ^ p ¼ F: LHS; LHS ¼ RHS:) Thus, F 1 ^ ¼ F p1 ; F ^ : N ! P: It remains to be showed Since F p : N ! P; and F ^ ^ ^ p is smooth. Since that F is smooth. Since F p ¼ F; and F is a smooth map, F ^ : N ! P is a map, and F ^ p is smooth, by Theorem 5.81, F ^ is smooth. F This completes the proof of existence part. Uniqueness: Let F1 : N ! P be a smooth mapping such that F1 p ¼ F; and let F2 : N ! P be a smooth mapping such that F2 p ¼ F: We have to show that F1 ¼ F2 : Since p : M ! N is onto, p p1 ¼ IdN : Since F1 p ¼ F; F p1 ¼ ðF1 pÞ p1 ¼ F1 ðp p1 Þ ¼ F1 IdN ¼ F1 : Thus, F1 ¼ F p1 : Similarly, h F2 ¼ F p1 : It follows that F1 ¼ F2 : This proves the uniqueness part. Theorem 5.83 Let M be an m-dimensional smooth manifold, N be an n-dimensional smooth manifold, and P be a p-dimensional smooth manifold. Let p1 : M ! N be a smooth submersion onto map. Let p2 : M ! P be a smooth submersion onto map. Let p1 be constant on ﬁbers of p2 ; and p2 be constant on ﬁbers of p1 : Then, there exists a unique diffeomorphism F : N ! P such that F p1 ¼ p2 : Proof Existence: We ﬁrst try to show that p2 p1 1 is a function. For this purpose, 1 let ðx; yÞ 2 p2 p1 ; and ðx; zÞ 2 p p : We have to show that y ¼ z: 2 1 1 Since ðx; yÞ 2 p2 p1 ; there exists z such that ðx; zÞ 2 p1 1 1 ; and ðz; yÞ 2 p2 : 1 1 Since ðx; zÞ 2 p1 ; ðz; xÞ 2 p1 : Since ðx; zÞ 2 p2 p1 ; there exists w such that 1 ðx; wÞ 2 p1 1 ; and ðw; zÞ 2 p2 : Since ðx; wÞ 2 p1 ; ðw; xÞ 2 p1 : Since ðw; zÞ 2 p2 ; p2 ðwÞ ¼ z: Similarly, p2 ðzÞ ¼ y; p1 ðwÞ ¼ x; and p1 ðzÞ ¼ x: Since p1 ðwÞ ¼ x ¼ p1 ðzÞ; and p2 is constant on ﬁbers of p1 ; p2 ðwÞ ¼ p2 ðzÞ: Since z ¼ p2 ðwÞ ¼ p2 ðzÞ ¼ y; y ¼ z: This proves that p2 p1 1 is a function. Since p1 : M ! N is onto, and p2 : M ! P is onto, dom ðp2 p1 1 Þ ¼ N; and 1 1 1 ran ðp2 p1 Þ P: Since p2 p1 is a function, dom ðp2 p1 Þ ¼ N; and 1 ran ðp2 p1 1 Þ P; we can write p2 p1 : N ! P: 1 1 Put F p2 p1 : Since F ¼ p2 p1 1 ; and p2 p1 : N ! P; F : N ! P: 1 1 Now, F p1 ¼ ðp2 p1 1 Þ p1 ¼ p2 ðp1 p1 Þ: Clearly, p2 ðp1 p1 Þ ¼ p2 : 1 (Reason: Let ðx; yÞ 2 LHS ¼ p2 ðp1 p1 Þ: So there exist u; v such that 1 ðx; uÞ 2 p1 ; ðu; vÞ 2 p1 1 ; and ðv; yÞ 2 p2 : Since ðu; vÞ 2 p1 ; ðv; uÞ 2 p1 : It follows

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that p1 ðvÞ ¼ u; and p1 ðxÞ ¼ u; and hence, p1 ðvÞ ¼ p1 ðxÞ: Since p1 ðvÞ ¼ p1 ðxÞ; and p2 is constant on ﬁbers of p1 ; p2 ðvÞ ¼ p2 ðxÞ: Since ðv; yÞ 2 p2 ; p2 ðvÞ ¼ y: Since p2 ðvÞ ¼ y; and p2 ðvÞ ¼ p2 ðxÞ; p2 ðxÞ ¼ y; and hence, ðx; yÞ 2 p2 ¼ RHS: Hence, LHS RHS: Next, let ðx; yÞ 2 RHS ¼ p2 : Since ðx; yÞ 2 p2 and p2 : M ! P; x 2 M; and y 2 P: Since x 2 M; and p1 : M ! N; there exists u 2 N such that 1 ðx; uÞ 2 p1 : Since ðx; uÞ 2 p1 ; ðu; xÞ 2 p1 1 : Since ðx; uÞ 2 p1 ; ðu; xÞ 2 p1 ; and ðx; yÞ 2 p2 ; ðx; yÞ 2 p2 ðp1 1 p1 Þ ¼ LHS: Hence, RHS LHS: Since LHS RHS; and RHS LHS; LHS ¼ RHS:) Thus, F p1 ¼ p2 : 1 Since p2 p1 1 : N ! P; and F ¼ p2 p1 ; F : N ! P: Now, we want to show 1 that F is 1–1, that is, p2 p1 is 1–1, that is, if ðx; zÞ 2 p2 p1 1 ; and 1 ðy; zÞ 2 p2 p1 , then x ¼ y: For this purpose, let ðx; zÞ 2 p p ; and ðy; zÞ 2 2 1 1 1 p2 p1 : We have to show that x ¼ y: 1 Since ðx; zÞ 2 p2 p1 1 ; there exists u such that ðx; uÞ 2 p1 ; and ðu; zÞ 2 p2 : 1 1 Since ðx; uÞ 2 p1 ; ðu; xÞ 2 p1 : Since ðy; zÞ 2 p2 p1 ; there exists v such that 1 ðy; vÞ 2 p1 1 ; and ðv; zÞ 2 p2 : Since ðy; vÞ 2 p1 ; ðv; yÞ 2 p1 : Since ðv; yÞ 2 p1 ; p1 ðvÞ ¼ y: Similarly, p1 ðuÞ ¼ x; p2 ðuÞ ¼ z; and p2 ðvÞ ¼ z: Since p2 ðuÞ ¼ z ¼ p2 ðvÞ; and p1 is constant on ﬁbers of p2 ; p1 ðuÞ ¼ p1 ðvÞ: Since x ¼ p1 ðuÞ ¼ p1 ðvÞ ¼ y; x ¼ y: Thus, we have shown that F : N ! P is 1–1. Now, we want to show that F : N ! P is onto. For this purpose, let us take any c 2 P: We have to ﬁnd y 2 N such that ðy; cÞ 2 Fð¼ p2 p1 1 Þ: Since p2 : M ! P is onto, and c 2 P; there exists z 2 M such that ðz; cÞ 2 p2 : Since z 2 M; and p1 : M ! N; there exists y 2 N such that ðz; yÞ 2 p1 ; and hence, 1 1 ðy; zÞ 2 p1 1 : Since ðy; zÞ 2 p1 ; and ðz; cÞ 2 p2 ; ðy; cÞ 2 p2 p1 : Thus, we have shown that F : N ! P is onto. Now, we want to show that F : N ! P is smooth. Since F p1 ¼ p2 ; and p2 is smooth, F p1 is smooth. Since p1 : M ! N is a smooth submersion onto map, and F p1 is smooth, by Theorem 25, F is smooth. Lastly, we have to show that F 1 : P ! N is smooth. Since F p1 ¼ p2 ; and F is 1–1 onto, F 1 p2 ¼ p1 : Since F 1 p2 ¼ p1 ; and p1 is smooth, F 1 p2 is smooth. Since p2 : M ! P is a smooth submersion onto map, and F 1 p2 is smooth, by Theorem 5.81, F 1 is smooth. Thus, we have shown that F : N ! P is a diffeomorphism such that F p1 ¼ p2 : This completes the proof of existence part. Uniqueness: Let F1 : N ! P be a smooth mapping such that F1 p1 ¼ p2 ; and let F2 : N ! P be a smooth mapping such that F2 p1 ¼ p2 : We have to show that F1 ¼ F2 : 1 Since p1 : M ! N is onto, p1 p1 1 ¼ IdN : Since F1 p1 ¼ p2 ; p2 p1 ¼ 1 1 1 ðF1 p1 Þ p1 ¼ F1 ðp1 p1 Þ ¼ F1 IdN ¼ F1 : Thus, F1 ¼ p2 p1 : Similarly, F2 ¼ p2 p1 1 : It follows that F1 ¼ F2 : This proves the uniqueness part. h

5.8 Properly Embedded

395

Note 5.84 Before going ahead, let us recall the deﬁnition of open submanifold of M: Let M be an m-dimensional smooth manifold, whose topology is O; and differential structure is A: Let G be a nonempty open subset of M. Let O1 be the subspace topology over G inherited from M, that is, O1 ¼ fG1 : G1 G; and G1 2 Og: Since M is an m-dimensional smooth manifold, M is a Hausdorff space. Since M is a Hausdorff space, and G is a subspace of M, G with the subspace topology is a Hausdorff space. Since M is an m-dimensional smooth manifold, M is a second countable space. Since M is a second countable space, and G is a subspace of M, G with the subspace topology is a second countable space. Put AG fðU; uÞ : ðU; uÞ 2 A; and U Gg: Now, we shall try to show that AG is an atlas on G, that is, 1. [fU : ðU; uU Þ is in AG g ¼ G; 2. all pairs of members in AG are C 1 compatible. For (1): By the deﬁnition of AG ; it is clear that [fU : ðU; uÞ is in AG g G: So, it remains to be proved that G [fU : ðU; uÞ is in AG g: For this purpose, let us take any p in G. Since p is in G, and G is a subset of M, p is in M. Since p is in M, and M is an m-dimensional smooth manifold, there exists ðU; uÞ in A such that p is in U. Since ðU; uÞ is in A; U is an open subset of M. Since U is open, and G is open, U \ G is open. Thus, U \ G is an open neighborhood of p: We shall try to show that ðU \ G; ujðU\GÞ Þ is in AG ; that is, ðU \ G; ujðU\GÞ Þ 2 A; and U \ G G: Since U \ G G; it remains to be showed that ðU \ G; ujðU\GÞ Þ 2 A: Since ðU; uÞ is in A; ðU; uU Þ is an admissible coordinate chart of M. Now, since U \ G is a nonempty open subset of U, by the Lemma 1, ðU \ G; ujðU\GÞ Þ is an admissible coordinate chart of M, and hence, ðU \ G; ujðU\GÞ Þ 2 A: For (2): Let us take any ðU; uÞ; ðV; wÞ 2 AG : We have to show that ðU; uÞ andðV; wÞ are C1 compatible. Since ðU; uÞ 2 AG , by the deﬁnition of AG ; ðU; uÞ 2 A: Similarly, ðV; wÞ 2 A: Since ðU; uÞ 2 A; ðV; wÞ 2 A; and A is a differential structure, ðU; uÞ and ðV; wÞ are C 1 compatible. Thus, we have shown that AG is an atlas on G. Hence, AG determines a unique smooth structure on G. Thus, the open subset G of smooth manifold M becomes an m-dimensional smooth manifold. Here, G is called an open submanifold of M: Note 5.85 Let M be an m-dimensional smooth manifold with smooth structure A: Let G be a nonempty open subset of M. In the open submanifold G of M; the smooth structure of G is determined by the atlas fðU; uÞ : ðU; uÞ 2 A; and U Gg of G. Thus, G becomes an m-dimensional smooth manifold.

396

5 Immersions, Submersions, and Embeddings

We shall try to show that the inclusion map i : G ! M (that is, for every x in G, iðxÞ x) is a smooth embedding. Since i : G ! M is the identity mapping IdG ; and IdG is smooth, i : G ! M is smooth. Now, we want to show that i : G ! M is a smooth immersion, that is, for every p 2 G; the linear map dip : Tp G ! TiðpÞ M is 1–1. For this purpose, let us ﬁx any p 2 G: We have to show that the linear map dip : Tp G ! TiðpÞ M is 1–1. Clearly, Tp G ¼ Tp M; and TiðpÞ M ¼ Tp M: So we have to show that the linear map dip : Tp M ! Tp M is 1–1. Here, i ¼ IdG ; so we have to show that the linear map dðIdG Þp : Tp M ! Tp M is 1–1. Since dðIdG Þp ¼ IdTp G ¼ IdTp M ; and IdTp M : Tp M ! Tp M is 1–1, dðIdG Þp : Tp M ! Tp M is 1–1. Thus, we have shown that i : G ! M is a smooth immersion. Also, clearly i : G ! M is 1–1. Since G is open in M, i : G ! M is an open map. Since i : G ! M is a 1–1 immersion, and open, by Note 5.69, i : G ! M is a smooth embedding. Conclusion: Let M be an m-dimensional smooth manifold. Let G be a nonempty subset of M. If G is open in M, then there exists an m-dimensional smooth structure on G such that the inclusion map i : G ! M is a smooth embedding. Now, we want to prove its converse part: Let M be an m-dimensional smooth manifold. Let G be a nonempty subset of M. Let G be endowed with the subspace topology of M. Suppose there exists an m-dimensional smooth structure A on G such that the inclusion map i : G ! M is a smooth embedding. We shall try to show that G is open in M. Since i : G ! M is a smooth embedding, i : G ! M is a smooth immersion. Since G; M are m-dimensional smooth manifolds, and i : G ! M is a smooth immersion, by Theorem 5.46, i : G ! M is a local diffeomorphism. Since i : G ! M is a local diffeomorphism, by Note 5.38, i : G ! M is an open map. Since i : G ! M is an open map, Gð¼ iðGÞÞ is an open subset of M. Deﬁnition Let M be an m-dimensional smooth manifold. Let S be a nonempty subset of M. Let S be endowed with the subspace topology of M. Let k be a nonnegative integer. If there exists an (m − k)-dimensional smooth structure A on S; such that the inclusion map i : S ! M is a smooth embedding, then we say that S is an embedded submanifold of M with codimension k. Here, we also say that M is an ambient manifold for S. From the Note 5.85, we can write: Let M be an m-dimensional smooth manifold. Let S be a nonempty subset of M. Let S be endowed with the subspace topology of M. S is open in M if and only if S is an embedded submanifold of M with codimension 0. In this sense, we say that the open submanifolds of M are exactly the embedded submanifolds of M with codimension 0.

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397

Note 5.86 Let M be an m-dimensional smooth manifold with smooth structure A, N be an n-dimensional smooth manifold with smooth structure B, and F : N ! M be a smooth embedding. Let FðNÞð MÞ be endowed with the subspace topology of M. Since F : N ! M is a smooth embedding, F : N ! M is a topological embedding, and hence, F : N ! FðNÞ is a homeomorphism. Since F : N ! FðNÞ is a homeomorphism, N is homeomorphic to F(N). Since N is an n-dimensional smooth manifold, N is an n-dimensional topological manifold. Since N is an n-dimensional topological manifold, and N is homeomorphic to F(N), F(N) is an n-dimensional topological manifold. Let us take any ðU; uÞ 2 B: We shall try to show that ðFðUÞ; u ðF 1 ÞÞ is a coordinate chart of F(N). Clearly, F(U) is open in F(N). (Reason: Since ðU; uÞ 2 B; and B is a smooth structure on N, U is open in N. Since F : N ! FðNÞ is a homeomorphism, and U is open in N, F(U) is open in F(N).) Here, ðU; uÞ 2 B; and B is a smooth structure on N, uðUÞ is open in Rn : Since u : U ! uðUÞ; U N; and F : N ! M is 1–1, u ðF 1 Þ is a function whose domain is FðUÞ: Since u : U ! uðUÞ is onto, U N; and F : N ! M is 1–1, the range of u ðF 1 Þ is uðUÞ: Since u : U ! uðUÞ is 1–1, U N; and F : N ! M is 1–1, u ðF 1 Þ : FðUÞ ! uðUÞ is 1–1. Thus, we see that u ðF 1 Þ is a 1–1 mapping from open subset FðUÞ of FðNÞ onto open subset uðUÞ of Rn : Now, we shall try to show that u ðF 1 Þ : FðUÞ ! uðUÞ is continuous. Since F : N ! M is a topological embedding, F is a homeomorphism from and 1–1. Since F 1 : N onto F(N), and hence, F 1 : FðNÞ ! N is continuous 1 FðNÞ ! N is continuous and 1–1, its restriction ðF Þ FðUÞ : FðUÞ ! U is contin uous. Since F : N ! M is 1–1, ðF 1 ÞFðUÞ ¼ ðFjU Þ1 : Since ðF 1 ÞFðUÞ ¼ ðFjU Þ1 ; and ðF 1 ÞFðUÞ : FðUÞ ! U is continuous, ðFjU Þ1 : FðUÞ ! U is continuous. Since ðFjU Þ1 : FðUÞ ! U is continuous, and u : U ! uðUÞ is continuous, their composite u ððFjU Þ1 Þ : FðUÞ ! uðUÞ is continuous. Further, since u ððFjU Þ1 Þ ¼ u ðF 1 Þ; u ðF 1 Þ : FðUÞ ! uðUÞ is continuous. Now, we shall try to show that F ðu1 Þ : uðUÞ ! FðUÞ is continuous. Since ðU; uÞ 2 B; and B is a smooth structure on N; u1 : uðUÞ ! U is continuous and 1–1. Since F : N ! M is a smooth embedding, F : N ! M is continuous. Since F : N ! M is continuous, and u1 : uðUÞ ! U is continuous and 1–1, F ðu1 Þ : uðUÞ ! FðUÞ is continuous. Also, ðu ðF 1 ÞÞ1 ¼ F ðu1 Þ: It follows that ðu ðF 1 ÞÞ1 : uðUÞ ! FðUÞ is continuous. Thus, we have shown that ðFðUÞ; u ðF 1 ÞÞ is a coordinate chart of F(N). Conclusion: Let M be an m-dimensional smooth manifold with smooth structure A, N be an n-dimensional smooth manifold with smooth structure B, and F : N ! M be a smooth embedding. Let FðNÞð MÞ be endowed with the subspace topology of M. If ðU; uÞ 2 B; then ðFðUÞ; u ðF 1 ÞÞ is a coordinate chart of FðNÞ:

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5 Immersions, Submersions, and Embeddings

Note 5.87 Let M be an m-dimensional smooth manifold with smooth structure A, N be an n-dimensional smooth manifold with smooth structure B, and F : N ! M be a smooth embedding. Let FðNÞð MÞ be endowed with the subspace topology of M. Since M is an m-dimensional smooth manifold, M is a Hausdorff space. Since M is a Hausdorff space, and F(N) is a subspace of M, F(N) with the subspace topology is a Hausdorff space. Since M is an m-dimensional smooth manifold, M is a second countable space. Since M is a second countable space, and F(N) is a subspace of M, F(N) with the subspace topology is a second countable space. Put AF

FðUÞ; u F 1 : ðU; uÞ 2 B :

From the Note 5.86, we see that AF is a collection of coordinate charts of F(N). Now, we shall try to show that AF is an n-dimensional atlas on F(N), that is, 1. [fFðUÞ : ðU; uÞ 2 Bg ¼ FðNÞ; 2. all pairs of members in AF are C1 compatible. For 1: Let us take any ðU; uÞ 2 B: Since ðU; uÞ 2 B; and B is a smooth structure of N; U N: Since U N; and F : N ! M; FðUÞ FðNÞ: Hence, [fFðUÞ : ðU; uÞ 2 Bg FðNÞ: Now, it remains to be showed that FðNÞ [fFðUÞ : ðU; uÞ 2 Bg: For this purpose, let us take any FðpÞ 2 FðNÞ; where p 2 N: Since p 2 N; and N is an ndimensional smooth manifold with smooth structure B, there exists ðV; wÞ 2 B such that p 2 V: Since p 2 V; FðpÞ 2 FðVÞ [fFðUÞ : ðU; uÞ 2 Bg: Hence, FðNÞ [fFðUÞ : ðU; uÞ 2 Bg: For 2: Let ðFðUÞ; u ðF 1 ÞÞ 2 AF ; and ðFðVÞ; w ðF 1 ÞÞ 2 AF ; where ðU; uÞ 2 B; and ðV; wÞ 2 B: Let FðUÞ \ FðVÞ 6¼ ;: We have to show that ðw ðF 1 ÞÞ ðu ðF 1 ÞÞ1 is smooth. Here,

1 w F 1 u F 1 ¼ w F 1 F u1 ¼ w F 1 F u1 ¼ w IdN u1 ¼ w u1 :

Since F : N ! M is 1–1, FðU \ VÞ ¼ FðUÞ \ FðVÞ 6¼ ;; and hence, U \ V 6¼ ;: Since U \ V 6¼ ;; ðU; uÞ 2 B; ðV; wÞ 2 B; and B is a smooth structure on N; w u1 ð¼ ðw ðF 1 ÞÞ ðu ðF 1 ÞÞ1 Þ is smooth, and hence, ðw ðF 1 ÞÞ ðu ðF 1 ÞÞ1 is smooth. Thus, we have shown that AF is an n-dimensional atlas on FðNÞ; and hence, F (N) is an n-dimensional smooth manifold, whose differential structure is determined by the atlas AF :

5.8 Properly Embedded

399

Note 5.88 Let M be an m-dimensional smooth manifold with smooth structure A, N be an n-dimensional smooth manifold with smooth structure B, and F : N ! M be a smooth embedding. Let FðNÞð MÞ be endowed with the subspace topology of M: From the Note 5.87, FðNÞ is an n-dimensional smooth manifold, whose differential structure is determined by the atlas AF fðFðUÞ; u ðF 1 ÞÞ : ðU; uÞ 2 Bg: Now, we want to show that F : N ! FðNÞ is a diffeomorphism, that is, 1. F : N ! FðNÞ is a homeomorphism, 2. F : N ! FðNÞ is smooth, and F 1 : FðNÞ ! N is smooth. For 1: Since F : N ! M is a smooth embedding, F : N ! M is a homeomorphism. For 2: Let us take any p 2 N: Let ðU; uÞ 2 B; where p 2 U: Let ðFðVÞ; w ðF 1 ÞÞ 2 AF ; where ðV; wÞ 2 B; and FðpÞ 2 FðVÞ: We have to show that ðw ðF 1 ÞÞ F u1 and u F 1 ðw ðF 1 ÞÞ1 are smooth. Here, ðw ðF 1 ÞÞ F u1 ¼ w ððF 1 Þ FÞ u1 ¼ w IdN u1 ¼ w u1 ; and u F 1 ðw ðF 1 ÞÞ1 ¼ u F 1 ðF w1 Þ ¼ u ðF 1 FÞ w1 ¼ u IdN w1 ¼ u w1 : Since FðpÞ 2 FðVÞ; there exists q 2 V such that FðpÞ ¼ FðqÞ: Since FðpÞ ¼ FðqÞ; and F is 1–1, p ¼ qð2 VÞ; and hence, p 2 U [ V: Since p 2 U [ V; U [ V 6¼ ;: Since U [ V 6¼ ;; ðU; uÞ 2 B; ðV; wÞ 2 B; and B is a smooth structure on N, w u1 ð¼ ðw ðF 1 ÞÞ F u1 Þ; and u w1 ð¼ u F 1 ðw ðF 1 ÞÞ1 Þ are smooth, and hence ðw ðF 1 ÞÞ F u1 ; and u F 1 ðw ðF 1 ÞÞ1 are smooth. Thus, we have shown that F : N ! FðNÞ is a diffeomorphism. Note 5.89 Let M be an m-dimensional smooth manifold with smooth structure A, N be an n-dimensional smooth manifold with smooth structure B, and F : N ! M be a smooth embedding. Let FðNÞð MÞ be endowed with the subspace topology of M. From Note 5.87, F(N) is an n-dimensional smooth manifold, whose differential structure is determined by the atlas AF fðFðUÞ; u ðF 1 ÞÞ : ðU; uÞ 2 Bg: Also, by Note 5.88, F : N ! FðNÞ is a diffeomorphism. Now, we want to show that the n-dimensional smooth manifold FðNÞ; with atlas AF ð¼ fðFðUÞ; u ðF 1 ÞÞ : ðU; uÞ 2 BgÞ; is an embedded submanifold of M, that is, the inclusion map i : FðNÞ ! M is a smooth embedding. Clearly, i ¼ F F 1 : From the Note 5.88, F : N ! FðNÞ is a diffeomorphism, so F 1 : FðNÞ ! N is a diffeomorphism. Since F 1 is a diffeomorphism, and F : N ! M is a smooth embedding, their composition F F 1 ð¼ iÞ is a smooth embedding, and hence, ι is a smooth embedding. Thus, we have shown that F(N) is an embedded submanifold of M with codimension m − n.

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5 Immersions, Submersions, and Embeddings

Note 5.90 Let M be an m-dimensional smooth manifold with smooth structure A; and N be an n-dimensional smooth manifold with smooth structure B: Here, the collection fðU V; ðuU wV ÞÞ : ðU; uU Þ 2 A; ðV; wV Þ 2 Bg is an atlas of the ðm þ nÞ-dimensional product manifold M N: Let us ﬁx any p 2 N: Now, consider the mapping F : M ! M N deﬁned as follows: For every x 2 M; FðxÞ ðx; pÞ: We shall try to show that F : M ! M N is a smooth embedding. For this purpose, it sufﬁces to show that 1. F is 1–1, 2. F is a closed map, and 3. F is a smooth immersion. For 1: It is clear that F is 1–1. For 2: Let A be a closed subset of M. We have to show that F(A) is closed in M × N. Here, FðAÞ ¼ A fpg ¼ ðM NÞ ððAc NÞ [ ðM fpgc ÞÞ: Since A is a closed subset of M, Ac is open in M, and hence, Ac × N is open in M × N. Since N is an n-dimensional smooth manifold, the topology of N is Hausdorff. Since the topology of N is Hausdorff, and p 2 N; fpg is a closed subset of N, and hence, fpgc is open in N. It follows that M fpgc is open in M N: Since M fpgc is open in M N; and Ac N is open in M N; ðAc NÞ [ ðM fpgc Þ is open in M × N, and hence, ðM NÞ ððAc NÞ [ ðM fpgc ÞÞð¼ FðAÞÞ is closed in M × N. Thus, F(A) is closed in M × N. For 3: We have to show that F : M ! M N is an immersion, that is, for every q 2 M; the linear map dFq : Tq M ! TFðqÞ ðM NÞ is 1–1. For this purpose, let us ﬁx any q 2 M: We have to show that the linear map dFq : Tq M ! TFðqÞ ðM NÞ is 1–1. Since q 2 M; and M is an m-dimensional smooth manifold, there exists an admissible coordinate chart ðU; uÞ of M such that q 2 U: Since p 2 N; and N is an n-dimensional smooth manifold, there exists an admissible coordinate chart ðV; wÞ of N such that p 2 V: For every r 2 U; o o o ; ; . . .; m ox1 ox2 ox r

r

r

is the coordinate basis of Tr M corresponding to ðU; uÞ; where ! o 1 o d u uðrÞ : oxi r oxi uðrÞ

5.8 Properly Embedded

401

Here, ðU V; ðu wÞÞ is an admissible coordinate chart of M N such that FðqÞ ¼ ðq; pÞ 2 U V: For every ðs; tÞ 2 U V; ! o o o o o ; ; . . .; m ; mþ1 ; . . .; mþn oy1 ðs;tÞ oy2 ðs;tÞ oy ðs;tÞ oy oy ðs;tÞ ðs;tÞ is the coordinate basis of Tðs;tÞ ðM NÞ corresponding to ðU V; ðu wÞÞ; where ! o o d ðu wÞ1 : ðuwÞðs;tÞ oyi ðs;tÞ oyi ðuwÞðs;tÞ Observe that for every u 2 U;

ðu wÞ F u1 ðuðuÞÞ ¼ ððu wÞ F ÞðuÞ ¼ ðu wÞðFðuÞÞ ¼ ðu wÞðu; pÞ ¼ ðuðuÞ; wðpÞÞ:

Hence, the matrix representation of linear map dFq is the following ðm þ nÞ m matrix: 2

1 60 6 6 6 6 60 6 60 6 60 6 4

0 0 1 0 .. . 0 0 0 0 0 0 .. .

0

0 0

3 0 07 7 7 7 7 17 7: 07 7 07 7 5 0

Clearly, 2

1 60 6 6 6 6 60 rank 6 60 6 60 6 4 0

0 1

0 0 .. .

0 0 0

0 0 0 .. .

0

0

3 0 07 7 7 7 7 17 7 ¼ m; 07 7 07 7 5 0

Hence, dFq : Tq M ! TFðqÞ ðM NÞ is 1–1. Thus, F is a smooth immersion.

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5 Immersions, Submersions, and Embeddings

Since F : M ! M N is a smooth embedding, by Note 5.89, FðNÞð¼ M fpgÞ is an embedded submanifold of M × N with codimension ðm þ nÞ n ð¼ mÞ; and F (N) is diffeomorphic to M. Thus, we have shown that M fpg is an embedded submanifold of M × N with codimension m, and M fpg is diffeomorphic to M. In words, we say that a “slice” of a product manifold M × N is an embedded submanifold of M × N, and each slice is diffeomorphic to M. Note 5.91 Let M be an m-dimensional smooth manifold with smooth structure A; and N be an n-dimensional smooth manifold with smooth structure B: Here, the collection fðU V; ðu wÞÞ : ðU; uÞ 2 A; ðV; wÞ 2 Bg is an atlas of the ðm þ nÞ-dimensional product manifold M N: Let G be an open subset of M, and f : G ! N be a smooth map. Put Cðf Þ fðx; f ðxÞÞ : x 2 Gg (called the graph of f). Now, consider the mapping F : G ! M N deﬁned as follows: For every x 2 G; FðxÞ ðx; f ðxÞÞ: Clearly, FðGÞ ¼ Cðf Þ: We shall try to show that F : G ! M N is a smooth embedding. For this purpose, it sufﬁces to show that 1. 2. 3. 4.

F is 1–1, F 1 : Cðf Þ ! G is continuous, F : G ! M N is a smooth map, and F is a smooth immersion. For 1: It is clear that F is 1–1. For 2: Clearly, F 1 ¼ p1 Cðf Þ ; where p1 : M N ! M is deﬁned by p1 ðx; yÞ x; for every ðx; yÞ 2 M N: Since p1 : M N ! M is continuous, its restric 1 tion p Cðf Þ ð¼ F 1 Þ is continuous, and hence, F 1 : Cðf Þ ! G is continuous.

For 3: Let us take any p 2 G: Since p 2 G; and G is an open submanifold of manifold M, there exists ðU; uÞ 2 A satisfying p 2 U G: Here, p 2 G; and f : G ! N; so f ðpÞ 2 N: Since f ðpÞ 2 N; and N is an n-dimensional smooth manifold with smooth structure B; there exists ðV; wÞ 2 B satisfying f ðpÞ 2 V: Thus, ðU V; ðu wÞÞ is an admissible coordinate chart of ðm þ nÞ-dimensional product manifold M N satisfying FðpÞ ¼ ðp; f ðpÞÞ 2 U V: We have to show that ðu wÞ F u1 is smooth. Let us observe that, for every q 2 U; ððu wÞ F u1 ÞðuðqÞÞ ¼ ðu wÞðFðqÞÞ ¼ ðu wÞðq; f ðqÞÞ ¼ ðuðqÞ; wðf ðqÞÞÞ ¼ ðuðqÞ; ðw f ÞðqÞÞ ¼ ðuðqÞ; ðw f u1 ÞðuðqÞÞÞ: Thus, for every q 2 U; ððu wÞ F u1 ÞðuðqÞÞ ¼ ðuðqÞ; ðw f u1 ÞðuðqÞÞÞ: Now, since f : G ! N is a smooth map, w f u1 is smooth, and hence, ðu wÞ F u1 is smooth. For 4: We have to show that F : G ! M N is an immersion, that is, for every q 2 G; the linear map dFq : Tq M ! TFðqÞ ðM NÞ is 1–1.

5.8 Properly Embedded

403

For this purpose, let us ﬁx any q 2 G: We have to show that the linear map dFq : Tq M ! TFðqÞ ðM NÞ is 1–1. Since q 2 Gð MÞ; and M is an m-dimensional smooth manifold, there exists an admissible coordinate chart ðU; uÞ of M such that q 2 U: Since q 2 U; f ðqÞ 2 N: Since f ðqÞ 2 N; and N is an n-dimensional smooth manifold, there exists an admissible coordinate chart ðV; wÞ of N such that f ðqÞ 2 V: For every r 2 U; o o o ; ; . . .; m ox1 ox2 ox r

r

r

is the coordinate basis of Tr M corresponding to ðU; uÞ; where ! o 1 o d u uðrÞ : oxi r oxi uðrÞ Here, ðU V; ðu wÞÞ is an admissible coordinate chart of M N such that FðqÞ ¼ ðq; f ðqÞÞ 2 U V: For every ðs; tÞ 2 U V; ! o o o o o ; ; . . .; m ; mþ1 ; . . .; mþn oy1 ðs;tÞ oy2 ðs;tÞ oy ðs;tÞ oy oy ðs;tÞ ðs;tÞ is the coordinate basis of Tðs;tÞ ðM NÞ corresponding to ðU V; ðu wÞÞ; where ! o o 1 : d ðu wÞ ðuwÞðs;tÞ oyi ðs;tÞ oyi ðuwÞðs;tÞ Observe that for every u 2 U; ðu wÞ F u1 ðuðuÞÞ ¼ ððu wÞ F ÞðuÞ ¼ ðu wÞðFðuÞÞ ¼ ðu wÞðu; FðuÞÞ ¼ ðuðuÞ; wðFðuÞÞÞ ¼ uðuÞ; w F u1 ðuðuÞÞ : Hence, the matrix representation of linear map dFq is the following ðm þ nÞ m matrix: 2

1

60 6 6. 6. 6. 6 60 6 6 oðp ðw F u1 ÞÞ 1 6 ðuðqÞÞ 6 ox1 6 6 6 oðp2 ðw F u1 ÞÞ 6 ðuðqÞÞ 6 ox1 6 6. 6 .. 6 4 oðp ðw F u1 ÞÞ n ðuðqÞÞ ox1

0

0

1 .. . 0 oðp1 ðw F u1 ÞÞ ðuðqÞÞ ox2 1 oðp2 ðw F u ÞÞ ðuðqÞÞ ox2 .. . oðpn ðw F u1 ÞÞ ðuðqÞÞ ox2

7 7 7 7 7 7 7 1 7 7 oðp1 ðw F u1 ÞÞ 7 ð uðqÞ Þ 7: oxm 7 7 1 7 oðp2 ðw F u ÞÞ 7 ð uðqÞ Þ 7 oxm 7 7 .. 7 . 7 5 1 oðpn ðw F u ÞÞ ð uðqÞ Þ m ox 0 .. .

...

3

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5 Immersions, Submersions, and Embeddings

Clearly, 2

1 60 6 6. 6. 6. 6 60 6 6 oðp ðw F u1 ÞÞ 1 6 ðuðqÞÞ rank6 ox1 6 6 1 6 oð p 2 ð w F u Þ Þ 6 ðuðqÞÞ 6 ox1 6 6. 6 .. 6 4 oðp ðw F u1 ÞÞ n ðuðqÞÞ ox1

0 1 .. . 0 oðp1 ðw F u1 ÞÞ ðuðqÞÞ ox2 1 oð p 2 ð w F u Þ Þ ðuðqÞÞ ox2 .. . oðpn ðw F u1 ÞÞ ðuðqÞÞ ox2

0

3

7 7 7 7 7 7 7 1 7 7 1 oð p 1 ð w F u Þ Þ 7 ð uðqÞ Þ 7 ¼ m; oxm 7 7 1 7 oð p 2 ð w F u Þ Þ ðuðqÞÞ 7 7 oxm 7 7 .. 7 . 7 5 1 oð p n ð w F u Þ Þ ð uðqÞ Þ m ox 0 .. .

Hence, dFq : Tq M ! TFðqÞ ðM NÞ is 1–1. Thus, F is a smooth immersion. Thus, we have shown that F : G ! M N is a smooth embedding. Since F : G ! M N is a smooth embedding, by Note 5.90, FðGÞð¼ Cðf ÞÞ is an embedded submanifold of M N with codimension ðm þ nÞ mð¼ nÞ; and F(N) is diffeomorphic to M. Thus, we have shown that Cðf Þ is an embedded submanifold of M × N with codimension n, and Cðf Þ is diffeomorphic to G. Deﬁnition Let M be an m-dimensional smooth manifold. Let S be a nonempty subset of M. Let S be endowed with the subspace topology of M. Let k be a nonnegative integer. Let S be an embedded submanifold of M with codimension k. If the inclusion map i : S ! M is a proper map, then we say that S is properly embedded. Note 5.92 Let M be an m-dimensional smooth manifold. Let S be a nonempty subset of M. Let S be properly embedded. We shall try to show that S is a closed subset of M. Since M is an m-dimensional smooth manifold, M is a Hausdorff locally compact space. Since S is a nonempty subset of M, and the topology of M is Hausdorff, the subspace topology of S is Hausdorff. Since S is properly embedded, i : S ! M is a proper map. Since S is a Hausdorff topological space, M is a Hausdorff locally compact space, i : S ! M is a continuous mapping, and ι is a proper map, by Note 5.67, i : S ! M is a closed mapping, and hence, iðSÞð¼ SÞ is closed in M. Thus, S is closed in M. Note 5.93 Let M be an m-dimensional smooth manifold. Let S be a nonempty closed subset of M. Let S be an embedded submanifold of M. We shall try to show that S is properly embedded, that is, the inclusion map i : S ! M is a proper map, that is, for every compact subset C of M, i1 ðCÞð¼ C \ SÞ is compact in S. For this purpose, let us take any compact subset C of M. We have to show that C \ S is compact in S. Let fGi gi2I be any open cover of C \ S: Here each Gi is open in S; there exists an open subset Hi of M such that Gi ¼ Hi \ S: Since for each i 2 I; Gi ¼ Hi \ S; and S is a closed subset of M, fHi gi2I [ fSc g is an open cover of C in M.

5.8 Properly Embedded

405

Since fHi gi2I [ fSc g is an open cover of C in M, and C is a compact subset of M, there exist i1 ; . . .; ik 2 I such that C Hi1 [ [ Hik [ Sc ; and hence, C \ S ðHi1 [ [ Hik [ Sc Þ \ S ¼ ðHi1 \ SÞ [ [ ðHik \ SÞ [ðSc \ SÞ ¼ ðHi1 \ SÞ [ [ðHik \ SÞ ¼ Gi1 [ [ Gik : This shows that C \ S is compact in S. Thus, we have shown that S is properly embedded. Note 5.94 Let M be an m-dimensional smooth manifold. Let S be a nonempty compact subset of M. Let S be an embedded submanifold of M. We shall try to show that S is properly embedded. Since M is an m-dimensional smooth manifold, the topology of M is Hausdorff. Since the topology of M is Hausdorff, and S is a nonempty compact subset of M, S is closed in M. Since S is closed in M, by the Note 5.92, S is properly embedded. Note 5.95 Let M be an m-dimensional smooth manifold with smooth structure A; and N be an n-dimensional smooth manifold with smooth structure B: Here, the collection fðU V; ðu wÞÞ : ðU; uÞ 2 A; ðV; wÞ 2 Bg is an atlas of the ðm þ nÞ-dimensional product manifold M N: Let f : M ! N be a smooth map. Here, the graph of f is given by Cðf Þ fðx; f ðxÞÞ : x 2 Mg: By Note 5.91, Cðf Þ is an embedded submanifold of M N with codimension n; and Cðf Þ is diffeomorphic to M. Now, we want to show that Cðf Þ is a closed subset of M N; that is ðCðf ÞÞc is open in M N: For this purpose, let us take any point ðx; yÞ 2 ðCðf ÞÞc : Since ðx; yÞ 2 ðCðf ÞÞc ; ðx; yÞ 62 Cðf Þ ¼ fðx; f ðxÞÞ : x 2 Mg: It follows that y 6¼ f ðxÞ: Here, y; f ðxÞ are distinct points of N, and in the topology of N, there exist an open neighborhood V1 of y, and an open neighborhood V of f(x) such that V1 and V are disjoint. Since f : M ! N is a smooth map, f : M ! N is a continuous map. Since f : M ! N is a continuous map, and V is an open neighborhood of f(x), there exists an open neighborhood U of x such that f ðUÞ V: Here, U is an open neighborhood of x, and V1 is an open neighborhood of y, U × V1 is an open neighborhood of (x, y) in M × N. Clearly, U × V1 is disjoint with Cðf Þ: (Reason: If not, otherwise, let ða; bÞ 2 U V1 ; and ða; bÞ 2 Cðf Þ: Since ða; bÞ 2 Cðf Þ; f ðaÞ ¼ b: Since ða; bÞ 2 U V1 ; a 2 U; and b 2 V1 : Since f ðaÞ ¼ b 2 V1 ; f ðaÞ 2 V1 : Since a 2 U; f ðaÞ 2 f ðUÞ V; and hence, f ðaÞ 2 V: Since f ðaÞ 2 V1 ; and f ðaÞ 2 V; V1 and V are not disjoint, a contradiction.) Since U × V1 is disjoint with Cðf Þ; the open neighborhood U × V1 of (x, y) contained in ðCðf ÞÞc : Thus, (x, y) is an interior point of ðCðf ÞÞc : Hence, ðCðf ÞÞc is open in M × N. Thus, Cðf Þ is a closed subset of M × N. Since Cðf Þ is an embedded submanifold of M × N, and Cðf Þ is a closed subset of M × N, by Note 5.93, Cðf Þ is properly embedded. Conclusion: Let M; N be smooth manifolds. Let f : M ! N be a smooth map. Then, Cðf Þ is properly embedded in M N:

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5 Immersions, Submersions, and Embeddings

5.9 Regular Level Sets Deﬁnition Let M be a ﬁve-dimensional smooth manifold, andlet ðU;uÞ be an 5 types: admissible coordinate chart of M. Any set among the following 10 ¼ 2 u1 ðfðx1 ; x2 ; x3 ; x4 ; x5 Þ 2 uðUÞ : x3 ¼ const; x4 ¼ const; x5 ¼ constgÞ; u1 ðfðx1 ; x2 ; x3 ; x4 ; x5 Þ 2 uðUÞ : x2 ¼ const; x4 ¼ const; x5 ¼ constgÞ; u1 ðfðx1 ; x2 ; x3 ; x4 ; x5 Þ 2 uðUÞ : x2 ¼ const; x3 ¼ const; x5 ¼ constgÞ; etc. is called a two-slice of U. Observe that u1 ðfðx1 ; x2 ; x3 ; x4 ; x5 Þ 2 uðUÞ : x3 ¼ const; x4 ¼ const; x5 ¼ constgÞ 6¼ u1 ðfðx1 ; x2 ; x3 ; x4 ; x5 Þ 2 uðUÞ : x2 ¼ const; x4 ¼ const; x5 ¼ constgÞ: (Reason: Since uðUÞ is a nonempty open subset of R5 ; fðx1 ; x2 ; x3 ; x4 ; x5 Þ 2 u ðUÞ : x3 ¼ const; x4 ¼ const; x5 ¼ constg 6¼ fðx1 ; x2 ; x3 ; x4 ; x5 Þ 2 uðUÞ : x2 ¼ const; x4 ¼ const; x5 ¼ constg: Now, since φ is 1–1, u1 ðfðx1 ; x2 ; x3 ; x4 ; x5 Þ 2 uðUÞ : x3 ¼ const; x4 ¼ const; x5 ¼ constgÞ 6¼ u1 ðfðx1 ; x2 ; x3 ; x4 ; x5 Þ 2 uðUÞ : x2 ¼ const; x4 ¼ const; x5 ¼ constgÞ:) Also, if ðc3 ; c4 ; c5 Þ 6¼ ðd3 ; d4 ; d5 Þ; then fðx1 ; x2 ; x3 ; x4 ; x5 Þ 2 uðUÞ : x3 ¼ c3 ; x4 ¼ c4 ; x5 ¼ c5 g 6¼ fðx1 ; x2 ; x3 ; x4 ; x5 Þ 2 uðUÞ : x3 ¼ d3 ; x4 ¼ d4 ; x5 ¼ d5 g: Now, since u is 1–1, u1 ðfðx1 ; x2 ; x3 ; x4 ; x5 Þ 2 uðUÞ : x3 ¼ c3 ; x4 ¼ c4 ; x5 ¼ c5 gÞ 6¼ u1 ðfðx1 ; x2 ; x3 ; x4 ; x5 Þ 2 uðUÞ : x3 ¼ d3 ; x4 ¼ d4 ; x5 ¼ d5 gÞ; etc. Deﬁnition Let M be a ﬁve-dimensional smooth manifold, and let S be a nonempty subset of M. If for every p in S, there exists an admissible coordinate chart ðU; uÞ of M such that p 2 U; and U \ Sð UÞ is equal to a two-slice of U, then we say that S satisﬁes local two-slice condition (see Fig. 5.2).

Fig. 5.2 Admissible coordinate chart

5.9 Regular Level Sets

407

Note 5.96 Let M be a ﬁve-dimensional smooth manifold. Let S be a nonempty subset of M. Suppose that S satisﬁes local two-slice condition. Let S has the subspace topology inherited from M. We shall try to show that S is a two-dimensional topological manifold. Since M is a smooth manifold, its topology is Hausdorff. Since the topology of M is Hausdorff, and S has the subspace topology inherited from M, the topology of S is Hausdorff. Since M is a smooth manifold, its topology is second countable. Since the topology of M is second countable, and S has the subspace topology inherited from M, the topology of S is second countable. Next, let us take any p ∊ S. Since S satisﬁes local two-slice condition, there exists an admissible coordinate chart ðU; uÞ of M satisfying p 2 U; and U \ S is a twoslice of U. Since ðU; uÞ is an admissible coordinate chart of M, U is open in M, and hence, U \ S is open in S. Since ðU; uÞ is an admissible coordinate chart of ﬁvedimensional smooth manifold M; uðUÞ is an open subset of R5 : Since U \ S is a two-slice of U, for deﬁniteness, let U \ S ¼ u1 ðfðx1 ; x2 ; x3 ; x4 ; x5 Þ 2 uðUÞ : x3 ¼ a3 ; x4 ¼ a4 ; x5 ¼ a5 gÞ; where a3 ; a4 ; a5 are constants. It follows that uðU \ SÞ ¼ fðx1 ; x2 ; x3 ; x4 ; x5 Þ 2 uðUÞ : x3 ¼ a3 ; x4 ¼ a4 ; x5 ¼ a5 g: Let us deﬁne p : R5 ! R2 fa3 g fa4 g fa5 g as follows: For every ðx1 ; x2 ; x3 ; x4 ; x5 Þ 2 R5 ; pðx1 ; x2 ; x3 ; x4 ; x5 Þ ðx1 ; x2 ; a3 ; a4 ; a5 Þ: Clearly, p is an open mapping. Since uðUÞ is an open subset of R5 ; and p : R5 ! R2 fa3 g fa4 g fa5 g is open, pðuðUÞÞð¼ fpðx1 ; x2 ; x3 ; x4 ; x5 Þ : ðx1 ; x2 ; x3 ; x4 ; x5 Þ 2 uðUÞg ¼ fðx1 ; x2 ; x3 ; x4 ; x5 Þ 2 uðUÞ : x3 ¼ a3 ; x4 ¼ a4 ; x5 ¼ a5 g ¼ uðU \ SÞÞ is open in R2 fa3 g fa4 g fa5 g; and hence, uðU \ SÞ is open in R2 fa3 g fa4 g fa5 g: Let us deﬁne a function j : R2 fa3 g fa4 g fa5 g ! R2 as follows: For every ðx1 ; x2 ; a3 ; a4 ; a5 Þ 2 R2 fa3 g fa4 g fa5 g; jðx1 ; x2 ; a3 ; a4 ; a5 Þ ðx1 ; x2 Þ: Clearly, j is a diffeomorphism. Since j : R2 fa3 g fa4 g fa5 g ! R2 is a diffeomorphism, j : R2 fa3 g fa4 g fa5 g ! R2 is a homeomorphism. Since j : R2 fa3 g fa4 g fa5 g ! R2 is a homeomorphism, and uðU \ SÞ is open in R2 fa3 g fa4 g fa5 g; jðuðU \ SÞÞð¼ ðj uÞðU \ SÞÞ is open in R2 ; and hence, ðj uÞðU \ SÞ is open in R2 : Clearly, ðj uÞðU \ SÞ ¼ fðx1 ; x2 Þ : ðx1 ; x2 ; a3 ; a4 ; a5 Þ 2 uðUÞg: (Reason: ðj uÞðU \ SÞ ¼ jðuðU \ SÞÞ ¼ jðfðx1 ; x2 ; x3 ; x4 ; x5 Þ 2 uðUÞ : x3 ¼ a3 ; x4 ¼ a4 ; x5 ¼ a5 gÞ ¼ jðfðx1 ; x2 ; a3 ; a4 ; a5 Þ : ðx1 ; x2 ; a3 ; a4 ; a5 Þ 2 uðUÞg ¼ fjðx1 ; x2 ; a3 ; a4 ; a5 Þ : ðx1 ; x2 ; a3 ; a4 ; a5 Þ 2 uðUÞgÞ ¼ fðx1 ; x2 Þ :ðx1 ; x2 ; a3 ; a4 ; a5 Þ 2 uðUÞg:) Now, we shall try to show that j ðujU\S Þ is a homeomorphism from U \ S onto ðj uÞðU \ SÞ; that is, 1. 2. 3. 4.

j ðujU\S Þ is 1–1 ðj ðujU\S ÞÞðU \ SÞ ¼ ðj uÞðU \ SÞ; ðj ðujU\S ÞÞ : U \ S ! ðj uÞðU \ SÞ is continuous, ðj ðujU\S ÞÞ : U \ S ! ðj uÞðU \ SÞ is an open mapping.

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5 Immersions, Submersions, and Embeddings

For 1: Since j is a homeomorphism, j is 1–1. Since ðU; uÞ is an admissible coordinate chart of M; u : U ! uðUÞ is 1–1, and hence, ujU\S is 1–1. Since ujU\S is 1–1, and j is 1–1, their composite j ðujU\S Þ is 1–1. For 2: Here, LHS ¼ j ujU\S ðU \ SÞ ¼ j ujU\S ðU \ SÞ ¼ jðuðU \ SÞÞ ¼ ðj uÞðU \ SÞ ¼ RHS: For 3: Since j is a homeomorphism, j is continuous. Since ðU; uÞ is an admissible coordinate chart of M, u : U ! uðUÞ is continuous, and hence, ujU\S is continuous. Since ujU \ S is continuous, and j is continuous, their composite ðj ðujU\S ÞÞ is continuous. For 4: Since j is a homeomorphism, j is open. Since ðU; uÞ is an admissible coordinate chart of M, u : U ! uðUÞ is open, and hence, ujU\S is open. Since ujU\S is open, and j is open, their composite ðj ðujU\S ÞÞ is open. Thus, the ordered pair ðU \ S; j ðujU\S ÞÞ is a coordinate chart of S. Also, p 2 U \ S: Hence, S is a two-dimensional topological manifold. Put A

U \ S; j ujU\S : ðU; uÞ is an admissible coordinate chart of M :

We want to show that A is an atlas for S. For this purpose, let us take any ðU \ S; ðj1 uÞjU\S Þ; ðV \ S; ðj2 wÞjV\S Þ 2 A; where ðU; uÞ; ðV; wÞ are admissible coordinate charts of M, and ðU \ SÞ \ ðV \ SÞ 6¼ ;: We have to prove that

1 : j1 ujU\S ððU \ SÞ \ ðV \ SÞÞ j2 wjV\S j1 ujU\S ! j2 wjV\S ððU \ SÞ \ ðV \ SÞÞ

is smooth, that is, 1 j2 wjV\S j1 ujU\S : j1 ujU\S ððU \ V Þ \ SÞ ! j2 wjV\S ððU \ V Þ \ SÞ is smooth, that is, 1 j2 wjV\S j1 ujU\S : ðj1 uÞððU \ V Þ \ SÞ ! ðj2 wÞððU \ V Þ \ SÞ is smooth, that is, 1 j2 wjV\S ujU\S ðj1 Þ1 : ðj1 uÞððU \ V Þ \ SÞ ! ðj2 wÞððU \ V Þ \ SÞ

5.9 Regular Level Sets

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is smooth, that is,

j2

1 wjV\S ujU\S ðj1 Þ1 : ðj1 uÞððU \ V Þ \ SÞ

! ðj2 wÞððU \ V Þ \ SÞ is smooth, that is, j2 w u1 uððU\V Þ\SÞ ðj1 Þ1 : ðj1 uÞððU \ V Þ \ SÞ ! ðj2 wÞððU \ V Þ \ SÞ is smooth. Here, ; 6¼ ðU \ SÞ \ ðV \ SÞ U \ V; so U \ V 6¼ ;: Since U \ V 6¼ ;; and ðU; uÞ; ðV; wÞ are admissible coordinate charts of M, w u1 : uðU \ VÞ ! wðU \ VÞ is smooth, and hence, the restriction ðw u1 ÞuððU\VÞ\SÞ is smooth. ; ðj1 Þ1 are smooth, their composite Since j2 ; ðw u1 Þ uððU\VÞ\SÞ

j2 w u1 uððU\V Þ\SÞ ðj1 Þ1 : ðj1 uÞððU \ V Þ \ SÞ ! ðj2 wÞððU \ V Þ \ SÞ is smooth. Thus, we have shown that A is an atlas for S. Let i : S ! M be the inclusion map of S. We want to show that i : S ! M is a smooth embedding, that is, (a) i : S ! M is smooth, (b) i : S ! M is a topological embedding, and (c) i : S ! M is a smooth immersion. For a: Let us take any p 2 S: So there exists ðU \ S; j ðujU\S ÞÞ 2 A; where ðU; uÞ is an admissible coordinate chart of M, and p 2 U \ S: Since p ¼ iðpÞ 2 M; there exists an admissible coordinate chart ðV; wÞ of M satisfying p ¼ iðpÞ 2 V: We have to show that 1 1 1 1 w i j ujU \ S ¼ w j ujU \ S ¼ w ujU \ S j 1 1 1 ¼ w ujU \ S j1 ¼ w ujU \ S j ¼ w u1 uððU \ V Þ \ SÞ j1 is smooth, that is, ðw u1 ÞuððU\VÞ\SÞ j1 is smooth. Here, U \ V 6¼ ;: Since

U \ V 6¼ ;; and ðU; uÞ; ðV; wÞ are admissible coordinate charts of M; w u1 :

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5 Immersions, Submersions, and Embeddings

uðU \ VÞ ! wðU \ VÞ is smooth, and hence, its restriction ðw u1 ÞuððU\VÞ\SÞ is smooth. Since ðw u1 ÞuððU\VÞ\SÞ is smooth, and j1 is smooth, their composite ðw u1 ÞuððU\VÞ\SÞ j1 is smooth.

For b: Since S has the subspace topology inherited from M; i : S ! M is a topological embedding. For c: Let us take any p 2 S: We have to show that the linear map dip : Tp S ! TiðpÞ M is 1–1, that is, the linear map dip : Tp S ! Tp M is 1–1. Since p 2 S; there exists ðU \ S; j ðujU\S ÞÞ 2 A; where ðU; uÞ is an admissible coordinate chart of M, and p 2 U \ S: Here, ðU; uÞ is an admissible coordinate chart of M satisfying iðpÞ 2 U: For every q 2 U \ S; ! o o ; ox1 q ox2 q is the coordinate basis of Tq S corresponding to ðU \ S; j ðujU\S ÞÞ; where ! 1 o o d j ujU\S : oxi q oxi ðjðujU\S ÞÞðqÞ ðjðujU\S ÞÞðqÞ For every r 2 U; o o o o o ; ; ; ; oy1 r oy2 r oy3 r oy4 r oy5 r

is the coordinate basis of Tr M corresponding to ðU; uÞ; where ! o 1 o d u uðrÞ : oyi r oyi uðrÞ Observe that for every q 2 U \ S; 1 u i j ujU\S j ujU\S ðqÞ ¼ uðiðqÞÞ ¼ uðqÞ ¼ ujU\S ðqÞ ¼ j1 j ujU\S ðqÞ : Also clearly, j1 : R2 ! R2 fa3 g fa4 g fa5 g is such that for every ðx1 ; x2 Þ in R2 ; j1 ðx1 ; x2 Þ ¼ ðx1 ; x2 ; a3 ; a4 ; a5 Þ: Hence, for every q 2 U \ S; the matrix representation of linear map diq is the following 5 × 2 matrix:

5.9 Regular Level Sets

411

2

1 60 6 60 6 40 0

3 0 17 7 07 7: 05 0

Since the 2

1 60 6 rank 6 60 40 0

3 0 17 7 07 7 05 0

is 2, the linear map dip : Tp S ! Tp M is 1–1. Thus, we have shown that i : S ! M is a smooth embedding. Hence, S is an embedded submanifold of M. Since for every q 2 U \ S; the matrix representation of linear map diq is the matrix 2

1 60 6 60 6 40 0

diq

3 0 17 7 07 7; 05 0

! ! ! ! o o o o ¼1 þ0 þ0 ox1 q oy1 iðqÞ oy2 iðqÞ oy3 iðqÞ ! ! o o o o þ0 þ0 ¼ 1 ¼ 1 : oy4 iðqÞ oy5 iðqÞ oy iðqÞ oy q

Thus, for every q 2 U \ S;

diq

! o o ¼ 1 : ox1 q oy q

! o o : ¼ ox2 q oy2 q

Similarly, for every q 2 U \ S;

diq

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5 Immersions, Submersions, and Embeddings

Now, since

diq

! !! o o ; diq ox1 q ox2 q

is a basis of the real linear subspace ðdiq ÞðTq SÞ of Tq M; ! o o ; oy1 q oy2 q is a basis of ðdiq ÞðTq SÞ; while ! o o o o o ; ; ; ; oy1 q oy2 q oy3 q oy4 q oy5 q is the coordinate basis of Tq M: Conclusion: For every q 2 U \ S; if ! o o o o o ; ; ; ; oy1 q oy2 q oy3 q oy4 q oy5 q is the coordinate basis of Tq M; then ! o o ; oy1 q oy2 q is a basis of ðdiq ÞðTq SÞ: Note 5.97 As in Note 5.96, we can prove the following result: Let M be an m-dimensional smooth manifold. Let S be a nonempty subset of M. Suppose that S satisﬁes local k-slice condition. Let S has the subspace topology inherited from M. Then, S is a k-dimensional topological manifold. Also, there exists a smooth structure on S such that M becomes a k-dimensional embedded submanifold of M. Deﬁnition Let M be an m-dimensional smooth manifold, N be an n-dimensional smooth manifold, and U : M ! N be a map. Any set of the form U1 ðcÞ; where c 2 N; is called a level set of U: Note 5.98 Let M be a three-dimensional smooth manifold, N be a four-dimensional smooth manifold, and U : M ! N be a smooth map. Let 2 be the rank of U: Let c 2 N: We shall try to show that the level set U1 ðcÞ satisﬁes local two-slice condition.

5.9 Regular Level Sets

413

By the constant rank Theorem 5.48, for every p 2 U1 ðcÞðthat is; UðpÞ ¼ cÞ there exist admissible coordinate chart ðU; uÞ in M satisfying p 2 U; and admissible coordinate chart ðV; wÞ in N satisfying UðpÞ 2 V; such that UðUÞ V; and for every ðx1 ; x2 ; x3 Þ in uðUÞ;

w U u1 ðx1 ; x2 ; x3 Þ ¼ ðx1 ; x2 ; 0; 0Þ:

Now, we shall try to show that U1 ðcÞ \ U ¼ u1 ðfðx1 ; x2 ; x3 Þ 2 uðUÞ : x1 ¼ ðp1 wÞðcÞ; x2 ¼ ðp2 wÞðcÞgÞ; that is, u U1 ðcÞ \ U ¼ fðx1 ; x2 ; x3 Þ 2 uðUÞ : x1 ¼ ðp1 wÞðcÞ; x2 ¼ ðp2 wÞðcÞg: Let q 2 U1 ðcÞ \ U: It follows that UðqÞ ¼ c; and q 2 U: Since q 2 U; uðqÞ 2 uðUÞ: Put uðqÞ ðx1 ; x2 ; x3 Þ: Hence, ðx1 ; x2 ; x3 Þ ¼ uðqÞ 2 uðUÞ: Also, ðx1 ; x2 ; 0; 0Þ ¼ w U u1 ðx1 ; x2 ; x3 Þ ¼ w U u1 ðuðqÞÞ ¼ wðUðqÞÞ ¼ wðcÞ ¼ ððp1 wÞðcÞ; ðp2 wÞðcÞ; ðp3 wÞðcÞ; ðp4 wÞðcÞÞ: Hence, x1 ¼ ðp1 wÞðcÞ; and x2 ¼ ðp2 wÞðcÞ: Thus, U1 ðcÞ \ U u1 ðfðx1 ; x2 ; x3 Þ 2 uðUÞ : x1 ¼ ðp1 wÞðcÞ; x2 ¼ ðp2 wÞðcÞgÞ: Next, let q 2 u1 ðfðx1 ; x2 ; x3 Þ 2 uðUÞ : x1 ¼ ðp1 wÞðcÞ; x2 ¼ ðp2 wÞðcÞgÞ: We have to show that q 2 U1 ðcÞ \ U; that is, UðqÞ ¼ c; and q 2 U: Since q 2 u1 ðfðx1 ; x2 ; x3 Þ 2 uðUÞ : x1 ¼ ðp1 wÞðcÞ; x2 ¼ ðp2 wÞðcÞgÞ U; q 2 U: Since uðqÞ 2 uðUÞ; ððp1 wÞðcÞ; ðp2 wÞðcÞ; ðp3 wÞðcÞ; ðp4 wÞðcÞÞ ¼ wðcÞ ¼ wðUðqÞÞ ¼ ðw U u1 ÞðuðqÞÞ ¼ ððp1 uÞðqÞ; ðp2 uÞðqÞ; 0; 0Þ; and hence, ðp3 wÞðcÞ ¼ 0 ¼ ðp4 wÞðcÞ: Since q 2 u1 ðfðx1 ; x2 ; x3 Þ 2 uðUÞ : x1 ¼ ðp1 wÞ ðcÞ; x2 ¼ ðp2 wÞðcÞgÞ; uðqÞ 2 fðx1 ; x2 ; x3 Þ 2 uðUÞ : x1 ¼ ðp1 wÞðcÞ; x2 ¼ ðp2 wÞðcÞg; and hence, ðp1 uÞðqÞ ¼ ðp1 wÞðcÞ; and ðp2 uÞðqÞ ¼ ðp2 wÞðcÞ: Since ðp1 uÞðqÞ ¼ ðp1 wÞðcÞ; ðp2 uÞðqÞ ¼ ðp2 wÞðcÞ; ðp3 wÞ ðcÞ ¼ 0 ¼ ðp4 wÞðcÞ; and wðUðqÞÞ ¼ ððp1 uÞðqÞ; ðp2 uÞðqÞ; 0; 0Þ; wðUðqÞÞ ¼ ððp1 wÞ ðcÞ; ðp2 wÞ ðcÞ; 0; 0Þ ¼ ððp1 wÞðcÞ; ðp2 wÞðcÞ; ðp3 wÞ ðcÞ; ðp4 wÞðcÞÞ ¼ wðcÞ: Since wðUðqÞÞ ¼ wðcÞ; UðqÞ ¼ c: Thus, uðU1 ðcÞ \ UÞ ¼ fðx1 ; x2 ; x3 Þ 2 uðUÞ : x1 ¼ ðp1 wÞðcÞ; x2 ¼ ðp2 wÞ ðcÞg: Hence, the level set U1 ðcÞ satisﬁes the local 2-slice condition. Now, by Note 5.97, if U1 ðcÞ has the subspace topology inherited from M; then U1 ðcÞ is an embedded submanifold of M. Since U : M ! N is a smooth map, U : M ! N is continuous. Since U : M ! N is continuous, and fcg is a closed subset of N; U1 ðcÞ is a closed subset of M. Since U1 ðcÞ is a closed subset of M, and U1 ðcÞ is

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5 Immersions, Submersions, and Embeddings

an embedded submanifold of M, by Note 5.93, U1 ðcÞ is a properly embedded submanifold of M, and its codimension is 2. Note 5.99 The following result can be proved as in Note 5.98: Let M be an m-dimensional smooth manifold, N be an n-dimensional smooth manifold, and U : M ! N be a smooth map. Let r be the rank of U: Let c 2 N: Then, the level set U1 ðcÞ is a properly embedded submanifold of M, and its codimension is r. Note 5.100 Let M be an m-dimensional smooth manifold, N be an n-dimensional smooth manifold, and U : M ! N be a smooth submersion. Let c 2 N: We shall try to show that the level set U1 ðcÞ is a properly embedded submanifold of M, and its codimension is n. Since U : M ! N is a smooth submersion, U : M ! N is a smooth map, and rank U ¼ n: Since U : M ! N is a smooth map, and rank U ¼ n; by Note 5.99, the level set U1 ðcÞ is a properly embedded submanifold of M, and its codimension is n. Deﬁnition Let M be an m-dimensional smooth manifold, N be an n-dimensional smooth manifold, and U : M ! N be a smooth map. Let p 2 M: If dUp : Tp M ! TUðpÞ N is onto (that is, the rank of U at p is n), then we say that p is a regular point of U: If p is not a regular point of U; then we say that p is a critical point of U: Note 5.101 Let M be an m-dimensional smooth manifold, N be an n-dimensional smooth manifold, and U : M ! N be a smooth map. Clearly, U is a smooth submersion if and only if every point of M is a regular point of U: Note 5.102 Let M be an m-dimensional smooth manifold, N be an n-dimensional smooth manifold, and U : M ! N be a smooth map. Let m < n. We shall try to show that every point of M is a critical point of U: If not, otherwise, let there exists p ∊ M such that p is not a critical point of U: Since p is not a critical point of U; p is a regular point of U; and hence, the linear map dUp : Tp M ! TUðpÞ N is onto. Since the linear map dUp : Tp M ! TUðpÞ N is onto, n ¼ dim N ¼ dim TUðpÞ N dim Tp M ¼ dim M ¼ m: Thus, m¥n; a contradiction. Thus, every point of M is a critical point of U: Note 5.103 Let M be an m-dimensional smooth manifold, N be an n-dimensional smooth manifold, and U : M ! N be a smooth map. Let G be the set of all regular points of U: We shall try to show that G is an open subset of M. For this purpose, let us take any p ∊ G. Since p ∊ G, p is a regular point of U: Since p is a regular point of U; the linear map dUp : Tp M ! TUðpÞ N is onto. Since the linear map dUp : Tp M ! TUðpÞ N is onto, by Theorem 4.7, there exists an open neighborhood W of p such that for every q ∊ W, the linear map dFq : Tq M ! TFðqÞ N

5.9 Regular Level Sets

415

is onto, and hence, the open neighborhood W of p is contained in G. It follows that p is an interior point of G. Thus, we have shown that G is an open subset of M. Deﬁnition Let M be an m-dimensional smooth manifold, N be an n-dimensional smooth manifold, and U : M ! N be a smooth map. Let G be the set of all regular points of U: Let c 2 N: If U1 ðcÞ G; then we say that c is a regular value of U; and U1 ðcÞ is a regular value set. If c is not a regular value of U; then we say that c is a critical value of U: If c is a regular value of U; then U1 ðcÞ is called a regular level set. Clearly, if U1 ðcÞ ¼ ;; then c is a regular value of U: Note 5.104 Let M be an m-dimensional smooth manifold, N be an n-dimensional smooth manifold, and U : M ! N be a smooth map. Letc 2 N: Let U1 ðcÞ be a nonempty regular level set. We shall try to show that U1 ðcÞ is a properly embedded submanifold of M, and its codimension is n. Let G be the set of all regular points of U: We know that G is an open subset of M. Let O1 be the subspace topology over G inherited from M, that is, O1 ¼ fG1 : G1 G; and G1 2 Og: Let A be the differential structure of M. Put AG fðU; uÞ : ðU; uÞ 2 A; and U Gg: We know that AG is an atlas on G. Thus, G becomes an m-dimensional smooth manifold (called open submanifold of M). Also, G is an embedded submanifold of M with codimension 0. Since U : M ! N is a smooth map, and G is an open submanifold of M, UjG : G ! N is a smooth map. Let us take any p 2 G: We want to show that the linear map dðUjG Þp : Tp G ! TðUjG ÞðpÞ N is onto. Since p 2 G; p is a regular point of M, and hence, the linear map dUp : Tp M ! TUðpÞ N is onto. Since p ∊ G, and G is an open submanifold of M, TpM is equal to TpG. Also, ðUjG ÞðpÞ ¼ UðpÞ: It follows that the linear map dðUjG Þp : Tp G ! TðUjG ÞðpÞ N is onto. Now, we want to show that U1 ðcÞ is an embedded submanifold of M. Since for every p 2 G; the linear map dðUjG Þp : Tp G ! TðUjG ÞðpÞ N is onto, UjG : G ! N is a smooth submersion. Since UjG : G ! N is a smooth submersion, and c 2 N; by Note 5.100, the level set ðUjG Þ1 ðcÞ is a properly embedded submanifold of G, and its codimension is n. Clearly, ðUjG Þ1 ðcÞ ¼ U1 ðcÞ: (Reason: Let p 2 ðUjG Þ1 ðcÞ: So p 2 G; and ðUjG ÞðpÞ ¼ c: Since p 2 G; c ¼ ðUjG ÞðpÞ ¼ UðpÞ; and hence, p 2 U1 ðcÞ: Thus, ðUjG Þ1 ðcÞ U1 ðcÞ: Next, let p 2 U1 ðcÞ: Since p 2 U1 ðcÞ; UðpÞ ¼ c: Since U1 ðcÞ is a regular value set, U1 ðcÞ G: Since p 2 U1 ðcÞ G; p 2 G; and hence, ðUjG ÞðpÞ ¼ UðpÞ ¼ c: It follows that p 2 ðUjG Þ1 ðcÞ: Thus, U1 ðcÞ ðUjG Þ1 ðcÞ:)

416

5 Immersions, Submersions, and Embeddings

Since ðUjG Þ1 ðcÞ ¼ U1 ðcÞ; and the level set ðUjG Þ1 ðcÞ is a properly embedded submanifold of G with its codimension n, U1 ðcÞ is a properly embedded submanifold of G with its codimension n. Since U1 ðcÞ is an embedded submanifold of G, the inclusion map i1 : U1 ðcÞ ! G is a smooth embedding. Since G is an embedded submanifold of M, the inclusion map i2 : G ! M is a smooth embedding. Since i1 : U1 ðcÞ ! G is a smooth embedding, and i2 : G ! M is a smooth embedding, their composite i2 i1 : U1 ðcÞ ! M is a smooth embedding. Since i2 i1 ¼ i; where idenotes the inclusion map from U1 ðcÞ to M; and i2 i1 : U1 ðcÞ ! M is a smooth embedding, the inclusion map i : U1 ðcÞ ! M is a smooth embedding. This shows that U1 ðcÞ is an embedded submanifold of M. Now, we want to show that U1 ðcÞ is a properly embedded submanifold of M. For this purpose, let us take any compact set K in M. We have to show that ði2 i1 Þ1 ðKÞ is compact in U1 ðcÞ: Here, ði2 i1 Þ1 ðKÞ ¼ ðði1 Þ1 ði2 Þ1 ÞðKÞ ¼ ði1 Þ1 ðK \ GÞ ¼ ðK \ GÞ \ U1 ðcÞ ¼ K \ U1 ðcÞ: Since K is compact in M, and U1 ðcÞ is closed in M, K \ U1 ðcÞ is compact in M. Since K \ U1 ðcÞ is compact in M, and K \ U1 ðcÞ U1 ðcÞ; K \ U1 ðcÞð¼ ði2 i1 Þ1 ðKÞÞ is compact in U1 ðcÞ; and hence ði2 i1 Þ1 ðKÞ is compact in U1 ðcÞ: Thus, U1 ðcÞ is a properly embedded submanifold of M with codimension n.

5.10 Smooth Submanifolds Deﬁnition Let M be an m-dimensional smooth manifold and S be a nonempty subset of M. If there exists a topology O over S with respect to which S becomes an n-dimensional topological manifold, and there exists a smooth structure on S with respect to which the topological manifold S becomes an n-dimensional smooth manifold such that the inclusion map i : S ! M is smooth immersion, then we say that S is an immersed submanifold (or smooth submanifold) of M with codimension m − n. Note 5.105 Let M be an m-dimensional smooth manifold and S be a nonempty subset of M. Let k 2 f0; 1; . . .; m 1g: Let S be an embedded submanifold of M with codimension k. We shall try to show that S is an immersed submanifold of M with codimension k. Put O fG \ S : G is open in Mg: We know that O (called the subspace topology of S) is a topology over S. Since S is an embedded submanifold of M with codimension k, O is a topology over S with respect to which S becomes a (m − k)dimensional topological manifold. Also, there exists a smooth structure A on S, with respect to which the topological manifold S becomes an (m − k)-dimensional smooth manifold such that the inclusion map i : S ! M is a smooth embedding, and hence, i : S ! M is a smooth immersion.

5.10

Smooth Submanifolds

417

Hence, S is an immersed submanifold of M with codimension kð¼ m ðm kÞÞ: This completes the proof. In short, every embedded submanifold is a smooth submanifold. Note 5.106 Let M be an m-dimensional smooth manifold with smooth structure A, N be an n-dimensional smooth manifold with smooth structure B, and F : N ! M be a 1–1 smooth immersion. Put O fFðVÞ : V is open in N g: Clearly, O is a topology over FðNÞ; and F is a homeomorphism from N onto F(N). Since F is a homeomorphism from N onto F(N), N is homeomorphic onto F(N). Since N is an n-dimensional smooth manifold, N is an n-dimensional topological manifold. Since N is an n-dimensional topological manifold, and N is homeomorphic onto F(N), F(N) is an n-dimensional topological manifold. Since N is an ndimensional smooth manifold with smooth structure B; and F is a homeomorphism from N onto F(N), fðFðVÞ; w ðF 1 FðVÞ ÞÞ : ðV; wÞ 2 Bg is a smooth structure on F(N). Thus, F(N) becomes an n-dimensional smooth manifold, and F is a diffeomorphism from N onto F(N). It is clear that w ðF 1 FðVÞ Þ ¼ w ðF 1 Þ: It follows

that the smooth structure on F(N) is fðFðVÞ; w ðF 1 ÞÞ : ðV; wÞ 2 Bg: Now, we want to prove that the inclusion map i : FðNÞ ! M is a smooth map. Clearly, i ¼ F F 1 : Since F : N ! FðNÞ is a diffeomorphism, F 1 : FðNÞ ! N is a diffeomorphism. Since F 1 is a diffeomorphism, and F : N ! M is a 1–1 smooth immersion, their composition F F 1 ð¼ iÞ is a 1–1 smooth immersion, and hence, i is a smooth immersion. It follows that F(N) is a smooth submanifold of M with codimension m − n. Conclusion: Let M be an m-dimensional smooth manifold with smooth structure A, N be an n-dimensional smooth manifold with smooth structure B, and F : N ! M be a 1–1 smooth immersion. Then, F(N) is a smooth submanifold of M with codimension m − n. Note 5.107 Let M be an m-dimensional smooth manifold and S be a nonempty subset of M. Let S be a smooth submanifold of M with codimension 0. We shall try to show that S is an embedded submanifold of M with codimension 0. Since S is a smooth submanifold of M with codimension 0, by its deﬁnition, there exists a topology O over S with respect to which S becomes an m-dimensional topological manifold, and there exists a smooth structure on S with respect to which the topological manifold S becomes an m-dimensional smooth manifold such that the inclusion map i : S ! M is a smooth immersion. Here, S is an m-dimensional smooth manifold, M is an m-dimensional smooth manifold, and the inclusion map i : S ! M is a 1–1 smooth immersion, so by Note 5.73, i : S ! M is a smooth embedding. Since i : S ! M is a smooth embedding, the inclusion map i : S ! M is a topological embedding, and hence, i is a homeomorphism from S onto iðSÞð¼ SÞ; where iðSÞ has the subspace topology inherited from M. It follows that O is equal to the subspace topology of S inherited from M. Since S is endowed with

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5 Immersions, Submersions, and Embeddings

the subspace topology inherited from M, and the inclusion map i : S ! M is a smooth embedding, and the smooth structure on S is m-dimensional, by the deﬁnition of embedded submanifold, S is an embedded submanifold of M with codimension 0. Note 5.108 Let M be an m-dimensional smooth manifold and S be a nonempty subset of M. Let S be a smooth submanifold of M with codimension k. Let the inclusion map i : S ! M be proper. We shall try to show that S is an embedded submanifold of M with codimension k. Since S is a smooth submanifold of M with codimension k; by its deﬁnition, there exists a topology O over S with respect to which S becomes an (m − k)dimensional topological manifold, and there exists a smooth structure on S with respect to which the topological manifold S becomes an ðm kÞ-dimensional smooth manifold such that the inclusion map i : S ! M is a smooth immersion. Here, S is an (m − k)-dimensional smooth manifold, M is an m-dimensional smooth manifold, the inclusion map i : S ! M is a 1–1 smooth immersion, and the inclusion map i : S ! M is proper, so by Note 5.71, i : S ! M is smooth embedding. Since i : S ! M is a smooth embedding, the inclusion map i : S ! M is a topological embedding, and hence, i is a homeomorphism from S onto iðSÞð¼ SÞ; where ι(S) has the subspace topology inherited from M. It follows that O is equal to the subspace topology of S inherited from M. Since S is endowed with the subspace topology inherited from M, the inclusion map i : S ! M is a smooth embedding, and the smooth structure on S is (m − k)-dimensional, by the deﬁnition of embedded submanifold, S is an embedded submanifold of M with codimension k. Note 5.109 Let M be an m-dimensional smooth manifold and S be a nonempty subset of M. Let S be a smooth submanifold of M with codimension k. Let S be compact. We shall try to show that S is an embedded submanifold of M with codimension k. Since S is a smooth submanifold of M with codimension k, by its deﬁnition, there exists a topology O over S with respect to which S becomes an (m − k)dimensional topological manifold, and there exists a smooth structure on S with respect to which the topological manifold S becomes an (m − k)-dimensional smooth manifold such that the inclusion map i : S ! M is a smooth immersion. Here, S is an (m − k)-dimensional smooth manifold, M is an m-dimensional smooth manifold, the inclusion map i : S ! M is a 1–1 smooth immersion, and S is compact, so by Note 5.72, i : S ! M is a smooth embedding. Since i : S ! M is a smooth embedding, the inclusion map i : S ! M is a topological embedding, and hence, ι is a homeomorphism from S onto iðSÞð¼ SÞ; where ι(S) has the subspace topology inherited from M. It follows that O is equal to the subspace topology of S inherited from M. Since S is endowed with the subspace topology inherited from M, the inclusion map i : S ! M is a smooth embedding, and the smooth structure on S is (m − k)-dimensional, by the deﬁnition of embedded submanifold, S is an embedded submanifold of M with codimension k.

5.10

Smooth Submanifolds

419

Note 5.110 Let M be an m-dimensional smooth manifold and S be a nonempty subset of M. Let S be a smooth submanifold of M with codimension k. Let p ∊ S. We shall try to show that there exists an open neighborhood U of p in S such that U is an embedded submanifold of M with codimension k. Since S is a smooth submanifold of M with codimension k, there exists a topology O over S with respect to which S becomes a (m − k)-dimensional topological manifold, and there exists a smooth structure on S with respect to which the topological manifold S becomes a (m − k)-dimensional smooth manifold such that the inclusion map i : S ! M is a smooth immersion. Since i : S ! M is a smooth immersion, and p ∊ S, by Theorem 5.75, there exists an open neighborhood U of p in S such that the restriction ijU : U ! M is a smooth embedding. Since ijU : U ! M is a smooth embedding, ijU : U ! M is a topological embedding, and hence, ι|U is a homeomorphism from U onto ðijU ÞðUÞð¼ UÞ; where (ι|U)(U) has the subspace topology inherited from M. It follows that the topology over U is the subspace topology inherited from M. Here, S is a (m − k)-dimensional smooth manifold, and U is an open neighborhood of p, so U is a (m − k)-dimensional smooth manifold. Since U is a (m − k)-dimensional smooth manifold, the topology over U is the subspace topology inherited from M, and the inclusion map ijU : U ! M is a smooth embedding, U is an embedded submanifold of M with codimension k. Note 5.111 Let M be an m-dimensional smooth manifold, N be an n-dimensional smooth manifold, and F : M ! N be a smooth map. Let S be a nonempty subset of M. Let S be a smooth submanifold of M. We shall try to show that the restriction FjS : S ! N is smooth. Let the codimension of S be k. Since S is a smooth submanifold of M, there exists a topology O over S with respect to which S becomes an (m − k)-dimensional topological manifold, and there exists a smooth structure A on S with respect to which the topological manifold S becomes an (m − k)-dimensional smooth manifold such that the inclusion map i : S ! M is a smooth immersion. Since i : S ! M is a smooth immersion, i : S ! M is smooth. Since i : S ! M is smooth, and F : M ! N is a smooth map, their composite F i : S ! N is smooth. Since F i : S ! N is smooth, and FjS ¼ F i; FjS : S ! N is smooth. Note 5.112 Let M be an m-dimensional smooth manifold, N be an n-dimensional smooth manifold, and S be a nonempty subset of M. Let S be a smooth submanifold of M with codimension k. Let F : N ! M be a smooth map, and FðNÞ S: Let F : N ! S be continuous. We shall try to show F : N ! S is smooth. Let us ﬁx any p 2 N: Since p 2 N; FðpÞ 2 FðNÞ S; and hence, FðpÞ 2 S: Since S is a smooth submanifold of M, there exists a topology O over S with respect to which S becomes an ðm kÞ-dimensional topological manifold, and there exists a smooth structure A on S with respect to which the topological manifold S becomes an (m − k)-dimensional smooth manifold such that the inclusion map i : S ! M is a smooth immersion. Since the inclusion map i : S ! M is a smooth immersion, by Theorem 5.46, i : S ! M is a local diffeomorphism. Since i : S ! M is a local diffeomorphism, and FðpÞ 2 S; there exists an open neighborhood U of

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F(p) in S such that ι(U) is open in M, and the restriction ijU : U ! iðUÞ is a diffeomorphism, and hence, ðijU Þ1 : iðUÞ ! U is smooth. Since U is an open neighborhood of F(p) in S, and F : N ! S is continuous, there exists an open neighborhood V of p in N such that F(V) ⊂ U. Since V is open in N, V is an open submanifold of N, and hence, V is an embedded submanifold of N. Since V is an embedded submanifold of N, V is a smooth submanifold of N. Since V is a smooth submanifold of N, and F : N ! M is a smooth map, by Note 5.111, FjV : V ! M is a smooth map. Since FjV : V ! M is a smooth map, ðijU Þ1 : iðUÞ ! U is smooth, and ðFjV ÞðVÞ ¼ FðVÞ U ¼ iðUÞ; their composite ððijU Þ1 Þ ðFjV Þ : V ! U is smooth. Since ððijU Þ1 Þ ðFjV Þ : V ! U is smooth, and ððijU Þ1 Þ ðFjV Þ ¼ FjV ; FjV : V ! U is smooth. Since FjV : V ! U is smooth, and V is an open neighborhood of p in N, F : N ! S is smooth at the point p. It follows that F : N ! S is smooth. Note 5.113 Let M be an m-dimensional smooth manifold, N be an n-dimensional smooth manifold, and S be a nonempty subset of M. Let S be an embedded submanifold of M with codimension k. Let F : N ! M be a smooth map, and FðNÞ S: Let F : N ! S be continuous. We shall try to show F : N ! S is smooth. Since S is an embedded submanifold of M with codimension k, by Note 5.105, S is a smooth submanifold of M with codimension k. Since F : N ! M is a smooth map, F : N ! M is continuous. Since S is an embedded submanifold of M, the topology over S is the subspace topology inherited from M. Since F : N ! M is continuous, FðNÞ S M; and the topology over S is the subspace topology inherited from M, F : N ! S is continuous. It follows, from Note 5.112, that F : N ! S is smooth. Note 5.114 Let M be an m-dimensional smooth manifold. Let S be a nonempty subset of M. Suppose that S satisﬁes local k-slice condition. Since S satisﬁes local k-slice condition, by Note 5.97, S is an embedded submanifold of M. Since S is an embedded submanifold of M, the topology over S is the subspace topology inherited from M, and there exists a smooth structure A on S with respect to which S becomes a k-dimensional smooth manifold such that the inclusion map i : S ! M is a smooth embedding. Now, we want to show that such a differential structure A is unique. For this purpose, let A1 be a smooth structure on S with respect to which S becomes a kdimensional smooth manifold such that the inclusion map i : S ! M is a smooth embedding. Next, let A2 be a smooth structure on S with respect to which S becomes a k-dimensional smooth manifold such that the inclusion map i : S ! M is a smooth embedding. We have to prove that A1 ¼ A2 : Let O be the topology over M; and let B be the smooth structure of M: Since the inclusion map i : S ! M is a smooth embedding from smooth manifold ðS; A1 Þ to ðM; BÞ; the inclusion map i : S ! M is a smooth map from smooth manifold ðS; A1 Þ to ðM; BÞ: Since the inclusion map i : S ! M is a smooth map from smooth manifold ðS; A1 Þ to ðM; BÞ; ðS; A2 Þ is an embedded submanifold of M; and i : S ! S is continuous, by Lemma 5.113, i : ðS; A1 Þ ! ðS; A2 Þ is smooth. Similarly,

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421

i : ðS; A2 Þ ! ðS; A1 Þ is smooth. Since i : ðS; A1 Þ ! ðS; A2 Þ is smooth, and ði1 ¼Þi : ðS; A2 Þ ! ðS; A1 Þ is smooth, i : ðS; A1 Þ ! ðS; A2 Þ is a diffeomorphism. Hence, A1 ¼ A2 : Note 5.115 Let M be an m-dimensional smooth manifold. Let S be a nonempty subset of M. Suppose that S is a smooth submanifold of M. Since S is a smooth submanifold of M, there exists a topology O over S, and a smooth structure A on S with respect to which S becomes a k-dimensional smooth manifold such that the inclusion map i : S ! M is a smooth immersion. Now, we want to show that for ﬁxed O; the differential structure A is unique. Let us ﬁx the topology O over S. Next, let A1 be a smooth structure on S with respect to which S becomes a k-dimensional smooth manifold such that the inclusion map i : S ! M is a smooth immersion. Next, let A2 be a smooth structure on S with respect to which S becomes a k-dimensional smooth manifold such that the inclusion map i : S ! M is a smooth immersion. We have to prove that A1 ¼ A2 : Let O be the topology over M, and let B be the smooth structure of M: Since the inclusion map i : S ! M is a smooth immersion from smooth manifold ðS; A1 Þ to ðM; BÞ; the inclusion map i : S ! M is a smooth map from smooth manifold ðS; A1 Þ to ðM; BÞ: Since the inclusion map i : S ! M is a smooth map from smooth manifold ðS; A1 Þ to ðM; BÞ; ðS; A2 Þ is a smooth submanifold of M; and i : S ! S is continuous, by Lemma 5.113, i : ðS; A1 Þ ! ðS; A2 Þ is smooth. Similarly, i : ðS; A2 Þ ! ðS; A1 Þ is smooth. Since i : ðS; A1 Þ ! ðS; A2 Þ is smooth, and ði1 ¼Þi : ðS; A2 Þ ! ðS; A1 Þ is smooth, i : ðS; A1 Þ ! ðS; A2 Þ is a diffeomorphism. Hence, A1 ¼ A2 :

5.11 Tangent Space to a Submanifold Note 5.116 Let M be an m-dimensional smooth manifold. Let S be a nonempty subset of M. Let S be an embedded submanifold of M. Since S is an embedded submanifold of M, the inclusion map i : S ! M is a smooth embedding, and hence, i : S ! M is a smooth immersion. Since i : S ! M is a smooth immersion, for every p 2 S, the linear map dip : Tp S ! TiðpÞ M is 1–1. Now, since iðpÞ ¼ p, for every p 2 S, the linear map dip : Tp S ! Tp M is 1–1, and hence, the image set ðdip ÞðTp SÞ is a subspace of the real linear space Tp M. Also, since dip : Tp S ! Tp M is 1–1, dimððdip ÞðTp SÞÞ ¼ dimðTp SÞ ¼ dimðSÞ. That is why, we adopt the convention of not to distinguish between ðdip ÞðTp SÞ and Tp S. Lemma 5.117 Let M be an m-dimensional smooth manifold. Let S be a nonempty subset of M. Let S be an embedded submanifold of M. Let p 2 S. Let v 2 Tp M. If there exists a smooth curve c : ðe; eÞ ! M whose image is contained in S, which is also smooth as a map into S, such that cð0Þ ¼ p, and c0 ð0Þ ¼ v, then v 2 Tp S.

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5 Immersions, Submersions, and Embeddings

Proof Let c : ðe; eÞ ! M be a smooth curve whose image is contained in S, which is also smooth as a map into S, such that cð0Þ ¼ p, and c0 ð0Þ ¼ v. We have to show that v 2 Tp S. Since c : ðe; eÞ ! S is a smooth map from one-dimensional smooth manifold ðe; eÞ into smooth manifold S, and 0 2 ðe; eÞ, dc0 : T0 ðe; eÞ ! Tcð0Þ S. Now, since d Tcð0Þ S ¼ Tp S, dc0 : T0 ðe; eÞ ! Tp S, and hence, v ¼ c0 ð0Þ ¼ ðdc0 Þð Þ 2 Tp S. h dt 0 Lemma 5.118 Let M be an m-dimensional smooth manifold. Let S be a nonempty subset of M. Let S be an embedded submanifold of M. Let p 2 S. Let v 2 Tp M. If v 2 Tp S, then there exists a smooth curve c : ðe; eÞ ! M whose image is contained in S, which is also smooth as a map into S, such that cð0Þ ¼ p, and c0 ð0Þ ¼ v. Proof Let v 2 Tp S. Here, S is an embedded submanifold of M, so S is a smooth manifold. Let k be the dimension of smooth manifold S. Since p 2 S, and S is a smooth manifold, there exists an admissible coordinate chart ðU; uÞ of S such that p 2 U, and uðpÞ ¼ 0 2 Rk . For every q 2 U; let ! o o ; . . .; k ox1 q ox q be the coordinate basis of Tq S corresponding to ðU; uÞ, where ! o 1 o d u uðqÞ : oxi q oxi uðqÞ Here, ! o o ; . . .; k ox1 p ox p is the coordinate basis of Tp S, and v 2 Tp S, so there exist real numbers v1 ; . . .; vk such that v¼v

1

! ! ! o o o k i þ þ v ¼v : ox1 p oxk p oxi p

Since ðU; uÞ is an admissible coordinate chart of S, φ is a mapping from U onto uðUÞ, where uðUÞ is an open neighborhood of uðpÞð¼ 0Þ in Rk . Since t 7! ðtv1 ; . . .; tvk Þ is a continuous map from R to Rk , and uðUÞ is an open neighborhood of uðpÞð¼ ð0; . . .; 0Þ ¼ ð0v1 ; . . .; 0vk ÞÞ in Rk , there exists e [ 0 such that

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423

for every t 2 ð0 e; 0 þ eÞ, we have ðtv1 ; . . .; tvk Þ 2 uðUÞ, and hence, for every t 2 ðe; eÞ, we have u1 ðtv1 ; . . .; tvk Þ 2 U. Now, let us deﬁne c : ðe; eÞ ! Uð S MÞ as follows: For every t 2 ðe; eÞ; cðtÞ u1 ðtv1 ; . . .; tvk Þ. Here, ðU; uÞ is an admissible coordinate chart of S, so U is open in S. Here, cð0Þ ¼ u1 ð0v1 ; . . .; 0vk Þ ¼ u1 ð0Þ ¼ u1 ðuðpÞÞ ¼ p. Thus, cð0Þ ¼ p. We shall try to show that γ as a function from ðe; eÞ to smooth manifold S is smooth. Since ðU; uÞ is an admissible coordinate chart of S, φ is a diffeomorphism from U onto uðUÞ, and hence, u1 is smooth. Since u1 is smooth, and t 7! ðtv1 ; . . .; tvk Þ is smooth, their composite t 7! u1 ðtv1 ; . . .; tvk Þð¼ cðtÞÞ is smooth, and hence, γ as a mapping from ðe; eÞ to S is smooth. Now, we want to show that γ is smooth as a map from smooth manifold ðe; eÞ to smooth manifold M. For this purpose, let us take any t0 2 ðe; eÞ. Here, cðt0 Þ ¼ u1 ðt0 v1 ; . . .; t0 vk Þ 2 U. Since S is an embedded submanifold of M, there exists a smooth structure A on S such that the inclusion map i : S ! M is a smooth embedding (i.e., i : S ! M is smooth, and smooth immersion.). Let ðW; wÞ be any admissible coordinate chart of M such that cðt0 Þ 2 W. We have to show that w c ðIdðe;eÞ Þ1 is smooth. Since i : S ! M is smooth, ðU; uÞ is an admissible coordinate chart of S satisfying cðt0 Þ 2 U, and ðW; wÞ is any admissible coordinate chart of M such that ðiðcðt0 ÞÞ ¼Þcðt0 Þ 2 W, w u1 is smooth. Further, since t 7! ðtv1 ; . . .; tvk Þ is smooth, t 7! ðw u1 Þðtv1 ; . . .; tvk Þ is smooth. For every t 2 ðe; eÞ,

1 1 w c Idðe;eÞ ðtÞ ¼ ðw cÞ Idðe;eÞ ðtÞ ¼ ðw cÞðtÞ ¼ wðcðtÞÞ ¼ w u1 tv1 ; . . .; tvk ¼ w u1 tv1 ; . . .; tvk ;

and t 7! ðw u1 Þðtv1 ; . . .; tvk Þ is smooth, so w c ðIdðe;eÞ Þ1 is smooth. For every t 2 ðe; eÞ, we have ðpi ðu cÞÞðtÞ ¼ pi ðuðcðtÞÞÞ ¼ pi u u1 tv1 ; . . .; tvm ¼ pi tv1 ; . . .; tvm ¼ tvi ¼ vi t; so d ð pi ð u c Þ Þ ð0Þ ¼ vi : dt Now, we shall try to show that c0 ð0Þ ¼ v: Here, ! ! ! d ð p ð u c Þ Þ o o o i ¼ v: ð0Þ ¼ vi ¼ vi c0 ð0Þ ¼ dt oxi cð0Þ oxi cð0Þ oxi p Hence, c0 ð0Þ ¼ v.

h

424

5 Immersions, Submersions, and Embeddings

Lemma 5.119 Let M be an m-dimensional smooth manifold. Let S be a nonempty subset of M. Let S be an embedded submanifold of M. Let p 2 S. Let v 2 Tp S. Let f 2 C 1 ðMÞ satisfying f jS ¼ 0. Then, vðf Þ ¼ 0. Proof Since f 2 C 1 ðMÞ, f jS 2 C 1 ðSÞ. (Reason: Since f 2 C 1 ðMÞ, f : M ! R is smooth. Here, S is an embedded submanifold of M, so the inclusion map i : S ! M is smooth. Since i : S ! M is smooth, and f : M ! R is smooth, their composite ð f jS ¼Þf i : S ! R is smooth, and hence, f jS 2 C 1 ðSÞ.) Since, by Note 5.116, we do not distinguish between ðdip ÞðTp SÞ and Tp S, and v 2 Tp S, v 2 ðdip ÞðTp SÞ, and hence, there exists w 2 Tp S such that ðdip ÞðwÞ ¼ v. Now, vðf Þ ¼ ððdip ÞðwÞÞðf Þ ¼ wðf iÞ ¼ wð f jS Þ ¼ wð0Þ ¼ 0. h Lemma 5.120 Let M be an m-dimensional smooth manifold. Let S be a nonempty subset of M. Let S be an embedded k-dimensional submanifold of M. Then, S satisﬁes the local k-slice condition. Proof Let us take any p in S. Since S is an embedded submanifold of M, there exists a smooth structure A on S such that the inclusion map i : S ! M is a smooth embedding (i.e., i : S ! M is smooth, and smooth immersion). Since i : S ! M is a smooth immersion, and p 2 S, by Theorem 4.13, there exist admissible coordinate chart ðU; uÞ in S satisfying p 2 U and admissible coordinate chart ðV; wÞ in M satisfying ðp ¼ÞiðpÞ 2 V such that uðpÞ ¼ 0; ðwðpÞ ¼ÞwðiðpÞÞ ¼ 0, and for every ðx1 ; . . .; xk Þ in uðUÞ,

wu

1

0 1 ðx1 ; . . .; xk Þ ¼ w i u1 ðx1 ; . . .; xk Þ ¼ @x1 ; . . .; xk ; 0; . . .; 0A: |ﬄﬄﬄ{zﬄﬄﬄ}

mk

Further, we may assume uðUÞ ¼ Cek ð0Þ, and wðVÞ ¼ Cem ð0Þ for some e [ 0. Since ðU; uÞ is an admissible coordinate chart in S satisfying p 2 U, U is an open neighborhood of p in S, and hence, there exists an open neighborhood W of p in M such that W \ S ¼ U. Since W is an open neighborhood of p in M, and V is an open neighborhood of p in M, their intersection V \ W is an open neighborhood of p in M. Since V \ Wð VÞ is an open neighborhood of p in M, and ðV; wÞ is an admissible coordinate chart in M satisfying p 2 V, ðV \ W; wjV\W Þ is an admissible coordinate chart in M satisfying p 2 V \ W. It sufﬁces to show that 10 1 08 0 1 < @ @x1 ; . . .; xk ; 0; . . .; 0A:@x1 ; . . .; xk ; 0; . . .; 0A ðV \ W Þ \ S ¼ wjV \ W |ﬄﬄﬄ{zﬄﬄﬄ} |ﬄﬄﬄ{zﬄﬄﬄ} : mk mk 2 wjV \ W ðV \ W Þ ;

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Tangent Space to a Submanifold

425

that is,

80 10 1 < wjV \ W ððV \ W Þ \ SÞ ¼ @x1 ; . . .; xk ; 0; . . .; 0A:@x1 ; . . .; xk ; 0; . . .; 0 A |ﬄﬄﬄ{zﬄﬄﬄ} |ﬄﬄﬄ{zﬄﬄﬄ} : mk mk 2 wjV \ W ðV \ W Þ ;

that is, 80 9 1 0 1 < = wððV \ W Þ \ SÞ ¼ @x1 ; . . .; xk ; 0; . . .; 0A : @x1 ; . . .; xk ; 0; . . .; 0A 2 wðV \ W Þ ; |ﬄﬄﬄ{zﬄﬄﬄ} |ﬄﬄﬄ{zﬄﬄﬄ} : ; mk

mk

that is, 80 9 1 0 1 < = wðU \ V Þ ¼ @x1 ; . . .; xk ; 0; . . .; 0 A : @x1 ; . . .; xk ; 0; . . .; 0 A 2 wðV \ W Þ : |ﬄﬄﬄ{zﬄﬄﬄ} |ﬄﬄﬄ{zﬄﬄﬄ} : ; mk

mk

Let wðqÞ 2 LHS, where q 2 U \ V. Since q 2 U \ V, q 2 U, and hence, uðqÞ 2 uðUÞ. Put uðqÞ ðy1 ; . . .; yk Þ. Thus, ðy1 ; . . .; yk Þ 2 uðUÞ, and hence, 0 1 1 wðqÞ ¼ w u ðy1 ; . . .; yk Þ ¼ w u1 ðy1 ; . . .; yk Þ ¼ @y1 ; . . .; yk ; 0; . . .; 0A: |ﬄﬄﬄ{zﬄﬄﬄ} mk

Since q 2 U \ V, q 2 V. Since q 2 U, and W \ S ¼ U, q 2 W. Since q 2 W, and q 2 V, q 2 V \ W, and hence, 0

1

@y1 ; . . .; yk ; 0; . . .; 0A ¼ wðqÞ 2 wðV \ W Þ: |ﬄﬄﬄ{zﬄﬄﬄ} mk

Thus, 80 1 < LHS ¼ wðqÞ ¼ @y1 ; . . .; yk ; 0; . . .; 0 A 2 @x1 ; . . .; xk ; 0; . . .; 0A : |ﬄﬄﬄ{zﬄﬄﬄ} |ﬄﬄﬄ{zﬄﬄﬄ} : mk mk 9 0 1 = @x1 ; . . .; xk ; 0; . . .; 0A 2 wðV \ W Þ ¼ RHS: |ﬄﬄﬄ{zﬄﬄﬄ} ; 0

1

mk

426

5 Immersions, Submersions, and Embeddings

Thus, LHS RHS. Next, let 0

1

LHS ¼ wðqÞ @y1 ; . . .; yk ; 0; . . .; 0A 2 RHS; |ﬄﬄﬄ{zﬄﬄﬄ} mk

where q 2 V \ W. We have to show that wðqÞ 2 wðU \ VÞ. It sufﬁces to show that q 2 U. Here, q 2 V \ W, so 0

1

@y1 ; . . .; yk ; 0; . . .; 0 A ¼ wðqÞ 2 wðVÞ ¼ C m ð0Þ e |ﬄﬄﬄ{zﬄﬄﬄ} mk

¼ ðe; eÞ ðe; eÞ ðe; eÞ ðe; eÞ ; |ﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄ{zﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄ} |ﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄ{zﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄ} k

mk

and hence, ðy1 ; . . .; yk Þ 2 ðe; eÞ ðe; eÞ ¼ Cek ð0Þ ¼ uðUÞ: |ﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄ{zﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄ} k

Since ðy1 ; . . .; yk Þ 2 uðUÞ, u1 ðy1 ; . . .; yk Þ 2 U, and 0 1 1 w u ðy1 ; . . .; yk Þ ¼ w u1 ðy1 ; . . .; yk Þ ¼ @y1 ; . . .; yk ; 0; . . .; 0 A ¼ wðqÞ: |ﬄﬄﬄ{zﬄﬄﬄ} mk

Now, since ψ is 1–1, q ¼ u1 ðy1 ; . . .; yk Þ 2 U.

h

Lemma 5.121 Let M be an m-dimensional smooth manifold. Let S be a nonempty subset of M. Let S be an embedded k-dimensional submanifold of M. There exists a unique smooth structure over S with respect to which S is an embedded k-dimensional submanifold of M. Proof Since S is an embedded submanifold of M, the topology over S is the subspace topology inherited from M, and there exists a smooth structure A on S with respect to which S becomes a k-dimensional smooth manifold such that the inclusion map i : S ! M is a smooth embedding. By Lemma 5.120, S satisﬁes the local k-slice condition. Since S satisﬁes the local k-slice condition, by Note 5.114, the smooth structure A is unique. h Lemma 5.122 Let M be an m-dimensional smooth manifold. Let S be a nonempty subset of M. Let S be an embedded k-dimensional submanifold of M. Let p 2 S. Let v 2 Tp M. Let ðf 2 C1 ðMÞ satisfying f jS ¼ 0Þ implies vðf Þ ¼ 0. Then, v 2 Tp S. Proof By Lemma 5.121, there exists a unique smooth structure A over S with respect to which S is an embedded k-dimensional submanifold of M. Thus, the

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427

topology of S is the subspace topology of M, and ðS; AÞ is a k-dimensional smooth manifold. Since S is an embedded k-dimensional submanifold of M, by Lemma 5.120, S satisﬁes the local k-slice condition. Since S satisﬁes the local k-slice condition, and p 2 S, there exists an admissible coordinate chart ðV; wÞ of M such that p 2 V, and V \ Sð VÞ is equal to a k-slice of V. For deﬁniteness, let V \ S ¼ w1 ðfðx1 ; . . .; xk ; xkþ1 ; . . .; xm Þ 2 wðVÞ : xkþ1 ¼ 0; . . .; xm ¼ 0gÞ; that is, wðV \ SÞ ¼ fðx1 ; . . .; xk ; xkþ1 ; . . .; xm Þ 2 wðVÞ : xkþ1 ¼ 0; . . .; xm ¼ 0g; that is, 80 9 1 0 1 < = wðV \ SÞ ¼ @x1 ; . . .; xk ; 0; . . .; 0A : @x1 ; . . .; xk ; 0; . . .; 0A 2 wðVÞ : |ﬄﬄﬄ{zﬄﬄﬄ} |ﬄﬄﬄ{zﬄﬄﬄ} : ; mk

mk

Since ðV; wÞ is an admissible coordinate chart of M such that p 2 V, for every q 2 V, ! o o o o ; . . .; k ; kþ1 ; . . . m oy1 q oy q oy oy q q is the coordinate basis of Tq M corresponding to ðV; wÞ, where ! o 1 o d w wðqÞ : oyi q oyi wðqÞ By Note 5.96, there exists a smooth structure B on S such that S becomes a kdimensional embedded submanifold of M, and ! o o ; . . .; k oy1 q oy q is a basis of ðdiq ÞðTq SÞ, where i : S ! M denotes the inclusion map of S. Now, since A is unique, A ¼ B. Since p 2 V, ! o o o o ; . . .; k ; kþ1 ; . . . m oy1 p oy p oy oy p p

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5 Immersions, Submersions, and Embeddings

is a basis of the real linear space Tp M. Now, since v 2 Tp M, there exist real numbers v1 ; . . .; vk ; vkþ1 ; . . .; vm such that ! ! ! ! o o o o 1 k kþ1 m v¼v þ þ v þv þ þ v : oy1 p oyk p oykþ1 p oym p We shall try to show that vkþ1 ¼ 0. Here, w : V ! wðVÞð Rm Þ is smooth, so the ðk þ 1Þth projection map pkþ1 w : V ! R of ψ is smooth. Here, V is an open neighborhood of p in M, and M is a smooth manifold, there exists an open neighborhood W of p such that p 2 W W V. By Lemma 4.58, there exists a smooth function v : M ! ½0; 1 such that 1. for every x in W ; vðxÞ ¼ 1, 2. supp v V. If x 2 ðM ðsupp vÞÞ \ V, then x 2 ðM ðsupp vÞÞ, and hence, by the deﬁnition of support, vðxÞ ¼ 0. It follows that if x 2 ðM ðsupp vÞÞ \ V, then ðvðxÞÞððpkþ1 wÞðxÞÞ ¼ ð0Þððpkþ1 wÞðxÞÞ ¼ 0. This shows that the following function f : M ! R deﬁned as follows: For every x 2 M, f ðxÞ

ðvðxÞÞððpkþ1 wÞðxÞÞ if 0 if

x2V M ðsupp vÞ;

is well deﬁned. Since V is open, x 7! vðxÞ is smooth, and x 7! ðpkþ1 wÞðxÞ is smooth, their product x 7! ðvðxÞÞððpkþ1 wÞðxÞÞ is smooth on V. Also, the constant function x 7! 0 is smooth on the open set M ðsupp vÞ. Now, from the deﬁnition of f : M ! R; f is smooth, that is, f 2 C 1 ðMÞ. From the deﬁnition of f : M ! R, we can write: For every x 2 M, ðvðxÞÞððpkþ1 wÞðxÞÞ if x 2 V f ðxÞ 0 if x 62 V: Clearly, f jS ¼ 0. (Reason: Let us take any x 2 S. We have to show that f ðxÞ ¼ 0. For this purpose, it sufﬁces to show that ðpkþ1 wÞðxÞ ¼ 0 whenever x 2 V. For this purpose, let x 2 V. Since x 2 V, and x 2 S, x 2 V \ S. Since x 2 V \ S, 80 9 10 1 < = wðxÞ 2 wðV \ SÞ ¼ @x1 ; . . .; xk ; 0; . . .; 0A:@x1 ; . . .; xk ; 0; . . .; 0A 2 wðVÞ ; |ﬄﬄﬄ{zﬄﬄﬄ} |ﬄﬄﬄ{zﬄﬄﬄ} : ; mk

mk

and hence, there exist real numbers x1 ; . . .; xk , such that 0

1

wðxÞ ¼ @x1 ; . . .; xk ; 0; . . .; 0 A 2 wðVÞ: |ﬄﬄﬄ{zﬄﬄﬄ} mk

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It follows that 0

1

ðpkþ1 wÞðxÞ ¼ pkþ1 ðwðxÞÞ ¼ pkþ1 @x1 ; . . .; xk ; 0; . . .; 0A ¼ 0: |ﬄﬄﬄ{zﬄﬄﬄ} mk

Thus, ðpkþ1 wÞðxÞ ¼ 0.) Since f 2 C1 ðMÞ, and f jS ¼ 0, by the assumption, vðf Þ ¼ 0. Since W V, for every x 2 W, we have x 2 V, and vðxÞ ¼ 1. Hence, for every x 2 W, f ðxÞ ¼ ðvðxÞÞððpkþ1 wÞðxÞÞ ¼ ð1Þððpkþ1 wÞðxÞÞ ¼ ðpkþ1 wÞðxÞ. Thus, f jW ¼ ðpkþ1 wÞjW . Now, since W is an open neighborhood of p, 0 ¼ 0ðpÞ ¼ ðvð f ÞÞðpÞ ! ! ! o o o 1 k kþ1 ¼ v þ þ v þv oy1 p oyk p oykþ1 p !! ! ! ! o o 1 ðfÞ þ þ vm ð f Þ ðpÞ ¼ v oym p oy1 p ! ! ! ! o o k kþ1 þ þ v ðfÞ þ v ðfÞ oyk p oykþ1 p ! ! !!! ! o o m 1 þ þ v ðfÞ ðpÞ ¼ v f jW oym p oy1 p ! ! ! ! o o k kþ1 þ þ v f jW f jW þv oyk p oykþ1 p !!! ! ! ! o o m 1 þ þ v ðpkþ1 wÞjW f jW ðpÞ ¼ v oym p oy1 p ! ! ! ! o o k kþ1 þ þ v ðpkþ1 wÞjW ðpkþ1 wÞjW þv oyk p oykþ1 p !!! ! o m þ þ v ðpkþ1 wÞjW ðpÞ: oym p

430

5 Immersions, Submersions, and Embeddings

Now, since ! o ðpkþ1 wÞjW ¼ i oy p ¼ ¼

!! o ðpkþ1 wÞjW i oy wðpÞ ! ! o o ðpkþ1 wÞjW w1 ¼ ðpkþ1 Þ oyi wðpÞ oyi wðpÞ

d w1 wðpÞ

0

if

i 6¼ k þ 1

1

if

i ¼ k þ 1;

0 ¼ v1 ð0Þ þ þ vk ð0Þ þ vkþ1 ð1Þ þ þ vm ð0Þ ðpÞ ¼ vkþ1 ð1ðpÞÞ ¼ vkþ1 1 ¼ vkþ1 : Thus, vkþ1 ¼ 0. Similarly, vkþ2 ¼ 0; ; vm ¼ 0. Hence, v¼v

1

¼ v1

! ! ! ! o o o o k þ þ v þ0 þ þ 0 oy1 p oyk p oykþ1 p oym p ! ! ! o o o o k ; . . .; k ¼ dip Tp S : þ þ v 2 span 1 k 1 oy p oy p oy p oy p

Thus, v 2 ðdip ÞðTp SÞ. Now, since we do not distinguish between ðdip ÞðTp SÞ and Tp S, we can write v 2 Tp S. h

Chapter 6

Sard’s Theorem

The root of Sard’s theorem lies in real analysis. It is one of the deep results in real analysis. How this result can be generalized into the realm of smooth manifold theory is only a later development. Its importance in manifold theory is for a deﬁnite reason. Upon applying Sard’s theorem, Whitney was able to prove a startling property about smooth manifolds: For every smooth manifold, ambient space can be constructed, etc. For pedagogical reasons, we have given its proof in step-by-step manner using Taylor’s inequality. Largely, this chapter is selfcontained.

6.1 Measure Zero in Manifolds Lemma 6.1 Let A be any nonempty compact subset of R4 : For every k ¼ 1; 2; 3; 4; let lk denote the k-dimensional (Lebesgue) measure over Rk . If for every real c; l3 ðfðy; z; wÞ : ðc; y; z; wÞ 2 AgÞ ¼ 0, then l4 ðAÞ ¼ 0. Proof Since A is compact in R4 , by the Henie-Borel theorem, A is bounded and closed. Since A is bounded, there exist real numbers a; b; a2 ; b2 ; a3 ; b3 ; a4 ; b4 such that a\b; a2 \b2 ; a3 \b3 ; a4 \b4 , and A ½a; b ½a2 ; b2 ½a3 ; b3 ½a4 ; b4 ½a; b R R R ¼ ½a; b R3 : Thus, A ½a; b R3 . d1 Let us ﬁx any d1 [ 0: Put d 2ðbaÞ ð [ 0Þ: It sufﬁces to ﬁnd a ﬁnite collection of four-dimensional rectangles that covers A; and sum of their l4 -measures is less than d1 : Next let us ﬁx any real c 2 ½a; b: Clearly, fðy; z; wÞ : ðc; y; z; wÞ 2 Ag is a compact subset of R3 . (Reason: Let us deﬁne a function f : R4 ! R3 as follows: For every ðx; y; z; wÞ in R4 ; f ðx; y; z; wÞ ðy; z; wÞ: Clearly, f : R4 ! R3 is continuous. It is clear that fcg R3 is a closed subset of R4 : Since fcg R3 is a closed subset of R4 ; and A is a closed subset of R4 ; their intersection A \ ðfcg R3 Þ is a closed subset of the compact set A; and hence, A \ ðfcg R3 Þ is compact. Since A \ ðfcg R3 Þ is © Springer India 2014 R. Sinha, Smooth Manifolds, DOI 10.1007/978-81-322-2104-3_6

431

432

6 Sard’s Theorem

compact, and f : R4 ! R3 is continuous, the f -image set f ðA \ ðfcg R3 ÞÞ ð¼ ff ðc; y; z; wÞ : ðc; y; z; wÞ 2 Ag ¼ fðy; z; wÞ : ðc; y; z; wÞ 2 AgÞ is compact in R3 ; and hence, fðy; z; wÞ : ðc; y; z; wÞ 2 Ag is compact in R3 ). Since the three-dimensional measure l3 fðy; z; wÞ : ðc; y; z; wÞ 2 Ag ¼ 0, and d [ 0; by the deﬁnition of measure, there exists a countable collection fC1 ; C2 ; C3 ; . . .g of open cubes in R3 such that fðy; z; wÞ : ðc; y; z; wÞ 2 Ag C1 [ C2 [ C3 [ ; and l3 ðC1 Þ þ l3 ðC2 Þ þ l3 ðC3 Þ þ \d: Since fðy; z; wÞ : ðc; y; z; wÞ 2 Ag is a compact subset of R3 ; and fC1 ; C2 ; C3 ; . . .g is an open cover of fðy; z; wÞ : ðc; y; z; wÞ 2 Ag; there exists a positive integer kc such that fðy; z; wÞ : ðc; y; z; wÞ 2 Ag C1 [ C2 [ [ Ckc ; and l3 ðC1 Þ þ l3 ðC2 Þ þ þ l3 ðCkc Þ l3 ðC1 Þ þ l3 ðC2 Þ þ l3 ðC3 Þ þ \d: We claim that there exists a positive integer n such that A \ ððc 1n ; c þ 1nÞ R3 Þ ðc 1n ; c þ 1nÞ ðC1 [ C2 [ [ Ckc Þ: If not, otherwise, let for every positive integer n; A \ ððc 1n ; c þ 1nÞ R3 Þ 6 ðc 1n ; c þ 1nÞ ðC1 [ C2 [ [ Ckc Þ: We have to arrive at a contradiction. Thus, for every positive integer n; there exists ðxn ; yn ; zn ; wn Þ 2 A \ ððc 1n ; c þ 1nÞ R3 Þð AÞ and ðxn ; yn ; zn ; wn Þ 62 ðc 1n ; c þ 1nÞ ðC1 [ C2 [ [ Ckc Þ: Since fðx1 ; y1 ; z1 ; w1 Þ; ðx2 ; y2 ; z2 ; w2 Þ; ðx3 ; y3 ; z3 ; w3 Þ; . . .g is a sequence in the compact set A; there exists a subsequence fðxn1 ; yn1 ; zn1 wn1 Þ; ðxn2 ; yn2 ; zn2 ; wn2 Þ; ðxn3 ; yn3 ; zn3 ; wn3 Þ; . . .g of fðx1 ; y1 ; z1 ; w1 Þ; ðx2 ; y2 ; z2 ; w2 Þ; ðx3 ; y3 ; z3 ; w3 Þ; . . .g such that fðxn1 ; yn1 ; zn1 wn1 Þ; ðxn2 ; yn2 ; zn2 ; wn2 Þ; ðxn3 ; yn3 ; zn3 ; wn3 Þ; . . .g is convergent, and hence, there exist real numbers a1 ; a2 ; a3 ; a4 such that limi!1 ðxni ; yni ; zni wni Þ ¼ ða1 ; a2 ; a3 ; a4 Þ: It follows that limi!1 xni ¼ a1 ; limi!1 yni ¼ a2 , limi!1 zni ¼ a3 , limi!1 wni ¼ a4 . Since each ðxn ; yn ; zn ; wn Þ 2 A \ ððc 1n ; c þ 1nÞ R3 Þ; each xn 2 ðc 1n ; c þ 1nÞ; and hence, limn!1 xn ¼ c: Since limn!1 xn ¼ c; a1 ¼ limi!1 xni ¼ c: Since a1 ¼ c; and limi!1 ðxni ; yni ; zni ; wni Þ ¼ ða1 ; a2 ; a3 ; a4 Þ; limi!1 ðxni ; yni ; zni ; wni Þ ¼ 1 n

1 n

ðc; a2 ; a3 ; a4 Þ: Since ðxn ; yn ; zn ; wn Þ 62 ðc ; c þ Þ ðC1 [ C2 [ [ Ckc Þ; and xn 2 ðc 1n ; c þ 1nÞ; ðyn ; zn ; wn Þ 62 ðC1 [ C2 [ [ Ckc Þ: Since limi!1 ðxni ; yni ; zni wni Þ ¼ ðc; a2 ; a3 ; a4 Þ; limi!1 ðyni ; zni wni Þ ¼ ða2 ; a3 ; a4 Þ: Since each ðyni ; zni wni Þ is in the closed set ðC1 [ C2 [ [ Ckc Þc ; ða2 ; a3 ; a4 Þ ¼ ðlimi!1 ðyni ; zni wni ÞÞ 2 ðC1 [ C2 [ [ Ckc Þc ; and hence, ða2 ; a3 ; a4 Þ 62 ðC1 [ C2 [ [ Ckc Þð fðy; z; wÞ : ðc; y; z; wÞ 2 AgÞ: It follows that ða2 ; a3 ; a4 Þ 62 fðy; z; wÞ : ðc; y; z; wÞ 2 Ag; and hence, ðc; a2 ; a3 ; a4 Þ 62 A. Since limi!1 ðxni ; yni ; zni wni Þ ¼ ðc; a2 ; a3 ; a4 Þ; and each ðxni ; yni ; zni wni Þ is in the closed set A; ðc; a2 ; a3 ; a4 Þ ¼ ðlimi!1 ðxni ; yni ; zni wni ÞÞ 2 A; which is a contradiction. So our claim is true. Thus, for every c 2 ½a; b; there exist a positive integer nc and an open subset Gc ð C1 [ C2 [ [ Ckc Þ of R3 such that A \ ððc n1c ; c þ n1c Þ R3 Þ ðc n1c ; c þ n1c Þ Gc : If we shrink ðc n1c ; c þ n1c Þ into another open interval J; that is, J ðc n1c ; c þ n1c Þ; then clearly A \ ðJ R3 Þ J Gc . (Reason: Let

ðx; y; z; wÞ 2 LHS ¼ A \ ðJ R3 Þ. It follows that x 2 J and ðx; y; z; wÞ 2 A: Since x 2 J ðc n1c ; c þ n1c Þ; x 2 ðc n1c ; c þ n1c Þ: Since x 2 ðc n1c ; c þ n1c Þ; and

6.1 Measure Zero in Manifolds

433

ðx; y; z; wÞ 2 A; ðx; y; z; wÞ 2 A \ ððc n1c ; c þ n1c Þ R3 Þ ðc n1c ; c þ n1c Þ Gc ; and hence, ðy; z; wÞ 2 Gc : Since x 2 J; and ðy; z; wÞ 2 Gc ; ðx; y; z; wÞ 2 J Gc ¼ RHS: Thus, LHS RHS). Here, l3 ðGc1 Þ ¼ l3 ðC1 [ [ Ckc1 Þ l3 ðC1 Þ þ þ l3 ðCkc1 Þ\d: Thus, l3 ðGc1 Þ\d: Similarly, l3 ðGc1 Þ\d; etc. Since ½a; b is compact, and fðc n1c ; c þ n1c Þ : c 2 ½a; bg is an open cover of ½a; b; there exist ﬁnite many c1 ; . . .; cl in ½a; b such that ½a; b ðc1 n1c ; c1 þ n1c Þ [ [ ðcl n1c ; cl þ n1c Þ: From the above observation, there 1

exist

1

open intervals

l

J1 ; . . .; Jl

l

such

that

J1 ðc1 n1c ; c1 þ n1c Þ; . . .; Jl 1

1

ðcl n1c ; cl þ n1c Þ; ½a; b J1 [ [ Jl ; and l1 ðJ1 Þ þ þ l1 ðJl Þ 2ðb aÞ: l

l

Further, since A ½a; b R3 , so A ¼ A \ ½a; b R3 A \ ðJ1 [ [ Jl Þ R3 ¼ A \ J1 R3 [ [ Jl R3 ¼ A \ J1 R3 [ [ A \ Jl R3 ðJ1 Gc1 Þ [ [ ðJl Gcl Þ:

Further, since J1 Gc1 ¼ J1 ðC1 [ [ Ckc1 Þ ¼ ðJ1 C1 Þ [ [ ðJ1 Ckc1 Þ and J1 C1 ; . . .; J1 Ckc1 are rectangles in R4 ; J1 Gc1 is a ﬁnite union of rectangles. Similarly, J1 Gc2 is a ﬁnite union of rectangles, etc. Now, since A ðJ1 Gc1 Þ [ [ðJl Gcl Þ; A is covered by a ﬁnite collection of rectangles in R4 : Here, l4 ðJ1 Gc1 Þ þ þ l4 ðJl Gcl Þ ¼ l1 ðJ1 Þ l3 ðGc1 Þ þ þ l1 ðJl Þ l3 ðGcl Þ \l1 ðJ1 Þ d þ þ l1 ðJl Þ d ¼ ðl1 ðJ1 Þ þ þ l1 ðJl ÞÞd 2ðb aÞ d ¼ 2ðb aÞ d1 h 2ðbaÞ ¼ d1 : Note 6.2 The Lemma 6.1 can be generalized as follows: Let A be any nonempty compact subset of Rn : For every k ¼ 1; . . .; n; let lk denote the k-dimensional measure over Rk : If for every real c; ln1 ðfðx2 ; . . .; xn Þ : ðc; x2 ; . . .; xn Þ 2 AgÞ ¼ 0, then ln ðAÞ ¼ 0: Lemma 6.3 Let A be a nonempty compact subset of R: Let f : A ! R be continuous. For every k ¼ 1; 2; let lk denote the k-dimensional measure over Rk : Then, l2 ðfðx; f ðxÞÞ : x 2 AgÞ ¼ 0: In short, the graph of a continuous function has measure 0 in R2 : Proof A is a compact subset of R1 ð¼ RÞ: We have to prove that l2 ðfðx; f ðxÞÞ : x 2 AgÞ ¼ 0: We want to apply the result of 6.2. Clearly, fðx; f ðxÞÞ : x 2 Ag is a compact subset of R2 : (Reason: First of all, we shall try to show that fðx; f ðxÞÞ : x 2 Ag is a closed subset of R2 : For this purpose, let us take a convergent sequence fðxn ; f ðxn ÞÞg in fðx; f ðxÞÞ : x 2 Ag: Let limn!1 ðxn ; f ðxn ÞÞ ¼ ða; bÞ: We have to show that ða; bÞ 2 fðx; f ðxÞÞ : x 2 Ag; that is, a 2 A and f ðaÞ ¼ b: Since limn!1 ðxn ; f ðxn ÞÞ ¼ ða; bÞ; limn!1 xn ¼ a; and limn!1 f ðxn Þ ¼ b: Since fðxn ; f ðxn ÞÞg is in fðx; f ðxÞÞ : x 2 Ag; for each positive integer n; xn 2 A: Since A is compact, A is closed. Since A is

434

6 Sard’s Theorem

closed, for each positive integer n; xn 2 A; and limn!1 xn ¼ a; a ¼ ðlimn!1 xn Þ 2 A; and hence, a 2 A: Since fxn g is a convergent sequence in A; limn!1 xn ¼ a 2 A; and f : A ! R is continuous, b ¼ limn!1 f ðxn Þ ¼ f ðaÞ: Thus, we have shown that fðx; f ðxÞÞ : x 2 Ag is a closed subset of R2 : Clearly, fðx; f ðxÞÞ : x 2 Ag A f ðAÞ: Since A is compact, and f : A ! R is continuous, f ðAÞ is compact. Since A is compact, and f ðAÞ is compact, their Cartesian product A f ðAÞ is compact. Thus, fðx; f ðxÞÞ : x 2 Ag is a closed subset of the compact set A f ðAÞ; and hence, fðx; f ðxÞÞ : x 2 Ag is compact.) For every real c; fx2 : ðc; x2 Þ 2 fðx; f ðxÞÞ : x 2 Agg ¼ ; or ff ðcÞg; so l1 ðfx2 : ðc; x2 Þ 2 fðx; f ðxÞÞ : x 2 AggÞ ¼ l1 ð;Þð¼ 0Þ or l1 ðff ðcÞgÞð¼ 0Þ; and hence, l1 ðfx2 : ðc; x2 Þ 2 fðx; f ðxÞÞ : x 2 AggÞ ¼ 0. Now, we can apply Lemma 6.1. It follows that l2 ðfðx; f ðxÞÞ : x 2 AgÞ ¼ 0: h Lemma 6.4 Let A be a nonempty compact subset of R2 : Let f : A ! R be continuous. For every k ¼ 1; 2; 3; let lk denote the k-dimensional measure over Rk : Then, l3 ðfðx; y; f ðx; yÞÞ : ðx; yÞ 2 AgÞ ¼ 0: Proof We have to prove that l3 ðfðx; y; f ðx; yÞÞ : ðx; yÞ 2 AgÞ ¼ 0 (see Fig. 6.1). Clearly, fðx; y; f ðx; yÞÞ : ðx; yÞ 2 Ag is a compact subset of R3 : (Reason: First of all, we shall try to show that fðx; y; f ðx; yÞÞ : ðx; yÞ 2 Ag is a closed subset of R3 : For this purpose, let us take any convergent sequence fðxn ; yn ; f ðxn ; yn ÞÞg in fðx; y; f ðx; yÞÞ : ðx; yÞ 2 Ag: Let limn!1 ðxn ; yn ; f ðxn ; yn ÞÞ ¼ ða; b; cÞ: We have to show that ða; b; cÞ 2 fðx; y; f ðx; yÞÞ : ðx; yÞ 2 Ag; that is, ða; bÞ 2 A: Since limn!1 ðxn ; yn ; f ðxn ; yn ÞÞ ¼ ða; b; cÞ; limn!1 xn ¼ a; limn!1 yn ¼ b; and limn!1 f ðxn ; yn Þ ¼ c: Here, fðxn ; yn ; f ðxn ; yn ÞÞg is a sequence in fðx; y; f ðx; yÞÞ : ðx; yÞ 2 Ag; so each ðxn ; yn ; f ðxn ; yn ÞÞ 2 fðx; y; f ðx; yÞÞ : ðx; yÞ 2 Ag; and hence, each ðxn ; yn Þ 2 A: Since A is compact, A is closed. Since limn!1 xn ¼ a; and limn!1 yn ¼ b; limn!1 ðxn ; yn Þ ¼ ða; bÞ: Since each ðxn ; yn Þ 2 A; limn!1 ðxn ; yn Þ ¼ ða; bÞ; and A is closed, ða; bÞ ¼ ðlimn!1 ðxn ; yn ÞÞ 2 A; and hence, ða; bÞ 2 A: Since each ðxn ; yn Þ 2 A; limn!1 ðxn ; yn Þ ¼ ða; bÞ 2 A; and f : A ! R is continuous, c ¼ limn!1 f ðxn ; yn Þ ¼ f ða; bÞ: Since ða; bÞ 2 A; ða; b; cÞ ¼ ða; b; f ða; bÞÞ 2 fðx; y; f ðx; yÞÞ : ðx; yÞ 2 Ag: Thus, we have shown that fðx; y; f ðx; yÞÞ : ðx; yÞ 2 Ag is closed. Here, fðx; y; f ðx; yÞÞ : ðx; yÞ 2 Ag A f ðAÞ: Since f : A ! R is continuous, and A is compact, f ðAÞ is compact. Since A is compact, and f ðAÞ is compact,

Fig. 6.1 fðx; y; f ðx; yÞÞ : ðx; yÞ 2 Ag is a compact subset of R3

6.1 Measure Zero in Manifolds

435

A f ðAÞ is compact in R3 : Since fðx; y; f ðx; yÞÞ : ðx; yÞ 2 Ag is a closed subset of the compact set A f ðAÞ; fðx; y; f ðx; yÞÞ : ðx; yÞ 2 Ag is compact.) Now, by 6.2, it sufﬁces to show that for every real c; l2 ðfðs; tÞ : ðc; s; tÞ 2 fðx; y; f ðx; yÞÞ : ðx; yÞ 2 AggÞ ¼ 0: For this purpose, let us ﬁx any real c. Since A is a nonempty compact subset of R2 ; clearly fs : ðc; sÞ 2 Ag is a compact subset of R: (See the proof of Lemma 6.1) Let us deﬁne a function fc : fs : ðc; sÞ 2 Ag ! R as follows: for every s 2 fs : ðc; sÞ 2 Ag; fc ðsÞ f ðc; sÞ: Since fc is the composite of continuous function s 7! ðc; sÞ; and continuous function f : A ! R; fc : fs : ðc; sÞ 2 Ag ! R is a continuous function. Also, the domain of fc ; that is, fs : ðc; sÞ 2 Ag is compact. It follows, by Lemma 6.3, that l2 ðfðx; fc ðxÞÞ : x 2 fs : ðc; sÞ 2 AggÞ ¼ 0: Now, it sufﬁces to show that fðs; tÞ : ðc; s; tÞ 2 fðx; y; f ðx; yÞÞ : ðx; yÞ 2 Agg ¼ fðx; fc ðxÞÞ : x 2 fs : ðc; sÞ 2 Agg: Let us take any ðg; 1Þ 2 LHS. It follows that ðc; g; 1Þ 2 fðx; y; f ðx; yÞÞ : ðx; yÞ 2 Ag; and hence, ðc; gÞ 2 A; and 1 ¼ f ðc; gÞ ¼ fc ðgÞ: Since ðc; gÞ 2 A; g 2 fs : ðc; sÞ 2 Ag; and hence, ðg; 1Þ ¼ ðg; fc ðgÞÞ 2 RHS: Thus, LHS RHS: Next, let us take any ðg; 1Þ 2 RHS: It follows that there exists a real x such that ðg; 1Þ ¼ ðx; fc ðxÞÞ: Since ðg; 1Þ ¼ ðx; fc ðxÞÞ; g ¼ x; and 1 ¼ fc ðxÞ ¼ f ðc; xÞ: Since 1 ¼ f ðc; xÞ; and f : A ! R; ðc; xÞ 2 A; and hence, ðc; g; 1Þ ¼ ðc; x; 1Þ ¼ ðc; x; f ðc; xÞÞ 2 fðx; y; f ðx; yÞÞ : ðx; yÞ 2 Ag: Since ðc; g; 1Þ 2 fðx; y; f ðx; yÞÞ : ðx; yÞ 2 Ag; ðg; 1Þ 2 LHS: Thus, RHS LHS: Since RHS LHS; and LHS RHS; LHS ¼ RHS: h Note 6.5 The Lemma 6.4 can be generalized as follows: Let n 2 f2; 3; 4; . . .g: Let A be a nonempty compact subset of Rn1 : Let f : A ! R be continuous. For every k ¼ 1; . . .; n; let lk denote the k-dimensional measure over Rk : Then, ln ðfðx; f ðxÞÞ : x 2 AgÞ ¼ 0. Lemma 6.6 Let X be a second countable, locally compact Hausdorff space. Then, there exists a sequence fKn g of compact subsets of X such that 1. fKn g is a cover of X; 2. K1 ðK2 Þo K2 ðK3 Þo K3 ðK4 Þo K4 : Proof Since X is a locally compact Hausdorff space, there exists a basis B of X such that if U 2 B then U is compact. Since B is a basis of X, B is an open cover of X. Since B is an open cover of X; and X is a second countable space, there exists a countable collection fU1 ; U2 ; U3 ; . . .g of members of B which also covers X. Since fU1 ; U2 ; U3 ; . . .g covers X; and for each positive integer n; Un ðUn Þ , fðU1 Þ ; ðU2 Þ ; ðU3 Þ ; . . .g also covers X: Put K1 ðU1 Þ : Since U1 2 B, ðU1 Þ ð¼ K1 Þ is compact, and hence, K1 is compact. Since K1 is compact, and fU1 ; U2 ; U3 ; . . .g is an open cover of K1 , there exists a positive integer n1 [ 1 such that K1 U1 [ [ Un1 . Here, ðU1 [ [ Un1 Þ ¼ ðU1 Þ [ [ ðUn1 Þ . Since ðU1 Þ ; . . .; ðUn1 Þ are compact, ðU1 Þ [ [ ðUn1 Þ ð¼ ðU1 [ [ Un1 Þ Þ is compact, and hence, ðU1 [ [ Un1 Þ is compact. Put K2 ðU1 [ [ Un1 Þ .

436

6 Sard’s Theorem

Since ðU1 [ [ Un1 Þ ð¼ K2 Þ is compact, K2 is compact. Since U1 [ [ Un1 is an open set, and ðU1 [ [ Un1 Þ ðU1 [ [ Un1 Þ ¼ K2 , K1 ðU1 [ [ Un1 Þ ðK2 Þo K2 . Thus, K1 ; K2 are compact sets satisfying K1 ðK2 Þo K2 . Since K2 is compact, and fU1 ; U2 ; U3 ; . . .g is an open cover of K2 , there exists a positive integer n2 [ n1 such that K2 U1 [ [ Un1 [ [ Un2 . Here, ðU1 [ [ Un2 Þ ¼ ðU1 Þ [ [ ðUn2 Þ . Since ðU1 Þ ; . . .; ðUn2 Þ are compact, ðU1 Þ [ [ ðUn2 Þ ð¼ ðU1 [ [ Un2 Þ Þ is compact, and hence, ðU1 [ [ Un2 Þ is compact. Put K3 ðU1 [ [ Un2 Þ . Since ðU1 [ [ Un2 Þ ð¼ K3 Þ is compact, K3 is compact. Since U1 [ [ Un2 is an open set, and ðU1 [ [ Un2 Þ ðU1 [ [Un2 Þ ¼ K3 , K2 ðU1 [ [ Un2 Þ ðK3 Þo K3 . Thus, K1 ; K2 ; K3 are compact sets satisfying K1 ðK2 Þo K2 ðK3 Þo K3 , etc. This proves 2. Since 1\n1 \n2 \n3 \ , U1 K1 , U1 [ [ Un1 K2 ; U1 [ [Un2 K3 , etc., and h fU1 ; U2 ; U3 ; . . .g covers X, fK1 ; K2 ; K3 ; . . .g is a cover of X: This proves 1. Lemma 6.7 Let n 2 f2; 3; 4; . . .g. Let A be a nonempty open subset of Rn1 . Let f : A ! R be continuous. For every k ¼ 1; . . .; n; let lk denote the k-dimensional measure over Rk . Then, ln ðfðx; f ðxÞÞ : x 2 AgÞ ¼ 0. Proof Since Rn1 is a second countable space, and A is a nonempty subset of Rn1 , A is a second countable space. Since Rn1 is locally compact Hausdorff space, and A is a nonempty open subset of Rn1 ; by Note 5.54, A is a locally compact Hausdorff. Since A is a second countable, locally compact Hausdorff space, by Lemma 6.6, there exists a sequence fKn g of compact subsets of X such that fKn g is a cover of X. Hence, by Note 6.5, ( 0 ln ðfðx; f ð xÞÞ : x 2 AgÞ ln ¼ ln ¼

1 X

1 [

fðx; f ð xÞÞ : x 2 Kn g

n¼1

ln

ðx; f ð xÞÞ : x 2 !

1 [

)! Kn

n¼1 1 X

ln ðfðx; f ð xÞÞ : x 2 Kn gÞ

n¼1

1 n o X x; f jKn ð xÞ : x 2 Kn 0 ¼ 0: ¼

n¼1

Hence, ln ðfðx; f ðxÞÞ : x 2 AgÞ ¼ 0:

n¼1

h

Lemma 6.8 Let n 2 f2; 3; 4; . . .g: Let A be a nonempty closed subset of Rn1 : Let f : A ! R be continuous. For every k ¼ 1; . . .; n; let lk denote the k-dimensional measure over Rk : Then, ln ðfðx; f ðxÞÞ : x 2 AgÞ ¼ 0: Proof Since Rn1 is a second countable space, and A is a nonempty subset of Rn1 ; A is a second countable space. Since Rn1 is locally compact Hausdorff space, and A is a nonempty closed subset of Rn1 ; by Note 5.55, A is a locally compact Hausdorff. Since A is a second countable, locally compact Hausdorff space, by Lemma 6.6, there exists a sequence fKn g of compact subsets of X such that fKn g is a cover of X: Hence, by Note 6.5,

6.1 Measure Zero in Manifolds

437

( 0 ln ðfðx; f ð xÞÞ : x 2 AgÞ ln ¼ ln ¼

1 X

1 [

ðx; f ð xÞÞ : x 2

ln

)! Kn

n¼1

! fðx; f ð xÞÞ : x 2 Kn g

n¼1

1 [

1 X

ln ðfðx; f ð xÞÞ : x 2 Kn gÞ

n¼1

1 n o X x; f jKn ð xÞ : x 2 Kn 0 ¼ 0: ¼

n¼1

n¼1

Hence, ln ðfðx; f ðxÞÞ : x 2 AgÞ ¼ 0:

h

Deﬁnition Let V be any n-dimensional real linear space. Any subset of V of the form a þ S; where S is a linear subspace of V; and a 2 V; is called an afﬁne subspace of V parallel to S and passing through a: Note 6.9 In real linear space R3 ; let a þ S be an afﬁne subspace of R3 parallel to S and passing through a: Let the dimension of the subspace S be 2. We want to show that l3 ða þ SÞ ¼ 0: Since the dimension of the subspace S is 2, there exist ðb1 ; b2 ; b3 Þ; ðc1 ; c2 ; c3 Þ 2 R3 such that S ¼ fsðb1 ; b2 ; b3 Þ þ tðc1 ; c2 ; c3 Þ : s; t 2 Rg. Let a ða1 ; a2 ; a3 Þ: Hence, a þ S ¼ ða1 ; a2 ; a3 Þ þ fsðb1 ; b2 ; b3 Þ þ tðc1 ; c2 ; c3 Þ : s; t 2 Rg ¼ fða1 ; a2 ; a3 Þ þ sðb1 ; b2 ; b3 Þ þ tðc1 ; c2 ; c3 Þ : s; t 2 Rg ¼ fðx; y; zÞ : x ¼ b1 s þ c1 t þ a1 ; y ¼ b2 s þ c2 t þ a2 ; z ¼ b3 s þ c3 t þ a3 g: If ðx; y; zÞ 2 a þ S; then there exist real numbers s; t such that 8 < b1 s þ c 1 t ¼ x a 1 b s þ c 2 t ¼ y a2 : 2 b3 s þ c 3 t ¼ z a3 : From the ﬁrst two equations, we have b1 c 1 s ¼ x a1 c1 and b2 c 2 y a2 c 2

b1 c 1 t ¼ b1 x a1 : b2 c 2 b2 y a2

Now, from the third equation, x a1 c 1 b1 x a1 b1 c 1 : þ c3 ¼ ð z a3 Þ b3 y a2 c 2 b2 y a2 b2 c 2 Clearly, this takes the form A1 x þ A2 y þ A3 z ¼ A4 ; and not all A1 ; A2 ; A3 are zero. For deﬁniteness, let A3 6¼ 0: It follows that

438

6 Sard’s Theorem

A4 A1 A2 2 aþS¼ x; y; x y : ðx; yÞ 2 R : A1 A3 A3 Now, we can apply Lemma 6.8. It follows that A4 A1 A2 2 x; y; x y : ðx; yÞ 2 R ¼ 0: l 3 ð a þ S Þ ¼ l3 A1 A3 A3

h

We can easily generalize the above result as the following: Lemma 6.10 Let n 2 f2; 3; 4; . . .g: Let A be an afﬁne subspace of Rn such that A 6¼ Rn : For every k ¼ 1; . . .; n; let lk denote the k-dimensional measure over Rk : Then, ln ðAÞ ¼ 0: Proof Its proof is similar to the proof given in the Note 6.9.

h

Deﬁnition Let M be an m-dimensional smooth manifold, and N be an n-dimensional smooth manifold. Let A be a nonempty subset of M: Let F : A ! N be any map. If for every p 2 A; there exist an open neighborhood W of p in M; and a ~ : W ! N such that for every x 2 W \ A; FðxÞ ~ ~ :W !N function F ¼ FðxÞ; and F is a smooth function, then we say that F : A ! N is smooth on A: Lemma 6.11 Let A be a nonempty subset of R3 : For every k ¼ 1; 2; 3; let lk denote the k-dimensional measure over Rk : Let l3 ðAÞ ¼ 0: Let F : A ! R3 be smooth on A: Then, l3 ðFðAÞÞ ¼ 0: Proof Let us take any p ðp1 ; p2 ; p3 Þ 2 A: Since F : A ! R3 is smooth on A; and p 2 A; there exist an open neighborhood Wp ðp1 ep ; p1 þ ep Þ ðp2 ep ; p2 þ ep Þ ðp3 ep ; p3 þ ep Þ of p in R3 ; and a function Fp : Wp ! R3 such that for every x 2 Wp \ A; Fp ðxÞ ¼ FðxÞ; and Fp : Wp ! R3 is a smooth function. Since A is a nonempty subset of R3 ; and R3 is a second countable space, A is a second countable space. Since A is a second countable space, and 1 1 1 1 p1 e p ; p1 þ e p p2 e p ; p2 þ e p 2 2 2 2

1 1 p3 e p ; p3 þ e p : p ð p1 ; p2 ; p3 Þ 2 A 2 2 is an open cover of A; there exists a countable collection of points ðnÞ ðnÞ ðnÞ pðnÞ p1 ; p2 ; p3 2 Aðn ¼ 1; 2; 3; . . .Þ such that

6.1 Measure Zero in Manifolds

439

1 1 1 1 ðnÞ ðnÞ ðnÞ epðnÞ ; p1 þ epðnÞ p2 epðnÞ ; p2 þ epðnÞ 2 2 2 2

1 1 ðnÞ ðnÞ p3 epðnÞ ; p3 þ epðnÞ : n ¼ 1; 2; 3; . . . 2 2 ðnÞ p1

is an open cover of A: It follows that 1

A [

n¼1

1 1 1 1 ðnÞ ðnÞ ðnÞ ðnÞ p1 epðnÞ ; p1 þ epðnÞ p2 epðnÞ ; p2 þ epðnÞ 2 2 2 2 1 1 ðnÞ ðnÞ p3 epðnÞ ; p3 þ epðnÞ ; 2 2

and hence, 1 1 1 1 1 ðnÞ ðnÞ ðnÞ ðnÞ A¼A\ [ p1 epðnÞ ; p1 þ epðnÞ p2 epðnÞ ; p2 þ epðnÞ 2 2 2 2 n¼1 1 1 ðnÞ ðnÞ p3 epðnÞ ; p3 þ epðnÞ 2 2 1 1 1 1 1 ðnÞ ðnÞ ðnÞ ðnÞ ¼ [ A \ p1 epðnÞ ; p1 þ epðnÞ p2 epðnÞ ; p2 þ epðnÞ 2 2 2 2 n¼1 1 1 ðnÞ ðnÞ p3 epðnÞ ; p3 þ epðnÞ : 2 2 Hence,

1 1 1 1 ðnÞ ðnÞ ðnÞ ðnÞ FðAÞ ¼ F [ A \ p1 epðnÞ ; p1 þ epðnÞ p2 epðnÞ ; p2 þ epðnÞ 2 2 2 2 n¼1 1 1 ðnÞ ðnÞ p3 epðnÞ ; p3 þ epðnÞ 2 2 1 1 1 1 1 ðnÞ ðnÞ ðnÞ ðnÞ ¼ [ F A \ p1 epðnÞ ; p1 þ epðnÞ p2 epðnÞ ; p2 þ epðnÞ 2 2 2 2 n¼1 1 1 ðnÞ ðnÞ p3 epðnÞ ; p3 þ epðnÞ : 2 2 1

440

6 Sard’s Theorem

It follows that

1 1 ðnÞ ðnÞ 0 l3 ðFðAÞÞ ¼ l3 [ F A \ p1 epðnÞ ; p1 þ epðnÞ 2 2 n¼1 1 1 1 1 ðnÞ ðnÞ ðnÞ ðnÞ p2 epðnÞ ; p2 þ epðnÞ p3 epðnÞ ; p3 þ epðnÞ 2 2 2 2 1 X 1 1 ðnÞ ðnÞ l3 F A \ p1 epðnÞ ; p1 þ epðnÞ 2 2 n¼1 1 1 1 1 ðnÞ ðnÞ ðnÞ ðnÞ p2 epðnÞ ; p2 þ epðnÞ p3 epðnÞ ; p3 þ epðnÞ 2 2 2 2 1 X 1 1 ðnÞ ðnÞ ¼ l3 F A \ p1 epðnÞ ; p1 þ epðnÞ 2 2 n¼1 1 1 1 1 ðnÞ ðnÞ ðnÞ ðnÞ p2 epðnÞ ; p2 þ epðnÞ p3 epðnÞ ; p3 þ epðnÞ 2 2 2 2 1

Hence, it sufﬁces to show that for every n ¼ 1; 2; 3; . . ., 1 1 1 1 ðnÞ ðnÞ ðnÞ ðnÞ l3 FpðnÞ A \ p1 epðnÞ ; p1 þ epðnÞ p2 epðnÞ ; p2 þ epðnÞ 2 2 2 2 1 1 ðnÞ ðnÞ p3 epðnÞ ; p3 þ epðnÞ ¼ 0: 2 2 For this purpose, let us take any d [ 0. Since ðnÞ ðnÞ ðnÞ ðnÞ FpðnÞ : p1 epðnÞ ; p1 þ epðnÞ p2 epðnÞ ; p2 þ epðnÞ ðnÞ ðnÞ p3 epðnÞ ; p3 þ epðnÞ ! R3 is a smooth function, by a result similar to Theorem 3.33,

0 ðnÞ ðnÞ ðnÞ ðnÞ FpðnÞ : p1 epðnÞ ; p1 þ epðnÞ p2 epðnÞ ; p2 þ epðnÞ ðnÞ ðnÞ p 3 e p ð nÞ ; p 3 þ e p ð nÞ ! L R3 ; R3

6.1 Measure Zero in Manifolds

441

is continuous. Since 3 3 3 3 ðnÞ ðnÞ ðnÞ ðnÞ p1 epðnÞ ; p1 þ epðnÞ p2 epðnÞ ; p2 þ epðnÞ 4 4 4 4 3 3 ðnÞ ðnÞ p3 epðnÞ ; p3 þ epðnÞ 4 4 is a compact subset of the domain of the continuous function ðFpðnÞ Þ0 , its image under ðFpðnÞ Þ0 is compact in the metric space LðR3 ; R3 Þ and hence is bounded. Thus, there exists a positive real number C such that

n 0 3 3 3 3

ðnÞ ðnÞ ðnÞ ðnÞ sup FpðnÞ ð xÞ : x 2 p1 epðnÞ ; p1 þ epðnÞ p2 epðnÞ ; p2 þ epðnÞ 4 4 4 4 3 3 ðnÞ ðnÞ C: p3 epðnÞ ; p3 þ epðnÞ 4 4

Since FpðnÞ

3 3 3 3 ðnÞ ðnÞ ðnÞ ðnÞ : p1 epðnÞ ; p1 þ epðnÞ p2 epðnÞ ; p2 þ epðnÞ 4 4 4 4 3 3 ðnÞ ðnÞ p3 epðnÞ ; p3 þ epðnÞ 4 4 ! R3

is a smooth function, by a result similar to Theorem 3.28, for every x; y in

3 3 3 3 ðnÞ ðnÞ ðnÞ epðnÞ ; p1 þ epðnÞ p2 epðnÞ ; p2 þ epðnÞ 4 4 4 4 3 3 ðnÞ ðnÞ p3 epðnÞ ; p3 þ epðnÞ ; FpðnÞ ð yÞ FpðnÞ ð xÞ jy xjC: 4 4

ðnÞ p1

Since A\

1 1 1 1 ðnÞ ðnÞ ðnÞ ðnÞ p1 epðnÞ ; p1 þ epðnÞ p2 epðnÞ ; p2 þ epðnÞ 2 2 2 2 1 1 ðnÞ ðnÞ A; p3 epðnÞ ; p3 þ epðnÞ 2 2

and 0 l3 A \

1 1 1 1 ðnÞ ðnÞ ðnÞ epðnÞ ; p1 þ epðnÞ p2 epðnÞ ; p2 þ epðnÞ 2 2 2 2 1 1 ðnÞ ðnÞ p3 epðnÞ ; p3 þ epðnÞ l3 ðAÞ ¼ 0; 2 2 ðnÞ p1

442

6 Sard’s Theorem

so 1 1 1 1 ðnÞ ðnÞ ðnÞ ðnÞ l3 A \ p1 epðnÞ ; p1 þ epðnÞ p2 epðnÞ ; p2 þ epðnÞ 2 2 2 2 1 1 ðnÞ ðnÞ p3 epðnÞ ; p3 þ epðnÞ ¼ 0: 2 2 Since 1 1 1 1 ðnÞ ðnÞ ðnÞ ðnÞ l3 A \ p1 epðnÞ ; p1 þ epðnÞ p2 epðnÞ ; p2 þ epðnÞ 2 2 2 2 1 1 ðnÞ ðnÞ p3 epðnÞ ; p3 þ epðnÞ ¼ 0; 2 2 there exists a countable collection of open cubes Bdk ðk ¼ 1; 2; 3; . . .Þ of side length dk ð [ 0Þ such that A\

1 1 1 1 ðnÞ ðnÞ ðnÞ epðnÞ ; p1 þ epðnÞ p2 epðnÞ ; p2 þ epðnÞ 2 2 2 2 1 1 1 ðnÞ ðnÞ p3 epðnÞ ; p3 þ epðnÞ [ Bdk ; 2 2 k¼1

ðnÞ p1

and 1 X

d l3 ðBdk Þ\ pﬃﬃﬃ 3 : 2 3C k¼1

Here, for every k ¼ 1; 2; 3; . . ., diam

FpðnÞ ðBdk Þ ¼ sup js tj : s; t 2 FpðnÞ ðBdk Þ ¼ sup FpðnÞ ð yÞ FpðnÞ ð xÞ : x; y 2 Bd k

supfjy xjC : x; y 2 Bdk g ¼ C supfjy xj : x; y 2 Bdk g ¼ C diam Bdk ¼ C

pﬃﬃﬃ 3dk ;

so, 3 l3 FpðnÞ ðBdk Þ 2 diam FpðnÞ ðBdk Þ pﬃﬃﬃ 3 pﬃﬃﬃ 3 pﬃﬃﬃ 3 2C 3dk ¼ 2 3C ðdk Þ3 ¼ 2 3C l3 ðBdk Þ:

6.1 Measure Zero in Manifolds

443

It follows that 1 1 1 1 ðnÞ ðnÞ ðnÞ ðnÞ 0 l3 FpðnÞ A \ p1 epðnÞ ; p1 þ epðnÞ p2 epðnÞ ; p2 þ epðnÞ 2 2 2 2 1 1 1 ðnÞ ðnÞ p3 epðnÞ ; p3 þ epðnÞ l3 FpðnÞ [ Bdk 2 2 k¼1 X 1 1 pﬃﬃﬃ 3 X 1 ¼ l3 [ FpðnÞ ðBdk Þ l3 FpðnÞ ðBdk Þ 2 3C l3 ðBdk Þ k¼1

1 pﬃﬃﬃ 3 X ¼ 2 3C l3 ðBdk Þ\d:

k¼1

k¼1

k¼1

It follows that 1 1 1 1 ðnÞ ðnÞ ðnÞ ðnÞ l3 FpðnÞ A \ p1 epðnÞ ; p1 þ epðnÞ p2 epðnÞ ; p2 þ epðnÞ 2 2 2 2 1 1 ðnÞ ðnÞ p3 epðnÞ ; p3 þ epðnÞ ¼ 0: 2 2

h

We can easily generalize the above result as the following: Lemma 6.12 Let A be a nonempty subset of Rn : For every k ¼ 1; . . .; n; let lk denote the k-dimensional measure over Rk : Let ln ðAÞ ¼ 0: Let F : A ! Rn be smooth on A: Then, ln ðFðAÞÞ ¼ 0: Proof Its proof is similar to the proof given in Lemma 6.11.

h

Deﬁnition Let M be an n-dimensional smooth manifold. Let A be a nonempty subset of M: For every k ¼ 1; . . .; n; let lk denote the k-dimensional measure over Rk : If for every admissible coordinate chart ðU; uÞ for M; ln ðuðA \ UÞÞ ¼ 0; then we say that A has measure zero in M: Lemma 6.13 Let M be an m-dimensional smooth manifold. Let A be a nonempty subset of M: For every k ¼ 1; . . .; m; let lk denote the k-dimensional measure over Rk : Let fðU S i ; ui Þ : i 2 Ig be a collection of admissible coordinate charts for M such that A i2I Ui ; and for every i 2 I; lm ðui ðA \ Ui ÞÞ ¼ 0: Then, A has measure zero in M: Proof Let us take any admissible coordinate chart ðU; uÞ for M: We have to prove that lm ðuðA \ UÞÞ ¼ 0: Since M is an m-dimensional smooth manifold, M is a second countable space. Since M is a second countable space, and A is a nonempty subset of M; A is a second countable space. Since A is a second countable space, and fUi : i 2 Ig is an open cover of A; there exists Sa countable subcollection hence, A ¼ fUn : n ¼ 1; 2; 3; . . .g of fUi : i 2 Ig such that A 1 n¼1 Un ; and S1 S1 S U Þ ¼ ðA \ U Þ: It follows that uðA \ UÞ ¼ uðð A\ð 1 n n¼1 n¼1 n¼1 ðA \ Un ÞÞ\ S1 S1 n UÞ ¼ uð n¼1 ðA \ Un \ UÞÞ ¼ n¼1 ðuðA \ Un \ UÞÞ; and hence, 0 lm ðuðA\

444

6 Sard’s Theorem

S P1 UÞÞ ¼ lm ð 1 n¼1 ðuðA \ Un \ UÞÞÞ n¼1 lm ðuðA \ Un \ UÞÞ: Now, it sufﬁces to show that for each positive integer n; lm ðuðA \ Un \ UÞÞ ¼ 0: If not, otherwise, let there exists a positive integer n such that 0\lm ðuðA\ Un \ UÞÞ: We have to arrive at a contradiction. Since 0\lm ðuðA \ Un \ UÞÞ; uðA \ Un \ UÞ 6¼ ;; and hence, A \ Un \ U 6¼ ;: It follows that Un \ U 6¼ ;: Since Un \ U 6¼ ; and ðUn ; un Þ; ðU; uÞ are admissible coordinate charts for M; u

ððun Þ1 Þ : un ðUn \ UÞ ! uðUn \ UÞ is smooth. Here, A \ Un \ U 6¼ ;; so un ðA \ Un \ UÞ is a nonempty subset of un ðUn \ UÞð Rm Þ: Since un ðA \ Un \ UÞ is a nonempty subset of un ðUn \ UÞ; and u ððun Þ1 Þ : un ðUn \ UÞ ! uðUn \ UÞ is smooth, the restriction ðu ððun Þ1 ÞÞjun ðA \ Un \ UÞ is smooth. Since ðUn ; un Þ 2 fðUi ; ui Þ : i 2 Ig, 0 lm ðun ðA \ Un \ UÞÞ lm ðun ðA \ Un ÞÞ ¼ 0, and hence lm ðun ðA \ Un \ UÞÞ ¼ 0. Now, we can apply Lemma 6.12. It follows that lm ðuðA \ Un \ U ÞÞ ¼ lm ¼ 0; which is a contradiction.

u ðun Þ1

un ðA\Un \U Þ

ðun ðA \ Un \ U ÞÞ

h

Lemma 6.14 Let M be an m-dimensional smooth manifold. Let fAn g be a collection of nonempty subsets of M: For every k ¼ 1; . . .; m; let lk denote the kS dimensional measure over Rk : Let each An has measure zero in M: Then, 1 n¼1 An has measure zero in M: Proof For this purpose, let us take S any admissible coordinate chart ðU; uÞ S1for M: A Þ \ UÞÞ ¼ 0: Here, 0 l ðuðð We have to prove that lm ðuðð 1 n n¼1S n¼1 An Þ S P1 m 1 ðA \ UÞÞÞ ¼ l ð ðuðA \ UÞÞÞ l ðuðA \UÞÞ ¼ lm ðuð 1 n n \ UÞÞ; m n¼1 n n¼1 n¼1 m so it sufﬁces to show that for each positive integer n; lm ðuðAn \ UÞÞ ¼ 0: If not, otherwise, let there exists a positive integer n such that 0\lm ðuðAn \ UÞÞ: We have to arrive at a contradiction. Since An has measure zero in M; and ðU; uÞ is an admissible coordinate chart for M, lm ðuðAn \ UÞÞ ¼ 0; a contradiction. h Lemma 6.15 Let M be an m-dimensional smooth manifold. Let A be a nonempty subset of M: For every k ¼ 1; . . .; m; let lk denote the k-dimensional measure over Rk : Let A has measure zero in M: Then, Ac ð M AÞ is a dense subset of M: Proof We have to show that Ac is dense in M; that is, ðAc Þ ¼ M; that is, ððAc Þ Þc ð¼ Ao Þ ¼ ;: If not, otherwise, let Ao 6¼ ;: We have to arrive at a contradiction. Since Ao 6¼ ;; there exists a point p 2 A such that p is an interior point of A: Since p is an interior point of A; there exists an open neighborhood G of p in M such that p 2 G A: Since p 2 A M; and M is an m-dimensional smooth manifold, there exists an admissible coordinate chart ðU; uÞ for M such that p 2 U: Clearly, U \ Gð UÞ is an open neighborhood of p: Since U \ Gð UÞ is an open neighborhood of p; and ðU; uÞ is an admissible coordinate chart for M; ðU \ G; ujU\G Þ is an admissible coordinate chart for M: Here, uðpÞ is an interior point of the subset ðujU\G ÞðU \ GÞð¼ uðU \ GÞÞ of Rm ; and hence, lm ðu

6.1 Measure Zero in Manifolds

445

ðU \ GÞÞ [ 0: Since ðU \ G; ujU\G Þ is an admissible coordinate chart for M; and A has measure zero in M; 0 ¼ lm ujU\G ðA \ ðU \ GÞÞ ¼ lm ðuðA \ ðU \ GÞÞÞ ¼ lm ðuðU \ ðA \ GÞÞÞ ¼ lm ðuðU \ GÞÞ: This is a contradiction.

h

Lemma 6.16 Let M be an n-dimensional smooth manifold, N be an n-dimensional smooth manifold, F : M ! N be a smooth map, and A be a nonempty subset of M: Let A has measure zero in M: Then, FðAÞ has measure zero in N: Proof We have to prove that FðAÞ has measure zero in N: For this purpose, let us take an admissible coordinate chart ðV; wÞ for N such that ðFðAÞÞ \ V 6¼ ;: We have to show that ln ðwððFðAÞÞ \ VÞÞ ¼ 0; where ln denotes the n-dimensional measure over Rn : Clearly, FðA \ F 1 ðVÞÞ ¼ ðFðAÞÞ \ V: (Reason: LHS ¼ FðA \ F 1 ðVÞÞ ðFðAÞÞ \ ðFðF 1 ðVÞÞÞ ðFðAÞÞ \ V ¼ RHS: So, LHS RHS: Next, let us take any q 2 RHS ¼ ðFðAÞÞ \ V: It follows that there exists p 2 A such that FðpÞ ¼ q 2 V: Since FðpÞ 2 V; p 2 F 1 ðVÞ: Since p 2 F 1 ðVÞ; and p 2 A; p 2 A \ F 1 ðVÞ; and hence, q ¼ FðpÞ 2 FðA \ F 1 ðVÞÞ ¼ LHS: Hence, RHS LHS). Since M is an n-dimensional smooth manifold, by Lemma 4.47, there exists a countable collection fðU1 ; u1 Þ; ðU2 ; u2 Þ; ðU3 ; u3 Þ; . . .g of admissible coordinate charts for M such that fU1 ; U2 ; U3 ; . . .g is a basis of M: Let us take any p 2 A \ F 1 ðVÞ: Since p 2 A \ F 1 ðVÞ; p 2 F 1 ðVÞ; and hence, FðpÞ 2 V: Thus, V is an open neighborhood of FðpÞ in N: Since F : M ! N is a smooth map, F : M ! N is continuous. Since F : M ! N is continuous, and V is an open neighborhood of FðpÞ; F 1 ðVÞ is an open neighborhood of p in M: Since F 1 ðVÞ is an open neighborhood of p in M; and fU1 ; U2 ; U3 ; . . .g is a basis of M; there exists a positive integer kðpÞ such that p 2 UkðpÞ F 1 ðVÞ; and hence, A \ UkðpÞ A \ F 1 ðVÞ: Since p 2 A \ F 1 ðVÞ; p 2 A: Since p 2 A; and p 2 UkðpÞ ; p 2 A \ UkðpÞ : It follows that for every p 2 A \ F 1 ðVÞ; there exists a positive integer kðpÞ such that p 2 A \ UkðpÞ A \ F 1 ðVÞ, and hence, A \ F 1 ðVÞ ¼ [fA \ UkðpÞ : p 2 A\ F 1 ðVÞg: Now, ðFðAÞÞ \ V ¼ FðA \ F 1 ðVÞÞ ¼ Fð[fA \ UkðpÞ : p 2 A \ F 1 ðVÞgÞ ¼ [fFðA \ UkðpÞ Þ : p 2 A \ F 1 ðVÞg; and hence wððFðAÞÞ \ VÞ ¼ wð[fFðA \ UkðpÞ Þ : p 2 A \ F 1 ðVÞgÞ ¼ [fwðFðA \ UkðpÞ ÞÞ : p 2 A \ F 1 ðVÞg ¼ [fðw FÞðA [ U kðpÞÞ : p 2 A \ F 1 ðVÞg. Thus, wððFðAÞÞ \ VÞ is the union of countable many sets ðw FÞðA \ UkðpÞ Þ; where p 2 A \ F 1 ðVÞ: Here, fkðpÞ : p 2 A \ F 1 ðVÞg f1; 2; 3; . . .g; so fkðpÞ : p 2 A \ F 1 ðVÞg can be written as fk1 ; k2 ; k3 ; . . .g: It follows that Ukð pÞ : p 2 A \ F 1 ðV Þ ¼ fUk1 ; Uk2 ; Uk3 ; . . .g;

446

6 Sard’s Theorem

and hence, 1 [ ðw F Þ A \ Ukj : [ ðw F Þ A \ Ukð pÞ : p 2 A \ F 1 ðV Þ ¼ j¼1

It follows that ln ðwððFðAÞÞ \ V ÞÞ ¼ ln [ ðw F Þ A \ Ukð pÞ : p 2 A \ F 1 ðV Þ ! 1 1 [ X ¼ ln ð w F Þ A \ U kj ln ðw F Þ A \ Ukj j¼1

¼

1 X

j¼1

1 ln w F ukj ukj A \ Ukj :

j¼1

Now, it sufﬁces to show that each 1 w F ukj ln ukj A \ Ukj is 0: Since F : M ! N is a smooth map, w F ðukj Þ1 is smooth. Also, since A has measure zero in M; ln ððukj ÞðA \ Ukj ÞÞ ¼ 0: Now, we can apply Lemma 6.12. It follows that 1 w F u kj ln ¼ 0: h ukj A \ Ukj Note 6.17 In the following discussion, if a ða1 ; . . .; an Þ 2 Rn ; and e [ 0; then 1 1 1 1 n 1 1 n a e; a þ e a e; a þ e 2 2 2 2 will be denoted by Ce ðaÞ; and

1 1 1 1 a1 e; a1 þ e an e; an þ e 2 2 2 2

will be denoted by Ce ½a: Here, Ce ðaÞ will be called an open cube of side length e; and Ce ½a will be called a closed cube of side length e: Let G be a nonempty open subset of Rn : Let 0\e: We shall try toSshow that there exists a countable collection fUn g of open cubes such that 1 n¼1 Un ¼ S1 G; n¼1 ðUn Þ ¼ G; and side length of each Un is strictly less than e: Let us take any a ða1 ; . . .; an Þ 2 G: Since a 2 G; and G is open in Rn ; there exists a positive rational number r such that Cr ðaÞ G; and r\e: Now, since the

6.1 Measure Zero in Manifolds

447

collection fs : s 2 Rn ; and all coordinates of s are rational numbersg is dense in Rn ; there exists s ðs1 ; . . .; sn Þ 2 Rn such that s 2 C2r ðaÞ; and all coordinates of s are rational numbers. Since s 2 C2r ðaÞ; for every i ¼ 1; . . .; n; i a s i ¼ s i ai \ r ; 4 and hence, a 2 C2r ðsÞ: Now, clearly C2r ½s Cr ðAÞ: (Reason: Let us take any x ðx1 ; . . .; xn Þ 2 C2r ½s: It follows that for every i ¼ 1; . . .; n; jxi si j 4r : Now, for every i ¼ 1; . . .; n; i x ai xi si þ si ai r þ si ai \ r þ r ¼ r ; 4 4 4 2 and hence, x 2 Cr ðaÞ:Þ Thus, for every a 2 G; there exist a positive rational number r; and s 2 Rn such that all coordinates of s are rational numbers, and a 2 C2r ðsÞ C2r ðsÞ ¼ C2r ½s Cr ðAÞ G: Now, since the collection of all open cubes of the form C2r ðsÞ; where r is a positive rational number, and all coordinates of sð2 Rn Þ are rational numbers, is aScountable set, there exists a countable collection fUn g of open cubes such that 1 n¼1 Un ¼ S ðU Þ ¼ G; and side length of each U is strictly less than e: h G; 1 n n n¼1 Note 6.18 Let G be a nonempty open subset of Rn : Let 0\e: We shall try to show that there exist a countable collection fUn g of open a countable colS1 cubes, and S 1 Un ¼ G, lection fVn g of open cubes such that n¼1 n¼1 ðUn Þ ¼ G, S1 n¼1 Vn ¼ G, Un ðUn Þ Vn ðn ¼ 1; 2; 3; . . .Þ, and side length of each Vn is strictly less than e: Let us take any a ða1 ; . . .; an Þ 2 G: Since a 2 G; and G is open in Rn ; there exists a positive rational number r such that Cr ðaÞ G; and r\e: Now, since the collection fs : s 2 Rn ; and all coordinates of s are rational numbersg is dense in Rn ; there exists s ðs1 ; . . .; sn Þ 2 Rn such that s 2 C4r ðaÞ; and all coordinates of s are rational numbers. Since s 2 C4r ðaÞ; for every i ¼ 1; . . .; n; i a s i ¼ s i ai \ r ; 8 and hence, a 2 C4r ðsÞ: Now, clearly C4r ½s C2r ðaÞ: (Reason: Let us take any x ðx1 ; . . .; xn Þ 2 C4r ½s: It follows that for every i ¼ 1; . . .; n; jxi si j 8r : Now, for every i ¼ 1; . . .; n; i x ai xi si þ si ai r þ si ai \ r þ r ¼ r ; 8 8 8 4 and hence, x 2 C2r ðAÞ:) Also, it is clear that C2r ½s Cr ðAÞ: (Reason: Let us take any x ðx1 ; . . .; xn Þ 2 C2r ½s: It follows that for every i ¼ 1; . . .; n; jxi si j 4r : Now, for every i ¼ 1; . . .; n;

448

6 Sard’s Theorem

i x ai x i s i þ s i ai r þ s i ai \ r þ r \ r ; 4 4 8 2 and hence, x 2 Cr ðAÞ:) Thus, for every a 2 G; there exist a positive rational number r; and s 2 Rn such that all coordinates of s are rational numbers, and a 2 C4r ðsÞ C4r ðsÞ ¼ C4r ½s C2r ½s Cr ðAÞ G: Now, since the collection of all open cubes of the form C4r ðsÞ; where r is a positive rational number, and all coordinates of sð2 Rn Þ are rational numbers, is a countable set, there exist a countable collection S fUn g of open S cubes, and a countable S1 col1 U ¼ G; ðU Þ ¼ G, lection fVn g of open cubes such that 1 n n n¼1 n¼1 n¼1 Vn ¼ G, Un ðUn Þ Vn ðn ¼ 1; 2; 3; . . .Þ, and side length of each Vn is strictly less than e: h

6.2 Taylor’s Inequality Note 6.19 We want to prove the following result: Let p 2 U R2 : Let U be an open set. Let S U; and S be star shaped with respect to point p ðp1 ; p2 Þ 2 S. Let f : U ! R be C 1 on U: Let x ðx1 ; x2 Þ 2 S: Then, ! of f ðxÞ ¼ f ð pÞ þ x p ðð1 tÞp þ txÞ dt ox1 0 ! 2 Z1 of 2 ðð1 tÞp þ txÞ dt þ x p ox2 0 0 1 1@ X i i x p ððDi f Þð pÞÞA ¼ f ð pÞ þ 1! i2f1;2g 0 1 X i 1@ Z1 x pi x j p j ð1 tÞDij ðð1 tÞp þ txÞ dtA þ 1! ði;jÞ2f1;2gf1;2g 0 0 1 1@ X i i x p ððDi f Þð pÞÞA ¼ f ð pÞ þ 1! i2f1;2g 0 1 1@ X i x pi x j p j Dij f ð pÞ A þ 2! ði;jÞ2f1;2g2 0 1 j k 1 1@ X i 2 i j k Z þ x p x p x p ð1 tÞ Dijk ðð1 tÞp þ txÞ dtA 3! 3 0

1

1

ði;j;k Þ2f1;2g

¼

Z1

6.2 Taylor’s Inequality

449

Proof Let us take any t satisfying 0 t 1: Now, since S is star shaped with respect to point p; and x 2 S; ð1 tÞp þ tx is in S; and hence, f ðð1 tÞp þ txÞ is a real number. Here, d d f ðð1 tÞp þ txÞ ¼ f p1 þ t x1 p1 ; p2 þ t x2 p2 dt dt 1 of d 1 1 p ¼ ð ð 1 t Þp þ tx Þ þ t x p ox1 dt 2 of d 2 2 p þt x p þ ðð1 tÞp þ txÞ ox2 dt of ¼ ðð1 tÞp þ txÞ x1 p1 1 ox 2 of þ ð ð 1 t Þp þ tx Þ x p2 ; 2 ox

Hence, 2 1 of of 2 Z ð ð 1 t Þp þ tx Þ dt þ x p ð ð 1 t Þp þ tx Þ dt ox1 ox2 0 0 1 2 Z1 of of 1 2 dt þ ð ð 1 t Þp þ tx Þ x p ð ð 1 t Þp þ tx Þ x p ¼ ox1 ox2 0

x1 p 1

Z1

¼ f ðð1 tÞp þ txÞjt¼1 t¼0 ¼ f ð xÞ f ð pÞ:

Now, ! of f ð x Þ ¼ f ð pÞ þ x 1 p ðð1 tÞp þ txÞ dt ox1 0 ! 2 Z1 of 2 þ x p ðð1 tÞp þ txÞ dt : ox2 0

1

Z1

On applying integration by parts, we get f ðxÞ ¼ f ð pÞ þ x1 p1

! Z1

ððD1 f Þðð1 tÞp þ txÞÞdt

0

þ x p 2

2

Z1

0

!

ððD2 f Þðð1 tÞp þ txÞÞdt

¼ f ð pÞ þ x1 p1 ðtððD1 f Þðð1 tÞp þ txÞÞÞjt¼1 t¼0 Z1 d ð1 tÞp1 þ tx1 t ððD11 f Þðð1 tÞp þ txÞÞ dt 0

450

6 Sard’s Theorem

d 2 2 ð1 tÞp þ tx þððD12 f Þðð1 tÞp þ txÞÞ dt dt t¼1 þ x2 p2 ðtððD2 f Þðð1 tÞp þ txÞÞÞjt¼0 dt Z1 d t ððD21 f Þðð1 tÞp þ txÞÞ ð1 tÞp1 þ tx1 dt 0 d 2 2 ð1 tÞp þ tx þððD22 f Þðð1 tÞp þ txÞÞ dt dt Z1 ¼ f ð pÞ þ x1 p1 ððD1 f Þð xÞ t ððD11 f Þðð1 tÞp þ txÞÞ x1 p1

0

þððD12 f Þðð1 tÞp þ txÞÞ x p2 dt Z1 þ x2 p2 ððD2 f Þð xÞ t ððD21 f Þðð1 tÞp þ txÞÞ x1 p1 2

0

þððD22 f Þðð1 tÞp þ txÞÞ x2 p2 dt ¼ f ð pÞ þ x1 p1 ððD1 f Þð xÞÞ þ x2 p2 ððD2 f Þð xÞÞ 2 Z1 x1 p1 tðððD11 f Þðð1 tÞp þ txÞÞÞdt 0

2 x1 p

1

x 2 p2

Z1

tððD12 f Þðð1 tÞp þ txÞÞdt

0

2 Z1 x 2 p2 tðððD22 f Þðð1 tÞp þ txÞÞÞdt 0

¼ f ð pÞ þ

x1 p1

Z1 ðD1 f Þð pÞ þ x1 p1 ðD11 f Þðð1 tÞp þ txÞdt 0

Z1 þ x2 p2 ðD12 f Þðð1 tÞp þ txÞdt

!

0

Z1 þ x2 p2 ðD2 f Þð pÞ þ x1 p1 ðD21 f Þðð1 tÞp þ txÞdt 0

Z1 þ x2 p2 ðD22 f Þðð1 tÞp þ txÞdt

!!

0

x1 p

1 1 2Z

tðððD11 f Þðð1 tÞp þ txÞÞÞdt

0

Z1 2 x1 p1 x2 p2 tððD12 f Þðð1 tÞp þ txÞÞdt 0

x2 p

1 2 2Z 0

tðððD22 f Þðð1 tÞp þ txÞÞÞdt

6.2 Taylor’s Inequality

451

¼ f ð pÞ þ x1 p1 ðD1 f Þð pÞ þ x2 p2 ðD2 f Þð pÞ 2 Z1 þ x1 p1 ððD11 f Þðð1 tÞp þ txÞ tðððD11 f Þðð1 tÞp þ txÞÞÞÞdt 0

Z1 þ 2 x1 p1 x2 p2 ððD12 f Þðð1 tÞp þ txÞ tððD12 f Þðð1 tÞp þ txÞÞÞdt 0

þ x p 2

1 2 2Z

ððD22 f Þðð1 tÞp þ txÞ tðððD22 f Þðð1 tÞp þ txÞÞÞÞdt

0

¼ f ð pÞ þ

of 2 1 1 2 of x p1 ð p Þ þ x p ð p Þ 1! ox1 ox2

! 1 1 1 o2 f 1 2Z þ x p ð1 t Þ ðð1 tÞp þ txÞ dt 1! oðx1 Þ2 0 Z1 o2 f þ 2 x1 p 1 x2 p 2 ð1 tÞ 1 2 ðð1 tÞp þ txÞ dt ox ox 0 ! ! 2 2 1 o f 2 2Z þ x p ð1 tÞ ðð1 tÞp þ txÞ dt oðx2 Þ2 0 0 1 1@ X i ¼ f ð pÞ þ x pi ððDi f Þð pÞÞA 1! i2f1;2g 0 1 X i 1@ Z1 x pi x j p j ð1 tÞDij ðð1 tÞp þ txÞ dtA: þ 1! ði;jÞ2f1;2gf1;2g 0

Similarly, 0 1 1@ X i f ð x Þ ¼ f ð pÞ þ x pi ððDi f Þð pÞÞA 1! i2f1;2g 0 1 1@ X i þ x pi x j p j Dij f ð pÞ A 2! 2 ði;jÞ2f1;2g 0 1 j k 1 1@ X i 2 i j k Z þ x p x p x p ð1 tÞ Dijk ðð1 tÞp þ txÞ dtA 2! 3 0 ði;j;k Þ2f1;2g

¼

This result can be generalized in Rn ; and is known as Taylor’s theorem.

452

6 Sard’s Theorem

Note 6.20 Let p 2 U Rn : Let U be an open set. Let S U; and S be star shaped with respect to point p ðp1 ; . . .; pn Þ 2 S. Let f : U ! R be C 1 on U: Let x ðx1 ; . . .; xn Þ 2 S: Let all the third-order partial derivatives of f be bounded in absolute value by a constant M on S: Then, 8 0 1 < X f ðxÞ f ð pÞ þ 1 @ xi pi ððDi f Þð pÞÞA : 1! i2f1;...;ng 0 19 = i 1@ X þ x pi x j p j Dij f ð pÞ A ; 2! ði;jÞ2f1;...;ng2 0 1 X 1 i Z1 2 i j j k k @ A ¼ x p x p x p ð1 tÞ Dijk ðð1 tÞp þ txÞ dt 2! ði;j;kÞ2f1;...;ng3 0 X Z1 1 i ð1 tÞ2 Dijk ðð1 tÞp þ txÞ dt x pi x j p j xk pk 2! 3 0 ði;j;k Þ2f1;...;ng X i 1 1 x pi x j p j xk pk Z ð1 tÞ2 Dijk ðð1 tÞp þ txÞ dt ¼ 0 2! 3 ði;j;k Þ2f1;...;ng

¼ ¼ ¼

1 2! 1 2! 1 2! 1 2! 1 2! 1 2!

X

i 1 x pi x j p j xk pk Z j1 tj2 Dijk ðð1 tÞp þ txÞdt 0

ði;j;k Þ2f1;...;ng3

X

i 1 x pi x j p j xk pk Z 1Dijk ðð1 tÞp þ txÞdt 0

ði;j;k Þ2f1;...;ng3

X

i 1 x pi x j p j xk pk Z Mdt 0

ði;j;k Þ2f1;...;ng3

X

i x pi x j p j xk pk M

ði;j;k Þ2f1;...;ng3

X

jx pjjx pjjx pjM

ði;j;k Þ2f1;...;ng3

X

M j x pj3 ¼

ði;j;k Þ2f1;...;ng3

1 3 n M j x pj 3 : 2!

1 ðn n n Þ M j x pj 3 3!

6.2 Taylor’s Inequality

453

Thus 8 0 1 < X 1 f ð xÞ f ð pÞ þ @ xi pi ððDi f Þð pÞÞA : 1! i2f1;...;ng 0 19 = i 1@ X i j j A Dij f ð pÞ x p x p þ ; 2! 2 ði;jÞ2f1;...;ng

1 n3 jx pj3 M; 2!

etc. Conclusion: Let p 2 U Rn : Let U be an open set. Let S U; and S be star shaped with respect to point p ðp1 ; . . .; pn Þ 2 S. Let f : U ! R be C 1 on U: Let x ðx1 ; . . .; xn Þ 2 S: Let all the third-order partial derivatives of f be bounded in absolute value by a constant M on S: Then, 8 0 1 < X 1 i i f ð xÞ f ð pÞ þ @ x p ððDi f Þð pÞÞA : 1! i2f1;...;ng 0 19 = X 1 þ @ xi pi x j p j Dij f ð pÞ A ; 2! 2 ði;jÞ2f1;...;ng

1 n3 jx pj3 M: 2!

In the following discussion, we shall use this inequality.

6.3 Sard’s Theorem on Rn Lemma 6.21 Let G be an open subset of Rð¼ R1 Þ: Let F : G ! R3 be a smooth function. Let F ðF1 ; F2 ; F3 Þ: For every k ¼ 1; 2; 3; let lk denote the k-dimensional measure over Rk : Then, 91 1 0 08 2 3 ðD1 F1 Þð xÞ = < l3 @F @ x : x 2 G; and rank4 ðD1 F2 Þð xÞ 5 ¼ 0 AA ¼ 0: ; : ðD1 F3 Þð xÞ Proof Put

9 3 ðD1 F1 Þð xÞ = C x : x 2 G; and rank4 ðD1 F2 Þð xÞ 5 ¼ 0 : ; : ðD1 F3 Þð xÞ 8 <

2

454

6 Sard’s Theorem

We have to show that l3 ðFðCÞÞ ¼ 0: Here, 9 8 2 3 ð D 1 F 1 Þ ð xÞ > > = < 6 7 C ¼ x : x 2 G; and rank4 ðD1 F2 Þð xÞ 5 ¼ 0 > > ; : ð D 1 F 3 Þ ð xÞ ¼ fx : x 2 G; and ðD1 F1 Þð xÞ ¼ 0; ðD1 F2 Þð xÞ ¼ 0; ðD1 F3 Þð xÞ ¼ 0g ¼ fx : x 2 G; and ðD1 F1 Þð xÞ ¼ 0g \ fx : x 2 G; and ðD1 F2 Þð xÞ ¼ 0g \ fx : x 2 G; and ðD1 F3 Þð xÞ ¼ 0g ¼ ðD1 F1 Þ1 ðf0gÞ \ ðD1 F2 Þ1 ðf0gÞ \ ðD1 F3 Þ1 ðf0gÞ: Since F : G ! R3 is a smooth function, D1 F1 : G ! R is continuous. Since D1 F1 : G ! R is continuous, and f0g is closed, ðD1 F1 Þ1 ðf0gÞ is closed in G: Similarly, ðD1 F2 Þ1 ðf0gÞ; ðD1 F3 Þ1 ðf0gÞ are closed in G: It follows that ðD1 F1 Þ1 ðf0gÞ \ ðD1 F2 Þ1 ðf0gÞ \ ðD1 F3 Þ1 ðf0gÞð¼ CÞ is closed in G; and hence, C is closed in G: Put C1 fx : x 2 G; and 0 ¼ ðD1 F1 Þð xÞ ¼ ðD1 F2 Þð xÞ ¼ ðD1 F3 Þð xÞg: Clearly, C1 ¼ C; and C1 is closed in G: Put C2 fx : x 2 G; and 0 ¼ ðD1 F1 Þð xÞ ¼ ðD1 F2 Þð xÞ ¼ ðD1 F3 Þð xÞ ¼ ðD11 F1 Þð xÞ ¼ ðD11 F2 Þð xÞ ¼ ðD11 F3 Þð xÞg: Clearly, C2 C1 ; and C2 is closed in G: Similarly, we deﬁne C3 fx : x 2 G; and 0 ¼ ðD1 Fi ÞðxÞ ¼ ðD11 Fi ÞðxÞ ¼ ðD111 Fi ÞðxÞfor every i ¼ 1; 2; 3g; etc. Clearly, C4 C3 C2 C1 ¼ C; and each Ci is closed in G. Now, we shall try to show that l3 ðFðC1 ÞÞ ¼ 0: Since G is an open subset S of R; there exists a countable collection fUn g of open bounded intervals such that 1 n¼1 Un ¼ G; S ðU Þ ¼ G: Now, FðC Þ ¼ Fðfx : x 2 G; and 0 ¼ ðD F ÞðxÞ ¼ ðD1 F2 Þ and 1 n 1 1 1 n¼1 S ðU Þ ; and 0 ¼ ðD F ÞðxÞ ¼ ðD ðxÞ ¼ ðD1 F3 ÞðxÞgÞ ¼ Fðfx : x 2 1 n 1 1 1 F2 ÞðxÞ ¼ n¼1 S fx : x 2 ðU Þ ; and 0 ¼ ðD F ÞðxÞ ¼ ðD F ÞðxÞ ¼ ðD1 F3 Þ ðD1 F3 ÞðxÞgÞ ¼ Fð 1 n 1 1 1 2 n¼1 S1 ðxÞgÞ ¼ n¼1 Fðfx : x 2 ðUn Þ ; and 0 ¼ ðD1 F1 ÞðxÞ ¼ ðD1 F2 ÞðxÞ ¼ ðD1 F3 ÞðxÞgÞ S1 P1 S ¼ 1 n¼1 FðC1 \ ððUn Þ ÞÞ; so 0 l3 ðFðC1 ÞÞ ¼ l3 ð n¼1 FðC1 \ ððUn Þ ÞÞÞ n¼1 l3 ðFðC1 \ ððUn Þ ÞÞÞ: Hence, it sufﬁces to show that each l3 ðFðC1 \ ððUn Þ ÞÞÞ ¼ 0: Let us ﬁrst try to show that l3 ðFðC1 \ ððU1 Þ ÞÞÞ ¼ 0: Since D11 F1 : G ! R is continuous, and ðU1 Þ is compact, ðD11 F1 ÞððU1 Þ Þ is compact, and hence, ðD11 F1 ÞððU1 Þ Þ is bounded. Similarly, ðD11 F2 ÞððU1 Þ Þ; ðD11 F3 ÞððU1 Þ Þ are bounded. It follows that ðD11 F1 ÞððU1 Þ Þ [ ðD11 F2 ÞððU1 Þ Þ [ ðD11 F3 ÞððU1 Þ Þ is bounded, and hence, there exists a positive number K such that ðD11 F1 ÞððU1 Þ Þ [ ðD11 F2 ÞððU1 Þ Þ [ ðD11 F3 ÞððU1 Þ Þ is contained in ½K; K: Case I: when C1 \ ððU1 Þ Þ ¼ ;: In this case,

6.3 Sard’s Theorem on Rn

455

l3 ðF ðC1 \ ððU1 Þ ÞÞÞ ¼ l3 ðF ð;ÞÞ ¼ l3 ð;Þ ¼ 0: Thus, l3 ðFðC1 \ ððU1 Þ ÞÞÞ ¼ 0: Case II: when C1 \ ððU1 Þ Þ 6¼ ;. Now, let R be the length of closed interval ðU1 Þ : Let us ﬁx any positive integer N [ 1: Let us subdivide the closed interval ðU1 Þ into N closed intervals E1 ; E2 ; . . .; EN such that each Ei is of length NR : Thus, ðU1 Þ ¼ E1 [ E2 [ [ EN : It follows that C1 \ ððU S1 Þ Þ ¼ C1 \ ðE1 [ E2 [ [ EN Þ ¼ Þ [ [ ðC1 \ EN Þ ¼ fC1 \SEk : C1 \ Ek 6¼ ;g: Hence, F ðC1 \ E1 Þ [ ðC1 \ E2S ðC1 \ ððU1 Þ ÞÞ ¼ Fð fC1 \ Ek : C1 \ Ek 6¼ ;gÞ S ¼ fFðC1 \ Ek Þ : C1 \ Ek 6¼ ;g; P and hence, 0 l3 ðFðC1 \ ððU1 Þ ÞÞÞ ¼ l3 ð fFðC1 \ Ek Þ : C1 \ Ek 6¼ ;gÞ fl3 ðFðC1 \ Ek ÞÞ : C1 \ Ek 6¼ ;g: Now, it sufﬁces to show that if C1 \ Ek 6¼ ;; then l3 ðFðC1 \ Ek ÞÞ ¼ 0: For this purpose, let C1 \ Ek 6¼ ;: Since C1 \ Ek 6¼ ;; there exists a 2 C1 \ Ek : Since a 2 C1 \ Ek ; a 2 Ek : Now, by Taylor’s theorem, for every x 2 Ek ; jF1 ð xÞ F1 ðaÞj ¼ jF1 ð xÞ fF1 ðaÞ þ 0gj

2 ! 1 R K : ¼ jF1 ð xÞ fF1 ðaÞ þ ðD1 F1 ÞðaÞgj 1! N

Thus, for every x 2 Ek ; 2 ! 1 R K : jF1 ð xÞ F1 ðaÞj 1! N Similarly, for every x 2 Ek ; 2 ! 1 R K jF2 ð xÞ F2 ðaÞj 1! N

2 ! 1 R K and jF3 ð xÞ F3 ðaÞj : 1! N

It follows that for every x 2 Ek ; jF ð xÞ FðaÞj ¼ jðF1 ð xÞ; F2 ð xÞ; F3 ð xÞÞ ðF1 ðaÞ; F2 ðaÞ; F3 ðaÞÞj ¼ jðF1 ð xÞ F1 ðaÞ; F2 ð xÞ F2 ðaÞ; F3 ð xÞ F3 ðaÞÞj jF1 ð xÞ F1 ðaÞj þ jF2 ð xÞ F2 ðaÞj þ jF3 ð xÞ F3 ðaÞj 2 ! 2 ! 2 ! 1 R 1 R 1 R K K K þ þ 1! N 1! N 1! N 2 !! 1 R K ¼3 : 1! N

456

6 Sard’s Theorem

Hence, for every x 2 Ek ; 2 !! 1 R K : jF ð xÞ FðaÞj 3 1! N Thus, F ðC1 \ Ek Þ F ðEk Þ B 1 3

1!

K ðNR Þ

2

ðFðaÞÞ;

and hence, 0 l3 ðF ðC1 \ Ek ÞÞ l3 B 1 3

1!

2 !!!!3 1 R K ðFðaÞÞ 2 3 2 K ðNR Þ 1! N

! 0 as N ! 1: This shows that l3 ðFðC1 \ Ek ÞÞ ¼ 0: Hence, l3 ðFðC1 \ ððU1 Þ ÞÞÞ ¼ 0: Thus, in all cases, l3 ðFðC1 \ ððU1 Þ ÞÞÞ ¼ 0: Similarly, l3 ðFðC1 \ ððU2 Þ ÞÞÞ ¼ 0; l3 ðFðC1 \ ððU3 Þ ÞÞÞ ¼ 0; etc. Thus, l3 ðFðCÞÞ ¼ l3 ðFðC1 ÞÞ ¼ 0: h Note 6.22 Similar to Lemma 6.21, we can prove the following result: Let G be an open subset of Rð¼ R1 Þ: Let F : G ! Rn be a smooth function. Let F ðF1 ; . . .; Fn Þ: For every k ¼ 1; . . .; n; let lk denote the k-dimensional measure over Rk : Then, 9 11 0 08 3 2 ðD1 F1 Þð xÞ > > = < B B CC 7 6 ln @F @ x : x 2 G; and rank4 ... ¼ 0 AA ¼ 0: 5 > > ; : ðD1 Fn Þð xÞ Lemma 6.23 Let m; n be positive integers satisfying m n: Let G be a nonempty open subset of Rm : Let f : G ! Rn be a smooth function, and f ðf1 ; . . .fn Þ: Let l denote the Lebesgue measure over Rn : Let C be the collection of all critical points of f ; that is, 9 3 ðD1 f1 Þð xÞ ðDm f1 Þð xÞ > = .. . 7 6 C x : x 2 G; and rank4 ... . .. 5\n : > > ; : ðD1 fn Þð xÞ ðDm fn Þð xÞ 8 > <

2

Then, lðf ðCÞÞ ¼ 0: m Proof Since G is a nonempty open subset S1 of R ; there S exists acountable collection m fCn g of open cubes in R such that n¼1 Cn ¼ G, 1 n¼1 ðCn Þ ¼ G; and for each n ¼ 1; 2; 3; . . .; the side length Rn of Cn is \1: Now,

6.3 Sard’s Theorem on Rn

457

91 08 3 2 ðD1 f1 Þð xÞ ðDm f1 Þð xÞ > > > > =C B< 7 6. .. . C B 7 6 . . \n .. f ðC Þ ¼ f @ x : x 2 G; and rank4 . A 5 > > > > ; : ðD1 fn Þð xÞ ðDm fn Þð xÞ 91 08 3 2 ðD1 f1 Þð xÞ ðDm f1 Þð xÞ > > > > =C 1 [ B< 7 6. .. . B 7 6 . . ¼ f@ x : x 2 \n C .. ðCn Þ ; and rank4 . A 5 > > > > n¼1 ; : ðD1 fn Þð xÞ ðDm fn Þð xÞ 91 0 8 3 2 ðD1 f1 Þð xÞ ðDm f1 Þð xÞ > > > > =C < 1 B[ 7 6. .. . C 7 6 . . ¼ fB \n . x : x 2 ð C Þ ; and rank n . A @ > 5 4. > > n¼1> ; : ðD1 fn Þð xÞ ðDm fn Þð xÞ 91 08 3 2 ðD1 f1 Þð xÞ ðDm f1 Þð xÞ > > > > 1 = < [ B C 7 6. .. . C 7 6 . . ¼ \n . fB x : x 2 ð C Þ ; and rank n . A @> 5 4. > > > n¼1 ; : ðD1 fn Þð xÞ ðDm fn Þð xÞ 1 [ ¼ f ðC \ðCn Þ Þ: n¼1

Since f ðCÞ ¼

S1 n¼1

f ðC \ ðCn Þ Þ;

0 lðf ðC ÞÞ ¼ l

1 [ n¼1

!

f ðC \ ðCn Þ Þ

1 X

lðf ðC \ ðCn Þ ÞÞ:

n¼1

Hence, it sufﬁces to show that each lðf ðC \ ðCn Þ ÞÞ ¼ 0: We ﬁrst try to show that lðf ðC \ ðC1 Þ ÞÞ ¼ 0: Since f : G ! Rn is a smooth function, each ðDij fk Þ : G ! R is a smooth function, and hence, each ðDij fk Þ : G ! R is continuous. Since ðDij fk Þ : G ! R is continuous, and ðC1 Þ is a compact subset of G; ðDij fk ÞððC1 Þ Þ is compact, and hence, ðDij fk ÞððC1 Þ Þ is bounded. Since ðDij fk ÞððC1 Þ Þ is bounded, there exists a real Mijk [ 0 such that for every z 2 ðC1 Þ , jðDij fk ÞðzÞj Mijk : Put M maxfMijk : 1 i m; 1 j m; 1 k ng: Since f : G ! R is a smooth function, each Di fj : G ! R is a smooth function, and hence, each Di fj : G ! R is continuous. Since Di fj : G ! R is continuous, and ðC1 Þ is a compact subset of G; ðDi fj ÞððC1 Þ Þ is compact, and hence, ðDi fj ÞððC1 Þ Þ is bounded. Since ðDi fj ÞððC1 Þ Þ is bounded, there exists a real Mij [ 0 such that for every z 2 ðC1 Þ ; jðDi fj ÞðzÞj Mij : Put M0 maxfMij : 1 i m; 1 j ng: Let us ﬁx any positive integer N [ 1: Let us subdivide the closed cube ðC1 Þ into N m closed cubes E1 ; E2 ; . . .; EN m such that each Ei is of length RN1 : Thus, ðC1 Þ ¼ E1 [ E2 [ [ EN m : It follows that C \ ðC1 Þ ¼ C \ ðE1 [ E2 [ [ EN m Þ ¼ ðC \ E1 Þ [ ðC \ E2 Þ [ S[ ðC \ EN m Þ ¼ S fC \ Ek : C \ Ek 6¼ ;; 1 k N m g: Hence, f ðC \ ðC1 Þ Þ ¼ f ð fC \ Ek : C \

458

6 Sard’s Theorem

S Ek 6¼ ;; 1 k N m gÞS¼ ff ðC \ Ek Þ : C \ Ek 6¼ ;; 1 k N m g, P and hence, 0 l ðf ðC \ ðC1 Þ ÞÞ ¼ lð ff ðC \ Ek Þ : C \ Ek 6¼ ;; 1 k N m gÞ flðf ðC \ Ek ÞÞ : C \ Ek 6¼ ;; 1 k N m g. Let us take any positive integer k satisfying 1 k N m ; and C \ Ek 6¼ ;. Now, by Taylor’s theorem, for every x ðx1 ; . . .; xm Þ 2 Ek , and y ðy1 ; . . .; ym Þ 2 Ek ; jf ð yÞ ðf ðxÞ þ ðf 0 ð xÞÞðy xÞÞj 2 3 02 3 2 32 31 f 1 ð yÞ ðD1 f1 Þð xÞ ðDm f1 Þð xÞ f1 ð xÞ y 1 x1 6 . 7 B6 . 7 6. 76 . 7C .. . 6 6 6 6 7 7 7 7 B C . . . . . .. ¼ 4 . 5 @4 . 5þ 4. 54 . 5A f n ð yÞ y m xm fn ð xÞ ðD1 fn Þð xÞ ðDm fn Þð xÞ 2 3 02 3 2 3 1 f 1 ð yÞ ððD1 f1 Þð xÞÞðy1 x1 Þ þ þ ððDm f1 Þð xÞÞðym xm Þ f1 ð xÞ 6 . 7 B6 . 7 6. 7 C 7 B6 . 7þ 6. 7 C . ¼ 6 4. 5 @4 . 5 4. 5 A f n ð yÞ fn ð xÞ ððD1 fn Þð xÞÞðy1 x1 Þ þ þ ððDm fn Þð xÞÞðym xm Þ 2 3 f1 ð yÞ ðf1 ð xÞ þ ððD1 f1 Þð xÞÞðy1 x1 Þ þ þ ððDm f1 Þð xÞÞðym xm ÞÞ 6 . 7 7 . ¼ 6 5 4. fn ð yÞ ðfn ð xÞ þ ðððD1 fn Þð xÞÞðy1 x1 Þ þ þ ððDm fn Þð xÞÞðym xm ÞÞÞ jf1 ð yÞ ðf1 ðxÞ þ ððD1 f1 ÞðxÞÞðy1 x1 Þ þ þ ððDm f1 ÞðxÞÞðym xm ÞÞj þ þ jfn ð yÞ ðfn ð xÞ þ ðððD1 fn Þð xÞÞðy1 x1 Þ þ þ ððDm fn Þð xÞÞðym xm ÞÞÞj ! 1 2 pﬃﬃﬃﬃ R1 2 1 2 pﬃﬃﬃﬃ R1 2 1 2 pﬃﬃﬃﬃ R1 2 m M þ þ m M¼n M m m m m N N N 1! 1! 1! |ﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄ{zﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄ} n

1 ¼ nm M 2 ðR1 Þ2 N 3 1 \ nm M 2 : N 3

Thus, for every x; y 2 Ek ; 1 jf ð yÞ ðf ð xÞ þ ðf 0 ð xÞÞðy xÞÞj\ nm3 M 2 : N Since C \ Ek 6¼ ;; there exists a 2 C \ Ek 6¼ ;; and hence, for every y 2 C \ Ek , 1 jf ð yÞ ðf ðAÞ þ ðf 0 ðAÞÞðy aÞÞj\ nm3 M 2 : N Again, by Taylor’s theorem, for every x ðx1 ; . . .; xm Þ; y ðy1 ; . . .; ym Þ 2 Ek ;

6.3 Sard’s Theorem on Rn

459

2 3 2 3 2 3 f 1 ð yÞ f1 ð xÞ f1 ð yÞ f1 ð xÞ 6 . 7 6. 7 6 . 7 7 6. 7 ¼ 6 . 7 jf ð yÞ f ð xÞj ¼ 6 5 4. 5 4 . 5 4 .. f n ð yÞ fn ð xÞ fn ð yÞ fn ð xÞ jf1 ð yÞ f1 ð xÞj þ þ jfn ð yÞ fn ð xÞj pﬃﬃﬃﬃ R1 1 pﬃﬃﬃﬃ R1 1 1 1 m1 m m M0 þ þ m1 M0 0! 0! N N |ﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄ{zﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄ} n

¼ nm

3=2

1 1 M0 R1 \nm3=2 M0 : N N

Hence, for every y 2 C \ Ek , jf ðyÞ f ðaÞj\nm3=2 M0 N1 : It follows that f ðC \ Ek Þ is contained in an open ball of radius nm3=2 M0 N1 with center f ðaÞ: Since a 2 C \ Ek ; so a 2 C; and hence, 3 2 ðD1 f1 ÞðaÞ ðDm f1 ÞðaÞ . 7 6 . . .. rank4 ... 5\n: . ðD1 fn ÞðaÞ ðDm fn ÞðaÞ So, the range space of f 0 ðaÞ : Rm ! Rn is contained in an ðn 1Þ-dimensional subspace, say V; of Rn : It follows that for every y 2 C \ Ek , f ðaÞ þ ðf 0 ðaÞÞðy aÞ contained in f ðaÞ þ V: Since for every y 2 C \ Ek , 1 jf ð yÞ ðf ðaÞ þ ðf 0 ðaÞÞðy aÞÞj\ nm3 M 2 ; N so all points of f ðC \ Ek Þ is at most ðnm3 MÞ N12 distance “above” or “below” the afﬁne hyperplane f ðaÞ þ V: Thus, f ðC \ Ek Þ contained in a “box” whose height is 1 2 nm3 M 2 ; N and ðn 1Þ-dimensional base area is n1 1 3=2 ; 2 nm M0 N and hence, n1 ! 3 1 1 3=2 2 nm M0 lð f ð C \ E k Þ Þ 2 nm M 2 N N 1 3ðnþ1Þ ¼ 2n nn m 2 ðM0 Þn1 M : N nþ1

460

6 Sard’s Theorem

It follows that X flðf ðC \ Ek ÞÞ : C \ Ek 6¼ ;; 1 k N m g 1 3ðnþ1Þ N m ! 0 as N ! 1; 2n nn m 2 ðM0 Þn1 M N nþ1

lðf ðC \ ðC1 Þ ÞÞ

because m n: Thus, lðf ðC \ ðC1 Þ ÞÞ ¼ 0: Similarly, lðf ðC \ ðC2 Þ ÞÞ ¼ 0; etc. Thus, lðf ðCÞÞ ¼ 0: h Lemma 6.24 Let m; n be any positive integers. Let G be a nonempty open subset of Rm : Let f : G ! Rn be a smooth function, and f ðf1 ; . . .fn Þ: Let C be the collection of all critical points of f ; that is, 9 3 ðD1 f1 Þð xÞ ðDm f1 Þð xÞ > = .. . 7 6 .. C x : x 2 G; and rank4 . . .. 5\n : > > ; : ðD1 fn Þð xÞ ðDm fn Þð xÞ 8 > <

2

Let for every i ¼ 1; 2; 3; . . .; Ci fx: x 2 G; and all 1st; 2nd; . . .; ith order partial derivatives of f at x are zerog: Then, 1. C4 C3 C2 C1 C G;. 2. each Ci and C are closed in G: Proof 1. This is clear from the deﬁnitions of Ci and C: 2. Since f : G ! Rn is smooth, each Dj fi : G ! R is continuous. Since each Dj fi : G ! R is continuous, and f0g is closed in R; each ðDj fi Þ1 ðf0gÞ is closed in G: It follows that \

1 Dj fi ðf0gÞ ¼ x : x 2 G; and 0 ¼ Dj fi ð xÞ for every j 2 f1; . . .; kg;

j 2 f1; . . .; k g i 2 f1; . . .; ng and for every i 2 f1; . . .; ngg ¼ C1 Þ

is closed in G; and hence, C1 is closed in G: Similarly, C2 ; C3 ; . . . are closed in G: Here,

6.3 Sard’s Theorem on Rn

8 > > <

461

9 3 ðD1 f1 Þð xÞ ðDk f1 ÞðxÞ > > = 6. 7 .. . 7\n . . C ¼ x : x 2 G; and rank6 . . . 4 5 > > > > : ; ðD1 fn Þð xÞ ðDk fn ÞðxÞ 8 9 2 3 ðDi1 f1 ÞðxÞ ðDin f1 ÞðxÞ > > > > < = \ 6. 7 .. . 6 7 x : x 2 G; and det4 .. . .. ¼ 0 ¼ : 5 > > > : ; i1 ; . . .; in 2 f1; . . .; k g; > ðDi1 fn ÞðxÞ ðDin fn ÞðxÞ i1 ; . . .; in are distinct 2

Now, since each Dj fi : G ! R is continuous, each 9 8 3 2 ðDi1 f1 Þð xÞ ðDin f1 Þð xÞ > > = < .. . 7 6 .. . x : x 2 G; and det4 . .. 5¼0 > > ; : ðDi1 fn Þð xÞ ðDin fn Þð xÞ is closed in G; and hence, 8 > < \

9 3 ðDi1 f1 Þð xÞ ðDin f1 Þð xÞ > = .. . 7 6 . .. x : x 2 G; and det4 ... 5¼0 > > ; i1 ; . . .; in 2 f1; . . .; k g; : ðDi1 fn Þð xÞ ðDin fn Þð xÞ i1 ; . . .; in are distinct 2

h

is closed in G: This shows that C is closed in G:

Lemma 6.25 Let m; n be any positive integers. For every k ¼ 1; . . .; n; let lk denote the k-dimensional measure over Rk : Let G be a nonempty open subset of Rm : Let f : G ! Rn be a smooth function, and f ðf1 ; . . .; fn Þ: Let C be the collection of all critical points of f ; that is, 9 8 3 2 ð D 1 f 1 Þ ð x Þ ð D m f 1 Þ ð xÞ > > = < .. . 7 6 .. . C x : x 2 G; and rank4 . .. 5\n : > > ; : ð D 1 f n Þ ð x Þ ð D m f n Þ ð xÞ Let, for every i ¼ 1; 2; 3; . . .; Ci fx : x 2 G; and all 1st; 2nd; . . .; ith order partial derivatives of f at x are zerog: Then,

ln f C½m ¼ 0; n

where ½mn denotes the greatest integer less than or equal to

m n:

462

6 Sard’s Theorem

Proof Since G is a nonemptyS open subset ofSRm ; there exists a countable collection 1 fUl g of open cubes such that 1 l¼1 Ul ¼ G, l¼1 ðUl Þ ¼ G; and side length of each Ul is strictly less than 1: It follows that 1 1 C½m ¼ C½m \ G ¼ C½m \ [ ðUl Þ ¼ [ C½m \ ðUl Þ ; n

n

l¼1

n

l¼1

n

and hence,

f C½m ¼ f n

1 [ l¼1

C½m \ ðUl Þ n

! ¼

1 [ f C½m \ ðUl Þ : n

l¼1

Thus, ! 1 1 [ X f C½m \ ðUl Þ ln f C½m \ ðUl Þ ; ¼ ln 0 ln f C½m n

n

l¼1

l¼1

n

and therefore, it sufﬁces to show that each ln f C½m \ ðUl Þ is zero: n

For this purpose, let us ﬁx a positive integer l: We have to show that ln f C½m \ ðUl Þ ¼ 0: n

Let us ﬁx any positive integer N [ 1: Let us subdivide the closed cube ðUl Þ into N m closed cubes E1 ; E2 ; . . .; EN m such that each Ei is of length 1 N ðside length of Ul Þ: For simplicity, let the side length of Ul be 1. Thus, ðUl Þ ¼ E1 [ E2 [ [ EN m : It follows that C½m \ ðUl Þ ¼ C½m \ ðE1 [ E2 [ [ EN m Þ n n ¼ C½m \ E1 [ C½m \ E2 [ [ C½m \ EN m n n n o [n ¼ C½m \ Ek : C½m \ Ek 6¼ ;; 1 k N m : n

n

Hence, [n o f C½m \ ðUl Þ ¼ f C½m \ Ek : C½m \ Ek 6¼ ;; 1 k N m n n n o [n ¼ f C½m \ Ek : C½m \ Ek 6¼ ;; 1 k N m ; n n

6.3 Sard’s Theorem on Rn

463

and hence, [n o f C½m \ Ek : C½m \ Ek 6¼ ;; 1 k N m 0 ln f C½m \ ðUl Þ ¼ ln n n n o Xn m : C½m \ Ek 6¼ ;; 1 k N : ln f C½m \ Ek n

n

Let us take any positive integer k satisfying 1 k N m ; and C½mn \ Ek 6¼ ;: Let all the ð½mn þ 1Þ th order partial derivatives of f be bounded in absolute value by a constant M on the compact set ðUl Þ : Now, by Taylor’s theorem, for every x ðx1 ; . . .; xm Þ 2 C½mn \ Ek ; and y ðy1 ; . . .; ym Þ 2 C½mn \ Ek ; jf ð xÞ f ð yÞj ¼ jðf1 ð xÞ; . . .; fn ð xÞÞ ðf1 ð yÞ; . . .; fn ð yÞÞj ¼ jðf1 ð xÞ f1 ð yÞ; . . .; fn ð xÞ fn ð yÞÞj jf1 ð xÞ f1 ð yÞj þ þ jfn ð xÞ fn ð yÞj ( ) 1 1 1 ¼ f1 ð xÞ f1 ð yÞ þ ð0Þ þ ð0Þ þ þ m ð0Þ þ 1! 2! ! n ( ) 1 1 1 þ fn ð xÞ fn ð yÞ þ ð0Þ þ ð0Þ þ þ m ð0Þ 1! 2! ! n m m m m 1 1 m m½ n þ1 jx yj½ n þ1 M þ þ m m½ n þ1 jx yj½ n þ1 M n ! n ! |ﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄ{zﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄ} n ! m m 1 ¼ n m m½ n þ1 jx yj½ n þ1 M ! n ! m 1 ½mnþ1 pﬃﬃﬃﬃ 1 ½ n þ1 m n m m M : N n !

It follows that f ðC½mn \ Ek Þ is contained in a closed ball of radius ! m 1 ½mnþ1 pﬃﬃﬃﬃ 1 ½ n þ1 n m m m M ; N n ! and hence, m pﬃﬃﬃﬃ 1 ½ n þ1 m 1 m M ln f C½m \ Ek 2 n m m½ n þ1 n N n !

!!!n :

464

6 Sard’s Theorem

Thus, 0 ln f C½m \ ðUl Þ N m n

! 0 as N ! 1;

!!!n ! m 1 ½mnþ1 pﬃﬃﬃﬃ 1 ½ n þ1 m 2 n m m M N n !

because m\

hmi n

þ1 n

that is;

m hmi \ þ1 : n n

Thus, ln f C½m \ ðUl Þ ¼ 0:

h

n

Lemma 6.26 Let m; n be any positive integers. For every k ¼ 1; . . .; n; let lk denote the k-dimensional measure over Rk : Let G be a nonempty open subset of Rm : Let f : G ! Rn be a smooth function, and f ðf1 ; . . .; fn Þ: Let C be the collection of all critical points of f ; that is, 9 3 ð D 1 f 1 Þ ð x Þ ð D m f 1 Þ ð xÞ > = .. . 7 6 C x : x 2 G; and rank4 ... . .. 5\n : > > ; : ð D 1 f n Þ ð x Þ ð D m f n Þ ð xÞ 8 > <

2

Then, ln ðf ðCÞÞ ¼ 0: Proof (Induction on m): By Lemma 6.23, the theorem is true for m ¼ 1: Case I: when m n. By Lemma 6.23, ln ðf ðCÞÞ ¼ 0: Case II: when n\m: For i ¼ 1; 2; 3; . . .; put Ci fx : x 2 G; and all partial derivatives of f at x upto ith order are zerog; and C0 C: By Lemma 6.24, C4 C3 C2 C1 C G: It follows that C ¼ ðC C1 Þ [ ðC1 C2 Þ [ ðC2 C3 Þ [ [ C½m1 C½m [ C½m ; n n n and hence, f ðCÞ ¼ f ðC C1 Þ [ ðC1 C2 Þ [ ðC2 C3 Þ [ [ C½m1 C½m [ C½m n n n ¼ f ðC C1 Þ [ f ðC1 C2 Þ [ f ðC2 C3 Þ [ [ f C½m1 C½m [ f C½m : n

n

n

6.3 Sard’s Theorem on Rn

465

It follows that 0 ln ðf ðCÞ ¼ ln ðf ðC C1 Þ [ f ðC1 C2 Þ [ f ðC2 C3 Þ [ [ f C½m1 C½m [ f C½m n

n

n

ln ðf ðC C1 ÞÞ þ ln ðf ðC1 C2 ÞÞ þ ln ðf ðC2 C3 ÞÞ þ þ ln f C½m1 C½m þ ln f C½m : n

n

n

Since, by Lemma 6.25, ln ðf ðC½mn ÞÞ ¼ 0; it sufﬁces to show that 1. 0 ¼ ln ðf ðC1 C2 ÞÞ ¼ ln ðf ðC2 C3 ÞÞ ¼ ¼ ln ðf ðC½mn1 C½mn ÞÞ; 2. ln ðf ðC C1 ÞÞ ¼ 0: For 1: For simplicity of discussion, here we shall assume 3 for n: Since Rm is a second countable space, there exists a countable basis B of Rm : SinceC2 is closed in the open set G; G C2 is open in Rm : Since C2 C1 G, C1 C2 G C2 G: Let us take any a 2 C1 C2 : It follows that a 2 C1 ; and a 62 C2 : Since a 2 C1 ; for every i 2 f1; . . .; mg; and for every j 2 f1; 2; 3g, ðDi fj ÞðaÞ ¼ 0: Further, since a 62 C2 ; there exist i1 ; i2 2 f1; . . .; mg; and j1 2 f1; 2; 3g such that ðDi1 i2 fj1 ÞðaÞ 6¼ 0: For simplicity, let ðD11 f1 ÞðaÞ 6¼ 0; that is, ðD1 ðD1 f1 ÞÞðaÞ 6¼ 0: Observe that D1 f1 : G ! R is a smooth function. Let us deﬁne a function ðF 1 ; F 2 ; . . .; F m Þ F : G C2 ! Rm as follows: For every x ðx1 ; x2 ; . . .; xm Þ 2 G C2 ;

F 1 x1 ; x2 ; . . .; xm ; F 2 x1 ; x2 ; . . .; xm ; . . .; F m x1 ; x2 ; . . .; xm ¼ F x1 ; x2 ; . . .; xm 1 ðD1 f1 Þ x ; . . .; xm ; x2 ; . . .; xm :

Clearly, FðC1 C2 Þ f0g Rm1 : (Reason: For every x ðx1 ; x2 ; . . .; xm Þ 2 C1 C2 ð G C2 GÞ; we have x 2 C1 ; and hence, FðxÞ ¼ ððD1 f1 ÞðxÞ; x2 ; . . .; xm Þ ¼ ð0; x2 ; . . .; xm Þ 2 f0g Rm1 :) Since ðD1 f1 Þ : G ! R is a smooth function, F : ðG C2 Þ ! Rm is smooth. Here, 2

ðD1 ðD1 f1 ÞÞðaÞ 6 0 detðF 0 ðaÞÞ ¼ det6 4 ... 0 ¼ ðD1 ðD1 f1 ÞÞðaÞ;

3 ðD2 ðD1 f1 ÞÞðaÞ ðDm ðD1 f1 ÞÞðaÞ 7 .. 0 1 7 . .. .. 5 . . 0

1

mm

which is nonzero. So, by Theorem 4.1, there exists an open neighborhood Ua of a such that Ua is contained in G C2 ; FðUa Þ is open in Rm ; and F has a smooth inverse on FðUa Þ: Thus, F : Ua ! FðUa Þ is a diffeomorphism, and F 1 : FðUa Þ !

466

6 Sard’s Theorem

Ua is a diffeomorphism. Since F 1 : FðUa Þ ! Ua is a diffeomorphism, Ua G C2 G; and f : G ! R3 is smooth, their composite f ðF 1 Þ : FðUa Þ ! R3 is a smooth function. Since Ua is an open neighborhood of a in Rm , there exists an open set Va 2 B such that a 2 Va ðVa Þ Ua , and ðVa Þ is compact. Since ðVa Þ Ua G C2 ; C2 \ ðVa Þ ¼ ;: Let us observe that ðC1 C2 Þ \ ðVa Þ is compact. (Reason: Since C1 G; C1 is closed in G; and G is open in Rm ; C1 [ ðGc Þ is closed in Rm : Since C1 [ ðGc Þ is closed in Rm ; and ðVa Þ ð Ua GÞ is compact, their intersection ðC1 [ ðGc ÞÞ \ ððVa Þ Þð¼ C1 \ ððVa Þ ÞÞ is compact, and hence, C1 \ ðVa Þ is compact. Since C1 \ ðVa Þ is compact, and ðC1 C2 Þ \ ðVa Þ ¼ ðC1 \ ðVa Þ Þ ðC2 \ ðVa Þ Þ ¼ ðC1 \ ðVa Þ Þ ; ¼ C1 \ ðVa Þ ; ðC1 C2 Þ \ ðVa Þ is compact). Since f : G ! R3 is a continuous function, and ðC1 C2 Þ \ ðVa Þ is compact, f ððC1 C2 Þ \ ðVa Þ Þ is compact. We shall try to show that l3 ðf ððC1 C2 Þ \ ðVa Þ ÞÞ ¼ 0: Now, since f ððC1 C2 Þ \ ðVa Þ Þ is compact, by Lemma 6.1, for every real c; l2 ðfðy1 ; y2 Þ : ðc; y1 ; y2 Þ 2 f ððC1 C2 Þ \ ðVa Þ ÞgÞ ¼ 0. For this purpose, let us ﬁx any real c. We have to show that l2 ðfðy1 ; y2 Þ : ðc; y1 ; y2 Þ 2 f ððC1 C2 Þ \ ðVa Þ ÞgÞ ¼ 0; that is, l2 ðfðf2 ðx1 ; . . .; xm Þ; f3 ðx1 ; . . .; xm ÞÞ : ðx1 ; . . .; xm Þ 2 ðC1 C2 Þ\ ðVa Þ ; andf1 ðx1 ; . . .; xm Þ ¼ cgÞ ¼ 0. Let us take any u ðu1 ; u2 ; . . .; um Þ 2 FðUa Þ: Since F 1 : FðUa Þ ! Ua is a diffeomorphism, the rank of ðF 1 Þ0 ðyÞ for every y 2 FðUa Þ is m. Now, since u 2 FðUa Þ; the rank of ðF 1 Þ0 ðuÞ is m. Since the rank of ðF 1 Þ0 ðuÞ is m, and

0 1 0 ðD1 f1 ; f2 ; f3 Þ F 1 ðuÞ ¼ ðD1 f1 ; f2 ; f3 Þ0 F 1 ðuÞ ð uÞ ; F

rank

0 ðD1 f1 ; f2 ; f3 Þ F 1 ðuÞ ¼ rank ðD1 f1 ; f2 ; f3 Þ0 F 1 ðuÞ :

Since u 2 FðUa Þ; and F is 1–1 over Ua ; there exists a unique x ðx1 ; x2 ; . . .; xm Þ 2 Ua such that FðxÞ ¼ u; and hence, F 1 ðuÞ ¼ x: Also, since ððD1 f1 Þðx1 ; x2 ; . . .; xm Þ; x2 ; . . .; xm Þ ¼ Fðx1 ; x2 ; . . .; xm Þ ¼ FðxÞ ¼ u ¼ ðu1 ; u2 ; . . .; um Þ; ðD1 f1 ÞðxÞ ¼ ðD1 f1 Þðx1 ; x2 ; . . .; xm Þ ¼ u1 ; x2 ¼ u2 ; . . .; xm ¼ um : Now, ðD1 f1 ; f2 ; f3 Þ F 1 u1 ; u2 ; . . .; um ¼ ðD1 f1 ; f2 ; f3 Þ F 1 u1 ; u2 ; . . .; um ¼ ðD1 f1 ; f2 ; f3 Þ F 1 ðuÞ ¼ ðD1 f1 ; f2 ; f3 Þð xÞ ¼ ððD1 f1 Þð xÞ; f2 ð xÞ; f3 ð xÞÞ ¼ u1 ; f2 ð xÞ; f3 ð xÞ ¼ u1 ; f2 F 1 ðuÞ ; f3 F 1 ðuÞ ¼ u1 ; f2 F 1 ðuÞ; f3 F 1 ðuÞ ¼ u1 ; f2 F 1 u1 ; u2 ; . . .; um ; f3 F 1 u1 ; u2 ; . . .; um : Hence, for every ðu1 ; u2 ; . . .; um Þ 2 FðUa Þ;

6.3 Sard’s Theorem on Rn

467

ðD1 f1 ; f2 ; f3 Þ F 1 u1 ; u2 ; . . .; um ¼ u1 ; f2 F 1 u1 ; u2 ; . . .; um ; f3 F 1 u1 ; u2 ; . . .; um :

It follows that for every u ðu1 ; u2 ; . . .; um Þ 2 FðUa Þ; 0 rank ðD1 f1 ; f2 ; f3 Þ0 F 1 ðuÞ ¼ rank ðD1 f1 ; f2 ; f3 Þ F 1 ðuÞ 2 3 1 0 0 . 6 7 ¼ rank4 ðD1 ðf2 ðF 1 ÞÞÞðuÞ ðD2 ðf2 ðF 1 ÞÞÞðuÞ . . ðDm ðf2 ðF 1 ÞÞÞðuÞ 5: ðD1 ðf3 ðF 1 ÞÞÞðuÞ ðD2 ðf3 ðF 1 ÞÞÞðuÞ

ðDm ðf3 ðF 1 ÞÞÞðuÞ

Now, let x ðx1 ; x2 ; . . .; xm Þ 2 ðC1 C2 Þ \ Ua : It follows that x 2 ðC1 C2 Þ \ Ua ðC1 C2 Þ; and hence, FðxÞ 2 FððC1 C2 Þ \ Ua Þ FðUa Þ: Thus,

3 ðD1 ðD1 f1 ÞÞð xÞ ðDm ðD1 f1 ÞÞð xÞ .. 7 6 .0 3 [ 1 rank4 0 5 0 0 33m 2 ðD 1 ðD 1 f 1 ÞÞð x Þ ð D m ðD 1 f 1 ÞÞð x Þ .. 7 6 . ðDm f2 Þð xÞ ¼ rank4 ðD1 f2 Þð xÞ 5 ðD1 f3 Þð xÞ ðDm f3 Þð xÞ 3m ¼ rank ðD1 f1 ; f2 ; f3 Þ0 ð xÞ ¼ rank ðD1 f1 ; f2 ; f3 Þ0 F 1 ðF ð xÞÞ 3 2 0 0 1 . 7 6 ¼ rank4 ðD1 ðf2 ðF 1 ÞÞÞðF ð xÞÞ ðD2 ðf2 ðF 1 ÞÞÞðF ð xÞÞ . . ðDm ðf2 ðF 1 ÞÞÞðF ð xÞÞ 5: ðD1 ðf3 ðF 1 ÞÞÞðF ð xÞÞ ðD2 ðf3 ðF 1 ÞÞÞðF ð xÞÞ ðDm ðf3 ðF 1 ÞÞÞðF ð xÞÞ 2

It follows that 3 2 1 0 0 7 6 0 ¼ det4 ðD1 ðf2 ðF 1 ÞÞÞðF ð xÞÞ ðD2 ðf2 ðF 1 ÞÞÞðF ð xÞÞ ðD3 ðf2 ðF 1 ÞÞÞðF ð xÞÞ 5 ðD1 ðf3 ðF 1 ÞÞÞðF ð xÞÞ ðD2 ðf3 ðF 1 ÞÞÞðF ð xÞÞ ðD3 ðf3 ðF 1 ÞÞÞðF ð xÞÞ ðD2 ðf2 ðF 1 ÞÞÞðF ð xÞÞ ðD3 ðf2 ðF 1 ÞÞÞðF ð xÞÞ ¼ det ; ðD2 ðf3 ðF 1 ÞÞÞðF ð xÞÞ ðD3 ðf3 ðF 1 ÞÞÞðF ð xÞÞ 3 2 0 0 1 7 6 0 ¼ det4 ðD1 ðf2 ðF 1 ÞÞÞðF ð xÞÞ ðD2 ðf2 ðF 1 ÞÞÞðF ð xÞÞ ðD4 ðf2 ðF 1 ÞÞÞðF ð xÞÞ 5 ðD1 ðf3 ðF 1 ÞÞÞðF ð xÞÞ ðD2 ðf3 ðF 1 ÞÞÞðF ð xÞÞ ðD4 ðf3 ðF 1 ÞÞÞðF ð xÞÞ ðD2 ðf2 ðF 1 ÞÞÞðF ð xÞÞ ðD4 ðf2 ðF 1 ÞÞÞðF ð xÞÞ ¼ det ; ðD2 ðf3 ðF 1 ÞÞÞðF ð xÞÞ ðD4 ðf3 ðF 1 ÞÞÞðF ð xÞÞ

etc. Hence, for every x ðx1 ; x2 ; . . .; xm Þ 2 ðC1 C2 Þ \ Ua ;

468

6 Sard’s Theorem

ðD2 ðf2 ðF 1 ÞÞÞðF ð xÞÞ ðD3 ðf2 ðF 1 ÞÞÞðF ð xÞÞ . . ðDm ðf2 ðF 1 ÞÞÞðF ðxÞÞ \2: . rank ðD2 ðf3 ðF 1 ÞÞÞðF ð xÞÞ ðD3 ðf3 ðF 1 ÞÞÞðF ð xÞÞ ðDm ðf3 ðF 1 ÞÞÞðF ðxÞÞ

Thus, for every u ðu1 ; u2 ; . . .; um Þ 2 FððC1 C2 Þ \ Ua Þ;

ðD2 ðf2 ðF 1 ÞÞÞðuÞ ðD3 ðf2 ðF 1 ÞÞÞðuÞ . . ðDm ðf2 ðF 1 ÞÞÞðuÞ . rank \2: ðD2 ðf3 ðF 1 ÞÞÞðuÞ ðD3 ðf3 ðF 1 ÞÞÞðuÞ ðDm ðf3 ðF 1 ÞÞÞðuÞ Since FðUa Þ is open in Rm ; fðy2 ; . . .; ym Þ : ð0; y2 ; . . .; ym Þ 2 FðUa Þg is open in Rm1 : Let us deﬁne a function fc : fðy2 ; . . .; ym Þ : ð0; y2 ; . . .; ym Þ 2 FðUa Þg ! R2 as follows: For every ðy2 ; . . .; ym Þ satisfying ð0; y2 ; . . .; ym Þ 2 FðUa Þ; fc y2 ; . . .; ym ¼ f2 F 1 0; y2 ; . . .; ym ; f3 F 1 0; y2 ; . . .; ym : Since ðy2 ; . . .; ym Þ 7! ð0; y2 ; . . .; ym Þ is smooth, and ðf2 ðF 1 ÞÞ : FðUa Þ ! R is smooth, their composite ðy2 ; . . .; ym Þ 7! ðf2 ðF 1 ÞÞð0; y2 ; . . .; ym Þ is smooth. Similarly, ðy2 ; . . .; ym Þ 7! ðf3 ðF 1 ÞÞð0; y2 ; . . .; ym Þ is smooth. It follows that fc : fðy2 ; . . .; ym Þ : ð0; y2 ; . . .; ym Þ 2 FðUa Þg ! R2 is smooth. For every u ðu2 ; . . .; um Þ satisfying ð0; u2 ; . . .; um Þ 2 FðUa Þ; we have ðD1 ðfc Þ1 Þðu2 ; . . .; um Þ ¼ ðD2 ðf2 ðF 1 ÞÞÞð0; u2 ; . . .; um Þ; ðD2 ðfc Þ1 Þðu2 ; . . .; um Þ ¼ ðD3 ðf2 ðF 1 ÞÞÞð0; u2 ; . . .; um Þ; etc., and hence, 0

ðfc Þ0 ðuÞ ¼ ðfc Þ u2 ; . . .; u ¼

m

" ¼

D1 ðfc Þ1 ðuÞ D1 ðfc Þ2 ðuÞ

# D2 ðfc Þ1 ðuÞ . . Dm1 ðfc Þ1 ðuÞ . D2 ðfc Þ2 ðuÞ Dm1 ðfc Þ2 ðuÞ

ðD2 ðf2 ðF 1 ÞÞÞð0; u2 ; . . .; um Þ ðD3 ðf2 ðF 1 ÞÞÞð0; u2 ; . . .; um Þ ðD2 ðf3 ðF 1 ÞÞÞð0; u2 ; . . .; um Þ ðD3 ðf3 ðF 1 ÞÞÞð0; u2 ; . . .; um Þ . . ðDm ðf2 ðF 1 ÞÞÞð0; u2 ; . . .; um Þ . : ðDm ðf3 ðF 1 ÞÞÞð0; u2 ; . . .; um Þ

Clearly, every point of fðy2 ; . . .; ym Þ : ð0; y2 ; . . .; ym Þ 2 FððC1 C2 Þ \ Ua Þg is a critical point of fc : (Reason: Take any u ðu2 ; . . .; um Þ 2 fðy2 ; . . .; ym Þ : ð0; y2 ; . . .; ym Þ 2 FððC1 C2 Þ \ Ua Þg: It follows that ð0; y2 ; . . .; ym Þ 2 FððC1 C2 Þ \ Ua Þ FðUa Þ; and 0 ðD2 ðf2 ðF 1 ÞÞÞð0; u2 ; . . .; um Þ ðD3 ðf2 ðF 1 ÞÞÞð0; u2 ; . . .; um Þ rank ðfc Þ ðuÞ ¼ rank ðD2 ðf3 ðF 1 ÞÞÞð0; u2 ; . . .; um Þ ðD3 ðf3 ðF 1 ÞÞÞð0; u2 ; . . .; um Þ . . ðDm ðf2 ðF 1 ÞÞÞð0; u2 ; . . .; um Þ . \2: ðDm ðf3 ðF 1 ÞÞÞð0; u2 ; . . .; um Þ

6.3 Sard’s Theorem on Rn

469

Now, by induction hypothesis, l2 ðfc ðfðy2 ; . . .; ym Þ : ð0; y2 ; . . .; ym Þ 2 FððC1 C2 Þ \Ua ÞgÞÞ ¼ 0: Now, since ðVa Þ Ua ; 0 l2 f2 F 1 0; y2 ; . . .; ym ; f3 F 1 0; y2 ; . . .; ym : 0; y2 ; . . .; ym 2 F ððC1 C2 Þ \ ðVa Þ Þ l2 f2 F 1 0; y2 ; . . .; ym ; f3 F 1 0; y2 ; . . .; ym : 0; y2 ; . . .; ym 2 F ððC1 C2 Þ \ Ua Þ 2 ¼ l2 fc y ; . . .; ym : 0; y2 ; . . .; ym 2 F ððC1 C2 Þ \ Ua Þ ¼ 0; and hence, l2

f2 F 1 0; y2 ; . . .; ym ; f3 F 1 0; y2 ; . . .; ym : 0; y2 ; . . .; ym 2 F ððC1 C2 Þ \ ðVa Þ Þ ¼ 0:

It sufﬁces to show that 1 f2 x ; . . .; xm ; f3 x1 ; . . .; xm : x1 ; . . .; xm 2 ðC1 C2 Þ \ ðVa Þ ; and ðD1 f1 Þ x1 ; . . .; xm ¼ c f2 F 1 0; y2 ; . . .; ym ; f3 F 1 0; y2 ; . . .; ym : 0; y2 ; . . .; ym 2 F ððC1 C2 Þ \ ðVa Þ Þg: Let us take any ðf2 ðx1 ; . . .; xm Þ; f3 ðx1 ; . . .; xm ÞÞ 2 LHS; where ðx1 ; . . .; xm Þ 2 ðC1 C2 Þ \ ðVa Þ ; and f1 ðx1 ; . . .; xm Þ ¼ c. Now, since ðx1 ; . . .; xm Þ 2 ðC1 C2 Þ\ ðVa Þ ; 2 0; x ; . . .; xm ¼ ðD1 f1 Þ x1 ; . . .; xm ; x2 ; . . .; xm ¼ F x1 ; . . .; xm 2 F ððC1 C2 Þ \ ðVa Þ Þ; and hence, F 1 ð0; x2 ; . . .; xm Þ ¼ ðx1 ; . . .; xm Þ: It follows that 1 f2 x ; . . .; xm ; f3 x1 ; . . .; xm ¼ f2 F 1 0; x2 ; . . .; xm ; f3 F 1 0; x2 ; . . .; xm ¼ f2 F 1 0; x2 ; . . .; xm ; f3 F 1 0; x2 ; . . .; xm 2 RHS: Thus, LHS RHS: We have shown that for every a 2 C1 C2 ; there exists an open neighborhood Va 2 B of a such that l3 ðf ððC1 C2 Þ \ ðVa Þ ÞÞ ¼ 0; and a 2 Va ðVa Þ S G: Here, [fðC1 C2 Þ \ ðVa Þ : a 2 ðC1 SC2 Þg ¼ ðC1 C2 Þ; so f ðC1 C2 Þ ¼ f ð fðC1 C2 Þ \ ðVa Þ : a 2 ðC1 C2 ÞgÞ ¼ ff ððC1 C2 Þ \ ðVa Þ Þ : a 2 ðC1 C2 Þg: Since each Va 2 B; and B is a countable collection, ff ððC1 C2 Þ \ ðVa Þ Þ : a 2 S ðC1 C2 Þg is a countable collection, and hence, l3 ðf ðC1 C2 ÞÞ ¼ l3 ð ff ððC1 C2 Þ\

470

6 Sard’s Theorem

ðVa Þ Þ : a 2 ðC1 C2 ÞgÞ

P countable

Thus, l3 ðf ðC1 C2 ÞÞ ¼ 0: Similarly,

l3 ðf ððC1 C2 Þ \ ðVa Þ ÞÞ ¼

P countable

0 ¼ 0:

0 ¼ l3 ðf ðC2 C3 ÞÞ ¼ ¼ l3 f C½m1 C½m : n

n

For 2: For simplicity of discussion, here, we shall assume 3 for n. Since Rm is a second countable space, there exists a countable basis B of Rm . Since C1 is closed in the open set G; G C1 is open in Rm : Since C1 C G; C C1 G C1 G: Now, since f : G ! R3 is a smooth function, f jðGC1 Þ : ðG C1 Þ ! R3 is smooth. Clearly, the collection of all critical points of f jðGC1 Þ is C C1 ; that is, 8 <

9 3 ðD1 f1 Þð xÞ . ðDm f1 Þð xÞ = C C1 ¼ x : x 2 G C1 ; and rank4 ðD1 f2 Þð xÞ . . ðDm f2 Þð xÞ 5\3 : : ; ðD1 f3 Þð xÞ ðDm f3 Þð xÞ 2

Let us take any a 2 C C1 : Since a 2 C C1 G; a 2 G: Now, since a 62 C1 ¼ fx : x 2 G; and all first order partial derivatives of F at x are zerog; there exists j 2 f1; 2; . . .; mg; and i 2 f1; 2; 3g such that ðDj fi ÞðaÞ 6¼ 0: For simplicity, let i ¼ j ¼ 1: Thus, ðD1 f1 ÞðaÞ 6¼ 0: Let us deﬁne a function ðF 1 ; F 2 ; . . .; F m Þ F : ðG C1 Þ ! Rm as follows: For every x ðx1 ; x2 ; . . .; xm Þ 2 G C1 ;

F 1 x1 ; x2 ; . . .; xm ; F 2 x1 ; x2 ; . . .; xm ; . . .; F k x1 ; x2 ; . . .; xm ¼ F x1 ; x2 ; . . .; xm f1 x1 ; . . .; xm ; x2 ; . . .; xm :

Since f : G ! R3 is a smooth function, f1 : G ! R is a smooth function, and hence, F : ðG C1 Þ ! Rk is smooth. Here, 2

ðD1 f1 ÞðAÞ 6 0 detðF 0 ðAÞÞ ¼ det6 4 ... 0

3 ðD2 f1 ÞðAÞ ðDm f1 ÞðAÞ 7 1 .. 0 7 . .. .. 5 . . 0

1

¼ ðD1 f1 ÞðAÞ; mm

which is nonzero. So, by Theorem 4.1, there exists an open neighborhood Ua of a such that Ua is contained in ðG C1 Þ; FðUa Þ is open in Rm ; and F has a smooth inverse on FðUa Þ: Thus, F : Ua ! FðUa Þ is a diffeomorphism, and F 1 : FðUa Þ ! Ua is a diffeomorphism. Since F 1 : FðUa Þ ! Ua is a diffeomorphism, Ua ðG C1 Þ G; and f : G ! R3 is smooth, their composite f ðF 1 Þ : FðUa Þ ! R3 is a smooth function. Since Ua is an open neighborhood of a in Rm ; there exists an open set Va 2 B such that a 2 Va ðVa Þ Ua ðG C1 Þ; and ðVa Þ is compact. Let us observe that ðC C1 Þ \ ðVa Þ is compact.

6.3 Sard’s Theorem on Rn

471

(Reason: Since C G; C is closed in G; and G is open in Rm ; C [ ðGc Þ is closed in Rm : Since C [ ðGc Þ is closed in Rm ; and ðVa Þ ð Ua ðG C1 Þ GÞ is compact, their intersection ðC [ ðGc ÞÞ \ ððVa Þ Þð¼ C \ ððVa Þ ÞÞ is compact, and hence, C \ðVa Þ is compact. Since C \ðVa Þ is compact, and ðC C1 Þ \ ðVa Þ ¼ ðC \ ðVa Þ Þ ðC1 \ ðVa Þ Þ ¼ ðC \ ðVa Þ Þ ; ¼ C \ ðVa Þ ; ðC C1 Þ \ðVa Þ is compact.) Since f : G ! R3 is a continuous function, and ðC C1 Þ \ ðVa Þ ð ðVa Þ GÞ is compact, f ððC C1 Þ \ðVa Þ Þ is compact. We shall try to show that l3 ðf ððC C1 Þ \ ðVa Þ ÞÞ ¼ 0: Now, since f ððC C1 Þ \ ðVa Þ Þ is compact, by Lemma 6.1, it sufﬁces to show that for every real c; l2 ðfðy1 ; y2 Þ : ðc; y1 ; y2 Þ 2 f ððC C1 Þ \ ðVa Þ ÞgÞ ¼ 0. For this purpose, let us ﬁx any real c. We have to show that l2 ðfðy1 ; y2 Þ : ðc; y1 ; y2 Þ 2 f ððC C1 Þ \ ðVa Þ ÞgÞ ¼ 0, that is, l2 ðfðf2 ðx1 ; . . .; xm Þ; f3 ðx1 ; . . .; xm ÞÞ : ðx1 ; . . .; xm Þ 2 ðC C1 Þ \ ðVa Þ ; and f1 ðx1 ; . . .; xm Þ ¼ cgÞ ¼ 0: Let us take any u ðu1 ; u2 ; . . .; um Þ 2 FðUa Þ: Since F 1 : FðUa Þ ! Ua is a diffeomorphism, the rank of ðF 1 Þ0 ðyÞ for every y 2 FðUa Þ is m. Now, since u 2 FðUa Þ; the rank of ðF 1 Þ0 ðuÞ is m. Since the rank of ðF 1 Þ0 ðuÞ is m, and

0 1 0 F f F 1 ðuÞ ¼ f 0 F 1 ðuÞ ð uÞ ;

rank

0 f F 1 ðuÞ ¼ rank f 0 F 1 ðuÞ :

Since u 2 FðUa Þ; and F is 1–1 over Ua ; there exists a unique x ðx1 ; x2 ; . . .; xm Þ 2 Ua such that FðxÞ ¼ u; and hence, F 1 ðuÞ ¼ x: Also, since ðf1 ðx1 ; x2 ; . . .; xm Þ; x2 ; . . .; xm Þ ¼ Fðx1 ; x2 ; . . .; xm Þ ¼ FðxÞ ¼ u ¼ ðu1 ; u2 ; . . .; um Þ; f1 ðxÞ ¼ f1 ðx1 ; x2 ; . . .; xm Þ ¼ u1 ; x2 ¼ u2 ; . . .; xm ¼ um : Now, ðf ðF 1 ÞÞðu1 ; u2 ; . . .; um Þ ¼ f ðF 1 ðu1 ; u2 ; . . .; um ÞÞ ¼ f ðF 1 ðuÞÞ ¼ f ðxÞ ¼ ðf1 ðxÞ; f2 ðxÞ; f3 ðxÞÞ ¼ ðu1 ; f2 ðxÞ; f3 ðxÞÞ ¼ ðu1 ; f2 ðF 1 ðuÞÞ; f3 ðF 1 ðuÞÞÞ ¼ ðu1 ; ðf2 ðF 1 ÞÞðuÞ; ðf3 ðF 1 ÞÞðuÞÞ ¼ 1 1 ðu ; ðf2 ðF ÞÞðu1 ; u2 ; . . .; um Þ; ðf3 ðF 1 ÞÞðu1 ; u2 ; . . .; um ÞÞ: Hence, for every ðu1 ; u2 ; . . .; um Þ 2 FðUa Þ;

f F 1 u1 ; u2 ; . . .; um ¼ u1 ; f2 F 1 u1 ; u2 ; . . .; um ; f3 F 1 u1 ; u2 ; . . .; um :

It follows that for every u ðu1 ; u2 ; . . .; um Þ 2 FðUa Þ; rank f 0 F 1 ðuÞ ¼ rank f 2 1 6 ¼ rank4 ðD1 ðf2 ðF 1 ÞÞÞðuÞ ðD1 ðf3 ðF 1 ÞÞÞðuÞ

0

F 1 ðuÞ 0 . ðD2 ðf2 ðF 1 ÞÞÞðuÞ . . ðD2 ðf3 ðF 1 ÞÞÞðuÞ

3 0 7 ðDm ðf2 ðF 1 ÞÞÞðuÞ 5 ðDm ðf3 ðF 1 ÞÞÞðuÞ

472

6 Sard’s Theorem

Now, let x ðx1 ; x2 ; . . .; xm Þ 2 ðC C1 Þ \ Ua : It follows that x 2 ðC C1 Þ \ Ua ðC C1 Þ; and hence, FðxÞ 2 FððC C1 Þ \ Ua Þ FðUa Þ: Thus, 3 ðD1 f1 Þð xÞ ðDm f1 Þð xÞ . 7 6 3 [ rank4 ðD1 f2 Þð xÞ . . ðDm f2 Þð xÞ 5 ¼ rankðf 0 ð xÞÞ ¼ rank f 0 F 1 ðF ð xÞÞ ðD1 f3 Þð xÞ ðDm f3 Þð xÞ 3m 2 3 0 1 0 . 6 7 ¼ rank4 ðD1 ðf2 ðF 1 ÞÞÞðF ð xÞÞ ðD2 ðf2 ðF 1 ÞÞÞðF ð xÞÞ . . ðDm ðf2 ðF 1 ÞÞÞðF ð xÞÞ 5: ðD1 ðf3 ðF 1 ÞÞÞðF ð xÞÞ ðD2 ðf3 ðF 1 ÞÞÞðF ð xÞÞ ðDm ðf3 ðF 1 ÞÞÞðF ð xÞÞ 2

It follows that 2

1

0

0

3

7 6 0 ¼ det4 ðD1 ðf2 ðF 1 ÞÞÞðF ð xÞÞ ðD2 ðf2 ðF 1 ÞÞÞðF ð xÞÞ ðD3 ðf2 ðF 1 ÞÞÞðF ð xÞÞ 5 ðD1 ðf3 ðF 1 ÞÞÞðF ð xÞÞ ðD2 ðf3 ðF 1 ÞÞÞðF ð xÞÞ ðD3 ðf3 ðF 1 ÞÞÞðF ð xÞÞ ðD2 ðf2 ðF 1 ÞÞÞðF ð xÞÞ ðD3 ðf2 ðF 1 ÞÞÞðF ð xÞÞ ¼ det ; ðD2 ðf3 ðF 1 ÞÞÞðF ð xÞÞ ðD3 ðf3 ðF 1 ÞÞÞðF ð xÞÞ 3 2 0 0 1 7 6 0 ¼ det4 ðD1 ðf2 ðF 1 ÞÞÞðF ð xÞÞ ðD2 ðf2 ðF 1 ÞÞÞðF ð xÞÞ ðD4 ðf2 ðF 1 ÞÞÞðF ð xÞÞ 5 ðD1 ðf3 ðF 1 ÞÞÞðF ð xÞÞ ðD2 ðf3 ðF 1 ÞÞÞðF ð xÞÞ ðD4 ðf3 ðF 1 ÞÞÞðF ð xÞÞ ðD2 ðf2 ðF 1 ÞÞÞðF ð xÞÞ ðD4 ðf2 ðF 1 ÞÞÞðF ð xÞÞ ¼ det ; ðD2 ðf3 ðF 1 ÞÞÞðF ð xÞÞ ðD4 ðf3 ðF 1 ÞÞÞðF ð xÞÞ

etc. Hence, for every x ðx1 ; x2 ; . . .; xm Þ 2 ðC C1 Þ \ Ua ; rank

ðD2 ðf2 ðF 1 ÞÞÞðF ð xÞÞ ðD3 ðf2 ðF 1 ÞÞÞðF ð xÞÞ . . ðDm ðf2 ðF 1 ÞÞÞðF ðxÞÞ . \2: ðD2 ðf3 ðF 1 ÞÞÞðF ð xÞÞ ðD3 ðf3 ðF 1 ÞÞÞðF ð xÞÞ ðDm ðf3 ðF 1 ÞÞÞðF ðxÞÞ

Thus, for every u ðu1 ; u2 ; . . .; um Þ 2 FððC C1 Þ \ Ua Þ; rank

ðD2 ðf2 ðF 1 ÞÞÞðuÞ ðD3 ðf2 ðF 1 ÞÞÞðuÞ . . ðDm ðf2 ðF 1 ÞÞÞðuÞ . \2: ðD2 ðf3 ðF 1 ÞÞÞðuÞ ðD3 ðf3 ðF 1 ÞÞÞðuÞ ðDm ðf3 ðF 1 ÞÞÞðuÞ

Since FðUa Þ is open in Rk ; fðy2 ; . . .; ym Þ : ðc; y2 ; . . .; ym Þ 2 FðUa Þg is open in R : Let us deﬁne a function fc : fðy2 ; . . .; ym Þ : ðc; y2 ; . . .; ym Þ 2 FðUa Þg ! R2 as follows: For every ðy2 ; . . .; ym Þ satisfying ðc; y2 ; . . .; ym Þ 2 FðUa Þ; m1

6.3 Sard’s Theorem on Rn

473

fc y2 ; . . .; ym ¼ f2 F 1 c; y2 ; . . .; ym ; f3 F 1 c; y2 ; . . .; ym : Since ðy2 ; . . .; ym Þ 7! ðc; y2 ; . . .; ym Þ is smooth, and ðf2 ðF 1 ÞÞ : FðUa Þ ! R is smooth, their composite ðy2 ; . . .; ym Þ 7! ðf2 ðF 1 ÞÞðc; y2 ; . . .; ym Þ is smooth. Similarly, ðy2 ; . . .; ym Þ 7! ðf3 ðF 1 ÞÞðc; y2 ; . . .; ym Þ is smooth. It follows that fc : fðy2 ; . . .; ym Þ : ðc; y2 ; . . .; ym Þ 2 FðUa Þg ! R2 is smooth. For every u ðu2 ; . . .; um Þ satisfying ðc; u2 ; . . .; um Þ 2 FðUa Þ; ðD1 ðfc Þ1 Þðu2 ; . . .; um Þ ¼ ðD2 ðf2 ðF 1 ÞÞÞðc; u2 ; . . .; um Þ; ðD2 ðfc Þ1 Þðu2 ; . . .; um Þ ¼ ðD3 ðf2 ðF 1 ÞÞÞðc; u2 ; . . .; um Þ; etc., and hence, " # D1 ðfc Þ1 ðuÞ D2 ðfc Þ1 ðuÞ . . Dk1 ðfc Þ1 ðuÞ 0 0 2 m . ðfc Þ ðuÞ ¼ ðfc Þ u ; . . .; u ¼ D1 ðfc Þ2 ðuÞ D2 ðfc Þ2 ðuÞ Dk1 ðfc Þ2 ðuÞ ðD2 ðf2 ðF 1 ÞÞÞðc; u2 ; . . .; um Þ ðD3 ðf2 ðF 1 ÞÞÞðc; u2 ; . . .; um Þ ¼ ðD2 ðf3 ðF 1 ÞÞÞðc; u2 ; . . .; um Þ ðD3 ðf3 ðF 1 ÞÞÞðc; u2 ; . . .; um Þ . . ðDk ðf2 ðF 1 ÞÞÞðc; u2 ; . . .; um Þ . : ðDk ðf3 ðF 1 ÞÞÞðc; u2 ; . . .; um Þ Clearly, every point of fðy2 ; . . .; ym Þ : ðc; y2 ; . . .; ym Þ 2 FðC \ Ua Þg is a critical point of fc . (Reason: Take any u ðu2 ; . . .; um Þ 2 fðy2 ; . . .; ym Þ : ð0; y2 ; . . .; ym Þ 2 FððC1 C2 Þ \ Ua Þg. It follows that ð0; y2 ; . . .; ym Þ 2 FððC C1 Þ \ Ua Þ FðUa Þ; and rank ðfc Þ0 ðuÞ ðD2 ðf2 ðF 1 ÞÞÞðc; u2 ; . . .; um Þ ðD3 ðf2 ðF 1 ÞÞÞðc; u2 ; . . .; um Þ ¼ rank ðD2 ðf3 ðF 1 ÞÞÞðc; u2 ; . . .; um Þ ðD3 ðf3 ðF 1 ÞÞÞðc; u2 ; . . .; um Þ . . ðDk ðf2 ðF 1 ÞÞÞðc; u2 ; . . .; um Þ . \2:Þ ðDk ðf3 ðF 1 ÞÞÞðc; u2 ; . . .; um Þ Now, by induction hypothesis, l2 ðfc ðfðy2 ; . . .; ym Þ : ðc; y2 ; . . .; ym Þ 2 FððC C1 Þ \ Ua ÞgÞÞ ¼ 0: Since ðVa Þ Ua ; f2 F 1 c; y2 ; . . .; ym ; f3 F 1 c; y2 ; . . .; ym 0 l2 : c; y2 ; . . .; ym 2 F ððC C1 Þ \ ðVa Þ Þ l2 f2 F 1 c; y2 ; . . .; ym ; f3 F 1 c; y2 ; . . .; ym : c; y2 ; . . .; ym 2 F ðC \ Ua Þ 2 ¼ l2 fc y ; . . .; ym : c; y2 ; . . .; ym 2 F ððC C1 Þ \ Ua Þ ¼ 0; and hence,

474

6 Sard’s Theorem

l2

f2 F 1 c; y2 ; . . .; ym ; f3 F 1 c; y2 ; . . .; ym : c; y2 ; . . .; ym 2 F ððC C1 Þ \ ðVa Þ Þ ¼ 0:

It sufﬁces to show that 1 f2 x ; . . .; xm ; f3 x1 ; . . .; xm : x1 ; . . .; xm 2 ðC C1 Þ \ ðVa Þ ; and f1 x1 ; . . .; xm ¼ c f2 F 1 c; y2 ; . . .; ym ; f3 F 1 c; y2 ; . . .; ym : c; y2 ; . . .; ym 2 F ððC C1 Þ \ ðVa Þ Þ : Let us take any ðf2 ðx1 ; . . .; xm Þ; f3 ðx1 ; . . .; xm ÞÞ 2 LHS; where ðx1 ; . . .; xm Þ 2 ðC C1 Þ \ ðVa Þ ; and f1 ðx1 ; . . .; xm Þ ¼ c: Now, since ðx1 ; . . .; xm Þ 2 ðC C1 Þ \ðVa Þ ;

c; x2 ; . . .; xm ¼ f1 x1 ; . . .; xm ; x2 ; . . .; xm ¼ F x1 ; . . .; xm 2 F ððC C1 Þ \ ðVa Þ Þ;

and hence F 1 ðc; x2 ; . . .; xm Þ ¼ ðx1 ; . . .; xm Þ: It follows that 1 f2 x ; . . .; xm ; f3 x1 ; . . .; xm ¼ f2 F 1 c; x2 ; . . .; xm ; f3 F 1 c; x2 ; . . .; xm ¼ f2 F 1 c; x2 ; . . .; xm ; f3 F 1 c; x2 ; . . .; xm 2 RHS: Hence, LHS RHS: Thus, we have shown that for every a 2 ðC C1 Þ; there exists an open neighborhood Va 2 BSof a such that l3 ðf ððC C1 Þ \ ðVa Þ ÞÞ ¼ 0; and a 2 Va ðVa Þ G: SHere, fðC C1 Þ \ ðVa Þ : a 2 ðC C S1 Þg ¼ ðC C1 Þ; so f ððC C1 ÞÞ ¼ f ð fðC C1 Þ \ ðVa Þ : a 2 ðC C1 ÞgÞ ¼ ff ððC C1 Þ \ ðVa Þ Þ : a 2 ðC C1 Þg: Since each Va 2 B; and B is a countable collection, ff ððC C1 Þ \ðVa Þ Þ : a 2 ðC C1 Þg is a countable collection, and hence, [ l3 ðf ðC C1 ÞÞ ¼ l3 ff ððC C1 Þ \ ðVa Þ Þ : a 2 ðC C1 Þg X X l3 ðf ððC C1 Þ \ðVa Þ ÞÞ ¼ 0 ¼ 0: countable

Thus, l3 ðf ðC C1 ÞÞ ¼ 0:

countable

h

6.4 Sard’s Theorem on Manifolds

475

6.4 Sard’s Theorem on Manifolds Theorem 6.27 Let m; n be positive integers such that m\n: Let M be an m-dimensional smooth manifold, and N be an n-dimensional smooth manifold. Let f : M ! N be a smooth mapping. Then, f ðMÞ has measure zero in N: Proof For simplicity, let n ¼ m þ 1: Let us deﬁne a mapping F : M R ! N as follows: For every p 2 M; and t 2 R; Fðp; tÞ f ðpÞ: Clearly, F is smooth. (Reason: We know that the mapping ðp; tÞ 7! p from product manifold M R to M is smooth. Since ðp; tÞ 7! p from M R to M is smooth, and f : M ! N is smooth, their composite ðp; tÞ 7! f ðpÞð¼ Fðp; tÞÞ from M R to N is smooth, and hence, F is smooth). Observe that fðp; 0Þ : p 2 Mgð M RÞ is of measure zero in M R: (Reason: Let ðU R; ðu IdR ÞÞ be an admissible coordinate chart of the product manifold M R; where ðU; uÞ is an admissible coordinate chart of M; and ðR; IdR Þ is an admissible coordinate chart of R: We know that M R is an ðm þ 1Þ-dimensional smooth manifold. We have to show that lmþ1 ððu IdR Þðfðp; 0Þ : p 2 Mg \ ðU RÞÞÞ ¼ 0; where lmþ1 denotes the ðm þ 1Þ-dimensional Lebesgue measure over Rmþ1 : Here, lmþ1 ððu IdR Þðfðp; 0Þ : p 2 M g \ ðU RÞÞÞ ¼ lmþ1 ððu IdR Þðfðp; 0Þ : p 2 U gÞÞlmþ1 ððuð pÞ IdR ð0ÞÞ : p 2 U Þ ¼ lmþ1 ððuð pÞ; 0Þ : p 2 U Þ lmþ1 ðRm f0gÞ ¼ 0Þ: Since F : M R ! N is smooth, and fðp; 0Þ : p 2 Mgð M RÞ is of measure zero in M R, by Lemma 6.16, Fðfðp; 0Þ : p 2 MgÞð¼ fFðp; 0Þ : p 2 Mg ¼ ff ðpÞ : p 2 Mg ¼ f ðMÞÞ has measure zero in N; and hence, f ðMÞ has measure zero in N: h Theorem 6.28 Let M be an m-dimensional smooth manifold, and N be an ndimensional smooth manifold. Let f : M ! N be a smooth mapping. Let C be the collection of all critical points of f . Then, f ðCÞ has measure zero in N. Proof Let us take any admissible coordinate chart ðV; wÞ for N: We have to show that ln ðwððf ðC ÞÞ \ V ÞÞ ¼ 0; where ln denotes the n-dimensional Lebesgue measure over Rn : Since M is an mdimensional smooth manifold, by Lemma 4.49, there exists a countable collection fðU1 ; u1 Þ; ðU2 ; u2 Þ; ðU3 ; u3 Þ; . . .g of admissible coordinate charts of M such that fU1 ; U2 ; U3 ; . . .g is a basis of M: Here, C M ¼ U1 [ U2 [ U3 [ ; so

476

6 Sard’s Theorem

0 ln ðwððf ðCÞÞ \ V ÞÞ ¼ ln ðwððf ððC \ U1 Þ [ ðC \ U2 Þ [ ðC \ U3 Þ [ ÞÞ \ V ÞÞ ¼ ln ðwðððf ðC \ U1 ÞÞ [ ðf ðC \ U2 ÞÞ [ ðf ðC \ U3 ÞÞ [ Þ \ V ÞÞ ¼ ln ðwðððf ðC \ U1 ÞÞ \ V Þ [ ððf ðC \ U2 ÞÞ \ V Þ [ ððf ðC \ U3 ÞÞ \ V Þ [ ÞÞ ¼ ln ðwððf ðC \ U1 ÞÞ \ V Þ [ wððf ðC \ U2 ÞÞ \ V Þ [ wððf ðC \ U3 ÞÞ \ V Þ [ Þ ln ðwððf ðC \ U1 ÞÞ \ V ÞÞ þ ln ðwððf ðC \ U2 ÞÞ \ V ÞÞ þ ln ðwððf ðC \ U3 ÞÞ \ V ÞÞ þ : Since 0 ln ðwððf ðCÞÞ \ V ÞÞ ln ðwððf ðC \ U1 ÞÞ \ V ÞÞ þ ln ðwððf ðC \ U2 ÞÞ \ V ÞÞ þ ln ðwððf ðC \ U3 ÞÞ \ V ÞÞ þ ; it sufﬁces to show that for each k ¼ 1; 2; . . .; ln ðwððf ðC \ Uk ÞÞ \ V ÞÞ ¼ 0: Let us try to show that ln ðwððf ðC \ U1 ÞÞ \ V ÞÞ ¼ 0: Since u1 is a diffeomorphism from open set U1 onto open set u1 ðU1 Þ; and w is a diffeomorphism from open set V onto the open set wðVÞ; the collection of all critical points of f in U1 is equal to ðu1 Þ1 ðthe collection of all critical points of ðw f

ðu1 Þ1 Þin u1 ðU1 ÞÞ. Since f : M ! N is a smooth mapping, w f ðu1 Þ1 is smooth. Since w f ðu1 Þ1 is smooth, by Lemma 6.26, 0 ¼ ln ððw f ðu1 Þ1 Þ ðthe

collection of all critical points of ðw f ðu1 Þ1 Þ in u1 ðU1 ÞÞÞ ¼ ln ððw f Þ ððu1 Þ1 ðthe collection of all critical points of ðw f ðu1 Þ1 Þin u1 ðU1 ÞÞÞÞ ¼ ln ððw f Þðcollection of all critical points of f in U1 ÞÞ ¼ ln ðwðf ðcollection of all critical points of f in U1 ÞÞÞ ¼ ln ðwððf ðC \ U1 ÞÞ \ VÞÞ: Thus, ln ðwððf ðC\ U1 ÞÞ \ VÞÞ ¼ 0: Similarly, ln ðwððf ðC \ U2 ÞÞ \ V ÞÞ ¼ 0; ln ðwððf ðC \ U3 ÞÞ \ V ÞÞ ¼ 0; etc. Theorem 6.28 is known as Sard’s theorem.

h

Chapter 7

Whitney Embedding Theorem

This is the smallest chapter of this book, because it contains only two theorems which are due to Whitney. These theorems have three serious reasons to study. Firstly, in its proof the celebrated Sard’s theorem got an application. Secondly, the statement of Whitney embedding theorem was contrary to the common belief that a smooth manifold may not have any ambient space. Thirdly, in its proof, Whitney used almost all tools of smooth manifolds developed at that time. Fortunately, in this chapter we have all the prerequisite for its proof in the special case of compact smooth manifolds. For the general case, which is more difﬁcult, one can ﬁnd its proof somewhere else.

7.1 Compact Whitney Embedding in RN Lemma 7.1 Let M be an m-dimensional smooth manifold. Let M be compact. Then, there exists a positive integer N such that M is embedded in the Euclidean space RN : Proof Since M is an open cover of M, by Lemma 4.52, there exists a countable collection ðU1 ; u1 Þ; ðU2 ; u2 Þ; ðU3 ; u3 Þ; . . . of admissible coordinate charts of M such that 1. 2. 3. 4.

U1 ; U2 ; U3 ; . . . covers M, U1 ; U2 ; U3 ; . . . is locally ﬁnite, Each un ðUn Þ is equal to the open ball B3(0), 1 1 u1 1 ðB1 ð0ÞÞ; u2 ðB1 ð0ÞÞ; u3 ðB1 ð0ÞÞ; . . . is an open cover of M.

1 1 Since u1 1 ðB1 ð0ÞÞ; u2 ðB1 ð0ÞÞ; u3 ðB1 ð0ÞÞ; . . . is an open cover of M, and M is 1 compact, there exists a positive integer k such that u1 1 ðB1 ð0ÞÞ; . . .; uk ðB1 ð0ÞÞ is 1 an open cover of M. For every i ¼ 1; . . .; k; ui is continuous, and B1 ½0 is com1 pact, so u1 i ðB1 ½0Þ is compact. ui ðB1 ½0Þ is compact in M, and M is Hausdorff, so 1 ui ðB1 ½0Þ is a closed subset of M. Since u1 i ðB1 ½0Þ is a closed subset of M,

© Springer India 2014 R. Sinha, Smooth Manifolds, DOI 10.1007/978-81-322-2104-3_7

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7 Whitney Embedding Theorem

1 1 u1 i ðB2 ð0ÞÞ is an open subset of M, and ui ðB1 ½0Þ ui ðB2 ð0ÞÞ, by Lemma 4.58, there exists a smooth function wi : M ! ½0; 1 such that

a. for every x in u1 i ðB1 ½0Þ; wi ðxÞ ¼ 1; b. supp wi u1 ðB 2 ð0ÞÞ: i Here, u1 : U1 ! u1 ðU1 Þð Rm Þ; so let u1 ðu11 ; . . .; u1m Þ; where each u1i : U1 ! R: Similarly, let u2 ðu21 ; . . .; u2m Þ; where each u2i : U2 ! R; etc. c Take any p 2 U1 \ ðu1 1 ðB2 ½0ÞÞ : Clearly, w1 ðpÞu11 ðpÞ ¼ 0: (Reason: Since 1 supp w1 is contained in u1 ðB2 ð0ÞÞ; w1 ðxÞ ¼ 0 for every x 62 u1 1 ðB2 ð0ÞÞ: Since c 1 1 1 p 2 U1 \ ðu1 ðB2 ½0ÞÞ ¼ U1 ðu1 ðB2 ½0ÞÞ U1 ðu1 ðB2 ð0ÞÞÞ; w1 ðpÞ ¼ 0; and hence, w1 ðpÞu11 ðpÞ ¼ 0 u11 ðpÞ ¼ 0: This shows that the following deﬁnition of function is well-deﬁned. ^ 11 : M ! R as follows: for every p 2 M; Let us deﬁne a function u ^ 11 ð pÞ u

w1 ð pÞu11 ð pÞ if p 2 U 1 c 0 if p 2 u1 1 ðB2 ½0Þ :

^ 11 : M ! R is a smooth function. Since ðU1 ; u1 Þ is Now we want to show that u an admissible coordinate chart of M; u1 : U1 ! u1 ðU1 Þ is a diffeomorphism, and hence, u1 : U1 ! u1 ðU1 Þ is smooth. Since ðu11 ; . . .; u1m Þ u1 : U1 ! u1 ðU1 Þð Rm Þ is smooth, u11 : U1 ! R is smooth. Since u11 : U1 ! R is smooth, w1 : M ! ½0; 1 is smooth, and U1 is an open subset of M, their product p 7! w1 ðpÞu11 ðpÞ is smooth over open set U1 : Also the constant function p 7! 0 is c ^ 11 is well-deﬁned, u ^ 11 : smooth over the open set ðu1 1 ðB2 ½0ÞÞ : Now since u ^ 11 : M ! R is well-deﬁned, for every M ! R is smooth. Since the function u p 2 M; ^ 11 ð pÞ u

1 w1 ð pÞu11 ð pÞ if p 2 u 1 ðB2 ½0Þ c 0 if p 2 u1 1 ðB2 ½0Þ :

^ 12 : M ! R deﬁned as follows: for every p 2 M; Similarly, the function u ^ 12 ð pÞ u

1 w1 ð pÞu12 ð pÞ if p 2 u 1 1ðB2 ½0Þ c 0 if p 2 u1 ðB2 ½0Þ

^ 21 : M ! R is deﬁned as follows: for every is smooth, etc. Further the function u p 2 M; ^ 21 ð pÞ u

1 w2 ð pÞu21 ð pÞ if p 2 u 2 1ðB2 ½0Þ c 0 if p 2 u2 ðB2 ½0Þ

^ 22 : M ! R is deﬁned as follows: for every is smooth. Similarly, the function u p 2 M;

7.1 Compact Whitney Embedding in RN

^ 22 ð pÞ u

479

1 w2 ð pÞu22 ð pÞ if p 2 u 2 1ðB2 ½0Þ c 0 if p 2 u2 ðB2 ½0Þ

is smooth, etc. Now let us deﬁne a function 0

1

! !C B B C B R R C F: M ! B |ﬄﬄﬄﬄﬄﬄﬄﬄ{zﬄﬄﬄﬄﬄﬄﬄﬄ} R R |ﬄﬄﬄﬄﬄﬄﬄﬄ{zﬄﬄﬄﬄﬄﬄﬄﬄ} C @ A m m |ﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄ{zﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄ} 0 1 k @|ﬄﬄﬄﬄﬄﬄﬄﬄ{zﬄﬄﬄﬄﬄﬄﬄﬄ} R R A ¼ Rkðmþ1Þ k

as follows: for every p 2 M; ^ 1m ð pÞÞ; . . .; ðu ^ k1 ð pÞ; . . .; u ^ km ð pÞÞ; ðw1 ð pÞ; . . .; wk ð pÞÞÞ: ^ 11 ð pÞ; . . .; u F ð pÞ ððu Since each component function of F is smooth, F : M ! RN is smooth, where N kðm þ 1Þ: We shall try to show that 1. F : M ! RN is 1–1, 2. F : M ! RN is a smooth immersion, that is, for every p 2 M; the linear map dFp : Tp M ! TFðpÞ RN ð¼ RN Þ is 1–1. For 1: Let FðpÞ ¼ FðqÞ; where p; q 2 M: We have to show that p ¼ q: Here, ^ 1m ðpÞÞ; . . .; ð^ ^ km ðpÞÞ; ðw1 ðpÞ; . . .; wk ðpÞÞÞ ¼ FðpÞ ¼ ðð^ u11 ðpÞ; . . .; u uk1 ðpÞ; . . .; u ^ 1m ðqÞÞ; . . .; ð^ ^ km ðqÞÞ; ðw1 ðqÞ; . . .; wk ðqÞÞÞ; FðqÞ ¼ ðð^ u11 ðqÞ; . . .; u uk1 ðqÞ; . . .; u so ðw1 ðpÞ; . . .; wk ðpÞÞ ¼ ðw1 ðqÞ; . . .; wk ðqÞÞ; and for each i ¼ 1; . . .; k; ^ im ðpÞÞ ¼ ð^ ^ im ðqÞÞ: ui1 ðqÞ; . . .; u Since p 2 M ¼ u1 ð^ ui1 ðpÞ; . . .; u 1 ðB1 ð0ÞÞ 1 [ [ uk ðB1 ð0ÞÞ; there exists a positive integer l such that 1 l k; and p 2 u1 and hence, wl ðpÞ ¼ 1: Since 1 l k; and l ðB1 ð0ÞÞ; ðw1 ðpÞ; . . .; wk ðpÞÞ ¼ ðw1 ðqÞ; . . .; wk ðqÞÞ; 1 ¼ wl ðpÞ ¼ wl ðqÞ: Since wl ðpÞ ¼ wl ðqÞ; and wl ðpÞ ¼ 1; wl ðqÞ ¼ 1: Since wl ðpÞ 6¼ 0; and supp wl is contained in 1 1 1 u1 l ðB2 ð0ÞÞ; p 2 ul ðB2 ð0ÞÞð ul ðB2 ½0ÞÞ; and hence, p 2 ul ðB2 ½0Þ: Since 1 ^ lm ðpÞ ¼ wl ðpÞulm ðpÞ ^ l1 ðpÞ ¼ wl ðpÞul1 ðpÞ ¼ 1ul1 ðpÞ ¼ ul1 ðpÞ; . . .; u p 2 ul ðB2 ½0Þ; u ^ l1 ðqÞ ¼ ul1 ðqÞ; . . .; u ^ lm ðqÞ ¼ ulm ðqÞ: Since ¼ 1ulm ðpÞ ¼ ulm ðpÞ: Similarly, u ^ lm ðpÞÞ ¼ ð^ ^ lm ðqÞÞ ¼ ul ðpÞ ¼ ðul1 ðpÞ; . . .; ulm ðpÞÞ ¼ ð^ ul1 ðpÞ; . . .; u ul1 ðqÞ; . . .; u ðul1 ðqÞ; . . .; ulm ðqÞÞ ¼ ul ðqÞ; ul ðpÞ ¼ ul ðqÞ: Since ul ðpÞ ¼ ul ðqÞ; and ul is 1–1, p ¼ q: For 2: Let us take any p 2 M: We have to show that the linear map dFp : Tp M ! RN is 1–1. For this purpose, let ðdFp ÞðvÞ ¼ 0; where v 2 Tp M: We have to show that v ¼ 0:

480

7 Whitney Embedding Theorem

1 Since p 2 M ¼ u1 1 ðB1 ð0ÞÞ [ [ uk ðB1 ð0ÞÞ; there exists a positive integer l such that 1 l k; and p 2 u1 l ðB1 ð0ÞÞ: It follows that for every x 2 1 ^ lm ðxÞÞ ¼ u1 ðB ð0ÞÞð u ðB ½0ÞÞ; we have wl ðxÞ ¼ 1; and ð^ ul1 ðxÞ; . . .; u 1 2 l l ðwl ðxÞul1 ðxÞ; . . .; wl ðxÞulm ðxÞÞ ¼ ð1ul1 ðxÞ; . . .; 1ulm ðxÞÞ ¼ ul ðxÞ: For simplicity, ^ 1m ðxÞÞ ¼ u1 ðxÞ: u11 ðxÞ; . . .; u let l ¼ 1: Thus, for every x 2 u1 1 ðB1 ð0ÞÞ; we have ð^ u1 Þp ÞðvÞ; . . .; ðdð^ uk Þp ÞðvÞ; ðdðw1 Þp ÞðvÞ; . . .; ðdðwk Þp ÞðvÞÞ ¼ Here, 0 ¼ ðdFp ÞðvÞ ¼ ððdð^ ððdðu1 Þp ÞðvÞ; . . .; ðdð^ uk Þp ÞðvÞ; ðdðw1 Þp ÞðvÞ; . . .; ðdðwk Þp ÞðvÞÞ; so ðdðu1 Þp ÞðvÞ ¼ 0: Since u1 is a diffeomorphism, dðu1 Þp is an isomorphism. Since dðu1 Þp is an isomorphism, and ðdðu1 Þp ÞðvÞ ¼ 0; v ¼ 0: Since F : M ! RN is 1–1, F : M ! RN is a smooth immersion, and M is compact, by Note 5.72, F : M ! RN is a smooth embedding. h

Note 7.2 The above Lemma 7.1 is known as the compact Whitney embedding in RN :

7.2 Compact Whitney Embedding in R2mþ1 Lemma 7.3 Let M be an m-dimensional smooth manifold. Let M be compact. Then, there exists an embedding F : M ! R2mþ1 : Proof By Lemma 7.1, there exists a positive integer N such that M is embedded in the Euclidean space RN : Since M is embedded in the Euclidean space RN ; there exists a smooth embedding f : M ! RN : h Case I when N\2m þ 1. Let us not distinguish between ðx1 ; . . .; xN Þð2 RN Þ and 0

1

B C @x1 ; . . .; xN ; 0; . . .; 0 A 2 R2mþ1 : |ﬄﬄﬄﬄﬄ{zﬄﬄﬄﬄﬄ} |ﬄﬄﬄ{zﬄﬄﬄ} N

ð2mþ1ÞN

Thus, RN R2mþ1 ; RN is an N-slice of R2mþ1 ; and RN satisﬁes local N-slice condition. It follows, by Note 5.97, that RN is an N-dimensional embedded submanifold of R2mþ1 ; and hence, there exists a smooth embedding g : RN ! R2mþ1 : Since f : M ! RN is a smooth embedding, and g : RN ! R2mþ1 is a smooth embedding, their composite map g f : M ! R2mþ1 is a smooth embedding. Case II when 2m þ 1\N. Since M is an m-dimensional smooth manifold, and R is a 1-dimensional smooth manifold, the product M M R is a ð2m þ 1Þ-dimensional smooth manifold. Let us deﬁne a mapping a : M M R ! RN as follows: for every p; q 2 M; and t 2 R,

7.2 Compact Whitney Embedding in R2mþ1

481

aðp; q; tÞ tðf ð pÞ f ðqÞÞ: Clearly, a is smooth. (Reason: Since ðp; q; tÞ 7! p is smooth, and f : M ! RN is smooth, their composite ðp; q; tÞ 7! f ðpÞ is smooth. Similarly, ðp; q; tÞ 7! f ðqÞ is smooth. Since ðp; q; tÞ 7! f ðpÞ is smooth, and ðp; q; tÞ 7! f ðqÞ is smooth, ðp; q; tÞ 7! ðf ðpÞ; f ðqÞÞ from M M R to RN RN is smooth. Since ðp; q; tÞ 7! ðf ðpÞ; f ðqÞÞ from M M R to RN RN is smooth, and ðx; yÞ 7! ðx yÞ from RN RN to RN is smooth, their composite ðp; q; tÞ 7! ðf ðpÞ f ðqÞÞ from M M R to RN is smooth. Since ðp; q; tÞ 7! t from M M R to R is smooth, and ðp; q; tÞ 7! ðf ðpÞ f ðqÞÞ from M M R to RN is smooth, ðp; q; tÞ 7! ðt; ðf ðpÞ f ðqÞÞÞ from M M R to R RN is smooth. Since ðp; q; tÞ 7! ðt; ðf ðpÞ f ðqÞÞÞ from M M R to R RN is smooth, and ðt; xÞ 7! tx from R RN to RN is smooth, their composite ðp; q; tÞ 7! tðf ðpÞ f ðqÞÞ ð¼ aðp; q; tÞÞ from M M R to RN is smooth, and hence, a : M M R ! RN is smooth.) Since a : M M R ! RN is smooth, and dimðM M RÞ ¼ 2m þ 1\N ¼ dimðRN Þ, by Theorem 6.27, lN ðaðM M RÞÞ ¼ 0; where lN denotes the Ndimensional Lebesgue measure over RN : Let us deﬁne a mapping b : TM ! RN as follows: for every ðp; vÞ 2 ‘ TMð¼ r2M Tr MÞ; where p 2 M; and v 2 Tp M; bðp; vÞ ¼ ðdfp ÞðvÞ: Recall that TM, the tangent bundle of M, is a 2m-dimensional smooth manifold. Clearly, a is smooth. (Reason: Since f : M ! RN is smooth, by Theorem 5.24, ‘ ‘ ðp; vÞ 7! ðf ðpÞ; ðdfp ÞðvÞÞ from TM to TRN ð¼ r2RN ðTr RN Þ ¼ r2RN RN ¼ fðr; xÞ : r 2 RN and x 2 RN g ¼ RN RN Þ is smooth. Since ðp; vÞ 7! ðf ðpÞ; ðdfp ÞðvÞÞ from TM to RN RN is smooth, and ðx; yÞ 7! y from RN RN to RN is smooth, their composite ðp; vÞ 7! ðdfp ÞðvÞð¼ bðp; vÞÞ from TM to RN is smooth, and hence, b : TM ! RN is smooth.) Since b : TM ! RN is smooth, and dimðTMÞ ¼ 2m\2m þ 1\N ¼ dimðRN Þ, by Theorem 6.27, lN ðbðTMÞÞ ¼ 0: Since aðM M RÞ RN ; bðTMÞ RN ; lN ðaðM M RÞÞ ¼ 0; and lN ðbðTMÞÞ ¼ 0; 0 lN ððaðM M RÞÞ [ ðbðTM ÞÞÞ lN ðaðM M RÞÞ þ lN ðbðTM ÞÞ ¼ 0 þ 0 ¼ 0; and hence, lN ððaðM M RÞÞ [ ðbðTMÞÞÞ ¼ 0: This shows that RN ððaðM M RÞÞ [ ðbðTMÞÞÞ is nonempty, and hence, there exists a 2 RN such that a 62 aðM M RÞ; a 62 bðTMÞ; and jaj 6¼ 0: Let Ha ðf v : v 2 RN ; v a ¼ 0gÞ denotes the orthogonal complement of a. We know that Ha is a “hyperplane” of RN ; so we shall not distinguish between Ha and RN1 : Let us deﬁne a function Pa : RN ! RN1 as follows: for every v 2 RN ;

482

7 Whitney Embedding Theorem

Pa ðvÞ v

ðv aÞ jaj2

! a

is the orthogonal projection of v on the hyperplane Ha : Clearly, Pa : RN ! RN1 is linear, and smooth, and therefore, dðPa Þv ¼ Pa for every v in RN : Since f : M ! RN is smooth, and Pa : RN ! RN1 is smooth, their composite ðPa f Þ : M ! RN1 is smooth. Now we shall try to show that 1. ðPa f Þ : M ! RN1 is 1–1, 2. ðPa f Þ : M ! RN1 is an immersion, that is, for every p 2 M; the linear map dðPa f Þp : Tp M ! TðPa f ÞðpÞ ðRN1 Þð¼ RN1 Þ is 1–1, that is, for every p 2 M; ðdðPa f Þp ÞðwÞ ¼ 0 implies w ¼ 0: For 1: Let ðPa f ÞðpÞ ¼ ðPa f ÞðqÞ; where p; q 2 M: We have to show that p ¼ q: We claim that p ¼ q: If not, otherwise, let p 6¼ q: We have to arrive at a contradiction. Since p 6¼ q; and f : M ! RN is a smooth embedding (and hence, f is 1–1), f ðpÞ 6¼ f ðqÞ: Since Pa ðf ðpÞÞ ¼ ðPa f ÞðpÞ ¼ ðPa f ÞðqÞ ¼ Pa ðf ðqÞÞ; and Pa : RN ! RN1 is linear, ð f ð pÞ f ð qÞ Þ

ð f ð pÞ f ð qÞ Þ a jaj2

a ¼ Pa ðf ð pÞ f ðqÞÞ ¼ 0:

Since ð f ð pÞ f ð qÞ Þ

ð f ð pÞ f ð qÞ Þ a jaj2

a ¼ 0;

and f ðpÞ 6¼ f ðqÞ; ðf ðpÞ f ðqÞÞ a 6¼ 0; and hence, j aj 2 jaj2 a¼ ðf ð pÞ f ðqÞÞ ¼ a p; q; ðf ð pÞ f ðqÞÞ a ð f ð pÞ f ð qÞ Þ a

!

2 aðM M RÞ: Thus, a 2 aðM M RÞ; which is a contradiction. So our claim is true, that is, p ¼ q: Thus, ðPa f Þ : M ! RN1 is 1–1. For 2: Let us take any p 2 M: Let ðdðPa f Þp ÞðwÞ ¼ 0 where w 2 Tp M: We have to show that w ¼ 0: If not, otherwise, let w 6¼ 0: We have to arrive at a contradiction. Since f : M ! RN is a smooth embedding (and hence, f : M ! RN is an immersion), and p 2 M; the linear map dfp : Tp M ! Tf ðpÞ ðRN Þð¼ RN Þ is 1–1. Since the linear map dfp : Tp M ! RN is 1–1, and 0 6¼ w 2 Tp M; ðdfp ÞðwÞ 6¼ 0: Here,

7.2 Compact Whitney Embedding in R2mþ1

dfp ðwÞ

dfp ðwÞ a 2

j aj

483

a ¼ Pa dfp ðwÞ ¼ Pa dfp ðwÞ d ðPa ÞF ð pÞ dfp ðwÞ ¼ d ðPa f Þp ðwÞ ¼ 0; so dfp ðwÞ dfp ðwÞ a a: ¼ jaj2 ¼

Since

dfp ðwÞ ¼

dfp ðwÞ a jaj2

a;

and ðdfp ÞðwÞ 6¼ 0; ððdfp ÞðwÞÞ a 6¼ 0; and hence, ! jaj2 j aj 2 dfp ðwÞ ¼ dfp w a ¼ dfp ðwÞ a dfp ðwÞ a ! 2 j aj ¼ b p; w 2 bðTM Þ: dfp ðwÞ a Thus, a 2 bðTMÞ; which is a contradiction. Hence, w ¼ 0: Since ðPa f Þ : M ! RN1 is 1–1, ðPa f Þ : M ! RN1 is an immersion, and M is compact, by Note 5.72, ðPa f Þ : M ! RN1 is a smooth embedding. Continuing this construction repeatedly for ﬁnite number of times, we get a map h F : M ! R2mþ1 ; which is an embedding. Note 7.4 The Lemma 7.3 is known as the compact Whitney embedding in R2mþ1 :

Bibliography

Apostol, T.M.: Mathematical Analysis, 2nd edn. Addison-Wesley, Reading (1974) Boothby, W.M.: An Introduction to Differentiable Manifolds and Riemannian Geometry, 2nd edn. Academic Press, Orlando (1986) Lee, J.M.: Introduction to Smooth Manifolds, 2nd edn. Springer, New York (2013) Loring, W.Tu.: An Introduction to Manifolds, 2nd edn. Springer, New York (2012) Rudin, W.: Principles of Mathematical Analysis, 3rd edn. McGraw-Hill, New York (1976) Spivak, M.: Calculus on Manifolds. Westview Press, USA (1998)

© Springer India 2014 R. Sinha, Smooth Manifolds, DOI 10.1007/978-81-322-2104-3

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