Download - CourseD Lecture1 Hilbert Modules
-
8/3/2019 CourseD Lecture1 Hilbert Modules
1/59
-
8/3/2019 CourseD Lecture1 Hilbert Modules
2/59
Contents
Introduction and definition
Basic properties
Operators on Hilbert C*-modules
Hilbert C*-modules
http://goforward/http://find/http://goback/ -
8/3/2019 CourseD Lecture1 Hilbert Modules
3/59
(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
http://goforward/http://find/http://goback/ -
8/3/2019 CourseD Lecture1 Hilbert Modules
4/59
(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
http://goforward/http://find/http://goback/ -
8/3/2019 CourseD Lecture1 Hilbert Modules
5/59
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
http://goforward/http://find/http://goback/ -
8/3/2019 CourseD Lecture1 Hilbert Modules
6/59
-
8/3/2019 CourseD Lecture1 Hilbert Modules
7/59
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
E l
http://goforward/http://find/http://goback/ -
8/3/2019 CourseD Lecture1 Hilbert Modules
8/59
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
E l
http://goforward/http://find/http://goback/ -
8/3/2019 CourseD Lecture1 Hilbert Modules
9/59
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
E l
http://goforward/http://find/http://goback/ -
8/3/2019 CourseD Lecture1 Hilbert Modules
10/59
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
N
http://goforward/http://find/http://goback/ -
8/3/2019 CourseD Lecture1 Hilbert Modules
11/59
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
http://goforward/http://find/http://goback/ -
8/3/2019 CourseD Lecture1 Hilbert Modules
12/59
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
http://goforward/http://find/http://goback/ -
8/3/2019 CourseD Lecture1 Hilbert Modules
13/59
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
http://goforward/http://find/http://goback/ -
8/3/2019 CourseD Lecture1 Hilbert Modules
14/59
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
http://goforward/http://find/http://goback/ -
8/3/2019 CourseD Lecture1 Hilbert Modules
15/59
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
http://goforward/http://find/http://goback/ -
8/3/2019 CourseD Lecture1 Hilbert Modules
16/59
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
http://goforward/http://find/http://goback/ -
8/3/2019 CourseD Lecture1 Hilbert Modules
17/59
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
http://goforward/http://find/http://goback/ -
8/3/2019 CourseD Lecture1 Hilbert Modules
18/59
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
http://goforward/http://find/http://goback/ -
8/3/2019 CourseD Lecture1 Hilbert Modules
19/59
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
http://goforward/http://find/http://goback/ -
8/3/2019 CourseD Lecture1 Hilbert Modules
20/59
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
http://goforward/http://find/http://goback/ -
8/3/2019 CourseD Lecture1 Hilbert Modules
21/59
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
http://goforward/http://find/http://goback/ -
8/3/2019 CourseD Lecture1 Hilbert Modules
22/59
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
http://goforward/http://find/http://goback/ -
8/3/2019 CourseD Lecture1 Hilbert Modules
23/59
Proof (continued)
-
8/3/2019 CourseD Lecture1 Hilbert Modules
24/59
( )
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)
http://goforward/http://find/http://goback/ -
8/3/2019 CourseD Lecture1 Hilbert Modules
25/59
( )
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
http://goforward/http://find/http://goback/ -
8/3/2019 CourseD Lecture1 Hilbert Modules
26/59
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
http://goforward/http://find/http://goback/ -
8/3/2019 CourseD Lecture1 Hilbert Modules
27/59
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)
http://goforward/http://find/http://goback/ -
8/3/2019 CourseD Lecture1 Hilbert Modules
28/59
( )
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)
http://goforward/http://find/http://goback/ -
8/3/2019 CourseD Lecture1 Hilbert Modules
29/59
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)
http://goforward/http://find/http://goback/ -
8/3/2019 CourseD Lecture1 Hilbert Modules
30/59
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
http://goforward/http://find/http://goback/ -
8/3/2019 CourseD Lecture1 Hilbert Modules
31/59
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
http://goforward/http://find/http://goback/ -
8/3/2019 CourseD Lecture1 Hilbert Modules
32/59
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
http://goforward/http://find/http://goback/ -
8/3/2019 CourseD Lecture1 Hilbert Modules
33/59
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
http://goforward/http://find/http://goback/ -
8/3/2019 CourseD Lecture1 Hilbert Modules
34/59
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
http://goforward/http://find/http://goback/ -
8/3/2019 CourseD Lecture1 Hilbert Modules
35/59
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
http://goforward/http://find/http://goback/ -
8/3/2019 CourseD Lecture1 Hilbert Modules
36/59
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
http://goforward/http://find/http://goback/ -
8/3/2019 CourseD Lecture1 Hilbert Modules
37/59
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
http://goforward/http://find/http://goback/ -
8/3/2019 CourseD Lecture1 Hilbert Modules
38/59
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
http://goforward/http://find/http://goback/ -
8/3/2019 CourseD Lecture1 Hilbert Modules
39/59
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)
http://goforward/http://find/http://goback/ -
8/3/2019 CourseD Lecture1 Hilbert Modules
40/59
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
http://goforward/http://find/http://goback/ -
8/3/2019 CourseD Lecture1 Hilbert Modules
41/59
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
http://goforward/http://find/http://goback/ -
8/3/2019 CourseD Lecture1 Hilbert Modules
42/59
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
http://goforward/http://find/http://goback/ -
8/3/2019 CourseD Lecture1 Hilbert Modules
43/59
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
http://goforward/http://find/http://goback/ -
8/3/2019 CourseD Lecture1 Hilbert Modules
44/59
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
http://goforward/http://find/http://goback/ -
8/3/2019 CourseD Lecture1 Hilbert Modules
45/59
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
http://goforward/http://find/http://goback/ -
8/3/2019 CourseD Lecture1 Hilbert Modules
46/59
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)
http://goforward/http://find/http://goback/ -
8/3/2019 CourseD Lecture1 Hilbert Modules
47/59
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
http://goforward/http://find/http://goback/ -
8/3/2019 CourseD Lecture1 Hilbert Modules
48/59
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
http://goforward/http://find/http://goback/ -
8/3/2019 CourseD Lecture1 Hilbert Modules
49/59
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
http://goforward/http://find/http://goback/ -
8/3/2019 CourseD Lecture1 Hilbert Modules
50/59
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
http://goforward/http://find/http://goback/ -
8/3/2019 CourseD Lecture1 Hilbert Modules
51/59
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)
http://goforward/http://find/http://goback/ -
8/3/2019 CourseD Lecture1 Hilbert Modules
52/59
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)
http://goforward/http://find/http://goback/ -
8/3/2019 CourseD Lecture1 Hilbert Modules
53/59
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
http://goforward/http://find/http://goback/ -
8/3/2019 CourseD Lecture1 Hilbert Modules
54/59
Corollaries
-
8/3/2019 CourseD Lecture1 Hilbert Modules
55/59
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
http://goforward/http://find/http://goback/ -
8/3/2019 CourseD Lecture1 Hilbert Modules
56/59
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
http://goforward/http://find/http://goback/ -
8/3/2019 CourseD Lecture1 Hilbert Modules
57/59
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)
http://goforward/http://find/http://goback/ -
8/3/2019 CourseD Lecture1 Hilbert Modules
58/59
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)
http://goforward/http://find/http://goback/ -
8/3/2019 CourseD Lecture1 Hilbert Modules
59/59
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
http://goforward/http://find/http://goback/