CourseD Lecture1 Hilbert Modules

8/3/2019 CourseD Lecture1 Hilbert Modules

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Contents

Introduction and definition

Basic properties

Operators on Hilbert C*-modules

Hilbert C*-modules
http://goforward/http://find/http://goback/


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(some) References

Original papers

W. L. Paschke. Inner product modules over B-algebras. Trans.

Amer. Math. Soc. 182 (1973), 443468.

W. L. Paschke. The double B-dual of an inner product moduleover a C-algebra B. Canad. J. Math. 26 (1974), 12721280.

M. A. Rieffel. Induced representations of C-algebras. Adv.

Math. 13 (1974), 176257.

G. G. Kasparov. Hilbert C-modules: theorems of Stinespring

and Voiculescu. J. Operator Theory 4 (1980), 133150.

Hilbert C*-modules


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(some) References

Textbooks

E. C. Lance. Hilbert C-modules a toolkit for operator

algebraists. London Math. Soc. Lecture Note Series 210.

Cambridge Univ. Press, Cambridge, 1995.

N. E. Wegge-Olsen. K-Theory and C

-Algebras. A friendlyapproach. Oxford Univ. Press, Oxford, 1993.

V. M. Manuilov, E. V.Troitsky. Hilbert C-modules. Transl. Math.

Monographs 226. Amer. Math. Soc. Providence, 2005.

Bibliography on Hilbert C-modules

by M. Frank:

http://www.imn.htwk-leipzig.de/mfrank/mlit.pdf

Hilbert C*-modules


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Introduction

Let A be a C

-algebra.

Idea:

Take the definition of a Hilbert space and replace the field of

scalars by A.

This was done independently by W. Paschke in 1973 and M.

Rieffel in 1974.

Note that scalars play a two-fold role for Hilbert spaces:

Hilbert spaces are linear spaces (modules) over scalarsInner product takes values in scalars

Hilbert C*-modules


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Pre-Hilbert modules

Let M be a right module over a C-algebra A. An action of

a A on M we denote by x a, x M.

Definition

A pre-Hilbert A-module is a (right) A-module M equipped with a

sesquilinear form , : M M A satisfying the following

properties:x, x is a positive element in A for any x M;

x, x = 0 implies that x = 0;

y, x = x, y for any x, y M;

x, y a = x, ya for any x, y M and any a A.

The map , is called an A-valued inner product.

Hilbert C*-modules

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Examples

Example

Let J A be a right ideal. Then J can be equipped with thestructure of a pre-Hilbert A-module with the inner product ofx, y J defined by x, y := xy.

In particular, if J = A, A is a pre-Hilbert C-module over itself.

Example

Let J1, . . . , Jn be right ideals of a C-algebra A and let M be the

linear space of all n-tuples (x1, . . . , xn), xi Ji. Then Mbecomes a right A-module if the action of A is defined by

(xi) a := (xia) for (x1, . . . , xn) M, a A, and becomes apre-Hilbert A-module if the inner product of elements

(x1, . . . , xn), (y1, . . . , yn) M is defined by(x1, . . . , xn), (y1, . . . , yn) :=

ni=1 x

i yi.

Hilbert C*-modules

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Examples

Example

Let {Ji}iN be a countable set of right ideals of a C-algebra A.

Then their direct sum M =

iN Ji is a right module over A. It

becomes a pre-Hilbert A-module if the inner product of

elements (xi)iN, (yi)iN M is defined by(xi), (yi) :=

iN x

i yi.

The sum in the last example converges because it consists of a

finite number of summands, which makes the last exampleunsatisfactory from the point of view of analysis.

Hilbert C*-modules

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Examples

Our next example comes from topology.

Example

Let X be a compact metric space, : E X a (locally trivial,finitedimensional) vector bundle over X.

Let (E) be the set of all continuous sections of E, i.e. maps

s : X E such that s = idX.Then (E) is a module over the C-algebra C(X) of continuousfunctions on X.

A metric on E is a sesquilinear form E E C(X) such thatits restriction to each fiber is an inner product.

Any metric on E makes (E) a pre-Hilbert C-module overC(X).

Hilbert C*-modules

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Norm

Definition

Let M be a pre-Hilbert A-module, x M. Put

xM := x, x1/2.

We usually skip the subscript M when it does not lead to

confusion of norms.Proposition

M is a norm on M and satisfies the following properties:

1 x aM xM a for any x M, a A;

2 x, yy, x y2Mx, x for any x, y M;3 x, y xM yM for any x, y M.

(1) is obvious; (3) follows from (2), so let us prove (2).

Hilbert C*-modules

Proof of the Cauchy inequality


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Proof of the Cauchy inequality

Let be a positive linear functional on A. Then (, ) is a

usual (scalar-valued) inner product on M, possibly degenerate.Applying the usual Cauchy inequality for this inner product, we

obtain, for any x, y M,

(x, yy, x) = (x, y y, x)

(x, x)1/2 (y y, x, y y, x)1/2

= (x, x)1/2 (x, yy, yy, x)1/2

(x, x)1/2 y, y1/2 (x, yy, x)1/2.

Thus, for any positive linear functional , we have(y, xx, y) y2M (x, x), hencex, yy, x y2Mx, x.

Hilbert C*-modules

Hilbert C modules


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Hilbert C-modules

Definition

A pre-Hilbert A-module M is called a Hilbert C

-module if it iscomplete with respect to the norm M.

If M is a pre-Hilbert C-module over A then the action of the

C-algebra A and the A-valued inner product on M can be

extended to the completion of M, which thus becomes a HilbertC-module.

Consider some examples:

Example

If J A is a closed right ideal then the pre-Hilbert module J iscomplete with respect to the norm J = .

In particular, the C-algebra A itself is a free Hilbert A-module.

Hilbert C*-modules

Examples


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Examples

Example

If M1, . . . , Mn be Hilbert C

-modules over A then one can definetheir direct sum M =

ni=1 Mi. The inner product on M is given

by the formula x, y :=n

i=1 xi, yi, wherex = (xi), y = (yi) M.The direct sum of n copies of a Hilbert module M we denote by

Mn or by Ln(M).

Example

If {Mi}iN is a countable set of Hilbert C-modules over A then

let M be the space of all sequences x = (xi) : xi Mi such thatthe series

iN xi, xi is norm-convergent in the C

-algebra A.

We define the inner product on M by

x, y :=

iN xi, yi for x, y M.

Hilbert C*-modules


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Let us check that the series

iN xi, yi is convergent.Since the series

iN xi, xi and

iN yi, yi are convergent,

for any > 0 there exists a number N such that, for all n N,

we have N+ni=N

xi, xi < , N+n

i=Nyi, yi

< .Then, by the generalized Cauchy inequality,

N+ni=N

xi, yi2 N+n

i=Nxi, xi

N+ni=N

yi, yi < 2.

This proves that the inner product on M is well defined.

Exercise

Prove that the module M is complete with respect to the norm

M.

Hilbert C*-modules

Standard Hilbert C-module


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Standard Hilbert C -module

If each Mi equals A then the construction of the last Examplegives the Hilbert C-module over A that consists of all

sequences (ai)iN, where ai A, i N, such that the series

iNai ai

is convergent in A.

This module is usually denoted by l2(A) or HA and is called thestandard Hilbert C-module over A.

If A is the field C of complex numbers, then l2(A) is the usualHilbert space l2.

Hilbert C*-modules

Standard module in the commutative case


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Standard module in the commutative case

In the commutative case there is the follwing description of

l2(A).Let X be a locally compact Hausdorff space, A = C0(X). Letl2 = l2(C) be usual Hilbert space with the norm topology and letC0(X; l2) be the space of continuous l2-valued functions on Xvanishing at infinity.

Proposition

The map

j : l2(A) C0(X, l2), j(f)(x) := (f1(x), f2(x) . . . ),

where f = (f1, f2, . . . ) l2(A), f1, f2, . . . C0(X), is an isometricisomorphism.

The proof reduces to verification of surjectivity of j.

Hilbert C*-modules

Submodules and orthogonal complements


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Submodules and orthogonal complements

Definition

Let N M be a closed submodule of a Hilbert C-module M.

We define the orthogonal complement N by

N = {y M : x, y = 0 for all x N}.

Then N is a closed submodule of the Hilbert C-module M.

However, the equality M = N N

does not always hold, asthe following example shows.

Example

Let A = C[0, 1] be the C-algebra of all continuous functions on

[0, 1]. Consider, in the Hilbert A-module M = A, the submoduleN = C0(0, 1) of functions that vanish at the end points of [0, 1].Then, obviously, N = 0.

In such a case we say that N is not orthogonally

complementable in MHilbert C*-modules

Generators and basis


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Generators and basis

A Hilbert C-module M is called finitely generated if there exists

a finite set {xi} M such that M equals the linear span (over C

and A) of this set.A Hilbert C-module M is called countably generated if there

exists a countable set {xi} M such that M equals thenorm-closure of the linear span (over C and A) of this set.

Let now A be unital.

An element x of a Hilbert C-module M is called non-singular if

x, x A is invertible.

The set {xi}i of elements of M is called orthonormal ifx

i, x

j =

ij.

It is called a basis of M if finite sums of the form

i xi ai,ai A, are dense in M.

E.g. the set e1 = (1, 0, 0, . . .), e2 = (0, 1, 0, ), etc. is a basis inl2

(A).

Hilbert C*-modules

DuprFillmore lemma


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Dupr Fillmore lemma

Since Hilbert C-submodules may be not orthogonally

complementable, we cannot imitate the GramSchmidtorthogonalization. The following result is its weak version.

Lemma (DuprFillmore)

Let M be a Hilbert A-module, e1, . . . ,

en

M are orthonormal,

x M, > 0.If y M satisfies y, y = 1 and y {x, e1, . . . , en}, then thereexists en+1 M such that

1 e1, . . . , en, en+1 are orthonormal,

2 en+1 SpanA(e1, . . . , en, x, y),3 dist(x, SpanA(e1, . . . , en+1)) .

Hilbert C*-modules


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Proof

Let x = x n

i=1 eiei, x, x = x + y.

Then x, x = x, x + 2 1 2 1 > 0.

Therefore the element x is nonsingular. Put

en+1 = x x, x1/2, then

en+1 SpanA(x, y) {e1, . . . , en}.

Therefore the elements e1, . . . , en, en+1 are orthonormal.Since we have taken x SpanA(x, e1, . . . , en) anden+1 SpanA(x

, y), we obtain (2). Finally, put

w = en+1x, x1/2 +ni=1 eiei, x SpanA(e1, . . . , en+1),then the equality w x = x x = y = proves(3).

Hilbert C*-modules

Kasparov stabilization theorem


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Kasparov stabilization theorem

The following theorem shows that the standard Hilbert

C

-module absorbs any countably generated HilbertC-module over A just as the usual l2 absorbs any other

separable Hilbert space.

Theorem (Kasparov)

Let A be a C-algebra and M a countably generated Hilbert

A-module. Then

M l2(A) = l2(A).

Proof

We start by proving the theorem for the case, where A is unital.

Hilbert C*-modules


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Proof (continued)


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( )

By the DuprFillmore lemma, there exists an element

en+1 SpanA(e1, . . . , en, xn+1, em) ()

such that the elements e1, . . . , en, en+1 are orthonormal and

dist(xn+1, SpanA(e1, . . . , en+1)) 1

n+1 .

It follows from () that

{e1, . . . , en+1} SpanA(x1, . . . , xn+1, e1 . . . , em).

By setting m(n+ 1) = m, we complete the step of induction.

Thus, an orthonormal sequence en satisfying the properties (1)and (2) has been constructed. But the property (2) means that

this sequence generates the whole module M l2(A), soM l2(A) = l2(A).

Hilbert C*-modules

Proof (end)


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( )

Thus, Theorem is proved for unital C-algebras.

Let A be a non-unital C-algebra and let A+ be its unitalization.

Defining the action of A+ on the Hilbert C-module M over A by

the formula

x (a, ) := x a+ x, x M, (a, ) A+, C,we equip M with the structure of a Hilbert C-module over A+.

Consider the C

-module l2(A+

) over A+

and denote by l2(A+

)Athe closure of the linear span of all the elements of the form

x a, x l2(A+), a A, in l2(A

+).It is easy to see that l2(A

+)A = l2(A).The isomorphism M l2(A

+) = l2(A+) implies the isomorphism

M l2(A) = MA l2(A+)A = (M l2(A

+))A = l2(A+)A = l2(A)

(we use here that M = MA, which can be proved by using anapproximate unit in A).

Hilbert C*-modules

Projective modules


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j

Definition

A Hilbert A-module M is finitely generated projective if there

exists a Hilbert A-module N such that M N = An for some n.

The following two theorems of Dupr and Fillmore show that

finitely generated projective submodules lie in Hilbert

C

-modules in the simplest way.

Theorem (DuprFillmore)

Let A be a unital C-algebra and let M be a finitely generated

projective A-submodule in l2(A). Then

1 the nonsingular elements of M are dense in M;

2 l2(A) = M M;

3 M = l2(A).

Hilbert C*-modules

Proof


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We begin with the case, where M = An.Let g1, . . . , gn be an orthonormal basis in M. Fix > 0 and foreach m put

em = em n

i=1 gigi, em.

Then em M and

em, em = 1

ni=

1 em, gigi, em.

Since x, em 0 for each x l2(A), as m , we concludethat em, e

m 1. So there exists a number m0 such that for

any m m0, the element em is nonsingular.

Then one can define

em = emem, em1/2

with em, em = 1. Let x M

. Then

em, x = em, e

m

1/2em, x = em, e

m

1/2em, x 0.

Hilbert C*-modules

Proof (continued)


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( )

Choose a number m m0 such that em, x < and set

x

= x + e

m.It is easy to see that x x = . (1)Let us check that the element x is nonsingular. Put

u = x emem, x, v = em(em, x + 1).Then u v (since u em), and x

= u+ v. Therefore

x, x = u, u + v, v

= u, u + (em, x + 1)(em, x + 1),

and the right hand side is invertible, since em, x < .Therefore x, x is invertible too.Together with the estimate (1), this proves density of

nonsingular elements in M.

Hilbert C*-modules

Proof (continued)


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Let {xn} be a sequence, in which each element em is repeated

infinitely many times. Put x = x1 ni=1 gigi, x1.Then (taking = 1) one can find gn+1 M such thatgn+1, gn+1 = 1 and dist(x, gn+1A) 1. Thereforedist(x1, SpanA(g1, . . . , gn+1)) 1. At the next step we replacethe module M by SpanA(g1, . . . , gn+1), x1 by x2, and = 1 by

= 1/2.

Going on with this procedure, we obtain an orthonormal

sequence {gk}kN, extending the basis g1, . . . , gn of M, suchthat

dist(xi, SpanA(g1, . . . , gn+i)) n} is abasis of M. Hence l2(A) = M M

and M = l2(A).

Hilbert C*-modules

Proof (end)


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We pass now to the case of an arbitrary finitely generated

projective module M. Let M N = An.By Kasparov stabilization theorem, N l2(A) = l2(A), hence

An = N M N l2(A) = l2(A).

Therefore, if K is the orthogonal complement to the submodule

N M in the module N l2(A), then K = l2(A) and

N M K = N l2(A).

But K = M is obviously the orthogonal complement to thesubmodule M in the module l2(A).

Hilbert C*-modules

Projectivity implies complementability


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Theorem (DuprFillmore)Let A be a unital C-algebra and let M be a finitely generated

projective Hilbert submodule in an arbitrary Hilbert C-module

N over A. Then N = M M.

Proof

As in the previous theorem, the proof can be reduced to the

case, when M is a free module, M = An. If {g1, . . . , gn} is thestandard basis of M, then put, for x N, x = x

n

i=1gigi, x.

Then x M and x x M, hence N = M M.

Hilbert C*-modules

Dual modules


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For a Hilbert C-module M over a C-algebra A let us denote

by M the set of all bounded linear A-module homomorphisms

from M to A.The structure of a vector space over C is given by the formula

( f)(x) := f(x), where C, f M, x M.The formula (f a)(x) := af(x), a A, introduces thestructure of a right A-module on M.

M is obviously complete with respect to the normf = sup{f(x) : x 1}.M is called the dual (Banach) module for M.

The elements of M are called functionals on M.

Note that there is an obvious isometric inclusion M M, whichis defined by x x, = x.Definition

A Hilbert module M is called self-dual if M = M.

Hilbert C*-modules

Self-dual modules


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Here is some information about self-dual modules (without

proof).

If A is unital, An is self-dual.

Any finitely generated projective Hilbert C-module over a unital

C-algebra is self-dual.

The condition of self-duality is very strong there are quite afew self-dual modules:

Any Hilbert module over a C-algebra A is self-dual iff A is

finitedimensional.

Self-dual Hilbert C-modules behave quite like Hilbert spaces.

Let N be a Hilbert C-module and let M N be a self-dualsubmodule. Then N = M M.

Hilbert C*-modules

Dual modules


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For an arbitrary Hilbert C-module M, M may be strictly greater

than M, moreover, the Banach module M may not admit the

structure of a Hilbert C-module at all.Description of the dual module for the standard Hilbert module

l2(A) is given by the following statement.

Proposition

Consider the set of sequences f = (fi)iN, fi A, such that the

norms of partial sumsNi=1 fi fi are uniformly bounded.

If A is a unital C-algebra, then this set coincides with l2(A);

the action of f on elements of the module l2(A) is given by

f(x) = iN fi xi,where x = (xi) l2(A), and the norm of f is given by

f2 = supN

Ni=1 fi fi .Hilbert C*-modules

Bidual modules


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Let M be the second dual (bidual) module for M.

For x M, f M, putx(f) := f(x).

The map x x is an isometric map from M to M:

x = sup{f(x) : f M, f 1} f x x ;

x 1

x x(x) =1

x x, x = x .For F M, define the functional F M by

F(x) := F(

x).

Identifying M with M = {x : x M} M, we obtain that F isthe restriction of F onto M M.Note that (x)= x for all x M.It is clear that the map F F is an A-module map from M toM and

F F. (In fact, this map is an isometry.)

Hilbert C*-modules

Inner product on bidual modules


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Define the inner product , : M M A by

F, G := F(G), F, G M.It can be directly checked that F a, G = aF, G for a A.Besides, for x, y M, one has

x, y = x((y)) = x(y) = (y(x)) = y, x = x, y.Therefore the inner product defined above is an extension of

the inner product on M.

Theorem (Paschke)M is a Hilbert C-module with respect to this inner product. In

particular, F, F 0 and F, F = F2.

Hilbert C*-modules

Further duals. Reflexivity


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We may consider the further duals, M etc., but, unlike the

Banach spaces case, they stabilize: M = M, so we have onlythree different modules

M M M,the first two of which are Hilbert modules, and all possible

variants can be realized:

M = M = M; M = M = M; M = M = M; M = M = M.

DefinitionM is called reflexive if M = M.

Reflexive modules are more frequent than self-dual ones. In

the commutative case there is a criterium

TheoremLet X be a compact Hausdorff space, A = C(X).l2(A) is reflexive iff X contains a copy of the StoneCechcompactification N of N.

Hilbert C*-modules

Operators on Hilbert C-modules


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Definition

Let M, N be Hilbert C-modules over a C-algebra A. A

bounded C-linear A-module homomorphism from M to N iscalled an operator from M to N.

Let HomA(M, N) denote the set of all operators from M to N. IfN = M, then EndA(M) = HomA(M, M) is obviously a Banach

algebra. However, we shall see soon that there is no naturalinvolution on this algebra.

Definition

Let T HomA(M, N). We say that T is adjointable if there

exists an operator T HomA(N, M) such thatTx, y = x, Ty for all x M, y N.

It is easy to check that if an adjoint operator T exists then it is

unique.

Hilbert C*-modules

Non-adjointable operators


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The main difference between Hilbert spaces and Hilbert

C-modules is that a bounded operator is not necessarily

adjointable, as the next example shows.

Example

Let A = l be the C-algebra of bounded sequences,

an = (0, . . . , 0, 1, 0, 0, . . .) A with 1 at n-th place.

Note that the series nN an is bounded by 1, but notconvergent.

Let M = l2(A), {en}nN an orthonormal basis in M.

Let T EndA(M) be defined by Ten = e1 ak. Take x l2(A),x =

nN en an, where

nN x

nxn is norm-convergent in A.

Then Tx = e1

nN anxn.

Hilbert C*-modules

Example (continued)


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The series

nN anxn is convergent: by the Cauchy inequality

for Hilbert C-modules,mn=k anxn

2 mn=k a2n mn=k xnxn = mn=k xnxnand the latter series is convergent by the definition of l2(A), soT is well defined. Also

Tx, Tx = nN anxn x, x ,so T is bounded.

If T would exist as a bounded operator then

T

ek, en = ek, Ten = 0, k = 1;en an, k = 1.But the series

nN a

nan is not norm-convergent in A, so T

e1is not well defined.

Hilbert C*-modules

Operators on Hilbert C-modules


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Denote by HomA(M, N) the set of all adjointable operators fromM to N.

The algebra EndA(M) = HomA(M, M) is an involutive Banachalgebra. Moreover, it is a C-algebra: this follows from the

estimate

TT supxB1(M)

{TTx, x} = supxB1(M)

{Tx, Tx} = T2 ,

where B1(M) denotes the unit ball of the module M.

Example

Let M = N N. Define P : M M to be the projection ontoN along N. Then P is bounded, P = 1, and P = P, henceP EndA(M).

Hilbert C*-modules

A non-adjointable projection


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Example

Let J A be a closed ideal such that the equality Ja= 0,a A, implies a= 0 (such ideals are called essential).

Put M := A J, N := {(b, b) : b J} M.

Then N = {(c, c) : c J}. Therefore N N = J J = M.

However, the submodule L = {(a, 0) : a A} M is atopological (non-orthogonal) complement to N in M.

Let P be a projection on N along L, P(a,j) = (aj, 0), where(a,j) M, and P 2.

Then P, if defined, should satisfy P(a,j) = (a, a), whichdoesnt lie in M, so P EndA(M), but P / End

A(M).

Hilbert C*-modules

Positivity


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Positivity for operators can be understood in two different ways.

Lemma

For T EndA(M) the following conditions are equivalent

1 T is positive in the C-algebra EndA(M);

2 m, Tm is positive in A for each m M.

Proof. If (1) holds then (selfadjoint) T1/

2

exists, hencem, Tm = T1/2m, T1/2m 0.

If (2) holds then m, Tm is selfadjoint, henceTm, m = m, Tm and polarization shows that

Tn, m = n, Tm for any m, n M, hence T is selfadjoint.Therefore, T = R S, where R, S EndA(M) are positive andRS = 0. Then m, S3m = Sm, TSm 0. But S3 0,hence m, S3m = 0. So S3 = 0, hence S = 0 and T = R 0.

Hilbert C*-modules

Compact operators


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Let M, N be Hilbert C-modules over A, x N, y M.Define the operator x,y : M N by

x,y(z) := xy, z, z M.

Operators of this form are called elementary operators. They

clearly satisfy the equalities

1 (x,y) = y,x;

2 x,yu,v = xy,u,v = x,vu,y for u M, v N;3 Tx,y = Tx,y for T HomA(N, L);4 x,yS = x,Sy for S Hom

A(L, M).

We denote the closed linear span of the set of all elementary

operators by K(M, N).The elements of K(M, N) are called compact operators.In the case N = M, the equalities (1)(4) mean that the algebraK(M) = K(M, M) is a closed two-sided ideal in the C-algebraEndA(M).

Hilbert C*-modules


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Compact operators acting on Hilbert modules are not compact

operators in the usual sense, when one considers them as

operators from one Banach space to another.

However, they are a natural generalization of compact

operators on a Hilbert space.

Proposition

Let l2

(A) be the standard Hilbert module over a unitalC-algebra A and let Nn l2(A), Nn = A

n be the free

submodule generated by the first n elements of the standard

basis.

An operator K EndA(l2(A)) is compact if and only if the norms

of restrictions of K onto the orthogonal complements Nn of the

submodules Nn vanish as n .

Hilbert C*-modules

Proof


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Denote by Pn the projection in l2(A) onto the submodule Nn .

Then, for any z Nn one has

x,y(z)2 = x,y(z), x,y(z) = y, z

x, xy, z

x2 y, z2 = x2 Pny, z2

x2 Pny2 z2 .

Since Pny tends to zero, we have x,y|Nn 0.

If K K(l2(A)) then it is a limit of finite linear combinations of

the form i ixi,yi, hence K|Nn 0 as n .

Hilbert C*-modules

Proof (end)


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Let us assume now that for some operator K, one has

K|Nn 0.Then, since

nm=1 Kemem, z = 0 for any z Nn, one has, for

z 1 and z Nn,

supz Kzn

m=1 Kemem, z = supz Kz 0 ()as n .

If z Nn, then Kz =n

m=1 Kemem, z.

This means that (

) still holds if the supremum is taken over the

unit ball of the whole module l2(A). Therefore the operator K isthe norm limit of the operators Kn =

nm=1 Kem,em.

Hilbert C*-modules

Compact operators


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Proposition

K(A) = A.

Proof. If A is unital, this is obvious (and K(A) = EndA(A)).In general, identify a,b with the left multiplication by ab

. This is

an isometric -homomorphism and its range contains a denseset {ab : a, b A} A, so it is an isomorphism,

Corollary

K(A) = EndA(A) iff A is unital.

Proof. EndA(A) is unital.

CorollaryK(An) = Mn(A); K(l2(A)) = K A.

Proof. The first assertion is obvious; to obtain the second one,

one has to pass to the limit n .

Hilbert C*-modules

Description of operators in the commutative case


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In the commutative case there is the following description for

EndA(l2(A)) and for K(l2(A)).

Let X be a locally compact Hausdorff space, A = C0(X).Let K = K(l2) be the space of (usual) compact operators on theHilbert space l2 with the norm topology.

Let C0(X;K) be the space of continuous K-valued functions onX vanishing at infinity.

Let B = B(l2) be the space of all bounded operators with the-strong topology.Let Cb(X;B) be the space of bounded continuous B-valuedfunctions on X.

Theorem (Frank)There are canonical identifications K(l2(A)) = C0(X;K) andEndA(l2(A))

= Cb(X;B).

Hilbert C*-modules

Non-adjointable compact operators


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One can get rid of the adjointability condition in the definition of

elementary operators and define them as

f,x : y x f(y), x N, y M, f M.

The closed linear span of all f,x, f M, x N, is a

non-adjointable (not necessarily adjointable) analog forK(M, N).

When N = M, we get a closed two-sided ideal of suchnon-adjointable compact operators in the Banach algebra

EndA

(M).Although these algebras are not C-algebras, in some aspects

they behave like the usual compact and bounded operators.

Hilbert C*-modules

When Kernel and Range are complemented


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Theorem (Mishchenko)

Let M, N be Hilbert A-modules and T HomA(M, N) an

operator with closed range. Then

Ker T is an orthogonally complemented submodule in M,

Ran T is an orthogonally complemented submodule in N.

ProofLet Ran T = N0 and let T0 : M N0 be an operator such thatits action coincides with the action of T. By the open mapping

theorem, the image of the unit ball, T0(B1(M)), contains some

ball of radius > 0 in N0. Therefore, for each y N0, one canfind some x M such that T0x = y and x 1 y. One

has T0 y2 = y, T0T0 y y T0T0 y ,Hilbert C*-modules

Proof (continued)


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hence

y2

= T0x, y = x, T0 y x T0 y 1 y y1/2 T0T

0 y

1/2 ,

Whence

y 2 T0T0 yfor any y N0.Let us show that the spectrum of the operator T0T

0 does not

contain the origin. Suppose the opposite, i.e. that

0 Sp(T0T0 ). Let f be a continuous function on R such that

f(0) = 1 = f , f(t) = 0 if |t| 12 2.

Using functional calculus in the C-algebra EndA(M), we definethe operator S EndA(M) by the formula S = f(T0T

0 ).

Hilbert C*-modules

Proof (continued)


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Then S = 1 and T0T0 S 12 2.Choose x M such that x = 1 and Sx > 12 . Then theinequality

T0T0 Sx

1

22 < 2 Sx

contradicts the assumption (with y = Sx).

So 0 / Sp(T0T0 ).

Therefore T0T0 is invertible and, in particular, surjective, so, for

any z M, one can find w N0

such that T0

z = T0

T0

w.

Then z T0 w Ker T.

Hilbert C*-modules


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Corollaries


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Corollary

If P End

A(M) is an idempotent, then Ran P is an orthogonallycomplemented submodule in M.

Corollary

Let M, N be Hilbert A-modules and let F : M N be a

topologically injective (i.e. Fx x for some > 0 and forall x M ) adjointable operator, then F(M) F(M) = N.

Corollary

Let M be a Hilbert A-module and let J EndA(M) betopologically injective and selfadjoint. Then J is an

isomorphism.

Hilbert C*-modules

More Corollaries


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Lemma

Let M be a finitely generated submodule in a Hilbert C-module

N over a unital C-algebra A. Then M is an orthogonal directsummand in N.

Proof

Let x1, . . . , xn M be a finite set of generators. Define theoperator F : An N by F(ei) = xi, where e1, . . . , en An is

the standard basis.

It is easy to see that F is adjointable with the adjoint

F : N An acting by the formulaF(x) = (x1, x, . . . , xn, x), where x N.

By the previous Theorem, Ran F = M is an orthogonal directsummand.

Hilbert C*-modules

When a projection is adjointable


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Proposition

Let A be a unital C

-algebra, l2(A) = M N, P : l2(A) M aprojection and N a finitely generated projective module. Then

l2(A) = M M if and only if P is adjointable.

Proof

If P exists, then (1 P) = 1 P exists as well. Therefore,Ker(1 P) = M is the range of a selfadjoint projection.

To prove the converse, let us verify first that l2(A) = N + M.

By the Kasparov stabilization theorem, one can assume,without loss of generality, that

N = SpanAe1, . . . , en, N = SpanAen+1, en+2, . . . .

Hilbert C*-modules

Proof (continued)


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Let gi be the image of ei under the projection onto M:

e1 = f1 + g1, . . . , en = fn + gn, fi M, gi M.

Since the projection induces an isomorphism of A-modules

N = M, the elements g1, . . . , gn are free generators andgk, gk > 0 (i. e. the spectrum of this positive operator is

separated from 0 and hence it is invertible).So, if fk =

i=1 f

ikei, then ek f

kk ek =

i=k f

ikei + gk.

On the other hand,

1 = ek

, ek

= fk

, fk

+ gk

, gk

, 1 (fkk

)(fkk

)

gk

, gk

> 0,

i.e. the spectrum of the element 1 fkk is separated from theorigin, hence this element is invertible in A,

Hilbert C*-modules

Proof (end)


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ek = (1 fkk )

1

i=k fikei + gk

N + M (k = 1, . . . , n),

thus N + M = l2(A).

Let x N M. Since any element y l2(A) = M N hasthe form y = m+ n, one has x, y = x, m + x, n = 0; inparticular, x, x = 0, thus x = 0. Therefore l2(A) = N

M.

Consider the map Q = 1 on N0 on M

, which is a bounded

projection, since l2(A) = N M.

Let x + y M N, x1 + y1 N M. Then

P(x + y), x1 + y1 = x, x1 + y1 = x, x1,x + y, Q(x1 + y1) = x + y, x1 = x, x1.

Therefore P exists and equals Q.

Hilbert C*-modules

CourseD Lecture1 Hilbert Modules

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Transcript of CourseD Lecture1 Hilbert Modules