A HILBERT C ∗ -MODULE ADMITTING NO FRAMES

HANFENG LI

Abstract. We show that every infinite-dimensional commutative

unital C ∗ -algebra has a Hilbert C ∗ -module admitting no frames.
In particular, this shows that Kasparov’s stabilization theorem for
countably generated Hilbert C ∗ -modules can not be extended to
arbitrary Hilbert C ∗ -modules.

1. Introduction
Kasparov’s celebrated stabilization theorem [4] says that for any C ∗ -
algebra A and any countably generated (right) Hilbert A-module XA ,
the direct sum XA ⊕ HA is isomorphic to HA as Hilbert A-modules,
where HA denotes the standard Hilbert A-module ⊕j∈N AA (see Sec-
tion 3 below for definition). This theorem plays an important role in
Kasparov’s KK-theory.
There has been some generalization of Kasparov’s stabilization the-
orem to a larger class of Hilbert C ∗ -modules [7]. It is natural to ask
whether Kasparov’s stabilization theorem can be generalized to arbi-
trary Hilbert A-modules via replacing HA by ⊕j∈J AA for some large
set J depending on XA . In other words, given any Hilbert A-module
XA , is XA ⊕ (⊕j∈J AA ) is isomorphic to ⊕j∈J AA as Hilbert A-modules
for some set J?
An affirmative answer to the above question would imply that XA
is a direct summand of ⊕j∈J AA , i.e, XA ⊕ YA is isomorphic to ⊕j∈J AA
for some Hilbert A-module YA .
In [2] Frank and Larson generalized the classical frame theory from
Hilbert spaces to the setting of Hilbert C ∗ -modules. Given a unital
C ∗ -algebra A and a Hilbert A-module XA , a set {xj : j ∈ J} of ele-
ments in XA is called a framePof XA [2, Definition 2.1] if there is a real
constant C > 0 such that j∈J hx, xj iA hxj , xiA converges in the ul-
traweak operator topology to some element in the universal enveloping

Date: November 28, 2008.

2000 Mathematics Subject Classification. Primary 46L08; Secondary 42C15.
Partially supported by NSF Grant DMS-0701414.
1
2 HANFENG LI

von Neumann algebra A∗∗ of A [10, page 122] and

X
(1) C hx, xiA ≤ hx, xj iA hxj , xiA ≤ C −1 hx, xiA
j∈J

for every x ∈ XA . It is called a standard frame of XA if furthermore

P
j∈J hx, xj iA hxj , xiA converges in norm for every x ∈ XA . Frank and
Larson showed that a Hilbert A-module XA has a standard frame if and
only if XA is a direct summand of ⊕j∈J AA for some set J [2, Example
3.5, Theorems 5.3 and 4.1]. From Kasparov’s stabilization theorem
they concluded that every finitely generated Hilbert A-module has a
standard frame. However, the existence of standard frames for general
Hilbert A-modules was left open. In fact, even the existence of frames
for general Hilbert A-modules is open, as Frank and Larson asked in
Problem 8.1 of [2].
The purpose of this note is to show that the answers to these ques-
tions are in general negative, even for every infinite-dimensional com-
mutative unital C ∗ -algebra:
Theorem 1.1. Let A be a unital commutative C ∗ -algebra. Then the
following are equivalent:
(1) A is finite-dimensional,
(2) for every Hilbert A-module XA , XA ⊕ (⊕j∈J AA ) is isomorphic
to ⊕j∈J AA as Hilbert A-modules for some set J,
(3) for every Hilbert A-module XA , XA ⊕YA is isomorphic to ⊕j∈J AA
as Hilbert A-modules for some set J and some Hilbert A-module
YA ,
(4) every Hilbert A-module XA has a standard frame,
(5) every Hilbert A-module XA has a frame.
In Section 2 we establish some result on continuous fields of Hilbert
spaces. Theorem 1.1 is proved in Section 3.
Acknowledgements. I am grateful to Michael Frank, Cristian Ivanescu,
David Larson, and Jingbo Xia for helpful comments.
2. Continuous fields of Hilbert spaces
In this section we prove Proposition 2.4.
Lemma 2.1. There exists an uncountable set S of injective maps N →
N such that for any distinct f, g ∈ S, f (n) 6= g(n) for all but finitely
many n ∈ N, and f (n) 6= g(m) for all n 6= m.
Proof. Take an injective map T from ∪n∈N Nn into N. For each x : N →
N define fx : N → N by fx (n) = T (x(1), x(2), . . . , x(n)) for all n ∈ N.
If x 6= y, say, x(m) 6= y(m) for some m ∈ N, then fx (n) 6= fy (n)
 A HILBERT C ∗ -MODULE ADMITTING NO FRAMES 3

for all n ≥ m. Now the set S := {fx ∈ NN : x ∈ NN } satisfies the

requirement. 
We refer the reader to [1, Chapter 10] for details on continuous fields
of Banach spaces. Let T be a topological space. Recall that a con-
tinuous field of (complex) Banach spacesQover T is a family (Ht )t∈T of
complex Banach spaces, with a set Γ ⊆ t∈T Ht of sections such that:
Q
(i) Γ is a linear subspace of t∈T Ht ,
(ii) for every t ∈ T , the set of x(t) for x ∈ Γ is dense in Ht ,
(iii) for every x ∈QΓ the function t 7→ kx(t)k is continuous on T ,
(iv) for any x ∈ t∈T Ht , if for every t ∈ T and every ε > 0 there
exists an x0 ∈ Γ with kx0 (s) − x(s)k < ε for all s in some
neighborhood of t, then x ∈ Γ.
Lemma 2.2. For each s ∈ [0, 1] there exists a continuous field of
Hilbert spaces ((Ht )t∈[0,1] , Γ) over [0, 1] such that Ht is separable for
every t ∈ [0, 1] \ {s} and Hs is nonseparable.
Proof. We consider the case s = 0. The case s > 0 can be dealt with
similarly. Let H be an infinite-dimensional separable Hilbert space.
Take an orthonormal basis {en }n∈N of H. Let S be as in Lemma 2.1.
For every f ∈ S and every t ∈ (0, 1] set vf,t ∈ H by
1/n − t π 1/n − t π
vf,t = cos( · )ef (n) + sin( · )ef (n+1)
1/n − 1/(n + 1) 2 1/n − 1/(n + 1) 2
for 1/(n + 1) ≤ t ≤ 1/n. Set Ht to be the closed linear span of
{vf,t : f ∈ S} in H for each t ∈ (0, 1]. Let H0 be a Hilbert space with
an orthonormal basis {e0f }f ∈S indexed by S. Then Ht is separable for
each t ∈ (0, 1] while H0 is nonseparable. Q
For each f ∈ S, consider the section xf ∈ t∈[0,1] Ht defined by
xf (t) = vf,t for t ∈ (0, 1], and xf (0) = e0f . Then xf (t) is a unit vector
in Ht for every t ∈ [0, 1], and the map t 7→ xf (t) ∈ H is Qcontinuous on
(0, 1]. Denote by V the linear span of {xf : f ∈ S} in t∈[0,1] Ht .
We claim that the function t 7→ P ky(t)k is continuous on [0, 1] for
every y ∈ V . Let y ∈ V . Say, y = m j=1 λj xfj for some pairwise distinct
fP1 , . . . , fm in S and some λ1 , . . . , λm in C. Then the map t 7→ y(t) =
m
j=1 λj xfj (t) ∈ H is continuous on (0, 1]. Thus the function t 7→
ky(t)k is continuous on (0, 1]. When t is Psmall enough, xf1 (t), . . . , xfm (t)
m
are orthonormal and hence ky(t)k = ( j=1 |λj |2 )1/2 . Thus the function
t 7→ ky(t)k is also continuous at t = 0. This proves the claim.
Since V satisfies the conditions (i), (ii), (iii) in the definition of con-
tinuous fields of Banach spaces (with Γ replaced by V ), by [1, Proposi-
tion 10.2.3] one has the continuous field of Hilbert spaces ((Ht )t∈[0,1] , Γ)
4 HANFENG LI
Q
over [0, 1], where Γ is the set of all sections x ∈ t∈[0,1] Ht such that
for every t ∈ [0, 1] and every ε > 0 there exists an x0 ∈ V with
kx0 (t0 ) − x(t0 )k < ε for all t0 in some neighborhood of t. 
Lemma 2.3. Let Z be an infinite compact Hausdorff space. Then
there exists a real-valued continuous function f on Z such that f (Z) is
infinite.
Proof. Suppose that every real-valued continuous function on Z has
finite image. Take a non-constant real-valued continuous function h
on Z. Say, h(Z) = B ∪ D with both B and D being nonempty finite
sets. Then at least one of h−1 (B) and h−1 (D) is infinite. Say, h−1 (D)
is infinite. Set W1 = h−1 (B). Then W1 and Z \ W1 are both nonempty
closed and open subsets of Z, and Z \ W1 is infinite.
Since every real-valued continuous function g on Z \ W1 extends
to a real-valued continuous function on Z, g must have finite image.
Applying the above argument to Z \ W1 we can find W2 ⊆ Z \ W1
such that both W2 and Z \ (W1 ∪ W2 ) are nonempty closed and open
subsets of Z \ W1 , and Z \ (W1 ∪ W2 ) is infinite. Inductively, we find
pairwise disjoint nonempty closed and open subsets W1 , W2 , W3 , . . . of
Z. Now define f on Z by f (z) = 1/n if z ∈ Wn for some n ∈ N and
f (z) = 0 if z ∈ Z \ ∪∞n=1 Wn . Then f is a continuous function on Z and
f (Z) is infinite, contradicting our assumption. Therefore there exists
a real-valued continuous function on Z with infinite image. 
Proposition 2.4. Let Z be an infinite compact Hausdorff space. Then
there exist a continuous field of Hilbert spaces ((Hz )z∈Z , Γ) over Z, a
countable subset W ⊆ Z, and a point z∞ ∈ W \ W such that Hz is
separable for every z ∈ W while Hz∞ is nonseparable.
Proof. By Lemma 2.3 we can find a continuous map f : Z → [0, 1] such
that f (Z) is infinite. Note that f (Z) is a compact metrizable space.
Thus we can find a convergent sequence {tn }n∈N in f (Z) such that its
limit, denoted by t∞ , is not equal to tn for any n ∈ N. For each n ∈ N
take zn ∈ f −1 (tn ). Set W = {zn : n ∈ N}. Then W is countable and
f (W ) = {tn : n ∈ N} 3 t∞ . Take z∞ ∈ W with f (z∞ ) = t∞ . Then
z∞ 6∈ W .
By Lemma 2.2 we can find a continuous field of Hilbert spaces
((Ht0 )t∈[0,1] , Γ0 ) over [0, 1] such that Ht0 is separable for every t ∈ [0, 1] \
{t∞ } while Ht0∞ is nonseparable. Set Hz = Hf0 (z) for each z ∈ Z. Then
Hz is separableQ for every z ∈ W while Hz∞ is nonseparable. For each
γ ∈ Γ set xγ ∈ z∈Z Hz by xγ (z) = γ(f (z)) ∈ Hf0 (z) = Hz for all z ∈ Z.
0

Then V := {xγ ∈ z∈Z Hz : γ ∈ Γ0 } is a linear subspace of z∈Z Hz

Q Q
satisfying the conditions (i), (ii), (iii) in the definition of continuous
 A HILBERT C ∗ -MODULE ADMITTING NO FRAMES 5

fields of Banach spaces (with Γ replaced by V ). By [1, Proposition

10.2.3] one has the continuous field of Hilbert
Q spaces ((Hz )z∈Z , Γ) over
Z, where Γ is the set of all sections x ∈ z∈Z Hz such that for every
z ∈ Z and every ε > 0 there exists an x0 ∈ V with kx0 (z 0 ) − x(z 0 )k < ε
for all z 0 in some neighborhood of z. 

3. Proof of Theorem 1.1

In this section we prove Theorem 1.1.
Recall that given a C ∗ -algebra A, a (right) Hilbert A-module is a right
A-module XA with an A-valued inner product map h·, ·iA : XA ×XA →
A such that:
(i) h·, ·iA is C-linear in the second variable,
(ii) hx, yaiA = hx, yiA a for all x, y ∈ XA and a ∈ A,
(iii) hy, xiA = (hx, yiA )∗ for all x, y ∈ XA ,
(iv) hx, xiA ≥ 0 in A for every x ∈ XA , and hx, xiA = 0 only when
x = 0,
(v) XA is complete under the norm kxk := k hx, xiA k1/2 .
Two Hilbert A-modules are said to be isomorphic if there is an A-
module isomorphism between them preserving the A-valued inner prod-
ucts. We refer the reader to [5, 6, 8, 11] for the basics of Hilbert C ∗ -
modules.
We give a characterization of frames avoiding von Neumann algebras.
Proposition 3.1. Let A be a unital C ∗ -algebra and let XA be a Hilbert
A-module. Let {xj : j ∈ J} be a set of elements in XA . Then {xj : j ∈
J} is a frame of XA if and only if there is a real constant C > 0 such
that
X
(2) Cϕ(hx, xiA ) ≤ ϕ(hx, xj iA hxj , xiA ) ≤ C −1 ϕ(hx, xiA )
j∈J

for every x ∈ XA and every state ϕ of A.

Proof. Suppose that {xj : j ∈ J} is a frame of XA . Let C be a constant
witnessing (1). Then every state ϕ of A extends uniquely to a normal
state of A∗∗ , which we still denote by ϕ. Applying ϕ to (1) we obtain
(2). This proves the “only if” part.
Now suppose that (2) is satisfied for every x ∈ XA and every state ϕ
of A. Let x ∈ XA . Note that hx, xj iA hxj , xiA = (hxj , xiA )∗ hxj , xiA ≥ 0
for every j ∈ J. For any finite subset F of J, from (2) we get
X
ϕ( hx, xj iA hxj , xiA ) ≤ ϕ(C −1 hx, xiA )
j∈F
6 HANFENG LI

for every state ϕ of A, and hence

X
hx, xj iA hxj , xiA ≤ C −1 hx, xiA .
j∈F
P
Thus the monotone increasing net { j∈F hx, xj iA hxj , xiA }F , for F be-
ing finite subsets of J ordered by inclusion, of self-adjoint elements in
A∗∗ is bounded above. Represent A∗∗ faithful as a von Neumann alge-
bra on some Hilbert space H. Then we may also represent A∗∗ naturally
as a von Neumann algebra on the Hilbert space H ∞ = ⊕n∈N H. By [3,
Lemma 5.1.4] the above net converges in the weak operator topology
of B(H ∞ ) to some element a of A∗∗ . Since the weak operator topology
on B(H ∞ ) restricts to the ultraweak operator topology on A∗∗ , we see
that the above net converges to a in the ultraweak operator topology.
Then (2) tells us that
Cϕ(hx, xiA ) ≤ ϕ(a) ≤ C −1 ϕ(hx, xiA )
for every normal state ϕ of A∗∗ . Therefore, C hx, xiA ≤ a ≤ C −1 hx, xiA
as desired. This finishes the proof of the “if” part. 
Let ((Hz )z∈Z , Γ) be a continuous field of Hilbert spaces over a com-
pact Hausdorff space Z. We shall write the inner product on each Hz as
linear in the second variable and conjugate-linear in the first variable.
By [1, Proposition 10.1.9] Γ is right C(Z)-module under the pointwise
multiplication, i.e.,
(xa)(z) = x(z)a(z)
for all x ∈ Γ, a ∈ C(Z), and z ∈ Z. By [1, 10.7.1] for any x, y ∈
Γ, the function z 7→ hx(z), y(z)i is in C(Z). From the conditions
(iii) and (iv) in the definition of continuous fields of Banach spaces in
Section 2 one sees that Γ is a Banach space under the supremum norm
kxk := supz∈Z kx(z)k. Therefore Γ is a Hilbert C(Z)-module with the
pointwise C(Z)-valued inner product
hx, yiC(Z) (z) = hx(z), y(z)i
for all x, y ∈ Γ, and z ∈ Z. In fact, up to isomorphism every Hilbert
C(Z)-module arises this way [9, Theorem 3.12], though we won’t need
this fact except in the case Z is finite.
Lemma 3.2. Let ((Hz )z∈Z , Γ) be a continuous field of Hilbert spaces
over a compact Hausdorff space Z. Suppose that there are a countable
subset W ⊆ Z and a point z∞ ∈ W \ W such that Hz is separable for
every z ∈ W while Hz∞ is nonseparable. Then Γ as a Hilbert C(Z)-
module has no frames.
 A HILBERT C ∗ -MODULE ADMITTING NO FRAMES 7

Proof. Suppose that {xj : j ∈ J} is a frame of Γ. By Proposition 3.1

there is a real constant C > 0 such that the inequality (2) holds for
every x ∈ XA and every state ϕ of A. For each z ∈ Z denote by ϕz the
state of C(Z) given by evaluation at z. For any z ∈ Z and any vector
w ∈ Hz , by [1, Proposition 10.1.10] we can find x ∈ Γ with x(z) = w.
Taking ϕ = ϕz in the inequality (2), we get
X
(3) Ckwk2 ≤ | hxj (z), wi |2 ≤ C −1 kwk2 .
j∈J

For each z ∈ Z let Sz be an orthonormal basis of Hz . For each

w ∈ Sz , from (3) we see that the set Fw := {j ∈ J : hxj (z), wi 6= 0}
is countable. Note that the set Fz := {j ∈ J : xj (z) 6= 0} is exactly
∪w∈Sz Fw . For each z ∈ W , since Hz is separable, Sz is countable and
hence Fz is countable. Then the set F := ∪z∈W Fz is countable.
Since Hz∞ is nonseparable, we can find a unit vector w ∈ Hz∞ or-
thogonal to xj (z∞ ) for all j ∈ F . If j ∈ J \ F , then xj (z) = 0 for all
z ∈ W , and hence by the condition (iii) in the definition of continu-
ous fields of Banach spaces in Section 2 we conclude that xj (z∞ ) = 0.
Therefore hxj (z∞ ), wi = 0 for all j ∈ J, contradicting (3). Thus Γ has
no frames. 
For any C ∗ -algebra A, A as a right A-module is a Hilbert A-module
with the A-valued inner product ha, biA = a∗ b for all a, b ∈ A. Given
a family {Xj }j∈J of Hilbert A-modules,
Q their direct
P sum, denoted by
⊕j∈J Xj , consists of (xj )j∈J in j∈J Xj such that j∈J hxj , xj iA con-
verges
P in norm, and has the A-valued inner product h(xj )j∈J , (yj )j∈J iA :=
j∈J hxj , yj iA .
We are ready to prove Theorem 1.1.
Proof of Theorem 1.1. (1)⇒(2): Suppose that A is finite-dimensional.
Then A = C(Z) for a finite discrete space Z. For each z ∈ Z denote by
pz the projection in C(Z) with pz (z 0 ) = δz,z0 for all z 0 ∈ Z. Let XA be
a Hilbert A-module. For any z ∈ Z and any xpz , ypz ∈ XA pz , one has
hxpz , ypz iA ∈ Apz = Cpz . Thus hxpz , ypz iA = λpz for some λ ∈ C. Set
hxpz , ypz i = λ. Then it is easily checked
Q that XA pz is a Hilbert space
under this inner product, ((XA pz )z∈Z , z∈Z XA pzQ ) is a continuous field
of Hilbert spaces over Z, and XA is isomorphic to z∈Z XA pz as Hilbert
A-modules. Take an infinite-dimensional Hilbert space H such that the
Hilbert space dimension of H is no less than that of XA pz for all z ∈ Z.
Then XA pz ⊕ H is unitary equivalent
Q to H as Hilbert spaces. It is
readily checked that Q((H)z∈Z , z∈Z H) Q is a continuous field ofQ Hilbert
spaces over Z, and ( z∈Z XA pz ) ⊕ ( z∈Z H) is isomorphic to z∈Z H
as Hilbert A-modules. Let J be an orthonormal basis of H. Then
8 HANFENG LI
Q
it is easy to see that z∈Z H and ⊕j∈J AA are isomorphic as Hilbert
A-modules. Therefore XA ⊕ (⊕j∈J AA ) and ⊕j∈J AA are isomorphic as
Hilbert A-modules. This proves (1)⇒(2).
The implications (2)⇒(3) and (4)⇒(5) are trivial.
The implication (3)⇒(4) was proved in [2, Example 3.5]. For the
convenience of the reader, we indicate the proof briefly here. Suppose
that XA and YA are Hilbert A-modules and XA ⊕ YA is isomorphic
to ⊕j∈J AA for some set J as Hilbert A-modules. We may assume
that XA ⊕ YA = ⊕j∈J AA . Denote by P the orthogonal projection
⊕j∈J AA → XA sending x + y to x for all x ∈ XA and y ∈ YA . For
each s ∈ J denote by es the vector in ⊕j∈J AA with coordinate 1A δj,s
P j ∈ J. Set xj = P (ej ) for each j ∈ J. For any x ∈ XA , say,
at each
x = j∈J ej aj with aj ∈ A for each j ∈ J, one has
X X X
hx, xiA = a∗j aj = hx, ej iA hej , xiA = hP x, ej iA hej , P xiA
j∈J j∈J j∈J
X X
= hx, P ej iA hP ej , xiA = hx, xj iA hxj , xiA .
j∈J j∈J

Therefore {xj : j ∈ J} is a standard frame of XA . This proves (3)⇒(4).

The implication (5)⇒(1) follows from Proposition 2.4 and Lemma 3.2.


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Department of Mathematics, SUNY at Buffalo, Buffalo, NY 14260-

2900, U.S.A.
E-mail address: hfli@math.buffalo.edu