extension theory of hilbert modules over...

Math. Nachr. 246-247 (2002), 83 – 105

Extension Theory of Hilbert Modules over Semigroups

By Kunyu Guo∗) of Shanghai

(Received April 10, 2001; accepted April 25, 2002)

Abstract. Hilbert modules over semigroups are essentially the same thing as uniformly boundedrepresentations of semigroups on Hilbert spaces. We develop extension theory of Hilbert modulesover semigroups and mainly apply them to study questions from operator theory and representationsof nonamenable groups.

1. Introduction

Let H be a Hilbert space, and G a semigroup with the identity. We call H a HilbertG–module if there is a uniformly bounded representation πH : G→ B(H) of G. Thismeans that

π(1) = I ,

π(st) = π(s)π(t) , for all s , t ∈ G ; andsupt∈G‖π(t)‖B(H) < ∞ .

By the preceding definition, the only difference between uniformly bounded represen-tations of a semigroup G and Hilbert G–module is one of perspective. Each has itsadvantages and we shall pass from one to the other when it is convenient. One reasonfor studying Hilbert modules over semigroups is that many problems from operatortheory are not subsumed in the language of Hilbert modules over function algebras[DP]. An example is that there exists a power bounded operator T on a Hilbert spaceH , which is not polynomially bounded [Fo, Le]. Therefore H becomes a Z+–Hilbertmodule via the representation

πH : Z+ −→ B(H) ; πH(n) = Tn .

However, it seems impossible to make H by the operator T as a Hilbert module overthe disk algebra A(D). Another reason arose in the context of representations of

2000 Mathematics Subject Classification. 46J15, 46H25, 47A15, 47D25.Keywords and phrases. Hilbert modules, extensions, semigroups, representations.∗) [email protected]

c© 2002WILEY-VCHVerlagGmbH &Co. KGaA,Weinheim, 0025-584X/02/246-24712-0083 $ 17.50+.50/0

84 Math. Nachr. 246-247 (2002)

groups, especially representations of nonamenable groups. Other reasons will becomeapparent later.

In [DP], Douglas and Paulsen initiated a systematic study of Hilbert modules overfunction algebras. In the language of Hilbert modules over function algebras, manyconcepts and problems from operator theory are subsumed into the suggestive topicsof module theory. In [CC1], Carlson and Clark introduced Ext–functor into thestudy of Hilbert modules over function algebras. The importance of such a homologicalinvariant is based on the fact that the study for it actually reveals intrinsic relationshipbetween operator theory and complex analysis [CC2, CCFW, CG, GC, Guo, Fer].Parallel to the theory developed by Douglas and Paulsen [DP], Muhly and Solel

[MS] have developed the theory of Hilbert modules over operator algebras and inaddition they succeeded in applying this theory to study incidence algebras and theirgeneralizations.

In the present paper, we introduce extension functor, Ext, into the study of Hilbertmodules over semigroups. Suppose that H and K are Hilbert G–modules. We wantto consider all possible Hilbert G–module J such that H is a submodule of J and Kis a quotient module, i. e., K ∼= J/H (= J �H). We then have a short exact sequence

E : 0 −→ Hα−→ J

β−→ K −→ 0

of Hilbert G–modules; such a sequence is called an extension of K by H , or J is calledan extension of K by H . By definition, ExtG(K,H) is the extension group of K by H ,whose elements are equivalence classes of extensions of K by H . The group operationsand equivalence relation will be defined in Section 2.

In Section 2, we provide a general framework for extension groups. Most of thismaterial are parallel to those of Hilbert modules over function algebras [CC1, Section2]. Thus many of the concepts they introduced, and results they proved, are quiteserviceable for semigroups that are not necessarily abelian. For the readers who arefamiliar with extension theory of Hilbert modules over function algebras, they mayskip Section 2.

In Section 3, we first prove that if G is left amenable, then each unitary HilbertG–module is G–projective. This gives a useful way of describing some extensiongroups. Section 3 also prove the following: let G be an abelian semigroup with can-cellation law, then

ExtG(l2(G), l2∗(G)

) �= 0

if and only if G is not a group. In particular, if G is an abelian group, and G+ asubsemigroup of G which is not a subgroup of G. Then ExtG+

(H2(G), H2

(G))

isnontrivial. We provide several examples in Section 3 to show the relationship betweenHilbert G–modules and Hilbert modules over function algebras, and applications inoperator theory.

In Section 4, we consider extensions of Hilbert modules on nonamenable groups. Inthis case, extensions are closely related to representations of nonamenable groups. LetH, K be Hilbert modules over a nonamenable group G. If both πH(G) and πK(G) areamenable, we show ExtG(K,H) = 0. Furthermore, if G contains F2 as a subgroup,we prove that ExtG

(l2(G), l2(G)

) �= 0.

Guo, Hilbert Modules over Semigroups 85

In the present paper we emphasize the role of homology theory. We assume alwaysthat semigroups and groups appearing in this paper are discrete.

2. Homological preliminaries

In studying Hilbert modules over semigroups, as in studying Hilbert modules overfunction algebras, the standard procedure is to look at submodules and associatedquotient modules. The extension problem then appears quite naturally: given modulesH,K (over a fixed semigroup G) what module J may be constructed with submoduleH and associated quotient module K ? Since the categories of Hilbert modules oversemigroups lack enough projective and injective objects, it is impossible to define theExt–functor as the derived functor of Hom as in [HS, Hel]. However, the standardhomological construction from classical homological algebra enable us to classify allextensions in the category H(G) of all Hilbert modules over a fixed semigroup G. Firstof all, let us review necessary homological constructions and introduce some notionand results. Results and their proofs in this section are parallel to those for extensionsof Hilbert modules over function algebras (see [CC1] or [Hel]). Therefore in this sectionwe often state results without proofs.

The symbol H(G) will denote the category of all Hilbert G–modules and HilbertG–module homomorphisms. Briefly, a Hilbert G–module H is a Hilbert space Htogether with a uniformly bounded representation of G, π : G→ B(H),

π(1) = I ,

π(st) = π(s)π(t) , for all s , t ∈ G ; andsupt∈G‖π(t)‖B(H) < ∞ .

A Hilbert G–module homomorphism α : H → K is a bounded linear operator betweenthe underlying Hilbert spaces, and a G–module homomorphism (α(th) = tα(h)). LetJ be a Hilbert G–module and H a submodule of J . Then we form the quotient moduleJ/H , which is a Hilbert G–module. The quotient space J/H is a Hilbert space sinceof course it can be identified with J�H . One easily check that module action is givenby t · h = P (th) for t ∈ G and h ∈ J �H , where P is the orthogonal projection fromJ onto J �H .

Now suppose that H and K are in H(G). We want to consider all possible HilbertG–module J such that H is a submodule of J and K is an associated quotient module,i. e., K ∼= J/H (= J �H). We then have a short exact sequence

E : 0 −→ Hα−→ J

β−→ K −→ 0

of Hilbert G–modules; such a sequence is called an extension of K by H , or J iscalled an extension of K by H . Let S(K,H) be the set of all extensions of K byH . We call two extensions E, E′ to be equivalent if there exists a Hilbert G–modulehomomorphism θ such that the diagram

E : 0 −→ Hα−→ J

β−→ K −→ 0‖ θ ‖

E′ : 0 −→ Hα′−→ J ′ β′

−→ K −→ 0

86 Math. Nachr. 246-247 (2002)

commutes. The set of equivalence classes of S(K,H) under this relation is defined tobe the extension group, ExtG(K,H). The zero element of ExtG(K,H) is the splitextension:

0 −→ H −→ H ⊕K −→ K −→ 0

where the middle term is the (orthogonal) direct sum of the two modules. The firstthing is to show that ExtG(−,−) is a bifunctor from H(G) to the category of abeliangroups. It is contravariant in the first and covariant in the second variable. In fact,using the method completely similar to that for Hilbert modules over function algebras,one can establish the functoriality of ExtG(−,−). For the sake of reader, we sketchthe outline.

A pullback diagram in the category H(G) is a diagram of Hilbert G–modules andhomomorphisms of the form

H1α1−→ K

α2←− H2 .

The pullback of the diagram is a Hilbert G–module J together homomorphismsβi : J → Hi such that

α1β1 = α2β2 ,

and if J ′ is another Hilbert G–module with homomorphisms γi : J ′ → Hi such thatα1γ1 = α2γ2, then there exists a unique homomorphism θ : J ′ → J with βiθ = γi fori = 1, 2. The definition of a pushout is the dual statement, obtained by reversing allof the arrows.

Completely similar to that of Hilbert modules over function algebras, we see theexistence of pullbacks and pushouts in the category H(G) and hence one can establishthe functoriality of ExtG. Let [E] ∈ ExtG(K,H) be the equivalence class of anextension of K by H

E : 0 −→ Hγ−→ J

δ−→ K −→ 0

and if α : K′ → K, then α∗[E] is defined to be the equivalence class of Eα which isthe upper row of the following diagram

Eα : 0 −→ Hγ′−→ J ′ δ′−→ K′ −→ 0

‖ ψ αE : 0 −→ H

γ−→ Jδ−→ K −→ 0

where {J ′, δ′, ψ} is the pullback of the diagram in the lower right corner and γ′ isgiven by γ′(h) = (γ(h), 0), for h ∈ H .

Likewise, if β : H → H ′, then βE is obtained by taking the pushout of E along β.Set [ β∗E ] to be the equivalence class of βE.

As done in [CC1], the reader easily verify that if E and E′ are equivalent exten-sions of K by H , then Eα and E′α are equivalent, and βE and βE′ are equivalent.Furthermore, we have induced homomorphisms

β∗ : ExtG(K,H) −→ ExtG(K,H ′) ,α∗ : ExtG(K,H) −→ ExtG(K′, H)


where β∗[E] = [βE], and α∗[E] = [Eα]. By careful verifying, one can show that thehomomorphisms α∗, β∗ satisfy condition α∗β∗=β∗α∗. Furthermore, if α′ : K′′ → K′,β′ : H ′ → H ′′ are G–module homomorphisms, then

(αα′)∗ = α′∗α∗ , (β′β)∗ = β′∗β∗ .

Moreover, when G is abelian one may naturally make ExtG(K,H) as two–sideG–module. In fact, let t ∈ G, and t : K → K, t : H → H be given by the ac-tions of t. Then we have homomorphisms t∗, t∗ from ExtG(K,H) to itself. A carefulverifying shows that both actions of G are the same and they are compatible with theaddition on ExtG(K,H). When G is nonabelian, it is impossible to make ExtG(K,H)as a G–module. As indicated above, we have shown the following

Proposition 2.1. ExtG(−,−) is a bifunctor from the category H(G) to the categoryof abelian groups. It is contravariant in the first and covariant in the second variable.In particular, when G is abelian, ExtG(−,−) is a bifunctor from the category H(G)to the category of G–modules.

The following Hom–Ext sequences are valid in the category of Hilbert G modules.Their proofs are similar to those of Hilbert modules over function algebras [CC1] ormodules over rings [HS].

Proposition 2.2. Let E : 0→ H1α→ H2

β→ H3 → 0 be an exact sequence of HilbertG–modules. Let H be a Hilbert G–module. Then we have the following Hom–Ext–sequences

0 −→ HomG(H,H1)α∗−→ HomG(H,H2)

β∗−→ HomG(H,H3)δ−→ ExtG(H,H1)

α∗−→ ExtG(H,H2)β∗−→ ExtG(H,H3) ,

where δ is the connecting homomorphism and is given by δ(θ) = [Eθ] for θ : H → H3,and

0 −→ HomG(H3, H) β∗−→ HomG(H2, H) α∗−→ HomG(H1, H)

δ−→ ExtG(H3, H) β∗−→ ExtG(H2, H) α∗−→ ExtG(H1, H) ,

where δ(θ) = [θE] for θ : H1 → H.

Let H, K be Hilbert G–modules, and B(K,H) all bounded linear operators from Kto H . Suppose a map D : G→ B(K,H) satisfies

D(st) = D(s)πK (t) + πH(t)D(s) , for all s , t ∈ G ,we say that D is a G–derivation. For a G–derivation D, if there exists a boundedlinear operator T : K → H such that

D(t) = DT (t) = TπK(t)− πH(t)T

for any t ∈ G, we say that D is an inner G–derivation. Let Deru(K,H) denote alluniformly bounded G–derivations, i. e.,

Deru(K,H) ={D : sup

t∈G‖D(t)‖ <∞

}

88 Math. Nachr. 246-247 (2002)

and Inn(K,H) all innerG–derivations. As in the case of Hilbert modules over functionalgebras [CC1], one see that extension is closely related derivation problem. For anextension of K by H

E : 0 −→ Hα−→ J

β−→ K −→ 0 ,

then there exists an uniformly bounded G–derivation D ∈ Deru(K,H) such that E isequivalent to the following extension ED defined by D

ED : 0 −→ Hi−→ H⊕K π−→ K −→ 0 ,

where H⊕K is Hilbert space direct sum of H and K with the G–module structuredefined by

t(h, k) = (th +D(t)k, tk) , h ∈ H , k ∈ K .

Finally, two extensions ED1 and ED2 are equivalent if and only if D1 −D2 is inner.

Proposition 2.3. Let H, K be Hilbert G–modules. Then

ExtG(K,H) = Deru(K,H)/Inn(K,H) .

3. Extensions of Hilbert modules over semigroups

In this section we study extensions of Hilbert modules over semigroups, especiallyconcentrate attention on amenable semigroups.

Let H, K be Hilbert G–modules, and let kerπH , and kerπK be kernels of homomor-phisms

πH : G −→ πH(G) ; πK : G −→ πK(G)

respectively. Then H, K are Hilbert G–modules in the obvious way, where G denotesthe quotient semigroup G/(kerπK ∩ kerπH). The next theorem seems intuitive, andit will serve the study of extensions over nonamenable groups.

Theorem 3.1.

ExtG(K,H) = ExtG(K,H) .

Proof . Let D : G → B(K,H) be a uniformly bounded derivation, and let x ∈ker πK ∩ kerπH . Then by induction,

D(xn) = nD(x) .

This impliesD(x) = 0 because D is uniformly bounded. Now define D : G→ B(K,H)byD(g) = D(g) for all g. Now we check that this is well defined. Firs note that g1 = g2if and only if there exists an element x ∈ kerπK∩kerπH such that g1 = g2 x. It followsthat

D(g1) = D(g2 x) = D(g2)πK(x) + πH(g2)D(x) = D(g2) ,


and hence D is well defined. Since

D(g1g2) = D(g1 g2)= D(g1)πK(g2) + πH(g1)D(g2)= D(g1)πK(g2) + πH(g1)D(g2)

we conclude that D is a uniformly bounded derivation from G to B(K,H). Obviouslythe corresponding D �→ D is one–one and onto from derivations over G to those overG, and D is inner if and only if D is inner. By Proposition 2.3, we see ExtG(K,H) =ExtG(K,H), completing the proof. �

Recall that a semigroup G is called left amenable if there exists a left invariantmean on G. This means there exists a positive linear functional µG on L∞(G) with‖µG‖ = |µG(1)| = 1, and for any f ∈ L∞(G) and t ∈ G, µG(ft) = µG(f), whereft(s) = f(ts), for all s ∈ G. Right amenability and (two–side) amenability are definedin the similar way. When G is a group, it is easy to see that left and right amenabilityare equivalent, and hence in the case of groups, “left, right” are often omitted. Latewe shall denote by G∗ the semigroup with the underlying set G, and the semigroupoperation being given by s◦ t = ts for s, t ∈ G. It is easy to see that G is left amenableif and only if G∗ is right amenable. A basic fact is that each abelian semigroup isamenable [Pa].

For a Hilbert G–module H , we say that H is G–projective if for each pair H1, H2 ∈H(G), and every pair homomorphisms ψ : H → H2 and φ : H1 → H2 with φ onto,there exists a homomorphism ψ : H → H1 such that ψ = φψ. Also, for I ∈ H(G), Iis called G–injective if for every pair H1, H2 ∈ H(G) and every pair homomorphismsψ : H1 → I, and φ : H1 → H2 with φ one–to–one and having closed range, there existsa homomorphism ψ : H2 → I such that ψ = ψφ. Using Proposition 2.2 we see that aG–module H is projective if and only if for each G–module K, ExtG(H,K) = 0, andG–module I is injective if and only if for each G–module J , ExtG(J, I) = 0. Let G beleft amenable. The next theorem says that if a G–module H is similar to a unitarymodule, then H is G–projective. It is a basic tool which enable us calculate someextension groups. For the function algebra version of Theorem 3.2, see our papers[CG, Guo]. Recall that a Hilbert G–module is said to be unitary (isometric) if πH(t)is a unitary (isometric) operator for each t ∈ G.

Theorem 3.2. Suppose G is left amenable. If a G–module K is similar to a unitarymodule, then K is G–projective. Similarly, if G is right amenable, and a G–moduleH is similar to a unitary module. Then H is G–injective.

Proof . To showK is projective, we need to prove ExtG(K,H) = 0 for anyG–moduleH . By the functoriality of ExtG, we may suppose K is unitary. From Proposition 2.3,our task is to show that each uniformly bounded derivation D ∈ Deru(K,H) is inner.To do this, we write B1(H,K) for all trace class operators from H to K, and identifyB(K,H) with B∗

1(H,K) by setting

〈T, C 〉 = tr(TC) , T ∈ B(K,H) , C ∈ B1(H,K) .

Let µ be a left invariant mean of G. Defining T ∈ B(K,H) = B∗1(H,K) by setting

〈T, C 〉 = µt (〈D(t)π∗K(t), C 〉) ,

90 Math. Nachr. 246-247 (2002)

that is, 〈T, C〉 is the mean of the bounded complex function

t �−→ 〈D(t)π∗K (t), C 〉

on G. For each s ∈ G, we have

〈πH(s)T − TπK(s), C 〉 = 〈T, CπH(s) − πK(s)C 〉= µt (〈D(t)π∗

K (t), CπH(s) − πK(s)C 〉)= µt (〈πH(s)D(t)π∗

K (t)−D(t)π∗K(t)πK(s), C 〉)

= µt

(⟨(D(st) −D(s)πK (t)

)π∗

K(t)−D(t)π∗K(t)πK(s), C

⟩)= µt (〈−D(s), C 〉) + µt (〈D(st)π∗

K (t), C 〉)−µt (〈D(t)π∗

K(t)πK(s), C 〉)= 〈−D(s), C〉 + µst (〈D(st)π∗

K (st)πK(s), C 〉)−µt (〈D(t)π∗

K(t)πK(s), C 〉)= 〈−D(s), C 〉

for all C ∈ B1(H,K), so that D(s) = TπK(s)−πH (s)T . We see that D = DT is inner,and hence ExtH(K,H) = 0 for each G–module H . It follows that K is projective. IfG is right amenable, then G∗ is left amenable. Let π∗

H be the conjugate representationof G on H , i. e., t �→ π∗

H(t). This just is the representation of G∗ on H . Clearly,this representation is unitary, and hence H is G∗–projective. This implies that H isG–injective. This completes the proof of Theorem 3.2. �

Theorem 3.2 provides one kind of method for calculating of some extension groups,especially associated with isometric modules. The following proposition gives a crite-rion for judging isometric modules.

Proposition 3.3. Let G be a right amenable semigroup, and H a G–module. ThenπH is similar to an isometric representation if and only if πH is uniformly belowbounded, that is, there exists a positive constant c such that for any h ∈ H,

inft∈G‖πH(t)h‖ ≥ c ‖h‖ .

Equivalently, a Hilbert G–module H is similar to an isometric module if and only ifπH is uniformly below bounded.

Proof . It is easy to see that necessary condition is obvious. We prove the oppositedirection. Let µG be a right invariant mean on G. Then for any h1, h2 ∈ H , we set

〈〈h1, h2〉〉 = µG (〈πH(t)h1, πH(t)h2〉) .Since

c′ ‖h‖ ≥ ‖πH(t)h‖ ≥ c ‖h‖ ,it follows that 〈〈 , 〉〉 is a Hilbert inner product on H . Consequently, if we equip Hwith the new Hilbert inner product, by the invariance of µG, πH becomes isometric.Finally, by the Riesz representation theorem, there exists a bounded linear operatorA such that

〈〈h1, h2〉〉 = 〈Ah1, h2〉 for all h1 , h2 ∈ H .


It is easy to check that A is positive and invertible. We thus conclude that

t �−→ A12 πH(t)A− 1

2

is an isometric representation, completing the proof. �

Now let H be an isometric G–module, and H a unitary extension of H . Note thatif G is abelian, then an isometric G–module can always be extended to a unitaryG–module. We thus have an exact sequence of G–modules

EH : 0 −→ Hi−→ H

π−→ H �H −→ 0 .

Following Propositions 2.2, and Theorem 3.2, for any G–module K we have

Corollary 3.4. Suppose G is left amenable. We have

ExtG(K,H) ∼= coker{π∗ : HomG

(K, H

)→ HomG

(K, H �H)} .

The correspondence is given by δ(θ) = [δθ] for θ ∈ HomG

(K, H � H

), where δθ

(derivation) is defined by δθ(t) = PHπH(t)θ, PH is the orthogonal projection from Honto H.

Below we give some concrete examples to show the relationship between HilbertG–modules and Hilbert modules over function algebras, and applications in operatortheory.

Example 3.5. Assume G is an abelian group (written as addition), and G+ asubsemigroup of G with the identity. Let G be the Pontryagin dual group of G, andlet m be the normalized Haar measure on G. For s ∈ G, we let ξs denote the evaluationhomomorphism from G to the circle group T defined by ξs(x) = x(s) for x ∈ G. It iswell known that the family of functions {ξs : s ∈ G} forms the canonical orthogonalbasis of the Hilbert space L2

(G,m

). When no confusion rises, we denote the Hilbert

space with the orthogonal basis{ξs : s ∈ G+

}by H2

(G). Now we consider H2

(G)

as a G+–module by

ρ(s)ξt = ξs+t .

Similarly, we set L2(G,m

)as a G+–module in the same way. Then L2

(G,m

)is

G+–projective. Let H2(G)⊥ be L2

(G,m

)�H2(G)

with the natural quotient modulestructure. Below we will assume that G = G+ −G+. Then

α : H2(G) −→ L2

(G,m

)is a homomorphism if and only if there exists φ ∈ L∞(G,m) such that α(f) = φf .Let P+ denote the orthogonal projection from L2

(G,m

)onto H2

(G). For each f ∈

L2(G,m

), a Hankel operator Hf :H2

(G)→H2

(G)⊥ is defined by Hfh=(I−P+)(f h)

with the domain H∞(G). Clearly, Hf is dense defined. We will denote by B(G+)

all those f such that Hf is bounded. It is easy to check that a bounded operator

92 Math. Nachr. 246-247 (2002)

α : H2(G) → H2

(G)⊥ is a homomorphism if and only if there exists an f ∈ B(G+

)such that α = Hf . From Corollary 3.4, we have

ExtG+

(H2(G), H2

(G))

= B(G+)/(L∞(G)+H2

(G)).

We say that G+ is pure if G+ does not contain nontrivial subgroup of G. It is easyto see that G+ is pure if and only if G+ ∩ ( − G+

)= {0}. If G+ is pure, and

G = G+ ∪ (− G+), then we make G as a order group in the natural way. This means

that s ≤ t if t−s ∈ G+. From [YCG, Theorem 1.1], we see that a Hankel operator Hf

is bounded if and only if there exists a φ ∈ L∞(G) such that f − φ ∈ H2(G). Hence

when G is a order group with the positive cone G+,

ExtG+

(H2(G), H2

(G))

= 0 .

In this case, a trivial example is that G = Z and G+ = Z+. Another class of examplesare that G = Z2, and G+ = {[m, n] : βm+n ≥ 0}, here β is a fixed irrational number.

Now let G = Z2, and G+ = Z2+. From [CS], one knows that there exists a bounded

Hankel operator Hf which does not admit a bounded symbol. Therefore we have

ExtZ2+

(H2(T 2), H2

(T 2)) �= 0 .

Example 3.6. Let FS2 be the free semigroup with two generators a, b. This meansthat each element x in FS2 is uniquely of the form am1bn1 . . . amkbnk, here mi, ni ≥ 0for i = 1, 2, . . . , k. It is well known that FS2 is not left amenable, and not rightamenable. Let C be the set of all complex numbers. We consider C as a HilbertFS2–module with the trivial action of FS2 on C. Let the action of FS2 on l2(FS2)be given by

(τ (s)h)(t) = h(ts) .We claim

ExtFS2

(C, l2(FS2)

) �= 0 .

It follows that C, as a Hilbert FS2–module with the trivial action of FS2 on C, is notprojective, although C is a unitary FS2–module.

To prove the claim, we construct an uniformly bounded derivation

D : FS2 −→ B(C, l2(FS2)

)which is not inner. For each t ∈ FS2, we define δt as the function δt(s) = 1 if s = t,and δt(s) = 0 if s �= t. Now let χa denote the characteristic function on the setof all nontrivial elements of FS2 whose reduced words begin in an with n ≥ 1. Astraightforward verifying shows that

D(t) = (τ (t)− 1)χa = χa(t)δe

is a uniformly bounded derivation from FS2 to B(C, l2(FS2)

), here e is the identity

of FS2. If D is inner, then there exists an f in l2(FS2) such that D(t) = (τ (t)− 1)f .So,

τ (t)(f − χa

)= f − χa ,


for any t ∈ FS2. It follows that f = χa + constant. This is impossible, and henceD(t) = χa(t)δe is not inner. The claim is proved. In the similar method, we can prove

ExtFS2

(l2(FS2), C

) �= 0 .

This implies that C, as a trivial Hilbert FS2–module, is not injective.

Example 3.7. Let H be Hilbert modules over the polydisk algebra A(Dn), and letH be isometric. If we consider H as Hilbert Zn

+–module in the obvious way, i. e.,

πH([m1, m2, . . . , mn])h = zm11 zm2

2 . . . zmnn h .

Then by Corollary 3.4 and [CG], for any Hilbert module K over A(Dn),

ExtZ+n(K,H) = ExtA(Dn)(K,H) .

This says that if the following operators are commutative,(πH(zi) Ai

0 πK(zi)

), i = 1 , 2 , . . . , n ,

then they are polynomially bounded if and only if they are jointly power bounded.Recall that operators {T1, T2, . . . , Tn} are jointly power bounded if

sup[m1,m2,··· ,mn]∈Zn

+

∥∥Tm11 Tm2

2 . . . Tmnn

∥∥ < ∞ .

Example 3.8. In [CCFW], they introduced the full subcategory C of the categoryof all Hilbert modules over the disk algebra A(D). By definition C consists of allHilbert modules H over A(D) with the property that the action by z on H is similarto a contraction. By the remarkable work of Pisier [Pi2], the category C actually isproper subcategory of all Hilbert modules over A(D). In [CCFW], they proved thatthe category C has enough projective objects and injective objects, and a module His projective if and only if H is similar to an isometric module. Let G be a semigroup.We use CG to denote the category of all those G–modules H which are similar tocontract G–modules. It is easy to see that CG is a full subcategory of H(G). Nowtake G = Zn

+, and let e1 = [1, 0, . . . , 0], . . . , en = [0, 0, . . . , 1]. Suppose that T be apolynomially bounded operator on H which is not similar to a contraction [Pi2]. Ifthe action of Zn

+ on H is defined by

e1h = Th , e2h = h , . . . , enh = h , h ∈ H .

Then H is not similar to a contract Zn+–module, and hence CZn

+is a proper subcategory

of H(Zn+

). By [CCFW], in the category CZ+ , a Hilbert Z+–module is projective if and

only if it is similar to a isometric module. In what follows we will show that H2(Tn) isnot projective in the category CZn

+if n > 1. For simplicity , we consider n = 2. From

Example 3.5, we see ExtZ2+

(H2(T 2), H2

(T 2)) �= 0. Therefore there exists a nontrivial

extension

E : 0 −→ H2(T 2) i−→ H2

(T 2)⊕H2

(T 2) π−→ H2

(T 2) −→ 0

94 Math. Nachr. 246-247 (2002)

where the action of Z2+ on H2

(T 2)⊕H2

(T 2)

is given by

[m, n] ◦(h1

h2

)=

(Tzmwn D

([m, n]

)0 Tzmwn

)(h1

h2

)for some uniformly bounded derivation D : Z2

+ → B(H2(T 2), H2

(T 2))

. By Propo-sition 3.3, a careful verifying shows that H2

(T 2)⊕H2

(T 2)

is similar to a isometricZ2

+–module, and hence a contract module. Consequently, H2(T 2)

is not projective inthe category CZ2

+. By means of A. Bernard theorem [Gar, Chapter V,Th.2.3], one

can prove that H2(Tn) (n > 1) is not projective in the category C(A(Dn)), whoseobjects are those Hilbert modules H over A(Dn) which are similar to contract mod-ules. Recall that A. Bernard theorem says that the unit ball of A(Dn) is the closedconvex hull of inner functions in A(Dn).

Example 3.9. Let α : G′ → G be a group homomorphism. For any HilbertG–module H , then H becomes a Hilbert G′–module under the action g′ · h = α(g′)h.If α is surjective, then by the proof of Theorem 3.1

ExtG′(K,H) = ExtG′/ker α(K,H) = ExtG(K,H) .Set α : FS2 → Z2

+ by

α(am1bn1 . . . amkbnk) = [m1 +m2 + . . .+mk, n1 + n2 + . . .+ nk] .

Then α is a surjective homomorphism, and hence the Hardy space H2(T 2)

becomesa FS2–module under the action

x · h = zm1+m2+...+mkwn1+n2+...+nk h ,

here x = am1bn1 . . . amkbnk. So,

ExtFS2

(H2(T 2), H2

(T 2))

= ExtZ2+

(H2(T 2), H2

(T 2)) �= 0 .

Now we return to general case. Let G be a semigroup. As in [Pi1], one uses T1(G)to denote all the functions f : G→ C which admit the following decomposition: thereexist functions f1, f2 : G×G→ C such that

f(st) = f1(s, t) + f2(s, t)and

sups∈G

∑t∈G

|f1(s, t)| < ∞ , supt∈G

∑s∈G

|f2(s, t)| < ∞ .

One equips this space with the norm

‖f‖T1(G) = inf

{sup

s

∑t

|f1(s, t)|+ supt

∑s

|f2(s, t)|}

where the infimum runs over all such decompositions. We have clearly

l1(G) ⊆ T1(G) ⊆ l∞(G)


and T1(G) is a Banach space [Pi1].For a Hilbert G–module H , we shall denote by H∗ the Hilbert G∗–module with the

underlying Hilbert space H , and the action of G∗ on H being defined by t h = π∗H(t)h.

When G is abelian, H∗ also is a Hilbert G–module.Below we will regard l2(G) as a Hilbert G–module with the module action defined

by shifts, i. e.,

ρ(s)δt = δst .

Theorem 3.10. Let G be an abelian semigroup with cancellation law. Then


) �= 0

if and only if G is not a group.

Combining Proposition 2.3 with Theorem 3.10, we immediately obtain that

Corollary 3.11. Let H be a Hilbert space. Let G be an abelian semigroup withcancellation law which is not a group. Then

ExtG(l2(G)⊗H, l2∗(G)⊗H) �= 0 .

Assume G is an abelian group (written as addition), and G+ a subsemigroup of Gwhich is not a subgroup of G. Then we have

Corollary 3.12.

ExtG+

(H2(G), H2

∗(G)) �= 0 .

To prove Theorem 3.10, we need the following lemma.

Lemma 3.13. Assume that G be an abelian semigroup, and T1(G) �⊆ l2(G). Then


) �= 0 .

Proof . Based on the idea of [Bo] or [Pi1], we give the details of the proof. Us-ing Proposition 2.3, we shall construct a uniformly bounded derivation D : G →B(l2(G), l2∗(G)

)which is not inner. Let f ∈ T1(G), and f /∈ l2(G). Consider the small

Hankel operator hf : l2(G) → l2∗(G) which admits [f(st)]s,t∈G as its representativematrix. Note that f has a decomposition

f(st) = f1(s, t) + f2(s, t)with

sups∈G

∑t∈G

|f1(s, t)| ≤ C , supt∈G

∑s∈G

|f2(s, t)| ≤ C

for some constant C. If we write hf = A1+A2, where Ai has the representative matrix[fi(s, t)]s,t∈G for i = 1, 2. Then

‖A1‖B(l∞(G)) ≤ C , and ‖A2‖B(l1(G)) ≤ C .

96 Math. Nachr. 246-247 (2002)

Since hfρ(t) = ρ(t)∗hf ,

A1ρ(t) − ρ(t)∗A1 = −(A2ρ(t) − ρ(t)∗A2) .

Because ρ(t) and ρ(t)∗ are of norms less than 1 both on l1(G) and l∞(G), and itfollows that

‖A2ρ(t)− ρ(t)∗A2‖B(l2(G)) ≤ 2C

by the Riesz–Thorin interpolation theorem (see [Pi1]). Now let

D(t) = A2ρ(t) − ρ(t)∗A2 .

Then D : G→ B(l2(G), l2∗(G)

)is a uniformly bounded derivation. If D is inner, then

there exists a bounded linear operator A ∈ B(l2(G), l2∗(G))

such that

D(t) = A2ρ(t) − ρ(t)∗A2 = Aρ(t) − ρ(t)∗A .This induces

〈D(t)δe, δe〉 = 〈A2δt, δe〉 − 〈A2δe, δt〉= f2(e, t)− f2(t, e)= 〈Aδt, δe〉 − 〈Aδe, δt〉= 〈δt, A∗δe〉 − 〈Aδe, δt〉 .

It follows that f2(e, t)− f2(t, e) ∈ l2(G), and hence

f(t) = f2(e, t) − f2(t, e) + f1(e, t) + f2(t, e) ∈ l2(G) .

This contradiction shows that D is not inner, and hence


) �= 0 .

The proof is complete. �

Let G be an abelian semigroup with cancellation law (written as addition). Set

G0 = {s ∈ G : there exists a t such that s+ t = 0} .Then G0 is a group, and it is the maximal group which is contained in G. Let

G+ = {t ∈ G : t = 0, or t /∈ G0} .It is easy to see that G+ is a subsemigroup of G, and G = G0+G+. For each t ∈ G,we write t = t− + t+ for the unique decomposition of t. Moreover, we define a partialordering � on G as follows: s � tmeans that there exists a t′ ∈ G+ such that s = t+t′.Clearly, � is linear, that is, if s � t, then s+ t′ � t+ t′ for any t′ ∈ G.

Proof of Theorem 3.10. If G is a group, then clearly


)= 0


by amenability of G. Now assume G is not group. Using the above notation, then G+

is not trivial. This implies that there is a nonzero element t0 in G+. Set

S = {3nt0 : n = 0, 1, 2, . . .} .We use χS to denote the characteristic function of S. Obviously, χS /∈ l2(G). Belowwe shall show χS ∈ T1(G). Consider the decomposition

χS(s+ t) = χ{s+t∈S; s+�t+} + χ{s+t∈S; s+ ��t+} .

Assume that s is given (hence s+ is given), such that s + t ∈ S and s+ � t+. Weclaim that there exists at most such one t. To reach contradiction, suppose that thereexist t1, t2 such that s+ � t+i and s + ti ∈ S for i = 1, 2. By s + ti ∈ S, we sees− + t−i = 0, and hence s+ ti = s+ + t+i ∈ S. It follows that there exist nonnegativeintegers n1 < n2 such that s+ + t+1 = 3n1t0 and s+ + t+2 = 3n2t0. Since

2 s+ � s+ + t+2 = 3n2 t0 = 3n2−n1 3n1 t0 = 3n2−n1(s+ + t+1

),

there is a g ∈ G+ such that

3n2−n1(s+ + t+1

)+ g = 2 s+ .

Since G is cancellative, we see

s+ +(3n2−n1 − 3

)s+ + 3n2−n1t+1 + g = 0 .

This implies that s+ ∈ G0, and similarly, t+1 ∈ G0. So, 3n1t0 ∈ G0, and hence t0 ∈ G0.This is impossible. The claim is proved. Hence

sups

∑t

χ{s+t∈S; s+�t+} ≤ 1 .

Now let t be fixed (and hence t+ is fixed) such that s + t ∈ S and s+ �� t+. Weclaim that there exists at most such one s. To reach contradiction, suppose that thereexist s1, s2 such that s+i �� t+ and si + t ∈ S for i = 1, 2. By si + t ∈ S, we sees−i + t− = 0, and hence si + t = s+i + t+ ∈ S. It follows that there exist nonnegativeintegers m1 < m2 such that s+i + t+ = 3mit0 for i = 1, 2. Since

s+2 + t+ = 3m2 t0 = 3m2−m1 3m1 t0 = 3m2−m1(s+1 + t+

),

this gives that

s+2 = 3m2−m1 s+1 +(3m2−m1 − 2

)t+ + t+ .

Since m2 > m1, we see s+2 � t+. This is a contradiction, and hence there exists atmost such one s such that s+ t ∈ S and s+ �� t+. Therefore

supt

∑s

χ{s+t∈S; s+ ��t+} ≤ 1 .

From the above discussion, one sees that χS is in T1(G). By Lemma 3.13, we obtainthe desired conclusion. The proof is completed. �

98 Math. Nachr. 246-247 (2002)

Remark 3.14. When G = Z+, the analogous construction in the proof of Theorem3.10 have appeared in [Fo, Bo], also see [Pi1, Lemma 2.13].

Example 3.15. By Corollary 3.12, we see

ExtZn+

(H2(Tn), H2

∗ (Tn)) �= 0 .

In particular, we have

ExtZ+

(H2(T ), H2

∗ (T )) �= 0 .

By [CCFW], for each extension

E : 0 −→ H2∗ (T ) α−→ H

β−→ H2(T ) −→ 0 ,

if H is similar to a contract Z+–module, then the extension E is split. Therefore, theextension group ExtZ+

(H2(T ), H2∗(T )

)consists of those equivalent extensions which

are not similar to contractions. Equivalently, under similarity, ExtZ+

(H2(T ), H2∗(T )

)is the equivalent classes of all power bounded operators with the following form whichare not similar to contractions, (

Tz A

0 Tz

).

Note that H2∗(T ), as a contract Hilbert module over A(D) is given by

f · h = P (f(z)h) ,

where P is the orthogonal projection from L2(T ) onto H2(T ). Then clearly byProposition 2.3, the extension group ExtA(D)

(H2(T ), H∗(T )

)is embedded into

ExtZ+

(H2(T ), H2∗(T )

)as subgroup. If one can prove

ExtA(D)

(H2(T ), H∗(T )

) �= 0 ,

then by [CCFW], a negative answer for Halmos problem [Pi2] is possible by differentmethod from [Pi2], although Pisier [Pi2] have produced examples of polynomiallybounded operators which are not similar to contractions.

4. Extensions of Hilbert modules over nonamenable groups

It is well known that a classical problem in the context of representation theory ofgroups asks whether all uniformly bounded representation of a group are similar tounitary representations. For nonamenable groups, this problem remained open fora while until, in 1955, Ehrenpreis and Mautner [EM] gave counterexample forG = SL2(R). Later it was realized that rather simpler counterexamples be describedon the free groups with at least two generators (see [Pi1]). In the positive direction,the most general results seems to be Dixmier’s theorem [Pi1, pp. 6] which says thatif G is amenable then the preceding problem has an affirmative answer. Actually, oneeasily modifies the argument [Pi1, pp. 6] to yield that if πH(G), as discrete group, is


amenable, then the representation πH of G on H is similar to a unitary representation.In the context of Hilbert modules, the preceding problem asks whether each HilbertG–module is similar to a unitary module. In this section, we study extensions of Hilbertmodules over nonamenable groups, which is closely related to the above problem. Inwhat follows we always assume that G is a group.

Let π : G→ B(H) be a representation of G on H . We say that π is unitarizable ifπ is similar to a unitary representation of G. In other word, a Hilbert G–module H isunitarizable if H is similar to a unitary module.

Lemma 4.1. Assume that H, K are unitarizable. If J is an extension of K by H

0 −→ Hα−→ J

β−→ K −→ 0 .

Then J is unitarizable if and only if the above extension is split.

Proof . To prove sufficiency, we may suppose H, J, K are unitary G–modules. Letξ ∈ J � αH and ζ ∈ H . Since

〈πJ(t)ξ, α(ζ)〉 = 〈ξ, π∗J(t)α(ζ)〉

=⟨ξ, πJ

(t−1)α(ζ)

⟩=⟨ξ, α

(πH(t−1

)(ζ))

⟩= 0 ,

this says that J � α(H) is invariant under G, and hence β : J � α(H) → K is aisomorphism of G–modules. It follows that the above extension is split. Conversely, ifthe above extension is split, it is obvious that J is unitarizable, completing the proof.

�

By Dixmier’s theorem [Pi1, pp. 6], if G is amenable, then each Hilbert G–module isunitarizable. Combining Dixmier’s theorem with Lemma 4.1, we immediately obtain

Proposition 4.2. Let G be amenable. Then for any Hilbert G–modules H, K, wehave

ExtG(K,H) = 0 .

According to Proposition 4.2, we shall concentrate attention on extensions of Hilbertmodules over nonamenable groups. Now suppose H, K are unitarizable over a non-amenable group G. Lemma 4.1 thus indicates that ExtG(K,H) just is the set ofextensions of K by H under similarity, which are not unitarizable.

For nonamenable groups, the simplest example is the free group F2 on two gen-erators. It is obvious that if a group G contains F2 as a subgroup, then G is nec-essarily nonamenable. Recall that the famous V. Neumann’s (or Day’s) conjec-ture says that if a group G is nonamenable, then it contains F2 as a subgroup [Pa].Evidence in favor of the conjecture is based on a series of examples, for example,SL(n, C); SU(n, C); PSL(2, R) and the isometric group of Rn, Gn (see [Pa]). How-ever, A. Yu. Ol’shanskii (see [Pa, pp. 168]) has constructed a group G0 which isnonamenable, but does not contain F2 as a subgroup, so that the conjecture is false.

100 Math. Nachr. 246-247 (2002)

Lemma 4.3. Both πK(G) and πH(G) are amenable if and only if G/(kerπK ∩

ker πH

)is amenable.

Proof . Now suppose G/(kerπK ∩ kerπH

)is amenable. Since kerπK/

(ker πK ∩

ker πH

)is a normal subgroup of G/

(kerπK ∩ kerπH

), Third Isomorphism Theorem

[Rob] implies that the quotient group is isomorphic to G/kerπK. From [Pa], onesees that amenability for discrete groups is preserved by forming quotient groups, andhence G/kerπK is amenable. Note that

G/kerπK∼= πK(G) .

So, πK(G) is amenable. In the same method, πH(G) is amenable. Now let us reachthe opposite direction. By Second Isomorphism Theorem [Rob],

kerπK/(ker πK ∩ kerπH

) ∼= (kerπH kerπK

)/kerπH ,

and (ker πH kerπK

)/kerπH

∼= πH

(ker πK

).

Since amenability for discrete groups is preserved by forming subgroup [Pa],πH

(ker πK

)is amenable, and hence kerπK/

(kerπK ∩ kerπH

)is amenable. Now

Third Isomorphism Theorem [Rob] implies that G/(kerπK ∩ kerπH

)is a group ex-

tension of ker πK/(ker πK ∩ kerπH

)by G/kerπK (∼= πK(G)). Since amenability for

discrete groups is preserved by forming group extensions [Pa], we thus conclude thatG/(kerπK ∩ kerπH

)is amenable. The proof is completed. �

Theorem 4.4. Let both πK(G) and πH(G) be amenable. Then

ExtG(K,H) = 0 .

Proof . Applying Theorem 3.1, Lemma 4.3 and Proposition 4.2. �

Now we consider the complex number field C as a Hilbert G–module with the trivialaction of G on C. For a Hilbert G–module H , a bounded function d : G → H withthe property

d(st) = d(s) + πH(s)d(t)

is called a bounded derivation from G into H . If there exists some h ∈ H such that

d(t) =(πH(t)− 1

)h ,

then the derivation d is said to be inner. From Proposition 2.3 and Theorem 4.4, wesee that if πH(G) is amenable, then each bounded derivation is inner. Now let Tn bethe n–circle group, and γ : G → Tn a homomorphism. Set the action of G on Cn

being given by

t · h = γ(t)h , t ∈ G , h ∈ Cn .

Since γ(G) is abelian, each bounded derivation d from G to Cn is inner, that is,

d(t) = (γ(t) − 1)h


for some h ∈ Cn.Moreover, by applying Proposition 2.3, we have the following

Proposition 4.5.

ExtG(C,H) = DerG(H)/InnG(H) ,

where DerG(H) and InnG(H) denote the sets of all bounded derivations, all innerderivations from G into H, respectively.

Now we equip l1(G) with the left regular representation of G, that is,

(λ(t)h)(s) = h(t−1s

),

and define a G–module homomorphism ρ : l1(G)→ C by

ρ

(∑t

atδt

)=∑

t

at .

The kernel of ρ, denoted by l10(G), is called the augmentation ideal of G. This yieldsthe canonical projective presentation of the complex field C in the category of allBanach G–modules,

0 −→ l10(G) −→ l1(G) ρ−→ C −→ 0 .

Because l1(G) is G–projective in the category of all Banach G–modules, applyingProposition 2.2, we have

Proposition 4.6. A bounded function d : G → H is a derivation if and only ifthere exists a unique bounded homomorphism α ∈ HomG

(l10(G), H

)such that

d(t) = α(δt − δe) .The derivation d is inner if and only if α extends to a bounded homomorphism α :l1(G)→ H.

Consider l2(G) as a Hilbert G–module by the left regular representation. Given abounded homomorphism α ∈ HomG

(l10(G), l2(G)

), then injectivity of l∞(G) implies

that α extends to a homomorphism from l1(G) to l∞(G). This implies that there isan h ∈ l∞(G) such that

α(δt − δe) = (λ(t) − 1)h .

From the above discussion and Proposition 4.6, one sees that for each bounded deriva-tion d : G→ l2(G), there exists a h ∈ l∞(G) such that

supt‖(λ(t)− 1)h‖2 < ∞

and

d(t) = (λ(t) − 1)h .Set

102 Math. Nachr. 246-247 (2002)

S(G) ={h ∈ l∞(G) : sup

t‖(λ(t)− 1)h‖2 <∞

}.

If we equip the seminorm on S(G) with

‖h‖G = supt‖(λ(t) − 1)h‖2 ,

then S(G)/C is a Banach G–module in the obvious way. A careful verifying showsthat there exists a positive constant C0 such that

C0 ‖h‖2 ≤ supt‖(λ(t)− 1)h‖2 ≤ 2 ‖h‖2

for each h ∈ l2(G). It follows that l2(G) is a closed submodule of S(G)/C in the normof S(G)/C. However, we not know if S(G)/C and l2(G) are equal. If we take G = Fn,the free group on n generators, this problem is equivalent to: for h ∈ l∞(Fn), if thereexists a fixed constant C such that∑

s

|h(ts)− h(s)|2 ≤ C , for all t ∈ Fn ,

does the limit lim|s|→∞ h(s) exist? where |s| denotes the length of reduced word s. Inany case, we have

Proposition 4.7.

ExtG(C, l2(G)

)= S(G)/

(l2(G)+C

).

Based on the ideas of [Pi1, Th.2.1] (proved originally in [BF]), the next theoremstrengthens [Pi1, Th.2.1]. Its proof is similar to that of [Pi1, Th.2.1].

Theorem 4.8. Assume T1(G) �⊆ l2(G). Then

ExtG(l2(G), l2(G)

) �= 0 .

Proof . For the sake of reader, we give the outline of proof. By Proposition 2.3, wewill construct a uniformly bounded derivation D : G → B

(l2(G), l2(G)

)which is not

inner. For each t ∈ G, let λ(t) act on CG by left translation by t. It is easy to verifythat a linear operator A on CG commutes with all λ(t) if and only if there exists afunction f on G such that the representation matrix of A is

[f(s−1t

)]s,t∈G

. Indeed,since λ(t)A = Aλ(t), for all t ∈ G, we see

Aδs = f(s−1 · )

where f = Aδe. This induces

〈Aδs, δt〉 = f(s−1t

).

The opposite direction is easy. Now let f ∈ T1(G), and f /∈ l2(G). Consider adecomposition of f by f

(s−1t

)= f1(s, t) + f2(s, t) with

sups∈G

∑t∈G

|f1(s, t)| ≤ C , supt∈G

∑s∈G

|f2(s, t)| ≤ C


for some constant C. Let A, A1, A2, be linear operators on CG whose representationmatrices are [

f(s−1t

)]s,t∈G

,[f1(s, t)

]s,t∈G

,[f2(s, t)

]s,t∈G

respectively. Then clearly by the conditions of f1, f2, we have,

‖A1‖B(l∞(G)) ≤ C , and ‖A2‖B(l1(G)) ≤ C .

Since Aλ(t) = λ(t)A,

A1λ(t) − λ(t)A1 = −(A2λ(t) − λ(t)A2) .

Because λ(t) is of norm 1 both on l1(G) and l∞(G), it follows that

‖A2λ(t) − λ(t)A2‖B(l2(G)) ≤ 2C

by interpolation [Pi1]. Now let

D(t) = A2λ(t) − λ(t)A2 .

Then D : G→ B(l2(G)

)is a uniformly bounded derivation. If D is inner, then there

exists a bounded linear operator B ∈ B(l2(G))

such that

D(t) = A2λ(t) − λ(t)A2 = Bλ(t) − λ(t)B .

So,〈D(t)δe, δe〉 = 〈A2δt, δe〉 − 〈A2δe, δt〉

= f2(e, t)− f2(t, e)= 〈Bδt, δe〉 − 〈Bδe, δt−1 〉= 〈δt, B∗δe〉 − 〈Bδe, δt−1 〉 .

It follows that f2(e, t)− f2(t, e) ∈ l2(G), and hence

f(t) = f2(e, t) − f2(t, e) + f1(e, t) + f2(t, e) ∈ l2(G) .

This contradiction implies that D is not inner. We thus conclude that

ExtG(l2(G), l2(G)

) �= 0 .

�

Corollary 4.9. If G contains F2 as a subgroup, then ExtG(l2(G), l2(G)

) �= 0.

Proof . If G contains F2 as a subgroup, then G contains an infinite free set E. ByLemma 2.7 (ii) in [Pi1], we know that the characteristic function of E∪E−1 is in T1(G)not in l2(G). From Theorem 4.8, the conclusion follows. �

Let SU(n) be n× n unitary matrices, and γ : G→ SU(n) a group homomorphism.Cn, as a Hilbert G–module, is defined by

t · h = γ(t)h , h ∈ Cn ,

104 Math. Nachr. 246-247 (2002)

and denote this module by Cnγ . It is easy to check that when γ = γ′,

ExtG(Cn

γ , Cnγ′)

= 0 .

How to characterize ExtG(Cn

γ , Cnγ′), even more generally, ExtG

(Cn

γ , Cmγ′)

seems dif-ficult because this will involved with some deep representation theory of groups, andgeometry of groups. We will continue this in the forthcoming paper.

Acknowledgements

The author is partially supported by NNSFC(10171019), Shuguang project in Shanghai,

and Young teacher Fund of higher school of National Educational Ministry of China.

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Department of MathematicsFudan UniversityShanghai, 200433P.R. ChinaE–mail:[email protected]

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