FUNDAMENTALS OF THE THEORY O F OPERATOR ALGEBRAS SPECIAL TOPICS
VOLUMEIV
Advanced Theory-An
Exercise Approach
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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
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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
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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
COMPARISON THEORY OF PROJECTIONS
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
COMPARISON THEORY OF PROJECTIONS
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
COMPARISON THEORY OF PROJECTIONS
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
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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
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COMPARISON THEORY OF PROJECTIONS
(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.
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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
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COMPARISON THEORY OF PROJECTIONS
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
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~~~~
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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
COMPARISON THEORY OF PROJECTIONS
.
.
(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),
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+
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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
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COMPARISON THEORY OF PROJECTIONS
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
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( { 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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COMPARISON THEORY OF PROJECTIONS
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
COMPARISON THEORY OF PROJECTIONS
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
COMPARISON THEORY OF PROJECTIONS
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 $0 be Cp, E(ilej. Let y be a vector in ‘H and let gj be Ei .y. From Theorem 5.4.3, there is ? = y j for j in {l?.. . ,n}. We an operator A in 2l such that AEllej have Solution.
n
j=l n
j=1 n
n
j=1 n
j=l
300
COMPARISON THEORY OF PROJECTIONS
6.9.39. Let G be a (discrete) group and F be the set of elements in G whose conjugacy classes are finite. Show that the center C of .CG (and RG)is precisely the set of elements L, in &G such that x is constant 011 the conjugacy class (go) for each go in F and x(g) = 0 if g 4 F. Solution. From the computation of the second paragraph of the proof of Theorem 6.7.5, if L, is in C, x is constant on each conjugacy class. Since x E l2(G), x vanishes on each infinite conjugacy class. Thus 2 has the form described if L, E C. Conversely, if L, E LG and x has the form described, then LXL,, = Lx*x, = Lx,*x = Lx,Lx for each g in G from the same computation. Since {Lxg : g E G } generates LG,L, E C. 6.9.40. Let G be a (discrete) group and F be the net of finite subsets of G partially ordered by inclusion. For x in Z,(G) and IF in F,define zr(g) to be x(g) if g E IF and 0 if g 4 IF. (i) Show that limp LxFzg= Lxxg in lz(G) for each g in G. (ii) Suppose L,: E &G and y in l2(G) is such that limrL,,y exists. Show that this limit is Lxy.
Solution. (i) Since L,xg x * x g = Ro,x, we have IlL,xg
- Lx~gl12= I I R X , ( X P
= xp * xg = R z g x r and L x x g =
- 4112= 1121 - ./I2
=
c
1.(h)I2.
her As x E lz(G), given a positive E , there is a set IF1 in 3 such that Chgp 1x(h)I2 < E if Fl .'lE Thus limr Lxpzg= Lrxg in Iz(G). (ii) Since (xs)*(g)= xr(g-1) = z(g-1) = z*(g) if g E F-l, where B-' = { g - l : g E F), and (zr)*(g) = 0 if g 4 F-', we have that (Q)* = (z*)r-~. Thus for each g in G, (LXY,2g) = (Y,LX*XB) = li$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
COMPARISON THEORY OF PROJECTIONS
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
COMPARISON THEORY OF PROJECTIONS
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
COMPARISON THEORY OF PROJECTIONS
(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
COMPARISON THEORY OF PROJECTIONS
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‘<_G-E, M+N
rn
7.6.20. Let R be a von Neumann algebra with center C acting on a Hilbert space ‘Ifand , let z be a vector in 3t. Show that (i) the range of the support of w,lC is [AA’x : A E R,A’ E R’]; (ii) the support of w,(C is CE where E is the support of w,lR.
Solution. (i) From Remark 7.2.6, the range of the support of w,IC is [C’z : C’ E C‘]. Now C = R n R’so that the von Neumann algebra S generated by R and R‘ is contained in C’, and an operator in S’ is in R and in R’ - hence in C. Thus S’ C and S C’, so that S = C‘. Hence w,lC has support [AA’z : A E R,A‘ E 2’1. (ii) From Remark 7.2.6, the range of E is [A’z : A’ E R’]. From Proposition 5.5.2, the range of CE is [AA’z : A E R,A’ E R’]. Hence, from (i), CE is the support of w,lC. rn
328
NORMAL STATES
7.6.21. Let R be a von Neumann algebra acting on a Hilbert space ‘H and {z,} be a sequence of unit vectors in 7-1 tending in norm to 20. Let E , be the support of w,, IR and P, be the support of wZnIC, where C is the center of R. Show that, in the strong-operator toPologY, (i) {E,Eo} tends to Eo; (ii) {P,Po} tends to PO. Solution. (i) The set consisting of vectors of the form A’z0 with A’ in R’ and vectors y orthogonal to the range of EOhas closed linear span H , from Remark 7.2.6. Of course, E,Eoy = 0 + 0 = Eoy. Now EoA’zo = A’zo and E,A’z, = A’z,, so that IIE,EoA’zo - EoA’soII = IIE,A’xo - A’z011 = “(1- E,)A’~oll = “(1- &)(A‘$,
L IIA’IIIIzn - zoII
-A’zo)~~ +
0.
(ii) The stated result follows from (i), with C in place of R.
w
7.6.22. Let R be a von Neumann algebra acting on a Hilbert space ‘H, {z,} be a sequence of unit vectors in H tending (in norm) to 20, and E, be the support of w,,IR. Does { E n } tend to Eo in the strong-operator topology? Proof? Counterexample? Solution. The sequence { E n } need not tend to Eo. Let R be the operators in f?(‘H), with H a separable Hilbert space, represented by (bounded) diagonal matrices relative to an orthonormal basis { e n } for H . Let 2, be C:=, a,,e,, where am, = 6 3 ( ~ m n ) when -~ m = 2 , 3 , . . . and a l , = [ ( n 2- l)n2 t 6 ] f ( ~ n ) - With l. these choices (5,) converges to el (= zo) in norm and each zn is generating for R (since a, # 0 for all m and n ) . Thus (in this case) each E , ( n # 0) is I and Eo is the (minimal) projection with (one-dimensional) range [ell. w Let R be a von Neumann algebra acting on a Hilbert 7.6.23. space ‘H, and let z and y be vectors in H such that w,lR = w,lR. Show that (i) the mapping Az -, Ay extends to an isometry of [ R z ]onto
[Ryl; (ii) there is a unique partial isometry V’ in R’ with initial space [ R z ]and final space [Ry]such that V‘Az = Ay for each A in R.
329
EXERCISE 7.6.24
Solution. (i) Since
for each A in R,the mapping Ax + Ay of {Rs}onto {Ry} is a well-defined linear isometry. By uniform continuity, this isometry extends uniquely t o an isometric linear mapping of [Rs]onto [Ry]. (ii) There is a unique partial isometry V‘ with initial space [Rs] such that V’Ax = Ay from (i). We note that V’ lies in R‘. Suppose B E R. If y’ is orthogonal to [Rz], then 0 = V’By’ = BV’y’ since (By’, A s ) = (y’, B * A s ) = 0 for each A in R. Moreover, for each A in R, V’BAz = B A Y = B V ’ A s . Hence the bounded operators V‘B and B V ’ agree on subsets that span 3-1. Hence V ’ B = B V ’ for each B in R and V’ is in 72’.
7.6.24. Let w and w’ be normal positive linear functionals on a von Neumann algebra 72 acting on a Hilbert space H . Suppose that the union of the support projections of w and w’ is a cyclic projection G in R. Show that (i) there are normal positive linear functionals w1 and w2 on R with orthogonal supports contained in G such that
and that there is a vector z in H such that w,lR = w t o 2 = w‘+wl; (ii) there are operators H’ and K‘ in R’ such that 0 5 H’ 5 I, 0 5 Ir“ 5 I , w = w ~ l , l R ,and w‘ = w1pzlR; (iii) J ) z- H’z1I2 5 IJw2]1,) ) z- ~ ’ ~ 51 J 1J w~l J J ; (iv) there are vectors 5 and s’ in H such that w,lR = w, w,lIR = w’, and 112 - 2/11
5 PlIw - w
I
1111 1 2
*
Solution. (i) From Theorem 7.4.7, w - w’ = w1 - w2, where w1 and w2 are normal positive linear functionals on R with orthogonal supports M and N , respectively, such that
330
NORMAL STATES
Since the supports of w and w' are subprojections of G, w(GTG) = w ( T ) and w'(GTG) = w'(T) for each T in R. From Exercise 7.6.19(ii), M N 5 G. It follows that
+
+ w ~-)G) ( I= + ~ l ) ( l G- ) = 0, so that the support E of w + w2 (= w' t w1) is a subprojection of (0'
(0
G. By assumption, G is cyclic, so that E is cyclic from Proposition 5.5.9. There is a vector z in H such that w,lR = w t w2 = w' 01 from Proposition 7.2.7. (ii) Since w 5 wEIR,there is an operator H' in R' such that 0 5 H' 5 I and w = W H ~ from ~ ~ Proposition R, 7.3.5. Similarly, w' 5 w,JR and there is an operator K' in R' such that 0 5 11' 5 I and w' = wlpzlR. (iii) As (I5 I - HI2, we have
+
11% - H'#
= {(I - H 1 ) 2 2 , Z ) 5 ((I- H ' 2 ) 2 , Z ) = ( z , z ) - (H'z,H'z) = (WE - W H ' Z ) ( I ) = IIw2II.
Similarly, IIz - I P z ~ I5~llwl 11. (iv) From (iii), IlK'z
- Z1l2 + 1. - H'z112 I llw1II t llw2II = IIw - 01 '1
and llri'z
- H'2.11 5 IlK'z - ZII + 112 - H'zll 5 [211w - w I 1111/2 .
We complete the solution by choosing H'a as s and 11'2 as s f . 7.6.25. Let w be a normal positive linear functional on a von Neumann algebra R acting on a Hilbert space 3-1, and let be a vector in 'H. Suppose that the union of the supports of w and wvlR is cyclic in R. Show that (i) there are vectors s and z' in 'H such that
w,lR = w , U,lIR = wvlR, 11s - 2'11 5 [211w - wulRll]1'2; (ii) there is a partial isometry V' in R' with initial space [Rs'] and final space [Rv]such that V's' = v and that if W' is a parand [Rs'] (in tial isometry in R' whose initial space contains [Rz] with V', particular if W' is an isometry) and W' agrees on [Rx'] then w = w,lR, (*) where u = W'x.
-4 15 [2110 - wvlRl111/2,
IJU
331
EXERCISE 7.6.26
Solution. (i) If we use w,lR in place of w' in Exercise 7.6.24, we conclude that there are vectors x and x' with the properties stated. (ii) From Exercise 7.6.23(ii), there is a partial isometry V' in R' with initial space [Rx'] and final space [Rv] such that V ' A d = Av for each A in R. If W'is as described, then W'x'= v and W'*W'x= x so that w,(R = w,lR = w , where u = W'z.Moreover, llu
- 011 = IIW'X - v11 = IIW'X - w'x'11 = JIX - 5'11 5 [2(JLd- W , J R J J ] 1 / 2 .
m
7.6.26. With the notation and assumptions of Exercise 7.6.25, and P be the central projeclet E' be the projection with range [Rv] tion in R (whose existence and uniqueness is guaranteed by Proposition 6.3.7) such that P 5 C,y, PE' is either properly infinite or 0 (in which case P is 0), and ( I - P)E' is finite. Show that there is an isometry U' in R' such that U'([Rx'] = V'([Rx'] and conclude that there is a vector u satisfying (*) if either of the following two conditions holds: (i) PE' = 0 [Hint. Use Exercise 6.9.10.1; (ii) R'P is countably decomposable and both PE' and P ( I - E') are properly infinite with central carrier P .
Solution. If there is a U' as described, Exercise 7.6.25(ii) applies with U' in place of W' and (*) is satisfied with U'x for u. (i) If PE' = 0, then P = 0 and E' is finite as is the projection F' with range [Rx']. From Exercise 6.9.10, we can find a unitary operator U' in R' such that U'F' = V'. (ii) With F' as in (i), P ( I - F ' ) is countably decomposable since RIP is. By assumption, P ( I - E') is properly infinite with central carrier P. Since PV' is a partial isometry in R'P with initial projection PF' and final projection PE', there is an isometry UA in R'P such that UAPF' = PV' from Exercise 6.9.10(ii). Now ( I - P)E' is finite and (I - P)V' is a partial isometry in R'(I - P ) with initial projection ( I - P)F' and final projection ( I - P)E'. Again, from Exercise 6.9.10(i), there is a unitary operator Ui in R'(I - P ) such that U{(I- P)F' = ( I - P)V'. Let U' be the (unique) linear operator whose restriction to P ( X ) is UA and to ( I - P)(7-t)is U i . Then U' is an isometry in R' and
U'F' = U'(PF' -t ( I - P ) F ' ) = UhPF'
+ U i ( I - P)F' = V'.
332
NORMAL STATES
7.6.27. With the assumptions of Exercise 7.6.25 and the notation of Exercise 7.6.26, show that there is a vector u satisfying w J R = 0,
llu - v11 _< 211w - W V l R ( 1 * ' 2
when R' is countably decomposable. [Hint. In case PE' is properly infinite and P ( I - E') is not, express PE' as a countably infinite sum of projections in R' equivalent to PE' and use these projections to replace v by a close approximant v' such that PG' and P ( I - GI) are properly infinite, where G' has range [Rv']. Remember to prove that the union of the supports of w and w ~ I JisRcyclic in applying Exercise 7.6.26( ii) .]
Solution. In view of the results of Exercise 7.6.26, it remains to study the case in which PE' is properly infinite and P ( I - El) is not properly infinite with central carrier P. As in the proof of Theorem 6.3.4, construct a countably infinite (orthogonal) family {PEL} of projections equivalent to PE' with sum PE'. Given a, positive E , there is an n'(# 1) such that IIPEk,vll < E . Let v' be ( I - P ) v t (&, PEL)v. Then ((21'
C PEA). - PE'vII = IIPEh,vll <
- ~ 1 =1 I[(
E.
n#n'
The range of ( I - P)E' -t (PE' - PEL,) (= GI) is
(A(I- P)
[Rv'] since
+ B P ) d = ( I - P)Aw+ (PE' - PEL,)Bv
-<
for each A and B in R. As PE; 5 G' and PEI PE', we have G' so that that PCEI 5 CGI. At the same time, ( I - P)E' ( I - P)CEI 5 CQ. Thus CEI5 C G ~But . G' 5 E l , SO that CQ 5 C E I .Hence CEI = CGJ.Since P 5 CEI = C Q , PG' = & ', PEL, and ( I - P)G' = ( I - P)E', we have that P is the unique central subprojection of CGIsuch that PG' is properly infinite and ( I - P)G' is finite. In the present case, P ( I - GI)(> 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
< <
T, I. With u we have from Proposition 4.2.8(iii) that 0 5 Tntl a given unit vector in H and m large enough ThI2u is close to T112u
342
NORMAL STATES
since the mapping A + All2 is strong-operator continuous on the unit ball of B(7f)+ from Proposition 5.3.2. (In fact, this mapping is strong-operator continuous on 8(7f)+from Exercise 5.7.36(iii).) Now when n I m, n
( T A / 2 ( C ( A k- A ~ - I ) + ) T A /U~)u , k= 1 m
5 ( T A / 2 ( C ( A-~A ~ - I ) + ) T A / ~ u , u ) k=l m
= ( T m ( X ( A k- Ak-l)+).,.)
I (I.,.);
k=l
and as m -+
00,
k= 1
Thus T 1 / 2 ( C ; = l ( A k- Ak-1)+)T1l25 I for each n in N. (iii) From (ii), for each j in { 1,.. ,n }
.
n
n
Thus { T 1 / 2 ( C ; = j ( A k- Ak-1)+)T1/2}is an increasing sequence (over n ) of operators in 2.l bounded above by I. Its limit Cj lies in 2.l and 0 5 Cj 5 I. Since n
n
for each j and all n, {Cj}is decreasing. In addition, for each n and m, m
n
m
k=l
343
EXERCISE 7.6.36
n k=l
t Cn+1}is monotone (iv) It follows from (iii) that {T1/2AnT1/2 decreasing. Since l
l
~
~
/
+ Cn+lII ~ ~ I n I I~ T I I I I A~~ I I / +~IICn+lII 5 2,
+
{T1/2AnT1/2 Cn+l}is bounded below by -21 and has a strongoperator limit B in U. At the same time T1l2AT1l2 C is a weakoperator limit point of {T1/2A,T1/2t Cn+1} where C (in U) is the strong-operator limit of {Cj}. Thus T1l2AT1I2t C = B and ~ 1 / 2 ~ ~ (= 1 /B 2E 8. Since T E U, R ( T ) E Q from Exercise 7.6.35(ii). Now (v) ( T ( z +CE"=,Ak- Ak-i)+z), u ) (= ( z + C E 1 ( A k- Ak-i)+Z, T U ) ) is approximated closely by:
+
c)
((2
n
n
k= 1
k=l
+ c ( A k - Ak-l)+z, [ I + X ( A k - Ak-l)+]-'u) = ( z , u )
for large n, where u is any preassigned vector in 7-l and z is either x or y. (From (i), converges to some vector in 'H when z is either x or y.) Thus T ( z - Ak-l)+z) = z , and x , y E R(T)('H). (vi) Suppose A is a maximal abelian self-adjoint subalgebra of U. If {A,} is a bounded increasing net in A, its strong-operator limit lies in 2l and commutes with A. Hence that limit lies in A. It follows that A satisfies the same condition as U, so that what we have proved thus far about U applies to A. In other words, for this part of the exercise, we may assume that U is abelian. With that = AT E U. Since x is in the range of T and assumption, T1I2AT1I2 As = x , x is in the range of AT. In addition AT (= T A ) is selfadjoint and TAy = 0 since Ay = 0. Thus R ( T A ) € 8, R(TA)x = x , and R(TA)y = 0. From Exercise 7.6.35(v), (vi), U = U- (that is, A = A - ) in this case.
ELl(&
+
344
NORMAL STATES
7.6.37. With the notation and assumptions of Exercise 7.6.36, show that (i) M A N lies in U, where M and N are spectral projections for T corresponding to bounded intervals with positive left endpoints; (ii) M,AF and FAMm are in U, where F = R ( T ) and { M m } is a sequence of spectral projections for T corresponding to bounded intervals with positive left endpoints such that EmM , = F [Hint. Consider (MmAMn i- Mn)(MnAM, M n )and suitable monotone limits .]; (iii) F A F A F E U; (iv) F A F A F z = z and F A F A F y = 0; (v) ?a=%-.
+
Solution. (i) Let S be a bounded interval with a positive left endpoint and let g ( t ) be t - l for 1 in S and 0 for t in R\S. From Exercise 7.6.36(vi), a maximal abelian subalgebra A of U containing T is weak-operator closed in B ( H ) and, therefore, contains g ( T ) . From Theorem 5.2.8, g(T)T = M , where A4 is the spectral projection for T corresponding to S . Since T A T E U from Exercise 7.6.36(iv), M A T ( = g ( T ) T A T ) E U. Similarly, M A N E U, where N is another spectral projection for T corresponding to a bounded interval with positive left endpoint. (ii) Since (MrnAMn -t lvn)(MnAMm t Mn) = MmAMnAMm MmAMn
+
+ MnAMm + Mn
and C,”=l M n ( = .F) E U, we have that
MmAFAMm
+ MmAF + F A M , + F E 2l
and also MmAFAMm E U. (Both operators are strong-operator convergent sums of positive operators in U.) Thus M , A F + F A M , E U, and M,AF MmAMm(= Mm[MmAF F A M , ] ) E U. Since MmAMm E U, we have that MmAF E U. (iii) From (ii)
+
+
( M , A F ) * ( M m A F(= ) F A M m A F ) E U. Again, 00
F A F A F (=
C FAM,AF)
m=l
E U.
EXERCISE 7.6.38
345
From Exercise 7.6.36(v), F x = x and F y = y , so that FAFAFx = x and FAFAFy = 0. (iv)
(v) Combining the conclusions of Exercise 7.6.35 with what we have proved thus far, we see that for each cyclic projection E in 2lwith generating unit vector x and each unit vector y in ( I - E ) ( R ) , there is a self-adjoint operator FAFAF in 2l such that FAFAFx = x and FAFAFy = 0. The conditions of Exercise 7.6.35(v), (vi) are fulfilled, and 2i = 2l-. m[56]
Let U be a C*-algebra with the property that each 7.6.38. bounded monotone-increasing net in 2l has a least upper bound in % and suppose 2l has a separating family of states (if each state of the family is 0 at some positive A in 9,then A is 0) whose limits on such nets are their values at the least upper bounds. We call such states of !? normal l and refer to a C*-algebra satisfying these conditions as a W*-algebra. Show that a C*-algebra is * isomorphic to a von Neumann algebra if and only if it is a W*-algebra. [Hint. Observe that A + T*AT is an order isomorphism of %h onto itself for every invertible T in 9.3
Solution. Suppose 2l is * isomorphic to a von Neumann algebra,. The * isomorphism transforms increasing bounded nets onto such nets, least upper bounds onto least upper bounds, and normal states onto normal states. Thus % is a W*-algebra in this case. Suppose 2l is a W*-algebra and cp is the mapping CaEA en,(,), where {q(u) : a E A} is the family of normal states of 2l. Since { q ( a ) } is separating for U, cp is a * isomorphism. Write xa for xq(,) and suppose {cp(&) : b E Ira} is a bounded increasing net in cp(2l) with strong-operator limit B . Then {Ab} is a bounded increasing net in %. By assumption { & } has a least upper bound A in 2l and {q(a)(Ab)}tends to q(a)(A)for each a in A. Thus {(V(Ab)Za,2,)) tends to ( v ( A ) x a ,xa). But {(q(Ab)xa,2,)) tends to ( B x a , x a )as well. Thus ((cp(A)- B ) x a , x a )= 0 for each a in A. With T as in the hint, {T'AbT} has T*AT as least upper bound and cp(T*At,T) has cp(T)*Bcp(T)as strong-operator limit. Thus ((cp(A)- B ) q ( T ) x acp(T)x,) , = 0 for each a in A and each invertible T in 2l. With S in 2l, S nI is an invertible element of 2l for all
+
346
NORMAL STATES
large positive integers n. Thus
when n is large. But ((cp(A)- B ) x , , x , ) = 0 and the inner product ((cp(A) - B)cp(S)xa,cp(S)xa)is independent of n. It follows that ((cp( A ) - B)cp(S)x,, cp( S ) x a )= 0 for each S in 2l. Since x, is a cyclic vector for the representation ~ ~ and( cp is~ the1 direct sum of ( ~ ~ :( ~ 1 a E A}, p(A) = B . Hence cp(2l) satisfies the conditions of Exercise 7.6.37, and cp(U) = cp(U)-. Thus U is a W*-algebra. m[56] Let 2l be a C*-algebra acting on a Hilbert space. 7.6.39. Suppose that each bounded increasing sequence in U has its strongoperator limit in 2l and that each Orthogonal family of non-zero projections in 2l is countable. Show that 2l= 2l-. [Hint.Note that the only use of nets (as opposed to sequences) in Exercise 7.6.35-37 is to establish that arbitary unions of projections in U lie in U and prove this under the present assumptions.] Solution. As indicated in the hint (and noting the argument of the solution to Exercise 7.6.35(iv)) it suffices to establish that the union of an increasing net of projections in 2l is in U under the present assumptions. We establish that the union F of an arbitary family {Fa : a E A} of projections in U lies in U. Let {Eb : b E B} be a maximal orthogonal family of (non-zero) projections in 2l such that Eb 5 F for each b. By assumption B is countable (possibly finite) so that we can denote the family {Eb} by { E l , E2,. ..}. Since { E l , El E2,. .} is an increasing sequence of projections in a, its strong-operator limit En(= E ) is in 2l by assumption. We assert that E = F . Since E 5 F , E V Fa 5 F for each a in A. The range projection of E -t Fa is E v Fa and is the strong-operator limit of the increasing sequence { [ $ ( Et Fa)]lIn}.Thus E V Fa lies in U as does E V Fa - E . If E V Fa - E # 0 , it can be adjoined to { E l , E z , .. .} to form a larger orthogonal family of non-zero projections in U contained in F - contradicting the maximality of { E l , &, . . }. Thus E V Fa = E , Fa 5 E for each a in A, and F = E E 2l. m[56]
+
.
En
.
347
EXERCISE 7.6.41
Show that a C*-algebra U is * isomorphic to a count7.6.40. ably decomposable von Neumann algebra R if and only if each bounded increasing sequence in U has a least upper bound in U, there is a separating family of (normal) states of U whose limits on such a sequence are their values at the least upper bound, and each orthogonal family of non-zero projections in 2l is countable.
Solution. Suppose U is * isomorphic to a countably decomposable von Neumann algebra R acting on a Hilbert space 3-1. Then bounded increasing sequences in U map onto such sequences in R under the isomorphism. The least upper bound of the image sequence in R is the image of an element of U that is the least upper bound of the sequence in U. Vector states of R composed with the isomorphism are normal states of U, and the set of such form a separating family for R. An orthogonal family of non-zero projections in U maps onto such a family in R. Since R is countably decomposable, the family of projections is countable. The argument of Exercise 7.6.38 applies to a C*-algebra U satisfying the given conditions, with sequences and Exercise 7.6.39 used in place of nets and Exercise 7.6.37. Thus U is * isomorphic to a von Neumann algebra R. From the condition on orthogonal families of projections in U, R is countably decomposable. w[56] 7.6.41. Let U be a C*-algebra and UzL, be a Banach space such that U is (isometrically isomorphic to) the (norm) dual space of U#. (i) Show that an element A in U is a self-adjoint element in the ball (U)Tof radius r in U with center 0 if and only if IIA in11I2 5 r2 n2 for each integer n. (ii) Show that the set of self-adjoint elements in (U), is weak * closed in U. (iii) Show that the set (a+),of positive elements in (U), is weak * closed in U.
+
+
Solution. (i) Note that for each A in U
Thus if A is a self-adjoint element in (U),,,
348
N O R M A L STATES
for each integer n. Suppose A = A1 t i A 2 with A1 and A2 self-adjoint elements in 2. If A2 # 0, then some non-zero b lies in sp A2. Now b t n E sp(A2 t n l ) , so that
But with In1 large and nb positive, T2
+ n2 < b2 + 2bn -t n 2 ;
+
whence r2 n2 < [ [ At in11I2 for such n. (ii) We think of the elements of U, as linear functionals on U by means of their isometric injection into the norm dual of U (the second dual of %#:I. If 77 is such an element, we write q ( A ) for the value of A at q. With this notation,
Thus, for each integer n,
and these sets are weak * closed. (This could be deduced from Theorem 1.6.5(i), as well.) From (i),
is weak * closed. (iii) From Lemma 4.2.1, if IlAll 5 T and A is self-adjoint, A is positive if and only if [[A- ~1115 T . Thus A E (a+),if and only if A E (%h)r and Iq(A - T I )5[ T for each 77 in (%#)I.That is so that
(!&)r
From (ii), each of the sets { A E ( U h ) r : Iq(A - T I )5[ T } is weak closed. Thus (2l+), is weak * closed.
*
EXERCISE 7.6.42
349
7.6.42. Adopt the notation of Exercise 7.6.41, and let F be the family of subsets of M whose intersection with every (a), is weak * closed, where n is a positive integer. With S a subset of U, and a a positive number, denote by M(S, a ) and Mo(S,a ) , respectively, the subsets
L a , q E S } , { A E a : Iq(A)l < a,q E S); let 2l,(S,a) and Ut(S,a) denote the sets U(S,a) n (U), and Mo(S,a ) n respectively. F is the family of closed sets of a topology (the Show that (i) {A E
: Iq(A)l
“F-topology”) for M. (ii) Let 0 be an F-open subset of U. Show that for each A in M, A 0 is an F-open set. Conclude that the mapping B + A B of 3 onto itself is an F-homeomorphism. (iii) Let { q j } be a sequence of elements of 8, tending to 0 in norm. Show that UO({vj},a) is F-open for each positive a. (iv) Let 0 be an F-open set containing 0 and n be a positive integer. Show that there is a finite subset S of U, such that U,(S, 1) is contained in On(%),, and that there is a finite subset 7 of (M),1, such that % L +(S l u 7 , 1 ) c 0 n (U)n+l* [Hint.Assume the contrary and establish the finite intersection property for the family
+
+
{U,+l(S U 7 , l )n (8\0) : 7 a finite subset of show that an element of the intersection of this family lies in (U), and deduce a contradiction.] (v) Let 0 be an F-open set containing 0. Construct a sequence {S,} of finite sets S, such that Sn+l C_ for n in { 1,2,. . .} and such that for n in El, U,(S1 u . . - u S,, 1) c 0 n (a),. (vi) For a given F-open set 0 containing 0, construct a sequence { q j } in 31, tending to 0 in norm such that % ( { q j } ,1) C 0. (vii) Given an 3-open set S in M containing 0, find an F-open set 00containing 0 such that 0 0 t 0 0 C 0 and conclude (using (ii)) that addition is F-continuous on 8. (viii) Show that the mapping (a,A)-aA: CXM-U is F-continuous. Conclude (using (vi) and (vii)) that F provides U with a locally convex linear topological structure.
350
NORMAL STATES
Solution. (i) If { F a } is a subset of 7 ,then
((7 F a ) n (a), = n ( F an (a),) a
a
and each Fa n (a),is weak * closed by definition of T . Thus is in 3. If { F l , . . . ,F,} is a finite subset of F,then m
(UFj)n (U), =
u
n,
Fa
m
(Fjn (%d;
j=1
j=1
and Uj”=, Fj E T . In addition, 8 and U are in F as is each weak * closed subset of U. Thus F is the family of closed subsets for a topology on U, the T-topology. (ii) We show that ( A t 0 )n (U), is a (relative) weak * open subset of (U),. Suppose C = A t B with B in 0 and llCll 5 n (so that C E ( A t 0 ) n (U),). Then IlBll = IIC - All n t llAll 5 n t m, where llAll 5 m. Thus B E 0 n (a),+, and 0 n (%),+m is a (relative) weak * open subset of (U),+, since 0 is F-open. It follows that there is a weak open set 01 in U such that 0 n (U),+m = 01n (U),+m. Now A t 01 is weak * open, so that ( A t 01) n (U), is a weak * open subset of (U), containing C (= A B ) . Moreover, ( A t 01)n (U), E ( A t 0 )n (U),. Thus (A -t 0 )n (U), is a weak * open subset of (U), for each positive integer n, and A -t0 is T-open. Since the mappings B --$ A t B and B -+ - A -tB are inverse to each other and both are F-open mappings of U onto U, both mappings are F-homeomorphisms. (iii) We show that U:({qj},u) is a weak * open subset of (U), for each positive integer n. In fact, since Jlqjll + 0, there is a positive integer n~such that llqjll < ? when j > no. Thus ( q j ( A ) ( llqjllllAll < a when A E (U), and J > no. It follows that U:({qj},a) = U:({ql, ...,q,,,}, a ) andU:({qj},a)is aweak * opensubset of (U),. (iv) Since 0 n (U), is a weak * open subset of (U), containing 0, there is a weak * open set 00containing 0 such that 0 n (U), = 00n (U),. Since 00is weak * open, there is a finite subset S of elements of U, such that U(S, 1) C 00,whence
<
*
+
<
%(S, 1) C
00
n (U),
= 0 n (U),.
Suppose that for each finite subset 7- of (U,),,, ant1
(Su 7-71) g 0 n (Q )n+l.
351
EXERCISE 7.6.42
Then for each such 7,
.
Let 3,. . , 7 k be finite subsets of (2l#)1ln, and let '& be their union. Then & is a finite subset of (U,)l/n and
Now %,+1(SU~, l)n(%\O)# 8;and the collection of sets described in the hint has the finite intersection property. Since each set of the form 2l(S U 7,l)is weak * closed a6 is (%\0) n (%)n+l(for 8\0 is 7-closed), and is weak * compact, the entire collection of sets has a non-empty intersection. Let A0 be an element of the intersection. For each q in ( % # ) * i n ,
so that )v(Ao)J5 1. Since !2l is the dual of
Thus A0 E
a,,
(a)*.By choice of A0 and S,
But A0 E %\O - a contradiction. Thus there is a finite subset 7 of (a,),/, such that
(v) From (iv), there is a finite subset S1 of
such that % I ( & , 1)
g 0 n (%)I and a finite subset S2 of (%,)I such that %2(& U S2,l) 2 0 n(2i)2. Similarly, there is a finite subset S3 of (8,);such that % 3 ( S 1 US^ US^, 1) On(%)3. Continuing in this way, we construct the sequence, S l , & , S 3 ,
. ,. , with the stated properties.
352
NORMAL STATES
(vi) The sequence { q j } constructed by enumerating the elements of the sequence constructed in (v), 81,Sz,$3, . . . , successively, tends to 0 in norm in U,, and 1) c U,(Sl u . * . u &, 1) G 0 n (%)a
8,({77j}jC,M,
for each positive integer n. With A in U({qj}, l), choose an integer n such that IlAll < n. Then
A E %({qj}, 1) E 0 n (U), and A E 0. Thus U({qj},l) c 0. (vii) Let { q j } be the sequence constructed in (vi). From (iii), 0 0 containing 0. If A and B are in 00,then Iqj(A)I < !j and 1qj(B)l < f, whence 1qj(A t B)I < 1 and A t B € Uo({qj},l)C 0.Thus 00t 00C 0. Given Ao, Bo in U and an F-open neighborhood 01 of A0 t Bo, -(A0 t Bo) t 01 is an F-open neighborhood 0 of 0 from (ii). As in the preceding paragraph, find an 3-open neighborhood 00of 0 such that 00t 00C_ 0. Again from (ii), A0 t 00and Bo t 00are F-open neighborhoods of A0 and Bo, respectively, and Uo({qj},$) is an F-open set
Ao t 0
0
t Bo t 0
0
= Ao
+ Bo t
0 0
t 0 0 C Ao t Bo t 0 = 01.
Thus addition is F-continuous on U. (viii) Given a0 in C, A0 in U, and an 7-open neighborhood 0 of 0, we can find a sequence { q j } of elements of U# tending to 0 in norm such that U((qj},l) C (3 from (vi). Let 6 be the smallest of 1, $(1u01 t 1)-l, and $(Iqj(Ao)( t l)-' (for j in N). F'rom (iii), UO({qj},b) is an F-open neighborhood 00of 0. If la - a01 < 6 and A E A0 t 0 0 , then Iqj(aA - aoAo)l I lqj(aA - aAo)l t 177j(aAo - aoAo)l
I laIl%(A - A011 -I- la - aoll~.i(Ao)l I (1.01 t 116 t bI~j(Ao)lI f $ I 1.
+
Thus a A E aoAo t 0, and the mapping (a,A)+aA:
CXU+U
is F-continuous. It follows from this and (vii) that 7 imposes a linear topological structure on U. As noted, given the 7-open neighborhood 0 of 0, U({qj}, 1) G 0 and Uo({qj},1) is a convex 7-open neighborhood of 0. Thus F imposes a locally convex topology on U. I
EXERCISE 7.6.43
353
7.6.43. With the notation and terminology of Exercise 7.6.42: (i) identify the F-continuous linear functionals on U [Hint.Use Exercise 1.9.15.1; (ii) show that a convex subset of 2f is weak * closed if and only if it lies in F ; (iii) conclude that the sets of self-adjoint and positive elements in U are weak * closed.
Solution. (i) We show that a linear functional p on U is Fcontinuous if and only if p is weak * continuous. If p is 7-continuous and a positive E is given, there is an F-open neighborhood V of 0 such that Ip(A)I < E when A E V . Now V n (U)2 is a (relative) weak * open subset of (a), containing 0. Thus p is weak * continuous a t 0 on (a),,and by translation, p is weak * continuous on (%)I. From Exercise 1.9.15, p is weak * continuous on 2l. If p is weak * continuous on U, then p is F-continuous since the F-topology on 24 is stronger than the weak * topology. (ii) Since each weak * closed set in 24 is F-closed, it suffices to show that each F-closed convex subset K of 24 is weak * closed. From Exercise 7.6.42, the F-topology is a locally convex topology on U. Thus if A0 E U\K, there is an F-continuous linear functional q on 2l and a real b such that, for each A in K , Req(Ao) > 6 2 Req(A), from Corollary 1.2.12. Now 7) is weak * continuous on U from (i). Thus K is the intersection of the weak * closed half-spaces containing it, and K is weak * closed. (iii) From Exercise 7.6.41, (a+), is weak * closed for each integer n ; that is, Uf is F-closed. From (ii), U+ is weak * closed since U+ is convex. Similarly, the set of self-adjoint elements in U is weak * closed. 7.6.44. With the notation and assumptions of Exercise 7.6.42, let Uh, and Uh be the real-linear spaces of hermitian elements in 3, and Q, respectively. (i) Suppose T E a\!&. Show that there is an q in U i such that Imq(T) # 0. [Hint.Use Exercise 7.6.43(iii) and a Hahn-Banach separation theorem.] (ii) Suppose A is a non-zero element of 2 l h . Show that there is an q in Uhy such that q ( A ) # 0 and that ah, separates U.
354
NORMAL STATES
= U,. [Hint. Show that U: is norm (iii) Show that U: t i%: closed and use Exercise 1.9.5.1 Show that there is a state q of U in (iv) Suppose A E ah\%+. %# such that q ( A ) < 0. (v) Deduce that, with A and B in ah, A I B if and only if q ( A ) 5 q ( B ) for each state q of % in %#.
Solution. (i) From Exercise 7.6.43(iii), Uh is weak * closed. Since T E %\%h, there is an element q’ in U, and a real b such that, for each A in %h, Req‘(A) 5 b < Req’(T). For each integer n!,nA E %h so that n Re q’(A) = Re q’(nA) I b. Hence R e q ’ ( A ) = 0 for each A in ah. Let q be iq‘. Then q ( A ) is real for each A in ?&- that is, q is a hermitian functional on U in %#. Since Req‘(0)= 0 I b < Req‘(T), we have Im q(T) = Im iq’(T) > 0.
(ii) From (i), there is a hermitian linear functional q on % in U# such that Im q(iA:)# 0 since i A # ah. But q( A ) = Re q( A ) = Im iq(A) = Im q( i A ) # 0.
It follows that the real linear subspace Uh, separates U, for if T = A1 iA2 with A l , A2 in and T # 0, then at least one of A l , A2 is different from 0 and there is an q in Uh, such that q ( T ) = q(A1)t irl(A2) # 0. (iii) If % in %# is a norm limit of elements in ah,,then % ( A ) E B for each A in %h; and qo E % .: Thus %: and :%i are closed, reallinear subspaces of %#. Now %: t i%h, is a complex-linear subspace of %# that separates the elements of %. As U is the dual of %#, %: i%: is norm dense in U# from Corollary 1.6.3. Suppose ql and qz are elements of norm 1 in Uh,. There is an A in (%)I such that < 1q1(A)I. Multiplying A by a suitable scalar of modulus 1, we may assume that 171(A)I = q1(A) = q l ( [ A4- A * ] / 2 )and, hence, that
+
+
a
EXERCISE 7.6.45
355
A E ah. In this case, q1(A) is real and Z%(A) is purely imaginary, so that 1 ;i < I17l(A) - 2172(A)I L I1171 - 2rlZII.
+
From Exercisel.9.5, ah, 2%: is norm closed in U,. Hence U, = Uh, 2%.: (iv) From Exercise 7.6.43(iii), is weak * closed. Since A E %\a+,there is an 170 in U, and a real b such that for each H in UU+,
+
In particular, with 0 for H , we have that b 5 0. From (iii), 171 2172 with 71 and 772 in U i . Thus
+
=
since A and H are in a h . Hence ql(A) < b 5 0. If r,q(H) < 0 for some H in U+, then n q ~ ( H=) r ] l ( n H ) < 6 for a suitably large positive integral n and nH E U+. But this contradicts the property of 171 just established. Thus 0 I q l ( H ) for each H in a+. Since ql(A) < 0, 71 # 0 and some positive scalar multiple 17 of 171 is a state of U in 3, with the desired properties. (v) If A 5 B and 17 is a state of U, then q(A) q(B). On the other hand, if A $ B , then B - A E Uh\U+ and q(B - A) < 0 for some state 7 of U in Q,. For this state 17, q ( B ) < q(A). Thus, if q ( A ) 5 q ( B ) for each state 17 of U in U,, A 5 B . 7.6.45. With the notation and assumptions of Exercise 7.6.41, show that (i) each monotone increasing net in U with an upper bound has a least upper bound in U; (ii) U is a W*-algebra in the sense of Exercise 7.6.38. Solution. (i) Let be an increasing net of operators in U that is bounded above. To show that { A , } has a least upper bound in U, it will suffice to show that the cofinal subset {A,}+,, has a least upper bound in U; we may assume that {A,} C_ (U),. for some positive r and that {A,} has a first element A,,. Since the mapping T + T A is an order isomorphism of U onto itself for each self-adjoint A in U, it will suffice to show that { A , - A,,} has a least upper bound in U; we may assume that { A , } (U+)..
+
356
NORMAL STATES
Since (at),is weak * compact, { A , } has a cofinal subset { A , , } convergent to some A in (a+),. With q a state of U in U,, {q(A,t)} converges to q ( A ) . As {A,,} is monotone increasing and 77 is a state of U, ( q ( A a t ) }is monotone increasing to q(A). Thus q(A,j) _< q ( A ) for each a' and each state q in a,. If A - A, # 21t for some a in A, then A - A,, # Ut when a' 2 a and there is a state q of U in U, such that q ( A - A,#) < 0 from Exercise 7.6.44(iv). But as just noted, q(A,!) 5 q ( A ) - a contradiction. Thus A, 5 A for each a in A; A is an upper bound for {A,}. Suppose B is an upper bound for {A,} in U. Then q ( A , # ) 5 q ( B ) for each state q of M and, in particular, for each such q in U,. But with q in U,, { q ( A , l ) } converges to ? ( A ) . Thus q ( A ) 5 q ( B ) for each state q of B in a,. From Exercise 7.6.44(v), A 5 B ; hence A is the least upper bound of {A,} in U. (ii) From (i), each state of 2l in U, is normal in the sense of Exercise 7.6.38. If A E Ut and q ( A ) = 0 for each normal state q of U, then q ( - A ) 2 0 for each such q and -A 2 0 from Exercise 7.6.44(v). Hence A = 0; the set of normal states of U is separating. Combining this with the result of (i), we see that the conditions of Exercise 7.6.38 are fulfilled. Thus 2l is a W*-algebra. w[92] 7.6.46. Let R be a von Neumann algebra. Prove that (i) R has a faithful normal semi-finite weight; (ii) R has a faithful normal state if and only if R is countably decomposable.
Solution. Let {E, : a E A} be an orthogonal family of non-zero cyclic projections in R, with sum I. For each a in A let z, be a unit vector such that [R'z,] is the range of E,. The set {z, : a E A} is generating for R',and is therefore separating for R. (i) We can define a weight p on R by p(B) = E ( R z a , z a )
( B E R+).
aEA
It is clear that p is normal, and p is faithful since the set {z,} is separating for R. For each index a, p(Ea) = 1. For each finite subset F of A, p ( E ( F ) )< 00 and thus E(F) E Fp, where E(F) = C a E F E a . Moreover E(ILr)RE(lF)E Fp, for each R in R+,since 0
I W ) R E ( F )I II~IIE(F)E Fp.
EXERCISE 7.6.47
357
The linear span M , of F, contains UE(IF)RE(IF) (where IF runs through all finite subsets of A). Thus M ; = R, and p is semi-finite. (ii) If R is countably decomposable, the index set A is countable, and can be replaced by { 1 , 2 , 3 , . . .} or by { 1 , 2 , 3 , . .. ,k} for some k. The equation w(R)= Cn-2(RZn,Zn),
c
where c = (C n-2)-1, defines a faithful normal state w on R. Conversely, suppose that R has a faithful normal state w. Given any orthogonal family {Gb} of non-zero projections in R,we have
and hence the index set { b } is countable. Thus R is countably decomposable. This part of the exercise also follows by applying Exercise 7.6.13 rn to I in R, together with Exercise 7.6.6.
7.6.47. Find an example of a normal semi-finite weight p on a von Neumann algebra R and an operator A in R such that (i) A E N , but A* $! N,; (ii) p is faithful, A E N,, but A* # N,.
Solution. For R we choose B('H), where 'H is a separable Hilbert space and { e n } is an orthonormal basis for 'H. (i) For A we choose the operator V * , where V is the isometry of 3-1 that maps en onto e,+l. Let p be wel. Then p ( A * A ) = p ( V V * ) = ( V * e l , V * e l ) = 0 ; while p ( A A * ) = p(V *V ) = p ( I ) = lle11I2 = 1. Thus A E N , but A* # N , . Since p is a vector state of R, p is a normal semi-finite weight on R. (ii) For this example, we choose for A* the partial isometry that maps e2n-1 onto eZn and e2n onto 0 for n in N. For p we choose 00 we,,-, C,"==, 2-nwe2n, a faithful normal semi-finite weight on B(3-t). Then p ( A * A ) = Cr=12-nlle2n112 = 1, while p ( A A * ) = C,"=,lle2n-1112 = 00. Thus A E N, but A* # N , . w
+
7.6.48. Let M be the subspace of C([O,11) consisting of those f such that f / i is bounded on (0,1], where &(A) = X for each X in [0,1]. Show that (i) M is a proper ideal in C([O,11) and if 91 5 g 5 92 with g 1 , g ~ in M , then g E M ;
358
N O R M A L STATES
(ii) the real-linear space Mh of real-valued functions in M is an archimedian partially ordered vector space with order unit L (see pp. 212,213, 297); (iii) po is a positive linear functional on Mo, where p o ( b ~ - f ~= ~ ) 6 and Mo = ( 6 ~ f L 2 : 6 E R, f E C([O,l],R)}; (iv) po extends to a positive h e a r functional p1 on Mh; (v) p is a weight on C([O,l]),where p ( f ) = pl(f) for each posi1])\Mh; tive f in Mh and p(g) = t o o for each positive g in c([O, ) 1 (with (vi) L E Fp M P = M , p ( ~ =~ 0,) L E N,, and p ( ~ = the notation of Section7.5). SoEution. (i) If f 1 , f i E M , there are constants Ml and M2 such that Ifi(X)/Xl I MI and Ifi(X)/Xl 5 M2 for all A in (0,1]. Then I(fi t fi)(X)/Xl I MI t M2, so that f1 t f2 E M. With k in C([O,I]), I(rCfi)(X)/Xl 5 11rC11M1 for each X in (0,1]; so that rCf1 E M . Finally, L E M and 1 4 M . Thus M is a proper ideal in C([O,11). Ifgi I g I g2, then 0 I g - gl I g2 - 91. With gl,g2 in Mh, g - gl E Mh if and only if g E Mh. Moreover, g2 - g1 E Mh. Thus it will suffice to show that g E Mh when 0 5 g 5 f and f E Mh. In this case,
for some constant and all X in (0,1] since f E Mh. Thus g E Mh. (ii) With the usual ordering, C([O,l],R) is an archimedian partially ordered vector space with order unit 1. Thus the subspace Mh is a partially ordered vector space with the induced ordering. If f E Mh and M is a constant such that If(A)/Al 5 M for each X in (0,1], then - M L 5 f I M L ;so that L is an order unit for Mh. If g E Mh and g 5 aL for each positive a, then g(X) 5 0 for each X in [0,1]. Thus Mh is an archimedian partially ordered vector space with order unit L. (iii) If b~ - f L 2 = 0 for some f in C([O,11,R) and some real scalar b, then f ( A ) = b/X for each X in (0,1]; and f is not bounded unless 6 = 0. If 6 = 0, then f = 0. Thus b~ - f~~= 0 only if b = 0 and f = 0. It follows that po is a well-defined linear functional on Mo. Suppose bL - f L 2 2 0. Then Xf(X) 5 6 for each A in (0,1] and Xf(X) + 0 as X 3 0. Hence 6 2 0 and p o ( 6 ~- f ~ = ~6 2) 0. It follows that PO is a positive linear functional on Mo.
EXERCISE 7.6.49
359
(iv) From Exercise 4.6.49, (i), (ii), and (iii), po extends to a positive linear functional p1 on Mh. (v) With p as defined and f , g positive functions in C ( [ O , l ] ) , we show that p(f g) = p(f) p ( g ) . If f and g are in Mh, then f+gEMhand
+
+
+ 9) = Pl(f + 9) = Pl(f) + Pd9) = P(f) + P ( d . If either or both of f and g are not in Mh, then f + g # Mh from (i) since 0 _< f _< f t g and 0 5 g 5 f + g . Thus = P ( f + 9) = P(f) + P ( d . P(f
0O
Of course af $ Mh i f f # Mh and a > 0; so that 00 = p ( a f ) = a p ( f ) in this case. Iff E Mh and a > 0, then a f E Mh and p(af) = P1 ( a t ) = Wl (f)= ap(f)*
Hence p is a weight on C([ 0 , 1 ] ) . (vi) By construction of p , Mh+ = F, so that M, = M. By definition of M, L E Mh+.Since i 2 = 0 - L - (-1)b2, p ( i 2 ) = 0. Thus L E N , ; but L = 1 - L - 0 . i 2 ,so that p ( ~ = ) 1. 7.6.49. Let R be a von Neumann algebra acting on a Hilbert space 'H and xo be a generating trace vector for R. For A in R,let JoAxo be A*xo. Show that (i) JO extends to an isometric conjugate-linear mapping J of 7.t onto 'H such that J 2 = I ( J is an involution); (ii) JA*J = A ' ( € R') for each A in R, where A -+ A' is the * anti-isomorphism described in Theorem 7.2.15. Solution. (i)
Since for each A in R,
IIJoAzol12 = ( A * X O , A * X=O () A A * x ~ , x=: ~( A) * A ~ o , s o=) ( ( A ~ o l ( ~ , JOis a well-defined conjugate-linear mapping of Rso onto Rso. Thus Jo has a (unique) isometric conjugate-linear extension J mapping 'H(= [Rso]) onto 3-1. As JiAx0 = JOA*sO = Azo, J 2 - I annihilates 72x0. But J 2 - I E f?('H) and (Rzo]= 'H. Thus J 2 = I . (ii) As JA*JBCxo = JA*C*B*xo = BCAxo = BJA*JCxo and [Rso]= 'H, JA*JB = B J A * J , where A , B , C E R. Thus JA'J E R'. Since JA*Jxo = JA*JIxo = JA*XO= A X O we , have that JA*J = A'. rn
360
NORMAL STATES
7.6.50. Let { e n : n E N} be an orthonormal basis for a (separable) Hilbert space H o , and let H be the (Hilbert space) countable direct sum of HO with itself. Suppose T is the operator on H with domain consisting of all vectors {zk} such that k2)1xk112 < oa and T ( { t k } )= {ksk}, B is the operator on 'H that maps {yk} to { C E 1 k - l y k , O , O ,...}, $0 is {2-1e1,2-2e2 ,...}, and 20 is {ZE, 2 - " k , 0 , 0 , . . .}. Show that 63 I)' (= 72); (i) T = T* and T q (f?(Ro) (ii) B E R; (iii) B T z o = zo; (iv) Szo # zo if S q R;(But see Exercise 8.7.60(vi).) (v) t o is generating and separating for R.
xEl
Solution. (i) Let e j k be the vector in R whose k th coordinate is ej and all of whose other coordinates are 0. Then { e j k : j , k E N} is an orthonormal basis for 3-1. Let f j k be the function on N x N that is 0 at all points except (j,k)where it takes the value 1. Then { f j k : j , k E N} is an orthonormal basis for l2(N x N). Let U be the unitary transformation of R onto 22(N X N) that maps e j k onto f j k for all (j,k)in N x N. Then U T U - l = M f , where f ( j , k ) = k for all j in N. From Theorem 5.6.4, M f is a self-adjoint operator affiliated with the multiplication algebra d of 22(N x N). Let U j j ' k be the unitary operator on /2(N x N) that interchanges f j k and f j l k and leaves each other frs fixed. Then U j j t k M f U j ; ! k = M f so that M f q A n { U j j l k : j , j ' , k E N}' (= do). Thus T is a self-adjoint operator affiliated with U - l A o U . Now U-'doU consists of those elements of B ( H > that are direct sums of scalar operators on 3-10, From Lemma 6.6.2, U - l d o U C R. Hence T*= T q R. (ii) If { y k } is a unit vector in H , then
k= 1
k=l
Thus B E f?(H). As B = C ~ l $ k - l Iwhere ~ , 10 is the identity operator on H o , B E R from Lemma 6.6.2. (iii) Since C?=,k2112-kekl12 = xT=1k22-2k < 00, 20 E Z)(T); and Txo = {k2-'ek}. Thus BTzo = {Ck=l 2 - ' e k , 0 , 0 , . . .} = 20.
361
EXERCISE 7.6.51
Let A0 be the operator in (iv) Suppose S q R and Szo = LJ(’H0) that maps ek onto 2kel for k in {l,. .,n) and e j onto 0 when j > n. Let A’ be A0 @ A0 @ Then A‘ E R’ so that A’S E SA’. Since xo E D(S),A’zo E D(S) and A ’ q = A’SXo = SA’zo. Now A’s0 = { e l , . ,e1,0,0,.. ,} and A’zo = (nel,O,O,. .}. Thus
.
. . a .
.
..
..,e1,0,0,. ..}) = {el,O,O,. ..}. But l[n-l{e, ,...,e1,0,0,...}112 = + 0, so that S is not closed. S(n”(e1,.
Thus no (closed) operator affiliated with 72 maps 50 onto to. (v) Since the sequence of vectors 2‘kek in HO is 12-independent and generates ?lo, 20 is generating and separating for R from Exercise 5.7.42. 7.6.51. Let SObe the set of faithful normal states of a von Neumann algebra R acting on a sepmable Hilbert space H. Show that (i) SO satisfies some one (and hence all) of the conditions (i), (ii), (iii), and (iv) of Theorem 4.3.9, where M of that theorem is replaced by R; (ii) H is self-adjoint if w ( H ) is real for each w in So. Solution. (i) We establish (ii) of Theorem 4.3.9 for our choice of So. Suppose H is a self-adjoint operator in R. From Corollary 4.3.10, there is a sequence {yn} of unit vectors in ‘H such that ((Hy,,y,)l + IlHll. Let ( e k } be an orthonormal basis for 3c with y, as el. Let X I be (1 - 2-n)1/2el and zk be 2-(n+k-1)/2ek for k in (2,3,...). With these choices, C z l w,, IR is a faithful normal state w, of R,for if A is a positive operator in 73 and on(A) = 0, then 0 = Azk = Aek for each k in {1,2,. ..}, whence A = 0. Note that 00
w n ( H ) = (1 - 2-n)(~y,?yn)
+ C2-(n+k-1)(~ek,ek), k=2
so that
IlHll 2 IWn(H)I 2 (1 - 2 - n ) l ( ~ 9 n , y n ) l - 2-nHall
--$
IlHll.
It follows that 1 1 8 1 1= sup(lw(H)I :w E SO}. From (i) and Theorem 4.3.9, SO) = S, where S is the (ii) p ( A ) is family of all states of R . Since w ( A ) is real for each w in SO, real for each p in SO), that is, for each state p of R. Hence A is self-adjoint from Theorem 4.3.4tii). w
362
NORMAL STATES
7.6.52.
Let
A0
be a closed operator acting on a Hilbert space
X,and suppose (Aox,x) 2 0 for each 2 in D(A0). (i) Show that A0 and A0 t I are closed symmetric operators on 3c. [Hint. Use Proposition 2.2.7.1 (ii) Suppose A is a positive self-adjoint extension of Ao. Note that A t 1 is a positive self-adjoint extension of A0 + I and that A +I is a one-to-one linear transformation with range 1-I. Show that the inverse B of A t I is in (8(3-1))+. (iii) With B as in (ii), y in 3-1, and 2 in 'D(A0) (= D(A0 t I)), show that
and that By E D(A,").
Solution. (i) From the second relation in Proposition 2.1.7, with 2 and y in D(Ao), A02 in place of u and Aoy in place of w, we deduce 2.4(3) (p.102) with A0 in place of T. Since each of the inner products on the right-hand side of 2.4(3) is real (when T is replaced by Ao), the vector entries can be interchanged in each of these inner products, yielding the right-hand side of the second relation of Proposition 2.1.7 with 2,y, &z, Aoy, replacing u, w, x,y, respectively. Thus (Aoz,y) = (x,Aoy), when s,y E D(A0). Hence A0 & A,". Since ((A0t l ) x , x) 2 0 for each x in ' D ( A 0 )(= D ( A o t I ) ) ,Ao +I is also symmetric. If 2, E 'D(Ao), x, + 2, and (A0 t I)z, -+ y, then A0zn + y - 2. Since A0 is closed, 2 E D(A0) and Aox = y - x. Hence x E D(Ao t I ) and (A0 t 1)x = y. It follows that A0 t I is closed. (ii) Since A t l i s closed (asjust argued for A o t I ) , A t 1 = A T l , and A t I is selEadjoint. Moreover,
-+
+
for each x in D ( A ) ( = D(A I ) ) ; A I is a positive self-adjoint operator with null space (0). Hence A I is a one-to-one linear transformation with range dense in 3-1 (from Exercise 2.8.45, the closure of the range of A + I (= (A I)') is the orthogonal complement of the null space of A I). From Lemma 2.7.9, A t I has closed range. Hence H is the range of A I.
+
+
+
+
363
EXERCISE 7.6.53
+ I)z and B is the inverse mapping to A + I, then 0 IllBY1I2= 11. 2 5 ( 4 4 + 0 4 = (BY,Y)I llBYllllYll
If y = (A
so that B E ( B ( X ) ) f . B is self-adjoint. By definition of the (iii) Since B E (l?(?l))$, operator B, B(A0 I)" = x (for each z in D(A0 t I)).Thus
+
with x and y as described in the statement of this exercise. Thus By E D((A0 + I ) * ) and (A0 I)*By= y. We show that (A0 +I)*= AZ+I- more generally, that (T+S)*= T*+S* when S is bounded. Suppose v E D((T S ) * ) and u E D(T) (= D(T S)). Then
+
+ + (Tu,v) = ((T + S)u,v) - (Su,v) = (u, (Tt S)*v - S*v),
so that v E D(T*)and T*v = (T + S)*v - S*v. It follows that v E D(T* S*)and (T*+ S*)v = (Ti-S)*v. Hence (T+ S)*C T*+ S*. Since the reverse inclusion T* S* C (T S)* is valid, in general, ( T i - S)* = T* -t S* when S is bounded. It follows that By E D(&) (= D(A,' I ) = D((A0 + I ) * ) ) .
+
+
+-
+
Let A0 be a closed operator acting on a Hilbert space 7.6.53. 'H, and suppose ( A o z , x ) 2 0 for each x in D ( A 0 ) . Define (u,v)' for each pair of vectors u,v in 'D(A0) to be ((A0 I)u,v) and let D' be the completion of D ( A 0 ) relative to the definite inner product (u,v) + (u,v)' on 'D(A0). (i) Show that the "identity" mapping of D(A0) onto itself has a (unique) bounded extension L mapping D' into 31, L is one-to-one, and 1 1 ~ 1 1 5 1. [Hint.Choose zn in D(A0) tending to z' in D'. If ~ ( z '= ) 0, show that llznll 3 0 and (z',zm)' = 0 for each m.] (ii) With y in 'FI, show that x + (x,y) (x E D ( A 0 ) ) extends to a bounded linear functional on 2)' of norm not exceeding [[yll. (iii) Show that there is a vector By in D(A,') satisfying (*) of Exercise 7.6.52(iii). [Hint. Find z' in 2)' such that (z, y) = (z,z')' for all 2 in D(A0) and let By be ~ ( z ' ) . ] (iv) Show that B E (B(7-f)):. [Hint. Use (i) and (ii) to show for each y in that llBll 5 1. Use the relation (x,y) = (z, ~-l(By))' 31 and each x in D(A0) (implicit in the hint of (iii)) to show that (BY,!/) = (L-l(BY),L-l(BY))'.l
+
364
NORMAL STATES
(v) Show that B is a one-to-one mapping and that its inverse A1 is a self-adjoint extension of A0 t I . [Hint. With y not 0, choose x in 'D(A0)such that (z,y) # 0. Use the equality (z,y) = ( x , ~ - ' ( B y ) ) l . Apply the discussion following Theorem 7.2.1. With a: and u in D(Ao), show that (u,s)' = ( u , ~ - l ( B ( Ai-o I)x))' and conclude that = B(Ao +I)z.] (vi) Show that A1 - I ( = A ) is a positive self-adjoint extension of Ao, and D ( A ) ~ ( 2 ) ' ) .
Solution. (i) Note that with
5
in D(Ao),
so that the identity mapping of D(A0) onto itself has a (unique) bounded extension L mapping D 1into H,and 11~115 1. To show that L is a one-to-one mapping, we proceed as in the hint. Since
we have that (lxn(l+ 0 when
~(2')=
(2',xm)' = lim(x,,x,)l n
0. Thus, for each m,
= Sm((A0 n i-I ) x , , x , )
(A0 i-I ) * x , ) = 0, = lim(x,, n
+
since x, E ' ~ ( A o I )
so that Z' = 0 and
L
E D((A0 t I ) * ) from Exercise 7.6.52(i). But
is one-to-one.
(ii) Since I(x,y)l I 11zIIIIyII I 11x11'11yll from (i), when 2 E DD(A0) and y E R,we see that the functional x + (5,y) on D(A0) has bound not exceeding llyll relative to the norm z + 11x11'. This functional extends (uniquely) to a functional of norm not exceeding llyll on '0'. (iii) From (ii) and Riesz's representation theorem (Theorem 2.3.1), there is a (unique) vector ZI in 2)' such that (5,y) = (x,z')' for each 5 in D(A0). Let By be L ( z ' ) . Choose 2, in D(A0) so that (2,) tends to z' (in '0'). Then
365
EXERCISE 7.6.53
Thus (z,y) =
( 5 , ~ ’ )=’
lim(z,zn)’ = lim((A0 n
n
+- I ) x , z n )
= ((A0 t I)z,~(z‘)) = ((A0 t I).,By).
(iv) From (i), (ii), and (iii) (and with the notation of the solution t o (iii)), IlBYll = l l L ( 4 1 1 2 llz’ll‘ 2 IlYll. Hence IlBll 5 1. Choose z n in D(A0) so that llzn - ~ ’ 1 1 ‘ tends to 0. Then from (i),
11zn - BY(( =
llL(5n -
z’)ll
I (1zn - ~ ‘ 1 1 ’
+
0,
and from (iii), ( B y , y ) = lim(z,,y) = lim(zn,z‘)’ = lim(zn,i-l(By))’ n
n
n
= (z’,L-’(By))‘ = IIL-1(By)llr2 2 0,
or note that limn(zn,z’)’ = ( z ’ , ~ ’ ) ’2 0. Thus B E (B(‘FI))t. (v) If y is a non-zero element of ‘FI, then
0#
(BY))’ for some x in (the dense manifold) D(A0). Hence ~ - l ( B y ) # 0. From (i), L is a one-to-one mapping, whence B y # 0. From the discussion following Theorem 7.2.1, with B in place of T (and now, the null space is (0)), the inverse A1 to B is a self-adjoint operator with domain contained in L ( D ’ ) , If z, u E D(Ao), then from (iii) and Exercise 7.6.52(i), (11,~)’ =
(2,Y)
=
(2, L - l
((A0 t I ) u , x )= (u,(Ao t 1 ) ~=) ( u , ~ - l ( B ( A ot I).))’.
Since D(A0) is dense in D’,z = ~ ( x=) B(A0 t I)., whence A ~= x AiB(A0
L - ~(B(Ao
t I).).
From (i), z =
+ I)z = (A0 -+ I)z.
Thus A1 is a self-adjoint extension of A0 t I . (vi) As in the solution to Exercise 7.6.52(ii), A1 - I is self-adjoint. Since A0 4-1C Al, A0 C A1 - I (= A). Since 0 5 B 5 I , B ( I - B ) 2 0. Hence (ABY,BY) = (AlBY,BY) - (BY,BY) = (Y,BY) - (BY,BY) = ((1- B)Y,BY) = ( B ( I - B)Y,Y) 2 0. But D(A) = D(Al), and D(A1) is the range of B. Thus 0 2 A and D(A) C L(D’).
366
NORMAL STATES
With the notation of Exercise 7.6.53, show that A is 7.6.54. the unique positive self-adjoint extension of A0 whose domain is contained in ~ ( 2 ) ‘ ) .(This extension is known i18 the fiiedrichs extension of Ao.) [Hint. Use Exercise 7.6.52(ii) and (iii) and an argument of the type indicated in the hint to Exercise 7.6.53(i) to show that L is one-to-one.]
Solution. From Exercise 7.6.52(ii) and (iii), a positive selfadjoint extension A’ of A0 gives rise to an operator B’ with the properties proved for B in that exercise. Thus ((A0 I)s,(B B‘)y) = 0 for each x in D(A0). If, in addition, we assume that the domain of A‘ is contained in L(D’),then there is a vector u’ in 2)‘ such that ~ ( u ’ = ) ( B - B’)y. Let {xn} be a sequence of vectors in D(A0) tending to u‘ (in D‘). Then
+
and (zn,~‘)’= lim(xn,xrn)’ m = lim((A0 m
-tl ) x n , z m )
= ((A0 iI ) z ~( ,B - B’)y) = 0. Hence ( u ’ , ~ ’ ) ’= l.imn(xn,u’)‘ = 0, and u’ = 0. It follows that ~ ( u ‘= ) ( B - B’)y = 0 and B = B’. Since B and B‘ are the inverse mappings to A and A’, respectively, A = A’. 434(p. 480)] 7.6.55. Let R be a von Neumann algebra acting on a Hilbert space 3-1 and A0 be a symmetric operator affiliated with R. Suppose (Aos,x) 2 0 for each x in D(A0). Show that the F’riedrichs extension of A0 (see Exercise 7.6.54) is affiliated with R.
Solution. From Exercise 7.6.52(i), A0 is symmetric and by definition of “affiliation,” A0 is closed. Exercises 7.6.53 and 7.6.54 guarantee the existence and uniqueness, respectively, of the F’riedrichs extension A of Ao. Let U’ be a unitary operator in R‘. Then U’AU“ is a positive self-adjoint extension of U‘AoU’* and D(U‘AU’*) E U’(L(D’)).Since A0 77 R, U’AoU’* = Ao. From uniqueness of A (Exercise 7.6.54), it remains to show that U’(L(D’)) L(D’). Suppose z E ~ ( 2 ) ’ )and ~ ( z ’ )= z (with z‘ in D’). Then {zn} tends to z‘ for some sequence (2,) in ’P(A0). Since A0 R,
EXERCISE 7.6.55
367
U ' ( V ( A o ) )= 'D(A0)and U'x, E V ( A o ) .NOW
as n,rn -+ 00 since {z,} converges in V'. Thus {U'x,} converges in V' to some u' and { U ' x n } converges in H to ~ ( u ' ) Since . {z,} tends
to z in
H,{U'z,}
U'(L(V')) L(V').
tends to U'z in 'H. Thus U'z = ~ ( dE )~(27') and
CHAPTER 8 THE TRACE
8.7.
Exercises
8.7.1. Let T be the center-valued trace on a finite von Neumann algebra R, and define a bounded linear operator v : R -+ R by v ( A ) = A - T ( R ) . (i) Prove that v is a projection, and identify its range and null space. (ii) By appeal to Lemma 6.5.6, or otherwise, show that 11v1[ = 2 when R is of type 111. (iii) By use of Theorem 7.3.1, or otherwise, show that 1 1 ~ 1 1 = 2n-l(n - 1) when R is a factor of type I, (with n < 00). Solution. (i) Since T is a linear mapping from R into C (the center of R) and r ( C ) = C for each C in C, it follows that T is a projection with range C. Its null space can be described as the set of operators with trace zero in R. Hence Y is the complementary projection, with null space C and range the set of operators with trace zero in R. (ii) Suppose that R is of type 111. Since llrll = 1, by Theorem 8.2.8(ii), (v), it follows that 1 1 ~ 1 15 2. From Lemma 6.5.6, for each positive integer n, we can choose equivalent projections E l , . . . ,E , in R, with sum I . If
A = El - (Ez -t E3 t * - . t E n ) then
and
Y ( A )= A
- T ( A )=
n
2
- -(E2 4-E3 + * n
*
- + En).
EXERCISE
8.7.1
369
Since IlAll = 1 and Ilv(A)II = 2n-'(n-l), we have 1 1 ~ 1 12 2n-'(n-l). When n + 00, we obtain 1 1 ~ 1 12 2; the reverse inequality has already been proved, so 1 1 ~ 1 1= 2. (iii) Suppose that R is a factor of type I,. Then R is a finitedimensional normed space, its unit ball (R)1 is a compact convex set, and thus (R)1 is the (norm-) closed convex hull of its extreme points. By Theorem 7.3.1, the extreme points of (R)1 are precisely the unitary elements of R. Each A in (R)1 can be approximated in norm by a convex combination C ajUj of unitary elements, v ( A ) is then close to C ajv(Uj), and
It follows that
From the isomorphism of
R with M,(@), each unitary element U in
R has the form U = AlEl t
i- A,E,,
where {El,.. . ,E , } is an orthogonal family of minimal projections in R,with sum I , and XI, . . . ,A, are complex numbers with modulus 1. We have .(U) = X I = A(E1+ * * * $- En), where A = n-'(A1
+ . t A,).
Thus
.(U) = (A1 - A)El t
' . *
t (A, - A)&,
and Ilv(U)ll = max{lAj - A1 : 1 5 j _< n}. Upon renumbering, we may suppose that
< l - -1t - n- - 1 - 2(n-1) n n n (with equality when A 1 = 1,Aa = with (l),shows that llvll = 2n-'(n
... = A, - 1).
= -1). This, together w
370
THE TRACE
8.7.2. Suppose that U1 and !?i2 are self-adjoint subdgebras of a finite von Neumann algebra R,r is the center-valued trace on R, and cp is a * isomorphism from 0 1 onto 2 l 2 such that r((p(A))= r ( A ) for each A in M I . For j = 1,2,let f3j be the set of all operators of the form CO A1 C 1 f A2 C2 t An Cn,with {Co,. ,C,} a finite subset of the center C of R and A1,. . .,An in Mj. (i) Show that Bj is a self-adjoint subalgebra of 72 that contains C U 2 l j , and that Q extends to a * isomorphism (p from B1 onto BZ such that p(C) = C and r ( $ ( B ) ) = T ( B )whenever C E C and
+ -
+
..
B E Bi. (ii) Let ( Q k } be an orthogonal family of cyclic projections in C with sum I, and for each k let zk be a unit vector such that Qk has range [c’zk]. Show that the equation pk(R) = (T(R)sk,zk)defines a faithful norrnd state p k on the von Neumann algebra R Q k , and prove that the restriction p’(BlQkis a * isomorphism from & Q k onto B z Q k . (iii) By using the result of Exercise 7.6.7, show that glB1Qk extends to a * isomorphism from B r Q k onto BTQk. Deduce that p extends to a * isomorphism from f?; onto B i . Solution. (i) It is apparent that Bj is a linear subspace of R that contains C U 2lj. Since
(AiCi)*= C,*A; = A;C;,
Ci(A2C2)= (A2C2)Ci = A2(C2Ci),
and (AlCl)(AZC2) = (AlAz)(C,C2), whenever Cl,C2 E C and A l , A2 E 2 l j , it follows that Bj is a self-adjoint subalgebra of ‘R. Suppose that CO,. ..,Cn E C, Al,. ..,A, E Q1, and let
With A0 = Bo = I , and Bj = p ( A j ) when 1 5 j 5 n, we have
n
EXERCISE 8.7.2
371
Accordingly, since 7 is faithful, it follows that S = 0 if and only if T = 0. From the preceding paragraph, there is a one-to-one linear mapping (p from 0 1 onto &, defined by the requirement that
whenever Co, . . . ,C,E C and A 1 , . . . ,A, E 2 l 1 . It is not difficult to verify that (p is multiplicative and adjoint-preserving, and is therefore a * isomorphism from L31 onto B 2 . It is apparent that p(C) = C when C E C . Moreover
r ( p ( B ) )= r ( B )for each B in B1. (ii) Since X k is a generating vector for the von Neumann algebra C ' Q k , it is a separating vector for C Q k . Thus w,, l c Q k is a faithful normal state of C Q k . Since r l R Q k is an ultraweakly continuous linear mapping from R Q k into C Q k , and T ( R ) > 0 when R > 0, it now follows that P k (= ox,o 7 1 R Q k ) is a faithful normal state of so
RQk.
Since ( p ( B Q k ) = p ( B ) p ( Q k ) = p ( B ) Q k , for d l B in &, it follows that the restriction C p I B l Q k is a * isomorphism from & Q k Onto & Q k . (iii) When B E f ? l Q k ( & &),
By applying the result of Exercise 7.6.7 to the faithful state p k of and the subalgebras & Q k and f?2Qk of R Q k , we deduce that pl&&k extends to a * isomorphism ?+!Jkfrom f?,Qk onto f?;Qk. The equation $(B) = x $ k ( B & k ) ( B E Br) R Q k
now defines a tension of (p.
* isomorphism m[85]
$ from f?,
onto S,, and $ is an ex-
372
THE TRACE
8.7.3. Let T be the center-valued trace on a finite von Neumann algebra R acting on a Hilbert space ‘H. (i) Prove that, for each 5 in ‘H, there is a sequence (91, y2,. . .} of elements of ‘H such that C llyn112 < w and
(++,
4 = C ( A Y n , Yn)
(A E R).
(ii) Let S be the set of all vectors y in ‘H with the following property: there is a vector 5 in ‘H such that
I ( 7 . ( 4 5 , 4 (A E R+). Show that S is a separating set for R. (AY,Y)
(iii) Suppose that {Ra} is a bounded net of elements of R and that the net {T(RXR,)}is weak-operator convergent to 0. Show that llRaA’yll + 0 for each A’ in R’ and y in S, and deduce that { R a }is strong-operator convergent to 0.
Solution. (i) Since T is ultraweakly continuous by Theorem 8.2.8(vi), the positive linear functional w, o T on R is ultraweakly continuous (hence, normal). The stated result now follows from Theorem 7.1.12. (ii) Suppose that A E R, and Ay = 0 for each y in S. Given any z in ‘H, choose y l , y 2 , . . . as in (i). It is apparent that y, E S, whence Ay, = 0, for each n. Thus (T(A*A)z,x) = C(A*AY~,Y,J = 0. Since the above argument applies to each x in ‘H, it follows that T(A*A) = 0, and A = 0 since T is faithful (Theorem 8.2.8(iii)). Hence S is a separating set for R (and is, therefore, generating for
R’) . (iii) Given {Ra}, A’, and y, as in the statement of (iii), choose 5 (corresponding to y) with the property set out in (ii). Then
Since the set of all vectors of the form A‘y, with A‘ in R’ and y in S, has closed linear span N , these vectors suffice to determine the strong-operator topology on bounded subsets of R. Hence { R a }is H strong-operator convergent to 0.
3 73
EXERCISE 8.7.4
8.7.4. Suppose that R is a van Neumann algebra with center C, cp is a * homomorphism from R onto a C*-algebra U, and U(U) is the unitary group of U. Prove that (i) for each A in U, the norm-closed convex hull of the set { U A U * : U E U(U)} meets q(C); (ii) q ( C ) is the center of 9.
Solution. (i) Each A in U has the form cp(R),with R in R. By the Dixmier approximation theorem, c o ~R)= ( contains an element C of C. Given any positive E , we can choose a finite set Vl,. . . ,Vn of unitary operators in R, and positive scalars a l , . . .,a, with sum 1, such that n
j=1
With U j the element cp(V,) of U(U), n
n
Thus cp(C)lies in the norm closure of co{UAU* : U E U(U)}. (ii) If A lies in the center of U, the set { U A U * : U E U(2l)) consists of the single point A , and the same is true of its normclosed convex hull. By (i), A € y(C). This shows that p(C) contains the center of U, and the reverse inclusion is apparent. 8.7.5. Suppose that R is a von Neumann algebra with center C , and Al, A z E R. Show that (with the notation of Section 8.3) the set cox(A1 Az)' nC is contained in the norm closure of the set
+
+
SoZution. Suppose that C E CO.R(AI Az)= n C. We have to show that, given any positive E , it is possible to choose Cj in coz(Aj)= n C ( j = l , 2 ) so that IIC - (Cl C2)ll< E . With the notation introduced at the beginning of Section 8.3, we can choose a in D so that
+
374
THE TRACE
By Proposition 8.3.4, there exist Cl,C2 in C and a sequence in D such that
{am}
Then, Cj E co.~(Aj)=n C ( j = 1,2) and
8.7.6. Let R be a von Neumann algebra with center C. With the notation of Section 8.3, show that if P is a projection in C and a E D, there is an element a p of D for which
aP(A) = Pa(A) t (I- P)A
(A E R).
Hence show that if P I ,. .. ,P, are projections, with sum I, in C and E D, there is an element P of D for which
PI,. . . ,P,
Deduce that for all A in R c o . ~ ( A ) ' n C = {C1+...+Cn : C j ~ c o . ~ ( P j A ) ' n C ( j =1,..., n)}, and
nC}. co.~(PjA)' n C = {PjC : C E CO.R(A)= Solution. There exist unitary operators U1,. . . ,u k in R and positive scalars a l , . . , a h with sum 1 such that
.
We can define an element a p of D by
375
EXERCISE 8.7.6
where I$ is the unitary operator P U j A simple calculation shows that aP(A) = P a ( A )
+ I - P in R, for j = 1,. . . ,k.
+ (I- P ) A
( A E R).
Upon replacing a by P j and P by Pj, it follows that (in place of a p ) there is an element y j of D for which yj(A) = PjPj(A) t (1- Pj)A = Pj(PjA)
+ (I- Pj)A
An easy induction argument shows that, for
(A E R). T
= 1,... ,n,
+
C,, where Cj E c o ~ ( P j A ) =n C Suppose that C = C1 t .. (and, hence, Cj = PjCj = PjC) for j = 1,... ,n. Given E (> 0) we can choose PI,. . . ,Pn in D such that
With ,f3 constructed from
PI,. . . ,,Onas above,
PjP(A) = PjPj(A) = Pj(PjA), IIPj(P(A) - C)ll = llPj(PjA) - Cjll
<E
( j = 1,.* * ,n ) ,
and thus M A ) - CII < E . It follows that C E coR(A)'nC. From this, each Cj in coR(PjA)'nC has the form PjC, for some C in CO.R(A)=n C. Conversely, suppose that C E c o ~ ( A ) ' n C. To complete the proof, it now suffices to show that PjC E co,(P'A)= n C, since C = PIC + PnC. Given any E ( > 0), we can choose a in D so that IIC - a(A)II < E . Then
+
IlPjC - a(PjA)II = IIPj(C - ~ ( A ) ) l < l E. Hence PjC E c o . ~ ( P j A ) =n C , as required.
376
T H E TRACE
8.7.7. Suppose that R is a von Neumann algebra with center C , P is a projection in C that is properly infinite relative to R, and EOis a subprojection of P in R such that EO P - Eo P (see Lemma 6.3.3). (i) By expressing P - Eo as the sum of n projections in R, each of which is equivalent to Eo, and by arguing as in the proof of Proposition 8.3.9, show that N
N
( n = 1,2,. . .).
Deduce that cP E c o ~ ( E o ) =fl C for all c in [0,1]. (ii) Prove that co~(E0)' n C contains each operator C of the form c j p j , where c1,. . , , c k are scalars in [O,11 and P I , . . . ,Pk are mutually orthogonal projections in C with sum P. (iii) Deduce that
c,",,
Solution. (i) By repeated application of the halving lemma we can construct projections El,. . .,En in R such that
For each j = 0,.
. . ,n, P - EO Ej and N
and (by using partial isometries that implement these equivalences) we can construct a, unitary element Uj of R such that UjEoUj* = P - Ej. Also,
Eo
N
Ej,
P - Eo
N
P - Ej,
and there is a unitary element Vj of R such that VjEoVj* = Ej.
377
EXERCISE 8.7.8
Since n
.
.
n
1 - -[(n
n+l
+
n + 1)P - P ] = P, n+l
+
it follows that n(n l)-'P, ( n l)-'P E c o ~ ( E 0 n) C. By taking limits as n + 00, it follows that co~(E0)' nC contains P and 0, and thus (by convexity) contains CP for all c in [0,1]. (ii) Suppose we are given an operator C of the form CjPj, as in the statement of (ii), and let PObe I - P. Then PO,... ,Pk are projections with sum I in C, and PoEo = PoPEo = 0. Since PjP = Pj, and c j P E c o . ~ ( E o ) = n C by (i), it follows from the results of Exercise 8.7.6 that
x;=l
O E co~(PoE0)"n C,
~ j PEj CoR(PjE0)'
nC
xF=l
( j = 1,.. . ,k), and thus C j P j E COR(EO)' n C. (iii) Since 0 5 EO 5 P, we have 0
< UEoU* 5 UPU* = P
for each unitary operator U in R. It follows that c o ~ ( E 0 ) "is contained in the (norm-closed convex) set { A € R : 0 A P}. Hence
< <
(1)
coR(Eo)' n C
{C E C : o 5 C 5 P}.
At the same time, c o ~ ( E o ) =fl C is norm-closed and contains the cjPj considered in (ii), and these form a norm-dense operators subset of the set {C E C : 0 5 C 5 P}. Hence the inclusion of (1) is an equality.
xi=l
8.7.8. Let R be a von Neumann algebra with center C. Show that the following two conditions are equivalent. (i) R is finite. (ii) If A, B E R and X E c o ~ ( A ) ' n C, Y E CO.R(B)=n C, then x Y E coR(A B)= n C.
+
+
378
THE TRACE
Solution. First, suppose that 72 is finite, and let T : 72 -, C be the center-valued trace. Given A, B , X , Y as in condition (ii), it follows from Theorem 8.3.6 that X = T(A), Y = T ( B ) ,and
X t Y = T ( At B ) E coR(A+ B)= nC. Thus (i) implies (ii). Now suppose that R satisfies condition (ii). Given any A in R, let B be I - A. Since A + B = I and c o ~ ( I ) n =C consists of the single element I, it follows that X Y = I whenever X E c o ~ ( A ) =n C and Y E c o ~ ( B ) =n C. This implies that c o ~ ( A ) =n C (and, also, c o ~ ( B ) =n C) contains just one point. Since this applies to each element A of R, it now follows that R is finite; for otherwise, we could find P and EOwith the properties set out in Exercise 8.7.7 and, by the results of that exercise, c o ~ ( E o ) =n C would not reduce to a single point. Thus (ii) implies (i). m
+
8.7.9. Suppose that R is a properly infinite von Neumann algebra with center C, and E is a projection in R. (i) Suppose that F 1 , . . . ,Fk are projections in R,P is a projection in C, Fj 5 E for j = l,...,k,and 3'1 V F 2 V . . . V Fk = P. Prove that P E N P. (ii) Suppose that C is a non-zero element of c o . ~ ( E ) =n C, and unitary operators Ul,...,u k in 'R and positive scalars a l , . ,ak with sum 1 are chosen so that
..
Prove that V l l UjQEUJ = Q for some non-zero projection Q in C. (iii) Prove that, if c o ~ ( E ) =nC # {0}, then Q E N Q for some non-zero projection Q in C. (iv) Prove that, if COR(E)=nC # {CE}, then Q(CE- E ) N Q for some non-zero projection Q in C. (v) Show that there are central projections Q1 and 9 2 maximal with respect to the properties that Q I E N 41and Qz(1- E ) Q 2 . Prove that N
coR(E)=nC = {I-Q2 t C : C E C, 0 5
C 5 QlQ2).
379
EXERCISE 8.7.9
[Hint. By use of (i) and the comparison theorem, show that (I - Q1)(1- Q 2 ) = 0 , so that I = (I-
Q i )t
(I-
t Q1Q2.
Q2)
Apply (iii) t o ( I - Q l ) E , (iv) to ( I - Q z ) E , Exercise 8.7.7 to QIQzE, and Exercise 8.7.6 to E.] Interpret the result of (v) in the case of an infinite factor.
Solution. (i) If Q E is finite, for some projection Q in C, then Q F j ( 5 Q E ) is finite for each j = 1, ..., k, and hence so is Q P (= QFlVQF2V-..VQFk).SinceRisproperlyinfiniteandQPEC, it follows that QP = 0. Thus QE is infinite when Q is a non-zero central subprojection of P , Hence either P = 0 (in which case the result of (i) is obvious) or P E is properly infinite (with central carrier PI. With P E properly infinite, we can choose projections E l , . . ., Ek in R so that
El
N
E2
N
Let Gr be F1 V F2 V
N
Ek
N
+ E2 +
P E = El
t Ek.
... V Fr, for r = 1,.. . ,k, and note that
GI = F1 = PF1
5 PE
Gr+1 - Gr = Gr V Fr+1 - Gr
N
N
El,
FT+1 - Gr A Fr+l
I Fr+1 = PFr+1 4 PE
N
when 1 5 r
N
Er+1,
< k. Thus
P = G k =G1+(G2 - G 1 ) + * . . + ( G k - G k - l )
5 El + E2 +
+ Ek = P E 5 P,
and P P E . (This can also be proved by using Exercise 6.9.4; with P E properly infinite, since PFj 5 P E we have P E P E V PF1 P E V PF1 V PF2 * * - P E V PF1 V * * - VPFk = P E V P = P.) (ii), (iii) Suppose that 0 # C E c o . ~ ( E ) =n C, and choose U 1 , . . . , Uk and a l , . . . , a k as in the statement of (ii). Then > 0. With E ( > 0) defined by N
N
N
N
N
c
k
IlC((= 2~ t (IC - C a j U j E U ; ( I , j=1
380
THE TRACE
for some non-zero projection Q in C. Thus k
k
j=1
j=l
2 QC - (IlCll- 2&)Q2 E Q . Now
V:=l
UjQEUJ 5 Q. Also, for each unit vector x in the range # 0 for some j, since
of Q , UjQEU;x
k
C aj(UjQEUJX,x)2 & ( Q x , x )=
E.
j=1
-
Hence V j=l UjQEU; = Q . From (i), with'UjQEU? (- Q E 5 E ) in place of Fj and Q in place of P, it follows that QE E . (iv) For each unitary operator U in R, we have the equality, U(CE- E)U* = CE - UEU*. Thus
CO.R(CE - E)= n C = {CE - C : C E CO.R(E)=n C}, from which the condition co.~(E)' n C # { C E } is equivalent to CO.R(CE - E)= n C # (0). From (iii), with CE - E in place of E , this entails Q(CE - E ) Q for some non-zero projection Q in C. (v) The existence of central projections Q1 and Q 2 with the stated properties follows from the comparison theorem (6.2.7), applied first to the projections E , I in R, then to the projections I - E , I . Again by the comparison theorem there are projections Pi and P2 in C such that Pi P2 = ( I - Q l ) ( I - Q 2 ) and Pl(I - E ) 5 P I E , P2E 5 & ( I - E ) . Since P I E V P l ( I - E ) = PI,it follows from (i) that P I E PI. Hence ( Q 1 + P l ) E Q1 + P I , and the maximality assumption concerning Q 1 now implies that PI = 0. A similar argument shows that P2 = 0; so ( I - Q 1 ) ( I - Q 2 ) = PI P2 = 0, and
-
+
-
N
+
(1)
+
I = (1 - Q i ) ( I - Q2)
+ QiQz.
381
EXERCISE 8.7.10
If con((I - Q , ) E ) = n C # {0}, it follows from (iii) that Q ( I - Q I ) E Q ( N Q ( I - Q 1 ) )for some non-zero projection Q in C, and this contradicts the inaximality assumption concerning Q 1 . Hence N
C O R ( (I Q 1 ) E ) =n C = (0).
(2)
From the niaximality assumption concerning Q2, P ( I - E ) -i P (whence P E # 0) whenever I' is a non-zero subprojection of I - Q2 in C . From this, it follows both that ( I - Q 2 ) E has central carrier I - Q? and also (by applying (iv) to ( I - Q 2 ) E ) that C O ~ ( ( IQ- ? ) E ) =n C = { I
(3)
-
Q2).
Since QIQ2E ccr Q1Q2 N Q1Q2(1- E ) , it follows from the resuIt of Exercise 8.7.7 that
(41
coR(QIQ?E)' n C = { C E C : 0 5 C 5 Q1Q2).
In view of ( l ) ,(a),( 3 ) , (4), and the result of Exercise 8.7.6, we have
C O ~ ( E ) = ~{ C I - =Q 2 + C : C ~ C , 0 1 C I Q 1 Q 2 } . When R is an infinite factor, the projections Q1 and Q 2 are either 0 or I , and are not, both 0 since ( I - Q1)(I - Q ? ) = 0. Thus three possible cases occur. If E N I ( I - E ) (the case i n which Q 1 = Q? = I ) , coR(E)=nC is { a 1 : 0 5 a 5 I}. If E -i I N I - E (the case in which Q1 = 0 and Q2 = I ) , coa(E)=n C is (0). If E N I + I - E (the case in which Q1 = I and Q2 = O ) , c o n ( E ) =nC is { I } . m[90]
-
Suppose that 'R is an infinite factor and { E x } is the 8.7.10. spectral resolution of a self-adjoint element A of R. Let C denote the center { z l : z E C} of R , and define real numbers CL,6 by a = sup{X E R : E,,
b = inf{X E R : I - E A 4 I}.
Show that -1IAll F
fl
I b IIlAll
and coR(A)= nC = { e l : c E R,
[Hint. Use the results of Exercise 8.7.9.1
5 c 5 b}.
382
THE TRACE
Solution. The set {A E R : Ex 4 I } contains the interval -11A11) and does not meet the interval [11A11, oo), since Ex is 0 when X < -(IAJI and is I when X 2 IlAll. Thus -1lAll 5 a 5 ((All. We assert also that Ex 4 I whenever X < a; for given such A, we can choose p(> A ) so that E, 4 I , and then Ex 5 E, 4 I . A similar argument shows that -1lAll 5 b 5 IlAll, and that I - Ex 4 I whenever X > b. If it were the case that b < a, we could choose X so that b < X < a, and deduce that (-00,
Since Ex V ( I - E x ) = I, and either Ex 5 I - Ex or I - Ex 5 Ex (because R is a factor), (1) contradicts the conclusion of Exercise 8.7.9(i). Hence a 5 b. Since c o ~ ( A ) =is a bounded norm-closed convex set of selfadjoint operators, and C consists of scalar multiples of I , there are real numbers a0 and bo such that C O ~ ( A )n=C = {cI : c E R, a.
(2)
I c 5 bo}.
We have to show that a = a0 and b = bo. Suppose that 0 < E 5 1, and let S be (11A11 t la1 t 3)-l&. Then
Ea+6 I , and (from the result of Exercise 8.7.9(v))I E con(E,+s)=. With the notation of Section 8.3, we can choose a0 in D so that ((ao(Ea+6) - Ill < 6. Then, N
Thus h ( A ) E COR(A)and pao(A) 5 ( a t & ) I for , all ,O in D. From this, B E c o ~ ( A ) =and B I ( a i-& ) I ,for all B in coa(ao(A))'. By ) the family choosing an element CO (necessarily of the form C O I in cox(ag(A))= n C, we conclude that c 0 1 E coa(A)= n C for some co satisfying co I a t E . From (2), a0 5 a t E .
EXERCISE
8.7.10
At the same time,
Ea-a 4 I , and (from the result of Exercise 8.7.9(v)) we have t h a t C O ~ ( E , - ~ )fl' C = (0). Since a 0 1 E coa(A)', we can choose cu in D so t h a t J J a o I- a(A)II < 6. Then
Now c o ~ ( ~ ( E , - a ) ) fl = c" E c o ~ ( E a - ~ )n=C = {O), so we can choose P in D such t h a t lIPa(Ea-s)(( < 6. Thus
2 a - E. We have now shown that a - E 5 a0 5 a E , for every positive E . It follows t h a t a0 = a. We show t h a t b = bo by a similar argument. This time, we choose (IIAll Ibl 3 ) - ' ~for S and note that and
a0
+
+ +
Now I - Eb-6 I , and (from the result of Exercise 8.7.9(v)) we have t h a t I E c o R ( I - Eb-6)=. We can choose a in D so that N
Thus p a ( A ) E: con(A) and P a ( A ) 2 (b - E ) I , for all P in D. From = B 2 (6 - E)I for all B in con(a(A))=. By this, B E C O ~ ( A ) and
384
THE TRACE
choosing an element C1 (necessarily of the form c l I ) in the family c o ~ ( a ( A ) ) =n C, we conclude that c l I E c o ~ ( A ) =n C for some c1 satisfying c1 2 6 - E . From (2), 6 - E 5 60. At the same time,
1 - E b + 6 4 1, and (from the result of Exercise 8.7.9(v)) n C = (0). Since 601 E c o ~ ( A ) = we , can choose c o ~ (I &+a)’ a0 in D so that ( ( 6 0 1- ao(A)II < 6. Then
+
6 0 1 I ao(A)t 6 1 5 (6 t 6)I - ( 6 6 - IIAll)ao(I- Eb+6) t 61 = (6 t 2 6 ) I - (6 t 6 - IIAll)ao(I - Eb+6). Now c o ~ ( a o ( 1 -&+6))= n C C coR(I - Eb+C)= n C = {0}, can choose /3 in D such that ((/3ao(I-Eb+~)ll< 6. Thus
56 + ~ . We have now shown that 6 - E 5 60 It follows that 60 = 6 .
and E.
SO
we
60
5 6 t E,
for every positive
8.7.11. Show that if A is a self-adjoint element of a countably decomposable type I11 factor R, and C is R n R’(={zI : z E C}), then coR(A)= nC = {cI:c E R, m 5 c 5 M } where [m,M ] is the smallest interval containing sp(A).
Solution. The spectral resolution { E x } of A satisfies
Ex = O O<Ex
(X<m),
(m
M).
From this, and since every non-zero projection in R is equivalent t o I (Corollary 6.3.5), it follows that the interval [a,6] described in Exercise 8.7.10 coincides with [m,MI. Hence the result to be proved in the present exerise is a consequence of Exercise 8.7.10.
385
EXERCISE 8.7.12
Suppose that R is an infinite factor, { E x } is the 8.7.12. spectral resolution of a self-adjoint element A of R, C is the center {zI : z E C} of R,Ic is the largest proper (norm-closed) two-sided ideal in R (see Exercise 6.9.51), and p : R -+ R/Ic is the quotient mapping. By using the result of Exercise 8.7.10, show that C O ~ ( A )nC = = where
[VL, M
{cI : c E w, vz 5 c 5 M } ,
] is the smallest interval containing sp(p(A)).
Solution. Let Po be thr. set of all projections E in R for which E 4 I , and let Z be the set of all operators in R with range projections in Po. From Exercise 6.9.51, Z is a. two-sided ideal in R and Z= = Ic. A projection E i n R lies in Ic if and only if it lies in Z (Exercise 6.9.46), and this occurs if and only if E < I . From this, together with Exercise 8.7.10,
cor,(A)’ n C = { c l : c E R, u 5 c
5 b},
where a = sup{X E
Iw : E X E A:},
6 = inf{X E R : I - Ex E Ic}.
It remains to identify [ u , b] nu, the smallest interval [m,MI containing SP(Y(A) 1. If X < a 5 b < p , then Ex E Ic and I - E , E K , so cp(Ex) = 0 a.nd p(E,) = I . Since
it follows that XI 5 q ( A ) 5 P I . Accordingly s p ( ~ ( A ) )C [ X , p ] whenever X < a 5 b < ji, and thus sp(y(A)) [a, b]. Since [m,M ] is the smallest interval containing sp(p(A)),
and mI 5 cp(A) _< M I . When A > a, p(E1) is a non-zero projection in p(R) (since Ex 4 h”] and commutes with p ( A ) . Thus
386
THE TRACE
Hence m 5 X whenever X > a. From this, together with (l), m = a. Similarly, when p < b, cp(I - E,) is a non-zero projection in cp(R) (since I - E, 4 K) and commutes with cp(A). Thus Pcp(I - E,) = C PW- E,)) s cp(A(I - E,)) = cp(A)cp(I- E,) 5 Mcp(I- E,). Hence p 5 M whenever p < b. From this, together with (l), A4 = b and [a,b]= [ m , M ] . 8
8.7.13. Interpret the results of Exercise 8.7.12 and 8.7.10 in the case of a countably decomposable factor R of type I, or 11, (in particular, for the factor B ( H ) , with ‘H a separable Hilbert space), taking into account the information given in Corollary 6.3.5, Theorem 6.8.7,and Exercise 2.8.25. Solution. The result of Exercise 8.7.12 is unchanged in the cases now under consideration, except that we have additional information about the ideal K . In these cases, K is the unique proper norm-closed two-sided ideal in R,and is the norm closure of the set of all operators in R that have finite range projections (Theorem 6.8.7). When R is B(H)(with 3.t separable), “finite” means “finite-dimensional,” and K is the ideal of dl compact linear operators (Exercise 2.8.25). If E is a projection in R,E N I if E is infinite (Corollary 6.3.5), so E 4 I if and only if E is finite. Hence, in Exercise 8.7.10, a = sup{X E Ha : EA is finite}, b = inf{X E W : I - Ex is finite}. When the factor R is B ( H ) , “finite” can again be replaced by “finite-dimensional.” 8
8.7.14. Show that if Z is a norm-closed two-sided ideal in a von Neumann algebra R, and R E Z, then COR(ARB)= Z for all A and B in R. Solution. Since R E Z, we have ARB E 1,and n
C UjUjARBU; j=1
EZ
387
EXERCISE 8.7.16
..
.
for all choices of scalars a l , . ,a, and unitary operators U1,. . ,Un in R. Hence co*(ARB) c Z, and con(ARB)' C_ Z since 2 is norm closed. Suppose that R is a von Neumann algebra with center 8.7.15. C, and K is a norm-closed ideal in C. Let j ( K ) be the norm-closed of the set RK (= {RK :R E 32, K E K}),and let linear span [RK]
J ( K ) = {R E R : CO.R(ARB)=n C C_ K ( A , B E R)}. Show that j ( K ) and J ( K ) are norm-closed two-sided ideals in 72, and j(K)nC= J(K)nC=K. Prove also that, among all norm-closed two-sided ideals Z in R such that Z n C = K,j ( K ) is the smallest and J(K) is the largest.
Solution. Since B(AK) = (BA)K and (AIC)B = (AB)K whenever K E K (C C) and A,B E R, the linear span of RK is a two-sided ideal in R, and j ( K ) (= [RK]) is a norm-closed twosided ideal in R. It is apparent that j ( K ) contains Ic, and is the smallest normclosed two-sided ideal in 72 that contains K , We prove next that j(K) n C Ic; from this, it will follow that j ( K ) n C = K , and that j ( K ) is the smallest norm-closed two-sided ideal in R that intersects C exactly in K. Suppose, then, that C E j ( K ) n C. We have to show that C E K . To this end it suffices (since K is norm closed) to show that, for every positive E , C is at distance less than E from K . Since C E [RK], there are finite sets {Rl,. . ,Rn}in R and { K1,. ..,K,} in K such that
.
llRlK1-I
.
* *
+ RnK, - Cll < E .
By Proposition 8.3.4, there exist C1,. ..,Cn in C and a sequence {a,} in D, such that ( j = 1,. ..,n).
lim Ilam(Rj) - CjII = O
m+oo
NowClK1 + . - . + C n K , ~ K , a n d
IlC& t
*
'
t C,K, - Cll
= lim ))am(R1)K1+ * - . m-oo
= lim Ilam(R1K1 t * m-mo
5 llRlK1 +
* - -
+ R,Kn
*
+ am(Rn)K,
- C(l
+ R n K n - C)ll
- Cll < E .
388
THE TRACE
This completes the proof of the stated results concerning j ( K ) , We now consider J ( K ) . Suppose that R1,Rz E J ( K ) and A1 ,A2, B1, B2 E R. It is apparent from the definition of J ( K ) that
for all A and B in 'R. Since K is a norm-closed linear space, it now follows from the result of Exercise 8.7.5 that
for all A and B in R. Thus AlRlBl t AzRzBz E J(Ic), and J(K) is a two-sided ideal in R. We assert next that J ( K ) is norm closed. For this, suppose that R lies in the norm closure J(K)=of J(K).We have to show that, for each A and B in 12, COR(ARB)=n C E K ; equivalently (since K is norm closed) we must prove that, for each C in coa(ARB)' nC and each positive E , IlC - Coil < E for some COin Ic. Since R E J ( K ) = and C E coa(ARB)', we can choose Ro in J ( K ) and o in D so that
We then have
For each p in D,
so
Since Ro lies in the two-sided ideal J ( I c ) , the same is true of a(AR0B); so coa(a(ARoB))= n C C Ic. From (l),with D replaced by an element COof coR(a(ARoB))'nC, we have ~ ~ C - C < EOand ~~ CO E K. As already noted, this completes the proof that J ( K ) is a norm-closed two-sided ideal in R.
EXERCISE 8.7.16
389
We now prove that J ( K ) n C = K. If Ir' E K,then K E j ( K ) , and A K B E j ( K ) for all A and B in R. Since j ( K ) is a norm-closed twosided ideal a n d j ( K ) n C = K ,it now follows that con(AKB)= E j ( K ) and coa(AKB)' n C j ( K ) n C = K.
c
J ( K ) n C. Conversely, if K E Accordingly, A' E J ( K ) ; so K J ( K ) n C, it follows from the definition of J ( K ) that K contains CO.R(K)=nC (= { K } ) ;in other words, K E K. Thus J ( K ) nC = K. Let Z be a closed two-sided ideal in R,for which Z n C = K. Given R in Z, we have
and
C O ~ ( A R B )n= c
z nc = K ,
for all A and B in R ;so R E J(K).Thus Z C J ( K ) , and J ( K ) is the largest norm-closed two-sided ideal in R that intersects C exactly in K. ~ [ 2 7Chap.III,Par. ( 5,Sec. 2,pp. 254-2571 8.7.16. With the notation of Exercise 8.7.15, let K1, K2, K, A) be norm-closed ideals in C , and let Z be a two-sided ideal in R. Prove that (i) J ( K 1 ) J ( K 2 ) if and only if K1 K 2 ; (ii) na)= JW; (iii) Z is a maximal two-sided ideal in R if and only if Z = J(K) for some maximal ideal K in C ; (iv) Z is the intersection of a family of maximal two-sided ideals in R if and only if Z = J ( K ) for some proper norm-closed ideal K in (a E
waEAnaEA
c
C. Solution. (i) It is apparent, from the definition of J ( K ) , that J(K1) C J ( K 2 ) if K1 C K 2 . Conversely, if J ( K 1 ) J(K2),then K1 = J(&) nC c J ( K 2 )nC = K2. (ii)
An element R of R lies in J ( n a E AK,) if and only if
390
THE TRACE
However, (1) is satisfies if and only if COR(ARB)=n C G K a
(A,B E R)
for each a in A; that is, if and only if R E J ( K a ) for each a in A. (iii) Suppose first that Z is a maximal two-sided ideal in R. Then Zis norm closed, Z n C is a norm-closed ideal K in C, K # C since I 4 Z and J ( K ) # R since J ( K ) n C = K # C. From the final assertion of Exercise 8.7.15, Z 5 J ( K ) ; so Z = J ( K ) , since Z is maximal. If K E KO # C, for some ideal KO in C, then K 5 KO G K r # C. By (i), J ( K ) C J ( K r ) # R. Since J(K)(= 2 ) is a maximal two-sided ideal in R, we have J(K) = J ( K g ) . By (i), K = Kc; so K = KO. Thus K is a maximal ideal in C, and (as noted above) Z = J ( K ) . Conversely, suppose that Z = J ( K ) , where K is a (necessarily norm-closed) maximal ideal in C. If Z C 10 # R, for some normclosed two-sided ideal 10in R, then
K
= J ( K ) n C = Z n C 5 2,
n C # C,
and ZOnC = K since K is maximal. By the final assertion of Exercise 8.7.15,Zo 5 J ( K ) = Z. Thus Z = 30,and Z is maximal. (iv) Suppose first that Z = J ( K ) , where K is a proper normclosed ideal in C. By Theorem 3.4.1, and since C C ( X ) for some compact Hausdorff space X, K = n K a for some family { K a } of maximal ideals in C. By (ii), Z = J(K) = J ( K a ) ; so by (iii), Z is the intersection of a family of maximal two-sided ideals in R. Conversely, suppose that Z = where each Z a is a maximal two-sided ideal in R. By (iii), Z a = J ( K a ) for some maximal ideal I c a in C. From (ii),
=
n
n&,
where K is the proper norm-closed ideal
nK, in C.
8.7.17. With the notation of Exercise 8.7.15, suppose that the von Neumann algebra R is finite. Prove that
J ( K ) = { R E R : T(R*R)E K}, where
T
is the center-valued trace on R.
391
EXERCISE 8.7.18
Solution. For each R in R, the set co*(R)= n C consists of the single point r ( R ) , by Theorem 8.3.6. Hence the definition of J ( K ) given in Exercise 8.7.15 can be expressed, in the present case, as follows:
J ( K ) = { R E R : r ( A R B ) E IC ( A , B E R ) } . It therefore suffices t o show that, if R E R and r ( R * R ) E IC, then r ( A R B ) E K for all A and B in R. Suppose, then, that A , B , R E R and r(R*R) E IC. From Theorem 3.4.1, and since C 2 C ( X ) for some compact Hausdorff space X ,there is a set Po of pure states of C such that (1)
x: = {C E c : p(C) = 0
(p
E Po)}.
When p E Po,p o r is a state u of R, and u(R*R) = 0. By the Cauchy-Schwarz inequality, 0 = u ( B A R ) = p ( 7 ( B A R ) )= p ( 7 ( A R B ) ) .
It now follows from (1) that r ( A R B ) E
K.
8.7.18. Suppose that for each positive integer n, R, is finite von Neumann algebra with center C,, and r, is the center-valued trace on R,. Let R be the finite von Neumann algebra C $R,,let C be the center C $C, of R, and let K be the norm-closed ideal
in C. With the notation of Exercise 8.7.15, show that
By considering suitable projections in R, show that j ( K ) # J ( K ) if either each R, is of type 111 or R1 is of type I1 , R2 is of type 12 , R3 is of type 13, and so on.
392
THE TRACE
Solution. The equation
defines a semi-norm p on R, and p is norm continuous since p ( R ) 5 llRll. In addition,p(ARB) 5 llAllp(R)llBll for all A , B , R in R. From this, the null space p-'(O) of p is a norm-closed two-sided ideal in R. Now
and ~ ~ ' ( n 0 C) is the norm-closed ideal K in C. We have to prove that p-'(O) = j ( K ) . From the assertion in the final sentence of Exercise 8.7.15, j ( K ) C p - ' ( 0 ) . To establish the reverse inclusion, suppose that R = C$R, E p-'(O), so that llRnll -+ 0 as n + 00. For each positive integer n, let C, be I(R,III and let
Then C $A, is an element A of R, $C, is an element C of K , and R = AC E RK j ( K ) . Hence p-'(O) = j ( K ) . F'rom Exercise 8.7.17,
J ( K ) = { R E R : r(R*R)E K } , is the center-valued trace on R. If R is an element
where T of R,we have
so r(R*R)E K: if and only if llrn(RiRn)ll+ 0 as n -+
00.
$R,
Thus
A projection E in R has the form C $En, where En is a projection in R,, for n = 1 , 2 , . . . . From the descriptions of j ( K ) and J(K)just obtained, it follows that E E j ( K ) if and only if llEnll 0 (equivalently, E E j(K)if and only if E , = 0 for all but a finite set of values of n), and that E E J(K) if and only if ]lrn(En)ll 0. From Theorem 8.4.4; in both of the cases to be considered (R, is either of type 111 or of type I,) there is a projection E, in R, such that 7 ( E n )= n-'I. Thus llEnll= 1, 11~(E,)ll= n-', and $En is an element of J ( K ) \ j ( K ) . --f
--f
EXERCISE 8.7.19
393
8.7.19. With the notation of Exercise 8.7.15, suppose that the von Neumann algebra R is properly infinite, and that Ic = (0). Let EO be a projection in R. (i) By using the results of Exercises 8.7.15 and 6.9.51, show that Eo E J ( ( 0 ) ) if and only if PEO 4 P for each non-zero projection P in C. (ii) Give a second proof of the assertion in (i), by using the definition of J ( { 0)) given in Exercise 8.7.15 and extending the reasoning indicated in Exercise 8.7.9. Solution. From the final assertion of Exercise 8.7.15, J ( ( 0 ) ) is the largest norm-closed two-sided ideal in R that has intersection (0) with C. It now follows from parts (ii) and (iii) of Exercise 6.9.51 that J ( ( 0 ) ) is the ideal Z= occurring in that exercise. From Exercise 6.9.46, Eo lies in I= if and only if it lies in the ideal Z. It is clear from Exercise 6.9.51(i) that EOE Zif and only if EO E PO (with PO defined as in that exercise). Accordingly, EO E J ( ( 0 ) ) if and only if PEO 4 P for each non-zero projection P in R. For our second proof of this result, suppose first that Eo E J ( { O } ) . This implies that c o ~ ( E 0 ) ' nC = {0}, and hence that the projection Q 1 occurring in Exercise 8.7.9(v) is 0. Hence PEO 4 P for each non-zero projection P in C . Now suppose that EO 4 J ( ( 0 ) ) ; we have to show that QEo Q for some non-zero projection Q in C. From the definition of J ( { O } ) in Exercise 8.7.15 it follows that there exist A , B in R and C(# 0) in C such that C E COR(AEOB)'.We can choose unitary operators u 1 , . . . ,u k in 2 and positive scalars a l , . . . , a k with sum 1 such that N
k j=1
With
E(>
0) defined by k
llCll = 2~ -I-IIC - C a j U j A E o B U j t l ( , j=1
let Q be the spectral projection for (the normal operator) C , corresponding to the Bore1 set (A E C : 1x1 > llCll - E } . Then Q is a non-zero projection in C, and
IlcQ~ll 2 (IlCll - ~ > l l Q ~ I l ,
394
THE TRACE
k
IlCQx
- C a j U j A E o B u ; Q z ( l 5 (IlCll-
2~)11QzII,
j=1
for each vector x in the Hilbert space on which R acts. Accordingly, for each non-zero vector x in the range of Q, there is at least one j in (1,. . ,k } for which E0BV;Qx # 0. For j = 1,.. . ,k , let Fj be the range projection of the operator QUjB*Eo. Then
.
(this last, by Proposition 6.1.6), and I-Fj has range the null space of EoBUJQ, by Proposition 2.5.13. We can interpret the last sentence of the preceding paragraph as follows: if z = Qx # 0 , then Fjx # 0 for at least one value of j in (1,. . . ,k}. From this, and since Fj 5 Q for each j , we have
It now follows from Exercise 8.7.9(i) that QEo
N
Q.
8.7.20. With the notation of Exercise 8.7.15, suppose that the von Neumann algebra R is properly infinite and countablly decomposable, K is {0), and Eo is a projection in R. By using the result of Exercise 8.7.19, show that EO E J ( ( 0 ) ) if and only if EO is finite. Deduce that J ( ( 0 ) ) is (0) when R is a countably decomposable type I11 von Neumann algebra.
Solution. If EO is finite, then PEO 4 P (because P is infinite) for every non-zero projection P in the center C of R. If EO is infinite, there is a non-zero projection Q in C such that QEo is properly infinite and has central carrier Q,and QEo N Q by Corollary 6.3.5. It now follows from Exercise 8.7.19 that EO E J ( ( 0 ) ) if and only if EO is finite. If R is a countably decomposable type I11 von Neumann algebra, there are no non-zero finite projections in R. Thus J ( ( 0 ) ) contains no non-zero projections, and J ( ( 0 ) ) is (0) by Exercise 6.9.47(i). 8.7.21. With. the notation of Exercise 8.7.15 suppose that the von Neumann algebra R is properly infinite. Let E be a projection
395
EXERCISE 8.7.21
in R, and let Q be the largest projection in C such that Q E Q. Show that ( I - Q ) E E J ( { O ) ) , and that E E J ( K ) if and only if Q E K. Deduce that N
and that J ( K ) = j ( K ) when R is a countably decomposable type I11 von Neumann algebra. Is (1) always satisfied when K is norm-closed ideal in the center C of a finite von Neumann algebra? Solution. From the maximality of Q, we have the relation P ( I - Q ) E 4 P for every non-zero projection P in C. It therefore follows from Exercise 8.7.19 and 8.7.16(i) that
Hence E E J ( K ) if and only if Q E E J(K). However, QE E J ( K ) if and only if Q E J ( K ) (from Remark 6.8.2, because Q E Q) and Q E J ( K ) if and only if Q E K (because Q E C and J(K)n C = K). Thus E E J ( K ) if and only if Q E K. We have already noted that J ( ( 0 ) ) C J ( K ) , and j ( K ) C J(K) by the final assertion of Exercise 8.7.15. Hence N
Moreover, the two-sided ideal j ( K ) t J ( ( 0 ) ) is norm closed, by Exercise 4.6.63. In order to prove ( l ) , it now suffices (in view of Exercise 6.9.47(i)) to show that j ( K ) t J ( ( 0 ) ) contains each projection in J ( K ) . Given such a projection E , and with Q chosen as in the statement of the present exercise, we have ( I - Q ) E E J({O)), Q E K C j ( K ) , and thus
E = Q E t ( I - Q ) EE j ( K )t J((0)). This proves (l), in the case of a properly infinite von Neumann algebra. In the case of a finitevon Neumann algebra, it follows from Exercise 8.7.17 that J ( ( 0 ) ) is {0}, since the center-valued trace is faithful. However, there are examples in which R is finite and J ( K ) # j ( K ) (Exercise 8.7.18), and in these instances (1) is not satisfied. m[89]
396
THE TRACE
Suppose that R is a properly infinite von Neumann 8.7.22. algebra with center C, Z is a norm-closed two-sided ideal in R, and (with the notation of Exercise 8.7.15) J ( ( 0 ) c Z. Show that Z = J ( Z n C).
Solution. With K: the norm-closed ideal Z n C in C, it follows from the final assertion of Exericse 8.7.15 that j(K:)
c z c JW).
Since J ( { O } ) C Z, we have
dK:>t J ( ( 0 ) ) c 1; that is (Exercise 8.7.21), J ( K ) c Z. Thus Z = J(K:). Suppose that R and S are von Neumann algebras 8.7.23. acting on a Hilbert space H,and R C S. Let @ be a positive linear mapping from S into R, such that
@(RlSRz)= Rl@(S)Rz @(I) = I, whenever R1, Rz E R and S E S. For all S in S, (i) prove that @(S*)= @(S)*; (ii) by considering the element [S- @(S)]*[S- @(S)]of S, show that (P(S)*@(S)5 @(S*S); (iii) deduce that Il@(S)ll 5 IlSll. [A mapping @ with the properties set out in this exercise is described as a conditional expectation from S onto R.] Solution. (i) Each self-adjoint element H of S has the form HI - H 2 , where H 1 , H z E S+.Since 9 is a positive linear mapping, @ ( H I ) ,@ ( H z ) E R+ and @ ( H ) ( = @ ( H I )- @ ( H 2 ) ) is self-adjoint. Each element S of S has the form H + iK,where H and K are self-adjoint elements of S. From the preceding paragraph, @ ( H )and @(K)are self-adjoint. Thus @(S)*= [ @ ( H )t i@(K)]*= @ ( H )- i @ ( K )= @(S*). (ii) When R E R(c S ) and S E S, we have (S- R ) * ( S - R ) 1 0 , and thus 0 5 9 ( ( S - R)*(S - R ) )
+ R*R) = (P(S*S) - R*@(S)- Q(S)*R+ R*R.
= @(S*S- R*S - S*R
397
EXERCISE 8.7.24
When R is the element @ ( S )of R,we obtain 0 5 @ ( S * S) @(S)*@(S), so @ ( S ) * @ ( S5) @(S*S). (iii) From (ii), and since S*S
5 llS((21,we have
@(S)*@(S)F @ ( S * S F ) IIS1l2@(I)=
11~1I2L
Let U be the unitary group in a von Neumann alge8.7.24. bra R acting on a Hilbert space H . Suppose that c o ~ ( T ) -meets the commutant R', for each T in B(3-1), where coa(T)- denotes ) the set the weak-operator closure of the convex hull c o ~ ( T of {UTU* : U € 2 4 ) . Let M be the set of all positive linear mappings cp : B(3-1) + B(7-f)such that
whenever T E B(3-1) and R i , R!, E R'. Let D(C_M ) be the set of all mappings a : B ( H ) -, B ( H ) that can be defined by an equation of the form k
a ( T )=
ajUjTUj*, j=l
where u1,,. . ,u k E U and a l , . . . , a k are positive scalars with sum 1. (i) Prove that cp(R') = R' whenever cp E M and R' E R', and that cp1 o 9 2 E M whenever c p 1 , c p ~E M . (ii) Show that M can be viewed as a closed subset of the product X T , where XT is c o ~ ( T ) -with the weaktopological space operator topology. (iii) Suppose that TOE B(3-1) and Ah E co~(T0)-n R'. Show by a compactness argument that Ah = $(TO)for some in M . (iv) Show that if T I , .. . ,T, E B ( H ) , there is an element cp of M such that cp(Tl),. . . ,cp(Tn)E R'. [Hint. Use (i) and (iii) to give a proof by induction on n.]
nITEB(31)
+
398
THE TRACE
(v) For each finite subset
IF of B ( H ) , let
Mp = {cp E M : cp(T) E R' whenever T E F}. Show that the family of all such sets M s has non-empty intersection. (vi) Deduce that there is a conditional expectation @ from B(3-1) onto R',with the property that @ ( T )E cox(T)- n R' for each T in B(7-f). Prove also that, if TOE B(3-1)and Ah E co~(T0)-n R',then @ can be chosen so that @(TO) = Ah. Solution. (i) If 9 E M and R' E R',we have
cp(R') = cp(R'1.I) = R'cp(I)I = R'. If 91,cp2 E M , then cp1 o 92 is a positive linear mapping from B(7-f) into B ( H ) , and ( p 1 ( 9 2 ( 1 ) ) = cpl(I) = I . Also,
whenever R i , Rb E R' and T E B(3t). In order to complete the proof that (PI ocpg E M , we have to show that cpl(cp2(T))E COR(T)for each T in B(7-f). Since cpz(T) E COR(T)- and cpl(cpz(T)) E co~(cpz(T))-,it suffices to show that cox(T1)- C COR(T)- when 21' E c o ~ ( T ) - .Now it is apparent that U S U * E COR(T)whenever U € U and S € cox(T). From this, together with the weakoperator continuity of the mapping A + UAU', it follows that USU* E COR(T)- whenever U E U and S E c o ~ ( T ) - . In particular, con(T)- contains the set {UT1U* : U E U},and so contains its weak-operator closed convex hull c o ~ ( T l ) - ,when TI E c o ~ ( T ) - . This completes the proof that 91 o 9 2 E M when p1,(p2 E M . (ii) The product space ll is the set of all mappings 71 : B(3-1) + B(7-f) such that n ( T ) E c o ~ ( T ) for - each T in B('H), with the coarsest topology that makes each of the "coordinate mappings" n + ?r(T): II + a(%) continuous relative to the weak-operator topology on B(3-1). Since the bounded closed set XT is weak-operator compact, it follows from Tychonoff's theorem that II is compact. The set M consists of those elements cp of II that satisfy the conditions cp(1)- I = 0,
cp(aS
+ bT) - acp(S) - b ( T ) = 0,
399
EXERCISE 8.7.24
whenever H E B ( H ) + , S,T E B ( H ) , R’,,Rb E R’, and a,b E @. Since each of the mappings
7r
+
~ ( 1-)I , x T ( U S 6T) - U T ( S ) ba(T), + n(R‘,TRL)- Rin(T)RL, 7r + 7r(H),
K -+
-+
from II into B ( H ) (with the weak-operator topology) is continuous, M is the intersection of a family of sets, each one of which is the inverse image of a closed set (either {0} or B(3-1)+) under a continuous mapping. Hence M is a closed subset of II, and is therefore compact in the relative topology. (iii) If TO E B ( H ) and Ah E c o ~ ( T ) -fl R‘, there is a net { S j } in c o ~ ( T 0 )that is weak-operator convergent to Ah. For each index j , there exists an element aj of D such that Sj = a j ( T 0 ) . Since M is compact, the net {aj} has a subnet { a j k }that converges to an element $ of M. Since the “coordinate mapping” a -+ 7r(To)is continuous,
$(TO)= limaj,(To) = limSj, = limSj = A;. k
k
3
(iv) Given 2’1 in B ( H ) , let A’, be any element of c o ~ ( T 1 ) -n R‘. By (iii), A’, = cp(T1)for some cp in M. This proves the stated result in the case in which n = 1. Suppose that T is a positive integer, and the stated result has been proved in the case in which n = T . Accordingly, given T I , .. . ,T,+1 in B(‘H), there is an element cpo of M such that cpo(Tl),. . . ,cpo(T,.)E R’. Choose A‘ in c o ~ ( c p ~ ( T ~ + ln) )R‘. - From (iii), $(cpo(T,.+l)) = A‘ (E R’) for some 1c, in M. Moreover, $(cpo(Tj)) = cpo(Tj) (E 72’) when 1 5 j 5 T , by (i). With cp the element 1c, o cpo of M, we have cp(Tj) E R’ ( j = 1,. . . ,T 1). This completes the proof by induction of the result stated in (iv). (v), (vi) For each T in B ( H ) ,the subset { c p E M : cp(T) E R‘} of M is closed, since R’ is (weak-operator) closed in B ( H ) and the mapping cp + cp(T) : M + B ( H ) is continuous. Thus M a (a finite intersection of sets of the type just considered) is closed in M, for each finite subset IF of B ( H ) . Moreover, M r is not empty, by (iv). Since, also, Mal n Ma, = Malua,,
+
the family {Ma} of all such sets has the finite intesection property. Since M is compact, the intersection of all the sets Ma is not empty.
400
THE TRACE
With 9 in this int,ersection, 9 E M and 9 ( T ) E R' for each T in B(3-t). Thus 9 is a conditional expectation from B(3-t) onto R',with n R' for each T in B(3-t). the property that @ ( T )E CO.R(T)Given TOin B(3-1) and Ah in co~(T0)-n R',choose $ as in (iii). Then 9 o $ E M , and 9 o $ maps B(3-t) into R' (because 9 does so). Hence 9 o $ is a conditional expectation from B(3-t) onto R', (9 o $ ) ( T ) E c o . ~ ( T ) -i l R' for each T in B(%), and
(9o $)(To) = @($(To))= @(Ah)= Ah.
.[99]
8.7.25. Let R be a von Neumann algebra of type I, acting on a Hilbert space 3-t. (i) Show that R satisfies the hypothesis of Proposition 8.3.11. [Hint.Use Theorem 6.6.5 and the continuity properies of * isomorphisms (Remark 7.4.4) to show that it suffices to consider the case in which R is n @ C , where C is an abelian von Neumann algebra. Let K be a set with cardinality n, so that each element of n @ C is represented by a matrix [ C j , k ] j , k E ~with entries in C. Let A be the set of all pairs (lF,d),in which Ii' is a finite subset of K and d is a finite-dimensional * subalgebra of C. When a = (F,d)E A, let R, be the set of all elements of n@C with matrices [ C j , k ] such that Cj,k E d for all j and k in K and Cj,k = 0 unless j , k E F.] (ii) Use (i) and Exercise 8.7.24 to show that there is a conditional expectation 9 from B(3-t) onto R', with the property that Q ( T ) E CO.R(T)-n R' for each T in B(3.1).
Solution. (i) Consider first the case in which R is n @ C , and define A and R,(u E A) as in the hint. It was noted in the proof of Corollary 8.3.12 that the set of all finite-dimensional * subalgebras of an abelian von Neumann algebra C is directed by the inclusion relation C, and has union norm-dense in C. It follows that the set A is directed by the relation 5 , in which (IF1,dl)5 (F2,da)if and and dl 5 42, and that { R , : a E A} is an increasonly if IF1 C ing net of finite-dimensional * subalgebras of n 8 C. For each finite subset F of K,let E(F)be the projection in n @ C whose matrix has 1 in the (k,k) position when k E F and has 0 in all other entries. When R E n @ C, E(F)RE(F)has a matrix (Cj,k]in which Cj,k = 0 unless j , k E F, and Cj,k E C when j , k E F. Each of the finite set of non-zero elements Cj,k can be approximated in norm, as closely
EXERCISE 8.7.26
401
as we please, by an element Aj,k of some finite-dimensional * subalgebra d j , k of C. Then, E(P)RE(IF)is approximated in norm by an element of R,,where a = ( F , d ) with d a finite-dimensional * subalgebra of c that contains U j , k E F d j , k . It follows that the norm doand also the weak-operator closure (UaElh Ra)-, sure (UaEARa)=, contains E(P)(n @ C)E(F) for each finite subset IF of K. Since V E(F) = I , we have (UaE- Ra)-= n 8 C. Given any type I, von Neumann algebra R,by Theorem 6.6.5 there is an abelian von Neumann algebra C and a * isomorphism 'p from n C onto R. With R a ( a E A) constructed as in the preceding paragraph, {'p(R,) : a E A} is an increasing net of finitedimensional * subalgebras of R. Now cp is isometric, and gives rise to a homeomorphism between the unit balls (n 8 C)l and (R)1 in the weak-operator topology, by Remark 7.4.4. By the Kaplansky density theorem, (UaEARa)lis weak-operator dense in ( n @I C)1; so (UaEA cp(Ra))lis dense in (R)1,and UaEd'p(R,)is dense in R. R', for each (ii) From (i) and Proposition 8.3.11, c o . ~ ( T )meets T in B(3-1). It follows from the result of Exercise 8.7.24 that there is a conditional expectation 9 from B ( H ) onto R', with the property that @ ( T E ) c o . ~ ( T ) -n R' for each T in B(7-l). 8.7.26. Show that the results of Exercise 8.7.25 remain valid for every type I von Neumann algebra R acting on a Hilbert space 3-1.
Solution. There is an orthogonal family { Q k : k E R} of central projections in R,with sum I , and (for each k in K) a cardinal n ( k ) such that RQk is of type In(k). By the result of Exercise 8.7.25, there is a family 3 k of finite-dimensional * subalgebras of RQk that is directed upward by the inclusion relation c , and has union weakoperator dense in RQk. Given any finite subset {k(l), .. .,k ( m ) } of K and any choice of Rj in Fk(j) (for each j = 1, . . . ,m ) , the linear span R1 +- . - - R, of Uj"=, Rj is a finite-dimensional * subalgebra of R. The set 3of all such algebras 721 +-. . .f R, is directed upward by C , and its union is weak-operator dense in R. Just as in Exercise 8.7.25, it now follows from Proposition 8.3.11 and Exercise 8.7.24 that there is a conditional expectation 9 from B(3-1)onto R', with the property that Q ( T )E COR(T)-nR' for each T inB(3-1).
+
402
TH E TRACE
8.7.27. Suppose that R is a von Neumann algebra acting on a Hilbert space 'H, and A is an abelian von Neumann subalgebra of R. (i) By adapting part of the proof of Lemma 8.2.3, show that the von Neumann algebra A' is of type I. (ii) By using the result of Exercise 8.7.26, show that there exist conditional expectations, @ from B(7-l) onto A, and I€' from B('H) onto A'. (iii) Prove also that there exist conditional expectations, CPO from R onto A, and @O from 72 onto R n A'. Solution. (i) Let {Pa}be an orthogonal family of cyclic projections in A, with sum I. For each index a, choose a vector x, such that Pa has range [d'z,], and let EL (in A') be the projection with range [Axa].The abelian von Neumann algebra AEL acts on the Hilbert space EL(3-I) and has z, as a generating vector. By Corollary 7.2.16, AEL = (AE;)' = ELA'EL; so ELA'E; is abelian, and EL is an abelian projection in A'. Note, also, that the central carrier of EL (relative to A') coincides with the central carrier Pa of Pa (relative to A), by Proposition 5.5.13. Thus C EL is an abelian projection in A' with central carrier C Pa(= I),and A' is of type I. (ii), (iii) Since A' is of type I by (i), it follows from the result of Exercise 8.7.26 that there is a conditional expectation @ from B(3-1) onto A"(= A). The restriction @lR is a conditional expectation from R onto A. Since A is of type I (in fact, of type 11), it follows from the result of Exercise 8.7.26 that there is a conditional expectation @ from B('H) onto A'. Since COA(T)- meets A', for each T in B ( Z ) (Corollary 8.3.12), I€' can be chosen in such a way that
Q ( T )E c o ~ ( T ) -n A' (1) for each T in B('H), by Exercise 8.7.24(vi). Let U(E R) be the unitary group of A. When R E R, R contains the set {WRU* : U E U},and therefore contains the weakoperator closed convex hull COA(R)-of that set. Accordingly, it follows from (1) that (2) I€'@) E R n A' ( R E R). Since I€' is a conditional expectation from a(%)onto A', it follows easily from (2) that the restriction @\Ris a conditional expectation 90from R onto R n A'.
403
EXERCISE 8.7.28
Suppose that R and S are von Neumann algebras 8.7.28. such i, that R E S and S has a faithful acting on a Hilbert space ? normal tracial state T . (i) Show that if H E S+,there is an element HO of Rt such that T ( H R ) = T ( H o R ) for each R in 72. [Hint. Apply Theorem 7.3.13 to suitable positive normal linear functionals on 72.1 (ii) Show that, for each element S of S , there is a unique element cp(S) of R such that T ( S R )= .r(cp(S)R)for each R in R. (iii) Prove that the mapping cp : S R defined by (ii) is an ultraweakly continuous conditional expectation from S onto R, and is faithful in the sense that cp(S) # 0 when 0 # S E St. --f
Solution. (i) If H E St,the equation p(R) = T ( H R )defines a positive normal linear functional p on R (for positivity, note that p ( R ) = T ( H ~ / ~ R H ~With / ~ ) )H. replaced by llHllI - H, it follows that the mapping
<
11H11~. From is a positive linear functional on R. Thus 0 5 p Theorem 7.3.13, there is an element KO of (R+)1such that
for each R in R. This proves (i), with Ho the element llHllK0 of
R+.
(ii) Suppose that S E S. Since S can be expressed as a linear combination of four positive elements of S, it follows from (i) that there is an element SO of R such that
@ R ) = T(SoR)
(1)
( R E R).
If S1 (in the von Neumann algebra R) has the property just ascribed to SO,then r ( ( S 1 - So)R) = 0 for each R in R;in particular, T((S1
Since
T
- So)(&- So)') = 0.
is faithful, S1 = So. Hence there is a unique element So of
R that satisfies (l),and we define cp(S) to be So.
404
THE TRACE
(iii) Suppose that S, S1, S2 E S , R1, R2 E R, and each R in R,
u1, a2
E @. For
T(V(alS1 t a2S2)R) = m l S 1 t a2S2)R) = UlT(S1R) t UzT(S2R) = U l M S 1 ) R ) t azr(cp(S2)R) = .r((Ulcp(Sl)t azcp(Sz))R), T(cp(RlSR2)R)= T(RlSR2.R)= T(SRZRR1) = 7(cp(S)R2RRl)= 7(Rlcp(S)R2R), and
T(cp(I)R)= T ( R ) = T ( I R ) . These equations, together with the uniqueness clause in (ii), imply that cp(a1S1 t a2S2) = alcp(S1) t a2#2), cp(RlSR2) = Rlcp(S)R2 and cp(1)= I. Moreover, if H E S+, it follows from (i) that cp(H) is an element HO of R+.Thus cp is a conditional expectation from S onto 72. Suppose that S E S+ and cp(S) = 0. Since T is faithful and T ( S )= ~ ( c p ( S )=) 0, it follows that S = 0. Hence cp is faithful. In order to prove that cp is ultraweakly continuous, we have to show that w o cp E S, whenever w E R,. Since R# is the linear span of its positive elements (see Remark 7.4.4),we may assume that w is a positive normal linear functional on R. Then, w o cp is positive, and we have to prove that w o cp is normal, Suppose that {H,} is a monotone increasing net of self-adjoint elements of S, with least upper bound H in S. Since cp is a positive linear mapping, the net {cp(H,)} in R is monotone increasing, has cp(H) as an upper bound in R,and therefore has a least upper bound K(5 cp(H)) in R. Since T is normal, T ( K )= limr(cp(H,)) = limT(H,) = T ( H ) = r ( p ( H ) ) ; a
so cp(H) - K 2 0 and ~ ( c p ( H-) K )= 0. Since T is faithful, cp(H) = K . Since w is normal, it now follows that limw(cp(H,)) = w ( K ) = w(cp(H)).
Thus w o cp is normal (and cp is ultraweakly continuous).
m
EXERCISE 8.7.29
405
8.7.29. Suppose that G is a discrete group with unit element e . Adopt the notation of Section 6.7, and recall that the von Neumann algebra CG has a faithful tracial state T defined by T ( A )= ( A x e ,xe). When z E lm(G), let M , (in B(lZ(G))be the operator of multiplication by z. (i) Suppose that CP is a conditional expectation from B ( / z ( G ) ) onto CG,and define a linear functional p on lW(G)by
Show that p is an invariant mean on G (in the sense explained in Exercise 3.5.7). (ii) Suppose that p is an invariant mean on G. Show that p is a state of the C*-algebra Zm(G). Suppose that T E B(l2(G)).Given z and y in lz(G),define z , , ~ in 1,(G) by
Show that there is an element AT of B(22(G))such that
and prove that
AT E CORJT>-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 <_ E V F . For each index a , E , 5 E, V F 5 G, so E, 5 G; that is, E 5 G. Also, F 5 E, V F 5 G; so E V F 5 G. Thus G = E V F , and { E , V F } converges to E V F . (ii) Since the net { E , A F } increases and is bounded above by E A F , it converges to a projection G in R, and G 5 E A F . For our example, suppose that 3-1 is L2([0,1]) with respect to Lebesgue measure, and let R be B ( H ) . For each real number t such that 0 5 t < 1, let Et be the operator of multiplication by the characteristic function of the interval [O,t]. Then { E t } is an increasing net of projections in R, with limit I ( = E ) . Let F (in R) be the projection from H onto the one-dimensional subspace that consists of constant functions on [0,1]. Then E A F = I A F = F , but Et A F = 0 when 0 <_ t < 1. Thus
v
lim Et A F = 0
t+l-O
< E A F,
in this example. We now consider again the general situation, in which E , A F -+ G 5 E A F , but now suppose that R is finite. From Exercise 8.7.33(i), the center-valued dimension function A on R is weak-operator continuous. By (i), {Ea V F } converges to E V F . From Exercise 8.7.31,
A ( G ) = lim A ( E , A F ) = lim[A(E,) iA ( F ) - A ( E aV F ) ]
+
= A ( E ) A ( F )- A ( E V F ) = A ( E A F ) . Since E A F - G is a projection in R and A ( E AF - G ) = 0 , it follows that G = E A F ; so { E , A F } converges to E A F , in the finite case. (iii) Now suppose that E and F are projections in R, and E is the limit of a decmusing net { E,} of projections in R. We can apply (i) and (ii), with I - E , I - F , { I - Ea}, in place of E , F , {Ea}, respectively. It follows that
( I - Ea) V ( I - F ) + ( I - E ) V ( I - F ) , that { ( I - E,) A ( I - F ) } converges to a projection I - G ( 5 ( I - E ) A ( I - F ) ) in R , that I - G < ( I - E ) A ( I - F ) in
412
THE TRACE
some instances, but that I - G = ( I - E ) A ( I - F ) if R is finite. From the equations
I - ( I - E ) V ( I - F ) = E A F , I - ( I - E ) A ( I - F ) = E V F, and the corresponding relations with E, in place of E , it now follows that E, A F + E A F , that { E , V F } converges to a projection G ( 2 E V F ) in 2, that G > E V F in some instances, but that G = E V F if R is finite. Suppose that E , F , and G are projections in a finite 8.7.35. von Neumann algebra R, E and F are the limits of increasing nets { E , } and { F , } , respectively, of projections in R, and G is the limit of a decreasing net {Ga} of projections in R (the index set being the same for the three nets). From the relation (1) E , A G , - E A G = ( E , A G, and by (i) (ii) (iii)
- E, A
G) - ( E AG - E , A G)
using the results of the preceding three exercises, show that the net { E , A G,} is strong-operator convergent to E A G; { E , V G,} is strong-operator convergent to E V G; { E , A Fa}is strong-operator convergent to E A F .
Solution. (i) Note that both the bracketed terms on the righthand side of (1) are projections, and { E A G - E , A G } is strongoperator convergent to 0, by Exercise 8.7.34(ii). From Exercises 8.7.32 and 8.7.33(i), 0
5 (A(E,
A G,
- E , A G ) x , ~5) (A(G, - G)x,x)
-P
0,
for each vector 2. Thus A(E,AG,-E,AG) -+ 0 in the weak-operator topology, and { E , A G , - E , A G} is strong-operator convergent to 0 by Exercise 8.7.33(ii). It now follows from (1) that { E , A G,} is strong-operator convergent to E A G. (ii) With E, and G, replaced by I - G , and I-E,, respectively, the argument of (i) shows that
( I - G,) A ( I - Ea) -+ ( I - G) A (I- E ) . Thus
E , v G, = I - ( I - G,)A ( I - E,)
-+
I - ( I - G) A ( I - E ) = E v G.
EXERCISE 8.7.36
413
(iii) The net {E,AF,} is increasing and therefore has a projection M as a strong-operator limit and least upper bound. Since E , A Fa I. E A F for each a , M 5 E A F . With a' fixed the net { E , A Far} has E A F,I as its strong-operator limit from Exercise 8.7.34(ii). Since E , A F,I 5 E , A Fa when a 2 a', E A F,I 5 M for each a', Again, from Exercise 8.7.34(ii), { E A Fa} has E A F as its strong-operator limit (and least upper bound). Thus E A F 5 M . Hence M = E A F - that is, { E , A F a } converges to E A F . Show that a normal tracial weight p on a factor R is 8.7.36. either faithful or identically zero.
Solution. If p is not faithful, there is a non-zero element A of that p ( A ) = 0. For a suitable spectral projection E of A and positive scalar c, 0 # E 5 c A , and thus p ( E ) = 0. Let { E , } be an orthogonal family of projections in R, maximal subject t o the condition that 0 < Ea 5 E for each index a . Since R is a factor, it follows from the comparison theorem that either E 5 I - C E , or I - C E , 5 E . In the first case, I - C E , has a subprojection equivalent t o E , and this can be added to the family { E , } , contrary to the maximality assumption. In the second case, I - C E , can
R+ such
itself be added t o the family (again giving a contradiction) unless C E , = I . This shows that C E , = I , and hence
since p is normal. However, for each index a there is a partial isometry V, in R such that E , = V,*V,, VaV,*5 E , and
It follows that p(E,) = 0, for each index a ; so p ( I ) = 0, and p is identically zero. 8.7.37. Show that a weight p on a von Neumann algebra R is tracial if and only if p( U H U * ) = p ( H ) whenever H E R+ and U is a unitary operator in R.
Solution. We noted, near the beginning of Section 8.5, that tracial weights satisfy the stated condition.
414
THE TRACE
Now suppose that p ( U H U * ) = p ( H ) for all H in R+ and all U in the unitary group U of R. Upo'n replacing H by A*A, where A E R, we obtain p((AU)*(AU))= p(A*A). Hence AU E N,, whenever A E Np and U E U ,where N, is the left ideal {S E R : p(S*S) < 00) in R. From this, and since R is the linear span of U ,it follows that N, is a two-sided ideal in R, as is M , (= N,*N,). Now M , is the linear span of the set {H E R+ : p ( H ) < 0 0 ) (= F,), and p ( U H U * ) = p ( H ) whenever H E F, and U E U. Hence
p(UAU*) = p(A)
( A E M,, U E U),
where (as usual) the linear extension of plF, to M , is also denoted by p. Upon replacing A (in M P )by AU (also in M,), we obtain p ( U A ) = p(AU);since R is the linear span of U,it now follows that (1)
p ( B A )= p(AB)
( A E M p , B E R).
Now suppose that A E R and p(A*A) < decomposition V H , then
H 2 = A*A E Fp 5 M,,
00.
If A has polar
AA* = VH2V*.
From (l),with V H 2 and V* in place of A and B , respectively, we obtain
p(AA*) = p ( V H 2 V * )= p(V*VH2)= p ( H 2 ) = p(A*A). We have now shown that p(A*A) = p(AA*) whenever A E R and p(A*A) < 00. Upon replacing A by A*, we obtain the same equality when A E R and p(AA*) < 00. Hence p(A*A) = p(AA*) ( 5 00) for all A in R,and p is a tracial weight. 8.7.38. Let p be the unique (faithful, normal) tracial state on a finite factor R. Show that there is a finite orthogonal family { E l , . . . ,Ek} of equivalent cyclic projections in R with sum I . Prove that, for j = 1,. ..,k, there is a vector zj in the range of Ej such that p(EjAEj) = ( A ~ j , z j ) ( A E R ) . Deduce that k
p(A) = C ( A z j , z j ) j=1
( A E R).
EXERCISE 8.7.39
415
Solution. Let F be a non-zero cyclic projection in R,and let E be a monic subprojection of F in R (Proposition 8.2.1). Since R is a factor, CE = I , so there is a finite orthogonal family { E l , . . . ,E k } of projections in R such that
-
Since Ej E 5 F and F is cyclic, Ej is a cyclic projection in R (Proposition 6.2.9). The von Neumann algebra Z’Ej, acting on the range of E j , has a generating vector, and this is a separating vector for E j R E j . By Theorem 7.2.3, the positive normal functional plEjREj has the form wZj IEjREj for some vector zj in the range of E j . Thus
p ( E j A E j ) = ( E j A E j z j , z j ) = (Azj,zj) for each A in R, and k
k
k
8.7.39. Suppose that p is the unique (faithful, normal) tracial state on a finite factor R, and define a norm 11 112 on R (as in Remark 8.5.9) by 11.1112 = [ P ( A * A ) ] ’ /By ~ . using the result of Exercise 8.7.38, show that the norm topology associated with (1 112 is coarser (weaker) than the strong-operator topology, but that these two topologies coincide on bounded subsets of R.
Solution. From Exercise 8.7.38, there is a finite family { X I , . . . ,z k } of vectors such that k
Note that j=1
416
T H E TRACE
Since p is faithful, the set (21,.. ,,xk} is separating for R, and is therefore generating for R’. Given A0 in R a n d E ( > 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<*Iill*/2 = ( ( I < (5 ( ((&(I;
5 ll6ll < 00. Given Iil and
Il.U(I(l>
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
<
Thus r(GTG):CG is bounded and
<
Hence ro is a linear mapping of GRG into CG. If 0 GTG, then 0 r(GTG) so that 0 5 r(GTG):CGand 0 5 TO(GTG).With A and B in GRG,
<
Finally, with C in C ,
ro(CG) = (r(CG)'CG)G= (Cr(G):CG)G = ( C : CG)G= CCGG = CG. Thus TO as defined is the (unique) center-valued trace on GRG.
8.7.59. Let R be a von Neumann algebra of type 111 acting on a Hilbert space R,and let A be a maximal abelian (self-adjoint) subalgebra of R. Suppose H is a positive element in the unit ball of the center C of R. Show that (i) there is a projection E in A such that A ( E ) = H , where A is the center-valued dimension function on R [Hint. Use Exercise 6.9.27.] ; (ii) E (in (i)) can be chosen as a subprojection of a given projection G in d provided H A(G).
<
EXERCISE 8.7.60
449
Solution. (i) From Theorem 8.4.4(ii), there is a projection F in R such that A ( F ) = H . Since I E A and F 5 I, it follows from Exercise 6.9.27(iii) that there is a projection E in A such that E F . From Theorem 8.4.3(iii), A(E) = A(F) = H . (ii) If G is a projection in A and H 5 A(G), we can again choose a projection F in R such that A ( F ) = H , and F 5 G since A ( F ) 5 A(G). Thus F N E 5 G, and A ( E ) = A ( F ) = H , for some rn projection E in A by Exercise 6.9.27(iii). N
8.7.60. Let V H and W K be the polar decompositions of S and T , respectively, and suppose, S,T 7 R, where R is a finite von Neumann algebra acting on a Hilbert space 'If. Let E , and F, be the spectral projections for W and I<, respectively, corresponding to the interval [-n,n] for each positive integer n. Show that (i) TF, is a bounded everywhere-defined operator T, in R and {G,} is an increasing sequence of projections in R with strongoperator limit I , where G, has range F,(E) n Ti1(En(E)) [Hint. Use Exercises 6.9.55 and 8.7.35(iii).]; (ii) ST has dense domain; (iii) ST is preclosed and has a unique closed extension S :T affiliated with R [Hint. 1Jse Theorem 2.7.8 and Exercise 6.9.54.1; (iv) S T has a dense domain; (v) S T is preclosed and has a unique closed extension S4T affiliated with R; (vi) if 50 is a unit generating vector for R and zo is any other vector in E , there is an operator T affiliated with R such that T x o = ZO. (Compare Theorem 7.2.1', Exercise 7.6.50(iv), and Theorem 5.6.15.)
-+ -+
Solution. (i) By choice of F,,KF, is a bounded everywheredefined self-adjoint operator in R and T, = T F , = WKF,, so that T , is a bounded everywhere-defined operator in R. From Exercise 6.9.55, the projection M , with range {x : T,x E &('If)}is in R and E , 5 M , . Now { E n } is an increasing sequence with strong-operator limit I (from the spectral theorem), so that A(&) = r(E,) T r ( I )= I in the strong-operator topology. Since {M,}is an increasing seI . Hence {M,} has strongquence and r(E,) 5 r ( M , ) , r ( M , ) operator limit I . Again from the spectral theorem, {F,} is an increasing sequence of projections with strong-operator limit I . From Exercise 8.7.35(iii), { F , A M,} (= {G,}) is an increasing sequence
450
THE TRACE
with strong-operator limit I. (ii) From (i), U2=rGn(H)is dense in 7-l. If x E Gn(H), then T,x E E,(H) so that T,x E D ( H ) = D(S). At the same time, x E Fn(H)so that x E D ( K ) = D(T) and T x = TF,x = T,x. Thus x E D(ST). It follows that ST has a dense domain. (iii) Since S and T are affiliated with R, S* and T* are affiliated with R. From (ii), T*S* is densely defined. Now T*S* C_ ( S T ) * ,so that (ST)*is densely defined. From Theorem 2.7.8, ST is preclosed. The closure S : T of ST is the smallest closed extension of ST. If U' is a unitary operator in R' and x E Z)(ST), then x E D(T), T x E V(S),U'x E D(T),TU'x = U'Tx E D(S),and
STU'X = SU'TX = U'STX. From Remark 5.6.3, S T q R since D(ST)is a core for S T . If A is a closed extension of ST affiliated with R , then S:T C A, whence S:T = A from Exercise 6.9.54. (iv) Since En t I and F, t I, we have that E, A F, t I from is dense in 3.1. If x E Exercise 8.7.35(iii). Thus U,"==l(E,A Fn)(7-l) (EnA F,)('H), then x E D ( H ) r l D(K).Hence x E D(S -t T ) . It follows that S T has a dense domain. (v) Since S rl R and T 77 R , S* q R and T* q R. From (iv), D(S* t T * )is dense in 'H. Since S* t T* C_ ( S T ) * ,D ( ( S T ) * )is dense in 7-t. From Theorem 2.7.8, S + T is preclosed. As with S : T in (iii), S$T 77 R and S i T is the only closed extension of S -i-T affiliated with R. (vi) From Theorem 7.2.1', there are operators Bo in R and TO affiliated with R such that BoTozo = Z O . From (iii), BOTOhas a closure T (= Bo :TO)affiliated with R. For this choice of T , we have m[76(pp. 221-229)] that Txo = Z O .
+
+
+
CHAPTER 9 ALGEBRA A N D C O M M U T A N T
9.6.
9.6.1. only if R is commutant.
Exercises Prove that a von Neumann algebra R is of type I if and
* isomorphic to a von Neumann algebra with abelian
Solution. If R is of type I, then so is R', by Theorem 9.1.3. Let E' be an abelian projection in R' with central carrier I. The mapping A 3 AE' is a * isomorphism from R onto the von Neumann algebra RE' by Proposition 5.5.5, and RE' has abelian commutant
E'R'E'.
Conversely, suppose that R is * isomorphic to a von Neumann algebra S such that S' is abelian. Then S' is of type I (in fact, of type II), and S (= S")is of type I by Theorem 9.1.3. Since type is m preserved by * isomorphism, R is of type I.
9.6.2. Suppose that R is a finite von Neumann algebra acting on a Hilbert space H,with a separating and generating vector x. Suppose y E H and T E R. Prove that (i) y is a generating vector for R if and only if y is a separating
vector for R; (ii) Tx is a separating and generating vector for R if and only if the range projection of T is I.
Solution. (i) Since y is separating for R if and only if y is generating for R', we have to show that [Ry] = 'H if and only if [R'y]= H. By Lemma 9.1.1, R' (as well as R) is finite. Thus [Ry] = H if and only if [Ry] 'H (that is, [Ry] N [Rz]) relative to R', and [R'y] = 3.1 if and only if [R'y] N 7i (that is, [R'y] N [R'x]) relative to R. Accordingly, we have to show that [Ry] N [Rz] if and N
452
ALGEBRA AND COMMUTANT
only if [R'y] [R'x],and this follows from Theorem 7.2.12. (ii) If T has range projection E (# I ) , then I - E E R,I - E # 0 and ( I - E)Tx = 0 ; so T x is not a separating vector for R. Now suppose that T has range projection I . If A E R and ATx = 0, then AT = 0 (since AT E R and x is a separating vector for a). Thus A(T(31))= {0}, and A = 0 since T ( H ) is dense in H. This shows that Tx is a separating vector for R;by (i), it is also a generating vector for R. N
Let R be a finite von Neumann algebra acting on a 9.6.3. Hilbert space 7.1, with a separating and generating vector. Observe, from Lemma 9.1.1 and its proof, that R' is finite and R has a generating trace vector 5 0 . Denote by cp the * anti-isomorphism from R onto R' described in Theorem 7.2.15, so that cp(A)xo = Azo for each A in R. Let r and T' be the center-valued traces on R and R', respectively. (i) Show that cp(C)= C when C E R fl 72'. (ii) Prove that r ( A ) = ~ ' ( c p ( A for ) ) each A in R. (iii) Let E be a projection in R. Show that E has range [R'Exo] and cp( E ) has range [REzo]. (iv) Let E and E' be projections in R and R',respectively. Show that r ( E ) = r'(E') if and only if there is a vector y such that E has range [R'y] and E' has range [Ry].[Hint. Use (ii),(iii), and Theorems 7.2.12 and 8.4.3(vi).]
Solution. (i) when C E R fl R' we have
c - Cp(C)E R',
[C - (P(C)Izo= 0,
and since xo is a separating vector for R',we have C - cp(C) = 0. (ii) From (i), and since cp is a * anti-isomorphism from R onto R', it follows that r' o cp is a positive linear mapping from R into the common center R fl R' of R and R', and has the properties ascribed in Theorem 8.2.8(i), (ii) to the center-value trace 7. From the uniqueness clause in that theorem, T' o cp = r. (iii) Since $0 is a generating vector for both R and R', by Lemma 7.2.14, E has range [ER'xo] (= [R'Ezo])and p(E) has range equal to [cp(E)Rxo](= [Rcp(E)xo]= [RExo]). ) (ii), it follows from Theorem (iv) Since T ( E ) = ~ ' ( c p ( E )by 8.4.3(vi) that T ( E ) = T'(E') if and only if E' w cp(E). Hence we
453
EXERCISE 9.6.4
have to show that E' N cp(E) if and only if there is a vector y with the properties set out in (iv). If there is such a vector y, then [R'Ezo] = [R'y] (since both are the range of E). By Theorem 7.2.12, [RE201 [Ry]; that is, N
cp( E) N E'.
Conversely, suppose that cp(E) E'. Let V' be a partial isometry in 12' with (o(E) and E' as its initial and find projections, respectively, and let y be V'Exo. If 'H is the Hilbert space on which R acts, we have N
E'(31) = V'lp(E)(31)= [V'REZO] = [RV'Eso] = [Ry], E ( H ) = [R'Eso]2 [R'V'Eso]= [R'y] = [R'V'cp(E)zo]2 [R'V'*V'cp(E)ZO] = [R'cp(E)zo]= [R'Ezo]= E ( X ) , and thus E(7t) = [R'y]. w 9.6.4. Let R be a finite factor with a generating vector $0 so that R' is finite by Proposition 9.1.2. Let E be the projection in R with range [R'zo], and denote by T , T ' , 7 1 , the unique tracial states of the factors R,R', ERE, R'E, respectively. Let co be the positive real number T ( E ) . (i) Prove that
COT*(ERE)= + W E ) ,
T'(R')
= Ti(R'E)
for all R in R and R' in R'. (ii) Suppose that F and F' are projections in R and R', respectively. Prove that r ( F ) = c o ~ ' ( F ' )if and only if there is a vector y such that F has range [R'y] and F' has range [Ry]. [Hint.Show that it is sufficient to consider the case in which F 5 E, and in that case apply the result of Exercise 9.6.3 to the projections F (in ERE) and F'E (in R'E).]
Solution. (i) The restriction TIERE is a numerical trace on
ERE,and c l l r J E R Eis a tracial state. Thus 7-1
= c;'71ERE,
COT1
= TIERE.
The mapping cp : R' + R'E is a * isomorphism from R' onto R'E, so 7-i o cp is a tracial state on R',and so coincides with 7'.
454
ALGEBRA AND COMMUTANT
(ii) We show first that, if either r ( F ) = cor'(F') or there exists a vector y with the properties stated in (ii), then F 5 E . For this, note first that the equation T ( F ) = cor'(F') implies that r ( F ) 5 co = r ( E ) ;from this, it follows that F 5 E , because otherwise we would have E 4 F (since R is a factor) and r ( E ) < r ( F ) . Note also that, if there is a vector y with the stated properties, we have [ R y ] E [Rzo]because zo is a generating vector for R, and hence [R'y] 5 [R'zo](that is, F 5 E ) by Theorem 7.2.12. We may assume henceforth that F 5 E. Let V be a partial isometry in R, with initial projection F and final projection a subFo, we have r ( F ) = T(Fo),so the projection Fa of E . Since F is equivalent to r(F0) = QT'(F'). It is easy relation T ( F )= COT'(F') to verify also that the existence of a vector y such that F has range [R'y]and F' has range [Ry]is equivalent to the existence of a vector yo such that FO has range [R'yo] and F' has range [Ryo]. Indeed, it follows from the type of argument used at the end of the solution to Exercise 9.6.3 that, given such a vector y , we can take V y for yo; and conversely, given such a vector yo, it suffices to take V*yo for y. Upon replacing F by Fo, we may assume henceforth that F 5 E . With the assumption that F 5 E , we prove next that a vector y has the properties set out in (ii) if and only if y = Ey and y is a generating vector, both for the projection F in the von Neumann algebra E R E acting on the range of E , and also for the projection F'E in the von Neumann algebra R'E. For this, suppose first that y has the properties stated in (ii). Then y = F y , whence y = E y , and [R'Ey] (= [R'y])is the range of F (E E R E ) ; moreover, F'E (= E F ' ) has range [ E R y ](= [ E R E y ] ) . Conversely, suppose that a, = E y , and the projections F (in E R E ) and F'E (in R'E) have ranges [R'Ey] and [ E R E y ] ,respectively. Let G' (in R') be the projection with range [Ry].Since G'E (= EG') has range [ E R y ] (= [ E R E y ] ) ,it follows that G'E = F'E. Thus G' = F' (since the mapping R' + R'E : R' + R'E is a isomorphism), and F' has range [Ry];moreover F has range [R'y],since E y = y . The finite factor E R E and its (finite) commutant R'E have xo as a separating and generating vector. From Exercise 9.6.3, the projections F (in E R E ) and F'E (in R ' E ) have a common generating vector if and only if q ( F ) = r{(F'E);from (i), this occurs if and . the preceding paragraph, the projeconly if r ( F ) = c ~ T ' ( F ' )From tions F (in R) and F' (in R') have a common generating vector y if and only if r ( F ) = QT'(F'). m[76(Chap.X, pp. 179-183)] N
*
EXERCISE 9.6.5
455
Suppose that R is a finite factor with finite commutant T and T’ the unique tracial states of R and R’, respectively. Prove that there is a positive real number c with the following property: if F ( 6 R) and F ‘ ( 6 R’) are projections, then T ( F ) = c.r’(F‘) if and only if there is a vector y such that F has range [R’y]and F’ has range [Ry].[We call c the coupling (or linking) constant of 72.1 [Hint. Use Corollary 8.2.9, Proposition 5.5.18, and Exercise 9.6.4.1 9.6.5.
R’, and denote by
Solution. The center {uI : a E C} of R is countably decomposable, so R is countably decomposable by Corollary 8.2.9. Since the central projection occurring in Proposition 5.5.18 is either 0 or I , either R or R’has a generating vector. Upon exchanging the roles of R and R’ if necessary, we may assume that R has a generating vector. The stated result now follows from Exercise 9.6.4. m[76]
Let R be a von Neumann algebra acting on a Hilbert 9.6.6. space ‘H, with center C and with a generating vector $0, and suppose that both R and R‘ are finite. Let E be the projection in R with range [R’zo];denote by T , T ’ , 71,T: the center-valued traces on the finite von Neumann algebras R, R‘, E R E , R’E, respectively; and let CO be ~ ( 3Note, ) . form Propositions 5.5.13 and 5.5.5, that CE = I and the mapping cp : R’ + R‘E is a * isomorphism from R’ onto
R’E. (i) Prove that CO has range dense in 3-1. (ii) Prove that CO~l(ERE =)r ( E R E ) E ,
c p ( ~ ’ ( R ’= ) ) ~i(cp(R’))
for all R in R and R‘ in R’. [Hint. To prove the first relation, let Qn be the spectral projection for Co corresponding to the interval [n-’, oo),for n = 1,2,. ... Show that it suffices to prove the equation for R in RQn,and so reduce to the case in which COhas a bounded inverse.] (iii) Suppose that F and F’ are projections in R and R’, respectively. Prove that T ( F )= Cor’(F’) if and only if there is a vector y such that F has range [R‘y] and F’ has range [Ry]. [Hint. Upgrade the argument needed to solve Exercise 9.6.4(ii).] Solution. (i)
Let P be the range projection of Co, so that
456
ALGEBRA AND COMMUTANT
P E C and ( I - P)Co = 0. Since T((1-
P)E)= ( I - P ) T ( E )= ( I - P)Co = 0,
it follows that E = PE. Thus P 2 CE = I , and so P = I . (ii) If 1c, is a * isomorphism from R' onto another von Neumann algebra S, then 1c, o T' o 1c,-' is the (unique) center-valued trace on S, since it inherits from r' the properties of a center-valued trace (as set out in Theorem 8.2.8). In particular, cp o T' o 'p-' = ri; so cp 0 7' = r; 0 $3. For n = 1,2,. .., define Qn as in the hint. Since COis a positive operator with range dense in 3-1, V F ! l Q n = I . Accordingly, in order to prove the first relation in (ii), it suffices to show that
Q n C o q ( E R E )= Q,r(ERE)E
( n = 1,2,. . .)
for each R in R. This last equation is equivalent to
Upon replacing R, 2 0 , E , CO by RQn,Qnzo, QnE, QnCo, respectively, we reduce t o the case in which COhas a bounded inverse. In this case, it is easy to verify that the mapping
ERE
+
C<'r(ERE)E : E R E
+ CE
is a center-valued trace on E R E , and so coincides with T I . Thus COq(ERE)= T ( E R E ) E ,for all R in R. (Compare Exercises 8.7.57, 8.7.58. Careful use of these exercises and the algebra N(C) will produce another proof of this last relation.) (iii) If T ( F ) = Cor'(F'), we have T ( F ) 5 CO= r ( E ) , and thus F 5 E. If there is a vector y with the properties set out in (iii), then
and hence (Theorem 7.2.12) [R'y] 5 [R'zo]; that is (again) F 5 E . We may assume henceforth that F 5 E . By reasoning as at the corresponding stage of the solution to Exercise 9.6.4(ii), we can simplify further to the case in which F 5 E. With the assumption that F 5 E , we can show (exactly as in the solution to Exercise 9.6.4(ii)) that there is a vector y with the
EXERCISE 9.6.7
457
properties set out in (iii) if and only if the projections F (in E R E ) and F’E (in R ’ E ) have a common generating vector. This occurs if and only if q ( F ) = r;(F’Z),by Exercise 9.6.3(iv), since E R E has a separating and generating vector $0. Since the range of CO is dense in N , and Co E C, the condition r1(F) = rl(F’E) is equivalent to rl(F)Co = ri(F’E)Co, and hence to
From (ii) COT@)
= .r(F)E = cp(r(F)),
Cor;(F’E) = COr;(cp(F‘))= Cocp(r’(F’)) = CoT’(F’)E = cp(Cor’(F‘)). From this, and since cp is one-to-one, (1)is equivalent to the equation
9.6.7. Let R be a von Neumann algebra acting on a Hilbert space 7-1, with countably decomposable center C, and suppose that both R and R’are finite. Denote by r , r‘ the center-valued traces on R, R’, respectively. (i) Deduce from Corollary 8.2.9 and Proposition 5.5.18 that there is a projection Q in C such that the von Neumann algebras RQ and R’(I - Q) have generating vectors q and 5 2 , respectively. (ii) Let E (in RQ) and E’ (in R’(1-Q))be the projections with ranges [R‘q] and [RQ], respectively. Define C1 in CQ and C2 in C(I - Q) by C1 = r ( E ) , C2 = r’(E). Suppose that F and F’ are projections in R and R‘, respectively. Deduce from Exercise 9.6.6 that F and F‘ have a common generating vector if and only if
r ( F ) Q = Clr’(F‘)Q,
C ~ T ( F ) (-I Q ) = T‘(F‘)(I- Q ) .
(iii) Prove that there is a unique invertible element CO, in the algebraN(C) of operators affiliated to C , with the following property: if F and F‘ are projections in R and R’, respectively, then r ( F ) = Co?(F‘) if and only if F and F’ have a common generating vector. (We call CO the coupling (or linking) operator of R) (iv) Show that a projection F in R is cyclic if and only if r ( F ) 5
CO -
458
ALGEBRA AND COMMUTANT
Solution. (i) By Corollary 8.2.9, R is countably decomposable, so the stated result follows from Proposition 5.5.18. then Q F has range [R’Qy] and ( I - Q ) F (ii) If F has range [R‘y], has range [R’(I - Q)y]. Conversely, if Q F has range [R’yl] and ( I - Q ) F has range [R’yz], then F has range [R‘(yl y z ) ] (see Proposition 5.5.10 and its proof). Similar comments apply to the projections F‘, QF’, ( I - Q)F’ in R’.it follows that the projections F and F’ have a common generating vector if and only if each of the pairs ( Q F , QF’) and ( ( I - Q ) F , ( I - Q ) F ’ ) has a common generating vector. From the result of Exercise 9.6.6(iii), applied to R Q , Q F and QF’ have a common generating vector if and only if T ( F ) Q = CIT1(F’)Q. By the same exercise, applied to R‘(I - Q ) , ( I - Q ) F and ( I - Q)F’ have a common generating vector if and only if the equality C27(F)(I- Q ) = T’(F‘)(I- Q ) holds. Thus F and F’ have a common generating vector if and only if
+
(I)
T ( F ) Q = ClT‘(F‘)Q, C27(F)(I - Q ) = T’(F’)(I - Q ) .
(iii) By applying the result of Exercise 9.6.6(i) to both R Q and R’(I - Q ) , it follows that the bounded (positive) operators C1 and C2 have range projections Q and I - Q , respectively. Thus C1 lQ(3-1) and Cz [(I- Q ) ( W )are one-to-one positive linear operators and have positive (densely defined, unbounded) inverses 21 and 2 2 . The operator C1 & ( I - Q ) is a positive element Co of N(C),and has an inverse 21Q CZ in N(C). The relations (1) are equivalent to the single equation T ( F ) = C07‘(F’). Since the operator T’(F’) is bounded, the composite mapping COT’(F’) is a closed linear operator, and so coincides with the product CO:T‘(F’) formed in N(C). It now follows from (ii) that F and F‘ have a common generating vector if and only if T( F ) = CO T‘(F’). Suppose that Ci is another element of N(C),with the property just attributed to Co. Since the projections E (in R) and Q (in R’) have a common generating vector 2 1 , and the projections I - Q (in R ) and E‘ (in R‘)have a common generating vector 22, it follows that CAQ = CA:T’(Q)= T ( E )= C1, CACz = CA :T‘(E’) = ~ ( 1 Q- ) = I - Q.
+
+
+
Each x in ’D(C0)has the form Q y C ~ Zfor, some y and z in ‘FI, and Cox = C 1 y + ( I - & ) z z . From the relations CiQ = C1, CiC, = 1-4,
459
EXERCISE 9.6.8
it follows that x E D(C4) and c;x = CAQy
+ c;c2. = cly + (I - Q ) z = cox.
A extends CO,and so CA = CO by Theorem 5.6.15(vii). Thus C (iv) If F is cyclic, then for some cyclic projection F' in R', T(F)
= co :T'(F') 5
co.
Conversely, suppose that T ( F ) CO.Then
Thus Q F 5 E , and Q F is cyclic since E is cyclic. At the same time, ( I - Q ) F is cyclic since ( I - Q ) F 5 I - Q and I - Q is a cyclic projection in 72. Since Q F and ( I - Q ) F are both cyclic, it follows m[37,38] (as noted in the proof of (ii)) that F is cyclic.
z
the same Let 'H be a Hilbert space, and denote by 9.6.8. set 3.1 with the conjugate Hilbert space structure (see the discussion preceding Definition 2.6.3). Suppose that A E B(3.1), and recall that the same mapping A can be viewed as an element A of B(R). (i) Show that the mapping A 4 A is a multiplicative adjointpreserving conjugate-linear isometry from B(3-t) onto B(7T). : X E sp(A)}. (ii) Show that sp(A) = Suppose that A is self-adjoint. Prove that A is self-adjoint, (iii) and sp(A) = sp(A). Show also that, for each f in C(sp(A)), f(A) is the same mapping as f ( A ) , where f is defined by f(t) = (iv) Give an example of an element f of C(sp(A)) such that f ( A ) is not the same mapping as f(A).
{x
m.
Solution. We use the notation introduced in the discussion preceding Definition 2.6.3. (i) Since A is the same mapping as A and addition in 'H is the same as addition in it follows that the mapping A 4 A preserves sums and products of operators. When c E C and x E 'H (the same the product E A (defined in B ( H ) ) satisfies set as
z,
z),
(EA)x = E;(Ax) = ET(Az) = c(A5) = (cA)x.
460
ALGEBRA AND COMMUTANT
Thus C A (in B ( H ) ) is the same mapping as cA (in B(7-1)),and ?A. Also, when x, y E 'H, we have
=
and (A)*= ( F ) .Since the norm in 7-1 coincides with the norm in 7-1, it follows that IlAll = IIAll. Finally, we observe that if B E B(H), then the same mapping B can be viewed as a bounded linear operator A acting on 7-1, and then A E B(7-1), A = B . A are the same (ii) When A E B(7-1)and X E C, X I - A and mapping, by (i); so X E sp(A) if and only if E sp(A). (iii) If A is self-adjoint, sp(A) G R, and from (i) and (ii) it follows that A is self-adjoint and sp(A) = sp(A). Suppose that p is a polynomial with complex coefficients,
xr-
p(X) = a0
+ a1X + a2X2 + + * * *
When X E sp(A) ( S R),
Thus
p(A) = a o I + a l A + a 2 A 2 + . . . + ~ , A n ,
p ( A ) = sol+ & A
+ a2A2 + . . . + &&An,
and p(A)= p(A) by (i). Thus p ( A ) is the same mapping as p(A). Given f in C(sp(A)), we can choose a sequence { p n } of polynomials such that Ilf'-pnll + 0, and we have Ilf-pnll --+ 0. From the preceding paragraph ,
f ( A ) x = lim pn(A)x = lim p,(A)x = f(A)x 71-03
n+co
for each x in 7-1. Thus f(A) is the same mapping as f(A). We note an alternative proof that f ( A ) and f(A) are the same mapping, when f E C(sp(A)). The mappings
f
--t
f
: C(sp(A)) + C(sp(A)),
T
-+
T
:
B(7-1) + B ( R )
are both conjugate-linear isometries that preserve adjoints and products, while the function calculus f + f(A) : C(sp(A)) B(7-1) is --f
461
EXERCISE 9.6.9
an isometric * isomorphism. From this, and since sp(A) = sp(A), it follows that the mapping
f + If0 : C(sp(A)) -+ a(%) is an isometric * isomorphism that sends the constant function 1onto and the identity mapping L on sp(A) to A. From the uniqueness of the (continuous) function calculus for A, we have
f ( 4= f(4. Upon replacing f by f, we deduce that f ( A )= fo; so f ( A )is the
same mapping as f ( A ) . (iv) I f f is the constant function with value i on sp(A), then has constant value -i. Hence
f
f(A)a: = is, f(8)a:= f(A)a: = -ia: (= i - 3 ) for each a: in H (= V),and f ( A ) is not the same mapping as
ft4.
8
9.6.9. With the notation of Exercise 9.6.8, let A (C f?(H)) be an abelian von Neumann algebra. (i) Show tht the mapping A -+ A : B ( H ) S ( R ) is a homeomorphism with respect both to the weak-operator topologies and to the strong-operator topologies. Deduce that the set { A : A E A } is an abelian von Neumann algebra A acting on %. (ii) Suppose that T is a closed densely defined linear operator on 'H. Show that the same mapping T can be viewed as a closed densely defined linear operator T on R. Prove that T is self-adjoint, or positive, if and only if T has the same property, and that T ' q A if and only if T qd. (Note that A does not indicate the closure of A in the notation of this exercise.) - (iii) Show that the restriction to N(d) of the mapping T -+ T is a one-to-one mapping from N(d) onto N(d);is conjugatelinear, multiplicative, and adjoint-preserving relative to the * algebra structure on N ( A ) (see Theorem 5.6.15); and is an isomorphism for the order structure on N ( A ) (see the discussion preceding Lemma 5.6 -22). (iv) Suppose that T q A and T is self-adjoint. Show that sp(T) = sp(T) and that f ( T ) is the same mapping as f(T) when f is a Bore1 function on @. [Hint.Apply Theorem 5.6.27 to the mapping --f
?+)!
:
f
-
f(T)
:
B"
-+
N(d).]
462
ALGEBRA AND COMMUTANT
Solution. (i)
The weak-operator topology on B(31) is the lo-
cally convex topology determined by the semi-normsP , , ~(2,'y € %), where p,,,(A) = I(Az,y)l. The strong-operator toplogy is determined by the semi-nome p, (z € R),where p,(A) = llAxL. Similar comments apply to the corresponding topologies on B(H). From this, and since
it follows that the mapping A 4 A : 8 ( H ) + B ( q ) is a homeomorphism, for each of these topologies. It follows that { A : A E A } is a closed subset A of S(R), and from Exercise 9.6.8(i), A is a self-adjoint abelian subalgebra of S ( R ) and contains 1. (ii) Suppose that T is a claed linear operator with domain 'D a dense linear subspace of 'H and with range in ?i. Then P can be and the mapping T can viewed as a dense lineax subspace of be viewed as aa operator T with domain and range in R. Since addition in R coincides with addition in ?i, T is additive. When c E C and z E 5 (the same set aa D),
z,
Hence T : + is linear. Since the product norm topology on H x coincides with the product norm topology on 'H x 8,and
the graph of T is the same set as the graph of T, it follows that T is closed. B y the argument set out in the introductory discussion to Subsection 9.2, A first approach to modular theov, the adjoint of T is the same mapping as the adjoint of T;so (F ) = (T)*. In particular, therefore, T is seAf-adjoint if and only if ?4 is self-adjoint. Since (T2,z)-= (T2,z)-= ( 2 , T S ) for each s in 2),T is positive if and only if T is positive. It is apparent from Exercise 9.6.8(i) that the mapping A + 2 : B(%) S ( R ) carries A' onto (A)', and that d is unitary if and only if A is unitary. If Tqd,then T = UTU" for each unitary operator U in A'. Since T and are the same mappings as T and U,respectively, i t now follows that T = VTV-' for each unitary
463
EXERCISE 9.6.10
operator Tin (A)',and so Tqd. The last argument can be reversed, t o show that T 7 A if and only Tr,d. (iii) It follows from (ii) that the restriction to N(d)of the mapping T -, T is one-to-one and adjoint-preserving, carries N ( A )onto N(d),and carries the positive cone of N ( A )onto the positive cone of N(A).Suppose that S, T E N(d),and let R be the element SPT of N(d).When z E D ( S ) n D(T) (= D(3)n D(T)),we have
-
( S + T ) Z= Z?z+Tz = SZ + T z = RZ = Rz.
+
z,
Hence 55' T and therefore 55'PT C 3. (Recall that 3+-?5 is the closure of the operator S -lT, and the superscript does not -,,denote closure in the present context.) Since S+T, -..-R E N(d) - and y+T C_ 'EE, it follows from Theorem 5.6.15(vii) that S+T = R. Hence the mapping T + T : N(A) -+ N(d)is additive, and a similar argument shows that it is multiplicative, relative to the algebraic operations and 7 in N ( A ) and N(A). The argument used in the solution of Exercise 9.6.8(i) shows that cT = fl when T E N(d) andcEC. (iv) With T self-adjoint in N(A),T is self-adjoint in N(d),and both T and T have real spectrum. Moreover X (E W) lies outside sp(T) if and only if XI - T has an inverse R in the algebra N(d), and R E A. (See the discussion preceding Proposition 5.6.20.) Since the mapping A A carries N(d)o n t o N ( d ) , A onto d, and XI-T onto X I - Ti, A lies outside sp(T) if and only if X I - ??has an inverse R in the algebra N(d),and fi: E d (that is, if and only if X 4 sp(T)). Thus sp(T) = sp(T). The mapping f f ( T ) : B, + N(d) is a a-normal homomorphism that maps the constant function 1 to I, and the identity mapping L on C to T. From this, together with the result of (iii), homomorphism the mapping II, (described in the hint) is a a-normal from f3, into N(A)that carries 1 to I and I to (T') (= T). F'rom Theorem 5.6.27, +(f) = f ( T ) for all f in S,. Thus --$
--$
and f ( T ) is the same mapping as
f(7).
9.6.10. Use the equation A-l = J A J and the result of Exercise 9.6.9 to prove that f(A-l) = Jf(A)J for every Bore1 function f on C! (see the discussion preceding Proposition 9.2.3).
464
ALGEBRA AND COMMUTANT
Solution. We can view the mapping J as a unitary transformation from H onto From this standpoint, the equation A-l = J A J can be rewritten in the form of a unitary equivalence
z.
-
between linear operators A (on H ) and A-1 (on equivalence preserves function calculus, we have
z).Since unitary
f ( F=)~ f ( h ) J * for each Bore1 function f on C. It now follows from Exercise 9.6.9(iv) that f ( A - l ) = f ( F=)J f ( A ) J .
Suppose that R is a von Neumann algebra acting on H , with a generating trace vector 20, and refer to Lemma 7.2.14, Theorem 7.2.15, and Exercise 7.6.49 for information including the following: there is an involution J of 'FI and a * antiisomorphism cp from R onto R' such that 9.6.11.
a Hilbert space
JAXO= A*xo,
(P(A)xO= A X O ,
p ( A ) = JA*J
for each A in R. Suppose that HV is the polar decomposition of an element T of R that has a bounded inverse, and u = T X O . (i) Prove that u is a separating and generating vector for R. (ii) Let A,, J , , and { o t } be the modular operator, involution, and modular automorphism group associated with the separating and generating vector u of R. Prove that
and deduce that
J , = JV*cp(V*)= cp(V)VJ,
A , = cp(V*H-2V)H2,
EXERCISE 9.6.11
465
Solution. (i) Since T has a bounded inverse (necessarily in R), RT = R and [Ru] = [RTzo]= [Rzo]= 'H.
If A E R and Au = 0, then AT E R and ATxo = 0; since zo is separating for R (Lemma 7.2.14), it follows that AT = 0, and A = (AT)T-' = 0 . Thus u is a separating and generating vectcor for R. (ii) The mapping SO : AU + A*u : R u + Ru is a preclosed densely defined conjugate-linear operator on 'H, and its closure S, has polar decomposition J,A:l2. When A E R, we have Ax0 = AT-lu, and thus SoAzo = (AT-')*u = T*-lA*u = T*-'A*Tzo = T*-lJT*A$o. From this, and since Rzo coincides with the domain Ru of SO,we have So T * - l J T * , and thus S, T * - l J T * . It follows that S, is bounded, as well as being closed and densely defined; so S, has domain 'H, and s, = T * - ~ J T * .
c
Now
c
T * - JT* ~ =H-~VJV*H = JJH-'VJV*H = J~(V*H-')V*H = JV*(~(V*)(~(V*H-'V)H
(for the last step, note that V * commutes with (p(V*H-l) since V * E R and (p(V*H-') E R', and
cp(V*)(p(V*H-'V)= 'p(V*H-') since V is unitary and cp is an anti-isomorphism). We have that the mapping JV*(p(V*)is a conjugate-linear isometry from 'H onto 'H, and (p(V*H-'V)H is a positive bounded operator with bounded inverse since it is the product of two such commuting operators. If we now consider S, as a densely defined closed operator from 'H into 'H, it has two polar decompositions,
J,A;'~,
(~v*(p(v*))((p(v*~-lv)~).
466
ALGEBRA AND COMMUTANT
From the uniqueness of the polar decomposition,
Thus Ju= J V *J V J = (p(V)VJ, and
Let K be (p(V*H-'V), so that H and K are bounded commuting positive operators with bounded inverses, and A, = K 2 H 2 . Let 31 be the commutative C*-algebra generated by H and K , so that there is a * isomorphism $ from 2l onto C(X) for some compact Hausdorff space X. Then $(H) = h, $(K)= k, and $(Au) = k2h2, where h and k are continuous functions taking strictly positive values throughout X. For each continuous complex-valued function f on the interval (0, oo), it follows from Proposition 4.4.7 and Example 4.4.9 that $(f (H)) = f(+(H)) = foh, +(f(K)) = f ok, +(f(Au)) = fo(h2k2). In particular, when t E W and f(s) = eit, we obtain
,)(Hit) = h",
$(Kit) = kit,
Thus A: = H2it112". Since H E 'R and Ir' E 'R', we have at(A)
-~
t ~ ~ , -' ~i 2t i t ~ 2 i t ~ ~ - 2 i t ~ - 2 i t
- H 2 i t ~ ~ - 2 i t (A E R). 9.6.12. Suppose that u is a separating and generating vector for a von Neumann algebra R acting on a Hilbert space 1-I, A' E R', and T > 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
ALGEBRA AND COMMUTANT
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
ALGEBRA AND COMMUTANT
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
ALGEBRA AND COMMUTANT
(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
ALGEBRA AND COMMUTANT
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
ALGEBRA AND COMMUTANT
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
ALGEBRA AND COMMUTANT
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
ALGEBRA AND COMMUTANT
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
ALGEBRA AND COMMUTANT
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
ALGEBRA AND COMMUTANT
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
ALGEBRA AND COMMUTANT
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
ALGEBRA AND COMMUTANT
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
ALGEBRA AND COMMUTANT
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
ALGEBRA AND COMMUTANT
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
ALGEBRA AND COMMUTANT
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
ALGEBRA AND COMMUTANT
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
ALGEBRA AND COMMUTANT
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
ALGEBRA AND COMMUTANT
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
ALGEBRA AND COMMUTANT
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
ALGEBRA AND COMMUTANT
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
ALGEBRA AND COMMUTANT
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
ALGEBRA AND COMMUTANT
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
ALGEBRA AND COMMUTANT
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
ALGEBRA AND COMMUTANT
(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
ALGEBRA AND COMMUTANT
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
ALGEBRA AND COMMUTANT
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
ALGEBRA AND COMMUTANT
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
ALGEBRA AND COMMUTANT
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
ALGEBRA AND COMMUTANT
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
ALGEBRA AND COMMUTANT
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
ALGEBRA AND COMMUTANT
(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
ALGEBRA AND COMMUTANT
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
ALGEBRA AND COMMUTANT
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
ALGEBRA AND COMMUTANT
[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
ALGEBRA AND COMMUTANT
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
ALGEBRA AND COMMUTANT
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
SPECIAL REPRESENTATIONS OF C*-ALGEBRAS
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
SPECIAL REPRESENTATIONS OF C*-ALGEBRAS
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
{
$Q(HAj - AjH) : H E C p } . j=l
556
SPECIAL REPRESENTATIONS OF C*-ALGEBRAS
Hence show that there exists
H in C, such that
llHAj - AjHll 5
-1
( j = 1, ...,n).
(iii) Let r be the set of all triples (IF,p, H ) in which IF is a finite subset of Q, p E A, H E C,, and [ [ H A- A H [ [ 5 n-l for all A in F, where n is the number of elements in IF. Show that r is directed by the binary relation 5 , in which (Fl,pl,Hl)5 ( F 2 , p 2 , H 2 ) if and only if F1 C IF2,pl p2, and H1 5 H2. K as follows: H , is H if y (iv) Define a mapping y + H , : I' is (IF, p, H ) . Show that { H , } is an increasing two-sided approximate identity for K with the additional property that
<
--$
for each A in U. (An approximate identity with this additional property is described as quasi-central.) Solution. (i) Since {Vx} is an increasing net in K and 0 5 Vx 5 I , { @ ( V , ) }is an increasing net in @ ( K )(& l?('&,)) and 0 5 @(Vx)5 I . It follows that {@(Vx)}is strong-operator convergent to an element F of @ ( K ) - . For each K in K ,{VxK} is norm convergent to I(; so { @ ( V x K ) } (that is, { @ ( V x ) @ ( K ) }is) norm convergent to @(K)and strongoperator convergent to F @ ( K ) .Thus @(K)= F @ ( K ) ,and a similar = @ ( K ) F . Since F S = S F = S for argument shows that @(A') each S in @ ( K ) ,the same is true for all S in @(K)-,since operator multiplication is (seperately) ultraweakly continuous. Hence F is the unit element E of @ ( K ) - (= @(U)-E). (ii) From (i), and since E @ ( A j )- @ ( A j ) E= 0, the net
C:=,
is strong-operator convergent to of operators acting on 0. Thus 0 lies in the strong-operator closure of the subset n
{ j=l @@(VxAj- AjVx) : X E A,
X 2p}
EXERCISE 10.5.6
557
of C;=,@@(a), and so lies in the strong-operator closure of the (larger, convex) subset n
{ C @ @ ( H A j- A j H ) : H E C,}. j= 1
Hence 0 lies in the weak-operator closure of this last set. Since the weak-operator topology on C;=, $@(a) coincides with the weak topology in the Banach space sense (Corollary 10.1.2), it now follows from Theorem 1.3.4 that 0 lies in the norm closure of the set in (1). Accordingly, there is an element N of C, such that
Since Qi is an isometry,
IlHAj - A j H ] l I
.-*
( j = 1,.. . ,n).
(iii) Given any finite subset IF of !2l and any p in A, it follows from (ii) that there is an element H of C, such that (IF, p , H ) E I?. The relation _< on r is reflexive and transitive. Suppose that (IF1, p1, H ) , (IF2, p2, H 2 ) E r; we shall show that there is an element (IF, p , H ) of r such that (IFj,CLjJj)
I (F,P,H)
(j=
thus completing the proof that I’ is directed by I.To this end, let IF be IF1 U IF2. For each j = 1,2, we can choose an expression for H j as a finite convex combination of elements V,, with X 2 p j . The suffices A, that occur in the chosen expression for H I , or H2, or both, form a finite subset A0 of A, and there is an element p of A such that p 2 X for each X in A0 (and thus, p 2 p 1 , p 2 pa). Since
VXIVKIV”
(XEho, P I V E A ) ,
we have
I V, 5 H
( H E C;,
j = 1,2). By (ii), we can choose H in C, in such a way that (F,p, H ) E I?, and we then have Hj
558
SPECIAL REPRESENTATIONS OF C*-ALGEBRAS
(iv) Given Ii' in Ic, A in U, and E (> 0), we can choose po in A such that IIK - VAKII < E , IIK - Ii'Vxll < & whenever X E A and X 2 PO. Since the open ball in U with center 0 and radius E is convex, it follows that
Let IF0 be a finite set consisting of no elements of U, one of which is A , where no > 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<,
KH,
+
K, H,A - AH,
+ 0,
in the norm topology. It is apparent from the definition of the relation 5 on I' that { H , } is an increasing net. Since Vx E Ic, 0 5 Vx 5 I , and H , is a finite convex combination of elements Vx, it follows that H, E Ic and 0 5 H , 5 I. Thus { H , } is an increasing two-sided approximate identity for Ic (and is quasi-central). ~[3,12] Suppose that 3-1 is a Hilbert space, A1 ,...,A , E 10.5.7. B('H)+, and A1 * * A , 5 I. (i) Show that 11 C;=, Ajzj1I2 5 CY=i(Ajzj,zj) whenever zl,. ..,zn E 3-1. [Hint.Define an inner product on 1.1 @ 1.1 @ -.@ 3-1 ( n copies) by
+ +
-
559
EXERCISE 10.5.7
and apply the Schwarz inequality to the vectors { z , z , . n { z 1 , 2 2 , . . . ? zn}?where z = Cj=l Ajzj .] (ii) Deduce that
for all complex numbers
a1
.. , z }
and
,. . .,a,.
Solution. (i) By applying the Schwarz inequality as indicated in the hint, we obtain
where the norms on the right hand side are those associated with the inner product in the hint. The inequality can be rewritten in the form
Hence
and this is the required result.
560
SPECIAL REPRESENTATIONS OF C*-ALGEBRAS
(ii) We have to prove that
equivalently
112
ajAjYllz
j=1
I
2
laj12 (AjY, Y > ,
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
SPECIAL REPRESENTATIONS OF C*-ALGEBRAS
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
SPECIAL REPRESENTATIONS OF C*-ALGEBRAS
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
SPECIAL REPRESENTATIONS OF C*-ALGEBRAS
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
SPECIAL REPRESENTATIONS OF C*-ALGEBRAS
(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 <
EXERCISE 10.5.16
571
From the maximality of the family {Gb}, we deduce that p(F1) 5 w,(Fl) whenever Fl is a projection in R and F1 5 F . Since p and
w, are bounded, and each element of ( F R F ) + can be approximated in norm by positive linear combinations of projections F1 of the type just considered, it now follows that plFRF 5 w,JFRF = ~ F , J F R F . From Proposition 7.3.5, p J F R F = w,(FRF for some vector y in the range of F . Since
the vector state wylRis singular or zero by Corollary 10.1.16, and is hence zero by Theorem 10.1.15(iii). Thus
p ( F ) = W y ( F ) = 0.
.[2,110]
10.5.16. Suppose that R is a von Neumann algebra, p is a faithful state of R, and pu t ps is the decomposition of p into its ultraweakly continuous and singular parts (see Theorem 10.1.15(iii)). Show that pu is a faithful normal positive linear functional on R. [Hint.Use Exercise 10.5.15.1
Solution. It follows from Theorem lO.l.l5(iii) that pu is a normal positive linear functional on R, and it remains to show that pu is faithful. If 0 # A E R+,we can choose a projection E in R and a positive scalar c such that A 2 c E > 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
SPECIAL REPRESENTATIONS OF C*-ALGEBRAS
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
SPECIAL REPRESENTATIONS OF C*-ALGEBRAS
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
SPECIAL REPRESENTATIONS OF C*-ALGEBRAS
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
SPECIAL REPRESENTATIONS OF C*-ALGEBRAS
(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
SPECIAL REPRESENTATIONS OF C*-ALGEBRAS
(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
SPECIAL REPRESENTATIONS OF C*-ALGEBRAS
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
SPECIAL REPRESENTATIONS OF C*-ALGEBRAS
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
SPECIAL REPRESENTATIONS OF C*-ALGEBRAS
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
SPECIAL REPRESENTATIONS OF C*-ALGEBRAS
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
SPECIAL REPRESENTATIONS OF C*-ALGEBRAS
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
SPECIAL REPRESENTATIONS OF C*-ALGEBRAS
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(%),
<
( q ( ~=)@ 6 r1 o@o
o @ - l ( ~= ) A,
and ( o fj is ultraweakly continuous on @(%)-. Hence t o fj is the identity mapping on @(a)-.Similarly, fj o c i s the identity mapping on 9(8)-. It follows that ij is an ultraweak homeomorphism and a linear isomorphism of a(%)- onto P(S)-. ~ (ii) If q is a Jordan * isomorphism of ?2ionto 8,then P 0 9 0 @ -is a Jordan isomorphsim of @(a) onto g(13). F'rom Exercise 10.5.28, fj is a Jordan * homomorphism as is (. F'rom (i), fj is a Jordan * isomorphism. m
*
EXERCISE 10.5.30
595
Use the pattern of the construction of the Cantor 10.5.30. set to construct a closed nowhere-dense subset S' of [0,1] having Lebesgue measure f (remove the centered open interval of length f from [0,1] and so on). Let S be [0,1] \ S', 'H be L2(0, l), and K be L2(S) (relative to Lebesgue measure), and let 2l and B be the (abelian) C*-algebras consisting of all multiplication operators corresponding to continuous functions (on [0,1]) on 'H and K , respectively. Let cp be the mapping on U arising from restricting a continuous function on [0,1] to S. Show that (i) cp is a * isomorphism of U onto B; (ii) cp extends to an ultraweakly continuous * homomorphism (p of 2l- onto B - ; (iii) (p is not an isomorphism (compare this conclusion with those of Exercise 10.5.29); (iv) the * isomorphism cp-' from B onto 2l is not ultraweakly continuous (see the discussion following Corollary 10.1.11).
Solution. (i) If f and g are continuous functions on [0,1] and fo and go are their restrictions to S, then
Moreover, if 0 = cp(Mj) (= M j , ) , then fo is 0 almost everywhere on the open dense subset S . Since fo is continuous on S, fo is 0 on S ; the continuous function f is 0 on a dense subset of [0,1]. Thus f is 0 and M f is 0. It follows that cp is a * isomorphism of 2l onto B. (ii) From Example 5.1.6, 2l- and a- are the algebras of multiplications by all bounded measurable functions on [0,1] and S , respectively, and each is maximal abelian. Again, the restriction mapping applied to the algebra of bounded measurable functions on [0,1] is a * homomorphism (p of 2l- onto f?-, and (i? extends 9.Let zo and yo be vectors in K and x and y be the vectors in 'H corresponding to the ( L z - ) functions defined as 0 on [0,1] \ S and xo and yo on S , respectively. With A in 2l-, ( A x , y ) = (P(A)xo,yo). Thus wzo,yo o (i? = ~ ~ , ~ ( 2 Itl -follows . that (p is continuous as a mapping from 2l- with its weak-operator topology to B- with its
596
SPECIAL REPRESENTATIONS OF C*-ALGEBRAS
weak-operator topology. Since %- and f?- are abelian von Neumann algebras acting on separable Hilbert spaces, the ultraweak and weakoperator topologies coincide on 2l- and on B- (see Exercise 7.6.4). Thus p is an ultraweakly continuous * homomorphism of the algebra %- onto the algebra B-. (iii) The characteristic function f of [0,1] \ S is not 0 almost everywhere since [O,11 \ S has measure Thus M I is a non-zero element of 2l-. But the restriction o f f to S is 0; hence ( P ( M f )= 0. Thus (p is not an isomorphism. (iv) Suppose that cp-l : 8 -+ % is ultraweakly continuous. From Lemma 10.1.10, cp-' extends to an ultraweakly continuous * homomorphism $ from B- onto %-. Since $09 : %- -+ 2l- is ultraweakly continuous, and
i.
it follows that ($ o p ) ( A ) = A for all A in %-. This implies that (P is one-to-one, contradicting (iii). Hence 9-l is not ultraweakly continuous. 10.5.31.
Let 2l and
B be C*-algebras, and let 7 be a Jordan
* isomorphism of U onto B .
Show that (i) 7 is a linear order isomorphism of % onto 7 - l are positive linear mappings); (ii) q is bounded; (iii) 7 is an isometry.
B (that is, q and
Solution. (i) Since 7-l is a Jordan * isomorphism of B onto %, it will suffice to show that 7 is a positive linear mapping. If A E %+, then q ( A ) = q ( A 1 / 2 ) 22 0; hence 7 is a positive linear mapping. (ii) From (i) and Exercise 10.5.10, 7 is bounded. (iii) Let 9 and il! be the respective universal representations of U and B. Since 7 is bounded, from Exercise 10.5.29(ii), Q o 7 0 has a unique extension i-7 that is a Jordan isomorphism of @(a)onto !P(B)- and is an ultraweak homeomorphism. From Exercise 10.5.26, there are central projections PI, P2 in @(a)and Ql,Q2 in !P(B)such that PI t P2 = 1, Q 1 Q 2 = 1, rl(P1) = Q I , rl(p2) = 9 2 , ijI@(%)-Plis a * isomorphism from @(%)-PI onto ! P ( B ) - Q l , and ijl9(%)-9is a * anti-isomorphism from @(2l)-P2 onto ! P ( B ) - Q 2 . From Theorem 4.1.8 (the proof of that result applies without change
*
+
597
EXERCISE 10.5.32
when cp is a * anti-isomorphism rather than a * isomorphism), the restrictions of f j to @(U)-Pj,j E { l , 2 } are isometries. With A in
a(%)-
9
m[53,108]
Thus f j is an isometry.
10.5.32. (i) Show that a linear order isomorphism of one C*algebra onto another that maps I onto I is a Jordan * isomorphism. [Hint. Use Exercise 10.5.9.1 (ii) Find an example of a positive linear mapping of one C*algebra onto another that is a linear isomorphism but whose inverse is not a positive linear mapping. [Hint.Consider C2 for each of the C *-algebras .] Solution. (i) Let q be a linear order isomorphsim of one C*algebra U onto another C*-algebra B such that q(1) = I . From Exercise 10.5.9, q( A)2 5 q( A2) for each self-adjoint A in U. Since the same considerations apply to 7-l and q(A) is a self-adjoint element in B, A2 = [rl-1(77(A))12 5 v-1(7(A)2)*
Thus q(A2) 5 ~ ( q - l ( q ( A ) ~ = ) ) v ( A ) ~ . Hence q(A2) = v ( A ) ~ )for each self-adjoint A in U. With B and C self-adjoint elements of U,
+
+
Hence q(BC C B ) = q(B)q(C) q(C)q(B)when B and C are selfadjoint elements of %. This same equality follows now for arbitrary B and C in U (by expressing B and C as linear combinations of self-adjoint elements of U). Hence q is a Jordan * isomorphism. (ii) The matrix $ transforms the C*-algebra C2 linearly
[
$1
7 T
+
+
onto itself ( ( u , 6 ) is mapped to (f(u 6),+(3a 6 ) ) ) . Since the determinant of this matrix is not 0, the transformation is a linear
598
SPECIAL REPRESENTATIONS OF C*-ALGEBRAS
isomorphism. The positive elements of C2 are those ( a , b ) with a and b real non-negative numbers. The isomorphsim is therefore a positive linear mapping. However, (+l,-l), a non-positive element of C2, is mapped to (0, ti),a positive element in C2. Thus the inverse of the transformation is not a positive linear mapping. m[54,108] 10.5.33. If q is a linear order isomorphsim of one C*-algebra U into another C*-algebra B and q(1) = I , must q be a Jordan * homomorphism? Proof? Counterexample?
Solution. Let U be any non-commutative C*-algebra (for instance, the algebra of complex 2 x 2 matrices) and q be the function representation of U in C(S(U)), where S(U) is the state space of U (see pp.263, 264). Then q is a linear order isomorphism of U into C(S(U)) and q(1) -- 1. We show that q is not a Jordan * homomorphism. Suppose the contrary. Then q ( A 2 )= v ( A ) ~ for each A in U, and d A 2 )= 7(A2)(P)= v ( 4 2 ( P ) = [9?(A)(P)l2= P W 2 for each state p of U. In particular, each state p of U is definite on each self-adjoint A in U. From Exercise 4.6.16, p(AB - B A ) = 0 for each such A and all B in U. Since S(U) separates 24,each self-adjoint A in U lies in the center of U, and U is commutative-contrary to hypothesis. Thus q is not a Jordan homomorphism.
*
Let 3-1 and K be Hilbert spaces, U and B be self10.5.34. adjoint subalgebras of B(3-1) and B ( K ) , respectively, and S,(U), S,(B) be the (convex) families of weak-operator continuous states on U and B,respectively. into Sw(f?) (i) Show that an affine mapping 'p' of S,(U) ('p'(awl+ (1- a)w2) = acp'(w1) (1- a ) c p ' ( q )for each a in [0,1] and w1, w2 in S, (U)) has a unique linear extension cp mapping Us, the linear space of all weak-operator continuous linear functionals on U, into 8:. (ii) With cp as contructed in (i), show that there is a weakoperator continuous, positive linear mapping a of B- into U- such that w ' ( a ( B ) ) = cp(w)'(B)for each B in B- and each w in Ut, where w' is the (unique) weak-operator continuous extension of w from U to 8 - ; and such that a ( I )= I. (iii) When 'p' is an affine isomorphism of Sw(U)onto Sw(B),show that Q (in (ii)) is a Jordan * isomorphism off?- onto 2l-.
+
599
EXERCISE 10.5.34
Solution. (i) By polarization,
for each pair of vectors z and y in 7-1. Each element of !2lz is a linear combination of vector functionals ~,,~l!2l. Thus is the linear CkWk (= p ) in this linear space span of Sw(!21). For c1w1 ( c j E @, Wj E sw(!21)), define cp(p) to be Cicpl(wi)-k . * * 4-ck(p'(wk). To establish that cp is well defined and linear, it suffices to show that cp(p) = 0 if p = 0. Assume p = 0. Expressing each c j in terms of its real and imaginary parts (and evaluating at self-adjoint elements in each of !2l and l?), we may assume that c j is real. In this case 0 = p = alwlI -t . a,wL - (blwy . . . b,w$), where all aj and b j are positive. We must show that
+
+
+ +
+
Evaluating each side of (*) at I , we have that ( a =) b , (= b ) . Since cp' is affine,
bl
+ +
a1
+
t a, =
The equality (*) follows by multiplying each side of (**) by a (= b ) . w1 defines an isometric linear mapping (ii) The mapping w of !2l$ onto (!2l-)$. The composition of the inverse of this mapping with cp and then with the corresponding mapping on l?: produces a linear mapping (which we denote by cp again) of (!?l-)$ into (l?-)$. Let z and y be vectors in 7-1 and B be an operator in B - . The mapping (z, y) + cp(wZlyIU-)(B) is a conjugate-bilinear functional on 7-1. Since cp extends c p l linearly and maps states (in Sw(21))to states (in Sw(l?)),IIcp(w)ll = llwll when w is a positive element of !2l$. From the polarization identity noted above, it follows now, that ---f
when z and y lie in the unit ball of 'H. From Theorem 2.4.1, there H such that (cr(B)z,y) =
is a (unique) bounded operator a ( B ) on
600
SPECIAL REPRESENTATIONS OF C*-ALGEBRAS
c p ( ~ , , ~ l U - ) ( Bfor ) all z and y in IH. If B 2 0, then ( a ( B ) z , z )= cp(w,,,lU-)(B) 2 0 since cp maps positive linear functionals onto positive linear functionals. Thus a ( B ) 2 0. Since
C) = a a ( B )t a(C). With A’ in U’, WAI,,~IU= we have that a(aB iW , , A I * ~ ~whence U, ( a ( B ) A ’ z , y )= ( A ’ a ( B ) s , y )for all x and y in IH. Hence a ( B ) E 9’’ (= U-), and a is a positive linear mapping of B- into U-. Ry linearity of cp, w’(a(B)) = cp(w)‘(B) for each w in U; and each B in f?-. With w in Sw(U),~ ( w E) SW(O)and 1 = cp(w)(I)= w ( a ( 1 ) ) . Hence a ( I ) = I . (iii) With cp’ an affine isomorphism of Sw(U) onto Sw(f?),the argument of (ii) applies to the inverse of Q’. The extension cp is a linear isomorphism of !2l$ onto f?f and there is a positive linear mapping P of U- into B- such that w’(P(A))= cp-’(w)’(A) for each w in f?$ and each A in U-, and such that p(1) = I . Combining this conclusion with the result of (ii), we have
for each A in U- and each w in B$. Since the elements of (!2l-)$ separate 8 - , a ( P ( A ) ) = A for each A in U-. Similarly, Pa(B) = B for each B in 0-.Thus a is a linear order isomorphism of B- onto U- such that a ( I ) = I . From Exercise 10.5.32(i), a is a Jordan * isomorphism of B-’ onto U-. m[59] 10.5.35. Let U acting on a Hilbert space ?-I be the universal representation of the C*-algebra U. With w in U#,let w’ denote the (unique) weak-operator continuous extension of w from 24 to U-. For B in U-, let 6 denote the function on U# that assigns w’(B) to w . For a given B in U-, assume that glS(U) is weak * continuous. Show that (i) 5lC is weatk * continuous, where
c = {up : a E [0,1],
p E S(U)};
(ii) g[(U#)lis weak * continuous [Hint. Recall that the hermitian elements of (%#)I are contained in C - C.]; (iii) B E U. [Hint. Use Exercise 1.9.15.1
601
EXERCISE 10.5.35
Solution. (i) Suppose po in C and a positive E (< 1) are given. If po = 0, and p1 in C is such that (IIp11I =) p l ( 1 ) < ~ / ( 1 11Bl1), then Il&o> - &l)I = IPl(B)l IIlPlII llBll < E .
+
h
Thus BIC is continuous at 0. We may assume that po = uoro, 0 < a0 5 1, and 70 E S(2l). Since S(2l) is weak compact and glS(2l)is weak * continuous, E(S(2l)is uniformly continuous with respect to the weak * uniform structure on S(2l). Thus there is a weak * open neighborhood V of 0 in a", say V = (7 E 2l# : Iq(Aj)) < 1, j = 1 ,...,n} for some A1 > . .. ,A, in U, such that ( ( p l( B )- p2(B)1 < ~ / if 2 p1- p2 E Y and pi,p2 E S(M). Let A0 be I and let rn be max{11Aoll,. .. ,~ ~ A , l ~ , ~ I B ~ ~ } . Suppose p1 E C (say, p1 = a171 with TI in S(2l) and a1 in [O,l]) is such that I(p0 - p l ) ( A j ) (< aoc/2m for j in (0,. , .,?I). Then
*
In particular, when j = 0, we conclude (from the first inequality) that
la0
- a11 < a o ~ / 2 mThus .
.
for j in (1,. . ,n}. By choice of A l , , . . , A , , I.ro(B)- ri(B)I < ~ / 2 . Hence
SIC is weak * continuous. (ii) Since C is weak * compact, B^IC is uniformly continuous in
Thus
the weak * uniform structure on C. Given a positive E , we can find a weak * open neighborhood Vo of 0 in such that \ ( P I -pz)(B)J< E if p1,p2 E C and p1 - p2 E YO.Thus is weak * continuous at 0 on C - C. Now C - C contains the closed unit ball in the (real) linear space 2.4; of hermitian functionals in 2l# from Theorem 4.3.6. It follows that is weak * continuous at 0 on (%;)I and hence, on (U$)2. Translating by elements of (%:)I, we now have that is weak * continuous on ( 2 l f ) l .
602
SPECIAL REPRESENTATIONS OF C*-ALGEBRAS
Each p in (%#)I has a unique expression as p 1 - ip2 with p1 and and the mappings p p1, p -+ p2 are weak * continuous. Hence 5 is weak * continuous on (U#)l. (iii) From Exercise 1.9.15, there is a Bo in U such that w‘(B) = g ( w ) = w ( & ) for each w in U#. In particular, (Bx,y) = (Box,y) for all 2 and y in 3.1. Thus b = Bo E U. p2 in (%;)I
10.5.36. Let cp be a weak * continuous affine isomorphism of the state space S(U) of one C*-algebra 31 onto the state space S ( B ) of another C*-algebra B. Show that there is a Jordan * isomorphism ty of B onto U such that p(ty(B)) = cp(p)(B)for each p in S(U) and each B in B. [Hint. Use the universal representations of U and B and the results of Exercises 10.5.34 and 10.5.35.1
Solution. Suppose U acting on ‘H and B acting on K are the universal representations of U and B. From Proposition 10.1.1, and a$ (of Exercise 10.5.34) coincide. From Exercise 10.5.34(iii), there is a Jordan * isomorphism a of 8- onto U- such that w’(a(A)) = cp(w)’(A)for each w in %$ and each A in B-. When A E B, cp(w)’(A)= cp(4W Suppose A E B . Then, with the notation of Exercise 10.5.35, a ( A ) is the composition of cp and A. By assumption, p is weak * continuous on S(U). By definition, A is weak * continuous on B#. Thus a&i) is weak * continuous on S(U). From Exercise 10.5.35, a ( A ) E U. The symmetric argument shows that t y - l maps U into B. Thus ty is a Jordan * isomorphism of B onto U such that p ( a ( B ) )= cp(p)(B)for each p in S(U) and each B in B. m[59] Let M be a factor acting on a separable Hilbert 10.5.37. space 3-1. (i) Suppose that p is a normal pure state of M . Show that M is of type I. (ii) If M is of type I, is each pure state of M normal? Proof? Counterexample?
Solution. (i) Since r p ( M )acts irreducibly on ‘H, the weakoperator closure of r P ( M )is B(‘H,). From Corollary 7.1.7, r , ( M ) is weak-operator closed, so that K, is a * homomorphism of M onto B(’H,). If r , is an isomorphism, then M is of type I. If K,, is not an
EXERCISE 10.5.38
603
isomorphism, its kernel is the unique norm-closed ideal J' in M and M is of type 1,or .,1 (See Theorem 6.8.7 and Corollaries 6.8.4, 6.8.5.) From Proposition 7.1.15, x p is ultraweakly continuous, and in case x p is not a * isomorphism, J contains an orthogonal family { E n } of projections with sum I . But then
n
n
-a contradiction. Thus x p must be a * isomorphism, and M is of type I. (ii) From Exercise 4.6.69(i), there is a pure state p of B(3-1) that annihilates the ideal K of compact operators. Since I is the sum of one-dimensional projections (each of which lies in K), no such pure state is normal. 10.5.38. Let 9 be the universal representation of a C*-algebra % on %a. Suppose { p n } is a sequence of states of % tending in norm to po and P, is the support of P,lC, where pn is the unique ultraweakly continuous linear extension of pn o @-' from @(%) t o @(%)and C is the center of (a(%)-. Show that { ~ , P Ois} strong-operator convergent to PO.[Hint.Use Exercises 7.6.14, 7.6.28, and 7.6.21.1
Solution.
From Proposition 10.1.1,
Choose xo in 3-10 such that po = w,,J@(2l)-. From Remark 10.1.3, each normal state of a(%)- is a vector state. Thus, from Exercise 7.6.14, each countably decomposable projection in a(%)- is cyclic. Hence the result of Exercise 7.6.28 applies, and there is a vector 2, in 7i0 such that uz,l@(%)-= p n and ((5,- x01) 5 2JJp,-, 0. From Exercise 7.6.21(ii), {P,Po} is strongoperator convergent to PO. 10.5.39.
Let Qr be the universal representation of a C*-algebra
2l on 3-1a. With p a state of %, we say that p is of a certain type I, III,II,, or 111) when xp(%)- is a von Neumann algebra of that type. Let P,, P,,,P,,, P,, be the maximal central projections in @(a)-such that @(%)-P,, @(U)-P,,, @(%)-PCm, and @(U)-P,, are (0) or of types I, IIl,II,, and 111, respectively. Show that
604
SPECIAL REPRESENTATIONS OF C*-ALGEBRAS
(i) a state p of U is of type I,, 111, 11,, or I11 if and only if p(P,), p(Pc,),?(ELw), or p(Pm) is 1, respectively, where p is the ultraweakly continuous extension of p o 9-' to @(a)-; (ii) the set of states of U of a given type is a norm-closed convex subset of '2".
Solution. (i) F'rom the definition of @ (as a direct sum of the representations engendered by the states of U), there is a projection E' in @(a)' such that @,rp : A -+ 9(A)E' ( A E U) is (unitarily) Q ( A ) C p extends to a * isomorequivalent to T,,. Now 9 ( A ) E ' phism of 9(U)-E' onto 9(U)-Cp from Proposition 5.5.5. It follows that p is of a given type if and only if CEI is a subprojection of the central projection corresponding to the summand of @(U)-of that type. There is a unit vector yp in the range of E' (corresponding to z, through the unitary equivalence of T,, and 9p)such that PO@-' = wy,(9(U). Hence, when p is of a given type, p = wy,~9(U)and 1 = p(E') 5 ~ ( C E 5' )p(Pb),where b is one of n , c1, c,, or 00. Conversely, if p(&) = 1 for b one of n, c1, c, or 00, then Pbyp = y, and [@(U)'@(U)-y,,], the range of Cjp from Proposition 5.5.2, is contained in P b ( 7 - l ~ )It. follows that p has the same type as @(U)-Pb. (ii) Since @-' is a * isomorphism of 9(U) onto U, (@-')#, the Banach adjoint of 9-', is an isometric isomorphism of U# onto @(a)# that carries the set of states of U onto the set of states of @(a).The composition of (9-')#and the mapping described in Proposition 10.1.1, of 9(U)#onto @(a);is an isometric isomorphsim of U# onto a(%), that carries the set of states of U onto the set of normal states of @(a)-. Now the set of normal states of (a(%)- that take the value 1 at pb is a norm-closed convex subset of @(a);;the set of states of U of a given type is a norm-closed convex subset of !2l#. m[65] --f
Find an example of a countably generated (norm10.5.40. separable) C*-algebra whose set of states of type I is not weak * closed in U#.
Solution. From Theorem 10.2.3, T,,(U>- = 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
SPECIAL REPRESENTATIONS OF C*-ALGEBRAS
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
SPECIAL REPRESENTATIONS OF C*-ALGEBRAS
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
SPECIAL REPRESENTATIONS OF C*-ALGEBRAS
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
SPECIAL REPRESENTATIONS OF C*-ALGEBRAS
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
SPECIAL REPRESENTATIONS OF C*-ALGEBRAS
(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
SPECIAL REPRESENTATIONS OF C*-ALGEBRAS
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
SPECIAL REPRESENTATIONS OF C*-ALGEBRAS
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
SPECIAL REPRESENTATIONS OF C*-ALGEBRAS
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
SPECIAL REPRESENTATIONS OF C*-ALGEBRAS
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
SPECIAL REPRESENTATIONS OF C*-ALGEBRAS
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
SPECIAL REPRESENTATIONS OF C*-ALGEBRAS
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
SPECIAL REPRESENTATIONS OF C*-ALGEBRAS
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
SPECIAL REPRESENTATIONS OF C*-ALGEBRAS
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
SPECIAL REPRESENTATIONS OF C*-ALGEBRAS
(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
SP(P) = sP(exP6)
c sp{expiu : 1~ E P, lzll 5 211Hll)
c { z E C : 121 = 1, z # -1). (iii) Let f be the function “log” defined on C8,and let g be the function “exp” defined on the strip
SZ = {ZE C!: --K < I m z <
T}.
From (ii), SP(6) c_ a, SP(P) G @*; so (with the notation of Theorem 3.3.8),
E W), f E W). Note also that P r= exp S = g(6), and f o g is the identity mapping on 0. From Theorem 3.3.8, 9
f ( P ) = f ( m >= (f o d ( 6 ) = 6; that is, logp = 6. (iv) The function “log” is holomorphic on the open set C8,and (by (ii)) sp(P) C8 and C \ sp(P) is connected. It follows from Exercise 10.5.65(iii) that (logP)(%) & 2l; that is, 6(2l) U.
c
EXERCISE 10.5.67
10.5.67. Hilbert space
633
Suppose that V is a unitary operator acting on a
H,Q is the * automorphism of B(3.1) defined by a ( A ) = VAV*
( AE B(H)),
and z is a unit vector in ‘H. / ~ , E is (i) Show that Il(VE - EV)zll = (1 - l ( V ~ , z ) l ~ ) *where the projection from ‘H onto the one-dimensional subspace containing 5.
(ii) By considering [a(2E- I)- 2 E
+ I ] V z ,prove that
where L is the identity mapping on a(%). Solution. (i)
Let k be (Vz,.). Since
( V E - E V ) s = va:- EV2 = v z - (Vz,z)z = v z - kz, we have
(ii) Since 2 E - I is a (self-adjoint) unitary operator, we have
634
SPECIAL REPRESENTATIONS OF C*-ALGEBRAS
Suppose that V is a unitary operator acting on a 10.5.68. Hilbert space 3c, a is the * automorphism of B(3-1) defined by
a ( A )= V A V *
.
( A E B(X)),
A1,. . ,A, are distinct elements of sp(V), and c = alA1 +.--+a,A,, where 01,. . ,a, are positive real numbers with sum 1. (i) Suppose that E > 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
a j= ~E.
j=1
(ii) It follows from (i) that, for each positive integer m, there is a unit vector zm such that
By Exercise 10.5.67(ii),
By taking limits as m + 00, we obtain
636
SPECIAL REPRESENTATIONS OF C*-ALQEBRAS
10.5.69. Suppose that 3.1 is a Hilbert space, ct is a * automorphism of f3(3.1), and Ila - < 2, where I is the identity mapping on
f3(W
(i) By using Corollary 9.3.5, show that there is a unitary operator V acting on 'H such that
a ( A ) = VAV*
( A E B(3.1)).
(ii) Deduce from Exercise 10.5.68 that
for all c in the (closed) convex hull of sp(V). (iii) Let co be the point closest to 0 in the (closed) convex hull of sp(V), and let U be the unitary operator c,'(co(V. Show that
a ( A ) = UAU*
( A E B(3.1))
and sp(U)
E
{z E
C : It1 = 1, Rez 2 i(4 - 110
- 6112 )112 }.
[This condition on sp( U )can be interpreted geometrically as saying that sp(U) lies in the arc of the unit circle symmetric about 1 with endpoints midway (on the circle) between 1 and the points on the circle at (straight-line) distance .1 - from 1. In view of the assumption that IIQ - LII < 2, it follows that sp(U) is contained in the "open right half-plane" { z E C : Re t > 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<. In particular,
- 6112 )112
R e z 2 lcol L $(4 - I.,
(and 1.1 = 1) for each t in sp(U). Since U = ZOV, where zo (= c;'lcol) is a complex number of modulus 1, it follows that UAU* = VAV* = a ( A ) , for each A in
B('H). 10.5.70. Suppose that {'Ha : a E A} is a family of Hilbert spaces, 'H is C$N,, U is a C*-algebra of operators acting on R, and Iu- is C$B(R,). Let a be a * automorphism of Iu such that Ila - L I ~ < 2, where L is the identity mapping on Iu. (i) By using the result of Exercise 10.5.14, with ?r : U B(7-f) the inclusion mapping, show that there exist automorphisms & of U- and a, of B(R,), for each a in A, such that 15 extends a, ---f
IIaa
- ball 5
-
= .1
- Lll < 2
(where i, L , are the identity mappings on 21-,B(7-fa), respectively), and for each element C $A4 of 21- (= C $B(7-fH,)). (ii) Show that there is a unitary element U of
a ( A )= UAU* and
( A E U)
such that
638
SPECIAL REPRESENTATIONS OF C*-ALGEBRAS
Solution. (i) From Exercise 10.5.14, with A : % 3 B('H) the inclusion mapping, cy extends to a * automorphism & of U-. From ultraweak continuity of & (Remark 7.4.4) together with Theorem 5.3.5 (Kaplansky density), it follows that
II& - L'll = 1. - LII < 2 (see the argument used in proving that IIfl - ill = - LII, in the solution to Exercise 10.5.14(i)). The center C of U- consists of all bounded operators of the form C $a,I acting on 7-l (= C $'Ha). If P, denotes the projection from 'H onto (the canonical image of) Flaythen {Pa : a E A} is the family of all minimal projections in C. If &(Pa)# Pa, then &(Pa)= Pb for some element b of A \ { a } , and II&(P,) - Pall = 1; so
a contradiction. Thus &(Pa)= Pa,for each a in A. When A E %-,
& ( A )=
&( A)P, =
&(APa);
moreover, for each a in A, the restriction &l%-P, is a * automorphism of %-Pa, and %-Pa is naturally * isomorphic to B(%,). It follows that there is a automorphism a, of B(H,) such that
*
By considering elements whenever C $A, E %- (= C eB(3-1,)). C $A, of U' in which only one of the direct summands A, is nonzero, it follows that ]la,- ball
5 It& - L'll = 110 - 611 < 2
(a E
A).
(ii) For each a in A, it follows from Exercise 10.5.69(iii) that there is a unitary operator U, acting on H,,such that
a , ( A ) = U,AU,*
( A E f3('Ha))
639
EXERCISE 10.5.71
whenever
C $A,
E U-; in particular, a ( A ) = UAU*
( A E 2l).
Moreover, SP(U) =
[ U ~ ~ ( u-a ) ] aEA
c ( 2 E C :1.1
= 1, R e z 2 i(4 - /la- L
II~)~~~},
by Exercise 3.5.26(ii) (although that exercise is formulated for the direct sum of a sequence of operators it remains valid for direct sums m with an arbitrary index set). 10.5.71. Suppose that U is a C*-algebra acting on a Hilbert space 'FI, a is a * automorphism of U, and U is a unitary operator acting on 3.t such that
a ( A ) = UAU* and sp(U) C
( A E U)
{t E C :1.1 = 1, R e z > 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
SPECIAL REPRESENTATIONS OF C*-ALGEBRAS
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
SPECIAL REPRESENTATIONS OF C*-ALGEBRAS
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< I<*) - c l , then H is self-adjoint,
+
6(A)= i ( H A - A H ) ( A E B('H)), sp(H) contains both IlHll and -llHll, and IlHll = #llSll. [Hint. For the last assertion, suppose that E > 0, and let E,, Fc be the spectral projections for H, corresponding to the intervals [IlHIl - E , 11H11], [-llHll, -1lHll E ] , respectively. Consider 6(Vc),where V, is a nonzero partial isometry such that VzK 5 E,, VcV: 5 F,.]
+
Solution. Let c be the midpoint of the smallest interval [u,bJ that contains s p ( i ( K IL'*)), and let T be $ ( b - u). Then [ - r , r ] is the smallest interval that contains sp(H), where H is the self-adjoint operator :(I< I<*)- c l . Thus IlHll = T , and r, -r E sp(H). When
+
+
A E B(X),
6 ( A )= i ( K A - A K ) , b ( A ) = 6(A*)*= [ i ( K A *- A * K )]*= i(l<*A- A K * ) . Thus
-+
6 ( A ) = :i[(X K * ) A - A ( K
+
-+ I{*)]
= i[(!j(K I<*) - c1)A - A ( i ( X + I < * ) - d)] = i(HA- AH).
Given any positive real number E , choose E,, F,, V,, as in the hint. (Note that EE # 0 # F,, because 11H11,-11Hl1 E sp(H), and hence there exists a non-zero partial isometry V, with the stated proper ties. ) Since
IIHE, - 4 1 1 5 E ,
IIHK -t T C l l I E,
v, = F,V,E,,
it follows that 11611 2
ll~(V€)Il= IlHVe - KHII = I l H F C V E - V€E,HII 2 11 - rF,V, - r&E,II - 2~ = 211TVell - 2 E = 2(r - E ) = 2(11Hll - E l .
646
SPECIAL REPRESENTATIONS OF C*-ALGEBRAS
Since this has been proved for every positive E , we have llSll The reverse inequality is evident, so llSll = 211Hll.
2 211Hll.
10.5.77. Suppose that M is a simple C*-algebra (containing I) acting on a Hilbert space 'H, 6 is a * derivation of !2l, R is a self-adjoint element of B('H), and
'llHll =
illSll,
b ( A ) = I(HA - A H )
( A E %).
(i) Let F' be a non-zero projection in a'. Show that the mapping A + AF' is a faithful representation of % on the Hilbert space F'(31). Prove that
S(A)F' = i(F'HF'A - AF'HF')
( A E %),
and deduce that IIF'HF'II = 11H11. (ii) Suppose that E' is a projection in U', and let S be (I - E')HE'. Show that S E M'. Deduce from (i) that S = 0. [Hint. If S # 0, let F' be the spectial projection for S*S,corresponding to the interval [E,m), where 0 < E < IlS*Sll. Deduce from (i) that
((F'HE'HF'II 2 llH1I2 2 llF'R2F')) = Ip'HE'HF' 4- F'S'SF'II, and hence obtain a contradiction.] (iii) Show that H E U'. (iv) By considering the universal representation of the C*-algebra B generated by Q and H, and using Proposition 10.1.4, show that H E !2l. Solution. (i) It is apparent that the mapping A + AF' is a representation cp of !2l on F'(H). The kernel of cp if a two-sided ideal in Q, not the whole of M, and is therefore (0) (since M is simple). It follows that cp is faithful. For each A in 24,
6(A)F' = F'G(A)F' = ~ F ' ( H A- AHIF' = i(F'HF'A - AF'HF').
647
EXERCISE 10.5.77
Thus
IlWII = I l V ( W ) I I = IlW)F'II
= IIF'HF'A - AF'HF'JI
5 2llF'HF'IIllAlI
( A E a),
and
211HIl = IlSll 5 2IlF'HF'II. It follows that I(F'HF'I1 = 11H11. (ii) When A E U, we have H A - AH E U, so 0 = ( I - E')(HA - AH)E'
= (I- E')HE'A - A ( I - E')HE' = S A - AS. It follows that S E a'. If S # 0, choose E and F' as in the hint. Then F' is a non-zero projection in a', E' 2 F', and
llS((*E'2 F'S*SF' = S*SF' 2 EF'.
H(I - E')HF
a contradiction. Thus S = 0. (iii) If E' is a projection in a', then ( I - E')HE' = 0, by (ii). With I - E' in place of E', we obtain E'H(I - E') = 0. Thus
HE' = E'HE' = E'H,
648
SPECIAL REPRESENTATIONS OF C*-ALGEBRAS
and H E U" = U-. (iv) Let B be the C*-subalgebra of B(3-1) generated by U and H, and let @ : B --+ B ( H * ) be the universal representation of B. Then @(a)is a simple C*-algebra of operators acting on Ha,iP6a-l is a * derivation of a(%),@(IT) is a self-adjoint element of B(Ha),
Il@(H)lI = IPII = +llSIl = ~ll@~@-lll~ and
( @ b @ ) - l ) ( T=) W ( A ) = i @ ( H A- A H ) = i(@(H)T- T @ ( H ) )
for all T (= @(A)) in @(a).We can now apply the result of (iii), and @ ( H ) in place of Q and H, respectively, to conclude with @(a) that @ ( H )E @(a)-. Thus
@(R)E @(B)n @(a)'-. Since @(a)is a norm-closed convex subset of @(f?), and @ is the universal representation off?, it follows from Proposition 10.1.4 (with B and K replaced by B and @(a), respectively) that
@(B)n @(a)= @(a). Thus
@(H)E @(a), and H E U.
4641
10.5.78. Suppose that U is a simple C*-algebra (with unit), and 6 is a derivation of 2.Prove that 6 is inner. [Hint. Show that it suffices to consider only * derivations. By taking an irreducible representation of U, reduce to the case in which 3 C B ( X ) and 2l- = B ( X ) , for some Hilbert space 'FI. In this case, use Exercises 7.6.15, 10.5.76,and 10.5.77.1 Solution. The equation 6*(A)= 6(A*)* defines a derivation S* of ?U. If 61 = t 6.) and 62 = #i(S* - 6), then 61 and 62 are * derivations of U, and 6 = 61 t i62. If both 61 and 62 are inner, then so is 6. We assume, henceforth, that 6 is a * derivation. Let cp be an irreducible representation of %. Since (p-l(O) (# %) is a two-sided ideal in a, and B is simple, it follows that cp-l(O) = (0); so Q is faithful. Upon replacing U and 6 by cp(?2l)and cpbv-',
i(6
649
EXERCISE 10.5.79
respectively, we may suppose that U acts irreducibly on a Hilbert space 'H. Thus U E B(3-1), a- = B(3-1). From Exercise 7.6.15, 6 extends to a derivation $ of B('H),and llsll = IlS((; moreover, 8 is weak-operator continuous on bounded subsets of B(H). Since
$(A*) = 6(A*) = b(A)* = $(A)*
( A E U),
it follows (from the continuity of 6 just noted, together with the Kaplansky density theorem) that $(A*) = $(A)* for all A in B ( H ) (= 24-). Thus 8 is a * derivation of B(3-1). From Exercise 10.5.76, there is a self-adjoint operator H acting on 8,such that
It follows that
so H E 2l by Exercise 10.5.77(iv). Thus 6 is inner.
~(96,971
10.5.79. Find an example of a derivation of a C*-algebra such that the derivation is not inner.
Solution. Let 'H be an infinite-dimensional Hilbert space and K be the ideal of compact operators on X. Let M be the C*-algebra { a l } K . Then each T in B(3-1) induces a derivation (adT)IM of Q defined by (adT)IM(A) = AT - TA for each A in U. If S is another operator in B ( K ) such that (ad S)lM = (ad T)IQ, then (S - T ) A = A ( S - T) for each A in U. Since K: and, hence, M act irreducibly on 'H, S - T = uI for some scalar a. It follows that S E 2l if and only if T E Q. Thus, if we choose T in B('H) \ 2l (say, T is chosen as a projection such that T and I - T have infinite-dimensional ranges), then (adT)IU is a derivation of U that is not inner.
+
10.5.80. Let 2l0 be an abelian C*-algebra acting on a Hilbert space H such that 2l0 n K = (0), where K is the ideal of compact operators on 'H, and let U be the C*-algebra Uo K. Show that (i) the center C of U consists of scalar multiples of I;
+
650
SPECIAL REPRESENTATIONS OF C*-ALGEBRAS
(ii) co%(A)=fl C = 0 for each A in 2l not of the form a 1 + B with B in K , where co%(A)= is the norm-closed convex hull of {UAU* : U E U) and U is the group of unitary operators in 2l (compare this conclusion with Theorem 8.3.5) [Hint. Consider the quotient mapping of 2l onto 9lIIC.J; (iii) there is an %o and an A in % not of the form a1 + B with B in IC. Solution. (i) From Proposition 10.4.9, M is a C*-algebra. Since
IC acts irreducibly on 31, % acts irreducibly on 3-1. Thus C E 2l' = ( a 1 : a E C).
(ii) Since the quotient mapping cp of 2l onto %/K is a * homomorphism of 2l onto %/K,each unitary element in % has a unitary element in %/K as its image. Thus Q maps com(A)into comlK(cp(A)). As Q is norm continuous, cp maps com(A)' into conlK(~(A))'. But Q/K: is isomorphic to the abelian algebra !Mo since ?2l = go K and !& n K: = (0). Thus coglK(cp(A))= = {cp(A)). If C O ~ ( A )n=C # 0, then cp(aI) E {cp(A)) for some scalar a, whence A = a 1 B with B in K. (iii) Let E be a projection in B(7-l) such that E and I - E have infinite-dimensional ranges and let 2b be the (abelian) twodimensional C*-algebra generated by E and I. Choose E for A.
+
+
10.5.81. Suppose that 2l is a C*-algebra. A subset P of ?2l is described as a primitive ideal if P is the kernel a-'(0) of some irreducible representation a of %. (This implies that P is a closed two-sided ideal, and P # %.) The set of all primitive ideals in % is denoted by prim(%). (i) Suppose that Z is a closed two-sided ideal in a, and Z # 2l. Show that 1 = n { P : P E prim(%), Z C P). [Hint. Apply Corollary 10.2.4 to the C*-algebra %/I.] (ii) For each P in prim(!M), let cpp : 2l -' % / P be the quotient mapping. Show that
[Hint. Consider the mapping 511 -+ CPEprim(m) $cpp(A) from a into ) (see Exercise 3.5.3).] the algebra C ~ E p r i m ( %$O/P
651
EXERCISE 10.5.81
(iii) Show that a maximal (proper) two-sided ideal in % is primitive. (iv) Suppose that 1 1 and ZZ are closed two-sided ideals in %, P E prim(%), and A1 A2 E P whenever A1 E 2'1 and A2 E ZZ. Show that P contains at least one of 1 1 and 12.[Hint. Let n : % + B ( X ) be an irreducible representation that has kernel P , and consider the closed subspace K of 3-1 generated by the set (n(A2)x : AZ E ZZ, x E X}.] Solution. (i) It is apparent that ZC
n(P : P
E prim(%), Z
c P}.
To establish the reverse inclusion, we have to prove the following assertion: if A E % and A 4 Z, there exists P in prim(%) such that I c P and A 4 P . If A E % and A 4 Z, we have cp(A) # 0, where cp : % --t %/Z is the quotient mapping. From Corollary 10.2.4, no(cp(A)) # 0 for some irreducible representation no of %/I.Now no o cp is a representation K of U, and n is irreducible since .(a) = xo(%/Z). Moreover, .(I) = (0) and n(A) # 0. The kernel P of K is a primitive ideal in %, and Z C P, A 4 P. (ii) With the notation of Exercise 3.5.3, &prim(a) @%/P is a C*-algebra, and the mapping
is a
*
homomorphism. Since cpF has kernel P, the kernel of cp is
n{P : P E prim(%)}, and this is (0) (by (i), with (0) in place of
1).It follows that cp is a * isomorphism, and is therefore isometric. Thus
IlAll = IIdA)II = IIC@cpP(A)ll = suP{llP,(A)II : P E prim(%)),
for each A in %. (iii) If Z is a maximal (proper) two-sided ideal in %, then Z is closed (Proposition 3.1.8), and it follows from (i) that Z C P for some P in prim(%). Since Z is maximal, we have Z = P;so Z is primitive.
652
SPECIAL REPRESENTATIONS OF C*-ALGEBRAS
(iv) Choose A : % -, D('H), and define the closed subspace K: of 'H, as in the hint. When A2 E Z2 and x E 'H, we have
It follows that
Since A is irreducible, K: is 'H or (0). If K: is 'H, n(A1) = 0 for each A1 in 1 1 , and 11 C A - ' ( O ) = P . If K: is {0}, n(A2)z = 0 for all A2 in 1 2 and x in 'H; so A ( & ) = 0, and Z2 C P. m[71] 10.5.82. Suppose that U is a C*-algebra and prim(%) is the set of all primitive ideals in %. Given any subset S of 24, the hull h ( S ) of S is defined by
h ( S ) = {P E prim(%) : S
P}.
Show that (i) { h ( S ): S E U} is the family of all closed sets in a topology on prim(%); (ii) if 3 C prim(%) and 10= n{P : P E F},then the closure F- of F is given by F- = h(Z0); (iii) if P E prim(%), then the one-point set {P} is closed in prim(%) if and only if P is a maximal (proper) two-sided ideal in U; (iv) if P1,P2 E prim(%) and Pi # P2, there is an open set in prim(%) that contains just one of Pl,P2; (v) prim(%) is compact in the following sense: if F a = 0, where each is a closed subset of prim(%), then naEpra = 0 for some finite subset F of A; (vi) prim(%) has just two elements, and the topology described in (i) is not Hausdorff, if % = B('H), where 'H is a separable infinitedimensional Hilbert space. [The topology in (i) is called the Jacobson topology on prim(%); with this topology, prim(%) is called the primitive ideal space, or primitive spectrum, of 0.The result of (iv) can be expressed as the assertion that prim(%) is a To-space.]
ra
naEA
653
EXERCISE 10.5.82
Solution. (i) Given an arbitrary family {S,} of subsets of %, it follows a t once from the definition of the term “hull” that
Suppose that S1,Sz 9, and let Zj be the closed two-sided ideal of % generated by S j . Given any primitive ideal P in U, Sj C P if and only if Zj C P ; so h ( S j ) = h(Zj). If P E h(Z1) U ~ ( Z Z )then , P contains at least one of Z1,22, whence P 2 ZIZZand P E h(Z1Zz). Conversely, suppose that P E h(Z1Zz); that is, P E prim(%) and P 2 2 1 2 2 . By Exercise 10.5.81(iv), P contains at least one of Z1,Zz; so P E h(Z1)U h(Z2). We have now shown that
h(S1)u h(S2) = h(Z1)u h(Z2) = h(Z1Zz). Observe next that h(%)= 0, and h({O})is prim(%). It follows from the preceding three paragraphs that the family { h ( S ): S C a} of subsets of prim(%) contains 0 and prim(%), and admits finite unions and arbitrary intersections. Accordingly, this family is the collection of all the closed sets, for some topology on prim( %) . (ii) Each closed set in prim(%) has the form h ( S ) , where S %. Moreover, h ( S ) 2 F if and only if S P for each P in F (equivalently, if and only if S C_ TO),and this implies that h ( S ) 2 h(Z0). Hence h(Z0) is the smallest closed set that contains F ;that is, F- = h(Z0). (iii) From (ii), the closure of { P } is the set of all primitive ideals that contain P . Thus { P ) is closed if and only if no primitive ideal properly contains P ; from Exercise 10.5.81(i), this occurs if and only if P is a maximal (proper) two-sided ideal in 8 . (iv) Since PI # P z , either PI g PZ or PZ g PI; we may suppose that PI P2. The closed set h(P1) contains PI but not P2, so its complement is open and contains P2 but not P I . (v) From (ii) and since F a is closed, = h(Z,), where Z, = P. If Z is the linear span of {Za}aE~, then Z is a two-sided ideal in 8. If Z P E prim(%), then 1, & P and P E h(Za) = for all a in A. But I?, = 0. Thus no primitive ideal contains Z. From Exercise 10.5.81(i), the norm closure of Z, and hence Z itself, are %. Hence there is a finite subset { ~ ( l ).,, .a.( n ) } (= F) of A such that 1 = A1 . - . A , , where A1 E Z,(l),. . . , A , E If
c
npEr,
ra
c
naEA
+ +
ra
654
SPECIAL REPRESENTATIONS OF C*-ALGEBRAS
+
E naEFra, then Aj E iTa(j) G P; and I = A1 t ..* An E P-a contradiction. Thus naEFra = 8. Let K be the set of a l l compact linear operators acting on (vi) the (separable, infinite-dimensional) Hilbert space 3-1. As noted in the discussion preceding Propsition 10.4.1, B(H)has just two normclosed two-sided ideals (other than B('H)), (0) and K. Since the identity mapping on B(3-1)is an irreducible representation of B ( H ) on 'H, its kernel (0) is a primitive ideal of B('H). By Exercise 10.5.81(iii), K: is a primitive ideal of B(3-1). Thus
P
prim(B(3-1)) = { (01, n}.
From (iii), the set consisting of the single element (0) is not closed in prim(B('H)); so prim(B(3-1)) is not a Hausdorff space. 10.5.83. Suppose that % is a C*-algebra, and f : prim(%) + [0,1] is a continuous function on the primitive ideal space prim(%). For each P in prim(%), let ' p p : U U/P be the quotient mapping. The following results are subsumed in Exercise 10.5.84, but are required for the solution of that exercise. (i) Define open subsets 00,.. ,Onof prim(%) by
.
Oj =
{P E prim(%) : $+ < J ( P ) < +},
and let ~j
= n { P : P E prim(%) \ Oj).
Show that
Oj = ( P E prim(%) :Zj If P } ,
10
t -..t 1, = %,
(ii) Suppose that A E at. By using (i) and Exercise 4.6.64, show that there exist Ao, . . .,A, such that
A=Ao+...+A,,
Aj
EZ~+
(j=O,...,72).
Prove that IlcpP(B>- f(P)cp,(A)lI
< ;llAll
where B (E U) is defined by n
B=
C $Aj. j=O
(PE prim(%>>,
655
EXERCISE 10.5.83
Solution. (i) Since f is continuous and takes values in [0,1], . .,Onform an open covering of prim(2l). By applying the sets 00,. Exercise 10.5.82(ii) to the closed set prim(%) \ Oj, we obtain
prim(%) \ Oj = h(Zj) = {P E prim(%) : Zj C P } , where Zj is defined as in the statement of (i). Thus
Oj = { P E prim(%) : Zj P } . A straightforward argument by induction on n, based on Exercises - In is closed, and 4.6.63 and 4.6.64, shows that the ideal Zo
+- + +
Given any P in prim(%), we can choose j in (0,. P E Oj; then, Zj P , so
. . ,n}
such that
+ -
Since 10 - t 1, is not contained in any primitive ideal of %, it follows from Exercise 10.5.81(i) that
+ +
(ii) Given A in 2l+ (= (To ... I,)+),the existence of the elements A o , . .. ,A, in 2l with the stated properties follows from (1). Since 0 5 A j 5 A, we have IlAjll 5 IlAll. If P E prim(%), we can choose an integer k such that
From (2) and the definition of Oj, P j # b; so by (i),
ZjSP
4 Oj if 0 5 j 5 n and k - 1 #
(OSjIn, k-l#j#k).
Since Aj E Zj,
pp(Aj) = 0
(0 I j
I n, k - 1 # j # k).
656
SPECIAL REPRESENTATIONS OF C*-ALGEBRAS
With B defined as in the statement of (ii), we have n
n
where C' denotes summation over the (one, or two) values of j such that 0 5 j 5 n and Ic - 1 5 j 5 k. For these values of j,
10.5.84. Suppose that U is a C*-algebra, A E % and f is a (bounded) contiiiuous complex-valued function on the primitive ideal space prim(%). For each P in prim(%), let pp : U + U / P be the quotient mapping. (i) Suppose that E > 0. By using the result of Exercise 10.5.83(ii), show that there is an element B, of Q such that
(ii) Deduce that there is an element B of % such that
[The result in (ii) is known as the Dauns-Hoffman theorem.]
Solution. (i) From Exercise 10.5.82(v), f is bounded. We can express f and A in the form 4
4
j=1
k=l
EXERCISE 10.5.84
657
where each f j is a continuous mapping from prim(%) into [0,1], each A k lies in %+, and u l , . ..,u 4 , b 1 , ... , b 4 are scalars. From Exercise 10.5.83(ii), with n a sufficiently large positive integer, we can find elements B j k ( j ,k = 1,. . . ,4)of 2l such that
for all P in prim(%). Thus
for all
P in prim(%), and summation over j and k in (1,. . . ,4} gives
where
(ii)
From (i), there is a sequence { B n } in U such that
so {Bn}is a Cauchy sequence in %. With B the limit of follows from (1) that
{Bn}, it
658
SPECIAL REPRESENTATIONS OF C*-ALGEBRAS
10.5.85. Let B be a C*-subalgebra of the C*-algebra U, and let cpo be an idempotent (cpo o cpo = 90)linear mapping of 2f onto B such that IIcpoll = 1. Suppose U acting on 3-1 is the universal representation of 2l and, in this representation, E is a projection in B - . Show that (i) cpo is a positive linear mapping of U ! onto B such that cpo(I) = I [Hint. Compose cpo with states of B and use Theorem 4.3.2.1; (ii) tpo extends uniquely to an ultraweakly continuous idempotent linear mapping cp of 8' onto B- such that llcpll = 1, and cp is a positive linear mapping; (iii) wzo cp is a state of U- definite on E when z is a unit vector in either E(3.C) or (I- E ) ( H ) (see Exercise 4.6.16); (iv) E p ( E A ) E = Ecp(AE)E = E y ( A ) E , Ecp(EAE)E = Ecp(A)E, and (I- E)cp(EA)(I- E) = ( I - E)cp(AE)(I- E ) = 0 for each A in U-; (v) cp(EAE) = E d A ) E for each A in 8' [Hint. Recall that, with A self-adjoint, -11A11E 5 EAE 5 llAllE.1; (vi) cp( EA( I-E)) = (I-E)cp( EA(I- E ) ) EtEcp( EA( I - E ) ) ( I E) for each A in Q- . Solution. (i) Since I E B, and cpo is idempotent with range B, tpo(I) = I. If p is a state of B, then ( p o cpo)(I) = 1. Since llp o cpolf 5 IIpllllmll = 1, p o cpo is a state of 2l by Theorem 4.3.2. If H is in U+, p(cpo(H))2 0 for each state p of B. Since cpo(H) E B, cpo(H) E B+ by theorem 4.3.4(iii). Thus cpo is a positive linear mapping of 2l onto B. (ii) By assumption, cpo is a bounded linear mapping of 2l onto B . If w is an ultraweakly continuous linear functional on B , then wocpo is a bounded linear functional on 2l and hence is ultraweakly continuous from Proposition 10.1.1. Thus cpo is ultraweakly continuous and extends uniquely to an ultraweakly continuous linear mapping cp of U' into B' such that IIpII = Ilcpoll = 1. Since B- C_ U-, cp o cp is defined, ultraweakly continuous, and coincides on 0 with cpo o cpo (= cpo = cplU). The ultraweakly continuous mappings cp o cp and cp agree on the ultraweakly dense subset 2l of IZL- so that they agree on U-. Hence cp is idempotent. Since the unit ball of B is contained in the unit ball of U and IIcpII = 1, cp maps the ultraweakly compact unit ball of !? onto lan ultraweakly compact (hence closed) subset of Bthat contains ( B ) l . From Kaplansky density theorem (B); = (B')I. Hence, cp(U-) = B'. From (i), cp is a positive linear mapping.
EXERCISE 10.5.85
659
(iii) Since cp(I) = cp,-,(I) = I , (w, o cp)(l)= 1. From (ii), cp is a positive linear mapping of 2l- onto 17- so that wx o cp is a state of '2-. As E2 = E , the states p of U- that are definite on E are those such that p ( E ) = p(E2)= P ( E ) ~that ; is, the states definite on E are precisely those that take the value 1 or 0 at E . Since E E 13- and cp is idempotent with range B - , (w, o cp)(E) = w,(E). When x E ( I - E)('H), (w, o cp)(E) = 0, and when x E E ( X ) , (w, o cp)(E) = 1. Thus w, o cp is definite on E when x is a unit vector in either E(3-1) or (I - E)(?l). (iv) From (iii) and Exercise 4.6.16, when x is a unit vector in E ( X ) or in ( I - E ) ( H ) ,
for all A in U-. Thus, with x a unit vector in E ( X ) , (cp(EA)x,x)= (cp(A)x,x). This same equality holds for all x in E ( X ) , so that
Ecp(EA)E = Ecp(A)E. With 2 a unit vector in ( I - E)('H),we have (cp(EA)x,x)= 0. This same equality holds for all x in (I - E)(3-1). It follows that ( I - E ) v ( E A ) ( I- E ) = 0. In the same way, we have that Ecp(AE)E = Ecp(A)E and that ( I - E)cp(AE)(I- E ) = 0 for all A in '2-. Thus Ecp(EAE)E = Ecp(AE)E = Ecp(A)E. (v) From the hint and since cp is a positive linear mapping (by (ii)),
Hence cp(EAE) = Ecp(EAE)E = Ecp(A)E, from (iv). (vi) From (iv),
cp(EA(I - E l ) = Ecp(EA(I - E ) ) E t ( I - E ) p ( E A ( I - E ) ) E E p ( E A ( I- E))(I - E ) t ( I - E)q(EA(I- E))(I- E ) = ( I - E ) p ( E A ( I - E ) ) E -I-E v ( E A ( I - E ) ) ( I - E ) ,
+
for each A in !2l-.
660
SPECIAL REPRESENTATIONS OF C*-ALGEBRAS
With the notation and assumptions of Exercise 10.5.86. 10.5.85, show that (i) J J E T ( IE - )t( I - E)SEJ)= max{ JJET(I-E)JI, 11 ( I - E)SEII} for all T and S in B(9-l); (ii) ( I - E ) ( p ( E A ( I - E ) ) E = 0 [Hint.Suppose the contrary and show that, for each A in U- and all large integers n,
nll(1- E)cp(EA(J- EPII
= IIp(EA(I - E ) ) t ( n - 1)(1- E)(p(EA(I- E))E(I
(use (i) and Exercise 10.5.85(vi)), and deduce a contradiction from this equality.]; (iii) (p(EA)= &(A) and cp(AE) = cp(A)E for each A in 2l-; (iv) cp( B A ) = By( A ) and 'p( A B ) = cp( A)B for each A in U- and each B in 0-; (v) cp is a conditional expectation from ' l2 onto B- in the sense of Exercise 8.7.23; (vi) (po is a conditional expectation from 2l onto B (that is, the conditions on ip in Exercise 8.7.23 are fulfilled for po with yo,%, and B,in place of ip, S , and R).
Solution. (i) With
2
a unit vector in E ,
IIET(I - E). + ( I - E)SE.1I2 = IIET(I - E)z(I2-t ll(I - E)SEz(12
I IIEW - E)1I211(~ - E>4I2-t ll(I- ~)s~l1211~412 I max { (lET(I- E>1I2,ll(I- E)SEIl2}* (The last inequality follows from the fact that its left side is a convex On the other hand, combination of I(ET(I-E)1I2 and JJ(I-E)SEIJ2.)
IIET(I- E)ll = IIE[ET(I - E ) t (I - E)SE]II 5 IJET(I- E ) -t ( I - E)SEII. Similarly, [](I- E)SEII 5 IIET(I- E )
(IET(I- E ) t ( I - E ) S E ( (=
-+ ( I - E)SEII, so that
{(IET(I- E)ll, ll(I - E)SEJJ}.
(ii) Suppose (I - E)cp(EA(I - E ) ) E positive integers n ,
#
0. Then for all large
IlE9(EAV - E W - E>llI 1141- E ) V ( E W - JWEII,
EXERCISE 10.5.86
661
so that from (i) and Exercise 10.5.85(vi) and since 'p is an idempotent with range B- and norm not exceeding 1,
a contradiction. Thus ( I - E)'p(EA(I - E ) ) E = 0. (iii) From (ii) and Exercise 10.5.85(vi),
(*>
(p(EA(I - E ) ) = E'p(EA(I - E ) ) ( I - E ) .
Thus for each A in U-, from (*) and Exercise 10.5.85(v),
so that
E'p(A) = E q ( E A E ) E = 'p(E A E ) = 'p(EA).
+
+ E y ( E A ( I- E ) ) ( I- E )
'p(E A ( I
-E))
Similarly, 'p(AE)= ' p ( A ) Efor each A in U-. (iv) Let B be a self-adjoint element in 8- and A be in 24-. From Theorem 5.2.2(v), given a positive E , there is a (finite) orthogonal family { E l , . . . , E n } of projections in B- and (real) scalars a l , . . . , a ,
662
SPECIAL REPRESENTATIONS OF C*-ALGEBRAS
such that llB - C;==,ajEjII
< ~/(211All).From (iii), we have
Thus q(BA) = B'p(A) and similarly, cp(AB) = (p(A)B for each B in B- and each A in U- . (v) From Exercise 10.5.85(ii), cp is a positive linear mapping of U- onto B- and cp(I) = I. From (iv), cp(BAC) = Bcp(A)C for each A in U- and B,C in B - . Thus cp is a conditional expectation from U- onto B- (in the sense of Exercise 8.7.23). (vi) Since 90 maps U onto B and is the restriction of cp to U, by (v) it is immediate that 'po is a conditional expectation from U onto
B.
8
10.5.87. Suppose the C*-algebra 2l is (linearly isomorphic and isometric to) the norm dual of a Banach space a+,and let q be the natural injection of U+ into 2l". (i) Let u be an element of U##. Show that u o q = A for a unique A in U (viewed as linear functionals on a#). (ii) Let U acting on 3-1 be the universal representation of U, and let A + A be the (isometric linear) isomorphism (of Proposition 10.1.21) between U- and U##.Let cp(A) be the element in U (obtained in (i)) such that A o q = cp(A), where A E 24-. Show that cp is an idempotent mapping of U- onto U and llcpll = 1. Deduce that 'p is a conditional expectation from 2l- onto U. Show that K is a weak-operator-closed (iii) Let K be cp-'(O). two-sided ideal in 3-. (iv) Let P be the central projection in U- (of Theorem 6.8.8) such that K = U-P. Show that % - ( I - P) = %(I- P). [Hint.Note that A - cp(A) E U-P for each A in U-.] (v) Show that U(I - P) is * isomorphic to U and conclude that U is a W*-algebra. (Compare this exercise with Exercises 7.6.417.6.45.)
EXERCISE 10.5.8s
663
Solution. (i) Suppose t E (%,)I. Then, since q is an isometry, I(. 0 q)(t)Il 5 11.11119(~)11 IIIVII, and * 0 9 is a bounded finear functional on U,. By assumption, Q is the norm dual of 8,. Thus there is an A in % such that Y o q = A, and A is unique. (ii) Let A be an element of Q (in U-). We show that cp(A) = A. Since cp is a linear mapping of U- onto ?2l, this will show that cp is an idempotent mapping of U- onto 2l. With t in U,,
cp(A)(O = ( A 0 N ) = V ( t ) ( A )= A W Thus cp(A) = A. At the same time, if B E (%-)I, then and IIcp(B)(OII= 11@ O rI)(Oll 5 11~(011 = 11tIL
E (U##)I
Thus llcp(B)II 5 1. It follows that llcpll 5 1, and from Exercise 10.5.86, cp is a conditional expectation from U- onto Q. (iii) We note first that Ic is weak-operator closed. We have that A E K, if and only if ( A o q)(() = 0 for all t in U+. Now q(t) E U# and 2l acting on 'H is the universal representation of U, so that there (~)JU. are vectors z(() and y(t) in 'H such that q(() = W ~ ( C ) , ~Thus A E Ic if and only if U~(.CC,~(~)(A) = 0 for all ( in U,. It follows that K, is weak-operator closed. Since cp is a conditional expectation from U- onto U (see Exercise 8.7.23), cp(BAC) = Bcp(A)C for each A in 0- and B,C in 3. Thus, if A E K ,0 = Bv(A)C = cp(BAC),and BAC E Ic. By weakoperator continuity of left (and then right) multiplication, BAG' E K when A E Ic and B, C E Q-. Hence K: is a weak-operator-closed twosided ideal in %-. (iv) Since cp is idempotent, A - v ( A ) E K for each A in U-. Thus A - cp(A) E U-P and A - cp(A) = [A - (p(A)]P.It follows that A ( I - P) = cp(A)(I - P ) E %(I - P). Hence %-(I - P) = %(I - P). (v) If A E % and 0 # A (= cp(A)),then A 4 K so that A 4 %-P. Thus A # A P and A ( I - P) # 0. Since P commutes with Q, the mapping A -, A ( I - P)of U onto %(I - P) is a * homomorphism and from the foregoing, this mapping is a * isomorphism. From (iv), % ( I - P ) = U - ( I - P ) , so that U is * isomorphic to the von Neumann algebra %-(I- P) (acting on (I- P)('M)).Hence Q is a W*-algebra in the sense of Exercise 7.6.38. .[92,113(pp. 130-133),114]
10.5.88. Let 2l be the uniformly matricial (the CAR) algebra of type (23}, and let {Uj} be a generating nest of type (2j). (The
664
SPECIAL REPRESENTATIONS OF C*-ALGEBRAS
existence of such an algebra U is proved in Proposition 10.4.18.) With S a subset of U, let S‘ be the set of those T in U that commute with every S in S,and let Bj be n Uj for j in {1,2,. . .}, where %I) = { e l : c E C}. (i) Show that each Bj is a C*-subalgebra of U (containing I) * isomorphic to the algebra of 2 x 2 complex matrices. (ii) Show that BjBk = BkBj when Bj E B j , Bk E Bk, and j # k; 2lj is the 1in.ear span of products B1. * * Bj with Bk in Bk; U is the norm closure of the linear span of products B1 B , with Bk in f3k and 7~ in {1,2,. .}. (iii) In each B j , choose a 2 x 2 self-adjoint system of matrix units and let a?), u p ) ,a?’ be the elements of Bj whose matrix representations relative to the chosen matrix units are
.
.-
(These are the “Pauli spin matrices.”) Show that generate Bj as an algebra and that {a,( j ) ,ay( 8,az (j) : j = 1,2,. . .} generates U-as a C*-algebra. ) Show that (iv) Let Cj be a:’) * . . a ~ - ’ ) ( a-~iaf))/2.
[The equations (*) are referred to as the “canonical anticommutation relations”-abbreviated “CAR”-and the set of elements { C j } is said to “satisfy the CAR.”] (v) Show that
Conclude that the set {C,, . . . ,C j } generates Uj as a (finite-dimensional) C*-algebra and that {Cl,C2, ...} generates M as a C*algebra.
EXERCISE 10.5.88
665
(vi) Let W be the set of products C:(l) *.-C:(h)Cn(l).s.C,(~), where 0 < m(1) < m(2) < ... < m(h) 5 j , 0 < n(1) < < n ( k ) 5 j , and h, k E (0,. . . , j } . (Such a product is said to be “Wick ordered.” If h = 0, the product is C,(1) * - . C , ( k ) . If k = 0, the product is Cz(ll. Ck(,). If h and k are 0, the product is I.) Show that W is a (linear-space) basis for 31j.
Solution. (i) By definition, I E Uj and of course I commutes with each element of 2lj-1. Thus I E B j . By assumption, there is a * isomorphism of 31j onto B ( Z j ) , where Z j is a Hilbert space of dimension 23. This isomorphism carries 31j-1 onto a subfactor M of B ( Z j ) of type Izj-1 and carries Bj onto M’. Now M’ is itself a factor; it remains to show that M’ is of type 12. Since M contains a family of 2j-l orthogonal (equivalent) minimal projections with sum I , each of these 2j-l minimal projections has dimension 2. Let E be one of these minimal projections. Then M’ is * isomorphic to M’E (from Proposition 5.5.5) and M’E is B(E(7-l))(for E M E , its commutant, consists of scalar multiples of E , from Proposition 6.4.3). Thus M’ is of type 1 2 and f3j is * isomorphic to the algebra of all complex 2 x 2 matrices. (ii) Suppose j < k. Then Bj E 2lj U k - 1 when Bj E B j . If Bk E Bk, then B k E Ui-, and BjBk = BkBj. We show, by induction on j, that each element of 31j is a (finite) sum of products of the form B1 . - -Bj with Bk in ,131,. By definition of (310 and) 01, B, = !2l1 so that our assertion is evident when j is 1. Suppose the assertion is established for j - 1 (2 1). The algebra N generated by Uj-1 and Bj consists of (finite) sums of products ABj with A in Uj-1 and Bj in B j . From our inductive hypothesis, each such product ABj is a (finite) sum of products B1 - - Bj with B k in B k . If we use the * isomorphism (of (i)) between Uj and B ( Z j ) together with the double commutant theorem, we see that the image NO of N in B ( 7 i j ) has commutant consisting of scalars. It follows that N’; = B(7fj). Since No (and B(7fj)) are finite dimensional, NO = N;. Thus N = Uj, and each element of 2lj is a sum of products of the form B1 Bj with B k in Bk. This completes our induction. It follows from the preceding argument that the linear span of the products B1 - .B, (where BI, E Bk and n = 1,2,. . .) is Uj”=, Uj. Thus the norm closure of this linear span is 31.
666
SPECIAL REPRESENTATIONS OF C*-ALGEBRAS
(iii) Note that ~ ( ( U ? ’ ) ~ & Q $and ) ) 50, 1 (j) is the chosen system of 2 X 2 matrix units for Bj. Thus Q ~ ’ ) , & ) , &) generate B j as an algebra. As noted in (ii), { B j :j = 1 , 2 , . . .} generates U as a C*-algebra. (Indeed, we have shown that {up),&’ : j = 1,2,. .} generates U as a C’dgebra.) (iv) Since &),@,up) E for all j
.
At the same time, (&) = 0, so that Cj” = 0 (and CjCj CjCj = 0 ) . Again from (w), when j < k,
Thus, when k < j, ckc; = -c;ck, and taking adjoints, -C;Cj. Finally, observing that
+
cjc; =
+
we have that CjC; CTCj = I, for all j. (v) Noting that C;Cj = ~(QL” - @ ) * ( o x( j ) - i@) and computing with the 2 x 2 matrix representing the right side of this equation, we have that 2C;Cj - I = up). From the relations
it follows that
Thus { C l , .. . , C j } generates the same C*-algebra as { B l , . . . , f ? j } , which from (ii), is 2lj. Again from (ii), {C1,C2,.. .} generates U as a C*-algebra.
667
EXERCISE 10.5.89
.
(vi) From (v), {Cl,. . ,Cj} generates the finite-dimensional C*algebra Uj. From (iv), Cj = 0 so that Uj is the linear span of the set of (finite) products of distinct elements of {C,, , . . ,Cj , C i , . . . ,C,t}. By using (iv) (the CAR), we see that each such product is a linear combination of elements of W . Thus !?lj is the linear span of W . It remains t o show that W is linearly independent. Suppose
Among the terms of the sum on the left side of (***) with nonzero coefficients, choose one with h k as small as possible. Let ~ , ( ~ ) . . . ~ ( h ) ; ~ ( l ) , , , ~(f ( k )0) be its coefficient. Let A be the element q p ( 1 ) - q y j - h ) and B be Cn,(l) Cn,(j-k), where
+
---
{m’(l),. ..,m‘(j - h ) } = {l,..., j } \ (rn(1) ,”. ,m(h)} (n’(1) ,...,n y j - k ) } = (1 ,...,j}\{n(l) ,.”, .(k)}.
Multiplying both sides of ($*$)on the left by A and on the right by B and using (*), we conclude that Um(1). . . , ( ~ ) ; ~ ( l ) . . . ~ ( kC,*C2 )C~C * ..Cj’Cj l
= 0.
From (v), C,*Ckis a non-zero element of D k . If we view U in a representation on a Hilbert space, each f?k is a factor of type 12 and the algebra generated by {Djh : h # k} is contained in Bb. Theorem 5.5.4 now tells US that CiCI C,*C2,CiC1 C,*C2C;C, ,...,CTC1 * * * C;Cj are non-zero. It follows that 0 = ~ , ( l ) . . . ~ ( ~ ) ; n ( l ) . . . ~ (#k ) 0 - a contradiction. Thus W is a basis for Uj. The linear independence of W can also be proved by showing that W has 2 2 j elements since W generates Uj linearly and Uj has H linear dimension 22j. With the notation of Exercise 10.5.88, let KO be a 10.5.89. representation of U on the Hilbert space 3t. Then {ro(Cj)}is a family of operators on ‘H that satisfy the CAR. We say that a family of operators acting on a Hilbert space and satisfying the CAR is a “representation of the CAR.” Thus each representation of the CAR algebra 21 gives rise to a representation of the CAR. Show that each representation of the CAR arises, in this way, from a representation of the CAR algebra.
668
SPECIAL REPRESENTATIONS OF C”-ALGEBRAS
Solution. Let {A(Cj)} acting on the Hilbert space 3.1 be a representation of the CAR. Fkom Exercise 10.5.88(vi),
defines a linear mapping rj of rUj into the C*-algebra CUo generated by {n(Cl), n(Cz), .}. We show that rj is a * homomorphism. For this, it will suffice to show that nj(AB) = Aj(A)Aj(B) for each pair of element A and B in W. Now rj(AB) is computed by expressing A B as a linear combination of elements of W, and this is effected by means of the CAR applied to the juxtaposed terms A and B, and then replacing each Ch by ?r(Ch). Precisely the same sequence of anticommutation operations can be applied to the oper) r(Ck)* in the juxtaposed terms A ~ ( A and ) ?rj(B) ators T ( C ~and since {Aj(Cj)} acting on H satisfy the CAR (by assumption); the result will be nj(AB). Thus rj(AB) = nj(A)nj(B) and nj is a * homomorphism of Mj into Mo. But M j is a finite factor and is therefore simple by Corollary 6.8.4. Thus A j is a * isomorphism and, by Theorem 4.1a8(iii), is an isometry. Since A j and ~ j + agree l on l on the (algebraic) generators {el,..,Cj} for Mj, T j and ~ j + agree Mj. It follows that the family of mappings {rj} defines an isometric * isomorphism of Ug121jinto !& which extends (uniquely) to an (isometric) * isomorphism A of M into G.By construction, r(Cj) is the image of Cj under the * isomorphism A so that A(%) = 210 and {T(Cj)} is the representation of the CAR arising from the representation K of the CAR algebra 2l. .[33(pp. 269-275)]
..
.
10.5.90. With the terminology of Exercise 10.5.89, we call a representation (A(Cj)} of the CAR on a Hilbert space 3.1 “irreducible” when no closed subspace of 31 other than {0} and 3t is invariant under all the operators in (~(Cj),r(Cj)*}.Show that there are an uncountable infinity of (unitarily) inequivalent irreducible representations of the CAR.
Solution. Suppose
{~(cj)} is a representation of the CAR on
H. By virtue of Exercise 10.5.89, we can “extend” A to a representa-
.
: j = 1,2,. .} generates M as a tion of a on 7-1. Since {r(Cj),~(Cj)*
EXERCISE
10.5.91
669
Banach algebra, {n(Cj), n(Cj)*} has no closed invariant subspaces other that (0) and ?-l if and only if the same is true of ~(9). Thus {7r(Cj)}is an irreducible representation of the CAR if and only if 7r is an irreducible representation of 2l. Moreover, two representations n1 and 7r2 of 2l on H Iand ’H2, respectively, are (unitarily) equivalent if an only if there is a unitary transformation U of ‘R1 onto ‘Hz such that UrI(Cj)U* = 7r2(Cj) for all j in { 1 , 2 , . . .} - that is, if and only if {nl(Cj)} and {7r2(Cj)} are (unitarily) equivalent representations of the CAR. From Example 10.4.19, there are an uncountable infinity of inequivalent irreducible representations of 2l and these give rise to an uncountable infinity of inequivalent irreducible representations of the CAR. With the notation of Exercise 10.5.3, show that 10.5.91. coU(R) # (R)1 if R is not finite. Conclude that coU(R) = (R)1 if and only if R is finite. [Hint. Use Theorem 7.3.1.1
Solution. If R is not finite, there is a projection E in R different from I and equivalent to I . Let V be a partial isometry in R such that V*V = I and V V * = E (# I ) . From Theorem 7.3.1, V is an extreme point of (R)1. Thus V # coU(R) unless V E U ( R ) . But V is not a unitary operator. Thus coU(R) # (R)1when 72 is not finite. From Exercise 10.5.3(iv), coU(R) = (R)1 when R is finite. Thus coU(R) = (R)1 if and only if R is finite. Let 2l be a C*-algebra, U(2l) its unitary group, 10.5.92. coU(2l) the convex hull of U(U), and S an element of 2l such that llSll < 1. Show that (i) each invertible element of (2l)l is the midpoint of two elements of U(2l) [Hint. Use the proof of Theorem 4.1.7 and “polar decomposition.”]; (ii) for each U in U(U), ( S + U ) / 2 is the midpoint of two elements of U(2l); (iii) for each U in U(U), U ( n - 1)s = C;=,u k for some U1,. . . ,U , in U(U); (iv) S = n-l u k for some U1,. . . , U , in U(U) when llSll < 1 - 2n-1 (we say that S is the mean of the n unitary elements U 1 , . . . ,Un)[Hint. Note that ( n - l)-’(nS - I ) has norm less than 1 and use it in place of S in (iii)-use I in place of U.]; (v) the open unit ball of U is contained in coU(U) and conclude
+
670
SPECIAL REPRESENTATIONS OF C*-ALGEBRAS
again (see Exercise 10.5.4) that [coU(U)]= = (a),; (vi) each T in U is some positive multiple of a sum of three unitary elements of 31.
Solution. (i) If T is an invertible element of (%)I and V H is its polar decomposition, then H = (T*T)lI2 E U and V = T(T*T)-’l2E U(U). The element H i ( l - H2)’I2 is in U(U) and H = ( U f U * ) / 2 , where U = H i ( l - H 2 ) ’ I 2 . Thus T = (VU V U * ) / 2 with V U and VU* in U(U). (ii) Since (S U ) / 2 = U(U*S 1 ) / 2 and IlU*Sll = llSll < 1, (U’S t 1 ) / 2 and (S t U ) / 2 are invertible elements of (%)I. From (i), (S -/- U ) / 2 is the midpoint of two elements of U(U). (iii) From (ii), there are elements U‘ and U“ of U(U)such that U+S = U’+U’’. Thus, for each positive integer n,there are elements U 1 , . . . ,U n and h,. . . ,Vn-1 (= U n ) of U ( U ) such that
+ +
+
+
+
u t (.
- 1)s= u1 t v1
+ ( n- 2 ) s
= u, t u, t v, t ( n - 3 ) s n
k=l
(iv) If llSll
< 1 - 2 n - l , then
If we substitute ( n - l ) - ’ ( n S - I ) for S and I for U in (*), we have that S = n-l U k , with Uk in U(U). (v) If S is in the open unit ball of U, then llSll < 1 - 2n-’ for all sufficiently large n. Thus S is a mean of n elements of U(U)from (iv), and S E coU(U). Since the open unit ball of U is norm dense in (a), and (%)I is norm closed, [coU(U)]= = (%)I. (vi) If llSll < (= 1 - 2(3)-’) then S is a mean of three unitary elements of U, from (iv). If T E U and E is a positive real number, then 11(311Tll t 3&)”TIJ < so that T = (lITll t E ) ( U It U2 t U 3 ) for some U1,U z , U3 in U(U). m[68]
3
10.5.93. With the notation of Exercise 10.5.92, show that (i) +[u(u) t U(U)]= [&(a) t (1 - ~)u(U): o Ia I I];
671
EXERCISE 10.5.94
(ii) when 2l is a von Neumann algebra, the sets in (i) coincide with { U H : U E U(U), 0 5 H 5 I , H E U}.
[Hint. Note that, with U1, U2 in U(U), U1+ U2 = U l ( I t U i U 2 ) and I + UiU2 is a normal operator.]
Solution. (i) It suffices to show that the convex combination aU1 t (1 - a)U2 is the midpoint of two elements of U(2l) when < a < 1 and U1, U2 E U(2l). In this case,
and
Ila-l(l - a)U;U211 = a - l ( l - a ) < 1.
+
Thus I a - l ( l - a)U,*U2 and aU1 t (1 - a)U2 are invertible. In addition, we have aU1 + ( 1 - a)U2 E (%)I. From Exercise 10.5.92(i), aU1 t (1 - a)U2 is the midpoint of two elements of U(U). (ii) As noted in the solution to Exercise 10.5.92(i), with H in (2l)tand U in U(U),both H and U H can be expressed as midpoints of two elements of U(U). Conversely, (U1 t U 2 ) / 2 = U l ( I t U i U 2 ) / 2 . Now ( I + U i U 2 ) / 2 is a normal element of U and generates an abelian von Neumann subalgebra A of U. The polar decomposition Vh' of ( I U i U 2 ) / 2 has its components V and K in A. In particular V*V and V V * are the same projection E of A. Hence I - E V is a unitary operator W in A U) and ( I t U i U 2 ) / 2 = W K . Thus rn (U1 U 2 ) / 2 = UlWIP, U1W E U(U),and 0 5 K 5 I .
+
+
+
(c
10.5.94. Let 2l be a C*-algebra. Show that each convex combination of 2" or fewer elements of U(U) is a mean of 2" elements of U(U). [Hint.Use Exercise 10.5.93(i) and argue by induction.]
xi:1
Solution. Suppose that T = ajUj, where 0 5 a j 5 1, aj = 1, and U j E U(U).If n = 1, then T is the mean of two elements of U(2l) by Exercise 10.5.93(i). We argue by induction on n . Suppose that n 2 2 and that our assertion is known for n - 1. Let 2"-1 a be Cjz1 a j . We may assume that a # 0 # 1 - a , for otherwise we are dealing with a convex combination of 2"-' or fewer unitaries and our inductive hypothesis applies. (Note that r - l EL=,Vk =
xi:,
672
SPECIAL REPRESENTATIONS OF C*-ALGEBRAS
+
From Exercise 10.5.93(i), UWj (1 - u)Vj = (Ujl+Uj2)/2 for some Ujl and Uj2 in U(%).Hence T = 2-, cjl;'(Ujl t Ujz), and T is a mean of 2n unitary elements of 2l.
10.5.95. Let 3.1 be a Hilbert space and V be an isometry of 'H (V*V = I ) such that VV* = E < I . Show that (i) s p V is the closed unit disk in C [Hint. Let eo be a unit vector in (I - E ) ( ? t ) , e k be Vkeo for Ic in {1,2,. . .}, and argue, as in Example 3.2.18, that X is an eigenvalue for V * when 1x1 < 1.1; (ii) if U 1 , . . ,U n are unitary operators acting on 'H, then
.
c n
IIV - n-l
U,ll 2 2n-'
k=l
[Hint. Assume the contrary, study IlUiV - n-'IIl, and use (i).]; (iii) S, is a mean of n, but not fewer than n, unitary elements of R when V lies in the von Neumann algebra R and S, = a,V, where 1 - 2(n - 1)-l < a, < 1 - 2n-*
[Hint. Use Exercise 10.5.92(iv) and (ii).]; (iv) there is no number n such that each element of (R)?can be expressed as a convex combination of n or fewer elements of U(R), when R is an infinite von Neumann algebra and
(R)?= { A E R : IlAll < 1). Solution. (i) With the notation of the hint, we have that V*eo = 0 since eo is orthogonal to the range of V . Also V*ek = V*Vek-l = e k - 1 when C E {1,2,. . .}. If 1x1 < 1, then
k=Q
k=O
EXERCISE 10.5.96
673
x
Thus X E spV' and E spV. Since s p V is closed and llVll 5 1, sp V is the closed unit disk in @. u k ( l < 2n-l. Then (ii) Suppose IIV - n-l
xizl
But (UiV)*(UiV) = V*V = I and (UiV)(UiV)* = U,*VV*Ul < I, so that UiV is a "non-unitary" isometry. Thus, from (i), we have -(1 n-') E sp(U,*V - n-'I), and 1 n-' 5 IlU,'V - n-'IIl - a contradiction. Thus llV - n-l Ukll 2 2n-'. Since IISn(( = a , < 1 2n-l, S, is a mean of n unitary (iii) elements of R from Exercise 10.5.92(iv). Suppose Sn = T - ~ Uk with U 1 , . . . ,U,. unitary elements of R. Then
+
xi=,
+
from (ii). Hence T 2 2(1 - a,)-'. But 1 - 2(n - l)-' < a,, by assumption, so that n - 1 < 2(1 - u,)-l 5 T . Since T is an integer, n 5 T , and Sn is a mean of no fewer than n unitary elements of R. (iv) Since R is an infinite von Neumann algebra, it contains a non-unitary isometry V. With S, as in (iii), we see that S, E (R)? and S, is the mean of no fewer than n unitary elements of R. 10.5.96. Let U be a C*-algebra, U(U) be its unitary group, and A be a self-adjoint elelement of 2l. Let S, be [-1, -(1 - 2a)] u [(l- 2a),1],
where 0 5 u 5 f. Show that (i) sp A C_ S, if A = aU1 (1 - u)U2 for some UI and U2 in U(U) [Hint. With X in sp A , use Exercises 4.6.16 and 4.6.31 to find a state p of 2l, definite on A , such that p ( A ) = X.]; (ii) ( is a continuous mapping of S, into Cf such that
+
+
11 u ( l - a)-l((t)l = (t(1- u)-lI,
674
SPECIAL REPRESENTATIONS OF C*-ALGEBRAS
where Ct is the set of complex numbers of modulus 1 with nonnegative imaginary part,
for t in S,, d ( r ) is the (unique) element of Ct with real part T , and g is the inverse mapping to t + 11 a(1 - a)-le(t)l of [-1.11 onto the closed interval [(l- 2a)(l - u)-',(l - a ) - ' ] ; (iii) fi and f2 are continuous mappings of S, into the complex numbers of modulis 1 and af1 ( t ) (1 - a)f2 ( t )= t for t in S,, where fl(t) = t i ( l - t2)'12 and f 2 ( t ) = t - i ( l - t2)l12 when a = 51 , f2(t) = t(1 - u ) - l ( l + a(1- a)-l[(t))-l and fl(t) = fi(t)r(t) when
+
+
+
a#
4;
(iv) A = aU1 SPA C S a ; (4
+ (1 - u)V2
for some U1 and
U2
in U ( U ) when
&(a) + (1 - a)U(U) = {UH : whenOiaS (vi)
u E U(U),H
= H* E U, s p a E S,}
i;
d ( U ) t (1 - .)U(U)
c U(U) t (1 - b)U(U)
when 0 5 a 5 b 5 $; in particular, each convex combination of two unitary elements of U is a mean of two unitary elements of U. (See Exercise 10.5.93(i).)
Solution. (i) If A = aU1 t (1 - u)U2 with U1 and U2 in U(U), then IlAll 5 1. Suppose X E SPA. From Exercise 4.6.16 and 4.6.31, there is a state p of U, definite on A, for which p(A) = A. We have
Hence spA c Sa. (ii) Since 11 t a(1 - a)-ld(t)l is the distance from the point -1 in C to the point a ( 1 - u)-'d(t) (whose real part is a ( 1 - a ) - l t ) on the circle with center 0 and radius a(1 - a ) - ' , the mapping
t + 11
+
- a)-le(t)l
675
EXERCISE 10.5.96
is a continuous (increasing) one-to-one mapping (hence, homeomorphism) of [-1,1] onto [(l- 2 a ) ( l - u ) - ' , ( l - a)-']. If t E Sa, then (1 -a)-'. 2 It(1-
.)-'I 2 (I - 2 a ) ( l -
a)-'.
Thus g(lt(1 - a)-'l) E [-1,1] and ( ( t ) = e(g(lt(1 - .)-'I)) Moreover, t is continuous and
E
Ct.
3,
(iii) When a = Sa = [-1,1] and t = [ j l ( t )t f 2 ( t ) ] / 2for each t in [-1,1], with f1 and f2 as defined. If a # 1 f 2 ( t ) , as defined, is a complex number of modulus 1, from (ii), for t '.in S,. Moreover, fi is continuous on S,. Also, Ifl(t)l = IA(t)((t)l = 1 and f1 is continuous on S,. In addition, u f l ( t ) (1 - a)f2(t) = t for each t in S,. (iv) If sp A C Sa, fl(A) and f2(A) axe unitary elements of !2l such that A = a f l ( A ) t (1 - a ) f z ( A ) ,where f1 and fz are the functions defined in (iii), from Theorem 4.1.3 and the properties of f1 and f2. We may take f l ( A ) for U1 and fz(A) for U2. (v) If T = U H with U in U(8)and H a self-adjoint element of !2l with the spectrum in S O , then H = aU1+ (1- a ) & from (iv), and
+
T = U H = u U U ~+ (1 - a)UU2 E &(a) t (1 - a>U(%). If 0 5 a < 1/2 and S = a& + (1 - a)% with & and V, in U(%),then V,+S = (1- a)(a(l - a)-'V,*Vl + I ) . Since 0 5 a < 1/2, a(1 - a)-' < 1, whence v,'S and S are invertible. Hence S =
U(S*S)1/2with U in U(2l) and (S*S)'/Z = u*s = aU*V1+ (1 - a)U*V2.
It follows from (i) that sp(S*S)'/'
C S a and
S E ( U H : U E U(%),H = H * E 8, sp H
c S,}.
(vi) If 0 5 a 5 6 5 1/2, S, C sb so that, from (v),
676
SPECIAL REPRESENTATIONS OF C*-ALGEBRAS
Suppose T is an element of the unit ball of a C*10.5.97. algebra U with unitary group U ( U ) and inf{llT - Ull : U E U ( U ) } 5 2a, where 0 5 a I Show that T E d(U) t (1 - a)U(U). [Hint. Use Exercises 4.6.1.6 and 4.6.31 to establish that sp H C [l - 2b, 11 when a < b < where H = (T*T)'/2.]
3.
3,
Solution. If a < b < !j, there is an element U in U ( U ) such that [IT - UII < 2b < 1. Thus 111 - U*TII < 1 so that U*T and T are invertible. Hence T = VH, where V E U(U)and H = (T*T)1/2E U. If X E sp H , there is a state p of U, definite on H , such that p ( H ) = A, from Exercises 4.6.16 and 4.6.31. We have
[ I - p(U*V)Xl = I p ( l - U*VH)I 5 111 - U*VHJI= IIU - TI1 < 2b and 1-2b I Xlp(U*V)l I A. Thus sp H C [1-2b, 11. Since this inclusion holds for each b greater than a, sp(H) [l - 2a, 11. From Exercise 10.5.96(iv), H and, hence, T are in d ( U ) t ( l - a ) U ( U ) . m[68]
10.5.98. Let U be a C*-algebra and S be an element of U such that llSll 5 1 - E , where 0 < E < ( n t 1)-'. Let blV1 t t b,V, be a convex combination of unitary elements V1,. . ,V, of U such that
.
11s - (blK t ..' t b,V,>ll < 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
SPECIAL REPRESENTATIONS OF C*-ALGEBRAS
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 <_ a2, 0 5 a3 < E (5 1/3), and a1 t a2 t a3 = 1. If a1 = 0, then S = azUz(I uz1u3U;U3) and a3 < 1/3 < a2, so that S E U i n v . If 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
<:
727
EXERCISE 12.4.2
(i) we can replace A by a self-adjoint operator in (%)I [Hint. Consider (A A*)/2.]; (ii) llA2 - All < 2a [Hint.Note that A - A2 = A ( I - A) A(I - E ) A(I - E ) - E(I - E).]; (iii) each s in sp A is such that 0 < s2 - s t 2a, 0 < s - s2 t 2a, that s lies in the interval [f - t 2a)'12, f t 2a)'I2], and s is not in [f - - 241/2 ' 21. t - 2(p],
+
+
(i +
(i
(a
(i
(iv) there is a projection F in U such that
I(E - FIJ < a + f -
(a - 2a)'?
[Hint.Consider f(A), where f is 0 to the left and 1 to the right of the second interval described in (iii).] Solution. (i) Since IIE - All < a, we have that IIE - A*[/ < a. Thus I(E - A t E - A*ll = 112(E - ;(A A*))" < 2 ~ .
+
-+
Since ( A A*)/2 E ( U h ) l , we can replace A by (A (ii) Proceeding as in the hint, we have
[[A2- All
+ A*)/2.
I IIA(I - A) - A ( I - E)ll t llA(I- E ) - E ( I - E)ll L IlAllllE - All 4-IIA - EllllI - Ell 5 2a.
(iii) From Proposition 3.2.10, s2 - s E sp(A2 - A) when s E sp A. From Theorem 3.2.3 and (ii), Is(Is -
11 = Is2 - $1 5 "A2 - A/( < 2a.
+
+
Thus 0 < s2- s 2a and 0 < s - s2 2a. Studying the polynomials s2 - s t 2a and s - s2 t 2a, we see that the endpoints of the intervals described are roots and that s lies in one but not the other interval. (iv) With f as described in the hint, f(A) is a projection F in U (assuming, as we may from (i), that A is self-adjoint) from Theorem 4.1.3, and
[ I F-
+
5 maxi(+ -+ 2a)'12 - f, - (i - 2a)'12) = 1 2 - (1 4 - 2a1112.
Hence
[IF - Ell 5 [IF - All t [ [ A- Ell < a
+ + - (i - 2a)'12.
4671
728
APPROXIMATION BY MATRIX ALGEBRAS
12.4.3. Let U be a C*-algebra and S be a subset that generates a dense linear subspace of Q. Suppose that for each finite set of elements A l , .. .,A , in S there is a state p of U satisfying
Show that U has a tracial state.
Solution. With IF a finite subset of S, let 55 be the set of states p of U such that p ( A B ) = p ( B A ) for all A and B in IF. By assumption, 7r is non-null. Moreover, 55 is a weak * closed (hence, compact) subset of U#. With l F 1 , . . . ,IF, a finite collection of finite subsets of S , Irl"..."Fm c Trlfl ' * fl 7rm. Hence Thus
(7i :
IF finite, lF E S} has the finite intersection property.
n{%IF :
finite,
IF c S)+ 0.
Suppose p is in this intersection. Then p ( A B ) = p(BA) for all A and B in S and, hence, for all A and B in the linear span of S. By assumption, this linear span is norm dense in U. Since p is a state, p is norm continuous. Hence p ( A B ) = p ( B A ) for all A and B in U. Thus p is a tracial state of Q. a 12.4.4. Let !Z be a C*-algebra and {U, : a E A} be a family of C*-subalgebras of U totally ordered by inclusion and such that U a E ~ Uisa norm dense in U. Show that (i) U is a simple C*-algebra if each U, is simple [Hint. Consider the quotient mapping modulo an ideal and recall that * isomorphisms are isometric.]; (ii) U admits a trace if each 8, admits a trace [Hint. Use Exercise 12.4.3.1; (iii) U has a unique trace if each U, has a unique trace; (iv) 2l is simple and has a unique trace if each U, is a factor of type 111; (v) U is not uniformly matricial if each U, is a factor of type 111.
Solution. (i) If Z is a two-sided ideal in U distinct from U, then Z=, the norm closure of 1,is distinct from U and the quotient
EXERCISE 12.4.5
729
mapping cp of 2l onto the C*-algebra %/Z' is a * homomorphism. Since each Ma is simple, Q I 2 t a is a * isomorphism and (from Theorem 4.1.8(iii)) an isometry for each a. Since Q is (norm) continuous on 0 and isometric on the norm-dense subalgebra U%, Q is isometric on 2l. Thus I= = (0), and 2l is simple. (ii) Let S be U?21aand let A * ,..., A , be a finite subset of S. Since {a,} is totally ordered by inclusion there is some 2l, containing A1, ... ,An. By assumption, Ma has a tracial state Tam Let p be a state extension of 7, to 2l. Then p ( A j A k ) = p(AkAj) for all j and k. Since S is norm dense in I#, the conditions of Exercise 12.4.3 are fulfilled, and M admits a tracial state. (iii) Let T~ be the unique tracial state of a,. Note that each trace r' on 2l coincides with r, on a,. By continuity of r' and 7,r' = T , where T is the trace on 24 constructed in (ii). (iv) From Corollary 6.8.4, each factor of type 111 is simple, and from Theorem 8.2.8 each such factor has a unique tracial state. Applying (i) and (iii), we have that U is simple and has a unique tracial state when each a, is a factor of type 111. (v) As noted in the comments preceding Theorem 12.1.9, each matricial C*-algebra is norm separable. It will suffice to show that Ma is not norm separable. Since %, is of type 111, 8, contains a countably infinite orthogonal family {E,} of non-zero projections. With A a subset of N, let E A be '&*En. Then ( E A } is an uncountable family of projections in 3, and l l E ~- EL,((= 1 when A # A'. Thus 21a is not norm separable and 2l is not matricial. 12.4.5.
Let M be a factor of type 111, E a projection in M, an operator in (N)1 such
JV a finite type I subfactor of M, and A that
llA - Ell2 < b < Show that there is a projection M in JVsuch that
[Hint.See the proof of Lemma 12.2.3.1 Solution. The beginning of the proof of Lemma 12.2.3 applies without change (but replace Mo,there, by M and delete references to B).
730
APPROXIMATION BY MATRIX ALGEBRAS
12.4.6. Let M be a factor acting on a Hilbert space 3-1. (i) Show that a finite orthogonal family of projections { E l , . ,Em} with sum I lies in a finite type I subfactor of M if and only if each Ej is a finite sum of monic projections in M . (See the introduction to Section 8.2. We say that Ej is a rational pmjection in M . ) [Hint. When M is infinite, use Exercise 6.9.4 to show that each Ej I . When M is of type 111, use Lemma 6.5.6.1 (ii) Suppose N i s a finite type I subfactor of M , E is a projection in M , and B is an operator in N such that IIE - Bll < 1/8. Show that E is rational. [Hint. Use (i) and Exercises 12.4.1 and 12.4.2.1 (iii) Show that a normal operator in M lies in a finite type I subfactor of M if and only if it is a finite linear combination of orthogonal rational projections with sum I . (We call such an operator rational. )
..
N
Solution. (i) Suppose N is a factor of type I, and Ej E N M for each j. Then I is the sum of n minimal projections in N . Now all minimal projections in N are equivalent (in N, hence, in M ) , from Corollary 6.4.7. Thus each minimal projection in Af is monic in M . From Corollary 6.5.5, each Ej is a finite sum of minimal projections in N . Hence each Ej is rational in M . Suppose, now, that each Ej is rational in M . If we replace each E j by a finite sum of monic projections, it will suffice to show that the total family of monic projections so obtained lies in a finite type I subfactor of M . We may assume, therefore, that each Ej is monic. Let I be E !i E ( j ,Ic), where E ( j ,Ic) Ej for all j . If M is a finite factor of type I, there is nothing to prove. If M is infinite, then each Ej is equivalent to I since a finite number of projections in M equivalent to Ej have sum I . To see this, note that if E ( j , 1) is finite, then E ( j ,2),. . . ,E ( j , n ( j ) )are finite from E ( j , n ( j ) ) )is finite, from Proposition 6.3.2,and I (= E ( j , 1) t Theorem 6.3.8- contradicting our assumption. Thus Ej is infinite. Using the result of Exercise 6.9.4, we see that N
---+
Ej
N
E ( j , 1)
N
E ( j , l)+E(j,2)
N
E ( j , l)+E(j, 2)+E(j, 3)
N
* * * N
1.
It follows that, when M is infinite, all the Ej are equivalent. From Lemma 6.6.4,there is a self-adjoint system of m x m matrix units for which the projections E l , . . , , E m are the principal units. The linear span of these matrix units is a type I, subfactor of M containing El,. . .,Em.
731
EXERCISE 12.4.7
There remains the case where M is a factor of type 111. Let be n ( l ) . . . n ( m ) and ~ ( jbe) ~ / n ( j ) . From Lemma 6.5.6, Ej = F ( j ,k), where F ( j , 1) F ( j , ~ ( j ) )If. T is the (unique) tracial state on M , then
x?:!
1 = T(I) =
c
T ( E ( j ,k)) =
k= 1
whence
T(E j )
N
c
-
T
N
T(Ej),
k=l
= n ( j ) - l . Hence
and
Hence { F ( j , k ) : j E { 1 , . . . ,m } , k E (1,. ..,~ ( j )is} an orthogonal family of equivalent projections in M with sum I. Again, from Lemma 6.6.4, they lie in a finite type I subfactor of M as do E l , . .., E m . (ii) From Exercise 12.4.2 there is a projection F in N such that [ ( E- F(I < 5 / 8 . From Exercise 12.4.1, there is a unitary operator U in M such that UEU' = F . From (i), F is rational. Hence E is rational . (iii) If N is a finite type I subfactor of M and A is a normal operaa j E j , where { E l , . . . ,E n } is an orthogonal tor in N , then A = family of projections in N with sum I (since JV is * isomorphic to the algebra of complex T x T matrices for some integer T ) . From (i), each Ej is a rational projection in M . Suppose A = ~ ~ bjFj = for1 some orthogonal family { F l , . . . ,Fm} of rational projections with sum I. From (i), F1,. . . ,Fm lie in some finite type I subfactor of M . Of course, A is in this subfactor.
x;==,
12.4.7. Let M be a factor and { E l , . . . , E m } be an orthogonal family of rational projections (see Exercise 12.4.6) in M with sum I. Let
732
APPROXIMATION BY MATRIX ALGEBRAS
be a self-adjoint system of n ( r ) x n ( r ) matrix units in E,ME, such that Cyc,' E ( j , j ,T ) = E,. (i) Suppose M is infinite. Show that each E ( j , j , r ) I and that there is a finite type I subfactor of M containing all E ( j , C , r ) . [Hint. Use Exercise 6.9.4 and Lemma 6.6.4.1 (ii) Suppose M is finite. Show that there is a finite type I subfactor of M containing all E ( j , k, r ) . [Hint.When M is of type 111, use Lemma 6.5.6 to express E( 1, 1, T ) as a sum of equivalent orthogonal projections such that all are equivalent as T varies. Use Lemma 6.6.4 to replace { E ( j ,C, r ) } by a system of matrix units in E,M E, containing the subprojections of E( l , l , r ) for each r. Now use the argument for (i) with the new matrix unit systems.] N
Solution. (i) Since M is infinite, each rational projection in M is equivalent to I (as noted in the third paragraph of the solution
to Exercise 12.4.6(i)). It follows that E ( j , j , r ) are equivalent for all j and r. Let V, be a partial isometry in M with initial projection E(1, 1, 1) and final projection E(1,1, T ) for r in (2,. .. ,m } and let V1 be E ( l , l , 1). Then
is a system of projections and partial isometries satisfying the conditions of Lemma 6.6.4 and is, therefore, contained in a self-adjoint system of matrix units whose linear span is a finite type I subfactor N of M . By construction of this system of matrix units for N (as indicated in Lemma 6.6.4), each E ( j , k , r ) is in the system. Hence E ( j ,k, r ) E N for all j , C, and T . (ii) If M is finite and of type I, there is nothing to prove. We may suppose that M is of type 111. Let T be the (unique) tracial state on M . By assumption, T(E,) = p ( r ) / q ( r ) ,where p ( r ) and q ( r ) are (positive) integers for each r in ( 1 , . ..,m}. Hence ~ ( E ( j ,rj),) = p( r ) / q (r ) n ( r )for each j in { 1,. . . ,n( T ) } and each T in { 1,. . . ,m}. Let s be the least common multiple of q( l)n( l ) , . . . ,q(m)n(m)and let t ( r ) be sp(T)/q(r)n(r). From Lemma 6.5.6, there are t ( r ) equivalent subprojections F ( 1, 1, r ) ,. . ,F( 1, t ( r ) ,r ) of E( 1,1, T ) in M with sum E ( l , l , r ) and each has trace p ( r ) / q ( r ) n ( r ) t ( r (= ) 8 - I ) . Let V ( j ,T ) be a partial isometry in M with initial projection F ( 1,1, r ) and final projection F ( 1, j, T ) for each j in (2,. . . ,t ( r ) } and let V(1,r) be F ( l , l , r ) . Then E(k,l,r)V(j,r)E(l,C,r) is a partial
.
733
EXERCISE 12.4.8
isometry V ( k , j ,T ) with initial projection F ( k , 1, T ) and final projecMoreover, tion F ( k , j , r ) subprojections of E ( k , k , r ) in M . F ( k , j , T ) = E ( k , k,T ) . The system of projections and partial isometries
cfl'!
satisfies the conditions of Lemma 6.6.4 and is contained in a finite self-adjoint system of matrix units that generates a (finite type I) subfactor of &ME,. Moreover, E ( j ,b, T ) is in this subfactor for j , k in ( 1 , . . . , n ( r ) } . If we replace the system { E ( j , k ,T ) } by the system of matrix units (in E T M E Tjust ) constructed for all T in ( 1 , . . . ,m}, we are in the situation treated in (i); all the principal matrix units are equivalent (they each have trace s - ' ) and their sum is I . Again, there is a finite type I subfactor nl of M containing all these system of matrix units and, hence, containing all E ( j , k, T ) . 12.4.8. Let E l , . . . ,En be a commuting family of projections in a factor M of type 111 acting on a Hilbert space 'H. Let 2 1 , . . . ,x, be vectors in 'H and E be a positive number. Show that (i) there is an orthogonal family { F l , . . ,F,} of projections in M such that each Ej is a finite sum of projections in { F l , . . . ,F T } ; (ii) with {Fl,.. . ,FT} as in (i), there is a family of rational projections (see Exercise 12.4.6) { G I , . . ,G T }in M such that Gj Fj and
.
<
ll(Fj - G j ) x k ( ( < E / T (iii) N1,.
( j E { l , . . . , ~ } k, E { l , - - - , m } ) ;
there is a finite type I subfactor in N such that N j Ej and
. . ,N n
<
N of M and projections
Solution. (i) Suppose we can find the family {F;, . . . , F l } for {El,.. . ,E k } , where k < n, such that each Fj' commutes with each E l , . . . , E n . When b = 1, { F i } is such a family, where F: = E l . Then
734
APPROXIMATION BY MATRIX ALGEBRAS
is such a family for { E l , . . . ,Ek+l}. In this way, we construct the familyfor { E l , . . . , & } . (ii) Let r be the (unique) tracial state on M . Since M is a factor of type 111, there is an increasing sequence {Fjk} of subprojections of Fj in M such that r ( F j k ) is rational for each k and r ( F j k ) --+ 7 ( F j ) as k --+ 00. From Exercise 8.7.3(iii), { F j k } is strong-operator convergent to Fj. Thus, we can choose k’ so that
Let Gj be Fjk) for each j in {l,...,T } . Since r ( G j ) = r(Fjkt) and r(F’k) is rational for all j and k , { G I , . .. , G r } is an orthogonal family of rational projections in M . G j ( t ) ,where E j is (iii) With the notation of (ii), let N j be CtZ1 4 j ) Fj ( t ) . (Recall that each Ej is a sum of projections in { P I , .. . ,Fr} from (i).) Then N j 5 Ej and
for j in (1,. . . ,n} and k in (1,. Thus n
r(I -
E
Nj)
j=1
.. ,m}. Each Nj is rational in M . n
= 1-
C7(Nj) j=1
and I - C;==,Nj is rational. From Exercise 12.4.6, we have that { N l , . ,N n , I - Cj”=, N j } is contained in a finite type I subfactor
..
ofM. 12.4.9. Let M be a factor of type 111 acting on a Hilbert space ‘H. Let { E l , .. , E n } be an orthogonal family of projections in M and W j ,k,T ) : j , k E (1,. , n ( W
.
..
be a self-adjoint system of n ( ~x )n ( r ) matrix units in ETME, such E ( j , j , T ) = E,.. With E a positive number and 21,.. . ,x, that 3. vectors in ‘H, show that there are a finite type I subfactor N of M and operators F ( j ,k,T ) in N such that
xvc)
735
EXERCISE 12.4.10
.
for all j, k in ( 1 , . . . , n ( r ) } ,h in ( 1 , . . , m } , and
T
in { I , . . . ,n}, and
is a self-adjoint system of n ( r )x n ( ~matrix ) units in F,.MF,, where F,. = CyF! F(j,j,.) and {F,. : T E ( 1 , . . . , n } } is an orthogonal family of projections in N . [Hint. Use Exercise 12.4.8 to find rational subprojections F(1, 1, T ) of E(1,1,.). Construct F ( j ,k , T ) from E ( j , 1 , ~ and ) F ( l , l , r ) . Use Exercise 12.4.7 to find N.]
Solution. By applying the result of Exercise 12.4.8, choose a rational subprojection F(l, 1, T ) of E(1,1, T ) in M for each T in { 1 , . . . ,n } such that
for T in ( 1 , . .. ,n } , b in ( 1 , . . . ,n ( r ) } , and h in ( 1 , . . . ,m}. Let F ( j , 1, T ) be E ( j , 1, r)F(1, 1, T ) and F ( j , b, T ) be F ( j , 1, r ) F ( k ,1, T ) * for j , k in ( 1 ,..., n ( ~ )and } T in ( 1 ,..., n}. Then { F ( j , k , ~ )is} a self-adjoint system of n ( r ) x n ( r )matrix units and Cy$) F ( j , j , T ) is a rational subprojection F,. of E,. It follows from the result of Exercise 12.4.7 that there is a finite type I subfactor N of M containing all F ( j , k , T ) . (Note, for this application of Exercise 12.4.7, that I-C F,. is a rational projection in M . ) Moreover
for j , b in { l , .. . , n ( r ) } ,T in ( 1 , .
. . , n } , and h in { l , .. . ,m}.
12.4.10. Let R be a type I, von Neumann subalgebra of a factor M of type 111, where n is a finite cardinal, and let E be a positive real number. (i) With ( A l , . . . , A m } a finite set of operators in ( R ) 1 , show that there are a finite type I subfactor N of M and operators B1,. . ,B , in (N)1such that
.
736
APPROXIMATION BY MATRIX ALGEBRAS
[Hint. Use the (unique) tracial state on M to represent M on a Hilbert space 3-1 so that for some unit (trace) vector xo in 7-1, llA112 = (IAxoll when A E M . Choose matrix units { E j k } for R such that Ell is an abelian projection in R. With Ah expressed where C ( h , j , k ) is in the center of R, use as Cpk-lC(h,j,k)Ejk, the spectral theorem to find a finite family (91,... ,Q,.} of central projections in R and complex scalars X(h,j, k,t ) such that I
-
Apply the result of Exercise 12.4.9.1 (ii) With A a normal operator in M , show that there are a finite type I subfactor JU of M and a B in hl such that llA - Bll2 < E . (iii) If A1 and A2 are normal operators in M , are there a finite type I subfactor JV of M and operators B1 and B2 in N such that llAj - Bill2 < E ( j E {1,2})? Proof? Counterexample? [Hint. Consider the factor t~~ studied in Theorem 6.7.8, and note that the set of operators in M commuting with a given finite type I subfactor of M is a subfactor of M of type 111.1
Solution. (i) Let C be the center of the von Neumann algebra
{ E j k : j , Ic E ( 1 , . . . ,n } } be a self-adjoint system of n x 7~ matrix units in R such that Cj”=, E j j = I and Ell is an abelian projection in R. (With these choices, E j j is an abelian projection in R since all E j j are equivalent and each has central carrier I since they have sum I.) From Theorem 6.6.5 (especially its proof and that of Lemma 6.6.3), each A in R is (uniquely) expressible as a sum Cyk-lCjkEjk with Cjr, in C. Suppose A h = C,”,-, C ( h ,j , k)Ejk for each h in (1,. . .,m}. From Theorems 3.4.16 ind 4.4.3, there is a finite orthogonal family { Q l , .. . ,Q,.} of projections in C with sum I and complex scalars p t ( C ( h , j , k ) ) (= X(h,j,k,t)),where p t ( Q t ) = 1 and pt is a pure state of C, such that for all h , j , k , and t
R and
I
-
It follows that IlAhQt
- BitII < ~
j,k=l
/ for 2 all h and t , where
EXERCISE 12.4.10
From Example 11.1.7, llBLtI( 5 l l A h Q t l ( 5 1. Letting B i , we have that 11Bi11 5 1 and
737 BLt be
Now { E j k Q t : j,k E (1,. . . ,n}} is a self-adjoint system of n x n matrix units in QtMQt and {Ql,. . . ,Q T } is an orthogonal family of projections in M . Applying the result of Exercise 12.4.9 to M in the trace representation (as described in the hint), we find a finite type I subfactor Jz/ of M and operators Fjkt in N such that { F j k t } is a self-adjoint system of n x n matrix units for each t in (1,. . . ,r}, F j j t } is an orthogonal family of projections in N , and
{X;==,
and Bh E (N)1 for each h in (1,. . . ,m}. (ii) Since A is normal, A , A * ,and I , generate an abelian (that is, type 11) von Neumann subalgebra Ro of M . If we apply the result of (i) to A and Ro in place of A 1 , . . . ,A , and R, we obtain the desired conclusion.
738
APPROXIMATION BY MATRIX ALGEBRAS
(iii) No! Let M be the factor .C32(of type 111) considered in Theorem 6.7.8 and r be the (unique) tracid state on M. It was noted in the proof of Theorem 6.7.8 that there is no unitary operator U in M such that r ( U ) = 0 and
r([u&,j - LajU]*[ULxaj - LxojU])’” < 1/25
( j E {1,2}).
With Lo,, and LXa2for our normal operators A1 and Az, suppose llAj - Bji2 < 1/50 ( j E {1,2}) for some B1 and B2 in a finite type I subfactor N of hi. Let {F’k} be a self-adjoint system of matrix units for N such that CF’j = I. From Lemma 6.6.3, N’ n M is * isomorphic to F11MF11. F’rom Exercise 6.9.16(iv), F11MF11 is a factor of type 111. Hence N‘ n M is a factor of type 111. The restriction ro of r to N’n M is the unique tracial state on N’ n M. If E is a projection in N’ n M such that ro(E) = 1/2, then 2 E - I is a unitary operator U in N’ n M such that ro(U) = r(U) = 0. Moreover, U B j = B j U ( j E {1,2}). Thus
- a contradiction.
I
12.4.11. Let E and F be projections on a Hilbert space 3-1, and let R be the von Neumann algebra generated by E , F , and I. Show that (i) (E - F)2 commutes with E and F; (ii) ERE = CE, where C is the center of R [Hint.Use (i) and consider (E - F)2E.]; (iii) E and I - E are abelian projections in R; (iv) R is either abelian, of type 12, or the direct sum of an abelian von Neumann algebra and one of type 12. Solution. (i) Note that
E ( E - F ) 2 = E ( E - E F - F E + F ) = E - E F - EFE = E - EFE = ( E - EF - F E + F)E = ( E - F)2E. By symmetry, F commutes with (E - F ) 2 .
+ EF
EXERCISE 12.4.12
739
(ii) From (i), (E- F)2is an element C of C and C E = E-EFE. Thus E F E = ( I - C ) E . It follows now that if A lies in the subalgebra of R generated (algebraically) by E , F , and I , then E A E E CE. Hence ERE = C E . (iii) From (ii), E is an abelian projection in R. Since R is generated (as a von Neumann algebra) by I - E , F , and I , we now have that I - E is abelian in R. (iv) With Q a non-zero central projection in R, Q is the sum of the two abelian projections Q E and Q ( I - E ) . From Theorem 6.5.2, R has the desired form. 12.4.12. Suppose % is generated as a C*-algebra by two projections and I . Let x be an irreducible representation of % on a Hilbert space H . Show that 3-1 has dimension at most 2. [Hint. Use 12.4.11.1
Solution. If E and F are projections in % such that E , F , and I generate %, then x ( E ) and x ( F ) are projections in B(3-I). Moreover, x ( E ) , x ( F ) , and I generate B ( H ) since x(%) acts irreducibly on H and x ( E ) ,x ( F ) ,and I generate x(%) as a C*-algebra. (See Theorem 5.4.1.) From Exercise 12.4.11, B(3-1) is either abelian or is a von Neumann algebra (factor) of type 12. Hence 'H has dimension at w most 2. 12.4.13. Let R be a von Neumann algebra that has no minimal projections. With E a projection in R and w a normal state of R such that w ( E ) > 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<) 1/2. Then 1/2 < w‘(E - Ea/2) and E - Ea/2 has a subprojection GOin E R E such that 0 < ”(Go) < 1/2 - ~ ’ ( E a / 2 ) (by applying the result proved in the preceding paragraph with 1 / 2 w‘(Ea/2)in place of E ) . Hence {Mb,GO}bEl E S,which contradicts the maximality of { M b } b E B . It follows that w’(Ea/2) = 1/2 and w(Ea/2) = a/2. Next, we apply this same procedure to the restrictions of w t o Ea/2RE,12 and ( E - E a / 2 ) R ( E- Ea/2)to produce projections Ea/4 and Fa/4,respectively, in R such that Ea/4 < Ea/2, Fa/4 < E - Ea/2, and w ( E + ) = w(Fa/4)= a/4. Let E3a/4be Ea/2+Fa/4.Continuing in this way, we construct a family { E a r }of projections in ERE such that W(Ear) = ar and Ear 5 Ear, when T 5 T ’ , where T and T’ are dyadic rationals in [0,1]. With s in [0,1], let Ea, be the union of all projections Ear such that T 5 s and T is a dyadic rational in [0,1]. Then w(E,,) = as since w is normal. The family {Ex : 0 5 X _< w ( E ) } has the desired properties.
741
EXERCISE 12.4.14
12.4.14. In solving (i), (ii), and (iii), make the following (inductive) assumption: (*) If R is a von Neumann algebra that has no minimal projections, then for each set of n normal states w 1 , . . . ,w, of R such that w1 has support I , there is a projection E in R such that w l ( E ) = . - - = w n ( E )= (i) Show that if F is a projection in R such that w l ( F ) = - . = w n ( F ) ( = a ) , where w 1 , . . . ,wn are as described in (*), then there is a family { F A : 0 5 X 5 a } of projections FA in R such that Fo = 0, Fa = F , w l ( F x ) = = on(Fx)= X for each X in [O,a], and FA 5 FA, when X 5 A'. [Hint.Use (*) to construct Fa, for each dyadic rational T in [0, I] by considering the restrictions of 0 1 , . . . ,on to FRF.] w(Fx) is continuous (ii) With the notation of (i), show that X on [O,a] for each normal state w of R. [Hint.Consider the faithful representation of R corresponding to w1, in which w1 = o,lR with 2 a generating and separating vector for R. Note that w = w,lR and that y = lim ALz for some sequence { A ; } in R'.] (iii) With w1,. . . ,wn,w normal states of R such that w1 has support I , show that there is a projection M in R such that u l ( M ) = ... = w n ( M ) = u ( M ) = f. [Hint.With E and I - E in place of F , and { E x } and { F A } ,respectively, the families of projections in R described in (i), note that the function X -, w ( E ~ / z F ( 1 - ~ ) / 2on ) [0,1] is continuous and takes the value (iv) Conclude that (*) holds for all n in N. . . ,wn are normal states of R (do not as(v) Show that if q,. sume that w1 has support I ) , there is a family {Ex : X E [O,l]} of projections in R such that w l ( E x ) = * .. = w n ( E x ) = X for each X in [0,1] and Ex 5 Ex, when X 5 A'.
a.
-
0
.
-
--f
3.1
+
Solution. (i) We may assume that a # 0. Then a-'wl,. . . , a-'w, are normal states when restricted to F R F . Since F R F has no minimal projections, the assumption (*) applies, and there is a projection EO in F R F such that Wl(E0) =
* * *
= wn(Eo) = w1(F)/2 = 4 2 .
If we apply this conclusion to each of Eo and F - Eo, we find projections El and E2 in EoREo and ( F - E o ) R ( F - Eo),respectively, such that, with j in {1,2},
742
APPROXIMATION BY MATRIX ALGEBRAS
Let FA be 0, E l , Eo, EOt E2, and F , when X is 0, a/4, a/2, 3a/4, and a, respectively. Continuing in this way, we construct Fa, when T is a dyadic rational in [0,1]. For an arbitrary s in [0,1], let Fa, be the union of Fa,, where T 5 8 and T is a dyadic rational in [0,1]. Since w1,. .. ,wn are normal states of R, {FA : 0 5 X 5 a } has the desired properties. (ii) Proceeding as in the hint, we may assume that R acting on H is such that w1 = w,lR and that 2 is a separating and generating unit vector for R. (The GNS representation corresponding to w1 is faithful since w1 is faithful from Exercise 7.6.6. From Corollaries 7.1.7 and 7.1.16, that representation is a normal isomorphism.) I t follows that w = w,lR for some unit vector y in H from Theorem 7.2.3. Let a posithe E (< 1) be given and choose A’ in R’such that [[A’$- 311 < ~ / 2 .(Note that A‘ # 0 since E < 1.) Let G be FA- F y where 0 5 X - A’ < (~/211A’11)~.Then
Thus X + w(Fx) is continuous on [O,a]. (iii) We proceed as in the hint. From (*), there is a projection E = w n ( E ) = 1/2. When X is 0, in R such that w l ( E ) =
When X is 1,
Since 1/2 is in the closed interval with endpoints w ( E ) and w ( 1 - E ) (= 1 - w ( E ) ) ,and X + w ( E ~ / t 2 F(l-X)/2)is continuous (from (ii)), there is a A’ in [0,1] such that w ( M ) = 1/2, where M is EX$/2 F ( ~ - x I ) By / ~ choice . of { E x } and { F A } ,for j in ( 1 , . . . ,n} and X in [O,11, W j ( E X / 2 t F(1-X)/2) = X/2 t (1 - X)/2 = 1/2.
+
Thus u l ( M ) =
* * *
= wn(M) = u ( M ) = 1/2.
743
EXERCISE 12.4.15
(iv) Combining Exercise 12.4.13, which starts the induction, and (iii), which establishes the inductive step, we have that (*) holds for all positive integers n. (v) From (iv), (*) is valid; from (i), with I in place of F , there is a family { E x } having the properties described provided w1 has support I . In general, let E be the union of the supports of the given states w l , .. . ,w, and let the restriction of these states to ERE be is ofERE w I2 , . . . , w ; + ~ . Then n - ' ( ~ ~ + . . - + w & ~ )anormalstatewi with support E . Apply the conclusion just reached t o w i , . . . ,w;+, (asnormal states of E R E ) to produce the desired family { E x } (without the assumption that w1 has support I ) . m[43,75]
12.4.15. (i) Let R be a von Neumann algebra that has no minimal projections and w1,. . . ,a, be normal states of R. Show that if El is a projection in R and A1 is in [0,1], then there is a projection E in E l R E l such that w j ( E ) = wj(X1E1) for each j in (1,. . . ,n}. [Hint.Restrict each w j to EIRE1 and use Exercise 12.4.14(v).] (ii) Assume (inductively) that, (**) if { E l , . . . , E k } is a family of k mutually orthogonal projections in a von Neumann algebra R that has no minimal projections, ( 0 1 , . . . ,wn} is a finite set of normal states of R, and X I , . . . ,Ak are in [0,1], then there is a projection E in F R F such that w j ( E ) = w j ( A ) for each j in (1,. . . ,n } , where A = XlEl t * X k E k and F = El Ek. Show that if { E l , . . . , E k + l } is a family of k 1 mutually orthogonal projections in R, {wl,. , . ,wn} is a finite set of normal states of R,and X 1 , . . . ,Xk+1 are in [0,1], then there is a projection E in GRG such that w j ( E ) = w j ( B ) for each j in (1,. . . ,n } , where B = Xk+lEk+l and G = El t '. t E k + 1 . [Hint.Restrict XIEl t Ek and w j to each of F R F and Ek+lREk+l, where F = El t apply (**) and (i).] (iii) Conclude that (**) is valid for each k in N. (iv) Show that the set P of projections in a von Neumann algebra R that has no minimal projections has weak-operator closure (72):. (Compare Exercise 5.7.8(i).)
+ + *
+
aj
*
a
-- +
+
+
Solution. (i) Let a j be w j ( E 1 ) and w i be uT1wjlEIRE1 when > 0. (Let w: be 0 on E1RE1 when aj = 0.) Then each w i is either
0 or a normal state of E l R E l . From Exercise 12.4.14(v), there is a
744
APPROXIMATION BY MATRIX ALGEBRAS
projection E in El72E1 such that u i ( E ) = A1 when aj > 0. Hence
for e a c h j in {1,...,n}. (ii) Proceeding as in the hint, we let bj be w j ( F ) and C j be w j ( E k + l ) . When bj > 0, let w i be 6;'wjJFRF. (Let w i be 0 on F R F when bj = 0.) When C j > 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
APPROXIMATION BY MATRIX ALGEBRAS
.
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
APPROXIMATION BY MATRIX ALGEBRAS
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
APPROXIMATION BY MATRIX ALGEBRAS
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
APPROXIMATION BY MATRIX ALGEBRAS
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
APPROXIMATION BY MATRIX ALGEBRAS
(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
APPROXIMATION BY MATRIX ALGEBRAS
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 )<~&/cm( j E { 1 , . . ,n } , k E { 1,.. . ,m ( j ) } ) . With this choice of S1 and S2, we have that
c
m(j)
IPj(A1 @ A2 - $1 @ S2)I
5
ClOjk(A1
@ A2 - s 1 @ S2)l
k=l
< Cm(j)&/CmI & for each j in { 1,.. . ,n}. Thus A1 @ A2 is in the ultraweak closure of s 1 @SZ. 12.4.27. Let M be a factor of type 111 and T be the (unique) tracial state on M Q M . Recall from Exercise 11.5.25 that the mapping that assigns B @ A to A @ B , for A and B in M , extends to a * automorphism a of M Q M . (i) Suppose U is a unitary operator in M Q M that implements a. Show that r ( U ( E @ F ) ) = 0 when E and F are projections in M . [Hint. Use Exercises 12.4.16(iii), 12.4.25(ii), 12.4.24(iv), and 12.4.21(iii) .] (ii) With A and B in ( M ) : , show that A @ B is in the weakoperator closure of P @ P, where
P @ P = { E @F : E , F E P } and P is the set of projections in M . [Hint.Use Exercises 12.4.15 and 12.4.26.1 (iii) With A and B in ( M ) : and U as in (i), show that r(U(A 8 B ) ) = 0. (iv) Show that a is not inner.
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APPROXIMATION BY MATRIX ALGEBRAS
Solution. (i) We may suppose that M acts on a Hilbert space 3-1 and 20 is a trace vector for M in 3-1 (by constructing the GNS representation corresponding to the tracial state on M , for example). Then for each T in M Q M , 7.P)=
(T(z0 €3 201,50 €3 zo).
From Exercise 12.4.16(iii), given a positive E , there are projections M and N in a finite type I subfactor No of M such that
ll(E - M)zoII < 4 2 ,
ll(F- N)zoII < 4 2 .
Hence
ll(E€3 F - M
€3 N ) ( z o €3 .o)ll
5 ll(E €3 ( F - N ) ) ( z o€3 .o)ll t ll((E- M ) €3 N)bo €3 z0)ll 5 I)Ezollll(F- N).oll t ll(E - M)zollllN~011< E . From Exercise 12.4.25(ii), NOis contained in a matricial subfactor of M . Since M is of type 111, N is of type 111. From Exercise 12.4.24(iv), N admits an outer * automorphism. From Exercise 12.4.21(iii), T ( U ( M €3 N)) = 0. Thus
N
~ T ( U ( €3E F))I = )T(U[E€3 F - M €3 N])l = I((E€3 F - M €3 N ) ( z o €3 zo), U*(zo €3 z0))l i ll(E€3 F - M €3 N>@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
APPROXIMATION BY MATRIX ALGEBRAS
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
APPROXIMATION BY MATRIX ALGEBRAS
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
APPROXIMATION BY MATRIX ALGEBRAS
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
APPROXIMATION BY MATRIX ALGEBRAS
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
APPROXIMATION BY MATRIX ALGEBRAS
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
APPROXIMATION BY MATRIX ALGEBRAS
(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
APPROXIMATION BY MATRIX ALGEBRAS
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
APPROXIMATION BY MATRIX ALGEBRAS
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 )
<
< j ( n ) }(= Y )
spans a dense linear submanifold of H?). From (iii), Y is an orthogonal set. From (ii), (ej(1) A
* * * Aej(n),ej(l)A “ ‘ A e j ( n ) ) =det((ej(,),ej(,))) = 1.
Thus Y is an orthonormal basis for Ht). (vi) We may assume that x is a unit vector an orthonormal basis for 3t. The mapping
,...,~ , ) + e l A x l A . . . A x ,
(XI
($1
el
and that {em} is
,...,x,
EN)
is multilinear, alternating, and bounded, since [ [ e l A x1
A
A
x,11 = \I[(. t 1)!]’/2~;+l(el 8 x1 8 I [(nt 1)!]1’2 11x1 I1* * * 112nll.
8 xn)ll
With u a vector in 1-1$+1, from Bessel’s inequality and (v)
c
\ ( e l A ej(1) A
. . * A ej(n),u)I2 I 11u1I2.
j(l)<--<j(9l)
Thus ( X I , . .. ,x,) + el A 21 A * . A 5, is an alternating, multilinear, From weak Hilbert-Schmidt mapping of 1-I x * - x H into (iv), there is a (unique) bounded linear mapping an(x)* (= a ( e l ) * ) of H(n“)into such that
%til.
-
l-Itil
Un(X)*(21 A
.
for all XI,.. ,x, in 7i.
*
* . A x,) = x A 2 1 A *
A Xn
EXERCISE 12.4.40
779
In the notation of Exercise 12.4.39, let 3-1$? be 12.4.40. C:=, $'If?), where Xg) is a one-dimensional space generated by a unit vector 20. (The space 'If$)is referred to as antisymmetric Fock space and 20 is the Fock vacuum.) Let { e m } be an orthonormal basis for 3-1. Show that (i) C,"=o@ a n ( e l ) * ( =a ( e l ) * )is a partial isometry with initial space K spanned by
and final space H$)8K , where ao(el)*(cso)= cel [Hint. Use Exercise 12.4.39(iii) and (v).]; (ii) the mapping x -+ a(z)* is linear, and
for all 2 and y in X ,and conclude that { a ( e j ) * }is a representation of the CAR in the sense of Exercise 10.5.89 [Hint. Use (i) for the first relation of (*) and polarize (+) for the second.]; (The representation of the CAR described here is referred to as the Fock representation.) (iii)
and a ( z ) z ~ = 0 for all 2 , 2 1 , ..., x n in 'H [Hint. Use Exercise 12.4.39(ii) and expand the determinant expression for (2 A y2 A * * * A y n , q A . A 5,) in terms of its first row.]; (The operators a(.) and a(z)* are referred to as annihilators and creators, respectively.) (iv) { a ( z ) , a ( z ) * : z E 'If} generates the CAR algebra24 (on 3-12') [Hint. Use Exercise 10.5.89.1; (v) the self-adjoint operator algebra !210 that is generated by {a(s),a(z)* : 2 E X} consists of linear combinations of I and products a ( q ) *...a ( . , ) * a ( y l ) . . . a ( y m )with all creators to the left and
780
APPROXIMATION BY MATRIX ALGEBRAS
all annihilators to the right ( Wick-ordered monomials) and is norm dense in the CAR algebra U; each annihilator is in the left kernel of WO(= w,,IU) and the null space of wol!Uo is the linear span of the Wick-ordered monomials (other than scalar multiples of I ) [Hint. Use (iii) and (iv).]; (The state wo is referred to as the Foclc vacuum state.) (vi) if p is a state of U such that p 5 two for some positive real t , then p = wo; and conclude that wo is pure and the Fock representation is irreducible. [Hint. Note that each annihilator is in the left kernel of p and that the restrictions of p and wo to UOhave the same null space. Show that 20 is generating for UO.] Solution. (i) We proceed as indicated in the hint. From Exercise 12.4.39(v), the set
is orthonormal, spans a (closed) subspace
K, of l i p ) , and
‘lit)
is an orthonormal basis for 8 Kn. From Exercise 12.4.39(iii), an(el)* annihilates %(,“I 8 K, and maps the orthonormal basis X, 00 for K, onto Y,+l. Now K = Cn=O$Kn, where KO = [zo]. Thus, a ( e l ) * maps K isometrically onto X(78) @ K and annihilates 31$) ex. It follows that a(e1) is a partial isometry with initial space ‘FI(78)8 K and final space K. Hence
(ii) From Exercise 12.4.39(iv), (vi),
+-
Thus a(cz y ) * = ca(z)* + a( y) * . If z is a unit vector in ‘FI, we may choose 2 as el in an orthonormal basis {em} for 7f. From (i),
781
EXERCISE 12.4.40
then, a(x)* is a partial isometry whose initial and final spaces are orthogonal complements of one another. In this case, a(z)a(z)*
+ a(z)*a(z) = I
When x is an arbitrary (non-zero) vector in ‘Ifand y = 11~11-~x,
and the first relation of (*) is established. By polarization of this first relation (that is, substituting z y, z - y, z iy, and z - iy for x, successively, and then combining the results), the second relation of (*) is established. The third relation of (t) is a consequence of Exercise 12.4.39(iv), since
+
+
+
for all 51,.. . ,x, in If. Thus a(y)*a(x)* a(z)*a(y)* and its adjoint are 0. With the basis elements e j and el, in place of x and y in (*), and C j of Exercise 10.5.88(iv) replaced by a ( e j ) , we have that { a ( e j ) } is a representation of the CAR in the sense of Exercise 10.5.89. (iii) Proceeding as in the hint,
for all 51,.. . ,x, and y2,. . . ,yn in IFI and n in { 1,2,. . . }. In addition, since 50 is orthogonal to the range of a(x)*,a(z)xo = 0. Thus ).(a satisfies the relation described in (iii). (iv) From (ii) and Exercise 10.5.89, { a ( e j ) , a ( e j ) * } generates the CAR algebra U. With x in IFI, x = Cj”=,c j e j where, :C I c j I 2 < 0. From (ii), n
n
j=1
j=1
782
APPROXIMATION BY MATRIX ALGEBRAS
At the same time, Ilyll-’a(y)* is a (non-zero) partial isometry for each (non-zero) y in ‘H. Thus Ila(y)*ll = llyll for each y in ‘H, and
j>n
j>n
It follows that a(.)* and ).(a are in 2l. The possibility of “Wick ordering’’ each monomial in annihi(v) lators and creators (resulting in a linear combination of Wick-ordered monomials but not necessarily a monomial) is a consequence of the relations (*) of (ii). It follows that the set of linear combinations of the Wick-ordered monomials is a self-adjoint operator algebra U o with norm closure a, the CAR algebra (from (iv)). Since a(z)x~ = 0, we have that wo(a(z)*a(z)) = 0, and a(.) is in the left kernel of 00 for each x in H. As
each element of !20,other than scalar multiples of I , is in the null space of W O . , have that (vi) Since 0 5 p 5 ~ W Owe
for each x in 3-1. Thus each Wick-ordered monomial (other than scalar multiples of I ) is in the null space of p. Since p ( I ) = wo(I) = 1, plU0 = WO~UO.But 2lo is norm dense in 2l and p and 00 are norm continuous. Thus p = 00 and WO is a pure state of 2l. As a(q)* a(x,)*z~ = z1 A A zn, the vacuum xo is generating for 2l. Now uo = w,,I2l so that ?2lacting on R$)(the Fock representation of U) is irreducible from Corollary 4.5.4 and Theorem 10.2.3. “481
...
CHAPTER 13 CROSSED PRODUCTS
13.4.
Exercises
13.4.1. Suppose that M is a von Neumann algebra acting on a Hilbert space 'H, G is a discrete group with unit e , a : g + ag is an automorphic representation of G on M that is implemented by a unitary representation g + U ( g ) of G on 'H, and R ( M , a ) is the (implemented) crossed product of M by a. Recall, from Definition 13.1.3 and the discussion following it, that R ( M , a ) acts on the Hilbert space 'H 8 K', where K' is /2(G), and that an element R of f?(H8 K ' ) lies in R ( M , a ) if only if R is represented by a matrix of the form [U(pq-')A(pq-')Ip,q€G for some mapping g + A ( g ) from G into M . Given such an element R of R ( M , a ) ,let @'(R)be the element A ( e ) 8 I of M@&. Show that (i) ip' is a conditional expectation from R ( M ,a) onto M@'@KL (see Exercise 8.7.23); (ii) ip' is faithful, in the sense that @'(R) # 0 when 0 # R E R ( M ,a)+; (iii) 9'is weak-operator continuous on the unit ball of R ( M ,a). Solution. (i) From Definition 13.1.3, R ( M ,a) contains M ~ C K , If . R (in R ( M , a ) ) has matrix [U(pq-')A(pq-')], then @'(R)(= A ( e ) 8 I ) has matrix [Sp,qA(e)]; so (1)
ip'(R) =
EgREg
( R E R(M,a)),
g€G
where E , (in B('H @I K ' ) ) is the projection represented by the matrix that has I in the ( g , g ) position and 0 at every other entry. In other words, @'(R ) is the "diagonal part" of R . From ( l ) , ip' is a positive linear mapping (from R ( M ,a) into M@IB@K,). Moreover, @'(I) = I , since C g E G E g= I . Given R in
784
CROSSED PRODUCTS
R ( M , a )with matrix as above, and S1,Sz in M@&, we may set S1 = A1 8 I , S2 = A2 8 I , with A1 and A2 in M . Since Sj has matrix [S,,,Aj], the matrix of S1 RS2 has AlA(e)A2 at each diagonal position; so
a'(SlRS2) = (AlA(e)A2)8 I = (A1 8 I ) ( A ( e )8 I)(Az 8 I) = Sl@'(R)S2. Thus @' is a conditional expectation from R ( M , a ) onto M @ @ p . (ii) If 0 # R E R ( M , a ) + ,then R112 # 0, so R1i2Eh # 0 for some h in G. Thus
o<(R~/~E~)*(R'/~E~) = EhREh
5
E,RE, = @'(R). S€G
(Exercise 4.6.11 can be used to effect this last argument.) (iii) From Exercise 8.7.23(iii), @' maps the unit ball of R ( M , a ) into the unit ball of M @ C p . The weak-operator topology on any bounded subset of B('H 8 K ' ) is determined by the set of all linear fun ct ionals W,,y : s (SX, Y ) +
where the vectors x , y satisfy x = Ehx, y = Eky for suitably chosen elements h and k of G. To establish weak-operator continuity of @' on the unit ball of R ( M , a ) ,it suffices to show that the linear functional u , , ~o @' is weak-operator continuous for vectors x and y as above, and this follows from the fact that
EXERCISE 13.4.2
785
13.4.2. Suppose that M is a type I11 von Neumann algebra, G is a discrete group, and a : g -t ag is an automorphic representation of G on M . Show that the crossed product von Neumann algebra R ( M , a ) is of type 111. (Note that this is in sharp contrast with the situation for continuous crossed products - see, for example, Theorem 13.3.7.) [Hint. Use Exercise 13.4.1 and Proposition 11.2.25.1
Solution. From Proposition 13.1.2 (see also the third paragraph of the introduction to Section 13.1) we may assume that cr is unitarily implemented, and interpret R ( M ,a ) as an implemeted crossed product. From Exercise 13.4.1, there is a faithful conditional expectation 9'from R ( M , a ) onto its type I11 von Neumann , Ic' is /2(G)), and 9' is weak-operator subalgebra M ~ C K :(where continuous on the unit ball of R ( M , a ) . By Proposition 11.2.25, R ( M ,a)is of type 111. 13.4.3. Suppose that G is a countable locally finite i.c.c. group, LG is the factor of type 111 acting on l2(G) described in Section 6.7, K is a separable infinite-dimensional Hilbert space, and M is the von Neumann algebra l ~ @ l ? ( I c ) Show . that (i) M is a factor of type 11, with commutant M' of type 111; (ii) there is a * automorphism of M that is not unitarily implemented. [Hint. Use Propositions 12.2.7, 13.1.10, and Exercise 9.6.33.1
Solution. (i) From Theorems 6.7.2(ii) and 6.7.5, LG is a factor of type 111 with commutant L& (= R G )also of type 111. Since D ( K ) is a factor of type I,, it follows (see Corollary 11.2.17 and the table preceding Proposition 11.2.19) that M (= Lc@f?(Ic))is a factor of type 11,. Moreover, M' (= L& @ C K )is * isomorphic to L&, and is therefore of type 111. (ii) By Proposition 12.2.7, LG is the finite matricial factor. Suppose that 0 < X < 1. From Proposition 13.1.10, there is a * automorphism 8 of M such that T o 8 = AT, where T is a faithful normal semi-finite tracial weight on M . From Exercise 9.6.33, M has a generating vector z, and 8 is unitarily implemented if and only if 8(E) E , where E (in M ) is the projection with range [ M ' z ] . Since M' is finite, the projection I with range [ M z ]is finite in M ' , so E is finite in M by Proposition 9.1.2. It follows from Proposition N
786
CROSSED PRODUCTS
8.5.2(ii) that 0 < r ( E ) < 00. Hence
+q # M E ) = M E ) ) , thus E
+ 8 ( E ) ,and therefore 8 is not unitarily implemented.
13.4.4. Suppose that R is a factor of type 111, K is a separable infinite-demensional Hilbert space, M is the factor R&?(K) of type 11,, q is the unique tracial state on R, and 7, is a faithful normal semi-finite tracial weight on M . Let c be a real number such that 0 < c 5 1, and let E be a projection in R such that q ( E ) = c (see Proposition 8.5.3). Show that R is * isomorphic to E R E if and only if there is a * automorphism 8 of M such that rooo 8 = cr,. [Hint. Most of the necessary ideas can be found in the proof of Proposition 13.1.10.1
Solution. There is a self-adjoint system { F j k : j , k = 1,2,. ..} of matrix units in M such that C F j j = I and F11MF11 is * isomorphic to R. Thus F11 is a non-zero finite projection in M , 0 < r,(F11) < OCI by Proposition 8.5.2(ii), and we may assume that 7, has been normalized so that r,(F11) = 1. It follows that r1, FlIMF11 is the faithful normal tracial state on FlIMF11. Suppose that R is * isomorphic to E R E . Let $ be a * isomorphism from R onto F11MF11, and note that r1 o $-* is a tracial state on F11MF11, whence 71 o $-l = roo[F11MF11. If we define E l l to be the projection $ ( E ) in F11MF11, then ~Co(El1) = Tl(?qEll))
= n ( E ) = c;
moreover, there is a * isomorphism cp from F11MF11 (= $(R)) onto E11MEll (= $(ERE)). We now follow the argument used in proving Proposition 13.1.10. The projection El1 can be embedded in a self-adjoint system { E j k : j , k = 1,2,. . .} of matrix units in M (with C Ejj = I ) . We can identify M with both No 8F11MF11 and No @ E11ME11; the * isomorphism No 8 cp from No 8 F11MF11 onto No 8 EllMEll then gives rise to a * automorphism 8 of M such that 8(F11)= ,311, and 7, o 8 = cr,. Conversely, suppose that there is a * automorphism 8 of M such that r, o 8 = cr,. Choose $, and define a projection Ell in F11MF11 (C M ) , as in the preceding paragraph. Then K?@ll)
= c = C%O(Fll)= G O ( ~ ( ~ l 1 ) ) .
EXERCISE 13.4.5
787
F'rom this, the projections $11 and B(F11) in M are equivalent. Let V be a partial isometry in M such that V*V = O(F11) and V V * = $11. Then V can be considered as a unitary transformation from the range of 8(F11) onto the range of E l l , and as such implements a untary equivalence relation between the von Neumann algebras B(F11)M9(F11)and E11ME11. Since 8 I F11MF11 is a * isomorphism from F ~ I M Fonto ~ I B(F11)M8(F11),it now follows that there is a * isomorphism 81 from F I I M F I Ionto E I I M E I I ,and $-l o 81 o $ is a * isomorphism from R onto ERE.
13.4.5. Suppose that R is a factor of type 111, K: is a separable infinite-dimensional Hilbert space, M is the factor Re,B(IC)of type II,, and T~ is a faithful normal semi-finite tracial weight on M. Suppose $so that c is a positive real number, n ( 2 c ) is a positive integer, and E is a projection in the type 111 factor n 8 R such that rn(E) = c / n , where rn is the unique tracial state on n 8 R. Show that R is * isomorphic to E ( n @ R)E if and only if there is a * automorphism 8 of M such that r, o 8 = cr,. [Hint. Adapt the argument required in solving Exercise 13.4.4.1 Solution. There is a projection F in M, necessarily non-zero and finite, such that F M F is * isomorphic to [email protected] may assume that T~ has been normalized so that rm(F) = 1 , so that 7,IFMFis the unique tracial state of F M F . We can choose projections El and G1 in F M F such that rm(G1) = l / n and G l M G l is * isomorphic to R, while T , ( E ~ )= c/n and ElMEl is * isomorphic to E(n @ R ) E . We have to show that G l M G l is * isomorphic to ElMEl if and only if there is a * automorphism B of M such that rm o B = cr,. Suppose that G l M G l is * isomorphic to E1MEl. By identifying M with both No @ GlMGl and No @ ElMEl (see the solution to Exercise 13.4.4) we can obtain a * automorphism 8 of M such that B(G1) = E l . Since the faithful normal semi-finite tracial weight rooo 8 on M is a positive scalar multiple of ,T (from Proposition 8.5.5), and Go(fl(G1)) = .r,(El)
= c / n = CTm(Gl),
it follows that rooo 8 = CT,. Conversely, suppose there is a * automorphism 9 of M such that r, o 8 = cr,. Then T,(O(G*)) = cr,(G1) = c / n = rm(E1).
788
CROSSED PRODUCTS
Thus @ ( G I )
E l , the von Neumann algebras &ME1 and e ( G , ) M O ( G l ) (= B(G1MG1)) are unitarily equivalent, and thus E l M E l and G l M G l are * isomorphic. N
13.4.6. Suppose that R is a factor of type 111, Ic is a separable infinite-dimensional Hilbert space, M is the factor R@B(Ic)of type 11, and r, is a faithful normal semi-finite tracial weight on M. Let TQ(R)be the set of all positive real numbers c with the following property: there is a * automorphism 8 of M such that r, 0 8 = crW. (Note that Exercise 13.4.5 provides an alternative characterization of the elements of 3Q('R).)Show that (i) 3 Q ( R )is a subgroup of the multiplicative group of positive real numbers (we refer to FG(R) as the fundamental group of R); (ii) 3 0 ( R ) = 3 Q ( E R E ) = 3Q(n8 12) when E is a non-zero projection in R and n is a positive integer.
Solution. (i) By taking for 8 the identity automorphism of
M, we see that 1 E FQ(R). If cI,c2 E FQ(R),we can choose automorphisms 6$,f?, ofM such that T,O& = c ~ T , , rooo82 = ~ 2
* 7
Since
it follows that c1c2,c11 E 3 Q ( R ) .HenceFG(R) is a (multiplicative) subgroup of the positive real numbers. (ii) Since FG(R)is defined solely in terms of algebraic properties of R@f3(IC), and R@B(lc)is * isomorphic to No @I R, it suffices t o prove that the type 11, factors
No €3 n, are
No 63 E R E ,
No @I ( n @I 12)
* isomorphic.
The algebra No@Rcontains a self-adjoint system of matrix units j, k,= 1,2,. .} such that C Fjj = I and Fll(No @I R)Fi1 is * isomorphic to R. If Ell (in F11(No@R)F11)is the image of E (in R) under a * isomorphism between these algebras, then E11(No@ID)E11 is * isomorphic t o E R E . Moreover, by the argument used in proving Proposition 13.1.10, El1 can be embedded in a self-adjoint system {Ejk : j , L,= 1,2,...} of matrix units in No 8 R, with C Ejj = I . {Fjk :
.
~
.
EXERCISE 13.4.7
789
In this way, No 8 R can be identified, up to * isomorphism, with No C3 E11(No 8 R ) E 1 1 , and hence with No 8 E R E . Since a (bounded) No x No matrix of n x n matrices over R can be viewed as a (bounded) N o x No matrix over R, and conversely, there is a natural * isomorphism between No 8 ( n 8 R) and No 8 R. w 13.4.7. Suppose M is a factor of type 11, with commutant M' of type 111 acting on a separable Hilbert space X. Let r, be a faithful normal semi-finite tracial weight on M , E be a projection in M such that r,(E) = 1, and Mo be EME. Show that (i) the mapping, f : 8 + T , ( B ( E ) ) , is a homomorphism of the group a u t ( M ) of * automorphisms of M onto .FG(Mo) (defined in
Exercise 13.4.6). [Hint.Use Proposition 8.5.5.1; (ii) the kernel of the mapping f (defined in (i)) is the subgroup aut,(M) of a u t ( M ) consisting of those automorphisms of M that are implemented by unitary operators on 7-1 [Hint.Use the result of Exercise 9.6.33.]; (iii) the mapping, f : Baut,(M) -+ f ( B ) , is an isomorphism of the quotient group aut(M)/aut,(M) onto FG(M0).
Solution. (i) From Proposition 8.5.2(ii), E is a finite non-zero projection in M , and the same is true of B(E) for each B in aut(M). It follows that f maps a u t ( M ) into the multiplicative group B+ of positive reds. With B in a u t ( M ) , r, o B = c r , for some positive real c from Proposition 8.5.5. From the result of Exercise 13.4.6, the set of such c is the subgroup .FG(Mo) of B+. Now, by choice of E ,
Thus, for each projection F in M ,
With
81
and
02
in aut(M),
Thus f is a homomorphism of a u t ( M ) onto FG(M0). (ii) Suppose B E aut(M). From (*), f ( B ) = 1 if r,(B(F)) = r,(F) for some finite non-zero projection F in M and only if r,(B(G)) = r,(G) for each projection G in M. It follows from
790
CROSSED PRODUCTS
the result of Exercise 9.6.33 that f(0) = 1 if and only if 8 is implemented by a unitary operator on 1-1. (iii) Since aut,(M) is the kernel of the homomorphism f and FG(Mo) is the image of f, the mapping .f is an isomorphism of m[55] aut(M)/aut,(M) onto FG(M0).
13.4.8. With the notation of Exercise 13.4.6, determine 36(R) when R is the finite matricial factor. Solution. Suppose that 0 < X < 1. By Proposition 13.1.10, there is a * automorphism 0 of M such that rm o 8 = Xr, (the notation being that of Exercise 13.4.6); so X E FG(R). Since FG(R) is a subgroup of the multiplicative group {c E R : c > 0), and contains the open interval (0, l ) , it follows that
,FQ(R)= {C E B :
c
> 0).
.[55]
a
13.4.9. With the notation of Theorem 13.1.15, suppose a, + as T + 00 and C,"=, 6: < 00. Identify U with np(U),as in the proof of Theorem 13.1.15, so that U- is a factor and p = w,I U for some cyclic unit vector 5 for 31. Suppose
H i = +[u;'E!;) -t (1 - up)-'E$;)] K, = H i H i . . .H:, rh(A) = w,(AKn)
(T
= 1 ? 2 , .. .),
( n = 1 , 2 , . ..)?
( A E U-).
Show that (i) I(, is in the centralizer of wzl U-; (ii) r; is a vector state of U-; (iii) [Irk - rAl12 5 p ( ( K m (iv) the sequence {r;} converges in norm to a normal state r' of U-; (v) r' is a tracial state of U-; (vi) U- is a (by Theorem 12.2.1, is the unique) finite matricial factor. Solution. (i) Suppose that 1 5 n 5 Ic and A1 E U1, A2 E U2, . .., A,, E U k . Since the algebras % I , . . . ,%k are mutually commuting,
EXERCISE 13.4.9
791
p is the product state @p,, and H i is in the centralizer of P r , we have
Since 2l- is the weak-operator-closed linear span of the set of all products of the form A1A2 . . . A k (with A l , A 2 , ... , A k as above), it now follows that w z ( l i n A )= w z ( A l i n ) for each A in %-. Thus the lies in the centralizer positive operator K, (and hence, also, of wzI 2l-. (ii) From (i),
where y = Kk/2z.Moreover
792
CROSSED PRODUCTS
(iv) Suppose that 1 5 m < n. Since the algebras 2l1,%2,. mutually commuting, we have
(I(, - Hi2Hi2
I I H L ( I - H,+1 Hm+2 . * * H;)12 H:(I - 2HL+, H;+, * . HA t H:+, H:+, Hh2).
= [H:H i
-
.. are
* ' *
---
Since p is the product state @ p r , and p,(HC) = 1, it follows that p((Km -
= p 1 ( H i 2 ) ~ 2 ( H i *2 .)* p m ( H : ) [ l - t t T I , n
m
r=l
r=l
- 1 &a;2 -
-t ( 1 - a,)( 1 - ar)-2]
= [4a,( 1 - a,)]-'
Since 0 < a, 5
= [l - ( 1 - 2 ~ ~ ) ~ l - l .
3 and 6, = log(aF'(1-
a,)),
we have
for sufficiently large T . The convergence of the series CE162, now entails convergence of the series C z l ( l- 2 ~ and, ~ hence, ) ~of the infinite products 00
00
00
r=l
r= 1
r= 1
It now follows from ( 1 ) that the sequence { T : } of normal states of !2lis a Cauchy sequence, relative to the norm topology in the predual
EXERCISE
13.4.10
793
Banach space (U-)#, and so converges in norm to a normal state of 2l-. (v) We note first that
T
where T, is the unique tracial state of U,. To verify this, observe that 2HCHP = I (with H , as in the proof of Theorem 13.1.15), whence
p,(AH:) = 2r,(H,AHC) = rr(2HCH,A) = T,(A). Suppose that m is a positive integer and A,,B, E 2l, ( T = 1 , . , . ,m ) . Whenever n 2 m, we have the following chain of equations (in which ( 2 ) , and the fact that p,(HC) = 1, are used near the end);
By taking limits as n * 00, we obtain
Since 2l- is the weak-operator-closed linear span of the set of all operators of the form A1 A2 - . A , or B1B2 . . - B,, it now follows that T' is a normal tracial state of M- . (vi) From Proposition 8.5.2, the normal tracial state T' of U- is faithful; so the matricial factor U- is finite. m[18] With the notation of Theorem 13.1.15, suppose (this is equivalent to the assumption that C z l e - b r < 00, since 0 < a, = (1 + ebr)-'). Identify U with rP(U), as in the proof of Theorem 13.1.15, so that 2l- is a factor and p = w,lU for 13.4.10.
C,"=,a , < 00
794
CROSSED PRODUCTS
some cyclic unit vector state of U, given by
2
for U. For
T
= 1 , 2 , . , ., let w, be the pure
and let H , be the element (1- a,.)-1/2E$i)of U,. Define a sequence {a,} of states of U by
Show that where K, = HlH2 H,; (i) u, = W,<,,~~U, (ii) I I K , ~ : - ~ ~ 2 1 = 1 2[1 2 - n : = m + l ( ~ - a,)1/2] (1 5 rn < n ) ; (iii) the sequence { K , x } converges in norm to a unit vector y; (iv) w,lU is a pure state of U; (v) U- is a type I, factor.
Solution. (i) Since p = w,I U, we have to show that
For this, it is sufficient to consider the case in which A is a product AlA2 - . Ah, where k 2 n and A1 E U1, A2 E U2, . .., Ak E U h (since 0
U is the norm-closed linear span of the set of all such products). In this case,
since the algebras U1, U2,. . . are mutually commuting. Thus
while
EXERCISE 13.4.10
Thus, in order to prove (I), it suffices to show that
Now A, has the form
xj,k=, CjkEj;), and 2
This proves (2) and hence, also, (1). (ii) Since p = w,l2l,
Now p is the product state @p,, and we have that p,(H,2) = 1, pT(Hr) = (1 - a,)'I2. Thus
From (i), l i n x is a unit vector. Since 0 < a, 5 and C,=, a, < oo, the infinite product - a,) converges (with non-zero limit). Thus (iii)
nE1(l
00
It follows from (ii) that, when 1 _< m
< n, 00
l l l i - n ~-
l ~ m5~ 2lI2{1 ll - [ (1 - a,)] 112} 1 1 2 . r=m+l
796
CROSSED PRODUCTS
The right-hand side of this last inequality has limit 0 when m -, 00, by (3); so { K n z }is a Cauchy sequence of unit vectors, and converges in norm to a unit vector y. (iv) From (i) and (iii), = lim
n+w
with convergence in the norm topology of the Banach dual space '2". Suppose that A1 E U1, A2 E U2,. . ., Ak E U k . When n 1 k, it follows from the definition of Q, as a product state that
an( A1A2 * * * Ah) = W( A 1 ) ~ 2A2) ( * * uk(Ak)* By taking limits as n uy(
-+
00,
we obtain
AlA2 . * Ak) = W (A 1 ) ~ 2 ( A 2 ) 'uk(Ak). *
* *
Since this last equation is satisfied for all products AlA2 Ak (with A, in U,, 1 5 T 5 k), w y l U is the product state mu,, and is a pure state of U by Proposition 11.4.7. (v) Let E' be the projection, in the factor a', that has range [Uy]. Since uylU is a pure state, the representation A + AE' of U (that has y as a cyclic vector) is irreducible. Thus E' is a minimal projection in U', U' is a type I factor, and hence %" (= U-) is a type I factor. Since U is infinite dimensional, as a linear space, the same is true of U-; so W- is of type I,. m 13.4.11. a, -,0 and
With the notation of Theorem 13.1.15, suppose that
czla, diverges.
e - b r diverges. (i) Show that b, -+ 00 and C,"=, Suppose that { c , } and {d,} are sequences of positive real (ii) numbers such that C z l c, diverges and d, + 00, and a,b are real numbers with a < b. Show that the set 00
{ t E W : a < t < b,
c, sin2($,t)
diverges}
r=l has positive Lebesgue measure. [Hint. Suppose the contrary. Upon deleting a finite number of terms from the sequences {c,.}, {d,}, reduce to the case in which 4d;l < b - a for T = 1 , 2 , . . . . Consider
1
fn(t>dt,
EXERCISE 13.4.11
797
where
(iii) Deduce from (i), (ii), and Theorem 13.1.15 that rp(U)- is a type I11 factor, and the complement in R of the set T(rp(U)-) meets each non-empty open subset of R in a set of positive Lebesgue measure.
Solution. (i) Since
and C,"=,a, diverges, it follows that c z l e e - b r diverges. At the same time, since a, + 0, we have
(ii) Assume that the assertion made in the statement is false, so that C,"=,c, sin2 $d,t converges for almost all t in the open interval ( u , b ) . We shall see in due course that this assumption leads t o a contradiction. The assumption remains unaffected if we delete a finite number of terms at the start of the series, and (since d, co) we may therefore suppose that --f
4dF1 Since
(T=
1,2, ...).
C,"=,c,
diverges, our assumption implies that the sequence {fn(t)} defined in the hint converges to 0 for almost all t in ( a , b ) . Since, also,
0
I f n ( 4 5 1,
it follows from the dominated convergence theorem that
798
CROSSED PRODUCTS
However,
r=l
= 4(b - u),
for all positive integers n. This contradicts (1). Hence the assertion in (ii) is established. (iii) From Theorem 13.1.15, the complement in R of T ( r p ( U ) - ) is the set
c 00
{t E R :
e-br sin2(!&t) diverges}.
T=l
From (i) and (ii), this set meets every non-empty open interval (hence, meets every non-empty open subset of W) in a set of positive Lebesgue measure. In particular, therefore, T ( r p ( U ) - )# R, and the factor rp(!21)-is of type I11 (see the discussion following the proof of Theorem l3.1.9). With the notation of Theorem 13.1.15, show that 13.4.12. T ( r p ( U ) - )= { O } if an =
1 n+l
-
(72
= 1 , 2 , . . .).
[Hint.Note that 6,, = log n, and that it is necessary to show that the series 00 n-l sin2( log n ) diverges for all non-zero real numbers t . When t > 0, define c(> 1) by the condition tlogc = A , and for k = 1 , 2 , . . ., let nk be the largest positive integer such that
it
EXERCISE
13.4.12
799
Consider the sum [cnkl
C
$ sin2($tlogn),
n=nk+l
where [cnk] denotes the largest integer not exceeding cnk.] Solution. We have
b, = log and
1 2
1 - a, (7) = log n,
e-bn sin2(-b,t) =
sin2(itlogn).
From Theorem 13.1.15, the assertion that T ( T ~ ( % )=- )(0) is equivalent to the statement that the series
n= 1
diverges for all non-zero real numbers t , and it suffices to consider the case in which t > 0. Let c = exp(n/t) (> l ) , so that t log c = n. With nk defined as in the hint, the sequence {nk} is non-decreasing and diverges to +oo. Thus, for sufficiently large b,
Whenever
Hence
nk
< n 5 [cnk],it follows from the definition of n k that
800
CROSSED PRODUCTS
and
1
cnk
- 1 - nk
2cnk c-1 (> 0) 2c as k + 00. Since nk -+ 00, it follows that the series 00 n-l sin2( log n) does not satisfy the Cauchy criterion for convergence, and therefore diverges.
it
13.4.13.
With the notation of Theorem 13.1.15, show that
T(np(21)-)= (0, f 2 n , f 4 n , . ..} if bn = [logn] ( n = 3,4,. . .), where [z] denotes the largest integer not exceeding z. [Hint. When t is a multiple of 2n, each term of the e-bn sin2(+b,t) is 0, since b, is an integer. In view of series this, and from Theorem 13.1.15, it suffices to prove that the series diverges when t is not a multiple of 2n. For such t , by considering the cases in which t/2n is rational and irrational, show that there is a strictly increasing sequence { m k } of positive integers and a positive real number b, such that
EF=,
Let nk be the largest integer for which lognk
and consider
< mk. Show that
2n~.
n=ni+l
Solution. If (t/2n) is irrational, the set
{g-[g]
1
: m = 1,2, ...
EXERCISE 13.4.13
801
is dense in the interval [0,1], by Dirichlet's theorem. Hence there is a strictly increasing sequence {mk} of positive integers such that
Thus
Upon deleting a finite number of terms from the sequence {mk}, we may suppose that
1 sin( +mkt)l 2 a
( b = 1 , 2 , . . .).
3.
In this case, we take 6 to be If ( t / 2 ~is) rational but not an integer, say p/q in lowest terms (with q > 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
since a is an isometry. Thus i(R#), with its strong topology, is a topological group. From (i), the mapping a --+ a+ is an antiisomorphism. Thus aut(R), with its bw-topology, is a topological group. (iii) Let u be a unit generating and separating vector for R. From Exercise 9.6.65, for each 1c, in aut(R), there is a unique unitary operator U, on H that implements 1c, and maps V, onto itself. Moreover, the mapping 1c, UQ is an isomorphism of the group aut(R) into the group U(3-I)of unitary operators on ‘H. We show that this mapping is continuous from aut(72) with its bw-topology to U(31)in the strongoperator topology. For this, it suffices to prove that IlU,u - vII < E for a preassigned positive E and a given v in V,, provided 1c, lies in a suitably chosen bw-neighborhood of the identity automorphism L of R, since V , spans ‘H (see Exercise 9.6.60). If (3 =
{a E aut(R) : llwv o a - w,I
then 0 is a bw-(open) neighborhood of (3 =) UGv E V , (by choice of U,) and
IlU,v
- vI(2 = 11u;v Il
l4
L
~ 1 <1 c2},
in aut(72). With 11, in (3,
- vII2 = 112 - 0112 - wvl
Rll 5 llwJ0 II, - U v l Rll < E 2 ,
from Exercise 9.6.60(iv). 13.4.20.
With the notation of Exercise 13.4.19, suppose g
+
U, is a strong-operator continuous unitary representation of a topological group G on H such that U,AU,* (= a,(A)) E R for each g in G and A in R. Show that g -+ as is a continuous homomorphism of G into aut(R) provided with its bw-topology.
Solution. Thus
With g and g‘ in G, U,,, = U,U,t by assumption.
“,,#(A) = U,,#AU,*,# = U,(U,#AU;)U,* = ag(ag,(A)), for each A in R. Hence a,,# = a g a g and ~ , the mapping g homomorphism of G into aut(R).
--t
a, is a
810
CROSSED PRODUCTS
To prove continuity of this homomorphism, we must show that IIw o as - wII < E for a preassigned positive E and a given w in R,, provided g is in a suitably chosen neighborhood of the unit element e in G. Since functionals of the form wZIYlR (5,y E %) span a norm-dense linear subspace of R, and the mapping w -+ w o ag is an isometry of R#,we may assume that w is U ~ , ~ ILet R .b be (1 1 1 ~ 1 1t Ilyll)-'. Choose a neighborhood 0 of e in G such that
+
when g E 8.Suppose g E 0 and A E
Thus IIw o ag - wII < E , and g
--f
(%?)I.
Then
ag is continuous.
13.4.21. Let R be a von Neumann algebra and f?(R)be the linear space of bounded linear transformations of R into itself equipped with the weak topology induced by the family
of linear functionals on f?(R).Show that (i) an automorphic representation a of topological group G on R is continuous in the sense of Section 13.2 if and only if the mapping a is continuous from G into B(R); (ii) the mapping that assigns to each unitary operator U in R the automorphism A -+ UAU* ( A E R ) is a continuous mapping from U ( R ) ,the unitary group of R equipped with the weak-operator topology, into B(R). [Hint. Use Exercise 5.7.5.1
Solution. (i) By definition, cx is continuous if and only if
is continuous for each A in R and w in R#.From Proposition 1.3.2, this is precisely the condition for a to be continuous from G into
B(R)*
EXERCISE 13.4.22
811
(ii) Again from Proposition 1.3.2, it will suffice to show that the map ping f : U + w(UAU*) (U E U ( R ) ) is continuous on U(R)in the weak-operator topology for each A in R and w in R,. From Exercise 5.7.5, the strong-operator and weakoperator topologies coincide on U(R).The mapping U + UAU* is strong-operator continuous from U ( R ) to R. Since w is strongoperator continuous on bounded subsets of R,f is continuous. 13.4.22. Let R be a von Neumann algebra and a be an automorphic representation of R on R (not assumed continuous). Suppose that each a ( t )is inner and G is the family of unitary operators in R that implement some a ( t ) . (i) Show that G is a subgroup of the group of unitary operators in R and that U and V in G implement the same automorphism of R if and only if V = CU for some unitary operator C in the center of R. (ii) Suppose U and V in G implement a ( t ) and a ( s ) , respectively. Show that UV = VU if t / s is rational [Hint. Suppose t / s = n / m with n and m integers. Choose W in G implementing a ( t / n )and note that W" implements a(t). Conclude that U = CW" for some central unitary operator C , and that WU = UW. Then note that WmU = UWm.] (iii) Suppose a is continuous. Show that G is abelian.
Solution. (i) If U and V in G implement a ( t ) and a ( s ) , respectively, then
Thus UV-' E G, and G is a group. At the same time, if t = s, UV-'A(UV-')-' = cr(O)(A)= A for each A in R, and UV-' is a unitary operator C-' in the center of R. Thus V = CU. If U and V are in G and V = CU for some unitary C in the center of R, then for each A in R,
812
CROSSED PRODUCTS
so that V and U implement the same a(t).
(ii) By assumption, a ( t / n ) is inner; thus there is a W in G implementing cr(t/n). Both W" and U implement cr(t). Hence U = CW", with C a central unitary operator in R, from (i). Thus
wu = WCW" = CW"W = uw, and UW" = WmU. As t / n = s / m , we have that W implements a ( s / m ) . Hence both W m and V implement cr(s). It follows that V = C'W" for some central unitary operator C' in R. Thus
uv = UC'W"
= C'W"U =
vu.
(iii) Suppose U and V in G implement a ( t ) and o(s), respectively. If t / is ~ raiional, then UV = VU from (ii). If t / s is not rational, then we can choose a sequence { t n } tending to t such that t , / s is rational for each n. With 5 and y vectors in the Hilbert space H on which R acts,
as n tends to 00, for each A in R. Choose Un in G implementing U,V = VUn from (ii). Thus
& ( i n ) . Then
Y)
( ~ 5 ,=
(UnVui'z, Y) = (a(tn)(V)s, Y)
+
(UVu -lz, Y)
for all 5 and y in H.It follows that UV = V U ,and G is abelian. m[45] Let cr be a continuous automorphic representation of M by inner automorphisms, G be the family of unitary operators in M that implement some cr(t), U be the C*-subalgebra of M generated by G and p be a pure state of U. Show that (i) p restricts to a character ( of G (norm-, but not necessarily, strong-operator-) continuous on G [Hint.Use Exercise 13.4.22.1; (ii) the kernel H of ( and {cI : IcJ = 1) (= C) generate G, and H n C = {I}; (iii) the mapping that assigns to each U in H the automorphism that U implements on M is an isomorphism of H onto {cw(t): t E R}, conclude that there is a homomorphism of R into the unitary group of M that implements a. (in general, this homomorphism will not be st rong-operator continuous. ) 13.4.23. R on a factor
EXERCISE 13.4.24
813
Solution. (i) From Exercise 13.4.22(iii), G is abelian. Hence !2l is abelian. From Proposition 4.4.1, p is a homomorphism of U into C such that p ( 1 ) = 1. Thus is a character of G continuous in the norm (but, in general, not in the strong-operator) topology. (ii) With U in G, we have that CU E ‘H, where [ ( U ) = c, so that H and C generate G. If a1 E H , then 1 = a = < ( a I ) . Thus
<
H n C = {I}. (iii) If U E H and U implements the identity automorphism of M , then U is in the center of M . Since U is unitary, U E C. From (ii), we have that U = I . Let 7 be the mapping that assigns to each U in H the automorphism of M that U implements. We have just noted that 17 is an isomorphism of H into { a ( t ) } .With t in R, there is a V in G that implements a ( t )since a ( t )is inner. From (ii), V = cW, with W in H . It follows that W implements a ( t ) ,so that 17 is an isomorphism of H onto { a ( t ) } .Let Ut be q-’(a(t)). Then t + Ut is a homomorphism of R into the unitary group of M , and Ut implements a ( t ) for each t in R. 13.4.24. Let R be a von Neumann algebra acting on a Hilbert space W ,u be a separating and generating unit vector for R, (J,A) be the modular structure corresponding to ( R , u ) , and t + ut be the modular automorphism group of R implemented by t + A”. Suppose that each ut is an inner automorphism of R. Show that (i) there is a maximal abelian subalgebra A of R such that o t ( A )= A for each A in A [Hint. Use Exercise 13.4.22.1; (ii) with A as in (i), A is contained in the centralizer of w,lR; (iii) there is an ultraweakly continuous conditional expectation 9, (see Exercise 8.7.23) mapping R onto A such that 9, o ut = 9, for all t in R. [Hint. With H in (R+)1,define @,-,(A)to be w,(HA) for A in A. Show that 0 5 wo 5 w , ( A and use Theorem 7.3.13 to construct 9,. Show that W A , o 9, is strong-operator continuous at 0 on R and use Lemma 7.1.3 to prove that 9, is ultraweakly continuous .] ; (iv) w, o 9, = w,lR, a, is uniquely defined by this equality, and Q U is faithful (see Exercise 8.7.28(iii)).
Solution. (i) Since t + ot is implemented by t + A i t , the automorphic representation t + ot of R on R is continuous. By assumption, each ut is inner. From Exercise 13.4.22(iii), the family G of unitary operators in R that implement some ut is an abelian group.
8 14
CROSSED PRODUCTS
From Zorn's lemma, we have that G is contained in a (self-adjoint) maximal abelian subalgebra A of R. Since ot is implemented by some U in A,
at(A) = UAU-l = AUU-l = A for each A in A. (ii) From Theorem 9.2.13 and Proposition 9.2.14(ii), the centralizer of o,l R contains A. (iii) With H in (R+)1,let wo be the element of dl defined by wo(A) = w,(HA) ( A E A). If A E A+, then All2 E A+ and from ( 47
wo(A) = w,(HA) = U , ( H A ' / ~ A ~=/ W ~ ), ( A ' / ~ H A ' / 2~ )0 . At the same time, since 0 5 H 5 I ,
o 5 w o ( ~=)W , ( A ~ / ~ H A=~(HA%, /~) A%) 5 (A1i2u,A'/'u) = w,(A). Thus 0 5 wo 5 w,J A. From Theorem 7.3.13 (applied to the simple case of the abelian von Neumann algebra A), there is a positive Ho in (A)1 such that wo(A) = w,(HoA) for all A in A. Moreover, Ho is unique, for if w,(KoA) = w,(HoA) ( A E A ) for some KO in A, then 0 = w,((Ko
- H o ) ( K o - Ho)*) = ll(IC0
- Ho)*u112.
Since u is separating for R, KO = Ho. We may now extend the mapping H -+ HOlinearly to R by expressing each element of R as a (finite) linear combination of (four) elements of (R+)1. Note for this that if C ajHj = 0 with H j in (R+)1( j E (1,. ..,n } ) and K j is the element of (d+)1obtained from H j , then
As before, CajICj = 0. Thus the extended mapping a, is well defined and linear. As defined, 9, is positive and @,(I) = I . In addition, w,(@,(T)A) = w,(TA) ( A E A ) for each T i n R, and with Ao in A,
wU(9,(TAo)A)= w,(TAoA) = w,(@,(T)AoA)
( A E A).
EXERCISE 13.4.24
815
From the uniqueness established before, we have that
Since 9, is a positive linear mapping, 9, is hermitian (adjoint preserving); thus
from the preceding equality. Hence 9, is a conditional expectation from R onto A. (This part of the solution can also be effected by using Exercises 10.5.85, 10.5.86, and 10.5.10.) To show that 9, is ultraweakly continuous, we must show that w o 9, is weak-operator continuous on (R)1for each w in A,. It will suffice to prove this for each w in a norm-dense subset of the set of states in A#. We prove that
is such a subset. Since u is separating for R, u is separating for A. Let E’ be the projection (in A’) with range [Au]. The mapping A + AE‘ (A E A ) is a * isomorphism of A onto AE’. Since AE’ has a separating vector u , each normal state of AE’ (and of A ) is a vector state corresponding to some unit vector in [Au] from Theorem 7.2.3. It follows that S is dense as asserted. We prove, now, that W A , o 9, is weak-operator continuous on (R)1for each A in A. From Lemma 7.1.3, it will suffice, for this, to show that W A , o 9, is strong-operator continuous at 0 on (R)1.This continuity follows from the fact that
for each T in R. Note that
9,(O*(T)) = @ , ( U T U * ) = U@,(T)U* = 9,(T)UU* = @ , ( T ) for each T in R,where 9, for all t in R.
01
is implemented by U in A. Thus 9,001 =
816
CROSSED PRODUCTS
(iv) Since u U ( a U ( T ) A=) w,(TA) for each T in R and all A in d, the equality w, o 9, = w , J R results from replacing A by I . Conversely, given that w, o 9, = w,lR for some conditional expectation !PU of 'R onto A, we have that
for each T in
R and all A in A. Thus
and Q,(T) = 9 , ( T ) since w,lR is faithful. It follows that the condition w, o 9, = w,l R determines 9, uniquely. If H E R+ and 9,(H) = 0, then from the formula just proved,
w,(H) = WU(9,(H)) = 0. Again,
H = 0 since w,I R is faithful. Hence 9, is faithful.
13.4.25. Let M be the (left) von Neumann algebra C F corre~ sponding to the free group 3 2 on two generators a and b (see Example 6.7.6) and let u be the unit trace vector for M corresponding to the function that takes the value 1 at the unit element of F2 and 0 at all other elements. (i) Identify the modular group of M corresponding to ( M , u ) . (ii) Identify the group of unitary elements in M that implement some element of the modular group found in (i). (iii) Let A be the maximal abelian subalgebra of M corresponding to the generator a. (See Exercise 6.9.42.) Note that A is contained in the centralizer of w u ( M ,and let 9 be the conditional expectation of M onto A described in Exercise 13.4.24(iii). How does 9 compare to cp of Exercise 8.7.28(iii) (where A and M replace R and S,respectively)? (iv) With L , in M , show that Q(L,) = L,,, where $ ' ( a n ) = x ( a n ) for each n in Z and x'(c) = 0 when c 4 { a n : n E 2). [Hint. Recall from Exercise 13.4.24, that w, o 9 = w,lM.]
Solution. (i) Since o,lM is a faithful normal tracial state of M , the centralizer of w,I M is M . From Theorem 9.2.13 and Proposition 9.2.14, each element of M is mapped onto itself by each automorphism of the modular automorphism group corresponding
EXERCISE 13.4.25
817
to ( M , u ) . This modular automorphism group consists just of the identity automorphism of M in this case. (ii) From (i), a unitary operator in M implements one of the automorphisms of the modular group for (M, u ) if and only if it lies in the center of M. The group of such unitary operators is therefore {cl : Icl = 1) since M is a factor. (iii) In this case, 9 = Q for w, o 9 = w, o Q = w,I M, from Exercise 13.4.24(iv ). (iv) Since w, o @ = u,I M and L,u = 2, we have
( z , A u ) = (A*L,u,u) = w,(@(A*L,)) = u,(A*@(L,)) = ( @ ( L x ) uAu) ,
= (Y,AU), for each A in A, when L , = ~(L,)(E A ) . From Exercise 6.9.41(ii), y(c) = 0 if c # an for all n in Z. If z is 1 at an and 0 at other elements of F2,then L , E A and
It follows that y = z' and 9 ( L , ) = LXt.
CHAPTER 14 DIRECT INTEGRALS A N D DECOMPOSITIONS
14.4.
Exercises
If 7-l is the direct integral of the Hilbert spaces {'H,} over (X, p ) and cp, 'p' are two decomposable representations of a C*algebra U on 'H, show that cp = 'p' if cpp = cpl, almost everywhere. 14.4.1.
Solution. Since cp and cp' are decomposable, for each A in U, cp(A)(p) = cp,(A) and cp'(A)(p) = cpl,(A) almost everywhere. But QP - cpl, almost everywhere, so that cp(A)(p) = cp'(A)(p) almost rn everywhere. From Remark 14.1.7, cp(A) = cp'(A). Thus cp = 9'. 14.4.2. If 'FI is the direct integral of the Hilbert spaces {'H,} over (X,p) and p --+ p ' , p --+ cpl, are decompositions of the representation cp of the norm-separable C*-algebra U, prove that Q, = cpl, almost everywhere.
Solution. Let { Aj} be a countable norm-dense subset of U. By hypothesis, there is a null set N j , such that
for each p in X \ N j . Let N be UE1Nj. Then N is a null set and cpp(Aj) = cpl,(Aj) for all j when p E X \ N . Since ' p p and cpl, are norm continuous and {Aj} is norm dense in U; cp,(A) = cpl,(A) for rn all A in U, when p 4 N . Thus cpp = cpl, almost everywhere. In the five exercises that follow, we outline some basic results of the theory of locally compact, abelian groups. Our goal, attained in Exercise 14.4.10, is a strengthening of the result of Exercise 13.4.23(iii). Our starting point is Haar measure on such groups [H:
EXERCISE 14.4.3
819
pp. 250-2631. We indicate the group operations additively (the group “product” of s and t is s t and the group “inverse” to t is -t), the group identity by 0, and the integral o f f (in LI(G)) relative to Haar measure by J f ( t ) d t .
+
14.4.3. Let G be a a-compact, locally compact, abelian group. (i) Examine the discussion beginning with Definition 3.2.21 and through to the statement of Proposition 3.2.23, with G in place of W, and conclude that it remains valid. (ii) Show that the non-zero linear functional p on L1(G) is multiplicative if and only if there is a character x of G (that is, a continuous homomorphism of G into “1) such that, for each f in L1(G),
[Hint.See the proof of Theorem 3.2.26.1 (iii) Let {U, : a E A} be a neighborhood base of 0 in G such that each U, has compact closure, and let u, be a positive function in L1(G) such that llualll = 1 and u,(t) = 0 when t 4 24,. Show that 11 f * u, - fll, -+ 0 over the net A (directed by inclusion of the neighborhoods U,)for each f in L,(G). [Hint. See Lemma 3.2.24.1 (iv) Show that, for each r in G, different from 0, there is a character x of G such that ~ ( r ) 1. [Hint. Use the pattern of the construction in the comments following the proof of Lemma 3.2.24, in conjunction with (iii), to obtain an abelian C*-algebra S ( G ) (acting on Lz(G)). The pure states of S ( G ) give rise to characters of G through the construction indicated in the proof of Theorem 3.2.26.1
+
Solution. (i) Lebesgue measure is Haar measure on B. Definition 3.2.21, the comments following it, and Proposition 3.2.22 remain valid, as do their proofs, when G replaces B and Haar measure on G replaces Lebesgue measure on B. (ii) Following the pattern of the proof of Theorem 3.2.26, we note that if p, as defined in (ii), is 0, then x is 0 almost everywhere on G, which is absurd. With ~ ( tfor ) expitr and f(x) for f(r) (and the other corresponding replacements), the equation of 3.2(4) remains valid, so that p is multiplicative on L1(G). Assume, now, that p is multiplicative, and adjoin a unit to Ll(G) as described in Remark 3.1.3, if one is not present. We extend p linearly to the algebra with unit adjoined, if necessary, and
820
DIRECT INTEGRALS AND DECOMPOSITIONS
conclude, using Proposition 3.2.20, that p is a bounded linear functional on L l ( G ) . The argument, in the proof of Theorem 3.2.26, defining ( and showing that ( is a homomorphism establishes that ( is a homomorphism of G into T I when R is replaced by G. To show that (, as constructed, is continuous on G, we make use of the assumption that G is a-compact. It follows from this assumption that the set of continuous functions on G that vanish on the complements of compact sets is dense in Ll(G). Thus there is a continuous function f vanishing on the complement of a compact subset K of G such that p ( f ) # 0. Let V be a compact neighborhood of 0 in G . Then V K ( = {w k : w E V , k E K } ) is a compact subset of G (as the continuous image of the compact subset V x K of G x G under the mapping (g,g') 4 g 9'). If T E V ,
+
+
+
and this integral is small for T in a sufficiently small neighborhood of 0 by uniform continuity of f on V K . (iii) Assume, first, that f is continuous on G and vanishes outside the compact subset K of G. With E positive, there is a U, such that If(s) - f ( t ) l < E when s - t E U,. If ub C U,,
+
In addition, f
* U b - f vanishes outside u b + I<. Hence 1. [If* ub - flip 5 &CP 7
where c is the (Haar) measure of ub t K (which is finite since Ur t K is compact), and ua - fll, + 0 over the net A. For an arbitrary f in L,(G), we can choose {fm} so that - f m I l p -, 0, llfmllp 5 I l f l l P , where each fm is continuous and vanishes outside a compact subset of G. Then
[If *
[If
821
EXERCISE 14.4.4
Choosing m large enough, for all b, we have
From what we have proved, there is a U, such that if Ub C
U,,then
llfm * 21b - fmllp 5 &/3. Hence [If * ub - flip 5 6 for dl such b.
(iv) Replacing R by G in the comments following the proofs of Proposition 3.2.22 and Lemma 3.2.24, we define dI(G) and 2ll(G). From (iii), Lf = 0 if and only if f is 0 almost everywhere for f in LI(G). Thus the mapping f -+ L j of LI(G) onto dI(G) is a * isomorphism. Let %(G) be the linear space generated by I and 2ll(G). Then &(G) is an abelian C*-algebra. Let po be a pure state of &(G) and, with f in Ll(G), let p ( f ) be po(Lf). Then p is a multiplicative linear functional on L l ( G ) . If pol%l(G) # 0, then p gives rise to a character ( on G as described in (ii). Moreover,
If each character of G takes the value 1 at a given element T of G, then p ~ ( L j , = ) p o ( L j ) for each f in LI(G) and each pure state po of %(G). In this case, Ljr = L j and fr = f for each f in Ll(G). If T # 0, we can choose a neighborhood V of 0, with compact closure, such that V and V - T are disjoint. Let f be a continuous function on G that vanishes outside of V and takes the value 1 at 0. Then fr vanishes on V so that fT and f are not equal almost everywhere. Thus T is 0 if [ ( T ) = 1 for each character [ of G. 14.4.4. Let G be a a-compact, locally compact, abelian group. Show that (i) the mapping t + f t ( t E G,f E L,(G)) is a continuous mapping from G into the normed space Lp(G) [Hint.Establish this first, when f is continuous and vanishes outside a compact set.]; (ii) f * g is continuous when f E LI(G) and g E L,(G); (iii) S-S(= ( 8 - 8 ' : 8,s' E S}) contains an (open) neighborhood of 0 when S is a measurable subset of G having positive measure. [Hint.Use (ii) with f and g replaced by the characteristic functions of S and -S.]
Solution. (i) Suppose f is continuous and vanishes outside the compact subset K of G. Let V' be a compact neighborhood of 0 in
822
DIRECT INTEGRALS AND DECOMPOSITIONS
G and let b be the (finite) measure of K + V'. By uniform continuity of f , there is an open neighborhood V of 0 such that V C_ V' and
when t - t' E V . If t - t' E V , then
llft - ft4;
J If(s - t ) - f(s - t')lPds = J If(s t t' - t ) ds
=
f(s)lP
Since the continuous functions that vanish outside of compact subsets of G form a norm-dense subset of Lp(G),given a function g in L p ( G ) ,we can choose a sequence {f(")} of continuous functions - slip + 0. vanishing outside compact sets in G such that Il f (") Then
Given a positive E , we can choose m so large that 119 - f(m)llp < ~ / 3 . With this choice, and from what we established in the preceding paragraph, there is an open neighborhood V of 0 in G such that - fj?)llp < ~ / when 3 t - t' E V . With t - t' in V , we have, then, that 11gt - gttllP < E . It follows that the mapping t + ft is (uniformly) continuous from G into Lp(G). (ii) Note, first, that for each s in G,
[If!")
Thus (f * g)(s) converges for each s in G and f t g is defined on all
823
EXERCISE 14.4.5
Given a positive E , there is, from (i), an open neighborhood V of 0 such that l l f p - frjII1 < E/llglloo when T - T’ E V. Thus I(f * g)(s) - (f * g)(s‘)l < E when s’ - s E V , and f * g is (uniformly) continuous on G. (iii) Since G is a-compact and compact subsets of G have finite Haar measure, S has a measurable subset of finite, positive measure. Replacing S by this subset, we may suppose that S has finite, positive measure. In this case, f E &(G) and g E L,(G), where f and g are the characteristic functions of S and -S, respectively. From (ii), f * g is continuous on G. Now t --+ f ( t ) g ( r - t ) is the characteristic function of S n ( T 5’) so that (f * g ) ( T ) is the Haar measure of Sn(r+S). Let 0 be the set of points T in G such that (f*g)(T) > 0. Then 0 is an open subset of G containing 0 (since S is assumed to have positive measure). If T E 0, then S n ( T S) has positive measure. In particular, S n ( T t S) has a non-empty intersection, and T + s = s’ for some s and s’ in S. Thus T E S - S when T E 0, and 0 S - S.
-+
+
14.4.5. A mapping cp from a measure space into a topological space X is said to be measurable when cp-l(O> is a measurable set for each open subset 0 of X . Let G be a a-compact, locally compact, abelian group, and let cp be a measurable homomorphism of G into a topological group H that contains a countable dense subset. Show that cp is continuous. [Hint.Use Exercise 14.4.4(iii).]
Solution. Let V‘ be an open neighborhood of the identity in H , V be an open neighborhood of that identity such that VV-I C V’, and { h l , h2,. . .} be a countable dense subset of H . With h in H , V - l h is an open set containing h. Thus some hj lies in V-’h
824
DIRECT INTEGRALS AND DECOMPOSITIONS
and h E V h j . It follows that { V h l , V h z , ...} covers H . Let Sj be p-'(Vhj). Then S j is a measurable subset of G and UGip"=ISj = G. Hence some Sj has positive measure. F'rom Exercise 14.4.4(iii), there is an open neighborhood U of 0 in G such that U 5 Sj - Sj. Since cp is a homomorphism
It follows that cp is continuous at 0. Hence cp is continuous on G. Let E be a topological group, C be a closed normal 14.4.6. subgroup of E, x be a continuous, idempotent (x(x(g)) = x(g)) homomorphism of E onto C , H be the kernel of x, and cp be the quotient mapping of E onto E / C . Show that (i) x(c) = c for each c in C ; (ii) s = hc for each s in E , where
(iii) cp(U) = cp(0)for each (relative open) subset U of H , where 0 = (-'(U); (iv) cp restricts to a topological group isomorphism (homeomorphism and group isomorphism) of H onto E / C provided with the quotient topology (the open subsets of E / C are the images under cp of the open subsets of E ) . Solution. (i) Since x maps E onto C , c = x(s) for some s in E , when c E C. Thus x(c) = x(x(s)) = x(s) = c, since x is assumed to be idempotent. (ii) With s in E and e the identity of E ,
from (i), since x(s)-' E C. Thus ([(s) =)sx(s)-' E H ;and s = s~(s)-~x(s= ) hc, where h = sx(s)-' E H and c = x(s) E C. (iii) Note, first, that if u E 24, then
((u)= ux(u)-l = ue-' = u E U ,
EXERCISE 14.4.7
825
whence u E [-'(U)= 0. Thus U c 0 and cp(U) cp(0). If s E 0, then ((s) E U and s = ((s)c with c in C from (ii). Thus ~ ( s= ) cp([(s))cp(c) = cp([(s)) since cp(c) is the identity of E / C . It C_ cp(U). ) Thus cp(U) = cp(0). follows that cp(s) E cp(U), whence ( ~ ( 0 (iv) If s = hc with h in H and c in C , then cp(s) = cp(h). Since each s in E has a decomposition of the form hc, from (ii), cp maps H homomorphically onto E / C . If cp(h) = cp(h'), with h and h' in H , then h-'h' E C. From (i), h-'h' = x(h-'h') = x ( h ) - l x ( h ' ) = e-'e = e ,
and h = h'. Since cp is continuous on E , the restriction of cp to H is a continuous isomorphism of H onto E / C . Note that [ is continuous since ((s) = sx(s)-' and x is assumed to be continuous. With U and 0 as in (iii) and U a (relative) open subset of H , we have that 0 (= (-'(U)) is open in E . Thus cp(0) is open in E / C . From (iii), cp(U) (= ~ ( 0 is) open ) in E / C . Thus cp is also an open mapping of H onto E / C . It follows that cp is a rn homeomorphism and group isomorphism of H onto E / C . [E. Kehlet provided a special case of this exercise.]
14.4.7. Let E be a topological group, C be a compact normal subgroup of E , and cp be the quotient mapping of E onto E / C . Suppose E / C is locally compact. (i) With V a compact subset of E / C containing the unit C of E / C and (0, : a E A} an open covering of cp-l(V) (= U ) , let {O1(g), . . . ,0,(g)} be a finite subcovering of gC for g in U . Let 0 be Uy=lOj(g). Find an open neighborhood 0, of the unit e of E such that gC0, c 0. [Hint. For each h in gC,find an open neighborhood Vh of e such that hVhVh 0.1 (ii) With the notation of (i), show that (cp(g0,) : g E U } is an open covering of V . (iii) With the notation of (i) and (ii), let {cp(g(1)0,(1))7'*- 7 cp(g(m)0,(m))l
be a finite subcovering of V . Show that
{ ~ l W ) ) , *- *9aL(l)(g(1)),** .,0 l ( g ( m ) ) , . . . , 0 ~ ( m ) ( g ( m ) ) } is a covering of U and a finite subcovering of (0, : a E A}. (vi) Conclude that U is compact and that E is locally compact.
826
DIRECT INTEGRALS AND DECOMPOSITIONS
Solution. (i) Note, first, that for g in U ,
since cp(g) E V. As gC is compact, the covering (0,) of U admits the finite subcovering {Ol(g),...,0n(g)} of gC. Let 0 be U;=lOj(g). Then 0 is open and gC C 0. For each h in gC,there is a neighborhood Vh of e in E such that hVhVh c 0. From the open covering {hvh : h E gc},we can select a finite subcovering {h(l)V,(l), . ..,h(m)Vh(,)}of gC. Let 0,be ny=lVh(j). If h E gC, then h E h(j)Vh(j) for some j in (1,.. . , m } . Thus
and gC0, C_ 0. (ii) Since cp(U) = V and is open and
'p
vc
is an open mapping, each set 'p(g0,)
u cp(g0,).
g€U
Thus (cp(g0,) : g E U } is an open covering of V. (iii) With g in U , we have that 'p(g) E cp(g(j)O,(j)) for some j in { 1,. . . ,m } . From (i) and since C is a normal subgroup of E ,
Hence {01(9(1)),* * * 7 &(l)(g(1))9 * * ' 9 01(g(m)), ' * * 7 on(m)(g(m>)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
DIRECT INTEGRALS AND DECOMPOSITIONS
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
DIRECT INTEGRALS AND DECOMPOSITIONS
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
DIRECT INTEGRALS AND DECOMPOSITIONS
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
DIRECT INTEGRALS AND DECOMPOSITIONS
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
DIRECT INTEGRALS AND DECOMPOSITIONS
(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
INDEX OF VOLUMES 111 AND IV
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
INDEX OF VOLUMES 111 AND IV
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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