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MATH EMATI CAL
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Hubert V. M. Manuilov E. V. Troitsky
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Translations of
MATH EMATI CAL
MONO GRAPHS Volume 226
Hubert V. M. Manuilov E. V. Troitsky
American Mathematical Society
Translations of
MATH EMATI CAL
MONO GRAPHS Volume 226
Hilbert V. M. Manuiov E. V. Troitsky Translated by the authors
American Mathematical Society Providence, Rhode Island
EDITORIAL COMMITTEE AMS Subcommittee Robert D. MacPherson
Grigorii A. Margulis
James D. Stasheff (Chair)
ASL Subcommittee Steffen Lempp (Chair) IMS Subcommittee Mark I. Freidlin (Chair) B. M. E. B.
FI4JIbBEPTOBbI MOIIYJIM MOCKBA, 2001 This work was originally published in Russian by Faktorial Press under the title "C ©2001. The present translation was created under license for the American Mathematical Society and is published by permission.
Translated from the Russian by V. M. Manuilov and E. V. Troitsky. 2000 Mathematics Subject Classification. Primary 46L08; Secondary 46Lxx.
For additional information and updates on this book, visit
www.ams.org/bookpages/mmono-226
Library of Congress Cataloging-in-Publication Data Manuilov, V. M. (Vladimir Markovich). 1961[C* Gil'bertovy moduli. English] Hilbert / V.M. Manuilov, E.V. Troitsky ; translated by V.M. Manuilov and E.V. Troitsky. p. cm. (Translations of mathematical monographs, ISSN 0065-9282 v. 226) Includes bibliographical references and indexes. ISBN 0-8218-3810-5 (acid-free paper) 1. 2. Hilbert algebras. I. Troitskii, E. V. (Evgenii Vadimovich) II. Title. III. Series. QA326.M3413 2005 2005042811
512'.556—dc22
Copying and reprinting. Individual readers of this publication, and nonprofit libraries acting for them, are permitted to make fair use of the material, such as to copy a chapter for use in teaching or research. Permission is granted to quote brief passages from this publication in reviews, provided the customary acknowledgment of the source is given. Republication, systematic copying, or multiple reproduction of ally material in this publication is permitted only under license from the American Mathematical Society. Requests for such permission should be addressed to the Acquisitions Department, American Mathematical Society. 201 Charles Street. Providence. Rhode Island 02904-2294. USA. Requests can also be made by e-mail to reprint-permission@ams .org. © 2005 by tile American Mathematical Society. All rights reserved. The American Mathematical Society retains all rights except those granted to the United States Government. Printed in the United States of America.
® The
paper used in this book is acid-free and falls within the guidelines established to ensure permanence and durability. Visit the AMS home page at http://www.ams.org/
10987654321
100908070605
Contents Preface
vii
Chapter 1. 1.1. 1.2. 1.3. 1.4. 1.5.
Basic Definitions
1
C-algebras
1
Pre-Hilbert modules Hubert The standard Hubert module HA and strong Morita equivalence Hubert
3 4 8 11
Chapter 2. Operators on Hubert Modules 2.1. Bounded and adjointable operators 2.2. Compact operators in Hubert modules 2.3. Complementable submodules and projections in Hubert C-modules 2.4. Full Hilbert 2.5. Dual modules. Self-duality 2.6. Banach-compact operators 2.7. operators and index. Mishchenko's approach 2.8. Representations of groups on Hilbert modules 2.9. Equivariant Fredholm operators
15 15 18
Chapter 3.
55 55 58
3.1. 3.2. 3.3. 3.4. 3.5. 3.6. 3.7.
Hubert Modules over
Inner product on dual modules Hilbert and dual Banach spaces Properties of Hilbert Wt-modules Topological characterization of self-dual HUbert F'redholm operators over Dupré — Fillmore theorem for Hilbert modules over finite
22 24 27 32 33 42 52
61
62 65 66 69
Chapter 4. Reflexive Hilbert 4.1. Inner product on bidual modules 4.2. Ideals and bidual modules 4.3. Reflexivity of Hilbert modules over V 4.4. Reflexivity of modules over C(X) 4.5. Hilbert modules related to conditional expectations of finite index
75 75 79 82
Chapter 5. Multipliers of A-Compact Operators. Structure Results 5.1. Extension of a Hilbert by the enveloping 5.2. Multipliers and centralizers 5.3. Multipliers of A-compact operators 5.4. Quasi-multipliers of A-compact operators
99 99
V
84 86
101
107 110
vi
CONTENTS
Strict topology Multipliers and Hubert modules. The commutative case Inner products on Hubert
114 118 126
Chapter 6. Diagonalization of Operators over 6.1. Problem of diagonalizing operators in Hubert 6.2. Quadratic forms on related to selfadjoint operators 6.3. Diagonalizing operators in the 6.4. Continuity of "cigenvalues" 6.5. Case of infinite of real rank zero 6.6. Case of 6.7. Case of continuous fields of trace 6.8. Schrödinger operator as an operator acting on a Hubert 6.9. Example: A continuous field of rotation algebras
133 133 136 138 143 145 146 148 154 157
Chapter 7. Homotopy of Groups of Invertible Operators 7.1. Technical lemmas 7.2. Proof of the Cuntz—Higson theorem 7.3. The case A C 7.4. Some other cases 7.5. Dixmier-Douady Theorem for 12(A) 7.6. Some generalizations 7.7. Neubauer type homotopy
159 159 164
Chapter 8. Hilbert Modules and KK-Theory 8.1. Tensor products 8.2. Main definitions 8.3. Cuntz's approach 8.4. Generalized Kasparov bimodules 8.5. Classifying spaces for some K- and KK-groups
181 181 182
Bibliography
193
Notation Index
199
Index
201
5.5. 5.6. 5.7.
166
169 173 175 176
184 188 189
Preface Hubert provide a natural generalization of Hubert spaces arising This generalizawhen the field of scalars C is replaced by an arbitrary tion, in the case of commutative appeared in the paper [591 of 1. Kaplansky; however the noncommutative case seemed too complicated at that time. The general theory of Hilbert (i.e., for an arbitrary serving as 'scalars') appeared 25 years ago in the pioneering papers of W. Paschke [100] and M. Rieffel [108]. This theory has proved to be a very convenient tool in the theory of operator algebras, allowing one to obtain information about by studying Hilbert over them. In particular, a series of results about some classes of (like and monotone complete
was obtained in this way [38]. An important notion of Morita equivalence for [109, 18]. This was also formulated in terms of Hubert notion also has applications in the theory of group representations. It turned out to be possible to extract information on group actions from Hubert arising from these actions 1102, 1101. Some results about conditional expectations [2] were of finite index [5, 133] and about completely positive maps of also obtained using Hilbert The theory of Hilbert may also be considered as a noncommutative generalization of the theory of vector bundles [33, 69]. This was the reason Hilbert modules became a tool in topological applications — namely in index theory of
elliptic operators, in K- and KK-theory [92, 90, 62, 63, 64, 65, 128] and in noncommutative geometry as a whole [24, 29]. Among other applications, one should emphasize the theory of quantum groups [135, 136], unbounded operators as a tool for Kasparov's KK-theory [3, 4] (also Section 8.4) and some physical applications [72, 80]. Alongside these applications, the theory of Hilbert itself has been developed as well. A number of results about the structure of Hubert modules and
about operators on them have been obtained [74, 42, 88, 77, 81, 127]. Besides these results, an axiomatic approach in the theory of Hubert modules based on the theory of operator spaces and tensor products was developed [12, 11]. A detailed bibliography of the theory of Hubert can be found in [43]. Some results presented here were only announced in the literature or the proofs were discussed rather briefly. We have tried to fill such gaps. We could not discuss
all the aspects of the theory of Hilbert here, but we tried to explain in detail the basic notions and theorems of this theory, a number of important examples, and also some results related to the authors' interest.
Vt'
PREFACE
viii
A major part of this book formed the content of the lecture course presented by the authors at the Department of Mechanics and Mathematics of Moscow State University in 1997.
We are grateful to A. S. Mishchenko for introducing the theory of Hubert to us. Together with Yu. P. Solovyov, lie has acquainted us with the circle of problems related to its applications in topology. modules
While working in this field and in the process of writing the present text, a significant influence oii us was niade by our friend and co-author M. We have discussed a number of problems of the Hubert C*_mnodule theory
with L. Brown, A. A. Irmatov, G. G. Kasparov, R. Nest, G. K. Pederseii and 'N. Paschke. Some applications were considered also with J. Cuntz, A. Ya. Heleinskii, J. Kaminker, V. Nistor, J. Rosenberg, K. Thomsen, B. L. Tsygan and others. Our research was partially supported by a series of subsequent Russian Foimdation of Basic Research grants.
CHAPTER 1
Basic Definitions 1.1.
For basic information about we refer to the books [31, 104, 123, 58]. We present here only some results on C*_algebras, which will be necessary for our purpose. Recall that an involutive Banach algebra A is called a if the equality IIa*aIl
11a112
for each element a E A. Any can be realized as a norm-closed subalgebra of the algebra of bounded operators 13(H) on a Hilbert space H. We do not assume existence of the unit element in By A+ we denote the obtained from the A by unitalization. As a linear space with involution, coincides with A C and multiplication in is given by the formula (a, z)(b, w) = (ab + zb + aw, zw), a, b e A, z, w E C. Any pair (a, z) E A+ defines an operator A —p A by b '—+ ab + zb and the norm of (a, z) is the norm of this operator. The spectrum of an element a of a unital is the set Sp(a) of complex numbers z such that a — z• 1 is not invertible. If a C*_algebra A has no unit, then the spectrum of an element a E A is defined as its spectrum in the A+ D A. The spectrum is a compact subset of C. An element a E A is called positive (we write a > 0) if it is Hermitian, i.e., if it satisfies the condition a* = a, holds
and if one of the following equivalent [31, 1.6.1] conditions holds:
(i) Sp(a) C I0,oo);
(ii) a = (iii) a =
b*b
for some b E A; for some Hermitian h E A. The set of all positive elements (A) forms a closed convex cone in A and P+ (A) fl = 0. Among the Hermitian elements h defined in (iii) there exists a h2
unique positive element, which is called the positive square root of a (we write h
a1!2).
A linear functional : A C is called positive if > 0 for any positive element a E A positive linear functional is called a state if = 1. We where a 0 and the supremurn is taken over all states. have hail = sup A of an algebra A into the 13(H) of all bounded
operators on a Hilbert space H is called a representation. A vector
E H is
called cyclic for the representation ir : A —÷ 13(H) if the set of all vectors of the form a E A, is dense in H. The vector H is called separating for the representation ir : A —+ 13(H) if the equality = 0 implies a 0. To each positive linear functional w on a A we can associate a unique (up to the unitary equivalence) representation of the algebra A on some
1. BASIC DEFINITIONS
2
for all a e A Hubert space and a vector such that w(a) = E is cyclic. The construction of such a representation is called the and the vector GNS-construction. A is an increasing net ea E A, E A. An approximate unit of a has an such that Ileall 1 and urn la — aeall = 0 for any a E A. Each approximate unit ea such that ea 0 and e0 e0 for [311. DEFINITION
possessing a countable approximate unit is
1.1.1. A
called a-unital. DEFINITION 1.1.2. An element h E A is called strictly positive if for any positive nonzero linear functional (or, equally, for any state) one has cp(h) > 0.
REMARK 1.1.3. Existence of a strictly positive element is equivalent to existence
0. Then h := of a countable approximate unit. One can assume that := hut is a countable approximate unit. Any is strictly positive. Conversely, separable is a-unital. The details can be found in [1041. We will often use the following statements. LEMMA 1.1.4 ([104, Lemma 1.4.4]). Let x, A such that a 0 and
x*x Tn(n) such it follows from (i) that
is equal to e3 or
}. Since em' ..L{ei,. .
. ,
.
.
By Lemma 1.4.3, there exists an element (1.3)
E
such that the elements
,
.
.
. ,ën,xn÷i,em')
are orthonorinal and —
SpanA(el,..
—
Cfl.44))
1
It follows from (1.3) and from condition (i) that
{e1,. ..,efl+1} C SpanA(xl,.
..
By setting m(n + 1) Tn', we complete the induction step. Thus, an orthonormal sequence satisfying properties (i) and (ii) has been constructed. But property (ii)
means that this sequence generates the whole module M HA, so M HA HA. Thus, Theorem 1.4.2 is proved for unital Let A be a nonunital and let A+ be its unitalization. Defining tile action of on the Hilbert A-module M by the formula x (a, A) M with the structure of a Hilbert At-module. Consider the Ak-module "A+ and denote by "A+ A the closure of the linear span of all the elements of the form x a, x HA+, a A, in HA+. It is easy to see that HA+A = HA. The
isomorphism M HA+
HA+ implies the isornorphism
M HA = MA
HA+A = (M
HA+)A
= HA.
0 DEFINITION 1.4.4. A Ililbert A-module M is called a finitely generated projective A-module if there exists a Hilbert A-module .N such that M for some n. The following two theorems of Dupré and Fillmore show that finite-dimensional projective submodules lie in Hilbert C*_modules in the simplest way. THEOREM 1.4.5 (Dupré — Fillmore, [36]). Let A be a unital a finite-dimensional projective A-submodule in the standard Hilbert A-module
HA. Then (i) (ii) (iii)
the nonsingular elements of the module M1 are dense in M';
__
1. BASIC DEFINITIONS
11)
PROOF. We begin the proof of the theorem with the case where M be an orthonormal basis in M. Fix > 0. For each m. put
L71(A).
Let 91'...
gj(gj,em).
—
Thcii
E M± and =
Since (x, Cm)
0
for each x e HA, we conclude that
there exists a number m0 such that for any m Then one can define , / ,
Cm —
=
with
1.
Let x
x) =
Choose a number rn
,
1. Therefore is nonsingular.
m0, the element ,
M1. Then ,
em)
—1/2
,
x) =
'rn0 such that
,
(em,
—1/2
,
x)II
0. By (2.4) one can find a sufficiently large n such that .
E
lix
—
0 in M such that T0x = y and lxii each y E IIyii. One has
=
0 and for all x e Al) adjointable A-homomorphism. Then F(M) F(M)1 = Al. COROLLARY 2.3.6. Let M be a Hilbert A-module and let J : M M be a selfadjoint topologically injective A-homomorphism. Then J is an isomorphism. LEMMA 2.3.7 Let M be a finitely genemuted Hubert submodule in a Hubert module Al over a unital C* -algebra. Then M is an orthogonal direct summand in Al. E M be a finite set of generators. Define the operator where by the formula = i = 1,... ,n, is e the standard basis. It is easy to see that F is adjointable with the adjoint F* acting by the formula F*(x) = ((x1,x),. .. where x E Al. By Theorem 2.3.3 the module Tm F = M is an orthogonal direct summand. 0
PROOF. Let x1
F:
,
LEMMA 2.3.8. Let A be a unital HA = projection and Al a projective module. Then HA = M adjointable.
M if and only
a
p is
2. OPERATORS ON HILBERT MODULES
24
PROOF. If exists, then (1 — =1 — exists as well. Therefore, by Theorem 2.3.3, Ker(1 — p) M is the image of a selfadjoint projection.
To prove the converse, let us verify first that HA =
jV' + M'. By the
Kasparov stabilization theorem, one can assume, without loss of generality, that .Af = spanA(el,.. = spanA(en+1, efl+2,...). Let gi be the image of under the projection onto M': . ,
Since the projection induces an isomorphism of A-modules Al Mi-, the elements > OA (i.e., the spectrum of this positive gi,. . , gn are free generators and (g,, operator is separated from 0 and hence it is invertible). So, if .
then i=1
On the other hand, 1—
1 = (ek,ek) = (fk,fk) + (gk,gk),
i.e., the spectrum of the element 1
> 0,
is separated from the origin, hence this
—
element is invertible in A, 1
ek= 1
ThusAl1
+ M' (k= 1,...,n).
E
k
fk
= HA.
has the form y = m+n, one has (x, y) = (x, m)+(x, n) =
thus x =
0.
Therefore HA =
0;
in particular, (x, x) = 0,
Consider the map q
which is a bounded projection since HA =
= Let x+y E
{
x1 +yl E
Then
(p(x+y),x1 +yi) = (x,x1 +yi) = (x,xi), (x+y,q(xi +yi)) = (x+y,xi) = Therefore
0
exists and equals q.
2.4. Full Hubert Let M be a Hilbert A-module. Recall that it is full if the closure (M, M) C A of the linear span of all (x, x), x E M, coincides with A. One can always consider any Hilbert module as a full Hilbert module over the (M, M). The following statement shows that full modules are easier to handle. a full THEOREM 2.4.1 ([71, 36]). Let A be a cy-unital Hubert A-module. Then: (i) There exists a Hubert A-module Al such that 12(M) HA Ef3A(. If A is unital, then there exist a number n and a Hubert A-module Al' such that (ii) If the module M is countably generated, then 12(M)
HA.
2.4. FULL HILBERT CS-MODULES
25
PROOF. Consider the set
s= {cEA:
EM}.
1cM
To prove the theorem, two following lemmas are required. LEMMA 2.4.2 ([16]). For any a E A, a such that 11(1 — c)aII
0 and any c > 0, there
exists
PROOF. Since the module M is full, one can find a finite set of elements m
cES E
M
such that —
k
<E/2.
(2.5) Put
=
yj(E2
(yj,yj))"2,
i =
1,... ,k;
b=>2(yj,yj). Then IIcII
hence c E S. Let 1(t) :=
+ b)"2b(E2 +
= E4t2(E2
0 one can find a number N such that the estimate N-f-n
N-f-n
N+n
N+n 111112
holds for all n > 0, the convergence of the series (2.8) follows.
< If 112 Li
2.5. DUAL MODULES. SELF-DUALITY
29
Note that for the functional f =
from Example 2.1.2, the partial sums are uniformly bounded, however the corresponding series is not conver-
gent.
Let us describe an interesting example of a dual module. EXAMPLE 2.5.6 ([44]). Let A = 13(H) be the algebra of all bounded operators on a separable Hilbert space H. Consider pairwise orthogonal projections E A, i E N, such that the series converges to 1 E A, and each projection as an orthogonal sum of Hilbert pj is equivalent to 1. We can consider H = spaces isomorphic to H, : H —+ H2 being isometries, so that 1 = p, =
As was shown above (see Proposition 2.5.5),
{{ai}
l2(A)' =
A with respect to the inner product
a
:=
The maps :
12(A)' —' A,
: {a,} —+
define an isometric isomorphism of A and 12(A)'.
be a positive linear functional on A. If M is a Hilbert A-module and is a if = {x E M : c.p((x,x)) = O} is a linear subspace of M, then given pre-Hilbert space with the inner product (., by the formula Let
x,yEM. and the We denote the norm defined by this scalar-valued inner product by Hubert space obtained by completion of with respect to this norm by Let f E M'. By Theorem 2.1.4 and Corollary 2.1.6, we have
f(x)*f(x)
If
then
for all x E M. Therefore, if x E
=
0
=
Hence, the map
x+
(2.13)
defines a linear functional on
Jj 2C/3. Since Y is compact, one can pass to a convergent subsequence yk = Yi(k). For a convergent sequence, by induction, we can choose Ek such that the corresponding Uk of yk are pairwise disjoint. Then we can choose Yk I
inside these neighborhoods so close to yk that IUi(k)I 2 C/2. choose functions bk such that bk (Yk) =
6=C/2.
1
and supp bk C Uk \
Finally one can It remains to take D
COROLLARY 2.5.12. Suppose that X is a compact connected separable Hausdorff
is MI if and only if X/G has at least two separated points.
C-space. Then
is a quotient space of X/G with PROOF. The Gelfand spectrum V of respect to the equivalence of nonseparated points. Since X is connected, Y is connected too. So, by Theorem 2.5.11, GG(X) is MI if and only if Y is not reduced D to one point.
2.6. Banach-compact operators DEFINITION 2.6.1. Let
be Hilbert A-modules and let M' he the dual
module for M. Consider the closure BK(M ,
E M, y E of the set .r
f
of the linear span of all operators
= y . 1(x) M', in the Banach space HomA(M,J'f). We call the elements Banach-compact operators.
hi the case JV' = M the set BK(M , N) is equipped with the natural structure of a Banach algebra. If T e EfldA(M) is a (not necessarily adjointable) operator, then the equalities
Tx = y f(Tx) = 9y,foT(X), TOv,i(x) Ty. f(x) = OTy.f(X) show that BK(M) is a two-sided ideal iii the Banach algebra EndA(M). we give hi the case of the standard Hilbert module over a unital one more (geometric) description of compact and Banach-compact operators. Let
2.7. C-FREDHOLM OPERATORS AND INDEX. MISHCHENKO'S APPROACH
33
C HA be a bounded set. We call it A-pre-compact if for each E > 0 there exists a free finitely generated A-module N N C HA, such that dist(S,N) <e. S
PRoPosITIoN 2.6.2. Let T e EndA(HA) (resp. T E Then the following conditions are equivalent: (i) T E 13K(HA) (resp., T E (ii) the image T(Bl(HA)) of the unit ball Bl(HA) is A-pre-compact. PROOF. If statement (i) holds, it is sufficient to prove that one can find an approximating module N for a finite set of elements in HA and this can be easily done by the Dupré - Fillmore method, as in the proof of Theorem 1.4.5. So suppose that (ii) is fulfilled. Then for any E > 0, one can find elements b1, . , bk E HA, generating the module N C HA, such that (b2, b3) öjj and diSt(T(Bl(HA)),.Af) <E, where B1(HA) is the unit ball of HA. Denote by PAr the projection onto N and consider the operator PET. It can be decomposed as . .
= bi(fi,x) + ...
(2.15) where Since x E IITx — bII <E. Therefore
E Bl(HA), one can find an element b
N such that
IITx—PgTxII=IITx—b+b—PgTxII
(.216 )
= IITx - b[I + IIPAr(b - Tx)II <E +
E = 2E,
hence IT — PNTII n such that if
= F(Lm) +M, is the projection on
and
=
along
:
HA —' HA
then
= where P2 is a projective finitely generated A-module, and where
a1,. ..
, ak
is a projective finitely generated A-module. Indeed, HA = are generators of the module ..N, then — I II a3—a3+a3,
I
TI
•_i
I!
For m —+ oo we have la/Il —' 0 since a7 = and x onto —p 0 as m
If
oo.
where x is arbitrary, Then for m sufficiently large,
we have
=
fl
(the proof repeats the proof of Theorem 2.7.5). Similarly
= Since m n, we have Lm
1S4.
fl
N1
where
is a finitely generated projective
A-module. From the equalities
F
C M1
that F :
Therefore we have the following equalities in K(A):
= = [P2], + [Pfl = [LmI, [Pfl = + [22] = VI;] + = and we have proved that the index is well defined. —
[JV1] +
Thus
[.N1]
—
[..V2] =
0
1This projection is well defined since C Mi form n and hence FIL.L is an isomorphism, whence it follows that HA is a closed A-module, L'm fl = 0, L'm + = HA. Therefore is a direct sum of closed A-modules and Q'm is bounded and A-linear. HA =
2. OPERATORS ON HILBERT MODULES
38
LEMMA 2.7.10. Let an operator F : HA —4 HA be A-Fredholm. Then there exists a number e> 0 such that any adjointable operator D satisfying the condition IF — DII >Tf(g) (rs + (h5 0
(2n +
j=1 s=1
2)a)
=
=
= j=1 s=1
t=1
3j s=1 E
t=1
j=1 s1
___________________
2.8. REPRESENTATIONS OF GROUPS ON HILBERT MODULES
45
is C-invariant. Let us estimate the distance by taking z' =
Thus
so that
=
-
p(z, z') =
p(z,
(2n ®
= 1
1
—
(2n+2)
1/2
k
+1
(2n+2)
Therefore
z) + p(z,
p(yn+i, (W1 +••• +
D,,(g),
2. OPERATORS ON HILBERT MODULES
48
=
which, being a trace, satisfies the relation P8
One also has
=dsfxS(g)Tgdg,
T9P8 = dsTgj X8(g')Tg' dg' =
=
d8
f
dg'T9 = P8T
LEMMA 2.8.9. Define
M8
(2.27)
M' :=
P8(M),
where the sum is supposed to be completed either as a Hubert sum. or as a closure in M of the algebraic sum (which is the same). Then
M' = M.
(2.28)
PROOF. Assume that a C-linear functional f on M vanishes on M and that x E M is an arbitrary vector. Then, for any set of indices, we have E M', so that
0=
f
dg.
Therefore, by the Peter—Weil Theorem 2.8.5, f(T9 (x)) =
0
holds almost everywhere
= f(x) =
and, by continuity, it vanishes everywhere. In particular, Hence, by the Hahn—Banach theorem, M = M.
0.
0
THEOREM 2.8.10 ([91]). Let M be a Hubert B-module equipped with a strongly
continuous unitary representation of C and let the group act trivially on B. Now let {V8} be the complete collection of pairwise nonequivalent unitary representations of C and let
M8 := Hornc,c(V8,M) C Homc(V8,M)
®M
be a Hilbert B-module with the inner B-valued product defined by the formula :=
orthonormal basis for V8. i,j=1
Then, for the Hubert sum, we have an equivariant B-linear isomorphism
vEV8,çoEM8, and
r(V8®M8) =M5, where M8 is defined by (2.27).
PROOF. To begin with, let us note that the r8 are algebraically injective. Indeed, let
o=r,
2.8. REPRESENTATIONS OF GROUPS ON HILBERT MODULES
49
Since, by the Schur lemma, is either an isomorphism or 0, the above equality can be true only if either = 0 or = 0. But then ® p = 0. By Lemma 2.8.9, it is sufficient to prove only that r5 maps ®M5 bijectively onto M8. := P8(M) we obtain, by relation (2.20), Note that, by setting that the operators induce isoinorphisms
M is a sum of isomorphic modules. } be the orthonormal basis of with respect to which the were defined. Define the homomorphism
Thus M8 = Let { matrix elements . . .
,
(2.29)
® x) =
[Mfl —* M5,
:
in square brackets to emphasize that there is no action of G on it. By the properties of the operators the map V is an isomorphism. where we have put
Since, by (2.21),
®x) = T9Pf1(x) = and
V
®
=
V
map
V
—,
:
W5(x)(v) :=
V(v 0 x).
Then
Ox) = V(v ® x). Since we have an isomorphism on the right-hand side and since is algebraically injective, r5 is an isomorphism (see Lemma 2.8.12), whence V is an isomorphism. In particular, the image of coincides with M8 and these images are orthogonal to each otl1er. Hence r is topologically injective and its image coincides with M. 0 o
OWS)(v
REMARK 2.8.11. Let the G-A-module M belong to the class P(A) of projective finitely generated modules. Then obviously M8 = Honxc(V3,M) e P(A). Let us show that only a finite number of summands do not vanish in the sum Denote generators of M by a1,.. . , a8. Using the Mostow lemma [93], choose C-periodic vectors b1,. . so close to a1,.. . , a8 that they generate M as an A-module (see Lemma 2.7.3). By decomposing the linear span of the orbit Gb3, which is a finitedimensional C-C-module, into irreducible modules, let us find a new system of generators c1,.. . now belonging to irreducible G-C-modules. Then it is evident that the number of nonzero summands does not exceed N. .
,
,
LEMMA 2.8.12. Let F: L M, T : N spaces, S = FT an isomorphism and Ker F =
0.
L be continuous maps of Banach Then F is an isomorphism.
PROOF. Since S is an isomorphism and F is bounded, T is topologically micetive and its image T(N) is closed in L. Suppose it does not coincide with L. Choose a vector 0 x E L\T(N). Then 0 F(x) FT(N). Indeed, let F(x) = FT(y) for some yEN. Since z = Ty E T(N). z—x 0, while F(z—x) = FT(y).—F(z) = 0.
50
2. OPERATORS ON HILBERT MODULES
We get a contradiction to the condition KerF =
0. Hence, T is a topologically injective epirnorphism, i.e., isomorphism, as well as F = ST'. U
Let us recall some facts about integrating operator-valued functions [62]. Let and p: C(X) —p A an involutive homomorX be a compact space, A a A be a continuous map and let, for each phism of unital algebras. Let F : X In this case the integral x e X, the element F(x) commute with the image of EA
be an open covering and let can be defined as follows. Let X = = 1 be a subordinate partition of unity. Choose points e and form the integral sum
= If the limit of such an integral sums exists, then it is called an integral. If X is a Lie group G, it is natural here to use a Haar measure dg and to define =
C(X)
fQ(g)dg := A is viewed as for a norm continuous map Q G —p 8(H), where the a subalgebra in the algebra 8(H) of bounded operators on a Hubert space H. If Q:G C 8(H), then, since
>0,
fG
we obtain that and
e
JQ(g)dg
(the positive cone P+ (A) is convex and closed). Hence we have proved the following lemma.
be a continuous function. Then, for the LEMMA 2.8.13. Let Q: C inteqral in the sense of [62], the following inequality holds:
0. THEOREM 2.8.14 ([128]). Let GL = GL(A) be the full general linear group, i.e., the group of invertible operators in End 12(A), and suppose that for the group C, a representation g (g C. T9 E GL) is given, and that the map
Cx
12(A),
(gu)
T9u
is continuous. Then there exists an A-valued inner product on l2(A) equivalent to the initial one (i.e.. generating an equivalent norm) and such that the representation g '—* is unitary with respect to this new product.
2.8. REPRESENTATIONS OF GROUPS ON HILBERT MODULES
51
PROOF. Let (, )' be the initial inner product. For any u, v E 12(A) there exists A, x Define the new product by the formula
a continuous map C
(u,v) =
I
JG
where the integral can be considered in the sense of any of the two definitions in [62], since
the map is norm continuous. It is easy to see that this new product gives a
sesquilinear map 12(A) x 12(A) —+ A satisfying properties (i)—(iv) of Definition 1.2.1
and that, by Lemma 2.8.13, (u, u) 0. Let us show that this map is continuous. Fix an arbitrary u E 12(A). Then x i—p 12(A) is a continuous map C defined on a compact space, thus the set the uniform boundedness principle ['7, v. 2], (2.30)
lim
I
x E G} is bounded. Therefore, by
=
0
-=
const
uniformly in x e C. If u is fixed, then
and, by equality (2.30), one has (u, v) 11=1
f(Tx (u), .
dxli
vol C sup IITx(v)II —40
(v —40).
ccEG
We
have obtained the continuity at the point 0, hence on the whole space 12(A) x = (ai(x),a2(x),...) e 12(A), the equality (u,u) = 0 takes the form
12(A). For
=
J
0.
C
Let A be viewed as a subalgebra of the algebra of bounded operators on a Hubert space L with an inner product (, )j.,. For each p e L we have
0=
=
j
dx
=
f
dx
(cf. [62]). Therefore = 0 almost everywhere, hence, by continuity, for all x and = 0. In particular, u = 0. Since each operator is an automorphism, we obtain
f
=
0
= (u,v).
Now we show the equivalence of the two norms, which, in particular, implies the continuity of the representation. There is a number N > 0 such that N
2. OPERATORS ON HILBERT MODULES
52
for any x E G. Hence, by [62], we have 11u112
= II(u.u)IIA =
if I
\XEG
=
=
. .
,
3. HILBERT MODULES OVER W-ALGEBRAS
60
It follows from Proposition 2.5.7 that
= But
= = w(a*f(x)) =
=
= ((f. for any x E M. Since the subspace
x+
a)(x))
= ((f
+
is dense in H,1,, the equality
=
= ((f.
a,g)) = ((f = holds for any p E P. hence a*(f,g) = (f a,g). Passing to adjoints, we also obtain (f,g)a = (f,g a). The obtained inner product on M' is an extension of the inner product from M. Indeed, if x,y EM, E P, then .
=
=
=
= (x,y). Further, if f EM', then
Hence
=
ço((f,x)) =
= f(x). Let us show that M' is complete with respect to the norm defined by the inner product constructed above. There also exists the norm I'M' on M' defined as the norm of linear maps from M to A, with respect to which the space M' is complete. Let us prove that II•I'M' = 1.11. Since therefore (f,
2
1(x)* f(x) = (x,f)(f,x) —
we
obtain
If
that 11111
Ill
But since
=
=
(x,x), for each w E
11111
E P, IIwII
:
= 1}
111112,
and
= we have proved that M' is a Hubert A-module. It remains to verify that it is self-dual. Let F E (M')'. The restriction of F to M C M' is an element of the module M', hence one can find a functional I E M' such that F(s) = f(x) for So,
all x E M. Let us define the functional F0 E (M')' by Fo(g)
= F(g)
—
(f,g),
gE
M'.
for all x E M. We should verify that Fo(g) = 0 for all in converging to g E M'. Let Since F0 is bounded, we can find a number K such that Fo(h)*Fo(h) K(h, h) for
It is clear that
=
0
E P. Choose a sequence {yn +
allhE M'. For alln= 1,2,... one has =
—
—
i,,))
Kço((g
—
—
3.3. HILBERT W-MODULES AND DUAL BANACH SPACES
61
But since —
yn,g —
—
—
we have
—
0.
—
+(yn +
+
—
=
+
—
+
+
Therefore
=
(3.1)
0.
Since the equality (3.1) is valid for any normal functional Fo(g) = 0.
3.3. Hubert
E P, we obtain that D
and dual Banach spaces
PROPOSITION 3.3.1. Let M,H be Hubert over a A and let T : M —9 H be a bounded operator, T E HornA(M,AI). Then there exists a unique extension of the operator T to an operator T : M' PROOF. Define the operator T# : by (T#y)(x) := (y,Tx), x E M, the operator T# is bounded, IIT#yII y E f'I. Since II(T#y)(x)II lxii iTilliyii. ForanyaEAwehave (T#(y . a))(x) = (y. a,Tx) = a* (y, Tx) ((T#y) a)(x): .
therefore the map T# is A-linear. Define the map T : M' by (Tf)(y) = Since T (T#)#, the map T is also a bounded where y E H, f E M'. (f, A-linear map. The equality = (Tx5(y) shows that the operator T is an extension of the operator T.
Let us show uniqueness of this extension. Let S M'
H' be a bounded x E M} c M'.
A-linear map that coincides with T on the submodule M = Then their difference V = T — S vanishes on M. Since the module M' is self-dual, the operator V has an adjoint operator : H' M'. If g E H', x E M. then (V*g)(x) = = = 0, V=
i.e.,
0, hence S
U
T.
COROLLARY 3.3.2. Let M be a Hubert A-module. Then the map T '—* T defines a monomorphism = EndA(M'). Let as
the
Hilbert of operators acting on them.
us show that self-dual
are dual Banach spaces,
PROPOSITION 3.3.3 (1100]). Let M be a self-dual Hilbert M is a dual Banach space.
as well
Then
PROOF. Let us introduce the notation M° for the Hubert module M viewed a Banach space with multiplication by scalars given by the formula A x := Ax, x E M°. Consider an algebraic tensor product over C, where is a 0 pre-dual space of normal functionals on A. Equip the space 0 M° with the maximal cross-norm and define for x E M the linear functional on 0 M° by the formula as
=
3. lULBERT MODULES OVER W-ALGEBRAS
62
where E A,, well defined. Since . .
M°. It is easy to see that this functional is
.
®
lii)
lxii
Ilwiil
< it follows from the definition of the maximal cross-norm [661 that be a sequence Let us show that in fact the equality = lxii holds. Let of functionals of norm 1 in A, such that —' iixii2. For each element of the form ® ® x E A, ® M° we have lxii and iixii < Therefore the map x '—' defines an isometric inclusion hence lxii iixii2, M C (A. ® M°)'. To prove the statement it is sufficient to check that the set : x E M} is closed in (A, ® Mo)* with respect to the weak* topology, =
since it would mean that M is isometric to the dual space of some quotient space of be a net in M that converges to some element F E (A, ®M°)' A, ®M°. Let with respect to the weak* topology. For y E M, let us define the linear functional on A. by = ® y), where E A,. The functional is bounded,
there exists a unique element f(y) E A satisfying the properties f is obviously and = linear. Let us show that it is A-linear as well. Let y E M, a, b E A, E A,. Define = ço(a'b). Then it follows from the equalities the normal functional E A, by liFil iiyii, hence
iiFii
cp(f(y a)') =
= =
F(p® (y .a)) =
.
=
a)) =
.
=
= p(a'f(y)') = cp((f(y)a)'),
E A, that f(y . a) = f(y)a. Since the module M is self-dual, hence we can find an clement x0 E M such that 1(y) = (x0, y). Thus F = 0 is closed in (A, ®M°)'. which hold for any
Consider the weak* topology on the dual Banach space M. A net } in M obviously converges to an element x E M with respect to this topology if for every A, and for every y EM. Some modification of the previous argument allows us to also obtain the following statement. PROPOSITION 3.3.4 (11001). Let M be a self-dual Hubert W'-module. Then is a W'-algebra. the C'-algebra
3.4. Properties of Hubert W'-modules The elements of self-dual Hilbert W'-modules admit the following convenient representation (an analog of the polar decomposition). Any PROPOSITION 3.4.1 ([100]). Let M be a self-dual Hubert element x EM can be written as x = z• (x,x)'12, where z EM is such that (z,z) is the projection onto the image of (x, x)V2. Such a decomposition is unique: if
x = z' a, where a 0, and if (z', z') is the projection onto the image of a, then = z and a = (x,x)hh/2.
3.4. PROPERTIES OF HILBERT W-MODULES
PROOF. For x E M and
n E
N
63
put
= ((x,x) + 1/n)'!2,
x
< 1. Let y eM be an accumulation Since = (x,x)((x.x) + point for the sequence in the weak* topology (which exists due to the com} pactness of the unit ball). Since ha,, — (x,x)"2h1 x, 0 and (x,x)'!2. Let p be the projection onto the image of x= Then
p(x, x) 1/2 = (x, hence
x = y.p(x,x)"2
x)
1/2
1/2
p = (x, x)
and
(x,x)
(x,x)'!2p(y,y)p(x,x)'!2.
Therefore
(x,x)'!2(p
=
0.
1, we have r — p(y,y)p 2 0. Therefore
Since
0,
whence it follows that p(p — p(y, y)p)1/2 =
0,
p(y, y)p. Put z
hence p
Thenz.(x,x)'!2=y.p(x,x)'!2=x,(z,z)=p(y,y)p=pandz.p=z. In
y p.
order to prove the uniqueness of the decomposition suppose that x = z'
a,
a 2 0, and that (z', z') is the projection onto the image of a. Then (x, x) = a(z',z')a a2, hence a (x,x)'!2 and (z',z') =p. Since (z'—z'.p,z'—z'.p) = 0, we obtain z' = z'p. Also one has where
(z,x) = (x,x) 1/2 = (z,z , )(x,x) 1/2 i.e.,
p
—
(z,
z') =
0.
Now
it
= can
0. whence be easily
we obtain that (p—(z,z'))p=p—(z,z'.p)
seen that (z — z',
z —
z') = 0, hence
z' = Z.
0
be a set of projections in a A. For each of them the C A has the natural structure of a singly generated projective paA Hilbert A-module. Similarly to the definition of the standard Hilbert module we can define the module as the set of sequences (ma), ma C Mc, C A, Let
set M0 =
such that the series is norm-convergent in A. The dual filbert module is called the ultra weak direct sum of the modules Mc,. For self-dual (sc, Ma)' Hilbert W*_modules we have the following structural result. THEOREM 3.4.2 ([1001). Let M be a self-dual Hubert over A. Then there exists a set {pc, } of projections in A such that the module M is isomorphic to the ultra weak direct sum of the modules pc,A.
PROPOSITION 3.4.3. Let Al C HA be a Hilbert submodule over a W* -algebra A. If Jsf-'- = 0, then the dual module 15/' coincides with
PROOF. Let j Al HA be an inclusion of modules. The restriction of defines the map j' dual to j. If fig, f E functionals f is such that = 0, then f±Isf and, by assumption, one has f = 0. fE Therefore the map j' is injective. Consider the composition of maps :
i=j'ooj
'—+
3. HILBERT MODULES OVER
64
If n E JV, then i(n) = j'(j(n)) = = Therefore the map i coincides with the inclusion map Al '—p /'f'• The dual map (after identification of the first and second dual modules) (i' o i)' = i'
o
i"
: .IV' = .Af" —,
—+.N'•'
is an isomorphism, hence the map i' should be surjective; then the map j' is sur-
0
jective as well.
PROPOSITION 3.4.4. Let A be a and 1Z C HA an A-submodule without orthogonal complement, i.e., 1?)- = 0 in HA. Then R! =
PROOF. It is sufficient to show that if the orthogonal complement to a submodule 1?. in HA is equal to zero, then the orthogonal complement to 7?. in is equal to zero too. Let us assume the contrary. Suppose that there exists a such that f(r) (f, r) 0 for some r E 7?.. Since the series functional f * is norm-convergent in A, there is a number n such that f (n) (z) 0 for .
.
(Ii,...
.
.
.).
.
But, as f(Th) E 11A, we get a contradiction.
0
Even Hilbert over commutative behave differently from Hubert spaces. The following example shows what kind of pathology can occur.
Let A = be the of bounded sequences and let co C its subalgebra of sequences vanishing at infinity. Consider the Hubert
M = l2(co)
be
over A.
PROPOSITION 3.4.5 ([85]). There is a submodule .iV C M such that it coincides with its bi-orthogonal module, but is not complementable.
PROOF. Put where fk E is
c0 for
//= {(f1,f2,f3,...),(f1,2f2,3f3,...)}, all k
eN
and the series
k2IfkI2
is convergent in c0. It
easy to see that the submodule Al C M is closed and coincides with its bi-
orthogonal complement, ftf-'--'- = Al. Let us show that Al is noncoinpiementable.
Assume the contrary: let M = .)VEB/C for some submodule K C M. Then there exists an A-linear (nonorthogonal) bounded projection P : M —' Al. Since P commutes with the action of A, it can be written as a sequence (Pk)kcN, where Pk is a projection acting on the k-th coordinate. More precisely, if Pk E A is the projection onto the k-th coordinate, then Pk = PkP. Since P is bounded, the norms of the projections Pk should be uniformly bounded. But any projection Pk projects onto where N : 12 —4 is an unbounded operator, N(aj,a2,...) = (a1,2a2,. . .). For each k there is a decomposition Mk = 12 =ft4 DKk, where Pk implies JCk = pkJC, j'4 = that the complements JCk should consist of the pairs {(x, y) : x, y l2} such that for some > 0 independent of k. > Consider an arbitrary element C0, M, where 12,.. .), (gi' 92,...) for any i. If then this element can he written a.s the sum = gi
M
For a fixed sequence
n,
we use a superscript to denote the number of an element in the For sufficiently large n the elements of A = =
the submodule K satisfy
—
>
—
But the left-hand side of this
3.5. CHARACTERIZATION OF SELF-DUAL W-MODULES
65
inequality vanishes as k oo (when ii is fixed) and the right-hand side does not vanish, since e c0 (it is sufficient to take the constant sequences = $ 0). This contradiction shows that the submodule .N is not complementable. D
3.5. Topological characterization of self-dual Hubert Let M be a Hubert over a A and let P E
be the set of normal states on A. Let us define (see [42]) two topologies on M as follows. Denote by r1 the topology given by the system of seminorms E P, and denote by r2 the topology given by the system of serninorms y E M, E P. In the case where A = C and M is a Hilbert space, the topology r1 is the norm topology and the topology r2 coincides with the weak topology, so in general these two topologies do not coincide. THEOREM 3.5.1 ([42]). Let M be a Hubert Then the following conditions are equivalent: (i) the module M is self-dual; (ii) the unit ball B1 (M) is complete with respect to the topology r1; (iii) the unit ball Bi(M) is complete with respect to the topology r2. PROOF. Let us prove the implication (i) (ii). Assume, for this purpose, that the unit ball B1(M) is not complete with respect to the topology r1. Let us denote
by L the linear span of the completion of B1(M) with respect to the topology Ti. For the extensions of seminorms from M to L we use the same notation. By assuniption, there exists an element r E L \ M and a bounded net E A, such that for any p E P and for any e > 0 there exists some (1 E A, for which — 0 and define the elements E M-'- by =
em
—
Theii 1
The inecluality
r((gj,gj)) shows that the series T((gj, em)) is convergent, hence there exists a number m0 such that for all m> nz0, one has E * I \ (Yuiem)) < > 1—
To proceed wit.h the proof, the following lemma is required.
LEMMA 3.7.2. If SE Hk, lxii = 1 and if r((x,x)) > 1— then there exists a projection p e A such that r(p) > 1 — E and the operator p(x, x)p is invertible in the W* -algebra pAp. PROOF. Let us denote by dP(A) positive operator a (x,x) E A, a = A
where
a
projection-valued measure defined by the
f1
AdP(A). Put
Jo, AA0, Ao e [0; 1]. Then f(a) = p is a projection. Denote dr (P(A))
a C-valued measure on [0; 1] and, by [96], one has
r(a)=f
Adp(A).
Then (3.3)
1—
Since P(1) =
1,
(3.4)
1 —
2(1
1.
by dp(A).
It
is
3.7. DUPRE - FILLMORE THEOREM
71
Choosing in an appropriate way a number A0 $ 0, we get follows from the definition of the function f(A) that
r(p) = r (f(A)) =
=
([Ao; 11)
>
1 — e.
It
/ f(A) dp(A)
fdi(A)
-E.
Now consider an operator pap E pAp. From the spectral theorem we obtain that A I (A) dP(A).
pAp lies within the Therefore the spectrum of the element pap of the segment [Ao; 1], hence it is separated from zero, so this operator is invertible in 0 pAp. Now we can proceed with the proof of Theorem 3.7.1. Let —1/2
an
— ' em—em
=
pbp
=
b
E
pAp
en
.
= P. = Let us now take an element y E M1 such that y $ 0,
Since the series
1.
converges, for any e >0 we have
/
*
,,
T((em,y)*2 b (em,y))
,,
11b112T((em,y)*(em,y))
and
lxii 1.
Let us come back to the proof of Theorem 3.7.1. Let contains each an infinite number of times. Put y = Yi — for e1 = 1 one can find an element y' E such that Ily'M (3.8) and a projection pi with T(pi) >
}
be
a sequence that gk(gk. y1): Then
1 and
— y'llr <E', such that the operator pi(y'. y')pi is invertible in —1/2
the algebra p1Ap1. Then, taking h1 = y'b', where b' = E p1Ap1, we obtain from (3.8) the inequality < = distT(y,y') < ei, where B1 denotes the unit ball of the Hubert module in the initial norm. Therefore
distT(yl.Bj(SpanA(M,hl))) <Er.
3.7. DUFRE - FILLMORE THEOREM
73
can find an element h2 E (SpanA(M, h1))-'- such that is a projection, r(p2) > and P2 <e2. Repeating this procedure and taking Ek = we obtain a set of mutually orthogonal elements E M-t- such that the operator (hi, = pj is a projection, T(pj) > and for which one has
Further, taking 62 = the operator (h2, h2) =
we
hk))) k
PiZ.
= i>k
i>k 1x112)
i>k
(>PiZ). i>k
1, where qj are some projections. Applying the construction described above to these generators, we can construct, by induction, a set of elements such that they satisfy the equality (hr, hr) = 1 and generate the module N. Therefore } is a basis in N and the module H is isomorphic to HA. It remains to prove that Al' = M'. Since the module H C M' is closed in the initial norm, the restriction f E (M1)' belongs to H'. Note that the module M-'- is self-dual, (M-'-)' = M', because of the self-duality of the modules and M. Suppose that f = 0. Since the submodule H is dense in with respect to the trace norm, one has f = 0 on due to the continuity of the map —* A with respect to this norm because of the inequality f: .
.
.
,
(3.12)
T((f(y))*f(y))
11f112.r((y,y)),
where y E M'. This proves injectivity of the map M' Now let 0 E H'. This functional can be extended up to a map from to A. If C H is a with respect to the trace norm, then sequence converging to the element y E put 0(a) = It is well defined due to the estimate (3.12) with 0 instead of f. Thus, the A-modules and H' coincide and the theorem is proved since 0 are isomorphic. the modules H' and
CHAPTER 4
Reflexive Hubert 4.1. Inner product on bidual modules For a Hubert C5-module M over a C5-algebra A we define the bidual Banach right A-module M" as the set of all bounded A-homomorphisms from the dual module M' into A. It turns out that the A-valued inner product on M can always be extended to the bidual module, as opposed to the dual module, which admits an extension of the inner product only in the case of W5-algebras. Let x E M, f E M'. Put ±(f) The map x '—p ± is an isometric map from the A-module M into the A-module 1±0
=
lxii
For
l._ilx(x) Ii
a functional F E
we
fe
1} 11111 IIxIl
11111
lxii
1
=
II (x, x) II = lixil
define the functional F e M' by the formula
F(x) := Identifying M and M = x e M} C M', we obtain that F is the restriction for all x G M. It is clear that the map of F onto M c M'. Note that F '—i F is an A-module map from M" to M' and IFil iF. We shall check soon that this map is an isometry. Let us define the inner product (•,•) M" x
F,GeM".
(F,G):=F(G),
(4.1)
—+ A by the equality
It can be directly checked that (F . a,G) = a5(F,G) for a E x,y E M, one has
A.
Besides, for
= (y,x)5 = = (x,y). = the inner product defined by formula (4.1) is an extension of the inner
=
Therefore
product on M. To check the properties of an inner product we require the following construction.
Consider the right A-module A x M. In addition to the natural direct sum inner product (., defined by the formula ((A, x), (b, = a5b + (x, y), where a,b E A, x,y E M, let us consider another inner product on Ax M. Take I E M', f 0, and a number t> 11111 and put (4.2)
((a, x), (b, Y))f,t :=
t2 a5b
+ as 1(r) + f(x)5b + (x, y). 75
4. REFLEXIVE HILBERT C-MODULES
76
Properties (iii) and (iv) of Definition 1.2.1 obviously hold. Let us check properties (i) and (ii). The first one is valid due to the inequality ((a, x), (a,
=
t2 a*a
+ a*f(x) + f(x)*a + (x, x)
+ a*f(x) + f(x)*a +
(4.3)
+ a*f(x) + f(x)*a +
> t2
(4.4)
(ta+
= (ta+
>0.
Suppose that ((a, x), (a, X))ft = 0. Then equality should be reached at each step in (4.3)—(4.4). Subtracting line (4.3) from line (4.4), we obtain (IIfli_2 — t_2)f(x)*f(x) = 0,
hence 1(x) = 0. Therefore t2a*a + (x, x) = 0 and we can conclude that a = 0 and x = 0, so we have checked the validity of property (ii). Thus, the module A x M with the inner product defined by formula (4.2) is a Hilbert A-module. We denote the Hubert module A x M equipped with this inner product by (A x M)f,t and let the norm on this module corresponding to this inner product be denoted by Note that lxii. For x,y E M, a E A. we have (f a + ii = Iia*f(y) + (x, y) ii 1 ((a, x), (0, Y))1,t ii (a, x)
11(0, y)
=
1 (a, x)
Therefore II(f a + £)iI Ii(a,
(4.5)
PROPOSITION 4.1.1 ([1011). Let 1sf c M' be a submodule containing the module
M. Then the norm of any functional i,b E H' satisfies the equality
=
PROOF. Without loss of generality we assume that = 1. Define the functional f E M' by the formula 1(x) := x E M. Then if ii ( 1. It is necessary to prove the inverse inequality, if ii 1. Take g E H such that < 1 and put
ForaE AandxEM wehave iica+f(x)ii = (the last inequality follows from (4.5)), hence the map
is a bounded modular map, (4.6)
1. Therefore x)*fc(a, x)
((a, x), (a, x))91
for all a E A, x E M. From the estimate (4.6) we get A*c*ca + a*c*f(x) + f (x)*ca + f(x)*f(x) a*a + a*g(x) + g(x)*a + (x, x). Taking a = —2g(x), we obtain 4g(x)*c*cg(x) + f(x)*f(x) D consider the inner product (•, Since IF, a for all (a,x) E A x M, the map
Ax M—'A, is bounded and its norm does not exceed D (we mean here the norm defined by the Therefore inner product (•,
(ca+F(x))*(ca+F(x))
(4.7)
Inequality (4.7) holds for all t > D and, taking the limit t —* D, we obtain (ca + F(x))*(ca + F(x)) D2(D2a*a + F(x)*a + a = —D2F(x), we get — 1)*(D_2c
—
D2(_D_2F(x)*F(x) + (x, x)),
1)?(x)
hence
Suppose that D2
1)*(D_2c_ 1) + D2(x,x). Sp(c). Then one can find a number 0 such that —
for all x E M. But then
1)*(D_2c
-
-
—
D2
F(x)*F(x) < —(x,x), whence
D2=
- 1+o
2D22 -
1+8
The obtained contradiction shows that D2 E Sp(c). But since Ilcil =
IIF'II
we have hlchl = D2, hence II(F,F)hI =
hIF1I2
=
hE'll2
= D2,
and II(F,F)ll E Sp((F,F)). For an
arbitrary element a E A we have IIacaII =
= hI(F'a,F'a)II E Sp((F'a,F.a)) = Sp(aca).
4. REFLEXIVE 1-IILBERT
78
The following lemma concludes the proof. LEMMA 4.1.3 (1101]). Let an element c E A be such that the inclusion Ilacall
E
0
0. Then c> 0.
Sp(aca) holds for any a E A, a
Proposition 4.1.2 shows that the inner product defined on M" satisfies conditions (i) and (ii) of Definition 1.2.1. It remains to check condition (iii). Notice
that
(F+iG,F+iG)0,
(F+G,F+G).>0,
hence these expressions are selfadjoint. Then
((F,G)+(G,F))* = (F,G)+(G,F), _i((F,G)_(G,F))* =i((F,G)—(G,F)). Therefore (F,G) = (G,F)*. The module M" is a Hilbert A-module, since the operator norm on M" coincides (by Proposition 4.1.2) with the norm defined by the inner product. Thus, we have proved the following theorem. x THEOREM 4.1.4 ([1011). The map (.,.) —f A defined by (F, G) = F(G), F, G E M", is an A-valued inner product on M". The norm defined by this inner product coincides with the operator norm on M". The map F '—p F is an :
isometric inclusion M" C M'. Let us pass now to dual modules of higher order. Let '1" E (M")'. Define the M' by the formula
functional
xcM. Further, let f E M'. Define
E (M")' by the formula
:= (F(f))*, ;—
The maps
f i—
composition
are
E
A-module inorphisms. Consider their
F
M'
(4.8)
F e M".
M we have
=
= 1(x) the composition (4.8) is the identity map, whence the map M' —* M'" is an isometric inclusion and the map M" M' is an epimorphism. Let us show that the last map is also monomorphic. For that purpose, apply Proposition 4.1.1 for the case E M' is the = M". Let E I'!' = M". Then the functional restriction on M of the functional Suppose that = 0. Then, = by Proposition 4.1.1, we have = 0, hence the map F41(x) =
is monomorphic. Thus this map is an isometric isomorphism. COROLLARY 4.1.5. For a Hubert (Jv1tI)II So,
M one has (M")' = M' and
_A4".
the series of dual modules M, M',... stabilizes on the third step and the
inclusions
M c M" = M" c M' = M"
4.2. IDEALS AND BIDUAL MODULES
79
unlike the module M', which is. generally speaking, only a Banach module. Let us illustrate by examples that all possible variants can be realized:
are isometric. Thus M and M11 are Hubert
(i) Let A be a unital a free A-module with n generators. Then the module M is self-dual, hence M = M" = M'. By Theorem 3.2.1, for any Hilbert A-module M (ii) Let A be a its dual module M' is a self-dual Hilbert module, hence M = M'. (iii) {44] Let A = C0(O, 1] be the (nonunital) of functions on a segment [0, 1] vanishing at zero, M = A. Then M' = C[0, 1], M" = C0(0, 1] and M = M'. (iv) [44] Consider the module C0(0, 1) of functions on the segment [0, 1] vanishing at the end points, over the A = C0(0. 1]. In this case one has M' = CEO, 1], M" = Co(0, 1], that is, M M'.
M is called reflexive if M" = M.
DEFINITION 4.1.6. A Hubert
Later we shall discuss other examples of reflexive Hilbert C*_modules.
4.2. Ideals and bidual modules A. The ideal J is equipped with the Let J C A be a right ideal in a natural structure of a pre-Hilbert A-module with the inner product (x, y) = x*y for x, y E J. In this section we describe the bidual module J". Note that the inclusion map J c A is bounded and A-linear, hence it defines an element of the dual module J'. Denote this element by e J' and put
J:=
:F
J"}.
Obviously .J is a linear subspace of A. Since for a E A and for F e
F(ft)*a = (a*F(fj))* =
one has
((Frn
= x holds for any x C J, we get
a right ideal in A. Since the equality
J ç J. PROPOsITION 4.2.1 ([101]). If J C A is a right ideal in a A, then (i) the ideal J is closed; (ii) the map i : F '—* F(f1)*, where F C J", is an isomorphism of Hilbert A-modules J and J". PROOF. The map i is obviously a homomorphism of A-modules. This map is also bounded, since = 1. Since the equality (f1
.
x)(y) = xfL(y) =
holds for any x, y C J, one has
= iIPIIj' = =
= Therefore
x= :
x C J, lxii
1}
x x
C
J,
1}
=
i an isometry from J" C .J". Hence Iii(F)ii = onto J. Applying condition (ii) of Theorem 2.1.4 to the map i, we obtain that
4. REFLEXIVE HILBERT C-MODULES
80
(i(F), i(F)) < (F, F). Changing the map i by its inverse, we get (i(F), i(F)) = (F,F), i.e., (F,F) = (F(fL)*,F(f,)*) = It is easy to derive from here that (F, C) = for F, C E J". Since the module J" is complete with respect to the norm
J is closed.
0
LEMMA 4.2.2 ([1011). In a A consider the set of right ideals K, K C A, with the following property: any bounded A-module homomoTphism from the ideal J to A admits a unique extension to an A-module homomorphism
Jc
from K to A. There is a unique maximal element in this set of ideals and it coincides with J.
PROOF. Let K, J c K C A, bea right ideal in A such that any element f E J' can be uniquely extended to a map f E K'. For a E K define F E J" by the formula F(f) = f(a)*. Due to uniqueness, the map fL has to be the inclusion map K C A, hence = fL(a) = a. Therefore K C J. On the other hand, Corollary 4.1.5 shows that any element of J' admits a unique extension to an element in J'. 0 Note that J J C K are two right ideals in a A, then the inclusion J c K holds. Indeed, for any F E J" the map g from K' to A belongs to the module K". Consider now the commutative case. Let X be a locally compact Hausdorif space and let S C X be its subspace. Denote the of continuous functions on X vanishing at infinity by Co (X), the of bounded continuous functions on S by C(S) and the ideal of functions vanishing on S by C0(X, S) C C0(X). If W C X is an open subset, then it is easy to see that there exists a unique maxi-
mal open set W, WC W C X, such that any function in C(W) extends uniquely to a function in C(W). There is an obvious inclusion W c W, where W denotes the closure of W. If X is a metric space, then one can show that W = W for any open subset W. This is not true in general. For example, if X is a Stone space and if W C X is a dense open subset, then one can show that W = X [34].
Let E C X be a closed subspace and let J = C0(X, E) be an ideal in A = C0(X). Put U = X \ E and denote by Y the space of all bounded functions oii X vanishing on E such that their restrictions onto the subset U are continuous on U. The set Y is equipped with the structure of an A-module (with the pointwise multiplication), and products of functions in Y by functions in J belong to J. For g E Y define f9 J' by the formula f9(x) = gx, x E J. The following proposition describes the dual module J'. PRoposiTioN 4.2.3 ([101]). The map g between the Banach A-modules J' and Y.
is an isometric isomorphism
PROOF. The map g '—p f9 is obviously an isometric A-homomorphism. We
have to show surjectivity of this map, i.e., to show that for any f E J' there is an element g E Y such that f = f9. Without loss of generality we can assume x E J, i.e., If(x)(t)I Ix(t)I holds that If III' < 1. Then f(x)f(x) < at each point t E X. For any point t U choose a function Xj e J that satisfies = 1. Define the function h : U —' C by h(t) = f(xt)(t). Note that the estimate Ih(t) < 1 holds for all t E U. For each y E J at any point t E U one has (xty — y)(t) = 0, hence f(xty — y)(t) = 0. Therefore f(y)(t) = f(xty)(t) = f(xt)(t)y(t) = h(t)y(t).
4.2. IDEALS AND BIDUAL MODULES
81
Since f maps J into A, the function hy should be continuous on the set U for any function y E J. Now one can take g = h. Then f = 0 After describing the dual module .1' we can describe the bidual module 5 = J". Let Xu be the characteristic function of the set U. Theii the map is the inclusion : F E Y'}. map J C A and J = PROPOSITION 4.2.4 ([101]). The sets J and C0(X, X \ U) coincide. PROOF. Let D C X be the intersection of the zero sets of all functions F E Y'. By the closedness of the ideal J one has J = Go(X. D). Since J c J, One has D c E. Put V= X \ D. It is easy to see that V c U. Let. us show that in fact the equality V = U holds, i.e.. that any function in C(U) extends uniquely to a function in C(V) and that if W 3 U is an open set such that any function in C(U) uniquely extends to a function in C(W). then 14' c V.
Note that for all F E Y', g E V. t U. the equality F(g)(t) = holds. Indeed, take a function x E J with x(t) = 1. Note that gx
E
A. Since
gx = XUgx, one has
F(g)(t) = F(g)(t)x(t) = F(gx)(t) = = =
Let h E C(U); let us show that the function h can be extended to a bounded continuous function on V (uniqueness of such an extension follows from V c U). Let g E V be a function such that its restriction coincides with h. Let 14' be a precompact set that lies in V together with its closure. Choose a functional F E V' such that = land IIF(xu)II = 1 (by Proposition 4.2.1, IIF(xu)M = IlFik.,). Define the function h on W by h = F(g)(t), t E W. The function h is continuous and hi with EU = lihil. For each t E W and for any net and limo = t, one has, for sufficiently large a,
=
=
=
h(t) t Since h(t) any point t E V is contained in some neighborhood of the form W, one can use this argument to define the function h on the whole V. Now let W 3 U be an open set with the property that any continuous function
in C(U) extends uniquely to a function in C(W). Let us show that then W c V.
Take a point t E W and a function a E A such that a(t) =
1
and aix\w =
0.
For a function g e Y, let denote the unique bounded continuous extension of the function to W. Define F V —' A by
F(g)(s) =
if s E W,
(0
It is clear that F E V'. Since F(xu)(t) = t
= 1, it is easy to see that D, hence t e V, whence 14' ç V and the proof is finished. 0
COROLLARY 4.2.5 (1101]). Let E be a closed subset of a metric locally compact Hausdorff space X. Then the Hilbert Co(X)-modnle C0(X, E) is reflexive.
4. REFLEXIVE HILBERT C-MODULES
82
4.3. Reflexivity of Hubert modules over In this section we describe the reflexivity results from [124]. Let AC he the of compact operators on a separable Hubert space H, and let AC+ be the of operators of the form a = A + K, where A E C, K E AC. THEOREM 4.3.1 ([124]). Any countably generated Hubert
is reflex-
ive.
PROOF. By the stabilization Theorem 1.4.2, any countably generated Hubert module is a direct summand in the standard module 12(J(+). Therefore it is sufficient to prove reflexivity of the module Proposition 2.5.5 gives a description of the dual module:
= LEMMA
4.3.2. 1ff =
= (fi): e 12(ACj' and jfK e
AC,
then f.K = (f2K) e
PROOF. Since the operator K can be approximated by finite-dimensional operators, it is sufficient to prove the lemma in the case where K is finite dimen= sional. Notice that the operator is a positive operator, the kernel of which contains Ker K and the image of which is contained in Im K*. Since dim Irn K* aiid codim Ker K are finite, the norm convergence of the series follows from its weak convergence. 0
Let FE 12(ACj". Put F1 =
where is the standard basis of E Since M" C M' for any Hilbert module M. the sequence F = (Fi) is an element of the module 12(AC+)'. Let us prove that the series is convergent in the to the element F(F) = (F, F). Let K E AC be a finite-dimensional operator in H. By Lemma 4.3.2, E
(F,F)K = (F,F. K) = hence
K*(F,F)K = and
= for any
e H, where (•,•) is a scalar-valued inner product on H. Let 'i e B1(H),
where B1 (H) is the unit ball of H. Choose an element dimensional operator K' such that mj =
Then
= Therefore
(4.9) for any q.
ij)
II (F, F) 11
IIiiII2
E H and a finite-
4.3. REFLEXIVITY OF HILBERT MODULES OVER
LEMMA 4.3.3. Let f =
and
E
83
for all
E
i
E N. Then
F is continuous and the algebra K is closed, one can assume are finite dimensional operators. Denote by that all c H the image of the operator k, dim < oo. Let H = H of two closed infinite-dimensional subspaces. Assume first that each of a the subspaces lies in one of the siibspaces H1 or H2. Let F(f) = K + A, K E A C. Choose a compact operator k with an image L C H2 such that dim L = Replace by zeros those terms in the sequence (k1, k2,. .), for which C H1, and denote the obtaimied sequence by Then (k1k, k2k,...) = 14k....) .). because of the equivalence of the conditions = 0 awl Tm I link. Thus .
.
.
i.e.,
= = oc, one has 14,...) = K' + A for some K' E IC. Interchanging the subspaces H1 and H2, one can construct a sequence 14',...) such that = K"+A for some K" E IC. Thus Since dim L
(1.
\—I
1.
ui..'I
\
'1'
—
2
hence F(k1, k2,
. .
.)
= K' + K" + 2A.
Therefore A = 0. In the case of arbitrary subspaces V1 one can find finite-diumensional operators n1 such that
k1 + = ni, + n7, (Iii 12,.. .)
E 12(IC+),
(nil, rn2,.. .), (ni, 112....) E l2(1C1).
and the image of each of the operators H2.
in one of the subspaces H1.
0
K + A, where
Put F,
F*F = LEMMA 4.3.4.
K E IC, A,, A E C. Then
+
+
= A.
PROOF. At first let us show that the series
convergent. Suppose
lAd2 is
thecontrary. PutM = Choose
Choose,
A112
> 0 to satisfy the estimate
further, a vector
e H with
=
1
to satisfy the inequalities <E.
i=
Then the inequalities
M+
1
N.
> M+1.
REFLEXIVE HILBERT C -MODULES
84
4.
contradict (4.9). So,
0C
2
N. It follows from the inequalities (4.10), (4.11) that at the point to (where A(to) =
1)
we have
IF(f)(to)
-
(t )f(t)I
IIFII
.
for all N' > N. Thus, the series
converges to a continuous function Moreover, we shall see soon that this series is uniformly convergent, hence it converges to a continuous
F(f)(t) at all the continuity points to of the sum
function, which should coincide with the function F(f)(t) due to the property L of X. The uniform convergence of this series would follow from that of the series (4.12)
because of the Cauchy-Bunyakovskii inequality (Proposition 1.2.4). So we have to prove the uniform convergence of (4.12) on X. Since the series (4.12) converges at each point and it.s sum coincides C with the continuous function (F, F)(t) = F(F)(t) at each point of continuity. Denote by E C X the set of the continuity points for the series (4.12). Now let t0 be some point of discontinuity for the series (4.12). Without loss of generality (multiplying by some continuous function, if necessary) one can assume that there is a sequence of points E E converging to the point to and satisfying the conditions
4. REFLEXIVE HILBERT C-MODULES
=
(F,
1
and
1F1(to)12
K(E(;). Then OflV caii choose a sufhciently small neighborhood of thc point .r such that fl = 0 (I j) for the set
= 1, 92
}
= G/GJ.. Let f 1
a colit iiiuous liulinegative function with
support iiiside LI1. Then
K(Ec:)EG(f)x =
Contradiction with the definition of K (Es) completes the proof.
0
THEOREM 4.5.6 (combination of Proposition 4.5.1, Leninia 4.5.4, and Theorem 4.5.5). Suppose a discrete gn.up C acts on a locally compact Hausdorff space X in such a way that k := max{#(Gx) : x E X} < +00. Then E0 is well defined. If X is a noriTial space. then K(EG) = k. Hence, by Proposition 4.5.1, C(X) a Hubert over GG(X).
THEOREM 4.5.7. Let X be a compact Hausdorff space and let C be a group acting uniformly continuously on X. If all orbits of the action of C have the same finite number of points, then the conditional erpectation
E(f)(x)
= #(Gx)
is well defined on C(X), and the Hubert generated and projective.
>
f(g1.r)
{C(X). E((.. .))} is finitely
PROOF. The idea of the proof is contained in [133J and requires two technical lemmas.
LEMMA 4.5.8. Let X he a compact Hausdorff space with a uniformly continuous action of a group C and let all orbits contain an equal finite number of points. Then, for any point x E X and any element g E C1. one can find an open neighborhood U1 of a point x, on which g acts identically.
PaOOF. Let us denote the cardinality of orbits #(Gx) by n. Let x1 E X be the orbit of the point a: and let E C be elements such that = x1. Choose and assume that each neighborhood U1 of the point x contains some point E satisfying the y E U1 such that g0y y. Fix neighborhoods U1, of the points condition U.r, fl Us.., = 0 for i j. Then we can find a neighborhood c U1 of the point x such that c Ut,. Since the group C acts uniformly continuously, one can find a neighborhood W1 c V1 of the point x such that g(W1) ç V1 for each g e C1. If y W1 and goy y, then the orbit Cy of this point contains at
4.5. CONDITIONAL EXPECTATIONS
least n+ 1 different points {h7y E contradiction proves the lemma.
:
i=
89
E
1
The obtained
0
4.5.9. Under the hypotheses of Lemma 4.5.8. for any point x E X one of this point such that the action of the subgroup G1 on can find a neighborhood V1 is trivial.
from Lemma 4.5.8 can be put
PROOF. We should show that a neighborhood choseii for all g E simultaneously. For each g E
U1(g)={yEX:gy=y}. Suppose the contrary, i.e., that the set U1(g) does not coiitain a neighborhood of .r. It means that any neighborhood U1 of x contains some point z such that for some e C1 we have g:z :. Consider a neighborhood of a: and neighborhoods {Uh.r} for fixed representatives {h1 e} e G of cosets in C/C1 such that their intersections are empty pairwise and c U,,,.. As in the proof of Lemma 4.5.8. we find a neighborhood W,. c Va. of the point x such that g(W.E) ç V,. for each g e C1. Put. U.,. = U9EG, G(W1) c Va. It is a G1-invariant open neighborhood of the point x E X. The assumption g:z z for some z E U,., means that the orbit of the point z consists of at least ii + 1 points. 0 E C Va be a neighborhood of the point x such that the action of C,. on Let V,. is trivial. Then one can find a function f,. E C(X) such that supp f1 C V1 and = 1. For each g E G one has either (gV,.) fl V,. = 0 or gV,. 1/,.. Therefore —
JQ(f)2
if g.r = x.
10
where denotes the action of C on functions. = f(g1.r). Consider a finite covering {U,. , } of the space X by sets of the above form. Put
> 1,
v=
Note that if we take one element map
in each coset C/Ga., theim, by Lemma 4.5.4, the
Ec(f)(x) is well defined for all a: E X. f Moreover, one has
= v"2(f,.)112 e C(X).
= #(Gx)
C(X) and it is a conditional expectation omm C(X).
= f, E
C(X); hence the set {ui,
. .
.
,
is a basis of the Hilbert
CG(X)_module {C(X), EG((., .))}. Therefore this Hilbert module is finitely generated and projective. 0 Theorem 4.5.7 generalizes results of [133) and shows that if all orbits consist of an equal finite number of points, then the corresponding conditional expectation is of algebraically finite iiidex. Being finitely generated and projective, the Hilbert module A = {C(X), E((., .))} is self-dual. In the case of a finite index (when
4. REFLEXIVE HILBERT C-MODULES
90
< cc and the pre-Hilbert module A is complete) we cannot expect this module to be self-dual. However, sometimes this module is reflexive, i.e., A" = A, where A' is the dual Banach Cc (X )-module of bounded Cc (X)-homomorphisms from A into CG(X). THEOREM 4.5.10 ([47]). Let the group G act uniformly continuously on a compact Hausdorff space X. Suppose that all orbits consist of not more than ii points, and that the number of points for which the length of their orbit is less than ii is finite. Then the Hubert {C(X), .))} is reflexive.
PROOF. Describe first the dual Banach A'. Let x1,. . the points with cardinality of orbits smaller than n. One can choose open neighborhoods U1,.. . , Urn of these points in such a way that each neighborhood and if, for some U2 is invariant with respect to the action of the subgroup U3. Denote by V the C-invariant compact h E C, one has hx2 = x3, then hU2 functional on the module set X \ (U1 U . . U Urn). Let F E A' be a .
,
be
A. Consider its restriction on the Hubert {C(Y), .))}. For a by the formula function g C(Y) we take its extension E C(X) and define This definition does not depend on the choice of an extension F]y(g) = If V' V is another compact G-invariant subspace not containing the points then (Fly')ly = F]y. Since the orbit of each point of the set V has is the same cardinality, by Theorem 4.5.7, the C'(Y)-rnodule {C(V), Ec((., finitely generated and projective, hence self-dual. Denote by C(X \ {xi,. }) the set of continuous functions on the noncompact space X \ {xi,. ,Xm}. The restriction onto this space defines the map . .
,
. .
A' —b C(X
(4.15)
\ {x1,..
.
,Xrn}).
It is easy to verify that the map (4.15) is injective. Let us study local properties of functionals from A' in a neighborhood of the such points x1 Xrn. Let Xo be one of these points. It has a neighborhood The group contains a normal subgroup that if gx0 = x0, then = Choose a C0 of the elements that do not move points from the neighborhood representative 9j in each coset G/GXO. Then outside the point x0 the action of the functional F E A' can be written as
F(f)(x) =
(4.16)
>
.
and this action can be continuously extended to the point x0. Consider the orbit of the point x0. Then one can write the sum (4.16) in the = x0,x',.. .
form
F(f)(x) =
k—i f .
jrO Passing
(
to the limit (which exists by assumption), we obtain
F(f)(xo)
i =
F*(gjx)
j0 ( i:g,x0=xJ
.
4.5. CONDITIONAL EXPECTATIONS
hence there exists (for f
91
1 E C(X)) the limit F(g2x)
for any x E X \ {x1,. . Recall that the function F(x) is defined only outside the point x0. If we would like the action F on the Hubert {C(Y), .))} to be of the form (4.16) on the whole X, it would be necessary to define the function F(x) at the point x0 by . ,
F(xo) =
(4.17)
lim
F(g2x).
n x—'xo
To complete the proof we need the following lemma. LEMMA 4.5.11. The module A' is isomorphic to the module of all bounded func-
tions F(x) on X that are continuous on X \ {x1,. . , Xm} and satisfy the condition .
(4.17).
PROOF. We have to show that the image of the monomorphism (4.15) consists of bounded functions. Suppose the contrary. Then there exists a point such that A'. > n.IIFII, where IIF1I is the norm ofF in the dual Banach of the point such that Moreover, one can choose a neighborhood =0 for those elements of the group G, for which ±. Consider a function f E C(X) Then it follows from the equality (4.16) that such that f(±) = 1 and supp f C
=
.
f(±),
and the inequality
= gives
=
> IIF1I
0
a contradiction.
Having the above description of the dual module A', we can describe the bidual
module A". Since one has the canonical inclusion A" C A', and since the inner product on A can be naturally extended to an inner product on A" (see Theorem it is sufficient to verify to which functions in 4.1.4) making it a Hilbert A' one can extend the inner product. Consider a function F from A' fl A". Adding to it (if necessary) a continuous function from A, we can assume that F(xo) = 0. Then the C°(X)-valued inner product of F by itself is an element of CG(X) of the form (4.18)
(F, F)(x) = E(IF(x)12) =
for all x E X. But since (F, F)(xo) = 0, it follows from the assumption F E A" that lim F(g2x) = 0 x—.xo
for each summand of the equality (4.18). Therefore we obtain from (4.17) that the
function F is continuous at the point x0, hence the module A is reflexive.
0
Later in this section we will prove a final version of Theorem 4.5.10 due to V. Seregin.
92
4.
REFLEXIVE HILBERT C-MODULES
CONSTRUCTION 4.5.12. Consider a uniformly continuous action of G on a Haus-
dorif compact space X with bounded cardinality of orbits. Let us develop the construction used in the proof of Theorem 4.5.10. For any point x0 E X let be a neighborhood of x that is invariant under the action of the stabilizer of the given point and is disjoint from its own images under the action of the group elements not belonging to Such a neighborhood exists due to uniform continuity: indeed, take a neighborhood with all these properties except that of invariance; then choose, by the definition of uniform continuity, a neighborhood := and put Wxo with CxoWxo C In this case one can take representatives g3 (j = #Gx°) for the cosets for the subgroup C and the neighborhoods that appear in this construction are of the form = = g3(x°), where is the orbit of x0. Let X —* [0, 1], E H, be a continuous function such that . .
= Io, 11,
.
,
x=x.
the cardinality of orbits is bounded, E is well defined and one can assume is Suppose so Then C = := is invariant, equals 1 on the orbit x0 and is supported in U, U1o. All these properties, except the invariance, are obvious. If g C and y U1o. then and h E Gxo, so y gjz, where z E and gg, = gkh for some Since
that
pu(gy) =
= pxo(z) = Pj(Y)
=
Similarly, we can choose a neighborhood with the same properties as and contained in together with its closure. Then, taking a function instead of such that it equals 1 on and has support in we obtain the corresponding function THEOREM 4.5.13 ([118]). Let X be a compact Hausdorff space and let G be a discrete group acting uniformly continuously on X. If the cardinalities of all orbits are uniformly bounded by some number N, then the Hubert Cc(X)-inodule H {C(X), is reflexive.
PROOF. Denote by B(X) the space of all bounded functions on X and by H the space of all maps from H to B(X). Choose and Px for any x as in Construction 4.5.12 and define the maps H' B(X) and 71: B(X) H as follows: (4.19)
=
and (4.20)
ij(f)(h)(x) =
#
Xc
F E H' and #Gx denotes the number of points in the orbit of x, as before. Let us verify that is well defined, i.e., that the function f = is bounded: f(x)I 0 we have h
—
= IIF(hipu)(x)
J
-
>
i
<sup + 2
11111
(
>
/
-
+ instead of x°),
Hence, by (4.22) (with
ij(f)(h)(x) — ii(f)(h)(x°)I =
0.
Additivity of 77(f) follows immediately from (4.20). Now let r E C'(X) be an invariant function. Then 77(f)(hr)(x) =
=
>
x,EGx
x,EGx
= 77(f)(h)(x)r(x) and ri(f) is a For any h E H one has 77(f)(h)(x) =
>
>
x,EGx
11111
.
IIhII,
so 77(f) is a continuous functional. Hence, 11(f) E H'.
Conversely, let us prove that condition (4.22) is satisfied for any function f from the image of For f, there exists a functional F E H' such that f = Namely, F = 17(f). Consider an arbitrary point x0 and choose for it a neighborhood (see 4.5.12). Take the function (see 4.5.12) and evaluate F at this Vso c function. Then, for x E Vxo, the function defined by the formula
F(4)(x) =
=
>
z
is continuous because it lies in the image of the functional F. This is equivalent to the condition f(x1) x,
=
f(z) zEGx°flV,,o
Thus, (4.22) holds for f and Lemma 4.5.14 is proved.
0
4.5. CONDITIONAL EXPECTATIONS
95
Let us continue the proof of the theorem and let us give a functional description B(X) —* H, where H is the space of H". Define the maps ii: H" .8(X) and of all maps from H" to B (X), by the formulas
v(F)(x) = and
= x (4.23)
E
h=
X, h =
Note that
I((h)(coz))(x)I
IIhII
lihil.
Then
F(h)(x) = F (h —
+>
(x)
x,EGX
=F
(h -
h(x1)fx) (x) + F (
(x)
x,EGx
= F(hi)(x) + x, EGx
= F(hi)(x) + x, EGx
= F(hi)(x)
+>
and invariance of the functions F(f1,) was used. h — >xEGx Further, for any point y E Gx, one has
= h(y)
hi(y)
—
x,EGx
= h(y)
—
h(y) = 0,
so the function F(h1) equals 0 on this orbit. so small that Indeed, choose, as in Construction 4.5.12, the neighborhoods < e for the corresponding invariant Ihil < r holds on their union. Then function
and
=
<Elif II.
= x,EGx
So we obtain IIF(hi)(x)II =
= IIF(hipu)(x)II 0, hence
(4.24)
pv=Id.
4. REFLEXIVE HILBERT C-MODULES
96
Let us prove now that p. is injective. Suppose that f point x0 E X, we have f(x°) 0. Then
0 E B(X),
so,
for some
=
=
0,
=
0 and Kerp = 0. Suppose that f Im ii. Then there is a functional F E H" such that f = Since p is injective, by (4.24), one has F = p(f). For an arbitrary point x choose (see Construction 4.5.12). The function is continuous and one and
hence 1u(f)
has
=
=
y, EGY
=
=
of x. As was already shown, condition (4.22) is satisfied in the neighborhood for 1. Thus I is bounded due to the estimate =
If(x)I
0 and any x c X there
(Lyapunov)
p(gx, gy) the
for each
g E
C
provided
metric space X is called exists S >
0 such that
p(x, y) Oforanyb1
LEMMA 5.1.2 ([123. Lemma IV.3.21). Let B be a Ct -algebra,
1,...,n. A matrix[cjjl e
is positive
i,j = b,1 E
B.
Since for any al
the matrix
E A, one has
is positive. Therefore the clement is positive for any b1 e Att, hence [z, z) 0 for any z E M ® Att. Set E
.N={ZEM®Att :[z,z]=0}. is an Att-submodule in M 0 A*t and the quotient module M 0 At*/Ar is a pre-Hilbert Att-module. Denote by M# the Hilbert Att-module obtained by completion of M 0 Att/A( with respect to the norm given by the inner product [., .]. We call this module the extension of the module M by the algebra Att. The Att always contains the unit element and for any x E M, a E A, we have (x. a) ® 1 — x 0 a E .A/. Therefore the A-module map x x ® 1 + .ftf, is well defined. Since [xO 1 1-i-N] = (x,y), this map is an Then
isometric inclusion.
Denote by HomA(M, Att) the set of all bounded A-linear maps from M to At*. Introduce on this set the structure of a vector space over C by the formula 99
5. MULTIPLIERS
100
(Ao)(x) := where A E C. x E M. E HomA(M.A**). and also with the structure of a right A**_module by the formula b E A**. For a b)(x) functional I E (M#)' we can define the map fR E HomA(M, A**) as the restriction of f onto M. namely. fR(.r) := f(x ® 1 + 1V). Obviously 11111. THEOREM f
M. the map
5.1.3 ([100]). For any A and for any Hubert A-module fR is an isometry of (M#)' onto Ho1nA(M, A**).
>0 PROOF. Let. a matrix satisfy the condition e for any e A. Let us show that it is sufficient to prove positivity of the For this purpose, it is sufficient to show that matrix >0
(5.1)
E A**. Without loss of generality one can assume that the for any b1.. . , elements are in the unit ball Bi(A**) of the W*_algebra A**. Since the unit ball .
B1 (A) of the C*_algehra A is dense in B1 (A**) with respect to the strong* topology. oiie can find iiets E A, A E A, converging to the elements E A** with respect to the strong* topology. Then the net converges to with respect to the weak topology (see [123, whence the inequality (5.1) follows. We have to show that any map e HO1I1A(M, A**) with < 1 can be extended up to a unique functional f E (M#)' with If II 1. Consider the functional A** given by the formula fo : M A** ®ai) =
fo
map. For
Obviously fo is an
E A, one has
. aj)*Ø(xj . a3)
=
ai)
;r7
=
*
i=l
i=1
Therefore for any
i.j
E A**, the following inequality holds:
i.e.,
fo(z)*fo(z) < for any z E M ® A**. Therefore the functional f : the formula xi ® and
+
:=
satisfies the inequality f(g)*f(y) y f e (M#)'. It follows from the equality f(x ® 1 +
extension of 0.
A** is well defined by
= 0(x) that f iS all
0
5.2. MULTIPLIERS AND CENTRALIZERS
101
COROLLARY 5.1.4. Let A be a a Hubert A-module. inner Then an A-valued inner product on M can be extended up to an product on the set HomA(M, A**) making this set a Hubert COROLLARY 5.1.5.
Let A be a
A-module. Then the Hilberi
a self-dual Hubert
M# is self-dual too.
As one more corollary we shall present the following characterization of self-dual Hubert modules.
THEOREM 5.1.6 ([42, 40]). Let A be a ments are equivalent: (i) the Hilbert A-module HA is self-dual; A is finite-dimensional. (ii) the
Then the following state-
PROOF. Note that each condition of the theorem implies the existence of a uiiit in the C*_algebra A. Indeed, if A is finite—dimensional, then 1 E A. If the A defined by module HA is self-dual, then the bounded A-module map f HA E A, i e N, has to be an element of the the formula 1(a) = a1. where a = module HA itself. This means in particular that End.4(A) = and so the identit.y mapping A —* A is identified with the unit element of A. In what follows, let us use the description of the dual module H'4 as the set of all of the series sequences b = (bk), E A, such that the partial sunis = are uniformly bounded. Self-duality means that for any increasing sequence (efl) of positive elements of the C*_algebra A, its boumidedness is equivalent to its norm-convergence. But, for finite-dimensional C*_algebras, the monotone bounded sequences are convergent, and this proves the implication (ii) (i). To prove the = opposite implication we pass to the Hubert module over the enveloping 147*_algebra A**. This module is self-dual by Corollary 5.1.5. Since any monotone bounded sequence in a I'V *_algcbra is convergent, any positive linear functional on = (A**)*. Suppose that the W*_algebra A** is A** has to be normal, i.e., infinite—dimensional. Then it contains an infinite collection of mutually orthogonal projections Pk E A** such that Pk = 1. Hence there exists an inclusion of the commutative of bounded sequences into A**. Let p be a Let us extend it to a positive linear positive linear functional on the algebra functional on the larger algebra A**. According to time assumption, is normal. is normal too. Hence we have Therefore its restriction = to the algebra = This contradiction shows that the obtained a false conclusion A is finite-dimensional as algebra A** is finite-dimensional, hence the well and this proves the implication (i) (ii). 0 An interesting general theory of extensions of Hubert. C*_moclules can be foimd in [6].
5.2. Multipliers and centralizers While writing this section we extensively used results of [104, 134]. Let 8(H) be the algebra of all bounded operators on a Hubert, space H and let A be a algebra.
DEFINITION 5.2.1. A two-sided closed ideal J C A is called essential if JflJ' 0 for any nonzero ideal J' C A.
102
5.
MULTIPLIERS
REMARK 5.2.2. Au ideal J C A is essential if and only if
J± :={aeAIaJ=O}=O. 13(H) is called nondegenerate if DEFINITION 5.2.3. A representation p : A for any ii E H there exists an element a e A such that p(a)h 0.
REMARK 5.2.4. For an arbitrary representation p, we can consider its restric-
tion onto the orthogonal complement H' to the invariant subspace H p(A)h = 0}, which is invariant as well. The new representation p' : A —+ 13(H') is then nondegenerate. Thus, roughly speaking, we lose nothing working only with nondegenerate representations.
LEMMA 5.2.5. A representation is nondegenerate if and only if p(A)(H) is dense in H. PROOF. Let a representation p be nondegenerate and let h±p(A)(H), i.e., for any f E H and any aE A, 0= (h,p(a)f) (p(a*)h,f) holds. Then p(b)h = 0 for any b e A, hence h 0. Conversely, let p(A)(H) = H and let h e H be an arbitrary nonzero vector. Without loss of generality one can assume that lihil = 1. Since p(A)(H) is dense in > 1/2, H, one can find g e H and a E A such that Ih—p(a)gII (h — p(a)g,h — p(a)g) = 1— + 1/4, p(a*)h (g,p(a*)h) + (p(a*)h,g) > 1,
u 13(H) be a faithful nondegenerate representaDEFINITION 5.2.6. Let p: A tion, so we can assume A c 13(H). An operator x e 8(H) is called a (two-sided)
multiplier of A if
axEA for each a e A. Denote by M(A) the set of all multipliers. It is easy to see that M(A) is au involutive unital algebra. REMARK 5.2.7. It seems that the definition of multiplier depends on the choice
of a (nondegenerate faithful) representation. In Theorem 5.2.11 we shall see that this is not the case. PROPOSITION 5.2.8. The set M(A) is a unital C A
an essential ideal in M(A). If A has no unit, then
C M(A).
PROOF. Three statements are nontrivial: 1) that M(A) is norm-closed, 2) that the ideal A is essential, and 3) that there exists an inclusion into the bidual space. x with respect to the norm, Xe. E M(A), x E 13(H). Then 1) Let x0a xa and ax for any a. Since A is closed, xa E A and ax E A for any a, i.e., x e M(A) by definition. 2) Let J be an ideal in M(A), JflA = 0, and let x E J be an arbitrary element. Then xa E A (since x is a multiplier) and xa E J (since a E A C M(A) and J is an ideal) for any a E A. Therefore xa E J fl A = 0, xa = 0 for any a E A. Then x = 0 by Lemma 5.2.5.
5.2. MULTIPLIERS AND CENTRALIZERS
103
A** (cf. remark after Theorem 3.1.3), it is sufficient to For this purpose, note that for any x E M(A) and for any
3) Since A!!
prove that M(A) C
weakly convergent net aA E A, we have
x w-limaA = w-lim(xaA) AEA
.XEA
E
since xaA E A. where [AIW is the weak closure of A in 8(H) and where w-Iim = A!! due to denotes the limit with respect to the weak topology. One has = A!!. Since the nondegeneracy of the representation, hence XA!! = C
1EA",onehasxEA1'.
D
The next definition historically precedes Definition 5.2.6. DEFINITION 5.2.9. A pair (L, R) of maps L:A
R: A
A,
A,
R(a)b = aL(b) for each a, b E A,
is called a double centralizer of A. Let us denote the set of all double centralizers of A by DC(A). PROPOSITION 5.2.10. Let (L,R) e DC(A). Then (i) L(ab) = L(a)b and R(ab) = aR(b);
(ii) L and R are linear; The
(iii) L and R are bounded and IILIt = IIRII. set DC(A) with the operations
(Li,Ri)+(L2,R2):zr(Lj+L2,Ri+R2),
zEC,
z(L,R)=(zL,zR),
(L1,Ri)(L2,R2) := (L1L2,R2R1), (L,R)* := L*(a) (L(a*))*, R*(a) := (R(a*))*,
a
A,
is a normed involutive algebra with respect to the norm (L, R) II :=
= II RD.
PROOF. 1) Let a,b E A, z E C and let ea (ck E A) be an approximate unit of A. Then eaL(ab) = R(ea)ab = eaL(a)b, + zb) = CQL(za + zb) = = = + L(za + zb) = z(L(a) + L(b)).
L(ab) = L(a)b,
+ R(ea)zb + L(b))),
+
2) Thus L is a linear operator on the Banach space A and, to prove its continuity, it is sufficient to check that its graph is closed. Let a and —p b. Then IIv(L(a)
— b)II
IIvL(a) —
+
—
= IIR(v)(a — an)jj + IIvL(an) IIR(v)II . ha —
+ hivhl
—
vbII
vblI
hhL(an) — bhI
0
for any v E A, hence vL(a) = vb, whence b = L(a), since v was taken arbitrarily. Thus the graph of L is closed, so L is continuous. Properties of R can be verified in the same way.
5. MULTIPLIERS
104
3) Let us compare ILl! and IL!!2
=
IIRII:
IIL(a)*L(a)II = sup IIR(L(a)*)aII
IIL(a)112 =
IIulI=1
IIaII=1
< sup IIRII IIL(a)*II Mall < sup IIRII
IILII
11a112
=
IIRII
IILII.
IIaII=1
IIaII=1
IIRII. A similar calculation gives the opposite inequality. The remaining statements are obvious, one has only to verify that
whence IILII
R2(R1(a))b = R1(a)L2(b) = aL1(L2(b)).
0 THEOREM 5.2.11. The map
DC(A),
p : M(A)
x
(Li,
= ax.
=
is an isometric *-isomorphism between M(A) and DC(A). Therefore DC(A) is a and M(A) does not depend on our choice of a nondegenerate representation.
so the pair (Lx, PROOF. First of all one has aL1(b) = axb = belong to DC (A). The linearity of the map p is clear. Also,
= (xy)a = x(ya) = = a(xy) = (ax)y = = (LXL,J.RYRX) =
does
= =
thus p. is a honiornorphisin of algebras. It is involutive,
= (Lxa*)* = (xa*)* = = (a) = (Rxa*)* Since IlxalI
*
= =
(a).
IIxlI hall, one has 11L111 lIxIt. Conversely, since the representation is x with respect to the strong topology, where
is nondegenerate,
an approximate unit of A. Indeed, for any a E A, the net
a is norm-
xa with respect to the norm. From the nondegeneracy we obtain the strong convergence x on the dense set AH, and, due to the houndedness, lIxM, the strong convergence takes place everywhere on H. Let > 0 be taken arbitrarily. Choose 11. e H such that IIhM = 1 and xh and one can find o such that Ia'!! !IxhhI +e/2. Since .r = <E/2. Then lIxlI +&. Therefore hlxh — convergent, whence
I
= and
lxii, since
IlxII —
is arbitrary. Hence p is an isornetry.
It remains to verify that Imp = DC(A). Consider an arbitrary element are bounded and by the weak and compactness of the unit ball 8(H), they have points of accumulation .rL E A!!, respectively, with respect to the weak topology. Passing, if necessary, to XR e sub-nets, we can suppose without loss of generality that
(L. R) C DC(A). Then the sets
=
L(eQ) and
XR =
5.2. MULTIPLIERS AND CENTRALIZERS
105
Then ZR = XL. Indeed, for any a,b E A we have axLb =
a
=
axRb =
a
=
= aL(b) = R(a)b,
a
= R(a)b,
whence XL = ZR (cf. the proof of the second part of Proposition 5.2.8). Let us denote x XL = ZR. Then
= xa = Lr(a).
L(a) =
R(a) = w-liinR(ae0) = ax = in particular, x E M(A). Tile above equalities show that p.(x) = (L, R).
0
EXAMPLE 5.2.12. 1) The equality M(A) = A holds if and only if A is unital. 2) For a commutative algebra A = c0(X) one has
c(Iix).
M(c0(X)) = Cb(X)
where Cb(X) is the C*_algebra of all bounded functions with the sup norm and I3X
is the Stone-Cech compactification of X. of compact operators one has 3) For the algebra = = 8(H). The proof of 1) and 2) can be found, for in [1341 and 3) will be proved below in a more general situation (see Theorem 5.3.1). DEFINITION 5.2.13. Let p: A 8(H) be a faithful non(legenerate representation, so we can assume A c 8(H). An operator x E 8(H) is called a left multiplier of A if Xe E A
for ally a E A. Denote by LM(A) the set of all left multipliers. It is clear that they form a unital algebra. Similarly one can define right multipliers, RM(.4). An operator x E 8(H) is called a quasi-multiplier of A if axb E A
for any a, b E A. Denote by QM(A) the set of all quasi-multipliers. It is clear that they form an involutive linear space. DEFINITION 5.2.14. A linear map A A A(ab) = A(a)b,
A is called a left centralizer if
for each a. b
E
A.
Similarly one defines a right centralizer. Denote the spaces of left and right Cciitralizers by LC(A) and RC(A) respectively.
DEFINITION 5.2.15. A linear map q: A x A —* A is called a quasi-centralizer if A0 E LC(A), where A0 : b '—' q(a, b),
for any a, b E A,
PbERC(A),
foranva, bE A.
and
In other words,
q(ca,bd) = cq(a.b)d, LEMMA 5.2.16. If p E RC(A), then
for any a, b. c, d E A. E LC(A).
5. MULTIPLIERS
106
(p(a*))*. Then PROOF. Recall that p' is defined by p*(a) p*(ab) = (p((ab)*))* = (p(b*a*))* = (b*p(a*))* = p*(a)b
D LEMMA 5.2.17 ([104, Lemma 3.12.21). Each right centralizer, each left centralizer and each quasi-centralizer is bounded.
PROOF. Let p E RC(A). Suppose that it is unbounded. i.e., there exists a < 1/n and E A such that > n. Then the element there exists an is well defined. By Proposition 1.1.5, for each a := Then the equality element = e A such that IIa"611 and sequence
IIp(xn)II = Ilunp(a"3)II
gives a contradiction, hence p is bounded. In a similar way one can prove that any left centralizer is bounded too. Thus, q E QC(A) is continuous separately in each variable as a map A x A —i A. By the uniform boundedness principle, such an D operator is jointly continuous (see [35]). PROPOSITION 5.2.18 ([104, Prop. 3.12.3]). Let A
13(H) be a nondegenerate
faithful representation. Then there exists a one-to-one isometric linear correspondence between left, right and quasi-multipliers and left, right and quasi-centralizers, respectively. In the first two cases this correspondence is a homomorphism of algebras. in the third one it is a homomorphism of involutive spaces.
PROOF. The correspondence for the left and right multipliers, together with its properties, were already described in Theorem 5.2.11. Let q E QC(A) and let x E A!! be an accumulation point (with respect to the weak topology) of the bounded directed net {q(ea, eQ)}, where {eQ} is an approximate unit for A. Passing, if necessary, to a sub-net, we can. as before, assume that x = w-limQ q(eQ, ea). Then A
=
q(a,b) =
axy
for any a. b e A, so x e QM(A) C A". The required properties can be verified
0
exactly as the similar properties in 5.2.11.
B. PROPOSITION 5.2.19. Let A be a (closed two-sided *-) ideal of a B M(A) such that its restriction
Then there exists a unique homomorphism to A is the identity map.
PROOF. Put := (L,,, Rb), i.e.. Lb(a) = ba, Rb(a) = ab, where we identify E DC(A). M(A) = DC(A). Since A C B is an ideal, ba E A and ab E A, so Thus, obviously : A '—p M(A). Now assume that in addition to 'y, there exists a homomorphism ö: B —' M(A)
that coincides with 'y on A. Then ö(b)a = ö(b)ö(a) = ö(ba) =
ba,
'y(b)a = 'y(b)y(a) = y(ba) =
ba
for any b e B and a E A, i.e., -r(b) and ö(b) coincide as multipliers of A, hence
0
COROLLARY 5.2.20. Let p: A 13(H) be a faithful representation of A and let A C B be an ideal. Then there exists a representation of B that extends p.
5.3. MULTIPLIERS OF A-COMPACT OPERATORS
107
PROPOSITION 5.2.21. Let A and B be
a a morphism
be
M(A)/A —, M(B)/B, which makes the diagram
a morphism A
H1
B
commutative. If
M(A)
M(A)/A
M(B)
M(B)/B
is an isomorphism, then
and
PROOF. Let (L,R) e DC(A). Define L,R: B
are isomorphisms too. B by
:= p(R(a)). b e B, b = Let us show that these maps are well defined. Let ec, be an approximate unit of the A and let b = Then 1(b) := ço(L(a)),
cp(L(a) —
L(a')) = lim
e0L(a')) = lim ço(R(e0)) ço(a — a')
—
A similar equality holds for right multipliers. Since for b1 = one has
=
p(a2) =
=
=
0.
and b2 =
= 131L(b2),
DC(A) —i DC(B) by ço"(L, R) = (1, ñ). Then it is a This map induces a map for quotients. that extends Indeed, if (x — y) e A, x, y E M(A), then — y) = — y) e B. We have obtained the desired commutative diagram. = If now is an isomorphism, then Hence, is an isomorphism and the inverse map is defined by
(1,
E
DC(A). Define
:
*-morphism of
(L,R) By the five-lemma,
Row).
0
is also an isomorphism.
REMARK 5.2.22. The homomorphism of
o
o
coincides with the canonical extension
to Au restricted onto M(A).
5.3. Multipliers of A-compact operators THEOREM 5.3.1 ([63]). Let M be a Hubert A-module. Define the map
DC(K(M)),
T '—' (T1,T2),
0x,Ty.
0Tx,y,
Then q5 is an isomorphism.
PROOF. First of all, note that OTx,yZ = Tx(y,z)
=
Ox,T*yZ =
x(T*y,z) = x(y,Tz) =
Osy
oT(z),
so that T1 and T2 can be defined in an equivalent way (and for all compact operators simultaneously) by the formulas
T1(k):=Tok,
T2(k):=koT,
108
5.
MULTIPLIERS
From these equalities we obtain at once that T1 and T'2 are well defined as maps is a two-sided ideal) and are bounded
IC(M) —+ IC(M) (since AC(M) C by 11Th. Since
T2(k1)k2 = k1Tk2 = k1T1(k2),
k1, k2 E
(T1,T2) E DC(AC(M)). Since
(TS)1(k) = TSk = T1(S1k),
(TS)2(k) =
kTS = S2(T2k),
is a homomorphism of algebras. It also respects the involution:
=
=
=
=
=
=
= (T*)2,
O.r.yT*,
= (T*)i.
is algebraically injective. Indeed, let T1 = 0 and T2 = 0. Then for = T.r(Tx,Tx) holds, whence (Tx, Tx)3 0 and any x E M, 0 = Tx = 0. Hence T = 0. To prove the surjectivity of 0. we construct an inverse continuous map 0. Let (T1, 7'2) be an clement of DC(IC(M)) and let x E M. Consider the limits
The map
(5.2)
T(x) := limT,1(x),
(5.3)
T*(x) =
:=
:=
+ 1/ri]', [T2(O.t.x)]*(x)[(x,x) + 1/n]'.
Let us prove that these limits exist. By Theorem 5.2.11, (T1, T'2) = (LF, Rp), where
F E M(IC(M)). Then (T1(k))*Ti(k) = (Fk)*Fk = k*F*Fk < IIF!12k*k = T2(k)(T2(k))* = kF(kF)* = kFF*k* llFlI2kk* =
where the inequalities are the operator inequalities in (T,1(x) —
lIT1 l12k*k,
)
Then
T71(.r) —
= {[(x,x) + 1/rij' — [(x,x) + {[(.r, x) + 1/n.j_i — [(.r, x) + 1/rnj_1}
< {[(x,.r) + 1/ni_i — [(x.x) + 1/fliil}((Tl(ozX))*TI(OXX)(x),x)
x {[(x,x) + 1/n]' — [(x,x) + 1/m]'} {[(x,x) + 1/n]' x
—
[(x,x) +
{[(x. x) + 1/nj' — [(x, x) + 1/m]1}
= l1T1I12{[(x,x) + 1/nj' — [(x,x) + x {[(x,x) + 1/n]' — [(x,x) + 1/rn]1} }2 = ItT1 ll2(x, x)3{[(x, x) + 1/n]' — [(x. x) + Thus, the Cauchy criterion of convergence for (5.2) is the same as for the limit from
Lemma 1.3.9. Therefore the convergence is proved. The convergence of (5.3) can he proved similarly. We obtained the maps T and T* defined everywhere on M.
5.3. MULTIPLIERS OF A-COMPACT OPERATORS
109
Also, by Lenuna 1.3.9,
(x,T*y) =
[(y,y) + 1/ri]')
[(y,y) + 1/n]')
= lim
+
= lirn
+ 1/nr',y. [(y,y) + 1/n]')
=
= (Tx,y).
+ 1/n]1,09.9(y) [(y,y) +
Em
It remains to verify that
Hence, by Lemma 2.1.1, T, T* E (T1,T2). Put
/
+
=
iV' —)
and note that, by Lemma 1.3.9,
= x(y, z)
(5.4)
lim x(x,
(y, z) = lim
Therefore
=
=
=
= T(x)(y,z) =
0
A similar argument for 7'2 completes the proof.
It is easy to obtain the following extension of this theorem. THEOREM 5.3.2 ([73, Theorem 1.5]). There exists an isometric isomorphism of Banach algebras
0: EndA(M) —+ LM(K(M)) that extends the homomorphism 0 from Theorem 5.3.1. PROOF. As usual, unitalizing, if necessary, we can assume that the A is already unital. As before, define 0 by the formula
0(T)(k) = Tk, k e JC(M), so that it extends 0 from Theorem 5.3.1. Then the calculations presented in the proof of 5.3.1 for T1 show that 0 is an algebraically injective homomorphism of 1. To prove that it is surjective, define the continuous inverse algebras and map
for 0 similarly to 5.3.1 by
:= Em
:=
S E x E M. By the same argument as in Theorem 5.3.1, the above limit exists. It defines an A-linear map. The boundedness of the operator : M —+ M and the continuity < 1, we have (see [100, 3.11] of can be verified as follows. For any x e M, and Proposition 3.4.1)x = n(x,x)'/2, whereu E (M#)', and one has u(x,x)0 EM for any /3 > 0. Put
y := u(x, x)3
/2
EM,
EM,
z:
Then
(x,y) = (x,x)3G,
(z,z) =
1/2112
Passing to the limit E —i
(y, z) II *
I
)1*
(x', x') 1/2
1
I
.
II
. II
(y, z) 112
(x, x) —E (y, z) (z, y) (x, x)
.
(x, x) —E
0, we obtain IIWE
11Th
(x, x)
lIz 112.
y)112 II
and
1
(x', xi') 1/2(1/ 1/) 1/211 II (x, x)
Therefore ftC(M). Since
1/211
hizIl.
Thus, cb(T) AC(M)xK(M) :
cb(T)
0(T) (Ok'x',y', Ox.ky)
) —
we
—
have 0(T) E QC (K(M)). Moreover, the previous calculation shows that
is
continuous: 11011 < 1.
Let us show the algebraic injectivity of
i.e.,
that Ker = 0. Let T
0. This
means that there exists a vector x E M such that T(x) is a nonzero functional, T(x)(y') 0 for some (nonzero) y' E M. Then (T(x)(yI))*T(x)(ij) IIT(x)h12 (y', (y', 1/) Therefore (1/ . (T(x) (1/) )* (T(x) (1/))
=
.
(T(x) (1/) )* (T(x) (y')))
(1/, }3
IIT(x)1h2{ (T(x)(yI))*T(x)(yI)
0
and Ox,y'.(T(x)(y'))
.
.
(T(x)(yI))*(T(x)(yI))I)
= (Oy'.(T(x)(y')) (T(x)(y')).y'(T(x)(y')) (T(x)(y')) [1/ . (T(i) (ye) (T(x) (1/))],
= (y'.
(1/.
1/. (T(x)(yI))*(T(x)(1/)) . (1/. (T(x) (T(x) 1/ . (T(x) (y') ) * (T(x)
1/ .
0.
Thus, 0(T) 0 and the algebraic injectivity of 4' is proved. To prove that the map 4 is surjective and isometric, it is sufficient to define, as in the previous theorem, a map i,1': QC(K(M)) —+ End(M,M'),
= S,
In the last expression we used the presentation of quasi-multipliers in the form of quasi-centralizers. On the other hand, y' = v'. (v',v'), x = v• (v,v) and S(Ox',y',
= = Ox',v'9S(O,,, ,,, ,8,,,,)v.y
6x'.y (S(8,,,,,, 8,,,, )v.v').
Statement 4 is proved, and this completes the proof of the theorem.
0
5.5. Strict topology DEFINITION 5.5.1. Let A 8(H) he a nondegenerate faithful representation of a A. The strict topology on 13(H) is the topology satisfying one of the following (obviously, equivalent) conditions: (i) it is the weakest topology, for which the maps ra : 13(H)
13(H),
Ia : 13(H) —÷ 13(H),
ra : x i—+ xa, Ia : x
x E 13(H), a E A,
ax,
are continuous; (ii) it is the topology generated by the system of seminorms IIxaII,
IIaxII.
Usually this topology is denoted by /3 analogously with the Stone-tech compactifi-
cation (cf. 5.2.12). For example, we denote by
the space of the maximal ideals
for the closure of the algebra C(X) in 13(H) with respect to the strict topology, and we denote by /3 -lim the limit with respect to the strict topology. PRoPosITIoN 5.5.2. The set M(A) is strictly closed, [Al0 c [M(A)]0 = M(A).
5.5. STRICT TOPOLOGY
115
PROOF. Let the net {XQ}aEA c M(A) be strictly convergent to x E 13(H). Then for any a E A there exist the norm-limits
R(a) :=
L(a) := defining the maps L, R: A
A. Since
=
aL(b) = a
b = R(a)b,
the pair (L, R) is an element of DC(A), i.e., a double centralizer. Identifying double centralizers with multipliers, by Theorem 5.2.11 we obtain that y E M(A). Then y. Indeed, 13
ya — xaa
= L(a)
—
xaa —p 0,
ay — axa
= R(a)
—
axa
0
0
with respect to the norm topology. So, M(A) is fl-closed. LEMMA 5.5.3. The net M(A) 1 = fl-lima
is
an approximate unit for A if and only if
PROOF. The result follows immediately from the definition.
0
PROPOSITION 5.5.4. (i) The conjugation in M(A) is fl-continuous.
(ii) Multiplication in M(A) by a fixed element is fl-continuous. (iii) Multiplication in M(A) is jointly fl-continuous on bounded sets. PROOF. Let x E M(A) and let
xa,
x. Then
ax
axa
for any a E A,
whence, after conjugation,
(b=a*),
foranybEA i.e.,
—f-'
Now let y e M(A) be a fixed element and xe,, —i x. Then 13
I(xay)a — (xy)aII
—
— a(xy)Il
x(ya)Il —p 0
— axII .
for any a E A. This means that
xy.
(3 for any for any a E A and —' y, 'y E r. Then, for any a E A and for any > 0, there exists a pair (ao, 'yo) such that for any pair (cl,-y) > and > 'ye) one has e A x r (i.e., for a >
Now let XQ
13
x,
IIXaD
II(xaYy)a — (xy)aIl
— —
—
a(xy)II
— —
axII
yalJ
+ + + +
xyall
—
— x(ya)II <E, —
axyll
—
(ax)yII
0
This means that (xy) =
THEOREM 5.5.5. The algebra of multipliers M(A) coincides with the fl-closure
ofA inB(H), M(A) =
[A]13.
5. MULTIPLIERS
116
PROOF. By Proposition 5.5.2, it is sufficient to prove that M(A) C
Let
be an approxiniate unit for A. By Lemma 5.5.3, 1 E M(A). Since x is a Let x E M(A) be an arbitrary element. Consider the net multiplier, the elements of this net belong to A and, by Proposition 5.5.4, the net is 8-convergent. Hence, y = But y = x. Indeed, for any e a E A, one has
ay =
a
yb =
=
= ax,
=
=
xb,
so x is the same multiplier as y.
D
DEFINITION 5.5.6. Let A '—* 8(H) be a nondegenerate faithful representation A. The left strict topology on 13(H) is the topology satisfying one of a of the following equivalent conditions: (i) it is the weakest topology for which the maps
xEB(H), aEA,
ra
are continuous; (ii) it is the topology generated by the system of seminorms
:= lixall. For the left strict topology, one can prove the following analog of Theorem 5.5.5.
THEOREM 5.5.7. The algebra of left multipliers LM(A) coincides with the closure of A in A!! with respect to the left strict topology.
PROOF. This statement can be obtained by the same method as in 5.5.5, it is D sufficient to use only the "left half" of the argument. For Hubert modules, it is natural to consider the following two topologies on the space of bounded homomorphislns. DEFINITION 5.5.8. Let M be a Hilbert A-module. The strong module topology on End(M) is the topology generated by the system of seminorins sh(x) := IIx(h)II, x E End(M), {sh}hEM
and the *-strong module topology on End*(M) is the topology generated by the system of seminorms {sh;
sh(x) := IIx(h)II,
:=
x E End*(M).
PRoPosITIoN 5.5.9. The strong topology is not weaker than the *-strong module topology on bounded sets in IC(M) (hence, by Theorems 5.5.5 and 5.3.1, everywhere
on End*(M)). The left strong topology is not weaker than the strong module topology on IC(M) (hence, by Theorems 5.5.7 and 5.3.2, everywhere on End(M)). The corresponding topologies coincide on bounded sets in
PROOF. We will verify the equivalence of the appropriate seminorms. We have sh(x) = IIx(h)II = IIxk(g)II IIxkII = .
for some k e 1C(M), g e M (see Lemma 2.2.3) and
= jx*(h)II = IIx*k(g)II
. ugh = IIk*xhI
=
(x)
5.5. STRICT TOPOLOGY
117
for some k e K(M), g e M. Conversely, let k e K(M) be an arbitrary element and let 0 with respect to the strong module topology. Assume that the net < c. Then for any E > 0 there exist vectors h1,.. . and Xa is bounded: from M such that 91,... ,
and there is
such that
= 1,... ,fl,
for any
=
IIXakII
c•
.
+
=
IIxaII + >
c.
E+
+ PROPOSITION
5.5.10. Let A and B be
a morphism such that B C
PROOF. Let
= MgiII
Then
is strictly continuous.
x in M(A) and let b be an arbitrary element of B. Then
for some a E A. The nets
b=
and axa are norm-convergent in A. does not increase the norm. = Therefore = and are Cauchy nets, so they are -L y E M(B). norm-convergent. Since b was arbitrary, this means that Thus, for any b' e B, b' = one has The map
yb' =
being a homomorphism of
=
=
= i1(xa') = D
REMARK 5.5.11. In particular, the map from Proposition 5.2.21 is the extension by continuity of the morphism thereby, is unique. A similar theory can be developed for quasi-multipliers, if one uses the following definition.
5. MULTIPLIERS
118
DEFINITION 5.5.12. Let A 13(H) be a nondegenerate faithful representation of a A. The quasi-strict topology on 8(H) is the topology satisfying one of the following equivalent conditions: (i)
it is the weakest topology for which the maps
xEB(H),a,bEA, are continuous; (ii) it is the topology generated by the system of seminorms {Vab}a.bEA,
l/ab(X) := IaxbII.
5.6. Multipliers and Hubert modules. The commutative case The following results describing the modules 12(C0(X, A)) and spaces of operators on them in terms of spaces of maps are obtained by combination and small modification of [41, 1].
DEFINITION 5.6.1. Denote by C0(X,M) (resp. by A0(X)) the space of continuous maps X M (rcsp. X A) vanishing at infinity.
Note that Ao(X) is a with respect to the sup-norm and C0(X, M) is a Hilbert Ao(X)-module with the inner product given by (f,g) = where f,g E C0(X,M), XE X. DEFINITION 5.6.2. Let X be a locally compact Hausdorif topological space and let M be a Hilbert A-module. Let us call the pair (X, M) compatible if the following conditions are satisfied:
(i) the j : M ®c c0(X) —, c0(X,M),
j(m® f)(x) := f(x)m, m EM, I E c0(X),
is an isometric isomorphism;
E C0(X,M) be such that p(x0) = 0 for some x0 E X and let F E EndA0(x)(Co(X,M)) be an arbitrary operator; then (Fp)(xo) = 0.
(ii) let
REMARK 5.6.3. Here and further, by a tensor product we always mean the projective tensor product. In the case of C*_algebras it is also called spatial or minimal. However in (i) (e.g. if M = A) we use the tensor product with a commuon this tensor product coincide. tative algebra, which is nuclear, so all For more details see [94, §6.3].
REMARK 5.6.4. Generalizing [41], we shall prove in the following two lemmas that the pair (X, 12(A)) is compatible (in [41] the case of compact X is considered). Note that a weaker compatibility condition for an arbitrary pair (namely, if in (ii) we replace End by End*) follows from [1]. Thus, the second part (the one concerning adjointable operators) of Theorem 5.6.8 [41] below can be deduced from [1] using identification of (M) with multipliers for LEMMA 5.6.5. The map
j : 12(A0(X))
C0(X, 12(A)),
I = (11,12,...) is an isometry.
j(f)(x) := (fi(x), 12(X),...), E 12(A0(X)),
5.6. COMMUTATIVE CASE
119
PROOF. It is clear that j is an isometric inclusion. Let us show that it surjective. Let p E C0(X, 12(A)) be an arbitrary element. Then, since
0 and let K C X be a compact set such that the inequality <E holds for any yE Y := X\K. For each point x E K choose a number n(x) such that we have
< i—n(x)
Since the map
is continuous, for each x E K one can find an open neighborhood that
in K such
<E 7=n(x)
for each z
Due to the compactness of K one can choose a finite subcovering Take ii := m > n,
sup xEX
<max
, sup
}
maxIsup I
<max
xEY
*e} =
3=1
E.
By the Cauchy criterion the series is convergent.
U
LEMMA 5.6.6. Let E c0(X,12(A)) and let FE EndA0(x)(Go(X,12(A))) be an arbitrary operator. Suppose that co(xo) = 0 for some x0 E X. Then (Fço)(xo) = 0.
PROOF. By 2.1.4, one has (Fço,Fço)
(Fp(xo),Fp(xo)) = (Fp,Fp)(xo)
0 an arbitrary number and m E M an arbitrary (fixed) element. One can find an open neighborhood U of the point x such that —
<E
5. MULTIPLIERS
122
holds for any y E U and for some map of the form IS(T)*(x)m — S(T)*(y)mII =
Then one has
<E
—
0
foranyyEU. GL*(A))
Denote by 13(X, GL*(A)) C (pointwise) inverse. COROLLARY 5.6.9. The
the set of functions with bounded
homomorphism J defined above gives an
isometric iso-
morphism of the groups GL(A0(X))
8.(X,GL(A)),
8(X,GL*(A)).
GL*(Ao(X))
PROOF. Since J is a homomorphism of algebras, the statement immediately
0
follows from its unitality.
Note that one has to be cautious when identifying different classes of operators in the standard Hilbert module over a commutative with continuous sets of operators of the same class on a Hilbert space. For example, although an operator of finite rank on the Hilbert module Hc(x) defines a continuous set of operators of finite rank on a compact space X, the inverse statement is not true, as can be seen from the following example obtained by D. Kucerovsky [67]. It is interesting that this example is of topological origin. Denote by the standard tautological vector bundle over the complex projective space CP(n). For information about vector bundles and their characteristic classes we refer the reader, for example, to the books [49, 60]. Let F(ert) be the Hilbert C(CP(n))-rnodule of sections of the bundle
5.6.10 ([67]). Let K be a compact operator with an algebraically nimage on the Hilbert C(CP(n))-module r(en). Then there exists a point x0 e CP(n) such that K(xo) = 0, where K(x) E C(CP(n),K.) is the ftC-valued function defined by the operator K. LEMMA
generated
PROOF. The operator K has the form sk(rk,-), where 8k, rk are continuous sections of the bundle Let E = be the vector bundle equal to the direct sum of ii copies of the bundle Then si Sn is a section of the bundle E. Let us calculate the higher order Chern class of the bundle E. We have .
Cn(E)
because 0. The nonvanishing of cn(E) means that any section of the bundle E vanishes at some point. In particular, the section s1 ED.. . ED vanishes at
some point x0 E CP(n), hence all the sections x0 too.
i=
1,
. .
.
, n,
vanish at the point
0
EXAMPLE 5.6.11 ([67]). Let
X =JJCP(n) be the disjoint union of complex projective spaces and let X+ be the one-point corn-
pactification of X. Define a Hilbert fl as a direct sum of spaces of sections of the bundles The rncdule fl is countably genfl = erated, so it can be realized as an orthogonally complemented submodule of the
5.6. COMMUTATIVE CASE
123
standard Hubert module Hc(x+). Define the compact operator K on the module fl by the formula
where s = E fl. The operator K defines a continuous family of rank one operators over the space X+. However, since the set K(x) does not vanish at any
point of X, by Lemma 5.6.10 the operator K is not an operator of a finite rank on the module fl. At the same time, since the set K(x) is continuous, K E K(fl). Extending the operator K by zero, one can obtain a compact operator on the module Hc(x+) possessing the same property. Now we consider nonadjointable operators in the commutative case [54]. Let A = C(X), where X is a compact Hausdorif space. We will use the shortcut ICA for the space IC(HA) of A-compact operators on the Hilbert module HA = 12(A). We also will use the characterization of compact operators from Proposition 2.2.1. As before, let be the module generated by the first n vectors of the standard basis and let be the orthogonal projection onto that module (for A = C we use the notation pg). The space of all operators F: 12(A) —' 12(A) satisfying the conditions of Definition 2.7.4, except, possibly, the condition to be adjointable, will be denoted by TA. THEOREM 5.6.12. An operator K E End HA is compact if and only if, for the map S introduced in the proof of Theorem 5.6.8, S(K) maps X to K and S(K) X
/C
is norm continuous.
PROOF. Let us prove that 8(K) is norm continuous if K E ftCA. Choose an
arbitrary point x E X and a number c > 0. There exists a number n E N such that IIK(1 — Pn)II < e/4. Note that S(K) is strongly continuous. Hence, = is norm continuous. Thus, we can choose a neighborhood x such that —S(Kp71)(x)II <e/2 for each y E Then, for these y, the following inequality holds: IS(K)(y) — S(K)(x)II
pn))(X)II +
+ E
IIS(K(1 —
++ E
=
—
E.
Now let us show that S(K)(x) E ftC for each x E X. For an arbitrary e > 0 choose a number n such that IIK(1 — Pn)II 0 an integer and the space of rank n projections in H. The strong topology and the norm topology coincide on
PROOF. Take Po E and e> 0. Let (fl,... be an orthonormal basis of po(H). If a projection p E is sufficiently close to P0 with respect to the strong topology, the norms are arbitrarily small and i = 1, n, is a basis — of pH. Hence, there exists a neighborhood (with respect to the strong topology) V of P0 in such that the inequalities . .
.
,
i= 1,...,n, hold for the orthogonalization an arbitrary element with iieii
of the basis . .. , 1. Then, for p E V. one has
=
—
.
Let
H be
—
EXAMPLE 5.6.15 ([54]). The following example shows that there exists a strong-
1c such that J(F) Let us identify 12 and L2[0, 1]. Let be the set of all functions f E L2[0, 1] such that 1(x) = 0 for x t. Let us denote by the orthogonal projection L2 —, il'. Define the following family of isometries [33]:
ly continuous map F: X
L2[0, 11, t E (0,1], f E (Utf)(x) := Ut: Let R E 1c be a Fredhoim operator with index R = 1. Suppose also that RR* = 1 and R* R = 1 — p', where p' is a one-dimensional projection. Define a strongly continuous map F: [0, 1] 1c by the formula
F(t) Since F*(t) =
(1
—
+
—
+ UI1RUtPt
fort> 0.
fort=0. one has
F*(t)F(t) = 1-Pa + F(t)F*(t) = 1
= 1+ K1,
P(t) + P(t) = 1, where K1 is some rank one operator. Hence, F(t) E 1c. We have —
indexF(0) = 0 and, by Theorem 5.6.13, we see that J(F)
1
indexF(1) A = C[0,
1].
5. MULTIPLIERS
126
Thus, an arbitrary strongly continuous map F : X
'1c does not define an Nevertheless, one can define a new topology on cIC (independent
operator in
of X!) such that the continuity of F: X
with respect to this new topology
implies J(F) E cIC(X) and vice versa. DEFINITION 5.6.16 ([54]). An F-topology is the topology on the following pre-base: Ue,ai
generated by
< i = 1,..., n}; a,,,D = {R E Fc I II(R — = {R E FcI there exists an operator T GL(l2),
where a1,.. .
such that T(V) C V and IITR — DII <E}, are some vectors from 12 and V is a finite-dimensional subspace
of 12.
We complete this section with the following two theorems. The proofs can be found in [54]. THEOREM 5.6.17 ([54]). The set of all F-continuous maps f : X —* cides with
coin-
THEOREM 5.6.18 ([54]). The spaces (tIC, norm) and (cIc, F-topology) are weakly homotopy equivalent.
5.7. Inner products on Hilbert We have already encountered the case where several inner products on the same Hilbert module were used. In this section we will study the relations between different inner products more thoroughly. Let M be a module over a A, on which two sesquilinear maps (•) are defined in such a way that M is a pre-Hilbert module with respect and to each of these maps. DEFINITION 5.7.1. Two inner products, (., and (, if the norms defined by these inner products are equivalent. is
are
called equivalent
Note that if the inner products are equivalent to each other and if {M,, (., is a Hilbert C*_module too. a Hilbert C*_module, then {M,, Let us consider first the case of different inner products on self-dual Hilbert
modules.
over a is another inner product
PROPOSITION 5.7.2 ([42]). Let M be a self-dual Hilbert
If
A with the inner product (,
equivalent to the given one, then there exists a unique invertible positive operator such that (x,y)1 = (Sx,Sy)2 for all x,y EM. S
PROOF. For x E M consider the functional on M defined by the formula Since the module M is self-dual, there exists an element Bx E M
y i—' (x, such that
(x.y)2 = (Bx,y)1 the for all y M. The map x '—i Bx is an A-homomorphism. Denote by i = 1,2. By our assumption, there exist norm defined by the inner product (., constants k,l > 0 such that I
5.7. INNER PRODUCTS ON HILBERT C-MODULES
127
for any x E M. Then 12 II
= II(x, Bx)211
IIBxII2
lixiL IIBxII1
11x01, i.e., the map B is bounded. The equality (Bx,x)1 = (x, x)2 0 implies that the operator B is positive with respect to the initial inner product. The inequality
Therefore IBxIJ1
= k2 (x, x)2 =
k2
(Bx, x)
< k2 IIBxII 1 lxiii
shows that the estimate iiBxiI1 IixiI1 holds, from which we obtain, by [92] (cf. the proof of Theorem 2.3.3), invertibility of the operator B. To complete the proof it remains to put S = B—h1'2. 0 Let M be a Hubert module with an inner product PROPOSITION 5.7.3 Let be another inner product equivalent to the initial one. Then the (•, given by the formula x (x, )2' x E M, defines an invertible map M
positive element in QM(K(M)) C
Conversely, any invertible positive (which can be identified with an element T of the set EndA(M,M)) defines an inner product (x,y) T(x)(y). x,y E M. element in
C
PROOF. Let EndA(M, M') —p QM(AC(M)) be the isometric isomorphism defined in Theorem 5.4.1, = Ox,t.T(z)(y), x,y,z,t e M. The map x (x, •)2 is bounded, hence it defines a map T M —+ M' and the element E QM(1C(M)) is defined by the formula
(note that the elementary operators of the form are considered with respect to the initial inner product Then for s E M one has
= (x (x. (y,y)2,s)1,s)1 .
= (x
0.
Since linear combinations of elementary operators are dense in the algebra the operator is positive. Let us show that it is invertible. Let us pass to A**. Both inner prodthe Hubert module M# over the enveloping ucts can be extended to the module M# and to the self-dual module (M#)'. By
((M#)') such Proposition 5.7.2, there exists an invertible operator S E that these extensions of the inner product are related by (x, y)1 = (Sx, Sy)2 for any x, y E (M#)'. But the image of the operator under the inclusion ((M#)') obviously coincides with the product S*S. Since QM(K(M)) c the operator S is invertible, the spectrum of the operator ç)(T) is separated from zero, hence 4(T) is invertible. In the opposite direction the statement can be proved similarly. 0
(.,
COROLLARY 5.7.4 ([44]). Let M be a Hilbert The following conditions are equivalent:
with an inner product
(i) any other inner product equivalent to the initial one is defined by an invertible operator S E EndA(M) and is given by the formula (x, y)2 = (Sx,Sy)1,
x,y EM;
5. MULTIPLIERS
128
(ii) each positive invertible quasi-multiplier T E QM(ftC(M)) can be decomposed into a product T = S*S for some invertible left multiplier Se THEOREM 5.7.5 ([44], see also [17]). Let M be a countably generated Hubert with an inner product (., Then for any inner product (•, equivalent to the initial one, there exists an invertible operator S e EndA(M) such that (x,y)2 = (Sx,Sy)1.
By assumption, the K(M) is a-unital, hence it contains a strictly positive element H E K(M). It is sufficient to show that each positive invertible quasi-multiplier admits a decomposition T = S* S with some left multiplier S. Put K = (HTH)1/2 E ,kC(M), = K (H2 + H e ,kC(M). Then and the sequence converges to K with respect to the norm. IVnII is norm-convergent to KK'. Then, for any K' E H . ftC(M), the sequence Since H is dense in K(M), we conclude that the sequence (Va) is convergent to some element S e LM(K(M)) with respect to the left strict topology and SF! = K. Therefore HS*SH K*K = HTH and, finally, = T. D PROOF.
As we can see from the following example of a nontrivial inner product, the requirement for Hilbert to be countably generated is essential in Theorem 5.7.5.
EXAMPLE 5.7.6 ([17, 44]). Let H be a nonseparable Hilbert space. Consider the space
equipped with the weak topology and the standard Hilbert C(X)-module Hc(x). Let us show that there exist inner products on Hc(x) equivalent to the standard one, which cannot be written as (S., 5.) for any operator S E Endc(x)(Hc(x)). For this purpose it is sufficient to find a quasi-multiplier T E QM(AC ® C(X))
that cannot be represented as T = S*S, S E LM(K ® C(X)). Let us use the identification of LM(ftC ® C(X)) (resp. QM(K ® C(X))) with the set of bounded maps from X to 5(H) continuous in the strong (resp. weak) topology (it is discussed in detail in Section 5.6). Define the new inner product on the module Hc(x) by the formula (5.9)
(y,
=
(y,
x(x))(x),
where y,z E Hc(x), x E X. it is easy to see that
(y,y)0 (y,y). This inner product defines a positive invertible quasi-multiplier T. Suppose that T = S*S for some S E ® C(X)). Let us show that one can choose a separable infinite-dimensional Hilbert space HT C H such that T(X)HT C HT and T1(X)HT C HT for all x e X. Let {ei,... ek, .} be a basis of some separable subspace H0 C H. Since X is compact, the sets T(X)ek and T'(X)ek are compact subsets in H for each number k. Therefore they generate a separable T'(X)Ho C Hi. Further, Hilbert subspace H1 C H such that T(X)H0 C H such that by induction we find separable subspaces C Ha), i.e., the closure of the union of Finally, set HT := C ,
all
.
.
5.7. INNER PRODUCTS ON HILBERT C-MODULES
Let us denote by Xo C X the subset 3/4 r
1/4,
129
ris linear
The restriction of the operator S onto the subspace X0 has the form S1
0
53)' 1/ 23/ 82
*
IA
*
*
* s1s2=r,
with respect to the decomposition H = HT
Since the subspace is invariant under the action of the operator s1 e B(HT) is invertible and the is continuous in the strong operator is unitary. Since the map u topology on the group of unitary elements, we conclude that is continuous on X0. Therefore the map r = is also strongly continuous as a map from X0
Thus, the assumption that T = S*S implies that any weakly
to
continuous bounded (by 1/4) linear map r : HT —4 turns out to be strongly continuous. But, since the strong and the weak topologies do not coincide on the ball of radius 1/4 in B(H#, HT), the obtained contradiction shows that the inner product (5.9) is not related to the standard inner product on the module Hc(x) by any invertible bounded operator S e EndA(Hc(x)).
If one considers various equivalent inner products on a Hilbert the problem of whether an operator is adjointable depends on the concrete inner of adjointable (resp. we denote the product. By operators (resp., compact operators) with respect to the inner product i= 1, 2. The adjoint operator for the operator T with respect to this inner product is denoted by T(j). Let M be a Hubert A-mod'ule over a A PRoPosITIoN 5.7.7 with the inner product (., Let S e EndA(M) be an invertible operator defining the inner product = (S., S.)1. Then the operator S is adjointable with respect to the first inner product if and only if it is adjointable with respect to the second
one.
*(1)
If S is
then the sets EndA true for IC(1)(M) and ic(2)(M).
*(2)
and EndA
coincide and the same is
PROOF. Let the operator S be adjointable with respect to the inner product Then one has
(Sx,y)2 = (S2x,Sy)1 =
=
S with respect for any x, y e M, and the operator to the second inner product. The converse statement can be proved similarly. Let us assume now that S e Let B e Then
(Bx, y)2 = (SBx, Sy)1 = (Sx,
= (x, for any x,y E M, hence BE statement about compact operators can be proved in the same way.
The
0
However, if the inner product is defined with the help of a nonadjointable operator S, then operators adjointable with respect to one of the equivalent inner
5. MULTIPLIERS
130
products need not be adjointable with respect to another. Thus a problem of whether a functional on M can be represented as an inner product by elements from M arises. More precisely, we define the set F C M' by
F= U
I3EB; yEM
where B is the set of all inner products (•, equivalent to the initial one. The functional f E M' is called representable if f e F. We study the set F for the standard Hilbert HA. Denote the extension of the initial inner product from the module HA to = HA.. and to its adjoint module still by (., •). It is easy to see that C is representable, then there exists an element z PROPOSITION 5.7.8. If I E in the module HA, for which the operator inequality
a(z,z) 2 unitary operator U —' such that (6.17) U maps the A-modules generated by the
into the modules generated by for some unitary elements u2 E A. Put "eigenvectors"
=
i.e.,
n(a) =
=
U,
cl}. m>ri
and it follows from (6.16) that 1(Q) changing a, we obtain that Then
(6.18)
—
< e. But,
—
a=
1
(then i = K
we obtain
1), —
—
I
(r) am
—
—
(r)
mn( 1) It
follows from (6.17) and (6.18) that —
(6.19) (r)
=
(au)
=
—
mn(1)
__(1) to satisfy the conditions A1(1)x1 = we get the estimate
Choosing
.
(6.20)
.
.
—
_(1)
x1
(1)
A1
(2) and A1 x1(2) =
(2)
x1
(2)
A1
<E.
In the same way we can obtain the estimates (6.19), (6.20) for other i.
THEOREM 6.4.2. Let K,. : HA —p HA, operators and let IlK1 — K211 Then
r=
1,2, be compact strictly positive
(i) there exists a unitary operator U in Hk such that it maps the "eigenvectors" of K2 into the "eigenvectors" of K1 and IIU*KIU — K2jj < e; (ii) the "eigenvalues"
satisfy the inequality
of the operators K,. (r = 1,2) can be chosen to —
1,
e
m(i,j) = 0,
i
g=diag(u,1,1,1,...)=diag(u,u'u,l,u'u,l,...).
We would like to define a hornotopy gj For
G, t
[0, -ir], in such a way that
this purpose, put
mj(1, 1) = u for t
(
[0,
ir/2],
mt(2i,2i+1) mt(2i+1,2i+1)
(cost —sint'\(u 0'\( cost cost
0 0
1
7.1. TECHNICAL LEMMAS
161
for i> 1, and
mt(r,s) =
for the remaining r, s.
0
For t e [ir/2, ir] put
(
rnt(2i — 1,2i—
1)
mt(2i — 1,2i) mt(2i,2i)
1)
(cost —sint'\(u'
0'\(
sint'\(u 0'\
cost
1)'
0
cost
and mt(r,s)
=
0
for remaining r, s.
0 LEMMA 7.1.5. Let W C
be defined by
W={g EGIgIH' =IdH'}, where
H' ® H1,
12(A)
H'
H1
12(A).
Then W is contractible inside G to
V = {g e = IdH', g(H1) = H1}. PROOF. With respect to the decomposition 12(A) = H' hornotopy by
(1
.,
I
/3(s)(1-t)
7(8)
) be the inverse for (
Let the operator (
H1 we define the
). Then
eRo
1)'
(1
7
1
(1
(1
(1
(1
(1—t).0 1
Hence, the homotopy lies in G.
0
7. HOMOTOPY TRIVIALITY OF GROUPS OF INVERTIBLE OPERATORS
162
and
EndAl2(A), M(KA)
Let 1CA, LM(KA)
EndA(l2(A),l2(A)')
QM(/CA)
denote the of A-compact operators on 12(A). the algebra of the left miiitipliers, the C*_algebra of multipliers and the space of quasi-multipliers, respectively (see [17, 73, 44, 63, 92, 1001 and Sections 5.3 and 5.4).
Let ci be a strictly positive element (see, e.g., Section 1.1) in a u-unital algebra A and let := (ci) be a countable approximate unit, where the functions i = 1,2,..., are defined by the following graph:
forz3andw2=ci2 1/2
1/2
.
Then
j=i+2,i+3,...,
71
,...,
13
32
—
If A is not unital, the notion of the "standard basis" {e1} for the module 12(A) makes no sense. Nevertheless, the elements ej'y are well defined for any -y E A, namely,
Denote by Q, the orthogonal projections onto these one-dimensional submodules LEMMA 7.1.6. The X
12(A) defined by
injection i: A
k(1) < k(2) < k(3)
ek(i)wix,
is adjointable and preserves the inner product. In particular, its image Tm i is defined by the selfadjoint projection of the form
p=
(7.2) PROOF.
First of all,
(ix, iy)
ek(i)wiY)
= = >X*wjwjy = Consider
the operator t
:
12(A) —+ A
t(z) :=
= (x,y). of
the form
= >WjZk(j).
ek(i)wiy)
7.1. TECHNICAL LEMMAS
163
This series satisfies the Cauchy criterion: if tl1e number in is so large that
i=m+1
then r
1/2
r
1/2
r
i(1) in such a way that 1 r N. —pj(I))Frei(l)alIj < 21.21' r = 1 4
Find i(2) > j(1) such that (by Lemma 7.1.8) 1
i(2) such that 11(1 —
and
t + s by the construction. Therefore one has (7.7)
<E,
IIP'F'rPII
r = 1,. .. , N.
As was shown in Lemma 7.1.2 and Lemma 7.1.5, it suffices to know how to construct a homotopy of any piecewise linear map with the image in a finite polyhedron in GL8 with vertices F1,. . , FN into a map taking values in a compact set {D(x)} C GL8 such that .
PD(x)=D(x)P=P VxES. For this purpose we can apply a homotopy of Neubauer type (see Section 7.7). By (7.7), we have to verify only that there exists an operator H0 : P'(12(A)) P(12(A))
such that operators H0P' and
P are adjointable. Let us assume H0 =
7. HOMOTOPY TRIVIALITY OF GROUPS OF INVERTIBLE OPERATORS
166
= JJl* where Jl*
Then H0P' =
is
an isomorphism P'(12(A))
12(A)
and J: 12(A) P(12(A)). We have proved the following statement. Then GL*(A) is con-
THEOREM 7.2.2 ([30]). Let A be a ci-unital tractible with respect to the norm topology.
7.3. The case A C AC Let a A be a subalgebra of the algebra AC of compact operators on a separable Hilbert space H. Under these restrictions we can prove the following statement. LEMMA 7.3.1. Let a, b E A,
Then
(fit 12,...) e
Ilafibli
(i
0
co).
PROOF. Since a* E AC, for any E > 0 we can find a number N = N(e) and a basis h1, h2,... in H such that
HN=spanc(hl,...,hN),
n. Then, since the sequence increases, lB — Bmll
lI(B —
+ lp(B
+ llp(B — Bm)pII
—
+ 11(1 — p)(B
—
Bm)(1
—
+ 11(1 —p)B(1 —p)II
where
k(1, 1); k(1,2), k(2, 1); k(1, 3), k(2, 2), k(3, 1);...
is some increasing sequence, is an adjointable embedding preserving the inner prod-
uct. If we denote H1 := ImJ1, then, by (7.12), IIF1IH1II
m(1) := 1 be so large that 11(1 — Pm(2))F!J(1)II < e/2, where
Denote by F2 the restriction of the m(2)-th line of the
Yi := matrix F onto G1
12(A) and find, by the same algorithm, a new embedding J2 such that its image equals H2 and 11F21H211
m(2) be so large that Let
11(1
Pm(3))F'YiII < 11(1
E
.2i' Pm(3))YiII
.
1,2,
i
0 such that for any C: X satisfying the inequality IIG—Fil <E, the submodule still satisfies all restrictions and the projection onto (x, is close to the projection onto CF. 0 THEOREM
8.5.5 ([126, 89, 30]). The map index defines an isomorphism index:
K(X,A).
PROOF. It is evident that this map is an epimorphism, since, for any bundles E and F, there exists a family of operators making the following diagram
8.5. CLASSIFYING SPACES
commutative:
191
(10) 00
>XXHA
XXHA
Here we have used Lemma 8.5.2. To prove injectivity, one has to construct a homotopy connecting an arbitrary map F: X with X —p IdHA, where is the set of operators of index zero. For F, one has [NFl — [CFI = 0, i.e., after taking sums of these bundles and some trivial bundle X x one obtains an isomorphism
J:Npe(X x x e La)) on the right-hand side. Hence, taking ii + k instead of n and writing J fiberwise as J(x), define, for
t E [0, 1], the following family of maps:
E
F
for h E
Since the fiberwise perturbation is compact, the map takes values in F* and the index is constant, hence, zero. For t = 1 one obtains a map to GL* (A), which is homotopic (in the class of maps to GL* (A)) to a map into a single point, according to Theorem 7.2.2. El REMARK 8.5.6. Using Theorem 7.2.2 it is proved in [89, 30] that
K(C(X) 0 A)
[X, F*I
K(C, C(X) ® A) (the last identification in fact is by definition) Thus, K(X; A)
K(C(X) ® A)
K(C, C(X) 0 A).
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Notation Index
a112
a>
a0
4
1
LC(A) LM(A)
24
0
1
M'
I
A"
105 105
27
56
M 19
1
/3-urn
M(A)
114
102 19
Co(X) 80 DC(A) 103 GL GL
6
50 52
43
29
E:A—.B EndA(M)
pVq pAq
6
15 16
71 71
P(A) 29
RC(A) RM(A)
6
flB 42 HornA(M,H) index F /C
RR(A) s-tim 55 Sp(A) 1
15 16
36
tsr(A)
19
18 IC(M) IC(M,Jsf) 18 ICA
K(A) Ko(A)
12(A)
12(M)
18
w-lim
32
La
33 33 33
KK(A, B)
149
65 65
123
8K(M,Jtf) K(E) 87
105 105 147
3 114
Z(A) 182
5
e
6
6
199
55
27
5
57
Index
equivalent projections, 58 essential ideal, 101 F-topology, 126 factor, 58 functional, 27 nonsingular, 130 normal, 56 positive, 1 representable, 130 G-B-module, 42
A-orthobasis, 8 A-valued product, 3 equivalent, 126 A, B-bimodule, 182 A-B-equivalence bimodule, 11 algebra 1
divisibly infinite (DI), 31 module-infinite (MI), 30 u-unitai, 2 universal enveloping, 56 von Neumann, 56
42
GNS-construction, 2 group, Grothendieck, 33 linear, full general GL, 50 general GL*, 52 inner product, equivalent, 126 Kasparov bimodule, 182 unbounded, 189 KK-groups, Cuntz definition, 188 Kasparov definition, 182 lemma of Dixmier-Douady, 173 module dual, 27 Hilbert, 4 countably generated, 8
56
continuous, 57 finite, 57 of type 1, 57 of type II, 57
of type lii, 57 of type 1100, 57
of type III, 57 properly infinite, 57 semifinite, 57 u-finite, 57 approximate unit, 2 bicommutant, 55 Cauchy-Bunyakovsky inequality, 4 centralizer double, 103 left, 105 quasi-, 105 commutant, 55 compatible pair, 118
4
conditional expectation, 6 faithful, 6 of finite index, 87
full, 11 reflexive, 79 self-dual, 27
algebraically, 86 characteristic number of, 87 element of hermitian, 1 positive, 1
standard, 6 pre-Hilbert, 3 projective finitely generated, 9 multiplier, 102
strictly positive, 2 element of module nonsingular, 8 i-orthogonal submodules, 171
loft, 105 quasi-, 105 right, 105 operator homotopy, 184 201
202
operator on module, 15 A-Fredholm, 34 index of, 36 adjoint, 15 Banach-compact, 32 compact, 18 elementary, 18 orthogonal complement, 6 polarization equality, 4 positive square root, 1 pre-Hilbert A-B-bimodule, 11 property (E), 169 property (K), 169 representation of an algebra, 1 nondegenerate, 102 universal, 56 set A-precompact, 33 space pre-dual, 56
spectrum of an element, 1 state, 1 strong Morita equivalence, 12 submodule, complemented, 22 symmetrization, 33 tensor product exterior, 181 interior, 181 theorem of Cuntz-Higson, 166 of Dixmier-Douady for modules, 174 of Dupre-Fillmore, 10 of von Neumann about bicommutant, 55
INDEX
topology,
left strict, 116 quasi-strict, 118 strict, 114 strong, 55 strong*, 55 strong module, 116 strong* module, 116 weak ,55 ci(E, E)-, 56
o(E, Em)-, 56 0-strong 55 tl_strong* 55 or-weak, 55
trace, 57 faithful, 57 finite, 57 normal, 57 semifinite, 57 triple, degenerated, 182 ultra weak direct sum, 63 unitalization, 1 vector cyclic, 1
separating, 1
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