Von Neumann Algebras

North-Holland Mathematical Library Board of Adasory Editors:

M. Arlin, H. Bass, J. Eei Is, W. Feit, P. J. Freyd, F. W. Gehring, H. Halberstam, L. V. Hi5rmander, M. Kac, J. H. B. Kemperman, H. A. Lauwerier, W. A. J. Luxemburg, F. P. Peterson, 1. M. Singer and A. C. Zaanen

VOLUME 27

itm

NORT1-1-HOLLAND PUBLISHING COMPANY AMSTERDAM • NEW YORK • OXFORD

Von Neumann Algebras J acques DIXMIER Université Paris 6

Paris, France

1931

N.H 1981 PC

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Library of Congress Cataloging in Publication Data

Dixmier, Jacques. Von Neumann algebras.

(North-Holland mathematical library ; y. 27) Translation of : Les a lgebres d opérateurs dans 1' espace hilbertien (a lgébres de Von Neumann) . Bibliography: p. Includes indexes. - 1. Von Neumann algebras. 2. Hilbert space. I. Title. II. Seri,es. QA326.D513 81-16835 512'.55 ISBN 0- 11- 114-86308-7 AACR2

PRINTED IN THE NETHERLANDS

CONTENTS

PREFACE TO THE ENGLISH EDITION: RECENT DEVELOPMENTS IN THE THEORY OF VON NEUMANN ALGEBRAS, by E. C. Lance 1. Modular theory 2. Connes' classification of type III factors 3. Structure theory for type III factors 4. Examples 5. Classification of injective factors PREFACE TO THE SECOND EDITION INTRODUCTION

xi xi xv xviii xxiii xxix xxxv xxxvii

PART I. GLOBAL THEORY CHAPTER 1.

1.1. 1.2. 1.3. 1.4. 1.5. 1.6. 1.7.

DEFINITION AND BASIC PROPERTIES OF VON NEUMANN ALGEBRAS Commutant and bicommutant Hermitian operators in a von Neumann algebra . . Unitary operators in a von Neumann algebra . . Cyclic projections in a von Neumann algebra . Homomorphisms Ideals in von Neumann algebras Maximal abelian von Neumann algebras

1 1 3 4 4 7 9 12

CHAPTER 2.

2.1. 2.2. 2.3. 2.4.

ELEMENTARY OPERATIONS ON VON NEUMANN ALGEBRAS Induced and reduced von Neumann algebras Product of von Neumann algebras Operators in a tensor product of Hilbert spaces Tensor products of von Neumann algebras

CHAPTER 3. DENSITY THEOREMS 3.1. Topologies on L(H) 3.2. The above topologies compared 3.3. Linear foi-ms on L(H) 3.4. The von Neumann density theorem 3.5. Kaplansky's density theorem

17 17 20 22 25 33 33 37 38 44 47

vi

CONTENTS

CHAPTER 4. POSITIVE LINEAR FORMS 4.1. Positive linear forms on a *-algebra of operators 4.2. Normal positive linear forms on a von Neumann algebra 4.3. Normal positive linear mappings 4.4. Structure of normal homomorphisms Isomorphisms of tensor products 4.5. Application: 4.6. Support of a normal positive linear form 4.7. Polar decomposition of a linear form 4.8. Decomposition of a hermitian form into positive and negative parts CHAPTER 5.1. 5.2. 5.3. 5.4. 5.5.

5. HILBERT ALGEBRAS Definition of Hilbert algebras The commutation theorem Bounded elements in Hilbert algebras Central elements in Hilbert algebras Elementary operations on Hilbert algebras

CHAPTER 6. TRACES 6.1. Definition of traces 6.2. Traces and Hilbert algebras 6.3. Trace-elements 6.4. An ordering in the set of traces isomorphisms of standard 6.5. An application: von Neumann algebras 6.6. Normal traces on L(H) 6.7. A first classification of von Neumann algebras 6.8. Classification and elementary operations 6.9. The commutant of the tensor product of two semi-finite von Neumann algebras 6.10. The space Ll defined by a trace 6.11. Trace and determinant

53 53 56 59 61 62 63 65 71 77 77 79 81 83 85 93 93 97 101 102 104 105 109 114 116 117 119

CHAPTER 7.1. 7.2. 7.3.

7. ABELIAN VON NEUMANN ALGEBRAS Basic measures Existence of basic measures Structure of abelian von Neumann algebras

127 127 129 132

CHAPTER 8.1. 8.2. 8.3. 8.4. 8.5.

8. DISCRETE VON NEUMANN ALGEBRAS A second classification of von Neumann algebras Abelian projections Discrete algebras and elementary operations Definition of types Complete Hilbert algebras and type I factors

137 137 138 141 141 142

CHAPTER 9. EXISTENCE OF DIFFERENT TYPES OF FACTORS . 9.1. A lemma 9.2. Construction of certain von Neumann algebras .

147 147 148

CONTENTS

Examples taken from measure theory Existence of different types of factors

9.3. 9.4.

vii

151 153

PART II. REDUCTION OF VON NEUMANN ALGEBRAS

CHAPTER 1.1. 1.2. 1.3. 1.4. 1.5. 1.6. 1.7. 1.8.

1. FIELDS OF HILBERT SPACES Borel spaces, measures Fields of vectors Measurable fields of Hilbert spaces Basic properties of measurable fields of Hilbert spaces Square-integrable vector fields Basic properties of direct integrals Measurable fields of subspaces Measurable fields of tensor products

161 161 162 164 166 168 170 173 174

CHAPTER 2. FIELDS OF OPERATORS 2.1. Measurable fields of linear mappings 2.2. Examples 2.3. Decomposable linear mappings 2.4. Diagonalisable operators 2.5. Characterisation of decomposable mappings 2.6. Constant fields of operators

179 179 180 181 185 187 188

CHAPTER 3.1. 3.2. 3.3.

195 195 196

3.4. 3.5. 3.6.

CHAPTER 4.1. 4.2. 4.3. 4.4. 4.5. 4.6.

3. FIELDS OF VON NEUMANN ALGEBRAS A preliminary theorem Measurable fields of von Neumann algebras . . Relations between a decomposable von Neumann algebra and its components Constant fields of von Neumann algebras Reduction of discrete or continuous von Neumann algebras Measurable fields of homomorphisms 4. FIELDS OF HILBERT ALGEBRAS Measurable fields of Hilbert algebras Decomposable Hilbert algebras Involution and von Neumann algebras associated with 1.1 Elements bounded relative to LI Central elements relative to ti Uniqueness and existence of the decomposition

CHAPTER 5. FIELDS OF TRACES 5.1. Measurable fields of traces 5.2. Decomposition of traces 5.3. Uniqueneqs of the decomposition 5.4. Reduction of properly infinite, purely infinite, finite and semi-finite von Neumann algebras . . .

198 201 204 207 211 211 212 213 215 217 218 223 223 225 228 230

CONTENTS

viii

DECOMPOSITION OF A HILBERT SPACE INTO A DIRECT INTEGRAL Posing the problem Existence theorems Uniqueness theorems

CHAPTER 6. 6.1. 6.2. 6.3.

233 233 233 237

PART III. FURTHER TOPICS 1. COMPARISON OF PROJECTIONS Comparison of projections A theorem on comparability Cyclic projections of A and cyclic projections of A' Applications: I. Properties of cyclic and separating elements Applications: II. Characterisation of standard von Neumann algebras

243 243 245

CHAPTER 2. CLASSIFICATION OF PROJECTIONS 2.1. Definitions 2.2. Cyclic projections of A and cyclic projections of A' 2.3. Finite projections 2.4. Semi-finite projections 2.5. Properly infinite projections 2.6. Purely infinite projections 2.7. Comparison of projections and dimension

259 259

CHAPTER 3. MORE ON DISCRETE VON NEUMANN ALGEBRAS . 3.1. Structure of discrete von Neumann algebras . . . Isomorphisms of discrete von Neumann algebras 3.2.

269 269 270

CHAPTER 1.1. 1.2. 1.3. 1.4. 1.5.

CHAPTER 4.1. 4.2. 4.3. 4.4.

4. OPERATOR TRACES Definition, Traces on Z1Relations between scalar traces and operator traces Existence and uniqueness theorems for operator traces

248 249 252

260 260 262 263 264 265

275 275 276 277 280

CHAPTER 5. AN APPROXIMATION THEORY 5.1. The approximation theorem 5.2. An application: two-sided ideals of A and ideals of Z 5.3. Characters of finite von Neumann algebras

285 285

CHAPTER 6. THE COUPLING OPERATOR 6.1. The coupling operator 6.2. Properties of the coupling operator

297 297 299

288 291

CONTENTS

6.3. 6.4.

Applications: I. Comparing the strong and ultra-strong, weak and ultra-weak topologies . . Applications: II. Conditions for an isomorphism to be spatial

CHAPTER 7. HYPERFINITE FACTORS 7.1. Factors contained in a finite von Neumann algebra 7.2. Existence and uniqueness of continuous hyperfinite factors 7.3. Some inequalities 7.4. A new definition of hyperfinite factors 7.5. Hyperfinite factors and elementary operations 7.6. Further examples of finite factors 7.7. Existence of finite, non-hyperfinite, factors

301 304 307 307 308 310 313 318 319 321

CHAPTER 8. 8.1. 8.2. 8.3. 8.4. 8.5. 8.6. 8.7.

ANOTHER DEFINITION OF FINITE VON NEUMANN ALGEBRAS Statement of the theorem Fundamental projections Weights on the set of fundamental projections Construction of a trace to within E The proof of the theorem concluded Consequences of the theorem More on tensor products

ix

329 329 330 331 334 337 338 343

CHAPTER 9. 9.1. 9.2. 9.3. 9.4.

DERIVATIONS AND AUTOMORPHISMS OF VON NEUMANN ALGEBRAS Derivations of algebras Derivations of C*-al4ebras: continuity, extension Derivations of von Neumann algebras Automorphisms of von Neumann algebras

349 349 350 352 355

APPENDIX I 1. Spectrum 2. Spectral measures 3. Extending the Gelfand isomorphism

359 359 360 360

APPENDIX II

363

APPENDIX III 1. Support of an operator 2. Partial isometries 3. Polar decomposition of an operator

365 365 365 366

APPENDIX IV

367

CONTENTS

APPENDIX V 1. Borel sets 2. Polish spaces 3. Souslin sets 4. Measurability of Souslin sets 5. Existence of measurable mappings

371 371 371 372 373 374

SOME REMARKS CONCERNING THE TERMINOLOGY

377

INDEX OF NOTATION

381

INDEX OF TERMINOLOGY

383

REFERENCES Journal articles General texts Books, monographs and conference proceedings on operator algebras

387 387 434 435

PREFACE TO THE ENGLISH EDITION: RECENT DEVELOPMENTS IN THE THEORY OF VON NEUMANN ALGEBRAS by E. C. Lance In his introduction to the companion volume on C*-algebras, J. Dixmier makes the comment that at the time when he wrote this book, the major part of the theory of von Neumann algebras seemed to him to have reached a more or less definitive form. His belief is vindicated by the fact that during the twelve years which elapsed between the preparation of the first and second editions of the book, few major developments took place. By the time the second edition appeared in 1969, however, it was already clear that fundamental new advances had been made in the theory, and the following half-dozen years brought a succession of important results which carried the theory of von Neumann algebras to an altogether new level of richness and interest. Now, twelve years after the appearance of the second edition, the theory again seems to have reached a period of stability and it is possible to view in perspective these advances in the theory, due principally to Tomita, Takesaki and Connes, and to see how they develop and illumine the basic theory as presented in this book. The following account presupposes some familiarity with the material covered in the book. References to the body of the book are given by means of triples such as (I, 2; 3), meaning chapter 2, section 3 in part I. Bibliographic references are given at the end of each section. These are not intended to be complete: I have made no attempt to attribute each result to its original source, but merely to provide an orientation to the student who wishes to become familiar with the modern theory of von Neumann algebras.

1.

Modular theory.

The structure of a semifinite von Neumann algebra is described by Dixmier as folloWs. Let A be a semifinite von Neumann algebra. Then there is a faithful normal semifinite trace ci5 on A+ (I, 6, 7). There is a natural correspondence between such pairs (A, cp) and full Hilbert algebras (I, 6, 2), from which one sees that every semifinite von Neumann algebra is isomorphic to a

PREFACE

xii

standard algebra (I, 5, 5). Furthermore, if A, acting on a Hilbert space H, is standard then there is an involution J on H such that JAJ = A' and JCJ = C* for every C in the centre of A (III, 1, 5), and these conditions characterize standard von Neumann algebras. With this structure theory it is quite easy to obtain, for example, the commutation theorem for tensor products of semifinite algebras (I, 6, 9). In seeking to extend these results to general von Neumann algebras, one is guided by the motivating example of the algebras defined by a locally compact group (I, 5, Exercise 5). If G is a unimodular group then the algebra 1.1 of continuous functions on G with compact support is a Hilbert algebra whose associated semifinite von Neumann algebra has a faithful normal semifinite trace (i5 satisfying

(P(Ux ) = x(e)

(X E

(

i)

where e is the identity element of G. If G is not unimodular then U is no longer a Hilbert algebra and cj) is not a trace. So one searches for generalized versions of the concepts of Hilbert algebra and trace which are appropriate to this example. The generalization of the trace is easy to define, and is already anticipated by Dixmier at the end of (I, 4, 2). It is obtained by omitting condition (iii) from the definition of a trace in (I, 6, 1). Such functionals , or forms, are called weights. So a weight on a von Neumann algebra A is a function 0 A+ [0, OE] satisfying (i)

+ T) = c(S) + c(T)

(ii) (XS) = X(S)

(S, T E

A+ )

(S E A+ ,

X

0) .

Faithfulness, normality, finiteness and semifiniteness are defined for weights exactly as for traces (I, 6, 1). We shall refer to a faithful normal semifinite weight as an fns weight. The problem about normality raised at the end of (I, 4, 2) is answered by the following result of Haagerup.

For a weight cp on a von Neumann algebra A, the following conditions are equivalent: (i) (I) is normal, THEOREM [756].

cp is completely additive (that is, (1)(Zxi ) =Z(I)(xi ) for any set {xi} ce for which Zxi is defined), (iii)(1) is ultraweakly lower-semicontinuous. (iv)(1)(x) = Zcp i (x) (x E A± ), where {cp.} is a set of positive normal functionals.

RECENT DEVELOPMENTS

It is straightforward to prove that, for any von Neumann algebra A, there is an fris weight (1) on M. The set

nqb

= {TE A:

4)(T*T) < oe}

is a dense left ideal in A. The space nq5nn$ = U an associative algebra over the complex numbers C, with a scalar product given by

(S, T) = 11)(T*S) and an involution T T*. But it is not a Hilbert algebra since the involution is in general not isometric (or even bounded) as a conjugate-linear operator on the scalar product space. An attempt to generalize the concept of Hilbert algebra was made by Dixmier, who introduced the idea of a quasi-Hilbert algebra (I, 5, 1). This is suitable for describing the algebra of continuous functions with compact support on a locally compact group (I, 5, Exercise 5), but not the algebra U(p described above. The crucial breakthrough was made by Tomita, who observed that, in the algebra 4, the involution is closable as a conjugatelinear operator on the scalar product space. (Tomita's work was never published, but an amplified version of it was given by Takesaki [[Ee]].) Define a left Hilbert algebra to be an associative linear algebra U over C, with a scalar product (x, y) which makes it a prehilbert space (whose Hilbert space completion will be denoted H) and an involutive antiisomorphism X x0 , such that

(i)the map x x0 is closable as a densely-defined conjugate-linear operator on H,

(ii)(xy, z) = (y, x°2) (x, y, z E U), (iii)for each x in U, the map y 4- xy is continuous, (iv)the set of elements xy (x, y E LI) is total in U. These axioms are satisfied both by group algebras (with

x # -- x*, ) and by the algebras U0 described above.

As in (I, 5, 1), operators Ux in L(H) are defined by Uccy = xy (y E U). These operators form an involutive algebra whose weak closure U(U) is called the left von Neumann algebra associated with U. Generalizing the definition in (I, 5, 5), we say that a von Neumann algebra A acting on a Hilbert space H is standard if there is a left Hilbert algebra U which is dense in H such that A = U(U). The closure of the mapping x x# on H is called the sharp Operation and is denoted by S. As a closed conjugate-linear

PREFACE

xiv

operator on H, S has a polar decomposition. This is written as S - Jel, where J is an involution (conjugate-linear isometry) on H and A is a positive self-adjoint linear operator (unbounded, in general) on H. Tomita's main theorem can be expressed as follows.

Let U be a left Hilbert algebra and define J, A as above. Then THEOREM [[Ee]].

(i)JU(U),J = U(U)',

(ii)AltU(U) A-it = U( U)

(t e R, the real numbers),

(iii)for c in the centre of U(U), JCJ = C* and Aitcrit = C. This difficult and powerful theorem gives a structure theory for arbitrary von Neumann algebras which generalizes the theory for semifinite algebras described at the beginning of this section. Thus every von Neumann algebra is isomorphic to a standard algebra, and a standard algebra is antiisomorphic to its commutant. This gives a positive answer to the question at the end of (I, 6, 8). Also, the commutation theorem holds for tensor products of arbitrary von Neumann algebras, giving a positive answer to the question at the end of (I, 2, 4). In addition to solving these problems, Tomita's theorem introduces a quite new and unexpected element into the theory, namely the one-parameter automorphism group at given by

at (S) = A

it

SA

-it

(S E

U(U) , t

E

R) ,

which is called the modular automo phism group of U(U). We shall see that modular automorphism groups furnish a powerful tool for analyzing von Neumann algebras. Suppose 11) is an fns weight on a von Neumann algebra A. We construct the left Hilbert algebra (4 and note that U(4) is naturally isomorphic to A. By means of this isomorphism we can transport the modular automorphism grou of U(Up) to a group of automorphisms of A which we denote by a. We shall call this group the modular automorphism group of (A, cp). The group (e) t is characterized by an important property which we shall now describe. With A, (1) as above, suppose that (at ) is a strongly-continuous one-parameter group of automorphisms of A which leave ci5 invariant. We say that (15 satisfies the KMS condition with respect to (at ) if for each pair of elements S, TE 4 there exists a complex-valued function F which is continuous and bounded on the strip {Z E C:

irrl Z

1}

RECENT DEVELOPMENTS

XV

and analytic in its interior, such that F(t) = Cu t (S)T) ,

F(t + i) = cP(Ta t (S))

(t ER)

.

(This condition first arose in physics; the initials stand for Kubo, Martin and Schwinger.)

If cp is an fns weight on a von Neumann algebra A and (at ) is a strongly-continuous one-parameter group of automorphisms of A which leave cP, invariant then cP satisfies the KMS condition with respect to (at) if and only if at = (t E R). THEOREM [543], [[Ee]].

Thus, given an fns weight on a von Neumann algebra, we have a characterization of the associated modular automorphism group which is intrinsic in the sense that it is formulated purely in terms of the von Neumann algebra, without any reference to left Hilbert algebras. In fact, the whole modular theory can be developed intrinsically, and the best expositions of modular theory use an intrinsic approach. However, the formulation in terms of left Hilbert algebras, apart from being the first historically, clearly fits in best with Dixmier's approach to the theory of semifinite von Neumann algebras. References : [543], [659], [696], [857], [890], [[Ee]].

2.

Connes' classification of type III factors.

It will be convenient at this point to change notation to bring it into line with modern usage. From now on, von Neumann algebras will be denoted by capital letters (typically M, N, R) and their elements by lower case letters (x, y, e, ...). Let Aut(M) denote the group of all *-automorphisms of the von Neumann algebra M, with the topology of simple ultraweak convergence. Let G be a locally compact abelian group. By an automorphic action of G on m we mean a continuous homomorphism from G into Aut(M). Suppose that a is an automorphic action of G on M. For f in L 1 (G) we can define a bounded linear transformation oc(f) on M by

u(f)x = f f(s)a s(x)ds G

(x E M),

where ds denotes Haar measure on G. Let G denote the dual group and for f in L 1- (G) let f denote its Fourier transform. Define a closed subset Sp(u) of G by Sp(a) .= {y Ea: f(y) = 0 whenever u(f) = 01.

The set Sp(a) is called the Arveson spectrum of a [685]. Perhaps the best intuitive understanding of Sp(u) is obtained from the following proposition, which describes Sp(a) as the set

PREFACE

xvi

of approximate eigenvalues for the action a.

Let a be an automorphic action of G on 1,4 and let y E É. Then y E Sp(a) if and only if there is a directed net {x 1 } of elements of 14 of norm one such that PROPOSITION [633].

s

(x) -(s, y) x

H

o

uniformly on compact subsets of G. Even more useful than the Arveson spectrum in analyzing automorphic actions is a subset r(a) of Sp(a) called the Connes spectrum of a, which is defined as follows. Let Ma denote the fixed-point subalgebra of M, m

a

= {xEM:

(X)

= X

(s E

G) }.

For each projection e in Ma, form the reduced algebra Me (I, 2, 1). For xe in Me and s in G, define a:(xe ) to be the restriction of as (x) to the range of e. This is well-defined, since e E Ma , and gives an automorphic action of G on Me . Define the Connes spectrum of a by F(a) =

n

Sp(ae ) ,

the intersection being taken over all projections in Ma. THEOREM [633]. The Connes spectrum r(a) is a closed subgroup of G. Let a, 13 be automorphic actions of G on M. We say that a and (3 are outer equivalent if there is a strongly-continuous mapping s us from G to the group of unitary elements of M such that

(i) (35 (x) = us as (x)u: (ii)u = u s as (u t ) s +t

(X E M, S E G)

(s,

tE

G) .

It is straightforward to check that this is indeed an equivalence relation. Notice that simple algebraic considerations dictate that a map s us satisfying (i) cannot be a group homomorphism, but should rather satisfy the "twisted" multiplication relation (ii); a map satisfying the condition is called a

unitary cocycle. THEOREM [633]. If a and 13 are outer equivalent autômorphic actions of G on M then F(a) = We now 'apply these concepts to the special case of modular automorphism groups. Let (1) be an fns weight on a von Neumann algebra M. The modular group (4)gives an automorphic action of R on M. In computing the spectrum of this action it is convenient to identify the dual group with the multiplicative group R positive reals, the duality being given by

xvii

RECENT DEVELOPMENTS

(

5

, y ) = y is

(S E

R, y

E R±)

The first important result is the following, known as the uni-

tary cocycle theorem. THEOREM [633]. If cP, II) are ins weights on the von Neumann algebra M then a cl) and 0 are outer equivalent. It follows that the closed subgroup F(a) of B):. is independent of (1) and is therefore an algebraic invariant of the von Neumann algebra M. We shall use this invariant to refine the Murrayvon Neumann classification of von Neumann algebras into types I, II and III (I, 8, 4). The classification is mainly of interest in the case where M is a factor, although one can obtain some results about general von Neumann algebras on a separable Hilbert space by the use of direct integral theory ([556], [866]). Before describing Connes' classification of factors, however, we want to discuss the relationship between semifiniteness of M and the modular automorphism groups of M. An action a of a locally compact abelian group G on a von Neumann algebra M is said to be inner if a is outer equivalent to the trivial action (in which each element of G acts as the identity automorphism of M). The action a is said to be pointwise inner if each automorphism as is inner. Clearly an inner action is pointwise inner. In general, the converse result is false ([633]), but in the particular case where G = R and M is a factor with separable predual it is true that pointwise inner implies inner (see [844]). THEOREM [[Ee]]. and only if cyci5 is

The von Neumann algebra m is semifinite if inner (for any fns weight cp on m).

It follows from the above that if M is any factor of type III with separable predual then M must have outer automorphisms. This considerably extends the remark at the end of (III, 9, 4). It also follows that if M is any semifinite von Neumann algebra then r(c14 ) = {1}. As we shall see below, the converse is false: it is possible for a von Neumann algebra of type III (even a factor with separable predual) to satisfy F(e) = {1}. Suppose now that M is a factor of type III. Since the Connes spectrum of its modular group is a closed subgroup of 4, it is specified by a unique number X in [0, If] such that

if X = o ,

{1) F(a (1) )

{Xn: n

Rx

E Z)

if o < X < 1, if X = 1.

The factor M is then said to be of type III. We shall see in Section 4 that for each X in [0, 1] there exist factors of type III A with separable predual (equivalently, acting on a separable Hilbert space).

xviii

PREFACE

The first important feature of Connes' invariant is that it can be effectively computed. Sometimes this can be done directly from the definition of the Connes spectrum, but there is another description of it which is often useful. We shall discuss this for the case of a factor with separable predual, since this is the most important case and since the description If M is such a factor then there then becomes somewhat simpler. is a faithful normal finite weight, in other words a faithful normal positive linear functional, on M. This can be normalized to take the value one at the identity and is then called a faithful normal state. Let (I) be a faithful normal state on M. The fixed point algebra of the modular group 04 is a von Neumann subalgebra of M denoted by M4 and called the centralizer of (1) because of the following result. PROPOSITION [[Ee]].

0 = {x

If (1) is a faithful normal state on m then

E M:

(xy) = cP(yx)

(y E m) } .

Let Z(M) denote the centre of M. Let A ci) be the modular operator associated with (15, so that A(1) is an unbounded selfadjoint operator which is the infinitesimal generator of a unitary group which implements the modular group (4, and denote its spectrum by sp(y. For each projection e in 0, the restriction of (1) to the reduced algebra Me is a faithful normal state cPe of Me . Define

S o (M) = » ( p(Acpe ), where the intersection is taken over all projections in Z(0). THEOREM [633, 635]. Let (1) be a faithful normal state on a type III factor . Then 0 E s o (m) , and S o (M) {0} - F(a4) . Another important description of Connes' invariant is given as follows. For any factor M, define S(M) =nsp(A cp), the intersection being taken over all fns weights THEOREM [633].

(i) 0 E S (M)

of M.

if and only if m is of type

(ii) F(a) = S(m) `-{0 } . References : [633], [685], [M], [[Aa]]. 3. Structure theory

for type III factors.

Let U be an automorphic action of a locally compact abelian group G on a von Neumann algebra M. We are going to construct a new von Neumann algebra called the crossed product of M by a. (In fact, the construction works equally well when G is nonabelian, but parts of the subsequent application to duality theory will require G to be abelian.) If G is discrete and M is abelian, this construction reduces to the classical group

RECENT DEVELOPMENTS

xix

measure space construction of Murray and von Neumann described by Dixmier in (I, 9, 2). Suppose M acts on a Hilbert space H and let L2 (G, H) be the Hilbert space of all (equivalence classes of) square-integrable H-valued functions on G. For x in M and s in G we define bounded operators ff(x), us on L2 (G, H) as follows: (71- (x)C) (t) = a

((t)) -

(EEL

2

(G, H) ,

t E G) .

(us C)(t) = C(t - s) The crossed product of M by a, denoted by M xa G or M, is the von Neumann algebra generated by all the operators ff(x), us . Notice that TT is a faithful representation of M as a subalgebra of M and that u Tr(x)u* = Tr(ct (s)) s s

(X E M,

S E

G) .

Thus if we identify M with its image ff(M), we can regard the automorphisms as of M as extending to inner automorphisms of M. If 0 is an isomorphism from M onto a von Neumann algebra N and a i8 an action of G on M then we define an action 0 a of G on N by as = Oa3 0 -1 (S E G). Given actions a, 13 of G on M, N respectively, we say that a and 13 are outer conjugate if there exists an isomorphism 0: M N such that 0 a is outer equivalent to

PROPOSITION [674]. If a is isomorphic to N )( G.

and 13 are outer conjugate then M x a G

Given an action a of G on M, construct the crossed product M on L2 (G, H) and, for y in consider the unitary operator Vy on L2 (G, H) defined by

a,

(V

Y

(s)

= (s, y)C(s)

(E E L 2 (G, H)

S E

G).

It is easy to check that conjugation by Vy leaves Ti(s) fixed, and has the effect on us of multiplying it by (s, y). So we obtain an automorphism ay of satisfying

A

ay (Tr (x) ) =

(x) ,

Y

(u ) = (s, y) us s

(x cm, s

G)

This gives an action & of G on M which is said to be dual to the action a. The dual action has many nice properties, for instance the following. PROPOSITION [933].

ff(m).

The fixed-point subalgebra

WI

is equal to

PREFACE

XX

The Connes spectrum of a is equal to the kernel of the restriction of the dual action to the centre of THEOREM [835].

F(a) = { y E G : a (x) = x

(x E Z(A)) ).

Y

Given an fns weight on M, on M, but this will require First, recall that if (15 and 04) are outer equivalent, so fut : t E RI with

A:

we want to construct a dual weight some preliminary considerations. 11) are fns weights on M then 04 and that there is a unitary cocycle

a(x) = u t atcl) (x)u*

(

x EM

)

.

We write ut = ( DIP : DOt and call (DIP : D(1)) the Radon -Ni kodym cocycle of If) with respect to (I). NOW let a be an action of G on M. Define K(G, M) to be the space of all *-ultrastrongly continuous functions from G to M with compact support and define operations on K(G, M) as follows:

( C" ) ( s)

ott C(st)T1( - t)dt

= G

el (s) = u_ s C(-s)* C.x(s)

C,

= C(s)x

E K(G, NI)

X

EM,

S

E G.

Under the first two of these operations, K(G, M) becomes an involutive algebra, and if we define

p(C) =f u Tr(C(s))ds G

s

then p defines a *-homomorphism from this algebra into a weakly dense subalgebra of Given an fns weight (15 on M, define 13 (1) to be the linear subspace of K(G, M) spanned by elements of the form

A.

C.x THEOREM

[886].

(C E

M)

X E

1)(X * X) < c° )

-

There is a unique fns weight CI) on m such that

( i) $(11 (e * C)) = (P((e * C)(e))

(CEN))

where e is the identity element of G, modular group 01 satisfies olt ( ff(x)) = Tr(d(x)) (x E M) uTr((r4o : DC1)) t ) t s)

(ii) the

(S E

G,

t

R) .

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The weight $ is said to be dual to cP. Let a be an action of G on M. Then we have an ction a of on M and so we can form a second crossed product A = A )( a G. Takesaki proved the following fundamental duality theorem.

a

THEOREM [674]. The algebra m is isomorphic to m 0 L(L 2 (G)). In fact, there is a canonical outer conjugacy relation between {A, CO and {m 0 L(L2 (G)), a 0 II. Suppose now that M is properly infinite and G is separable. It follows from the results of (III, 8, 6) that M 0 L(L2 (G)) is isomorphic to M by an isomorphism which transforms a 0 I to a. So in this case {A, 0 is outer conjugate to {M, a), which justifies calling the above result a duality theoreM. Now we apply these results to the case where M is a factor of type III. Let cp be an fns weight on M, so that 04 is an acti9n of R an M. We form M and look at the dual weight- $. Since e leaves (I) invariant, it is apparent from the defining properties of (1) that e is induced by the strongly continuous unitary group 'Cut : t ER) in M and is therefore inner. Thus M is semifinite, and in fact is a von Neumann algebra of type Moe . If we use the duality theorem (tIlis time identifying R with R under the usual , e 25t.;, we obtain Takesaki's basic structure duality (s, theorem for type III factors.

If m is a factor of type III then there is a von Neumann algebra N of type TI Œ, an fns trace T on N and a one-parameter group fO t : t ER) of automorphisms of N such that THEOREM [674].

TO

e

-t

and m is isomorphic to N xe, R. outer conjugacy.

T

(t E R)

The action 0 is unique up to

The fact that M is a factor implies that the restriction of O' to the centre of N is ergodic. We shall see in Section 5 that this ergodic flow on Z(N) sometimes determines M completely. For the present, we observe that, since we already know that Mach is the kernel of the flow, we can determine the type of M in terms of the flow. Thus it is easy to see that M is of type III 1 if and only if N is a factor (and so the flow is trivial). With more work, one can see that if M is of type III x (0 < X < 1) then elZ(N) is periodic, and in fact is isomorphic to the action of R)1! by translation on Loe (4/F(04 ));if M is of type 1110 then the flow is aperiodic and can be described in terms of a virtual subgroup of R)1 [835]. !

The structure theorem gives a great deal of information about type III factors, but it does have limitations: crossed products by continuous groups are not easy to work with, and it is advantageous where possible to describe a type III factor M

PREFACE

xxii

as the crossed product of a semifinite von Neumann algebra by a single automorphism e (more precisely, by the group Z acting as powers of e). If M is of type Til l then such a discrete decomposition may not be possible ([693j), but for a factor of type ITTA (A < 1) with separable predual it can be done. In order to describe the discrete decomposition we introduce some ideas which are interesting in their own right. Suppose for the rest of this section that M is a type III factor with separable predual. Given an fns weight 4) on M, define

= ftER : 4 is inner}. By the unitary cocycle theorem, T(M) is independent of 4), and it is evident that T(M) is a subgroup of R. THEOREM

t

(i)

[633].

For t E R, the following are equivalent:

T(M),

(ii)there is a faithful normal state p of m with 4 = 1. THEOREM [633]. If m is of type 111 A with 0 < A < 1 then T(m) = TZ, where T = -2ff/logX. Suppose M is of type ITIA with 0 < A < 1. With T as above, let p be a faithful normal state of M such that 4 = 1. Let F be the factor of type Z. with separable predual and let tr be its usual fns trace (I, 6, 6). Since M F is isomorphic to M, we can identify p 0 tr with an fns weight 4) on M with the following properties: (

i

)

O.

= 1,

(ii) (pa) = co, an fns (iii)the restriction of cp to its centralizer M trace (such a weight is called strictly semifinite [543]). An fns weight 4) on M satisfying these three properties will be called a lacunary weight. THEOREM

[633].

(i) sp(N) =

If (1) is a lacunary weight on m then

F(A,

(ii)0 is a factor of type TiŒ0 (iii)there is a unitary u in m such that

(u*xu) = 4(x) for all x >

0

in M.

Write N = 0, and T for the restriction of 4) to N. Conjugation by u gives an automorphism e of N such that Te = AT (we say that

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O scales the trace by A). This gives the existence part of the following structure theorem. THEOREM [633]. Let m be a factor with separable predual of type III A (0 < A < 1). There exist a factor N of type 11. and an automorphism 0 of N scaling the trace by A such that m is isomorphic to N x e Z. Moreover, { N, 0} is unique up to outer con-

jugacy. To conclude this section, we describe briefly what happens when M is of type 1110. This time, one defines an fns weight cP on M to be lacunary if

isolated point of sp(A cp), mci5 is a properly infinite von Neumann algebra,

(i) 1 is an

(ii)

(iii)cP, is strictly semifinite. The algebra N = 0 is of type II co but is not a factor, in fact it has nonatomic centre. To prove the existence of a lacunary weight and of a suitable automorphism 0 of N is much more complicated than in the IITA. case, and we refer to [633] for details. The structure theorem is as follows.

Let M be a factor of type 111 0 with separable predual. There exist a von Neumann algebra N of type Ii,. with nonatomic centre and an fns trace T, and an automorphism 0 of N which acts ergodically on Z(N) and strictly decreases T in the THEOREM [633].

sense that there exists IS < 1 such that Te(X)

< 6T(X)

for all x > 0 in m,

such that m is isomorphic to N xe Z. If R, g also satisfy these conditions then there exist nonzero projections e in z(N) and é in z(R) such that the restrictions of 0 and to the reduced algebras Ne and fie are outer conjugate. References : [633], [674], [835], [M].

4. Examples. At the time when this book was written, only finitely many nonisomorphic factors of type II with separable predual were known, and the construction by Powers [411] of a continuous family of type III factors was too recent to be included (see the Preface to the second edition, and the remarks at the end of (I, 9, 4)). A continuum of type Ill factors was discovered almost immediately afterwards by McDuff [479, 480]. With the use of crossed products and infinite tensor products it is now possible to construct an enormous number of factors to illustrate the theory. Infinite tensor products of von Neumann algebras are not discussed in Dixmier's book, although they were introduced by

PREFACE

xxiv

von Neumann himself [77]. For a more modern and readable account, see A. GUICHARDET, "Produits tensoriels infinis et représentations des relations d'anticommutation" (Ann. Sci. gcole Norm. Sup. 83 (1966) 1-52). We shall describe the infinite tensor products only of factors of finite type I. Write Fn for the the algebi. a of p X trace on Flo . If p positive element h that

p < co), isomorphic to factor of type I (2 p matrices, and denote by tr the normalized is a state on F then there is a unique in Fp (called tie density matrix for p) such

p (x) = tr (hx)

(X

E

.

2) and, for each i, let Let (pi) be a sequence of integers (pi pi be a state on FP . with density matrix hi. For n = 1, 2, ...

the tensor product

0 F = F 0 0 F is a C*-algebra in a • P. P P1 n 7, =1 way, natural since it is isomorphic to F p , where P = p 1p 2 pn .

Define a state

i=1

The algebras

0 p i on 0 F i=1 i=1

= tr ((h 1 0 ... 0

2

0 F • 1.=1

h) x)

(x E

F ) .

an inductive system with respect to the -

inclusion mappings

n+1

x

OF

i=1 Pi

+OF,

i=1 Pi

and we can form the inductive limit, which is a normed algebra CO

whose completion

.

0 F

is the infinite C*-tensor product of the

P. n

0 pi commute with the inclusion mappings, co i=1 they define a state 0 p. on this C*-algebra. We denote by . 2 2=1 0 {Fn , p.} the weak closure of the image of 0 F in the G NS • P. co i=1 7,=1 representation given by the state 0 p.. The von Neumann algebra i=1 0 {F , p.} is actually a factor, and we call such factors i=1 Pi F .

Pi

Since the states

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Araki-Woods factors [423] (in the literature they are sometimes called ITPFI factors). Consider now the special case in which pi = 2 (all i) and each pi is equal to a state p of F 2 whose density matrix haT eigenvalues p and 1 - p for some p in [1/2, 1]. Write A = p - 1 00

1. Then 0 IF2, p) is the type Lm factor F if i=1 it is the hyperfinite type II factor R (III, 7, 2) if A = 1; A=0; and if 0 < A < 1 it is of type ITIA and is called the Powers factor RA. Tt is shown in [423] that RA is the unique ArakiWoods factor of type IIIA, also that there are unique Arakitypes IL Woods factors Ro l and R 1111 respectively. However, there are uncountably many nonisomorphic Araki-Woods factors of type III, distinguished from each other by an invariant which turns out to be equivalent to the invariant T(M) mentioned in the previous section. For details, see [423] and [633]. so that 0

A

The remaining constructions in this section all involve crossed products by groups of automorphisms. We begin by constructing a family of automorphisms of the hyperfinite factor R which have very special properties.

Let u be an automorphism of a factor M. The set

{n €Z

n.is inner} .

is a subgroup of Z and is therefore equal to pZ for a unique p O. We call p the outer period of a and write p = p(a). If If p > 0 then no nonzero power of u is inner then p = O. up to a scalarmultiple which is unique for some u in M ad u aP = since M is a factor. It is easy to check that a(u) = yu for some y E C such that yP - 1. We call y the obstruction of a and write y = y(u). The condition y(a) = 1 is necessary and sufficient for the existence of an automorphism outer equivalent to a whose pth power is the identity. For an automorphism a with p(a) = 0 we define y(u) = 1. Notice that p(a) and y(a) depend only on the outer conjugacy class of a. For each pair (p, y) we construct an automorphism 3J of R with outer period p and obstruction y. If p = 0 we can t e s o1 to be the shift automorphism on the doubly-infinite tensor product

e

be an auto0 {F 2 , tr}. Suppose that p > 0 and yP = 1. Let i=_co morphism of an abelian von Neumann algebra inX, p) whose powers act freely and ergodically (I, 9, 3). For example, (X, p) could given by a rotation be the unit circle with Haar measure, with of the circle through an irrational multiple of TE. The crossed product

e

xxvi

PREFACE

Lm (X, eP Z by the automorphism OP is a factor isomorphic to R([300]) which is generated by elements ff(x) (X E L (X, p)) and a unitary u corresponding to the generator OP of Z. Define

sY (ff(x)) =

sY (u) = yu.

One checks that SY extends to a trace-preserving automorphism on the algebra of elements of the form X Tr(x.)u i , hence to an

i—n automorphism of R, which has the required properties. We will now use the automorphisms sY to sketch the result of Connes [742] that for any A in (0, 1) there is a factor of type III which is not antiisomorphic to itself. This solves negatively the problem raised at the end of (I, 1, 5). If M is a von Neumann algebra, we denote by M ° the opposite, or reversed, algebra, in which the *-linear structure is the same as in M, but the order of multiplication is reversed. We wish to construct a factor M such that M° is not isomorphic to M. Before embarking on the construction, we need to define some special subgroups of Aut(M). Let M be a factor with separable predual. We give Aut(M) the topology of simple norm convergence on M. For a factor of type H i this coincides with the topology of simple ultraweak convergence used previously, but in general it is stronger [757]. We denote by Int(M) the normal subgroup of inner automorphisms, by Out(M) the quotient group Aut(M)/Int(M) and by E : Aut(M) Out(M) the quotient map. Let Int(M) denote the closure of Int(M). A bounded sequence (x71 ) in M is called a central sequence if co for each p in M* , xn p - pxn ± 0 as n where xp, px are the functionals given by

xp(y) = p(yx) ,

px(y) = p(xy)

(x, y

EM)

An automorphism a of M is called centrally trivial if 0 *-strongly as n a(xn ) - xn co, for any central sequence Denote by Ct(M) the set of all such automorphisms: it is n ). a normal subgroup of Aut(M). Now let Px be a factor of type IIIA and let P A = N A x e Z be its discrete.decomposition as in Section 3. Let to be a complex cube root of unity, let a be the automoprhism L0 0 of R 0 N A and let Q A be the crossed product (R 0 N A ) xci Z. Then Qx is a factor of type III A since a scales the trace of R 0 N A by A. Also, R 0 N A and are determined up to outer conjugacy by Qx because of the uniqueness of the discrete decomposition. It turns out that if P is Pukanszky's factor of type III A ([171]) — then 3sW01EInt(RONx), I00ECt(R0Nx) and Int(R 0 N A ) nCt(R 0 NA) = Int(R 0 N A ). Also, E(Ct(R 0 NA)) and

—

xxvii

RECENT DEVELOPMENTS

E(Int(R 0 NA) ) are commuting subgroups of Out(R 0NA) . factorization

Thus the

co a = ( s 3 o 1) (1 0 e) (with the first factor in Int and the second in Ct) is uniquely determined, and it follows that w = Y(sW 1) is an isomorphism invariant _ of QA. If we replace QA by Q0 then w will get reA placed by w. This shows that Q° is not isomorphic to Q as A required. We now turn our attention to finite factors, and use some of the above ideas to obtain a powerful isomorphism invariant for factors of type il l . The factors with which we shall be concerned are those which are isomorphic to their tensor product with R, and we begin by observing that many factors have this property. Indeed, if M is a factor of the form N 0 R then M a'MO R since ROR -74 R (III, 7, 5). IfM is any factor of type III, with trace tr then we can form the infinite tensor CO

product re = 0 {M, tr} by a process like that described above i=1 for finite type I factors. We can write M = N 0 F2 for some factor N, and we then have Mx = le 0 FO R, so that 2 = Mx 0 R. We now make some definitions which will help in the analysis of such factors. A central sequence (xn ) in M is called hypercentral if xnyn - ynxn 0 *-strongly as n oo, for any central sequence 0 *-strongly as n (yn ). It is called trivial if xn - Al for some bounded sequence (A n ) of scalars. We say that M is ultrafactorial if every hypercentral sequence in M is trivial. This terminology is motivated by the following considerations. Let w be a free ultrafilter on the positive integers. Form the

productIlm.,whereeachM.is an isomorphic copy of M, and let 2 i=1 Jui = {(x.) E UM. : lim tr(xtx.) = O.

Then J is a maximal two-sided ideal in 11Mi and the quotient MW is a factor of type Ill ([124], [160], [517E) called the ultraproduct of M. There is a natural embedding of M in Mw given by

x

(x, x, x, ..J/J

w

.

The relative commutant of M in Mw is called the asymptotic centralizer of M and denoted by Mw.

xxviii

PREFACE

THEOREM [517]. The type ii factor m is ultrafactorial if and only if mto is a factor.

For a type Il l factor m, the following conditions are equivalent: THEOREM [517, 745].

(i) M

M 0 R,

(ii)mw is noncommutative, (iii) E(Int(M)) is not abelian, (iv) Tnt(M) is not contained in Ct(M). When these conditions hold, E(Ct(M)) is the centralizer of E(Int(M)) in Out(M). In general, Int(M) is not closed in Aut(M) (in fact, it is closed if and only if M fails to have property r (III, 7, 7) [693], [721], and M is then called a full factor), and so Out(M) is not a topological group. But it is a Borel space with the quotient Borel structure from Aut(M). For a type Ill factor M (with separable predual, as always) we define x(M) = E(Int(M) nCt(M)). This abelian group is a Borel space and, whenMZMOR, it is equal to the centre of E(Int(M)). THEOREM [743]. If M is ultrafactorial then the Borel structure on x(m) is countably separated (in other words, x(m) is a standard Borel space (11, 1, 1)). For certain factors constructed as crossed products, the invariant x(M) can be computed, and in this way one obtains the following examples. A.) Let Fn denote the free group on n generators and let U(Fn ) be the associated factor (III, 7, 6). Let M be the CO

0 {U(F 2 ), tr} by the action of Z2 given by i=1 exchanging the generators of F 2 in each component. Then Z4, so M is a type Ill factor which is Z 2 but x(M 0 M) X(M) not isomorphic to its tensor product with itself. crossed product of

B.) Given p and y with yP = 1, p > 2, one can construct an automorphism of U(Fn ) with outer period p and obstruction y, where n = (p - 1) 2 . Let w be a complex cube root of unity and let e be the .automorphism s(i 0 c4L. on R 0 U(F 4 ). Then p(e) = 3, outer conjugate to 0, y(e) = 1, so there is an automorphism Z9 . with 0 = 1. Let M = (R 0 U(F 4 )) X Z 3 . Then x(M) Given M, one can reconstruct the dual action of Z 3 since it is given (up to outer conjugacy) by the unique subgroup of Z9 of order 3. Thus one can retrieve R 0 U(F 4 ) and the group {1, 6(e), £(0 2 , and deduce that M is not isomorphic to M° . This gives an example of a type III factor not antiisomorphic to itself. )}

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C.) The Borel space of 'any compact abelian group appears as x(M) for a suitable M. The above results were all announced by Connes in [743] but details of the proofs have not been published. The interested reader should consult the two excellent survey articles by Connes in [[I]]. We conclude this section with one more extraordinary example due to Connes [950]. Let M be a factor of type III and let tr be an fns trace on N = M 0 F, normalized so that tr(T 0 j) = 1 when f is a minimal projection in F. Call a projection e EN fundamental if eNe M. Murray and von Neumann [67] defined the fundamental group of M to be the subgroup {tr(e)

e is fundamental in N}

of R. As noted in [633], it is equal to the group of positive numbers A for which there exists an automorphism of N scaling the trace by A. The fundamental group of R is R. Let G be a countable discrete group (such as SL(3, Z)) with the property that the trivial representation of G is an isolated point in the dual space. Then U(G) is a full factor, so that Out(U(G)) is a topological group. But the rigidity property of G implies that Out(U(G)) is discrete, and since it is a Polish space it is separable. So it must be countable. It follows that the fundamental group of U(G) is countable. Hence there exists an uncountable family (M 1 ) of nonisomorphic type Ill factors such that the factors M 1 0 F are all isomorphic. The factors M 1 can be represented on Hilbert spaces in such a way that their commutants are all isomorphic to each other. References : [423], [693], [742], [743], [745], [950].

5.

Classification of injective factors.

The proliferation of examples, seen in the previous section, makes a complete classification of factors seem hopeless. Indeed, Woods [679] has shown that the Borel space of isomorphism classes of factors with separable predual is not countably separated. However, there is a class of factors for which an almost complete classification is possible, and which contains the factors which arise in many applications of von Neumann algebra theory. These are the AFD factors which we shall now describe. Murray and von Neumann [67] called a factor of type III

approximately finite if it is the inductive limit (in the strong topology) of an increasing sequence of finite type I factors. Since the factors are finite, Dixmier (III, 7, 2) considers the term "approximately finite" inappropriate and calls them "hyperfinite." We shall be interested in infinite factors

PREFACE

XXX

possessing the same property, and for these the term "hyperfinite" is equally inappropriate. So we shall follow current usage and call them approximately finite-dimensional (AFD). Thus a factor M with separable predual is an AFD factor if it contains an increasing sequence of subfactors Mk Fpk (in fact, one can take p k = 2 1< unless M is finite-dimensional) such that CO

U is strongly dense in M. Notice that any Araki-Woods k=1 k factor is AFD. The first result is a generalization of Theorem 3 in (III, 7, 4):

For a factor m with separable predual, the following are equivalent: THEOREM [793].

(i)M is AFD,

xrz in m and a *-ultrastrong neighbourhood (ii)given x l, V of o in m, there exist a finite-dimensional *-subalgebra G of i n). M and y l, yn in G such that xi - yi EV (1 The structure of AFD factors is tight enough to ensure that any two AFD factors of type Ill are isomorphic (III, 7, 2). However, it is often very difficult to show that a factor is AFD. For instance, it is not at all obvious that its commutant should be AFD. For this reason, it is useful to introduce an apparently weaker notion called injectivity. A von Neumann algebra M acting on H is called injective if there is a projection of norm one (sometimes called a conditional expectation) from L(H) onto M. Let G be a finitedimensional von Neumann algebra, and let du denote Haar measure on the (compact) unitary group U of G. The map

t

utu*du

(t E L(H))

is a projection of norm one onto G I , so G' is injective. It is not hard to prove that the intersection of a decreasing sequence of injective algebras is injective, and that the commutant of an injective algebra is injective. Hence every AFD factor is injective.

Any injective factor of type II I is AFD (and hence isomorphic to R). THEOREM [790].

The proof of this astonishing theorem, Connes' greatest single contribution to von Neumann algebra theory, uses all the gadgetry mentioned in the previous section (the automorphism groups Int(M) and Ct(M), ultraproducts) and much more besides. We shall use the theorem to analyse injective factors of other types.

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Suppose M is an injective factor of type II. We can express M in the form N 0 F where N is a type II factor. It is easy to see that N must be injective, so by the theorem it is isomorphic to R, and therefore M must be isomorphic to the Araki-Woods factor R0 1 . Next, suppose M is an injective factor of type In the discrete decomposition M N x e Z, N must and is therefore isomorphic to R ol . In order to is uniquely determined, we need to know how many of R 01 can scale the trace by A. The surprising Connes is as follows.

III A (0
THEOREM [745, 833]. (i) Any automorphism of R is outer conjugate to one of the automorphisms s (see Section 4); (ii) for

0 < A < 1 there is, up to conjugacy, one and only one automorphism of Rol which scales the trace by A. It follows that the Araki-Woods factor RA is the only injective factor of type IIIx• For factors of type 1110, the situation is more complicated. First, we already know that there are uncountably many nonisomorphic Araki-Woods factors of type 1110. Second, an example due to Krieger and Connes ([633]) shows that there is an AFD factor of type 1110 which is not an Araki-Woods factor. In fact, there is no explicit description of all type 1110 AFD factors, but their classification turns out to be equivalent to two well-known classification problems in ergodic theory, as we shall now see. Let {Z,v} be a Lebesgue measure space (up to Borel isomorphism, {Z, V} is the unit interval with Lebesgue measure). A transformation of Z is a bijective mapping T on Z such that both T and T-1 are measurable and preserve null (negligible) sets. Let G be a separable locally compact abelian group with Haar measure A. An action of G on Z is a homomorphism s T s from G to the transformations of Z such that the map (z, s)

T z : {Z x G, v x A} s

{Z, VI

is measurable. An action of R is called a flow. An ergodic action is said to be strictly ergodic if the orbits T z = {T z : S E G s

GI

(z E Z)

are null sets. (See G. MACKEY, "Ergodic theory and virtual groups," Math. Ann. 166 (1966) 187-207.) Two actions T, T 1 of G are said to be equivalent if there is a transformation U of Z such that

xxxii

PREFACE

T l z = UT U s

-1

(s

V-almost everywhere

2

E G) .

They are said to be weakly equivalent if there is a transformation V of Z which preserves orbits, in other words T

G

2= VT V

G

-1

2 V-almost everywhere.

Two transformations T, T 1 are said to be weakly equivalent if they generate weakly equivalent actions of Z. A

Given an action T of G on Z, we can define an action c. of G on = tm (Z, V) by -1

a (x)(z) = x(T s s

Lrn (Z, V), ZEZ, S E G).

(X

Conversely, any action of G on A comes from an action of G on Z (cf. Appendix IV). Let T be a transformation of Z, generating If the action is free and an action of the integers on Z. corresponding crossed product of A by Z is a ergodic then the Factors of this factor (I, 9, 3), which will be denoted W*(T). form are called Krieger factors. THEOREM [554, 807]. (i) For an ergodic transformation T, the factor W*(T) is AFD; (ii) if T, T 1 are ergodic transformations then w*(T) is isomorphic to w*(Ti) if and only if T and T 1 are

weakly equivalent. For a factor following are equivalent: THEOREM [740].

rel

with separable predual, the

(i)m is a Krieger factor, (ii)there is a sequence of projections of norm one 'Irk from

m

CO

onto finite-dimensional subalgebras Nk, with U N k weakly dense k=1 in m, which is coherent in the sense that ffori = ffm, where m = min{k, I}. Notice that, in this theorem, we cannot require the Nk to be factors. In fact, if we strengthen condition (ii) by requiring each Nk to be a factor then we get a characterization of ArakiWoods factors (this follows from the results in [566]). This shows that any Araki-Woods factor is a Krieger factor.

Now let M be an injective factor of type 1110, and form the discrete decomposition M N X0 Z. The restriction of to Z(N) is an ergodic automorphism of a nonatomic commutative von Neumann algebra and so gives rise to an ergodic transformation T of a Lebesgue measure space. Using the above characterization of Krieger factors and the uniqueness of the type IL AFD factor, Connes [740, 790] showed that M W*(T). Thus every injective factor of type III ° is a Krieger factor.

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xxxiii

RECENT DEVELOPMENTS

We could also have formed the continuous decomposition to Z(N) gives rise M N xe R. This time, the restriction of to a strictly ergodic flow. A useful model for this flow was constructed by Connes and Takesaki [835] by means of an equivalence relation on a set of fns weights of M, and it is called the flow o f weights of M. Using methods of ergodic theory (representing a flow as a flow built under a ceiling function), Krieger showed in [807] how to construct from this flow a transformation T of a Lebesgue measure space such that W*(T) = M. Thus the classification of type 1110 AFD factors is equivalent to the classification (up to equivalence) of strictly ergodic flows.

e

Finally, we need to consider the case of a type 1111 factor. Connes showed in [790] that any injective factor of type 1111 is AFD. Krieger [478] showed that there is only one Krieger factor of type 1111, namely the Araki-Woods factor R. This remains the only known example of a type liiiAFD factor. To summarize: we have been considering four classes of factors, as follows. (A- W) = {Araki-Woods factors}, (K) = {Krieger factors}, (AFD) = {AFD factors}, (Inj.) = {Injective factors}. We have the relations (A-W) c (K) c (AFD) = (Inj.). There are no known factors in (AFD) but not in (K), and any such factor would have to be of type 1111. There exist factors of type III() (but not of any other type) in (K) but not in (A-W). There is exactly one isomorphism class of AFD factors of each of the types In (1 1 n I aa), III, hoe, IIIx (0 < A < 1). There is one known isomorphism class of AFD factors of .type M i . There are uncountably many isomorphism classes of AFD factors of type III ° , and their classification is equivalent to the classification of ergodic transformations (up to weak equivalence) or of strictly ergodic flows (up to equivalence). In Mackey's terminology, this last problem is the same as that of finding all the virtual subgroups of R. References : [554], [745], [790], [807], [833], [835], [896].

May, 1981

E. C. LANCE

•

PREFACE TO THE SECOND EDITION

The changes compared with the first edition are as follows:

1 0 The following have been inserted: results of Kadisun, Ringrose and Sakai on derivations and automorphisms; of Sakai and Tomita on the polar decomposition of linear forms; a result of Niiro on the disintegration of traces; a result of J. Schwartz on the types of factors obtained in a central disintegration; results of Sakai on the non-commutative LebesgueNikodym theorem, and the type of a tensor product, and a result of Hugenholtz and Stormer on the type of certain factors. Some other results have been incorporated in the exercises. It has not been possible to keep count of numerous recent theorems due to theoretical physicists, notably to Araki, Woods and Powers. 2 ° The theory of complete Hilbert algebras, useful in group theory, has been added. 3 ° In the reduction theory, Mackey's point of view has been adopted by taking a Borel space as base space. We thus have formulations which are more suitable for applications to C*algebras. 4 0 Various corrections and detailed improvements have been made, for which thanks are due to those colleagues who communicated their observations to me. The terminology has been updated and the bibliography completed

I

INTRODUCTION

In this book, we study, under the name of von Neumann algebras, those algebras generally known as "rings of operators" or "W*-algebras." The new terminology, suggested by J. Dieudonn4, is fully justified from the historical point of view. Certain of the results are valid for more general algebras. We have, however systematically avoided this kind of generalization, except when it would facilitate the study of von Neumann algebras themselves. Parts I and II comprise those results which at present appear to be the most useful for applications, although we do not embark on the study of those applications. Part III, which is more technical, is primarily intended for specialists; it is virtually independent of Part II. To keep this book to a reasonable length, many interesting questions have been put to one side. Some are sketched in exercises. The references given in the course of the text will assist the reader to complete his studies. These references are not intended to imply any order of priorities. Most of Part II is inspired by a forthcoming book by R. Godement on spectral theory.

* *

We use, without giving any references, elementary results from algebra and general topology. The reader of this book is also assumed to have a reasonably thorough knowledge of topological vector spaces and particularly of Hilbert space; on these topics, see [[3]] to [[ 15]] ( 1 ) and the Appendices. As

( 1 ) The numerals in brackets refer to the bibliography of journal articles. Those in double brackets, to the bibliography of general texts, in which we have inserted, for the convenience of the reader, a list of books at least partly devoted to Hilbert space.

xxxviii

INTRODUCTION

far as spectral theory is concerned, we only assume knowledge, in principle, of the Gelfand-Neumark representation theorem, for which see [[9]]; nevertheless, the reader should preferably have a little acquaintance with the details of this theory. A reading of Part II, as well as of certain passages of Parts T and TTT, assumes knowledge of integration theory. Finally, a few of the exercises appeal to other theories.

** Any reference in which the part of chapter is not specified refers to the current part, or chapter. The symbol L signals the end of a proof. It is suggested that the reader begins by looking over the index of terminology, located at the end of the book.

PART I GLOBAL THEORY

•

CHAPTER 1. DEFINITION AND BASIC PROPERTIES OF VON NEUMANN ALGEBRAS.

Commutant and bicommutant. Let H be a complex Hilbert space and L(H) the set of continuous linear operators of H into H. This set possesses 1.

the structure of an algebra over the field of complex numbers. In agreement with common practice Let M be any subset of L(H). in algebra, we shall name the commutant of M, to be denoted by M', the set of those elements of L(H) that commute with all the elements of M. For example, (L(H)) 1 is the set of scalar operators; indeed, a linear map of H into H which commutes with all rank-1 operators must leave all 1-dimensional subspaces of H invariant, and thus, as is well known, must be a homothetic map; we will be able to give an alternative proof of this fact before long.

We shall put (M')' = M" (the bicommutant of M), (W) ' = M " , It is clear that M' is a subalgebra of L(H) containing the identity operator on H (this operator will be denoted by T H , or simply I when no confusion is possible). We have MC M". The inclusion AWN implies M'DN', and thus M"C:N"; in particular, M' D (M") ' = M"; since, moreover, M' C (M' ) " = M", we see that

. . . .

M I = Ml" = M (5) = M C M" = M

(4)

,

=.

There is an adjoint operation in L(H). If S EL(H), we will always denote the adjoint of S by S*. We have

(S + T)* = S* + T*,

(XS)*

5k-S*,

(ST)* = T*S*,

S** = S

(X denotes a complex number, A its complex conjugate). We thus see that L(H) can be regarded as an involutive or *-algebra. Every subalgebra of L(H) which is stable with respect to the adjoint operation is called an involutive or *-subalgebra of L(H), or a *-algebra of operators. Let MCL(H); if M is adjoint-stable, then M' is a *-subalgebra of L(H).

PART I, CHAPTER 1

2

A von Neumann algebra in H is a *-subalgebra A of L(H) such that A=A". DEFINITION 1.

The set of scalar The algebra L(H) is a von Neumann algebra. operators on H, often denoted by CH, is a von Neumann algebra. Every von Neumann algebra in H contains cH. Let M be an adjoint-stable subset of L(H). The set M P is a von Neumann algebra. The set M" is a von Neumann algebra containing M; and, if A is a von Neumann algebra such that MCA, then M"C=A" = A; thus M" is the smallest von Neumann algebra containing M. Now if M is any subset of L(H) , put N = MUM*, where Mt is the image of M under the adjoint operation; the von Neumann algebras containing M are just those containing N; hence there is a smallest von Neumann algebra containing M, namely N", called the von Neumann algebra generated by M. PROPOSITION 1.

Let (A i ) iEl be a family of von Neumann alge-

bras. Then A = n Ai is avon Neumann algebra, and A' is the iEI Von Neumann algebra generated by the A. To say that TE A is equivalent to saying that T commutes with the Al, hence with LiAi; hence with (UAi)", i.e. with the von Neumann algebra generated by the Ai. 0

Proof.

The centre Z of a von Neumann algebra A is AnA'. This is also the centre of A'. Its commutantZ P is the von Neumann algebra generated by A and A'.

DEFINITION 2. A factor is a von Neumann algebra whose centre only contains the scalar operators. Let A be a von Neumann algebra. To say that A is a factor is equivalent to saying that A' is a factor, or again that the von The algebra Neumann algebra generated by A and A' is L(H). L(H) is itself a factor. At the opposite extreme from factors, it is appropriate to consider abelian von Neumann algebras. It is clear that every operator of an abelian von Neumann algebra is normal (i.e. commutes with its adjoint). On the other hand, let (Ti)i EI be a family of normal operators; suppose that Ti commutes with Tx and T;kc for any iEI and xEI; let M be the set of the Ti and of the Tt. we have MCM P , hence RCM" , from which it follows that M" is abelian; in other words, the von Neumann algebra generated by the T i is abelian. In chapter 3, section 4, we will give topological definitions of von Neumann algebras. References : [65], [73].

3

GLOBAL THEORY

Hermitian operators in a von Neumann algebra. Let A be a *-subalgebra of L(H). Every TE A can be uniquely

2.

expressed in the form T = T 1 + iT 2 , where T 1 and T2 are hermitian elements of A. Indeed,

T

1 = —(T + T*)EA 1 2

and

T

2

= — (T* - T)EA. 2

Let A be a von Neumann algebra, and T an hermitian element of A. For each continuous function f of a real variable, f(T) belongs to A. The spectral projections of T belong to A. PROPOSITION 2.

(Throughout this book, the word "projection" is taken to mean "orthogonal projection.")

Proof. We know (cf. Appendix 1) that f(T) and the spectral projections of T commute with every operator that commutes with T, and in particular with the operators of A P . Hence f(T) and the spectral projections of T belong to A" = A. 0 Since the only projections in H that commute with L(H) are plainly 0 and I, we rediscover the fact that (L(H)) 1 = C. Let us adopt, once and for all, the following convention: if M CL(H), M-1- will denote the set of hermitian elements of M which are > O.

Let A be a von Neumann algebra and let TE A. EA+ if and only if T can be written in the form s*s for

COROLLARY 1.

Then T some s EA.

Proof. The condition is obviously sufficient. Conversely, if T EA+, we have 1(2 EA, and T = (T 1/) * T 1/2 . 0

Let A be a von Neumann algebra and M the set of its projections. Then A is the von Neumann algebra generated by M. COROLLARY 2.

Proof. We have MilcA" = A. We show that ACM". Let SE A, TE M f , and let us show that S and T commute. It is enough to consider the case of hermitian S, and it then suffices •to prove that the spectral projections of S commute with T (cf. Appendix I). Now these spectral projections belong to A, and therefore to M. 0 A von Neumann algebra A is said to be a-finite if every family of non-zero pairwise orthogonal projections of A is countable. In a separable Hilbert space, every von Neumann algebra is a-finite.

PART I, CHAPTER 1

4

Another term for G-finite, as applied to von Neumann algebras, is "countably decomposable." The theory of von Neumann algebras generalises integration theory to a certain extent [101]; it is well known that in this latter theory hypotheses of countability on the underlying space are sometimes useful; similar remarks apply to the study of a-finite von Neumann algebras. References : [73].

Remark.

Unitary operators in a von Neumann algebra. PROPOSITION 3. Let A be a von Neumann algebra. Then every element of A is a linear combination of unitary elements of A. 3.

Proof. It is enough, by section 2 above, to consider the case of an hermitian operator TE A such that 11T11 <1. We then — have I - T 2 > 0. Putting U = T + we have UE

- T 2 ) 11 ,

A, and

Utilr = U*U = (T + i(I - T 2 ) 11 )(T - i(I - T 2 ) 1/2 ) = T 2 + I - T 2 =I, so that U is unitary. Moreover, T =

1

+ U*).

0

Let TE L (H ). Then TE A if and only if u'rru i-1 =T for every unitary operator u'E A'. COROLLARY.

Proof. The condition expresses the fact that T commutes with the unitary operators of D tors of A'.

A', hence (prop. 3) with all the opera-

The corollary will often be used, without explicit reference being made, in the following form: an operator built up from operators of A by a "unitarily invariant" procedure is again in A. For example, let F(7 A+ be an upper-bounded increasing net and T its supremum in L(H) (cf. Appendix II); we then have TE At Another example: if S = UISI is the polar decomposition of an element S of A (cf. Appendix III), then U and ISI belong to A. The support of S (cf. Appendix III) also belongs to A. References : [65], [66].

Cyclic projections in a von Neumann algebra. Let A be a *-subalgebra of L(H) and X a closed linear subspace of H. To say that X is invariant under A is equivalent to 4.

saying that Pxc A P (Px denotes, here and throughout this book, the projection onto X). In fact, if PxE A t , we have, for TE A and x EX, Tx = TPxx = P xTx, so that Txc X. Conversely, suppose that X is invariant under A. Then, for each TEA, we have

5

GLOBAL THEORY

TP X (H) C: X, hence

P TP X X

= TP

X'

hence P XT = (T*P X )* = (PXT*PX )* = PXTPX = TPX . If M is any subset of subspace of

H, we denote by 4 the closed linear

H generated by the Tx(T E Ax E M) .

plainly invariant under

A.

This subspace is

Hence the corresponding projection,

A

denoted by E m , is a projection of A / . one element, x, we use the notations

XA x

If M consists of just and EA; a projection of

the form E A is called a cyclic projection of

A'

Let A be a 4 -subalgebra of L(H), and M a subset of H. We say that M is cyclic feir A if the union of the T(M), TEA, is total in H, i.e. if 4 - H. We say that M is separating for A if the conditions TEA, T(M) = 0 imply that DEFINITION 3.

T = O. If M = fxl, where xEH, we say that X is a cyclic (resp. separating) vector for A. This means that the TX, TE A are everywhere dense in H (resp. that the conditions TE A, TX = 0, imply that T = 0). PROPOSITION 4. Let A be a *-subalgebr of L(H) containing 1 H, and M an arbitrary subset of H. Then, X is the smallest among the closed linear subspaces N of H such that Mc:hi and

PN

A' A Proof. We have MICXm because 1HE A. Morerer, if MCN and if PN E A' , then N is invariant under A and X C: N. D

PROPOSITION 5. Let A be a *-subalgebra of L(H) containing 1H, and M any subset of H. Then M is cyclic for A if and only if M is separating for A'. Proof. If M is separating for A', the obvious equality (1 - E) (M)= 0 implies that I - E A = 0, and so AuA = H. Suppose, conversely, that X A - H; then, the conditions T I E A', T' (M) = 0 imply, for every TE A, that T I T(M) = TT' (M) = 0, hence that T' (X) = 0, hence T 1 = 0, from which it follows that M is separating for

A'

0

6

PART I, CHAPTER 1

Let A be a von Newmann algebra. Then M is separating for A if and only if M is cyclic for A'. Proof. Interchange the roles of A and A' in proposition 5. 0 PROPOSITION 6. Let A be a von Neumann algebra. Then A is a-finite if and only if A possesses a countable separating set. COROLLARY.

Proof. Let M be a countable separating set for A; let (E i ) ie , be a pair-wise disjoint family of projections of A ; for every xE M, we have Ex = 0 for all but countably many of the indices i; hence E(M) = 0 and consequently Ei = 0 for all but countably many indices i; this shows that A is G-finite. Suppose, on the other hand, that A is G-finite; let (y x ) xE K be a maximal family of non-zero elements of H such that the A' r A' E are pair-wise orthogonal; we have L E = I because the Yx xEK Yx family ( Yx ) XEK is maximal, and K is countable because A is G-finite; hence the set of the y x 's is a countable separating set. Let A be a von Neumann algebra, Z its centre, and T an element - of A. For every projection G of Z such that TG = T, we have T*G = (TG)* = T*. Hence the infimum F of the projections of Z majorizing the support of T (cf. Appendix III), which is the smallest projection of Z majorizing the support of T, is also the smallest projection of Z majorizing the support of T*. We have TF = T, and every projection E of Z such that TE = 0 is orthogonal to F. We say that F is the central support of T. PROPOSITION 7.

Let M be a subset of H,

and

A N = X.

Then we

A' have X N = X„ . Proof. We have NC X

hence X CX . Since X,, is the M M ' N M closed linear subspaces of H containing M and smallest of the N

A' E 4, / in other words such that P c Z it suffices to show that E N ' N invariant is under both that X N A and A'. It is clear that X N is invariant under A'. Now let xEN, T P E A', TE A; we have T (T 'X)

(as N is invariant under

T(X

)C:X Al N..

A),

= T 1 (TX) E T 1 (N)

and so T(A'x)CX

A' , N

hence

0

COROLLARY 1. A Let M be a closed linear subspace of H such that P E A. Then EM the central support of PN. N

7

GLOBAL THEORY

A'

A

A

Z'

Since XN = N, we have XN = XN (prop. 7), hence XN is the smallest of the closed linear subspaces Z of H containing 0 N and such that P Z E Z. COROLLARY 2. Let M be a subset of H. Then the projections I A' A EE and A' have the same central support, namely E Z . E A E m

Proof.

M

M

Proof. 1.

This follows at once from proposition 7 and corollary

0

The following conditions are equivalent: (i) is a factor; (ii) for any non-zero elements R, S of A, there exists TE A such that RTS X O. COROLLARY 3.

A

(ii) (i): if A is not a factor, there exists in Z a projection P such that P / 0, 1 - P / 0 (cor. 2 of prop. 2). For every TE A, we have PT(I - P) = P(I - P)T = O.

proof.

(i) ' (ii): suppose that there exist R, SC A such that R / 0, S / 0, and RTS = 0 for every TE A. Let N be the closure of S(H), and Let P = 4. By corollary 1, P is a non-zero projection of Z. Moreover, for every TE A, R vanishes on TS(H), hence on T(N); hence R vanishes on X, from which it follows that RP = O. Hence P X I and A is not a factor. 0 Proposition 4 can evidently fail if I 4A (take A = 0). The corollary to proposition 5 can fail if A is an arbitrary *-subalgebra of L(H) (exercise 3). The question of the existence of separating or cyclic elements will be studied further on several occasions (and even in chapter 2, section 1). References : [65], [73].

5.

Homomorphisms.

Let A and B be von Neumann algebras. A mapping is called a homomorphism, if it is linear, if

(D (ST) = (D(S)(D(T)

for

SE

A,

TE

(D of A into B

A,

and if

1) (S*) = cD(S)*

(

for

SE

A,

in other words, if (D is a homomorphism for the *-subalgebra structures of A and B. A mapping (D' of A into B is called an antihomomorphism if it is linear, if

(D' (ST) =

(T) (D I (S)

for

SE

A, T E A,

and if (D

I (S*) = ( D' (S)*

for

SE

A.

8

PART I, CHAPTER 1

PROPOSITION 8. Let A and B be von Neumann algebras and (1) a homomorphism or an antihomomorphism of A into B. Then

(i) (I)(A+) (= B+ ; (ii)If E is a projection of A, (1)(E) is a projection of B; if (1) is injective, (iii)For each SEA, we have I) (S) we even have (S) S (iv)If S is an hermitian operator of A, and if f is a (complex-valued) continuous function of a real variable such that f(0) = 0, (D(S) is an hermitian operator of B, and (1)(f(S)) = f(4)(S)). Proof. We consider the case where (I) is a homomorphism. The

I

I ( II < IISI ;

I = I I;

case of an antihomomorphism is treated analogously. If SEA+ , we have S = T*T for some TEA, hence (1) (S) = (1) (T) *(1) (T) €13+ ,

from which (i) follows. If E is a projection of A, we have E = E 2 = E*, and so

(1)(E) = (1)(E) 2 = so that (1)(E) is a projection of B. For each S E A, we have S*S <

(S) * (5)

< H

Whence (ii).

H s H2. I, S

and so

H 2w) < H

S H 2 .T

[because (I) is a projection]. Hence

H 4)(s) H 2 < HsH 2 / which proves the first assertion of (iii). Assertion (iv) is clear if f is a polynomial (without constant term), and we may pass to the general case by continuity (thanks to the relation I)( s) < s

I (

II

I II).

We finally show that, if (I) is injective, we have

H (1)(s) H = HsH. 2 As HS H =HS*S SE A4- . If we had

+

H,

it is enough to consider the case where

for some SE A , there would exist a continuous complex-valued function f of a real variable such that f(0) = 0, with f(S) / 0 arid 43.(f(S)) = f(4)(S)) = 0, which is a contradiction. 0

9

GLOBAL THEORY

If (I) is a bijective homomorphism (resp. antihomomorphism) of A onto B, then 4)-1 is a homomorphism (resp. antihomomorphism). We then say that (I) is anisomorphism (resp. antiisomorphism) of A onto B, or sometimes an algebraic isomorphism (resp. antimorphism); we also say that A and B are isomorphic (resp. antiisomorphic) or algebraically isomorphic (resp. antiisomorphic). Let H and K be two complex Hilbert spaces, U an isomorphism of H onto K (i.e. a linear isometry of H onto K) and A a von Neumann algebra on H. The mapping S USU-1 is an isomorphism of A onto a von Neumann algebra B on K. Such an isomorphism is said to be spatial; we say that A and B are spatially isomorphic. Now let V be a bijective mapping of H onto K such that

(VxlVy) = (y1x)

for x,y

E

H,À,p

E

C.

The mapping S VS*V-1 is an antiisomorphism of A onto a von Neumann algebra C on K. Such an antiisomorphism is said to be spatial; we say that A and C are spatially antiisomorphic. Let He the Hilbert space conjugate to H, i.e. the space H Xx, (x,y) x + y and the endowed with the operations (X,x) scalar product (x,Y) (171x). The identity mapping of H onto He of the mapping V above. Thus, the mapping play the role can S S* is a spatial antiisomorphism of A onto a von Neumann algebra Ac on He . Every antihomomorphism of A into a von Neumann algebra D is the composition of this antiisomorphism and a homomorphism of Ac into D. This fact reduces most problems relating to antihomomorphisms to problems concerning homomorphisms. The structure of isomorphisms and, to some extent, of homomorphisms, will be studied in detail from time to time in the sequel, beginning in chapter 4. We do not know of any antiisomorphic von Neumann algebras which are not in fact isomorphic. References : [65], [66], [67], [79].

6.

Ideals in von Neumann algebras.

Let A be a von Neumann algebra. If M is a left (resp. right) ideal of A, then M* (the set of S* as S runs through M) is a right (resp. left) ideal. Hence a self-adjoint left (resp. right) ideal of A is two-sided. Conversely, a two-sided ideal M of A is self-adjoint; for let SE M, and S = WISI be its polar decomposition; we have WE A

and

S* = ISIW* = W*SW*E M.

Let M be a linear subspace of A. For M to be a left (resp. right) ideal of A, it is necessary and sufficient that, for

PART 1, CHAPTER 1

10

each S E M and each unitary operator U E A, we have US E M (resp. SUE M). The necessity is obvious, and the sufficiency follows from proposition 3. Let ni be a left ideal of A. Let S E A. Then S E ni if and only In fact, let S = WISI be the polar decomposition of if ISI E M. S. We have I SI = W*S, and W E A_ PROPOSITION 9. Let m be a two-sided ideal of A. Then every element of m is a linear combination of elements of mt.

m. We have S = S l + iS 2 , with S l E Y11, S2 E ni, S i and S 2 hermitian. It thus suffices to consider the case where S is hermitian. There exist spectral projections E, F of S such that Let

Proof.

SE

ES > 0,

E+F=

Then, E

E

A, ES

FS < O.

F E A, and so E M

+

—FS

,

E M

and S = ES + FS.

D

LEMMA 1. Let mo be a subset of A+ possessing the following properties: (1) If S E M o and if u is a unitary operator of A, we have trisu E Mo ; (ii)If S Ern ° and if T is an operator of Al- majorized by s, we have T E Mo; (iii)If S E M 0 and T E M O' we have s +TEM 0

Then, the set n of the s E A such that ss* ideal of A, and mo = (n 2 ) -1- .

E

mo is a two-sided

(The symbol n2 denotes, here and henceforth, the two—sided ideal which is the square of the ideal n in the usual algebraic sense.) Proof. have

Let

SE

n, and let U be a unitary operator of A. (SU) (SU)* = SUU*S* = SS* (US) (US)* = U (SS*)

and so SU E n; US E n. S + T E n; in fact,

Moreover, if

U -1 E

SE

EM

o'

0'

n and T

E

(S + T) (S + T)* < 2SS* + 2TT* E M

so that (S + T) (S + T)* E Mo .

n, we have

0'

We

11

GLOBAL THEORY

Thus, n is a two—sided ideal of

A.

C (n 2 ) +. Con— If s c m 0 ' we have S 1/2 E n, hence S E n 2 ; thus ,71 0 versely, let T E VL. 2 ; T is the sum of operators of the form AB*, with A c Vi, B E n; if T is hermitian, the identity (1)

4AB* = (A + BHA + B)* — (A — 13)(A — B)* + i(A + iB)(A + iB)*—i(A — iB)(A — iB)*

n shows that T is majorized by an operator of the form where Ci are in n, i.e. by an operator of M ; 0 have T E M 0'• thus, (n2 ) + CM0 . Li

*

if T — > 0, we thus

Let s and T be elements of A+ such that S T. There exists a unique operator A E L(H) such that: 1 s 1/2 =AT1/2 ; 2 the support of A is majorized by that of T. We have A E A. Proof. For every x e H, we have LEMMA 2.

m

s 1/2x

11 2

= (sxix) < (rxix) =

H ri.lx

p.

In particular, T1/2x = 0 implies S 1/2x = O. The mapping T1/2x ± S ilx is a continuous linearmapping Cof T 1/2 (H) into H. Let B be the unique continuous linear mapping of T1/2(H)= T(H) into H which extends C. (Here and henceforth, -F4 denotes the closure of M). We have S 1/2 = BT 1/2 . The existence and uniqueness of A follows at once. The uniqueness implies in addition that, if LP is a unitary operator of A P , we have U'AU I-1 = A. Hence A E A (corollary to proposition 3). D

Let mo be a subset of A . Then there exists a two-sided ideal m of A such that mo . m+ if and only if mo possesses properties (i), (ii) and (iii) of lemma 1. PROPOSITION 10.

Proof.

Conditions (1), (ii) and (iii) are sufficient by lemma 1. Conditions (i) and (iii) are clearly necessary. Finally, suppose that TEm+ , S E A+ and S _ < T. By lemma 2, there exists an AEA such that

S = S 11 (S 1/2 ) * = AT I1T 1/2A * = ATA * , SO that S E m.

0

Let m be a two-sided ideal of A. (i)The set of s E A such that ss*EM is a two-sided ideal n of A such that n2 = m; (ii)If 8 is a linear map of m into a complex vector space such that 0(ST) = 0(TS) for SE M and TE A we have e(ST) = (TS) for SE n and TE n. PROPOSITION 11.

12

PART I, CHAPTER 1

Proof. Assertion (i) follows immediately from lemma 1 and proposition 10. (1) and

(1')

We prove assertion (ii).

By the identities

4B*A = (A + B)*(A + B) - (A - B)*(A - B) + i(A + iB)*(A + iB)

-

i(A

-

iB)*(AiB),

it is enough to show that e(S*S) = e(SS*) for sc n. S = WISI is the polar decomposition of S, we have s*s =

1s1 2 = w*wls1 2 ,

SS* = WISI 2W*

and

Now, if

1 5 1 2 E M,

hence

2 2 8(S*S) = 8(W*W1S1 ) = e(wISI W*) = e(SS*).

E

The ideal n of proposition 11 will be denoted by m1/2 . One can define ma for any a > 0 [14]. References : [4], [12], [14], [25], [97].

7.

Maximal abelian von Neumann subalgebras.

Let A be a von Neumann algebra. If we partially order the abelian von Neumann subalgebras of A, we may consider the notion of a maximal abelian von Neumann subalgebra in A. The following proposition proves the existence of such subalgebras.

Every abelian von Neumann subaigebra of A is contained in a maximal abelian von Neumann subalgebra of A. PROPOSITION 12.

Proof.

It is enough to show that the set of abelian von Neumann subalgebras of A is inductively ordered. Now if (Ai)iEI is a totally ordered family of such subalgebras, the von Neumann algebra B generated by the Ai is abelian by what we observed at the end of section 1, and is contained in A; it is clear that B is an upper bound of the A. in the set of abelian von Neumann subalgebras of A. 0 PROPOSITION 13. For a von Neumann subalgebra

B of A to be

maximal in A it is necessary and sufficient that it possess one of the following equivalent properties. (i)BP n A B; (ii)The von Neumann algebra generated by B and A' is B'. Proof. The equivalence of conditions (i) and (ii) follows from proposition 1. Now let B be a von Neumann subalgebra of A. To say that B is abelian is the same as saying that BC:B P , hence that Bc:Ivr)A. Now suppose that this condition is satisfied. To say that B is not maximal abelian in A is the same as saying that there exists an SE A, S4 B, commuting with S* and with B,

13

GLOBAL THEORY

and hence that B / B' A. Thus, to say that B is maximal abelian in A is the same as saying that B = B 7 r)A. We see, in particular, that a maximal abelian von Neumann subalgebra in A contains the centre of A (which was evident a

priori). The notion of maximal abelian von Neumann subalgebra will be useful in chapter 9 and in part II, chapter 3. References : [28], [57], [78], [100].

1. Let A be a von Neumann algebra on H, Z its centre, and E a projection of A such that the relations T A, ET = 0 imply that TE = O. Show that EE Z. [Let SE A. Putting T = SE - ESE, show that SE = ESE. Deduce that E(H) is invariant under A] [53]. 2. Let A be a von Neumann algebra on H, and Al its unit ball. a. Show that I is an extreme point of Al. [If I = 1/2(A + B), AE A i, BE A lt let C = 1/2(A + A*), D = 11(B + Bt), whence I = 11(C + D), CE Al , DE Al. Deduce from this that C = D = I. Then A = I + iA hermitian and AE A l' imply that A = I. 1 -; A 1 Similarly B = I.J

Exercises.

An element UE A such that U*U = I is an extreme point of [If u = 1 (A + B), AE A l , BE A l , deduce from a that U*A = U*B = I. Let F = UU*. Show that A and B map H isometrically onto F(H), and then that A = B = U.] Similarly, an element VE A such that VV*= I is an extreme point of Al. (Use the isometry T T* of A onto A.). b.

A

Let T be an extreme point of A l . Show that T*T is a projection (cf. chapter 4, lemma 8). [If not, there exists CE A+, commuting with T*T, with T*TC / 0, and C.

I = II T*T(I + C)2 II = II T*T(I - C)2

H.

Then, T(I + C) E A 1 , T (I T(I + C) = C) E A 1 , whence T whence TC = 0]. Let E = T*T, F = TT*. Show that (I - F)A(I E) = O. [If AE (I

-

F)A(I

-

E),

H

A

H<

-

C),

1,

we have T ± A < 1, henceT=T+A=T- A]. IfAis a factor, show that we have T*T = I or TT* = I. (Use corollary 1 of proposition 7.)

d.

The extreme points of er) A i are the projections of

A.

[If E = 1/2(A + B) is a projection of A, with A > 0, B > 0, we have A EEAE, BE ZAE. Apply a to EAE] [34], [174], [2. 8 9], [322].

3. Let H be the Hilbert space where elements are the complexvalued functions defined on the interval [0, 1] which are

14

PART I, CHAPTER 1

square-integrable with respect to Lebesgue measure. For every essentially bounded measurable complex-valued function g on [0, 1], let T be the operator f f4 on H. Let A be the set of the T a as g runs through the set of polynomials with complex coefficiénts. Let h be the characteristic function of [1/2, 1]. Show that h is separating for A, but is not cyclic for A'. (Show that Th is a strong limit of operators of A, hence that T'h = T'Thh = ThT'h for each T'EA'.)

4. Let A be a *-algebra of operators on H. If T is an invertible element of L(H) such that the map A 4- TAT-1 is an automorphism of A (for the *-algebra structure), there exists a [The unitary operator U on H such that UAU-1 = TAT-1 for every A E A. equality (TAT)* - TA*T-1 shows that T*TE A'. Let T = ulTI be the polar decomposition of T. We have ITI E A', so that UAU-1 for AE A]. TAT 5. Let A be a complete normed algebra over the field C and B a *-algebra of operators on the complex Hilbert space H. Let (I) be an isomorphism of the algebra A onto the algebra B. For x E B, let II x II be the norm of x in B and II x II the norm obtained by using (1) to transfer the norm of A. Show that A-lx*x is x* 11 111 X Hl. (If A > H x*x Ill' I x 11 2 < invertible, hence A > x* x b. Show that the involution of B is continuous for the norm lx H. (Using a, show that the hypotheses Hxy, - Y Hi 0 and imply 2 = Y * ; then apply the closed graph 14 - 2 ill theorem.)

a.

H

H

-

11

11 .)

-

c.

Show that (I) is continuous.

(Use a and b).

d. Let T be a homomorphism of the algebra A onto the algebra B. Show that T is continuous. [The condition T(x) = 0 is equivalent to T(x) E M for any maximal regular right ideal M of B, which is semi-simple; now T-1 (M) is a maximal regular right ideal of A, and is therefore closed; deduce from this that the kernel N of T is closed; then A/N is a complete normed algebra; apply e to the isomorphism obtained from T by passage to the quotient.] e. Deduce from e that, if B is a norm-closed *-algebra of operators on H, then every automorphism of the algebra B is continuous. Reference : C. E. RICKART, The uniqueness of norm problem Banach algebras (Ann. Math., v. 51, 1950, p. 615-628).

in

6. Let A be a von Neumann algebra. For each two-sided ideal m of A, let re be the two-sided ideal generated by the projections of m and m° the two-sided ideal which is the norm-closure of m. We say that M is restricted if M =

15

GLOBAL THEORY

a. Let T E Mt . For A > 0, let EA be the greatest spectral projection of T such that TEA < AE A . Show that I - E E M. E tik.) (We have T - EA = TTA for some b.

Deduce from a that M ° is the norm-closure of Mm .

c. Show that every projection E of M° is in m. [Let (T,) be a sequence of elements of M such that II Tn E II ÷ o ; if 16 S

ETnE,

Rn

Sn Sn'

we have II S 1 Hence — E
E II ÷ °,

II R

E II

O.

for sufficiently large n, and then E = RR

d. Deduce from b and C that the maps M 4- Mc° , M 4- M0 define mutually inverse one-to-one correspondences between the set of restricted two-sided ideals of A, and the set of norm-closed [12], [124]. two-sided ideals of A. 7. A von Neumann algebra with no non-trivial two-sided ideals is a 0-finite factor. (If the von Neumann algebra A is not a-finit?, the set M of the TEA whose support is of the form EA with M countable, is a non-trivial - two-sided ideal of A) [4], [94]. 8. Let H be a complex infinite-dimensional Hilbert space with a countable basis, and mi (resp. M2) the set of compact (resp. finite-rank) operators on H. a. Show that Mi and M2 are two-sided ideals of L(H) and that, if M is any two-sided ideal of L(H) other than {0} and LUI) , we have

m2 CmCm i . b. Let T be the set of (infinite) sequences of non-negative real numbers converging to zero. A subset S of T is called an ideal of T if it possesses the following properties: (i) if (An ) E S and if 1T is a permutation of {1, 2, ...), then (A ff ( n )) E S ; (ii) if (An ) E S and (pn )ES, then (An 1.171 ES; (iii) if (An ) E SI (1171 )E T and pn < An for every n, then (pn ) E S. Let m be a two-sided ideal of L(H) containing m For each 1 TE and each orthonormal basis (en ) of H consisting of eigenvectors of T, the sequence (An of the corresponding eigenvalues of T is an element of T; let S(T) be the subset of T thus associated with T when (en ) varies, and S= U S(T). Then, S is TE m+ an ideal of T and the mapping m 4- S is a bijection of the set of )

.

e

)

16

PART I, CHAPTER 1

two-sided ideals of L(H) containing Mi onto the set of ideals of (To show that every ideal of T yields a two-sided ideal miCim i , use proposition 10, and the maximum-minimum definition of

T.

eigenvalues) [4]. Let A be a von Neumann algebra on H, M a total subset of H, (Tn ) a sequence of elements of A, and T an element of L(H) such that Tx and Tx converge weakly to Tx and T *x for any XE M as n -I- 03 . Show that T E A. [Show that, if SE A', XE M, y E M, we have (TSx1y) = (STx1y)] [28]. 9.

a.

b.

H.

Let (el, e2, ...) be an orthonormal basis of Tri E L(H) be defined by 3 Ten = -ne 2 + — 2 ne3,

Tn e 2 = 2e2 - 3e3,

Te i = 0

Let

for

i 2,n.

Let TE L(H) be defined by Te2 = 2e 2 - 3e3,

Te. = 0

for i / 2.

Show that Te iconverges strongly to Te i , for each i, but that T does not belong to the von Neumann algebra A generated by the Tn . [ Let S be the hermitian operator in L(H) defined by Se i = e l +

1

1 e2 + -3- e 3 +

,

Se i

1

el

for j > 1,

show that S commutes with the Tn , but that (STe 1le 2 ) X(TSelle2).] Cf. chap. 3. 10. Let A be a von Neumann algebra on H, and T a (not necessarily continuous) linear mapping from a linear subspace of H into H. We say that T is affiliated to A, and we write TnA, if UlTU 1-1 = T for every unitary operator U' of A'. a.

If TE L(H), the conditions TnA, TE A are equivalent.

b.

If TnA, and if T* exists, we have T*nA.

c. Suppose that T tion is dense in H. Then TnA if and only only if the spectral

[101].

is closed, and that its domain of definiLet T = UITI be its polar decomposition. if UE A and trrI n A. Further, trrInA if and projections of ITI belong to A. [65],

CHAPTER 2. ELEMENTARY OPERATIONS ON VON NEUMANN ALGEBRAS

1.

Induced and reduced von Neumann algebras.

Several procedures exist for constructing new von Neumann algebras from given ones. One of these "elementary operations" is already known to us, namely the operation of taking the commutant of a given von Neumann algebra. In this section, we shall define some other elementary operations. Let H be a complex Hilbert space, E = Px a projection on H and TE L(H). We denote by Tx, or TE , the restriction of ET to X, which is an element of L(X). We have Tx = (TE)x = (ET) x= (ETE)x. If M is a subset of L(H), we denote by Mx, or ME , the set of all the Tx for TE M. Consider the following two cases:

1 ° M is a *-algebra of operators A, and EE A. Let B be the set of the TE A such that TE = ET = T, in others words the set of the TE A such that T(X) C=X and T(X1 ) = O. The set B is a *-subalgebra of A; we have B = EBEC=EAE, and EAECB, so that B = EAE. The map T -Y- TE is a *-isomorphism of B onto AE . 2 ° M is a *-algebra of operators A, and EE A'. Every operator TE A leaves both X and XI invariant. The mapping T 4- TE is a *-homomorphism of A onto AE . PROPOSITION 1.

Let A be a von Neumann algebra and E = Px a

projection in A. AE and (A') E are von Neumann algebras and we have (A 1 ) E (AE ) P .

(ii)If M is a set, closed under multiplication and the adjoint operation, which generates the von Neumann algebra A, then ME generates the von Neumann algebra AE . (iii)If N is a set which generates the von Neumann algebra A', then NE generates the von Neumann algebra (AE )'.

17

18

PART I, CHAPTER 2

Proof. We can assume that N is closed under the adjoint operation, replacing N by NUN* if necessary. We can also assume that I EM, replacing M by MIJ{I} if necessary. The *-algebras AE and (A l ) E clearly commute with each other. On the other hand, let T EL(X) be an operator commuting with Let S = ToE € L(H). It is the operators rq, where T 1 EN. immediate that S commutes with N; thus SEN' = A" . A and consequently T = S E EA,E . Setting N A', we see that A E = (( A I ) E ) ? is a von Neumann algebra. Returning to the case of arbitrary N, we again see that NE generates (AE )'. Now, let T' E L(X) be an operator commuting with ME . We shall A r show that T' E (A') E This will establish that (A') E = (mE ) , (A') E is a vontha Neumann algebra, and that ME generates AE. The proof will thus be achieved.

It is enough to consider the case where T' is unitary (Chap. Proposition 3). For xl, x2, , Xn E X and T 1 , T2, , T HEM, we have 2 n n x. (T.ET'x.tT.ET l x.) = (ET *.T.ET'x.IT'x.) = X 2 2 2 2 J 1, J

i,j=1

i=1

= i,j=1

i,j=1

(T I ET.T.Er.IT'x.) * =(ET.T.Ex.tx.) j 2 2 2 j i,j=1 j 2

T .x i=1 " Let Z be the closed linear subspace of H generated by the ele-

ments

y-

A T.x . by proposition 4 of chapter 1, we have P - E X' i -

'

by corollary 1 of proposition 7 of chapter 1, P z is therefore the central support of E. There exists a unique operator S' E L(H) such that

S'

T.x.I =

S'P = P s' = s'. 7

T.T x•

For every TE M, we have

T.x.) = S I .(

S'T( i=1

TT• x 2

2

-

TT.T'x. = TS P ( 2 2=1 .

T.x ) 2 i ' 2=1 2

19

GLOBAL THEORY

and PT = TP z , so that S'T = TS'. Hence S' EM I = for S EX, we have S'S = T'x; thus T' = S E (A 1 ) E. E

A'.

Finally,

0

,Instead of writirlig (A')E or (AE ) 7 we shall henceforth write A E. The algebra AE is called the von Neumann algebra induced by A' in X. The homomorphism T' ± T of A' onto AE is called the induction of A l on A. Every von Neumann algebra of the form AE is called a reduced von Neumann algebra of A. We have an obvious transitivity property: projections of A such that X 1 DX 2 then

A

-A

if Px, P v are

1 '2

A' -A' )

X1X2' X2 X1 X2 X2 Let A and B be von Neumann algebras, (D a homomOrphism of A into B, E a projection of A and F = (D(E). The restriction of (D to EAE is a homomorphism of the * - algebra EAE into the *-algebra FBF, which in an obvious way defines a homomorphism (D E of AE into BF, called a reduced homomorphism of (D. If (I) is an isomorphism of A onto B, then (D E is an isomorphism of AE onto BF . We have analogous results for antihomomorphisms and antiisomorphisms.

Let A be a von Neumann algebra, E a projection of A and F its central support. Then the induction of A' on is an isomorphism if and only if F = I. PROPOSITION 2.

Proof.

The condition is necessary since (I - F) E = O. Conversely, suppose that F = I. Putting X = E(H), the T(X), TE A, generate H (chapter 1, corollary 1 of proposition 7). If T'E A' is such that rq = 0, we have T'(X) = 0, hence T'T(X) = TT 1 (X) =0 for every T E A, hence T' = O. 0

Let A be a von Neumann algebra, Z its centre and E a projection of A. The centre of AE is Z E . In particular, if A is a factor, then AE and A4 are factors. COROLLARY.

Proof.

It is clear that Z E is contained in the centre of AE . On the other hand, let T be an element of AE r) A. We have T = T' for some T' EA ' . Let F be the central support of E. Replacing T' by FT', we can suppose that T' = FT'. Proposition 2 applied to AF then shows that T /1„ is in the centre of A. Hence T' EZ, and consequently T E 7 E . 0 PROPOSITION 3. Let A be a cl-finite von Neumann algebra and Z its centre. There exists a projection G of Z such that AG possesses a cyclic element and AI _G a separating element.

Proof. Let (xi)iE I be a maximal family of non-zero elements of H possessing the following properties: 1 0 the E =EY1., are pairwise orthogonal; 2 ° the Ei = E. are pairwise orthogonal. Let

20

PART I, CHAPTER 2

E = X Ei

E

A,

E =

iEI

E'•

F = I - E,

€

F' = I - E l .

2E'

(resp. Fi) be the central support of F (resp. F l ). If F 1 F 1/ / 0, we have (F') Fl 0, hence (F') F / 0 (prop. 2), i.e. F l F / 0; let y be a non-zero element of F(H) r1 F l (H); then is orthogonal to the Ei, and E 1 is orthogonal to the g which contradicts the maximality of 'the family (xi)iE I . Hence F iFI = 0. Putting G = F l , we have Let

E =I-F>I- G

E l = I - F' > I - F1 > G.

and

Moreover, since A is 0-finite, I is countable; multiplying the xi by suitable scalars, we can suppose that xi E H. Since xi =Eix=Eil c°7 let, then x X 11 xi 11 2 < iEl

iEl

A majorizes the 4, hence E l , and that Ef majorwe see that E x izes the Ei, and therefore E. Hence Gx is cyclic for AG and U (I - G) x is cyclic for AG. COROLLARY.

A G-finite abelian von Neumann algebra

possesses a separating element. Proof. Retaining the abov7 notation, I - G is majorized by A' . Hence ExA' , and G is majorized by hence a fortiori by Ex

EA' = I and x is separating for A.

D

Proposition 1 (ii) fails if M is an arbitrary subset generating the von Neumann algebra A (exercise 1). Proposition 3 fails without the countability hypothesis: if A is abelian, if A G possesses a cyclic element and AI _G a separating element, then A possesses a separating element and is therefore 0-finite. References : [12], [65], [100], [117].

2.

Product of Iron Neumann algebras.

Let (Hi) iE , be a family of complex Hilbert spaces and H their direct sum, namely the set of families (x.) :EI where xi E Hi and 7, 7, . 4For every family (Ti)i ei , where TiE L(Hi) and xi 2 X

H

iEI

H < . H Ti H < 4- .,

we can clearly define a TE L(H) by the formula T((x.)i EI ) = ( Tx) €1 . We will denote this operator by (Ti)/ whia should not cause any confusion. It is plain that sup

I, (T 2. + S.(Ti) ) = (T- + ( -), 2

(T -S .) = (T •) (S 0 ,

(T4:)=(T.) * .

Let E. be the projection PH. in the space H. For an operator of L(H) to be of the form (Ti) 2: it is necessary and sufficient that it commute with the E.

21

GLOBAL THEORY

This established, let, for each iEI, Ai be a von Neumann algeLet A (resp. B) be the set of the operators (Ti) bra in H.. 2 such that Ti €A1 (resp. Ti €A,) for each ( and sup HTid < + 0D)• The * algebras of operators A and B in H commute with one another. We shall show that A' = B, A = B', which will imply in particular that A and B are von Neumann algebras. Let T be an operator of L(H), commuting with B. Since it is clear that Ei ES, T commutes with the Ei, and hence T is of the form (Ti). any iEI, we must have For every operator (T.2 ), where T- E A, —

(T.T 1.) = (T.) (T) = (T'.) (T.) -- (T I.T.) 2

2 2

i

hence T-Tt 2 = A2• ' for any i E I. Thus 2 2 = T I-T • and thus T •2 E Av T E A which shows that B' = A. Interchanging thé roles of the and the At we see similarly that A' = B.

A

is called the product von Neumann algebra of the von Neumann algebras Ai , and is denoted by H A. We have thus established the formula ( H = H A. The von Neumann algebra

iEi

ici

ici

(If we are dealing with a finite family A l , A 2 , , An , we also use the notation A l x A 2 x x An .) The centre Z of A is the set of operators ( T1 ) such that Ti belongs, for every iEl,tothecentreZ.ofklyin others words, Z = H In i' iEI particular, the E i belong to the centre of A. On the other hand, let A be a von Neumann algebra, (Ei) EI a family of pairwise disjoint projections of the centre of A, such that E. 2 = I. We shall see that such a family gives rise in a iEI canonical way to a family (Ai)i E , of von Neumann algebras and a spatial isomorphism of A onto H A.. Indeed, let H. = E.(H). iEi

The space H is the direct sum of the H. Let Ai = AE .. We have A. = A' Ai. . For every T EA, put T.2 = TEi and form the 2operator (T i ) E H A.. The mapping T 4- (Ti) is an isomorphism of A onto

iEl H A.. in fact, if (T.) 2

H A

there exists an operator TE L(H) i' jET EI which induces Ti in Hi for every i ET, and T commutes with every element of A' , hence T E A" = A. E

The operation of forming the above product is associative in an obvious sense. Moreover, let, for each i, Fi be a projection of A.2 ' Let F = (F.), which is a projection of A. We have 2

AF =

.

H (A.) F ., EI

A F' =

H

. 2-E I

2

F.

PART I, CHAPTER 2

22

Finally, let, for each j E I, (Di be a homomorphism of von Neumann algebra

Bi.

put

Ai

into a

If, for each T = (Ti) iei E . 11 Ai , we 2EI

E H

(D(T) = ((D.(T.)). y

B.,

yEI . y yEI

(D is a homomorphism of H Ai into H B. There is an analogous iEI iEI property involving antihomomorphisms. : [118].

Reference

3.

Operators in a tensor product of Hilbert spaces.

Let H 1 and H 2 be two complex Hilbert spaces. We construct their algebraic tensor product Ho, a complex linear space. There is a unique pre-Hilbert space structure on Ho such that

(x 10 X2Iy10y2)

(X ilYi ) (X 2IY2 )

for x ,y E H 1 1 11

X21y2

E

H . 2

This leads to a metric on Ho . The completion of Ho in this metric is a Hilbert space called the Hilbert tensor product of H1 and H 2 and is denoted by H i 0H 2 , with a slight abuse of notation. Let TlE L(Hi ), T2E L(H 2 ). The algebraic tensor product T 0 of T 1 and T 2 operates in Ho and is continuous; in fact, it is enough to show this when T1 = I and then when T 2 = I; suppose n . for example that T 2 = I; let 1 xi"1 Oxi2 be an element of H • we 0' i=1 can take the x to be orthonormal; then

2

n

T

o

i=1

2 .1

0 xil 2

i

= i =l

T ixi 0

2=1

11

H2


—

X

2

rho: i=1

m

i 2

=HT

n . 2 1 X x1 i=1

2 x

2

With a slight ablve of language, the continuous extension of T o H is called the tensor product of T 1 and T 2 , and is denoted to by T 1 0 T 2 . We see immediately that T 1 0 T 2 is a bilinear function of T 1 and T 2 , that

23

GLOBAL THEORY

and that

(T

1

0 T )* = T* 2 1

T*. 2

In an analogous way, we define the Hilbert tensor product H 1 0 H 2 ® ... ® Hn of Hilbert spaces H1 , H2, , Hn and the tensor product

T

1

®T

of operators Ti

2

E

e... 0 T E n

L(H

1

0 H

2 e

••• 0 Hn )

L(Hi).

We are going to study H 1 0 H2 in an unsymmetrical way. Let (e i )-7-€ be an orthonormal basis of H 2 using which we may H 2 canonically with the space L 2identfy (I) of families of complex numbers such that (X.)The I .1 2 <

iEI

mapping x l xi 0 ei is a linear isometry Ui of H 1 onto a complete and thus closed linear subspace H1-- of H; the Hi are pairwise orthogonal; the linear subspace of H o generated by the Hi is dense in H o ; hence H is the direct sum of the H. It follows from this that every element x of H can be uniquely expressed in the form is some family 11 0 • where (xi1 ) iEI •

iEI

of elements of H 1 such that

/ iEI

xl;

11x11 2 =

11 2

< .;

furthermore

11 2 . j El

Conversely, let the complex Hilbert space H be the direct sum of a family (H i ) ET of closed pairwise orthogonal linear subspaces; let Hi be another Hilbert space, and for each i ET, let Ul be an isomorphism of the Hilbert space H 1 onto the Hilbert space Hi. Then, putting H2 = L6(1) [a Hilbert space possessing a canonical orthonormal basis (ei)], we may use the above to define a canonical isomorphism of Hi 0 H 2 onto H under which the element xi ee l: of H 1 0 H2 corresponds to the element Ui(x1) of Hi. We will often identify H 1 0 H 2 and H by means of this isomorphism. Retaining the above notation, LI is a linear mapping of H into Hi, such that ul(H e Hi) = 0, and which maps HI isometrically onto H 1 1 - 2utui is the identity operator on Hi, and Uiall is the projection of H onto Hi. For every T EL(H), we have UiTUx EL(H i ), and T is completely determined by knowledge of the U*Tux = Tx, or of the matrix (Tix). By abuse of notation, we sL11 write T = (Tix ). For example, suppose that H 1 is the one-dimensional Hilbert Space C. Then, the mapping I 0 x2 4- x2 enables us to identify

PART I, CHAPTER 2

24

H = Hi 0 H2 canonically with H 2 = L(I). The operators Tix in H I may be identified canonically with complex numbers A. We

easily see that Xix = (Texlei). Returning to the general case, the representation of T by the matrix (Ti) thus extends the usual matrix representation of operators. If S = (Si x ) and T =(Ti), we immediately have

S + T - (S.

AS = (AS. ),

2X

2X

+ T. ),

S* = (St.),

2X

ST= (R. ) 2X

with, for every X1 E Hp xi

Riz i

1

US

UatIlorL

TUx xi=LISTUX =

°LEI the series converging strongly in H 1 .

X

S iaax TXl ,

DEI

[

Examples.

Let T l E L(Hi), T 2 E L(H2), and put (T2e 5 lei) = Aix , so that (Ai x ) is the matrix representing T 2 with respect to the basis (ei) in the usual sense. Then the matrix (Tix) of T = T 1 0 T 2 is given by Tix = XixT i ; in fact, for Xi E Hp we have

UV T 1 0 T 2 )Uxx1 = 11(T 1 0 T2) (x 1 0 ex ) = 11(T 1X1 0 T 2 gx )

( r = 2,atILTx0Xe iax = ctc , 11

XX axUtUTx= 2 a 1 1

acI

X.Tx. 2x 1 1

In particular, the matrix representing T1 0 I is (6ixT 1 ), where cSix is the Kronecker delta, and the matrix representing I 0 T2 is (X. IH ). 2 X'

1

LEMMA 1. If TE L(H) commutes with the uiul form T 1 01, for some T1 E (Ht) .

Proof.

then T is of the

Choose a particular element a of I. We have T.

2x

= UtTU = U*11 UtTU = U*TU UtU . 2 X U Ca 2 X x a a

Now, IlUx = 0 if i / some T 1 E L(H i ) . 0

x, and UtU = IH I . i 7

,

Hence Ti x 8. T„ for 2X

For every subset M of L(H 1 ), let NM be the set of the T = (Tix) EL(H) such that Tix EM or every i El and every X E I; further, let Pm be the set of the T = (Ti)c L(H) such that: 1° T. = 0 for i / x; 20 T. is the same element of M for all i. le then have (PW Nml, (Pm)" =PM» ; and, if M contains 0 and 1H 1, (NM)' = . (NM)" = Proof. Let R = (Rix ) E LW), S = (six ) E L(H), and suppose Six = 0 for i / x, Sii being an element T of L(H i ) not dependLEMMA 2.

ing on i.

The condition RS = SR immediately translates into RixT = TRi x for any i, S E T. It follows from this that

25

GLOBAL THEORY

NW.

Replacing M by M', we deduce from this that NmuDNM, and hence that PmIC:(Nm)'. We show that C:Pml (NW if 0, TH EM; let R = (R-2 x) E (Nm)'; we have at once 1 M' for any i, x E I; consethat (NA4)'(:(Pm) 1 = Nmi, and so P quently, we have U 17131 E NM, because all the elements of the matrix representing tJtJ are equal to 0 or to IH 1 ; hence R comand lemma 1 shows that RE Pm'. We have mutes with the thus shown that (NW P = Pm', if 0, IH E M. Since 0, TH i E M P 1 = Pmu for every M. Finally, for any M, we have (Pm)" = = Nmu. if 0, IH i E M, we have (NW" =

(Pm) 1 =

In particular, the operators of L(H) commuting with the operators of the form I 0 T 2 , where T2E L(H 2 ), are just the operators of the form T1 0 I, where TiE L(4 1 ). These lemmas will be useful, not only in section 4, but also in chapter 3, lemma 6. References : [65]. 4.

Tensor products

of von Neumann algebras.

We retain the notation of section 3. Furthermore, let A l be a von Neumann algebra in H 1 and A 2 a von Neumann algebra in H2. + T1 0 T 2 , The operators of the form R1 0 R2 + S 1 0 S 2 + where R , S , 1 1

' T1

E

A

1

and

R , S , 2 2

T E 2

A 2'

form a *-algebra of operators Aci , containing IH, in H. The von Neumann algebra generated by Ao is called the tensor product von Neumann algebra of A l and A2 and is denoted by A l 0 A 2 , with a slight abuse of notation. We define analogously the tensor product Al 0 A2 0 ... 0 An of von Neumann algebras Al, A 2 , ..., An If A 2 = CH.) , Ao is the set of operators T1 0 IH 2 for T I E Al. By lemma 2, A o is a von Neumann algebra since All = Al . Hence the mapping T 1 4- T 1 0 IH2 is an isomorphism of A l onto A l 0 CH 2 , which is called the ampll,ation of A l onto A l 0 CH 2 . Similarly, the mapping T2 IH 1 0 T2 is an isomorphism of A 2 onto CH 1 0 A2. Moreover Al 0 A 2 is the von Neumann algebra generated by Al 0 CH 2 and CH 1 0 A. PROPOSITION 4. (i) The operators of A, cH 2 are the operators representable by a matrix (Tix ) such that Tix = 0 for i / x, and such that Tii is an element of Al independent of i. (ii) The operators of Al 0 L(H 2 are the operators representable by a matrix (Tix ) such that Tix E AT for any i, xE T. Cf-f 2 = (Al 0 L(H 2 )) . (iii) We have Al )

26

PART I, CHAPTER 2

Proof.

(i) follows immediately from the preceding remarks.

The algebra (Al 0 L(H2)) / is the intersection of (A10CH 2 ) 1 and

(C

H, 0 L(H 2 )) 1 = L(H 1 ) 0 C H2 .

Now for an operator of the form T 1 0 Iti o to commute with A'1 0 cu , it is necessary and sufficient that T I E A l . Whence

(iii).

112

Finally,

(ii) is a consequence of (i), (iii) and of lemma 2. E

Still in the notation of section 3, the projections Ei = belong to CH 1 0 L(H2), and therefore a fortiori to

P Hi

Al 0 L(H2) = (Al 0 CH2 ) / . Similarly, the partial isometries UxU1 = Uix , which have Ei as initial projection and Ex as final projection, belong to Al 0 L(H2) = (Ai 0 CH 2 )'. This established, we have: PROPOSITION 5. (i) (A 1 0 L(H2)) E . is spatially isomorphic to

Al , and (A10 cH 2 ) E2. is spatially isomorphic to A. (ii) Let A be a von Neumann algebra in a complex Hilbert space H, and (Ei)iEl a family of pairwise disjoint projections of A with sum equal to 1; suppose that there exists, for every i ET and x ET, a partial isometry u having initial projection Ei and final projection Ex. Let a be a particular element of T, H 1 = E a (H), H 2 = Then, A is spatially isomorphic to the von Neumann algebra AH, 0 L(H2) in H 1 0 H2, and A' is spatially isomorphic to All 1 te- cH2 .

Proof.

We will actually prove something more precise, as the proof will furnish certain canonical isomorphisms. We prove (i). The isomorphism Ui of H i onto Hi maps A to (A'1 0 n2 ) Ei by proposition 4 (i), and t5erefore maps Al to

(A1 ° L(H 2 ) ) Ei • Now (ii). Let Ui be the isomorphism of H1 onto Hi = E(H) defined by Uai. The Ui may be used to define a canonical isomorphism of H onto H1 0 H 2 (section 3), by means of which we shall identify these two spaces. For T EA, the matrix Tix representing T is given by

T

ix

= UtTUX = (U*.TUax) Ea

and so A (=AH L (H2 ) . ing T 1 is given by

For T'

E

A',

E

AH ' 1

the matrix (T) represent—

27

GLOBAL THEORY

= (U*.T I U )

ax E a

az,

= (T'Ut U )

-ba ax

=

Ea

I EA!, , 6.2,x T Ea ti1

®C 1 2 . Hence D AH 0 L(H2) , and finally and so A' (= y 1 A -= AH L(Hir, A = AH ® Ca . 1

1 H2 Thus, passing from A l to A 1 ® L(H 2 ) (resp. to Al CH 2 ) and passing from a von Neumann algebra to a reduced ( resp. induced) The algebra are, to a certain extent, inverse operations isomorphism inverse to an ampliation is an induction. For i = 1, 2, n, let 14,77 be a complex Hilbert space, Mi a subset of L(Hi), and Ai the von Neumann algebra generated by Mi. Let A be the von Neumann algebra generated by the operators PROPOSITION 6.

T, I

H1

T

...

2

...

I LI

"2

-I-

(T E M )

Mil 2

H n (T 1 • • •

2 '

I

M

1 '

H

o Tn (Tn E Mn ) .

H2

We have

A = A

l

ED

A2

o An .

If we replace Mi by mi utvit, the Ai and A stay the I suppose that Mi = M. Then, Ai = same.Wcnthrfo Lemma 2 shows that Ai 0 CH 2 ... CH is the von Neumann algebra generated by the operators Ti 0 IH 2 IH , where T i E M i

Proof.

n

A l oc

- H2

Henc

®..®C CA.

Similarly,

C

H1

0 ...

A- ®

®C

Hy,

CA

for

1

i

n,

and hence

A1 oA 2 o

o AC A.

The reverse inclusion is obvious. The operation of forming the tensor product of von Neumann algebras is associative in the following sense: let H1, H2 , , Hn be complex Hilbert spaces, H = H 1 0 H2 ..... Hn , K --= H10 H2 ... 03) The Hilbert space H may be canonically H P identified with K 0H p +i 0 - - Hn . Let Ai be a von Neumann algebra in Hi ,

PART I, CHAPTER 2

28

A

=

Ai o

A2

e

An ,

A

is generated by the algebras

The von Neumann algebra

c

Hl

øA

2

® ... C

C

C

• " '

Hn '

H1

H2

An .

0 ...

The same is true of BokoA o o •" p+1 p+2

in virtue of proposition 6, since B generated by

A, øcv

oc

"2

c

H '

H

oA o 2 "'

An

is the von Neumann algebra

Hp '

c oc

'

H1

H2

"

oA

Hence

p+ 1 Let A1 (resp • A 2 ) be a von Neumann algebra in H 1 (resp. H2) and E1 PK 1 (resp. E 2 = PK 2 ) a projection of Ai (resp. A2) . The operator E i 0 E2 is a projection of A 1 0 A2, and (E1 ®E2) (Hi 0 H 2 ) may be identified with K i K2 . Having said this, we have

(A 1

0 (A ) 0 A ) = (A ) 2 E2 . 2 E 10E2 1 E1

In fact, A 1 0 A2 is generated by the operators R

1

OR

2 +S1

OS

2

+ . +T

1

0T

2

(R1, S1 ..., T1 c A1, R2, S2, ..., T2 E A2) , and hence [prop. (ii)] (Ai0A2) El0E2 is generated by the operators NE

1

0E ) (R OR + S 0 S + 1 2 1 2 2

= (E R E ) 11 1 E

1

0 (E R E ) 2 2 2 E

+ T

+ 2

1

OT ) (E 0E ) ) 1 2 E103)E2 2

+ (E T E ) l 1 1 Ei

(E T E ) E 2 2 2 E2

0 (A ) E • (A ) 2 1 E1

hence

(A l ®A )

)

C (A ) 1 El

(A) ; 2 E2

1

P"

29

GLOBAL THEORY

the reverse inclusion is obvious. We see analogously that

(A'1

0

A') 2 El0E 2

(A') 1 El

=

(A') 2E2

0

Finally, let, for each iE I (resp. XE K), Ai (resp. Bx) von Neumann algebra in Hi (resp. Kr ). Let A = H Ai in

be a

iEI H=

e.

B = II

H

x in K

e

xEK canonically identified with

A

08 =

e

xEK

K.„.. The space HO

7,(H•

(2• , x)EIxK

Kx )

.

K may be

We then have

AI,- 0 B xi- we leave the job of verifying this to H (i,x)EIxK

the reader. Proposition 4 also has the following application: PROPOSITION 7. Let H be a Hilbert space, A a von Neumann algebra in H, Z = ARA', Tii(i,j = 1, n) elements of A, and

T-i- ti- (i " n) elements of A'. , j = 1, tions are equivalent: (i) EkTikr4j = 0 for i,j = 1,

Then the following condin.

(ii)There exist elements zij(i,j = 1, 7,( Tikzkj = 0,

Proof.

1 EkZikTki = T ij

for

n) of Z such that i,j = 1,

n.

If there exist Zii satisfying (ii), we have

ZkT ikT kj = E kTiJI ZUT ij = /. (Z ICT ikZkI )T Zj = O. Suppose that EkTikTki = 0 for i,j = 1, n. Let K be the direct sum of n copies of H. We shall identify K with HO L, where L is an n-dimensional Hilbert space. The elements of L(K) may be identified with nxn matrices whose entries are elements of L(H) n E Ao L(L). Let T = (TO

T' =

L(L). We have TT' = O. Let E be the set of projections E' of A' @ L(L) such that TE' O. Let E (') be the supremum of the elements of E. The projection onto T (K) is an element of E since TT / = 0 and T 1 EA' 0 L(L); hence . _

EA'

@

= T / . Furthermore, it is clear that TE6 = O. Hence, if we put E / = (Z..). T. Z . = 0, , we have Z..E A l , 0 2j 2,j=1,-., n 2,7 k = Ili . It remains to prove that ZijE A. It is enough kZikTJ to show that, for every hermitian R P E A', we have Zie / = 2j Let S' = R'€) I A P OCL, which is hermitian. We have TS'E = S / TE 0 / = 0, and so the projection onto S / E;1 (K) is an element of E ; hence S'E l (K)CE I (K), and consequently 0 0

PART I, CHAPTER 2

30

E 0P S'E' 0 =S P E 01 '. taking adjoints, we get E;S P E I; = ES', whence S P E 0P = ES' ,whence Z..R ' 1.,„7 1 = R 1 Z.-.

D

Certain simple properties of the tensor product will not be obtained until later (chapter 4, proposition 2; chapter 6, corollary to proposition 14). Problem: do we have (Ai® A2) ' = Al®Ap (cf. exercise 5, chapter 6, proposition 14, and [443]). The elementary operations, applied to the abelian von Neumann algebras, yield all the discrete von Neumann algebras (chapter

8). References : [31], [50], [65], [77], [89], [128], [190]. Exercises. 1. Let H be a complex Hilbert space, H1 and H 2 two complementary orthogonal subspaces of H of the same dimension, and M the set of operators TE L(H) such that T(Hi) (=H 2 , T(H2) = 0. Show that the von Neumann algebra generated by M is L(H), but that the von Neumann algebra generated by MH1 is CH i . Let Z, A, B be von Neumann algebras, with ZC:AC:B, Z being contained in the centre of B.

2.

a. Let TE B. Show that, in the set M of projections EE such that TEE A, there exists a projection greater than all the others. (Consider a maximal element of M.) Deduce from this that, if Ti A, there exists a non-zero projection FE Z such that, for every non-zero projectionG ofZ majorized by F, we have TG A. b. Suppose that, for every non-zero projection E of Z, we have A E / BE. Show that there exists a projection TE B such that, for every non-zero projection E of Z, we have TE 4 AE . (Let N be the set of projections FE Z such that there exists a projection SE BF possessing the following property: for every non-zero projection G of Z majorized by F, we have Set AG . Consider a maximal family of non-zero pair-wise orthogonal projections of N and show, with the help of a, that their sum is I.)

3. Let A and B be von Neumann algebras. For A 0 B to be G-finite, it is necessary and sufficient that A and B be a-finite. If A and B are G-finite, let (xi) and (Yi) be separating sequences for A and B. Then the double sequence (xi® y) is cyclic for A' B', hence for (A 0 B)', and therefore separating for A (31 B.] Let A and B be von Neumann algebras in H and K. For Ao B to be a factor, it is necessary and sufficient that A and 13 be factors. [If A and B are factors, show that A® B and A' B' generate L(H® K) .]

4.

31

GLOBAL THEORY

Let A (resp. B) be a von Neumann algebra in H (resp. K) . Let T E (A® B) ' . 5.

Show that (A0 B) --= (A' 0 L (K) ) n ( L (H) that A 0 13 is generated by AO ci< and CH 0 B.) a.

B') .

(Use the fact

There exists an element S in the *-algeLet x€ H, y E K. b. bra of operators Co generated by A' ® cK and CH° B' such that is arbitrarily small. [Show that (T - S) (x0 y)

T(x 0 y)

= (E

A'

n

c.

Let (x )

yn+.x ,

)r(x ®

y)

;

A' Br, X.E X , y.€ X J. x y

where

be a sequence of elements of H such that

m

m2 <

1H x m m (yn )

B'

is a limit of elements of the form

deduce from this that T (xO y) X1 0 y 1

E

a sequence of elements of K such that

H Yn n There exists RE

C 0 such iri,n

H2

<

c° .

that

H

(T

-

R)(xm 0 yn ) H 2

is arbitrarily small. [Let 141 , K1 be infinite-dimensional Hilbert spaces; identify (xm ) with an element x of H 1 ® H, (yn ) with an element y of K 0 K 1 ; form, in the space

H1

®

(

H 0 K)

0 K 1 = (H 1 0 H) 0 (K 0 K 1 ),

the operator IH 1 0 TO IK 1 , which commutes with

(L(H 1 ) 0 A) 0 (B® L(K 1 )) and apply b to it.] This result makes it likely that (A® B) ' --= A' o B P (cf. however, chapter 1, exercise 9b).

Suppose that A is a factor in K and that T 1 , T2, ...,Tn E are linearly independent (over C). Let Ti, T. TE A'. 6.

Then the relation

1 T.T'. 2, 7,

0 implies that the T vanish (use

i=1 proposition 7).

Let B be the *-algebra of operators generated

A

32

by

PART I, CHAPTER 2

A

and

A',

i.e. the set of operators

y

Rj R, where

i==1

R.E

A,

R I.E A 7 ; show that the mapping X R.R: ÷

R•OR'. defines

2=1 2=1 an isomorphism of the *-algebra 8 onto a *-algebra of operators in KOK [65]. 7. Let A be a von Neumann algebra in H. If T1, ..., TE A and x l , xn eH are such that T ixi + + Txn = 0, there exist Tije A (2,j = 1, ..., n) such that xi = EiTijxj, 0; in other words H is a tame A-module. [Introduce K 7 and L as in the proof of proposition 7. Let x = (x l , ,xn) K. Let E be the cyclic projection of A 0 L(L) defined by x. Put E = (Tii).] [333]. ,

CHAPTER 3.

1.

Topologies on

DENSITY THEOREMS

L(H).

The norm defines a topology on L(H) called the norm topology. This is the topology of uniform convergence over bounded subsets of H, where H is endowed with the strong i.e. its norm topology.

Norm topology.

Strong topology. Let xE H. The function T ± H Tx H is a seminorm on L(H). The collection of all these seminorms determines a Hausdorff locally convex topology on L(H) called the topology of strong pointwise convergence or just the strong topology for short. A base of neighbourhoods of zero for this topology is obtained by taking, for each finite set (xi, x2, ..., xn ) of elements of H, the set of TE L(H) such that 1 for 1 < i < n. We can also define the strong topolTxogy Is the coarsest topology on L(H) for which the maps T Tx of L(H) into H (with its strong.topology) are all continuous.

d

H<

This topology is compatible with the linear space structure of L(H), but not with its algebra structure in general; in other words, the map (S, T) ST is not strongly continuous in general (exercise 2). Nevertheless, if Li(H) denotes the unit ball of L(H), the equality ST - S oT o = S(T - T o ) + (S - S o )To shows that the map (S, T) ST of L i (H) X L(H) into L(H) is ST, T strongly continuous. The (partial) maps S ST are strongly continuous (for fixed T and S respectively). When H is infinite-dimensional, the map T T* is not strongly continuous. Indeed, let (el, e2, ...) be an orthonormal sequence in H, and let Un be the operator x 4- (xlen )el in H: then Un converges strongly to zero; however we have (Upz e l Ix) = (e l l(xlen )e l ) (en lx) for every X EH, hence Up i = en and q does notconverge strongly to zero. The unit ball Li(H), endowed with the uniform structure associated with the strong topology, is a complete space ([[ 3]], chapter III, section 3, theorem 4). If H is separable, this

PART I, CHAPTER 3

34

space is in fact metrisable and has a countable base ([[ 3 ]], chapter III, section 3, proposition 6), from which it follows that every von Neumann algebra can then be generated by a countable family of elements.

Let A be a a-finite von Neumann algebra, and A l its unit ball endowed with the strong topology. Then, Al is metrisable. PROPOSITION 1.

Proof.

Let M = {x l , x2,...}(=H be a countable separating subset for A. Let N be the set of elements of the form

Tixi + `*2±...+Thxn , where the 1°-2 are in A'. Then, N is dense in H. Since A l is equicontinuous, the strong topology on Al is the same as the topology of pointwise convergence over N; since T(T / x + T P x + 1 1 22

+ T 1 x) = L T.Tx. nn • E

for

T EA,

=1

this topology is also that of pointwise convergence over M. Hence the mapping

T 4- (Tx is a

homeomorphism

1

,

Tx , ...) 2

of Al onto a subset of the product space

H x H x ..., which is metrisable. Let A be a a-finite von Neumann algebra, M a bounded subset of A, and T a point in the strong closure of M. Then there exists a sequence of elements of M which converges strongly to T. COROLLARY.

Weak topology. Let x, y EH. The function T 1(Tx1y) I is a seminorm on L(H). The collection of all these seminorms defines a Hausdorff locally convex topology on L(H) called the topology of weak pointwise convergence or just the weak topology, for short. The seminorms T 1(Tx1x)1 are enough to define this topology, because of the equality 4(Tx1y) = (T(x + y) Ix + y) - (T(x - y) 1X - 3) + i(T(x+iy) Ix + iy) - i(T(x - iy)

- iy) .

We obtain a base of neighbourhoods of 0 for this topology by taking, for each finite set (xl, y1), (x2, y 2 ), •.., (xn , yn ) of pairs of points of H, the set of T EL(H) such that

1(Txilyi) < 1

for

1 < i < n.

35

GLOBAL THEORY

We can also define the weak topology as the coarsest topology on L(H) for which the maps T (TXIY) of L(H) into C are continuous, or alternatively, for which the maps T -›- Tx of L(H) into H, endowed with its weak topology, are continuous. This topology is compatible with the linear space structure of L(H), but not with its algebra structure in general (exercise 2). Nevertheless, the maps S 4- ST, T ST are weakly continuous (for fixed T and S respectively). The mapping T T* is weakly continuous due to the equality

I (Tx101

I (T* Y1 x) I

=

'

The unit ball Li(H), endowed with the weak topology, is a compact space ([[3]], chapter IV, section 2, theorem 1, corolIf, further, H is separable, let (xl, x2, ...) be a lary 3). dense sequence in H. Then the weak topology on L1(H) is defined ((S - T) xilxj), and is thus by the sequence of écarts (S, T)

metrisable and possesses a countable base. Ultra-strong topology. Let (xi, x2, ...) be a sequence of Co

H

elements of H such that

i=1

x.2

11 2 <

w.

For each T EL (H ), we

have co

H

Txi

H 2
ill The function T

Co

i=1

i=1

d

Txi

H2

is a seminorm on L(H).

The

collection of all these seminorms defines a Hausdorff locally convex topology on L (H) called the ultra-strong topology. We obtain a base of neighbourhoods of 0 for this topology by taking, for 00

every sequence (xi) of elements of H such that 00

the set of T

EL(H) such that

1

i=1

H Tx.2 H 2

< 1.

/ II x.2 11 i=1

2

<+°°,

This topology is

also the coarsest for which the mappings T ÷ (Txl, Tx2, ...) of L(H) into the direct sum H e H e ... are continuous, (xi) being Co

H

any sequence of elements of H such that

i=1

x

H2. <_ ..

This topology, compatible with the linear space structure of L(H), is not compatible with the algebra structure in general ST of L 1 (H) X L(H) (exercise 2). Nevertheless, the map (S, T) (for fixed T and S into L(H) and the maps 5 ST, T ST

36

PART I, CHAPTER 3

respectively) are ultra-strongly continuous. The map T T* is not ultra-strongly continuous in general (cf. section 2).

Ultra-weak topology.

(xl,

Let

x2, ...), (yi, y2, ...) be two

H such that

sequences of elements of CO

2 <

op,

i=1 L (H) ,

For each T

x I (Tx.i y .)1 <

i=1

H Yi 11 2 < + m.

we have

i=1

H

Tx-

H- H

Y.

H

00

r-

<

The function T

Y II

II xi 11 . 11

< II T II

MTH i=1 / II

X • II 2 2

I(Tx . ly). X

i=1

H

11 21 1/2 <

is a seminorm on L(H).

The

i=1

collection of all these seminorms defines a Hausdorff locally convex topology on L(H) called the ultra-weak topology. The

seminorms T

are enough to define it. I • / (Tx.lx.)I 2 2

We obtain

=1

a base of neighbourhoods of 0 for this topology by taking, for every finite family (4, (4 ) , (xi), (yi), (4), (4) of sequences of elements of H such that 00

00

L.

H

x].-

i=1 the set of T

H2 <

m,

H2 <

co,

X

H

H2

such that

E (H)

L (Tx.kly k.) 1-

<1

for

1 <

k < n.

37

GLOBAL THEORY

This topology is also the coarsest for which the linear forms 00

T ÷

(y.) are any two X (Tx.ly.) are continuous, where (x.), 2 2 2 2

i=1 sequences of elements of

H such that

H x.H 2

+

HY. H

i=1

2

i=1

This topology is not compatible with the algebra structure of ST, T ST (for The mappings S L(H) in general (exercise 2). fixed T and S respectively) and T + T* are ultra-weakly continuous. The ultra-strong and ultra-weak topologies have the interesting property of being isomorphism invariants (chapter 4, theorem 2, corollary 1). The ultra-strong topology is also sometimes known as the "strongest" topology. References : [15], [73], [74], [234].

The above topologies compared.

2.

We immediately have the following table, where the symbol < means "finer than": Norm topology < Ultra-strong topology < Strong topology A

A

Ultra-weak topology < Weak topology When H is infinite-dimensional, the symbol < may even be taken to mean "strictly finer than" (exercise 1, and part III, chapter 6, propositions 7 and 8). The strong and ultra-strong topologies coincide on bounded subsets of L(H). In fact, let W be an ultra-strong neighbourhood of O. There exists a sequence (x ) of elements of H such i 03 that / II x• i=1

11

2 <+

00, and such that the inequality

Tx . 112 _< 1 implies that TE W. H 00 i=1

H

such that i=1

X.

12

1 2'

Let

W1

Let n be a positive integer be the strong neighbourhood of

n-1 0 defined by the inequality tions

H

T

H<

1 and T EW

1

imply

H

Tx.y

H2<- 12'

Then the condi-

38

PART I, CHAPTER 3

n-1

H i=1

Tx.

m 2 _< . H 2-1

Tx.

12

T

H2

x. m

2

1 — 2 2 < 1

hence TEW, which proves our assertion, We may see similarly that the weak and ultra-weak topologies coincide on bounded subsets of L(H). For T to converge strongly to 0, it is necessary and sufficient that T*T converge weakly to 0, as is shown by the equality (T*Tx1x). Tx Similarly, for T to converge ultra-strongly to 0, it is necessary and sufficient that T*T converge ultraweakly to O.

m

12,

References : [15], [73], [74].

3.

Linear forms on L (H) .

The space L(H) = L is a complex Banach space. The set of norm-continuous linear forms on L constitutes the dual L* of L, which is itself a complex Banach space as well. The linear forms on L which are continuous for one among the strong, weak, ultra-strong and ultra-weak topologies constitute subspaces of L* that we are going to study. For x, y EH, we shall denote by wx y the weakly continuous linear form on L defined by wx i y(T) ='= (Txly). The set L- of all weakly continuous linear forms on L coincides with the set of all strongly continuous linear forms, and with the set of finite sums of forms Wx,y (H311, chapter IV, section 2, proposition 11). This set is a linear subspace of L*, not closed for the norm of L* in general (part III, chapter 6, propositions 7 and 8); let L * be its norm-closure in L*. In what follows, it will be convenient to use the following terminology. Let E, F be two complex normed linear spaces, E*, F* their topological duals, and (x, y) 4- (3(x, y) a continuous bilinear form on E X F. Each y EF defines the linear form fu :x -›- f3(x, y) on E, and the mapping y -›- 4 of F into E* is linear and continuous. If this mapping is bijective and isometric, we shall say that F is the dual of E for the form (3. If this is the case, we have

HY

H

=

sup

Wx,y)

I

for every y E

F,

and sup

II 11<1

3(x,y)1

for every

x

E

E.

39

GLOBAL THEORY

canonical bilinear form on L* x L induces on L- x L a bilinear form for which L is the dual of the normed linear space L-. LEMMA 1.

The

Proof. Let E be the dual of the normed linear space L-. Each element of L defines a continuous linear form on L-, and thus an element of E. Let 8 be the linear mapping, clearly injective, of L into E thus obtained. This mapping is continuous if E is endowed with the weak *-topology a(E, L-) and L the topology L-) which is the same as the weak operator topology. Let L 1 and E l be the unit balls of L and E. The ball L1 is weakly compact and so 0(L 1 ) is weakly compact. Finally, 8(L 1 ) and El have the same polar in L-, namely the unit ball of L-. Hence 0(L1) = E l , which shows that 8 is an isometry of L onto E. E

L with

Lemma I also enables us to identify complex Banach space L.

the dual of the

LEMMA 2.

The ultra-strongly continuous linear forms on L coincide with the ultra-weakly continuous linear forms, i.e. Co

X i=1

with the linear forms

.

where

co

H

x.

H

co

2 <

1

co

i=1

i=1

H

H2 <

They belong to L. Proof. Let

= i=1 / xi,Yi l

with

X i=1

H x- H

2

co

2 H Yi H iz=1

<

X

< + 00.

It is clear that (1) is the uniform limit over the ball L1 of the forms

, and hence ckL * . 1 w i=1 xi,yi

The form (I) is ultra-weakly

continuous, and every ultra-weakly continuous form is ultrastrongly continuous. Finally, let ci) be an ultra-strongly continuous linear form. Consider the Hilbert space H = 07, 1Hi, where H - H 1 = H We have

14)(T)

<

i=1

I TXi I

2

2

(xi) being a sequence of vectors of H such that

40

PART I, CHAPTER 3

CO

H x. II

2

<+

co,

7,==i

R.

T

The sequence (xi) is an element x of Let be the operator of L(R) defined by T((2i)) = ( Tzi). We have 1(1)(T)I < Tx and so there exists an element g = ( yi ) of such that

m H

R

0.0

( T) = (Tx@ = i-1

LEMMA 3.

Proof.

(Tx.ly.). 0 2 2

The 'linear forms of L. are ultra-weakly continuous We set out by establishing the following result:

For each linear form (1)E L , there exist two orthonormal systems (e l , e 2 , en ), (ei, e, qi ) in H, and numbers Xi 0(1 n) such that "

We have cp =

el

• •. j=1 XJ' Y J

be other elements of

Let x'1' x ' 2"

, Yl , v2 ,

Yq

H such that the two finite-rank linear

(YIY .)x. and y

mappings y

x ql

k=1

( Y14)Xj< of H into itself

are in fact the same. For each pair (2 1 , 22) of vectors of H, let {z i , 22} denote the linear mapping of rank 1, z ÷( 212 1 )22. We have ( {

j=1

2

1 , 2 2)xilYj) =

(xj1 2 1)(z2W) = (

j

= (

x. ( z 214)xf< l zi)

g

(z2lYj)xj121)

( zi , 2 2 } xj< 1 4) , {

hence by linearity and continuity

(Tx.IY

j=1

( Tx;.

=

kl

14)

k

i.e. w

xj• ,y j• - k=

for every TE

L,

k=1

GLOBAL THEORY

41

We carry out the polar decomposition of the operator ? ÷

i=1

and then apply spectral theory to the finite(YIY•)x't a J

rank hermitian operator thus obtained. We thus see that we can choose for (xk) and (4) two orthogonal systems. Whence, immediately the existence of orthonormal systems (e.), (e'.) and 2 2 scalars X. 2

0 such that cp =

X.(1) i=1

I CH T) I

7/1

T 11 . 11

i-1

e i' e i

ei

II

II' II

H (t) H

H (P H

„Y

,Y X., i=1 2

-< 1

I

-k -< 2

for

X.,

14)(T) =

and

i=1

which proves our intermediate result.

This established, let (Pc

and II cl) k

i=1

X.. Moreover, for T = 1 {e.,e 1.1., we have 2 2 i=1 2 i=1

II T II hence

T II 71 X.

n

n and so

We then have

1.

k

L.

2, and

We have cti =

k=1

(pk , with (1)E

nk so cp k = vL Akwe k e lk , k=1 i

with

nk

H

H = li

- 1,

e.

X.1,

k

0,

i=1

Whence

/ nk k1/2k (I) = Xç s e. k=i i=1 (X.) co

Finally,

k 1/2 i

y

•

-k

42

PART I, CHAPTER 3

and, similarly,

tnk

k

, 2\ (x • ) 2e . k 11 <

X

k=1 i=1

hence 4) is ultra-weakly continuous.

0

Let M be an ultra-weakly closed 'linear subspace of L(H), M* the dual of the Banach space M, Mr the ball H T H r of M, and (1) a linear form on M. (i) The following conditions are equivalent: THEOREM 1.

(i 1) (I) is weakly continuous; (i 2) 4) is strongly continuous; (i 3) 4)

xi,yi

(ii) The folloiwng conditions are equivalent: (ii 1) (I) is ultra-weakly continuous; (ii 2) (I) is ultra-strongly continuous; (ii 3) 4) =

, with

w

xi , Yi

i=1

H x. H 2 <

yi

00 ,

m2 <

co,

i=1

The restriction of cp to M 1 is ultraweakly (resp. weakly) continuous; (ii 4) Et-esp. (ii 5)].

The restriction of (1) to Ml is ultrastrongly (resp. strongly) continuous. (iii)Let g, (resp. M* ) be the set of weakly (resp. ultraweakly) continuous linear forms on M. Then, M* is the normclosure of il/L, and M is the dual of M* for the bilinear form induced on M* x M by the canonical form on M* x M. (ii 6) [resp. (ii 7)].

(iv)Let K be a convex subset of M. The following conditions are equivalent: (iv 1) K is ultra-weakly closed; (iv 2) K is ultra-strongly closed; (iv 3) [resp.

(iv 4)] K

closed for every r;

n Mr i.s ultra-weakly (resp. weakly)

(iv 5) [resp. (iv 6)] K n Mr is ultra-strongly (resp. strongly) closed for every r. Proof. The implications (i 3) (i 2) are clear. (i 1) Moreover, if 4) is strongly continuous on M, cp extends to a strongly continuous linear form on L, and hence

43

GLOBAL THEORY

(PI

(1)

, which shows that (i 2) =0. (i 3).

Whence part (i)

i=1 xi'Yi

of the theorem. Similarly, use of lemma 2 proves the equivalence of (ii 1), (ii 2), (ii 3). The equivalence (ii 4) (ii 5) ]resp. (ii 6) 0. (ii 7)] follows from the fact that the topologies induced on M 1 by the weak and ultra-weak (resp. strong and ultra-strong) topologies are the same. The implications (ii 6) are clear. To prove part (ii) of the (ii 1) (ii 4) theorem, it thus only remains to prove the implication (ii 6) (ii 1). Moreover, lemma 2 implies that M is normdense in M Let MI be the set of elements of L * orthogonal to M; this is a norm-closed linear subspace of L. Since M is ultra-weakly closed (a hypothesis which has not hitherto been used), M is just the linear subspace of L orthogonal to MI . The Banach space L is the dual of the Banach space L* for the bilinear form (q), T) (1)(T) on L* X L. If II) belongs to the Banach space L * /MI and if TE M, let (11), 1 ) denote the value at T of any representative of 11) in Ç. By duality theory, the Banach space M is the dual of the Banach space L * /MI for the form (11), T) 4-(tp, T). Moreover, the restriction map L * M* has kernel MI ; its range is the set of (linear) forms on M which can be extended to ultra-weakly continuous forms on L (lemmas 2 and 3), i.e. M . We therefore have a canonical bijection e of L /MI onto M * ; if WE M * and if II) = e - (w) , we have (, T) = w(T) for every TE M and so

114

sup E Mt

k1 T)I =

sup Mf

HT)

I = 11 (1) 11

Thus, M* is norm-complete, and M is the dual of M * for the w(T). This proves (iii). The equivabilinear form (w, T) lences (iv 1) 0' (iv 2), (iv 3) •• (iv 5) follow from the equiva(iv 4) [resp. lence (ii 1) 40. (ii 2). The equivalence (iv 3) (iv 5) 40. (iv 6)] follows from the fact that the topologies induced on Mr by the weak and ultra-weak (resp. strong and ultra-strong) topologies are the same. The equivalence (iv 1) 0' (iv 3) follows from (iii) and a property of Banach spaces (][3]], chapter IV, section 2, theorem 5). Thus, (iv) is proved. We can now finally establish the implication (ii 6) 0' (ii 1). Let cp be a linear form on M whose restriction to M 1 is ultra-strongly continuous, and let KCI'M be the hyperplane of cp-1 (0); the set KniM is ultra-strongly closed; hence, 1 by (iv), K is ultra-weakly closed, from which it follows that (41) is ultra-weakly continuous. We call M the predual of M. Thus, M may be canonically * identified with the Banach dual of its predual. The ultra-weak topology on M is the coarsest topology for which the linear forms comprising M * are all continuous.

PART I, CHAPTER 3

44

References : [7], [15], [67], [68], [73], [74].

4.

The von Neumann density theorem.

LEMMA 4. Let A be a *-algebra of operators in H. Let X be the closed linear subspace of H generated by the Tx, TEA, x E H. Let X' be the closed linear subspace of H consisting of those x EH such that Tx = 0 for every T E A. Then, X and X' are orthogonal complements in H, and, for every TEA, we have TPx = PxT = T.

Proof. For each T E L(H) let NT denote the set of x E H such that Tx = O. We know that the closure of T(H) is the orthogonal complement of NT . Hence the closed linear subspace generated by the T(H) , TEA, has orthogonal complement equal to the intersection of the NT , T E A. Moreover, for T E A, we have TPx? = 0, hence T = T(I - Pxr) = TPx and similarly T* = T*Px, whence, taking adjoints, T = PxT. E LEMMA 5. Let A be a *_algebra of operat9rs in H such that XA = H. Then, for each y E H, we have y E

A Proof. Let X = Xu , y' = Pxy, y" = y - y ' . For each TEA, we have Ty E X by definition of X, and Ty' E X because X is invariant under A, hence Ty " E X; hence ( T*Ty " I y") = 0, i.e. Ty" O. Therefore y" = 0 (lemma 4) , i.e. y E X. The following lemma constitutes the crucial step.

Let A be a *-algebra of operators in H such that XH = H. Then every operator of A" is in the ultra-strong closure of A.

A LEMMA

6.

Proof. Let SE A", xE H. variant under S, and x, X

A

We have E A E A ' , so that X is in(lemma 5) x, hence Sx E XxA . x

This established, let (xi)i ci be a family of elements of H such that X 11 xi, 11 2 < + 00 , and let us show that there exists a jE I TE A such that / 11 (S - T)x• 112 is arbitrarily small. To this iEI

7

,

end, let K be the direct sum cf a family (Ki)i Ei of pairwise orthogonal closed linear subspaces, such that there exists a linear isometry U. of H onto Ki. Let y i =UX. Since 2 < + c, 7, we can construct y = yi E K. For each 11 yi 11 iEI iE I RE L. (H) let rg. denote the operator of L(K) represented by the matrix (R 5 ), where Rix = 0 for i x, and R.2, 2• • = R for each i (chapter 2, section 3) . As T runs through A, T runs over a *-algebra of operators A in K. By lemma 2 of chapter 2, we have (A) As the r r i"g(T E A, K) plainly generate K, the begin5 E ning of the proof shows that Sy is a strong limit of elements Ty, TEA, i.e. that ,

".

45

GLOBAL THEORY

r4_,Tyll=

D (s _ ri)x.H 2 2 iEI

can be made arbitrarily small.

Let A be a *-algebra of operators in H, and A l the A. of ball unit (i)The following eight conditions are equivalent: THEOREM 2.

(i 1) [resp. (i 2)]

A (resp. A l ) is weakly closed;

(i 3) [resp. (i 4)] A

(resp.

A l ) is strongly closed;

(i 5) [resp. (i 6)] A

(resp.

A l ) is ultra-weakly closed;

(i 7) [resp. (i 8)]

A (resp. A l ) is ultra-strongly closed . A

(ii)Suppose the conditions (i) are satisfied. Let X = X. Then P Av is the greatest projection of A. For every TE A, we have , P X T = TP X „ - T. The operators of A" are the operators T+

(TEA,

A.

A

C) .

Proof.

We already have the equivalences (i 1) (i 3) (i 2) '4* (i 4) <=> (i 5) •#' (i 6) <=> (i 7) (1 8) [theorem 1, (i) (i 2). Suppose from now and (iv)] , and the implication (i 1) on that A is ultra-strongly closed. Let X' = X1 . By lemma 1, each operator of A is reduced by X and X' , and induces 0 in X'. The *-algebra of operators B = Ax in X is ultra-strongly closed. Lemma 6 applied to B shows at once that ix E 8, hence that Px is The operators of A' are indeed the greatest projection of A. reduced by X and X', induce in X' an operator of L(X') , and in X an operator from B'. Hence the operators of A" are reduced by X and X', induce a scalar operator in X', and in X an operator of B". By lemma 6, B" = B. The operators of A" are thus indeed the operators T + X.IH, (TE A, X E C) . Finally, if S E L(H) is in the weak closure of A, we have S E A", hence S = T + X. it/ with a T E A; moreover,

S -= SP thus, (i 7)

X

= TP

X

+XP

X

E

A•

A is weakly closed, which proves the implication (i 1). E

COROLLARY 1. Let A be a *-algebra of operators in H such that XA = H (for example, such that IHE A). Then A", i.e. the von eumann algebra generated by A, is the closure of A in any one of the weak, strong, ultra-weak or ultra-strong topologies. be the closure of A for the weak (resp. ultra-weak) topology, which is a *-algebra of operators. We have A CA 1 CA" , hence A" CA 1 C A", i.e. AY = A". Moreover, A 1 = A by theorem 2. Hence A 1 = A". We see similarly

Proof.

Let A l (resp. A 2 )

46

PART I, CHAPTER 3

that A 2 = A". Finally, Al (resp. A 2 ) is also the strong (resp. ultra-strong) closure of A, by theorem 1. E

A COROLLARY 2. Let A be a *-algebra of operators in H such that XH = H, and A l its unit ban. For A to be a von Neumann algebra, it is necessary and sufficient that A, or Al , be closed in any one of the weak, strong, ultra-weak or ultra-strong topologies.

Pr oof. For every set MCL(H), M' is weakly closed.

The conditions are therefore necessary. They are sufficient by theorem

2 A norm-closed *-subalgebra of L(H) is called a c*-aigebra of operators in H, or a sub-C*-algebra of L(H). Von Neumann algebras are particular examples of C*-algebras.

Let A be a von Neumann algebra and m an ultraweakly closed left ideal of A. Then m is weakly closed. There exists exactly one projection E of A such that m is the set of TEA satisfying T = TE. If m is a too-sided ideal, E belongs to the centre of A. COROLLARY 3.

Let n = minim*, an ultra-weakly closed *-algebra of operators. Let E be the greatest projection of n, and M' the (weakly closed) set of TE A such that T = TE. We have EEM, and so the relation TE M r implies TE M; consequently, M'Cm. On the other hand, let TE M and T = WITI be its polar decomposition; we have ITI EM I hence ITI En, hence ITIE = ITI and consequently,

Proof.

TE = WITIE = WITI = T,

from which it follows that TE fi'. Thus M = M' . If F is a projection of A such that m = AF, we have E = EF, hence E F and similarly, F E, whence E - F. Finally, if M is a two-sided ideal, then M and hence E also, are invariants with respect to the unitary operators of A, and so EE A', from which it follows that E belongs to the centre of A. E

If A is a factor, every non-zero too-sided ideal of A is ultra-strongly dense in A. Proof. If m is a non-zero two-sided ideal of A, its ultraCOROLLARY 4.

strong closure is an ultra-weakly closed non-zero two-sided ideal, and the central projection E of corollary 3 is necessarily equal to I since A is a factor.

Let A be a von Neumann algebra, m a too-sided ideal of A, and ni its weak closure. For each Te 070 4", there exists an increasing filtering set Fcie such that T is the supremum of F. COROLLARY 5.

47

GLOBAL THEORY

Proof.

Let (Ti)i ci be a maximal family of non-zero operators ofesuchthat r for every finite subset J of I. The iEJ

operatorsYT.form an increasing filtering set FCM -4- whose iEJ supremum S is an element of (We- majorized by T. Let R = T - S

- + E ( M)

As AE m converges weakly to the greatest projection E of M, 1/2 R 2AR converges weakly to R ,ER 2 = R7 hence, if R / 0, we have OAR 1/2 / 0 for some AE M, hence for some A E M± such that I. But we then have A 0 R 1/2 AR 1/2

R,

R 11AR

Em,

and this contradicts the maximality of the family (Ti) iEIHence R = O. Li The method used in the proof of lemma 6 is used in an analogous way in the proof of Jacobson's algebraic density theorem. The equivalence (i 1) (i 5) of theorem 2 does not hold if A is an arbitrary linear subspace of L(H) . It is easy, in fact, to construct ultra-weakly continuous linear forms on L(H) which are not weakly continuous (when H is infinite-dimensional). References

[7], [12], [67], [73], [74].

5. Kaplansky's density theorem.

Let A, B be *-algebras of operators in H, with AC'S, and suppose that A is strongly dense in B. Let M (resp. N) be the set of hermitian elements of A (resp. B). Then, the unit ball of A (resp. M) is strongly dense in the unit ball of (resp. N). Proof. We can suppose that A and B are norm-closed, because the unit ball of A (resp. M) is norm-dense in the unit ball of the norm-closure of A (resp. M). THEOREM 3.

Let SE N. Then, S is in the weak closure of A. However, the mapping A 1.1(A + A*) is weakly continuous, leaves S fixed, and maps A into M. Hence S is in the weak closure of M and therefore in the strong closure of M since M is convex (theorem 1).

1. The function x Suppose, further, that 2x(1 + is strictly increasing from -1 to +1 in [-1, +1]. In view of the fact that B is uniformly closed, there thus exists an S / E N such that S = 2S I (I + S 12 ) -1 . Let T / be an arbitrary element of 1 for M, and put T = 2T I (I + T /2 ) -1 . Since 12x(1 + x 2 ) -1 1 IITO 1, every x, we have and

48

PART I, CHAPTER 3

(1)

1

- (T 2

- S) = (I+T

12 -1 r

) LT (I+S /2 ) - (I+T /2 )S 1 hI+S /2 ) -1

= (I+T/2)-

1(TI _sp)(i±sp2)-1

+ (I +T /2 ) -1T' (S' -T')S t (I +S' 2 ) -1 = (I+T /2 ) -1 (T / -S P ) (I +S /2 ) -1 + 1 T(S / -T I )S. 4 As T / converges strongly to S I (which is possible by the above), we see that T converges strongly to S, in view of the inequalities

(1

T /2 ) -11

1,

II T II

We thus see that the unit ball of M is strongly dense in that of

N. To complete the proof, construct the Hilbert space

R = H1 e

H2 ,

where

Hi = H 2 = H,

and represent each operator T E L(R) by a matrix (Ti where Tis E L(H) (chapter 2, section 3) . Let X (res. B) be the set of T E L(R) such that TU E A (resp. Tis E B) for i,j - 1,2. It is immediate that X and B are *-algebras of operators in R and that X is strongly dense in g. This established, let SE B,

with H s H 5_ 1.

Define g E L(R) by the matrix

0 S*

S 0

.

We see

at once that H s H < 1 and that S is hermitian. Hence there exist hermitian operators R= (Ris) of X such that H R H < 1 which converge strongly to S. Then, R21 converges strongly to 0 1. S, and H R21 H

Let H be a Hilbert space, A a c*-algebra of operators in H, At the set of elements of norm 5_ 1 in A+, and (1) a linear form on A. If (HAI is strongly continuous at 0, then ci5 is ultra-weakly continuous. COROLLARY.

Prof. Let Ah be the set of hermitian elements of A, and A l , A7 the unit balls of

A, Ah.

The mappings T 4- T+, T ÷ T- of

All into AI are strongly continuous at 0, because

m Txh H --< H Tx k

m‹H H

and il TX for every x E H. Hence Tx Ah Ai is strongly continuous at O. By linearity, c1)12t11 fl strongly continuous at 0, hence cdiq is strongly cong_nuous. Let D be a closed convex subset of C. Then 4 -1 (D) ()A7 is a strongly closed convex subset of hA. Each point of L(H) in the weak closure of this supset is in its strong 4osure [theorem 1 (iv)]. Hence (1) -1 (D) ()A7 is weakly closed in A7. Each open square D' of C has complement equal to the union ok four closed h half-planes; hence (15-1 (D / ) r),q is weakly open in A l . Hence flAl

49

GLOBAL THEORY

is weakly continuous. Since T

1 -(T + T*) and T 2

h

1 — (T - T*) 2i

A, (I) IA is weakly are weakly continuous mappings of Al into A each r 0, Let B be the weak closure of A. For each continuous. closed ball in A (resp. B) of centre 0 let A r (resp. Br ) be the and radius r. Then Ar is weakly dense in Br (theorem 3). Hence extends to a weakly continuous function 14, on Br . For r < r', tPr' extends 1P r . We thus have a function 1P on B which extends all the 1P r 's. It is immediate that 4) is a linear form on B whose restriction to B 1 is weakly continuous. Hence 1P is ultra-weakly continuous [theorem 1 (ii)].

(Or

References : [47],

Exercises.

1.

[ 67],

[367].

Let H be a complex infinite-dimensional

Hilbert space. a. Let (el, e2, ...) be an orthonormal system in H, and Un the partial isometry x (xl e n )el. Show that U71,. is the operator that Un converges strongly to zero, that does x± (xlel)en, not converge strongly to zero, and that Un does not converge to zero in norm.

WI

Deduce from a that the mapping T T* is not strongly (resp. ultra-strongly) continuous. b.

c. Deduce from b that the strong (resp. ultra-strong) topology is strictly finer than the weak (resp. ultra-weak) topology. d. Deduce from a that the norm topology is strictly finer than the ultra-strong topology [74].

2.

Let

H

be a complex infinite-dimensional Hilbert space.

a. Let W be an ultra-strong neighbourhood of zero in LU-I). Deduce from exercise ld that there exists a 2E H such that sup II T °a" (Suppose the opposite, and apply to W theorem =

TEW 2 of [[ 3]], chapter III, section 3.)

b. Deduce from a that, if L(H) x L(H) is endowed with the product ultra-strong topology and L(H) with the weak topology, then the mapping (S, T) 4- ST is not continuous. Let x, y be non-zero elements, and (xi)i c i a family of elements of H such that Let W be the set of TE L(H) such that x. 112 < iEI

Ts

be such that sup m m Choose 1.-El TEW SEW such that Sx = Xz, X / 0; then choose TEW such that TSx = I finally choose a unitary operaz tor U such that I(UTSx1y)1 > 1. We have S EW and UT E W] [ 74].

II Txill2

< 1.

Let

ZE

H

XI II T II > 1 II 11';

3. Let (e l , e 2 ...) be an orthonormal basis of the complex Hilbert space H. Let Pn be the projection onto the line Ce n . Let Tm,n = Pm + mPn . Show that zero is in the ultra-strong

50

PART I, CHAPTER 3

closure of the set M of the Tm,n , but that no sequence of elements of M converges weakly to zero. (If Tm n converges V' y Visbounde,hcm is bounded and weakly to zero, II Tmy , ny j I takes the same value an infinite number of times.) Deduce from this that L(H) is not metrisable for any one of the ultrastrong, ultra-weak, strong and weak topologies. (Since there would exist a sequence ultra-strongly dense in L(H), if L(H) were metrisable, L(H) would have a countable base) [73],[74]. 4. We adopt the notation of theorem 1. Show that, if H is separable, the Banach space M * (endowed with the norm induced by (If M is a dense sequence in H, the that of M*) is separable. linear combinations of the cox,y , where x, y E M, with rational complex coefficients, are dense in M* ,) 5. Let A be a von Neumann algebra, M and Mi two-sided ideals of A, and n the product ideal (in the usual algebraic sense). Let m, mi , n be the strong closures of m, M i , n. Show that Mflm i . [Let E, El be the greatest projections of M, M in= . There exist elements S (resp. Sl) of M (resp. Mi ) which converge strongly to E (resp. El) and remain bounded; then SSiE n converges strongly to EF, so that EFE — n, and — mn — m i cTil 1. 12]. Let H be a complex Hilbert space, A a *-algebra of operators in H, and B the weak closure of A. Suppose that I E B. Suppose also that every norm-continuous linear form 41 on A is Let lp be the ultra-strongly continultra-strongly continuous. uous extension of (15 to B. Show that cl) ip is a linear isometry of the dual A* of A onto the space B* of ultra-strongly contin(Use theorem 3.) uous linear forms on B. Deduce from this that the bidual of A is canonically isomorphic to B, and that the canonical mapping of A into its bidual may be identified with the identity mapping of A into B. (Use theorem 1.)

6. a.

Show that if we take for A the set of compact operators in H, the conditions of a are satisfied, and B = L(H). [Let 41 be a norm-continuous linear form on A. To show that 4) is ultrastrongly continuous, reduce the question to the case where OS*) - OS) for every SE A. For x, y c H, let fx, 0 be the operator 2 4- (21x)y. Show that

b.

0{x, 0) = (Aolx),

where AE L(H) depends only on (I) and is hermitian. Show that, for every orthonormal system (e l , e 2 , ...) in H, we have

< by observing that

IXO({ei,

-,

converges for every sequence

(X-2 ) of real numbers converging to zero.

Deduce from this that

51

GLOBAL THEORY

Ad) is compact and that, if (pi) is the family of eigenvalues of Ad) corresponding to the orthonormal basis (Ei) consisting of •Conclude that (15 = eigenvectors, we have E. 2 <

1

is ultra-strongly continuous.] c. In the dual L* of the Banach space L = L(H) , let L be the 11 subspace of ultra-strongly continuous linear forms, and A the annihilator of the space A of compact operators. Deduce from b that the space L* is the direct sum of the subspaces L * and 1 Show that, if 4) 1 E L * and cp 2 E A , we have

AI

H

h

4) 2

[Construct elements S, TE

H= L

H

ch H

4-

H 4) 2 H

such that S is of finite rank,

H s H < 1,

Ich(s)1 > H (1)1 H — c,

II T II

ICI) 2 (T) I

< 1#

.

>

E;

(P2

then, adding to T a suitable finite-rank operator, ensure < 1] [7], [111], [112], [113], [[13]]. further that

1 s T II

7. Let subset of

A be a A, and

von Neumann algebra in

H, M

a strongly dense

T' a closed operator in H.

a- For T'A' (chapter 1, exercise 10) to hold, it is necessary and sufficient that for every SE M, T 1 commutes with S, i.e. that T'S extends ST'. Suppose that T' is self-adjoint and > 0, and let M be its domain of definition. Then T'A' if and only if, for every S E M, M is invariant under S and S* and for x, y EAU

b.

(T'S*xlT l y) = (T'xIT'Sy). [Taking x and y in the domain of definition

N

of T' 2 , we have

(T' 2 x1Sy) =(S*x1T 12 y) = (xIST I2 y), whence Sy E

A1

and T' 2 Sy = ST' 2y.

Hence, using a, we have

•

POSITIVE LINEAR FORMS

CHAPTER 4.

Positive linear forms on a *-algebra of operators. Let A be a *-algebra of operators in H containing 'H. 1.

A linear form cP on A is said to be positive if 41(T) > 0 for every TE A+. We then have cP(T*) = OT) and cP(T*T) > 0 for every TEA. Hence

1(S*T)1 2 < (121(S*S)14)(T*T) for any

SEA,

TEA (Cauchy-Schwarz inequality).

In particular

1(T)12 < (1)(I)(T*T) < '1 )(1 ) 2 11 T *T II = 1)(i)211T 112, from which it follows that ci5 is continuous and has norm equal to 41(I). For each To E A, the linear form T 4- OT0TT6c) is positive, because T > 0 implies T0TT8 >• 0. We say that (15 is faithful if the conditions ci5(T) = 0 and TE A+ imply that T = 0. Let (1) and lp be two linear forms on A. We say that (I) majorizes q) , and we write 4 > if (I) - lp is positive. If xE H, the linear form wx,x on A is positive; it is denoted simply by co x .

Let X E H, and let ip be a positive linear form on A I T' majori2ed by wx. There exists a T F E A' such that 0 and tp LEMMA

Proof.

1.

For

S E

A 12

1(S*T)

and T

E

A, we have

(s*s)(T*T)

H sx

11 2 1 Tx 11 2.

Hence, putting ((Tx1Sx)) = OS*T), we define on the subspace Ax of H a unique form which is sesquilinear, clearly hermitian, Positive and of norm < 1. There then exists a hermitian operator T o > 0 in the space Ax = X such that 11)(S*T) = (Tx1T0Sx) and

54

PART I, CHAPTER 4

1.

11 To 11

For RE A, s

E A, TEA,

we have

( Rx1 T 0TS x ) = 0 ( T S ) *R) = IP ( S* (T *R )) = ( T *Rx IT 0 Sx) = (RxITT 0 Sx)

from which it follows that ToT = TT 0 on the space X. Hence ToPx is a hermitian operator of A' which is ?_ 0 and of norm Let T' be its square root. We have, for TE A

1.

,

4)(T) = (Tx1T 0 x) = (Tx1T /2x) = (TT' xITPx)" LEMMA 2. If wx,y is, on A, a positive linear form, there exists a zEH such that wx,y =co2 on A.

Proof. 4w

x,y

We have, for each T

(T) =2w

x,y

(T) + 2w

x,y

E A+,

(T*) = 2(Tx1y) + 2(Tylx)

= (T(x+y)lx+y) - (T(x-y)Ix-y)

(T(x+y)lx+y),

hence 4 wx,y _< wx-hy,x+y . Lemma 2 then follows from lemma 1. LEMMA 3. Let B (resp. B 1 ) be a *-algebra of operators containing I in the Hilbert space K (resp. K 1 ). Let (I) (resp. (D i )

LI

be a *-homomorphism of A onto B (resp. 13 1 ). Suppose there exists an element xE K (resp. x l E K 1 ) whlch is cyclic for B (resp. B 1 ), such that (0(T)xlx) = (4)1(T)xl1x 1 ) for every TEA. Then there exists an isomorphism of the Hilbert space K onto the Hilbert space K 1 which transforms B into B 1 and (I) into (D i . Proof.

For every TEA, we have

110(T)x11 2 =

(T)*(p(T)xix) = (4)(T*T)xix) = (cD i (r*T) x l ,xl )

= H (1) 1 (T)x 1 H 2 .

Hence there exists a linear isometry U of Bx onto Bix i such that WT)xl = U(D(T)x for every TEA. The mapping U extends to an isomorphism V of K onto K 1' For S, TEA, we have

1 (S) V(D(T )x =

1 (S)cD 1 (T)x 1 =

hence cloi(S)V = \JCS).

1 (ST)x 1 = U(D(ST)x =

0

The above lemma is a uniqueness result. The next lg,mma will be an existence result. The meaning of the adverb "canonically" in the statement of the lemma will be made precise in the course of the proof.

Every positive linear form (I) on A canonically defines a Hilbert space K, a linear mapping r of A onto a dense linear subspace of K, and a norm-decreasing homomorphism of A into L(K), such that, if we put x = F(Iti)EK, we have LEMMA 4.

55

GLOBAL THEORY

and 0(T) -= (T(T)xlx) for each TE A. Furthermore, 4)(Iti) - 'K. If 0 is faithful, (Di is an isomorphism of A onto I(A), and x is separating for (NA). Proof. For SE A and TE A, put (SIT) = 0(T*S). Then A becomes F(T) = ,T(T)x

a pre-Hilbert space. By virtue of the inequality 10(T*S)1 2 < 0(5*5)0(T*T),

A

such that 0(S*S) = 0 is also the set of SE A suchthat0(T*S) = 0 for every TE A; it is therefore a left ideal of A. The quotient space A/m is a Hausdorff pre-Hilbert space. Let K be its completion. The canonical mapping F of A onto A/m is thus a linear mapping of A onto a dense linear subspace of K. Moreover, for S E A, the operator in A of leftmultiplication by S defines, by passing to the quotient, a belinear operator g in A/m; this operator has norm cause for TE A, we have the set m of the

SE

H s H,

(STIST) = 0(T*S*ST) <

H

SS 110(T*T) =

Hs

H 2 (T 1 T)

hence g extends to a continuous linear operator (D(S) in K. We easily see that T is a homomorphism such that T(IH) IK [for example, for RE A, SEA, TE A, we have

(r(s)IT(R)r(T)) = (sIRT)

(T(R)*r(s)Ir(T))

(T*R*s) = (R*sIT) = (Tm*msdr(T)) ; hence (R)* =T(R*)].

Finally,

T(T)x

T(T)F(I)

r(T)

and ((T)xlx)

(4)(T)rmIr(i)) = (TII)

o(r).

If 0 is faithful, we have m F(T) = 0, hence TEM, hence

= 0; the condition T(T)x - 0 implies O. E T LEMMA 5. Let 0 be a norm-continuous linear form on A. If 0(i) H 0 H, then 0 is positive. Proof. We can suppose that 0(I) = 11 11 = 1. Let TE A+ , and suppose that the number 0(T) is not > O. There then exists a closed disc 1 2 - 201 < p in C which contains the spectrum of T without containing 0(T). The spectrum of the normal operator T - z o is contained in the disc 121 < p, whence H T - z o H < P. Hence

10 ( T) - zol = 10(T) -z o oid - 10(T

-

20) 1

11 ci) 11.11 T-

II

1:1,

56

PART I, CHAPTER 4

which is absurd.

0

The argument in lemma 5 is due to Phelps [The range of Tf for certain linear operators T (Proc. Amer. Math. Soc., 16, 1965, pp. 381-382)].

2.

Normal positive linear forms on a von Neumann algebra.

Let A be a von Neumann algebra. A positive linear form (I) on A is said to be normal if% for every increasing filtering family FC=A+ with supremum T E A+, cP(T) is the supremum of cp(F) . DEFINITION 1.

Every positive linear form majorized by a normal positive linear form is normal. If (1) is normal and if To E A, the positive linear form T 0 TT0* ) is normal. LEMMA 6. Let (1) be a linear form on A defined by cP= w

i=1 xi'Yi (resp. (1) =

, in which case assume that

w co

H

xi

H2

00 xil y 4 2 < +

<

co).

i=1

i=1

Let K be an n-dimensional (resp. separable, infinite-dimensional) Hilbert space. Let 4) be the ampliation T T 0 IK of A onto Aoc K" Then there existxEHoKandyEHoKsuch that ( T) ((T)xly). If yi =xi for each i, we can suppose that y = x. Proof.

Suppose, for example, that

Co

-

with

w

i=1

i=1 xi3Yi

H

x.

H2 <

i=1

Let (e.) be an orthonormal basis of K. 00

H

yi

H2 <

Put CO

x = X • 0 e.

E

H 0 K,

i=1

Y =1Ye-E 14 0 K. i=1

We have 00 M

(Tx i 0

ei

00 0

i=i

ei) =

(Txi ly i ).

57

GLOBAL THEORY

Let A be a von Neumann algebras R an element of e, and (I) and 1p two normal positive linear forms on A such that (R) < 1P(R). Then there exists a non-zero element s of e majorized by R, such that 4, (T) < 11)(T) for ever? non-zero T of A+ majorized by s. LEMMA 7.

Proof. Let (Ri )j 1 be a totally ordered family of operators let Ro be its supremum; we have

R0

Ro

E

and

R

cP(R0 ) = sup c(R)

sup 'OR)

Zorn's lemma yields the existence of a maximal operator R i c A+ R and cP(R 1 ) Ip(R 1 ). Let S such that R i R - R 1 . We have R, S / 0; and, if T E A+ is non-zero and majorized by SE At S S, we have OT) < OT), for otherwise, R i would not be maximal. El THEOREM 1. Let A be a von Neumann algebra and (I) a positive linear form on A. The following conditions are equivalent 7 (i)(I) is normal; (ii)4 is ultra-weakly continuous;

(iii) =

w X" with

i,1 /

i=1

H xi ..

H2

<

Every ultra-weakly continuous linear form on A is a linear combination of normal positive linear forms. Proof.

The implications (iii) Moreover, the equality

(1)

(ii)

(i) are immediate.

4(Tx1y) = (T(x + y) Ix +y) - (T(x - y) Ix

+

y)

i(T(x + iy)Ix + iy) - i(T(x - iy)lx - iy)

shows that every ultra-weakly continuous linear form on A is a linear combination of positive ultra-weakly continuous linear forms. We prove the implication (ii) (iii). Let cl) be a positive ultra-weakly continuous linear form on A. With the notation of lemma 6, we have

cp(T) = ( 11) (T)xly). By lemma 2, there exists zcHeKsuch that

(1)(T) = ((T)2I2); whence the existence of elements zi of

H such that

58

PART I, CHAPTER 4

CO

0T)

y (rzi k i ).

We prove the implication (i) (ii). Let (1) be a normal positive linear form on A. Let (Ri)i e , be a totally ordered family of operators of A+, such that Ri 5_ 1, and such that the linear formsT on A are ultra-strongly continuous; let R be of the Ri; we have RE A+, R 5 I; moreover, for the supremum every element T of the unit ball Al of A, we have 14)(T(R

R.))1

2

(T(R - R)T*)(R - Ri)

(I)OR - Ri);

since OR - Ri) can be made arbitrarily small, we see that OTR) is the uniform limit of (TR) over A l is therefore ultrastrongly continuous on Al and consequently ultra-strongly continuous on A (chapter 3, theorem 1). This proves the existence of a ma-Timal operator S in A+ such that S 5 I and such that T (ST) is ultra-strongly continuous. Suppose that S I. We will arrive at a contradiction, which will complete the proof of the theorem. Let S P = I - S, and let 2 be an element of H such that OS P ) < (S P 212). By lemma 7, there exists S i E Al- such that S1 5_ S P , S1 / 0, and such that OT) < (T212) for every non-zero T of A+ majorized by S i . Then, by the Cauchy-Schwarz inequality, we have, for each TE A, ,

,

I(TS1) 1 2 5 14)(I)0S 1 T*TS 1 ) 5 0I)(SiT*TS121g) =

(1)(I) II TS iz 11 2 ,

since S T*T5 1

5

11 2 2 S

„2, 5 11T 11 H s i 11S 1 .

Hence the linear form T (TS 1 ) is strongly continuous and consequently T OT(S + S i )) is ultra-strongly continuous. This contradicts the maximality of S. Instead of "normal," the term "completely additive" is also used. For an alternative definition of normal positive linear forms (which is in fact the classical definition), cf. exercise 9. Lemma 7 is the analogue of a well-known lemma of integration theory used in one of the proofs of the Lebesgue-Nikodym theorem. Let A be a von Neumann algebra in H. Let T OT) be a function on A+ possessing the following properties: 1

0

20

3

0

0 <.(T)

< + 00;

(T 1 + T2) =

(T1)

( XT)

for X ?_ 0;

= X(T)

OT2) ;

If F is an increasing filtering set in A+ with supremum A, we have OT) = sup OF). TE 4

GLOBAL THEORY

59

Problem: does there exist a family (xi )i ci in H such that (Txilxi)? cp(T) iEI

We shall see (chapter 6, corollary to proposition 2) that this is true in one special case, anyway. Let A be a complete normed *-algebra with identity such that , i.e. a normed *-algebra isomorphic to a C*x x*x = algebra of operators containing I in a complex Hilbert space. We can take for the elements > 0 in A the elements of the form x*x. Suppose that every increasing filtering family that is bounded above in A+ has a supremum. This does not imply that A is isomorphic to a von Neumann algebra, even if A is abelian. Suppose, further, that for every non-zero T of A+, there exists a normal positive linear form cti on A such that cp(T) O. Then, by [166], A is isomorphic to a von Neumann algebra.

II II 1 2

II

References : [9], [15], [17], [19], [42], [43], [111], [112], [113], [166].

3.

Normal positive linear mappings.

Let A and B be von Neumann algebras. A linear of A into 13 is said to be positive if (A) C8 [which mapping T implies that CT*) = 11(T)*]. We say that T is normal positive if, further, for every increasing filtering set FC=A+ with supremum TE A+, UF) has supremum cD(T). DEFINITION 2.

Let A and B be two von Neumann algebras, and cl) a normal positive linear mapping of A into B. Then cl) is ultraweakly continuous, hence the restriction of T to bounded subsets of A is weakly continuous. If, further, there exists a contant k > 0 such that T(Tt)T(T) s k(TT) (a condition which is always satisfied if B is abelian), then T is ultra-strongly continuous, and hence the restriction of T to bounded subsets of A is strongly continuous. THEOREM 2.

Proof.

For each normal positive linear form-cl) on B, OT is a normal positive linear form on A. Hence for each ultra-weakly continuous linear form cp l on S, (j)'0(1) is ultra-weakly continuous (theorem 1). Hence T is ultra-weakly continuous. Now, if T converges ultra-strongly to zero, T*T converges ultra-weakly to zero, and hence so does T(T*T), hence also T(T*)T(T) if

IT(T*)(1)(T) < kT(T*T); hence T(T) converges ultra-strongly to zero, which proves that is ultra-strongly continuous. Finally, if B is abelian, we shall show that, for S, TE A, we have

T(T * S)T(T*S)* < T(T*T)T(S*S);

T

60

PART I, CHAPTER 4

it is enough to show that, for every character X of B, we have

X(0(T * S))X(0(T * S) * )

X(0(T * T))X( 0 (S * S)).

Now this is just the Cauchy-Schwarz inequality, as X00 is a positive linear form on A. El COROLLARY 1. Let A and B be two von Neumann algebras, and 0 an isomorphism of A onto B. Then, 4) is bicontinuous for the to ultra-weak and ultra-strong topologies. The restriction of bounded sets is bicontinuous for the weak and strong topologies.

Proof.

Since (1) is an order-isomorphism, is positive and normal. Moreover, cl) (T) *IT (T) = (T*T) for every T E A. C1

Let A and B be von Neumann algebras, and 0 a normal homomorphism or antihomomorphism of A into B such that 0(1) = 1. Then, 0(A) is a von Neumann algebra. The algebra A may be identified with the product Al x A2 of two von Neumann algebras, 0 vanishing on A l and injective on A2. If M is a subset of A, and if C is the von Neumann algebra generated by M, (C) is the von Neumann algebra generated by 0(M). Proof. By theorem 2, 0 is ultra-weakly continuous. The COROLLARY 2.

kernel of 0 is an ultra-weakly closed two-sided ideal of A. Then (chapter 3, theorem 2, corollary 3), A may be identified with a product A l x A 2 of von Neumann algebras, 0 vanishing on Al and being injective on A 2 . We are thus led to the case is injective and consequently isometric [chapter 1, where proposition 8 (ill)]. Let, then, 1) be a von Neumann algebra contained in A. The unit ball of 0(D) , which is the image of the unit ball of -V under 0, is ultra-weakly compact. Hence (chapter 3 , theorem 2), 0(D) is a von Neumann algebra. This proves the remaining assertions of the corollary. El We saw in section 1 the close relationship that exists between positive linear forms and homomorphisms. Now let A and B be von Neumann algebras in the complex Hilbert spaces H and K, 0 a normal homomorphism of A onto B, and x E K; then the positive linear form T (0(T)xlx) on A is normal. Conversely:

Let A be a von Neumann algebra, cp a normal positive linear form on A, and 1 the canonical homomorphism defined by (1) (lemma 4). Then, 1 is normal, and 0(A) is a von Neumann algebra. PROPOSITION 1.

Let K be the Hilbert space in which 0(A) acts, and r the canonical mapping of A into K. Let F be an increasing filtering set in A+, with supremum TEA+ . Then, 0(F) is an increasing filtering set in 0(A), majorized by cio(T). Furthermore, for each SE A, (4)(T)r(s)1r(s) = fls*Ts) is the supremum of ((F)r(S)IF(S)) = (p(s*Fs) by the normality of (P. Since the F(S)'s are dense in K, 0(T) is the supremum of 0(F) in L(K).

Proof.

61

GLOBAL THEORY

Thus, II) is normal, and 0(A) is a von Neumann algebra by corollary 2 of theorem 2. A homomorphism of one von Neumann algebra onto another is not (cf. chapter 8, exercise 5). always normal.

References : [7], [15], [19], [31], [79], [89], [100], [136].

4.

Structure of normal homomorphisms.

We are already acquainted with three particular types of normal homomorphisms between von Neumann algebras: spatial isomorphisms (chapter 1, section 5), inductions (chapter 2, section 1), and ampliations (chapter 2, section 4). Having said this, the following theorem now gives the general form of normal homomorphisms, and thus, in particular, of isomorphisms:

Let A and B be two von Neumann algebras, and a normal homomorphism of A onto B. There exists an ampliation 1 of A onto a von Neumann algebra C, an induction 0 2 of C onto a von Neumann algebra D., and a spatial isomorphism 0 3 of 1) onto = 03 0 0 2 0 0 1' B such that Proof. Let H and K be the Hilbert spaces in which the algebras A and B act respectively. Suppose, to begin with, that there exists a cyclic element y E K for B. Put, for TE A, OT) = (0(T)yly). The positive linear form cp on A is normal, THEOREM 3.

hence (theorem 1 and lemma 6) there exist a Hilbert space H P , a von Neumann algebra C in H', an ampliation 01 of A onto C, and an element x of H', such that (1)(T) = (01 (T) x I x). Let H" . X. The mapping S -›- se is an induction 02 of C onto the von Neumann algebra D = C H “, and we have

(0 (T) y I y) = cp(T)

( (02 0 0 1 ) (T)xlx).

Thus, 0 2 04) 1 is a homomorphism of A onto of A onto B, XD =

H",

XB = K

and

1), 0 is

a homomorphism

(0(T)y1y) = ( (0 2 0 0 1 ) ( T)xlx).

By lemma 3, there exists an isomorphism of H" onto K which transforms 02001 into 0. This isomorphism defines a spatial isomorphism 0 3 of 1) onto B, and we have 0 =. 0 3 04)20(D i . We now pass to the general case. Consider a maximal family (yi)i ci of non-zero vectors of K such that the closed linear subspaces X. = K. are pairwise orthogonal. Because of the maximality oi the family(yi), we have K = ei ci Ki. Let 01- be the normal homomorphism T (0(T))Ki of A onto BKi• By the first part of the proof, we can define, for each 2€I, the following objects: a family (Hix ) XEIi of Uilbert spaces, copies of H, with direct sum H:, an ampliation 0 1: of A onto a von Neumann 2

62

PART I, CHAPTER 4

algebra Ci in 14,:, a close c;1 linear subspace Hy of HI invariant under C i' the induction e2 of C.2 onto D. =„. (C.)e, and a spatial isomorphism (6 of Di onto BK., such that V- = IGIT0V2" (1) (D i . Let H' be the direct sum of the for I, i.e. of the H for xEIi and iEI; the direct sum H" of the is a closed linear subspace of H'. For every TE A, let (1)1(T) bé the element of L(H 1 ) which induces (DI(T) in H! for each iEI; it is clear that (D i is an ampliation of A onto a von Neumann algebra C in H'. The subspace H" is invariant under C; let 11 2 be the induction of C onto D = CH". The isomorphisms of H3 onto Ki define an isomorphism of H" = eiE ,H7 onto K = ei ci Ki, and hence a spatial isomorphism of 0 onto B. Moreover, we have (31) = (D30(1)20(4. D

k

COROLLARY. Let A and B be two von Neumann algebras, and 11 an isomorphism of A onto B. Then there exist a von Neumann algebra C and two projections E', F' of C', with central support 1, such that one can identify A with C E P, B with CF I, and 4) with the isomorphism TE , 4- TF (T E C) .

Proof.

With the preceding notation, (1) 2 is an injective induction T TF , of C onto CF, and 11 1 may be identified with an injective induction T 4TEl of C onto CE P. D Then we can Remark. Suppose that B' is CY-finite. suppose that the ampliation of A onto C is of the form T T 0 1. is a separable Hilbert space. Indeed, I is 17 where countable since the PK. are non-zero orthogonal projections of B'. Moreover, the Ii an be assumed countable by lemma 6.

13

References : [17], [31], [89].

Application:

Isomorphisms of tensor products. PROPOSITION 2. Let Al, A2, B1, B2 be von Neumann algebras, (D i a normal homomorphism of A l onto B 1, and cD 2 a normal homomorphism of A2 onto B2. There exists a unique normal homomorphism cp of A l ® A2 onto B 1 0 B2 such that (D(T1 0 T2) = 11 1 (T 1 ) 04) 2 (T 2 ) for T 1 E Al, T2 E A 2 . If l and (1.2 are isomorphisms (resp. spatial isomorphisms) then 4) is an isomorphism (resp. a spatial isomorphism). 5.

Proof.

The uniqueness of (11 is immediate since (I) must be ultra-weakly continuous and the *-algebra generated by the operators T1 0 T 2 , where T I E Ai and T 2 E A 2 , is ultra-weakly dense in Al 0 A 2 . To prove the existence of (1), it suffices, in virtue of theorem 3, to consider the following two cases: a. (1) 1 is the identity mapping of A l , and c1 2 is the ampliation T 2 T 2 0 IK of A2 onto A 2 o cK, K being a complex Hilbert space; b. (1) 1 is the identity mapping of A l and (1)2 is the induction T 2 (T2) E l, E' being a projection of A.

GLOBAL THEORY

Case a. Let

63

(D be the ampliation of A l 0 A 2 onto

(A

A2)

1

0 C

K

= A l o (A 2 o cv) N

(associativity of the tensor product). we have 11(T i 0 T 2 ) = (T 1 0 T 2 )

Case b.

(8) I K = T1 o

For T l

E

Al

and T 2

E

A 2,

o 1K = T1 o

T2

A 1 o A 2 onto A2 ) , = A o (A 1 2 ) E ,'• IOEI

Let (I) be the induction of (A

1

0

For T I E A i and T 2 E A 2 , we have (D(T

1

, 0 T ) = (T 2 1 0 T 2) IOE'

T l o (T 2 ) E / = T l

(11 2 (T 2 ) •

Suppose that (Di and 1T 2 are isomorphisms, and let us prove that (I) is an isomorphism. It again suffices merely to consider here cases a and b. Now, in case a, IT is an ampliation, and therefore an isomorphism. In case b, E / has central support T. Let H1, H 2 be the Hilbert spaces in which Al and A 2 act, and

=

E l (H 2 ).

We have

,A 2P A

= fr2 (chapter

1, proposition 7,

CH1 0A

corollary 1), hence X

H l0H 2

= H1

o H 2' from which it follows

that I 0 E / [regarded as a projection of (A 1 0 A2 ) P] has central support I. Hence 4) is an isomorphism. Finally, it is clear that if (D1 and (D 2 are spatial isomorphisms, then (D is a spatial isomorphism. Il References : [67], [113], [128], [133].

Support of a normal positive linear form. PROPOSITION 3. Let A be a von Neumann algebra, and cp a normal positive linear form on A. Among the projections GE A such that 6.

there is one, which we will denote by all the others. We have (P(G) = 0,

(TF) =

(FT)

=o

for every T

E

F,

greater than

A.

Let M be the set of TE A such that (1)(T*T) = O. By the Cauchy-Schwarz inequality, M is also the set of TE A such that Hence M is a left ideal, ultra(p(S*T) = 0 for every S E A. weakly closed by theorem 1. The existence of F then follows from chapter 3, corollary 3 of theorem 2. Finally, for TE A, we have

Proof.

PART I, CHAPTER 4

64

111)(TF) 1 2

WFT)1 2

(1)(TT*)(1)(F) = 0,

OT*TWF) = O.

D

Let F be the projection defined by proposition F is called the support of (P. Two normal 2. The projection I positive forms on A are said to be mutually singular if their supports are disjoint. DEFINITION 3.

-

Let E cp be this support. We have (p(T) = (1)(E(1) TE1) ) for every TE A, and (I) defines a faithful normal positive linear form on AE(p . In particular, to say that (I) is faithful is the same as saying that Eq) = I. Let (I) =

u

, where

y,H x.2 i=1

112 < + 00 ;

a projection G of

A

is

i=1 x i such that (I)(G) = 01f and only if Gxi = 0 for every i; hence the support of (I) is EA 1 , where M denotes the set of the xi. In particular, (I) is faithful if and only if M is separating for A. For a von Neumann algebra to be a-finite, it is necessary and sufficient that there exists on A a faithful normal positive linear form (I); the condition is clearly sufficient; on the other hand, if A is a-finite, there exists (chapter 1, proposition 6) a separating sequence (xi) CO

for

(1) =

A;

y

i=1

we can suppose that

11 2 Y H x.2 H < + co, and then take i=1

wx,.•

PROPOSITION 4. Let A be a von Neumann algebra, Al its unit ball, cl) a normal positive linear form on A, and Eq) its support.

On Al, the convergence of 14)(T*T) to zero is equivalent to the strong convergence of TE (1) to zero.

Proof.

The form (I) canonically defines a Hilbert space K, a mapping r of A into K, and a homomorphism (I) of A onto a von Neumann algebra B = (D(A) in K (lemma 4, proposition 1). Put x = r(i). Suppose, to begin with, that (11 is an isomorphism. Let F = 4 1 , F i = I F and El the projection of A such that F 1 = (1)(E1). Then, F 1 is the greatest projection of A such that 0 = (4)(E1)xlx) = (1)(E 1 ). Hence Eq) = I - E l , 0(E(p) = F. This established, for TEA, to converge strongly to zero, it is necessary and sufficient, if TE Al, that (P(TE 41 0 ) = cP(T)F converge strongly to zero (theorem 2, corollary 13- , hence that -

H cT)Tlx H 4. 0

for every T' E B',

hence that

H (T)x

1 2 = (4)(T*T)xix) = (p(T*T)

65

GLOBAL THEORY

converge to zero. is an We now pass to the general case. The kernel of ultra-weakly closed two-sided ideal of A ; by chapter 3, corollary 3 of theorem 2, A may be identified with the product C x D of two von Neumann algebras, (I) and 0 inducing 0 on C, and the to D being an isomorphism. It then merely restriction of suffices to apply the result of the first part of the proof. D

Let A be a von Neumann algebra, (I) and 1P two Eli) their supports. normal positive linear forms on A, and E The following conditions are equivalent: PROPOSITION 5.

(i)(I)(T) = 0 and

TE

A+ imply ip (T)

0;

(ii)E -

(in) On the unit ball Al of A, the topology defined by the seminorm [p(T*T)] 1/2 is finer than that defined by the seminorm [11)(T*T)] 1/2 .

Proof.

The implication (ii) => (iii) follows from proposition (i) is immediate. Finally, suppose The implication (iii) that condition (i) is satisfied;- we have ( ( I - E(p) = 0, hence 11)(I I - Eq) , hence E = 0, hence I - E Ecp. 0

4.

Reference : [19].

Polar decomposition of a linear form. Let A be an algebra. If f is a linear form on A and if x€A, we define the linear forms x.f and f.x on A by means of the 7.

formulas

(x. f) (y) = f(yx) , for any y

E A.

( f. x)

(y) =

f (xy)

We have

x1 .(x2 .f) = (x ix 2 ).f,

(f.x 1 ) .x 2 = f. (x 1 x 2 ).

If A is a *-algebra, and if f is a linear form on A, we define the linear form f* by f*(y) = f(y*). We have (x.f)* = f*.x*. We say that f is hermitian if f = f*. If A is a normed algebra and if f is continuous, x.f and f.x are continuous and If A is a von Neumann algebra and if f is ultra-weakly continuous, x.f f.x and f* are ultra-weakly continuous.

H x.fli 114- 11f1l1 II f-x 11 11f11 -11x11

,

LEMMA 8. Let H be a Hilbert space, A a C*-algebra of operators in H containing 1, Al the unit ball of A, and a an extreme point of A l . Then T*T is a projection. = 1. Let B be the Proof. We have H T H . 1, hence H T*T

C*-algebra of operators generated by I and T*T; it is

PART I, CHAPTER 4

66

commutative. Let 0 be its spectrum, which is compact (Appendix 1). Let B' be the Banach *-algebra of continuous complexvalued functions on S-2, and cp the Gelfand isomorphism of B' onto B. Put f = (1)-1 ((T*TO) , and g = f2 . We have f 0 , g 15_ 1. Suppose there exists an co c 0 such that 0 < g(w) < 1. There then R, continuous and 0, such that exists a function h:0 sup g(1 + 7 ) 2 = sup

0,

gh

g(1

h) 2 = 1

(it suffices to take the support of h in a sufficiently small neighborhood of w, and to take h sufficiently small). Let U = (p(h) E S. We have U = U* and

Il

T*T(I + U) 2

H=H

T*T(I

H = 1.

U) 2

Then T(I + U) E A1, T(I - U) E Al. The equality T = 1/2(T(I + U) + - U)) implies, since T is extremal, that T = T(I + U) = T(I U), whence TU - 0 and gh - 0, which is absurd. Hence g can only take the values 0 and 1, from which it follows that T*T is a projection. D LEMMA 9. Let A be a von Neumann algebra, E and f an element of the predual of A.

(1) 11 f11 2 > II f-E 11 2 ± 11

f.

a projection of A

(I — E) 11 2 -

"-I-) If 11 f 11 = 11 f.E II , we have f = f. E. Proof. Assertion (ii) follows from (i).

We prove (i). Let H be the Hilbert space in which A acts. Let L* be the predual of L(H) AI the annihilator of A in L * , and E > O. As we have seen in the proof of chapter 3, theorem 1, the predual of A may be identified with L* / Al . Hence there exists gE L* , extending f, such that II g II II f II -I- E. If we show that ,

H g12

H g.E 112

H g.(1

_ E) 112 ,

we will deduce from this that ( 11f11 ± E) 2 11f- E 11 2+ 11f- (I- E)11 2, whence (i), since E > 0 is arbitrary. We are thus reduced to the case where A = L(H) . Since the weakly continuous linear forms on L(H) are norm-dense in L * , we can take f to be weakly continuous. There then exist two orthonormal systems 4) in H, and numbers (e l , ..., e n ), (el, X1 > 0, Xn 0 such that

f

= X we e + 1, 1

+X w

11

nen,enI ,

(chapter 3, proof of lemma 3). (f.E)(T) =

Jell

xn

=x 1

Hence, for each

TE

YX i (TeilEe'27 ),

(f.(I - E))(T) = P.(Te.1(I 2-

2-

E)e!),

L(H),

67

GLOBAL THEORY

whence

Il f.E 11 2

f.(i - E) 11 2 Eel:1

n = X X‘:(11.Eei. 2.,11 2 + 11 2 +2

E)e , 1 02

10 2 ± (Dti ii (I

-

E)e .i 112)

Y

X.X.(11Ee:11.11Ee:11 a 1
y

YX.+2

x.x j =

i=1li<jn

11 (1 - E)e II .11(1 - E)e :11 ) x.)2 =

-L=1

il

f 11 2 .

D

LEMMA 10. Let H be a Hilbert space, A a *-algebra of operators in H containing i, f a continuous linear form on A, (I) a positive form on A and S an element of A such that II s II < 1, 0 d»S=f3 11 (i) 11 = 11 f 11 If T E A satisfies 11 T 11 <13 f. T and 11 f• T 11 = fil then f.T = 4). Proof. We can suppose that 11 (I) 11 = 11 f 11 = 1. Apply lemma 4 to A and (1), yielding a Hilbert space K, a vector x E K, and a homomorphism 4) of A into L(K). We have

Il

(xl c(T*S*)x )

3

l

= (4)(ST)x x) =(ST) = f(T) = (f.T) (I) = II f.T II = 1.

= 1 and 11.T*S* II < 1, we deduce from this Since (xlx) = (limiting that 0(T*S*)x = x case of the Cauchy-Schwarz inequality). Hence for each T E A, (f.T)(R) = f(TR) = 4)(STR) = (cD(STR)xlx) = ((R)x1(1)(T*S*)x) = (4)(R)xlx) = 1)(R), whence

f.T = 4).

0

Let A be a von Neumann algebra and f an element of the predual of A. THEOREM

(i)

4.

There exists a pair (q), u) with the following properties:

4) is a normal positive form on A and II 1)11 = 11 f 11 b. u is a partial isometry of A whose final projection is equal to the support of (1); c. f = cp.u, (;) = f.u*. a.

68

PART I, CHAPTER 4

(ii) Let $' be a normal positive form on A, and u' a partial isometry of A whose final projection is majorized by the support of cp', with f = $' f.u'*. Then $' = $, u' = u. Proof. We can suppose that II f = 1. Let Al be the unit ball of A, and A 2 the set of TE A l such that f(T) = 1. Since Al is ultra-weakly compact and f ultra-weakly continuous, there exist T E A1 such that If(T)1 = 1; multiplying T by a suitable scalar, we see that A2 / 0. Furthermore, A2 is convex and ultra-weakly compact; hence there exists an extreme point V in A2. This point is also extreme in Al, because, if V= 1/2(S +T) with S, TE A1, we have If(S)1 < 1, If (T) 1 <— 1, and 1 = f(V) = 1/2(f(S) + f(T)) -, whence f(S) = f(T) = 1, hence S, T E A2, and S = T = V as V is extreme in A 2 . By lemma 8, VV is a projection, hence V is a partial isometry.

We have II 4) II 1. Since II f 11 -11 V 11 cp(I) = f(V) = 1, 4) is positive (lemma 5), of norm 1, and normal because ultra-weakly continuous. Let E be the support of cp. We have (P(V*V) = f(VV*V) = f(V) = 1, hence V*V E. Put U = EV*. Then UU* = E(V*V)E = E, hence U is a partial isometry with final projection E. For every TE A, we have Put $ = f.V.

( ET)

(f .u*) (T) = f (u*T) = f(VET) = and so (1) = f.U*.

Put U*U = F.

We have

= (T)

H f.F H <

1, and

(f.F)(U*) = f(U*UU*) = f(U*) = f(VE) = ( 1)(E) = 1, whence H f.F H = for every TE A, ($.U) (T)

hence f = (p.U.

1 =

whence f.F = f [lemma 9 (ii)].

Then,

OUT) = f(U*UT) = f(FT) = (f.F)(T) = f(T) We have thus proved (i).

u' have the properties of (ii). We have H uf*H < 1, 0 and II (1)? II = II f' U?* II Il çb' 11 II f 11 = 11$ 1.u' hence H f.ut* H = By lemma (applied with 10 S = U, f T = u'*), we have (1) = f.U P * = (p P . We show that U = U P . Let X = UU P *. Then Let (p P ,

f.u , *

Il I

$(x) = ($.u) (u'*) = f(u'*) = (f.u'*)

(I)

= d(I) = 1

hence (,(X*) = 1, 1 = $(X) 2 < (p(XX*) < 1, hence (1)(XX*) = 1, and - X) (E - X)*) = 1- 1- 1 + 1 = 0.

Now EXE = X since the final projections of U and U P are majorized by E. Since cp is faithful on EAE, we see that X = E.

GLOBAL THEORY

Let H be the Hilbert space in which have

H

U(U P*x) H =

H

EX

H = H x H,

A

69

acts.

hence

For

X E

E(H), we

H u(u'*x) H

H u'*x11;

hence U P E F(H); since U maps F(H) isometrically onto E(H), the equality UU I *x = x = UU*x implies that U f *x = U*x. Moreover, U* and U f * vanish on E(H) I , hence U P * = U*, U f = U. D

With the notation of theorem 4, we say that (I) is the absolute value of f and we denote it by If I. The rela tion f = IfI.0 is called the polar decomposition of f% PROPOSITION 6. We retain the notation of theorem 4. Then If*I = u*.Ifl.u, and the polar decomposition of is = ft Proof. Since In has support UU*, we have DEFINITION 4.

(1)

f*

(1fl.u)* = u*.Ifl = u*.Ifl.uu*.

It is clear that U*.IfI.0 is a normal positive form on G is a projection of A, we have (u*.Ifl.u)(G) =

IfIcuGu*) =

A.

If

o.

Since the support of UGU* = UG(UG)* is majorized by the support of the above condition is equivalent to UG = 0, hence to U*UG = O. It follows from this that

In,

(support of U*). Ifl.0 = U * U-

(2) Finally, by (1)

f*.0 = u*.Ifl.u.

(3)

The proposition follows from (1), (2), (3) and theorem 4

(ii) .

D

LEMMA 11. Let H be a Hilbert space, A a c*-aZgebra of operators in H, $ a continuous positive form on A, and S an element of A such that the form 4).s is hermitian. Then ($.S) (T) I II s llc (T) for every T E A. Proof. By hypothesis, (.S) (T*) = (4.S) (T) for every TE A, i.e. $(ST*) = $(ST) = $(T*S*). Icp(sT)1 = H(sT 1/2T1/2 )

Hence, if

osTs*OOTO = $(S 2 T) 1/2 (T) 11 .

Similarly, (4)(S 2 T1 < $(S 4T) 1/2 (T) 1/2 , and, more generally, $(52n+LT)1/20T)1/2. We thus have, working step by step, (s 2nT)

70

CHAPTER 4

PART I,

1$(8T)1 < $(S2nT) l/2(T) 1/2

110 1/271H

T

1

$ (T),1/4 .

1/211 S

-4(T)

1/2 n

II 4) (T) 1/2 + 1/4 +

. 1/2 n

+ co,

whence, letting n

l(sT)1 _< II s 11$(T)

D

THEOREM 5. Let A be a von Neumann algebra, and cp, ip normal positive forms on A such that 4) tp. There exists an element T o of A such that 0 T0 I and such that 4) (T) = p (T 0 TT () ) for each T

A.

E

Proof. Suppose, to start with, that 1P is faithful. Replacing A by an isomorphic von Neumann algebra, we can suppose that there exists a vector x, separating for A, such that i4(T) = (Txlx) for every T E A (lemma 4 and proposition 1). By lemma 1, there exists T6 E A' such that 0 .L T(') I and (4)

$(T) = (TT;x14x)

For every T'

E

A'

,

for every

TE A.

put

$' (T') = (T'xIx),

(5) (6)

fq (T') = (T l x1T(') x) = (T6*T'xix) = ($'.4*) (T').

Let (7)

f'

= If' 1.u(')

be the polar decomposition of f', so that If'l = f'.uP0 * =

(8)

By lemma 11 applied to

If' I

H T6*U6*

therefore exists To

every T'

E

A'

A',

(1,' and T (') *U(') *, we have (pl. By lemma 1 applied to A' there A such that 0 To I and such that, for ,

E

,

Ifil(TP) =(TorrxITx).

(9)

By (5), (8) and (9) (10)

(T' ILJ6T (') x) =

(T*U(') *T'xIx) =

I

fq I

(T')

= (T P T101x1T 011x) = (T'xIT0x)

GLOBAL THEORY

71

by (5), (8), (7) and (6),

(11)

(T P xIU P* U P T P x) = (T"U /* U P T P x1x) 0 0 0 0 0 0 = If =

Since x is cyclic for

(12) (13)

( Uo'rr ' ) =

(If' I

.u0') (T') =f'

(T / )

(T P xIT P x). 0

A', (10)

and (11) imply

UPTIx = 0 0

o

x,

U P* U l T l x = Tx. 0 0 0 0

Then, for every T E A, we have, by (4) , (13)

and (12) ,

(j)(T) = (TT P xIT P x)= (TU P* U P T P xIT P x) 0 0 0 0 0 0 = (TU P T P xIU P T / x) = (TT xIT x) = (T TT xlx) =4)(T TT ). 0 0 0 0 0 0 0 0 0 0 If i» is not necessarily faithful, let E be its support. By the above, there exists To E A such that 0 To 5_ I, ToE = ET0 =To , and (I) (T) = tp(T 0 TT0) for T E EAE. Since 0 _5(1)(1 -E) _5'4)(1 -E) =0, the support of cp is majorized by E. Then, for every T EA, we have

( T) = (ETE) = ip(T0 (ETE)T0 ) = 11)(To TT0 ).

D

References : [34], [214], [239], [258], [289], [301], [343].

8.

Decomposition of a hermitian form into positive and negative parts. THEOREM 6. Let A be a von Neumann algebra, and f an ultraweakly continuous hermitian linear form on A.

(1) There exists exactly one pair (q)4 1 ) of mutually singular normal positive forms on A such that f.= cl) (ii)We have IfI = (1) + (iii)If E, E P are the supports of cl), (I) P , then LI has support E + E P ; we have cp = f.E 3 l = f.E P ; moreover, f = Ifl.(E - E P ) is the polar decomposition of f% Proof. Let f = IfI.0 be the polar decomposition of f. Since f = f*, proposition 6 at once proves that U = U*. Now U*U is a -

projection, hence U 2 is a projection, hence the spectrum of U is contained in {-1, 0, 1), from which it follows that U = E l E2 where E l , E2 are two disjoint projections of A. Proposition 6 also yields

72

PART I, CHAPTER

(1)

1f 1= u.1f1. 0 = (E I

4

E 2 )-1f1.(E 1

E 2 ).

But Ifl has support U 2 = E l + E 2 , hence

(2)

1f1 = (E

+ E ) .1f1. (E + E ). 2 1 2 The equalities (1) and (2) yield, on addition, 1

131 = E 1 .1f1 .E 1 whence E

E 2 .1f1.E 2 ,

. E l' E2. Ifl 13° 1 •E2 =E2. If' •E Ifl .E1 = E1 2' .(E 1 E 2 ) fi.E 1 E 2 . Ifl .E 2 . f = 1 fl .0 = 1 fl =

and

Now E l. 1 fl E l and E2. If' .E2 are mutually singular normal positive forms on A. Now let 4), (1) / be mutually singular normal positive forms on A such that f = - cpf. Let E, E / be the supports of 4), 4) / . For a projection D of A to be such that 4 + 4) / )(D) = 0, it is necessary and sufficient that 4)(D) = 4) / (D) = 0, i.e. that DE F DE ' = O. Hence cp + 4) / has support E + E f . Moreover, E is a partial isometry with initial and final projection equal to E + E t . We have, for every TEA,

(4

+

4) / ).(E — E')) (T) = (4) + cp / ) (ET — E / T) 4)(ET) —4) / (E / T) =4)(T) — 4)'(T) =f(T),

(f.(E — E 1 )*)(T) — 4 — 4) / )(ET

E P T)

= (1)(ET) +(1) / (E / T) =4)(T) +4) / (T) =

+0 (T) .

By theorem 4 (ii), we thus have (1)1, U = E E / . The equalities f = Ifl v show that the pair (q), is uniquely determined by f. Hence cp = E1.1f1.E1, (P P = E2.1f1. E2, E = El, E / = E 2 and all the assertions of the theorem are now clear. ]

If

7. Let A be a von Neumann algebra, and 4) and V two normal positive forms on A. Then for (1) and cp' to be mutually singular, it is necessary and sufficient that PROPOSITION

cl) -

'

H (1) H

(Pf

H.

Proof. If (I) and (1) / are mutually singular, we have 1,1) — (1) / 1 = 11) + (I) / by theorem 6, hence

H cl)

(1) ?

H = H =

(I)

H = (4) +

(I)

=

v) (1)

II 4) I ± II

gb'

.

73

GLOBAL THEORY

Suppose that 11 $ - $l 11 = 11 4) 11 ± 11$ ' II. Since $ - $/ is weakly continuous on the unit ball of A, and this unit ball is 1 and weakly compact, there exists TE A such that H T H (4) - $') = II 4) 11 ± 11 4)1 11- Replacing T by 1/2(T + T*), we can suppose that T is hermitian. Then

+ $'(T)

OT - )

$'(T -4- ) =

- $')(T) =

H$H

H $'

H.

Now

H

$(T 4-)

H,

$1(T-)

- 0T-) - $

$'

1(T-4-) 5

0,

whence

(1)(T ± ) =

II (I) I

II

(4)' (T- ) =

if, '

Il.

Let E, E' be the supports of Ts" and T. We have

$(T +)

$(E)

5H

H.

Hence $(1 E) = 0, and E majorizes the support of $. Similarly, E l majorizes the support of $/. Since EE' = 0, $ and cp / are mutually singular. 0 Comparison of positive forms, initiated in sections 2, 6, 7, 8 can be developed further (exercise 8; part III, chapter 7, exercise 14). The analogy with integration theory will be apparent. References : [92], [111], [112], [113],

[188], [190], [239],

[258], [289].

Let A be a von Neumann algebra, and E a projection of A. Deduce from corollary 2 of theorem 2 a new proof of the fact that (A / ) E is a von Neumann algebra.

Exercises.

1.

Let A be a von Neumann algebra, El, E2, ..., En pairwise + En , and $ a disjoint projections of A, E = El + E2 + positive linear form on A. The positive linear form T $(ETE) is majorized by a scalar multiple of the positive linear form

2.

T4- X $(EirrEi) .

[When E = I, we have, for T

E

At

i=1

$(T1 T1 Ei)

OT) = i=1

$(T) 1/2 X [$(EiTEi)j1 1/2 , i=1

whence

OT) 5

—2 n X [$(EiTEi) ] 1/2 i=1

k X $(EiTEi)

i=1

[19].

74

PART I, CHAPTER 4

3. Let A and B be two von Neumann algebras, Au (resp. Bu ) the group of unitary operators of A (resp. B), (resp. 1 ) a faithful positive linear form on A (resp. B), and an isomorphism of the group Au onto the group Bu such that 00(U)) = OU) for every UE A. Show that has a unique extension to an isomor(Endow A and B with the pre-Hilbert strucphism of A onto B. tures defined by (1) and Then, is an isometry. Use propo1) of chapter [20], 3 sition [150], [170], [182]. 4. Let A be a *-algebra of operators, (1) a norm-continuous linear form on A. Show that (1) is a finite linear combination of norm-continuous positive linear forms on A. [Let A 0 be the set of hermitian elements of A, a real normed space. Its dual At is the set of continuous linear forms 11) on A such that We can suppose that (I)E A. The set K of positive 0 forms of norm < 1 in A* is convex and weakly compact; the set of 0 T E A o such that lip(T)I 1 for every 1P E A is the unit ball of the unit ball of hence A.I'; is the set of X 19 1 - X 2q) 2 , where Ao; X1, X2 > 0, Xi + X 2 = 1, 14, 11)2E K] [111], [188 ] . 5. Let A be a *-algebra of operators in the complex Hilbert space H, such that 1 = H, A the weak closure of A, a von Neumann algebra, (4)i)i E I the family of norm-continuous positive linear forms on A. For each iEI, let be the homomorphism of A defined by cp.2 ' and Hi the Hilbert space in which (D(A) acts. Let H be the direct sum of the H, 0 the homomorphism of A into L(H) defined by the a. Show that 0 is an isomorphism of A onto S = 0(A), and that every norm-continuous linear form on A is weakly continuous. (The (1)i 00-1 are weakly continuous by construction of (1). Then apply exercise 4.) b. There exists a canonical isomorphism of the bidual of the normed space A onto the weak closure 13 of B, which transforms the canonical mapping of A into its bidual into the mapping 0. (Use exercise 6 a of chapter 3). c. The isomorphism 0 -1 of B onto A is weakly continuous and extends to a weakly continuous homomorphism of -8- onto Tl [111], [112], [113], [188]. Let A 1 (resp. A 2 ) be a von Neumann algebra, and 0 1 (resp. 02) an ultra-weakly continuous linear form on Ai (resp. A 2 ). There exists exactly one ultra-weakly continuous linear form on A = A l 0 A 2 such that

6.

OT 1

0 T2 )

= 1 (T1)(1) 2

(T2)

for

T 1 e A l' T 2

E

A 2'

If 41_ and 0 2 are positive, so is 0; if 01 and 02 are faithful,

75

GLOBAL THEORY

se is 4).

[If 4) 1 = X w

i=1 can take 4) =

y. w xioz.

yi

and 4) 2

z

•=1

show that one

t j' j

00

iD

if (1) = X w and 4) 2 = X w 1 • z' i=1 x i j=1 is separating for A and (z) separating for (x.) 2 1 •) is separating for Al 0 A2.]

2,j=1

are faithful, A 2' hence (x•

x.

•

jiY2 j

7. Let A be a von Neumann algebra, Z its centre, 4) a normal positive linear form on A, 4) the canonical homomorphism defined by 4), M the kernel of (I), which is a strongly closed two-sided ideal of A, F the projection of Z such that m = AF, and E support of 4). Show that I - F is the central support of E. [Let F' be this central support; we have cD(I - F') = 0, hence I F' < F; moreover OF) = 0, hence FE4) = 0, hence FF' = 0.] 8. Let A be a von Neumann algebra in H, and 4) a normal positive linear form on A. For S, TE A, put (SIT) (1) = OT*S). Let A4) be the Hausdorff pre-Hilbert space obtained by passing to the quotient, A completion. Let II) be another normal positive linear form on A. a. The identity mapping of A defines, by passage to quotients, a linear mapping of Au onto A , if and only if E, < Y,

y

-

Y.

b. The identity mapping of A defines, by passage to quotients, a continuous linear mapping of Au onto A and only if 4) is majorized by a scalar multiple of tp. c. The identity mapping of A defined, by passage to quotients, a linear mapping of Alp onto A a closed extension if and only if the conditions 11)(TV n ) ÷ 0, (1)((Tn -T )*(Tn Tru )) 0 imply that (1)(T*T ) 0. We then say that 4) is a most domnated n n by -

Let 4)1, 4)2, ..., be normal positive linear forms on

d.

A,

03

almost dominated by II), and such that 00

(Pi=

X

X 4).(I) < + co. Show that i=1 on,

(1). is normal and almost dominated by II). (Show that

AA

i=1

may be identified with a subspace of the direct sum of the X .) Let 4) be a normal positive linear form on A, z an element of H, and II) = w 2 . Then 4) is almost dominated by II) if and only (chapter 1, exercise 10) if there exists a closed operator with dense domain such that 4) = z . [To show that the condition is necessary, reduce the question to the case where z is cyclic and separating for A. Then, Alp may be identified with A z (=H. The identity mapping of A, regarded as a mapping of Az into A 0 : has a least closed linear extension of the form WT', 0 in H, and where where T' is a self-adjoint operator e.

76

H

PART I, CHAPTER 4

WT t x

=

H

T t x 11 whenever x belongs to the domain M of T t .

Show that M is invariant under every operator of S, T, TIE A, we have

A.

For

(T t ST121T t T2) = OT*ST 1 ) = (T t T1zIT t S*T2); deduce from this that (1° SxIT t y) = (T t xIT 1 S*y) for any x,y E M. Then use exercise 7b of chapter 3] [19]. 9. Let A be a von Neumann algebra, and (1) a positive linear form on A. For (I) to be normal, it is necessary and sufficient that, for every family (E)ici of pairwise disjoint projections cf A, wehave 14) X Ei = XflEi). [Adapt the proof of the impli-

cation (i)

iEI iEI j (ii) of theorem 1] [15].

10. Let H be a Hilbert space, A a von Neumann algebra in H, and f an ultra-weakly continuous linear form on A. There exists an ultra-weakly continuous linear form on L(H), extending f, with H fll = g [If f > 0, there exist xi E H such that < 0.0 and such that f(T) = X(Txilxi) for each TE A. Put g(T) = X(Txilxi) for each TE L(H); whence H g H = g(I) = f(I) = r 11 • In the general case, use the polar decomposition of f.j 1.214j, [289].

H.

H xi H'

fill

11. Let A be a von Neumann algebra, (1) and (p t two mutually singular normal positive forms on A, and f = (1) (p t . We have -

sup

TE

A+ , II

T II

f(T) ,

II cp t II

=

sup

—f(T) .

T E A+ ,11 T 11.1

[If TE A+ and H T H 5_1, we have f(T) H (1) H; letting (1)(T) E denote the support of (I), we have, moreover, f(E) =(;b(E) = 110.]

CHAPTER 5.

1.

HILBERT ALGEBRAS

Definition of Hilbert algebras.

Let U be an associative algebra over the field C of complex numbers, endowed with a scalar product (xly) which makes it a Hausdorff pre-Hilbert space. Let H be the corresponding completion of U. We assume given: .TA of U onto U; we will 1° A bijective linear mapping x denote by x 4- xv the inverse of this mapping;

2° An involutery antiautemerphism x x* of U, i.e. a bijective mapping of U onto U such that (Xx + py) * =

x _- x. (xy)* = y*x*,**

+ Tly*,

(this antiautemerphism makes U a *-algebra). DEFINITION 1. We say that U is a quasi-Hilbert algebra if the following axioms are satisfied:

(i) (xIY) = (Y*Ix*) for x EU, y E (xylz) = (yix*Az) for x E U, y E U, z E (ii) (ill) For each x E U, the mapping y xy is continuous; (iv) The set of elements xy , where x EU, y EU, is total in U; (NI) If a and b are two elements of H such that (alxy) = (blx^y^) for every x E U and every y E U, there exists a sequence (x n ) in U such that x n 4- b and x a.

We say that U is a Hilbert algebra if, further, x A = x for each x E U. Axioms (i) and (iii) imply that the mapping y is continuous. Axioms (i) and (ii) imply ( 1)

(x ylz)

yx = (x*y*) *

(z*Iy *x*) = (y A z* Ix*) = (x12 yA*),

which re-establishes symmetry between left- and rightmultiplication. We further have

(xylz)

= (ylx* A 2) = (x* A * Ay12) hence (xlzyA*) =(x*A*AlzyA*); 77

PART I, CHAPTER 5

78

since y A * is an arbitrary element of U, axiom (iv) implies that = x*A*A, whence

X

(2)

x v * = x *A,

(3)

xA* = x*v.

Relation (1) and axiom (ii) can therefore also be written

(4)

(xy12) = (xlzy*v),

( 5)

(xy12) = (ylxv*z).

In the case of Hilbert algebras, axiom (v) satisfied.

is automatically

By axiom (i), the mapping X X* extends uniquely to an involution J of H, i.e., we recall, to a bijective mapping J of onto H such that J

2

= I.

(JalJb) =

(bla) ,

J(Xa +

= XJa +

H

T-1-31).

H canonically defined

The mapping J is called the involution of

by U. The mappings y xy, y yx extend to unique elements Ux and We immediately have

Vx of L(H) .

(6)

U

Xx+Ily =XU +X 1.1U ,

(7)

V

Xx+Ily

= XV

x + 1.1V ,

xy = U xU y

Ux*A = Uc;

V

xy =VV, y x

VxA* =V* ;

UV = VU; x y y X

(8) (9)

U

JUx J

Vx* ,

JV J = x

U.

The Ux (resp. Vx ) form a *-algebra of operators in H. By chapter 3, corollary 1 of theorem 2, and axiom (iv), the weak closure of this algebra is a von Neumann algebra U(U) [resp. called the von Neumann algebra left- (resp. right-) associated with U. Hence the xy, where xEU, yEU are dense in U. The algebras U(U) and V(U) commute with one another by (8), and JU(U)J = V(U), .JV(U)J = U(U) because of (9). The mappings X Ux , X ÷ Vx are called the canonical mappings of U into U(U) and V(U). The Hilbert algebras are also sometimes called unitary algebras. They constitute, as we shall see, a powerful tool in the study of von Neumann algebras. References : [1], [2], [11],

[

13]

,

[

19]

,

[27], [29], [30],

79

GLOBAL THEORY

[72], [87], [90], [95], [101], [115], [117], [127], [189], [261], [[9]]. [66],

The commutation theorem. DEFINITION 2. An element aEH is said to be left- (resp. right-) bounded if there exists a continuous operator ua (resp. Va ) of L(H) such that uax = Va (resp. Vax = Uxa) for xeU. 2.

The elements of U are left- and right- bounded and the notations Ua , Va are consistent with the previous notations Ux , Vx when aEU. Moreover, the equality Uax = Va (resp. Vax = Uxa) shows, if we let Vx (resp. Ux ) converge weakly to I, that aEUa (H) [resp. aEVa (H)]. It follows from this, in particular, that the mappings a 4- Ua , a Va are injective.

If a is left-bounded and TEV(U) 1 , then Ta is leftbounded, and TUa = UTa; the ua form a left ideal m of V(U)'. If a is right-bounded and TEU(U) ', then Ta is right-bounded, and TVa = VT a ; the va form a left ideal n of U(U)'. LEMMA 1.

Proof.

Let xEU, yEU and a be left-bounded. We have UaVxy = Ua (yx) =Vyxa = Va = VxUay,

so that U a commutes with the Vx, Ua EV(U)'.

If TEV(U) P , we have

TUax = TVxa = VxTa, and so Ta is left-bounded and UTa = TUa • applies if a is right-bounded. ]

A similar argument

Let m* (resp. n*) be the image of M (resp. mapping T T*. LEMMA 2.

Let mi =mnm*, ni = nnn*. P11 " = V(U) 1 , 1

n) under the

We have

n" = 1

Proof. By lemma 1, M IP 'cV(U) P , ecU(U) P .

We shall show that V(U) P cm7. The proof that U(U) / c4 is similar. Therefore, let TEV(U) P , TiEmi, and let us show that TT 1 = T IT. Lemma 1 implies immediately that, for xEU, x P EU, we have 4c1TUxEMi . Hence UkciTuxTi=T iqr,TUx , and it suffices to let Ux and then Ux l converge weakly to I. D

ni commute. Proof. Let U a EM 1' Vc En . We have U*a = ub, V = vd with a and left-bounded, and c and d right-bounded. For each xEU and LEMMA 3. Mi and

b each yEU, we have

80

PART I, CHAPTER 5

(alxy) = (alv x) = (vyA* alx) = (U ayA*Ix) =

(YA* 1UbX) = (yA *IVxb) = (VxA*Y A* 1b) = ((xA y A )*Ib) = (Jblx AyA).

By axiom (v), there exists a sequence (xn ) in U such that a. A similar calculation yields the existence of a sequence (yn ) in U such that c, yivi. 4- d. Then

x* b and xA

(U v xly) = (vxlu y) = (u clv b) = lim(U acc c y b

*IV x*) n y n xAlu yv) = lirri(sY*Ix*y) = lim(x A xlyy v ) = lim(v xn n n yn = (vxaluy d) = (UaxlVaiy) = (VU c a xly),

hence UaVe = VerJa .

0

THEOREM 1. U(U) P = V(U), V(U) / = U(U).

Proof.

We already know that U(U) cV(U) / .

V(U)' =

c

Moreover,

= U(U)" =

hence U(U) = V(U) / and consequently U(U) / = V(U). PROPOSITION 1.

Let U be a Habert algebra, and LP a *-

subalgebra of U. (i)U' is a Hilbert algebra. (ii)If the ux, where xEU', are strongly dense in U(U), then U' is dense in U. (iii)If U' is dense in U, we have U(U 1 ) = U(U), V(U 1 ) = V(U). Proof. (i) It is clear that U / satisfies axioms (1), (ii) and (iii). If 2E17 is orthogonal to the xy (xcU', yeU'), we have so that Ux* 2 = 0 (because (Ux *21y) = (zIos) = 0 for x, U x *2 is a limit of elements of U / ); for every tEU, we therefore have (Vt21x) = (2IXt*) = (Ux* 2It*) = 0, from which it follows that Vtz is orthogonal to U / ; now 2 is a limit of elements of the form Vtz, hence is orthogonal to U / and finally 2 = 0. (ii) Suppose the condition in (ii) is satisfied. Let a, beU. There exist ux (xEUI) which converge strongly to Ua . Hence there exist xEU' such thatxb is arbitrarily close to ab. Then, by a similar argument, there exist yEU P such that xy is arbitrarily close to xb. Hence U P is dense in the set of ab(a, bEU), hence in U. (iii) If (1 / is dense in U, U and U / have the same completion. We clearly have U(U') c u(U) V(U P ) c V(U) , hence U(W) = . Whence U(W) = U(U) , V (U l ) = V(U) . D TRU) I = D

81

GLOBAL THEORY

Throughout the remainder of the book (with the exception of exercise 5), only Hilbert algebras will appear. We have nevertheless proved theorem 1 for quasi-Hilbert algebras, because it is useful in certain applications, and because the proofs just for Hilbert algebras are no shorter. References : [13], [19], [27], [29], [30], [53], [67], [90], [98], [101], [115], [123], [255], [277].

3.

Bounded elements in Hilbert algebras.

For the rest of this chapter, we suppose that U is a Hilbert algebra. PROPOSITION 2. For an element a EH to be left-bounded, it is necessary and sufficient that it be right-bounded. The element Ja then has the same properties, and we have: U

Ja

= U* = JVJ, a a

V

Ja

v* = JU J. a a

For xEU and aEH, we have U JX a = JVx Ja • If a is rightbounded, we deduce from this that JVaJx = VxJa, hence Ja is left-bounded, and Uja = JVaJ• If Ja is left-bounded, we deduce from the same equality that Ujxa = JUjax, where Uya = JUjaJy for yEU, hence a is right-bounded and Va = JUjaJ.

Proof.

Suppose again that a is right-bounded. We have, for xEU and

yEU, (U Jaly) = (Jalx*y) = (y*xla) = (xIU a) = (xlVay) = (V * xly), x a hence UxJa = V. 4 x.

Hence Ja is right-bounded and Vja = V. a

We therefore have: a right-bounded .14. Ja left-bounded =."Ja right-bounded a left-bounded. Interchanging the roles of a and Ja, we see that all these conditions are equivalent. The formula Vja = VI = JUaJ follows from the above. We deduce from it that U

Ja

= JVJ = JV* J U*. a Ja a

D

For every element C of the common centre of U(U) and V(U), we have JCJ = C*. COROLLARY.

Proof.

For xEU, we have, by lemma 1, Cx* =

CUx* = (c*u )* = u* C*x'

hence, by proposition 2, CJx

= U

CJx = JC*x,

CJ = JC*.

D

82

PART I, CHAPTER 5

DEFINITION 3. An element aE H having the properties of proposition 2 is said to be bounded relative to U.

Remark.

Let ad-I. For x, yEU, we have (xyla) = (xlvu lca) . Hence, to say that a is bounded amounts to the same thing as saying that the bilinear form (x, y) 4- (xyla) is jointly continuous in the two variables x, y. We then see that it comes to the same thing to say that a is bounded relative to U, or relative to every Hilbert subalgebra of U which is dense in U. PROPOSITION 3. The Ua (resp. Va ), for a bounded, form a twosided ideal of U(U) [resp. V(U)]. We have, for TEU(11)3 TU

= U

Ta5

UT = U JT*Ja' a

for T 1 E V(U)3

T t V =V , T a a

VaT P = V JTI*Ja .

Proof. The formulas TUa = UTa , T r Va = VTla have been established in lemma 1. Moreover, UT = (T*U*)* = (T*U )* = vic = U T*Ja JT*Ja' a a Ja V T I = (T I *V*)* = (T i *V )* = V*, =V, . T *Ja JT *Ja a Ja

Da

PROPOSITION 4. For an element a E H to be bounded, it is necessary and sufficient that there exist a sequence (xn ) of elements ofUsuch that Hxn -aH 4-0and sup Huxn II < + Then, u converges strongly to Ua . xn Proof. Suppose there exists a sequence (xn ) of elements of U such that Hx - a H 0 and sup IlUxn H= m < + II Then, for n every yEU, we have

H vyx,,I II=Huxny

milyH,

hence

mHy

H.

Hence a is bounded. Furthermore, Uxny = Vyxn converges strongly to Va = Ua y, hence Uxn converges strongly to Ua . Conversely, suppose that a is bounded, and let E > O. There E, H vx H 1 (chapter 3, exists an xEU such that H Va - a H theorem 3). Then there exists a yEU such that E, Hu H< H u a H (chapter 3, theorem 3). II Y Since Uax = Va , we see that H yx - a H 5 2E. Furthermore, H Uyx

H

H yy H.H Us

= H uy H. H vs H

H ua H.

Whence the existence of a sequence (xn ) with the properties of the proposition. 0

83

GLOBAL THEORY

DEFINITION 4.

We say that U is full if every bounded element

of H belongs to U. Let U be a Hilbert algebra. Let BU be the linear space of bounded elements of H. As a subspace of H, B is endowed with a pre-Hilbert structure. For aEB and bES, we have uab = vba; indeed, let (xn ) be a sequence of elements of U converging strongly to b; for each yEU, we have

(uaxn ly) = (Vxn aly) =

(alyx)

= (aNyx)1

hence, in the limit,

(92 bly) = (aluy ib) = (alVjby) = (Vbcity), Put, for aEB and bEB, ab =uab =vba. which proves our assertion. We thus define on B a multiplication which extends that of U and which makes B an associative algebra, because, for aEB, bEB, cEB, we have

a(bc) = u v b = vc u b = (ab)c. a c a Finally, for aEB, put a* = Ja. We have a*b* = U Ja Jb = JVa b = J(ba) = (ba)*; hence B becomes a *-algebra and U is a *-subalgebra of B. We easily check that B is a Hilbert algebra, called the Hilbert algebra of bounded elements. An element of H, bounded relative to B, is in B. Hence the Hilbert algebra of bounded elements

is full. By proposition 1 (iii), we have

U(U) = U(B),

V(U) = V(B).

The above work enables one, almost always, to reduce problems relating to Hilbert algebras to problems relating to full Hilbert algebras.

Instead of "full," we sometimes say "maximal." References : [13], [27], [29], [30], [90], [101], [115].

4.

Central elements in Hilbert algebras.

PROPOSITION 5. For an element aE H, the following conditions are equivalent: (i) (ax) = (alyx) for xEU, 1/EU; (ii) uxa = Va for xELI; (iii)Ta = Yr*Ja for TEU(U) [hence for TEV(U)].

84

PART I, CHAPTER 5

Proof.

For x, yEU, we have

(alxy) = (ux* aly),

(alyx) = (Vx*aly),

whence the equivalence of conditions (i) and (ii). Moreover, Vx = JU*J, so condition (iii) implies condition (ii). Finally, condition (ii) implies condition (iii) since every TO(U) is a weak limit of operators U. D

If aEH satisfies the conditions of proposition 5, we say that a is central relative to U. DEFINITION 5.

If aEU, we recover the usual algebraic notion, by condition (ii) of proposition 5. The set Z of central elements of H is a closed linear subspace of H, which is unaltered if we replace U by a *-subalgebra of U which is dense in U [since U(U) is unaltered]; in particular, the central elements are the same relative to U and relative to the Hilbert algebra of bounded elements. PROPOSITION 6.

Let Z = U(U) n V(U) be the common centre of

U(U) and V(U): (i)J(Z) = Z; (ii)p zE 7 1 ; U) = E V MEZ: (iii)For aEZ ' EU( a a (iv)EU(U) = EV ( U) EZ. Proof. If a E z and TEU(U), we have T(Ja ) = J(JTJa) = JT*a = (JT * J)(Ja), whence (1).

hence JaEZi

If Ra, we have

T(Ra) = R(Ta) = R(JT*Ja) = (JT*J)(Ra), whence (ii). implies

hence RaEZ,

The equality E U(U) = E v(U) is immediate, and

a

E

UM)

a

E

a

U(U) n V(U) = Z

Finally, (iii) implies (iv).

•

E

DEFINITION 6.

The projection 4 (11) = E V(U) of proposition 6 is called the characteristic projection o U. In particular, if U has an identity element, the characteristic projection of U is I.

For a bounded element a of H to be central, it is necessary and sufficient that u a EZ, or that va EZ• We then have ua . va . PROPOSITION 7.

85

GLOBAL THEORY

Proof.

If the bounded element a is central, we have, for

Ux = Va = Ua = V x, a x x a from which it follows that Ua EZ. we have, for xEU,

Uax

u Uxa ,

= U U = U U a x x a

and consequently a is central.

U

hence

a

xEU,

V ,

a

on the other hand, if Uo E Z,

U a = U x x a

hence

vx a,

We argue similarly if Va EZ.

Important related material is given in exercise 4. References : [13], [29], [90].

5.

Elementary operations on Hilbert algebras.

Let U be a Hilbert algebra. Without changing either the preHilbert structure or the involution of U, replace the multiplication (x, y) xy by the multiplication (x, y) yx. We check at once that we thus obtain a Hilbert algebra U P which is called the reversed Hilbert algebra to U. We have

U(U P ) = V(U),

V(U P ) = U(U).

Let (Ui)i E , be a family of Hilbert algebras. Let Hi be the completion of Ili. Let U be the direct sum of the Ui: an element of U is a family (xi)i ci , where xiEUi, and where all but a finitenumberofthe.are zero. Define an algebra structure and a pre-Hilbert structure on U in the usual way, and put We check immediately that U is then a Hilbert (x.)* = (xt). 2 2 algebra, called the direct sum Hilbert algebra of the U. The completion U of U is the direct sum of the Hi. We have

U(U) = H U(Ui),

V(U) = H V(Ui);

iE I

iEI

in fact, it is clear that, for every

U

x (E. II U(1-1.) ,

xEU,

U(U) c H U(U.);

hence

'1-El

similarly V(U) c H V(U.) and consequently iEI

U(U) =

D

=

whence our assertion. U(U) n V(U) , and

If E. = P

H U(U); iEI

iEI

Hi'

the E. are projections of

86

PART I, CHAPTER 5

U(U.) = (U(U)) E. , 2 2

V(U.) = (V(U)) E- . 2 2

Conversely, let U be a Hilbert algebra, and let (Ei) icI be a family of pairwise disjoint projections of u(u) n V(U), with sum I. Let Hi = Ei (H), Ui = U n Hi. Since Hi is invariant under U(U) and V(U), we have Oicili, Liiiicui. IfxEUi, we have x*EUi, because E.2 Jx = JE.x 2 = Jx (corollary of proposition 2). Suppose is full. For zEU, Eiz is bounded, hence EizEU; it folU that lows from this that U. = E.(U) is dense in H 2.. Thus, Lii pos2 2 sesses a Hilbert algebra structure, its completion being Ho . It is immediate that the Ui are full. We at once have U(U)

c

(U(11)) E

V(U) c (V(U)) Ei ,

and therefore

U(U) = (U(U)) Ei

V(U) = (V(U)),.•

The direct sum Hilbert algebra of the U is a dense *-subalgebra of U. PROPOSITION 8.

Let U be a Hilbert algebra, and H the comple-

tion of U.

(i) Let I be a two-sided ideal of U, and

its closure in H.

EU(U) n V(U). (iif Let E be a projection of U(U) n V(U). Then K = E(H) is a two-sided *-ideal of U, which is dense in E(H) if U is

Then

(iii) Let E l be another projection of U(U) n V(U), and K' = E' (H) n U. Suppose that U is All. Then KK i = 0 i f and only if E and E t are disjoint.

Proof. (i) I, hence I also, are closed with respect to left and right multiplication by elements of U; hence P—€ U(U)

n V(U)

U(U) n V(U)

(ii) Assertion (ii) has been established previously. (iii) If EE / = 0, we have KK'c KnK' = O. Conversely, suppose that KR' = O. Let xEK, yEK I ; we have x*EK, hence, for every zEU, (2x1y) = (Iyx*) = 0; it follows from this that (xly) = 0; consequently, K and K t , and thus E and E', are disD joint.

Remark. Putting, for X, .J.EC, (X41) =Xj, X* = X, we clearly endow C with a 1-dimensional Hilbert algebra structure. Constructing direct sums of such algebras, we see that there exist for every Hilbert space H, Hilbert algebras dense in H. Let U1 , U2 be Hilbert algebras. a *-algebra when we put

The algebra U1 0 U 2 becomes

GLOBAL THEORY

.\*

In

v i L x1 0 '2=1

2/

n

87

x 1- * .

. 1 2-1

We know besides that there exists on 11 1 0 112 a unique preHilbert structure such that (x1 0 x 2 1y 1 0 y 2 ) - (x1 1y 1 )(x2 Iy 2 ). We immediately check that U = 11 1 0 U2 is then a Hilbert algebra, called the tensor product Hilbert algebra of the Hilbert algebras U 1 an d U 9 . Let H, H1 , H 2 be the completions of U0U 2, UU 2 . We have H = H i 0 H 2 . 1, 1 PROPOSITION 9.

(i)ucul 0 112 )

u(u1 )

0 U(U 2 ) ; v(u i 0 U2 ) = V(U 1 ) 0 v(U 2 ) ;

(ii)If the characteristic projections of 111 and U2 are equal to 1, the characteristic projection of U is equal to I. Proof. (

For X 1 , ylE 11 1 and x 2 , y 2 E 11 2 , we have

UxiOU x2)(Y1

Y2) ' ( Ux1Y1 )

(Ux2Y2 )

= x ly i 0 x 2 y 2 = (x 1 0 X 2 ) (y 1 0 y 2 )

= u, 1 - 2(Y1

Y 2),

-

hence Uxi 0 Ux2 = U x10x2 . Since U(U 1 0 U 2 ) is generated by the operators Ux100c2 , we see that U(U

1 0

U 2 ) CU(U 1 ) 0 U(U 2 ).

Similarly, V(U

1

q) U

2

) c

V(U

1

) (0

V(U ) . 2

Hence U(U

1

0 U ) = V(U 0 U )' 2 1 2

V (U )

1

el V(U 2 )

U(U ) 1

O

U(U ) . 2

Whence (i). Now let a l be a central element of H 1 , a2 a central element of H 2 . For x l , y i E Up x2, y 2 E U 2 , we have

(a l 0 a 2 I (x 1 0 x 2 ) (y 1 0 y 2 )) = (a l 0 a 2 lx 1y 1 0 x 2y 2 ) =

(a l loy ])

=

(a

1

0

(a 2 1x 2y 2 ) =

a 1 (y 0 y ) (x 0 x )), 1 2 2 1 2

(a 2 1y 2x 2 )

88

PART I, CHAPTER 5

hence a l 0 a2 is a central element of H. This established, if the characteristic projections cf U 1 and U 2 are equal to I, the elements Tial (resp. T2a2), where T l E U(U1) [resp. T2E U(112)] and where a l (resp. a 2 ) is a central element of H1 (resp. H 2 ) , form a total set in H1 (resp. H2). Hence the elements T ia l 0 T2a2 = (T1 0 T2) (a1 0 a2) form a total set in H. Since T1 0 T2E U(U1 0 U2), we see that the characteristic projection of U is equal to I. 0

Let H be a complex Hilbert space, and A a von Neumann algebra in H. We say that A is standard if there exists a Hilbert algebra U dense in H such that A = U(U). DEFINITION 7.

The results of this section then imply the following proposition: PROPOSITION 10.

(i)If A is standard, A' is standard; (ii)If A and B are standard, A o B is standard; (iii)Let A = H Ai; for A to be standard, it is necessary iEI and sufficient that the Ai be standard. Later on (chapter 6; and part III, chapter 1), we will find out how to characterise completely standard von Neumann algebras. For another "elementary operation,"

cf.

exercise 3.

References : [90], [101], [116].

Exercises. 1. Let U be a Hilbert algebra. If Z = U(U) n V(U) is G-finite, there exists a bounded element which is separating for Z. [Let (xi)i ci be a maximal family of non-zero bounded elements such that theErEZx iP arepairwise disjoint. Show that

X E. = I, that I is countable, that we can suppose 2 iEI

H xi 11 2

< + co and supi 11 u .11 < + on, and that x =

icI then a bounded element, separating for Z.]

xi is iEI

2. Let U 1 and U2 be two Hilbert algebras, al (resp. a 2 ) an element bounded relative to U 1 (resp. U2). Then, a l 0 a2 is bounded relative to U 1 0 U 2 , and

a l0a2

=U

al 0

U

a2 1

V

al

oa

2

=V

al

0

a,

.

3. Let U be a full Hilbert algebra, H its completion, and J the involution of H canonically defined by U.

89

GLOBAL THEORY

For TE U(U) [resp. T P E V (U ) (U)] is a T(U) [resp. a. right (resp. left) ideal of U, and T(U) n JTJ(U) is a *-subalgebra of U. b.

Taking for T a projection E of U(U), and putting E l = JEJ, E(U) n E P (U) = EE' (U)

is a full Hilbert algebra U

1,

dense in

E(H) n E I (H) = EE'

(H) ,

and

U(U 1 ) = 4.

(U U) )

E EE

V(U ) = (V( U) ) ) p 1 E EE

Let U be a Hilbert algebra, H

[116].

its completion,

J the in-

volution of H canonically defined by U, and H the group of automorphisms of U. Let G be the group of unitary operators of U(U) . When U a. runs through G, UJUJ runs through a group G1 CH of unitary operators of L(H) , and U UJUJ is a homomorphism of the (nontopological) group G onto G1 . The kernel of this homomorphism is the group of unitary operators of the centre of U(U) . For an element a of H to be central, it is necessary and sufficient that UJUJa = a for every U E G. b.

c. For each element a of H, let Ka be the convex set generated by the UJUJa (U€ G) , and Kai the closure of Ka in H. Show that K a1 is invariant under G1, hence that the point b of Ka' closest to zero is invariant under G1, hence that b E Z (the set of central elements) . Let, moreover, c = P a; show that K al + z , hence that b = e.

d- Let a be a bounded element of H, (U 1 , U2, ..., Un ) elements of G, (X1, X2, ..., Xn ) real numbers 0 such that

X xi = 1, b X XjUiJUiJa. Show that ub = XjUjuu j=1 i=1 i=1 Deduce from this that II Vxo II < II U II for every x E U and every c E K a , hence for every e e K. Conclude from this that all elements of IC, and in particular P are bounded. Za' ae. Show that the set of bounded central elements is dense in Z. [29], [106]. Let G be a -locally compact group, and U the algebra of 5. continuous complex-valued functions on G of compact support (multiplication being given by convolution of functions). Let dg be left-invariant Haar measure on G. Let A(g) be the modular function of G, such that d(g 1 ) = A(g) -1 dg. For XEU, put

PART I, CHAPTER 5

90

x*(g) = x(g 1 )

and

xA(g) = A(g) - x(g).

For x, y E U, put

(xly) = fx(g)y(g)A(g) -1/2dg. Show that U is a quasi-Hilbert algebra. [To prove axiom (iii), show that, for x, y, 2 E U, we have a.

(xy I z)I

II y 11. 11 z Hflx(g) IA(g) -Aidg ,

hence

H xy H

H y HfIx(g) tA(g) -1/4dg.

To prove axiom (r), show that, if a and b are square-integrable functions on G for the measure A(g)dg, the equality (alxy) (blxAyA) for x, y E U implies equality of the measures a(g)dg and A(g) -12b(g)dg.] b.

For g

E

G and x

E

H (the completion of U) , put

(Ugx)(g') = x(g -lg')A(g) 14 ,

(Vgx)(g l )

x(g 1 g)A(g).

Show that the U (resp. Va ) are unitary operators in H which g V01. generate u(U) rLresPWe will return to this example in part III, chapter 7, section

6 References : [1], [2], [13], [27], [29], [53], [56], [67],

[ 90 ] , 6. a E H, uax =

[9 8 ], [99], [[9]]Let U be a Hilbert algebra and H its completion. For define linear mappings u , v of U into H by the formulas a a Va, V ax = 0xa.

The operators u*, v* extend uj a , v aa respectively. Hence a a admit least closed extensions U al = u**, Ni z - va** . Morev 'ta ' a a over, we put a.

u a = u* Ja

va

Ja

The operators U a , Va are closed, extend U' a' VaP respectively, and U' = u* ,

u a = W .* , Ja

via

v* , Ja -

va = v i* . Ja b. I-J -11U(11), UaTIU(U), VTIV(U), VaTIV(U) (chapter 1, exercise 10). (Show that uaVxy = VxUay for x, y E U; deduce from this that U al Vai) = Vx *for X E U and b belonging to the domain of 14.) a

Ja -

-

91

GLOBAL THEORY

c. If y E H, there exists an increasing family of projections + 00, such (E(r)) r> 0 of V(U), converging strongly to I as r that E(r)y is bounded for each r. [Let U = WT be the polar O. Take for E(r) decomposition of L.Jy, with T self-adjoint the spectral projections of T.] d. Let B be the set of bounded elements. Then B is contained in the domain of U. Let y be in the domain of U a . Let (E(r)) have the properties of c. We have U f E(r)y = E(r)Uay.

e.

We have U a = u ar ' Va = V. a

(Use d.)

[90], [99].

=

CHAPTER

6.

TRACES

Definition of traces.

1.

DEFINITION 1. Let A be a von Neumann algebra. A trace on A+ is a function cp defined on A+, taking non-negative, possibly infinite, real values, possessing the following properties:

(i)

If

S E A+

and T

E

A+ , we have (j)(s + T) = (p(s) +

(ii) If S E A + and if X is a non-negative real number, we have d(XS) = 4(s) (with the convention that O. + co = 0); (iii) If s E A+ and if u is a unitary operator of A, then cp(uSu -1 ) = Os). We say that (p is faithful if the conditions S E A+ , Os) = 0, imply that s = O. We say that (1) is finite if OS)

< + 00

for each

SE

At

We say that (I) is semi-finite-if, for each SEA +, OS) is the supremum of the numbers (1)(T) for those T E Al- such that T < S and (p(T) < + 00.

We say that (i) is normal if, for each increasing filtering set F c A+ with supremum S E A+ , q(S) is the supremum of (1)(F) .

Let A be a von Neumann algebra, and cp a trace on At The set of the T € A+ such that (1)(T) < + co is the positive part of a two-sided ideal rn of A. There exists exactly one linear form -(1) on m coinciding with cp on ni+, and we have ep (ST) =ep (TS) for s E ni T E A. Finally, let s E ni; if ci) is normal, the linear form T ÷ .(1)(ST) on A is ultra-weakly continuous. PROPOSITION 1.

,

Proof.

The set of TE A+ such that OT) < + 03 satisfies the conditions of proposition 10 of chapter 1, and is therefore the positive part of a unique two-sided ideal M of A. Every element of M is a linear combination of elements of MI- (chapter 1, proposition 9), and the properties of cp immediately imply that there exists a unique linear form "(;b on m coinciding with cp on Mt . If SEM and if U is a unitary operator of A, we have ip(uSu-1 ) = .0S), by the properties of (I) if SEMI- , and by linearity in the general case; hence CP (SU) =11)(USUU-1 )

93

94

PART I, CHAPTER

6

thanks to proposition 3 of chapter 1, we deduce from this that . (ST) =14)(TS) for SE M and TE A. Finally, let SE M, and put 11)(T) = ',4)(ST) for TE A. We show that 11) is ultra-weakly continuous if (I) is normal. We can confine attention to the case where SE M-4- 1 and we shall show that .(I) is then positive and normal; this, together with theorem 1 of chapter 4, will effect the proof. Now, if TE At we have

11)(T) = cp(ST'T') = (I)(T'ST') > 0

(since T'ST > 0),

and so ip is positive. Now let F c A+ be an increasing filtering set with supremumi TE A+. Then, i S 1/2 FS 1/2 is increasing filtering, is mOorized by SiTS 1/2 , and S 1TS 2 is in the strong closure of t S 1/2 FS'2 ; hence STS 2 is the supremum of S 1/2 FS. Observe that S E m, and that, for RE A, S 1/2 (S 1/2 )*

(RS) (RS)* = RSR*

E M.

Then (chapter 1, proposition 11), we have 2

FS 2 C

S 1/2 TS 1/2

EM,

and 11)(T) = cp(S 1/2 S 11T) = O STS) is the supremum of

(p(sF's 1/2 ) = by normality of (I).

L_ 1

cs 2 s

=

(F)

0

By an abuse of language, the name trace is sometimes given to the linear form ep on m. If cp is finite, .(1) is a positive linear form on A, and the results of chapter 4 are applicable. However, it would not be adequate for what follows to consider only finite traces.

Let (I) be a function taking non-negative, possibly infinite, values, defined on A+, such that O. OS + T) = ,4)(S) + cp(T), (1) (XS) = X(S) for SE A, TE A+, X Then for cp to be a trace, it is necessary and sufficient that (I) ( R * R) = d( RR*) for every R E A. COROLLARY 1.

Proof. Suppose, to begin with, that cp is a trace. Let RE and R = UIRI be its polar decomposition. We have 1 2 2 and RR* - UIRI U*= I R IR* = U*U I R 12

A,

If m denotes the two-sided ideal of definition of el), the condition R*R E M implies that RR*E M and the two conditions are therefore equivalent; furthermore, if it is full, we have (RR *)

= "OUIRI 2U*) = "OU*UIR1 2 ) = OR*R).

95

GLOBAL THEORY

We therefore have (P (RR*) = (P(R*R) for each RE A. On the other hand, suppose that (p (RR*) = (P(R*R) for each RE A. Let S E A+, and let V be a unitary operator of A. We have

(1) (VSV 1 ) = (I) ( hence cp is a trace.

( VS 2 )

(

VS 2 ) * ) = (1)«VS-1)*(VS 2)) =

D

2. Let cp be a normal trace on canonically identified with the product of algebras Al, A2, A3 such that cp induces on ,races (pi, (1)2, (P3 possessing the following 1 0 h is faithful and semi-finite; 2 ° (1) 2 for every non-zero S of COROLLARY

A+ .

Then, A may be three von Neumann AI, A Ali normal properties: ,

= 0; 3 ° (1) 3 (S) = +

By proposition 10 of chapter 1, the set of the TE A+ such that (p(T) = 0 is the positive part of a two-sided ideal n of A, clearly contained in the ideal of definition M of .(1). Let E (resp. F) be the greatest projection in the strong closure of m (resp. n). We have E F, and E, F are projections of the centre of A (chapter 3, theorem 2, corollary 3). Put

Proof.

A1 =A E-F'

A2

=

A F,-

A3

=

A I-E .

Then A may be identified with the product of A l , A2, A 3 and the follow at once from chapter 3, corolproperties of (pi, lary 5 of theorem 2, and the normality of (p. D This corollary reduces the study of normal traces to that of normal traces which are also faithful and semi-finite. The projection I - F is called the support of (1). When (1) is finite, the support of 4) is the same as the support of .(1) in the sense of definition 3 of chapter 4.

Let (I) be a normal trace on Al". For (1) to be semifinite, it is necessary and sufficient that every non-zero element of A+ majorize a non-zero element T of Al- such that ( T) < + COROLLARY 3.

Proof. The condition is obviously necessary. Moreover, in the notation of corollary 2, it implies that A 3 = 0, and hence that (1) is semi-finite. 0 PROPOSITION 2. Let A be a von Neumann algebra, 0 a normal trace on A+, and (Ti)i ci a family of elements of A- such that X T. = I (in the sense of the weak topology). For each TE

tE l we have 1

( T) = X

tEl

L2

(T.TT.). 2 •2-

96

PART I, CHAPTER 6

Proof.

For every subset

J

of I, let T = X T.. Let F be the j iEJ

set of finite subsets of I. If T

E ?71+ ,

we have, for J

(TT) =

Let M be the ideal of definition of E

F,

.(I) (TT.) = 2

iEJ

iEJ

2 2

moreover (appendix II) I is the ultra-strong limit of the Tj , hence (proposition 1) d(T) is the limit of the .OTTJ ); hence (I) (T1TT1.1 ). We now pass to the general case. Thanks to ( T) = iEI

corollary 2 of proposition 1, it suffices to study the case where (I) is semi-finite. Let G be an increasing filtering set in m+, with supremum T (chapter 3, theorem 2, corollary 5). We have 1,

( T) = sup OS) = sup( SEG

=

SEGEI

2

1-

(T :ST :) ) = sup

sup SEG,JEF( iEJ

X 11) (T 1!TT I!) ) =

JEF (iEJ

(1) (T l!TT I!) .0 iEI

2 2

COROLLARY. Let A be a von Neumann algebra in H, and cp a normal trace on A+ . There exists a family (xi)iE l of vectors of H such that (I) = wx . on A+ .

y

iEI

By corollary 2 of proposition 1, it is enough to study 1° OT) = + for every non-zero T of the following two cases: A+ ; 2 0 (1) is semi-finite. In the first case, we can take for the family (xi) 1-E' the family of all the vectors of H. Let us study the second case. In proposition 2, we can suppose that T. E ni for every i (chapter 3, theorem 2, corollary 5). Then, 'OTTO .(1)(TIITT) is normal and positive each linear form T on A, and it is enough to apply theorem 1 of chapter 4.

Proof.

In certain treatises, the term "trace" refers to normal or faithful traces. Instead of "semi-finite," one sometimes says "essential." Instead of "trace," the term "pseudo-trace" is sometimes used. Let A be a von Neumann algebra, m a two-sided ideal of A, and (1) a positive linear form on M such that OST) = d(TS) for S E M In certain cases, we may then extend the restriction and TE A. of (I) to to a normal trace on A+ (cf. part III, chapter 1, exercise 11).

e

On an abelian von Neumann algebra, it is easily seen that in general there exist non-normal finite traces. In the case of

97

GLOBAL THEORY

factors, cf. exercise 6, and part III, chapter 2, corollary of proposition 15. References :

[12], [14], [19], [65], [66], [70], [78],[80], [89], [101], [117].

Traces and Hilbert algebras.

2.

Let U be a Hilbert algebra, and H the (Hilbert space) completion of U. For SE U(U) + Eresp. sE V(U) 41, put: THEOREM 1.

(S) = (ala)

if s = ua (resp. s 1/2 = Va ) for some bounded a

E

f-f;

d(S) = + co otherwise.

Then, (1) is a faithful, semi-finite normal trace on U(U)+ Eresp. V(U) -11. The two-sided ideal of the TE U(U) [resp. V(U)] which can be expressed in the form u, (resp. v,) for some bounded a, coincides with the two-sided ideal Of the TE U(U) [resp. V(U)] such that cp(T*T) < + m. If a and b are bounded elements, then

Eresp. "cp(Tra ) = (alb)]

"(1)(qua) ' (alb)

We will only consider the case of U(U). It is clear that cp is a positive-real-valued function on U(U) + , possessing property (ii) of definition 1. Moreover, for TE U(U) to be of the form Ua for some bounded a, it is necessary and sufficient that IT] be of the form ub for some bounded b [since the U form a two-sided ideal of U(U)], and hence that a

Proof.

1)( IT1 2)

4)(T * T )

(

Let

prove property We prove (i) of definition 1. U (U ) +,

RE

SE

U (U ) + ,

T

R + S.

We have R = AT 1/2 , S = BT 1/2 , A and B being elements of U(U) which vanish on T(H) 1 (chapter 1, lemma 2). Put C = A*A + B*B; we have 1

* 2 T=R + S I1 S = T 1/2A * AT 1/2 + T1/2B BT1/2 = 1

and hence (CT 1xIT 2X) = (7 1 x1T 1x) for every xE H, hence 1 THTI , we see (Cu lu) = (u lu) for every U ETTRT; as C vanishes on 7 that C is the projection on T(H) , whence r1 2- = CT Ii. This established, if T 1/2 = Ua for some bounded a, we have = AU a

and, moreover, a

OR)

E

=

U , Aa

S 1/2 = BUa = U

Ba

T(H) , so that Ca = a; then

+ d(S) = (AalAa) + (BalBa) = (cala) =

(ala) = OT).

98

PART I, CHAPTER 6

If, on the other hand, q(T) = + 00, we have OR) = + 00 or OS) = + 00 because, if we had R1/2 = ub and S I/ = Uc with b and bounded, we would have `I‘ - =A*AT1/2 +13*ET1/2 =

hence

+ E*5 1/2 =

(1)(T) < + co.

Hence OR + S) = OR) + q(S) in every case. Now let T be any element of U(U) and let us show that (TT*) = OT*T), which will establish that cp is a trace on Let T = W T be the polar decomposition of T. Suppose U(U) ± . that 0T*T) < + 00. We therefore have tTi = Ua for some bounded furthermore, aE ITI(H) , hence a, and 0T*T) H a Moreover, wa H =

I I

11 2 ;

=

H

Ham.

T* = (WU )* = UW* = U JWa' a a

and so

Jwa

E T*

(H),

and consequently

H

This

H

H = H Wall = H a H . established, since IT*1 = WT* = Uwjwa , we have 2 0TT*) = H WJWa H H a H 2 cp(T*T). WJWa

H=

JIAJa

The condition cp(T*T) < + 03 implies that 1 (TT*) < + 00 and the two conditions are therefore equivalent; furthermore, we have (1)(T*T)

cp (TT*)

for every

At the same time, if we put T = OUtUb

)=

0T*T) =

ub =

I

a

T

E

U(U).

Uva , we have shown that

11 2

(bib);

using polarisation, we deduce from this that .4) (ugub ) any bounded elements a and b.

for

(I)

is faithful. For, if q(T) = 0 for a TE U(U) + , we and (ala) = O. hence a 0, T 1/2 = 0, T = 0. a We show that (1) is normal. Let (Ta ) be an increasing filtering set in 11(.1) -4- , with supremum Tc U(U)+. We have merely to show that OT) < sup OTa ). This is obvious if sup (Ta) = + 00 . Suppose henceforth, therefore, that q(T) < 1 < + 00. Then, T1/2 = U , with aa a

The trace have T1/2 = U

-

(aa laa ) = OTa )

11.

Since T converges strongly to T while remaining majorized by a T, 712 converges strongly to T 1/2 . [Indeed, if p(x) is a polynomial which approximates x1/2 very closely on the interval p(T) and p(Ta ) are very close to T and T T the

[0,11

99

GLOBAL THEORY

sense of the norm; and, moreover, p(Ta ) converges strongly to 0(T).] Hence, for X, yE U, we have

Il (T1/2y1x) = lim (Taylx) = lim (U aa ylx) = lim (aa lxy*). As the xy* are dense in H and the ac all lie within a fixed ball, aa converges weakly to an a E H which satisfies

(Aix) = (alxy*) = (vyalx);

1,

hence a is bounded, T' = Ua , and

OT) = (ala) -.ç. sup (a la ) = sup OT ). a a a In fact, let TE U(U) + , T / O. The trace (p is semi-finite. For each bounded a, we have UaT1/2 = ub for some bounded b, hence

L 4(T 2u*u T 2 ) < + co. a a Now,

L, _,_ T -U*U T` 5_ II

a a

U 11„ 2 T. a

I

Moreover, we cannot have U a TI = 0 for every bounded a, because, making Ua converge weakly to I, we would then deduce that T = O. We can therefore find U a such that

Il T U*U T

2 /

0,

T 2 u*u

L

T 2 _..ç T,

a a

a a

which proves our assertion.

0

DEFINITION 2. Given a Hilbert algebra U, the traces defined by theorem 1 are called natural traces on TRW+ and V(U) 4". These traces are unaltered if U is replaced by the Hilbert algebra of bounded elements. Let A be a von Neumann algebra, (I) 1a trace on A+, m the ideal of definition of '(;b. For SE M12 , TE M-2 we have T*S E M; put

(SIT) = '4)(T*S). It is plain that a pre-Hilbert space structure is thus defined on m½. We will always employ the notation (SIT) in the above sense (when no confusion regarding (I) is possible) and we will denote by S 11 2 the corresponding semi-norm (SIS) 12 (to distinof S). guish it from the usual operator norm H S

H

H

Let A be a von Neumann algebra, w a semi-finite faithful normal trace on A+, and m the ideal of definition of THEOREM 2.

100

PART I, CHAPTER 6

di. Endowed with the scalar product (SIT) = (1)(T*S), a full Hilbert algebra. Let K be the corresponding completion of el, and J the involution of K canonically defined by m 1/2 . For RE A, the mapping S RS (resp. s SR) of rn into m 1/2 extends by continuity to an operator cD(R) [resp. T(R)] of L(K). The mapping (I) (resp. T) is an isomorphism (resp. antiisomorphism) of A onto U(m1/2 ) [resp. V(m1/2 )] which extends the canonical mapping of the Hilbert algebra m into U(m) [resp. V(m 1J)], and we have T(R1 = J4(R*)J. Finally, let cp and ip be the natural traces on U(m1 ) -4- and V(m1/2 ) +; for RE Al- , we have w(R) = 01, (R)) =

Proof. For R, S, TE mil, we have, in view of proposition 11 of chapter 1, (1)

(T*IS*) = Co(ST*) = (70(T*S) = (SIT),

(2)

(RSIT) = (1)(T*RS) = (I)((R*T)*S) = (SIR*T),

(3)

(RSIRS) = Co(S*R*RS) < II R* R II 63(S*S)

R 11 2 (SIS).

Furthermore, (S(I-T*) IS(I-T*)) = CJNI-T)S*S(I-T*))

when T converges ultra-strongly to I, (I - Tt) (I - T) converges ultra-weakly to zero, and hence (proposition 1) (S -ST* IS -ST*) converges to zero, which proves axiom (iv) for Hilbert algebras. For RE A, sE m1/2 , inequality (3) still holds, and yields the existence of the continuous extension T(R). It is immediate (resp. T) is a homomorphism (resp. antihomomorphism) of that A into L(K), which extends the canonical mapping of ml into U(M1/2 ) [resp. V(M1/2 )] and that T(R) = J(R*)J. Moreover, T and T are injective; because if for example (D(R) = 0, we have, for every SE M1/2 RS = 0, hence R = 0, letting S converge strongly to I. It is clear that (31)(A) and T(A) commute, hence ,

T (A)

c

V(m1/2 ) P = U(m1/2 ) ,

T(A)

c

U(mlY =

;

since T(A) DUM 1/2 and T(A) , the von Neumann algebras generated by (1)(A) and T(A) respectively are U(Mil) and V(M11 ). Finally, as in the proof of proposition 1 of chapter 4, we see that T and T are normal. Hence )

1/2 T(A) = U(m ),

T(A) = V(m 1/2 )

(chapter 4, theorem 2, corollary 2).

GLOBAL THEORY

101

For S E rfl, we have w(S) = (S 1/2 1S 1/2 ) =

cl)(4)(S)) =

If S is now an arbitrary element of A+, let Fcm+ be an increasing filtering set with supremum S. We have 00(S)) = sup 14)(4) (F)) = sup W(F) =

and similarly tp(4)(S)) = w(S). Finally, if a is a bounded element of K, we have, by theorem 1, OUllUa < + co; hence Ua = 0(S), with w(S*S) < + 00, hence SE r0; thus, Ua = U from which it follows that a = S e M1/2 : the Hilbert algebra M1 is full. 0 References : [13], [19], [29], [30], [66], [99], [101], [116], [117], [120].

Trace-elements. DEFINITION 3. Let A be a von Neumann algebra in H. An element a of H is called a trace-element for A if wa is a trace on A, in other words if (TiT 2a la) = (T2T1al a) for any T1, T2 E A. 3.

Let U be a Hilbert algebra and H its (Hilbert space) completion. Every central element of H is a traceelement for U(U) and V(U). PROPOSITION 3.

Let T1, T2E U(U). Put T1 = JTVE V(U) (J being the involution on H canonically defined by U). We have Tia = Ta, Ta = T1_ *a since a is central, and therefore

Proof.

(T T 2 1 ala)

= (T 2 T 1I ala) = (T 1I T 2 ala) = (T 21 aIT*a) = (T 1 T 2 ala), 2 alT"a) 1

from which it follows that a is a trace-element for 11(1). proof for V(U) is similar. 0

The

Let A be a von Neumann algebra and a a trace-element for A. Then EA', which is the support of co a (chapter 4, section 6) is a a projection of the centre Z of A. Since EA and 4 1 have the a same central support (chapter I, proposition 7, corollary 2), EA admits 4 1 for central support. In particular, if a is a cyclic for A, then a is also separating for A. Put, then, for TE A, 0(T) = Ta: 4) is a bijective mapping of A onto U and we have the following proposition: PROPOSITION 4: (i) If the *-algebra structure of A is transferred to U by means of T, U becomes a full Hilbert algebra. (ii) We have A = U(U), A' = V(U);

102

PART I, CHAPTER 6

(iii) The element a is an identity element for U (and is thus central). Proof. Transfer the scalar product of [lc H to A by means of 11 -1

.

For T l E A, TE A, we have

(T 1 IT 2 ) = (T 1 alT 2 a) = wa(T*T 2 1 ). Then, A becomes a full Hilbert algebra (theorem 2). Whence (1). Similarly, using (I) to transfer the results of theorem 2, we see that A = U(U). Hence A' = V(U).. Finally, a is the image under (11 of the identity element of A, whence (iii).

COROLLARY 1. If a is a trace-element and is cyclic for A, then a is a trace-element and is cyclic for A'.

Proof.

This follows immediately from propositions 3 and 4.

Let A be an abelian von Neumann algebra. there exists a cyclic element for A, we have A' = A. COROLLARY 2.

0

If

Proof. Each element of H is a trace—element for A. By proposition 4, A' is antiisomorphic to A, hence is abelian. Hence A c A' c A" = A. For a direct proof of corollary 2 of proposition 4, cf, exercise 5.

References : [29], [66], [89], [100]. 4.

An ordering in the set of traces.

This section settles uniqueness questions regarding traces. Sections 6 and 7 will be devoted to existence questions.

DEFINITION 4. Let A be a von Neumann algebra, cp and (1) 1 two traces on At We say that (1) majorizes cp', and we write (1) > if cp(T) > (T) for every T E At THEOREM 3. Let Z be the centre of A, and cp a semi-finite normal trace on A+. For each SE Z such that 0 S I, the function T cp(ST) on A+ is a normal trace (1)s majorized by (I), and every normal trace majorized by (1) is of this type. If (1) is (p s is injective. faithful, the mapping S

Proof.

For each SE Z such that 0 < S I, it is clear that on the function T 4(ST) A+ is a normal trace majorized by cp. Conversely, let (P I be a normal trace majorized by (1), and suppose to begin with that 4) is faithful. Let M and M / be the ideals of definition of .(1) and '(1) / ; we have McW. By theorem 2, "4) canonically defines a Hilbert alebra structure on M1/, two von Neumann algebras U(M 1/2 ) and 7(0) in the completion K of M1/, an isomorphism (11 of A onto U(M 1/2 ), and an antiisomorphism T of A onto V(m1/2 ). For TE M1/2 , TIE m1/2 , put ((T I T 1 )) = V (T I T). As (p i < cp, there exists a unique hermitian operator Al: L(K), such

103

GLOBAL THEORY

that 0 for TE M -2 (4)

I, p.nd such that ((TIT1)) = (TIA(T i )). ,

TIE Ml,

RE

We have,

A:

(T14)(R)A(T 1 )) = (R*TIA(T i )) = ((R*TIT 1 )) = .0 P (TtR*T).

(5) (TINT(R)(T 1 )) = (TIA(RT 1 )) = ((TIRT 1 )) = '0'(TIR*T), (6) (TIT(R)A(Ti )) = (TR*IA(T 1 )) = ((TR*IT 1 )) = .0 1 (TITR*), (7) (TIAT(R)(T 1 )) = (TIA(T i R)) = ((TITO)) = .0 P (R*TIT) = .0 / (TITR*). Hence AE U(M 2 ) n V(m1/2 )P, and consequently A = T(S), for some S E Z, O < S < I. It follows that, for T E M1/2 and T 1 E M`, we have

V(TIT) = (TIT 1 S) = ep(STIT); hence .0 / (T) = .0(ST) for TE M± . Then, T d(ST) 0 / (T) and T are two normal traces on A+ which agree on and hence are identical (chapter 3, theorem 2, corollary 5). The uniqueness of S follows from the above proof together with the fact that T is injective. Finally, if (I) is not assumed to be faithful, A may be identified with the product of two von Neumann algebras Al, A 2 , where 0 induces normal traces 01, 0 2 on AI, A; which possess the following properties: 1° 01 is faithful and semi-finite; 2 ° 02 = 0. Let 0i, (1) be the traces induced by O P on At, A. We have 0 11(T) = 0 1 (S 1T) for some element Sl of the centre of A l such that 0 _< S1 I. Let S be the element of Z defined by the element S 1 of A l and the element 0 of A2. We clearly have 0 I. for every TE A+, and 0 s: S (P I (T) = c(ST)

On a factor, two semi-finite faithful normal traces are proportional. COROLLARY.

Let (1) and cp 1 be two such traces. The trace 0" = is semi-finite faithful and normal and majorizes 0, 0'. Hence 0' = X'cp" for some constants X > 0 and X' > O. Hence = 0 0 = xxr-1 0 i .

Proof.

Let A be a von Neumann algebra, i a trace on A+, CP I a positive linear form on A. We say that 0 majorizes 0' if 0(T) ?- 0 / (T) for TE A The method of proof of theorem 3 then also yields the following result:

t

Let cp be a semi-finite normal trace on A+, and m the ideal of definition of cp. For every S E m such that 0 < S I, the function T -.0(S IITS 11 ) = '0(ST) on A is a normal positive 'linear form cp s majorized by 4), and every normal PROPOSITION 5.

104

PART I, CHAPTER 6

positive 'linear form majorized by cp is of this type. If ct) is faithful, the mapping s cps is injective. Proof. Let SE M be such that 0 S I. It is clear that (;bs is a normal positive linear form; furthermore, for TE A+, (ST) =

OTI2 ST1/2 ) 5

(TIIT1/2 ) =

and so cps is majorized by 4) Conversely, let (1) 1 be a positive linear form majorized by cp. As in the proof of theorem 3', we introduce K and A; equations (4) and (5) still hold, and hence A E U(M12 so that A = (S) for some S E A, 0 S 5_ I; the proof is then concluded in exactly the same way as that of theorem 3. 0 .

)

References : [19], [65], [66], [92], [101].

5.

An application: isomorphisms of standard von Neumann algebras.

LEMMA 1. Let A be a standard von Neumann algebra in H, and cp a semi-finite faithful normal trace on A+. There exists a Hilbert algebra U, dense in H, such that: 1° A = 2° (p is the corresponding natural trace on M. Proof.

There exists a full Hilbert algebra B, dense in H, such that A = U(S). Let ip be the corresponding natural trace on A+, and x = (1) + Tp. The traces cp, p, x are semi—finite, faithful and normal and ip < x, < x. Hence there exist operators S, S P in the centre Z of A such that O < S < I,

O < S P < I,

.q)(T) = x(ST),

cp(T) = x(S P T)

for TE A+ (theorem 3). The operators S and S P have support I since 4) and ip are faithful. Using the Gelfand isomorphism of Z onto an algebra of continuous functions, we see that there exist projections El, E 2 , ... of Z, pairwise disjoint, with sum I, and numbers Xi > 0, such that SE. > X.E., 2

S P E. > X.E..

It is enough to prove the lemma for each algebra AE. . We can therefore suppose henceforth that

S > X > 0, Let Z = 5 1-1/2 5 1/2

Z >

E

S P > X > 0.

Z; we have

> O

and

(1)(T) = Ip(Z -2T)

for

TE

A+ .

We now define a Hilbert algebra U: the underlying pre—Hilbert space of U is the same as that of B; the involution of U is that of B; the multiplication of U is defined by

GLOBAL THEORY

105

z(xy) = zuxy = uxzy = x(zy) = ZVx =

(x,y)

V 7.x = (Zx)y.

It is immediate that we have thus defined a Hilbert algebra structure (particularly on account of chapter 5, corollary of be the canonical mappings of x proposition 2). Let x U into U(U) and V(U). We have U P = ZUT , V P = ZVx . This shows x that V(U) c

U(U) c U(B),

V(B),

U(U)

and so

=

U(B)

=

A.

Moreover, an element of H bounded relative to U is bounded relative to B, and is therefore in B = U, from which it follows that U is full. Let, then, (I) P be the natural trace on U(U)+. Let TE U(U) -4We have $ 1 (T) = + 00 if and only if 11)(T) = + 00, i.e. if and only if (I)(T) = + 00. Moreover, if T1/2 = Ux for some XE U, we have .

—1 1 1T½ ZT=ZU=U,

x

X

Hence $ = $ P .

hence

(I)(T) = 'q)(Z

-2

T) = II x II

2 $ ( T) •

0

Let A and A l be standard von Neumann algebras. Then every isomorphism of A onto A l is spatial. THEOREM 4.

Let 4) be an isomorphism of A onto A l . Let U be a full Hilbert algebra such that A = U(U), and let $ be the corresponding natural trace on A+. Transferring $ to At by means of 4), we obtain a semi—finite faithful normal trace $1 on At. By lemma 1, there exists a full Hilbert algebra U 1 such that A l = U(U1) and Ux , such that $ 1 is the corresponding natural trace. Let 2:x be the canonical mappings of UintoA of U 1 and q into and Q i :x the Hilbert algebra isomorphism of an is A l . Then, Qi 1 0 (I) oS2 U onto the Hilbert algebra U1 which extends to an isomorphism W of the completion H of U onto the completion H 1 of U l . For X, yE U, we have

Proof.

—1 -1 —1 W (11(U x )Wy = W U P Wy = W Wx

((Wx)(Wy))

= W

-1

W(xy) =

xy

= Uxy,

and therefore (D(Ux ) = WUxW-1 . Thus, the isomorphisms (I) and T WTW-1 agree on a two—sided ideal of A which is ultra— strongly dense in A. Since these isomorphisms are ultra— strongly continuous (chapter 4, corollary 1 of theorem 2), they are the same. 0

References : [101]. 6.

Normal traces on

L(H) .

Let H be a complex Hilbert space, and (e) E l an orthonormal basis of H. For TEL(H) +, put THEOREM 5.

PART 1, CHAPTER 6

106

4)(T) = X (Teilei). iEI

Then (1) is a semi-finite faithful normal trace on L(H)+, which is independent of the choice of orthonormal basis (e ) ic1 If E is a projection, (1)(E) is the Hilbert dimension of E(H). .

(In this last assertion, it is convenient to identify all infinite cardinals with + 00.)

Proof.

It is clear that (I) possesses properties (i) and (ii) of definition 1. Now, let TE L(H) and let (ell x' XEI\- be any orthonormal basis of H. We have X (T*Tede

iEI

2 2

=

iEI

Te. 11 2 = =

X 1(Tede 2 xl )1 2 jEI xEK XEK iEI

I( Ted 2, 2 el)1 X

X (TT*e xl lex) ) = XII T*e t 1 2 = X X1 XEK XEK XEK iEI

2 12 x le-)

(Te'

X 1 (Tede = X 2 Xl) 1 2 * XEK iEI This shows at one and the same time that OT*T) = 4)(TT*), hence that cp is a trace, and that this trace is independent of the orthonormal basis chosen. Since each positive linear form wei is such that is normal, 4) is normal. If TE L(111 2 , we have T½ = 0, and therefore T = 0: 0 = d(T) = X )

1-

I TI-2ei

iEI (1) is faithful. Now let E be a projection; we can choose the eu in such a way that Eei = 0 or Eei = eu for every i; cp(E) is then the number of the ei such that Eei = ei , i.e. the Hilbert dimension of E(H) (with the indicated convention regarding infinite for cercardinals). This shows in particular that 4)(T) < + cp is therefore a factor, tain non-zero T of L(H) ± . As L(H) is semi-finite (corollary 2 of proposition 1). 0 By the corollary of theorem 3, every normal trace on L(H)+ is proportional to the trace cp above, or is identically equal to infinity on the non-zero operators of 1.(H) -4- . COROLLARY.

The set n of the TE L(H) such that

2 X11Te-11 = iEI

is a two-sided ideal of

Y I.(Te i le

)1 2 < + 00

L(H), independent of the orthonormal basis (ei)iEI, and consisting of compact operators. Every

107

GLOBAL THEORY

I2

< + 00 represents, with X Iti,xEI respect to (ei)jEI , an operator of n. If we put, for TE n and T' E n, matrix (tix ) such that

P

(TIT')

t. t. , X (Te.IT F e.) = =-bx -b x i,xEI

iEI

we define on n a Hilbert space structure independent of the orthonormal basis (e.). The too-sided ideal 6 of the finiterank operators is dense in n in the sense of this Hilbert space structure.

n is a two-sided ideal of L(H) independent of the basis (ei)i E I by theorem 5. The inequality Proof.

The set

H Te. H 2

H T H' < iEI

2 X 11 Te 2. 11 iEI is independent of (ei)] shows that every element of n is the [which follows immediately from the fact that

limit, in the sense of the norm topology, of operators represented by matrices having only finitely many non-zero entries, and a fortiori of finite rank. Hence each T En is compact. Now consider a matrix (tix) such that

2 1 it. < + op;

X i,xEI putting

T( X X.e.) = 7-El

( X t ix Xx)e i , icI xEI

we define, due to the inequality

ic I

X t.2X X

XE I

2 (X 2, XE

It' I

1 2 ) ( .X Ix 21 2\ 2x 2E 1

an operator T E L(H), and we have TE n. The pre-Hilbert structure of n may be obtained either directly or by application of is dense in theorem 2. The fact that n is complete and that n is now immediate. 0

6

The operators of

n are known as the Hilbert-Schmidt opera-

tors. Let H I be the Hilbert space conjugate to H, that is, we Tx, recall, the space H endowed with the operations (X, x) (y1x). For the x + y, and the scalar product (x, y) (x, y)

PART I, CHAPTER 6

108

x H', H' may be identified identify H with its dual in is of the form element T of n -> Y y. 0 x. allows us to

bilinear form (x, y) ÷ (xIY ) on H (We will not with the dual of H. the rest of this section). Every n Y ÷ Y ( YIY J - ) xJ- and the mapping T

6

J j=1 '7 identify the vector space 6 with the vector space which is the algebraic tensor product of H' and H. We can, furthermore, suppose that the y- are orthonormal and the x. orthogonal, in J J which case

j=1

n 2 (TIT) = Y 11 x • II = j=1 J

2

n

y

y. 0 x.

j=1 J

this shows that the above mapping is an Hilbert structures of the vector spaces It follows from this that it extends to of the Hilbert space n onto the Hilbert

;

'3 isometry for the preunder consideration. a canonical isomorphism space H' ® H. Let S E n.

GO

y

Since S is compact, we have Sy =

(ylu.)v., the u. being

co ti J J orthogonal; then (SIS) = X H V. H 2 ;

j =1

orthonormal and the y

i

J.1

we thus see that S is the limit in the sense of the Hilbert n space structure of n of the operators y ÷ (ylu.)v., which we <7 a j-1 n operators the u. 0 Y.. Finally, S may identified with have J j=1 J

y

y

CO

be identified with the element

y

u. 0 Y. of H' 0 H.

• <7 j=1

<7

mapping T 4- T* of n onto n may be identified with the linear isometry of H' 0 H onto H' 0 H which maps u 0 v to v 0 u. (ii) If SE L(H), the mapping T -).- ST of n into n may be identified with 1 H' 0 S, and the mapping T ->- TS of n into n with s* ® Proof. If T is the operator y ± (y1u)v, we have PROPOSITION 6. (i) The

= ((y1V)uly i ), (T* YIY I) = ( YI TY ') = ( Y I u)(Ylv) and so T* is the operator y -> (ylv)u, which proves (i). moreover, for SE L(H), we have STy = (yu)Sv, and so ST may be identified with

GLOBAL THEORY

u 0 sv = (1

H'

0

109

S) (u o v),

which proves the first assertion of (ii). follows from the first together with (i).

The second assertion

0

COROLLARY. If H is a Hilbert space, the von Neumann algebras L(H) 0 cH and cH o L(H) in H 0 H are standard von Neumann alge-

bras. Proof.

This follows from theorem 2, proposition 6 and the fact that the Hilbert spaces H and H', having the same (Hilbert) dimension, are isomorphic. D

References : [ [ 13]].

A first classification of von Neumann algebras. DEFINITION 5. A von Neumann algebra A is said to be finite (resp. semi-finite) if, for every non-2ero T of A+, there exists a finite (resp. semi-finite) normal trace on Al- such that d(T) / O. The algebra is said to be properly infinite (resp. purely infinite) if the only finite (resp. semi-finite) normal trace on A+ is 0. 7.

A non-finite von Neumann algebra is also said to be infinite. This classification is invariant with respect to isomorphisms and antiisomorphisms.

There are various logical connections between these different notions, as indicated in the following diagram (where A < B means that the properties A and B are incompatible, unless

H = 0): Purely infinite

Semi-finite

4

Properly infinite

>

< Finite

We will see in chapter 9 that there exist purely infinite von Neumann algebras. By theorem 5, L(H) is semi-finite, and properly infinite (resp. finite) if H is infinite- (resp. finite-) dimensional. An abelian von Neumann algebra A is finite; in fact, for every xE H, w is a finite normal trace on A+ Every von Neumann subalgebra of a finite von Neumann algebra is finite. On the other hand, nothing can be said about the finiteness of the von Neumann subalgebras of a semi-finite von Neumann algebra, since L(H) is itself semi-finite. PROPOSITION 7. Let (Ai) ici be a family of von Neumann algebras, and A the product von Neumann algebra. For A to be finite (resp. properly infinite, purely infinite),

PART I, CHAPTER 6

110

it is necessary and sufficient that each Ai be finite (resp. semi-finite, properly infinite, purely infinite). Every finite (resp. semi-finite) normal trace (P on A+ induces on each Ai a finite (resp. semi-finite) normal trace (pi; and, if (I) is non-zero, at least one of the (pi is non-zero. Conversely, let aEI, and (1) be a non-zero finite (resp. semifinite) normal trace on A-0{-1; for T = (Ti) E A+, put (p(T) =4) 10, (T01 ): then, cp is a finite (resp. semi-finite) normal trace on At The proposition follows immediately from these remarks.

Proof.

PROPOSITION 8. Let A be a von Neumann algebra, and Z its centre. Among the projections of Z, there exists a greatest projection, say El (resp. E2, Fl, F2) such that AE1 (resp. AE2' AF l , F2 A. ) is finite (resp. semi-finite, properly infinite, purely -tnftn-Lte). We have

E F 1 1

0, E +F =I; 1 1

E F =0, E +F 2 2 2 2

1; E

E2 , F1

F2 .

Proof . Let (Gi)i ei be a maximal family of non-zero projections of Z, pairwise disjoint, and such that the AGi are finite (resp. semi-finite, properly infinite, purely infinite). Let G = X G. By proposition 7, G is the greatest projection of Z iEI such that A G is finite (resp. semi-finite, properly infinite, purely infinite). Whence the existence of El, E2, F l , F 2 . It is clear that

E

1 F 1 = 0,

E 2 F 2 = 0;

E1

F 1 ->F 2 .

2'

To show that E l + F 1 = I (resp. E2 + F2 = I), it is enough to show that AI_E l (resp. AI_E 2 ) is properly infinite (resp. purely infinite); now, if there existed on AI_E l (resp. AI_E2 ) a nonzero finite (resp. semi-finite) normal trace, its support É would be a non-zero projection of Z such that AE is finite (resp. semi-finite), and so E l (resp. E2) would not be the greatest projection of Z such that AE 1 (resp. AE2 ) is finite 0 (resp. semi-finite). COROLLARY 1. A von Neumann algebra is canonically isomorphic to the product of a finite von Neumann algebra, a purely infinite von Neumann algebra, and a semi-finite properly infinite von Neumann algebra.

Proof.

We have

A = A El x A F2 x AE2F1.

0

2. A factor may be finite, purely infinite, or infinite and semi-finite. Proof. If A is a factor we have E l = I or F 2 = I or COROLLARY

E2F1 = I.

0

111

GLOBAL THEORY

Let

PROPOSITION 9.

A be a von Neumann algebra, and

Z its

centre. (1) The following conditions are equivalent:

(i 1) There exists a faithful semi-finite normal trace on

A4-;

(i 2)A is semi-finite. (ii)The following conditions are equivalent: (ii 1) There exists a faithful finite normal trace on (ii 2) A is finite and a-finite; (ii 3)

A is finite and

A+;

Z is a-finite;

(iii)Every finite von Neumann algebra is thé product of finite a-finite von Neumann algebras. Proof.

The implications (i 1) (i 2), (ii 1) (ii 3) are immediate. Now, suppose that A is semi-finite (resp. finite). Let (ci)ici be a maximal family of non-zero semifinite (resp. finite) normal traces on A+, whose supports Ei, which are non-zero projections of Z, are pairwise disjoint, Ei, and let us show that E = I; if E Let E = I, there iEI exists a semi-finite (resp. finite) normal trace (p on A+ such (T(I that (I)(I - E) / 0; the trace T E)) is non-zero, seminormal, and its support is majorized finite (resp. finite), and by I E, which contradicts the maximality of the family (Wi ET . Hence Ei = I. This established, 11) = cpi is a iEI iET normal trace on A+, whose support is E. = I, and is therefore 2E1 faithful; also, ip is semi-finite; for let TE A+, T 0; there existsanEisuch that TE-2 / 0; let SE A+, S / 0, be such that S 5 TE iand d(S) < + 03; we then have S T and (S) = (pi(S) < + 03, which proves our assertion. Thus, the im (i 1) is established. If now the plication (i 2) are finite, the AE . are finite and a-finite, which proves If, fuither, Z is G-finite, I is countable; (i). multiplying the (pi by suitable scalars, we can suppose that (p ( i) < 00 ; then, ip is finite, which proves the implication iEI (ii 3) = (ii 1) .

y

y

y

y

y

For a von Neumann algebra to be isomorphic to a standard von Neumann algebra, it is necessary and sufficient that it be semi-finite. COROLLARY.

This follows from theorem 1, theorem 2 and part (i) of D proposition 9.

Proof.

PART I, CHAPTER 6

112

It is more difficult to find conditions for a von Neumann algebra to be spatially isomorphic to a standard von Neumann algebra. This problem will be extensively considered in part III (chapter 1, section 5; chapter 6, section 2).

Let A be a finite von Neumann algebra, Z its centre, and cp a semi-finite normal trace on For every non2ero T of Z -1-, there exists a non-zero T 1 of Z -1- majored by T such that q(T1 ) < + oe. PROPOSITION 10.

Proof. Let 11) be a finite normal trace on Al- such that OT) / O. Replacing (1) by (1) + i we can suppose that tp There exists (theorem 3) a SE Z -4- such that i(R) = cp (SR) for ,

every RE M. We have ST / 0 since 11)(T) / 0, hence there exists a non-zero T 1 of Z -4- majorized by T, and a number X > 0, such that ST 1 XT 1 . Then

0T 1 )

X

-1

cp(ST 1 ) = X

-

< + oe.

D

THEOREM 6. Let U be a Hilbert algebra, F its characteristic projection, Z U(U) n V (U ), and E the greatest projection of Z such that (U(U)) E [resp. (V(U)) E ] is finite. Then we have E =F.

Proof. We first show that F E, i.e. that (U(U)) F is finite. To this end, let T be a non-zero element of U(U) -1- such that T = TF. There exists a central element a E H such that 11 T 1/2a e = (Tala) / 0; because the equality Ta = 0 for every central a would imply that TT l a = TITa = 0 for every T E V(U) and every central a, and hence TF = O. Then, wa is a finite normal trace on A+ (proposition 3), which is non-zero on T. Hence (U(U)) F is finite. We now show that E F. It will be enough to establish the following: for every non-zero projection G of Z majorized by E, there exists a non-zero central element a such that a E G(H) Now, let cp be the natural trace on U(U) -4- . By proposition 10, there exists a non-zero projection G I of Z majorized by G such that c(G P ) < + 03. We have G / = U a , for some non-zero, central, bounded a (theorem 1, and chapter 5, proposition 7), and

a

E

G 1 (H)

0

On a semi-finite factor F, there exists a faithful, semifinite, normal trace 0 let Mo be the ideal of definition of Cp. Every normal trace on F-4- is of the form X (1) where 0 X + 00, co with the convention that O. + = 0 (corollary 2 of proposition 1, and the corollary of theorem 3). In particular, /14 only depends on F and not on the choice of cp. We say that the elements of Mdo (resp. M) are the trace elements (resp. the Hilbert-Schmidt elements) relative to F.

GLOBAL THEORY

113

THEOREM 7. Let F be a semi-finite factor in a Hilbert space H. Let G be a group of unitary operators in H such that uFU-1 = F for every UE G. Suppose that there exists a vector

E7 , H possessing the following properties: (i)E is separating for F; (ii)eE is the set of vectors of H which are invariant under G. Then F is a finite factor, and E is a trace element for F. Each UE G defines the automorphism T UTU -1 of F. For every TE F, we put w(T) = (Ti). By (ii), W is invariant under G.

Proof.

a.

We choose once and for all a semi-finite, faithful, normal trace ip on Ft and a projection E of F such that O . < tj(E) < + on. We have EE / 0, as E is separating for F. We can therefore suppose that H EE = (EEIE) = 1.

b.

H

c.

For every UE G, there exists a number

X(U) > 0 such that

(UTU -1 ) = X(U)i(T) for every TE F -4- (corollary of theorem 3). It is clear that X is a homomorphism of G into the group of real numbers > O. We show that this homomorphism is trivial. If not, there exists aUEG such that X(U) < 1. Put En = UnEU-n for n = 1, 2, 3, .... For every projection F of F such that i(F) < + œ, the linear form T (TF) = 11)(FTF) on F is positive and normal (proposition 1), and its support is F; since ip(E*E F) =

nn

FE

n n

)) = X(U)n(E) n

0,

we deduce from chapter 4 (proposition 4) that En F converges strongly to zero; now the union of the F(H), for all the projections F of F such that i(F) < + 00, is total in H, and hence En converges strongly to zero. But

H EnE H = H unEE H = H EE H,

hence

= 0,

which is a contradiction. We have thus shown that 11) is invariant under G. Let TE F-4- . Let KT cF+ be the convex hull of the set of the UTU -1 for UE G. Let KTI cF-4- be the weak closure of KT . Then Kri, is weakly compact, and UK 1 U-1 = for every UE G. Hence the KI" which is lower semi-continuous for the function S II S E on weak topology, attains its minimum at some point So of Kr. If U E G, we have H u-lE H H s oE H, and 11(S 0 + US 0U-1) E 0 hence

d.

us

H s oE H

H 1/2(s o

us 0 u-1 )E

I

II soc I -

PART I, CHAPTER 6

114

Thus,

H s o H = H 1/2(s o

US0U-1)E

H•

As the norm in H is strictly convex, we deduce from this that S

0

E = US

-1

0U

E = US

0

E.

By hypothesis (ii), S o EE C. Since E is separating for F, So is scalar. Thus Krt, contains a scalar operator 0. Furthermore, the function w is constant on KT, and therefore on KT1 . In particular, it takes the value 1 on K. Hence KE contains a scalar operator A > O. The function 11) is lower semi-continuous for the weak topology (corollary of proposition 2), and finite and constant on KE by c, hence finite on K. In particular, tp(X) < + 03 , and so F is finite.

e.

We can now suppose that 11)(I) = 1. w are constant on KT, hence on 4. By operator 1_1E14. Then

Let TE F. Then ip and d, there exists a scalar

w(T) = w( p ) = pw(I) = 4)(p)w(I) = Ip(T)w(I) from which it follows that w is a trace. In part III, chapter 8, we will be able to reformulate definition 5 in an entirely different way. If

U is a quasi-Hilbert algebra, U(U) can be purely infinite.

References : [10], [13], [19], [29], [42], [65], [66], [90], [117], [123], [400], [416].

8.

Classification and elementary operations.

Let A be a semi-finite (resp. finite) von Neumann algebra, and E a projection of A or of A'. Then AE is semi-finite (resp. finite). PROPOSITION 11.

Suppose, first, that E E A. For every T E A, we have T . Put, for every normal trace (I) on A+, cpE(T) = cp (T o E) . It is immediate that cp E is a normal trace on A+ E , semi-finite if cp is semi-finite, finite if q) is finite. This established, for every non-zero T of A, there exists a semi-finite (resp. finite) normal trace (I) on A+ such that cpE(T) = 0 E) / O. Hence AE is semi-finite (resp. finite). If now EE A', AE is isomorphic to an algebra AF , where F is a projection of the centre of A (chapter 2, proposition 2); hence AE is semi-finite (resp. finite) by the first part of the proof.

Proof. 0 EE

115

GLOBAL THEORY

The tensor product of two semi-finite (resp. finite) von Neumann algebras is a semi-finite (resp. finite) ,:on Neumann algebra. PROPOSITION 12.

-

Proof.

If two von Neumann algebras are replaced by two isomorphic von Neumann algebras, their tensor product is replaced by an isomorphic von Neumann algebra (chapter 4, proposition 2.) By the corollary of proposition 9, we can suppose that the given von Neumann algebras A l , A 2 are the algebras U(U1), U(U 2 ) corresponding to Hilbert algebras U 1 , U 2 . Then Atl i ) 0 U(U 2 ) = U(U 1 0 U 2 ) [chapter 5, proposition 9 (1)], and If, further, Al and A 2 are so U(U1) 0 U(U2) is semi-finite. finite, the characteristic projections of U 1 and U2 are equal to I (theorem 6), and hence the characteristic projection of U 1 0 U2 is equal to I [chapter 5, proposition 9 (ii], hence ] U(U1) 0 U(U 2 ) is finite (theorem 6).

Let A be a semi-finite von Neumann algebra. There exists a von Neumann algebra Al antiisomorphic to A, a Hilbert space K, and a projection E of Al ® L(K), such that A' is spatially isomorphic to (A1 0 L(K)) E . PROPOSITION 13.

There exists an isomorphism (D of A onto a standard von Neumann algebra B (corollary of proposition 9). Now, -1 (1° = (D34 2 0(4, where (Dl is the ampliation of S onto C = B 0 cK (K being a sùitable Hilbert space), where (1) 2 is the induction of C onto V = CE (E being a suitable projection of C') and where (I0 3 is a spatial isomorphism of V onto A. Then A' is spatially isomorphic to

Proof.

V'

L(K)) ,

= C' = (B'

and it suffices to put B' =

Al .

0

The commutant of a semi-finite von Neumann algebra is a semi-finite von Neumann algebra. COROLLARY 1.

Proof.

This follows from propositions 11, 12 and 13.

0

COROLLARY 2. Let A be a von Neumann algebra, and Z its centre which is also the centre of A'. The greatest projection E of Z such that AE is semi-finite is equal to the greatest projection F of Z such that A;, is semi-finite. By corollary 1, M. is semi-finite, hence E E. D terchanging the roles of A and A P , we have F

Proof.

F.

In-

Let A be a purely infinite von Neumann algebra. is purely infinite.

COROLLARY 3.

Then

A'

Proof.

The projection E of corollary 2 is equal to O.

[I

COROLLARY 4. Let A be a purely infinite von Neumann algebra, and E a projection of A or of A'. Then AE is purely infinite.

PART I, CHAPTER 6

116

Since A' is purely infinite by corollary 3, it is enough to consider the case where E E A'. There then exists a projection F of the centre of A such that AE is isomorphic to As A = A F x A i _ F , A F is purely infinite (proposition 7). AF .

Proof.

D

The commutant of a finite (resp. properly infinite) von Neumann algebra is not necessarily finite (resp. properly infinite), as exemplified by the case of L(H) and cH. (cf. nevertheless part III, chapter 2, exercise 2). For some additional material, part III, chapter 8, theorem 2.

cf. exercises

10 and 11, and

Problem: does proposition 13 hold for an arbitrary von Neumann algebra? References : [40], [65], [67], [89].

9.

The commutant of the tensor product of two semi-finite von Neumann algebras. Let

PROPOSITION 14.

Al

and

be semi-finite von Neumann

A2

algebras. We have

(A 1

o

A2 )

A'1

/ =

0 A. 2

Proof.

There exist Hilbert algebras U1 , U2 such that A l is isomorphic to U(U1 ) and A2 to U(U2). Hence (chapter 4, theorem 3), we can suppose that

(U(11 1 )

Al =

0 C K1 ) E 1,

A 2 = (U(U 2 ) 0 C K ) E p,

2 2 with E IE

Putting E' = E

A 1 oik 2

(U(U1 )

0 C v )',

E 2 = (U(U 2 ) 0 9( 2 ) 1 .

0 E, we then have

= (u(U) oc

K

ou(U ) OC ) = (U(U oU )oct, v ) I. 1 2 Om E 2 K2 E1 1 2

Moreover

A'1

= (u(U ) o c

A2 =

1

(u(U)

0 C

)' K1 E '

(V(U

K 2) E2s =

(v(U 2 ) 0 L(K 2)) E 2 '

1)

L(K 1 )) E /, 1

hence

AI

(V(1/ 1 )

L(K i ) 0 v(U 2 ) 0 L(K 2 )) E 1

(V(Ul o U2 ) 0 =

L(Ki o K 2 )) E i

(u(U1 o U2) o cKloK 2 q 1 ' (A 1 o A2):*

0

117

GLOBAL THEORY

Let A l and A2 be arbitrary von Neumann algebras, Z i and Z2 their centres, and A = A l 0 A2 . The centre Z of A is e qual to Z 1 0 Z2. In particular, if Al and A2 are factors, A l 0 A2 is a factor. COROLLARY.

Proof. Let Hi and H2 be the spaces in which A l and A 2 act, respectively. We have Z

1

0 C

Z

H2

'

CHi

0 Z

2

cZ,

and hence

Z

1

0 Z

2

cZ

•

Moreover,

A

now, the (chapter Hence Z 1 Z1 0 Z =

1

0

c, c A c Z' ,

A'

c, c A' cP ;

0

1 "2 "2 von Neumann algebra generated by Al and Ai is Zi; hence 2, proposition 6) Zi 0 CH 2 cZ / ; similarly, CH i glZcZ P . 02 Z' CZ ' . Now (proposition 14), we have (Z 1 0 Z 2 ) Hence Z 1 0 Z 2 D Z. 11

The last assertion of the corollary is very elementary (chapter 2, exercise 4). Recall, moreover, the problem posed at the end of section 4 of chapter 2.

References 10.

: [65],

[128],

[443].

The space L 1 defined by a trace.

Let A be a von Neumann algebra, and (I) a trace defined on the two-sided ideal M of A.

Let SEA,

LEMMA 2.

TEni.

We have

1 (1)(ST)I _< [(I) Proof.

S

and T. chapter 1,

Let S = We have

(I) (ST) =

VISI, T ITI Ern,

4)(

Is I IT* I

] 1/2 .

VITI 1 be L

the polar decompositions of I T P E "12, and so, by proposition 11 of

=

( ITI 1/2U I 5 IVI T I 1/4) = OPIT SI NI 1/2 )(ITI 1/2"ISI 1/2 ) * ].

As ITN'S 1 1/2 E M1/2 and equality yields

ITI 1/2V* IS I 1/2 E M1/2 F

the Cauchy-Schwarz in-

14)(ST) 1 2 4[ ( ITI 1/2 u1 s1 )(IT I 1/2 u1s 1 1/2 )*M( ITI 1/4 v*Isl 1/2 )(I TI 1/2v *Isl 1/2 )*] I I )(1)(ISIVITIV*)* =CP(UISI1U*,T, Now 1S*I = UISIU* and 1T*I = VI T IV* *

D

Let A be a von Neumann algebra, cl) a trace defined on the too-sided ideal m of A, s EA and TEM. We have THEOREM 8.

118

PART I, CHAPTER 6

1 4)(ST)I < Proof.

(

ISTI)

< 11 s 11(a(ITI).

Suppose, to begin with, that S > 0 and T > O. We

have

T 12 ST 12 < IIH S IIH

T,

1

.

cp (ST) = (4) (T,1/2 )

hence

.

II SII(T)•

If now SE A and TE M are arbitrary, we have

whence,

(1)(15*I .ITI) H I s* I 11(1T1) = H s 11(P(ITI), (1)(1s1.1T*1) I ii Isl 1101T*1) = H s b(ITI), by lemma 2, 1 ,4)(ST)I I H s 114)(ITI)• Consequently, I(ST)I = I (4)(STI) I H H(lsT1) = (i(lsT1),

which is the first inequality of the theorem. Moreover, let ST = WISTI be the polar decomposition of ST. We have

= OW*ST) <

H w*s H(ITI)

ils H(ITI),

which is the second inequality of the theorem.

D

COROLLARY 1. Let TEM. The Teal number 01TI) is the supremum of the real numbers i(ST)1 when S varies over the set of operators of A such that Il S < 1. This supremum is attained.

Proof. T.

AS

usual, let T = VIT be the polar decomposition of

We have (

COROLLARY 2.

and

I T I ) = 1(V*T)I

The function T

II V* II

WTI)) defined

seminorm.

1.

on m, is

Proof. This function is the supremum of a collection of seminorms by corollary 1. 0 Suppose, from now on, that cp is normal, faithful and semifinite. The space M, endowed with the norm T ± WTI), is a complex normed vector space. We will denote the compiex Banach space obtained by completing this normed space by L 1 ( ) , or more briefly by L l , and m will be identified with a dense vector subspace of Ll. The norm of an element fEL 1 will be denoted by II f II 14 or by f l

il

Let TEM. The linear form S d(ST) on A is an ultra-weakly continuous (proposition 1) linear form cpT , and therefore, with the notation of chapter 3, section 3, T€ A * . Furthermore,

119

GLOBAL THEORY

HT

II 1 , by corollary 1 of theorem 8. The I (I) T II = (])( I T I ) = mapping T (1)T is therefore a linear isometry of M onto a vector subspace of A. We shall show that this subspace is dense in A in the sense of the norm. As A may be identified with the dual of A* [chapter 3, theorem 1 (iii)], it is enough to show that, for every SE A, there exists a TE fl such that cpT(S) / 0; now, let S = u1s1 be the polar decomposition of S; there exists (because R can converge strongly an RE M + such that RI sl R while remaining in M+ , majorized by I); then to I (PR2u*

( s ) = (1)(R2u*s) = (P (R lsI R) / (31,

whence our assertion.

Hence the mapping T (pT extends in

exactly one way to an isomorphism of the Banach space L 1 (4)) onto the Banach space A* of ultra-weakly continuous linear forms on A. Using theorem 1 (iii) of chapter 3 again, we obtain the following result:

Let A be a von Neumann algebra, and 4) a semifinite faithful normal trace defined on the two-sided ideal m of A. The normed spaces A (endowed with the norm and m (endowed with the norm in duality with respect to the bilinear form (s, T) (ST), are such that A is the dual of m. References : [15], [32], [76], [101], [[13]]. THEOREM 9.

HH)

H H 1)

11.

Trace and determinant.

Let A be a von Neumann algebra, T an invertible element of A, and T = ulTI its polar decomposition. Then, U is unitary, and ITI is invertible. We can therefore form logITI which is a hermitian element of A.

Let (I) be a finite trace on A such that We call the determinant associated with 4) the func-

DEFINITION 6.

= 1. tion A defined on the set of invertible elements of formula

A by the

A(T) = exp[fllogITI)]. It is immediate that A(T)

LEMMA 3.

= A ( I T I ) = [A (

IT1 2 )]

= [ A(T * T)] 1/2 .

Let f(X) be a function which is analytic in an open subset of the complex plane, and F a Jordan curve in this set. Let t S t be a differentiable (in the sense of the norm of A) mapping of an open interval T of the real line into A. Suppose that the spectrum of st lies within the interior of r for tE T. Then, f(St) EA is a differentiable function of t, and

120

q

PART I, CHAPTER 6

hf(S t )] = c[g(S t )S1],where

g(X) = Proof.

dS

df(X)

q

and

dx

=

t

t

—

dt

We have, for tE 1,

f(S t ) =

TTT

f f(x) (X - s t ) -1 a.

In view of the equality

(X - S t' )

- (X - S t )

= (X - S t' )

(S t' - S t )(X -

we therefore have

t ,_ t [f(s t ,)

- f(s t

277, r

f(X) (X - S t' ) -1 ti1-t (s ti -s t )(X-s t )1dX.

When t' t, (X - St') -1 converges, in the sense of the norm of A, to (X - St) -1 , uniformly with respect to X in r; and 1 (S t' S t ) converges to S tl '- hence f(S ) is differentiable t'-t t respect to t, and with

1 Ll-f(S ) = . f f(X) (X dt t 27 -/-

S

-1 t)

Si(X

S ) t

t

-1

dX.

Moreover, the equality

- S t)

(X'

yields, when

- (X - S t ) X' X

=

,-1

d

dX ‘A

S t)

(X'

'')t)

- (X

(X - X')(X

S t)

S ) -2 t

Hence

d

Ef(X)(X

-1 S)]

= g(X)(X

St)

f(X)(X

s t ) -2

and, consequently, g(S ) =

t

Then

1

27/, g(X)(X - S t ) 1dX =

1 277:

f(X)(X -

St)

-2

a.

121

GLOBAL THEORY

d

f

10 /Q )1

j

=

27i j

=

1 1 -2Tri

r

JkA4L0k

r

- s t) l s t/(x -

st

)

l

]dx

-2 SlcDk=cP[g(S )S 1 ]. t t t

t

D

LEMMA 4. Let S and T be elements of A, with T hermitian. have A (exp S* exp T exp S) = exp OS* + T + S).

We

For t real, put

Proof.

S t = exp(tSflexp T exp(tS)E A. Each St is positive hermitian and invertible. Furthermore, St is a differentiable function of t. There exist numbers a > 0, a / < + co such that

<msm‹ t

HS t

a <

a,

for every

tE

[

0,

1].

We now apply lemma 3, taking for fn.) the principal value of loin in the complex plane with the negative real axis deleted, and for r a suitable rectangle containing a and a' within its interior. As S' = S*St + S tS ' we have

dt

Wog

St)

= cp(S t S't ) = OS t S*S t + S) = OS* + S),

and hence

Olog S i )

-

Wog S o )

= OS* + S),

i.e. logA(exp S* exp T exp S) - d(T) = OS* + S).

The function A has the following properties:

THEOREM 10.

(i) A(AI) = 'XI for A / 0; (ii) A(T) = A(T) = A(T*T) 1/2 for TE A, T invertible; (iii) A(ST) = A(S)A(T) for S

E

A,

TE

(iv) A(exp T) = lexpcp(T) I for T (v)

Proof.

A(T)

_<

lim H

Tnen for T

E

E

A,

S

and T invertible;

A;

A, T invertible.

n4-4,0 (i) For A / 0, A(AI) = exp(logIXII) = exp logl

Al

= lx1-

(ii) For T EA, T invertible, it has already been observed that A(T*T) = A(T) 2 . Moreover, TT* = U(T*T)U -1 for some unitary operator U EA and so log(TT*) = U log(T*T)U -1 ,

hence

A(T*T) = A(TT*).

122

PART I, CHAPTER 6

(iii) Let S and T be invertible elements of A, and S = u1s1 , their polar decompositions. In view of lemma 4 T=vI together with (ii), we have A(ST) = A(UISIVITI)

= A(ITIV* ISIU * UISIVITI) ½ = A(ITIv * IsI 2vITI) = [exp(logITI + log (V*I5I 2 V) + log1T1)] 1/2 = exp(logISI + logITI) = A(S)A(T). (iv) Lemma 4 yields, in view of (ii) and denoting by RX the real part of a complex number X, A(expT) = [A(expT*expT)] 1/2 = [expflT*+T)] 1/2 =expR(1)(T)

(y) For TE A, T invertible, we have ITI logITI

log H T

H,

hence

A(T)

II T II

= lexpcp (T) I .

and therefore

H.

exp log HTM=HT

Consequently, in view of (iii), A(T) =A(Tn) l/n every integer n. D

111/n II T n

COROLLARY. For every TE A, cp(T) belongs to the convex the spectrum of T.

for

hull

of

Proof. It suffices to show that (I)(T) lies in each closed half-plane which contains the spectrum S(T) of T. [Recall that, since S(T) is a compact subset of the complex plane, its convex hull is closed.] Moreover, if we replace T by aT + a l , where a and a' are two complex numbers, OT) is replaced by 04(T) + a' and S(T) by LS(T) + a'. This allows us to suppose that the half-plane considered is the half-plane RX O. Then, S(exp T), which is the same as exp S(T), is contained in the interior of the circle 1X1 = 1. Hence

lexP by theorem 10,

(T)

I

(iv) and (v).

=

(

exp T)

I,

Consequently, ROT)

O.

Cf. part III, chapter 2, exercise 6. In proving the results of this section (results which will not be used in the sequel) we have appealed to theories which do not appear among the list set out in the introduction. Nevertheless, the reader could consider analytic functions of an operator as defined by the Cauchy integr, and the necessary knowledge may of [[ 9]]. then be found in theorem 24 References : [23], [24].

123

GLOBAL THEORY

Exercises. 1. Let A be a von Neumann algebra, and 11) a trace on A+. For 4) to be normal, it is necessary and sufficient that q) be lower semi-continuous for the weak topology (use the corollary of proposition 2). If this is the case, and if T o E A+ is such that 4)(T0 ) < + OD, the restriction of 4) to the set of elements of A+ majorized by To is weakly continuous. 2. Let A be a 0-finite von Neumann algebra in H, and q) a normal trace on A+. Show that there exists a sequence

00 ...) of elements of H such that 4) = X x • . [If (x 1 , x 2' i=1 ( T) = + m for every non-zero T of A+, use a countable separating set for A. If q5 is semi-finite, form a sequence (E.) of 00

pairwise disjoint projections of A such that

X E. = I and such i=1 that cp(E.) < + co for every i; then use proposition 2.]

3. Let A be a von Neumann algebra, (I) a faithful trace on A+, m the ideal of definition of '4), and S and T hermitian operators of m 1/2 . If . (1)(ST) 5 . 4)(SUTU-1 ) for every unitary operator U of A, then S and T commute. [Let R be an hermitian operator of A. Show that the function of a real variable

4)( se itR -itR )

t

itR

-itR

is differentiable, with derivative (1)(iSRe Te Set this derivative equal to zero for t = 0, and make R = i(ST - TS)]. There is an identical result if . 4)(SUTU -1 ) for every unitary operator U of A [15], [76]. . 4)( ST) 4. Let U be a Hilbert algebra having an identity elelemt e. The trace w e on TI(U) -4- is the natural trace. 5. Show directly that, if an abelian von Neumann algebra A in H possesses a cyclic element x, we have A = A'. [Let T / E A'; we need to show that T / is normal. There exists -a sequence (Tn ) in A such that Tn x ± T l x. Since Tm - Tn is normal, the TX form a Cauchy sequence, and have a limit y; for every TE A,

(Y1Tx) = lim (T;',Icx1Tx) and hence y = T*x .

lim (xITTnx) = (xITT l x) = (T s*XITX),

Therefore, for every TE A,

T(T l x) = T I (Tx)

T (Tx)

and

T*(Tx)

T(T I*x) =T"(Tx);

the equality It T(Tx) II = II T(Tx) H yields, in the limit, T/*2 T/Z for every rii , *( hence T' (TX) =

HE

H

Tx) 11;

H

H

H

H

PART I, CHAPTER 6

124

Let H be an infinite-dimensional complex Hilbert space, M a two-sided ideal of L(H) , and (I) a trace on M [i.e. a linear form on m, positive on m+ , such that (ST) = (TS) for TE ni and

6.

E L(H)].

= 0. [Show that, if H1 and H 2 are two a. If M = L(H), (orthogonal) complementary subspaces of H of the same dimension, we have

°P H1 )

= °P H2 ) =

= (" PH 1 ) 4- (1)(P H 2 "

and hence that 01) = 0.] b. Let 6 be the two-sided ideal of finite-rank operators. Suppose mD 6 and 06) = O. Let S, S' be compact operators of m+, (Xi(S)) the sequence of eigenvalues > 0 ofS arranged in decreasing order, each eigenvalue being repeated a number of times equal to its multiplicity (the sequence terminating in 0, 0, 0, ... if S is of finite rank). If Xi(S) = 0(Xi(S)) as i + co, we have OS) = 0. [Reduce to the case where S and S have the same eigenvalues. Then, adding to S and S' operators of which does not alter d(S) nor OS'), show that 0 OS) EOS'),whereE > 0 is arbitrarily small.]

6,

c. Let ip be the trace on L(H) +defined in theorem 5, n the ideal of definition of i. Let (I) be a trace on n. Show that (ti is proportional to 4). [First show that cl) is proportional to II) on observing that (1) takes the same value on all rank-1 projections. Replacing (1) by (I) + )4, reduce to the case where 06) =0. Then show that On+) = 0, using b.]

6,

d.

Let

p be

X (S) + 1

the set of compact operators S

+ X(S) = 0(log i)

0 such that

as

There exists a two-sided ideal M such that M n and M+ = a non-zero trace (I) on m such that On) = 0 [359].

p,

and

Let H and K be complex Hilbert spaces. A given orthonormal basis (ei)i E , of K canonically defines isomorphisms Ui of H onto subspaces of H 0 K. Let A be a von Neumann algebra in H, (I) a normal trace on A+, and B = A o L(K) . For T E B+ , we put

7.

OT) = X OU*TU.). z, iEI that II) is a normal trace on B+, faithful (resp. semi(Argue as in finite) if (I) is faithful (resp. semi-finite) . theorem 5.) Let H be a complex Hilbert space, cl) the trace on fined by theorem 5, and M the ideal of definition of .(1).

8.

L(H) l- de-

GLOBAL THEORY

a . If H

TE

125

m, we have, for every orthonormal basis

(e.). 7,, "/„Ei7

of

,

< 1-w X 1 (Te 7,d"' 7, jEI

(1)(T) = X

and

(Te

iEI

.1

2

e•)•

2

(Write T as a linear combination of elements of m -4- .)

If T

b.

L (H) and if X 1 (Te d ei 2, 2, iEI

E

< I- 00 for eVery

l

ortho-

[Writing T=T 1 +iT 2 , normal basis (ei)i E , of H, we have TE m. with T 1 = 1/2(T + T*), T 2 = 1/2i(T - T*), reduce to the case where T is hermitian. Choosing a suitable orthonormal basis, show that TI- Em + and T- E Mt] c. some

Show that we can have Ti iv and

• 1"1 < + co for X 1 (T e 2-V iEI orthonormal basis (e0i EI of H. [Let Hl, H 2 be (ortho-

gonal) complementary subspaces of H of the same dimension; take H i u H 2 for every i.]

T unitary such that T(H i ) = H2, T(H2) = H i , and ei E

If H is infinite-dimensional and if TE L(H) is such that X 1 (Te 2,• e x) 1 < + 00 for every orthonormal basis (ei ) i EI d.

show that T = O. (First show that, for every xE H, Tx is proportional to x.) e•

If T e L(H) is such that

1

(Te.

7, e x )

I

< + 00 for some

i, x orthonormal basis (ei)i E I Of H, we have T E M. [Let (fx)x EL be another orthonormal basis of H. Putting fx = X u xi ei , we iE ] have

l(T fA IA ) I --

X l u xi lduxx 1.1( T eilex ) I

i,xEI

and so

X I crfx 1fx)

XEL

Then use

, 1/2 1 -ç

[1(Tellex ) 1( X luxi1 2 ) XEL i,xEI

X

(

2 1/2 X luxx l ) I XEL

b.]

Problem: If TE M, does there exist an orthonormal basis (ei) iE , of H such that

X 1(T e.2 1ex ) 1 < + cc? ? i,xEI

w.

PART I, CHAPTER 6

126

If a finite von Neumann algebra A possesses a countable cyclic set M, then A is of countable type. [M is separating for the centre of A; use proposition 9 (ii).] 9.

Let

10.

H

and

A

and B be two von Neumann algebras in the spaces

K.

a. If A is properly infinite, A ® B is properly infinite. [Let E be a projection of the centre of A 0 B such that (A 0 B)E is finite. Then, (A 0 CK) E is finite, hence E = 0.] b. If E (resp. F) is the greatest projection of the centre of A (resp. B) such that AE (resp. BF ) is finite, G - E 0 F is the greatest projection of the centre of A 0 B such that (A 0 B) G is finite. [We have

A

0 B =

(A

0 B

F

)

x

(A

E

® BI-F ) x (AI-E x (AI-E 0 BI-F), 0 BF)

apply a and proposition 12.] c. If A 0 B is properly infinite, one of the algebras is properly infinite (use b).

A, B

Let A and B be two von Neumann algebras. If A 0 B is purely infinite, one of the algebras A, B is purely infinite. [If there exists a projection E (resp. F) in the centre of A (resp. B) such that AE (resp. BF ) is semi-finite, then AE 0 B F is semi-finite.] 11.

The converse is false (part III, chapter 8, theorem 2). Let U 1 , U 2 be two Hilbert algebras, E l , E2 their characteristic projections. The characteristic projection of U1 0 U 2 is E l 0 E 2 . (Use theorem 6 and exercise 10 b.) 12.

13. Let A be a von Neumann algebra, cp a finite trace on such that cl)(I) = 1, and A the associated determinant.

a.

A

Show that A is continuous for the norm topology.

b. Show that, if T i and T 2 are invertible positive hermitian operators of A such tn.at T 2 T i , we have A(T 2 ) A(T 1 ). [Note that Ti 1/2 T2Ti 1/2 _< I , hence that A(TVIT2Ti 1/2 ) 1. ] [23], [24].

ABELIAN VON NEUMANN ALGEBRAS

CHAPTER 7.

1.

Basic measures.

Let H be a complex Hilbert space, and y an abelian C*-algebra of operators in H. Let Z be the spectrum of y, Lœ(Z) the set of continuous complex-valued functions on Z vanishing at infinity, f -›- Tf the Gelfand isomorphism of L(Z) onto y, and vx y the spectral measure on Z defined by the pair (x, y) of elements of H. (Concerning all this material, cf. Appendix I, sections 1 and 2.)

A positive measure y on z is said to be basic if it possesses the following property: for a subset of Z to be locally y-negligible, it is necessary and sufficient that it be locally v , -negligible for any XE H. DEFINITION 1.

(Throughout this book, when we speak of a measure on a locally compact space, we will always mean Radon measure; cf. [ [4]].) If there exists a basic measure v on Z, every basic measure is a positive measure equivalent to v, and conversely. If y is a basic measure, every measure Vx x possesses a V-integrable density with respect to v. By'formula (6) of Appendix I, there exists, for every measure vx,y , a v-integrable function h on Z such that Vx,y = hx,y V. The h x,y possess X,11 the followlog properties, which areconsequences of the properties of the V x,y

(1)

' x ,y h Xx+X'xs,y = Xh x,y + Xh

(3)

h

E

C) ,

h

T X,T

g

ly

=

0,

x,x

ggrh x,y

(g, g'

g

except on V-negligible sets which depend on

g

Xt

h y,x = hx,y'

(2)

(4)

(X,

i

, .

127

E

Lco(Z)) .

x, x', y, X, X', g,

PART I, CHAPTER 7

128

Since the union of the supports of the Vx x is dense in Z, the support of a basic measure V is the whole of Z. Then, Loo (Z) may be identified, along with its *-algebra and Banach space structures, with a subspace of Ifc3 (Z, V) (the space of essentially bounded V-measurable complex-valued functions on Z, in which two functions, equal locally almost everywhere, are identified.) PROPOSITION 1. There exists a homomorphism of the *algebra L:w .c (z' v) onto a *-subaZgebra of L(H), which extends the Gelfand -bsomo phism and which is continuous when *1 (Z, V) is endowed with its weak topology as the dual of Q(z, v) and L(H) with the weak operator topology. This homomorphism is unique. It is an isometry. Its image is the weak closure of Y. For fc Lœc (z, v), xcH, y H, we have, again denoting by f T f the extended isomorphism:

(Tfxly) =fr(C)dVx,y (C).

Proof.

The equality

(TfXIY) = which holds for f€1,00(Z), shows that the Gelf and isomorphism is continuous for the stated weak topologies. Now the unit ball of L(H) is weakly complete; moreover, every point of IZ(Z, v) is in the weak closure of a bounded subset of 1.00 (Z); [indeed, if f E IZ(Z, v) is such that II f 1, f is the weak limit of funcg 1; tions g of I_Z(Z, v) of compact support, such that then g is the limit in mean of continuous functions h with sup1; and port contained in a fixed compact set, such that I h g is, a fortiori, the weak limit of h]. Hence ([ [ 3]], part III, chapter 2, proposition 8), the Gelfand isomorphism extends to a unique weakly continuous linear mapping of 4(Z, V) into LU-t) , whose image Z is contained in the weak closure of Y. We again denote this mapping by f Tf . The formulas

H H

(Tf.xly)

-ff(C)hx,y (C)c/V(C),

T = T2,

Tfg = TfT6 ,

valid for f, g € 4.(Z), remain true by continuity for f, g E v) [for the last formula, it is necessary to make f, then g converge separately to elements of Lc (Z, V)].

H Tf 11 = 11 11.

We show that Firstly, T = TaT*, and so the f gg homomorphism f Tp is increasing for the natural_ grderings. To the function 1, cof. responds an idempotent self-adjoint operator, i.e. a projection E. If 0 < X, we therefore have 0 < T. < XE and consequently this shows that for f O. Suppose now that the function f 0 Tf II f II majorizes the number p > 0 on the non-locally-V-negligible measurable set X; then, Tf majorizes pE r , where E' is the

H Tf 1

x;

129

GLOBAL THEORY

projection corresponding to the characteristic function of X; now, X is non-negligible for at least one spectral measure Vx,x and we then have

(E'xIx)

= f dy (c) / o, so X 5,5 H Tf H p. Finally, H

E' / 0;

that

= HfH

this shows that Tf H any f for (by considering ff). and consequently

for

f

0

00 The unit ball Z 1 of Z is thus the image of the unit ball of Lc (Z (Z, V) which latter is weakly compact. Hence Z l is weakly compact and consequently Z is weakly closed (chapter 3, theorem 2). D This extension of the Gelfand isomorphism should be compared with the extension defined in Appendix I, section 3. Cf. also exercise 2 of Appendix I. References : [9], [18], [28], [45], [70], [100], [[9]].

2. Existence of basic measures. H is cyclic for Y' y x,x is a basic

If x

PROPOSITION 2. measure on Z.

Let xE H. We need to show that V/,, ,y is absolutely continuous with respect to Vx,x . Firstly, if y is of the form T / x, for some T / E Y', we have, for every positive function f E L(Z)

Proof.

v YY(f) =

'

(T

f

Trxlvx) = IITIT f 1/2

x11 2

2 II TI II Vx,x (f):

II 2(TfX1x )

HT'

T / 112V5 x, which proves our assertion. We pass hence V Y Y to the general case. We will use criterion 5 of [[id], chapter 0 Let f, a function V, section 5, corollary 5 of theorem 2. of 40 (Z), and 6 > 0, be fixed. First, we can find y / E Y's such that

HY

yi

H

Then, for every h

IVy,y (h) - Vy , ,y/ (h)1

ill -1 H Y H -1

,

H Y: H

H Y H.

0 of Lœ (Z) majorized by f, we have I (Thy ly) - (Thy ly I) I + 1 (T hy ly 1 ) - (Thy' ▪ II

Th 11.11 y 11.11 y - y'

ly l )

+ II Th 11.11 y -y'

.L 211 f 11.11 y 11.11 y - y' II 5_ Then, by the first part of the proof, there exists 6 > 0 such that v5 ,5 (h) 5_ 6 implies vy r ,u ( h) E/2. Then, h E Loo(Z) r h
II

130

PART I, CHAPTER 7

For there to exist a bounded basic measure on z, it is necessary and sufficient that there exist a cyclic element for yl. Proof. The condition is sufficient by proposition 2. ConPROPOSITION 3.

versely, suppose that there exists a bounded basic measure v on Z. We will show that V" is cl -finite; this, together with chapter 2, corollary of proposition 3, will prove the proposition. We can confine attention to the case where Y" is the weak closure of Y. Let (Ei)i ei be a family of non-zero pairwise disjoint projections of Y". Let Zi be a v-measurable subset of Z whose characteristic function corresponds to Ei under the isomorphism of proposition 1. The intersection of two distinct Z'S is negligible; for every finite subset J of I, we thus have 1 V(Z) V(Z) < + 00; hence the inequality

1

iEJ

a positive integer) only holds for a finite number n (n of indices i; moreover,v(Z i ) > 0 for every i since Ei / 0; hence D I is countable.

v(Z)

Let Z be an abelian von Neumann algebra in H. (i) There exists a sub-c*-algebra Y of Z, weakly dense in Z, whose spectrum z carries a basic measure. (ii)If H is separable, Y can be so chosen that, in addition, z is compact and second-countable. (iii) Suppose that Z is cr-finite (which is the case if H is separable). For every sub-c*-algebra y of Z, the spectrum of y carries a bounded basic measure. PROPOSITION 4.

Proof.

If Z is cl-finite, there exists a cyclic element x for Z / . Then, for any sub-C*-algebra y of Z we have Y I D Z I , and so s is cyclic for yl. This, with proposition 2, proves (iii). If H is separable, Z is the von Neumann algebra generated by a sequence (Ti). Taking for y the sub-C*-algebra of Z generated by I and the Ti, the spectrum of y is compact and secondcountable (Appendix I, section 1), and carries a basic measure by (iii). Whence (ii). We finally consider the general case. 10

There exist:

a family (Hi)i c i of Hilbert spaces;

2 ° in each Hi , a cl-finite von Neumann algebra Z. such that H may be identified with the direct sum of the Hi's and Z with the product of the Z's [chapter 6, proposition 9, (ill)]. Let Ei =H• and let Z / be the spectrum of Z. To the projection Ei, correiponds a continuous function on Z / which is equal to its own square, i.e. the characteristic function of a clopen set Z. As EiEx = 0 for i / x, the Zi are pairwise

131

GLOBAL THEORY

Consider the sub-C*-algebra Y of Z consisting of the T = (T 7 )j 1 possessing the following property: for any E > 0, is finite. The C*the set of the iEI such that II T i II algebra y is weakly dense in Z. Every non-zero character of Z lying in one of the Zi defines, by restriction to Y, a non-zero character of Y. We thus have a continuous mapping 0 of U Z. iEI Conversely, let x be a non-zero into the spectrum Z of Y. character of V; we have disjoint.

x(E)x(E) = x(EiE x) = 0

for

i / x;

hence x(E) = 0 for each i of a subset J of T whose complement has at most one element; if we had J = I, we would have x = 0 by the norm-continuity of x; hence I J = (i 0 1 and x(E 0 ) = 1. For every TE Z, we have TEi o E Y; put

)( 1 ( 7 ) = )(CIE- ) l'O we see immediately that the character x of Y is thus extended to a non-zero character x' of Z which belongs to Z•l . We thus 'O have a continuous mapping e' of Z into U Z.. We easily verify

2E1 that

e'oe

and 000' are the identity mappings of

U Z. and Z j ET

respectively. In short, Z may be identified (along with its topology) with the union of the Zi's. Similarly, Zi may be identified with the spectrum of Zi. Now, by (iii), there exists a basic measure vi on Zi. The vi define on Z a positive measure V, which induces vi on each Z. Moreover, if x = X x. E H, with

iEI E x. E H., and T = (T.) 2EI • 2 Z. (TX

y,

we have

I X) = X iEI

(T.X.IX.),

2

2

and so v , xi . This established, on Zi the measure v for a sub set Y of Z to be locally negligible for every Vx,x , it is necessary and sufficient that each set Yn zi be negligible for every vx . i.e. that each Y n zi be negligible for Vi n , in other works, that Y be locally negligible for V. Hence û is a basic measure on Z. 0 It can happen that the spectrum Z of an abelian C*-algebra of operators y carries no basic measure, even if Z is compact and second-countable (exercise 3 e). References : [9], [18], [45], [71], [89], [100], [101].

PART I, CHAPTER 7

132

3.

Structure of abelian von Neumann algebras. THEOREM 1. Let H be a complex Hilbert space, and Z an abelian von Neumann algebra in H. There exists a locally compact space Z, a positive measure y on Z, with support z, and an isometric isomorphism of the normed *-algebra Z onto the normed *-algebra V). If H is separable, we can require z to be compact and second-countable. Let Y be a sub-C*-algebra of Z, weakly dense in Z, whose spectrum Z carries a basic measure V [proposition 4, (i)]. By progosition 1, there exists an isometric isomorphism of Z onto Lc (Z, V). If H is separable, we can suppose that Z is compact and second-countable [proposition 4, (ii)]. D

Proof.

Conversely: THEOREM 2. Let Z be a locally compact apace, y a positive measure on Z, and H = 1,(Z, V). For fE 1, c (z, v), let T f E L(H)

be the operator defined by the formula Tfg = fg(gEH) be the set of the Tf's.

.

Let Z

(1) Z is a von Neumann algebra such that Z = Z'. (ii) The mapping f Tf is an isometric isomorphism of the normed *-algebra r.,(z, Y1 onto the normed *-algebra Z. Proof. It is clear that the mapping f T 4. is a *-homomorphism of ff(Z y) onto Z, and that m T. If I.f () I jf X < c on a non-y-negligible measurable set Y, -we have, with xy denoting the characteristic function of Y, ITfXYI Xx y , hence For g, h c H and f E L(Z, V) II II X. Hence H Tf = f

I

we nave

(5)

(Tfglh) =

I I

f(Og()h( .0dV(0

—E 1 and gh Lc(Z, V); hence the mapping f Tf is weakly continuous (for the weak topologies already considered). The unit ball of Z, the image of the unit ball of q(Z, v), is weakly compact, and so Z is an abelian von Neumann algebra. Let

U = Hn

v).

It is immediate that U is an abelian Hilbert algebra which is dense in H. Then, U(U) c U(U) = V(U) (chapter 5, theorem 1); by symmetry

U (U ) = V(U) = U(U) / = V(U) r . For f E U, uf = vf = Tf E Z, hence

U(U) c Z c Z i ,

and so

Zc U(U)

= U(U),

GLOBAL THEORY

and finally Z = U(U) = Z l .

133

D

Remark.

If y has support Z, 1_00(Z) may be isometrically identified with a *-subalgebra of q(Z, y). The image y of 1(Z), under the isomorphism f Tf, is a sub-C*-algebra of Z. The spectrum of y may be identifiet with Z, in such a way that the Gelfand isomorphism may be identified with the isomorphism f T F of 14,0 (Z) onto Y. If g, h EH, it follows from (5) that V g h is the measure with density (Radon-Nikodym derivative) gTi, wi h respect to V. Now, when q and h vary through L(Z, V), g h runs through the whole of 1,- (Z, v), and hence V is -a basic measure on Z. The mapping f 4- Tt of q(Z, V) onto Z is the weakly continuous extension of the Gelfand isomorphism defined by proposition 1.

t

In theorem 1, the specification of Z does not determine Z and V uniquely. Cf., nevertheless, Appendix IV. References : [9], [45], [71], [89], [100], [101]. 1. Let H be a complex Hilbert space, and y a commutative C*-algebra of operators in H. Let Z be the spectrum of Y, and V a positive measure on Z. For the Gelfand isomorphism to be continuous [14,0 (Z) being endowed with the topology induced by the weak topology of L'i (Z, v), and L(H) with the weak operator topology], it is necessary and sufficient that every measure Vx,x be absolutely continuous with respect to V. If this is the case, the Gelfand isomorphism admits a weakly continuous extension to Ifô C (Z V), and, for V to be basic, it is necessary and sufficient that this extension be injective.

Exercises.

2. Let Z be a locally compact space, y a positive measure on Z, and f Tf an isomorphism of the *-algebra f(Z, V) onto an abelian von iqeumann algebra Z. Let X be a measurable subset of Z, and E the projection of Z corresponding to the characteristic function of X. For Z E to be G-finite, it is necessary and sufficient that X be,up to a locally negligible set, the union of countably many integrable sets (i.e. measurable and of finite measure). (To see the sufficiency, reason as for proposition 3. To see the necessity, consider families of non-zero projections of Z, pairwise disjoint, majorized by E, corresponding to integrable subsets of Z, and apply Zorn's lemma to them.) 3. a. The property of a vOn Neumann algebra having a countable family of generators is invariant with respect to isomorphism (use chapter 4, theorem 2, corollary 2). b. If a von Neumann algebra A is isomorphic to a von Neumann algebra acting in a separable space, then A is G-finite and is generated by a countable family of elements. c. Let G-finite,

A be a von Neumann algebra, and Z its centre. If Z is and if A is generated by a countable family

134

PART I, CHAPTER 7

of elements, then A is isomorphic to a von Neumann algebra acting in a separable space. (Let x be a separatin element for Z; is isomorphic to the algebra induced by A in V; show that the A Hilbert space 4 is separable.) d. Adopting the notation of theorem 2, suppose v bounded. Show that the constantly-1 function on Z is separating and cyclic for Z. If, further, there exists a countable family M of elements of L(H) generating the von Neumann algebra Z, H is separable. (The images of the function 1 under the operators of M form a total set in H). Obtain from this an example of a 0-finite abelian von Neumann algebra which is not generated by a countable family of elements. (Take for Z a suitable product of copies of the real interval [0, 1], and for y the product of Lebesgue measure, in such a way that L6(Z, V) is not separable.) e. Adopting the notation of theorem 2, take for Z the interval [0, 1], endowed with the discrete topology, and for V the measure defined by assigning mass + 1 to each point. Show that Z is not a-finite, but is generated by a single element (for example, by the operator T where f is the function Let y be the sub-C*-algebra of Z generated by this operator Tf and I. Show that the spectrum y is compact and second-countable, but carries no basic measure. .

f. Let Z be an abelian von Neumann algebra in a separable Hilbert space. Show that Z is generated by a single hermitian element. [Form a sequence (E l , E2, ...) of projections generating Z. Obtain from it a family (F r) of projections of Z, generating Z, where r varies over the set of rational numbers in [0, 1] with the following properties: F o = 0, F 1 = I, r r l implies F < F I, lim F r = Fro. Construct an hermitian r r

r÷ro,rro

operator T such that F r is the greatest spectral projection for which F r T < rF r ' Show that T generates Z] [73]. g. Let H be a separable Hilbert space, A and B abelian von Neumann algebras in H, and C the von Neumann algebra generated by A and B. Then C is generated by a single element. (Use f.) be a family of weakly closed *-algebras in H, comh. Let D.) a muting pairwise, and D the von Neumann algebra generated by the If each DJ is generated by a single element, so is D. V. (Use !l and g) E313]. i. Let be a von Neumann algebra in H, the product of a family (EJ) of von Neumann algebras. For E to be generated by a single element, it is necessary and sufficient that each Ebe generated by a single element. (Use h). We do not know whether every von Neumann algebra in a separable Hilbert space can be generated by a single element, or even

GLOBAL THEORY

135

(Cf. nevertheless, part III, by a finite number of elements. chapter 3, exercise 8 and chapter 8, exercise 12.) 4. We adopt the notation of theorem 2. Every hE L(Z, V) defines a linear form (ph on Z by means of the formula cp (T ) = ff(c)h()dv(C). h f Show that the forms (Ph are just the forms Wx y on Z, as well as (Use being the ultra-weakly continuous linear forms on Z. theorem 1 of chapter 3). Compare with theorem 8 of chapter 6, and with part III, chapter 1, corollary of theorem 4, and chapter 6, section 3, remark 1 [19], [89], [100], [109]. 5. Let Z be an abelian von Neumann algebra. There exists a locally compact space Z and a positive measure V on Z with support Z possessing the following properties: a. the normed *-algebra Z is isomorphic to the normed *-algebra 1.,(Z, V); b. every function of q(Z, V) agrees locally almost everywhere with a continuous function. [Show that the space Z and the measure V constructed in the proof of proposition 4, (i) possess property b] [9], [45], [100], [110]. 6. Let U be an abelian full Hilbert algebra. There exists a locally compact space Z, and a positive measure v on Z with support Z, such that U is isomorphic to the Hilbert algebra T.,(Z, v) nLœ c (z, V). (Use theorem 1, and chapter 6, theorems 1 and 3).

■

CHAPTER 8.

DISCRETE VON NEUMANN ALGEBRAS

1.

A second classification of von Neumann algebras. DEFINITION L. Let A be a von Neumann algebra, and Z its centre. We say that A is discrete if it is isomorphic to a von Neumann algebra whose commutant is abelian. We say that A is continuous if there exists no non-zero projection E of Z such that the von Neumann algebra AE is discrete. The above classification is invariant with respect to isomorphisms and antiisomorphisms. A factor may be discrete or continuous. A discrete factor A is isomorphic to the algebra of all operators in a suitable Hilbert space. In fact, transforming A by means of an appropriately chosen isomorphism, we can suppose that A', which is, like A, a factor, is abelian. Then, A' reduces to the scalar operators, which proves our assertion. PROPOSITION 1. Let (Ai) i e i be a family of von Neumann algebras, and A the product von Neumann algebra. For A to be discrete (resp. continuous), it is necessary and sufficient that each Ai be discrete (resp. continuous).

Proof.

iC

2

are abelian, then

A'

=

R Al. is abelian. . 2 2E1 Moreover, if the Ai are replaced by isomorphic von Neumann algebras, A is replaced by an isomorphic von Neumann algebra. This shows that, if the Ai are discrete, then A is discrete. Conversely, if A is discrete, an isomorphism of A onto a von Neumann algebra B such that B' is abelian defines an isomorphism of each Ai onto a von Neumann algebra Bi such that 14: is abelian; hence the Ai's are discrete. If the

If an algebra Ai is not continuous, it is the product of a non-zero discrete von Neumann algebra and another von Neumann algebra; thus A is the product of von Neumann algebras of which one is discrete and non-zero; consequently, A is not continuous. On the other hand, if A is not continuous, there exists a nonzero projection E of its centre such that AE is discrete; let

PART I, CHAPTER 8

138

E = (Ei)iEI, where Ei is a projection from the centre of Ai; we • (at least) one of the (A.) have A = II (A ) is discrete i Ei ' 2 Ei i I E and non-zero, and hence one of the A's is not continuous. 0

In the same way that, from proposition 7 of chapter 6, we deduced proposition 8 and its corollary 1, we may deduce from proposition 1 the following results:

Let A be a von Neumann algebra, and Z its centre. Among the projections of Z, there exists a greatest projection E (resp. F) such that AE (resp. 747 ) is discrete (resp. continuous). We have EF 0, E + F = I. COROLLARY 1.

COROLLARY 2. A von Neumann algebra is canonically isomorphic to the product of a discrete von Neumann algebra and a continuous von Neumann algebra. PROPOSITION 2.

A discrete von Neumann algebra is semi-finite.

Let A be a discrete von Neumann algebra. Considering an isomorphic image of A if necessary, we can suppose that A' is abelian. Then, A t is finite, and hence A is semi-finite (chapter 6, proposition 13, corollary 1). LI

Proof.

COROLLARY 1. Let A be a von Neumann algebra, Z its centre, and E (resp. F) the greatest projection of Z such that AE (resp. AF ) is semi-finite (resp. discrete). We have E F. COROLLARY 2.

A purely infinite von Neumann algebra is con-

tinuous. Remark. Since a discrete factor is isomorphic to an algebra L(H), we already know from very elementary considerations (chapter 6, theorem 5) that a discrete factor is semi-finite. In chapter 9, we will make use of the following result:

Let A be a factor, and ()) a faithful normal trace on At If there exists a strictly decreasing infinite sequence of projections En of A such that ci(En) < + 00 for every n, then A is continuous. PROPOSITION 3.

Proof. Suppose that A is discrete. Replacing A by an isomorphic algebra if necessary, we can suppose that A = L(H), where H is a suitable Hilbert space. By theorem 5 of chapter 6, the E n (H)'s are finite-dimensional, from which it fol lows that the infinite sequence (En ) cannot be strictly decreas ing. 0 References : [10], [42], [43], [65], [100], [117].

2.

Abelian projections.

Let A be a von Neumann algebra, and E a non-zero projection of To say that a von Neumann algebra reduces to just the scalar operators amounts to the same thing as saying that it contains

A.

GLOBAL THEORY

139

no projections apart from 0 and I; therefore, to say that AE reduces to the scalar operators is the same as saying that the only projections of A majorized by E are 0 and E.

Let A be a von Neumann algebra, and E a projection of A. We say that E is minimal (relative to A) if E / 0 and if every projection of A majorized by E is equal to 0 or to DEFINITION 2.

E. We will generalise definition 2 as follows:

Let A be a von Neumann algebra, and E a projection of A. We say that E is abelian (relative to A) if the algebra AE is abelian. DEFINITION 3.

Let Z be the centre of A. For every projection E of A, the centre of AE is ZE (chapter 2, corollary of proposition 2). Since a von Neumann algebra is generated by its projections, we see that a projection E of A is abelian if and only if every projection F of A majorized by E is of the form EG, where G is a projection of Z (that we can then take to be equal to the central support of F). It follows from this that an abelian projection E is minimal among the projections of A which have the same central support as E. In particular, if of A is minimal.

A

is a factor, a non-zero abelian projection

Let A be a von Neumann algebra in H. ing six conditions are equivalent: THEOREM 1.

(i) [resp. (ii)]

The follow-

A (resp. A') is discrete;

[resp. (iv)] Every non-zero projection of A (resp. A') majorizes a non-zero abelian projection of A (resp. A'); (iii)

(y) [resp. (vi)] There exists an abelian projection of (resp. A') whose central support is I.

A

Proof.

(iii) We are going to establish the implications (i) (iii) (v) (ii) =*. (iv) (vi) (i) . The implications (i) (ii) (iv) [resp. (iii) =0. (v) and (iv) =0. (vi), (v) and (ii) can be seen by interchanging are and (vi) (i)] equivalent, as the roles of A and A'. It thus suffices to establish the implications (i) = (iii) (v) = 0' (ii). (i) = (iii): We suppose that A' is abelian and we need to show that every non-zero projection E of A majorizes a non-zero projeqion of A. Let x be a non-zero element of E(H), and F = E A; we have F E. The algebra A. is abelian and possesses the cyclic vector x, and so (chapter 6, proposition 4, corollary 2) AF is abelian. Hence F is an abelian projection of A majorized by E. (iii) (V): Let Z be the centre of A ; let (Ei) i ci be a maximal family of non-zero, pair-wise disjoint, projections of Z,

140

PART I, CHAPTER 8

possessing the following property: for every iEI, there exists an abelian projection Fi of A whose central support is Ei; if every non-zero projection of Z majorizes a non-zero abelian projection of A, we have X E . = I, and so F = F is a projeci iEI iEI tion of A with central support I. Moreover, A = 11 A F iEI F• and hence F is abelian. ,

(ii): Let F be an abelian projection of A with central (v) I. support Then, AF is abelian, hence A. is discrete, hence A l , isomorphic to i/q, (chapter 2, proposition 2), is discrete.

If A is continuous, then A' is continuous.

COROLLARY 1.

If A' is not continuous, there exists a non-zero projection E of Z = An A' such that 7tk is discrete. Then, AE is Proof.

discrete, and so COROLLARY 2.

Proof. If crete.

A

A

is not continuous.

0

An abelian von Neumann algebra is discrete. is abelian,

A'

is discrete, and hence

COROLLARY 3. Let A be a factor in H. ditions are equivalent:

is dis-

The following five con-

A (resp. A') is discrete; [resp. (iv)] A (resp. A') possesses minimal projections;

(i) [resp. (iii)

A

0

(ii)]

(17) There exist Hilbert spaces H1, H2 and an isomorphism of H onto H 1 0 H2 which transforms A into L(H 1) 0 cH 2 and A' into cH l 0 L(H 2 ). Proof. Conditions (i), (ii), (iii) and (iv) are equivalent by theorem 1. Condition (v) plainly implies that A and A' are discrete. Finally, suppose that A is isomorphic to an algebra L(H1 ). There exist (chapter 4, theorem 3) a Hilbert space K and a projection E / E (L(4 ) 0 C ) 1 = c 1

K

H1

0 L(K)

such •that A is spatially isomorphic to (L(H i ) 0 CO E '. We have E l = I 0 E, where E is a projection of L(K). Putting H2 = E(K), we see that A is spatially isomorphic to OH]) 0 CH 2 . 0 Condition (iii) of corollary 3 thus furnishes an "internal" characterisation of the von Neumann algebras isomorphic to algebras L(H). Corollary 2 expresses the fact that an abelian von Neumann algebra A is isomorphic to an abelian von Neumann algebra B such that B' is abelian, i.e. such that B l = B, something we also know from theorems 1 and 2 of chapter 7.

GLOBAL THEORY

141

Instead of "abelian" projections, we sometimes speak of "irreducible" projections. References : [6], [10], [42], [43], [65], [100], [117].

3.

Discrete algebras and elementary operations.

We already know that the commutant of a discrete (resp. continuous) von Neumann algebra is a discrete (resp. continuous) von Neumann algebra, and that a product of discrete (resp. continuous) von Neumann algebras is a discrete (resp. continuous) von Neumann algebra. PROPOSITION 4. Let A be a discrete (resp. continuous) von Neumann algebra, and E a projection of A or of A!. Then, AE is discrete (reap. continuous). As A' is discrete (resp. continuous), it is enough to consider the case where Ee A t . There then exists a projection F of the centre of A such that A E is isomorphic to AF . Now, A = AF x Ai_ F , and hence AF is discrete (continuous) by proposition 1. 0

Proof.

PROPOSITION 5.

bras.

Al

Then,

®

Let Al and A2 be discrete von Neumann algeA2 is discrete.

If we replace A l and A 2 by isomorphic von Neumann algebras, A l 0 A 2 is replaced by an isomorphic von Neumann algebra (chapter 4, proposition 2). We can therefore suppose that Ai and A are abelian. Then Ai 0 A is abelian. Now

Proof.

(A Hence

A1

0

1 0

A2 ) A2

1 =

A'1

0

A2

is discrete.

(chapter 6, proposition 14).

0

In part III, chapter 3, we will undertake a deeper study of the structure of discrete von Neumann algebras (Cf. also part III, chapter 8, corollary of theorem 2). Reference

4.

: [65].

Definition of types.

The purely infinite von Neumann algebras are called type III algebras. The continuous semi-finite von Neumann algebras are also called type II algebras, and, more precisely, type Ill if they are finite, or type Moo if they are properly infinite. The discrete algebras are also called type I algebras. For (a. von Neumann algebra A to be type I (resp. type II, III) it is necessary and sufficient that A' be type I (resp. II, III). Let A be a discrete factor, K a complex Hilbert space such that A is isomorphic to L(K), and n the dimension of K. Then, n is equal to the cardinality of every family of pairwise disjoint

142

PART I, CHAPTER 8

minimal projections of A of sum I. Hence n is an algebraic invariant of A. We say that A is type In . If n is an infinite cardinal, we also say that A is type I. We will generalise these terms in part III, chapter 3. References : [ 42], [43], [65].

Complete Hilbert algebras and type I factors. LEMMA 1. Let U be a (non-zero) complete Hilbert algebra. Then U, U(U) and V(U) possess minimal non-zero hermitian idem5.

potents.

Proof. a. Let E be the set of non-zero hermitian idempotents of U. We show that E is non-empty. There exists in U a nonzero hermitian element a. Then Ua is non-zero hermitian. There exists TE U(U) such that TU0, is a non-zero spectral projection of U . Since TU a = UTa , Ta is a non-zero hermitian idempotent of U. b. The mapping (x, y) 4- xy of U x U into U is separately continuous. As U is complete, this mapping is continuous (r[3]], chapter III, section 4, proposition 2). There therefore exists a constant M > 0 such that Hxy II mil x H. H y H for any x, y E U. If e E U is a non-zero idempotent, we have 1I e 2 11 /4-1 . e II M il e 1 2 , hence li e H c. Let a We have a inf 11 e H 2 M -2 . There exists an H

•

ecE e E E such that H e 11 2 < 2a. e1e2 = e 2 e 1 = 0, then 2 (e 1 1e 2 ) -- (e 1 le 2 )

If e = e l

(e 1 1e 1 e 2 ) =0, hence 2a

e2 with e l , e 2 E E,

Ile 1 II 2 + I1e 2 11 2 Ilell 2 < 2a,

a contradiction. Hence e is a minimal element of E. Every nonzero projection of U(U) majorized by ue is of the form ue , for some e / EU; then, e' is an element of E majorized by e, whence e e and Ue l = Ue •' hence U e is a minimal projection of U(U). 0

Let U be a full Hilbert algebra such that The following conditions are equivalent: (i) U is complete; (ii) U(U) is type 1; (iii) following multiplication of the norm of U by a constant, U is isomorphic to the Hilbert algebra of Hilbert-Schmidt operators in a suitable Hilbert space. PROPOSITION 6. U(U) is a factor.

GLOBAL THEORY

Proof.

(iii)

(i) (ii): theorem 1. of

143

(1): obvious. this follows from lemma 1, and from corollary 3

(ii) (iii): since U is full, a -*Ua is an isomorphism of the Hilbert algebra U onto the Hilbert algebra of the HilbertSchmidt elements of U(U) endowed with the natural trace (chapter If U(ü) is type I, there exists a Hilbert space 6, theorem 1). K and an isomorphism of U(U) onto L(K). This isomorphism transforms the natural trace on U(U) -1- into a trace on L(K) + , which is a multiple of the trace defined in theorem 5. Hence, after multiplication of the norm of U by a constant, the Hilbert algebra U is isomorphic to the Hilbert algebra of Hilbert-Schmidt operators in K. 0

Remark.

It follows from considerations of dimension that the Hilbert space of proposition 6 (iii) is completely determined by U, up to isomorphism.

Let ü be a complete Hilbert algebra. Let (Ei) be the family of minimal projections of U(ü) n V(U). (i)The Hilbert space U is the direct sum of the üi = E(U). PROPOSITION 7.

(ii)The Ui are self-adjoint two-sided ideals of ü which annihilate one another pairwise. (iii)After multiplication of the norm of Ui by a constant, Ili the Hilbert algebra of Hilbert-Schmidt opera- isomrphct tors in a suitable Hilbert space. (iv)If U(U) and V(U) are finite von Neumann algebras, each Ui is finite-dimensional. Proof.

It is clear that the E.2 are pairwise disjoint. Let 0, B = F(U) is a non-zero comE =E.2' and F = I - E. If F possesses a minimal projecU(B) plete Hilbert algebra. Hence tion G (lemma 1). The central support G / of G is a minimal projection of U(B) n V(8) . Hence there exists in U(U) n V(U) a minimal projection disjoint from the E's, which is impossible. Hence F = 0, which proves (i). Assertion (ii) follows from chapter 5, proposition 8. Assertion (iii) follows from proposition 6. Suppose that U(U) is finite. For every i, U(U) is a finite factor, is type I (proposition 6) and hence is type In for some n < + co, and is therefore a finite-dimensional vector space; this proves (iv). 0

The closed two-sided ideals of U are direct sums of some of the Ui's. The minimal (resp. maximal, different from LI) non-zero closed two-sided ideals are the Ui 's (resp. the U e Ui's). Proof. The direct sum of a selection of the U_.'s is plainly COROLLARY.

a closed two-sided ideal of U.

On the other hand, if I is a

144

PART I, CHAPTER 8

closed two-sided ideal of U, we have P I E U(U) nV(U) (chapter 5, proposition 8); hence, for every i, PI majorizes El: or is disjoint from it; hence I is the direct sum of those Ui which it contains. The rest of the corollary is obvious. 0 References :

[1], [148], [[9]].

Exercises. 1. Let A be a continuous von Neumann algebra in K, and E a non-zero projection of A. Show that E(K) is infinte-dimensional. [If dim E(K) < + 00, AE is discrete.] 2. Give an alternative proof of the implication (i) (v) in corollary 3 of theorem 1. [If A is isomorphic to L(K), there exists in A a family (Ei)i cI of minimal projections, pairwise disjoint, such that • X E= I, and U-2x such that U* 2x = E-2' 2xU2E1 2 U.2x U* 2x = E. use proposition 5 of chapter 2.] 3. Let A be a discrete factor in H, and B a von Neumann algebra contained in A'. If the von Neumann algebra C generated by A and B is L(H), we have B = A'. [If S / A', there exist in A', which is isomorphic to an algebra L(K), non-scalar operators commuting with B, and therefore with C] [65]. (Cf. part III, chapter 7, exercise 13 d. ) 4. Let A be a von Neumann algebra, Z its centre, E a minimal projection of A, and F its central support. Show that F is a minimal projection of Z, and that AF is a discrete factor. Deduce from this that the supremum of the minimal projections of A is a projection G of Z, and that AG is a product of discrete factors.

5. a. Let Z be an abelian von Neumann algebra, and x a normal character of Z. Show that the support of x is a minimal projection of Z. Let Z be the interval [0, 1], V Lebesgue measure on Z, H V), and Z the von Neumann algebra of theorem 2 of chapter 7 . Show that no character of Z is normal (use a) [7 o].

b.

6. Let A be a von Neumann algebra possessing the following property: (M) Every decreasing sequence of projections of A is stationary. a. Show that every non-zero projection of minimal projection.

A

majorizes a

b. Deduce from a that I is a finite sum of minimal pairwise disjoint projections. c. Show that A is the product of a finite number of factors possessing property (M). (Apply b to the centre of A.) d. Show that a factor possessing property (M) is a finitedimensional factor. (use b.)

GLOBAL THEORY

145

e. Show that (M) is equivalent to the following property: (Use c and d.) the vector space A is finite-dimensional. 7. a. Let A be a von Neumann algebra in H, (En ) = (Px n ) an infinite sequence of non-zero, pairwise disjoint projections of A. Let xn be an element of Xn such that II xn II = 1. Let U be a non-trivial (i.e. free) ultrafilter on the set of non-negative integers. For TE A, put OT) = lim u (Txn lxn ). Show that cp is a linear form on A which is continuous for the norm topology but not for the ultra-strong topology. [Consider the projection Fn which is the supremum of En+ i, En+ 2, ...; Fn converges ultrastrongly to zero and cp(Fn ) = 1.]

b.

Show that a von Neumann algebra which is an infinite(use dimensional vector space is a non-reflexive Banach space. a, exercise 6 and chapter 3, theorem 1.)

M

CHAPTER 9.

EXISTENCE OF DIFFERENT TYPES OF FACTORS

1.

A lemma. LEMMA 1. Let A be a von Neumann algebra, B a von Neumann subalgebra of A, maximal abelian in A, cp a faithful normal trace on For Tc m 1/2, let KT be A+, and m the ideal of definition of the convex hull in A of the set of operators uTu-1, where u varies over the set of unitary operators of B. Let Kr be the weak closure of KT . Then, Kr c m and K n B is non-empty. Proof. Recall that .(1) defines a pre-Hilbert structure on 0, and that II • 112 denotes the corresponding norm. We can clearly confine attention to the case where II T 1. < 1 and H T 11 2 Then, for every unitary operator U of B, we have

H

II

uTu-1 H

1

and

H

UTU-1 11 2

1.

If we denote by Al the unit ball of A (for the norm H . 11), and by K ,the unit ball of the pre-Hilbert space 17115 , we thus have KT c Ai n K. Moreover, (I) = wx . (chapter 6, corollary of propoiEI sition 2). Hence cp(S*S) = 11Sx.112 is a function of S which iEI

is lower semicontinuous for the weak topology on A. Thus i, K is weakly closed; hence iC c A K, and, in particular, KT' c T

1

The set K is invariant under the group G of transformations 1 T S uSU- A-, where U varies over the set of unitary operators of B; hence the same is true forThe set WI, is weakly compact, attains its minimum over q at and hence the function S an element TnE This element is unique because, as cp is Lu T faithful, ni 2 ris strictly convex for the norm 11 . 11 2' Hence T 0 is a fixed point for G, i.e. To commutes with the unitary operators of B; thus, T 0 Br n A = B (since B is maximal abelian) and finally T o E K, n B.

Ç.

s r2

References : [15], [78].

147

148

PART I, CHAPTER 9

Construction of certain von Neumann algebras. Let H be a complex Hilbert space and A an abelian von Neumann algebra in H. Let G be a discrete group with neutral element e, 2.

and let s Us be a unitary representation of G in the space H. Suppose that UV-Aus = A for every s E G. It follows from this that, for every s E G, the mapping T Us-1TU5 = TS is an automorphism of the von Neumann algebra A. For every s E G, let Hs be a Hilbert space of the same_dimension as H, and Js a linear isometry of H onto H. Let H be the direct sum of the Hs 's for s c G. In the manner of chapter 2, section 3 , we will represent every element R of L(H) by the matrix (R s, t)SEG,tEG, where

s, t

= J*RJ EL(H). s t

For every T e A let (1)(T) be the element of LA defined by the following matrix (Rs, t) : R s, t = 0 if s ti Rs .tp = T for every S E G; in other words, 4)(T) is the element of L(H) reduced by the H s 's such that J(T)J5 T for every B E G. The mapping (1) is an isomorphism of the von Neumann algebra A onto an abelian von Neumann algebra X c L(H) . For every y E G, let Uy be the element of LU-1) defined by the following matrix (Rs, 0 Rsa, t = 0 if st -1 / y; Ryt, t = Uy for every t G; in other words, Uy is a unitary operator which permutes the H s 's among themselves, and which, more precisely, maps Hs into Hys , with Jps Uy Js = Uy. Thus, U2 Uy maps Hs to H2y s, with J*

U

ZyS 2

y

J =J* U J J* U J ZyS s ys ys y s S

=

UU

2 y

=U . 2y

Hence ".1

U U =U z y zy

(1)

Moreover, U7,4(T)U„ is reduced by each Hs , and a a -

J*(u

s

-1

y

(1)(T)175 )J =(J*17J

Y

8

S

I J )(J* (D(T)J )(J* U J ) =U _ 1 TU = TY; ys ys y s

Y -j- YS YS

thus --1

(2)

U

(1) (T)U =

The operators of the form

(1)(T )u 1 yl

+ cD(T,)u

+ Y2

constitute a *-algebra Bo of L(T1).

+ cD(T)U n Yn Indeed,

149

GLOBAL THEORY

(4)

( T*) =

(T)Z )* =

and IT(T )Z (1)(T )77 = (I)(T )0(TY 1 1 )Z rL7 1 yi 2 y2 1 2 Y1 Y2

= 4)(T

-1 )u 12

Let B be the von Neumann algebra generated by B o , i.e. the weak closure of B. 0 The algebra A is an abelian von Neumann subalgebra of B.

We compute the matrix (R

st -y, and R

= J* yt

t

s ,t ) of

We have R

4)(T)ZT .

= 0 if

j = ( J* (D(T)J )(J* rLJ J ) = TU . y t yt yt yt y t

There thus exists a family (Ty ) _.2iEG in A such that R s, t Tst-lu s t-1. This property is therefore also possessed by the operators of Bo , and is preserved under passage to weak limits: every

element S

B is represented by a matrix of the form (T3t_1U5t-1)' with Ty E A for every y E G. In particular, for S E B+, we have E

, T = J*SJ E + .

e e

Let 4) be a serrti-finite faithful normal trace on A+, which is invariant with respect to G. For every S = (Tst_ iu E B+, put 11)(S) = cp (T e ) . Then, IP is a semifinite faitgrul normal trace on B. This trace is finite if and only if (1) is finite. For TE A+, 11)(4)(T)) = Proof- For S E B ±, S i E 13+ , X 0, we immediately have PROPOSITION 1.

11)(S + S ) = 4(S) + Ip(S ), 1 1 We show that 4(SS*) = tp(S*S) We have

i(XS) = X1P(S).

for S c B. Let S = (T -1U _1) •

st

S* = ( U* _1T* _1).

ts

ts

Put SS * =

(R

st

_1 U

st

_1 ) ,

S *S =(R'lu_1) .

st

st

We have R e = X T 1U t 1U

tEG

R

X Unr tU t , tEG

lT

1 = X T tal,

tEG

st

150

PART I, CHAPTER 9

where the series converge in the sense of the strong topology. All the terms of these series a -Je in A+ . Hence

(S*S) = OR') = Y (U*T*T U ) = t t t

tEG

e

y tEG

OT tT*) t

=

e

) --tp(SS*).

Hence 'q) is a trace, and the normality of cp immediately implies that of 1p. We show that 11) is faithful; let S = (T and suppose that i(S) = p(Te) = 0; since cp is faitftul, ° it follows from this that T e = 0; for every xe Hs , we therefore have

I

L-

2

= (Sxlx) = (SJ J*xl J o,Thcx) = (T J*0IJ*x) = 0, 0 e s s 1 1 hence Six = 0; thus, Si = 0 and, consequently, S = 0. We show that 11) is semi-finite; the operator 'HE A is the supremum of an increasing filtering family of operators Tie A+ such that (1)(Ti) < + co; the formula 4)(0(T)) = (1)(T) being obviously valid, the operator IRE S is supremum of the increasing filtering family (0(Ti)), and 11)( (D(Ti)) < + 00 . Finally, it is clear that tp is finite if and only if cp is finite. E S 2 x II

PROPOSITION 2. Suppose that X is a maximal abelian von Neumann subalgebra of B. For B to be semi-finite, it is necessary and sufficient that there exist on A+ a semi-finite faithful normal trace on A+ which is invariant with respect to G.

Proof.

The condition is sufficient (proposition 1). Conversely, suppose that B is semi-finite. Let ip be a semi-finite faithful normal trace on 13+. This trace induces a faithful normal trace on A+. There exists a faithful normal trace (I) on A+ such that OT) = 012(T)) for TE At This trace is invariant with respect to G; in fact, for TE A+ and yE G, we have, in view of (2) -

1

)=

-1 (TY) = 11)(4)(TY)) = Ipai -10(T)Zi ) =00(T)) -

All the difficulty lies in showing that (1) is semi-finite. To achieve this, we will first establish the following result, if S = (T _iu -1) E B is such that cp(s*S) < + we have ot st (1)(TVe ) < + 00 . In fact, for every T1 = 0(T) E X and every S E G, we have

J*T SJ = TJ*SJ = TT = TT = J*SJ T = J*ST J s 1 s s s e s s s 1 s e (using for the first time, in an essential way, the fact that A is abelian). In particular, if T1 is a unitary operator of A, we have, replacing S by STi l in the above equality, -1 J*(T ST )J = J*SJ . 0 1 1 0 s 0

151

GLOBAL THEORY

The operator JARJs thus remains the same when R runs through the convex set Ks generated bx the operators T1ST-1 (for T 1 a variable unitary operator of A), and therefore when R runs through the weak closure K' of K3 . Now, A is a maximal dbelian von Neumann subalgebra of B. By lemma 1, KL contains an element S o of X such that 11)(S5S 0 ) < + 00. The relation

J*S s Js = J*SJ s s = Te implies that So = (D (Te), which proves our assertion. The operator Iv is a strong limit of operators SE B such that V(S*S) < + co. ft S = (T 4. _1U Te = JpJ e converges • < q 'co. By chapter 6 (proposition 1, strongly to and (1)(W:Y corollary 2), this proves that (I) is semi-finite. 0

Suppose that A is a maximal abelian von Neumann subalgebra of L(H). Suppose, moreover, that An Au = 0 for y / e. is a maximal abelian von Neumann subalgebra of B. Then, Proof. Let S = (T * _1U _1) be an element of B which commutes LEMMA 2.

ag

with A. For every TsErA, thus have S(T) = 4)(T)S, whence, a matrix calculation yields, TT t Since A T s -1t.jst -1 = T st lu • is m ximal abelian, this implies rnat Ty Uy E m, i.e.a Ty E Au for every y E G. Since, moreover, Ty E A, the second hypothesis of the lemma implies that Ty = 0 for y / e. We then have S = 4)(Te ) E. D LEMMA 3. To the hypotheses of lemma 2, add the further hypothesis that the only elements of A invariant with respect to G are the scalar operators. Then, B is a factor. Proof. Let S be an element of the centre of B. We have SEX by lemma 2, and hence S = (1)(T) every y E G, we have

for some TEA. Moreover, for

IT(T) = ri5-111)(T)U = UTY),

hence T = TY.

By the hypothesis of the lemma, T is a scalar operator, and hence S = (I0(T) is a scalar operator. 0 References : [65], [78], [210], [211], [224], [231], [232], [235], [236], [237], [246], [247], [275], [276], [290], [392], [441].

3.

Examples taken from measure theory.

We retain the notation of section 2. We are now going to put ourselves in a situation where the conditions of lemmas 2 and 3 are satisfied. Let Z be a locally compact space, countable at infinitx, and V a positive measure on Z. Take for H the Hilbert space T.,(Z V). C

152

PART I, CHAPTER 9

For every fE IZ(Z, V), let Tf be the element of L(H) defined by Tf(g) = fg(gE H). The set of the Ts is a maximal abelian von Neumann subalgebra A of L(W (chap'ter 7, theorem 2). Let, moreover, G be a discrete group which acts on the left on For every SE G, suppose that the mapping C sC(CE Z) is a homeomorphism of Z onto Z which transforms V into an equivalent measure. (Thus, v is "quasi-invariant" with respect to G.) For rs (C) on Z, every SE G there therefore exists a function C which is > 0, finite, locally V-integrable, and such that Z.

dv(sc) = rs We have, for

s, t

E

(c )dv( c )

G,

rst (c)dv(c) = dv(stc) = r s (t)dv(tc) = r s (tc)r t (C)dv(C), hence rs t(C) = rs (tOrt(c) almost everywhere on Z. Setting t = s -1 , we deduce from this that I = r5 (s -1C)r 8 -1 (C) almost everywhere on Z. For g

H and s

E

E

G, put

(u s g)(c) = r s _1(0 1/2g(s -1 0. We have

f and hence U

2 -1 2 i(U g)(C)I dV(C) = fr 5 -1(C)Ig(s )1 dV(C) s 2 -= flg(s 1 )1 2dv(s 1 0 =fig(C)I dV(C), is an isometry; furthermore,

-1-1 (usu tg)(c) =r5_1(c) 1/2(u tg)(s ) =r 5 _1() 1/2rt_1(s 1 0 1/21 g(t s 0 _ _ =r 5 t _ 1 () 1/2g(t ls = (u g )(c); st so that s

f Elf°c (Z,

Us is a unitary representation of G in H. V )„ we have, moreover,

(U 1T u5 g) s f

=

r

1/2 (Tfusg)

For

(sc) = rs (c) 1/2 f(sc) (u sg) (s)

= r 5 (0 1/2f(s)r 5- 1(s0 1/2 00 = f(s)g(), hence, putting f5 (r) = f(sC),

-1

Us TfUs = Tfs . Thus, the conditions of the beginning of section 2 are satisfied, and therefore so is the first hypothesis of lemma 2. We now make the following additional hypothesis:

GLOBAL THEORY

153

(4 ) For every element s / e of G, and every non-negligible measurable subset z' of z, there exists a non-negligible measurable subset z" z', such that z" n sz" = c. We shall show that then An Au = 0 for every y E G different Y from e. An element of An Au is of the form Tf, with tE Lœc (z, v), and of the form TorUy , with gE LZ, V). Let Xy denote the characteristic funaion of a subset Y of Z. Let X be the set of ce Z such that PC) / O. If X is non-negligible, there exists a measurable non-negligible subset X' of X of finite measure such that X' nyX 1 = cp. We have

A Xf

= Tfxx , = TUX, = g(Uy Xx l)

almost everywhere; now Uy xxlvanishes almost everywhere on X I ; hence no( ' = 0 almost everywhere, which is a contradiction. Hence X and consequently f are negligible, so that Tf = O. Suppose, finally, that:

(**) G is ergodic in Z. Then, let Tf be an element of A invariant with respect to G. For every se G, we have Tfs = U-S-1TfU5 = Tf, hence f = fs almost everywhere. By the ergodicity of& G, there exists a constant equal to f almost everywhere. Hence Tf is a scalar operator. Thus, when the hypotheses (*) and (**) are satisifed, lemmas 2 and 3 are applicable, and B is a factor. The hypothesis that Z is countable at infinity merely serves to simplify the discussion somewhat. References : [65], [78].

4.

Existence of different types of factors.

We retain the notation of sections 2 and 3, and we suppose that hypotheses (*) and (**) are satisfied.

Suppose that V is invariant with respect to G, V((0) = 0 for every cc Z, and 0 < V(Z) < + [resp. V(Z) = + 00 ]. Then., S is a type II (resp. II,„) factor. Proof. For every element T of A+ , put PROPOSITION 3.

f

(Tf.) = ff(C)dV(C). It is clear that cp is a faithful trace on A. The invariance of V with respect to G implies the invariance of cp with respect to G. The trace (I) is normal, because the isomorphism f Tf is compatible with the natural orderings in V) and A; and, if (fa) is an increasing filtering family of positive functions in f(Z, v) with supremum fe Lœ c (Z, V) [in the sense of the natural

154

PART I, CHAPTER 9

ordering in el) (Z, V)], we know that ff(c)dv(c) is the supremum of ffla (c)dV(c). If V(Z) < + 00 , it is clear that (I) is finite. If v(Z) = + 00, the function 1 on Z is the supremum [in the sense of Ti(Z, V)] of an increasing filtering family of positive functions fEq(z, V) such that Lf(C)dV(C) < + 00 , and hence 4) is semifinite but not finite. Then, there exists on 8+ a semi-finite faithful normal trace, which is finite if and only if v (Z) < +00 (proposition 1). Since B is a factor, we see that B is semi-finite, and finite if and only if v(Z) < + 00. To complete the proof of the proposition it therefore suffices to show that B is not discrete. Now, the hypotheses V(Z) > 0, v((0) = 0 for every C EZ, imply the existence of a decreasing sequence (Zn ) of measurable subsets of Z such that + 00 > V(Z i ) > V(Z2) > ... (consider a point in the support of V, and open neighbourhoods of this point whose measures converge to zero). The xz define strictly decreasing projections En of A such that On ) < + on. The (10 (En) form a strictly decreasing sequence of projections of B such that

11)(11 (E )) < + co. n

Hence-B is not discrete (chapter 8, proposition 3). THEOREM 1.

There exist type Il l factors and type I'm factors.

Proof.

Take for Z the one - dimensional torus (resp. the real line) with its usual topology, for v the Haar measure on Z, and for G a dense subgroup of Z (G being regarded as a discrete group). Make G act on Z by translation. The measure v is invariant with respect to G, we have 0(0) = 0 for every CE G, and 0< v(Z) < + co [resp. v(Z) = + co]. It is a classical result that G is ergodic in Z. Finally, let y be an element of G different from e, and let Z' be a non-negligible measurable subset of Z. There exists a neighbourhood V of e in Z, so small that V n y v = . Then, if s is an element of G such that Z" = Z' nsv is non-negligible, we have Z u nyZ" = (I). The factor B is then type II I (resp. We return to the general situation of section 3, with the hypotheses (*) and (**).

If the subgroup Go of G consisting of the elements which leave y fixed is distinct from G and ergodic in z, then B is a type III factor. Proof. For every element Tf of A+, put PROPOSITION 4.

$(Tf) = As in the proof of proposition 3, we see that cp is a semifinite faithful normal trace on A+ which is invariant with respect to Go . Arguing by contradiction, we shall suppose that

GLOBAL THEORY

155

13 is semi - finite. There then exists on A+ a semi-finite faithful normal trace (I) / which is invariant with respect to G' (proposition Put w = 0 0'. There exist elements Tg , Tg , of A+ such 2). that

f

)= w(TfT ), g

0 / (Tf.) = w(T T I ) f g

for every TfE A+ (chapter 6, theorem 3).

For

SE

Go, we have

-1 ) = 4)(T (T ) =w(T T ) w(T T ) =(.)(T U T U ) =w(UT u 1 T) = Orr f 5-1 fs gs s fs g f f g f gs for any TfE A+, hence gs = g almost everywhere. As Go is ergodic, we conclude from this that g is equal almost everywhere to a constant > 0; similarly for g l . Hence 4) is proportional to (1) t and, consequently, invariant with respect to G, which is contrary to the hypotheses. D THEOREM 2.

There exist type iii factors.

Proof. Take for Z the real line, for V Lebesgue measure, and for G the group of transformations of Z of the form C± aC + b, where a and b are rational, and a > 0. The subgroup G o is obtained on setting a = 1; it is ergodic in Z. We easily see that the hypothesis (*) is satisfied. It then suffices to apply proposition 4. 0

Remark.

The type 111, II III factors that we have just constructed act in separable Hilbert spaces, under the condition that a countable group G is taken in the proof of theoremj. There exist other procedures for the construction of factors. (Cf. part III, chapter 5, exercise 8, part III, chapter 7, section 6; and [77], [78], [296], [315], [316], [317].) The theory of C*-algebras also furnishes examples. References : [13], [65], [78]. We shall see (part II, chapter 6, section 2) that the study of arbitrary von Neumann algebras reduces, to some extent, to the study of factors. The type I factors are well known (chapter 8, corollary 3 of theorem 1). The type II w factors are tensor products of type 111 fadtors and type I factors (part III, chapter 8, exercise 11). Unfortunately, a classification of type II or type III factors is not available.: herein lies one of the main outstanding problems of the theory. In separable Hilbert spaces, nine pairwise non-isomorphic type II factors (part III, chapter 7, section 7, theorem 4; and [309], [441], [443], [ 4 6 3 ], [474]), and continuous families of pairwise nonisomorphic type III factors ([171], [310], [411], [435]) are known to exist.

Exercises. cD t (T), U t and

1. W be

We adopt the notation of section 2. Let the elements of L(P) defined by the matrices

156

PART I, CHAPTER 9 3 -1

), (6 - 1

(6 s,t T

4TH) and (6

Kronecker deltaY.

s,t

-1U ), where 6 8,t is the s

a. Show that y ÷ U P* is a unitary representation of G in that W is a unitary operator such that W

2

= 1,

W( T )W = (13.' ( T) ,

P,

wu w = u t* ,

and that CP(T) E

Bt

and

b. Suppose henceforth that A' = A. Let SE L(P) commute with the t^7 1 and (1) 1 (T). Show that S and WSW are represented by the matrices (T 1 ) and (Us Ts _ it u; 1 ), where T is for every st--Lu Y y E G, an element of A. Show that two elements of L0-7), one commuting with the (D(T) and the Uu , the other commuting with the (1) 1 (T) and the U / Deduce from this that B' is the von Neumann comute(sbl. algebra generated by the 11 / ( T) and the L.J, and that B / = WBW [13], [65], [78]. C.

2. We adopt the notation of section 2, and suppose that the conditions of lemma 2 are satisfied. Show that, if T is a minimal projection of A, (1)(T) is a minimal projection of B. [A projection of B majorized by (1)(T) commutes with X.] Use this to obtain some examples where the construction of section 3 leads to type I factors [65]. 3. Show that there exist *-algebras of operators B, C in a complex Hilbert space, isomorphic for the *-algebra structure, whose weak closures are respectively finite and properly infinite. (Take for B and C the algebras of exercise 6 b of chapter 2, A and A' being finite factors in an infinite-dimensional Hilbert space.) 4. Show that there exists a type 111 factor possessing a cyclic element and acting in a non-separable space. (In the proof of theorem 1, take G to be uncountable. The cyclic element is the transform by one of the Js 's of the function lE H.) Show that this factor is G-finite (like every finite factor) but is not generated by a countable family of elements. (A von Neumann algebra generated by a countable family of elements which possesses a countable cyclic set acts in a separable space.) 5. In chapter 7 (exercise 3), we studied the following properties that a von Neumann algebra is capable of possessing: (i) of being G-finite; (ii) of being generated by a countable family of elements. We now propose to study some related properties. Let

A

be a von Neumann algebra in

H, and let Z be its centre.

GLOBAL THEORY

157

(i) For every a. The following conditions are equivalent: projection E of Z such that ZE is of countable type, AE is (ii) A is the product of 0-finite von Neua-finite; (iii) For every decomposition Z = II Z E . such mann algebras; iEI that the ZEi are 0-finite, (iii) nite. [Show that (i) is o-finite over its centre.

the AEi are 0-fi(ii) (i).] We then say that

A

b. If A is the von Neumann algebra generated by Z and a countable family of elements, A is 0-finite over its (Reduce to the case where Z admits a separating element centre. x. Then, A is isomorphic to the algebra induced in 4; we can therefore suppose that x is cyclic for A. By the hypothesis on A, there exists a sequence T1, T 2 , ... of elements of A such that the TT1x, TE Z, form a total set in H. Then, the T1,x form a separating set for Z'. A fortiori A c Z' is 0-finite. c. Show that there exist von Neumann algebras cf-finite over their centre which are not generated by their centre and a countable family of elements (use exercise 4). (Cf., nevertheless, part III, chapter 3, exercise 1.)

M

PART II REDUCTION OF VON NEUMANN ALGEBRAS

=

CHAPTER I.

1.

FIELDS OF HILBERT SPACES

Borel spaces, measures.

A Borel space is a set endowed with a set B of subsets of E, possessing the following properties: CpE B, B is closed under countable unions and the taking of complements (and hence under countable intersections). The elements of B are called the Borel sets of E. Let E, F be two Borel spaces. A mapping f of E into F is said to be a Borel map if the inverse image under f of every Borel set of F is a Borel set of E. Let E be a Borel space, and E P a subset of E. The intersections with E / of the Borel sets of E define a Borel structure on E / , said to be induced by the Borel structure of E. Let E be a topological space. The Borel sets of E for the topology define a Borel structure on E said to be subordinate to the topology. A Borel space E is said to be discrete if every subset of E is a Borel set. A Borel space is said to be standard if its Borel structure is subordinate to a Polish space topology (for example, the topology of a second-countable locally compact space). It is clear that, if E is a countable Borel space, saying that E is standard is equivalent to saying that E is discrete. If E is an uncountable standard Borel space, E is isomorphic to the Borel space subordinate to the topological space [0, 1]. A Borel set of a standard Borel space is a standard Borel space (Cf.K.KURATOWSKI, Topology, I, Metrisable spaces, complete spaces, 2nd. edition, p. 358, remark 1). Let Z be a Borel space, and B the set of Borel sets of Z. In this book, we call a positive measure on Z a mapping V of B into [0, + 00] such that: 1 0 if Z i , Z2, ... are pairwise disjoint elements of B, we have v(Zi u Z2 U ...) = V(Z1) +V(Z2) + ...; 2 0 Z is the union of a sequence of Borel sets Y i , Y 2 , ... such that v(Yi) < + co for every i. A subset of Z is said to be Vnegligible if it is contained in a set Y EB such that V(Y) = O.

161

162

PART II, CHAPTER 1

A subset of Z is said to be V-measurable if it is of the form X u N, with X E B and N V-negligible. The set M of V-measurable sets is closed under countable unions, countable intersections and the taking of complements; if we put V(XuN) = V(X), we obtain an extension of V to M which still satisfies the axioms for a measure. We will assume known the theory of measurable and integrable functions, etc. with respect to V. A positive measure V on Z is said to be standard if there exists a y-negligible subset N of Z such that the Borel space Z N is standard. If Z is a locally compact space, countable at infinity, a positive (Radon) measure on Z, regarded as a function on the set of Borel sets of Z, is a positive measure in the above sense. When Z is second-countable, this measure is standard.

Fields of vectors

2.

Let Z be a Borel space, and y a positive measure on Z. Whenever it is possible without risk of confusion, we will speak of "measurable," "integrable," etc. instead of "V-measurable," "V-integrable," etc.

We call a field of complex Hilbert spaces over Z a mapping c H(C), defined on Z, such that H(C) is, for every CE Z, a complex Hilbert space. Then F = II H(c) is a complex vector CEZ space; an element x of F is a mapping C± x(C) defined on Z such that x(C) EH(C) for every CE Z; such a mapping is called a vector field over Z. If Y is a subset of Z, an element of IT H(c) CEY is called a vector field over Y. LEMMA 1. Let xl, x2 , ... be a sequence of vector fields such that the functions c (xm (C)Ixn (C)) are measurable. For CE Z, let X(c) be the complex vector space generated algebraically by the xn (c), and d(c) the (algebraic) dimension of X(c). Then, the set z of the cEz such that d(c) = p(p = 0, 1, 2, ..., is measurable. There exists a sequence y l, y 2 , ... of vector fields such that:

(i)for every c z, y i (c), y 2 (c), •.. algebraically generate

X(c); (ii)if d(c) = No, y l (c), y 2 (0, ... form an orthonormal system; if d(c) < 80, yi(c), y2(c), yd (c )(c) form an orthonormal system, and y(r) = o for n > d(C); (iii)for each field yn, there exists a covering of z by disjoint measurable sets z l , z 2, ... possessing the following property: on each zk, yn can be put in the form c 4- Lic0c)xi(c),

163

REDUCTION OF VON NEUMANN ALGEBRAS

where the fi are measurable complex-valued functions, which are identically zero for i sufficiently large. Proof. We first define, by induction, a sequence zl, 22, ... of vector fi'elds in the following way: let CE Z; we take for zk(C) the first of the vectors xl(C), X2(C), ... which is linearly independent of the zi() for i < k, if such a vector exists; otherwise, we put zk(C) = O. Then 1° for every CE Z, generate the vector space X(C); 2 ° if z (C) z (C) d(C) = then zi(C), z2(C), ... are linearly independent; 0 3 if d(C) < No , then Zi(C) = 0 for i > d(C), and the zi(C) with indices < d(c) are linearly independent. We prove the following assertion, where j denotes an integer > 0: (Ai) There exists a sequence (Y1, Y2, ...) of disjoint measurable subsets of Z, such that, for every i < j and every k = 1, 2, ..., we have, either zi(C) = 0 for any CE Yk or zi(C) = xni (C) / 0 for any cE Yk with an index ni independent of Suppose that (As) is established for j < /, and let us prove (A/). Let Z' be a measurable subset of Z such that, for i < 1, we have, on Z I , z(r) = xni(C) 0 for some index ni independent 0 of C. Then, the set Xq of the E Z ? such that zi(C) = xq (C) is defined by the vanishing of the Gram determinants A(xyli (C), x n2 (C), xn 1 _ 1 (C), xm (C)) of

xn (c) = 1

1

(c),

x (Ç) = n2

2

2

(C) ,

x 7/1_1 (C)

=Z 1 _ 1 (C),

X(C) m

for m < (1, and by the condition A(xn2 (C), xvi 1 _ 1 (C), xg (C)) is measurable, whence we immediately Aave (Al).

O.

Hence

X

On each of the sets Y1, Y2, ... of (Afl, d(C) is, either constant or > j. The set of c for which onc) < j is therefore measurable, whence the first assertion of the lemma. Now, at each point c EZ io , orthonormalize the sequence of non-zero zi(C)'s using the Gram-Schmidt process. We obtain vectors Yi() If p < 86, put yi(c) = 0 for i > p. The fields Yi clearly possess properties (i) and (ii) of the lemma. Suppose that property (iii) is established for the indices n < j, and let us prove it for yi, using (AJ). If zj(C) = 0 on Yk, we have 0 on Yk, we have yi(c) = 0 on Yk. If zi(c) = xnj (C)

H -1 Yi ( C) = II u ( C ) IIU (), with

u( c ) = zi ( c ) -

y

n<j

(zi (ol yn (o) yn (c),

164

PART II, CHAPTER 1

and our induction hypothesis, together with the measurability of the functions C± (x (C)Ix ()), implies property (iii) for the index j. 0

Remark- The statement of the lemma and its proof are still valid if the word "measurable" is everywhere replaced by the word "Borel." References : [29], [80].

3.

F

Measurable fields of Hilbert spaces.

Let c H(c) be a field of complex Hilbert spaces over Z, and the vector space of the vector fields.

DEFINITION 1. We say that the H(c) form a v-measurable field of complex Hilbert spaces if there is given a linear subspace S of F possessing the following properties:

(1) For every x€ S, the function c able;

11 x(c)

H

is y-measur-

(ii)If y E F valued function

is such that, for ever? xE S, the complex(x (C) 1Y ()) is v-measurable, then, y ES; (iii)There exists a sequence (xl , x 2 , ...) of elements of S such that, for every cE z, the xn (c) form a total sequence in H(c).

The vector fields belonging to S are then called v-measurable vector fields. A sequence (x l, x 2 , of v-measurable vector fields possessing property (iii) is called a fundamental sequence of v-measurable vector fields. Property (iii) implies that the H() 's are separable. If x and y are measurable vector fields, (x(C) which is a linear combination with constant coefficients of 11 (x + y) (C) 11 2, 11 (x - y)(C) 11 2 , 11 (x + iy)(C) 11 2 and 11 (x - iY)(C) 11 2, is a measurable function of c. Then, by property (ii), the product of a measurable vector field with a measurable complex-valued function is a measurable vector field; the limit of a sequence of measurable vector fields which converges weakly at each point of Z is a measurable vector field.

Remarks.

1.

When multiplying the fields xi of definition 1 by measurable functions, we can suppose that each function ÷ II x(r) II is bounded and vanishes outside a set of finite measure. By adding to the xi's their linear combinations with rational complex coefficients, we obtain a sequence (y1 , y2, of measurable vector fields such that, furthermore, for every C E Z, the yn (C)'s are dense in H(C).

2.

If V1 is a positive measure on Z equivalent to V, the field of the H() 's, together with the space S, is a

REDUCTION OF VON NEUMANN ALGEBRAS

165

V 1-measurable field of Hilbert spaces. In other words, the notation of V-measurable field of Hilbert spaces only involves the class of the measure V.

Examples. 1. Suppose that Z is discrete. Every complexvalued function on Z is measurable. The only subspace S of possessing the properties of definition 1 is F itself.

F

2. Let Ho be a separable complex Hilbert space. The constant field corresponding to H o over Z is the V-measurable field defined as follows: a. H(c) = Ho for every Cc Z7 b. the v-measurable vector fields are the V-measurable mappings of Z into H 0 ' (As H 0 is separable, the measurable mappings of Z into Ho are the same whether Ho is endowed with the strong or the weak topology.) Property (i) of definition 1 is satisfied. Moreover, the constant mappings of Z into Ho are measurable; hence, if a vector field c x(C) satisfied property (ii) of definition 1, (x(C) la) depends measurably on C for every a E Ho, and hence x is measurable. Finally, if (an is a total sequence in H 0' the constant fields C an form a fundamental sequence of measurable vector fields. )

Let y be a measurable subset of Z. A vector field x over Y is said to be measurable if it can be extended to a measurable vector field over Z. For this to be possible, it is necessary and sufficient that, for any measurable vector field y over Z, the function c (x( ) ly ()) (defined over Y), be measurable. The condition is plainly necessary. Conversely, suppose that it is satisfied, and extend the field x to Z by putting x(C) = 0 for c el Y. The field thus extended is measurable by definition 1, (ii). The measurable vector fields over Y define a structure of measurable field on the field (H(C))c E y; this field is denoted by EIY (if E denotes the given field over Z) and is called the field induced by E over Y. If Z is the union of a sequence (ZI, Z2, ...) of measurable sets, a vector field over Z is measurable if and only if its restriction to each Zi is measurable. Let Z, Z' be Borel spaces; v, v' positive measures on Z, z'; and E = (( H(c)), S), E' = (( H'W)), S'), V-measurable and V'measurable fields of Hilbert spaces. Let n : Z Z' be a Borel isomorphism transforming v into v'. An 11-isomorphism of E onto E' is a family (V(C)) ccz possessing the following properties: is an isomorphism of the Hilbert space E Z, V(C) (i) for every H(C) onto the Hilbert space W(r)(C)); (ii) for a field C x(C) €H(r) to belong to S, it is necessary and sufficient that the field n(C) V(C)X(C) €H' (n()) belong to S'. If Z = Z' and v = v', an isomorphism of E onto E' is an 11isomorphism, where fl is the identity mapping. A field isomorphic to a constant field is said to be trivial.

166

PART II, CHAPTER 1

Remark 3.

If, in definition 1, we replace "V-measurable" by Borel throughout, we obtain the definition of a Borel field of Hilbert spaces over Z. (This definition only involves the Borel structure of Z, and not the measure v.) The definitions and properties of this section carry over if "measurable" is replaced by "Borel" throughout. Let E = ( ( 4( ) ), S) be a Borel field of Hilbert spaces over Z, and let v be a measure on Z. A vector field C x(C) cH(C) over Z is said to be v-measurable relative to E if, for every y ES, the function C (x(C)Iy(C)) is v-measurable. Let S' be the set of y-measurable vector fields over Z. We easily see that S' possesses properties (i), (ii), (iii) of definition 1. Hence ((H(C)), S') is a v-measurable field of Hilbert spaces, which is said to be deduced from E. Every element of S' is equal almost everywhere to an element of 5 (this follows immediately from the analogous property for scalar-valued measurable functions). If condition (iii) of definition 1 is suppressed, obstacles are encountered as soon as one attempts to prove the non-trivial results of the theory. References : [26], [28], [48], [52], [80], [100], [117], [118], [145], [193], [194], [205], [273], [329], [375].

4.

Basic properties of measurable fields of Hilbert spaces.

Let Z be a Borel space, y a positive measure on Z, and C H(C) a v-measurable field of Hilbert spaces over Z.

The set z, of the cE z such that the dimension d(c) of H(c) is equal to p is measurable. PROPOSITION 1. (i)

(ii) There exists a sequence

(y i, y 2 , ...) of measurable

vector fields possessing the foLlowing properties: a. of

if d(c) = 80 , (yi(C), y 2 (

)

, ...) is an

orthonormal basis

H(c);

b. if d(c) < No , (y i ( c), y 2 ), yd (c) (0) is an orthonormal basis of H(c), and yi(c) = O for i > (

Proof.

Let (x l , x 2 , ...) be a fundamental sequence of measurable vector fields. We note that d(c) is also the algebraic dimension of the vector space generated by the xi()'s. It then suffices to apply lemma 1 to the xi. The yi obtained are measurable by part (iii) of lemma 1. D DEFINITION 2. A sequence (y 1, y 2, ...) of measurable vector fields possessing the properties or proposition 1 (ii) is called a measurable field of orthonormal bases. PROPOSITION 2.

measurable fields.

Let (xl, x 2, ...) be a fundamental sequence of For a vector field x over z to be measurable,

REDUCTION OF VON NEUMANN ALGEBRAS

167

it is necessary and sufficient that the functions (x(C)Ixi (0) be measurable. Proof. The condition is plainly necessary. Now suppose that the functions C (x(C)Ixi(C)) are measurable. Let (y 1 , y2, ...) be the sequence of vector fields deduced from the sequence (xi) by application of lemma 1. The sequence (Yi) is a measurable field of orthonormal bases. By part (iii) of lemma 1, the functions C (x(C)Iyi(C)) are measurable. For every measurable vector field y over Z, the number (x() ly(C)), which is equal to CO

•

(x(c) lYi(c))

(y(c) ly i (c)) ,

depends measurably on C. Hence x is measurable.

0

For p = 1, 2, ..., Nb, let Z be the set of cE z such that d(c) = p, and Hn be a complex pEdimensional Hilbert space. For every p, t6 field induced by (H(c)) cez over z is isomorphic to the constant field defined by H. PROPOSITION 3.

Let (y 1 , y 2 , ...) be a measurable field of orthonormal bases, and (ell) an orthonormal basis of Hn . For CE Zn , let U(C) be the isomorphism of H(C) onto H which maps y 11C) to e, y2(C) to e, ..., etc. By propositionP 2, to say that the vector field x over Z is measurable comes to the same thing as saying that, for every p, the functions

Proof.

C.+ (x(C) lYi(C))

= (U(C)x(C)Ie 19.)

on Zn are measurable, that is to say that, for every p, the mappIng C U(C)x(C) of Z into H is measurable. D Proposition 3 very often allows one to reduce the study of measurable fields of complex Hilbert spaces to the study of constant fields.

Let z be a Borel space, y a positive measure on z, and c H(c) a field of Hilbert spaces over z. Let (x 1, x 2 , ...) be a sequence of vector fields possessing the following properties: 1° the functions c (xi(C)Ixj ( ) ) are measurable; 20 for every CE Z, the xi (c) form a total sequence in H(c). Then, there exists exactly one measurable field structure on the H(c)'s such that the fields xi are measurable. PROPOSITION 4.

Proof. The uniqueness follows at once from proposition 2. We prove existence. Apply lemma 1 to the fields xi, and let Yi' y 2 , ... be the fields obtained. Let S be the set of vector fields x such that the functions C (x(C)Iyi(C)) are measurable. We show that S satisfies conditions (i), (ii), and (iii) of definition 1. If SE S,

PART II, CHAPTER 1

168

00

Ilx(011 2 =11(x( 1y.(c»1 2 )

i= 1

depends measurably on Y. Moreover, we have Yi ES, y 2 ES, ...; hence, if a vector field y is such that (x(C)Iy(C)) depends measurably on c, for every X E S, we have yE S. Finally, for every Cc Z, the yn (C)'s form a total system in H(C). Hence specifying S endows the H() 's with a measurable field structure. Finally, the functions c -4- (xi()W()) are measurable by part (iii) of lemma 1. Hence xi E S for every All the results of this section remain valid when Remark. "Borel" replaces "measurable" throughout. This remark also In applies to many (but not all) of the results which follow. to this. we will not draw explicit attention future, References : [28], [48], [80 ],..

5.

Square-integrable vector fields.

Let c H(C) be a y-measurable field of complex Hilbert spaces Over Z. A vector field x over Z is said to be squareintegrable if it is measurable and if

J ii x(c) edv(c)

<

w.

The set of square-integrable fields is a complex vector space K. For x E K, y c K, (x(c) IOC)) is an integrable function of c. Put

(xIY) The space structure.

K

f(x(c) ly(c))dv(c).

is thus endowed with a complex pre-Hilbert space We have, for x E K,

11

x

11 2 =

Jii x(c)

2

dv(c)

The X E K such that II x II = 0 are just those x's which vanish almost everywhere. We will identify two elements of K which are equal almost everywhere; in other words, we consider the Hausdorff pre-Hilbert space H associated with K. In what follows, the elements of H will be regarded as vector fields, whenever this will not cause confusion. For X E H, we will therefore speak of the values x(c) EH() of x. It should be noted, nevertheless, that the x(C) are only determined to "within negligible sets." This explains the exceptional negligible sets that appear in most of the statements which follow. Frequently, these sets do not cause any difficulty, and we will sometimes even refrain from mentioning them at all, preferring

169

REDUCTION OF VON NEUMANN ALGEBRAS

statements which, although strictly speaking are incorrect, are easy to understand. On the other hand, such an omission can sometimes be a source of serious errors.

H is a Hilbert space.

PROPOSITION 5. (i)

(ii) If a sequence (X n ) of elements of H converges to an element x of H in the sense of the metric of H, there exists a subsequence of (xn ) which converges to x almost everywhere. Let (xi, x2, ...) be a Cauchy sequence in H. It is enough to prove that some subsequence (xyzk ) converges almost everywhere to an element x of H, and that Hxnk x H 4- O. Now, taking a subsequence if necessary, we can suppose that

Proof.

cc

H xn+1 n=1

xn

H<

00

Then,

H x +1 (r) 1=1

.

H<

x (r)

for every C not belonging to

a certain negligible set N ([[4]], chapter IV, section 3, theoCO

(x n+1 (C) - x n (C)) conn=1 verges, in the sense of the metric of H(C), to an element x(C) of H(C) such that rem 1).

For C N, the series xl(C) +

cc

H x(c) H

H x l ( c)

xn+1 (c) H n=1

H

sn (c)

H.

Put x(c) = 0 for EN. The vector field C x(C) is ')-measurable, being the limit almost everywhere of the measurable fields By [[ 4]], chapter IV, section 3, theorem 1, we have x (hence x E H) , and fil x(c) H2dv(c) < cc

Il s

x

H

Dix n+1

x n H,

which proves that x is the limit of the x 's in the sense of the metric of H. 0

The space H it called the direct integral of the H(o's and is denoted by f H(c) dy (C). The space H actually depends on the choice of the space S and DEFINITION 3.

of the measurable vector fields.

Sr notation

When necessary, we can use the

H(C)dV(C).

If x is a square-integrable vector field, we will also sometimes denote the corresponding element of H, by analogy with the

170

PART II, CHAPTER 1

above notation, by f

x(C)dV(C).

Let Y be a measurable subset of Z. The square-integrable vector fields which vanish off Y form a closed linear subspace of H , denoted by e H(C)dV(C), which may be canonically identiY fied with the diri ct integral of the induced measurable field c .+ H(C), C E Y-

Examples. 1. Suppose that Z is discrete, and that each point of Z has measure 1 (cf. section 2, example 1). Then K is just the space of vector fields x such that X 11x ( C ) 11 2 < ±`°, and ceZ H = K is the direct sum of the H(C)'s. In this case, we can canonically identify the H() 's with subspaces of this is not the case in general.

H.

However,

2. Let Ho be a separable complex Hilbert space and let c ± H(c) be the corresponding constant field over Z. Then, the square-integrable vector fields are the square-integrable mappings of Z into Ho, and

2 H(C)dV(C) = L " (Z, V). "0 Proposition 5 may thus be viewed as a generalization of the Riesz-Fischer theorem.

Remark. Let v l be a positive measure equivalent to v. Put p(c) is a v-measurable function on Z such V1 = pv, where c that 0 < p(c) < co for every C. Then, the mapping, which, to the square-v-integrable vector field x assigns the vector field

c

p(c) - x(c),

H(c)dv i (c)

is an isomorphism of

H(r)d(r) onto

since

f ilocrlx(c)112dv1(c)=filx(c)112p(c)--ip(c)dv(c) =filx(c)112dv(c). For fixed V and 'al' this isomorphism does not depend on the choice of p. It is called the canonical isomorphism of

fp

f4(c)dv(C)

onto

rH(c)dv (c). 1

References : [26], [28], [48], [52], [80], [100], [117], [118].

6.

Basic properties of direct integrals.

PROPOSITION 6.

orthonormal bases.

Let (y 1, y 2 , ...) be a measurable fi eld of

171

REDUCTION OF VON NEUMANN ALGEBRAS

(i)Let x be a vector field. Then XE H if and only if the (x(c)Iyi(c)) are square-integrable and the functions c r i=1 (ii)

If x

(

E

H and y

xIY ) =

E

H, we have

i=1

(iii)If xE H, then x is the Limit in the sense ofthemetric 14,ofthevectorfieldsC±1(x""""(C) i=1 Proof. Let x be a vector field. We have, for

as n

+ co.

every

C,

Whence we immediately have (1), in view of proposition X € H and y E H, we have, for every c,

2.

II

2

x () 11

of

X (x (0 1Y" K»1 2 " i=1 If

00

( x(

)

1 0 0)

=

(

x()I yi (c)) (0 o l yi (c»

i=1

and

r i=1

.1

clyi ( 0)1dv(c)411x(c01.11y(c)Ildv(c)<+., 1(x(01Yi (c )) 1-1(y(H

whence

(ii).

In particular, co

Mx M 2

=

i=1

fl(x()I y .( c»1 2 dv(c);

hence, if we put

xn (c)

= i=1

r 12dv( il(x(ol y .(c»c),

we have

Mx - xn M2 =

i=n+i

172

PART II, CHAPTER 1

and hence 11 x COROLLARY.

x n .4. 0 as n + 00 ; whence (iii) If y is standard, H is separable.

Proof. By the hypothesis on V, there exists a se9ence (fl , f2 , ...) of complex-valued functions dense in Lc (Z, V). Then, the vector fields of the form

C

fill (C)y i (C) + frz2 (0y 2 (C) +

+ frip (C)Yp (C)

are dense in H [proposition 6, (iii)], and the set of these fields is countable. D

Let (xl, x 2 , ...) be a fundamental sequence of measurable vector fields such that the functions c 4- 11 x i:(C) 11 2 are bounded. Let M be the set of vector fields obtained - by multiplying the xi by bounded measurable scalar-valued functions which vanish off a set of finite measure. Then M is a total set in H. PROPOSITION 7.

Proof. It is clear that M c H. Let x be an element of H orthogonal to M. For every bounded measurable function f on Z which vanishes off a set of finite measure, and for every i = 1, 2, ..., we have

f

f(C)(xi (C)Ix(C))dV(C) = O.

Since f is arbitrary, it follows from this that (xi(C) lx( ) ) except on a negligible set Ni. Let N be the (negligible) union of the Ni's. For C 4N, we have (Xi(C)Ix(C)) = 0 for every i, hence x(C) = O. Hence Hs O. 0

H=

Let (x i , x2 , ...) be a sequence of measurable vector fields. Let M be the set of square-integrable vector fields obtained by multiplying the x. by measurable scalarvalued functions. If M is total in 4, the xi(c) form, almost everywhere in z, a total sequence in H(). Proof. Let X(c) be the closed vector subspace of H(c) generated by the xi(c) l s. We need to show that X(C) = H(C) almost PROPOSITION 8.

everywhere in Z. Now, let (yi, y 2 , ...) be a fundamental sequence of square-integrable vector fields (section 3, remark 1). Then yi is the limit, in the sense of the metric of H, of a sequence (21, 22, ...) of finite linear combinations of fields of M. Taking a subsequence, if necessary, we can suppose (proposition 5) that the 2n (C) converge to y(r) except on a negligible set Ni. Now for fixed c, each 2 n (C) is a finite linear combination of the x (cs. Hence, for P Ni, yi(C) EX(C) . The union P N of the Ni's i s negligible. For c N, we have yi(C) EX(C) for every i, and hence X(C) = H(C). D References : [28], [80].

REDUCTION OF VON NEUMANN ALGEBRAS

173

Measurable fields of subspaces. PROPOSITION 9. Let c -4- H(c) be a v-measurable field of complex Hilbert spaces over Z. For every c E z, let K(c) be a closed linear subspace of H(C), and E(C) the projection PK (c) . Let S be the set of measurable vector fields x such that x(c) E K(c) for every c. The following conditions are equivalent: K(c), endowed with S, is (i)The field of Hilbert spaces c a v-measurable field; 7.

(ii)There exists a sequence (xi) of measurable vector fields [for the field c H(C)] such that, for every C, (x l (C), x2(C), ...) is a total sequence in K(r);

(iii)For any measurable vector field x [for the field c H(c)], the field C E(c)x() is measurable. Proof.

(ii): obvious. (i) (iii): suppose such a sequence (xi) exists. Applying

(ii) lemma 1, we can replace the xi by measurable fields y i such that, for every Cc Z, the non-zero yi(C)'s form an orthonormal basis of K(C). Then, for every measurable vector field x,

c

E(C)x(C) = Xfx(C)Cyi(C))yi(C)

is a measurable field.

(iii) (i): suppose condition (iii) is satisfied and let us show that S satisfies properties (i), (ii) and (iii) of definition 1. This is obvious for property (i). Now, let x be a vector field such that x(C) E K( ) for every C E Z, and such that, for every y ES, (X(C) IOC)) depends measurably on c; then, for H(C)], the any measurable vector field z [for the field C E(c)2(c) belongs to S, and hence (x(C)I2(C)) = field c (x(c)IE(c)z(c)) depends measurably on c; hence x S. Finally, if (xi ) is a fundamental sequence of measurable vector fields for the field c E(C)xi(C) are in S and, H(C), the fields C for every cE Z, the E(C)xi(c) form a total sequence in K(C). Under the conditions of proposition 9, we say that the K() 's form a measurable field of subspaces. Then,

.1.

K(c)dv(C) is a complete, and thus closed, subspace of H(c)dv(C) Reference : [59].

PART II, CHAPTER 1

174

Measurable fields of tensor products. PROPOSITION 10. Let c H(c), c K(c) be two v-measurable fields of complex Hilbert spaces over Z. There exists on the field c H(c) 0 K(c) exactly one y-measurable field structure possessing the following property: for any two measurable vector fields c x(c)E H(c) and c -›- y(C)6 K(c), the field c x(c) 0 y(c) is measurable. 8.

Let (xi) [resp. W)] be a fundamental sequence of measurable vector fields for the H() 's [resp. for the K(C ) ' s]. The fields c xi(C) 0 yi(c) possess the properties that allow us to apply proposition 4. There thus exists on the H(C)0K(C)'s a measurable field structure such that the fields C xi(C) 0 yi(c) form a fundamental sequence of measurable vector fields. If c x(C)e H(C), C y(c)E K(C) are measurable vector fields,

Proof.

(c)

y (c)

® y (c) ) =

depends measurably on c, and so the field C x(C) 0 y(c) is measurable. Whence the existence of the structure in the statement of the proposition. Moreover, for every measurable field structure on the H( ) 0 K(C ) ' s possessing the property of the statement, the fields c xi(C) 0 y•(c) form a fundamental sed quence of measurable vector fields. Whence the uniqueness. 0

Let c H(c) be a v-measurable field of complex Hilbert spaces over z, and c K(c) the constant field over z corresponding to a separable Hilbert space Ko . There exists a PROPOSITION 11.

unique isomorphism of the Hilbert space

H(c)dy(C))

Ko

onto the Hilbert space f (H(c) ø K())dv(c) which, for every

H(c)dv(c) and every y E

x6

Ko , transforms

x el y into the field

c -4- x(c) 0 y.

e

Proof. If x l , x2, ..., xn are elements of f H(C)dV(C) and yi, y2, ..., yn elements of Ko , we have

n

I

y

x.(

)

.0 yi

2

dV(C)

=

f

n i,j=1

i=1 1"

n

-=

c )) (y .i y . ) y Ocx,.(c)ix.(c),civ( 1- d d (x .(x .) (y

.IY 0

i,j=1 1- d 1. d

'

i1 x.

. i=1 " " 7

REDUCTION OF VON NEUMANN ALGEBRAS

175

and hence there exists a linear isometry of the algebraic tensor

e

e

Ko into f (H(C) 0 K(C))dV(C) which

product of f H(C)dY(C) and transforms

n X x. 0 y.2 into the field yi .

i=1

e

The range of the isometric extension

of this mapping to

H(c)dv()) o K o is the whole of

indeed, let (zi) be a fundamental sequence of measurable vector fields for the H() 's, such that the 's are bounded; let (2 1.) be a total sequence in Ko ; then, the fields d o 2:1 form a fundamental sequence of measurable vector c .4- zi(c) fields for the H(C) 0 K(C)'s; hence the fields C 4 f(C)Zi(C) 02 ,1i, where f is a bounded measurable scalar-valued function vanishing off a set of finite measure, form a total set in

H 2i (c) H

(H(c) 0 K(C))dV(C)

(proposition 7); now these fields lie in

e,

the range of whence our assertion. property of the statement is obvious.

Finally, the uniqueness

D

We will henceforth identify

o K(c))dv(c)

and

(Je H()dv())®Ko

by means of the isomorphism of proposition 11.

K(c) Let Ko be a separable Hilbert space, and c unique the corresponding constant field over z. There exists a Ko onto the Hilbert isomorphism of the Hilbert space r.,(z, y) space K(c)dy(c) which, for every fEL(z, v) and every y€ K0, COROLLARY.

f

transforms f 0 y into the field c f(C)y. Proof. Take for the field C 4 H(C) the constant field corresponding to the one-dimensional Hilbert space C. Then,

f9

H(C)dV(C) =

v)

(section 5, example 2). Moreover, H(C) 0 K(C) may be canonically identified with K(C). 0 We will henceforth identify

176

PART II, CHAPTER 1

K(c)dv(c)

2

= 1_,K0

(z, V)

2 L(z, v) 0 K o c

with

by means of the isomorphism of the corollary. Reference : [53].

Exercises.

1.

Let C

H(C) be a v;measurable field of com-

plex Hilbert spaces over Z, and

H =

H(C)dV(C).

Let

(x l , x2, ...) be a sequence of measurable vector fields. Let M be the set of square-integrable vector fields obtained by multiplying the xi's by the characteristic functions of measurable sets. If M is strongly dense in H, the x() 2 's are strongly dense in H(C) almost everywhere. [Argue as for proposition 8, replacing X(c) by the strong closure of the set of the xi(C) ' s, and y i by a sequence of measurable vector fields such that, for every CE Z, the y i (c)'s are dense in H(C) and supllYi ( C ) II <+°°]cEZ 2. Let c H(c) be the constant field corresponding to the one-dimensional Hilbert space C, and

H =

f

e

H(C)dV(C).

Show that

the weak convergence of a sequence (xn ) of elements of H to zero does not necessarily imply the weak convergence of the xn (c) to zero almost everywhere. 3. Let C H(c) be a v-measurable field of complex Hilbert Z, and (xi) a fundamental sequence of measurable spaces over vector fields. For every measurable vector field C x(C), there exists a sequence of vector fields of the form

f(c)x (c),

C 4'

where the f. are measurable complex-valued

i=1 functions on Z, which converge to x(C) almost everywhere on Z. [Reduce to the case where y(Z) < + 00 and where II x ( C ) II and the II xi ( C ) I. I s are bounded. Then use proposition 7 and proposition

5 (ii)]

28].

4. Let Z be a Borel space, V a positive measure on Z, C÷H(C) a field of complex Hilbert spaces over Z, F = H H(C), and Q the CEZ set of linear subspaces S of V which possess the following property: for every x€S, the function C II x ( C ) II is v-measurable.

H(c) is a y-measurable field, the set of If the field c V-measurable vector fields is a maximal element of Q. a.

b. Let S be a maximal element of Q. Suppose that there exists a sequence (xi) of elements of S such that, for every Ce Z, the xi(C) form a total sequence in H(C). Then, the field c H(c), endowed with S, is v-measurable. [48], [52].

177

REDUCTION OF VON NEUMANN ALGEBRAS

Let Z be a locally compact space, C -4- H(C) a field of complex Hilbert spaces over Z, and F = TI H(C). We say that

5.

CEZ c -4- H(c) is a continuous field of Hilbert spaces if there is given a linear subspace S of F possessing the following properdepends continuously on C; ties: (i) for every x ES, x( de(ii) if y c F is such that, for every x ES, pends continuously on c, then yE S; (iii) for every CE Z, the x(C) I s, xE S, form a total set in H(C). The fields of S are

H

)

H

H 00 - x(c) H

then called continuous vector fields. a.

If x, yE

S, (X(C)

IY(0) depends continuously on

c.

b. Property (ii) can be replaced by: (ii) if y € F is such that II y ( C ) II depends continuously on c and for every x E S, (x() y()) depends continuously on c, then y E S (Cf. j). c. If X E S and if f is a continuous complex-valued function on Z, the field C .4- f(c)x(c) is continuous. (Use b). d. Let (xn ) be a sequence of continuous vector fields, and X E F. If 11X(C) - x(C) II ÷ 0 uniformly over Z, we have x E S. e. For x E F to be continuous, it is necessary and sufficient that, for every C E Z and every 6 > 0, there exist a continuous field y such that 6 in some neighborhood of C

H x(cf) - y (c) H ‹

f. Let (xi) 17 E' be a family of continuous vector fields such that, for every C EZ, the x(C)'s form a total set in H(C). Let x E F. If the functions C ÷ IA x ( c) II and C .4- (x(C)Ixi(C)) are [ Use (ii) and (ii i ) together.] continuous, we have x E S. g. For every C E Z and every a E H (c ) , there exists an xE S such that x(c) = a. [Form a sequence of fields X n E S such that X+1 ( C ) - x74 ( C 1 ) II < 2 -n on Z and such that X(C) converges strongly to a.j

I

h. Let n be an integer > O. The set of dim H(c) > n is open.

CE

Z such that

be the set of x E F such that II x(C) II depends continuously on c. For a linear subspace of F to satisfy (i) and (ii), it is necessary and sufficient that it be maximal among the linear subspaces of V contained in F0 . i.

Let

F0

j. If H(c) is, for every cc Z, a copy of a fixed Hilbert space H o , the set of strongly continuous mappings of Z into Ho (i), (ii) and (iii). However, a mapping g of Z intosatife H 0 such that (7s(C)Ix(C)) depends continuously on C for every strongly continuous mapping x of Z into Ho is not always strongly continuous. k.

If S has the property:

178

PART II, CHAPTER 1

(iv) There exists a sequence (xn ) of elements of S such that, for every C E Z, the xn (C) 's form a total sequence in H(C) f then there exists exactly one Borel field structure on the field C ± H(C) such that the X E S are the Borel vector fields [28], [48].

CHAPTER 2.

1.

FIELDS OF OPERATORS

Measurable fields of linear mappings:

Let Z be a Borel space, v a positive measure on Z, and H I (C) two V-measurable fields of complex Hilbert C H(C), C spaces over Z. For every CE Z, let T(C) be an element of L(H(C), H'(C)), i.e. a continuous linear mapping of H(C) into T(C) is called a field of continuous H' (c). The mapping C H I (C) linear mappings over Z. Usually, the measurable field C H(C) and we will then speak of will be the same as the field C a field of operators.

The field of continuous linear mappings c T(c) is said to be measurable if, for every measurable vecx(c) EH(), the vector field c T(c)x(c) EH ' () tor field c is measurable. DEFINITION 1.

T(C) is measurable, II T ( C ) II depends measurIf the field c ably on C. In fact, let (xl, x2, ...) be a sequence of measurable vector fields with values in the H(C)'s such that, for every cE Z, the xi(C)'s are dense in H(C) (chapter 1, section 2, remark 1). Then, II T ( C ) II is the supremum of the numbers (with the convention that 0. -f-co = 0), xi (c) and the numbers depend measurably on C; whence our assertion.

H Tcoxi ( c ) H .H

H -1

be two fundaLet (xl, x2, ...),(xl, x;, mental sequences of measurable vector fields with values in the T(c) to be H() 's and H'(c)'s respectively. For the field c functions sufficient that the and measurable, it is necessary (T(c)xi(c) (0) be measurable. c PROPOSITION 1.

The condition is plainly necessary. Conversely, suppose that the functions

Proof.

C

(T(C)xi ( C ) Ixii( C ))

(xi(C)IT(C) *xii (C))

T(C)*xelj(C) are are measurable. Then, the vector fields C measurable (chapter 1, proposition 2). Hence, for every measurable vector field x with values in the H() s, the functions

179

PART II, CHAPTER 2

180

c

(T(c)x(c) x 1.(

)

) =

ix(C)IT(C)*x l.(0)

are measurable. Hence (chapter 1, proposition 2) the vector field C + T(C)x(C) is measurable. 0 If c + T(c) and c + T' (c) are measurable fields of operators, the fields c + T(c) + T'(C), C + T(c)T'(C) are clearly measurable. The field C + T(C)* is measurable by virtue of proposition 1. Analogous properties hold for measurable fields of linear mappings. Let Y be a measurable subset of Z. A field of linear mappings c + T(C) over Y is said to be measurable if it extends to a measurable field over Z, or, equivalently, for every measurable vector field x over Y, the vector field C + T(C)x(C) over Y is measurable. If Z is the union of a sequence (Z I , Z 2 , ...) of measurable subsets, a field of linear mappings over Z is measurable if and only if its restriction to each Z is measurable. Let C + T(C) c L(H(C)) be a measurable field of operators over Z, and C + K(C) a measurable field of subspaces, such that PK (c) T(c)PK(c) for every cE Z. Then the field of operators c + T(c)K( c ) is measurable [relative to the field c + K(C)], as follows immediately from proposition 9 of chapter 1. If c + H 1 (c) is another y-measurable field of Hilbert spaces over Z, and if c + Ti(c)E L(Hi(c)) is a measurable field of operators, then the field c + T(c) T1(C) is measurable [relative to the field c + H(C) 0 H1(C)]. In fact, let x(C) E H(C)

4- X 1 (C) E

be measurable vector fields. Then, the field C

(T(C) 0 T 1 ()) (x() 0 x l (C)) = T(C)x(C) 0 T i (C)x l (C)

is measurable, and our assertion then follows easily from proposition 1. References : [28], [48], [52], [80], [100], [117],

[118].

2. Examples. 1. Suppose that Z is discrete. A measurable field of continous linear mappings is just any mapping c + T(C) such that T(C) E

L(H(c), H' (

)

).

2. Let c T(c) E L(H(C), H 1 (0) be a measurable field of continuous mappings. Suppose that T(c), for every C E Z, is an isomorphism of the Hilbert space H(c) onto the Hilbert space H'(c). Then, the field c + T() -1 = T( ) * is measurable, from which it follows that (T(c)) ccz is an isomorphism of the measurable field (H(C)) cc z onto the measurable field (H' ())cEZ.

181

REDUCTION OF VON NEUMANN ALGEBRAS

3. Let Z be a locally compact space, countable at infinity, a V positive measure on Z, H o a separable Hilbert space, and C 4 H(C) the corresponding constant field over Z. Let C 4 T(C) be a field of operators, i.e. a mapping of Z into L(H0 ). The following hypotheses are equivalent: a. The field C 4 T(C) is measurable; b. the functions C 4 (T(C)Xilxj) [where (xi) is a total family in Ho ] are measurable; c. the mapping C 4 T(C) of Z into L(H0 ) endowed with the strong topology is measurable. In fact, a and b are equivalent by proposition 1. It is clear that implies b. Finally, suppose that hypothesis a is satisfied; let Y be a compact subset of Z; since the function C 4 II T() H is measurable, there exists a compact subset YiCY such that II T(C) II is bounded on Yi such that V(Y`, Y 1 ) is arbitrarily small; moreover, (yi, y 2 , ...) being a dense sequence in H o , there exists a compact subset Y 2 C Yi such that the mappings C 4 T(C)yi are continuous on Y2 (Ho being endowed with the strong topology), and such that V(Y 1 `, Y2) is arbitrarily small; then, v(Y , Y2) is arbitrarily small, and, on Y2, the mapping c T(c) y is continuous for any E H o (Ho being endowed with the strong topology); hence hypothesis e is satisfied. 3.

Decomposable linear mappings.

Let c H' H(c), c Hilbert spaces over Z.

H =

(c) be two v-measurable fields of complex Let

H' =

H(c)dv(c),

H' (c)dv

.

A measurable field

(C) E L(H (C)

(c))

is said to be essentially bounded if the essential supremum A of the function c 11 T( C ) 11 is finite. If this is the case, for every square-integrable vector field x, the vector field x l defined by s' (c) T(C)x(C) is square-integrable, and we have s' H < AH s H. Hence the mapping x 4 x' is an element T of L(H, H') such that T A.

H

H H<

PROPOSITION

2.

We

H

have

T

H

A.

Proof. For every square-integrable vector field x and every essentially bounded measurable complex-valued function f on Z, we have

f if(c)1211 T(c)x(c) edv(c) T(fx)

11 2

11 T

11 2 11

fx 11 2 = 11 T

and therefore, since the function

11 2 f1f ( C ) 1 2 IIs ( C ) 11 2C1V(),

f is arbitrary,

PART II, CHAPTER 2

182

T(c)x(c)

(1)

I

I T II

x(C)

II

almost everywhere. This established, let (xn ) be a sequence of square-integrable vector fields such that, for every E Z, the xn (C)'s are dense in H(C). Almost everywhere, we have

H Tcoxn ( c ) hence

H

I

T(c)

H T II -II xn ( C ) II

I

I T II •

for every n,

Hence X < II T II •

If two essentially bounded measurable fields of linear mappings define the same element of L(H, H'), they are COROLLARY.

equal almost everywhere. Proof.

The difference C the zero mapping, and hence

H

T(C) of these two fields defines T( c = 0 almost everywhere. 0

)H

A mapping T L(H, H I ) is said to be decomposable if it is defined by an essentially bounded measurable field T(c). We then write DEFINITION 2.

T

If the fields c

T (c )dv )

H'

H(C), C

are the same, we speak of

(C)

decomposable operators. The T()'s are defined, by the corollary of proposition 2, "up to negligible sets." In particular, given a point cE Z of measure zero, T(c) can be chosen arbitrarily. T(c) be an isomorphism of the measurable Example. Let c field (H(c)) cE z onto the measurable field (H' ( ))CEZ. Then,

T(c)dv(C)

T =

and

T-1 =

H' onto H. be decomposable operators. If

are mutually inverse isomorphisms of PROPOSITION 3.

Let

T1, T2

T i (C)dV(C)

T1 =

T(C) -1dV(C)

and

H onto H'

and of

T 2 (C)dV(C),

T2 =

we have T

1

+ T

2

=

(T (c) + T 2 1

AT

Proof.

1

= -

(C))4W,

kr (C)dv(C), 1

x

E

T 1 (C)T

2 (C)dv(C),

I

We prove, for example, the last assertion. Let TI 1 = 1

For

T* = 1

T 1T 2 =

H, y

E

H, we have

183

REDUCTION OF VON NEUMANN ALGEBRAS

(T l x1y)4(T (c)*x(C)Iy(C))dv(c)=f(x(c)IT (c)y(c))dv(c)=(xIT y), 1 1 1 1 hence T / = T*.

1

LI

Analogous properties are possessed by decomposable linear mappings. Let

6 T =

T(C)dV(C)

be a decomposable operator. By propositions 2 and 3, T is unitary (resp. hermitian, a partial isometry, etc.) if and only if T(c) is unitary (resp. hermitian, a partial isometry, etc.) almost everywhere. Let c

E(c) be a measurable field of projections. Let E =

E(;)dV(C),

which is a projection of L(H).

Let

K(c) = E(c)(H(c)),

K = E(H).

For a field x H to be in K, it is necessary and sufficient that x = Ex, i.e. that x(C) = E(C)x(C) almost everywhere. We thus see that K is the direct integral of the measurable field of subspaces c + K(c) (chapter 1, proposition 9.) PROPOSITION 4. T. =

2

Let

T. (C)dV(C) (i = 1, 2, ...)

2

and

T =

T(C)dV(C)

be decomposable operators.

(i)If Ti converges strongly to T, there exists a subsequence (Till» such that, almost everywhere, Tnk (c) converges strongly to T(c ) . (ii)If, almost everywhere, Ti(c) converges strongly to T(c), and if sup ll Ti H < ., then Ti converges strongly to T. Proof.

Suppose that Ti converges strongly to T. Let

a = supil T.

Too

H <

..

We have H H ‹ a for every i, except on a negligible set N. Let (x.i) be a fundamental sequence of measurable vector fields such tnat• xd E H for every j. We have H Tixi - Tx,..7, H .4- 0 as i + + co. Hence (chapter 1, proposition 5) there exists a negligible set Ni and a subsequence (Tnk ) such that

184

H

PART II, CHAPTER 2

Tnk ()xj

(c)

Tcoxi (c) H

0 as k + + co, for C Ni. Let N' be the union of the N-'s, which _s negligible. The sequence (n l , n 2 1 ...) can be dsupposed, by applying a diagonal process, independent of j. Then, for N u , we have

-

H Tnk (C)x.(C) d + 00, for any

as k

- T(C)x•(C) d

II

O

j, and

H Tnk (c) H

a,

so that T nk (C) converges strongly to T(C). Suppose now that, almost everywhere, Ti(C) converges strongly to T(C), and that a = supll T . H < .. We have, except on a

H

(c) H _

Ti negligible set, < a for every i, and hence II T(C)II-<-a. We keep the above notation. For every j,

H

T.(C)x.(C) -T(C)x. (c) 112 d d converges to zero almost everywhere as i -4- + 00 ; furthermore, ]

H T i (C)x j (C) - T(C)x.(C) 11 2 < [24 x (C) 1 2 , d j and the function c x.(c) 2 is integrable. Hence

H

11

d

H T.x. - Tx H .4. 0 d j

as i

+

For every measurable complex-valued function f on Z such that

suplfR) CEZ

I

= b

',

we have (T i

- T)(fti)

as i -4- +

b II (Ti - T)xi II ÷

H

(c) H

Now, if we suppose that each function C is xi bounded, the fX form a total set in H ( chapter 1, proposition j 7). Hence T- converges strongly to T. D PROPOSITION 5.

T

1 =

There exists a sequence T 1 (C)dV(C)1

T

2 =

of decomposable operators in H such that, for every ce z, L(H(0) is the von Neumann algebra generated by the Ti( ) 's. Proof. It is enough to prove the proposition when the field C

H(C) over Z is the constant field corresponding to a

REDUCTION OF VON NEUMANN ALGEBRAS

185

separable Hilbert space Ho . Then, L(H0) is the von Neumann algebra generated by a sequence S i , 5 2, ... of elements, and we can take for the fields C Ti(C) the fields C .4' Si. fl References : [28], [48], [52], [57], [80], [100], [119], [123].

Diagonalisable operators. 00 Let Lc (Z, V) be the set of essentially bounded measurable complex-valued functions on Z, in which we identify two functions that are equal almost everywhere. If Z is a locally compact space, countable at infinity, let Ito (Z) be the set of complex-valued functions on Z which are continuous and vanish at infinity. We endow 1.,(Z, V) and I..03 (Z) with their usual *algebra and Banach space structures. If tE Lœ c (Z, V) or if f(C)IE L(H(C)) is measurafE 14,0 (Z), the field of operators C ble and essentially bounded. We will denote by Tf the corresponding operator of L(H). 4.

DEFINITION 3. The operators of the form Tf., where fc IZ(z, v) [resp. fE Ift (z)] are said to be diagonalisable [resp. continu-

ously diagonalisable]. By proposition 3, the mapping f Tf is a homomorphism of the *-algebra V) (resp. Lm (Z)) onto the *-algebra Z (resp. of diagonalisable (resp. continuously diagonalisable) operators. These homomorphisms are said to be canonical.

Remark.

Let Z' be the (measurable) set of the CE Z such that H(C) X 0, xz l its characteristic function, and v i = xz /V. We see at once that the field C H(C), with the same measurable vector fields, is also V'-measurable, that

H(c)dv(c)

1-1(c)civ 1

(c),

that the measurable fields of operators, the decomposable operators and the diagonalisable operators are the same for v and V'. The canonical homomorphism of Ifl (Z, v) onto Z is the composition of the obvious canonical homomorphi'sm of V) onto IZ(Z, V) and the canonical homomorphism of erp,(Z, V') onto Z, which is an isomorphism by proposition 2. For the canonical homomorphism of IZ(Z, v) onto Z to be an-isomorphism, it is therefore necessary and sufficient that H(C) X 0 almost everywhere (in which case v = v'). Moreover, for the canonical homomorphism of L(Z) into L'(Z, v) to be injective, it is necessary and sufficient that the support of v be Z. Thus, when H(C) X 0 almost everywhere and the support of v is Z, Z may be identified with the spectrum of y in such a way that f(C) = C(Tf) for CE Z and

f E 40 (Z) .

186

PART II, CHAPTER 2

Suppose that Z is locally compact and countable at infinity, H(c) X 0 almost everywhere, and V has support Z. Then y is a basic measure on the spectrum Z of Y. PROPOSITION 6.

Proof.

If

XE

H and f

40(Z) , we have

(TfX1x) = ff(C)(x(C)Ix(C)) dV(C), which shows that (with the notations of part I, chapter 7)

H x( c ) H 2 dv( c ). Hence the measure Vx is absolutely continuous with respect to V. Conversely, let N be a subset of Z which is negligible for all the measures Vx,x• Then, for every x€ H, x(C) = 0 V-almost everywhere on N. Now there exists a field X E H such that x(C) X 0 V-almost everywhere on Z, in view of the fact that 0 H(C) X 0 almost everywhere. Hence N is V-negligible. PROPOSITION 7. (i) The algebra Z is an abelian von Neumann

algebra. If z is locally compact and countable at infinity, then Z is the weak closure of V. (ii)The canonical homomorphism of IZ(z, v) onto Z is continuous when T.;(z, v) is endowed with its weak* topology as the dual of 1, 1 (z, y) and Z with the weak operator topology.

(iii)Z' is a-finite. Proof. We prove (ii). E

H,

Let

y E H,

fc Lœc (z, v).

Then,

(Tfsly) = ff(C)(x( )ly (C))dv(C) depends continuously on f when 1_,(Z, V) is endowed with its weak* topology as the dual of *z, V), since the function C (x(C)Iy()) is integrable. We now prove (i). In view of the remarks preceding proposition 6, we can suppose that H(C) X 0 almost everywhere. The unit ball of 1.,(Z, V) is weak*-compact. Hence the unit ball of Z is weakly compact and consequently Z is a von Neumann algebra (part I, chapter 3, theorem 2). Suppose that Z is locally compact and countable at infinity. Since Lm (Z) is weak* dense in (Z, v), y is weakly dense in Z.

q

We now prove (iii). The fields xi of proposition 7 of chapter I can be taken to be in H. By this proposition, the xi's form a cyclic set for Z, and hence a separating set for V 0

REDUCTION OF VON NEUMANN ALGEBRAS

187

In chapter 6, we will prove the converses of the above results. References : [28], [80], [100].

5.

Characterisation of decomposable mappings. H(C) and C W(C) be two V-measurable As always, let C fields of complex Hilbert spaces over z, H =

Hs(c)dy(c).

H' =

H(c)dv(C),

Every function fE I,c (Z, V) defines a diagonalisable operator L(H) and a diagonalisable operator T _I FE L(H 1 ). If TEL(H, H I ) is decomposable, it is clear that TT = for every f f E Loe Conversely; C Z ' V).

Let T be a continuous linear mapping of H into H' such that, for every fE 1,7(z, y), we have TI' TTf.. Then, T is decomposable. THEOREM 1.

Proof. Let (xl, x 2 , ...) be a fundamental sequence of measurable vector fields over Z (relative to C 4. H(C)) such that sup il xi(C) < . and xi E H for every i; we can put

CEZ Yi = Txi

E HI . For every finite sequence p = (P1, P2, ..., P n ) of rational complex numbers, put

x For every fc q(z,

flf()1 2

y (

p =

i=1

p 2-x2 ,

Yp =

2. = 1

P -Y "

I = T fy I p , and so V), we have TTfx p = T fTxp

c) 2i dV(C)

I

H

H

H T eflf() 1

2

m x (c) I 2 dV(C),

so that, since f is arbitrary, yp(c)

< 011.11x p (c)

except on a negligible set N p . Let N be the negligible set which is the union of the N 's as p varies over the set of finite sequences of rational complex numbers. For C N, we have, for any rational complex numbers p l , p2, p n

Hence there exists a continuous linear mapping T(C) of H(C) into H' (C) such that T(C)x.(C) = y.(C) for every i; moreover

2

2

188

PART II, CHAPTER 2

H

H T(c) H

TH. Put T(C) = 0 for CE N. The field C T(C) is measurable and therefore defines a decomposable mapping T 1 of H into H'. We have T ixi = yi = Txi for every i, and hence T1 (T) = TfiT ixi =Tf`Txi =T(Trxi )

for every f€

H (chapter

Now, the vectors Ti form a total system in proposition 7). Hence T = T

1

Ç(Z,

V).

1,

Tmdv(c). D

=

For an operator TE L(H) to be decomposable it is necessary and sufficient that it commute with the diagonalisable operators. COROLLARY.

Remark. Suppose that Z is locally compact and countable at infinity. Let TE L(H, H / ) be such that TJ.T = TTf for every Then T is decomposable. and proposition 7.

fE 40 (Z).

This follows from theorem 1

References : [28], [80], [117].

6.

Constant fields of operators.

Let c H(c) be a y-measurable field of come K(c) the plex Hilbert spaces over z, H = H()dy(c), and c constant field over z corresponding to a separable Hilbert space K0. Let PROPOSITION 8.

S

=

S(C)C1V(C) E L(H)

and

T

E

L(K

0

Then, identifying

(f

H(c)dv(c))

o

Ko

and

f (H(c) 0 KW)dy(C),

we have

S Proof.

Let xE

0 T =

H()dy(c)

identified with the field C

(S(C) 0 T)dV(C).

and y E K o .

x(C)

(S 0 T)(x 0 y) =

0 y.

Then, x 0 y may be We have

(Sx) 0 (Ty),

which may be identified with the field C

Hence

(Sx)(C)

0 Ty =

S(C)x(C)

0 Ty = (S(C) 0 T)(x(C) 0 y).

REDUCTION OF VON NEUMANN ALGEBRAS

T) (x

(S

y) =

189

(s() o T)dv(C))(x0y).

Let Ko be a separable complex Hilbert space, c K(c) the corresponding constant field over z, and COROLLARY.

K =

K(c)dv(c) = 1, 2 (z, v) 0 K.

For every T E L(K0 ), we have IoT=fT(c)dv(c),

with T(c) = T for every

E Z.

Proof. Apply proposition 8 to the case where C H(C) is the constant field corresponding to the one-dimensional Hilbert 0 space C, and where S(c) = 1 for every C.

Let Ko be a separable complex Hilbert space, and U the set of unitary operators of L(K0 ). For the weak topology, and a fortiori for the strong topology, U is a Borel set in LEMMA 1.

L(K0 ). 0, the set of the For every xe Ko and every A TE L(K0 ) such that H Tx A is weakly closed. Hence the set of the TE L(K0) such that H Tx H = A is Borel for the weak A, topology (being defined by the conditions H Tx H . Now, if (xi) is = 1, 2, ...) a dense n for A 1/n H Tx H> sequence in Ko, U is the set of the TE L(K0 ) such that H Tx, H = H xi H, H Tx i H = H xi H for every i, and is thus Borel for the weak topology. fi Proof.

-

Before stating lemma 2, observe that, if Y is any subset of Z, there exists a V-measurable subset X of Z containing Y and possessing the following property: if X f is another V-measurable subset of Z containing Y, then X f contains X to within a negligible set. (It suffices to take X a Borel set containing Y and of measure equal to the outer measure of Y.) The set X is determined by Y, to within a neglibible set; we will say that X is the measurable envelope of Y.

Let Y be a subset of z, x its measurable envelope, E:c H(c) and E' :c H' (c) v-measurable fields of complex Hilbert spaces over z, and J a countable set. For every i E J, let c Tg(c) L(H(c)) (resp. c TA(c) E L(H t (C))) be a measura(resp. E'). Suppose that, ble field- of operators relative to for every c E Y, there exists an isomorphism u(c) of H(C) onto H t (c) such that u(c)Tg(c)u(c) -1 = Tii(c) for every :7 E J. Suppose that y is standard. Then, altering E and Et on negligible sets if necessary, there exists an isomorphism c v(c) of Elx onto E' Ix such that v(c)Ti(c)v(C) -1 = V.(C) for every jEJ and every LEMMA 2.

EX.

PART II, CHAPTER 2

190

Proof. We can suppose that X = Z, and that the given set of fields of operators is adjoint-stable. We have dim H(C) = dim H P (C) everywhere on Y and therefore almost everywhere on Z. We are thus reduced to the case where E and E' are both the constant field over Z defined by a separable Hilbert space Ko . Let (x ) be a total sequence in K0 . The mappings C Tj(C)xi, C T i-(C)xi (jEJ; i = 1, 2, ...) are measurable. Moreover, we can take Z to be compact metrisable, so that Z is the union, to within a negligible set, of a sequence of compact subsets such that the restrictions to each of these subsets of the mappings

C 4- T -(C)x., 2 d

C -÷

d

(j E J; i = 1, 2, ...)

1,

are continuous (for the strong topology of K0 ). It is enough to construct the V(c)'s in each of these compact subsets. Suppose henceforth, therefore, that the mappings C .4. T.( C)X., 2 d

C ÷ T t.(C)x., 2 d

(i E J; i = 1,

2, ...)

are continuous on Z for the strong topplogy of Ko . Let L1 be the unit ball of L(K0 ); endowed with the weak topology, L1 is a compact metrisable space. For every C EZ, let Mr be the set of the TE L i satisfying the following conditions: if Ti() =TTi(C) for every j E J; 2 ° T belongs to the set U of unitary operators of K o . For C EY, we have U(C) E M c , hence Mr (I). Let A l (resp. A 2 ) be the set of pairs (C, T) E Z x L 1 which satisfy condition 1° (resp. 2 ° ). Then, A l is the set of pairs (C, T) E Z X L 1 such that for jE J, j, k = 1, 2,

(Txi T is (C)*xk ) = (T j ( C) xi I T *Xid since the functions

(Txi lTii (C)*xk ),

(C, T)

(TJ (C)xi tT*xk )

(C, T)

are continuous on Z x Ll, Al is closed in Z x L l ; by lemma 1, A 2 =ZxUis Borel inZxL 1 ; hence A 1 nA 2 is Borel inZxL 1 . From this follows (Appendix V) the existence of a V-measurable mapping C V(C), defined on X, with values in 1.1, such that V(c) EM for CE X. Then, V(C) is unitary, and we have

T'.(c)V(C) = V(C)T.(C) d

for

d

j

E

J.

D

Let Ko be a separable complex Hilbert space, and H(c) be a vof operators of L(K0 ). Let c family (Si ) ici a measurable field of complex Hilbert spaces over z, THEOREM 2.

H

= f H(c)dv(c),

191

REDUCTION OF VON NEUMANN ALGEBRAS

and for every ic I, C Ti(C) an essentially bounded measurable field of operators over z; let T. = 1,

T.(C)dV(C) E 1,

1.(14).

Suppose that V is standard and that there exists, for every cEZ, an isomorphism u(c) of the Hilbert space H(c) onto the Hilbert space Ko such that

= T.(c)

for every i.

Then, there exists an isomorphism of H onto 1,6(z, v) Which, for every i E 1, transforms Ti into I 0 S.

Ko

The spaces H(C) all have the same dimension as Ko. We are thus led to the case where C H(C) is the constant field over Z corresponding to K o . Let J be a countable subset of I such that the family (Si)i Ej is weakly dense in the family (Si)i ci . By lemma 2, there exists, for every CE Z, a unitary operator V(C) in Ko , depending measurably on C such that

Proof.

1.1(0 -1 5.1.1(C

)

for every ieJ and thus for every i c I. Then

v()dy(c) = is a unitary operator in H = 1.,(Z, V) 0 K o which transforms Ti into the operator defined by the constant field C Si, i.e. 0 Si (corollary of proposition 8). 0 THEOREM 3. For i = • 1, 2, ..., let Ki be a separable complex Hilbert space, and ( 517 )_4, 1 a family of operators of 1.(K). Let H(c) be a v-measuraNe field of complex Hilbert spaces over z,

H

H()dy(c), and, for every

E 1, c

Tj(c) an essen-

=

tially bounded measurable field of operators over z; let

=

T codV(C) E 1.(H) . i

Let z l, z 2 , ..., be a sequence of subsets of z with union equal to z. Suppose that V is standard and that there exists, for every CE zi, an isomorphism ui() of H(c) onto Ki transforming each j into S. Then, there exists (if V X 0) an isomorphism of one of the K's onto a closed linear subspace H' of H reducing the Ti 's which transforms, for every je I, slci: into (Ti)Hf. .

192

PART II, CHAPTER 2

Restricting to a non-negligible measurable subset of Z on which the dimension of H(C) is constant, we are at once led to the case where c H(C) is a constant field over Z, corresponding to a Hilbert space Ko which we can then suppose to be the same as all the Ki's. Arguing as in theorem 2, we can suppose I countable. By lemma 2, we are consequently led to the are measurable and where U-(C) case where the Z's depends 2 2 measurably on C for CE Z. Then, one of the Z's, Z l for example, is non-negligible; hence, restricting to a non-negligible measurable subset of Z l , we are reduced to having to prove the conclusion of the theorem under the hypotheses of theorem 2 (whose notation we adopt henceforth), together with the condition v X 0. Now, let f be an element of T.,(Z, V) such that II f = 1. Then, the mapPingx-›-foxis an isomorphismW of K 0 onto a subspace H' of L2 (Z, V) ® H 0'- since

Proof.

(I 0 S.)(f 0 x) = f 0 S.x, and (I 0 S.)*(f a x) = f 0 s*.x,

,l

J

we see that H' reduces the I 0 S transforms Sj 's' and that W into (I 0 S )Hf; theorem 3 is then a consequence of theorem 2. 0 -

d

References : [57], [58], [91]. Thanks to [492], the hypothesis that V is standard can be suppressed in theorems 2 and 3.

Exercises. tors in

J

e

1.

Let Z be the algebra of diagonalisable opera -

H(c)dv(c).

For Z' = Z, it is necessary and sufficient

that the dimension of H(c) be 0 or I almost everywhere.

2.

e a.

Let T =

e I

H()dy(c).

f

T(C)dV(C) be a decomposable operator in

Let E be the support of T, which is decomposable.

e

E(c)dv(c). Show that almost everywhere, E(C) is the f support of T(c). [Let (xi) be fundamental sequence of square-

Let E =

a

integrable vector fields. Let K(c) = E(c)(H(c)). The T P T*Xi, where T' runs through the algebra of diagonalisable operators, form a total system in E(H); hence, almost everywhere, the T(c)*xi(c) form a total system in K(c).]

h.

Let T = WS be the polar decomposition of T,

W=

f

e

w(c)dv(c),

s=f

e

s(c)dv(c).

Show, using a, that T(c) = W(c)S(c) is almost everywhere the polar decomposition of T(c).

193

REDUCTION OF VON NEUMANN ALGEBRAS

C.

Let c

T i (C) be a measurable field of operators. T

Let

1 (C) = W 1 (C)S 1 (C)

be the polar decomposition of T1 (C). Show that the fields 4 Sl(C) are measurable. [Reduce to the case where C 4 the field C 4 T 1 (C) is essentially bounded, and use b.] 3. Let Z be a locally compact space, countable at infinity, V a measure > 0 on Z, and C 4 H(C) a v-measurable field of complex Hilbert spaces over Z. Let Z 1 be the set of CEZ such that H(C) X 0 and v' = xz , V, where xz , denotes the characteristic function of Z'. If the support of V' is Z, the canonical homomorphism of Lw (Z) onto the algebra y of continuously diagonalisable operators is an isomorphism, from which it follows that Z may be identified with the spectrum of Y. But, if Z'-Z' is not V-negligible, V is not a basic measure on Z. 4. Let Z be the algebra of diagonalisable operators. Show afresh that, if TE T is decomposable, by the following procedure: reduce to the case where C 4 H(C) is the constant field corresponding to a Hilbert space Ho ; then

H

2 L (Z, V) 0 H o C

=

and

Z = Z 0 C , 1 HQ

Z 1 being the algebra of multiplication operators, by the functions of q(Z, V), in L6(Z, V); then, Z' = Z1 0 L(H 0 ); furthercr-finite, so that T is the strong limit of more, Z' is a sequence of operators of the form T

l

®S

1

+T

2

0 S

2

+

+ T

n

0 S , n

dewhereTiEZ1andsiEL(H0);showthatanoperatorT-0.is 2 composable, and deduce from this that T is decomposable. 5. Let c -4- H(c) be a v-measurable field of complex Hilbert spaces over Z, and Y a metrisable topological space. For every CE Y, let c TOO be a V-measurable field of operators, with

H for any c E Z and y

E

Y.

Ty (C)

H<

m < 4- co

Let

T = Y

Y

If, for every CE Z, Tu (c) is a strongly continuous function of y T is a strongly continuous function of y. [Use proposition 4, Y. 1 (ii)j. 6.

Let Z = [0, 1], and let v be Lebesgue measure on Z.

PART II, CHAPTER 2

194

a. For CE Z, let Ic be the set of fc 1.(Z, V) possessing the following property: there exists a y-measurable subset Y of Z, of density 1 at c, such that f(C / ) converges to zero as C s converges to c, while remaining on Y. Show that I c is a *-ideal of the *-algebra Ti(Z, V). b. Let Jr be a maximal ideal of T__:(Z, V) containing I. Let (p c be the cfiaracter of I;(Z, V) corresponding to J. For f E TVZ, y), put f / (c) = cp c (f). Show that the function c --)- f4 () coincides almost everywhere with f. [Observe that, for almost every cE Z, f(c 1 ) converges to f(c) as c' converges to c on a suitably chosen set of density 1 at c.] c.

Let C -4- H(C) be a y-measurable field of finite-dimensional

complex Hilbert spaces over Z,

e H = H(c)dv(C),

and Z the alge-

bra of diagonalisable operators in H. Show that, for every TE 7 1 , we can choose an essentially bounded V-measurable field

e of operators c ,-). T(c) such that:

1° T = f

T(C)dV(C); 2 ° the

mapping T -4- T(c) is a homomorphism of Z' onto L(H(C)) for every c. (Use b) [57]. Cf. also J. DIEUDONNE, Sur le théoréme de

Lebesgue-Nikodym (Iv), J. Ind. Math. Soc., 15, 1951, 77-86. 7.

Let Z = [0, 1], and y be Lebesgue measure on Z,

H = LC2 (Z,

e V) =

where c ÷ H(c) is the constant field corresponding to the onedimensional Hilbert space C. Let T be the identity mapping of onto H. Show that there exists no field c ± T(C) c L(H, H(C)) such that, for every X E H, we have

e Tx = f (T(C)x)dV(C) • [Let xn E H be the function c -4- e 2iTrnC-; T(c) may be identified with a continuous linear form f; on H such that, almost everywhere on Z, we have f(x) =

e 2iiTnC12 /1 n

e 227TnC for every n.

Now

H

CHAPTER

1.

A

3.

FIELDS OF VON NEUMANN ALGEBRAS

preliminary theorem.

will continue to denote a Borel space, V a positive measure on Z, C H(C) a V-measurable field of complex Hilbert spaces over Z,

H

=f

e

H(c)dv(c),

and Z the algebra of diagonalisable operators.

For every iE 1, let c -4- Ti(c) be an essentially bounded measurable field of operators, and THEOREM 1.

e Ti =

Let A(c) be the von Neumann algebra in H(c) generated by the T() 's, and A the von Neumann algebra in H generated by the e Ti's and by Z. Let s = s(c)dv(c) be a decomposable operator. (i)

If

f

s E A, we have s(C)

E A(C) almost everywhere.

(ii)If s(c) E A(C) almost everywhere, and if I is countable, we have SEA.

(i) is obvious if S is one of the Ti', of if S E Z, hence if S is in the *-algebra B generated by Z and the TI-is. Proof.

Moreover, Z' is G-finite, [chapter 2, proposition 7, (iii)], hence A is G-finite, hence, if SEA, s is the strong limit of a sequence of elements of B (part I, chapter 3, proposition 1 and theorem 3). Using proposition 4 of chapter 2, 'we see that S(C) E A(C) almost everywhere. Suppose that S(c) EA(c) almost everywhere, and that I is countable. Let T 1 E A'. As ZcA, we have T' EV, and therefore

e T 1 = f T s (c)dv(c)

(chapter 2, theorem 1).

195

196

PART II, CHAPTER 3

and with the Ts, s' T' (r) commutes AsT / commuteswiththeT 2 with the TI-. (c)'s and with the Ti(C)* / s, for any i, almost everywhere. Hence T' (ri) commutes with S(C) and with S(C)*, almost everywhere, from which it follows that T / commutes with S and S. Thus, SE A" = A. 0

For every iE I, let c T i (c) be an essentially bounded measurable field of operators, and COROLLARY 1.

T. = 7,

2

Let A(c) be the von Neumann algebra in H(c) generated by the Ti(o's and B the von Neumann algebra in H generated by the Ti's. (i)If Z is a maximal abelian von Neumann subalgebra of IV, we have A(c) = L(H(c)), almost everywhere. (ii)If A(c) L(H(c)) almost everywhere, and if i is countable, Z is a maximal abelian von Neumann subalgebra of B'. Proof. If Z is a maximal abelian von Neumann subalgebra of B', the von Neumann algebra A generated by Z and the Ti / s, or by Z and B, is Z / (part I, chapter 1, proposition 13). Let S1

=

s 1 (c)dv(c),

S2 =

S 2 (C)dV(C),

be a sequence of operators of Z / such that, almost everywhere, L(H(c)) is the von Neumann algebra generated by the Si(C) I s (chapter 2, proposition 5). By theorem 1, (i), we have S v-(c) EA U ) almost everywhere, hence A(C) = L(H(C)) almost everywhere. Conversely, suppose that I is countable and A(C) = L(H(C)) almost everywhere. Then [theorem 1, (ii) ], every decomposable operator is in A, so that Z' A and hence Z is a maximal abelian von Neumann subalgebra of 13' (part I , chapter 1, proposition 13) .

The von Neumann algebra V is generated by Z and by a countable family of elements. COROLLARY 2.

This follows from theorem 1, (ii) and from proposition 5 of chapter 2. 0.

Proof.

Without the countability hypothesis on I, theorem 1, (ii) and corollary 1, (ii) are no longer true (exercise 2). References : [28], [55], [57], [100], [118].

2.

Measurable fields of von Neumann algebras.

For every CE Z, let A(C) be a von Neumann algebra in H(C). The mapping C A(C) is called a field of von Neumann algebras over Z.

197

REDUCTION OF VON NEUMANN ALGEBRAS

DEFINITION 1. A field of von Neumann algebras C 4- A(C) over z is said to be measurable if there exists a sequence c T i m, c T 2 (c), ... of measurable fields of operators such that, almost everywhere, A(c) is the von Neumann algebra generated by the Ti() '• Replacing Ti(C) by H Ti ( c ) H - 1 Ti(C) at the points C such that 0, we can suppose that the fields C -4- Ti(C) are Too

H

H

essentially bounded; they then define operators of P. CH( ) is measurable. The field Examples. The field c C -4- L(H(C)) is measurable (chapter 2, proposition 5). Let Y be a measurable subset of Z. Let C A(C) be a field of von Neumann algebras over Y. This field is said to be measurable if it extends to a measurable field of von Neumann algebras over Z. It comes to the same thing to say that there exists a sequence

c 4- T ( ) , 1

C

T 2 (C),

•••

of measurable fields of operators over Y, such that, almost everywhere on Y, A(C) is the von Neumann algebra generated by the Ti()'s. If Z is the union of a sequence (ZI, Z2, ...) of measurable subsets, a field of von Neumann algebras over Z is measurable if and only if its restriction to each Zi is measurable. PROPOSITION 1.

Let c

A(c) be a measurable field of von Neu-

mann algebras. (i)The set A of decomposable operators T =

T(C)dV(C)

such that T(c) E A(C) almost everywhere is a von Neumann algebra in H such that ZcAc Z', generated by Z and a countable formiLy of elements.

(ii)If c B(c) is a measurable field of von Neumann algebras defining the same von Neumann algebra A in H, we have A(C) =13(C) almost everywhere. Assertion (i) is an immediate consequence of theorem 1. We prove (ii). Let

Proof.

T1 =

T(C)dV(C),

T

2

•••

be a sequence of decomposable operators such that A(C) is, almost everywhere, the von Neumann algebra generated by the T.() 's. We have Ti EA, hence -(C) EB(C) almost everywhere, and so A(c) cB(C) almost everywhere. Similarly, B(C) A(C) almost everywhere. 0

PART II, CHAPTER 3

198

A von Neumann algebra A in H is said to be decomposable if it is defined by a measurable field c A(c) of von Neumann algebras. We then write DEFINITION 2.

A = By proposition 1, (ii), the A() 's are defined by negligible sets.

A,

to within

For a von Neumann algebra A in H to be decomposable, it is necessary and sufficient that it be the von Neumann algebra generated by Z and a countable family of decomposable operators. THEOREM 2.

Proof. The condition is necessary by proposition 1, (i). the other hand, if A is generated by Z and by a sequence T

1

- f T (C)dV(C), 1

T

2 =

On

T 2 (C)dV(C),

of decomposable operators, let A(C) c L(H(C)) be the von Neumann algebra generated by the Ti()'s. The field C -›- A(C) is measurable, and

A = .

A(c)dv(C)

E

by theorem 1.

Remark. If V is standard, H is separable; to say that A is generated by Z and a countable family of decomposable operators amounts to the same thing as saying that ZcAcZ' Problem; is it the same in general? References : [28], [48], [49], [80], [100], [338].

3.

Relations between a decomposable von Neumann algebra and its components. Suppose A and A' are decomposable.

THEOREM 3.

A

A(c)dv(C),

A' =

Let

A'Wdv(C).

(i)A(c) and A' (c) commute almost everywhere.

(ii)If2 is the centre of A (and therefore of A'), A(C) and A' (c) are, almost everywhere, factors which generate the von Neumann algebra L(H(c)). (iii)Conversely, if A(c) and A'(c), almost everywhere, generate the von Neumann algebra L(H(c)), Z is the centre of A andA'. Proof.

Let

199

REDUCTION OF VON NEUMANN ALGEBRAS

T

1

=

T (C)dV(C), 1

(resp. T 1s = f T 1I (C)dV(C),

T

=

2

T (C)dV(C), 2

T 21 =

• • •

T 2/ (C)dV(C),

...)

be decomposable operators which, together with Z, generate A (resp. A / ). Almost everywhere, A(C) is generated by the TU) 's, and A / (C) by the T() 's. As Tv• commutes with T1 and T /-* T-(C) commutes with T 41 (C) and T /-(C)* almost everywhere, hence A(C) and A / (C) comae almost everywhere. Whence (i). We will prove (ii) and (iii) together. To say that A n = Z is the same as saying that A and A' generate the von Neumann algebra Z / , hence that Z, the Ti and the T generate the von Neumann algebra V, hence that, almost everywhere, the Ti(C) and the T 41 (C) generate L(H(C)) (theorem 1), hence that, almost everywere, A(C) and A / (C) generate L(H(C)). Finally, if A(C) and A / (C) generate L(H(C)), we have

A(c) n A(c) 1 c (A' (c) u A(c)) 1 = L(H(C))' from which it follows that A(C) is a factor and similarly

A / (C).

0 Let

PROPOSITION 2.

Al =

A l (c)dv(c),

A2

• • •

be a sequence of decomposable von Neumann algebras, and A the von Neumann algebra which they generate. Then, A is decomposable, and, if

A =

A(c) is almost everywhere the Ai(c). Proof.

Let

e Ti = 1

.

the von Neumann algebra generated by

i fe i

T2 T-1-1 (C)dV(C), be operators which, with Z, generate A. Ai(c) is generated by Ti(c), T(C), by Z and the T7.:(i = 1, 2, ...; j = 1, 2, composable, an, almost everywhere, A(C) T1'.(c), that is, by the Ai(C)'s. D LEMMA 1. Suppose that y is standard. Neumann algebras c -4- A(c) is measurable, also measurable.

•

•

Almost everywhere, Then, A is generated ...), hence A is deis generated by the

If the field of von A(c)' is the field c

200

PART II, CHAPTER 3

Proof.

It is enough to prove the lemma when C H(C) is the constant field over Z corresponding to a separable Hilbert space Ho , and Z is compact and metrisable. Let L1 be the unit ball of L(H 0 ), endowed with the strong topology. For i = 1, 2, ..., let Ti(C) be an operator of Ll, depending measurably on C, and such that, for every C Z, A(C) is the von Neumann algebra generated by the Ti(C ) ' s. The space Z is, up to a negligible set, the union of a sequence of compact subsets Yi such that the restrictionstoeachy-oftherroppingsc4— T1, (c) , C 4- T-(C)* are con2 tinuous. Since it suffices to prove that the restriction of the field c 4- A(C) / to each Yi is measurable, we henceforth restrict attention to the case where the -mappings C Ti(C), C 4- Ti(C)* are continuous. Let M be the subset of Z x L 1 consisting of pairs (c, T') such that T / EA(C) / , i.e. such that

T'T.() = T.(c)T I ,

T'T.(c)* = T.(c)*T s ;

since the mappings (C. T') of Z x

T'T.() - T i (C)T i .

(C, T 1 ) 4- T I T.(?)*

T i (C)*T /

L1

into L(H ) are continuous, M is closed. Let o (M1, m 2 , ...) be a base for the topology of M. The Mi are Borel sets in the space Z x L l , which is a separable, complete metric space. Let Zn be the (V-measurable) projection of Mn Z. There exists (appendix V) a y-measurable mapping C T(c) of Zyl into L 1 such that (c, T7/1 (C))E Mn for every cE 211 ; extend this field to Z by assigning the value 0 on Z , Z , and again denote by c Th(c) the field obtained, which is vIlmeasurable. It is enough to show that, for every cE Z, the Th(c) generate A(C) ° . Now, let T / be an operator of A(c) / such that II T 1 II •< 1 - W e Let (M ni, Mr1 ,) , ...) be a base of neighbourhave ((,, T / )E hoods of (c, T / ) in M. Tften, tc, T / ) is the limit of (C, Th (C)) in Z x L l , i.e. T / is the strong limit of the Th (C)'s. 0 P

onto

M.

THEOREM 4.

(i)If A = We have

Suppose y is standard.

I

Awciv(c) is a decomposable von Neumann algebra,

A' =

A(

) l dv(c).

(ii)Let

B1 -

e B i (c)dV(C),

B2 =

be decomposable von Neumann algebras, and let B be their inter413) section. Then, B is decomposabLe, and, if B = I B(c)dv(c), B(C)

201

REDUCTION OF VON NEUMANN ALGEBRAS

is almost everywhere the intersection of the Bi(c)'s. Proof. We prove (1). Let T

1

=

T2 =

T (C)dV(C), 1

• • •

be decomposable operators which, with Z, generate A. Almost everywhere, A(c) is generated by the Ti(C) I s. For a decomposa-

e ble operator T 1 =

T' ()d\()to be in

A',

it is necessary and

f

sufficient that it commute with the Ti's and with the T's, hence that, almost everywhere, T(c) commute with the T() 's and with the Ti(C)"s, hence that, almost everywhere, T(C) € A(C) 1 • Now, the field C -›- A(C) 1 is measurable (lemma 1). Hence

e

A' We now prove (ii).

= f A(C)'dv(C).

For a decomposable operator

T =

T(C)dV(C)

to be in all the Bi's, it is necessary and sufficient that, almost everywhere, T(C) be in all the 8j() 's, i.e. in their intersection B(C). To prove (ii), it is therefore enough to prove B(c) is measurable. By lemma 1, it suffices that the field c B(C) 1 is measurable. Now, B(C) 1 is to prove that the field C the von Neumann algebra generated by Bl(C) / , B2(C) 1 , ... (part I, chapter 1, proposition 1). It then suffices to reapply lemma 1, and proposition 2. 0 References : [28], [80], [100], [117], [118]. Thanks to [338], we can suppress, in lemma 1 and theorem 4, the hypothesis that V is standard.

Constant fields of von Neumann algebras. PROPOSITION 3. Let c H(c) be a v-measurable field of come plex Hilbert spaces over z, H = H(c)dv(c), Ko a separable com4.

plex Hilbert space, c K(c) the corresponding constant field over z, A = A(c)dv(c) a decomposable von Neumann algebra in H, and B a von Neumann algebra in Ko . Then, identifying H ® K0 with f (H() 0 Ko)dv(C), we have

A Proof.

Let

® B =

(A(c) o B)dv(c).

202

PART II, CHAPTER 3

s1 =

s 1 (c)dv(c),

5

2 =

•••

be decomposable operators in H which, with Z, generate A. Let T 1 , T2, ... be operators in K0 which generate B. Then, the al,.(c) 0 Td.'s generate A(C) o B almost everywhere, and so the field c 4- A(C) 0 B is measurable. Furthermore, as S. 0 T. = I -1, ,7 i

J

1e (A(c) 0 B)dv(c) is

(chapter 2, proposition 8),

0 T.)dV(C)

d

the von Neumann algebra generated by the

Si 0 Ti's and by the diagonalisable operators of

H

® Ko , which

e form the algebra Z 0 CK 0 . generated by

A

J B,

Hence

0 ci(0 and CH 0

(A(C) 0 B)dV(C), which is

is equal to

A

®

B.

0

Let Ko be a separable complex Hilbert space, c 4- K(c) the corresponding constant field over z, and COROLLARY.

e

K=f

K(c)dv(c) =

L2c (z, y) o K o .

For every von Neumann algebra B in Ko, we have

e Z013=18(c)dv(C), with B(c) = B for every E Z. Proof. Apply proposition 3 to the case where C constant field c 4. C.

H(C) is the

0

LRMMA 2. Let Y be a subset of z, x its measurable envelope, Ko a separable complex Hilbert space, Ao a von Neumann algebra in Ko, and A(c) a von Neumann algebra in Ko generated by a sequence of operators depending measurably on c. Suppose that, for every c EY, there exists a unitary operator u(c) in Ko such

that

u(c)

-

o u(c) = Acc).

Suppose, moreover, that v is standard. Then there exists, for almost every c E X, a unitary operator v(c) in Ko depending measurably on c, such that

v(c) Proof.

-

o v(c)

Let (x.2 ) be a total sequence in K o , (Si) [resp. (q)] a sequence of operators in K0 generating A0 (resp. Al) ), c .4. T.(c) [resp. c 4- TOC)] a sequence of measurable mappings of . 2 Z into L(K0 ) such that, for every CE Z, the T() 's [resp. T 2s.(r) , I s] generate the von Neumann algebra A(C) [resp. A(C) ° ].

203

REDUCTION OF VON NEUMANN ALGEBRAS

Ti()* is a mapping We can suppose that, for every i, C T ( C). The mappings c Tk(c), and c T(c)* is a mapping C TI(C)xj are measurable. We can suppose that Z Ti(c)xj, C is compact metrisable. Then Z is, to within a negligible set, the union of a sequence of compact subsets such that the reTi(C)x.0-, strictions to each of these subsets of the mappings C T,t(c)x_i are continuous (for the strong topology of K0). We can hencefôrth suppose that the mappings C Ti(C)xj., 11(c)x:7 are continuous on Z for the strong topology of Ko . Let L 1 be he unit ball of L(K 0 ) endowed with the strong topology. Let M cZ x L 1 be the set of pairs ((,, T) such that T is unitary and T -1 A0T = A(C). To say that (C, T) EM is the same as saying that T is unitary, that T*SiT commutes with T:i(C) and that T*StT commutes with T-(C) for any i and j; or, equally, d that T is unitary, that (S.TT s.(C)x ITx ) 1 2 d k

= (S.Tx k ITT /.( C) * x 1 y

and that

(s s.TT.(c)x k ITx) = ( S 2/.Txk 2

d

ITT.(C) * X,)

d

for any i, j, k, 1. Hence M is Borel in Z x L 1 (cf. chapter 2, lemma 1). Furthermore, for every CEY, (C, U(C)) CM, hence the projection of M onto Z contains Y. Hence (appendix V) there exists a measurable mapping C V(C) of X into L(K0) such that (C, V(C)) EM for almost every C EX. D

Let Ko be a separable complex Hilbert space, Ao a von Neumann algebra in Ko, c -4- H(c) a v-measurable field of complex Hilbert spaces over z, PROPOSITION 4.

H A(c) cL(H(c)) a Z the algebra of diagonalisable operators, c v-measurable field of von Neumann algebras over z, and

A =

A(c)dv(c).

there exists, for every cE Z, an isomorphism u(c) Ko such that u(c) -1 A0u(c) = A(c). Suppose, moreis standard. Then, there exists an isomorphism of v) o K o which transforms A into Z ® Ao . are immediately led to the case where C .4- H(C) is the constant field corresponding to Ko . By lemma 2, there exists, for every cE Z, a unitary operator V(C) in K0 depending measurably on c such that V(c) -1 A0V(C) = A(C). Then,

Suppose that of H(c) onto over, that y H onto 1,(z, Proof. We

fe

V(c)dV(C) is a unitary operator in

H = T.,(z, y) ® Ko

which

204

PART II, CHAPTER 3

transforms A into the von Neumann algebra defined by the constant field C 4- A 0 , i.e. Z 0 A 0 (corollary of proposition 3). 0

Let K o be a complex Hilbert space, A o a von Neumann algebra in Ko, c H(c) a y-measurable field of complex A(c) c LU-((r)) a v-measurable Hilbert spaces over z, and c field of von Neumann algebras over z. If y is standard, the set of the cc z such that A(c) is spatially isomorphic to A o is measurable. D Proof. This follows at once from lemma 2. PROPOSITION 5.

Reference : [80]. Thanks to [338] and [492], we can suppress, in propositions 4 and 5, the hypothesis that V is standard.

5.

Reduction of discrete or continuous von Neumann algebras.

PROPOSITION 6.

Let A = A(c)dv(c) be a decomposable von Neu-

mann algebra, and E = JE()() a projection of A (resp. A'), E(c) being, for every CE z, a projection A(C) [resp. A(c)1. Let c K(c) be the measurable field of subspaces corresponding to the E() 's, and

K = E(H)

K(c)dv(C).

Then, A Proof.

K

= f A(c) K(c) dv(C).

Let

T

1

= f T 1 (C)dV(C),

T2 =

T 2 (C)dV(C),

be operators of A such that the T() 's generate A(C) almost everywhere. We can suppose that the sequence (Ti) is closed with respect to the taking of adjoints and to multipli cation. Then, the (ETiE)K and ZK generate AK (part 1, chapter 2, proposition 1). Now

ET.E =

E(c)T.()E(c)dv(c) 1

hence

(ETiE) K As the (E(C)Ti(C)E(C)),(( c ) generate A(C)K( c ) almost everywhere, we have

AK =

A(c) K(c) dvc. ()

0

205

REDUCTION OF VON NEUMANN ALGEBRAS

For E = E(C)dv(C) to be an abelian projection of

COROLLARY.

A()civ(c), it is necessary and sufficient that E(c) be, almost everywhere, an abelian projection of A(c).

A =

Proof. To say that E is abelian is the same as saying that AK is abelian, and therefore that the A(c)K ( c ) 's are abelian almost D everywhere, hence that E(C) is abelian almost everywhere. LEMMA 3.

Let A =

algebra, and c

Ac() be a decomposable von Neumann K(c) a measurable field of subspaces, E(C) = E A(c) , K(C)

K =

A

E.= E K.

Then E =

Proof. Let K 1 = E(H) and C

K i (C)dV(C).

subspaces such that K 1 =

T1 =

K(r) be the measurable field of

T i (C)dV(C),

T2 =

Let

T 2 (C)dV(C)

be operators of A such that 1 ° the Ti(C)'s generate A(C) almost everywhere; 2 ° the Ti's are strongly dense in A. Let (xi) be a fundamental sequence of square-integrable vector fields. Then, the TTiPiq c) x i-(TEZ; i = 1, 2, ...; j = 1, 2, ...) form a total sequence in Ç. Hence the Ti(C)PK( c )xj(C)'s form, almost everywhere, a total sequence in K l (C). Hence, almost everywhere,

P K1(C) = E( C ) " PROPOSITION 7.

Let A = A(c)dv(c) be a decomposable von Neu-

mann algebra. (i) If A is discrete, A(c) is discrete almost everywhere. (ii)If A is continuous, and ify is standard, A(c) is continuous almost everywhere. Proof.

Suppose that

A

is discrete. G =

Let

G(C)dV(C)

be an abelian projection of A of central support I (part 1, chapter 8, theorem 1). Almost everywhere on Z, G(C) is an abelian projection of A(c) (corollary of proposition 6) of central support I (lemma 3). Hence A(C) is discrete almost everywhere.

PART II, CHAPTER 3

206

Suppose V is standard. Let Y be the set of the CE Z such that A(c) is not continuous. We shall show that there exists an abelian projection G

G(C)dV(C)

of A, with G(c) / 0 for CE Y. If A is continuous, we will conclude from this that Y is negligible, hence that A(C) is continuous almost everywhere.

To prove our assertion, we may, by the usual methods, reduce it to the case where a. C H(C) is the constant field corresponding to a Hilbert space Ho; b. z is compact metrisable and there exist continuous mappings

1

2

1

•

of Z into the unit ball L1 of L(H0) endowed with the strong topology, such that, for every C E Z, the T() 's generate A(C) and the T() 's generate A(C) / . We can further suppose that, . for every - 1, 2, ..., there exists a j = 1, 2, ... such that T.(c)* = T.(c),

for any that:

C E Z.

1°.(C)* = 1°.(C)

Let M be the set of pairs (c, T) E

ZX

L 1 such

10 TT!2 ( C) = T! ( C) T for i = 1, 2, ... ; 2

2° T is a projection; 3 ° (TTi(C)T) (TTi(C)T) = (TTi(C)THTTi(C)T)

for i, j = 1, 2, ..4

4° T X 0. Conditions 1 0 , 2 ° , 3 ° define closed sets in Z x L l , while condition 4 0 defines an open set, and hence M is Borel. Moreover, for C EY, the set of T such that (C, T) EM is non-empty. Then (appendix V), there exists a V-measurable mapping C -4- G(C) defined on a \)-measurable subset X of Z containing Y such that (C, G(C)) EM for every C E X. The G(C) are projections (condition 2 0 ), belong to A(C) (condition 1 ° ), are abelian (condition 3 ° ), and non-zero (condition 4 ° ) . Put G(C) = 0 for C E Z •"- X, and

e

'

let G

=

f

G(C)&(C).

Then, G is an abelian projection of

A,

and

G(C) / 0 on X, and therefore on Y. 0 COROLLARY 1. Suppose that v is standard. Let E (resp. F) be the greatest projection in the centre of A such that AE (resp.

e AF ) is discrete (resp. continuous). Let E = E(C)dV(C), e F(C)dV(C). Then, almost everywhere, E(C) [resp. F(C)] is F = 1

207

REDUCTION OF VON NEUMANN ALGEBRAS

the greatest projection in the centre of A(c) such that Am E(c) [resp.A(c) F(c) ] is discrete (resp. continuous). Proof. Almost everywhere, E(C) and F(C) are disjoint projections, with sum I, in the centre of A(C) (theorem 4), A(c) E(r) is discrete and A(C) F(c) is continuous (propositions 6 and 7Y. 0

Suppose that V is standard. For A to be discrete (resp. continuous), it is necessary and sufficient that A(c) be discrete (resp. continuous) almost everywhere. COROLLARY 2.

Proof.

This follows from corollary

1.

D

References : [10], [80], [117].

6.

Measurable fields of homomorphisms.

Let C 4 H(C) and C 4 K(C) be two V-measurable fields of complex Hilbert spaces over Z,

H =

K

H(c)dv(c),

K(c)dv(c).

Let

c

A(c)

c

L H(c)), (

c

B(c)

c

L(K(c))

be measurable fields of von Neumann algebras over Z,

A =

B =

A(c)dv(c),

B(c)dv(c).

For every C E Z, let be a homomorphism of A(C) into B(C). The mapping C 4 (D c is called a field of homomorphisms. This field is said to be measurable if, for any measurable field of operators

the field C

is measurable.

(T(C)) E13()

Put then, for T =

(1)(T) =

T(C)dV(C) EA,

(T(c))dv(c)

ES.

It is immediate that (I) is a *-homomorphism of the *-algebra into the *-algebra B. PROPOSITION 8.

A

If the (I) 's are normal, (I) is normal.

Proof. Let (Tx) be an increasing filtering set in A-4- with supremum Tc A+ . Since A is 0-finite, there exists an increasing sequence (T 1 , T 2 , ...) of operators Tx with supremum

208

PART II, CHAPTER 3

T (part I, chapter 3, corollary to proposition 1). T. =

7,

Put

T

T.(C)dV(C), 2

with €A()

T-(C)

for every

E Z.

Taking a subsequence, if necessary, we can suppose that, almost everywhere, the T() 's form an increasing sequence converging strongly to T(C) [chapter 2, proposition 4, (1)]. Hence, almost everywhere, I c (T(c)) is the strong limit of the increasing sequence formed by the (T) 's [chapter 2, proposition 4, (ii)], and is consequently the supremum of the (TA) 's. 0 PROPOSITION 9. Let c 0 c , c Tc be two measurable fields of

normal homomorphisms of A(c) into B(c), defining the same homomorphism of A into B. Then, O c = T c almost everywhere. Proof. Let T

1

=

T

T (C)dV(C), 1

=

2

T (C)dV(C), 2

be operators of A such that, almost everywhere, the Ti(C)'s generate A(C). We have

f

T (T.(C))dv(C), C

0 (T.())dv(C) = C

and hence, almost everywhere,

C 2

= T (T.(c)) C

for every i.

Hence, almost everywhere, Tc and I c coincide on a *-subalgebra of A(C) which is ultra-weakly dense in A(C). As Ic and Tc are ultra-weakly continuous (part I, chapter 4, theorem 2), we have 0 0C = T C almost everywhere.

A normal homomorphism 0 of A into B is said to be decomposable if it is defined by a measurable field c 0 c of normal homomorphisms of A(c) into B(c). We then write DEFINITION 3.

=

C

By proposition 9, the I c are defined by I to within negligible sets.

If . 0 is almost everywhere, an isomorphism of A(c) onto B(c), -bs an isomorphism of A onto B. Proof. Let PROPOSITION 10.

T1 =

T 1 (C)dV(C),

T2 =

T 2 (C)dV(C),

REDUCTION OF VON NEUMANN ALGEBRAS

209

be operators of A such that, almost everywhere, the Ti(C)'s generate A(C). Then, almost everywhere, the (Ti()) 's generate B(C). Hence B is generated by the (T) 's and the algebra of diagonalisable operators in K. Now, )(Ti) e (A), and every diagonalisable operator in K is the image under (I) of a diagonalisable operator in H. Hence (1)(A), which is a von Neumann algebra (part I, chapter 4, theorem 2, corollary 2), is identical with B. Besides, if

T = rT(C)dV(C)

A,

E

the relation (1)(T) = 0 implies that (TOT(C)) = 0 almost everywhere, hence T(C) = 0 almost everywhere, hence T = O. 0

c H(c)

Let plex Hilbert spaces over PROPOSITION 11.

z,

H = c

be a v-measurable field of com-

H(Ody(C),

A(c) e L(H(c)) a v -measurable field of von Neumann algebras,

and

A =

A(c)dv(c).

Let K be a Hilbert space, B a von Neumann algebra in K such that B' is a-finite, and (I) an isomorphism of A onto B . Suppose that v is standard. Then there exists a v-measurable field c K(c) of Hilbert spaces over z, a v-measurable field c B(c) e L(K(c)) of von Neumann algebras, a v-measurable field c cDc of isomorphisms of A(c) onto B(r), and an isomorphism of

K(c)dv(c) onto K which transforms

fe,T

feB(c)dv(c)

into B and

(c)dv(c) into (1.

Proof. By theorem 3 of part I, chapter 4, it suffices to study the following two cases:

a.

Ho being a Hilbert space, we have

K=HoH 0

,

B

A

CH

O and (10 is the isomorphism T ÷ T 0 I u . Furthermore (part I, HO chapter 4, section 4, remark), we can suppose that Ho is.separa-

ble.

Then, K may be identified with

1, proposition 11),

fe

(A(c)

Ea Ci(o )dV(C)

A ®

(H(C) 0 Ko )dV(C)

(chapter

CH0 may be identified with

(proposition 3), and ITI may be identified

210

PART II, CHAPTER 3

e with

(I) dV(C), where (I) denotes, for every C , the isomorphism

C C S ÷ S 0 IH0 of A(C) onto A(C) 0 CK0 (chapter 2, proposition 9). b. Denoting by E' a projection of A' of central support I, we have K = E' (H), B = AE, and 4) is the isomorphism T .÷ TE E. f

Let

E' =

E'(C)dV(C),

B(C) = A(C) E , ().

K(C) = E'(C)(H(C)),

Almost everywhere, E'(C) is a projection of A(C) of central support I by lemma 3 [in view of the fact that the field

c

A(c)'

is measurable and

dard]. Hence the mapping S A(c) onto B(c). We have

A' =

A(C)'dV(C), since V is stan-

S E (C) is an isomorphism

K =

K(C)dV(C)

(chapter 2, section 3),

B =

B(C)dV(C)

(proposition 6)

(11

of

and

= fe4) dv(c).

0

1.

Show that theorem 1, (ii) is false if one suppresses the hypothesis that T is countable. [Choose the fields C Ti(C) in such a way that the Ti's are diagonalisable, but that, for every E Z, the T() 's generate

Exercises.

2.

Let

c

A(C) cL(H(C)),

C

B(C) cL(K(C))

be two measurable fields of von Neumann algebras over Z. a. The set Y of the measurable.

E Z

such that A(C) and B(C) commute is

b. If V is standard, the set of the or such that A(C) cB(C), is measurable.

such that A(C) =B(C), (Use lemma 1 and a.)

CE Z

V is S'tandard, the set of the E Z such that A(C) is a factor is measurable. [To say that A(C) is a factor is the same as saying that A(C) nA(c)' commutes with L(H(C)).] C. If

d.

Suppose that A(C) cB(C) and that A(C) B(C) for every C. Show that there exists a measurable field C E(C) of projections over Z such that E(C) E B(C), E(c) A(c) for every CE Z. (Use part I, chapter 2, exercise 2 b) [80], [338], [360].

CHAPTER 4.

1.

FIELDS OF HILBERT ALGEBRAS

Measurable fields of Hilbert algebras.

Let C H(C) be a V-measurable field of complex Hilbert spaces over Z. For every CE Z, let U(C) be a Hilbert algebra whose U(C) is called a field of completion is H(C). The mapping C Hilbert algebras.

U() of Hilbert algebras over z is A field c said to be measurable if the following conditions are satisfied: (i)If c x(c) is a measurable vector field such that x(c) €U(r) for every c, the field c x(c)* is measurable. x(c), c -4- y(c) are measurable vector fields such (ii)If c that x(c) E U(C) and y(C) c U() for every c, the field c x(c)y(c) is measurable. (iii)There exists a fundamental sequence (xi) of measurable vector fields such that xi(c) EU(c) for every i and every c. The measurable vector fields x such that x(C) E WC) for every C form a *-algebra Uo. If xc Uo , the field of operators DEFINITION 1.

C

Ux ( c ) is measurable; in fact, the vector fields C

LJ

= 2 x(C) x.(C)

are measurable, and it suffices to apply proposition 1 of chapVx(c) is measurater 2; similarly, the field of operators C ble. We can suppose, when convenient, that the fields xi of defini1 0 the tion 1, (iii) possess the following further properties: 0 are essentially bounded; 2 the funcfunctions C 0 tions C are essentially bounded; 3 the set of II Ux-(c) II 2 these fields i s closed under multiplication and the taking of adjoints. Properties 1 0 and 2 0 can be ensured by multiplying -'s by suitable scalar-valued functions. Property 3 ° is the obtained by enlarging the sequence of the xi 's.

H xi ( c ) H

211

212

PART II, CHAPTER 4

PROPOSITION 1. Let c U(c) be a field of Hilbert algebras over Z. Suppose that there exists a fundamental sequence (x,4 )

of measurable vector fields possessing the following properties: 1 0 xi(C) EU() for every c; 2 ° the fields c xi(c)* are measurable;

3 0 the fields c 4- xi(C)xj(C)* are measurable. U() is measurable. Then, the field c Proof. If C x(C) is a measurable vector field such that x(C) EU) for every EZ,

(x(C)*Ix.()) = (x.(C) * Ix()) 2 2 depends measurably on C, and so the field C x(C)* is measurable, which is property (i) of definition 1. Moreover,

(x( C) * Ix.(C)x-(C) * ) d

x(C)*xi(C) is depends measurably on C, and so the field C measurable; then, if C -4- y(C) is another measurable vector field such that y(c) E U(C) for every C c z, (x(c)y(C) lx i ()) depends measurably on C, and hence the field C x(c)y(c) is measurable, which is property (ii) of definition 1. Finally, D property (iii) of definition 1 is plainly satisfied. References : [13], [29], [86], [106], [116], [117]. 2.

Decomposable Hilbert algebras.

Let

H =

fkodv(c),

and let U be the set of fields C x(C) of U0 nH such that the field of operators C 4 - Ux(c) is essentially bounded (hence also the field C Vx(r) ). Then, U is a *-subalgebra of U0 . For, if xe U and y e U, we"'have

I (xy)(C) h

mu y (c) h

H ux(c) 00 h

almost everywhere, where M denotes the essential supremum of the function c and hence xy E H ; moreover

H ux(c) H,

(xy) (C)

=

ux(C) uy (C)

,

213

REDUCTION OF VON NEUMANN ALGEBRAS

from which it follows that the field of operators C U( xy ) (c ) is essentially bounded; hence xy EU; moreover, II x*( C) II = II x(C) II

and

H ux*(c) 11 = I ux(c) II

so that x* c U, which establishes our assertion. Moreover, as a subspace of have: PROPOSITION 2.

Proof. If xeU,

H, U is a pre-Hilbert space. We

U is a Hilbert algebra which is dense in H. yE

U, 2 E U, we have

(xIY) = f(x()ly())dv(C) = f(y*(C) lx*(c))dv(C) = ( Y * Ix *), (XY1

2)

(X(C)Y(C) 12 (C))dV(C) =

Hxy 11 2 = fll

J

(() lx*(C)2(C))dv(C) =(Ylx *2 ) ,

x(C)Y(C) 112dV(C) -- m 2 f11 00

12

= 14 2 11 y 11 2

where M denotes the essential supremum of the function ± II ux(C) II. Finally, let (xi) be a fundamental sequence of measurable vector fields such that x,-(C) EU(c) for every C and such that the functions C xi () c H uxi (c) H, are bounded. The xi(C)x_i(C) form, for every E Z, a total system in H(C). Hence the fields C f(C)g(C)xi(C)x4(C), where f and g are bounded measurable scalar-valued functions which vanish outside a set of finite measure, form a total system in H (chapter 1, proposition 7). Now fxie U and gxj E U. This shows at the same time both that U is dense in H and that U is a Hilbert 0 algebra.

A Hilbert algebra U in H is said to be decomU(c) of Hilposable if it is defined by a measurable field c bert algebras. We then write DEFINITION 2.

U = References : [13], [29], [86], [106], [116], [117].

3.

Involution and von Neumann algebras associated with U.

Let J [resp. J(C)] be the involution of H [resp. H(C)] defined canonically by U [resp. U(C)]. Let C x(C) be a measurable vector field. With the notation of definition 1,

(J(C)x(C) lxi(C)) = depends measurably on C, and so the field C J(C)x(C) is measx(C) is in H, the field C urable. If the field C J(C)x(C)

214

PART II, CHAPTER 4

e is in

H.

Denote by

f

J(C)dV(C)

the mapping of

H

fined; this is an antilinear isometry of with J on U. Hence

H

into

into

H,

H

thus de-

which agrees

0 J

= f

J(C)dV(C).

For xc U, the field of operators C essentially bounded. We have

ux = reu indeed, for y

.

{

(uxy)(C) U

x(C)

x(c)

E U,

e and so

ux(c) is measurable and

dV(C)

= (xy) () =

coincides with U

x

ux(c) y(C); on U, and therefore on

H.

Similarly,

vx

=

(C)

dV(C).

PROPOSITION 3. The von Neumann algebras U(U), V(U) are decomposable, and we have U(U) =

U(U(C))dv(C),

V(U)

=

V(U(C))dv(C).

Proof. For every XE U, we have Ux Z', Vx E where Z denotes the algebra of diagonalisable operators. Hence ,

U(U) cZ',

v(U) c

Z'.

Consequently,

Zc u( U )s = V( U ),

Z c v(U)' = u(U).

Let (x. ) be a fundamental sequence of measurable vector fields 2 such that x-(C) E U(c) for every C and every i, and such that the 2 functions c are bounded. Let

H, c

T. =

c)

U

dV(C). x(c) 2By part I, chapter 5, proposition 1, U(U(0) is the von Neumann algebra generated by the Ux .( cl 's. Using theorem 1 of chapter 3 we see that we will have pr8vea the equality

2

U(U) =

U(U(c))dv(c)

if we show that U(U) is the same as the von Neumann algebra A generated by Z and the Tt 's. Now, ZcU(U), and the Tt's, which

215

REDUCTION OF VON NEUMANN ALGEBRAS

commute with the Vis, X E U, are in U(U), and hence A c U(U) Consequently, among the operators TiTf (where Tf is the diagonalisable operator defined by a bounded measurable scalar-valued function f) appear the operators Uy , where the y form a total set in U (by proposition 7 of chapter 1); now, these operators generate U(U) (part I, chapter 5, proposition 1). Hence A = U(U) and, consequently,

u(U(c))dv(c).

U(U) = We see similarly that

V(U) =

V(U(c))dv(c).

0

For Z to be the centre of U(U) and V(U), it is necessary and sufficient that, almost everywhere, U(U(c)) and V(U(c)) be factors. COROLLARY.

Proof.

This follows from proposition 3, and from chapter 3, theorem 3. 0 References : [13], [29], [86], [106], [116], [117].

4.

Elements bounded relative to U.

If C a(C) is a measurable vector field such that, for every €Z, a(c) is bounded relative to U(c), the fields c -4- ua(c) , c va(c) are measurable. Proof. Let (xi) be a fundamental sequence of measurable vecPROPOSITION 4.

tor fields such that xi(C) E

c4u

U(c)

for every

c.

The fields

x.(c) = V x-(c) a(C) 2

a(C) 2

are measurable, and so (chapter 2, proposition 1) the field c 4 Ua ( c ) is measurable. Similarly, the field C 4 Va ( c )-is measurable. 0

Let a E H. For a to be bounded, it is necessary and sufficient that a(c) be bounded almost everywhere, and that the fields c ua(c), c va(C) be essentially bounded. We then have PROPOSITION 5.

U

Proof. that

a

=

U

a

(c)dv(c),

va

Suppose that a(c) is bounded almost everywhere, and M almost everywhere; we have, for every X E

d ua(c) d H vx(C) a(c) H = H

almost everywhere, hence

Va

u,

u a(C) x(c)

H

md X d;

mH

x(c)

H

hence a is bounded.

PART II, CHAPTER 4

216

Conversely, suppose that a is bounded. composable. Let

ua =

As Ua E

ua is de-

T(C)dv(C).

For every xE U, we have

uax

Vxa,

T(C)x(C) = V x(C) a(C)

hence

almost everywhere. Let (xi) be a fundamental sequence of measurable vector fields such that: 1 ° xi(c) e U(C) for every c; 2 ° the functions C H xi (c) H,uxi ( c ) are bounded; 3 ° the set of these fields is closed with respect to multiplication and the taking of adjoints. Let Y be a measurable subset of Z of finite measure, and yi the field equal to xi on Y and to 0 on Z`Y. Then, yi EU, hence T(c)yi(C) Vy (c)a(C) for every i, almost everywhere. Thus, almost everywheré on Y, a(c) is bounded relative to a dense *-subalgebra of U(c) and, consequently, bounded relative to (1(). Hence, almost everywhere on Z, a(c) is bounded relative to U(C). Furthermore, T(C)yi(c) = Ua ( c )Yi(C) almost everywhere, so that T(C) = Ua ( c ) almost everywhere on Y, hence T(C) = Ua ( c ) almost everywhere on Z. Thus, thefieldC .÷ Ua(c), is essentially bounded, and

H

u = a

U

Va

v

a(c) dv(c).

Similarly,

=

a(C) dV(C).

0

PROPOSITION 6. For every c E z, let B(c) be the algebra corresponding to U(c). Then, the field c measurable, and

B =

full

Hilbert

B(c) is

B(c)dv(C)

is the full Hilbert algebra corresponding to U. Proof. Let C a(C), C 4- b(c) be two measurable vector

fields

such that a(C) E B(C) E B(C) for every C, i.e. such that b(C) are for a(c) and bounded every C. Then, the field

c

a(c)b(c) = u a(c) b(c)

is measurable. Moreover, the field

C

a(C)* = J(C)a(C)

is measurable. As the field C -->- B(C) plainly satisfies property (iii) of definition 1, we see that this field is measurable.

REDUCTION OF VON NEUMANN ALGEBRAS

217

Thus,

B =

B(C)dv(c)

is a dense Hilbert algebra in H of which U is a *-subaIgebra. By the definition of B and proposition 5, every element that is bounded relative to U is in B. 0 Reference : [116]. 5.

Central elements relative to U.

Let Z [resp. Z(C)] be the subspace of H [resp. H(C)] consisting of the central elements relative to U [resp. U(C)]. PROPOSITION 7. and

The

Z(c) for

a measurable field of subspaces,

Let (xi) be a fundamental sequence of measurable vecEU(C) for every i and every C. For tor fields such that x-(C) 2 every CEZ, the xi(C)xj(C) - xj(C)xi(C) generate the same closed linear subspace Z'(C) of H(C) as the xy - yx, where xEU(C), y E U(?) The field of the subspaces Z' (C) is measurable [chapter 1, proposition 9, (ii)]. As Pz( r ) - I - Pv (c) , the field C Pz( c ) and, consequently, the field C Z(C), are measurable [chapter 1, proposition 9, (iii)]. Suppose that the fields x-2 are chosen in such a way that the functions

Proof.

H xi ( c ) H,

c

c

H uxim H

are bounded, the set of the xi being, furthermore, closed with respect to multiplication and the taking of adjoints. Let a E H. If a(c) Z(c) almost everywhere, we have

U

* a(C) = J(C)U x-(C) J(C)a(C) x-(C) 2 2 almost everywhere. Putting T.

2 we therefore have Ta = JT*Ja and, consequently, for every 2 2 bounded measurable scalar-valued function f vanishing outside a set of finite measure, T.T. = T AJT I:Ja = JT-T 4fJa = J(T fT.) * Ja. 1. Y 1,.1The linear combinations of the operators TfTi form a *subalgebra of U(U), which, as we observed when proving proposition 3, generates U(U). Passing to the limit, we therefore

218

PART II, CHAPTER 4

have Ta = JT*Ja for every Te U(U), and, consequently, a E Z. Conversely, if a€ Z we have u. (C) =J() u. ()J ()a() , for every i, almost everywhere. Hence, almost evenNWere, a(C) is central relative to a dense *-subalgebra of U(C), from which it follows that a(C) E Z(C)

Let E be the characteristic projection of U, and E(c) the characteristic projection of UM. Then, COROLLARY.

E =

E(C)dV(C).

U(U) ' E(C) = E U(U(C)) . It is therefore Z Z(C) enough to apply proposition 7 and chapter 3, lemma 3. D Proof.

We have E = E

References : [29], [117].

6.

Uniqueness and existence of the decomposition.

PROPOSITION 8.

Suppose that

U =

U(c)dv(C)

Then, if the Li(c)'s and the U' () 's are full almost everywhere, we have U(C) = W(C) almost everywhere. Proof. Let (xi) be a fundamental sequence of measurable vec1 ° xi(C) E U(r) for every C; 2 ° the functor fields such that: xi tions C uxi(c) are bounded; 3 ° the set of these fields is closed under Multiplication and the taking of adjoints [multiplication and the adjoint operation being defined by the U(C ) ' s]. Let Y be a measurable subset of Z of finite measure. Let yi be the field equal to xi on Y, and to 0 on Z•,Y. We have yiE U. Hence yi(C) EU I (C) almost everywhere, and, consequently, xi(C) EU I (C) almost everywhere on Y. Furthermore, almost everywhere on Y, the products xi(C)xj(C) and the adjoints xi(C) * , calculated in U(C) and in (1) (C), are the same. This shows that, almost everywhere, there exists a *subalgebra, common to U(C) and U / (C), which is dense in H(C). Since U(C) and U / (C) are full almost everywhere, we have U(C) = U / (C) almost everywhere (part I, chapter 5, section 3, D remark) (

c )

c

.

THEOREM 1.

For a full Hilbert algebra U in H =

H(Ody(c)

to be decomposable, it is necessary and sufficient that the von Neumann algebra U(U) be decomposable. Proof. We already know that the condition is necessary (proposition 3). Conversely, suppose that U(U) is decomposable. Then, the algebra Z of diagonalisable operators is contained in

219

REDUCTION OF VON NEUMANN ALGEBRAS

U(U), and U(U) cZ'.

Let J be the involution of

H

associated

with U. There exists a countable cyclic set for Z. As each element of this set is the limit of a sequence of elements of U, there exists a sequence of elements of U which is cyclic for Z. Suitably augmenting this sequence, we finally arrive at a sequence (Yi) of distinct elements of U, which is cyclic for Z, and which is closed under addition, multiplication, the taking of adjoints, and multiplication by rational complex numbers. For each i, choose a field C yi(C), a representative of yi. For every E Z, let B(C) be the set of the Yi()'s, j =1, 2, .... Almost everywhere, we have

= (r 1 y + r 2 y d.)

r 1 y.() + r 2y 2,

,

for any i, j and rational complex numbers rl, r2 . we shall endow the B() 's with structures of algebras over the field of rational complex numbers, by putting

= (y .y .) 1, cy

.

This definition of the multiplication does not make sense if belongs to one of the setsNijki defined by the conditions

yi (C) = y k (C),

C

= y i (C),

However, this set is negligible; in fact, let Y be the measurable set of the E Z such that

Yi() = y l (C), and let E be the corresponding projection of Z; we have

E Y v =, Ey k ,

Ey. = Ey i ,

hence (in view of the fact that EE Z),

E(y.y.) = EU y. = U Ey. = U Eyi d Eyk Ey = E(y ky i ), y d Yi j and so (yiyi )(C) = (yky)(c) almost everywhere on Y; whence our assertion. Hence, outside a negligible set, we have welldefined a multiplication on B(C), such that yi(C)yj(C) = (yiy.)(C) and, discarding another negligible set, Ehe fact that the `.'s form an algebra over the field of rational complex numbers implies that the B() 's are algebras over this field. In an analogous way, using the fact that J(Eyi) = E(Jyi) for every projection E of Z, we see that, putting y.(C)* = we have defined on the B(C)'s *-algebra structures over the

PART II, CHAPTER 4

220

field of rational complex numbers, with

(y.() y.()) = (y i (C)* yi(C)*) (provided we discard yet another negligible set). Consider the complex linear subspace U 1 (C) of H(C) generated by B(C). There exists on U l (C) exactly one structure of *-algebra over C that induces on B(C) the *-algebra structure that we have just constructed. We show that U l (C) is a dense Hilbert algebra in H(C). The equality

(y i (Otyi (0) = for any i and X,

y E U i (C) .

j implies that (xly) = (y*Ix*) for any Put

Yi

-_... 16 u

La . (C)dv(c).

-

We have

=

(y i y o.)(C)dv(C) =-Y =U y. 2 d Yi d

=

u

Yi

(c)y-(C)dv(C), d

hence, almost everywhere,

U .(C)y j•(C)

y .(C)y d.(C)

Similarly, almost everywhere,

u( )* y k (C)==u 4,(Uy.(C)=.(C)*y.(C). Yi 0 Y* d Hence

(y i ( c)yi (C)Iy k (c))= ( yi c)Iy i (c)*y k (c)) (

for any i, j, k, almost everywhere. Hence (xYlz) = (Ylx"), for any x, y, 2 E U 1 (c), almost everywhere. We see simultaneously that, almost everywhere,

I y i () y i ( c) II

II uyi II II

yi (c)

IL

and hence the mapping y xy in U 1 (r) is continuous. Finally, the Ty i ls, TE Z, form a total set in H, contained in U; every element of H can be approached arbitrarily closely by products of two elements of U and, consequently, by linear combinations of elements of the form (Tyi)(T lyi )( TEZ, TIE Z); now

221

REDUCTION OF VON NEUMANN ALGEBRAS

(Ty)(T l

) = UTyiT lyi = TT l uyi yi = (TT 1 )(y iyi ),

from which it follows that the sequence of the yiyi's is cyclic for Z; hence, almost everywhere, the yi(C)yj(C)'s form a total set in H(C) (chapter 1, proposition 8); we have therefore established that, except on a negligible set N, U 1 (r) is a Hilbert algebra that is dense in H(C). We redefine U l (C) on N in an arbitrary manner, except for the requirement that U1(r) be a Hilbert algebra that is dense in H(C) (cf. part I, chapter 5, section 5, remark). The field C (11(C) is measurable by proposition 1. Put

U

1 .

Since

Uyi (C)yi (C) = Uy i(o yi (C) almost everywhere, we see that U .(C) = U almost everywhere. Denoting by T the diagonalisable operator corresponding to the function f of (Z, V), we have .()

(Ty.)() =.f()y i (C),

H

hence

U (Tfyi)(c) =f(C)0yi(c) =f(C)0yi (C),

)(c H

and so is essentially bounded, hence Tfyi E U1. u(T fyi If f1EL °3 (Z, , the product (Tfyi) (Tfiyi) , calculated in U 1 , is the field c f(C)yi(C)fi(C)yi(C), i.e. )

TfTfi (y iyi ) = (rJ).)(T iv,

fr y . ) calculated in U. Hence there exists a *-subalgebra common to U and U l , that is dense in H. Since U is full, U is the full Hilbert algebra corresponding to U l . Let U(c) be the full Hilbert algebra corresponding to U l (C). Then, the field C U(C) is measurable, and U = I U(C)dV(C)

(proposition

6).

D

There is little likelihood of proposition 8 continuing to hold if the hypothesis that U(c) and U / (c) are full is suppressed. References : [13], [29], [86], [106], [116], [117].

Exercises. 1. Let Z be a Borel space, and V a positive measure on Z; for every EZ, let U(c) be a Hilbert algebra; then

B = H Li(c)

cEz

is a *-algebra. A *-subalgebra C of B, possessing the following properties, is given: (i) for everyx E C, the functionC 4. 11x()H is measurable; (ii) if y E B is such that, for every x E C, the function

222

PART II, CHAPTER 4

(x() l()) is measurable, then y E C; (iii) there exists a sequence (xi) of elements of C such that, for every CE Z, the -(C)'s form a total sequence in U(C). Let the Hilbert space H(C) be the completion of U(C). Show that there exists on the H(C) exactly one v-measurable field structure possessfield C ing the following property: for a vector field C 4 x(C) to be in C, it is necessary and sufficient that it be measurable and E Z. that x(C) e U(C) for every The field C 4 H(C) being endowed with this structure, show that the field C -4- U(C) is measurable.

C

H(C) be a \.-measurable field of complex Hilbert Let C spaces over Z, and C 4 U(C) c H(C) a V-measurable field of Hilbert algebras. If, for every C EZ, U(C) possesses an identity element I(C), the field C 4 I(C) is measurable. [Let (T-) be a fundamental sequence of measurable vector fields such that -(C) E U(C) for every i and every C. Then, 2.

(I(C)tXi(C)xj (C)) = depends measurably on

C.]

CHAPTER 5.

1.

FIELDS OF TRACES

Measurable fields of traces.

Let c H(c) be a y-measurable field of complex Hilbert spaces over Z,

H =

H(c)dv(c),

and

A

a decomposable von Neumann algebra. For every cE Z, let (p c be a trace on A(C) + . The mapping c -0- cp is called a field of traces over Z.

The field of traces c -4- (p c is said to be measurable if, for any measurable field of operators c -4- T(c) such ((T()) is measthat T(c) E A(c) 4- for every c, the function c urable. T(c) is essentially Let then, assuming that the field c DEFINITION 1.

bounded,

T =

T(C)dV(C) E

A.

• The number I (p c (T())dV(C) only depends on T; set it equal to

( T).

It is clear that (p is a trace on At which we denote by

C

(i) for (p to be finite, it is necessary and sufficient that f(p c (i)dv(c) < i-œ (which implies that qb c is PROPOSITION

1.

finite almost everywhere). (ii)If the (p c 's are normal almost everywhere, (p is normal. If, further, (1) is semi-finite, the (p c 's are semi-finite almost everywhere. (iii)If the (p c 's are faithful almost everywhere, (1) is faithful.

223

224

Proof.

PART II, CHAPTER 5

For cp to be finite, it is necessary and sufficient

that ,4)(I) < + 00.

Now, cp(I) = f (P c (I)dV(C); whence (i).

if Suppose that the (P c 's are faithful almost everywhere.

T =

T(C) (IV ( C) €

A,

and if

Then,

cp(T) = 0,

we have (1) (T(C)) = 0 almost everywhere, hence T(C) = 0 almost everywhere, hence T = 0, whence (iii). Let (Tx) Suppose that the cp c 's are normal almost everywhere. be an increasing filtering family in A+ with supremum TEA+. Since A is 0-finite [chaPter 2, proposition 7, (iii)], we can extract from the family of the TX's an increasing sequence T 1 , T 2 , ... with supremum T (part I, chapter 3, corollary to proposition 1). Put T. =

f Ti(c)dv(c).

T

f T(c)dv(C).

There exists an increasing sequence of integers (n 1 , n2, ...) such that Tnk (C) is an increasing sequence converging strongly to T(C) almost everywhere (chapter 2, proposition 4). Then, almost everywhere, 4) c (Tnk (C)) is increasing and converges to Hence

(1)(Tnk ) = 14)0Tnk (C))dv(C) is increasing and converges to

(p(T) If, further, cp is semi-finite, we show that is semi-finite almost everywhere. Since A is 0-finite, I is the supremum of an increasing sequence (Si) of operators of A+ such that OSi) < + 00. Put

s

i =

We have cp c (Si(C)) < + 00 almost everywhere, and, taking a subsequence if necessary, Si(C) is increasing and converges strongly to I, almost everywhere. Hence cp is semi-finite almost everywhere. Problems: If cp is normal, are the (1) 's normal? If (I) is faithful, are the cp c 's faithful? If the cp c 's are semi-finite, is cp semi-finite? References : [80], [117].

225

REDUCTION OF VON NEUMANN ALGEBRAS

2.

Decomposition of traces. THEOREM 1. Let U = U(c)dv(c) be a decomposable Hilbert algebra in H, (p the natural trace on U(U) +, and (p c the natural trace on U(U(C)) + . Then, the field of traces c (p c is measurable, and (I)

e Proof.

Let

a =

e - f (I) C dv(c).

a(c)dv(c)

f

be an element of

H,

bounded for U.

We have

(U*

U

C a(C) a(C)

) = (a(C)1a())

and

(u (c) ua( ) )dv(C). (P(U*U a a) = (cda) = f(a(C)1a(C))dy(C) = fcp c Now, let T be any element of U(U), and put T=

f T(c)dv(C),

with T(C)

E

for every

U(U(C)) ±

€Z.

chapter 3, corollary of proposition 1 and corollary 5 of theorem 2, there exists an increasing sequence (Ti) of operators of U(U) + , with ,supremum T, such that cp (Ti) < + on for every i. Put

U(U) is 0-finite. Using part I,

T. = f T.(C)dV(C). I-

2-

We already know that the functions C + (i) c (Ti(C)) and that

are integrable,

Moreover, taking a subsequence if necessary, the functions C + cp c (Ti(C)) form, almost everywhere, an increasing sequence converging to the function C + (I) (T(C)). Hence this function is measurable, and

1 THEOREM 2.

4) (T(C))dV(C) = limcp(Ti) = (1)(T).

Let A

= A(c)dv(c)

D

be a semi-finite decomposable

von Neumann algebra. Suppose that y is standard. (i) The A() 's are semi-finite almost everywhere.

226

PART II, CHAPTER

5

(ii)For A to be finite, it is necessary and sufficient that A(c) be finite almost everywhere. (iii)Let (1) be a semi-finite faithful normal trace on At Then, there exists a measurable field c 4 cp c of semi-finite dy(c). faithful normal traces on the A(c) -4- 's such that cp= Proof. There exists a Hilbert space K, a standard von Neumann algebra S in K, and an isomorphism (10 of A onto B. As A is a-finite, S, and consequently 8', Hence (chapter 3, proposition 11) there exist are G-finite. a V-measurable field c K(C) of Hilbert spaces over Z, a Vmeasurable field C -4- 8(C) c L(K(C)) of von Neumann algebras, and a V-measurable field C -4- (P c of isomorphisms of A(C) onto 8(C), such that one can identify

f

K with

8(C)dV(C) and (I) with I (D cdV(C).

K(C)dv(C), B with Let

4) be the normal trace on

S+ which is the transform of 4) by (D. There exists (part I, chapter 6, lemma 1) a dense Hilbert algebra U of K such that = U(U) and such that 'q) is the natural trace on U(U)+. Then, U(U) = B being decomposable, U is decomposable (chapter 4, theorem 1). Let e U=1 U(c)dv(C)We have

B = U(U) =

U(U())dv(c)

(chapter 4, proposition 3),

and so 8(C) = U(U(c)) almost everywhere. This already proves (i); furthermore, for A, or B, to be finite, it is necessary and sufficient that the characteristic projection of U be I (part I, chapter 6, theorem 6), hence that the characteristic projection of U(C) be I almost everywhere (chapter 4, corollary of proposition 7), hence that B(c), or A(c) be finite almost everywhere; whence (ii). Let 11) be the natural trace on U(U(C)) + . We have re tp = j 1p dV(C) (theorem 1).

C

Finally, let cp r be the normal trace on A(c) -1- which is the transform of tpc by 6V- . For every measurable field of operators

C 4' T(C) c A(C) + , = -00(D ( T(C))) depends measurably on C, hence the field c

(p c is measurable; furthermore, if

227

REDUCTION OF VON NEUMANN ALGEBRAS

T =

T(C)dV(C)

E A+ ,

we have

T(T) = hence

(I)(T)

= TP(4) (T)) =

flyclyT(C)))dV(C) = flyT(C))dV(C);

hence re

CI) = j (I) dV(C). COROLLARY.

that

Suppose

V

0

is standard.

Let

A = Amdv(c)

be a decomposable von Neumann algebra, and cp a normal (resp. faithful normal) trace on A+. There exists a measurable field (p r. of normal (resp. faithful normal) traces on the A(c) -1- 's such 3hat

re j (P cdv(C).

1)

(

There exist projections El, E 2 , E3 of the centre of A, with sum I, possessing the following properties: 1 0 the trace 4) 1 induced by (I) on A 1 semi-finite and faithful; 2 ° the trace (I) 2 induced by (I) on Ag is infinite for every non-zero operator; 3 ° the trace 0 induced by (1) on A4r.. 3 is zero. Let

Proof.

-

E

l

=

f

E 1 ()dV(C),

E

= f E

2

2

()dV(C),

E

3

=

f E 3 ()dV(C).

Almost everywhere, E l (c), E2(c), E3(C) are projections of the centre of A(C), which are mutually disjoint and have sum I. We have

A

AE2

E1

A(c) E1(c) dv(C)

=

-1 A(c) E2() dy(c), AE3 = f

A(C) E3() dV(C) (chapter 3, proposition 6).

1 Let C 4- (I) c be a measurable field of semi-finite faithful normal

e 1

traces on the A(C ) - 1() 's such that (1) 1 =

(1) cj-oN(C)

(theorem 2).

f

Let, for every C, 0 be the trace on A(C) E0(r) which is infinite for every non-zero Sperator, and (p c the tracé on A(C) -4- which induces 0 on A(c ) l(c ), 0 on A(C ) 2(c) , and 0 on A(C ) - (C) .

C

C

3

PART II, CHAPTER 5

228

.4- T(C)

For every measurable field

E

AU),

2 1 (1) (T(C)) = II) (T(C)El(C)) + 11) (T(C) E2(C)) c C depends measurably on C, and hence the field C + (I) c is measurable. Further, if T(C)dV(C) EA +,

T = we have

2 1 (T) = 4) (TE ) ± (i) (T ) 1 E2 2 = icp 1 cr(0 EiR )) dU(C)41) c (T(C) E2(c) )dv(C)=1(1) (T(C))dV(C), hence

re dVW C Finally, if (1) is faithful, E3 = 0, and so the cp I s are faithful. 0 I) =

(

References : [10], [80], [117], [123].

3.

Uniqueness of the decomposition.

The following lemma belongs with the global theory. LEMMA 1. Let A be a von Neumann algebra, w and w' two normal

traces on A+, and B a strongly dense *-subalgebra of A. Suppose that w(T*T) = w' (TT) < + 00 for every T E B. Then, w = w'. The set of the TE A+ such that w(T) < + co is the positive part of a two-sided ideal m of A. Let (1) be the linear form If S E M, the linear form on m which agrees with w on m+ . T 6J(ST) = (TS) on A is ultra-weakly continuous (part I, chapter 6, proposition 1). Similarly define m l and V. Let T 1 , T2 E B; then T 1T 2 is a linear combination of elements of the form R*R where RE B, hence T 1T 2 E ni nne and (i)(T 1 T 2 ) = (i) 1 (T iT 2 ); 6J P (TiT 2T) C(T 1 T 2 T) and T consequently, the linear forms T (where TEA) coincide on B, and therefore on A. Let SERInne ; by what we have just seen, the linear forms T 6 (TS) and T ci' (TS) (where TE A) coincide on 82 , therefore on B, and therefore on A; in particular, d(S) (.1) I (S).

Proof.

We have m nm' DB 2 , and hence m nm' is strongly dense in A. If RE A+, there exists an increasing filtering family (Ri ) of elements of (m n m' ) 1- with supremum R (part I, chapter 3, corollary 5 of theorem 2). Then w(R) = supw(Ri) = supw P (Ri) = w' (R). D THEOREM 3.

Let

A - I A(c)dv(C)

229

REDUCTION OF VON NEUMANN ALGEBRAS

be a decomposable von Neumann algebra. Let c two measurable fields of normal traces on the A(c) -"s.

(p cdv(c) and f dv(c) are both equal to the same semi-finite trace cp, we have (p c = Tp c almost everywhere. Proof. By chapter 2, proposition 7 (iii), A is U-finite. Let M be the ideal of the TE A such that cp(T*T) < + co. Since 4) is semi-finite, M is strongly dense in A. Let (Ti ) be a sequence of elements of A such that A is the von Neumann algebra generated by Z (the algebra of diagonalisable operators) and the T's. Since the unit ball of m is strongly dense in that of A d and the latter is metrisable for the strong topology, each T is the strong limit of a sequence of elements of M. We can therefore suppose that Ti E M for every j, that I is the strong limit of a subsequence of (Ti), and, further, that the Ti's form a *-algebra over the field of rational complex numbers. Put 0 T.

d

d

For every measurable subset Y of Z, we have, denoting by Ey the corresponding projection of Z.

C

(T

.(

dY

C)* T -(C ) )dv(C) =cp(T*.T.E y ) =f d d d Y C d

d

hence

C d

d

=

(T .(c)*T .()) < + C d d

co,

for every j, except on a negligible set N. Augmenting N if necessary, we can suppose that, for every EZ`N, the T() 's form a *-algebra over the field of rational complex numbers that is strongly dense in A(C). For EZN, we have

C

C

for any i and j, and so cp c (T*T) = lit(T*T) for any T in the *algebra over C generated by the Ti(C ) ' s. Hence (lemma 1)

$c

1Pc-

D

We will observe that lemma 1 is useless if the A(C) are factors. Theorem 3 is plainly incomplete. There again, the problems are not settled even if V is standard. References : [80], [306].

230

PART II, CHAPTER 5

Reduction of properly infinite, purely infinite, finite and semi-finite von Neumann algebras.

4.

We will very occasionally use results from part III in this section.

e

Let A

= A()dy(c)

be a decomposable von Neumann algebra. Suppose that y is standard. If A is purely infinite, A(c) is purely infinite almost everywhere. THEOREM 4.

Proof. Put H = H ( C)dv(C) . Let Y be the set of the C E Z such that A(C) is 1 ot purely infinite. We are going to show

e that there exists a projection G =

x

e

=f

x(C)dv(c)

EGA

G(C)dV(C) of A, and an

such that x(C) X 0 for

C EY

and x is a

trace-element for AG . We will deduce from this that x = 0, and hence that Y is negligible. To prove our assertion, we reduce it, by the usual technique to the case where: a. c ÷ H(C) is the constant field corresponding to a Hilbert space Ho; b. Z is compact metrisable and there exist continuous mappings

C

T (C), 1

C

T

2

(C),

C

T), (C) 1

C

T 2I (C),

of Z into the unit ball L1 of L(Ho ) endowed with the strong topology, such that, for E Z, the T() 's generate A(C) and the T() 's generate A(C) I . We can further suppose that, for every j = 1, 2, ..., there exists a j = 1, 2, ... such that T.()* = T.(C),

for any E Z. such that:

Let M be the set of the (c, T,

1 ° TT() = T P.(c)T for i = 1, 2

'1°1(0* = 1°.(C) 2

2

y) Z X

L1 X

H0

2, ...;

2 ° T is a projection; 3 ° Ty = y; 4 ° HTTi(c)T)(TTi(C)T)yly) = UTTi(c)T)(TTi(C)T)yly) j = 1, 2, ...;

for

5° H y H = 1. Then M is closed (i-f being endowed with the strong topology). For C E Y, the set of the (T, y) such that (C, T, y) E M is nonempty (part III, chapter 2, corollary 1 of proposition 7). Hence (appendix V) there exist measurable mappings c G(C), C x(C) defined on a measurable subset X of Z containing Y, such that (C, G(), x(C)) EM for every C EX. Put G(C) = 0, x(C) = 0 for E ZX. The G() 's are projections (condition 2 ° ) ,

231

REDUCTION OF VON NEUMANN ALGEBRAS

(condition 1 0 ), and

belong to A(C)

e

5 0 ).

x(C)

0 for

C EY

(condition

e

Put G -,- f G(C)dV(C),

x = f x(C)dV(C).

Then G is a projec-

tion of A, and X E G(H) (condition 3 0 ) • Let w be the normal positive form S + (Sxlx) on GAG. By condition 4 ° , we have W(S 1 S 2 ) = W(S2S1) when Sl, S 2 belong to the *-algebra generated

e by GZG and the G(1 Ti(C)dV(C))G.

As this algebra is weakly

dense in GAG (chapter 3, theorem 1) we see that x is a trace0 element for AG .

Suppose that y is standard. Let E (resp. F) be greatest projection in the centre of A such that AE (resp. the AF ) is semi-finite (resp. purely infinite). Let COROLLARY 1.

e

e

F(c)dv(c). Then, almost everywhere, E(c) [resp. F(c)] is the greatest projection in the centre of A(c) such that A(c) E(c) [resp. AF()] is semi-finite (resp. purely E(C)dV(C), F =

E

= f

infinite). Proof. Almost everywhere, E() and F(C) are disjoint projections with sum I, in the centre of A(C) (chapter 3, theorem 4), A(C)E/ c ) is semi-finite [proposition 6 of chapter 3, and theorem 2 (i) ] , and A(C)F(c) is purely infinite (proposition 6 of chapD ter 3, and theorem 4).

2.

Suppose that y is standard. For A to be semifinite (resp. purely infinite), it is necessary and sufficient that A(c) be semi-finite (resp. purely infinite) almost everywhere. COROLLARY

Proof.

This follows from corollary 1.

e

D

Let A -,- f A(c)dv() be a decomposable von Neumann algebra. If A is properly infinite, then A(c) is properly infinite almost everywhere. THEOREM 5.

Proof. There exist disjoint projections E, FE A, such that E —F—E+F=I (part III, chapter 8, corollary 2 of theorem

e 1) .

Let E .-,-

f

e E(C)dV(C), F', E(C)

F(C)

F(C)dV(C). E(C) +

We have

F(C) =

except on a negligible set N. Let CE Z' - N. For every non-zero projection G of the centre of A(C), we have GE(C) GF(C) — G, and so G is infinite (part III, chapter 2, proposition 4). Hence A(C) is properly infinite (part III, chapter 2, proposition 9). D

232

PART II, CHAPTER 5

Suppose that is standard. Let E (resp. F) be the greatest projection in the centre of A such that AE (resp. COROLLARY 1.

AF ) is finite (resp. properly infinite).

Let E =

E(C)dV(C),

F = F(C)dV(C). Then, almost everywhere, E(C) [resp. F ( c)] is the greatest projection in the centre of A(C) such that A(c) E(c) F(c) ] is finite (resp. properly infinite). [resp.A(c) Proof.

Argue as above, this time using theorem 2 (ii) and 5. 0

Suppose that is standard. For A to be finite (resp. properly infinite), it is necessary and sufficient that A(c) be finite ( resp. properly infinite) almost everywhere. COROLLARY 2.

Proof.

This follows from corollary 1.

0

References : [10], [80], [311]. Let C ÷ H(C) be a 'u-measurable field of complex Hilbert spaces over Z. For every CE Z, let (p c be the trace on L(H(C)) + defined in part I, chapter 6, section 6. Show that the field C 4- (p c is measurable. (Consider a measurable field of orthonormal bay.)

1.

Exercises.

2.

Let

A

A(C)dV(C) be a decomposable von Neumann algebra,

=

and C -4- (p c a measurable field of finite traces on the A() 's

such that (p c (IH (c) ) = 1 for every C.

Suppose that

dV(C) = 1.

e Then, (1) = f (PcdV(C)

is a finite trace on

A

such that (p(IH) = 1.

Let A (resp. A c ) be the determinant associated with (1) (resp.(p.

e Show that if T =

T(C)dV(C) EA is invertible, then T(C) is in-

vertible almost everywhere, and A(T) = explogA c (T(C))dV(C).

f

(Use exercise 2 3.

Let

A

eb

of chapter 2).

= f A(C)dV (C) be a decomposable ab von Neumann algebra.

Suppose that v is standard. The set of the C such that A(C) is type I (resp. Ill, II, III) is measurable. (Use chapter 3, corollary 1 of proposition 7, corollary 1 of theorem 4, and corollary 1 of theorem 5). [80], [311].

CHAPTER 6.

1.

DECOMPOSITION OF A HILBERT SPACE INTO A DIRECT INTEGRAL

Posing the problem.

As in chapters 1, 2, and 3, given: 1 ° a compact metrisable space Z; 2 ° a positive measure V on Z of support Z; H(C) of non-zero Hilbert spaces 3 0 a v-measurable field C over Z, we can construct canonically: 1 ° the separable Hilbert space

H =

H(c)dv(c);

2 ° the abelian von Neumann algebra Z of diagonalisable operators; 3 ° more precisely, the C*-algebra Y of continuously diagonalisable operators, whose weak closure is Z. We are now going to show that the order of these constructions can be reversed, in an essentially unique way. This will greatly add to the importance of the preceding chapters.

Existence theorems. THEOREM 1. Let H be a separable complex Hilbert space, y an abelian c*-algebra of operators in H, z the spectrum of y, and y a basic measure on Z. Suppose that I is in the weak closure of y. Then, there exists a v-measurable field c -4- H(c) of non-zero complex Hilbert spaces over z, and an isomorphism of H onto 2.

H(Ody(C), which transforms the Gelfand isomorphism into the canonical isomorphism of r„.(z) onto the algebra of continuously diagonalisable operators. (i) We will denote by f T. the weakly continuous isomorphism of LCm (Z (Z, V) onto the weak cliosure Z of y which extends the Gelfand isomorphism (part I, chapter 7, proposition 1); this since it contains I (part I, chapter weak closure is moreover

Proof.

233

234

PART II, CHAPTER 6

3, theorem 2). Let (xi, x2, ...) be a dense sequence in H. Adding to the xils their linear combinations, with rational complex coefficients, we can suppose that the xi's form a linear subspace H' of H over the field of rational complex numbers. For x, y E H, let hx,y be the Radon-Nikodym derivative of the spectral measure Vx,y with respect to V. By the formulas of part I, chapter 7, section 1, and the countability of H', there exists a v-negligible subset N of Z such that, for c 4N, the function (x, y) h 5, 4 (C) is, on H', a positive hermitian sesquilinear form. Let P(C) be the Hilbert space obtained from H' by passing to the quotient and completing, H' being endowed with this sesquilinear form, and let OC) be the canonical mapping of H' into H(C). Let N i Z "---N be the set of the €ZN such that H(C) = O. For C E Z N, the condition E N1 is equivalent to the condition "hx . x.(C) = 0 for every i and every j"; we therefore d see that Ni is ' \)-measurable; let f be its characteristic function; we have

(Tfcci Tfi) = (Tfcci I x i ) =

(c)dv 5 .

.(c) -

if( c)h.

.(c)dv(c)

=0,

hence Trxi = 0 for every i, hence T_E. = 0; we thus see that N1 is V-negli'gible. Choosing new spaces O(C) arbitrarily on N UNi, which is v-negligible, we can arrange that H(C) 0 for every -

EZ

(ii) We now endow the H(c)'s with a measurable field structure. For c N uN i , put xi(C) = cp(C)xi. The number (x.() x.())

d

= hx • ,x • (c) 2 d

depends measurably on C; and, for every C Et.N uNi, the X() 's form a total set in H(C) . There thus exists on the H(C) 's exactly one measurable field structure such that the fields C xi(C) are measurable vector fields (chapter 1, proposition

4). (iii) We are going to define an isomorphism of H onto

H(c) dv(c) . x

Let

=

T i-1 fi

We have

xi E H

[fl . f2 ,

fn E

co L

C (Zi V) ].

235

REDUCTION OF VON NEUMANN ALGEBRAS

l

(C)dV(c) f-(C)f-d (Oh == .[Cb.)x, x (C) =fh xix (C)dV(C) = x-' 2 jx i1 2

x ii

=1 1

j=i f-(C)f

•(C)(x.(c) lx.(c))dv(c)

f. ( C)x )x. (C) ()

i=1

This shows at once that the vector field

is square-integrable, and consequently that this field only depends (up to negligible sets) on the vector x and not on its

n representation in the form

1 T .4, x., and finally that the i=1 J i 1-

mapping U0 which, to the element

n

x

of

H,

assigns the field

e f*.(c)x.(c)

are dense in

of f H(C)dV(C) , is an isometry. These fields

f e H(r)d\.'()

(chapter 1, proposition 7).

Moreover,

n

X T x. are dense in H. Hence U0 i=1 fi 76 extends to an isomorphism U of H onto H(C)dV(C). the vectors of the form

on A function fc LC (Z, V) defines on the one hand an operator Tf in H, and on the other hand a diagonalisable operator T.F in

fe

H(c)dv(c).

With the above notation, we have

Tfx = Moreover, T fF U 0

x

i = 1 Tffi

x..

is the vector field

c i=1 We thus see that T? 0

f(c) f 2

x = U 0 T fx,

x 2(C) •

and hence that T F = UT?

-1

.

236

PART II, CHAPTER 6

Let H be a separable complex Hilbert space, and Z anabelian von Neumann algebra in H. Then, there exist a compact metrisable space Z, a positive measure v on z with support z, a v-measurable field c ÷ H(c) of non-zero complex Hilbert spaces THEOREM 2.

e over z, and an isomorphism of

H onto

H(c)dv(c) which transforms Z into the algebra of diagonalisable operators. 1

Proof. Let Y be a sub-C*-algebra of Z, weakly dense in Z, whose spectrum Z is compact metrisable and carries a basic measure v [part I, chapter 7, proposition 4]. There exist (theorem 1) a y-measurable field C 4- H(C) of non-zero

e Hilbert spaces over Z and an isomorphism of

H

onto

f

H(C)dV(C)

which transforms Y into the algebra of continuously diagonalisable operators, and hence Z into the algebra of diagonalisable operators [chapter 2, proposition 7, (i) ] . D

Let H be a separable complex Hilbert space, and A a von Neumann algebra in H. There exist a compact metrisable space z, a positive measure v on z of support z, a v-measurable field c 4- H(c) of non-zero complex Hilbert spaces over Z, a vmeasurable field c 4- A(C) of factors in the H(O's, and an COROLLARY.

e

isomorphism of

I

H onto

H(c)dv() which transforms A into

e A(c)dv(C). Proof.

Apply theorem 2 to the centre Z of

A.

We obtain Z, V,

e c÷

H(C), and an isomorphism U of H onto [ H(C)dV(C) which J

transforms Z into the algebra of diagonalisable operators. We have ZcAcV, hence UAU -1 is decomposable (chapter 3, theorem 2). Hence there exists a v-measurable field C ÷ A(C) of von

e Neumann algebras in the H() 's such that UAU-1 -

A(C)dV(C).

As the centre of UAU is the algebra of diagonalisable operators, the A(C)'s are factors almost everywhere (chapter 3, theorem 3). D The corollary to some extent reduces the study of von Neumann algebras to that of factors; this was one of the principal goals of "reduction theory." Nevertheless, we saw in part I that one can study general von Neumann algebras directly by methods which comprise the "global thoery."

References : [28], [49], [80], [100], [117], [145], [193], [194], [205], [206].

REDUCTION OF VON NEUMANN ALGEBRAS

237

Uniqueness theorems.

3.

Given H and y, we wish to show that Z, V and the field C -4' H(C) are essentially unique. We have already remarked (chapter 2, section 4) that Z may be canonically identified with the spectrum of Y in such a way that the canonical isomorphism of L(Z) onto Y may be identified with the Gelfand isomorphism. This establishes the uniqueness of Z (up to homeomorphism), and allows us to state the uniqueness theorem in the following way:

Let Z be a locally compact space, countable at Let y be a positive measure on z of support z, c .4- H(c) a y-measurable field of non-2ero Hilbert spaces over THEOREM 3.

infinity.

e

z,

H = f H(c)dy(c), Y the algebra of continuously'diagonalisable operators in H, and f -4- Tf the canonical isomorphism of 4,0 (z) onto Y. Define, analogously, v l, c ÷ H l (c), H1, yl, f -4- rr 3 .. Let u be an isomorphism of H onto H 1 transformng Tf into tj-. for every fE Loo (z). Then, y and y l are equivalent, anorthere exist after necessary modification of the H(c)b and the H 1 's on negligible sets, an isomorphism c ,4- v(c) of the field (H(r)) onto the field (H 1 (0), such that u = wv, where v is the

e

e

isomorphism f v(c)dv(c) of H onto f H (c)dv(c), and where w is

1 the canonical isomorphism of f H i mdv(c) onto H 1 . Proof. Identify Z with the spectrum of y and of Y1 6

in such a way that, for every function fE Loo (Z), Tf and T. - UTfU-1 are the elements of y and y l corresponding to f und6r the Gelfand isomorphisms. Then, if x, y E H, we have Vx ,u = VUx,Uy, hence are the same. Hence the basic measures defined on Z by Y and

y and vl are equivalent.

Let Y = / eHl(C)dV(C), W be the canoni-

cal isomorphism of if onto H 1 , and '.the image of yl under W-1 , which is the algebra of continuously diagonalisable operators in Y. Every function fE L(Z) defines an operator T . in Y, and we have 71'-'

= (W

-1

f

U)T (W f

-1

U)

-1

.

Hence W-1 U = V is a decomposable linear mapping of H into Y (chapter 2, theorem 1), from which it follows that there exists a measurable field c -4- V(c) of linear mappings of H(c) into

e with V = of

f

Finally, since V is an isomorphism

V(C)dv(c).

H onto if, we have V*V = I

H'

VV*

ly,

PART II, CHAPTER 6

238

hence V( ) * y( ) = I

H(c)

,

v( c ) v( c)* =

c)

almost everywhere, i.e. V(C) is almost everywhere an isomorphism of H(C) onto H1(C). 0

Let z be a Borel space, v a standard positive ) ) a v-measurable field of non-zero come plex Hilbert spaces over z, H = H(c)dv(c), and Z the algebra of diagonalisable operators in H. Define z l , v i, E l = (H1 (c 1 )), H1 , and Z I analogously. Let u be an isomorphism of H onto H 1 transforming Z into Z 1 . Then there exist: THEOREM 4.

measure on

z, E = (H(

1 ° a v-negligible Borel set N in z, and a v 1-negligible Borel set N 1 in z1; 2 ° a Borel isomorphism n of ZN onto z 1 ', N 1 which transforms v into a measure V 1 equivalent to v 1; 3°

an n-isomorphism (v(c)) of Elz , N onto El lz 1 N, N 1 , which

H 1 (c 1 )Lj1 (c 1 ) in

defines an isomorphism v of H onto H i =

such

a way that u = wv, where w is the canonical isomorphism of onto H 1 . Proof. We can suppose that Z and Z are compact metrisable. 1 The isomorphism U defines an isomorphism of Z onto Z1, and therefore an isomorphism / of q(Z, V) onto *° (Z i , V 1 ). Hence there exist N, N1, y l , n, with properties 1 ° and 2 ° of the theorem, n also defining the isomorphism / (appendix IV). For fE Loe c (Z, v) [resp. flc V 1 )], denote by Tf (resp. T1. 1 ) the diagonalisable operator of H (resp. 1'1) defin7d by f f1 ). If f and fi correspond under /, T and WTfW I corf respond under U, i.e.

WT, W = UT fU , or 4 -1 -1 Tf w u w UT

,

f' or

—1 T„ V = VT J

f'

putting

= W 1 u E L( H,

).

239

REDUCTION OF VON NEUMANN ALGEBRAS

It all, therefore, comes down to showing that V is "decomposable" in a generalised sense. We could have presented this generalisation in chapter 2, but it would have been cumbersome. We will briefly indicate how the arguments of chapter 2 may be extended, leaving the details to the reader, which should not, however, Let (x l , x2, ...) be a fundamental secause any difficulty. that Xi EH; we quence of measurable vector fields over Z`N such -2 can put yi = Vxi EYi . Making use of the equality TAV = VTf, we show, just as for theorem 1 of chapter 2, that

p.x.(C) y

i=1 "

i=1 "

almost everywhere on for any rational complex numbers pn . Hence there exists a continuous linear mapping P1 , p2, E < 1 V(C) of H(C) into H1(11(C)), for such that II V(C) and

V(Ç)x.() =y i (n(0) for every i, almost everywhere. Generalising proposition 1 of chapter 2, we conclude from this that the V() 's transform every measurable vector field over Z`-N into a measurable vector field over Z1`1\11.

Arguing similarly with V-1 , there exists, for CiE a continuous linear mapping V P (Ci) of H1(C 1 ) into H(11 -1 (C1)), such for every i, 1, and V i (C1)yi(C 1 ) = xi(11 that almost everywhere; and the V'(C 1 )'s transform every measurable N 1 into a measurable vector field over vector field over V(C) and V' (TI()) are, almost everywhere, Z`N. Consequently, inverse isomorphisms of H(C) onto H'(71()) and of H t (n(c)) onto H(C). After modification on a negligible set, (V(C)) cczv is an fl-isomorphism of EIZ•, N onto ElIZ1%--N1. The corresponding isomorphism of H onto H1 acts on elements of the form Tfxi in the same way as V, and is therefore equal to V.

H v?( c1 ) H _5_

Reference : [80].

Exercises.

1.

Let C

H(C) be

a v-measurable field of com-

H = 9 H()dy(c), Z the and A a factor contained

plex Hilbert spaces over Z,

algebra of

diagonalisable operators, in V. - Supvon Neumann subalgebra pose that Z is a maximal abelian in A'.

0 almost everywhere, there If A is discrete, and if H(C) a. 0 separable exist: 1 a Hilbert space Ko; 2 0 almost everywhere on Z, an isomorphism U(C) of H(C) onto Ko; 3 ° for every TE A, a unique operator TIE L(K0 ) such that

240

PART II, CHAPTER 6

6

T =

U(C)

-1

T

1

U(C)dv(C).

(Use chapter 2, proposition 8 and exercise 1, and theorem 3 of the present chapter). As T runs through A, T 1 runs through L(K0 ).

e

b.

In the general case, for every T = f T(C)dV(C) EA, T / 0

implies T(C) 0 almost everywhere. [Let Y be the measurable subset of Z consisting of the C such that T(C) = 0. Let E be the diagonalisable projection corresponding to Y. We have T E = 0, hence E = 0] [16], [57], [58]. Problem: What relations exist between the T() 's? [245], [265].

2. With the hypotheses of theorem 2, show that one can, with the notation of that theorem, take for Z a compact interval of the real line. (Thanks to part I, chapter 7, exercise 3 f, take Y to be generated by a single hermitian element. Then the spectrum of V is a compact subset of the real line) [80].

PART III FURTHER TOPICS

M

CHAPTER 1.

COMPARISON OF PROJECTIONS

Comparison of projections. DEFINITION 1. Let A be a von Neumann algebra, and E and F two projections of A. E and F are said to be equivalent (relative to A), and we write E F, if there exists an element u of A such that u*u = E, UU* = F. We write E F, or F E, if there exists a projection of A equivalent to E and majorized by 1.

F. If X = E(H) and y y, tions X y,

F(H), we will also make use of the notaX.

Two projections E and F of A are equivalent if there exists in a partial isometry with E as initial projection and F as final projection, or, equally, if there exists in A an operator whose restriction to E(H) is an isometry of E(H) onto F(H); this restriction then defines a spatial isomorphism of AE onto AF, and of A onto A.,. We see at once that the relation E F is an equivalence relation, and that the relation E -
A

Let E and F be two projections of A, and E' and F P their cenF' (by part I, chapter 1, tral supports. If E F, we have E' E', so that proposition 7, corollary 1), and similarly F' E' = F'. Consequently, if E -< F, we have E' < F'. Let (Ei)i ci [resp. (F.2 ) 2EI • ] be a family of pairwise disjoint projections of A, and let E = X E. (resp. F = X F.) . If

2 2 iEI icI Ei —F i for every iEI, we have E F; indeed, let Ui be a partial isometry of A having Ei as initial projection and Fi as final projection; there exists a unique partially isometric projection U having E as initial projection, F as final projection, and coinciding with Ui on E(H); this operator U is invariant with respect to the unitary operators of A', hence U e A, and our Fi for It follows from this that, if E assertion is proven. every icT, we have E -< F.

243

244

PART III, CHAPTER 1

Let (Ax)XEK be a family of von Neumann algebras, and let A be their product. For every x EK, let Ex and Fx be two projections of A. Let E = (Ex ), F = (Fx ), which are projections of F) to hold, it is necessary and suffiA. For E F (resp. cient that we have Ex —Fx (resp. Ex -‹Fx ) for every x€ K.

It is clear that the relation E —F implies E F and F Conversely: PROPOSITION 1. If E F and

FP Proof. We have E F' F and F E' imply F l E —F' —E". Let, then, U as initial projection, and

(2n)

u

n

F.

F and F —E l E. The relations E" with E" E' E; furthermore, be a partial isometry of A having E E" as final projection. Put

X' = E' ( 4),

X = E(H), X

E, we have E

E.

(X),

X" = E"(H),

X (2n4-1) = Un (X')

(so that X (C)) = X, X (1) . X I , X (2) . X"). The relations x(2n1 1)Dx(2n f 2) X D X I D X" imply that x(2n) hence the sequence -

- -

CO

X (i )

i s decreasing; let

Y =

v(i) (I A .

Then, X is the direct sum

i=0 of the orthogonal subspaces

Y,

e X(1),

x (1) e x (2) ,

x (2) e x(3),

X (3) e X (4)

• • •

and X / is the direct sum of the orthogonal subspaces y, X (2)

ex (3)

X (1) eX (2) , X

Now, U maps X (i) e X hence

(i+1)

(4)

eX

(5)

X

(3)

eX

(4)

isometrically onto X(i+2) e

(i+3) X (i) 0 X (i+ 1 ) ,, X (i+2) e X We thus see that X — X / and, consequently, E « E / — F. PROPOSITION 2.

Let Te A.

D

Then the supports of T and T* are

equivalent. Proof.

Let T . = WIT 1be the polar decomposition of T. The supports of T and T* are WW and WW. Moreover, WE A. ]

COROLLARY 1.

Let E = Px, F

Py be two projections of A.

Then Xe (X n YI )

Y e (V nX1)

245

FURTHER TOPICS

1 1 . The null space of FE is generated by X and X n e (X n Y 1 Hence the support of FE is the projection onto X Similarly, the support of EF is the projection onto Y e (Y n X I Now, EF = (FE)*. 0

Proof.

).

).

Let E let E l be the support of T 1 , x and y = T'x. Then, X' -‹ X'''. Furthermore, if E l .= = x, we have XA XAx' A is Proof. For every T E A, we have Ty = TT l x = T / Tx, and SO,A the closure of T' (X),i.e. of T'E(H). By proposition 2, A x COROLLARY 2.

is therefore equivalent to the closure of ( T l E A1 X'

* ( H) = E 4T 1 *(H)c XA , '

whence

XA

XA .

If E l x = x, X A is contained in E'04) = T l *(H), and henc9 the Ax *(U) is equal to X; we therefore have closure of ET'

0

If A is finite, every projection of A equivalent to I is equal to I. PROPOSITION 3.

If E —I, we have I = U*U, E = UU* with a Ue A, and hence, for every finite trace (I) on A, cp(I — E) = cp(U*U) 4)(UU*) = O. Hence I E = O. D

Proof.

The proof of proposition 1 is similar to a well known proof in the theory of cardinal numbers (the role of equipotence of sets is here played by equivalence of projections). In chapter 8, we will prove, not without some difficulty, the converse of proposition 3. In some texts, the notation E-‹ F means, "E F and E is not equivalent to F."

(in

our notation),

References : [42], [65].

2.

A theorem on comparability.

Let E l and E2 be projections of A, and F 1 and F 2 their central supports. If F1F 2 0, there exist two equivalent non-zero projections G l , G 2 of A, majorized by E l and E 2 respectively. LEMMA 1.

Proof. X

1

Let

= E (H), 1

X

2

= E

2

(H),

y1

F 1 (H) (H)

y 2 = F 2 (H).

By part I (chapter 1, proposition 7, corollary 1), Y 1 is the closed linear subspace generated by the T(X1)'s, as T runs through A; we can even (part I, chapter 1, proposition 3), restrict ourselves to taking only the unitary elements T of A, in which case T(X 1 ) X 1 . As F 1F 2 0, we can therefore suppose, replacing X1 and X 2 by equivalent subspaces if necessary, that

PART III, CHAPTER 1

246

X 1 and X2 are not orthogonal. Then, corollary 1 of proposition 2 furnishes two equivalent non-r.,ero subspaces contained in X 1 and X2 respectively. 0 THEOREM 1. Let A be a von Neumann algebra, Z its centre, and E and F projections of A. There exists a projection G of Z such that EG and E(I - G) < F(I - G).

Proof.

By Zorn's lemma, there exists a maximal family ((Et, Fi))i ci of pairs of projections of A possessing the following properties: the E's (resp. FiTs) are non-zero, pairwise disjoint, majorized by E (resp. F), and Ei — F. Let E ° = X E. iEI

E,

F ° = X F. iEI

F.

We have E ° F ° . Let E l = E - E ° , F 1 = F F ° . By lemma 1 and the maximality of the family ((Ei, the central supports 1 1 of E and F are disjoint. There therefore exists a projection G ,- Z such that

E

1

G,

1 F 5_ I - G.

This established, we have

E°G

F ° G,

E ° (I - G) rs-, F ° (I - G),

hence FG

F ° G E ° G + E i G = EG,

and

E(I - G)

E ° (I - G)

F ° (I - G) + F l (I - G) = F(I - G).

D

Let A be a factor, and E and F two projections of A. Then either F or F -‹ E. Proof. The projection G of theorem 1 is equal to 0 or to I. 0 COROLLARY 2. Let A be a von Neumann algebra, Z its centre and (Ei)i cl a family of non-zero projections of A, which are oairwise chsjoint and equivalent. There exists a projion G of Z ., and a family (Fx)xEK of non-zero projections of A, paiPwise disjoint and equivalent, and majorized by G, possessing the following properties: 10 I c K; 2 ° for xEl, Fx Ex G; 3° if We put COROLLARY 1.

FO = G - X F , x xEK we have F 0 Fx, and F 0 is not equivalent to the Fx 's. more, if i is 2-nfinite, we can suppose that Fo O.

Further-

247

FURTHER TOPICS

Let (E,..) xEK be a maximal family of projections of A extending the family (Ei)i ei (so that I cK), for which the Ex 's are pairwise disjoint and equivalent. Let E 0 = I - X E. Let

Proof.

XEK

G be a projection of Z such that

E G o

E G,

E (1 - G)

E (I - G)

(the choice of x in K being immaterial as the Ex 's are equivalent). Put F o = E G, F = ExG. The relation Fx F o would 0 x imply Ex -<E 0 , contrary to the maximality of the family (E_) x x€K . Hence, on the one hand Fx 0 and, on the other hand, F is not 0 equivalent to the Fx 's. We thus have

F =EG=G0 0

EGGxEK

x

XF x€K

x

Finally, if T is infinite, K is infinite; let K' be the complement in K of an arbitrarily chosen element of K; K' is equipotent with K; we have G = F

+IF xeK

F

x

hence (proposition 1) G —

o

+

IF -<1F , x x

XEK

G,

xEK

F. Replacing the F's by suitably xEK

chosen equivalent projections, we see that we can suppose that GF. xEK

Let A be a von Neumann algebra. For A to be continuous, it is necessary and sufficient that every projection of A be the sum of two equivalent disjoint projections. COROLLARY 3.

Proof. If A is not continuous, A possesses a non-zero abelian projection E; two equivalent projections of A majorized by E have the same central support (section 1), and are therefore equal (part I, chapter 8, section 2); hence A does not possess the property of the corollary. Now, suppose that A is continuous; let E be a non-zero projection of A ; we shall show that E majorizes two non-zero equivalent disjoint projections E l , E 2 of A; applying the same result to E - (E I + E 2 ), and using transfinite induction, we see that E will be the sum of two disjoint equivalent projections of A, and the proof will be completed. To prove our assertion, we can replace A by AE (which is continuous); it is therefore necessary to prove that if A is continuous and 0, there exist two non-zero, disjoint, equivalent projections of A. Now, A is not abelian; let El be a projection of A not belonging to the centre of A, and let Ei = I - El; applying theorem 1 to E l and Ei, we obtain two disjoint

248

PART III, CHAPTER 1

E 12 ; then E 2 -, E 3

projections E2, E 2f of A such that 0 / E2 D and E2, E 3 are disjoint.

E 21 ,

Applying corollary 2 of theorem 1 to the case of a factor, the reader will convince himself that this corollary plays the role of a "euclidean division." References : [6], [10], [42], [54], [65], [105], [114], [117]. 3.

Cyclic projections of of A ' .

A

and cyclic projections

Let A be a von Neumann algebra in H, and (T)i ci a family of elements of A. There exists a TEA such that: I° H TT H _< 1 for every iEl; 2° if y E H is such that Ty 11 2 < + 00, we have y E T(H) . LEMMA 2.

iCI

Let H' be a complex Hilbert space whose dimension is card I + 1. Let B be the algebra A 0 L(H') in H ® H'. Using an orthonormal basis of H', identify H 0 H' with H e (ei e ei), where H and the H.'s are pairwise orthogonal subspaces of Hoff'. For every j E 1, aere exists a partial isometry Ui in B having H and Hi as initial and final subspaces. The algebra A may be identified with BH (part I, chapter 2, proposition 5).

Proof.

Let X be the set of the x E H 18) H' such that PH.x = Ui T.PHx for every i e I. It is clear that X is a closed linear subspace of H 0 HP, invariant with respect to every unitary operator of B'. Hence Px E B. Furthermore, the conditions xe X and PHx = 0 imply that PH.x = 0 for every i E I, so that x = 0. Corollary 1 of proposition 2 then shows that Px PH relative to B: there exists a partial isometry U E B whose initial projection is majorized by PH and whose final projection is Px. Let be the restriction to H of PHU. We have Te BH = A, and T(H) = PH;X), from which it follows that T(H) is the set of the y c H such that

X iEI

II

T

H 2 = yH

UTy

1

2

< +

iEI

It follows from this that, for every z E H, the numbers M TI•Tz II are bounded. The numbers H TT H are therefore bounded part III, chapter 3, theorem 2). Multiplying T by a suitable scalar, we obtain the operator T of the lemma. D

Let A be a von Neumann algebra in H, x an element of H, and y an element of A. There exists an element z of H and elements S, T of A such hat x = Tz, y = sz, z = Ez (E denotes the support of T). Proof. Let (S 1 , S 2 , ...) be a sequence of elements of A such LEMMA 3.

that H snx - Y 4-n . Put 2n (Sn+ 1 s n ) = Tn , and let T be an operator of A which possesses the properties of lemma 2

249

FURTHER TOPICS

vis-.-vis the family (Tn ). We have (Sn+1 - Sn)T < 2 -n , hence the operators S nT converge, in the sense of the norm, to an operator S of A. Moreover,

II

CO

CO

11Tx11 2 = n=1

OD

4n11 S n=1

n+1 x -

nx

4n (2.4 -n ) 2 < +

<

co-,

n=1

hence there exists a zEH such that T2 = x, and we can plainly suppose that 2 is orthogonal to the kernel of T, i.e. E2 = z, where E denotes the support of T. Finally, the equality SnTz = Snx yields, in the limit, Sz = y.

Let A be a von Nevimann qlgebra io H, aç1 x and x l elements of H. The relations E xm l E2 and E' Em are x1 equivalent. A there exists a y E XxA and a partial Proof. If E A isometry U' E A' 1 such that x l = Ws. Apply lemma 3 to A, x and y, and let z be the element of H thus obtained. By corollary 2 of proposition 2 (where we interchange the roles of A and A'), THEOREM 2.

we have

E

A' xi

We see similarly that 4 ' COROLLARY.

lent Proof.

E A' -< E A' — E A' . z x Y

A -< E. BY implies that Ex1

1 The relations E xl1

p

x1 — EA' x and EA' x are equiva-

Apply theorem 2 and proposition 1.

I]

Reference : [65].

Applications: I. Properties of cyclic and separating elements. LEMMA 4. Let A be a von Neumann algebra, E and F projections of A, and E' and F' their central supports. Suppose that F' E'. If there exists a cyclic element x for AE and a separating element y for AF , then we have F E. Proof. Suitably identifying A with a product of two von Neu4.

mann algebras, we can, thanks to theorem 1, make do with studying the case where E F. Let E l be a projection of A such that E E F. There exists a cyclic element for AEl . We can therefore restrict attention to the case where E F. We then have E' = F'. The element x is separating for A, and therefore separating for A ' (since the inductions of A'r on A and A;, are F E isomorphisms), and therefore cyclic for AF . Thus,

X A F = F(H)

D XAF and,

y is cyclic for

AD!,,

4 X

consequently (theorem 2), X and so XA = F(H).

.

Now,

Moreover, E(H) )(x .

PART III, CHAPTER 1

250

Hence E(H)

F(H).

0

Let x be a separating qlement for A. exists a cyclic element for A, we have 4 — I. PROPOSITION 4.

If there

Let E P = E A A'. By part I (chapter 1, proposition 7, corollary 2), E P has I for central support. Moreover, x is E / and, conseHence (lemma 4), we have I cyclic for 1), (proposition I es-, quently E'. E

Proof.

If there exists a cyclic element and a separating element for A, there exists an element which is both cyclic and separating for A. COROLLARY.

Proof. Let x be a separating element for A. There e*ists (proposition 4) a partial isometry U / EA' having E / = E I as final projection. As x is separatintalprojec,nd ing and cyclic for AE /, U / x is separating and cyclic for A. D

Let A (resp. Al ) be a von Neumann algebra; suppose that there exists a cyclic and separating element for A (resp. A 1). Then, ever? isomorphism (11 of A onto A l is spatial. THEOREM 3.

Proof.

There exist (part I, chapter 4, corollary of theorem 3) a- complex Hilbert space K, a von Neumann algebra C in K, and projections E P , F / of C / , of central support I, such that A may be identified with CE!, A l with CF P and t with the isomorphism TF 1 (TE C). By the hypothesis of the theorem and lemma 4, TE! we have E / F / . Let U P be a partial isometry of C / having E / as initial projection, and F / as final projection. For every TEC, the restriction of U P to E' (K) transforms T E / into TF I. Hence t is spatial. 0 We will frequently discover (theorem 6; chapter 3, proposition 3; chapter 6, proposition 10; chapter 8, corollaries 8 and 9 of theorem 1) conditions which ensure that an isomorphism is spatial.

Let A be a von Neumann algebra in H possessing a separating element, and (I) a normal positive linear form on A. Then there exists an element x of H such that cp = w. THEOREM 4.

Let 2 be a separating elypent for A. Let E P = E 2 . The central support of E l is I (because E3 = I) , and hence the induction of A on Av. ' is an isomorphism and enables us to transfer the form cp to AE !. It suffices to prove the theorem for AE / and for this transferred form. We can thus suppose henceforth that z is cyclic for A. If the theorem holds for (I) + w z, it holds for (I) which is majorized by (I) + w2 (part I, chapter 4, lemma 1). We can therefore suppose henceforth that 14) > wx , and hence that (1) is faithful. Then (part I, chapter 4, lemma 4 and proposition 1), there exists a Hilbert space K, a von Neumann algebra B in K, a cyclic and separating element y for B and an isomorphism t of A onto B such that (p(T) = (II(T)y y) for TE A. Now, by theorem

Proof.

FURTHER TOPICS

3, IT, is spatial.

251

D

Let A be a von Neumann algebra. The following conditions are equivalent: (i) Every normal positive linear form on A is a form of the kind wx; (ii)For every projection E of A such that AE is a-finite, AE possesses a separating element. These conditions are satisfied in particular when A is abelian, and when A is standard. COROLLARY.

If A possesses property (i), it is clear that, for every projection E of A, AE possesses property (i). Now, if, further, AE is a-finite, there exists a faithful normal positive linear form on AE . This form is a form wx , and x is separating for AE .

Proof.

Suppose condition (ii) is satisfied. Let (I) be a normal positive linear form on A, and let E be its support; then AE is a-finite, and therefore possesses a separating element; by theorem 4 applied to AE , there exists an element x E E(H) such that (1)(T) = (TX Ix) for every Te EAE; then, for every TEA, we have

cp(T) =11)(ETE) = (ETExlx) = (Tx1x). If A is abelian, A satisfied condition (ii) by part I, chapter 2, corollary of proposition 3. Suppose that A is standard, and let E be a projection of A such that AE is a-finite. We show that AE possesses a separating element. By proposition 3 of part I, chapter 2, applied to AE , it is enough to consider a cyclic element x. Let F be the the case where A central support of E. The element x is separating for A, and therefore A'. As A is standard (part I, chapter 5, section 5), AF possesses a separating element y. Then, E is separating for AE 0

Cf. on this subject, chapter 6, section 3, and chapter 8, corollaries 10 and 11 of theorem 1. When a von Neumann algebra A possesses a cyclic and separating element, A and A' are, in certain respects, "equally large"; this idea will be made precise in chapter 6 (and even in the following section, to some extent). References : [17], [19], [31], [40], [62], [65], [66], [67], [70]. [89], [10.0], [119].

252

PART III, CHAPTER 1

Characterisation Applications: 11. standard von Neumann algebras. Let A be a a-finite von Neumann LEMMA 5. and Z its centre. If there exists an involution J muting with the projections of Z, such that Jetl' = possesses a cyclic and separating element. 5.

of algebra, of H, comA', then A

We have (part I, chapter 2, proposition 3) A = A i x A 2 , with A l possessing a cyclic element and A 2 a separating element. However, J commuting with the projections of Z, defines involutions which interchange A l and Ai, A2 and hence there exists a separating element for A l and a cyclic element for A2. Hence A possesses a cyclic element and a separating element, and consequently (corollary of proposition 4) a cyclic and separating D element.

Proof.

Let A be a a-finite von Neumann algebra. For A to be standard, it is necessary and sufficient that A be semi-finite and that A possess a cyclic and separating element. THEOREM 5.

Proof.

The conditions are necessary (lemma 5 and part I, chapter 6, corollary of proposition 9). Conversely, suppose that A is semi-finite and possesses a cyclic and separating element. There exists (part 1, chapter 6, corollary of proposition 9) an isomorphism (I, of A onto a standard von Neumann algebra B. As A is G-finite, the same is true of B, and so B possesses a cyclic and separating element (lemma 5). Then, (I) is spatial (theorem 3), and hence A is standard. D

Let A be a finite von Neumann algebra, possessing a cyclic and separating element. Then, every separating element for A is cyclic for A, and every cyclic element for A is separating for A. COROLLARY.

By theorem 5, A' is antiisomorphic to A, and is therefore finite. Hence the two assertions of the corollary are equivalent, and follow immediately from propositions 3 and 4.

Proof.

We are now going to attempt to free ourselves of the countability hypothesis of theorem 5.

Let A be a von Neumann algebra, (Ei)i ci an infinite family of projections of A, with supremum 1, such that the AE.'s are a-finite, and (F cc ) xEK a family of pairwise disjoint non-zeiio projections of A. Then, card K Card I. TpdAMA 6.

Proof.

Since AE . is G-finite, there exists a countable set al such aat Ei = E. Let Ki be the (countable) subset of K consisting of the XEK such that F(Mi) X O. We have K = U Ki, because, if an X E K were such that F(M) = 0 for iEI every i E I, Fx would be disjoint from all the non-zero Ei's. Hence

FURTHER TOPICS

Card K

253

(Card 1) 8 =Card I. 0

Every von Neumann algebra A is the product of von Neumann algebras B possessing one or other of the following two properties: (i) B is a-finite, (ii) there exists an uncountable family (Ei ) ici of pairwise disjoint, equivalent projections of. S with sum I, such that the 5Ei '5 are a-finite. Proof. Let E be a non-zero projection of A such that LEMMA 7.

AE is G-finite (for example a non-zero cyclic projection). Decreasing E if necessary, we can suppose (theorem 1, corollary 2) that there exists a family (E,.. a,)xEK of projections of A, equivalent to E, pairwise disjoint, and a projection G of the centre of A majorizing the Ex 's, such that F = G - X E x -‹ E. The x K algebra AF is G-finite. is countable, there If K exists a countable cyclic set for Ac, and hence AG is If K is uncountable, we can (theorem 1, corollary 2) G-finite. suppose that F = O. In both cases, AG possesses property (i) or property (ii) as stated in the lemma for B. It is now enough to repeat the argument for Ai _ G , and to use transfinite induction. THEOREM 6. Let A be a von Neumann algebra in H, and Z its centres suppose that there exists an involution of H such that JAJ = A' and JCJ = c* for every c e Z. Let A l , Z i, J 1 , acting in another Hilbert space H 1 , have the same properties. Then, every isomorphism q) of A onto Al is spatial.

Proof.

If

A =

H A., the iEI

A i 's

satisfy the same hypotheses as

A

(thanks to the fact that J commutes with the projections of Z). Moreover, if A is G-finite, theorem 6 follows from lemma 5 and from theorem 3. By virtue of lemma 7, we can therefore confine ourselves to the case where there exists in A an infinite family of pairwise disjoint equivalent projections, with sum I, the corresponding reduced algebras being There then exist in A', Al , Al similar families of G- finite. projections, these four families being equipotent. By part I (chapter 4, theorem 3, corollary), there exists a von Neumann algebra B and projections Px, Py of B', with central support I, such that A may be identified with Bx , Al with By, and (1) with the isomorphism Tx T y (TE 13). As in theorem 3, the proof will be complete if we show that X — Y. Theorem 1 enables us to restrict attention to the case X Y, Y X. Suppose, for example, that X Y. This leads us at once to the case where X c Y. There exists an infinite family (Px .) . [resp. (Py.) . ] of mutually disjoint, equivalent proljeHions of B', witR igPx(resp.Py),such that the Aki's (resp. AWs) are G-finite. Moreover, we can add to the family

254

PART III, CHAPTER 1

(Pxi )i ci non-zero projections -‹ Pxi in such a way that the family (1DXf17)xEK obtained consists of mutually disjoint projections of if', with sum P y (as theorem 1, applied transfinitely, yields immediately). Lemma 6, applied to 13 and to the families 1 , shows that I and K are equipotent. Hence (Xx)xEK' ( Y) p px 0 X. = Px and finally Y — X. Y xEK

y

Let A be a von Neumann algebra in H, and Z its centre. For A to be standard, it is necessary and sufficient that A be semi-finite and that there exist in H an involution J possessing the following properties: (i) JAJ = Al; (ii) JCJ = C* for every c E Z. Proof. The conditions are necessary (part I, chapter 5, COROLLARY.

corollary of proposition 2, and chapter 6, corollary of proposition 9). Now, suppose that A is semi-finite. There exists (part I, chapter 6, corollary of proposition 9) an isomorphism 11 of A onto a standard von Neumann algebra Al which satisfies the conditions of the corollary. If A satisfies them as well, (1) is spatial (theorem 6), and hence A is standard. 11 The conclusion of the corollary of theorem 5 no longer necessarily holds if the hypothesis that A is finite is suppressed (exercise 6).

Lemma 6 clearly enables one to associate with each von Neumann algebra a cardinal number which is invariant under algebraic isomorphism. The conditions of theorem 6 are satisfied by certain not necessarily standard von Neumann algebras (cf. [13]).

Theorem 6 generalises theorem 4 of part I, chapter 6. References : [17], [66], [101].

Exercises. 1. Let A be a von Neumann algebra, and E and F two projections of A. For E F, it is necessary and sufficient that there exist a projection GE A with G E, G — F. (Let E E l _< F. Apply the comparability theorem to F - E l and I-E. If F I - E, the proposition is obvious. If F-E 1 , we have F I and we can take G = I) [65]. 2. Let A be a von Neumann algebra, and Px and P y two projections of A. a.

Show that P

X+V

- P

Y

PX

P

XnV

.

(Use corollary 1 of

proposition 2).

b.

Deduce from this that, if (I) is a trace on Al- , we have

(P(Px)

(1:(P)

) ( Px+v)

(1

(1)(1D XnV )

[ 65].

255

FURTHER TOPICS

3. Let A be a finite von Neumann algebra, and S and T two elements of A such that ST = I. Then, TS = I. (The support of T is equal to I, and so, therefore, is that of T* by propositions 2 and 3. Show that T is invertible.) (This property characterises the finite algebras: 8, theorem 1.)

cf. chapter

4. Give all*the details of the following argument, which is a new proof of lemma 1: let A be a von Neumann algebra, and E l and E2 two projections of A with central support I; as the mapping T4 rr,) (T P EA ? ) is an isomorphism of ALI onto q 2 , the normal positive linear forms (1)1 on iq and (1) 2 on correspond 2 1 with one another under this isomorphism; decreasing (1) 1 and cp2 if necessary, ye can suppose that (1)1 =cox , (1) 2 =wx2 , with x i X0, x 2 O; A ; by part I, Chapter 4, lemma 3, there let X1 = XA x l , X 2 - X x2 exists an isomorphism of the Hilbert space X 1 onto the Hilbert space X 2 which transforms Tk into Tk for every T' E A'; con1 2 clude that Px — Px . 2 1 5. Let A be a von Neumann algebra, (I) a normal positive linear form on A, and E a non-zero projection of A. Show that there exists a non-zero projection F of A majorized by E such that the form T (1)(FTF), TEA, is a form wx . (Choose F of the form EA ' X ' and apply theorem 4). 6. Let the complex Hilbert space K be the direct sum of subspaces Kl, K2, ..., let H be another complex Hilbert space, and Un an isomorphism of H onto Kn (n = 1, 2, ...). These define an ampliation of L(H) onto a von Neumann algebra A in K. Let (e l , e2, ...) be an orthonormal basis of H, and e Une p . Show that x = e ll + 1/2 e 22 + 1/3 e33 + ... is cyclic and separating for A, but that y = e21 + 1/2 e 32 + 1/3 e43 + ... is separating, without being cyclic, for A. 7. Let H1, H2 be two 2-dimensional complex Hilbert spaces,

H= H 1 eH 2' A = L(U 1 ) x c uii2. A' = CH, x L(H 9 ) . There exists an involution

We have transforms

8.

Let

A

A

into

A',

But

A

of

H

which

is not standard.

be a von Neumann algebra, and in a two-sided ideal of

A E, we have F E in (rea. Let E be a projection of m. If F duce to the case where F E; then, F = UU*, E = U*U for some UE A, and U = UE E In, hence FEM). b. Let El, E 2 be projections of M, and E their supremum. have E cm. (Use a and exercise 2 a) [12].

We

9. Let A be a von Neumann algebra, and P the set of projections of A. We call an ideal of P every subset I of P

PART III, CHAPTER 1

256

possessing the following properties: (i) if E I and F E, we have F I; (ii) if E l and E2 are in I, their supremum is in I. a. Let I be an ideal of P, and M the two-sided ideal of A generated by I. Show that m is just the set M' of the TEA (Show that such that ET = TE = T for some (variable) E of I. M I is a two-sided ideal of A) . Deduce from this that the set of the projections of m is I.

b.

Obtain from this a bijective correspondence between the set of ideals of P and the set of restricted two-sided ideals (part I, chapter 1, exercise 6) of A. (Use a and exercise 8) [12], [124].

10. Let A be a von Neumann algebra in H, and m a restricted two-sided ideal of A (part I, chapter 1, exercise 6). a. Let TE At For an element S of L(H) to be in & and be majorized by T, it is necessary and sufficient that S = irT1T 1/2 T, show, for some T1 E Y71+ such that T 1 _< 1. (If S EM+ and S using exercise 9 a, and lemma 2 of part I, chapter 1, that S 1/2 = RT 1/2 for some REM such that H R H 1; then S = (S)*S 1/2 = T1/2 R*RT.) b. Show that the set F of the operators of M-1- majorized by T is increasing filtering. (Use a and exercise 9 a.) Problem: is the result true if m is not restricted? c.

If the strong closure of

m is

A, T is the supremum of F

[12].

Let A be a von Neumann algebra, M a two-sided ideal of A strongly dense in A, and (I) a linear form on m, positive on re, such that (ST) = OTS) for any SE ni, TEA, and possessing the following property: if Fc re is an increasing filtering set with supremum TEM+ , cl)(T) is the supremum of OF). Then, there exists exactly one normal trace ip on A+, which agrees with (I) on Mt , and 4) is semi-finite. [Let n be the restricted ideal generated by the projections of m (part I, chapter 1, exercise 6). For REAP, put Ip(R) = sup(FR ), FR being the set (increasing filtering by exercise 10) of the R 1 E n+ majorized by R. It comes down to proving the following: if F A+ is an increasing filtering set with Supremum R, and if X = supiP(F), we have Ip(R) = X. Now, let G be the increasing filtering set of the Sc n+ which are majorized by a (variable) operator of F; G is majorized by R, and every majorant of G majorizes F (by exercise 10 c), and thus R too; hence R is the supremum of G; let p = sup(1)(G); we plainly have p X Ip(R). We are going to show that, for RlE FR, we have OR1) p, which will imply that OR) p. There exists a projection EEM such that R1 = ER 1E; EGE is increasing filtering and majorized by ERE, and ERE is in

11.

257

FURTHER TOPICS

the strong closure of EGE, hence ERE is the supremum of EGE; hence (HERE) = sup0EGV; p majorizes OG), and therefore (EGE); hence p

OER 1 E) = OR 1 )]

d(ERE)

[12].

12. Let A be a von Neumann algebra, and T a closed operator, with dense domain, affiliated to A (part I, chapter 1, exercise 10). The supports of T and of T* are equivalent. (The support E of T is defined in the same way as when T is continuous: I - E is the projection onto the kernel of T) [65]. 13. Let A be a finite von Neumann algebra in H, and M a linear subspace of H. We say that M is essentially dense in (relative to A) if there exists an increasing sequence X 1 , X 2 , ... of closed linear subspaces of H contained in M, total in H, such that Px. EA for every i.

H

2

a.

The intersection of two essentially dense linear subspaces

M, M' is essentially dense. [Let (X 1 , X 2 , ...), [resp. Xi, X, ...)] be a sequence possessing the properties of the definition relative to M (resp. M I ); then, the Xi n Xi!: form an increasing sequence of closed linear subspaces; PX. n YtE A; for 2 -2 every finite normal trace cp on A,

1) (1) ( Pcx.nxt)l ) ' (1) ( Px 2 I, by exercise 2 b, hence OP are total in

(1) ( Pxtl"

, 1) -4-0, and hence the X.n

1.-

H.]

If T is a closed operator, with dense domain, affiliated to A (part I, chapter 1, exercise 10), T -1 (M) is essentially dense. [Let T = WITI be the polar decomposition of T, X(X)(0 A + co) the largest spectral subspace of ITI such that XPxo..) , and X1-- the set of the X E H such that ITIPX(X) TPx(i1xE Xi; the X(i) nXi are closed, increasing, and contained in T-1 (m); the corresponding projections are in A; for every on A, the support of finite normal trace b.

X2 ()

P)0-1

)* is majorized by Px1; , and the support of (p X frpX(i) X 21TPX(i) conclude from this that OP xii) -4- 0, and that the X(i) n X s are

P

total in H.] c. Let T 1 , T2 be closed operators, with dense domains, affiliated to A. Show that T 1 + T 2 and T1T 2 admit minimal closed (By a and b, extensions, with dense domains, affiliated to A. T1 + T2, T1T2, TI + T, Tri have dense domains). A , there exists a closed operator T, d. Let x€ H. If y EX1r with dense domain, affiliated to A, such that y = Tx (use e and lemma 3) [65][101].

PART III, CHAPTER 1

258

14. Let A be a finite von Neumann algebra in H, and 4) and 4) two normal positive linear forms on A, with supports E cp, Ell) such E. Show that there exists a closed operator TrIA that E 0 (part :, chapter 1, exercise 10), with dense domain, such that, ) where Tn is defined for every SEA, we have 4)(S) =lim4)(T*ST n n n÷l-co

in the following way: if T = WITI is the polar decomposition of T, and if E(X) is the greatest spectral projection of ITI XE(X), put Tn = wITIE(n) E A. (Reduce at such that ITIE(X) once, by considering AE?, to the case where 4) is faithful; then by means of an isomorphIsm onto A, to the case where 4) = wx , x being cyclic and separating for A ; then 4) = wu by theorem 4; apply to x and y exercise l3 d) [19], [92], [101].

15. Let Z be aBorel space, Va standard positive measure on Z, C H(C) a v-measurable field of complex Hilbert spaces over Z, C A(C) c L(H(0) a V-measurable field of von Neumann algebras, and

E

e

e

E(C)dV(C)

e

= I

two projections of

A =

f

and

A(c)dv(c).

F = f F(C)dV(C) For E

F, it is necessary

F(C) almost everywhere. [The condiand sufficient that E(C) tion is obviously necessary. To prove that it is sufficient, argue, for example, as for lemma 2 of part II, chapter 2.] [80].

CLASSIFICATION OF

CHAPTER 2.

1.

PROJECTIONS

Definitions.

Let A be a von Neumann algebra, and E a projecWe say that E is finite (resp. semi-finite, properly infinite, purely infinite, infinite) if the algebra AE is finite (resp. semi-finite, properly infinite, purely infinite, infinite). DEFINITION 1.

tion of A.

When we wish to be precise, we say that E is finite (resp. semi-finite, etc.) relative to A. If E is finite (resp. semi-finite, etc.) is finite (resp. semi-finite, etc.).

and if F

E, then F

A finite projection is semi-finite. A purely infinite projection is properly infinite. An abelian projection is finite.

Let (Ei) ici be a family of projections of A whose central supports are pairwise disjoint; let E = X E.. 2 PROPOSITION 1.

iEI

For E to be finite (resp. semi-finite, properly infinite, purely infinite), it is necessary and sufficient that each Ei be finite (resp. semi-finite, properly infinite, purely infinite). H A and so the proposition follows E' iEI from part I (chapter 6, proposition 7). 0

Proof.

We have

AE

.

,

Let E be a finite (resp. semi-finite, purely infinite) projection of A and F a projection such that F-<E. Then F is finite (resp. semi-finite, purely infinite). PROPOSITION 2.

Proof.

It is enough to study the case where F E. Then AF = (AE)F, and so the proposition follows from part I (chapter 6, proposition 11 and proposition 13, corollary 4). We remark that the projection 0 is simultaneously finite, semi-finite, properly infinite and purely infinite. This is actually somewhat more convenient than the usual convention under which 0 is only a finite and semi-finite projection. References : [10], [42], [65].

PART III, CHAPTER 2

260

2.

Cyclic projections of A and cyclic projections of A'.

LEMMA 1.

Let Px be a projection of A, Pxt a projection of A',

and V = X n X' . Put B = A , C = Ak. Then, By and Cy are both equal to the same von Neumann algebra D, and Cy and By are equal to D'. If Px andPxt have the same central sqpport, B is isomorphic to D and C to D'. If X = Xfand X' = X (in which case Px and Px , have the same central support), x is cyclic and separating for D and D'. Proof.

We have

B

Y

=

(

A ) = A = (A = C' XY V XI ) V Y'

which serves to define D; and

C = (CY Y Y

=

B V ' = (B )' = D'.

If Px and P have the same central support P z , the induction of A on Ak = ' is an isomorphism; hence the central support of Py (regarded as an operator of B') is Ix; hence B is isomorphic to By A T D. Similarly, C is isomorphic to C y = D'. Finally, if X = and X' = XA x , x is cyclic for Ak = B ' , and hence for B Y' -- t' ' and also for Ax, = C', hence for Cy = D.

Let x€ H. Then, 4, is abelian (resp. finite, semi-finite, properly infinite, pure -1y infinite) if and only if is abelian (resp. finite, semi-finite, properly infinite, purely infinite). PROPOSITION 3.

41

Proof. We have A = A l x A2, A' = Al x idk, with A l and Al semi-finite, A 2 and A'2 purely infinite; moreover, every projection of A 2 or of A'2 is purely infinite (part I, chapter 6, proposition 13, corollary 4). It therefore suffices to study the case where A is semi-finite. We adopt the notation of lemma 1. All the algebras that appear there are semi-finite. The algebras V and D' are standard (chapter 1, theorem 5) , hence Ax = B is antiisomorphic to Akt = C. Hence Ax is abelian (resp. finite, semi-finite, properly infinite) if and only if Akt is abelian (resp. finite, send-finite, properly infinite) . 0 References : [17], [19], [31], [40], [65], [100].

3.

Finite projections.

PROPOSITION 4. Let E be a finite projection of A. Every projection of A majorized by E and equivalent to E is equal to E. Proof. This follows from proposition 3 of chapter 1.

We will later (chapter 8, theorem 1, corollary 1) prove a converse of proposition 4.

261

FURTHER TOPICS

PROPOSITION 5. Let E l , E 2 , ..., En be projections of A, and let E be their supremum. If the Ei's are finite, so is E.

Proof.

It is enough to consider the case where

n = 2.

(i) Suppose to start with that E l and E2 are equivalent and disjoint. Then AE = AE1 0 L(K) where K is a 2-dimensional complex Hilbert space (part I, chapter 2, proposition 5). As the algebras AE and L(K) are finite, AE is finite (part I, chapter 6, propositio n 12). ,

(ii) Suppose that E l and E2 are merely disjoint. By theorem 1 of chapter 1, and proposition 1, it is enough to study the case where E 1 E 2 and the case where E l . Suppose, for example, that E2 E l . Let K be a complex Hilbert space, F a rank-1 projection in K, and identify every element T of A with the element T 0 F of B = A 0 L(K). Instead of showing that E is finite relative to A, we are going to show, which comes to the same thing, that E is finite relative to B. We can suppose, taking K to have dimension 2, that there exists in B a projection El equivalent to E l and orthogonal to E l . Then El + El + El, and E l + Ei is finite by part (i) of the proof. (iii) Finally, consider the general case. Let X1 = E l (H), X2 = E2(H), X be the closure of X1 + X 2 , i.e. X = E(H), and Y = X e Xl. An element of Y orthogonal to X 2 is orthogonal to X 1 + X2, and therefore to Y, and hence is zero. Hence X2 (chapter 1, proposition 2, corollary 1). Hence Py is finite and, consequently, Px = Px 1 + Py is finite by part (ii) of the proof. PROPOSITION 6. Let E, F be too equivalent finite projections of A, and G a projection of A majorising E and F. There exists a unitary operator u of A such that UEU-1 = F, UGU-1 = G. In particular, G - E — G - F.

Proof. have G 1 chapter and the with E l

Let the projection G 1 be the supremum of E and F. We G and G is finite (proposition 5). By theorem 1 of 1, it is enough to study the case where G1 - E G 1 - F case where G 1 - F G1 - E. If, for example, G1 -E—E 1 G1 - F, we have G

1

> F + E

1

E + (G

1

- E) = G 1 ,

hence F + El = G 1 (proposition 4), hence G 1 - F = E l — G 1 - E. Let V (resp. W) be a partial isometry of A having E (resp. G 1 - E) as initial projection, and F (resp. G 1 - F) as final projection. Let U be the unitary operator which agrees with V on E(H), with W on (G 1 - E)(H), and with I on (I - G 1 )(H). Then U c

A,

UEU

-1

= F,

UGU

-1

= G.

PART III, CHAPTER 2

262

Remark. The proof of the proposition as well as its conclusions are valid under the following hypotheses: E and F are equivalent projections of A with supremum G l , every projection of A majorized by G1 and equivalent to G 1 is equal to Gl, G is a projection of A majorizing E and F. We will use this remark in chapter 8. References : [10], [42], [65].

4.

Semi-finite projections.

Let E be a non-zero semi-finite projection of There exists a non-zero finite projection of A majorized by

PROPOSITION 7.

A. E.

There exists on A a faithful semi-finite normal trace q). Let T be a non-zero operator of A such that 4)(T) < + 03 , and F a non-zero spectral projection of T such that F AT. Then, (I)(F) < + 03 . The restriction of (I) to FAP defines on 4 a faithful finite normal trace, hence AF is finite. 0

Proof.

COROLLARY 1.

Let E be a semi-finite projection of A.

(1) There exists a family (Ei ) icI of pairwise disjoint finite projections of A such that E = X E.. 1-El

2

(ii) If E 0, there exists a finite projection F of ized by E such that AF admits a non-zero trace-element.

A major-

Proof. Let (Ei) i ci be a maximal family of pairwise disjoint, finite, non-zero projections of A majorized by E. Let E' = E E.. Then, E' is semi-finite and does not majorize iEI any non-zero finite projection, hence (proposition 7) E' = O. We prove (ii). By proposition 7, we can suppose that E is finite and no9Tzero. Let x be a non-zero vector in the range of E. Let F = Ex . There exists a non-zero finite normal trace (1) on AF . As x is separating for AF , (1) is of the form coy (chapter 1, theorem 4), and y is a trace-element for AF . 0

Let E be a semi-finite projection of A. There exists a finite projection F of A, majorized by E, which has the same central support as E. COROLLARY 2.

Proof.

Let (Ei) ic , be a maximal family of non-zero finite projections of A, majorized by E, whose central supports El are disjoint. 'Let F = E. E, which is finite (proposition 1). iEI Let F` (resp. E') be the central support of F (resp. E). We have F' < E l . If E / F' X 0, we have E(E' - F') X 0, hence there exists (proposition 7) a non-zero finite projection majorized by E and orthogonal to F', and therefore to the q's.

263

FURTHER TOPICS

This contradicts the maximality of the family (Ei)i ci .

Hence

F' = E'.

For A to be semi-finite, it is necessary and sufficient that A be isomorphic to a von Neumann algebra 13 such that B' is finite. Proof. The condition is sufficient by part I (chapter 6, COROLLARY 3.

proposition 13, corollary 1). Conversely, if A is semi-finite, A' is semi-finite, hence there exists (corollary 2) a finite projection E l of A / with central support I. Then, A is isomorphic to B = AE, and B' = tY is finite.

Let A be a semi-finite von Neumann algebra. For A to be continuous, it is necessary and sufficient that there exist a decreasing sequence (E l, E2, ...) of finite projections of A, with central support 1, such that En - En+1 En+ 1 for COROLLARY 4.

n = 1, 2, .... If A is continuous, the sequence (E.) exists by corollary 2, and chapter 1, corollary 3 of theorem 1. Suppose now that there exists a sequence (E i ) with the stated properties. It suffices to show that AE1 is continuous (because then A l is continuous, hence A', isomorphic to i,11/4 , is continuous, and hence A is continuous). We will therelore suppose henceforth that A is finite. Using part I, chapter 6, proposition 9 (ill), we are led at once to the case where A is finite and There then exists a faithful finite trace cp on A G-finite. [part I, chapter 6, proposition 9 (ii)]. For every abelian projection E of A, we have E < En for every n (by chapter 1, theorem 1 and the minimality property of abelian projections), hence 2 -n ( E 1 ), whence (I)(E) = 0, and hence E = O. cp (En) cp(E) Hence A is continuous.

Proof.

PROPOSITION 8. Let E be the greatest projection of the centre of A, semi-finite relative to A (part I, chapter 6, proposition 8). For a projection F of A to be semi-finite, it is necessary E. and sufficient that F E, F is semi-finite. (proposition 2). If F is Proof. If F

semi-finite, let G be its central support. Then AF is semifinite, hence A;, is semi-finite, hence A is semi-finite, hence AG is semi-finite, hence G E. Corollary 3 of proposition 7 provides a definition of semifinite algebras analogous to the definition of discrete algebras. References : [10], [42], [65].

Properly infinite projections. For a projection PROPOSITION 9. Let Z be the centre of A. E of A to be properly infinite, it is necessary and sufficient 5.

PART III, CHAPTER 2

264

that, for every projection F of A such that FE infinite.

0, FE

is

Proof.

The condition is necessary (proposition 1). Moreover, if E is not properly infinite, there exists a non-zero projection E l of the centre of AE such that AE1 is finite (part I, chapter 6, proposition 8). We have E l = FE , where FEZ. Then FE / 0, and FE is finite. The condition of the proposition is therefore sufficient. I]

If a projection E of A is properly infinite, its central support E' is properly infinite. COROLLARY 1.

If F is a projection of the centre of A such that FE' / 0, we have FE / 0, hence FE, and, a fortiori, FE', are infinite.

Proof.

COROLLARY 2. The greatest projection of the centre of A which is properly infinite relative to A is also the greatest properly infinite projection of A.

Proof.

This follows immediately from corollary 1.

El

Let (Ei)i c , be a family of projections of A, and E their supremum. If the Ei's are properly infinite, so is E. COROLLARY 3.

Proof.

This follows at once from proposition 9.

PROPOSITION 10. Let (Ei) ie , be an infinite family of pairwise disjoint, equivalent, non-zero projections of A, and E = X E.. iEI

Then, E is properly infinite. Proof.

Let J be the complement in I of an arbitrarily chosen For every projecelement in I. Then, I and J are equipotent. tion F of Z such that EF 0, we have EF E.F EF, so that iEJ EF is infinite (proposition 4), from which it follows that E is properly infinite (proposition 9).

We will later (chapter 8, theorem 1, corollary 2) prove a converse to proposition 10. References : [ 10], [42], [65].

6.

Purely infinite projections.

For a projection E of A to be purely infinite, it is necessary and sufficient that E majorize no non-zero finite projection. PROPOSITION 11.

Proof. The condition is necessary (proposition 2) . It is sufficient, because, if E is not purely infinite, E majorizes a non-zero semi-finite projection, and therefore a non-zero finite projection (proposition 7).

FURTHER TOPICS

265

Let E be the greatest projection of the centre of A which is purely infinite relative to A (part I, chapter 6, proposition 8). For a projection F of A to be purely infinite, it is necessary and sufficient that F E. PROPOSITION 12.

Proof.

If F E, F is purely infinite (proposition 2). If F is purely infinite, let G be its central support. Then, AF is is purely purely infinite, hence A!„ is purely infinite, hence E. 0 infinite, hence AG is purely infinite, hence G

Cf. also exercise 1, and chapter 8 (theorem 1, corollary 5). References : [10], [42], [65].

7.

Comparison of projections and dimension.

If A Let A be a factor, and (I) a faithful normal trace on is purely infinite, (I) is identically infinity on the set of nonzero operators of A+. If A is semi-finite, we require (I) to be semi-finite; then, (I) is uniquely determined up to multiplication by a constant X such that 0 < X < + 00 .

Let A be a factor, and M the set of its projections. A relative dimension on M is defined to be the restriction to M of a faithful normal trace on A+, assumed to be semifinite if A is semi-finite. DEFINITION 2.

This relative dimension D is uniquely determined up to multiplication by a constant X (0 < X < + co). If E F, we have D(E) =D(F) Let E and F be projections of A. (part I, chapter 6, proposition 1, corollary 1). Hence, if E D(F). F, we have D(E) PROPOSITION 13. Let A be a factor, tions, and D a relative dimension on

M the set of its projecM.

(i)If E is a finite projection of A, D(E) < + (ii)If E is an infinite projection of A, D(E) = + co. (iii)If E and F are finite projections such that D(E) = D(F) F). [resp. D(E) F (resp. D(F)], we have E Proof. Let (I) be a faithful normal trace on A+ , semi-finite if

A

is semi-finite, whose restriction to M is D. Let E be a projection of A. The algebra AE is a factor (part I, chapter 2, corollary of proposition 2). The trace tI) induced by (I) on A E is faithful normal, and semi-finite if (I) is semi-finite. As two faithful normal traces on a factor are proportional, we see that E is finite (resp. infinite) if and only if D (E) = (1) ( E) = OI E(H))

PART III, CHAPTER 2

266

is finite (resp. infinite), whence (i) and (ii). Now, let E and F be finite projections of A such that D(E) = D(F). SupF (chapter 1, theorem 1, corolEl pose, for example, that E D(F) < + co, hence D(F - E l ) = 0, hence lary 1); we have D(E 1 ) F E l = 0 since cl) is faithful, hence E —F. Finally, if two projections E, F of A are such that D(E) < D(F), we do not have F E, and we therefore have E F. PROPOSITION 14. Let A be a tions. D a relative dimension taken by D.

factor, M the set of its projecon M, and D the set of values

n finite (resp. n = + co), we can, mul(i)If A is type tiplying D by a suitab e scalar, arrange that D - {0, 1, ..., n1 (resp. D = {0, 1, 2, + œl). If A= L(H), D(E) is then the dimension of E(H) (agreeing to identify all infinite cardinals with + co). (ii)If A is type II I , we can, multiplying D by a suitable scalar, arrange that D is the interval [0, 1 . (iii)If A is type 11e , D is the interval [0, + oe]. ]

(iv)If A is type III, D

{0, +

is type In , we can suppose that A = L(H), H being n-dimensional; then, for an appropriate choice of D, D(E) is the dimension of E(H) (part I, chapter 6, theorem 5); whence (1). If A is type Il ' , we have D(I) = 1 for an appropriate choice of D. By chapter 1 (theorem 1, corollary 3), D then contains all numbers of the form p.2 n (p, n integers 0, p 2 n ), and is therefore dense in [0, 1]; moreover, let (X i , X 2 , ...) be an increasing sequence of numbers of D, with limit X; let Ei be a projection of A such that D(E 1 ) Xi; we construct, by induction, an increasing sequence of projections Fi of A such that Fi Ei, which is possible by propositions 6 and 13; let F be the supremum of the Fi's; we have D(F) = X, hence X E D; whence (ii). If A is type Ile , there exists an increasing filtering family of finite projections Ei of A with supremum I (proposition 7, corollary 1); then, the D(Ei)'s are arbitrarily large, and [0, D(E)] c D by (ii), hence D = [0, +(n]. Finally, (iv) is immediate. 0

Proof.

If

A

Let A be a finite factor, M the set of projections of A, and D a finite-valued function 0, defined on M, possessing the following properties: (i) if EE M and FEM are disjoint, D(E + F) = D(E) + D(F): (ii) if U is a unitary operator of A, and if E E M, then D(uEu -1 ) = D(E) . Then, if D / 0, D is a relative dimension. PROPOSITION 15.

A

Proof. If A is type In , for n finite, we can suppose that = L(K), K being an n-dimensional complex Hilbert space. On

every minimal projection, D takes the same value, X, say.

Hence,

267

FURTHER TOPICS

for every projection E of A, we have D(E) = Xdim E(K), which proves the proposition in this case. Suppose from now on that A is continuous. Suppose D(I) / 0, and let D P be a relative dimension such that D' (I) D(I). For every projection E such that D' (E)is of the form p.2-n (p, n integers 0), there exist projections E l , E2, E 2n of A, which are equivalent, mutu2n ally disjoint, and such that I = X E., E = E.. We have

2 nD(E 1 ) = D(I)

D' (I) = 2N D' (E1)

and so D(E 1 ) = D P (E 1 ), hence D(E) = D' (E). Now let F be any projection of A. There exists a projection F 1 F of A such that D' (F1)is of the form p.2-n and such that D' (F) - D' (F 1 )is arbitrarily small; as D(F) D(F 1 ) D P (F 1 ), we see that D(F) D I (F). Similarly, D(I - F) D' (I F), hence D(F) D' (F) and finally D(F) = D' (F).

Let A be a finite factor, and ip a real-valued function defined on the set N of hermitian elements of A, possessing the following properties: COROLLARY.

(i) i4(XT) = Xlp(T) for X ER,

TE

N;

(ii) 11)(T 1 + T 2 ) - 4)(71 1 ) + 11)(T 2 ) if T 1 , T 2 are commuting ele-

ments of N; 0 for TE A;

(iii) 11)(T)

(iv) OUTU-1 ) = 11)(T) if T

E

N and if U is a unitary element of

A. Let (1) be a faithful finite normal trace on striction of cp to N is proportional to i.

A.

Then, the re-

Proof. By properties (ii), (iii), (iv), there exists a constant X > 0 such that ip X(1) on M (proposition 15). Let T be any element of N. Let E > O. There exist mutually disjoint projections El, E 2 , ..., En of A, commuting with T, and real numbers p l , p2, pn such that p.E. - E.I •< T < . p.E. + E.I. 2 2 E E 2=1 In view of properties (i), (ii), (iii), we see that both i(T) and XOT) lie between

2

2

E1(I)

and

p.ip(E.) +

PART III, CHAPTER 2

268

Hence tP(T) = 4(T). Proposition 14, (i)

justifies the term "relative dimension."

Proposition 14, (i) and (ii), allow one to define a normalised relative dimension when A is discrete or finite. However, these two normalisations do not agree when A is type I n , 1 < n < + co. For G-finite factors, proposition 13, (iii) can be freed from the finiteness hypothesis (cf. chapter 8, theorem 1, corollary 5); for an extension of proposition 15, cf. chapter 8, theorem 1, corollary 7.

Reference : [65]. Exercises. 1. Let A be a von Neumann algebra, E l and E 2 projections of A, and F 1 and F 2 their central supports. If E l 2 properly infinite, and F 1 F2, we have E 1 E. isfnte,E (Apply the comparability theorem to E l and E2) [65]. 2. Let A be a finite von Neumann algebra. If there exists a finite cyclic set {xi , x2 , ..., xn for A, then A is finite. (The E A .'s are finite by proposition 3, and have supremum I) [65]. }

3. Let A be a von Neumann algebra. A projection E of A is said to be discrete (resp. continuous) if AE is discrete (resp. continuous). a. Let Z be the centre of A. Let F be the greatest projection of Z which is discrete (resp. continuous) relative to A. For a projection E of A to be discrete (resp. continuous), it is necessary and sufficient that E < F. b. If E is discrete, there exists an abelian projection of majorized by E and having the same central support as E.

A

4. Let A be a finite factor, N the set of invertible elements of A, and A a finite-valued function, > 0, defined on N and possessing the following properties: (1) If S, TEN, A (ST) = A (s) A (T) ; (ii) If s E N, A (s*) = A(s) ;

(iii)

A(XI) = X for X > 0;

(iv) A(S)

1 if 0

S

I.

Show that A is the determinant defined in part I, chapter 6, section 11. [Show first that A(u) = 1 for U unitary. For every hermitian element S of A, put w(5) = logA(exp S) and show, with the help of the corollary of proposition 15, that w, extended by linearity to A, is a faithful finite normal trace on A. To show 0), observe that the function that w(XS) =XW(S) (X real, S X w(XS) is additive, and is 0 for X > 0, and is therefore linear] [24].

CHAPTER 3.

MORE ON DISCRETE VON NEUMANN ALGEBRAS

Structure of discrete von Neumann dlgebras. DEFINITION 1. A von Neumann algebra A is said to be homogeneous if there exists in A a family of pairwise disjoint, 1.

equivalent abelian projections, with sum I. Such an algebra is discrete. Every von Neumann algebra isomorphic or antiisomorphic to a homogeneous algebra is homoA disgeneous. An abelian von Neumann algebra is homogeneous. crete factor is homogeneous. A homogeneous algebra is spatially isomorphic to an algebra of the form B 0 L(K) with B abelian; on the other hand, an algebra B 0 nK), with B abelian, is homogeneous (part I, chapter 2, It follows from this that the tensor product of proposition 5). two homogeneous algebras is homogeneous.

LEMMA 1. Let A be a von Neumann algebra, E and F projections of A, E l and F / their central supports. If E l F l and if E is abelian, we have E- F.

Proof.

Thanks to theorem 1 of chapter 1, it suffices to study where F E and, consequently, the case where F _< E. the case We then have F = EF 1 since E is abelian (part I, chapter 8, secE l _< F l , we deduce from this that F = E. tion 2). As E

Let A be a homogeneous von Neumann algebra in H. Let (Ei),-, ET [resp. (F x) xcic] be a family of pairwise disjoint, equivalent ahelian projections of A with sum I. We have Card I = Card K (provided that H 0). PROPOSITION 1.

If K is finite, A is finite (chapter 2, proposition hence I is finite (chapter 2, proposition 10). Thus I and K 5), are either both finite or both infinite. Support first that I and K are finite. The Ei's and the Fx 's have central support I, and hence E. Fx (lemma 1). Suppose, for example that E. —Y F , hence Card I = Card K l with IC cK. Then I = x xEK P icI F=I (chapter 1, proposition 3), and consequently K l = K, xEK /

Proof.

269

PART III, CHAPTER 3

270

Card I - Card K. If I and K are infinite, we can, by suitably decomposing A into a product, suppose that the AE .'s and the AE-2..'s an the AFx 's are AF's are G-finite ( because the X Card K and abelian). Then, lemma 6 of chapter 1 yields Card I 0 Card I. Card K Proposition 1 shows that n . Card I is an algebraic invariant of A. We say that A is type In (which generalises the terminology adopted in part I, chapter 8, section 4 when A is a factor). If A - B 0 L(K), with B abelian, n is the dimension of K. For A to be finite, it is necessary and sufficient that n be finite. If A is type I n , we also say that n is the multiplicity of A'. Nevertheless, n is not an algebraic invariant of

Every discrete von Neumann algebra A is of the where the A's are type In ., with the ni's mutually

PROPOSITION 2.

form H A. iEI

distinct. Proof.

Apply chapter 1 (theorem 1, corollary 2), taking for a family consisting of a single non-zero the family (E.). abelian projection of A. Adopting the notation of that corollary, the Fx 's are equivalent, mutually disjoint, non-zero, abelian, and Fo is abelian. If the central support of F o were equal to that of the Fx 's, we would have Fo Fx (lemma 1), contrary to the corollary which we are using. Hence there exists a projection E of the centre of A, majorized by G, orthogonal to F o , and such that the FxE's are non-zero. The equality F yields E F E. Hence AE is homogeneous and G - Fo x x xEK xEK non-zero. It then suffices to repeat the argument on A I-E (which is discrete) and so on transfinitely. Li COROLLARY.

H (

3,. 0

CH

)

Every discrete von Neumann algebra is of the for where the B.'s are abelian.

If A is discrete, so is A', and hence --=-11(C,.0014.)) with the C.'s abelian. Hence

Proof.

iEI

A =

H

(C /. 0 CH )

and it suffices to put

C /. = B..

0

References :110], [43], [61], [100].

isomorphisms of discrete von Neumann algebras. PROPOSITION 3. Let A, B be von Neumann algebras such that A', B' are homogeneous. Suppose that A and B are of the same multiplicity. Then every isomorphism (I) of A onto B is spatial. Proof. There exist (part I, chapter 4, corollary of theorem 3) 2.

a von Neumann algebra

C,

projections E / , F / of

C/ ,

with central

271

FURTHER TOPICS

support I, such that A may be identified with CEP, B with CF /, TF , (TE C). Let (E)i ci [resp. and 0 with the isomorphism TEl abelian, pairwise family of disjoint, equivalent be a ] (FY) x XEK projections of C', with sum E' (resp. F'). We have E,;: (lemma 1). Since A = CE, and B = CF I have the same multiplicity, F'. I and K are equipotent. Hence E' LI

COROLLARY 1. Let A and B be two von Neumann algebras such that A' and B' are abelian. Every isomorphism of A onto B is spatial.

Proof.

A

and S are of multiplicity 1. E

Let A be a discrete von Neumann algebra, Z its centre, and 0 an automorphism of A which leaves the elements of Z fixed. There exists a unitary operator uc A such that 0(T) - UTU-1 for every TEA. COROLLARY 2.

We can suppose, replacing A by an isomorphic von Neumann algebra if necessary, that A' is commutative, and hence equal to Z. By corollary 1, there exists a unitary operator U such that 0(T) = UTU-1 for every TEA. For TE Z, we have UTU -1 = 0(T) = T; hence

Proof.

U E Z I = A" = A. References : [ 43],

[44],

[ 61],

0

[125].

Let A be a discrete von Neumann algebra, and Z its centre. If A is a-finite over Z (part I, chapter 9, exercise 5), A is generated by Z and a countable family of elements. [Reduce to the case where A is homogeneous, by proposition 2, then to the case where, further, Z is Then, A = B ® L(K), with B abelian, and a-finite. A is a-finite. Deduce from part I (chapter 2, exercise 3) that K is separable.]

Exercises.

1.

(Cf. part II, chapter 3, theorem 1, corollary 2.) 2. Let A be a von Neumann algebra. For A and A' to be homogeneous type I n and I n! , it is necessary and sufficient that

A = B 0 L(K) ® C K "

A'

= S ® c K

0

L(K'),

an abelian von Neumann algebra such that B = B', and K (resp. K') being an n- (resp. n'-) dimensional complex Hilbert space. Deduce from this that the von Neumann algebra generated by A and A' is type I nn , [100].

B being

3. A ring is said to be binormal if, for any elements xl , x 2 , x 3 , x4 of the ring, we have Icaxaol xa(2) xa(3) xa(4) = 0, where G runs through the set of permutations of {1, 2, 3, 4 } ,

272

PART III, CHAPTER 3

and where EG (= ±1) is the sign of the permutation. a. The ring of 2 X 2 matrices over a commutative ring is binormal. It is enough, by linearity, to consider the case

(l O\ x l = (0 0)' 1

(o o'\ x 2 = ( o)'

(o l\ x 3 = (0 0)'

0 0) x4 = (0 1

b. The ring of 3 X 3 matrices over C is not binormal. consider the matrices

X3 =

100 000 0 0 0,

000 X 2 = 10 0 00

O 01 O 00, O 00

x

000 = 000 4 001

c. Deduce from b that a von Neumann algebra that contains three pairwise disjoint, non-zero, equivalent projections, is not binormal. d. Deduce from a and c that a von Neumann algebra is binormal if and only if it is the product of an abelian von Neumann algebra and a type 1 2 von Neumann algebra [3]. 4. A von Neumann algebra A such that A = A' is spatially isomorphic to a von Neumann algebra of the type defined in part I (chapter 7, theorem 2). (Use theorems 1 and 2 of part I, chapter 7, and corollary 1 of proposition 3 of the present chapter) [100].

5. A von Neumann algebra A in H is said to be uniform if there exists in A a family (Ei) icI of finite, pairwise disjoint, equivalent projections, with sum I. a. For a von Neumann algebra to be uniform, it is necessary and sufficient that it be spatially isomorphic to an algebra B 0 L(K), with B finite. b. Let A be a uniform von Neumann algebra. Let (Ei) iEI [resp. (F__x )xEK] be a family am of finite, pairwise disjoint, equivalent projections of A, with sum I. If I is infinite, K is infinite, and Card I = Card K (if H / 0). (Argue in somewhat the same way as for proposition 1.) The number Card I is then called the order of A.

c. Let A be a properly infinite von Neumann algebra, Z its centre, and E a non-zero finite projection of A. There exists a non-zero projection G of Z and a family (Ei) of pairwise disjoint projections equivalent to EG, with sum G. (Use corollary 2 of theorem 1, chapter 1).

273

FURTHER TOPICS

d. Every properly infinite semi-finite von Neumann algebra is the product of uniform von Neumann algebras of mutually differ(Use a, b, c.) ent orders. Every semi-finite von Neumann algebra A is of the form II(BitaC.), where the B's are finite. (Apply d to A'.) e.

iEI f. Let A be a uniform, properly infinite von Neumann algebra, Z its centre, and E a finite projection of A. There exists a family (Ei) of mutually disjoint projections equivalent to E, whose sum belongs to Z. (Use c.) g. Let A, B be von Neumann algebras such that A', B' are uniform, properly infinite, and of the same order. Every isomorphism of A onto B is spatial. (The proof is analogous to that of proposition 3, using b and f.) h. Let A be a uniform properly infinite von Neumann algebra. For A to be standard, it is necessary and sufficient that A' be (To show uniform, properly infinite and of the same order as A. is isomorphic to a A that is sufficient, note the condition that [37], [89]. standard von Neumann algebra B, and use g). 6. Let Z be a Borel space, y a positive measure on Z, C a y-measurable field of complex Hilbert spaces over Z,

H = C

H(C)

H(c)dv(C),

A(C) a v-measurable field of von Neumann algebras over Z,

and

A =

A(Ody(C).

If A is type In (n = 1, 2, ..., N6), A(C) is type I n almost everywhere. (Decompose mutually disjoint, equivalent abelian projections of A, with sum I.) The converse is true if y is standard [10], [80].

7. Prove theorem 2 of part II, chapter 6, by the following method: reduce to the case where Z' is homogeneous; then Z' = B 0 L(K), where B = B'; hence (exercise 4) there exists a compact metrisable space Z, a positive measure y on Z, such that B is the algebra of multiplication operators by the functions of If°C (Z' y) in H 0 = L(Z, V); then

H =H 0 0K=fH(c)dv(c), where c H(c) is the constant field over Z corresponding to and Z = B 0 cf.( is the algebra of diagonalisable operators of

K; H.

274

PART III, CHAPTER 3

8. Let A be a discrete von Neumann algebra in a separable Hilbert space H. Show that A is generated by a single element. [Thanks to part I, chapter 7, exercise 3 f, h, i, this may be reduced to the case where A is homogeneous, and then to the case where , A = L(H). Observe that L(H) can be generated by a compact hermitian operator with eigenvalues of multiplicity 1 and by a rank-1 projection.] [288].

CHAPTER 4.

1.

OPERATOR TRACES

Definition.

Let A be a von Neumann algebra, and Z a von Neumann algebra contained in the centre of A. Throughout this chpater, we choose, once and for all, a locally compact space Z, a positive measure V on Z, and an isomorphism of the normed *-algebra Lc (Z, V) onto Z (part I, chapter 7, theorem 1); also, we will identify LcCD (Z, V) and Z by means of this isomorphism; this identification is compatible with the natural orderings on the two algebras. Then, Z -F will be embedded in the set Z + , of meas0, finite or otherwise, on Z. (In 7 + , as in urable functions q(Z, V), we identify two functions equal locally almost everywhere.) Every increasing filtering setA majorized in Z -F , admits a supremum; it easily follows that, in Z -F, every increasing filtering set admits a supremum.

A trace associated with Z and defined on A+ or Z-trace on A+ ) is a mapping 0, defined on A+, with values in 7 +, possessing the following properties: (1) If S E A+ and T E A+ , we have 0(s + T) = T(s) + ; DEFINITION 1.

S E

A+ and

(

T E

Z

we have 0(TS)

= T(S) ;

If s E A+ and if u is a unitary operator of A, we have 0(usu-1 ) = 0(s). (iii)

We say that 0 is faithful if the conditions S E A+ , (1)(s) = 0 imply that s = O. We say that 0 is finite if 0(s) EZ + for every S E At We say that is semi-finite if, for every non-zero s of A', there exists a non-zero T of A+ majorized by s such that 0(T) E Z. We say that 0 is normal if, for every increasing filtering set F c A+ with supremum s,A -F , 0(s) is the supremum of 0(F). If Z is the set of scalar operators, a Z-trace on A+ is of the form S (I)(S)I, where (1) is a trace on At 275

PART III, CHAPTER 4

276

PROPOSITION 1. Let 4) be a Z-trace on At The set S EA + such that cD(S) cz + is the positive part of an

of the ideal m of A. There exists a unique linear mapping l) of m into Z agreeing with (I) on re, and we have 1(ST) = .0(TS) for Sent, TEA. Finally, let S c m; if (I) is normal, the linear mapping T .4) (ST) of A into Z is ultra-strongly and ultra-weakly continuous, and its restriction to the bounded subsets of A is strongly and weakly continuous. Proof. We work along similar lines to proposition 1 of part I, chapter 6. Instead of using theorem 1 of part I, chapter 4, we use theorem 2 of part I, chapter 4. We leave the details to the reader.

We sometimes call the mapping V a Z-trace on M, when there is no risk of confusion. References : [6], [12], [63], [101], [104], [105]. 2.

Traces on Z +

0, finite or otherwise, on Z + , will be called a A function w trace on 2+ if w(S + T) = w(S) + w(T) and w(XS) = Xw(s) for 0 (with the convention, once and for all, that S e2 + , TE 2+, x 0. + co = 0). By analogy with the definition of part I, chapter 6, we define, in the obvious way, faithful traces, semi-finite traces, and normal traces on Z. Let cp be a normal trace on Z+ ; for every S E Z + , put w(s) = sup cp (S.), where (S i ) is the family of functions of Z + majorized by S; it is clear that w is a normal trace on 2+, faithful (resp. semi-finite) if and only if ()) is faithful (resp. semi-finite); moreover, every normal trace w on 2+ is obtained in this way, (I) simply being the restriction of w to Z. We will henceforth identify w and (1). Let S Er.

S(C)dV(C) remains unchanged if S is

The number

f

modified on a locally negligible set. Put

w(s) = fs(c)dy(c). Then, w is a semi-finite, faithful, normal trace on

2+ .

Let w and w l be two normal traces on z+, w being and semi-finite. There exists exactly one element fai4hful QEZ + such that w 1 (S) = w(SQ) for every se Z. If w 1 is faithful semi-f-Lnite), we have Q(c) > 0 [resp. Q(C) < + .3] locally almost everywhere on z. LEMMA 1.

Proof. By part I, chapter 6, (proposition 1, corollary 2), there exists a projection E of Z (identified with the characteristic function of a measurable subset Y of Z) which possesses the following properties: (i) if S E Z + is not zero locally almost every on Z'-Y, w 1 (S) = + .; (ii) the trace wi induced by

277

FURTHER TOPICS

wl on Z is semi-finite. Let w' be the trace induced by w on and II) = w' + wl, which is faithful and semi-finite. There exist (part I, chapter 6, theorem 3) elements Q0, Q 1 of Z such that w' (S) = IP(SQ0), WI(S) = IP(SQ1) for S Z. Identify 7E with ZE, i.e. with the set of functions of Z vanishing on Z`■ Y• Then For Q 1 = Q1Q(3 1 is a function of 2+ that vanishes on Z"-Y. S € Z we have ,

W'1 (S) = 11)(SQ 1i)=IP(SQ,Q 0 1 Q 0 ) w being faithful, we have, locally almost everywhere on-Y, Q0 (C) > 0, hence Q0(C) -1 Q0() = I], and consequently w'1 (S) = 4(SQ 1Q0 ) = w l (SQ 1 ). Put Q(C) = Q 1 (C) for CEY, and E Z"-Y; we have co for Q(C) = + [because,

w (S) = w(SQ) 1

for every

SE

Z.

We prove the uniqueness of Q. Let Q' EZ , and suppose that w(SQ) = w(SQ') for every S E 2+. If Q X Q', there exists a nonlocally negligible measurable subset X of Z such that for example, Q(C) < Q ' (r) for Ce X. There thus exists an S E Z+ such that SQ < SQ', SQ X SQ I , SQEZ and W(SQ) < + co (thanks to the fact that w is semi-finite); then, w(SQ' - SQ) =w(SQ 1 ) - w(SQ) = 0

contradicting the faithfulness of w. Finally, if Q(C) = 0 [resp. Q(C) = + co] on a non-locally negligible set, w l is not faithful (resp. is not semi-finite), whence the last assertion of the lemma. 0

Relations between scalar traces and operator traces. PROPOSITION 2. Let A be a von Neumann algebra, Z a von Neumann algebra contained in the centre of A, (I) a normal Z-trace on A+ , and w a normal trace on Z+ . (i)(1) = wa is a normal trace on A+ . (ii)For (j) to be faithful, it is necessary and sufficient that w and be faithful. (iii)If w and iT are semi-finite, so is cp. (iv)If (1) is semi-finite and w is faithful, (I) is semi-finite. (NI) If 4) is semi-finite and 1 is faithful, w is semi-finite. Proof. (i) is immediate. 3.

(ii) If w and (D are faithful, so is cp. Suppose 4) is faithful. It is clear that (D is faithful. Now let T be a non-zero element

278

PART III,

CHAPTER 4

of Z + We have w(TCI)) = w(0(T)) = 11)(T) > O.

0(I) is the supremum of an increasing filtering family of elements ofZ + ,hencew(TS) >0 for certain elements SE Z + , and so w(T) > 0 !(1) is thus faithful. Now,

(iii) Suppose w and are semi-finite. Let S be a non-zero element of There exists a non-zero element Si of A+ such If 0(S 1 ) = 0, we have (1)(S 1 ) = 0, and S, 0(S1) EVthat Si (iii) is proved. If 0(S1) 0, there exists a non-zero element T of Z+ majorized by 0(S1) such that w(T) < + we have T = T1O(S 1 ), with a Ti E Z+ such that Ti I; let, then

e.

.

,+ S2 = S T € A . 1 1

0(S2) = T10(S 1 ) = T, we see that S2 0, S2 cp(S 2 ) = w(T) < + thus cp is semi-finite.

As

Si

S, and

(iv) Suppose 4) is semi-finite and w faithful. Let S be a nonzero element of A+. There exists a non-zero element Si of such that Si S,

e

(1)(S 1 ) = W(4(S 1 )) w is faithful, this necessitates 0(S 1 ) < + co locally almost everywhere. As the proof is finished if 0(S1) - 0, we can therefore suppose henceforth that 0 < 0( 5 1 ) X < + m on a non— locally negligible measurable subset of Z. There then exists a projection E of Z such that E0(S1) X 0, Ell(S1) E Z. Let S2 = ESi. As 0(S2) = E0(S1), we see that S2 / 0, S2 S, is semi-finite. EZ; thus 0(S2) As

(v) Suppose that 0 is faithful and w non-semi-finite. There exists a non-zero element T of Z+ such that W(Ti) = + co for every non-zero T i of Z -F majorized by T. As 0(T 1 ) = TOM vanishes, up to a locally negligible set, on the same set as T i , we have

q(T 1 ) = w(O(T thus (I) is not semi-finite.

1

)) = + co;

D

Let A be a von Neumann algebra, Z a von Neumann algebra contained in the centre of A, cp a nol:ial trace on A+, and w a semi-finite faithful normal trace on 2+. There exists exactly one normal Z-trace 0 on A+ such that (j)= w o(p. PROPOSITION 3.

,

Proof. Let S EA+ The mapping T OST), where T runs through Z -F, is a normal trace on Z -F , and therefore on Z -F . lemma 1, there exists a unique element 0(S) E 2+ such that .

By

279

FURTHER TOPICS

O( ST) = w(O (S) T) for every T EZ + . Let S S be elements of A+, U a unitary ele' 1 ment of A, and S2 and T elements of Z+. We have

w(O(S +S )T) = (ti ( (s +s 1 )T) = OST) + 1

1 T) = W(((S) +

w(O(SS )T) =c1)(SS T) = w(O(S)S T), 2 2 2 w(O(USU-1 )T) =flUSU-1T) = (P(SU-1TU) = 11)(ST) = is normal. Let (Si) be and so 0 is a Z-trace. We show that an increasing filtering set in A+ with supreTum S EA ; then, the O(Si)'s form an increasing filtering set in Z + with supremum Q; moreover, the equality cp(SiT) = w(O(Si)T) for T EZ + yields, in the limit, OST) = w(QT) , hence Q = O(S), which proves our assertion. Finally, the uniqueness of is immediate since the formula = wo0 implies, for S EA+ , w(O(S)T) = (1)(ST) for any T E Z± .

Let A be a von Neumann algebra, Z its centre, (I) a semi-finite, faithful, normal Z-trace on A+ , and 4) a normal trace on A+. There exists exactly one normal trace w on Z+ such that (1) = wo0. PROPOSITION 4.

Let w l be any semi-finite faithful normal trace on Z + Let (p 1 = 0) 1 00, which is a semi-finite, faithful, normal trace on A+ (proposition 2). By part I (chapter 6, proposition 1, corollary 2), there exists a projection E of Z (identified with the characteristic function of a measurable subset Y of Z) which possesses the following properties; (i) if SEA+ is not in AE, (p(S) = + co; (ii) the trace (p' induced by (1) on is semi-finite. Let (pi be the trace induced by (pi on A and let lp = P P which is faithful and semi-finite. There exists (part I, chapter 6, theorem 3) elements Q0, Q 1 of Z such that (OS) = (Pi(S) = Ip(SQ 1 ) for S c A. Let Q be the function of Z + equal to + co on Z ,,Y, and to Q0Q1 1 on Y; let w be the normal trace on Z + defined by w(T) = w i (TQ) for any TE Z. We show that OS) = w(O(S)) for every S E A. If SAE, 0(S) is non-zero locally almost everywhere on ZN.Y, and so (1)(S)Q is infinite on a non-locally negligible set, hence w(O(S)) = w 1 (0(S)Q) = ± °3= (1)(S). It is therefore enough henceforth to consider the case where S E /4E. Identify ZE with ZE, i.e. with the set of functions of Z vanishing on ZN,Y. Let (Qi)i ci be the increasing filtering family of functions of Z+ majorized by QI 1 . Then 4)(S)Q 0Q1 1 is the supremum of the 0(S)Q 0Qi's, hence

Proof.

g

(

-1

= s upw 1 ( 1( S ) Q 0Qi )

= supw i ( ( SQ 0Qi )) = sup()) (SQ 0 Qi ) =suptp (SQ 0Qi Q i ) =suf4 (SQ iQ l ) •

280

PART III, CHAPTER 4

Now, Ql(C) is only zero on a locally negligible subset of Y, since $i is faithful, hence Q 1 (V -1Q1(C) = I locally almost everywhere on Y. Hence the sufaemum of the QiQ i 's is E, from which it follows that w(0(S)) = $(SE) = $(S). We show that w, whose existence we have just proved, is unique. The elements of Z + of the form 0(S), where S EA + (elements for which the value of w is prescribed), form the positive part of an ideal M of Z. Let F be the greatest projection of the strong closure of M. Suppose that I - F X 0. As 11 is semifinite, there exists a non-zero S E A+, majorized by I - F, such that 0(S) EZ + . Then

0(S) = 0(S(I - F)) = 0(S)(I - F) = 0, which contradicts the faithfulness of (D. Hence F = I. By , normality, w is therefore unique on Z+ and consequently on Z -E . 0 It is easy to see that proposition 4 is false if Z is only contained in the centre of A (and for example if Z reduces to just the scalar operators). Similarly, the hypotheses of faithfulness or of semi-finiteness of propositions 3 and 4 are indispensable. References : [6], [12], [101].

Existence and uniqueness theorems for operator traces. THEOREM 1. Let A be a von Neumann algebra, and Z a von Neumann algebra contained in the centre of A. For A to be semifinite, it is necessary and sufficient that there exist a semifinite faithful normal Z-trace on A. Proof. Let w be a semi-finite faithful normal trace on 2+. 4.

Suppose that there exists a semi-finite faithful normal Z- trace on A+ . Then, w00 is a semi-finite faithful normal trace on A+ 0 (proposition 2), hence A is semi-finite. Conversely, if A is semi-finite, there exists on A+ a semi-finite faithful normal trace $, hence (proposition 3) there exists a normal Z-trace (1) on A+, such that $ = wo(1). By proposition 2, T is faithful and semi-finite. 0

Let A be a von Neumann algebra, Z its centre, and cti l and (D 2 two semi-finite faithful normal Z-traces on A+. There exists exactly one element Q of Z+ such that 0 1 (s) = ,T2(S)2 for every s E A+ . We have 0 < Q ( c) < + co locally almost everywhere on z. THEOREM 2.

Proof. Let w l be a semi-finite faithful normal trace on Z. Then $ = w 1 00 1 is a semi-finite faithful normal trace on A+ (proposition 2). By proposition 4, there exists a normal trace w2 on Z + such that 4 = w204) 2 , and w2 is faithful and semi-finite

281

FURTHER TOPICS

by proposition 2. By lemma 1, there exists an element Q E 2+ such that w2(T) = wi(TQ) for every TE Z+ , and such that O < Q(C) < + co locally almost everywhere on Z. Then, for every SE A+, we have W

1

(4) 1 (S)) = w 2 (0 2 (S)) =

By proposition 3, the mappings S 0 1 (S) and S 02(S)Q are the same. If now 10 1 (S) = 0 2 (S)Q',with an element Q' E 2+, put w (T) = w i (TQ I ) for every TE Z+ ; for every SE A+ , we have W

2

( r (S)) = 2

W

1

(0 1 (S)) =

W

1

(4) 2 (S)Q 1 ) = w 21 (02 (S)) •

By proposition 4, we therefore have w2 =

Q = Q'.

LL),

and, by lemma 1,

D

Let A be a finite von Neumann algebra, and let Z be its centre. There exists exactly one normal Z-trace on A+, such that I(T) = T for T E 7 + . This Z-trace is finite and faithTHEOREM 3.

PPoof. Let T be a semi-finite faithful normal Z-trace on A+ (theorem 1). Let Q = T(I) E 7+ . Since T is faithful, we have Q(C) > 0 locally almost everywhere. Suppose Q(C) = + co on a non-locally negligible measurable subset of Z; let w be a semifinite faithful normal trace on 2+; there would exist a non-zero T of Z+ such that, for every non-zero T 1 of Z+ majorized by T, we would have w(0(T')) = w(QT') = + co; as the trace wo0 is semifinite, this would contradict proposition 10 of part I, chapter 6. Hence 0 < Q(c) < + co locally almost everywehre. This established, the mapping S 4- T(S)Q -1 on A+ is a normal Z-trace cl such that 0(I) = I. We have cl(T) = = T for TE Z+ ; this property implies that I is finite, and also that 0 is faithful [because the set of the S E A+ such that cl(S) = 0 is the positive part of a two-sided ideal M of A, and the greatest projection E of the strong closure of M is such that EEL 0(E) = 0]. Finally, the uniqueness of 0 follows from theorem 2. Let A be a finite von Neumann algebra. The Ztrace defined on A, whose existence is proved by theorem 3, is called the canonical Z-trace of A. DEFINITION 2.

This mapping, often denoted by T 4- T 11, is therefore a normal pcsitive linear mapping of A +canto the centre LZ of A, such that • = T for T EZZ. We have (5*)4 = (S4)*, (ST)4 = (TS)61 for SE A, TE A, and (ST)4 = sliT for S EA, TE Z. By part I, chapter 4, theorem 2, the mapping T T9 is ultra-weakly and ultra-strongly continuous.. If A is a finite factor, the canonical mapping l; is of the form T q)(T)I, where cp is the unique normal trace on A such that 4)(I) = 1. This trace 4) is called the canonical trace on A.

PART III, CHAPTER 4

282

References : [6], [12], [101], [104], [105], and T. IWAMURA, On continuous geometries 1 (Jap. J. Math., 9, 1944, 57 - 71). 1. Let A 1.3e a continuous finite von Neumann algeS4 the canonical Z-trace of A, and T an bra, Z its centre, S element of Z such that CV T I. Show that there exists a projection EE A such that E ll = T. [Show first that the result is true if T is a step-function taking a finite number of values of the form p.2 -n , with p, n integers ?_ 0. For this, use chapter 1 (theorem 1, corollary 3). Then establish the general result by arguing as for chapter 2, proposition 14.] [89]; and T. IWAMURA, On continuous geometries I (Jap. J. Math., 9, 1944, 57-71) .

Exercises.

2. Let U be a Hilbert algebra, let H be its completion, and let Z be the set of central elEments of H. Suppose that A = U(U) is finite. Let T -4- TR be the canonical Z-trace of A (Z , the centre of A). Let a be a bounded element of H. We adopt the notation of part I (chapter 5, exercise 4) . a. Show that, for every b E Ka' , we have L.J1= U. (First show ub n it for b E Ka . Then, if bn b with bn E Ka and b E Ka' , stays bounded and ub n converges weakly to ub. Use the fact that the canonical Z-trace is ultra-weakly continuous.)

H

b.

Show that U

Up a

H

.

c. Show that A is a factor is and only if Z is onedimensional. (If dim Z > 1, A is not a factor by part I, chapter 5, exercise 4 e. If A is rt 1 factor, there exist bounded elements a, b of H such that U, qare linearly independent; then use b) [ 29], [106]. 3. Let A be a semi-finite von Neumann algebra, and Z its centre, identified with L7(Z, V). Let (I) be a semi-finite faithful normal Z-trace on At Show that the set of the TEA + such that (1)(T) is finite locally almost everywhere is independent of (I) and of the identification of Z with L:7(z, V), and that it is the positive part of a two-sided ideal M of A. Show that M is strongly dense in A, and is the union of the two-sided ideals of definition of the T's, as T runs through the set of faithful normal Z-traces on At Show that M = A if and only if A is finite [12]. 4. Let Z be a Borel space, V a positive measure on Z, C H(C) a v-measurable field of non-zero complex Hilbert spaces over Z,

H = f H(c)dv(c),

A =

A(c)dv(C)

a decomposable von Neumann algebra in H, c (p c a measurable field of traces on the A(C) +I s, and Z the algebra of diagonalisable operators, identified with L7(Z, V).

283

FURTHER TOPICS

e a.

For every T =

f T(C)dV(C)

e

A+ ,

let 0(T) be the function

c .4- (P c (T(C)) which is an element of Z. Show that 0 is a Ztrace on A+, normal if the (Pc's are normal. (Argue as for proposition 1 of part II, chapter 5.) b.

A+

Conversely, every semi-finite faithful normal Z-trace 0 on

is of this type.

[Let W be the trace S -4-

I S(C)dV(C)

on

Z.

Form the trace w00. Apply part 11, chapter 5, theorem 2, to obtain the field C 4' (pc. Then use a and the uniqueness result of proposition 3.]

•

CHAPTER

1.

AN APPROXIMATION THEOREM

5.

The approximation theorem. the complex Hilbert space H, let T

be a continuous hermitian operator, and E a non-zero projection commuting with T; in this section, we put In

ME (T) = sup!'

II

11 x H11=1,Ex=x (Txlx),

(Txlx ) ,

m (T) = infm

wE (T ) = ME (T)

mE (T).

11=1,Ex=x

When E = I, we simply write M(T) , m(T), w(T). We agree, moreover, that w0 (T) = O. If F is a family of projections commuting with T, we put w F (T) = suPEEFwt (T) '

Let A be a von Neumann algebra, Z its centre, and T an hermitian operator of A. There exist projections G, G F of Z ., disjoint and with sum 1, and a unitary operator U of A such that LEMMA 1.

3/4 w(T),

-1 w G /(1/2(T + UTU ))

3/4 w(T).

G

(1/2(T + UTU

-1

))

W

Proof. Let n(T) = 1/2(M(T) + m(T)).

There exist spectral projections E, F of T, disjoint and with sum I, such that ME (T) < n(T), mF (T) n(T). Let G, G P be projections of Z such that GG P = 0, G + G P = I, EG -4 FG, FG' EG P (chapter 1, theorem 1). There exists a partial isometry V (resp. W) of A, with initial projection EG (resp. FG'), and final projection G1 5 FG (resp. GI EG P ). Hence, there exists a unitary operator U EA which transforms EG(H) into G i (H), G1(H) into EG(H), FG' (H) into Gi(H), G(H) into FG P (H), and which reduces to the identity operator on every vector orthogonal to these subspaces. We show that G, G P , U satisfy inequality (1), for example. We have

285

PART III, CHAPTER 5

286

TG

m(T)EG + n(T)FG = m(T)EG + n(T)G

1

G 1 ),

+ n(T)(FG

hence (UTU

-1

m(T)G + n(T)EG + n(T)(FG - G). 1 1

)G

Adding these inequalities we have (dividing throughout by two) 1 (T + UTU

-1

2

)G

1 (m(T) + n(T))(EG + G )+n(T)(FG 2 1 -

1

- (m(T) + n(T))(EG + G

Since it is clear that 1/2(T -+ UTU -1 )G proved inequality (1). 0

1

+FG - G

1

-

G ) 1

3 )=(M(T) - -W(T))G.

M(T)G, we have indeed

Let T be an hermitian operator of A, and F a finite family of pairwise disjoint projections of Z., with sum I. There exists a finite family F' of pairwise disjoint projections of Z., with sum I, and a unitary operator U of A, such that LEMMA 2.

F

(,

2

(T + UTU

-1

))

4 r

Proof.

Once the lemma is proved for a finite family of von Neumann algebras, its truth for the product von Neumann algebra follows immediately. We need therefore only consider the case where F reduces to the single projection I. Lemma I then enables one to construct a family F' (consisting of two elements) and a U which possess the required properties. Given an element T of A, we will denote by K T the convex subset of A generated by the elements UTU -1 ,where U runs through the group G of unitary operators of A. We will denote by Kri, the closure of KT in A for the norm topology. Moreover, consider the set S of the functions U f(U) defined on G, with real 0, zero except on a finite set of points, and such that values f(U) = 1. For fE S and TE A, we will put f.T = f(U)UTU-1 ; UEG UEG when f runs through S, f.T runs through K T . It is immediate that, for g E S, we have g.(f.T) = (gf).T, denoting by gf the function W

f(U)g(V) UEG,VEG,VU=W

on G, i.e. the convolution of g and f, which is an element of (G is regarded as a discrete group).

S

LEMMA 3. Let T be an hermitian operator of A, and let E > O. There exists an fE S and an SE Z such that H f.T - s 5 E.

H

287

FURTHER TOPICS

Proof. For every integer p > 0, there exists a family F = (E 1 , E 2 , ..., En ) of pairwise disjoint p9jections of Z, this folwith sum I, and an feS, such that wT(f.T) -‹(--)Pw(T): 4 lows immediately from lemma 2 by induction on p. Suitably choosing real numbers ai, we therefore have

f.T Now, S =

a.E E Z, and

y

a.E. i=1 "

(1)pw(T). 4

/ 3\ p

-4- w(T) is arbitrarily small for p

i-1

D

sufficiently large.

Let T 1 , T 2 , ..., Tn be elements of A, and let E > O. There exist an f€S and elements S l „ s 2 , sn of Z such that LEMMA 4.

Il

f.TS h h

h = 1, 2, •.., n.

for

<- E

As every element of A is a linear combination of two hermitian elements of A, we can suppose that the Th's are hermitian. Arguing by induction on n, suppose that we have found a gES and elements Si, S2, ..., S n _i of Z such that E for h - 1, 2, ..., n-1. Let (lemma 3) g' be g* Th Sh an element of S and Sn an element of Z such that g / .(g.Tn ) - sn E. We have, for h = 1, 2, ..., n-1,

Proof.

H

H

H g'. (g .Th

g'. (g. Th ) - s h

It therefore suffices to put

sh) H

g'g = f.

H g . Th

sh

E.

D

LEMMA 5. Let Ti, T2, ..., Tn be elements of A. There exist elements s l , s 2 , sn of Z and a sequence (fl , f2 , ...) of o elements of S such that, for h = 1, 2, ..., n, Hfi .Th -s h H

as i

+ oe.

Proof.

Using lemma 4, and arguing by induction on i, we construct, for every integer i > 0, an element gi of S and. elements Sf, g 1 ).Th H _< 2 -i S;I:i of Z such that H (gigi _ i for h = 1, 2, ..., n. Put fi = _l g l . We have s h 11 = H

gi+1 *

. 1• = Hg 2+

(fi .Th ) - S (f• 2

.T -S i ) H

h h

Hf..T - s h H .‹ 2 ,

hence

H fi+1 .Th

fi .Th 11

2-i+1

h

288

PART III, CHAPTER 5

Hence (f i .Th, f2.Th, ...) is, for h = 1, 2, ..., n, a Cauchy sequence, whose limit Sh is also the limit of the S and, conse0 quently, belongs to Z.

As a particular case of lemma 5, we thereNote that sh E Krl. h . fore have the following result: THEOREM 1.

For every

T E A., KT " = Krl,

nZ is non-empty.

COROLLARY. Let A be a finite von Neumann algebra, Z its centre, and ,T the canonical Z-trace of A.

(1) For every T

E

A, Kr7, reduces to the single point OM .

(ii) Let T be a linear mapping of A into Z, continuous for the norm topology, such that T(T 1T 2 ) = T(T 2T 1 ) for T 1, T 2 EA, and such that T(T) = T for T E Z. Then, T = (D. Proof.

We shall show that q reduces to 'HT) . As cl) possesses the properties supposed true for 'I', this will simultaneously prove (i) and (ii) . Now T(UTU-1 ) = T(U-1 UT) = T(T) for any T E A and unitary operator U of A. Hence 4' is constant on Kr and, consequently, on KTI . Hence, for S E KT", we have S='Y(S) = T(T) . 0

Later, we will prove converses of this corollary (chapter 8, theorem 1, corollaries 3 and 4) . Reference

2.

: [6].

An application: ideals of Z.

two—sided ideals of A and

The algebra A is no longer supposed to be finite.

Let T 1 , T 2 be elements of A. Then Kr4 +1,) is contained in the closure of 1q 1 + q for the norm topo7ogy. 2 Proof. Let S E K g. and let E > O. There exists an f c S 1 + T2' H such that 11 f.(T i + T2) - S ii E. Then (lemma 5) there exist g c S and elements Si E Ky.. Ti c K r7. , S2 E K p l . rr 2 C Kr7 , such that 1 LEMMA 6.

a

■2

II

g.(f.T i ) - s 1 11

E,

As 11 g.(f.(T i + T 2 )) - S II

S 2 II -

E.

6, we see that

il s which proves the lemma.

II g.(f.T 2 ) -

(

s 1 + s 2) H <- 3E,

Il

Let T E A, and s E Z. Then KgT is contained in the closure of sKr'l for the norm topology. LEMMA 7.

Let R E Kgrr , and 6 > O. There exists an f E S such that II f. (ST) - R 11_< E. Then (theorem 1) there exists a gES and an R1 E Kt I . rr c 4 such that II g. (f.T) - R 1 II E. These inequalities lead o the inequalities

Proof.

FURTHER TOPICS

s( g . (f.T))

-RH.

H

E(i

H f. (ST) -

g.(f.(sT) - R) H _<

H s( g .(f.T)) Hence 11R - sR i H

289

sR i

EH

s

R 11

E,

H.

s H), which proves the lemma.

E

LEMMA 8. Let m be a two-sided ideal of A, closed for the norm topology. Then, mn Z is the union of the KTs, when T runs through m.

Proof. As KTFF = {T} for T E Z, it is clear that m n Z is contained in the union of the KTFFT s, TEM. Moreover, for T E M, we have KT c Pr, hence Kii,c m, hence qcmnZ. LI LEMMA 9. Let ni be a two-sided ideal of A, and Tr its closure for the norm topology. Then, — m n Z is the closure of m n Z for the norm topology.

Proof. Let n be the closure of ni n Z for the norm topology. It is clear that ncnin Z. Suppose that there exists an element S of M n Z which does not belong to n. We are going to arrive at a contradiction, which will prove the lemma. Let Z be the spectrum of Z, Loo (Z) the algebra of continuous complex-valued functions on Z, and T fr the canonical isomorphism of Z onto Loo(Z) . Then n is the set of the T E Z such that fr vanishes on a certain closed subset Y of Z. There exists EY such that f5 (r) X 0. Hence there exists a spectral projection EE Z of S such that fE (C) = 1 and such that the restriction of S to E(H) is invertible. Moreover, as S is the norm limit of elements of there exists a TEM such that the restriction of T to E(H) is also invertible. Then, for a suitably chosen element T1 of A, we have TT]. E, from which it follows that E c ni. This contradicts the inequality fE (C) = 1. LEMMA 10. Let n be an ideal of Z ., closed for the norm topology, and ni the set of the T E A such that Ky. iTT ,,c n for any T1 E A, T2 E A. Then, m is the largest two-s-z-de6 ideal of A such that ninZcn, m is closed for the norm topology, and we have ni

n Z = n.

Proof. If T E M and T F E M, we have T + T F E ni by lemma 6. It is therefore clear that M is a two-sided ideal of A. If TEninZ, we have {T} = Kru c n, hence Tcn, from which it follows that MnZc n; if now T E n, we have, for any T 1 , T2 in A, TK"TiT 2 cn , hence (lemma 7) K TiTT FF 2 = K T" TiT2 c n , so that TEmnZ. Thus, n = m n Z. Finally, if ffl F is a two-sided ideal of A such that M t n Z c n, let m" be its closure for the norm topology; we have renZcn (lemma 9) , hence, for every T e K c n (lemma 8) , and therefore m" c m. PROPOSITION 1. Let n be an ideal of Z which is closed for the norm topology. There exists a greatest two-sided ideal m of A

290

PART III, CHAPTER 5

such that mnZcn, and we have MnZ = n. There also exists a smallest two-sided ideal m' of A such that m' n Z = n. Proof. The existence ofMand the fact thatMnZ=nfollow from lemma 10. Moreover, let m' be the two-sided ideal of A generated by n. We have m' n Z Dn., and M / CM, hence m' n Z = n, from which it follows that m' is the smallest two-sided ideal of A such that m' nZ = n. 0

1. The mapping 1V1.4- MnZ isa bijection of the set of maximal two-sided ideals of A onto the set of maximal ideals of Z. COROLLARY

Proof.

Let M be a maximal two-sided (hence norm-closed) ideal of A. Let n' be a norm-closed ideal of Z containing m n Z and different from Mn Z. The greatest two-sided ideal M P of A such that m' n Z = n' contains M and is different from M, and is therefore equal to A; hence n' = Z, which proves that MnZ is a maximal ideal of Z. Moreover, if n is a maximal ideal of Z, it is clear that the greatest two-sided ideal M of A such that MnZ=nisamaximal two-sided ideal of A ; and if M i is another two-sided ideal of A such that m i nZ = n, we have Mi CM, hence Ml = M if M i is maximal. 0

Let m be the intersection of the maximal twosided ideals of A. We have mn Z = o. COROLLARY 2.

Proof. This follows from corollary 1 and from the fact that the intersection of the maximal ideals of Z is just zero. 0 COROLLARY 3. If A is a factor, the set of the two-sided ideals of A, differing from A, possesses a greatest element. To say that a two-sided ideal M of A is distinct from amounts to the same thing as saying that Mn Z = 0. 0

Proof.

A

When A is finite, lemma 8 and lemma 10 may be stated (by the corollary of theorem 1) in the following way:

Let A be a finite von Neumann algebra. (i)Let m be a two sided ideal of A, closed for the norm topology. Then, mn Z is the image of m under the canonical Ztrace of A. PROPOSITION 2.

-

(ii)Let n be an ideal of Z, closed for the norm topology. The greatest two-sided ideal muofAsuch that mnZ=nis the set of the TEA such that (TT 1 )1 e n for every T1 E A. COROLLARY 1. Let m be a two-sided ideal of A (supposed finite). If mn Z = 0, we have m = 0. Proof. Let M F be the greatest two-sided ideal of A such that m' n Zi. = 0. For every TE ni', we have TT* E M P , hence (TT*) Q1 E m' n Z = 0, hence T = 0. D

291

FURTHER TOPICS

The intersection of the maximal too-sided ideals of A (supposed finite) is zero. Proof. This follows from corollary 1, and from corollary 2 of D proposition 1. COROLLARY 2.

If Ais a finite factor, the only too-sided ideals of A are 0 and A. COROLLARY 3.

Proof.

This follows from corollary 1.

0

References : [29], [60], [63], [124]. 3.

Characters of finite wma Neumann algebras.

In this section, A denotes a finite von Neumann algebra, Z its centre, and T T9 its canonical Z-trace. Let L be the set of central linear forms on A [i.e. such that $(T1T 2 ) = $(T2T1) for any T1 E A, T 2 E A], continuous for the norm topology. Let L' be the set of linear forms on Z which are continuous for the norm topology. Then, the following proposition generalises, in the case of a finite von Neumann algebra and the canonical Z-trace, propositions 2 and 4 of chapter 4. PROPOSITION 3. Let $ L, and let cp' E L' be t. the restriction of $ to Z. We have, for every TE A, $(T) = $'(T4). The mapping cp' is a bijection of L onto L'. Proof. Since $ is central, $ is constant on KT . Since $ is continuous for the norm topology, $ is constant on K. Then, by the corollary of theorem 1, we have

cp(T) = $(`111:1) = This shows that the it is surjective, let linear form $ on A by ties of the canonical of $ to Z is $'. D

mapping $ $' is injective. To show that $' be an element of L' ,and define a the formula $(T) = $'(T4). By the properZ-trace, we have $E L, and the restriction

A character of the finite von Neumann algebra A is any norm-continuous central Linear form x on A, whose restriction to the centre Z of A is a character of Z. DEFINITION 1.

Recall that the characters of Z are the homomorphisms of Z onto the complex field. They are continuous for the norm topology. The set of characters, which is in (1-1) correspondence with the set of maximal ideals of Z, is endowed with a natural topology which makes it a compact space, namely the spectrum Z of Z, and Z may be identified with the algebra of continuous complex-valued functions on Z. By proposition 3, the characters of A may be identified with the points of Z. The characters of A are finite traces on A (non-normal, in general; cf. part I, chapter 8, exercise 5). For an element

PART III, CHAPTER 5

292

of L to be a character, it is necessary and sufficient that we have, for any TEA and SE Z, X(ST) = x(S)X(T). The condition is obviously sufficient. Conversely, if x is a character, we have X(ST) = x((ST) ) = X(ST) = x(S)x(TO) = X(S)X(T) .

As the elements of L' are simply the complex Radon measures on Z, proposition 3 implies the following results: PROPOSITION 4. There exists a canonical bijection (1) -4- v of L onto the set of complex Radon measures on z. This bijection is defined by the formula (1)(T) = f

x(T)dv(x)

for every

TEA.

For cp to be a trace, it is necessary and sufficient that y be positive. COROLLARY. Let T be the convex subset of L consisting of the finite traces on A such that cp(1) 1. Then, the non-zero extreme points of T are the characters of A. Proof. It is well known that the point-masses of mass 1 on Z are the non-zero extreme points of the convex set consisting of 0 such that V(I) 1. 0 the measures

Let x be a character of A. The set m of the TEA such that x(T*T) = 0 is a maximal two-sided ideal of A. The mapping x m is a bijection of the set of characters of A onto the set of maximal two-sided ideals of A. Proof. Observe, to begin with, that the conditions PROPOSITION 5.

x(T*T) = 0, on the one hand, and x(T1T) = 0 for every T1 E A on the other hand, are equivalent by the Cauchy-Schwarz inequality. This established, let x' be the restriction of x to Z, which is a character of Z and therefore defines a maximal ideal n of Z. The condition x(TiT) = 0 for every T 1 E A is equivalent to the condition x'((T T)q) 0 for every Ti E A, and therefore to the condition (T iT)01 E n for every T 1 E A. NOW (proposition 2) this condition defines a maximal two-sided ideal M of A. Moreover, n, n the mappings x -4- x', x' M are bijections between the following sets: the set of characters of A, the set of characters of Z, the set of maximal ideals of Z, the set of maximal two-sided ideals of A. D

References : [29] , [63] , [69], [ 7 0], [107] , [1 2 1], [123] , [124]. Let A be a finite von Neumann algebra, Z its centre, T an ment of A, and L q'the weak closure of KT . Show that K,Tn Z = I, where T -4- TR denotes the canonical Z-trace of A. (Argue as for the corollary of theorem 1, using the fact that the mapping T TR is ultra-weakly continuous) [6].

Exercises.

1.

FURTHER TOPICS

293

2. a. Let A be a von Neumann algebra, Z its centre, T1, T2, ..., Tn elements of A, and S1 E K 1 . There exists a sequence (f1, f2, •..) of elements of S suc1i that fi.Ti converges (in the sense of the norm topology) to S1 and fi.Th to an element of Km" for h = 2, 3, ..., n. (Choose g S in such a way that II g.Ti 2 -q; then apply lemma 4 to g.T i , g.T 2 , s 1 II g.Tn to determine f .)

q b. Let VI, be a norm-closed ideal of Z. Let S E A+ , T E A+. If K11! c n and S T, we have Ks" c n. (Show that every element of K s" is majorized by an element of KT", by using a.)

3. Let A be a von Neumann algebra, Z its centre, n a normclosed ideal of Z, and ni the largest two-sided ideal of A such that M n Z = n. Show that ni is equal to the set 171 of the T E A such that KT" *T c n. [First show that M M P and that M P C Z = Vi; it then suffices to prove that M P is a two-sided ideal of A ; for every unitary operator U of A and every T E we have UT E M I and TU E M P ; finally, let T1 E ni ' T 2 E ni ' we have ,

(T + T )*(T + T ) 1 2 1 2

;

2T*T + 2T*T 11 2 2'

and qtr 1 +Tr 2 is contained in the closure of Krkr i + qr 2 , and therefore in Yi; hence T1 + T2 E M P , by exercise 26 b1 If A is finite, m is the set of the T E A such that (T*T)4 E n. Rediscover this result by using proposition 5 [29], [124]. Let A be a finite von Netkmann algebra, and (i) a finite Ztrace i. on A. Show that 4)(T) = ci) (I) for every T E A, where T TQ! denotes the canonical Z-trace of A. (Use theorem 1.) Deduce from this that is normal [12]. 4.

5. Let H be a complex Hilbert space, H1 and H 2 two infinitedimensional complementary orthogonal subspaces of H, and A the set of operators of the form S = XiPH i + X2PH,) + T, where T E L(H) is compact, and where X 1 E C, E C. g how that A is a C*-algebra containing I, whose centre consists of just the scalar operators. Show that, if S is unitary, IX11 = 1X21 = 1 . Deduce from this that, if U 1 , U 2 , ... Un are unitary operators of A and p i , p 2 , 'An real numbers 0 with sum equal to 1, 1 .0 (P - P )U. is then of the form PH - PH + R, with 2 H1 H2 2 1 2 -

i=1 R compact, and cannot therefore converge, in norm, to an opera-

tor of the centre of A [6]. 6.

For every n of N (the set of integers 0), let Hn be an n-dimensional Hilbert space, An = L(Hn Zn the centre of An (which may be identified with C), and A the finite von Neumann algebra H A of centre Z = H Z ; Z may be identified with the n nEN n nEN algebra L(N) of bounded sequences (X 0 , X 1 , ...) of complex ),

,

PART III, CHAPTER 5

294

numbers. Let U be a non-trivial ultrafilter on N. Let n be the maximal ideal of Z consisting of the sequences (Xi) such that lim Xi = O. Let M be the greatest two-sided ideal of A such LL that ninZ = n. Let M' be the two-sided ideal of A generated by [Let n. Show that M is distinct from the norm-closure of M'. T = k (r n ) nc ivE A, where Tn is a rank-1 projection in H. Wekhave (TS)1 En for any S E A, because the sequence with which (TS) I is identified converges to zero. However, for any element R of M', we have H T - R 1.] Let A be a von Neumann algebra, Z its centre, M a normclosed two-sided ideal of A, 13 the *-algebra A/m, Y the centre of B, and (1) the canonical mapping of A onto B. Show that UTU-1 E m for T(Z) = V [If T E A is such that (1)(T) E Y, we have T every unitary operator U of A, hence T - T 1 E M for every T I E hence in particular for every T 1 E fq.]

7.

Let A be a finite von Neumann algebra, Z its centre, cp a trace on A, and m the two-sided ideal of A consisting of the The sesquilinear form TEA such that (1)(T*T) = O. (S, T) 4 (I) (ST*) on A defines, on passage to the quotient, a scalar product on U = A/m which makes U a Hilbert algebra with identity. Let H be the completion of U, Z the set of central elements of H, and (lo the canonical mapping of A onto U.

8.

a. 0(Z) is dense in Z. [Let TE A and let U be a unitary element of A. Show that (1)(T2 - T(U)0(T)T(U) -1 is orthogonal to Z. Deduce from this that (1)(T4) is the orthogonal projection of (1)(T) onto Z.] Problem: is U full? b. For U(U) and V(U) to be factors, it is necessary and sufficient that (I) be a character of A. [Using a and chapter 4, exercise 2 c, show that it is necessary and sufficient that Z/(mnZ) be 1-dimensional.] c. Take for A the example of exercise 6. Show that one can choose (I) in such a way that M is the ideal considered in exercise 6, and that u(U) is then a type II I factor. (By b, it is enough to prove that A/m is infinite-dimensional.) We thus have here a new proof of the existence of type II ]. factors [124]. Let A be a C*-algebra containing I, B a sub-C*-algebra containing I, and n a linear projection of A onto B such that 1. Then Tr is positive, n(ATB) = An(T)B for A, BE B and 11 Tr II and'n(T)*n(T) n(TT) for TEA [195]. A, TE

9.

a.

b. Let A be a von let HT be the weakly where U runs through that A has Schwartz'

Neumann algebra in H. For every TEL(H), closed convex hull of the set of the UTU-1 the set of unitary elements of A. We say property if, for every TE L(H), HT contains

FURTHER TOPICS

295

an element of A'. If this is true, there exists a linear pro1 and Tr(T) E HT for jection Tr of L(H) onto A' such that 11 Tr 11 every TE L(H) [309]. C. There exist type II1 factors which possess Schwartz' property and others which do not possess it [309].

d. Suppose there exists, in the unitary group of A, an amenable subgroup G such that A is the von Neumann algebra generated by G. Then A has Schwartz' property; this is the case if A is abelian [296], [319], [336], [399]. If A is the algebra of multiplication operators by bounded measurable functions, in L 2 ([0, 1]), there exist at least two distinct mappings TT : L(H) .4- A' = A possessing the properties of b [230]. e. Let A be an abelian von Neumann algebra, and Tr a positive linear mapping of L(H) into A l such that ff(T) = T for TE A. Then T1(ST) = ff(S)T and Tr(TS) = Pit(S) for every S E L(H) and every T EA [230]. f. Let A be a von Neumann algebra, (I) a semi-finite faithful normal trace on A+, m the ideal of definition of ep, B a von Neumann subalgebra of A, and n = mnB; suppose that n is strongly dense in B. Let 131 be the set of the TEA such that q(TS) = 0 for SE n. The sum B + Bi is direct and equal to A. Then Let Tr be the corresponding projection of A onto B. 1, Tr is faithful and positive, and ultra-weakly and 11 Tr II ultra-strongly continuous; if SEA and if TEAnB' = C, we have If B is the set of elements of A commuting with TI(ST) = ff(TS). C, then for every TEM, the weakly closed convex hull of the set fUTU -1 : U a unitary element of CI

meets B in exactly one point, which is none other than ff(T), and which belongs to n [15], [130], [134], [179], [241], [278], [293]. g. Let A be a von Neumann algebra in H, G a group, g -4- Ug a unitary representation of G in H such that U AU-1 = A for every gE G, UG the set of the Ug 's for gE G, S = Agn U G , and E 0 the projection of H onto the subspace of vectors invariant under Uu . We have E 0 G U" n u' u" BP Suppose that the central support of G G E 0 relative to 13 ' is I, in other words, that A'E o (H) is total in H. Then there exists a unique G-invariant normal linear projection Tr of A onto B, and Tr is faithful and positive. For TEA, ff(T) is the unique point of B in the weakly closed convex hull (g E G) , and Tr(T) is the unique element of A such of the U V E TE that Tr(T)E O 0 0 ' A normal positive linear form w on A is Ginvariant if and only if w = (w113) 0 T1 [370], [371], [462]. .

CHAPTER 6.

THE COUPLING OPERATOR

The coupling operator. LEMMA 1. Let A be a von Neumann algebra in H, Z its centre, and J an involution of H, commuting with the projectiyns of Z, 1.

and mph -Opt JAJ = A'. and X'X — JX .

Proof.

For every

When T I runs through

J(T I x) = T(Jx),

XE

H, we have J(4: ) =

A', T = JT I J runs through A, A J(X AI ) = X . hence

and

Jx

To prove the second aslertion of the lemma, it suffices to consider the case where Xf%-< XA Jx (thanks to theorem 1 of chapter 1, and to the fvt that J commutes with the projections of Z); we then have VI X4; (chapter 1, theorem 2); as J defines an antiisomorphism of A onto A', we deduce from this that

A' ) = X A ; J(X Jx

X A = J(XxA ' ) Jx

hence (chapter 1, proposition 1) X XA

X AJX'

COROLLARY. Let U be a Hilbert algebra, H the completion of U, J the involution of H canonically defined by U, Z = U(U) nV(U), 0 a normal Z-trace on U(U) +, and 0' the Z-trace on V(U) + obtained on transforming (I) by means of J. For every X E H, we

have

Proof.

U (U) ). (NE V(U) ) = () Y (E x U(U) V(U) — JEx J (lemma 1), whence We have Ex cP(E

Let A be chapter 4, space Z, a with L(Z, tions 0,

V(U)

) = O f (JE

V(U)

J) = V(E

U(U)

).

a von Neumann algebra in H, and Z its centre. As in we will choose once and for all a locally compact positive measure V on Z, and an identification of Z V), which embeds Z -4- in the set Z1- of measurable funcfinite or otherwise, on Z. 297

PART III, CHAPTER 6

298

Suppose that A is semi-finite. Let 0 (resp. 0 1 ) be a semi-finite faithful normal Z-trace on A+ (resp. A' 1 ) . There exists exactly one element c of 2+ such that, for every cO' (EA) . We have 0 < c(C) < + 0D locally x E H, we have (1)(4 ) almost everywhere. PROPOSITION 1.

(i) We prove the uniquenes of C. Let C 1 be an element of 2+ such that 0(EA ) = C 1 O t (4) for every XE H. Suppose that, for example, we hA. C(C) < Cl(C) on a non-locally negligible measurable subset Z t of Z; we shall arrive at a contradiction. Let G be the projection of Z corresponding to the characteristic function of Z t . Since 0 1 is semi-finite, there exists a non-zero projection E t of A' such that E t < G, / t (E t ) Making E t smaller if necessary, we can suppose Tt E t is of the form EA x . Since 0 t is faithful, we have 0 < 0 t (E x ) < + 00 on a non-locally negligible subset of Z t . Then,

Proof.

A W(E )

A 1 (1)'(E ),

which is impossible. (ii) When A is given, it is enough to confirm the existence of C and the last assertion of the proposition for a particular choice of 0 and of 0 1 . This follows from the relations that hold between two semi-finite faithful normal Z-traces (chapter 4, theorem 2). (iii) Suppose that the existence of C and the last assertion of the proposition are established for A and A'. Let Px (resp. PO be a projection of A (resp. A') with central support I. T.;( Adopt the notation of chapter 2, lemma 1. The induction T t is an isomorphism 8 of A' onto Br; this isomorphism allows us in particular to identify Z with Zx, the centre of B and B r ; transferring 0 / to B r+ by means of 0, we obtain a semi-finite faithful normal Z-trace 01 on Br + . Moreover, let M be the set of the TE A such that TPx = PxT = T; the restriction of 0 to M-4- defines a semi-finite faithful normal Z-trace 0 1 on S. For X E X, we have

X A' =X B'c X, x x

X A nX =P (X A ) = X B ; X x x

hence (E AI )

=

1

(E IV );

x

0 1 (EA ) = 0 t (0). 1 x x

The proposition is therefore valid for B and B r . Another application of the same argument shows that the proposition is also true for V and Dr. (iv) We are going to show that every semi-finite von Neumann algebra is spatially isomorphic to a von Neumann algebra such as

FURTHER TOPICS

299

A

being further supposed standard. This, with the corollary of lemma 1, will complete the proof. Every semi-finite von Neumann algebra is spatially isomorphic to a von Neumann algebra of the form (A1 0 CO E 11 1 where A l is a standard von Neumann algebra in Hi, and where EE (A 1 0 ciO' (part I, chapter 4, theorem 3, and chapter 6, corollary of proposition 9). Con0 (K 0 K), the von Neumann algesider, in the Hilbert space bra Al 0 (CK 0 L(K)); it is the tensor product of two standard von Neumann algebras (part I, chapter 6, corollary of proposition 6), and is therefore standard [part I, chapter 5, proposicK is spatially isomorphic to tion 9, (ii)] and Al L(K)) E (A1 cK E, being a suitably chosen projection of E A 1 o cK o L(K).

"0,

Let A be a von Neumann algebra such that A and A' are finite. Take for (I) and (13, ', in proposition 1, the canonical Z-traces of A and A'. The corresponding function of Z -4- is called the coupling operator between A and A' and is denoted by DEFINITION 1.

CA When A is a factor, CA may be identified with a finite real number > 0, the coupling number between A and A'. The term "coupling operator," sometimes called the "invariant," is english for the french term "fonction de liaison." Roughly speaking, CA measures the "relative size" of A and A' (cf. exercise 2). CA can, in fact, be defined for any A. References : [20], [31], [40], [65], [66], [67], [89].

2.

Properties of the coupling operator.

We plainly have CA' = C. Let Z' be a locally compact subset of Z, V' the measure induced by V on Z r , G the projection of Z associated with the characteristic function of z', and B = AG . The centre of AG is Z G , which may be identified with q(Z 1 , v'). We check immediately that CB is the restriction of CA to Z r . Now let E' be a projection of A' with central support I. The induction T TE , enables one in particular to identify the centre of AE I with Z. Let (I (resp. V) be the canonical Z-trace of A (resp. A'). Then: PROPOSITION 2.

CAE , is the product of CA and 4)'(E').

Proof. For every operator T' of E'A l E l , there exists exactly one function (to within locally negligible sets) of q(Z, V), let it be O(T'), such that (1)'(T') = Q(T 1 )(1) 1 (E'); the uniqueness is immediate, because, on account of the faithfulness of (D', the function l'(E') only vanishes on a locally negligible set; when kE' for some k > 0, hence T' is hermitian, we have -kE 1 < T'

PART III, CHAPTER 6

300

-k(1) 1 (E l )

V(T 1 )

ari (E f ),

whence the existence of O(T') in this case and the general case. The mapping T' -4- Q(T') is a EWE' into Z, and may therefore be identified mappingof into 7Er • It is clear that this tive. We have O(Ti:* = O(*1). If QEZ, we

0'(E'QE') = QV(E'),

by linearity in linear mapping of with a linear mapping is posihave

Q(E'QE') = Q.

hence

Hence (chapter 5, corollary of theorem 1) Q may be identified with the canonical ZE I-trace T' of A' A This established, for every xe E/(H), E x ' and E E, correspond TR , of A onto AEI ; hence, denoting by T under the isomorphism T the canonical ZEI -trace of AE I, we have

A'

A',

T(ExE ) = (1)(E x ).

Moreover,

A 0'(E x) = Q(EA)

I (E) = T'(E Apr

x

by the first part of the proof. Hence

A' I T(E E ) = c (1) 1 (E')T'(E A x x

).

]

Suppose that A and A' are finite and For A to possess a separating (resp. cyclic) elemen -4 a-finite. it is necessary and sufficient that CA (resp. CA PROPOSITION 3.

Proof.

If there exists a separating element x for

E A' = 1,

E

A

I,

hence

I = (D(E

A'

)

A,

we have

(D P (E A )

and, consequently, CA ?_ I. Conversely, suppose that CA ?_ I, and let us show that A possesses a separating element. It suffices (part I, chapter 2, proposition 3) to study the case y here there exis a separating element x for A'. We then have E= I. I, hence E henc 4)(4 ) I: the element x is separating for A. Interchanging the roles of A and A', we see that the existence of a cyclic element for A is equivalent to the condition cA I. 0

Suppose that A and A' are finite. For A to be standard, it is necessary and sufficient that CA = T. Proof. The condition is necessary (corollary of lemma 1). Conversely, suppose that CA = I and let us show that A is stanPROPOSITION 4.

dard. As a product of standard von Neumann algebras is standard, it suffices to study the case where A and A'

301

FURTHER TOPICS

are G-finite. Then (proposition 3) there exists a separating element for A and also a separating element for A'. Hence (chapter 1, theorem 5) A is standard. References : [20], [31], [40], [65], [66], [67], [89].

Applications: I. Comparing the strong and ultra-strong, weak and ultra-weak topologies. PROPOSITION 5. Let A be a von Neumann algebra in H, and n an integer > 0. The following properties are equivalent: (i) Every ultra-weakly continuous linear form on A is the sum of n forms wx,y . (ii)Every normal positive linear form on Ai s the sum of n form wx . (iii)The subsets of A defined by a condition. of the type 3.

X H Tx i

112

1 form a base of neighbourhoods of 0 in A for the

i=1

ultra-strong topology. (iv)Every projection E of A such that AE is a-finite is the supremum of n cyclic projections. Let K be an n-dimensional Hilbert space. To say that the integer n satisfy condition (i) [resp. (ii), (iii), comes to the same thing, as is immediately seen, as saying in H o K, A o cK and the integer 1 satisfy (i) [resp. (iii), (iv)]. It therefore suffices to study the case

Proof.

A

and (iv)] that, (ii),

n= 1 (i) (ii): Suppose that condition (i) is satisfied. Let (1) be a normal positive linear form. Then, (I), being ultra-weakly continuous, is a form wx,y , hence (part I, chapter 4, lemma 2) is a form w2 . (ii) that

(iii):

For every sequence (yi) of elements of

CO

X H Yi H

2

<

H such

00 ,

i=1

let y I be the ultra-strong neighbourhood of 0 in the colridition

Ty i 11 i-1

2 < 1.

A defined by

PART III, CHAPTER 6

302

We have 00

-E=1

11

11 2 = ci)(T*T)

00 $ being the normal positive linear form X w„.. If condition 02 i=1 is satisfied, we V(y i ) = V(X), whence have $ = wx hence , (ii) condition (iii). Let $ be an ultra-weakly continuous linear form on A. If condition (iii) is satisfied, there exists an XEH 1 implies 114)(T)1 _< 1. Hence $(T) is a consuch that 11 Tx tinuous linear function of Tx, from which it follows that there exists y EH such that $(T) = (Txly).

(iii)

(i):

H

(ii) <4. (iv):

This is the corollary of theorem 4, chapter

1. D

Remark 1.

The conditions of proposition 5 are satisfied, with n = 1, when A is abelian and when A is standard (chapter 1, corollary of theorem 4). We will encounter a further case of very broad scope in chapter 8 (theorem 1, corollary 10).

Suppose that A and A' are finite. For the conditions of proposition 5 to be satisfied, it is necessary and sufficient that CA 1/n. PROPOSITION 6.

Proof. Let K be an n-dimensional Hilbert space. Let B = A ® cK, acting in H 0 K. We have CS = ncA (proposition 2). As in the proof of proposition 5, we are led to the case where n = 1. It then suffices to apply proposition 3. 0

Let A be a von Neumann algebra. properties are equivalent: PROPOSITION 7.

The following

(i)Every ultra-weakly continuous linear form on A is the sum of a finite number of forms x,y w -, (ii)Every normal positive linear form on A is the sum of a finite number of forms wx; (iii)The weak and ultra-weak topologies coincide on A; (iv)The strong and ultra-strong topologies coincide on A. Proof.

The equivalence (i) s (iii) follows from the definition of the weak and ultra-weak topologies (part I, chapter 3, section 1). Since the weak (resp. ultra-weak) topology is the weak topology corresponding to the strong (resp. ultra-strong) topology (part I, chapter 3, theorem 1), we have the implication (iv) (iii). The implication (iii) (iv) follows from the relations between the weak and strong (resp. ultra-weak and ultra-strong) topologies given in part I, chapter 3, section 2. Since every ultra-weakly continuous linear form is a linear combination of normal positive linear forms, we have (ii) (i).

303

FURTHER TOPICS

Finally, the implication (i) = (ii) follows from lemmas 2 and 6 0 of part I, chapter 4.

Remark 2.

Let

A 1 , A 2, A =A 1

An xA

be von Neumann algebras, and

2 x

• x

An .

For the strong and ultra-strong topologies to coincide on A, it is necessary and sufficient that they coincide on A l , A2, ..., A. To decide whether the conditions of proposition 7 are satisfied, it therefore suffices to consider the following cases: 1° A' is properly infinite; 2 ° A' is finite and A is properly infinite; 3 ° A' and A are finite. The first case will be studied in chapter 8. The other two cases will be discussed now.

If the strong and ultra-strong topologies coincide on A, every projection of A which is the supremum. of a countable family of cyclic projections is majorized by the supremum of a finite family of cyclic projections. LEMMA 2.

Proof.

Let E be the supremum of the projections

00 2 E X, °D . Let V (xr) x2 ' '"' We can suppose that X 11 x 11 < r=1 be the ultra-strong neighbourhood of 0 in A consisting of the

A' E

E A'

CO

TEA such that

X r=1

k Txr M 2

1.

There exist vectors

c V fx ). Let the projection F be , Yi' y 2 , ..., y n such that (i t%ur , A' , r, thesupremmofei ,4 1 2 ,—,E,,For every number a > 0, we F)x r,= 0 for every r. F) E V (y) c V (Xr)* Hence (I have a(I I - E. Hence F Consequently, I F E. il

If A' is finite and A is properly infinite, the ultra-strong topology on A is strictly finer than the strong topology. Proof. For every xE H, E is finite, hence 4 1 is finite (chapter 2, proposition 3). Hence the supremum of a finite family of cyclic projections of A is finite (chapter 3, proposition 5). Now there exists an infinte family (that we can supPROPOSITION 8.

pose countable) of equivalent, disjoint, non-zero projections of A [such a family is furnished by chapter 1, corollary 2 of theorem 1, applied to a family (Ei) consisting of a single finite projection]; making these projections smaller, if necessary, we can suppose that they are cyclic; their supremum is infinite (chapter 2, proposition 10). The proposition then follows from lemma 2. D

304

PART III, CHAPTER 6

Suppose that A and A' are finite. For the strong and ultra-strong topologies to coincide on A, it is necessary and sufficient that ql be essentially bounded. Proof. If c-A 1 is essentially bounded, the strong and ultraPROPOSITION 9.

strong topologies coincide on A (proposition 6). If CA 1 is not essentially bounded, there exists, for every integer p > 0, a non-zero projection Ep of the centre Z of A such that CAEr 1/p. Making the E,'s smaller, if necessary, we can suppose thA the Z E 's are G-finite. Let E be the supremum of the Ep's. Th Cprojection E is the supremum of a countable family of cyclic projections. If the strong and ultra-strong topologies coincide on A, E is the supremum of a finite number q of cyclic projections (lemma 2); then, every projection of A majorized by E is the supremum of q cyclic projections, hence (proposition 6) 1/q, which is a contradiction. CAE The converse of lemma 2 is false (chapter 8, exercise 7). References : [15], [19], [31], [40], [70], [89], [117].

Applications: II. Conditions for an isomorphism to be spatial. Let A be a von Neumann algebra, and Z its centre, identified Z, V). Let with L(C (Z be another von Neumann algebra, Y its centre. If 0 is an isomorphism of A onto B, Q defines an isomorphism of Z onto Y, and hence V may be identified with 4.

LcciC (z v). We will employ this convention in the following assertion.

Suppose A, A', B, B' are finite. isomorphism of A onto B. If CA = CB, 0 is spatial. PROPOSITION 10.

Let

0

be an

Proof. Let 0 be an isomorphism of A onto B. There exist a von Neumann algebra C and projections E', F' of C', with central support I, such that one can identify A with CE, B with C F I, and Q with the mapping TE, TF, T running through C (part I, chapter 4, corollary of theorem 3). Let G / be the supremum of E' and F', which is a projection of C / . As A' = C, B' =C F1 1 are finite, E F and F' are finite projections of C', and hence G ' is finite. Replacing C by Cr, we see that we can suppose C I C, isomorphic to A and B, is finite. Thisfinte.Morv, established, denote by (D / the canonical Z-trace of C I ; we have (proposition 2)

C'(E')= CA C

C C '(F') -

C

B'

If CA = CB, we deduce from this that 0'(E') - (1) 1 (F'), whence E' — F / (by the comparability theorem for projections and the faithfulness of (D / ). Hence 0 is spatial. D

305

FURTHER TOPICS

Let A be a finite von Neumann algebra, Z its centre, and Q an automovphism of A which leaves the elements of Z fixed. Then, Q is spatial. COROLLARY.

Proof. There exists a family (E:i) of disjoint, finite projections of A' with sum I (chapter 2, proposition 7, corollary 1). Let F. be the central support of E. Then, Q induces an auto2 "Z AF ., and the induction TF . morphism Y TE ttransforms Ti into an auiomorphism (Di of AE. As AE. and Ag are finite, and 0,1: leaves the elements of thJcentre °I As ,. fAted, (101: is defined by a unitary operator Ui in E(H) (proposition 10). Let U be the unitary operator in H which coincides, for every i, with Ui It is clear that U defines the automorphism Q. 0 in E(H). References : [20], [31], [66], [67], [89].

Exercises. 1. Let U be a Hilbert algebra, H the completion of U, Z = U(U) n V(U), J the involution of H canonically defined by U, 0 a normal Z-trace on U(U) + , and 0 1 the normal Z-trace on V(U) + obtained by transforming 11 by means of J. a.

If xE

H is bounded, Eg (

is the support of qc, JE (U)J is

the support of JUp = Vx , and hence 0(E P T U )) = 0 / (E L(U) ). b. If x is any element of H, let xn be a sequence of bounded elements converging to x. Let

V(U)

V(U)

U(U) n n xn xn Replacing xn by EE'xn , we can suppose that En (resp. Eh) converges strongly to E (resp. E') with E n E (resp. E1.11 E l ). For every trace w on 2+, we have E = E

,

w(O(E )) n

E

r

=

E

U(U)

w(4)(E)),

,

E

= E

E

f

= E

w(0 1 (E 1 ))

(Use part I, chapter 6, corollary of proposition 2.) With the help of a, conclude that w(cD(E)) = W(0'(E')). Whence a new proof of the corollary of lemma 1.

e.

For any x EH, show directly that

0(E V(U) ) = arguing as in a, but using, in addition, part I, (chapter 1, exercise 10, and chapter 5, exercise 6). 2. Let A be a von Neumann algebra. Suppose that A and A' are homogeneous, finite and of multiplicities n and n' respectively. Show that the coupling operator is constant and equal to nn'-. 3.

Let

A

be a von Neumann algebra, Z its centre.

306

PART III, CHAPTER 6

a. Let x be a trace-element for A. If TEZ, Tx is a traceelement for A. Every element of X Z is a trace-element for A. Z 71 . The mapping T-4- ETE ofAinto Z' is b. LetE=Eez linear, positivf, and ESTE - ETSE for S E A, TEA. Suppose, further, that x is separating for Z. Then, for every TE A, there exists exactly one T' E Z such that ETE = T'E. (Observe that Zv possesses a cyclic element, hence that ZE = Show that T T is the canonical Z-trace of A (which is finite). C.

Suppose that A and A' are finite, and that CA I. There exists a projection FE Z 1 possessing the following property: for every Te A, there exists exactly one T' E Z such that (Reduce to ETE = T I E, and T T 1 is the canonical Z-trace of A. the case where A is a-finite. Then use proposition 3 and C) [66], [89].

d.

CHAPTER 7.

1.

HYPERFINITE

FACTORS

Factors contained in a finite von Neumann algebra.

LRMMA 1. Let A be a von Neumann algebra, cp a faithful finite normal trace on A, B a *-subalgebra of A containing I, T an element of A, and H - 112 the norm defined by 4) (part I, chapter 6, section 2). The following conditions are equivalent: (i)T is a limit, for the norm H . 11 2 , of elements of B; (ii)T is a limit, for the norm H - 1123 of elements s of B such that H s H is bounded; (iii)T is a limit, for the uZtra-strong topology, of elements of B. Proof. We have (ii) (iii) by part I (chapter 4, proposition 4). We have (iii) (i) because, if S converges ultra-strongly to zero, S*S converges ultra-weakly to zero, hence 11s 112 = 4)(s*s) 1/2 converges to zero. Finally, suppose that condition (i) is satisfied, and let us show that condition (ii) is then satisfied. The closure g of B for the norm H . 11 2 is a *subalgebra of A containing I, ultra-strongly closed by the above, and therefore a von Neumann algebra. By part I (chapter 6, theorem 2), B is canonically endowed with a Hilbert algebra structure. The element T is bounded relative to S, hence (part I, chapter 5, proposition 4) T is the limit, for the norm . H 2 of elements Tn of B such that the H UTn are bounded; then, the are Tn H's bounded since the mapping S Us (S is an isomorphism of von Neumann algebras. 0

I

H's

H

LEMMA 2. Let A be a finite von Neumann alg ebra, Z its centre, (1) a faithful finite normal trace on A, A 4- Pel the canonical Ztrace of A, and H . 11 2 the norm defined by 0. Let T E A, and 6 > 0, be such that 11 TT / - T / T 11 2 __ EH T / 1 for any T / E A. c' Then 11 T - TP1 112 Proof. Let n > 0. Let U 1 , U 2 , ..., Un be unitar y operators of

A

and X 1 , X 2 ,

Xn real numbers 307

0 with sum 1 such that

308

PART III, CHAPTER

, -1 X A.U.2 TU2. j=1 s

(chapter

7

5, corollary

of theorem

1).

We have

- T i=1 s

2

. X X.11 2=1

u.lTu.-T H2 '1-

n

n -

=

TU.-U.T 2 2 i1

2

. X 2-1- =1

C.

Hence

2

E

7 is arbitrary, proves the lemma. 0 PROPOSITION 1. Let A be a finite von Neumann algebra, (Ai) iEI a family, totally ordered under inclusion, of factors contained in A, and B the von Neumann algebra generated by the Ai 's. Then, B is a finite factor. Proof. Let So be an element of the centre of B. It is enough to show that, for every projection E of the centre Z of A such We can therefore that ZE is G-finite, (S0)E is scalar.

which, since

confine attention to the case where the centre of A is G- finite. Let, then, (1) be a faithful finite normal trace on A. For every 6 > 0, there exist (lemma 1) an iEI and an S Ai such that 11 S c, were h S 0112 H - H2 denotes the norm defined by (I) in A. For every T Ai , we have

11 ST - TS 11 2 = 11 (S

S o )T - T(S

So) 112

2611T 11.

Hence (lemma 2) 11 S - S 11 2 2, denoting by S SIL:1 the canonical trace of A. Then, 1 S o - SDI 11 2 3E. Since SR is scalar, limit S. of scalar operators, is also scalar. 0 0 which is a On the subject of proposition

References

2.

1, cf. exercise 1.

: [22], [67].

Existence and uniqueness of continuous hyperfinite factors. DEFINITION 1. A factor A is said to be hyperfinite if it satisfies one of the following conditions: (i) A is type 1p , p < .;

FURTHER TOPICS

(ii)

309

A is finite, and is the von Neumann algebra generated

by an increasing sequence of factors which are type I 2n , ... etc. 1 1 , 1 2 , 14, 1 8 , Every factor isomorphic or antiisomorphis to a hyperfinite factor is hyperfinite.

Let A be a type In (n finite) factor in H, and B a factor containing A which is either continuous and finite, or (p 1). There exists a type I 2n factor A l such that type 1 2 LEMMA 3.

Ac

A1

As A is isomorphic to L(Hn ), Hn being an n-dimensional Hilbert space, there exist n minimal projections El, E2, ..., En A, mutually disjoint, equivalent (relative to A, and there- of fore relative to B) and with sum I. Putting K = E l (H), C = BK, there exists a spatial isomorphism of B onto C 0 L(Hn ) which transforms A into C L(Hn ). Now, if B is finite and continuK ous (resp. type C is finite and continuous (resp. type 2Pn), 1 2p). In both cases, C possesses two disjoint equivalent projections F l , F 2 with sum I (in the first case, this follows from chapter 1, theorem 1, corollary 3). Let V be a partial isometry of C such that V*V = F l , VV* = F2. Then, the linear combinations of F l , F 2 , V, V* form a type 12 factor C 1 contained in C. Putting A l = C 1 0 L(Hn ) , we have A c A l c 13, and Al is type

Proof.

2n' THEOREM 1.

Every continuous finite factor contains a continuous hyper finite factor. Let A be a continuous finite factor. Lemma 3, applied inductively, shows that there exists an increasing sequence A 0 , A l , ... of factors contained in A such that Ai is of type I • Let B be the von Neumann algebra generated by the A ' s. proposition 1, B is a finite factor. We have dim B = + 00,By hence B is continuous. Finally, B is plainly hyperfinite. 0

Proof.

There exist continuous hyperfinite factors acting in a separable Hilbert space. COROLLARY.

Proof.

This follows from theorem 1 and from the existence of continuous finite factors acting in a separable Hilbert space (part I, chapter 9, section 4, theorem 1 and remark). 0 Given a finite factor A, we will always denote by H • 112 , in what follows, the norm defined by the canonical trace of A, and we will call the corresponding metric the canonical metric of A. THEOREM 2.

Two continuous hyperfinite factors are isomorphic.

Let A, B be two continuous hyperfinite factors, (Ai) [resp. (Bi)] an increasing sequence of factors contained in A (resp. B), where Ai, Bi are type I 2i, generating A (resp. B). Suppose constructed, for 1 < i < n, isomorphisms cDi of Ai onto Bi, each (Di being an extension of 4). for j < i. As + 1 is

Proof.

PART III, CHAPTER 7

310

L (H 2 ) and Brz± i to Bn ø L(H 2 ) , where H 2 is a isomorphic to An two-dimensional Hilbert space, there exists an isomorphism of An±i onto Bn± i extending (D n . By induction, there therefore exists, for every i , an isomorphism cDi of Ai onto Bi such that (D • extends (D • for j i. Let

C= UA.., i=1

V

= U B., i=1

Then, C (resp. D) is a *-subalgebra of A (resp. B), which is ultra-strongly dense in A (resp. B) ; furthermore, the an isomorphism (D of C onto D. The canonical metricdefin of A (resp. B) induces on Ai (resp. Bi) the canonical metric of Ai (resp, Bi). As (10i transforms the canonical trace of Ai into (1). is an isometry, and, finally, (D is an isometry that of (for the canonical metrics). Now C (resp. D) is dense in A (resp. B) for the canonical metric (lemma 1). The isomorphism (I) of the Hilbert algebra C onto the Hilbert algebra V extends to an isomorphism of the full Hilbert algebra which extends C onto the full Hilbert algebra extending D. Finally, these full Hilbert algebras are just A and B (part I, chapter 6, theorem 2.) ] The continuous hyperfinite factors are also called "approximately finite." This terminology clashes with the now generally accepted terminology of "finite factor."

Reference : [67].

3.

Some inequalities.

LEMMA 4, Let A be a finite factor, E a projection of T an hermitian element of A such that H T H 1.

(1) There exists a spectral projection

H (ii) If T Proof,

F

E 11

0, we have

H

2

9 11

T 1 - E 11 2

T - E

A, and

F of T such that

H 1/22 .

1 3 11 T - E 112.

We have

T

2

- T = (T

2

2 - ET) + (ET - E ) + (E

T),

hence

H T2 -T 11 2 11T H,HT-E11 2 -1-11E

H. H T-E 11 2+11 1

Let E be a real number such that 0 < E < greatest spectral projections of T such that

T - E11 2 3 11 T - E 11 2 Let F, F 1 be the

311

FURTHER TOPICS

(I - E)F

TF < CF . 1 1 2 1 E for X lying -2F F 1 . As 1X 2 - X1 E Let F2 = I outside the intervals [-E, +E], [1 - C, 1+E], we have

1

II

— Ell F 11 2 2 2 hence

(T

2

TF <_ F,

-EF

- T)F 2 11 2

II

6 -E11

11 F 2 11 2 Moreover,

11 TF - F 11

II

TF1

1

T

2

- T 11 2

3 11 T - E 11 2 '

E 11 2'

II

hence

6 11 T - F 11 2 -- 11 TF - F 11 2 + 11 TF 1 11 2 + 11 TF 2 11 2 _E +6 +6 11 T - E 11 2' -51 1 If 11 T - E 11 2-2- , the inequalities of the lemma are plainly true. If have

11 T - E 11 13 < 12 , we can take

C=

11 T - E 11I. We then _51

11 T - F 11 2

811 T - E 11

Whence

1/2

11 F - E 11 2 <8 11 T - EH 2 + HT-En 2

= 11 T

1/2 E 11 2 (8 + 11 T - E 11 -)

which proves (1) . Now suppose that T tion as above, we have

11 T 1/2 F

F 11

9 iT - El1,

O. With the same nota-

T 1/2 F 1 11

E

1/2

hence

11

- F 11 2

11 T 1/2 F - F 11 2 + 11 T 1/2 F 1 11 2 + 11 T 1/2 F 2 11 2 -Ei- E 1/2 + - 11 T - E 11 2 111 T- E 11(1+7.2 -2 ) _611 T - E 11 1/24 7 11 T - E 11 2-1- 11 T - E 11 2

Consequently,

11 T 1/2 - E 11 2

611 <

11

E

II 1/24

+ 911

1/2 - E II 2

1/4 E 11 2 (6 + 9.2)

1311 T

1-1

E 11 2 -

D

Let A be a finite factor, and E and F two projecThere exists a partial isometry w E A whose initial (resp. final) projection is majorized by E (resp. F), and is such that II w - E 112 14 11 F E LEMMA 5.

tions of A.

312

PART III, CHAPTER 7

Proof.

Let FE = WB be the polar decomposition of FE. We have 0 B I. The operator W is a partial isometry; its final projection is majorized by F; as (FE)* = EF, its initial projection is majorized by E. We have B

2

(FE)*(FE) -EFE,

" 2 HB -EH =HE(F-E)E H

hence

2

2 H F-E 11 2 '

By lemma 4, (ii), we have

HB

F

13 11 F - E

11 2

Then,

II

W

E

11 2

II

W -FE 11 2

+ II FE - E 11 2

H w(E-B) 11 2 + II (F -E)E 11 2 -S 11 E - B 11 2 + 11 F - E 11 2

11

F-EH 2 +HF-E

11 2

14 11 F

E

H2.

0

13

Let A be a finite factor, u a unitary operator of A, W a partial isometry of A, which coincides with u on its initial subspace. Then, LEMMA

6.

HU

211 w - I 11 2 .

w H2

Proof.

The initial subspace of W is the final subspace of W. Hence UW* = WW*. Taking adjoints, we get WU* = WW*. Hence

(u

- W ) ( U - w)* = uu* - uw* - wu* + ww* =

I

and, consequently, denoting the canonical trace on

H u - w H2 = 2

- WW * )

W)I + 14)(W(I

H w - 1 11 + II Wk 2

11 2 = 2 11 W

- ww* A

by

(I),

W * ))1

I 11 2.

Let A be a finite factor, and E and F two equivalent projections of A. There exists a unitary o pe rator UE A such that F- uEu -1 andu-IH2 H < 36H F E H /8 2 Proof. Let W be the partial isometry whose existe-nce is given LEMMA 7.

-

by lemma 5, and E l E and F / F its initial and final projections. We have E l ^, F l , hence, by proposition 6 of chapter 2, E El F F / . Let V be a partial isometry of A, with initial projection E - E l , and final projection F - F l . Arguing similarly with I - E and I - F, we construct partial isometries W1, V 1 of A, and the sum W + V + W 1 + V 1 is a unitary operator U of A such that F = UEU -1 . Now

313

FURTHER TOPICS

Hw

11 w 1 - (I -

E) 11 2

1 4 11 F

E 11 2

E

1/4 2'

1/4 1411 (1 - F) - (I -E) 11 = 1411F - E II . 2

Hence

11w + w 1 - I 11 2

2811F - E

1/4 2'

As u coincides with the partial isometry W + W1 on its initial subspace, we have (lemma 4)

H u - (w + w 1 ) H 2

2 1/2 HW+W

H 1/8 8 11 F-E 11 2 .

1/2 1 -IH 2

Hence

HU - I 11 2

36 11 F

E 11 12 /8 .

0

Reference : [67].

4.

A new definition of hyperfinite factors.

In lemmas 8, 9, 10, 11, A denotes a continuous finite factor possessing the following property:

(*) For any elements T1, T 2 , ..., Tm of A and real number > 0, there exist a * - subalgebra B of A of finite dimension over C, and elements Si, S 2 , ..., S m of B such that S h Th 112 for h = 1, 2, ..., m. We always denote the canonical trace on A by cp. LEMMA 8.

number E l >

For any elements T 1 , T 2 , ..., Tm of A and real 0, there exists a type I 2n factor B c A, and elements E l for , _h Th 112 Sm of B such that Hq

S l, s 2, h = 1 .: 2, ..., m.

Proof. Denote by C the finite-dimensional * sUbalgebra of A, and R 1 , R2' P m the operators of C, whose existence is Ey be pairwise disjoint/ guaranteed by (*). Let El, E2, minimal projections of the centre of C, with sum I (we can, if necessary, adjoin I to C). Each CE. is a finite-dimensional factor. Let El E4 , EPi be pairwise disjoint minimal projections of CEi, with sum I. Let W. be a partial isometry of CE, with initial projection E ]:, an. final prOqction E. Each Rh is a linear combination of the operators 1, 2, ..., p). We need therefore to 1, 2, ..., pi; i (j, k prove the existence of a type I 2n factor B contained in A and possessing elements Vj arbitrarily close to the Wi.'s in the 1. (In fact, we will sense of the norm 11:1112, with H 2 then have -

314

H

PART III, CHAPTER 7

H2

H 40(1* - vii *) H 2 + H II Wik 11 11 II.

- v:3;7 * 11 2 +11

-II11 2 +

V`I *

‘772 )v.12., *

H2

- vik 11 2

Il 11

11 2 )

w

Fqi be pairwise disjoint, equivalent projecLet Fl, 1, 1, 1 ...I I, tions of A majorized by E l , such that 1

2 = (1)(F.)

= flF1 17

1,

)

= 2 -n ;

give qi the largest value possible (n being fixed for the moment) such that

1

(F

1

+ F

2 + i

+

Fg ))

1 2 q1, by means of W., we obtain projecTransporting Fi, Fi, ..., Fi tions equivalent to those, which are pairwise disjoint, and majorized by E. Carrying out this construction for 2 d = 1, 2, ..., pi, we obtain what we will call "the projections F of index i." Making i range from 1 to p, we obtain a certain number of "projections F," of trace 2 -n , which are pairwise disjoint. Let us adjoin to them a sufficient number of new "projections F" of A, of trace 2 -n, in such a way that the projections F are all pairwise disjoint, and have sum I. We can regard the projections F as minimal projections of a type I 2n factor Bc A. Further, for a suitable choice of B, the partial isometries of B which have for initial projection F or F or ... or Fqi 2 and for final projection certain of the F's cA index i, coincide on their initial sUbspaces.with one of the Wc7. 1 s. . Hence there exist partial isometries V3 in B such that &i-11"7-4 is a partial isometry with initial projection E1 (F 122 + F + i i + Fi). Then, 2 2 H - v-i II2 = (E i - (F . + F + + Fgi)) _< 2 -n . 2 Taking n arbitrarily large, we therefore make 11 trarily small. D

2

11 2 arbi-

For any elements T 1 , T 2 , ..., Tm of A, projection E of A such that 11)(E) = 2 -n, and real number E 2 > 0, there exist a type I 2p factor BcA, with p Sm of B, n, elements s i , s 2 , and a projection F of B such that H Sh - Th IT E 2 for h = 1, 2, m, H F E 11 2 E 2, and cp(F) Proof. By lemma 8, there exists a type T v factor Bc A, and elements S i , S 2 , ... Sm , Sm+ 1 of B such that HSh Th 11 2 El for h . 1, 2, ..., M, HS71+1 - E 11 2 E l . By lemma 3, we can, augmenting B if necessary, suppose that p n, Replacing LEMMA 9,

315

FURTHER TOPICS

is hermitian. 5_4_1 by 12 ( Sm+1 + S* m+1), we can suppose that S m+1 Put

-1 2 . R = 2S m+1 (I + S m+1 )

H

We have

R

H _<

1, and, observing that E = 2E(I + E 2 ) -1 , 2

1

— (R - E) = (I +S m+1 ) 2

[cf.

-1

(S

m+1

E) (I + E)

-

-1

1 R(E - S )E + — 4 m+ 1

part I, chapter 3, formula (1)]; hence

5 R - E 11 2 _< -2-

H

HSmi.1

5

- E 11 2 5

Ci.

By lemma 4, (1), .tere exists a projection F 1 of B such that H F 1 E 11 2 15 C l . Then,

(1)(E)1 = 1(F 1 - EII)

1(F 1 )

1(1)(F 1 ) -

1

15 E.

Let F be a projection of B majorized by F i or majorizing F i such that (1)(F) = 2 - n (such an F exists since B is type I 2p, p n). We have

II

F 1 - F II

2 ( (

F

1

- F)

2

) -

(F)

(F)

1

I(I) (F 1 )

I

2

I

E.

1

Hence

HF-EH

2

__11F- F 11 +HF -EH 1 2 2 1

1

1

E2

0

for E 1 small enough.

For any elements T 1 , A such that of LEMMA 10.

-n cp(E) = 2 ,

15E 1/4 + 15 c i

ET

h

T 2 , . . . Tm

= T E = T h h

(h

of A, projection E

1, 2, . . . ,

in) ,

and real number E 3 > 0, there exist a type 12P factor Sc A, with p n, such that E E B, and elements s 1 , s 2, . . . ; Sm of B such that ES

h

=s E = S h h'

H

s

h

- T 11 h 2

E3

for

h = 1, 2, ..., m.

Furthermore, if E is a minimal projection of a type I 2n factor C contained in A, we can suppose that SC. There exist (lemma 9) a type I 9p factor B i c A (p n), elements R i , R2 , ... Rm of B i and a projection F of B i such that

Proof.

11 R

- Th 11 2 -c 2 (h = 1, 2, • • •

rn)

11 F -E 11 2

By lemma 7, there exists a unitary operator

UE

E2 ,

4)(T) =.

A such that

PART III, CHAPTER 7

316

E = U 1 FU Then, B = U E B. Put

-1

H

and

1/8 E H 2

H2

U

36 E

36H F

B iU is a type 1 2p factor contained in

A

1/8 . 2

such that

Rh = U 1 Rh U E B. We have

H Rh' -

H2 =

H

H

Rh -uThu-1 H 2 2

+

H

T

h

H

Th

- UT u* h

H2

Rh - Th 11 2 +

- UT 11 + h 2

H

UT

h

E 2 + lI Th 11.111

u II2+IITh II . Il I 1/8 E + 7 2 11 T HE . 2 h 2

uThu-1 11 2 E

u* 11 2

Taking the number E 2 of lemma 9 so that

E2 we - have

H

H

1 /8

7211 T h "2 Rh - Th 112

ER F E-T II = h h 2

for

63

h = 1, 2, ..., m.

E3, hence

E(R hI - T h ) E II 2 < II Rh _ Th II 2 -‹ C3 for h . 1,2, ... ,m.

therefore suffices to put s h ERhE. Now, let E l , E 2 , ,E n be pairwise disjoint minimal projections of C, with sum I, with El = E. These projections allow one to identify A with a 2n-dimensional Hilbert space), C itAE 0 L(H 2n ), (H self being identified with CE (H) 0 L(H 2 ) . Then, BE 0 L(H 2n) is a type I lying between C and A, containing E and the S h' 's. replacing B by this new factor, we have the last assertion of the lemma. 0

LEMMA 11. For any elements T1, T 2 , ..., Tm of A, type I 2 n factor B contained in A, and real number E 4 > 0, there exist a type 1 215, factor C, with p n, such that BcC c A, and elements S 2, sm of C such that H sh - Th 11 2 5_ E 4 . Proof. Let E 1 , E 2' E2n be pairwise disjoint, minimal

projections of B, with sum I, and let Wi be a partial isometry of B with initial projection El and final projection E. Put

T.

.

= WtT W

We have ET. . =. T. . E = T.. . 1 2,j .02 2,,3;12 1 2,d;h

FURTHER TOPICS

317

There exist (lemma 10) a type 1 2 (p n) factor C, such that BcCcA, and elements s1 'j. of such that c 3 for any 1., j, h. We have - T. • S• . 2,d;h 2, , d ;h II

2n 2 n T =

W .WtT W

E .T E. = 2=1 d=1

=

X W ,T „ Wt.

q-Jd

Put S

h

. W *. € W .S . i -i, d;h d

=

C.

We have

HS-TH - T. . 2 E,d;h h 2 -<X11W.(S.

)wtd

h

< X H

sS

—

i

T. .

H2

H2 22n E

3 < E

4

i,j

for E 3 sufficiently small.

D

Let A be a finite factor. conditions are equivalent: THEOREM 3.

The following three

(i)A is hyperfinite; (ii)A is the von Neumann algebra generated by an increasing sequence of finite-dimensional *-subalgebras; (iii) (a)

A is the von Neumann algebra generated by a count-

able family of elements;

(13) For any elements T 1 , T2, „., Tm of A, and real number > 0, there exist a finite-dimensional *-subalgebra B of A and elements Sl„ S2, •.., S m of B such that II C for Th 112 m, h = 1, 2, Proof. It is clear that (i) (ii). Now suppose that A is the von Neumann algebra generated by an increasing sequence A Q' A l' ... of finite-dimensional *-subalgebras. Then, condition (iii, a) is obviously satisfied, and condition (iii, 13) is satisfied by lemma 1. Finally, suppose conditions (iii, et) and (iii, (3,) are satisfied, Let R1, R 2 , ... be a sequence of elements of A generating A. Condition (iii, (3), which is none other than condition (*) from the beginning of this section, allows us to apply lemma 11. Applying this lemma inductively, we can construct an increasing sequence Bo, B1, „. of factors contained in A, with the following propertie: 1 ° Bi is type such that I2P 1; 2 0 Bi contains elements Si, S, 1 Then (le T 2, /. for h = 1, II Si - R h H 2 mm a 1) the h

PART III, CHAPTER 7

318

von Neumann algebra generated by the Bi's contains all the Rn 's, hence is equal to A. If Bi = Bi+1 = Bi +2 = A is type I 2p i . If pi + as i + co, we can, interpolating type I,)0, factors between the B's (which is possible thanks to lemma 1)', and renumbering the Bi's, suppose Bi to be type I 2 i. Hence, in this case also, A is hyperfinite, LII

References : [67], [144], [181], [197].

5. Hyperfinite factors and elementary operations. PROPOSITION 2. The tensor product of two hyperfinite factors is a hyperfinite factor. Proof. We already know that, if A and 13 are finite factors, A 0 B is a finite factor (part 1, chapter 6, proposition 12 and corollary of proposition 14). Moreover, suppose that A (resp. B) is the von Neumann algebra generated by an increasing sequence (Ai) [resp. (Si)] of finite-dimensional *-subalgebras. Then, A 0 B is the von Neumann algebra generated by the increasing sequence of the Ai 0 Bi's (part I, chapter 2, proposition 0 6), and hence is hyperfinite.

Let A be a hyperfinite factor, and E a projecThen, AE is hyperfinite.

PROPOSITION 3.

tion of A. Proof.

Let (I) be the canonical trace of

A.

(i) Let A 0 , A 1, ... be an increasing sequence of factors, where A1 is type I i , generating A. Suppose EE Ai. Then, the (Ai)E, j > i, form an increasing sequence of finite-dimensional *-algebras which generate AE, hence AE is hyperfinite. More generally, if (1)(E) = p.2 -n (p, n integers), E is equivalent to a projection belonging to one of the A.'s, and hence AE is again hyper finite.

(ii) We pass to the general case. The factor AE plainly satisfies condition (iii, a) of theorem 3. We show that it also satisfies condition (iii, (3). Let T1, T 2 , .,. Tm be elements of EAE, and E > O. There exists a projection F of A such that F < E and such that OF) is of the form p.2 -n, with, further, H E - F 11 2 We have E.

H T,- FT,F ,z, H2 ‹- H ET,2 E

FT,

H + H FT,E - FT,F H2 < 20 T. H. E 2 2

2 2

Moreover, by part (i) of the proof, there exist a finitedimensional *-subalgebra B of FAF, and elements Si, S2, ..., S m of B such that

H FT.F 2

Si H 2

(i = 1, 2, .„, m),

Then,

T. - S. H < 6 (1 + 2H T. H) 2 2 2 2

for

i = 1, 2, ..., m.

319

FURTHER TOPICS

Moreover, BcEAE. Finally, the canonical trace of A induces on AE a trace proportional to the canonical trace of AE . Hence, the norm II II 2 of A induces on AE a norm proportional to the norm H H 2 of AE.

Let A be a hyperfinite factor, finite factor, it is a hyperfinite factor. PROPOSITION 4.

If A' is a

Proof. There exist (part I, chapter 6, proposition 13) Hilbert spaces K and K l , a factor A i in K antiisomorphic to A, and a projection E of A l 0 L(K1), such that A' is isomorphic to (A1 0 L(K1)) E , The space K K 1 may be identified with the of subspaces possessing the foldirect sum of a family (Ki) . lowing properties: 1 ° PK.E2- 61( 0 L(K1); 2 ° the (A1 0 L(K 1 ))/‹.'s are isomorphic to A l , Let (I) be a semi-finite faithful normal. trace on (A1 0 L(K 1 ))+. As A' is finite, and A l 0 L(K i ) is a hen J is a factor, iEJ P K . (chapsufficiently large finite subset of I. Then, E iEJ ter 2, proposition 13). Replacing E by an equivalent projecP K „ Then, replaction, we see that we can suppose that E < iEJ ing K ® K 1 by Ki, we see, finally, that we can suppose K1 to ieJ be finite-dimensional. As A l is hyperfinite, A l 0 L(Ki) is hyperfinite (proposition 2), and so (A 1 0 L(K 1 )) E is hyperfinite (proposition 3). D

X

Reference : [67].

6.

Further examples of finite factors.

Let G be a discrete group with identity element e, and U the algebra of complex-valued functions on G, zero except at a finite number of points, the multiplication in U being defined by convolution of functions:

(x * x') (g) =

X

x(h)x 1 (h -lg)

hEG (the sum only includes a finite number of non-zero terms). For x€ U, define x* bY x*(g) = x(g-1 ). For x U, y E U, put (x IY ) = / x(g)y(g)- . It is immediate that U becomes a pre-

gEG

Hilbert space, its completion being the space H of complexvalued functions x on G such that lx(g)1 2 < + ... Let E E U

X

gEG be the characteristic function of the element g of G. PROPOSITION 5. (i) U is a Hilbert algebra, with identity ele-

ment ce,

320

PART III, CHAPTER 7

(ii)U(U) and V(U) are finites every element of U(U) [resp. V(U)] is of the form ua (resp. va ), a being a bounded element of H. (iii)For an element b of H to be central, it is necessary and sufficient that the function b be constant on the (conjugacy) classes of G. (iv)U(U) and V(U) are factors if and only if every class of G, other than {e}, is infinite. (y) If a and b are bounded elements of H, uab (resp. Val') is the convolution product a * b (resp. b * a). Proof.

An easy calculation shows that, for

(xly) = (y*lx*)

x, y, z

E

U,

(x * ylz) = (ylx* * z).

and

It is clear that ce is the identity element of U, and so the elements of the form x * y, XEU, y EU comprise all the elements of U. If we put

11x 11

II

= gEG

we have

Hence U is a Hilbert algebra. Since U possesses an identity element, its characteristic projection is equal to I, hence U(U) and V(U) are finite; furthermore, the two-sided ideal of U(U) [resp. V(U)] consisting of the elements Ua (resp. Va ), a bounded, contains TH, hence is the whole of U(U) [resp. V (U )]. Thus, (1) and (ii) are proved. We prove (y) . For a E U, b E U, and g c G, we have (Ua b) (g) = (Vba) (g) = (a * h) (g) . Moreover, for fixed g E G, (Uab) (g) (Vba) (g) and (a * b) (g) are separately continuous functions of a and b when a and b run through the set of bounded elements of H. Hence Uab = Vba = a * b for a and b bounded in H. We prove (iii). For bEH to be central, it is necessary and sufficient that Uc b = VE b for every g€ G, hence that E a * b b * g for every g€ G, hAce that the functions h -4- b(h) and h -4- b(g- -hg) on G be the same, i.e., finally, that b be constant on the classes of G. If there exists a finite class of G other than Ce}, there exists in U a central element not proportional to ce , hence a non-scalar central element in U(U): thus, U(U) is not a factor. On the other hand, if there exists a non-scalar central element Ua (a bounded) in U(U), then a is central and not proportional to L e , hence is constant and non-zero on a class of G distinct from fe}; as X la(g)1 2 < + c°, this class is finite. Whence

(iv).

gEG

321

FURTHER TOPICS

References : [1], [2], [13], [29], [30], [53], [56], [67], [ 90 ], [ 98 ], [ 99 ], [[ 9 ]] ,

Existence of finite, non-hyperfinite, factors. DEFINITION 2, Let A be a finite factor, and (1) the canonical trace on A. We say that A possesses property F if, for any elements T1, T2, ;,•, Tm of A and real number C > 0, there exists a unitary operator u of A such that OU) = 0 and 7.

H u- 1Thu -

Th 112 c for h = 1, 2, ..., M. Property r is invariant under isomorphism. PROPOSITION 6.

A continuous hyperfinite factor possesses

property F. Let A be a continuous hyperfinite factor, T 1 , T 2 , ..., Tm elements of A, and c > 0. Let (A0, Al, A2, ,..) be an increasing sequence of factors, of type Ii,12, 1 3 , ... etc. which generate A. There exist an integer i and elements 6/2 for Sl, S2, ..., Sm of Ai such that H Th - Sh 11 2 h = 1, 2, ..., m. Moreover, A1 + 1 is isomorphic to the tensor product of Ai and a type 12 factor. There therefore exists a type 12 factor B commuting with Ai and contained in A1 +1 . Let U be a unitary operator of B whose canonical trace (relative to B, and therefore relative to A) is zero (we can take U to be a symmetry). Then,

Proof.

11

u -1 (T

h

- s )u h

11 2

=HT

h

-

S

h 1 12

< 2

and

U S

h

U = Sh.

Hence

,, u-1 T U h

- T h

11 2

E.

D

Suppose that, in the group G of proposition 5, every class other than Ce} is infinite, and that there exists a subset F of G possessing the following properties: LEMMA 12.

(i) There exists an element g l of G such that

F u g iFg i l u {e} = G; (ii)There exists elements g 2 , g 3 of G such that F, g 2 Fg 1, g 3 Fg 1 are disjoint. Then, U(U) is a finite factor not possessing property F. Proof. Suppose that U(U) possesses property F. Let c > 0, and let U (y, a bounded element of H) be a unitary operator Y c for with canonical trace 0, such that H ugi - upugluy 112 We have i = 1, 2, 3 (we put U, = U,). y (U

Y

) = (U I I) = ( YIE ) = Y(e)-

PART III, CHAPTER 7

322

Moreover,

II 2

Uy* Ug. Uy

Ugi

= II U y

U* U U

gi

y

* Y * E

-

gi II 2 = II

gi

I

and

II

II Y II

Uy

II 2

1.

Hence

X

geF

log) 12

±

X

ly(g) 1 2

geg 1 Fg-1 1 7 12 2 y iy(g) 12 )(g) 12 = X IY(g)I + L I(E -1*Y* E g l geF 91 geF geF log) 1 2 x iy(g) 1 2 ± 7 1). x i y(g) 12 ± L -1 geF gEg3Fg7i1 gEg 2 Fg 2 =

+ 2E,

XI Y (g) I 2 4"geF X I( E -1 * y * E )(g)1 2 92 92

+ X 1(E -1 4 cy *E )(g)1 2 g3 g3 geF

geF 3 X

gEF

l() 1 2

- 4E.

Thus,

12 c X IY(g) 1

— 2

gEF

4

1

1-

which is impossible for E sufficiently small. THEOREM 4. There exist finite, ing in a separable Hilbert space.

E, 0

non-hyperfinite, factors act-

Let G be the free group on two generators gl, g2. Every element of G other than e may be written in a canonical way in the form ... gIgP2g14 ..., with non-zero exponents. Let

Proof.

a E G.

Then, the glagI n are all different, unless a is a power

of g l ; however, in this latter case, the gr2lag-2.n are all distinct, unless a = e. Thus, every class of G other than {e} is infinite. Moreover, let F be the set of the a / e of G whose canonical expression terminates in a g, n = ±1, ±2, .... Then, conditions (i) and (ii) of lemma 12 are satisfied, with g3-g. Hence U(U) is a. continuous finite factor which does not possess property r, and which is therefore not hyperfinite. 0

FURTHER TOPICS

323

Concerning the classification of type I l ]. factors, cf. the concluding remarks of part I, chapter 9, section 4. Reference : [67].

Exercises. 1. Let V be Lebesgue measure on [0, 2], H = L 2 ([0, 2], V). Every V-measurable set Z in [0, 2] defines a projection of H, namely the operator of multiplication by the characteristic function of Z. Let, in particular, E be the projection corresponding to [0, 1]. For n = 1, 2, ..., let Z1, ZY22, Z2n be a partition of [0, 1] into measurable sets with the following properties: n n n-1 =Z uz.; a. Z i

n b. v(z ) 2i1 n

c. V(Z.)

2i-1 2/, 3 n-1 = — v(z ); 4 i

n-1 ) .

1

= — v(Z 4 i

Let Y- be the subset of [1, 2] defined as follows: Y.) . (resp. is obtained from Z/5 7:, (resp. Z1) by means - ôf - the translation C -4- C + 1. Let XY.1 = 417 uY71, and E721, the projection of L(H) corresponding to X. For every n, E 71, E 2 , ..., Er2ln can be regarded as the minimal projections of a type 1 2 n factor An commuting with E. We can, further, choose the As to be increasing. Show that the von Neumann algebra A generated by the An is is not a factor. (Show that El + El + + 4'71_ 1 converges strongly to E as n 4- + co, hence that E belongs to the centre of A.) 2. Show that, if A is a finite von Neumann algebra, the set of factors contained in A, ordered by inclusion, is inductive. (Use proposition 1) [22]. 3. Let A be a finite von Neumann algebra, (Ai)i ci a family, totally ordered by inclusion, of von Neumann subalgebras of A, having the same centre Z as A. Then, the von Neumann algebra generated by the Ai 's has centre Z. (Argue as for proposition

1). 4. a. Let A be a von Neumann algebra, cp a finite faithful normal trace on A, and B a maximal abelian von Neumann subalgebra in A. Let TEA, and E > 0, be such that TT' - T'T 112 T' for any T' E B. Then, the distance of T from B, for the norm 11 11 2 , is E. (By lemma 1 of part I, chapter 9, there exists an element S EB in the strong closure of the convex set K generated by the UTU-1 , U a unitary operator of B. For every element R of K, MR-TM 2 E. Hence

m

11S

cm

T 112

E-)

324

PART III, CHAPTER 7

1 a family, b. Let A be a finite von Neumann algebra, (Bi) totally ordered under inclusion, of von Neumann subalgebras of A, and Ci a maximal abelian von Neumann subalgebra in Bi, with CicCx when Bi Bx . Let B (resp. C) be the von Neumann algebra generated by the Bi's (resp. Ci's). Show that C is a maximal abelian von Neumann subalgebraof B. (Argue as for proposition 1, using a.) 5. Let A be a type In (n finite) factor, and B a continuous (resp. type Ipn) finite factor containing A. For every integer q (resp. for every divisor q of p), there exists a type I n facc B. tor Al such that Ac (Argue as for lemma 3) [67]. q 6. Let A be a hyperfinite continuous factor, and (p l , p 2 , ...) a sequence of integers tending to + 00 such that pi divides pi +1 . Show that there exists an increasing sequence of factors (Ai), where Ai is type In., contained in A such that A is the von Neumann algebra generated by the A's. (Arguing as for theorem 1, show that there exists one continuous hyperfinite factor for which the assertion is true. Then use theorem 2) [67].

7. A hyperfinite factor is isomorphic to a factor acting in a separable Hilbert space (cf. part I, chapter 7, exercise 3 c). 8.

Let I be an uncountable set, and G the (discrete) group of the permutations of I which leave fixed all but a finite number of the elements of I. a.

Show that every class of G other than {e) is infinite.

b. Show that the factor U(U) of proposition 5 possesses [If T 1 , T2, ..., Tm are eleproperty (iii, (3) of theorem 3. ments of U(U), the Ti's may be approached arbitrarily closely, in the sense of the norm 2 , by operators of the form Si = Uxi , Xi € U. The xi vanish outside a subgroup of G of finite order. Hence the Si's generate a finite-dimensional *subalgebra of U(U).]

HH

c. Show that U(U) does not possess property (iii, a) of theorem 3. (If a von Neumann algebra is generated by a countable family of elements and possesses a cyclic element, it acts in a separable Hilbert space. Now G is not countable.)

9. Use the notation of section 6. If an element a E H is such that, for every b H, a * b (which is a bounded function on G) belongs to H, then a is bounded. (Show that the mapping b 4- a * b is continuous, by applying the closed graph theorem) [90]. 10. Let G be a 'discrete group, not equal to {e}, all of whose classes, other than {e}, are infinite.

FURTHER TOPICS

325

a. If G is the union of an increasing sequence of finite subgroups, the factor U(U) of proposition 5 is continuous hyperfinite. [Use theorem 3, (iii).] b. If, for every finite family 9. 1 , g 2 , gn of elements of G, there exists an element g / e of G commuting with the (Note that U c is unitary and then U(U) possesses property F. that its canonical trace is O.)

(ConC. Deduce from a and b a new proof of proposition 6. sider the group of permutations of an infinite countable set I which leaves all but a finite number of the elements of I fixed) [67]. 11. Let bert space.

A

be a continuous finite factor in a separable Hil-

a. Let (A77) 77 EI be a family, totally ordered by inclusion, of hyperfinite factors contained in A. Show that the von Neumann algebra generated by the Ails is a hyperfinite factor. [Use theorem 3, (iii).] b. For every hyperfinite factor 13 c A, there exists a maximal hyperfinite factor C such that Sc C c A (use a), and C is continuous (use lemma 3) [22].

12. Let G be a discrete group, and G1 a subgroup of G. the notation of section 6.

Use

a. Show that the U a 's, where a E H is bounded and vanishes outside G l , form a von Neumann subalgebra A of U(U). (Use lemma 1 ) b. For A to be a maximal abelian subalgebra of U(U), it is necessary and sufficient that G1 be abelian and that, for every g G such that g G i , the set of the g 1 gg 11 , where g l runs through G l , be infinite. c. For A' n U(U) to be just the scalar operators, it is necessary and sufficient that, for every gE G, the set of the g lggi l , where gl runs through Gl, be infinite. d. Let K be an infinite (commutative) field, the union of an increasing sequence of finite subfields (for example, the algebraic closure of a finite field). For CLEK, a / 0, and 13 EK, let (a, (3) be the bijection of K onto K defined by (a, I3)E = 4 + (3. The (a, Ws form a group G, the union of an increasing sequence of finite groups. Let G1 be the subgroup of G consisting of the (a, Ws for which °LEK ' , K', K' being a proper infinite subfield of K. Show that U(U) is a continuous, hyperfinite factor, that A / U(U), and that A' n U(U) is just the scalar operators. (Use c and exercise 10 a) [16] [67].

13. Let A be a von Neumann algebra, and Z its centre. For every subset M of A, we put MA = M' n A.

326

PART III, CHAPTER 7

If M is closed under the taking of adjoints, a von Neumann algebra containing Z and M. -

(MX)A is a

b. A

is said to be normal if, for every von Neumann subalgebra B of A containing Z, we have (13 1 ) ik = B. Show that, if A is discrete, then A is normal. [We can suppose that A' is abelian, hence A' = Z. Then, WA = B', (BA)Â = B" = B.] c. If A is a continuous finite factor in a separable Hilbert space, A is not normal. (If A is hyperfinite, use exercise 12 d. If A is not hyperfinite, consider, thanks to exercise 11 b, a maximal hyperfinite factor Sc A. Let C = B. Distinguish three cases: 1 ° C is not a factor; 2 ° C is just the scalar operators; 3 ° C is a factor, not just the scalar operators. Show that the third case contradicts the maximality of B. In the first two cases, show that (BA):4 / B).

Problem (Fuglede-Kadison):

can the first case actually occur?

d. A continuous semi-finite factor [Write A = A l 0 space is not normal. tinuous finite factor, and use c.]

A in a separable Hilbert L(K), where Al is a con-

Problem: is a continuous von Neumann algebra non-normal? [16], [22], [65], [78], [283]. 14. Use the notation of section 6. Let Go be the subgroup of G consisting of the finite classes of G. If G/Go is infinite, U(U) is continuous.

[Let Z = U(U) n V(U). Let (Gi) be the family of (left) cosets xGo of G with respect to Go. Let Ei be the operator of multiplication by the characteristic function of Gi in H. Show that the Ei's form an infinite family of pairwise disjoint, equivalent projections of V. Then apply exercise 2 of chapter 3] [56]. 15. Use the notation of section 6, taking for G the discrete free group generated by two elements a, b. The automorphism of G which interchanges a and b defines an automorphism SI of U(U) which is not inner (Kadison). [Suppose that (1) is defined by a unitary element Ux of U(U), X being a bounded element of H; we have, for every integer n, UxUcanUI = Ubn , hence X * can * x* = chn ; however, if y denotes an element of H with finite support very close to x, we have (y * c an * y*) (bu ) = 0 for n sufficiently large.] 16. Let A be a continuous finite von Neumann algebra in a separable complex Hilbert space H. Let Z be the centre of A. Suppose that there exists a *-subalgebra B of A possessing the following properties: (i) every finite subset of B generates a finite-dimensional *-algebra; (ii) Z and B generate A.

FURTHER TOPICS

327

a. Suppose further that A is standard. Show that A is spatially isomorphic to Z 0 A0 , A0 being a continuous hyperfinite factor acting in a separable Hilbert space. [Write

e

A =A(c)dv(c), Z fZ

being compact metrisable, and the A() 's

being factors. Show that the A(C)'s can be supposed standard and hyperfinite, and hence all spatially isomorphic to A0 . Then use proposition 4 of part II, chapter 3.] b. In the general case, show that A is isomorphic to Z 0 Ao. (Show that A is isomorphic to a standard algebra acting in a separable space) [40], [360].

M

CHAPTER 8.

1.

ANOTHER DEFINITION OF FINITE VON NEUMANN ALGEBRAS

Statement of the theorem.

Let A be a von Neumann algebra. For it to be finite, it is necessary and sufficient that every projection of A equivalent to i be equal to 1, in other words, that the hypotheses T E A, T*T = I, imply that TT* = I. THEOREM 1.

The condition is necessary (chapter 1, proposition 3). In the rest of the chapter (with the exception of sections 6 and 7), A denotes a von Neumann algebra such that every projection of A equivalent to I is equal to I. We intend to show that A is finite.

Remark 1.

Suppose that the following result has been proved:

(*) There exists (if H / (D) a non-zero projection F of the centre Z of A such that AF is finite. The theorem will then follow. Because, let (Fi) i ci be a maxi mal family of pairwise disjoint non-zero projections of Z such that the AF i 's are finite. Let G= X F I G' = I - G. Every

iEI projection of AG I equivalent to I G f is equal to ' G I, hence, if G' / 0, there exists a non-zero projection G" of Z majorized by G' and such that AO is finite, which contradicts the maximality of the family (Fi) i ci . Hence G' = 0, X Fi = I, and hence

iEI

A

=

H AF. is finite. iEI

We are going to prove the assertion (*) in what follows. LEMMA 1. (i) Let E and F be two equivalent projections of A; if E F, we have E = F. (ii) Let (Ei) i EI be a family of pairwise disjoint, non-zero, equivalent projections. Then I is finite. Proof. (i) We have I =F + (I -F)

+ (I

,

hence I =E + (I -F) , 329

E =F.

330

PART III, CHAPTER 8

(ii) If I is infinite, there exists a subset J of I, J I, which is equipotent with I. Then, E., in contradicE1 ieI ieJ tion to (1). Suppose that A is not continuous. There exists (chapter 3, proposition 2) a non-zero projection F of Z such that AF is homogeneous and type In , with n finite, by lemma 1, (ii). Hence AF is finite, which proves (*). We can therefore, throughout the rest of this chapter, make the following hypothesis:

Remark 2.

(**) U is continuous. References : [6], [65], [66].

Fundamental projections. DEFINITION 1. A projection E of A is said to be fundamental if there exist a projection F of Z and pairwise disjoint, equivalent projections El, E 2 , E 3 , E 2n of A, with sum F, 2.

with E 1 = E. Every projection equivalent to a fundamental projection is fundamental, as follows immediately from chapter 2, section 3, remark. LEMMA 2. Every projection E / 0 of A majorizes a non-zero fundamental projection.

Proof. There exist (chapter 1, theorem 1, corollary 2; and lemma 1) a projection F of Z, and pairwise disjoint projections E l , E2, •.., En of A, possessing the following properties: V 1 0 0 E1 E2 En_ 1 En ; 3 ° F = E i E2 + +En E1—F0;2 . Let p be an integer such that 2P > n. There exists, by corollary 3 of theorem 1, chapter 1, applied inductively, a family (G1, G 2 , ..., G 2p) of pairwise disjoint, equivalent projections of A, with sum F. Apply theorem 1 of chapter 1, to E l and We need to consider, separately, the case where G 1 -‹ E l and the case where El -‹ G1. If G 1 -4: El, the lemma is proved (because El EF / 0, hence F / 0, hence Gl / 0). If E l -4tG 1 , then F = X E. -‹ X G. / F, which is impossible (lemma 1). 2 . 2

0

Every projection E of A is the sum of a family of disjoint fundamental projections.

COROLLARY. mutually

Proof.

Let (Ei)i ci be a maximal family of pairwise disjoint, non-zero, fundamental projections majorized by E. By lemma 2, EEi = 0, whence E = E.. 2 iEI iEI Remark. Let pE H(X / 0), F a non-zero fundamental projection majorized by ExA , and (F1, F2, ..., F 2n ) equivalent, pairwise

331

FURTHER TOPICS

disjoint projections of A, whose sum G is a projection of Z, with F1 = F. Then, AG = AF 0 L(K), where K is a 2n-dimensiona1 If AF is Hilbert space (part I, chapter 2, proposition 5). finite, AG is finite and the proof is at an end (section 1, remark 1). It therefore suffices to show that AF is finite. Now, AF satisfies the same hypothesis as A (lemma 1), is continFinally, we can adjoin, to uous, and Fx is separating for AF . the previous hypotheses already made on A, the following:

(***) There exists a separating element x for A. References : [6], [65].

Weights on the set of fundamental projections. LEMMA 3. Let E be a non-zero fundamental projection, F a projection of Z., (El, E 2 , E 2 n) pairwise disjoint projections of A, equivalent to E, with sum F. Then, F and the integer n only depend on E. 3.

Every projection of Z which majorizes E majorizes the E i rs, hence F as well, from which it follows that F is the central support of E. Moreover, let ET, E, ..., ET: be pairwise disjoint projections of A, equivalent to E, with sum F. If p < 2n, we have

Proof.

F

E. / F, i=1

which is impossible (lemma 1); the impossibility of the hypothesis 2n < p is proved similarly. 0 DEFINITION 2. (D(E) = 2 -n F, (1)(E) = 0(a).

put put

LEMMA 4. cli(E) =

Let E

Let E be a fundamental projection. If E / 0, n and F being defined by lemma 3. If E = 0,

E, E' be E' are

fundamental projections. equivalent.

The relations

Proof.

Lemma 3 immediately shows that the relation E E' Conversely, suppose that implies the relation (1)(E) = cp(E'). (1(E) = 11(E')

and let us show that E E'. Using theorem 1 of chapter 1, we E', or E l -; E. are reduced to the case where we have either E E 2n ) Suppose, for example, that E E'. Let (El, E2, [resp. (Ei, En)] be disjoint projections equivalent to E (resp. E'), with sum F E Z. E < E for 1
i=1

2-

i=1

2.-

i=1

2

PART III, CHAPTER 8

332

Ef

= E I and finally Ei

0

E. ?-

5. Let E, E l be fundamental projections of A such that p. There exist disjoint pro(1)(E) = 2 -PF, cD(E l ) = 2 -qF, and q jections (Ei) 1
Proof.

Let

n

be disjoint projections, equivalent to

(El)l<2<2-1 .

E t , with sum F.

24-19

It is clear that E" =

and that (1)(E") = 2 -19F.

E l. is fundamental,

1;1 Hence (lemma 4) E"

E.

The equality

2q -19 E l. then implies the existence of the projections E. of

E" =

2

i=1 the lemma.

Let E l , E2, ..., E n , F l , F 2 , ..., Fp , F- be fundaprojections, the E's (resp. the Ft's) being pairwise mental LEMMA 6.

disjoint. Let E = .X E,

=

2 1

F..

Suppose that F E and

j=1

4)(F- ) +

(F

j=1

i=1

Then, there exists a projection F n+1 , equivalent to F-, orthogonal to F, and majorized by E. Proof. Consider the central supports of the Ei's, of the F.'s, and of Suitably expressing A as a product of von Neumann algebras, we are led to the case where all these supports are equal to 0 or to I. Since the zero projections play no part in the matter, we can suppose that all these supports are equal to I. We therefore write = 2-ri,

(F.)

=

(F-) = 2 -s .

Let q be an integer greater than the res, the si l s and s. Let G be a fundamental projection (which will serve as a "unit of measure") such that (1)(G) = 2 - q (chapter 1, theorem 1, corollary 3). Let E 1 , E 2 ..., E 2g-ri (resp. F l , F 2 , ..., FV -Si;

j j

i

j

?;2q-S) be disjoint projections, equivalent to G, with sum Ei (resp. Fj; ri (lemma 5). Then the E (1‹i
-s +

2 j=7

,3•

1 2 2 1=1

333

FURTHER TOPICS

or

( 1)

2g -13 i.

2 q-8 ,/

2q-8 +

i=1

j=1

This enables one to choose, in the set

M

of the

joint subsets M 1 , M2 , containing 2 q-8 and

x

X

, two dis-

2 q-8i elements

j=1 E M 1 is equivalent to

respectively. Then, the sum F t of the Ei F t are the sum F t of the E?z',- M2 is equivalent to F, and F t E F t --E F disjoint and majorized by E. We have F (chapter 2, section 3, remark), whence the lemma. E ,

with finite values 0, defined on the set of fundamental projections of A, is called a weight if it possesses the following properties: DEFINITION 3.

A function

E

4)(E),

(i) Let E be a fundamental projection, (Ei) EI a family of disjoint fundamental projections with sum E .; t en ( E) =

iEI

(ii)

If 0E) = 0, then E = O.

We say that cp is central if the relation

E l E2

implies

cp(E 1 )= 0E 2 ).

Let x be a separating element for A. For every fundamental projection E, put 0(E) = (4)(E)xlx). Then, cp is a central weight. LEMMA 7.

Proof.

E 2 implies 0E 1 ) =cp(E 2 ) By lemma 4, the relation E l is separating the relation 0E) = 0 implies for A, Since x 0(E) = 0, hence E = O. Finally, let E be a fundamental projection, and (Ei) i e , a family of disjoint fundamental projections with sum E. We are going to show, and this will complete the proof of the lemma,that cp(E) = X () ( EI: ), and to accomplish this

iEI that

X (1)(Ei) iEI

exists in

Z

and is equal to 11(E).

Firstly, for every finite subset

X

O(Ei)

J

of I, we have

(HE).

ieJ In fact, if this were not the case, we would be able, suitably expressing A as a product of von Neumann algebras, to confine X flE i ). Let G be a attention to the case where IT(E) + 2 - P iEJ

334

PART III, CHAPTER 8

fundamental projection such that 0(G) = 2 - P. By lemma 6, applied twice, there would exist two disjoint projections E', G P , equivalent to E and G respectively, majorized by X E.. iEJ Whence

E

X E. iEI 7-

which is impossible (lemma 1).

El + GI ,

X 0(E.) 0(E) . ConseiEJ 0(Ei) exists and is majorized by 0(E).

quenity,

Hence

iEI We now show that t(E)

X 0(E.). If this were not the case, iEI we would be able, suitably expressing A as a product of von Neumann algebras, to confine attention to the case where 2 - P + X 0(E.) iEI

Let E be a fundamental projection such that IT(E0) = 2 -P . As 0 the zero projections play no part in the matter, we can suppose that the E's are non-zero, and hence form a sequence (El, E2, ...) since A possesses a separating element. Then, by lemma 6 applied inductively, there would exist a sequence (E6, El, E, ...) of disjoint projections of A, majorized by E, E for i = 0, 1, 2, .... We would therefore have with E'

E = E. i I

E

and

i=1

E -

E' i=1

E

0

0,

which is impossible (lemma 1). References : [6], [65].

4.

Construction of a trace to within c .

LEMMA 8. Let (I) and ip be two weights on the set of fundamental projections, and let > 0. There exists a number 0 > 0 and a fundamental projection E 0 such that, for every fundamental projection F . < E, we have 0(F) 5 oF)

Proof.

0(1 + 6)d(F).

We can suppose that 4)(I) = Ip(I). Suppose that, for every fundamental projection G / 0, there exists a fundamental projection G 1 G such that cp(G 1 ) > Ip(G). There would then exist

FURTHER TOPICS

335

(Zorn's lemma and lemma 2) a family (Gi) of disjoint fundamental projections, with sum I, such that ( MG-) 2 > I CL.), 2

(

whence

(1)(I) = X(P(G.) > Xlp(G.) =

which is impossible. Hence there exists a fundamental projection G X 0 such that, for every fundamental projection G 1 5. G, we have (I)(G1) L 4)(G1) Let e be the supremum of the real numbers n such that, for every fundamental projection G 1 5 G, we have 0 < + co, and e(G i ) 5 tP(G1) for rith(G1) t4) (G1). We have 1 every fundamental projection G 1 5 G. Suppose that, for every non-zero fundamental projection G1 5. G, there exists a non-zero fundamental projection G 2 5. G1 such that 0(1 + C)CP(G2) ING 2 ). Arguing as above, we would deduce from this that 0(1 + 0,4)(G1) 1P(G 1 ), in contradiction to the definition of 0. Hence there exists a non-zero fundamental projection E < G such that, for every fundamental projection F E, we have OF) < 0(1 + E)4)(F). Moreover, 0 F)-E tp(F) since F 5. G. 0 (

Let E > O. A positive linear form cp on A is called a trace to within E if for every TE A+ and every unitary operator UE A, we have OuTu -1 ) < ( 1 + E)4)(T). DEFINITION 4.

If S is any element of

A, we deduce from this that

(1)(5S * )

(1 +

because we have S = US1, with U unitary and S 1 hermitian, UE A, S i E A (by virtue of the remark of chapter 2, section 3), hence 2. if 4SS*) SS* = US 21 U -1 ' S*S Conversely, = S A (1 + EWS*S) 1 for every SE A, we have, for every TE A+ and for every unitary operator U E A, cp(UTU

-1

) = (1)((UT 1/2 ) (UT1/2 )*)

(1+E)(1)((UT 1/2 )*(UT 1 ))

(1 +E)4)(T) .

For every E > 0, there exists a non-zero fundamental projection EE A and a non-zero normal positive linear form w on AE which is a trace to within E. Proof. Let 4) be a faithful normal positive linear form on A LEMMA 9.

[they exist by hypothesis (***)], and ip a central weight on the set of fundamental projections (lemma 7). By lemma 8, there exists a fundamental projection E 0 such that, for every fundamental projection F E, we have d(F) 5. (1 + E)11)(F)

(multiplying 4) by a suitable scalar if necessary). The restriction of 4) to EAE defines a non-zero normal positive linear form w on AE . Let T E A+ be such that ET = TE = T, and U a unitary

PART III, CHAPTER 8

336

operator of

A

commuting with E. We show that

(2)

cp(UTU -1 )

(1 +

which will establish that w is a trace to within E. If T is a fundamental projection (majorized by E), UTU-1 is a fundamental projection majorized by E and equivalent to T, hence

cp(UTU -1 )

(1 + E)1(UTU -1 ) = (1 + E )l)( T)

If T is any projection of

A

(1 + E )( T).

majorized by E, we have T =

IT., i

where the Ti's are disjoint fundamental projections (corollary of lemma 2); then

1 ), (1, )= 1 4)(7,s.),(p(uriu - 1 )= 1 ,4)=171J i from which it follows that (2) is again true in this case. Finally, if T is any element of A -1- such that ET = TE = T, T is thenormlimitofoperators

are projec-

tions of A majorized by E and the Xi's are scalars (2) follows on passing to the limit. 0

0, so that

For every E > 0, there exists a non-zero normal positive linear form (I) on A which is a trace to within E. LEMMA 10.

Let E and w be the projection and linear form whose existence is assured by lemma 9. Let El, E2, Ep be disjoint, equivalent projections with sum FEZ, with E l = E. Let Wi be a partial isometry having E as initial projection, and Ei as final projection. The form w may be identified with a linear form on the set B of the TE A such that ET = TE = T. For TEA,

Proof.

put cp(T) =

2=1

w(W*.TW.). 2 2

As

W

.Wt --

= F,

1:=1 0 ° Wri E B,

we have, putting T1 cp(T*T) =

w(w*T*w .W*.TW .) =

i,j (p (TT*) =

s

dJ s

w(WtTW.WtT*W.) = —sd d s s,d

w(Tt sd sd

=

1,i

sd sd

337

FURTHER TOPICS

whence 11)(T*T) 5. (1 + E)d?(TT*). Moreover, cp(E) = W(E) / 0.

0

References : [ 6], [66], [140], [262], [303]. 5.

The proof of the theorem concluded.

We use the notation LEMMA 11.

T' K" T of chapter 5.

S, K T'

For every

TE

A, K

reduces to a single point.

Proof. Suppose first that T is hermitian. We can plainly restrict attention to the case where 0 < T I. 'Let Si, S2 be elements of fq, and suppose that Si / S 2 . There exists a projection E / 0 of Z and a scalar a > 0 such that, for example, SiE - S 2E ûE. We will arrive at a contradiction. Replacing A by AE , it is enough to deduce a contradiction from the hypothesis Si - S2 > c.i. Now, let (I) be a non-zero positive linear form on A which is a trace to within E. For every fE .S, we have

1( f.T)

Ecp (T ).

cp( T)I

-

Hence

1

d(T)I

(S1)

I

EOT),

(

S2) - cp(T)I

E(T)

and, consequently, 2 E(T)

IOS 1

- S 2 )I

Whence the contradiction for E sufficiently small. If now TEA is arbitrary, write T = Ti + iT 2 , with T 1 , T 2 hermitian. Put Kr11, 1 = {Si }, K 2 = {S 2 . Let SE K. For every > 0, there exists an fES such that H f.T - s H E, whence E, and consequently H f.T* }

s* H _

H f-T, -

1 vs

+ s*) H

E

,

H

1

f.T 2 -- ,(s 2 - s*) H

E.

As E > 0 is arbitrary, we deduce from this that 4 (S + S*) =

1

-

- S*) = S 2 , hence S = S i + iS2. We will henceforth denote by T

1'

0

the sole member of the set

K.

he mapping T of A into Z is positive linear. LEMMA 12. We have (T 1T 2 )9 = (T 2T 1 ) for any T1 E A, T2 E A. Proof. The linearity follows from lemma 6 of chapter 5. If T 0, we have Ti > 0 for every T i E KI, hence T4 O. If TEA and if U is a unitary operator of A, we have K =K UT U(UT)U =KTU1

338

PART III, CHAPTER 8

hence (UT)E1 = . Since every element of A is a linear combination of unitary elements of A, we deduce from this the relation

(T T )E1 = (T T ) 1:1 1 2 2 1

for

T

1

E

A, T

2

E

A

.

We can at last prove theorem 1. Let 4) be a non-zero, normal positive linear form en A, which is a trace to within C. For TE A, put 4)'(T) = 4)( 11 ). By lemma 12, (1) 1 is a trace on A. We have cif (I) = 4)(I) / 0. Finally, let T E A+ ; 1. for every T E KT , we have 4)(T 1 ) (1 + 04)(T) , hence cp' (T) = (1 (T as 4) is normal, we see that 4) t is normal. Then the support F of 4)' is a non-zero projection of Z, and 4)' defines a faithful normal trace on AF , so that AF is finite. This completes the proof. 0

wig)

There is probably a shorter proof of theorem 1. remains, however, to be discovered.

o

);

This proof

References : [6], [66], [140], [262], [303].

6.

Consequences of the theorem.

COROLLARY 1. Let A be a von Neumann algebra, E a projection of A. For E to be finite, it is necessary and sufficient that every projection of A majorized by E and equivalent to E be equal to E.

Proof.

It suffices to apply theorem 1 to AE .

D

We have here another definition of finite projections. Then, propositions 9 and 11 of chapter 2 furnish an alternative definition of properly infinite and purely infinite projections. COROLLARY 2. Let E be a properly infinite proection of A. There exist pairwise disjoint projections (E l , E 2 , E 3 , ...) equivalent to E, with sum E.

Proof. Suppose merely that E is infinite. Let E2 E = El , with E 2 / E l , E 2 E l . Let U be a partial isometry of A having E l and E 2 as initial and final projections. By induction, we see that Ul is a partial isometry with initial projection E l and final projection E with El E2 .... Let F. = E.2 - 2+1* The restriction of U '1- to F 1 (H) maps F l ( H) isometrically onto Fi_i_ 1(H), hence 0 / F 1 F2 F3 .... By corollary 2 theorem 1, chapter 1, applied to AE and to the family (F 1 , F2, ...), there exists a projection F of Z and an infinite family (G4) :1- ci of pairwise disjoint, equivalent, nonzero projections (A A, with sum EF. We can write I = I l uI 2 u ..., the In 's being disjoint and equipotent with I. Then X G., X G., ... are pairwise disjoint, non-zero projecjEI1 jEI2 tions, equivalent to G.. Hence EF = G 1 + G 2 + •.., the jEI

339

FURTHER TOPICS

Gi being non-zero, pairwise disjoint projections of lent to EF.

A,

equiva-

Finally, suppose that E is properly infinite. Applying proposition 9 of chapter 2 and Zorn's lemma, we construct a family EFx and, (Fx ) xcK of disjoint projections of Z, such that E = xEK 1 2 equivalent.to EFx , disfor every xcK, projections Gx , Gx , The projections Ei = X Glx- are the joint, and with sum EFx .

xEK projections of the corollary.

0

Let A be a von Neumann algebra with centre If there exists a linear mapping 1 of A into Z, such that COROLLARY 3.

(D(T)

T

for

T E Z,

and such that 4)(S 1S 2 ) = cD(S 2 S 1 ) for s 1 E A, finite. Proof.

Let E be greatest projection of Z infinite relative to A. By corollary 2, we where E l , E 2 are disjoint projections of A, UU* for some UE A, hence Then E U*U, E l larly, 11 (E) = IT(E2). Hence 2E

whence E = O.

-

21(E )

-

Z.

S2 EA,

then

A is

which is properly have E = E l + E 2 , with E El — E 2 . (D(E) = (1)(E 1 ). Simi-

4)(E 1 ) + (D(E 2 ) -= 4)(E)

E,

0

COROLLARY 4. Let A be a von Neumann algebra with centre Z. 1-f, for every TE A, the set '4 of chapter 5 meets Z in a single

point, A is finite. The mapping (I) of A into Z is linear (chapter 5, lemma 6). We plainly have (1)(T) = T for TE Z. Moreover, if S1 E A and if S2 is a unitary operator of A, we have

Pr oof. Let (1)(T) be the point where K meets Z.

, hence cD(SS 1 2 ) = (D(S 2 S. 1 S 2 (S1S2)S2-1 = Kcb-J1S2 ) This equality also holds for arbitrary S2 in A by linearity. Corollary 4 then follows from corollary 3. 0 KS1S 2

K

Let A be a von Neumann algebra, E and F two projections of A, and E P and F' their central supports. Suppose that itlk is a-finite, that F is properly infinite, and F'. Then E -‹ F. that E In particular, in a a-finite factor, two infinite projections are equivalent. COROLLARY 5.

340

PART III, CHAPTER 8

Proof.

Consider a maximal family (Ei) i ci of non-zero, disjoint projections of A, majorized by E, and such that Ei F. Then E - X E. has a central support majorized by that of F, and icI chapter 1 shows that E - X E. = 0, E = 1 of lemma E.; since icI icI AE is G-finite, I is countable. Moreover, there exists (corollary 2) an infinite family (Fx) xEK of disjoint projections of A equivalent to F, with sum F; since I is equipotent with a subset of K, we have E= X E. iEI COROLLARY 6.

Let

Neumann algebra. set K' of chapter 5 zero point.

F .rEK

x

=F.

0

A be a purely infinite a-finite von Let T be a non-zero element of A. Then the meets the centre Z of A in at least one non-

Proof.

We can immediately reduce the problem to the case where T is hermitian and of norm 1. Changing T to -T if necessary, we can suppose that there exists a non-zero spectral projection E of A and an integer n > 0 such that TE n-l E, n-1E - (I whence T E). Let F be the central support of E. Writing A =AF X Ai _F , we are led to the case where F = I. If E majorizes a central projection of A, the corollary is immediate. We can therefore suppose that E majorizes no central projection of A, from which it follows that the central support of I E is I. By corollary 5, E I - E. By corollary 2, there exist pairwise disjoint projections El, E 2 , En+1 of.A, equivalent to E and with sum E. Put I - E = E 0 . There exist unitary operators U0, ..., Uni.1 of A such that UiEjU.77 1 = Ea . ( g ) , where the G. are the n+2 cyclic permutations of 0, 1, ..., 61. We have XU.h.U7 1 - I for every j. Put i 27 (n ± 2) LU.TU.-1 . S L

Z

Then

(n + 2)S

, -1 XU.kn (E + E + 2 1 2 = (n + 1)n

-

+ E

n+1)

- E )U.-1 0

-

I-1=n I.

n-1 (n + 2) -1 for every element Si of K. and therefore Hence S 1 for every element S 1 of q. Hence K n Z containsS a non-zero point. Now Ks c KT since S E KT , hence K; c KT' . D COROLLARY 7. Let A be a factor, M the set of projections of D a function defined on M, possessing the following prop-

A, and

erties:

341

FURTHER TOPICS

+ 00 for every non-zero E of

(i) 0 = D(0) < D(E)

(ii)if

M;

E EM and FEM are disjoint, D(E+F) =D(E) + D(F);

(iii)if u is a unitary operator of A, and if E D(UEU 1 ) = D(E).

Then, D is a relative dimension of A, or is identically infinite on the set of non-zero projections of A. Proof.

We first show that, if EEM is infinite, then D(E) = + 00. Let E E E be disjoint projections of A such 2' 3 that

E = E

1

+ E2 + E3 ,

E l , E 2 —E

E

3

(corollary 2).

We have E 1 -< E l + E 2 E l ; similarly, E —E 1 , hence El + E2 E2 + E3 Hence there exists (E I + E 2 ). E3, hence I - E l — I a unitary operator U of A such that UE1u -1 = E l + E2. Then

D(E

1

) = D(E

E 2 ) = D(E

1

1

) + D(E ) 2

and, consequently, D(E 1 ) = + 00 ; since D(E)

D(E 1 ), D(E) = + co.

We now show that, if D is finite for one finite projection E 0 , D is finite for every finite projection E. By chapter 1, (theorem 1, corollary 2), there exist pairwise disjoint projections E l , E 2 , ..., En of A with sum E, such that

E

EE —

1

2

—E

n -1

›- E . n

Then, in view of proposition 6 of chapter 2, we have D(E) = X D(E.) i=1

nD(E ) < + 00• 0

This established, the assertion of the corollary is immediate if A is purely infinite. If A is not purely infinite, let E, F be two finite projections of A, and D / a relative dimension of A. We are going to show, and this will complete the proof, that D(E)/D(F) = D I (E)/D P (F). Let G be the supremum of E and F, which is finite. The restriction of D to the projections of GAG defines, on the set of projections of AG , a function to which one can apply proposition 15 of chapter 2. We then see that D and D I are proportional on the projections of GAG. 0

Let A and B be two von Neumann algebras. Suppose that there exists in A' (reap. B') an infinite family (q)iEI [resp. (q) ici ] of pairwise disjoint, equivalent projections with sum 1, such that the A (reap. Bp I ) are a-finite (which is the case if A' and B' are a-finite and COROLLARY 8.

342

PART III, CHAPTER 8

properly infinite). spatial.

Then, every isomorphism (I) of A onto B is

Proof.

Let (Ix) xEK be a partition of I, the Is being countably infinite. Put E; = X E P.; the E" are properly infinite iEI x

(chapter 2, proposition 10), equivalent, pairwise disjoint, with sum I, and the ilku's are a-finite. Replacing the Eql:'s by the Ek's, we cEn therefore suppose henceforth that the E,;-,'s are properly infinite, and similarly that the F lys are properly infinite. There exist a von Neumann algebra C, and projections E l , F l , of C', with central support I, such that A may be identified with CE P, B with CF' and (I) with the isomorphism TE? TF/(TE C) We have E l. — (corollary 5), hence 2

.

E P = X E P. — X F P. = F l . . 2 iEI 2E' Hence cip is realised by a partial isometry of

CP .

D

COROLLARY 9. Let A be a von Neumann algebra, Z its centre, and (I) an automorphism of A which leaves the elements of Z fixed. If A' is properly infinite, 4) is spatial.

Proof.

If

A

=

H A., (I) induces automorphisms iEI

i

of the A 's

i

and it suffices to prove the corollary for Ai and (Di. Thanks to lemma 7 of chapter 1, we can then satisfy ourselves by studying the following two cases: 1° A' is a-finite (and properly infinite); 2 ° there exists in A' an infinite family (E)i e , of pairwise disjoint, equivalent projections, with sum I, such that the A's are a-finite. In both cases, corollary 8 proves that (I) is spatial. 0

Let A be a von Neumann algebra. If A' is properly infinite, every ultra-weakly continuous linear form on A is a form wx,y Proof. Let (Ei, E, ...) be a sequence of pairwise disjoint COROLLARY 10.

A', equivalent to 1, with sum I (corollary 2). The induction T TE i of A onto AE iis a spatial isomorphism (1). Let H1 = Ei(H). We can identify H with H1 0 K (K beir7 a separable infinite-dimensional Hilbert space), and (10 -1 with the ampliation TE !1 0 IK of AE ionto A = AE 0 cK. The corolTE i lary then follows from lemma 5, part I, chapter 4: n

projections of

COROLLARY 11. Suppose that A' is properly infinie. For A to possess a separating element, it is necessary and sufficient that A be a-finite. Proof. The condition is plainly necessary. Conversely, if A is a-finite, there exists

a

faithful

normal positive

FURTHER TOPICS

343

linear form on A. By corollary 10, this form is a form W. Then, x is separating for A. 0 References : [6], [10], [15], [31], [42], [62], [65], [66], [89], [404], [414].

More on tensor products.

7.

Let A on A+, and s trace mapping T .÷ ST* of bounded subsets of LEMMA 13.

be a von Neumann algebra, (I) a faithful normal an element of A such that 0s*s) < + co. The A into A is strongly continuous on the

A.

For every TE A, put Ip(T) = cp(S*ST); then 1P is ultraweakly continuous (part I, chapter 6, proposition 1). There exists a family (xi) of vectors such that cp = tox . on A-4- (part I, chapter 6, corollary of proposition 2). We hate, for every

Proof.

A,

H ST*x.2 "H2 =

Lw

.(TS *ST*) = (TS*ST*) = 'flS*ST*T) = 1P(T*T).

As T converges strongly to zero, staying bounded throughout, T*T converges ultra-weakly to zero, hence Ip(T*T) ÷0, hence ST*xi÷0 for every i. Now the xi's constitute a separating set for A since cl) is faithful. As ST* stays bounded, we see that ST* converges strongly to zero. Li

LEMMA 14. Let A be a von Neumann algebra, and E a non-zero properly infinite projection of A. The mapping T T* of EAE into EAE is not strongly continuous on the bounded subsets of

EAE

Proof.

There exists pairwise disjoint, equivalent projections El, E2, ... of A, with sum E (corollary 2 of theorem 1). By part I, chapter 2, proposition 5 (ii), AE is spatially isomorphic to AEl 0 UK), where K is an infinite-dimensional Hilbert space. Hence AE contains a von Neumann subalgebra isoT* of L(K) into L(K) is morphic to L(K). Now the mapping T not strongly continuous on bounded subsets (part I, chapter 3, section 1). 0

Let Al, A2 be two von Neumann algebras. If one of them is purely infinite, then A l o A 2 is purely infinite. THEOREM 2.

Proof.

Let H 1 , H 2 be the spaces in which A l , A2 act. We will suppose that A l is purely infinite. Choose an orthonormal basis (ei)i E I of H2, which enables us to identify H1 0 H2 canonically with H. (where Hi = H 1 for every i). The elements of

iEI A l 0 L(H2) are represented by matrices (Tij), where Tii E A 1 , for every i, xE 1. If x A i , the element X 01 of Al 0 L(H2) is represented by the matrix (6ixX) (part I, chapter 2, proposition 4). If T = (Tii) E A 1 0 L(H2) and X E A, then T. (X 0 I) is

344

PART III, CHAPTER 8

represented by the matrix (TX) (part I, chapter 2, section 3). If Ai 0 A2 is not purely infinite, there exists a faithful normal trace (1) on (A i 0 A2) -i- and an SE (A 1 0 A2) -4- such that S / 0 and cp(S) < + op. Let S = (Sig). If Sii were zero for every i E I we would have (Sx1x) = 0 for every X E Hi and every i, hence S'1 (Hi) = 0 for every i, hence S 1/2 = 0 and S = 0, which is a contradiction. Hence there exists an index i o such that Si oi o / O. For every T = (Tig) E A1 0 L(H 2 ), we will put Ti oi o = T. We have S" 0, n - / O. There therefore exists a non-zero projection E of A i and a real number X 0 such that s-x for every x E Hi

H Exil

xit

H

-

Consider the mappings Œ

Al

Al

0 A

2

+

Al

Y 0 A

2

+

Al

where a(X) = X 0 I, B(T) = ST*, y(T) = T". These mappings are strongly continuous on bounded subsets: the first by part I (chapter 4, corollary 1 of theorem 2); the second by lemma 13; the third--clearly. Now 1(f3(a(X))) = (S(X 0 I)*) - = S-X*. The mapping X + S -X* of Ai into Ai, and a fortiori the mapping X + EX*, are therefore strongly continuous on bounded subsets. Taking X E EA1E, this contradicts lemma 14 since E is a purely infinite projection of A i . 0

Let Al, A2 be two von Neumann algebras. them is continuous, then A l 0 A 2 is continuous. COROLLARY.

If one of

Write A i = B i x C i , A 2 = B 2 x C 2 , with Bi, B2 semifinite and C i , C 2 purely infinite. Then Ci 0 B 2 , B i 0 C 2 and C 1 ® C 2 are purely infinite (theorem 2) and therefore continuous. It suffices to prove that B i 0 B2 is continuous, assumingfor example, that B i is continuous. Now there exists a decreasing sequence (En ) of finite projections of Bi, with central support I, such that En En+1 for n 1 (chapter 2, corolEn+1 lary 4 of proposition 7). Put Fn = En 0 I E Bi 0 8 2. The Fn are decreasing projections of B i 0 B 2 , such that Fn Fn+1 Fn+1 for n 1. By part I (chapter 1, corollary 1 of proposition 7), the central support of Fn is I. Hence Bi 0 8 2 is continuous (chapter 2, corollary 4 of proposition 7). 0

Proof.

Reference : [192].

Exercises. mann algebra.

1.

A

be a purely infinite 0-finite von NeuLet Z be its centre. Let

a. Let M be a two-sided ideal of A. If m 0, we have n Z / O. (Let TEM+ , T / 0; exercise 6 of part I, chapter 1, yields a projection E / 0 in M; E is equivalent to its central support F; use exercise 8 of chapter 1.)

FURTHER TOPICS

345

b. The intersection of the maximal two-sided ideals of zero. (Chapter 5, proposition 1, corollary 2). c. A purely infinite 0-finite factor trivial two-sided ideals (use a).

has

no

A

is

non-

d. Let A be a purely infinite a-finite factor (they exist: part I, chapter 9, section 4, remark). Let K be a nonseparable Hilbert space. Then, A 0 L(K) is a purely infinite factor which possesses non-trivial two-sided ideals DA 0 L(K)) / = A' 0 CI( is purely infinite; L(K) is not 0-finite, hence A 0 L(K) is not 0-finite; use exercise 7 of part I, chapter 1] [25]. 2. Let A be a von Neumann algebra, and $ a faithful positive linear form on A. Suppose that there exists a real number 1 such that $(T*T) A$(TT*) for any TE A. Show that A is finite. [Let E be the greatest properly infinite projection of A; write E = El + E 2 + ..., where the E's are disjoint projections, equivalent to E; we have j(E) A - (E) for every i, hence $(Ei) = 0, $(E) = 0, E = O.] 3.

Let

A

be a von Neumann algebra in H.

a. Let E0 = Pxo , El = Px 1 , E2 = Px 2 , ... be pairwise disjoint, equivalent projections of A, with sum 1, Ui a partial isometry of A having Ei + 1 as initial projection, and Ei as final projection

(i =

0,

1, 2, ...).

Let T = I

U. (the series 2

i=0 converging strongly), and Y = T(X1 e X2 e ...). Show that Tx = 0 implies x = O. Deduce from this that YnX0 = O. Show that T*(H) nX0 = O. Deduce from this that Y is dense in H. b. Let E, F be two disjoint, equivalent, properly infinite projections of A, with sum I. Show that there exists aTEA 1 0 Tx = 0 implies x = 0; 2 0 TF(H) is dense in H; such that: 0 3 TF(H) nE(H) = O. (Apply to F corollary 2 of theorem 1, and use a.) c. If A is properly infinite, there exist two elements S 2 of A such that S i (H) nS2(H) = 0, Si(H) and S2(H) being dense in H. (Apply b twice, interchanging the roles of E and F.) Let x be a separating element for A. Show that y = Six and 2 - S2x are separating elements for A, and that the equality T'y = T's for a T P EA ? implies T'y = T / 2 = O. [We have T'y = T / 2 = SiT / x = S2T / x €S1(H) nS2(H).] Deduce from this that there exists no closed operator TA' (part I, chapter 1, exercise 10) • with dense domain, such that Ty = z. d. Suppose that A is properly infinite and 0-finite. Show that there exist on A two faithful normal positive linear forms $1, $2 such that every positive linear form majorized by

346

PART III, CHAPTER 8

(pi and cp2 is zéro. (Reduce to the case, by using an isomorphism on A, where A' is properly infinite; then, A admits a separating element x; with the notation of c, take 4) 1 = wy, (1)2 = w2, and apply lemma 1 of part I, chapter 4) [19]. 4. Let A be a von Neumann algebra, and E and F equivalent projections of A. There exist disjoint projections El, E2 (resp. Fl, F 2 ) of A, with sum E (resp. F), and unitary operators U 1 , U2 of A such that UlElUi l = F l , U 2E2U 1 = F2. (If E is finite, take El = F2 = 0, and apply proposition 6 of chapter 2. If E is properly infinite, choose El, E2, F l , F2 in such a way that E El « E2 « F F1 — F2 and show that I - El F l , I- E2 —I—I- F2) [10]. 5. Let A be a von Neumann algebra, and E and F two projections of A, AE being a-finite. For E < F, it is necessary and sufficient that OE) d(F) for every normal trace cp on A+. [To show that the condition is sufficient, E, then to the case where F < E. reduce to the case where F If F is finite, reduce further to the case where there exists on A+ a semi-finite faithful normal trace cp such that (F) < + 00 ; the relations d(E) - F) OF), cp(E) = d(F) + If F is properly infinite, apply corolthen yield E - F = 0. lary 5 of theorem 1.] -

6. Let A be a von Neumann algebra such that every normal positive linear form on A is the sum of a finite number of forms wx . Then, there exists an integer n such that every normal positive linear form on A is the sum of n forms w corollary 10 of theorem 1, and propositions 6, 7, 8, 9 of chapter 6) [19]. .

7. Prove the converse of lemma 2 of chapter 6. [If the strong and ultra-strong topologies do not coincide on A, either A is properly infinite with A' finite, or A and A' are finite and CV- is not essentially bounded. In the first (resp. second) case, use the proof of proposition 8 (resp. 9) of chapter 6] [19]. 8. Let A be a von Neumann algebra, Z its centre, and 0 a linear mapping of A into a vector space E over C, such that 0(ST) = 0(TS) for S EA, TEA, and such that, for every non-zero projection F of Z, we have 0(F) / 0. Then, A is finite. (Adapt the proof of corollary 3 of theorem 1). 9.

Let

A

be a von Neumann algebra.

Let Z be its centre.

a. Let .(10 be a finite Z-trace on A. Show that 0 is normal. (Reduce to the case where A is either properly infinite, or is finite. In the first case, show that = 0, using corollary 2 of theorem 1. In the second case, use exercise 4 of chapter 5) .

FURTHER TOPICS

347

Let M be a restricted ideal of A (part I, chapter 1, exercise 6). Show that a Z-trace on M is normal. (Use a) [12].

b.

10. Let A be a von Neumann algebra, and Z its centre. Suppose that A is finite and A' properly infinite. There exists a projection FE Z' possessing the following property: for every TEA, there exists exactly one T' E Z such that FTF = T'F, and T T 1 is the canonical Z-trace of A. (Reduce to the case where A is a-finite. corollary 11 of theorem Then use 1, and exercise 3 of chapter 6) 1. 89]. 11. Let A be a type hoe factor. Show that A is spatially isomorphic to an algebra B 0 L(K), where B is a type 111 factor and K is an infinite-dimensional Hilbert space. [By corollary 2 of theorem 1, there exists a sequence El, E 2 , ... of nonzero, equivalent, pairwise disjoint projections of A. By chapter 2, proposition 7, we can suppose that the E's are finite. By chapter 1, corollary 2 of theorem 1, we can suppose that I. Then use part I, chapter 2, proposition 5 (ii).] 12. Let H (resp. H2) be a countably-infinite- (resp. 2-) dimensional Hilbert space. a. Let A be a von Neumann algebra in H, (X1, X2, ...) a sequence generating A, and y a generator of L(H) (cf. chapter 3, exercise 9). Represent every element of A 0 L(H) by a matrix (xii) of elements of A. For k = 1, 2, ..., let 2k E A ® l(H), be the element represented by the matrix (xii) such that X11 = xk, xii = 0 for i / 1 or j / 1. Then the zk's and I 0 y generate A ® L(H).

b.

Deduce from a that

A ® L(H) is generated by two elements. A ® L(H) ® L(H 2 ) can be generated by

c. Deduce from b that two elements, one of which is unitary. d. Deduce from c that A ® L(H) generated by two unitary elements.

0 L(H 2 ) 0 L(H 2 ) can be

e. Deduce from d that, if B is a properly infinite von Neumann algebra in H, then B can be generated by a single element. [497]

-

CHAPTER 9. DERIVATIONS AND AUTOMORPHISMS OF VON NEUMANN ALGEBRAS

1.

Derivations of algebras.

Let A be an algebra. [a, b] - ab - ha.

For any a, b

E

A, we will put

We call a derivation of A a linear mapping 6 of A into A such that d(ab) = (6a)b + a(6b)

(1)

for any a, b

E

A.

We then have

b] + [a, 6b]

S([a, b]) =

(2) for any a, b

E A.

In fact,

(S([a, b]) = 6(ab) - d(ba) = (Sa)b + a(6b) - (6b)a - b(6a) = [6a, b] +

[a, 6b].

If 6 leaves invariant a subset M of A, then 6 also leaves invariant the commutant M f of M; because, if a E M i and b E M, we have [6a, b] 6([a, b]) - [a, (Sb] = O. In particular, 6 leaves the centre of A invariant. If A possesses an identity element I, we have SW = 6(1 2 ) = 6(1).1 + 1.6(1) = 26(I), hence 6(I) = O. If x E A, the mapping a .4- [x, a] of A into A is a derivation; indeed, x(ab) - (ab)x = (sa - ax)b + a(xb - bx). We call this derivation the inner derivation defined by x. Suppose that A is endowed with an involution x x*. Let 6 be a derivation of A, and put S*(a) = (S(a*)) * for every a € A. Then 6* is a derivation, because d*(ab) = (d(b*a*))* = ((db*)a* + b*(Sa*))* = a((S*b) + (6 4(a)b.

If 6 = 6*, the derivation 6 is said to be hermitian; in this case, da is hermitian for every hermitian a in A. If 6 is any 349

PART III, CHAPTER 9

350

+ (S*) + i derivation of A, we can write d 1 1 -0 + (S*), --,(6 - 6 49 are hermitian derivations of A. 22 2 2.

Derivations of C*-algebras:

(S *), and

continuity,

extension. LEMMA 1. Let A be a c*-algebra of operators, B a commutative sub-c*-algebra of A containing 1, cp a positive form on A whose restriction to B is a character of B, and (S a derivation of A. Then OH) = 0. Proof. Let T be an hermitian element of 8, and let us show that (I)(ST) = 0. We can suppose that OT) 0 (adding a suitable scalar operator to T). By considering the Gelfand transform of 8, we see that T can be written in the form U 2 - V2 , with U, V hermitian elements of B such that d(U) = cp(V) = 0. We have ST = UOU) + ((SU)U - (V(SV) - (6V)V. Now Icp(U((SU)) 1 2 (1)(U 2 )W6U) * (6U)) = 0, hence (1)(U(6U)) = 0. see similarly that cp(OU)U) = (1)(V((SV)) = (PHSV)V) = 0. 0

We

Let A be a c*-algebra of operators containing 1, C its centre, and (S a derivation of A. Then 6(C) = 0. LEMMA 2.

Proof. Since S(C) cC (section 1), we are reduced to the case where A is commutative. Then every character of 6(A) (lemma 1), whence 6(A) = O. 0

A

vanishes on

Let A be a c*-algebra of operators, 6 a derivation Then (S is norm continuous.

LEMMA 3.

of A.

Proof. If I Et A, (S extends to a derivation of A + C.I (which is a C*-algebra) vanishing at I. We can therefore suppose that I E A. Thanks to section 1, we are led to the case where d is hermitian. Suppose that 6 is not continuous. Its restriction to the subspace of hermitian elements of A is not continuous. By the closed graph theorem ( [[ 3]], part I, section 3, corollary 5 of theorem 1), there exists a sequence (S1, S2, ...) of hermitian elements of A converging to zero such that SSn 4- A, where A is a non-zero hermitian element of A. We can suppose that S n / 0 for every n. Multiplying the S n 's by a suitable constant, we can suppose that the spectrum of A contains a real number > 3. Replacing S n by S n + 2 Sn H (which leaves (SS n unaltered), we can suppose that S n L- 1/3 II S n There exists an hermitian element H of A such that

11

1

1, HAH H11 Put Tn = HSnH, B = HMI. We have ST

n

H.

2 3H .

= (Ws n H + HOS )H + HS ((S11), n n

FURTHER TOPICS

351

hence (ST

(1)

Moreover, 1/311 1/3H S n

H

H

Tn

sn 11H 2 H, whence

hence

(2)

n

-4- B.

< H i sn 11H2, h- lTn -1 B. Tn H Tn 1 Tn

whence 11-111 Sn

sn

02 . 3H 2 ;

This established, there exists for every n a character q),, of the commutative C*-algebra generated by I and Tn such that iTn ) = 1. As I is an interior point of the core of Tn elements 0 of A, Ipn extends to a positive form (pn on A a[3]], part II, chapter 3, proposition 6). By lemma 1, we have (Pn((STn) = O. B Y (1 " IWB)1 < 1 for n large enough. However, lTn ) = 1, and we have a contradicby (2), (pn(B) Tn tion. 0

i4)(11

H-

Let A be a c*-algebra of operators, B its weak closure, and 6 a derivation of A. Then 6 is ultra-weakly continuous and extends with no increase of norm to a derivation of LEMMA 4.

Proof. Let H be the Hilbert space in which A acts, and At the set of elements of A+ of norm _< 1. If T E At and x, y E H, we have (3) 1(6(T)x1Y)

=

1(6(T 1/2 )T 1/2xIY) + (T 1/2,5(T 1/2 )xly)

11 6 11 - 11

T1/2x

11.- 11

Y

II ± 11 6 11 -11 T 1/2 Y I II x II

II 6 II ( I x 112+11y 11 2 ) 1/2 ( I T1/2x 112 4" I T 1/2 Y 11 2 ) 1/2 = H 6 1111 x 11 2 HY 11 2 ) 1/2 ((Tx1x) + (Tyly)) 1/2 (with 11 6 11 < + ., by lemma 3). Let (xl, x2, (Y1 , y2 , be sequences of elements of H such that xi H- < oe, 2 Yi 11 < + co, and let E > O. There exists an integer N such

XII

I

that, for

+00

m

T

H

X

1, we have

(6(T)xilyi)

< E.

In view of

(3), there therefore exists a weak neighbourhood V of 0 in

such that

1,00 TE V

X

(6

(T)Xi I yi )

A +1

2E.

1 By part I, chapter 3, corollary of theorem 3, the linear form T X(d(T)xilyi) on A is ultra-weakly continuous. Hence 6 is ultra-weakly continuous.

352

PART III, CHAPTER 9

Every point of B is in the ultra-weak closure of a bounded subset of A (part 1, chapter 3, theorem 3). Moreover, the bounded, ultra-weakly closed subsets of B are ultra-weakly compact and therefore ultra-weakly complete. Hence ([[3]], part III, chapter 2, proposition 8) 5 extends to an ultra-weakly continuous linear mapping 6 of B into B. The relation T(ST) = (6S)T + S(6T) holds for S, TEA, and both sides are separately continuous functions of S and T for the ultra-weak topology. Hence -6 is a derivation of B. The fact that 6 6 II follows from part I, chapter 3, theorem 3. 0

II

References : [252], [367].

3.

Derivations of von Neumann algebras.

LEMMA 5. Let A be a von Neumann algebra, and 6 a derivation of A. (i)For every projection E of A, the mapping T E6(T)E is a derivation of EAE of norm II 6 (ii)Let Let F be an increasing filtering set of projections of A with supremum I. Suppose that, for every EE F, there exists TE E ErAE such that H TE H -5and E6(T)E = [TE , E] for every T E EAE. Then there exists To E A such that IIH T O 5 II and 6(T) = [T 0 , T] for every T E A.

H6H

'

H6

For every EE F, let FE be the set of E' EF, such that E' E. The set of the FE'S is a filter base B on F. As the closed ball of centre 0 and radius II 6 II in A is weakly compact, the mapping E TE admits a weak limit point Tn following B, such that II TO II 11 Moreover, if TEA, t5e mapping E E6(T)E converges strongly to 6(T) following B. We therefore have 6(S) = [To, T] if Then the derivation 6 : T S EA Q = U EAE. 6(T) - [To, T] of

Proof.

(i) is immediate. We prove (ii).

T6

EE F

A

vanishes on Ao.

For every TEA, we therefore have

E6 1 (T)E = 6'(ETE) - 6(E)TE - ET6'(E) = 0 for any EE F, and consequently 6'(T) = 0 onpassing to the strong limit. 0

Let A be a von Neumann algebra, F the set of projecA such that AE is of countable type. Then F is increasing filtering with supremum I. LEMMA 6. tions E of

Proof. This follows from dn. fact that the elements of F are the projections of the form E ti,‘1 where M is a countable set of vectors (part I, chapter 1, proposition 6). D Let A be a von Neumann algebra, and 6 a derivation of A. There exists a To EA such that II To II 5- II 6 II and 6(T) = [ T0 , T] for every T E A. THEOREM 1.

353

FURTHER TOPICS

1 ° Let U be the unitary group of Proof. define the mapping Au of A into A by A (T) = (UT + 6(U))U u

-1

A.

For U € U, we

.

If U, V E U, we have Auriv(T) = [U(VT + 6(V))V

+ 6(U)JU

-1 -1 -1 --1 + 6(U)U + U6(V)V U = UVTV 1U = (UVT + 6 (uv) ) v

-1 -1 u ,

hence

(1)

AA =A

Uv

UV

.

Let E be the set of the K c A possessing the following proper1 ° K is non-empty, convex and weakly compact; ties: 2 ° Au (K) c K for every U E U; 3 ° every element of K has norm (1). For let Mc A be the set of 6 (u)u -1 -- 11 6 11- Firstly, E for U E U. If U, V E U, we have Av(6(U)U -1 ) = (V6(U)U -1 + = 6 (vu) u- lv-1 , hence Av (M) c M. Consequently the weakly If (K,L) is a closed convex hull of M is an element of E. totally ordered family of elements of E, we have nKi E E. By Zorn's lemma, there exists in E a minimal element Ko . Suppose that it has been proved that Ko reduces to a single point To , in other words Ko - K o = fol. Then we have, for every U E U,

-1 2 -1 T = A (T ) = UT U + 1 o(U)U ID U 0 o hence 6(U) = [To, U] and by linearity 6(T) = [To, T] for every T E A. Note that Ko - Ko is convex and weakly compact, and that U(Ko - K 0 )U-1 = Ko - K o for every U E U, because

U(S - T)U

-1

= A (S) - A (T) U U

for S, TE Ko . 2 ° If A is the product of von Neumann algebras A1, A 2 , we have 6(A1) c Ai and 6(A2) c A2 by lemma 2, and it suffices to prove the theorem for A1 and A2. We can therefore suppose that A is semi-finite or purely infinite. If A is semi-finite, the set of projections E of A such that AE is finite is increasing filtering with supremum I (chapter 2, proposition 5 and corollary 1 of proposition 7) , hence it is enough to consider the Finally, by lemmas 5 and 6, case where A is finite (lemma 5) . it suffices to prove the theorem when A is finite and

354

PART III, CHAPTER 9

A

G-finite and when

is purely infinite and G-finite.

3 0 Suppose that A is finite and G-finite, and let S, TE Ko . There exists a faithful finite normal trace on A ; this trace defines a pre-Hilbert norm II 112 on A. Let a = sup R 11 2 < + 00 . Let H c K 0 be the set of the Au (1/2(S +T)) REK0 for UE a; let H P c K0 be the weakly closed convex hull of H; by (i), H is invariant under the Au 's, hence so is H', hence H I = Ko by the minimality of Ko . For every E > 0, there exists an R C K0 such that a - c < II R1[2; as the norm II II 2 is lower semicontinuous for the weak topology (part I, chapter 6, corollary of proposition 2), there exists aUEU such that

H

a-

E <

A

1 U

1

(S + T))

- (A

A (T)) u (S) + u

2 2 Since a and Au(S) 11 2 a and Au (T) 112 11 2 is a preHilbert norm, there exists U such that Au(S) - A u (T) H2 is arbitrarily small. Now

H

H

H Au (s) -

Au (T) 11 2 =

H

H

H u(s _

T)u-1

2 . 11

H S - T 11 2 .

Hence S = T. 4 0 Suppose that A is purely infinite and G-finite, and let S, TE Ko . Let floe a weakly continuous linear form on A. Let a = sup Lf(R)I < + (13 . Arguing exactly as in 3 ° : for RE K0 every E > 0, there exists U E U such that a - E 11/2(f(Au (S)) + f(Au(T))) I; as If(Au (S)) I a and If(AU ( T)) 1 .. a, there exists UE U such that If(Au (S) - Au (T)) I = I is arbitrarily small. If (u(s - T)U -1 ) If Ko - K o {0 } , we can choose S, TE K0 such that S - T is a non-zero element of the centre of A (chapter 8, corollary 6 of theorem 1); then

U(S - T)U -1 = S - T for every UE U, and the above work shows that f(S - T) = 0. Hence S - T = 0, which is a contradiction. 0

Let H be a Hilbert space, A a c*-algebra of operators in H, B the weak closure of A, and 8 a derivation of A. There exists To E B such that To and 6(T) = [To, T] for every T E A. COROLLARY.

H

Proof.

H

H6H

This follows from lemma 4 and theorem 1.

References : [367], [368], [376], [404], [472].

355

FURTHER TOPICS

4. Automorphisms of von Neumann algebras In this section, we will make extensive use of the properties of the holomorphic functional calculus ([[9]], and N.BOURBAKI, Theories Spectrales, parts I and II, Paris, Hermann, 1967). Let E be a complex Banach space. We denote by L(E) the set of continuous linear operators in E. The classical (operator) norm on L(E) makes L(E) a Banach space; the corresponding topology on L(E) is called the norm topology. The set GL(E) of the invertible elements of L(E) is norm open in L(E). Let A be the set of z EC such that - TI < Im(z) < TT, and A' the set of SEC which do not lie on the non-positive real axis, i.e. which are not real, O. Let log:A' -4- A be the principal branch of the logarithm funciton; this is a bijection of A' onto A, and the inverse bijection is the mapping exp of A onto A'. Let A (resp. A') be the set of the X E L(E) [resp. GL(E)] such that Spx cA (resp. SpxcA'). Then A (resp. A') is a norm open exp x, subset of L(E) [resp. GL(E)] and the mappings x homeomorphisms of A onto A I and y .4. log y are mutually inverse and of A' onto A. [The operators exp x, log y are defined by Cauchy integrals; for every x€ L(E), exp x can also be defined by the series

1 r x". ] n>On.

LEMMA 7. Let X E L(E) y = exp x, E E E. (i)If xE = 0, we have yE = C. (ii)If yC = C, and if Spx doès not contain any of the points ±2iff, ±4iff, ±6iff, ..., we have xE = O. Proof. (i) is immediate. Suppose that Spx satisfies the condition of (ii). Let f be the entire function such that exp(z) - 1 = zf(s). We have y - I = (exp x) - I = xf(x), and o f(spx) = Spf(x), hence f(x) is invertible. Consequently, the D kernel of y - I is equal to the kernel of x, whence (ii). LEMMA 8. Let X E L(E) and y = exp x. Let F be the Banach space of continuous bilinear mappings of E x E into E. Define X I E L(F) and y l E GL(F) by

(x s i) (E l,

2) = x(f(E l , E2)) — =

(Ci , E2 EE, fE F).

2)

f(El , xE2),

-1

-1

Eli Y E2 ))

Let f E F.

If x 1 f = 0, we have y' f = f. (11) If Y s f = f, and if every z E SpX satisfies lim(z)I < 2Tr 3' we have x' f = Q. (i)

356

PART III, CHAPTER 9

The mapping y y' is a morphism of the Lie group GL(E) into the Lie group GL(F). It is a classical result that the corresponding morphism of the Lie algebra L(E) of GL(E) into x'. As y = exp x, we the Lie algebra L(F) of GL(F) is x deduce from this that y' = exp x'. Using lemma 7, it now suffices to show that, if Spx satisfies the condition of (ii), we have ±2iff, ±4iff, 4Spx'. Now, define x 0' x 1, x 2 E L(F) by

Proof.

(x 0f) (El, E2) = x(f(Ei,

(xif) (El, (x2f) Then he

x'

2) =

-f(xE l ,

E2) = -f(E 1 , xE2) .

xo + x l + x 2 , the xi's commute with one another, and Spx 1 c Spx 0 + Spx 1 + Spx 2 .

If X e c is such that

x

-

X is invertible, it is clear that the

xi - X are invertible, hence ESpx s ,

we have 1Im(2)1 < 3.

LEMMA 9. y = exp X.

Spxi Spx. = 2ff.

3

Consequently, for every

D

Let E be a complex Banach algebra, XE L(E),

(i)If x is a derivation of E, y is an automorphism of the algebra E. (ii)If y is an automorphism of the algebra E, and if lim(z) I < for every Z E SpXj then x is a derivation of E. Proof. Apply lemma 8, taking for f the multiplication of E. 0 LEMMA 10. Let E be a complex Banach algebra, and y an isometric automorphism of the algebra E such that I I y - I II < /i. (i) Every z E Spy lies in the open angle - — 2Tr < arg(z) < 3 3 which enables one to form x = log y. (ii) x is a derivation of E. Proof. The conditions on y imply that every ZE Spy satisfies IzI = 1 and lz - 11 < /T, whence (i). We have Spx = log(Spy), and so (ii) follows from lemma 9 (ii). 0 THEOREM 2. Let A be a von Neumann algebra, and (I) an automorphism of A. Suppose that H - H < rT. There exists a unitary element u o of A such that VT) = u0Tu-6 1 for every TEA. Proof. Let d = log (D, which is a derivation of A (lemma 10). By theorem 1, there exists To E

A

such that

357

FURTHER TOPICS

6(T) = LT0 (T) - RT0 (T) for every TEA, where LT0 (resp. Rib) denotes left- (resp. right-) multiplication by To in A. Let So = exp To, which is an invertible element of A. We have

cD(T) = (exp(LT0 + R_To ))(T) = (exp LTo .exp R_To )(T) (

7 m,nn

m n

1

T ) LT R _(T) m.n. 0 0

= ( X 7.71 - Trroi) T ( X Y2,0

(-Te )

X

m,no

m T T(-T ) m!n! 0 0 1

n)

-1 = S TS 0 0 Moreover, for every T S oT*s -0

E

1

A, -1* = 11(T*) = (T)* = S o

hence S*S commutes with T*. Consequently, 'S o l belongs to the 0 0 center of A. Hence if S o = UolSol is the polar decomposition of S o , we have (1)(T) = U 0TU6 1 for every TEA. D References : [103], [200], [216], [369], [404], [405]. replaced Theorem 2 was firstproved in [404], with the bound by 2. The method of the text (due essentially to J.-P. Serre) is a little shorter but does not enable one to obtain the bound 2 There exist factors which possess outer automorphisms chapter 7, exercise 15).

(cf.

•

APPENDIX I

1.

Spectrum

For every locally compact space Z, we denote by Lm (Z) the set of continuous complex-valued functions on Z which vanish at infinity. This set is an abelian *-algebra (the involution being defined by taking the complex conjugate function of any function). If, for every fE 40 (Z), we put II f = sup If()

CEZ 1(Z) is at the same time a complete normed algebra.

Let H be a complex Hilbert space, and y an abelian C*-algebra of operators in H. Let Z be the set of characters of Y, i.e. of the homomorphisms of y onto C. Endowed with the weak* topology of the dual of the Banach space Y, this is a locally compact space. For fe Lm(Z), there exists a unique element Tf of Y such that C(Tf) = f(C) for every c€ Z, and f Tf is an isomorphism of the normed *-algebra Lm (Z) onto the normed algebra y which preserves the natural orderings. We say that Z is the spectrum Tf is the GeZfand isomorphism of Lm (Z) onto Y. of Y and that f If y possesses an identity element, 40 (Z) possesses an identity element which must be the function 1 on Z. Hence Z is compact. Let X 4- g(X) be a continuous complex-valued function of the complex variable X; if fELco (Z), we have gtife 40 (z), and we put g(Tf.) = Tgo f; this agrees with the usual definition if g is a polynomial. Suppose that Y has a countable base for the norm topology (equivalently, that Y is separable); in other words, suppose that Y is generated, as a sub-C*-algebra of L(H), by a countable family of elements. Then, Z has a countable base. Indeed, there exists a sequence (fi ) norm dense in 14),(Z); let Z i be the open subset of Z consisting of the e Z such that Ifi(c) > 1 ; we show that the Zi's form a base for the topology of Z; let C o E Z, and let Y be an open subset of Z containing C o ; let f be a function of Lm(Z) equal to 2 at Co and to 0 on Z''-Y; let i be an integer such that < 1. Then, C o E Zi c Y, which f proves our assertion.

I

f II

359

APPENDIX I

360

2.

Spectril measures

Let X E H, y E H. The mapping f L(Z) , and we have

(Tfxly) is a linear form on

I (Trzly) H 11 f Il. lix II. Il y II. Hence there exists a unique bounded measure Vx,y on Z, called a spectral measure, such that Vx y (f) = (Tfœly) for fc Ito (Z). We have, for x, , y E H, X, E C, g, g' e 49 (Z)

I

(1) v

(2)

I x II. II y II ,

vs, y 11

xx+X'x' ,y V

(3)

(4)

= Xv

v

x ,y

,

x,y

y,x

x,x

(5)

+ X'v „

x,y

Te,Tgly

0,

= nnfV " x,y

Let us prove formula (5), for example. T gX,T g fy

g (f) =(T fT gx1T g. f y) = (T g--

I y)

For fc 40 (Z), we have y (g g

= (g g r V x, y ) (f) .

Formulas (2), (3) imply

4v

(6)

v . - iv =v + iv . x,y x+y,x+y x-y,x-y x+1,y,x+2,y and f E 40(Z) .

Let S E L(H)

.

. .

We have

( S TfXIY ) =(TfX1S *0 =Vx,s*y(f)

and

(TiSx I y ) =

Hence S commutes with V if and only if .0-x,S*y = VSx,y for any x E f-f, y E H. Finally, for every positive function f / 0 of 4,(Z) , there exists an X E H such that (TrHx) > 0. We conclude from this that the union of the supports of the vx,x 's is dense in Z.

3.

Extension of the Gelfand isomorphism.

Let L be the set of bounded complex-valued functions on Z, which are measurable for all the spectral measures. Then, L is clearly a *-algebra, and 40 (Z) is a *-subalgebra of L. We norm = Pif() If f e L, fis integrable for L by putting

IfI

CEZ

L

all the spectral measures (which are bounded), and we have

Iff(c)dy

x,y

() 1

f I - 11 vx, y II

I fII.IlxIl.II y I

361

APPENDIX I

Hence

(x,

y)

ff(C)dV

(C) is a continuous sesquilinear form

x,y

on H, so that there exists exactly one operator Tf E L(H) such

that (Tfx1y) =

x,y

-- Hfil. When

We have H Tfdl f € Loo(Z) , the notation Tf is consistent with the previous notation, being defined in the same way by the Vx,y 1 s. We have, besides, the following properties:

XT f

r

(i)

T Xf+V

(ii)

T

(iii)

Tfp = TfTf p;

( iv)

TE

(y)

Property (i)

whence

E L;

X,

XI E

C)

= T14

y";

Tfx,y

(Tyx1y)

X I Tfl (f,

f\)

x,y

is obvious. We have

4 .xly), = P(C)dv x,y (C) = ff(C)dv y,x (C) = (Tfylx) = (T p

(ii) .

If S E

, we have

f xIS*y) = (STf.rly) , (TfSX1y) = ff(c)dv sx,y (C) = ff (C) dv x,S*y (C) = ( T hence TfS = STf, whence (iv) .

(TgTfxly) = ff ( C)C1Vx whence

(v).

Finally,

(T

xl y)

ff(C)r (C)dv

ff'

whence

x_,y

If

gy

g

(C)

E Imo( Z)

ff(C)g(C)dVx,y (C)

(C) = ff (C) dv Tf fx„y (C) =

ft

x124)

'

(iii) .

A subset Y of Z which is measurable for every spectral measure has for characteristic function a function Xy of L such that xy = xy = Hence, there corresponds to it a projection Ey Of . An operator of L(H) which commutes with all the Er 's commutes with V; in fact, if g E Lop( Z) g is the norm limit of finite linear combinations of functions Xy : hence Tg is the norm limit of finite linear combinations of projections Ey , which proves our assertion. When V is the c*-algebra generated by a single hermitian operator T of L(H) , the projections Ey are called spectral

362

APPENDIX I

projections of T. They commute with every operator commuting with T; also, every operator commuting with the spectral projections of T commutes with T.

References : [18], [25], [28], [45], [100], [[9]].

1. Let T be an hermitian operator of L(H), and the sub-C*-algebra of L(H) generated by T and I. Show that the mapping C C(T) is a homeomorphism of the spectrum Z of Y onto a compact subset of R, namely the spectrum of T in the usual sense, which maps the function on Z associated with T to the C. function C

Exercises.

2. Let H 1 be the Hilbert space of square-integrable functions on [0, 1] for Lebesgue measure, and Ti the operator in H 1 of multiplication by the function x x. Let H2 be the Hilbert space of complex-valued functions f on [0, l ] such that f(x) 1 2 cf°, and T2 the operator in H2 of multiplica,1 xEL0,1j tion by the function x x. Let H = H 1 e H 2 , T the continuous linear operator in H which extends T1 and T2, and Y the C*algebra of operators generated by T. show that the spectrum of Y may be identified with [0, 1] (cf. exercise 1), and that a function which is measurable for every spectral measure is Lebesgue measurable. Deduce from this that the extension of the Gelfand isomorphism studied in section 3 does not have range equal to the von Neumann algebra generated by Y.

1

Reference : [[is]] , p. 65 (Cf. part 1.)

1,

chapter 7, proposition

APPENDIX II

Let K be a complex Hilbert space, and M the set of hermitian operators of L(K). Recall that M is endowed with a natural T if (Sx)x) (Tx)x) ordering. [If S M and TcM, we write S the set of the every N TEM such for x€ K.] We will denote by I. T that 0 Let F c M. Suppose that F is increasing filtering, i.e. that, for every pair S l , S 2 of elements of F, there exists an S E F such that S Sl, S S2. Suppose, further, that F is majorized, i.e. that there exists an hermitian operator majorizing all the operators of F. Then:

The set F admits a supremum in M, which is in the strong closure of F. For every S E F, let Fs be the set of the TE F majorizing S. It suffices to prove our assertion for Fs instead of F. We can therefore suppose that F is minorized (bounded below) as well as majorized, and, consequently, we are led to the case where F c N. Moreover, the Fs form a filter base. As N is weakly compact, this filter base has a weak limit point To E M. For every S E F, the set of hermitian operators majorizing S is S. Thus, T o majorizes weakly closed and contains Fs , hence To F and is in the weak closure of F. If T1 is another hermitian operator majorizing F, T1 majorizes the weak closure of F, and therefore majorizes T o . Thus, T o is the supremum of F. Finally, if x l , x2, ..., X EK, and TE F, we have

H

(T0 - T)xi

h2

H (T 0 - TOx h, 2i = ((To - T)xilxi) H (T0 - T)xi 11 2 's can be made arbitrarily

for every i, hence the small, which proves that To is in the strong closure of

F.

In the course of the proof, we also established the following point: if a majorant of F is in the weak closure of F, this majorant is the supremum of F. Reference : J.-P. VIGIER, Étude sur les suites infinies d'opérateurs hermitiens, Thèse no. 1089, Geneva, 1946, p. 12.

M

APPENDIX III

1.

Support of an operator.

Let H be a complex Hilbert space, and TE L(H) . The set of Let X X E H such that Tx = 0 is a closed linear subspace of H. be its orthogonal complement. The projection E = Px is called Ey E X 1 , hence the support of T. For every y E H, we have y T(I - E)y = O. Consequently, T = TE. Conversely, let El be a projection of L(H) such that TE 1 = T; we have, for every s EH, T(i - El)g = 0, hence (I - E1)2 X I ; hence I —E i I E, Thus, E is the smallest of the projections E 1 of L(H) such that TEi = T. -

-

The closure of T(H) is a closed linear subspace Y of H. Let F = Py. It is clear that F is the smallest of the projections F1 of L(H) such that FlT = T or, which comes to the same thing, T*Fi = T*. Hence F is the support of T*.

2.

Partial isometries.

Let U E L(H), and E its support. We say that U is a partial isometry if U is isometric on X = E(H). Then, U(H) = U(X) is a closed linear subspace Y of H, and U maps X isometrically onto . Y. Let F = Py. We say that E (resp. X) is the initial projection (resp. the initial subspace) of U, and that F (resp. Y) is the final projection (resp. the final subspace) of U. Let

,r E

X, y = UX

E

V. For every

2E

14, we have

(xis) = (xlEs) = (UxIUEs) = (yIU2),

hence

x = U*y.

Thus, the mapping x Ux of X onto y has for its inverse (isometric) mapping the mapping y .4- U*y of Y onto X. Since,' furthermore, the support of U* is F, we see that U* is a partial isometry, with initial projection F, and final projection E. We also see that U*U = E, UU* = F. Conversely, let V E L(H) be such that V*V is a projection G. Then, for every xE H, we have

366

II vx 11

=

v*vx ix) = (Go: ix) = II Gx II ,

hence V is isometric on G(H) and zero on G(H) 1 , which proves that V is a partial isometry. Similarly, if WE L(H) and if WW* is a projection, W* is a partial isometry, and hence W is a partial isometry.

3.

Polar decomposition of an operator.

Let TE L(H), E the support of T, F the support of T*, X =E(H) ITI = (T*T) 12 . We have, for every x€ H, and Y = F(H). We put:

II

Tx

H2

= (T*Tx1x) = H ITlx

II.

Hence ITI has support E and, consequently, ITI(H) = X. Furthermore, the mapping ITlx 4- Tx is a linear isometry of ITI(H) onto T(H), and therefore extends to a linear isometry V of X onto Y. Let U be the partial isometry with support E which coincides with V on X; this partial isometry has E as initial projection, and F as final projection. We have T = UIT1, an equality called the polar decomposition of T. On the other hand, if we have an euqality T = UlTi, where T1 is positive hermitian and where U 1 is a partial isometry whose initial projection is the support of 2 hence T = 1T1, and then T l' then we have T*T = T 1 UU 1T 1 = Tr, 1 U 1 = U. We have

T*

ITIU* = U*(UITIU*);

the operator UITIU* is positive hermitian, with support F -; and U* is a partial isometry, with initial projection F, and final projection E. Hence the eugality T* = U*(UIT1U*) is the polar decomposition of T*. Thus,

1T*1 = UIT1U*,

T = U*IT*1U.

Reference : J. VON NEUMANN, Uber adjungierte Funktionaloperatoren (Ann. Math., 33, 1932, 294-310).

APPENDIX IV

If Z is a locally compact space and V a positive measure on Z, we will denote by L7(Z, v) the set of complex-valued functions on Z which are measurable and essentially bounded with respect to V, and we identify two functions which are equal locally almost everywhere; we will denote by 1_00(Z) the set of continuous complex-valued functions on Z which vanish at infinity. The following result is due to J. von Neumann [Einige Satze über messbare Abbildungen (Ann. Math., 33, 1932, 574-586)] in the more general case where Z and Z l are separable complete metric spaces.

Let z and z 1 be two second-countable locally compact spaces, a positive measure on z, v l a positive measure on z 1 , and (1) an isomorphism of the *-algebra r_;(z, v) onto the *-algebra 1.7(z i , v 1 ). Then, there exist:

1° a v-negligible set N in z, and a v 1 -negligible set N 1 in Z • 2 0 a Borel isomorphism of Z , N onto Zr-N i which transforms v

into a measure equivalent to v l , with ((f)) (fl()) = f(c) for every f E Lcc° (Z, V) and every EZ We begin with several comments: 1 0 (I) is isometric. For (10 maps positive funcitons, which may be written in the form gg, to positive functions; and is the least number a > 0 such f that a 2 ff > O. 2 ° We can replace v and v l by equivalent positive measures, hence in particular (since Z and Zl are countable at infinity) by bounded measures. Then, adjoining points at infinity to Z and Z l , which carry zero mass, we see that we are reduced to the case where Z and Zl are compact (and second-countable), as we henceforth suppose. 3 ° Let Y be the support of v, and Y1 the support of V 1 . It is enough to prove the proposition for the spaces Y, Y 1 and the measures induced on Y, Y1 by v and V1. We will therefore suppose henceforth that the supports of y and yl are Z and Zl, which allows us to identify and L(Z 1 ) with *-subalgebras of L7(Z, v) and 1.;,(Z1, v 1 ). 367

368

APPENDIX IV

c (Z, V) generated by Let A be the norm-closed *-subalgebra of Lœ L(Z) and 4)-1 (4,(Z 1 )). Let A1 = 4)(A), which is the closed *c (Z i , v 1 ) generated by 4,0 (Z 1 ) and 4)(L(Z)). subalgebra of Lœ Then, A and A 1 are norm-separable. Let Z r be the spectrum of the Banach algebra A, a second-countable compact space. Let 8 be the G-elfand isomorphism of L » (Z r ) onto A. The linear form

f 4. f fdl.) on A, transferred to L(Z r ) by means of

e-1 ,

is a posi-

tive measure v r on Z' with support Z'. The canonical injection of Loo (Z) into A defines, by taking the transposed operator, a continuous mapping of Z' into Z; for f € 49(Z) , fo 0 = 3-1 ( f) , so that 0 must be surjective. It is clear that O (v' ) = V. Every function f E A defines, on the one hand the function g = foe E Lc'c3' (Z r V') , and, on the other hand, the function h = 8-1 ( f) E L,D0 (Z f ) ; we shall see that g - h is y'-negligible; we already know that g = h if f € Lx,(Z); in the general case, let (fn ) be a sequence of functions of 49 (Z) such that

e

,

JIf

- fn I 2 dv -÷ 0, and let gn = fn oe. 2 12 n dv, j ig - gnI dv' = flf - fl

We have since

v=

We have

f

(h - gn )

(T1

g;1 )dY r = f(f - fn) (f -

41)dv

bx definition of y r . Hence g and h are both limits of gn in 1,(Z', V') , whence our assertion. Since Z and Z r are compact and second-countable and V = O (v' ) , there exists a disintegration X c of v r relative to part VI, chapter 3, theorem 1) . Let d be a metric on Z' cornvatible with the topology. Let p be an integer > O. We are going to show that the support of X c has diameter L 1/p, belonging to a negligible set Nn . except for Indeed, let h l , h2, hn be positive functions of L('), whose supports X1, X 2 , ..., Xn have diameter < l/p, and are such that

h. > 0 at every point of Z'.

Let

i=1 fi =e(h i ),gi =f1.00,7<.=inf(g i , The functions

g., h., k. are

hi ).

equal v'-almost everywhere. We

g . > 0 almost everywhere, hence, if Yi denotes the set i=1 of E Z such that fi(c) > 0, Z is the union of the Y's, to within a v-negligible set. Moreover, have

APPENDIX IV

f

369

dV(C)fki (OdyC r ) = fki(Odv f (C 1 ) = fgi (OdV I (C) = fdV(C)fgi (C)Ot c (C),

ki

hence (remembering that

gi )

fk.(C)dX (C) = fg .(C)Ot (C) 2c v-almost everywhere on

Z.

Now, for C' E

e-1 (Y) n (z'

, we

have

= 0

gi (c') > 0;

and

hence, almost everywhere on Z, Xc is concentrated on (Z' 0-1 (yi )) uxi ; if, further, C E Yi, X is concentrated on 0-1 (Y-); hence, almost everywhere on Y-, 2 X C is concentrated on X. and consequently its support has diameter - 1/p. This being true for every i, our assertion is proved. Let N be the v-negligible union of the Np 's. For c (IN, the support of Xc has diameter zero, and hence X c is a point-mass. There exists, for every CE ZN, N, a point 0 1 (C) E 0 -1 (fC)) such that X c = ce'(). As Xc depends measurably on c, the same is true of ef(c). The image of v under er is

f

E

e r (C)

dV(C) = fX dV(C)

= V I

.

The set N' = Z'N--0'(Z), whose inverse image under 0' is empty, is therefore v'-negligible. Thus, 0' and the restriction of 0 to Z I N.N' are mutually inverse mappings of ZN, N onto and of Z'N,N' onto Z'-N, which transform .9 into v' and '9' into v; also, the isomorphism of f(Z, .9) onto IZ(Z', v') which they define reduces to 0-1 on A. Furthermore, Z is the union of a sequence of pairwise disjoint subsets Zo, Zl, Z2, ..., with Zio and eflzi continuous for i > 1. Hence, enlargingv-neglib N and N' if necessary, we can suppose that 0 F IZN■ N is a Borel isomorphism of Z ■ N onto Z'N,N'.

Arguing similarly on A l , we can define, on the spectrum Zi of A l , a positive measure vf with support a v 1 -negligible set N 1 in Z l , a 91-negligible set NI in ZI, and inverse Borel isomorphisms el, e l of Z 1 '-N1 onto ZiN, NI and of Z1 N J onto Z1 ,...N 1 , which transform V1 into VI and Vi into V1, and which define an isomorphism of q(ZI, vi) onto q(Z i , v1) reducing to 11 on Al, where 0 0 1 denotes the Gelfand isomorphism of LD (ZI) onto Al. We are thus reduced to proving the proposition when A

=

A

1

= LD

(Z

1

).

370

APPENDIX IV

Then, the isomorphism of T(Z) onto L(Z1) induced by (D defines a homeomorphism n of Z onto Zl such that ((f)) (ri()) = f(r) for every fe Ito (z) and every Cc Z. Identifying Z with Zl by means of fl, we are led to the case where Z = Zl and where (D induces the identity mapping on 40 (Z); it then suffices to prove the following lemma, which holds without the countability hypothesis:

Let Z be a locally compact space, y and v l positive measures on z with support z, and (I) an isomorphism of 1(Z, y) onto IZ(Z, v 1 ) which reduces to the identity on 40 (z). Then, y and v l are equivalent [so that q(Z, V) = Lm(Z' V1)], and is the identity mapping. LEMMA.

Let f be a lower-semicontinuous bounded positive function on 00 Z. Then, f is the supremum in q(Z, y) and in L c (Z, V 1 ) of the functions of 1_.(Z) majorized by f. Hence (1)(fD = f. Now, let g be a positive function of 1(Z, yl) with compact support. There exists a decreasing sequence of bounded, lower semicontinuous functions gn g, such that

f

g ndy jgdY,

fg ndy i jgdy i .

Then, g is the infimum of the gn 's in IZ(Z, v) and IZ(Z, V 1 ). Hence: 1 ° if g is y-negligible, g is V1-negligible and conversely, which shows that V and V 1 are equivalent; 2 ° we have ((g) = g. Finally, every positive function hErj(Z, V) = V1) is the supremum in IZ(Z, V) of the positive functions with compact support of 1_,(Z, V) which it majorizes; hence (1)(h) = h. 0

APPENDIX V

1.

Borel sets.

Let E be a topological space. A set T of subsets of E is called a tribe if it satisfies the following conditions: a. the complement of every set of T belongs to T; b. every countable intersection of sets of T belongs to T. Every intersection of tribes on E is a tribe on E. There therefore exists a smallest tribe containing the set of closed sets of E; the sets of this If E is the Hausdorff, tribe are called the Borel subsets of E. the subsets of E of the form A1 n A2 n A3 n ..., where each Ai is a countable union of compact subsets of E, are Borel subsets of E called KG 6 sets in E.

2.

Polish spaces.

A topological space E is said to be polish if its topology is second-countable (i.e. has a countable base), and if there exists a metric compatible with the topology of E for which E is complete. Every countable product of polish spaces is a polish space. Every space E which is the (disjoint) sum of a countable family (Fn ) of polish spaces is polish. Indeed, for each n, let dn be a metric on Fn compatible with the topology of Fn and for which Fn is complete, such that dn (x, y) 1 in Fn X Fn ; we define a metric d on E compatible with the topology of E by putting d(x, y) = 1 if x and y do not belong to the same set Fn , and d(x, y) = dn (x, y) if x and y are both in F. We immediately check that every Cauchy sequence in E is convergent, and that E contains a countable dense set, whence our assertion. Let E be a polish space, and F a subspace of E. If F is closed in E, then F is polish. Suppose now that F is open in E. Let F' = E•■ F, and let d be a metric on E compatible with the topology of E for which E is complete. For x, y cF, put d(s, y) =

d(x, y)

Id(x , F 1 ) -1 371

d(Y F') -1 1

APPENDIX V

372

It is immediate that 6 is a distance on F compatible with the topology of F. Moreover, if (xn ) is a Cauchy sequence in F for 6, then xn converges in E for d to an element x; the numbers d(xn , F') are bounded below by a number > 0, and so x E F; also, xn converges to x for 6. Hence F is complete for 6, so that F is polish. The set N of natural numbers, regarded as a subspace of R, is a complete metric space. Hence NN, endowed with the metric

co

d( (m.) , (n .) ) =

2, 2 i=0

-

Im- - n-I +Im. - n .1 '

1,

2

is a complete metric space. If E is a polish space, there exists a continuous mapping of NN onto E. In fact, for every > 0, there exists a countable covering of E consisting of non-empty closed sets of diameter < E, since E is secondcountable. Consequently, there exists a mapping (no , n) F(no, flu ) of the set of finite sequences of integers ? 0 into the set of non-empty closed subsets of E, possessing the following properties: 1 ° F(cp) = E; 2 ° for any sequence (n o , np), we have

0'

np) = F (n 0'

, n

0) u F (n

0'

, n , 1) u ...; p

0

the diameter of each of the F(no, 2 -P for every np ) is p O. This established, for every x = (mo, mi, ...) ENN , the sequence of the F(mo, ..., mn ) for p = 1, 2, ..., is a Cauchy filter base on E, and therefbre converges to a point of E that we denote by f(x). Every point of E belongs, for any p 0, to a' set Fp = F(mo, mp ), and is therefore a limit of the filter base (F0, F l , ...); thus, f(NN) = E. If x = (Mo, rnp ...) and y = (no, ni, ...) are two elements of NN such that d(x, y) < 2 , we have mo = no , ml = nl, ..., mi - ni, hence f(x) and f(y). are both in F(mo , ..., mi) and, consequently, are distant < 2 -1' from each other. This proves that f is continuous.

3

3.

Souslin

sets.

Let E be a polish space. We say that a subset of E is

Sous lin is it is the image of a polish space P under a continuous mapping of P into E. The image of a Souslin subset of E under a continuous mapping of E into another polish space is Souslin. Let (An ) be a sequence of Souslin subsets of E. Then, the union A of the As is Souslin; for, let P n be a polish space and fn a continuous mapping of P into E such that fOlDn ) = An ; let the space P be the (disjoin4 sum of the Pn 's, which is

373

APPENDIX V

polish; the mapping f of P into E, equal to fn on Pn , is continuous, and f(P) = A, so that A is Souslin. Similarly, the intersection A r of the An 's is Souslin; for let P r be the product space of the Ps, which is polish; let Q be the subspace of P' consisting of the (xn )'s such that frn (xm) = fn (xn) for every pair of indices m, n; then Q is a closed subspace of P', and is therefore polish; if pn denotes the canonical mapping of P' onto Pn , the restriction f' of fop n to Q is independent of n, is continuous, andr(Q) = A', hence A' is Souslin. Thus, the set of subsets of E which, together with their complements, are Souslin is a tribe which, by section 2, includes the closed subsets of E. Hence every Borel set of E is Souslin. Let A be a relatively compact Souslin subset of E. We shall see that there exists a compact space F, a K0- 6-set in F, B, say, and a continuous mapping f of F into E such that A = f(B). Replacing E by A, we can suppose that E is compact. Let g be a continuous mapping of NN into E such that g(NN) = A. Let R be the one-point compactification of the real line. The space NN is a subspace of the compact space RN , and it is easy to see that NN is the intersection of a sequence Gl, G2, ..., of open -v. E, which is compact. Let B be the subsets of R" Put F = graph of g, B its closure in F, and f the canonical projection Moreover, as g is continuous, of F onto E. We have A = f(B). B is closed in NN X E, hence B = B n (NN X E). Now, i is compact, and

NN x E = (G x E) 1

n (G

2

E)

n ...;

finally, Gn is a countable union of closed and therefore compact subsets of RN, hence G X E is a countable union of compact subsets of F.

4.

Measurability of Souslin sets.

Let E be a second-countable locally compact space, V a measure 0 on E, and A a Souslin subset of E. We are going to show that A is V-measurable.

We can confine attention to the case where A is relatively compact. There then exists a compact space F, a continuous mapping f of F into E, and a family (Bn,p ) of compact subsets of F such that, putting

B

n =

B

uB

n, 2

u...

and

B = B

l

nB

2

n

we have A = f(B). We can further suppose that 5 n_t 1cB n, 2c... for every n. We shall show that for every a < V*(A), there a. As f(C) is exists a compact set CcB such that V(f(C)) compact, this will complete the proof.

374

We first show the existence of a sequence (p l , p2, ••.) of integers such that, if we put Cn = B n B 1,10, 1 n nBn . p , we have V * ( f(Cn )) > a for every n. Suppose 'tie Bi ,p . havg been determined for i < n. As Cn _i cE c Bn , we have 7' = (Cn _i n Bn, i ) u (Cn _ i n Bn,2 ) u..., hence V*(f(Cn_i n Bn, i) ) + co. Hence converges to V*(f(Cn _i)) as i V * ( f(Cn_inBni p n )) > a for pn chosen suitably, which proves our assertion. Then put C=C nC 1 2

. . = B

1 ,pi

n B 2P

As the sequence of the Bi p i n . nBn ,pn 'S is a decreasing sequence of compact subsets i with intersection C, f(C) is compact and equal to the intersection of the f(Bi ,pi n... nBn,pn ), hence

v(f(C)) = iiraV(f(B i,pi n

5.

n B

n,Pn

)

limV * ( f(Cn )) 1. a-

Existence of measurable mappings.

Since N is well ordered, we can endow NN with the lexicographic ordering. Then, NN is totally ordered. We show that every non-empty closed subset P of NN admits a least element. Let no be the smallest integer such that there exists an element of P of the form (no ,...). Let ni be the smallest integer such that there exists an element of P of the form (no , n l , ...); etc. Then, xo = ( no, n i , n 2 , ...) x for every x e P; moreover, xo is in the closure of P; hence Xo E P since P is closed. Let QcNN be the countable set of elements of NN all but a finite number of whose coordinates are equal to zero. Then, every open set V in NN is a union of intervals [2 1 , 2 2 E, where Indeed, let x = ( no, n l , ...) E.V. There exists 21 EQ, 22 EQ. an integer i such that every element of NN , whose coordinates with indices 0, 1, ..., i are equal to those of x, belongs to V. Let

2 1 = ( no , n l , ..., ni , ni+1 , 0, 0, ...),

22

=

( n o , n l , ..., n i , n i+1 + 1, 0, 0, ...).

We see at once that xE

[21, 22[ CV,

whence our assertion.

This settled, we shall now establish the following result: Let Z be a second-countable locally compact space, a positive measure on Z, Z' a subset of Z, and G a polish space. For every CEZ, let Gc be a subset of G, non-empty for C E Z i . Suppose that the setEof pairs (C, x)EZ)
APPENDIX V

375

Z containing Z', and a V-measurable mapping n of Z" into G, such E Z". that n(c) E G c for every We need to construct a v-measurable subset Z" of Z containing Z' and a V-measurable mapping n 1 of Z" into Z X G, with values in E, such that fffl i ( C) = C for CE Z", denoting by Ti the canonical mapping of Z x G onto Z. Regarding E as a continuous image of NN we are led to the following situation: we have a continuous mapping f of NN into Z, such that A = f(NN) DV, and we are going to show the existence of a V-measurable mapping g of A into NN such that fog is the identity mapping of A. This will complete the proof. Now, for every xE A, f-1 ({x}) is a closed subset of NN. Let g(x) be its smallest element, so that fog is indeed the identity mapping of A. We show that g is V-measurable. By what we have seen above, it suffices to show that, for every 2 ENN g-1 (1f- , 4), where If- , z[ = fXEN N : X < zl, is V-measurable. Now, we easily see that this set is equal to fd -f- , 4), hence is Souslin and, consequently, V-measurable. ,

References : [80]; and N. BOURBAKI, Topologie genérale, part IX, 2nd. edition, Act. Sc. Ind., No. 1045, Paris, Hermann, 1958.

M

SOME COMMENTS REGARDING TERMINOLOGY

Generally, our notation is in line with current usage, and usually is in agreement with that of N. Bourbaki. Below, we clarify certain points, in adequate detail. Nevertheless, the reader should not imagine that these explanations are enough to prepare him for reading the book, if he lacks the knowledge mentioned in the Introduction.

Set theory. A finite set is regarded as countable. Let A and B be two sets, and f a mapping of A into B. We say that f is injective if x / y implies f(x) / f(y). we say that f is surjective if the image f(A) of A under f is equal to B. We say that f is bijective if f is injective and surjective. The mapping f is also denoted by x 4- f(x). Let A be a set, and f a mapping of A into A. A subset A' of A is said to be stable (or invariant) for f if f(A') CA'. Let A, B, C be three sets, f a mapping of A into B, g a mapping of B into C. The composite mapping of A into C is denoted by gof or gf. We denote by N the set of integers 0, by R the set of real numbers, and by C the set of complex numbers. Let A and B be two sets, and f a mapping of A into B. If B=R (resp. C) we say that f is a real-valued (resp. complex-valued) function. If A = R, we that that f is a function of a real

variable. Topology. A topological space Z is said to have a countable base, or to be second-countable, if there exists a sequence (V 1 , V2, ...) of open subsets of Z such that every open subset of Z is the union of some of the V-2 1 s. When Z is metrisable, it comes to the same thing to say that Z contains a dense sequence. When Z is a

377

378

COMMENTS ON THE TERMINOLOGY

Hilbert spacd, it comes to the same thing to say that Z contains a countable orthonormal basis, and so the terminology cannot give rise to any confusion.

Algebra. A vector space E over the field of complex numbers is called a complex vector (or linear) space. Let E and F be two complex vector spaces. A mapping f of E into F is said to be linear if

f(Xx + py) = Xf(x) + pf(y) for any x, y E E, X, p E C. The vector subspace f(E) of F.

rank of

f

is the dimension of the

A linear mapping of E into E is also called a linear operator in E, or just an operator in E. An operator of the form x Xx (where X is a fixed element of C) is called a homothetic mapping, or a scalar operator; it is denoted by X when no confusion is possible. The operator 1 (or I) is called the identity

operator. Let E* be the algebraic dual of E, and (x', x) the value at a point x of E of the linear form X r E Et. Let A' be a vector subspace of E*. The restriction to E X A' of the mapping (x, x') is called the canonical bilinear form on (x, x')

E

X Ar .

An algebra over C is a complex vector space A, endowed with a bilinear mapping (X, y) -4- xy of A X A into A, such that (xy)2 = x(y2) for any x, y, 2E A (we only consider associative algebras). We say that A is abelian if any two elements x, y of A commute. In the general case, the centre of A is defined to be the set of the XE A which commute with all the elements of A. We call an involution in A a bijective mapping x -4- x* of A onto A such that

x** = x, (Xx)* =

(x + y)* = x* + y*,

(xy)* = y*x*,

for

x, y

E

A,

X EC.

An algebra over C endowed with an involution is called an involutive algebra or a *-algebra.

Topological vector spaces. Let E be a complex vector space. A seminorm on E is a function p defined on E, with values 0, such that p(Xx) = PIP(x) and p(x + y) p(x) + p(y) for any x, y EE, X E C. If, further, p(x) = 0 implies that x = 0, we say that p is a norm; we generally use the notation X in this case, and E is known as

H H

379

COMMENTS ON THE TERMINOLOGY

a normed complex vector space. The set of x e E such that H x H 1 is called the unit ball of E. In a complex topological vector space E, a set A is said to be total if the smallest closed vector subspace of E containing A is E itself. A complex pre-Hilbert space is a complex vector space H endowed with a scalar product [that is denoted by (xly) in this book], i.e. a mapping (x, y) .4. (xly) of H X H into C satisfying the conditions

(y1 x) = (x ly),

(Xx + pylz) = X(x12) + P(y12),

(x, x)

0

(x lx) 1/2 is a for any x, y, Z E H, X, J EC. Then, the function x seminorm on H, and we always endow H with the topology defined by this seminorm. If this topology is Hausdorff, i.e. if this seminorm is a norm, we have a complex Hausdorff pre-Hilbert space. If, further, H is complete, H is called a complex Hilbert space. Let H be a complex Hilbert space, T a continuous operator in 1. We H. We denote by HT H the supremum of H Tx H for H x H denote by T* the unique continuous operator in H such that (T*xly) = (xITy) for any x, y€ H. If T = T*, we say that T is hermitian. If TT* = T*T = I, we say T is unitary. If T maps the unit ball of H into a strongly compact subset of H, we say that T is compact. (Remember that the terminology employed by numerous authors is "completely continuous"). An involution in H is a bijective mapping J of H onto H such that

2 J x = x, for any x,

J(x + y) = Jx + Jy,

y

E

H, XE C.

J(Xx) = XJx,

(JxIJy) = (y Ix)

•

INDEX OF NOTATION

The reference numbers indicate the part, the chapter and the section (or, in a few cases, the exercise) , in that order.

L(H)

IH : 1

M' ppl

lz

cf m

PX A Em

I, 6, 2. LJ- ( ), L l : I, 6, 10.

11,s H2 :

:

1 1 1 1 1 1

,

11 f1114 ,

,

H

fui: 1 , 6, 10.

40(z) : I, 7, 1 • f(Z, v) : I, 7, 1.

,

,

e

,

J

,

H(C)dV(C) : II, 1, 5.

S e

,

:

II,

1, 5.

e

XA : I, 1, 4.

I extUdy(C)

Ex A : I, 1, 4.

f

y H(c)dv(c)

XA x : I, 1, 4. M 2 : I, 1, 6. M2 : I, 1, 6. M° : I, 1, exercise 6. I, 1, exercise 6. TrIA: I, 1, exercise 10. Tx : I, 2, 1. T : I, 2, 1. : I, 2, 1.

J

J

A(C)0N(C)

: II, 3, 2.

e iTcdy(C) U(C)dv(C)

E >-F X— Y X Y X y

2,-

T4 f. T CA 381

II, 2, 3.

:

E —F E -‹ F

.

1, 5.

T(C)dv(C)

e f (p c dv(c)

An

II,

e

fe

iEI Al x A 2 x ... x : 1, 2, 2. A 1 elA 2 O... ®A I, 2, 4. 3. wx., y : I 3 Wx : I, 4, 1. q5 ip : I, 4, 1. x.f, f.x, f*, Ifl: I, 4, 7. y x , vx : I, 5, 1. U(U), V(U) : I, 5, 1. ep : I, 6, 1. (SIT) : I, 6, 2.

:

11, 1, 5.

e

J

H Ai : 1, 2, 2.

:

:

II, 3, 6.

: II, 4, 2. :

II, 5, 1.

1, 1, 1, 1, 1, 1, 4, 4, 5, 6,

1. 1. 1. 1. 1. 1. 1. 4. 1. 1.

.

r

INDEX OF TERMINOLOGY

The reference numbers indicate, in succession, the part, the chapter and the section (or, occasionally, the exercise). Abelian projection I, 8, 2 Absolute value of a linear form I, 4, 7 Affiliated to a von Neumann algebra (operator) I, 1, exercise 10 Algebra Hilbert I, 5, 1 quasi-Hilbert I, 5, 1 von Neumann I, 1, 1 von Neumann, generated by a set I, 1, 1 von Neumann, 0-finite I, 1, 2 von Neumann, a-finite over its centre I, 9, exercise 5 Algebraically antiisomorphic I, 1, 5 isomorphic I, 1, 5 Almost dominated (positive linear form) I, 4, exercise 8 Ampliation I, 2, 4 Antihomomorphism I, 1, 5 Antiisomorphic I, 1, 5 Antiisomorphism I, 1, 5 algebraic I, 1, 5 spatial I, 1, 5 Associated, left- or right-, with a quasi-Hilbert algebra (von Neumann algebra) I, 5, 1

I, 7, 1 Basic (measure) Bicommutant I, 1, 1 Binormal III, 3, exercise 3 Borel III, 1, 1 Bounded relative to a Hilbert algebra I, 5, 3 right- or left- relative to a quasi-Hilbert algebra I, 5, 2 C*-algebra of orators I, 3, 4 Canonical mapping of U into U(U) or V(U), (U a quasiHilbert algebra) I, 5, 1 Central relative to a Hilbert algebra I, 5, 4 Central weight III, 8, 3 Character of a finite von Neumann algebra III, 5, 3 Characteristic projection I, 5, 4 Commutant I, 1, 1 Continuous (von Neumann algebra) I, 8, 1 Coupling operator III, 6, 1 Cyclic element I, 1, 4 projection I, 1, 4 set I, 1, 4

383

384

INDEX OF TERMINOLOGY

Decomposable Hilbert algebra II, 4, 2 homomorphism II, 3, 6 linear mapping II, 2, 3 operator II, 2, 3 von Neumann algebra II, 3, 2 Decomposition, polar I, 4, 7 and Appendix III Derivation III, 9, 1 Determinant associated with a trace I, 6, 11 Diagonalisable operator II, 2, 4 Dimension, relative III, 2, 7 Direct integral of a measurable field of Hilbert spaces II, 1, 5 Direct sum (Hilbert algebra) I, 5, 5 Discrete von Neumann algebra I, 8, 1 Equivalent projection III, 1, 1 Essentially bounded measurable field of linear mappings II, 2, 3 Essentially dense III, 1, exercise 13 Factor I, 1, 1 Faithful positive linear form I, 4, 1 trace I, 6, 1 Field Borel, of Hilbert spaces II, 1, 3 constant, of Hilbert spaces II, 1, 3 continuous, of Hilbert spaces II, 1, exercise 5 measurable, induced II, 1, 2. measurable, of Hilbert algebras II, 4, 1

measurable, of Hilbert spaces II, 1, 3 measurable, of homomorphisms II, 3, 6 measurable, of linear mappings II, 2, 1 measurable, of operators II, 2, 1 measurable, of orthonormal bases II, 1, 4 measurable, of subspaces II, 1, 7 measurable, of traces II, 5, 1 measurable, of von Neumann algebras II, 3, 2 measurable, vector II, 1, 3 of Hilbert spaces II, 1, 2 square-integrable vector II, 1, 5 vector II, 1, 2 vector, measurable on a measurable subset II, 1, 3 Finite projection III, 2, 1 trace I, 6, 1 von Neumann algebra I, 6, 7 Full Hilbert algebra I, 5, 3 Fundamental projection III, 8, 2 Fundamental sequence of measurable vector fields II, 1, 3 Hilbert-Schmidt operator I, 6, 6 Homogeneous von Neumann algebra III, 3, 1 Homomorphism I, 1, 5 Z, v) [resp. canonical of L(C (Z 40(Z)] into the algebra of diagonalisable (resp. continuously diagonalisable) operators II, 2, 4 Hyperfinite III, 7, 2

INDEX OF TERMINOLOGY

Induction I, 2, I Induced Borel structure II, 1, 1 von Neumann algebra I, 2, 1 Infinite projection III, 2, 1 von Neumann algebra I, 6,7 Involution canonically defined by a quasi-Hilbert algebra I, 5, 1 Isomorphic I, 1, 5 Isomorphism I, 1, 5 algebraic I, 1, 5

385

Partial isometry Appendix III Positive linear form I, 4, 1 Predual I, 3, 3 Product of von Neumann algebras I, 2, 2 tensor, of Hilbert algebras I, 5, 5 tensor, of von Neumann algebras 1, 2, 4 Projection, final or initial Appendix III Properly infinite projection III, 2, 1 von Neumann algebra I, 6, 7 III, 7, 7 Property Purely infinite projection III, 2, 1 von Neumann algebra I, 6, 7

r

canonical of f H(OdV(C) 0 onto f H(C)&1(C) II, 1, 5 Gelfand Appendix I of measurable fields II, 1, 3 spatial I, 1, 5 Measurable envelope I, 2, 6 Measure positive II, 1, 1 spectral Appendix I Minimal projection I, 8, 2 Multiplicity III, 3, 1 Mutually singular normal positive forms I, 4, 6 Norm topology I, 3, 1 Normal positive linear form I, 4, 2 positive linear mapping I, 4, 3 trace I, 6, 1 von Neumann algebra III, 7, exercise 13 Z-trace III, 4, 1 Order of a von Neumann algebra III, 3, exercise 5

Reduced homomorphism or antihomomorphism I, 2, 1 von Neumann algebra I, 2, 1 Restricted two-sided ideal I, 1, exercise 6 Reversed Hilbert algebra I, 5, 5 Semi-finite projection III, 2, 1 trace I, 6, 1 von Neumann algebra I, 6, 7 Separating element I, 1, 4 set I, 1, 4 Spatially antiisomorphic I, 1, 5 isomorphic I, 1, 5 Spectrum Appendix I Standard Borel space II, 1, 1 measure II, 1, 1 von Neumann algebra I, 5, 5 Strong topology I, 3, 1 Subordinate Borel structure II, 1, 1 Subspace, final or initial Appendix III

386

INDEX OF TERMINOLOGY

Support central I, 1, 4 of a normal positive linear form I, 4, 6 of a normal trace I, 6, 1 of an operator Appendix III Trace canonical . III, 4, 4 natural I, 6, 2 on A+ I, 6, 1 on a two-sided ideal I, 6, 1 on 2+ III, 4, 2 to within C III, 8, 4

Trace-element I, 6, 3 Type of a von Neumann algebra I, 3, 4 and III, 3, 1 Ultra-strong topology I, 3, 1 Ultra-weak topology I, 3, 1 Uniform von Neumann algebra III, 3, exercise 5 Weak topology I, 3, 1 Weight III, 8, 3 Z-trace, canonical Z-trace III, 4, 1 and III, 4, 4

REFERENCES

JOURNAL ARTICLES The first part covers the entire period up to around 1954. The articles are arranged in alphabetical order of authors' names. The second part more or less covers the period from 1954 to 1980. The articles are here classified by year. For some time, it has become more and more artificial to separate work devoted to von Neumann algebras from that dealing with C*-algebras. The distribution of references between this book and the second edition of my book on C*-algebras (Cahiers scientifiques, fascicule XXIX, English translation, NorthHolland 1977) is therefore somewhat arbitrary. Several papers from the field of theoretical physics are listed, which have to do with von Neumann alqebTas, but we have not attempted to give a complete list. FIRST PART

[1] W. AMBROSE, Structure theorems for a special class of Banach algebras,Trans.Amer.Math. Soc. 57(1945) 364-386. [2] W. AMBROSE, The L 2 -system of a unimodular group I, Trans. Amer. Math. Soc. 65 (1949) 27-48. [3] A. BROWN, The unitary equivalence of binormal operators, Amer. J. Math. 76 (1954) 414-434. [4] J. W. CALKIN, Two-sided ideals and congruences in the ring of bounded operators in Hilbert space, Ann. Math. 42 (1941) 839-873. [5] J. DIXMIER, Mesure de Haar et trace d'un opérateur, C. R. Acad. Sc. 228 (1949) 152-154. [6] J. DIXMIER, Les anneaux d'opérateurs de classe finie, Ann. Ec. Norm. Sup. 66 (1949) 209-261. [7] J. DIXMIER, Les fonctionnelles linéarires sur l'ensemble des op4rateurs bornés d'un espace de Hilbert, Ann. Math. 51 (1950) 387-408.

REFERENCES

388

[8] [9] [10] [11]

dans les anneaux d'opérateurs, J. DIXMIER, Application C. R. Acad. Sc. 230 (1950) 607-608. J. DIXMIER, Sur certains espaces consideies par M. H. Stone, Surma Brasil. Math. 2 (1951) 151-182. J. DIXMIER, Sur la réduction des anneaux d'opérateurs, Ann Éc. Norm. Sup. 68 (1951) 185-202. J. DIXMIER, Algèbres quasi-unitaires, C. R. Acad. Sc. 233 (1951) 837-839.

[12] [13]

J. DIXMIER, Applications 11 dans les anneaux d'opérateurs, Compos. Math. 10 (1952) 1-55. J. DIXMIER, Algébres quasi-unitaires, Comment. Math. Rely.

26 [14]

J. DIXMIER, Remarques sur les applications

3 [15]

(1952) 275-322. (1952) 290-297.

J. DIXMIER, Formes linéaries sur un anneau d'opérateurs,

Bull. Soc. Math. Fr. 81 [16] [17] [18]

Archiv Math.

(1953) 9-39.

J. DIXMIER, Sous-anneaux abéliens maximaux dans les facteurs de type fini, Ann. Math. 59 (1954) 279-286. J. DIXMIER, Sur les annea