Holomorphic Automorphism Groups in Banach Spaces: An Elementary Introduction

H0LOM 0R PHIC AUTO M0R PHISM GROUPS IN BANACH SPACES: AN ELEMENTARY INTRODUCTION

NORTH-HOLLAND MATHEMATICS STUDIES Notas de Matematica (97)

Editor: Leopoldo Nachbin Centro Brasileiro de Pesquisas Fisicas, Rio de Janeiro, and University of Rochester

NORTH-HOLLAND -AMSTERDAM

0

NEW YORK

OXFORD

105

HOLOMORPHIC AUTOMORPHISM GROUPS IN BANACH SPACES: AN ELEMENTARY INTRODUCTION

Jose M. ISIDRO Facultad de Matematicas Universidad de Santiago de Compostela Spain

and LMO

L. STACHO

Bolyai lntezet Szeged Hungary

1985

NORTH-HOLLAND -AMSTERDAM

NEW YORK

OXFORD

@

Elsevier Science Publishers B.V., 1984

All rights reserved. No part of this publication may be reproduced, stored in a retrieval system, or transmitted, in any form or by any means, electronic, mechanical, photocopying, recording or otherwise, without the prior permission of the copyright owner.

ISBN: 0 444 87657 X

Publishers:

ELSEVIER SCIENCE PUBLISHERS B.V. P.O. BOX 1991 1000 BZ AMSTERDAM THE NETHERLANDS

Sole distributors forthe U.S.A. and Canada:

ELSEVIER SCIENCE PUBLISHING COMPANY, INC. 52 VAN DER BILT AVENUE NEW YORK, N.Y. 10017 U.S.A.

Library of Congress Cataloging in Publlcstion Data

Isidro, Joe6 M. Holomorphic automorphism groups in Benach spaces. (North-Holland nrsthematics studies ; 105) ( b a s de matendtics ; 97) Bibliography: p. 1. HolomrpMc functions. 2. Automorphism. 3. Banach spaces. I. StSch6, Lhsz16 L. 11. Title. 111. Series. Iv. s e r i e s : b t a s de m a t e d t i c a (Amsterdam, lctherlends) ; 97. W . n 8 6 no. 97 tQA333 510 s t515.9'83 84-21164 ISBN 0-444-87657-X (U.S. ) PRINTED IN THE NETHERLANDS

PREFACE

Since the early 70's,there has been intensive development in the theory of functions of an infinite number of complex variables. This has led to the establishment of completely new principles (e.g. concerning the behaviour of fixed points) and has thrown new light on some classical finite dimensional results such as the maximum principle, the Schwarz lemma and so on. Perhaps the most spectacular advances occurred in connection with the old problem of the determination of the holomorphic automorphisms of complex manifolds. This book is based on the introductory lectures on this latter field delivered at the University of Santiago de Compostela in October 1981 by the authors. Originally, it was planned as a comprehensive postgraduate course relying on a deep knowledge of holomorphy in topological vector spaces and infinik dimensional Lie groups. However, seeing that some of the undergraduate students were mainly interested in the study of bounded domains in Banach spaces, the authors restricted their attention to these aspects. This proved to be a fortunate idea. We realized that by combining the methods of the theories developed independently by W. Kaup and J . P . Vigu6 with minor modifications, even the main theorems could be derived. This was achieved in a self-contained way from the most fundamental principles of Banach spaces (such as the open mapping theorem), elementary function theory and the pure knowledge of the Taylor series representation of holomorphic maps in this setting. It may often happen in teaching mathematics that avodding the introduction of strong tools leads to abandoning natural heuristics. Probably, this is not the case now. It is enough to V

vi

PREFACE

recall how deeply the early development of the theory of finite dimensional Lie groups and Lie algebras was inspired in Cartan's investigation of the structure of symmetric domains. Moreover, we think that this approach to the automorphism groups of Banach space domains may also serve as motivating and illustrative material in introducing students to the theory of Lie groups and complex manifolds. The text is divided into eleven chapters. In chapter 0 we establish the terminology, and some typical examples of later importance (e.g. the Mtibius group) are studied. In chapter 1 we show the main topological consequences of the Cauchy estimates of Taylor coefficients for uniformly bounded families of holomorphic mappings. These considerations are continued in chapter 2 and applied specifically to the case of the automorphism group, concluding with the topological version of Cartan's uniqueness theorem. The global topological investigations finish in chpater 3, where the Caratheodory distance is introduced to obtain the completness properties of the group AutD. In chapter 4 a completely elementary introduction to Lie theory begins by showing where one-parameter subgroups come from. Chapter 5 is devoted to a description of the Banach Lie algebra structure of complete holomorphic vector fields in order to lay the foundation of chpater 6, in which the Banach Lie groups structure of AutD is studied. In chpaters 7 and 8 we discuss the basic theory of circular domains and determine explicitly the holomorphic automorphism group of the unit ball of several classical Banach spaces. In chapter 9 we introduce the reader to another fruitfully developing branch of these researches by proving Vigue's theorem on the Harish-Chandra realization of bounded symmetric domains. Finally, in chapter 10 and elementary introduction of the Jordan approach to bounded symmetric domains is presented and the convexity of the Harish-Chandra realization is proved. We would like to express our sincere acknowledgement to Prof. L. Nachbin who suggested the idea of writing these notes

PREFACE

vii

and who, together with Prof. E. Vesentini, introduced the authors to infinite dimensional holomorphy and this fascinating branch of mathematics. Thanks are also due to M. Teresa Iglesias €or the careful typing. The authors, August 1984.

J.M. Isidro

Santiago de Compostela Spain.

Stach6 Szeged Hungary.

L.L.

This Page Intentionaiiy Left Blank

TABLE OF CONTENTS

PREFACE

V

CHAPTER

0. PRELIMINARIES.

CHAPTER

1 . UNIFORMLY BOUNDED FAMILIES OF HOLOMORPHIC

§l.

52. 13.

CHAPTER

51. 52. 53.

CHAPTER

51. 52.

53. 54. 55.

1

MAPS AND LOCALLY UNIFORM CONVERGENCE. Cauchy majorizations. Continuity of the composition operation. Differentiability of the composition operation.

5

9 13

2 . TOPOLOGICAL CONSEQUENCES OF THE GROUP

STRUCTURE OF THE SET OF AUTOMORPHISMS. The topological group Aut D Cartan's uniqueness theorem. Topological version of Cartan's uniqueness theorem.

.

17 19 20

3. THE CARATHEODORY DISTANCE AND COMPLETENESS

PROPERTIES OF THE GROUP OF AUTOMORPHISMS. The Poincar6 distance. The Caratheadory pseudometric. The Caratheadory differential pseudometric. Relations between the Carathgodory pseudometric and the norm metric on D.

32

Completeness properties of the group Aut D.

37

ix

29

33 35

TABLE OF CONTENTS

X

CHAPTER 51. 52. 53. 54. 55.

CHAPTER 51.

52.

53. 54. 55.

CHAPTER 51. 52. 53.

54.

55. 56. CHAPTER §I.

52.

53. 54. 55.

4. THE LIE ALGEBRA OF COMPLETE VECTOR FIELDS. One parameter subgroups. Complete holomorphic vector fields. The Lie algebra of complete holomorphic vector fields. Some properties of commuting vector fields. The adjoint mappings. 5. THE NATURAL TOPOLOGY ON THE LIE ALGEBRA OF COMPLETE VECTOR FIELDS. Cartan's uniqueness theorem for autD. Some majorizations on autD. The natural topology on autD. autD as a Banach space. autD as a Banach-Lie algebra.

6. THE BANACH LIE GROUP STRUCTURE OF THE SET OF AUTOMORPHISMS. The concept of a Banach manifold. The concept of a Banach-Lie group. Specific examples: the linear group and its algebraic subgroups. Local behaviour of the exponential map at the origin. The Banach-Lie group structure of AutD. The action of AutD on the domain D. 7. BOUNDED CIRCULAR DOMAINS. The Lie algebra autD f o r circular domains. The connected component of the identity in AutD. Study of the orbit (AutDIO of the origin. 0 The decomposition AutD=(Aut D)(AutoD). Holomorphic and isometric linear equivalence of Banach spaces.

43

49 54

58

61

65 66

69 70

74

77 83 87 101 108 111

113

120 124 126 128

TABLE OF CONTENTS 56. 57.

CHAPTER 51.

52. 53. 54.

55. 56. 57. S8.

CHAPTER §1.

52. 53. 54. 55. 56.

§7.

58. 59.

CHAPTER 51. 92. 53.

54 *

55.

96. 57.

xi

The group of surjective linear isometries of a Banach space. Boundary behaviour and extension theorems.

130 132

a.

AUTOMORPHISMS OF THE UNIT BALL OF SOME CLASSICAL BANACH SPACES. Some geometrical considerations. Automorphisms of the unit ball of LP(Q,p),2#p#m. Automorphisms of the unit ball of some algebras of continuous functions. Operator valued Mijbius transformations. J*-algebras of operators. Minimal partial isometries in Cartan factors. 0

0

Description of Aut B(F1) and aut B(F1). Description of Aut 0 B ( F k ) and aut0B(Pk).

139 142 148 157 164 169

178 183

9. BOUNDED SYMMETRIC DOMAINS.

Historical sketch. Elementary properties 3f svmetric lomains. The canonical decomposition of autD. The complexified Lie algebra of autD. The local representation of autD. The pseudorotations on autD. The pseudorotations on D. The construction of the image domain 8. The isomorphism between the domains D and 8.

191 193 199 20 1 203 207 213 221 224

10. THE JORDAN THEORY OF BOUNDED SYMMETRIC

DOMAINS. Jordan triple product star algebras. Polarization in J*-algebras. Flat subsystems. Subtriples generated by an element. JB*-triples and Hermitian operators. Function model f o r EC. (Ec,*) as a commutative Jordan algebra.

231 235 238

240 242 249 262

x ii

TABLE OF CONTENTS

98.

p o s i t i v e J * - t r i p l e s and t h e c o n v e x i t y of homogeneous c i r c u l a r d o m a i n s .

270

59.

Some p r o p e r t i e s of t h e t o p o l o g y of l o c a l uniform convergence.

280

L I S T OF REFERENCES AND SUPPLEMENTARY READING

285

CHAPTER

0

PRELIMINARIES

Throughout what follows, E and E l denote complex Banach spaces whose norms will be represented indistinctly by ]I I ] , and D is a bounded domain in E.

-

0.1. DEFINITION. A m a p p i n g f: D+E1 i s s a i d t o b e h o l o m o r p h i c i f , for e v e r y aeD, we h a v e f

(a+h)=

. . - ,h)

m

C fLn (h, n=O

i n a n e i g h b o u r h o o d o f a.

Here, for every ndN,

a"

1

(0.1)

f(a+tlhl+. . . + t h 1 n n

is a continuous n-linear operator from En into E l . Remark that, f o r ndN and hcE, we have (0.2)

The family of all holomorphic mappings from D c E into a set D 1 C E 1 is denoted by Hol(D,D1). When E = E 1 and 9=D1 we write H o l ( D ) instead of Hol(D,D ). 1

0.2. DEFINITION. A s u b s e t B c D i s s a i d t o b e c o m p l e t e l y i n t e r i o r t o D, a n d we w r i t e B C C D, if dist(B,aD) > O .

For feHol(D,D1) and B c c D we define

I / f 11

I/ f /I

B=: SUP xeB

IIf

by means of (XI

I1

0.3. DEFINITION. A n e t ( fI. ) , i n Hol(D,D 1 is s a i d t o JCJ c o n v e r g e locally u n i f o r m l y t o a m a p p i n g fcHol(D,E1) i f , f o r 1

CHAPTER

2

0

We denote by 7 the topology on Hol(D,D1) of local uniform convergence over D. If a net (f , ) , in Hol(D,D ) is locally 3

JCJ

1

uniformly convergent to fcHol(D,El), we write T= lim f . = f jeJ

'

0.4. EXERCISE. (a) Let E be the Banach space !Z1 and

function f: E-E is holornorphic on the whole space E and that f is not bounded on the open unit ball B ( E ) of E. Thus we may have 1 1 f / l B =m even if fcHol(D,El) and B c c D. (b) Is T a metrizable topology?. 0.5. DEFINITION. A m a p p i n g fcHol(D) is s a i d to b e an nutomorphism

oS D

if t h e r e exists gcHol(D) s u c h t h a t fg= id

n = gf

Here fg stands f o r the composite of the mappings f and g, and idD represents the identity mapping of D. The family of all automorphisms of D is represented by AutD. 0.6. EXERCISE, (a) Prove that a mapping f: D+E satisfies fcAutD if, and oniy if, f is a surjective bijection of D and, for every asD, the operator ':f is invertible.

invertibility of f

(b) Can the assumption concerning the be weakened?.

(c) Show that AutD with the usual law of composition is a group.

0.7. EXAMPLE. Let A be the open unit disc of E and, for k,ucQ with \ k l = 1 and l u l < l , let us define M as the restrick,u tion to A of the Mdbius transformation

PREL IMI N A R I €3S

3

Then, t h e f o l l o w i n g r e s u l t h o l d s : 0.8.

The g r o u p AutA is g i v e n b y

THEOREX.

P r o o f : F i r s t , l e t us o b s e r v e t h a t w e have

where

1w1
1,u

l+uu' and v=: 1 +Gut

w=:

' + u

~

s a t i s f y IvI= 1

and

1+ u i '

. !loreover , 14

1,-u

= i d - !I

A-

1,-u

M

and

1,u

M

k,O

M-k , O = i d A = f l k,O Mk , O

T h u s , t h e M6bius t r a n s f o r m a t i o n s f o r m a g r o u p w i t h r e g a r d t o c o m p o s i t i o n a n d e v e r y f4

w e have M

(an)=

kru

(A).

af4

L e t Scan and M

k,u f i x e d ; s i n c e / k / = ) L \ = 1, w e h a v e

that is,

ktu

M k ,u

w e have M

ktu

(2A)c

( < ) fa

an.

E.

i s b i h o l o m o r p h i c on

Thus a M

k,u

(A)caA.

k,u

Therefore

be a r b i t r a r i l y

On t h e o t h e r h a n d

f o r a l l ss:h , whence w e see t h a t M

k,u

is

(A)

a bounded o p e n s e t i n CC whose b o u n d a r y i s c o n t a i n e d i n aA. However, o n e e a s i l y c h e c k s t h a t t h e o n l y s e t w i t h t h e s e p r o p e r -

ties is A ,

i.e., M

kru

(A)= A

and M

keu

eAutA.

C o n v e r s e l y , l e t fsAutA b e g i v e n a n d w r i t e u = : f ( O ) , g = : M

1 ,-u

f.

Then g ( O ) = 0 so t h a t t h e Schwarz lemma may b e a p p l i e d t o g and

g-1

.

Then, f o r a n y CsA\{O},

w e have

Is(<) ISIL

and

1

lil=19- g ( C ) / ~ / g ( i )t/h,a t i s , / g ( s ) ) =151. T h e r e f o r e g i s l i n e a r , so t h a t g = PI f o r some kcaA. But t h e n f= M

krO

g= M

1 ,u

l,u

M

k,O

i s a M6bius t r a n s f o r m a t i o n .

#

4

CHAPTER

0.9.

0

DEFINITION. A domnin D is said to he h n r n o g e n a o u : :

i f , for2 e v e r y p a i r a , b c D , t h e r e i s a n autornorphisrn fcAutD s u c h t h n t f ( a )= b .

Lf D i:; homogeneous, t h e n w e s a y t h n t A u t D acts

i u > o n s i t ? : u a Z : 4 on D. I t f o l l o w s t h a t A i s a homoueneous domain. By R i e m a n n ' s famous

t h e o r e m , e v e r y p r o p e r domain D i n

e q ~ i v u L c n Lt o A ,

i s hihl.iZornor'phicnIZy

i . e . , t h e r e a r e f c H o l ( A , D ) a n d gcHol(D,A: s u c h

that

f(A)=

D,

q(D)= A,

f g = id

D'

gf= i d

A

Hence, f o r e v e r y p r o p e r domain D c i T , w e c a n d e s c r i b e e x p l i c i t e -

ly t h e q r o u p AutD by t h e f o r m u l a AutD= { f M f - ' ; McAutA}. L e t u s r e c a l l t h a t t h e mapping f c a n b e e x p r e s s e d i n terms of t h e Green f u n c t i o n of t h e domain D . domain o f

I n p a r t i c u l a r , each proper

a: i s homogeneous.

0.10.

EXAMPLE. L e t E b e a n y complex Banach s p a c e and

p u t D = : B ( E ) f o r t h e open u n i t b a l l of E .

I f f i s any s u r j e c t -

i v e l i n e a r i s o m e t r y of E l t h e n w e h a v e f I D e A u t D .

CHAPTER 1 UNIFORMLY BOUNDED FAMILIES OF HOLOMORPHIC MAPS AND LOCALLY UNIFORM CONVERGENCE

§I.-

Cauchy majorizations.

Let E and El be complex Banach spaces, D c E and D I C E l be bounded domains and asD be a given point in D. Write 6=: dist(a,aD). Then, PROPOSITION. (Cauchy estimates). We h a v e

1.1. ( 1 .I1

f o r a22 fsHol(D,D,) a n d neN.

Proof: By Cauchy's classical integral representation theorem, we have

whenever a+Ahl+...+AhncD. Therefore

...,hncE with

for all hl,

IIhllI,...,IIhnI/<6/n.By the continuity

of f;", this estimate also holds for l i h l l l = the n-linearity of f a(n,

5

...=

IIhnII= 6/n. By

6

CHAPTER

1

n" se n , whence t h e From t h e S t i r l i n g f o r m u l a we d e r i v e n! r e s u l t follows.

# EXERCISE. Show by examples i n E n t h a t , i n g e n e r a l , nn Ii n ( 1 . I ) c a n n o t be improved. t h e constant , 1.2.

1.3. REMARK. For the "symmetric" t e r m s f ( " ( h , ...,h ) w e have the sharper estimate

P r o o f : From ( 0 . 2 ) i t f o l l o w s t h a t

(1.1) we obtain

whenever a + A h c D . Then, a s i n t h e p r o o f o f

# From t h e Cauchy estimates w e c a n d e r i v e t h e l o c a l l y L i p s c h i t z i a n b e h a v i o u r o f t h e d e r i v a t i v e s o f f g i v e n by t h e following : 1.4.

PROPOSITION. Let a, b s D be g , i v e n a n d a s s u m e l h a t

the segment [ a , b ] = : { a + X ( b - a ) ; Xe[O,l]} d i s t ( [ a , b ] , aD)= p > O .

Then

We

Z i e s i n D uith

have

f o r aZZ fsHol(D,D1) a n d n m .

P r o o f : S e t h n + l = : b-a. w e have

(f:"-f:")

(hlI

- .. ,h,)

=

\:

[

By t h e Newton-Leibnitz

fa+th (n

n+ 1

.. . , h n ) ] d t =

(hl,

formula

BOUNDED FAXILIES AND UNIFORM CONVERGENCE

(hl

7

,.. . , h n ) ] d t =

...+t n + l h n i l ) d t w h i c h by

(0.1)

is equal t o

Then, u s i n g ( 1 . 1 ) w e d e r i v e

r ( n + l )(

pe

n+l

/I

/I fll,

hnil

It

whence t h e r e s u l t f o l l o w s .

1 . 5 . THEOREM. L e t ( f . ), a n d f r e s p e c t i v e l y be a n e t 1 IeJ and an e l e m e n t i n Hold(DID1) and d e n o t e b y B r c D a n y b a l l

c e n t e r e d a t aeD. T h e n , t h e f o l l o w i n g s t a t e m e n t s a r e e q u i v a l e n t : ( a ) The n e t

( f . ), 1

1eJ

is u n i f o r m l y c o n v e r g e n t t o f o n B.

i b i F o r a 2 2 k a , we h a v e l i m I I f ! k jEJ

-fikll

1 ,a

= 0.

Proof: The i m p l i c a t i o n a = > b i s obvius. L e t r b e t h e r a d i u s of B so t h a t O < r < 6 = : d i s t ( a , a D ) . G i v e n a n y

nm, w e h a v e f o r a l l hcE w i t h

w h i c h , by

(1.2)

[I

i s dominated by

hlI

rr

8

1

CHAPTER

N O W , g i v e n any E > O

t h e r e e x i s t s n CiN such t h a t 0

and, by h y p o t h e s i s t h e r e i s joeJ such t h a t

for a l l jeJ, ] > j

0'

T h i s completes t h e p r o o f .

# 1.6.

THEOREM. L e t t h e b a l l s B 1 and B2

be c o m p Z e t e l y

( f j ) j e J is a n e t in Hold(D,D1), t h e foZtowing

i n t e r i o r t o D. I f

a s s e r t i o n s are equivaZent:

l a ) fj+f r e l a t i v e t o

11 . I I B .

ib) f j + f r e l a t i v e t o

I) - / I B

1 I

2

P r o o f : Since D i s c o n n e c t e d , w e can f i n d b a l l s B ~ ~ B ; , . .,.B A + l such t h a t :

B'EB 0

Bi+l:

and

1

B'c:cD k

B2

f o r a l l k= 1 , 2 , . . . , n

t h e c e n t e r a ' o f Bi s a t i s f i e s

k= l , . . . , n + l . A s f +f r e l a t i v e t o j

a ; and t h e b a l l BA

\I

1)

,

"A i n s t e a d of

f j( ,r a i + f(ar,

a p l y i n g theorem 1 . 5 t o t h e p o i n t and D , w e g e t

for a l l

1

s o t h a t f.+f r e l a t i v e t o 7

/I. [IB,

rm

+f r e l a t i v e t o Asthe rolesof

11

- [IB

=> 1

I

. After

of t h e argument w e g e t f , + f r e l a i i v e t o I

f j

aicBL-l f o r

k

serveral reiterations

I( - I I B ,

n+ 1

f.+f r e l a t i v e t o 3

. Thus

I / . 11

B2

.

B1 and B2 may be changed, t h e proof i s c o m p l e t e .

#

BOUNDED

FAMILIES AND UNIFORll CONVERGENCE

9

1 . 7 . COROLLARY. T h e t o p o l o g y T o n Hol(D,D ) i s m e t r i z a 1 is a m e t r i c o n Hol(D,D1) //

b l e . F o r any baZZ B c c D ,

We h a v e Tlimf.= f

i n Hol(D,D1) 3 f!k +f(k f o r a l l ],a a kcPI, o r if and o n l y if t h e r e e x i s t s a b a l l B c c D s u c h t h a t

whose a s s o c i a t e d t o p o Z o g y i s 2'.

i f , and onZy if, t h e r e e x i s t s aeD s u c h t h a t

/ I fj-fll, 52.-

+o-

Continuity of the composition operation.

Let D,D1 and D2 be bounded domains in the Banach spaces E, and E 2 . As a first application of the previous theorem we show that the composition of mappings Hol (D,D1)XHol (D1,D2)+Hol (D,D2)

is continuous with regard to the topology of local uniform convergence. The way we shall follow is perhaps not the shortest possible but it provides information that turns out to be useful later. 1.8.

PROPOSITION. L e t feHol(D,D1) and qeHol(D1,D2) b e

h o Z o m o r p h i c m a p p i n g s whose r e s p e c t i v e T a y l o r ' s s e r i e s a t acD and b=: f (a)eD a r e

f(a+h)= f(a)+

...,h)

C fin(h, n=1

f o r a l l kCN.

Here, the detailed interpretation of ( 1 . 4 )

10

CBAPTER

v

1

1

+...+ vm>1

Pnoof: We have the following formal expansion for gf about the point asD (1.5)

gf (a+h)= g(b) +

?

m= 1

.

rjLm[f (a+h)-b;. .;f (a+h)-b]=

We point out that (1.5) is uniformly convergent in a suitable neiqhbourhood of a. Indeed, by ( 1 . 1 ) we have the majorizations

13 where 6=: dist(a,aD) and

E=:

dist(b,aD1

.

v +..+v

Hence

m

BOUNDED FAMILIES AND UlJIFORl4 CONVERGENCE

11

whenever

which i s s a t i s f i e d f o r s u f f i c i e n t l y s m a l l v a l u e s of f o r 1 1 hlI <6,. T h e r e f o r e ( 1 . 5 ) makes s e n s e f o r a l l moreover , w e may w r i t e

/I

I/ hll ,

say

h//<6,;

q f ( a + h )= q(b) + m

+ I

.. , h ) ;. . ; f

m

c

c V + . .f 1

m=l k=l

V

V 1 ,. . , V

m = g(b)+ C

m

=k

m

21

k

Z

k=l m=l V

( m ('1 (h, 9 b [fa

1

(h,.., h ) ] =

9,( m [ f l( V ( h r . - i h ) ; - . ; f

c +..+V

(Va

m

=k

Observe t h a t , f o r e a c h kdN, t h e k - t h t e r m o f t h i s s e r i e s i s a c o n t i n u o u s k-homogeneous p o l y n o m i a l of t h e v a r i a b l e h . By t h e uniqueness of t h e Taylor expansion, t h i s completes t h e proof. ?+

If E1,E2,..,En a n d E a r e complex Banach s p a c e s , w e w r i t e L(E1,..,E / E ) f o r t h e s p a c e of a l l c o n t i n u o u s n - l i n e a r mappings L: E l x E2 x..xEn+E. b e a n e t i n L(E1, ..,E,IE) with (L.) , 1 1eJ L *L a n d assume t h a t , f o r r = 1,2,. . , n , (K;) j e J a r e n e t s i n 1.9.

LEMMA. L e t

j

L(E:,..,E;

r

IE,)

w i t h Kr*Kr. T h e n we h a v e I

Lj (Ki r in

*.

rK))+L(K 1

*

- rKn)

CHAPTER

12

Proof: Let us write 'K similarly L . I

to

=:

L

1

K~

- *

j ,O-'

,

K~

K ~ - K and ~

=:

j,l

3

, L I., I-' -. L,-L. we may assume that 1 for all indexes j,r,s. Then

Given any

(u;Ulr..,

se can estimate

where the term in the right hand side of the inequality is majorized by

Thus, in any case we have

I / L~ ,u(Kj 1,ur**

r K mj

Therefore L ,( K1,, 1

1

,om

/I +O

(u;u~,..,~~)#(O;O,..~O)

.. , K m3, ) +L (K',. . ,Km ) 1

Lj,o( K j 1.10.

if

,ot *

THEOREM. L e t

since

- ,K;,~) = L ( K 1 ( f j ) jeJ

I . .

rKm)

.

a n d (gj)jcJ b e n e t s i n

Hol (D,D1) a n d Hol(D1,D2) such that

Tlim g . = g

Tlim f , = f I

jeJ

jcJ

3

T h e n we h a v e Tlim g.f, = gf in Hol(D,D2).

j e ~

J

'

Proof: Let us fix acD arbitrarily. By corollary 1.7 it suffices to show that ( g . f , ) ( m + (gf)m: for all m m . Observe 1 1 a that from proposition 1.1 it follows that for fixed m, the equilipschitzian in some neighbourhood of mappings ~ - + g (are ~ j r X

.

the point b=: f (a) Explicitly

r

we have

BOUNDED FAMILIES AND UIJIFORM CONVERGENCE

if E = : dis f (a)+b, there exists and index j j?j

0'

0

But then

13

ince such that f (a)cB for all ,

3

because of theorem 1.5 (applied to g , and the point b) and the 3 fact that f (a)+f(a) Therefore, by lemma 1.9 we have

.

,

3

#

U I , . . U m 21

It turns out from the proof of ( 1 . 6 )

1 .I 1

.

COROLLARY. If g f

3

that, in general, we have 3

I n p a r t i c u Z a r , gj , b ,+ g L k w h e n e v e r b.+b a n d Tlim g = g. 3

+gb

~ r b .

(k

3

(k

(k a n d b.+b t h e n g ( k

,b j+gb

jeJ

.

3

j

1.12. EXERCISE. (a) Prove the theorem directly by showing that given any acD, we have 11 g.f.-gfllB-+O for a 3 3 sufficiently small ball B c c D centered at a. (b) Let U c c D and kcN be given; prove that the family f (k=: x+fik , XCU, fsHol(D,D1), consists of holornorphic mappings and is uniformly bounded. What is ( f(

k)F?.

53.- ~Differentiability of the composition operation.

For later purposes, it is useful to investigate also 'the differentiability properties of the composition operation.

14

CHAPTER

1

1.13. 3EFINITION. L e t ICE b e a n i n t e r v a l a n d s u p p o s e t t h a t for e v e r y ts1 we a r e g i v e n a m a p p i n g f sHol(D,D1). We s h u l Z s a y t h a t t + f t i s " T - d e r i v a b z e " if we h a v e T l i m r; 1 (ftth-ft)= A h+O

f o r some AeHol(D,E1). A t t h i s p o i n t i f i s a h a r d q u e s t i o n w h e t h e r t h e mapping

t+ftgt i s d e r i v a b l e i n t h e T s e n s e i f so a r e t + f t a n d t + g However, it i s e a s y t o p r o v e a s l i g h t l y weaker s t a t e m e n t .

t

.

i n Hol(D,D ) is s a , i d 1.14. DEFINITION. A n e t (f.) 1 1 jeJ t o h e " w e u k l y T - c o n u e r g c n l " t o f if, f o r e v e r y xeD, t h e r e e x i s t s a b a l l B c e n t e r e d a t x s u c h t h a t ( 1 f . - f l I B+O. We w r i t e 1' l i m f . = f i n t h a t c a s e .

1

Wj€!J 3

A rriup t + f t from I i n t o Hol(D,D1) is s a - i d to b e " w e a k l y

T - d e r i v a b Z e " a t t O c I if we h a v e

f o r same ACHO~(D,E~).

1.15. LEMMA. If t h e maps I+Hol(D,D1) a n d I+Hol(D1,D2) y i v e n r e s p e c t i v e l y b y t+ft a n d t + g t a r e w e a k l y T - d e r i v a b l e a t t o , t h e n t + g t f t is U c a k Z y T - d e r i v a b l e a t t o . P r o o f : W e may assume t = 0 . L e t u s f i x xeD. W e can 0 0 d dt logt is

choose a b a l l B I C C DI c e n t e r e d a t f x s u c h t h a t L=: bounded when r e s t r i c t e d t o B1 a n d

L e t B 2 c E 2 be a b a l l c e n t e r e d a t 0 such t h a t L B 1 c C B 2 .

Furthermore, l e t B C c D be a b a l l c e n t e r e d a t x such t h a t f O ( B ) c c B1 a n d

BOUNDED F A I I I L I E S AND UNIF0R:I CONVERGENCE

where K=:

d dt

/,f

t

. Then, for

15

sufficiently small values of t we

have ft(BlcB,

t1

and

t

( g -g

o

l (Bl)CB2

Moreover

and applying proposition 1 . 1 we obtain

On the other hand, for ysB we have

11

g0(fty)-g0(foy+tKy)I /

611

go('IlB

1

11

fty-fOy-tKy//

Hence

0

From the Taylor series of go at f y we see that

Therefore

#

This Page Intentionaiiy Left Blank

CHAPTER

2

TOPOLOGICAL CONSEQUENCES O F THE STRUCTURE O F THE

51.-

GROUP

SET O F AUTOMORPHISIIS

The t o p o l o g i c a l g r o u p Aut D .

So f a r a l l o u t c o n s i d e r a t i o n s were v a l i d f o r a n y of t h e s p a c e s

Hol(D,D1). The p r i v i l e d q e o f Aut D w i t h respect t o them i s , f i r s t o f a l l , t h a t a n y mapping i n it a d m i t s a n i n v e r s e a n d t h a t t h e c o m p o s i t i o n o p e r a t i o n c a n be i t e r a t e d a n y number o f t i m e s i n it. H e n c e f o r t h w e r e s t r i c t o u r a t t e n t i o n t o Aut D . L e t u s c o n s i d e r f i r s t t h e c o n t i n u i t y o f t h e mapping f + f - '

w i t h r e g a r d to T.

S i n c e we have a l r e a d y e s t a b l i s h e d t h e c o n t i n u i t y o f t h e m u l t i p l i c a t i o n (1.e. t h e o p e r a t i o n ( f , q ) + q f i n Aut D , c f . Chap 1 , § 2 ) , it s u f f i c e s t o prove t h a t Tlim q.= i d

(2.1)

jeJ

3

-1

D

=> T l i m q . = i d D

jeJ

'

I n d e e d , l e t ( f ) j e J b e a n e t i n Aut D s u c h t h a t

i n Aut D. Tlim f = f jeJ j

a n d w r i t e q , = : f j f - l . Then w e h a v e 1

T l i m g . = T1im f , f - l = idD jeJ jeJ whence by a s s u p t i o n it follows

'

'

1

T l i m gT = T l i m f f T 1 = i d D jeJ jeJ

a n d t h e r e f o r e T l i m f T 1 = f-'. jeJ 3 To see ( 2 . 1 ) w e p r o v e t h e below e s t i m a t i o n :

2.1.

LEW4.4. L e t B , B ' c c D b e g i v e n b a l l s . T h e n t h e r e

e x i s t s a c o n s t a n t K s u c h t h a t we h a v e

17

18

CHAPTER

2

Proof:Let us fix xeB and set a = : g ( x ) . I f a , bcB' , t h e n by Cauchy m a j o r i z a t i o n s w e h a v e

Hence w e i m m e d i a t e l y o b t a i n t h e f o l l o w i n g : 2.2.

THEOREX. Aut D is a t o p o l o g i c a z g r o u p w i t h

regard

to t h e topology T. P r o o f : I t s u f f i c e s t o show ( 2 . 1 ) . L e t u s f i x a n y b a l l B c c D . S e t 6=: d i s t ( B , a D ) a n d d e f i n e

{xeD; d i s t ( x , B ) < 6 / 2 } . Then, f r o m T l i m g . = i d

BS/*=:

t h a t g +idD with r e s p e c t t o j

//

- /I

jeJ

; thus,

1

D

w e deduce

t h e r e i s a j o e J such

that

f o r a l l jljo a n d , i n p a r t i c u l a r , g j ( B ) c B A I 2 . A p p l y i n g lemma

2 . 1 t o g=: 9 .

3,

f=: id

n a n d B ' = : B 6 / a w e see t h a t

f o r a l l j > j o . Thus g T 1 + i d D w i t h r e g a r d t o 1

11

- /IB

and t h e result

#

f o l l o w s from t h e o r e m 1 . 6 . 2 . 3 . EXERCISE. Show by e x a m p l e s i n Aut A t h a t t h e c o n s t a n t K i n lemma 2 . 1 , B'

.

2.4.

i n g e n e r a l , must d e p e n d o n b o t h B a n d

REImRK. F o r a b e t t e r u n d e r s t a n d i n g of t h e s i t u a -

t i o n i n lemma 2 . 1

it i s i m p o r t a n t t o n o t e t h a t :

Given a n y b a l l B c c D , t h e r e i s a c o n s t a n t K s u c h t h a t w e have

19

TOPOLOGY ON THE GROUP OF AUTOMORPHISMS

f o r a l l f,gcAut D. Proof: L e t B ' = B

s/2

be d e f i n e d as above,where a g a i n

6=: d i s t ( B , a D ) , a n d c o n s i d e r a n y xsB. NOW, i f t h e p o i n t y = : g - l f ( x ) l i e s i n B ' , t h e n by p r o p o s i t i o n 1 . 4 w e h a v e

T h u s , by t h e a r b t r a r y n e s s o f xsB w e o b t a i n

I/ 2.5.

f-9

EXERCISE. U s i n g t h e f a c t t h a t c o n s t a n t m a p p i n g s

h a v e n u l l d e r i v a t i v e , show

f o r a l l f , gsAut D.

52.-

t h a t w e have

Hint: s h i f t D.

C a r t a n ' s uniqueness theorem. _____-

Next w e i n v e s t i g a t e t h e c o n s e q u e n c e s o f t h e € a c t t h a t , i n Aut D ,

t h e c o m p o s i t i o n c a n be i n f i n i t e l y i t e r a t e d . The p r o o f o f

t h e r e s u l t w e s h a l l o b t a i n s h a d s t h e f i r s t l i g h t on how t h e g e o m e t r y of D d e t e r m i n e s t h e a u t o m o r p h i s m s .

f:'=

9:'

2.6.

THEOREM.

L e t f , geAut D . I f for some acD

and

f ( ' = 9:'

then f = q.

we h a v e

P r o o f : L e t u s c o n s i d e r t h e map h = q - l f . W e h a v e h ( ' = (1

and h a = i d . T h e r e f o r e it s u f f i c e s t o p r o v e t h e s t a t e m e n t o f t h e theorem f o r h=: i d D . Suppose h f i d , .

Then t h e r e e x i s t s some R d N s u c h t h a t

a

CHAPTER

20

h',

C o n s i d e r t h e i t e r a t e d maps h ' ,

2

...

W e show by i n d u c t i o n t h a t

f o r a l l p a . Obviously a

( h p ) ('=

and

(hp)il= id

f o r a l l p m , and t h e a s s e r t i o n of

( 2 . 3 ) f o r p= 1 i s nothing but

t h e d e f i n i t i o n of k . Assume ( 2 . 3 ) h o l d s f o r p. By p r o p o s i t i o n 1 . 8 w e have

vl,.

C o n s i d e r t h e c a s e 2 g k i R . Then, from

f o r k = 1,2,...

(hp)ik=0

( h p ) ( 2 = '.-.=

. , v m 31

and

L e t u s compute ( h P t l ) i ' .

( h P f l ) ('=

(hp)

hJk= 0

w e d e r i v e ( hP + l ) (a= 0

From ( h p ) i 2 = . . = (hP);'-'=

. . , h a( 1] =

[hi'] + ( h p ):'[h;',

hi'cp

0

we get

hi'=

(p+l)h,('

which p r o v e s ( 1 . 3 ) . Hence w e have

lirn

11

(hp)i'II =

m

which c o n t r a d i c t s t h e Cauchy

P*m

majorizations

//

(Chapter 1 , p r o p o s i t i o n 1 . 1 ) ( h p ) ;'I\

6 (

e

)

2

sup(

11 X I \

i

xsD1

f o r a l l p a . I n f a c t w e have p r o v e d t h e f o l l o w i n g : 2.7.

COROLLARY.

L e t hsHol(D) be g i v e n and a s s u m e t h a t (1

t h e r e e x i s t s a p o i n t acD for w h i c h we h a v e h ( O = a , ha = i d . Then h= i d

D

.

e n e s s theorem. Roughly s p e a k i n g , C a r t a n ' s u n i q u e n e s s t h e o r e m s t a t e s t h a t ,

TOPOLOGY ON THE GROUP OF AUTOMORPHISMS

21

g i v e n a p o i n t a e D , t h e a u t o m o r p h i s m s of D d e p e n d o n l y on t h e i r 0 - t h a n d I-st d e r i v a t i v e s a t t h e p o i n t a , i . e . , a n y f s A u t D i s u n i q u e l y d e t e r m i n e d by t h e p a i r f (O a

t

f

('.

d

Is t h i s correspon-

d e n c e c o n t i n u o u s ? . The a n s w e r i s a f f i r m a t i v e . W e h a v e t h e f o l l o w i n g t o p o l o g i c a l v e r s i o n of C a r t a n ' s uniqueness theorem: 2.8.

THEOREM.

L e t asD b e f i x e d a n d a s s u m e t h a t

f , f . c A u t D, jsJ, s a t i s f y I Tlim f = f. I jeJ

,

f o r s= 0 ,1 . T h e n we h a v e

f ! ",a+f:s

P r o o f : E s s e n t i a l l y , w e c a r r y o u t t h e r e a s o n i n g s of t h e p r o o f o f C a r t a n ' s t h e o r e m i n a more g e n e r a l s e t t i n g ( w h e r e t h e r o l e o f f i s now p l a y e d b y f , ) . I D e f i n e h . = : f - ' f . . The r e l a t i o n h ! * = f - ' [ f j ( a ) ] + a 1 I I Observe t h a t also

is clear.

i n a c c o u n t of p r o p o s i t i o n 1 . 1 . N e x t w e s t a b l i s h t h e f o l l o w i n g a x u l i a r y stament LEMMA. I f

2.9.

h i o +a Ira

,

h!'

+id

1 la

and

i s a n e t in H o l ( D ) s u c h t h a t

(h,),

(kl IeJ

h . +O Ira

for k = 2 , .

.., R -I

t h e n we h a v e

An i n m e d i a t e c o n s e q u e n c e w e o b t a i n 2.10.

COROLLARY. F o r a n y n e t

( h . ) , c H o l ( D ) with I IeJ

I n d e e d : I f w e h a d h j f t i d D t h e n by t h e o r e m 1 . 5 t h e r e would be a n R > 2 s u c h t h a t h(.k

-to

3 ,a

Then, b y lemma 2 . 9

f o r 26kSR

and

22

CHAPTER

2

for all j c J and pm. Therefore

-

l i m //(hy);'l/

bpX

jeJ

for pm. But this contradicts the Cauchy estimates (proposition 1.1).

p r o v i n g the corollary. T h u s , in our case f-l f , - + idD whence Tlim f , = f. j

1

e

'

~

Thus, o u r only remainder task i s to prove the lemma. This requires a better overlook on the expansion of h p as a direct iteration of the formula given by proposition 1.8 would lead to very involved expressions. Instead, let us procceed a s follows: Start from

This a s s e r t s that, for f , gsHol(D) a n d for sufficiently small vectors xeE, (gf)(a+x) is the sum of all possible expressions

Let us write (2.4) i n the more visualizable form

/I\

x..x..x

...

/I\

x..x..x

...

/I\

x..x..x

TOPOLOGY ON THE GROUP O F AUTOMORPHISMS

23

I n s u c h a way, it seems t o b e i n t u i t i v e l y c l e a r t h a t f P ( a + x )

i s t h e sum of a l l p o s s i b l e e x p r e s s i o n s c o r r e s o n d i n g t o t h e g r a p h s of t h e form

.....

x

x

x

x

H e r e t h e symbol

7(

.....

.....

x

x

.....

x

x

x

x

x

x

c a n b e i n t e r p r e t e d a s t h e s i g n of s u b s t i t u -

t i o n . Now w e s t a b l i s h t h e p r e c i s e mathematical development of t h i s technique.

I n o r d e r t o be s e l f - c o n t a i n e d ,

w e s h a l l make

no r e f e r e n c e t o t h e u s u a l t h e o r y of t r e e g r a p h s .

2.11.

DEFINITION. L e t nCN b e a r b i t r a r i l y S i z e d . A n-up2.e A =

" t r e e of h e i g h t n r r i s a n

(ao,..,a

n- 1

1 of f u n c t i o n s

such t h a t t h e d o m a i n of a

11 f o r e v e r y p , dom

P

i s a segment

=tl,2,..dp(A)loflN. P r a n g e a,-1= { I } .

c1

31

" o n t o " dom

for Ogpgn-1 LX

Pf 1

I

.

41 for p = I , . . , n

a

P

i s a "monotone i n c r e a s i n g " mapping

we h a v e dom

c1P =

range ap - 1 *

The number d ( A ) i s c a l l e d t h e w i d t h o f A a t t h e h e i g h t p , and W e s a y t h a t d ( A ) i s t h e d e g r e e of

we s h a l l w r i t e d(A)=: d 0 ( A ) . A.

24

CHAPTER 2

2.12.

Note

C o n s i d e r t h e p l a i n graph

EXAMPLE.

t h a t t h e s e q u e n c e of t h e v e r t i c e s i s r e l e v a n t ! This c a n

b e i n t e r p r e t e d a s a t r e - of h e i g h t 3 a s follows:

0

1

a 1 (2)= 1

(I)= 1

a o ( l ) =1

... cx 0 ( 5 ) =

3

,.,

a1(5)= 2

...

~1

0

(10)=5

A t t h i s p o i n t w e c a n p r o v i d e a n e x a c t i n t e r p r e t a t i o n of

2.13.

(2.5).

L e t T r e e s ( n ) d e n o t e Ihe s e t o f a l l

DEFINITION.

t r e e : : o f h e i g h t n. G i v e n g e H o l ( D ) ue d e f i n e t h e " t r e e - d e r i v n t i ves

g i A of g at a c D a s foZZows:

F o r AcTrees ( 1 ) w e s e t g

(A

=:

(d(A) i n t h e u s u a Z s e n s e . ga

giH is a l r e a d y d e f i n e d for a l l BcTrees (n-1) and A=

( C X ~ ~ . ).€ ~ T rC e eX s n- 1

( n ) we s e t

If

TOPOLOGY ON THE GROUP OF AUTOMORPHISMS

25

-1

)sTrees(n-l), a 0 (k)=: {m: ao(m)= k } a n d Jf means cardinality. ljhere B= (al,..,cx

n- 1

2.14. PROPOSITION. T h e series 1 AeTrees (n)

I1

(A

4,

(xr ..rxI

I1

is uniformly convergente o n some neighbourhood o f the origin and we have

Proof: For n= 1 this formula is equivalent to the usual Taylor expansion of g. Remark that by the Cauchy estimates we have

where 6 = dist(a,aD)

and p ~ [ O , m ) .

NOW, assume we had proved (2.71

gn-' (b+x)= g"-' (b)+

C

BeTrees ( n )

gLE(x,* * r x )

for sufficiently small vectors x and

for all

pc[Orm),

where W

w

Y

(PI=

ep

1 [ v = 1 dist (y,aD)

1"

We prove (2.7) and (2.8) for n. Let us begin with (2.8)

CHAPTER

26

2

there exists a But given any BcTrees(n-I) and v l I . . , v d(B) unique (Y. such that (aolR)cTrees(n) and w k = ## a 0- 1 ( k ) f o r 0 k = l,..ld(B). Thus the second member of (2.9) is

which, due to Cauchy's majorizations, is dominated by m

c

c RcTrees(n-1)

V

1

+..tV

d

(R)

By the induction hypothesis the last sum is dominated by

Thus (2.8) is established. Now (2.8) is immediate

c g;* ( X I ,.x)= AeTress (n)

= g n (a+x)-gn

.

(a)

#

27

TOPOLOGY ON THE GROUP OF AUTOMORPHISM

Now w e c a n p r o v e t h e lemma. Proof o f lemma 2 . 9 :

I t f o l l o w s from p r o p o s i t i o n 2 . 1 4

f o r any p,kcW, acD and qcHol(D)

.

that

Consider any AsTrees(p)

with

d ( A ) = R s u c h t h a t A h a s a v e r t e x where t h e number o f e n t e r i n g e d g e s i s d i f f e r e n t from 1 and R , A=

)

( ~ 1 ~ , . . , ~ 1

that is, i f

t h e n t h e r e a r e s , udN w i t h

P-1 # c ~ - ~ ( u ) f ?W . e show t h a t , i n t h a t case, !h

# c ~ ~ ~ ( v and ) f l

+O. 3 ,a

Indeed, it i s

e a s y t o see t h a t

t h a t i s , 11 h! ,(fI/ i s n o t g r e a t e r t h a n t h e p r o d u c t of t h e norms of a l l t h o s e d e r i v a t i v e s t h a t o c c u r a t some v e r t e x o f A i f w e draw h ( A as j,a

I 1

h ( * . . . h a( *

...

I I (*

... h ( * . ..ha

By Cauchy m a j o r i z a t i o n s a l l t h e f a c t o r s o f

(2.10)

are bounded

by a c o n s t a n t i n d e p e n d e n t o f j . A t t h e same t i m e , by c o r o l l a r y 1 . I 1 we have

CHAPTER

28

2

and k f l e n t a i l s h ! k +O = i d ( k a n d h e n c e 3 ,a b e c a u s e h s ( a ) * a . The o n l y t r e e s A i n T r e e s ( p ) w i t h

since k= # a - ' ( v ) < L

-to

h(k

j ,hs ( a )

I t h e p r o p e r t i e s t h a t e a c h of t h e i r v e r t i c e s a d m i t a 1 o r Q e n t e r i n g edge and d ( A ) = L a r e

... ... ... ... ... I

I

I

. . . .. .. .. ..

p

I

1

L e t u s c a l l them AlI...,A

h. (A"

(e

,,a *ha

P

for

2 . 1 5 . EXERCISE. i n t h e g r a p h form.

1

I P

1

R

R

Therefore

/

p . . .

. Observe

V=

1I

.

that

. p , whence

( a ) Write t h e series of f n f n - 1 - .f 1 (b) Prove ( 2 . 1 0 ) .

CHAPTER

3

THE CARATHEODORY DISTANCE AND COMPLETENESS PROPERTIES OF THE GROUP OF AUTOMORPHISMS

51.-

The Poincar6 distance.

3.1. DEFINITION. We s a y t h a t a m e t r i c d o n a b o u n d e d d o m a i n D i s Aut D - i n v a r i a n t i f w e h a v e

f o r a l l x , ysD a n d a l l fcAut D . 3 . 2 . LE+W.The f u n c t i o n

d e f i n e s a n Aut A - i n v a r i a n t

m e t r i c o n A.

As usually, we write tanh(<)=:

for its inverse function.

eC-e-5 e'+e-'

,

<eC, and

tanh-' stands

Proof: Let us consider the mappings

It is straightforward to check that Tlim U j +m

t j=

Ut for t.+t and 3

that Ut+s= Ut U s for all t, sdR. That is, the mappinqlR+Aut A given by t+Ut is a continuous homomorphism. We have d (O,Ut(0))= t for all t30.

A

We introduce the expression di(" =:

m

infCCtk;tk>O,3<,,,..,5, 1

a)=:

, 3M1,..,MmsAut A 29

such that

CHAPTER

30

i 0=

m

P

,

I

ik-l

Mk(0)'

,

3

t MkU k ( 0 ) = t k

for

k= l,.m}.

satisfies the triangle inequality and is Aut A-in-

Trivial Y d variant .

We show that d i = dA. One key observation is that (3.1)

dA(o.,B)= infItp0;

3McAut A

M ( O ) = a,

t

MU ' ( 0 ) ~

because the set whose infimum is taken consists only of one element. Indeed, if M ( O ) = a and M ( p ) = B where pc[O,I), then M(O)= 0 whence, by the Schwarz lemma, there exists ks3A M 1 , -u such that M 1 ,-,M= 14k , O ' Therefore

Thus

k=

5 ,-a ( a )

h, (8)1

and

-0.

M= M 1 ,(Y "k,O

with this unique value of k. Then t

tanh(tl)= U

1

(O)=

T4

-1

( o ) = MkrO

M 1 ,-u

( a )=

so that t l = dA(n,B). From (3.1) it readily follows that d i < d A . Let us define n=: =:

min{m;3a, UcA, 3t1,..,tm?0, 3 < o l . . , ~ m l3M1,..,MmcAut A

s u that

co=

t

a,

b m = 0 , Mk(0)= < k - l , MkU k ( 0 ) = ik and

Choose

t 1 +..+tm
a,@ so that t

C1=

. .... .. .. in-1= Mn(0)

<,=

MIU '(0) t M U 2(0) 2

............ B = 5 n= M nu t n ( o ) .

,

f o r k=l,..,m

CARATHEODORY DISTANCE AND COMPLETENESS PROPERTIES

f o r some M1,..,MnsAut A

, t l , . . , t 20

31

a n d < o , ~ l , . . , < n e A By . the

d e f i n i t i o n of n w e h a v e * t

t + . . + t 2 d A ( a , < n - 1 ) ={ t 2 O : M U 1 n-1

(a)=

<,-1

where

*

M =:

Thus i f d i f d ,

M I ,a M

,

k,O

M1,-cl('n-l 1

w i t h k= :

lfi*l,-a(
)I

t h e only p o s s i b l e v a l u e o f n i s n = 2 s i n c e t h e r e

is t h e chain

Then w e c a n w r i t e

MIUL1(0)=

<

M2U

a

2(0)=

By t h e Schwarz lemma w e h a v e M 1 = XM

f o r some k s i n c e k,O M-l M 1 ( 0 ) = M-l(a)= r1-l M(O)= 0. T h e n , a g a i n by t h e Schwarz

l e m m a , we have

M2= M1 U

L1

M

k' , O t

MU tMk,,=M14

ktO

t2

U 'Mk,,oU

f o r s u i t a b l e k', k"c3A. However, (3.2)

Iu

t l(A)

ISUt'(IAI) t

f o r a l l AeA s i n c e t h e t r a n s f o r m a t i o n U

is circle-preserving

a n d , f o r g i v e n p > O , it s h i f t s t h e c i r c l e {A:

IAl= p)

d i r e c t i o n of t h e p o s i t i v e r e a l a x i s . T h e r e f o r e

whence t - < t l + t 2 .

in the

32

3

CHAPTER

EXERCISE. ( a ) I f M ( c / ) = N ( c x ) , t h e n t h e r e e x i s t s

3.3.

ke3A s u c h t h a t M M1,aMk,OM1,-a= N . ( b ) Give a d e t a i l e d p r o o f f o r

(3.2).

(c) P r o v e t h a t a l l Aut A - i n v a r i a n t m e t r i c s a r e of t h e form

where f :

[0,1)%3+

is a function t h a t s a t i s f i e s If ( 0 1 -f (a).! cf (

I

7 Li-u 1 ) 1-&I3

for a l l

n,~c[0,1). ( d ) Does a n y f :

property

(*)

d e f i n e a n Aut A - i n v a r i a n t

3.4.

REMARK.

[O,l))IR+

with t h e

metric on A?.

The p r o o f of t h e lemma h a s t h e f o l l o w i n g

h e u r i s t i c a l background: I f A i s c o n s i d e r e d a s t h e PoincarG

model1 of t h e Bolyai-Lobatchewsky

plane ( t h u s t h e s t r a i g h t lines

a r e r e a l i z e d a s c i r c l e s o r t h o g o n a l t o 3A and t h e g r o u p Aut A r e p r e s e n t s t h e g r o u p of e v e n c o n g r u e n c e s o f t h e p l a n e ) , t h e n

Ut i s a s h i f t o f t u n i t s i n t h e r i g h t d i r e c t i o n on t h e l i n e (-1 , I ) .

The d i s t a n c e d A i s c a l l e d t h e P o i n c a r e ' m e t r i c . §2.

-

The C_ a r a t_ h g o~ d o r y- p s~ e u d. ometric. _

Now w e a r e g o i n g t o c o n s t r u c t a n Aut D - i n v a r i a n t metric i n bounded domains D of a Banach s p a c e E .

3.5.

DEFINITION. L e t D be a b o u n d e d domain i n E . F o r

x , ycD we s e t

We c a l l d D t h e " C a r a t h d o d o r y p s e u d o m e t r i c " o n D. 3.6.

El.

THEOREM.

L e l D a n d D 1 b e b0undc.d domains i n E a n d

T h e n a n y f s H o l ( D , D ) is a c o n t r a c t i o n w i t h r e g a r d t o t h e 1

CARATHEODORY DISTANCE AND COMPLETENESS PROPERTIES

c o r r e s p o n d i n g C a r a t h Q o d o r y p s e u d o r n e t r i c s dD, dD

d

D

i s a n Aut D - i n v a r i a n t

1

.

33

In particular

pseudometric.

Proof: Given any mapping +eHol(D,A), the expression

is trivially a pseudometric on D since dA is a metric on A. Since the supremum of any family of pseudometrics is also a pseudometric, d is a pseudometric on D. If fsHol(D,D1) then, D for x, ycD, we have

3.7. EXERCISE. Show that for D=: A , the Caratheodory pseudometric coincides with the Poincarg metric on A. 53.- The Carathgodory differential pseudometric.

Next we compare the pseudometric dD with the natural metric induced on D by the norm of the space E. Before doing it we introduce an easily manageable relative of dA. 3.8. DEFINITION. For XCD and VsE ( 3 4)

6D (x,v)=: sup{\$;’(v)

1;

we s e t

@sHol(D,A), $(x)= 0 )

We c a l l A D t h e “ C a r a t h L o d o r y d i f f e r e n t i a 2 p s e u d o m e t r i c ” o n D.

Th s terminology is explained by the following lemma:

(a) T h e m a p p i n g (x,v)+~~(x,v), xcD, vCE,is lower s e m i c o n t i n u o u s a n d , f o r a n y f i x e d XCD, AD(x,-) is a seminorm on E. 3.9. LEMMA.

(b) I f X, ysD a n d t+xt i s a s m o o t h c u r v e j o i n i n g x w i t h y so t h a t xo= x, x = y , t h e n 1

CRAPTER

34

3

rl

P r o o f : The f i r s t s t a t e m e n t f o l l o w s d i r e c t l y from t h e

$eHol(D,A) , 6J,D ( x , v ) = : l$i'(v)

f a c t t h a t , f o r any f i x e d

I

has

t h e mentioned p r o p e r t i e s . Given @cHol(D,A) , we have ' d

-

d A [ $ ( y )I ~ ( x ) ] =

d a [ @ ( X t + srl @ ( x ) ] d t

0 ds'o

which, by t h e t r i a n g l e i n e q u a l i t y , i s dominated by

and, by s e t t i n g

+ t =:

1.1

l,-Uxt)

9 , t h e l a s t i n t e g r a l can be

w r i t t e n i n t h e form

j1 1

J, I

(x,)

X;

1 dt,c

0

' 6 D ( x t ,xi)d t

'0

s i n c e $teHol(D,A) and $ ( x ) = 0 t

t

for all t c ( 0 , l ) .

i&

The holomorphic mappings have t h e f o l l o w i n g c o n t r a c t i v e c h a r a c t e r i n terms of t h e Caratheodory d i f f e r e n t i a l m e t r i c : 3.10.

PROPOSITION. If f : D+D1 Cs u h o l o m o r p h i c i n u p p i n g ,

Lhcn Me h a v e

f o r a l l xsD a n d vcE.

P r o o f : W e have

CARATHEODORY DISTANCE AND COMPLETENESS PROPERTIES

3.11.

COROLLARY.

If

35

i s biholomorphic and

f : D+D1

f ( D ) = D 1 t h e n we h a v e

6

[f(x), f ' l ( v ) ] =

AD(X,V)

Dl

f o r a l l , xcD a n d vsV.

Proof:

94.-

R e l a t i o n s between t h e Carathgodory pseudometric and t h e

m e t r i c on D. norm ---_= Now w e a r e g o i n g t o d i s c u s s t h e r e l a t i o n s b e t w e e n t h e C a r a t h g o d o r y p s e u d o m e t r i c a n d t h e metric i n d u c e d on D by t h e norm o f t h e space E .

3.12.

PROPOSITION. I f B is t h e b a l l Of radius p w i t h

c e n t e r asD, we h a v e

(3.5)

6B ( a , v ) =

1 ; I / vll

and (3.6)

d B ( a , x ) =d ( A

1 P

/I

x-a/I , O ) =

tanh-l(

1

(1

x-a(/)

for all xsB a n d vsV. P r o o f : L e t veE and + s H o l ( B , A ) w i t h + ( a ) =0 b e g i v e n and c o n s i d e r t h e f u n c t i o n @ : have + ( O ) =

A+A d e f i n e d by @ I S ) =

0; t h u s , by t h e Schwarz lemma w e h a v e

+ ( a + j v ) .W e / $ ( j ) / c / < f/o r

a l l SsA, whence w e o b t a i n

1 Therefore 6 ( a , v ) < B

P

/I v / /.

On t h e o t h e r h a n d , g i v e n a n y

vsE\{O}, by t h e Hahn-Banach t h e o r e m t h e r e e x i s t s a c o n t i n u o u s

CHAPTER 3

36

l i n e a r f u n c t i o n a l $oeE

*

such t h a t

2.

Then, for t h e f u n c t i o n $ ( X I = : [ 'L

$(BIZ

1

;( x - a ) ]

2.

w e have $ ( a ) = 0 ,

A,

T h i s p r o v e s ( 3 . 5 ) . To show ( 3 . 6 ) i n t r o d u c e t h e same f u n c t i o n s

, fix $ as

xeB, s e t v = : x-a a n d b e f o r e . S i n c e $sHol(B,A) ,

by t h e o r e m 3.6 w e h a v e

1

I(

d , ( x , a ) C d A ( $ ( x l, $ ( a ) ( = d A ( ; x-a(I , O ) 'L

2.

On t h e o t h e r h a n d , by t h e d e f i n i t i o n o f d,,

3.13. P=:

THEOREM.

L e t K c c D be u convex s e t w i t h

d i s t ( K , aD). T h e n w e h a v e d D I K Q d i S t l l . I /

exactly 1

diam (D)

2.

dA(x,a)>dAl$

I/

x-YII

q ) ( X l Y ) $

p1 II

IK

'

or m o r e

x-YII

f o r a l l x,ycK.

P r o o f : Given yeX, l e t B d e n o t e t h e b a l l of c e n t e r y and r a d i u s diam(D)

.

S i n c e B T D , w e h a v e d B I D s d D .But

d B ( x , y )= t a n h

f o r a l l xsB b e c a u s e t h e f u n c t i o n t a n h - I .is c o n v e x on LO,=)

i t s d e r i v a t i v e a t t h e p o i n t 0 i s equal t o 1 , h e n c e t a n h for a l l t>O.

This proves t h e left-hand

1

and F,><

side inequality.

To see t h e s e c o n d i n e q u a l i t y , f i x x,ycK a r b i t r a r i l y and s e t

x t = : x + t ( y - x ) , B t =: t h e b a l l w i t h c e n t e r x t a n d r a d i u s p . W e have B c D f o r a l l t

tc[0,1].

Therefore

CARATBEODORY DISTANCE AND COMPLETENESS PROPERTIES

37

f o r a l l veE. Hence

# 3.14.

COROLLARY.

The t o p o l o g i e s i n d u c e d on D b y dD

a n d t h e norm o f E c o i n c i d e . I n p a r t i c u l a r d D i s n o t o n 2 3 a pseudometric but a metric. 55.-

C -o m p l e t e n e s s p r o p e r t i e s o f t h e g r o u p Aut D . _ ___

(f ) b e a T - C a u c h y s e q u e n c e in I I ~ J A u t D a n d a s s u m e t h a t for> some aeD we h a v e f , ( a ) + b s D . T h e n THEOREM. L e t

3.15.

there exists

fsAut D

such t h a t T l i m f . = f . I

I

JCJ

P r o o f : C l e a r l y w e h a v e a u n i q u e mapping f e H o l ( D , E ) w i t h Tlim f = f . T h u s , i f were f e A u t D , by t h e o r e m 2 . 2 we would 3 IeJ h a v e T l i m f T 1 = f-l. jeJ

1

We show t h a t -1

( f j )jcJ

(3.7)

i s a T-Cauchy

sequence.

L e t u s w r i t e b . = : f , ( a ) and c h o o s e a n y b a l l s 1

3

B c c D and

B ' c c D c e n t e r e d r e s p e c t i v e l y a t a and b s u c h t h a t f ( B ) c B ' .

From t h e o r e m 3 . 1 3 i t f o l l o w s t h a t

Hence w e c a n f i x 6>0 s o t h a t t h e set C=

i s contained i n B. L e t u s i n t r o d u c e a l s o Cf=: 1 W e may assume w i t h o u t l o s s o f g e n e r a l i t y t h a t d D ( b . , b ) <6 for I 2 a l l jsJ. T h e n , from t h e t r i a n g l e i n e q u a l i t y a p p l i e d t o d D a n d

theorem 3.6

we obtain

38

CHAPTER

3

f o r a l l jeJ. L e t u s c o n s i d e r a n y y e C ' . By t h e o r e m 3 . 6 w e h a v e

because f

t1 ( y )e C . 3

Therefore

But t h e n , by t h e o r e m 3 . 1 3 w e a l s o h a v e

S i n c e by c o r o l l a r y 3 . 1 4 , C' c o n t a i n s some o p e n b a l l , f r o m theorem 1 . 6 w e o b t a i n t h a t

for a n y b a l l B 1C C D. Thus ( 3 . 7 ) i s e s t a b l i s h e d a n d i t f o l l o w s t h e e x i t e n c e o f a u n i q u e gcHol(D,E) s u c h t h a t T l i m f T 1 = g . jeJ

The t h e o r e m i s now a c o n s e q u e n c e of

( 3 . 7 ) and ( 3 . 8 ) . Indeed,

( 3 . 8 ) i m p l i e s f - l ( C ' ) c C f o r all jcJ so t h a t g ( C ' ) C ? C C D . 1

T h e r e f o r e t h e mapping f g i s d e f i n e d on C ' a n d w e h a v e

f g = l i m f . f T 1 = i d on C ' jeJ

3 1

.

To c o m p l e t e t h e p r o o f w e must show

t h a t g ( D ) c D s o t h a t f g i s d e f i n e d on D and fg= i d D . L e t u s f i x a n y xsD. W e h a v e a l r e a d y s e e n t h a t , by s e t t i n g g ! = : f: 1

1 IC'

and f ! = : f . J

JIG

,

the r e l a t i o n

T l i m f'g'= f o g IC' j,keJ j k

h o l d s . Hence i t f o l l o w s (f . f - l ) ~ S - ( i d D ) f~o"r s= 0 , l . Theorem 2.7

i m p l i e s T l i m €.f-'= 3 k

by theorem 3.6 j '

3 k

idD. I n p a r t i c u l a r , f . f - l ( x ) + x 7 k

SO

that,

dD(f . f x , x ) + d D ( x , x ) = 0 I

k

Thus g i v e n a n y b a l l B c c D centered a t x , t h e r e i s joeJ t h a t ( f j o f -k1 X)k-j

0

i s a dD-Cauchy n e t c o n t a i n e d i n B. By

such

CARATHEODORY DISTANCE AND COMPLETENESS PROPERTIES

39

t h e o r e m 3 . 1 3 , t h i s means t h a t ( f , f i 1 x ) k 2 j o i s a l s o a Cauchy '0

n e t w i t h r e g a r d t o t h e norm, and h e n c e i t a d m i t s a l i m i t y e g . T h e r e f o r e w e have

g(x)= l i m f - l x = f - l l i m f . f i l x = f k

keJ

10 k e J

'0

-1

Y ~ D .

10

S i n c e ( 3 . 7 ) i s t r u e a n d f T 1 ( b ) + a , w e may a p p l y a l l o u r r e s u l t s -1

t o ( f . 1jeJ 3

fs l u t D

.

3

instead

( f j )j e J .

Of

Hence g f = i d D' t h a t i s ,

# 3.16.

REMARK.

I n g e n e r a l , Aut D i s n o t T-complete

as

t h e f o l l o w i n g example shows: L e t u s p u t D=:

A

a n d d e f i n e f =: Un by means o f

Then Un i s T - c o n v e r g e n t t o t h e c o n s t a n t mapping U :

1'1;

however

U$Aut A . T h i s phenomenon j u s t i f i e s t h e a s s u m p t i o n o f t h e e x i s t e n c e o f a p o i n t asD s u c h t h a t l i m f , ( a ) =bsD i n t h e t h e o r e m . jeJ

3

L e t u s n o t e , f u r t h e r m o r e , t h a t t h e boundedness of D i s a l s o of

c r u c i a l i m p o r t a n c e a s shown by t h i s e x a m p l e : L e t u s t a k e D=:

E and d e f i n e f cAut D by means o f

f n : x+ I x n

xeE,

nm.

Then f n i s T - c o n v e r g e n t t o t h e c o n s t a n t mapping f : x+O w h i c h c l e a r l y i s n o t i n Aut E . T h i s s e c o n d e x a m p l e s h a d s some l i g h t o n t h e u s e o f t h e C a r a t h E o d o r y metric i n t h e p r o o f . Roughly s p e a k i n g it s e r v e s t o e s t a b l i s h t h a t t h e F r g c h e t d e r i v a t i v e s of t h e f . t e n d t o o p e r I a t o r s which a r e bounded from b e l o w . 3.17.

EXERCISES. ( a ) P r o v e t h a t i f D = D , t h e n d >d. 1 D 1 / DD ( b ) Show t h a t g i v e n f s A u t D a n d xeD,

w e h a v e 6D(x,v)= G D [ f ( x ), f i l ( v ) ] f o r a l l v s E .

40

3

CHAPTER

( c ) Prove t h a t t h e l e f t uniform s t r u c t u r e o f T on Aut D i s c o m p l e t e . ( d ) On Aut D , t h e t o p o l o g i e s o f p o i n t w i s e c o n v e r g e n c e a n d u n i f o r m c o n v e r g e n c e on compact subsets of D c o i n c i d e .

( e ) Show t h a t , f o r a n y Banach s p a c e E, t h e open u n i t b a l l B ( E ) o f E i s a l w a y s d B ( E ) - c o m p l e t e . ( f ) P r o v e t h a t t h e domain o f d e f i n e d by D = :

Then D

A\{O}

i s n o t d -complete. D

3 . 1 8 . THEOREM. L e t D b e a b o u n d e d h o m o g e n e o u s d o m a i n . is a c o m p l e t e m e t r ~ i cs p a c e w ~ t hraegard t o d D .

c D be a Cauchy n e t w i t h r e g a r d t o t h e metric d D . F i x a n y p o i n t aeD a n d s e t 6=: d i s t ( a , a D ) . T h e n , i f B d e n o t e s t h e b a l l ( f o r t h e m e t r i c o f t h e norm on D ) c e n t e r e d a t a w i t h r a d i u s p < 6 , we h a v e B c D . By c o r o l l a r y 3 . 1 4 w e c a n Proof: L e t (x, )

,

7 JfJ

find a ball B

( f o r t h e m e t r i c d D ) c e n t e r e d a t n w i t h some

r a d i u s r such t h a t w e have

Since ( x . ),

1 1eJ

i s a d -Cauchy n e t , w e c a n f i x a n D

index j eJ so 0

that dD(xk,x. )
f o r a l l k s J , k>,j N o w , a s D i s homogeneous, t h e r e e x i s t s some 0' gcAut D s u c h t h a t g ( x . ) = a . T h u s , a s d D i s A u t D - i n v a r i a n t , w e get

'0

so t h a t

f o r a l l k e J , k > j o . By t h e o r e m 3 . 1 3 , e q u i v a l e n t on B; t h e r e f o r e ,

( g (x,) )

d

a n d t h e norm m e t r i c a r e

D

i s a Cauchy n e t f o r

~~

0

CARATHEODORY D I S T A N C E AND COMPLETENESS P R O P E R T I E S

/I - 1 1

and t h e r e e x i s t s some g e E c D such t h a t

from which we d e r i v e

a g a i n by theorem 3 . 1 3 . A s < = g ( q ) f o r some q c D and d A u t D-invariant,

w e have

so t h a t D i s d -complete. D

D

is

41

This Page Intentionaiiy Left Blank

4

CHAPTER

THE L I E ALGEBRA OF COMPLETE VECTOR FIELDS

P e r h a p s t h e most i n t e r e s t i n g q u e s t i o n c o n c e r n i n g Aut D i s how t h e a u t o m o r p h i s m s c a n be d e t e r m i n e d from t h e s h a p e of D . To t a c k l e t h i s p r o b l e m d i r e c t l y seems t o b e h o p e l e s s i n g e n e r a l .

It i s much m o r e f r u i t f u l , a s i t was f i r s t o b s e r v e d b y E . C a r t a n , t o b e g i n t h e c o n s i d e r a t i o n s w i t h t h e d e s c r i p t i o n of t h o s e a u t o m o r p h i s m s t h a t , i n some s e n s e , l i e v e r y n e a r t o t h e i d e n t i t y The b a c k g r o u n d of t h i s i d e a i s t h e e x p e c t a t i o n t h a t , i f t h e b o u n d a r y o f D i s a smooth h y p e r s u r f a c e a n d f s A u t D i s close t o i d D , t h e n , i f t h e v e c t o r f i e l d v=: x + x - f ( x )

c a n be continuously

e x t e n d e d t o a D , v must b e a l m o s t t a n g e n t t o a D . T h u s , intuitively, i f t h e d i f f e r e n c e between f and i d D i s i n f i n i t e s i m a l , t h e n f-id

D

i s a h o l o m o r p h i c v e c t o r f i e l d of i n f i n i t e l y s m a l l v e c t o r s

t h a t are t a n g e n t t o 2 D . such vector f i e l d s

I t c a n b e hoped t h a t t h e c a l c u l a t i o n o f

i s e a s i e r t h a n t h e s o l u t i o n of t h e o r i g i n a l

p r o b l e m . However, w e a l s o h a v e t o c o n s i d e r t h e c a s e of a non smooth b o u n d a r y a n d t h e c a s e i n w h i c h t h e e l e m e n t s o f Aut D c a n n o t b e c o n t i n u o u s l y e x t e n d e d t o aD. 51.-

One-parameter

subgroups.

I n what f o l l o w s w e s h a l l d e n o t e by [t]

t h e e n t i r e p a r t of t h e

r e a l number t d R .

*

c A u t D, ( t j ) jewcIR+ with 1 l a t , + O and AsHol(D,E) b e g i v e n , and a s s u m e t h a t t h e s e q u e n c e I 4.1.

PROPOSITION. L e t

(f ,)

1 tj

,

A,=: - (f.-idD) 7

7

s a t i s f i e s T l i m A , = A. Then, f o r e v e r y t d R , j-m

J

ftsAut D such t h a t 43

t h e r e e x i s t s a unique

4

CHAPTER

44

]+Cc

J

M o ~ c u u e r , i h e m u p p i n g IWAut D g i v e n b y t + f

t

is a T - c o n t i n u o u s

ii o mu rrio rp h i sm . P r o o f : The p r o o f p r e s e n t e d h e r e i s c e r t a i n l y n o t t h e s h o r t e s t , b u t it p r o v i d e s a d e e p i n s i g h t i n t o t h e b a c k g r o u n d of t h e t h e o r y .

We d e f i n e l i n e a r o p e r a t o r s

9 . a n d f i . on H o l ( D , E ) a s f o l l o w s : 1

1

(4.1)

O b s e r v e t h a t , i f we w r i t e if f o r t h e f u n c t i o n x + f i l A ( x ) , xCD, we h a v e

(4.3)

Proof .of ' ( 4 . 3 ) : L e t B C C D b e a b a l l a n d f i x 6 > 0 so

t h a t Bg=: B + G B ( E ) c U . Then, by t h e T a y l o r e x p a n s i o n of f , w e have

%,f ( X I = I

1 t { f [ f . (x)]-f ( X ) I = 1

1

-1

{ f [x+t.A.(x)]-f ( x ) } = 3

3 7

for a l l xeB. U s i n g t h e Cauchy e s t i m a t e s w e o b t a i n

Moreover, from t h e Cauchy e s t i m a t e s w e d e r i v e

II

f

' A , - f ( 1A [ ( , = 3

(1

f

COMPLETE VECTOR FIELDS

so t h a t T l i m f ( ' A , = f ( l A i n U . The r e l a t i o n ( 4 . 2 ) j+m

3

45

is obvious.

H e n c e f o r t h , l e t u s f i x a b a l l B c D a n d a number 6 > O w i t h t h e property that

/ I All

<m.

W i t h o u t l o s s of g e n e r a l i t y w e may assume

M=:

sup{ / I

B6

~

~

B6

1 ; jml<m 1

O b s e r v e t h a t , a s a n y h o l o m o r p h i c mapping i s l o c a l l y b o u n d e d ,

f o r e v e r y xsD t h e r e i s a b a l l U c e n t e r e d a t x s u c h t h a t

I]

All,


L e t us define

exptti)f=:

c

tk -k A f

k=O k!

W e s h a l l show t h a t t h e r e e x i s t s a number T > O s u c h t h a t , f o r e v e r y t c L O , T]

, we

have

(4.4) w i t h r e s p e c t t o t h e norm

(1 - // B '

and

(4.5) Then, w e c a n e a s i l y d e d u c e t h e p r o p o s i t i o n f r o m 4 4 . 4 )

and

( 4 . 5 ) . Indeed, i f t c [ O , r ] , we have

f:

3

un form y f o r xeB. By ( 4 . 5 )

SO

t h a t w e have

C

tk -.k k !(A idD)xcB6fC D

k=O

f o r a l l xcB. Then t h e o r e m s 3 . 1 5 a n d 1 . 6 e n s u r e t h a t ( t h i s i s t h e main p o i n t of t h e p r o o f ! ) :

CHAPTER

46

Tlim fi j+m

f o r a l l tc[O,i].

=

4

ftsAut D

J

NOW, t h e mapping [ O , T ] + H O ~ ( D , E ) g i v e n b y

t + f t i s a n a l y t i c i n t h e r e a l s e n s e i f w e endow H o l ( D , E ) w i t h t h e norm

11 - I I B

; h e n c e , by t h e o r e m 1 . 6

it i s a l s o 2'-continuous.

I n t h e same way a s i n e l e m e n t a r y a n a l y s i s w e c a n e s t a b l i s h t h e relation

for a l l s,te[O,r]

with s.+tc[Q,.r]. F i n a l l y , f o r t d R , we extend

t h e d e f i n i t i o n o f f t by s e t t i n g

One c h e c k s i n a s t r a i g h t f o r w a r d manner t h a t ( 4 . 6 ) h o l d s f o r a n y s,tdR. Then, by t h e o r e m 2 . 2 ,

w e have

Tlim f

=

ftcAut D

j +m

f o r a l l t a r a n d t h e mapping IR*AutD g i v e n by t + f t i s T-continuous s i n c e it i s so i n [o,T]. P r o o f of

11

gl/ K < m

(4.4):

L e t us c o n s i d e r a n y g s H o l ( D , E ) w i t h

where K i s b a l l c o n c e n t r i c w i t h B a n d

K Z E = : K + 2 € B ( E ) c B 6 . Assume t h a t t h e i n d e x j

have

11

fj-idD1l

Then, i f w e w r i t e v = : f j ( x ) - x

is l a r g e enough t o

<E

B6

f o r a n y xeK, w e h a v e

w h i c h , by t h e Cauchy e s t i m a t e s , i s d o m i n a t e d by

47

COMPLETE VECTOR FIELDS

Using t h i s o b s e r v a t i o n and t a k i n g

whenever n <

. 2Mt.

E=:

1 2n

6 we obtain

T h e r e f o r e w e may a p p l y ( 4 . 6 ) f o r e s t i m a t i n g

1

-

t h e b i n o m i a l e x p a n s i o n of

6 2M I .

A

f o r any t e . 0 ,

( I + t, A , ) 1 3

T h i s i s p e r h a p s t h e c r u c i a l p o i n t o f t h e whole s e c t i o n . L e t u s

6 2~

fix ts[O,

and w r i t e n , = :

)

3

[tti']. W e

have n . < 3 2Mt.

.

we

3

know t h a t

For f i x e d k w e have n.

k

(,j)t.=

3

1 -

k!

( n . t . )( n . - 1 ) tj . . . ( n . - k + l ) t * 3 7

3

3

j

tk k!

a n d , by ( 4 . 3 ) ,

Thus t h e p r o p o s i t i o n f o l l o w s f r o m t h e b e l o w e l e m e n t a r y lemma: 4.2.

LEMMA. I f

i n a Banach s p a c e

El,

m

we h a v e

w h e r e C / / a s / / < m , and t h e r e e x i s t S=l s u c h t h a t / l a j \1 6 w s f o r a 2 2 j , s = 1 , 2 ... w i t h

as+as (s= 1 , 2 , . . . I 3

w1,w2, m

c

w

s=l

...

<m

,

m

then

c

s=l

m

as J

+

c

as.

s=l

I n d e e d , it s u f f i c e s t o a p p l y t h e lemma t o t h e s p a c e E =: 1

{ h c H o l ( B , E ) ; h i s bounded on B ) w i t h t h e norm

t o the vectors as=: ( z j ) t ; i ; ( f ) 1

IB

,

t S

as=: s! i s ( f )

IB

11 * I I B

and

with the

48

CHAPTER

4

Proof of t h e lemma: W e h a v e

/I a s \ ($0'

for a l l sdN;

W

t h u s t h e series

C a s i s a b s o l u t e l y c o n v e r g e n t . Then s=l

c

m

for a l l Nc.IISI: : Given E > O ,

t h e r e e x i s t s N such t h a t

Once s u c h a n N h a s b e e n f i x e d , t h e r e e x i s t s j cTN s=1

11

0

as-aSI/<~/2 7

t-tf

such t h a t

for a l l j > j0 ' T h i s c o m p l e t e s t h e p r o o f .

4 . 3 . COROLLARY. t

us<€/2.

s=N+1

For on$ XCD,

L h c r n u p p i n g IR+D g i z ~ e nbg

(x) s a t i s f i c s d d t f t ( x ) = A[ft(x)]

o n L h e whole real l i n e .

4.4.

DEFINITION. A v e c t o r f i e l d x + A ( x ) , X E D ,

is sai'a?

to be " c o m p l e t e in D r r if the m a x i m a l s o l t i t i o n o f L h c initin2

value problem a d t

(4.10)

~t = A (t x ) , x t ~

, X0 =

D

X

i s d e f i n e d o n the w h o l e IR. If AeHol(D,E)

is a v e c t o r f i e l d w h i c h i s c o m p l e t e i n D , t h e n

we define exp A=: x+xl,

xsD

where x 1 i s i m p l i c i t e l y g i v e n by ( 4 . 1 0 ) . T h u s , what w e h a v e p r o v e d c a n be r e f o r m u l a t e d a s

4 . 5 . THEDREM. L e t

assume t '0, j

b e a s e q u e n c e i n Aut D and

(f )

j a t h a t , f o r some AsHol(D E ) a n d some zL)e

have Tlim

I -

j+cc

t j

f .-,idD) = A 3

( t . ).

COMPLETE VECTOR FIELDS

Then t h e v e c t o r f i e l d A

is c o m p l e t e in D a n d We h a v e

[ttj']

Tlim f , j+W

49

= exp(tA)

J

f o r a l l t d R . M o r e o v e r , t h e m a p p i n g IR+AutD g i v e n b y t + e x p ( t A ) i s

a T-continuous 4.6.

homomorphism. EXERCISES. ( a ) Give a d e t a i l e d p r o o f f o r ( 4 . 9 ) ( b ) Given a n y AeHol(D,E) a n d t o d R , l e t W

d e f i n e e x p ( t A) a s f o l l o w s : 0

dom e x p ( t o A )= {xeD; t h e r e i s a n i n t e r v a l JCR, [O,tO]cJ,

f o r which t h e i n i t i a l v a l u e p r o b l e m ( 4 . 1 0 )

h a s a s o l u t i o n on J}. exP t o A ) x = : xtO where xtO i s i m p l i c i t e l y d e f i n e d by (4.

0).

1 ) Check t h a t t h i s i s a n e x t e n s i o n o f t h e d e f i n i t i o n o f e x p A

as p r e v i o s u l y s t a t e d f o r complete v e c t o r f i e l d s and t 2 ) Show, a s i n t h e proof of (4.41,that we have T l i m ( I + on e a c h o p e n U s u b s e t U c c d o m e x p ( t A )

.

t

0

1.

=

A)"=exp(tA)

n+m

3 ) Show t h a t , f o r e a c h o p e n s u b s e t U w i t h U c c d o m e x p ( t A ) , w e n C n i m k = O k!

have T l i m

z k ( i d ) = exp(tA).

4 ) Use lemma 2 . 4

t o show t h e o r e m 1 . 6 .

52.-

D

Complete h o l o m o r p h i c v e c t o r f i e l d s .

The f o l l o w i n g q u e s t i o n r a i s e s n a t u r a l l y f r o m t h e p r e v i o u s t h e o r e m : Given a T - c o n t i n u o u s o n e - p a r a m e t e r g r o u p t + f Aut D ,

is italways possible t o find AeHol(D,E)

in

such t h a t

f t = e x p ( t A ) f o r a l l t d R ? I t i s t h e same a s a s k i n g w h e t h e r t h e 1 T l i m ;( f l / n - i d ) n e c e s s a r i l y e x i s t s , o r w h e t h e r t h e T - c o n t i D n+n u i t y o f t + f a l w a y s e n t a i l s i t s T-dif f e r e n t i a b i l i t y

.

Our n e x t t a s k w i l l b e t o p r o v e t h a t t h e a n s w e r i s a f f i r m a t i v e . This r e q u i r e s a c a r e f u l study of t h e p o s i t i o n of t h e p o i n t s fk/n ( x ) , k = 1,2,... i n comparison w i t h x + x ( f ' / " ( x ) - x ) f o r l a r g e n . I n t h i s s e c t i o n , w e p r o v e a lemma d u e t o H . C a r t a n w h i c h , i n

CHAPTER

50

4

many aspects, can be considered as a discrete version of the Bellman lemma in the theory of ordinary differential equations, 4.7. LEMMA. L e t fsAut D b e g i v e n a n d nssumc t h a t

B c c D is a b a Z l w i t h O
-XI

by the Cauchy estimates, we have

Proof: If p g n - I ,

=

I(

xp-yp/I+ ( / [fP-idD] (xl)-[fP-idD] (x,)

:: II Xp-YpIl

'1

d-6 II

+

,

Since x 0 = yo and x 1= y 1 by induction

d

I1 xl-xoII

the statement of the lemma follows

COROLLARY. I n t h e a b o v e conditions, we h a v e

4.8.

11

fP-idD I1

(fp-idD)-p(f-idD)] I B

C.

1

I[

11

f-id

PC1

11

fk-idDl/

k=l

whenever

/I

I/ 5

k

f -idDIIB <6

Bd

f o r k= 1,2,..,p-I.. In pariicular we

have

w h e n e v e r p~p(&),w h e r e

p(6)=: maxtp:

1)

fk-idDI/

<6

f o r a22

kcp}.

a

Proof: We have

x -y = [fP(x)-x]-p[f P P

(x)-x]

whence the assertions are immediately obtained by the arbitra-

51

COMPLETE VECTOR FIELDS

r i n e s s of xcB. 4.9.

REMARK.

Roughly s p e a k i n g , t h e lemma a n d i t s

corollary a s s e r t t h a t the triangle xo,xp,yp

i s a l m o s t symme-

t r i c and i t s a n g l e a t x 0 i s i n f i n i t e s i m a l i n c o m p a r i s o n w i t h 1 . max{ 11 f k-idD]IB , k= l,...,p}. F o r v l , v 2 s E , w e c a n d e f i n e

a

However, i n g e n e r a l , t h i s c o n c e p t i s n o t so u s e f u l a s i t i s i n t h e case of H i l b e r t spaces. I t i s i m p o r t a n t t o know t h e l i m i t s w i t h i n w h i c h t h e i n e q u a l i t y

can be a p p l i e d . Thus, w e g i v e upper and lower bounds f o r

(4.11)

p(6).

4.10.

LEMMA.

L e t fcAut D be g i v e n a n d assume t h a t

B c c D i s a b a l l w i t h d < d i s t ( B , a D ) . Then t h e number p ( 6 )

satisfies

6

(4.12)

II

f-id

-l
II

Bd+6

26

/I

f-idDIID

1 f o r a l l . 6 , 0<6< -d. 4

p= 0 , l

P r o o f : F i x a n a r b i t r a r y zOcBdand s e t z - = :f p ( z , ) f o r P By t h e d e f i n i t i o n o f p ( 6 ) w e have z cBdt6 i f k,
,...

P

T h u s , f o r p ~ p ( 6+)I w e o b t a i n

11

zp-zo

/I s /I

zo-z1

1 1 +. . .+ / /

II=

Z ~ - ~ - Z

P-1 C k=D

11

( f - i d D ) z k / / 6 P / / f - i dI/

Bd+6

whence, by t h e a r b i t r a r i n e s s o f z o i n B d , w e g e t (4.13)

f o r p < l + p ( 6 ) . But i n t h e case p= l + p ( 6 ) w e h a v e

To p r o v e t h e s e c o n d i n e q u a l i t y w e make u s e of

( 4 . 1 1 ) . W e have

CHAPTER

52

Thus, for 6 <

4

Td we have

Finally, from the first inequality or corollary 4 . 6 and ( 4 . 1 3 1 we derive the following estimate in terms of f-idD only:

for all p$l+p(6). t

THEOREM. L e t t-+f be a l ‘ - c o n t i n u o u s o n e - p r u m e t e r g r o u p i n Aut D. Then t h e r e e x i s t s a u n i q u e h o Z o m o r p h i c v e c t o r f i e Z d A i n D s u c h t h a t A i s c o m p l e t e i n D a n d we h a v e ft= exp(tA) for2 a 2 2 tdR, 4.11.

Proof: By theorem 4 . 5 it suffices to show that the sequence A =: n(fl’n-idD) , nEN, is T-convergent in Hol(D,E). Thus it n

suffices to prove that ( A n ) n m is a T-Cauchy sequence. Let B C C D be a ball and assume that E > O has been given. Choose 6 6 > 0 small enough to have d-fi < E . By assumption we can find a number cr>O such that we have

(1

fs-idD(/ <6 Bd

1 for all sc(-o,+o). Let m , n m be large enough to have m,n> 3. Observe that ( 4 . 1 1 ) can be applied to the automorphism fl’mn and to the number p= m,n because, for k&max{m,n], we have k/mn
(1

(fm’mn-idD ) -rn(fl’mn-idD) [ (

or, multiplying by n,

B < E m ( /f l’mn-

idD((

COMPLETE VECTOR FIELDS

II

An-Amn

11

BsE

I/

53

AmnII

In a similar manner we obtain

II

A,-AmnIIB<4

AmnlIB

By the triangle inequality,

I1

A,-A,I/

B<2EI/AmnII

Therefore, the proof will be complete if we establish the existence of a number T > O such that sup

(4.15)

/I

AtIIB

( ~ 0

where

A t= :

1

t (ft-idD)

1tls.T

Proof of ( 4 . 1 5 )

:

Let u s choose

so that we have

T

Fix te(--r/2, 1/2) arbitrarily and set n=: [~t-']

.

Then ( 4 . 1 1 )

yields

I/

nt

t

(f -idD)-n(f -idD lIB'c

n

2 11

f

t

-idDI/

whence by the triangle inequality we obtain nl/ft-id,II that is ,

since n t > ~ / 2and, finally

<21/fnt-idD/I

,

54

CHAPTER

4

COROLLARY. T h e v e c t o r f i e l d A i s b o u n d e d o n

4.12.

every ball B c c D . 4.13.

u

and l e t ( f , )

( a ) L e t U C C D be a n o p e n s u b s e t of

EXERCISE. ,

1 l a

c A u t D a n d ( t, )

s u c h t h a t , f o r some AeHol(U,E) ,

ClR:

jm w e have

1

,

tj+O,

be sequences

1 l i m - ( f .-id ) = A t I D j +W 1

w i t h r e s p e c t t o t h e norm

11 - 1 1 "

. Show

t h a t t h e n , A c a n be

h o l o m o r p h i c a l l y extended t o D and i t s e x t e n s i o n i s a v e c t o r f i e l d which i s c o m p l e t e i n D .

f

t

($I=

(eit$lle

Z i t

( b ) Examine t h e e x a m p l e t + f t

q2,..)

and $= (+,l$,...)ek

m

.

,

where

( c ) Look f o r a n unbounded c o m p l e t e h o l o m o r p h i c v e c t o r f i e l d i n some bounded d o m a i n . §3.- The L i e a l g e b.r -a of c o m p l e t e h o l o m o r-p h i-c v e c t o r f i e l d s . 4.14.

DEFINITION.

The f a m i l y o f a l l h o l o m o r p h i c vector

f i e l d s A o n D t h a i a r e c o m p l e t e i n D # i Z l be d e n o t e d b y a u t D .

LEMMA. L e t A s a u t D b e g i v e n . If # e w r i t e f =: e x p ( t A ) , t h e n # o h a v e f t e A u t D f o r a l l t d R . M o r e o v e r , t h e m a p p i n g t + f t i s T - c o n t i n u o u s a n d we h a v e t h e r e l a t i o n 4.15.

t

Tlim t+O

1 t

(ft-id

) = A.

P r o o f : I t i s w e l l known f r o m t h e e l e m e n t a r y t h e o r y of d i f f e r e n t i a l e q u a t i o n s t h a t , g i v e n any AsHol(D,E), t h e s e t

U=: { ( x , t ) e D X R ; t h e r e i s a n o p e n i n t e r v a l J w i t h 0 , t s J

for

which ( 4 . 1 0 ) h a s a s o l u t i o n o n J}

i s a domain i n DXR c o n t a i n i n g D x C O ? . The map f ( x , t ) =: x t , u n a m b i g u o u s l y d e f i n e d by ( 4 . 1 0 ) as a c o n s e q u e n c e of t h e l o c a l u n i q u e n e s s of t h e s o l u t i o n s o f d i f f e r e n t i a l e q u a t i o n s , i s r e a l a n a l y t i c on U a n d f ( . , t ) i s h o l o m o r p h i c f o r e a c h f i x e d t .

COMPLETE VECTOR FIELDS

55

Furthermore, we have f [f(x,t) , s ] = f(x,t+s) whenever (x,t), (f(x,t) ,s)cU. Hence fteHol(D,D) and id = f0= f-t ft= ft f-t, D

i.e. ftcAut D for all tdR. For the remainder part of the proof, let us fix any ball B c c D and any 6>0 such that B g c C D and 11 All This is possible BS<W. because holomorphic maps are locally bounded. For each xeB and and t d I R we have

Now we set t =: sup{t>O; fS(x)eB6 0

for all se[-t,t] and all xeB1

Given tl>tO arbitrarily, there exist xeB and sCt such that 1 fS(x)+Bg. Observe that we have s>tO necessarily. But then

.

By the arbitrarines; of tl>tO we have G C t O / / A ( / Thus tO>O and B6

for all t, 1 t I
S't

corollary 4.10 implies that A is bounded on every ball completely interior to D , (i.e., our previous considerations are valid for each ball B and each 6>0 with B g c c D ) . Then

for all xeB and

te[-toIt0]. Therefore

56

4

CHAPTER

f o r all xeB and t , Itl-
bounded on e v e r y i n t e r i o r b a l l , i s a l s o bounded on e v e r y i n t e r i o r b a l l . T h i s c o m p l e t e s t h e proof by t h e a r b i t r a r i n e s s of B C C D . d

With e a c h Acaut D w e a s s o c i a t e t h e o p e r a t o r

A: Hol(D,E)+Hol(D,E) g i v e n by Ag=:q(lA. 4.16.

COROLLARY.

For any g e H o l ( D , E ) , any baZZ B a n d

11

any d > O w i t h B g C C D a n d that c ~ e h a v e

gl(

,

<m

t h e r e is n n u m b e r r > O such

Bfi t

t k

m

Ftg(x)=:g[f ( x ) ] =

I;

*

(A g ) (x)

k=O

I t l s i . T h e s e r i e s c o n v e r g e : : uniformly o n

w h e n e v e r x8B and

1 -

P r o o f : For t > O w e s e t A =:

. Therefore

Tlim A = A t

t+O

4.17.

(fT-id 1. W e have

t

t h e s t a t e m e n t i s a consequence of L e t A l l A2eAut D b e g i v e n .

THEOREM.

B.

(4.4).

Then A1+A2

a n d A ( ' A ~ - A ; ~ A2 b e Z o n g t o a u t D . 1

= g[exp(tA

)]

(1

=: g+Ag g and F . = : g+qft= 1 where g s H o l ( D , E ) , te2R and j = 1 , 2 . C o n s i d e r t h e

Proof: L e t u s w r i t e

n e t ( gt ) c A u t D d e f i n e d by gt=: f : Tlim(ft-id,)= t+O

A +A 1

2'

A

t

g

f:.

W e show t h a t

I n d e e d , g i v e n any b a l l B E D ,

for

s u f f i c i e n t l y small v a l u e s o f t w e have 'tlB

=

gt2

?:(idg)=

[exp t

*

i22][exp

t A l l ( i d B )=

*

= i d + t ( A Z + A 1 ) ( i d B )+. .= i d + ( A + A 2 ) B B 1

Therefore

t1

(gt-id )

IB

+A +A2 i n 1

(1 - I /

+.

.

B. By t h e a r b i t r a r i n e s s

of B , t h e convergence h o l d s a l s o i n t h e t o p o l o g y T. Then theorem 4 . 5 e s t a b l i s h e s t h a t A +A2&aut D . 1

The proof of A i 1 A 2 - A 2 ( 1A s a u t D i s s i m i l a r 1

by c o n s i d e r i n g t h e

COMPLETE VECTOR FIELDS

n e t h =: t

fi

f:

1

T l i m - (h -id t+O t 2

fit )=

57

fYt and showing t h a t (1

A(1 A2-A2 1

4.18. DEFINITION. G i v e n A1 ,A2eHol(D,E), we d e f i n e (1

[A,,A,I=: A:~A2 - A ~A 1 T h e r e f o r e w e have

for xsD. The o p e r a t i o n A2

c,]

i s c a l l e d t h e L i e p r o d u c t of A1 and

*

For f i x e d AsHol(D,E) , t h e l i n e a r o p e r a t o r [A, .] i s c a l l e d t h e

a d j o i n t of A and w i l l be denoted by A

W '

4.19. PROPOSITION. F o r e v e r y AcHo~(D,E), t h e adjoint of A is d e r i v a t i o n o n a u t D, i . e . , we h a v e

f o r a l l A1,A2cHol(D,E). Proof: Since t h e Lie product i s c l e a r l y anticommutati-

ve, a l l w e have t o prove i s t h e J a c o b 3 i d e n t i t y

f o r a l l A1,A2,A3eHol(D,E). B u t

= A

(1 (1

1

(1 (1

(1

(1

A 2 A3-A1 A 3 A2-(A2 A3-AJ AZ) ('A 1 =

-

Summing up t h e s i m i l a r e x p r e s s i o n s f o r t h e c y c l i c p-errnutations of t h e i n d e x e s w e o b t a i n t h e d e s i r e d r e s u l t . iy

CHAPTER

58

4

4.20. DEFINITION. An a l g e b r a U i t h a p r o d u c t

[,I

is

caZZed a " L i e a l g e b r a " if [,] is a n t i c o m m u t a t i v e a n d sntisfies the Jacvbi i d e n t i t y .

Thus we have proved 4.21. THEOREM. T h e s e t aut D i s a r e a l L,Le a l g e b r a w i t h r e s p e c t t o t h e p r o d u c t IAIIA2]=: A 1( 1 A2-A2( 1A 1 . dk

4.22. EXERCISES. ( a ) Show that dtk

la

(exptAl)..(exptAn)

belongs to the Lie subalgebra of aut D generated by A1,..A,. (b) Prove that we can write

T=

-' in corollary 4 . 1 3 .

811 All

B6

54.- Some properties of commuting vector .fields. __- Now we turn to the investigation of holomorphic vector fields A,BsHol(D,E) with the property %= ii. In general,

i6X=

i ( X ( l B ) = (X ( ' B ) ('A=

2x( 2( B , A ) +x('B('A

and hence

that is,

..

[A,B] = BA-AB

Thus,

and

6

commute if and only if [A,B]= 0 .

Furthermore, we remark that if XcHol(D,E) is an arbitrary vector field then, using the argument leading to (4.7) we get (4.16)

for any open ball B with B 6 c C D , any faHol(D,E) and n m . Therefore we have (exptX)x=

-c

n=O

t" 3

(Xnid,)xeBg

COMPLETE VECTOR FIELDS

6

59

-1

I t \ <2e / / X [ I B and xeB s i n c e t h e series c o n v e r g e s u n i f o r m l y on B by ( 4 . 1 6 ) and i t c l e a r l y s a t i s f i e s ( 4 . 1 0 ) . For

whenever

t h e d e f i n i t i o n o f e x p ( t X ) see E x e r c i s e s 4 . 4 . 4 . 2 3 . LEMMA. L e t A,BcHol(D,E) w i t h a n d assume t h a t xcdom(expA).

0 be given

[A,B]=

T h e n We h a v e

e x p ( A + t B ) x = ( e x p t B ) (expA)x

tm.

f o r s u f f i c i e n t l y s m a l l v a l u e s of Proof: L e t us w r i t e t

h t ' '= : e x p ( t B + A )

g =: e x p t B

f S = : expsA

Consider t h e arc S=: { (expsA)x;

1 dist(S,aD) , 3

6=:

Since

&. and k=O

u=: s 6

1

E=:

and s e t

6 -

then

m

t RB^ k ) 1. d U R!

I

,

%

= C 7 1 (sA+tBInidU = n. 2 6 n=O 26

C

O0

R= 0

t RB^ R -

= C

m ( C

R= 0 '!

if

,

1 s 1 , 1 t 1 <E

commute, i f

sk A ^k ( k!

1

sc[O, I ]

k

k=O k!

ik\id u2 6

I s ] , I t / < €and ycU

26

Hence, f o r se[O,l]

and Is'I, l t ' l , I t l < E

f' (x)CU

,

ht

I s

(XI

= gtfS

w e have

( x )cUg

and 9 t + tI f s + s NOW,

I

(XI=

g

'ht

'IfS

( x ) = g t ' f S ' g t f S (x)= h t ' " g t f

(x)eU2&

l e t u s examine t h e s e t J=: {se[O,l];

g t f s ( x ) = htPs(x)

for

Itl<E)

W e knwn t h a t O s J . L e t s c J be f i x e d a n d t l , s ' e ( - - E , ~ ) b e s u c h t h a t s+s'[O,l]. Then, f r o m t h e a s s u m p t i o n SCJ w e o b t a i n

CHAPTER

60

ht,StS'

exp{

4

(x)= exp[ (s+s')A+tB]x=

7 s' [ (s+s')A+tB]}exp{ 7 S' [(s+s')A+tB]

s+s

- ht' , s Iht'', s t ' tt " s + s

= 4

)x=

S+S

I

where t'=: ts'/(s+s')

(XI= h

t a t s '

4

t"fs(x)=

t s+s' (XI= g f (XI

and

ti'=: ts/(s+s'), so that s+s'eJ.

This means J=I,[ J + ( - - E , E n )[ ]0 , 1 ] .

Whence J = [ 0 , 1 ] .

#

vith [A,B]= 0

4.24. COROLLARY. For a l l A,BcautD

we

h av e

.

exp (AtB)= (expB)(expA) Proof: For xeD, we define

J =: {t'm; exp(A+t'B)x= (expt'B) (expA)xj

Obviously O s J we obtain

X

. If t'eJx

,

applying lemma 4.23 to A+t'B and B

exp[A+ (t'+t)B]x= (exptB)[exp(A+t'B)]x= = (exptB)(expt'B)(expA)x= [exp(t+t')B] (expA)x

for sufficiently small values of t. Thus J x is open. However, Jx is a l s o closed since the mappings t'+(expt'B) (expA)x and t'-+exp(Att'B)x are continuous on IR. Therefore J =lR. 4.25. THEOREM. T h e L i e a l g e b r a autD

is " p u r e l y r e a l " ,

i . ~ (autD) . ni(autD)= {Of. Proof: Let Ae(autD) fl i(autD) be given. Then with (i% and, by corollary 4.23, we have

for all s,tdR. Hence

commutes

COMPLETE VECTOR FIELDS

61

f o r a l l id. T h u s , g i v e n a n y X B D , t h e mapping $x: &-+D d e f i n e d by <+(expr;A)x i s r e a l a n a l y t i c on t h e whole C.

On t h e o t h e r

hand, w e have

<

i f x s B c c D and

i s s u f f i c i e n t l y s m a l l , i . e . $x i s holomorphic

i n some n e i g h b o u r h o o d o f 0 a n d t h e r e f o r e , i n view o f i t s r e a l S i n c e D i s bounded, $x i s

a n a l y t i c i t y , i t i s h o l o m o r p h i c on C .

c o n s t a n t by L i o u v i l l e ' s t h e o r e m . Thus A x = x w a s a r b i t r a r y i n D , w e h a v e A= 0 .

di

10

+x(<)= 0.

Since

# 55.-

The a d j o i n t m a p p i n gs.

W e a r e g o i n g t o i n t r o d u c e a f a m i l y o f mappings o f f u n d a m e n t a l importance i n L i e theory.

4.26.

DEFINITION. L e t U b e a n y s u b d o m a i n o f D a n d

a s s u m e t h a t f : U-+D a n d X :

W E a r e , :-espectiveZy, a holomorphic

m a p p i n g a n d a h o l o m o r p h i c v e c t o r y i e l d o n U. In t h e s e q u e l we shaZl write

x=:

g

g f = : gfg-l

g (-1l x g - l

#

#

g

w i t h o u t a n y d a n g e r o f c o n f u s i o n . T h e maps f + g f a n d X+g X #

#

( d e f i n e d r e s p e c t i v e l y on h o l o m o r p h i c l o c a l t r a n s f o r m a t i o n s and l o c a l v e c t o r f i e l d s ) w i l Z b e c a Z l e d t h e " a d j o i n t m a p s N of g . By r e p e a t i n g t h e p r e v i o u s c o n s i d e r a t i o n s , it i s i n m e d i a t e t h a t , i f g , : D+D' g,(D)= D',

and

g 2 : D'+D"

g 2 ( D ' ) = D",

a n d , by s e t t i n g

XI=:

are b i h o l o m o r p h i c maps w i t h

t h e n w e have

g#X,

d o m ( e x p X ' ) = g [dom( expX) ( c f . Exercises 4 . 4 ) .

,

expX'= g t (expx)

62

4

CHAPTER

4.27.

EXERCISES. L e t g: D-tD'

mapping w i t h g ( D ) = D ' .

be a biholomorphic

Show t h a t :

(a) t h e a d j o i n t map g # : autD-tautD' is a s u r j e c t i v e a l g e b r a s a u t D and

c o n t i n u o u s isomorphism of t h e Banach-Lie

autD'

. (b) t h e a d j o i n t map g # : AutD-tAutD i s a s u r j e c t i v e

isomorphism of t h e t o p o l o g i c a l g r o u p s A u t D and A u t D ' endowed w i t h t h e i r r e s p e c t i v e t o p o l o g i e s of l o c a l uniform convergence on D and D ' . From now on w e r e s t r i c t o u r s e l v e s t o t h e c a s e i n which D = D ' . 4.28.

THEOREM. L e t u s s u p p o s e t h a t

( g j ) j e J and

( t j )j e J a r e , r e s p e c t i v e i ! y , n e t s i n A u t D a n d IR w i t h

l i m t . =0 j+m

T l i m ( g , - i d D ) = AeautD

3'

j-tm

J

Consider a n y h o l o m o r p h i c l o c a l t r a n s f o r m a t i o n f : U+D a n d a vector X: W E . T h e n , for a n y b a l l B s u c h t h a t B c c D , ue have 6

field

1

(g. f-f) j

I#

-t(Af-f('A) IB

IB

and

uniformly on B. P r o o f : Given B w i t h B 6 c c D , i f gilB'U.

Therefore

u n i f o r m l y on B . S i m i l a r l y , w e have

11

26 < t.A/( I B6 e

w e have

COMPLETE VECTOR FIELDS

u n i f o r m l y on B , where [A,X]= : 4.29.

63

[A

EXERCISE. Give a d e t a i l e d proof of t h e s e f o m l a s

H i n t : t h e e s s e n t i a l t o o l s have been developed i n t h e p r o o f of proposition 4.1

.

I n p a r t i c u l a r , i f gt=exptA and t i s so small t h a t g - t B c U ,

we

have

w i t h r e s p e c t t o t h e norm

I/ - 1 1

B

. Therefore

Y(t)=:

t

g X #

IB

s a t i s f i e s the d i f f e r e n t i a l equations

where t h e d e r i v a t i v e i s t a k e n i n t h e s e n s e of t h e norm Hence w e have

f o r s u f f i c i e n t l y s m a l l v a l u e s of t , because

f o r ndN. I n d e e d , i f B 1 ; , c C D and ZsHol(D,E) w e have

11

- 11

B '

64

CHAPTER

4

T h a t is, roughly s p e a k i n g w e have

l o c a l l y . W e s h a l l see i n t h e n e x t c h a p t e r i n which s e n s e t h e tn s e r i e s C 7A i X converges g l o b a l l y i f XcautD. n=O

CHAPTER

5

THE NATURAL TOPOLOGY ONTHE L I E ALGEBRA OF COMPLETE VECTOR FIELDS

§ I . C a r t a n ' s u n i q u e n e s s theorem f o r a u t D . THEOREM. L e t ACautD b e g i v e n a n d s u p p o s e that

5.1.

A ( * = A(l= 0 f o r some aeD. T h e n A=O.

Proof: S e t f for a l l tdR,

t

( x )= :

(exp t A ) x . Observe t h a t f t ( a ) = a

since

d dt

f t ( a ) = A[ft(a)]= 0

i s s a t i s f i e d i n t h i s c a s e . L e t u s compute ( f t )

d' .

By t h e

a n a l y t i c i t y of t h e mapping ( t , y ) + f t ( y ) w e have

d dt - - -a - at2

=

('

( f t ) (lh= x

a at l l o f

a __ a f at1 at, I*

a

t+tl

t+t

-

( x + t2 h ) = A[ft(x+t2hl]=

(x+t2h)=

at2

[(f');'h]

f o r a l l xsD. Thus w e have

for a l l t d R . In particular

65

10

66

CHAPTER

5

whence ( f t ) ( l = ( f t ) ( O = i d f o r a l l t m . T h e n , b y C a r t a n ' s t u n i q u e n e s s t h e o r e m , we h a v e f = i d D f o r a l l t m , whence A= 0 .

# 5.2.

COROLLARY. G i v e n a n y a t l D ,

t h e rnapp-ing

is i n j e c t i v e a n d continuous.

autD+ExL(EIE) d e f i n e d b y A+(ALo,A:l)

(S

(S

P r o o f : The c o n t i n u i t y i s c l e a r . I f A l , a = A 2 , a f o r s= 0 , l

,

t h e n A1-AlcautD

0

a n d (A1-A 2 a('=

f o r s= 0 , l .

ff

The p r e v i o u s c o r o l l a r y n a t u r a l l y r a i s e s t h i s q u e s t i o n : i n v e r s e mapping

( A ( ~ , A ' -+A ~ ) continuous?.

Is t h e

BY i t s l i n e a r i t y , i t s

c o n t i n u i t y means some k i n d of " l i p s c h i t z i a n i t y " . To o b t a i n a

r e s u l t i n t h i s d i r e c t i o n , w e go b a c k t o AutD. G i v e n a b a l l B t c D c e n t e r e d a t aeD, t h e r e e x i s t s a c o n s t a n t K s u c h t h a t we h a v e 5 . 3 . THEOREM.

for a l l f c A u t D . Proof: W e p r o c e e d by c o n t r a d i c t i o n . L e t ( f j ) j m \ r c A U t D be a s e q u e n c e w i t h f . # i d D f o r je3N s u c h t h a t 1

where p Since

,+a.

I

/I

1

f.-idDI/,sdiamD, 3

1

. Let

s=G

by t h e o r e m 2 . 7 , T l i m f , = i d j+m

define

'

p , = : l+maxEpeU?; J

11

w e have D

11

( f j - i d D ) : s \ l +O

,

i.e.,

u s f i x d>O s u c h t h a t B f i c C D a n d

k


fj-idDII d

From lemma 4 . 8 w e know t h a t p j a +

d 211 f j - i d D / l R

f o r a l l ] E N . F o r t h e s a k e o f s i m p l i c i t y w e may assume t h a t

67

TOPOLOGY ON VECTOR F I E L D S

w e have

Pj< -

(5.3)

d

I1 f j - i d D ! / B P.

f o r a l l jdN. By d e f i n i t i o n we have f . ] f . t h e norm

//

./ /

. However,

1

i d D with r e s p e c t t o

w e s h a l l show t h a t

Bd

P.

P.

(fjJ);O+

a

and

(fj7);'+

which i s a c o n t r a d i c t i o n by theorem 2 . 7 .

id

An a p p l i c a t i o n of

corollary 4.6 yields

1

L-

d-d/4

11

f o r a l l j C N and p$p

P- 1 f (a)-a(( C

I/

. Therefore,

by ( 5 . 3 ) ,

k=1

1

f k - i d ( /B d ( a ) L

for a l l psp,. 1

P.

Now,let u s c o n s i d e r t h e sequence (f..':)I 3

where

W e have

68

5

CHAPTER

f o r k= 1 , 2 , . . . , p ,

1

jm.

and

Therefore

J

P,

W e must p r o v e t h a t t h e r i g h t - h a n d

s i d e t e n d s t o 0 . To d o t h i s ,

l e t u s d e n o t e by r t h e r a d i u s o f B and c h o o s e j

0

such t h a t

2d/u < r / 2 f o r a l l j > j o . Then, by ( 5 . 4 ) w e have d

( a ) e B ' = : B,,2 f o r a l l j > j o and k = 1 ,

*

*

lPj

(a)

. Therefore,

by p r o p o s i t i o n 1 . 4

and ( 5 . 2 )

J

whenever j > j o and 1CkCp.. Taking i n t o a c c o u n t ( 5 . 4 ) and t h e I f a c t t h a t /I f,-idDII 0,we can w r i t e 3

B

-+

f o r some c o n s t a n t y ( i n d e p e n d e n t of j , k ) and a l l j ? j k= l , . . , p j .

0'

Hence,

(5.5) f o r ] > l o . I t i s w e l l known from e l e m e n t a r y a n a l y s i s t h a t

R

( l + a . ) I +I

I

whenever a

j

+

0 and

c1

j

P

-+

j

0 . B u t , by ( 5 . 3 1 ,

69

TOPOLOGY ON VECTOR F I E L D S

5.4. B’cCD

B,

COROLLARY.

F o r e v e r y p a i r of ba2l.s B and B ’ w i t h

t h e r e e z i s t s a c o n s t a n t K ’ s u c h t h a t we h a v e

for all f , gsAutD s a t i s f y i n g f ( B ) C B ’ and g ( B ) C B ’ . P r o o f : Given f , gsAutD w i t h f ( B ) c B ’ a n d g ( B ) C B I ,

by

t h e o r e m 5 . 3 a n d remark 2 . 4 w e h a v e

4K2

4CK

2

+K2

w h e r e , by p r o p o s i t i o n 1 . 4 ,

t h e right-hand

s i d e i s d o m i n a t e d by

f o r some c o n s t a n t s K 1 ’ K 2 , K 3 d e p e n d i n g o n l y on B a n d B ’ .

ff

5 . 5 . E X E R C I S E . L o o k f o r c o u n t e r e x a m p l e s t o show t h a t t h e c o n s t a n t K ’ i n c o r o l l a r y 5 . 4 must a c t u a l l y d e p e n d o n B a n d B’

.

53.-

The ~- n a t u r a l t o p o l o g y on a u t g .

From t h e s t r o n g s t a t e m e n t o f t h e o r e m 5 . 3 it i s a l r e a d y e a s y t o deduce a r e s u l t c o n c e r n i n g autD: 5.6.

THEOREM.

Given any b a l l B c c D c e n t e r e d a t a s D ,

t h e r e e x i s t s a c o n s t a n t K B s u c h t h a t we h a v e

70

CHAPTER

5

f o r a l l aeautD. I f B ' c c D is a n o t h e r b a l l c e n t e r e d a t a ' s D ,

ue have

on autD. P r o o f : Theorem 5 . 3 f u r n i s h e s a c o n s t a n t K

such t h a t

B

t

f o r a l l fsAutD. Hence, g i v e n AsautD and w r i t i n g f = : e x p t A r tdF, we have

tm.

But Tlim O't t o A a l s o i n t h e norms for a l l

I

t

( f - i d ) = A whence

11

- /IB

D

and

1

1 t

( 1 *:'I\

C

(ft-idD) tends

. This

proves (5.6).

s=o

To prove t h a t

w e need o n l y t o copy t h e proof of theorem 1 . 6 .

5.7.

* II

REMARK. The e q u i v a l e n c e of two norms on a v e c t o r

s p a c e i m p l i e s t h e e x i s t e n c e of p l

u1 I1 94.-

XI1

xi1

#

2 a l lX I 1

,

p 2 > 0 such t h a t w e have

1 for a l l x-

autD a s a Banach s p a c e .

A f t e r t h e p r e v i o u s theorem, t h e n e x t q u e s t i o n i s a t hand: Is autD endowed w i t h any o f t h e norms

I[ * \ I B

1

C 11 -:'I\ s=o c a u t D with or

a Banach

s p a c e ? . That i s , g i v e n a sequence ( A , ) 3 jm A!' +L(', s = 0 ,1 , does t h e r e e x i s t AeautD s u c h t h a t w e have la (

A,'=

L('

f o r s= 0,1? W e c a n prove a much s t r o n g e r r e s u l t t h a t

h a s c r u c i a l importance i n t h e t h e o r y of symmetric domains.

5.8.

THEOREM.

L e t t h e n e t s ( f j ) j e J C A u t D and

(tj), CIR: b e g i v e n a n d assume t h a t we c a n f i n d some acD, some JCJ such t h a t the n e t L(%E a n d some L ( ' ~ L ( E I E )

TOPOLOGY ON VECTOR FIELDS

A =:

1 t. 1

j

(f.-idD) I

s a t i s f i e s A ( s +L ( s f o r S = O , ? . jra t h a t we h a v e T l i m A , = A . jfJ

71

Then, t h e r e ezists AsautD

such

3

P r o o f : L e t u s f i x a b a l l B c c D c e n t e r e d a t acD and c h o o s e d>O s u c h t h a t B Z d c c D . We may assume

sup j E J

s=o

11

AiZll

<m

Then, by t h e o r e m 5 . 3 t h e r e e x i s t s M > O s u c h t h a t

11

fj-idDI/

6Mt 1

B2d

f o r a l l j c J . For 6
1)

f7-id

I

I/

<&I Bd

From lemma 4 . 8 w e see t h a t

J

W e s e t t = : 6 / M and n . = : [ t t y l J where w e k e e p t and 6 f i x e d I 3 f o r t h e moment.. N o t e t h a t n . < p ,(6) f o r a l l jsJ. T h e r e f o r e , by I

c o r o l l a r y 4 . 6 w e have (5.7)

/I

3

( f . ' - i d D ) - n . ( f . - i d D ) I / B'<

I

3

n.6

11

1

f j-idD

n.6 6

d/2

/I

6

M d/2

Mt.6 -6 t

I

Denote by r t h e r a d i u s of B . From ( 5 . 7 ) u s i n g t h e Cauchy

estimates w e o b t a i n

11

n.

( f j l - i d D ) ~ S - n3 . tI. Aj,a( S IS1m a x ( 1 , r )

/I

f o r s = 0 , l . N o w , t h e t r i a n g l e i n e q u a l i t y a p p l i e d t o t h e segrrents s u g g e s t e d by t h e below d i a g r a m

72

5

CHAPTER

n.

f '-idD

n f kk-idD

i

n t A = n (f - i d

n 1. t 1 . A1. = n 1. ( f j- i d D )

k k k

k

k

D

implies

Observe t h a t

f o r s= 0 , l .

Hence, t h e r e e x i s t s and i n d e x j , ( S )

for a l l j , k > j o ( 6 ) and

s= 0 , l .

n.

such t h a t

S i n c e n . S p . ( S ) , from t h e d e f i n i I

3

t i o n o f p . ( 6 ) w e see t h a t f J ( B ) c B d f o r a l l jeJ. Thus w e may 1 1 n. apply c o r o l l a r y 5 . 4 t o B ' = : B and f = : f j l , g= f n k k . Hence d

r o a l l j , k > j o ( 6 ) , where M4 i s i n d e p e n d e n t of j , k, 8 a n d t . Applying t h i s r e s u l t t o ( 5 . 7 ) , t h e t r i a n g l e i n e q u a l i t y y i e l d s f n, l. - i d D r /

f kn - i d D k

1

n J. t J. A3.

t h a t is,

nktkAk

.

TOPOLOGY ON VECTOR FIELDS

73

f o r some M4>0, a l l j, k?j0(6) a n d a l l t e ( O , G / M ) .

Prom ( 5 . 9 ) it

r e a d i l y follows(by taking t h e superior l i m i t i n j , k with fixed that

6,t)

l i m sup11 j, k

A 3, - A ~ I ~ -~

< ~ ~ 6

i s a Cauchy n e t w i t h r e g a r d t o t h e ( A j )j e J C o n s e q u e n t l y , w e h a v e ( A , ) ( s + ~ ( s ' x ) , s= O , I , f o r

for a l l 6>0, i.e., norm

.

.

11

3 x

some ~ ( 0 t X ) e Eand L ( l ' X ) c L ( E - l E ) whenever xeB. But t h e n

i s a Cauchy n e t w i t h r e g a r d t o t h e norm

I( . \ I B ,

(A,)

I jeJ

whenever

i s a b a l l c e n t e r e d a t a p o i n t x s B . By r e p e a t i n g t h e

B'CCD

a r g u m e n t , f r o m t h e c o n n e c t e d n e s s of D w e o b t a i n t h a t ( A , ) , 3 IeJ i s a Cauchy n e t w i t h r e g a r d t o t h e norm / I . / I f o r any b a l l B"

B"CC

D,

(Aj)jeJ

i.e.

i s a T-Cauchy n e t . Thus w e h a v e

T l i m A = A f o r some A e H o l ( D , E ) . B u t t h e n , t h e o r e m 4 . 3 e s t a u i s h 3

e$ J t h a t AeatuD. 5.9.

COROLLARY.

L e t B c c D a n y baZZ c e n t e r e d a t aeD.

/ I .I1 B

T h e n a u t D is B a n a c h s p a c e w i t h r e g a r d t o t h e norms s=o

/ I "I;.

and

*

Proof: L e t us suppose t h a t ( A , ) , is a Cauchy s e q u e n I l a ce i n t h e norm /I .;*\I Then A!' * L ( s f o r s= 0 , l .

e

.

s=o

Choose a s e q u e n c e

(E,)

,

I l a

1la

o f p o s i t i v e numbers w i t h

E,*

I

0.

S i n c e w e have

1 T l i m - (exptA.-id ) = A . t I D I t+O

we can p i c k t . > O such t h a t 1

t . < E .

1

1

and

where f . = : e x p t A Obviously we have I j j'

J

f o r s= 0 , l . Now t h e o r e m 5 . 8 e n s u r e s t h a t L('=

A('

f o r some

74

CHAPTER

5

AsautD.

# 5.10.

EXERCISE. Prove c o r o l l a r y 5 . 9 d i r e c t l y by u s i n g

t h e l o c a l uniform c o n t i n u i t y o f t h e s o l u t i o n s o f o r d i n a r y d i f f g r e n t i a l equations with regard t o t h e i n i t i a l values.

5.-

autD a s a Banach-Lie a_ l g e_ bra. ~

L e t u s f i x any b a l l B c c D and 6>0 s u c h t h a t B 6 c c D , and endow

/I .I / B .

We a l r e a d y know t h a t ( a u t D , /I i s a Banach s p a c e . L e t u s now c o n s i d e r i t s L i e - a l g e b r a s t r u c t u re. autD w i t h t h e norm

5 . 1 1 . LEMMA. For a l l AcautD, t h e m a p p i n g A # : X+[A,X]

is a b o u n d e d Z i n e a r o p e r a t o r o n a u t D . P r o o f : The l i n e a r i t y of A f t i s o b v i o u s . On t h e o t h e r hand, by t h e Cauchy e s t mates and t h e f a c t

11 . l \ B ~ \ l

we

have

11

[A,X]

I/

=

/I A'lX-X'lA 1

~ ~ I I A l I lI X

BS

f o r a l l XeautD and some M ( i n d e p e n d e n t of X ) .

5 . 1 2 . COROLLARY. T h e m a p p i n g

# ; A+A#

is a c o n t i n u o u s

Z i n e a r o p e r u t o r on autD. P r o o f : W e have M I

11

A # ( l 6 M ' I I All

f o r a l l AcautD

and some

>O. 5.13.

PROPOSITION. We h a v e exp ( A # ) = (expA)

f o r a l l AcautD.

P r o o f : L e t XcautD be a r b i t r a r i l y f i x e d . By lemma 5.11

w e have t h a t

TOPOLOGY ON VECTOR FIELDS

[exp(tA ) ] X = : #

75

c t k AkX k! #

k=O

'L

i s a w e l l - d e f i n e d element Y ( t ) o f autD. Moreover,

d dt

?(t)= lim 1 "[ Y . ( t + h ) -".Y ( t ) ] = A Y % ( t ) # h+O h %

f o r a l l t6B a n d Y ( O ) = X . But t h e norm c o n v e r g e n c e of 1 [? ( t u h )-Y'L ( t )] means i t s T-convergence i n v i e w of c o r o l l a r y 5.12.

'L

Thus t h e mapping t + Y ( t )

s a t i s f i e s t h e d i f f e r e n t i a l equa-

tion (5.10)

i n t h e Banach space ( a u t D , T). But w e h a v e s e e n i n 5 4 C h a p t e r

IV, t h a t t h e mapping Y ( t ) = ( e x p t A ) # X ,

tm,

satisfies this

e q u a t i o n , t o o , whence t h e r e s u l t follows.

5.14.

LEMMA. L e t

#

@ be a c o n t i n u o u s automorphisrn of t h e

Banach L i e a 2 g e b r a a u t D . Then we h a v e

for a 2 2 A , XcautD. P r o o f : S i n c e @ i s a n automorphism of t h e L i e a l g e b r a autD, w e have

a n d , by r e i t e r a t i n g t h e a r g u m e n t w e o b t a i n

f o r n a . A s 4 i s a c o n t i n u o u s l i n e a r o p e r a t o r on a u t D , by p r o p o s i t i o n 5.13,

#

This Page Intentionaiiy Left Blank

CHAPTER

6

THE BANACH L I E GROUP STRUCTURE O F THE SET O F AUTOMORPHISMS

We have s e e n t h a t A u t D i s a t o p o l o g i c a l group when endowed w i t h t h e t o p o l o g y T of l o c a l uniform convergence. Now w e a r e g o i n 9 t o c o n s t r u c t anokher t o p o l o g y T a on AutD such t h a t (AutD, T a ) c a r r i e s t h e s t r u c t u r e of a r e a l Banach-Lie group which a c t s a n a l y t i c a l l y on D . F i r s t we i n t r o d u c e some p r e p a r a t o r y m a t e r i a l . The concept ___

51.-

of a Banach ___-__ manifold.

L e t M and E be r e s p e c t i v e l y a Hausdorff s p a c e and a Banach space o v e r any of t h e f i e l d s I R o r

which w e i n d i s t i n c t l y r e p r e s e n t

by X .

6 . 1 . D E F I N I T I O N . A " c h a r t V o f M o v e r E is a p a i r

(u,u)

w h e r e U i s a n o p e n s u b s e t of M a n d u is a h o m e o m o r p h i s m of U o n t o a n o p e n s u b s e t o f E.

on M i s a c o l l e c t i o n o f c h a r t s (Ual~ol)aeI E s u c h t h a t t h e foZZozJing c o n d i t i o n s a r e s a t i s f i e d :

An " a n a l y t i c s t r u c t u r e ' '

of M o v e r MI:

The f a m u l y

(Ua)aeI

i s a n o p e n c o v e r o f M.

M ~ : For e a c h p a i r a , ~ e ~ t h ,e m a p i n g l i B v v i l :

I - 1 , ( ~ ~ n ~ , ) +Bl(LJ i a

nu,)

is a n a l y t i c . M3:

T h e c o l l e c t i o n ( U a , ~ a ) a e I is a maximal f a m i l y o f c h a r t s on M f o r w h i c h c o n d i t i o n s M a n d M2hoZd. 1

A

" B a n a c h m a n i f o Z d " i s a p a i r ( M I A ) w h e r e M is a H a u s d o r f f

s p a c e a n d A is a n a n a l y t i c s t r u c t u r e o n M o v e r some B a n a c h s p a c e E. I f t h e r e i s no danger of c o n f u s i o n , w e s h a l l r e f e r t o t h e Banach manifold M w i t h o u t any r e f e r e n c e t o i t s a n a l y t i c s t r u c t u r e A .

77

CHAPTER

78

6

A c c o r d i n g a s t h e f i e l d x i s 3R o r .'U w e s a y t h a t M i s a p e a 2 o r a complex manifold. REMARK. C o n d i t i o n M 3 w i l l o f t e n be cumbersome t o c h e c k

6.2.

i n s p e c i f i c i n s t a n c e s . I n f a c t , i f c o n d i t i o n s M 1 and M2 a r e s a t i s f i e d , t h e f a m i l y (Uct,ucr)aer can be e x t e n d e d i n a unique manner t o a l a r g e r f a m i l y o f c h a r t s f o r w h i c h c o n d i t i o n M3 i s s a t i s f i e d , t o o . Thus, M3 i s n o t e s s e n t i a l i n t h e d e f i n i t i o n o f a Banach m a n i f o l d . 6 . 3 . EXEMPLES. L e t U b e a non v o i d o p e n s u b s e t of a Banach s p a c e E . The p a i r ( U , i d u ) i s a c h a r t o f U o v e r E a n d d e f i n e s a n a l y t i c s t r u c t u r e on U .

The m a n i f o l d so c o n s t r u c t e d i s c a l l e d

t h e c a n o n i c a i : m a n i f o l d on U . L e t M and N b e ,Banach m a n i f o l d s

F respectively. If

o v e r t h e Banach s p a c e s E a n d

( U , u ) a n d ( V , v ) are c h a r t s o f M a n d N , t h e n

(UxV, uxv') , where uxv: ( x , y ) + ( u ( x ) , v ( y ) ) , i s a c h a r t of M x N o v e r ExF. The f a m i l y o f t h e p a i r s So c o n s t r u c t e d i s a n

the pair

a n a l y t i c s t r u c t u r e and t h e c o r r e s p o n d i n g m a n i f o l d i s c a l l e d t h e

p r o d u c t of M and N . L e t M be a Banach m a n i f o l d o v e r a . c o m p l e x Banach s p a c e E. Then E c a n be c o n s i d e r e d a s a r e a l Banach s p a c e ,

by

%.

too, which we denote

Any c h a r t ( U , u ) o f M o v e r E i s a c h a r t o v e r

%

and t h e

f a m i l y o f t h e s e c h a r t s d e f i n e s a r e a l a n a l y t i c s t r u c t u r e on M . The m a n i f o l d so c o n s t r u c t e d i s c a l l e d t h e u n d e r Z y i n g r e a l

m a n i f o l d of M. 6.4.

DEFINITION. L e t a B a n a c h m a n i f o Z d M a n d a p o i n t xcM b e

g i v e n , a n d c o n s i d e r t h e s e t of t h e p a i r s

[ ( U , U ) rh] w h e r e

is a c h a r t of M at x and hsE. Me s a y t h a t [ ( V , U ) , h , ]

(U,u)

and

[ ( V , V ) , h 2 ] a r e " e q u i v a l e n t " i f We h a v e

( v O u - l )( '

u(x)

.h = h 1

2

W e w r i t e T M f o r t h e q u o t i e n t s e t . The e q u i v a l e n c e c l a s s o f t h e

a IX

e l e m e n t [(U,u) rh] which i s d e n o t e d by h au t a n g e n t v e c t o r t o M a t x.

,

is called a

THE L I E GROUP OF AUTOMORPHISMS

79

L e t u s f i x any c h a r t (U,u) of M a t xcM. The mapping E+T M g i v e n

a au. I

by h+ h

X

i s a b i j e c t i o n by means of which w e can

t r a n s f e r theXBanach s p a c e s t r u c t u r e of E t o T M. W e say that TJ4 X

endowed w i t h t h i s Banach s p a c e s t r u c t u r e i s t h e t a n g e n t s p a c e t o M a t x. 6.5.

DEFINITION. L e t a Banach m a n i f o Z d M and a Banach

space F be g i v e n .

We s a y t h a t a mapping f : M+F i s " a n a Z y t i c a t

a p o i n t xcM"if t h e r e i s a c h a r t f0u-I:

(U,u)

o f M a t x such t h a t

u ( U ) + F i s a n a z y t i c . We s a y t h a t f i s " a n a l y t i c o n M " if

i t i s a n a Z y t i c at e v e r y p o i n t xcM and we c a l l

f 0 u - l a "ZocaZ

expression" o f f a t x. L e t f, g: M+F be a n a l y t i c mappings a t a p o i n t XCM,

by f0u-l: u ( U ) + F , yov

-1

and d e n o t e

: v(V)+F t h e i r l o c a l r e p r e s e n t a t i o n s i n

t h e c h a r t s (U,u) and ( V , v ) , r e s p e c t i v e l y . W e s a y t h a t f and 9 a r e e q u i v a Z e n t a t x i f t h e r e i s a neighbourhood W c U f l V of x -1 -1 F on W . W e d e n o t e by B X t h e q u o t i e n t s e t such t h a t f o u = gav and each e q u i v a l e n c e c l a s s i s c a l l e d an a n a l y t i c germ a t x. O F

i s endowed w i t h a v e c t o r s p a c e s t r u c t u r e i n an o b v i o u s manner. NOW, t a n g e n t v e c t o r s t o M a t x can be i n t e r p r e t e d a s d i f k r e n t i a l

o p e r a t o r s a c t i n g on a n a l y t i c germs a t x i n t h e following manner: for f s O F

and h

a au

I x

eT M w e s e t x

a+ au 6.6.

( x ) .h=:

DEFINITION.

h (feu-')'' u (x)

L e t M and N be Banach m a n i f o l d s o v e r

t h e Banach s p a c e s E and F, r e s p e c t i v e l y . We s a y t h a t a continuous mapping f : M-+N i s a " m o r p h i s m " o f Banach m a n i f o l d s i f , for e a c h p o i n t XCM,

t h e r e are charts

y= f ( x ) s u c h t h a t v 0 f o u - l :

(U,u) of M a t x and

(V,v) of N a t

u ( U ) + v ( V ) is a n a l y t i c .

Suppose t h a t f : M+N i s a morphism of Banach m a n i f o l d s . Then (V.

€0

u -1)

(1

u (x)

i s an element of L ( E , F ) and w e c a n d e f i n e a

c o n t i n u o u s l i n e a r mapping d f ( x ) : T x M + T f ( x ) N by s e t t i n g

80

CHAPTER

6

for h e E . I t i s e a s y t o check t h a t df(x) does n o t depend on t h e c h a r t s ( U , u ) and (V,v) w e have chosen. W e say t h a t d f ( x ) i s t h e

d e r i v a t i v e of f a t x and t h a t ( 6 . 1 )

is i t s locuZ e x p r e s s i o n

w i t h r e s p e c t t o t h e c h a r t s ( U , u ) and ( V , v ) . L e t M be a Banach manifold o v e r E and l e t U b e a n open s u b s e t

of M. W e s e t iTxM; xcu)

TU=:

( U , u ) i s a c h a r t of M , w e d e f i n e a mapping T : TU+ u ( U ) x E

If

U

by means of

a

h -

T : u

au

Ix

f

(u(x), h ) .

Then, w e have 6.7.

PROPOSITION. T h e r e e x i s t s

cz

u n i q u e topology on TM

s u c h t h a t Lhc f o l l o w i n g c o n d i t i o n s a r e s a t i s f i e d : ( a ) F o r eZ)Cray o p e n s u b s e t U of M, TU

Cn u n o p e n

:;uhsst o f TX.

(b) For1 e o c r y c h a r t ( U , u ) of M, T u : TU*u(U) X E is a h o m e o m o r p h i s m .

t h e mapping

W e l e a v e t h e proof a s an e x e r c i s e . I t i s c l e a r t h a t TM w i t h

t h i s topology i s a Hausdorff s p a c e . Moreover, if ( U , u ) i s a i s a c h a r t of TM o v e r t h e Banach space

c h a r t of M , t h e n (TU, Tu)

ExE, and w e have 6.8.

PROPOSITION. T h e f a m i l y { (TU,TU):

( u , u ) i s a c h a r t of M I

d e f i n e s a n d analytic s t r u c t u r e o n TM. The Banach m a n i f o l d so c o n s t r u c t e d on TM i s c a l l e d t h e tangent

-

b u n d l e t o M . Obviously, t h e c a n o n i c a l p r o j e c t i m s n 1' TM+M and TI TM-tE, g i v e n by

-

2'

a

nl: h au

Ix

+X

and

a v2: h -

au I x

+h

THE L I E GROUP O F AUTOMORPHISMS

a

81

cTM, are Banach m a n i f o l d morphisms. :x Moreover, i f f : M+N i s a morphism of Banach m a n i f o l d s , i t s

for h

I

d e r i v a t i v e d f : TM+TN i s a morphism o f t h e c o r r e s p o n d i n g t a n g e n t bundles.

6 . 9 . D C F I N I T ~ - O N . ~ ~ A ~ a n a l y t i c v e c t o r f i e l d ” o n a Banach m a n i f o l d M is morphism X: M+TM s u c h t h a t we have

I f X:

M+TM i s a n a n a l y t i c v e c t o r f i e l d on M , t h e n i t s v a l u e

a

Ix

X ( x ) a t xcM i s a t a n g e n t v e c t o r t o M a t x , X ( x ) = h ( x ) - c T M . au The l o c a l e x p r e s s i o n o f X w i t h r e s p e c t t o t h e c h a r t s (U,u) o f M a n d (TU,Tu) o f TM i s g i v e n by

where h: M-tE i s a n a n a l y t i c mapping on M . W e d e n o t e by T ( M ) t h e s e t o f a l l a n a l y t i c v e c t o r f i e l d s o n M .

6.10.

DEFINITION.Let X = :

f(x)

a ax

be a n a l y t i c v e c t o r f i e l d s o n M and l e t

l x and

Y= g ( x )

be g i v e n .

a

AX=: Af ( x ) au

a aulx

W e define

Ix

for xcM. It i s easy t o v e r i f y t h a t X+Y,

AX a n d [ X , Y ]

are elements of

T(M) a n d t h a t , i n t h i s way, T ( M ) becomes a L i e a l g e b r a . W e c a l l

it t h e L i e a l g e b r a of a n a l y t i c v e c t o r f i e l d s o n M . 6.11. manifolds.

DEFINITION. L e t @ :

Y c T ( N ) a r e “ r e l a t e d by

(6.2)

M+N be a morphism of Banach

W e say t h a t t h e a n a l y t i c v e c t o r f i e l d s XcT(M) and if we have

d$.X= Yo@

L e t u s t a k e c h a r t s (U,u) of M a t x a n d ( V , v )

o f N a t y = @ ( x ),

82

6

CHAPTER

and assume t h a t X = f ( x )

a au I x

and

a

Y= g ( y ) -

av IY

are t h e

c o r r e s p o n d i n g local e x p r e s s i o n s of X and Y . Then t h e e x p r e s s i o n

of

( 6 . 2 ) i s g i v e n by

6.12.

PROPOSITION.

L e t ip:

M+N be a m o r p h i s m o f Banach

m a n i f o l d s and assume t h a t X l I X 2 c T ( M ) a r e r e l a t e d b y (t, w i t h Y1,Y2cT(N),

r e s p e c t i v e l y . Then X1+X2,

r e l a t e d by $ w i t h Y 1 + Y z l h Y 1 and

AX1 and

are

[X1,X2]

[Y,,Y,].

W e leave t h e proof as an e x e r c i s e . 6.13.

DEFINITION.

L e t $: M+N b e a m o r p h i s m o f Banach

m a n i f o l d s . Then:

(a) We s a y thal ip i s an nirnmersion” i f , f o r e v e r y x c M , dip ( x ) : TxM*T

ip ( x )

N

i s i n j e c l i v e and t h e i m a g e d $ ( x ) . T M

c Z o s e d topologCcally c o m p l e m e n t e d s u b s p a c e o f T

is a

N.

(t, ( x )

( h ) We s a y t h a t @ i s a ” s u b m e r s i o n ” i f , for e v e r y x e M ,

dip(x) : T x M + T + ( x l N i s s u r j e c t i v e and t h e k e r n e l K e r d g ( x ) i s a ( o b v i o u s l y c l o s e d ) t o p o l o g i c a l l y complemented s u b s p a c e of T M. X

N o w we have ( s e e 12 I 5 5 ) . 6.14.

PROPOSITION. Let

4 : M+N be a m o r p h i s m of Banach

manifolds. Then the f o l l o u i n g statements are e q u i v a l e n t : (a) The mapping @ :

M-tN is a n i m m e r s i o n and a s u b m e r s i o n .

(b) For e a c h x c M , t h e mapping d i p ( x ) : TxM*T i p ( x j N is a

s u r j e c t i v e i s o m o r p h i s m of Banach s p a c e s .

( c ) For e a c h x c M , t h e r e a r e a n e i g h b o u r h o o d U of x in M and a n e i g h b o u r h o o d V o f y= $ ( x )

in N such t h a t

a n a l y t i c homeomorphism of U o n t o V. 6.15. ip:

D E F I N I T I O N . If a m o r p h i s m o f

@ IU

is a n

Banach m a n i f o l d s

M-tN s a t i s f i e s a n y o f t h e a b o v e c o n d i t i o n s , we s a y t h a t ip i s

a “ l o c a l i s o m o r p h i s m ” o f M and N.

83

THE L I E GROUP O F AUTOMORPHISMS

B y a n f ' i s o m o r p h i s m N o f Banach m a n i f o l d s we mean a b i j e c t i v e l o c a l isomorphism 6.16.

4:

M-tN.

PROPOSITION. L e t M , N and

4 be r e s p e c t i v e l y a

t o p o l o g i c a l s p a c e , a Banach m a n i f o l d o v e r E and a mapping

4 : M+N.

Then t h e f o l l o w i n g s t a t e m e n t s a r e e q u i v a l e n t ( a ) F o r e v e r y XCM,

t h e r e i s an o p e n n e i g h b o u r h o o d U o f

x i n M, t h e r e i s a c h a r t (V,v) o f y = : $ ( x ) in N and t h e r e is a c l o s e d t o p o l o g i c a l l y c o m p l e m e n t e d s u b s p a c e F of E s u c h t h a t v Q $ is a homeomorphism o f U o n t o F n vp# ( U ) ] . (b) T h e r e e x i s t s a Banach m a n i f o l d s t r u c t u r e o n M s u c h t h a t i t s u n d e r l y i n g t o p o l o g y i s t h e t o p o l o g y of M and

M+N

@:

i s an i m m e r s i o n . The m a n i f o l d s t r u c t u r e s a t i s f y i n g t h e s e c o n d i t i o n s i s u n i q u e and i t s c h a r t s a r e t h e p a i r s (V,V,I$

1") , where

U is as i n

(a)

.

W e c a l l it t h e @ - i n v e r s e image of t h e m a n i f o l d s t r u c t u r e i n N . 6.17.

DEFINITION. L e t N be a Banach m a n i f o l d and d e n o t e

by i : M-tN a t o p o l o g i c a l s u b s p a c e M o f N and t h e c a n o n i c a Z inclusion. I f the pair (M,i) s i t i o n 6 . 1 6 , we s a y t h a t M

s a t i s f i e s t h e c o n d i t i o n s of p r o p o endowed w i t h t h e i n v e r s e i m a g e

m a n i f o l d s t r u c t u r e o f t h a t in N is a s u b m a n i f o l d o f N . 52.-

The c o n c e p t o f a Banach-Lie

6.18.

group.

DEFINITION. A " B a n a c h - L i e " g r o u p is a s e t G w h e r e

we h a v e a g r o u p s t r u c t u r e t o g e t h e r Q i t h an a n a l y t i c s t r u c t u r e o v e r a Banach s p a c e E s u c h t h a t t h e mapping GxG+G g i v e n b y ( x , y ) + x y - l is a n a l y t i c . A c c o r d i n g a s E i s r e a l o r complex w e s a y t h a t G i s a r e a l o r a

compZex Banach-Lie

group.

I f e denotes t h e i d e n t i t y element o f G I we have

6.19.

PROPOSITION. L e t t h e s e t G

be endoved w i t h a

g r o u p s t r u c t u r e and an a n a l y t i c s t r u c t u r e o v e r E . T h e n G i s a Banach-Lie satisfied:

g r o u p i f and o n l y i f t h e f o l l o w i n g c o n d i t i o n s a r e

84

CHAPTER

6

L1: P'or nZli xOeG, t h e mapping G-tG g,Luen b y x+x x i s a n a l y t i c . 0

by

L 2 : For a1,l x0@G, t h c mapping G+G g i v e n

cinal'yt'ic .in a n o p e n neighbourhood

L3: T h e mapping GxG-+G o p e n n e i g h h o u r h o o d of

g i v e n by

x

+

x xx - 1

of e .

0

0

.is

(x,y)+xy-' i s a n a Z y t i c and

( e , e ).

P r o o f : I f G i s a Banach-Lie

group, t h e n t h e s e c o n d i t i o n s

a r e obviously s a t i s f i e d . Let

( x O t y 0 ) e G x Gb e g i v e n . Then w e h a v e

xy-i=

f o r a l l x,ysG.

~ x o y ; l ~ y o r ~ x ;( yl ox ~Y ) -1

-1

IY;'

Thus, t h e mapping ( x , y ) + x y - ' c a n be r e p r e s e n t e d

i n a n e i g h b o u r h o o d of

(x,,yo)

as a c o m p o s i t e of mappings o f t h e

t y p e s m e n t i o n e d i n c o n d i t i o n s L 1 , L2 a n d L

3'

whence t h e r e s u l t

follows.

7Y 6.20.

COROLLARY. L e t G b e e n d o w e d w i t h

a group s t r u c t u r e

and uri a n a 1 , y t i c s t r u c t u x a t . . T h e n G is a R a n u c h - L i e o n l y if t h e f o l l o w i n g c o n d i t i o n s a r e s a t i s f i e d :

group is a n d

L;:

T h e m a p p i n g G-tG g i v e n b y x+x-l i s a n a l y t i c o n G .

L;:

The mapping G x G + G g i v e n by

(x,y)+xy i s a n a l y t i c on GxG.

Proof: If ( x , y ) + x y - ' is a n a l y t i c , so a r e t h e m a p p i n g s y + ( e , y ) + e y - l and ( x , y ) - ( x , u - ' ) + x ( y - ' ) -

1

.

I f y-ty-l a n d

( x , y ) + x y a r e a n a l y t i c , so i s ( x , y ) + ( x , y - l ) + x y - l .

7Y 6.21. ~ P O U ~s tJ r

COROLLARY. L e t G b e a B a n a e h - L i e

group. T h e n the

u c t u r e o n G is c o m p a t i b Z e w i t h t h e t o p o Z o g y u n d e r Z y i n g

t h e a n a l y t i c s t r u c t u r e o$ G , 6.22.

i.e.,

G i s a topoZogicuZ group.

EXERCISES. Assume t h a t G i s a Banach-Lie

group.

85

THE LIE GROUP OF AUTOMORPHISMS

Show that the topological group G satisfies the following conditions: (a) G is metrizable. (b) Both the left and right uniform structures of G are complete. 6.23. DEFINITION. L e t G a n d H b e B a n a c h - L i e g r o u p s . We Q m a p p i n g f: G+H is a " m o r p h i s m " of B a n a c h L i e g r o u p s

say t h a t

if f is a m o r p h i s m of b o t h t h e g r o u p s t r u c t u r e s a n d t h e manifold s t r u c t u r e s of G a n d H. 6.24. PROPOSITION.Let G a n d H b e Banach-Lie groupsand denote b y f: G+H a g r o u p h o m o m o r p h i s m . T h e n f i s a morphism of

Banach-

L i e g r o u p s if a n d o n l y i f f: G-tH i s a n a l y t i c i n a n e i g h b o u r h o o d o f e.

Proof: Let x sG be given. If f: G+H is a group homomor0 phism, we have f(x)= f(xilx) for all xcG. Thus, by conditions L' and L; of corollary 6.20,if f is analytic in a neiqhbourhood 1 of e, it is analytic on G. The converse is obvious. 6.25. DEFINITION. L e t t h e B a n a c h - L i e g r o u p G a n d t h e e l e m e n t asG b e g i v e n . We d e f i n e t h e r r Z e f t " a n d " r i g h t t r a n s l a t i o n s " b y a a s t h e m a p p i n g s G+G g i v e n r e s p e c t i v e l y b y

La: x+ax

r

:

x+xa

,

xeG.

Obviously, La and ra are automorphisms of the analytic structure of G. Moreover, the mapping i

:

x+axa

-1

xsG

is a Banach-Lie group automorphism of G. 6.26. DEFINITION. L e t G b e a B a n a c h - L i e g r o u p . We s a y t h a t a s u b s e t H c G i s a " B a n a c h - L i e s u b g r o u p " of G is H is a s u b g r o u p a n d a s u b m a n i f o l d of G w i t h r e s p e c t t o t h e c a n o n i c a l i n c l u s i o n i: H-tG.

86

6

CHAPTER

6.27.

EXERCISES.

( a ) L e t G be a B a n a c h - L i e

t h a t t h e i d e n t i t y component of e i s a B a n a c h - L i e ( b ) L e t H be a B a n a c h - L i e G.

g r o u p . Show subgroup of G. s u b g r o u p of

Show t h a t H i s c l o s e d a n d t h a t t h e c a n o n i c a l i n c l u s i o n

i : H-tG i s a morphism o f B a n a c h - L i e

6.28.

DEFINITION.

groups.

L e t G be n B a n a c h - L i e

g r o u p . We sag

thal an a n a l y t i c u e c l v r f i e i ' d X c T ( G ) i s " l e f t i n v a r i a n t " ,if, for all acG, X i s r e l a l e d t o i t s e l f b y Ra , i.e., i f ZJC h a v e (6.3)

dQ,;X=

aeG

XoK.

a

is t h e local ( U , u ) i s a c h a r t of G a n d X = f ( x ) au / x e x p r e s s i o n of X I t h e n ( 6 . 3 ) i s e q u i v a l e n t t o If

W e d e n o t e by G ( G ) t h e s u b s e t of Y ( G ) c o n s i s t i n g of a l l l e f t

i n v a r i a n t a n a l y t i c v e c t o r f i e l d s on G. A s a n immediate co n s e-

we obtain

q u e n c e of p r o p o s i t i o n 6 . 1 2 , 6.29.

PROPOSITION. L e t G be u Banach-Lie g r o u p .

G ( G ) is a L i e s u b a l g e b r a of

6.30.

Then

T(G).

PROPOSITION. L e t

(Y:

G ( G ) + T e ( G ) b s t h e evaZuaLion

at I h e p o i n t eeG. T h e n a is a s u r j e c t i v e isomorphism of v e c t o r spuces. P r o o f : L e t (U,u) be a c h a r t o f G a t e ; t h u s

by X - + X ( e )

a au

le

f o r XcG(G)

b e c a u s e of d e f i n i t i o n 6 . 1 0 .

some X,YeG(G).

. Clearly

c1

is given i s a l i n e a r mapping c1

Assume t h a t w e h a v e X ( e ) = Y ( e ) f o r ( e )I w e

A s X a n d Y are l e f t i n v a r i a n t and a = R

have

for a l l acG.

Let h

a au

le

Thus X= Y and a

is i n j e c t i v e .

e T e ( G I b e g i v e n . Then w e d e f i n e X ( a ) =: dR (e) .h

a

au

le

87

THE L I E GROUP O F AUTOMORPHISMS

f o r asG a n d it i s i m m e d i a t e t o c h e c k t h a t dRe ( e ): Te ( G ) + T e ( G )

i s t h e i d e n t i t y mapping. Thus

Moreover, X i s a n a l y t i c . Indeed, s i n c e l e f t t r a n s l a t i o n s a r e a u t o m o r p h i s m s of t h e m a n i f o l d s t r u c t u r e o f G ,

(aU, u 0 R - l )

c h a r t o f G a t t h e p o i n t a and t h e l o c a l e x p r e s s i o n of R

a

is a is the

i d e n t i t y map. T h u s , X i s l o c a l l y r e p r e s e n t a b l e a s t h e c o n s t a n t mapping x-th

a au

(x

f o r xsaU a n d X i s a n a l y t i c . B e s i d e s , X i s

l e f t invariant since

X [ R a ( x ) ] = X ( a x ) = dR

ax

(e).h

a au

le

=

f o r a l l a,xsG. T h e r e f o r e , l e f t i n v a r i a n t v e c t o r f i e l d s on a Banach L i e g r o u p a r e a n a l y t i c a n d t h e y a r e u n i q u e l y d e t e r m i n e d by t h e i r v a l u e s a t the point esG. By means of t h e i s o m o r p h i s m a: G ( G ) + T e ( G )

we can t r a n s f e r t h e

Banach s p a c e s t r u c t u r e o f T e ( G ) t o G ( G ) , a n d i t i s i m m e d i a t e t o v e r i f y t h a t , i n t h i s way, G ( G ) becomes a Banach-Lie a l g e b r a . W e c a l l i t t h e Bannck-Lie

53.-

a Z g e b r a of G .

S p e c i f i c ____ e x a m p l e s : The l i n e a r g r o u p a n d i t s a l g e b r a i c subgroups.

L e t A b e a r e a l o r complex Banach a l g e b r a w i t h u n i t e . W e

i n d i s t i n c t l y d e n o t e b y M a n y o f t h e f i e l d s IR o r it.

6.31.

DEFINITION. We d e f i n e t h e " c o m m u t a t o r p r o d u c t " o n

A by m e a n s o f

[x, y] = : xy-yx

x,ycA.

CHAPTER

88

6

I t i s i m m e d i a t e t o see t h a t t h i s p r o d u c t s a t i s f i e s t h e c o n d i -

t i o n s o f d e f i n i t i o n 4 . 2 0 a n d t h a t t h e commutator p r o d u c t

[,I:

AxA-tA

i s c o n t i n u o u s . Thus A i s a Banach-Lie a l g e b r a .

L e t u s d e n o t e by G ( A ) t h e s e t o f r e g u l a r e l e m e n t s o f A ; G(A)

thus,

i s a g r o u p and a n o p e n s u b s e t of A . T h e r e f o r e , G ( A ) i s a

Banach m a n i f o l d i n a c a n o n i c a l manner ( c f . e x a m p l e s 6 . 3 ) .

Now

w e have

6 . 3 2 . LEMMA. W i t h ils c a n o n i c a l structures of g r o u p a n d g r o u p #hose B a n a c h - L i e

B a n a c h manifold, G ( A ) is a B a n a c h - L i e

algebra is A . p r o o f : I t i s i m m e d i a t e t o c h e c k t h a t c o n d i t i o n s L 1 and L

of p r o p o s i t i o n 6 . 1 9 a r e s a t i s f i e d . F o r yeA w i t h 2 we h a v e

)I

y-eI\ < I

m

Y-l=

[ e + ( y - e ) J -l;

x

1"

(-

n=O

t h e series b e i n g c o n v e r g e n t i n t h e norm of A . (x,y)+xy-l i s a n a l y t i c i n a neighbourhood of

T h u s , t h e mapping (e,e), i.e.,

c o n d i t i o n L 3 i s s a t i s f i e d , t o o , a n d G ( A ) i s a Banach L i e g r o u p o v e r t h e Banach s p a c e A. Let a c G ( A ) be f i x e d ; w i t h r e s p e c t t o t h e c a n o n i c a l c h a r t , t h e expression of t h e l e f t t r a n s l a t i o n R a i s R T h u s , i t s d e r i v a t i v e dR

a

( X I - ax f o r x c G ( A ) .

i s g i v e n by

I f X= X(x)

3 -

is a l e f t invariant au v e c t o r f i e l d on G ( A ) , by d e f i n i t i o n 6 . 2 8 w e have f o r a l l xeG(A) a n d hwl.

I

dR (x).X ( x ) = X [ R a ( X I By ( 6 . 4 )

]

t h i s is equivalent t o

f o r a l l a,xeG(A). Taking x= e we o b t a i n X ( a ) = a X ( e ) f o r a c G ( A ) o r , by c h a n g i n g t h e n o t a t i o n ,

THE LIE GROUP O F AUTOMORPHISMS

89

xcG(A)

(6.5)

I t i s e a s y t o see t h a t v e c t o r f i e l d s of t h e

where h = : X ( e ) c A .

form ( 6 . 5 ) a r e a c t u a l l y l e f t i n v a r i a n t . M o r e o v e r , f o r X=:

xh

a

1

z/ x

and

a

xh2

Y=:

[ X , Y ] ( x ) = ( x h h -xh h 2

1

lX

1 2

a au

w e have

Ix

= X[hl,h2]

a au lX

so t h a t t h e mapping T G ( A ) + A o b t a i n e d by e v a l u a t i n g a t e c G ( A ) i s a s u r j e c t i v e Banach-Lie

isomorphism between T G ( A ) and A .

# 6.33.

DEFINITION. We s a y t h a t t h e B a n a c h - L i e

group G ( A )

i s t h e "Zineari g r o u p " o f t h e Banach a l g e b r a A a n d d e n o t e i t b y GL(A;IK).

Assume t h a t A i s complex Banach a l g e b r a ; t h e n it i s a r e a l

.

Banach a l g e b r a , t o o , w h i c h i s d e n o t e by A Thus t h e l i n e = IR group G L ( A , E ) , with i t s underlying r e a l manifold s t r u c t u r e , i s a r e a l Banach-Lie

G.W e

group over

s a y t h a t it i s t h e underZying

reaZ l i n e a r g r o u p o f G L ( A , E ) .

6.34. DEFINITION. For t d R and x c A exptx=:

we d e f i n e

t n xn

C n! n=O

S i n c e A i s c o m p l e t e , w e h a v e e x p t x c A ; a c t u a l l y , e x p t x i s a reg u l a r e l e m e n t of A a n d ( e x p t x ) - I = e x p ( - t x )

,

so t h a t t h e mapping

e x p : I R x A + G L ( A J K ) g i v e n by ( t , x ) + e x p t x i s r e a l a n a l y t i c . F o r t = l

w e s i m p l y w r i t e e x p x i n s t e a d o f e x p l x . The mapping A + G L ( A , I K ) g i v e n by x+expx i s r e a l a n a l y t i c , too. 6.35. @:

PROPOSITION. L e t A , B b e Banach a Z g e b r a s and

G L ( A J K l + G L ( B , I K ) a m o r p h i s m of t h e c o r r e s p o n d i n g

Z i n e a r groups

Then t h e d e r i v a t i v e d @( e ) : TeGL ( A ,lK) +TeGL ( B & ) i s a homomorphism o f t h e B a n a c h - L i e a l g e b r a s A and B and we h a v e @ ( e x p x ) = e x p [d@( e )x]

CHAPTER

90

6

Proof: For any fixed xeA, the mapping f: I R + G L ( B ; I K ) given by t+f(t)=: b(exptx1 is real analytic. Moreover, since $ is a group homomorphism, by setting a=: d$(e)xsB, we have f ( 0 ) =$(e)=e and

=

CP (exptx)lim g1 [+(expsx)-el= $(exptx)a s+o

Thus,f is the solution of the initial value problem

in the Banach space B. Now, we consider the function g: I R + G L ( B , l K ) given by t+g(t)=: expta with a= d@(e)x. It is easy to see that g(0)= e and

Thus g is also the solution of (6.6) and we have f(t)= g(t), i.e. $(exptx)= exptd$(e)x for all tdIR and xcA. NOW, let x,ycA be given and consider the mapping F : I R + G L ( B J K ) given by F(t)=: $(exptx)$(expty), tcTR. A computation similar to the one above gives ( 6-7)

F' (t)= $ (exptx) (a+b)@ (expty)

where we have put a= d@(e)x, derivative of (6.7) at t= 0

b=: dQ(e1y. By taking the

F" ( 0 ) = aL+2ba+bL

Similarly, if G(t) = :

+ (expty)@ (exptx), we

have

THE L I E GROUP O F AUTOMORPHISMS

91

2 2 G " ( O ) = a +2ab+b

so t h a t Y ( t )= : F ( t ) - G ( t )

Also,

satisfies

a p p l y i n g twice t h e c h a i n r u l e a t t = 0 t o compute Y " ( 0 )

we o b t a i n

whence t h e r e s u l t f o l l o w s by comparing w i t h ( 6 . 8 ) .

6.36.

EXERCISE. L e t '4:

A+B b e a Banach-Lie

algebra

homomorphism. Show t h a t t h e r e e x i s t s a unique Banach-Lie homomorphism such t h a t d $ ( e ) = Y . 6.37.

PROPOSITION. L e t H b e a s u b g r o u p o f t h e l i n e a r

g r o u p GL(A,X) a n d l e t B b e a c l o s e d s u b s p a c e of A . A s s u m e t h a t t h e r e a r e a n e i g k b o u r k o o d U of e i n G L ( A , l K ) a n d a n e i g k b o u r k o o d V of

0 i n A s u c k t h a t t h e e x p o n e n t i a l m a p p i n g exp: V n B + U n H i s

a homeomorphism f o r t h e c o r r e s p o n d i n g i n d u c e d t o p o l o g i e s . T k e n B i s a B a n a c k - L i e s u b a l g e b r a of A a n d H i s a B a n a c k - L i e w h o s e L i e a l g e b r a i s B.

group

P r o o f : S i n c e exp ( '= i d , by t h e i n v e r s e mapping theorem 0

t h e r e i s no loss of g e n e r a l i t y i n assuming t h a t ( U , l o g

) IU

,

is a

c h a r t of G L ( A , I K ) a t e l where log d e n o t e s t h e i n v e r s e of exp. Thus ( U n H , 1 0 g l U n H i s a c h a r t of H o v e r t h e Banach s p a c e B . Now, l e t hcH be g i v e n . A s H i s a subgroup of G L ( A , I K ) t r a n s l a t i o n R h maps U l l H o n t o a s e t R , ( U n H ) c H

,

the l e f t

which i s a

neighbourhood of h i n H I and t h e p a i r (6.9) i s a c h a r t of H a t h . I t i s e a s y t o see t h a t t h e f a m i l y g i v e n by ( 6 . 9 )

f o r hcH i s an a n a l y t i c s t r u c t u r e o v e r B . Moreover, f o r

t h i s a n a l y t i c s t r u c t u r e , H i s a Banach-Lie p o s i t i o n 6.30,

B i s a L i e s u b a l g e b r a of A .

group; t h u s by pro-

#

CHAPTER

92

6.30. nach-Lie

6

REMARK. N o t i c e t h a t , i n g e n e r a l , H i s n o t a Ba-

s u b g r o u p o f GL(A,IK) b e c a u s e , a s a m a n i f o l d , H may f a i l

t o be a s b n i f o l d o f GL(AJK): i t s t a n g e n t space a t e is B , which i n g e n e r a l i s n o t a complemented s u b s p a c e o f A . 6.39.

DEFINITION. L e t A b e a Banach a l g e b r a o v e r I K and

l e t H he a s u b g r o u p of G L ( A J K ) . We s a y t h a t H is ::ithgroup of d e g r e e Cn of GL(A,IK) S

(I

IK-algebraic

if t h e r e e z , i s t s a n o n v o i d s e t

of c o n t i n u o u s v e c t o r - U a Z u e d I K - p o l y n o m i a Z s q : AxA-tE

cn w i l h q ( O , O ) =

of

degree

0 s u c h t h a t w e have

Of p a r t i c u l a r i n t e r e s t f o r u s , t h o u g h n o t i n c l u d e d i n t h e a b o v e d e f i n i t i o n , i s t h e s i t u a t i o n i n w h i c h w e h a v e a Banach a l g e b r a o v e r C and a s u b g r o u p H o f G L ( A , t ) , ' b u t t h e e q u a t i o n s ( 6 . 1 0 ) d e f i n i n g H a r e ' l l i - p o l y n o m i a l s q: AxA+E o n t h e u n d e r l y i n g El-struct u r e s of A x A

and E

Notice t h a t , i n a l l t h e s e c a s e s , H is closed i n GL(AJK).

Clear-

l y , a n y f i n i t e p r o d u c t a n d a n y i n t e r s e c t i o n of IK-algebraic s u b g r o u p s o f d e g r e e & n i s a M - a l g e b r a i c s u b g r o u p of d e g r e e Cn. The d e f i n i t i o n i n c l u d e s t h e c a s e i n which H i s d e f i n e d b y a s e t S o f M - p o l y n o m i a l s q : A+E d e p e n d i n g on a s i n g l e v a r i a b l e . A l s o ,

by t h e Hanh-Banach t h e o r e m , t h e p o l y n o m i a l s qcS can b e c h o s e n t o b e IK-valued. 6.40.

ussume t h a t H

THEOREM. L e t A b e a Bunach a l g e b r a o v e r I K and

i s aIK-algebraic

s u b g r o u p of d e g r e e hn of

G L ( A , I K ) . Then H is u Banach L i e g r o u p whose B a n a c h - L i e

algebra

Is

P r o o f : W e w r i t e w=: ( u , v ) f o r t h e e l e m e n t s o f AxAvihich i s a Banach a l g e b r a o v e r I K w i t h r e s p e c t t o t h e norm

/I w J / =:

M-Banach

, //

n

v / / I . L e t u s p u t P = : 9 ~ ~ ( a x . 4f )o r t h e k=l s p a c e o f continuouslK-polynomials p: AxAjlK of d e g r e e

max{// u l /

s n s u c h t h a t p ( O , O ) = 0 . Now w e d e f i n e a mapping

THE L I E GROUP O F AUTOMORPHISMS

@:

93

GL(AJK)+GL(P(P) ,K) by means of [ @ ( x ) p(]u , v ) = : p ( u x , x - l v )

where pcP, Y : A+L(P)

(u,v)eAxA and xcGL A S ) . A l s o , w e d e f i n e a mapping by [Y(x)p] ( u , v ) =:

where pcP,

(u,v)cAxA and xcA.

F i r s t , w e s t u d y some p r o p e r t i e s of 0 and Y . W e have

( a ) 4 i s a Banach-Lie group homomorphism and d + ( e ) = Y. The proof i s a n e x e r c i s e . Thus, by p r o p o s i t i o n 6 . 3 5 w e g e t ( b ) Y i s a Banach-Lie

a l g e b r a homomorphism and

[ @ ( e x p x ) p ] =[ ~ X P + ( X ) ] P

(6.11)

f o r a l l xcA and pep. ( c ) L e t z s A be g i v e n . Then e a c h of t h e s u b s p a c e s P k ( ~ x A ) k, = l , 2 , . . , n r

i s i n v a r i a n t by Y ( z ) , i . e . ,

w e have Y ( z ) c P k ( A x A ) . Moreover, i f z s A i s a r e g u l a r e l e m e n t of A , t h e n Y ( z ) i s a r e g u l a r e l e m e n t of L ( P ) . I n d e e d , l e t pcP(AxA) be g i v e n and suppose t h a t F c L k ( A x A , I K ) i s

its a s s o c i a t e d symmetric k - l i n e a r mapping s o t h a t w e have

Then, t h e mapping f : AxA-+AxA g i v e n by (6.12)

f ( z ) : w= ( u , v )

+

(uz,-zv)

s a t i s f i e s f ( z ) c L ( A x A ) . T h e r e f o r e , from t h e d e f i n i t i o n of y w e get

and Y ( z ) pcPk ( A K A ) .

Now, suppose t h a t zcA i s r e g u l a r . Then f

( 2 ) a s d e f i n e d by ( 6 . l a i s r e g u l a r i n L ( A x A ) and f ( z ) - l = f ( z - I ) . Thus, t h e r e s t r i c t i o n

94

6

CHAPTER

o f Y ( z ) t o each of t h e s u b s p a c e s P ( A x A ) i s a r e g u l a r element k o f L(Pk(AxA)) , t h e i n v e r s e image of pePk(AXA) b e i n g g i v e n by

Therefore Y ( z ) i s r e g u l a r i n L ( P )

,

too.

Next w e show t h a t B i s a c l o s e d L i e s u b a l g e b r a of A . S i n c e H i s a IK-algebraic subgroup of d e g r e e s n of G L ( A , X ) t h e r e i s a s e t of IK-polynomials

,

S c P such t h a t

H= I ~ ~ G L ( A ; I K ) ;q ( z , z - ' ) =

o

VqcS 1

W e d e f i n e a n o t h e r s e t o f polynomial Q c P by means of

(6.13)

Q=:

{pep;

p(h,h-')= 0

VhcH}

C l e a r l y , Q i s a c l o s e d M - s u b s p a c e of P and S c Q : t h u s , i n particular (6.14)

[zcGL(AJK), q ( z , z - ' ) = 0

VqcQ]=>

zeH

We c l a i m t h a t , f o r xeGL(AJK), w e have t h e e q u i v a l e n c e (6.15)

xeH < = > + ( x ) Q c Q

Indeed, l e t xeGL(A,X) be g i v e n and assume t h a t xsH. A s H i s a subgroup of G L ( A , l K )

,

we have Hx= H-lx= H . From ( 6 . 1 3 ) and t h e

d e f i n i t i o n of 4 w e o b t a i n

f o r a l l qcQ and hcH; t h u s @ ( x ) Q C Q by ( 6 . 1 3 ) .

Conversely, l e t

xeGL(A,X) be g i v e n and assume t h a t + ( x ) Q c a . By ( 6 . 1 4 )

it

1

s u f f i c e s t o show t h a t q ( x , x - ) = 0 f o r a l l qcQ. L e t qeQ be given; by assumption we have @ ( x ) q c Q t; h u s by ( 6 . 1 3 ) w e o b t a i n -1 - 1 i . e . , q ( h x , x h ) = 0 for a l l heH, and [ @ ( x ) q ] ( h r h - l ) =0 , t a k i n g h= e s H we g e t q ( x , x - l ) = 0 . Now w e c l a i m t h a t , f o r yeA, w e have t h e e q u i v a l e n c e

95

THE L I E GROUP OF AUTOMORPHISMS

y€B < = > Y ( y ) Q c Q

(6.16)

I n d e e d , l e t ysB be g i v e n . Then we have exptycH f o r a l l tdR a n d , by (6.15)

,

@ ( e x p t y ) Q c Q t; h e r e f o r e , from ( 6 . 1 1 ) w e d e r i v e

[ e x p Y ( y ) ] Q C Q .I f w e f i x any q c Q , t h e mapping IR-tP g i v e n by t-texpY(y)q t a k e s i t s v a l u e s i n t h e closedIK-subspace Q of P a n d , by t a k i n g i t s d e r i v a t i v e a t t = 0 , w e g e t Y!(y)qcQ, whence Y ( y ) Q c Q . Conversely,

l e t yeA be such t h a t Y ( y ) Q c Q . A s Q i s

a c l o s e d x - s u b s p a c e of P , w e have [ e x p t + ( y ) ] Q c Qf o r a l l

tm.

Thus, by ( 6 . 1 1 ) , @ ( e x p t y ) Q c Q ,whence exptysH f o r a l l tdR and t h e r e f o r e ycB. I n p a r t i c u l a r , a s d @ ( e ) =Y i s a Banach-Lie a l g e b r a homomorphism,

( 6 . 1 6 ) e n t a i l s t h a t B i s a c l o s e d L i e s u b a l g e b r a of A . Next, w e show t h a t H i s a Banach-Lie group.

a:

L e t A =: A Q i A & the complexified of (I:

A = A when

IK= a)

. For

xcA

(I:

,

t h e Banach a l g e b r a A ( t h u s

S p ( x ) 1s the spectrum of x

i n A'.

From

t h e s p e c t r a l t h e o r y we know t h a t t h e s e t s

and

a r e , r e s p e c t i v e l y , neighbourhoods of e i n G L ( A J K ) and 0 i n A . According t o t h e holomorphic f u n c t i o n a l c a l c u l u s ( c . f .

11

I ) , on

U w e can s e l e c t a holomorphic b r a n c h of t h e l o g a r i t m i c f u n c t i o n

L e t u s d e n o t e by l o g i t s p r i n c i p a l d e t e r m i n a t i o n . By t h e spectral

mapping theorem, l o g : U+V i s a complex b i a n a l y t i c map ( t h u s , a r e a l b i a n a l y t i c map, t o o , i n c a s e I K = I R ) whose i n v e r s e i s exp: V+U.

T h e r e f o r e , by p r o p o s i t i o n 6 . 3 7 ,

it s u f f i c e s t o show

t h a t we have exp ( V

nB ) c U

H

log(U n H ) c V f l B

NOW, l e t ycV f l B be g i v e n ; from ycV and ycB w e g e t expycu and

exptycH f o r a l l tm, t h u s expycU n H .

CHAPTER

96

6

Next, l e t xcUnH be g i v e n and p u t y=: l o g x . Thus, i n p a r t i c u l a r (6.17)

YCV

M o r e o v e r r f r o m XSU w e d e r i v e

a n d by t h e s p e c t r a l mapping t h e o r e m

W e c l a i m t h a t t h e s p e c t r u m Sp(TY(y)] of ' Y ( y ) i n t h e c o r n p l e x i f i e d algebra L(P)'

of L ( P ) ( o r i n L ( P ) whenlK= C) s a t i s f i e s

Indeed, w e p r o v e . t h a t f o r A r e g u l a r element of L ( P )

'.

d with

(imgX/2v

I

AI-Y(y) i s a

Now w e h a v e

so t h a t

f o r pePk(AxA), k = 1 , 2 , . . , n ,

a n d w= ( u , v ) e A x A . T h u s , it s u f f i c e s

A

i; I - f ( y ) i s r e g u l a r i n L ( A x A ) f o r k = 1 , 2 , . . , n . B u t , due t o t h e d e f i n i t i o n of f ( y ) ,

t o prove t h a t

S i n c e by ( 6 . 1 8 ) ,

I imgh I ? v / n

e n t a i l s h/kgSp ( y ) a n d A/kgSp ( - y ) A A e-y a n d i; e + y a r e r e g u l a r i n

f o r k= I , 2 r . . r n r t h e e l e m e n t s A a n d so i s

Since

x k

I-f ( y ) i n L (AxA)

4 ( x )= 4 ( e x p y ) =

expY ( y )

.

by t h e f u n c t i o n a l c a l c u l u s a n d

t h e s p e c t r a l mapping t h e o r e m i t f o l l o w s t h a t

97

THE L I E GROUP O F AUTOMORPHLSMS

By R u n g e ' s t h e o r e m , t h e r e i s a s e q u e n c e of p o l y n o m i a l s p k c C \ X / , kdN, s u c h t h a t w e have l o g h = limpk(X) u n i f o r m l y when XcSp I @ ( x ) 1

. Then

k+m

I

1 Y ( y ) = log$ ( x ) = [Xe-Q( x ) ]-IlogAdX= 2 ~ i JY

S i n c e xeH, by ( 6 . 1 5 ) w e have @ ( x ) Q C Q a n d , as Q i s a closed s u b s p a c e of PI w e g e t Y ( y ) Q = l i m p k [ + ( x ) ] Q C Q .Then, by ( 6 . 1 6 ) w e a

obtain

k+m

Y ( y )c B

(6.19)

F i n a l l y , from ( 6 . 1 7 ) and ( 6 . 1 9 ) w e deduce Y ( y ) c V n B .

ff 6.41. over

a:

REMARK.

T h e case i n which A i s a Banach a l g e b r a

and H i s a n n - a l g e b r a i c

included i n our considerations.

subgroup of GL(A,E)

c a n a l s o be

Indeed, we can c o n s i d e r t h e

u n d e r l y i n g = - s t r u c t u r e s of A and G L ( A , C ) and d e f i n e n P = @ P ( A x A ) t o b e t h e Banach s p a c e of c o n t i n u o u s l R - p o l y n o m i a l s k=1

P: AxA

k

-t

a:

of d e g r e e Sn w i t h p ( O , O ) = 0 . A s i n o u r c a s e t h e

polynomials defining H belong t o a subset of P I we a r e i n a s i t u a t i o n i n which t h e o r e m 6 . 4 0 i s a p p l i c a b l e . A number o f i n t e r e s t i n g examples o f X - a l g e b r a i c

s u b g r o u p s of

d e g r e e s n of GL(A,IK) are i n c l u d e d i n t h e f o l l o w i n g PROPOSITION. L e t X I Y and f s L ("X,Y)

6.42.

be r e s p e e t i v e z y

t w o Banach s p a c e s 0 v e r . X and a x - r n u Z t i Z i n e a r mapping f : Xx...xX-+Y

.

Let

m= 0 o r m= 1 and s u p p o s e t h a t X= Y when m= 1 .

Then, t h e s e t H o f t h e eZements clcGL(L(X),IK)

satisfying

98

CHAPTER

f

(6.20)

( a x 11 . .

,(YX

n

1 = am f

(XI

I

-

6

* lXn)

XI

1 . .

,x

cx

is u B a n a c h - L i e g r o u p ~ l h o s eBanach-Lie a l g e b r a B is t h e s e t of a1 1. 6 c L ( X ) s a t i s f y i n g f(6x

(6.21)

1

,..,x n ) +..+

f(xl,...,fixn ) = m d f ( x l ,.., x n )

Here am a n d mfi d e n o t e r e s p e c t i v e l y t h e i d e n t i t y a n d t h e z e r o t r a n s f o r m a t i o n on Y when m= 0 . P r o o f : O b v i o u s l y A = : L ( X ) i s a Banach a l g e b r a o v e r K a n d H I a s d e f i n e d by ( 6 . 2 0 ) GL ( L(X)JK)

. We

i s a subgroup of t h e l i n e a r group

c l a i m t h a t H i s a IK-algebraic

s u b g r o u p of d e g r e e

Cn. I n d e e d , f o r f i x e d x 1 , x 2 , . . x ex, t h e mapping : T,(X)+X

PXlI.. lxn

g i v e n by

i s a c o n t i n u o u s n-homogeneous IK-polynomial,

and H i s d e f i n e d by

t h e s e t S of e q u a t i o n s PXlI..,X

( a ) =0

f o r x l , . . , x n ~ X . M o r e o v e r , B a s d e f i n e d by ( 6 . 2 1 ) ,

Banach-Lie

is a closed

s u b a l g e b r a of A = L ( X ) . T h u s , it s u f f i c e s t o show

t h a t H and B are a s i n theorem 6 . 4 0 i . e . w e have B= ( n e L ( X ) ; e x p t a c H

Vtm]

NOW, s u p p o s e t h a t G e L ( X ) s a t i s f i e s e x p t 6 e H f o r a l l

R e p l a c i n g cx b y e x p t d i n ( 6 . 2 0 )

tm.

and t a k i n g t h e d e r i v a t i v e a t

t = 0 w e see t h a t 6 s a t i s f i e s ( 6 . 2 1 ) ; t h u s 6 c B . C o n v e r s e l y , s u p p o s e t h a t 6eB and d e f i n e l i n e a r mappings Fn: L("X,Y)+L(X,Y)

by means o f

THE L I E GROUP OF AUTOMORPHISMS

x1,..,xneX

for GsL("X,Y),

and k = 1 , 2 , . . , n .

99

L e t us w r i t e

F=: (F1+F2+..+Fn)-F0 S i n c e 6 s a t i s f i e s ( 6 . 2 1 ) w e h a v e F ( f ) = 0 and FO,F 1 I

-

*

IFn

commute. T h e r e f o r e

.

( e x p F 1 ) . ( e x p Fn ) f = e x p ( F 1 + . . + F n ) f = ( e x p F o ) ( e x p F , ) f = ( e x p F0 ) f = =

(exp6)mf

which shows t h a t exp6cH. S i n c e B i s a l i n e a r s p a c e , w e c a n d o t h e same a r g u m e n t w i t h t 6 i n s t e a d of 6 , whence w e c o n c l u d e t h a t expt6sH f o r a l l tc3R.

# EXAMPLES. L e t A b e a Banach a l g e b r a o v e r IK ( w h e r e

6.43.

A may h a v e no u n i t a n d f a i l t o b e a s s o c i a t i v e , f o r e x a m p l e , a n y

Banach-Lie

a l g e b r a o f a n y B a n a c h - J o r d a n a l g e b r a ) . Then w e c a n

a p p l y p r o p o s i t i o n 6-42 2

t o t h e case i n which n = 2 , m= 1 , X = Y = A

i s t h e m u l t i p l i c a t i o n on A , i . e . ,

and f c L ( X , X )

f ( x , y ) = x.y.

O b s e r v e t h a t t h e a u x i l i a r y Banach a l g e b r a L ( X ) a p p e a r i n g i n p r o p o s i t i o n 6 . 4 2 i s now L ( A ) which i s a s s o c i a t i v e a n d h a s u n i t e v e n i f A f a i l s t o b e so. T h u s , t h e o r e m 6 . 4 0 a n d p r o p o s i t i o n 6 . 4 2 a r e a p p l i c a b l e . Accordingly, t h e set

w h i c h i s t h e g r o u p of a u t o m o r p h i s m s of A , i s a Banach-Lie g r o u p i n t h e normed t o p o l o g y o f L ( A )

.

The Banach-Lie a l g e b r a of

t h i s group i s

which i s t h e a l g e b r a o f l K - d e r i v a t i o n s M-algebraic equations p

of A. Besides, H i s t h e

s u b g r o u p o f G L ( L ( A ) 3 ) of d e g r e e $ 2 d e f i n e d b y t h e XPY

(a)= 0 where

100

CHAPTER

6

and x,yeA. L e t X b e a complex H i l b e r t s p a c e a n d Y = 5

. Then

can apply

p r o p o s i t i o n 6 . 4 2 w i t h n = 2 , m= 0 t o t h e spaces X a n d Y a n d t h e 2

r e a l b i l i n e a r mapping f c L ( X , E ) X,

f ( x , y ) = ( x l y ) . A c c o r d i n g l y , i f a* d e n o t e s t h e a d j o i n t o f t h e

operator H=:

g i v e n by t h e s c a l a r p r o d u c t on

C X E( X~)

,

t h e set

{ ~ c G L ( ~ ( ,E); x )

(axlay)= -

which i s t h e u n i t a r a y g r o u p of H , i s a r e a l Banach-Lie Banach-Lie

g r o u p whose

algebra is ( 6 X l y ) + ( x l d y ) =0

B= { 6 C L E ( X ) ;

The u n i t a r y g r o u p o f X is a r e a l a l g e b r a i c s u b g r o u p of G L ( L c ( X ) ,lK)

of d e g r e e 6 2 d e f i n e d by t h e e q u a t i o n s p

XIY

( a )= 0

where PX,Y

(a)= ( a x l a y ) - ( x l y )

a n d x,ysX. Let

x

b e a H i l b e r t s p a c e over 6: a n d d e n o t e by Q a n y c o n j u g a -

t i o n on X . Take f e L ( 2 X l C ) t o b e t h e r e a l b i l i n e a r mapping g i v e n by f ( x , y ) = : ( Q x / y ), x , y c X . Then w e c a n a p p l y p r o p o s i t i o n 6 . 4 2 w i t h n = 2 , m= 0 . If a

t

d e n o t e s t h e t r a n s p o s e d of t h e o p e r a t o r

ac,!&(X) , t h e n , t h e s e t

which is t h e o r t h o g o n a l g r o u p of X , i s a r e a l Banach-Lie whose Banach-Lie B = {6CLc(X);

group

algebra is

( Q 6 x / y ) t ( Q X l f i y ) =0

YX,y€X)= { 6 C L E ( X ) ; 6 t 6 t = 0 )

The o r t h o g o n a l g r o u p of X i s a r e a l a l g e b r a i c s u b g r o u p of

THE L I E GROUP O F AUTOMORPHISMS

GL(L (X)

LT

,a)

of d e g r e e c 2 d e f i n e d by t h e e q u a t i o n s p

101

X,Y

(a)= 0

where

and x,ysX.

6.44. EXERCISES. L e t X,Y b e H i l b e r t s p a c e s o v e r (I. i s a c l o s e d complex subspace &of L L T ( X , Y ) such

A J*-algebra

t h a t w e have -*As acGL(L

2 whenever A , B s

a. A

J*-automorphism i s any

,(I) such t h a t w e have LT ( A )

for a l l A , B s A .

Prave t h a t t h e group of a l l J*-automorphisms

of '2 i s a r e a l a l g e b r a i c subgroup o f G L ( L L T ( & ) , C ) of d e g r e e

c 3 . Thus i t i s a r e a l Banach-Lie group. Prove t h a t i t s BanachL i e algebra i s the set

of a l l J * - d e r i v a t i o n s of L a ( h ) .

Now w e t u r n o u r a t t e n t i o n t o t h e c o n s t r u c t i o n of a r e a l BanachL i e group s t r u c t u r e on AutD.

m o u g h o u t t h i s s e c t i o n , B c D and 6 > 0 s t a n d f o r a f i x e d open b a l l and a r e a l number such t h a t B 6 c c D . W e know t h a t (autD,

/I .I(

)

B6

i s a r e a l Banach-Lie a l g e b r a . W e d e n o t e by Holm(B,E) t h e complex Banach s p a c e of holomorphic mappings f : B+E t h a t a r e bounded on B . Thus ( H o l m ( B , E ) ,

11

./IB)

i s a r e a l Banach s p a c e ,

too *

As autD and Holm(B,E) w i l l always be endowed w i t h t h e t o p o l o g i e s

.

.I I B ,

r e s p e c t i v e l y d e f i n e d on them by t h e norms I / /I and I ] B6 w e s h a l l omit any r e f e r e n c e t o t h e s e norms. However, we s h a l l c o n s i d e r s e v e r a l t o p o l o g i e s on AutD H o l m ( B , E ) ; t h u s , i n o r d e r t o a v o i d any p o s s i b l e confusicm,

whenever w e r e f e r t o A u t D w e

102

6

CHAPTER

s h a l l e x p l i c i t e l y m e n t i o n t h e t o p o l o g y w e are c o n s i d e r i n g on i t 6.45.

LEMMA. li'herc i s

neighbourhood

R

i n a u t D s u c h t h a t , f o r any A c M

,

M

of Lhe o r i g i n

the series

13 AnidD

-1; n=O

is c o n v e r g e n t t o (expA) i n t h e s p a c e Holw(BrE). T h e m a p p i n g IB HolB(B,E) g i v e n b y

exp: M

+

(6 . 2 2 )

P r o o f : L e t AeautD b e g i v e n . Then, f o r tCiR a n d ndN, w e

have

t" An A i d e H o l m ( B r E ) ;t h e r e f o r e , w e D n!

can d e f i n e a formal

power s e r i e s IR+Holm(B,E) by means o f

t-t

(6-23)

2;

n t"! i n i d D

n =O

w e have

A s i n t h e p r o o f of p r o p o s i t i o n 4 . 1 ,

f o r a l l ndN, where M = :

/I

M =:

i d D / I B< m i s i n d e p e n d e n t of n . N o w , {AcautD;

11

All,

s 6

i s a n e i g h b o u r h o o d of t h e o r i g i n i n a u t D a n d , f o r a n y f i x e d AE

M

,

t h e r a d i u s of c o n v e r g e n c e o f

( 6 . 2 3 ) is g r e a t e r t h a n 1 .

S i n c e Hol-(B,E) i s c o m p l e t e , m

f(t,A) lB=:

C n=O

i s convergent i n Holm(BrE) t o (exptA) Moreover, i t is e a s y t o see t h a t tn-l

tn

A

n! A n i d B IB

f o r all t c [ - l , + I ] .

f ( t , A ) = C -i n i d D = A[f ( t , a ) ] dt n=l ( n - l ) !

d

THE L I E GROUP OF AUTOMORPHISMS

103

and f ( O , A ) = i d D . T h u s , b y d e f i n i t i o n 4 . 4 ,

Now ( 6 . 2 2 ) d e f i n e s a f o r m a l power s e r i e s b e t w e e n t h e r e a l Ba-

t h i s series

n a c h s p a c e s a u t D a n d H o l m ( B I E ) . S i n c e f o r AEM

i s convergent, exp: M + H o l m ( B I E ) d e f i n e s a real a n l a y t i c mapping on M

. #

6.46.

REMARK. N o t i c e t h a t e x p : M + H o l m ( B , E ) t a k e s i t s

v a l u e s n o t o n l y i n t h e s p a c e H o l m ( B , E ) b u t i n t h e smaller s e t AutD. 6.47.

LEMMA. T h e r e a r e a n e i g h b o u r h o o d

and a n e i g h b o u r h o o d expM

+

N

N

of

idD i n Holm(B,E)

of 0 i n a u t D

M

such t h a t

i s a b i j e c t i o n . Moreover, b o t h exp: M

i n v e r s e log: N

-t

a r e Z i p s c h i t z i a n on M

M

P r o o f : By lemma 6 . 4 5 e x p : M

N

and i t s

.

Holm(BIE) i s a real

+

a n a l y t i c mapping on a n e i g h b o u r h o o d

and

N

+

M of 0 i n a u t D . I t s

d e r i v a t i v e a t t h e o r i g i n i s t h e e l e m e n t of L ( a u t D , H o l m ( B , E ) ) g i v e n by e x p h l A = f; i d D = A T h u s , by t h e i n v e r s e mapping t h e o r e m , t h e r e a r e a n i e g h b o u r h o o d

M ' o f 0 i n autD and a neighbourhood

M'

such t h a t exp:

+

N'

N'

o f i d i n Holm(B,E)

i s a b i a n a l y t i c mapping. By t h e

c o n t i n u i t y of t h e d e r i v a t i v e a t 0 , t h e r e i s a convex neighbourhood

Then, f o r A1,A2e

M

o f 0 i n a u t D i n which e x p '

MI'

w e have

i s bounded

104

CHAPTER

and e x p i s l i p s c h i t z i a n o n M " .

6

A similar a r g u m e n t a p p l i e s t o

i t s i n v e r s e l o g . T h e r e i s no loss of g e n e r a l i t y i n a s s u m i n g

t h a t MI'= M

and N = expM

.

Is N = : expM a T-neighbourhood o f i d

6.48.

QUESTION.

6.49.

EXERCISE. Show t h a t i f q u e s t i o n 6 . 4 8 h a s a n

D

in

AutD?.

a f f i r m a t i v e a n s w e r , t h e n by l e m m a 6 . 4 7 , e v e r y FcAutD a d m i t s a n e i g h b o u r h o o d t h a t i s homeomorphic t o M by A+F expA. However, a s w e s h a l l see i n c h a p t e r 8 , t h e a n s w e r i s n o t a l w a y s a f f i r m a t i v e . 'Thus, i n g e n e r a l , w e may o n l y e x p e c t t h a t f o r some g r o u p t o p o l o g y , w h i c h i s f i n e r t h a n T , t h e m a p p i n g s A-+expA, AcM

, are

l o c a l homeomorphisms o f AutD o v e r a u t D f o r a l l FcAutD.

To e s t a b l i s h t h e e x i s t e n c e o f s u c h t o p o l o g y w e s h o u l d know t h a t t h e c o m p o s i t e mapping expAloexpA2 c a n a l w a y s b e w r i t t e n i n t h e form e x p C f o r some CeautD, whenever A l l A2 a r e S u f f i c i e n t l y n e a r t o 0 i n a u t D . T h i s f a c t i s a s p e c i a l case of o n e of t h e t h e main g o a l s of t h e g e n e r a l L i e t h e o r y , known a s t h e CampbellH a u s d o r f f t h e o r e m ( c f . I3 I ) 6.50.

.

T h e r e a r e a n e i g h b o u r h o o d M of t h e o r i g i n

THEOREM.

-i.n a u t D a n d a r e a l a n a l y t i c m a p p i n g C : M+autD s u e h t h a i w e h a v e

f o r a l l A1,A2eM. By t h e c o n t i n u i t y of C a t t h e o r i g i n , w e c a n f i n d neighbourhccds M l c M and M Z C M of 0 i n a u t D s u c h t h a t

C(M1, M ~ ) C M and

(6.25)

6.51.

C(M2,M2) C M 1

REMARK. The e x p l i c i t f o r m of t h e mapping C i s a l s o

known. One c a n show ( c f . C(slA1,

13 1 ) t h a t g i v e n E c a D , w e h a v e

s2A2]lB= C ( s l A 1 , =

,.

s2A2 ) i dB=

l o g ( e x p s l i l , e x p s 2 5 2 ) id,

THE LIE GROUP OF AUTOMORPHISMS

105

in the sense that the formal power series

L k,R 2 0

s:

si X k t R ( A 1,A21 %.log [id+ (expslAl exps2i2-idl]

,

where Xk, (A1,A2)= : k+1

c

=:

(-1)

n+ 1

1

c

pl+..+pn=k, ql+..+qn=R pl!ql!

n=l

piqi30

1 .. pn!q,!

-p1 3 2 A1 A2

.-

Pi +qi>o

I

"pnAqn

..A1 A2 converges in the norm

11 . I I B

to id

B

whenever

11

slAl11

and

11 s 2 A 2 I l are sufficiently small. (This is not consequence of any majorization!). Then, we necessarily have

where

and the convergence is meant in the topology of autD. Since (A1,A2)eautD for all k,R because they are partial derivatiC k,k ves in the T-sense of the mapping ( s l l ~ 2 ) + C ( s l A 1 1 ~ 2at A 2 )0, Dynkin's identity yields

k+l =

c

n=l

n+l

(-1)

1

Pn - l A ' n - lAPnAqn - 1A P I 'I . .. 1!qn! A1#A2#.-A1# 2# 1# 2

P 1!ql 1

Pn

It would be interesting to have a direct proof for the formula expAl.expA2= exp [ C Ck, (A1,AZ)] kit

in the setting of AutD.

THEOREM.I'?icr~c exists a un-iquc H a u s d o r f f

6.52.

T

6

CHAPTER

106

o n AutD such that (AutD, T a )

topology

-is a topological g r o u p a n d

1 M ; n = 1,2,..} {exp -

n

is n f u n d a m e n t a l . s y s t e m o $ n e i g h b o u r h o o o d s of i d MOreQUcr, 1'

D

for T

a

.

22'.

P r o o f : From t h e g e n e r a l t h e o r y o f t o p o l o g i c a l g r o u p s , it s u f f i c e s t o p r o v e t h a t t h e s t a t e m e n t s ( a ) , (b), ( c ) a n d ( d ) below a r e s a t i s f i e d . m

1

n

exp M= {idgl. n=1 I n d e e d , l e t fCeXpM b e s u c h t h a t f f i d D . Then, t h e r e i s some AcM

(a) W e h a v e

w i t h A 4 0 f o r w h i c h expA= f ; t h e r e f o r e , w e c a n f i n d some nEJN s u c h t h a t At#

S i n c e t h e e x p o n e n t i a l mapping i s i n j e c t i v e on

M,

M , w e h a v e ft#exp

1 n

a,

thus

A!;

f4 n

n=1

exp

n1 M .

( b ) L e t n l a n d n ClN be g i v e n ; t h e n t h e r e e x i s t s some 1 2 1 1

mdN such t h a t exp

fii

M ) fl ( e x p

M c (exp

M)

2

1

I n d e e d , i t s u f f i c e s t o c o n s i d e r m=:

.

max(nl,n2).

( c ) L e t ndN b e g i v e n ; t h e n t h e r e e x i s t s some mdN s u c h t h a t (exp

1 n

M ) . (exp

I n d e e d , by ( 6 . 2 4 ) g i v e n ndN assume M

1 ; M) ' c e x p

1 M.

we have C ( O , O ) = 0 . A s C i s c o n t i n u o u s a t 0,

we can f i n d

mm

such t h a t C (

t o b e s y m m e t r i c , i . e . , M = -M (exp

iii1

M) ( e x p

1 M) - '=

1 E

MI

. Then

1 ; M ) c 1g

M.

We may

-1 M ) = ( e x p ;I;; M I . ( e x p -

m

( d ) L e t geAutD a n d ndN b e given; t h e n t h e r e e x i s t s some mdN such t h a t g . (exp ; 1 M ) .g-'Cexp 1 M.

I n d e e d , once gcAutD h a s b e e n f i x e d , by c o r o l l a r y 5 . 1 2 t h e a d j o i n t mapping gy':

autD+autD o f g - '

i s a n a u t o m o r p h i s m of t h e

107

THE L I E GROUP OF AUTOMORPHISMS

-1

a l g e b r a autD. T h e r e f o r e t h e s e t g # (

Banach-Lie

1 n

M)

is a

neiqhbourhood of 0 i n autD and w e may f i n d some m a such t h a t 1 m

-1

M c q #

so t h a t

1

(

M). Moreover, by p r o p o s i t i o n 5 . 1 3 ,

g.exp(

1

M)q

-1

cexp

I n

M.

I n o r d e r t o show t h a t T >,T it s u f f i c e s t o prove t h a t e v e r y T-neighbourhood of i d D c o n t a i n s a T -neighbourhood of i d

D

.

Now,

t h e f a m i l y of s u b s e t s o f AutD g i v e n by

f o r E > O i s a fundamental system of T-neighbourhoods of i d D . By

lemma 6 . 4 5 t h e mapping exp: M+Holm(B.E) i s c o n t i n u o u s a t t h e o r i g i n ; as

i s a neighbourhood of i d D f o r t h e t o p o l o g y induced by H m ( B , E ) on AutD, t h e r e e x i s t s some ndN such t h a t e x p ( 1 M ) c N ( E ) .

# 6.53.

DEFINITION.

We r e f e r t o t h e t o p o l o g y i n t r o d u c e d by

t h e o r e m 6 . 5 2 on A u t D a s t h e “ a n a Z y t i c t o p o l o g y “ o n AutD. By A u t O D w e d e n o t e t h e connected component o f i d D i n (AutD, T a ) .

6.54.

LEMMA. T h e r e i s a n e i g h b o u r h o o d M o f t h e o r i g i n i n

autD s u c h t h a t exp: M-texpM

i s a homeomorphism when b o t h M and

a r e endowed w i t h t h e i r r e s p e c t i v e t o p o l o g i e s a s s u b s p a c e s of autD and ( A u t D , T). expM

P r o o f : L e t M be as i n theorem 6 . 5 0 .

By lemma 6 . 4 7 ,

exp: M-+Holm(BIE)i s a homeomorphism of M o n t o a neighbourhood

expM of i d D i n H o l m ( B I E ) . Now it s u f f i c e s t o r e a l i z e t h a t exp t a k e s i t s v a l u e s n o t o n l y i n Holm(BIE) b u t i n t h e s u b s e t AutDcHol,(B,E)

and t h a t t h e t o p o l o g y induced by Holm(B,E) on

AutD i s p r e c i s e l y T .

ff

CHAPTER

108

6.55.

6

REMARK. Observe t h a t e x p : M+expM 1s a homeonmrphism,

t o o , f o r t h e t o p o l o g i e s i n d u c e d on M

a n d expM

by autD and

(AutD, T,). Thus, i n p a r t i c u l a r , T and 4 a g r e e on t h e s u b s e t expM o f AutD, b u t from t h i s f a c t w e c a n n o t c o n c l u d e t h a t T a n d I' a g r e e on t h e whole g r o u p AutD: w h e r e a s expM of i d D f o r T I it may f a i l t o be so f o r T . 55.-

i s aneighbourhood

The Banach-Lie o up s t r u c t u r e of AutD. ____ - - - -___-g r-

Now w e are g o i n g t o c o n s t r u c t a r e a l Banach-Lie g r o u p s t r u c t u r e on AutD whose u n d e r l y i n g t o p o l o g y is T For t h i s p m p o s e , l e t and M 2 b e a s i n t h e o r e m 6 . 5 0 and 6 . 5 2 so t h a t M ,M1

.

C ( M 1 , M I1

(6.26)

=M

C ( M 2 , M 2 )= M I

and exp: M-bexpM i s a homeomorphism f o r t h e t o p o l o g i e s i n d u c e d by a u t D and (AutD, T,). L e t u s d e n o t e by l o g : expM-tM i n v e r s e and w r i t e F=: I N ; N

its

open and O e i l r c M ]

Then, t h e f a m i l y IexpN; NcF} i s a f u n d a m e n t a l s y s t e m of n e i g h b o u r h o o d s o f i d D f o r T,. 6.56.

THEOREM. T h e r e is a u n i q u e r e a l a n a l g t i c B a n a c h

r n a n i f o Z d s t r u c t u r e on (AutD, T ) f o r w h Z c h t h e f a m i l y a

is a s y s t e m o f c h a r t s a t t h e i d e n t - i t y e l e m e n t id D' Proof:

L e t geAutD be g i v e n . The l e f t t r a n s l a t i o n

Lg: f + g o f , fcAutD, is a n automorphism o f t h e t o p o l o g i c a l -1 Now w e d e f i n e g r o u p (AutD, Ta) whose i n v e r s e i s ( L g ) - ' = Lg a s y s t e m of c h a r t s a t geAutD a s t h e family of p a i r s

.

C l e a r l y , c o n d i t i o n M1 of d e f i n i t i o n 6 . 1 is s a t i s f i e d b e c a u s e

THE L I E GROUP O F AUTOMORPHISMS

U

109

g.expN

NeF, g e A u t D i s an open c o v e r of A u t D f o r t h e t o p o l o g y Ta. Moreover, t h e s e

l o c a l c h a r t s are a n a l y t i c a l l y c o m p a t i b l e i n t h e r e a l s e n s e ,

i.e.,

t h e y s a t i s f y c o n d i t i o n M 2 , t o o . Indeed, assume t h a t

f o r some g ,g2eAutD and N 1 , N 2 c F . 1

AleN

1

Then, t h e r e are A l c N l

and

such t h a t gl.expA = f = g2.expA2 1

w e have

Thus, f o r AcNl n N 2

=

S i n c e A1,A2cM2

,

log [expAlexp ( -A2) expA]

from theorem 6.50 w e d e r i v e

expAlexp(-A2) = expC(A1 , - A 2 ) = expA 3 where, by ( 6 . 2 6 )

, A 3 =:

C ( A ,-A2) 1

i s a f i x e d e l e m e n t of M

1'

Then,

from theorem 6.50 w e d e r i v e l o g [expA exp (-A 1

2

) expA] = l o g (expA expA) = 3

Whence t h e t r a n s i t i o n homeomorphism c o r r e s p o n d i n g t o t h e c h a r t s (glexPN1 I

and (g2expN2' 1oglg2expN2

i s g i v e n by

A J C ( A ~ , A ) which i s a r e a l a n a l y t i c mapping.

# 6 . 5 7 . THEOREM. The m a n i f o l d

(AutD, T a ) is a r e a l Banach-

L i e g r o u p whose B anach- Li e a l g e b r a i s autD.

110

CHAPTER

6

P r o o f : By p r o p o s i t i o n 6 . 1 9 a n d 6 . 3 0 it s u f f i c e s t o p r o v e t h e s t a t e m e n t s ( a ) , (b) and ( c ) below. ( a ) F o r e v e r y f i x e d gcAutD, t h e mapping Lg: q+gf i s r e a l a n a l y t i c on AutD. I n d e e d , l e t gcAutD b e f i x e d . Choose a n y fcAutD 1

c h a r t ( f .expN, l o g L f ( g f e x p u , l o g L (gf)

and a n y l o c a l

1 of (AutD, T a ) a t f . Then

I gfexpN

i s a l o c a l c h a r t a t g f and t h e

e x p r e s s i o n o f Lg i s t h e s e c h a r t s i s g i v e n by

= l o g expA= A

for A e N . Thus, Lg i s a n a l y t i c . ( b ) For e v e r y f i x e d gcAutD, t h e mapping

Tg: AutD+AutD g i v e n by f + g f g - l i s r e a l a n a l y t i c . I n d e e d , l e t gcAutD be g i v e n . Choose any fcAutD

and a n y l o c a l

1

c h a r t (f.expN, logLf-

) a t f . Since t h e a d j o i n t !f.expN g # : autD+autD o f g i s c o n t i n u o u s , t h e r e i s some N C M 1

t h a t g# (iV,)CrV; t h e r e f o r e ,

mapping such

l o g e x p g # ( A )= g#A

€ o r A c N l . Then ( f . e x p N 1 , l o g L f - l l f e x p N ) a n d (gfg-lexPNl' logL(gf9

-1

1

-1

1

1 are local c h a r t s a t f

Igfg-lexpN 1

and g f g - l . Moreover, t h e e x p r e s s i o n of Tg i n t h e s e c h a r t s i s g i v e n by

for a&N

1'

Thus T g i s a n a l y t i c .

( c ) The mapping F:

(AutD)x(AutD)+AutD g i v e n by

111

THE LIE GROUP OF AUTOMORPHISMS

(f,g)+fg-' is analytic in a neighbourhood of ( idD,idD)

.

Indeed, let M2 be as in (6.26). Then (expM2, loglexpM21 is a chart of AutD at idD and its "Cartesian square" is a chart of AutDxAutD at (idD,idD) By (6.25) we have

.

for all A1,A2eM2. Then, it is easy to check that the expression of F in these charts is given by (A1'A2 which is real analytic. 56.- The action of AutD on the domain D. ~~

We endow the domain D with its underlying real analytic manifold structure and consider DxAutD as a product manifold. Then we define the action of AutD on D as the mapping JI: DxAutD+D given by (x,f)+f (x)

.

6.58. THEOREM. T h e mapping tic o n DxAutD.

+:

(x,f)+f(x) is r e a l a n a l y -

Proof: It suffices to prove its analyticity near the identity element idD. Now, let xcD be given and fix any ball B c c D centered at x. Starting with this ball B we can construct a neighbourhood M of 0 in autD as we did in 54. Then (B, idB)x(expM, log,expM)is a local chart of D AutD at (x,idD). Also, (D, idD ) is a chart of D at x. Thus, it suffices to show that the mapping

which is the expression of JI in these charts, is real analytic in BxM. By lemma 6.45, (y,A)+(y,expA) is real analytic in BxM with

112

CHAPTER

6

v a l u e s in B x H o l m ( B , E ) . O b v i o u s l y , t h e mapping B x H o ~ ~ ( B , +E E ) g i v e n by ( y , f 1 +f (y) is s e p a r a t e l y h o l o m o r p h i c ;

t h e r e f o r e , by H a r t o g ' s t h e o r e m , it is h o l o m o r p h i c a n d , i n p a r t i c u l a r , real a n a l y t i c . But ( 6 . 2 7 ) i s t h e c o m p o s i t e of ( y , A ) + ( y , e x p A ) and (y,f)+f ( y ) , whence t h e r e s u l t follows.

#

CHAPTER

7

BOUNDED CIRCULAR DOMAINS

I n t h i s c h a p t e r w e s h a l l s t u d y t h e group A u t D f o r domains D w i t h some p a r t i c u l a r g e o m e t r i c p r o p e r t i e s .

§I

.- The

L i e a l g e b r a autD f o r c i r c u l a r domains.

7.1.

DEFINITION. We s a y t h a t a b o u n d e d d o m a i n D

is

“ c i r c u l a r ” i f OeD a n d , f o r a l l xcD a n d a l l AcC w i t h I A / = 1

,

we h a v e AxeD. Throughout t h e whole c h a p t e r , D w i l l s t a n d f o r a bounded c i r c u l a r domain.

7 . 2 . LEMMA. L e t D b e a b o u n d e d c i r c u l a r d o m a i n . T h e n t h e v e c t o r f i e l d Z : x + i x i s c o m p l e t e i n D.

tm,

it

x. S i n c e D i s c i r c u l a r , w e have f c A u t D and t h e mapping t + f t i s a T-continuous one-parameter group. By theorem 4 . 5 i t s a s s o c i a t e d v e c t o r f i e l d , which i s o b v i o u s l y 2, i s complete i n D. P r o o f : For

w e d e f i n e f t : x+e

t

# W e c a l l Z t h e c i r c u l a r v e c t o r f i e l d and it w i l l p l a y an i m p o r t a n t r o l e i n t h e s t u d y of c i r c u l a r domains. S i n c e O c D , any (non n e c e s s a r i l y c o m p l e t e ) holomorphic v e c t o r f i e l d X i n D i s u n i q u e l y determined by i t s T a y l o r s e r i e s a t 0 . W e w r i t e Pn f o r t h e space of c o n t i n u o u s n-homogeneous CU

polynomials P: E+E, so t h a t w e have X= C P = : X ( nS P O

n=O n

for nm.

113

P n where

CHAPTER

114

z#=: [ Z , . ] be t h e a d j o i n t of 2 ; and , by r e i t e r a t i n g t h i s o p e r a t i o n

Let

7

t h e n w e may a p p l y Z #

to X

a n d t a k i n g l i n e a r combinations,

w e o b t a i n e x p r e s s i o n s of t h e form P(z#)x= (ao+a

z + . . + ar Z '#) X

1 #

where P ( A ) = a o + a l A t . . + a r A r

i s a polynomial i n t h e indetermi-

nate A .

1.3. LEMMA. LeL P(X)c(I:[A] b e a n y poZynumiaZ i n A a n d m wssume t h a l X = l' P ,is a h o l - o m o r p h i c v e c t o r J i e l d i n D. T h e n n=O n we have

P r o o f : F o r t h e homogeneous components P

of X w e h a v e

By r e i t e r a t i n g t h i s o p e r a t i o n a n d t a k i n g l i n e a r c o m b i n a t i o n s

we o b t a i n t h e r e s u l t .

# m

7.4.

LEMMA.

Assume t h a t X = 1 P n s a t i s f i e : : XeautD. Then

we h a v e P = 0 f o r a l l 1-113.

n=O

n

P r o o f : L e t XeautD b e g i v e n . S i n c e a u t D i s a r e a l L i e a l g e b r a and ZcautD, w e h a v e P ( Z # ) X s a u t D f o r a n y p o l y n o m i a l w i t h r e a l c o e f f i c i e n t s P(A)dR[A].

By t a k i n g P ( A ) = A 3 t A

and

a p p l y i n g lemma 7 . 3 w e o b t a i n

B u t now we h a v e P ( - i ) = P ( 0 ) = P (i) = 0 , so t h a t t h e T a y l o r series of P ( Z # ) X a t 0 i s

BOUNDED CIRCULAR DOMAINS

f o r k= 0 , l a n d , by C a r t a n ' s

[P(Z#)X] (k= 0

Thus, we have

115

0

u n i q u e n e s s theorem, P ( Z # ) X = 0 . However, P ( n i - i ) f O f o r a l l n > 3 ; therefore P = 0 for n>3. DEFINITION. For a n y b o u n d e d c i r c u l a r d o m a i n D , we

7.5.

set 0

a u t D =:

P1

n

aut D=:

(autD)

0

0

(autD)O= { X ( o ) ; XeautDj

E =:

Aut D = :

(POOP2)n autD

{FsAutD, F i s l i n e a r ] .

PROPOSITION. F o r b o u n d e d c i r c u l a r d o m a i n s D ,

7.6.

we

h a v e t h e t o p o Z o g i c a 2 d i r e c t sum d e c o m p o s i t i o n 0

autD= ( a u t o D )8 ( a u t D)

(7.1)

M o r e o v e r , a u t 0 D i s t o p o l o g i c a 2 2 y i s o m o r p h i c w i t h E0 ( c o n s i d e r e d 0 a s a r e a l l i n e a r s u b s p a c e of E l b y t h e m a p p i n g X - t X ( 0 ) a n d a u t D 0

c a n v i e w e d a s t h e L i e a Z g e b r a of Aut D. P r o o f : L e t XcautD be given: by lemma 7 . 4 w e have f o r some P k e P k , k = 0,1, 2 .

X = P +P + P 0

1

2

Applying lemma 7 . 3 t o

t h e polynomial P (1)= h 2 and t h e v e c t o r f i e l d X w e d e r i v e P(Z

#

)x=

2

2

2

1 P(ni-i)P = C (in-i) P = -(Po+P2) n n

n=O

n=O

so t h a t P0+PZ€autDa n d , t h e r e f o r e , P 1= X - ( P 0 + P 2 ) c a u t D . C l e a r l y 0

P + P 2 c a u t D and P l c a u t D ; t h u s autD a d m i t s t h e d i r e c t sum 0 0 0 decomposition autD= ( a u t D ) Q ( a u t o D ) By lemma 5.1 I , t h e c a n o n i -

.

2

c a l p r o j e c t o r s Z#

and

I-Z#

2

are c o n t i n u o u s .

Now, l e t c s E O be g i v e n . Then, t h e r e e x i s t s a unique symmetric b i l i n e a r mapping QccL( E x E I E ) such t h a t t h e v e c t o r f i e l d A:

x + c - Q c ( x , x ) , XCD, b e l o n g s t o autD. I n d e e d , t h e r e i s some

X = P + P +P cautD w i t h c = X ( 0 ) = P o . Then w e have c+P2cautD and 0

1 2

,

requirements. I f t h e r e i s

Q ( x , x )=: -P2

(XI

a n o t h e r Q:

i n t h e same c o n d i t i o n s , from c-QccautD and

I

xcD

c-QA cautD w e g e t Qc-Q:=

satisfies the (c-QA

)-(c-Qc)cautD; t h u s

CHAPTER

116

f o r k= 0 , l

(Qc-QA)Ak= 0

Q,=

7

and, by C a r t a n ’ s u n i q u e n e s s theorem,

QL-

Now, w e show t h a t Eo i s complete and t h a t t h e mapping

E o + a u t D g i v e n by c+c-Q i s a c o n t i n u o u s s u r j e c t i v e i s o m r p h i s m 0 of Banach s p a c e s . Indeed, assume t h a t Q = 0 f o r some c c E A s c-QccautoD, w e have ccautD. From ZeautD we o b t a i n

0’

[ Z , c ] = i c s a u t D . S i n c e autD i s p u r e l y r e a l , w e have c = 0 . Thus,

i s a n isomorphism o n t o t h e image subspace which i s o b v i o u s l y a u t D. NOW, l e t u s t a k e any b a l l B c c D c e n t e r e d a t 0 OeD. By theorem 5.6, t h e r e a r e c o n s t a n t s K 1 , K Z such t h a t w e CW-Q

have

f o r a l l XeautD. Applying t h i s t o t h e v e c t o r f i e l d X = c-Q e a u t D w e o b t a i n 0

for a l l c e E o . Thus c+c-Q

i s a homeomorphism. S i n c e w e know t h a t a u t D i s c l o s e d i n a u t D , Eo i s complete. A s f o r t h e

0 a s s e r t i o n c o n c e r n i n g a u t 0D , w e c a n r e p e a t t h e arguments of

theorems 6.56 and 6.57 r e s t r i c t i n g o u r s e l v e s t o t h e group 0 A u t D i n s t e a d of AutD.

I n t h e c o u r s e of t h e proof w e have e s t a b l i s h e d t h e f o l l o w i n g 7.7.

Eo i s a r e a l s u b s p a c e of E a n d , f o r e a c h a unCque QceL(EXEIE) s u c h t h a t t h e v e c t o r f i e l d

COROLLARY.

c e E O , t h e r e is

x*c-QC(xIx), XCD, b e l o n g s t o a u t D .

7.8.

DEFINITION. Ve r e s e r v e t h e n o t a t i o n Qc f o r t h e

symmetric biZCnear mapping d e s c r i b e d above. 7.9.

PROPOSITION.

For bounded c i r c u l a r dom ai ns D ,

h a ve 0

0

0

[aut D , a u t D] c a u t D ,

0

[aut D , a u t 0 D ] c a u t O D

we

BOUNDED CIRCULAR DOMAINS

0

117

0

[ a u t D , autoD] c a u t D

Mor.eover, 0

( a ) For a l l L c a u t D , c c E o a n d x c E , it holds L c c E O a n d

Q L C ( x , x ) = LQ ( x , x ) - ~ cQ( L x x) 0

( b ) For a l l c 1 , c 2 e E 0 , We h a v e Q

( . , c 2 ) + Q c( c l , . ) c a u t D 2

,1

( c ) For xcE a n d c 1 ,c2eE0 t h e following e q u a l i t y h o l d s

[Q,

Q, 1

( x , x ) , X I = Qc [Qc ( X , x ) , X I . 2

2

1

0 ( d ) For a l l FcAut D , c c E O a n d XEIE, we h a v e FccEO a n d QFc ( x , x ) = FQc ( F - l x ,

F-lx)

.

0

Proof: L e t L , L 2 c a u t D be 9 i v e n ; t h e n

so t h a t Banach-Lie

0

[aut D,

1

0 a u t o D ] c a u t D.

[L 1 , L 2 ]

is l i n e a r

I n p a r t i c u l a r , autoD i s a

s u b a l g e b r a of a u t D .

0 L e t L s a u t D a n d AeautOD b e a r b i t r a r i l y g i v e n ; t h e n w e h a v e

A ( x ) = c - Q c ( x , x ) , X C D , where c s E O a n d Q c i s a s y m m e t r i c b i l i n e a r mapping ExE-tE. An e a s y c o m p u t a t i o n g i v e s

S i n c e t h e mapping x+2Qc(Lx,x)-L[Qc~x,x))is a n e l e m e n t of P 2 , i t $0

2 ZOWS

[L,A] cautOD. B e s i d e s L c =

[L ,A] O s E o

and

which p r o v e s ( a ) . L e t A 1 , A 2 c a u t D b e g i v e n and assume t h a t 0

where c l , c 2 e E

0

follows t h a t

,

and Qc l

e L ( E x E 1 E ) a r e symmetric.

Q C2

It

7

CHAPTER

118

S i n c e t h e mapping x+QC ( x , Q c ( x , x ) 1 -Q, 1

2

1

(Qc ( X , X ) , X I 2

,

XED,

is

a n element of P 3 , by lemma 7 . 4 , it m u s t be i d e n t i c a l l y n u l l . I C ~ +Qc ) ( c l , . ) e a u t 0 D. T h i s p r o v e s ( b ) and

Thus [A1 ,A2] = Q c ( . (c)

.

1

2 0

F i n a l l y , l e t FcAut D and acEO be g i v e n . Then w e have F = expL 0

f o r XCD, Fx i s t h e v a l u e a t t = 1 of t h e

f o r some L s a u t D , i . e . ,

s o l u t i o n of t h e i n i t i a l v a l u e problem

i n t h e space E . By ( a ) w e have L ( E O ) C E o ; t h e r e f o r e , i f t h e i n i t i a l v a l u e i s some C c E o f l D , ( 7 . 2 ) can be i n t e r p r e t e d a s an i n i t i a l v a l u e problem i n E S i n c e E i s complete and t h e 0' 0 s o l u t i o n of ( 7 . 2 ) i s u n i q u e , w e have Fc= (expL)ccE f o r a l l 0 ccEOfl D. A s F is l i n e a r , FceEO for a l l c s E o . Moreover, F#[c-Q, ( x , x ) ] = F[c-Q,

( F - l x , F - ' x ) ] = Fc-FQ, (F-'x,F-'x)

f o r a l l xeE, so t h a t

Q,,

( x , x ) = FQ,

(F-'X,F-'~)

which shows ( e l .

#J 7.10.

COROLLARY.

T h e s u b s p a c e E o is i n v a r i a n t u n d e r t h e

g r o u p A u t 0D. In p a r t i c u l a r , E

0

is a c o m p l e x s u b s p a c e of E . T h e 0

m a p p i n g c+Q is c o n j u g a t e l i n e a r a n d w e h a v e Q , ( c , . ) e a u t D f o r C

a l l ceEo. 0

P r o o f : W e have F=: i d cAut D D

because D i s c i r c u l a r .

Applying (e) we g e t Q i c = -iQc. Then a p p l y (b) w i t h

BOUNDED CIRCULAR DOMAINS

7.11.LEMMA

.

We h a v e E o = t X ( c ) ; c c E O

119

I

XcautDl.

Proof: Let u s set ( a u t0 D ) E = : { L ( c ) ; c c E o l L s a u t 0 D)

I

( a u t o D ) E o = { A ( c ) ; ccEO,AcautoD}

0

F i r s t w e show t h a t (aut$)EocEo. I n d e e d , l e t c c E O be g i v e n and

t a k e any Lcaut 0D . Then A = : c-QccautD so t h a t

a t 0 w e g e t L ( c ) c E O and t h e r e f o r e

By e v a l u a t i n g [L,A] ( a u t0D ) E C E 0

[L,A]cautD. But

0

Now w e show t h a t ( a u t o D ) E o c E o . I n d e e d , l e t c c E O be g i v e n . Then A= c-Q cautD; s i n c e ZsautD, w e have A 2# ( Z ) . = [A,[A,Z]]cautD.

[A, [A,Z]]x=

4iQ (x,c)

But

xcD

S i n c e t h e mapping x + 4 i Q c ( x , c ) i s l i n e a r , w e must have 0 [A, [A,Z]]caut D and by t h e p r e v i o u s s t e p w e g e t Qc ( x , c )cEO

(7.3)

f o r a l l c e E O and xcE t h e symmetry of Q,

0'

I n t e r c h a n g i n g t h e r o l e s of x and c , by

we g e t Qx(c,x)cE0

(7.4)

f o r a l l x,ccEO. As c+Q,

is linear,

Thus, from ( 7 . 3 ) and (7.4) w e d e r i v e Q C ( x + c , x ) c0 ~ f o r a l l x , c e E O . Then, from QC(X+C,X)=

Q C t x , x ) + Q ( c , x ) e E0

and ( 7 . 3 ) we o b t a i n Q c ( x , x ) c E o f o r a l l xcEol SO t h a t

120

7

CHAPTER

( a u t o D )Eo c E O . by p r o p o s i t i o n 7 . 6

NOW,

( a u t o )E g = : { X ( c ) ; ccEO

i t follows

, X c a u t D ] c ( a u t0 D)E o + ( a u t o D )E O c E o .

The c o n v e r s e i n c l u s i o n i s o b v i o u s .

52.-

The c-o.n-n-~~ e c t e d component of t h e i d e n t i t y i n- AutD. -

--~

L e t cT b e a n y H a u s d o r f f t o p o l o g i c a l g r o u p a n d d e n o t e by W a n

o p e n s y m m e t r i c c o n n e c t e d n e i g h b o u r h o o d of t h e i d e n t i t y e l e m e n t e i n J . F o r nCN w e s e t

From t h e g e n e r a l t h e o r y of t o p o l o g i c a l g r o u p s , it i s known that

lY= :

U

W"

nEN

i s a c l o s e d n o r m a l s u b g r o u p of J a n d t h a t H i s t h e c o n n e c t e d

component of e i n J ; t h u s , H d o e s n o t d e p e n d o n t h e c h o i c e o f W.

S i n c e t h e mapping e x p i s a l o c a l homeomorphism a t 0 , i n

p a r t i c u l a r we get. 7.12.

LEMMA. L e t M be t h e n e i g h b o u r h o o d of 0 in a u t D

g i v e n b y Zernma 6 . 5 6 .

Then

is t h e c o n n e c t e d c o m p o n e n t o f i d D i n ( A u t D , T a ) . M o r e o v e r , Aut D i s a c l o s e d norvnul szibgroup of b o t h (AutD,T ) a n d 0 a (AutD,Y). 7.13.

DEFINITION. L e t S and

autD r e s p e c t i v e l y .

G

be s u b s e t s of D and

We d e f i n e t h e " o r b i t " of S b y

I t i s immediate t o check t h a t i f

G

G

by meuns o f

i s a s u b g r o u p of AutD, t h e n

121

BOUNDED CIRCULAR DOMAINS

w e have

GG ( S ) =

G(S).

L e t E o be t h e subspace o f E g i v e n by d e f i n i t i o n 7 . 5 . C l e a r l y E

0

n D i s a bounded open c i r c u l a r s u b s e t of t h e s p a c e E

however, E o n D

-

0'

may f a i l t o be connected

LEMMA. I f D i s a bounded c i r c u l a r domain of E,

7.14.

t h e n ue have

(AutoD) ( E o n D ) c E o f l D .

P r o o f : L e t gcAut D be g i v e n . By lemma 7 . 1 2 w e have 0

g= ( e x p A l ) o .

.O

(expA ) f o r some AkcM,

k = 1 , 2 , . . , n . Thus, i t

s u f f i c e s t o show t h a t

f o r a l l AcM. NOW, l e t AeM and xeEOn D be g i v e n . L e t us c o n s i d e r t h e i n i t i a l

v a l u e problem

(7.5)

i n t h e space E . I t s s o l u t i o n y ( t ) = (exptA)x s a t i s f i e s y ( t ) e D for a l l

tm.

S i n c e xcEO and by p r o p o s i t i o n 7 . 6 Eo i s c o m p l e t e ,

t h e i n i t i a l v a l u e problem ( 7 . 5 ) h a s a s o l u t i o n i n E o , t o o . A s t h e s o l u t i o n i s u n i q u e , w e have (exptA)xcEo f o r a l l t d R ; t h u s , (expA) ( E n~ D ) c E~ n D .

# 7.15. LEMMA. L e t D be a b o u n d e d c i r c u l a r domain of E .

T h e n , i f E n D i s c o n n e c t e d (in p a r t i c u l a r , i f D i s b a l a n c e d ) , 0

we h a v e

Proof: A s E

0

nD

i s assumed t o be c o n n e c t e d , E o n D - i s a

bounded c i r c u l a r domain o f t h e s p a c e E o and it makes s e n s e t o

122

7

CHAPTER

speak

.

of t h e group A u t ( E o n D)

L e t qsAutOD be g i v e n . Then g i s a biholomorphic b i j e c t i o n of D

o n t o D; t h e r e f o r e q

i s a biholomorphic b i j e c t i o n o f

lEon D E o n D o n t o i t s image g ( E o il D)

g(E0n

(7.6)

. From

lemma 7 . 1 4 ,

DICE^^

D

Thus, a p p l y i n g gT1eAut0D t o ( .6) and lemma 7 . 1 4 a g a i n , w e obtain

~~n

E o n Dcg-'

c~-' ( E o n D) = Eo

so t h a t

n D.

D)C

~ , nD

Then g ( E 0 n D ) = E o n D and

.

eAut(E,, fl D ) fl D Observe t h a t , i f D i s b a l a n c e d , t h e n E 'lEO

0

nD

i s balanced t o o ;

hence i t i s connected and t h e lemma h o l d s .

#

7.16. LEMMA. L e t D be a boun ded c i r c u l a r domain of E .

T h e n , if Eo n D is c o n n e c t e d ( i n p a r t i c u l a r , if D is b a l a n c e d ) , the set

( A u t o D ) O i s a n e i g h b o u r h o o d of 0 i n t h e s p a c e E o .

P r o o f : Consider t h e mapping

0:

Eo+aut D g i v e n b y ~ 4 c - Q ~ . 0

i s a neiqhbourhood of 0 i n autD, M n autDo i s a neighbourhood of 0 i n a u t D and, by p r o p o s i t i o n 7 . 6 , Since M

U=:

+-'Mn

0

autoD i s a neighbourhood of 0 i n E o . NOW, c o n s i d e r

t h e composite J of t h e mappings C+C-Q

-t

exp (c-qC)

+

"

By lemma 6.32 w e have J ( c ) = C

1

[exp ( c - ~ , )] o (i:idD)O f o r ceU, where w e

n=O

have p u t A c = :

c-Qc. Moreover, we have ( i : i d D ) = 0 f o r a l l n f l

-1 and (Acid,)O= c

,

so t h a t J ( c ) = c f o r ceU and J f U ) is a neighbourhood of 0 . Thus, by lemma 7 . 1 2 . OeU= J ( U ) =

(expAc)O; csU}c{expA)O;

a n d , by lemmas 7.15 and 7 . 1 4 ,

A e M I c (AutoD)0

BOUNDED CIRCULAR DOMAINS

123

whence t h e r e s u l t f o l l o w s .

# 7.17.

PROPOSITION. L e t D b e a ( n o n n e c e s s a r i z y c i r c u Z a r l

b o u n d e d d o m a i n i n E a n d l e t J b e a s u b g r o u p o f AutD s u c h that, for some XGD, t h e orbit J ( x ) o f x b y J i s a n e i g h b o u r h o o d of x. T h e n J ( x ) = D a n d t h e s u b g r o u p J acts t r a n s i t i v e Z y o n D. P r o o f : F i r s t w e show t h a t J ( x ) i s a n open s u b s e t of D . L e t y s J ( x ) be g i v e n . Then w e have g ( x ) = y f o r some g c J . A s J ( x )

i s assumed t o be a neiqhbourhood of x , t h e r e e x i s t s some open

s u b s e t W c D such t h a t xcWCJ(x) and, a p p l y i n g g w e o b t a i n

S i n c e q i s a homeomorphism, g(W) i s open; t h u s , by t h e a r b i t r a r i n e s s of y , J ( x ) i s open. Now w e show t h a t J ( x ) k a c l o s e d s u b s e t of D. L e t ycD b e any

-

p o i n t of t h e closure J ( x ) of J ( x ) i n D . Then, t h e r e i s a sequence ( y n )n C NJC ( x ) s u c h t h a t y,*y.

T h e r e f o r e , w e have

f o r some q n CJ and ndN. L e t dD be C a r a t h e o d o r y d i s t a n c e i n D . S i n c e w e have assumed t h a t J ( x ) i s a

y,= g,(x)

neighbourhood o f x and, by c o r o l l a r y 3 . 1 4 ,

d,

i n d u c e s t h e norm

t o p o l o g y on D ,

f o r some E > O . Moreover, a s y,+y

it f o l l o w s t h a t

f o r a l l n m 0 . Since dD is J - i n v a r i a n t , dD(x,q,ly)i'

so t h a t , by ( 7 . 7 )

,

from ( 7 . 8 ) w e o b t a i n and t h e r e f o r e

0

J ( x ) = J ( x ) . Thus J ( x ) i s c l o s e d i n D .

ycg, 0

S i n c e D i s c o n n e c t e d , w e have J ( x ) = D and D i s homogeneous

CHAPTER

124

7

under the action of J .

# 7.18. COROLLARY. L e t D b e a b o u n d e d c i r c u l . a r d o m a i n of .if E o n D i s c o n n e c t e d ( i n p a r t i c u l a r , if D i s baZancedJ ~e h a v e (AutoD)O= Eon D. E.

Then,

Proof: Consider the Banach space E the bounded domain 0' 7.15, Aut D is a 0 subgroup of Aut(Eo n D) ; by lemma 7.16 (AutoD)0 is a neighbourhood of 0 in E o n D. Then, proposition 7.17 gives the result. E o n D and the point OcEOn D. By lemma

ff 5 3 . - Study of the orbit origin. ___ (AutD)O of the-___-

In order to make a deeper study of the orbit (AutD)O of the origin we recall some properties of analytic sets. 7.19. DEFINITION. A s u b s e t R of domain D i s s a i d t o b e i n D if, f o r e v e r y p o i n t xcD, t h e r e i s a n e i g h b o u r h o o d U of x a n d t h e r e is a s e t F cHol(U,!l) of h o l o m o r p h i c f u n c t i o n s f: U+E s u c h t h a t we h a v e

complex-analytic

Roughly speaking, a subset D of D is analytic in D if, and only if, R can be locally represented a s the "joint kernel" of a set of holomorphic functions. For a study of the elementary properties of analytic sets see for example 1 4 5 1 p. 50. DEFINITION. L e t D b e a b o u n d e d c i r c u l a r d o m a i n in E. We d e n o t e b y R t h e s u b s e t o f D c o n s i s t i n g of t h e p o i n t s x c D for w h i c h (AutoD)x is a c o m p l e x - a n a l y t i c c l o s e d s e t i n D. 1.20.

LEMMA. L e t D b e a b o u n d e d c i r c u l a r d o m a i n i n E . T h e n , i f Eon D i s c o n n e t e d ( t h u s , i n p a r t i c u l a r , when D i s b a l a n c e d ) , u e h a v e Ll#$ a n d (AutD)R=R. 7.21.

BOUNDED CIRCULAR DOMAINS

125

P r o o f : S i n c e E n D i s assumed t o be c o n n e c t e d , by 0

c o r o l l a r y 7.18 w e have (AutoD)O=E

0

fl D .

Since Eo i s a c l o s e d

complex subspace of E l by t h e Hanh-Banach E

s e p a r a t i o n theorem,

f l D i s a complex-analytic c l o s e d set i n D;

0 and R f $ .

t h u s se have O c R

Let x c ( a u t D ) R ; t h e n t h e r e a r e ycR and gcAutD such t h a t gy= x . S i n c e by lemma 7 . 1 2 AutOD i s a normal subgroup of AutD, w e have ( a u t o D ) x = (AutoD)gy= g ( a u t o D ) y . B u t (AutoD)y i s a complexa n a l y t i c c l o s e d set i n D and t h e s e p r o p e r t i e s a r e p r e s e r v e d by gsAutD; t h u s , xcR and ( A u t D ) R c R . The o p o s i t e i n c l u s i o n i s obvious. 7.22.

# LEMMA. L e t D b e a b o u n d e d c i r c u l a r d o m a i n f o r

w h i c h E fl D i s c o n n e c t e d . 0

T h e n , f o r a l l xcll, we h a v e

P r o o f : I f x=O, t h e n t h e a s s e r t i o n i s t r i v i a l . L e t xcR be g i v e n w i t h x#O and p u t V = :

i

{Act; I A / < I /x / I - 1. S i n c e (AutoD)x

i s a complex-analytic c l o s e d s e t i n D ,

i s a complex-analytic c i r c u l a r group f

t

c l o s e d s u b s e t of V . A s D i s c i r c u l a r , t h e

(y)=: e

it

y , tCR, ysD, i s c o n t a i n e d i n AutoD;

therefore

e

it

xee

it

(AutoD)x= (AutOD)x

f o r a l l t d R , so t h a t { e i t ; t d R } c W .

S i n c e W i s an a n a l y t i c

i s connected (1431 p r o p o s i t i o n 1 page 5 0 ) . Then t h e u n i t d i s c o f t i s c o n t a i n e d i n W , whence t h e r e s u l t s u b s e t of V , V \ W follows.

# 7.23. PROPOSITION. L e t D b e a b o u n d e d c i r c u l a r d o m a i n i n E.

T h e n , if E o n D is c o n n e c t e d ( t h u s , i n p a r t i c u l a r , if D is

b a z a n c e d l , we h a v e

CHAPTER

126

7

(AutD)O= (AutoD)O= R = E fl D 0

Proof: First we show that Rc(AutoD)O. Let xcR be given. By lemma 7 . 2 2 we have Oe(AutO)x; thus, gx= 0 for some gsAut0 D. -1 -1 Then we have x= g Ocg (AutoD)O= (AutO)O and Rc(AutoD)O. Now we show that (AutD)ncQ. Obviously O c R ; by lemma 7.21, R is invariant under AutD, so that applying AutD to the relation OeR we get (AutD)0 c (AutD)R c R.

Thus , we have R c (AutoD)0 C (AutD)O c R and corollary 7 . 1 8 completes the p r o o f .

#

COROLLARY. L e t D b e a bounded c i r c u l a r dom ai n i n E. T h e n , i f E 0 fl D i s c o n n e c t e d ( i n p a r t i c u l a r , i f D i s balanced), t h e orbit (AutD)O is b a l a n c e d . 7.24.

Proof: Let xe(AutD)O and ACE, I A l S l , he given. B y proposition 7 . 2 3 we have xeR; then by lemma 7 . 2 2 we have hxc (AutoD)X C(AutoD)(AutD)O = (AutD)0 .

#

.

0

54 .- The decomposition AutD= (Auk D) (AutoD) I -

7.25. THEOREM. Let D c E and 8=$ be bounded c i r c u l a r d o ma in s i n t h e Banach s p a c e s E and 2 r e s p e c t i v e l y , and assum e % % t h a t f: D+ D is an a n a l y t i c i s o m o r p h i s m of D o n t o D s u c h t h a t f(O)= 0 . T h e n , t h e r e is a s u r j e c t i v e c o n t i n u o u s l i n e a r map FcL(E,2) s u c h t h a t FID= f.

Gt:

' L ' L

Proof: For tdR we define gt: E+E and E+ E by means t 'L t %t of g (x)=: eitx and g ( y ) = : city. Obvioysly g eAutD, GtcAutD 'L and, as f : D+D is a surjective isomorphism, the mapping h=: g -t f-1'Lt g f satisfies heAutD and h(O)= 0. From the chain rule 'Lt % and the fact that we have Lgt= g L for all LeL(E,E) , we derive -t ( 1

hA1= ( g

lo

(f

-1

( 1 'Lt ( 1 -

lo

(g

lo - g

-t

-1

(f

(1 (1 t

lo

f

0

= id

so that, by Cartan's uniqueness theorem, we get h= idD and

127

BOUNDED CIRCULAR DOMAINS

e i t f ( x ) = f ( ei t x ) for a l l

ttB

and xcD. By developi,ng

both

Gtf and

fgt i n t o t h e i r

Taylor series a t 0 we o b t a i n

s i n c e w e have assumed t h a t f ( O ; ) = 0 . Thus, by t h e u n i q u e n e s s of t h e T a y l o r s e r i e s , l e ( n - l ) i t - l ] f o( n ( x , . . , x ) = 0

f o r t d R , 1112, and xeD; t h e r e f o r e f i n = 0 . f ( x )= f ive .

(1

0

(XI

f o r xcD and f = F

It f o l l o w s t h a t (1

ID

'L

where F=: f o c L ( E , E ) i s s u r j e c t -

# 7 . 2 6 . COROLLARY. L e t D a n d

E and

8

8

b e t h e o p e n u n i t b a l l s of

'L

and assume t h a t f : D+D i s a h o l o m o r p h i c map of I) o n t o

'L

D s u c h t h a t f ( O ) = 0 . Then f i s an i s o m o r p h i s m if and o n l y i f we

h a ve f = F

f o r some s u r j e c t i v e l i n e a r i s o m e t r y F : E+8. ID

P r o o f : Assume t h a t f i s a n isomorphism. By theorem 7 . 2 5

w e have f = F

'L

f o r some s u r j e c t i v e F e L ( E , 2 ) . S i n c e f ( D ) = F(D)=D, ID

F i s an i s o m e t r y . The c o n v e r s e i s c l e a r .

# 7 ; 2 7 . DEFINITION. L e t D be a bounded c i r c u l a r domain i n E. We d e f i n e t h e " i s o t r o p y s u b g r o u p " o f t h e o r i g i n , I s o t D, by means of

Obviously I s o t D is a c l o s e d subgroup of A u t D for t h e t o p o l o g i e s T and T a . Moreover, from c o r o l l a r y 7 . 2 6 w e immediately o b t a i n 0

t h a t I s o t D = Aut D. The c i r c u l a r subgroup Z s a t i s f i e s

Z C (Aut'D)

n

(AutoD).

CHAPTER

128

7

7.28. THEOREM. L e t D b e a b o u n d e d circuZar9 domain in E . T h e n , if E IlD is c o n n e c t e d ( i n p a r t i c u l a r , if D i s b a Z a n c e d i , 0 ue h a v e

AutD= (AutOD) (AutoD)= (AutoD)(Aut0D) Proof: Let feAutD be given. From proposition 7.23 we

derive f(O)e(AutD)O= ( A u t o D ) O ; thus we have f(O)= g ( 0 ) for some gsAutoD. Then h=: g-lfcAutD and h(O)= g-'f(O)= 0, whence by theorem 7.25 we obtain h= F for some surjective F a L ( E , E ) , so 0 ID that heAut D. Thus we have f = gh with gaAutOD and hsAut0D. The other equality comes from the fact that AutOD is a normal subgroup of AutD. Remark that the factorization f= g.h, with g and h in the above conditions, is not unique as the circular subgroup satisfies Z c (AutoD)I7 (AutOD)

.

§5.-

ff

Holornorphic and isometric linear equivalence of~Banach___ spaces

.

'L

Let E and 3 be complex Banach spaces and D, D their respective open unit balls. 7.29. DEFINITION. (a) We s a y t h a t E a n d 2 a r e " h o Z o m o r p h i c a Z Z y e q u i v a Z e n t " if t h e r e is some s u r j e c t i v e a n a l y t r ?i ic i s o m o r p h i s m f: D+D. ?i

(b) We s a y t h a t E a n d E a r e " i s o m e t r i c a Z Z y Z,LnearZy e q u i v a l e n t " i f t h e r e e x i s t s some s u r j e c t i v e l i n e a r i s o m e t r y L:

E+$.

7.30. THEOREM. E a n d 2 a r e isomorphically e q u i v a l e n t if and onZy i f t h e y i s o m e t r i c a Z Z y Z i n e a r l y e q u i v a l e n t . Proof: The "if part" is obvious. Thus, let f: "DD be any surjective holomorphic isomorphism. Then, the mapping PI, f#: AutD + AutD given by g-tf-lgf is a surjective isomorphism o f these groups, so that we have Aut8= f-l(AutD)€ and, therefore,

BOUNDED CIRCULAR DOMAINS

,

f (Aut8)= (AutD)f

129

(AutD)= (Aut8)f

f

-'

Let us denote by Eo and 80 the Banach spaces associated with E and 2 by definition 7.5 , so that

8

(AutD)O= E0 n %

,

(AutD)O= E o n D

(7.9)

by proposition 7.23. We claim that (7.10)

2.

(AutD)O = f-' (AutD)0

(AutD)O = f (Aut8)0, 'L

Indeed, from f-'(AutD)f= AutD we get that [f-1 (AutD)f]O=

'L

%

(AutD)O = E o

'

L

nD

'L

is a complex-analytic closed set in D. Let us put < = : f ( 0 ) ; % then f-l(AutD)< is a complex analytic closed set in D and, applying f we obtain that (AutD)< is a complex analytic closed subset of D. Therefore, by proposition 7.23 we have < c E0 n D. Then, as the orbit E o n D of 0 is AutD-invariant,

E" 0 n 8= (nut&)O = =

[f-' (AutD)f] O =

f-l (AutD)CCf-'(AutD) ( E o

n D) c f - l

( E o n D)

so that, applying f we obtain

In a similar manner we get E~ I

t

D c f (80n

8)

.

%

Thus we have E o n D= f ( E o n D) , which is equivalent to (7.10) 'L From the second of these formulas we obtain f - (0)E: (AutD)0 ; 'L % 'L 'L thus we have f-'(O)= q ( 0 ) for some gcAutD. Then h=: fg is an 'L

analytic isomorphism of D onto D with h(0) = fg(0)= 0 the result follows by corollary 7.26.

whence i+

CHAPTER

130

7

The group ~. __ of surjective linear isometries of a Banach space.

56.-

-

Let E be a complex Banach space with unit ball D=: B ( E ) . The group Aut 0 D of all surjective linear isometries of E turns out to be a subgroup of both AutD and GL(L(E)) I the linear group 0 of E. We obtain some properties of Aut D by looking at it a s a subgroup of these two groups. h

Let autD= autUDOaut0 D be the decomposition of the Lie algebra autD given by proposition 7 . 6 . For aeGL(L (El , let a# be the adjoint of CL (cf. definition 4 . 2 6 ) . 7.31.

PROPOSITION. Assume t h a t D Cs h o m o g e n e o u s . Then 0

we have t h e foZZouing c h a r a c t e r i z a t i o n of Aut D a s a s u b g r o u p

of GL(L (El ) :

Proof: Let asGL(L(E)) be such that acAutOD. It is an immediate consequence of the proof of proposition 7 . 9 (d) that aut D is a#-invariant(even if D is not homogeneous). 0

In order to prove the converse statement, we show first that the relation a # (autoD)c autOD

implies a# (AutoD)c AutOD

Indeed, let feAutOD be given; by Lemma 7 . 1 2 we can find Ak eautoD, k= lr2r...,nr such that

Therefore, by proposition 5 . 1 3 we have a f= afcl-'= It

[a(expAl)a-']0.

.O

[a(expAn)a-']=

= a (expA1)a ..oa (expAn)= exp(a#Al),..,exp # #

(a A )

a n

BOUNDED CIRCULAR DOMAINS

131

From the assumptuion a#(autoD)cautOD we get a#AksautoD for k = 1,2,..,n so that, again by lemma 7.12, exp(aA )eAutoD and, # k finally a#f= exp(a#A1)

0 .

.oexp(a#Ao)eAut0D.

NOW, let acGL(L(E)) be such that a# (autoD)C aut D. Since D is 0 assumed to be homogeneous, by proposition 7.23 it follows that

so that cx (D)= [a(Aut D)1 0 c (AutoD)O = D 0

A similar argument with a-'cGL(L(E)) gives a(D)= D. Thus, a is a surjective linear isometry, i.e., mcAut 0 D. 79 7.32'. COROLLARY. If t h e u n i t baZZ D o f E i s h o m o g e n e o u s , 0 t h e n Aut D i s a r e a l a l g e b r a i c s u b g r o u p of d e g r e e 2 of GL(L(E)) In p a r t i c u l a r , Aut 0 D i s a B a n a c h - L i e g r o u p f o r t h e t o p o l o g y of u n i f o r m c o n v e r g e n c e on D.

.

Let AcautOD be given. By proposition 7.6 A(x) = c-Q, (x,x)I

xeD

for some csE. Then a#A has the expression (cf. definition 4.26) (a#A)x= a(c)-aQ ( C X - ~ X a-lx) , , xcD Thus, again by proposition 7.31, a#A belongs to autoD only if. mQ (a-'xp-'x) = Qa(cl(x,x)

VxcD

if, and

132

CHAPTER

7

which i s e q u i v a l e n t t o

NOW,

f o r f i x e d CEE and xeD, t h e mappings L ( E ) X L ( E ) + E g i v e n

re s p e c t i v e l y by

a r e o b v i o u s l y c o n t i n u o u s homogeneous polynomials of d e g r e e 2 and 1 , and ( * ) can be r e f o r m u l a t e d as

w i t h C C E , X E D and acGL(L(E)). T h i s a s e t of e q u a t i o n s d e f i n i n g 0

A u t D a s a r e a l a l g e b r a i c subgroup of d e g r e e 2 of GL(L(E)).

Then, theorem 6 . 4 0 c o m p l e t e s t h e p r o o f .

# 7.33. homogeneous.

EXERCISE. Assume t h a t t h e u n i t b a l l D of E i s 0

Show t h a t , on t h e group Aut D , t h e a n a l y t i c

topology Ta, t h e t o p o l o g y of l o c a l uniform convergence 9' and t h e topology TU of uniform convergence o v e r D c o i n c i d e . 57.-

Boundary behaviour and e x t e n s i o n theorems.

W e r e c a l l t h a t i f D i s a bounded c i r c u l a r domain, t h e n any

AeautD

a d m i t s a unique r e p r e s e n t a t i o n of t h e form

where ceE

0

, LeL(E)

and Q c : E

+

E i s a c o n t i n u o u s symmetric

b i l i n e a r mapping. I n p a r t i c u l a r , A i s and e n t i r e mapping. 7.34.

DEFINITION.

Let AsautD b e g i u e n ; t h e n ue seL

T h u s t h e numbers p and C d e p e n d o n t h e d o m a i n D a n d on t h e

v e c t o r f i e l d AcautD. For any f i x e d EcE and AcautD, w e c a n c o n s i d e r t h e i n i t i a l v a l u e

BOUNDED CIRCULAR DOMAINS

133

Problem d = A[y(t)], 2

(7.11)

y(O)=

5

+5

whose maximal s o l u t i o n $ ( t ) i s d e f i n e d i n a domain dom

5

IR. O f c o u r s e , i f 5eD t h e n w e have

+

5

of

- E. 5-

( t )= ( e x p t A j 5 and dom+

7.35. LEMMA. L e t AsautD a n d SeE b e g i v e n . T h e n : (a1 dom+5=(-c-110g[l + d i s t ( S, O ) -l]

,

-'I)

C-llog [I + d i s t ( 5 , ~ )

( b ) F o r a n y xeD and a n y tdR w i t h I t l < G - l l o g ( l + l l C-x/l - I ) we h a v e

/I

+5(t)-+x 1 1 (
.

11 + 5 ( t ) - + x/ I ( t )W e ( t ) ] .Moreover, from t h e 5 5 t h e o r y o f o r d i n a r y d i f f e r e n t i a l e q u a t i o n s , w e know t h a t t h e P r o o f : For tsdom+

know t h a t $ ( O ) =

5

and

we put 6 ( t ) = :

5, dt J15(t) = A[+

f u n c t i o n t + S ( t ) i s a b s o l u t e l y c o n t i n u o u s i n dam+<; t h u s , by L e b e s g u e ' s theorem, S i s d e r i v a b l e almost everywhere i n dom+ B e s i d e s , 6 h a s r i g h t and l e f t d e r i v a t i v e s a t any p o i n t o f

dam+

5'

5'

From

S(t,)-S(t,)=

we obtain

II

9 5 ( t l ) - + x ( t l ) II

-11

+ 5 ( t 2 ) - + xII( ct 2 )

CHAPTER

134

d dt

(7.12)

which i s v a l i d

7

s(t)sC[s(t)t62(t)]

a l m o s t everywhere i n dom$

6 ( t ) t 6 2 ( t ) f 0whatever i s tcdom$

5'

5'

For x#< w e have

and from ( 7 . 1 2 ) w e g e t

d log 6( t ) dt l + f i ( t ) sc a l m o s t everywhere i n domq

f o r a l l tcdomq

5'

whence w e e a s i l y d e r i v e

5'

On t h e o t h e r hand,

f o r tedomijl

F;'

L e t us set t*=:

sup{tER; tedom$

F,

1

From t h e e x t e n s i o n theorems f o r o r d i n a r y d i f f e r e n t i a l e q u a t i o n s we g e t

so t h a t w e have

l i m * 6 ( t )=

a.

Therefore

t+t

I n a similar way w e can prove t h a t

(C-'log(l+f,( 0 ) - I ) , O ] c

dom$ 5

BOUNDED CIRCULAR DOMAINS

S i n c e 6(0)=

I/

$5(0)-+x(O)

11

=

11

<-xi1

135

a n d xsE was a r b i t r a r y , by

making x r a n g e o v e r D w e g e t t h e f i r s t s t a t a n t ; t h e s e c o n d o n e h a s been proved i n ( 7 . 1 3 )

7.36.

.

#

DEFINITION. For rdR we s e t

D(T,A)=:

where $

E,

{ < s E ; .redom$

d e n o t e s t h e maximal s o Z u t i o n of

Thus, w e have D c D ( T , A )

51 17.111.

f o r a l l rdR

7 . 3 7 . LEMMA. T h e foZlowing r e Z a t i o n s haZd

( a ) {SeE; d i s t ( 5 , D ) < ( e C I T / - I ) - ' } C D ( T , A )

for a l l T m .

( b ) If AsautD, t h e n A , a D c a u t( a D ) and ( e x p t A ) 2D= a D

for aZZ

tm.

P r o o f : L e t T ~ Rb e g i v e n a n d assume t h a t 5 c E s a t i s f i e s

. We

d i s t ( 5 , D ) < ( e CI "-1)

may s u p p o s e t h a t < # D a s o t h e r w i s e From

t h e r e i s nothing t o prove

we o b t a i n

IT SO

I
t h a t , by lemma 7 . 3 5 w e o b t a i n Tsdom+

5

and < ~ D ( T , A ) .

The above r e a s o n i n g a l s o shows t h a t , f o r a l l
s o l u t i o n of t h e i n i t i a l v a l u e problem (7.111, (exptA)a D c E \ D f o r a l l tCR. W e s h a l l now show t h a t (exptA)a T h e r e f o r e , f o r t d R w e have

DC5

CHAPTER

136

7

I n d e e d , l e t u s f i x a n y S c a D a n d .rdIR. Then, f o r e v e r y E > O , t h e r e e x i s t s some xEeD s u c h t h a t

11

5-xE

/I

<

r

whence

A p p l y i n g lemma 7.35 t o x E a n d 'r w e g e t

we h a v e 11 <-xE)I+o: t h e r e f o r e ]I j j l E ( ~ ) - j j l X ( . s ) I l + O . S i n c e E $ x ( r ) c D , w e h a v e )I ( T ) ~ B . A s €,caD a n d T ~ I Rw e r e a r b i t r a r y , w e 5 E g e t ( e x p t A ) a D c 5 for t d R .

F o r E+O

t o (7.14) we o b t a i n

F i n a l l y , applying exp ( - t A ) aDc[exp(-tA)]aDcaD

so t h a t ( e x p t A ) a D = aD for a l l t d R .

ff A s u s u a l l y , w e w r i t e DF f o r t h e p a r a l l e l s e t D =: E

D+EB(E).

7 . 3 8 . LEMMA. L e t AeautD be g i v e n . T k e n , f o r e v e r y E > O , t h e r e e x i s t s v > O s u c h t h a t f = : expA a d m i t s a n i n j e c t i v e koZornorphic e x t e n s i o n F : D +E w i t h F(D ) c D c . i s bounded on D

n

rl

.

rl

I n paraticuzar, F

P r o o f : L e t AeautD b e g i v e n a n d d e n o t e b y p a n d C t h e numbers a s s o c i a t e d w i t h A by d e f i n i t i o n 7 . 3 4 .

W e may assume

t h a t C>O a s o t h e r w i s e t h e s t a t e m e n t i s o b v i o u s l y t r u e . By lemma 7 . 3 7 , o n c e rdR h a s b e e n f i x e d , w e h a v e

t h u s , by s e t t i n g

T=

1 and

B0UNBFrD CIRCULAR DOMAINS

137

we have D c d o m expA il

s o t h a t it makes s e n s e t o d e f i n e F: Dq+E b y means o f F ( S ) = : (expA)S; t h a t i s , F ( < ) i s t h e v a l u e a t t h e i n s t a n t t = 1

of t h e s o l u t i o n o f t h e i n i t i a l v a l u e p r o b l e m

From t h e t h e o r y of o r d i n a r y d i f f e r e n t i a l e q u a t i o n s w e know t h a t F: D +E i s a n i n j e c t i v e h o l o m o r p h i c mapping. O b v i o u s l y , 0 w e h a v e F I D =f . By a p p l y i n g lemma 7 . 3 5 t o a n y xCD, a n y < = x + u w i t h UCE, I / u I I < q , a n d t = 1 w e o b t a i n

Whence F ( D n ) c D E . The e x t e n s i o n F o f f= expA i s c l e a r l y u n i q u e .

# 7.39.

THEOREM. L e t fsAutD be g i u e n . T h e n , f o r e v e r y & > O ,

t h e r e e x i s t q > O such t h a t f admits and i n j e c t i v e holomorphic e x t e n s i o n F: Dq+ E w i t h F ( D q ) C D E . I n p a r t i c u l a r , F i s bounded o n D,,. P r o o f : L e t fsAutD a n d E > O b e g i v e n . By t h e o r e m 7 . 2 8 we h a v e f = goh f o r some gcAutOD a n d hcAutoD. Then, by t h e o r e m 7 . 2 5 and lemma 7 . 1 2

it follows t h a t f= G

ID

o(expA1) 0 . .

o

(expAn)

f o r some s u r j e c t i v e l i n e a r i s o m e t r y G : E+E a n d A 1 , . . A O b v i o u s l y g= G

ID

cautD.

admits a n i n j e c t i v e holomorphic e x t e n s i o n t o

t h e whole E . Moreover, a s G i s u n i f o r m l y c o n t i n u o u s on E , w e c a n e a s i l y f i n d a number q o > O

such t h a t

NOW, by lemma 7.38 w e c a n f i n d a number v1>O

h =: expA 1

1

such t h a t

a d m i t s an i n j e c t i v e h o l o m o r p h i c e x t e n s i o n H1

CHAPTER

138

7

with

By r e i t e r a t i n g t h e argument and s e t t i n g q = : m i n ( r l O , q l , . . , q n )

we get the result.

# 7.40.

EXERCISES.

( a ) Show t h a t , f o r a n y fsAutD, t h e r e i s

a p a r a l l e l negighbourhood D

rl

of D i n which f i s u n i f o r m l y

continuous.

(b) Show t h a t any feAutD maps a n y s u b s e t S c c D i n t o a s u b s e t f ( S ) cc D .

CHAPTER 8

AUTOMORPHISMS OF THE UNIT BALL OF SOME CLASSICAL BANACH SPACES

51.-

Some qeometrical considerations.

Let A be a convex non void subset of E. We recall that a function f: A - t I R is "convex" if we have

f [Xi (l-X)y]sAf (x)+ ( I - A ) f ( y ) for all As[O,I] and all x,ysA. We denote by Conv(A) the set of the functions f : A + E that are convex in A.

8.1

. DEFINITION. We

say that fsConv(A) is "subdifferentiable"

at a point asA if there exists a continuous real linear functional A: E-tIR such that we have

A (x)-A (a)Cf (x)-f (a) f o r all xcA.

The set Bf(a) of the functionals A in those conditions is called the subgradient of f at the point a. 8.2. PROPOSITION. Any lower semicontinuous function is subdifferentiable. In particular, the norm subdifferentiable

11 - 11

is

at any point asE, afO, a n d w e have

139

140

Here A:

8

CHAPTER

d e n o t e s t h e s e t of c o n t i n u o u s r e a l l i n e a r f u n c t i o n a l s

E a .

Proof: See { 2 8 1 theorem F page 3 0 . S i n c e a n y c o n t i n u o u s complex l i n e a r f u n c t i o n a l $cE* i s u n i q u e l y

a:

d e t e r m i n e d b y i t s r e a l p a r t As%

by t h e formula

$ ( x ) = A(x)-iA(ix), and

I/ $ 1 1

=

L e t D=:B(E)

11

All

, we

XCE,

a l s o have

be t h e open u n i t b a l l o f E . W e r e c a l l f r o m c h a p t e r

7 t h a t any XeautD c a n b e u n i q u e l y r e p r e s e n t e d i n t h e f o r m X(X)= C+L(X)-Qc(X,X)

where c e E O

,

XcE

LeT((E) and Q c L s ( E x E i E ) i s a c o n t i n u o u s symmetric

b i l i n e a r map. 8 . 3 . PROPOSITION. L e t aeaD and XcautD be g i v e n . T h e n , f o r a 2 2 4eE;

s u c h t h a t ReQeVlj

-

11

(a) ue have $ [ Q (~a , a ) ]=

R e @ [ L ( a ) ] =0 Proof: L e t u s f i x any

Om.

TE)

We h a v e eieacaD.

Consider t h e

i n i t i a l v a l u e problem

a n d d e n o t e by

.p ( t ) = :( e x p t X ) ei6 a i t s maximal s o l u t i o n . By

lemma 7 . 3 7 w e know t h a t X

IaD

dom q = IR

,

i s c o m p l e t e i n a D so t h a t g ( t ) a D = aD

f o r all t d R . Let u s d e f i n e $ ( t ) = : e-ie.exp(tX)eiea

AUTOMORPHISMS OF CLASSICAL BANACH

SPACES

141

f o r t d R . Then $(O)=

II

a,

for tm. T h u s , i f R e $ s V ( /

. (1

$ ( t ) II = 1

( a ),

I n a s i m i l a r manner, by s e t t i n g

+ ( t ) =e :- i e . e x p ( - t X ) e

i B

a

we obtain Re$

so t h a t w e have

Re

[+I

(011 30

[+I

(0) J

=

0 . S u b s t i t u t i o n of

gives

But t h i s i s v a l i d f o r a l l

Om, whence

t h e r e s u l t follows

##

immediately. 8.4. $(a)=

COROLLARY.

11 CpII 1 1 0

all

.

L e t XcautD, acE

and @EX

a:

be such t h a t

T h e n we h a v e

P r o o f : W e may assume a f 0 and $#O a s o t h e r w i s e t h e r e s u l t

i s o b v i o u s . From $ ( a ) =

11 $ 1 1

11

all

we obtain

CHAPTER

142

Then p r o p o s i t i o n 8 . 2 a p p l i e d t o result.

8

'

-and

/I @I1

~

a

gives the

II a II #

b2.-

Automorphisms of t h e u n i t b a l l of L ' ( Q , I I ) , -~ 2jCpfm. L -

1 1 1 1 -

Throughout t h i s s e c t i o n ,

(R,p)

r e p r e s e n t s a r e a l measure s p a c e

w i t h 0 < p ( 6 2 ) < w . For e a c h p w i t h l & p < a , Lp s t a n d s f o r t h e Banach s p a c e of

I / f 11

p=:

( c l a s s e s o f ) measurables f u n c t i o n s f : Q

(jR I f 1'

dp) " ' < a

.

+

a: s u c h t h a t

W e s h a l l make no d i s t i n c t i o n between

f a n d i t s c o r r e s p o n d i n g c l a s s . With q w e r e p r e s e n t t h e conjugate

1 1 + = 1 . The u s u a l c o n v e n t i o n s a r e a p p l i c a b l e f o r P q t h e c a s e p = m . F o r e a c h m e a s u r a b l e s e t S , x, d e n o t e s t h e

of p I i . e . -

c h a r a c t e r i s t i c function of S. 8.5

LEMMA. L e t p b e g i v e n w i t h 16pim a n d a s s u m e t h a t

dimPL>l. T h e n , t h e r e a r e t w o m e a s u r a b l e s e t s , A , B s u c h t h a t

P r o o f : W e s h a l l show t h a t t h e r e e x i s t s a m e a s u r a b l e s e t A s a t i s f y i n g O < p ( A ) < ~ ( i l ) .Then t h e p a i r A and B = :

R\A

s a t i s f i e s t h e r e q u i r e m e n t s of t h e lemma. W e p r o c e e d by c o n t r a d i c t i o n . Assume t h a t f o r e v e r y m e a s u r a b l e s e t A w e had

S i n c e p ( Q ) < w , w e have x Q s L p . A s dim L p > l , t h e r e e x i s t s some feLP which d o e s n o t b e l o n g t o t h e s u b s p a c e g e n e r a t e d by particular

O b v i o u s l y , w e have

x,;

in

AUTOMORPHISMS OF CLASSICAL BANACH SPACES

e x a c t l y o n e of t h e s u b s e t s i n t h e

Thus, by ( 8 . 1 ) and ( 8 . 2 ) , right-hand

s i d e of ( 8 . 3 ) h a s p o s i t i v e m e a s u r e . W e may assume

t h e c o r r e s p o n d i n g s u b i n t e r v a l t o be Io= ( 0 , 1 ] (8.4)

143

A 0 =: f-'(O,I]

,

so t h a t

O
NOW w e w r i t e A = f - l ( O r l / 2 ] U f- 1 (1/2,1]. A g a i n , e x a c t l y o n e of 0

t h e s e t w o s u b s e t s h a s p o s i t i v e measure. W e w r i t e I1 f o r t h e corresponding subinterval: thus

By i n d u c t i o n w e g e t a s e q u e n c e o f i n t e r v a l s s e q u e n c e of m e a s u r a b l e s e t s ( A k I k m

('k) k€!IN

and a

with

f o r a l l km. A s t h e s e q u e n c e s o f t h e e x t r e m e p o i n t s of t h e Ik a r e monotonous a n d bounded, t h e y d e f i n e a number <m and i t

i s e a s y t o check t h a t

Therefore,

and w e have f =

<x,

bL

almost everywhere; t h u s € belongs t o t h e

s u b s p a c e g e n e r a t e d by

x

R

l

which i s a c o n t r a d i c t i o n .

#

8.6. DZFINITION. L e t p w i t h 1 < p < m b e g i v e n . F o r fcLp, we d e f i n e f " i n t h e f o l l o w i n g manner

144

8

CHAPTER

*

-1 1

(a) If p= 1, t h e n f ( X I = E ( x ) f* ( x ) = 0 w h e n f(x)= 0. I f (XI I (b)

11 p f l ,

fl/

uhen f(x) 40

and

then f*= ? ( f ( p - 2 .

8 . 7 . LEMMA. For f e L P , t h e f u n c t i o n

f* s a t i s f i e s

( a ) f*eLq (b)

II f l l

;

=

II f * I /

;

Proof: First we c o n s i d e r the case p= 1. Obviously f * is essentially bounded and (1 f * ( (M = S=: { X C R ; f(x)= 0 1 we have

(1 f (1

. Moreover,

writing

Now we consider the case l i p < - . Then

Thus

Finally, since p f l ,

=

II flip ( 1 1 f l l p p ) l ’ q =

II f l I pII

f*II

#

AUTOMORPHISMS OF CLASSICAL BANACH SPACES

145

and be given. Then,for e a c h p a i r of m e a s u r a b l e s e t s A,B w i t h Afl B= 4 , A U B = R, t h e f u n c t i o n f=: xA + p eie x, belongs to Lp and i t s c o r r e s p o n d i n g * f i s g i v e n by 8.8

LEMMA. L e t p ,

f*=

Isp<-,

A

pZ0

e-it3

+

XB

if P f l

Proof: It is an easy consequence of the definitions and the facts

xA.x,=

0,

x A +xB =

1.

#

8 . 9 . THEOREM. L e t p b e g i v e n w i t h lsp<m, p f 2 and assume t h a t dim Lp>l. T h e n e v e r y v e c t o r field t h a t i s c o m p l e t e i n t h e u n i t b a l l of Lp i s l i n e a r .

Proof: Let XeautB(LP) be given. We recall that there exists some ccLp such that X admits the representation

First, we shall prove the following auxiliary statement. 8.10.

LEMMA. U n d e r t h e a s s u m p t i o n s o f t h e o r e m 8 . 9

,

for

e u e r y m e a s u r a b l e s e t A w i t h O
Proof: Once A has been given, we put B=: R\A.

In order

to simplify the notations, we set

f o r j,k,R= 1,2.

f=:

x 1 +P

e

it3

NOW, by lemma 8.8, for p ~ and 0 8dlR we have

x2cLp. By lemma 8.7 we have < f , f * > =

11

flI P I / f * / I

g'

146

CHAPTER

8

thus, applying c o r o l l a r y 8 . 4 t o t h e p a i r ( f , f * ) , we obtain

T h e r e f o r e , from t h e e x p r e s s i o n o f f * g i v e n by lemma 8 . 8 w e g e t

i n the case p f l

i0 =
e

i0

x2;

x2)

xl+p e

p-1 I

.-iO

xl+p

x2

A f t e r computing t h e i n t e g r a l s a p p e a r i n g i n t h i s f o r m u l a and r e a r r a n g i n g t h e t e r m s i n a c c o r d a n c e w i t h t h e powers of e

ie

,

we get

fit2 p 2 e 2 i 0 +[o,,

Pp+1+2a:2

+ [ & + 2 B 1 22

which i s v a l i d f o r a l l

P + ( U l + P 2 ! J 2 ) 2 / p Y- 2

PP+(II1+PP

p20

u 2 ) 2 / p Y- 1 l + B l2l

(8.5)

Besides, as B i l i s independent of

p

,

(8.6)

Also,

From

8.51,

lim

7 1 (ul+p

P+m

P

(8.6)

P q 2 / P Y- 1 =

and ( 8 . 7

pp-l .iB,

and 0 d R . T h u s , a l l c o e f f i c i e n t s

must be n u l l ; i n p a r t i c u l a r

(8.7)

s41+

we derive

Y- 1 u 2 / p

147

AUTOMORPHISMS OF CLASSICAL BANACH SPACES

AS

w e have assumed pf2 and B 2

d o e s n o t depend on p , t h e l a s t

12

equality entails

12

= 0 and

1

Therefore

= 0.

a s w e wanted t o show. The proof i n t h e c a s e p = 1 i s s i m i l a r .

Now w e w e p r o c e e d t o p r o v e t h e theorem. O b v i o u s l y , it s u f f i c e s t o show t h a t , f o r any m e a s u r a b l e s e t S,

T h i s i s c l e a r when u ( S ) = 0 . I f O
the result is also

Thus w e have t o c o n s i d e r t h e case

S i n c e w e have assumed t h a t dim L p > l , by lemma 8 . 5

t h e r e are measurable sets A , B w i t h A O < u ( A )
,

,

n B=

@,

A

u B=

R,

O < v ( B ) < v ( R ) . From lemma 8.10 a p p l i e d t o A and B

we get cdu= J S

[

Jn

cdp=

[

cdp+j

8 . 1 1 . COROLLARY. F o r l $ p < a

#

cdp= 0

B

'A

,

pf2 a n d dim L p > l w e have

(AutD)O= { O j a n d AutD={U ID

; U i s a s u r j e c t i v e l i n e a r i s o m e t r y of L p } .

P r o o f : W e r e c a l l from d e f i n i t i o n 7.8 and p r o p o s i t i o n 7.23 t h a t i f E o = { X ( O ) ; XcautD}, t h e n w e have (AutD)O= E o n D . From theorem 8 . 9

w e o b t a i n E0 = { O ) ,

t h u s (AutD)O= {O}.

Then,

f o r FsAutD w e have F ( 0 ) = 0 ; t h e r e f o r e , by c o r o l l a r y 7.25, f o r some s u r j e c t i v e l i n e a r i s o m e t r y U of L p . The F= U I D converse i s obvious.

#

148

8

CHAPTER

the

9 3 . - Automorphisms of

u n i t b a l l o f some a l g e__b r a s of

continuous functions. T h r o u g h o u t t h i s s e c t i o n R d e n o t e s a compact H a u s d o r f f s p a c e and C ( Q ) r e p r e s e n t s t h e Banach s p a c e of c o n t i n u o u s f u n c t i o n s f: R

+

/I

C w i t h t h e norm

sup I f ( x ) 1

flI =:

.

X€R

8 . 1 2 . THEOREM. L e t A b e c l o s e d c o m p l e x s u b a z g e b r a of

C(Q) w h i c h c o n t a i n s t h e unit e of C(Q). S u p p o s e t h a t T : A+A is a s u r j e c t i v e l i n e u p i s o m e t r y of A.

T h e n T is o f t h e

form Tf= ~ . $ f

wirere 1 c1

fcA

i s a n e l e m e n t o f A s u c h t h a t l a ( x ) I = 1 for a l Z

01

XCR,

a n d @ i s a s u r a j e c t i v e a l g e b r a a u t o m o r p h i s m of A. Moreover, if T e = e , t h e n T is m u Z t C p l i c a t i v i cA

P r o o f : W e c o n s i d e r A as a Banach s p a c e a n d w r i t e A* f o r i t s c o n j u g a t e s p a c e . T h u s , t h e u n i t b a l l B(A*) i s a c o n v e x s e t t h a t i s compact f o r t h e weak-*

t o p o l o g y of A*. By t h e K r e i n c l o s e d c o n v e x h u l l of t h e

Milman t h e o r e m B(A*) i s t h e weak-*

s e t e x t r B ( A * ) o f t h e extreme p o i n t s o f B ( A * ) . I n p a r t i c u l a r , e x t r B ( A * )# $

.

F i r s t w e show t h a t e v e r y extreme p o i n t L o f B ( A * )

f o r m L=

Adx

where heP, / X I = 7 , and Ax:

i s of t h e

A * i 2 is the evaluation

f + f ( x ) a t some p o i n t x e n . I n d e e d , l e t L b e an extreme p o i n t o f B(A*); t h e n

11

LI/ = 1

a n d , by t h e Hahn-Banach t h e o r e m , w e c a n

/I FII

extend L t o an element FcC(R)* such t h a t

= 1 . L e t S denote

t h e set of such extensions

S = : { F c C ( R ) * ; F I A = L,

11

FII =

It i s clear t h a t S i s a c o n v e x a n d weak-* C(R)*;

/I

LI/ = 11

compact s u b s e t of

t h u s , u s i n g a g a i n t h e Krein-Milman t h e o r e m , w e c a n choose

a n e x t r e m e p o i n t F of S . Then F i s a c t u a l l y a n e x t r e m e p o i n t of

t h e u n i t b a l l of C(n)*

.

Indeed, i f w e have

F=

1 2

( F +F ) 1

2

AUTOMORPHISMS OF CLASSICAL BANACH SPACES

with

I1 F k ] j < I , k =

and, i f we w r i t e L have

11

k

1,2,

149

then

f o r t h e r e s t r i c t i o n of F k t o A, t h e n w e

LkII $ 1 and L=

1 2

(L1+L ) 2

S i n c e L i s an extreme p o i n t of B(A*) w e g e t L = L = L . 1 2 T h e r e f o r e , each Fk i s a norm p r e s e r v i n g e x t e n s i o n of L ,

i.e.,

FkeS, and because F i s an extreme p o i n t of S , w e have F = F = F . Now, a s 1 2

C(R)*

i s t h e s p a c e of complex B a i r e measures

on R , t h e extreme p o i n t s of i t s u n i t b a l l are known ( c f .

1131)

t o be t h e measures of t h e form X b X where AcC., / X I = 1 and b x

is

t h e e v a l u a t i o n on C(R) a t some p o i n t xeR. I t f o l l o w s t h a t o u r extreme f u n c t i o n a l F h a s t h e form

and, r e s t r i c t i n g t o A , w e o b t a i n L ( f ) = hf

so t h a t L = A6

X

(XI

feA

.

A s a w e l l known consequence of t h e Hanh-Banach

theorem, w e have

Given fsA, t h e f u n c t i o n a l L - + / L f l i s convex and weak-* c o n t i n u o u s on A*. So, it a t t a i n s i t s maximum on t h e weak-* compact convex s e t E(A*) a t some e x t r e m a l p o i n t ( c f . 128 I )

Now w e c o n s i d e r t h e g i v e n Banach s p a c e i s o m e t r y T of A . The t r a n s p o s e d mapping T* o f T I d e f i n e d by

150

CHAPTER

(T*L) f = L ( T f )

8

LsA*

i s a s u r j e c t i v e i s o m e t r y of A * ;

,

fsA

Thus T* must c a r r y t h e s e t

extrB(A*) o n t o i t s e l f . L e t us d e f i n e R 0 =:

{xcR;

GxcextrB(A*)}

S i n c e d eextrB(A*) i f and o n l y i f AGxeextrB(A*), for a l l ACE,

/ A ] = 1,

I f xcRO

we have

,

t h e n T*6x must a l s o be a n extreme p o i n t of B ( A * ) .

T h i s a s s o c i a t e s w i t h x a complex number a ( x ) of modulus 1 and a p o i n t r ( x ) e n 0 such t h a t T*6 = X

U(X)

6

T (x)

or, equivalently,

f o r a l l feA and xen,.

Taking f = eeA i n t h e above, w e g e t

a ( x ) = ( T e ) (x) f o r a l l X E R

0

so t h a t a i s ( t h e r e s t r i c t i o n t o R o

o f ) a f u n c t i o n TeeA.

Now w e b e g i n t o u s e t h e r i n g s t r u c t u r e of A . L e t f , gsA and xcRo be g i v e n . Then w e have

and m u l t i p l y i n g by

(8.11)

a ( x ) we obtain

a ( x ) [T(fg)] ( X I = ( T f ) (XI (Tg) ( X I

From ( 8 . 8 ) and ( 8 . 9 ) w e d e r i v e t h a t , f o r any hcA,

AUTOMORPHISMS OF CLASSICAL BANACH SPACES

151

From (8.11) we see that, for f, gcA, the function

vanishes

on Go;

consequently

for all f, gcA. If we take f = g= T-'(eI , we see that 1 cA. Since both c1 and 1 have modulus 1 on Ro and each c1 a, attains its maximum on R , we see l a ( x ) / = 1 for all xcR. Then if we define @ : A-+A by

@ f = ol-lTf

f €A

it is clear that 4 is a surjective algebra automorphism of A.

Next we characterize the group of automorphisms of the unit ball D=: B(A) of the algebra A. Since D is bounded circular domain, there is a Banach subspace A. associated with A by definition 7.5. We recall from chapter 7 that any vector field XcautD can be uniquely represented in the form X(h)= f+L(hl-Qf (h,h)

heA

for some Lcaut0D, f-Q cautOD and fcAo. f

-8.13. THEOREM. L e t R be a c o m p a c t H a u s d o r f f s p a c e and d e n o t e by A a c l o s e d ) c o m p l e x s u b a l g e b r a o f C(0) s u c h t h a t e c A . Then

(a) A. is t h e c o m p l e x s u b a l g e b r a o f A g i v e n by A = : {acA; %A}. 0

152

CHAPTER

8

( b ) E v e r y a u t o m o r p h i s m FeAutD

admits a unique

0

r e p r e s e n t a t i o n F = LQM w h e r e LsAut D is a s u r j e c t i v e linear isometry

01A and M&AutOD is

t h e MSbius transformalion

€+a Mf=: e+af f o r some acAo,

11

fsD

all < 1 .

Proof: Let us write !?(A)=: {acA; a=

a)

and notice that

R ( A ) is a closed real subalgebra of A with e c R ( A ) . Moreover,

its complexified R ( A )

satisfies (I:

R(A) = {asA;

ZCAl

It is easy to see that, for aeRiA)' Mabius transformation f +a , Mf= 1G f

with

11

a l l '1,

the

f&D

is an element of AutD. Thus

R (A)

n

D C (AutD)0

and, by proposition 7.23

(8.12)

c R(A) c A o

Now let fcA be given. Then, the vector field Xf(h)=: f-Qf(h,h), 0 hcA, satisfies X eautD. If 6 x : A -t C denotes the evaluation on A at any point x e n , we have &xeA* and

AUTOMORPHISMS OF CLASSICAL BANACH SPACES

As esaD,

153

from p r o p o s i t i o n 8 . 3 w e o b t a i n

r

6 x [ Q f ( e , e ) ] =6 = 6X

(8.13)

r

so t h a t Q f ( e r e ) =

X€R

and t h e r e f o r e F s A . Thus

From ( 8 . 1 2 ) a n d ( 8 . 1 3 ) w e g e t

a:

A = R ( A ) = {aeA; 0

%A}.

The r e m a i n d e r p a r t o f t h e p r o o f f o l l o w s e a s i l y f r o m t h e o r e m 7.25.

ff A number o f i n t e r e s t i n g e x a m p l e s a r e i n c l u d e d i n t h e o r e m 8 . 1 3

8 . 1 4 . EXAMPLE. L e t A b e t h e whoZe a Z g e b r a C(Q). T h e n t h e

g r o u p AutD c a n b e r e p r e s e n t e d a s t h e s e t of t h e t r a n s f o r m a t i o n s F: f

+

Ff g i v e n b y

where T: R

+

R i s a homeomorphism o f R , a : R

+

(I:

i s a continuous

11

f u n c t i o n w i t h l a ( x ) I = 1 f o r a l l xcR a n d a e C ( R ) w i t h

a / /< I .

-

P r o o f : By t h e o r e m 8 . 1 3 w e h a v e A o = { f e C ( n ) ; f c C ( R ) ) = = C , ( R ) ; t h u s , b y p r o p o s i t i o n 7 . 2 3 , t h e o r b i t of t h e o r i g i n i s

(AutD)O= C(R) ll D = D T h e r e f o r e , a n y FeAutD c a n b e u n i q u e l y r e p r e s e n t e d i n t h e f o r m F= L o M where M i s a

Mobius t r a n s f o r m a t i o n a n d L i s a surjective

l i n e a r i s o m e t r y o f C(R)

.

But i t i s known ( c f .

I13 I ) t h a t a n y L

i n t h o s e c o n d i t i o n s admits a unique r e p r e s e n t a t i o n

f o r some homeomorphism T of R a n d some acC(R) w i t h l a ( x ) I = 1

CHAPTER

154

8

f o r a l l XCQ, whence w e o b t a i n ( 8 . 1 4 ) . The c o n v e r s e i s o b v i o u s .

# 8 . 1 5 . EXAMPLE. L e t R = :

h be the closed idnit d i s k of a n d d e n o t e b y A / , h e a Z g e b r a of all functions f : h C that are continuous o n h and hoZurrror7phic o n A. T h e n A u t D is the s e t of the transformations +

(8.15)

Ff=

CL.

f

0.1

+a

f eD

e+Z. ( f O T ) w h e r e rcAutA is a conformal map of the u n i t d i s k and a, a are complex numbers w i b h / a / =1 ,

lal
P r o o f : By t h e o r e m 8 . 1 3 w e h a v e A o = { f c A ; E e A } . F o r a f u n c t i o n f c H ( A ) , i t s c o n j u g a t e ? i s h o l o r n o r p h i c on A i f a n d o n l y i f f i s c o n s t a n t ; t h u s , i n o u r c a s e A = C and t h e o r b i t 0

of t h e o r i g i n i s

Then, a n y FcAutD c a n b e u n i q u e l y r e p r e s e n t e d i n t h e f o r m

F= LoM where M i s a t r a n s f o r m a t i o n

Mi=

f +a e+af

w i t h acA a n d L i s a s u r j e c t i v e l i n e a r i s o m e t r y o f A . By theorem 8.12,

such an

L c a n be u n i q u e l y r e p r e s e n t e d i n t h e

form L f = a.@f

feA

where acA i s a f u n c t i o n s a t i s f y i n g I c l ( x )

I=

1 f o r a l l xeA a n d $

i s a n a l g e b r a a u t o m o r p h i s m o f A . S i n c e i n o u r case CL i s holomorphic on A and h a s c o n s t a n t modulus, a i s c o n s t a n t . Thus, t h e r e s u l t f o l l o w s from t h e p r o p o s i t i o n below. 8 . 1 6 . PROPOSITION. L e t @ b e a surjective a l g e b r a

automorphisms of A . T h e n @ c a n be uniquely represented i n the form

AUTOMORPHISMS O F C L A S S I C A L BANACH SPACES

where

T:

A

+

155

A i s a conformal map of A o n t o i t s e l f .

P r o o f : Suppose 4 i s an a l g e b r a automorphism of A . L e t fsA and XsC be g i v e n . I t i s c l e a r t h a t X b e l o n g s t o t h e r a n g e f ( h ) of f i f and o n l y i f Xe-f

i s n o t i n v e r t i b l e i n A. Since 4

i s a n a l g e b r a automorphism, he-f i f $ ( A e - f ) = Xe-4f

i s invertible.;

i s i n v e r t i b l e i f and o n l y t h u s f and 4f have t h e same

r a n g e . Applying t h i s t o t h e i d e n t i t y f u n c t i o n idAeA w e g e t t h a t

4 i d A i s c o n t i n u o u s on h , holomorphic on A and

T=:

In particular, T

T

i s an open s e t . Thus

(A)

T

( h )= h .

i s n o t c o n s t a n t a n d , as it i s holomorphic, T

( A ) C A . Next l e t zeA and fcA

be

g i v e n . Then - r ( z ) s A and w e have f-f [ T ( z ) ] =

z[id*-~(z)]g

f o r some gcA. Applying $I w e o b t a i n

so t h a t $ f - f [ T ( Z ) ] v a n i s h e s a t z , i . e .

f o r a l l feA and zcA. Thus

4 i s t h e mapping $ f = f o - r . S i n c e

$ is

s u r j e c t i v e , by c o n s i d e r i n g i t s i n v e r s e 4 - l w e o b t a i n

4-1

= foo

,

fcA,

f o r some f u n c t i o n o which i s c o n t i n u o u s on h ,

holomorphic on A and a ( A ) c A . Then w e have f = f o ( T o 0 ) f o r a l l fsA. I n p a r t i c u l a r , f o r f = i d A w e g e t and

T

TOO=

i d A , whence ?-(A)= A

i s a conformal map of t h e u n i t d i s k .

#

By a p r o c e s s of c o m p a c t i f i c a t i o n , w e can a l s o c o n s i d e r a l g e b r a s of c o n t i n u o u s f u n c t i o n s on some non compact t o p o l o g i c a l s p a c e s .

8 . 1 7 . EXERCISES.

( a ) L e t X be a completely r e g u l a r

Hausdorff s p a c e and d e n o t e by B ( X ) t h e Banach a l g e b r a of a l l bounded c o n t i n u o u s f u n c t i o n s f : X

+

endowed w i t h t h e supremum

norm. Show t h a t , i f BX d e n o t e s t h e Stone-CEch c o m p a c t i f i c a t i o n

CHAPTER

156

8

of X, then there is a surjective isometric algebra isomorphism

J: B ( X ) + C(@X) which commutes with the natural conjugation of B ( X ) . (b) Show that, if A is any closed

complex subalgebra of B ( X ) such that ecA, then the subspace A. associated with A by definition 1 . 5 is given by A = { Z c A ; fcA}. Thus

and every FeAutD can be represented in the form F= LoM where L is a surjective linear isometry of A and M is a Mtjbius transformation

f +a Mf= e+Zf where aeAo,

11

feD

a \ /< I .

Some interesting examples are included in the above situation.

For instance (c) Let X=: A be the open unit disk of (I: and denote by H m ( A ) the algebra of all bounded holomorphic functions f: A (I:. By an argument similar to that of proposition 8.16, show that any surjective algebra automorphism $ of Hm(A) can be represented in the form $ f = f o ? , fCHmr for some rcAutA. Thus we have (AutD)O= A and any FcAutD can be written in the form -+

Ff=

~1

f o t +a e+a.fo-r

fcD

where rsAutA and a and a are complex numbers with l a / = 1 la(
,

(d) Assume that X is a discrete topological space. Show that for all FeAutD, there exists a unique permutation T of X and a function M: X -+ AutA such that

AUTOMORPHISMS OF CLASSICAL BANACH SPACES

(Ff)x= M(x f [T (XI]

xcX,

157

feD.

54.- Operator valued Mijbius transformations.

Let H be a fixed complex Hilbert space with scalar product < , > and write B=: B[L(H)]. The transformations defined, for 1 AcB, analogously as in C by M' (X)=:

(X+A)( 1+A*X)

,

M"(x)=: (I+xA*)-~ (x+A)

,

XcB

seem to be the natural candidates to non-linear elements of AutB. Unlike in the case of a", the situation MA. (X*)#M,'(X) * may occur; therefore, in general, these transformations cannot belong to AutB. However, as it was observed by Potapov 1481, a slight correction transforms them into elements of AutB as we shall see. 8.18 LEMMA. G i v e n X,YeL (H) a n d GcGL (H), we h a v e X>Y and o n t y if G"XG2G"YG.

8.19. THEOREM. F o r e a c h AcB, t h e m a p p i n g

is a b i h o l o m o r p h i c autornorphisrn of B a n d we h a v e MA'=

M-A

Proof: Consider the power series of MA M I (xi= (x+A) ? (-A*x)"= A + A

n=O

m

c

n=O

u

(-I)~(I-AA*)xA*xA* .A*X ( 2 n + l l terms

Similarly

.

if

158

CHAPTER

m

M;(x)=

A+

c

n=O

(-I,)"xA*xA*

8

...

L--I

A*X(I-A*A)

(2n+l) t e r m s

Therefore, MA (X) ( 1 -A*A) = ( 1 -AA*)M i (X)

X€D

that is,

MA (X)= (1-AA*) -'MA

(X) ( 1-A*A) '=

( 1 -M*)'Mi ( X ) ( 1 -A*A)

-'

it immediately follows that M A ( X ) * = MA*(X*) holds f o r all XeB. Now we can show that M A ( D ) C E , i . e .

Thusl by applying the previous lemma with G= : ( 1 -A*A)

(8.17) Since

-'

( 1 +A*X) I (8.16) is equivalent to

(x*+A*) ( I - A A * ) - ' ( X + A I S

(I+x*A) (I-AA*)-~ (I+A*x)

AUTOMORPHISMS OF CLASSICAL BANACH SPACES

X*A ( 1 -A*A)-

'-x* 1 -AA * (

-1

A= o=

'

( 1-A*A) - A*X-A* ( 7 -AA* ) -

159

'x

as it can be seen from the corresponding power series expansions, the difference between the right hand side and the left hand side in (8.17) equals l-X*X which is a positive operator whenever / I X 11 ( 1 . Thus (8.16) is established. Clearly, the Potapov-MBbius transformation M A is holomorphic. So it remains to prove that M A ~ M - A =id, and it suffices to see that MA[M-A(Y)]= Y for all Y in some neighbourhood of -A. Now, the relation M (X)= Y is equivalent to A

that is ,

'

Mi (Y)= Ml [ ( 1 -AA* ) 'Y

I/ Y' (1

whenever

<1

( 1 -A*A) -']

. But now

Y'A*= (I-=*) ~ Y ( I - A * A ) - ~ A * = 0)

=

(l-AA*)fY

c

AA*= (1-AA*) 'YA* ( 1-AA*)

n=O

terms

2n+l

and hence ( 1 -Y 'A*)- ' = n=O =

Therefore

'

( 1 -AA*) (YA*) (1-AA*)

( 1 -AA*)

'

( 1 -YA*) -

'

( 1-AA*)

-'= -'

160

M','

CHAPTER

'

( Y ' ) = ( 1 -=*I ( 1 -YA*)

( 1 -AA*)

8

-'

[ (1-AA* ) tY ( 1 -A*A) -'-A]

=

8.20. COROLLARY. T h e domain B is homogeneous; n a m e l y {MA(0); AC.B}= B . 8.21. LEMMA. L e t AcB be f i x e d . Then f o r a l l

X&B and

Z e L ( H ) , w e have

Proof: W e know that

M' (X)= A + ( 1 -AA*)

m

C. (-1 )

n=O

A*X= A + ( 1-AA*)X ( 1 +A*X)

nXA*XA*

L

(2n+l)

2

terms

Hence

= (1-AA*) [I-X (1+A*X) -'A*] Z ( 1 +A*X)

Since I - X ( I + A * X )- 1 A * = I-XA*+XA*XA*+

...+( - I ) ~ ( x A * )...+ ~+ =

( I + X A*

-1

and since

the proof is complete.

79 8.22. EXERCISE. Show that, if

11 A 11

.

< I , then $ (A*A)A*= A*$ (AA*)

+:

A 4 T is holomorphic and

AUTOMORPHISMS OF CLASSICAL BANACH SPACES

161

Next we are going to study the one-parameter groups of Potapov-M8bius transformations. 8.23. PROPOSITION. L e t A, BsB b e s u c h t h a t b o t h AB* a n d -~ A*B a r e s e l f a d j o i n t o p e r a t o r s . T h e n MAoMB= M MA(B)

'

Proof: By Cartan's uniqueness theorem, it suffices to show that the automorphisms MAoMB and MMA(B) have the same image and the same derivative at the origin. Obviously MAIMB(0)]= MA(B)= MMA(B)(0). On the other hand, from lemma 8.21 we get

In the course of the proof of (8.16) we have established that G*

[I -MA (X)*MA (X)] G= 1 -X*X

where

-'

G= : ( 1 -A*A)

(1 +A*X)

and hence

-'( 1 -B*B)

I - M (B) ~ * M (B) ~ = (I -A*A)f ( I +B*A)

( I +A*B)

(I -A*A)f

Observe that, from our assumption, the operators B*B , B*A= A*B and A*A commute. Therefore [I-M,(B)*M~(B)I+

=

(I-A*A)t (I+A*B)-~(I-B*B)+

[l-MA(B)MA(B)*] f = [l-MA* (B*)*MA*(B*)]

'=

(1-AA*) '(l+Al3*)-'(1-BB*)'

Thus

=

(I-AA*)~(I+BA*) (I+BA*)-~ (I-BB* t Z(I-B*B)t (I+AB*)-~(I-A*A)+

On the other hand, by the chain rule and lemma 8.21, we also have

CHAPTER

162

= (I-AA

8

* #(I+BA*)-’(I-BB* f Z ( I - B * B ) f (I+AB*)(I-A*A)#. # 8.24.

THEOREM. L e t t+MA

be a c o n t i n u o u s one-pararnctcr t

g r o u p of P o t a p o u - M d b i u s

t r a n s f o r m a t i o n s . Then t h e o p e r a t o r

is s e 2 f a d j o i n t f o r a 2 2 t , s m if a n d o n 2 y if, f o r some A:As DcL (H), we h a v e M s *

v h e r e D- D

= At

exp t(D-s D*

t m

)

s t a n d s f o r t h e v e c t o r f i e l d X+D-XD*X,

P r o o f : Let t+MA

t

Then

x~L(H).

be a continuous one-parameter group.

MA = exp tV

t m

t

holds for some VeautB

,

Thus, by setting D=: d-

At= tD+w(t) where

lim t+O

=

so that

(O)=

1 t w(t)=

d dt l o

At

,

we have

0, and

D-XD*X

Conversely, we shall prove that, given DsL(H), the mapping t+exp t(D-’D*), t m , is a group of Potapov-MGbius transformations.

AUTOMORPHISMS OF CLASSICAL BANACH SPACES

163

Let D= JlDI be the polar decomposition of D, i.e., ID1 = : (D*D) and J is a suitable isometry of the (closure of the) range of D into H. Remark that D"= JIDI" for ndN. Since the function tanh (hyperbolic tangent) is the solution of the differential equation

'

and tanh(t)e[O,l) for tdR, the function TR-tL(H) given by A = : Jtanh(t(D1) satisfies t

A = D-A D*A t t' dt t

At€B

A = 0,

for all tdR. Moreover, A:As=

tanh (tID1 ) otanh ( s /DI

,

A sA*= Jotank(t ID1 ) otanh ( s ID/)* t

for s,tdR. Thus, by propostion 8.23,

Therefore, it suffices to prove that (8.18)

MAt(AS)'

At+S

s,tm

because (8.18) implies that t+MA is a one-parameter group of d Potapov-M6bius transformations shch that (O)= D and therefore MA

=

t

Now we have

exp t(D-'D*)

tf3R

164

CHAPTER

8

Here

8.25. COROLLARY. W e have

autB= ID- s D*

;

DeL(H) 1

55.- J*-alqebras of operators.

In view of theorems 1 . 2 8 and 8.19, in order to achieve the complete description of the elements of AutB we need only to compute the subgroup Aut0 B which can be identified with the family of all surjective linear isometries of L ( H ) . T o solve this problem, it is a natural approach to look for certain elements of L ( H ) with rather specific geometrical properties that must be preserved by all surjective linear isometries of L ( H ) . Namely, one conjectures that the operators

165

AUTOMORPHISMS OF CLASSICAL BANACH SPACES

u@v*: h+u,

hcH

with u,vcH, meet this requirement. However, a direct geometrical characterization of them is rather sophisticated. On the other hand, their algebraic description is quite convenient as Corollary 8.25 enables us to extend the use of Potapov-MBbius transformations in a far reaching algebraic direction. 8.26. DEFINITION. A c l o s e d c o m p l e x s u b s p a c e E o f L(H) i s s a i d t o bje a J * - a Z g e b r a o n H i f X A * X c E w h e n e v e r X , A s E . We endow E w i t h t h e “ t r i l i n e a r p r o d u c t ” 2

(x,A*,Y)=: XA*Y+YA*X

With respect to this product, E for X , A , Y s E

I

is a t e r n a r y a l g e b r a . Indeed,

we have

2 (x,A*,Y)= ( x + Y ) A * (x+Y)-XA*X-YA*Y~E A

linear

m a p p i n g L: E

+

E

of E

i s a J*-automorphism

if

L( E ) = E a n d

L(x,A*,Y)= (LX, (LA)*, LY) h o l d s for a l l X , A , Y e E A J*-deriuation

of

E

. i s a l i n e a r m a p p i n g L: E

-f

E

such t h a t

L(x,A*,Y)= (LX,A*,Y)+(X,(LA)*,Y)+(x,A*,LY) h o Z d s f o r a l l X,A,Yc E

.

Many familiar spaces are J*-algebras. For instanceI every C*-algebra is obviously a J*-algebra with respect to the natural triple product. Thus, by the Gel’fand-Naimark theorem, every B*-algebra is a J*-algebra, too. Also any Hilbert space H can be identified with {xBe”; xeH) where e is any fixed unit vector of H; therefore H is a J*-algebra, too.

166

CHAPTER

8

C a r t a n factors, which can be considered as the natural infinite

dimensional generalizations of the spaces appearing in Cartan's classification of bounded symmetric domains (cf. Chap. 9 1 , are of particular interest among the J*-algebras on H. To define them, let us recall that a conjugation on H is a conjugate 2 linear mapping Q: H+H such that IIQ I / = 1 and Q = 1. Given any conjugation Q on H I we can find a complete orthonormal system S = { e . ; jcJl i n H such that 1

Qe.= e . 3

~ ( i e . )-ie. =

3

3

1

jcJ

Thus

and, with respect to S, the matrix of the operator QA*Q is the transposed of that of A , i.e. = , 3,

k

3

j ,kcJ

T

The operation A+A =: QA*Q, AcL(H), is called the transposition associated with Q. 8.27.

A compZex subspace F of L ( H ) is said

DEFINITION.

to be a Cartan f a c t o r of:

type I

if F = {XcL(H);

projectors P

X= P XP } for some f i x e d ortkogonaZ 2

1

PP2' T

type I1 if F = { X e L ( H ) ;

X = XI f o r some transposition o n H.

type I11 if F = tXsL(H);

'I

X = -X) for some transposition o n H.

type I V if F is a c7,ost.d c o m p l e x subspace of L(H) such that { X 2 ; X C P ) C C . ~ and { X * ; X e F I c F . 8.28.

J*-algebras.

EXERCISE.

Verify that Cartan factors are

AUTOMORPHISMS OF CLASSICAL BANACH SPACES

167

8.29. LEMMA. (a) A c l o s e d c o m p l e x s u b s p a c e E o f L(H) i s a J * - a l g e b r a if a n d o n l y i f AA*AcE w h e n e v e r AcE

.

(b) G i v e n a J * - a l g e b r a E o n H, a map LcL(E J * - i s o r n o r p h i s m o f E if a n d o n l y if L( E ) = E a n d L(AA*A)= LA(LA)*LA h o l d s f o r a l l Ac E

)

is a

.

(c) A m a p LcL( E) is a J * - d e r i v a t i o n if w e h a v e L (AA*A)= (LA)A*A+A(LA)*A+AA*(LA),

on

E if a n d o n l y

ACE

Proof: The statements are immediate consequences of the fact that, for X,A,YcL (HI ,

8.30. THEOREM. L e t E d e n o t e a J * - a l g e b r a e v e r y Potapov-Mdbius

transformations M

B( E ) o n t o i t s e l f a n d Aut B( E ) = tM oL; AsB( E ) A

aut B ( E ) = {A+L-'A*; s

w h e r e A+L- A

*

Ac E

A

on H.

Then

w i t h A c B ( E ) maps

, L is a J*-automorphism of E 1 , L is a J*-derivation of E }

d e n o t e s t h e v e c t o r f i e l d X+A+L(X)-XA*X, XcB( E ) .

Proof: Let AcB(E ) be arbitrarily fixed and let A= J / A I be its polar decomposition. By setting D=:

m

I:

n=O

'

(AA*)"A, we have 2n+l D=J

-C n=O

1 IA/2 n + 1 = J t a n h - l 2n+1

]A1

Therefore A= J tanhlDl and so MA= exp(D-',D*). From the power series expansion we see that D c E , whence it follows that D-XD*Xc E , whenever Xc E Thus, M A maps B ( E ) into itself.

.

CHAPTER 8

168

of MA maps B ( E ) into itself, so Similarly, the inverse M -A that MA[B( E ) ] = B( b').

Since {MA(0) ; AeB( E ) I = B( E ) and since {A-'A*;

A s F;)=

aut B ( E ) , it suffices to see that, given

L c L ( E ) , we have

0

0

is a J*-automorphism of E

(8.19)

LeAut B ( E ) < = > L

(8.20)

LeautOB( E ) < = > L is a J*-derivation on E 0

Proof of (8.19): Suppose that LcAut B ( E proposition 7.9(d) we see that

Thus

(LA)(LA)* (LA)= L (AA*A) for all ACE

)

. From

.

Conversely, suppose that L is a J*-automorphism. Then L( E ) = E . Moreover, for XeL(H) with X= JlXl, we have XX*X= J I X I ; therefore

so that

holds for all A c E . Thus 11 L 11 G I . But L-l is also a J*-automorphism of E and, by the open mapping theorem 1 L-leL( E ) Thus / I L- 11 G I and L is a surjective linear isometry of E .

.

.

Proof of (8.20) Suppose that Leaut0B( E ) , and define 0 G =: exptL for tdR. Then GtcAut B( E ) and s o , for each fixed A C E , we have t

AUTOMORPHISMS OF CLASSICAL BANACH SPACES

-

=

( -

=

(LA) A*A+A (LA) *A+AA* ( L A )

dt

lo

G~A)A*A+A(

dt

lo

G ~ A ) * A + A*A(

dt

lo

169

t G A)=

Conversely, let L be a J*-derivation on E and set again t 0 G =: exptL for tdR. We must show that, for tdR, GtcAut B( E ) , or equivalently, that Gt is a J*-automorphism of E For fixed A c E , we have

.

d dt

) G - (~G ~ A , ( G ~ A *,

G ~ A =) - L G ~ ( G ~ A , ( G ~ A *, ) G ~ A +)

*,

+ G - ~( L G ~ A , ( G ~ A )

) + G - ~( G ~ A , ( G ~ A *,

) G ~ A +) G - ~( G ~ A , ( L G ~ A *,

G ~ A +)

LG t A ) =

= - G - t ~ ( G ~ A , ( G ~ A *, )

G ~ A +) G - ~ L( G ~ A , ( G ~ A *) , G ~ A =)

o

for all A c E and tdR. Thus G -(~ G ~ A, ) ( G ~ A *, ) G ~ A =) G O ( G O A , 0

whence GtcAut B( E )

(GOA)

*,

GOA) = (A,A* , A )

. iy

8.31. EXERCISE. Let E l and E 2 be J*-algebras on H and suppose that L c L ( E l l E 2 ) is a bijective mapping. Then L is isometric if and only if it is a J*-isomorphism.

§6.-

Minimal partial isometries in Cartan factors.

Recall that an operator JcL(H) is called a p a r - t i a i ! isometry if, for some subspace H of H, the restriction J is an isometry 0

and J

1

= 0.

IHO

It is a well known consequence of the existence

IHo

of the polar decomposition that

CHAPTER

170

(8.21)

8

A i s a partial isometry < = > AA*A= A,

AeL(H)

From (8.21) we can easily obtain the following 8.32. LEMMA. LeL

h e a J * - a L g e b r a o n H. T h e n a n y

E

linear a u l o r n o r ~ p h i s mo f B ( E ) p r e s e r o e s t h e s e t o f partial i s o t n c t r i e s of

F

.

8.33. DEFINITION. Let.

E

b e n J * - a Z y e b r a and asrjtlme i h a t

c b' a r e p a r t ' i u l isometrics. L e t U S s e t Hk =: ixcH; I/ JkxII = Ilx 11 f u r k = 1,2. We s a y t h a t J 1 is a p a r a t uf J2 if H l c H 2 a n d J = J hie w r i t e

J ,J

1

2

21H1

J1<Jz

if J 1 is p a r t of J

lIH1

.

2'

Clearly, the relation < is a partial ordering on the set of non-zero partial isometries of E . The minimal elements with respect to are called m i r i i m a Z partial i s o m e t r i e s of E

.

We recall that a net ( A , ), in L ( H ) is said to be convergent I JCJ to A with respect to the weak operator topology if we have llm = for all x,yeH. We write Tw for the weak 3 1 operator topology on L(H). 8.34. THEOREM. If t h e J * - a Z g e D r a E is c l o s e d i n L(H) with r e s p e c t to T t h e n E i s t h e T W - c l o s u r e of t h e linear W' h u l l . of p( E ) and mp( E

)=

{AcE ;

AA*A=A+O,

A E*A= QAI

Proof: Let Aet' be arbitrarily fixed and suppose that

where A+P(A) is the spectral measure of l A / , is the polar decomposition of A (cf. I13 1 )

.

Consider the sequence Y =: n Y,+~= A Y ~ A,

n= 0,1,.

.

Since

AUTOMORPHISMS O F CLASSICAL BANACH SPACES

171

h o l d s f o r a l l ndN. T h e r e f o r e , i f t h e sequence of odd polyno-

mials ( p n I n m

i s bounded on [0,

p o i n t w i s e on LO,

1 1 A /I ]

11 A I / ]

and p,

converges

t o some f u n c t i o n J i , w e have

I n p a r t i c u l a r , JP[a,P]cE

f o r e a c h a,BdR

w e l l known t h a t each o p e r a t o r JP [a,P J

.

.

+

I

a,Pm,

However, it i s is a partial

i s o m e t r y , and t h a t T -

~i Bore1 function}= SpanCP[a,B]; a,BclR+}

W

T h e r e f o r e A b e l o n g s t h e T w - c l o s u r e of S p a n ( E ) . ?1

Suppose now t h a t J c m p ( E ) and l e t XeE A=:

% * ' L

be g i v e n . D e f i n e A by

'L

J X J . C l e a r l y A C E and k e r J c k e r A , whence

range(A1 c ( k e r A ) l c ( k e r 5 ) l = {xeH;

[I ?XI/

=

I / x [I 3

Thus, i f t h e p o l a r decomposition of A i s a g a i n A= J / A ] , t h e n

%

T h e r e f o r e w e have J < y J

f o r some yea,

'L

Iyl= 1 , and, a s J i s

2.

minimal, J = y J . But now f o r any s p e c t r a l p r o j e c t i o n P[a,P]

of

\ A \ , t h e o p e r a t o r J P { ~ , R J i s a p a r t i a l isometry c o n t a i n e d i n 'L

J= y J .

I t follows t h a t JP({II All

I)=

J . Therefore

#

172

8

CHAPTER

E

8.35. P R O P O S I T I O N . L e t T -elaced %w

Jcmp( E

)

be a n y J * - a l g e b r a

w h i c h is

a n d a s s u m e t h a t f o r any Jep( E ) t h e r e is s o m e %

B = (Span mp( E

sucrlz t h a t J < J . T h e n

)TTW.

Proof: L e t F be the family of the finite sums 2.

...+J,, ndN, of mutually orthogonal minimal partial isome'L tries Jkc m p ( E ) , lSk$n, i.e.

%

J1+

for all pairs k , & with k f k . Then, the linear hull of F is T -dense in E because, given any Jcp(E 1 , the net W is not empty and it is weakly convergent F(J)=: { ? e F ; J"<J} to J .

8.36. COROLLARY. In p a r t i c u l a r , if e v e r y Jcp( E 1 'L

'L

c o n t u i n s some Jcmp(E) s u c h t h a t dim range J < m

E

=

(Span mp( E 1 )-W

then

.

8.37. P R O P O S I T I O N . A l l Cartan f a c t o r s of L ( H ) a r e

T -cZosed. W

Proof: Given any operators R 1 , R l c L ( H ) and a conjugation Q on H , the mappings L ( H ) + L ( H ) given by

LR1R2

: X - + R ~ X R ~,

T

-

Q'

X+QX*Q

are T W -continuous. On the other hand, if Fk is any Cartan factor of type k, 1skS3, we have

F1= {XcL(H);

X=

XI

for some projectors P 1 , P 2 F 3 = { X s L ( H ) ; T X= -XI

Q

As for Cartan factors F4 page 334.

,

the proof can be found in 1 1 9 1 ,

ff

AUTOMORPHISMS O F CLASSICAL BANACH SPACES

8.38.

DEFINITION.

Given a J*-algebra

E

*

t h a t a n o p e r a t o r A c E i s m i n i m a l if A E A = EA.

for t h e s e t of m i n i m a l e l e m e n t s o f

173

o n H, we s a y

We w r i t e m ( E )

E .

It i s i m m e d i a t e t h a t the set m ( E )

J*-isomorphisms

8.39.

E

L of

THEOREM.

.

i s preserved by a l l Moreover, if E i s T w - c l o s e d , t h e n

L e t P1,P2 and Q b e , r e s p e c t i v e Z y ,

o r t h o g o n a l p r o j e c t o r s w i t h H . = : range P 1

c o n j u g a t i o n on H .

F = { X e L ( H ) ; P2XP1= X I

F3 = { X S L (H)

; QX*Q=

t h e corresponding Cartan f a c t o r s . C a r t a n f a c t o r of t y p e I V . m ( F ) = {f@e*; ecH1,

A2= 0 )

XI

-XI

Furthermore

,

Let F 4 be a n y

Then

fcHZ),

m(F,)= m ( F q ) = {AcF4;

( j = 1 , 2 ) and a

p 2 = I X C : L ( H ) ; QX*Q=

1

1

j

Denote b y

m(F ) = {e@(Qe)*; 2

{f@(Qe)*-e@(Qf)*;

eeH}

erfc.HI

provided t h a t dimF4>l.

Proof: I n g e n e r a l , f o r

e , f c H w i t h efoff, the t w o

relations

C a s e k= 1. From t h e -~

above remarks it i s e a s y t o check

t h a t , f o r e e H l and f e H 2 , w e have f P e * e m ( F 1 ) .

Conversely,

let

A c m ( F 1 ) w i t h AfO be g i v e n . T h e n A= P Z A P l and w e c a n choose some

CHAPTER

174

8

xsH such that P2APlx#0. In particular, the vectors e=: p l x and f=: P,Ae satisfy AefQ and A*ffO so that L

(Ae 0 (A*f)*#O Moreover f@e*cF1, and from the minimality of A we obtain

whence A= (XAe)@(A*f)* for some XeE.

Case -k=

2. Let ecH be given. Then Q[e@(Qe)*]*Q= eP(Qe)* so that e@(Qe)*cFZ and, as remarked at the begining of the proof e@(Qe) * e m ( F 2 )

.

Conversely, let Aem(F2) with AfQ be given, and choose eeH such that AefO. Since,F 2 is *-invariant, re0 (Qe)*] * C P z EA3A[eP(Qe) *JA= (Ae)@(A*Qe) * Moreover, from AcF 2 we derive A*Q= QA, whence A= f@(Qf)* with f = : XAe for a suitable A d . Case k = 3 . For the sake of shortness, we introduce the notation [v,u]=:

v@(Qu)*-LIB(Qv)*

v ,ucH

NOW, l e t f,eeH be given; it is easy to see that [f,eJcF3. For arbitrary AcL(H) Q[f,e]*Q= - [ f , e J , hence [f ,e]A* [f ,el= M-N holds with M= f@(Qe)*+eP(Qf)* N = fO(Qf *+e@(Qe) *

AUTOMORPHISMS OF CLASSICAL BANACH SPACES

175

Since F3 is a J*-algebra, for AeF3 we have [f,e]A*[f,e]eF3. Moreover, from AeF3 we obtain A*Q= -QA; it follows that M= [f ,el eF3 Therefore (8.22)

N= M-[f,e]A*[f,e]cF2 [f,e]A*[f,e]=

n F 3 = 101. Thus, for AeF3

[f,e]cC[f,e]

On the other hand, the relation [f,e]= 0 holds if and only if f,e are linearly dependent. By putting A=: [f,e] in ( 8 . 2 2 ) we obtain

[fie][f,e] * [f,e] =

A

[fie]

with A = /I ell 2 11 flI 2 -l<elf>121whence and [f,e]sm(F3).

[f,e]A*[f,e]=

C[fle]

Conversely, let Aem(F3) with AfO be given. For e,feH we have CA3A[f,e]*A=

(AQe)@(A*f)*-(AQf)@(A*e)"

Now we can choose e l f such that the vectors AQe and AQf are linearly independent. (Indeed, if we had dim range A= 1 , then A would have the form A= u@v* for some u,veH. But u@v*eF3 holds if and only if u@v= 0). Hence QAaf 'Be I *-e '@f * = for some independent couple f',e'eH

[f ' ,e '3 and A= A[f',e'].

# Case k = 4 . Suppose that AeF4 and A XeF4, we have

2

=

0. Then, given any

AX*A= (AX*+X*A)A= [(A+X*) 2-X*2]AeCA i.e., Aem(F4). Conversely, let Aem(F4) be given; then AF*A= CA and A 2 = a1 for some a d : .Thus, for some Bee, 4

CHAPTER

176

8

2 * 2

@A= (A*AA*)*A= A A A

= a2A*

If ct 2 f o r we have A*= yA for some yea. Since we have assumed that dim F4>l, we may fix X c F q \ Q A if afO. But then we get the contradiction

* * * *

2

2

QA3A(A X A ) A= A XA =

CI

2

X

Proof: Exercise 8.41. THEOREM. With t h e p r e c e d e n t n o t a t i o n s , 0

( a ) E v e r y e l e m e n t of Aut B(Pk)' k = 1,2,3, i s a c o n t i n u o u s

operator

w i t h r e s p e c t t o t h e T w - t o p o l o g y o n Fk.

( b ) We h a v e F k = (Span m p ( F k ) ) - T w f o r k= 1,2,3.

Proof: (a) Suppose that ( A ) is a net such that j jeJ TW - 1 i m A I, = A in F k I and let LcAutoB(Fk) and f,ecH be 1

arbitrarily fixed. We have to show that < L ( A , ) f ,e>-+ 3

Case k= 1: Write f'=: P 2 f and e l = : P 1e. Then f'Re'*em(F1) and we can find vectors f " e H 2 e"cH1' such that

For any pair vcHZ

ucHIr and any operator XcFl we have

AUTOMORPHISMS OF CLASSICAL BANACH SPACES

177

(V@U*)X* (V@U*)= VPU* Applying this, first to f'Pe'* and A, and then to fl'@e''* and L(A) , we obtain ,Ae ' >f '@e I *

(8.24)

( f 'Pe' * ) A* ( f 'Qe ' * ) = < f I

(8.25)

(f"@e"*)(LA)*(f"@e"*)= f"@e"*

,

As L is a J*-automorphism, from (8.23) and (8.24) it follows = c (f"Pe"*)(LA)* (fa1@eIt*)

whence, by cornpairing with (8.25) = = for all I jsJ. The result follows by letting t+m and taking into account that T-lim A , = A. The proof €or the cases k= 2 and W

j

7

k= 3 is quite similar. (b) Let Jsp(F,) be arbitrarily given. It suffices to 'L 'L find some Jsmp(F ) with dim range J<m. Let us choose k

es{xcH; / / Jx /I = / \ x / / } with /I e / / = 1 . Then, the following choices satisfy the requirement: If k = 1, J=: (Je)Pe*. If k = 2, we define P: H+H to be the projector onto the subspace 21 Span{e,QJe} and put J=: JP. (Indeed, then J(QJe)= Qe so that 21 21 {xsH; 11 JxII = 11 xi1 1 and range J are Q-invariant) Finally, % for k= 3, we can take again J=: JP where P is as before. N o w J (QJe)= -Qe.

.

# 0

8.42. -

COROLLARY.

F o r k= 1,2,3,

every element

LeAut B(F ) i s u n i q u e l y d e t e r m i n e d b y its r e s t r i c t i o n t o t h e k s u b s e t mp(Fk).

#

CHAPTER

178

8

57.- Description of Aut 0 B ( F 1 ) and aut0B ( F 1 ) . _ _ - .~ ___ _ _ _ _ __ _ _ _ _ _ _ _Let us set F 1 = : {XeL(H); P 2 X P 1 = XI where P 1 , P 2 are orthogonal projectors on H, and write H =: range P . ( j = 1 , 2 ) . Furthermore, 1 0' let L denote any fixed element of Aut B ( F 1 ) . 8.43. LEMMA. L e t feH2 --f o I. I. ow 1: n g s tu te me n ts ho i! d s :

I

ff0, be g i v e n . T h e n , o n e o f t h e

(lf) T h e ~ eis a v e c t o r f'&HZ and a s u r j e c t i v e Z i n e u r i s u m e t r y Uf : H1+H1 s u c h t h a t

( 2 f ) G i v e n a c o n j u g a t i o n Q, on H2

,

there i s a vector

f'sH2 and a s u r j @ c d i z ; e l i n e a r i s o m e t r y Uf: H1+H2 such that L(fQe*)= (Q2Ufe)@f'*

ecH

Proof: Fix any feH2 , f f 0 , and consider any finite n and X=: fQ C u; are set {uk; k = 1 , n)cH 1 ' Since fQu* k 1 minimal operators in F we can write

...,

1'

for suitable g,gkeH2 and vIvkeH , lcksn. Then, as both 1 range L (X) and range (LX)* are one-dimensional spaces, at least one of the sets Eg,}, { v k l must consist of mutually parallel vectors (i.e, g j I j g k V . or vj I/ vk v j l k ) . Irk

Now we proceed to prove the lemma. Let us fix any pair of independent vectors el,e2eH (We may assume dimH1>l as 1' otherwise F 1 would be isomorphic to H2 whose linear isometries are well known). Write

179

AUTOMORPHISMS OF CLASSICAL BANACH SPACES

There are two possibilities: (If) The situation u 1 I / u 2 does not hold. Then, as previously proved, glll g, must hold and by (8.25) we can write

for a fixed f' and suitable ei, Now, let ecHl be arbitrarily given and put L(f@e*)= gBu*. As {e;,ei,u) cannot consist of mutually parallel vectors, we must have 911 f' and therefore

holds for a unique e'eH1. Thus we can define a mapping Uf: H1+H1 by setting

It is easy to verify that U is a surjective linear isometry f of H1 (because L is a surjective linear isometry of F1). (2 ) The situation ul\lu2 does h o l d . The construction f of the required elements can be carried out in a similar manner. However, a conjugation Q, on H, must come up to compensate the fact that the variable factors e l e' of the * tensor products L(f@e ) and e'@f' appear in opposite places.

# 8.44. THEOREM. L e t

LcAut0 B(F1) b e g i v e n . T h e n o n e of

t h e f o l l o w i n g s t a t e m e n t s ho I d s : ( 1 ) T h e r e a r e s u r j e c t i v e l i n e a r isometrics U: Hl+H

V: H2+H2 (8.26)

such t h a t

L(A)= VP2A U P 1 ,

AcFl

( 2 ) G i v e n any c o n j u g a t i o n s Ql,Q2 on H1 a n d H,

,

1

and

180

8

CHAPTER

r e s p e c t i u e l y , t h e r e are surajecliue linear isornetries

U: H1+H2 a n d V: H2*H

(8.27)

1

s u c h that

L (A)= VQ1A*Q2UP

,

AsF

Proof: Let us fix an arbitrary unit vector f cH 0 2' Replacing L by L' : A+Q, (LA)*Q2P2 if necessary, we may assume L to satisfy condition ( 1 ) in lemma 8.43, i.e. f0

Again, replacing L by L": A+VOL(A)U P I fO V f'= f, we may assume that

if necessary, where

0 0

L (foBe*)= foBe*,

(8.29)

ecH

1

Furthermore, it suffices to establish (8.26) for the operators AcFl of the form A= fBe* where fsH, , esH1. too. Applying lemma 8.43 Now let us f i x a unit vector eOcH 1' (interchanging the spaces H 1 and H2) to the vector eo, we see that one of the following possibilities holds: There exist a unit vector eisH1 and a surjective linear isometry V : H2+H2 such that (1,

0

)

eO

(8.28')

L(fQeE)=

(v,0 f)@e;* ,

~

C

H

~

(2e ) Given a conjugation Q on H , there exists a 0 1 1 vector e'cH and a surjective linear isometry Ve : H2+H such 0 1 0 1 that

First we show that (8.28") is impossible. Suppose that (8.28") holds. Then we may a ssume eA= f o as it can be seen by applying (8.29) to eo and (8.28") to f o . Now as L is an

AUTOMORPHISMS OF CLASSICAL BANACH SPACES

181

isometry, we have

whence it follows

However, the right hand side of this latter equality is a continuously Frgchet real-derivable function of (elf)whereas the left hand side does not admit a Frechet real-derivative. Thus, we necessarily have (8.28'). Replacing L by

L"'

:

A+v;'L(A)

if necessary, we may assume that

0

L(foBe*)= fo@e*

,

,

L(fBe,*)= f@et

esH 1'

fsH2

.

To do this, consider It only remains to prove that L = idF 1 Since eo was arbitrarily fixed, the same any e 1sH1 , flsH2 argument that was used to establish that (8.28") was impossible shows that, €or some surjective linear isometry V el (and similarly) U ,

.

fl

L(fl@ey)= (V fl)@e:= el

fl@(Uf ey) 1

As the ranges and kernels of the tensor products on the right hand sides coincide, it follows that L(f 1 eel)= yfl@et for some ysQ, \ y / =1. Next we show that y does not depend on elf. Indeed, if e 1 eo and f l1 f, , then for some matrix ( a . ) Osj, ksl, we have

1

Ik

,

182

CHAPTER

8

Hence, i t r e a d i l y f o l l o w s t h a t y = 1 ( c f . e x e r c i s e b e l o w ) . T h i s f a c t can be i n t e r p r e t e d a s

,

whenever e l l e g and f l l f ,

aikeC, O X j ,

a r b i t r a r i n e s s of e l , f l ), L = i d

.

5

ksl,

t h a t i s (by t h e

#

8.45. EXERCISE. Prove t h e r e l a t i o n y = 1 . H i n t : t h e

[i i]

cannot be w r i t t e n a s ( a 1 a 2 ) Q ( a 3 a 4 ) * if y f l .

matrix

8.46. EXERCISES.(l) Show t h a t t h e s e t of t h e i s o r n e t r i e s L:

Fl+fF

1'

for which ( 8 . 2 6 ) h o l d s i s t h e i d e n t i t y component of

0 Aut B ( P 1 ) .

D e s c r i b e t h e o t h e r c o n n e c t e d components i f any.

( 2 ) L e t MeL(H ) and N e L ( H ) be s e l f a d j o i n t o p e r a t o r s and 2

t h e mapping

1

e i t M P 2 A eitN P1 f o r t e IR and A s F l . Show t h a t

w r i t e Lt(A)=:

IR+AutoB(F1) g i v e n by t + L t

i s a continuous

one-parameter group of i s o m e t r i e s of F1 whose a s s o c i a t e d v e c t o r field is

With t h e above n o t a t i o n s , show t h a t t h e v e c t o r f i e l d s f

MIN

and f n ,

have t h e same a s s o c i a t e d one-parameter

group i f ,

I"

and only i f , f o r some pe1R we have M'= M+pI

(3) L e t Her(Hk)=:

I

N'= N - p I

{XeL(Hk); X = X*}

h e r m i t i a n e l e m e n t s of L ( H k ) f o r k= 1,2,

d e n o t e t h e s e t of

and endow i H e r ( H k ) w i t h

i t s u s u a l Banach-Lie a l g e b r a s t r u c t u r e . Show t h a t i Her(H2,H )=: 1

{(iM,

iN);

MeHer(H2),

w i t h t h e p r o d u c t s t r u c t u r e and t h e norm

NsHer(H1))

183

AUTOMORPHISMS OF CLASSICAL BANACH SPACES

i s a Banach-Lie

a l g e b r a . Show t h a t t h e s u b s e t

i s a c l o s e d L i e i d e a l of i H e r ( H 2 , H 1 ) . 0

( 4 ) Prove t h a t t h e L i e a l g e b r a a u t B ( F 1) of t h e L i e 0

group A u t B ( F 1 )

i s isomorphic t o t h e q u o t i e n t i H e r ( H 2 , H 1 ) / J

endowed i t s q u o t i e n t s t r u c t u r e and t h e norm

9 8 . - D e s c r i p t i o n of Aut 0B ( F ) and a u t 0B ( F ) k

-

W e d e n o t e by

M=:

{

B

k

’) “2

:

01

j

k= 2 , 3 , 4 .

,Cclc} t h e J * - a l g e b r a

of

a l l 2 x 2 s y m m e t r i c m a t r i c e s w i t h complex e n t r i e s . A s u s u a l l y , whenever a n u l l e n t r y a p p e a r s i n a m a t r i x AcM

w e l e a v e a blank

i n t h e c o r r e s p o n d i n g p l a c e . Any element of M c a n be u n i q u e l y r e p r e s e n t e d i n t h e form

i . e . , a s a sum of a d i a g o n a l m a t r i x p l u s a scalar m u l t i p l e of a p a r t i c u l a r element of M

of

M

w e l o o k f o r t h e J*-automorphisms

t h a t preserve diagonal matrices. 8.47.

t r y of

form

. First

M

LEMMA. L e t K : M + M b e a s u r j e c t i v e Z i n a a r i s o m e t h a t p r e s e r v e s d i a g o n a l , m a t r i c e s . T h e n A is of t h e

CHAPTER

184

Proof:

Then, if Z =

8

We must have

a b -a b it follows that [ b l a2)= [b 1-a2) whence

On the other hand, we must have (cf. lemma 8.32 and theorem 8.39)

for suitable a , B e t . But then a,@€

111.

# Denote by F 2 = {XcL ( H ) ; X T = X } the Cartan factor of type I1 corresponding to a given conjugation Q on H, where x[= QX*Q. 0

8 . 4 8 . THEOREM. L e t LcAut B ( F Z ) be g i v e n . T h e n t h e r e

exists a u n i q u e unitary o p e r a t o r L ( A ) = UAU'

UcL(H) s u c h t h a t AeF2

Proof: From the characterization of the minimal elements of F 2 it follows that

AUTOMORPHISMS OF CLASSICAL BANACH SPACES

185

for some not necessarily linear mapping T: H+H. Observe that the values of T are determined up to a constant factor belonging to - I l l } ; furthermore,

Given any u,veH, we have

whence, by (8.30), it follows that ~(TU)@(QTU)* = [(Tu)@(QTu)*] [(TvI@(QTv) *]*[(Tu)@(QTu)*]= =

*(Tu)@(QTu)*

That is, 2 =

2



u vcH

1

and, in particular, the mapping T preserves orthogonality. As a consequence we obtain that

Indeed, we have Tf ~T(0ul+t!u2)whenever

f 1 uIru2

Observe that, if h l T u l r T u 2 then, for some fcH, we have Lrf@(Qf)*]=

h@(Qh)* 2

(because < f l u . > I

=

i.e.

Tfc{-h,h}

2

2

= = 0 1

7

and for

f lu1,u2 j = 1,2).

186

CHAPTER

a

Therefore h 1 T ((cu1+Cu2)whenever h 1 T u l ,Tu2I i.e.

1

{ T u l I T u 2 } lC [T(O~-I~+EU~)] and

SO

(8.32)

C T u l + C T u Z ~T(QUl+CU2)

Now

L [ul@ [Qu, 1 * + u 2 @( Q u ,1 *] c cSpanI L [ ( a l u l+a2u2 1 @Q ( a u + u 2 u 2 ) * J ;

Q

1, a 2 d C l c

CSpanI [T(nlul+a2U2)]@[QTCllul+a2u 2 I ] * ; whence (8.31) follows by

al,a2c@J

8.32).

Consider now any orthonormal couple { e l l e 2 } .From (8.31) we can see the existence of a linear mapping

of symmetric matrices such that

Since for any orthonormal couple { h l l h 2 }we have

it follows from lemma 8.47 that

AUTOMORPHISMS OF CLASSICAL BANACH SPACES

holds for some uc{-l,l}.

187

Thus, the isometry defined by

satisfies

Finally, let us fix a unit vector e cH. Observe that 0 Indeed, we whenever f 1 eo , 1 1 f 11 = 1 X I = 1

.

'e 0 ,Xf= 'e0,f

have dom U

e O ti f

=

dom U

eo,f

;

moreover, as T is pointwise

determined up to a constant ? I , u e o, h f is unambiguously (eo)= Te for determined by its linearity and the fact U 0 e o ,X f

Xca. Hence, we can define a mapping T' on H by means of

Also, if the subscript x satisfies

denotes the value at XCH, T'

for all f,xcH. The left han side of (8.33) is a linear mapping ofthe variable f. It follows that the mapping f+T'f is Linear,too. Indeed, given f,qcH, we can write f = ae0 +f0 , g= Peo+g0 for some a,BeE, and fo,gOcH with fo,gOl eo . Now, the orthogonal projection of the right hand side of (8.33) onto the subspace {eoI1 is equal to (T'fo)@(QT'eo): for any xeH, whence

188

CHAPTER

8

This implies T'f 0 + T u g = T'(fo+g 1 . On the other hand, 0 0 T'f= aT'e 0+ T ' f o Since

T ' g = 4T'eo+T'g0

for all feH, it follows that T ' is a

T'fe{-Tf,Tf}

l i n e a r isometry of H such that (8.30) holds with T' instead of

T. This completes the proof, because then

holds and, by theorem 8.41, the latter entails L ( A ) = T'AT"

AcF2

# 8.49. EXERCISES. system in H and ( a , ) ,

... ,hn 1

( 1 ) If Chl,h2,

j k j , k = I ,...,n

matrix, then

is an orthonormal is any complex symmetric

( 2 ) Show that the Lie group Aut'B that its L i e algebra is given by

aut0 B ( F 2 ) = {f

M'

X

-f

F2)

is connected and

i(MX+XMT); McHer

Hint: Look for continuous one-parameter groups s+Ls

of the form

Ls (A)= U,AU',

,

AeF2,

se IR

(cf. example 6.43). 8.50. REMARK. I1 wouZd b e i n t e r e s t i n g t o f i r i d r e p r e s e n t a t i o n f o r m u l a s f o r t h e e l e m e n t s o f t h e groups Aut 0 B(F3) a n d Aut0B(F4). I t s e e m s t h a t t h e m e t h o d p r e s e n t e d f o r 0

0

Aut B ( F 2 ) Can b e a d a p t e d t o t h e c a s e of Aut ( F 3 ) . H o u e v e r , t h e

AUTOMORPHISMS O F C L A S S I C A L BANACH S P A C E S

189

study of Cartan factors of type I V , a n d i n particular t h a t of 0

Aut B(P4)

,

r e q u i r e s a different approach (cf. 1 2 4 1 a n d 1 7 4 1

respectively).

This Page Intentionaiiy Left Blank

CHAPTER

9

BOUNDED SYMMETRIC DOMAINS

51.-

H i s t o r i cal sketch. DEFINITION. We s a y t h a t a b o u n d e d domain D i n a

9.1.

c o m p l e x Banach s p a c e E i s " s y m m e t r i c a t a point" a c D i f t h e r e 2

e x i s t s an a u t o m o r p h i s m ScAut(D) s u c h t h a t S = id, and a i s an i s o t a t e d f i x e d p o i n t f o r S. We s a y t h a t D is " s y m m e t r i c " if i t i s s y m m e t r i c at e v e r y p o i n t acD. Throughout t h e whole c h a p t e r , E a n d D d e n o t e r e s p e c t i v e l y a f i x e d complex Banach s p a c e and a bounded s i m p l y c o n n e c t e d symmetric domain D i n E . Symmetric domains i n C n where i n t r o d u c e d by E . C a r t a n i n 1935 who p o i n t e d o u t t h e v e r y d e e p c o n s e q u e n c e s o f t h i s d e f i n i t i o n by c l a s s i f y i n g c o m p l e t e l y a l l p o s s i b l e f i n i t e d i m e n s i o n a l symmetric domains. H i s p r o o f s w e r e b a s e d on t h e c o m p l e t e c l a s s i f i c a t i o n of f i n i t e dimensional semisimple L i e a l g e b r a s ( g i v e n a l s o by h i m s e l f p r e v i o u s l y ) , a t o o l whose u s e seems r a t h e r h o p e l e s s i n i n f i n i t e dimensions. Since w e are p r i m a r i l y i n t e r e s t e d i n i n f i n i t e dimensions, we s t a t e C a r t a n ' s theorem without proof. L e t E1,..,E

n

be f i n i t e d i m e n s i o n a l complex Banach s p a c e s of xl

d i m e n s i o n s r l l . . l r and w r i t e Ek fox t h e s p a c e n k=l n Vk= l , . . , n I 8 E k = : { ( X ~ ~ . . , X ~ X) k; C E k k =1 endowed w i t h t h e norm

191

CHAPTER 9

192

Then w e have: THEOREM ( C a r t a n ' s c l a s s i f i c a t i o n t h e o r e m ) . If dim E<m,

s u c h t h a t D is t h e n t h e r e a r e Banach s p a c e s E 1 , . . , E n b i h o l o r n o r p h i c a l l y e q u i v a l e n t t o t h e open u n i t b a Z l B( (3 of

n @ Ek k=1

, where e a c h of t h e Ek

k = 1 Ek)

-i-s e q u a l t o some of t h e s p a c e s

d e s c r i b e d below ( o r d e r e d i n 6 fundamental t y p e s ) : w h i t p , qdN

TYPE I :

L(Ep,Eq)

TYPE TI:

C A ~ L ( @ , E ~:) A ~ = A ) where p , q m and A

t

denotes

t h e Lransposedof A .

I n t h e s e c a s e s , E p l lCq a n d

L(Crp,iEq)

have,respectiveZy, their

e u c l i d e a n norm a n d Lhe o p e r a t o r norm T Y P E IV:

The s p a c e Ep

TYPE V:

A p a r t i c u Z a r Banach s p a c e of d i m e n s i o n 1 6 .

TYPE V I :

A p a r t i c u z a r Banach s p a c e o f d i m e n s i o n 2 7 .

w i e h t h e norm

Prom c h a p t e r 7 w e know t h a t t h i s r e p r e s e n t a t i o n of D i s unique n

up t o an i s o m e t r i c l i n e a r isomorphism. The domain B ( 0

k=l

E ) is k

c a l l e d t h e H ar i s h- C handr a r e a z i z a t i o n of D. If w e drop t h e c o n d i t i o n d i m E < m l t h e s i t u a t i o n seems t o become

e s s e n t i a l l y more c o m p l i c a t e d . So f a r , t h e s t r o n g e s t a n a l o g o u s g e n e r a l r e s u l t i s a v e r y r e c e n t theorem of W . Kaup ( 1 9 8 3 ) 1351: THEOREM. L e t D b e a b o u n d e d s y m m e t r i c domain of E. Then

D i s b ih o Z o mo r phi caZZy e q u i v a l e n t t o t h e u n i t o p e n b a l l o f some Ranach s p a c e w h i c h i s u n i q u e l y d e t e r m i n e d up t o i s o m e t r i c

193

BOUNDED SYMMETRIC DOMAINS

l i n e a r isomorphisms. I n t h i s c h a p t e r , w e prove o n l y a s l y g h t l y weaker s t a t e m e n t due t o J.P. Vigu6 ( 1 9 7 6 ) : THEOREM.

L e t D be a b o u n d e d s y m m e t r i c d o m a i n of E . T h e n ,

t h e r e e x i s t s a baZanced domain D ' i n E s u c h t h a t D and D ' a r e b i h o l o m o r p h i c a 1 Zy e q u i v a l e n t . S i n c e b i h o l o m o r p h i c a l l y e q u i v a l e n t b a l a n c e d domains are l i n e a r l y e q u i v a l e n t , Kaup's theorem shows t h e c o n v e x i t y of D ' . However, from t h e way of ViguG's c o n s t r u c t i o n t h i s f a c t cannot be d i s c o v e r e d . F i n a l l y , w e r e m a r k t h a t s e v e r a l i d e a s of Kaup's method go back t o a modern e l e m e n t a r y J o r d a n t h e o r e t i c approach of t h e s c h o o l of M . Koecher ( 1 9 6 9 ) t o f i n i t e d i m e n s i o n a l symmetric domains which w e recommend t o t h e i n t e r e s t e d r e a d e r s ,

(cf.,

1 4 2 1 and

(430.

52.-

Elementary p r o p e r t i e s of symmetric domains.

= -

The remaining p a r a g r a p h s c o n t a i n t h e proof of V i g u g ' s theorem d i v i d e d i n t o s t e p s t h a t might have some i n t e r e s t i n t h e m s e l v e s . 9.2. 2

S = id

LEMMA. G i v e n acD a n d ScAutD s u c h t h a t S = a

t h e r e e x i s t a n e i g h b o u r h o o d U of a a n d a

D ,

and

biholornorphic

map f : U-tE s u c h t h a t f ( a ) = 0 a n d f g S i s L i n e a r . P r o o f : W e may assume a = 0 . W r i t e L = : S:

and remark t h a t

L 2 = i d . For hsHol(D,E) I w e d e f i n e + ( h ) = : LohoS. Then t h e

+

operator i s l i n e a r and s a t i s f i e s $ 2 = i d . Hence, t h e mapping 1 g = : - ( S + $ ( S ) ) i s a f i x e d p o i n t of $, i . e . g = Logos. Furthermore 2 1 g ( O ) = 0 and g i l = 'z ( L + L ) = L . S i n c e gdl i s i n v e r t i b l e , t h e r e i s 3

neighbourhood V of 0 where g i s biholomorphic. P u t U = V n S ( V ) ;

t h e n U i s a neighbourhood of 0 such t h a t S ( U ) = U . By s e t t i n g g = : g1,

we have f ( O ) = 0 and f = LofoS, i . e . F S- foSof-1= L I f ( " ) .

t -

#

CHAPTER 9

194

9 . 3 . PROPOSITION.

a t a s a t , i s f i e s S ( a ) =a and

F o r e v e r y asD, a n y s y m m e t r ' y S of D S ( ' = -id. I n particuZnr, S u n i q u e

Proof: W e assume a = 0 . L e t u s d e n o t e by S any symmetry of D a t a . By lemma 9 . 2 w e have a l o c a l c o o r d i n a t e map f a t 0 such t h a t f ( O ) = 0 and L = : SA1 i s a l i n e a r c o n t i n u a t i o n of f # S . We show t h a t L = - i d .

-x

A s s u m e w e had Lxf

f o r some X C E , x f 0 .

Then, f o r s u f f i c i e n t l y s m a l l v a l u e s of t ( s a y /t1<6), t h e points xt=:

t ( x + L x ) l i e i n t h e neighbourhood where f i s defined. 2

Moroever, a s L = i d , w e have L ( x t ) = xt f o r a l l t s ( - 6 , + 6 ) . and xt#O

But

f - 1 (O)=O t+O isolated fixed point.

i s a f i x e d p o i n t f o r S . S i n c e l i r n f-'(x,)=

then f - l ( x , )

for t f O ,

S c a n n o t have an

By C a r t a n ' s uniqueness theorem, S i s u n i q u e l y dermined.

9.4.

a by

DEFINITION

#

We s h a l Z d e s i g n a t e t h e symmetry o f D at

Sa'

9 . 5 . PROPOSITION. T h e m a p p i n g D + A u t D g i v e n b y a+Sa is T-continuous. P r o o f : Given a s D , w e show t h a t

T l i r n S,+c

(9.1)

Sa= i d

c+o

D

Then, m u l t i p l y i n g on, t h e r i g h t by S i l = S a and a p p l y i n g theorem 2 . 2 we o b t a i n T l i r n Sa+c= Sa. c+o

By theorem 2 . 8 ,

i n o r d e r t o prove ( 9 . 1 ) it s u f f i c e s t o show

that

l i r n (S,+c C'O

S )'"=idik a a

for

k= 0 , l

Now w e have

=

s = +(~a + c )+ [s,+~ (a)-s,+~ ( a + c ) ]= a+c+ [sa + c (a)- s ~ (+a +~c ) ]=

BOUNDED SYMMETRIC DOMAINS

195

1

= a+.-!

('a.+c ) a + t c c d t 0

Take a b a l l B c c D c e n t e r e d a t a and p u t 6=: d i s t ( B , a D ) . From t h e Cauchy e s t i m a t e s w e have

Since S

a+c

( D ) c D and D i s bounded, we g e t (1

II ( s a + cS a ) a

-all

$11

CII

(I+

1

T

M)

f o r some M i n d e p e n d e n t of c . T h e r e f o r e ,

l i m (S,+c

Sa)Lo= a= id('.

On t h e o t h e r hand,

c+o

( S a + c Sa);'-id=

( S a + c 1 S( 'a ( a ) ) (2

( S a ) i l - i d = (Sa+c ( 1 ) (1 = a +('a+c a + c

c dt

From t h e Cauchy e s t i m a t e s ,

H e n c e f o r t h w e assume t h a t O c D a n d w r i t e S=: S o f o r t h e symmetry of D a t 0 .

9.6.

PROPOSITION. L e t acD b e g i v e n . T h e n , f o r e v e r y

CCE, t h e r e e x i s t s a unique A c a u t D s u c h that

CHAPTER 9

196

Namely s u c h a v e c t o r f i e Z d is g i v e n by

-

A = T lim 1 2 t %+ct O't

S-idD)

P r o o f : We may assume a = 0 . I n o r d e r t o u t i l i z e theorem

2t 1 ( S t c S - i d ) ,( k I k = 0 , 1 , W e s h a l l make u s e of t h e f o l l o w i n g T a y l o r t y p e f o a l a s

5 . 8 , we s t u d y t h e convergence o f for t + O .

for holomorphic maps F : (9.2)

F ( b + h ) = F ( b ) + F ( b + h ) - F ( b ) =Fb( 0+

I,

1

FbtTh (' hdT =

L e t u s t a k e any b a l l B c c D c e n t e r e d a t 0 and p u t

6=: d i s t ( B , a D ) . For s u f f i c i e n t l y small v a l u e s of t , w e have t a c B c D and a p p l y i n g ( 9 . 2 ) t o t h e f u n c t i o n F = : Stc p o i n t s b=: t c ,

h=:-tc,

and t h e

we obtain

Moreover, a s t ( l - < T ) c e B for a l l r , < e [ O , I ] , estimates we obtain

where M i s independent of t . T h e r e f o r e ,

f r o m t h e Cauchy

BOUNDED SYMMETRIC DOMAINS

so that

1 lim 2t t*O

(S

tc

197

S-idD ) O(O= c.

On the other hand, applying ( 9 . 2 ) to the operator-valued function f: x + ( S t c ) i 1 and the points b=: tc, h=-tc, we have

Again by the Cauchy estimates,

But now

Moreover, by proposition 9.5 we have T lirn St c = S which (2 (2 O't implies lim ( S t c ) t c = S o O't

.

Thus, theorem 5.8 entails that A =: T lirn t+O

1 2t

(S

tc

S-idD)

defines a vector field AcsautD which obviously satisfies the requirements of the statement.

#

198

CHAPTER 9

9.7.

DEFINITION. H e n c e f o r t h , f o r c e E , Ac wiZl d e n o t e

t h e v e c t o r f i e l d of a u t D u n i q u e l y d e t e r m i n e d b y t h e condition::

(9.3)

(Ac):'=

0

I t i s e a s y t o show t h a t I ): E

(Ac):'=

+

$ SL2(c,-)

a u t D g i v e n by $ = : c * Ac i s a

c o n t i n u o u s r e a l l i n e a r mapping. 9.8.

PROPOSITION. L e t D b e a b o u n d e d s y m m e t r i c domain.

T h e n D -Is h o m o g e n e o u s u n d e r t h e a c t i o n of P r o o f : By p r o p o s i t i o n 7 . 1 7 ,

t h e s u b g r o u p AutoD.

it s u f f i c e s

t o show t h a t

t h e o r b i t (Aut D ) O o f t h e o r i g i n by t h e s u b g r o u p Aut D i s a 0

n e i g h b o u r h o o d o f 0 . By lemma 6 - 4 5 a n d r e m a r k 6 . 2

, we

0

can f i n d

a n e i g h b o u r h o o d M o f t h e o r i g i n i n a u t D s u c h t h a t t h e mapping AcM * expAcAutD i s r e a l a n a l y t i c , a n d e x p M c A u t O D . S i n c e $:

c * Ac i s r e a l a n a l y t i c , w e c a n f i n d a n e i g h b o u r h o o d U o f 0

i n E s u c h t h a t @ ( U ) c M . Moreover, b y t h e o r e m 6 . 5 7 , fsAutD * f ( 0 ) e D is r e a l a n a l y t i c . T h e r e f o r e , t h e c o m p o s i t e

i s r e a l a n a l y t i c i n U and w e have

B e s i d e s , g ( O ) = 0 and

f o r all c c E , so t h a t g o( 1= i d . T h e n , by t h e i n v e r s e mapping t h e o r e m , g(U) i s a n e i g h b o u r h o o d of t h e o r i g i n .

BOUNDED SYMMETRIC DOMAINS

53.-

199

The c a n o n i c a____ l decomposition o f autD. -

By i n v e s t i g a t i n g t h e e f f e c t o f t h e a d j o i n t S# o f t h e symmetry a t 0 , w e o b t a i n a good p i c t u r e o f a u t D . We a l r e a d y know t h a t a u t D i s a r e a l Banach s p a c e and t h a t S # : a u t D

-t

2

c o n t i n u o u s l i n e a r o p e r a t o r . Moreover, a s S = i d

autD i s a

,

is a

S#

p r o j e c t o r , so t h a t it f u r n i s h e s a d e c o m p o s i t i o n o f a u t D i n t o t h e d i r e c t t o p o l o g i c a l sum autD= L

Ql

4

of t h e e i g e n s u b s p a c e s c o r r e s p o n d i n g t o t h e e i g e n v a l u e s -1

+ I o f S#

and

, L = : {AcautD;

S A-

&=:

S#A= A )

(9.4)

9.9.

{AcautD;

# - -A}

PROPOSITION. The s u b s p a c e s L and Q a r e ,

respectively,

t h e image and t h e k e r n e l o f t h e m a p p i n g s $: E

c

-f

+

autD

Q: a u t D

+

E

A

+

A(0)

Ac

P r o o f : L e t AcautD b e g i v e n . By d e f i n i t i o n o f S #

, we

have

f o r a l l xcD. By t a k i n g t h e f i r s t d e r i v a t i v e , w e g e t

Thus, e s p e c i a l i z i n g t h e s e r e l a t i o n s f o r x= 0 and t a k i n g i n t o account t h a t S(O)= 0 , (9.5)

( S # A) 0( O --

f o r a l l AsautD.

S"(O)=

-A('0

-id,

we obtain

A'=

(S#A)

AA'-SA'(A(O)

,. )

CHAPTER 9

200

N o w we prove L = I m $ .

L e t A c a u t D be s u c h t h a t S # A = -A

and

d e f i n e c=: (S#A)hl=

Ad" T h e n w e have ( S # A ) ( = -A 0( ' a n d , by ( 9 . 5 1 , AL1-SA2(c,.) so t h a t A. (1-01 S 2 ( c , . ) . H e n c e by

d e f i n i t i o n 9 . 7 , we get A= Ac. C O n V e r S e l y , l e t u s assume t h a t A= A

c and A:'= satisfies

A:"=

1 2 Sd2(c,.).

By (9.5;

f o r s o m e C C E , so t h a t t h e vector f i e l d S # A

B y C a r t a n ' s u n i q u e n e s s t h e o r e m w e have S A= -A.

#

The proof of Q= k e r g

is quite s i m i l a r .

# 9 . I D . COROLLARY.

We h a v e

I n particular, L is a c l o s e d L i e s u b a l g e b r a of a u t D . M o r e o u e r ,

for a l l A c e & Proof:

and LcL. Since S #

i s a L i e a l g e b r a a u t o m o r p h i s m , w e have

Furthermore, [L,Ac]O=

L A I A c O - ( A c ) ;'LO=

Lo( 1C.

201

BOUNDED SYMMETRIC DOMAINS

54.-

The c o m p l e x i f i e d L i e a l g e b r a of autD. ---____

For l a t e r u s e , w e i n t r o d u c e t h e c o m p l e x i f i e d C a u t D = : (autD)B i (autD) of t h e L i e a l g e b r a autD. S i n c e autD i s p u r e l y r e a l , i . e . (autD) fl i (autD) = { O )

,

t h i s r e p r e s e n t s EautD as a d i r e c t

t o p o l o g i c a l sum. Thus, though t h e v e c t o r f i e l d s AcEautD a r e no l o n g e r complete i n D ( a n d , i n p a r t i c u l a r , t h e y do n o t s a t i s f y C a r t a n ' s u n i q u e n e s s t h e o r e m ) , t h e y have a u n i q u e r e p r e s e n t a t i o n

of t h e form X = A l t i A 2 w i t h A l l AZcautD. 9.11.

D E F I N I T I O N . For ceE, w e d e f i n e

cc=: 71

Qc=:

(Ac-iAic)

1

(Ac+iAic)

T h e r e f o r e , w e have

Q,(o)=

Cc(0)= c

(9.6)

f o r a l l C B E . Moreover, t h e mappings E and c

-+

A = C + Q c c c

0

EautD g i v e n by c

-+

Qc a r e , r e s p e c t i v e l y , complex l i n e a r and complex

-+ cc

c o n j u g a t e l i n e a r . Both of them are c o n t i n u o u s . 9.12.

PROPOSITION. We h a v e

[Ac,Aic]fO

P r o o f : Suppose w e had [ A c , A i c ] =

f o r a 2 2 C B E , cfO.

0 f o r some C C E , c f 0 .

F i x any b a l l B c c D c e n t e r e d a t 0 and assume t h a t EautD h a s been endowed w i t h t h e norm

1 1 .I1

B. Then, t h e r e

e x i s t s E > O such

that

1

expXIB=

C - X k = O k!

^k

id

IB

i s well d e f i n e d on B whenever XeCautD and mappings A

-+

AAc

and

6 > 0 such t h a t w e have

A

+

CAc

11

XI1 <E. S i n c e t h e

a r e c o n t i n u o u s , t h e r e i s some

202

CHAPTER 9

for a l l ACE, IXIs6. T h e r e f o r e , w e c a n d e f i n e

From t h e a s s u m p t i o n [A A . ] = 0 , w e e a s i l y d e r i v e [Ax , A X c ] = 0 c 1c f o r a l l Act; t h u s by c o r o l l a r y 4 . 2 4 , w e have

But o b v i o u s l y ( - A A c + C h c ) O = -A

hC

( 0 ) + C X c ( 0 ) = 0 by ( 9 . 6 ) so t h a t

~ X ~ ( - A ~ ~ + A o ~ ~for ) O a=l l

1 ~ 1 6 6 ,i . e .

(9.7)

O=

(expAx

Moreover, s i n c e X

+

(eXPCxc)0

CXc i s complex l i n e a r a n d c o n t i n u o u s , w e

have " 1 k OD Xk k (expCXc)O= 1 - ( C X c i d B ) O = 1 - (Ccid,)O k! k! k=0 k=O

and t h e r e f o r e t h e mapping g: X

-+

(expCXc)O i s h o l o m o r p h i c i n

/ A 1 < 6 . We a l r e a d y know t h a t h : X (expAXc)O i s r e a l a n a l y t i c on t h e wohle C. By ( 9 . 7 ) , g and h c o i n c i d e i n a n open s u b s e t o f C; t h u s h i s a n e n t i r e mapping. B u t A X c i s a c o m p l e t e v e c t o r field i n D , so t h a t h ( X ) = (expAXc)OcD f o r a l l XcE. By Liouville's t h e o r e m w e have -+

for a l l XcC. But t h i s i s a c o n t r a d i c t i o n s i n c e

Let D b e a b o u n d e d symmetric domain and assume that AsautD satisfies [A,X]= 0 f o r all XcautD. T h e n 9.13.

PROPOSITION.

203

BOUNDED SYMMETRIC DOMAINS

we h a ve A= 0 .

Proof: Let AcautD be such that [A,X]= 0 for all XcautD and assume that we had AfO. Then, in the decomposition A= Ac+L with AccQ, LcL, we have two possibilites: LfO, L=O. Suppose L#O. As LcL entails LO= 0, by Cartan's uniqueness theorem, there must exist some C'CE such that LAlc'fO. Then, for X=: Ac,cautDIwe have

By corollary 9.10, we have [Ac,Ac,]cL

so that [AcIAc,]O= 0 and [L,Ac,]O= L 0( ' c ' f 0 . Therefore [A,X]#O which is contradictory.

Suppose L= 0. Then Ac#O and cf0. But then, by proposition 9.12, the vector field X=: A . cautD satisfies [A,xJ= [Ac ,Aic] # O 1c which is contradictory.

# 55.-

The local representation of autD.

9.14. LEMMA. The mapping c

+

(expCc) i s b i h o 2 o m o r p h i c

i n a n e i g h b o u r h o o d of 0 . P r o o f : It is easy to see that the mapping c

(expCc)O is holomorphic in a neighbourhood of the origin. On the other hand d dt l o (expCtc)O= ( ? c idB ) O =

+

C (O)= c

so that its Frdchet derivative at 0 is invertible. The result follows by the inverse mapping theorem.

# 9.15. DEFINITION. H e n c e f o r t h , U d e n o t e s a n e i g h b o u r h o o d of 0 s u c h t h a t c + (exp C 1 0 a d m i t s a h o l o m o r p h i c i n v e r s e o n U. M o r e o u e r , tle s e t

204

CHAPTER 9

We i n t r o d u c e t h e m a p p i n g J : V * U b y m e a n s of J:

(expCc)O * c

T h u s , we h a v e

U= J(V) a n d J ( E x p C c ) O = c f o r a l l ccU.

A s previously,

we

w r i t e L = : { L e a u t D ; LO= 0 1 . f o r aZZ xeU

PROPOSITION. We ha.ue ( J# L ) x = L 0' l x

9.16.

and LeL. t

Proof: L e t L s L be g i v e n and s e t G =: e x p t L f o r tdlR. W e t

h a v e T l i m G = i d D ; t h e r e f o r e , w e c a n f i n d a number 6>0 a n d a t+O

n e i g h b o u r h o o d W I C V of t h e o r i g i n s u c h t h a t G t ( W l ) C V f o r a l l

t , Itlc6. L e t u s s e t W 2 = :

J(W1),

i s d e f i n e d on

so t h a t J G t J - l

W2 f o r I t \ < & NOW, . l e t xeW2 b e g i v e n ; f r o m G-tO=

0 we d e r i v e

( j # G t ) x = ( J G t J - l ) x = JG t ( e x p C x ) O = J G t (expCx)G- t o =

JIGi(expcx)]o=

By t h e o r e m 4 . 2 8

w e have G

t

c

# x

where L # C = [ L , c ] = Y Y

= (exp t L ) # C ~ =

z

t" n !L

~

C

~

n= 0

c

(1

f o r yeE,

i.e.

Lo

f o r ndN. T h u s , ( J # G ~ ) ~ J[expC =

whence

lo=

exp t ~ A l x

( J # L ) x = L o( 1X . The r e s u l t f o l l o w s by t h e i d e n t i t y

principle.

#

205

BOUNDED SYMMETRIC DOMAINS

9 . 1 7 . LENMA. We h a v e [ C x , C y ] =

0 r

[QxrQy]=

0

f o r a22

x,ysE. P r o o f : For a n y U , V C E , c o n s i d e r

N U , v

= : [A, ,Av]

- [Aiu‘Aiv]=

[ C U + Q U , CV+QV] - [ i C U - i Q v

By c o r o l l a r y 9 . 1 0 w e have N

utv

r

iCv-iQ

eL, and therefore

since [ Q ~ , Q( 1~= I (~ Q( 1Q ~ , - Q ~( 1Q,);’=

and

i f s = 0 , l and c c E . I t follows t h a t

J#Nix,y

= 2[cix,Cy];1=

2i[Cx,C

] ( l = iJ#Nx

Y O

IY

i.e. Nix,y

by theorem 4 . 2 5 .

= iNx

Thus

IY

C L n(iL)c(autD) n ( i a u t D ) = {Ol

V

3=

CHAPTER 9

206

9.. 18

.,,PROPOSITION.

( a ) (J Cc)x= c

#

WE have

f o r all ccE a n d xcU

(b) F o r a l l ccE, t h e U e c t o r fieZd J#Q, is a c o n t i n u o u s h o m o g e n e o u s polynomial o f s e c o n d d e g r e e . Proof:

( a ) L e t c c E and xcU b e g i v e n . Then w e have

Whence , by lemma 4 . 2 3 w e d e r i v e

so t h a t J # C c i s a c o n s t a n t v e c t o r f i e l d of v a l u e c .

(b) L e t ceE, xcE and ycU be g i v e n . By t h e p r e v i o u s s t e p w e have x = ( J # C x ) y ; t h e r e f o r e

i s a v e c t o r f i e l d t o which p r o p o s i t i o n 9 . 1 6 a p p l i e s . Thus, w e

have I

.

Now, f o r x i n a neighbourhood of t h e o r i g i n , t h e segment [O,x] E by means of l i e s i n U and w e can d e f i n e + : [0,1] ( J # Q c ) t x . I t i s e a s y t o check t h a t + ( O ) = 0 , so t h a t by $(t)= ( 9 . 8 ) w e have -f

207

BOUNDED SYMMETRIC DOMAINS

which i s a c o n t i n u o u s homogeneous polynomial of second d e g r e e i n x . The r e s u l t f o l l o w s by t h e i d e n t i t y p r i n c i p l e .

56.- ______ The p s e u d o r o t a t i o n s on autD. ~

_

I

_

_

9 . 1 9 . D E F I N I T I O N . For t d R , we i n t r o d u c e t h e m a p p i n g s t

$ :

E autD

+

E autD in t h e f o l l o w i n g m a n n e r :

( a ) L e t AsautD b e g i u e n . T h e n A a d m i t s a u n i q u e r e p r e s e n t a t i o n A= A i L

w i t h Ace& a n d L c L a n d we d e f i n e L

$L:

A= A +L

+

A it +L e c

( b ) Now Qt may b e e x t e n d e d t o EautD b y c o m p l e x l i n e a r i t y

b e c a u s e we h a v e E a u t D = ( a u t D ) @ i ( a u t D ) , t h e sum b e i n g d i r e c t . I n o r d e r t o show t h a t $t i s a L i e a l g e b r a automorphism of a u t D , we introduce an a u x i l i a r y transformation.

9.20. R ~ :E

-+

by

E

For tdIR, we d e f i n e t h e m a p p i n g

DEFINITION.

R ~ :x

+

e

it

x.

L e t AcautD b e g i v e n and d e n o t e by J : V

+

U t h e neighbourhood U

of 0 and t h e isomorphism J g i v e n by d e f i n i t i o n 9 . 1 1 . Thus, A is u n i q u e l y determined by i t s r e s t r i c t i o n t o U and J # $tA i s a holomorphic v e c t o r f i e l d on U = J ( V ) . B e s i d e s , by p r o p o s i t i o n 9.18,

J # A i s a n e n t i r e holomorphic v e c t o r f i e l d ( a c t u a l l y , J # A

i s a polynomial of d e g r e e n o t g r e a t e r t h a n 2 ) so t h a t R t # J # A i s a l s o an e n t i r e holomorphic v e c t o r f i e l d and it makes s e n s e t o t

compare ( R # J # A ) I u

with

J#($tA)

Iu

. We

g e t t h e following

result 9 . 2 1 . PROPOSITION.

f a ) We h a v e R;J#A=

t

J#+ A for a l l tm

208

CHAPTER 9

a n d AeautD. ( b ) F o r all

td[R,

+t i s a Lie a l g e b r a a u t o m o r p h i s m o f

autD. ( a ) S i n c e any AcautD may be w r i t t e n i n t h e form

Proof:

A= Cc+Qc+Lfor some ceE and L c L ,

e q u a l i t y Rt J =J

#

#

#

+

t

it s u f f i c e s t o check t h e

i n these particular vector f i e l d s .

L e t c c E be g i v e n . By p r o p o s i t i o n 9 . 1 8 ,

JC C C i s c o n s t a n t ; t h u s

w e have

~ ~ c ~~ )# x( e=~

(J

~ c , e) ixt = (J

# c ~ )e ~ itc =

# and

Q, i s a homogeneous polynomial of second d e g r e e , w e have

As

(J#$

t Q,)X=

J#Q

it x= J# ( e e c

-it

Q ~ ) X =e

-it(~#~c)x

and t

t

( R # J # Q c ) x = (R

)

(1

-t J#Qc ( R - t x ) = eitJ#Q, ( e - i t x ) = e R x

= e

Since f o r LsL

,

-it

it - 2 i t

e

J# Q, ( x ) =

(J#Qc)x

J L i s l i n e a r , w e have

#

( b ) Obviously, $t i s an isomorphism of autD a s a v e c t o r s p a c e . By s t e p (a), f o r A l t A 2 s a u t D , w e have

BOUNDED SYMMETRIC DOMAINS

209

whence t h e c o n c l u s i o n f o l l o w s by t h e i d e n t i t y p r i n c i p l e .

ff L e t u s d e n o t e by

TI

j

,

j = 0,1,

the canonical projections

a s s o c i a t e d w i t h t h e d e c o m p o s i t i o n a u t D = L@Q.

9.22.

1. I

LEMMA. The norm

d e f i n e d o n a u t D by

\ A / = : max{II ( n . A ) h k l \ ; I

i s i n v a r i a n t under alZ t r a n s f o r m a t i o n s 4

j , k = 0,13 t

,

tdR.

Moreover, i t

d e f i n e s t h e n a t u r a l t o p o l o g y on autD. P r o o f : L e t u s s u p p o s e t h a t w e h a v e An TI,A

I n

\An]

+

+

/I

n.A

jA,l

+

0 i n a u t D , so t h a t

0 . Conversely, i f

I1

1

=

C

k=O

0 then

11

-+

0 i n autD; t h e n

(n.A )

(kll

I n 0

(kl\

" ( 71 j An o

j,k= 0 , l ; t h u s by t h e o r e m 5 . 6 , w e h a v e I T , A

+

I n

0; thus

-+

-+

o

for

0 ( j = 0 , l ) and

A = n A + n A + O . n O n 1 n

Moreover,

from

4 tA=

@

t

+L w e g e t (Ac+L)= A . lt e c

TI^^ t A =

t

4 noA

El$

t A=

4

t

TIA (k-

(k

f o r a l l a e a u t D . From d e f i n i t i o n 9 . 4 w e o b t a i n ( A X c ) O - k ( A c ) O f o r Ad!:,

which completes t h e p r o o f .

ff 9.23.

DEFINITION. L e t B b e a n e i g h b o u r h o o d of t h e o r i g i n

i n autD s u c h t h a t

( a ) t h e mapping A s N

-+

expAsAutD is i n j e c t i v e

(b) N i s i n v a r i a n t u n d e r a 1 2 t r a n s f o r m a t i o n s 4 Then we s e t G = : expN a n d d e f i n e yt: expA

-+

expQtA

t

,

tdR

.

CHAPTER 9

210

for> t m and AsN.

Observe that by lemma 6.47 and lemma 9 . 2 2 such a neighbourhood exists,

Our next task will be to extend the mappings Y ' : G +. G to the identity component AutOD of AutD. By lemma 7.15, any GcAut D 0 admits a representation of the f o r m G= G1G2..Gn with G = expA and AkcN for k = 1,2,..,n so that we could set k

k

.. (YtGn)

Y t G= : ( Y t GI)

The trouble is that the representation of G we have used is not unique. 9.24.

PROPOSITION. L e t G1,G2,..,GneG b e s u c k t h a t

.

G1o G20 . o G n= idD a T h e n fort a l l tdR.

we

have

t t t ( Y GI). (Y G2). ( Y G ) = idD

.

Proof: Let us write

for tdR. We begin with the following observation: Given any XcautD, we have G#tX= X

(9.9)

for a l l tdR. Indeed, by assumption, there are A k e N , k = 1,2,..p such that G = expA Write X = : @ tY where Y=: $-tXcautD and t is k k' kept fixed; then, by proposition 5 . 1 3 and lemma 5.14, we have

= $

t

(GI#..G,#)Y=

From ( 9 . 9 )

$

t

t

t

(G1..Gn ) # Y = @ (idD ) # Y = $ Y = X . t

we c a n deduce G = idD' Indeed, set

21 1

BOUNDED SYMMETRIC DOMAINS

Fh =: Gt+h(Gt)-l for t, hdR. B y (9.9) we have

Fix=

(9.10)

X

for all XcautD and h m . Now we show that the mapping hdIR FhcAutD h= 0. Let A= A +LcautD be fixed; since ccE continuous real-linear mapping, we have -f

d +tA= dt

1 lim ( + t + h ~ - + t ~lim ) = (A h

h+O

is Tderivable at A cautD is a

-A

c

h+O

+

e

it)=

i(t+h)

Moreover, by lemma 6 . 4 5 , the mapping A-texpA, AcN, is real analytic with regard to the T topology on AutD. Thus, considering the composed mapping t + + t +~ exp we get the T derivability of t

+

t

A

I

YLH with H= expA. Applying

t

..

this to each of the Y Gk = expGtAk, k= 1,2,. , n , by lemma 1 . I 5 we get the Tweak derivability of Gt. Therefore, for some neighbourhood B of 0, we have 1

(Gtfh-Gt) converges in the norm

(1 .( I B

or, equivalently, 1 (Fh-idD) converges in the norm

I / .I/ G~ ( B )

whence it follows that

1 (Ft-id ) = A t T lim h D h+O

CHAPTER 9

212

t

for some A eautD, s o that h

-+

Fh is T derivable at h= 0.

Then, theorem 4.28 entails 1

T lim

h (F#x-x)=

h+O

[A~,x]

for all XeautD, so that by (9.10) we have [At,X]= 0 for all XeautD. Thus, by proposition 9.13 O= A

t

=

T lim h+O

1

tth

[Gt+h(Gt)-l-Gt(Gt)-l]= T l i m h (G h+O

t -1

-Gt) (G )

d t whence Gt= 0 for all t d R , that i s , G is constant and Gt= G o= i dD' 9.25. COROLLARY. L e t G1,G2..GncG a n d H1,H2..H,&G g i v e n and a s s u m e t h a t G,oG20..oG = Hl0H o . . ~ H T h 8 n (Y~G,). ( y t c n ) = ( Y tH ~ ) . ( Y ~ H ln~o )r a l , l 2 tm. m

.

.

.

be

Proof: We need o n l y to observe that, if H= expA w i t h AeN , then t

(YtH)-l= (exp$tA)-l=exp(-$A ) = exp[lp =

t

t

t

(-A)]=

-1

Y exp(-A) = $ (H

)

. #

Yt:

9.26. DEFINITION. F o r tCR, w e d e f i n e t h e mapping AutOD AutOD by means of -+

Yt: Gl..Gn w h e n e v e r GI

..

(YtGl) (YtGn)

+

,..,GneG.

We know that AutOD=

u

G";

therefore, in view of the previous

n m

proposition, the mappings Y Moreover, we have

t

are well-defined on AutOD.

BOUNDED SYMMETRIC DOMAINS

(YtG)(Y-tG)= G

and

Yt(GH)= (YtG)( Y t H )

213

,

for G, HcAutOD and tdR. 9.27. EXERCISES. Consider the mapping RxAut0D given by (t,G)

+

Y'G.

+

AutOD

Show that t

(a) For fixed G, the application (t,G) -t Y G is a one-parameter group t

Y G is real analytic when AutOD is endowed with the analytic topology Ta Is it T continuous?. (b) The joint application (t,G)

-f

.

57.- The pseudorotations on D. - ____ We recall that, by proposition 9.8, D is homogeneous under the action of AutoD, so that D = {G(O);

GcAutoD}

D i n the 9.28. DEFINITION. For tdR, w e d e f i n e Tt: D f o l l o w i n g m a n n e r : L e t xcD be g i v e n ; t h e n we h a v e x= G(0) f o r some GcAutoD, a n d we s e t -+

In order to see that this definition makes sense we have to verify that, for G1,G2eAut D with G1O= G20, we have (YtG )0= (YtG2)0. By passing to G=: G I 1G2 we must prove that, 1

t for all GeAutoD, the relation GO= 0 implies (Y G)O= 0 for all

tdR. This will be our next task. 9.29. DEFINITION. We s e t IsotD=: {GcAut D; So

GO= 0 )

far, we have made no u s e of the assumption concerning the

s i m p l e c o n n e c t i v i t y of D. We shall apply it to prove the

following:

CHAPTER 9

214

9.30. PROPOSITION. A s s u m e t h a t t h e b o u n d e d s y m m e t r i c d o m a i n D i s s i m p Z y c o n n e c t e d . T h e n t h e s u b g r o u p IsotD is a r c w i s e c o n n e c t e d v i t h r e g a r d t o t h e t o p o Z o g y Ta.

Proof: It suffices to show that, for any GsIsotD, there continuous path :'I [OJ] IsotD such that r ( O ) = idD and l'(l)= G.

exists a T

-f

Let GeIsotD be given; then we have GeAut D

and

GO= 0.

Therefore, we can find A1,A2,...,A,cN

We divide the interval

k for te [ 5

, k+l

1

such that

[O,l] into n subintervals

and k = 0,l , , , ,n-I . Obviously,

f

is a

T a continuous path which connects id and G in the space AutoD. D

In order to connect them in the subspace IsotD, we project this path f : [ O r I ] + AutOD into D by applying each ? (t)= G to the origin 0, so that we get the path y: [0,1]

t

+

D defined by

Since G belongs to IsotD, y is a closed path: y(O)= id,(O)= 0 and y ( l ) = G ( 0 ) . Thus, as D is assumed to be simply connected, y is homotopic to the origin 0. Let us denote by R=: [O ,I] x [ O r 11 the unit rectangle and denote by f: (s,t)eR + f(x,t)eD a homotopy in D continuously deforming the path y into the origin 0, so that we have

BOUNDED SYMMETRIC DOMAINS

215

f(O,t)= y(t)= Gt(0)

f(l,t)= id,(O)=

f ( s , O ) = idD (O)= 0

f(s,l)=

0

(9.11) G ( O ) = 0.

We shall construct a lifting of f: R + D to AutoD, i.e., a T continuous function f: R + AutOD such that

for all (s,t)cR. Then we shall have F(l,t)O= f(l,t)= 0 for all tc[0,1] , so that F(l ,t)eIsotD for te[0,1] and, by writing

we obtain the path

r:

[0,1]

+

IsotD we were looking for.

Let U be the neighbourhood of the origin in E constructed in the proof of proposition 9.8; thus the mapping (9.12)

g: ceU

-+

(exp Ao)Oeg(U)

is an isomorphism and

Then, we have

for some carathgodorian open ball B E ( 0 ) centered at 0. As the homotopy f: R -+ D is continuous, the mapping

is uniformly continuous on RxR; therefore, there exists an m a such that

CHAPTER 9

216

Now we devide the horizontal side [ 0 , 1 ] into m subintervals

of the rectangle R

and construct recurrently the lifting F of f on each of the k+l subrectangles R =: [ k , ]X[O,l], k = O,l,..,m-l.

m

k

We claim that, for (s,t e R 1

, we have

G;

Indeed, as the caratheodorian distance is AutD-invariant, by

(9.11) we have

Therefore, by ( 9 . 1 2 ) and we define

it makes sense to apply J-l to G;lf(s,t)

c(s,~)=:J - 1 Gt-1 f(s,t)

for (s,t)eRo. Let us set F0(srt)=: GtexpA

c(s,t)

for (s,t)eRO. Then, it is easy to check that Fo is a lifting of f over R o . Now we proceed by induction on k. Assume we had already

constructed a lifting Fk of f over R ; thus

for all (srt)CRk. We claim that, for ( ~ , t ) e R ~,+we ~ have

217

BOUNDED SYMMETRIC DOMAINS

Indeed, by ( 9 . 1 3 )

and the induction hypothesis we have

Thus, it makes sense to apply J-l to Fk- 1 ( ;k;i ,t)f(s,t) and we define

for (s,t)CRk+l.If we set Fk+l(slt)=:Fk(

mk

,t)expA c(s,t)

for ( s , t ) ~ R ~, +then ~ it is easy to check that F k + l lifts f on R k + l . Moreover, Fk+l and Fk agree on the common border of their rectangles of definition:

so that Fk+l extends the previous partial lifting. This

completes the proof.

# Let U and V be the neighbourhoods of 0 in E constructed in definition 9.8 and put R ~ = :x

-t

e itx

for tdIR and xeE. We may assume that U is an open ball centered t at 0 , so that U is invariant under the transformations R By setting

.

it is easy to see that V is invariant under the transformations St. Finally, we recall that, by proposition 9 . 2 1 , we have

21 8

CHAPTER 9

€or a l l AcautD. 9.31.

LEMMA. L e t t d R b e g i v e n .

T h e n , t h e r e a r e a number

6 > 0 and a n e i g h b o u r h o o d W of 0 s u c k t h a t we h a v e

P r o o f : Take a n y 6 > 0 s u c h t h a t t h e b a l l B 2 & ( 0 ) w i t h c e n t e r a t 0 and r a d i u s 2 6 i s c o n t a i n e d i n V;

then

S t [ B 6 ( 0 ) ] C V i s a neighbourhood o f 0 and w e d e f i n e

The p a i r 6,W s a t i s f i e s o u r r e q u i r e m e n t s .

11

I n d e e d : L e t AcautD b e s u c h t h a t

All

and t a k e any x e B 6 ( 0 ) .

Consider t h e i n i t i a l v a l u e problem d dt

y ( t ) = A[y(t)]

I

y(o)= x

whose s o l u t i o n i s d e n o t e d by y ( t ) = ( e x p t A ) x , a n d s e t T(x)=:

inf{t>O;

(1

(exptA)x-xl(

W e claim t h a t , r ( x ) > l . Indeed, f o r OSt
If it w e r e r ( x ) < l f o r some x e B 6 ( 0 ) , by t a k i n g a w i t h r ( x ) < a < I ,

w e would o b t a i n

BOUNDED SYMMETRIC DOMAINS

11 for all tc [ O J (x)

219

(exptA)x-xll S & a < &

, so

that

which is contradictory. Thus, -r(x)?l for all xcB6(0) and we have (expA)W c (expA1B& ( 0 ) c B 2 6 ( 0 ) C V t

Therefore, it makes sense to apply S (expA)Smt to W and, by (9.14), we get

aZZ

9.32. PROPOSITION. We h a v e (YtG)O= 0 f o r aZZ GcIsotD a n d tm. Proof: We recall that the family {exp A;

11

A I ~ v<

1

, na)

is a fundamental system of Ta-neighbourhoods of idD in AutD; therefore, the family of subsets

is a bases for the left uniform structure on the topological group (AutD, r a ) . NOW, let GcIsotD and tt3R be given. By proposition 9.30 there is a T a continuous path r : [0,1] -t IsotD connecting idD and G in IsotD. Let & and W be determined as in lemma 9.31. Since r is uniformly continuous on [0,1 3 , there is a partition O = s o < s1< . . < s m-1 < sm = 1 of [0,1] such that we have

CHAPTER 9

220

and, by lemma 9.31

for all j = O,l,..,m. Since OcW

and

G s . CIsotD, we get 1

for all j, and finally

= Y t (Go -1

G,)Yt(GY1 G2)..Yt(G-' m - 1 G,)O=

0.

# 9.33. PROPOSITION. T h e mappingsTt: D + D of d e f i n i t i o n 9 . 2 8 a r e w e l l d e f i n e d o n D. M o r e o v e r , f o r aZZ t m , Tt i s a hoZornorphic e x t e n s i o n of St= J-l0RtoJ a n d t + Tt is a Tcontinuous o n e - p a r a m e t e r g r o u p IR + AutoD.

Proof: We have already shown that the Tt are well defined on D. Now we prove that they are holomorphic in D. Let xcD be given and fix any GsAutOD with GO= x . Since J: c -+ (expA ) O is real bianalytic in some neiqhbourhood V of 0, so are the mappings

F =: c

-+

Y t (GexpA ) O =

( Y tG) (expA it) 0

2

e 1

t

for t fixed. Therefore T

= F2F;

c

is real analytic in a

neighbourhood F 1 ( V ) of x . By the arbitrariness of x, T t is real analytic in D. In particular, for x= 0, G= idD and V = W (where W is the neighbourhood of 0 constructed in lemma 9.31), we have

BOUNDED SYMMETRIC DOMAINS

221

Tt (expA ) O = (YtexpAc)O = [St (expAc)S-l] O = St (expAc)0 for all ceW. This means that Tt is an extension of S t . Tt is real analytic in D and, in W, Tt coincides with St holomorphic, Tt is holomorphic in D. Obviously, we have Tt(D)cD for t definition of T , we can easily s,t&; thus TteAutOD and t Tt AutoD. To show its T-continuity -f

Since , whichk

all tdR. Moreover, from the obtain T S + t = TSTt for all is a one-parameter group in it suffices to prove that

T lirn T t = idD

(9.15)

t'O

Since T

St, we have

"=

t

t

lirn T = lim S = lim J-loRtoJ= idD

t'O

t'O

t

As the transformations T (9.15) by theorem 1.6. S8.-

uniformly on V.

O't

are automorphisms of D, this entails

Construction of the image domain.

Next, a bounded balanced domain D, which is biholomorphically equivalent to D, can be defined in terms of the transfomtions t T . 9.34. DEFINITION. We introduce t h e m a p p i n g F: D

a n d we set A

D=: F(D). 9.35. LEMMA. extension of J.

6

is bounded, and F is a h o Z o m o r p h i c

-t

E

by

222

CHAPTER 9

Proof: W e h a v e

The holornorphy o f F i s c l e a r s i n c e x f o r a l l tCIR a n d t

-f

Tt i s

L e t xeV be g i v e n . A s T

+

T t ( x ) i s holomorphic

Tcontinuous.

t 0

under t h e transformations S

J and V i s i n v a r i a n t

fl

# 9 . 3 6 . COROLLARY.

Given AeCautD,

t h e r e i s a unique

A

polynomial! A: E F('A=

-f

E o f d e g r e e t w o s u c h t h a t we h a v e

i~. P r o o f : L e t AcCtautD b e g i v e n . By p r o p o s i t i o n 9 . 1 8 ,

the

mapping A=: J # A i s a p o l y n o m i a l of d e g r e e two t h a t s a t i s f i e s

i n t h e n e i g h b o u r h o o d U o f 0;

t h u s , w e h a v e AF= F ( 1A i n D . Any

o t h e r A'cCtautD s a t i s f y i n g F ( l A = i ' J c o i n c i d e s w i t h J # A o n U ; therefore

i\

i s uniquely determined.

# 9.37.

DEFINITION. For AeCautD,

denotes t h e unique

polynomial o f degree t # o a s s o c i a t e d w i t h A by c o r o l l a r y 9 . 3 6 . From p r o p o s i t i o n 9 . 1 0 ,

2 are

g i v e n by

w e know t h a t t h e p o l y n o m i a l s

ec, 5 ,

and

BOUNDED SYMMETRIC DOMAINS

9.38.

PROPOSITION.

fi

i s a connected open circular s e t .

P r o o f : O b v i o u s l y , Oc6. have

223

Since

Flv=

J a n d T t l V = St, w e

F T t ( x ) = F S t ( x ) = J ( J - l R t J ) ( x ) =e i t J ( x ) = e i t F ( x )

f o r a l l xcV a n d t d l R . A s F a n d Tt a r e h o l o m o r p h i c on D , f r o m the identity principle we derive F Tt = ei t F f o r a l l tclR. Then e

so t h a t

6

itA it D = e F ( D ) = F T ~ ( D )F ( D ) =

6

is circular.

N e x t , l e t xcD be g i v e n a n d d e n o t e b y B a n y b a l l c e n t e r e d a t x w i t h r a d i u s G < d i s t ( x , d D ) . S i n c e t h e mapping c continuous, t h e r e e x i s t s E>O whenever

/I

c/I

<E.

such t h a t w e have

-+

Cc i s

11

C

11 c I I

<€i

W e c l a i m t h a t t h e i n i t i a l v a l u e problem

h a s a maximal s o l u t i o n d e f i n e d on t h e i n t e r v a l [O,l]

i s ccE w i t h

11

c B

<E.

I n d e e d , d e n o t e by y ,

the solution of

Assume t h a t f o r some c =

( 9 . 1 6 ) and set

5 w e h a d ~ ( € , ) < lThen .

whatever

CHAPTER 9

224

Moreover, for tcr(i;), we have

so that

which is a contradiction.

11

Therefore, for each CCE with [O,l] 6 by means of yc

--

cII

<E,

we can define a curve

-+

;,(t)=:

F(yc(t))

and we have

f o r all te[0,1]

and ceE with

11 cII

<E.

But then

*

y, (t)= tc+F (X)

A

which is a neighbourhood of F(x). As x was arbitrary in D, D is open. Obviously, 6= F(D) is connected.

ff 59.-

The isomorphism between the and 6. ~ domains _ _ _ _D . . _ _ _ -

We complete the proof of ViguE's theorem by establishing that 3 is balanced and that the mapping F: D 6 is injective. -+

As we already know that speak of aut6 and Aut6.

6

is a bounded circular domain, we may

BOUNDED SYMMETRIC DOMAINS

9.39.

225

LEMMA. We h a v e iicaut6 for a l l AcautD.

for e v e r y GeAutoD, t h e r e is a u n i q u e k A u t

6 0

Furthermore,

s u c h t h a t FG= 6F.

P r o o f : F i x AcautD and xcD a r b i t r a r i l y , and s e t y ( t ) = : (exptA)x,

G(t)=

Fy(t)

f o r t d R . Then we have

for a l l

tm;

t h u s , i f w e p u t ;=:

Fx, t h e c u r v e t

-t

G(t)

i s the

s o l u t i o n of t h e i n i t i a l v a l u e problem

t h a t is ,

$(t)= (exptff);

ii s

Thus, t h e v e c t o r f i e l d

complete i n

6, i . e . ,

i c a u t 6 , and

w e have F (exptA) x = Fy ( t )=

$ ( t )=

( e x p t i )=;

(exptfi)Fx)

f o r a l l xeD and t d R . T h e r e f o r e FexpA= expffF. Now, l e t GeAutOD be g i v e n . Then we can w r i t e G= (expA1) ..(expAn) f o r some A1,..,AneautD,

where

so t h a t

&: ( e x p i l ) . . (expffn)eAuts. B e s i d e s , s i n c e

F I V =J , from e F = F G w e d e r i v e

CHAPTER 9

226

t I u =( F G F - ~ so that

d

JGJ-~

is uniquely determined by G. *

9.40. DEFINITION. For GeAutOD we d e n o t e by G t h e u n i q u e e l e m e n t o f Auto6 s u c h t h a t FG= 6F. 9.41.

COROLLARY.

6

is a homogeneous b a l a n c e d d o m a i n .

Proof: From F(O)= 0

we derive G ( O ) = eF(O)= FG(0) and

so that 6 is homogeneous. From corollary 7.10 we know that any homogeneous bounded circular domain 6 is balanced.

Let 6 D and 66 denote the carathgodorian differential metrics on D and 6 ( c f . definition 3.81, respectively 9.42.

LEMMA. T h e r e is a c o n s t a n t M>O s u e h t h a t we have

Proof: Let xcD and veE be given. Fix any GcAutOD with x = GO and put u = : (GA')-'v; by corollary 3.11 we have (1 6 ( O , u ) = 6 (GO,GO u) whence D D (1

1

6, (x,v)= bDr0 , ( G ~ ) - v]

Moreover, we have FO= 0 and Fo( 1 = id; thus, the relation FG= GF entails Fx= GO and F"e"= Therefore, again by corollary x o

"'.

3.11

BOUNDED SYMMETRIC DOMAINS

227

whence

From definition 3.8 and the fact that J= F is an isomorphism JV between V and U, it is easy to derive

for some M>O independant of x and v .

# 9.43. COROLLARY. T h e m a p p i n g c for all

+

(expCc) i s w e 2 2 d e f i n e d

~€6.

Proof: First, we consider the mapping c + (expEc)O. It is easy to see that the maximal solution of the initialmlue problem

(t)= tc. As 6 is balanced, the segment [O,l]c is 2 contained in D whatever is cs8. Therefore, (expec)0 is well defined for all c d .

is given by

Consider now the maximal solution y

of

We claim that r(c)>,l for all c c 6 , so that (expCc)O is well defined for all cs6. Indeed, assume we had -r(c)
228

CHAPTER 9

lim t+T

y,(t)=

a

(C)

By lemma 3.9, and lemma 9 . 4 2 ,

for O
is As we have seen in the previous step, the segment [O,l]c contained in 6; therefore, for te[t1,t2] we can find a ball B(t) centered at tc and contained in 6.

By proposition 3.12 we have

From the assumption r(c)
(9.18)

II

CII

dist(tc,a6)

c

so

C

dist ([O,l]c,a6)

From (9.17) and (9.18) we finally get

f o r tl,t2 7't(C). However, as D is homogeneous, by theorem

3.18, D is complete with respect to the metric dD; thus, there exists some aeD such that lim y c ( t ) = a, which is a t+T (c)

229

BOUNDED SYMMETRIC DOMAINS

contradiction.

# NOW, Vigu6's theorem is at hand. 9.44. PROPOSITION. T h e map F: D

6 is

-f

biholomorphic.

Proof: It suffices to show that F is injective. NOW, c -+ Cc is a complex linear continuous mapping; therefore, it is holomorphic. From the theory of ordinary differential equations we know that, for c&, the solution of the initial value problem (9.19) depends holomorphically on the parameter c. By corollary 9.43 the maximal solution of (9.19) is well defined on [0,1] whatever is cc6, i.e.

is a holomorphic mapping on 6.By lemma 9.35, we have F I v= J where J: (expCc)O + c. Therefore we get FF= idv , and by the identity principle J?F= idD. Similarly F$= idD' A

#

This Page Intentionaiiy Left Blank

CHAPTER

10

THE JORDAN THEORY OF BOUNDED SYMMETRIC DOMAINS

51.-

Jordan _triple product star alqebras. _

___.

.

10.1 DEFINITION. Let L s (ExEI E) denote the Banach space of all continuous symmetric bilinear mappings ExE + E equipped with its usual norm and denote by *: E + Ls(ExEIE) a continuous conjugate linear application c + c*. For ccE and x,ycE we write * * xc y=: c (x,y). T h u s , for e a c h f i x e d CeE, (E,c*) c a n b e v i e w e d a s a " n o n n e c e s s a r i l y a s s o c i a t i v e " Banach a Z g e b r a and t h e

-+

o p e r a t i o n c : ExE

+

E g i v e n b y (x,y) + xc*y is c a l l e d t h e

"c*-muI t i p l i c a t ion I f . E g i v e n by (x,c,yl Clearly we have

The o p e r a t i o n ExExE " t r i p l e product

If.

+

where M is the norm of the map c The s t r u c t u r e

(E,*)

+

c

*

+

xc*y i s c a l l e d t h e

.

i s c a l l e d a " J o r d a n t r i p Z e p r o d u c t star

algebra", or simply a J*-tripZe,if

it s a t i s f i e s the axioms:

( J 1 ):

x(a a*b)*x= 2 (aa*x)b*x-aa* (xb*x)

(J2):

(xa*x)b*x= xa* (xb*x)

f o r e v e r y a,b,xeE.

10.2. EXAMPLE. Let D be a bounded balanced domain in a Banach space E and E = : (aut D ) O . For 0 Q cL (ExEIE) denote the unique symmetric bilinear c s that the vector field x + c-Q,(x,c), xeD, belongs 231

circular ceEof let mapping such to aut D.

232

CHAPTER 1 0

Then, according to proposition 7 . 9 ,

the triple star product

fulfills (J1) and ( J 2 ) , i.e. ( E , , * )

is a J*-triple.

In particular, by ViguB's theorem, to every bounded symmetric

*

domain D we may associate a J -triple on its supporting space in a natural way. 10.3. PROBLEM. How can be algebraically characterized those J*-triples ( E l * ) €or which there exists some bounded balanced domain D c E such that the vector fields of the form x c-xc*x, xeD, belong to aut D for all csE?. -+

In this chapter, our main purpose will be to answer this question.

In the sequel, we shall use the notations 'c*=: x ab*=: x + ab*x for a,b,c,xsE. 10.4. PROPOSITION. G i v e n a J * - t r i p Z e

(El*),

xc*c

-f

and

t h e cZosed

r e a l L i e s u b a l g e b r a fl g e n e r a t e d i n t h e L i e a Z g e b r a P(E) o f aZZ p o Z y n o m i a l v e c t o r f i e l d s by &=: {c-'c*; ceE) a d m i t s t h e t o p o Z o g i c a l d i r e c t sum d e c o m p o s i t i o n

w h e r e L i s t h e c Z o s e d r e a l s u b a Z g e b r a of L ( E ) g e n e r a t e d try t h e f a m i l y {ice*; ceE}. P r o o f : We have

where Q 1=: &

and

Qn+'=:

v

[Qk,QR]

for

n>2

k+k=n

k,k>I

Herev denotes the real linear hull operation, as usually. Since

233

JORDAN THEORY

and, by

axiom J2 we have -2xa* (b-xb*x)+2xb* (a-xa*x)= 2 (ab*-ba*)x

[a-sa*, b-'b*]x=

i for all xeE, by putting a=: 2 c icc*c{ [a-'a*,

and b=: c, we obtain

b-sb*];

a,bcE}cQ

2

for ccE. Therefore L+&CA

The sum in the left-hand side is topologically direct, because clearly Q and L are closed subspaces in P(E). Moreoverl by writing L =: v{icc*; 0

ceE}

we obtain

for all a,bcE, whence

Q Let us define L: that (10.1)

2

= Lo

for nCiN in the same manner as Qn. We claim

Q2 k -

CQ

and

Q2k

k

C

v

r=l

Li

for k = 1,2,... Indeed, we have already established (10.1) for k = 1. Now we proceed by induction on k. Assume we have

for all s= l12,.~.,,k.Then

CHAPTER 1 0

234

v

Q2k+1= I:Q",Q"I= m + n = 2 k +1 m,n>l

v

v

[ Q ~ ~ , Q ~c ~ + ~p;,d I

s+t=k sB1,t>O

rSk

and also

Thus,it suffices to show that rL;,Ql

=u

,

r = 1,2,

...

For r=l this relation means [iaa*, b-sb*]e&

,

a ,beE

which is indeed true, since (aa*) ( I = aa* and s o , by axiom J 1 we have

[iaa*, b-sb*]x= =

I f we have

iaa*b-iaa*(xb*x)t2xb*(iaa*x)= iaa*b-ix(aa*b)*x= i(aa*b-s(aa*b)*)x

[L;,Q]cQ

f o r s= 1 , 2 ,

...,r

then, from the

definition of Lo' and the Jacobi identity, we derive

This completes the proof.

2 35

JORDAN THEORY

10.5. COROLLARY. We h a v e [ L , Q ] c g . N a m e l y ,

ac-S (ac)* = i.e.

[a ,

c-”c*]

x (Rc)*x= R (xc*x)-2xc* (ax)

xcE

f o r a l l RcL and ccE.

Proof: Let R c L and ccE be given. Then R = lim

an€

L i k=1

.

where

n-tm

Hence [R,,

c-’c*]cg

[ a , c-’c*]=

for all ndN and

lim [ R ~ , C-~C*]CQ n+m

From the linearity of R it follows that

Thus, the vector field v=: RC-~(RC)*-[R, c-”c*] belongs to Q and vanishes at 0. Therefore v= 0.

# 10.6. EXERCISE. Prove that if (El*) is a *-triple such that L@Q (defined as in proposition 10.4) is a Lie algebra of vector fields, then (El*) is a J*-triple.

52.- Polarization in J*-algebras. Before proceeding to geometrical considerations, we investigate the basic algebraic consequences of the J*-triple axioms.

10.7. LEMMA. L e t E, F b e c o m p l e x v e c t o r s p a c e s and A1,A2: ExE + F r e a l b i l i n e a r m a p s . T h e n , if s = s (i.e., A1

A2

A 1 (x,x)= A 2 (x,x) f o r a l l xcE) and A1,AZ a r e e i t h e r s y m m e t r i c and c o m p l e x b i l i n e a r o r s e s q u i l i n e a r m a p s , t h e n we h a v e A 1 = A 2 ’

Proof: It suffices to note that if A: ExE symmetric complex bilinear, then

+

F is

CHAPTER 1 0

236

and, if A: E x E

10.8.

-t

F is sesquilinear, then

PROPOSITION. Axiom

J 1 is e q u i v a l e n t to t h e

f o l . l o w i n g e v a Z u a t i o n .formula: (J)

x1 (fla*f2)*x2= ( a f ~ x l ) f ; x 2 - a f ; ( x l f ~ x 2 ) + ( a f ~ x 2 ) f ~ x l

Proof: The implication (J) => ,(J1) is trivial. Conversely, assume we have (J1), i . e .

x (aa*f *x= (aa*x f*x+(aa*x)f*x-aa* (xf*x)

.

This means that , for fixed a, feE, the symmetric bilinear mappings A 1 (xl,x,)

=:

x 1 (aa*f)*x2

A ~ ( x ~ , x ~ )(aa*xl)f*x2+(aa*x2)f*xl-aa*(xlf*x2) =: coincide for x = x 2 , and hence everywhere by lemma 10.7. 1

Therefore, for any fixed x1 mappings A; (a,g)=:

,

x2, feE, the sesquilinear

x 1 (ga*f)*x,

coincide for a= g, and hence for arbitrary a, gcE by lemma 10.7

JORDAN THEORY

10.9. COROLLARY. Axiom ( J 2 ) is a

237

c o n s e q u e n c e of

(J1).

Proof: Axiom ( J 1 ) implies (J); thus in particular

x (ax*b)x= 2 (xa*x)b*x-xa* (xb*x) But we have ax*b= bx*a, i.e. the terms a,b can be interchanged in the right-hand side. Therefore 2 (xa*x)b*x-xa* (xb*x)= 2 (xb*x)a*x-xb* (xa*x)

and so

.

(xa*x)b*x= xa* (xb*x)

PROPOSITION. Axiom ( J 1 ) is a l s o e q u i v a l e n t t o

10.10.

(J'):

,

[ . a *

bb*] = a (ab*b)*- (bb*a)a* ,

a ,bcE

Proof : Indeed , (J) implies

b (aa*b)*x= (aa*b)b*x+ (aa*x)b*b-aa* (bb*x)= (aa*b)b*x+bb*(aa*x)-aa* (bb*x) Thus [&*, proving (J' )

aa*]x= b (aa*b)*x- (aa*b)b*x

.

If we suppose that (J') holds, the pofarization argument of lemma 10.7 yields

[a,a;, whence

bib;]=

a l (a2b;b2)*-(blb;al)a;

CHAPTER 1 0

238

a l ( a 2 b ; b 2 ) *x= [ala;,

blb;]x+(blb;al)a;x=

= a 1a * 2 ( b1 b*x)-b 2 1b * 2 ( a l a ; x ) + ( b1b*a 2 1 )a;x= =

which i

(blb:x) a ; a l + ( b 1b*a 2 1 a;x-blb;

(J) f o r a

=

1

xl,

a2= f l

I

b = aI

(ala;x)

d

b2= f 2

= x 2'

# 5 3 . - F l a t subsystems.

~ t . t(F,*) be a rcaZ triple. we say that a s e t S C F i s a " f l a t s y s t e m " if ab*= ba* for all a , b e S . 10.11.

DEFINITION.

We s a y t h a t S is " e o m r n u t a t i v e or' u b e l i u n " i f aa*-bb* aa* coinmutes w i t h bb*) for a l l a , b c S .

(i.e.

F i n a Z Z y , S is " a s s o c i a t i v e " if

(x1 a*x 2 ) b * x3 = x 1 a * ( x 2 b * x 3 ) for a l l x 1 1 x 2 1 x eS and a , b c S . 3

By p a s s i n g t o r e a l l i n e a r combinations and u s i n g t h e p o l a r i z a t i o n argument, it i s e a s y t o o b t a i n the f o l l o w i n g

10.12.

LEMMA. L e t

(F,*) b e a r e a l J * - t r i p l e

and S c F .

Then ( a ) S is f Z u t , i f a n d o n l y i f Vs is f l a t

(b) S is abelian ; f

a n d o n l y if VS is a b e l i a n , w h i c h

o c c u r s if a n d o n l y if we h a v e albT+bla;-a2b;+b

for a l l a l f b , ,

2 a* 2

a 2 ' b2&.

( c ) S is a s s o c i a t i v e i f and o n l y i f VS i s a s s o c i a t i v e .

JORDAN THEORY

239

If ( F , * ) i s a complex: J * - t r i p l e and S c F t h e n We a l s o h a v e . (b') S i s a b e l i a n i f and o n l y i f a:VS i s a b e l i a n , w h i c h o c c u r s i f and o n l y i f we have

f o r a l l a l l b l , a2, b2sS.

(c') s i s a s s o c i a t i v e i f and o n l y if Here 'VS

denotes the complex linear hull

VS

(I:

of

is a s s o c i a t i v e .

S.

10.13. EXERCISE. Prove the above lemma.

10.14. LEMMA. E v e r y e o m p l e x f l a t J * - t r i p l e

i s abelian.

Proof: Let a,beF; as F is flat we have ab*= ba* Since F is complex, iasF whence

and ab*= 0. Then a b*= a2b;= 0 1 1 F is abelian.

for all a l l b l , a2 b2eF

and

# 10.15. PROPOSITION. E v e r y r e a l f Zat J * - t r i p Ze i s associative.

Proof: Let x l , x2' x3eF and albeF be given. Applying axiom (J) and the fact that F is flat, we derive (xla*x2)b*x 2 3 +(ax*x 1 3 )x;b-ax*(bx*x 1 3 = b(x 1a*x 2 )*x3= (axTb)x*x 2 3) = =

=

x2 (axTb)*x3+(xla*x3) b*x2-xla*(x2b*x3 ) =

[ (xla*x2)b*x3+(X1a*x3)b*x2-x 1a* (x2b*x3)]+

CHAPTER 1 0

240

+ ( x1 a * x j ) b * x 2 -x 1 a*(x2b*x3)

S u b s t r a c t i n g ( x a*x ) b * x 3 we e a s i l y d e r i v e 1 2

( x1 a * x 3 ) b * x 2 = x l a * ( x 3 b * x 2 )

# b4.-

elemsnL.

S u_ b t_ ri_ p l e_s g e n e r a t e d by a n _ 10.16.

DEFINITION. L e t

(E,*)

be a J * - t r i p l e .

A subtriple

of (E,*) i s a c l o s e d s u b s p a c e F of E s u c h t h a t FF*FcF,

i.e., we have x y * z s F whenever x , y , z e F . A c c o r d i n g a s F i s a r e a l o r a complex s u b s p a c e of E , we s p e a k of a r e a l o r a c o m p l e x s u b t r i ple. Given any s e t S C E , t h e s r n a l Z e s t s u b t r i p 1 . e of

(E,*) c o n t a i n i n g is c a l l e d t h e J * - s p a n of S . The s m a 1 I e s t complex s u b t r i p l e c o n t a i n i n g S i s t h e 'J*-span of S . I t is c l e a r t h a t S

m

J-span

v

S=

S"

n=O

where

s 0 = : vs

and

S"+l=:

v

Sk(SR)*Sm

k+R+m=n

Similarly, 'J*-span

S=

H e n c e f o r t h w e assume t h a t ( E , * )

LT

V(J*-span S )

d e n o t e s a complex J * - t r i p l e .

G i v e n CCE, w e s h a l l w r i t e F C = : J*-span Furthermore, we set

{c}

,

EC=: 'J*-span

{c}

241

JORDAN THEORY

cn=: for n= 1,2,..., i.e., to c-multiplication.

c

n

n- 1c

(CC")

is the n-th power of c with respect

10.17. LEMMA. We have ck c R*

=

ck+R-1c*

f o r k i t = 1,2,..

Proof: The stament is obvious for k = R = 1. Suppose that, for some n, we have (10.2)

CkC

whenever k+R= r+s

R*

r s* = c c

and k , R , r, stn. Then, the system S

n

=:

{ck ;

k= 1,2,...,n~

is flat. Therefore Ck cR*,

Let k , Rsn derive

cr cs*

k,

r, s s n

R,

k

and consider the term c (cR+l)*. From axiom J we

ck(cE+l)*x= Ck(CC* c R ) * x= =

(CC*

ck)cL*x+(cc* X)CR* ck- cc*(ck cR* x)=

= ck+l cR*x

-

[ck cR*, Cc*]x=

k+l cR*x

c

f o r all xcE. Hence, for any rtn we have

which is the relation (10.2) with n+l instead of n.

10.18. THEOREM. F C is a f l a t s u b t r i p Z e of (E,*) a n d a,

FC=

v

n=l

cn

CHAPTER 1 0

242

Proof: By lemma 1 0 . 1 7 , the system S=: {en; n m } is flat, and hence so is its closed linear span. It is clear that v S C F c . On the other hand, from ( 1 0 . 2 ) we see

*

= c c for any k, R , n= 1,2,... subtriple; thus F c C V S .

c

k+R+m-2

That is, S

k+R+m-1

= c

S*SCS

whence VS is a

#

10.19. COROLLARY. EC is a n a s s o c i a t i u e a b e l i a n

s ~ t br i p Ze a n d m

EC=

v

n=1

§

5. - J&--triples 10.20.

cn

and- Hermitian operators. DEFINITION.

ue s a y that a ~ * - t r i p ~ (eE , * )

is

" b o u n d e d " , o r t h a t it i s a JB*-triple, if there e x i s t s a b o u n d e d o p e n n e i g h b o u r h o o d U o f t h e o r i g i n i n E s u c h that a l l v e c t o r f i e l d s c-s c * for ccE a r e cornpZete i n U .

Curiously, the letter B h e r e refers to Banach's name and not to the adjective "bounded", because these structures are closely related to the Jordan-Banach algebras. In the sequel we shall reserve the notations & , L , L o and A to designate the families of vector fields(cf. Proposition 1 0 . 4 ) Q = : {c-'c*;

ceE}

L=: Lie span(L 0 ) 10.21.

1. 1

,

,

L 0 =:

{ice*;

ccE)

A = : L @Q

LEMMA. If (El*) is a JB*-tripZe, t h e n we c a n find

o n E, which is equivalent to t h e o r i g i n a ~o n e sucli that, for every LcL, expL is a n isometry f o r 1 . 1 .

a norm

(1 (1 , *

JORDAN THEORY

243

Proof: Choose U in accordance with the above definition, i.e., c-S * c c autU for all ccE. Since autU is a Lie algebra Iu of vector fields, we have A ecautU for all AcA. In particular, lu L cautU for all LcL. Iu

Let us consider the functional defined on E by Iv/=: 6u(o,v)

v cE

From lemma 3.9 we know that 1 . I is a continuous seminorm on E. Moreover, if 0 < 6 and M are such that GB(E)cUcMB(E), then

Given any AcA,

we have exp(A

1u

)eautU

and therefore

for all VCE, i.e., (expA ) ( ' is a 1 . (-isometric linear IU 0 operator. To complete the proof it suffices to remark that every LcL is a linear operator, thus (expL) expL.

A'=

ff 10.22.

COROLLARY. For a l l ccE a n d tdR, exp(itcc*) is an

isometry w i t h r e s p e c t t o

/ .1 .

1 0 . 2 3 . EXERCISE. Prove lemma 10.21 elementarily by showing that the convex balanced hull co(AU) of U is an exp-invariant balanced convex bounded negihbourhood of 0, and

so its gauge can be taken as

1.1.

1 0 . 2 4 . DEFINITION. We say t h a t a l i n e a r o p e r a t o r AcL(E) is h e r m i t i a n if exp(itA) is a 11 I/ - i s o m e t r y f o r a22 tdR. We s e t Her(E) f o r t h e s e t o f h e r m i t i a n o p e r a t o r s o n E.

-

Thus, AcHer(E) if and only if iA

/B(E)

eautB(E)

that Her(E) is a Lie subalgebra of L(E).

. This fact shows

CHAPTER 1 0

244

10.25.

EXAMPLE. In a Hilbert space H we have A i s selfadjoint}= {A€L(H); (Axlx)dR

Her(H)= {AcL(H); where

VxcH}

is the scalar product on H.

<,> 10.26.

LEMMA. A Z i n e a r o p e r a t o r AeL(E) is h e r m i t i a n if

a n d onZy if <$,Ax>t-IR w h e n e v e r xcE and $€El a r g s u c h t h a t

($,x>=

II $ 1 1

/I XI/

*

Proof: Let u5 fix x 8 E and A e L ( E ) arbitrarily, and consider the function 9 : IR IR given by -f

g(t)=:

/I

Xt//

where x t = : exp(itA)x.

Since t + xt is analytic and y + I] y 11 is lipschitzian, 9 is locally lipschitzian. Therefore g'(t) exists except for a subset S ( A , x ) of IR whose Lebesgue measure i s null, and we have

(10.3)

g'(t)= Re< @ , i Axt>

I/ $11

Indeed, Let $ be in the above conditions and put For sdR we have

=
1 S

( X t + S - ~t) > =

Y=:

1

s+o

(Xt+s

II $ 1 1

Re

Thus cp'(t)= lim Re<'?, 1 s

\y .

-xt ) > = Re

JOBDAN THEORY

245

proving ( 1 0 . 3 ) . Now we prove the lemma. Let AcL(E) be such that

Fix any xcE and any tdR\S(A,x). By the Hanh-Banach theorem, there are functionals @eE', @ f O such that / I XtII NOW we can apply (10.3) to any of these <@,xt>= / I $ 1 1 funetionals t- to obtain

.

4'(t)=

Re<

'

-, I1 @ I 1

iAxt>

As A is assumed to satisfy (10.41, we have < @ , Axt>dlR whence cp' (t)= 0. Thus, is null almost everywhere in IR and q7 is constant. Then

9'

Since xcE and tdR were arbitrary, we get AcHer(E). Conversely, let AeHer(E) be given and assume that xcE and $eE' are such that <@,x>= / / [IxI / Then we have for all tcB , so that is derivable q(t) = 11 XtII = 11 x 11 everywhere in IR and f ' = 0. Applying (10.3) to t=O we get

.

9

whence < @,Ax>dR.

ff A fundamental fact concerning hermitian operators is the following result known as Sinclair's theorem. As usually, Sp(A) and p ( A ) denote respectively the spectrum and the spectral radius of A. 10.27. THEOREM. If AsHer(E) t h e n Sp(A)c7R a n d p(A)=

11

All

.

CHAPTER 1 0

246

Proof: Let A denote the closed complex subspace spanned 2 in L ( E ) by the family {I, A , A , . . . I . Then A is a commutative Banach subalgebra of L ( E ) . According to Gel'fand's theorem, there is a compact topological space K and a continuous algebra homomorphism g: A -+ C ( K ) such that

for all X e . A , where ran g(X) denotes the range of the function g ( X I We have

.

(I g ( X ) (I

IS!=

max

=

(1

P(XIC(~X

Sespx i.e., g is a contraction. Thus

g(exp itA)= g

[ n=OC (;:ln An] = c - (it)" n! g n=O

( A ) "= exp [its (A)]

Let us fix SeSpA arbitrarily. Then S = g(A)z for some z e K ; therefore, as g is a contraction, we have

I=

11

g(exp itA) ( 1 =

exp itAjI =

/I

exp itg(A) I( >(exp[itg(A)]zI=

lexp(it6) I = exp(-tImc)

€or all t m . Thus I m S = 0, whence SpAcIR. In order to prove

11

p(A) / I

=

11

All, we define

f o r rdR and observe that ArcHer(E). p(A,)=

It is easy to see that

Irl and we recall that 1

sin A r =: 2i [exp(i ~ ~ ) - e x p ( -Ai,)]

247

JORDAN THEORY

i.e.

(1

(10.5)

sin A

11

+ 2 2

6

=

for all rCR. Since

W

= I - (-1) 2n+l n=O

(-I/2)< 2 n + 1 for

n

,I]

EC[-I

n -1/2

) > O for all ndN, the theorem follows n immediately if we establish that the series

and since (-1) (

W

c - 2n+ (-l)n (-:I2) 1

(sin

n=O

converges in

L ( E ) for a l l r, Osr
-1

(10.6)

A--= sin L

Indeed, (10.5) and

II

m

ArlI c

c

n=O

1

I 2n+l

0sr
O~r
sin A r )

10.6) imply m

-1/2 1 11 n =

for

sum satisfies

(-1ln

sin A ~ Ic /c ~ 2n+l (

sin-' I = n / 2

n=O

-1/2

I=

248

CHAPTER 1 0

To prove (1O.6lf observe that

Since, by Gel'fand's spectral radius theorem, for BeL ( E )

,

p

(B)= lim n-fm

]bn11*'

the map

B

sin-'B=:

-f

(-11~ ( 2n+ 1

C

m

n=O

1/2)~2n+l n

is well defined and holomorphic on the domain I B C G ( E ) ; p(B)
=

{sin

E;

O
if

SCS~A.}={sin 5; <e[-rfr]Ic[-sin

r, sin r]

i.e., p(sin Ar)
For sufficiently small values of r, we have sin-'(sin A ) =

c

m

n=O

(-1 ) 2n+l (-;/2)

[ ; (2m+l)!

( - I ) ~ A 2 m t 1 j 2 n t 1 -r

m=O

t4

=

k c k=O ak Ar

where a = :

c n=O

(42) 2n+l n

(2ml+l1 t

c ..

..

t

n

(-1)

m j

m 12m +1) =k j T =1 1

m l , ,mk20 m

However, we have < = sin-'(sin


1 k =O

a 1= 1

and

a k= O

ak

Ek

for 1 5 1 < 1 , whence

for all k f l , which proves 1 1 0 . 6 ) .

JORDAN THEORY

249

56.- ____ Function model for E C .

*

DEFINITION. A J - t r i p l e ( E l * ) is s a i d t o b e " H e r m i t i a n " if cc*eHer(E) for a l l c c E . 10.28.

In this terminology, lemma 1 0 . 2 1 can be rephrased as 10.29.

LEMMA. E v e r y JB*-triple i s t o p o Z o g i c a Z Z y

i s o m o r p h i c to a h e r m i t i a n o n e .

Throughout this section, ( E , * ) will be a fixed hermitian * J -triple such that

for all x,y,zeE and some We know already that E =

ME. 'v

We also fix an element ceE, cf0. cn is a commutative and

n=l

associative complex subtriple of E. Therefore, by setting x.y=: xc*y

(x,yeE)I

the structure (Ec, .) is a commutative Banach algebra, F C is a closed real subalgebra of EC and cn is the n-th power of c in (Ec,.), so that the notation is not misleading. It is convenient to introduce the multipZication o p e r a t o r r e p r e s e n t a t i o n Rc: EC * L(Ec) given by

Rc: x Obviously, R

-f

xc*

I EC

is a continuous algebra homomorphism.

10.30. LEMMA. We h a v e FC (Fc)*cHer (E); i n p a r t i c u l a r R xeHer(EC) f o r aZZ xcFC. Proof: It suffices to see that cm c"*eHer(E) for m,n= 1,2,.. But we have 4icm cn*= [(icm)-s(icm)*

,

cn-s cn*Je:[Q,Q]CL= Lie spn{iaa*; aeE}C

CHAPTER 1 0

250

C

iHer (E)

since iHer(E) is a Lie subalgebra of

L(E)

.

ff

In view of Sinclair’s theorem, we have the following more precise picture of Gel’fand’s representation for (Ec,. ) : We set K=:

(thus

As

S~(CC*

I EC

)

CO(K)3C(K)

and

CO(K)=: {fcC(K); f ( O ) =

01

if O 4 K ) with

u s u a l l y , we write idK for the identity function on K.

THEOREM. T h e r e is a unique Banach a l g e b r a homomorphism G: E C + C O ( K ) such that 10.31.

and

acEC

(10.8)

P r o o f : Let A o C R [t] be the algebra of the polynomials N vanishing at 0. For p(t)= I: rk tk‘Ao let us define k =1 N

p(c)=:

c

k=l

yk

k

T h u s , if- denotes the closure in C O ( K ) , we have

JORDAN

THEORY

251

E C = { p ( c ) ; pcAOII f q ( t ) c A o h a s r e a l c o e f f i c i e n t s , t h e n R c q ( c ) c H e r ( E C ) a n d , by

S i n c l a i r ' s theoreml we have (10.9)

11

R c q ( c ) 11 = p[Rcq(c)]= P [ q ( R c ( c ) ) ] =

If p ( t ) eAO i s a r b i t r a r y , then

1 where p 0 ( t ) = t : Ip(t)

12.

From ( 1 0 . 9 ) w e o b t a i n

(10.8')

Moreover, w e

RC(

whence, a l s o

From ( 1 0 . 7 ' ) i t f o l l o w s t h a t t h e mapping g o : A O ( c ) * C O ( K ) g i v e n by

252

CHAPTER 1 0

g,b~c)]=: p

PCAO

is a well defined continuous algebra homomorphism and

II

( 1 0 .7,,1

g o b ) II

II

K=

cc* aa*

I EC II

aeAo (c

f

It is an imnediate consequence of ( 1 0 . 7 " ) that g admits a 0 continuous extension to E'= A*(c)- satisfying (10.7). NOW let ( P , ) ~ ~ C Abe ~ a sequence such that p (c) a. Then, by (10.9) n we have +

d

2

II id, /P,l -

1 q

lP,l

Thus, the sequence

.

(

2

II

K=

1

I1 RcC t

I P p

I 2- t1

I P p

2

I Ill=

1 p n ) n m converges uniformly to some ldK

element of Co (K) But we have p whence

=

g,[p

n

(c)]

-+

g(a) in C O ( K )

,

and (10.8) holds.

# COROLLARY. We h a v e g(ala;a3)== 1 g(al)g(a2)g(a3) K for a l l a l , a2, a3cEC. 10.32.

JORDAN THEORY

Proof: I n d e e d , c k c ktk-1 m ktL+m-1 = c .c = c

I

- - id,

R*

m

c = c

253

k+R-1 c x cm=

i.e.

( a i d k ) -( @ id,)R (y id:)= K

1 0 . 3 3 . EXERCISE. L e t SeK, < > O

and a , x o c E C

be g i v e n a n d

define

Show t h a t

Hint: 10.34.

DEFINITION. Ve s e t

11

all c = :

" t p(aalEc) f o r acEC

and Uc=: { a c E C ; /I a ] \ c < l } . F u r t h e r m o r e , for ad! a n d ~ d R \ { O l , we d e f i n e t h e f u n c t i o n q a 1 5 : C + E b y m e a n s of

1 0 . 3 5 . L i 3 l M A . L e t cm\{O} be g i v e n a n d s u p p o s e t h a t D c E i s a bounded open domain such t h a t q e a u t D for a 2 1 ale acc. T h e n we h a v e < > O and D = A.

ID

254

CHAPTER 10

Proof: C l e a r l y

w e g e t a u t D 3 t t R ; tdRl where R i s t h e v e c t o r f i e l d R:c+ ic. it Thus, e x p ( t R ) r = eitc and e D = D f o r t d R . L e t u s c o n s i d e r any < o ~ a Dand any act:, a # O . Then, w e c a n f i n d

a sequence ( < , ) n m ~ D

s u c h t h a t 5,

+

5,.

Write

e x p ( t q a l c ) c n f o r ndN and t d R . I t i s w e l l know t h a t ,

‘;n,t=:

f o r some 6 > 0 , w e have S o , tcE f o r

It I< p .

would imply

c,=

lim 5

n+m

I= Ico I .

se (-6

,6)

Since 5

for

On t h e o t h e r h a n d ,

exp ( - t q a ,

Lo ,

D n aD= $. It follows t h a t 5

I Go

=

I n d e e d , if

O,t

I co,

+

LO,,

mini

, we

0

c a D for I t [ < & and hence

I f 15,

I,

have

Ico I ,

for some m. But t h e n i.e. 5 0 , uE:

contradicting 5

0,u

eaD

I f w e had c0= 0 , then

colt=

0,

c o ,t&D b e c a u s e c o ,t e D

c D c o n t r a d i c t i n g 5 e 2 D and

such t h a t

n.s

I t / < & .Thus

‘tt

whence

t h e n we c a n c h o o s e

JORDAN THEORY

a contradiction. Thus

co=

2 55

0. Now we have

Since we may replace c 0 by e it c o for any chapter 8, proposition 8 . 4 , yields

the argument of

tdIR,

i.e.

5 = 150/2 for all COe2D. This is possible onlyif

D=

D since D is bounded.

10.36.

THEOREM. L e t

<>O

and

ff US

c o n s i d e r t h e following statements:

(a) E C is a J B * - s u b t r i p l e of (E,*). (b) Sp(cc* (c)

11 .I/

I EC ) C 7 R +

a n d aa*a#O f o r a l l a e E C \ { O ) .

is a norm o n EC a n d cc*

I EC

i s injective.

(d) Uc i s b o u n d e d . T h e n we have (a) = > (b) => (c) a n d (a)<=> (b)+(d).

Proof: (a) = > (b). Let U be a bounded neighbourhood of s * eautU for all acEC. Let us fix L@K\{O} lu arbitrarily and consider D =: {g(a)<; a & ) . The mapping 5 a + g(a)5 is a continuous linear functional on EC which is non identically null; consequently, D is a bounded open subset of 5 C. Given any x SU and acEC, consider the orbit 0 in EC such that a- a

0

x t =: exp t(a-sa*)x 0

tm

d xt= a-xta*x * therefore if we define We have dt t'

CHAPTER 1 0

256

i =: t

g(x,)E

and

a=: g(a)F,,

we have

As a consequence of the previous lemma, we have t ; > O and D = CA. E,

On the other hand, the relation aa*a= 0 entails

d dk

(ta)= a= a-(ta)a*(ta)

t m

i.e. {expt(a-sa*)0; tdR}= R a c U whence a= 0. (b) => ( c ) . We have

Thus, we have

11

all c < l

if and only if

whenever / g ( a ) < ( < c, i.e., g(a)
As

11

all c < 1 3 =

is a convex set in

(c,

n

(eK\{ 0 3

for a l l <EK\{O}.

That

{aeE ; g(a)Sc'JSA)

the linearity of g entails the

2 57

JORDAN THEORY

c o n v e x i t y of u'. The f u n c t i o n a l

/I . I (

i s o b v i o u s l y p o s i t i v e homogeneous, i . e .

l a ( (c = / A / ( 1 all C . Furthermore, g i v e n a c E C \ { O } aa*a#O by h y p o t h e s i s , and s o

(1

Thus

11 - I(

w e have

i s a norm on E C .

R c ( c ) a = 0 . L e t u s choose a sequence of polynomials ( p , ) n m C A O such t h a t p , ( c ) + a . By

NOW, assume t h a t cc*a= 0 , i . e . ,

setting q,(t)=:

t1

/p,(t)

12,

w e have

whence, by t a k i n g t h e l i m i t , we o b t a i n aa*a= 0 and a = 0 . ( d )+ ( b ) = > ( a )

.

I n d e e d , Uc i s an open neighbourhood of

0 i n EC s i n c e t h e map a

U c = {acEC;

11

+

aa*

i s c o n t i n u o u s and

I EC

a a * l E c / l< I > . L e t u s show t h a t a- a *

I uc

cautUC f o r

a l l a s E C . S i n c e polynomial v e c t o r f i e l d s a r e bounded on bounded

s e t s , t h i s means t h a t t h e maximal s o l u t i o n of d d t Xt=

(10 . l o )

* a-x t a

t

x = x 0

p a s s e s through t h e boundary of Uc f o r some xcUc. T h a t i s x

cauC

o r , equivalently,

f o r some toe%?,

while

CHAPTER 10

258

Let us write yt=: g(x,)<

as usually. From

and

a=: g(a)<

(10.10) it follows that t m

(10.10')

With the aid of chapter 8 proposition 8.4,it is easy to see that q is complete in JCA. Since yOcKA, it follows that a,< e e A , a contradiction.

yto

(a) = (d). Let U be a bounded open neighbourhood of 0 in EC such that a-sa* rautU for all acEC. We prove that

uccu.

lu

Given any a8ECI we have exp (a-S aX ) OeU. Let us define xt=: exp t (.a- s a* ) 0 and, once ( e K \ { O }

t m

has been fixed arbitrarily, set

Thus the functions t + xt and t + y, satisfy the differential equations (IO.lO), (10.10') with the initial values x = 0 and y= 0 , respectively. Since we have already established the implication (a) = > (b)I we know that < > O . In this case, the explicit solution of (10.10') is given by

Y,=

IaI l a / c1 - tanh -

V T

e

tm

259

JORDAN THEORY

That is, for asEC we have

(10.11)

s *

g [exp(a- a

01=

Jid,

tanh

-

rnK

To conclude U c c U we must show that, for every b d , there s * exists some asEC such that b= exp(a- a )O. Observe that, from the hypothesis (a), it follows that the Gel'fand representation g of EC is injective. Indeed, (a) = > ( b ) = > ( c ) whence, in particular,

whenever xcEC\ { O }

. Thus , fixing bsUC

arbitrarily, it suffices

to find asEC such that

i.e.

We have

Therefore

= cm n=O

1 2 n + l 9 [(bb*) nb]

CHAPTER 1 0

260

since

/g(b)I 6 11

rnK ..

implies also

11

blI c < l

bb*

I EC

for beUC. Moreover, the relation bcUC

I(

< I , whence we see that the mapping W

?(b)=:

1 n=O

2n+l (bb*)"b ,

bcUC

is well defined and we have

g(b)

Thus, the choice a=:

suits ( 1 0 . 1 2 )

. Consequently U c c u .

In the courSe of the proof (a) = > (b) we have seen that, for any < c K \ { O } , we have c > O and {g(a)(; aeU}= D =

5

C A

Hence, it readily follows that

i.e., U C U c . Thus U = Uc, which completes the prooE.

k! COROLLARY. If EC i s a J B * - s u b t r i p Z e o f (E,*), I\ * I [ i s an e q u i v a Z e n t norm o n E C , t h e u n i t baZZ of is Uc, and Uc is t h e u n i q u e bounded o p e n n e i g h b o u r h o o d 10.37.

then

/I . (1

s *

of 0 in EC in w h i c h e v e r y v e c t o r f i e l d a- a Moreover,

9:

b

+

$'(b)=:

C n=O

'

,

aeEC

(bb*)nb, 2n+l

,

i s complete.

bcUc

is a r e a l b i a n a Z y t i c mapping of Uc o n t o EC u h o s e i n v e r s e i s g i v e n by

JORDAN THEORY

26 1

We close this section with the following J o r d a n r e p r e s e n t a t i o n of the JB*-triples generated by a single element. 10.38. THEOREM. I f EC i s a JB*-subtripZe o f E, t h e n t h e m a p p i n g j: a

(E',*)

+

- g(a)

rnK

o n t o (c,,(K)

,*I.

i s a t o p o l o g i c a l J*-isomorphism

of

H e r e c ~ ( K ) is e n d o w e d with i t s n a t u r a l

t r i p l e p r o d u c t flflf3=: f

f

1 2 3'

Proof: We have

for all a l r a,, a3eE

.

Thus j is a J*-homomorphism. Furthermore,

.

Since the norm / / 1 1 is equivalent to 11 / I c on EC, it follows that the range of j is closed in C0 ( K ) and j is a topological J*-isomorphism of EC onto range (j)C C o ( K ) Observe that

.

Thus, in view of the Weierstrass-Stone theorem, we have range(j1 y{jb(c)];

peAO}- =

CHAPTER 10

262

COROLLARY. E v e r y J B * - t r i p l a g e n e r a t e d by a s i n g l e el’ement i s t o p o l o g i c a l l y J * - i s o m o r p h i c t o ( C O ( K ) ,*) f o r some 10.39.

compact s u b s e t K c l R & .

10.40. EXERCISES. (a) Let R be a compact space, acR and ca(sZ)=: I f c C ( R ) ; f(a)= 0 1 . Prove that (C,(Q),*) is a

JB*-subtriple of C(Q) with respect to the natural triple product. is an equivalent norm on C a ( n ) . (b) Assume that 1 1 I/ PrQve that ( C a ( i l ) , / / / / a , * ) is a hermitian JB*-triple if and only if /I / I a is a l a t t i c e n o r m , i.e.,

If

. 1w21= >

57.-

-

I l f Ila 41 f 2 1 1

a

a

( E c *)aS a commutative Jordan alqebra. I -

Next we extend o u r considerations from the associative algebra ( E C , * ) to the non necessarily associative algebra (E:*). As previously, ( E y * ) is a fixed hermitian J*-triple, ceEC\{O} is also fixed and we write xy=: xc*y ( x , y c E ) and X=: SP(CC*),~~ I

respectively. 10.41. DEFINITION. L e t A b e a c o m m u t a t i v e b u t n o n n e c e s s a r i l y a s s o c i a t i v e Banach a Z g e b r a . We s a y t h a t A i s a commutative J-algebra

if

2

w h e r e u = : uu. G i v e n a c o m m u t a t i v e J - a l g e b r a A, R: A

-+

the linear operator

L ( A ) g i v e n by

R(u)v=: uv

u ,v c A

i s t h e m u l t i p l i c a t i o n r e p r e s e n t a t i o n of A.

The importance of R derives from the fact that &(A) is an

associative algebra. The axiom ( J A ) can be stated as

263

JORDAN THEORY

(JA')

R(u2)-R(u)

ucA

.

R(u) commute) in terms of R.

(i.e., R(u?) and

If A is not associative, in general we do not have R(xy)= R(x)R(y). This difficulty can be overcome by sacrificing L(A) given by linearity. The quadratric representation P: A -f

) P(u)=: ~ R ( u ) ~ - R (2 u

ucA

is one of the most powerful tools in the study of J-algebras. It can be shown that P(xlP(y)= P(xy) for x,yeA; however, the usual proofs (see e.g. 161) are not elementary.

10.42. PROPOSITION. T h e algebra ( E l - ) ,i . e . , E with the c-multiplication, is a commutative J-algebra such that

(10.13)

2

R(u )R(u)= R(u)R(u

2

)=

2 7

3

R(u) +

1 5 R(u)~

u€E

Proof: Obviously, ( E l . ) is a commutative Banach algebra. To prove (10.13) we start from ( J 2 ) We have

.

(cu*clv*c= cu* (cv*c) u[(cu*c)v*c]*u=

u[(cu*(cv*c)]*u=

U[(CV*C)U*C]

= u[cv* (cu*c)]*u

for any fixed U,VCE since the u and v-multiplications are commutative. Using axiom (J) we obtain u [(cu*c)v*c] *u= 2 [V(CU*C) *u] -V(CUXC) * (uc*u)= =

2 [(uc*v) c*u+(uc*u)c*v-uc* (vc*u)]c*u-

- [(uc*v) c* (uc*u)+ (uc*(uc*u)) c*v-uc* (vc*(uc*u I ] = =

2 [ (uc*u)c*v]c*u- [(uc*u) c* (uc*v)+ (uc*(uc*u) c*v-

CHAPTER 10

264

( R ( u ) v )+

-UC*(Uc*u)C*v)]= 2R(u) [R(u2)v]-[R(u2) 3

2

+R(u ) v - R ( u ) ( R ( U )v]= {~R(U)R(U~)-R(U~)R(U)-R(U~) }v I n a similar manner, we have

u [(cv*c)u*c] *u= 2 ( u (cv*c) Xu) c*u-u (cv*c) *u2= = 2 [2

(vc*u)c*u-vc*u2] c*u- [ (vc*u) c*u2 +c*u2+ 3

2

2

+ ( V C * U)C*U-VC*U ] = 2R(u) [ ~ R ( u ) v - R ( u

- [R ( U 2

R ( u )v + R ( u ) R ( u 2 )v-R ( u 3 )v] =

= {4R(U)

Thus, if we set

2

3

2

2

-3R(U)R(u ) - R ( u ) R ( u ) + R ( u 3}v )

)v]-

265

JORDAN THEORY

2

B=: R(u)R(u

A=: R(u )R(u) C=: R(u)

3

D=: R(u

3

2

)

)

we have 3B-A-D= 3B+2A-2D-2C= 4C-3B-A+D= 2A-D whence, by substracting the last term, we obtain 3B-3A= 3B-D-2Cz 2D-3B-3A+4C=O Therefore, A= B and so 3A-D-2C= -2(3A-D-2C)= 0 2 C= 1 D. and finally A= B= 3 3

ff

10.43. PROPOSITION. If (A,-) i s a commutative J-algebra, then we have the foZZowing "polarization f o r m u l a t 1 R(uLz)= ~ R ( u z ) R ( U ) - ~ R ( U ) R ( Z ) R ( U+R(u')R(z) ) f o r a L 1 U,ZCA.

Proof: Given any vcA and any

-cm, by axiom (JA) we

have

whence, by developing and comparing the coefficients of both sides, we get R(u2)R(~)zt2R(uv)R(u)z= 2R(u)R(uv)Z+R(v)R(U 2 ) i.e. u2 (vz)+2 (uv)(uz)= 2u[(uv) This can be reformulated as

I.

2

+v(u z)

Z

T

in

CHAPTER

266

10

DEFINITION. L e t A be a n o n n e c e s s a r i l y a s s o c i a t i u e B u n u c h a l g e b r a . We d e f i n e t h e " u n i t a l e x t e n s i o n " 2 of A 10.45.

hY 'L

A=: EQA= {XBa; h c E , aeA}

II

Ixl+lla I I

XQalI = :

(XOa)(UQb)=:ALI@

(Xb+i.la+ab)

wh er e A , v e E and albeA.

we identify 0Qa with a, aeA, and X@O and we shall write A+a instead of XQa.

A s usually,

with A , AsE,

'L

It is easy to check that A is an algebra and 1 is the unique 'L

'L

element ecA such that ex= x = xe for all x c A . DEFINITION. We s a y t h a t u e A i s " i n v e r t i b l e " if u admits a unique inverse tlith respect t o the I m u l t i p l i c a t i o n . We s h a l l d e n o t e t h e i n v e r s e of u b y u-l 10.46.

10.47.

J-algebra

LEMMA.

The u n i t a Z e x t e n s i o n o f a com m ut at
is a c o m m u t a t i v e J - a Z g e b r a .

Proof: Let A be a commutative J-algebra. It is immedia'L te from the definition that A is a commutative J-algebra. On the other hand, we have z

2.

'L

'L

R(X+u)= R(X)+R(u)= Aid a: +R(u)

'L

2

%

2

z

2

2

z

2

R[(X+u) ] = R(u +2hu+h ) = R(u )+2XR(u)+A id, f o r Xc6 and 'L u e A , where we have written R to denote the multiplication 'L representation on A . Thus it suffices to show that and

'L

R ( u 2 ) - R(u)

ueA

267

JORDAN THEORY

I f pea and v e A ,

[$(u2),

then ? . 2 Q 2 2 2 $(U)] (ii+V)= [R(u ) , R ( u ) ] p + [ R ( u ) , R(u)]v= p u u-pu u+O= 0

10.48.

;Y

D E F I N I T I O N . T h r o u g h o u t t h e r e m a i n d e r p a r t of t h i s

s e c t i o n , A w i l l d e n o t e a f i x e d unital c o m m u t a t i v e B a n a c h a l g e b r a . We s e t

un=: R ( u ) " l , 10.49.

U=: { x e A ;

(1

R ( x ) - i d ( ( < I } , Au=: Spaniu"; n= 0,...,1}.

LEMMA. We h a v e u n t 2 = uu

ntl

2 n

= u u

f o r n= o , I , ~ , . .

and u c A . P r o o f : The case n = 0 i s t r i v i a l . I f w e a s s u m e t h e s t a t e m e n t t o be v a l i d f o r s o m e nm, t h e n

unc3= R ( u )

n+3

1= R(u)u

n+2-

- uu

R(u2)un+'= 10.50. R(u")

n+2-

2

U2Un+'.

PROPOSITION,

#

For a l l n = 0 , l

admits a representation as a

~ ( u ~ I) n . p a r t i c u l a r ,

2

- R ( u ) R ( u )u"= R ( U ) R ( u ) u " =

r e s u l t t o be v a l i d f o r some

a l l UCA,

polynomial on R ( u ) and

R(u") - R ( U ~ ) f o r

P r o o f : T h e cases n = 1 , 2 ,

,... a n d

m,n= 0,1,2,.

..

are t r i v i a l . I f w e a s s u m e the

nm,

from t h e p o l a r i z a t i o n f o r m u l a

we obtain R(un+')=

R ( u 2 u n-1 ) = 2 R ( ~ u " - ~ ) R ( u ) - P ( u ) R ( u ~ - 2~R)(=u n ) R ( u ) -

-P(u)R(u"-')=

2pn[R(u) ,R(u2 )]R(u)-[2R(u)

2

-

- R ( U ) ] P ~ - ~[ R ( u ) , R ( u 2 ) ]

f o r s u i t a b l e p o l y n o m i a l s pn - 1 r P n . 10.51.

PROPOSITION.

#

m+nm n We h a v e u - u u

n

for a l l u c A and

a l l n= o , I , ~ , . . I n p a r t i c u l a r , A ~ = : S p a n I u : n= a n a s s o c i a t i v e s u b a l g e b r a of A .

o,I,~,...I

i s

Proof: The s t a m e n t i s t r i v i a l f o r n + m < 3 . A s s u m e i t . t o be t r u e for s o m e p a i r n , m w i t h m + n = d ; t h e n

2 63

CHAPTER

10

10.52.LEMMA. For ueU, u is i n u c r t i b l e a n d cc R(u) -1 1 . T h u s , t h e m a p p i n g U A U g i v e n b y u+u -1 = C u-I 1 n=O is h o l o m o r p h i c o n U .

U-l=

P r o o f : F o r UCU w e h a v e m

i n v e r t i b l e and R ( u ) - l =

1

11

R u ) - i d l l < 1 1 so t h a t R ( u ) i s

[R(u)-id]".

Now, t h e r e l a t i o n u v = l

n=O

is equivalent t o ~ ( u ) v = 1I i . e l

v = ~ ( u- ')I .

ff F o r xcU, w e i n t r o d u c e t h e a u x i l i a r y f u n c t i o n

H ( x ) : U+L(A)

g i v e n by u + - ( u - ' ) ( ' .

-1 -1 1 0 . 5 3 . THEOREM. We have P ( x ) = R ( x ) R ( x . ) = H ( x ) - ' for

c A b e g i v e n . Then

-'.

H(x)= R ( x - ~ ) R ( x )

whence

o p e r a t o r of t h e f o r m

M o r e o v e r l H ( x ) commutes w i t h a n y

C a k R ( x I k with n, -nhkCm

m a

S i n c e x - l and x 2 b e l o n g t o A X l w e h a v e

x-1 x 2 = x -1 ( x x ) = ( x - l x ) x = x

X€U

T h e r e f o r e , t h e f u n c t i o n s f : U+A a n d g : U-tA g i v e n r e s p e c t i v e l y by z+z

-i

t h a t is,

and z+z2 s a t i s f y

JORDAN THEORY

269

so that 2

2

2

id= -R(x )H(X)+~R(X-~)R(X)= -R(x )H(X)+2R(x )H(x)= P(x)H(x)= = H(x)P(x).

10.54. PROPOSITION

.

We h a v e [P(x)y]-'=

P(x)-ly-'

ff

for

a12 x,yeU.

proof: Let XEU and YEA be given; then X-'P

(x)y= R(X-' ) R ( X - ~-'R(X)~= ) ~(x)y= xy

Now, let us fix any ysU and define the functions f: U+U and g: U+A by means of z + z - ' and z+P(z)y respectively. Then 9;ly-l is given by

(z(zy))-22y] A 1 y - L 2y-1(xy)+2 (x(y-ly)) -2 (xy-l)y= = 2R(y- 1 )R(y)x+2~-2R(y)R(y-')X= 2x : z+[2

so that

10.55. THEOREM. (Fundamental formula). F o r e v e r y x,yEA P [P (x)y] = P(X)P(Y)P(X) ~n particular P(X

2

I = P[P(x)I]=

P(x)P(I)P(x)= P(X)

2

,

XEA

Proof: As the expressions on both sides of the fundamental formula are polynomials of x,y, it suffices to prove it for x,yeU. Let x,ycU be given; from porposition 10.54 we get -lz-l

(1

IY

X,ZEU

whence, by the definition of H(x), We obtain -H [P ( X I y] P ( x ) = P ( x )

so that, by theorem 10.53

[-H (y)]

1

YEA

CHAPTER

270

10

p [P (XIy] - P (XI = P (x)- P ( y )-

’

whence the result follows. 10.56.

#

THEOREM. L e t wcA b e g i v e n a n d a s s u m e t h a t u,veAW. P (uv)= P (u)P (v)

Then

Proof: By proposition 10.51 , AW is an associative subalgebra of A; thus, for x r y c A W we have 2

2

2

P(x)y= 2 x y-x y = x y

and P(x)-P(y)

. Therefore, 2

2

P (x”) = P [P (XIy] = P (x)P (YIP(XI= P ( x ) P (y)= P ( x 1 P (y)

.

2

for all x,yeA Moreover, we have ( x 2 ) 1( l = 2id and 1 = 1 ; therefore, by the inverse function theorem, there exists a neighbourhood V c A W of 1 such that each vcV can be written in the form v= x2 for some x e A W . Thus P (vu)= P (v)P (u)

for all u,v in V. Since both sides of the last equality are polynomials on u,v, that relation holds for all pairs u ,v€AW.

#

1 0 . 5 7 . EXERCISES. ( 1 ) Let ( E l * ) be a J*-triple and put Q(a): x+ax*a for a,xcE. Show that Q[Q(a)b]= Q(a)Q(b)Q(a).

(2) By using the representation Q, prove that the

fundamental formula is valid in (E,c*) Hint: P(u)= Q(u)Q(c). §8.-

Positive J*-triples and the convexity of homoqeneous circular domains.

We begin with the following simple but crucial observation. 1 0 . 5 8 . LEMMA. L e t ( E , * ) b e a J B * - t r < p Z e a n d a s s u m e t h a t i s a b o u n d e d o p e n n e i g h b o u r h o o d of 0 i n E s u c h t h a t a-S a* eautU f o r a l l aeE. T h e n Iu

U

U = I c ~ E ; P(CC*) l E c ) < l I

JORDAN THEORY

a n d we h a v e Sp(cc*

I EC

271

) c l R + f o r a l l ccE.

Proof: Let ccE be given. It is easy to see that EC is a JB*-triple, too, and that Vc=: U Il EC is a bounded open neighbourhooi? of 0 in EC such that exp(b-sb*)Vc= V c

(10.14)

bcEC

By corollary 1 0 . 3 7 , the only bounded open neighbourhood of 0 in EC satisfying (10.14) is the set

wc=

{xeEC;p(xx*

I EC

) < I3

Thus,we have UnE

(10.15)

c

=

vc =

{xeEC;

P ( X X * , ~ ~ ) < ~ }

NOW, let ccE be such that ccU. Then ceUnEC and by ( 1 0 . 1 5 ) we have p (cc*1 E c ) (1. Conversely, let ceE be such that p (cc* E c ) > 1 By ( 1 0 . 1 5 )

.

we have ccUn EC and therefore ceU so that

On the other hand, for ceE, EC is a JB*-triple and from theorem 1 0 . 3 6 we obtain Sp(cc*~~,)cIR+.

i4 10.59. LEMMA. A s s u m e t h a t (El*) is a J B * - t r i p l e a n d l e t ceE a n d hcQ:\{O} b e g i v e n . T h e n We h a v e A+Sp(cc*) if a n d o n l y i f A-c i s a n i n v e r t i b l e ernent o f t h e c o m m u t a t i v e J - a l g e b r a (2",c")

.

Proof: Set K=: SP(CC*~~,)and consider the Jordan representation j : ' EC -+ C O ( K ) of (EC,*) (cf. theorem 10.38 and corollary 1 0 . 3 9 ) . The map j is a topological J*-isomorphism and we have

CHAPTER 1 0

272

Writing

. for the c-multipl cation in 2, (A-c)

- (p+a)=

Xu+ ha-pc-c .a)

Thus 1-c is invertible in

.)

paTI

aeEC

if and only if

Aa+A-lc-c.a=

o

f o r some a e E C l which by the isomorphism j is equivalent to hf+A-'&fK

-

idK f = 0

which is equivalent to X#K. 10.60.

PROPOSITION. If ( E l * ) a J B * - t r i p l e , t h e n we h a v e

p(cc*)=

p(cc*

I EC

and

SP(CC*)C~R+

f o r a l l csE.

Proof: Let us fix csE\{OI arbitrarily and consider the multiplication representation R on the commutative unital J-algebra (8,~").By corollary 10.19 we know that the subset {R(a); is a commutative family in L ( 2 ) . Therefore, t h e closed algebra R spanned in L ( 2 ) by {R(a); aegcl is a commutative unital Banach subalgebra of L ( 2 ) . Let n be its spectrum and

213

JORDAN THEORY

t h e G e l ' f a n d r e p r e s e n t a t i o n of R . NOW, l e t c s S p ( c c * ) b e g i v e n . We h a v e

~ ( c ) X:+ X

-f

X C + C C * X,

AeE,

X ~ E

Thus

i.e.,

t h e o p e r a t o r
2.

i s n o t i n v e r t i b l e i n L ( E ) , hence,

E

2,

it i s n o t i n v e r t i b l e i n R . T h e r e f o r e r e r a n g e g ( R ( c ) ) , i . e . , w e c a n c h o o s e some wcR s u c h t h a t

L e t us define , I ,

0 ' : 2

where P ( c ) = : 2 R ( c ) - R ( c e q u a t i o h z 2 - 2 5 z + n = 0,

2

g[P(c)]w

1 , a n d l e t ~ , , 5 , be t h e r o o t s o f t h e

i.e.

For j= 1 , 2 , w e h a v e

and t h e r e f o r e

C o n s e q u e n t l y , P ( c - c , ) h a s n o i n v e r s e i n R . I t follows t h a t

I c - 5 , i s n o t a n i n v e r t i b l e e l e m e n t of 1

9".

Indeed, i f we had

CHAPTER 1 0

274

( c - c . ) - ( a . + h . )1= 7

7

3

for some a . c E C and h . e C , then it would follow P(a.+A.)sR and I

3

1

3

~ ( c - i . ) ~ ( a . + h .P )( = a.+X.)~(c-c ) = P[(a.+A.) -(c-c.)]= P ( I ) = I 3

1

3

I

1

3

j

3

3

i.e., the invertibility of P(c-c.) in R. Thus, by lemma 10.59 7

we have

cjet01

u SP(CC*

I EC

j= 1,2

whence O$ 1 ( y +y ) = 2 1 2

since S ~ ( C C *

I EC

yc 1 P(CC* 2

I EC ) +

1 P(cc*,Ec)

)CJR+.

10.62. DEFINITION. L e t B be a Banach s p a c e and A e L ( E ) . We s a y t h a t A i s a " h e r m i t i a n p o s i t i v e o p e r a t o r " i f AcHer(E) and SPA-+ . We w r i t e Her+(E) f o r t h e s e t of h e r m i t i a n

p o s i t i v e o p e r a t o r s of L (El

.

L e t ( E , * ) be a J * - t r i p l e .

We s a y t h a t E i s a " h e r m i t i a n p o s i -

tive J*-triple

"if we haue cc*eHer+(E) for aZZ ccE.

F i n a l l y , for ccE we s e t

10.63. LEMMA. L e t E b e a Banach s p a c e , A c L ( E ) and

JORDAN THEORY

275

We h a v e ACHer+(E) if a n d o n l y if <4,x> >O f o r a l l

(4,x)cD.

Proof: Assume that AcHer+(E) and write r=: p ( A ) . r Observe that A- T idEsHer(E) and r Sp(A- 2 idE)c [- 3

, 7r ]

From Sinclair's theorem we obtain

// Thus , if

( @,x)cD

,

A-

5 idE//

S P

then

whence <@,Ax> >O. Conversely, assume that > O for all (Y,y)cD. Let xcE be arbitrarily fixed and define

xt =: exp(-tA)x

gw=:

I/

XtII

for tdR. Now we may apply ( 1 0 . 4 ) to the operator iA. It follows that g'(t)cO for almost all t d R and so the function 9 does not increase. Thus (1 xtI1 & ( I x / ( for tar whence by the arbitrariness of XCE, we obtain

(10.16)

/I exp(-tA) / I

SZ

t>O

By lemma 1 0 . 2 6 , A is hermitian; thus by Sinclair's theorem we have Sp(A)clR. Therefore, if we had Ac(SpA)\IR+ for some Ac6, then we would have A > O . However, in this case, from ( 1 0 . 1 6 ) and the spectral mapping theorem we would obtain

CHAPTER 10

276

which is a contradiction. Thus Sp COROLLARY. If AsHer

10.64.

I / All

=

SUP

I

($,x)eD

Proof: Let us write r 1= :

inf <@,Ax> (4,x)eD

Then A-rlidE and

r2idE -A

are positive hermitian operators,

whence Sp(A)c[rl,rz]. Therefore

(1

All= p ( A ) = max{lrll,lr2(l.

79 EXERCISES. ( a ) Show that Her+(E) is a closed convex cone in L ( E ) 10.65.

.

PROPOSITION. If ( E l * ) i s a h e r m i t i a n p o s i t i v e J * - t r i p Z e y t h e n I / * 11 is a continuous s e m i n o r m o n E . 10.66.

Proof: We have cc*cHer+(E) f o r ccE. Then, by the defiand Sinclair's theorem nition of / I 11

-

On the other hand, the functional c

-+I/

CC*

I/

'

is continuous.

Let u be as in lemma 1 0 . 6 3 and write

for (@,x)cD and a,beE. Then, by corollary 1 0 . 6 4 we have

JORDAN THEORY

271

<arb> , a,bcE, is a @tX semiinner product on E because, due to lemma 10.63, we have

For fixed (+,x)cD, the mapping (arb)

-+

Thus,c -+ is a seminorm on E for each (@,x)cD. The @tx supremum of a family of seminorms is always a seminorm.

# As a summary of the chapter we get the following result due to

Kaup

.

10.67 (a)

//

THEOREM. L e t

. 11

( E l * )b e a J B * - t r i p l e .

Then

is an e q u i v a l e n t norm o n E and i t s u n i t b a l l

Bm i s t h e o n l y bounded o p e n n e i g h b o u r h o o d of 0 in E in w h i c h

a l l v e c t o r f i e l d s c-'c*,

ceE , a r e c o m p l e t e .

(b) ( E l * ) i s a h e r m i t i a n p o s i t i v e J * - t r i p l e w i t h t h e norm / I * 11 and we h a v e

s

when endowed

*

(c) The mapping c + exp(c- c ) O i s a r e a l b i a n a l y t i c i s o m o r p h i s m o f E o n t o Bm whose i n v e r s e is

Proof: As (El*) is a JB*-triple, by proposition 10.60 we have p(cc")= P(CC*

I EC

)

ccE

Now, fix any cCE, cf0. Then, EC is also a JB*-triple, and, by

218

CHAPTER 1 0

theorem 10.36,

the map acEC

is a norm on EC. Thus, from the definition of

11 I /

we get

so that 11 - 1 1 is a norm which, by proposition 10.66, is continuous on E. In particular,

is an open neighbourhood of 0 in E. NOW, as E is a JB*-triple, there exists some bounded open neighbourhood U of 0 in E such that we have exm(c- s c* )U= U f o r all ccE. By lemma 10.58 such a neighbourhood must be

i.e., U is the open unit ball of E with respect to the norm 1 1 . 11 Therefore 1 1 11 and the norm 1 1 11 are equivalent. Thus, we have proved (a)

.

-

.

NOW, as U = Bm is a bounded domain, autBm is a Lie algebra. Since we have c-'c*eautBm for csE, we have cc*eautB,, too. Then, the operators cc* are all hermitian and, by proposition 1 0 . 6 0 , we have Sp(cc*)ClR, so that (El*) is a hermitian positive triple when endowed with the norm proposition 1 0 . 6 6 , we have

11 11

m.

Then, by

JORDAN THEORY

Moreover, from ( 1 0 . 8 )

/ j cc*c 1)

w=

p

279

and corollary 1 0 . 3 2 we obtain

[ (CC*C)(CC*C)* ]

f =

11

(CC*C)(CC*C)* j E C I I

+=

for ccE. Thus, we have prove (b). Finally, (c) is an immediate consequence of corollary 1 0 . 3 7 .

# 10.68.

COROLLARY. L e t E be a comp2ex Banach s p a c e and

assume t h a t D i s a ( s i m p Z y c o n n e c t e d ) b o u n d e d s y m m e t r i c domain

of E.' T h e n , t h e H a r i s h - C h a n d r a r e a Z i z a t i o n p a r t i c u l a r , t h e r e i s an e q u i v a l e n t n o r m

6

of D i s c o n v e x . In

1 1 . 11

on E s u c h t h a t

D i s b i h o l o m o r p h i c a l l y e q u i v a Z e n t t o t h e u n i t b a l Z Bm of (Er

I1 - II

,I

*

Proof: From chapter 9 we know that the Harish-Chandra realization 6 of D is a bounded balanced circular domain of E. Let us consider its associated J*-triple (E,*) as given by example 1 0 . 2 . Obviously (E,*) is a JB*-triple and is a bounded neighbourhood of the origin such that c-'c"eaut6 for all csE. By theorem 1 0 . 6 7 , b is the open unit ball of E with respect to the norm 1 1 I / w . In particular, 6 is convex and the Harish-Chandra realization gives a biholomorphic isomorphism of D onto 6. #

c

.

COROLLARY. L e t E b e a c o m p l e x Banach s p a c e , D c E a b o u n d e d b a l a n c e d c i r c u l a r domain and Eo=: (autD)O. T h e n : 10.69.

(a) The s e t (AutD)O= D n E 0 is convex. (b) If A,eautD i s t h e u n i q u e v e c t o r f i e l d w i t h t h e 0, t h e r e i s a reaZ a n a l y t i c c, (Ac):'= p r o p e r t i e s (Ac);'=

CHAPTER 1 0

280

i s o m o r p h i s m 4 : D nEo

+

E0 s u c h t h a t

Proof: From chapter 7 we know that D n E o is a bounded balanced circular domain. Let (E , * ) be its associated J*-triple. Then D n E O is a bounded neighbourhood of 0 in Eo such that (expAc)(Dn Eo)= D n Eo for all ccE ; thus ( E o , * ) is a JB*-triple and, by theorem 10.67, D n E O is the unit ball of Eo for a suitable norm; in particular, D n E o is convex. (expA ) O is a real bianalytic isomorphism of Eo Moreover, c onto D n E o ; thus, its inverse 4 : D n E O Eo is a real analytic isomorphism which satisfies +

+

(10.17)

for all acD n E o .

59.-

Some properties of the topoloqv of local uniform-con=gence.

Next we apply the previous results to the study of the topolosy T of local uniform convergence on the group A u t D for a bounded balanced circular domain D. 10.70.

PROPOSITION. The mapping T: AutD

f

+

Tf=:foexpA

-+

AutD g i v e n b y

fsAutD

@f-l(O) 0

t a k e s its va Z ues i n Aut D a n d is c o n t i n u o u s at id, when AutD is e n d o w e d w i t h t h e t o p o l o g y T .

Proof: By ( 1 0 . 1 7 ) we have (Tf)O = (fexpA )O= f f - l ( o ) = 0 bf-l(o)

281

JORDAN THEORY

so t h a t T f e A u t 0 D f o r fcAutD. M o r e o v e r , i f we endow A u t D w i t h t h e topology T

,

t h e mappings

AutD * D

n Eo

E

+

( 1 0 -1 8 ) f

-f

f-l(O)

-f

autD

+

0

@f-'(O)

*

A

@f-I(o)

are known t o be c o n t i n u o u s . W e know a s w e l l t h a t , f o r a s u i t a b l e n e i g h b o u r h o o d M Of 0 i n a u t D , t h e mapping M c a u t D * AutD

* expA

A

i s c o n t i n u o u s ( a c t u a l l y , a homeomorphism) when AutD i s equippec! w i t h t h e t o p o l o g y T

a

. In

particular,

since T >TI

a u t D +AutD (10.19)

i s continuous f o r t h e topology T composite T o f

on autD. T h e r e f o r e , t h e i s

(10.18) and (10.19) i s continuous a t i d

D

for

T.

# 10.71.

PROPOSITION. L e t

(

~ b e )a s~e q u e~n c e i n E

c

0

w i t h cn * 0 . Then t h e s e q u e n c e f = : expA n

converges t o idD uniformZy on D. P r o o f : F o r x e D n E O a n d ndN, l e t y ( t ; x , c n ) d e n o t e t h e maximal s o l u t i o n of t h e i n i t i a l v a l u e p r o b l e m

282

CHAPTER 1 0

Then, we have f n ( x ) = y ( l ; x , c n ) for all nClN and x c D n E

AS c

-f

0'

Thus

Qc i s a continuous real linear mapping, we have

for some constant K and all c e E o , y s E O . As D is bounded and y ( s , x , c ) s D n E O C D for s d R , xcDn Eo and ccEO, from (10.20) we

derive

whence the result follows.

#+ 10.72. PROPOSITION. L e t (fn)nm b e a s e q u e n c e in AutD s u c h t h a t T lim fn= idD. T h e n we h a v e lim f n = idD uniformZy o n D.

n+w

n+07

Proof: From T lirn fn= idD n+m

we obtain f - ' (0) + 0 and n

therefore $ f ~ l ( 0 )+ 0. Then, by proposition 1 0 . 7 1

we have

on D. Moreover, from T lirn f

=

id

n+m

D

and proposition 1 0 . 7 0 we derive

T lirn h n = T lim Tf = idD n+m n+m As the transformations hn

are linear, the latter entails

JORDAN THEORY

(10.22)

hn

+

idD

283

u n i f o r m l y on D

On t h e o t h e r hand, from t h e d e f i n i t i o n h n , gn and T w e g e t

h g = (Tf )expA n n n

= fnexpA

expA

=

fn

T h e r e f o r e , by ( 1 0 . 2 1 ) and ( 1 0 . 2 2 ) w e have

10.73. PROPOSITION. L e t D b e a bounded b a l a n c e d c i r c u -

l a r domain D o f a Banach s p a c e E . On t h e g r o u p AutD, t h e t o p o z o g y T c o i n c i d e s w i t h t h e t o p o l o g y TU o f uniform c o n v e r g e n c e o v e r D. be a sequence of A u t D such t h a t

P r o o f : L e t ( f n )n~

T l i m f n = f . I t s u f f i c e s t o show t h a t ( f n )n~

converges t o f

u n i f o r m l y on D . S i n c e A u t D i s a t o p o l o g i c a l group w i t h r e s p e c t

,

we have T l i m f - ' f n = i d D a n d , by p r o p o s i t i o n 1 0 . 7 2 , n-+sequence ( f - ' f n ) new c o n v e r g e s t o i d D u n i f o r m l y on D .

to T

the

On t h e o t h e r hand, by theorem 7 . 3 9 , f o r e v e r y E > O t h e r e e x i s t q>O

such t h a t f can be e x t e n d e d t o a bounded holomorphic

mapping F : DE

-+

D

rl

. If

M i s a bound f o r F on DE, from Cauchy's

i n e q u a l i t i e s we obtain t h a t F i s

- 1 i p s c h i t z i a n on D . Then

w e have

11

f,-f

11 D = 1 1

ff-'fn-f

IID

6

M

/I

f-'fn-idllD

-+

0

.

This Page Intentionaiiy Left Blank

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topology, Proc. Amer. Math. SOC. 9 1 ( 1 9 8 4 ) 2 2 3 - 2 2 4 . I82

I

VIGUE, J.P. : Sur la caractgrisation des automorphismes analytiques d'un domaine borne, preprint.

1831 VIGU6, J.P.: Points fixes d'applications holomorphes dans

un domaine born6 convexe de tn, preprint.

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