Fundamentals of the Theory of Operator Algebras Vol.4: Special Topics-Advanced Theory, an Exercise Approach (Pure and Applied Mathematics (Academic Press), Volume 100)

FUNDAMENTALS OF THE THEORY O F OPERATOR ALGEBRAS SPECIAL TOPICS VOLUMEIV Advanced Theory-An Exercise Approach This...

Author: Richard V. Kadison | John R. Ringrose

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FUNDAMENTALS OF THE THEORY O F OPERATOR ALGEBRAS SPECIAL TOPICS

VOLUMEIV

Advanced Theory-An

Exercise Approach

This page intentionally left blank

FUNDAMENTALS OF THE THEORY OF OPERATOR ALGEBRAS SPECIAL TOPICS VOLUME IV A duanced The0ry-A

n Exercise Approach

Richard V. Kadison

John R. Ringrose

Department of Mathematics

School of Mathematics

University of Pennsylvania

University of Newcastle

Philadelphia, Pennsylvania

Newcastle upon Tyne, England

COPYRIGHT 0 1992, BY RICHARD V. KADISON ALL PARTS RESERVED. NO PART OF THIS PUBLICATION MAY BE REPRODUCED OR TRANSMITTED IN ANY FORM OR BY ANY MEANS, ELECTRONIC OR MECHANICAL, INCLUDING PHOTOCOPY, RECORDING, OR ANY INFORMATION STORAGE AND RETRIEVAL SYSTEM, WITHOUT PERMISSION IN WRITING FROM RICHARD V. KADISON. THE STATEMENTS OF ALL EXERCISES (0ACADEMIC PRESS 1986) APPEAR WITH THE PERMISSION OF ACADEMIC PRESS. BIRKHAIJSER BOSTON 675 Massachusetts Avenue, Cambridge, MA 02 139-3309 Library of Congress Cataloging-in-Publication Data Kadison, Richard V., 1925Fundamentals of the theory of operator algebras. (Pure and applied mathematics ; 100-100.2 (QA3.P8 vol. LO)) Vol. 4 has imprint : Boston : Birkhauser. Vol. 3-4: lacks series statement. Includes bibliographies and indexes. Contents: v. 1. Elementary theory -- v. 2. Advanced thcory [etc.] -- v. 4. Special topics : advanced theory, an exercise approach. 1. Operator algebras. I. Ringrose, John R. 11. Title. 111. Series: Pure and applied mathematics (Academic Press) ; 100-100, 2. QA326.K26 1983 512l.55 ISBN 0-8176-3498-3 (v. 4)

82-13768

Printed on acid-free paper Printed by Quinn-Woodbine, Woodbine, New Jersey Printed in the U.S.A. ISBN 0-8 176-3498-3 ISBN 3-7643-3498-3 9 8 7 6 5 4 3 2 1

CONTENTS Contents of Volume III

ix

Exercise Groupings

xi

Chapter 6.

Comparison Theory of Projections - Exercises and Solutions

274

Chapter 7.

Normal States and Unitary Equivalence of von Neumann Algebras - Exercises and Solutions

312

The Trace

368

Chapter 8. Chapter 9.

- Exercises and Solutions Algebra and Commutant

451

- Exercises and Solutions

Chapter 10. Special Representations of C*-Algebras - Exercises and Solutions

546

Chapter 11. Tensor Products - Exercises and Solutions

680

Chapter 12. Approximation by Matrix Algebras - Exercises and Solutions

726

Chapter 13. Crossed Products - Exercises and Solutions

783

Chapter 14. Direct Integrals and Decompositions - Exercises and Solutions

818

Bibliography

842

index

847


PREFACE These volumes are companions to the treatise; ‘‘Fundamentals of the Theory of Operator Algebras,” which appeared as Volume 100 - I and I1 in the series, Pure and Applied Mathematics, published by Academic Press in 1983 and 1986, respectively. As stated in the preface t o those volumes, “Their primary goal is to teach the subject and lead the reader to the point where the vast recent research literature, both in the subject proper and in its many applications, becomes accessible.” No attempt was made to be encyclopzedic; the choice of material was made from among the fundamentals of what may be called the “classical” theory of operator algebras. By way of supplementing the topics selected for presentation in “Fundamentals,” a substantial list of exercises comprises the last section of each chapter. An equally important purpose of those exercises is to develop “hand-on” skills in use of the techniques appearing in the text. As a consequence, each exercise was carefully designed to depend only on the material that precedes it, and separated into segments each of which is realistically capable of solution by an attenti ve, diligent, well-moti vated reader. The process by which the exercises were designed involved solving each of them completely and then subjecting the solutions to detailed scrutiny. It became apparent, in the course of this operation, that the written solutions could be of considerable value, if they were made generally available, as models with which a reader’s solutions could be compared, as indicators of methods and styles for producing further solutions on an individual basis, and as a speedy route through one or another of the many special topics that supplement those in the text proper of “Fundamentals” (for the reader without the time or inclination to develop it as an exercise set). The present texts contain those written solutions; the first of these texts has the solutions to the exercises appearing in Volume I of “Fundamentals” and the second has the solutions to those appearing in Volume 11. The statements of the exercises precede their solutions, for the obvious convenience of the reader. In most instances, where an exercise or group of exercises de-

viii

P RE FAC E

velops a topic, the solutions have been given in what the authors feel is optimal form. Solutions are, of course, geared to the state of knowledge developed at the point in the book where the exercise occurs. Very occasionally, this necessitates an approach that is slightly less than optimal (for example, Exercise 2.8.10 occurs before square roots of positive operators, which could be used to advantage in its solution, are introduced). From time to time, an exercise reappears, with the task of finding a solution involving newly acquired information. Of course, knowledge of and long experience with the literature of the subject has had a major influence on both the content of the exercises and the form of their solutions. In many cases, there is no specific source for the exercise or its solution. In virtually no instance was the solution of am exercise copied directly from the literature; the solutions were constructed with the path and stage of development of “Fundamentals” where the exercise occurs, very much in mind. Often, where a set of exercises develops a special topic, the solutions present a new and simpler route to the results appearing in the set. References are placed after the solutions. As in “Fundamentals,” no attempt is made to be thorough in referencing. The references appearing are chosen with a few goals in mind: to supply the reader with additional material, closely related to the exercise and its solution, that may be of interest for further study, to provide a very sketchy historical context that may be deepened by consulting the papers cited and their bibliographies. Where a set of consecutive exercises is largely inspired by a single article, for the most part, the first of the set and/or the highpoints of the topic note the article. A guide to the topics treated by sets of related exercises follows this preface.

CONTENTS O F VOLUME I11 vii

Preface Contents of Volume

iV

Exercise Groupings

Chapter 1. Linear Spaces

ix xi 1

- Exercises and Solutions

Chapter 2. Basics of Hilbert Space and

40

Linear Operators

- Exercises and Solutions Chapter 3.

Banach Algebras - Exercises and Solutions

Chapter 4.

Elementary C*-Algebra Theory - Exercises and Solutions

Chapter 5. Elementary von Neumann

84

139 207

Algebra Theory - Exercises and Solutions Bibliography

263

Indez

268


EXERCISE GROUPINGS Algebras of affiliated operators 6.9.53-6.9.55, 8.7.60 Approximate identities in C*-algebras 4.6.35-4.6.37, 4.6.60, 10.5.6 p-compactification of N 3.5.5, 3.5.6, 4.6.56, 5.7.16-5.7.19 Canonical anticommutation relations 10.5.88-10.5.90, 12.4.39, 12.4.40 Characterizations of von Neumann algebras among C*-algebras (i) in terms of order structure 7.6.35-7.6.40 (ii) as dual spaces 7.6.41-7.6.45, 10.5.87 Compact linear operators 2.8.20-2.8.29, 2.8.37-2.8.39, 3.5.12, 3.5.14, 3.5.15, 3.5.17-3.5.20 Completely positive mappings 11.5.15-1 1.5.24 Conditional expectations 8.7.23-8.7.30, 8.7.50, 10.5.85-10.5.87, 13.4.1, 13.4.24, 13.4.25 The Connes 2'-invariant (i) calculation for certain matricial factors 13.4.9- 13.4.15 (ii) general properties 14.4.14-14.4.16, 14.4.19, 14.4.20 Coupling constant and operator 8.7.57, 8.7.58, 9.6.3-9.6.7, 9.6.30

xii

EXERCISE GROUPINGS

Derivations and automorphisms (i) continuity of derivations 4.6.65, 4.6.66, 7.6.15, 10.5.12, 10.5.13 (ii) inner and universally weakly inner derivations 8.7.51-8.7.55, 10.5.61-10.5.64, 10.5.71, 10.5.72, 10.5.7610.5.79, 12.4.38 (iii) inner and universally weakly inner au tomorphisms 10.5.1, 10.5.14, 10.5.60-10.5.75 Diagonalization of abelian self-adjoint subsets of n 8 R 6.9.18-6.9.35 Diximier approximation theorem 8.7.4-8.7.13, 10.5.2, 10.5.80 Extreme point examples (i) simple examples 1.9.19, 2.8.13, 2.8.14, 5.7.8, 5.7.12, 5.7.13 (ii) projections (see also Proposition 7.4.6) 12.4.13-12.4.15 (iii) unitary elements (see also Theorem 7.3.1) See the listing under “Unitary elements of C*-algebras,” sections (ii) and (iii). Extremely disconnected spaces 5.7.14-5.7.2 1 Flip automorphisms (i) algebras generated by two projections 12.4.11, 12.4.12 (ii) free action 12.4.17, 12.4.18 (iii) a Liapunov theorem for operator algebras 12.4.13-12.4.15 (iv) the flip in various algebras 11.5.25, 12.4.16, 12.4.19-12.4.27, 12.4.34, 12.4.35, 14.4.1114.4.13 Friedrichs extension 7.6.52-7 .6.55

EXERCISE GROUPINGS

xiii

Fundamental group (of a factor of type 111) 13.4.4-1 3.4.8 Generalized Schwarz inequality and applications 10.5.7-10.5.10, 11.5.23 Harmonic analysis on certain groups 3.5.33-3.5.42 Ideals (i) ideals in C*-algebras 4.6.41, 4.6.42, 4.6.60-4.6.64, 10.5.11 (ii) ideals in von Neumann algebras 6.9.46-6.9.51, 8.7.14-8.7.22 (iii) primitive ideals and the Dauns-Hofmann theorem 10.5.81-10.5.84 Isometries and Jordan homomorphisms 7.6.16-7.6.18, 10.5.21-10.5.36 Modular theory (i) states that satisfy the modular condition 9.6.14-9.6.17 (ii) dual cones 9.6.51-9.6.65, 13.4.19 Non-normal tracial weight 8.7.42-8.7.45 Relative commu t ants (i) commutation formulae for tensor products 12.4.34- 12.4.37 (ii) maximal matricial subfactors 12.4.29,12.4.30 (iii) non-normal factors 12.4.22, 12.4.23, 12.4.28-12.4.31, 14.4.17, 14.4.18 Represent ations (i) function representations of partially ordered vector spaces and C*-algebras 4.6.26, 4.6.27, 4.6.48-4.6.54 (ii) states, representations; quasi-equivalence, type 7.6.33, 7.6.34, 10.5.38-10.5.50

XiV

EXERCISE GROUPINGS

Stone-Weierstrass theorems for C*-algebras 4.6.70, 10.5.52-10.5.59 Strong continuity of operator functions 5.7.35-5.7.37, 12.4.32 Tensor products (i) center of a tensor product of C*-algebras 11.5.1-1 1.5.4 (ii) inductive limits and infinite tensor products 11.5.26-11.5.30 (iii) simple C*-algebras 11.5.5, 11.5.6 (iv) slice maps and relative commutant formulae for vori Neumann algebras 12.4.34-12.4.37 (v) type I C*-algebras 11.5.8, 11.5.9 Unitary implementation (i) * automorphisms and * isomorphisms 9.6.18-9.6.33, 13.4.3 (ii) continuous groups of * automorphisms 9.6.65, 13.4.19-13.4.23, 14.4.8-14.4.10 (iii) norm-continuous groups of * automorphisms See the listing under “Derivatives and automorphisms,” section (iii) (iv) trace-preserving isomorphisms 8.7.2, 9.6.35-9.6.39 Unitary elements of C*-algebras (i) exponential unitaries and connectivity 4.6.2, 4.6.3, 4.6.5-4.6.9, 4.6.59 (ii) Russo-Dye theorem 10.5.3- 10.5.5 (iii) convex combinations of unitary elements 10.5.91-10.5.100 Vectors and vector states 7.6.19, 7.6.21-7.6.28, 7.6.33, 9.6.54, 9.6.60, 9.6.63

FUNDAMENTALS OF THE THEORY OF OPERATOR ALGEBRAS

VOLUME I ELEMENTARY THEORY

VOLUME I1 ADVANCED THEORY

VOLUME I11 SPECIAL TOPICS

Elementary Theory-An Exercise Approach

VOLUME IV

TOPICS SPECIAL

Advanced Theory-An Exercise Approach

CHAPTER 6 COMPARISON THEORY OF PROJECTIONS

8.9.

Exercises

6.9.1. Suppose E and F are projections in a von Neumann algebra R, E is properly infinite, CE = I , and E 5 F . Show that F is properly infinite and CF = I .

Solution. If P is a non-zero central projection, then PE # 0 since CE = I . From Proposition 6.2.3, P E 5 P F . Thus P E G 5 P F . From Proposition 6.3.2, G is infinite since P E is infinite. Hence P F is infinite. In particular, P F # 0. It follows that F is properly rn infinite and CF = I . N

6.9.2. Let R be a von Neumann algebra and E and F be projections in R such that F 5 E and E is countably decomposable. Show that F is countably decomposable.

Solution. Let {Fa : a E A} be an orthogonal family of nonzero subprojections of F in R and let V be a partial isometry with initial projection F and final projection contained in E . Then the set { V F a V * : a E A} is an orthogonal family of non-zero subprojections of E . Since E is countably decomposable, A is countable. Hence F is countably decomposable. rn 6.9.3. Let E , F, M , and N be projections in a von Neumann algebra R such that E 5 it4 and F 5 N . (i) Suppose M N = 0. Show that E V F ;;IM t N . (ii) Is it true that E V F 5 M V N ? Proof-counterexample?

Solution. (i) From the Kaplansky formula (Theorem 6.1.7), EV F - F

N

E -EA F

5 M.

275

EXERCISE 6.9.4

Since ( E V F - F ) F = 0 and F

EV F = EV F

5N, -

F+ F

5 M + N.

(ii) No! Let 3-1 be a two-dimensional Hilbert space and R be B(3-1). Let E and F be orthogonal projections with one-dimensional ranges. Let M and N be E . Then E V F = I and M V N = E so that E V F M V N although E M and F rv N .

-

6.9.4. Let E be a properly infinite projection in a von Neumann algebra R. Let F be a projection in R such that F 5 E . Show that E E V F.

-

-

-

Solution. From the halving lemma (6.3.3), there is a projection G i n R s u c h that G < E and E N G E - G . Then F 5 E E - G . From Exercise 6.9.3(i),

E V F 5 G - k E - G = E.

Since E 5 E V F , E

-

E

V

F from Proposition 6.2.4.

6.9.5. Let E be a finite projection and F be a properly infinite projection in a von Neumann algebra R. Suppose CE 5 CF. Show that Q E 4 Q F for each non-zero central subprojection Q of C F .

+

Solution. Suppose Q E QF for some non-zero central subprojection Q of CF. Then from the comparison theorem (6.2.7), there is a non-zero central subprojection P of Q such that

P F = PQF

5 PQE = P E .

But P E is finite and P F is infinite - contradicting Proposition 6.3.2. Thus QE 4 QF for each such non-zero central subprojection Q O f CF.

-

Let E and F be equivalent projections in a finite von 6.9.6. Neumann algebra R. Show that I - E I - F .

Solution. Suppose I - E and I - F are not equivalent. From Theorem 6.2.7, there is a central projection P such that either

P ( 1 - E ) 4 P ( I - F ) or P ( I - F ) 4 P ( 1 - E ) .

276

COMPARISON THEORY OF PROJECTIONS

Suppose P ( I - E)

CIJ

G < P(I - F). Then, since P E

N

PF,

contrary to the assumption that R is finite. The symmetric argument w applies if P(I - F) 4 P(I - E). Thus I - E I - F. N

6.9.7. Let E and F be equivalent finite projections in a von Neumann algebra R. Show that I - E I - F . N

Solution. Since E and F are finite, the same is true of EV F , by Theorem 6.3.8. By applying the result of Exercise 6.9.6 to the projections E and F in the finite von Neumann algebra (EVF)'R(EVF), it follows that E V F - E ry E V F - F (in ( E V F ) R ( E V F ) , and hence in 2).Thus

I - E =(I -E v F)t(Ev F)-E N(I-EVF)+(EVF)-F=I-F.

-

Let E l , E2, PI, and F2 be finite projections in a von Neumann algebra R such that 0 = El& = F1F2, El FI, and El E2 F1 F2. Show that E2 F2. 6.9.8.

+

-

+

-

Solution. Let V be a partial isometry in 'R with initial projection El E2 and find projection F' t Fa. Let GI be V&V* and G2 be VE2V*. Since GI = VEI(VE1)' and El = (VEl)*VEI,we have that G1 El ( w F1). In the same way, G2 E2. Moreover,

+

N

N

From Theorem 6.3.8, F1 t Fa is finite in R;hence (FI t F2)R(F' t F2) is a finite von Neumann algebra. Since F1 and GI are equivalent projections in this finite von Neumann algebra,

from Exercise 6.9.6.

277

EXERCISE 6.9.9

6.9.9. Let E , F , M , and N be projections in a von Neumann algebra R. Suppose that M and N are finite, M N, 0 =ME = N F , and E 4 F . Show that E + M 4 F + N . N

+

Solution. From Proposition 6.2.2, E M 5 F t N . Suppose E+M F + N . From Proposition 6.3.7, there is a central projection P such that P 5 CE and either PE is properly infinite or P = 0, and ( I - P ) E is finite. From Theorem 6.3.8, ( I - P ) ( Et M ) is finite and hence ( I - P ) ( F + N ) is finite under the present assumption. N

Since

( I - P ) ( Et M )

( I - P ) ( F t N ) and ( I - P ) M

N

( I- P)N,

we have that ( I - P ) E ( I - P ) F from Exercise 6.9.8. Since E 4 F by assumption, P E 4 P F . In particular, P # 0 and P E is properly infinite. Now P 5 CE,so that CPE = P and CPF = P. It follows that PF is properly infinite. From Exercises 6.9.4, 6.9.5, and our present assumption, N

P E N P(E+ M )

N

P(F t N )

contrary to our earlier conclusion. Thus E

N

+M

PF, 4

F

+N .

rn

Let E and F be projections in a von Neumann algebra F and V is a partial isometry in R with initial projection E and final projection F . Show that (i) there is a unitary operator U in R such that U E = V if E is finite; (ii) there is an isometry W in R (that is, W*W = I ) such that W E = V if I - E is countably decomposable, I - F is properly infinite, and CI-E 5 C I - F .

6.9.10.

R. Suppose E

N

Solution. (i) From Proposition 6.3.2, F is finite; and from I - F . Let Vo be a partial isometry in R Exercise 6.9.7, I - E with initial projection I - E and final projection I - F . Let U be Vo V . Then U is a unitary operator in R and U E = V . (ii) From Theorem 6.3.4, I - E 5 I - F . Let V1 be a partial isometry in R with initial projection I - E and final projection a subprojection of I - F . Let W be V1 V . Then W is an isometry rn in R and W E = V . N

+

+

278


6.9.11. Let R be a von Neumann algebra. Show that the following two statements are equivalent: (i) R is finite; (ii) for each pair of equivalent projections E and F in R, there is a unitary operator U in R such that UEU* = F . What are some of the consequences of defining “equivalence” of projections E and F in R t o be “unitary equivalence” ( U E U * = F for some unitary operator Uin R)?

Solution. Since E is equivalent t o F in R, there is a partial isometry V in R such that V*V = E and V V * = F . Since R is finite, E is finite and, from Exercise 6.9.10(i), there is a unitary operator U in R such that U E = V . Thus

UEU* = UEEU’ = VV* = F If R is infinite and we take I for E , there is a projection F in R such that F < I and F N I . In this case,

UEU* = U I U * = UU* = I

#F

for each unitary operator U . If R is finite, the new “equivalence” would be the same as the equivalence of Definition 6.1.4 from (i). If R is infinite, there are projections E and F in R such that F < E ,

INE NF

N

I -E

N

I - F.

In this case, there is a unitary operator U in R such that U E U * = F . Thus E is ”infinite” in the new sense; but I is not ”equivalent,” in the new sense, t o a proper subprojection so that I is always “finite” in the new sense. At the least, the new “finiteness” and “infiniteness” are peculiar. In any event, the old and new equivalences are different when R is infinite. m 6.9.12. Show that, if R is a semifinite von Neumann algebra, there is an orthogonal family {Qa}of non-zero central projections in R such that C Q a= I and each Q a is the sum of an orthogonal family of mutually equivalent finite projections in R. [Hint. Look a t the proof of Proposition 6.3.12.1

EXERCISE 6.9.12

(*I

2 79

Solution. Suppose that we have proved the following assertion: given any non-zero central projection P in R, there is a central projection Q in R such that 0 < Q 5 P and Q is the sum of an orthog-

onal family of mutually equivalent finite projections in R. Let { Q a }be an orthogonal family of non-zero central projections in R,maximal subject t o the requirement that each Q a should be the sum of an orthogonal family of mutually equivalent finite projections in R. If C Qa # I , we can apply (*), with P the projection I-C Q a ; it follows that there is a central projection Q such that 0 < Q 5 I - CQa and Q is the sum of an orthogonal family of mutually equivalent finite projections in R. Then, Q can be added t o the set { Q a } , contrary to the maximality assumption. Hence C Q a = I . It remains t o prove (*). To this end, let P be a non-zero central projection in R. If P is not properly infinite, it has a central subprojection Q (# 0) that is finite in R, and (*) is (trivially) satisfied. We assume henceforth that P is properly infinite. Since R is semifinite, there is a finite projection E in R for which CE is I. Let {Eb : b E B} be an orthogonal family of projections in R,maximal subject to the condition that Eb P E for each index b. Then N

Eb 5 C E = ~ CPE = PCE = P, and (from the maximality assumption) P E P - C Eb. From the comparison theorem, there is a central projection Q in R such that Q ( P - Eb) 4 Q P E . Upon replacing Q by Q P , we have a central projection Q such that 0

< Q 5 P,

Q - X Q E b -i Q E .

The projections QEb(N Q E 5 E ) and (Q - x Q E b ) ( - i Q E 5 E ) are all finite; but their sum Q is infinite since P is properly infinite. Hence the index set B is infinite. With 60 in B, there is a one-to-one correspondence between the sets B and B \ { b o } . Thus

-

whence x b E a Q E b Q . If W is a partial isometry in R, from C b E B Q E bt o Q , and Fb is the range projection of WQEb, then {Fb : b E B} is an orthogonal family of mutually equivalent finite projections in R,and Fb = Q.

280


6.9.13. Let E be a properly infinite projection in a von Neumann algebra R and F be a cyclic projection in R such that CE 5 C F . Show that there is an orthogonal family { Q a }of central subprojections of CE with sum CE such that, for each a, Q a E is the sum of an orthogonal family of projections each equivalent to Q a F .

Solution. Suppose that we have proved the following assertion:

(*I

given any non-zero central subprojection P of CE there is a central projection Q in R such that 0 < Q IP and Q E is the sum of an orthogonal family of projections each equivalent to QF.

Let {Q,} be an orthogonal family of non-zero central subprojections of CE,maximal subject to the requirement that each Q a E should be the sum of an orthogonal family of subprojections each equivalent to Q a F . If C Q a # CE,we can apply (*), with P the projection CE - C Q a ;it follows that there is a central subprojection Q such that 0 < Q ICE - C Q a and Q E is the sum of an orthogonal family of projections each equivalent to QF. Then Q can be added to the family {Q,} contrary to the maximality assumption. Hence

C Qa

= CE.

It remains to prove (*). To this end, let P be a non-zero central subprojection of CE.Let {Eb : b E b} be an orthogonal family of subprojections of P E maximal subject to the condition that Eb P F for each index b. Then N

and (from the maximality assumption) PF ;d E - X E b . From the comparison theorem, there is a central projection Q in R such that Q ( E - C Eb) 4 Q P F . Upon replacing Q by Q P , we have a central projection Q such that

The projections QEb are each equivalent to QF. (There is at least one Eb, since CPF = CPE= P , PF is cyclic, and P E is properly inQEb is not properly infinite, there finite, from Theorem 6.3.4.) If is a non-zero central subprojection QO of Q such that Q o ( C Q E b )

281

EXERCISE 6.9.14

(= Z Q o E b ) is finite. From Proposition 6.3.2, QoE - XQoEb is finite; and from Theorem 6.3.8, QoE is finite, contrary to our assumption that E is properly infinite. Hence X Q E b is properly infinite.

QEb is properly infinite,

and

from Exercise 6.9.4. Let V be a partial isometry in R with initial projection C QEb and final projection QE. Then the family {QVEbV* : 6 E B} is an orthogonal family of subprojections of Q E with sum QE such that QVEbV* QF for each 6 in B. Thus (*) is proved.

-

Let R be a von Neumann algebra and { E l , . . . ,E n } , { F l , . . . ,Fn} be two finite sets of projections in R such that El * * * E n , F1 * - * F,, Xj"=,Ej = I , and Cj"=,Fj = I . Show that Ej F j .

-

6.9.14.

N

-

N

N

Solution. From Proposition 6.3.7, there is a central projection P in R such that PE1 is properly infinite or P = 0 , and (I- P)E1 is finite. From Proposition 6.3.2, ( I - P)E2,. . .,( I - P ) E , are finite. From Theorem 6.3.8, X,",l(I - P)Ej = I - P is finite. Hence ( I - P)F1,. . . , ( I - P)Fn are finite. If ( I - P)E1 is not equivalent to ( I - P ) & , there is a central projection Q in R such that either Q ( I - P)E1 4 Q ( I - P)F1 or Q ( I - P)F1 4 Q ( I - P ) E l . In the first case,

Q ( I - P)Ej

-

N

G j < Q ( I - P)Fj

( j E (1,.

.. , n } ) .

Hence Q ( I - P ) C G j < & ( I - P ) - contradicting the finiteness of & ( I - P ) . Thus ( I - P ) E I ( I - P)F1. Suppose P # 0. Then PE1 is properly infinite and from Exercise 6.9.4, N

PE1

-

-

- -C n

PE1 -t PE2

*

a

*

PEj = P

j= 1

since PE1 rv P E , . Now PF1 is properly infinite, for if PO is a non-zero central subprojection of P such that PoPFl(= PoF1) is

282


finite, then Cj”=,PoFj (= PO)is finite. But PoPEl(= PoEl) and, hence, Po are infinite since PE1 is properly infinite - contradicting the finiteness of PO. Since PF1 is properly infinite, as before, PF1 P. Thus PE1 PF1. It follows that El F1. N

N

N

Let E be a non-zero projection in a von Neumann 6.9.15. algebra R acting on a Hilbert space 3-1. Show that, in the von Neumann algebra E R E acting on E(3-1), (i) G is an abelian projection if and only if G is a subprojection of E abelian in R; (ii) each central projection has the form P E with P a central projection in R ; (iii) G is a finite projection if and only if G is finite in R and G is a subprojection of E ; (iv) G is properly infinite if and only if G is a subprojection of E properly infinite in R.

Solution. (i) We have that the projection G is abelian in E R E if and only if G E R E G (= G R G ) is abelian, in which case G is abelian in R. (ii) From Proposition 5.5.6, the center of E R E is CE, where C is the center of R. Thus, a projection in CE has the form PoE with PO in C. From Proposition 5.5.5, the mapping T’E -+ T’CE is a * isomorphism of R‘E onto R’CE. Now C C 72’ and PoE maps onto PoCE. Since PoE is a projection, POCEis a (central) projection P in R and P E = PoCEE = PoE. (iii) A subprojection of E in R lies in E R E and an equivalence of two projections of E R E relative to R is an equivalence relative to E R E . Thus G is finite in E R E if and only if it is a subprojection of E finite relative t,o R. (iv) We have G properly infinite in E R E if and only if PEG (= PG) is either 0 or is infinite in E R E for each central projection P in R from (ii). From (iii), PG is infinite in E R E if and only if it is infinite in R. Thus G is properly infinite in E R E if and only if it is (a subprojection of E ) properly infinite in R. 6.9.16. Let 92 be a von Neumann algebra acting on a Hilbert space ‘H, and let E be a non-zero projection in R. Show that E R E acting on E(3-1) (i) is finite if R is finite;

EXERCISE 6.9.17

(ii) (iii) (iv) (v) (vi)

283

is of type I if R is of type I; is of type I, if R is of type I, and E is properly infinite; is of type 111 if R is of type 111; is type 11, if R is of type 11, and E is properly infinite; is of type I11 if R is of type 111.

Solution. (i) From Exercise 6.9.15@), E is finite relative to E R E since it is finite relative to R . Thus E R E is finite if R is finite. (ii) Since R is of type I , there is an abelian projection F in R with central carrier I . From Proposition 6.4.8, there is an orthogonal family { E a }of abelian projections in R with sum E . From Exercise 6.9.15(i), each E, is abelian in E R E . Thus E R E is of type I. (iii) From Exercise 6.9.15(iv), E is properly infinite in E R E . From (ii), E R E is of type I. Thus E R E is of type I,. (iv) Since R is finite, E R E is finite from (i). If G is a nonzero abelian projection in E R E , then from Exercise 6.9.15(i), G is a non-zero abelian projection in R. On the assumption that R is of type 111, no such G exists. Thus E R E has no non-zero abelian projections, and E R E is of type 111. (v) Since R is of type 11,) there is a finite projection F in R with CF = I . Moreover, R has no non-zero abelian projections. Hence, from Exercise 6.9.15(i), E R E has no non-zero abelian projections. As E is properly infinite, C E F 4 E from Exercise 6.9.5. Thus C E F EO < E for some projection Eo in R, and EO is finite with central carrier CE in R. It follows from Exercises 6.9.l5(iii) and 5.7.38 that EO is a finite projection in E R E with central carrier E (relative to E R E ) . From Exercise 6.9.15(iv), E is properly infinite in E R E . Thus E R E is of type 11,. (vi) From Exercise 6.9.l5(iii), E R E has no finite projections other than 0 since R has none. As E # 0 , E R E is of type 111. N

6.9.17. Let E be a projection in a maximal abelian (selfadjoint) subalgebra A of a von Neumann algebra R. Suppose E is minimal in the family of projections in A with central carrier CE (relative to R). Show that E is abelian in R. Solution. From Proposition 5.5.6, the center of E R E (acting on E ( R ) , where R acts on 3-1) is CE, with C the center of R , and AE is a maximal abelian subalgebra of E R E . If F is a non-zero

284


projection in A E , then F E A and F 5 E. Thus F = C F F I: C F E . If F < C F E ,then ( I - C F ) Et F is a projection in A (since C A ) with central support C E . Moreover, ( I - C F ) E-t F < E . But this contradicts the minimality of E . Hence F = C F E . It follows that each projection in A E is in CE, so that A E = CE. Thus the center CE of E R E is a maximal abelian subalgebra A E of E R E , It follows that E R E = CE, E R E is an abelian von Neumann algebra, and E is an abelian projection in R. m[66] 6.9.18. Let A be a maximal abelian (self-adjoint) subalgebra of a von Neumann algebra R. Suppose A # R. Show that A contains two orthogonal non-zero projections E and F such that CE = CF and E 5 F .

Solution. If G is a projection in A such that CGCI-G = 0, then G = CG,for

It follows that either each projection in A is central in R, in which case the center of 72 coincides with A and A = R (since A is maximal abelian in R ) , or ( P =)CGCI-G# 0 for some projection G in A. Since A # R (that is to say, R is not abelian), the latter case applies. In this case, PG and P( I - G) have the same central carrier P . From the comparison theorem (6.2.7)) there is a non-zero central projection Q such that Q 5 P and either QG 5 Q ( I - G ) or Q ( I - G) 5 QG. In any event, one of QG', Q ( I - G) serves as E and the other serves as F , when R is not abelian. a

6.9.19. Let R be a von Neumann algebra with no abelian central summands, and let A be a maximal abelian (self-adjoint) subalgebra of R. Show that A contains a projection E such that C E = C I - E = I and E 5 I - E .

Solution. Let {E,} be a family of non-zero projections in A maximal with respect to the properties that {CE,} is an orthogonal family and E , 2 I - E, for each a. From Exercise 6.9.18, A contains non-zero orthogond projections Eo and F' with the property that Eo 5 Fo ( 5 I - Eo). Thus the family {E,} is non-null. Let E be C, E,. Then CE = C a C ~(=& P). If P # I , then R ( I - P ) is

285

EXERCISE 6.9.20

a non-abelian von Neumann algebra (since R is assumed to have no central summands that are abelian) and A(I - P ) is a maximal abelian subalgebra of R ( I - P ) . Again from Exercise 6.9.18, there is a non-zero projection El in A(I - P ) such that El 5 ( I - P ) - E l . If we adjoin El to the family { E a } ,we have a properly larger family than { E a } , contradicting the maximality of { E a } . Thus P = I . Since Ea = CE,Ea 5 CE,(I - Ea) = CE, - E , for each a , we have that

E =C E " a

5 CcE,- E ,

= I - E,

a

andCE=P=I. Let R be a countably decomposable von Neumann 6.9.20. algebra and A a maximal abelian (self-adjoint) subalgebra of R that contains no non-zero projection finite in R. Show that A contains n orthogonal projections with sum I equivalent in R for each positive integer n.

Solution. Note, first, that under the hypotheses on A, R has no abelian central summand, for if P is a non-zero central projection such that R P is abelian, then P is in A and P is an abelian, hence finite, projection in R. From Exercise 6.9.19, there is a projection E in A such that CE = C I - E = I . From Corollary 6.3.5, E , I - E , and I are equivalent in R, for all non-zero projections in A are properly infinite in R. Again E R E acting on E ( H ) , where R acts on the Hilbert space H,is a countably decomposable von Neumann algebra in which A E is a maximal abelian subalgebra that contains no nonzero finite projections. Applying Exercise 6.9.19 once again, we find a projection F in A E equivalent to E - F and E (in E R E , and hence in R). We now have three orthogonal equivalent projections, F , E - F , and I - E , in A (with sum I ) . Continuing in this way, we construct n orthogonal equivalent projections with sum I in A. m[66] Let R be a von Neumann algebra of type I with no 6.9.21. infinite central summand, and let Pn be a central projection in R such that RPn is of type I,. Let A be a maximal abelian subalgebra of R. Show that

286


(i) some non-zero subprojection of P, in A is abelian in R; (ii) A contains an abelian projection with central carrier I .

Solution. (i) If n = 1, then PI is a non-zero abelian projection in the center of R and hence in A. If n > 1, then RP, acting on P,('FI), where 3-1 is the Hilbert space on which R acts, is a von Neumann algebra !without abelian central summands and AP, is a maximal abelian subalgebra of it. From Exercise 6.9.19, AP, con~ P, and El 5 P, - El. Now tains a projection El such that C E = EIRE1 acting on EI(H)is a type I von Neumann algebra with no infinite central summand. (See Corollary 6.5.5 or Exercise 6.9.16.) Again, either d E 1 has a non-zero abelian projection F , in which case F R F = F E l R E l F is abelian and F is an abelian projection in R, or there is a non-zero projection E2 in dE1 such that E2 5 El - E2. Continuing in this way (we consider E2RE2 next), we produce either a non-zero abelian projection in A or a set of n non-zero projections E l , . . . ,En in RP, such that E j + l 5 Ej - Ej+l, El 5 P, - E l , and Ej+l < E j . (From the assumption that R has no infinite central summand, n must be finite.) If Q is the central carrier of En, then En, Q(En-1 - En), Q ( E n - 2 - En-l),.**,Q(& - E2), Q(Pn - El)

+

are n 1 orthogonal projections in RP, with the same (non-zero) central carrier, which contradicts the fact that RP, is of type I,. Thus the process must end with a non-zero abelian subprojection of P, for R in A before we construct E n . (ii) Let { E a }be a family of non-zero projections in A abelian for R and maximal with respect to the property that {CE,}is an orthogonal family. Let P be C, CE,. If P # I , then R ( I - P) is a von Neumann algebra of type I with no infinite central summand. From what we have just proved in (i), A(I - P ) , a maximal abelian subalgebra of R ( 1 - P), contains a non-zero abelian projection EO for R ( I - P). But then, adjoining EO to { E a }produces a family properly larger than { E a } of non-zero abelian projections for R in A with mutually orthogonal central carriers, which contradicts the maximal property of { E , } . Thus P = I . From Proposition 6.4.5, C, E, is an abelian projection for R, has central carrier I , and lies ind. 6.9.22. Let El be an abelian projection with central carrier I in a von Neumann algebra R of type I, with n finite. Show that

EXERCISE 6.9.23

287

(i) there is a set of n orthogonal equivalent projections with sum I in R containing El (so that each is abelian in R); (ii) ( I - E I ) R ( I - E l ) is of type I,-1.

Solution. (i) Let { E, : a E A} be a maximal orthogonal family of equivalent projections in R containing E l . Let E be CaEA E,. If I - E has central carrier I , then El 5 I - E from Proposition 6.4.6(ii) and there is a subprojection Eo of I - E equivalent to E l . In this case, { E o , E, : a E A} is a properly larger family than { E , : a E A} contradicting the maximality of { E , : u E A}. Thus I - E does not have central carrier I and there is a non-zero central projection P such that P ( I - E ) = 0. Thus CaEA P E , = P and RP is of type I, where m is the cardinality of A. From uniqueness of the type decomposition (Theorem 6.5.2), m = n. If E # I and Q is C I - E , then QE1 5 I - E from Proposition 6.4.6@). Let F be a subprojection of I - E equivalent to QE1. As before, we extend { F , QE, : a E A} to a maximal orthogonal family of projections {Fb : 6 E B} in RQ each equivalent to QE1 . Again, there is a non-zero central subprojection Qo of Q such that Q o Fb = Qo; and RQo is of type I k , where k is the cardinality of B. Now {Fb : b E B} contains { F , QEa : a E A}, which has n -t 1 elements; so that n t 1 ,< k contradicting the assumption that R is of type I,. Thus CaEA E, = I . If we relabel { E , : u E A} of (i) as { & , . . . , E n } , then (ii) I - El = E2 t .- En. F'rom Exercise 6.9.15(i), E2,. . . , E n are abelian projections in ( I - E1)R(I - E l ) . Thus (I- E l ) R ( I - E l ) is of type I,-1.

+

6.9.23. Let R be a von Neumann algebra of type I, with n finite, and let A be a maximal abelian subalgebra of R. Show that (i) there are n (orthogonal, equivalent) projections in A with sum I each abelian in R with central carrier I relative to R; (ii) A is generated algebraically by the n abelian projections of (i) and the center of R; (iii) A contains p orthogonal projections with sum I equivalent in R if n = pq (with p and q positive integers). Solution. (i) We proceed by induction on n. When n = 1 , R is abelian, R = A, and I is a projection in d abelian in R with central carrier I . Moreover, d (= R) is the center of R. Suppose n > 1 and

288


we have established our assertion when R is of type 11, with b less than n. Then R has no infinite central summands. Exercise 6.9.21(ii) applies and there is a projection El in d such that El is abelian and C E= ~ I. It follows from Exercise 6.9.22(ii) that (I- E l ) R ( I - El) acting on (I- E1)(3t)is a von Neumann algebra of type In-1, where R acts on the Hilbert space 71, and d ( I - El) is a maximal abelian subalgebra of it. The inductive hypothesis applies and I - El is the sum of n - 1projections E l , , ,E n in d ( I - El), each Ej is abelian in ( I - El)'R(I - E l ) (and, hence, in R), and has central carrier I - El in (I- E l ) R ( I - E l ) . As in the solution to Exercise 6.9.22, it follows that I = C q = * = CE, * (ii) Let C be the center of R. Since EjREj acting on Ej('H) is an abelian von Neumann algebra with center CEj and in which A E j is a maximal abelian subalgebra, we have that E j R E j = CEj = d E j for each j in (1,.. ,n}. If A E d,then

..

--

.

where Cj E C. (iii) With E l , . . .,En the projections constructed in (i), let Fj be E,+kp, where j E (1,. ., p } . Then F1,. ,Fp are orthogonal projections in A with sum I equivalent in R . 4661

.

~~~~

..

Let R be a countably decomposable von Neumann 6.9.24. algebra of type I,, and let A be a maximal abelian subalgebra of R in which I is the union of projections in A finite in R. Show that (i) d has a projection finite in R with central carrier I; (ii) d has a projection abelian in R with central carrier I ; (iii) some non-zero central projection Q in R is the sum of projections in A abelian in R with central carriers Q in R; (iv) there is a countable family of orthogonal projections in A with sum I each abelian with central carrier I; (v) there are n Orthogonal projections in A with sum I equivalent in R for each positive integer It. Solution. (i) Let {F,,} be a family of projections in A finite in 'R and maximal with respect to the property that {CF,}is an orthogonal family. If P = C bCF,and P # I, then I - P is a nonzero projection in A. If I P is orthogonal to all finite projections

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289

EXERCISE 6.9.24

of R in A, the union of these finite projections is not I , contrary to assumption. Thus there is a projection FO in A finite in R such that Fo(I- P) # 0. But then { & ( I - P),Fb} is a family of finite projections in A, properly larger than {Fb}, whose central carriers form an orthogonal family. This contradicts the maximal property of {Fb}. Thus P = I . From Lemma 6.3.6, F b is a projection F in A finite with central carrier I in R. (ii) With the notation of (i), F R F is a von Neumann algebra of type I with no infinite central summand and A F is a maximal abelian subalgebra of it. From Exercise 6.9.21(ii), A F contains a projection EOabelian in FRF (and hence, in R) with central carrier F i n FRF. Since F has central carrier I in R so has Eo. Thus Eo is an abelian projection in R with central carrier I and EO lies in A. (iii) Let { E a }be a maximal orthogonal family of projections in d abelian with central carrier I in R,and let E be C, E,. If E # I , then ( I - E ) R ( I - E ) is a von Neumann algebra of type I in which d(I - E ) is a maximal abelian subalgebra. Moreover, I - E is the union of projections in A ( I - E ) finite in ( I - E ) % ? ( I - E ) . From (ii), d(I- E ) contains a projection El abeiian with central carrier I - E in ( I - E ) R ( I - E ) . It follows that El is abelian with central carrier CI-E in R. If C I - E = I , we can adjoin El to { E a } contradicting the maximal property of {&}. Thus (Q =) I - CI-E # 0. Now Q ( I - E ) = 0 so that { Q E a } is a family of projections in A abelian with central carriers Q and sum Q in 72. (iv) Let (Q,}be a maximal orthogonal family of central projections in R each with the property of Q in (iii). If 0 # I - C,Qc (= Q O ) , then RQo is a von Neumann algebra of type I, and AQo is a maximal abelian subalgebra of it with the property that QO is the union of projections in AQo finite in RQo. Thus there is a non-zero central projection Q1 in RQo that is the sum of projections in AQo abelian with central carriers Q1 in RQo (hence, in R). Adjoining Q1 to {Q,} produces a family that contradicts the maximality of {Q,}. Hence CcQc = I. Since R is countably decomposable of type I,, the same is true of RQ,. We can index the set of projections in AQc with sum Qc (abelian in R with central carriers Q c )as E l , , Ez,, . . .. From Proposition 6.4.5, C, En, (= En)is an abelian projection with central in R for each positive integer n. Moreover, each carrier I (= C,Qc) En is in A and C,"==, En = CcQc= I . With the notation of (iv), if Fk = C g o E k + j n for k in (v)

cb

290


.

.

(1,. .,n}, then {Fl,. . ,F,} are n orthogonal projections in A, equivalent in R with sum I .

Let R be a countably decomposable von Neumann algebra of type I,. Show that each maximal abelian subalgebra A of R contains n orthogonal projections with sum I equivalent in R for each positive integer n. [Hint. Consider the union EO of all projections in A finite in R and apply the results of Exercises 6.9.20 and 6.9.24 as well as Proposition 6.3.7.1 6.9.25.

Solution. Let EO be as in the hint. If EO = 0, then A has no finite non-zero projections and Exercise 6.9.20 applies to complete the solution. If Eo = I , then Exercise 6.9.24 applies t o complete the argument. We may suppose that 0 < EO < I. In this case, ( I - Eo)R(I - Eo) acting on ( I where R acts on the Hilbert space 3-1, is a countably decomposable von Neumann algebra of type I and d ( I - Eo) is a maximal abelian subalgebra of it that contains no finite non-zero projections. From Exercise 6.9.20, there are n orthogonal projections E l , , . ,En in A ( I - E o ) with sum I - EO equivalent in ( I - Eo)R(I-Eo) (and hence, in R). From Proposition 6.3.7, there is a central projection PO in R such that PoEo is finite and either I - PO is 0 or ( I - PO)& is properly infinite. Suppose PO # 0. Then PO is properly infinite since R is of type I,, and PO= PoEo Po(I- E o ) with PoEo finite. Hence Po(I- Eo) is properly infinite from Theorem 6.3.8. Now Po(1- Eo) = Po('& Ej) and PoE1,. . . ,PoE, are equivalent, by choice of E l , . . . , E n . Thus PoEl, . ,PoE, are equivalent (countably decomposable) properly infinite projections in APo, and Po(E1 t Eo), Po&, . . ,FOE, are n equivalent (countably decomposable, properly infinite) projections with sum PO in APo. (Use Corollary 6.3.5 for this last.) It will suffice to locate n orthogonal equivalent projections in A ( I - PO)with sum I - PO. In effect, we may assume that EO is properly infinite with central carrier I (that is, that PO = 0). With this assumption, &RE0 acting on EO('H)is a countably decomposable von Neumann algebra of type I, and AEo is a maximal abelian subalgebra with the property that & is the union of projections in AEo finite in EoREo. From Exercise 6.9.24, there are n projections PI,.. . ,Fn in dEo equivalent in EOREO (hence in R) with sum Eo. The n projections El F1 ,.. . ,E , F, are equivalent in R, have sum I , and lie in A . 4661

&)(a),

.

+

..

.

+-

+

291

EXERCISE 6.9.26

Let R be a finite von Neumann algebra and E , Eo, 6.9.26. F , FO be projections in R such that EO 5 E , FO 5 F , EO Fo, and E 5 F. (i) Show that E - EO 5 F - Fo. (ii) Suppose E F . Show that E - Eo N F - Fo. N

N

Solution. (i) If E - EO 5 F - Fo, then there is a non-zero central projection P in R such that P ( F - Fo) 4 P ( E - Eo)from the comparison theorem. Thus P ( F - Fo) G < P ( E - Eo) for some proper subprojection G of P ( E - Eo). Now PEo PFo and PE 5 P F , so that N

N

from Proposition 6.2.2. It follows from Proposition 6.2.4 that P E is equivalent to the proper subprojection G 4- PEo, contradicting the assumption that R is finite. Thus E - EO 5 F - Fo. (ii) If E F , then E 5 F and F 5 E . From (i), E-Eo 5 F-Fo and F-Fo 5 E-Eo. From Proposition6.2.4, E-Eo F-Fo. N

N

Let R be a von Neumann algebra of type 111. Let A 6.9.27. be a maximal abelian subalgebra of R and E be a projection in A. (i) Show that there is a sequence { E n }of projections in A such that EO = E , CE, = C E ,En 5 En-l, and En 5 En-l - E n for n in {1,2,. . .}. (ii) Suppose F is a projection in R such that CECF# 0. Show that there is a non-zero projection G in A such that G 5 E and G 5 F. (iii) Suppose F is a projection in R such that F 5 E . Show that some subprojection El of E in A is equivalent to F . (iv) Show that A contains n orthogonal equivalent projections with sum I for each positive integer n.

Solution. (i) If E = 0, we choose 0 for each En. Suppose E # 0. Then E R E is a von Neumann algebra of type 111 from Exercise 6.9.16(iv), and A E is a maximal abelian subalgebra of it. In particular, E R E has no abelian central summands. From Exercise 6.9.19, AE contains a projection El with central carrier E relative to E R E and such that El 5 E - El. It follows that C E = ~ C E . If we repeat this construction with El in place of E , we locate a projection

292


E2 in A such that C E = ~ CE,E2 5 E l , and Ez 5 El - E2. In this way, we construct a sequence {E,} with the desired properties. (ii) If we replace R, A, E , and F , by RQ, A Q , E Q , and F Q , where Q = CECF.we may assume that CE = CF = I . With this assumption and with the notation of (i), if PEn j! P F for each nonzero central subprojection P of CF in R, then F 5 E , from the comparison theorem. If F 5 En for each n, then E,-1 - E , has a subprojection equivalent t o F for each n. In this case, R contains an infinite orthogonal family of projections equivalent t o F , which contradicts the assumption that R is finite. Thus, for some n and some non-zero central projection P , PEn 4 P F ; so that P E , will serve as G . (iii) Let S be the set of pairs consisting of orthogonal families {E,} and { F a } ,indexed by the same set A, of non-zero projections, where Ea Fa for all u in A, Ea E A, Fa E R,Ea 5 E , and Fa 5 F . Let 5 be the partial ordering of S for which

-

when

A 2 B. Applying Zorn’s lemma, we find a maximal element

-

( { E c } ,{F,}) in S . Let El be C, E , and F1 be CcF,. From Proposition 6.2.2, El F1. Now El E A, El 5 E , and F1 5 F . Since R is finite and F 5 E , F - F1 5 E - El from Exercise 6.9.26. If F - F1 # 0 , then E - El # 0 ; and F - F1 has a non-zero central carrier contained in the central carrier of E - E l . From (ii), there is a non-zero projection EO in A such that Eo 5 E - El and EO FO 5 F - F1. But then ( { E o , E c } , ( F ~ , F c }is) an element of S properly larger than ({E,},{F,}) - contradicting the maximal property of ({E,}, {F,}). It follows that El F1 = F. (iv) From Lemma 6.5.6, there are n orthogonal equivalent projections F1,. . . ,F, in R with sum I since R has no central portion of type I. From (iii), there is a projection El in A such that El FI. From Exercise 6.9.6, I - El I - F1 (= F2 t -..t F,). Again from (iii), there is a subprojection E2 of I - El in A such that E2 F2. Continuing in this way, we find E l , . . . ,E , in A such that E l - F l , ..., E , N F n a n d E 1 + . . . + E , ~ F l + . . . + F , = .I Since R is finite, El t . .. -t En = I .

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-

-

-

-

6.9.28. Let R be a countably decomposable von Neumann algebra and A a maximal abelian subalgebra of it. Suppose R has

EXERCISE 6.9.29

293

no central summand of type I. Show that each non-zero projection in A contains n non-zero orthogonal projections in A equivalent in

R. Solution. Let E be a non-zero projection in A. Then E R E acting on E ( H ) , where 72 acts on the Hilbert space ‘H, is a countably decomposable von Neumann algebra with no central summand of type I, from Exercise 6.9.16, and AE is a maximal abelian subalgebra of ERE. If we show that AE contains n orthogonal non-zero projections equivalent in ERE,then these n projections are orthogonal subprojections of E in A and are equivalent in R. It suffices to show that A contains n orthogonal non-zero projections equivalent in R. If A contains a non-zero finite projection F , then F R F acting on F ( ? f ) is a von Neumann algebra of type 111 and AF is a maximal abelian subalgebra of it, from Exercise 6.9.16. From Exercise 6.9.27(iv), AF contains n orthogonal subprojections equivalent in F R F with sum F . We may suppose now that A has 110 non-zero finite projections. In this case, Exercise 6.9.20 applies and A has n orthogonal projections with sum I equivalent in R.

Let R be a countably decomposable von Neumann 6.9.29. algebra with no central summand of type I, and let n be a positive integer. Show that each maximal abelian subalgebra of R contains n orthogonal projections with sum I equivalent in R.

Solution. Suppose A is a maximal abelian (self-adjoint) subalgebra of R. Let { E j } be a,n orthogonal family of projections in A, maximal subject to the condition that each E j can be expressed as a sum Ejl i- E j , of n projections in A that are equivalent in 72. If C E j # I , it follows from Exercise 6.9.28 that I - C Ej contains n non-zero projections in A equivalent in R,the sum of these projections can then be adjoined to the family { E j } , contrary to our maximality assumption. Hence

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294


6.9.30. that n 8 R

Let R be a von Neumann algebra with center C. Show has center C 8 I,,.

Solution. From Lemma 6.6.2, (n 8 R)' = R' 8 In. Thus the center of n @ R, ( n 8 2)n (R'8 In), has all of its non-zero entries on the diagonal, and all diagonal entries coincide with some one element of R R R' (= C). Thus C 63 1, is the center of TZ @ 12. rn

6.9.31. Let 72 be a von Neumann algebra and n be a positive integer. Suppose P is a central projection in n@Rsuch that ( n @ R ) P is of type I,. Show that m is divisible by n. Solution. The matrix of P has a central projection QOof R at each diagonal position and 0 at all others, from Exercise 6.9.30. Since (n872)P is of type I, the projection M in ( n 8 R ) Pwhose matrix has QOat the 1, 1entry and 0 at all others is a sum of projections abelian in ( n 8 72)P.Using Proposition 6.4.5, we can find a subprojection G of A4 abelian in (n@ R)P with the same central carrier P as M. The matrix of G has some projection Go in RQo at the 1, 1 entry and 0 at all others. Since G is abelian in ( n 8 R)P with central carrier P,Go is abelian in RQo with central carrier Qo. Thus RQo is of type I. Suppose MI,. .,Mk are orthogonal abelian projections in RQo each with the same non-zero central carrier Q. By placing each Mj at one of the diagonal positions and 0 at all other positions, we form nk orthogonal abelian projections in ( n 8 R)P each with central carrier 91, the matrix with Q at each diagonal entry and 0 at all others. Since ( n 8 R)P is of type I,; nk _< rn. It follows that each central summand of 72Qo is of type Ij, where j 5 $. Now RQo has a central summand of type 11, from Theorem 6.5.2 since it is of type I and each central summand of it is finite (as just noted). If we assume, in the preceding argument, that A41 $- -..t Mk = Q,then .. ,Mk have sum Q1. the nk abelian projections formed from MI,. Thus (n 8 R)Q1is of type I,I, as well as of type I,. By uniqueness of type decomposition (see the comments preceding Theorem 6.5.2), m = nk. Hence m is divisible by n, k = f , and RQo is of type

.

Ik.

EXERCISE 6.9.33

295

6.9.32. Let R be a von Neumann algebra and n be a finite cardinal. Show that (i) n @ R is finite if R is finite; (ii) n @ R is properly infinite if R is properly infinite; (iii) n @ R is countably decomposable if R is countably decomposable.

Solution. Let Ej be the element of n @ R whose matrix has I in the j , j entry and 0 at all others. Then f = Cy==, E j , where f is the identity of n 8 R. If F is a subprojection of Ej in n 8 R, then EkFEk = 0 when k # j , so that EkFEh = 0 for all h in ( 1 , . . . ,n } from Exercise 4.6.11. Thus the h, k entry of F is 0 unless h = k = j . If FO is the j , j entry of F , then F i = Fo, for F i is the j , j entry of F 2 (= F ) . Similarly F,*, the j , j entry of F’ (= F ) , is Fo. Hence FO is a projection in R. (i) If V is a partial isometry in n @ R with initial projection Ej and final projection F , then EjVEj = V . Thus V has 0 as its h , k entry unless h = k = j . Let VObe the j , j entry of V . Then V , h = I and VoV, = Fo. Since R is finite, FO = I and F = Ej. It follows that Ej is finite for each j in ( 1 , . . . ,n}. From Theorem 6.3.8, Cj”=,Ej ( = I ) is finite and n @ R is finite. (ii) From Exercise 6.9.30, n 8 R has center C 8 I,. Thus if Q is a central projection in n @ R,there is a central projection Qo in R such that Q has all its non-zero entries on the diagonal and each is Qo. If Q is non-zero, Qo is non-zero. Since R is properly infinite, Qo Eo < Qo for some projection Eo in R. If E is the projection in n @ R with all its non-zero entries on the diagonal and each equal t o Eo, then E < Q and Q E . Thus Q is infinite in n @ R and n @ R is properly infinite. (iii) Let {Fa : a E A} be an orthogonal family of non-zero subprojections of E j . Then Fa has as its only non-zero entry a projection E , in R in the j , j position. It follows that { E , : a E A} is an orthogonal family of non-zero projections in R. Since R is countably decomposable, A is countable. Hence Ej is a countably decomposable projection in n @ R. From Proposition 5.5.9, Ej is the union of an orthogonal family of cyclic projections in n @ R ,and this family is countable. Thus Cy==, E j (= f ) is the union of a countable family of cyclic projections in n @ R. From Proposition 5.5.19, R is countably decomposable. N

N

296


Let R be a von Neumann algebra and n be a cardinal. 6.9.33. Show that (i) R is finite if n @ R is finite; (ii) R is properly infinite if n is finite and n @ R is properly infinite; (iii) R is countably decomposable if n @I R is countably decomposable,

Solution. (i) From Lemma 6.6.2, R is * isomorpphic to R @ I n . Since n @ Ris finite, each von Neumann algebra, in particular, R B I n , contained in n @ Ris finite. From its definition, finiteness is preserved by * isomorphisms. Thus R is finite. (ii) If QO is a non-zero central projection in R and Q is the (central) projection in n @ R with QO at each diagonal entry and 0 at all others, then n @ (RQo)= ( n @ R)Q. Now ( n @ R)Q is properly infinite since n @ R is properly infinite. If RQo were finite, then n@(RQo) (= ( n @ R ) Qwould ) be finite from Exercise 6.9.32(i). Thus RQo is infinite and R is properly infinite. (iii) As in (i), 'R is * isomorphic to the subalgebra R @ I , of n 8 R. Since each von Neumann subalgebra of a countably decomposable von Neumann algebra is countably decomposable and * isomorphisms preserve countable decomposability, R is countably decomposable. Let R be a countably decomposable von Neumann 6.9.34. algebra and n a positive integer. Show that each maximal abelian subalgebra of n 8 72 contains n (orthogonal) equivalent projections with sum I .

Solution. Let A be a maximal abelian subalgebra of n 8 R and C be the center of n @ R. Then C A so that each of the central projections corresponding to the central type decomposition of n @ R lies in A. If P is a central projection in n @ R, then A P is a maximal abelian subalgebra of ( n @ R ) P . From Theorem 6.5.2, there are central projections Pw, PI,Pz,.. . , and Pc,with sum I such that either P, = 0 or (n@b)P,is a von Neumann algebra with no central summand of type I; and either Pm = 0 or ( n @ R)Pm is a von Neumann algebra of type I, for all m in { o o , ~2,. , . .}. From Exercise 6.9.32(iii), n @ R is countably decomposable since R is. From Exercise 6.9.31, Pm = 0 with m finite and m not divisible by n.

EXERCISE 6.9.35

297

Thus AP, contains n equivalent projections El,,. . . ,En, with sum P, from Exercise 6.9.29, and AP, contains n equivalent projections El,,. . . , E n , with sum P, from Exercise 6.9.25. If P, # 0 with m finite, then rn = kn for some positive integer b; and AP, contains n projections El,,.. . , E n , with sum P, equivalent in n @ R from Exercise 6.9.23(iii). Let Ej be Ej, Ejm -t C,"=lEj, for each j in (1,. . . ,n } , where E j , is defined t o be 0 if m is not divisible by n. Then { E l , . . . ,En}is a set of n equivalent projections in A with sum I . m[66]

+

Let S be an abelian self-adjoint subset of n @ R, 6.9.35. where n is a positive integer and R is a countably decomposable von Neumann algebra. Show that there is a unitary operator U in n @ R such that UAU-l has all its non-zero entries on the diagonal for each A in S.

Solution. Let A be a maximal abelian self-adjoint subfamily of n @ R containing S. (The existence of such an A is derived from an application of Zorn's lemma.) Since the algebra generated by A is an abelian self-adjoint family containing A, A coincides with this algebra (from the maximal property of A ) . Thus A is a maximal abelian self-adjoint subalgebra of n 8 R. From Exercise (3.9.34, A contains n orthogonal equivalent projections Fl,.. . , F, with sum I. Let Ejk be the element of n @ R with I at the j , k entry and 0 a t all others. Then { E l l , . . . , En,} is an orthogonal family of equivalent projections in n @ R with sum I . From Exercise 6.9.14, Ejj Fj for j in {1,..., n } . Let Vj be a partial isometry in n 8 R with initial projection Fj and final projection Ejj and let U be C;=, Then U is a unitary element in n @ Rand U F j U - ' = E j j . Since Fj commutes with every element in A and hence in S, Ejj commutes with UAU-l for all A in S. Thus UAU-' has all its non-zero entries on the diagonal when A E S. m[66] N

4.

6.9.36. With the notation and assumptions of Exercise 5.7.40 (iii), show that (i) the center C of 2l' is generated as a finite-dimensional linear space by its minimal projections; (ii) 9l'P is a factor of type I, with n finite and 2lP has linear dimension not less that n2 when P is a minimal projection in C.

298


Solution. (i) From Exercise 5.7.4O(iii), the C*-algebra U' is finite-dimensional. It follows that U' is a von Neumann algebra and that its center C is finite dimensional. From Theorem 4.4.3, C E C ( X ) , where X is a compact Hausdorff space. Since C is finite . ,Pk be the dimensional, X is a finite set {PI,.. . , p k } . Let PI,. projcetions in C corresponding to the characteristic functions (in C ( X ) ) of the one-point sets {PI},. . , { p k } . Then P I , . . . ,Pk are minimal projections in C and C is generated as a linear space by them. (ii) Since U' is finite dimensional, U' is strong-operator closed. Thus Q' is a von Neumann algebra and Ql'P is a von Neumann algebra with center C P . From (i), CP consists of scalar multiples of P , whence U'P is a factor. Since U'P is finite dimensional, U'P is a factor of type I, with n finite. If the linear dimension of U P is less than n2,then U P is a factor of type I, with m < n. Now Pso is a generating vector for UP. Thus the mapping AP -+ APso is a linear transformation of UP onto P(R), and P(3t)has dimension not exceeding m 2 ,the linear dimension of UP. But P Z Ois separating for U'P by Corollary 5.5.12. Thus the mapping A'P + A ' P q is a linear isomorphism of U'P into P(H),and P(3t) has linear dimension not less than n2 (> m 2 ) . We conclude from this contradiction that U P has linear dimension not less than n 2 . rn

.

.

6.9.37. Let R be a von Neumann algebra with center C. Show that R is of type I and C is the weak-operator-closed linear span of its minimal projections (we say C is totally atomic) if and only if R is the weak-operator-closed linear span of its minimal projections.

Solution. Suppose R is of type I and C is totally atomic. Let { Q a : a E A} be the set of distinct minimal projections in C and E be an abelian projection in R with central carrier I . From the comments following Proposition 6.4.3, RQa is a factor. Since Q a E is a non-zero abelian projection in RQa,RQa is a factor of type I. Now RQa is the weak-operator closed linear span of its family {&b : b E B} of minimal projections since each projection in RQa is the sum of an orthogonal family of minimal projections in RQa from Proposition 6.4.8 and the comment following Proposition 6.4.2 and since RQa is the norm-closed linear span of its projections from Theorem 5.2.2. Each Eab is minimal in R. Finally, each projection E in R is C EQa since C Qa = I by assumption.

EXERCISE 6.9.38

299

Suppose, now, that R is the weak-operator closed linear span of its family {Fb : b E B} of minimal projections. Let Qb be the central carrier of Fb in R. From Proposition 6.4.3,Q b is a minimal projection in C. Thus Q b Q b t = 0 unless Q b = 96,. Let { Q a : a E A} be the family of distinct Q b . If Q is a central projection orthogonal t o each Q a , then QFb = QQbFb = 0 for all b. By assumption, I is in the weak-operator closure of the linear span of {Fb : b E b}. Thus Q = Q . I = 0, C Q a = I, and C is totally atomic. Choose one Fa with central carrier Q a , then C Fa is an abelian projection in R with central carrier cQa(= I) from Proposition 6.4.5. Thus R is of type I and C is totally atomic. m

6.9.38. Let r2L be a C*-algebra acting on a Hilbert space ‘H. Suppose 2l‘ is a factor of type In with n finite and 3L has linear dimension at least n2. Show that there is a vector xo in ‘H such that ‘H= {AQ : A E 2l). Let {Eik}be a self-adjoint system of n x n matrix units for a’. Then Ei, is a minimal projection in 9.4‘ and 21”Ei1 acting on E{,(X) is a von Neumann algebra with commutant E{,Q’Ei,. From Proposition 6.4.3, E{,U’E;, consists of scalar multiples of E ; , , and from Theorem 5.4.1, 21Eil acts irreducibly on Ei,(H). From Proposition 5.5.5, the mapping (T =) TCE;* -, TE;, is a * isomorphism of 2l” onto 3L”E:,. Thus B(Ei1(%))has linear dimension at least n2 (from the assumption that U has linear dimension at least n’). It follows that E;,(H) has dimension at least n. Let { e l , . . . , e n } be an orthonormal set in Eil(‘H)and let $0 be Cp, E(ilej. Let y be a vector in ‘H and let gj be Ei .y. From Theorem 5.4.3, there is ? = y j for j in {l?.. . ,n}. We an operator A in 2l such that AEllej have Solution.

n

j=l n

j=1 n

n

j=1 n

j=l

300


6.9.39. Let G be a (discrete) group and F be the set of elements in G whose conjugacy classes are finite. Show that the center C of .CG (and RG)is precisely the set of elements L, in &G such that x is constant 011 the conjugacy class (go) for each go in F and x(g) = 0 if g 4 F. Solution. From the computation of the second paragraph of the proof of Theorem 6.7.5, if L, is in C, x is constant on each conjugacy class. Since x E l2(G), x vanishes on each infinite conjugacy class. Thus 2 has the form described if L, E C. Conversely, if L, E LG and x has the form described, then LXL,, = Lx*x, = Lx,*x = Lx,Lx for each g in G from the same computation. Since {Lxg : g E G } generates LG,L, E C. 6.9.40. Let G be a (discrete) group and F be the net of finite subsets of G partially ordered by inclusion. For x in Z,(G) and IF in F,define zr(g) to be x(g) if g E IF and 0 if g 4 IF. (i) Show that limp LxFzg= Lxxg in lz(G) for each g in G. (ii) Suppose L,: E &G and y in l2(G) is such that limrL,,y exists. Show that this limit is Lxy.

Solution. (i) Since L,xg x * x g = Ro,x, we have IlL,xg

- Lx~gl12= I I R X , ( X P

= xp * xg = R z g x r and L x x g =

- 4112= 1121 - ./I2

=

c

1.(h)I2.

her As x E lz(G), given a positive E , there is a set IF1 in 3 such that Chgp 1x(h)I2 < E if Fl .'lE Thus limr Lxpzg= Lrxg in Iz(G). (ii) Since (xs)*(g)= xr(g-1) = z(g-1) = z*(g) if g E F-l, where B-' = { g - l : g E F), and (zr)*(g) = 0 if g 4 F-', we have that (Q)* = (z*)r-~. Thus for each g in G, (LXY,2g) = (Y,LX*XB) = li$Y, L(,*),Xg) = lip(% (LXv-1)*zg) = li$L,,,

Y,xg)

= (v*),

EXERCISE 6.9.41

where u = limr L,y, since the sets F in a cofind subnet of F. Thus u = L,y.

301

F such that F = F-' form

6.9.41. Let Go be a subgroup of the (discrete) group G. Let R be .CG acting on /2(G), and let Ro be { L , E R : z(g) = 0, g $! Go). Show that (i) 720 is a von Neumann subalgebra of R; (ii) Ro is the weak-operator closure of the linear span of { L x g : g E Go}; (iii) Ro is * isomorphic to CG,, acting on 12(Go).

So6ution. (i) Suppose Lx and L , are in 720. Then 0 = s(g) = y(g) if g 4 Go, and (ax y)(g) = 0. Hence

+

and 720 is a linear space. Since Go is a subgroup of G,

Thus Lj. = L,. E Ro. With e the unit element of G, g # e if g 4 Go. T h u s z ( ) = O i f g # G o a n d I = L X e E R o .I f g # G o a n d h o E G o , e-f then gho # GO;so that

Thus L,L, = LZ*, E Ro. It follows that Ro is a self-adjoint subalgebra of R containing 1. We show, next, that 720 is weak-operator closed. Since R is a von Neumann algebra from Theorem 6.7.2, it will suffice to show that 20is weak-operator closed in R. Let L , in R be in the weak-operator closure of Ro and suppose g 4 Go. Then ( L y z e zg) , (= y(g)) is in the closure of

whence y(g) = 0. It follows that L, E Ro and 720 is a von Neumann subalgebra of R. (ii) Suppose L, E Ro and TLXg= LZgTfor each g in Go, where T E 13(/2(G)). We adopt the notation of Exercise 6.9.40 and denote by FO the intersection F n Go for each IF in F. Then Q = zipo since

302


L, E R& and x vanishes outside Go by definition of 720. Since L , is in the linear span of {LZg : g E Go}, TL, = L,T for each IF in 3. From Exercise 6.9.40(i), limrL,x, = L,xg for each g in G . Since T is bounded,

lim L,Txg F

= limTL,x, F

= TL,xg

for each g in G. With Tx, in place of y in Exercise 6.9.40(ii), we conclude that L,Tx, = TL,x, for each g in G . Thus L, commutes with the commutant of the strong-operator closure of the linear span of {Lxg : g E GO},a von Neumann algebra. From the double commutant theorem, L, is in this strong-operator closure. Thus Ro is the strong-operator closure of the linear span of {Lxg : g E G O } . (iii) Let E&be the projection in D(l?(G))with range spanned by {xg : g E GO}. Since the range of Eh is stable under each L,, with g in Go and {L,, : y E Go} generates I& from (ii), EL E RL. Now x, is a generating vector for R' (c RL),whence Eh has central carrier I in %?& from Proposition 5.5.2. F'rom Proposition 5.5.5, the mapping T 4 TEA of Ro onto ROE:,acting on Eh(Z2(G))is a * isomorphism. The mapping of 12(Go) onto EA(l2(G))that assigns xg to ys, where y, is the element of /2(Go)that assigns 1 to g and is 0 elsewhere on Go is a unitary transformation U of Iz(G0) onto Ei(lz(G))with the property that ULCG~U-' = R O ( E ; ( / ~ ( G )Thus ) . U implements a * isomorphism (unitary equivalence) of 'CG~and ROE;.It follows that Ro is * isomorphic to . C G ~ . 6.9.42. With the notation of Example 6.7.6, let Aj be the von Neumann subalgebra of LF,,generated by LXaj.Show that dj is a maximal abelian subalgebra of L F ~ .

SoZution. F'rom the proof of Theorem 6.7.5, L, in LF,,commutes with Lxa, if and only if z(ajgaf') = z(g) for each g in Fn. Thus x(uygujm) = s(g) for each integer rn and all g in Fn if L, commutes with Aj. If ud * * . u: is the reduced form of g and contains a generator other than a j , then by examining the various possibilities where k and h are and are not j, we see that { u ~ g u ~:"m E Z} is an infinite subset, of Fn.Since z E 12(Fn),it follows that x(g) = 0. F'rom Exercise 6.9.41, L, E d j . Thus Aj is maximal abelian. m

EXERCISE 6.9.43

303

6.9.43. Let G be a (discrete) group and a be an automorphism of G. (i) Show that the mapping L, -+ L,,, is a * automorphism of CG.

(ii) Let G be 3 2 and a be the automorphism of G that interchanges the generators a1 and a2. Show that there is no unitary ~ ~ that UL,U-' = L,,, for each L , in C F ~ . operator U in I C such (We say that the automorphism L , -+ L,,, is outer in this case.)

Solution. (i) This is apparent since CG is defined entirely in terms of the group G and the mapping of CG described is derived structurally from the automorphism a of G. Nevertheless, we perform some routine verivications. Note that

so that

(5

* y) o a = ( 5 o a ) * ( y o a).Note, too, that

(5* 0 a ) ( g )=

z ( a ( g ) - l ) = z ( a ( g - 1 ) ) = ( 5 0 a)*(g)

so that x* o a = (x o a)*. (ii) Suppose L , in C F ~is such that

Then Lysx,, - LXa2*,and y*x,, = xa2*y. It follows that y(ga,')

=

y(a;'g) for each g in G, and y(a2gal') = y ( g ) . Hence y(a;gaIn) = y ( g ) for each integer n. Now (aznga," : n E Z} is an infinite subset of 32for each g in 3 2 . Since y E 6 2 ( 3 2 ) , y ( g ) = 0 for all g in 3 2 and y = 0. In particular, no unitary operator L , in C32 is such that L,L,L;~ = L,,, for each L , in C F 2 .

304


Let Fm be the free group on a countable number of 6.9.44. generators a l , a2,. Let G, be the subgroup generated by a l , a,+l, a,+2,. Let R, be the von Neumann subalgebra of C3, generated by ( L , : g E G,}. Show that (i) each R, is a factor * isomorphic to L3,; (ii) nr=lR, = d1, where d1 is the (abelian) von Neumann subalgebra of CF, generated by La,,and that d1 is a maximal abelian subalgebra of C3,.

...

...

*

Solution. (i) Each G, is isomorphic to Fm so that LG, is isomorphic t o LF,. From Exercise 6.9.41, R, is * isomorphic to LG,, Since Fm is an i.c.c. group, C3, and R, are factors. (ii) If L , E n?i1Rn,then z vanishes outside each G, from Exercise 6.9.41. Thus s(g) = 0 if the reduced form of g contains any of az, a s , . . .. From Exercise 6.9.42, d1 is a maximal abelian subalgebra of L3,. From Exercise 6.9.41, L , E d1. Of course, d1 C n:=lR,. Thus d1 = nr=lR,.

.

6.9.45. (i) Show that GI @ G2 is an i.c.c. group if and only if G I and Gz are i.c.c. groups. (ii) Let G be 3 2 @ n. Show that LG is not isomorphic to C F ~ .

*

Solution. (i) Let el and e2 be the respective identity elements of G1 and Gz. Suppose G1 and G2 are i.c.c. groups and (g1,92) # (e1,e2). Say 91 # el. If g{ and gr are distinct conjugates of gl in GI, then (gi,g2) and (g;,gz) are distinct conjugates of (gl,g2) in G1$Gz. By hypothesis, g1 has an infinite set of conjugates in G1 so that (g1,gz) has an infinite set of conjugates in GI @ Gz. It follows that G1 @ Gz is an i.c.c. group. Suppose now that GI @ Gz is an i.c.c. group. If g1 # e l , then (g1,ez) has an infinite conjugacy class C in G1 @ G2. Each element of C has the form (9, e2), where g is a conjugate of gl in GI. Two such elements, ( g ' , e2) and (g", ez), are distinct if and only if g' # g". Thus gl has an infinite conjugacy class in GI. It follows that G1 and, symmetrically, Gz are i.c.c. groups. (ii) From (i), Examples 6.7.6 and 6.7.7, and Theorem 6.7.5, CG is a factor of type 111. In the proof of Theorem 6.7.8, we show that, for each pair of elements A1 and A2 in Ln and each positive E , there is a unitary operator U in Cn such that ( U s e , x e ) = 0 and

EXERCISE 6.9.46

305

II(VAj - AjU)zell < E for j in {1,2}, where e is the identity element of II. We showed that there is no such unitary operator in L32when E is &, A1 is LEal, and A2 is LIa2. It follows from this that .Cn and C32 are not * isomorphic. To show that LG and L F ~are not * isomorphic, it will suffice, precisely as in the last paragraph of the proof of Theorem 6.7.8, t o find an element ( h , g ) of 3 2 @ II (different from the identity) commuting with a given finite set of elements ( h l , g l ) ,...,(hntgn) in 3 2 @ II. Choose g in ll (different from e) commuting with each gk (as in the last paragraph of the proof of Theorem 6.7.8). Then (e’,g) commutes with each ( h k , g k ) , where e’ is the identity of 95. m[99] 6.9.46. Let Z= be the norm closure of a two-sided ideal Z in a von Neumann algebra R. Show that each projection in I= lies in Z. Solution. that llE - All

If E is a projection in Z= , we can choose A in Z so

< 1. Then, IIE - EAEll = IIE(E - A)EII < 1.

In the Banach algebra E R E , the element EAE lies within distance less than one from the identity element E, and so has an inverse B. Moreover, B E R and E = B E A E E Z. 6.9.47. Suppose that Z and J’ are two-sided ideals in a von algebra R,and 7 . is norm closed. Prove that (i) Z 2 J’ if each projection in Z lies in J’; (ii) Z E J’ if, for each projection E in Z, there is a projection F in ,? such that E 5 F. Solution. (i) Suppose that each projection in Z lies in J’. Given a self-adjoint element A of Z, the spectral resolution { E x } of A satisfies E - , E Z,I - E , E Z for every positive real number E , by Lemma 6.8.1; so E - c , I - E, E J , and A ( E - , -t I - E,) is an element A, of 3. Since 3 is norm closed and A, E 3,and

for each positive E , it follows that A E 3. Since Z and ,7 are selfadjoint, by Proposition 6.8.9, and J’ contains each self-adjoint element of 1,we have Z C 3.

306


(ii) With E a projection in 2,the assumption that E 5 F for some projection F in 3 entails E E 3, by Remark 6.8.2. It now follows from (i) that, Z 2 3. 6.9.48. Suppose that R is a factor. (i) Show that if Z and 3 are norm-closed, two-sided ideals in R, then either Z .7 or J C Z. [Hint. Use the result of Exercise 6.9.47(ii).] (ii) Prove that there is a norm-closed, two-sided ideal Z in R such that Z # R and Z contains every proper two-sided ideal in 12. [Z is (0) in some cases; see Corollaries 6.8.4 and 6.8.5.1

Solution. (i) If Z J , by Exercise 6.9.47(ii) there is a projection & in Z such that EO 2 F for each projection F in 3. Since the equivalence classes of projections in a factor are totally ordered (Proposition 6.2.6), we have F 5 EO for every projection F in 3. Again by Exercise 6.9.47(ii), J’ C_ Z. (ii) Let Z be the union of all norm-closed two-sided ideals properly contained in R. Since these ideals are totally ordered by inclusion, by (i), it follows that Z is a two-sided ideal in R. Moreover Z # R, since I $ Z. Given any two-sided ideal J’ properly contained in R, the norm closure 3=is another such ideal (Proposition 3.1.8), so 3 .7= C Z.In particular, Z=E 1,and Z is norm cIosed. 6.9.49. Let R be a von Neumann algebra. (i) Suppose that Z is a two-sided ideal in R and PO is the set of all projections in Z.Prove that (a) E V F E POif E, F E PO; (b) if E and F are projections in R and E 5 F E PO,then

E E Po. (ii) Suppose that a family PO of projections in R satisfies the conditions (a) and (b) set out in (i). Let ZObe the set of all operators in R with range projection in Po. Show that ZOis a two-sided ideal in R, and PO is the set of all projections in 10.Prove also that if 3 is a two-sided ideal in R,then POis the set of all projections in J’ if and only if ZO 3 1;.

Solution. (i) Suppose that POis the set of all projections in a two-sided ideal Z in R. From Remark 6.8.2, PO has property (b) set

EXERCISE 6.9.50

307

out in the exercise. If E , F E Po, then

E V F -E

N

F - E A F 5 F E PO,

and E V F - E E Po by (b). Hence E , E V F - E E Z, whence

EV F = ( E V F - E ) + E E 1. Accordingly, E V F E PO,and (a) is proved. (ii) Suppose that Po is a family of projections in R that satisfies conditions (a) and (b), and let ZObe the set of all operators in R that have range projection in PO.By arguing as in the first paragraph of the proof of Theorem 6.8.3, but using PO instead of the family of finite projections, it follows that ZOis a two-sided ideal in R. A projection E in R lies in ZOif and only if E E Po; that is, PO is the set of all projections in Zo. By Exercise 6.9.46, Po is the set of all projections in Zc. Suppose that J’ is a two-sided ideal in R. If ZO C J’ C Zc, then since 10 and Zr have the same projections (those in Po), it follows that these are the projections in J’. Conversely, suppose that PO is the family of all projections in J’. By Exercise 6.9.46, Po is the family of all projections in J’=, and by Exercise 6.9.47(i), the norm-closed ideals J’= and Z r have the same projections, and thus coincide. Hence J’ J’= = 1 : . On the other hand, if A E 1 0 , the range projection E of A is in PO,so E E J’ and A = E A E J’; whence, ZOC J’. rn Suppose that E , F, G are projections in a von Neu6.9.50. mann algebra R, such that G is properly infinite and P E < PG, P F 4 PG whenever P is a central projection in R for which PG # 0. Show that P E V PF 4 PG for each such projection P .

Solution. With P a central projection such that PG # 0, there is a central subprojection Q of P such that QPE = QE 5 Q F = QPF and ( P - Q ) F 5 ( P - Q ) E . From Proposition 6.3.7, there is a central subprojection Qo of Q such that QoF is properly infinite or Qo is 0, and (Q - Qo)F is finite. In this case, (Q - Qo)E and (Q - Qo)(E V F ) are finite. From Exercise 6.9.5,

308

C:OMPARI S 0 N THE 0 RY OF PROJECTIONS

provided ( Q - Qo)G # 0. From Exercise 6.9.4,

provided QO# 0. Thus, if QG # 0, Q(E V F ) 4 QG. Symmetrically,

if ( P - Q ) # 0. Thus

P ( E V F ) = P E V P F 4 PG. Let R be a properly infinite von Neumann algebra 6.9.51. with center C, and let PObe the set of all projections E in R such that P E P for each non-zero projection P in C. From the result of Exercise 6.9.50, with G = I , note that. E V F f ‘Po whenever E , F E PO.Prove that (i) the set Z of all operators in R with range projection in PO is a two-sided ideal in R; (ii) the norm-closed, two-sided ideal Z= in R satisifies Z= n C =

{Oh

(iii) if J’ is a two-sided ideal in R, and J’nC = (0), then J’ E I=. Interpret these results in the case in which R is an infinite factor. Solution. (i) Let E and F be projections in R.It is apparent, from the definition of PO,that E E ’Po if E 5 F f PO. When G is the properly infinite projection I in R,the result of Exercise 6.9.50 reduces to the assertion that E V F E POwhenever E , F E PO. Accordingly, from Exercise 6.9.49(ii), the set Z of all operators in R with range projection in Po is a two-sided ideal in R,and the projections in Z (equivalently, the projections in I=)are precisely those in PO. (ii) If P is a non-zero projection in C, then P 4 PO and thus P 4 F . Accordingly, the norm-closed ideals (0) and Z= n C in C contain the same projections (that is, just 0). By Exercise 6.9.47(i), Z= nC = (0). (iii) Suppose that J’ is a two-sided ideal in R, and J nC = (0). Let E be a projection in J . Given any non-zero projection P in C, we have P ,” (since J’ n C = (0)) but PE E J’, and hence

2

309

EXERCISE 6.9.52

+

P E P;so PE 4 P . Thus E E Pa (C_ T = ) ;since each projection in J lies in Z=, it follows from Exercise 6.9.47(i) that J E I=. When R is an infinite factor, PO is the set of all projections E in R such that E 4 I . To say that an ideal $' in R meets the center of R only at 0 amounts to the assertion that I 4 3, equivalently that 3 # R. Accordingly, the set Z of all operators in R that have range projection in PO is a two-sided ideal in R, the norm-closure Z= is a two-sided ideal properly contained in R,and Z= contains every (norm-closed) proper two-sided ideal in R. Thus I= is the (unique) largest two-sided ideal properly contained in R (see Exercise 6.9.48(ii)). rn 6.9.52.

Let R be a von Neumann algebra. Suppose T q R.

Show that (i) R ( T ) and N ( T ) are in R (see Exercise 2.8.45); (ii) R ( T * )= R ( T * T )= R ( ( T * T ) ~ ) ; (iii) R ( T ) and R(T*)are equivalent in R.

Solution. (i) From Exercise 2.8.45, z E N ( T ) ( % )if and only if z E D(T) and T z = 0. If U' is a unitary operator in R', then U'z E V ( T ) when z E V ( T ) and T U ' z = U'Tz. Thus TU'z = 0 when z E N ( T ) ( % ) ,and N(T)('FI)is stable under each unitary operator in R'. Hence N ( T ) E R. From Exercise 2.8.45, R ( T ) E R. (ii) We show t1ia.t N ( ( T * T ) $ )= N ( T * T ) so that (ii) follows from Exercise 2.8.45. If z is in the range of the operator N ( ( T * T ) h ) , then z E V ( ( T * T ) i and ) ( T * T ) * z= 0 from Exercise 2.8.45. Thus E V(T*T), 1 1 T'Tz = (T*T)Z(T*T)Zz = 0, and z is in the range of N(T*T). Suppose u is in the range of N(T*T). Then u E 'D(T*T)and T*Tu = 0. Thus u E D ( ( T a T ) * ) , 1

1

0 = (T*Tu,u)= ([(T*T)3]12u,u) = ll(T*T)2u1112,

and u is in the range of N ( ( T * T ) * ) . It follows that N(T*T) =

N((T*T)~). (iii) From Theorem 6.1.11, T = V ( T * T ) i ,where V is a partial isometry in 72 with initial projection R ( ( T * T ) j )and final projection R ( T ) . From (ii), R(T*)= R ( ( T * T ) i ) .Thus R ( T ) and R(T*) are equivalent in R. rn

310


Let S be a symmetric operator affiliated with a finite von Neumann algebra R. Show that S is self-adjoint. [Hint.Use Proposition 2.7.10,especially (i) and (ii).] 6.9.53.

Solution. Suppose S 17 R and S is symmetric (S C S*). From Exercise 6.9.52, R(S t il) N R([S4- iI]*).From Exercise 6.9.7,

I - R(S + iI) I - R([S+ iIj*). N

From Exercise 2.8.45,

+ iI) = I - R([St ill*), N ( [ S t ill*) = I - R(S t iI) so that N ( S + iI) N ( [ S -t ill*). Again from Exercise 2.8.45, the range of N ( S + il) consists of vectors z in D(S -t iI)(= D(S))such that Sz -+ iz = 0. Since S C S*,z E D(S*)and Sz = S*z, so that N(S

N

(Sz, 4 = ( z , S * a ) = ( z ,Sz) = ( S Z , 2)

and 0 = ( S z + i z , z ) = (Sz,z)

+ i(.,z).

Thus ( 8 , z ) = 0 and z = 0. Hence N ( S + i I ) = 0 and N ( [ S + i I ] * ) = 0. Similarly N ( [ S- i l l * ) = 0. From Proposition 2.7.10,S is self-adjoint (for (S f iI)* = S* 7 iI). 6.9.54. Let A and B be operators affiliated with a finite von Neumann algebra R, and let VH be the polar decomposition of B. Suppose A C B. Show that (i) V*A is a symmetric operator afliliated with R; (ii) A = B.

Solution. (i) Since A C B,

V * A E V * B = V*VH = H = H*

(V*A)*.

Thus V * A is symmetric. Suppose (5,) is a sequence of vectors in the domain of V * A that converges to some vector 5 and for which { V * A z , } converges to some vector y. As V* is isometric on the range of A,

1lA.n - A ~ ~ l (JV*AZ, l = - V * A ~ m l l - -0+

311

EXERCISE 6.9.55

as n,m -,

00,

so that { A x , } converges to some vector t and

V * A z , --t V * Z= 9. But since A is closed, x E D ( A ) = D ( V * A ) and z = Ax. Thus y = V * z = V * A x ,and V * A is closed. The preceding argument shows that V * A is closed, symmetric, and densely defined (for D ( V * A )= D ( A ) ) .If U' is a unitary operator in R',then U'*AU' = A so that

lJ'*V*AU' = V*U'*AU' = V * A since V * E R. Thus V * A1772. (ii) From (i) and Exercise 6.9.53, V * A is self-adjoint. Now V * A is contained in H and since self-adjoint operators are maximal symmetric, V * A = H . Hence

A = R ( B ) A = V V ' A = V H = B.

m[76(Lemma16.4.1,p.226)]

6.9.55. Let E be a projection in a finite von Neumann algebra R acting on a Hilbert space 3-1. With T in R, let F be the projection with range { x : T x E E(3-1)}. Show that F E R and that E 5 F .

Solution. With A' in R' and T x in E(3-1), TA'x = A'Tx E E(3-1) since A'E = EA'. Thus F(3-1) is stable under R',a self-adjoint family, and F E R" (= R). Note that T x E E(3-1)if and only if ( I - E ) T x = 0. Thus F(3-1) is the null space of ( I - E)T - that is, F = " ( I - E ) T ] . From Propositions 2.5.13 and 6.1.6,

(*> I - F = R [ T * ( I- E ) ] &[(I- E ) T ] 5 I - E . If E 2 F , then there is a central projection P in R such that P F 4 P E from the comparison theorem (6.2.7). Now P ( I - F ) 5 P ( I - E ) from (*) and Proposition 6.2.3 so that N

P(I - F )

N

EO 5 P ( I - E ) .

From Exercise 6.9.9,

P = PF

+ P(I-

F ) 4 PE

+ EO 5 P E + P ( I -

E)= P

since P ( 1 - F ) and EO are finite. But this is absurd. It follows that E 5 F. m[76]

CHAPTER 7 N O R M A L STATES A N D U N I T A R Y EQUIVALENCE O F VON N E U M A N N ALGEBRAS

7.6.

Exercise

Let R be a von Neumann algebra acting on a Hilbert 7.6.1. space 'H, and let w be a normal state of R. Show that (i) the support of w has range [ R ' x j : j = 1,2,. . .],where w = M &=I wxj IR; (ii) the support of w is the union of the projections E j , j E N, where Ej has range [ R ' x j ] .

Solution. Let E be the projection (in R)with range equal to : j = 1,2,. . .I. (i) Suppose F is a projection in R such that

[R'Xj

0 = w ( F ) = C ( F x j ,~

j ) .

j=1

Then F x j = 0 and 0 = A ' F x j = F A ' x j for each j in N and each A' in 72'. Thus F E = 0. Conversely, if G E = 0 for a projection G in R,then G x j = 0 for each j in W, and 0 = C g l ( G x j , x j ) = w ( G ) . It follows that E is the support of w. (ii) 03Since the ranges of E j , j E N span [ R ' x j : j = 1,2, ...I, E = Vj,l Ej and Ej is the support of w.

Vgl

Let w be a normal positive linear functional on a von 7.6.2. Neumann algebra R acting on a Hilbert space 3-1. Let E be the support of w and F' be a projection in R' such that E 5 Cp. Show that

3 13

EXERCISE 7.6.3

(i) there is a normal positive linear functional wo on RF’ such that wo(AF’) = w ( A ) for each A in R; (ii) the support of wo is EF’.

, (A)= Solution. (i) Since E is the support of w and E 5 C F I w ~ ( A C Ffor~ each ) A in R. From Proposition 5.5.5, the mapping cp,

AF’

--+

ACp

:

RF’

-+

RCp

is a * isomorphism (and hence a linear order isomorphism). Let be u ~ R C F IThen .

wo(AF’) = u ( A ) = w ( A C p ) = w l ( A C p ) = (

~

01

o 1cp)(AF’).

Thus wo(AF‘) = w ( A ) = 0 if AF‘ = 0 , and wo is a well-defined linear functional on RF’. Moreover, wo = w1 o cp. Since 01 is a normal positive linear functional, wo is. (ii) As w o ( ( I - E ) F ’ ) = o ( I - E ) = 0, the support of wo is a subprojection of EF’. If wo(G) = 0 for some projection G in RF’, then (p(G)F’ = G and 0 = wo(G) = w(cp(G)). Hence p ( G ) 5 1’ - E and G = cp(G)F’ 5 (I- E)F’. Thus EF‘ is the support of wo. Adopt the notation of Exercise 7.6.2. 7.6.3. Must EF’ be cyclic in RF‘ under F’R‘F‘ when E is cyclic (i) in R under R’? Proof? Counterexample? (ii) Must wo be a vector state of RF’ when w is a vector state of R? Proof? Counterexample?

Solution. (i) No! Let ‘Ho be a separable, infinite-dimensional Hilbert space and let ‘H be the (countably-) infinite direct sum of 3-10 with itself - we use the notation of Exercise 5.7.42. With { e n } an orthonormal basis for 3-10, the vector {el,ez/2,e3/3, ...} is a generating and separating vector for R. Choose 1 for E . Let F’ be the projection with range consisting of all vectors {x,O,O,. . .} (x E ‘Ho). Then F’R’F’ consists of all operators whose matrix representation has some scalar multiple of I0 at the ( 1 , l ) entry and 0 at all other entries, where I0 is the identity operator on ‘Ho. Thus with Q(# 0) in 3-10, F’R‘F’({2o,O,O,. . .}) = [{50,0,0,. . .}I, a one-dimensional subspace of F’(3-1). It follows that F’(= F’I = F ’ E ) is not cyclic under F‘R’F‘ although I ( = E ) is cyclic under

72‘.

314

NORMAL STATES

(ii) No! Since ‘R has a separating vector from (i), each normal state of R is a vector state of R. Let Yn be (2-n/2e,,0,0,. . .} and let w be C,”==,q,,.Then w is a normal state of R. The partial isometry in R‘ whose matrix representation has I0 in the ( j ,1)entry and 0 at other entries maps gn onto zn,where zn has 2-”I2en at the j t h coordinate and 0 at other coordinates. Thus the support of w is I from Exercise 7.6.1(i). At the same time C p = I from Exercise 6.9.30. Thus from Exercise 7.6.2(ii), the support of wo is F’ (= IF’), and as noted in (i), F‘ is not cyclic in RF’ under F‘R’F’. F’rom Proposition 7.2.7, wg is not a vector state of RF’. 7.6.4. Let A be an abelian von Neumann algebra acting on a Hilbert space H,and let w be a normal state of A. Show that (i) w is a vector state of d ; (ii) the weak-operator topology and the ultraweak topology coincide on A.

Cj”=,

w X j( A for some Solution. From Theorem 7.1.12, w = countable family of vectors {zj} in ‘H such that Cj”=,11zj112 < 00. From Exercise 7.6.1, the support E of w is the union of the family { E j } of projections Ej (in A), where Ej has range [d’zj]. From Proposition 5.5.19, E is countably decomposable in A. Thus AE is a countably decomposable von Neumann algebra on E(’H).

(i) Since AE is an abelian, countably decomposable von Neumann algebra on E(‘H),AAE has a separating vector, from Corollary 5.5.17. Hence oldE is a vector state of AE, from Theorem 7.2.3. Choose a unit vector 2 in E(’H) such that w x ( d E = w l d E . Then w,ld = w , since E is the support of w. (ii) As noted in Remark 7.4.4, the ultraweak topology on a von Neumann algebra is the weak topology induced by the normal states of that algebra. Thus it suffices to show that each normal state of A is weak-operator continuous on A. But this is immediate, since, from (i), each normal state of d is a vector state. 7.6.5. (i) Suppose (S,S,rn) is a a-finite measure space, {gj} is a sequence of functions in L 1 ( = Ll(S,m)),and 11gjll1 < 00. Show that the series C,”=,gjconverges almost everywhere to a

cEl

EXERCISE 7.6.5

315

function g in L1, and

for each f in L , (= L,(S,m)). (ii) How is the result of (i) related to Exercise 7.6.4?

Sobtion. (i) For each positive integer n, define h , and k, to be the L1 functions I=;==, gj and C;=,Igjl, respectively. Since the sequence { k n } is increasing, and n

(where K = ,C ; llgjlll < oo),it follows from the monotone convergence theorem that {k,} converges almost everywhere to a function k in L1; so C,”=, 1gjl converges almost everywhere to k. From this, it follows that C;, gj converges almost everywhere; the sum function g is measurable and satisfies 191 5 k, so g E L1. Thus, almost everywhere, { h n } converges to g and lhnl _< k. Given f in L,, { f h n } converges to f g and lfhnl 5 llfllOOk (E L l ) , almost everywhere. By the dominated convergence theorem, fgjdm = lim j=1

C J fgj dm

n-tcc j = 1

= lim

J f h n dm = J fg dm.

n-+w

(ii) Let A be the multiplication algebra for the measure space ( S , S , m ) and let w be a normal state of A. There is a sequence { Z j } in L2 (= L2(S, m)) such that w = CZ, /A. Let gj be )zjl2. Then gj E L1 and

j= 1

j=1

316

NORMAL STATES

Hence the conditions of (i) are satisfied, with g j 2 0, and we can choose g ( 2 0) as in (i). Let x be the element g1I2 of L2. For each f in L,,

This shows that w is the vector state wzld, and gives a measuretheoretic proof (for this particular von Neumann algebra d) of the result of Exercise 7.6.4(i). 7.6.6. Show that a normal state of a von Neumann algebra is faithful (see Exercise 4.6.15) if and only if its support is I. Solution. Let w be a normal state of the von Neumann algebra R. Suppose w is faithful. Then for each projection E in R, w ( E ) > 0, and from Definition 7.2.4, w has support I . If w is not faithful, then w ( A ) = 0 for some non-zero, positive operator A in R. As noted in the discussion following Definition 7.1.1, A 2 AEo for some non-zero spectral projection Eo of A and some positive A. Since EO E R, 0 5 Xw(E0) 5 w ( A ) = 0; and ~ ( E o=) 0. It follows that w does not have support I in this case. 7.6.7. Suppose that, for j = 1,2, pj is a faithful normal state of a von Neumann algebra Rj,and Uj is a weak-operator dense selfadjoint subalgebra of Rj.Let cp be a * isomorphism from U1 onto Uz, such that p l ( A ) = pz(cp(A))for each A in U1. Show that cp extends to a * isomorphism $ from R1 onto 7 2 2 , such that p1 ( R ) = p2 ( $ ( R ) ) for each R in R1. [Hint. Consider the repreesentation ‘ p j of Rj engendered by p j , and use Proposition 7.1.15.1 Solution. Let Ij denote the unit element of Rj.If I1 E 31,then is a unit for the dense subalgebra U2 of 7 2 2 , and by continuity is a unit for 7 2 2 ; so cp(I1)= 1 2 , and I2 E U2. From this, and a similar argument in which the roles of I1 and I2 are reversed, it follows that I1 E ‘ 2 1 if and only if I2 E U2. If I j $ Uj, for j = 1,2, let Bj be the algebra { d j -I- A : c E C, A E Uj} and let cpo be the * isomorphism cp(l1)

EXERCISE 7.6.7

317

+

c11 A --t cI2 t cp(A) from ,131 onto ,132. Upon replacing and q by ,131,172 and cpo, and noting that p1 If?, = p2 o 90, we may assume henceforth that 11 E 2l1 and 1 2 E ' 2 2 . Let cpj : Rj B(1-1j) be the representation of Rj engendered ---f

by the state

pj,

and let

xj

be a unit cyclic vector for cpj such that

If A E Rj and cpj(A) = 0, then

and A*A = 0 since p j is faithful. It follows that A = 0, so cpj is faithful (and, hence, isometric). By Proposition 7.1.15, with 3 = {zj}, cpj(Rj)is a von Neumann algebra and cpj is strong-operator continuous on the unit ball ( R j ) l of Rj. From this continuity, and since (2lj)l is strong-operator dense in ( R j ) l , it follows that cpj((2lj)l) is strong-operator dense in cpj((Rj)l) (= (cpj(Rj))l). Thus c p j ( 2 l j ) - = cpj(Rj),and cpj(2lj)zj is dense in 1-1j. When A E 2l1, we have

From this, and since yj(Uj)xj is dense in Xj, for j = 1,2, it follows that there is a unitary operator U from 1-11 onto 1-12 such that

When A, B E 2l1,

Since cp1(2ll)zl is dense in 1-11, we have Ucpl(A) = cp2(63(A))U, and

From this, cp~(U2)= Ucp1(2ll)U-', closures we obtain

and by taking strong-operator

318

NORMAL STATES

It follows from ( 2 ) and (3) that the mapping R --t cpT1(Ucpl(R)U-l) is a * isomorphism 11, from R1 onto R2 that extends cp. From ( l ) , and since I j E Uj, we have U z l = 2 2 . Thus, for all R in 721, PZ(+(R)) = (cp2(+'(R))z2952) = (Ucp1(R)U-lz2, .2)

.

= (cpl (R)Xl, .1> = Pl(R).

7.6.8. (i) Find an example of a factor (acting on a separable Hilbert space) for which there is no separating vector. (ii) Construct a. faithful normal state for the factor exhibited in (9. (iii) Conclude that a normal state of a factor need not be a vector state.

Solution. (i) Let 'H be an n-dimensional Hilbert space for some n exceeding 1 and let M be B('H). Then M is a factor (of type In), and there is no separating vector for M since 0 = ( I - E ) z when E is the projection in B(7-t)with range generated by z. (ii) Let { e l , . . . , e n } be an orthonormal basis for 'H and let T ( A ) be n - l C y = l ( A e j ,e j ) for A in M . Then T is the normalized trace of Exercise 4.6.18(i), and T is a faithful state of M , as noted in that exercise. (iii) If T were a vector state wx of M , then t would be a separating vector for M . (For if A z = 0, then r ( A * A ) = w x ( A * A ) = llAs112 = 0 so that A * A = 0 and A = 0.) Hence, from (i), T is not a vector state. Let {en : R. E N} be an orthonormal basis for the 7.6.9. separable Hilbert space 'FI. Define w ( A ) to be CF=l2 - n ( A e n , e n ) for each A in B('H). Show that (i) w is a faithful normal state of B(31); (ii) w is not weak-operator continuous on B(3-I); (iii) the weak-operator topology is strictly coarser than the ultraweak topology on B(7-l).

Solution. (i) From Theorem 7.1.12, w is a normal state of B(7-l). If A is a positive operator in B('H), then A = H 2 with H

319

EXERCISE 7.6.10

a positive operator in B(3-1). If w ( A ) = 0, then 0 = (Ae,,e,) = IIHenl12 for each n in N. Thus H e , = 0 and Ae, = 0 for each n in N. Since {e,} is a basis for 3-1, A = 0. Thus w is a faithful state of

B(3-1).

(ii) Since the weak-operator topology on B(3-1)is the weak topology induced by the linear span of the set { w , , ~ : x , y E 3-1) (= V ) of vector functionals on B(3-1),each weak-operator continuous linear functional wo on B(3-1) has the form a l w z ( l ) , y ( l ) . - - u,w,(,),~(,), where a l , . . . , a , are scalars and {s(l), ..., x ( n ) } , { y ( l ) , .. ., y ( n ) } are finite sets of vectors in 3-1 (by Theorem 1.3.1). Now w o ( I - E ) = 0, where E is the projection in B(3-1) with range [ x ( l ) ,. . . ,x ( n ) ] . Since 3-1 is infinite dimensional, I - E # 0. Thus wo is not faithful. (We have proved that no weak-operator continuous linear functional on B(3-1) is faithful when 3-1 is infinite dimensional.) From (i), it follows that w is not weak-operator continuous on B(3-1). (iii) By Definition 7.4.3, w (= CF=,w,,, , where x , = 2 - * e n ) is ultraweakly continuous on B(3-1). From (ii), w is not weak-operator continuous on B(3-1). Thus the weak-operator topology differs from the ultraweak topology on B(3-1). Since each weak-operator continuous linear functional on B(3-1) is ultraweakly continuous, the weakoperator topology on B(3-1) is strictly coarser than the ultraweak topology.

+ +

7.6.10. Let R be a von Neumann algebra acting on a Hilbert space 3-1, {Qa}aEA be an orthogonal family of central projections in R with sum I , and w be a normal state of R. Suppose that WlRQa = Z;zl w , ~ , I R Q ~where , n, 5 m, for all a in A. Show that m w = Cp1 W Z j IR.

Solution. We may assume that na = m for each a in A by defining xja to be 0 when n, < j 5 m. Replacing xj, by Qaxja, we may assume that each x j a E Qa(3-1). Since a

a

a

for each j in { I , . . . ,m } ,and for fixed j , { x j a } is an orthogonal family of vectors in 3-1; C a x j a converges to a vector xj in 3-1 for each j . With A in R, A = C, AQa and

320

NORMAL STATES

for each subset

B of A. Thus, by ultraweak continuity of w , a

a na

a j=1

j=1

m

Q

m

j=1 a

j-1

m

m

j=1

j=1

a

a

Thus w = C;='=, IR. 7.6.11. Find an example of a von Neumann algebra R acting on a Hilbert space 'If and a net of cyclic projections in R converging in the strong-operator topology to a projection in R that is not cyclic.

Solution. Let 7-l be a non-separable Hilbert space and let { E , :

A} be an orthogonal family of projections with sum I such that E , ( X ) is separable for each a in A. Let R' be the commutant of {Ea : a E A} and let R be R" (so that R is the (abelian) von Neumann algebra generated by {EQ : a E A}). Note that T' E R' if EaT'E, = T' for some a in A. Let 3 be the family of finite subsets of A partially ordered by inclusion. With IF in F, define Ep t o be C a E s E Q .Then Ep E R. We note that Ep is a cyclic projection in R.Indeed, with X a a preassigned unit vector in EQ(3-I), let z]p be CaEpza.Suppose a vector pa is given for each a in F. aE

CQTry,.Choose Ti (effectively in B ( E , ( X ) ) ) such that and Ti E B ( H ) . Then :'2 E R'. Let EaTiE, = Ta, T ~ z = , T' be C a E p T ; .Then T' E R' and T'xp = y, whence E p is cyclic in R. Now { E p : F E 3}is a (monotone) net of projections in R with strong-operator limit (union) I. However, I is not a cyclic projection in R since a generating vector fqr R' would be separating for R and R would be countably decomposable. But { E , : a E A} is an uncountable orthogonal family of non-zero projections in R (uncountable since 3-1 was chosen non-separable and each E a ( H ) was Let y be

chosen separable).

I

EXERCISE 7.6.12

321

Let R be a countably decomposable von Neumann 7.6.12. algebra acting on a Hilbert space H. Suppose that E is the strongoperator limit of a net of cyclic projections in R. Show that E is a cyclic projection in R.

Solution. Note that the sequential nature of {En}is used in the proof of Proposition 7.3.10 only in the second sentence of the third paragraph of that proof where Proposition 5.5.19 is adduced to conclude that E is countably decomposable. Under the given hypotheses, it follows that E is countably decomposable without the use of Proposition 5.5.19. With our present hypotheses, we may replace {En}by a net {E,} of cyclic projections in the proof of Proposition 7.3.10. 7.6.13. Show that a projection in a von Neumann algebra is the support of a normal state if and only if it is countably decomposable.

Solution. Let R be a von Neumann algebra acting on the Hilbert space H and let E be a projection in R. Suppose E is the support of a normal state w of R. Let { E , : a E A} be an orthogonal family of non-zero subprojections of E in R. Since E is the support of w and each E , is non-zero, w(E,) > 0. Since w is normal,

Thus w(E,) = 0 for all but a countable family of a in A. Hence A is countable and E is countably decomposable. Suppose E is countably decomposable. From Proposition 5.5.9, E is the sum of a (countable) family { E l , E2, . . .} of non-zero cyclic projections in R. Let z n be a unit generating vector for E , and let yn be 2-"I2x,. Then C F = l ~ Y n (=J R w ) is a normal state of R. If F is a subprojection of E in R and w ( F ) = 0, then F z , = 0 for all n. Hence F R ' z , = 0 and F E , = 0 for all TZ. But then o = F(C:==,En)= F E = F . Since

E is the support of w .

w

322

NORMAL STATES

Show that each normal state of a von Neumann alge7.6.14. bra 72 is a vector state of 'R if and only if each countably decomposable projection in R is cyclic.

Solution. F'rom Proposition 7.2.7, each normal state of R is a vector state of R if and only if all supports of normal states of R are cyclic projections i:n R. F'rom Exercise 7.6.13, the supports of normal states of R are precisely the countably decomposable projections in 72. Thus each normal state of R is a vector state of R if and only if rn each countably decomposable projection in 'R is cyclic. 7.6.15. Suppose that U is a C*-algebra acting on a Hilbert space 3.1. Let S be a derivation of %, and recall that S is norm continuous (Exercise 4.6.65(v)). By using Lemma 7.1.3, show that 6 is weak-operator continuous on bounded subsets of U. Deduce that 6 extends (uniquely) to a derivation 6 of !U-, and that llsll = IlSll. of

Solution. Suppose that 'P > 0, and that a net (A,} of elements is strong-operator convergent to 0. Given any z and y in 'FI,

(a):

(6(A,)z, y) = ((AtI26(Ati2)t 6(A:/2)A1,/2)z, y)

= ( s ( A ; / ~ A) ~' , ,/ ~ Pt) ( ~ ( A : / ~ ) A : / ~ Zy)., From this, and since {Ail2} is strong-operator convergent to 0 (by use of Proposition 5.3.2),

I ( W a $4I 5 II41IIAa/2 IIIlzIIIIA:/2 vll f 116II11Ail2IIIIAil2 2 I Illvll s 'P1~211~11(11~1111A~~2~ll t l l ~ : ~ 2 ~ l l l l v l l0;~ +

so {&(A,)}is weak-operator convergent to 0. The preceding argument shows that 6 is continuous at 0, as a mapping from (a): (with the strong-operator topology) into U (c B('H), with the weak-operator topology). From Lemma 7.1.3, 6 is weak-operator continuous on (U),, for d positive T . In proving that 6 extends (without increase of norm) to a derivation of Q-, we may assume that ll6ll = 1. For each positive T , (a), is weak-operator dense in (a-),, by the Kaplansky density theorem, and (U-), is weak-operator compact (hence, weak-operator complete). The restriction 61(2l), is a weak-operator continuous mapping from (%), into (%-),, this continuity is uniform by Proposition

323

EXERCISE 7.6.16

1.2.1(ii), and so 6((U), extends (uniquely) to a weak-operator continuous mapping 6, : (%-), (%-),. When s T , bs1(U-), is a weak-operator continuous extension of 6l(U),, and so coincides with 6,. It follows that there is a mapping 8 : U- + U- such that 8l(U-), = 6, for each positive T . Moreover, 8 extends 6, and is weak-operator continuous on bounded subsets of 2l- . Suppose that A , B E 2l- and c E @. From the Kaplansky density theorem, there are bounded nets { A a } , {Bb} in U that are weak-operator convergent to A , B , respectively. We have --+

>

From these relations, together with the last sentence of the preceding paragraph, it follows by taking limits (first, for { A a } ,then, for {Bb}) that

+

6(cA -+ B ) = d ( A ) 6 ( B ) , S(AB) = A 6 ( B ) -t 8 ( A ) B . Hence 8 is a derivation of U-. Since 8((U-),-) = 6,.((2l-),) C (a-),, we have 5 1 = IlSll, whence 11811 = ()6)) = 1. If 60 is another derivation of U- that extends 6, then the first two paragraphs of this solution (with 60 and 2l- in place of 6 and U, respectively) show that 60 (as well as 8) is weak-operator continuous on (U-), for all positive T . Since 8 - 60 is weak-operator continuous on (U-), and vanishes on the dense subset (U), of (U-),, it follows that 8 and 60 coincide on (au-), (and, hence, on U-). m[60]

llall

Let q be a linear mapping of a C*-algebra U into a 7.6.16. C*-algebra D. Suppose q ( I ) = I and 11q(T)ll = IlTll for each normal element T in U. Show that (i) q ( A ) is self-adjoint if A is a self-adjoint element of U [Hint. Consider p o q for an appropriate state p off?.]; (ii) q(B*)= q(B)*for each B in U.

Solution. (i) If q ( A ) = B -t iC with B and C self-adjoint and C not 0, there is a state p of 0 such that p(C) # 0, from Theorem 4.3.4(i). Let Uo be the (abelian) C*-subalgebra of U generated by A and I . Since each element of 2l0 is normal, qJMois an isometry (and q ( I ) = I > . Thus I I ( P 0 ~ ) I ~ O=I I 1. As ( P W ) ( I ) = P ( I ) = 1, ( P 077)IUo is a state of UO.But ( p o q ) ( A ) = p ( B ) t ip(C) and p(C) # 0 -

324

NORMAL STATES

contradicting the fact that p o q is hermitian and A is self-adjoint. Thus q(A)* = q ( A ) . (ii) If B = T t is with T and S self-adjoint in U, then q(B)*= q(T)*- iq(S)*= q ( T ) - i q ( S ) = Q(T = q(B*). rn

is)

7.6.17. Let q be a linear isometry of one C*-algebra U onto another C*-algebra B. (i) Show that U is a partid isometry in B, where U = q ( I ) . (ii) Let qo(T) be U*q(T)E for each T in U, where E = U*U. Show that l l ~ ~ ( A )=l lIlAll for each normal element A in 2l. [Hint.If 1 E s p ( A ) , there is a state p off? such that p([U f q(A)]*[Ut q ( A ) ] ) = 4.1 (iii) Show that 770 is hermitian. [Hint.Use Exercise 7.6.16.1 (iv) Show that U is a unitary operator in 0.[Hint.Consider T in such that q(T)= I - E . ] (v) Show that q ( V ) is unitary when V is a unitary element of

u.

Solution. (i) Since q is a linear isometry of U onto L3, q maps

each extreme point of (a), onto an extreme point of ( B ) 1 . From Theorem 7.3.1, I is an extreme point of (%)I. Thus q ( I ) ( = U ) is an extreme point of (B)1. Again from Theorem 7.3.1, U is a partial isometry in B such that ( I - F ) B ( l - E ) = (0), where U*U = E and UU' = F . (ii) To show that 770 is isometric on normal elements of 2l,it will suffice to show that [lqO(A)II = IlAll when IlAll = 1 E sp(A). For such an A , 2 E sp(1 t A ) and 111 t AIJ = 2 . Thus

F'rom Theorem 4.3.4(iv), there is a state p of

B such that

Since llU*q(A)ll = Ilq(A)*Ull 5 1 = Jlq(A)*q(A)JJ, we can conclude that p ( E ) = 1 = p ( U * q ( A ) ) .Thus p is definite on E (see Exercise 4.6.16) and 1 = P(U*V(A))P(E)= P(U*V(A)E). Hence I[U*q(A)EII = 1. It follows that the mapping qo of U into EBE is norm preserving on normal elements of U.

EXERCISE 7.6.18

325

(iii) Since % ( I ) = U * q ( I ) E = U * U E = E , and 170 is a linear mapping of the C*-algebra 2l into the C*-algebra EBE that is isometric on normal elements of U, % is hermitian from Exercise 7.6.16. (iv) As q maps U onto B there is a T in U such that q ( T ) = I - E . Let T be A t iB with A and B self-adjoint elements of a. We have

From (iii), qo(A) and qo(B) are self-adjoint. Thus 0 = % ( A ) = qo(B). From (ii), IlAll = 11w(A)ll = 0 and IlBll = 11770(B)ll = 0. Hence 0 = A = B = T , and E = I . In the same way, choosing S in 2l such that q ( S ) = I - F (recall from (i), that U U * = F ) , we conclude that F = I . Thus U is a unitary element of B. (v) The mapping T + q ( V T ) is a linear isometry of U onto B whed V is a unitary element of U. Hence q(V)is a unitary element of B from (iv). 7.6.18. With the notation and assumptions of Exercise 7.6.17, define q’(T) t o be U*q(T)for each T in U. Show that ~ each self-adjoint H in 24 [Hint. Consider (i) q ’ ( H 2 )= V ’ ( H >for q’(exp i t H ) in series form and use ( v ) of Exercise 7.6.17.1; for each A in 8; (ii) q ’ ( A 2 )= v ‘ ( A ) ~ q = Uq’ and q‘(AB t B A ) = q’(A)q‘(B)t q‘(B)q’(A)for (iii) all A and B in 2l. (We call 11’ a Jordan * isomorphism of 2l onto B. Compare Exercise 10.5.28.) Solution. ( i ) With H a self-adjoint element of U and t real, exp itH is a unitary element Ut of U. From Exercise 7.6.17(v), q‘(Ut) is a unitary element of 0.Now $ ( I ) = U*U = I and q’ is a linear isometry. From Exercise 7.6.16(ii), q’(Ut)* = q‘(U;) = q‘(U-t), Thus I = v’(U-t)q’(Ut), (-it)2 q’(i7-t) = I - i t q ’ ( H )t -q’(H2) tW3) 2!

so that

326

NORMAL STATES

and q‘(H2) = v ’ ( H ) ~ . (ii) With A in U, A = H elements of U. From (i),

9’(fq2

+ iK,where H and A’ are self-adjoint

+ v ’ ( o # w ) t 9’(K)9‘(H)

9‘(W2

= q’(H t K)2 = q‘([H t KI2)

+ HK t KH t K 2 ) = q’(H)2 + 9‘(HK + K H )t qq1q2.

= q’(A2

+

Thus $ ( H K t KH) = q’(H)q’(K) q‘(K)q‘(H)and

?’(A2) = q‘(R2t i ( H K t K H ) - K2) = q ‘ ( q 2 t i [ ? f ( H ) q ’ ( Kt) q’(K)q‘(H)]- V’(Jq2 = q’(A)2.

Uq’. With A and B in U,

(iii) By definition of q‘, we have 9 = we have from (ii),

q‘(A)2tq’(A)v’(B) t q’(B)q’(A)t v ’ ( B ) = ~ rl’(At BI2 = q’([At BI2) = v ‘ ( A ) ~t q‘(AB B A ) t q‘(B)’.

+

Hence $(AB t BA) = ~‘(A)Q‘(B) t v’(B)q’(A).

7.6.19.

4531

Let w and w’ be normal positive linear functionals on

-

-

a von Neumann algebra R. Suppose w w’ = w1 w2, where w1 and w2 are normal positive linear functionds on R with orthogonal supports M and N, respectively. Let G be a projection in R such

that w ( G T G ) = o ( T )and w ’ ( G T G ) = w’(T)for each T in 72, Show that (i) w1 and w2 are the unique positive linear functionals on R that are normal, have orthogonal supports, and whose difference is w - w’; (ii) M N 5 G.

+

Solution. (i) Since M and N are orthogonal projections in ((M - N ( (= 1. Now (w1

- w2)(M - N ) = Ul(M) t W 2 ( N ) = wl(M t I - M)

= l l 4 I + IIw2II.

+ W2(N t I - N )

R,

EXERCISE

7.6.20

327

The equality (*) has been established for each pair w1, w2 of normal positive linear functionals on R with orthogonal supports. If the difference of such a pair is equal to the hermitian functional w w’, the conditions of Theorem 4.3.6 are fulfilled. In particular, the uniqueness assertion applies, and w1, w2 are unique. (ii) From Theorem 7.4.7, there is a projection E in GRG such that the hermitian normal functional (w - w’)lGRG (= p ) attains its maximum on the unit ball of GRG at 2 E - G and p = p 1 - p2, where p1 and p2 are normal positive linear functionals on GRG and for each T i n GRG, p l ( T ) = p ( E T E ) and p2(T) = -p((G-E)T(G-E)). For j = 1,2, let pi be the positive normal linear functional T -+ pj(GTG) on R. Now

and p i ( I - E ) = pb(I - ( G -E )) = 0. It follows that w-U‘ = pi -pb, and the supports M’ of pi and N’ of pb satisfy M‘ 5 E , N’ 5 G - E (whence M’N’ = 0). From (i), pi = w1, pb = w2. Thus

M=M’IE,

N=N‘ PE;) is also properly infinite with central carrier P so that the assumptions of Exercise 7.6.26(ii) are fulfilled once we show that the union of the support G of w,lIR and the support M of w is cyclic in R. When this cyclicity is established, we can assert the existence of a vector u in 'FI such that w,lR = w and

IIu - vll(5 [2(lw - WvlIR11]1'2.

333

EXERCISE 7.6.28

Thus, in this case,

with the preceding inequality, we have

If ((w- w,lRII = 0, then w = w,lR; and we can choose outset. When 0 < IIw - w,IRII, we have

so that if

E

has been chosen suitably small,

E'

for u at the

will be small; and

It remains t o show that G V M is cyclic in R. Let E be the support of o,lR. From Remark 7.2.6, G(3-1) = [R'v'] and E('H) = [R'v]. From the definitions of w' and G', v' = G'v. Hence

G(3-1) = [R'v'] = [R'G'w]& [R'v] = E(3-1). It follows that G V M Hence G V M is cyclic.

E V M . By assumption E V M is cyclic. m

7.6.28. Let w be a normal positive linear functional on a von Neumann algebra R acting on a Hilbert space 3-1, and let v be a vector in 3-1. Suppose that the union of the supports of w and w,lR is cyclic in R. Show that there is a vector u in 3-1 such that

[Hint.Reduce to the case where R' is countably decomposable and apply Exercise 7.6.27.1

334

NORMAL STATES

Solution. With the notation introduced in Exercise 7.6.24 (and w,lR in place of u’), let N’ be the projection in R’ with range and [Rz’], where y is a generating spanned by [Ry], [Rv], [Rz], vector for G under R’. From Proposition 5.5.19, N ’ is a countably decomposable projection in R’, so that N‘R’N‘ is countably decomposable. The support M of w is [R‘z] from Remark 7.2.6 and [R’Rz]C C N ~ ( X whence ), M 5 C N : . From Exercise 7.6.2, it follows that there is a normal positive linear functional wo on RN‘ such that wo(AN‘) = w ( A ) for each A in R and that the support of wo is M N ‘ . In the same way, the support E of o,lR is a subprojection of C N,~whence EN’ is the support of w,lRN’. Since y is a generating vector for E V M ( = G) under R’and y E ”(‘If), “’R’N‘y]

= “’R’y] = ( E v M ) N ’ ( R ) = ( E N ’ v M N ‘ ) ( R ) .

The assumptions of Exercise 7.6.27 are now fulfilled with N‘(7-I) in place of R,RN‘ in place of R, N’R’N’ in place of R‘, wo in place of w , and u,lRN’ in place of w,lR; there is a vector u in “(R) such that I 112 . w,[RN’ == wo, (1’11- v11 I 2llWO - w,lRN 11 We note that (**) holds with this choice of u. If A E R,

w,(A) = ( A u , ~=) ( A N ‘ u , ~=) wo(AN’) = w ( A ) , so that w,JR = w . If IIAN’II 5 1, then

since IIAClvtII = llAN’ll _< 1 from Proposition 5.5.5 and Theorem 4.1.8(iii). Hence llwo - w,lRN’II 5 IIw - w,lRII and (**) follows. m[65] Let p be an ultraweakly continuous linear functional 7.6.29. on a von Neumann algebra R acting on a Hilbert space X. Let UO be any partial isometry in R such that Ip(Uo)l = llpll, and let U be d o , where la[ = 1 and ap(Uo) = Ip(U0)l.

335

EXERCISE 7.6.30

(i) Show that w is a normal positive linear functional on R, where w(A) = p(UA) for each A in R. [Hint. Use Theorem 4.3.2.1 (ii) Let F be UU*. Show that for each A in R, p(A) = p ( F A ) = w(U*A). [Hint.Suppose the contrary and choose A in R such that 11(I- F)AII = 1 and p ( ( I - F)A) > 0. Let 8 satisfy 0 < 8 < n/2 and t a n 8 = p ( ( 1 - F)A)/llpll, and consider p(U cos8 ( I - F)Asin8).]

+

Solution. (i)

Note that

so that w(1) = IIwII. From Theorem 4.3.2, w is positive. From the definition of w and the assumption that p is ultraweakly continuous, w is normal. (ii) By definition of w, w(U*A) = p(UU*A) = p(FA). We proceed as in the hint. If p((1-F)A) # 0, we can replace A by a suitable scalar multiple and assume that p((1-F)A) > 0 and II(I-F)AII = 1. There is nothing to prove if llpll = 0. If llpll # 0 and with 8 as indicated in the hint, we have

But with

5

a unit vector in 'H,

+

IIUcosBs t ( I - F)Asin8s1I2 = llFUcos8~11~Il(1- F)Asin8s112

5 COS' 8 + sin2 8 = I.

+

Hence (lUcos8 ( I - F)AsinBII 5 1 - contradicting (*).

w

7.6.30. Let R be a von Neumann algebra and p be an ultraweakly continuous linear functional on R. Show that (i) there are a normal positive linear functional wo on R and a partial isometry UO in R such that UOU; (= Eo) is the support of wo and

[Hint. Use Theorem 7.3.2.1;

336

NORMAL STATES

(ii) Uo = U1 and wo = w1 when U1 is a partial isometry in and w1 is a normal positive linear functional on R with support U1Ui (= E l ) such that p ( A ) = wl(U1A). [Hint.Show that llwlII = llpll and reduce to the case where llpll = 1. Prove that 1 = ~1 ( E l ) = wo(UoU;1*)= (Ui'y, Uzy), where R acting on 3.1 is the universal normal representation of R and y is a (unit) vector in 'H such that wo = wylR. Use the equality clause of the Cauchy-Schwarz inequality to conclude that UoUiy = y. Deduce that Eo = El and

R

uou; = Eo.]

Solution. (i) From Theorem 7.3.2,there are a normal positive linear functional wo on R and a partial isometry U in R such that

w(A) = p ( U A ) , p(A) = oo(u*A), ~

( u=)JJPJJ = llWoII = WO(EO).

Let E be U*U so that llw~II= p ( U ) = wo(U*U) = wo(E) and EO 5 E . Thus UEo and EoU* are partial isometries in R. Let UO be EoU". Then UoU; = EoU*UEo = EoEEo = Eo, and since EOis the support of W O ,

At the same time,

p(U,*A) = uo(UoU,*A) = wo(EoA) = wo(A)

( A E R).

(ii) We proceed as in the hint. Since, for each A in R, lP(A)l = IWl(U1A)I I Il4lllU~AIIi Il~lllllAll~

we have that llpll 5 ((w1((. On the other hand, IlWlII

= Wl(E1) = Wl(UlU,*) = P(U,*) 5 IIPllll~,*lls IlPllt

so that llpll = llwlII,. We may assume that llpll = Ilwlll = 1, R acting on 7t is the universal normal representation of 'R, and y is a (unit) vector in 3.1 such that wvJR= WO. Then

p ( U i A ) = wl(U1U;A)=w1(&A) so that

= Wl(A),

337

EXERCISE 7.6.31

From Proposition 2.1.3, UoU;y = y , whence llUTy\l = 1. Thus y E El(%). As [R’y] is the range of Eo, EO 5 El. By symmetry, El 5 Eo, and EO = El. Now

and EoUoU;EoY = UoU;y = y . Since y is separating for EoREo, UoU; = EoUoU;Eo = Eo. As U$ and U; are unitary transformations of Eo(7-l) onto their ranges with inverses UO and U 1 , respectively, the equality, Uo = U I , follows from the equality, Uo U; = Eo. Of course, then,

and wo = W I .

~[93]

7.6.31. Let E be a projection in a von Neumann algebra, and let P be the union of all projections in R equivalent to E . (i) Show that P = C E . (ii) Show that the support of w,, J Ris a central projection when 20 is a trace vector for R.

Solution. (i) If CE - P # 0, then there are equivalent non-zero subprojections EO of E and Fo of C E - P from Proposition 6.1.8. Thus E = E - Eo t Eo E - Eo t Fo, from Proposition 6.2.2. By definition of P , E - EO Fo 5 P , which contradicts the choice of Fo as a non-zero subprojection of CE - P . It follows that P = CE. I - F , then there (ii) Let F be the support of w,,lR. If G is a V in R such that V*V = G and V V * = I - F . Since F is the support of w,,(R and 20 is a trace vector for R, N

+

N

Thus w,,lR annihilates the union of all projections in R equivalent to I - F . From (i), this union is CI-F. But I - F is the union of all projections in R annihilated by w,,(R. Thus I - F and hence F are central projections in R.

7.6.32.

Let

be a trace vector for a von Neumann algebra wz0IC has support I , where C is the center of R. Show that 20 is separating for R. 50

R acting on a Hilbert space 7-l. Suppose that

338

NORMAL STATES

Solution. Suppose Ax0 = 0 for some A in R. Let E be R(A*A). Then E E R and Ezo = 0 from Proposition 2.5.13. Since xo is a trace vector for R,Fzo = 0 for each projection F in R equivalent to E. Thus the union P of all the projections equivalent to E annihilates 2 0 . From Exercise 7.6.31(i), P = CE and, in particular, P E C. Since uX,lC has support I, CE = 0. Thus E, A'A, and A are 0; and xo is a separating vector for R. a

7.6.33. Let 2l be a C*-algebra acting on a Hilbert space 8, and let {wen 1%) be a sequence of vector states of U tending in norm to p. Show that p is a vector state of 3L. Solution. Since p is the uniform limit on (%)Iof the ultraweakly continuous functions w X nI(%)l, p is ultraweakly continuous on (%)I. By uniform continuity, p has a unique ultraweakly continuous extension to the ultraweak closure of (%)I. From the Kaplansky density theorem, the ultraweak closure of (%)Iis (%-)I. It follows that p has a (unique) normal extension w to 2l-, and o is a normal state of %-. Again from the Kaplansky density theorem and ultraweak continuity of o - oZn I%-, we have IIW

- U s , 1%- II = IIp - wx, lull

-+

0.

From Theorem 7.3.11,w is a vector state of %-. Hence p (= +l) is a vector state of U.

m

7.6.34. Let ( p n } be a sequence of states of a C*-algebra % converging in norm to a state po. Let rn be the GNS representation corresponding to pn. Suppose that 11,12,. are equivalent to a single representation a of 0 on a Hilbert space 3-1. (i) Show that a0 is equivalent to the representation A a(A)E' of U on E'(3-I) for some projection E' in I(%)'. (ii) Find an example in which 10 is not equivalent to the representation a.

..

Solution. (i) Since xn is equivalent to a, the kernel K: of I is the kernel of R , for each n in W. Thus p1,p2,. annihilate K: as does their norm limit po. Hence there is a linear functional p on R(%) such that p o I = pa. From Exercise 4.6.23(ii), p is a state of I@). There are unit generating vectors x n in 3-1 for I(%) such that won o a = pn since rn is equivalent to A. Now a = cp o p, where

..

339

EXERCISE 7.6.35

p : U --t U/r-'({O}) is the quotient mapping and cp : U / K - ~ ( { O } ) K(U)is a * isomorphism (hence an isometry). (See Theorem 1.5.8 and Exercise 4.6.60.) From the definition of the quotient norm, p maps the open unit ball of 2l onto the open unit ball of % / r - ' ( { O } ) . Hence K (= cp o p ) maps the open unit ball of 24 onto the open unit ball of K(%), and --$

IIP -

WXn

I~(')ll

= IIPO- ~ n l l 0. +

From Exercise 7.6.33, p is a vector state w,l~(U) of .(a). Let E' be ' range [7r(U)x]. Then K O is equivalent to the projection in ~ ( 2 l ) with n(A)E' of 2l on E'('H). the representation A (ii) Adopt the notation of the solution to Exercise 7.6.22 and let 2l be the strong-operator dense C*-subalgebra of R consisting of all operators A in R for which { ( A e n , e n ) }converges. Let pn be w,,, 12l for n in N. As noted in the solution to Exercise 7.6.22, each 2, is generating for R and hence for 2l. From Proposition 4.5.3, rn (corresponding to p n ) is equivalent to the identity representation K of 2l on 'H. Let po be wel121. Then {xn} converges to el and { p n } converges to PO in norm since ---$

((wz, - we1 11

5

( ( x n-

+ 11.1 11)

el I I ( l I ~ n l l

5 2((xn- el 11.

Now KO (corresponding to po) is a representation of U on the onedimensional space [el] (= [%el]),and K is a cyclic representation of U on the infinite-dimensional space 7-f. Thus TO is not equivalent to T.

7.6.35. Let 2l be a C*-algebra acting on a Hilbert space 'H. Suppose that each increasing net of operators in U that is bounded above has its strong-operator limit in U. Show that (i) each decreasing net of operators in 2l that is bounded below has its strong-operator limit in 2l; (ii) the range projection of each operator in U lies in U; (iii) the union and intersection of each finite set of projections in U lie in 24; (iv) the union and intersection of an arbitary set of projections in U lie in U; (v) E E U, where E is a cyclic projection in 2l- with generating vector x, provided that for each vector y in ( I - E)('H) there is a self-adjoint A , in U such that A,z = x and A,y = 0; (vi) U- = U if each cyclic projection in U- lies in U.

340

NORMAL STATES

Solution. (i) If { A , ) is a decreasing net in 2.l that is bounded below with strong-operator limit A , then { -A,} is an increasing net in U with strong-operator limit -A. By assumption, - A E Iu, so that A E !2l. (ii) From Proposition 2.5.13, R ( A * ) = R(A*A), so that it suffices to show that R ( H ) E 2l for each positive H in Q. Of course R ( H ) = R(aH) for each positive scalar a. Thus we may assume that 0 5 H 5 I . In this case {H1/"} is a monotone increasing sequence in U and from Lemma 5.1.5, R(H)is its strong-operator limit. By assumption then, R ( H ) E M. (iii) From Proposition 2.5.14, E V F = R ( E t F ) E !2l, when E and F are projections in 8. Thus the union of a finite family of projections in ?2l is in Q. Since I - V,(I - E,) = A,E,, the intersection of a finite family of projections in U is in U. (iv) If { E , : a E A} is a collection of projections in U, the union of each finite subcollection lies in U from (iii). The family of such unions, indexed by the family of finite subsets of A directed by inclusion is an increasing net with strong-operator limit VaEAE,. By assumption then, VoEAE, E a. (v) From our assumption that A,a: = 2, R(A,)z = x. Since A,y = 0 and A, = A t , R(A,)y = 0. Thus GZ = 2 and Gy = 0 for each y in ( I - E ) ( H ) ,where G = / \ Y E ( I - E ) ( H ) R(A,). From (ii) and (iv), G E U. As E is cyclic under U' with generating vector x, E 5 G. As Gy = 0 for each y in ( I - E ) ( X ) , G 5 E . Thus

E=GE%. (vi) From Proposition 5.5.9 and (iv), each projection in 2l- lies in 2l. From Theorem 5.2.2(v), each self-adjoint operator in U- lies in U. Since U- is a self-adjoint algebra containing %,a-= 8. rn 7.6.36. Let U and 'H be as in Exercise 7.6.35. Suppose E is a cyclic projection in !2l- and z is a unit generating vector for E ( H ) under U'. With y a unit vector in ( I - E ) ( X ) , show that (i) there is a sequence { A n } in (!&)I such that A,$ --+ 5, Any + 0, ll(An - An-l)+zll < 21-n, and ll(An - An-l)+yll < 21-n, where A0 = 0; (ii) {T,} is a bounded monotone decreasing sequence of pos- Ak,l)+)-l, and itive elements of 8 , where T, = (I t C;=l(A~ T1/2(C~=l(Ak-Ak-1)+)T1/25 I for each n, where T is the strongoperator limit of {?',} (in %); (iii) for each j in 11,. ,n } , { T 1 / 2 ( C i = j ( A ~Ak-1)+)T1I2}is

..

341

EXERCISE 7.6.36

monotone increasing with n, bounded above by I , and if Cj is its strong-operator limit, then 0 5 C j 5 I , {Cj} is decreasing,

and

+

(iv) {T'/~A,T'/~ Cntl) is monotone decreasing and bounded and T ' / ~ A T ' / ~E IU, where A is a weak-operator limit point of {A,); ( v ) R(T) E U, R(T)x = x, R(T)y = y; (vi) each maximal abelian (self-adjoint) subalgebra of '2 is weakoperator closed. [Hint. Note that AT € U if U is abelian, and apply (i)-(v) and Exercise 7.6.35 to the subalgebra.]

Solution. (i) From the Kaplansky density theorem, there is a such that A,x -t x and Any 4 0 since sequence {A,} in E E (UC)l, E x = x, and Ey = 0. Passing to a subsequence of {A,) (using the Cauchy criterion on the convergent subsequences, {A,x) and {Any}), we can arrange that

For each self-adjoint A, At and A- have orthogonal ranges so that

Thus I(Atz([_< IIAxII. It follows that for each n in Pd,

(ii) Since

<
-n RL. (iii) From the results of (i), (ii), and Exercise 8.7.24, show that the following three conditions are equivalent: (a) there is a conditional expectation from B(lz(G))onto CG (= Rd); (b) there is an invariant mean on G; )Rd. (c) for each T in f ? ( l ~ ( G >c )o, ~ ~ ( T meets Solution. (i)

Suppose that CP is a conditional expectation from

B(12(G))onto CG,and p is defined as in the statement of part (i) of the exercise. Then p is a bounded linear functional on lm(G), and llpll 5 1, since 1 1 ~ 1 ) = 1, 11@11 5 1 by Exercise 8.7.23(iii), and ((M,II = llzlloo when z E lm(G). If u is the element of l,(G) that takes the value 1 throughout G, then p ( u ) = T(CP(M,)) = T ( Q ( 1 ) ) = T ( l ) = 1.

406

THE TRACE

When z E l,(G), y E IZ(G), and g, h E G, we have

and

LPMA:,

= Mz,Y

where zg (in I,(G)) is defined by z g ( h )= z(g-'h). In the notation of Exercise 3.5.7, zg is Tgz. Since L,, and L& lie in the range CG of the conditional expectation a, and r is a tracial state of LG,we have

Hence T f p = p for each g in G, and p is an invariant mean on G. (ii) Suppose that p is an invariant mean on G. Since I,(G) is a C*-algebra with unit u (defined in the solution to (i)), and p is a bounded linear functional on I,(G) satisfying llpll 5 1 = p(u), it follows (Theorem 4.3.2) that p is a state of Z(,G). Given T in B(Iz(G))and 2,y in Iz(G),the complex-valued function z , , ~defined in (ii) is an element of L , ( G ) , the mapping 5 + zZtyis linear for each fixed y, the mapping y + z , , ~is conjugate-linear for each fixed z, and ( I Z , , ~ ~ ( 5 ~ IIT1111~llllyll.It follows that the equation

defines a bounded conjugate-bilinear functional b on /z(G); corresponding to b, there is an element AT of B(lz(G))such that

If h E G, and

2,y

E

Iz(G), we have

EXERCISE 8.7.29

407

where u = Rxhxand v = Rxhy. Also, for each g in G,

Since p is an invariant mean, p ( ~ ~=,p ~( )~ ~ that , ~ ) is;

Thus R j , ATR,, = AT, for each h in G, and

-, is a weak-operator continuous linear If AT 4 c o ~ ~ ( T ) there functional w on B(lz(G))and a real number c such that

Rew(AT) > c 2 Rew(S)

(S E C O R ~ ( T ) ) .

In particular,

By expressing w as a finite sum of vector functionals w , , ~ and , using (1) and the definition of z ~ ,it~follows , that AT) = p(zw), where zW (in

I,(G)) is defined by

Since Rezu(g)5 c for all g in G, by (2),and p is a state of l,(G), we have c < R e o ( A T ) = Rep(%,)5 c, a contradiction. Thus AT E CO.R~(T)-. (iii) The assertions that (a) implies (b) and (b) implies (c) result from (i) and (ii), and (c) implies (a) by Exercise 8.7.24(vi). m[99]

408

THE TRACE

8.7.30. Suppose that G and H are discrete groups, and adopt the notation of Section 6.7. (i) Show that the von Neumann algebras CG and C H are unitarily equivalent if they are * isomorphic. (ii) Show that CG and C H are not * isomorphic if G has an invariant mean and H has no invariant mean. [Hint. Use (i) and Exercise 8.7.29(iii).] (iii) Re-prove Theorem 6.7.8 by showing that ll has an invariant mean and F2 has no invariant mean. [Hint. Let a and b be the two generators of F2, and let S be the set of reduced words in .& that begin with a non-zero power of b. Use the fact that F2 = S U bS and that S, US,a2S are disjoint .] (iv) Show that no group containing .F2 has an invariant mean. [Hint. Consider cosets of F2 and use the set S described in (iii).] Show that no two of C ~ , L Fand ~ , LG,are (v) Let G be TI@&. * isomorphic. [Hint. Use Exercise 6.9.45.1

Solution. (i) Since each of the von Neumann algebras CG and C H has a separating and generating vector (Remark 6.7.3), they are

unitarily equivalent if they are * isomorphic, by the unitary implementation theorem (7.2.9). (ii) If G has an invariant mean and H does not, it follows from Exercise 8.7.29(iii) that there is a conditional expectation from B(lz(G)) onto LG but there is no Conditional expectation from B ( / 2 ( H ) ) onto CH. Hence CG and LH are not unitary equivalent; by (i), they are not * isomorphic. (iii) The group II is locally finite (Example 6.7.7), and so has an invariant mean by Exercise 3.5.7. When X C 32,the characteristic function fx is in l m ( 3 2 ) , and fgX

=TgfX

(9 E F2),

where Tg is defined as in Exercise 3.5.7. The inequalities fS

t fbS 2

f32

2 fS t f a S t f a z S

can be written in the form fS

4-T b f S L f& L

fS

4-Tafs 4-

Ta2f.S.

If p is an invariant mean on F2, these inequalities imply that 2p( fs) 2 1 2 3p(fs) (recall, from Exercise 8.7.29(ii), that p is a state of lm(G)), which is impossible. Thus 3 2 has no invariant mean.

409

EXERCISE 8.7.31

It now follows from (ii) that the von Neumann algebras Ln and C F ~are not * isomorphic. 2 and { g k : k E IK} (iv) Let G be a (discrete) group containing 3 be elements of G such that the cosets 3 2 g k and &gk, are disjoint unless k = k' and such that G = U k E K 3 2 g k . Let SO be U k E n sg,. Then SO

ubS0

=

(u

Sg,)

k€K

u b(

u

sgk)

=

k EK

u ( sU b S ) g k =

U

k€X

k %t

&gk

=G

and SO,US,-,, u2So are disjoint. The argument given in (iii) applies now, with So in place of S, to show that G does not have an invariant mean. (v) Since 3 2 is a subgroup of G, G has no invariant mean from (iv). Hence CG is not * isomorphic to L n from (ii) and (iii). From the result of Exercise 6.9.45(ii), ,& and C F ~are not * isomorphic. From Theorem 6.7.8, Cn and C F ~are not * isomorphic. m[99] Let A be the center-valued dimension function on a 8.7.31. finite von Neuamnn algebra R. Prove that

A ( E V F ) t A(E A F ) = A ( E )t A ( F ) for all projections E and F in R. Deduce that E A F # 0 if A( E ) t A ( F ) > I , and that E V F if A ( E )t A ( F ) < I . Solution. Since E V F - F

E - E A F , we have A(E V F ) - A ( F ) = A ( E V F - F ) = A( E - E A F ) = A( E ) - A( E A F ) .

Thus

N

+

+

A ( E V F ) A(E A F ) = A ( E ) A(F). If A ( E )t A ( F ) > I , we have A(E A F ) = A ( E ) A ( F ) - A(E V F ) > I - A(EV F ) 2 0; so A ( E A F ) # 0, and E A F # 0 . If A(E) A ( F ) < I , we have A(E V F ) = A ( E ) A ( F ) - A(E A F ) 5 A ( E )t A ( F ) < I = A ( I ) , 8 and thus E V F # I .

+

+

+

#I

410

THE TRACE

8.7.32. Suppose that A is the center-valued dimension function on a finite von Neumann algebra R, E , F , and G are projections in R, and F 2 G. Show that

A ( E A F - E A G ) 5 A ( F - G). Solution. From Exercise 8.7.31, and since E V F 2 E V G, we have

+ +

A(E A F ) = A(E) A(F) - A(E V F ) < A(E) A ( F )- A(E V G), A(E A G) = A ( E ) -t A(G) - A(E V G).

Subtraction now gives the required result. Let A be the center-valued dimension function on a 8.7.33. finite von Neumann algebra R. (i) Prove that A is weak-operator continuous on the set P of all projections in R. (ii) Suppose that {E,} is a net of projections in R, and {A(&)} is weak-operator convergent to 0. Show that {E,} is strong-operator convergent to 0. [Hint. Use Exercise 8.7.3(iii).]

Solution. (i) Since A is the restriction to P of the centervalued trace T on R, it suffices to note that T is weak-operator continuous on bounded subsets of R. This follows from the ultraweak continuity of T (Theorem 8.2.8(vi)), since the utltraweak and weakoperator topologies concide on bounded subsets of R (Remark 7.4.4). (ii) Since llEall _< 1 and A(&) = T(E:E,), the stated result is a special case of Exercise 8.7.3(iii). 8.7.34. Suppose that E and F are projections in a von Neumann algebra R, and E is the (strong-operator) limit of an increasing net { E , } of projections in R. (i) Prove that the net {E, V F } converges to E V F . (ii) Prove that the net { E , A F } converges, and that lim(E, A F ) 5 E A F . Give an example in which lim(E, A F ) < E A F , and show that lim(E, A F ) = E A F if R is finite. (iii) State and prove the corresponding results for the case in which E is the limit of a decreasing net {E,} of projections in R.

EXERCISE 8.7.34

411

Solution. (i) Since the net { E , V F } increases and is bounded above by E V F , it converges to a projection G in R, and G 0), the 11 Il2-ball {A E R : IIA-Aoll2 < E } with center A0 and radius E contains the strong-operator neighbor~ . .,k)}. Hence the hood {A E R : 1lA.j - Aozjll < ~ k - (~j=/ 1,. norm topology derived from 11 112 is coarser than the strong-operator topology on R. It remains to prove that, on bounded subsets of R, the strongoperator topology is coarser than the topology associated with 11 112. This can be deduced from the result of Exercise 8.7.3(iii), but a simple direct proof is available in the present case. The set of vectors of the form A’xj, .with A‘ in R‘ and j in (1,. . . ,k}, has linear hull dense in the Hilbert space on which R acts, and so suffices to determine the strong-operator topology on a bounded subset B of R. If A0 E B,A’ E R‘,1 5 j 5 k, and E > 0, the sub-basic neighborhood {A E B : ll(A-Ao)A‘sjl( < E } of A0 (in the strong-operator topology on B ) contains the 11 112-ball {A E B : IIA - A0112 < EI[A’\I-~},since

Hence the two topologies coincide on f?.

m[77]

With the notation of Exercise 8.7.39, let 2l be a self8.7.40. adjoint subalgebra of R (not necessayily containing I ) . (i) Suppose that A , A l , A z , . . . are self-adjoint elements of R such that IlAll 5 1, A, E 2l for n = 1 , 2 , . . ., and IIA - A,ll2 -+ 0 as n + 00. Define a continuous function f : R -+ [-1,1] by f ( t ) = 2t(l -t t 2 ) - l , and note that the restriction fl[-l, 11 has a continuous inverse mapping g : [-1,1] + [-1,1]. Show that Ilf(A)- f(A,)llz + 0 as n -+ 00, and deduce that A (= g(f(A))) lies in the strongoperator closure %- of 31 in R. (ii) Deduce that the

11

Jl2-closureof U in R coincides with U-.

Solution. (i) We recall from Remark 8.5.9 that ( ( R S T ( ( 26 IIRIIIIS112IITII, for all R , S , and T , in R. For each self-adjoint T in

EXERCISE 8.7.41

417

R , / / ( It T2)-111 5 1 and Il2(1+ T2)-'Tll 5 1. From this, we have

as n

+ 00.

Since f(A), f(An) lie in the unit ball (R)1, it now follows from the result of Exercise 8.7.39 that the sequence {f(A,)} is strongoperator convergent to !(A). Since f(An) is the norm (hence, strongoperator) limit of a sequence of polynomials (without constant term) in A,, it follows that f ( A n ) E U-; so f(A) E 2l-. Since IlAll 5 1and (g o f ) ( t ) = t for all t in [-1,1], and g(0) = 0, a further application of the polynomial approximation argument just used gives

(ii) Since the topology derived from 11 112 is coarser than the strong-operator topology on R,it suffices to prove that each (1 112limit point of 2l lies in 3 - . Suppose that {An} is a sequence in 2l, A E R, and IIA - An1J2 -+ 0 as n + 00. We want to prove that A E U-. Since IIR112 = IIR*ll2 for each R in R,we have

so it suffices to consider the case in which A, A1 , A 2 , . . . are selfadjoint. Upon replacing A, A, by cA, cAn, respectively, for a suitable scalar c(# 0 ) , we may suppose also that llAll 5 1. In this case, it ~[78(Par.l)] follows from (i) that A E 2l-.

8.7.41. Suppose that G is a countably infinite discrete group in which each element, other than the unit e , has an infinite conjugacy class. As in Section 6.7, define an orthonormal basis {xg : g E G} of the Hilbert space Iz(G) and a factor LG of type 111, acting on Iz(G),for which each x g is a trace vector.

418

THE TRACE

(i) Show that G cannot be expressed as the union of a finite number of sets each of which is a right coset of some subgroup of G that has infinite index. [Hint. Suppose the contrary, and let { H I , ...,H , } be minimal set of subgroups of G, each with infinite index, such that G can be expressed as the union of a finite number of right cosets of H I , . . . ,H , . Choose some such expression for G, and by considering a right coset of H1 that does not appear in that expression, show that H1 (and hence, also, G) can be expressed as the union of a finite number of right cosets of H 2 , . . ,H , , contradicting the minimality assumption.] (ii) Suppose that h , k E G\{e}. Let

.

and note that N h is a subgroup of G with infinite index and N h , k is either empty or a right coset of N h . Deduce that if S is a finite subset of G\{e}, there is an element g of G such that g - l S g n S = 0. (iii) Let {S,} be an increasing sequence of finite subsets of G, with union G\{e}. For each positive integer n, choose an element g , of G such that gC1Sngn n S, = 0, and let U, (in LG)be the unitary operator defined by ( u n z : ) ( s= ) .(g,ls)

(9 E G ,

2

E 12(G)).

Show that lim ( U , A U ~ x , ,xh) = 0

n-+w

whenever A E Lc;, and g , h are distinct elements of G. Deduce that the sequence { U,.AU,*} is weak-operator convergent to T ( A ) I ,where T is the (unique) tracial state of Lc.

Solution. (i) With H I , . .. ,H , as in the hint, n > 1 since G is not the union of a finite number of right cosets of H I . Choose an expression for G as a union of a finite number of right cosets of H I , . . . ,H,, let H l g l , . .. ,H l g k be the right cosets of H1 that appear in this expression, and let H l g be a right coset that does not appear. Since H l g E G\ UFZlH l g j , H l g is contained in the union of the (finitely many) right cosets of H 2 , . . . ,H , that appear in the chosen expression. It follows that each of H l g l , . . ,H l g k is contained in the union of a finite number of right cosets of H 2 , . .., H , . These right cosets of H2,. . .,H , can be used to replace H l g l , . . . ,H l g k in

.

419

EXERCISE 8.7.41

the chosen expression for G . In this way, G can be expressed as the union of a finite number of right cosets of H 2 , . . . ,H , , contradicting the minimality assumption. This proves (i). (ii) The subgroup N h is the normalizer of h in G , and its right cosets are in one-to-one correspondence with the (infinitely many) conjugates of h, since gl'hgl = gF1hg2 if and only if Nhgl = N h g l . Thus N h has infinite index in G , and N h , k is either empty or a right coset of N h . With S a finite subset of G\{e}, the set { g E G : g-lSg n S #

0)

is the union U h , k E S N h , k of a finite number of right cosets of subgroups with infinite index, and is not the whole of G by (i). Thus g-lSg n S = 0 for some g in G. (iii) If A E CG,there is an element w of 12(G) such that A s = w * x for each x in /2(G). When g , h E G ,

(UnAU:xg,sh) = (AU:xg, U i x h ) = (Axg,lg,Zg,1h)

= (Axg;lg)(g;lh) = (w * x g ; I g ) ( g ; l h ) = w(g,'hg-lg,).

When g # h, and the finite set

E

> 0, we can choose m so large that S, contains

When n 2 m,

and lim (UnAU:xg,zh)= 0

n+o3

(g

# h).

Also, (UnAUZxg,zg)= w ( e ) = T ( A ) .Thus, for all g and h in G , lim (U,AU,*z,, xh) = T ( A ) ( xh). ~~,

n-+w

T H E TRACE

It follows that lim (U,AU,*u, v) = T(A)(u,v)

n+w

whenever u and v are finite linear combinations of the basis vectors x,. Since these linear combinations form a dense subspace of 3.1, the (bounded) sequence {U,AU,*) is weak-operator convergent to m[90] T(A)I. Let 3.1 be a separable infinite-dimensional Hilbert 8.7.42. space, and let po be the usual tracial weight on B('H) (see Remark 8.5.6). The purpose of this and the following three exercises is to show that B(3-1) has a semi-finite tracial weight T (neither faithful nor normal) that is not a multiple of pol even when restricted to the subset of 8(3.1)+ on which T takes finite values. Suppose that {yl, y2, y3,. . .) is an orthonormal sequence in 3.1, {A1, X2, X3, . . .} is a decreasing sequence of non-negative real numbers, and A (in f3(%)+) is defined by

Let {al,. . . ,z,} be a finite orthonormal system in 'H. Prove that (i) xr==,(Azk,.k) 5 A j = x y = l ( A ~ j~, j ) ; (ii) z ; I " , l ( A z k , ~ k2) Xj if the linear span [zl,. . . ,z,] contains yl,. . ,y,.

x7=1

xy=l

.

Solution. For each x in 3.1, AX,^) = CZl Xjl(x,yj)J2.Thus

where

Since

cj

5 l)yjll' = 1 (with equality if yj E [zl,. . .,zm]), and

421

EXERCISE 8.7.43

we have (3) M

Ccj 5 m.

(4)

j= 1

Thus m

m

m

00

j=l

k= 1

j=1 m

j=1

=

00

CXj(1

-Cj)

-

C

xjcj

j=m+l co

j=1 m

j=m+l

j=1 M ..

= Xm+l(m -

CCj)2 0. j=1

This proves the first relation in (i), and the second is apparent since (Ayj, yj) = X j . The inequality in (ii) is an immediate consequence of (1) and (3). a[29] Suppose that 3-1 is a separable infinite-dimensional 8.7.43. Hilbert space. For each positive compact linear operator A acting on 3-1, let {Xj(A)} be the sequence of non-zero eigenvalues of A, arranged in decreasing order and counted according to their multiplicities (as in Exercise 2.8.29(v)), and followed by a sequence of zeros if A has finite-diniensional range. Let

By using the results of Exercise 2.8.29 and 8.7.42, show that for all positive compact linear operators A and B , (i) sm(A) 5 sm(B)if A 5 B , (ii) s m ( A t B ) I s m ( A ) t s m ( B ) I ~ 2 m ( A B ) .

+

422

THE TRACE

Solution. Given a positive compact linear operator A acting on 1-I, it follows from Exercise 2.8.29 that there is an orthonormal sequence { y j } in 31 such that

j=1

where X j = Xj(A). From Exercise 8.7.42, m

k=l

.

for every orthonormal system (21,.. ,zm} in ‘H, with equality if zk = yk for each k = 1,... ,m. Moreover, m k=l

.

if the linear span [zl,. , ,zm] contains y1,. . . ,yn. In the arguments that follow, we apply these results to A , B , and A t B . (i) If A 5 B , we have m

m

k= 1

k=l

for each orthonormal system (21,.. . ,z m } . For an appropriate choice of this system, we obtain s m ( A ) 5 s m ( B ) . (ii) Let (21,.. . ,zm} be an orthonormal system consisting of “the 2 be an ~ orthonor) first rn eigenvectors” of A t B , and let ( ~ 1 , ... ,~ mal system whose linear span contains “the first m eigenvectors” of both A and B . Then m

s m ( A -k B ) = E ( ( At B)zk, zk) k=l m

m

k= 1

k=l

2m

2m

423

EXERCISE 8.7.44

8.7.44. Let p be a pure state of the C*-algebra 1, that vanishes on the ideal co (see Exercise 4.6.56). Define positive linear mappings C,D,S,T from I , into I,, each mapping the identity of I, onto itself, by

.} = ( 5 1 , f ( 5 l t 5 2 ) , i ( 5 1 t 5 2 D{zl, 5 2 5 3 , . *} = (51 5 1 5 2 7 5 2 9 5 3 , 5 3 *}, s { 5 1 , 5 2 , 5 3 , - .} = ( 2 2 , 5 3 7 5 4 , - * .}, T{xl,X2,53,.* = C{51,52,53,.

*

7

7

7

+

53),

-

.},

7 * *

*

a }

{ ~ l , ~ + (5 2~) , lf ( z l

t ' * * t ~ 4 ) , k ( t~*1' . + z 8 ) , . . . }

and let p be the state p o C o T of 1,. Prove that T = STD, that each of the operators C, D ,S , T maps co into co, and that C - CS and T - T S map I , into CO. Deduce that p vanishes on co and p =p o S =po

D.

Solution. For each element

of ,I

(5,)

T D { z n } = ($1 7 5 1 , f(51 t 5 2 ) , :(XI t * 5 4 ) , . - .}, and STD{z,} = T{s,}. Hence T = S T D . Moreover * *

{zn} -

S{%) = ( 5 1 - 5 2 7 5 2 - 2 3 , 5 3

C{Zn} - CS{Zn} = (21 - 5 2 , f ( 5 1

- 54,

*

- .},

- 53), i ( 5 1 - 54),

. . .}.

Thus the sequence (C - C S ) { x , } , and also its subsequence (T T S ) { x , } , converge to 0. In other words, C - CS and T - T S map I, into co. If (2,) E CO, the sequence {x,} converges to 0 and is thus Cesaro summable to 0. Thus C{Z,}, and its subsequence T{a,}, converge to 0 ; that is, C{z,}, T { s n } E CO. It is apparent that D{on},S { z n } E CO. Thus each of the operators C, D , S, T maps co into co. From the preceding two paragraphs, it follows that the operators CTD - CT (= CTD - CSTD = (C - CS)TD) and CT - C T S (= C(T - T S ) )both map 1, into CO. Since p vanishes on CO, P O

(CT - C T S ) = po(CT - CTD) = 0;

that is p = p o S = p o D . Moreover P(C0)

so p vanishes on

CO.

= p(CT(c0))G

P(C0)

= {O},

424

THE TRACE

8.7.45. With the notation of Exercise 8.7.43 and 8.7.44, let F be the set of all positive compact linear operators A acting on 3-1 for which the sequence {s,(A)/log(rn -t 1)},=1,2,... is bounded (and is therefore a positive element G ( A ) of lm). Define a mapping r : B('H)+ 3 [O,oo] by

Prove that (i) r ( a A ) = a r ( A ) when A E B(3-1)+ and a 2 0 (with the convention that 0 0;) = 0); (ii) U 3 U * = F,and G ( U A U * ) = G ( A ) , when U is a unitary operator acting on 7t and A E 7 ; (iii) r ( U A U * ) = r ( A ) when A E B(3-1)+ and U is a unitary operator acting on 3-1; (iv) if A E B(3-1)+, B E T , and A 5 B , then A E 7 [Hint. Deduce from the result of Exercise 4.6.41(iv) that A is compact.]; (v) if A , B E 7 ,then A B E T and

+-

for some Xo in the ideal co in 1,; (vi) r ( A B ) = r ( A ) r ( B ) when A , B E B(3-1)+; (vii) T is a semi-finite tracial weight on B ( W ) [Hint.Use the result of Exercise 8.7.371; (viii) T is not normal, and is not a multiple of the usual tracial weight on B(3-1)(see Remark 8.5.6) even when restricted to the subset of B(3-1)+ on which r takes finite values.

+

+

Solution. (i) When a > 0 and A E f3(3-1)+, a A E 7 if and only if A E T , and G ( a A ) = a G ( A ) when A E F. Thus r(aA) = a r ( A ) when a > 0 and A E B(3-1)+. It is apparent that this remains true (with the stated convention) when a = 0. (ii) If A E B(3-1)+ and U is a unitary operator acting on 'If, then UAU* is compact if and only if A is compact. When A is compact, its eigenvalues (and their multiplicities) are the same as those of U A U * , so s , ( A ) = s,(UAU*) for all rn = 1,2,. . . . From this, UAU* E F if and only if A E F ;and G ( U A U * ) = G ( A ) when A E 3.

425

EXERCISE 8.7.45

(iii) If A E B(1-l)+ and U is unitary, then from (iii) either A E 3, UAU* E 3 and

r ( A ) = p ( G ( A ) )= p(G(UAU*))= T ( U A U * ) , or A 4 F,UAU* 4 F l and T ( A )= r(UAU*)= 00. B , then B is compact; from (iv) If A E a(%)+,B E 3,and A I Exercise 4.6.41(iv) (with U the C*-algebra B(1-l) and C, the ideal of compact operators) it follows that A is compact. From Exercise 8.7.43(i), s m ( A )I s m ( B ) ;so

sm(A) sm(B) 0 5 log(m t 1) - log(m t 1)

*

Hence the sequence { s m ( A ) /log(m t l)}is bounded, and A E F. (v) If A , B E 3,we have 0 2

s m ( A t B ) < sm(A) t s m ( B ) log(m t 1) - log(m t 1)

by Exercise 8.7.43(ii). Hence the sequence { s m ( At B)/log(m t 1)) is bounded, A t B E 3,and

G(A

+ B ) 2 G ( A )t G ( B ) .

Again by Exercise 8.7.43(ii), s m ( A )t s m ( B )5 S ~ ~t B ( )A. Thus

G ( A )t G ( B ) I y, where Y (in lm) is the sequence {yn} given by Yn

=

s2n(A t B ) log(n t 1)

~2n(A tB)

(= log(2n t 2) - log2

> *

Accordingly,

D ( G ( A )t G ( B ) )5 DY = G ( A t B )

+ Xo,

where XO is the element DY - G ( A + B ) of E,. The proof of ( v ) will be complete when we show that X O E CO. Now XO = { z m } ,where 22n-1

X2n

= Yn -

= Yn -

s2n-1(A t B ) t B ) - s2n-1(A t B ) - ~2n(A log 2n log(n t 1) log% ' ~2n(t A B ) - s2n(A t B ) - ~ 2 n ( A t B) log(2n t 1)

log(n t 1)

log(2n

+ 1)

*

426

THE TRACE

-

1 hn(A t B) = s2n(A B){log(n t 1) log(2n t 1)) log(2n t 1) IJAt B(I < ~ 2 n ( At B)(log(2n t 1) - log(n t 1)) t l)log(2n t 1) log(2n t 1) < s2n(A f B ) log2 t llA+Bll + o - log(2n f 1) log(?& t 1) log(2n t 1) 1

as n -+ 00, since the sequence {sm(A t B)/log(m t 1)) is bounded. This shows that XOE co. (vi) Suppose that A, B E a(%)+.If A t B 6 F,then A, B E F by (iv), and we can choose XO in co so that the conclusion of (v) is satisfied. Since the state p of ,I vanishes on co, and p = p o D,it follows from the inequalities in (v) that

r(A)

+ r ( B )= P(G(A) t G ( B ) ) = P(D(G(A) t G(B)) - X o ) I p(G(A t B ) ) = r ( A t B ) I P(G(A) t G(B)) = r ( A ) t q);

so r ( A + B ) = r ( A ) + r ( B ) . If, however, A + B

4 7 ,then at least one

of A, B is not in F,by (v); so in this case, T ( At B) = T ( A )t r ( B ) = 00. (vii) From (i) and (vi), r is a weight on D('H). F'rom (iii) and Exercise 8.7.37, it is a tracial weight. In order to prove that T is semi-finite, we have to show that the linear span M , of the set T ( = {A E B('H)+ : r(A) < cm} = F7)is weak-operator dense in B('H). To this end, it suffices to prove that .F contains each element A of a(%)+that has finite-dimensional range. Such an operator A is compact and has only finitely many nonzero eigenvalues, so the sequence {sm(A)} is ultimately constant, G(A) (= {s,(A)/log(m f 1)))lies in C O ( ~lm), and A E T (with r ( A ) = p(G(A)) = 0). (viii) Let { E m }be an orthogonal sequence of projections, each having one-dimensional range in %, and let 00

A = x { l o g ( m t 1) -logm}E,. m=l

427

EXERCISE 8.7.46

Then the operator A is a positive compact linear operator, Xm(A) = log(m+l)-logm, s m ( A ) = log(rn+l), and A E 3;moreover, G ( A ) is the sequence { l , l , l ,...}, and T ( A ) = p ( G ( A ) ) = 1. However, T ( A , ) = 0 for n = 1,2,. . ., and A , A , where

From this, T is not normal (see the discussion preceding Proposition 8.5.1). It is not a multiple of the usual tracial weight po on B('H), even when restricted to 3, since po(A,) = log(n 1) and po(A) = 00. .[29]

+

Suppose that R is a finite factor acting on a Hilbert 8.7.46. space 'H, and that x and u are generating unit trace vectors for R. Let cp and $ be the * anti-isomorphisms from R onto R' determined (see Theorem 7.2.15) by the conditions

cp(A)x = A X ,

+ ( A ) u = AU

( A E R).

Let do be a maximal abelian von Neumann subalgebra of 72, and let d, and d$ be the abelian von Neumann algebras generated by do U t p ( d 0 ) and do U $(do),respectively. Prove that (i) there is a unitary operator U' in R' such that U'x = u; (ii) $ ( A ) = U'tp(A)U'* for each A in R; (iii) d+= U'd,U'*; (iv) d+ is a maximal abelian subalgebra of B(7-l) if and only if the same is true of d,. [In Exercise 8.7.48 and 8.7.49, we give examples to show that d$ can be, but need not be, a maxiaml abelian subalgebra of L?('H), in the circumstances set out in the present exercise.] Solution. (i) Since R has a unique tracial state, by Proposition 8.5.3, w,lR = wulR; SO

I(Au1I2 = w,(A*A) = w,(A*A) = llAs112 for each A in R. From this, and since x and u are generating vectors for R, there is a unitary operator U' acting on H such that U'Ax = Au for each A in R (in particular, U'x = u). Since U'ABx = ABu =

428

THE TRACE

AU'Bx for all A and B in R, it follows that U'A = AU' when A E R; so U' E 2'. (ii) When A E R , $ ( A ) u = Au =: AU'X = U'Ax = U'cp(A)a: = U'cp(A)U'*u. Since $ ( A ) , U'cp(A)U'* E R' and u is a separating vector for R',it follows that $ ( A ) = U'cp( A)U'* . (iii) Since U' E R' and do R, U'doU'" = do. From this, together with (ii), we have U'(d0 U cp(Ao))U'* = A0 U $(do); so U'dvU'* = d+. (iv) The stated result is an immediate consequence of (iii). 8.7.47. With the notation of Section 8.6, A n operator-theoretic construction, assurrie that conditions (a) and (b) are satisfied and that G acts ergodically on A. Suppose that uo is a unit vector in H , and U(g)uo = uo for each g in G . (i) Prove that uo is a separating and generating vector for A. in Ic (= CgEG$'Hg), where (ii) Let u be the vector CgEG$xg ze = uo and zg = 0 when g E G\{e}. Show that u is a generating trace vector for R. (iii) Let $ be the * anti-isomorphism from R onto R' determined (see Theorem 7.2.1Fj) by the condition

$(T)u = T u (T E R). Given that T (in R) has matrix [U(pq-')A(pq-l)],where A ( g ) E d for each g in G , find the matrix of $(T). (iv) Show that @ ( A )has matrix [6,,,A], and $ ( @ ( A ) )has matrix [G,,,U(p)AU(p)*]for each A in A. (v) Suppose that S E B ( K ) and S has matrix [S,,,]. Show that S E @(A)' if and only if Sp,s E d for all p and q in G , and that S E +(@(A))'if and only if U(p)*S,,,U(q) E A for all p and q in G. (vi) Deduce from (v) that the von Neumann algebra d+ generated by @(A)(a maximal abelian subalgebra do of R) and $(@(A)) (the maximal abelian subalgebra $(do)of R')is the maximal abelian $A of B(Ic). subalgebra CgEG Solution. (i) The projection E with range [Auo] lies in A'(= A). The projection U(g)EU(g)*has range U ( g ) E ( X ) ,and

W l ) E ( W = [U(g)duol = [U(g)dU(g)*uol= [Auol = E ( W ;

EXERCISE 8.7.47

429

so U ( g ) E U ( g ) *= E for each g in G . Since E E A, E # 0, and G acts ergodically on A, it follows that E = I . Hence uo is generating for A, and is also separating for A( = A’). (ii) Let po(A) = (Au0,uo) for A in A, so that po is a faithful normal state of d,and

for all A in d and g in G . From the proof of Lemma 8.6.3, there is a faithful normal tracial state p of R, defined by the equation

where T (in R) has matrix [T(p,q ) ] . Since p ( T ) = ( T u ,u),where u is the vector defined in (ii), it follows that u is a (separating) trace vector for R. Suppose that T’ E R‘ and T’u = 0. The matrix of T’ has the form [U(p)A‘(q-’p)U(p)*],where A’(g) E A for each g in G . Thus T‘u is the element CgEG $yg of X: (= CgEG $‘Ifg)given by yg = U(g)A’(g)U(g)*uo(= U(g)A’(g)uo).Since T’u = 0, each yg is 0; hence A’(g)uo = 0 for each g in G , and A‘(g) = 0 since uo is separating for A. It follows that T’ = 0. This shows that u is a separating vector for R’, and is therefore a generating vector for R. (iii) If T (in R ) has matrix [U(pq-’)A(pq-l)J,then we have that T u = CgEG $zg where

Now $ ( T ) (in 72’) has matrix of the form [ U ( p ) A ’ ( q - ’ p ) U ( p ) * ] ,where A’(g) E A for each g in G . Since $ ( T ) u = T u = CgEG $zg,we have

, the operSince uo is a separating vector for A (= U ( g ) d U ( g ) * )and ators A ( g ) , A’(g) E A, it now follows that A’(g) = A ( g ) . Thus the matrix of + ( T )is [ U ( p ) A ( q - ’ p ) U ( p ) * ] . $ A (E R ) , and @ ( A )has ma(iv) When A E d,@ ( A )= CgEG trix [&,,,A].This matrix can be written in the form [U(pq-’)A(pq-l], where A ( e ) = A and A ( g ) = 0 when g E G\{e}. From (iii), + ( @ ( A ) ) has matrix [ U ( p ) A ( q - ’ p ) U ( p ) * ]and , this is [Sp,,U(p)AU(p)*].

430

THE TRACE

(v) Simple matrix calculations show that, for each A in A, S commutes with @ ( A )if and only if S,,,A = AS,,, for all p , q in G. Thus S commutes with @ ( A )if and only if S,,, E A‘ (= A ) for all p , q in G. When A E A, $ ( @ ( A ) ) has matrix [G,,,U(p)AU(p)*] by (iv), and S commutes with $ ( @ ( A ) )if and only if ( P , q E G);

S , , , a ? ) A U ( q ) * = U(p)AU(p)*Sp,,q and this occurs if and only if

( P t q E GI.

U(P)*S,,,U(Q)A= A W * S , , , U ( q )

Hence S commutes with $ ( @ ( A ) )if and only if U(p)*S,,,U(q) E A’ (= A ) for all p , q in G. $A (vi) From the double commutant theorem, and since CgEG is a maximal abelian subalgebra of f ? ( K ) ,it suffices to show that

@(A)’n +(@(A))’=

(1)

$A. g€G

From (v), and since U ( g ) A U ( g ) * = A for each g in G, it is apparent that

@(A)’n $(@(A))’2

@A. g€G

It remains to establish the reverse inclusion. Suppose that S E @(d)’n$(@(d))‘. From (v), the matrix [SP,,] of S satisfies

s,

E A,

U(P)*sP,,U(q) E A

(P,!?E GI.

Thus

Sp,qE d n U ( p ) A U ( q ) *= d n U ( p q - l ) U ( q ) A U ( q ) * = dnU(pq-’)A. Since An U(g)A = (0) when g E G\(e}, by condition (b) of Section 8.6, it now follows that S,,, = 0 when p # Q. From this, and since SP,,E A, we have S=

$Sp,, E

$A.

EXERCISE 8.7.48

431

8.7.48. By using Exercise 8.7.47 and Example 8.6.12, give an example of a factor R of type 111 acting on a Hilbert space H, a generating unit trace vector u for R, and a maximal abelian von Neumann subalgebra do of R,such that the von Neumann algebra d$ occurring in Exercise 8.7.46 is a maximal abelian subalgebra of

w-0Solution. In Example 8.6.12, d is the multiplication algebra of the space L2 associated with Lebesgue measure m on the unit interval S(= [0, l)),and G consists of rational translations (modulo 1) of S. Each g in G leaves m invariant, and gives rise to a corresponding “translation unitary operator” U, acting on L2. The factor R of type 111 is obtained from the maximal abelian algebra d acting on L2 and the unitary representation g -+ U, of G on L2, by the process described in Section 8.6, An operutor-theoretic construction. Let uo be the unit vector in L2 defined by uo(s) = 1 for all s in S, and note that U,uo = uo for each g in G . The results of Exercise 8.7.47 now show that R has a maximal abelian von Neumann subalgebra @ ( A ) ( =do)and a generating unit trace vector u , such that the von Neumann algebra A+ generated by do U $(do)is maximal abelian in the algebra of all bounded operators (where 11, is the * anti-isomorphism from R onto R‘ associated with u). With the notation of Section 6.7, let G be the free 8.7.49. group on two generators, a and b. Recall that LG (= { L , : y E /z(G), L , E B(12(G))})is a factor of type 111 with commutant RG (= { R , : z E l2(G),R, E B ( l z ( G ) ) } ) and , 2, is a generating unit trace vector for CG.Let ?b, be the * anti-isomorphism from LG onto RG determined (see Theorem 7.2.15) by the condition $ ( T ) z e = Tx,. Let do be the von Neumann algebra generated by L,,, and recall (Exercise 6.9.42) that do is a maximal abelian subalgebra of

LG. (i) Show that $ ( L , ) = R, when y E l2(G) and L, E B(l2(G)). (ii) Show that $(do) is the von Neumann subalgebra of RG generated by R,, . (iii) Suppose that S E B(l2(G)),and let [S(g,h)]be the matrix of S, defined by

Show that S commutes with L,, if and only if S ( g , h ) = S(ag,ah)

432

THE TRACE

for all g and h in G, and that S commutes with R,, if and only if S ( g , h ) = S(ga,ha) for all g and h in G . (iv) Show that there is a partial isometry SO acting on /2(G), with matrix determined by the condition that So(g,h) is 1 if there exist integers m and n such that g = anban and h = ambuban, and S ( g , h ) is 0 otherwise. Prove that SO commutes with L,, and R E , . Deduce that the von Neumann algebra dq generated by do U $(do) is not maximal abelian in 17(l2(G)).

Solution. (i) Suppose that y E Iz(G) and L , E f?(lz(G)).Then L , E LG and $(L,) E RG;so $(L,) = R , for some z in lz(G). Moreover,

Thus $(L,) = R,. (ii) Let Go be the cyclic subgroup of G generated by a. From Exercise 6,9.41(ii)9

do = { L ,

E LG : x(g) = 0 when g

4 Go};

so

$(do)= { R , E RG :

z(g) = 0 when g

4 GO}.

This, together with the analogue of Exercise 6.9.41(ii) for the right translation algebra RG,shows that $(do)is the von Neumann algebra generated by R,, . There is an alternative proof of (ii), based on the fact that $(Lza) = R,,, together with the Kaplansky density theorem and the weak-operator continuity (on bounded sets) of $ and For this, note that the argument used in proving the weak-operator continuity (on bounded sets) of * isomorphisms between von Neumann algebras (Corollary 7.1.16) applies also to * anti-isomorphisms. (iii) S commutes with Lx, if and only if S = L;,SL,,; equivalently, if and only if

+-'.

Since L E , x h = x a h and Lxazg= zag,the last condition can be written in the form

EXERCISE 8.7.50

433

A similar argument shows that S commutes with Rx, if and only if S ( W , ha) = S(g, h )

(9, h E GI.

(iv) We can define disjoint subsets X and Y of G, and a one-toone mapping f from X onto Y , by

X = {urnbun : m , n integers}, Y = {umbubun : m,n integers}, f ( u m b u n ) = umbubun, Let E and F (in f?(/2(G)))be the projections of multiplication by the characteristic functions of X and Y , respectively. Given x in I2(G),we can define a vector Sox in the range of E by

sox =

c

x(f(s))xg,

SEX

and

F'rom this, and since it is easily verified that the range of SO is the whole of the range of E , it follows that SO is a partial isometry with F and E as (mutually orthogonal) initial and final projections, respectively. Since (Soxh,xY)= 1 if g E X and h = f(g),

= 0 otherwise. SO has the matrix described in the statement of (iv). From the criteria set out in (iii), SOcommutes with Lxa and Rza (and, hence, with their inverses LEa, R;J. It follows that SO commutes with A0 and $(do);so SO E dk. Since S;So # SoS;, A; is not abelian, hence dG # A $ , and A$ is not maximal abelian in D(12(G)).

8.7.50. With the notation of Section 8.6, An opemtor-theowtic construction, assume that conditions (a) and (b) of that section are satisfied, and let W be the unitary group of the maximal abelian subalgebra @(A) of R.Prove that (i) for each T in R, the weak-operator closed convex hull c o @ p ( ~ ) ( Tof) - the set (WTW* : W E W } meets @ ( A )in a single point [Hint. See the proof of Lemma 8.6.2.1; (ii) there is a unique ultraweakly continuous conditional expectation from R onto @(d),and D! is faithful.

434

THE TRACE

Solution. (i) Since T E R, @ ( A ) R, and R is weak-operator closed, it follows that C O ~ ( A ) ( T )C_- R. From Corollary 8.3.12, c o a ( ~ ) ( T ) -meets @(A)’. Thus c o g ( ~ ) ( T ) -meets R n @(A)‘ (= @(A))* Since T E R, T has a matrix of the form [17(pq-~>A(pq”)], where A(g) E A for each g in G. Each W in W has a matrix of the form [b,,,Wo], where WOis a unitary operator in A. Thus WTW* has matrix [WoU(pq-’)A(pq-l)W,’]; since A is abelian, each diagonal entry in the matrix of WTW”is A ( e ) . From this, it follows that each operator in C O ~ ( A ) ( T has ) - a matrix with A ( e ) at each , that diagonal position. In particular, if SOE C O ~ ( A ) ( Tn) -@ ( A )so SO(= @ ( A o ) )has matrix [Sp,,A0] for some A0 in A, then the only ) just possible value of A0 is A ( e ) . Hence c o a ( ~ ) ( T ) -meets @ ( A at one point, the operator @ ( A ( e ) )with matrix [6,,,A(e)]. (ii) Given T in R, [U(pq-’)A(pq-l)] is the form of the matrix for T , where A ( g ) E A for each g in G. Let P(T) be @ ( A ( e ) ) ;that is, Q(T)is the element of @ ( A )that has matrix [b,,,A(e)]. Simple calculations show that Q is a conditional expectation from R onto @ ( A )It . was noted in the proof of Lemma 8.6.2 that A ( e ) # 0 when T > 0, so Q is faithful. Since the matrix of T has A ( e ) at the ( e , e ) entry, and the * isomorphism Qr :A --t @ ( A is ) ultraweakly continuous (Remark 7.4.4),it follows that Q is ultraweakly continuous. Suppose that 9 0 is an ultraweakly continuous conditional expectation from R onto @(A).Given T in R and W in W . we have

Q‘o(WTW*)= WQ‘o(T)W* = @@), since W,Po(T) E @ ( A )(an abelian algebra). From this, together with the linearity and ultraweak continuity of g o , it follows that Qo(T) = Po(S) for each S in CO~(A)(T)-.We can take, for S, the unique element SO of c o a ( ~ ) ( T ) -n @(A),to obtain Qo(T) = Q o ( S 0 ) = SO. The same argument can be applied to Q , so Po(T) = Q ( T )and 80 = Q . Thus Q is the only ultraweakly continuous conditional expectation from R onto @(A). 8.7.51. Suppose that R is a von Neumann algebra with center C, H E R, and d = inf{llH - Cll : C E C}. Show that the equation

b ( A ) = H A - AH

( AE R)

defines a derivation 6 of R. (The term “derivation” was defined in Exercise 4.6.65. When 6 is obtained in the above manner from an

EXERCISE 8.7.52

435

element H of R, it is described as an inner derivation of R . We shall see, in Exercise 8.7.55, that every derivation of a von Neumann algebra is inner.) Prove also that d 5 ll6ll 5 2d. [Hint. Consider 6 ( U ) U * , where U is a unitary element of R.]

Sobution. When A , B E R,we have A6(B) + S(A)B = A ( H B - B H )

-+ ( H A -

AH)B = H A B - ABH = d ( A B ) .

From this, the (linear) mapping 6 : R When C E C,

-+

R is a derivation of R.

Thus IlS(( I 2 ( ( H- C ( (for , each C in C; so ll6ll 5 2d. For each U in the unitary group U of R,

Hence ll6ll 2 I(SI(for each element S of the norm-closed convex hull S of the set { H - UHU' : U E U}.Now

s = { H - Ii : K

E coa(H)=},

and CO.R(H)=contains an element COof C, by the Dixmier approximation theorem. Thus

8.7.52. Let 6 be a derivation of a von Neumann algebra R, and recall from Exercises 4.6.65 and 7.6.15 that 6 is bounded, and is weak-operator continuous on bounded subsets of R. (i) Suppose that P is a projection in the center C of R. By using the relation 6 ( P ) = 6 ( P 2 ) ,show that 6 ( P ) = 0. Deduce that the restriction 61RP is a derivation d p of the von Neumann algebra RP. Prove also that if the derivations 6 p of RP and 61-p of R ( I - P) are both inner, then 6 is inner.

436

T H E TRACE

(ii) Let E be a projection in R. Show that the equation

S,y(EAE) = EG(EAE)E

( A E R)

defines a derivation SE of the von Neumann algebra E R E , and llSEll

5 611*

(iii) Suppose that { E , : a E A} is an increasing net of projections in R,and VaEAE4 = I. Suppose also that, for each a in A, there is an element H , of &RE, such that \[Hall5 llSll and

SE,(A) = H,A - A H ,

( A E E,RE,).

Show that the net { H a } has a subnet that is weak-operator convergent t o an element H of R. Prove also that

6 ( A ) = H A - AH

( A E R).

[Hint. Consider first the case in which A E EbREb for some b in Pa.] Solution. (i) Since S ( P ) = S ( P 2 )= P S ( P ) -t- S(P)P = 2PS(P), we have P S ( P ) = 2PS(P);so P b ( P ) = 0, and S(P) = 0. For each A in R, S(AP) = AS(P) t 6(A)P= S(A)P. Thus S maps R P into R P , and from this it is apparent that 61RP is a derivation S p of R P . If the derivations 6 p and 61-p are both inner, we can choose H in R P and K in R(1- P ) so that

S(AP) = H A P - APH = H A - A H , S(A(1 - P ) ) = K A ( I - P ) - A ( I - P ) K = K A - A K , for all A in R. Thus

6 ( A ) = ( H t K ) A - A(H and S is inner.

+K)

( A E R),

437

EXERCISE 8.7.53

(ii) It is clear that S E , as defined in the statement of (ii), is a bounded linear mapping from E R E into E R E , with llb~ll5 IlSll. When A , B E R,we have

S(EAE * E B E ) = EAES(EBE) + S(EAE)EBE. Upon multiplying by E on both left and right, throughout this last equation, we obtain

-

+

GE(EAE E B E ) = EAEGE(EBE) GE(EAE)EBE. Thus 6~ is a derivation of E R E . (iii) The net { H a } has a weak-operator convergent subnet, with limit H in R satisfying llHll 5 IlS((, since the ball (R)llall is weakoperator compact. Suppose that A € R and b € A. Whenever a € A and a 1 b, we have E EaREa and

By taking limits over the appropriate subnet of { H a } and the corresponding subnet of { E a } ,and noting that the latter subnet is strongoperator convergent to I , we obtain

Since 6 is weak-operator continuous on bounded subsets of -+ A , it now follows that 6 ( A ) = H A - A H .

R,and

EbAEb

Suppose that 6 is a derivation of a countably decomposable finite von Neumann algebra R. For each U in the unitary group U of R , define an affine mapping uu : R R by

8.7.53.

--f

(i) Prove that auv = au o uv for d1 U and V in U. (ii) Show that S is inner if and only if there is an element H of R such that a u ( H ) = H for all U in U. (iii) Let K be the weak-operator closed convex hull of the set {S(U)U* : U E U}. Show that K is weak-operator compact and au(X:)E X: for all U in U.

438

THE TRACE

(iv) Let F be the family of all non-empty weak-operator compact convex subsets of Ic that are invariant under each of the mappings au (V E U). Show that F,partially ordered by inclusion, has a minimal element KO. (v) Show that 12 has a faithful normal tracial state p. (vi) Let M = sup(llKll2 :K E KO},where 11 112 is the norm on 7Z defined by llR112 = [p(R*R)l1I2. Given Ir'l and K2 in KO,show that Il.V(&)

- av(h;)ll2 = ll& - K2112

(U E w,

and that

for each K in the weak-operator closed convex hull Ic1 of the set (av(g(K1 t K2)) : u E U). (vii) &om the results of (vi) and the minimality property of KO, deduce that KO consists of a single element H of K . Prove that IlHll I ll4l and

6 ( R )= HR - RH

(RE R).

Solution. (i) When U,V E U and R E R,

a v v ( R ) = S(UV)V*U*t UVRV*U* = US(V)V*U* S(U)U* t UVRV*U* = 6(U)U* 4- U[S(V)V* VRV*]U*

+

+

= av(av(-l[l)). (ii) 6 is inner if and only if there is an element H of R sucxti hat (1)

( R E a).

S(R) = H R - RH

Since R is the linear span of 2.4, (1) is equivalent to

6(U)= HU and hence to 6(U)U* = H can be written in the form (2)

- UH

(U E U),

- UHU* (U

av(H)= H

E

2.4). This last condition

(U E 24).

439

EXERCISE 8.7.53

Thus (1) and (2) are equivalent, and 6 is inner if and only if (2) is satisfied for some H in R. (iii) The ball ( R ) l l b l l is convex and weak-operator compact, and contains the set {6(U)U* : U E U } (= S); so it contains the weakoperator closed convex hull K of S , and K is weak-operator compact. Since 6 ( U ) U * = au(0) and auv = au o a v , the set S is invariant under each of the mappings au; the same is true of K , since au is affine and continuous. (iv) The family T is partially ordered by inclusion. If a subfamily TOof .F is totally order by inclusion, it has the finite intersection property (since each finite intersection of sets in F ,coincides with one of those sets, and is therefore non-empty). From this, and since K is compact, the intersection K1 of all sets in FO is not empty. It follows easily that K1 E F, and that K1 is the greatest lower bound of F-,in 3. Since each totally ordered subfamily 30of .F has a greatest lower bound in 3, Zorn's lemma implies that 3 has a minimal element KO. (v) From Exercise 7.6.46(ii), R has a faithful normal state w . Thus w o T is a faithful normal tracial state of R, where 7 denotes the center-valued trace on R. . (vi) For each Ii in KO ( c K: G (R)lpll), (lIi-112=

so M

[p(K*li)]'/2

5 (lI

K2

in KO and U in

U,

- .u(I(2)112 = IIU(& - K2)U*l12 = ll& - K2ll2.

Since the norm

11

112

on R is derived from an inner product,

+

( ( a u ( W .u(Kz)Ilq t (lau(K1) - au(h'2)ll; = 211au(I~1)(1,2 t 2llau(I~2)(1;I 4M2.

From this and since au is f i n e ,

lIau(+(K1 t Ii-2))11; = I l + a u ( I h ) t qau(lr2)ll; 5 M2 - + l l U U ( K 1 ) - uu(Ii2)ll; = M2 - flllil - Ii'211;. Note also that i(K1 t I i 2 ) E K O ,and thus au(i(Ir'1t K2)) E KOfor each U in U.

440

T H E TRACE

Let

So = {R E

R

: I J R ( J 25

[M2 - tllIi.1 - 1121121112.

Thus So is a convex subset of R,and the preceding paragraph shows that (U E aU(+(& t A',)) E So n K o

u).

We assert also that SO is weak-operator closed. Indeed, since p is normal, there is a sequence ~ 1 , x 2 , ~ 3 , .. .of vectors such that C ll~j11~ < 00 and p(R) = C ( R ~ j , s for j ) each R in R. Hence

and n

So = { R E

R

:

C J J R z 5~ JM2~ ~-

fllli.1

- 11211; ( n = 1 , 2 . . .)}.

j=1

From this, SO is strong-operator closed, and is therefore weakoperator closed by Theorem 5.1.2. It results from the preceding paragraph that the weak-operator closed convex hull K 1 of the set {UU($(k'l

is a subset of So n KO. Thus subset of KO,and (IK((2

t li.2)) K1

:

u E U)

is a weak-operator compact convex

< [ M 2 - tllA-1 - Ii.21I2]2 112

(K E

Kl).

(vii) Since auv = a U o a V and each aU is affine and weak-operator continuous, the set K 1 is invariant under au for each U in U ;so IC1 E F. From our assumption that KO is minimal in F,and since K 1 E KO,it now follows that K I = KO.Thus

M = sup{llIi.112 : K E KO} = sup{l/llr'l(2 : K E K,}

5 [M2 - fllli.1 - K211;]1/2,

= 0, K1 = K 2 . We have now shown that any two elements K 1 and K 2 of K O are equal, so KO consists of a single element H of K. Thus llHll 5 ))6)), H satisfies (2) since au(K0) 2 KO, and hence H satisfies (1).

and

( ( K l- K 2 ( 1 2

441

EXERCISE 8.7.54

Suppose that S is a derivation of a countably decom8.7.54. posable type 111 von Neumann algebra R. Let C and U denote the center and unitary group, respectively, of R, and note that the results of Exercise 8.7.53(i)-(iv) remain valid in the present case. (i) Show that the difference set

is convex, weak-operator compact, and invariant under each of the mappings R + U R U * R -+ R, where U E U. (ii) Suppose that K O does not consist of a single point. By use of (i) and Proposition 8.3.9, prove that there exsist K1, I i 2 in KO and C in C such that l i l - X 2 = C # 0. Deduce that acl(li1) -

au(K2) = c

( U E U).

Let w be a weak-operator continuous linear functional on R such that Rew(C) = 1, and let M = sup{Rew(li) : K E K O } . Prove that Rew(Ii) I M - 1 Ii E KO,

K in the weak-operator closed convex hull K1 of the set (au(Ii2): u E U}.

for each

(iii) From the results of (ii) and the minimality property of K O , deduce that KO consists of a single element H of K . Prove that IIHII I IlSll and

S(R) = H R - RH

( R E R).

Solution. The arguments used in the solution to Exercise 8.7.53 show that assertions (i)-(iv) listed in that exercise remain valid in the present case. We now turn to the additional assertions of the present exercise. (i) The set KO- KO is convex since KO is convex, and is weakoperator compact since it is the image of the compact set KO x KO under the continuous mapping

442

THE TRACE

Suppose that H E K O - KO,and choose h'l,K2 in KO so that H = K1 - K 2 . Then UHU* = U ( K 1 - Ii-z)U* = [S(U)U*4- U K 1 U * ]- [6(U)U*f U K 2 U * ]

= ar/(li-1)- a v ( K 2 ) E KO - KO ( U E U), since au(K0) C KO. Thus KO - KO is invariant under the mapping R + URU* : R + 2,for each U in U. (ii) If KO does not consist of a single point, then KO- K O contains a non-zero element H of R. By (i), C O ~ ( H ) 'C KO - KO, and by Proposition 8.3.9, c o ~ ( H ) =contains a non-zero element C of C. Hence C E KO - K O ,and Ir'l - Ir'2 = c # 0

KO. For each U in U , au(K1) - au(K2) = U ( K 1 - Ir'2)U* = ucu* = c.

for suitable

K 1

and IC2 in

Since C # 0, there is a weak-operator continuous linear functional w on R such that Rew(C) = 1. Since K O is weak-operator compact M ( = sup{Rew(K) : K E KO})< 00. For each U in U ,a u ( K 2 ) E K O and R e w ( a ~ ( K 2 ) )= Rew(au(K,)) - Rew(C) 5 M

- 1.

Thus Ir' E KO and Rew(K) 5 M - 1 for all Ir' in the weak-operator closed convex hull K1 of the set ( a u ( K 2 ) : U E U}. (iii) We continue with the assumption that KO does not consist of a single point, and follow up the conclusions obtained from this assumption in (ii). Since auv = aU o av and aU is affine and weakoperator continuous, K1 is invariant under au for each U in U ;so K1 E T . From (ii), K1 5 KO,and K1 # K O since sup{Rew(K) : K E K1} 5 M - 1 < M = sup{Rew(K) : K E K O } . This contradicts our assumption that KO is minimal in T . It follows that KO consists of a single element H of K . Just as in Exercise 8.7.53, IlHll 5 ll6ll and 6 ( R ) =H R - R H (RER). H

EXERCISE 8.7.55

443

8.7.55. (i) Suppose that 6 is a derivation of a von Neumann algebra R . By using the results of the three preceding exercises, show that 6 is inner. (ii) Suppose that M is a (?-algebra acting on a Hilbert space 3, and 6 is a derivation of 3. By using (i) and Exercise 7.6.15, show that there is an element H of 2L- such that

6 ( A ) = H A - AH

( A E U).

Solution. (i) Let P be the largest projection in R such that ( P = 0, or) the von Neumann algebra RP is type 111. Then R ( I - P) is semi-finite, and by Exercise 8.7.52(i) it suffices to prove that the von Neumann algebras RP and R ( I - P ) have only inner derivations. We may assume henceforth that R is either type I11 or semi-finite. Consider first the case in which R is type 111, and let & be the family of all non-zero countably decomposable projections in R. Then El V Ez E € whenever E l , E2 E I (see, for example, Exercise 5.7.45). From this, and since I is the sum of an orthogonal family of cyclic projections in 72,it follows that V { E : E E €} = 1. We now adopt the notation, and apply the results, of Exercise 8.7.52(ii) and (iii), taking for the net { E a } the set & (indexed by itself, with its natural ordering). For each E in I ,E R E is a countably decomposable type I11 von Neumann algebra (from Exercise 6.9.16(vi)), and the derivation S E of E R E satisfies l l S ~ l l5 IlSll. By Exercise 8.7.54(iii) there is an element H E of E R E such that

It now follows from Exercise 8.7.52(iii) that S is inner. Now consider the case in which R is semi-finite, and let & be the family of all finite countably decomposable projections in R. From Theorem 6.3.8 and Exericse 5.7.45, ElVEz E & whenever E l , E2 E &. From this, since I is the sum of an orthogonal family of finite projections in R (see, for example, Exercise 6.9.12), and each member of that family can be expressed as a sum of (necessarily finite) cyclic projections in R, it follows that V { E : E E 8) = I . The proof that 6 is inner now follows the same pattern as the argument used in the type I11 case, except that we appeal to Exercises 8.7.53(vii) and 6.9.15(iii) in place of 8.7.54(iii) and 6.9.16(vi).

444

T H E TRACE

(c

(ii) From Exercise 7.6.15, the derivation 6 of U B(3.t)) extends to a derivation 8 of U-. By (i), there exists H in U- such that

8 ( A ) = H A - AH ( A E U-). In particular, therefore, 6 ( A ) = H A - AH when A E U. [ 11,51,60,64,95] 8.7.56. Let, R be a von Neumann algebra of type In with n finite, C be the center of R,and q be a center state of R. Let { E j k } be a self-adjoint system of n x n matrix units for R,and Akj be q(Ejk). Let A be the element AjkEjk of R, so that A corresponds to the matrix [Ajk] under the * isomorphism of Theorem 6.6.5 between R and n @I C. (i) Prove tha,t q ( B ) = n7(AB) for each B in R, where 7 is the center-valued trac,e on R, and deduce that 7 is ultraweakly continuous. (ii) Show that A is self-adjoint and C;', Ajj = I . (iii) By expressing A as A+ - A - and considering q ( A - ) , prove that A 2 0. (iv) Prove that there are n equivalent projections G I , .. . ,G , abelian in R wit:h sum I , and n positive operators C 1 , . . . ,C, in C with sum I , such that n

for each B in R, where q j ( B )is the (unique) element of C such that G j B G j = q j ( B ) G j . [Hint. Use Exercises 6.9.23 and 6.9.35, together with the * isomorphism between R and n 8 C . I

Solution. (i) From Example 8.1.3, ~ ( E j j=) n-'I and T(Ejk) = 0 when j

# k. Since

n

we have n

p= I n

p= 1

n

445

EXERCISE 8.7.56

Since q is a center state, for each C in C

q(CEjk) = C q ( E j k ) = n C r ( A E j k ) = n r ( A C E j k ) . By using the * isomorphism between R and n @ C , it follows that each element B of R has the form Cj”,k=l C j k E j k , where Cjk E C. From this, and since q and r are linear, q( B ) = nr( A B ) for each B in R. The ultraweak continuity of T (Theorem 8.2.8(vi)) entails the same continuity for q. (ii) From (i),

j,k=l

j=1

Since q is a positive linear mapping, q is hermitian; so

Akj = q ( E j k ) = q(Ekj)* = Aj*k, the matrix [Ajk] is a self-adjoint element of n @ C, and A is selfadjoi nt . (iii) We can express A in the form A+ - A - , as in Proposition 4.2.3(iii), and

0 5 q ( A - ) = n r ( A A - ) = - ~ T ( ( A - ) 5~ 0. ) Thus T ( ( A - ) ~=) 0, ( A - ) 2 = 0 since T is faithful, A - = 0, and A = A+ 2 0. (iv) We apply Exercise 6.9.35, with S consisting of the single positive element [Ajk] of n @ C ; note that the assumption that R is countably decomposable is not needed in the present case, since we can cite Exercise 6.9.23 in place of 6.9.34 in the solution to Exercise 6.9.35. It follows that there is a unitary element UO of n @ C such that Uo[Ajk]U,*is a diagonal matrix, with a positive element Cj of C in the j t h diagonal position, for j in (1,. . . ,n}. Hence there is a unitary operator U in R such that UAU’ = Cj”=,CjEjj, and n

446

THE TRACE

where G j = U*EjjU. Now GI,. . .,G , are equivalent abelian projections in R with sum I (so T(Gj) = n-'I), and each has central carrier I. It follows from Proposition 6.4.2 that G j R G j = CGj, and from Proposition 5.5.5 that the mapping A' -+ A'Gj is a * isomorphism from R' onto R'Gj; upon restriction of this last mapping, we obtain a * isomorphism C + C G j from C onto CGj. From these assertions, if B E R there is a unique element qj(B) of C such that G j B G j = qj(B)Gj, for j in (1,. . . ,n}. Moreover, n

q ( B ) = ~ T ( A B=)72

C T(CjGjB) j=1

n

=n

C CjT(GjB) j=1

n

= 12

C CjT(GjBGj) j=1

n

n

n

j=1

j=1

When B = I, qj(B) = I for each j , and we obtain n

I = q(1)= c c j q j ( I ) = j=1

ccj. n

rn

j=1

Let, R be a finite von Neumann algebra acting on a Hilbert space 'H and T be its center-valued trace. Suppose G is a projection in R. Show that (i) 7( G )I CG; (ii) CT( G ) = CG; (iii) there is a self-adjoint element CG affiliated with the center of R such that CG :T(G) = CG; (iv) for some choices of R and G, CG is not bounded, 8.7.57.

447

EXERCISE 8.7.58

Solution. (i) From Theorem 8.2.8(iv), CGT(G)= ~ ( C G G=) r(G). Since T is a positive linear mapping, 0 5 T(G)5 I. Hence r(G) I C G . (ii) From (i), C T ( ~5)CG. If P = CG - C T ( ~then ) , 0 = PC,(G) so that 0 = Pr(G) = T(PG). Now 0 5 PG so that 0 = PG from Theorem 8.2.8(iii). Thus PCG = 0. But P 5 CG so that P = 0 and CT(G) = cG* (iii) From Theorem 5.2.1, the center C of R is isomorphic to C ( X )for some extremely disconnected compact Hausdorff space X . If f and e are the functions in C ( X ) representing r ( G ) and CG, respectively, then 0 5 f I 1, e is the characteristic function of a clopen set Xo, and ef = f. If e' is the characteristic function of a clopen set X1 contained in Xo such that e'f = 0, then e' = 0 (and XI = 0) from (ii). Thus the (closed) set X , on which f vanishes meets X Oin a (closed) nowhere-dense subset 2 of X . Define a function g on X\Z by: 9(P) = 0

(P E X\XO),

g ( P ) = W P ) (P E XO\Z).

Then g is a self-adjoint function on X (that is g E S ( X ) ) . Since g: f = e , C G : r ( G )= CG from Theorem 5.6.19, where CG in S(C) corresponds t o g under the isomorphism of N(C) with N ( X ) . (iv) Let M be a factor of type 111 on a separable Hilbert space 3.10 (for instance, let M be Cn of Example 6.7.7), and let { E n } be a sequence of projections in M such that E, has trace (dimension) 2-,. Let R be the countably infinite direct sum of M with itself (acting on 'H, the countably infinite direct sum of 3.10 with itself). Let G be El @ Ez Then T ( G )= C,"==, @2-"1 and CG = I. If 2, is a vector in 3.1 with n th coordinate a unit vector in 'Ho and all other coordinates 0, then r ( G ) z , = 2-,x, 0. Hence T ( G )does not have a bounded inverse, so that CG is not bounded in this case. From these observations, C G x , = 2,x,. The domain of CG consists of all those vectors {z,} in 3.1 such that {2,2,} E 3.1. 9.

---f

Let R be a finite von Neumann algebra acting on a 8.7.58. Hilbert space 3.1, T be the center-valued trace on R, and G be a non-zero projection in R. Show that TO is the center-valued trace on GRG, where ro(S) = ( r ( S ) : C G ) Gfor S in GRG and CG is as in Exercise 8.7.5 7(iii ) .

Solution.

From Exercise 6.9.16(i), GRG acting on G('H) is a

448

THE TRACE

finite von Neumann algebra. Thus GRG has a center-valued trace 70 uniquely characterized by the conditions that it is a positive linear mapping of GRG into its center CG, where C is the center of R, such that ro(AB) = TO(BA) for each A and B in GRG and ro(C) = C for each C in CG. We show that these conditions are satisfied for TO (as described in the exercise); but first we must see that TO maps GRG into CG (rather than N(CG)). If T E R, then -llTllG 5 GTG 1IT IIG, whence

0. Define linear functions w, wc (C E R ) on R by

w ( R ) = (RA1u,u), wc(R) = (CRu,u)+r(RCu,u) and let

K: = {we : C E R , I(C((5 k), where k = i r - * (~A'II.

( R E R),

467

EXERCISE 9.6.12

(i) Show that

K

is a convex subset of the Banach dual space

R# of R,and is compact in the weak * topology on RW. (ii) Show that w E K . [Hint.Suppose the contrary, and deduce that there is an element B of R such that

Let B have polar decomposition VK, and obtain a contradiction by taking C = kV*.] (iii) Deduce from (ii) that there is an element A of R such that IlAll I k and

(RA'u, u) = (ARu,u) t r(RAu,u )

( R E R).

+

(iv) Prove that (A rI)-'A'u = Au, where A is the modular operator associated with the separating and generating vector u for

72. Solution. (i) The mapping C + w c : R + R y is linear, and is continuous relative to the weak-operator topology on R and the weak * topology on 72". This continuity assertion follows from the fact that, for each R in R, the linear functional

C

--+

wc(R) = (CRu,u)

+ r(Cu,R*u)

on 72 is weak-operator continuous. Since the ball ( R ) & in R is convex' and weak-operator compact, its image K under the mapping C + w c is convex and weak * compact. (ii) Suppose that w 4 K. Since K is convex and weak * compact, it follows from the Hahn-Banach theorem that there is a weak * continuous linear functional p on R# such that

There exists B in 12 such that p ( q ) = w l ( B ) for each w1 in RW,and we have

468

ALGEBRA A N D COMMUTANT

By taking C = obtain

kV*,where B has polar decomposition VK, we

Since this contradicts our earlier assertion that Reo(B) > 1, it now follows that w E K. (iii) From (ii), w = W A for some A in ( R ) k . (iv) In the argument that follows, the symbols SO,S, F, A have the meanings attributed to them in Subsection 9.2, A first approach to modular theory. From the equation in (iii), with R* in place of R , we have (SORU,A*U) = (R*u,A*u) = (AR*u,u) = (R*A’u,u)- T(R*Au,u) = (A‘u - TAU,Ru),

for each R in R. Since SO has domain follows that A*u E D(F) and

Ru,and F = S,*, it now

A’u - TAU= FA*u = FSAU = AAu. Thus A‘u = (A t rI)Au, and Au = (A + rI)-’A‘u.

m[22]

Suppose that D is an invertible (not necessar9.6.13. ily bounded) positive self-adjoint operator on a Hilbert space H, r > 0 , and B E B(’H). (i) Show that there is an element C of B ( H ) such that

469

EXERCISE 9.6.13

for all y1 and y2 in 'H. Prove also that llCll 5 $r-1/211BII. (ii) By an argument analogous to the proof of Lemma 9.2.8, show that

Solution. (i) Since { D i t : t E P} is a (strong-operator continuous) one-parameter unitary group, and multiplication of operators is jointly continuous on bounded sets in the strong-operator topology, the function (DitBD-"tyl,yz) o f t is continuous on P;its absolute value does not exceed I1BllllylllllyzI1. It follows that the integral in (i) is absolutely convergent, and has absolute value not greater than

Now ert

J

dt e-*t = r (e*t)2 .IreTtdt 1 =1 K [ar~tan(e*~)]y~ -I

+

+

- 2 .

From the preceding paragraph, the equation

defines a bounded conjugate-bilinear functional c on 'H, and llcll 5 ~r-1/211Bll. Corresponding to c, there is an element C of B ( X ) that satisfies the conclusions of (i). (ii) The argument is divided into three stages. Consider, first, the case in which D has the form C:, a j F j , where ~ 1 ,. .,am are positive real numbers and { F l , . . . ,F,} is an orthogonal family of projections with sum I. In this case, it follows from a calculation set out in the proof of Lemma 9.2.8 that the operator C defined in (i) has the form

.

470

ALGEBRA AND COMMUTANT

Moreover, since D a = Cy=, agFj (s = hi),we have

It now follows from (1) that

This completes the proof of (ii) in the first (very special) case. We next consider the case in which D is bounded and has bounded inverse, and choose positive real numbers a , b such that a1 5 D _< b1. As in the proof of Lemma 9.2.8, D is the limit in norm of a sequence { D , } of operators, each of the type considered in the preceding paragraph and satisfying aI 5 D , 5 b1; moreover, D" is the norm limit of the sequence {Di}, for each complex z. From the preceding paragraph,

for all

51

and

x2

in 'H, where C, (in B(3-1)) is defined by

Since

it follows from the dominated convergence theorem that

EXERCISE 9.6.13

471

as n --+ 00; that is, {C,}is weak-operator convergent to C. Now the sequences {Cn}, { D A / 2 } ,{D,1/2} are bounded, and

as n + 00. A similar argument applies when the positions of the exponents are exchanged. By taking limits in (2), it now follows that

for all z1and x2 in H. This completes the proof of (ii) in the second case. Finally, we consider the general case, in which D is unbounded. For each positive integer n, let E , be the spectral projection for D corresponding to the interval [n-', n]. Since D is positive and invertible, the increasing sequence {En}is strong-operator convergent to I . For a given choice of n, let DO,Bo, CO be the restrictions to En('H)of the operators D,E,BEn, E,CEn, respectively. Then DO, Bo, CO E B(E,('H)), Do being a positive operator with a bounded inverse. When y1, y2 E E,('H),

472


The fact that EnDit I En(3-I) = Ddt is a consequence of Corollary 5.6.31. It now follows from the preceding paragraph that, for all $1 and x2 in En(3-I), (Bx1,xz) = (Box1,xz)

= (coD;/2x1, Do - 1 / 2 x 2 ) t ~ ( c ~ D ;~ ~i // ~~ ~x ~~ ,)

= (C D 1/ 2~ D 1,- ~ / ' X Z-t) r(CD-1/2x1,D 1 / 2 ~ 2 ) . When 51, x2 E O(D1/2)n D(D-1/2), we can apply the preceding equation, with E n q , Enx2 in place of 11, 5 2 . This gives

(3)

( B E d l ,E n Q ) = (CD'/2Enx1, D-1/2Enx2) 4- r(CD-%,z1,

D1/2Enx2).

Since D S E n x j = E n D s x j ---t Dsxj as n + 00, for s = A$, it follows by taking limits in (3) that

( B x ~~ ,2 =) ( C D 1 / 2 ~D-1/2x2) l, -+ T ( C D - ~ / D1/'x2) ~X~,

.

.[23]

9.6.14. Suppose that {ot} is a one-parameter group of * automorphisms of a von Neumann algebra R; p, p1, pz, p 3 , . . . are positive linear functionals on R;IIp - pnll + 0 as n + 00; and {ot} satisfies the modular condition relative to pn for each n = 1 , 2 , 3 , . . .. Prove that {ot} satisfies the modular condition relative to p.

Solution. Suppose that A , B E R. For each positive integer n , there is a complex-valued function f n that is bounded and continuous on the strip R = { z E C : 0 5 Imz 5 l}, is analytic on the interior of R, and has boundary values

( t E W). = pn(at(A)B), fn(t t i) = pn(Bot(A)) Since fm - fn is bounded and continuous on R and analytic on the interior of Q, and fn(4

(1) I f m ( 4 - fn(z>I 5 llPm - ~nllllAIlIIBII for each z in the boundary of R, it follows from (a variant of) the maximum modulus principle that (1) is satisfied for all z in 0. Fkom this, {fn(z)} converges uniformly on R. The limit function f is bounded and continuous on R, is analytic on the interior of $2,and has boundary values

f ( t ) = p(at(A)B), f(t ti)= p(Bot(A)) ( t E W). Hence { o t } satisfies the modular condition relative to p.

EXERCISE 9.6.15

473

9.6.15. Suppose that R is a von Neumann algebra with center C and {ot} is a one-parameter group of * automorphisms of R.

(i) Suppose that {ot}satisfies the modular condition relative to a positive linear functional p on R, C E C+, and po is the positive linear functional on R defined by

po(A) = p ( C 4

( A E R).

Show that {ot} satisfies the modular condition relative to po. (ii) Suppose that {ot} satisfies the modular condition relative to q I R , where the vector u lies in the domain D ( C ) of an operator C affiliated to C. Show that ( 0 2 ) satisfies the modular condition relative to w c U I R . [Hint.Use (i) and Exercise 9.6.14.1

Solution. (i) When A, B E R

po(at(A)B)= p(Cat(A)B)= p(ot(A)CB), po(Bat(A)) = p(CBat(A)). Since A , C B E R and {ot} satisfies the modular condition relative to p, there is a bounded continuous complex-valued function f,on the strip R = { z E C : 0 5 Imz 5 l}, that is analytic on the interior of R and has boundary values

f(t> = p ( o t ( A ) C B )= po(at(A)B), f(t

+ i) = p(CBot(A))= po(Bot(A)).

Thus {ot} satisfies the modular condition relative to po. (ii) Suppose that C has polar decompostion V H , so that V E C, H is a positive operator affiliated to C, and u E D ( H ) . Let H have spectral resolution { E x } , and define

( n = 1,2,. . .).

un = VE,Hu

Then {u,} is norm convergent to Cu, and hence

Since u, = Gnu, where C, = V H E , E C ,

wUn(A)= (AC,u,Cnu)

w U ( C ~ C ~ A ) ( A E R).

1

From (i), { a t } satisfies the modular condition relative to w U nIR. It now follows from Exercise 9.6.14 that {ot} satisfies the modular m[112] condition relative to wcU I R.

474


9.6.16. Let {at} be the modular automorphism group corresponding to a faithful normal state w of a von Neumann algebra R. Suppose also that {ot} satisfies the modular condition relative to a positive normal linear functional wo on R, and 00 5 w. (i) Prove that, if H , K E R+ and w ( H A H ) = w ( K A K ) for each A in R, then H = K. [Hint.By using the first part of the discussion following Corollary 9.2.15, reduce to the case in which w = wU1R, where u is a separating (and generating) vector for R. In this case, use Exercise 7.6.23 to show that K u = V ' H u and H u = V'*Ku for some partial isometry V' in R'. Prove that

Deduce that V'H'/'u = H1/2u and that K u = Hu.] (ii) By the Sakai-Radon-Nikodfm theorem (7.3.6) there is a positive operator H in the unit ball of R such that

wo(A) = w ( H A H )

( A E R).

Show that wo(A) = w ( a t ( H ) A a l ( H ) ) and , deduce that at(H)= H,

w(AH)= w(HA)

( A E R, t E a).

[Hint.Apply Proposition 9.2.14 to w and wo.] (iii) Suppose that A , B E R . By using the modular condition, first for A , B , wo and then for A , B H 2 , w , show that w ( ( H 2 B - B H 2 ) A . ) = 0. (iv) Deduce that H 2 is a positive element C in the unit ball of the center of R and wo(A) = w ( C A ) for each A in R, (Compare this with Exercise 9.6.15 (i) .)

Solution. (i) The GNS construction, applied tow, gives rise to

a

* isomorphism cp from R onto a von Neumann algebra cp(R),and

cp(R)has a separating (and generating) vector u such that w = wuocp. for these facts, we refer to the discussion following Corollary 9.2.15. Since w ( H A H ) = w ( K A K ) for each A in R, we have

It suffices to prove that cp(H) = cp(K). Upon replacing R,w , H , K by cp(R),wu I cp(R),cp(H), cp(K),respectively, we reduce to the case

475

EXERCISE 9.6.16

in which w = ouI R, where u is a separating vector for R. In this case

From Exercise 7.6.23 there is a partial isometry V' in R',with initial space (RHu] and final space [ R K u ] ,such that V'AHu = A K u for each A in R. In particular,

V'HU = K u , H U = V'*V'Hu = V'*Ku. Thus

and

( I I P=U ( v 'J H ~~ / ~H~ u~, I ~ u=) I

) H ~ ./ ~ u ~ ~ ~

From equality in the Cauchy-Schwarz inequality, it now follows that V'H'12u = H1l2u.Thus

and H = K since u is a separating vector for R. (ii) From Proposition 9.2.14(i), w = w o at and wo = wo o at for all real t. Hence

for all A in R and t in for all real 2 , and

R. It now follows from (i) that a t ( H ) = H

w(HA)= w(AH) by Proposition 9.2.14(ii).

( A E R)

476


(iii) There exist complex-valued functions f and g, bounded and continuous on the strip fl = { z E C : 0 5 Imz 5 l}, analytic on the interior of 0, and such that

From the last conclusion of (ii),

for all real t. Thus f = g, and

w ( B H 2 A )= g ( i ) = f(i) = wo(BA) = w ( H B A H ) = w(H2BA). Hence w ( ( H 2 B- B H 2 ) A )= 0. (iv) Since w js faithful, it follows from (iii) (by taking A = ( H 2 B- B H 2 ) * )that

H2B = B H 2

( B E R).

Hence H E R n R',0 5 H 2 5 I since 0 5 H 5 I, and

wo(A>= w ( H A H ) = w ( H 2 A )= w ( C A ) where C = H 2 .

( A E R),

~[ll2]

9.6.17. Suppose that R is a von Neumann algebra with center C, acting on a Hilbert space 'FI, {at} is the modular automorphism group corresponding to a separating and generating vector u for

R,

w is a positive normal linear functional on R, and {ot}satisfies the modular condition with respect to w (as well as wulR).

+

(i) Prove that w w u l R = w v l R for some separating and generating vector w for R.

477

EXERCISE 9.6.17

Use Exercise 9.6.16 to prove the existence of elements S, T of C+ such that w,lR = wsvIR, w = w ~ , l R ,and S has null space (0) and range dense in 3-1. (iii) Deduce from Exercise 7.6.23 that u = V ' S v for some unitary operator V' in R'. (iv) Let S-' : S(1-1) + 31 be the inverse of the mapping S : 'H + S('H). Show that S-', T , and T:S-' (= C) are positive elements of the algebra N(C) (see Theorem 5.6.15). Prove also that u E D(C) and w = oc,lR. (Compare this with Exercise 9.6.15(ii).) (ii)

+

Solution. (i) The positive normal linear functionals w w, I R and cs, I R both have support I , and [Ru] = 'H. From Lemma 7.2.11, there is a vector v in 7-f such that [Rv] = 1-1 and w -tw, I R = w , I R. Moreover, since wv I R has support I , [R'v]= 'H. (See Remark 7.2.6.) (ii) The one-parameter group { o t ) satisfies the modular condition relative to each of the positive linear functionals w,w, I R, and hence also relative to their sum wvI R. Since w, I R 5 w, I R and w 5 w,, 1 R, it now follows from Exercise 9.6.16(iv) that there exist positive operators SOand TOin C such that

w,(A) = w,(SoA),

w ( A ) = ov(ToA)

( AE R).

Let S = $ I 2 , T = Then S, 7' E C+, w, 1 R = w s v 1 R, and w = w r v I R. If Q is the range projection of S, then

ll(I - Q>u1l2= wu(I - Q ) = WSV(I- Q ) = ( ( I - Q ) S V ,S V ) = 0 ; so Q = I , since u is a separating vector for R. Thus S has range dense in 'H, and null space (0). (iii) Since w, I R = wsv I R,[Ru]= H , and

[RSV]= [ S R v ]= [S('H)]= 'H, it follows from Exercise 7.6.23 that there is a unitary operator V' in R' such that V ' A S v = Au for each A in R;in particular, V ' S v = u. (iv) The operator S-' has domain S(1-1) dense in 'H, and its graph ((Sz, x) : x E 1-1) is closed in 'H x 'H, since S is bounded. For each unitary operator U in C', S = U * S U and thus S-l = U * S - ' U ; so S-l q C. Since S, T E C+, S-' and T are positive elements of

JW >.

478


We assert that the element C (= T:S-') of N(C)is positive. For this, note that C is the closure of the operator TS-' (with domain S('FI)) and, since ST E C+, (CSZ, S5) = (TS-lSs, S X )

= (Tz,SZ) =(STZ,~) 30 for each z in 'H. Since S ( H ) is a core for C, it now follows that a positive element of N(C). Since u = V'Sv = SV'v, we have u E S(7i) E D(C)and

C is

CU = TS-'U = TV'V = V'TV. From this

o c u ( A )= (AV'Tv, V'Tv) = (V'ATV,V'TU) = (ATv,Tv)= WT,,(A)= w ( A ) for each A in R.

41121

9.6.18. With the notation used in the development of modular theory in the context of a von Neumann algebra R and a faithful normal semi-finite weight p on R (starting in the paragraph preceding Proposition 9.2.39), suppose that U is a unitary operator in the center C of It. Prove that

where SOis the conjugate-linear mapping A + A* : U2+ U2.Hence show that ? r ( U ) J x ( U )= J, and deduce that

J n ( C ) J = .(C*)

(CE C).

Solution. We recall that U = Npn A$', Npis a left ideal in R, AE and N; is a right ideal. When A E U, we have A E and thus U A = AU E Npf l Ni = U .

JV",

EXERCISE 9.6.18

479

It follows that U A E 2l whenever A E IU, and hence that U A E U2 whenever A E 212. From this, and since the same argument applies with U* in place of U , we have 3' = { U A : A E U2}= { r ( U ) A : A E a'}).

When A E 212,

Since !212 = D(So), it now follows that

Son(U) = r(U*)So,

n(U)Son(U)= so.

Suppose that $0 E D(S).Since S is the closure of SO,there is a sequence { A n } in 212 such that

From the preceding paragraph, we have

and n(U)A, + r(U)zo. Thus

n(U)zo E D(S),

S n ( U ) x o = n(U*)Szo,

whenever $0 E D ( S ) . From this, and since the same argument applies with U* in place of U , we have 20 E D(S)if and only if r ( U ) z o E D ( S ) , and S r ( U ) = n(U*)S. Thus

r ( U ) S r ( U )= s . From the uniqueness of the polar decomposition of S, and since

we have r ( U ) J n ( U )= J

480


The last equation can be written in the form

Js(U)J = r(U*). Each element C of C can be expressed as a linear combination of unitary operators ajUj in C . We have

'&

4

T(C)J = x a j ~ ( U j ) J , j=1

and the conjugate-linearity of J now implies that

Suppose that R is a von Neumann algebra with center C, acting on a Hilbert space H, and J : 'H + H is a conjugate-linear isometry such that J 2 = I , J R J = R',and J C J = C* for each C in C. Let E be a non-zero projection in R, and let E' be the projection J E J in R'. Prove that (i) if y E 3.1 and E has range [R'y],then E' has range [ R J y ] ; (ii) CE = C p ; (iii) EE' is a non-zero projection and its range is invariant under

9.6.19.

J;

(iv) x = J x = E x for some unit vector x in 3c. Solution. (i) If E has range [R'y],then

E'(3c) = J E J ( H ) = JE(3.1) = [JR'y]= [ J R ' J J y ] = [RJy]. (ii) If P is a projection in C, we have

J P E J = J P J J E J = P * E ' = PE', so P E = 0 if and only if PE' = 0. Hence CE = Cp4

481

EXERCISE 9.6.20

(iii) Since CE = # 0 . Also,

CE,# 0, it follows from Theorem

5.5.4 that

EE'

JEE' = JEJJE'JJ = E'EJ = EE'J, and EE'(H) is invariant under J . (iv) Let y be a non-zero vector in the range of EE', and let z = i(y - J y ) . From (iii), z E EE'('H). Also,

and z # 0 unless y = Jy. In either case, there is a non-zero vector z (and we can assume that 11x11 = 1) such that x = Ex = E'x = J x .

Suppose that R is a von Neumann algebra with center

9.6.20.

C,acting on a Hilbert space H,and J : H H is a conjugate-linear isometry such that J 2 = I, J R J = R',and J C J = C* for each C in C. Let E (# 0) be a countably decomposable projection in R. Prove --f

that there is a vector x in 3-1 such that Jx = 2, E has range [R'z], and J E J has range [Rx].[Hint. By using Exercise 9.6.19, show that there is a (finite or infinite) sequence { z 1 , 2 2 , . .} of unit vectors in 'H such that z, = J z , = E z , for each n, the subspaces [E'z,] (n = 1,2,. .) are mutually orthogonal, and V[R'zn]is the range of E. Prove that the subspaces [Rzn](n = 1,2,. ..) are mutually orthogonal, and V[Rx,] is the range of J E J . Follow the reasoning used in the first paragraph of the proof of Proposition 5.5.18.1

.

.

Solution. Let {z, : a E A} be a family of unit vectors in 'H, maximal subject to the condition that x , = J x , = E x , for each a in A and the subspaces [R'x,] ( a E A) are mutually orthogonal. From Exercise 9.6.19(iv), the family {z,} is not empty. If E, denotes the projection with range [R'z,], then { E , : a E A} is an orthogonal family of non-zero subprojections of E in R. Since E is countably decomposable, the index set A is finite or countably infinite; so the family { x , } can be relabelled as a (finite or infinite) sequence (21,$ 2 , . ..}. Let J' be the projection E - C En in R . If F # 0, it follows from Exercise 9.6.19(iv) that x = J x = Fx for some

482


unit vector x in H,and x can be added to the family ( x , } , contradicting the maximality assumption. Thus F = 0, and E = C En. Since J x , = x , and E, has range [ R ’ x , ] , it follows from Exercise 9.6.19(i) that [ R x , ] is the range of the projection J E , J in R’. Moreover, the projections J E , J (n = 1,2,. . .) are mutually orthogonal, and have sum J E J . If x is the vector C n - l ~ , ,the reasoning set out in the first paragraph of the proof of Proposition 5.5.18 shows that E has range [R‘o]and J E J has range [Rx].Moreover,

JZ =

C n-* J o , = C

7 1 - l ~ = ~o

.

9.6.21. Suppose that R is a von Neumann algebra with center C. Show that there is an orthogonal family {Q,}of projections in C such that C Q a = I and each Q a is the sum of an orthogonal family of equivalent cyclic projections in R. Prove also that, if both R and R’ are properly infinite, the family {Qa} can be chosen in such a way that each Q , is the sum of an orthogonal family of infinitely many equivalent properly infinite cyclic projections in R. (Compare Exercises 6.9.12 and 6.9.13.)

Solution. By expressing R as the direct sum of a finite von Neumann algebra and a properly infinite von Neumann algebra, we may assume that R falls into one of these two classes. In each of the two cases so obtained, in order to prove the first assertion of the exercise, it suffices to establish the following result:

(*)

each non-zero projection P in C has a non-zero subprojection Q in C such that Q is the sum of an orthogonal family of equivalent cyclic projections in R.

Indeed, given this result, let {Q,} be a maximal orthogonal family of projections in C, each of which can be expressed as the sum of an orthogonal fanlily of equivalent cyclic projections in R, and let P be I - CQ,.If P # 0, it follows from (*) that the family {Q,} can be augmented by the addition of a non-zero subprojection Q of P,contradicting the maximality assumption. Thus P = 0, and

CQa=I.

We now prove (*) under the assumption that R is finite. Let x be a unit vector in the range of P and let E (in R) be the cyclic projection with range [R‘o].By Proposition 8.2.1, there is a monic

EXERCISE 9.6.21

483

projection El in R such that El 5 E ( 5 P). Thus C E ~ ( 5 P) is the sum of a finite orthogonal family { E l , . . .,Ek} of projections in R such that Ej El ( 5 E) for j = 1,. .,k. Since E is cyclic, each Ej is cyclic, and (*) is proved in the finite case. Suppose next that R is properly infinite. In this case, the nonzero projection P in C is properly infinite relative to R,and can therefore be expressed as the sum of an infinite orthogonal sequence {GI,G2,...) of projections in R, each equivalent to P. If El is a non-zero cyclic subprojection of GI in R, then Gj ( N G I ) has a subprojection Ej (in R ) equivalent to E l , for each j = 1 , 2 , , . . Let {Eb : b E B} be a maximal orthogonal family of projections, in R and each equivalent to E l , that contains the infinite sequence { E l , & , . .). From the maximality assumption, El 2 P - C E b ; so N

.

.

.

for some projection Q in C such that 0 < Q 5 theorem. Since

P,by the comparison

and the index set B is infinite, the usual argument (see, for example, the last paragraph of the solution to Exercise 6.9.12) now shows that Q is the sum of an orthogonal family {Fb : b E B} of projections in R, each equivalent to QEl ( 5 E l ) and therefore cyclic in R. This proves (*) in the properly infinite case, and so completes the proof of the first assertion in the exercise. (This case can also be established by applying Exercise 6.9.13.) Suppose now that both R and R' are properly infinite. The final assertion of the exercise will follow from the argument set out above (excluding the second paragraph), provided that the non-zero projection P in C contains a properly infinite cyclic projection F in R. Indeed, if this is the case, then G1 ( N P) contains a properly infinite cyclic projection, and this can be used as El in the preceding paragraph. If El is properly infinite, the same is true of the cyclic projections Fb (b E B ) constructed above. It remains to prove that P contains a properly infinite cyclic projection F in R. Since P is properly infinite in both R and R', it has the form

484


where {Gj} and {Gg}are orthogonal infinite sequences of projections in R and R‘,respectively, and G j P (in R),Gi P (in R’), for each j = 1 , 2 , . , . With Ej chosen as in the preceding paragraph, Ej has range [R‘yj] for some vector y j (= Pyj).Let 5‘ be a partial isometry in R’,with initial and final projections P and Gi, respectively. Then Ej has range [R‘zj],where xj = l$”yj, and [Rzj]is the range of a subprojection E$ of G) in 2’.Since { E j } and {Eli} are orthogonal families of projections, and z j is a common generating vector for Ej (in R) and Ej (in R’) for each j = 1 , 2 , . . . , the reasoning used in the first paragraph of the proof of Proposition 5.5.18 shows that ,C ; Ej is a cyclic projection F in R. Clearly F 5 P , and F is properly infinite since Ej El # 0 ( j = 1,2,. . .). N

.

N

N

9.6.22. Suppose that R and S are von Neumann algebras acting on Hilbert spaces 31 and K , respectively, and ‘pis a * isomorphism from R onto S. Suppose also that { Q a : a E A} is an orthogonal family of central projections in R with sum I , and, for each a in A, the * isomorphism cplRQ, from R Q a onto S v ( Q a ) is implemented by a unitary transformation from Qa(31)onto cp(Qa)(IC). Prove that cp is implemented by a unitary transformation from 3-1 onto Ic.

Solution. For each a in A, let Ua be a unitary transformation from Q a ( H ) onto cp(Qa)(K)that implements the * isomorphism cpIRQ,. Since C Q a = I , we have C c p ( Q a ) = I ; so there is a unitary transformation U from 7-1 onto K: such that U I Qa(31)= Ua ( a E A). When R E R,

Summation over a in

A gives

p(A)U = U A ,

cp(A) = U A U * .

IU

Suppose that 3.1 and K are Hilbert spaces, R (C D(7-1)) and S (E D ( K ) ) are von Neumann algebras, R has a generating vector 3 and E (in R) is the projection with range [R’z],S has a generating vector y and F (in S) is the projection with range [S’y], cp is a * isomorphism from R onto S, and cp(E) = F . Show that

9.6.23.

485

EXERCISE 9.6.23

the * isomorphism (plERE from E R E onto F S F is implemented by a unitary transformation UOfrom E ( H ) onto F(K).Deduce that cp is implemented by a unitary transformation U from 3-I onto K. [Hint. Let { E , : a E A} be an orthogonal family of projections in R, maximal subject to the condition that 0 < E, $ E for each a in A. Let V, be a partial isometry in R, with initial projection E, and find projection a subprojection of E . Show that C E , = I , and define U by requiring U ( E , ( H ) = cp(V,*)UoV, for each a in A.]

Solution. Since cp(E) = F , the restriction cp I E R E is a * isomorphism from the von Neumann algebra ERE (acting on E ( H ) ) onto FSF (acting on F ( K ) ) . Moreover, E R E and FSF have separating and generating vectors 5 and y, respectively. By Theorem 7.2.9, cp 1 E R E is implemented by a unitary transformation UOfrom E('H) onto F(K). Choose { E a } and {V,} as in the hint. Since [Rz] = H and E has range [R's],CE = I by Proposition 5.5.13. From maximality of ( E , } , the projections E and I - C E , do not have Ron-zero equivalent subprojections in R. From this, and since CE = I , it follows from Proposition 6.1.8 that C E , = I . Thus Ccp(E,) = 1. Let G, ( 5 E ) be the final projection of V,. Then q(V,) is a partial isometry in S, with initial projection cp(E,) and find projection cp(G,) ( 5 cp(E) = F ) . Since G, is in ERE, y(Ga) = UoG,U,*, and UoG, is a unitary transformation from Ga(3-I)onto cp(G,)(K). Thus cp(V,*)UoV, (= cp(V;)UoG,V,) maps E,(H) isometrically onto cp(Ea)(K),and U (as defined in the hint) is a unitary transformation from 3.1 onto /c such that U E , = cp(E,)U = cp(V,*)UoV, for each a in A. Suppose that A E R. Given a , b in A, we have V,AV< E ERE, so

v(v4 )V( A)v(Q*)uO = y( vQ A ' i ) Uo = uo

Summation over a and b in UAU* for each A in R.

A vb*-

vQ

A now yields (p(A)U = U A , so cp(A) =

486

ALQEBRA AND COMMUTANT

9.6.24. Suppose that 'H and K are Hilbert spaces, R (E B(7-l)) and S (E B ( K ) ) are von Neumann algebras, x E Z, and y E Ic. Let E (in R ) , E' (in R'),F (in S), and F' (in S') be the projections [S'y], and [Sy], respectively. Suppose that with ranges [R'z], [Rz], there exist orthogonal families of projections, {EL : a E A} and {FL : a E A}, indexed by the same set A and both having sum I, such that

E: E R',

EL

N

E',

Fi E S',

Fi

F'

( a E A).

Show that, if cp is a * isomorphism from R onto S, and cp(E) = F, then cp is implemented by a unitary transformation U from H onto K . [Hint. By using Exercise 9.6.23, show that there is a unitary transformation UO from E'(7-l) onto F'(K) such that cp(A)F' = UoAE'U,' for each A in R. Choose partial isometries V,l in R' and Wl in S', that implement the equivalences E' EA and F' N F;, respectively, and define U by the condition UEL = W;VoVa'+.]

-

SoCution. The stated conditions imply that E' and F' each has central carrier I. Hence the mappings R -+ RE': R + RE' and S -+ SF': S SF' are * isomorphisms, and the mapping

$ : AE'+ cp(A)F'

is a * isomorphism from RE' onto SF'; moreover, $(EE') = FF'. Now RE' has a generating vector z and [(RE')'%](= [E'R'E'z]) is the range of the projection EE' in RE'. Also, SF' has a generating vector y, and [(SF')'y] is the range of the projection FF' in SF'. From Exercise 9.6.23, $ is unitarily implemented. This proves the existence of an operator Uo with the properties set out in the hint. With V,l and WL chosen as in the hint, W;UoVk* is a unitary transformation from Ei('H) onto F;(K), for each a in A. Since C EL = I and C Fi = I, there is a unitary transformation U , from 'H onto K , defined as in the hint. When A E 72,

~ ( A ) F ; U= ~ ( A ) W ; F ' U ~ V ; * = Wicp(A)F'UoV,'* = W;U~AEV,," = WiUtjV,'*E:A = UELA.

EXERCiSE 9.6.25

487

Summation over all a in A now yields cp(A)U = U A , so

cp(A)= UAU*

(AER).

Suppose that, for j = 1,2, Rj is a von Neumann 9.6.25. algebra with center C j , acting on a Hilbert space H j , and J j : H j + H j is a conjugate-linear isometry such that J j = I , J j R J j = 723, and J j C J j = C* for each C in C j . Let cp be a * isomorphism from R1 onto 7 2 2 . Prove that cp is implemented by a unitary transformation from onto H 2 . [Hint. By using Exercises 9.6.21 and 9.6.22, reduce to the case in which R1 contains and orthogonal family { E , } of equivalent cyclic projections with sum I . Define Fa = cp(E,), EL = J1 E , J1, Fi = 52 FaJ2, and use Exercises 9.6.20 and 9.6.24.1

Solution. From Exercise 9.6.21, there is a family { Q , } of projections in C 1 , with sum I , in which each Q , is the sum of an orthogonal family of equivalent cyclic projections in R1. From Exercise 9.6.22, it suffices to prove that the * isomorphism ( P I R ~ Q , from RIQaonto R2cp(Q,)is unitarily implemented, for each a in A. Since JIQaJl = Q , , the range of Q a is invariant under J1; similarly, the range of cp(Q,) is invariant under & . If we now replace H I by Q a ( H 1 ) and R1 by RlQa, H2 by ~ ( Q a ) ( 3 - 1 2 ) and R2 by R29(Qa), J 1 by J 1 1 Q a ( W 1 ) and J 2 by J 2 1 d Q a ) ( N 2 ) , and by CP I RlQa, the assumptions set out in the exercise remain valid and, in addition, the identity in R1 is the sum of an orthogonal family of equivalent cyclic projections. In view of the preceding paragraph, we may now assume that R1 contains an orthogonal family { E , : a E A} of equivalent cyclic projections with sum I . Define Fa, Ek, and FA, as in the hint. Since cp is a * isomorphism, and the mappings A + JjAJj : Rj --.f 72; ( j = 1,2) are * anti-isomorphisms, each of the orthogonal families {Ea}, { E L } , {F,}, {F:} consists of equivalent countably decomposable projections in the appropriate von Neumann algebra, and has sum I . From Exercise 9.6.20, there exist vectors 2, in ‘FI1 and y a in 3 1 2 such that E,, EL, Fa,F,: have ranges [R:4 [ R 1 4 ,[ R i y a l , [ R 2 Y a ] , respectively. Since cp(E,) = Fa, it now follows from Exercise 9.6.24 that cp is unitarily implemented. [27(Chap. 111, Par. 1, Sec. 5 , Thm. 6, p. 225)l

488


9.6.26. Suppose that p1 and p2 are faithful normal semifinite weights on a von Neumann algebra R, and n1 and 7r2 are the corresponding representations of R,constructed as in Theorem 7.5.3. By using Theorem 9.2.37 and Exercises 9.6.18 and 9.6.25, show that 7r1 and 7r2 are equivalent.

Solution. From Theorem 7.5.3, n j is a faithful representation and xj(R)is a von Neumann algebra, for j = 1,2. We have to show that the * isomorphism R2 0 n;'

: 7rl(R)+ 7r2(R)

is unitarily implemented. This follows from Exercise 9.6.25; the conditions required in that exercise are satisfied (with n j ( R ) in place of Rj),by Theorem 9.2.37 and Exercise 9.6.18.

9.6.27. Suppose that R is a finite von Neumann algebra with center C, acting on a Hilbert space 3-1, cp is a * automorphism of R,and cp(C) = C for each C in C. Prove that there is a unitary operator U acting on 3-1 such that cp(A) = UAU* for each A in R. [Hint.By using Exercises 9.6.21 (applied to R') and 9.6.22, reduce to the case in which R' contains an orthogonal family of equivalent cyclic projections with sum I . Show that T O cp = T , where r is the center-valued trace on R,and deduce that cp(E) E for each projection E in R. Use Exercise 9.6.24.1 N

Solution. Let {Qa} be an orthogonal family of projections in C (the center of R') with sum I , such that each Qa is the sum of an orthogonal family of equivalent cyclic projections in R' (Exercise 9.6.21). Since cp(Q,) = Qa, the restriction cp 1 RQa is a * automorphism of the finite von Neumann algebra RQa; moreover, cplRQ, acts as the identity mapping on the center of RQa. If cplRQ, is implemented by a unitary operator acting on Qa(3-1), for each index a, then cp is implemented by a unitary transformation of 3-1 (Exercise 9.6.22). Upon replacing R by RQa and cp by cp I RQa, we may now assume that R' contains an orthogonal family {EL} of equivalent cyclic projections with sum I . Since cp(C) = C for each C in C, it is apparent that the mapping T o cp : R C inherits from r the properties, set out in Theorem 8.2.8, that characterize the center---f

489

EXERCISE 9.6.28

valued trace; so T O Y = T . Since r((p(E))= T ( E ) ,it follows (Theorem 8.4.3(vi)) that cp(E) N E, for each projection E in R. Let E' be any member of the family {EL}. Choose a vector z such that [Rz] is the range of E l , let E (in R) be the projection F, F has range with range [R'z],and let F be cp(E). Then E [ R ' V z ] and , E' has range [ R V z ]where , V (in R) is a partial isometry that implements the equivalence of E and F. It now follows from Exercise 9.6.24 (with y = Vz, F' = El, and Fi = EL) that cp is unitarily implemented. N

9.6.28. Suppose that R is a von Neumann algebra with center C, acting on a Hilbert space 'H, and both R and R' are properly infinite. Let cp be a * automorphism of R such that p(C) = C for all C in C. Prove that there is a unitary operator U acting on 7-l such that y ( A ) = UAU* for each A in R. [Hint.By use of Exercises 9.6.21 (applied to R') and 9.6.22,reduce to the case in which R' contains an orthogonal family of infinitely many equivalent properly infinite cyclic projections with sum I. Use Exercise 9.6.24 together with Corollary 6.3.5.1 Solution. By Exercise 9.6.21, there is an orthogonal family {Q,} of projections in C (the center of R') with sum I, such that each Q, is the sum of an infinite orthogonal family of equivalent properly infinite cyclic projections in R'. By reasoning as in the first paragraph of the solution to Exercise 9.6.27,we reduce to the case in which R' contains an infinite orthogonal family {EL} of equivalent properly infinite cyclic projections with sum I. Note that each EA has central carrier I. Let E' be any member of the family (EL}, and choose z, E, and F , as in the final paragraph of the solution to Exercise 9.6.27. Since E' is properly infinite and has central carrier I (relative t o R'), E is properly infinite and has central carrier I (relative t o R), by Propositions 9.1.2and 5.5.13;moreover E is countably decomposable (relative to R), since E is cyclic. It now follows that F (= cp(E)) is properly infinite and countably decomposable, and has central carrier I, relative to R. By Corollary 6.3.5, E N F. The argument set out in the last two sentences of the solution to Exercise 9.6.27 applies again, in the present context, and completes the proof that cp is unitarily implemented. 8

490


9.6.29. Suppose that R is a von Neumann algebra with center C , acting on a Hilbert space ‘H, and there is no projection P in C such that the von Neumann algebras RP and R’P are of types 11, and 111, respectively. Let cp be a * automorphism of R such that cp(C)= C for each C in C. By using Exercises 9.6.27 and 9.6.28 and Corollary 9.3.5, show that there is a unitary operator U acting on H such that cp(A) = UAU* for each A in R. (The exclusion of algebras that have a direct summand of type I1,with commutant of type 111 is necessasry, in the formulation of this exercise. In Exercise 13.4.3, we exhibit a factor 12 of type I1,with commutant of type 111, and a * automorphism of R that is not unitarily implemented. Exercise 9.6.33 provides additional information about * automorphisms of an algebra of type I1,with commutant of type 111.) Solution. There exist projections PI,. . .,P4 in C, with sum I, such that RP1 is of type I unless PI = 0, RP2 is of type 111

unless P2 = 0, RP3 is of type 11, unless P3 = 0, and RP4 is of type I11 unless P4 = 0. Since R has no central summand of type 11, with commutant of type 111, it follows that R’P3 is of type 11, if P3 # 0. For each j , cp I RPj is a * automorphism of RPj and leaves the center CPj of RPj elementwise fixed. Thus cp I RPj is unitarily implemented-by Corollary 9.3.5 in the case j = 1, by Exercise 9.6.27 in the case j = 2, and by Exercise 9.6.28 in the cases j = 3, 4. It now follows from Exercise 9.6.22 that cp is unitarily implemented.

9.6.30. Suppose that, for j = 1,2, Rj is a von Neumann algebra with countably decomposable center C j , acting on a Hilbert space Hj, both Rj and 125 are finite, Cj (in N(Cj))is the coupling operator of Rj (see Exercise 9.6.7),and T j is the center-valued trace on Rj. Let cp be a * isomorphism from R1 onto R2. (i) Prove that cp o TI = 72 o cp. (ii) By using Theorems 3.4.3 and 5.6.19, prove that the * isomorphism cpIC1, from C1 onto C2, extends to a * isomorphism II, from N(C1)onto N(C2). (iii) Suppose that a projection Q in C1 is the sum of an orthogonal family {,Ti,. ,EL} of equivalent cyclic projections in 72;. Choose 21 in ‘FI1 so that Ei has range [Rlzl], let E (in 721) be the projection with range [R:sl],and let F be cp(E). Show that, if +(C1) = C2, there is an orthogonal family {F:, . , , ,Fj!} of equivalent cyclic pro-

..

49 1

EXERCISE 9.6.30

jections in 72; with sum cp(Q), and a vectorx 2 2 in X 2 such that F has range [ 7 2 ; 2 2 ] and Fi has range [ 7 2 2 2 2 ] . (iv) By using (iii), together with Exercises 9.6.21, 9.6.22, and 9.6.24, show that cp is implemented by a unitary transformation from 3-11 onto 7t2 if and only if $ ( C l ) = C2.

Solution. (i) Since cp is a * isomorphism from 7 2 1 onto 7 2 2 , it carries C1 onto C 2 . It is apparent that the mapping cp-' o 7 2 o cp : 721 -+ C1 inherits from 7 2 the properties (set out in Theorem 8.2.8) that characterize a center-valued trace. Thus cp-' o r2 o cp = 71. (ii) We can express cp 1 C I as a composition c p ~ ol cpo o cp1 of * isoC(X1), cpo : C(X1) C(X2), and cp;' : morphisms cp1 : C1 C ( X 2 ) C 2 , where X1 and X2 are extremely disconnected compact Hausdorff spaces. From Theorem 3.4.3 there is a homeomorphism 7 from X2 onto X 1 such that cpo(f) = f o 7 for each f in C ( X 1 ) . If g E N(X1), then g is a normal function defined on a dense open subset X1 \ 21 of X I , and g o 7 is a normal function defined on the dense open subset X2 \ q-'(Z,) of X 2 . The mapping g + g o 7 is a * isomorphism Po from N(X1) onto N ( X 2 ) , and extends cpo. At the same time, cpj extends to a * isomorphism (i'j from n/(Cj) onto n / ( X j ) ( j= 1,2), by Theorem 5.6.19. Thus cp 1 C1 extends to a * isomorphism t,b (= $5;' o $30 o PI) from N(C1)onto N(C2). (iii) Since the cyclic projections E (in 7 2 1 ) and Ei (in 72;) have a common generating vector 2 1 , it follows from the defining property of the coupling operator C1 (see Exercise 9.6.7) that --$

--f

--f

T1(E)

= c1:T;(E;),

where T; denotes the center-valued trace on 72;. Since the equivalent projections E i , . . . ,EI, in 72; have sum Q in C 1 , we have T ~ ( E [=) Ic-'Q. Thus q ( E ) = K'(C1 q), and T2V)

= 7 2 ( c p ( W = (P(Tl(E)) = $( k-l(Ci

Q)) = k-l ($(Ci) :$(Q))

=~ - l ( + ( ~ l ) w ? b If $ ( C l ) = C2, we have T ~ ( F=) k-l(C2 :cp(Q))( 5 (72). By Exercise 9.6.7(iv), F is a cyclic projection in 722. Let 2 2 be a generating vector for F , and let F: (in 72;) be the projection with range [R222].Then

k-'(C2 ^p(Q))= T ~ ( F=) C2 : ~ i ( F i ) .

492


Since Cz is an invertible element of the algebra N(C2),it now follows that r l ( F i ) = k-lcp(Q). Thus Fi is contained in an orthogonal family { F i , . . . ,F i } of equivalent (and, necessarily, cyclic) projections in RL,with sum cp(Q). (iv) If cp is implemented by a unitary transformation U from W1 onto 712, then cp I C1 and $ are also implemented by U . Since all the constructs used in defining the coupling operator are preserved by unitary equivalence,

cz = uc1u-'

= $(Cl).

Conversely, suppose that $(Cl) = C2. From Exercise 9.6.21, there is an orthogonal family { Q a } of projections in C1 (the center of Ri) with sum I , such that each Q a is the sum of an orthogonal family (necessarily finite, since Rl,is finite) of equivalent cyclic projections in 72;. Choose a particulr index a, and let Qa

= Ei + * * * + E L ,

.

where E i , . . ,EL are equivalent cyclic projections in Ri. We can now choose 2 1 , E , F , 2 2 , {F;, . . . , P i } as in (iii), using Qa in place of Q. Since cp(E) = F , it follows from Exercise 9.6.24 that the * isomorphism cp I RQa from RIQa onto R2cp(Qa)is unitarily implemented. This conclusion holds for each index a, so cp is unitarily m[37,38] implemented, by Exercise 9.6.22. Suppose that, for j = 1,2, Rj is a finite von Neumann algebra acting on a separable Hilbert space Wj,and 72: is properly infinite. Prove that (i) Rj has a separating vector x j [Hint.Proposition 5.5.18.1; (ii) 72: contains an (infinite) orthogonal sequence of projections, with sum I , each equivalent to the cyclic projection with range [Rjxj] [Hint.Use Proposition 6.3.12.1; (iii) if cp is a * isomorphism from R1 onto R2, then cp is implemented by a unitary transformation from H Ionto 712. 9.6.31.

Solution. (i) From Proposition 5.5.18, there is a central proand I - Qj has range jection Qj in Rj such that Qj has range [R)-c:j] [Rjyj], for some vectors x j and yj. Since Rj is finite, the projection with range [Riyj] is finite in Rj;so I - Qj is finite in Ri,by

493

EXERCISE 9.6.32

Proposition 9.1.2. Since 72; is properly infinite, it now follows that I - Q j = 0 , Q j = I , and x j is a generating vector for RS. (ii) Since the projection, I, with range [Rgxj] is finite in Rj, the projection E[i with range [ R j z j ]is finite in by Proposition 9.1.2. From Proposition 5,5.13,Ei has central carrier I relative to RS. It now follows from Proposition 6.3.12(ii) that the (countably decomposable, properly infinite) projection I in R: is the sum of an orthogonal family (necessarily, countably infinite) of projections in RS each equivalent to the finite projection E(i. (iii) In view of (ii), the stated result follows from Exercise 9.6.24, with R, S, x , y, E , F replaced by R1,R2,21, 52, I , I , respectively, and with the set of positive integers as the index set A.

"5,

Suppose that, for j = 1,2, Rj is a von Neumann 9.6.32. algebra acting on a separable Hilbert space Hj,and R; is properly infinite. Let cp be a * isomorphism from R1 onto R2. By using Exercise 9.6.31 and Remark 7.2.10, show that cp is implemented by a unitary transformation from 7 i 1 onto 'H2. Solution. Let 91 be the largest central projection in 2 1 such that RlQl is finite, and note that cp(Q1) is the corresponding central projection in R2. Then cp I RlQl is unitarily implemented by Exercise 9.6.31. If Q1 # I, the von Neumann algebras R1(I - 91) and &(I - cp(Q1)) are properly infinite, and cpIRl(I - Ql) is unitarily implemented, by Remark 7.2.10. It now follows from Exercise 9.6.22 that cp is unitarily implemented.

9.6.33. Suppose that, for j = 1,2, Rj is a properly infinite von is Neumann algebra acting on a separable Hilbert space Hj,and finite. Note that Rj has a generating vector xj (Exercise 9.6.31(i)), and let Ej be the projection in Rj with range [%?>xi]. Show that a * isomorphism cp from R1 onto R2 is implemented by a unitary transformation from 'HI onto F f 2 if and only if cp(E1) E2. (For an example in which the condition p(E1) Ez is not satisfied, we refer to Exercise 13.4.3.)

"5

N

N

Solution. If cp is implemented by a unitary transformation from H1 onto X2,then cp(E1) (= UE1 U")has range

U

494


Since

[R2Uz1]= [UR121]= U(X1) = X2 = [R222],

-

it follows from Theorem 7.2.12 that [RiUzl] w [ R i q ] ;that is, cp(E1) E2. Conversely, suppose that E2 cp(El), and let V (in 7 2 2 ) be a partial isometry that implements this equivalence. Then

-

and [ R ~ V Q(= ] [VRiz2] = V E 2 ( X 2 ) ) is the range of cp(E1). It now follows from Exercise 9.6.23, with R,S,2, y, E , F replaced by R1,R2, 21, V 2 2 , E l , cp(El), respectively, that cp is unitarily implemented. I

9.6.34. Suppose that m and n are cardinals, A is a maximal abelian von Neumann algebra acting on a Hilbert space Ic, and R is the von Neumann algebra m @ (d@ I,). Prove that R is of type I, and R' is of type I, (in other words, prove the conwerse of Theorem 9.3.2).

Solution. We prove first that k @C is of type I k whenever k is a cardinal and C is an abelian von Neumann algebra. To this end, let A be a set with cardinality k, and let {&,b : a,b E A} be the selfadjoint system of matrix units in k @ C determined by the condition that the matrix of Ea,b has I in the (a,6) entry and 0 at each other entry. Then Ea,b is a partial isometry in k 8 C with initial projection E b , b and find projection From this, and since Ea,a(k@ C)&,, is * isomorphic t o the abelian von Neumann algebra C, it now follows : a E A} is a family of k equivalent abelian projections that in k @ C, with sum I. Thus k @ C is of type I k . With m and A @ I, in place of k and C, we deduce from the preceding paragraph that R (= m @ ( A @ I,)) is of type I,. Also, n 8 A is of type I,, and hence the same is true of the * isomorphic algebra ( n @ d)@ I, (= R').(See Lemma 6.6.2.) 9.6.35. Suppose that R is a von Neumann algebra of type I,, with commutant R' of type I,. Show that the center C of R has commutant C' of type Imn.

EXERCISE 9.6.36

495

Solution. In view of Theorem 9.3.2, we may suppose that R is m @I ( A @ I,), where A is a maximal abelian von Neumann algeI,) 8 I,. bra acting on a Hilbert space K. In this case, C is (d 18 The Hilbert space 7.1 on which R (and C) act is CaEA ${CbEB ex}, where the index sets A and B have cardinality m and n, respectively. An element 2 of 'H is a family { X a : a E A} of vectors in CbE-$K such that C Q E A [ (=~ ,1 (1 (~ 121(< ~ OO), and for each a in A, xa is a family { x Q , b : b E B} of vectors in K such that 1 1 ~ ~ =1 ) CbEl ~ ) ) ~ , , b ) ) ~Thus . the family {x,,b : ( a , b ) E A x B} is an element U s of CcEAXl$K. It is easy to check that U , as just defined, is a unitary transformation from 7.1 onto CcEAxl $K. The $(CbEl $ A ) , for some general element C of C has the form CaEA A in A, and U C U * = CcEAxl $ A . Since A x B has cardinality mn, it now follows that UCU* is A 8 I,,. Thus UC'U' (= (VCU*)') is mn 8 A;so UC'U* and C' are of type I,,. 9.6.36. Let R and S be finite type I von Neumann algebras both acting on the same Hilbert space 7.1 and having the same center C, and let cp be a * isomorphism from R onto S such that cp(C) = C for each C in C. Prove that there is a unitary operator U in C' such that cp(A) = UAU* for each A in R. [Hint.Use Exercise 9.6.35 and the discussion preceding Theorem 9.3.1.1

Solution. Since R is a finite type I von Neumann algebra acting on the Hilbert space 7.1, there is an orthogonal family

{P,,,

:

1 L m < No, 15 n 5 dim'H}

of projections in C, with sum I , such that RP,,, and R'P,,, types I, and I,, respectively (unless P,,, = 0). Let

are of

be the corresponding family for the von Neumann algebra S. If we it follows show that Q,,, = P,,, (= cp(P,,,)), for all pairs (m,n), from the discussion preceding Theorem 9.3.1 (and from Theorem 9.3.4) that cp is implemented by a unitary operator U acting on 'H; moreover, U E C' since

ucu* = cp(C) = c

(C E C).

496


It remains to show that

Suppose that (1) is not satisfied. In this case, since {Pm,n} and are orthogonal families of projections in C, each with sum I , it follows that there exist distinct pairs, ( m , n )and ( T , s ) , such that PmlnQTIa is a non-zero projection E in C. The von Neumann algebras RE, R’E,SE, S’E all have center CE, and their types are I,, In, I,, I,, respectively. From Exercise 9.6.35, (CE)’ has type both I,, and I,,; so mn = T S . At the same time, since cp(E) = E, cp I RE is a * isomorphism from RE onto SE, and thus m = r . Since 1 5 m = T < N o and mn = T S , it now follows that n = 8, contrary to our assertion that ( m ,n ) and ( T , 8) are distinct pairs. Thus (1) is satisfied; as already noted, this suffices to prove the assertions in the exercise. {Qm,n}

Let R be a finite type I von Neumann algebra with 9.6.37. center C and let B be a von Neumann algebra such that C E B 5 R. Prove that there is an orthogonal family {F’} of projections in B, with sum I , such that FjBFj = CFj for each index j . [Hint.Use Corollary 6.5.5 .]

Solution. By a familiar exhaustion argument, it suffices to prove that each non-zero projection E in B has a non-zero subprojection F in f3 such that FBF = CF. To this end, let PE be the set of all projections G in B such that 0 < G 5 E . Given G in P E , R contains an abelian projection EOwith the same central carrier as G, since R is of type I. By Corollary 6.5.5 (and since R is finite), there is an orthogonal family (QI(G),Q2(G),Qa(G), . . .} of projections in C, with sum CG,such that GQj(G) is the sum of j equivalent abelian projections in R. Let k be the smallest positive integer with the property that, for some projection G in P E ,Qk(G) # 0. Then m

n=k

and we can choose Go in PE such that Qk(G0) is a non-zero projection Q in C. Let E’ be QGo, so that F E PE and F is the sum of k abelian projections in R, each having central carrier Q.

EXERCISE 9.6.38

497

To complete the solution, we shall show that FBF = CF. Since F B F , CF are von Neumann algebras and CF C_ FBF, it suffices to show that CF contains each non-zero projection G in FBF. Given such a projection G , we have G E FBF C B and 0 < G 5 F 5 E ; SO G E PE.Thus a3

F 1G =

CGQ~(G). n=k

Now FQ,(G) 2 GQn(G),FQn(G) is the sum of k abelian projections each with central carrier Q n ( G ) ,and GQn(G)is the sum of n such projections. It follows from Proposition 6.4.6(iii) and the assumption that R is finite that FQk(G) = GQk(G),and Q , ( G ) = 0 when n > k. Thus

9.6.38. Suppose that R is a finite type I von Neumann algebra with center C and for j = 1,2, dj is an abelian von Neumann algebra such that C C dj C R . Let y be a * isomorphism from d1 onto dz such that y(C) = C for each C in C, and suppose also that T ( ~ ( A= ) )T ( A )for each A in d l , where T is the center-valued trace on R. Show that there is a unitary operator V in R such that cp(A) = V A V * for each A in d1. (Note that this exercise prepares for, and is subsumed in, the one that follows.) [Hint.Apply Exercise 9.6.37, with d1 in place of B, so obtaining an orthogonal family { F j } of projections in A1. Show that Fj q ( F j ) (in R) for each index j . Let V be C Vj, where Vj is a partial isometry in R that implements the equivalence of Fj and q(Fj).] N

Solution. By Exercise 9.6.37, there is an orthogonal family { F j } of projections in d1,with sum I, such that

dl Fj = Fjd1 Fj = CFj for each index j . Since ~ ( q ( F j ) = ) r ( F j ) , it follows (Theorem 8.4.3(vi)) that Fj q ( F j ) (in R ) ,so there is a partial isometry Vj in R with initial and find projections Fj and q ( F j ) ,respectively. Since C Fj = I , we have C q ( F j ) = I , and Vj is a unitary operator V in R. N

498


Given any A in ~ 4 1 we , can find a family {Cj}of elements of C such that AFj = CjFj. We have

= C V A F j V * = VAV* . Suppose that !& and 2l2 are self-adjoint subalgebras 9.6.39. of a finite type I von Neumann algebra R,and cp is a * isomorphism from 2l1 onto 2l2 such that r(cp(A))= r ( A ) for each A in U1, where r is the center-valued trace on R. Show that there is a unitary operator U in R such that 'p(A) = UAU* for each A in %I. (Hint. By use of Exercise 9.6.1, reduce to the c m in which R' is abelian. Then use Exercise 8.7.2 to reduce to the case in which ?2l1 and 2 l 2 axe von Neumann algebras that contain the center C (= R') of R, and p(C) = C for each C in C. By applying Exercise 9.6.38, with Aj the center of %j, reduce further to the case in which 2l1 and 8 2 have the same center. Complete the argument by using Exercise 9.6.36.1 Solution. The property to be established is preserved by a * isomorphism from R onto another von Neumann algebra. From this, and in view of Exercise 9.6.1, we may assume that R' is abelian. Thus R' C R" = 72, R' = R f l12' = C, and 73 = C'. For j = 1,2, let Bj be the * subalgebra of R generated by CU2lj. From Exercise 8.7.2, cp extends to a * isomorphism 8 from B1 onto B2,such that p(C) = C and r ( p ( B ) )= r ( B ) whenever C E C and B E B1. Moreover, 8 extends to a * isomorphism 8 from t?; onto B,; of course, @ leaves C elementwise fixed, and (by the ultraweak continuity of r and q), r o 8 = r . If $3 is implemented by a unitary element of R,then so is cp (= @ I%I). Upon replacing 241, %2, cp by S,, Si ,8, respectively, we may henceforth make the additional assumptions that %I and 242 are von Neumann algebras containing C, and cp(C) = C for each C in C. The center dj of 2lj contains C, and 'p 1 A1 is a * isomorphism from d1 onto A2 that satisfies the hypothesis of Exercise 9.6.38. From that exercise, there is a unitary operator V in 72 such that cp(A) = VAV* for each A in d1. Upon replacing 2 l 2 by V*2l2V, and Q by the * isomorphism

A --+ V*v(A)V : %1

--+

V*?2l2V,

499

EXERCISE 9.6.40

we may henceforth make the additional assumption that IzLl and 2f2 have the same center, and cp leaves this center elementwise fixed. After the preceding reductions, it suffices to prove that cp is implemented by a unitary operator in R when (in addition t o the assumptions listed in the exercise) 2l1 and M 2 are von Neumann algebras with the same center CO,R' = C E CO,and cp leaves CO elementwise fixed. For k = 1,2, we c m apply Exercise 9.6.37 with 2 l k in place of B; the family { F j } so obtained consists of abelian projections in U k , and has sum I. Thus U k is a type I von Neumann R. From Exercise 9.6.36, cp is algebra, and is finite since U k implemented by a unitary operator in CA (C_ C' = R). 4851 Suppose that { P I ,P2, P 3 , . . .} is a sequence of pro9.6.40. jections in an abelian von Neumann algebra d. Let BO be the C*-subalgebra of d generated by the positive operator Ao, where A0 = x:=13-nP,. Show that P, E % (n = 1,2,3,...). [Hint. Let A0 be the C*-algebra generated by (Ao, PI, P2, ..}, and let X be a compact Hausdorff space such that do S C ( X ) . Use Theorem 4.1.8(ii) and Example 4.4.9 to find a continuous function f such that Pl = f(Ao).I

.

Solution.

e2,.

With the notation of the hint, suppose that fo, e l ,

. ., are the functions in C ( X )representing Ao, P I ,P2, . . ., respec-

tively. Then en is the characteristic function of some (clopen) subset Xn of X. Now Ifo(p)l 2 when p E XIand Ifo(p)I I 3-" = Q when p 4 XI.Let f be a continuous function on R that takes the value 1 on [f,oo) and 0 on (-a, Then f o fo = e l . From Theorem 4.1.8(ii) and Example 4.4.9, f ( A o ) = PI and f ( A 0 ) is the norm limit of polynomials in A0 (without constant term). Hence PI E !&. Since

4

81.

03

C 3 - " P n = A0 - 3-lPl E %, n=2

P2 E !& by a similar argument. Continuing in this way, we have that P, E 00( n = 1,2,3,...). m 9.6.41. Suppose that A is an abelian von Neumann algebra acting on a separable Hilbert space. Show that A contains a positive operator that generates A as a von Neumann algebra. [Hint. Use part of the proof of Theorem 9.4.1, together with Exercise 9.6.40.1

500


Solution. Since d acts on a separable Hilbert space, d is countably decomposable, and so has a separating unit vector 5, by Corollary 5.5.17. Let P be the set of all projections in A. By reasoning as in the third paragraph of the proof of Theorem 9.4.1, but with A an arbitrary element of the cornmutant A' instead of the algebra A itself, it follows thai P has a countable strong-operator dense subset { P I ,P2, P3, . ..}. The von Neumann algebra generated by this subset contains P , and is therefore the whole of A. Define A0 in A+ by

n=l

By Exercise 9.6.40, the von Neumann algebra generated by tains each Pn, and so coincides with d . ~[80]

A0

con-

By noting that every abelian von Neumann algebra 9.6.42. acting on a separable Hilbert space is * isomorphic to one of the algebras described in Theorem 9.4.1, give another proof of the result of Exercise 9.6.41.

Solution. It follows from the first two paragraphs of Section 9.4 that each abelian von Neumann algebra acting on a separable Hilbert space is * isomorphic to a maximal abelian von Neumann algebra acting on a separable Hilbert space, and is therefore * isomorphic to one of the algebras described in Theorem 9.4.1. Since a * isomorphism between von Neumann algebras is a homeomorphism relative to the ultraweak topologies (Remark 7.4.4), it now suffices to prove that each of the algebras in Theorem 9.4.1 is generated (as a von Neuniann algebra) by a single positive element. We recall that A, is the multiplication algebra relative to standard Lebesgue measure m on the unit interval [0,1]. When 1 5 k 5 No, the algebra d k in Section 9.4 can be viewed as the multiplication algebra of a measure space (Sk,mk) in which sk has k points, to each of which mk assigns unit mass, and every subset of s k is measurable. Let A , (in d ; ) be the operator M , of multiplication by the identity mapping L on [0,1], and let A,+ (in be the operator MI< of multiplication by a function Ir' : S k -+ [2,3] that takes distinct values at the points of sk. Note that, for both L and Ir', the essential range (as defined in Example 3.2.16) contains the range. We assert that the von Neumann algebras A,, d k , and A, $Ak have positive generators A,, Ak, and A,$Ak, respectively. To prove

di)

EXERCISE 9.6.43

501

this, suppose that T, E A, and T k E d k . Then T, is the operator of multiplication by some bounded Borel function g c on [0,1], and Tk is the operator of multiplication by a bounded function g k on Sk. We can find a bounded Borel function g on R such that

From Exercise 5.7.27, T, = g ( A , ) and Tk = g ( A k ) ; so, by an easy application of the uniqueness of the bounded Borel function calculus (Theorem 5.2.9),

Since Tc = g(A,), T, lies in the von Neumann algebra generated by A,; this has been proved for all T, in A,, so A, generates A, as a von Neumann algebra. Similarly, Ak (respectively, A, @ Ak) generates rn d k (respectively, A, @ dk) as a von Neumann algebra.

9.6.43. Let A be the abelian von Neumann algebra generated by a bounded normal operator A acting on a separable Hilbert space H , and let B(sp(A)) be the algebra of all bounded Borel functions on sp(A). Recall from Remark 9.5.13 that the (continuous) function calculus for A is a representation ( P A : f + f(A) of C(sp(A)) on 'H. Let ( P A : L o o ( v ~-,) B(7-l) be the extension of 9~ described in Proposition 9.5.3. Show that ( P A ( g ) = g ( A ) for each g in B(sp(A)), and deduce that

Solution. We recall that L,((PA) is the algebra of all Borel functions on sp(A) that are bounded on the complement of some set in N W Athe , null ideal of ( P A . Thus

where N ( P A )is the algebra (ideal, in & ( ( P A ) ) of all Borel functions on sp(A) that vanish on the complement of some set in NVA. The range C of ' P A is the C*-algebra generated by A (Theorem 4.4.5), so C- = A. By Proposition 9.5.3, (PA : L o o ( q ~ + ) B(R) is a * homomorphism with kernel N ( c p ~and ) range C- (= A). From this,

502


together with ( l ) , it follows that (PA JB(sp(A)) is a * homomorphism from B(sp(A)) into 8(3-1),and has range A. It remains to prove that (PjA(g) = g ( A ) for each g in B(sp(A)). By Theorem 5.2.9, it will suffice to prove that the * homomorphism ( P A I f?(sp(A)) is a-normal, since it is apparent that (with the notation of that theorem) cp~(1)= I and ( P A ( & ) = A. Suppose that { g n } is an increasing sequence of elements of B(sp(A)), converging pointwise on sp(A) to an element g of B(sp(A)). From the monotone convergence theorem (and with the notation used in Proposition 9.5.3)

( P A ( g n ) Z ,x) =

lp(A) gn d p x

Lp(A)

gdpx

= (qA(g)xc, Z ) as n + 00, for each x in 3-1. It follows that the increasing sequence { ( P A ( g n ) } of self-adjoint operators has least upper bound @ A ( g ) ; so m[80(Satz 6, p. 213)] ( P A I B(sp(A)) is a-normal.

9.6.44. Suppose that {A, : a E A} is a commuting family of bounded self-adjoint operators acting on a separable Hilbert space ‘FI. Show that there is a bounded positive operator H acting on X, and a family { g , : a E A} of bounded Bore1 functions on sp(H), such that ( a E A). A, = g a ( H ) Solution. Let A be the abelian von Neumann algebra generated by {A, : a E A}. By Exercise 9.6.41, A is generated (as a von Neumann algebra) by an element H of A+. By Exercise 9.6.43,

A = M H ) : 9 E B(SP(H))I ‘ Hence, for each a in A, A , = g , ( H ) for some g , in B(sp(H)).

rn

where X is a 9.6.45. Let U be the abelian C*-algebra C(X), compact Hausdorff space. Show that %is generated (as a C*-algebra) by a single self-adjoint element if and only if X is homeomorphic to a compact subset of R. Deduce that the statement of Exercise 9.6.41 does not remain true if the term “von Neumann algebra” is replaced by “C*-algebra” at both occurrences.

EXERCISE 9.6.46

503

Solution. Suppose first that U is generated, as a C*-algebra, by a self-adjoint element Ao. Since 2l is C ( X ) , and the C*-algebra generated by A0 is * isomorphic to C(sp(A0)) (Proposition 4.1.4 and the discussion preceding it), it follows that C ( X ) 2 C(sp(A0)). By Theorem 3.4.3, X is homeomorphic to the compact subset sp(A0) of

lw. Conversely, suppose that X is homeomorphic to a compact subset Y of R. Since polynomials are dense in C ( Y ) ,by the Weierstrass approximation theorem, the identity mapping on Y is a self-adjoint element fo of C ( Y )and generates C ( Y )as a C*-algebra. From this, and since C ( X ) E C ( Y ) , it follows that C(X)has a self-adjoint generator. If D is the unit disk { z E C : Izl 5 I}, then D is not homeomorphic to a compact subset of R (because D\ {ZO} is connected, for each zo in D). From the preceding discussion, C(D)is not generated, as a C*-algebra, by a single self-adjoint element. Finally, note that C(D)is * isomorphic to an abelian C*-algebra of operators acting on a separable Hilbert space-for example, C(D)acts as an algebra of multiplication operators on L2(D, m ) , where m is Lebesgue plane measure-and this algebra has no self-adjoint generator. 9.6.46. Suppose that S is a set, and for each bounded complexvalued function f on S, denote by Mj the bounded linear operator, acting on l z ( S ) , of multiplication by f . Let d be the abelian von Neumann algebra { M j : f E lm(S)}.Prove that (i) if the cardinality of S exceeds that of R, there is no selfadjoint element A0 of A that generates A as a von Neumann algebra (compare this with Exercise 9.6.41); (ii) if S is the interval [0,1], L is the identity mapping on S, and A0 is the self-adjoint element M , of A, then A0 generates A as a von Neumann algebra, but

(compare this with Exercise 9.6.43). Solution. (i) Each self-adjoint element A0 of A has the form M,, with g a bounded real-valued function on S. Since the cardinality of S exceeds that of lw, there are distinct elements $1 and $2 of S. such that g(s1) = g(s2). Thus A0 lies in the C*-algebra do (a

504


proper subset of d)defined by

In fact, do is weak-operator closed, and is therefore a von Neumann algebra; for

where xj (in 44s)) takes the values 1 at sj and 0 on S \ {sj}. Since A0 lies in the von Neumann algebra do,and do is a proper subset of d,A0 does not generate d as a von Neumann algebra. We have now proved that d is not generated by a single self-adjoint operator. This shows that the statement of Exercise 9.6.41 does not remain true if the word “separable” is deleted. (ii) The operator A0 (= M L )is self-adjoint, and sp(A0) = [0,1] = S. We assert that

In order to prove ( l ) ,it suffices (by Theorem 5.2.9) to show that the * homomorphism

is o-normal, since (with the notation of that theorem) cp(1) = I and V ( L )= A o . Suppose that g , 91,g2,. . . E D(sp(Ao)), and the sequence { g n } is increasing and pointwise convergent to g . The increasing sequence {(p(gn)}of self-adjoint operators in A is bounded above by cp(g), and so has a least upper bound B (= B * ) in A, and B = Mh for some bounded real-valued function h on S. Given any s in S (= sp(Ao)), let x, (in I2(S)) be the characteristic function of the one-point set {s}. Then

= lim g n ( s ) n-+m

= ds)

(s E S).

Thus h = g , B = M h = M g = V ( g ) , cp is o-normal, and (1) is proved.

EXERCISE 9.6.46

505

From (l),together with Theorem 5.2.9, it follows that

where A0 is the von Neumann algebra generated by Ao. Each element of A has the form M j , where f is a bounded complex-valued function on S. We shall show that M j E A0 (whence A = do,and A is generated as a von Neumann algebra by Ao). To this end, it suffices to show that if E > 0 and 11,. .. ,I, are vectors in / 2 ( S ) ,there exists A in A0 such that

(3)

I l M r ~ j- Azjll < E

( j = 1,...,n ) .

Now

so there is a countable subset SOof S such that

( S E S \ S o , j = l ,. . . , n ) .

q(s)=O

We can define a bounded Borel function g on S (= sp(A0)) by

We have MgXj

=M p j

( j = 1,. . . ,n ) ,

and M g E A0 by (2). Thus (3) is satisfied, with A the element M g of d o . We have now shown that d is generated, as a von Neumann algebra, by Ao. With f a bounded complex-valued function on S that is not a Borel function, let A be the element M f of A. When g E B ( s p ( A o ) ) ,f(s) # g ( s ) for some s in S, and

Thus

A E A,

A

4

M A o ) : g E B(SP(Ao))I*

Hence the final assertion of Excercise 9.6.43 does not remain valid without the assumption that 7-l is separable.

506


9.6.47. Let S be the compact subset {O,l,i,i,. . .,i,.. .} of R, and let SO be S \ (0). Let cp and cpo be the representations of C(S) in which the elements of C(S) act by pointwise multiplication on the separable Hilbert spaces 4 ( S ) and 62(So), respectively. (i) Show that cp and cpo are faithful. (ii) Prove that the weak-operator closures of the C*-algebras cp(C(S)) (= U) and cpo(C(S)) (= Uo) are maximal abelian von Neumann algebras, and there is a unitary transformation V from I2(S) onto 62(So) such that VU-V' = 3,. (iii) Determine the null ideals of cp and cpo, and deduce that cp and cpo are not equivalent. (iv) With the notation of Proposition 9.5.3,determine the mappings (p and PO,and observe that (p is one-to-one but (po is not. (v) Let Q be an automorphism of C(S),and let 77 be the homeomorphism of S such that a(f)= f o 7 for each f in C(S) (see Theorem 3.4.3). Show that ~ ( 0 )= 0, and deduce that there exist unitary operators IJ and UO acting on l2(S) and 22(S0),respectively, such that

cp(Q(f)) = Uco(f)U*,

(f E W)).

cpo(a(f))= UocPo(f)Uo*

(vi) Show that there is no unitary transformation W from / 2 ( S ) onto Iz(S0) such that WUW* = 3 0 . (vii) Deduce thah the abelian C*-algebras U (acting on I z ( S ) ) and UO(acting on I z ( S 0 ) ) are * isomorphic, their weak-operator closures are unitarily equivalent, there is a * isomorphism from U onto U o that extends to a * homomorphism from U- onto U,, but there is no * isomorphism from U onto % that extends to a * isomorphism from 2.l- onto q-.

Solution. (i) When s E S, let z8 (in I2(S)) be the function that takes the value 1 at s and vanishes elsewhere on S. I f f E C(S) and cp(f) = 0, then 0 = (cp(f)zs)(s)= f(s)zs(s) =

f(4

(8

E 8);

f = 0, and

cp is faithful. When s E So, let ys (in lz(S0)) be the function that takes the value 1 at s and vanishes elsewhere on SO. If f E C(S) and cpo(f) = 0, then

so

0 = (cpo(f)Ys)(4 = f(sbs(4 = f(s)

(8

E

30);

507

EXERCISE 9.6.47

= 0, since f vanishes on the dense subset SO of S, and 90 is faithful. (ii) Let A, acting on 22(S), be the maximal abelian algebra {Mg: g E Im(S)},where M gis the operator of pointwise multiplication by the bounded function g on S. Let A0 by the corresponding algebra acting on Iz(S0). Given any one-to-one mapping h from So onto S, the mapping 2 --+ z o h from Iz(S) onto 12(So) is a unitary transformation V such that VdV* = do. It now suffices to show that rzL- = A and %- = do. Since v(f) is M , when f E C(S), it is apparent that 24 A, whence U- C A. Each element A of d has the form M,,with g in 6m(S). For k = 1,2,..., let gk be the element of C ( S )defined by so f

gk(i) gk(

(1 L n

=

k) = g k ( 0 ) = g(0)

I k),

( n > k).

Since llgkll 5 llgll and g ( s ) = l i m k , , g k ( $ ) for each s in S, it follows that A = M g = lim M,, = lim v ( g k ) k-rm

k+w

in the strong-operator topology (see the end of Example 2.5.12). Thus A E U-, and A = 8 - . A similar argument shows that do = 3,. (iii), (iv). Given 5 in Iz(S) and g in Zoo(S),we have (Mgz,

4= C(Mgz)(s)Zo = SES

c

g(s)l.(s)l2

*

SfS

Since S is countable, each y in I,(S) is a Borel function, and it follows from the preceding equations that

where p2 is the regular Borel measure on S that assigns mass 1z(s)I2 to s, for each s in S. In particular,

(2)

(V(f b,4 =

J,f d P I

.( E /2(S),

f E W)),

so pZ is the measure occurring in equation 9 4 5 ) and the ensuing discussion.

508


We can choose z in /2(S) so that z(s) # 0 for all s in S. Then, z is a separating vector for d (= q(C(S))-), and the only p,-null subset of S is the empty set 0. It follows that the null ideal JV~ consists of 0 only. Since Np = (0) and every complex-valued function on S is a Borel function, it follows that L,(cp) is 1,(S) and N ( 9 ) (C L,(cp)) consists of the zero function only (see the discussion preceding Proposition 9.5.3 for the definitions of L,(cp) and N((c)). Since N(cp) = {0}, the mapping p occurring in Proposition 9.5.3 is a * isomorphism. In fact, when g E L,(cp) (= l,(S)), p(g) is the unique bounded operator acting on Iz(S) such that

and it follows from (1) that

The preceding three paragraphs can be amended, in a fairly obvious way, so as to apply to cpo rather than q;we give a sketch of these arguments. The weak-operator closure do of cpo(C(S)) can be expressed in the form

where Mi,acting on /z(So), is the operator of pointwise multiplication by the bounded function g I SO.Given y in SO), (3)

where u, is the regular Borel measure on S defined by V,({O})

In particular,

= 0,

VY((4)

= lV(4l2

(8

E So).

EXERCISE 9.6.47

509

We can choose y in SO) so that y(s) # 0 for all s in SO. Then, y is a separating vector for do (= cpo(C(S))-),and the only v,-null subsets of S are 0 and (0). It follows that the null ideal Nq0 consists of these two sets. Since N , # N,,, cp and cpo are not equivalent. Since N,, = {0,{0}}, and every complex-valued function on S is a Bore1 function, it follows that L,(cpo) is I,(S) and N(cq0) (c L,(cpo)) is the one-dimensional subspace consisting of all functions on S that vanish on SO.From Proposition 9.5.3,together with (3), GJo(9)= Mg”

(9 E L ( S ) = ~ c & o ) ) ,

and 90 is not one-to-one. (v) Since 0 is the only point of S that is not an isolated point of S , it is a fixed point of any homeomorphism of S;so q(0) = 0, whence SO) = SO. The equations

define unitary operators U and UO acting on Iz(S) and I2(So), respectively. When f € C(S),we have

for each s in S, and hence

This proves that

and a similar calculation shows that

The results of (v) can be obtained by another method, that we now outline. We have to show that the representations cp o a and yo o a of C(S) are equivalent to p and PO,respectively. For this, it suffices to prove that

510


since all these representations have uniform multiplicity 1, by Proposition 9.5.9. Arguments similar to those used above in the solution to (iii) and (iv), taking into account the fact that ~ ( 0 )= 0, show that Nvoa = (0) = Nq, N p o o a = (0,(0)) = Nqo 0

(vi) Suppose there is a unitary transformation W from /2(S) onto Zz(S0) such that WUW' = 2i0. Since Q and cpo are faithful, and have ranges U and 2l0, respectively, we can define an automorphism a of C(S) by

4 f ) = Qil(Wcp(f)W*)

(f E C(S)).

With UO chosen as in (v), we have

and cp and cpo axe equivalent, contradicting (iii). Hence there is no such unitary transformation W. (vii) From (iv) and (i), cpo o cp-' is a * isomorphism from 2l onto !&,and extends t o a * homomorphism cpo o 9-l from 2l' onto h-. From (ii), U- and U i are unitarily equivalent. Suppose that 11, is a * isomorphism from U onto UOthat extends to a * isomorphism $ from ?2l- onto !&-. Since U- and K- are maximal abelian von Neumann algebras, it follows from Remark 7.2.10 (or Theorem 9.3.1) that 4 is implemented by a unitary transformation W of the underlying Hilbert spaces. Thus

WUW* = $(U) = +(a)= 2 0 , contradicting (vi). Hence there is no envisaged. ~[57]

* isomorphism II, of the type

9.6.48. The purpose of this exercise is to examine the conclusions of Proposition 9.5.3 as they apply to the representation considered in Example 9.5.10. Suppose that S is a compact Hausdorff space, and cp is the representation of C(S) constructed (as in Example 9.5.10) from a null ideal sequence {Nj} of separable type. Let C be the C*-algebra cp(C(S)),and let p : L,(cp) 4 C- be the * homomorphism associated with cp as in Proposition 9.5.3. With the notation used in Example 9.5.10, prove that 0) Lco(cp) = ~ , ( S , W ;

511

EXERCISE 9.6.48

(ii) when g E L,((o) and { 2 1 , 5 2 , . . .} is an element of the Hilbert space C,9"=, $'Hj (= H)on which p(C(S)) acts P ( g ) { z l , 22, - * = {921,952, .}; if Y is a Bore1 subset of S and E(Y) is the projection occur(iii) ring in Proposition 9.5.3, then a }

w ) { s l , 22,. *

whenever

.

( ~ 1 ~ x 2 .} , .E

.} = { P l , P

27.

- .}

3-1, where q is the characteristic function of

y; (iv) Q j = E(Yj).

Solution. (i) Near the end of Example 9.5.10, we define a separating vector u for C- , and note that (df)Qju,Qju) = L f e j d P for j = 1,2,. so

s,

. . . Now

Q1

is I and

((o(f)% 4=

el

(f E C ( S ) )

is the constant function 1 on

J,f

(f E C ( S ) )

*

Thus pu = p , and from the discussion preceding Proposition 9.5.3, L,(cp) is L,(S,S,p). (ii) When g E L,(S,S,p), we can define $(g) in B(3-1) as follows: ' d ' ( g ) { ~ l , x ' 2 , - -= . }{ 9 2 1 , g 2 2 , . . . }

whenever { 2 1 , 5 2 , . . .} E 3-1 (= Cj"=, $ Z j ) . From the definition of cp, $(f) = cp(f) when f E C(S);we want to prove that $ = (p. Let 5 be an element { 2 1 , 2 2 , . . .} of 3-1. For all g in L m ( S , S , p ) , ($(g)x,

2)

= ( { g z l , 9x2 7 - * 00

j=1

00

a},

{ 51 7 22

7

* *

.})

512


where k, is the element of L l ( S , S , p ) defined by

The change of order of summation and integration, in the preceding manipulation, is justified by the dominated convergence theorem, since the partial sums of the series C g(s)lzj(s)I2are dominated by the L1 function Ilgllwlcx(s). From the preceding paragraph, together with equation 9.5(5), it follows in particular that

for all f in C(S). Since the measures p and p x are regualr, this implies that p x is absolutely continuous with respect to p , and

for all g in L,(S, S,p ) . It now follows, from the preceding paragraph and Proposition 9.5.3, that

and hence that p(g) = $ ( g ) , for all g in L,(S,S,p) (= L,(cp)). (iii) The stated result is an immediate consequence of (ii) since, as noted in Proposition 9.5.3, E(Y)is cp(q). (iv) In Example 9.5.10 we have noted that { e l , O,O, .. .} is a separating vector u for C-, and Q j u = { e j , O , O , . . .}, where e j is the characteristic function of Yj. It now follows from (iii) that

E(Yj)u = E ( Y j ) { e l ,O,O,. . .} = { e j e l , O , O , . ..} = { e j , O , O , . . .} = Q j u , and hence that E ( Y j ) = Q j .

EXERCISE 9.6.49

513

9.6.49. (i) Suppose that S is a compact subset of the complex plane C,p is a (regular) measure defined on the a-algebra S of Borel subsets of S , and ( 5 )is a decreasing sequence of Borel subsets of S, with Y1 = S. Let ‘H be Cj”=, $ ‘ H j , where ‘ H j is the closed subspace of L2 (= L 2 ( S , S , p ) )consisting of the functions in L2 that vanish almost everywhere on the complement of Yj. Let L be the identity mapping on S , and let A0 be the element of B(’H) defined as follows:

whenever { 5 1 , 5 2 , . . .} E Cj”=, $‘Hi. Show that A0 is normal. (ii) Suppose that A is a bounded normal operator acting on a separable Hilbert space ‘ H A . By use of Remark 9.5.11 and the first paragraph of Remark 9.5.13, show that it is possible to choose S (= sp(A)), p , and {Yj}, satisfying the conditions set out in (i), in such a way that A is unitarily equivalent to the operator A0 described in (i). Solution. (i) The equation

defines a representation cp of the abelian C*-algebra C ( S ) on the Hilbert space ‘H (compare this with Example 9.5.10). Hence ( P ( L ) (= Ao) is normal, since it is an element of the abelian C*-algebra

cp(W)).

(ii) It has been noted in Remark 9.5.13 that the (continuous) function calculus for A is a representation ( P A of C(sp(A)) on ‘ H A , and A = c p ~ ( ~ From ). Remark 9.5.11, c p is~ equivalent to a representation tp of the type considered in Example 9.5.10 (with S = sp(A), and with suitably chosen p and {Yj} satisfying the conditions set out in (i)). Note that ( P ( L ) is the normal operator A0 occurring in (i). There is a unitary transformation U such that

By taking

L

for f , we obtain A = UAoU*.

9.6.50. Suppose that A is a bounded normal operator acting on a separable Hilbert space ‘H. We say that A has uniform multiplicity n , where 1 2 n 5 No, if the (continuous) function calculus for

514


A has uniform multiplicity n when viewed (as in Remark 9.5.13) as a representation V A of C(sp(A)) on H. (i) Show that A has uniform multiplicity n if and only if the abelian C*-algebra U generated by A, A*, and I has commutant 31' of type I,. (ii) Suppose that S is a compact subset of the complex plane @, p is a (regular) measure on the a-algebra S of Bore1 subsets of S, L is the identity mapping on S,M , is the bounded operator acting on L2 (= L 2 ( S , S , p ) ) of multiplication by L , and K is a Hilbert space of dimension n. Show that the bounded operator M , 8 I, acting on L2 @ K , is normal and has uniform multiplicity n. (iii) Show that, if A has uniform multiplicity n and S = sp(A), then p can be chosen in such a way that the conditions set out in (ii) are satisfied and A is unitarily equivalent to the operator M , 8 I described in (ii).

Solution. (i) From Theorem 4.4.5, the C*-algebra U is the range, cp~(C(sp(A))),of the function calculus QA of A. By Proposition 9.5.9, with Y = S, a representation cp of @(S) has uniform multiplicity n if and only if the range C of cp has commutant C' of type I,. It follows that U' is of type I, if and only if ( P A(equivalently, A) has uniform multiplicity n. (ii) From Theorem 9.5.12, the mapping $ : f

+

M , 8 I : C(S) + B(L2 8 K )

is a representation of the abelian C*-algebra C(S), and has uniform multiplicity n. Thus A0 is normal, where A0

= $(L) = M, @ I ;

moreover, +(C(S)) has commutant of type I,. Since C(S) is generated (as a C*-algebra) by L , E, and the constant function 1, $(C(S)) is generated by Ao, A,*, and I . It now follows from (i) that A0 has uniform multiplicity n. (iii) Suppose that A has uniform multiplicity n. The function calculus ( P A of A is a representation of @(sp(A)) on H , and has uniform multiplicity n , and is therefore equivalent to a representation $ of the type described in Theorem 9.5.12 (with sp(A) in place of S). Thus A (= ( P A ( L ) )is unitarily equivalent to M , @ I (= $(I,)). w

515

EXERCISE 9.6.51

9.6.51. Let R be a von Neumann algebra with center C, acting on a Hilbert space H , and let J be a conjugate-linear isometry of H onto 3-1 such that J 2 = I, J R J = R',and JCJ = C* for each C in

C.

(i) Show that A -+ JA* J is a * anti-isomorphism of R onto R'. be a * anti-isomorphism of R onto R'. Show that (ii) Let there is a unitary operator U on H such that (the conjugate-linear isometry) J U implements +:

+

JUA*(JU)*= +(A)

( A E R).

[Hint. Use Exercise 9.6.25.1

Solution. (i) Note, first, that since J* is the mapping of H into 'H obtained from J by using the adjoint of J when J is viewed as a linear mapping of 3-1 into 72 and that J so viewed is a unitary transformation of 3.1 onto R, we have J = J* (as mappings of % into 3-1) for both J and J* are the mapping inverse to J on 3-1. Note, too, that with A in B(3.1), (JAJ)* = JA*J, for JAJ E B ( H ) and, with z , y in H,

( ( J A J ) * z , y )= ( x ,J A J y ) = (AJy,J z ) = ( J y ,A*J z ) = (JA*J x , y) . Thus

cp(A*)= JA**J = JAJ = (JA*J)*= cp(A)*, where cp(B) = JB*J for B in

R.In addition, with A , B in R,

+

cp(aA t B ) = J(aA B)*J = JizA*J t J B * J = aJA*J JB*J

+

= w ( A )+ cp(B>, and

(p(AB)= J(AB)*J = JB*JJA*J = cp(B)cp(A). With A' in R', there is, by assumption, an A* in R such that cp(A)= JA*J = A', so that cp maps R onto R'. Finally, since J is an isometry of 3-1 onto H,given an z in 'H there is a y in 'H such that J y = Z . If 0 = p ( A ) = JA*J for some A in R , then 0 = JA*Jy =

516


J A * z , and A*z = 0 for each 2 in 'FI. Thus A* = 0 and A = 0. It follows that cp is a * anti-isomorphism of R onto R'. (ii) With cp as in (i), Iet 7 be the * isomorphism cp-l o $ of 7Z onto R. From Exercise 9.6.25, there is a unitary operator U such that 9 ( A ) = UAU* for each A in R. Hence $ ( A ) = (p(UAU")= J(UAU*)*J = JUA*(JU)*

-

( A E R)

9.6.52. Let R be a von Neumann algebra acting on a Hilbert space ')I with generating and separating vector u. Let So, S , Fo, F, J, and A be the operators defined in Section 9.2. Suppose J' is a conjugate-linear isometry of 3;1 into H such that J'u = u , Jt2 = I ,

J'RJ' = R', (AJ'AJ'u,u) 2 0

( A E R).

Let HoAu be J'A*u ( A E R ) and U be J'J. Show that (i) 0 _< ( A J A J u , u ) (A E R); (ii) HO = J'S", and Ho has closure J'S(= H); (iii) ( H z , s ) 2 0 for each 2 in D ( H ) ( = D(AlI2)) and H is symmetric [Hint.Use Exercise 7.6.52(i).]; (iv) H is self-adjoint [Hint.Note that H* = S:J' and that J'R'u (= Ru) is a core for H* as well as for H . ] ; (v) H is positive and H = UA1I2; (vi) J = J', and J is the unique operator with the properties assumed for J'. Solution. (i) With A in R, we have ( A J A J U ,U ) = ( A J S A * U U, ) = ( A ~ / ~ AA **U~) ,2 0, since All2 >, 0. (ii) By definition D ( H 0 ) = D(J'S0) = D(S0) = Ru,and

Hence HO= J'S,. Now (5,y) is in the closure of the graph of J'S0 if and only if there is a sequence (2,) in the domain of SO,tending to s , such that {J'&sn} tends to y, which occurs if and only if (Soz,} tends to J'y. Thus, (s,y) is in the closure of the graph of J'So if and only if (x,J'y) is in the closure of the graph of SO,that is, if

EXERCISE 9.6.53

517

and only if x E D(S)and S x = J'y (equivalently, J ' S x = y). Thus H = J'S. (iii) It follows from (ii) that D ( H ) = D(S)= D(A112). By hypothesis, with A in R,

( H A u , A u ) = (A*J'A*u,u)= (A*J'A*J'u,u) 2 0; hence ( H z ,E) 2 0 for each z in D ( H ) since Ru is a core for H . From Exercise 7.6.52(i), with A0 in place of H , H C H * . (iv) From (ii), HO = J'S0 and So = J'Ho. Thus

S l J ' C _ H," = H * ,

H,"J' C_ S,*,

H " = H," C_ S,*J'.

Hence, H * = S;J' = FJ'. Now R'u is a core for F , so that J'R'u is a core for FJ'. But H * = FJ' and J'R'u = J'R'J'u = Ru. Thus Ru (= D ( H 0 ) ) is a core for H and for H * . From (iii), H C H * , so that H = H * . (v) From (iii), ( H x , ~2) 0 for each x in D ( H ) . From (iv), H is self-adjoint. Hence H is positive. From (ii),

H = J'S = J' JA'I2 = UA'I2. (vi) Since H 2 0 and UA112is a polar decomposition for H , we have that I = U = J'J. Hence, J = J', and J is the unique operator with the properties assumed for J'. "61 9.6.53. Let R be a von Neumann algebra acting on a Hilbert space H,u be a separating and generating vector for R, and S, F , J and A have the meanings attributed to them in Section 9.2. With z a vector in H , let cp,(A) be ( A u , s ) for each A in R, and cpk(A') be ( A ' u , z ) for each A' in R'.Show that (i) z E D(S)(=D(F,*) = D(A'12)) and that Sz = z for a vector x in 3-1 if and only if the (normal) linear functional cpk on R' is hermitian; symmetrically, y E D ( F ) and F y = y if and only if 'py is hermitian; (ii) cp; 2 0 if and only if x = H u for some positive H affiliated with R [Hint. Use L , in combination with Exercise 7.6.55, and Lemma 9.2.28.1; (iii) the set of vectors x in 3-1 such that cpk 2 0 is a (norm-)closed cone V t in 3-1 (see p. 212) and conclude (by symmetry) that the same is true of the set V;I2 of vectors z in H such that cpz 2 0;

518


(iv) V t and V:l2 (of (iii)) are dual cones, that is, w E V t if and only if (20, w) 2 0 for each v in V i / 2 ,and w E VA12 if and only if (w, w ) 2 0 for each w in Vto; (v) V t is the norm closure of Rtu and Vi12 is the norm closure of R ' f u ; (vi) A112Rtu = R ' f u , A-'12R'+u = R+u,and deduce that V;'2, Vt are the norm closures of A'12Rtu, A-'/2R'tu, respectively. Solution. (i) If z E D(S)and Sz = z, then for each self-adjoint A' in R',

whence cp:(A') is real and cp: is hermitian. (See Corollary 9.2.30) Suppose, now, that cp; is hermitian and T' E R'. Then

(FT'u,s) = (T'*u,x)= cpk(T'*)= cp',(T') = ( T ' u , ~=) (z,T'u), whence z E D(F;) and s = F,'z = Sz. (ii) Suppose H is a positive operator affiliated with R such that u E D ( H ) and 2 = Hu. Let { E x } be the resolution of the identity for H (see pp. 310, 311), and let H , be HE,. Then H , E R and E,H & H,. Hence

H,u = E,Hu

+

Hu = x

( n+ m)

since E , tends to I in the strong-operator topology. Now (A'u, H,u) = (H,A'u,u) 2 0, when A' E R'+,since H , and A' are commuting positive operators. Thus 0 5 lim(A'u, H,u) = (A'u, H u ) = (A'u, z),

and 9; 2 0. Suppose z in 3-1 is such that 9: >_ 0. Then, in particular, pi is from (i). From Lemma 9.2.28, hermitian, and x E D ( S ) (= D(F,*)) L , 7 R (and Ltu = z). By definition of L,, L,T'u = T'z for each T' in R', so that

(L,T'u,T'u) = (T'z,T'u) = (T'*T'u,z) 2 0

EXERCISE 9.6.53

519

since T'*T' E R'+ and cpk 2 0. Thus (L,y, y ) 2 0 for each y in R'u, a core for L,, and ( L , z , z ) 2 0 for each z in D ( L , ) . From Exercises 7.6.54 and 7.6.55, L , has a positive self-adjoint extension (the F'riedrichs extension) H affiliated with R. As L x u = 2, H u = 2. (iii) If A' E R'+,then ( A ' u , a z + y ) 2 0 when a 2 0 and z , y E V t . Thus ax y E Y:. If v and -v are in Vz, then ( A ' u , v ) = 0 for each A' in R'+.Since each operator T' in R' is a linear combination of = 0. As [R'u] = 3-1, v = 0. Thus (four) operators in R'+,(T'u,v) V: and, symmetrically, Vi12 are cones in 3-1. If (5,) is a sequence of vectors in V t tending to z in norm and A' E R'+,then 0 5 ( A ' u , ~ , )-, ( A ' u , ~ ) .Hence and, symmetrically, V;l2 are (norm-)closed cones in 'H. (ivj If v E Vt'2, then ( A u , v ) 2 0 for each A in R+.If 1u E V:, then w = H u for some positive H affiliated with R from (ii). With H , as in the solution to (ii),

+

0 5 (H,u,v)+ ( H u , v ) = (w,v).

If (w,v) 2 0 for each v in then 0 5 (w,A'u) = ( A ' u , w ) for each A' in R'+,since A'u E U;12 for such A' (from (ii) applied with R' in place of R). Hence cp', 2 0 and w E V:. Thus w E V t if and only if (w, v) 2 0 for each v in U ; l 2 . Symmetrically, E VA12 if and only if (w, v) 2 0 for each w in V t . (vj From (ii), R+u C U:. If z E V:, there is a positive H affiliated with R such that z = Hu. With the notation of the solution to (ii), H,u E R+u and H,u + H u = 2. Thus V t is contained in the norm closure of Rtu and, hence, coincides with this norm closure. Symmetrically, V;12 is the norm closure of R'+u. (vij Let @ ( A )be JA*J for A in R . From the discussion at the beginning of Section 9.2, @ is a * anti-isomorphism of R onto R'. Hence @(R+) = R'+.With A in R+,Au E D ( S ) = D ( A 1 1 2 )and A 1 I 2 A u= J S A u = JA*u = J A * J u = @ ( A ) u . Thus A 1 / 2 R +=~R'+u. Since R'+u is norm dense in VAl2 from (v), V;12 is the norm closure of A 1 l 2 R + u . Symmetrically, with A' in R'+,A'u E 'D(F)= D(A-'l2) and

A-'I2A'u = JFA'u = JA'*u = JA'*Ju = @ - ' ( A ' ) u . Hence A-'/2'R'fu = R+u,and V t is the norm closure of the cone A - l2R'+u. m[ 1121

'

520


9.6.54. With the notation and assumptions of Exercise 9.6.53 and with w a normal state of R, show that (i) there is a vector w in V t such that w,lR = w [Hint.Use Theorem 7.2.3 to find a vector z such that w = w,lR and then use Theorem 7.3.2 to study cp’,.] (ii) JIw --uII = inf{[lz - u11 : w,lR = w } , where w is as in (i) [Hint. Use Exercise 9.6.53(ii) to express w as H u , with H a positive operator affiliated with R; Use Exercise 7.6.23(ii) to prove that if w,lR = w for some vector t, then

Re(z,

.>

5 ( H u ,4.1;

(iii) the vector ‘u in (i) is unique. [Hint.Prove that Re(z,u) = ( H Z L , ~with ) , the notation of the preceding hint, only when z = Hu and then use the formula of (ii).]

Solution. (i) Since u is separating for R,there is a unit vector z in ‘H such that w = w, 1 R from Theorem 7.2.3. From Theorem 7.3.2, there is a partial isometry V’ in R’ such that w‘ is a positive normal linear functional on R‘, where u’(A‘) = cp’,(V’A’) for each A’ in R’,and such that cp‘,(A’)= w’(V’*A’). Now

w’(A’) = cp’JV‘A‘) = (V‘A’u,z) = (A’u,V‘*z), so that (w =) V‘*z E

U:. In addition,

(A’u,Z ) = cp’,(A’)= ,’(,’*A’)

= cp’,(V’V’*A’) = (A’u, V’V’*).

Since u is generating for R’, z = V‘V’*z,whence wv I R = w, I R = w . If H is a positive operator affiliated with R and u E V ( H ) , (ii) then u E V ( H 1 1 2 ) H1I2u , E V ( H 1 1 2 ) , H1I2H1l2u = Hu (from 5.6.(18)), and A‘H’12u = H112A’ufor each A‘ in R’.Thus, if V‘ is a partial isometry jn R’,

((V’Ru,u)I = I(V’H’I2u,H’I2u)J

and (2)

R e ( V ’ H u , u ) 5 ( H u ,u ) .

521

EXERCISE 9.6.55

Suppose z is a unit vector in 3-t such that wz I R = W H I~R. From Exercise 7.6.23(ii), there is a partial isometry W' in R',with initial space [RHu],such that W'Hu = z . From (2),

Re(a,u)= Re(W'Hu,u)5 ( H u , ~ ) , so that

(3)

llHu - .]I2 = 2 - 2 R e ( H u , u ) 5 2 - 2Re(z,u) = 112 - u1I2.

From Exercise 9.6.53(ii), there is a positive operator H , affiliated with R such that w = Hu. From (3),

(iii) If w' is another vector in V t such that wvf I R = w , then

and v' = V ' H u for some partial isometry V' in R' with Hu in its initial space (where H is as in (ii)). Hence

Re(V'Hu,u) = Re(v',u)= Re(v,u)= ( H u , ~ ) , and the inequality of (2) is equality in the present case. It follows that

( V ' H u , u ) = I(V'Hu,u)l = Re(V'Hu,u) = ( H u , u ) , so that

Thus V ' H 1 / 2 u= H112uand w' = V'Hu = Hu = v.

41121

Let R be a von Neumann algebra acting on a Hilbert 9.6.55. space 'H, u be a separating and generating vector for R, and S, F , J and A have the meanings attributed to them in Section 9.2. With a in [0,3], let V t be the norm closure of {AOAu : A E R+}. (The notation Y t and V:l2 of Exercise 9.6.53 is in agreement with

522


the definition of V: by virtue of Exercise 9.6.53(v), (vi).) Let !j - a. Show that (i) V," is a (closed) cone and J A ~ A U= A"'AU ( A E R+),

(ii) W:, the real-linear span of of All2 and

a'

be

JV,O= v$

e,is contained in the domain

[Hint.Establish the first equation for y in Rtu;use the facts that J is an isometry and A1i2 is closed.]; (iii) llHz11 5 llrCzll when 2 E D ( H ) n D ( K ) and H and K are self-adjoint operators affiliated with an abelian von Neumann algebra such that H 2 5 Ka [Hint.Use a common bounding sequence of projections for H,K, H 2 , K2.]; (i.1 IlA"Yll L 21/211Yll ( 8 E w2 [Hint. Express Aa as Aa(I - E)+AaE, where E is the spectral projection for A corresponding to [0,1]. Note that 1lA"E)I 1 and that Aa(I - E) 5 A1l2(1- E). Use (iii).]; (v) A"V: is dense in V," [Hint.Use (iv).]; (vi) V," and V,.' are dual cones; in particular V;l4 is self-dual [Hint.If (y,z)2 0 for each x in V:, use Theorems 3.2.30 and 5.6.36 to show that h,(ln A)y is in the d u d cone to V: (with h, as defined in the proof of Theorem 3.2.30). Prove that (gn =) h,(lnA)y + 9, yn E 'D(A-af), and Ah-"'ynE V t . Use (v).]

Solution. (i) Since All2 = Aa'A" (from D(A'I2)

5.6.(18)),

Ru C

D(Aa). With A, B in R+ and b a positive number,

A'Au t bAaBu = Aa(A t bB)u E {A"Ku : K E Rt}. When H and K are in Rt,there is a K' in 'R'+ such that 1

( A ~ H U , A O ' K=~() H ~ L , A ~=K(HU,KIU) ~) 2 0, from Exercise 9.6.53(vi). Thus, with 2 in V t and y in V t ' , (2, y) 2 0. If x and -2 are in V,", then (z,Aa'Ku) = 0 when K E 32+. Since

EXERCISE 9.6.55

523

each operator in R is a linear combination of four operators in R+, x is orthogonal to Aa'Ru. We note that Aa'Ru is dense in 3-1, from which x = 0. Suppose that y E D(Aa'). Let E , be the spectral projection 1

for A, corresponding to the interval [O,n]. Then E,y E D(A2) 1

D(Aa'). Since Ru is a core for AT, there is a sequence {A,} such that

C

in R

Now Aa' is everywhere defined and does not increase norm on El('FI), while Aa does not decrease norm on D(Aa) n (I- El)(X). Thus, with xm for Amu - Eny, we have that

Thus

Now limn E,Aa'y = Aa'y, whence (x,Aa'y) = 0 for all y in D(AQ'). Since Aa' is self-adjoint and one-to-one, its range is dense in 'FI, so x = 0. Thus {AaKu : K E R+}and its closure :fC are cones. If A E R+,then from Exercise 9.6.10,

JAaAu = A - a J A ~= A-a JSAu = A - ~ J J A ~ =~A~ ~A' A ~ ~ .

Hence J maps a dense subset of VE onto a dense subset of VE' . Since J is an isometry, JV: = Y;'. (ii) From(i) (when a = 0, a' = f),with A in R+, A ~ ~ = ~ JAA UU .

524


Suppose x E V t . Then x is the limit of a sequence { A n u } for some sequence { A n } of operators in R+.Since J is an isometry

A ' / ~ A ,=~ JA,U

-+ J X .

As All2 is closed, x E D(A'/2) and A'12x = J x . Thus W t & 'D(A112), and A1j2y = J y for each y in W:. It follows that llYll = IlJYll = IIA1/zYll

for each y in WE. (iii) From Theorem 5.6.15(i), there is a common bounding sequence { E n } for H, K ,H2, K 2 . By assumption,

(W2E,x, Enx) 5 (K2Enx,Enx),

and

KE,x = E,Kx

-+

Kx.

Hence llHx11 5 llKx11. (iv) Proceeding as indicated in the hint, we have that A E is everywhere defined, bounded, and llAEll 5 1. Also, (I - E ) A C_ A ( l - E) (and A(1- E) is closed and self-adjoint). By passing to the function representation, we see that Aa = A a ( I - E)+A"E, IIAuEll 5 1, and [Aa(I - E)I2 5 [A'l2(I - E)I2. If y E 'D(A1I2) (C D(Aa)), then y E 'D(A'i2(1 - E ) ) C_ D(Aa(l - E ) ) ,

and

A'I2(I

- E)y = (I- E)A1/2y, A a ( I - E)y = (I- E)A"y .

From (iii), we have

EXERCISE 9.6.55

525

Thus, from (ii), if y E WE,

(v) Suppose x E V:. Then z is the limit in 8 of {A,s} for some sequence {A,} of operators in R+. From (ii), A,u - 2 E W t E D(A112);from (iv)

so that Aax E V t . Hence AaV: C V:. Since {A"Au : A E R+}is dense in V z , A"* is dense in V:. (vij From the solution to (ij, (x,y) 2 0 when x E V: and y E V:'. Thus Ut and V;' are contained in the d u d cones of one another. Suppose, now, that y is in the dual cone to V: (that is, (y, 2) 2 0 for each 2 in V:). Let h n ( p ) be (2n)-6(1- ilpl) when ]pl 5 n, and let h,(p) be 0 when n < IpI, where n is a positive integer. (See the beginning of the proof of Theorem 3.2.30.) Then, by Theorem 3.2.30 calculations, h,(t) = (1-cos nt)/7rnt2when t # 0 and h,(O) = n/(27r). Since both h, and A, are continuous and in &(a)n Lm(R), and h,(p) = h,(-p), h, is the Fourier transform of k, (either from Theorem 3.2.30 or direct calculation). From the equation noted in the statement of Theorem 5.6.36, (h,(ln A)y,x:) =

1

h n ( t ) ( A i f y ,dt z ).

Since Aitu = u, from Remark 5.6.32, and AitAA-it E 'R+ when A E R+,AitAu = A"AA-"u and AitAaAu = AaAitAu E Vt. Hence A"Vt = V t for each real t, and the unitary operator hit maps the dual cone of V z onto itself. Thus 0 _< (A"y,x), and since A,(t) 2 0 for each r e d t, h,(h A)y is in the d u d cone of V t from ( t ) . Since {h,} is monotone increasing with pointwise limit the constant function 1 on R and h + h(ln A) is a a-normal homomorphism of f? into the abelian von Neumann algebra generated by A (see Theorems

526


5.2.8 and 5.6.26), h,(In A) is strong-operator convergent to I. Thus (Yn =) h,(lnA)y Y. We show, now, that y, E YE'. Since hn vanishes outside a finite interval, y, E D(Af) for each real t . To see this, paw to the abelian von Neumann algebra generated by A and I and to the representing function algebra C ( X )for this von Neumann algebra. Then h,(ln A ) is represented by h, o f in C ( X ) ,where f represents In A in n / ( X ) . If q is a point in X at which the function representing A takes a value outside the interval [exp -n,expn], then (h, o f ) ( q ) = 0. It follows that At h,(ln A) is bounded for each real t . Since h,(ln A ) is a bounded, everywhere-defined operator, Ath,(ln A) is closed, and densely defined. Thus A'h,(In A) = At h,(ln A). In particular, y, E D(At) for each real t . Thus, with A' in 12'+ and A equal to .--)

JA'J,

o 5 (A=A~,Y,)= ( A - = ' A ~ / ~ A ~ , * , ) = (JAu,A-a'y,) = ( J A J u , A-"yn) = (A'U,A-~'Y,),

e,.

from (ii). Hence A*-a'ynE From (v), 9, E Aa'Vt E V,f. It follows that y E V;' and that V t is the dual cone to V:. In particular, V;14 is its own dual (we say that Y;l4 is self-dual). m[7,19,41] We adopt the notation of Exercise 9.6.55, but write 9.6.56. V , in place of Vt14. Let !240 and f30 be the (strong-operator-dense) * subalgebras of R and R', respectively, consisting of elements in reflection sequences. (See Subsection 9.2, Tomita 's theorem-a second approach.) Show that (i) AoJAoJu E V,(Ao E %o) [Hint. Recall that JAou = A1I2A$u, and use the fact that A-'l4AoA1l4 has a (unique) extension in R, by Theorem 9.2.26 and the discussion following it.]; (ii) AJAJu E V , ( A E R ) [Hint. Use (i) and the Kaplansky density theorem.]; (iii) { A 1 / 4 A 8 ~: A0 E (JZ(O)h} is dense in V, [Hint. Use Corollary 5.3.6 (especially its proof) and Exercise 9.6.55(iv).]; (iv) { A J A J u : A E R} = {A'JA'Ju : A' E a'},and this set is dense in V , [Hint. Use (ii) and (iii), and prove that BJBJu = A114A8u, where B (in 12) is the extension of A1/4AoA-114 and A0 E (a0)h

-1;

(v) AJAJV,

C V,

(AE

R).

527

EXERCISE 9.6.56

Solution. (i) Then

Let B in R be the extension of A - 1 / 4 A ~ A ' / 4 .

A ~ J A ~ J=UA ~ J A = ~U A ~ J S A ; U= A ~ A ~ / ~ A ~ * U = (A-'/4AoA1/4)( A-1/4AOA1/4)*u

= A1l4BB*uE V , . (ii) It will suffice to show that AJAJu E V , for each A in (R)1, Since 2l0 is a strong-operator-dense * subalgebra of R, A is in the strong-operator closure of (2lo)l. Hence AJAJ is in the strongoperator closure of {AoJAoJ : A0 € (&,)I}. Thus AJAJu E U, from (i). (iii) Suppose 2 in V , and a positive E are given. Choose A in R+ such that 1. - A1/4Aul(< $E. From (the proof of) Corollary . 5.3.6, there is a B in (!&)h such that I(Au - B2u1)< r 3 l 2 &From Exercise 9.6.55(iv), since Au and B2u are in W t ,

( ( A 1 / 4 A-u

B2ull < 2 1 / 2 2 - 3 / 2 ~ .

Hence 1. - A1/4B2u1( < E . It follows that { A 1 / 4 A i u: A0 E (?&),,} is dense in V,. (iv) With A in R,(A' =) JAJ E R',JA'J = A. Thus

AJAJ = JA'JA'= A'JA'J, and

{ A J A J : A E R} = ( A ' J A I J : A' E R')

I

/ 4 a (unique) extension B in R. With A0 in (!dO)h, A 1 / 4 A ~ A - 1has We have, B J B J u = BJBu = BJSB*u = BA1l2(A-'l4 A0 A' I4)u

= BA1j4Aou= A 1 / 4 A t ~ . F'rom (ii) and (iii), it follows now that {AJAJu : A E R } is a dense subset of V,. (v) With A, B in R,

A J A J B J B J u = ABJAJJBJu = ABJABJu E V u , from (ii). Since AJAJ is continuous, { B J B J u : B E R} is dense in V,, and V, is closed, AJAJV, C V,.

528


9.6.57. We adopt the notation of Exercise 9.6.56. Suppose x E V,. Show that (i) J x = x [Hint. Use Exercise 9.6.55(i).]; (ii) J E = E'J and JEE' = EE'J, where E and E' are the projections with ranges [ R ' x ] and [Rx],respectively; (iii) x is separating for R if and only if x is generating for R; (iv) if x is separating for R and J' is the modular conjugation corresponding to x , then J' = J . [Hint. Use Exercise 9.6.52(vi).]

Solution. (i) Note that JA114Au= A1I4Au,when A E Rt , from Exercise 9.6.55(i). Since J is continuous and V , is the norm closure of { A 1 / 4 A u: A E R+},J x = x . (ii) From (i),

JTx = JTJx E R'x,

JT'x = J T ' J x E R x ,

when T E R and T' E R'. Thus J maps R x isometrically onto R's; whence J maps { R x } * isometrically onto { R ' X } Hence ~.

E'Jy = E'JEy

+ E'J(I - E)y = JEy

for each y in 3-1, and E'J = J E . Thus

JEE' = E'JE' = E'EJ = E E ' J . (iii) Note that x is separating for R if and only if [R'x] is 'H, which occurs if and only if E is I . From (ii), E is I if and only if I = J E J = El. Thus, x is separating for R if and only if [ R x ]is 'H, that is, if and only if x is generating for R. (iv) Since x is separating for R, it is generating for R, from (iii), and there is a modular structure associated with x . From (i), J x = s; and of course J 2 = I and J R J = R'..Moreover, A J A J x E V, for each A in 72, from Exercise 9.6.56(v). Thus ( A J A J x , x ) 2 0, since V , is self-dual (Exercise 9.6.55(vi)). From Exercise 9.6.52(vi), J = J'.

We adopt the notation of Exercise 9.6.56. Let x , y, 9.6.58. and v be vectors in V , and suppose that 0 = (z,y)= ( u , ~ ) Let . E and E' be the projections with ranges [R'z]and [Rx],respectively. Show that

529

EXERCISE 9.6.5s

(i) v = 0 [Hint.Choose A, in R+ such that { A 1 / 4 A i u }tends to v and show that {IIAnull} tends to 0 . Choose B' in f ? and ~ use the fact that A1/4B'A-1/4 has a (unique) extension in R' to prove that (v,B'u) = 0.1; (ii) JEE'(= J') is the modular conjugation for EREE' (= S) acting on EE'(H) with generating and separating vector z [Hint. Use Exercises 9.6.57(i), (ii) and 9.6.52.1; (iii) VL E V,, where VL is the self-dual cone for {S,z} (corresponding to U, for ( 7 2 , ~ )[Hint. ) Use Exercise 9.6.56(iv), (v).]; (iv) EE'y = 0. [Hint. Show that EE'y E Ui by using (iii) and the fact that VG and U, are self-dual. Then use (i).] Solution. (i) Proceeding as indicated in the hint, we have

llAn~112= ( A i u , ~=) ( A ' / 4 A i ~ + , ~ ()v , u ) = 0 . Thus, with C' the extension of A1/4B'A-'/4 in

(A'/4A:u,B'u)

--t

R',

(w,B'u)

and

(A1i4A2,u,B'u) = ( A i u ,A'/4B'A-1/4u) = (&u, A,C'u) = (A,u,C'A,u) +

0.

Hence, (v, B'u) = 0. Since Dou is dense in SFt, v = 0. (ii) Let K be EE'(3-I) and J' be JEE'. Then J' is a conjugate linear isometry of K onto K , from Exercise 9.6.57(ii). Moreover,

( J ' ) 2 = JEE'JEE' = J2(EE')2= EE', so that J' is involutory. Of course J'a: = 2, and

J'EE'AEE'J' = EE'JAJEE'

( A E R),

whence J'SJ' = S'.Finally, with A in R,

(EE'AEE'J'EE'AEE'J'x,

X)

= (AJ'Ax,X ) = ( E ' E J A ~A, * ~ ) = (JE'Az, E'A*z) = (J A z , A*z) 2 0,

530


from Exercise 9.6.56(v) and self-duality of V, and since Ax and A*x are in El(%). It follows from Exercise 9.6.52(vi) that J' is the modular conjugation for { S , x } . (iii) From Exercise 9.6.56(iv),

{EE'AEE'J'EE'AEE'J'x : A E R} is dense in VL. F'rom Exercise 9.6.56(v),

E E ' A E E ' J ' E E ~ A E E ~ J=' XJ'AJ~EE'AEE'X = JEE'AJEE'A~ =JEAE'JEA~ = JEAJEEAx = EAJEAJx E Vu, since x E Vu. As Vu is closed, VL Vu. (iv) If w E VL, then w E V , from (iii), so that

(EE'Y, 4 = (Y,EE'w) = (y, w)2 0, since y, PO E V , and V, is self-dual from Exercise 9.6.55(vi). As VL is self-dual, EE'y E V i , But

so that, choosing EE'y for v and the vector x for u in (i), we conclude that EE'#=O.

9.6.59. With the notation of Exercise 9.6.58, let F and F' be the projections with ranges [R'y] and [Ry], respectively. Let z be Ey,let z' be E'y, and let M , M',N , and N' be the projections with ranges [R'z],[Rz], [R'z'],and [Rz'], respectively. Show that JM = N'J, JN = M'J, CM = CMI = CN = CNI,M 5 E, (i) and N' 5 E' [Hint. Note that J z = z', and that J P J = P for each central projection P in R by Exercise 9.6.18.1; (ii) if t # 0, there is a non-zero partial isometry U in R such that U*U 5 M ,UU* 5 N,and U * U z # 0; (iii) if z # 0 and G' is the projection with range [RU*Uz],in the notation of (ii), there is a non-zero partial isometry V' in R' such that V'*V' 5 G' 5 M',V'V'* 5 N', and V'*V'U*Uz # 0;

531

EXERCISE 9.6.59

(iv) if z # 0, then UV'z is a non-zero vector in NN'(3-I), and there is an A in N R N such that 0

< (UV'z,Az') = - i ( B J B J y , y ) ,

where B = A*U - JV'J E R , in the notation of (iii) [Hint. Note that AN'z' = Az' and z' is a generating vector for N R N N ' acting on NN'(3-I). By using E'V' = V ' , UE = U , and JE' = E J , prove that

(UV'y,Ay) = (UV'z,Az'), (y,V'JV'y) = 0 , (UJA*Uy,Ay)-L 0.1; (v) Ey = E'y = 0 and EF = E'F' = 0. [Hint. Note that 0 5 ( B J B J y ,y ) and use the conclusions of (ii), (iii), and (iv).] Solution. (i) From Exercise 9.6.57(ii), J E = E'J. Thus

J z = JEy = E'Jy = E'y = Z'

.

Hence JA'z = JA'Jz', and J maps [R'z]isometrically onto [ R z ' ] . Hence J M = N ' J . Similarly, JA'z' = JA'Jz, and J N = M ' J . Since A -+ JA*J is a * anti-isomorphism of R onto R', this mapping preserves central carriers. Moreover, JPJ = P (as in the hint). Hence CN = CMI and CM = CNI. From Proposition 5.5.13, CN = C N l an d CM = C M I .Finally, M 5 E and N' 5 E' since

(ii) If z

# 0, then M

and N are non-zero projections in R. Since

C M = C N ,M and N have equivalent, non-zero subprojections from the comparison theorem (Theorem 6.2.7). Thus, there is a partial isometry U in R such that 0 < U*U 5 M and UU* 5 N . Moreover, U'Uz # 0 since

[R'U'Uz] = [U*UR'z] = U*U(3-I)# (0). (iii) If z # 0, there is U as in (ii), and U * U z # 0. Since [ R U * U z ] [Rz], G' 5 M ' . Thus 0 # CQ 5 CMI = C N , , and there is a non-zero partial isometry V' in R' such that V'*V' 5 G' and V'V'* 5 N'. Moreover, V'*V'U*Uz # 0 since

[RV'*V'U*Uz]= [V'*V'RU*Uz]= V'*V'(X)# ( 0 ) .

532


(iv) If z # 0, then V'*U*UV'z # 0 from (iii), whence UV'z Since U has range in N(31) and V' has range in N'(31),

# 0.

NN'UV'z = NUN'V'z = UV'z. Since N and N' have ranges [R'z'] and [Rz'],respectively, z' is generating and separating for N R N N ' acting on "'(31). Hence there is an A in N R N such that AN'z' (= Az') i s n e x UV'z. Multiplying A by a suitable scalar, we may assume that (UV'z,Az') > 0. With B as defined,

B J B J = A * I J J A * U Jt JV'JV' - A*UV - JV'A*UJ, and -~ 2 ( B J B J = ~ , ~ ) - 5l(UJA*Uy,Ay)- i(y,V'JV'y) t Re(UV'y,Ay). Now V' = N'V' = E'N'V' = E'V' and U = U M = UM E = UE from (i), (ii), and (iii). Thus

(UV'y,Ay) = (UEE'V'y, Ay) = (UV'Ey,AE'y) = (UV'z,Az'), (y,V'JV'y) = ( y , E'V'JE'V'y) = (E'y,V'EJV'y) = (EE'y,V'JV'y) = 0 , and

from Exercise 9.6.58(iv). Thus 0

< (UV'z,Az') = (UV'y,Ay) = Re(UV'y,Ay) = - $ ( B J B J y , y ) .

V , from Exercise 9.6.56(v), and Since y E V,, BJBJV, U, is self-dual, we have that 0 5 ( B J B J y, y) . This inequality contradicts the conclusion of (iv), if z # 0. Thus Ey = z = 0 and E'y = z' = J z =: 0. It follows that [R'y]2 (I - E ) ( H ) and m [Ry] ( I - El)(%),whence E F = E'F' = 0. (v)

533

EXERCISE 9.6.60

With the notation of Exercise 9.6.56, let 'H, be 9.6.60. { x : J x = x}. Show that (i) 'H, is a real Hilbert space relative to the structure imposed by 'H; (ii) each element of H has a decomposition 3, t ixi with 2, and xi in 'H, and this decomposition is unique; (iii) each element of 'H, has a unique decomposition x+ - x-, where x+ and x- are orthogonal vectors in V , [Hint.Use Proposition 2.2.1 to find x+, and recall that V , is self-dual.]; (iv) if x,y E V,, then 112

- Y1I2 I I I W z l R - %lRII 5

IIZ -

Yllllx

+ Yll,

and conclude that u,lR = u,(R if and only if x = y [Hint.Use (iii) to express x - y as T J - w ,where TJ and w are in V , and (w, w)= 0. Let E and F be the projections with ranges [R'v] and [R'w], respectively. Consider (0, - u y ) ( E- F ) , and use the facts that E F = 0 (Exercise 9.6.59(v)) and 0 I (w,x), 0I (v,y).]; { u , l R : x E V , } is norm closed in R#. (v) Solution. (i) If x and y are in 'H, and a is real, then J ( a x + y ) = a J x J y = a x t y; thus 'H, is a linear space over R. Since J is continuous, 'H, is a closed (real-linear) subspace of 'H; hence 1-1, is complete. Finally,

+

b,Y)

= ( J x ,J Y ) = (Y,J * J x ) = (y,x),

and 'H, is a real Hilbert space. (ii) Let x, be (x t J x ) / 2 and let xi be (x - J x ) / ( 2 i ) . Then z = x, t i x i , J x , = x,, and J x i = xi. Thus 2, and xi are in 'H,. Suppose x = x: t ixf with x: and xf in 'H,. Then 2

+ J X = X: + ixc[+ x: - izf= 22:

and 5

- J X = Z:

+ i ~ -i 2, + i I

I

~I =i 2 i ~ c [ .

Thus xi = x, and xf = xi. (iii) Suppose x E 'H,. From Proposition 2.2.1, there is an element x+ in V , such that

534


Since U, is a cone, ay E V , for each positive a when y E U,. Thus (y,x - x+) I 0 for each y in V,. Since V , is self-dual (Exercise 9.6.55(vi)), (3- =)x+ - x E U,. Hence x = x+ - x-, and with 0 in place of y in (*), 0 2 -(x+,x-) = (x+,x - x+) 2 (0,x - s + ) = 0 .

Thus x+ and x- are orthogonal vectors in V,. If x = 5; - xL,where x i and xL are orthogonal vectors in V,, then (2-,x-)

I

I

I

= ( x - , x - ) - (z+,x->I (x,,x-)

I I 11x-11112-11.

Thus 11x-11 5 11xL11 and, by symmetry, llx!-ll 5 11x-11. Hence 112 -

q l=1 1 a= 1 1 4 1= 115 - 2+11

*

By uniqueness of x+ (as the vector in U, nearest x), we have that x+ = xi. It follows that x- = 3.: (iv) We proceed as in the hint. Since E F = 0, from Exercise 9.6.59(v), IIE - 3’11 5 1. Thus, since V , is self-dual,

Iluc I R - WY I RII 2 I(% - w,)(E - F)I = I(@ - F ) x , x ) - ( ( E - F ) Y , Y ) l = +I(@ - Y),Z t Y) t (x t Y,(E - F ) ( x - Y))l = $I(@ - F ) ( v - w),x t y) t .( t y,(E - F ) ( v - w))l = +I(. t w,2 t y) t (x t Y,V t w)I

* y)

= (v t w,x

= ( wt) ( w , 4 t (w) t (W,Y) 2 (v) - ( U V )- (w) tb y ) = (v - w,x- y) = 112 - Y1I2

-

With A in R,

I(&

-Wy)(A)I

= I(Ax,x) - (AY,Y)l

t Y),Z - Y) t ( 4 s - Y > 4 t Y)I 5 llA1111~t Y1111~- YO,

= +I(A(x

535

EXERCISE 9.6.61

+

so that llWI I R - w y I Zll 5 1. Yllllo - Yll. IR} is Cauchy convergent. By (v) Suppose o(n) E V , and virtue of the first inequality of (iv), ( ~ ( n )is}now Cauchy convergent. Hence {z(n)} tends to some y in V,. The second inequality of (iv) yields that { W = ( ~ I] R} converges to w y 1 R. Thus {usI R : z E V,} is norm closed in Ry. m[7]

Let H be a positive invertible (possibly unbounded) 9.6.61. operator on a Hilbert space N. Show that (i) H1I4(1+H1/2)-1 is a bounded, everywhere-defined operator on 'H and is equal to (H1/4-&H-1/4)-1; (ii) with z and y in 7f, 1/4

~-1/4)-'z,

((H -F:

y) =

(ezt + e - r t ) - l ( H ' t / 2 z

7

Y) dt

Jim

[Hint. Use Lemma 9.2.7 and argue as in the proof of Lemma 9.2.8.1; (iii) A1j4(1+L I ' / ~ ) - ' V ; C V; for each a in [0, with the notation of Exercise 9.6.55. [Hint. Use (ii), Exercise 9.6.55(vi), and note that AaV; C_ VE for each real t.]

31,

Solution. (i) Passing to the function representation of the abelian von Neumann algebra generated by H ,we have that the operators ( I + H 1 j 2 ) - ' and H1i42 (I+H1/2)-1 are bounded, everywheredefined operators. But H1/4(I is closed since H1j4is is closed and (I H'I2)-' is bounded. Thus H 1 i 4 ( I H1I2)-' bounded and everywhere defined. Again, from the function representation,

+

+

@1/44H4/4)--1 =

H1/4(1+@)-'

+

.

(ii) The argument is divided into three stages. Consider, first, the case in which H has the form C?=, ajFj, where a l , . . ,am are positive real numbers and { F l , . .,F,} is an orthogonal family of projections with sum I. In this case, from Lemma 9.2.7,

.

.

536


so that

=J, ( He r"t /+* xe,-yr)tdt We next consider the case in which H is bounded and has a bounded inverse, and choose positive real numbers a , b such that a1 5 H 5 b1. As in the proof of Lemma 9.2.8, W is the limit in norm of a sequence {H,} of operators, each of the type considered in the preceding paragraph and satisfying a1 5 H , 5 bI; moreover, H" and iW-1/4)-1 are the norm limits of the sequences { H : } , for each complex z, and {(HA/4+H;1/4)-1}, respectively. From the preceding paragraph,

Since I(Hf'x,y)Y 5 llzllllyll and JR(e*t t eVrt)-' dt is absolutely integrable, it follows from the dominated convergence theorem that

Finally, we consider the general case, in which H is unbounded. for each positive integer n, let E, be the spectral projection for H corresponding to the interval [n-', n]. Since H is a positive invertible operator, the increasing sequence { E n } is strong-operator convergent to I . For a given choice of n, let Ho be the restriction to En(%)of H . Then HO is in B(E,('H))+ and has a bounded inverse. When x , y E En(%), from Corollary 5.6.31 and the preceding paragraph,

537

EXERCISE 9.6.62

For general x and y in H ,the preceding equality applies with Ens and E,y in place of x and y. Now

and

(H'%,s,

E,y) = (H'+X, E,y)

+

(H't/2x,y).

From the dominated convergence theorem,

(iii) Suppose x E V z and y E V,"'. Then, as in the solution to Exercise 9.6.55(vi), Aitx E V," for each real t and (Aitx,y) 2 0. From (i) and (ii),

+

(A1l4(1 A112)-lx,y) = ((A1/4)$A-1/4)-'x,y)

Since

V t is the dual cone to VE', A1l4(1+ A'/2)-1z E Vz. Hence A'/'(1 -t A1I2)-'V" u C - V" u .

fl

9.6.62. With the notation of Exercise 9.6.56, let w be a normal linear functional on R such that 0 5 w 5 w,lR. From Proposition 7.3.5, there is an operator H' in (7Zf+)1 such that w = w , , ~ t , l R . (i) Suppose x E D(A-l12) f l Yu and

Show that 2 = 2 ( 1 + A1/2)-1H'u. [Hint.Note that x E D ( F ) and FX= A 1 i 2 J s= A 1 / 2 ~ . ] (ii) With s as in (i), show that

538


[Hint. Use Exercise 9.6.61(i).] (iii) With x defined by ( H ) , show that A-114H'~E V, and that x E V,. [Hint. Recall that JA-'I4J = All4, JH'J E R,and JV, = V,. Use Exercise 9.6.61(iii).] (iv) Define 5 by (**). Show that x E 'D(A-1/2)nV, and satisfies (*). [Hint. Note that x = 2(1 A1/2)-1H'u and that (1 t Al/2)-1A-l/2 C A-1/2(1+ A1/2)-1.] (v) With x as in-(iv), show that

+

u - 2 = 2A'I4(I

+ A1/2)-1A-''4(I - H')u E V,.

[Hint. Note that (I t A112)u = 221 and proceed as in (iii) (with I - H' in place of HI).] Solution. (i) By assumption and since x

E D(A-l12) = D(F),

for each A in R,

2w(A) = ( A u , ~-I-) ( A z , u ) = ( A u , s )t ( 5 , S A u )

= ( A u , x )t (Au,Fx) = ( A u , ~t) (Au,A'/2Jx). But x E V,, so that Jx = 2 from Exercise 9.6.57(i). Thus

%(A) = (Au,(It All2)%). By choice of H',w ( A ) = (Au,H'u), Since u is generating for R, 2H'u = (If A112)z and

x = 2(1+ A1/2)-1H'u.

+

- (I All2)-' and (ii) Note that A1i4(It A1/2)-1A-114 C

R'u c D ( F ) = D(A-'12) = D(A-1/4A-'/4) From Exercise 9.6.61(i), A114(1 where defined, so that

Thus

+

C_

D(A -1/4 ).

is bounded and every-

539

EXERCISE 9.6.62

(iii) From (ii), H'u E D ( A - 1 / 4 ) .Now

where H = J H ' J E R+. By definition, A114HuE V,. From Exercise 9.6.57(i), JA1I4Hu= A1/4Hu.Thus

A h - ' / 4 H '= ~ A'I4Hu E V,

.

By definition of z and Exercise 9.6.6l(iii),

(iv) From (iii), z E V,. Now

J:

= 2(1 t A'/2)-1H'u, and

Since H'u E D ( A - 1 / 2 ) ,H'u E D(A-'l2(I t A 1 I 2 ) - l ) . It follows that z E D(A-'l2). At the same time, ( I t A1f2)s= 2H'u. Thus

2 4 A ) = (Au,2H'u) = (Au,(I t All2)")

t ( A U , all2J Z ) = (Au,J : ) t (Au,Fx) = ( A u , z )t (z,A*u) = (AU,

J:)

= ( G , Z iWI,ZL)(A), for each A in R. (v) Since 0 5 H' 5 I , I - H' E R'+,and the argument of (iii) applies with I - H' in place of H' to show that

As ( I t A'12)u = 2u, u = 2(1 t A'/')-'u and u

- z = 2 ( 1 + A1/2)-'(I - H')u E V, .

.[40]

540


9.6.63. With the notation of Exercise 9.6.62, show that (i) there is a y in V , such that u - y E V , and

- w = fcwu,,t q J R

[Hint.Use Exercise 9.6.62(iv) and (v).]; (ii) u - +y

(= z ) E V,,

w,)R - w = W ( l / q y ( R

and

ll%P - 41 = (%Y) where y is as in (i) [Hint.Recall that V , is a cone and is self-dual.]; (iii) with y and z as in (ii) ((%lR- 41 I +llwulR - 41

[Hint.Note that 0 5 (y,y) 5 ( u , ~ ) . ] ; R and , E' the projections (iv) with x' in V , such that w _< w ~ J I E respectively, and Ro the von whose ranges are [R'z'] and [Rz'], Neumann algebra E R E E ' acting on EE'(7-l) (= YO), we have that z' is generating and separating for Ro,V,I C V,, the equation

wo(EAEE') = w ( A )

( A E 72) defines a positive normal linear functional wo on Ro,and there is a vector z' in V,, , such that

IIW In- wII Ii l l w d In- wII, and w 5 wZl(R[Hint.Use Exercise 9.6.58(ii) and (iii) and apply the result of (iii) t o wo and (Ro,~').]; (v) there is a sequence { u ( n ) }in V , with u as u(0) such that, w i %(n)lR,

- 41, and { u ( n ) }converges to some w in V, such that w,lR = w [Hint. Use z of (ii) as u(1) and apply the conclusion of (iv), inductively, to construct u(n). Use Exercise 9.6.60(iv) to show that { u ( n ) }is Cauchy convergent .] (vi) the set of (normal) linear functionals w' on R such that 0 5 uw' 5 w,lR for some positive a is a norm-dense subset of the set of all vector functionals on R,and each positive normal linear functional on R has a representation as W,,J IR for a unique w' in V,. [Hint.Note that OAI,IRis such an w' for each A' in R'. Use Exercise 9.6.60(iv) and (v).] IIWu(n)lR

- all I+llwu(n-l)IR

54 1

EXERCISE 9.6.63

Solution. (i) Since 0 5 w 5 w, I R,we have 0 5 w, 1 R - w 5 w, I R,and we may apply the conclusions of Exercise 9.6.62(iv) and (v) t o w, I R - w (in place of w).Hence there is a y as described. (ii) Since i y and u - y are in V , and V , is a cone, ( z = ) u - l2 y - ‘u. - y

+ $3 E V , .

Note that

w, I R - w = w, I R = w, IR

+ w1, I R 2

+ w1 zy

+ wy,u) IR - w I R - w, IR + w - w = w$ I R 2 &,y

by choice of y, and that llwu I R - wII = (%

I R - W )= +((U,Y)

+ ( Y d 4 = (%Y)

since u,y E V, and ( u , y ) = ( y , u ) 2 0. (iii) Since y and u - y are in V , and V , is self-dual,

But from (ii), llwz

I R - wII = W 2l , ( I )

= i(Y, Y)

L

+(.u.,Y> =

+llWU

IR - wII *

(iv) In the present case, 2’ takes the place of 5 in Exercise 9.6.58; we conclude from (ii) and (iii) of that exercise that d is generating and separating for Ro acting on 3-10 and V,I V,. (We use V,I in place of VL of Exercise 9.6.58(iii).) Since 0 5 w 5 W,I 172, the support of w is contained in E , from Remark 7.2.6, so that w ( A ) = w ( E A E ) for each A in R. Now E‘E has central carrier E (relative t o R ’ E ) , from Proposition 5.5.13, so that the mapping

EAE

+

EAEE’

( AE R)

is a. * isomorphism of E R E onto EREE’ from Proposition 5.5.5. Thus the equation wo(EAEE’) = w ( A ) ( A E R ) defines a positive normal linear functional on EREE‘. In addition, wo _< w,’ 1720. By applying the conclusion of (iii) t o wo,Ro, d ,and V,,, we see that there is a vector z’ in V,t such that

542 Thus, with H in


R+,since EE’z’ = z‘,

and w 5 wZl I R. In addition, since w 5 wxl

I R,

(v) Let u(0) be u and u(1) be z (of (iii)). Suppose we have found u(O), .. . ,u(n) with the properties described in the statement of this exercise. Then u(n) E V , and w 5 w,(?) I R. From (iv), with u ( n ) in place of d,there is a u(n t 1) (replacing z’) in V , such that 0 5 W,(,+l) I R and

The sequence { u ( n ) }is constructed by this inductive process. It follows that { u , ( ~I R} ) converges to w . From Exercise 9.6.60(iv), { u ( n ) }is Cauchy convergent and therefore tends to a vector w in V,. Again, from Exercise 9.6.60(iv),{wu(,) I R} tends to w, I R. Thus w = w, I R, and from Exercise 9.6.60(iv), w is the only such vector in V,. (vi) If A’ E R and H E Rt, then 0 5 A’*A‘H 5 llA’1I2H and

Since {A’u : A’ E R} is dense in ‘M, the set S of positive (normal) linear functiionals w’ such that aw’ 5 w, I R for some positive scalar a is norm dense in the set of all vector functionals on R. Since u is separating for R,all positive normal linear functionals on R are vector functionals, from Theorem 7.2.3. Now each element of S has the form w,,~I R for some w‘ in V,, from (v), and the set of positive (normal) linear functionals on R that are representable in this form is a norm-closed subset of R# from Exercise 9.6.60(v). Thus each positive normal linear functional on R has the form w,,~I R for some w‘ in V , , and w’ is unique from Exercise 9.6.60(iv). ~[5,7,19,40,41]

EXERCISE 9.6.64

543

Let R and S be von Neumann algebras acting on 9.6.64. Hilbert spaces H and K with separating and generating unit vectors u and v,respectively, and let cp be a * isomorphism of R onto S. Let V , and V , be the respective self-dual cones for R and S corresponding to u and w. (See Exercise 9.6.55(vi).) (i) With z in V , a separating or generating vector for R, show that V , = V , . [Hint.Use Exercises 9.6.58(iii), 9.6.57(iv), 9.6.6O(iii).] (ii) Show that there is a unique unitary transformation U of 3-1 onto K such that UAU-' = cp(A) and Uu' = v for some vector u' in V , [Hint.Use Exercise 9.6.63(vi) with the functional (w,lS) o q and apply Exercise 7.6.23.1 (iii) With U as in (ii), show that U V , = V,. [Hint.Use (i).] (iv) With w and w' normal states of R and S,respectively, denote by u, and w,t the (unique) vectors in V , and V , whose corresponding vector states are w and w', respectively. Show that Uu,rOv = V,I for each normal state o' of S, where U is as in (ii). [Hint.Note that U-lw,t E V , from (iii).] (v) Suppose R = S and 3-1 = K. Show that there is a unique unitary operator U' in R' such that U'u, = w, for each normal state w of R. (vi) With the assumption of (v), J the modular conjugation for (R,u)and J' the modular conjugation for (R,v), show that there is a unitary operator V in R such that V A V * = JJ'AJ'J for all A in R. [Hint.Use Exercise 9.6.51(i). Find u' in V, such that w,rIR = w,lR. Note that U'u' = v and U'JU'* = J', where U' is as in (v). Use Exercises 9.6.63(vi) and 9.6.57(iv) for this.]

Solution. ( i ) From Exercise 9.6.57(iii), 5 is both generating and separating for R when 2 E V , and it is either generating or separating for R. Thus, from Exercise 9.6.58(iii), V , E V,, where V, is the self-dual cone corresponding to (72,~).Exercise 9.6.57(iv) assures us that the modular conjugations corresponding to 5 and u are the same. Thus u is a "real" element relative to V, (that is, is in 'H, relative to the modular conjugation for (R,z)) in the sense of Exercise 9.6.60. From (iii) of that exercise, u = u+ - u-,where u+ and u- are orthogonal elements of V , (and, hence, of V , ) . But u - 0 is another decomposition of u as a difference of orthogonal elements of V,. The uniqueness clause of Exercise 9.6.6O(iii) allows us to conclude, now, that u- = 0 and u = u+ E V,. From Exercise 9.6.58(iii), again, V , C V,. Hence V , = V,.

544

ALGEBRA A N D COMMUTANT

(ii) Since v is separating for S , w, IS is faithful (on S) and I S ) o cp is faithful on R. From Exercise 9.6.63(vi), there is a (unique) vector u' in V , such that wuc I R = (w, I S)ocp. As wUr I R is faithful, u' is a separating vector for R.From Exercise 9.6.57(iii), u' is generating for R.From Exercise 7.6.23, the mapping Au' -+ cp(A)v ( A E R ) extends to a unitary transformation U of 3-1 onto X: such that, for each A in R,UAU-' = cp(A). Choosing I for A, we have that Uu' = v. Suppose, now, that U' is a unitary transformation of 3-1 onto X: such that U'AUI-' = p(A) (A E R) and U'x = v for some vector 5 in V,. Then, with A in R, (0,

U'AX = U'AU'-'U's = c p ( A ) ~ , A s = U'-'v(A)v, Now [(w,

1 S)0 p ] ( A )= (cp(A)v,V) = (AU'-'w,

5

= U'-'V.

U'-'v)

= ( A s , z )= w , ( A ) . But u' is the only vector in V, whose corresponding functional is (w, IS) 0 cp. Thus u' = x and U'Au' = (p(A)v = UAu' ( A E R). Hence U = U'. (iii) From (i), with u' as in (ii), V,, = V,. Since U is a unitary transformation of If onto Ic taking u' onto v such that URU-' = S, we have that uv, = UV,I = v,. (iv) Let U be as in (ii) and note that U-'vW~E V, from (iii). Now, with A in R,

But u w ~ oisc pthe only vector in Vu whose corresponding functional is w' o cp. Thus U ~ I ,=, ~U-lv,' and UU,I~,+, = vwt. (v) Under the present assumption, let U' be the (unique) unitary operator of (ii) corresponding to the identity automorphism of R. Then U'AU'-' = A ( A E R); whence U' E R'. Moreover, from (iv), U'u, = v, for each normal state w of R.

545

EXERCISE 9.6.65

(vi) Since A -, J'A'J' and A' + JA'* J are * anti-isomorphisms of R onto R' and R' onto R, respectively, and (J'A*J')* = J'AJ', the mapping

A

+ JJ'AJ'J

( AE R)

is a * automorphism of R. Let U' be the unitary operator in R' described in (v) and (from Exercise 9.6.63(vi)) let u' be the (unique) vector in V, whose corresponding vector state on R is wv I R. From (v), U'u' = v. From Exercise 9.6.57(iv), the modular conjugation for (R, u') is J. (Of course u' is separating and generating for R since v is.) As U'RU'* = R and U'u' = v, we have that U'JU'* = J', Let V be J U ' J . Then V is a unitary operator in R,and for each A in

R, VAV* = JU'JAJU'' J = J(U'JU'')U"U'*(U'JU'*)J = J J W A U ~ * J ~= J JJ~AU~U'*J'J = JJ'AJ'J. Hence the mapping A -+ JJ'AJ'J ( A E R ) is an inner phism of R (implemented by V). m[5,7,19,41]

* automor-

Let R be a von Neumann algebra acting on a Hilbert 9.6.65. space 7-f with a generating and separating vector u. Let $ be a * automorphism of R and U+ be the (unique) unitary operator on 7-f (described in Exercise 9.6.64(ii) and (iii)) such that U$V, = V , and

?)(A) = U,AU;

( A E R).

With ?)'another * automorphism of R, show that U++I = U + U p and conclude that the mapping $ -+ U+ is a unitary representation of the group of * automorphisms of R. Solution. Note that

Now U Q U + : V ~= U+V, = V,, from Exercise 9.6.64(iii). Thus, from the uniqueness assertion of Exercise 9.6.64(ii), U++: = U+U+f, and U+ is a unitary representation of the automorthe mapping $J phism group of R on 7-1. 4411 ---f

CHAPTER 10 SPECIAL REPRESENTATIONS OF C*-ALGEBRAS

10.6.

Exercises

10.5.1. Suppose that /3 is a * automorphism of a C*-algebra M, S is the state space of 8, @ : 8 + B(3-19)is the universal representation of rU, and T,, :8 - B(’H9)is the representation engendered by p, when p E S. Show that (i) the mapping p + p o ,4 :S + S is one-to-one and has range S;

(ii) for each p in S, the representations r po p and 7rpop of % are equivalent; (iii) there is a unitary operator U acting on ‘Ha such that U@(A)U* = @(P(A))for each A in M; (iv) the * automorphism @,P@-lof @(a) extends uniquely to a * automorphism p of the von Neumann algebra @(a)-. Solution. (i) If p E S, then p o /3 is a positive linear functional on U, and (P 0 P ) ( q = P ( P ( 0 = P ( 4 = 1; so p o ,4 E S. The mapping p -+ p o p : S -+ S is one-to-one and has

range S, since there is an inverse mapping p 3 p o P’l. (ii) With the usual notation associated with the GNS construction, we have ((TP 0

P)(A)x,,,4= bP(P(A))% = P(P(A)) = (POP)(A)

4 ( A E a, P E S).

Moreover x,, is a cyclic vector for 7rp o p , since “TP O

P)(w5l = [ 1 7 P ( 9 w = 31.,

547

EXERCISE 10.5.1

From Proposition 4.5.3 (the essential uniqueness of the GNS construction), ?r, o p is equivalent to r p o p . (iii) By (ii), for each p in S there is a unitary operator Up from Rp0ponto H ' , such that

$3-1,, Since 3-10 = CpES

it follows from (i) that the equation

defines a unitary operator U from 3.10 onto 3-19 (essentially, U permutes the terms of the direct sum). When A is in M and y = $9, is in Ha,we have

and thus @ ( P ( A ) )= U @ ( A ) U * . (iv) From (iii), @(a)= @@(a)) = U@(%)U*, so U@(U)-U*, and the equation

p(B)= UBU* defines a

@(a)-=

( B € @(a)-)

* automorphism p of @(a)-.When B = @(A) E a(%),

(@P@-')(B)= @ ( P ( A ) )= U @ ( A ) U *= U B U * = p ( B ) ; extends @,P@-l. Since every * automorphism of @(a)is ultraweakly continuous (Remark 7.4.4), two such automorphisms that coincide on @(U)are to a * automorphism equal. Thus is the unique extension of @Pap-' of @(a)-. so

p

548

SPECIAL REPRESENTATIONS OF C*-ALGEBRAS

10.5.2. Suppose that 2.4is a C*-algebra, U is the unitary group of ?U, and Ue (C U )is the set of all exponential unitary elements of 24;that is, Ue={expiH:H=H*EQ}.

Suppose .that A E U, and let K and ice (C K ) be the norm-closed convex hulls of the sets (UAU* : U E U} and {UAU* : U E U e } , respectively. (i) By using Theorems 5.2.5 and 5.3.5 and Proposition 5.3.2, prove that Q(L4.e) is strong-operator dense in cp(U) for each faithful representation Q of 2f. (ii) By using (i) and Proposition 10.1.4, prove that K e = K .

Solution. (i) Given U in U ,'p( U )is a unitary operator in cp(U) (C cp(U)-), and it follows from Theorem 5.2.5 that Q ( V ) = expiKO for some self-adjoint operator KOin 'p(U)-. By the Kaplansky density theorem (5.3.5), KOis the strong-operator limit of a bounded net {K,} of self-adjoint elements of cp(!2l). By Proposition 5.3.2, the net {exp iK,} is strong-operator convergent to exp iKo (= cp(U)). Let H, be the self-adjoint element cp-'(K,) of %. Then expiH, E Ue and, by Theorem 4.1.8(ii), cp(exp iH,)= exp icp(H,) = exp i l i a + cp( U ) in the strong-operator topology. The preceding paragraph shows that each element of cp(U) is the strong-operator limit of a net in p(Ue). Thus cp(Ue)(C cp(U)) is strong-operator dense in cp(U). (ii) It is apparent that I c e C K , so we have to prove the reverse inclusion. Since K is the norm-closed convex hull of the set of operators {UAU*: U € U},and Ke is a norm-closed convex set, it suffices t o show that UAU* E Ice for each U in U. The universal representation @ of M is faithful, hence isometric, so @ ( K e )is a norm-closed convex subset of @(a). By Proposition 10.1.4

Given U in U, @ ( V ) is the strong-operator limit of a net {V,} in @(Ue),by (i). Since

EXERCISE 10.5.3

549

for each vector 5, {V:} is strong-operator convergent to @ ( U * ) . Thus the net { V , @ ( A ) V , }in @(Ke)is strong-operator convergent t o @ ( U A U * ) . From (l),

9(UAU*)E

@(Ke)-

so UAU* f ice,and Ec, = Ec.

n @(a) = @(Ke);

a[9]

10.5.3. When R is a von Neumann algebra, let [coU(R)]= denote the norm closure of the convex hull coU(R) of the unitary group U ( R )of R. (i) Show that A*, AB E [coU(R)]' whenever A, B E [coU(R)]', that [coU(R)]' E (R)1,and that coU(R) contains each self-adjoint element of (R)1. [Hint. For the last result, see the proof of Theorem 4.1.71 (ii) Suppose that P is a central projection in R. Show that

A

-+ ( I - P),P + B , A + B E [coU(R)]=

whenever A E [coU(RP)]' and B E [coU(R(I - P))]=. (iii) Show that coU(R) contains each partial isometry V in R such that I - V*V I - V V * . (iv) Deduce from (i) and (iii) that coU(R) = (R)1when R is finite. (v) Suppose that 12 is properly infinite, V E R,and V is an isometry (that is, V*V = I ) . By choosing projections E l , . . ., E n in R such that

-

El

N

Ez

N

* * .

N

En I = El t E2 - I - + *En, -+ N

.

and considering the partial isometries V1,. . ,V,, where Vj = V ( I Ej),deduce from (iii) that V E [coU(R)]=. (vi) Suppose that R is properly infinite, V is a partial isometry in R,and I - V*V 5 I - V V * . Show that there is a partial isometry W in R such that V W and V - W are isometries. Deduce from (v) that V E [coU(R)]=. (vii) Suppose that R is properly infinite and V is a partiel isometry in R. By applying the comparison theorem to I - V*V and I - V V * , and using (vi), (i), and (ii), show that V E [coU(R)]=. Deduce that [coU(R)]' = (R)1. (viii) Show that [coU(R)]= = (R)1for every von Neumann algebra R.

+

550

SPECYAL REPRESENTATIONS OF C*-ALGEBRAS

Solution. (i) The norm-closed convex set (R)1 contains U(R), and so contains [coU(R)]=. The proof of Theorem 4.1.7 shows that each self-adjoint element of (R)1 has the form i ( U U*),with U in U ( R ) ,and so lies in coU(R). If A, B E coU(R), we can express A and B as convex combinations m n

+

A = c a j u j , B = XbkVk k=l

j=1

of elements

U1,.

. .,Urn, V1,. . .,Vn of U ( R ) . Since m

j=1

m

n

j=1 k=l

it follows that A*,AB E U ( R ) . If A, B E [coU(R)]=, we can choose sequences {An} and {B,} in coU(R) that converge in norm to A and B, respectively. Since A*, AB are the norm limits of the sequences { A : } , {A,&}, respectively, and A;, A,B, E coU(R) by the preceding paragraph, it now follows that A*, AB E [coU(R)]=. (ii) If A E coU(RP), we can express A as a convex combination Cj"=,ajVj, where %,. ..,V, are unitary elements of RP. Since m

-+

where Uj is the unitary element vj (I- P) of R, it follows that A t ( I - P) E coU(R). Suppose that A E [coU(RP)]=, B E [coU(R(I - P))]=. There is a sequence {A,} in coU(RRp) that converges in norm to A, and A, t (I- P) E coU(R) by the preceding paragraph. Thus A

+ (I- P) = lim[A, t (I- P)]E [coU(R)]=.

+ B E [coU(R)]=; from (i), A -+ B = [ A + (I- P ) ] [ P+ B] E [coU(R)]=.

A similar argument shows that P

(iii) If W is a partial isometry in 2,with initial projection I V*V and final projection I - VV*, then V W and V - W itre unitary operators in R,and

+

v = +(Vt W )+ $(V - W) E coU(R).

551

EXERCISE 10.5.3

(iv) Suppose that 72 is finite, and A E (R)1.Then A has polar decomposition V H , where H is a self-adjoint element of (R)l and V is a partial isometry in R. By (i), H E coU(R). Sinve V*V V V * and 72 is finite, I-V*V N I-VV* by Exercise 6.9.6; so V E coU(R), by (iii). It now follows from the second paragraph of the proof of (i) that A = VH E coU(72). The preceding paragraph shows that (%!)I coU(R), and the reverse inclusion has already been noted. (v) For each integer n ( 2 2), the existence of a set of projections E l , . ..,En with the stated properties follows by repeated application of the halving lemma (6.3.3). For each j h)

I I

- %*Vj = Ej I , - VjV; = I - V ( I - Ej)V* N

+

= I - VV* VEjV* 2 VEjV* Ej I ; N

N

so I - T V j N I thus

N

I - VjV. From (iii), V1,... ,Vn E coU(R), and n-yv1 t - * *

+ Vn)E coU(R).

Now

so

(1 - +)V E c o ~ ( R ) , V = lim(1-

+)vE [cou(R)]'.

(vi) If W is a partial isometry in R, with initial projection I V*V and find projection a subprojection of I - V V * , then V t W and V - W are isometries in 32. From (v), V W, V - W E [coU(R)]=,

+

so

v = $(V + W )+ $(V - W )E [coU(R)]'.

(vii) By the comparison theorem (6.2.7), there are central projections P and Q in R,with sum I , such that

P ( I - V * V )5 P ( I - V V * ) , Q(I - V V * )2 Q ( I - V * V ) .

552


From (vi), applied to the partial isometries V P in R Q , we have

RP and V*Q in

V P E [coU(RP)]=, V*Q E [coU(R&)]=. It now follows from (i) and (ii) that VQ E [coU(RQ)]=, and that

v = V P t VQ E [CoU('R)]=. Since [coU('R)]"contains each partial isometry V in R and (see (i)) each self-adjoint element H of (R),, we can deduce, as in the proof of (iv), that [coU(R)]= = (R)1. (viii) Let P be the largest central projection in R such that RP is finite, and let Q = I - P. Given A in (R)1, it follows from (iv) and (vii).that

BY

A = AP

so [cou(R)J= = (R)1. 10.5.4.

-+ AQ E [coU(R)]=;

I

Suppose that U is a C*-algebra and Ueis the set {expiH: H = H * E U}

of all exponential unitary elements of U. Show that the norm-closed convex hull [coU,]" of 2.4, is the unit ball (U)l. (This result is known as the Russo-Dye theorem. Further information on this subject can be found in Exercises 10.5.91-10.5.100.) [Hint. It suffices to prove the corresponding result for @(U),where @ is the universal representation of U. To this end, use Propositions 10.1.4 and 5.3.2 and Theorems 5.2.5 and 5.3.5.1 Solution. Let K: be [co@(Ue)]', the norm-closed convex hull of the set @(Ue)of all exponential unitary elements of @(a). If U is a unitary operator in @(a)',U has the form expiH for some selfadjoint element H of @(%)-, by Theorem 5.2.5. By the Kaplansky density theorem (5.3.5), H is the strong-operator limit of a bounded net {H,}of self-a.djoint elements of a(%). From Proposition 5.3.2,

EXERCISE

553

10.5.5

U (= expiH) is the strong-operator limit of the net {expiH,} in 9(Ue).Thus

where U ( @ ( O ) - )is the unitary group of 9(%)-.Since Ic- is convex and (strong-operator, hence) norm closed in @(%)-, it now follows from Exercise 10.5.3(viii) that (@(%)-)I & Ic-. By Proposition 10.1.4, since Ic is a norm-closed convex subset of a(%),

Since 9 is an isometry, evident. m[91]

(a), E [coUJ.

The reverse inclusion is

10.5.5. Suppose that U is a C*-algebra, A E U, and IlAll < 1. Let D be the disk { A E C : 1x1 IlAll < l}, and define a holomorphic function f : D --* U by

Show that (i) A g ( A * A ) = g ( A A * ) A for every continuous complex-valued function g on [0,1]. [Hint. Consider first the case in which g is a polynomial .]; (ii) if 1x1 = 1,

(I+ XA*)(XI + A)-' = A* t ( I - A*A)(XI + A)-', (XI

+ A ) - ' ( ( I + XA*) = A* + ( X I f A ) - ' ( I - AA');

(iii) f(0) = A, and f ( X ) is unitary when

1x1 = 1;

Deduce that the norm-closed convex hull of the unitary group U(U) of M is the unit ball (%)I. (Compare this with Exercises 10.5.4 and 10.5.92(v).)

554


Sotution. (i) For each positive integer n,

A(A*A)"

A(A*A)(A*A)* - * ( A * A ) = (AA*)(AA*) * (AA*)A= (AA*)nA.

Thus Ap(A*A) = p(AA*)Afor every polynomial p. Each element g of C([O,l]) is the uniform limit on [O,l] of a sequence { p n } of polynomials, and

(ii) When

1x1 =: 1 the operator X I +

A is invertible, since

We have

( I t XA*)(XIt A)-' = [A*(XIt A ) t ( I - A*A)](XIt A)-' = A* t ( I - A*A)(XIt A)-', and a similar argument yields the second equation in (ii). (iii) Let B be ( I - AA*)'j2 and C be ( I A*A)'I2. From (i),

-

f(0) = B-lAC = B-lBA = A. When

1x1 = 1, f(A)

is a31 invertible element of rU, and

fiom (ii), together with (i) applied to A*, we obtain

f(A)'*

= C"[A* t C2( XI t A)-']B = C-'A*B C(XI A)-'B = A*B'lB C(AI -t A)-lB2B-' = A* C[-A* ( A 1 A)"(I AA*)]B-' = A* - CA*B-' C ( I A"'A)-'(A-lI A*)B-' = A* - A*BB-l t C(I XA)-l(J;I A*)B"

-+

+

+

+

+

+

+

+ +

= A* - A* t f(A)* = f(A)*.

+

+

+-

555

EXERCISE 10.5.6

(iv) Let r be the unit circle {expie : 0 5 0 5 27r). By Cauchy's integral formula, 1

A = f(0)= 2ni

dt

&1

2n

=

f(e")dO.

The integral exists as the norm limit of approximating Riemann sums. Since (27r)-l J" : de = 1, these approximating sums are convex combinations of values taken by f on r, and so lie in the convex hull of U(U), by (iii). Thus A lies in the closed convex hull, [coU(U)]=. The argument just given shows that [coU(U)]= contains the open unit ball of U, and so contains (%)I; the reverse inclusion is m[46] evident. 10.5.6. Suppose that K is a norm-closed two-sided ideal in a C*-algebra U and {Vx : X E A} is an increasing two-sided approximate identity for K (see Proposition 4.2.12 and Exercises 4.6.35 and 4.6.36). Let Q : U + B ( 7 - l ~be ) the universal representation of U, so that Q ( K ) is a norm-closed two-sided ideal in @(a), and (see Proposition 10.1.5) let E be a projection in the center of a(%)- such that

Q ( K ) = @(a) n Q(U)-E = a(%)n Q(U)E,

Q(K)- = Q(U)-E.

(i) Show that the increasing net {Q(Vx)}is strong-operator convergent to E . (ii) Suppose that p E A and A 1 , . . . , A n E U. Let Jp be the set {VA: X E A, X 2 p } , and let C, be the set of all finite convex combinations of elements of Jp. Show that 0 lies in the strongoperator closure of the set

of operators acting on the Hilbert space C;=, $?fa. By reasoning as in the proof of Proposition 10.1.4, deduce that 0 lies in the norm closure of the set n

{

$Q(HAj - AjH) : H E C p } . j=l

556


Hence show that there exists

H in C, such that

llHAj - AjHll 5

-1

( j = 1, ...,n).

(iii) Let r be the set of all triples (IF,p, H ) in which IF is a finite subset of Q, p E A, H E C,, and [ [ H A- A H [ [ 5 n-l for all A in F, where n is the number of elements in IF. Show that r is directed by the binary relation 5 , in which (Fl,pl,Hl)5 ( F 2 , p 2 , H 2 ) if and only if F1 C IF2,pl p2, and H1 5 H2. K as follows: H , is H if y (iv) Define a mapping y + H , : I' is (IF, p, H ) . Show that { H , } is an increasing two-sided approximate identity for K with the additional property that

E - ' . By (ii), we can choose HO in C,, so that (IFo,po,Ho) is an element yo of r. Now suppose that +yo 5 y = (F,p,H ) E r. Since H , = H is in C, C,,,, it follows from (2) that

Since A E

FO C IF and (IF,p,H) E l?, we have

where n is the number of elements of IF. Thus

H,IC

+

I ,

j=l

for all y in 7t. This follows from (i), with ajy in place of

xj.

10.5.8. Suppose that R and S are von Neumann algebras, T,I is a positive linear mapping from R into S , and 11q(I)115 1. (i) Let { E l , . . . ,En} be an orthogonal family of projections in R, with sum I . By using the result of Exercise 10.5.7(ii), with q(Ej) in place of Aj, show that

for all complex numbers a l , . . . ,a,. (ii) Deduce that q(A)*q(A) 5 q(A*A) for each normal element A of R.

Solution. (i) We have to show that

where Aj is v(Ej). Since q is positive, Ej 2 0, and E i + * . . E t n = I, we have Aj 2 0 and

Hence the required result follows from Exercise 10.5.7(ii). (ii) For each self-adjoint element H of R,

EXERCISE 10.5.9

561

IlmlI.

and thus 117i(H)ll I IlHll Thus 1177(A>llL 211??(1>11l l 4 for all A in R,and 77 is bounded. Suppose that A is a normal element of R; we want to show that q(A)*q(A) 5 q(A*A). In view of the result of (i), and since q is bounded, it is sufficient to note that A can be approximated arbitrarily closely in norm by operators of the form C;==,a j E j , where a l , . . . ,an are scalars and { E l , , . . ,En}is an orthogonal family of projections in R with sum I . This follows, for example, from Theorem 5.2.8, since the identity mapping L on sp(A) (extended to be 0 on C! \ sp(A)) can be uniformly approximated by a Bore1 function that takes only finitely many values a l , . . . ,a,. Suppose that 3 and B are C*-algebras, q is a pos10.5.9. itive linear mapping from 2L into B, and 11q(1)11 5 1. Prove that q(A)*q(A) 5 q(A*A) for each normal element A of Q. (This result is known as the generalized Schwarz inequality.) [Hint. Let @ be the universal representation of 2L, and let cp be a faithful representation of B. Show that the mapping

770 = ' P o 770 @ - l :

@(a) + cp(B)

extends to a positive linear mapping 170 : @(a)-+ cp(B)-, and Ilqo(I)ll 5 1. Apply the result of Exercise 10.5.8(ii) to ij0.1

Solution. With the notation used in the hint, @ and cp are is a positive positive linear isometries (as are their inverses); so linear mapping from @(a) into cp(B), and Ilqo(l)(( = 11q(l)11 5 1. Given any vector 2 in the Hilbert space H, on which p(B) acts, the positive linear functional w, o qo on @(a) is weak-operator continuous by Proposition 10.1.1; so 770 is weak-operator continuous. By Lemma 10.1.10, q-,extends to a (bounded and) ultraweakly contin& p(B) uous linear mapping ~0 : a(%)- -, B(X,). Since $(@a(%)) and @(a(%)+) C cp(B+), the ultraweak continuity of % entails %(@(a)-) C cp(f3)- and ijo((@(U)-)+) (p(B)-)+; in this connection, note that the ultraweak density of @(a)+ in (@(a)-)+ follows from Corollary 5.3.6. From the preceding paragraph, fjo is a positive linear mapping from @(a)-into cp(B)-, and ll@(I))(5 1. It follows from Exercise 10.5.8(ii) that ijo(B)*%(B)5 %(B*B)for each normal element

562


and thus V(V(A*A)- rl(A)*17(A))= cp(q(A'4) - cp(q(A))*cp(V(A)) = @ ( B * B )- ijo(B)*ijo(B)2 0. Since cp is a faithful representation of the algebra B, it now follows that q(A*A) - r](A)*q(A)2 0 . m[54]

10.5.10. Suppose that 121 and B are C*-algebras, and 17 is a positive linear mapping from M into €3. Show that q is bounded and 11q11 = 11q(1)11. [Hint. Use Exercises 10.5.4and 10.5.9.1 Solution. From the argument used at the beginning of the solution to Exercise 10.5.8(ii), q is bounded and 11q11 2 11q(I)ll. It is evident that 11q(1)11 5 11q11, and it remains to prove that (1)

llv(A>ll 5 1177(~)11

( A E (Wl).

We may assume that q # 0, whence q(1) # 0; so upon replacing q by llq(l)ll-l q , we may suppose that llq(1)ll = 1. From the Russo-Dye theorem (Exercise 10.5.4), in order to prove (l), it suffices to consider only the case in which A is replaced by a unitary element U of U. In this case (and with Ilq(I)ll = l), it follows from the generalized Schwarz inequality (Exercise 10.5.9) that V ( U ) * W )5 T ( U * U ) = 9(1) 5 1177(1)ll1 = I . ~ h u IMWII s

5 1 = IW)II.

4911

10.5.11. (i) Suppose that M is a C*-algebra of operators acting on a Hilbert space 31 and V H is the polar decomposition of an element A of 121. Show that A(H 4- n-l1)-ll2 E 121 for each positive integer n, and deduce that VH1I2E %. (ii) Suppose that X: is a norm-closed left ideal in a C*-algebra U. By taking M in its universal representation, deduce from (i) and Proposition 10.1.5 that each element of K can be expressed in the form BC, with B in K and C in K+.

563

EXERCISE 10.5.11

Solution. (i) Since A E U and H = (A*A)'/' 2. 0, it follows that H E 2l+,and ( H n-'I)-'/' E U for each positive integer n; so A ( H n-'I)-'j2 E Q. In order to show that V H 1 / ' E U, it now suffices to prove that

+

+

lim I(VH1/' - A(H nhoo

+ n- ' I ) -I/' 11 = 0.

This follows from the fact that

(ii)

We may suppose that U is given, acting on a Hilbert space

H ,in its universal representation. From Proposition 10.1.5 (with CP the identity mapping on U) there is a projection E in U- such that

If V H is the polar decomposition of an element A of

K, then

and thus H1/' = H'/'E. From (i), VH112E U; so

V H 1 j 2 = V H ' / ' E E U n UE = K. It now suffices to take VH'/' for B and H112 for C. We note a second solution to (ii), that does not use Proposition 10.1.5. By taking a faithful representation, we may suppose

564


that U is given acting on a Hilbert space 3-1. If VH is the polar decomposition of an element A of Ic, then A = BC, where B = V H ' I 2 and C = H1I2. Moreover, C = (A*A)lI4 E Ic+, and B E U by (i). Since the range projection of H1I2coincides with that of H , and is the initial projection of V, it follows that VH1I2 is the polar decomposition of B. By applying (i), with B in place of A , we obtain V H 1 I 4 E U. Since C E Ic+ and H1I4 = C ' I 2 , we have H'I4 E K and B = V H 1 I 2 = VH1/4H'/4 E = K. rn

10.5.12. Suppose that U is a C*-algebra acting on a Hilbert space 3-1, X" is the Banach dual space of a Banach space X, 77 : U + X# is a norm-continuous linear mapping, and 77 is continuous also relative to the ultraweak topology on U and the weak * topology on X#. Show that (i) for each x in X, the equation

defines llP4 I (ii) on U-, (iii)

an ultraweakly continuous linear functional pz on 8, and 117711 1141; p3: extends to an ultraweakly continuous linear functional px and Iliizll = IIPzll; for each A in U-, the equation

defines an element ?(A) of X"; (iv) i j is a bounded linear mapping from U- into X#, ij(A) = q(A) when A E U, llfjll = Ilqll, and i j is continuous relative to the ultraweak topology on U- and the weak * topology on EX.

Solution. (i) Since the pair of linear mappings 7 : U -+ X# and f : o -+ o(z) : X# -+ C! are continuous relative to the ultraweak topology on U, the weak * topology on X", and the usual topology on C!, their composition p3: is an ultraweakly continuous linear functional on U. Since

565

EXERCISE 10.5.13

(ii) This follows from Corollary 10.1.11. (iii) It is apparent, from the definition of pz, that

for all z,y in X and a , b in @. Thus

since these ultraweakly continuous linear functionals on 2l- coincide on U. It follows that @(A), as defined in (iii) when A E 2l-, is a linear functional on X. Since

I(@(A))(xc)l= IPx(A)I F llPxl1 IIAII = IlPzll IlAll 5 117711 11x11 IlAll 7 the linear functional G(A)on X is bounded (that is, q(A) E X"), and (1)

Ilii(A>II I 117711 IlAll

(A E

u->.

(iv) From (iii), f j is a mapping from U- into X". Since the mapping px : 8- + @ is a linear functional when x E X, the mapping A + (fj(A))(s) : UC is linear, and 77 : 8- + X# is linear. It now follows from ( 1 ) that f j is a bounded linear mapping, and 117711 L 117711. When A E 2l, ---f

77 extends 7, whence 11fj11 2 Ilqll, and (since the reverse inequality has already been noted) 11fj11 = 11q11. Since the linear functional A + (ij(A))(x) on 2l- (that is, &) is ultraweakly continuous for each x in X, 77 is continuous relative t o the ultraweak topology on U and the weak * topology on X#. so q(A) = q ( A ) , in this ca.se. Thus

10.5.13. Suppose that U is a C*-algebra acting on a Hilbert spave 'FI, X is a Banach space, and the Banach dual space X" is a Banach U-module in the sense of Exercise 4.6.66. We describe X# as a dual %-module if, for each A0 in 3, the linear mappings

566


are weak mappings

* continuous.

If, further, for each po in X", the linear

Apo, A --t poA : U + X" are continuous relative to the ultraweak topology on U and the weak* topology on X#,we say that X# is a dual normal %-module. Now suppose that X# is a dual normal %--module (hence, by restriction, a dual normal U-module). Let 6 :U + X# be a derivation, and recall from Exercise 4.6.66 that 6 is necessarily norm continuous. Let 9 be the universal representation of U, and choose P and cy as in Theorem 10.1.12, with T : U + B('H) the inclusion mapping. Show that (i) X" becomes a dual normal @(U)--module when the action of $(a)- on X" is defined by

A

+

S . p = a ( S P ) p , p.S=pCw(SP)

( S E @(a)-,P E E " ) ,

and that P p = p . P = p for each p in XW; (ii) the equation 6 p ( S ) = S(a(SP)) defines a derivation

@(a) + X+;

Sp :

(iii) 6 p extends to a derivation 8 p : @(a)- Xx that is continuous relative to the ultraweak topology on @ and the ( weak % * I) topology on X+; (iv) i p ( P ) = 0, and 6 ( A )= Z p ( a - l ( A ) ) for each A in U; (v) 6 is continuous relative to the ultraweak topology on U and the weak * topology on X#, and extends to a derivation 8 p o cy-l : U- + 2". --$

Solution. (i) Since X# is an %--module and onto 8-, the equations

s

p = a(SP)p, p

*

s = pcy(SP)

cy

maps @(U)-P

( S E @(U)-, p E X")

define left and right actions of @(a)-on X". Since the * isomorphism cy is isometric, and is a homeomorphism relative to the ultraweak topologies on @(U)-P and U- (Remark 7.4.4), the mappings (p, S ) +

s

*

p,

(p,S) + p

inherit from the mappings

*

s : X" x a(%)- X" ---f

567

EXERCISE 10.5.13

all the properties required to ensure that X# becomes a dual normal @(IU)--module. Since P is the unit element of is(U)-P, a ( P ) is the unit element I of 2l- and

P ‘ p=I p =p =p l = p P *

( p E X”).

(ii) Since Q maps is(U)P onto U, it is apparent that dip, a defined in (ii), is a linear mapping from @(a) into X”. When &,S2

E

@(W,

-

so c5p : a(%)+ X# is a derivation. ( b p ( S ) ) ( z )on (iii) For each z in X, the linear functional S is(%) (bounded, by Exercise 4.6.66) is ultraweakly continuous, by Proposition 10.1.1; so 6 p is continuous relative to the ultraweak topology on @(a)and the weak topology on X#. Using Exercise 10.5.12, we have that t5p extends to a bounded linear mapping 6p : @(a)-+ X# that is also ultraweak-weak continuous. Given S and T in @(a)-, we have S = lim S, and T = lim Tb, in the ultraweak topology, for suitable nets {S,} and {Tb} of elements of @(%). Since 6pJ@(U) is the derivation b p ,

*

*

By taking limits over the net {S,}, and using the ultraweak-weak* continuity of 6 p and the fact that X# is a dual normal is(%)-module, we obtain

Upon taking limits over the net

so 6 p is a derivation.

{Tb}, we

have

568

SPECIAL REPRESENTATIONS OF CCALGEBRAS

(iv) From (iii) and (i), &(P) = Sp(P2) = P * Sp(P) S p ( P ) * P = 2Sp(P),

+

so S p ( P ) = 0. Given A in 8 ,

6 ( A )= S(a(O(A)P)= ) 6 p ( @ ( A )= ) Zp(@(A)). By (i), and since z p ( P ) = 0, we have S ( A ) = S p ( @ ( A ) *) P @ ( A ) Sp(P) = S p ( @ ( A ) P )= S p ( a - l ( A ) ) .

+

-

(c

@(a)-)and 6 p : (v) The mappings a-l : 8- --+ @(%)-I' @(a)--, X# are continuous relative to the ultraweak topologies on 8- and @(a)-and the weak * topology on X#. Hence the same is by (iv)). Finally, true of 8 p o a-l : 8- + X# and S (= 6 p o we assert that 8 p o a-1 is a derivation. For this, let A , B E 2l-, and let S be a - I ( A ) and T be a-I(B). Then S , T E @(U)-P, so A = a ( S ) = a ( S P ) , B = a ( T )= a ( T P ) . Thus

S p ( a - l ( A B ) ) = Sp(S7') = S * Sp(T) $p(S) T

-

+

=~ ( s P ) & ( ~ ( B t )S )p ( a - l ( ~ ) ) a ( ~ ~ ) = A S ~ ( C ~ - ~ ( SB ~) () ( Z - ~ ( A ) ) B . This shows that 6 p o a-l is a derivation from Q- into X" (and extends the derivation 6). ~[50]

+

10.5.14. Suppose that 8 is a C*-algebra, p is a * automor- L J J < 2, where L is the identity mapping on U. phism of U, and Let @ : U + B ( 7 - i ~be ) the universal representation of 8 , and let fi be the * automorphism of @(a)-occurring in Exercise 10.5.1. Let 1~ : U + B ( H , ) be a faithful representation of 8, and choose P,a as in Theorem 10.1.12. Show that - Lll = IIP (9 (ii) p ( P ) = P ; (iii) the restriction pl@(U)-P is a * automorphism of the von Neumann algebra @(U)-P; (iv) the * automorphism ~rpn-lof T ( U )is a homeomorphism of .(a) with its ultraweak topology, and extends to a * automorphism apa-1 of 7r(8)-.

Ip

41;

p

EXERCISE

So/ution. (i) Since the

10.5.14

* isomorphisms @ and @-'

569 are isome-

tries,

p

Since and Lare ultraweakly continuous (Remark 7.4.4) and (a(%)), is ultraweakly dense in (@(%)-)I by the Kaplansky density theorem (5.3.5), we have

Thus ll,6 - Lll 5 - ~ 1 1 ; the reverse inequlity is evident from (l), since - L extends @(/3 - L)@-'. and ,8 is a * auto(ii) Since P is a central projection in @(a)-, morphism of @(%)-, P ( P ) is a central projection Q in @(%)-. Thus P and Q are commuting projections, and

p

are projections with norm not exceeding IIP - QII. If we show that (IP - Qll < 1, it will then follow that

P - PQ = Q - PQ = 0, whence P = PQ = Q = p ( P ) . Now

Thus (iii)

p( P ) = P . This is evident from (ii).

570


(iv) When A E U, it follows from Theorem 10.1.12 that

Hence the * automorphism 7rPr-l of r(U) extends to the * automorphism of .(a)-, and the ultraweak continuity of these mappings and their inverses is a consequence of Remark 7.4.4. m[69]

10.5.15. Suppose that p is a singular positive linear functional on a von Neumann algebra R and E is a non-zero projection in R. Show that there is a non-zero subprojection F of E in R such that p ( F ) = 0. [Hint. Choose a vector 2 such that 2 = Ea: and 1 1 ~ 1 1>~ p ( E ) . Let G be X G b , where {Gb} is an orthogonal family of projections in R,maximal subject to the conditions that Gb 5 E and W,(Gb) 5 p(Gb) for each index b. Show that w,(G) 5 p(G), and deduce that E - G is a non-zero projection F in R such that plFRF 5 w,lFRF. Use Proposition 7.3.5 and Corollary 10.1.16 to show that p ( F ) = 0.1

Solution. Choose {Gb : b E B}, G, and F , as in the hint. For each finite subset IF of the index set B,

Thus (from the final paragraph of Section 1.2)

Since p ( E )

F 5 E.

< 1 1 ~ 1 1=~ w,(E), it follows that G # E and that 0
0. From Exercise 10.5.15, p s ( F ) = 0 for some non-zero subprojection F of E in R. Since p is faithful, 0 < p ( c F ) = plI(cF) t p&F)

= plI(cF) and pu is faithful.

41101

10.5.17. Suppose that 2.l is an infinite-dimensional C*-algebra, and let @ be the universal representation of 3. (i) Prove that @(a)-contains an infinite orthogonal sequence { E l , Ez, . . .} of non-zero projections. [Hint.Use Exercise 4.6.13.1 (ii) Prove that @(a)-has a norm-closed subspace that is isometrically isomorphic to the Banach space 1,. Deduce that, as a

572


Banach space, @(a)-is not reflexive. [Hint. Use (i) and Exercises 1.9.11 and 1.9.24.1 (iii) Prove that, as a Banach space, U is not reflexive. [Hint.Use (ii), Proposition 10.1.21, and Exercise 1.9.12.1

Solution. (i) From Exercise 4.6.13, there is an infinite sequence {Al,AZ,. ..} of non-zero elements of 8+ such that A j A k = 0 when j # k. It suffices to take for Ej the range projection of @(Aj). (ii) With each element of I,, that is, bounded complex sequence X = { q , x 2 , . . .}, we can associate the operator 03

j=1

where Ej is the sequence of projections occurring in (i). Since T is an isometric linear mapping from I, into @(U)-,and I, is a complete metric space, the range M of T is a norm-closed subspace of @(a)(and is isometrically isomorphic to I,). From Exercise 1.9.24, 1, is not reflexive; so M is not reflexive, and it follows from Exercise 1.9.11 that @(a)-is not reflexive. (iii) Since @(a)-is not reflexive, by (ii), it follows from Propois not reflexive. By two applications of sition 10.1.21 that @(a)## Exercise 1.9.12, @(a)is not reflexive, and (since @ is an isometric isomorphism) U is not reflexive. 10.5.18. Suppose that 8" is the Banach dual space of a C*algebra U and @ is the universal representation of U. When p E Uw, let ji be the unique ultraweakly continuous extension to @(a)-of the linear functional p o 9-l on @(U)(see the discussion preceding Proposition 10.1.14). When S E @(U)-,define a linear functional ? on U# by the equation

S"(d= iw mapping S ,? is

(P E

w.

Show that the --t an isometric isomorphism from a(%)- onto 2"". [Hint.Note that is an isometric isomorphism onto U, and use Proposition 10.1.21.1 from @(a)

Solution. The second (Banach) adjoint operator (@-')## is an isometric isomorphism from @(U)## onto a##,and

@(a)## = {s^ : s E @(a)-}

EXERCISE

573

10.5.19

in the notation of Proposition 10.1.21; so it suffices to show that

(1)

(@-*)##(?)=

s

(S E

@(a)-).

When p E a#,p is the unique ultraweakly continuous extension to @(a)-of the linear functional P O @ - ' on @(a); that is, 3, = p o @ - l , in the notation of Proposition 10.1.21. Thus

and (1) is proved.

.[102,109]

10.5.19. With the notation of the discussion preceding Proposition 10.1.14, prove that

whenever S,T E

@(a)-and p E 2l".

Solution. Suppose that S E @(a)-and p E 2P. In this case, Sp,pS E U#, and the linear functionals p , Sp, pS on @(a)-are ultraweakly continuous. If we show that the stated equations are satisfied when T E @(a), it follows by continuity that they remain valid for all T in @(a)-. We assume henceforth that T = @(A), where A E U. Then -

Y

574


Suppose that 9 is the universal representation of a 10.5.20. C*-algebra U, and define mappings ( p , S ) -+ s p ,

(p, S ) -+ p s : %# x

@(a)-+ a#

as in the discussion preceding Proposition 10.1.14. Show that (i) U# is a Banach @(%)--module, and a dual @(%)-modulein the sense of Exercise 10.5.13; (ii) if Xo is a norm-closed subspace of U# with the property that Bp,pB E XOwhenever B E @(U)and p E XO,then Sp,pS E XO whenever S E a(%)- and p E XO. [Hint.By the Hahn-Banach theorem it suffices to show that R(Sp) = 0 = fl(pS) whenever p E XO, S E @(a)-,R E a##,and ill& = 0. Use Exercises 10.5.18 and 10.5.19.1 Solution. (i) With the notation used in the discussion preceding Proposition 10.1.14,

for all p in U#,S in

@(a)-,and A in a. Thus

and the bilinear mappings (1)

(p,S ) + sp,

( p , S ) + p s : %" x

@(a)-+ %#

are bounded. When p E Ux,

(Ip)(A) = P(@(A)I) = ( p 0 @-')(@(A)) = p(A)

( AE

a),

so I p = p. A similar argument shows that p I = p. We assert next that the associative law holds for each of the three possible types of triple product, STp, SpT, pST. Indeed, from Exercise 10.5.19,

whenever A E U, p E %#, and S,T E @(a)-; so S(Tp) = (ST)p. The other two associative laws follow from similar arguments.

EXERCISE 10.5.20

575

F'rom the preceding paragraph, %# is a Banach @(%)--module and (by restricting the mappings in (1) to 214 x @(2l)) a Banach @(%)-module. When A1, A2 E 2l, we have

( @ ( A l ) P ) ( A 2= ) P(@.(A2)9(Al)) = P(@(A2Al)) = P(A2Al),

and (similarly) ( p @ ( A l ) ) ( A 2= ) p(AlA2), for each pin %#. It follows that, for all A1 and A2 in %, the linear functionals P

on 2421"are weak

+

(@(Al)P)(A2), P

+

(P@(Al))(A2)

* continuous. Accordingly, the mappings

are weak * continuous, for each A1 in %, and U# is a dual @(%)module. (ii) Suppose that p E &; we want to prove that Sp,pS E Xo whenever S E @(a>-. To this end, it suffices (Corollary 1.2.13) to show that 52(Sp) = R(pS) = 0

(2)

whenever 52 E U##and R(X0 = 0. Given such an 52, we can choose T in (a(%)- such that 52 = F , by Exercise 10.5.18. Since Bp,pB E XO for each B in @(%), it follows from Exercise 10.5.19 that

0 = R(Bp) = 'i;(Bp)= (Fp)(T) = P(TB), 0 = 52(pB) = 'i;(pB)= (p?j)(T)= P(BT) for each B in

a(%).Ultraweak

continuity of jj now entails

P(TS) = P(ST) = 0

( S E 9(2l)-).

Thus Q(Sp) = T ( S p ) = (G)(T) = P ( W ) = 0,

R(pS) = F ( p S ) = (p?Sj(T)= P(STj = 0,

for all S i n (iij.

@(a)-.This proves (21, and

so completes the proof of

576


10.5.21. Suppose that 2l and B are C*-algebras and 77 : U + B is a linear mapping such that

v(A*)= rl(A)*, v(AB t B A ) = rl(A)rl(B)+ rl(B)rl(A) for all A and B in U. (We refer to such a mapping 7 as a Jordan * homomorphism from U into B; compare Exercise 7.6.18.) When A, B E a, denote by [A,B ] the commutator AB - BA. Establish the identities

A B A + BAB = (A+B)3-A3-B3-(A2B+BA2)-(B2A+AB2), ABC + C B A = ( A -+ C ) B ( A-+ C ) - A BA - CBC, [[A,B ] ,C] = ,4BC + C B A - (BAC t C A B ) , [A,B12 = A ( B A B )+ ( B A B ) A- AB2A - BA2B, for all A,B,C in U, and show that (i) q(An)= v ( A ) (~n = 1,2,. . .); ( 4 rl(ABA) = rl(A)rl(B)rl(A);

(4

+

+

rl(ABC C B A ) = rl(A)v(B)rl(C) rl(C)rl(B)rl(A); (iv) rl("A, B1,CI) = "rl(A),v(B)1,v(C)I; (4 rl([A,BI2) = [rl(A),7l(B)I2*

Solution. (i) The stated result is apparent in the case n = 1. If it has been proved for a particular value r ( 2 1) of n, q( AT+1) = $Q( AA'

+- A'A)

= ${rl(A)v(A')t rl(A')v(A)) = +{rl(A)rl(A)'-t rl(A)'rl(A)I = rlW'+l*

Hence the stated result is valid for all (ii) Since

TI

= 1,2,.

. ..

( A t B)3 = ( A + B ) 2 ( A-t B ) = ( A 2-+ AB + B A + B 2 ) ( A+ B ) = A3 + B3 + ( A 2 B+ B A 2 )+ ( B 2 A+ AB2)+ A B A + BAB, we have

A B A + BAB = ( A+ B)3 - A3 - B3 - ( A 2 B+ BA2)- ( B 2 At AB2) .

EXERCISE 10.5.21

577

From this identity, and since v is linear and preserves squares, cubes, and the “Jordan product” AB t B A , it follows that

With - A in place of A , we obtain

and addition yields q( A B A ) = q( A)q(B)q(A ) . (iii) Since

+

( A t C ) B ( A C) = ABC t C B A t A B A t CBC, we have

+

A B C t C B A = ( A t C ) B ( A C ) - A B A - CBC. From (ii)

(iv) We have

[ [ A ,81,C] = ( A B - BA)C - C ( A B - B A ) = ABC C B A - (BAC t C A B ) .

+

From (iii)

578


(v) We have

[A,B]' = ( A B - B A ) ( A B - B A ) = { A ( B A B ) ( B A B ) A } - A B 2 A - BA'B.

+

From (i) and (ii)

10.5.22.

With the notation of Exercise 10.5.21, suppose that

Bo is the smallest norm-closed subalgebra of D that contains q(Q) (we do not assume that I E DO),and let C and CO be the centers of U and (i) (ii) (iii)

Do, respectively. Show that Bo is a self-adjoint subalgebra of D; if A , B E Q and [ A , B ]= 0, then [q(A),q(B)]E CO; if A , B E U and [ A , B ] = 0, then [ q ( A ) , q ( B ) ]= 0, and 17(AB) = rl(A)@); (iv) q(C) G CO; (v) q(1)is a projection in B and is the unit element of So; (vi) if E is a projection in U, then q ( E ) is a projection in Bo; (vii) if E , F are mutually orthogonal projections in U, then q ( E ) , 7 ( F ) are mutually orthogonal projections in Bo; (viii) if P is a c,entral projection in U, then q ( P ) is a central projection in &, and q ( A P ) = q(A)q(P)for each A in %.

Solution. (i) Since q is hermitian, q(U) is a self-adjoint subset off?, and BO is a self-adjoint subalgebra of B. (ii) If A , B E B and [A,B] = 0, then, from Exercise 10.5.21(iv),

for all C in U. This shows that the element [q(A),q ( B ) ] of 80commutes with q(C) for each C in U, and so commutes with every element of Bo; so [ q ( A ) , q ( B ) E ] CO.

579

EXERCISE 10.5.23

(iii) Again suppose that A , B E 2l and [ A , B ] = 0. By (ii), [ q ( A ) , q ( B ) ]E CO,so [ q ( A ) , q ( B ) ]is normal. At the same time, by Exercise 10.5.21(v),

[ r l ( A ) , ~ ( B= ) 1rl([A7B12) ~ = 0, and it follows (Corollary 4.1.2) that [ q ( A ) , q ( B )= ] 0. Thus

77(A)V(B)= ${rl(A)rl(B)-t- 77(B)V(A)) = i q ( A B+ BA) = q(AB). (iv) If C E C, we have [ A , C ]= 0, and hence [q(A),q(C)]= 0 by (iii), for each element A of 2l. Thus q(C) E Bo,q(C) commutes with each element of q(U) (and hence, with each element of Bo); so q(C) lies in the center CO of Bo. ~ = q(I), so ~ ( 1is) (v) Since I2 = I* = I , we have ~ ( 1=)q(I)* a projection in B . From (iii),

V(A)rl(I)= rl(AI) = 77V)= rl(W = 7 7 ( M A ) for each A in 2l, and this implies that B q ( I ) = B = q ( I ) B for each element B of the norm-closed algebra B,-, generated by q(U). Since, also, q(1)E Bo, it follows that q(I) is the unit element of Bo. (vi) If E is a projection in U, then E2 = E* = E ; so v ( E ) ~= q(E)* = q ( E ) , and q ( E ) is a projection in Bo. (vii) If E , F are mutually orthogonal projections in U, then q ( E ) , q ( F ) are projections in Bo, by (vi), Since EF = F E = 0 , it follows from (iii) that 77(E)rl(F)= rl(EF) = 0. (viii) If P is a projection in C, then q ( P ) is a projection in CO,by (vi) and (iv). For each A in U, AP = P A , and it follows from (iii) that 7l(AP) = 1?(A)77(P).

.

10.5.23. With the notation of Exercise 10.5.22, suppose that n 2 2, {Ejk : j , k = 1,.. . ,n} is a self-adjoint system of matrix units in U, and Cy==, Ejj = I . Let I0 be the unit element q(1)of Bo, and define

580


(i) Show that F k j = Fk’ ( j , k = 1,.. . , n ) , and that G k j = G;k, H k j = Hj;F when j # k. (ii) By using the relation E j k = E j j E j k E k k t E k k E j k E j j , show that Fjk

in

= Gjk + H k j

( j ,k = 1,. . . ,n; j

# k).

(iii) Show that F l l , . . . ,Fnn are mutually orthogonal projections BO with sum 10,and that

(iv) Suppose that j, k , 1 are distinct elements of { 1,2,, . . ,n}. Show that FjjFkl = 0, and

Prove SO that H j k H k l = Hj1. (v) Show that GjkGkj = GjlGlj and H j k H k j = H j l H I j if j , k , l are distinct elements of {1,2,. . . ,n}. (vi) For j = 1.,...,71, define Gjj and H j j to be GjkGkj and H j k H k j , respectively, where k # j (and note that, from (v), the definitions are independent of the choice of k ) . Show that Gjj

= FjkFkjFjj,

Hjj

= FkjFjkFjj

and deduce that

Fjj = Gjj t H j j . (vii) Show that

GjkHkj

= 0 = H j k G k j , when j

# k.

Deduce that

(viii) Show that {Gjk : j , k = 1,. . . ,n} and { H j k : j , k = 1,. . . , n } are self-adjoint systems of matrix units in Bo, and

581

EXERCISE 10.5.23

Sohtion. (i) Since E k j = E f k and 7j is hermitian, Fkj

( j ,k = 1,.

= q(Ekj) = q(Ejk)* = Fjk

.. ,

'72).

Gkj'= F k k F k j F j j = ( F j j F j k F k k ) * = G;k, and similarly H k j = H * , when j # k. $i) When j # k, it follows from Exercise 10.5.21(iii) that

Thus

Fjk

= q ( E j k ) = q(EjjEjkEkk

+

EkkEjkEjj)

= q(Ejj )q(Ejk ) q ( E k k ) -k ' d E k k ) q ( E j k) q ( E j j ) = FjjFjkFkk 4-

= Gjk

FkkFjkFjj

+Hkj.

(iii) Since E l l , . . . ,En, are mutually orthogonal projections with sum I, and q(1)= 10,it follows from Exercise 10.5.22(vii) Fl1,. . . ,Fnn are mutually orthogonal projections in Bo,with l o . From this, together with the definitions of G j k and H j k , apparent that

when k

FjjGjk

= G j k F k k = Gjk,

FkkGjk

= GjkFjj = 0,

FjjHkj

= H k j F k k = O,

FkkHkj

= HkjFjj = H k j ,

#j.

in U that sum it is

By addition, and by use of (ii), we obtain

FjjFjk

= F j k F k k = Gjk,

FkkFjk

= FjkFjj = H k j .

(iv) Since E j j & = EklEjj = 0 (when j , k , l are dl different), it follows from Exercise 10.5.22(iii) that FjjFk1

= v ( E j j ) v ( E k l )= q(EjjEk1) = 0.

Fkom this, together with (iii), GjkGkl

= GjkFkkFkl = FjkFkkFkI = Fjj Fjk F k l = Fjj(FjkFkI

+

FklFjk).

Since FjkFkl

-k F k l F j k = q(Ejk)q(E k l ) -k q(EkI)q(Ejk)

= q(EjkEk1 = q(Ej1)

= Fjl,

+

EklEjk)

582


it now follows that G j k G k l = FjjFj1 = Gjl. Since EkjEu = E1IEkj = 0, it fouows from Exercise 10.5.22(iii) that FkjFll = q(Ekl)q(Ell)= q(EkjE11) = 0.

From this, together with (iii),

(v) From (iv), when j , k, I are all different we have

and similarly, H j k H k j = HjlH1j. (vi) Given j in {1,2,.. . ,n } , choose k distinct from j . Then

Thus

(vii) When j

# C, from (iii) and Exercise

10.5.21(ii) we have

583

EXERCISE 10.5.23

If 1 # j , we have (whether or not 1 = k)

so, in fact GjkHkl

= 0,

HjkGkl

=O

for all j , k,1 in {1,2,. . . n } such that j # k. Upon premultiplying the first of these equations by Gkj, and the second by H k j , it follows that they remain valid also when j = k. It remains to prove that G j k H l m = 0 = HjkGlm for d l j , k,1, m such that k # 1 , and this is apparent from the fact that

(viii) From (i), together with the definition of G j j and it is apparent that

Hjj

in (vi),

From (iii) and (vi)l

Since

Gjk

GjkGlm

=

FjjGjkFkk

and

= 0 = HjkHim when

Hjk

=

FjjHjkFkk,

it iS apparent that

k # 1.

We now have to prove that

for all j, k,1 in { 1 , 2 1 . .. n}. From (iv), and from the definition of G j j and H j j in (vi)l these relations are satisfied when j # k # 1. It remains t o prove that

for all j and k. From (vii) and (vi)

and similar arguments establish the other parts of (1).

584


10.5.24.

With the notation of Exercise 10.5.23,let

and let D be the set of all elements of U that commute with all the matrix units E j k ( j ,k = 1,.. . ,n). Show that (i) each element A of U can be expressed uniquely in the form

j,k=l

with all the coefficients D j k in D,and (compare Lemma 6.6.3) Djk = EljAEkl; (ii) for each D in D, q ( D ) commutes with all the elements Fjk,

c;"=,

Gjk, H j k ;

(iii) If A E U, and A is expressed as in (i), then

(iv) G,H E CO,and G t H = 10; (v) i f A , B E D a n d j # k , t h e n

(vi) for all A and B in U,

(vii) the equations

define a * homomorphism 71 : U + BOG (C f?) and a homomorphism q~ : U + BoH (& f?), and q = ql t 772.

*

anti-

EXERCISE

10.5.24

585

Solution. (i) Suppose first that A E U, and define Djk, for all j , k in (1.2.. . . , n ) , by Djk = Cy=l EljAEkl. Then

and Djk E D. Also, DjkEjk = EjjAEkk, and

since C;', Ejj = I. Conversely, suppose that Djk E D for all j and k, and

Then

n

EljAEkl =

DrsEljErsEkl = DjkEll, r,s=l

and Djk = Cy=IDjkEll = Cy=tEljAEkl. (ii) When D E D, D commutes with Ejk; so q(D) commutes with q(Ejk) (= Fjk) and

by Exercise 10.5.22(iii). Moreover, q(D) commutes with Gjk and Hjk, since both Gjk and Hjk are finite products of elements of the form FTs. (iii) The stated result follows from (1). (iv) It follows from (iii) that Do is the norm-closed subalgebra of B generated by the elements q(D) (with D in D) and Fjk (with j , k in {1,2,. . . ,n)). Now G, H E Bo, since Gjj7H j j E Bo for each j in {1,2,. . . , n ) . Thus, in order to prove that G and H lie in the center Co of Bo, it suffices to show that they commute with each q(D) and each Fjk. From (ii) and the definitions of G and H , each q(D) commutes with both G and H . Moreover

586


by Exercise 10.5.23(ii), (vi), (vii), (viii). From (viii) of Exercise 10.5.23, G t H = 10. (v) If A, B E D and j # k,

This, together with (l), gives

Since F,?k = q ( E j k ) 2 = q ( E j Z k ) = 0 , and (similarly) F i j = 0, it now follows from (ii) that

by definition of G j k and Exercise 10.5.23(ii), (vi), (vii), (viii). Hence, upon postmultiplying throughout (2) by G j k , we obtain

Similarly, by postmultiplying throughout (2) by

Hjk,

we obtain

(vi) Given A and B in U, we can choose coefficients A j k , B j k in D so that n n

A=

C

j,k=l

AjkEjk,

B=

C

j,k=l

BjkEjk.

EXERCISE 10.5.24

587

Then by (l),

n

j,k=1

and we have corresponding expressions for v(B)G, q(B)H. Moreover, since

AB =

n

n

j,k=l

1=1

C (C

Aj1Blk)Ejkr

and it follows from (v) and (ii) that n

1

n

588


Similarly

j,k=l n

I=1

n

= Q( B)Hq(A)H.

(vii) The stated results follow from (vi), since q is a hermitian linear mapping from U into Bo and G,H are central projections with sum lo in Bo. m[49] 10.5.25. Suppose that R and S are von Neumann algebras and q : R + S is an ultraweakly continuous Jordan * homomorphism. Let ,130 be the smallest norm-closed subalgebra of S that contains q(R),let SObe B;, and let I0 be q(1).Denote by C, CO, and 20,the respectively. centers of R,Do, and SO, Show that Bo and So are self-adjoint subalgebras of S,and (i)

c,

20.

(ii) Show that I0 is a projection in S,and is the unit element of BO and SO. (iii) For each positive integer n, let P, be the largest projection in C such that RP, is of type I , unless Pn = 0, and let PO be I P,. Let Q n be q(Pn) for n 2 0. Show that { Q n } is an orthogonal family of projections in 2 0 , and Qn = l o . Note that, for n 2 2, the von Neumann algebra RP, contains a self-adjoint system of n x n matrix units, in which the n diagonal elements have sum Pn. Prove that RPo contains a self-adjoint system of 2 x 2 matrix units, in which the two diagonal elements have sum PO. Deduce from Exercise 10.5.24 that, for all n 2 0, there are projections G , and H , in 20,with sum Qn, such that the mapping A v(A)Gn : RP, SoG, is a * homomorphism and the mapping A q ( A ) H n : RPn + SOH, is a * anti-homomorphism. (iv) Prove that there are projections G and H in 2 0 such that G H = I0 and the equations

c;=,

--f

c:=,

--f

--+

+

% ( A ) = q(A)G, 772(A)= q ( 4 H

( A E R)

589

EXERCISE 10.5.25

define a * homomorphism q1 : R 3 SoG ( C S ) and a * antihomomorphism 7 2 : R SOH (& S) for which ql 72 = q. S is a hermitian linear (v) Conversely, suppose that [ : R mapping, and there exist projections 10, G, H in S such that G+H =

+

--$

--f

I0 7

for each A in R, the mapping A + t ( A ) G : R -, S is a * homomorphism, and the mapping A e ( A ) H : R + S is a * antihomomorphism. Show that 6 is a Jordan * homomorphism. --f

Solution. (i) From Exercise 10.5.22(i), BO is a self-adjoint subalgebra of S, and it follows that SO (= &-) is self-adjoint. If C € CO ( C 130 SO),then C B = BC for each B in Bo,and by continu20,and ity CS = SC for each S in SO;so C E 20. Thus CO c, & 2, = 2 0 . (ii) From Exercise 10.5.22(v), 10 is a projection in S, and is the unit element of 130 So).Since loB = BIo = B for each B in BO, it follows by continuity that 10s= S l o = S for all S in SO,so l o is the unit element of SO. (iii) From Exercise 10.5.22(vii), (viii), {Q,} is an orthogonal family of projections in Co 2 0 ) . Since q is ultraweakly continuous,

(c

(c

03

03

When n 2 2 and P, # 0, RPn is of type I n ; so Pn is the sum of n equivalent abelian projections in RP,, and these abelian projections are the diagonal elements in a self-adjoint system of n x n matrix units in R P , (Lemma 6.6.4). The von Neumann algebra RPo has no central summand of finite type I , and is therefore (unitarily equivalent to) the direct sum of an algebra of type 111 and a properly infinite algebra. From Lemmas 6.3.3 and 6.5.6, POis the sum of two equivalent projections in RPo, and these two projections are the diagonal elements in a self-adjoint system of 2 x 2 matrix units in

RPo. For each n ( 2 0), we have

590


by Exercise 10.5.22(viii). It follows that the restriction q(RPnis a Jordan * homomorphism from RP, into SoQn. Since Q , is a projection in the center Co of 130, the algebra B0Qn is norm-closed and has center CoQn. Moreover, B0Qn is generated, as a norm-closed subalgebra of SoQn, by the range q ( R ) Q n of q(RPn(since BO is generated by ~ ( 7 2 ) )When . n # 1, the existence of projections Gn and H , with the stated properties now follows from Exercise 10.5.24(iv), (vii), applied to q(RPn (since CoQn G 2oQn G 20 and BO SO).Since R S is abelian, q1RPl is a * homomorphism by Exercise 10.5.22(iii), and we can take G I t o be Q1 and H I to be 0. G,, H = C =;, H,. From (iii), G and H are (iv) Let G = Cr=o projections in 20,and

n=O

n=O

For a l l A in 72, q(A) = C,"=,q(APn),by ultraweak continuity of q, and

q(APn)G = q(APn)QnG = q(APn)Gn* Since each G , (E 20)commutes with the range of q, it now follows from (iii) that, when A , B E R,

Thus w, as defined in (iv), is a * homomorphism, and a similar argument shows that 772 is a * anti-homomorphism. For all A in R, q ( A ) E SO,and

591

EXERCISE 10.5.26

Thus ( is a Jordan

* homomorphism.

10.5.26. Suppose that R and S are von Neumann algebras and q is a Jordan * isomorphism from R onto S. Show that there exist central projections PI, P2 in R and Q 1 , Q 2 in S such that PI+P2 = I , Q1 Q2 = I , q(P1) = Q1, q(P2) = Q2, q1RP1 is a * isomorphism from RP1 onto SQ1, and qlRP2 is a * anti-isomorphism from RP2 onto SQ2. (Since R and S are unitarily equivalent to RP1 @RP2 and SQ1 @ S Q 2 , respectively, this result can be stated in the form that a Jordan * isomorphism from one von Neumann algebra onto another is a direct sum of a * isomorphism and a * anti-isomorphism.)

+

Solution. When A E

q(A)=

R+,q(A112)is self-adjoint, and

all^)^) = v ( A ' / ~ 2) ~0;

so q : R -, S is a positive linear mapping. The same is true of q-' : S + R, since q-' is a Jordan * isomorphism. It follows that q is an isomorphism for the order structures on R and S, and so preserves least upper bounds. From Exercise 10.5.22(v), q ( I ) is the unit element of S. If w is a normal state of S, it follows from the preceding discussion that w o q is a normal state of R. Thus q is ultraweakly continuous. From Exercise 10.5.25(iv), there are central projections Q1 and Q2 in S, with sum I , such that the equations

q i ( A ) = q ( A ) Q i , 72(A)= q(A)Q2

( A E R)

define a * homomorphism 171 : R + SQ1 and a * anti-homomorphism 72 : R + SQ2. From Exercise 10.5.22(vii), (viii), q - l ( Q j > is a projection Pj in the center of R,and

Pi

+ P2 = q-l(Q1 +

Q2)

= q-l(I) = I ;

592


moreover

*

It now follows that qlRP1 (= qllRP1) is a isomorphism from RP1 onto SQl, and q1RP2 (= Q lRP2) is a * anti-isomorphism from RP2 onto SQ2. m[53] Suppose that R is a factor, S is a von Neumann 10.5.27. algebra, and q is a Jordan * isomorphism from R onto S. Show that S is a factor and q is either a * isomorphism or a * anti-isomorphism.

Solution. With the notation of Exercise 10.5.26, the central projection PI in R is either I or 0. If PI = I , q (= qlRP1) is a * isomorphism. If PI = 0, then P2 = I , and q (= qlRP2) is a * anti-isomorphism. In both cases, q induces a * isomorphism from the (trivial) center of R onto the center of S, whence S is a factor. m[53] 10.5.28. Suppose that 24 is a C*-algebra, B is a C*-algebra of operators acting on a Hilbert space 'N, and q is a hermitian bounded linear mapping from 24 onto B. Let @ be the universal representation of 24, and let fj : @(a)- B(3-1) be the unique ultraweakly : @(U) B B(7-l) continuous linear mapping that extends q o (see Theorem 10.1.13). Show that 51 has range B-, and that q is a Jordan * homomorphism if q is a Jordan * homomorphism. By applying the result of Exercise 10.5.25 to q, deduce that q is a Jordan * homomorphism if and only if there is a projection P in the center of 8- such that ---f

---f

for all A and B in 24. (Note that, in the case of a Jordan * isomorphism from U onto B, the present exercise augments the information obtained in Exercises 7.6.16-7.6.18 about isometries between operator algebras.)

Solution. A mapping fj : @(a)-+ B(R) extends q o @-' if and only if q ( A ) = q ( @ ( A ) )for each A in U; so by Theorem 10.1.13, q0G-l extends uniquely to an ultraweakly continuous linear mapping q : a(%)- -+ B(7-l). Since q o @-' is a bounded linear mapping

593

EXERCISE 10.5.28

with range B,it follows from the open mapping theorem (1.8.4) that ( q o @-1)((Q(21))1) contains the ball ( B ) r , for some positive T . Now (@(%)-)I is ultraweakly compact, hence the same is true of its image under f j , and

so q((Q(w)l)

2w

r

r

= (B-)r.

This shows that the range of i j contains B - . From ultraweak continuity of fj,

so fj has range B - . If 77 : 2l+ B is a Jordan * homomorphism, then so is 77 o Q(2l) + f?, and (since fj extends q o Q - ' ) (1)

@-'

:

fj(A*)= q(A)*, q(AB t B A ) = q(A)ij(B)t ij(B)fj(A)

for all A and B in @(a).Since fj and the mapping A -+ A* are ultraweakly continuous, and operator multiplication is seperately ultraweakly continuous, it follows that (1) remains valid for all A and B in @(a)-(in the case of the second equation in (l),the extension from @(a) to @(a)is made one variable at a time). It follows that f l is a Jordan * homomorphism if q is a Jordan * homomorphism. If there is a central projection P in B- with the stated properties, then for all A and B in %,

and addition yields

Thus 7 is a Jordan * homomorphism if there is such a projection P . Conversely, suppose that 7) is a Jordan * homomorphism. Then the same is true of i j : Q(2l)- -+ B-, Since i j has range B-,it follows

594

SPECIAL REPRESENTATIONS OF C'-ALGEBRAS

from Exercise 10.5.25(iv) that there are central projections G and H in .O-, with sum I, such that the equations

-

% ( A ) = q(A)G, 92(A)= fj(A)H ( A @(a)-) define a * homomorphism 171 : @(a)- 8 - G and a * anti-homomorphism Q : @(rzL)- + B - H . When A , B E %, we have q(AB)G = fj(@(AB))G= q ( @ ( A ) @ ( B ) ) G = 4( @ ( A ))Gii(@P) )G = 17(A)Grl(W = rl(A)rl(B)G, and (similarly) q(AB)H = q(B)q(A)H. This proves the required m[59,108] result, with P the central projection G in B'. Let 17 be a bounded linear isomorphism of one C*10.5.29. algebra 2l onto another C*-algebra B and let @ and P be the universal representations of rzL and B, respectively. (i) Show that il! o q o W1 extends to a linear isomorphism of @(a)-onto q(13)' that is an ultraweak homeomorphism. (ii) Deduce that fj is a Jordan * isomorphism when 7 is a Jordan * isomorphism.

Solution. (i) By the open mapping theorem, q has a bounded inverse 6 (mapping 13 onto 24). If w is an ultraweakly continuous linear functional on q(13), then w o !@ o q o @-l is a linear functional on a(%)that is bounded and hence, ultraweakly continuous. Thus !@ o 770 @-l is ultraweakly continuous and, from Lemma 10.1.10, has a unique ultraweakly continuous linear extension fj mapping @(a)into Q ( B ) - . In the same way, @ o [ o iv-' has a unique ultraweakly continuous linear extension mapping o(f3)- into @(a)-.With A in a(%),

- = f?(R,,)when p is a pure state of the C*-algebra U. In particular, each pure state of 24 is of type I. As noted in Exercise 10.5.39(ii), the set of states of U of type I is convex. Now the weak * closed convex hull of the set of pure states of U is the set of all states of U. Thus, in order to show

605

EXERCISE 10.5.41

that the set of states of type I of a C*-algebra ‘2 is not a weak * closed subset of 2l# it will suffice to produce just one state of U that is not of type I. With the notation of Section 6.7, let 2 l 2 be the norm closure of the algebra of finite linear combinations of the unitary operators L Z g (g E F2). The vector 5 , ( e the unit of F 2 ) is generating for 2 l 2 ; whence n p is (unitarily) equivalent to the identity representation of !2l2 on / 2 ( & ) , where pis the vector state wZe(‘22, from Corollary 4.5.4. Now !2lT = C32 and C32is a factor of type 111, from Theorems 6.7.2, 6.7.5, and Example 6.7.6. Thus p is a state of 9 2 of type 1 1 1 , and the set of states of 2 l 2 of type I is not weak * closed. 10.5.41. Let T I , 7 r 2 , and 7r3 be representations of a C*-algebra U such that is quasi-equivalent to a subrepresentation of 7r2 (we say that 7r1 is quasi-subequivalent to 7r2 in this case, and write 7rl n2) and 7r2 is quasi-equivalent to 7r3. Show that n 1 7r3.

zq

zq

Solution. By assumption, there is a projection E’ in n2(2l)’ such that nl is quasi-equivalent to the representation A --+ nZ(A)E’. From Proposition 5.5.5, the mapping 7r2(A)E‘ + K Z ( A ) C Eextends I to a * ismorphism of n2(!2l)-E’ onto n2(2l)-C~1. Thus n1 is quasiequivalent t o the representation A T ~ ( A ) C Eand I there is a * isomorphism a of TI(%)- onto n 2 ( U ) - C ~ l such that a ( r l ( A ) ) = 7r2 ( A)CEIfor each A in U. Since n2 is assumed to be quasi-equivalent to n3, there is a * isomorphism p of n2(U)- onto n3(U)- such that P ( n 2 ( A ) )= n3(A) for each A in U. It follows that p o a is a * isomorphism of T I ( % ) - onto 7r3(2l)-p(Cp)and that ( p o a ) ( n l ( A ) ) = P ( n 2 ( A ) C p ) = ~ ~ ( A ) P ( C EThus I ) . n 1 is quasi-equivalent to the subrepresentation A + ~ ~ ~ ( A ) P (of C n3, E I and ) 7r1 n3. --.$

zq

10.5.42. Let 7r1 and 7r2 be representations of a C*-algebra 2l. Let be the universal representation of 2l on I&, and let F‘1 and P2 be the central projections in @(a)-corresponding to 7r1 and 7 r 2 , respectively, as described in Theorem 10.1.12. Show that 7r2 if and only if PI 5 P 2 ; (i) nl (ii) the set of quasi-equivalence classes of representations of U is partially ordered by if (iii) the quasi-equivaleiic,e class of n is minimal relative t o and only if 7r is primary.

zq

zq;

zq

606


Solution. (i) From Remark 10.3.2, with j in {1,2}, A j is quasiequivalent to the representation @pi ( A 4 @ ( A ) P j ) . If PI I P2, then @pl (= ( @ p 2 ) p l )is a subrepresentation of @pz. In this case, ~1 Z q @pz and @ p 2 is quasi-equivalent to 7 r 2 . Applying the result of Exercise 10.5.41, we have that ~1 Z q ~2 when PI 5 P2. Suppose now that ~1 Z q 7 r 2 . By transitivity of quasi-equivalence, we have @pl Z q 7 r 2 . Applying Exercise 10.5.41 again, we have @plZq @ p 2 . Thus there is a projection E' in @(U)'P2such that @pl is quasi-equivalent to @ E I (= ( @ p Z ) ~ l )From . Theorem 10.3.3(ii), P1

= cp, = CE' 5 P2.

(ii) Of course A Zq A' for quasi-equivalent representations A and A' of %. Let ~ 1 7 r, 2 , and 7r3, be representations of U with corresponding central projections P I , P2, and P 3 , respectively, in (a(%)-. If ~1 Zq ~2 and 7r2 Z q 7r3, then PI 5 P2 and P2 5 P 3 from (i). TI, Thus PI 5 P 3 and ~1 Z q 7r3 from (i). If A I Z q 7r2 and 7r2 then PI 5 P2 and P2 5 PI from (i). Hence PI = P2. Now ~1 is quasi-equivalent to @pl and 7 2 is quasi-equivalent to @pa (= Qip,). From the comments following Definition 10.3.1, quasi-eqivalence is an equivalence reladtion. Thus A I is equivalent to ~ 2 It. follows that Zq is a partial ordering of the set of quasi-equivalence classes of representations of "u. (iii) The class of II is minimal if and only if each subrepresentation of A is quasi-equivalent to A . Thus, from Proposition 10.3.12(i), the class of A is minimal if and only if A is primary.

zq

10.5.43. Let x1 and 7r2 be representations of a C*-algebra 2.4 quasi-equivalent to the representations A ; and A ; , respectively. Show that ~1 and 7r2 are disjoint if and only if A ; and A; are disjoint.

sq

Solution. If 7~ is a subrepresentation of n l , then A ~1 in the terminology of Exercise 10.5.41. Thus A Zq A: from that exercise; that is, A is quasi-equivalent to a subrepresentation AO of A : . Similarly, if A' is a subrepresentation of 7 r 2 , 7r' is quasi-equivalent to a subrepresentation A; of A ; . From transitivity of quasi-equivalence, A is quasi-equivalent to A' if and only if AO is quasi-equivalent to A ; . Thus 7r1 and 7r2 have quasi-equivalent subrepresentations if and only if A: and A: have quasi-equivalent subrepresentations. From Corollary 10.3.4(i), therefore, ~1 and 1 2 are disjoint if and only if 7r; and A; are disjoint,.

EXERCISE 10.5.45

607

Let 11 and 7r2 be representations of a C*-algebra 10.5.44. U, 9 be the universal representation of U, and PI and P2 be the central projections in a(%)- corresponding to x1 and 7r2 (as in Theorem 10.1.12). Show that (i) 7rl and 7r2 are disjoint if and only if PI P2 = 0; (ii) Q(U),the set of quasi-equivalence classes of representations of U partially ordered by Z q (we adopt the terminology and notation of Exercise 10.5.41 and include the 0 mapping of 2l as an element of Q(U)although this mapping has been excluded as a representation of U), is a lattice isomorphic to the lattice of projections in the (through the mapping that assigns PI to the quasicenter of @(a)equivalence class of TI); (iii) the quasi-equivalence classes of 7rl and 7r2 have the 0 mapping as their greatest lower bound in Q(U)if and only if T I and 7r2 are disjoint.

Solution. (i) From Remark 10.3.2, x1 and 7r2 are quasi-equivalent to 9 p , and 9 p 2 ,respectively. From Exercise 10.5.43, xl and 7r2 are disjoint if and only if 9pl and 9p, are disjoint. Since PI = Cp, and P2 = Cp,, 9 p , and 9 p z are disjoint if and only if PIP2 = 0, from Theorem 10.3.3(iii). Thus nl and x2 are disjoint if and only if PIP2 = 0. (ii) From Exercise 10.5.42(i) and Theorem 10.3.3(ii), the mapping from the quasi-equivalence class of 7r1 to PI is a well-defined, order isomorphism of Q(U)onto the lattice of projections in the center of a(%)-.Thus Q(U)is a lattice (relative to the partial ordering Z q of the representations of U). (iii) If x Zq xl and n Z q 7r2, then x1 and x2 have quasi-equivalent subrepresentations unless n is the 0 mapping. Thus xl and 7r2 are disjoint if and only if their quasi-equivalence classes have the 0 mapping as greatest lower bound in Q(2l). rn 10.5.45. Let p and 77 be states of the C*-algebra 9. Write 7 and p 7 when x p w Q 7rq ( 7 r p and xo are quasi-equivalent) respectively 7 r p Zq T,,( 7 r P is quasi-subequivalent to 7rq in the terminology of Exercise 10.5.41). Let { p n } be a sequence of states of a C*-algebra U and suppose that pn+l Zq pn for n in {1,2,. . .} and that { p n } tends in norm to po. Show that (i) po Zq p n for each n in {1,2,. . .}; (ii) po is a factor state quasi-equivalent to each pn when each pn p

Nq

sq

608


is a factor state; (iii) po need not be quasi-equivalent to pn when

Solution. (i) Using the notation of Exercise 10.5.38, we have that {P,Po} is strong-operator convergent to PO. On the other hand, P,+1 5 P,, from Exercise 10.5.42(i). Thus { P , } is strongoperator convergent to n,P, (= P ) . It follows that {P,P,} is strong-operator convergent to PPo and PPo = PO.Thus Po 5 P 5 P,, and po Zq pn for each n in { 1,2, . . .}. (ii) If each p, is a factor state, then the quasi-equivalence class of rp, is minimal in Q(U)from Exercise 10.5.42(iii), and the hypothesis, pn+l p,, becomes, pn+l p n (that is, pn+l and pn are quasieqivalent); the conclusion of (i) becomes, po pn for each n in {1,2,. . .}, in particular, po is a factor state of U. (iii) In the solution to Exercise 7.6.34(ii), we found an example of a C*-algebra U and states pn equivalent to one another (that is, with r p neqivalent to one another), hence quasi-equivalent to one another, tending in norm to to a state po with TO(%) one dimensional. Each nn(!Z) is infinite dimensional. Thus po is not quasi-equivalent to pn for n in N. m[65]

zq

Nq

N~

10.5.46. Let po be the norm limit of a sequence {p,} of factor states of a C*-algebra 2l. Show that po is a factor state quasiequivalent to all but a finite number of {p,}. Deduce that the set of factor states of U is norm closed.

Solution. Choose N such that lip, - pmll < 2 when n,rn 2 N . For such n and rn, r P nand xp, are not disjoint (from Corollary 10.3.6). Since pn and pm are factor states, they are either disjoint or equivalent from Proposition 10.3.12(ii). Thus if n,m 2 N , pn w q p m . The ass-umptions of Exercise 10.5.45(ii) are fullfilled for the sequence { ~ N , ~ N + I.},with . norm limit PO, and po is a factor , , .. It follows that a norm limit state quasi-equivalent to p ~pN+1,. of factor states is a factor state and the set of factor states of 2l is norm closed. m[1.6,65]

.

10.5.47. Use Exercise 7.6.34 to prove that a norm limit of a sequence of pure states of a C*-algebra is a pure state equivalent

EXERCISE 10.5.48

609

to all but a finite number of the pure states in the sequence. (By “equivalence” of states, we mean equivalence of the corresponding representations .)

Solution. Let{p,} be a sequence of pure states of a C*-algebra U converging in norm to a state po of 3. There is an integer N such that llpn - pmll < 2 when n , m 2 N . Let rn be the GNS representation constructed from pn. From Corollary 10.3.8, all r n with n larger than N are equivalent to a single representation T. From Theorem 10.2.3, T is irreducible, so that .(a)’ consists of scalar multiples of I . But from Exercise 7.6.34(i), there is a projection E’ in .(a)‘ such that TO (corresponding to po) is equivalent t o the representation A + n(A)E‘ of U on E’(3-1). Thus E’ must be I and TO is equivalent t o T and hence to T N , T N + ~ , .. .. m[65] 10.5.48. Suppose U is a C*-algebra acting irreducibly on the Hilbert space 3-1, x and y are unit vectors in 3-1, and 3-10 is the twodimensional subspace of 3-1 generated by x and y. Let E and F be the one-dimensional projections in B(3-10) with ranges [x]and [y], respectively. (i) Show that l l % l ~- 0,lq = 11% - wyll = I[(& - wy)lB(3-1o)ll = tr(lE - PI),where “tr” denotes the (non-normalized) trace on B(3-10)(viewed as the algebra of 2 x 2 matrices). (ii) Show that

[Hint. Note that x and y are eigenvectors for ( E - F ) 2 and deal with the cases (2, y) = 0 and (x,y) # 0 separately.] Solution. (i) Since U acts irreducibly on 3-1, 3- = B(3-1);and from the Kaplansky density theorem, ~ ~ w , ~U w,JUII = llwI - wyll. Since (w, - w y ) ( B )= (w, - w y ) ( Q B Q ) ,where Q is the orthogonal projection of 3-1 onto 3-10, IIw, - w y J J= II(w, - wy)lB(3-10)ll. Since EAE = ( A x , s ) E = w , ( A ) E and F A F = (Ay,y)F = w,(A)F, we have tr[(E - F)A] = tr(EAE) - t r ( F A F )

= w,(A) tr(E) - wy(A) t r ( F ) = u,(A) - w,(A).

610


If we view S('H0) as complex 2 x 2 matrices relative to diagonalizes E - F and assume that E - F has matrix

a

basis that and A

[ :,]

[: :],

has matrix then (w, - wy)(A) = Xu t X'd. If IlAll 5 1, then I ul -< 1 and [dl 5 1; so that

Choose a and d of modulus 1 such that Xu = 1x1 and X'd = IX'l. Let 6 and c be 0. Then IlAll = 1 and tr[(E' - F ) A ] = 1x1 t IX'I = tr(lE

- FI),

from which (i) follows. (ii) As in the hint, ( E - F ) 2 commutes with E and F , so that z and y are eigenvectors for ( E - F ) 2 . If (z,y)= 0, (ii) becomes: Ilw,(U - wYl2lll = 2. If E' and F' are the projections in B(1-I) with ranges [z]and [y], then I(w,-wy)(E'-F')I = 2, and 2 5 llwZ-wyll 5 IIwzll -t llwyI) = 2. If (5, y) # 0, then z and y correspond to the same eigenvalue for ( E - F ) 2 since ( E - F ) 2 is self-adjoint. But 1-Io is two dimensional. Thus ( E - F ) 2 is a scalar. Since

( z , ( E- F)%) = (z,z- E F z ) = 1 - I(z,y)12, IE - FI is the scalar [l - l(~,y)1~]'/~. Thus

from (i). 10.5.49. Use Exercise 10.5.48 to prove again that a norm limit of pure states of a C*-algebra is a pure state (and hence that the family of pure states is norm closed).

Solution. Let { p n } be a sequence of pure states of a C*-algebra U converging in norm to a state po of U. As in the solution to Exercise 10.5.47, there is an integer N and a representation n of ?2l on 1-I such that A" is equivalent to n when n 2 N . Passing t o a subsequence, we may assume that IIpn+l - pnll 5 2-" and nn is equivalent to A for all n in N. By choosing unit vectors successively

611

EXERCISE 10.5.50

in 'H, we can arrange that wxn o r = pn and that From Exercise 10.5.48 then, we have that

(zn+l,zn)

2 0.

It follows that (5,) is a Cauchy sequence in H and converges to a unit vector zo in H. Hence wx:,Ir(21)is the norm limit of wx,lr(21) and w,, o r = po. Since A ( % ) acts irreducibly on H,zo is generating for r(0) and A is equivalent to the GNS representation constructed from po by Proposition 4.5.3. Thus the GNS representation constructed from po is irreducible and po is a pure state of 2l from Theorem 10.2.3. 10.5.50. Use Theorem 10.2.3 and the result of Exercise 10.5.48 to show once again (see Exercise 4.6.26(ii)) that a C*-algebra 2l is abelian if there is a positive real number 6 such that llp1 - p 4 2 6 whenever p1 and p2 are distinct pure states of 2l.

Solution. Let p be a pure state of 2l and r pbe the irreducible representation of 2l obtained from p by means of the GNS construction. (See Theorem 10.2.3.) If l i phas dimentsion greater than 1, we can find linearly independent unit vectors z and y in lipsuch ' / ~6. From the result of Exercise 10.5.48, that 2[1 - l ( z , ~ ) 1 ~ ] < IIw, - wyll < 6. Thus llpl - p2II < 6, where p1 = wz o r p and p2 = w y o r,. From Corollary 10.2.5, p1 and p2 are pure states of a. Since r p ( U )acts irreducibly on H,,rp(A)z = z and r,(A)y = 0 for some A in 8. (Use Theorem 10.2.1 for this and recall that z and y are linearly independent.) For this A, p l ( A ) = 1 and p2(A) = 0. Hence p1 # p2, contradicting the choice of 6. It follows that H, is one dimensional, whence rp(21)= @I,for each pure state p of 2l. In particular rp(21)is abelian for each pure state p of 2l. If B and C are in 8 , then r,(BC - CB)= 0; so that p(BC - CB) = 0 for each pure state p of 2l. From Theorem 4.3.8(i), BC = CB. Hence 2l is abelian.

612


10.5.51. Suppose that 'H is a Hilbert space, K (C a(%))is the ideal of all compact linear operators, U is a C*-algebra such that K 24 C_ a(%),and p (# 0) is a bounded linear functional on U that vanishes on K . Show that p is singular, in the sense of the discussion preceding Theorem 10.1.15. [Hint.By considering the linear functional induced by p on U/K, show that p is a linear combination of states of U that vanish on K. Then use Proposition 10.1.17.1

Solution. Since p is a bounded linear functional on U and vanishes on K , there is a bounded linear functional q on U/K such that p = q o cp, where y : 24 --+ U/K is the quotient mapping (Theorem 1.5.8). By Corollary 4.3.7, q is a finite linear combination of states ~ 1 , .. .,qn of U/K. If 1 5 j 5 n, q j o cp is a state p j of U that vanishes on K , and p is a linear combination of p 1 , . .. ,p,. If w is an ultraweakly continuous linear functional on U, and 0 5 w 2 p j , then 0

5 w ( K ) 5 p j ( K ) = 0, w ( K ) = 0,

for each K in Kt. Since the linear span K of K+ is ultraweakly dense in U, it follows that w = 0. From Proposition 10.1.17, pj is I singular (for each j in { 1,. , . ,n}), and so is p. 10.5.52. Let U be a C*-algebra acting on a Hilbert space 'H, and let K be the ideal of compact operators. Suppose p is a positive linear functional on U such that llpll = llpl% n K11. Show that (i) there is an increasing sequence { E n } of finite-dimensional projections En in IU such that { p ( E , ) } tends to llpll; (ii) llp - pall 0, where p,(A) = p(E,AE,). -+

Solution. (i) Since p is hermitian (and positive on U and UnK is a self-adjoint subspace of a(%)),

From Exercise 2.8.29, each positive H in K has a finite or infinite set of positive eigenvalues A, corresponding to an orthogonal family of finite-dimensional projections G,, {A,} --t 0 when {A,} is infinite and H = CA,G,. If H E % as well, then and A 1 > A2 > f ( H ) E U for each continuous function f on [O,A1]. If we choose this f so that f(A,) = 1 and f(0) = f ( A j ) = 0 when j # n, then e m . ,

6 13

EXERCISE 10.5.53

G, = f ( H ) E U. Given a positive E , we can choose H in (U n K);' such that llpll- ;E 5 p ( H ) . If the eigenvalues {A,} of H form a finite set and G = C G , ( 2 H ) , then G E 2l and p(G) 2 p ( H ) 2 llpll- ; E . If they form an infinite set, then for m large enough, Am < llpll and IIE;=m+lX,Gn(I < g llpll. Thus P (x;=m+l AnG,) < $ 7 and

Thus ( p ( 1 ) =) ((pII = sup{lp(E)I : E E F},where -F is the set of projections (necessarily, with finite-dimensional range) in U n K. Note that if E , F E F ,then E t F E U n K and the (finitedimensional) range projection EV F of E t F is a polynomial function (with 0 constant term) of E t F . Thus E V F E T . Suppose now tha.t we have chosen a sequence { F , } in F such that {p(F,)} tends to llpll. Let E , be F1 V ... V F,. Then E , E F and { p ( E , ) } tends to llpll (= P ( W * (ii) With E a projection in U, from the Schwarz inequality lP(A) - P(EAE)l

JWu

- W E ) I + I P ( J w I - WI + I P ( V - E))I < p ( -~ E ) ' / ~ ~ ( E A * A E )t' ~/ ~( E A A * E ) ~ -/ ~q~1 (l 2I

I

-t p ( I - E ) ' i 2 p ( ( I- E ) A * A ( I - E))1/2 < E

when IlAll 5 1 and p ( I - E ) < (9 l l p l l ) - ' ~ ~ .With {En}as in (i), p ( I - E n ) (= llpll - p ( E , ) ) tends to 0, and IIp - pnJI tends to 0. 10.5.53. Let 2l be a C*-algebra, Z be a norm-closed, two-sided ideal in U, and p be a positive linear functional on 2l. Show that (i) p = p1 t p2 with p1 and p 2 positive linear functionals on U such that llplII = llpllZll and p21Z = 0; (ii) the decomposition of (i) is unique. [Hint. Use Proposition 10.1.5.1 Solution. We may assume that U acting on H is the universal representation of U. In this case, from Proposition 10.1.5, there is a vector x in H and a central projection P in 24- such that p = ~ ~ and Z- = U-P.

1 %

614


(i) Let xi be P,, 2 2 be (I- P)., p1 be wzl (24,and p2 be w,, From the Kaplansky density theorem,

1%.

and 0 = wx,121-P = wx,IZ- so that p2lZ = wx,IZ= 0. For each A in 2l,

Thus p = pi f p2. (ii) Suppose p = pi t pa, where pi and pa are positive linear functionals on 2l such that IIpiII = llpilZll and pal1 = 0. Then pi = wI; (24and pi = wz; 1% for some vectors z{ and zi in 7-l. Since

we have that Pxi = zi. Now pi t pi = p1 t p2 so that pi p2 - p i . Thus

- p1

=

Since P q = x1 and Px: = xi, ( w , ~- wz:)l%-(I - P) = 0. Thus (wE1- wX;)IU- = 0 and p1 = wZ11% = w,; 1% = p i . It follows that m[36(Lemma 3, p. 218)] p2 = p i . 10.5.54. Let 7-1 be an infinite-dimensional Hilbert space and K be the ideal of compact operators on 3-1. Show that (i) the vector state space (that is, the weak * closure of the set of vector states) of B(7-1) coincides with the pure state space (that is, the weak * closure of the set of pure states) of B(3-1); (ii) the set of states of B(7-1) that annihilate K is a non-null weak * compact convex subset K$ of a(%)#whose extreme points are pure states of O(3-1).

EXERCISE 10.5.55

615

Solution. (i) From Corollary 4.3.10, the vector state space contains the pure state space of B(7-1). From Corollary 10.2.5, each vector state of B('H) is a pure state. Thus the vector state space is contained in the pure state space of B(7-1). Hence the vector state space and the pure state space of B(7-1) coincide. (ii) Since the set of states that annihilate a single operator is a weak * closed convex set and K$ is the intersection of such sets, K$ is a weak * closed convex set. From Exercise 4.6.69(i), there is a state (in fact, a pure state) that annihilates K. Thus K$ is non-null. From the Krein-Milman theorem, K$ is the weak * closed convex hull of its extreme points. Let p be such an extreme point. If p = apl (1 - a)p2 with a in (0,l) and p1,p2 states of B(7-1), then p1 and p2 annihilate T*T for each compact operator T since p(T*T)= 0. Thus 0 = p l ( T ) = p2(T) and p1,p2 E K$. Since p is an extreme point of K;, p = p1 = p2. Hence p is a pure state of

+

B(7-1). 10.5.55. Let 7-1 be a Hilbert space and K the ideal of compact operators on 7-1. Let p be in K$, the set of states of B ( X ) that annihilate K. Show that (i) if p is a pure state of B ( X ) and E is a projection such that ( I - E)(7-1) is finite dimensional, for each finite set of operators A 1 , . . . ,A , in B(7-1) and each positive E , there is a unit vector z in E(7-1)such that

[Hint. Use Corollary 4.3.10.1; (ii) p is a weak * limit of vector states of B(7-1) and p is in the pure state space of B(7-1); (iii) aw, (1- a)p is a weak * limit of vector states of B ( X ) and is in the pure state space of B(7-l). [Hint. Use Exercise 4.6.69(ii).]

+

Solution. (i) Replacing E if necessary, we may assume that max{lIAjII : j E {O,. . . ,m } } 5 1, where A0 = I . As p ( I - E ) = 0 by assumption, p ( E A E ) = p(A) for each A in B(7-1). From Corollary 4.3.10, each pure state of B(7-1) is the weak * limit of vector states of B(7-1). Let 2' be a unit vector in 7-1 such that

616


and let zo be Ex'. Then (*) can be rewritten as

In particular (with 0 for j ) , 1- llx011~< 2 so that 1- 11z0)1< 2. Let x be xo/llzoll (from the preceding, with E less that 1, 20 # 0). Then

Combining this last inequality with (**) (and recalling that IlAjII 5 l),we have that luz(Aj)-P(Aj)l < E for j in { 1,.. . ,m } and E z = z. (ii) Let A l , . . . , A , in B(3-1) and a positive E be given. By adjoining the adjoints if necessary, we may assume that A1 = I and { A l , ... , A m } is a self-adjoint subset of B(7-f). From Exercise 10.5.54(ii), there is a finite convex combination alp1 t.* t anpn (= P O ) of pure states P I , . .. , p n of B(3-1) in K$ such that

From Corollary 4.3.10, we can find a unit vector

z1

such that

Let El be the projection with range [ A j q : j E {l,... , m } ] . From (i), there is a unit vector 1 2 in ( I - E1)(3-1) such that

Let Ez be the projection with range [ A j q , A j z z : j E ( 1 , . . . , m } ] . Again from (i), there is a unit vector 2 3 in ( I - E2)(3-1) such that

Continuing in this way, we construct unit vectors property that

and

Xk

is in

..

21,. ,z,

with the

617

EXERCISE 10.5.56

Suppose

T, s

E (1,. . . ,n} and

T

< s. Then

and since { A l , . . . ,Am} is a self-adjoint set, 0 = ( A ~ z , , z , ) . Thus (Ajz,, z,) = 0 whenever T and s are distinct elements of { 1,. , ,n} and j E {1,...,m}. Let z be U ; / ~ Zt~ -t- a,1/2 I,. Then 2 is a unit vector and for all j in (1,. . , m } ,

.

.-.

.

Thus

.

and (w,(Aj) - p ( A j ) ( < E for all j in (1,. . , m } . It follows that p is a weak * limit of vector states. Fkom Exercise 10.5.54(i), p is in the pure state space of D(‘H). (iii) From (ii), p is in the pure state space of B(7-f). Fkom Exercise 4.6.69(ii), awx (1 - a)p is in the pure state space of f?(H). From Exercise 10.5.54(i), therefore, aoX (1- a)p is a weak * limit of vector states.

+

+-

10.5.56. Let 2l be a C*-algebra acting on a Hilbert space ‘H, and let K be the ideal of compact operators on ‘H. Show that (1 - u)p is a weak * limit of vector states of U for (i) uw,lM each unit vector I in ‘H and each a in [0,1], where p is a state of 2l that is 0 on 2l n K [Hint.Use the quotient mapping of B(’H) onto f?(‘H)/Kand use Exercise 10.5.55.1; (ii) each weak * limit of vector states of M has the form uw,IB+ (1 - u)p, where z is a unit vector in ‘H, p is a state of 2l that is 0 on M n Ic, and a E [0,1]. [Hint.Use Exercises 10.5.52 and 7.6.33.1

+

Solution. (i) Since 0 n K is a norm-closed, two-sided ideal in M, U/M f l K is a C*-algebra that is * isomorphic to (a -t K ) / K . (See Corollary 10.1.9.) From Exercise 4.6.23(ii), there is a state po of %/an K such that p is the composition of po and the quotient

618


mapping of U onto M/M n Ec. Through the isomorphism of 0 / M n K: with (U t K)/Ec, there is a state p1 of (a t K ) / K that corresponds to PO. From Exercise 4.6.23(i), there is a state pf of 0 4- Ec, the composition of p1 with the quotient mapping of UtK:onto (!2l+-Ec)/K:, whose restriction to 0 is p and that is 0 on Ec. Let q be a state extension of pf from M Ec to B(3-1). From Exercise 10.5.55(iii), aw, (1 - a ) is~ a weak * limit of vector states of B(H). Hence aw,(M t (1- a)p is a weak * limit of vector states of 2l. (ii) Suppose po is a weak * limit of vector states of U. From Exercise 10.5.53, po = apl t (1 - a)&, where p1 is either 0 or a state of M for which 1 = ((pl(0n lcll and pz is either 0 or a state of M for which ~ 2 1 %t l Ec = 0. If p1 = 0, then we can choose a to be 0 and p to be p2. Assume p1 is a state of U. From Exercise 10.5.52, there is an increasing sequence {En: n = 2,3,. ..} of projections in U f l K such that { p l ( E , ) } tends to 1 and {lip1 - pnI(} tends to 0, where p,(A) = p1(E,AE,) for each A in M. Now polUnlC (= apl(UnK) is a weak * limit of positive vector functionals on %nK. Thus pl(Mnlc is a weak * limit of vector states of M n K (that is, positive vector functionals whose restrictions to U n K have norm 1);say {WZblM n K}b,o tends to pllu n K in the weak * topology on (U n K ) # . Then the net {uZb IEnOEn)bE~ tends to p1IEnUE, in the weak * topology on (E,ME,)#. Since E,UE, acting on En(%)is a finite-dimensional C*-algebra (von Neumann algebra), this convergence is in the norm topology on (E,UE,)f. As wz,IEnQEn = WE,,^^ IEnME,, p1IE,ME, is a positive vector functional, uy[EnMEn,on E,ME, (from Theorem 7.3.11, for exand {IIpl - u~,,~lMlI}tends to 0. ample). Thus pn = WE,~~U, From Exercise 7.6.33, p1 is a vector state w,lU of 0. Thus po = awZ[21t (1 - a)p2. If p2 = 0, then a = 1, po = oz121, and we can take for p, the restriction to U of any state of B('F1)that is 0 on K , If p2 # 0, we let p be p2. In any event, po = awzl?2l t (1 - a)p, where a, 2, and p, are as described. m[36]

+

+

10.5.57. Let M be a C*-algebra acting on a Hilbert space H. (i) Suppose 31 is two dimensional and B is a maximal abelian subalgebra of B(3-1). Show that M does not separate the set of vector states of B(3-1). (ii) Suppose 0 acts irreducibly on 3-1. Show that 2l does separate the set of vector states of B ( H ) . (iii) Suppose M separates the set of vector states of B('F1). Show

619

EXERCISE 10.5.58

that 2l acts irreducibly on 'H. (iv) Suppose 0 is a C*-algebra containing 2f and acting irreducibly on 'H. Show that 2l acts irreducibly on X if 2l separates the set of vector states of 8.

Solution. (i) Choose an orthonormal basis { e l , e2) consisting of (simultaneous) eigenvectors for 2l. Let u be (el +e2)/2112 and D be (el +ie2)/2112. Then wu # wv since u and D are linearly independent. But (Au, u ) = ( u t 6)/2 = (Av, v) for each A in 9, where Ae1 = uel, Ae2 = 6e2, so that w,lU = w,, 1%. (ii) Let x and y be linearly independent unit vectors in 'H and let A be an operator in B('H) such that Ax = x , Ay = 0. Since 2l acts irreducibly on 'H, 2l- = f?('H) and there is an operator A0 in 2l such that IIAox - 211 and llAoyll are small. Then I(Aoz,x) - 11 and I(Aoy,y)l are small. In particular, oz(Ao) # wy(Ao). Thus U separates the set of vector states of f?(X). (iii) Let x be a unit vector in 7-1. If [ax]' is at least one dimensional, choose a unit vector y in [Uz]' and let u be (x y)/2112 and D be ( x i - iy)/2'I2. Then w,(A) = ( ( A x , x ) t (Ay,y))/2 = w,(A) for each A in 2l since 24 is self-adjoint. However, w, # w, since u and D are linearly independent. Thus [2lx] = 'H for each unit vector x in 'H, and 2l acts irreducibly on 'H. (iv) Since f? acts irreducibly on 'H, 8 separates the set of vector states of B(7-1) from (ii). As 2l separates the set of vector states of f?, 2l separates the set of vector states of O(7-l). From (iii), 2l acts irreducibly on 3-1. rn

+

10.5.58. Let 2l be a C*-algebra acting irreducibly on a Hilbert space 'H, K be the ideal of compact operators on 'H, and f? be a C*subalgebra of 2l that separates the pure state space of U. Show that (i) f? acts irreducibly on 'H; (ii) q(B) = cp(2l), where q is the quotient mapping of 2l onto %/afl K: [Hint.Use Exercises 4.6.70 and 10.5.56.1; (iii) 2l = f3 if 2l n K = (0); (iv) U = f ? + K : i f K C 2 l ; (v) 2l = f?. [Hint.Note that if 0 n K = (0) and K C 2l, then wzlB extends to a state of U that annihilates K.]

Solution. (i) Since the vector states of 2l are pure states (by Corollary 10.2.5), f3 separates the set of vector states of 2. From

620


Exercise 10.5.57(iv), B acts irreducibly on 3-1. (ii) Since U acts irreducibly on R,either U n K = (0) or K G 2i from Proposition 10.4.10. In either case, each state p o cp of U, where p is a state of U/2i t l K , is in the pure state space of U, since p o cp is a weak * limit of vector states from Exercise 10.5.56(i), and vector states are pure since U acts irreducibly. (In fact, the pure state space and the vector state space of U coincide.) Thus, if p1 and p2 are distinct states of a/%nK ,p1 o cp and p2 o cp are distinct elements of the pure state space of U. By assumption, there is a B in B such that pl(cp(B))# pz(cp(B)). Thus cp(f3) separates the states of cp(U). From Exercise 4.6.70, cp(U) = cp(B). (iii) If 2l n K = (0), cp is an isomorphism onto. From (ii), cp(U) = cp(B), whence U = 13. (iv) If K G U, then B t K = U since cp(B) = cp(U) and cp is the quotient mapping of U onto U/K. (v) If K c B, then B = B t K = U from (iv). Since B acts irreducibly on 3-1 from (i), if Ic B , then B n Ic = (0) from Proposition 10.4.10. Similarly, U n Ic = (0) unless K c U. If U f l K = (0), then U = B from (iii). We may assume that K C U and B n K = (0). Let 2 and y be orthogonal unit vectors in 3-1 (the case in which 3-1 is one dimensional needs no discussion). Let p1, p2 be the states of U (= 8 t K) defined as ( B s , z ) ,(By, y), respectively, at B t C, where B E 13 and C E K. That p1 and p2 are states follows by noting that p1 = wx 0 (cplfl)-'

0 (0,

pz = uy

0

(cplf31-l

0

cp,

91.13is a * isomorphism, and cp(U) = cp(B). From Exercise 10.5.56(i), 1 p(p1 tua,IU)and i ( p 2 +w,(U) are elements of the pure state space of U that take the value and 0, respectively, at the one-dimensional projection with range [y], but whose restrictions to B are the same state, +(usf uy)lB. Thus B fails to separate the pure state space of 2f in this situation. It follows that U and B either both contain K or both meet it in (0). In both cases, U = 8. 1[36]

4

10.5.59. Let U be a simple C*-algebra and B be a C*-algebra contained in U that separates the pure state space of U. Show that

u = B.

Solution. Let 1~ be an irreducible representation of U. (See Corollary 10.2.4.) Since 3 is simple, ?r is a * isomorphism of U

621

EXERCISE 10.5.60

onto .(a), and n(B) separates the pure state space of rn Exercise 10.5.58,A ( % ) = ~ ( f ? ) .Thus U = B.

A(%).

From

10.5.60. Suppose that p is a * automorphism of a C*-algebra U, and !lj : U + B('FI*) is the reduced atomic representation of U. Given any representation cp : U + B('FI,) of U, we can consider also the representation cp o P : 2l+ f?(W+,). Show that (i) if cp and II, are equivalent representations of U, then cp o p and ?I, o ,d are equivalent representations; (ii) if cp is an irreducible representation of U, then cp o /? is an irreducible representation; (iii) there is a unitary operator U acting on 'HQ such that !€'(/?(A))= U @ ( A ) U *for each A in U; (iv) the * automorphism of Q(U)extends uniquely to a * automorphism p of the von Neumann algebra @(a)-. Solution. (i) Since cp and .1c, are equivalent, there is a unitary ' , onto 'FI$ such that transformation V from H

$ ( A ) = Vcp(A)V*

( A E U).

With P ( A ) in place of A , we obtain

so cp o p and II, o p are equivalent. (ii) The C*-algebra (cp o P)(U) coincides with cp(U), and acts

irreducibly on 'Hp (since cp is an irreducible representation). (iii) The reduced atomic representation !P is CaE-$7ra, where { A :~ a E A} is a family of irreducible representations of U that contains exactly one member of each equivalence class of irreducible representations of a. Given a in A, o P is an irreducible representation of U, by (ii), and is therefore equivalent to nf(,) for just one element f ( a ) of the index set A. In this way, we obtain a mapping f : A + A. Similarly, there is a mapping g : A -+ A such that, for each a in A, the representations A, o p-' and rg(,) are equivalent. From (i) A, (= ( A , 0 P - l ) o p ) is equivalent to rg(,)o p , and hence to A I ( ~ ( , ) ) . Thus f ( g ( a ) ) = a , and similarly g ( f ( a ) ) = a; so f has an inverse mapping g , and f is a one-to-one mapping from A onto A.

622


Let 'FI, be the Hilbert space on which a4(U) acts, so that 'HQ is CoEA @'Ha. Since x 4 o P is equivalent to ~ j ( ~there ) , is a unitary transformation U , from 'Hj(a)onto 'FI, such that

The equation

defines a unitary operator U acting on an element C e x 4 of Nu,we have

XQ.When A E U and x

is

Thus Q ( @ ( A ) ) U= U Q ( A ) ,for each A in %. (iv) From (iii), Q(U) = Q(P(U)) = UQ(U)U*; so Q(U)- = Uq(U)-U*, and the equation

p ( B ) = UBU" defines a * automorphism

(B E

*(a)-)

of !€!(a)-.When B = * ( A ) E U(U),

(QPQ-')(B)= Q ( P ( A ) )= U I ( A ) U * = UBU* = p ( B ) ; so

p extends QPQ-'.

Since every * automorphism of Q(U)- is ultraweakly continuous (Remark 7.4.4), two such automorphisms that coincide on g(U) are equal. Thus is the unique extension of QPQ-' to a * automorphism Of g(%)-.

p

EXERCISE 10.5.61

623

Suppose that 2l is a C*-algebra, 6 is a derivation 10.5.61. of 8, and & ( A * )= 6(A)* for each A in 2l (in these circumstances, we refer t o 6 as a * derivation of U). Recall, from Exercise 4.6.65, that 6 is bounded. Let Q be exp6, in the sense of the holomorphic function calculus for 6 in the Banach algebra B(2l) of all bounded linear operators acting on 2l, and note that Q = C~=o(n!)-'6n, by Theorem 3.3.5. (i) Establish the "Leibniz formula," n

P ( A B )=

C (:)sr(A)an-.(B)

( A , B E U; n = 1 , 2 , . . .),

r=O

where (3) is the binomial coefficient n!/r!(n- T ) ! . (ii) Prove that a is a * automorphism of 2l. (iii) Suppose that 2l acts on a Hilbert space 'H, H is a self-adjoint element of B('H), and

6(A)= i(HA- AH)

( A E 3).

Prove that n

P ( A ) = in E ( - l ) r ( r ) H n - r A H r

( A E 2l, n = 1 , 2 , . . .),

r=O

and that

Q ( A )= UAU*

( A E a),

where U is the unitary operator exp iH. [In Exercise 10.5.72, we prove that a * automorphism a of a C*algebra 2l has the form exp6, where 6 is a * derivation of U, if [la- LII < 2, where L is the identity mapping on 2l;Exercises 10.5.6510.5.71 lead up to this result. From Exercises 10.5.63 and 6.9.43(ii), it follows that a * automorphism of a von Neumann algebra cannot necessarily be expressed as the exponential of a * derivation.]

Solution. (i) We prove the result by induction on n. In the case n = 1, the required formula becomes (1)

6(AB)= A6(B)t 6(A)B,

and this is satisfied since 6 is a derivation.

624


Suppose that m is a positive integer, and the required formula has been proved in the case n = m; that is, m

Sm(AB)=

C (Y)Sr(A)Gm-"(B). r=O

By applying S to both sides of this equation, and using (l),we obtain

c (Y) m

Gm+'(AB) =

[Sr(A)Sm-"+'(B)t 6r+1(A)Sm-'(B)]

r=O

= (:)AS"+'(B)

=

c

(m~')6r(A)6m+'-r(B),

r=O

since

This shows that the required formula is satisfied when n = m -t 1, and so completes the inductive proof. (ii) Since eZe-" = e-'eZ - 1 for a l l z in @, it follows that (exp 6) c) (exp(-6)) = (exp(-S))

0

(exp 6) = L ,

where L denotes the identity mapping on U. Thus (Y (= exp6) has an inverse mapping exp(-S), and is therefore a one-to-one linear mapping from U onto U. Note also that, since S(A*) = S(A)* for each A in U, we have P ( A * ) = P ( A ) * ,and

a ( A * )=

C $ P ( A * ) = C &S"(A)* = a ( A ) * . n=O

n=O

It remains t o prove that a(A)cu(B)= a ( A B ) for all A and B in U. By formal manipulation of infinite sums, and by use of (i), we

625

EXERCISE 10.5.61

obtain M

O

O

n=O r=O n

=

C $ P ( A B ) = a(AB). n=O

These formal manipulations are justified because 0 0 0 0

M

O

O

and all the sums occurring above exist in the "unordered" sense discussed in Section 1.2. (iii) With H and U as described in (iii), we have

c O0

U = expiH =

n=O

c 00

in

-Hn,

U* = exp(-iH) =

n!

n=O

(4)"

-Hn, n!

and 00

n=O

n=O

Formal manipulation of infinite sums (justified, because the sums

626


exist in the "unordered" sense) gives O0

O0

jT+S

UAU* =

-(-1)"H'AH" T

n

in

c

H n-' AH'

._

00

=

r!s!

d S=o

00

n

~ ( - l ) ' ( ~ ) H n - 8 A H ' ( A E U). I".

n=O

s=o

If we prove that n

P ( A ) = in E ( - l ) T ( : ) H n - T A H r

(2)

( A E U),

T=O

it then follows from the preceding chain of equations that 00

UAU* =

C $P(A)

= (exp 6 ) ( ~=) o.(A).

n=O

It remains to prove (2). When n = 1, (2) reduces to the given equation 6(A) = i(HA - AH). Suppose that m is a positive integer, and (2) is satisfied in the case in which n = m; that is m

P ( A ) = im C ( - l ) T ( Y ) H m - T A H r . T=O

Then P + l ( A)

= 6( bm ( A ) ) = i[HP(A) - P(A)H] m

= jm+l ~ ( - l ) r ( ~ ) [ H m + l - r A H - Hm-'AHTfl]. r T=O

Thus 6 m + l (A)

{

= imfl Hm+',4

4- (-l)"+'AHmfl}

-pfl

C (- 1)' ( m r + l ) ~ m + l - r ~ ~

m+l

r.

r=O

627

EXERCISE 10.5.62

It follows that (2) is satisfied when n = rn t 1. This completes the inductive proof of (2). Suppose that 6 is a * derivation of a C*-algebra U 10.5.62. and a is the * automorphism exp 6 of U. Show that a is "universally weakly inner" in the following sense: if cp is a faithful representation of U, the * automorphism cpacp-' of cp(U) is implemented by a unitary operator in cp(M)-. [Hint.Use Exercises 8.7.55(ii) and 10.5.61.1

Solution. The C*-algebra cp(2l) acts on a Hilbert space R+,, ~ 6 c p -is~a * derivation of cp(U), and cpcrcp-l is a * automorphism of cp(U). It is apparent that c p ~ 5 ~ c p - = l ( c p 6 ~ - ' ) ~for , n = 0,1,2,. ..; by expressing exp 6 and exp (p6cp-l in power series form, it follows that 'pacp-' = cp(exp 6 ) ~ - ' = exp(cp6cp-'). From Exercise 8.7.55(ii), there is an element HO of cp(U)- such that

(cpS(o-l)(B)= HOB - BHo

( B E cp(U)).

Since, also

we have

(cpScp-')(B) = +[(HO- H,*)B- B ( H 0 - H i ) ] = i ( H B - B H ) , where H is the self-adjoint element ki(H,* - H o ) of cp(U)-. It now follows from Exercise 10.5.6l(iii) that

where U is the unitary element expiH of cp(U)-.

m[69]

10.5.63. Show that a * automorphism cr of a von Neumann algebra R is inner if and only if a has the form exp 6, where 6 is a * derivation of R.

628


Solution. If LY is inner, it is implemented by a unitary operator U in R, and U =: expiH for some self-adjoint element H of R (Theorem 5.2.5). The equation 6(A) = (HA - AH)

( A E R)

defines a * derivation 6 of R, and it follows from Exercise 10.5.61(iii) that (exp6)(A) = UAU* = a(A) (A E R). Thus a = exp6. Conversely, suppose that a = exp6 for some * derivation 6 of R. Since the identity mapping on R is a faithful representation of R, and R = R-,it follows from Exercise 10.5.62 that a is implemented w by a unitary operator in R; so a is inner. 10.5.64. Suppose that ?2lis a C*-algebra, 6 is a bounded linear operator from 2l into a, and expt6 is a * automorphism at of 2l for each real number t. Show that 6 is a * derivation of 2l. Solution. Suppose that A, B E U and t E W. Then O = al(A*) - at(A)* =

-

tn C 7[bn(A*) - J"(A)*], n.

n=O

and

The manipulation carried out in the last block of equations is justified, because

EXERCISE 10.5.65

629

(see the solution to Exercise 10.5.61(ii)), and the sum

exists in the "unordered" sense discussed in Section 1.2. Thus

O = t [ 6 ( A * )- 6 ( A ) * ]-t

C Itnn.I S n ( A * )- Sn(A)*] O'

n= 2

0 = t[G(AB)- A(6( B ) - 6( A)B]

t

Oo

n=2

2

t7 n [S"(AB)(n)Sr(A)Sn-r(B)], n. T r=O

for all real t . It follows (see the discussion preceding Theorem 3.3.1) that

S(A*) - S(A)* = 0, S(AB) - AS(B) - S(A)B = 0, and 6 is a

* derivation.

Suppose that B is a C*-algebra, L is the identity 10.5.65. mapping on B, p is a bounded linear operator from B into B, and cC\ sp(p) is connected, where sp(p) denotes the spectrum of /3 in the Banach algebra of all bounded linear operators acting on B. Let 2l be a norm-closed subspace of B, and suppose that p(U) C_ U. (i) Show that ( Z L - p ) - l ( 2 l ) & 2l when IzJ> IIpII. (ii) Suppose that A E 2l and p is a bounded linear functional on B that vanishes on U. Prove that

and deduce that ( I L - p)-l(U) 2l for all z in C \ sp(p). (iii) Suppose that f is a complex-valued function holomorphic on an open set containing sp(p), and let f ( p ) be the bounded linear operator on B that corresponds to f in the holomorphic function calculus for p. Show that (j(/3))(2l) U.

630


Solution. (i) When A E U and 1.1

and (ZL

> IIpII, we have ~

~ z< -1, l ~ ~ ~

- @)"(A) = z - ~ ( -L z"@)"(A) M

=

p"( A).

~-"-l

n=Q

The series on the right-hand side is norm convergent, and its sum lies in 8,since @"(A)E 2l and %is norm-closed. Thus ( z L - ~ ) ' ~ ( % )2l. (ii) Let g be the complex-valued function defined on C! \ sp(p) by g ( 4 = P ( ( Z L - p>-'(A)). The argument used in proving Theorem 3.2.3 shows that g is holomorphic on C \ sp(@). When z E C and 1.1 > IIpII, we have z E C \ sp(p); moreover, g ( z ) = 0 by (i), since A E % and p vanishes on %. We have now shown that g is holomorphic on the connected open set C \ sp(P), and vanishes on a non-void open subset of C \ sp(@). Thus g ( z ) = 0 for all z in C!\ sp(P). It follows from the preceding paragraph that if A E U and z E C\sp(P), then p ( ( z ~ - / l ) " ( A ) )= 0 whenever pis a continuous linear functional on B that vanishes on U. By the Hahn-Banach theorem (see Corollary 1.2.13), ( Z L - @)-'(A)E 8. Thus ( Z L - P)-'(U) E U. (iii) From the definition of the holomorphic function calculus (see the discussion preceding Theorem 3.3.5)

where C is the union of a finite collection of piecewise smooth curves in C\sp(p); the integral exists as the limit of approximating Riemann sums, relative to the norm topology on the algebra of bounded linear operators on B. When A E 2l,

and this integral exists as the limit (in the norm topology on B ) of approximating Riemann sums. Since rZ1 is a norm-closed subspace of f3, and ( z i - @ ) - l ( A )E 2l for all z on C by (ii), it follows that ( f ( P ) ) ( AE) a* Thus ( f ( P ) Wc

.

631

EXERCISE 10.5.66

10.5.66. Suppose that B is a C*-algebra, H is a self-adjoint Define bounded linear operators LH, element of B, and IlHll < RH, 6, and p, acting on B, by

3..

L H B = HB,

R H B = BH,

6(B) = i ( H B - B H )

( B E B),

and p = exp6. (For these operators, “functions” are to be interpreted in terms of holomorphic function calculus within the Banach algebra of all bounded linear operators acting on B, and “sp” will denote spectrum relative to that algebra.) (i) Prove that SP(LH) c_ SPB(H), SP(RH) G SPB(H). (ii) Deduce from (i) that sp(6)

c { i ( s - t ) : s , t E sp,(H))

14 I 2 l l H l l } , SP(P) c {expiu : 21 E R, 1.1 I211HII) c { z E c :1.1 = 1, z # -1). G

{ill

: 21 E B,

[Hint.Use Exercise 3.5.24 and the spectral mapping theorem (3.3.6).] (iii) Let “log” denote the principal value of the logarithm in

cs= { z E c : z # +I}, the plane slit along the negative real axis; that is log re’” = logr

+ iu.

( r > 0,

-T

< u < T).

Prove that 6 = logp. [Hint.Use Theorem 3.3.8.1 (iv) Suppose that U is a norm-closed subspace of 8,and p(U) E CU. By using Exercise 10.5.65(iii), show that 6(U) U.

Solution. (i) Suppose that X E @I \ spB(H), so that X I - H has an inverse S in B. The equation L s B = SB defines a bounded linear operator L s from B into B, and

Ls(XL- LH)B = Ls(XB - HB) = S(XI - H)B = B, ( X L - LH)LsB = XLsB - H L s B = (XI - H)SB = B, for all B in B. Thus the bounded linear operator X i - LH on B has a bounded inverse Ls. Similarly, X i - RH has a bounded inverse R s , where R s B = B S . Hence X $ s p ( L ~U) s p ( R ~ )and , SP(LH) u SP(RH)

c SPB(H).

632


(ii) When B E B, LHRHB = H ( B H ) = ( H B ) H = RHLHB, and

6 ( B )= i ( H B - B H ) = i(LHB - RHB). Thus LH and RH are commuting elements, in the Banach algebra of all bounded linear operators acting on B, and 6 = ~ ( L H- R H ) . From Exercise 3.5.24, applied with ~ L H and - ~ R Hin place of A and B, we have

- t ) : S E s p ( L ~ ) ,t E SP(RH)}. From (i), and since the self-adjoint element H of B has its spectrum Sp(6) = Sp(iLH - iRH)

{i(S

in the interval [-l\Hll, llHll], it now follows that sp(6) c_ { i ( s - t ) : s , t E SPB(H)}

5 {iu : 21 E R,1.1 L 211Hll). From the spectral mapping theorem, and since 211Hll

0; 0 1 , . . ,Onare disjoint open sets in C, each with diameter less that E , such that X j E Oj;and Ej is the spectral projection for V corresponding to Oj. Show that

.

.

Ej#O, Deduce that if z = range of E j , then

llVEj-XjEjll

C;=,

< E

( j = 1 ,...,7~).

a,!/2yj, where yj is a unit vector in the

(ii) By using Exercise 10.5.67 and (i), show that

Solution. (i) Note first that Ej is ej(V), in the sense of the bounded Bore1 function calculus for V (see Theorem 5.2.8), where e j is the characteristic function of Oj. Since Oj is open and meets sp(V) at A j , there is a continuous function fj : C + [0,1] such that f j ( A j ) = 1 and fj vanishes on C \ Oj;moreover,

0 # fj(V) = (fjej)(v)= f ( V ) E j , 0 #

Ej.

Since

I(t - Aj)ej(t)l 5 d i m O j < E

( t E C),

it follow^ that Il(V - Xj1)Ejll < E . When j # k, we have Oj n Ok = 8, and thus EjEk = 0. With y1,. . ,yn and z as in the statement of (i), (yj,yk) = 0 when j # k, and n.

.

EXERCISE 10.5.68

635

Moreover, since Ej commutes with V,

Thus

n

n

n

O}.]

Solution. (i) Since a(I)= I,it follows that a(C) = C for each element C of the center { z I : z E C!} of the type I factor B(7-l). By Corollary 9.3.5, a is implemented by a unitary element V of B(3.1). (ii) From Exercise 10.5.68,

for each c in the convex hull of sp( V); from continuity of the modulus function, this remains true for all c in the closed convex hull. (Although it is not used in the solution to this execise, we note the fact that the word "closed" can be omitted in the statements of (ii) and (iii), since the convex hull of sp(V) is already closed. This follows

637

EXERCISE 10.5.70

from an easy compactness argument, given the fact (from elementary plane geometry) that each element of the convex hull of sp( V) is a convex combination of at most three elements of sp(V).) (iii) With co and U defined as in the statement of (iii), lcol is the point closest to 0 in the closed convex hull K of sp(U); moreover, (1)

lkl 2 lcol 2 4(4 - /la- Lily > 0

(k E K ) .

If kl E K and Rekl < Icol, then K contains the line segment with endpoints at lcol and k l , and so intersects the interior of the disk with center 0 and radius lcol in @, contradicting (1). Thus Rek: 3 1 ~ 0 1 , for each k in I 0).

Prove that (i) there is aself-adjoint element H of B(7-l) such that llHll and expiH = U ; (ii) the equations

< fn

d ( B )= U B U * , 6 ( B )= i ( H B - B H )

( B E B(X)) derivation 6 of B('H), such that

define a * automorphsim d and a * ti = exp8 and & I % = a ; (ii) 6(U) C U, and 81% is a * derivation 6 of 2l such that exps = a. [Hint.Use Exercise 10.5.66(iv).]

640

SPECIAL REPRESENTATIONS OF C"-ALGEBRAS

Solution. (i) The equation fo(eit) = t

(-+r

< t < $r)

defines a continuous real-valued function fo on the arc in which the for unit circle intersects the open right half-plane, and Ifo(.z)l < each t in sp(U). The restriction fol sp(U) is a real-valued function f in C(sp(U)), [If11 < !pr, and expif(z) = z for each z in sp(U). If H is j ( U ) , in the sense of the continuous function calculus for the normal element U of B ( H ) (Theorem 4.4.5), then H is self-adjoint, llHll < $r, and expiH = U by Theorem 4.4.8. (ii) It is apparent that d is a * automorphism of B ( H ) , 6 is a * derivation of B ( X ) , and d l 8 = a. From Exercise 10.5.61(iii), with B(W) in place of M, it follows that 6 = exp8. (iii) We apply Exercise 10.5.66(iv), with B ( H ) , 8, 6, in place of B, 6, /?,respectively. Since &(a)= a@) = 8,it follows that $(a)C 5%. Hence 61% is a * derivation 6 of M, and by expressing exps in power series form, we have

7ir

a=

= (exp

= exp(81U) = exp 6.

10.5.72. Suppose that 0 is a C*-algebra, a is a * automorphism of 0, and Ila - LII < 2, where L is the identity mapping on U. Show that there is a * derivation 6 of Q such that (Y = expb. [Hint. It is sufficient to consider the case in which 0 is given, acting on a Hilbert space H,in its reduced atomic representation. In this case, use Exercises 10.5.70 and 10.5.71.1 Solution. Since the reduced atomic representation Q of !2i is it suffices to prove an isometric * isomorphism from U onto @(a), the stated result with Q(U)in place of M. Accordingly, by Proposition 10.3.10, we may suppose that U is a C*-algebra of operators acting on a Hilbert space C$Ha,and U' = C$f?('H,). From Exercise 10.5.70(ii), there is a unitary operator U in U' such that

& ( A ) = UAU*

( A E a)

and sp(U) {t E C! : Iz1 = 1, Rez > 0). It now follows from Exercise 10.5.71(iii) that a = expb for some derivation 6 of U. 4691

*

641

EXERCISE 10.5.73

Suppose that a is a * automorphism of a von Neu10.5.73. mann algebra 72, and IIa - < 2, where I is the identity mapping on R. Show that a is an inner automorphism of R.

Solution. From Exercise 10.5.72, (Y = exp6 for some tion 6 of 72; so a is inner, by Exercise 10.5.63. m[69]

* deriva-

10.5.74. Suppose that 2.4 is a C*-algebra and aut(2l) is the set of all * automorphisms of U. (i) Show that aut(2l) is a subgroup of the (multiplicative) group of invertible elements in the Banach algebra a(%)of all bounded linew operators from U into 2l. Deduce that aut(U), with its (relative) norm topology as a subset of a(%),is a topological group. (A set G that is both a group and also a Hausdorff topological space is descibed as a topological group if the mappings

( g , h ) - g h : G x G - t G and g - + g - ’ :

G-G

are continuous.) (ii) Show that 11~1- /311 _< 2 for all a and /3 in aut(U). (iii) Suppose that a E aut(U) and IIa - < 2, where L is the unit element of aut(2l). Deduce from Exercises 10.5.72 and 10.5.62 that a lies on a (norm-continuous) one-parameter subgroup of aut(%), and is universally weakly inner. (By a one-parameter subgroup of a topological group G, we mean a continuous homomorphism t + gt from the additive group R into G; we refer to “the one-parameter subgroup { g t } of G.”) (iv) Let autl(U) be the subgroup of aut(2l) generated (algebraically) by the set {a E aut(2l) : IIa - ~ 1 < 1 2). Show that autc(U) is a connected open subgroup of aut(U), and deduce that autL(U)is the connected component of I in aut(%). [Hint. By considering cosets of autl(21), show that aut,(U) is closed as well as open.] (v) Show that aut,(U) is the subgroup of aut(%) generated (algebraically) by the one-parameter subgroups of aut(U), and each element of autb(!21)is universally weakly inner.

Solution. (i) Each * automorphism of U is an isometric linear mapping from !2l onto U, and so lies in the group N of invertible elements of B(%). When a and ,i3 are * automorphisms of 2l, so are 0-l and a o p; so aut(%) is a subgroup of N . It was noted in

642


Section 3.1 that multiplication is jointly continuous, and inversion is continuous on the set of invertible elements, in any Banach algebra. This applies, in particular, t o B(%). It follows that JV and aut(U) are topological groups (each with the relative norm topology as a subset of f?(U)). (ii) When a,p E aut(U) (c a(%)),we have

(iii) If a E aut(%) and 11a-~ll< 2, it follows from Exercise 10.5.72 that a = exp6 for some * derivation 6 of U. By Exercise 10.5.62, a is universally weakly inner. For each real number t, t6 is a * derivation of U, expt6 is a * automorphism at of U, and {at} is a (norm-continuous) one-parameter subgroup of aut(U) containing (Y

(= 4. (iv) For each a0 in aut(%), the continuous mapping a aao from aut(%) onto aut(U) has a continuous inverse mapping a -+ and is therefore a homeomorphism of aut(U). Let 0 denote the open neighborhood {a E aut(U) : IIa - LII < 2) of L in aut(2.l). Since SZ C aut6(U),each element a0 of the subgroup aut,(%) has a neighborhood Qao contained in aut,(%). Thus autb(%) is an open subgroup of aut(U). It now follows from the preceding paragraph that every right coset of aut,(%) is open in aut(U). Hence aut(%) \ autL(U), a union of such cosets, is open in aut(%), and aut,(%) is closed (as well as open) in aut(%). If a E 0, we have --f

(since a is an isometric linear mapping from U onto U); so a-l E R. In view of this, each element of autL(U)can be expressed in the form a 1 a 2 . . * a , , w h e r e a l , a 2 ,..., a,ESZ. F o r e a c h j i n { 1 , 2 ,...,n}, L is connected to aj by a continuous arc (in fact, a one-parameter subgroup) rj in aut(U), by (iii). Since right translations are homeomorphisms of aut(%), r j c y j + l a j + z ...a, is a continuous arc in aut(u), connecting aj+laj+2 a , to ajaj+.l,.- a,. It follows that

---

rn u rn-lanu

u - - .u rla2a3. . .an

is a continuous arc 'l in aut(%), and connects L to ala2 .--a,. Since aut,(U) is clopen in aut(U), and L E autl(U), we have I' autL(U). Thus aut,(%) is (arcwise) connected.

EXERCISE 10.5.75

643

Since aut,(Zl) is clopen, connected, and contains L , it follows that autb(!21)is the connected component of L in aut(U). (v) Let G denote the subgroup of aut(U) generated (algebraically) by the (continuous) one-parameter subgroups of aut(U). Each of these one-parameter subgroups lies in the connected component aut,(%) of L in aut(%), so G C autl(U). On the other hand, G contains R, and so contains the subgroup aut,(a) generated (algebraically) by 51, since each element of R lies on a one-parameter subgroup of aut(%), by (iii). It follows that G = aut'(2l). In proving (iv), we noted that each element a of aut'(2l) has the form a1a2 a n , where a 1 ,a2,...,a n E 0. Since a 1 ,@a, . ,a, are universally weakly inner, by (iii), so is a. 4691

--

..

10.5.75. Suppose that U is a C*-algebra and aut(%) is the topological group considered in Exercise 10.5.74. Show that (i) aut(U) is discrete if and only if U is abelian; (ii) aut(U) is connected if 3 is a type I factor; (iii) aut(U) is neither discrete nor connected if U is the type 111 factor L3z considered in Exercise 6.9.43(ii). Solution. (i) Suppose first that U is abelian; we give two proofs that aut(2l) is discrete in this case. For the first proof, we note that rU is * isomorphic to C ( X ) , for some compact Hausdorff space X ,and it suffices to show that aut(C(X)) is discrete. Suppose that L # a E aut(C(X)). By Theorem 3.4.3, there is a homeomorphism 9 of X such that a(f)= f o 17 for each f in C(X). Since a # L, it follows that ~ ( z o#) Z O , for some zo in X. Let fo : X + [-1,1] be a continuous function such that fo(z0) = -1, fO(rl(Z0)) = 1. Then fo E C ( X ) ,llfoll = 1, and

IIa

- 41 L

Ila(f0)- foll = llfo 0 D - foll

2 IfO(V(Z0)) - fO(Z0)l = 2. From the preceding paragraph, Ila - ~ 1 1= 2 whenever L # a E aut(C(X)). Thus

- Pll = ll(aP-' - LIP11 = 1l.P-l - L l l = 2 whenever Q and p are distinct elements of aut(C(X)), and aut(C(X)) 1 1 0

is discrete.

644


We now give a second proof that aut(M) is discrete when M is abelian. By taking a faithful representation of U, we may suppose that M acts on a Hilbert space 'H. If a E aut(%) and Ilct - L I ~ < 2, it follows from Exercise 10.5.74(iii) that a is implemented by a unitary element U of M'; that is,

a ( A )= UAU*

( A E U).

Since U- is abelian, Q = L. From the preceding paragraph, IIa - L [ l = 2 whenever L # a E aut(2l). This implies that IIa -PI1 = 2 whenever a and p are distinct elements of aut(2l); so aut(U) is discrete. Conversely, suppose that aut(2l) is discrete. Given any selfadjoint element H of %, define a * derivation 6 of M by

6(A)= i(HA - AH)

( A E U),

and let at be exp t6 for all real t . Since {at} is a (continuous) oneparameter subgroup of the discrete group aut(U), it follows that at = L for all real 1. Given any element A of U,

-

A = at(A)

=

n=O

in n!

-6"(A)

= A +it(HA - A H ) t

Oo

n=2

tn -P(A) n!

( t E W).

Thus H A - AH = 0; this has been proved for all A and self-adjoint H in U, so 2l is abelian. (ii) Suppose that U is a type I factor. If a E aut(2l), then a is inner by Corollary 9.3.5. From Exercise 10.5.63, a = exp6 for some * derivation 6 of %. It follows that Q is connected to L by the continuous arc (one-parameter subgroup) {at}in aut(%), where at = expts. Thus aut(M) is connected. , is not discrete, by (iii) When 2l is the type 111factor C F ~aut(%) (i). Also, ?2lhas an outer * automorphism a,by Exercise 6.9.43(ii). Since 2!! is weak-operator closed in its given representation (acting on 62(&)), it follows that a is not universally weakly inner. F'rom Exercise 10.5.74(v), cr does not lie in the connected component aut'(2l) rn of L in aut(U). Hence aut(U) is not connected.

645

EXERCISE 10.5.76

*

10.5.76. Suppose that 'H is a Hilbert space and 6 is a derivation of B(7i). Let K be an element of B('H) such that 6 ( A ) = i ( K A - AK) for each A in B ( X ) (see Exercise 8.7.55(i)). Show that, if c is a suitably chosen real number and H = ;(I< Ill < E2(1 -&)-I and let T be b-l[S - (1 - ~)(blV1t ... t bn-lVn-l)], where b = t (1 - ~ ) b , . Show that (i) IlTll I 1; (ii) 112' - Vnll 5 2 ~ b - l< 1, where n-l I b,; (iii) T = (1 - ~ b - l ) U t , &b-lUn+l for some U , and Un+l inU(U) [Hint.Use Exercise 10.5.97.1; (iv) S is a convex combination alU1 t . - .t a,U, &Un+1 of unitary elements U 1 , . ..,Un+l of U; (v) (U)?n(co,U(U))' = (U)?ncon+ U ( U ) ,where n is a positive integer, conU(%) is the set of convex combinations of n (or fewer) elements of U ( U ) , co,+U(%) is the set of convex combinations of n 1 elements of U(U)in which the last coefficient may be chosen less than a preassigned positive E , and (U)? = { A r~ U : IlAll < 1).

E

+

+-

Solution. (i) We have

+

llTll 5 b-l[(l - E)llS - (blV1 5 b--"E2 t (1 - E ) b , t E(l

* * .

t bnVn)ll t (1 - E)bn + Ellslll

- E)] = 1.

677

EXERCISE 10.5.98

(iii) From (i), (ii), and Exercise 10.5.97, there are unitary ele~ 6 - Un+l. l ments U , and Un+l in 24 such that T = (1 - &b-')U, of T and (iii), we have From the definition (iv)

+

, a,-l = ( l - ~ ) b , - l , a , = b - ~ ( = (l-~)b,), whereal = ( l - - ~ ) b l ..., and U1 = V1,. . . ,Un-l = V,-l. Now

+ + +

+

+

so that a1 a, E = 1, and alU1 4-... a,U, EU,+~ is a convex combination of U1, . . . ,Un+l. (v) If 71 = 1, (COnU(24))= = (U(U))' = U(U) = con+U(U), since a norm limit of unitary elements of the C*-algebra 24 is in U(24). If S' E (24)pflcon+ U(%)and a positive E' is given, there are elements Ul,. .. ,Un+l in U(U) and non-negative real numbers a l , . ..,a,+l with sum 1 such that a,+l < E' and S' = alU1 a,+lU,+1. We have

+ +

11s'- (UlUl +

*.*

+ an-lUn-l + ( a , + U,+l)U,)II

5 2an+1 < 2 8 .

Hence S' E (a)?n (co,U(U))=. Suppose, now, that S' E (24)f' fl (co,U(U))=. For small (positive) E , llS'll < 1-E. Since S' E (conU(21))=, S' fulfills the conditions imposed on S, and from (iv), S' E con+ U(24).Thus

678


10.5.99. Let M be a C*-algebra and Qinv be its group of invertible elements. Show that the following statements are equivalent: (i) U i n v is norm dense in U; (ii) i(U(2l) U(2l))is norm dense in (%)I; (iii) (a)? c 0 2 + U ( ~ ) .

c

+

Solution. (i) --t (ii) If S E (%)I, there is a sequence {S,} in tending to S. Let an be (max(1, llSnll})-l. Then {anSn} tends to S and anSn E (%)I n !2linv* F'rom Exercise 10.5.92(i), a n S n E i ( U ( U ) U(0)). Hence $(U(U) U(2l))is dense in (%)I. (ii) + (iii) By assumption, (a)?= (a)?t l (copU(U))=. F'rom Exercise 10.5.98(v), (U)? n (COZ U(U))= = (a)?n coz+ U(U).Thus (a)?c co2+U(%>. (iii) -+ (i) It will suffice to show that the norm closure of U i n v contains (a)?.Let a positive E ( 5 1/3) be given. By assumption, each S i n ( % ) ? has theformalU~+u2Uz+a3U3,where Ul,U2,U3 E U(U), 0 5 a1 0, let E' be min{e,al/2} and let SO be (a1 - E')U~ (a2 t a3 -+- E')UZ.Then [IS- Sol1 5 2(e' $- a3) < 4e and Uinv

-+-

+

+

+

Thus S is in the norm closure of

Uinv.

m[68]

Let T be an element that is not a convex combi10.5.100. nation of fewer than n unitary elements of U, where n 2 3. Suppose that we have T = alU1 ... anUn, where u1, ...,un are non-negative real numbers with sum 1, and 171,. . ,U, are unitary elements in U. Show that (i) ai 5 U j ak if j # k; (ii) ( n - I)-' 5 aj ak if j # k; (iii) aj 5 2(n 1)-l for all j .

+ +

+ +

.

-+

By renumbering, we may assume that the ordering a1 5 uz 5 * . . 5 an holds. (i) If an > a1 a2, then Solution.

-+-

679

EXERCISE 10.5.100

Hence, ( ~ i + ~ 2 + ~ , ) - ' ( a i Ut ia z U z t ~ , U , ) E Uinvn(M)i, WhereUinv is the group of invertible elements of U; and from Exercise 10.5.92(i),

for V1 and V2 in U(U). This provides a convex decomposition of T in terms of n - 1 elements of U ( U ) - contradicting our assumptions. Thus ai

provided j # k. (ii) When j

F

a,

I a1 t a2 I a j t a k

# k, (ii) follows from (i) and

(iii) We show that a, 5 2 ( n t 1)-l, whence aj 5 a, 5 2(n+l)-' for all j. Suppose that 2(n t 1)-l < a,. Then

1 = a1 t . * . t an-1 t a, > ( n - l)a1 -t 2(n i- 1)-1, and

a1

and ( n

< (n + l)-'. Hence, if k

+ l)-'

+ 1,

< ak. But then, 1 = (a1 t a2) t (a3 t . * .t U

M )

t a,

t ( n - 3 ) ( n+ 1)-l -I- a, > 4(n t 1)-l t ( n - 3 ) ( n + 1)-1 = 1 2

-a

contradiction.

a,

CHAPTER 11 TENSOR PRODUCTS

1 1.6.

Exercises

11.5.1. Suppose that U is an abelian C*-algebra, and B is a C*-algebra with center C. Identify 2l 8 C, in the usual way, as a C*-subalgebra of U 8 B (see the final paragraph of Subsection 11.3, The spatial tensor product). Show that B 8 C is the center of 2l 8 B. [Hint. Identify 2l with C(S) for some compact Hausdorff space S , and B 8 B with C(S,B).]

Solution. There is a * isomorphism 11, from U onto C(S), for some compact Hausdorff space S. With L the identity mapping on B, 9 8 L is a * isomorphism from U 8 B onto C(S)8 B, and carries 2l8 C onto C(S) 8 C. We assume, henceforth, that 2l is C(S). There is a (unique) * isomorphism cp from C(S)@Bonto C(S,B) with the following property: when f E C(S) and B E B, cp(f 8 B ) is the continuous mapping f( ) B : s + f ( s ) Bfrom S into B (see the discussion following the proof of Proposition 11.3.2). In particular, therefore, cp(f @I C) is f( )C when f E C(S) and C E C. As C(S)@ C is generated (its a C*-subalgebra of C(S) @I B ) by simple tensors f @ C, and C(S,C) is generated (as a C*-subalgebra of C(S, a)) by functions f ( )C,it now follows that cp carries C(S)@Conto C(S,C). Accordingly, it suffices to show that C(S,a) has center C(S,C). Suppose first that F E C(S,C). Given any G in C(S,B),

( F G ) ( s )= F ( s ) G ( s )= G ( s ) F ( s )= ( G F ) ( s ) for each s in S, since G(s) E B and F ( s ) lies in the center C of B. This shows that FG = G F , for all G in C(S,B), so F lies in the center of C(S,B). Conversely, suppose that F lies in the center of C(S,B). Given

681

EXERCISE 11.5.2

any B in 8 , F commutes with the element of C ( S , B ) that has constant value B throughout S; so

F(5)B = B F ( s ) ( B E 8,s E S). Thus F ( s ) lies in the center C off?, for each s in S, and F E C(S,C). From the two preceding paragraphs, C(S, B) has center C(S,C); so !2f @ B has center U @ C. 11.5.2. Suppose that R is a von Neumann algebra with center 2 and B is a C*-algebra with center C. (i) Show that 2 8 B contains the center of R 8 B. [Hint. Suppose that S lies in the center of R B B . Given any positive E , approxRj 8 Bj of R 0 B , imate S in norm to within E by an element Cj”=, and deduce from Proposition 8.3.4 that S lies at distance not more than E from 2 0B.] (ii) By using (i) and Exercise 11.5.1, show that R @ Bhas center 2 @C.

Solution. (i) We use the notation introduced in the hint. Suppose that U1,. . . ,Un are unitary operators in R, al, . , a , are positive scalars with sum 1, and

..

n

Tj =

C akUkRjU;

.. ,m).

( j = 1,.

k=l

Since S commutes with uk 8 I, m

k= 1 n

j=1 m

682

TENSOR PRODUCTS

In terms of the notation introduced at the beginning of Section 8.3, the result of the preceding chain of inequalities can be expressed as follows: m

(1)

11s- C a ( R j ) 8 ~ j5 El ~ (aE D). j=1

From Proposition 8.3.4, there exist 21,. . . ,2, in 2 and a1 ,a2 ,a3, ... in D such that

Upon replacing a by a, in ( l ) , and taking limits as obtain m (IS-

c

z jc3 Bjll

T + 00,

we

I E.

j=1

We have shown that S lies at distance not more than E from B (for every positive E ) , and so lies in the norm closure 2 @ B

2 of20.13. (ii) Suppose that S lies in the center of R 8 B. From part (i), S E 2@B;so S lies in the center of 2@B,and this is 2 @ C by Exercise 11.5.1. It follows that 2 @ C contains the center of [email protected] establish the reverse inclusion, it suffices to note that every element of 2 0C commutes with every element of R O B , whence (by continuity) every rn element of 2 8 C commutes with every element of R @ B. 11.5.3. Suppose that, for j = 1,2, Cj is a C*-subalgebra of an abelian C*-algebra 2j. Let C be the C*-subalgebra (Cl @ 2 2 ) n (21 60'2) of 21 8 2 2 . (i) Show that each pure state of C has the form p1 @p2 I C, where p1 and p2 are pure states of 2 1 and 2 2 , respectively. (ii) Suppose that p1 and 71 are pure states of 2 1 , p2 and 72 are pure states of 2 2 , and p 1 @ p2 I C 1 @ C2 = 71 @ 72 I C1@C 2 . Prove that p1€3p2 I C = 71 8 7 2 IC. (iii) From (i), (:ii), and the Stone-Weierstrass theorem, deduce that C = C1 @ C2.

Solution. At the outset, we note the inclusions C 1 @ c2

ccc 21 @2 2

683

EXERCISE 11.5.3

among three abelian C*-algebras. 1 €3 2 2 (i) Each pure state p of C extends to a pure state p of 3 (Theorem 4.3.13(iv)), and pis multiplicative on 2 1 8 2 2 (Proposition 4.4.1). The equations

define multiplicative linear functionals (that is, pure states) p1 and p2 on 21 and 2 2 , respectively. Since

p is the product state p1 €3 p2 of 2 1 €3 2 2 . (See Proposition 11.1.1.) Thus p = PIC = Pl €3P2 (C. (ii) When A E C1, we have pi(A) = (PI 8 p2)(A @ 1 ) = (71 8 r2)(A @ 1)= 7i(A). Accordingly, if Al,.

..,A k E C1 and B1,. .. ,B k E 2 2 , we have k

k

k j=l

This shows that the product states coincide on C1 0 2 2 ; by continuity, P1

€3 P2 ICl 8 2

A similar argument shows that

p1

8 p2 and

2 = 71 @ P2

71

8 p2 of 21 8 2 2

ICl €3 2 2 .

684

TENSOR PRODUCTS

It follows that p1 €4 p2, TI €4 p2, TI @ 7 2 all have the same restriction to (Cl €4 Z2)n (zl@ C2) (= C). (iii) We can identify C with C(S), for some compact Hausdorff space S, and C1 8 C2 then corresponds to a norm-closed self-adjoint subalgebra A of C(S)that contains the constant functions. Two distinct points s1 and s2 of S give rise to two distinct pure states of C(S)( 3 C); by (i) and (ii), the corresponding pure states of C do not coincide on C 1 @ C2 (Z A). Hence some function in A takes different values at s1 and 82. By the Stone-Weierstrass theorem, A = C(S) and C1 €4 C2 = c. 8 11.5.4. Suppose that U 1 and are C*-algebras with centers and C2, respectively. Show that U1@%2 has center C1 8C2. [Hint. We may assume that ?2l1 and U2 act on Hilbert spaces H 1 and 1-12, respectively. Let 2j denote the center of a.; Show that the center of %1@U2 is contained in the intersection of the centers of U 1 @ U , and U, 8 U2, and use the results of Exercise 11.5.2(ii) and 11.5.3(iii).]

C1

Solution. It is apparent that each element of C1 @C2 commutes with each element of 3 1 0U2. By continuity, each element of C1@C2 (C U 1 €4 U2) commutes with each element of 2l1 €4 2l2. Hence the center C of U 1 €4 UZcontains C1 €4 C2. Suppose that C E C. Then C E U 1 8 U2 E 3 1 €4 U,. Moreover, C commutes with each element of U 1 @ UZ,and hence with each element of U 1 8 U, (E (U1 @ U2)-). Thus C lies in the center of 2l1 8 U,, and this is C1 €4 2 2 by Exercise 11.5.2(ii). A similar argument shows that C E 2 1 €4 C2. From the preceding paragraphs,

It now follows from Exercise 11.5.3(iii) that C = C1 €4 C2.

m[8,47]

11.5.5. Suppose that '2 and B are simple C*-algebras. Let be an irreducible representation of U €4 0 on a Hillbert space 7-1. (i) Show that the mappings

O(H), : B + T ( I @B ) : B + B ( H ) , : A + T ( A @ I ) : U -+

T2

7r

EXERCISE 11.5.5

685

are faithful representations of U and B, respectively. Prove also that q(U)- and ~ ( 0 )are - factors, and 7r2(B)- C T I ( % ) ' . Suppose that A 1 , . . . , A , E U, B1,. . . , B , E f?, and (ii)

Show that C;==, A j Q9 Bj = 0 . [Hint. Apply Theorem 5.5.4 (with the factor TI(%)- in place of R) and Proposition 11.1.8.1 (iii) Show that the mapping K 1% 0 B is an isometry, and deduce that ?r is a faithful representation of U 8 B. [Hint. Use Theorem 11.3.9.1 (iv) Deduce that U 8 f? is a simple C*-algebra. Solution. (i) Since ~1 is a representation of U and nl(1) = I, the kernel rF1(0)of ~1 is a two-sided ideal in U, and is not the whole of !24. Since U is simple, ~ c ' ( 0 )is (0) and RI is faithful. A similar argument shows that ~2 is faithful. Since A @I I commutes with I @ B whenever A E U and B E f?, ? r l ( A ) ( = R ( A@I I ) ) commutes . T I ( % ) - commutes with q ( S ) - , and with r z ( B ) ( = R ( I @ B ) ) Thus

If P is a projection in the center of T I ( % ) - , it follows from 8 (= rq(B)) the preceding paragraph that P commutes with ~ ( 1 B) for each B in B. Of course, P commutes also with R ( A @ I) (= q ( A ) E T I ( % ) - ) for each A in U. Hence P commutes with r ( A @ B ) (= R I ( A ) T ~ ( Bfor ) ) all A in U and B in B, whence

P E 7r(U 63 B)'. Since R is irreducible, P is either 0 or I. This proves that T I ( % ) - is a factor, and a similar argument shows that K,(B)- is a factor. (ii) Under the stated conditions, and with 'R the factor r1(U)-, we have 0=

n

n

j= 1

j=1

Cr(Aj @ B j ) = C

~1(Aj)~2(Bj),

r i ( A i ) ,. . . , x i ( & ) E R, ~ 2 ( B 1 .) ., . , r 2 ( B n )E R'.

686

TENSOR PRODUCTS

From Theorem 5.5.4, there is an n x n matrix [ c j k ] of scalars (scalars, because R is a factor) such that n

Since T I and

7r2

(k = 1,...,n),

are faithful representations, it follows that n

( j = 1,...,n ) , k=l

and this implies that C;=,A j 8 B j = 0 (see Propos,..m (iii) Since T is a representation of 2l @I B,

B From (ii), the mapping T -+ 11r(T)ll : % @ B. It follows from Theorem 11.3.9 that

+W

1. ,8).

is a C*-norm on

bearing in mind that %@If3is the spatial tensor product corresponding to the spatial C*-norm CY on % 0 B. From (1) and (2), the mapping T I 2l@ B is an isometry. By continuity, A is isometric on 2l @I L?, and is therefore a faithful representation. (iv) Suppose that % @ B is not simple, and let Z be a closed ideal of U 8 B that is neither (0) nor U 8 D. By composing an irreducible representation of the C*-algebra ( 2 l @ B ) / Z with the quotient mapping from % Q9 D onto (U 8 f?)/Z, we obtain an irreducible representation of % @ D that has non-zero kernel, and this contradicts 1[111] the conclusion of (iii). Thus % 8 B is simple.

EXERCISE 11.5.6

687

11.5.6. Suppose that rzL and 0 are C*-algebras and a is a C*norm on U @B. Show that the identity mapping on 31 0 B extends to a * homomorphism from M Bo B onto the spatial tensor product I#@B. Deduce that U @a B is simple if only if both and 8 are simple and a is the spatial C*-norm on M a B. Solution. If 0 denotes the spatial C*-norm on U @ B, we have a ( T ) 2 a ( T ) for all T in U @B, by Theorem 11.3.9. The identity mapping on M 0B can therefore be viewed as a norm continuous * isomorphism from a dense subset U0S of M@,,Bonto a dense subset B). This mapping extends by continuity to 0 @ B of M @ B (= a * homomorphism cp from U B into U @ B. Moreover, the range

of 9 is the whole of 3L @ B because it is closed (Theorem 4.1.9) and contains U @ B. If ? and ? Blare simple, and a is the spatial C*-norm on M @ B, then M @a B (= U @B ) is simple, by Exercise 11.5.5. Conversely, suppose that 0 @a 5 is simple. In this case, the * homomorphism c p : rzL@.,B+M@B, constructed above, has kernel (0); so cp is a * isomorphism, and is therefore norm preserving. Since cp I M 0B (the identity mapping on U @ B) is isometric, a ( T ) = a(T)for each T in U @B. Thus a is the spatial C*-norm on M 0B, when U @a 8 is simple. It remains to show that M and B are simple when M @ B is simple. If0 is not simple, let Z be a closed ideal in M such that (0) # Z # U. With 6' : M + M/Z the quotient mapping, and L the identity mapping on f3, the * homomorphism

has a kernel different from (0) and U@B. Thus % @ Bis not simple when M is not simple; so simplicity of U @ B entails simplicity of U (and, similarly, of B). Show that the tensor product B ( H ) @ B(K) of 11.5.7. the (represented) C*-algebras B(7t) and B ( K ) is not the whole of S(3-I@ K ) when both the Hilbert spaces H and K are infinite dimensional. [Hint. Let (yb : b E B} be an orthonormal basis of K , and define a unitary transformation U from CbEl $'H onto H @ K: by z b 8 yb. Represent an element TOof $zb) = the equation

V(c

688

TENSOR PRODUCTS

B(3-I @ K ) by the matrix [Tab]of the bounded operator U‘lT~U acting on EbeB ex,as in the discussion preceding Proposition 2.6.13. Show that, when TOE B(7-l) @ B(K),there is a (norm-) compact subset of B(3.1) that contains all the entries Tab in the matrix representing To.] Solution. We recall that, if A E B(3-I) and B E B ( K ) , then A @ I has matrix [&bA] and I @ B has matrix [ s a d ] ,where [sab] is the (complex) matrix of B relative to the orthonormal basis { Y b } of K . Thus A @ B has matrix [sabA]. It follows that an element C;==,A j @ Bj of B(1-1)@ B(K)is represented by a matrix in which each entry is a linear combination of A l , .. . ,A,. Suppose that [Tab]is the matrix of an element TOof B ( X ) @ B ( K ) , and let S be the norm closure, in B(R),of the set of all the entries Tat,. In order to show that S is compact, it suffices to prove that

S is ‘‘totally bounded” in the following sense: given any positive E , S is contained in the union of a finite collection of balls in B(3-I) with radius E . To this end, choose an element SO of B(7-l) 0 B(K)

- Sol1 < $ E ,

and let [&b] be the matrix of SO. Now IITab - SabII < !j& for d l a , b in B,and d l the entries sab lie in the ball (M)lpll in a finite-dimensional subspace M of B(3-I). Since (M)lpll is compact, it is covered by a finite collection of balls in B(l-l) with radius $ E . Upon doubling the radius of each ball in that collection, we obtain a finite set of balls of radius E whose union contains each Tab, and so contains s.Hence S is compact. Let N be the norm-closed subspace of B(7-l) generated by a linearly independent sequence {A1 ,A2 ,. . .} of operators. Since N is an infinite-dimensional separable Banach space, the unit ball (N)1 is not compact but has an everywhere-dense sequence { R1, Rz ,.. . }. The index set B is infinite, so there is a mapping f from b onto the set of all positive integers, and the diagonal matrix [S,bRf(b)] represents an element Ro of B ( X @ K). The closure (N)l of the set of all entries in this matrix is not compact, so (by the result of the previous paragraph) Ro 4 B(31) @ B ( K ) . such that llT0

11.5.8. Suppose that U and B are C*-algebras and (Y is a C*norm on U O B . When A E U and B E B , write A @ B and A @ * B to denote the corresponding simple tensors in the C*-algebras U8 B and U 0,respectively. Let 7r be a representation of U @a B on a

689

EXERCISE 11.5.8

Hilbert space 3-1, and define representations

a1

of U and

n2

of B by

Suppose that TI(%)- is a factor of type I. Show that (i) there exist Hilbert spaces 3-11 and 3-12, and a unitary transformation U from 3-1 onto 311 @ 312, such that

(ii) there exist representations 'Hz) such that

whenever A l , . . . , A , E U and spatial C*-norm on 2l0 B .

B1,.

cp1

(of U on 3-11) and

QZ

(of B on

. . ,B , E B , where u denotes the

Solution. (i) Since .I(%)- is a type I factor, it follows (see Example 11.2.5) that there exist Hilbert spaces 311 and 3-12 and a unitary transformation U from 3-1 onto 3-11 @ 3-12 such that

Note also that the von Neumann algebra tensor product, on the right-hand side of the last equation, coincides with the corresponding B commutes C*-algebra tensor product (Example 11.2.1). Since with A @a I , whenever A E Iu and B E 8,it follows that n2(B) (= n(1 ma B ) ) commutes with nl(A) (= n(A @ a I)).Thus q ( B ) TI(%)', and Un2(B)U* Ulr$l)'U*

c

c

= (Un(%)-U*)' = (B(3-11) @ @ % ) I

= @'HI @ B(312).

690

B

TENSOR PRODUCTS

(ii) From (i), there exist mappings 91 : 2l B ( H 2 ) such that

3

B(X1) and cp2 :

--f

From the fact that; ~1 and n2 are representations, it is apparent that cpl and cp2 are representations; moreover, from the equality, A @a B = (A I)(I B), we have

Un(A B n B)U* = (cpi(A) @ I ) ( I @ cp2(B))= cpi(A) @ cp2(B). (iii) We can form the tensor product representation cp1 @ cp2 of the spatial tensor product U @ B ,and cp1 892 does not increase norms (Theorem 4.1.8(i)). Thus

n

j=1 n j=1

11.5.9. Let 2l be a C*-algebra with the following property: for each representation 'p of U, the von Neumann algebra cp(U)- is of type I, (In these circumstances we describe 2l as a type I C*-algebra) Prove that 2l is nuclear. [Hint. Suppose that B is a C*-algebra and a, is a C*-norm on 2l@ B. Given any irreducible representation t of 2l @a B, define representations ~1 of 2l and 1 ~ 2of B as in Exercise 11.5.8. Prove that q(2l)- is a factor (necessarily of type I). By using the result of Exercise 11.5.8(iii), show that a, coincides with the spatial C*-norm u on U @B.]

691

EXERCISE 11.5.10

Solution. With the notation introduced in the hint, the reasoning used in the solution to Exercise 11.5.5(i) shows that r ~ ( % ) is a factor. Since 24 is a type I C*-algebra, TI(%)- is a type I factor. From Exercise 11.5.8(iii) n

n

j=1

j=1

whenever A l , . . . , A n E U and B1,. . . ,B, E B. Since this inequality holds for every irreducible representation K of U @a B, and since the direct sum representation of all the irreducible representations of % @a B is faithful (Corollary 10.2.4) and is therefore isometric, it now follows that n

n

The last inequality amounts to the assertion that a ( T )I a ( T )whenever T E % @ 13, and this, together with Theorem 11.3.9, shows that Q = a. Thus % is nuclear. ~[21,42,74,111] 11.5.10. Suppose that {eI,ez} and {fl,f2} are orthonormal bases of two-dimensional Hilbert spaces 'H and K, respectively. Let P be the projection from 1-1 @ K onto the one-dimensional subspace that contains e l @ f1 t ez @ f ~ so , that

By considering

show that P does not lie in the norm closure (equivalently, the weakoperator closure) of the set of all operators of the form Aj@Bj, where A l , . . . , A h E B('H)+ and B1,. . . ,Bk E B ( K ) + .

xt=l

Solution.

The range of P is generated by the unit vector

2 - 1 / 2 ( e 1 8 f ~ + e z @ f 2 )(= zo), and el@fz,ez@fI are bothorthogonal t o zo; so P(e1 @ fz) = P(ez @ f 1 ) = 0.

692

TENSOR PRODUCTS

Moreover,

k

Similarly,

Also,

k

EXERCISE 11.5.11

693

From the Cauchy-Schwarz inequality, together with the preceding estimates, we now obtain

llP - Sll k

and ((P - S((2 f . From the preceding paragraph, P cannot be approximated within norm distance less than by operators S of the stated form, and therefore does not lie in the norm closure of the set of all such operators s. The norm closure of this set coincides with the weak-operator closure, since the containing space B(H 8 K) is finite-dimensional.

a

11.5.11. Give an example of C*-algebras M and €3 such that the state space of Q 8 B is not the norm-closed convex hull of the set of all product states of U 8 B . [Hint. With the notation of Exercise 11.5.10, let !2l = B(3.1), B = B ( K ) , and let p be the state w, of B(3-18K ) (= B ( H ) 8 B ( K ) ) , where 2 is the unit vector 2-4(el 8 f1 t e 2 8 f2). Show that each convex combination po of product states of B(1-I) 8 B ( K ) can be expressed in the form

where

21,.

.., X I

E 'H and PI,... ,y~ E K . Let

and identify elements of B(1-I) and B ( K ) with their matrices relative to the orthonormal bases { e l , e 2 } and {fi, f2}, respectively. By considering the values taken by p and po at

694

TENSOR PRODUCTS

Solution.

Suppose that PO is a convex combination, m

n=l

of product states of B(3.1)8 f3(K:). Since 3.1 and K: are finite dimensional, each up) is normal, and can be expressed as a finite sum of vector functionals (see, for example, Theorem 7.1.12(a)). Thus PO can be expressed in the form stated in the hint. We now adopt all the notation used in the hint. When

we have

that is,

When A1=[;

;I,

we have p(A1 €3 B1) = 0. Thus

B1=[;

;I,

EXERCISE 11.5.11

695

From this, and a similar calculation involving A2 €3 82,where

we have

we have

p(A3

@8

3 )

=

f and

From the Cauchy-Schwarz inequality and (2), it now follows that

Thus IIP - POI] 2

- Po(& €3 8 3 ) ) - Po(& €3 & ) I

b(A3

€3

B3)

If 2 3€3 8311 L f - IIP - Poll,

=

IPo(A3

and JIP- Poll 2 f . It follows that the state p of B(7-l)€3 B ( K ) lies at norm distance at least f from the convex hull of the set of product states. rn

696

TENSOR PRODUCTS

11.5.12. Give an example of von Neumann algebras R and S and a normal state w of RG‘S that is not in the weak * closed convex hull of the set of all normal product states of R g S . [Hint. Proceed as in Exercise 11.5.11.1

Solution. We use the notation introduced in the hint to Exercise 11.5.11, and take

R = B(W), Since 3-1 and

s = B(K>,

w = wx.

X: are finite dimensional,

the set of product states coincides with the set of normal product states, and the norm-closed convex hull of this set coincides with the weak * closed convex hull. It follows from the result obtained in the solution to Exercise 11.5.11 that w does not lie in the weak * closed convex hull of the set of all normal product states. 11.5.13. In Example 11.1.7, we identify (in effect) the C*algebra C(S,U) of norm-continuous mappings (provided with pointwise operations and supremum norm) of a compact Hausdorff space S into a C*-algebra U with U 8 C(S). Show that such a mapping represents a positive element of U 8 C(S) if and only if its value at each point of S is a positive element of U,

Solution. Suppose F ( s ) 2 0 for each s in S, where F is in C(S,U). Then F = G2, where G(s) = F ( s ) i , and G E C(S,U)h. Hence F 2 0 (in U 8 C(S)). Suppose F 2 0. Then F = G2 for some G (= G*) in C(S,U), w so that F ( s ) = G ( s ) ~2 0 for each s in S. 11.5.14. Suppose U1,%2,Bl,B2 are C*-algebras and 771,772 are bounded linear mappings of 8 1 into B1 and U2 into B2, respectively. satisfying Must the (unique) linear mapping 770 of U1@82 into qo(A1 8 A2) = ql(A1) 8 772(A2) for all A1 in U1 and A2 in U2 be bounded when U1 0 U2 and B1 0 B2 are provided with C*-norms? Proof? Counterexample? [Hint. Consider Example 11.3.14.1

EXERCISE

11.5.15

697

Solution. No, qo need not be bounded. Let .F be the free 1 and Bl be the C*group on two generators a1 and u2. Let 9 algebra LO generated by {I,, : g E T } ;let 242 and B2 be the C*algebra Ro generated by { R , : g E F}.(See the notation of Example 11.3.14 and Section 6.7.) We write 2l1 02 l 2 for the algebraic tensor product of LO and Ro provided with the C*-norm 0 and B 1 0 B2 for this algebraic tensor product with the C*-norm a it acquires from the algebra generated by LO and Ro acting on L2(F). Choose the identity mappings (on LO and Ro)for each of q1 and N. Suppose 70 is bounded. Then the (unique) bounded extension q of 170 mapping 241 @ 8 2 into B1 @a B2 is a * homomorphism since the restriction 770 of q to the norm-dense * subalgebra 241 0 U2 of 9 1 @ 242 is a * isomorphism. From Theorem 4.1.8(i), llq(T)lla 5 llTll. for each T in 3 1 @ 8 2 . But from Theorem 11.3.9, llTlld _< llTlla (= ))q(T)lla) when T E 241 0 !2l2 (= LO 0 Ro).(We are writing llTlld and ~ ~ T rather than a ( T ) and a ( T ) to avoid confusion with 7.) Hence a and a coincide on Lo 0 Ro,contrary to the conclusions of Example rn 11.3.14. It folIows that qo is not bounded. 11.5.15. Let q be the mapping of M n ( C ) into itself that assigns to each matrix [ u j k ] its transpose matrix (whose (j,k)entry is a k j ) . Show that (i) q is a * anti-automorphism of M,(@); (ii) q is a positive linear mapping of M,(@) into itself (see the discussion preceding Lemma 8.2.2); (iii) when n 2 2, the (unique) linear mapping 7) @ L of M,(@)@MZ(@)into itself that assigns q ( A ) @ Bto A @ Bis not a positive linear mapping. [Hint. Express an element T of M , ( @ ) @ M z ( @ ) as a 2 x 2 matrix with entries from M,(@) and note that ( q @ L ) ( T ) has representing matrix obtained from that of T by transposing each block of the 2 x 2 matrix. Choose for T a positive matrix that has 0 at all entries not in the upper, principal, ( n 1) x ( n 1) block and a non-zero scalar at the (n,n t 1) entry. Recall that if a positive matrix has 0 at some diagonal entry, then 0 is at each entry of the corresponding row and column - see Exercise 4.6.11.1

+

+

Solution. (i) Since matrices are multiplied by a scalar and added on an entry-by-entry basis, q is a linear mapping of Mn(@) onto itself. The (j,k) entry of q([ujk][bjk]) is the ( k , j ) entry of [ a j k ] [ b j k ]namely , C,"=,u k r b r j , which is the (j,k)entry of the matrix

~ ~

698

TENSOR PRODUCTS

and q is an anti-automorphism of &In(@). Finally, the ( j , k ) entry of q([ajk]*) is the (k,j)entry of [ a j k ] ' , which is Zjk; while the ( j , k ) entry of q([ajk])*is the complex conjugate of the ( k , j ) entry of ~([ajk]), namely Zjk. Thus q is a * anti-automorphism of Atn(@). (ii) Each * anti-homomorphism cp of a C*-algebra CU is a positive linear mapping; for with A a positive element of M,

cp(A) = v ( ( A : ) ~ = ) ~ p ( A i2) 0~. In particular, q is a positive linear mapping, (iii) With Example 11.1.5 in mind, each element of the algebra Mn(@)8 M2(@)has a representation as a 2 x 2 matrix with entries from Mn(@) - the matrix representing A 8 B is

where B =

[iii

:I,

and that of (q 8 L ) ( A8 B ) (= q(A) 8 B ) is

That is, the effect of q @ on ~ the 2x2 matrix over Mn(@)representing A @ B is to transpose each n x n entry. Since this same process applied to dl elements of M n ( @ ) @ M 2 ( @is ) a linear mapping, it is the linear mapping q @ L. Choose for T the matrix with 0 at each entry except for the entries (1, l),.. . ,(n t 1,n t l), (n,nt l), and ( n t 1,n) at which entries the value 1 appears (2' is a 2n x 2n matrix). Then T 2 0, but ( 7 8 L ) ( T has ) 1 at the (1,2n) entry and 0 at the (2n,2n)entry; so that ( q 8 L ) ( Tis ) not positive, and q 8 L is not a positive linear mapping.

EXERCISE 11.5.16

699

11.5.16. A positive linear mapping q of a C*-algebra 2l is said to be completely positive when, for each positive integer n, q 8 2 , , the (unique) linear mapping whose value at A 8 B is q ( A ) 8 B for each A in 2l and each B in M , ( @ ) , is positive. Show that (i) q is completely positive when q is a * homomorphism; (ii) q is completely positive when q ( A ) = TAT* for each A in U, where rU acts on the Hilbert space 3-1 and T is a given bounded linear transformation of 3-1 into another Hilbert space K ; (iii) q is completely positive when q is a composition of completely positive mappings; (iv) q is completely positive when q ( A ) = T v ( A ) T * , where cp is a * homomorphism of 2l into B(1-I) and T is a bounded linear transformation of the Hilbert space 3-1 into the Hilbert space Ic; (v) not each positive linear mapping of a C*-algebra is completely positive. [Hint. See Exercise 11.5.15.1 Solution. (i) We show that q 8 L , is a * homomorphism of 2l8 M n ( @ )for each positive integer n when q is a * homomorphism. To see this, it suffices to show that

for all R and S in some set of linear generators for 2l@ for all A l , A2 in 2l and B1, B2 in Mn(@),

&In(@). Now,

Hence q @ L, is a * homomorphism and is, therefore, a positive linear mapping. Thus q is completely positive when q is a * homomorphism. (ii) Each element A of 2l@ M,(@) has a representation as an n x n matrix [Ajk] with entries Ajk from 2l, and ( q 8 L , ) ( A ) has [q(Ajk)]as its representing matrix (by an argument similar to that employed in the solution to (iii) of Exercise 11.5.15). Thus, if [Ajk] is

700

TENSOR PRODUCTS

positive and 7 arises from the linear transformation T as described in the statement of this exercise, (77 €3 L,)(A) has [TAjkT*] as its representing matrix. But

.

where Tjj = T for j in (1,. . ,n } and T j k = 0 when j # k. Thus ( q 8 L ~ ) ( A2 )0 w'hen A 2 0; and q is a completely positive mapping in this case. (iii) If 7 = r,q 0772, then, employing the n x n matrix representation of U @M,(@)as in (ii), we see that for each positive integer n,

77 €3 b, = (771 8 40 (172 €3 4. Since the composition of positive linear mappings is a positive linear mapping, 17 8 L, is positive when each of 771 and 772 is completely positive. (iv) From (i) and (ii), 77 is the composition of completely positive mappings. Hence, from (iii), 77 is completely positive. (v) The mapping 7 described in Exercisell.5.15 provides an example of a positive linear mapping 77 of a C*-algebra such that 7 is not completely positive. 11.5.17. Let 7 be a completely positive mapping (see Exercise 11.5.16) of a C*-algebra U into B(3-1) for some Hilbert space 1-1 and let {ea}aEAbe an orthonormal basis for 'H. Denote by ?% the linear space of functions from A to U that take the value 0 at all but a finite number of elements of A, where is provided with pointwise addition and scalar multiplication (so that ?% is the restricted direct sum of U with itself over the index set A). Show that (i)

a

a,

defines an inner product (see p.75) on where A = { A a } a Eand ~ A' = { A k l } a ~[Hint. E ~ To show that (,&A) 2 0, use the fact that the n x n matrix whose ( j ,k ) entry is AJAk is a positive element of CLI 8 Mn(C),where A l , . . .A , are the non-zero coordinates of (ii) 0 = ( A , = (B, for each fi in ?%, when (A, = 0, 2 is a linear space, where

n)

A)

L = {A E a : (A,A)= O},

A)

A.];

701

EXERCISE 11.5.17

and

( A t 2,h + .c)o

= (A,@

defines a definite inner product on KO,the quotient space m / E ; (iii) 0 5 (B,B) 5 llA112(&A), where A = { A a } , E ~and B = { A A a } a E ~and , conclude that cpo(A)is a bounded linear mapping of KO into KO,where

(Hint.Let T be the n x n matrix whose non-zero entries are in the first row, and this first row consists of the non-zero coordinates of A. Let R and S be the n x n matrices whose only non-zero entries are their ( 1 , l ) entries, and these are A*A and llA1121,respectively. Use the fact that T*RT 5 TuST in conjunction with the complete positivity of q.]; (iv) cp is a representation of 2.l on K , where cp(A) is the (unique) bounded extension of * ( A ) from KO to K , and K is the completion of KO relative to (, )o; (v) {fa L}=,=Ais an orthonormal set in K , when q(1) = I, where ja is the element of with I at the a coordinate and 0 at all others; (vi) Vucp(A)V = q ( A ) ( A E a), when ?(I) = I, where V is the (unique) isometry of H into K such that V e , = la 2 for each a in A.

+

~

+-

a,

Sohtion. (i) With A and A’ in there are at most a finite number of indices in A, say, 1,., ,n, at which 2 or A’ have non-zero coordinates. Thus

.

and there is no question of the convergence of the sum defining (A,A’). With B (= {B,),EA) another element of % and b a scalar,

702

TENSOR PRODUCTS

Since is a positive linear mapping, it is hermitian. Hence ?(A*) = ?(A)* for each A in !2l, and

With T as described in the hint to (iii), T*Tis the n x n matrix whose (j,k)entry is Aj*Ak. Since 77 is assumed to be completely positive, the matrix S whose (j,k) entry is r](Aj*Ak)is positive. With e thevector { e l , ...,e,}in'H$...$'H,

a.

Hence (A,A') + (A,d') is an inner product on (ii) By applying the Cauchy-Schwarz inequality to the inner product defined in (i), we see that

0 = (A,@ = @,A), when (A,A) = 0 and time, the equation

B E !%

Hence 2 is a linear space. At the same

(AtE , B t i),= (A,B) defines (unambiguously) an inner product on KO.If (atE , B t2)o = 0, then = 0 and E 2, whence B t 2 = 0 t 2. Hence (, )o is a definite inner product on KO. (iii) We proceed as in the hint. Since A*A 5 llA1121, R 5 S and T" RT 5 T'ST. With A l , . .,A , the non-zero coordinates of A, the ( j , k ) entries of T*RT and T*ST are, respectively, A:A*AAk and ((All2 Aj'Ak. Since 7 is completely positive, the matrix whose ( j , k) entry is q(AjA*AAk)is less than or equal to the matrix whose ( j , h )

(B,B)

a

.

703

EXERCISE 11.5.17

entry is l l A l 1 2 ~ ( A ~ A kApplying ). the vector state corresponding to the vector { e l , . . . , e n } to these matrices, we have

+

A

It follows that cpo(A)(A 2) = 0 if E 2, so that cpo(A) is a welldefined linear mapping of KO into itself. From the same inequality, we have that IIcpo(A)II IIlAll; hence cpo(A) has a (unique) bounded extension cp(A) mapping the completion K of KO into itself. (iv) To see that cp is linear, note that, with A and B in rU, b a scdar, and { A a } a ~ ~A() =in !%,

cp(A where

+ bB)(k t f?) = B t 2,

B = { ( A + b B ) A a } , E ~= { A A , } a E ~+ b { B A a } , E ~Hence .

+ 2) t bcp(B)(A+ L),

cp(A t bB)(At 2) = cp(A)(A

+

and the bounded operators cp(A bB) and cp(A) t bcp(B) agree on the dense subset K O of Ic. Hence cp(A bB) = cp(A) bcp(B). Note, too, that p ( A B ) ( A 2) = 6 t 2,

+

+

+

where

c = { A B A a } , € h . Thus

+ 2) = cp(A)cp(B)(A+ f?),

cp(AB)(A

and the bounded operators cp(AB)and cp(A)cp(B)agree on the dense subset KO of Ic. It follows that y ( A B ) = cp(A)cp(B). Finally, when A = { A a } aand ~ ~B = { B a } a ~ ~ ,

704

TENSOR PRODUCTS

so that cp(A)* = cp(A*).Hence cp is a representation of U on (v) Under the assumption that q ( I ) = I , we have

K.

(lat i , L t i ) o = = (q(I)ea,eal= ) (ea,eaf)) so that {fa t i } a E ~is an orthonormal set in K. (vi) Since q ( I ) = I , {fa t l } aisE an ~ orthonormal set in K from ( L , L I )

(v), and there is a unique isometry V mapping H into K such that V e , = j a t 2 for all a in A. For all a and a' in A and A in U, we

-

= (q(I*A)ea eal ) .[lo41 Thus V*cp(A)V = q ( A ) for all A in U.

Adopt the notation and assumptions of Exercise 11.5.18. 11.5.17 (exclusive of the assumption that q ( I ) = I ) . Let Ho be the dense linear manifold in 31 consisting of finite linear combinations of {ea}aEA, and let To(CaEA0 T a e a ) be CaEAo ra(fa t 2) for each finite subset & of A. Show that [Hint.Note that (i) To is a bounded linear transformation t t 2) = ( q ( ~ ) e ~ , e ~ l ) . ] ; (ii) T*cp(A)T = q ( A ) ( A E U), where T is the (unique) bounded extension of TOfrom Ho to 3-1; (iii) when q ( I ) = I , there is a Hilbert space 'H' containing H and a representation cp' of U on 3t' such that Ecp'(A)E = q ( A ) E for each A in U, where E is the projection of 'H' on H. [Hint. Let I C 1 be the range of V in K ,Kz be the orthogonal complement of K1 in K,Z' be H @ Kz,and U ( z ,y) be V x t y for x in 3.1 and y in Kz.Identify 7-f with {(x,O) : x E H} and q(A)(x,O)with ( q ( A ) s , O ) . Define $ ( A ) to be U*cp(A)U.]

(L

Solution. (i) As in the solution to Exercise 11.5.17(v), (lat &la) t i ) o = ( q ( I ) e a , e a l ) , whence

705

EXERCISE 11.5.19

Thus IlTOll IIlll(r)llt. (ii) For all a and a' in A and all A in 2.4,

+

(T*cp(A)Te,,e,#)= (cp(A)(i, i ) , f a g = (q(A)e,, ear).

+ Z)O

Hence T*'p(A)T = q ( A ) for all A in %. (iii) From Exercise 11.5.17(vi), V is a unitary transformation of 31 onto K1 and V*(p(A)V = q ( A ) for all A in M. Thus U ,as defined in the hint, is a unitary transformation of 'HI onto K, and 'p' is a representation of on 7 i ' . With A in Q, x in 'H, and y in K2,let cp(A)Vx be u + v , where u E K1 and v E Kz.Then, from Proposition 2.5.13, V'u = 0, and

Ecp'( A)E(x,y) = Ev'( A)(s, 0) = EU*cp(A)U(5,0 ) = EU*(p(A)Vx = EU*(u t v) = E(V*u,v) = (V*(u+v),O) = (V*cp(A)Vx,0) = ( q ( A ) z ,0) = v ( A ) E ( z ,Y). Hence E y ' ( A ) E = q ( A ) E .

.[lo41

11.5.19. Let q be a positive linear mapping of a C*-algebra 2l into B ( ? f )for some Hilbert space 31; let cp and cp' be representations of U on Hilbert spaces K and K', respectively; and let T and T' be bounded linear transformations of 'H into K and K', respectively, such that T*(p(A)T= q(A) and T'*p'(A)T' = q ( A ) for each A in 2l (as described in Exercises 11.5.17 and 11.5.18). Let KO and Kh be the closure of the ranges of T and T', respectively, and let E and E' be the projections of K and K' onto KO and Kb,respectively. Show that there is a untary transformation U of KO onto K,; such that, for each A in 0,

T' = UT,

E 4 A ) E = U*E"p'(A)E'U.

Solution. Define UoTx to be T'x for each x in H. Then T' = UoT. With A in 2.4,

( v ( A ) T sTY) , = (T*Y(A)TX,Y) = (rl(A)x,Y) = (T'*v'(A)T's, y) = (cp'( A)T's, T' y)

706

TENSOR PRODUCTS

for all x and y in 3.1. Letting A be I, we conclude that

(Tx,T y ) = (T'x,T'y) = (UOTX, UoTy). Thus Vo is well defined, linear, and extends (uniquely) to a unitary transformation U mapping KO onto Kh, and T' = U T . At the same time, (EP(A)ETa:,TY)= ( P ( A ) T s , T y ) = ((p'(A)T'x,T'y)

= (U*E'(p(A)E'UTz, T y ) for all z and y in 1-1. Since the range of T is dense in KO and the operators E(p(A)Eand U*E'y(A)E'U are bounded operators on K O , we have that E v ( A ) E = U*E'(p(A)E'U.

11.5.20. Let !2l be a C*-algebra acting on a Hilbert space 'H. Show that (i) the matrix [ ( ~ k r ~ j whose )] j,k entry is ( x k , x j ) is positive, where 21,. . , x , are vectors in H [Hint. Let { e l , . . . , e n } be an orthonormal basis for an n-dimensional subspace of H containing 2 1 , . .. , x , . Note that [ ( x k , x j ) ] is the matrix of T*T relative to { e l , . . . , e n } , where Tej = xj.]; (ii) the matrix [AJAk] whose j,k entry is A;Ak is in M,(U)+ for each set of n elements { A l , . . . ,A,} in B and conclude that the matrix all of whose entries are a given positive A in U is in M,(U)+ [Hint.Consider the matrix whose first row is A1, . . . ,A, and all of whose other entries are 0.1; (iii) each positive element of Mn(21) is a sum of matrices of the form [AJAk] [Hint. Express a positive element of M,(B) as T*T with T in M,(Q).]; (iv) [Ajk] 4 [ ( A j k z k , z j ) :] M,(B) + M,(@) is a positive linear mapping for each set of n vectors { X I , .. . , x , } in H and conclude that [ ( A j k x , ~ )2] 0 for each z in 'H when [Ajk] 3. 0 [Hint. From (iii), it suffices to show that [(AJAkxk,xj)]2 0 for each set of n elements A1,. . . ,A, in U. Use (i).]; (v) if [Ajk]is in M,(U)+ and [Bjk]is in M,(U')+, then [AjkBjk] 2 0. [Hint.Use (iii) and (ii).]

.

EXERCISE

11.5.20

707

Solution. (i) With { e l , . . . ,en} and T as in the hint, the j,k entry of the matrix for T*T relative to { e l , . . . ,en} is (T*Tek,e j ) = (Tek,Tej) = ( z k , z j ) . Thus [ ( z k , z j ) ]2 0. (ii) If A is the element of Mn(21) described in the hint, then A*A = [Aj’Ak]E Mn(!21)+. If A E 2l+ and each Aj is A t , then each Aj’Ak is A. Hence the n x n matrix all of whose entries are A is in Mn (a)+* (iii) Each element of Mn(21)+ has the form

where [Bjk]E M,(U). If Aj = B r j , then [B:jBrk] = [A;Ak]. (iv) The mapping described is clearly linear; hence, from (iii), it suffices to show that the value of this mapping at each matrix of the form [Aj*Ak]is an element of M , ( @ ) + . But, from (i),

If each of the vectors x j is the same vector z in H,then our mapping becomes [Ajk]-+ [ ( A j k z , ~ ) Hence ]. [(Ajkz,z)]2 0 when [Ajk]2 0(v) From (iii), [Ajk]is a sum ~ ~ = l [ A ~ )with * Aeach ~ ) A:) ] in U. Thus Ajk = A (jP I * A,( 4 and

xy=l

Hence, it suffices to show that [Aj*AkBjk]2 0 for each subset { A l , . . . ,A,} of 2l. By the same argument, it now suffices to show that [A*AkB;Bk] 2 0 for each subset { A l , .. . ,A,} of 2l and each subset { B l , . . . ,B , } of 3’.As Aj and A; commute with Bk and B; for all j and k, Aj*AkB?Bk = (AjBj)*(AkBk).Thus, from (ii),

708

TENSOR PRODUCTS

11.5.21. Define an n-state of a C*-algebra 2l to be a matrix [pjk] of linear functionals on 2l such that [pjk(Ajk)] 1 0 when [Ajk] E Mn(21)+ and p j j ( I ) = 1 for j in (1,. .,TI}. Show that (i) if U acts on a Hilbert space 31 and ( 2 1 , . . . ,zn}is a set of n unit vectors in H, then [wx, 1 a] is an n-state of !2l [Hint. Use Exercise 11.5.20(iv).]; (ii) a linear mapping q of 2l into a C*-algebra B , such that q(I) = I, is completely positive, if and only if [ p j k o q] is an nstate of U for each n-state [ p j k ] of B [Hint. Note that ( [ B j k ]3,Z) = ( [ ( B j k z k , z j )ii, ] ii), where 3 = ( 5 1 , . . . ,zn} and ii = {1,1,...,l}.]; (iii) [pjk] is an n-state of 9 when each p j k is the same state p of U [Hint. Use the GNS construction and (i).]; (iv) a positive linear mapping of 2l into an abelian C*-algebra B is completely positive. [Hint. Note that [ B j k ] E M,(B)+ if and only if [ p ( B j k ) ]2 0 for each pure state p of B . Then use (iii).]

.

Solution. (i) If

[Ajk]

E Mn(21)+,then

0 I [(AjkXk,Xj)l = [&,,Zj(Ajk)l

131 is an n-state of 2l. from Exercise 11.5.20(iv), so that [ushrZj (Since x j is a unit vector, w x j , Z j ( I = ) 1.) q is completely positive and [Ajk] E Mn(2l)+. Then Suppose (ii) [ ~ ( A j k )E] Mn(B)+ and [ ( p j k 0 q)(Ajk)] E M,(C)+ for each n-state [pjk] off?. Thus [ p j k o q] is an n-state of 2l when [ p j k ] is an n-state of U. (Since q ( I ) = I and p j j is a state, ( p j j o q ) ( I )= 1.) Suppose, now, that [ p j k o 54 is an n-state of 2l for each n-state [pjk] of B. Suppose B acts on a Hilbert space K and (21,.. . ,zn} is a set of n vectors in /c. Choose unit vectors y1,. . . ,yn in K and non-negative (real) scalars a l , . . . ,a, such that a j y j = Xj for each j . From (i), [wyk,vjI B] is an n-state of f?, so that [wyb,yj o 171 is an n-state of 2l by hypothesis. Thus, with [Ajk] in M,(U)+,

rd 1 0. [(V(Ajk)Yk?3 From Exercise 11.5.20(ii) and (v), [ajak] E A&(@)+

and

[(q(Ajk)2krXj)] = [(dAjk)Yk,3lj)ajakI 2 0. Proceeding as in the hint (with q(Ajk) in place of B j k ) , we have that ([dAjk)]z,$) = ([q(Ajk)zk, X j ) ] % 6 ) 2 0,

EXERCISE 11.5.22

709

where E = ( 2 1 , . . . ,z,} and ii = {l,.. . ,1}. Thus [q(Ajk)] EM,(B)+ and q is completely positive. (iii) Let A be the GNS representation of U corresponding t o the state p. From Exercise 11.5.16(i), A is completely positive so that [ ~ ( A j k )2 ] 0 when [Ajk] E M,(U)+. In this case, [ ( A ( A ~ ~ ) s , s )2] 0 for each vector 2 in the representation space for A, from (i), and in particular, for a generating unit vector SO for K ( U ) such that w,, o A = p. Thus

and [pjk] is an n-state of U, when p j k = p for all j and k in (1,. . . ,n}. (iv) Suppose q is a positive linear mapping of U into an abelian C*-algebra B. Let [Ajk] be an element of M,(U)+. From Exercise 11.5.13, [q(Ajk] E M,(B)+ if and only if [ ( p o q)(Ajk)] 2 0 for each pure state p of B since f? is abelian. Now p o q is a positive linear functional on U, so that [ ( p o q ) ( A j k ) ] 2 0 from (iii). Thus [ q ( A j k ) ] is in Mn(B)+ and q is completely positive. [ 10(Prop.1.2.2,~.148).106(Lemma 6.1,p. 262)] 11.5.22. Let U be an abelian C*-algebra and [ p j k ] be an n x n matrix of linear functionals on U. (i) Find a representation A of U on a Hilbert space 3-1 with a cyclic vector u and a matrix of operators H j k in .(a)- such that pjk(A) = (a(A)Hjku,.u) for each A in U. [Hint. Express each pjk as q j k i r j k with q j k and T j k hermitian, and let p be c,”,k,l(q$k t qyk t T& t T ~ ; ) . Use the GNS representation corresponding to p as K together with Proposition 7.3.5 and Corollary 7.2.16 .] (ii) Suppose [ p j k ( A ) ] 2 0 for each A in U+. Show that [ H j k ( p ) ] 2 0 for each p in X , where A(%)- 2 C ( X ) , and we denote by the same symbol an element of A(%)- and the function representing it. Conclude that ( H j k ] 2 0. [Hint. Suppose the contrary and find a non-null clopen subset XO (with corresponding projection E ) of X and a vector {q,. . . , u , } (= ii) in @” such that ([Hjk(p)]ii,ii) < 0 for each p in Xo. Deduce the contradiction 0 > C,”,,=, UjakEHjk and

+

c n

0>

j,k=l

UjUk(EHjkZL,U)

= ( [ ( E H j k U , U ) ] i i , i i ) >_ 0

710

TENSOR PRODUCTS

- the first inequality uses Exercise 11.5.21(iii); the last inequality follows from the choice of Hjk and the present hypothesis.] (iii) Show that [pjk] is an n-state of M (abelian)if and only if [pjk(A)]2 0 for each A in U+ and p j j ( 1 ) = 1. [Hint. Recall Exercise 11.5.20(ii), (v), and (iv).] (iv) Show that each positive linear mapping of U (abelian) is completely positive.

Solution. (i) We proceed as in the hint. Let uo be a unit generating vector for n(M). Since n(2i)- is abelian with a generating vector, n(2l)- is maximal abelian by Corollary 7.2.16. Choose uo and a multiple u of uo so that p(A) = (n(A)u,u ) for each A in U. As qTk, qJTk,T $ , rJ<are positive linear functionals on M dominated by p, they induce positive linear functionals on n(U) dominated by w,, I n(U) from Exercise 4.6.23(ii). Thus, from Proposition 7.3.5, there are operators ATk, A;k, B$, BJTkin n(%)- such that

for each A in U. It follows that, for each A in U,

where H j k = ATk - AYk -t i(B:k - BG) (E ~(24)’). (ii) From Theorem 5.2.1, n(U)- S C ( X ) for some extremely disconnected compact Hausdorff space X . If [ H j k ( p o ) ] 2 0, there i ) 0. By is some { a l , . . . ,an} (= ii) in Cn such that ( [ H j h ( p ~ ) ] i i , i < continuity of all H j k , there is a clopen subset Xo of X containing po such that ( [ H j k ( p ) ] i i , i i ) < 0 for each p in Xo. Let E be the characteristic function of X O . With p in Xo,

whence

EXERCISE 11.5.23

Since u is separating for

711

.(a)-,

by assumption. Hence, by strong-operator continuity anc hznsity,

0 5 [(EHjku,u)] contradiction. Hence [ H j k ( p ) ]2 0 for each p i n X and [ H j k ] 2 0 from Exercise 11.5.13. (iii) If b j k ] is an n-state of U, then [pjk(A)]2 0 for each A in U+ since the n x n matrix with each entry equal to A is in Mn(U)+ from Exercise 11.5.20(ii). Suppose [pjk(A)]2 0 for each A in 2l+. Then with H j k as in (ii), [ H j k ] E M n ( r ( 2 l ) - ) + . If [Ajk] E Mn(U)+, then [ ~ ( A j k )is] in Mn(r(21)-)+ from Exercise 11.5.16(i). Thus [r(Ajk)Hjk]2 0 from Exercise 11.5.20(v), and -a

0

I [(r(Ajk)Hjku,.)I

= [~jk(Ajk)]

from Exercise 11.5.20(iv) and by choice of H j k . Hence [ p j k ] is an n-state of U in this case. (iv) From the solution to Exercise 11.5.21(ii), the assumption that 24 is abelian, and the solution to (iii), it will suffice to show that [ ( p j k o ? ) ( A ) ]_> 0 for each A in U+ and each n-state [ p j k ] off?, the C*-algebra into which 77 maps. But with A in U+, q ( A ) E f?+ and the matrix all of whose entries are q ( A ) is in Mn(f?)+. Thus [ ( p j k o q ) ( A ) ] 2 0 for each n-state [ p j k ] of f?, and q is completely m[ 1041 positive. 11.5.23. Let q be a positive linear mapping of a C*-algebra 24 into a C*-algebra B such that ~ ( 1=)I . Show that (i) E T * E T E 5 E T * T E when E , T E f?(K) for some Hilbert space X: and E is a projection; (ii) q ( A ) * q ( A ) I v ( A * A ) for each normal operator A in U. [Hint. Use Exercises 11.5.18(iii) and 11.5.22(iv) with (i).] (This provides another approach to Exercise 10.5.9.)

712

TENSOR PRODUCTS

Solution. (i) For each x in

K,

(ET*ETEz,X) = I I E T E x ~I( ~I I T E z ~=~ (~E T * T E s ,z ) . Hence ET*ETE 5 ET*TE. (ii) Let 70 be vIUO,where UOis the C*-subalgebra of U generated by A , A * , and I . Then 770 is a positive linear mapping, is abelian. From Exercise 11.5.22(iv), ~0 is q o ( 1 ) = I , and completely positive. Suppose L3 acts on a Hilbert space ‘H. From Exercise 11.5.18(iii), there is a Hilbert space K: containing 3-1 and a representation Q of Uo in B ( K ) such that

for each B in UO,where E is the projection of K onto ‘H. Thus, from (i), we have that

11.5.24. Let { A n } be a sequence of positive operators on a Hilbert space ‘H with CT=,A , weak-operator convergent to I . (i) Find a positive linear mapping 7 of C ( X ) into B(’H) such that ~ ( 1 )= I and q(fn) = A,, where X is the compact subset (0, : n = 1,2,. . .} of W and fn takes the value 1 at and 0 at other points. (ii) Find a Hilbert space K containing ‘H and a sequence {En} of projections on /C with sum I such that EEnE = A n E for each n , where E is the projection in f?(K) with range ‘H. [Hint. Use Exercises 11.5.18(iii) and 11.5.22(iv).]

k

Solution. (i) Since C : = l ( A n ~ , ~converges ) for each x in ‘FI, with f i n C ( X ) , 00

n= 1

00

n=l

-

n=l

713

EXERCISE 11.5.25

Hence, by polarization, Cr=lf( + ) A n converges in the weak-operator topology to some ~ ( f ) and , 17 is a positive linear mapping with the desired properties. (ii) From Exercise 11.5.22(iv), the mapping 7,constructed in (i), is completely positive and V( 1) = I . Thus, from Exercise 11.5.18(iii), there is a Hilbert space K containing 'H and a representation cp of C ( X )on K such that E v ( f ) E = q ( f ) E ,where E is the projection of K onto 3-1. Since { f n } (as in (i)) is an orthogonal family of idempotents in C ( X ) ,(p(fn)}is an orthogonal family { F , } of projections F,, then on K and EFnE = AnE for all n. If F = C,"==,

EFE =

00

00

00

n=l

n=l

n= 1

C EFnE = C AnE = (CA n ) E = E .

Thus E 5 F and I - F 5 I - E . Let El be F1 t I - F and En be Fn for n in {2,3,. ..}. Then C,"==, En = I and EEnE = AnE for all n. 11.5.25. Let R be a von Neumann algebra acting on a Hilbert space 3-1. Show that y 8 z (z,y E 'H) extends to a unitary (i) the mapping z 8 y operator U on 3.1 8 'H; (ii) U (in (i)) is self-adjoint; (iii) the mapping A 8 B -, B 8 A ( A , B E R ) extends to a * automorphism of R %R. ---$

Solution. (i) With

21,.

. . ,z n , y1,. . . ,yn vectors in 3-1,

C;=,

In particular, C;=, zj 8 yj = 0 if and only if yj 8 xj = 0. Thus the mapping z 8 y + y 8 z (z,y E 'H) extends (uniquely) to a well-defined linear mapping UOof the everywhere-dense subspace 'Ho generated by simple tensors onto itself. From (*), UOis an isometric mapping of 'Ho onto Ho. Hence UO extends (uniquely) to a unitary operator U on 'H 8 'H.

714

TENSOR PRODUCTS

(ii) Note that U 2 ( x €3 y) = 5 €3 y for each x and y in 'H. Thus U 2 = I and U = U * . (iii) Since

U ( A€3 B ) U ( z @I y) = U ( A€3 B)(y €3 x) = U ( & @ B z ) = Bs €3 Ay = ( B @I A ) ( x €3 Y), for all x and y in 'H, we have that U ( A18 B)U = B €3 A for all A and B in B('H). Thus the mapping

A €3 B

+

B €3 A

( A , B E B('H))

extends to a * automorphism a of B('H)%B('H) (implemented by U ) . The restriction of a to RB'R is a * automorphism. m[98]

11.5.26. Let { H j : f E F} be a family of Hilbert spaces in which the index set F is directed by 5. Suppose that, whenever f,g E F and f 5 g, there is an isometric linear mapping U g j from ' H j into Hg, and UhgUgf = U h f whenever f,g,h E F and f 5 g 5 h. (i) Prove that U j j is the identity mapping on H j . (ii) Construct a Hilbert space 'H and, for each f in F, an isometric linear mapping U j from 'Hi into 'H, in such a way that U j = U g U g f whenever f , g E F and U { U j ( ' H j ) : f E F} is everywhere dense in 'H. [Hint. Let X be the Banach space consisting of all families {Zh : h E F} in which zh E Hh and SUp{llZhll : h E F} < 00 (with pointwise-linear structure and the supremum norm). Let XObe the closed subspace of X consisting of those families {Zh : h E F} for X/Xo which the net {115hll : h E F} converges to 0, and let Q : X be the quotient mapping. Adapt the argument used in proving Proposition 11.4.1(i).] (iii) Suppose that K is a Hilbert space, Vf is an isometric linear mapping from 'Hi into K , for each f in F, V j = VgUgj whenever f , g E F and f 5 g, and U(Vj(7if): f E IF} is everywhere dense in K. Show that there is a unitary transformation W from 'H onto K such that Vj = WUj for each f in IF. Note. In the circumstances set out above, we say that the Hilbert spaces ' H j (f E F) and the isometries Ugf(f,g E F,f 5 g) together constitute a diwcted system of Hilbert spaces; the Hilbert space 'H occurring in (ii) (together with the isometries U j (f E F)) is the inductive limit of the directed system. The effect of (iii) is t o show that the constructs in (ii) are unique up to unitary equivalence. --$

EXERCISE 11.5.26

715

Solution. (i) Since U j j is an isometric linear mapping and

it follows that Ufj= I . (ii) We use the notation introduced in the hint. Given f in IF, we define an isometric linear mapping U; from 3-11 into X as follows: when x E 3-11,U j x is the family { X h : h E I}in which

Note that X/Xo is an isometry, (a) the linear mapping QU; : 3-1f (b) QU; = QULU,f when f _< g. For this, suppose that x E Xj. To prove (a), let {yh : h E F} be an element y of Xo. Given any positive real number E , it results from the definition of XO that there is an element fo of lF such that IlYhll < E whenever h E F and h 2 fo. Since IF is directed, we can choose g in F so that g 2 f and g 2 fo. Since U;x is the family { x h } defined by ( l ) ,we have --$

Thus IIUjx - yII 2 11x11. It follows that the distance 11QUjxll from U i x to & is not less than 11x11. The reverse inequality is apparent, and (a) is proved. For (b), note that x E Xf and U , f x E 3.1,; we want to show that U ; x - U;U,fx E &. Now U j x - U;U,jx is an element { z h : h E F} of X, and we have to prove that the net {11zhll : h E F} converges to 0. In fact, we have the stronger result that 11zhll = 0 when h 2 g(> f ) , since

The range of the isometric linear mapping Q U ; is a closed subspace Kj of the Banach space X/Xo, and the norm in K f inherits

716

TENSOR PRODUCTS

from the Hilbert space H f the "parallelogram law" (see (6) in Proposition 2.1.8). When f 5 g ,

From this inclusion, and since IF is directed, it follows that the family { K f : f E F} of subspaces of X/Xo is directed by inclusion. Thus u { K f : f E P} is a subspace 3-10 of X/Xo, its norm satisfies the parallelogram law, its closure is a Banach subspace 3-1 of X/Xo, the parallelogram law remains valid in 3-1 by continuity, and thus 3-1 is a Hilbert space. The proof of (ii) is now complete if we take, for U f , the isometric linear mapping QU; from ' H f into 3-1. (iii) Under the conditions set out in (iii), the mapping VfUy' is a linear isometry from Uf(3-1r)onto Vf('Hj).When f 5 g , VgUi1 extends VfU,', since, for x in 3-15,

V,U,-'(UfX) = vgu,-'ugu,fx = V,U,fX = VfX = vfu;'ufx. From this, and since the family (Uf(3-1f) : f E IF} is directed by inclusion, there is a linear isometry WOfrom U U f ( ' H j ) onto u V f ( H f ) , such that WOextends VjU,' for each f in IF. Moreover, WOextends by continuity to a unitary transformation W from 'H onto K , W extends VfU;' for each f , and thus V f = W U f . m 11.5.27. Suppose that IF is a directed set, and the Hilbert : spaces 3-15 (f E IF) and isometric linear mappings Us! 3-15 + 3-1, ( f , g E F,f 5 g) together constitute a directed system of Hilbert spaces. Construct the (inductive limit) Hilbert space 3-1 and the isometric linear mappings U f : 'Hf + 3-1, as in Exercise 11.5.26(ii). Suppose also that A f E B(3-1j) for each f in P, sup{IIAfll : f E IF} < 00, and there is an element fo of IF such that A,U,j = U,jAf whenever f , g E IF and fo 5 f 5 g. Show that there is a unique element A of B(3-1) such that AUf = U f A f whenever fo 5 f E IF. (We refer to A as the inductive limit of the family { A f : f E IF} of bounded operators.) Solution. When f E F, U;Uf is the identity operator on 3-1f, U f A f U j E B(3-1), and IIUjAfUjI) 5 M , where M is the supremum

717

EXERCISE 11.5.28

of

{llAhll :

h E F}. When f , g E F a n d fo 5 f 5 g , we have

Thus UgAgU,"and U j A j U i have the same restriction to the space Uf (3-15) ( C 'HI, when f o 5 f 5 g . The family { U f ( ' H f ): fo 5 f E IF} of subspaces of 'H is directed by inclusion, and its union 'Ho is a dense subspace of 'H. From the final assertion of the preceding paragraph, it follows that there is a linear operator A0 : 'Ho 3 'Ho such that

A0 is bounded, and llAoll 5 M , so A0 extends by continuity to an element A of B('H). Since

It is apparent that

it follows that

AUj = U j A j V i U j = U j A j

(fo 5

f E IF).

So far, we have proved the existence of a bounded operator A with the stated properties. Suppose that A' is another such operator. Then

wherever fo 5 f E IF, hence A' I 'Ho = A I 'Ho, and A' = A by continuity. This establishes the uniqueness of A . Suppose that IF is a directed set, and the Hilbert 11.5.28. 'Hg spaces X j ( f E F) and isometric linear mappings U g f : H f ( f , g E F,f 5 g ) together contitute a directed system of Hilbert --f

718

TENSOR PRODUCTS

spaces. Suppose also that the C*-algebras !2lj (f E * isomorphisms

F) and the

together constitute a directed system of C*-algebras. Suppose, finally, that ~f is a representation of '#f on 'lit, for each f in r, and

whenever f,g E F and f 5 g. Construct the inductive limit Hilbert space 3.1 and the isometric linear mappings U f : Hf -+ 3.1 as in Exercise 11.5.26@), and construct the inductive limit C*-algebra 2l and the * isomorphisms ~f : Uf 4 U as in Proposition 11.4.1(i). Prove that there is a unique representation K of 121 on 'li such that

for each f in B. (We refer to K as the inductive limit of the family (91 : f E F} of representations.) Solution. The argument used to establish the existence of a representation A with the stated properties is divided into two stages. First, we use the result of Exercise 11.5.27 to construct a representation $h of Uh on H,for each h in F. Then, we view vh as a * isomorphism from U h onto the C*-algebra ~ h ( U h ) ( c a), and by considering the representation $h o 'pi1 of (Ph(%(h)on 3.1, for each h in IF, we construct the required representation K of a. In the first instance, choose (and fix) an element h of IF. Given A in a h , define a family { A j : f E F}, with A j in B('Hj), as follows:

When f , g E F and h 5 f 5 g, we have

EXERCISE 11.5.28

719

Since, also, llAfll 5 IlAll for each f in F,the family { A f : f E F) has the properties set out in Exercise 11.5.27. We denote by $h(A) the inductive limit of this family, and note that

In order to verify that t+hh is a representation of M h on 'H, suppose that A, B E ?21h and a, b E @, and note that

From these relations, it follows that

$h(aA

+ 6B)Pf = (a$h(A) + b$h(B))Pf,

$h(A*)Pf = Pf $h(A)*Pf, where Pf(= U fU j ) is the projection from 'H onto Uf('HI). Since F is directed, U f(H f ) Ug('Hg)when f 5 g, and the union u { U f(H/ ) : h 5 f E F) is dense in the Hilbert space H , it follows that the net { P f : f E F) is strong-operator convergent to I. By taking strong-operator limits in ( 3 ) ,we obtain

720

TENSOR PRODUCTS

Moreover, in the case in which A = I ( € ah),we have that A , = I (E B ( ' H j ) ) when h 5 f, from (1); so it follows from (2) that

and hence $ h ( I ) = I ( € B(7.1)). Thus t)h is a representation of 2 l h on 3-1. Now suppose that h , k E F and h 5 k. When k 5 f E P, we have

h ( A ) U f = UfKf(@fh(A)) = UJrf(@fk(@kh(A))) = $k(@kh(A))Uj ( AE ah). Since U { U f ( X j ) : k

5f

E

IF} is dense in 'H, it follows that

If we consider ' p h as * isomorphim from U h onto ' p h ( 2 l h ) (E U), we can form the inverse * isomorphism pi1 : 'ph(Uh) + U h and the representation '$h o 'phi of yh(!&) on 3-1. If h , k E,'!l h 5 k, and A E 'ph(Uh), we can substitute 'phl(A) for A in (4),to obtain

since

( ~= h ( ~ ok @ k h .

Thus

The C*-subalgebras ( ~ h ( U h ) ( h E P) of U are directed by inclusion and have union dense in %, and each of the representations $h o 'pi1 has norm 1. Accordingly, there is a * homomorphism TO of norm 1 from Uph(Uh) into B ( X ) that extends $h o ' p i 1 for each h in IF, and KO extends by uniform continuity to a representation K of 24 on 'H. When A E a h , we have

and from (2),

EXERCISE 11.5.29

So far, we have proved the existence of a representation

721 K

of

2l into B(3-1) that satisfies the stated conditions. Note that these conditions imply that

and so uniquely determine r ( B ) P f when B E (~f(Uf). Since cpf(Uf) C_ cpg(Ug)when f 5 g , the stated conditions uniquely determine r ( B ) P gwhenever B E ( ~ f ( U fand ) f 5 g , and so determine K ( B ) (= 1imgr(B)Pg). Since U f E ~ ~ o f ( Uisf )dense in U, K is uniquely determined by the stated conditions. 11.5.29. Suppose that {?la : a E A} is a family of Hilbert spaces and ua is a unit vector in 7-fa for each a in A. Note that the family IF of all finite subsets of A is directed by the inclusion relation When F is an element {al,. . . , a , } of IF, define a Hilbert space 3 - 1 ~and a unit vector U F in 3 - 1 ~by

c.

(i)

Suppose that F, G E P and

Show that there is a unique isometric linear mapping UGF 7 - l ~+ 3 - 1 ~such that

whenever x a j E H a j for j in { 1,. . . ,n}. (ii) Show that the Hilbert spaces 3 - 1 ~( F E P) and the isometric linear mappings UGF ( F , G E F , F C_ G) together constitute a directed system of Hilbert spaces. (iii) Suppose that the Hilbert space 3-1, together with the isometric linear mappings UF : 7 - l ~-+ 3-1 ( F E IF), is the inductive limit of the directed system occurring in (ii). Show that there is a unit vector u in 3-1 such that UFUF = u for each F in IF. (We refer to the inductive limit Hilbert space 3-1 as the tensor product of the family {‘Ha : a E A}, and denote it by @ , + , ? f a . Strictly speaking, we should use a nomenclature and notation that recognizes the dependence of this construction on the choice of ua.)

722

TENSOR PRODUCTS

Solution. (i) From Proposition 2.6.5 there is a unitary transformation V from 'HG onto 'HF @ 'HG\F such that

V(xa1 8 * . * @ %a,) = (xa1 8 * * . 8 xan) 8 (xan+, 8 *..

Zam)

whenever z a j E 'Haj for j in (1,. , , ,m}. Since UG\F is a unit vector in 'HG\F, and V* is a unitary transformation, we can define an isometric linear mapping UGF : 'HF 'HG by

UGFX= V*(x @ UG\F)

(Z

E 'HF).

If zaj E 'Ha, for j in { 1,...,n } ,

U G F ( Z@~ *~* * @ ~ a ), = V*((2a1€3 *..@ xan) €4 (Ua,+, 8 * . * 8 %am 1) = X a 1 @ - - . @ ' a , €3'a,+, @-**€3%,. This establishes the existence of a mapping UGF with the stated properties; the uniqueness of UGF follows from its continuity and linearity, since the linear span of the simple tensors is dense in ' H F . (ii) We have to show that UHGUGF = UHF whenever F,G, H are in IF and F G C H . For this, we may suppose that

F = {~l,*--,a,}, G = { U ~ , * . . , U , } , where n 5 m 5 k. If xaj E 'Haj ( j = 1,.

H = {~l,**-,ak} a ,

n ) , we have

-

U H G U G F (X~ @ ~* * xa, ) = U H G ( S 61 ~ ~. . . @ xa, @ u ~ , , +@~. .- 8uam) =zal 8 * * * 8 %an 8 ua,+l 8 - . €3 214, 8 ~ a , + l €3 * * * €3 Uah = U H F ( Z , ~€3 . . . @ x,,).

-

Hence UHGUGF= U H F . (iii) From the definition of UGF it is apparent that UGFUF= UG when F,G E IF and F G. Given any two elements, F and G , of IF, we can choose H in F such that F H , G H (for example, we could take H to be F U G). Then

UFUF= UHUHFUF= UHUH= UHUHGUG= UGUG. Thus the unit vector UFUF in 'H is independent of the choice of F in IF. ~[81]

723

EXERCISE 11.5.30

With the notation of Exercise 11.5.29, suppose that 11.5.30. is a representation of a C*-algebra 2l, on the Hilbert space N, for each a in A. Construct the directed system of C*-algebras consisting of C*-algebras UF (F E F) and * isomorphisms QGF : IZLF + 9 ! ~ (F,G E F,F C G), as in the discussion following Proposition 11.4.2, so that the inductive limit of the system is a C*-algebra B (= BaGA %), together with * isomorphisms Y F :BF -+ B (F E F)satisfying conditions analogous to those set out in Proposition 11.4.1(i). When F E F, the tensor product of the family { A , : a E F} is a representation AF of MF on NF. (i) Prove that A,

TG(@GF(A))UGF= U G F A F ( A ) ( A E %) whenever F,G E 8' and F E G. (ii) Deduce that there is a unique representation @aEA%LO) on 'H (= Q P a E A X H asuch ) that R((PF(A))UF

= UFAF(A)

A

of 2l (=

(AE a F )

for each F in F. (iii) Prove that wzco A o V F = wUFo X F ( F E IF). (iv) When a E A and F = { a } E F, we write pa in place of ( P F , so that cp, is a * isomorphism from 8, into %. Define states p of 2l and pa of (P, (21a)(G %) by p = w, o T , pa = w,, o A, o cp,'. Show pa. that p is the product state (v) Show that u is a cyclic vector for A if, for each a in A, u, is a cyclic vector for A,.

BaE-

724

TENSOR PRODUCTS

Since linear combinations of simple tensors form everywhere-dense subsets of 'HF and %F, it now follows that ~G(@GF(A))UGF = U G F ~ F ( A ) ( A E %F).

(ii)

In view of (i), (ii) is an immediate consequence of Exercise

11.5.28.

(iii) When F E B' and A E % F ,

(iv) Let a l , . . - , a , be distinct elements of A and Aj be in for j in ( l , - - . , n } . With { a l , . . - , a n }for G and { a j ) for F i n the formula (PF = (PGo Q G F , we obtain

whence

Thus, with w for utCl8

*

8 us,

725

EXERCISE 11.5.30

Hence p is the product state BaEApa. (v) When A1 E U,, ,-.. ,A, E Ua, and G = {al, . . - , a , } ,

If each u, is cyclic for r,, then

[ K ( ( p G ( % G ) ) u ] is

UG(%G). Hence

CHAPTER 12 APPROXIMATION B Y MATRIX ALGEBRAS

12.4.

Exercises

12.4.1. Let E and F be projections in a von Neumann algebra R such that llE - Fll < 1. Show that (i) E N F [Hint.See the proof of Lemma 12.1.5.1; (ii) there is a unitary operator U in R such that UEU* = F.

Solution. (i) From Proposition 2.5.14, the range projection R ( E F ) of EF is E - ( E A ( I - F)). But E A ( I - F ) = 0, for otherwise, there is a unit vector x in the ranges of E and I - F so that

1 = 11x11 = ll(E - F)xll

< llE - Fll < 1.

Thus R(EF) = E and, symmetrically, F = R(FE) N R((FE)*). Hence E and F are equivalent in R (see Proposition 6.1.6). (ii) Note that Il(I - E) - ( I - F)ll = llF - Ell < 1. From (i), I - E N I - F . Let V and W be partial isometries in R such that V*V = E, VV* = F , W*W = I - E , and WW* = I - F. Let U be V W . Then

+

U*U = V*V+V*W+W*V+W*W= V*V+W*W = E+I-E = I . Symmetrically, U U * = F -I-I - F = I. Thus U is a unitary operator in R and

UEU* = VEV* + VEW'

+ WEV*+ WEW* = VV* = F.

8

12.4.2. Let E and 2l be a projection and a C*-algebra, reand spectively, acting on a Hilbert space %. Suppose 0 < a llE - All < a for some A in (%)I. Show that

0, show that there is a family {Ex : 0 5 X I w ( E ) } of projections Ex in R such that EO = 0, E w ( ~=)E , w ( E x ) = X for each X in [O,w(E)], and Ex 5 Ex, when X _< A'. [Hint. Let w ( E ) be a and w' be a-lwlERE. Use a maximality argument on orthogonal families of projections in E R E such that the values of w' at their unions does not exceed $ to produce a projection Ea/2 in E R E for which w'(Eap) = f. Now find Ear such that w'(Ea,) = r for each dyadic rational T in [0,1].]

Solution. We proceed as in the hint, so that w' is a normal state of E R E . Partially order by inclusion the set S of orthogonal families {Ga}aeA of projections in E R E such that a

a

740

APPROXIMATION BY MATRIX ALGEBRAS

The union of a linearly ordered subset of S is in S, since w’ is normal, and is an upper bound in S for that linearly ordered subset. By Zorn’s lemma, there is a maximal element { M b } b E l in S. We shall Mb is a projection Ea12 in E R E for which w’(E,lZ) show that = 1/2. We prove, first, that if G is a projection in E R E such that o’(G) # 0 and a positive E is given, then there is a subprojection Go of G in E R E such that 0 < w’(G0) < E . It will suffice to show that each such G has a subprojection G1 in E R E such that 0 < w’(G1) 5 w’(G)/2, for then we can repeat this process to find a subprojection G2 of G1 in E R E such that 0 < w’(G2) 5 w‘(G1)/2 5 w‘(G)/4. Continuing in this way, we find a projection G, in E R E such that G, 5 G and 0 < w‘(G,) 5 2-,w’(G). To find G I , it will suffice to locate a projection N in E R E such that N < G and 0 < w ’ ( N ) < w’(G), for in this case, either 0 < w‘(N) 5 w‘(G)/2 or 0 < w’(G - N ) 5 w’(G)/2. Let M be the support of w’lGRG. Since o‘(G) > 0, M # 0. Since R has no minimal projections, M has a subprojection N in R such that 0 < N < M 5 G. As M is the support of w’lGRG, 0 < o ‘ ( N ) and 0 < w’(M - N ) . Thus

cb

0 < w‘(N) < u ’ ( M ) 5 w’(G). Since { M b } b e i E 8,W‘(Ea/2)5 1/2. Suppose ~ ’ ( E a / 2 0, let wy be Cj'wjIEk+lREk+l. (Let w[i' be 0 on Ek+lREk+l when cj = 0.) Then w i and w r are either 0 or normal states. From the (inductive) assumption (w), there is a projection M in F R F such that u g ( M ) = wg(A) and, hence, w j ( M ) = w j ( A ) for each j in { 1,.. . ,n}, where A = A1 El . . + X k E k . From (i), there is a projectian N in Ek+lREk+l such that w y ( N ) = u y ( X k + l E k + l ) and, hence, w j ( N ) = w j ( A k + l E k + l ) for each j in (1,.. . ,n}. It follows that w j ( E ) = w j ( B ) for each j in {l,,. . , n } , where E is M N and B is XlEl ... iXk+lEk+l. (iii) From (i) and (ii), it follows, by induction on k, that (**) is valid for all k in N. (iv) By polarization, it suffices to make our approximations on (finite sets of) vector states of R. We prove, first, that XlEl -..t A k E k (= A ) is in the weak-operator closure P- of P, where { E l . . ,Ek} is an orthogonal family of projections in R and X I , . . . , Xk are in [O,l]. Suppose that we are given a positive E and unit vectors $ 1 , . . ,z,. Let w j be the vector state of R corresponding t o x j . From (iii), there is an E in P such that w j ( E ) = o j ( A ) , and of course, J w j ( E )-w,(A)I < E for all j in { 1,., , ,n}. Thus P- contains each such A in R. From Theorem 5.2.2(v), each operator in (72)' is a norm limit of operators in R having the form of A. Thus (72): is contained in P-. Since (72); is weak-operator closed and P (72); , we have that P- 5 (72); and P- = (R)?. m[30]

+.

+

+

+

.

.

12.4.16. Let R be a type I von Neumann subalgebraof a factor M of type 111 and let {Pn}be the family of central projections in R such that RP, is of type I , or P, = 0 and C, P, = I. Suppose P, = 0 when n > rn for some finite cardinal m.

(i) With { A l , . . . , A , } a finite set of operators in (R)1 and E a positive number, show that there are a finite type I subfactor n/ of M and operators B1,. .,B, in (N)1 such that

.

745

EXERCISE 12.4.16

[Hint.Use Exercise 12.4.10 and 12.4.9.1 (ii) With E and F projections in M and E a positive number, show that there are a finite type I factor N and operators A and B in (N)1 such that IIB - F((2< E IIA - El12 < E , [Hint. Use (i) and Exercise 12.4.11.1 (iii) With the notation of (ii), show that A and B can be chosen to be projections. [Hint.Use Exercise 12.4.5.1 (iv) With H1 and H2 in (M);' and 241 and 242 ultraweakly open sets in ( M ) I containing H I and H2, respectively, show that there are a finite type I subfactor N of M and projections El and E2 in N such that El E 241 and E2 E 242. [Hint. Use (iii) and Exercise 12.4.15(iv).]

Solution. (i) Let A,.h be AhP, for T in { 1 , . . . ,m } and h in ( I , . . . ,n}. If P,.# 0, then RP,.is a type I, von Neumann subalgebra of the factor P,MP,. and A h , E (RP,.)1for each h in { 1 , . . ,n}. From Exercise 12.4.10, there are a finite type I subfactor N,. of P,MP, and operators Ilkl,.. . ,BC, in N,. such that llBChll 5 1 and

.

( h E ( 1 , . . .,4). B312 < E/2m } a self-adjoint system of Let { E ( j ,k,T ) : j,k E ( 1 , . . . ,n ( ~ ) }be n ( ~x )n ( ~matrix ) units for N,. such that E y $ ) E ( j , j , ~ = ) P, and let B:h be X ( j , k , r , h ) E ( j , k , ~ )If. P, = 0, let Bih be 0. (In this case, Arh = 0.) From Exercise 12.4.9, we can find a finite type I subfactor N of M and operators F ( j , k,T ) in N such that { F ( j ,k,T ) } is a self-adjoint system of n ( r )x n ( ~matrix ) units, F(j,j,T ) } is an orthogonal family of projections in N , and 1IA.h -

x$!l

{ E ; : ;

llE(j,k , T ) - F ( j ,k,79112 < E/2mc.(T)2 when j , k E (1,. .. ,T Z ( T ) } and T E ( 1 , . . . , m } , where c > max(lx(j, k, @)I}, with j , k in { 1 , . . . ,n ( r ) } ,T in ( 1 , . .. ,m } , and h in B,h

is

xy!l.lX(j, k,

T,

h ) F ( j ,k, T ) , then

B,.h

( 1 , . . . ,n}}. If E ( N ) l and

4,)

IB:,

- ~ r h 1 1I2 c

C IIE(.~,~,T)-~ ( j , k , r ) 1 1 2

j,k=l

< Cn(T)2&/2mcn(T)2= &/2m

746


.

for all T in (1,. . ,m}and h in (1,. ..,n}. Thus llArh-Brhl(2 < ~ / m . Let B h be C;='=, Brh. Then B h E (N)1 and

(ii) From Exercise 12.4.11, the von Neumann subalgebra Rc,of M generated by E and F has the properties prescribed for R (where

m = 2). From (i), there are Af, A , and B , as described. (iii) Adopt the notation of (ii). From Exercise 12.4.5, there are projections M and N in N such that IIE - MI12 provided

< c,

IIF - Nil2 < E

N , A, and B,are chosen such that

with E' sufficiently small. (iv) From Remark 7.4.4, the ultraweak and weak-operator topologies concide on the unit ball in each of M and B ( H ) , where M acts on the Hilbert space 31. Applying the result of Exercise 12.4.15(iv), there are projections A41 in U1 and M2 in 242. Since the weakoperator topology is coarser (weaker) than the strong-operator topology on ( M ) 1 ,there are vectors 21,.. . ,z m in 3.t and a positive E such that, for b in {1,2},

{ A E (MI1 : ll(A - Mk)zjI( < E , j E (1, * * * 3 m}}(=wk)c uk From Exercise 8.7.3(iii), the strong-operator and 11 112-metric topologies coincide on (M)1 (since convergence of nets in ( M ) 1 is the same for each of these topologies). From (iii), there are a finite type I subfactor N of M and projections El and E2 in N such that Ek E (wkE) u k ( b E (1,2})12.4.17. Let oy be a * automorphism of a von Neumann algebra R. Suppose that there is an A in R such that CA = I and Aoy(B) = B A for each B in R. With VH the polar decomposition of A, show that (i) H is in the center of R [Hint.Prove that H 2 a ( B )= oy(B)H2 for each B in R. Use the fact that a ( B * )= a ( B ) * . ] ;

EXERCISE 12.4.18

747

(ii) CH = I ; (iii) a ( B ) = V*BV for each B in R [Hint. Note that H ( a ( B )- V * B V ) = 0 and use Theorem 5.5.4.1; (iv) V is a unitary operator in R;conclude that a is inner. [Hint. Consider I - R( V ).]

Solution. (i) For each B in R, V H a ( B )= B V H . Thus (1)

H a ( B ) = V * V H a ( B )= V'BVH.

From (l), we have

a(B)H2 = [Ha(B*)]*H= (V*B*VH)*H = HV*BVH = H 2 a ( B ) .

Since a maps R onto R, H 2 E R'. Hence H , the positive square root of H 2 , is in R'. But A E R; whence H and V are in R . Thus H is in the center of R. (ii) Since C H A = VCHH = VH = A , I = C A 5 C H . Thus I = CH. (iii) From (1) and (i), H ( a ( B ) - V * B V ) = 0 . From Theorem

5.5.4,

a ( B )- V*BV = CH(Q(B)- V * B V )= 0. (iv)

From (iii), I = a ( I ) = V*V and

I - a ( R ( V ) )= a ( I - R ( V ) )= V * ( I- R(V))V = 0 . Thus V is an isometry such that R ( V ) = I . It follows that V is a unitary operator (in R). Hence a is inner. m[70]

12.4.18. Let a be a * automorphism of a von Neumann algebra R. When there is no element A in R, other than 0, such that A a ( B ) = B A for each B in R, we say that a acts freely (on R). (i) Suppose A E R and A a ( B ) = B A for each B in R. Show that ~ ( C A=)CA.[Hint. Consider C A and ~ - ' ( C Ain) place of B.] (ii) Show that either a acts freely on R or there is a non-zero central projection Q in R such that a(&) = Q and alRQ is inner. [Hint. Use Exercise 12.4.17.1 (iii) Show that there is a central projection P in R uniquely defined by the conditions: a ( P ) = P , P is 0 or alRP is inner, and alR(I - P) acts freely. [Hint. Use (ii) and a maximality argument.]

748


Solution. (i) Note that

~ ( C A )=AA ~ ( C A=) C A A = A. Thus ~ ( C A 2) CA.In addition,

whence CA 5 CY-l(CA).

Hence CA 2 ~ ( C A and ) , CA= ~ ( C A ) . (ii) If a does not act freely on R,then there is a non-zero A in R such that A a ( B ) = BA. From (i), ~ ( C A=)C A so that a induces a * automorphism ~ ~ R of C RCA. A From Exercise 12.4.17, ~ I R C A is inner. Since A # 0, CA # 0; we may use C A for Q. (iii) If a acts freely on R, let P be 0. If a does not act freely, let { & a : a E A} be a maximal orthogonal family of non-zero central projections such that a ( Q a ) = Q a and alRQa is inner. From (ii), { & a } is not empty. Let Ua be a unitary operator in RQa such that UaBU,' = a ( B ) for each B in RQa. Since { Q a } is an orthogUa converges in the strong-operator topology to onal family, CaEA a unitary operator U in RP, where P = CaEAQa, and with B in

RP,

Now a ( P ) = P since a ( & , ) = Q a for each a in A (from ultraweak continuity of a - see Remark 7.4.4). Thus

UBU* = a ( B ) P = a ( B ) a ( P )= a ( B P ) = a ( B ) , and alRP is inner. Suppose P # I and alR(I - P ) does not act freely. From (ii), there is a non-zero central projection Q in R ( I - P ) such that alRQ is inner. Then Q is a non-zero central projection in R orthogonal to each Q a . Thus { Q a , Q } is a family that contradicts the maximality of { & a } . It follows that alR(I - P) acts freely.

EXERCISE 12.4.19

749

To this point, we have established the existence of P with the stated properties. Suppose P‘ is a central projection in R with these same properties. Then a(P‘(I - P ) ) = P’(I - P) . Suppose P‘(I - P ) # 0. Then, since alRP’ is inner, alRP‘(I- P ) is inner. Thus, there is a unitary operator V in RP’(I-P) such that V a ( B )= BV for each B in RP’(I - P ) . With T in R ( I - P ) , we have that v # 0, v E RP’(1- P ) c R ( I - P ) , and

V a ( T )= VP’(I - P ) a ( T )= Va(TP’(1- P ) ) = TP’(I - P)V = TV, contradicting the fact that alR(I - P ) acts freely. It follows that P’(1- P ) = 0; whence P‘ 5 P . Symmetrically P 5 P’, and P is the unique central projection in R with the stated properties. ~[70]

12.4.19. With the notation of Exercise 11.5.25, suppose that the automorphism (Y of (iii) of that exercise is inner. Show that R is a factor. [Hint.Assume that the center of R contains an operator that is not a scalar and use Proposition 11.1.8.1

Solution. Suppose A in the center of R is not a scalar multiple of I . If A @ I = I @A , then from Proposition 11.1.8, there are scalars c11, c12, c21, c22

such that

From the first two equations and the assumption that A is not a scalar multiple of I , we conclude that c11 = c21 = c12 = c22 = 0. From the third equation, we then have the absurdity, 0 = I . Thus

a(A@I)=I@A#A@I. But A @ I is in the center of inner. ~[98]

R@R. Thus, R is a factor if a is

750


12.4.20. With the notation of Exercise 12.4.19, let P be a * automorphism of R and suppose that a is inner. Show that (i) /3&3-' is inner [Hint.Consider ( P & ) a ( P @ ~ ) - l awhere , L is the identity automorphism of R, and note that the set of inner * automorphisms of R is a normal subgroup of the set of all * automorphisms .]; ) ( A @ I ) @ , ( U o for ) each A in R and each (ii) @ , ( U o ) ( P ( A ) @ I = unit vector z in H,where UO is a unitary operator that lies in Rein and implements P@Ip-', and 9, is as in Proposition 11.2.24; (iii) P is inner. [Hint.Use (ii), Exercise 12.4.17, and Proposition 11.2.24 extended so that the condition that T be positive is removed.]

Solution. (i) With A and B in R, note that ( P @ L ) - ~= P - l & and that

It follows that (p@L)a(p@L)-'a= pep-'.

If 7 is an inner * automorphism of R and V is a unitary operator in R that implements 7 , then with A in R,

Thus P7P-l is inner, and the set of inner * automorphisms of R is a normal subgroup of the group of all * automorphisms of R. It follows that ( p @ ~ ) a ( P @ ~is) -inner; 'a hence Pep-' is inner. (ii) F'rom (i), there is a unitary operator UOin R@R that implements [email protected]'. With A in R,

Hence, for each unit vector z in 3-1,

from the (conditional-expectation) properties of 9,. (iii) By Exercise 12.4.19, R is a factor. If 9,(Uo) is Q,(Uo) 8 I (as in the proof of Proposition 11.2.24), then from (ii),

EXERCISE 12.4.21

751

It will follow from this equation and Exercise 12.4.17 that /3 is inner once we show that QZ(Vo)(equivalently, @ z ( U ~ is ) ) not 0 for some unit vector z in ‘H. We prove this by extending the conclusion of Proposition 11.2.24 to assert that @ , ( T )# 0 for some z if T # 0 and T E R@S (without the requirement that T be positive). Suppose that @ , ( T )= 0 for each unit vector z in K. Then, as in the proof of Proposition 11.2.24, i#,(T) = 0 and (T(z@z),y@z)= 0 for all z and y in 1-I and z in K. Define a conjugate-bilinear functional on K by (21, Y)o = (T(w8 u ) , 8 9). The second identity of Proposition 2.1.7 applies to ( , )o (although an inner product is treated in that proposition). Replacing v by y and z by u in that (polarization) identity, we have 4 ( ~ , Y ) o = ( u + Y , u + Y ) o - (u-Yy,u-Y)o +i(u iy, u iy)o - i(u - iy, 11 - iy)o.

+

-+

Since ( z , z ) o = ( T ( w @ z ) , w ’ @ z = ) O ( z E K),we have that (u,y)o = 0 for all u and y in K. Hence ( T ( z @ u )y@v) , = 0 (2,y E ‘H, u, v E K). It follows that T = 0. m[98] 12.4.21. Let N be a subfactor of a factor M of type 111 and * automorphism of M @ M (described in Exercise 11.5.25) that a.ssigns B @ A to A 8 B for all A and B in M . Let 7 be the (unique) tracial state on M @ M and cp be the conditional expectation described in Exercise 8.7.28, mapping M @M onto N @ Nin this case. Suppose U is a unitary operator in M @ M that implements cr be the

crIN@N. (i) Show that cp(U)cr(T)= T+o(U)for each T in N @ N . Suppose N admits an outer * automorphism. (ii) Show that cp(U) = 0. [Hint.Use Exercises 12.4.2O(iii) and 12.4.17.1 (iii) Show that T ( U T )= 0 (T E N @ N ) .

Solution. (i) By assumption,

U*TU = cr(T)

(2’ E N @ N ) .

Since cp is a conditional expectation from M@Monto N @ N ,

752


when T E N Q N . is outer, (ii) Since N admits an outer * automorphism, CYJNQN from Exercise 12.4.2O(iii). Since NQN is a factor, cp(U) has central carrier either 0 or I . From (i), the fact that crIN&V is outer, and the result of Exercise 12.4.17, cp(V) has central carrier 0. Hence cp( U )= 0. (iii) By choice of cp (see Exercise 8.7.28),

(T E N Q N ) .

T ( U T )= ~ ( c p U( ) T ) = ~ ( 0 = )0

~[98]

12.4.22. Let M be the factor of type 111 (described in Example 8.6.12) constructed from the interval [0,1)(= 5') with Lebesgue measure m and the group G of translations modulo 1 by rationals. With Ej the projection in M corresponding to the characteristic function of the interval [(j- l)n-',jn-') for j in (1,. . .,n} and Vn the unitary operator in M corresponding to translation by n-l, let M , be the von Neumann subalgebra of M generated by ( E j } and V,. Show that (i) M , is a factor of type I , [Hint. Consider VLjEl and use Lemma 6.6.4.1; (ii) M , M , when m is divisible by n; (iii) U?=,M,! is strong-operator dense in M [Hint. Consider subintervals of [0,1) with rational endpoints.]; (iv) M is the matricial factor of type 111.

+ +

Solution. (i) Note that El * * * En = I , V,*EjVn = Ej+l for j in (1,. .. ,n - l},and V,*E,V, = El. Thus V;lEj is a partial isometry with initial projection Ej and final projection Ej+l.Hence V,-'E,-1V,-'Ej4

* *

.V,-IEl

= V;(j-l)El,

and V;jE1 is a partial isometry with initial projection El and final projection Ej+l for j in (1,. . ., n - 1). From Lemma 6.6.4, ( E l , . .. ,En} and {V;jE1 : j E (1,. . . ,n-1}} generate afactor M , (of type In). Since V;jElV;-l is a partial isometry (in M , ) with initial projection E;. and final projection Ej+1 for j in (1,. , , ,n - l}, while ElV,"-' (= (V[(n-l)E1)*)is a partial isometry with initial projection En and final projection E l ,

v;'E~t V;2E1V,, t V ; ~ E ~ V t :

t V,'-nE1V:-2 t E~v,"-'

753

EXERCISE 12.4.23

is a unitary operator U in M,. As V;" = I , V, = I!:-",

and

Thus V, = U' E M,. It follows that El and V, generate M , (as do { E j } and Vn). (ii) Let Fj be the projection in M corresponding to the characteristic function of [ ( j - l)m-*,jm-*). By assumption, m = nk for some integer k. Thus, since the correspondence between elements of G and unitary operators in M is a representation of G (from the discussion preceding Proposition 8.6.1),

and, from (i), M , & M,. (iii) From (ii), M,! & M(,+1)! so that U,",lMn! is a * subalgebra 2l of M . Let E be the projection in M corresponding to the characteristic function of the interval [ T , T ' ) , where T and T I are rationals. If T = p / q and T I = p'/q', with p , p ' , q , and qI integers, then E E Mqqt C M(qqt)! from (ii). Since @ is a * isomorphism of A onto @ ( A )@ , is a homeomorphism between A and @ ( A )in their ultraweak topologies. Now projections corresponding to intervals such as [ T , T ' ) in A generate an algebra whose ultraweak closure is A. Thus projections such as E generate an algebra whose ultraweak closure is @ ( A ) It . follows that the ultraweak closure of 2l contains @(A).As V;;-1)! - V, and Vnmis the unitary operator in M corresponding to translation by the rational number m / n , V T E M,! 5 2l. Hence the ultraweak (and strong-operator) closure of M is M . (iv) Since M is a factor of type 111 and is the strong-operator closure of an algebra that is the union of a sequence of finite type I rn subfactors of M , M is the unique matricial factor of type 111. 12.4.23. Adopt the notation of Exercise 12.4.22, and let Go be the set of translations modulo 1 by dyadic rationals. Show that (i) Go is a subgroup of G; (ii) Go acts ergodically on S [Hint.Modify the argument of the last paragraph of Example 8.6.12.1;

754


(iii) the von Neumann subalgebra Mo of M generated by the operators corresponding to the multiplication algebra A of (S,m) and the unitary representation of G restricted to Go is a factor of type 111 distinct from M; (iv) Mb n M = @I. [Hint. Note that a maximal abelian subalgebra of M is contained in Mo.] Solution. (i) The dyadic rationals in [0, 1) form the subgroup Go of G generated by {2-n : n E N}. (ii) The argument of the last paragraph of Example 8.6.12 applies to show that Go acts ergodically, when we replace ‘G’ by ‘Go,’ ‘rational’ by ‘dyadic rational,’ and ‘n’ by ‘2n,’ and note that the a-algebra S is generated by the subintervals of [0,1) with dyadic rational endpoints, of lengths 1,1/2,1/4,. . . . (iii) From (ii) Proposition 8.6.9, and by an argument closely analogous to the last paragraph of the proof of Proposition 8.6.1, it follows that Mo is a factor. (Note that Proposition 8.6.1 itself does not give this result, because the present construction of Mo, from A and the unitary representation of Go, is not quite the one envisaged in that proposition. Indeed, the operators that generate Mo, @(A) (A E A ) and V ( g )( g E Go), act not on the Hilbert space associated with A and Go, but on the larger space associated with A and the ) maximal abelian (in M, hence in full group G. Nevertheless, @ ( A is M o ) , equation (1) preceding the proof of Proposition 8.6.1 remains valid, and the argument of the final paragraph of that proof can be used in the present context.) Since M is of type 111 and Mo is an infinite-dimensional subfactor of M, Mo is a factor of type 111. The matrix of the unitary operator V in M corresponding to translation modulo 1 by 1/3 on [0,1) has the unitary operator (on Lz([O, l), m)) corresponding to this translation as its (1/3,0)-entry. (See the discussion preceding Proposition 8.6.1.) The matrix of the unitary operator (in M o ) corresponding to an element of Go has 0 as its (1/3,0) entry. Thus each element of the strong-operator-dense subalgebra of Mo generated by the operators @(A) (A in A) and the unitary representation of Go has 0 as its (1/3,0) entry. Hence the matrix of each operator in Mo has 0 a6 its (1/3,0) entry, and V 4 M o . It follows that M Ois distinct from M. (iv) By construction, @ ( A ) Mo,where A is the multiplication algebra of ( S , S , m ) . From Proposition 8.6.1, @(A)is a maximal abelian subalgebra of M. If T E Mb n M, then T E @(A)‘n M and

EXERCISE

12.4.24

755

T E @ ( A )C Mo. Thus T E Mo n ML = C I , since Mo is a factor (from (iii)). Hence ML n M = @I. 12.4.24. With the notation of Exercise 12.4.23, show that (i) Mo is matricial and is therefore, the matricial factor of type 111 [Hint. Use Exercise 12.4.22.1; (ii) V*MoV = Mo, where V is the unitary operator in M corresponding to translation (modulo 1) by in [0, 1) [Hint. Note that G is abelian and use 8.6(1) and 8.6(2).]; (iii) the mapping T + V*TV(T E Mo), with V as in (ii), is an outer * automorphism of Mo [Hint. Use Exercise 12.4.23.1; (iv) each matricial factor of type 111 admits an outer * automorphism. (The fact that Mo is matricial, noted in (i), is also a consequence of a general result [20: Corollary 21: Each subfactor of type I11 of the matricial factor of type 111 is matricial.)

3

Solution. (i) From Exercise 12.4.22, M2n Mzn+l,and V2T is the unitary operator in M corresponding to translation by ~ n / 2 ~ . Thus U?=lM2n (= 90)contains the unitary operators in M corresponding to elements of Go. At the same time, arguing as in the solution to Exercise 12.4.22(iii), the projections in A corresponding to subintervals of [0, 1) with dyadic rational endpoints generate an ultraweakly dense * subalgebra of A, and the image of these projections under @ generate an ultraweakly dense * subalgebra of @ ( A ) . Each of these image projections lies in !2l,-~. Thus, the ultraweak closure of UOcontains @ ( A ) this , closure is Mo, and Mo is matricial. From Exercise 12.4.23(iii), Mo is a factor of type 111, whence Mo is the matricial factor of type 111. (ii) Since G is abelian, V*VznV = V p for each integer n (from 8.6(1)). From 8.6(2), V'@(A)VE @ ( A ) Thus . V*MoV = Mo. If there is a unitary operator W in Mo such that W*TW = (iii) V'TV for all T in Mo, then V W * E ML n M . From Exercise 12.4.23(iv), V W * = aI for some scalar a. Hence V = aW E Mo - contradicting one of the conclusions reached in the solution to Exercise 12.4.23(iii). Thus T + V*TV (T E M o ) is an outer * automorphism of Mo. (iv) From Theorem 12.2.1, each matricial factor of type 111 is * isomorphic to Mo. Since Ma admits an outer * automorphism, each matricial factor of type 111 admits an outer * automorphism.

756


12.4.25. Let R be a von Neumann algebra of type 111. (i) Suppose { N j } is an ascending sequence of distinct type I subfactors of R (each containing I ) . Show that the ultraweak closure of ujNj in R is a matricial type 111 subfactor of R. (ii) Show that each type I subfactor of R is contained in a matricial type 111 subfactor of R.

Solution. (i) Let U be the norm closure of UjAfj. Then U is a matricial C*-algebra and its ultraweak closure Af is a matricial von Neumann subalgebra of R. Since R is finite, JVis finite. From Corollary 12.1.3, N is a factor of type 111. (ii) Let N 1 be a type I subfactor of R. Since R is finite, N 1 is finite and is the linear span of some finite self-adjoint matrix unit system { E j k } . From Lemma 6.6.3, d !{ fl R is * isomorphic to E 1 1 R E 1 1 . It follows from Exercise 6.9.16(iv) that Afi n R is a von Neumann algebra of type 111. From Lemma 6.5.6, we have that N; n R contains a subfactor M 1 of type 12. The subalgebra N 2 of R generated by n/l and M 1 is a (finite) type I subfactor of R. Moreover, NI is contained properly in N 2 . Repeating this argument with N 2 in place of A f l , we construct N 3 , a type I subfactor of R containing N 2 properly. In this way, we arrive at an ascending sequence {Nj} of distinct type I subfactors of R. From (i), UjNj and, hence, N 1 are contained in a matricial subfactor of R of type 111. 12.4.26. Let, R 1 and R 2 be von Neumann algebras acting on Hilbert spaces 'HI and 'H2, and let S 1 and S 2 be subsets of ( R 1 ) 1 and ( R z ) ~respectively. , Suppose A 1 and A 2 are in the ultraweak closures of S 1 and S 2 , respectively. Show that A 1 8 A 2 is in the ultraweak closure of S1 8 S 2 , where s 1 8 S 2

= {SIB s 2 : s 1 E Sl,

52

E SZ}.

[Hint. Use Proposition 11.2.8.1

.

Solution. Let p 1 , . . ,pn be elements of the predual of R I BR 2 and let E be a positive number. We want to find S 1 in S 1 and S 2 in S 2 such that Ipj( A 1

B A 2 - $1 8 S2)1 < E

( j E { 1,.

-

,n } ) .

From Proposition 11.2.8, each pj is in the norm closure of the linear span of normal product states of R 1 8722. Let p i , in this linear span,

757

EXERCISE 12.4.27

be such that llpj - pill < ~ / 4 .Since Sk is contained in the unit ball of Rk (Ic E {1,2}) and this unit ball is ultraweakly closed, we have that IlAkll I 1. If we find S k in Sk such that Ipi(A1 @ Az - Si C 3 Sz)1 < ~ / 2

( j E (1,. . , n } ) ,

then Ipj(A1 @ A2 - S1@ S Z )I~Ipi(A1 @ AZ - S1@&)I t 211pj - pi11 < 8.

x:;)

We may assume that each pj = product state of R1 @ 7 2 2 . Let c be

ajkwjk,where w j k

is a normal

max{lajkl : j { l ~, . . . , n } , k ~{I,... , m ( j ) } } , and m be C;=, m ( j ) . Since A1 and A2 are in the ultraweak closures of SI and S2, respectively, and each wjk is a normal product state, there are operators S1 in S1 and S2 in S2 such that Iujk(A~@ AZ- S i @ S Z )@O €3 z0)ll < &, and T ( U ( E8 F))= 0. (ii) From Exercise 12.4.15(iv), A and B are in the ultraweak closure of P. From Exercise 12.4.26, A €3 B is in the ultraweak closure of P €3 P. (iii) From (i), the ultraweakly continuous mapping T 4 T ( U T ) vanishes on P @ P. Hence

T ( U ( A€3 B ) ) = 0, from (ii). (iv) If a is inner, there is a U as in (i). From (iii) and by linearity and ultraweak continuity of the mapping T + T(UT)on M Q M , T ( U T )= 0 for each T in M 6 M . In particular, 1= T ( 1 ) = T ( U U * ) = 0 -a

contradiction. Thus

a!

is outer.

m[98]

EXERCISE 12.4.20

759

Suppose R is a countably generated, countably de12.4.28. composable von Neumann algebra. Show that (i) (R)1 has a countable, strong-operator-dense subset [Hint. Use the Kaplansky density theorem.]; (ii) each von Neumann subalgebra of R is countably generated. [Hint.Use Exercise 5.7.46 to find a metric on (R)1 whose associated metric topology is the strong-operator topology. Then use separability arguments in conjunction with (i).] Solution. (i) Let (A1 ,A2,. . .} be a countable (self-adjoint) set of generators for R and let Ro be the (self-adjoint) algebra of is (non- commutative) polynomials in these generators. Then ~ Z O strong-operator dense in R. From the Kaplansky density theorem, (Ro)l is strong-operator dense in (R)1. Hence the set of polynomials in {Al, Az,. . .}, with (complex) rational coefficients, in ( R o ) ~is a countable strong-operator-dense subset of (R)1. (ii) From Exercise 5.7.46, (R)l admits a metric d whose associated metric topology coincides with the strong-operator topology on (R)1. From (i), this metric topology on ( R ) Iis separable. If S is a von Neumann subalgebra of R,the restriction of d to (S)1 yields a separable metric topology that coincides with the strong-operator topology on (S)1. A countable strong-operator-dense subset of (S)1 generates S as a von Neumann algebra. Thus S is countably genera ated.

12.4.29. Let R be a von Neumann algebra of type 111. (i) Show that the ultraweak closure of the union of a family of subfactors of R of type 111 totally ordered by inclusion is a subfactor of type 111. [Hint.Use Exercise 12.4.4 and Proposition 12.1.2.1 Suppose R is countably generated (as a von Neumann algebra) with countably decomposable center (for example - R acts on a separable Hilbert space). Show that (ii) each matricial subfactor of R is contained in a maximal matricial subfactor of R [Hint.'Use Corollary 8.2.9 and Exercise 12.4.28 to show that each von Neumann subalgebra of R is countably generated. Use Theorem 12.2.2.1; (iii) each finite type I subfactor of R is contained in a maximal matricial subfactor. [Hint.Use Exercise 12.4.25 and (ii).]

Solution. (i) Let 2l be the norm closure of UM,, where the

760


family { M a : a E A}, consisting of subfactors of R of type 111, is totally ordered by inclusion. From Exercise 12.4.4(iv), U is a (simple) C*-algebra with a, unique trace. Since U R and R is finite, the ultraweak closure ‘u- of U is finite. From Proposition 12.1.2,Z.i- is a factor. Since M a C U and M , is of type 111, U- has infinite linear dimension. Thus U- is a factor of type 111. (ii) Let N be a matricial subfactor of R. Partially order the family .F of matricial subfactors of R containing N by inclusion, and let { M , : a E A} be a totally ordered subset of 7 . Each M , is of type 111. From (i), the ultraweak closure Mo of U,M, is a factor of type 111. From Corollary 8.2.9, R is countably decomposable. From Exercise 12.4.28, Mo is countably generated. The criterion of Theorem 12.2.2 tells us, now, that Mo is matricial. Thus Mo is an upper bound for { M a : a E A} in 7 . From Zorn’s lemma, there is a maximal matricial subfactor of R containing N . (iii) From Exercise 12.4.25(ii), each finite type I subfactor of R is contained in a matricial subfactor of R. Thus, each finite type I subfactor of R is contained in a maximal matricial (type 111) subfactor w[35] of R (from (ii)). 12.4.30. Let R be a von Neumann algebra of type 111, and let M be a maximal matricial von Neumann subalgebra of R. Show that (i) M is a factor of type 111; (ii) M’ n R is a factor if (M’ n R)‘ n R = M ; (iii) (M’ n R)’ n R # M unless R is a matricial factor.

Solution. (i) Since R is finite, M is finite. From Corollary 12.1.3, M is a factor of type 111. (ii) If T is in the center of M ’ n R, then ( M ’ n R)’ n R contains T . If (M’ n R)’ n R = M , in addition, then T E M . But T E M‘. Since M is a factor (from (i)), T is a scalar. Thus M ’ n R is a factor. (iii) Since M’ n R is a von Neumann subalgebra of R and R is finite, M’ n R is finite. If (M‘ n R)’ n R = M , then M’ n R is a factor from (ii), and contains a type I, factor N with n greater than 1 unless M’ n R consists of scalar multiples of I . In this latter case,

R = ( M ’ n R ) ’ n R= M and R is a matricial factor. In the other case, M and N generate a matricial subfactor of R properly larger than M, contradicting the

EXERCISE

12.4.31

maximality of M . Thus (M’ n R)’ n R matricial factor.

#M

761 unless R(= M ) is a

12.4.31. We say that a von Neumann algebra R is normal when each von Neumann subalgebra S of R coincides with its own relative double commutant (that is, (S’ n R)’ n R = 8).Show that (i) R is a factor if R is normal; (ii) R is normal if R is a type I factor; (iii) R is not normal if R is a type 111 factor. [Hint. Use Exercises 12.4.23 and 12.4.30.1; (iv) R is not normal if R is a factor of type 11,. [Hint. Use Theorem 6.7.10.1

Solution. (i) Let C be the center of R. We have

({@I}’ n R)’n R = C = @I if and only if R is a factor. (ii) If R is a type I factor, then R is * isomorphic to f?(‘H), for some Hilbert space ‘H, from Theorem 6.6.1. The property of being normal is clearly preserved under * isomorphism between von Neumann algebras. The fact that B(‘H) is normal is the double commutant theorem. Thus R is normal when R is a type I factor. (iii) From Exercise 12.4.23(iv), the matricialII1 factor is not normal. From Exercise 12.4.3O(iii), R is not normal when R is not matricial. Thus, no factor of type 111 is normal. (iv) From Theorem 6.7.10, R is * isomorphic to n 8 M for some factor M of type 111. From (iii), M is not normal. Let S be a von Neumann subalgebra of M such that (S’n M)‘ n M # S. Then, from Lemma 6.6.2,

( n 8 S)’n ( n 8 M ) = (S’ 8 In)n ( n 8 M ) = (S’ r l M ) 8 I,. Thus [ ( n8 S)’ n ( n 8 M)]’n ( n 8 M )= ( n 8 (S‘n M ) ’ ) n ( n @ M )

= R. 8 [(S’ n M)’ n M ]

#n@S. It follows that R is not normal.

m[35]

762


12.4.32. Let M be a factor of type 111. (i) Show that IlU(H) - U ( H 0 ) l l ~5 211H - H0112 for each pair of self-adjoint operators H and HOin M , where U ( H ) is the Cayley transform appearing in Lemma 5.3.3. [Hint.Use the trace representation and Remark 8.5.9.1 (ii) Suppose h is a continuous red-valued function vanishing at 00 on W. Let { H , } be a sequence of self-adjoint operators in M such that {llHn - Ho112} tends to 0 as n tends to 00. Show that { h ( H n ) } is strong-operator convergent to h ( H 0 ) . [Hint.Use Exercise 8.7.3, the proof of Theorem 5.3.4, and (i).] (iii) With HO a self-adjoint operator in M , h as in (ii), and E a positive number, find a positive 6 such that Ilh(H) - h(Ho)ll2 < E provided that llH - Holl2 < S and H r~ M . (iv) Suppose M is countably generated and for each finite subset { A l , . .. ,A,} of M and positive E there are a finite type I subfactor JVof M and operators B1,. . ,B, in N such that

.

( ( A j- Bjll2 < E Show that

M

(J' E { ~ , * * * , Z J } ) *

is matricial. [Hint.Use (iii) and Theorem 12.2.2.1

Solution. (i) We may suppose that M acting on the Hilbert space 'H is the trace representation of M . In this case, there is a generating trace vector xo in 3-t for M and llAll2 = IIAzoll for each A in M . As in the proof of Lemma 5.3.3,

Il(U(H>- U(HO)).Oll 5 2ll(H - HOMO

+ iq-l~oll.

Thus, from Remark 8.5.9,

+ + ir)-lll

IlU(H) - U ( H O ) l l 2 5 211(H - Ho)(Ho - 211H - Holl2ll(Ho 5 2 1 p - HOll2.

W 1 l 1 2

(ii) Under the assumptions and from (i),

IlU(Hn) - U(HO)ll2

0.

From Exercise 8.7.3(iii), { U ( H , ) } is strong-operator convergent to U ( H 0 ) . Arguing as in the proof of Theorem 5.3.4 (with the notation of that proof), we have W n )

= fW(Hn))

+

f(U(H)) = h ( H )

763

EXERCISE 12.4.32

in the strong-operator topology as n tends to 00. (iii) From (i), IlU(H) - t Y ( H o ) l l 2 -+ 0 as IIH - Holl2 + 0 where H E M h . Thus, from Exercise 8.7.3(iii), the mapping H + U ( H ) is continuous at Ho from M h in the (1 Ila-metric topology to M in its strong-operator topology. Again, arguing as in the proof of Theorem 5.3.4 (with the notation of that proof), we have that

H --t U ( H ) --t f ( U ( H ) ) = h ( H ) is continuous at Ho from M h in the 11 112-metric topology to M in its strong-operator topology. Hence there is a positive 6 with the desired property. (iv) To apply the criterion of Theorem 12.2.2, we choose a finite subset { T I , .. . ,T p } of ( M ) 1 and let f be the function (defined in the second paragraph of the proof of Theorem 5.3.5) that assigns t to t in [-1,1] and t-' to each other real t . Then f is continuous on R and vanishes at 00. Let pj be the operator in 2 @ M with matrix

*;[ 21 and let

be a positive real number. Then f'j is self-adjoint and IIpjll 5 1. Thus f(f'j) = F j . From (iii), there is a positive 6 such that Ilf(Fj) < E provided llf'j - f i l l 2 < 6 and I? is a self-adjoint operator in 2 @ M . By assumption, there are a finite type I subfactor N of M and operators 5'1,. . . ,S, in N such that llTj - S j l l p < 6 for each j in (1,. . . , p } . Let $ j be the operator in E

f(a)ll,

Then

llf'j-$jll; = $(I(Tj-Sjlli - k ~ ~ T ' ' - S =~ llTj-Sj11;, ~~~)

whence

llTj - f ( 3 . d l l 2 = llf(p;.>- f ( 3 j > l l 2 < E ( j E (1,. . ., p } ) . Now f(3j)is a (self-adjoint) operator in (2 @ N ) , . Let

;:[ 31 be its matrix. Then

Rj

E

(N)1and

IIT~- RjII%s i(IIfiIIi + 2 1 I ~-j fijII4 + IIsII%)= IIpj - f<sj>IIg< E' for each j in (1,. . . , p } . Thus Theorem 12.2.2 applies, and M is matricial. .[78(Lemma 1.5.3, pp. 726-728)]

764


12.4.33. Let M be a countably generated factor of type 111, and let S be a self-adjoint subset of ( M ) l that generates M as a von Neumann algebra. Suppose that for each positive E and each finite subset (5’1,. . ,S,} of S,there are a finite type I subfactor N of M and operators T I , .. . ,T, in (N)1 such that llSj - Tjllz < E for each j in { 1,...,n}. Show that M is matricial.

.

Solution. I-f A1,. . , , A , are in ( M ) 1 , there are operators SI,. . . ,S, in S and (non-commutative) polynomials p l , .. . , p m in n variables such that

since S is self-adjoint and generates M . By strong-operator continuity of p j on ( M ) 1 and Exercise 8.7.3(iii), there is a positive 6 such that if llSj - Tjll2 < 6 for some T I , .. . ,T, in ( M ) 1 ,then

By assumption, there is a finite type I subfactor erators 2 - 1 , . . ,T, in (N)1 such that llSj - Tjll2 (1,. . ,n } . Let BI, be p k ( T 1 , . . . ,T,). Then

.

.

N of M and op< 6 for each j in

and BI, E N , for each k in (1,. . . ,m}. From Exercise 12.4.32, M is matricial. 12.4.34. Let N be a type I subfactor of a factor M , E be a minimal projection in N , and Mo be the set of elements in M that commute with all elements of N . Show that (i) E M E = MoE [Hint. Use Lemma 6.6.3.1; (ii) ( E 8 E ) ( M @ M ) ( E8 E ) = ( M o @ M o ) ( E8 E ) ; (iii) $ = MoGMo, where N is the set of elements in M @ M that commute with N @ N . [Hint. Consider the mapping f -+

F ( E 8 E ) (F E #).I

Solution. (i) Since N is a factor of type I and E is a minimal projection in N , there is a self-adjoint system ( E a , b ) a , b E ~ of matrix

EXERCISE 12.4.34

units for N such that Of course,

x b ( = @Eb,b =

765

I and E = Ea,afor some a in B.

MoE = EMoE C E M E . Suppose T E E M E . From Lemma 6.6.3, in Mo. Hence,

T =TE =

z b ( = B Eb,aTEa,b

(= R ) is

(r

Eb,aTEa,b)Ea,a = RE E MoE.

b€B

It follows that E M E C MoE, whence E M E = MoE. (ii) Since Mo@Mo C M G M and E 8 E in M Q M commutes with MoBMo, we have that

(MoQMo)(E€3 E ) C ( E 8 E ) ( M @ M ) ( E8 E ) . On the other hand, for each A and B in M , ( E 8 E ) ( A€3 B ) ( E €3 E ) = ( E A E ) CB ( E B E ) E ( E M E ) @ ( E M E = ) (MoE)@(MoE), and ( E €3 E ) ( M G M ) ( E@ E ) is generated by { ( E 8 E ) ( A8 B ) ( E €3 E ) : A , B E M } as a von Neumann algebra. Now ( M o E ) Q ( M o E )is generated as a von Neumann algebra by operators of the form T E €3 S E (= ( T 8 S ) ( E€3 E ) ) with T and S in Mo, and (Mo@,Mo)(E8 E ) is generated as a von Neumann algebra by operators of the form (7’8S ) ( E 8 E ) with T and S in Mo. Thus

( E 8 E ) ( M @ M ) ( E8 E ) = ( M o E ) @ ( M o E=) (Mo@Mo)(E8 E ) . (iii) Since T 8 S E # when T and S are in Mo, Mo@Mo C N . Suppose F E N . Then F ( E €3 E ) = ( E 8 E ) F ( E 8 E ) E ( E8E)(M@M)(E 8 E ) = (Mo@Mo)(E8 E ) .

Thus, there is

in Mo@Mo such that

(F-

S ) ( E @E ) = 0. But - 3 E ( N @ N ) ’E, 8 E E N @ N ,and N @ N is a factor. From Theorem 5.5.4, T - = 0 (since E 8 E # 0 by choice of E ) . Hence T = E MoGMo and JV Mo@Mo. It follows that # = Mo@/Mo.

s

s

766


12.4.35. Suppose M is a von Neumann algebra of type 11, and cr is the * automorphism of M @ M (described in Exercise 11.5.25) that assigns B 8 A to A @I B for each A and B in M . (i) Suppose (Y is inner. Show that M has the form n 8 M o , where n is an infinite cardinal and Mo is a factor of type 111. [Hint. Use Exercise 12.4,19 and Theorem 6.7.10.1 (ii) Show that if a is inner, then cr restricted to (Mo 8 I,)@(Mo 8 I,) is inner. [Hint. Use (i), Corollary 9.3.5, and Exercise 12.4.34.1 (iii) Conclude that Q is outer (that is, not inner). [Hint.Use the result of Exercise 12.4.27.1 Solution. (i) From Exercise 12.4.19, M is a factor. From Theorem 6.7.10, M is * isomorphic to n @ Mo for some infinite cardinal n and some factor Mo of type 111. (ii) From (i), we may suppose that M is n @I Mo. Let N be the type I, subfactor of n @I Mo consisting of those matrices each of whose entries is a scalar multiple of I . Then Mo @ I , is the set of elements of M that commute with N . With N @ N considered as a von Neumann subalgebra of M Q M , Q restricted to N @ N is inner, from Corollary 9.3.5, since N @ N is a factor of type I. Let V be a unitary operator in N63N that implements crI(NQN)and let U be a unitary operator in M @ M that implements a. Then UV-l is a unitary operator in M @ M that commutes with NQN. From Exercise 12.4.34,

and UV-' implements crl(M0 8 I,)&(Mo 8 I,). (iii) Since Mo@I, is afactor of type 111, ~ ~ I ( M O @ I ~ ) & ( M O @ I , ) is outer from Exercise 12.4.27 - contradicting the conclusion of (ii). m[98] Hence, cr is outer. 12.4.36. Let R and S be von Neumann algebras and p and u be non-zero elements of R, and S, respectively. Show that (i) there is a unique element p @ (T of (R@S), such that ( p 8 a ) ( R8 S) = p(R)o(S) for each R in R and S in S and that IIp 8 611 = llpllllall [Hint.Use Theorem 11.2.10 to consider R and S in their universal normal representations. Then apply Corollary 7.3.3.];

767

EXERCISE 12.4.36

(ii) there are unique operators @,,(F) and Q p ( F )in respectively, corresponding to each r;?l in R @ S ,satisfying p ’ ( @ u ( T ) ) = (p’

63 u ) ( f ) ,

R and S,

u ’ ( Q p ( f ) ) = ( p 63 a’)@.)

for each p’ in R, and each u’ in S, [Hint. Recall that R and S are the norm duals of R, and S,.]; (iii) aU and Q p (as defined by (ii)) are ultraweakly continuous linear mappings of R@S onto R and S,respectively, satisfying

@ , ( ( A 8 I ) F ( B @ I ) ) = A@,(F)B, Q p (8(C I ) T ( I 63 D))= C Q p ( F ) D for each 5? in R@S,A , B in

R,and C, D in S,and that

@,(R63 S) = u ( S ) R ,

Q p ( R63 S) = p(R)S

when R E R and S E S; (iv) Qju(5?) E Ro and q p ( F )E SOiff E %@SO, where Ro and SO are von Neumann subalgebras of R and S, respectively [Hint. Consider, first, the case where F = Ro @ SOwith Ro in Ro and SO in SO.Then use (iii).]; (v) p E Ro@& if @ u ! ( f ) E Ro and Q p t ( pE) SOfor each u‘ in S, and each p‘ in R,. [Hint. With A‘ in Rb,show that

((A‘ 63 I ) f ( z63 y),

8 v ) = ( f ( A ‘ 63 I ) ( z 8 y ) , u 63 v)

for all z and u in 7-1 and y and TJ in K . Let u‘ be W ~ , ~ p’[ Sbe, w ~ , A I * and ~IR p”, be W A I ~ for , ~ this. ~ R Use , Theorem 11.2.16.1

Solution. (i) Proceeding as indicated in the hint, Theorem 11.2.10 assures us that we may consider R and S in their universal normal representations on Hilbert spaces 7-1 and Ic, respectively, without loss of generality. In this case, there are vectors z and y in 7-1 of length llp11*/2 and u and v in K of length lla111/2such that p = o , , ~ ~and R u = w,,,IS from Corollary 7.3.3. The equation ( p 8 u ) ( T )= ( f ( x 8 u),y

8 v)

(* E R @ S )

defines an ultraweakly continuous linear functional p 8 u on R @ S . With R in R and S in S , we have ( p @ a ) ( R @ S= ) ( ( R @ S ) ( z @ u ) , y @= v ) (Rx,y)(Su, v) = p(R)u(S).

768 If R E

A.PPROXIMATION BY MATRIX ALGEBRAS (R)1and

S E (S)1, then RC?JS E ( R @ S ) 1and

Hence llpllllall L I(p 8 011. we have

On the other hand, with

5? in (R C?J S)l,

Thus Il@~gllI Ilpllllall, and IIpQ9all = Ilpllllull. Since operators of the form R 8 S, with R in R and S in S, generate an ultraweakly dense linear submanifold of RBS,there is at most one linear functional on R@S with the properties prescribed for p Q9 u. (ii) From the uniqueness clause of (i), (up t p’) 63 0 = “ ( p €3 u ) t p’ Q9 6.

Thus the mapping : p’

--f

(p’ €3 a)@)

(p’ E R,)

is a linear functional on R x . Since

@,(f),as defined, is an element of (R,),.Thus @,(5?) Theorem 7.4.2. Symmetrically, (iii) Since

E R from

Q P ( F E) S.

p ’ ( 9 ( u F t F‘)) = (p‘ @ fJ)(uFt F‘) = a ( p ’ 8 o)(F)t (p’ Q9 a)(F’)

= .p’(@u(F)) t p’(%(F’)) = p’(a@,(F) t a@’))

for each p’ in R,, 9, is a linear mapping. Moreover,

F + p ’ ( @ , ( F ) ) = (p’ C?J u ) ( F )

(i;E R @ S )

EXERCISE 12.4.36

769

is continuous from R @ S with its ultraweak topology to C for each p’ in 72,. Hence 9, is an ultraweakly continuous linear mapping from R@S into R. With A , B , and R in R,S in S, p’ in R,, and p“ the element of R, whose value at C in R is p’(ACB),we have

p ’ ( @ u ( ( A8 I ) ( R8 S ) ( B 8 I ) ) )= (p’ 8 .)((ARB) 8 S ) = p‘(ARB)a(S) = p”( R).( S )

= (p” 8 .)( R 8 S ) = p”(%(R 8 S ) )

= p’(A@,(R 8 S ) B ) .

Thus

@ , ( ( A 8 I ) ( R8 S ) ( B8 I ) ) = A @ , ( R 8 S ) B . Now the mappings

are ultraweakly continuous linear mappings of R@S into R that agree on generators of an ultraweakly dense linear submanifold of [email protected] they agree on [email protected] symmetric argument applies to Q,,and the first relations set out in (iii) are established. With R in R,S in S, and p‘ in R,, we have p’(@,(R 8 S ) ) = (p’ 8 .)(R 8 S ) = p’(.(S)R).

Thus 9,(R 8 S) = o ( S ) R and, symmetrically, Q J R 8 S) = p(R)S. It follows that 9, maps onto R and Q , maps onto S. (iv) From the last relations established in the solution to (iii), 9 , ( T ) E Ro when p = Ro @I S with Ro in Ro and S in S. Since operators of the form & 8 SO (& E Ro,SOE SO)generate an ultraweakly dense linear submanifold of Ro@Soand 9, is an ultraweakly continuous linear mapping, 9, maps Ro@So into Ro. Symmetrically, Q, maps Ro@& into So. (v) Suppose the operator 5? in R@S is such that 9,,l(p) E Ro and !J!,,,(f’) E SO for each u’ in S, and p’ in R+. Proceeding as

770


outlined in the hint, with the notation introduced there, we have

((A' 8 I ) F ( z 8 Y ) , U 8 v) = (p' 8 Q')(F)= ~ ' ( @ ~ l ( f ' ) ) = ( A ' @ o l ( F ) z , ~=) ( ! b o l ( F ) A ' ~ , ~ ) = p"(iDo,(F)) = (p" 8 Q ' ) ( f ' ) = (F(A'z 8 y ) , 8 ~ u) = @(A' 8 I ) ( z 8 y), u 8 v).

f' commutes with RI, 8 @ I . Symmetrically, f' commutes with @[email protected] f' E (RI,@Sh)'.From Theorem 11.2.16 and the double Thus

commutant theorem,

Hence

f' E Ro@So.

~[115,116,117]

12.4.37. Let R and S be von Neumann algebras acting on Hilbert spaces 'FI and K , respectively. Suppose Ro and So are von Neumann subalgebras of R and S, respectively. Show that (i) (Rhn R)@(SA n S)= (Ro@So)'n(R@S)[Hint. Use Exercise 12.4.36(v).]; (The result of Exercise 12.4.34(iii) is a special case of this formula.) (ii) A@Bis a maximal abelian subalgebra of RIBS when A and B are maximal abelian subalgebras of R and S,respectively; (iii) C@D is the center of R@S when C is the center of R and D is the center of S by using (i); (iv) Rh n R = @I and S A n S = @I if and only if we have that (Ro@So)' n (REIS)= @ I .

Solution. (i) If R E RI, n R and S E Sh n S, then R@S commutes with [email protected]

(2; n R)@(s: n S)G ( R O @ ~nO(R@S). )' Suppose f' E (ROBSO)' n (RIBS),A E Then, from Exercise 12.4.36(iii),

Ro,p

E R w , and

Q

E

&.

EXERCISE 12.4.38

771

Hence AQ?,(F)= @ , @ ) A and a,@) E Rb n R for each 0 in S,. Symmetrically, ?Z,,(T)E SA n S for each p in R#. From Exercise 12.4.36(v),

f E (R;n R)@(s; n s).

Hence

(&@so)’n (1263s)c (Rbn a)@(&, n S).

Combining this with the reverse inclusion, noted earlier, we have the desired formula. (ii) Since A and B are maximal abelian in R and S, respectively, A’ n R = A and B’n S = B. Thus

(d6W)’n (Rt?~S)= (A’ n 7Z)B(f3’f~S ) = A@!?, from (i), and d @ Bis maximal abelian in R@S. (iii) From (i),

But (RQS)’n (‘RQS) is the center of R@S. (iv) If the tensor product of two von Neumann algebras is the algebra of scalar multiples of I, then each of the von Neumann algebras is the algebra of scalar multiples of I. From (i), then, each of Rb n 72 and Si n S is CI if and only if

12.4.38. Let R be a matricial von Neumann algebra, { M n } a generating nest of finite type I factors for R,and X # a dual normal Rmodule (as described in Exercise 10.5.13).Choose a compatible selfadjoint system of matrix units for M,, and let 6, be the set (finite group) of unitary operators in M, whose matrix representations relative t o these matrix units have only 1 and -1 as non-zero entries. Let Q be the locally finite group U,Q, and p an invariant mean on Q (as described in Exercise 3.5.7). (i) Show that the linear span of G is U,M,; conclude that this linear span is ultraweakly dense in 72. (ii) Let X be the predud of X# and fz(U) be [U*S(U)](z)for each U in 6 and z in X, with 6 a derivation of R into X#, Show

772

that

APPROXIMATION BY MATRIX ALGEBRAS fx

is bounded on Q and that

2

---t

p(fz) is an element po of X#.

[Hint.Use Exercise 4.6.66.1 (iii) With the notation of (ii), show that ( V E Q).

6(V) = vpo - p o v

[Hint.Define g z ( U ) to be [V*U*S(U)V](o)and fz(U)to be fz(UV). Use the fact that 6 is a derivation and p is an invariant mean t o establish the desired relation.] (iv) Show that 6 ( A ) = Apo - poA for each A in R. [Hint.Use (iii) to establish this for each A in the linear span of 9. Then use (i) and Exercise 10.5.1.3.1 (v) Note that the result of (iv) is valid when R is assumed just to have an amenable group (one having an invariant mean) of unitary operators whose linear span is ultraweakly dense in R. Show that R has such a group when it is the ultraweak closure of an ascending family of finite-dimensional self-adjoint subalgebras, and in particular, when R.is abelian and acts on a separable Hilbert space. Solution. (i) Each matrix unit in M , of the compatible system has the form ( U V)/Z for two elements of G,. Thus the linear span of 9, is M , and the linear span of 0 is U,M,. By assumption, u,M, is ultraweakly dense in R. (ii) Since U*6( U ) E X#, fz is a complex-valued function on G. From Exercise 4.6.66, 6 is bounded (continuous). By definition of “Banach R-module,” the mapping

+

(T,p)

--$

Tp : R

X

X”

--t

X#

is a bounded bilinear mapping. Let b be its bound. Then, for each U in 9, If&U = I[U*S(Vl(4l5 ~ 1 1 ~ 1 1 1 1 ~ 1 1 , and fx is a bounded function on Q. With po as described,

EXERCISE 12.4.38

773

(iii) Adopt the notation of the hint. Then ~(f,)= p(fJ). From the assumption that X# is a dual R-module, the mapping p

--*

v*pv:

X”

+ X’

is weak * continuous. From Theorem 1.3.1, there is an element z’in X such that p ( z ’ ) = ( V * p V ) ( z for ) each p in Xy. Thus, for each U in 6, g,(U) = [V*U’6(U)V](Z) = [U*S(U)](z‘)= f,l(U), and gz = f,~. It follows that

Let c, be the constant function on 6 whose value at each element is [V*G(V)](s). With IJ in 6, we have that

Hence :f

= c,

+ g,,

and po = V * S ( V )iV*poV. It follows that

6 ( V )= VPO - pov

(V E 6).

(iv) Since the mappings 6 and A -, Apo - poA are linear mappings of R into X’ that agree on 6, these mappings agree on U,M,, the linear span of 6. At the same time, both mappings are continuous from R in its ultraweak topology to X” in its weak * topology. (That 6 is continuous is a consequence of Exercise 10.5.13(v); the mapping A + Apo - poA is continuous by virtue of the assumption that X# is a dual ~ r m R-module.) d As these mappings agree on an ultraweakly dense subset of R, they agree on R.

774


(v) Nothing in the reasoning of the solutions to (ii), (iii), and (iv), requires more than the assumption that R has an amenable group Q of unitary operators whose linear span is ultraweakly dense in 72. If {an} is an ascending sequence of finite-dimensional selfadjoint subalgebras M, of R and Un%, is ultraweakly dense in R, then each 2ln is a finite direct sum of finite type I factors from Proposition 6.6.6. We can choose self-adjoint systems of matrix units for the factor summands of On so that the system for %,+I is compatible with the system for a,. Again, if we form the finite group 8, of unitary elements of 8, each of whose factor components are unitary elements of those factors with matrix representations whose only non-zero entries are 1 and -1, then 8, E Qn+1 and 8, has linear span a,. Thus U,8, is locally finite, has an invariant mean, and has linear span ultraweakly dense in R. If 'R is abelian and acts on a separable Hilbert space, Exercise 9.6.41 and Theorem 5.2.2(v) allow us to express 'R as the ultraweak closure of an ascending union of finite-dimensional (abelian) C*-subalgebras. 4501 12.4.39. Let 3-1 be a separable Hilbert space and 3-1, be the n-fold tensor product 3-1 8 .@ 'Ifof 3-1 with itself. Show that (i) there are operators U,,and S, on 3-1, such that

-.

.

for all 21,., .,xn in 3-1, where 0 is a permutation of (1,. . ,n } and x(a)is its sign ( ~ ( 0=) 1if a is even and x(a) = -1 if a is odd), with U,,a unitary operator and S; a projection on 3-1, [Hint. Establish that ( $ 1 , . . . , Z n ) --+ x,,(1) 8 * @ xu(,) is a weak Hilbert-Schmidt mapping of 3-1 x . x 3-1 into H, and apply Theorem 2.6.4. Note that V,,tU, = U,,,t, ~(aa') = x(a)x(a'),and x(0-l) = ~(a).]; (ii) ($1 A . * . A x n , y l A...Ayn)=det((zj,Yk)),

-

-

where x l A - - + A x n= ( n ! ) ' / 2 S , - ( x l @ . ' . @ x , ) ; (The vector 2 1 A * - - A z n in Nnis referred to as the exterior or wedge product of 21,. .,x,.) (iii) x 1 A * . A zn = 0 if and only if {q . , ,x,) are linearly dependent and ( X I A A x,, y1 A A pn) = 0 with x1 A * * - A x , a non-zero vector in 3-1, if and only if there is a non-zero vector in [q,.., x , ] orthogonal to [yl,. . ,y,] [Hint. Use (ii).]; a

.

--

-

.

.

.

775

EXERCISE 12.4.39

(iv) ( 5 1 , . . . ,5,) + 2 1 A - .. A 5, is an alternating multilinear mapping A of 7-f x ... x 'H into the range 'H?) of S; (that is, A is linear in each coordinate and for each permutation u of { 1,. . . ,n } A(51,-*,%)

=X(M&7(1)

€3*..€3'5u(,)),

and there is a (unique) bounded linear mapping 6 of 'H?) into K such that a = 6 o A when a is a weak Hilbert-Schmidt, alternating, into a Hilbert space K [Hint.Use multilinear mapping of 'H x . . Theorem 2.6.4 to express a as iiop with & a bounded linear mapping of H ' , into K . Show that &S; = &, and let 6 be ( ~ t ! ) - ' / ~ i i . ] ; < j ( n ) }is an orthonormal (v) {ej(l) A ... A ej(,) : j(1) < basis for the range 'H(,") of S;, where {em} is an orthonormal basis for 'H [Hint.Use (iii) and (iv).]; (vi) there is a unique bounded linear mapping a,(s)* of 'H(,")into 'H(,il that assigns x A 21 A - A 2, to 21 A - . A 2, for all 21,... ,x, in 7.1. [Hint.Let 2 be e l , with { e m } an orthonormal basis for 'H, and show that ( x l , ...,2,) ---+ x A x 1 A . - . A x , is a weak Hilbert-Schmidt, alternating, multilinear mapping of 3-1 x . . - x 'H into Use (iv).] e 3 - 1

+

x?il.

Solution. (i)

For each permutation u of (1,. ..,n } (%..*,%

(21,...,2,)~2,(1)€3...€3~,(,)

is a bounded multilinear mapping L of 'H x ll"u(1)

€3

* * *

€3 %(n)ll =

E'H)

x 'H into H ' , since

*..

11~111

IIZnll.

With {em} an orthonormal basis for 'H and u a vector in H ',

C

I(ej(,,(l)) B

*

- B ej(,(,)), u>12= IIuI12,

j(l),... A,)

by Parseval's equation. Hence L is a weak Hilbert-Schmidt mapping. From Theorem 2.6.4, there is a unique bounded linear transformation U, of H ' , into H ' , such that L = U, o p, where p(51,.

. ,2,) *

= 21 €3

* . .€3 2,

(51,.

. .,x, E 'H).

Thus U , ( a 1 € 3 * . - € 3 2 2 ,= ) Zu(1) € 3 " " € 3 ' " u ( n ) .

776


Since U, maps the. orthonormal basis (ej(1) €3 U,,is a unitary operator. Note that, with XI,. . ,x, in R,

- - - @ ej(n)}onto itself,

.

Thus U,IU, = U,,I. Note, too, that x ( 0 - l ) = ~ ( 0 )It. follows that

~ ( 0 0 '= )

U

x(a)x(a'), and that

0

and that (S,)2

= ( n ! ) - 2~ X ( 0 ) X ( 0 ' ) U 0 U u I

= (n!)-1

c s, U

=s .,

Hence S , is a projection on X,. (ii) Using the fact that S; is a projection, we have

-

-

(iii) From (ii), (21 A A x,, y1 A . A yn) = 0 if and only if there are scalars c1,. .. , c n , not all 0, such that n

n

j=1

j=1

777

EXERCISE 12.4.39

(that is, if and only if, the rows of the matrix ((sj,yk)) are linearly dependent). In particular, with zk in place of yk, we have n

n

n

n

C;==,Cjzj = 0, if and only if $1 A * . z1 A - - - A x, = 0 if and only if {q,...,z,}

= 0. Therefore are linearly dependent. At the same time, if 51 A . A 2, and y1 A . - A yn are orthogonal cjxj in and x1 A - A x, # 0, then there is a non-zero vector C:==, [XI . . . ,~ c , orthogonal ] to [yl, . . ,yn]. Let T be the permutation of { 1,. . .,n} that interchanges the (iv) (distinct) elements j and k and leaves fixed all other elements. We have, from (i), that and

-

*

A

Zn

.

Thus A is an alternating multilinear mapping. Since a is, in particular, a weak Hilbert-Schmidt mapping, there is a bounded linear mapping i5 of H ' , into K such that (Y = i5 o p from Theorem 2.6.4. Since a is alternating

Thus &U17= x(o)&and

Now A = (TZ!)~/~S; o p. Thus, if & = (n!)-lI2ii, we have

778


If & o A = &I o A for some bounded linear mapping &’ of Ic, then & = &I since { X I A * * . A x , : x i , . ..,x,

into

€31)

generates a dense linear submanifold of H?). 8 ej(,) span a dense (v) Since vectors of the form ej(l) @I linear manifold in H,, the vectors ej(1) A ... A ej(,) span a dense linear manifold in H?).From (iv), ej(1) A ’

--A

ej(,)

= fek(1) A

*

-

*

A ek(,)

. .. ,j ( n ) }= {k( l),. . .,k( n)}. Hence

when { j (l),

{ejp) A

... A ej(,)

: j(1) < j ( 2 )

l), let mk be kq 1. Then

+

In this case, we take 6 to be lsin(pn/q)I. So far we have shown that, if (t/2r) is not an integer, we can choose {mk} and b with the properties set out in the hint. With nk defined as in the hint, it is apparent that nk 2 2 and the sequence {nk} is non-decreasing and diverges to 00. Moreover lognk < mk 5 log(nk

+ 1)

log(2nk) = lognk log2

+

<mkt1,

b, = [logn] = mk

(nk

+ 1 I n 5 2nk).

Note also, from the above inequalities, that emk < 2nk.

802

CROSSED PRODUCTS

Thus

Since nk ---t 00, it now follows that the series CFGle m b n sin2($b,t) does not satisfy the Cauchy criterion for convergence, and so diverges. 13.4.14. With the notation of Theorem 13.1.15, suppose that, for n = 1,2,. . ., b, = n!

when [en!]< T 5 [e(ntl)!],

where [z]denotes the largest integer not exceeding z. Show that T(7rp(!21)-) contains each rational multiple of 27r but is not the whole of R.

Solution. If s is a rational number, n!s is an integer for all sufficiently large n, and thus b,s is an integer for all sufficiently large T . Hence the series

converges, since it has only finitely many non-zero terms. By Theorem 13.1.15, 27rs E T(7rp(!21)-). When [en!]< T 5 [e(n+l)!],we have

+

so CEl e - b r diverges. Since a, = (1 e b r ) - l , it follows easily that a, 0 but ~ ~ l diverges. u , From Exercise 13.4.11(iii), --$

T ( A P ( W )

#

’

803

EXERCISE 13.4.15

13.4.15.

With the notation of Theorem 13.1.15,suppose that

r=l

r=l

Show that the factor xP(21)- is of type 11,.

[Hint. Let

r=l 00

r=l

r=l

Identify rU,with rP,(210) and 2le with rpe(21,) (see the first paragraph of the proof of Theorem 13.1.15) so that

for some cyclic unit vectors, 2, for ?2ioand 5, for Z e , and 2ll, U, are factors of types I,, Ill, respectively, by Exercises 13.4.10 and 13.4.9. Show that there is a * isomorphism cp from 2l onto 121,@ & such that P = W X , @ X , 0 cp. Deduce that the C*-algebra ~ ~ ( is unitarily 3 ) equivalent to 210@U,.]

Solution. We use the notation introduced in the hint, and define C*-subalgebras Br ( T = 1,2,...) of rU,63 2le by

The equations

.

define * isomorphisms 6, from Mr onto Br, for T = 1,2,.. . Since 2Lo @I 21e is generated by the commuting family (a,} of finite type

804

CROSSED PRODUCTS

I factors, it follows from Corollary 11.4.10 that there is a * isomorphism $1 from !2& 63 2& onto @Fp=,B,, with the property that $1(B,) is the canonical image of f3, in @FIBr, for r = 1,2,. . . Similarly, with the property there is a * isomorphism $2 from U onto @El%, By means of that &(a,)is the canonical image of Ur in @,*“=1Ur. and U, 8 U e the mapping $1 and $2, we can identify U with @El%, with @1Fp=~f3,.With these identifications, it follows from Proposition 11.4.5(ii) that there is a * isomorphism cp from 31 onto U, 8 Ue, such that cp 1 U, = 8, for r = 1,2,. .. When A1 E U1,. .., Azr E U2r, we have

.

.

and

Since U is the norm-closed linear span of the set of all operators of the form A1 A2 -..A2,-1A2, (with Aj in Uj for each j, and r any positive integer), it now follows that

Since zo 8 ze is a cyclic vector for M, 8 Me (= cp(U)), it now follows that cp is a representation of U unitarily equivalent to T,, (see Proposition 4.5.3). Thus rp(?21) is unitarily equivalent to cp(2l) (= U, @ a,,), and rp(U)- is unitarily equivalent to the von Neumann tensor product U; @ U;. From the table preceding Proposition 11.2.19, U; I$IU; is of type 11,, so the factor r p ( U ) - is of type 11., rn 13.4.16. Suppose that M is a countably decomposable semifinite von Neumann algebra and ( 0 1 ) is the modular automorphism

805

EXERCISE 13.4.17

group corresponding to a faithful normal state w of M . Show that the continuous crossed product R ( M ,u ) is unitarily equivalent to M @ A , where A is the multiplication algebra corresponding t o Lebesgue measure on R. [Hint. By Theorem 9.2.21, there is a continuous unitary representation t + U ( t ) : R -+ M such that U ( t ) implements ut. Interpret R ( M ,0)as an implemented continuous crossed product (Definition 13.2.6) and use Theorem 13.2.8 and Proposition 13.2.7(i).]

Solution. The implemented crossed product R ( M ,u ) (which is unitarily equivalent to the abstract crossed product) is generated as a von Neumann algebra by the operators

( A E M , t E a), A €91, U ( t )@ It the notation being that of Definition 13.2.6 with U ( t ) as in the hint. Since U ( t ) E M and

I @ 11 = ( U ( t ) *8 I ) ( U ( t )€9 I t ) , it follows that R ( M , a ) is generated as a von Neumann algebra by the operators

A@I,

I@lt

( AE M , t E

a).

Let V be I €9 T , where T is the unitary operator (Fourier transform) occurring in Theorem 13.2.8. By that theorem, Tl-tT* = wt; moreover, the operators { w t : t E R} generate the multiplication algebra A, by Proposition 13.2.7(i). Thus V R ( M , a ) V *is generated as a von Neumann algebra by the operators

A@I,

I@wt

( A E M , t E W),

and so coincides with the von Neumann algebra tensor product MGA. m[24(Prop.4.2, p.51)] 13.4.17. Let R be a von Neumann algebra acting on a Hilbert space H,t + at be an automorphic representation of R on R implemented by a strong-operator-continuous one-parameter unitary group t + Ut with Ut in R for each real t , and S be the crossed product of R by a. Show that (i) S is * isomorphic to R @ d where A is the multiplication algebra of Lz(R); (ii) S is of type I, 11, or I11 when R is of type I, 11, or 111, respectively.

806

CROSSED PRODUCTS

Solution. (i) By virtue of the discussion preceding Definition 13.2.6, we may assume that S is the “implemented” crossed product of R by a. In this case, S acts on H @ Lz(W) and is generated as a von Neumann algebra by the operators A @ I and Ut @ It with A in R and t in R. Since Ut E R and

we have that I @ It E S . Thus S is generated as a von Neumann algebra by the operators A @ I and I @ l t with A in R and t in R. From Proposition 13.2.7(i), the operators wp generate A. From Theorem 13.2.8, the operators 11 generate an algebra unitarily equivalent t o A. It follows from Theorem 11.2.10 that S is * isomorphic to R@A. (ii) Since A is of type 11, S is of type I, 11, or I11 when R is of type I, 11, or 111, respectively, from (i) and Table 11.1. Let R be a von Neumann algebra acting on a separable Hilbert space X,u be a separating and generating unit vector for R, and J be the modular conjugation corresponding to (R,u). (i) Find a norm-continuous one-parameter unitary group t + Ut on H that generates a maximal abelian subalgebra of R. [Hint. Use Exercise 9.6.4:l.I (ii) Show that UtJUtJ (= &) is a norm-continuous oneparameter unitary group and that VtAv,“ = UtAU,“ (= crt(A)) for each A in R and t in R. (iii) Show that R and {& : t E R} generate a type I von Neumann algebra 7.[Hint. Note that { J U t J } generates a maximal abelian subalgebra of R‘ and use Exercise 9.6.1.1 (iv) With at as in (ii) and 7 as in (iii), show that the crossed product of R by a is not isomorphic to 7 when R is of type I1 or 111. [Hint. Use Exercise 13.4.17.1 13.4.18.

Solution. (i) Applying Zorn’s lemma, we find a (self-adjoint) maximal abelian subalgebra A0 of R. Since H is separable, there is a self-adjoint H that generates do as a von Neumann algebra (from the result of Exercise 9.6.41). Let Ut be expitH. Then t -+ Ut is a norm-continuous one-parameter unitary group (from the first paragraph of the proof of Theorem 5.6.36) and each Ut is in do. With t small and positive, sp(eitH) { c E

c : IcI = 1,Il-

cI

< 1) (= s).

EXERCISE 13.4.19

807

There is a continuous function f on S with range in ( - n , x ) such that if c E S, c = expir, and T E (-T,T),then f(c) = T. In this case, f ( U t ) = t H . Hence H is in the C*-algebra generated by { U t } and { U t } generates do. (ii) Since J R J = R', for all t and t' in R, we have

Moreover, with A in R, we have

At the same time, V: = JU;JU; = JU-tJU-t = V-1, so that Vt is a unitary operator. In addition,

so that t + Vt is a norm-continuous one-parameter unitary group. (iii) From (i), { Ut : t E R} generates a maximal abelian subalgebra do of R. Since A -+ J A * J is a * anti-isomorphism of R onto R', { J U t J } generates the maximal abelian subalgebra J d o J of R'. As Ut E R and UtJUtJ = Vt, each JUtJ E 7 and J d o J C 7. At the same time, each Vt is in the von Neumann algebra generated by R and J d o J . Thus 7 coincides with the von Neumann algebra generated by R and J d o J . If T E 7',then T E R' n ( J d o J ) ' = J d o J since J d o J is a maximal abelian subalgebra of R'. Thus 7'is abelian. From Exercise 9.6.1, 7 is of type I. (iv) Since a is a continuous automorphic representation of R implemented by a (continuous) one-parameter unitary group in R, the crossed product of R by a is of type I1 or type I11 when R is of type I1 or type 111, respectively, from the result of Exercise 13.4.17(ii). In this case, the crossed product is not isomorphic to 7 (of (iii)) since 7 is of type I.

13.4.19. Let R be a von Neumann algebra acting on a Hilbert space 'H, aut(R) be the group of * automorphisms of R, and i(R,) be the group of isometries of the predual R, of R onto itself. The strong

808

CROSSED PRODUCTS

topology on i(R,) has a basis open neighborhood of a, determined by a finite set ( ~ 1 , . ,w,} of elements of R, and consists of those a in i(R,) such that Ila(wj) - ao(wj)lJ< 1 for each j in (1,. ,. ,n}. (i) Show that, each a in aut(R) is the (Banach space) adjoint of a (unique) a, in i(R,) and that the mapping a + a, is an anti-isomorphism of the group aut(R) with its image in i(R,). The mapping of (i) transfers the strong topology on i(R,) t o a topology on aut(R) we shall call the bounded weak-operator (bw-) topology. (ii) Show that i(R,) is a topological group in the strong topology, and conclude that aut(R) is a topological group in the bwtopology. (iii) Suppose R has a generating and separating vector. Show that there is a (strong-operator-) continuous unitary representation of aut(R) (provided with its bw-topology) on 3-1 that implements (the identity representation of) aut(R). [Hint. Use Exercises 9.6.65 and 9.6.6O(iv).]

..

Solution. (i) From Remark 7.4.4, w o Q E 72, when w E R# and a E aut(R). Let a,(w) be w o a. Since a is an isometry of R onto R, a, is an isometry of R, onto R, and a is its adjoint. If a maps R, into R, and a = a # , then w ( a ( A ) )= ( a ( w ) ) ( A )for each A in R and each w in R,. Hence a,(w) = w o a = a(w) for each w in R,,and a, = a. With a and a' in aut(R) and w in R,,

w((aa':)(A))= a , ( w ) ( a ' ( A ) ) = ( a : Q # ( w ) ) ( A ) for each A in R. Thus w o (aa')= (aka,)(w) for each w in R,, and (aa'), = a:a,. It follows that the mapping a a, is an anti-isomorphism of aut(R) with its image in i(R,), (ii) Given a, and a!, in i(R,) and w in R,, suppose a and a' in i(R,) are such that II(a - ao)"a:>(wlII

< 1,

Il@' - a:)(b)ll < 1-

Then, since a is ail isometry, lI(aa'

- ~oa:,)(w>ll I I b a ' - ad)(w)ll iII(aa:, - ~ 0 4 > < ~ > l l L 1i(a7- a x w ) i i + I I ( ~ a0)[a:,(w)iii < 1.

809

EXERCISE 13.4.20

If 11(a - uo)[a;*(w)]ll < 1, then

l(.O1

>J%(lI

- ~-1)( =411

- 41 = II(a - ~ o > [ a O ’ ( ~ > l)l covers U . (iv) It follows from (i), (ii), and (iii) that the inverse image of a compact subset of E / C containing C is a compact subset of E . Since E / C is locally compact, we can choose V to be a compact neighborhood of C. With this choice, U is a compact neighborhood of e in E . Thus E is locally compact.

14.4.8. Let M be a factor, a be a continuous automorphic representation of R on M by inner automorphisms, G be the group of unitary operators in M that implement the automorphisms a ( t ) ( t E It), C be {cI : c E TI},and ~ ( tbe) UC,where U in G implements a(t).

EXERCISE 14.4.8

827

(i) Show that 77 is well defined and is a homomorphism of R onto G/C. Suppose 77 in (i) is contimuous, where G is provided with its weak-operator topology and G/C with its quotient topology (see Exercise 14.4.6(iv)). Let E be the topological group

{ ( U , t ) : U E G, t E R, U implements ( ~ ( t ) ) as a subgroup of G @ R with the product topology on G x R, CObe the (closed) subgroup ( ( c 1 , O ) : c E TI} of E , and n(U,t) be t for each ( U , t ) in E . Show that (ii) n is a continuous, open homomorphism of E onto R with kernel CO,and conclude that E/Co is isomorphic and homeomorphic to R [Hint. To show that ‘lr is open, note that a basis for the open sets of E consists of sets of the form

{ ( U , t ) : a < t < b , ~ ( t=) UC, U E 0, 0 open in G} and use the assumption that 77 is continuous.]; (iii) E is a 0-compact, locally compact, abelian group [Hint. Use Exercises 13.4.22 and 14.4.7.1; (iv) each character of COis the restriction of a character of E [Hint. Use Exercise 3.5.38 to show that the group of characters of Co is Z. With the aid of Exercise 14.4.3, note that the subgroup of Z consisting of restrictions of characters of E “separates” points of CO.Conclude that this subgroup is Z.]; (v) there is a continous, idempotent homomorphism x of E onto Co and a closed subgroup H of E that is homeomorphic and isomorphic to R [Hint. Use (iv) to extend the identity mapping on COto a homomorphism x with the desired properties. Use Exercise 14.4.6.1; (vi) there is a (continuous) one-parameter unitary group t + Ut that implements a such that Ut is in M for each t in R.

Solution. (i) If U and V in U ( M ) both implement a(t),then UV-l E C so that UC = VC and 77 is well defined. In addition, ~ ( t + s )= UVC = UCVC = q ( t ) v ( s ) ,where U implements a ( t )and V implements cr(s). Finally, if U E G, then U implements some a ( t ) and q ( t ) = UC. Thus 77 is a homomorphism of P onto G/C. (ii) The projection of G @ R onto R is a continuous homomorphism. Its restriction n to E is a continuous homomorphism of E onto R. If ‘lr(U,t) is 0, then t is 0 and U implements a(O), the

828

DKRECT INTEGRALS AND DECOMPOSITIONS

identity automorphism of M . In this case, U is in the center of M and ( U , t ) = ( c l ,0) for some c in 11'1. Of course ( c l ,0) E E and n( ( c l ,0)) = 0. Thus COis the kernel of K. To show that n is an open mapping, it will suffice to show that n maps each open set of a basis for the open sets of E onto an open set in R. Sets of the form 0 x ( a , b ) constitute a basis for the open sets in U ( M )x R, where 0 is a weak-operator open subset of U ( M ) and a,b E B. Thus sets of the form E n (0 x ( a , b ) ) constitute a basis for the open sets of E. Now, if 0' = 0 n G

E n ( 0 x ( a , b ) )= { ( U , t ) : u < t < 6 , ~ ( t=) UC, U E 0') and, with q the quotient mapping of G onto G/C,

Since q is an open mapping and 7 is assumed to be continuous, K is an open mapping. Let $ be the isomorphism of E/Co onto R determined by K. If 'p is the quotient mapping of E onto E f Co, then K = $ o 9.With S an open subset of R, $-l(S) = 'p(~-l(S)). As 'pis an open mapping and K is continuous, $-l(S) is open. Thus 11, is continuous. Suppose U is an open subset of E/Co. Then $(U)= K ( ' ~ - ~ ( U As ) ) . 'p is continuous and K is an open mapping, $(U)is open. Thus II, is an isomorphism and a homeomorphism of E/Co onto R. (iii) From Exercise 13.4.22, G and hence E are abelian. From Exercise 14.4.7, E is locally compact since COis compact and E/Co is isomorphic and homeomorphic to R. From this same exercise, E/Co is a-compact since R is the union u?=,[-n, n] of the compact neighborhoods [-n,n] of 0. Again, from Exercise 14.4.7, E is 6compact. (iv) Since COis isomorphic and homeomorphic to "1, the character group of COis Z - as described in Exercise 3.5.38. Restrictions t o COof characters of E correspond to a subgroup ZOof Z.From Exercise 14.4.3(iv), for each c in "1, there is a character of E that takes a value different from 1 at (cl,O) when c # 1. Thus the characters of "1 in ZOseparate 11'1. Let q be the smallest positive element of ZO. With m in ZO,we have that m = n q t r , where 0 _< r < q and n E Z ; T = m - n q E 250. By choice of q, r = 0. The character of "1 corresponding to m is t -, zn = ( ~ q ) Thus ~ . each character of "1 corresponding to the

EXERCISE 14.4.9

829

elements of Zo takes the value 1 at all the qth roots of unity. From the result of the preceding paragraph, q must be 1 and ZOmust be Z. It follows that each character of Co is the restriction of a character of E . (v) Let x o be a character of E such that X O ( ( C I , O )=) c for each c i n %I, and let x ( ( U , t ) )be (xo((U,t))l,O) for each ( U , t ) in E . Then x is the desired idempotent. From (ii) and Exercise 14.4.6 (iv), T restricts to a homeomorphism and isomorphism of H onto R. From (v), the mapping t -+ (vi) Let ( U t , t ) in H be ( T I H)-'(t). Ut is a (continuous) one-parameter unitary group that implements the automorphic representation a.

14.4.9. Let R be a von Neumann algebra acting on a separable Hilbert space 1-1. Show that (i) (R)1is a csm space when (R)1is endowed with its strongoperator * topology (determined by the semi-norms T + IITxlI -t llT*x11) [Hint. Introduce a metric of the type described in Exercise 2.8.35. Use the argument of Proposition 2.5.11.1; (ii) the restriction of the strong-operator * topology to the unitary group U ( R ) of R coincides with the strong- (and weak-)operator topology on U ( R ) [Hint. Use Remark 2.5.10 and Exercise 5.7.5.1; (iii) U ( R )is a closed subset of (R)1and conclude that U ( R ) is a csm space when U ( R )is provided with its strong- (or weak-)operator topology.

Solution. (i) Proceeding as Exercise 2.8.35, we let { yI,y2, . . .} be an orthonormal basis for 'H and define d( S, T ) by

c 00

2-n(llSyn - TYnII t IIS*yn - T*ynll)*

n=l

Arguing as in the solution to that exercise, we conclude that d is a (translation-invariant) metric on B(1-1) and the associated metric topology restricts to the strong-operator * topology on bounded subsets of D(7-l). If {T,} is a d-Cauchy convergent sequence in (R)1, the argument of Proposition 2.5.11, applied to {Tn}and to {T;} shows that {Tn}converges to some T in ( B ( ' H ) ) l in the strong-operator * topology. It follows that T is the strong-operator closure of (R)1so that T E (R)1. Thus (R)1is complete relative to the metric d. We show that (R)1has a countable dense subset. It suffices to show that this is the case for (B(1-1))1in its strong-operator *

830

DIRECT INTEGRALS AND DECOMPOSITIONS

topology. Note, for this, that the family F of operators in (B('H))l whose matrix representations with respect to the orthonormal basis { yn} for 'H have all their non-zero entries complex-rational numbers and only a finite number of these, is a self-adjoint, countable subset. Moreover, the self-adjoint operators in F are strong-operator dense in the set of self-adjoint operators in B ( ' H ) ) l . If To E (B('H))1 and To = T -t iT' with T and T' self-adjoint operators in ( B ( H ) ) l ,then there are sequences {Tn} and {TA} of self-adjoint operators in 3 that are strong-operator convergent to T and T', respectively. It follows that {T, iTA} is a sequence of operators in 7 that is strongoperator * convergent to TO. Thus (R)1is separable and complete in the metric d, that is, (R)1is a csm space. (ii) From Remark 2.5.10,the * operation is strong-operator continuous on the set of normal operators in B(31).It follows that the sequence {Un} in U ( R ) converges in the strong-operator topology to U in U ( R ) if and only if { U ; } converges to U*. Thus { U , } is strong-operator convergent to U if and only if it is strong-operator * convergent to U . Hence the strong-operator and strong-operator * topologies coincide on U ( R ) . Both topologies coincide with the weak-operator topology on U ( R )by the result of Exercise 5.7.5. (iii) Suppose {Un} in U ( R )is strong-operator * convergent to U in (R)1.Then {Un} and { U ; } converge to U and U * , respectively, in the strong-operator topology. Hence U and U* are isometries and U is a unitary operator. It follows that U E U(R)and U ( R ) is a closed subset of the csm space (R)1.Thus U ( R ) is a csm space in the metric it inherits from the metric don (R)1 defined in (i).

+

14.4.10. Let M be a factor acting on a separable Hilbert space 'H and a be a continuous automorphic representation of R by inner automorphisms of M . (i) With E as in Exercise 14.4.8,show that E is a closed subset of U ( M ) x R, where U ( M ) , the unitary group of M , is endowed with its strong-operator topology. (ii) Conclude from (i) and Exercise 14.4.9 that U ( M )x R is a csm space and E is an analytic subset of it. (iii) Show that there is a measurable mapping t + Vt of R into U ( M ) such that ( & , t ) E E for each t in R, [Hint.Use Theorem 14.3.6.] (iv) With q as in Exercise 14.4.8,show that 77 is continuous. [Hint. Use Exercise 14.4.5.1

EXERCISE 14.4.10

831

(v) Show that there is a (continuous) one-parameter unitary group t +. Ut that implements a and such that each Ut is in M . is a sequence in E tending to Solution. (i) Suppose {(Un,tn)} ( U , t ) in U ( M )x R. Then {in} tends fo t and {Un} tends to U in the strong-operator topology. Thus, for each A in M and all x , y in

If, we have that ( a ( - t n ) ( A ) x , y ) = (U;AUnxc,y) = (AUnxr U n y ) -+ ( A U x , U y ) = ( U * A U x , y ) . But by continuity of a , ( a ( - t , ) ( A ) x , y ) + ( a ( - t ) ( A ) z , y ) . Thus a ( - t ) ( A ) = U*AU for each A in M ; whence U implements a ( t ) and U is in M . It follows that ( U , t ) E E and E is closed in U ( M )x R. (ii) From Exercise 14.4.9, U ( M ) is a csm space in its strongoperator topology. It follows that U ( M )x R is a csm space in its product topology, when U ( M )is endowed with its strong-operator topology; and E is an analytic subset of it from (i). (iii) If we apply Theorem 14.3.6, with R and Lebesgue measure in place of ( X , p ) , U ( M ) in its strong-operator operator topology in place of Y , E in place of S,and the projection p of U ( M )x R (= Y x X ) onto R in place of K , then A ( = p ( E ) = n ( S ) )becomes R, and there is a measurable mapping t i Vt of R into U ( M )such that (K, t ) E E for each t in ]Re. ) be defined as V,C for each (iv) From Exercise 14.4.8 (i), ~ ( tcan t in R, and 7 is then a well-defined homomorphism of R into G/C. Since U ( M ) has a countable dense subset relative to its strongoperator topology, the same is true of G and of the topological group G/C (in its quotient topology). If (3 is an open subset in G/C, its inverse image in G is open and is therefore the intersection of a strong-operator open subset (3' of U ( M )with G. Now ?)-1( =0{) tE

R : v, E O'}.

Since t +. fi is a measurable mapping of R into U ( M ) , v-l(O)is a measurable subset of R. It follows that 7 is a measurable homomorphism of R into the (separable, topological) group G/C. From Exercise 14.4.5, 7 is continuous. (v) From (iv) and Exercise 14.4.8 (vi), there is a (continuous) one-parameter unitary group t -+ Ut that implements cy and such that Ut is in M for each t in W.

832


14.4.11. Let M be a factor acting on a separable Hilbert space 3-1 and u be a separating and generating unit vector for M. Let t ot be the modular automorphism group corresponding to ( M , u ) . Suppose each ot is inner. Show that (i) M is semi-finite [Hint. Use Theorem 9.2.21 and Exercise 14.4.10.); (ii) R ( M , a )is * isomorphic to M @ A ,where A is the multiplication algebra corresponding to Lebesgue measure on W [Hint. Use Exercise 13.4.17.1; (iii) R ( M , a ) is semi-finite. .--)

Solution. (i) Since t + ot is a continuous automorphic representation of B on A4 by inner automorphisms, the result of Exercise 14.4.10(v) yields that o is implemented by a one-parameter unitary group t + Ut with each Ut in M. From Stone's theorem, there is a self-adjoint operator Ir' (affiliated with the abelian von Neumann algebra generated by (Vt} and hence with M) such that ut = ,itK = H i t where H = exp K 17 M and H is positive. Since

o t ( A )= H i t A H - i t

( A E M , t E a),

M is semi-finite by Theorem 9.2.21. (ii) From (i) and Exercise 13.4.17, R ( M , o ) is

Mad.

* isomorphic t o

(iii) From (i) and Exercise 13.4.17, R ( M , o ) is semi-finite since

M is semi-finite.

14.4.12. Let M be a factor of type I11 acting on a separable Hilbert space. Show that (i) M has a separating and generating vector [Hint. Use Proposition 9.1.6.1; (ii) M admits an outer automorphism. [Hint. Use Exercise 14.4.11.1

Solution. (i) From Theorem 9.1.3, M' is a factor of type 111. Thus, from Proposition 9.1.6, M has a joint generating and separating unit vector. (ii) From Exercise 14.4.11(i), some ot must be an outer automorphism of M.

EXERCISE 14.4.13

833

14.4.13. Let R be a von Neumann algebra of type I11 acting on a separable Hilbert space and let (Y be the * automorphism of R@R described in Exercise 11.5.25(iii). Show that a is outer. [Hint. Use Exercises 12.4.19, 12.4.20, and 14.4.12.1 Solution. If (Y is inner, then R is a factor, from Exercise 12.4.19, and each * automorphism of R is inner, from Exercise 12.4.2O(iii). But each factor of type I11 acting on a separable Hilbert space admits an outer automorphism, from Exercise 14.4.12(ii). Thus (Y is outer.

14.4.14. Let X and Y be csm spaces, f be a continuous mapping of X into Y , and A be an analytic subset of Y . Show that f - l ( A ) is an analytic subset of X. [Hint.Choose V a csm space and g a continuous mapping of V onto A. Study n ( B ) , where B , is the inverse image of the diagonal in Y x Y under the mapping ( x , v ) + (f(x),g(v)) of X x V into Y x Y and K is the projection of X x V onto X.]

Solution. Proceeding as in the hint, with the notation introduced there, we note that the mapping ( 5 , ~--$) (f(x),g(v)) is continuous and the diagonal in Y x Y is closed. Thus B is a closed subset of X . x V . Hence B is a csm space. Since K is continuous, K ( B )is an analytic subset of X . We complete the exercise by showing that n ( B ) = f-'(A). To see that n ( B ) = f - l ( A ) , observe that x E n ( B ) if and only if there is a v in V such that ( x , v ) E B , that is, if and only if there is a v in V such that f ( x ) = g(v). But g maps V onto A, so that there is such a v if and only if f ( z ) E A. Thus n ( B ) = f-'(A), and f - l ( A ) is an analytic subset of X . 14.4.15. Let R be a von Neumann algebra acting on a separable Hilbert space 7-l. Show that (i) {UU' : U E U ( R ) , U ' E U ( R ' ) } ( = Uj(R))is an analytic subset of the (csm) space U(7-l), where U ( R ) ,U(R'), and U(7-f)are the groups of unitary operators in R, R',and B('H), respectively, each provided with its strong-operator topology [Hint.Use Exercise 14.4.9, and consider the mapping (U,U') -+ UU' of U ( R ) x U ( R ' ) into U('H).]; (ii) an automorphism of R implemented by a unitary operator

834


V on 9 f is inner if and only if V E U i ( R ) ; (iii) T ( R )is an analytic subset of R. [Hint. Use the result of Exercise 14.4.14 in conjunction with (i) and (ii).]

Solution. (i) From Exercise 14.4.9(iii), U(R), U ( R ' ) , and U(8) are csm spaces in their strong-operator topologies. Thus the product space U(R)x U(R')is a csm space. The mapping (U,U') + UU' of U ( R )x U ( R ' ) into U(3-I)is (strong-operator) continuous and has U i ( R ) as its range. Thus Ui(R)is an analytic subset of U(3-I). (ii) Suppose that the unitary operator V implements a * automorphism (Y of R. Then (Y is inner if and only if there is a U in U ( R ) such that UAU* = VAV* for each A in R,that is, if and only if there is a U in U(R)such that U-'V = U' for some U' in U(R'). (iii) Let w be a faithful normal state of R, (J,A) be the associated modular structure, and t + ut be the associated modular automorphism group. Then ut is inner if and only if Ait E Ui(R) from (ii). It follows that T ( R )is the inverse image of U i ( R ) under the (strong-operator-continuous) mapping t + A". From (i), U i ( R ) is an analytic subset of U('H),and from Exercise 14.4.14, T ( R )is an a analytic subset of R. 14.4.16. Let M be a factor of type I11 acting on a separable Hilbert space 9 f . Show that (i) T ( M ) is a subset of R having Lebesgue measure 0 [Hint. Use Exercise 14.4.15(iii), Theorem 14.3.5, Exercise 14.4.4(iii), and Exercise 14.4.11(i) .I; (ii) M admits an automorphism a such that an is outer for all [n-l T ( M ) ] positive integers n. [Hint. Use (i) and note that has Lebesgue measure 0.1

Ur=l .

Solution. (i) From Exercise 14.4.15(iii), T ( M ) is an analytic subset of R, and from Theorem 14.3.5, T ( M )is a measurable subset of R. Now T ( M ) = T ( M ) - T ( M ) , so that T ( M ) contains an open neighborhood of 0 if T ( M )has positive measure (from Exercise 14.4.4(iii)). In this case, T ( M ) = R and each automorphism of a modular group for M is inner. It follows, from Exercise 14.4.11(i), that M is semi-finite when T ( M ) has positive measure. Thus T ( M ) has measure 0 when M is a factor of type I11 acting on a separable Hilbert space. (But see Exercise 14.4.20(vi).) (ii) From (i), Ur=,[n-l T ( M ) ] (= S) has measure 0. Choose s

-

EXERCISE 14.4.17

835

in R \ S. Then ns 4 T ( M )for each non-zero integer n. Let t + at be the modular group corresponding to some faithful normal state of M. Then a,,(= a,")is an outer automorphism of M for each non-zero n in Z. For a,we can choose as. m 14.4.17. Let M be a factor of type I11 acting on a separable Hilbert space H,and let a be a * automorphism of M implemented by a unitary operator U on H. Suppose that an(= a,) is outer for each non-zero integer n. Show that (i) R ( M , a )is a factor of type I11 [Hint.Use Proposition 13.1.5 and Exercise 13.4.2.1; (ii) @(M)'n R ( M , a ) = { c l : c E C} [Hint.Recall the matrix ) of R ( M ,a). Use Exercise descriptions of the elements of @ ( M and 12.4.17(iv).]; (iii) @(M)'is * isomorphic to M' [Hint.Use Proposition 9.1.6 and Theorem 7.2.9.1; (iv) R ( M , a ) ' is a proper subset of @(M)'[Hint. Show that U @I I1 is in R ( M , a ) and not in @ ( M ) . ] ; (v) @(M)'is not normal in the sense of Exercise 12.4.31.

Solution. (i) From Proposition 13.1.5(ii), R ( M ,a) is a factor since M is a factor and a, is outer when n # 0. F'rom Exercise 13.4.2, R ( M ,a)is of type I11 since M is of type 111. Thus R ( M ,a) is a factor of type 111. (ii) With A in M, @(A) has A at each diagonal entry in its matrix representation and 0 at all other entries. Those bounded operators with matrix representations having each entry in M' are the elements of cP(M)'. At the same time, the elements of R ( M ,a)have matrix representations of the form [U(n-m)A(n-m)], where U ( n )= U" and n --+ A(n) is a mapping of Z into M. (See the discussion following Definition 13.1.3.) Thus, an element of @(M)'n R ( M , a ) has as the (n,O)entry of its matrix representation an element A(n)' of M' of the form U(n)A(n) with A(n) in M. For each B in M and each n in Z, we have that a-,(B)A(n) = U(n)*BU(n)A(n) = U(n)*BA(n)' = U(n)*A(n)'B= A(n)B.

From Exercise 12.4.17 (iv), either A ( n ) = 0 or a(-.) is inner. Since a(-.) is outer for each non-zero n in 2,A(n) = 0 when n # 0. It

836


follows that the matrix representation of an element of the intersection @(M)'n R ( M , a ) has each diagonal entry the same element A ( 0 ) in M n M' and all other entries 0. Thus

*

) M acts on (iii) Since @ is a isomorphism of M onto @ ( M and a separable Hilbert space 'H, both M and @ ( M )are countably decomposable factors of type 111. As @ ( M )acts on H @I &(Z) and 'H is separable, H 8 L2(Z)is separable. Thus M' and @(M)'are countably decomposable factors of type I11 (see Theorem 9.1.3). From Proposition 9.1.6, M and M' have a joint separating and generating unit vector, and the same is true for @ ( M )and @(M)'.It follows from Theorem 7.2.9, that @ is unitarily implemented; in particular, M' is * isomorphic to @ ( M ) ' . (iv) Since @ ( M )C R ( M , a ) ,we have that R(M,a)' C @(M)'. Moreover, R (M ,a) ' = @(M)'if and only if @ ( M )= R ( M , a ) . Thus, it will suffice to show that @ ( M is ) a proper subset of R ( M ,a). For this, we note that U @I I1 (= U(1) @I II) is in R ( M , a ) but not in @ ( M ) .Indeed, the matrix representation of U @I 11 has 0 at each ) a proper diagonal entry and U at each n -t 1, n entry. Thus @ ( M is subset of R ( M , a ) . (v) The relative commutant of R(M,a)' in @(M)'is R ( M , a ) n @ ( M ) ' .From (ii), this relative commutant consists of scalars and its commutant relative to @(M)'is 9 ( M ) ' . From (iv), R ( M , a ) ' is a proper subset of @ ( M ) 'hence, , not equal to its own relative double commutant. Thus 9 ( M ) ' is not normal. rn 14.4.18. Show that a von Neumann algebra acting on a separable Hilbert space is normal (in the sense of Exercise 12.4.31) if and only if it is a factor of type I. [Hint.Use Exercises 12.4.31, 14.4.16, and 14.4.17.1

Solution. From Exercise 12.4.31, a normal von Neumann algebra is a factor and is not of type 11. It remains to show that a factor of type I11 acting on a separable Hilbert space is not normal. We make use of the result of Exercise 14.4.17, and denote our factor of type I11 by M'. From Exercise 14.4.16(ii), M' admits an automorphism a such that an is outer for all positive (and, hence, all non-zero) integers n. The conditions of Exercise 14.4.17 are fulfilled

837

EXERCISE 14.4.19

with M‘ in place of M . Thus @(M’)’is not normal and @(MI)’is * isomorphic to M (= M”). It follows that M is not normal. w 14.4.19. Let R acting on a Hilbert space 3-1 be a von Neumann algebra, S be a von Neumann subalgebra of R, and t -+ V, be a (continuous) one-parameter unitary group on 3-1 that implements one-parameter groups t + oi and 2 -+ ot of automorphisms of R and S, respectively, where t -+ ut is the modular automorphism group of S corresponding to a faithful normal state w of S. Let W be a faithful, ultraweakly continuous conditional expectation of R onto S and 9 be a group of unitary operators in 72 such that 9 and S generate R as a von Neumann algebra. Suppose that V S V * = S, u ( V A V * ) = w ( A ) ( A E S ) , oi(V) = V , and @’(V)= 0, for each V ( # I ) in 9. Show that VAA, : Vi E E, Aj E S } (= 2l) is a self-adjoint (i) (ViA1 t ..-t subalgebra of R and 8- = R; (ii) t + 0: is the modular automorphism group of R corresponding to w o @’.[Hint.Use (i) and Lemma 9.2.17.1

Solution. (i) Since VdSVd* = S for each

V A W B = VWW’AWB

&‘ in 9, we have that

(V,W E 9, A , B E S ) ,

with V W in 9 and W’AWB i n S. Thus 2l is a self-adjoint subalgebra of R containing S and 9. By assumption, S and 9 generate R as a von Neumann algebra. Hence 2l- = R. (ii) Note that w o a‘(= w’) is a faithful normal state of R since w and 9‘ are faithful and ultraweakly continuous. From (i) and Lemma 9.2.17, it will suffice to show that t --+ 0 ; satisfies the modular condition for each pair of elements of 2l relative to w’. It will, in fact, suffice to establish this for a pair of elements of the form V A and W B in 2l, where V,W E (3’ and A , B E S. Note, for this, that

Thus

w‘(a:(VA)WB)= 0 , W # V * w ’ ( o ~ ( V A ) V *= B )w(o~(V/AV*)B).

838


Similarly, w'( WBa;(V A ) )= W ( @'(WBW*at( WVAV*W * ) W V ) )

= w(WBW*at(WVAV*W*)@'(WV)). Hence

w'(WBai(VA))= 0, W # V* w'(V*Ba;(VA))= o ( V * B a t ( V A V * ) V= ) w(Bat(VAV*)). Since t +. at is the modular automorphism group of S corresponding to w , there is a bounded, continuous, complex-valued function f on { z E C! : 0 5 I m z 5 1) (= a) analytic on the interior of Q such that

f ( t ) = w(at(VAV*)B)= w'(a;(VA)V*B) f ( t + i) = w ( B a t ( V A V * ) ) =w'(V'Bal(VA))

( t E R) ( t E R).

If W # V * ,we use the constant function 0 on R for f to fulfil the modular condition for V A and W B . Thus t + a: is the modular rn automorphism group of R corresponding to w'. 14.4.20. Let M , acting on the Hilbert space 3-1, be the factor of type I11 constructed in Exercise 13.4.12, wo be a faithful normal state of M , ( J , A ) be the modular structure and 2 + at be the modular automorphism group of M corresponding to 00. Thus t + Ait (= U ( t ) ) implements t +. at. Let R ( M , o )be the (implemented) crossed product of M by a considered as an automorphic representation of R (as a discrete group) on M . With the notation of Definition 13.1.3, let R be the von Neumann algebra (which acts on 3-1 @ Za(1w)) generated by (S = ) @ ( M and ) { V ( t ) : t E G}, where G is a given subgroup of 118 (as a discrete group). Show that (i) R is * isomorphic to R ( M , a ) ,where a = a(G [Hint.Show that the projection of EfcR $3-1 onto EgEG $3-1 commutes with R, and the restriction of R to CgEG $3-1 is a * isomorphism of R onto R ( M ,a). Consider generators and matrix representations.]; (ii) R is a factor of type 111. [Hint.Use Exercises 13.4.2, 13.4.12, and Proposition 13.1.5(ii).] Let a:(T) be V ( t ) T V ( t ) *( t E R,T E R). With w the faithful normal state of S such that w o @ = wo, and 9' the conditional

EXERCISE

14.4.20

839

expectation of R ( M , a ) onto S described in Exercise 13.4.1, let w’ be (w o ia’)lR. Show that (iii) w’ is a faithful normal state of R and t -+ a: is the modular automorphism group of R corresponding to u’ [Hint. Use the Exercises 13.4.1 and 14.4.19.1; (iv) S ‘ n R ( M , a ) = { c I : c E C } [Hint. Use Exercise 12.4.17(iv).]; (v) G = T ( R ) [Hint. Use (iii) and (iv). Study the matrix representation of V ( t ) . ] ; (vi) there is a countably decomposable factor of type I11 for which the modular automorphism group consists of inner automorphisms. (Compare the result of Exercise 14.4.16(i).) Solution. (i) Let E‘ be the projection on 3-1 8 Z,(W) whose matrix has p , p entry I when p E G and all other entries 0. The $3-1, which is isomorphic to) H 8 /2(G), the range of E’ is (CgEG space on which R ( M ,a) acts. This range consists of functions from W into 3-1 that vanish off G (and are 12-convergent). Suppose g , h E G and z is a function from P into H that vanishes except at h. Then ( V ( g ) z ) ( p= ) 4J,g+hW(z(h)),

since the matrix for V ( g )is [6,,,+,U(g)]. Thus ( V ( g ) z ) ( p is ) 0 unless p = g h , and in this case, its value is U ( g ) z ( h ) .It follows from this that V ( g )maps the range of E’ into itself for each g in G, and E’ commutes with each V ( g ) . In addition, V ( g ) E ‘ , acting on the range of E’, is the generator in R ( M , a ) corresponding to g in G. Of course @ ( A )commutes with E‘ and Q(A)E’,acting on the range of E’, is the generator in R ( M ,a ) corresponding to A in M . Thus E‘ E R’and RE’, acting on the range of E‘, is R ( M , a ) . It remains to show that T in R is 0 if TE’ = 0. Note, for this, that if p - q = g E G, then the p , q entry of the matrix for T is V ( p- q)A(p- q ) = U ( g ) A ( g ) where , A(g) E M , which is the 0, - g entry of T . Since the matrix of E’ has I as its -9, - g entry and 0 for each other p , - g entry, TE‘ has the same 0, -g entry as T has. Thus the p , q entry of T is 0 when p - q E G , T E R , and TE’ = 0. We complete the argument by showing that the p , q entry of each T in R is 0 when p - q 4 G. This is true for V ( g )(the p , q entry, 6p,g+qU(g) is 0 unless p - q = g ) and, hence, for Q ( A ) V ( g ) when , A E M and g E G. It is true, therefore, for each element of the linear span of { i a ( A ) V ( g ) } a, dense, self-adjoint subalgebra of R, and, hence, for each T in R. It follows that R is * isomorphic to R ( M , a ) .

+

840


(ii) From Theorem 9.2.21, M is of type I11 since T ( M ) = (0) (from the result af Exercise 13.4.12). Thus R(M,cu) is of type I11 from Exercise 13.4.2. From Proposition 13,1.5(ii), Z ( M , a )is a factor since each cut (= ut) is outer when t E G \ (0). From (i), R is a factor of type 111. (iii) Since 9' and o are faithful and ultraweakly continuous, w' is a faithful normal state of R. (See the properties of 9' described in Exercise 13.4.1.) Now t -+ V ( t ) is a (continuous) one-parameter unitary group on 'H 63 12(R). From 13.1(8), V ( t ) implements ot transferred to 9(M) by a, and t --+ ut transferred by is the modular automorphism group of S(= 9(M)) corresponding to w . With V ( G )for 0 in Exercise 14.4.19 and V ( t )for G,we have that w ( V ( t ) A V ( t ) *= ) w ( A ) for each A in S and t in G (indeed, for each t in R) from Theorem 9.2.13 and Proposition 9.2.14(i). Moreover,

and a ' ( V ( 9 ) ) = 0 since Go,J(g) = 0, when g # 0. Applying the is the modular result of Exercise 14,4.19(ii), we have that t + automorphism group of R corresponding to w'. (iv) Each entry in the matrix of an element T in S' lies in M'. If T also lies in R ( M ,0 ) and has matrix [ U ( p - q)A(p - q ) ] , then U ( t ) A ( t )E M' and A ( t ) E M for each t in W. Let U ( t ) A ( t )be A(t)'. With B in M, we have

a-t(B)A(t)= U ( t ) * B U ( t ) A ( t= ) U(t)*BA(t)' = U(t)*A(t)'B= A ( t ) B . From Exercise 12.4.17(iv), A ( t ) = 0 when t # 0 since 0 - t is outer ( T ( M )= (0)) and M is a factor. Thus the only non-zero entries are on the diagonal and all these are equal to A(0)(= A(O)'),an element of M n M ' . It follows that

S'n R ( M , a ) = {cI:c E C}. (This may also be proved using Proposition 13.1.5(i) in place of Exercise 12.4.17( iv ).) (v) From (iii), t + 0; is a modular automorphism group for R and it is implemented by t -, V ( t ) .By construction, V ( t )E R when t E G. Thus G T ( R ) .

EXERCISE 14.4.20

84 1

Suppose W is a unitary operator in R that impelements a:. Then V ( t ) W *E S’ n R ( M , o ) . Thus W is a scalar multiple of V(2) (from (iv)), and V ( t )E R. Now the matrix for V ( t )has U ( t ) as its 2,0 entry. As proved in the last paragraph of the solution to (i), the p , q entry of the matrix for an element of R is 0 unless p - q E G. Thus t (= t - 0) E G. Hence T ( R ) G, and T ( R )= G. (vi) The special case where G = P yields a factor R ( M , a ) of type I11 for which T ( R )= P - each element of the modular automorphism group of R ( M ,0 ) is inner. As w’ is a faithful normal state of R ( M , a ) , R ( M , a ) is countably decomposable. But note that R ( M ,o) acts on Ctcl$N,a non-separable Hilbert space. This indicates the limits to extending the result of Exercise 14.4.16(i).

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INDEX OF VOLUMES 111 AND IV

A Abelian projection, 282,283 Abelian self-adjoint subset of n 8 72, 297 Affiliated operator, 309,310, 360,361,449, 450 Algebra abelian C*, see C*-algebra abelian von Neumann, aee von Neumann algebra of bounded operators, 188-192, 205-206 C*, ace C*-algebra CAR (Fermion), ace CAR algebra countably decomposable, ace Countably decomposable von Neumann algebra Fermion (CAR), bee CAR algebra finite-dimensional C*, 147-149

/l(Z), 115-117,

120-121

L ~ ( l k )128-129 ,

Ll(T~,rn), 121-128,

129131

maximal abelian, see Maximal abelian algebra multiplication, 231, 245-246, 248-250 nuclear C*, aee Nuclear C*-algebra quotient, 196-197 semi-simple, 106 simple, ace Simple C*-algebra UHF, aee Uniformly matricial algebra of unbounded operatom, 449 uniformly matricial, ace Uniformly matricial +bra von Neumann, ace von Neumann algebras

W*,345-346,

662-663

Amenable group, ace also Invariant mean; 772, 774 Annihilator, 779 Anti-homomorphism, 113, 584, 588, 589, 590 Anti-isomorphism, 359, 427, 428, 431, 464-465, 515, 591, 592 Approximate identity (in C*-algebras), 166-168, 196-197 quasi-central, 556-558 Approximation theorem, StoneWeierstrass, 138 Archimedian partially ordered vector space, 182 Automorphic representation, 809. 811-813, 826-831 Automorphism. 545. 546-547, 568-570, 621-629, 631-643, 785-786 flip, see Flip automorphism free action of, 747-749 group, 641-644 inner, 627-628, 631-637, 641, 839, 841 outer, bee Outer automorphisni universally weakly inner, 627, 641-643

848

INDEX O F VOLUMES 111 A N D IV

B Banach lattice, 184186 Banach module, 201

Banach-Orlicz theorem, 33 Bansch space non-separable, 30-31 reflexive, a e e Reflexive (Bansch space) separable, 30-31,35-36 Banach space, examples of C, 14,15, 17,20, 21, 166-189 CO. 14,15, 16, 17,20, 21, 188-189 I,, 14,19,20, 21, 188-189 11, 16,17 19,20, 22 Ii(Z) (as a Banach algebra), 115-117,120-121 I,, 24-26, 115-117 L m ,30-32 LI , 31-32 L1 (R)(as a B a n d algebra), 128-129 Li (Ti,m) (as a Banach algebra), 121-128,129-131 L,, 26-30 8-compactification, 89-90,188-189, 222-224 Bidual (of a C*-algebra), 572 Bore1 function calculus for bounded multiplication operatom, 232-233 for unbounded multiplication operators, 249-250 Bounded weak-operator topology, 808-810 Boundedly complete lattice, see Lattice C

C*-algebra abelian, 160-162,502-503,611 finite-dimensional, 147-149 generated by two projections, 739 monotone closed, 339-345 monotone sequentially closed, 345 nuclear, 690491 reflexive, 571-572 simple, see Simple C*-algebra singly generated, 502-503 type I, 690 uniformly matricial, 728-729 Canonical anti-commutation relations, aee CAR CAR, 664 algebra, see CAR algebra irreducible representations of, 668-669,77S782 representations of, 667-869, 779-782 CAR algebra, 663-669 factor representations of, 790-804

INDEX O F VOLUMES I11 AND IV

product states of, 790-804 Cayley transform, 762-763 Center state, 444 Center-valued trace, 44G448 Central carrier, 240, 337 Character of a locally compact group, 819-821 of TI,123-126 of Z.117-120 Closed subspace, 4 Compact h e a r operator, 55. 56, 58-66, 72, 95, 98, 99, 100-104, 190, 192, 204

Compact self-adjoint operator, 5 9 4 6 Completely positive mapping, 699-705, 708-71 I Complexification of a real B a n d algebra, 85-86 of a real Hilbert space, 41 of a real linear space, 4-5 of a real normed space, 4-5 Conditional expectation, 396-407, 433-434, 660-662, 783-784, 813-817

faithful, 403-404, 433434, 783-784, 813, 816 Cone dual, 517-545 self-dual, 522-545 Continuous crossed product, 805-806 Convolution, 115-117, 121-123, 819-821 Countably decomposable von Neumann algebra, 247, 285, 288 290, 292, 293, 295, 296, 321,347, 394, 437, 441, 759, 804

Countably decomposable projection, 246, 274, 321, 322 Coupling (or linking) constant, 455 Coupling (or linking) operator, 457 Creator, 779 Crossed product continuous, 805-806 discrete, 783, 785 by inner automorphisms, 805-806 by the modular group, 804-805 Cyclic projection, 246, 280-282, 313, 32&322, 452-459

D Daun-Hofmann theorem, 656 Definite state, 150-151, 165-166 Derivation of a C*-algebra, 2OC-203, 322-323, 443-444, 56e568, 623-629.631-632, 639-640, 6 4 6 4 4 9 inner, 434-443, 631-632, 645-649

849

850

INDEX O F VOLUMES 111 AND IV

outer, 649 of a simple C*-algebra, 646-649 of a von Neumann algebra, 434-443, 641, 771-772 * derivation, 623-629, 631-632,639-641,645-649 Diagonalization (of an abelian self-adjoint set), 297 Dilation, 712 Dimension function, 409-410, 448 Direct sum of B a n d algebras, 86-87 of operators, 51-52 restricted, 700 Directed system of bounded operators, 716-717 of C*-algebras, 718-721 of Hilbert spaces, 714-716 of representations, 718-721 Discrete crossed product, 783, 785 Discrete group algebra, 300-305, 405-409 Disjoint representations, 606-607 Dixmier approximation theorem, 373-386, 548-549, 650 Dual cones, 517-545 Dual group of Ti,123 of Z,117

Dual module, 565 Dual normal module, 566, 771 Dual space (of a C*-algebra), 572-575 bidual, 572

E Eigenvalue, 229 Equivalence (of projections:), 275, 309 Equivalence (of representations), 506 Essential range (of a measurable function), 24s250 Exponential unitary, 140-145, 548, 552 Extension of pure states, 178-180 of states, 178-180 Extension problem, 506, 510 Exterior product, 774-782 Extreme point, 49-50, 180, 217-218 Extremely disconnected space, 90, 218-221, 223-224, 225, 227, 228-229


F Factor, 413, 826-829, 830-832,834-837 discrete (i.c.c.) group examples, 300-305, 405409, 417-420, 431-433 finite, 414-420, 427-433, 453-455 groupme.%uFe-spaceexamples, 428-431, 433-434, 725-755 matricial, s e e Matricial factor normal, 761,835-837 semi-finite, 832 type 111, 797-802, 832, 834-836 Factor (primary) state, 607-608 Faithful conditional expectation, r e e Conditional expectation Faithful state, 150,316-318, 356-357, 361, 571 Faithful weight, 356-357 Ferrnion algebra, s e e CAR algebra Finite von Neumann algebra, 275, 282,291,295, 296, 368-372, 390-392, 403-404, 409-410, 412413, 446450, 451453, 455-459, 488-489, 49C493 Finite projection, 275, 276, 277,282 Flip automorphism, 713-714, 749-752, 757-758, 766, 833 unitary, 713-714 Fock representation (of the CAR), 779 Fock space (antisymmetric), 779 Fock vacuum, 779 Fock vacuum state, 780 Free action (of a

automorphism), 747-749

Friedrichs extension, 366 Function calculus, g e e Bore1 function calculus Function representation (of a Banach lattice), 184-186 Fundamental group, 788-790

G Generalized nilpotent, 95-99, 103-104 Generalized Schwarz inequality, 561-562, 711-712 Generating vector, 241-242, 243-246, 36C361, 451-452 Group amenable, 772 locally compact abelian, 818-829 topological, 641

H Hahn-Jordan decomposition, 158,326-327 Hilbert-Schmidt operator, 71-73, 103-104

851

852

INDEX O F VOLUMES 111 A N D IV

Holder's inequality, 24, 27

* homomorphism, 595-596 I Ideal, 86-88, in a C*-algebra, 166-171, 172, 196-200, 562-564, 613414 of compact operators, 612-613, 614518 in a factor, 306,308-309 maximal, 306, 389-390 primhive, 650-657 in a von Neumann algebra, 305,306-307, 308-309, 38&396 Idempotent mapping between C*-algebras, 658-662 Inductive limit of bounded operators, 716-717 of C*-algebras, 718-721, 723 of Hilbert spaces, 714-716 of representations, 718-721 Infinite tensor product of C*-algebras, 723-725 of Hilbert spaces, 721-'725 of representations, 723-725 Inner autoniorphism, see Automorphism Inner derivation, see Derivation Invariant mean, 91,405-40!3 Invariant T(R),834-835, 839-841 for certain crossed products, 83-841 for certain infinite tensor products, 797-802 Involution (of a Hilbert space), 359 Isometry between C*-algebras, 323-326, 592, 596-597 J

Jacobson topology, 652 Jordan * homomorphism, 576-591, 592-594, 598 Jordan

* isomorphism, 325-326, 591-592, 594,596-600, 602 K

KMS boundary condition, 8ec Modular condition

L Lattice, 186

Banach,184-186 boundedly complete, 218-220, 223-224, 225,227 sublattice, 137 bindependent, 243-245 Linear functional, 2

INDEX O F VOLUMES 111 AND IV

multiplicative, see Multiplicative linear functional positive (on a partially ordered vector space), 177-182 singular, 570-571, 612 Linear operator, aee Operator Linear order isomorphism, 596-598 Linear transformation, aee Operator Linking (coupling) constant, 455 Linking (coupling) operator, 457 Locally compact abelian group, 818-829 Locally finite group, 91

M Matricial factor, criterion for, 762-765 finite, 752-756, 762-765, 79S793 subfactor, 755, 756, 75S761 of type [I,, 803-804 of type 111, 797-802 Matrix units, 731-733 approximation of, 733-735 compatible system of, 771 Maximal abelian algebra, 283-293, 296-297, 302, 427-434, 448 in a factor, 302, 304, 427-434 in a tensor product, 770-771 Maximal ideal in a factor, 306 in a von Neumann algebra, 306, 389-390 Meager set, 224-227 Modular condition, 472478 Modular theory, 464468, 472-481, 488, 515-545, 832 Monotone closed C*-algebra, 339-345 Monotone sequentially closed C*-algebra, 346 Multiplication algebra, 231, 245-246, 248-250 Multiplication operator bounded, 50-51, 231, 232-233 unbounded, 24&250 Multiplicative linear functional on l m ( Z ) , 87-88 on ll(Z), 118-120 on L1 (G), 81S820 on L I ('TI,m), 124-127 Multiplicity (of an eigenvalue), 64, 65, 101 Multiplicity theory (for a bounded normal operator), 513-514 Multiplicity, uniform, 513-514

853

854

INDEX OF VOLUMES I11 AND IV

N Non-separable B a n d space, 30-31 Hilbert space, 45 Norm topology, 5-6 Normal operator, 108-109,513-514 Normal state, 227,312-320, 321-322, 32S334, 347, 361 Normal von Neumann algebra, 761,836-837 n-state, 708-711 Nuclear C*-algebra, 690-691 Null space (of an unbounded operator), 77-78 0

One-parameter subgroup (of a topological group), 641 One-parameter unitary group, 827,829,831 Operator affiliated, ace Affiliated operator compact, see Compact linear operator compact self-adjoint, 59-66 Hilbert-Schmidt, 71-74, 103-104 multiplication, aee Multiplication operator normal, 108-109, 513-514 rational, 730 symmetric, 362-367 tensor product, 74-75 unbounded, 75-78 Operator-monotonic increasing, 173-176 Order unit, 177-184 Outer automorphism, 303,751-752, 755, 757-758, 766,832,834-835

P Partially ordered vector space, 177-186 archimedian, 182 Pauli spin matrices, 664 Plancherel's theorem, 120,127 Point spectrum, 229 Polar decomposition approximate, 171-172 of invertible elements, 172-173 of normal functionals, 334-337 Positive linear functional (on a partially ordered vector apace), 177-184 Positive linear mapping, 560-562, 705-706 completely, aee Completely positive mapping Positive normal functional, 537-542

INDEX OF VOLUMES I11 AND IV

Positive square root, 64,66 Primary (factor) state, 607-608 Primitive ideal, 650-657 Primitive ideal space, 652457 Primitive spectrum, 652 Principle of uniform boundedness, 38 Product state, 693-696 of the CAR algebra, nee CAR algebra Project ion abelian, 282,283 countably decomposable, 246, 274, 321,322 cydic. see Cyclic projection equivalence of, 275,309 finite, 275,276, 277, 282 properly infinite, 274, 275, 280, 282 rational, 730-735 Projection of norm 1 (in a C*-algebra), 658-662 Properly infinite projection, 274, 275, 280, 282 Properly infinite von Neumann algebra, 295, 296,308,3718-381, 393-396, 482-484, 489, 492-494 Pure state of D ( H ) , 203-205, 615-617 of a C*-algebra, 608611 extension of, 178-180 Pure state space, 614-617,614-621

Q Quasi-central approximate identity, 556-558 Quasi-equivalence of representations, 605407 of states, 607408 Quasi-subequivalence of representations, 605-607 of states, 607408 Quotient C*-algebra, 196-197

R Radical (of a commutative Banach algebra), 106 Range projection, 77,78 Rational operator, 730 Rational projection, 730-735 Reduced atomic representation, 621-622,640 Reflexive (Banach space), 8-10, 20-21,24-25, 26, 28-29,36,572 Regular open set, 224-226

855

856


Representation, automorphic, 809, 811-813, 826-831 of D('H), 190-192 of C ( S ) ,506-512 disjoint, 606-607 equivalent, 488, 506-510 quasi-equivalent, 6 0 5 4 0 7 quasi-subequivalent, 605-607 reduced atomic, 621-622,640 unitary, 545, 808-810,812-813, 827-829 universal, see Universal representation Restricted direct sum, 700 Russo-Dye theorem, 552-555

S Self-dual cone, 522-545 Semi-finite von Neumann algebra, 278-279,804-805,832 Semi-norm, I Semi-simple, 106 Separable Banach space, 30-31, 35-36 Separating vector, 243-245, 318, 360-361, 451453 u-ideal, 224 Simple C*-algebra, 235-236, 620-621, 646-649, 684-687, 728-729

derivation of, 646-649 tensor product of, 664-4387

Singdar functional, 570-571, 612 S h e maps, 766-770 Square mot in a Banach algebra, 132-134, 137 positive (for a compact operator), 64, 66 State

of the CAR algebra, see CAR algebra center state, 444 definite, 150-151, 165-166 extension of, 178-180 factor, 607-608 faithful, see Faithful state normal, ace Nonnal state n-state, 708-711 primary (factor), 607-608 product, ace Product state pure, see Pure state tracid, 337-338, 414-420 of type I,, 111, II,, 111, 603-605 vector, Lee Vector state Stone's theorem, 251, 260 StontWeierstrass theorem, 138 Strong-operator continuity (of functions), 237-240

INDEX OF VOLUMES 111 AND IV Strong-operator topology, 207-209, 212, 214,215, 216, 247 Sublattice, 137 Sublinear functional, 1 Subspace, closed, 4 Support of a normal state, 312-313, 316,321,326 of a vector state, 327,328 Support functional, 1 Symmetric operator, 362-367

T Tensor product center of, 680-684, 770-771 infinite, see Infinite tensor product of operators, 74-75 of simple C*-algebras, 684487 of states, bee Product state Topological group, 641 Topology bounded weak-operator, 808-810 Jacobson, 652 norm, 5-6 strong (on i(R*)), 807-808 strong-operator, see Strong-operator topology ultraweak, 314,318-319 weak, 5-6 weak-operator, 207-217, 314,318-319 Totally disconnected, 220, 223 Trace, in reduced algebra, 447448 Trace norm, 415-417 Trace, normalized, 152 Trace vector, 337-338 generating, 359 Tracial state, 414-420 Tracial weight, 413-414 on U ( ? f ) , 420-427 non-normal, 42 4427 Transformation, linear, s e e Operator Translation invariant subspace in Ll(R), 128-129 in Ll(T~,rn), 129-131

U Ultraweak topology. 314,318-319 Unbounded operator affiliated, s e e Affiliated operator multiplication, 248-250

857

858

INDEX OF VOLUMES 111 AND 1V

symmetric, 362-367 Uniform boundedness principle, 38 Uniform multiplicity (for a bounded nonnd operator), 513-514 Uniformly convex Banach space, 10, 41 Uniformly matricial algebra, 728-729 Unitary element (in a C*-algebra) convex combinations of, 549-555, -79 exponential, 14Cb145, 548,552 Bip, 713-714 Unitary equivalence of bounded normal operators, 513-514 of representations of an abelian C*-algebra, 506-510 of von Neumann algebras, 484494,495499,543-644 Unitary group, 142-143 oneparmeter, 827,829,831 Unitary implementation of automorphic representations, 812-813, 8264327, 830-831 of * isomorphisms, 484-494,495-499,543-544,545, 785-786 Unitary representation of aut(R), 545,808-809 of 1,830-831 universal representation, 546-549.552-553, 555-558, 561-564, 566-570,871-575,594,603404,646,648,658

Universally weakly inner automorphiam, 627, 641-643

V Vector generating, 360-361, 451-452 separating, .we Separating vector Vector state, 152, 187-188, 204, 313-316, 318, 322, 327-334, 338-339, 615-619

Vector state space, 614-615,617418 von Neumann algebra, see r h o Factor abelim, 499-502, 503-505 characterized as dual spaces, 347-356, 662-663 charecterieed as W*-algebras, 345-347 countably decomposable, aee Countably decomposable von Neumann algebra discrete group examplen, 300-305,405-409 finite, a t e Finite von Neumann algebra generated by a bounded normal operator, 499-502, 503-505 generated by two projections, 738-739 maximal abelian, tee Maximal abelian algebra normal, 761, 836-837 properly infinite, see Properly infinite von Neumaan algebra semi-lhite, 278-279, 804-805, 832 of type I, 283, 285, 298, 401, 402, 451, 495-499, 738, 744, 805 of type I,, 285,286.294, 297,299,400,444,494,735 of type I,, 283, 288, 290

INDEX OF VOLUMES I11 AND 1V

of of of of

type type type type

11, 805 111, 283, 291, 391, 448, 756, 759, 760

IIm, 283, 766 111, 283, 785, 805

W W*-algebra, 345, 355-356, 662-663 Weak convergence, 36-38, 43 Weak

* convegence, 6, 31-32,

37-38

Weak * continuous linear functional, 12 linear operator, 12 Weak topology, 5-6 Weak-operator topology, 207-217, 314, 318-319 Wedge product, 774-782 Weight, 356, 357-359 faithfd, 356 normal, 356,357 semi-finite, 356, 357 trecial, see Tradal weight Wick-ordered monomial, 780 Wick-ordered product, 665 Wiener's Tauberian theorems. 128

859


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