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The Projectivity of some BanachAlgebras

Martin S G Jones

Thesis submitted for the degree of

Master of Philosophy

*School of Mathematics and StatisticsUniversity of Newcastle upon Tyne

Newcastle upon TyneUnited Kingdom

May 2006

Abstract

This dissertation describes conditions for some Banach algebras to be pro-

jective. We concentrate mainly on operator algebras, and review some of the

many results obtained by mathematicians in this field up to the present time.

In each case full proofs and explanations are provided.

We begin by defining projectivity, describing its role and some of its appli-

cations, and mentioning some important results obtained in relevant areas.

In the second chapter we follow Lykova’s proof [14] that if X is a Banach

space with an elastic basis then the Banach algebra of compact operators on

X is right projective; furthermore it is also left projective if X is reflexive.

We end the chapter with a simple example of a Banach algebra which is not

projective: the space of sequences which are power p summable, for p > 1.

In the third chapter we consider some biprojective Banach algebras. It

will be proved that unital modules over biprojective algebras are projective

and also that direct sums of biprojective algebras are biprojective.

Chapter four concerns Banach algebras of continuous functions on a lo-

cally compact Hausdorff space which vanish at infinity. We follow Helem-

skii’s proof [7] that in the case of complex-valued functions, such an algebra

is projective if and only if its spectrum is paracompact, and use this to show

that such a result also holds when the range is the algebra of compact oper-

ators on a separable Hilbert space. Finally we prove that in the case when

the range is an arbitrary Banach algebra A, then it is necessary for A to be

projective.

In final chapter we consider C*-algebras and expand on Lykova’s proof [15]

that a commutative C*-subalgebra of a biprojective C*-algebra is necessarily

biprojective.

Contents

1 Introduction 1

1.1 Background . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1

1.2 Banach modules and projectivity . . . . . . . . . . . . . . . . 2

2 Banach algebras which are left or right projective 13

2.1 Preliminary results . . . . . . . . . . . . . . . . . . . . . . . . 13

2.2 On the projectivity of the compact operators . . . . . . . . . . 23

2.3 Lack of projectivity of `p when p > 1 . . . . . . . . . . . . . . 31

3 Banach algebras which are biprojective 35

3.1 Examples of biprojective algebras . . . . . . . . . . . . . . . . 35

3.2 Modules over biprojective algebras . . . . . . . . . . . . . . . 38

3.3 Direct sums of biprojective Banach algebras . . . . . . . . . . 39

4 On the projectivity of C(Ω, A) algebras 42


4.2 On the projectivity of C0(Ω, K(H)) . . . . . . . . . . . . . . . 48

4.3 A necessary condition for C0(Ω, A) to be left projective . . . . 62

5 Projectivity of commutative C*-subalgebras of C*-algebras 64


5.2 Commutative C*-subalgebras of biprojective C*-algebras . . . 69

Bibliography 81

Chapter 1

Introduction

1.1 Background

Projectivity is a concept which can be considered in any category and essen-

tially concerns the possible solutions to lifting problems. Given objects X,

Y , P , morphism ϕ : P → Y and surjective morphism σ : X → Y , does there

exist a morphism ψ : P → X so that σ ψ = ϕ? The extension problem

is the same situation with the arrows reversed and i an injective morphism.

Both are represented below.

X

σ

P

ψ>>

ϕ// Y

Lifting problem

Xψ

J Yϕ

oo

i

OO

Extension problem

The object P is called projective if the lifting problem above can be solved

for each choice of X, Y , ϕ and σ. J is said to be injective if the extension

problem has a solution for any X, Y, ϕ and i.

Projectivity has been studied in categories such as groups and modules over

algebras in the purely algebraic setting. In an article [9] published in 1945

and another in 1946, Hochschild initiated its study in the context of abstract

1

1.2. Banach modules and projectivity 2

algebra. The research of projectivity in the category of modules of Banach

algebras was introduced by Helemskii in a paper of 1970, and has proved to

be a fruitful method of analysing Banach algebras. The study of homology

and cohomology deals with the extensive range of problems associated with

lifting and extension problems, and projectivity is a major tool in this study.

Interested readers are referred to Helemskii’s comprehensive book on the

subject [6].

A number of mathematicians have obtained results in this topic. Selivanov

has produced some important results on biprojectivity of C*-algebras, for

example in his 1976 article [21]. Dales & Polyakov published several results

in their paper [2] on the projectivity of Lp(G) algebras where G is a locally

compact group. In particular L∞(G) is projective if and only if G is finite,

and Lp(G) is projective (where 1 < p <∞) if and only if G is compact.

1.2 Banach modules and projectivity

In this section, A will denote a Banach algebra, not necessarily with an

identity. We describe the main properties of projectivity, which can be found

in [6] and other literature. We shall give the definitions and results for Banach

left modules over A, although in each case analogous results will hold for right

modules and bimodules.

1.2.1 Projectivity

A Banach left module over the Banach algebra A is a pair (X,T ) where X

is a Banach space and T : A ×X → X is a bounded bilinear operator such

that T (a, T (b, x)) = T (ab, x) for all a, b ∈ A and x ∈ X. In future we will

use the notation a · x in place of T (a, x).

If X and Y are Banach left modules over A, then a bounded linear map

γ : X → Y is a morphism of Banach left modules over A if γ(a ·x) = a · γ(x)for all a ∈ A and x ∈ X.

2


Definition 1.1. Projectivity

Let P be a Banach left module over A. Then P is projective if for every

• pair X and Y of Banach left modules over A;

• surjective morphism σ : X → Y of Banach left modules over A such

that there exists a bounded linear map α : Y → X with σ α = 1Y ;

• morphism ϕ : P → Y of Banach left modules over A;

Pψ

~~

ϕ

@@@

@@@@

Xσ

++Y

α

kk

there exists a morphism ψ : P → X of Banach left modules over A such that

σ ψ = ϕ.

The question may arise of why we consider only those maps σ which have

a right inverse α (in the category of Banach spaces), since this was not

stated in the general lifting problem on the previous page. One can see

that in this case, in the category of Banach spaces the lifting problem

has a trivial solution ψ = α ϕ. It turns out, that by considering only

lifting problems which have a positive solution in this category, we obtain

a more useful and natural notion of projectivity. If we do not do so,

then the corresponding notion is strict projectivity, and research has shown

that the class of strict projective modules is restrictively small and less useful.

We will often consider a Banach algebra A as a module of itself, by defining

the module operation A × A → A to be the multiplication of the algebra,

i.e. a · b = ab for a, b ∈ A. In this case, we will write that A is left projective

to mean that A is projective as a Banach left module over A, and similarly

for right projective. A Banach algebra is biprojective if it is projective as a

Banach bimodule over itself. If A is commutative then the notions of left and

right projectivity coincide, and so we will write simply that A is projective.

Notation. Category theory

To ease notation we shall use A-mod to denote the category of Banach left

modules over A. Similarly mod-A, A-mod-A, and A-mod-B are, respectively,

3


the categories of Banach right modules over A, Banach bimodules over A,

and Banach bimodules over A on the left and B on the right.

1.2.2 Projective tensor products

If X and Y are Banach spaces we may form their projective tensor product

X ⊗Y as follows. Firstly, let X Y denote the space of finite formal linear

combinations of E ×F over C, and let x y denote the element (x, y). Then

X Y =

n∑i=1

λi(xi yi) : λi ∈ C, xi ∈ X, yi ∈ Y, n ∈ N

.

We define a subspace M to be the linear span of elements of the following

forms:(x1 + x2) y − x1 y − x2 y;x (y1 + y2)− x y1 − x y2;

(λx) y − λ(x y);x (λy)− λ(x y);

where x1, x2 ∈ X, y1, y2 ∈ Y and λ ∈ C. We now define the algebraic tensor

product X ⊗ Y as the quotient of X Y by M , and use the notation x ⊗ y

for x y +M . Thus each element u ∈ X ⊗ Y can be written

u =n∑i=1

xi ⊗ yi

where n ∈ N, xi ∈ X and yi ∈ Y . We now define a norm on X ⊗ Y .

Definition 1.2. Projective norm

The projective norm π(u) of an element u ∈ X ⊗ Y is defined by

π(u) = inf

n∑i=1

‖xi‖ ‖yi‖ : u =n∑i=1

xi ⊗ yi

.

We define the projective tensor product X ⊗Y as the completion of X ⊗ Y

with respect to the projective norm; this can be shown to be a Banach space.

The following useful theorem is proved on page 94 of [20].

4


Theorem 1.3. Let X and Y be Banach spaces. For u ∈ X ⊗Y there exists

a representation u =∑∞

i=1 λixi ⊗ yi such that (xi)i∈N and (yi)i∈N are null

sequences (i.e. they converge to zero) and∑∞

i=1 |λi| <∞.

If A is a Banach algebra it will on occasion be useful to adjoin an identity e

to A. Let

A+ = A⊕ Ce = a+ λe : a ∈ A, λ ∈ C.

With the following operations and norm, this is a unital Banach algebra:

Addition: (a+ λe) + (b+ µe) = (a+ b) + (λ+ µ)e

Scalar multiplication: µ(a+ λe) = µa+ µλe

Continuous multiplication: (a+ λe)(b+ µe) = ab+ λb+ µa+ λµe

Norm: ‖a+ λe‖ = ‖a‖+ |λ|, where a, b ∈ A and λ, µ ∈ C.

Definition 1.4. Unital module

Let A be a Banach algebra with a unit e and X be a module in A-mod. Then

X is unital if e · x = x for all x ∈ X.

If X is a module in A-mod then we can extend the module operation to A+

by defining e · x = x for each x ∈ X. Thus X can be regarded as a unital

module in A+-mod. Similarly a morphism σ in A-mod is also a morphism

in A+-mod since σ(e · x) = σ(x) = e · σ(x).

Suppose P is a module in A-mod. Then A+ ⊗P is also a module in A-mod

by defining a module operation A×A+ ⊗P → A+ ⊗P on elementary tensors

as follows

b · (a+ λe)⊗ p = (ba+ λb)⊗ p

where b ∈ A, a + λe ∈ A+ and p ∈ P , and then extending to the whole of

A+ ⊗P . It can be checked that this is a bounded bilinear operator. Modules

of this form are called free in A-mod. Similarly, modules of the form P ⊗A+

and A+ ⊗P ⊗A+ are free in mod-A and A-mod-A respectively.

Lemma 1.5. Free Banach modules are projective.

Proof. We prove the case of left modules: that A+ ⊗P is projective in A-

mod. Let X and Y be modules in A-mod; σ : X → Y be a surjective

5


morphism in A-mod such that there exists a bounded linear map α : Y → X

with σ α the identity on Y ; and ϕ : A+ ⊗P → Y be a morphism in A-

mod. As described above we will regard these modules as unital modules

in A+-mod, and similarly for the morphisms. We define a linear operator

ψ : A+ ⊗P → X by

ψ((a+ λe)⊗ p) = (a+ λe) · α ϕ(e⊗ p)

on elementary tensors and extending. It is easily shown that ψ is a morphism

in A-mod. We now show that σ ψ = ϕ:

σ ψ((a+ λe)⊗ p) = σ((a+ λe) · α ϕ(e⊗ p))

= (a+ λe) · σ α ϕ(e⊗ p)

= (a+ λe) · ϕ(e⊗ p) = ϕ((a+ λe)⊗ p).

Therefore A+ ⊗P is projective in A-mod.

Notation. Tensor map

If f : X → X ′ and g : Y → Y ′ are bounded linear operators then f ⊗g : X ⊗Y → X ′ ⊗Y ′ denotes the map defined on elementary tensors by

(f ⊗ g)(x ⊗ y) = f(x) ⊗ g(y). It can be shown that f ⊗ g is also a bounded

linear operator.

If H1 and H2 are Hilbert spaces, we may form their Hilbert tensor product,

which is also a Hilbert space, in the following manner. An inner product is

defined on their algebraic tensor product H1 ⊗H2 by

〈η1 ⊗ η2, ξ1 ⊗ ξ2〉 = 〈η1, ξ1〉〈η2, ξ2〉

on elementary tensors and extending linearly. The Hilbert tensor product

H1 ⊗Hil H2 is obtained by completing H1 ⊗ H2 with respect to the norm

arising from this inner product. The projective tensor product H1 ⊗H2 can

be embedded in H1 ⊗Hil H2 via a map denoted by ϕ which can be shown to

have operator norm 1. More details on Hilbert tensor products may be found

in Section 2.6 of [10].

6


1.2.3 Criterion for projectivity

Here we describe a necessary and sufficient condition for a module to be

projective. All modules and morphisms considered in this section are in the

category of Banach left modules over the Banach algebra A, unless otherwise

stated.

Notation 1.6. The canonical morphism π.

If P is a module in A-mod, then the natural map π : A+ ⊗P → P defined by

π((a+ λe)⊗ p

)= a · p+ λp

is a surjective morphism in A-mod.

It may be noted that there is a bounded linear operator α : P → A+ ⊗Pdefined by α(p) = e ⊗ p, such that α ρ is the identity on P . However α is

not generally a morphism in A-mod. A morphism is called a retraction if it

has a right inverse in the same category.

The criterion we shall prove is now stated. Two lemmas will be proved as

part of the proof.

Theorem 1.7. Let A be a Banach algebra and P a module in A-mod. Then P

is projective in A-mod if and only if the canonical morphism π is a retraction

in A-mod.

Lemma 1.7.1. Let P and X be modules in A-mod and suppose that P be

projective in A-mod. If σ : X → P is a surjective morphism in A-mod and

there is a bounded linear operator α : P → X such that σ α is the identity

on P , then σ is a retraction in A-mod.

Proof. The identity operator 1P : P → P is trivially a morphism in A-mod.

Pψ

~~

1P

@@@

@@@@

Xσ

++P

α

kk

7


Therefore, from the definition, there exists a morphism ψ : P → X in A-mod

such that σ ψ = 1P .

Lemma 1.7.2. Let P and S be modules in A-mod and suppose that P is

projective in A-mod. If there is a morphism β : P → S in A-mod which is a

retraction, then S is also projective in A-mod.

Proof. Let X and Y be modules in A-mod; σ : X → Y be a surjective

morphism in A-mod such that there exists a bounded linear map α : Y → X

with σ α the identity on Y ; and ρ : S → Y an arbitrary morphism in

A-mod. Let γ : S → P be a morphism in A-mod which is a right inverse to

β.

P

ψ

β))S

γii

ρ

???

????

Xσ

++Y

α

kk

Then ρβ : P → Y is a morphism in A-mod and so, by the projectivity of P ,

there exists a morphism ψ : P → X in A-mod such that σ ψ = ρ β. Thus

we have a morphism ψγ : S → X in A-mod such that σψγ = ρβγ = ρ.

Hence S is projective in A-mod.

Conclusion of the proof of Theorem 1.7.

(⇒) Suppose P is projective in A-mod. Then A+ ⊗P and P are modules

in A-mod, π : A+ ⊗P → P is a surjective morphism in A-mod and

α : P → A+ ⊗P defined by α(p) = e⊗ p is a bounded linear map such that

π α is the identity on P . Hence by Lemma 1.7.1 there exists a morphism

ρ : P → A+ ⊗P in A-mod such that π ρ = 1P .

(⇐) If π : A+ ⊗P → P is a retraction in A-mod, then since A+ ⊗P is

projective in A-mod, Lemma 1.7.2 implies that P is also projective in A-

mod.

For a subset S of a Banach space we shall write S for the topological closure

of S. If A is a Banach algebra and B is a subalgebra of A then we shall use

8


the notation AB for the linear span of the set of elements of the form ab

where a ∈ A and b ∈ B. Thus

AB =

n∑i=1

aibi : n ∈ N, ai ∈ A, bi ∈ B

.

In this fashion we shall write A2 for AA, etc. We call A idempotent if A2 = A.

Corollary 1.7.3. If A is left projective and A 6= 0 then A2 6= 0.

Proof. Let us suppose, for a contradiction, that A is projective, there exists

0 6= a ∈ A, and A2 = 0. Then there exists a morphism ρ : A → A+ ⊗Ain A-mod such that π ρ is the identity map on A. Let us write ρ(a) =∑∞

i=1(bi + βie)⊗ ci, where bi, ci ∈ A and βi ∈ C. Then

a = π ρ(a) = π

(∞∑i=1

(bi + βie)⊗ ci

)=

∞∑i=1

(bici + βici) =∞∑i=1

βici.

Thus ρ(a) =∑∞

i=1 bi ⊗ ci + e⊗ a. Hence

0 = ρ(a2) = a · ρ(a) =∞∑i=1

abi ⊗ ci + a⊗ a = a⊗ a

which implies that a = 0, a contradiction.

1.2.4 Criterion for biprojectivity

We now prove a relatively simple criterion for a Banach algebra to be bipro-

jective. This result can be found in [6], page 359. Let

π1 : A+ ⊗A→ A,

π2 : A ⊗A+ → A,

π : A ⊗A→ A,

denote the canonical morphisms with the respective domains.

We start by proving a few lemmas.

9


Proposition 1.8.1. A biprojective Banach algebra is idempotent.

Proof. Let A be a biprojective Banach algebra. The map π1 can easily be

shown to be a surjective morphism in A-mod-A, and there is a bounded linear

operator α : A→ A+ ⊗A defined by α(a) = e⊗ a which is a right inverse to

π1. Therefore, by an analogous version of Lemma 1.7.1, π1 is a retraction in

A-mod-A. Let ρ : A→ A+ ⊗A be a morphism in A-mod-A which is a right

inverse to π1. Suppose, for a contradiction, that A2 is a proper subspace of

A, and let a ∈ A\A2. By the Hahn-Banach theorem, there exists a bounded

linear functional f ∈ A∗ such that f |A2 = 0 but f(a) 6= 0. Let f+ be a linear

extension of f to A+. Let us write, using Theorem 1.3,

ρ(a) =∞∑i=1

λi(bi + βie)⊗ ci

such that λi ∈ C such that∑∞

i=1 |λi| < ∞ and βi ∈ C, bi, ci ∈ A such that

bi + βie→ 0 and ci → 0 as i→∞. Then∥∥∥∥∥∞∑i=1

λiβici

∥∥∥∥∥ ≤ maxi∈N

|βi|maxi∈N

‖ci‖∞∑i=1

|λi| <∞.

Hence d =∑∞

i=1 λiβici ∈ A. Similarly it can be shown that u =∑∞

i=1 λibi ⊗ci ∈ A ⊗A and we can write

ρ(a) =∞∑i=1

λibi ⊗ ci + e⊗∞∑i=1

λiβici = u+ e⊗ d.

Applying π1 to both sides yields a = π1(u) + d and hence d = a − π1(u).

Therefore ρ(a) = u+ e⊗ (a− π1(u)) = u+ e⊗ a− e⊗ π1(u). It follows that

ρ(a2) =

a · ρ(a) = a · u+ a⊗ a− a⊗ π1(u),

ρ(a) · a = u · a+ e⊗ a2 − e⊗ π1(u)a.

It may be noted that a ·u ∈ A2 ⊗A and a⊗π1(u), u ·a, e⊗a2, and e⊗π1(u)a

lie in A+ ⊗A2. Thus applying the bounded linear functional f+⊗ f to these

two equations gives

10


(f+ ⊗ f) ρ(a2) =

f+(a)f(a)

0

which is a contradiction.

Lemma 1.8.2. If X is a projective module in A-mod and Y is a projective

module in mod-A, then X ⊗Y is a projective module in A-mod-A.

Proof. Since X and Y are projective, by Theorem 1.7 there exist morphisms

ρX : X → A+ ⊗X in A-mod and ρY : Y → Y ⊗A+ in mod-A such that

πX ρX and πY ρY are the identity maps on X and Y respectively. We define

a map ρ : X ⊗Y → A+ ⊗X ⊗Y ⊗A+ by ρ = ρX ⊗ ρY and check that it is a

morphism in A-mod-A and a right inverse to π : A+ ⊗X ⊗Y ⊗A+ → X ⊗Y .

Firstly,

ρ(a · x⊗ y) = ρX(a · x)⊗ ρY (y) = a · ρX(x)⊗ ρY (y) = a · ρ(x⊗ y)

and similarly ρ(x⊗ y · a) = ρ(x⊗ y) · a. Also,

π ρ(x⊗ y) = πX ρX(x)⊗ πY ρY (y) = x⊗ y.

Hence π ρ is the identity map on X ⊗Y and by Theorem 1.7, X ⊗Y is seen

to be projective in A-mod-A.

Theorem 1.8. A Banach algebra A is biprojective if and only if π : A ⊗A→A is a retraction in A-mod-A.

Proof. (⇒) Suppose A is biprojective. Then, as in the proof of Propo-

sition 1.8.1 above, there exists a morphism ρ : A → A+ ⊗A in A-mod-A

which is a right inverse to π1. By the same proposition, for any a ∈ A there

exist sequences (bi)i∈N and (ci)i∈N in A such that a =∑∞

i=1 bici. Therefore

ρ(a) =∞∑i=1

bi · ρ(ci) ∈ A · (A+ ⊗A) ⊂ A ⊗A.

Thus the image of ρ lies in A ⊗A. Since π = π1|A b⊗A we see that ρ is also a

11


right inverse to π.

(⇐) Let ρ be a morphism in A-mod-A which is a right inverse to π. Since

π1|A b⊗A = π = π2|A b⊗A then ρ is also a right inverse to π1 and π2. That π1 is a

retraction in A-mod implies that A is projective in A-mod. Also since A+ is

isomorphic to the free module C ⊗A+ in mod-A we see from Lemma 1.5 that

A+ is projective in mod-A. Hence by Lemma 1.8.2 we have that A ⊗A+ is

projective in A-mod-A. Since π2 is a retraction in A-mod-A, it follows from

the corresponding version of Lemma 1.7.2 that A is biprojective.

12

Chapter 2

Banach algebras which are left

or right projective

In this chapter we let X denote an infinite dimensional Banach space with an

elastic basis and K(X) denote the Banach algebra of compact operators on

X. It will be proved that K(X) is right projective. Furthermore we will show

that if X is reflexive, then K(X) is left projective as well. We then prove

that K(X) is not biprojective. Finally we show that `p is not projective for

any p > 1.

2.1 Preliminary results

2.1.1 The operators Sn and Rn

In this section we summarise the results which appear in section 28 of [12].

Let X be a Banach space. A sequence (ek)k∈N in X is called a Schauder

basis for X if for every element x ∈ X there is a unique sequence of scalars

(x(k))k∈N such that x is the limit of∑n

k=1 x(k)ek as n tends to infinity, with

respect to the norm of X. In this section we shall not assume that the basis

is normalised1.

1However the basis can easily be normalised if necessary by using (ek/‖ek‖).

13

2.1. Preliminary results 14

For such a Banach space X and for n ∈ N, we define operators Sn : X → X

and Rn : X → X by their action on the element x =∑∞

k=1 x(k)ek as follows:

Sn(x) =n∑k=1

x(k)ek and Rn(x) =∞∑

k=n+1

x(k)ek.

These are proved in [12] to be bounded linear operators. Moreover there

exists ζ ∈ R such that ‖Sn‖ < ζ for all n ∈ N.

Notation. For a Banach space X we let B(X) denote the Banach algebra of

bounded linear operators on X with composition multiplication and operator

norm. If X has a Schauder basis then for i, j ∈ N we can define eij ∈ B(X)

by

eij(ek) =

ei if j = k,

0 otherwise,

on the basis vectors of X and extend to the whole of X.

Observe that∑n

k=1 ekk is the bounded linear operator Sn.

Lemma 2.1. There exists D ∈ R such that for any m,n ∈ N with 1 ≤ m ≤ n

we have ∥∥∥∥∥n∑

k=m

ekk

∥∥∥∥∥ ≤ D.

Proof. If m ≥ 2 then∥∥∥∥∥n∑

k=m

ekk

∥∥∥∥∥ =

∥∥∥∥∥n∑k=1

ekk −m−1∑k=1

ekk

∥∥∥∥∥ = ‖Sn − Sm−1‖ ≤ ‖Sn‖+ ‖Sm−1‖ ≤ 2ζ.

(Notice that if m = 1 we have ‖∑n

k=1 ekk‖ = ‖Sn‖ < ζ.) Thus the condition

is satisfied for all m as stated by setting D = 2ζ.

Remark. We find that ‖Sn‖ ≥ 1 for all n ∈ N since

‖Sn‖ ≥∥∥∥∥Sn( e1

‖e1‖

)∥∥∥∥ =1

‖e1‖

∥∥∥∥∥n∑k=1

ekk(e1)

∥∥∥∥∥ =‖e1‖‖e1‖

= 1.

14


Theorem 2.2. Criterion for compactness

Let X be a Banach space with a Schauder basis. A closed subset W of X

is compact if and only if it is bounded and Rn(x) converges to 0 as n → ∞uniformly for x ∈ W .

Proof.

(⇒) It is clear that a compact set is bounded. The second statement will

now be proved. Let ε > 0 be arbitrary and set ε0 = ε/(2 + ζ) where ζ is, as

defined earlier, such that ‖Sn‖ ≤ ζ for all n ∈ N. Construct a finite ε0−net

x1, . . . , xm for W . Then for any x ∈ W there exists xk ∈ W for some

k = 1, . . . ,m such that ‖x− xk‖ < ε0. Therefore, for each n ∈ N,

‖Rn(x)‖ = ‖x− Sn(x)‖ ≤ ‖x− xk‖+ ‖xk − Sn(x)‖

≤ ‖x− xk‖+ ‖Sn(xk)− Sn(x)‖+ ‖Rn(xk)‖

< (1 + ζ)ε0 + ‖Rn(xk)‖.

For each k = 1, . . . ,m there exists Nk such that ‖Rn(xk)‖ < ε0 for all

n > Nk. Let N = maxN1, . . . , Nm. Then for all n > N and all x ∈ W we

have ‖Rn(x)‖ < (1 + ζ)ε0 + ε0 = (2 + ζ)ε0 = ε.

(⇐) We show that W is totally bounded, i.e. for an arbitrary ε > 0 we shall

construct a finite ε-net for W . Now W is bounded by assumption, so let

M ∈ R be such that ‖x‖ < M for all x ∈ W . Also by assumption, there

exists n ∈ N such that ‖Rn(x)‖ < ε/2 for all x ∈ W . Consider the following

subset of X:

Sn(W ) = Sn(x) : x ∈ W.

It is finite dimensional for it is spanned by e1, . . . , en. It is also bounded,

since for x ∈ W ,

‖Sn(x)‖ ≤ ζ‖x‖ < ζM.

Therefore Sn(W ) is relatively compact and so has a finite ε/2-net

x1, . . . , xm. This is also an ε-net for W , for suppose x ∈ W . Then there

15


exists xk ∈ Sn(W ) from the net such that ‖Sn(x)− xk‖ < ε/2 and then

‖x− xk‖ ≤ ‖Sn(x)− xk‖+ ‖Rn(x)‖ <ε

2+ε

2= ε.

Thus W is closed and totally bounded, and so compact.

2.1.2 Elasticity

We follow the definition used in [14] by calling the bases defined below elastic

bases. In fact, the modern name for them is an unconditional basis. The

norm defined in this section is sometimes known as the bounded multiplier

unconditional norm. More details on these definitions and results are to be

found in [16] Section 4.2 or [24] Section II.D.

Definition 2.3. Elastic basis

Let X be a Banach space with a Schauder basis (ek)k∈N. This basis is elastic

if every x =∑∞

k=1 x(k)ek ∈ X is such that for any bounded complex sequence

γ = (γ(k))k∈N the series∑∞

k=1 γ(k)x(k)ek converges, and additionally we have

sup|γ(k)|≤1 ∀k

∥∥∥∥∥∞∑k=1

γ(k)x(k)ek

∥∥∥∥∥ <∞.

Notation. For convenience we shall use the notation T∞ to denote the set

of complex sequences γ = (γ(k))k∈N such that |γ(k)| = 1 for all k ∈ N.

Let Y be the space of all complex-valued sequences y = (y(k))k∈N such that∑∞k=1 y

(k)ek converges in X. It is clear that Y is a linear space. We introduce

a norm for Y and show it is complete with respect to this norm. For y ∈ Y ,

define

‖y‖Y = supγ∈T∞

∥∥∥∥∥∞∑k=1

γ(k)y(k)ek

∥∥∥∥∥X

.

That this function is well defined follows directly from the definition of an

elastic basis.

Claim. The conditions for a norm are satisfied.

16


Clearly y = 0 implies ‖y‖Y = 0. Now suppose ‖y‖Y = 0. This implies

that ‖∑∞

k=1 γ(k)y(k)ek‖X=0 for any γ ∈ T∞. Now for any n ∈ N we use

the following argument to show that y(n)=0. Then it will follow that y = 0.

Let γn = (−1, . . . ,−1, 1,−1, . . .) with the 1 in position n, and let γ0 =

(1, 1, 1, . . .). Both of these are elements of T∞. Thus

0 =

∥∥∥∥∥∞∑k=1

γ(k)n y(k)ek

∥∥∥∥∥X

=

∥∥∥∥∥2y(n)en −∞∑k=1

y(k)ek

∥∥∥∥∥X

and 0 =

∥∥∥∥∥∞∑k=1

γ(k)0 y(k)ek

∥∥∥∥∥X

=

∥∥∥∥∥∞∑k=1

y(k)ek

∥∥∥∥∥X

.

Therefore ‖2y(n)en‖X ≤ ‖2y(n)en −∑∞

k=1 y(k)ek‖X + ‖

∑∞k=1 y

(k)ek‖X = 0

implying that y(n) = 0. It is easily checked that the required property of

scalar multiplication is satisfied and that the triangle inequality holds.

Claim. Y is complete with respect to this norm.

Proof. Let (yp)p∈N be a Cauchy sequence in Y . For every ε > 0 there exists

N ∈ N such that

‖yp − yq‖Y = supγ∈T∞

∥∥∥∥∥∞∑k=1

γ(k)(y(k)p − y(k)

q )ek

∥∥∥∥∥X

< ε

for all p, q > N . Thus ‖∑∞

k=1 γ(k)(y

(k)p − y

(k)q )ek‖X < ε for every γ ∈ T∞. In

particular it holds for γ = γn and γ = γ0 as defined above. Therefore, for

any n ∈ N we have∥∥∥∥∥2(y(n)p − y(n)

q )en −∞∑k=1

(y(k)p − y(k)

q )ek

∥∥∥∥∥X

=

∥∥∥∥∥∞∑k=1

γ(k)n (y(k)

p − y(k)q )ek

∥∥∥∥∥X

< ε

and

∥∥∥∥∥∞∑k=1

(y(k)p − y(k)

q )ek

∥∥∥∥∥X

=

∥∥∥∥∥∞∑k=1

γ(k)0 (y(k)

p − y(k)q )ek

∥∥∥∥∥X

< ε.(2.1)

Thus ‖2(y(n)p − y

(n)q )en‖X < 2ε and hence |y(n)

p − y(n)q | < ε/‖en‖. Therefore

(y(n)p )p∈N is a Cauchy sequence in C and hence converges to some y(n) ∈ C.

17


Let y = (y(n))n∈N.

We now show that y ∈ Y . Let xp =∑∞

k=1 y(k)p ek. By considering the in-

equality (2.1) above, (xp)∞p=1 is seen to be a Cauchy sequence in the Banach

space X, and so converges to some x ∈ X. It will now be shown that

x =∑∞

k=1 y(k)ek which implies y ∈ Y by definition. By inequality (2.1)

above, there exists N ′ such that ‖∑n

k=1(y(k)p − y

(k)q )ek‖X < ε for all n > N ′

and p > N . Taking the limit as q → ∞ yields ‖∑n

k=1(y(k)p − y(k))ek‖X ≤ ε

for all n > N ′, p > N . Thus∥∥∥∥∥∞∑k=1

y(k)p ek −

∞∑k=1

y(k)ek

∥∥∥∥∥X

= limn→∞

∥∥∥∥∥n∑k=1

(y(k)p − y(k))ek

∥∥∥∥∥X

≤ ε (2.2)

for all p > N . This implies x = limp→∞ xp = limp→∞∑∞

k=1 y(k)p ek =∑∞

k=1 y(k)ek as claimed. Clearly inequality (2.2) above implies that (yp)p∈N

converges to y.

There is a natural isomorphism T : Y → X given by T (y) =∑∞

k=1 y(k)ek.

The following shows that T is bounded with ‖T‖ ≤ 1:

‖T (y)‖X =

∥∥∥∥∥∞∑k=1

y(k)ek

∥∥∥∥∥X

≤ supγ∈T∞

∥∥∥∥∥∞∑k=1

γ(k)y(k)ek

∥∥∥∥∥X

= ‖y‖Y .

By the open mapping theorem, there exists a bounded linear operator T−1 :

X → Y such that T T−1 and T−1 T are the identity operators on X and

Y respectively. Therefore

supγ∈T∞

∥∥∥∥∥∞∑k=1

γ(k)x(k)ek

∥∥∥∥∥X

= ‖T−1(x)‖Y ≤ ‖T−1‖ ‖x‖X = C‖x‖X

where C = ‖T−1‖. This result easily generalises to the following:

Corollary 2.4. For any m,n ∈ N such that 1 ≤ m ≤ n,

supγ∈T∞

∥∥∥∥∥n∑

k=m

γ(k)x(k)ek

∥∥∥∥∥X

≤ C

∥∥∥∥∥n∑

k=m

x(k)ek

∥∥∥∥∥X

.

18


Proof. Let y(k) =

x(k) m ≤ k ≤ n,

0 k < m or k > n.Then clearly y =

∑∞k=1 y

(k)ek ∈

X since only finitely many terms are non-zero. Applying the main result of

this section yields

supγ∈T∞

∥∥∥∥∥n∑

k=m

γ(k)x(k)ek

∥∥∥∥∥X

= supγ∈T∞

∥∥∥∥∥∞∑k=1

γ(k)y(k)ek

∥∥∥∥∥X

≤ C

∥∥∥∥∥∞∑k=1

y(k)ek

∥∥∥∥∥X

= C

∥∥∥∥∥n∑

k=m

x(k)ek

∥∥∥∥∥X

as required.

2.1.3 The dual basis

For m ∈ N define a functional e∗m on X as follows. For an arbitrary element

x ∈ X let y = T−1(x) so that x =∑∞

k=1 y(k)ek. Then we define e∗m(x) = y(m).

It follows easily that e∗m is linear. Observe that

|e∗m(x)| = |y(m)| = ‖y(m)em‖‖em‖

=

∥∥∑mk=1 y

(k)ek −∑m−1

k=1 y(k)ek

∥∥‖em‖

≤2 supn∈N

∥∥∑nk=1 y

(k)ek∥∥

‖em‖=

2‖y‖‖em‖

=2‖T−1x‖‖em‖

≤ 2‖T−1‖‖em‖

‖x‖

which shows that e∗m is bounded with ‖e∗m‖ ≤ 2‖T−1‖/|em‖.

Notation. We use the notation δij =

1 i = j

0 i 6= j.

A sequence of functionals (fi)i∈N and vectors (xj)j∈N are said to be biorthogo-

nal if fi(xj) = δij for all i, j ∈ N. We observe that the functionals e∗m defined

above are biorthogonal to the basis vectors ek. Note that, by the definition

of the functionals e∗k,

x =∞∑k=1

y(k)ek =∞∑k=1

e∗k(x)ek.

19


Then, for any bounded linear functional f ∈ X∗ we have

f(x) =∞∑k=1

e∗k(x)f(ek) =∞∑k=1

cke∗k(x)

where ck = f(ek). Thus it can be written f =∑∞

k=1 cke∗k, and it will be seen

that when X is reflexive, this converges uniformly.

Definition 2.5. Reflexive

Let X be a Banach space and X∗∗ its second dual space with operator norm.

There is a natural bounded linear map J : X → X∗∗ defined by

J(x)(ϕ) = ϕ(x)

where x ∈ X and ϕ ∈ X∗, and it follows from the Hahn-Banach theorem

that J is isometric. The Banach space X is said to be reflexive if J is an

isomorphism.

Theorem 2.6. If a Banach space X is reflexive then (e∗m)m∈N is a Schauder

basis for X∗.

This will be proved in the following two lemmas. The first is stated without

proof and appears in [16], page 399.

Lemma 2.6.1. (e∗m)m∈N is a Schauder basis for Spe∗m : m ∈ N.

Lemma 2.6.2. If X is reflexive then X∗ = Spe∗m : m ∈ N

Proof. Let S denote the set Spe∗m : m ∈ N. Suppose S is a proper subspace

of X, for a contradiction. By the Hahn-Banach theorem, there exists a non-

zero bounded linear functional Λ ∈ X∗∗ which is zero on S. Let e∗∗k denote

the element of X∗∗ corresponding to ek in X. Since X is reflexive, (e∗∗k )k∈N

is a Schauder basis for X∗∗ so we may write Λ =∑∞

k=1 λke∗∗k . Observe that,

for any i, k ∈ N, e∗∗k (e∗i ) = e∗i (ek) = δik. Thus (e∗i )i∈N and (e∗∗k )k∈N are

biorthogonal. It follows that, for all i ∈ N,

Λ(e∗i ) =∞∑k=1

λke∗∗(e∗i ) =

∞∑k=1

λkδik = λi.

20


Since Λ is zero on S we know that Λ(e∗i ) = 0 for all i ∈ N. Therefore λi = 0

for all i and so Λ = 0. This is a contradiction.

2.1.4 Approximate identities

Let A be a Banach algebra and (J,) a directed set. A left approximate

identity is a net (uα)α∈J such that for all x ∈ A the net (uαx) converges

to x. Similarly, a right approximate identity is a net (uα)α∈J such that

for all x ∈ A the net (xuα) converges to x. A net which is both a left

and a right approximate identity is called a two-sided approximate identity.

An approximate identity which is a sequence is said to be sequential. A

left/right/two-sided approximate identity (uα)α∈J is said to be bounded if

there exists M ∈ R+ such that ‖uα‖ < M for all α ∈ J .

The question of whether K(X) has any of the various approximate identities

described above has been studied by various mathematicians. In particular,

in his paper of 1986 [5], P.G. Dixon proved necessary and sufficient conditions

on a Banach space X for K(X) to have a bounded left approximate identity.

We now prove a weaker version.

Example 2.7. If X is a Banach space with Schauder basis (en)n∈N then

K(X) has a bounded left approximate identity. If X is also reflexive then it

has a two-sided approximate identity.

In this section we shall denote by B1(X) the closed unit ball

x ∈ X : ‖x‖ ≤ 1 of the Banach space X.

For n ∈ N and x =∑∞

k=1 x(k)ek ∈ X, recall the bounded linear operator

Sn : X → X defined by Sn(x) =∑n

k=1 x(k)ek. Since Sn has finite rank n, it

is compact. Let a ∈ K(X).

Step 1. (Sn)n∈N is a left approximate identity

a(B1(X)) is compact so by Theorem 2.2, for any ε > 0 there exists N ∈ N

21


such that ‖Rn(x)‖ < ε for all n > N and x ∈ a(B1(X)). Note that

a(y) : y ∈ X, ‖y‖ ≤ 1 ⊂ a(B1(X)).

Thus, for all n > N ,

‖a− Sna‖ = supy∈X‖y‖≤1

‖Rna(y)‖ ≤ supx∈a(B1(X))

‖Rn(x)‖ < ε

as required.

Step 2. If X is reflexive then (Sn)n∈N is a right approximate identity

It is known that the norm of a bounded linear operator is equal to the norm

of its adjoint. Thus,

‖a− aSn‖ = ‖aRn‖ = ‖(aRn)∗‖ = ‖R∗na∗‖. (2.3)

By Section 2.1.3, since X is reflexive, the dual space X∗ has a Schauder basis

(e∗n)n∈N. For f =∑∞

k=1 cke∗k ∈ X∗ and x =

∑∞k=1 e

∗k(x)ek ∈ X we observe the

action of R∗n(f) on x:

R∗n(f)(x) = f(Rn(x)) = f

(∞∑

k=n+1

e∗k(x)ek

)

=∞∑

k=n+1

e∗k(x)f(ek) =∞∑

k=n+1

cke∗k(x).

Therefore R∗n(f) =∑∞

k=n+1 cke∗k. We see that R∗n acts on X∗ in an identical

way to how Rn acts on X, with their respective bases. By Schauder’s the-

orem, (a proof of which can be found in [25], page 282), a bounded linear

operator is compact if and only if its adjoint is compact. Therefore a∗ is

compact. By applying Theorem 2.2, for any ε > 0 there exists N ∈ N such

that ‖R∗n(f)‖ < ε for all n > N and f ∈ a∗(B1(X∗)). Then for all n > N ,

‖R∗na∗‖ = supg∈X∗

‖g‖≤1

‖R∗na∗(g)‖ ≤ supf∈a∗(B1(X∗))

‖R∗n(f)‖ < ε.

22

2.2. On the projectivity of the compact operators 23

By (2.3) above, ‖a− aSn‖ < ε. Since, for all n ∈ N, we have ‖Sn‖ ≤ ζ where

ζ is the constant described in Section 2.1.1, we see that (Sn)n∈N is a bounded

two-sided approximate identity for K(X).

2.2 On the projectivity of the compact oper-

ators

This section follows the work of Lykova [14].

2.2.1 Right projectivity of K(X)

We first show that if X is a Banach space with an elastic basis (ek)k∈N then

K(X) is right projective. Let us prove a simple lemma:

Lemma 2.8. If ε is a primary nth root of unity and j ∈ Z then

n∑k=1

(εj)k =

n if j is a multiple of n,

0 otherwise.

Proof. The sum is a geometric progression. If j is a multiple of n then∑nk=1(ε

j)k =∑n

k=1 1 = n. Otherwise,

n∑k=1

(εj)k = εj(εj)n − 1

εj − 1= 0

since (εj)n = (εn)j = 1j = 1, the numerator is zero, but εj 6= 1 so the

denominator is nonzero.

We recall that in Sections 2.1.1 and 2.1.2 two constants C and D were intro-

duced, satisfying ‖∑n

k=m ekk‖ ≤ D and ‖∑n

k=m γkxkek‖ ≤ C‖∑n

k=m xkek‖for any sequence (γk)k∈N such that |γk| = 1 for all k ∈ N, and x =∑∞

k=1 xkek ∈ X and 1 ≤ m ≤ n.

23


Proposition 2.9. For any m,n ∈ N such that 1 ≤ m ≤ n and a ∈ K(X),

the following inequality holds∥∥∥∥∥n∑

i=m

eii ⊗ eiia

∥∥∥∥∥ ≤ C2D

∥∥∥∥∥n∑

i=m

eiia

∥∥∥∥∥.

Proof. Let a ∈ B(X) and consider the element u ∈ K(X) ⊗K(X) defined

by

u =n∑k=1

1

n

(n∑

i=m

εk(i−m)eii

)⊗

(n∑

j=m

ε−k(j−m)ejja

)(2.4)

where ε = e2πi/n. Using the bilinearity of the tensor product we have

u =n∑k=1

1

n

n∑i=m

n∑j=m

εk(i−m)eii ⊗ ε−k(j−m)ejja =n∑k=1

1

n

n∑i=m

n∑j=m

εk(i−j)eii ⊗ ejja.

Rearranging gives

u =1

n

n∑i=m

n∑j=m

(n∑k=1

εk(i−j)

)eii ⊗ ejja.

By Lemma 2.8,∑n

k=1 εk(i−j) is zero unless i − j is a multiple of n (in which

case it is n). Since −n < i − j < n then i − j is a multiple of n if and only

if i− j = 0, i.e. i = j. Therefore

u =1

n

n∑i=m

n∑j=m

neii ⊗ ejja i = j

0 i 6= j

=

1

n

n∑i=m

neii ⊗ eiia =n∑

i=m

eii ⊗ eiia.

(2.5)

We will need the following two results.

Lemma 2.10. Let k ∈ Z and a ∈ B(X). Then∥∥∥∥∥n∑

i=m

ε−k(i−m)eiia

∥∥∥∥∥ ≤ C

∥∥∥∥∥n∑

i=m

eiia

∥∥∥∥∥.

Proof. Let x =∑

i∈N xiei ∈ X and let us write a =∑

i,j∈N aijeij. Then, by

24


applying Corollary 2.4,∥∥∥∥∥n∑

i=m

ε−k(i−m)eiiax

∥∥∥∥∥ =

∥∥∥∥∥n∑

i=m

ε−k(i−m)

(∞∑j=1

aijxj

)ei

∥∥∥∥∥≤ C

∥∥∥∥∥n∑

i=m

∞∑j=1

aijxjei

∥∥∥∥∥ = C

∥∥∥∥∥n∑

i=m

eiiax

∥∥∥∥∥ .It may also be noted, that by substituting a = I the identity operator, and

k for −k, into the lemma above, the result below follows easily.

Corollary 2.11. For k ∈ Z we have,∥∥∥∥∥n∑

i=m

εk(i−m)eii

∥∥∥∥∥ ≤ C

∥∥∥∥∥n∑

i=m

eii

∥∥∥∥∥.

By applying the property of the norm of the tensor product, i.e. that ‖a⊗b‖ ≤‖a‖ ‖b‖, on the element u described in (2.4) we have,

‖u‖ ≤n∑k=1

1

n

∥∥∥∥∥n∑

i=m

εk(i−m)eii

∥∥∥∥∥∥∥∥∥∥

n∑i=m

ε−k(i−m)eiia

∥∥∥∥∥.

We now use the above to estimate this norm

‖u‖ ≤n∑k=1

1

nC

∥∥∥∥∥n∑

i=m

eii

∥∥∥∥∥C∥∥∥∥∥

n∑i=m

eiia

∥∥∥∥∥= C2

∥∥∥∥∥n∑

i=m

eii

∥∥∥∥∥∥∥∥∥∥

n∑i=m

eiia

∥∥∥∥∥ ≤ C2D

∥∥∥∥∥n∑

i=m

eiia

∥∥∥∥∥since ‖

∑ni=m eii‖ ≤ D by Corollary 2.1. Therefore, by (2.5) we have,∥∥∥∥∥

n∑i=m

eii ⊗ eiia

∥∥∥∥∥ = ‖u‖ ≤ C2D

∥∥∥∥∥n∑

i=m

eiia

∥∥∥∥∥as claimed.

We will now use this inequality to show that, when a ∈ K(X), the sequence

25


(An)n∈N in K(X) ⊗K(X)+ where An =∑n

i=1 eii⊗eiia, is a Cauchy sequence.

Recall that∑n

i=1 eii is the operator Sn described in Section 2.1.1. Thus∥∥∥∥∥n∑

i=m

eiia

∥∥∥∥∥ =

∥∥∥∥∥n∑i=1

eiia−m−1∑i=1

eiia

∥∥∥∥∥ = ‖Sna− Sm−1a‖.

Let ε > 0. Since (Sn)n∈N is a left approximate identity for K(X) there exists

N ∈ N such that ‖Sna− a‖ < ε/2C2D for all n > N . Therefore

‖Sna−Sm−1a‖ = ‖Sna−a+a−Sm−1a‖ ≤ ‖Sna−a‖+‖Sm−1a−a‖ <ε

C2D

for all n,m > N + 1. Hence

‖An − Am‖ =

∥∥∥∥∥n∑

i=m

eii ⊗ eiia

∥∥∥∥∥ < C2D · ε

C2D= ε

for all n,m > N + 1 and so (An)n∈N is a Cauchy sequence in the complete

space K(X) ⊗K(X)+ and so converges.

Define ρ : K(X) → K(X) ⊗K(X)+ by ρ(a) =∑∞

k=1 ekk⊗ekka. We now show

that ρ is a morphism in mod-K(X). Firstly, it is easily shown that ρ is linear.

It is also bounded for ‖ρ(a)‖ = ‖∑∞

k=1 ekk ⊗ ekka‖ ≤ C2D ‖∑∞

k=1 ekka‖by Proposition 2.9. Then

∑∞k=1 ekka = limn→∞ Sna = a and so

‖ρ(a)‖ ≤ C2D‖a‖ implying ‖ρ‖ ≤ C2D. Finally, note that ρ(ab) = ρ(a) · b.Thus ρ is a morphism in mod-K(X).

It now remains to show that π ρ is the identity operator on K(X) where

π : K(X) ⊗K(X)+ → K(X) is the canonical morphism described in Def-

inition 1.6. Then K(X) is proved right projective by Theorem 1.7. Let

a ∈ K(X). Then

π ρ(a) = π

(∞∑k=1

ekk ⊗ ekka

)=

∞∑k=1

π(ekk ⊗ ekka) =∞∑k=1

ekkekka.

Now ekkekk = ekk and since∑n

k=1 ekk = Sn where Sn is the operator described

26


earlier, we have

π ρ(a) =∞∑k=1

ekka = limn→∞

Sna = a.

2.2.2 Left projectivity of K(X)

In this section we will show that if X is a reflexive Banach space with an

elastic basis (ek)k∈N then K(X) is left projective.

Proposition 2.12. For any m,n such that 1 ≤ m ≤ n and a ∈ K(X), the

following inequality holds∥∥∥∥∥n∑

j=m

aejj ⊗ ejj

∥∥∥∥∥ ≤ C2D

∥∥∥∥∥n∑

j=m

aejj

∥∥∥∥∥where C and D are the constants introduced in Sections 2.1.1 and 2.1.2.

This result is proved analogously to Proposition 2.9 in the previous section,

with the following lemma in place of Lemma 2.10

Lemma 2.13. Let k ∈ Z, ε be the primary nth root of unity, and a ∈ B(X).

Then ∥∥∥∥∥n∑

i=m

εk(i−m)aeii

∥∥∥∥∥ ≤ C

∥∥∥∥∥n∑

i=m

aeii

∥∥∥∥∥where C is the constant identified in Section 2.1.2.

Proof. Let x =∑∞

i=1 xiei ∈ X and write a =∑

i,j∈N aijeij. For each i ∈ N we

define yi = εk(i−m)xi. Since the basis is elastic we have y =∑∞

i=1 yiei ∈ X,

and by Section 2.1.2 we have

‖y‖ =

∥∥∥∥∥∞∑i=1

εk(i−m)xiei

∥∥∥∥∥ ≤ C‖x‖.

27


Thus ∥∥∥∥∥n∑

i=m

εk(i−m)aeiix

∥∥∥∥∥ =

∥∥∥∥∥n∑

i=m

εk(i−m)

∞∑j=1

ajixiej

∥∥∥∥∥=

∥∥∥∥∥n∑

i=m

∞∑j=1

ajiyiej

∥∥∥∥∥ =

∥∥∥∥∥n∑

i=m

aeiiy

∥∥∥∥∥≤

∥∥∥∥∥n∑

i=m

aeii

∥∥∥∥∥ ‖y‖ ≤ C

∥∥∥∥∥n∑

i=m

aeii

∥∥∥∥∥ ‖x‖.2.2.3 Lack of biprojectivity of K(X)

Let X be an infinite dimensional Banach space with a Schauder basis. In this

section we prove that K(X) is not biprojective. Suppose, for a contradiction,

that there exists a morphism ρ : K(X) → K(X) ⊗K(X) in K(X)-mod-

K(X) such that π ρ is the identity on K(X). Let us write

ρ(e11) =∞∑i=1

λiai ⊗ bi

where λi ∈ C such that∑∞

i=1 |λi| <∞ and ai, bi ∈ K(X) such that ai, bi → 0

as i→∞ (possible by Theorem 1.3). In accordance with [12], page 205, we

may write ai =∑∞

j,k=1 a(jk)i ejk for each i ∈ N and similarly for each bi, where

each of these series converge in the strong operator topology, i.e. the weakest

topology such that the map a 7→ ‖a(x)‖ is continuous for all x ∈ X, where

a ∈ K(X). Observe that for each n ∈ N,

ρ(enn) = ρ(en1e11e1n) = en1 · ρ(e11) · e1n

=∞∑i=1

λi

(en1

∞∑j,k=1

a(jk)i ejk

)⊗

(∞∑

j,k=1

b(jk)i ejke1n

)

=∞∑i=1

λi

(∞∑k=1

a(1k)i enk

)⊗

(∞∑j=1

b(j1)i ejn

). (2.6)

Recall from earlier in this chapter that Sm =∑m

k=1 ekk and there exists a

constant D ∈ R+ such that ‖Sm‖ ≤ D for all m ∈ N. The following lemma

28


is stated without proof.

Lemma 2.14. Let a =∑∞

j,k=1 a(jk)ejk and b =

∑∞j,k=1 b

(jk)ejk be elements of

K(X). Then for any j,m ∈ N,

m∑i=1

a(ji)b(ij)ejj = ejjaSmbejj.

Since (Sm)m∈N is a left approximate identity forK(X) we obtain the following

easy corollary:∞∑i=1

a(ji)b(ij)ejj = ejjabejj. (2.7)

Step 1. For each m ∈ N we have∑∞

i=1

∑mp=1 λia

(1p)i b

(p1)i = 0.

For each p ∈ N, we define a map Vp : K(X)×K(X) → C by

Vp(a, b) =∞∑j=1

a(jp)b(pj)

where a =∑∞

j,k=1 a(jk)ejk and b =

∑∞j,k=1 b

(jk)ejk. The map Vp is immediately

seen to be bilinear and we now show that it is also bounded. Using the lemma

above,

|Vp(a, b)| =

∣∣∣∣∣∞∑j=1

a(jp)b(pj)

∣∣∣∣∣ =1

‖epp‖

∥∥∥∥∥∞∑j=1

a(jp)b(pj)epp

∥∥∥∥∥ ≤ ‖epp‖ ‖a‖ ‖b‖.

By a theorem on page 22 of [19] for each p ∈ N there exists a bounded linear

functional Vp : K(X) ⊗K(X) → C such that Vp(a ⊗ b) = Vp(a, b) for all

a, b ∈ K(X). Now for any j, k, p, n ∈ N,

Vp(a(1k)enk ⊗ b(j1)ejn

)=

a(1p)b(p1) if j = k = p,

0 otherwise.

Applying Vp to (2.6) for any p, n ∈ N, yields

Vp(ρ(enn)) =∞∑i=1

λi

∞∑k=1

∞∑j=1

Vp

((a

(1k)i enk ⊗ b

(j1)i ejn

))=

∞∑i=1

λia(1p)i b

(p1)i .

29


Then we have, for all p, n ∈ N,

‖Vp‖ ‖ρ‖D ≥ ‖Vp(ρ(Sn))‖ =

∥∥∥∥∥n∑k=1

Vp(ρ(ekk))

∥∥∥∥∥ = n

∥∥∥∥∥∞∑i=1

λia(1p)i b

(p1)i

∥∥∥∥∥which implies

∑∞i=1 λia

(1p)i b

(p1)i = 0 for all p ∈ N.

Step 2. The series∑n

i=1

∑mp=1 λia

(1p)i b

(p1)i → 0 as n → ∞ uniformly for

m ∈ N.

Let ε > 0 be arbitrary. Since∑∞

i=1 |λi| ‖ai‖ ‖bi‖ <∞ there exists N ∈ N such

that∑∞

i=n+1 |λi| ‖ai‖ ‖bi‖ < ε/D‖e11‖ for all n > N . Now for all p, n ∈ N we

have ∣∣∣∣∣n∑i=1

λia(1p)i b

(p1)i

∣∣∣∣∣ ≤

∣∣∣∣∣∞∑i=1

λia(1p)i b

(p1)i

∣∣∣∣∣︸︷︷︸=0

+

∣∣∣∣∣∞∑

i=n+1

λia(1p)i b

(p1)i

∣∣∣∣∣ .Hence, by making use of Lemma 2.14, for all n > N and m ∈ N,∣∣∣∣∣

n∑i=1

λi

m∑p=1

a(1p)i b

(p1)i

∣∣∣∣∣ ≤

∣∣∣∣∣∞∑

i=n+1

λi

m∑p=1

a(1p)i b

(p1)i

∣∣∣∣∣ ≤∞∑

i=n+1

|λi|

∣∣∣∣∣m∑p=1

a(1p)i b

(p1)i

∣∣∣∣∣≤ D‖e11‖

∞∑i=n+1

|λi| ‖ai‖ ‖bi‖ < ε.

Step 3. We have∑∞

i=1

∑∞p=1 λia

(1p)i b

(p1)i = 1.

Since π ρ is the identity on K(X) we have

∞∑i=1

λiaibi = π(ρ(e11)) = e11.

Therefore, by Lemma 2.14,

∞∑i=1

λi

∞∑p=1

λia(1p)i b

(p1)i e11 = e11

∞∑i=1

λiaibie11 = e11

30

2.3. Lack of projectivity of `p when p > 1 31

which implies the claimed result.

Step 4. We obtain a contradiction.

It follows from Fubini’s Theorem that we may interchange the summation

signs to obtain

1 =∞∑i=1

∞∑p=1

λia(1p)i b

(p1)i =

∞∑p=1

∞∑i=1

λia(1p)i b

(p1)i = 0,

a contradiction, which completes the proof.

Remark. If X is a Banach space of finite dimension n then it is isomorphic

to the Euclidean space Cn. In this case K(X) is the algebra of square matrices

Mn(C) which will be shown to be biprojective in the next chapter.

2.3 Lack of projectivity of `p when p > 1

In this section it will be shown that for 1 < p < ∞, the Banach algebra

`p with componentwise operations and power p norm, is not projective. A

stronger version of this result can be found in Lykova’s paper [14].

2.3.1 `p is not projective for 1 < p ≤ 2

We will prove that `p is not left projective. This is sufficient because `p is

commutative. Note that if p ≤ 2 then q = 11− 1

p

≥ 2. In particular we have

q ≥ p.

Lemma 2.15. x = (1/n)n∈N ∈ `p for all p > 1.

Proof. Note that ‖x‖pp =∑∞

n=11np . We compare this series with the area

under the curve y = xp between x = 1 and infinity:∫ ∞

1

dx

xp=

[x−p+1

1− p

]∞1

=1

p− 1.

Thus ‖x‖pp < 1 + 1p−1

= pp−1

<∞.

31


Remark. It is well known that x /∈ `1.

Lemma 2.16. If x = (xk)k∈N ∈ `p then x ∈ `q for all q ≥ p.

Proof. Since x is a null sequence there exists N ∈ N such that |xk| < 1 for

all k > N . Thus, for any q ≥ p,

∞∑k=N+1

|xk|q ≤∞∑

k=N+1

|xk|p ≤∞∑k=1

|xk|p = ‖x‖pp.

Therefore∞∑k=1

|xk|q ≤N∑k=1

|xk|q + ‖x‖pp <∞

which implies x ∈ `q.

Let 1 < p ≤ 2 and suppose, for a contradiction, that there exists a morphism

ρ : `p → `p+ ⊗ `p in `p-mod such that π ρ(x) = x for all x ∈ `p where

π : `p+ ⊗ `p → `p is the canonical morphism described earlier. By Lemma 2.15,

x = (1/k)k∈N is an element of `p. We consider the action of ρ upon it.

Let ρ(x) =∑∞

i=1 λiyi ⊗ zi be a representation satisfying the properties of

Theorem 1.3. Then

x = π ρ(x) = π

(∞∑i=1

λiyi ⊗ zi

)=

∞∑i=1

λiyizi.

Let us use the notation yi =(yki)k∈N for each i ∈ N and similarly for each

zi. Making use of Lemma 2.16 we see that, for each i ∈ N, we have zi ∈ `q

where 1p

+ 1q

= 1. Thus, by Holder’s inequality, for each i ∈ N and for any

n ∈ N,

n∑k=1

|λiyki zki | ≤∞∑k=1

|λiyki zki | ≤ |λi|

(∞∑k=1

|yki |p) 1

p(

∞∑k=1

|zki |q) 1

q

= |λi| ‖yi‖p‖zi‖q

Note that since (zi)i∈N is a null sequence in `p and ‖zi‖q ≤ ‖zi‖p for each i

32


then it is also a null sequence in `q. Therefore, for any n ∈ N,

n∑k=1

|xk| =n∑k=1

∣∣∣∣∣∞∑i=1

λiyki z

ki

∣∣∣∣∣ ≤n∑k=1

∞∑i=1

∣∣λiyki zki ∣∣=

∞∑i=1

n∑k=1

∣∣λiyki zki ∣∣ ≤ ∞∑i=1

|λi| ‖yi‖p‖zi‖q

which, by Theorem 1.3, is finite and does not depend on n. Thus we have

the contradiction x ∈ `1. Therefore no such ρ can exist.

2.3.2 `p is not projective for p > 2

We will again prove that `p is not left projective.

Lemma 2.17. If p > q and ε ∈ R is such that 0 < ε < p/q − 1 then the

sequence

xε =(k−

1+εp

)k∈N

lies in `p but not in `q.

Proof. It follows from the Integral Test that xε ∈ `p for all ε > 0. For

ε < p/q − 1 we have (1 + ε)q/p < 1. Therefore for any n ∈ N,

n∑k=1

|xkε |q =n∑k=1

(k−

1+εp

)q=

n∑k=1

k−(1+ε) qp >

n∑k=1

k−1

which tends to infinity as n→∞. Thus xε /∈ `q.

As a corollary to this lemma, we note that for any ε ∈ (0, 1) we have xε ∈ `p

but xε /∈ `p2 .

Lemma 2.18. If y = (yk)k∈N ∈ `p and z = (zk)k∈N ∈ `q then yz ∈ `r where

r = pqp+q

and ‖yz‖r ≤ ‖y‖p‖z‖q.

Observe that when 1p

+ 1q

= 1 this is simply Holder’s inequality.

33


Proof. Let p′ = pr

and q′ = qr. It can be checked that 1

p′+ 1

q′= 1.

Define y′ = (yrk)k∈N and z′ = (zrk)k∈N. The following shows that y′ ∈ `p′ .

∞∑k=1

|y′k|p′=

∞∑k=1

|yrk|pr =

∞∑k=1

|yk|p <∞

and similarly z′ ∈ `q′ . Thus we may employ Holder’s inequality on y′ and z′,

giving that y′z′ ∈ `1 with ‖y′z′‖1 ≤ ‖y′‖p′‖z′‖q′ . Raising both sides to the

power 1/r yields

(∞∑k=1

|y′kz′k|

) 1r

≤

(∞∑k=1

|y′k|p′

) 1p(

∞∑k=1

|z′k|q′

) 1q

and the result follows from resubstituting y and z.

We fix ε ∈ (0, 1) and let x be the xε defined in Lemma 2.17 above. As noted

above we see that x ∈ `p but x /∈ ` p2 . Suppose, for a contradiction, that there

exists a morphism ρ : `p → `p+ ⊗ `p in `p-mod such that π ρ is the identity.

On following the method in the previous section we obtain the equation

x =∞∑i=1

λiyizi

for some λi ∈ C and yi, zi ∈ `p. Observe that, by Lemma 2.18 above we

have yizi ∈ `p2 for each i. Since `

p2 is complete, this leads to the contradiction

x ∈ ` p2 . Therefore, no such ρ : `p → `p+ ⊗ `p exists and `p is not left projective.

34

Chapter 3

Banach algebras which are

biprojective

3.1 Examples of biprojective algebras

We start by reproducing two well-known examples of biprojective Banach

algebras:

• the algebra `1 of absolutely summable complex-valued sequences with

componentwise operations and absolute sum norm;

• the algebra Mn(C) of square n × n complex matrices with the usual

matrix operations and operator norm, i.e. for M ∈Mn(C),

‖M‖Mn(C) = supx∈Cn

‖Mx‖2

‖x‖2

where ‖ · ‖2 denotes the Euclidean norm of Cn.

35

3.1. Examples of biprojective algebras 36

3.1.1 Biprojectivity of `1

Let (en)n∈N be the standard basis of `1. We note that for any x =∑∞k=1 xkek ∈ `1 and any n ∈ N, the following holds:∥∥∥∥∥

n∑k=1

xkek ⊗ ek

∥∥∥∥∥ ≤n∑k=1

‖xkek‖ ‖ek‖ =n∑k=1

|xk|. (3.1)

Since the limit of the right term exists as n tends to infinity, so does the limit

of the term on the left. Thus we may define ρ : `1 → `1 ⊗ `1 by

ρ(x) =∞∑k=1

xkek ⊗ ek

where x =∑∞

k=1 xkek. It will now be shown that ρ is a morphism in

`1-mod-`1.

Firstly, it is clear that ρ is linear. Inequality 3.1 above implies that

‖ρ(x)‖ ≤ ‖x‖ for any x ∈ `1. Hence ρ is bounded with ‖ρ‖ ≤ 1. Note that

for x, y ∈ `1, we have x ·ρ(y) = x ·∑∞

k=1 ykek⊗ek =∑∞

k=1 xkykek⊗ek = ρ(xy)

since xek = xkek for each k. Similarly ρ(xy) = ρ(x) ·y. Thus ρ is a morphism

in `1-mod-`1.

It remains to show that π ρ is the identity on `1 where π : `1 ⊗ `1 → `1 is

the canonical morphism described in Definition 1.6. Let x =∑∞

k=1 xkek ∈ X.

Then

π ρ(x) = π

(∞∑k=1

xkek ⊗ ek

)=

∞∑k=1

π(xkek ⊗ ek) =∞∑k=1

xkek = x.

Hence `1 is biprojective by Theorem 1.8.

3.1.2 Biprojectivity of Mn(C)

We will show that Mn(C) is biprojective by constructing a map

ρ : Mn(C) → Mn(C) ⊗Mn(C) which is a morphism in Mn(C)-mod-

36

3.1. Examples of biprojective algebras 37

Mn(C), such that π ρ is the identity.

We recall that eij is the operator whose matrix has value one in row i column

j and zero elsewhere. Under matrix multiplication

eijekl =

eil if j = k,

0 otherwise.

Note that ‖eij‖ = 1 for any 1 ≤ i, j ≤ n, for if x = x1e1 + . . . + xnen ∈ Cn

has norm at most one, then this implies |x1|2 + . . . + |xn|2 ≤ 1 and hence

‖eij(x)‖ = ‖xjei‖ = |xj| ≤ 1.

Clearly eij : i, j = 1, . . . , n is a basis for Mn(C). We define ρ on this basis

as follows

ρ(eij) = ein ⊗ enj

and extend linearly to the whole of Mn(C). Thus for an arbitrary matrix

M =∑n

i,j=1Mijeij we have

ρ(M) =n∑i=1

n∑j=1

Mijρ(eij) =n∑i=1

n∑j=1

Mijein ⊗ enj.

The map ρ is certainly linear and since Mn(C) is finite dimensional it follows

that ρ is bounded. For the final properties it will suffice to consider the basis

elements: they will then extend to the whole of Mn(C) by linearity. Note

the following equalities:

eij · ρ(ekl) = eijekn ⊗ enl =

ein ⊗ enl if j = k

0 otherwise

ρ(eijekl) =

ρ(eil) = ein ⊗ enl if j = k

ρ(0) = 0 otherwise

and ρ(eij) · ekl = ein ⊗ enjekl =

ein ⊗ enl if j = k

0 otherwise.

37

3.2. Modules over biprojective algebras 38

Therefore eij ·ρ(ekl) = ρ(eijekl) = ρ(eij)·ekl. Hence ρ is a morphism inMn(C)-

mod-Mn(C). Finally we observe that π ρ(eij) = π(ein⊗ enj) = einenj = eij.

This concludes the proof that Mn(C) is biprojective.

3.2 Modules over biprojective algebras

It will be shown in this section that any unital Banach left module over a

biprojective unital Banach algebra is projective.

Notation. If X is a module in A-mod then there is a natural map

∆ : (A ⊗A)×X → A ⊗X

defined on elementary tensors by ∆(b⊗ c, x) = b⊗ (c ·x) and extended to the

whole of (A ⊗A)×X.

We will use π : A ⊗A → A and π′ : A ⊗X → X to denote the canonical

morphisms with the respective domains. The map ∆ can easily be shown to

have the following properties, for every a ∈ A, u ∈ A ⊗A, and x ∈ X:

1. ∆ is bounded and bilinear;

2. a ·∆(u, x) = ∆(a · u, x);3. ∆(u · a, x) = ∆(u, a · x);4. π′ ∆(u, x) = π(u) · x.

Theorem 3.1. Let A be a biprojective Banach algebra with unit e and let X

be a unital module in A-mod. Then X is projective in A-mod.

Proof. By Theorem 1.8 there exists a morphism ρ : A→ A ⊗A in A-mod-A

such that π ρ is the identity on A. We define a map ρ′ : X → A ⊗X by

ρ′(x) = ∆(ρ(e), x).

Since ∆ is bounded and bilinear it follows that ρ′ is bounded and linear.

Since ρ is a morphism in A-mod-A we have that a · ρ(e) = ρ(a) = ρ(e) · a for

all a ∈ A. Thus by properties 2 and 3,

a·ρ′(x) = a·∆(ρ(e), x) = ∆(a·ρ(e), x) = ∆(ρ(e)·a, x) = ∆(ρ(e), a·x) = ρ′(a·x).

38

3.3. Direct sums of biprojective Banach algebras 39

Finally we check that π′ ρ′ is the identity operator on X using property 4:

π′ ρ′(x) = π′ ∆(ρ(e), x) = (π ρ(e)) · x = e · x = x

for all x ∈ X. Therefore X is projective in A-mod.

3.3 Direct sums of biprojective Banach alge-

bras

If A1, . . . , An are Banach algebras then their direct sum

A =n⊕k=1

Ak

with componentwise operations and norm defined, for a = (a1, . . . , an) ∈ A

by:

‖a‖A =n∑k=1

‖ak‖Ak

is a Banach algebra. For 1 ≤ k ≤ n we denote by Φk the natural monomor-

phism Ak → A defined by Φk(ak) = (0, . . . , 0, ak, 0, . . . , 0) where ak ∈ Ak

and is in position k on the right side. It is obvious that ΦkΦj = Φj if k = j

otherwise it is the zero morphism.

Theorem 3.2. If A1, . . . , An are biprojective Banach algebras then their di-

rect sum A =⊕n

k=1Ak is also biprojective.

Since Φk is bounded and linear for each k = 1, . . . , n, by a result in [19], the

tensor map (Φk ⊗ Φk) : Ak ⊗Ak → A ⊗A is also bounded and linear.

Since for each k = 1, . . . , n the Banach algebra Ak is biprojective, there

exists a morphism ρk : Ak → Ak ⊗Ak in Ak-mod-Ak such that πk ρk is

the identity of Ak, where πk : Ak ⊗Ak → Ak is the canonical morphism as

described in Definition 1.6. Define a map ρ : A → A ⊗A on an arbitrary

39


element a = (a1, . . . , an) of A by

ρ(a) =n∑k=1

(Φk ⊗ Φk)(ρk(ak)).

Linearity of ρ follows from that of each Φk ⊗ Φk and ρk. We check that ρ is

bounded:

‖ρ(a)‖ ≤n∑k=1

‖Φk ⊗ Φk‖ ‖ρk‖ ‖ak‖ ≤ maxk=1,...,n

‖Φk ⊗ Φk‖ ‖ρk‖

‖a‖.

We now show that for any ρ(a) · b = ρ(ab) for any a, b ∈ A.

Claim. For any k = 1, . . . , n and a, b ∈ A we have

(Φk ⊗ Φk)(ρk(ak)) · b = (Φk ⊗ Φk)(ρk(akbk)).

Proof. Note that Φk(ak)b = Φk(akbk). It follows that (Φk ⊗Φk)(ρk(ak)) · b =

(Φk ⊗ Φk)(ρk(ak) · bk). The final step follows from the property of ρk.

Hence

ρ(a) · b =n∑k=1

(Φk ⊗ Φk)(ρk(ak)) · b =n∑k=1

(Φk ⊗ Φk)(ρk(akbk)) = ρ(ab).

It follows similarly that a · ρ(b) = ρ(ab) as well. Thus ρ is a morphism in

A-mod-A. It remains to be shown that π ρ is the identity operator on A.

Claim. For any k = 1, . . . , n we have π (Φk ⊗ Φk) = Φk πk.

Proof. The result will be proved for an arbitrary elementary tensor x⊗ y ∈Ak ⊗Ak:

π (Φk ⊗ Φk)(x⊗ y) = π(Φk(x)⊗ Φk(y)

)= Φk(x)Φk(y)

= Φk(xy) = Φk πk(x⊗ y).

It then holds in the whole of Ak ⊗Ak by linearity and continuity.

40


Observe that

π ρ(a) =n∑k=1

π (Φk ⊗Φk)(ρk(ak)) =n∑k=1

Φk πk ρk(ak) =n∑k=1

Φk(ak) = a.

This concludes the proof that⊕n

k=1Ak is biprojective.

41

Chapter 4

On the projectivity of C(Ω, A)

algebras

In this chapter we will let Ω denote a locally compact Hausdorff space, and

K(H) the Banach algebra of compact operators on a Hilbert space H with

composition multiplication. We consider the projectivity of the Banach al-

gebra C0(Ω, K(H)) of continuous functions from Ω to K(H) which vanish at

infinity, with pointwise operations and supremum norm. It will be proved

that C0(Ω, K(H)) is left projective if and only if Ω is paracompact. Finally

we show that for C0(Ω, A) to be left projective it is necessary for A to be left

projective.


Many of these preliminary results have been adapted from Munkres [17]. An

exception is Proposition 4.5 which appears in [7].

4.1.1 Paracompactness

We first define some terms. Let Ω be a topological space. Ω is said to be

locally compact if it can be covered by open sets, each of which is contained

in a compact subset of Ω. A subset F ⊂ Ω is relatively compact if its closure

F is compact in Ω.

42


Definition 4.1. Support

For a function φ : Ω → R we define the support of φ as the closure of the set

of elements which are not mapped to zero. Thus

Supp φ = x ∈ X : φ(x) 6= 0.

Let B and T be collections of subsets of Ω. We say that T is locally fi-

nite if each point in Ω has a neighbourhood which intersects only finitely

many members of T . T is countably locally finite if each point in Ω has a

neighbourhood which intersects only countably many members of T . B is a

refinement of T if each member of B is contained in a member of T .

Definition 4.2. Paracompact

Ω is paracompact if it is Hausdorff and for every open covering T there exists

a locally finite open1 refinement B that covers Ω.

It can be noted that any compact space is also paracompact. We recall that

a topological space Ω is regular if for any closed subset F ⊂ Ω and point

t /∈ F there exist disjoint neighbourhoods U of t and V of F . The following

result can be found in [17], page 254.

Lemma 4.3. Let Ω be a regular topological space. Then Ω is paracompact

if and only if every open covering of Ω has a countably locally finite open

refinement that covers Ω.

Definition 4.4. Partition of unity

If U = Uαα∈J is an open cover for a topological space Ω, for some index set

J , then a family of continuous functions φα : Ω → [0, 1]α∈J is a partition

of unity dominated by U if

1. (Supp φα) ⊂ Uα for each α ∈ J ;

2. The collection Supp φαα∈J is locally finite;

3.∑φα(t) = 1 for each t ∈ Ω.

1Locally compact Hausdorff spaces are automatically regular and so by Theorem 28,Chapter 5 of [11] it is sufficient for every open covering to have a locally finite refinement.

43


Proposition 4.5. Let Ω be a paracompact locally compact topological space.

Then there exists an open cover U of Ω such that each point in Ω has a

neighbourhood which intersects no more than three sets of U .

As part of the proof of this proposition, we use the following two lemmas,

the first of which is stated without proof.

Lemma 4.5.1. If B1, . . . , Bn are relatively compact subsets then their finite

union B =⋃nk=1Bk is also relatively compact.

Lemma 4.5.2. Let U be a locally finite open cover of Ω, and K be a compact

subset of Ω. Then K intersects only a finite number of sets from U .

Proof. Since U is locally finite, every t ∈ K has an open neighbourhood Vt

which intersects only a finite number of elements (say Ut,1, . . . , Ut,mt) from

U . Therefore Vt : t ∈ K is an open cover of K and so has a finite subcover

Vt1 , . . . , Vtn for some t1, . . . , tn ∈ K. Thus K intersects only the following

elements of U :

Utk,j : 1 ≤ k ≤ n, 1 ≤ j ≤ mtk.

Proof of Proposition 4.5. Take any covering of Ω and let B be a locally finite

open refinement of it. Since closed subsets of compact sets are compact, we

see that the sets in B are also relatively compact. Let t ∈ Ω. We define,

inductively, the subsets Sn(t) by setting S0(t) = t and for n ≥ 1,

Sn(t) =⋃U ∈ B : U ∩ Sn−1(t) 6= ∅.

It will now be shown that Sn(t) is relatively compact for all n ≥ 1. Clearly

the singleton set S0(t) is compact. Let us assume that Sn−1(t) is relatively

compact for some n ≥ 1. Then by the lemma above, there are only a finite

number of sets from B which intersect the compact set Sn−1(t) and hence

also Sn−1(t). Thus Sn(t) is a finite union of relatively compact sets, and by

the above lemma, relatively compact.

44


We now define, for each t ∈ Ω, the subset S(t) by

S(t) =∞⋃n=1

Sn(t).

We will now show that for any t, t′ ∈ Ω, either S(t) = S(t′) or S(t)∩S(t′) = ∅.For suppose S(t) ∩ S(t′) 6= ∅.

Claim 1. t ∈ S(t′)

In the following reasoning sets of the form Un and V will denote open sets

in our cover B. Our assumption S(t) ∩ S(t′) 6= ∅ implies that

Un ∩ S(t′) 6= ∅ for some Un ⊂ Sn(t) and n ∈ N

⇒ Un ∩ V 6= ∅ for some V ⊂ Sm(t′) and m ∈ N

⇒ Un ⊂ Sm+1(t′) ⊂ S(t′).

By the definition of Sn(t), there exists Un−1 ⊂ Sn−1(t) such that Un−1 ∩Un 6= ∅. Therefore Un−1 ∩ S(t′) 6= ∅ and by the same argument as above

Un−1 ⊂ S(t′). We may continue inductively to show that U1 ⊂ S(t′) for some

U1 ⊂ S1(t) and hence that t ∈ S(t′).

Claim 2. S(t) ⊂ S(t′)

From above we see that t ∈ Sm(t′) for some m ∈ N, or in other words

S0(t) ⊂ Sm(t′). We prove inductively that Sn(t) ⊂ Sm+n(t′). Assuming that

Sn−1(t) ⊂ Sm+n−1(t′) for some n ≥ 1, we have

U ⊂ Sn(t) ⇒ U ∩ Sn−1(t) 6= ∅ ⇒ U ∩ Sm+n−1(t′) 6= ∅ ⇒ U ⊂ Sm+n(t

′).

Therefore for all n ≥ 1, we have Sn(t) ⊂ Sm+n(t′) ⊂ S(t′) implying that

S(t) ⊂ S(t′). By a symmetrical argument the reverse inclusion holds. Thus

S(t) = S(t′). We have proved that the sets S(t) and S(t′) either coincide or

are disjoint; thus we have defined an equivalence relation ∼ on Ω by

t ∼ t′ ⇐⇒ S(t) = S(t′).

45


We define, using the axiom of choice, a subset T of Ω so that it includes

exactly one member of each equivalence class. Thus Ω =⋃t∈T S(t) and for

any two distinct points t, t′ ∈ T we have S(t)∩S(t′) = ∅. We now define, for

each t ∈ T and n ≥ 0 a subset Vt,n of Ω by

Vt,n =

S1(t) n = 0

Sn+1(t) \ Sn−1(t) n ≥ 1

and claim that U = Vt,n : t ∈ T, n ≥ 0 is our desired cover of Ω.

Claim 3. U is a cover for Ω.

Any s ∈ Ω must lie in S(t) for some t ∈ T . Let n be the lowest integer such

that s ∈ Sn(t). If n = 0 then we have s = t ∈ S1(t) = Vt,0. Otherwise n ≥ 1

and s ∈ Sn+1(t) \ Sn−1(t) = Vt,n.

Claim 4. The sets in U are open and relatively compact.

Note that each Sn(t) is open, and for n ≥ 1 we have Vt,n = Sn+1(t) ∩(Ω \

Sn−1(t))

an intersection of two open sets. Also, we know that each Sn(t) is

relatively compact, and it is easily verified that, for all t ∈ T and n ≥ 0, Vt,n

is a closed subset of the compact set Sn+1(t) and hence compact.

Claim 5. Every s ∈ Ω has a neighbourhood which intersects no more than

three sets of U .

Observe that Vt,n ∩ Vt′,m 6= ∅ only if t = t′ and |n−m| ≤ 1. Thus, if s ∈ Vt,nthen Vt,n is a neighbourhood of s which intersects only itself, Vt,n−1 and Vt,n+1.

This concludes the proof.

The next important theorem is stated without proof and can be found in

[17], page 259.

Theorem 4.6. Let Ω be a paracompact Hausdorff space and suppose U =

Uαα∈J is an open cover of Ω. There exists a partition of unity on Ω dom-

inated by U .

46


4.1.2 C0(Ω) algebras

For a locally compact Hausdorff space Ω, C0(Ω) denotes the Banach algebra

of continuous functions which vanish at infinity, i.e. for any f ∈ C0(Ω) and

ε > 0 there exists a compact subset K ⊂ Ω such that |f(t)| < ε for every

t /∈ K.

We assert that every compact Hausdorff space is normal (a proof of this may

be found on page 202 of [17]). The following useful result can also be found

in [17], page 207:

Lemma 4.7. The Urysohn Lemma

If X is a normal space, and S and T are disjoint closed subsets of X, then

there exists a continuous function f : X → [0, 1] such that f(S) = 0 and

f(T ) = 1.

We may use this lemma to show the existence of functions in C0(Ω). For

example, for any t ∈ Ω there exists f ∈ C0(Ω) such that ‖f‖ = 1 and

f(t) = 1.

Theorem 4.8. C0(Ω) has a bounded approximate identity.

Proof. Let (Kλ)λ∈Λ be the compact subsets of Ω ordered by inclusion, where

Λ is some index set. For each λ ∈ Λ we construct (by using the Urysohn

lemma) eλ ∈ C0(Ω) such that ‖eλ‖ = 1 and eλ(t) = 1 for all t ∈ Kλ. Clearly

(eλ)λ∈Λ is bounded and we claim it is also an approximate identity of C0(Ω).

Let f ∈ C0(Ω) and ε > 0 be arbitrary. Let K ⊂ Ω be a compact subset such

that |f(t)| < ε for all t /∈ K, and ν ∈ Λ such that K ⊂ Kλ for all λ ν.

Then for all λ ν and t ∈ K we have eλ(t) = 1 and so f(t)eλ(t)− f(t) = 0.

And for all λ ∈ Λ and t /∈ K we have |f(t)eλ(t) − f(t)| ≤ |f(t)| < ε. Hence

‖feλ − f‖ < ε for all λ ν.

47

4.2. On the projectivity of C0(Ω, K(H)) 48

4.2 On the projectivity of C0(Ω, K(H))

In a paper [7] of 1970, Helemskii proved that the Banach algebra C0(Ω)

was projective if and only if its spectrum Ω is paracompact. Here we use

his methods to prove a similar result for C0(Ω, K(H)): that for a separable

Hilbert spaceH the Banach algebra C0(Ω, K(H)) is left projective if and only

if Ω is paracompact. We observe that the result for C0(Ω) is a special case

of that for C0(Ω, K(H)) since K(C) is isomorphic to C and so C0(Ω, K(C))

is isomorphic to C0(Ω). In fact, the result may be strengthened further: it

can be shown to hold for certain algebras defined by continuous bundles of

Banach algebras. More information on this can be found in [13].

4.2.1 Proof of sufficiency

Theorem 4.9. Let Ω be a paracompact locally compact Hausdorff space and

H be a separable Hilbert space. Then C0(Ω, K(H)) is left projective.

Notation. For a set Λ we shall let N(Λ) denote the set of finite subsets of

Λ, ordered by inclusion. This can be seen to be a directed set with order where λ1 λ2 ⇔ λ1 ⊂ λ2 for λ1, λ2 ∈ N(Λ).

By Lemma 4.5, there exists an open cover U of Ω such that each point in Ω

has a neighbourhood which intersects at most three sets in U . Let us write

U = Uµ : µ ∈ Λ where Λ is an index set. By Theorem 4.6 there exists a

partition of unity hµµ∈Λ dominated by U . For µ ∈ Λ and k ∈ N we define

an element Hµ,k ∈ C0(Ω, K(H)) by

Hµ,k(t) =√hµ(t)ekk.

For F ∈ C0(Ω, K(H)), λ ∈ N(Λ) and n ∈ N we define

uF,λ,n ∈ C0(Ω, K(H)) ⊗C0(Ω, K(H)) by

uF,λ,n =∑µ∈λ

n∑k=1

FHµ,k ⊗Hµ,k.

48


Let us write λ = µ1, . . . , µm and suppose n1, n2 ∈ N such that n1 > n2. Let

η and ξ be primary mth and nth1 roots of unity respectively. Using an identical

method to that used in Chapter 2, it can be checked that uF,λ,n1 − uF,λ,n2 =

1

mn1

m∑j=1

n1∑k=1

(m∑i=1

n1∑l=n2+1

ηjiξklFHµi,l

)⊗

(m∑i=1

n1∑l=n2+1

η−jiξ−klHµi,l

).

Now for any t ∈ Ω, there are at most three values of i, say i1, i2, i3 for which

hµi(t) 6= 0. Therefore∥∥∥∥∥

m∑i=1

n1∑l=n2+1

ηjiξklF (t)Hµi,l(t)

∥∥∥∥∥ =

∥∥∥∥∥3∑p=1

ηjipF (t)√hµip

(t)

n1∑l=n2+1

ξklell

∥∥∥∥∥≤ 3 max

µ∈λ

∥∥∥∥∥√hµ(t)F (t)

n1∑l=n2+1

ξklell

∥∥∥∥∥ (4.1)

for all t ∈ Ω, where ‖ · ‖ denotes the operator norm of K(H). Note that the

result also holds when the F (t) is absent in each expression. We recall from

Chapter 2 the existence of real constants C and D such that for any n1 > n2,

a ∈ B(H) and sequence (γk)k∈N such that |γk| = 1 for all k, we have∥∥∥∥∥an1∑

k=n2+1

γkekk

∥∥∥∥∥ ≤ C

∥∥∥∥∥an1∑

k=n2+1

ekk

∥∥∥∥∥ and

∥∥∥∥∥n1∑

k=n2+1

ekk

∥∥∥∥∥ ≤ D.

For h ∈ C0(Ω) and F ∈ C0(Ω, K(H)) we shall use hF to denote the element

of C0(Ω, K(H)) defined in the natural way: hF (t) = h(t)F (t).

Lemma 4.10. For any λ ∈ N(Λ) and n ∈ N we have

‖uF,λ,n‖π ≤ 9C2D2 maxµ∈λ

∣∣∣∣∣∣∣∣∣√hµF∣∣∣∣∣∣∣∣∣

where ‖ ·‖π denotes the projective norm of C0(Ω, K(H)) ⊗C0(Ω, K(H)), and

|||·||| the supremum norm of C0(Ω, K(H)).

49


Proof. It follows from the inequality (4.1) above that∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣m∑i=1

n∑l=1

ηjiξklFHµi,l

∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣ ≤ 3 max

µ∈λ

∣∣∣∣∣∣∣∣∣√hµF∣∣∣∣∣∣∣∣∣ ∥∥∥∥∥

n∑l=1

ξklell

∥∥∥∥∥ ≤ 3CDmaxλ∈µ

∣∣∣∣∣∣∣∣∣√hµF∣∣∣∣∣∣∣∣∣ .

By applying the property of the tensor product norm,

‖uF,λ,n‖π ≤ 1

mn

m∑j=1

n∑k=1

∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣m∑i=1

n∑l=1

ηjiξklFHµi,l

∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣m∑i=1

n∑l=1

η−jiξ−klHµi,l

∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣

≤ 1

mn

m∑j=1

n∑k=1

3CDmaxµ∈λ

∣∣∣∣∣∣∣∣∣√hµF∣∣∣∣∣∣∣∣∣ 3CDmax

µ∈λ

∥∥∥√hµ

∥∥∥C0(Ω).

Since 0 ≤ hµ(t) ≤ 1 for all µ ∈ Λ and t ∈ Ω, the result follows.

Lemma 4.11. For any n1, n2 ∈ N with n1 > n2 we have

‖uF,λ,n1 − uF,λ,n2‖π ≤ 9C2D supt∈Ω

∥∥∥∥∥F (t)

n1∑l=n2+1

ell

∥∥∥∥∥.

Proof. It follows from the inequality (4.1) above that for any t ∈ Ω,∥∥∥∥∥m∑i=1

n1∑l=n2+1

ηjiξklF (t)Hµi,l(t)

∥∥∥∥∥ ≤ 3 maxµ∈λ

∣∣∣√hµ(t)∣∣∣ ∥∥∥∥∥F (t)

n1∑l=n2+1

ξklell

∥∥∥∥∥≤ 3C

∥∥∥∥∥F (t)

n1∑l=n2+1

ell

∥∥∥∥∥.

The result follows by applying the property of the tensor product norm in a

similar way to the previous lemma.

Lemma 4.12. For any ε > 0 and F ∈ C0(Ω, K(H)) there exists a compact

subset K ⊂ Ω and λ ∈ N(Λ) such that

• hµ(t) = 0 if t ∈ K and µ /∈ λ; and

• ‖F (t)‖ < ε for all t /∈ K.

Proof. Let K ⊂ Ω be a compact subset such that ‖F (t)‖ < ε for all t /∈ K.

Now each t ∈ Ω has a neighbourhood Vt which intersects at most three sets

50


in U . Vt : t ∈ K is an open cover of K and so has a finite subcover

Vt1 , . . . , Vtn for some t1, . . . , tn ∈ K. Thus we see that there are at most

3n sets in U which intersect K. Since (Supp hµ) ⊂ Uµ for each µ ∈ Λ, there

are at most 3n values of µ for which hµ(t) 6= 0 for some t ∈ K. Let λ be the

set of these values of µ. This satisfies the conditions of the lemma.

We obtain the following straightforward corollary to this lemma:

µ /∈ λ =⇒∣∣∣∣∣∣∣∣∣√hµF

∣∣∣∣∣∣∣∣∣ < ε.

If t ∈ K then by the first property we have ‖√hµ(t)F (t)‖ = 0. For t /∈ K

we have by the second property ‖√hµ(t)F (t)‖ < ‖F (t)‖ < ε.

Lemma 4.13. For every ε > 0 and F ∈ C0(Ω, K(H)) there exists N ∈ Nsuch that ∥∥∥∥∥F (t)

n1∑k=n2+1

ekk

∥∥∥∥∥ < ε

for all t ∈ Ω and n1 > n2 > N .

Proof. By the lemma above there exists a compact subset K ⊂ Ω and

λ ∈ N(Λ) such that hµ(t) = 0 if t ∈ K and µ /∈ λ; and ‖F (t)‖ < ε/D for all

t /∈ K.

Now (∑n

k=1 ekk)n∈N is an approximate identity for K(H) so for all t ∈ K

there exists Nt ∈ N such that ‖F (t)∑n1

k=n2+1 ekk‖ < ε/2 for all n1 > n2 > Nt.

Since F is continuous, each t ∈ K has a neighbourhood Ut such that s ∈ Utimplies ‖F (t)− F (s)‖ < ε/2D. Thus, for all n1 > n2 > Nt and s ∈ Ut,∥∥∥∥∥F (s)

n1∑k=n2+1

ekk

∥∥∥∥∥ ≤

∥∥∥∥∥(F (s)− F (t))

n1∑k=n2+1

ekk

∥∥∥∥∥+

∥∥∥∥∥F (t)

n1∑k=n2+1

ekk

∥∥∥∥∥≤ ‖F (s)− F (t)‖

∥∥∥∥∥n1∑

k=n2+1

ekk

∥∥∥∥∥+

∥∥∥∥∥F (t)

n1∑k=n2+1

ekk

∥∥∥∥∥≤ ε

2DD +

ε

2= ε.

51


The set Ut : t ∈ K is an open cover of K and so has a finite subcover

Ut1 , . . . , Utm for some t1, . . . , tm ∈ K. Setting

N = maxi=1,...,m

Nti

we have ‖F (t)∑n1

k=n2+1 ekk‖ < ε for all t ∈ K and n > N . For t /∈ K we have

‖F (t)‖ < ε/D and so∥∥∥∥∥F (t)

n1∑k=n2+1

ekk

∥∥∥∥∥ ≤ ‖F (t)‖

∥∥∥∥∥n1∑

k=n2+1

ekk

∥∥∥∥∥ < ε.

The result follows.

Note that N(Λ) × N is a directed set by defining (λ1, n1) (λ2, n2) if and

only if λ1 ⊆ λ2 and n1 ≤ n2.

Proposition 4.14. For any F ∈ C0(Ω, K(H), the net (uF,λ,n)(λ,n)∈N(Λ)×N

converges.

Proof. We show that the net is Cauchy. Let ε > 0 and F ∈ C0(Ω, K(H) be

arbitrary. By Lemma 4.12 there exists λ ∈ N(Λ) such that µ /∈ λ implies∣∣∣∣∣∣√hµF∣∣∣∣∣∣ < ε/27C2D2. By Lemma 4.13 there exists N ∈ N such that

supt∈Ω

∥∥∥∥∥F (t)

n1∑l=n2+1

ell

∥∥∥∥∥ < ε

27C2D

for all n1 > n2 > N . Note that if λ1 ⊃ λ then λ1 is the disjoint union of

λ and λ1 \ λ. Then by Lemmas 4.11 and 4.10, (λ1, n1) (λ2, n2) (λ, n)

implies

‖uF,λ1,n1 − uF,λ2,n2‖π = ‖uF,λ,n1 + uF,λ1\λ,n1 − uF,λ,n2 − uF,λ2\λ,n2‖π≤ ‖uF,λ,n1 − uF,λ,n2‖π + ‖uF,λ1\λ,n1‖π + ‖uF,λ2\λ,n2‖π

<9C2Dε

27C2D+

9C2D2ε

27C2D2+

9C2D2ε

27C2D2= ε.

Thus the net is Cauchy and hence converges in the complete space

C0(Ω, K(H)) ⊗C0(Ω, K(H)).

52


We can therefore define a function ρ by

ρ(F ) = lim(λ,n)→∞

uF,λ,n.

We now show that ρ has the desired properties. Firstly it is clear that ρ

is linear, for uF+G,λ,n = uF,λ,n + uG,λ,n for all F,G, λ and n. Secondly, one

obtains immediately from Lemma 4.10 that for any F ∈ C0(Ω, K(H)),

‖uF,λ,n‖π ≤ 9C2D2 |||F ||| .

Hence ‖ρ‖ ≤ 9C2D2 by continuity. It follows from the definition of u that

uFG,λ,n = F · uG,λ,n for all F,G, λ and n. Hence ρ(FG) = F · ρ(G) for all F

and G, and ρ is shown to be a morphism in C0(Ω, K(H))-mod. Since

π ρ(F ) = lim(λ,n)→∞

∑µ∈λ

n∑k=1

FH2µ,k,

to show that π ρ is the identity on C0(Ω, K(H)) it is sufficient to show that

the net (Eλ,n)(λ,n)∈N(Λ)×N is a right approximate identity for C0(Ω, K(H))

where

Eλ,n =∑µ∈λ

n∑k=1

H2µ,k.

Let ε > 0 and F ∈ C0(Ω, K(H)) be arbitrary. By Lemma 4.13 there exists

N ∈ N such that ‖∑n

k=1 F (t)ekk − F (t)‖ < ε for all n > N and t ∈ Ω. Let

K ⊂ Ω be the compact subset described in the proof. It is claimed that for

all (ν, n) (λ,N) and t ∈ Ω,

‖F (t)Eν,n(t)− F (t)‖ =

∥∥∥∥∥∑µ∈ν

n∑k=1

F (t)hµ(t)ekk − F (t)

∥∥∥∥∥ < ε. (4.2)

Claim 1. (4.2) is satisfied for t ∈ K.

53


It is known that for ν ⊃ λ we have∑

µ∈ν hµ(t) = 1. Therefore∥∥∥∥∥∑µ∈ν

n∑k=1


∥∥∥∥∥ =

∥∥∥∥∥n∑k=1

F (t)ekk − F (t)

∥∥∥∥∥ < ε

by the above.

Claim 2. (4.2) is satisfied for t /∈ K.

Since ‖F (t)‖ < ε/D we have∥∥∥∥∥∑µ∈ν

n∑k=1


∥∥∥∥∥ ≤

∣∣∣∣∣∑µ∈ν

hµ(t)

∣∣∣∣∣ ‖F (t)‖

∥∥∥∥∥∞∑

k=n+1

ekk

∥∥∥∥∥< 1 · ε

D·D = ε.

Hence ‖∑

µ∈ν∑n

k=1 F (t)hµ(t)ekk − F (t)‖ < ε for all t ∈ Ω and so

|||FEν,n − F ||| < ε for all (ν, n) (λ,N). This concludes the proof.

4.2.2 Proof of necessity

Theorem 4.15. Let Ω be a locally compact Hausdorff space and H be a sep-

arable Hilbert space. If C0(Ω, K(H)) is left projective then Ω is paracompact.

Let us suppose that C0(Ω, K(H)) is projective. Then there exists a morphism

ρ : C0(Ω, K(H)) → C0(Ω, K(H))+ ⊗C0(Ω, K(H))

in C0(Ω, K(H))-mod such that π ρ is the identity. In this section we prove

that Ω is paracompact.

Proposition 4.16. There exists a continuous function Ξ : Ω× Ω → C such

that

1. Ξ(s, s) = 1 for all s ∈ Ω; and

2. For any compact subset K ⊂ Ω, Ξ(s, t) → 0 as t → ∞ uniformly for

s ∈ K.

54


Proof. We define a function i : C0(Ω, K(H)) ⊗C0(Ω, K(H)) → C0(Ω ×Ω, K(H)) in a natural way by

i(F ⊗G)(s, t) = F (s)G(t)

on elementary tensors and extending. For f ∈ C0(Ω) we define Qf ∈C0(Ω, K(H)) by

Qf (t) = f(t)e11 (4.3)

where t ∈ Ω. Note that Qf = Q√fQ√

f where√f ∈ C0(Ω) is

the function defined by√f(t) =

√f(t). This ensures that ρ(Qf ) ∈

C0(Ω, K(H)) ⊗C0(Ω, K(H)), and we may define the function Ξ by

Ξ(s, t) = ‖i(ρ(Qfs))(s, t)‖

where fs ∈ C0(Ω) is such that fs(s) = 1.

Lemma 4.16.1. Ξ is well defined.

Claim 1. i maps into C0(Ω× Ω, K(H)).

Proof. Let F,G ∈ C0(Ω, K(H)) be arbitrary. We will prove the result on

the elementary tensor F ⊗ G: it will be shown that i(F ⊗ G) is continuous

at the arbitrary point (s0, t0) of Ω × Ω. Let 0 < ε < 1. Since F and G are

continuous there exist neighbourhoods U of s0 and V of t0 such that

‖F (s)− F (s0)‖ <ε

2(1 + ‖G(t0)‖)for all s ∈ U, (4.4)

and ‖G(t)−G(t0)‖ <ε

2(1 + ‖F (s0)‖)for all t ∈ V.

By (4.4) we have ‖F (s) − F (s0)‖ < 1 and hence ‖F (s)‖ < 1 + ‖F (s0)‖ for

55


all s ∈ U . Thus for any (s, t) ∈ U × V ,

‖F (s)G(t)− F (s0)G(t0)‖

≤ ‖F (s)G(t)− F (s)G(t0)‖+ ‖F (s)G(t0)− F (s0)G(t0)‖

≤ ‖F (s)‖ ‖G(t)−G(t0)‖+ ‖F (s)− F (s0)‖ ‖G(t0)‖

<‖F (s)‖ε

2(1 + ‖F (s0)‖)+

ε‖G(t0)‖2(1 + ‖G(t0)‖)

<ε

2+ε

2= ε.

Since this holds on elementary tensors it will hold on the whole space.

Claim 2. Ξ is independent of the choice of fs.

Proof. Suppose f and g are two elements of C0(Ω) such that f(s) = g(s) = 1.

Note that Qf and Qg commute. Furthermore, since∑n

k=1 ekke11 = e11 for all

n ∈ N it can be checked, by defining

Qf,n(t) = f(t)n∑k=1

ekk ∈ C0(Ω, K(H))

where n ∈ N and f ∈ C0(Ω), that we have QfQg = Qf,nQg for all n ∈ N.

Then, by the properties of the maps ρ and i,

i(ρ(QfQg))(s, t) = i(Qf,n · ρ(Qg))(s, t) = Qf,n(s)i(ρ(Qg))(s, t)

=n∑k=1

ekk i(ρ(Qg))(s, t).

Since this holds for all n ∈ N and (∑n

k=1 ekk)n∈N is an approximate identity

for K(H) we see that i(ρ(QfQg)(s, t)) = i(ρ(Qg))(s, t). Then by symmetry

we obtain

i(ρ(Qf ))(s, t) = i(ρ(QgQf ))(s, t) = i(ρ(QfQg))(s, t) = i(ρ(Qg))(s, t).

Thus Ξ is independent of the choice of fs.

Lemma 4.16.2. Ξ is continuous.

56


Claim. For each s ∈ Ω there exists a neighbourhood Us of s and Qfs ∈C0(Ω, K(H)) such that Ξ(r, t) = ‖i(ρ(Qfs))(r, t)‖ for all r ∈ Us and t ∈ Ω.

Proof. By a theorem in [17] page 202, every compact Hausdorff space is

normal (and hence regular), so there exists a neighbourhood Us of s such

that Us ⊂ Ω. Let fs ∈ C0(Ω) be such that q|Us= 1. Then for every

(r, t) ∈ Us × Ω we have that Ξ(r, t) = ‖i(ρ(Qs))(r, t)‖.

Since i(ρ(Qs)) is continuous by Claim 1, the above shows that Ξ is continuous

at any point (s, t) ∈ Ω× Ω.

Lemma 4.16.3. Ξ(s, s) = 1 for all s ∈ Ω.

Proof. Observe that, for u =∑∞

k=1 Fk ⊗Gk we have π(u) =∑∞

k=1 FkGk and

so, for any s ∈ Ω,

i(u)(s, s) =∞∑k=1

Fk(s)Gk(s) =

(∞∑k=1

FkGk

)(s) = π(u)(s).

Therefore,

Ξ(s, s) = ‖i(ρ(Qfs))‖ = ‖π(ρ(Qfs))(s)‖ = ‖Qfs(s)‖ = 1

for all s ∈ Ω.

Lemma 4.16.4. For any compact subset L ⊂ Ω, Ξ(s, t) → 0 as t → ∞uniformly for s ∈ L.

Let L ⊂ Ω be compact and ε > 0 be arbitrary.

Claim. For each s ∈ L, there exists a neighbourhood Us of s such that

Ξ(r, t) → 0 as t→∞ uniformly for r ∈ Us.

Let s ∈ L and let Us ⊂ Ω and Qfs ∈ C0(Ω, K(H)) be as described in

Claim 4.2.2. Let ρ(Qfs) =∑∞

k=1 λs,kFs,k ⊗ Gs,k be a representation which

satisfies the conditions of Theorem 1.3. Then

Ms :=∞∑k=1

|λk| |||Fs,k||| =∞∑k=1

|λs,k| |||Fs,k|||

57


is a positive real number. Since Gs,k → 0 as k → ∞ we can take N ∈ N so

that |||Gs,k||| < ε/M for all k > N . And for each k = 1, . . . , N there exists a

compact subset Ks,k ⊂ Ω such that ‖Gs,k(t)‖ < ε/M for all t /∈ Ks,k. Setting

Ks =⋃Nk=1Ks,k gives a compact subset of Ω satisfying ‖Gs,k(t)‖ < ε/M for

all t /∈ Ks and k ∈ N. Then, for all r ∈ Us and t /∈ Ks,

Ξ(r, t) =

∥∥∥∥∥∞∑k=1

λs,kFs,k(r)Gs,k(t)

∥∥∥∥∥≤

∞∑k=1

|λs,k| ‖Fs,k(r)‖ ‖Gs,k(t)‖ < M · εM

= ε.

Conclusion of the proof. Us : s ∈ L is an open cover of L and so has a

finite subcover Us1 , . . . , Usn for some s1, . . . , sn ∈ K. Then for all s ∈ L

and t /∈ K where

K =n⋃i=1

Ksi

we have Ξ(s, t) < ε.

This concludes the proof of the proposition.

We now present Helemskii’s proof that the existence of such a function implies

that Ω is paracompact. We define Ψ : Ω× Ω → [0, 1] by

Ψ(s, t) = minΞ(s, t), 1 ·minΞ(t, s), 1.

From the properties of Ξ, it can be checked that Ψ is continuous and satisfies

the following conditions:

1. For any compact subset L ⊂ Ω,

• Ψ(s, t) → 0 as s→∞ uniformly for t ∈ L;

• Ψ(s, t) → 0 as t→∞ uniformly for s ∈ L;

2. Ψ(s, s) = 1 for all s ∈ Ω;

3. sups∈Ω Ψ(s, t) = 1 for all t ∈ Ω.

58


Lemma 4.17. For any subset S ⊂ Ω, the function m : Ω → [0, 1] defined by

m(t) = sups∈S

Ψ(s, t)

is continuous.

Proof. Let S ⊂ Ω. We show that m is continuous at the arbitrary point

t0 ∈ Ω.

Step 1. m is lower semi-continuous.

For an arbitrary ε > 0,m(t0)−ε/2 is not an upper bound for Ψ(s, t0) : s ∈ Sso there exists s0 ∈ S such that Ψ(s0, t0) > m(t0)− ε/2. By the continuity of

Ψ there exists a neighbourhood V of t0 such that |Ψ(s0, t)−Ψ(s0, t0)| < ε/2

for all t ∈ V . Then for all t ∈ V we have

m(t) > Ψ(s0, t) > Ψ(s0, t0)−ε

2> m(t0)−

ε

2− ε

2= m(t0)− ε

as required.

Step 2. m is upper semi-continuous.

Let us suppose, for a contradiction, that there exists ε > 0 and a net (tλ)λ∈Λ

in Ω such that tλ → t0 but that for every λ ∈ Λ there exists µλ λ such

that m(tµλ) > m(t0) + ε.

Claim 1. There exists a net (t′λ)λ∈Λ lying in a compact subset of Ω such that

t′λ → t0 and m(t′λ) > m(t0) + ε for all λ ∈ Λ.

Let L be a compact neighbourhood of t0, and κ ∈ Λ such that tλ ∈ L for all

λ κ. Then the net (t′λ)λ∈Λ defined by t′λ = tν where ν µλ and ν κ

satisfies the claimed properties.

Claim 2. There exists a convergent net (s′λ)λ∈Λ in Ω such that Ψ(s′λ, t′λ) >

m(t0) + ε.

The above implies that there exists a net (sλ)λ∈Λ in S such that

Ψ(sλ, t′λ) > m(t0) + ε for all λ ∈ Λ. Since L is compact we have

59


Ψ(s, t) → 0 as s → ∞ uniformly for t ∈ L. Hence there exists a compact

subset K ⊂ Ω such that Ψ(s, t) < m(t0) + ε for all t ∈ L and s /∈ K. This

implies sλ ∈ K for all λ ∈ Λ, and so (sλ)λ∈Λ has an accumulation point

s0 ∈ K and a subnet (sλ)λ∈Λ′ for some Λ′ ⊂ Λ, such that sλ → s0. We

now define the net (s′λ)λ∈Λ. For each λ ∈ Λ there exists µλ ∈ Λ′ such that

mλ λ. We define s′λ = sµλ. Then we have Ψ(s′λ, t

′λ) > m(t0) + ε for all

λ ∈ Λ, and s′λ → s0 as λ→∞.

It follows that Ψ(s0, t0) > m(t0) + ε which is a contradiction since s0 is an

accumulation point of S and so Ψ(s0, t0) ≤ m(t0). This concludes the proof

that m is continuous on Ω.

Lemma 4.18. There exists a set fµ : µ ∈ Λ of continuous functions

Ω → [0, 1] such that

• fµ(t) → 0 as t→∞ for all µ ∈ Λ; and

•∑

µ∈Λ fµ(t) = 1 for all t ∈ Ω.

Proof. We arrange the points of Ω into a transfinite sequence (sµ)µ∈Λ. For

each µ ∈ Λ we define the function fµ : Ω → [0, 1] by

fµ(t) = supσ∈Λσµ

Ψ(sσ, t)− supσ∈Λσ≺µ

Ψ(sσ, t)

where t ∈ Ω. Since 0 ≤ supσ≺µ Ψ(sσ, t) ≤ supσµ Ψ(sσ, t) ≤ 1 we see that

0 ≤ fµ(t) ≤ 1 for all µ ∈ Λ and t ∈ Ω. By the previous lemma, each fµ is

the sum of two continuous functions and hence continuous.

We now check that fµ : µ ∈ Λ satisfy the desired properties. Firstly, since

fµ(t) ≤ Ψ(sµ, t) for every µ ∈ Λ and t ∈ Ω, and Ψ(sµ, t) → 0 as t → ∞, we

have limt→∞ fµ(t) = 0. Also for any t ∈ Ω,∑µ∈Λ

fµ(t) = sups∈Ω

Ψ(s, t) = 1

as required.

60


We are now ready to prove that Ω is paracompact, by constructing a count-

able locally finite refinement of an arbitrary open cover B of Ω. For µ ∈ Λ

and n ∈ N we define a subset Uµ,n of Ω by

Uµ,n =

s ∈ Ω : fµ(s) >

1

n

where fµ is as described in the previous lemma. For n ∈ N we define

Un = Uµ,n : µ ∈ Λ, Uµ,n 6= ∅.

Claim. Un is locally finite for any n ∈ N.

Proof. Let n ∈ N and t ∈ Ω. Since∑

µ∈Λ fµ(t) = 1 there exists a finite subset

λ ∈ N(Λ) such that

0 ≤ 1−∑µ∈λ

fµ(t) <1

2n.

Since it is the sum of a finite number of continuous functions, the function

g : Ω → [0, 1] where

g(s) =∑µ/∈λ

fµ(s) = 1−∑µ∈λ

fµ(s)

is continuous. Therefore there exists a neighbourhood U of t such that |g(s)−g(t)| < 1/2n for all s ∈ U . Then for all s ∈ U we have g(s) ≤ g(t) + |g(s)−g(t)| < 1/n and so for any µ /∈ λ we have fµ(s) ≤ g(s) < 1/n. Hence U does

not intersect Uµ,n unless µ is an element of the finite set λ.

Since fµ(t) → 0 as t → ∞, for each µ ∈ Λ and n ∈ N, Uµ,n is contained

in a compact subset of Ω and so is relatively compact. Since it is covered

by B it has a finite subcover Bµ,n,1, . . . , Bµ,n,mµ,n for some mµ,n ∈ N and

Bµ,n,k ∈ B. We define

Cµ,n,k = Uµ,n ∩Bµ,n,k

and for each n ∈ N we define Cn = Cµ,n,k : µ ∈ Λ, 1 ≤ k ≤ mµ,n, Cµ,n,k 6= ∅.It can be seen that Cn is a refinement of B and is locally finite because Un is.

Note that for every t ∈ Ω there exists µ and n such that t ∈ Uµ,n. (Let µ be

61

4.3. A necessary condition for C0(Ω, A) to be left projective 62

such that fµ(t) > 0 and let n > 1fµ(t)

. Then we have t ∈ Uµ,n.) Therefore

C =⋃n∈N

Cn

is a countably locally finite refinement of B. Thus, by Lemma 4.3, Ω is shown

to be paracompact.

4.3 A necessary condition for C0(Ω, A) to be

left projective

Theorem 4.19. Let A be a Banach algebra and Ω a locally compact topo-

logical space. If C0(Ω, A) is left projective then A is left projective.

Let C0(Ω, A)+ and A+ denote the adjoinment of identities E and e to

C0(Ω, A) and A respectively, as described in Section 1.2. Suppose there

exists a morphism ρ : C0(Ω, A) → C0(Ω, A)+ ⊗C0(Ω, A) in C0(Ω, A)-mod

such that π ρ = 1. We fix t0 ∈ Ω and define a map Υ+ : C0(Ω, A)+ → A+

by Υ+(F + λE) = F (t0) + λe and Υ its restriction to C0(Ω, A). Then Υ and

Υ+ are easily shown to be linear and bounded with norm one. Therefore the

tensor map

Υ+ ⊗Υ : C0(Ω, A)+ ⊗C0(Ω, A) → A+ ⊗A

is bounded and linear. Let f ∈ C0(Ω) be such that f(t0) = 1 and ‖f‖ = 1.

Then for each a ∈ A, we define an element Ff,a ∈ C0(Ω, A) by Ff,a(t) = f(t)a.

We now define a map ρ′ : A→ A+ ⊗A by

ρ′(a) = (Υ+ ⊗Υ)(ρ(Ff,a)).

It may be noted that Ff,a+b = Ff,a+Ff,b. Thus the linearity of ρ and Υ+⊗Υ

ensures that ρ′ is linear. Also, since ‖Ff,a‖ = ‖a‖ we have

‖ρ′(a)‖ = ‖(Υ+ ⊗Υ)(ρ(Ff,a))‖ ≤ ‖Υ+ ⊗Υ‖ ‖ρ‖‖Ff,a‖ ≤ ‖ρ‖ ‖a‖

62

4.3. A necessary condition for C0(Ω, A) to be left projective 63

which shows that ρ′ is bounded with ‖ρ′‖ ≤ ‖ρ‖.

Claim. ρ′ is a morphism in A-mod.

By Theorem 4.8 there exists a bounded approximate identity (eλ)λ∈Λ of C0(Ω)

and also ν ∈ Λ) such that eλ(t0) = 1 for all λ ν. Note that Feλf,ba =

Feλ,bFf,a. Let a, b ∈ A and let us write ρ(Ff,a) =∑∞

k=1(Gk + λkE) ⊗ Hk.

Then for all λ ν,

(Υ+ ⊗Υ) ρ(Feλf,ba) = (Υ+ ⊗Υ)

(Feλ,b ·

∞∑k=1

(Gk + λkE)⊗Hk

)

= Feλ,b(t0)∞∑k=1

(Gk(t0) + λke)⊗Hk(t0)

= b · (Υ+ ⊗Υ) ρ(Ff,a) = b · ρ′(a).

By the continuity of Υ+ ⊗ Υ and ρ, taking the limit λ → ∞ gives us the

required result: ρ′(ba) = b · ρ′(a).

Claim. π′ ρ′ is the identity on A.

Let a ∈ A be arbitrary. Then using the same notation as above we see that

π′ ρ′(a) = π′

(∞∑k=1

(Gk(t0) + λke)⊗Hk(t0)

)=

∞∑k=1

(Gk(t0) + λke)Hk(t0)

= (π ρ(Ff,a))(t0) = Ff,a(t0) = a.

Hence π′ ρ′ is the identity.

63

Chapter 5

Projectivity of commutative

C*-subalgebras of C*-algebras

It is proved in this chapter that commutative C*-subalgebras of biprojective

C*-algebras are also biprojective.


An involution on a Banach algebra A is a function A → A that is an anti-

isomorphism, i.e. for all a, b ∈ A and λ ∈ C, the following hold:

• (a+ b)∗ = a∗ + b∗,

• (λa)∗ = λa∗,

• a∗∗ = a,

• (ab)∗ = b∗a∗.

A C*-algebra is a Banach algebra A with an involution such that ‖a∗a‖ =

‖a‖2 for all a ∈ A.

5.1.1 Representations

If A and B are C*-algebras, then a linear map φ : A → B is a *-

homomorphism if it is multiplicative and φ(a∗) = φ(a)∗ for all a ∈ A. It

64


is a fact (see [3] page 14) that a *-homomorphism is contractive, i.e. has

norm at most 1.

Definition 5.1. Representation

Let A be a C*-algebra A and H a Hilbert space. A representation of A on H

is a non-zero *-homomorphism π : A→ B(H).

An element h ∈ H is a cyclic element for a representation π if

π(a)(h) : a ∈ A = H

and a representation is called cyclic if such an element exists. A representa-

tion is said to be irreducible if every non-zero element of H is cyclic. Two

representations π : A→ B(H) and π′ : A→ B(H ′) are said to be equivalent

if there exists an isomorphism U : H ′ → H satisfying,

• ‖U(h′)‖ = ‖h′‖ for all h′ ∈ H ′

• π′(a) = U−1 π(a) U for all a ∈ A.

We denote by A the set of equivalence classes of irreducible representations

of A. With a topology, to be defined later, this will be the spectrum of A.

The following result and its proof can be found in [4] page 37.

Proposition 5.2. Let A be a C*-algebra and H,H ′ be Hilbert spaces. Sup-

pose π, π′ are representations of A on H,H ′ (respectively), and η, η′ are cyclic

vectors. If

〈π(a)(η), η〉 = 〈π′(a)(η′), η′〉

for every a ∈ A (where 〈 , 〉 represents the inner product), then π and π′ are

equivalent.

5.1.2 Positive functionals and the GNS construction

The spectrum of an element a in a Banach algebra A with unit e is defined

by

Sp(a) = λ ∈ C : a− λe is invertible.

65


The spectrum of an element in a non-unital algebra A is defined to be its

spectrum in the unitisation A+ described in Chapter 1. An element a is said

to be positive if Sp(a) ⊂ R+. The set of positive elements of an algebra A

shall be denoted by A+.

Definition 5.3. Positive linear functionals

Let A be a C*-algebra. A linear functional f is positive if f(a∗a) ≥ 0 for

every a ∈ A.

We now state the famous theorem by Gelfand, Naimark, and Segal, the proof

of which is known as the GNS construction.

Theorem 5.4. Let A be a C*-algebra and f a positive linear functional on

A. There exists a representation π of A on a Hilbert space H and a cyclic

element h ∈ H such that ‖h‖2 = ‖f‖ and

f(a) = 〈π(a)(h), h〉

for all a ∈ A.

A full proof of this theorem may be found in [3], page 29.

5.1.3 The spectrum of a C*-algebra

Here we briefly describe how the spectrum A of a C*-algebra A is defined

both in the commutative and noncommutative cases.

The spectrum of a commutative Banach algebra is defined to be its charac-

ter space endowed with the weak *-topology. Commutative C*-algebras are

characterised by the famous theorem of Gelfand:

Theorem 5.5. Gelfand Theorem

Let A be a commutative C*-algebra. Then A is isometrically isomorphic to

C0(A).

It is a fact that the spectrum of a commutative Banach algebra is a

locally compact Hausdorff space. The characters of C0(Ω) are precisely the

66


evaluations vt for t ∈ Ω, where vt(f) = f(t) for f ∈ C0(Ω). The character

space of C0(Ω) can be identified with Ω.

We now turn to the noncommutative case. We first describe the primitive

ideal space and endow it with a certain topology (the Jacobson topology).

An ideal I of a C*-algebra A is called a primitive ideal if it is the kernel of

an irreducible representation. We denote the set of all primitive ideals of A

by Prim(A). Two operations are now defined on subsets of A and Prim(A):

h, the hull, and k, the kernel. For a subset S ⊂ A we define

h(S) = I ∈ Prim(A) : S ⊂ I

and for a subset F ⊂ Prim(A),

k(F ) =⋂I ∈ F.

We now define a closure operation on subsets of Prim(A) by

F =

∅ if F = ∅,h(k(F )) otherwise.

This can be shown to satisfy the Kuratowski closure axioms, which encap-

sulate the properties of set closure in topological spaces. By a theorem in

[22], page 98, the class of complements of subsets F such that F = F defines

a topology on Prim(A). This topology on Prim(A) is called the Jacobson

topology.

Recall that A denotes the set of equivalence classes of irreducible represen-

tations of A. There is a natural surjection Θ : A → Prim(A) by defining

Θ(π) = kerπ. We now define, using the topology on Prim(A) and the map

Θ, a topology on A:

T is open in Prim(A) ⇔ Θ−1(T ) is open in A.

67


Definition 5.6. Spectrum of A

The spectrum of a C*-algebra A is the set A endowed with the inverse image

of the Jacobson topology under the map Θ described above.

It transpires that when A is commutative, this definition of A coincides

with the earlier definition of the spectrum of A. The character space of A

may be identified with A via a map φ 7→ πφ where πφ : A → B(H) is the

representation defined by πφ(a) = φ(a)I, where a ∈ A.

5.1.4 Enveloping algebras of C*-algebras

Let A denote a C*-algebra. An enveloping C*-algebra of A is a pair (B, θ),

where B is a C*-algebra and θ : A→ B is a *-homomorphism such that for

any representation π : A → B(H) of A there exists a unique representation

π′ : B → B(H) such that π = π′ θ.

For a Hilbert space H, there are numerous topologies that one can consider

on B(H). One is the weak operator topology, which is defined as the weakest

topology such that the sets W (T, x, y) are open for every T ∈ B(H) and

x, y ∈ H, where

W (T, x, y) = S ∈ B(H) : |〈(T − S)(x), y〉| < 1.

Definition 5.7. Von Neumann algebra

A von Neumann algebra is a C*-subalgebra of B(H) (where H is some Hilbert

space), which is closed in the weak operator topology, and contains the identity

operator.

A proof of the following result can be found in Chapter 12 of [4].

Theorem 5.8. Let A be a C*-algebra. Then there exists an enveloping von

Neumann algebra (EA, σ) of A.

68

5.2. Commutative C*-subalgebras of biprojective C*-algebras 69

5.1.5 Perfect maps and paracompactness

The following results appear in the additional material of the Russian

version of [11].

We recall that a topological space X is said to be regular if for every point

x ∈ X and neighbourhood U of x, there exists a closed neighbourhood V of

x such that V ⊂ U .

Theorem 5.9. Let X be a paracompact space and Y be a regular space.

Suppose f : X → Y is a continuous, surjective, closed map. Then Y is

paracompact.

Definition 5.10. A map f : X → Y is said to be perfect if

• f is continuous,

• f is closed,

• f−1(y) is a compact subset of X for every y ∈ Y .

Proposition 5.11. Let X be a compact space and Y be a Hausdorff space.

Then any continuous map f : X → Y is perfect.

Proposition 5.12. Let X and Y be topological spaces and f : X → Y be

perfect. Suppose Y ′ ⊂ Y and X ′ = f−1(Y ′). Then the map f |X′ : X ′ → Y ′

is perfect.

Theorem 5.13. Let X and Y be topological spaces with Y paracompact. If

there exists a perfect map f : X → Y then X is paracompact.

5.2 Commutative C*-subalgebras of bipro-

jective C*-algebras

The following theorem follows directly from [8], page 195:

Theorem 5.14. A commutative C*-algebra is biprojective if and only if its

spectrum is discrete.

69


In this section we build on this theorem and use it to prove the result, a

proof of which can be found in [15].

Theorem 5.15. Let A be a biprojective C*-algebra and B be a commutative

C*-subalgebra. Then B is biprojective and its spectrum is discrete.

By Theorem 5.8 there exists an enveloping Von Neumann algebra EA of A,

and canonical morphism σ : A → EA. Let EB be a maximal commutative

C*-subalgebra of EA containing σ(B). From a result in [4], page 375, EB is a

von Neumann algebra. By a result in [23], page 139 there exists a morphism

α : EA → EB in EB-mod-EB such that

1. α(x) = x for all x ∈ EB,

2. α(E+A ) ⊂ E+

B ,

where, as stated earlier, the notation A+ is used to denote the set of positive

elements of A. We extend σ to σ+ : A+ → (EA)+ in the obvious way. Let

Θ : (EB)+ → B+ be the dual map of the function σ+|B+ , i.e. for t ∈ (EB)+

we define Θ(t) = t σ+. This is a well-defined map, for Θ(t) is clearly linear

and multiplicative, and to see that it is non-zero on B+, let b ∈ B and λ ∈ Cbe such that λ 6= −t σ(b). Then b+ λe ∈ B+ and

Θ(t)(b+ λe) = t σ+(b+ λe) = t σ(b) + λ 6= 0.

We note that Θ is continuous by a result in [18], page 116. Furthermore

B+ is Hausdorff and (EB)+ is compact. Therefore by Proposition 5.11, Θ is

perfect. Let us use the notation Θ for the restriction of Θ to EBσ(B).

Proposition 5.16. Θ is a perfect map.

Proof. We show that EBσ(B) = Θ−1(B). Then Θ is perfect by Proposi-

tion 5.12. The following diagram may clarify the situation:

Bσ //

Θ(t)''PPPPPPPPPPPPPPPP EBσ(B)

t

C

70


Firstly, let us show that EBσ(B) ⊂ Θ−1(B). If t ∈ EBσ(B) then it is non-

zero on EBσ(B) and hence also on σ(B). Therefore Θ(t) = t σ 6= 0 and so

Θ(t) ∈ B.

Now let us prove Θ−1(B) ⊂ EBσ(B). Suppose t ∈ Θ−1(B). This implies

that t σ = Θ(t) ∈ B; hence t is non-zero on the image of σ|B. Since

σ(B) ⊂ EBσ(B) we see that t ∈ EBσ(B).

Since A is biprojective, by Theorem 1.8 there exists a morphism ρ : A →A ⊗A in A-mod-A which is a right inverse to π. We define ρ : AB →EA ⊗EAσ(B) by

ρ = (σ ⊗ σ|AB) ρ|AB.

Note that the image of ρ|AB lies in A ⊗AB for if c ∈ AB then c =∑∞

k=1 akbk

for some ak ∈ A and bk ∈ B. For each k ∈ N we have ρ(akbk) = ρ(ak) · bk ∈A ⊗AB ⊂ A ⊗AB. Hence the limit ρ(c) =

∑∞k=1 ρ(akbk) lies in the closed

subset A ⊗AB. In a similar way we find that the image of σ|AB is contained

in EAσ(B).

Claim 3. σ ⊗ σ|AB is a morphism in A-mod-B.

Proof. Note that AB is module in mod-B. Let a′, a ∈ A, b ∈ B and x ∈ AB.

Then σ ⊗ σ|AB(a′ · a⊗ x) = σ(a′a)⊗ σ|AB(x) = a′ · σ(a)⊗ σ|AB(x) and

σ ⊗ σ|AB(a⊗ x · b) = σ(a)⊗ σ|AB(x · b) = a′ · σ(a)⊗ σ|AB(x).

Since ρ|AB is a morphism in A-mod-A, then it is also a morphism in A-mod-

B. Therefore the composition ρ = (σ ⊗ σ|AB) ρ|AB is also a morphism in

A-mod-B.

Claim 4. π ρ = σAB.

Proof. Let x ∈ AB and suppose ρ(x) =∑∞

k=1 ak ⊗ xk where ak ∈ A and

xk ∈ AB for each k ∈ N. Then (σ ⊗ σ|AB) ρ|AB(x) =∑∞

k=1 σ(ak)⊗ σ(xk).

Therefore,

π ρ(x) = π

(∞∑k=1

σ(ak)⊗ σ(xk)

)= σ

(∞∑k=1

akxk

)= σ π ρ(x) = σ(x)

71


as claimed.

We have proved that the following diagram is commutative:

ABρ|AB //

σ|AB

ρ

''OOOOOOOOOOOO A ⊗ABσ⊗σ|AB

EAσ(B) EA ⊗EAσ(B)π

oo

As EB is commutative, it is isomorphic to C0(EB) where EB is the maximal

ideal space of EB. Therefore, for each s ∈ EB we can define a linear functional

fs on EA by

fs(x) = α(x)(s)

where x ∈ A and α : EA → EB is the morphism introduced earlier.

Claim. fs is positive for any s ∈ EB.

Proof. Let x ∈ EA be positive. By property 1 of α we know that α(x) is

positive. Thus

α(x)(s) = s(α(x)) ∈ Sp(α(x)) ⊂ [0,∞)

where Sp denotes the spectrum.

By the GNS construction, for each s ∈ EB, there exists a representation

πls : EA → B(H ls) for some Hilbert space H l

s and a cyclic vector ηls ∈ H ls such

that

fs(x) =⟨πls(x)(η

ls), η

ls

⟩for all x ∈ EA. We define a subset K l

s ⊂ H ls by K l

s = πls(EB)(ηls).

Remark. By property 2 of α, for any x ∈ EB we have

⟨πls(x)(η

ll)⟩

= fs(x) = α(x)(s) = x(s).

Remark. ηls ∈ K ls

72


Proof. Since EB is a von Neumann algebra, it has an identity e. Then

πls(e)(ηls) = ηls implying ηls ∈ πls(EB)(ηls) ⊂ K l

s.

Consider the representation πls|Kls

: EB → B(K ls) defined by

πls|Kls(x) = πls(x)|Kl

s

where x ∈ EB.

Remark. ηls is a cyclic vector for πls|Kls

because πls|Kls(EB)(ηls) = πls(EB)(ηls)

which is dense in πls(EB)(ηls).

Let Ts : EB → B(C) be the representation Ts(x) = x(s)I, where I is the

identity of B(C), and note that

〈Ts(x)(1), 1〉 = x(s) =⟨πls|Kl

s(x)(ηls), η

ls

⟩for all x ∈ EB. Now Ts(EB) = EB(s)I = CI and hence 1 is seen to be a

cyclic vector. By Proposition 5.2, πls|Klsis equivalent to Ts via an isomorphism

U : C → K ls which maps 1 to ηls.

Claim. πls(x)(k) = x(s)k for all x ∈ EB and k ∈ K ls.

Proof. By the equivalence of the representations, for each x ∈ EB, the fol-

lowing diagram is commutative:

K ls

πls|Kl

s(x)

//

U−1

K ls

CTs(x)

// C

U

OO

Hence πls(x)(k) = πls|Kls(x)(k) = U Ts(x) U−1(k) = U x(s)I U−1(k) =

x(s)k as claimed.

We now define a linear operator Qls : EA → H l

s by

Qls(a) = πls(a)(η

ls)

73


where a ∈ EA.

Remark. Qls is bounded.

Proof. Since any *-homomorphism is contractive we have

∥∥Qls(a)

∥∥ =∥∥πls(a) (ηls)∥∥ ≤ ∥∥πls(a)∥∥ ∥∥ηls∥∥ ≤ ‖a‖

∥∥ηls∥∥which shows that ‖Ql

s‖ ≤ ‖ηls‖ = 1.

We now copy the above technique on the opposite algebra EopA , which is de-

fined as having the same elements as A but with the multiplication reversed,

i.e. ab in Aop is equal to ba in A. Let πrs : EopA → B(Hr

s ) be a representation

and ηrs ∈ EopA a cyclic vector such that

〈πrs(x)(ηrs), ηrs〉 = fs(x) = x(s)

for all x ∈ EB. Define Krs = πrs(EB)(ηrs) ⊂ Hr

s and the representation πrs |Krs

of EB on Krs by πrs |Kr

s(x) = πrs(x)|Kr

s, where x ∈ EB.

This is also equivalent to the representation Ts described above, via an

isomorphism C → Krs which maps 1 to ηrs . Let Qr

s : EA → Hrs be the

bounded linear operator defined by Qrs(a) = πrs(a)(η

rs), where a ∈ EA.

We note that, for each s ∈ EBσ(B) we have Θ(s) ∈ B by proposition 5.16.

Therefore there exists xs ∈ B such that σ(xs)(s) = 1. It may also be noted

that B ⊂ AB because A has a bounded approximate identity; hence xs ∈AB. We define a map Φ : EBσ(B)× B → R+ by

Φ(s, t) =∥∥ϕs,t ((Qr

s ⊗Qlt

)(ρ(xs))

)∥∥2

74


where s ∈ EBσ(B) and t ∈ EB,

xs ∈ B (as discussed above),

ρ : AI → EA ⊗EAσ(B) is the map introduced earlier, and

ϕs,t : Hrs ⊗H l

t → Hrs ⊗Hil H

lt is the embedding of the projective

tensor product into the Hilbert tensor product of the

Hilbert spaces Hrs and H l

t .

Lemma 5.17. Φ is independent of the choice of xs.

Proof. Let x, y ∈ B such that σ(x)(s) = σ(y)(s) = 1. Then,

Qrs(σ(x)) = πrs(σ(x)) (ηrs) = σ(x)(s)ηrs = ηrs . (5.1)

Similarly, Qrs(σ(y)) = ηrs as well. Note that, by the reverse multiplication of

EopA and by (5.1) we have for any a ∈ EA,

Qrs(σ(y)a) = πrs(σ(y)a) (ηrs) = πrs(a)π

rs(σ(y)) (ηrs) = πrs(a)Q

rs(σ(y))

= πrs(a) (ηrs) = Qrs(a).

Thus for any ρ(x) =∑∞

i=1 ai ⊗ bi where ai ∈ EA and bi ∈ EAσ(B) we have

(Qrs ⊗Ql

t

)ρ(yx) =

(Qrs ⊗Ql

t

)(y · ρ(x))

=(Qrs ⊗Ql

t

) ∞∑i=1

σ(y)ai ⊗ bi =∞∑i=1

Qrs(σ(y)ai)⊗Ql

t(bi)

=∞∑i=1

Qrs(ai)⊗Ql

t(bi) by the above reasoning

=(Qrs ⊗Ql

t

)( ∞∑i=1

ai ⊗ bi

)=(Qrs ⊗Ql

t

)ρ(x).

Since xy = yx we have, by symmetry,

(Qrs ⊗Ql

t

)ρ(y) =

(Qrs ⊗Ql

t

)ρ(xy) =

(Qrs ⊗Ql

t

)ρ(x)

as claimed.

Lemma 5.18. Φ is continuous on EBσ(B)× EB.

75


For u ∈ EA ⊗EA let Φu : EBσ(B) × B → R+ be the function defined by

Φu(s, t) = ‖ϕs,t((Qrs ⊗Ql

t)(u))‖2.

Step 1. For each fixed w in the algebraic tensor product EA⊗EA the function

Φw is continuous.

Proof. Observe that, for any a, b ∈ EA we have

fs(ab∗) =

⟨πrs(ab

∗)(ηls), ηls⟩

=⟨πrs(b

∗)πrs(a)(ηls), ηls⟩

=⟨πrs(a)

(ηls), πrs(b)

(ηls)⟩

and similarly ft(b∗a) =

⟨πlt(a)(η

lt), π

lt(b)(η

lt)⟩. Therefore, for any s, t ∈ EB

and w =∑n

i=1 ai ⊗ bi ∈ EA ⊗ EA we have

∥∥ϕs,t ((Qrs ⊗Ql

t

)(w))∥∥2

=

⟨∞∑i=1

Qrs(ai)⊗Ql

t(bi),∞∑j=1

Qrs(aj)⊗Ql

t(bj)

⟩

=∞∑

i,j=1

〈πrs(ai) (ηrs) , πrs(aj) (ηrs)〉

⟨πlt(bi)

(ηlt), πlt(bj)

(ηlt)⟩

=∞∑

i,j=1

fs(aia∗j)ft(b

∗jbi) =

n∑i,j=1

α(aia∗j)(s)α(b∗jbi)(t).

Thus we see that the function Φw, as a sum of products of continuous func-

tions, is continuous on EB × EB.

Step 2. For each fixed u ∈ EA ⊗EA the function Φu is continuous.

Proof. By Theorem 1.3 every element u ∈ EA ⊗EA can be written in the

form

u =∞∑i=1

λiai ⊗ bi

where λi ∈ C such that∑∞

i=1 |λi| < ∞ and (ai)i∈N and (bi)i∈N

are null sequences in EA. Therefore there exists n ∈ N so that

‖∑∞

i=n+1 λiai ⊗ bi‖ < ε/2‖ρ‖. Let us write u1 =∑n

i=1 λiai ⊗ bi and

u2 =∑∞

i=n+1 λiai ⊗ bi.

76


By the above, for any ε > 0 there exists a neighbourhood R of (s0, t0) such

that for any (s, t) ∈ R we have

|Φu1(s, t)− Φu1(s0, t0)| <ε

2.

Also, for any (s, t),

Φu2(s, t) ≤ ‖ϕs,t‖ ‖Qrs‖ ‖Ql

t‖ ‖ρ‖ ‖u2‖ <ε

4

since ‖Qrs‖ < 1, ‖Ql

t‖ < 1 and ‖ϕs,t‖ < 1. Therefore |Φu2(s0, t0)−Φu,2(s, t)| <ε/2. Hence |Φu(s0, t0)− Φu(s, t)| < ε.

Step 3. The function Φ is continuous.

Proof. Consider a point (s0, t0) ∈ EBσ(B) × EB and x0 ∈ B such that

σ(x0)(s0) = 1. Let U ⊂ EBσ(B) be the neighbourhood of s0 defined by

U = s ∈ EBσ(B) : σ(x0)(s) 6= 0.

Note that for any s ∈ U if we set xs = x0/σ(x0)(s) then we have σ(xs)(s) = 1.

Then, for all (s, t) ∈ U × EB,

Φ(s, t) =∥∥ϕs,t ((Qr

s ⊗Qlt

)ρ(xs)

)∥∥2=

∥∥ϕs,t ((Qrs ⊗Ql

t

)ρ(x0)

)∥∥2

|σ(x0)(s)|2.

Thus on the neighbourhood U × EB of (s0, t0), we see that Φ is the quotient

of two continuous functions, and hence continuous at that point.

Lemma 5.19. Φ(s, t) = 0 for all s ∈ EBσ(B) and t /∈ EBσ(B).

Proof. Let s and t be as described. Note that EBσ(B) is a closed left ideal

of EB, and so EBσ(B) ⊂ EB. Now σ(B) ⊂ EBσ(B) so x(t) = 0 for all

x ∈ σ(B). Hence, for all a ∈ EA and x ∈ σ(B),

Qlt(ax) = πlt(ax)(η

lt) = πlt(a)π

lt(x)(η

lt) = πlt(a)x(t)η

lt = 0.

So for any u ∈ EA ⊗EAσ(B) it must be that (Qrs ⊗Ql

t)(u) = 0.

77


Lemma 5.20. Φ(s, s) 6= 0 for all s ∈ EBσ(B).

Proof. Let s ∈ EBσ(B). Note that, for x ∈ N ls, since x∗x is positive we have

fs(x∗x) ∈ R so

fs(x∗x∗∗) = fs((x

∗x)∗) = fs(x∗x) = fs(x∗x) = 0

which implies that x∗ ∈ N rs . We can therefore define a map j : H l

s → Hrs by

j(x+N ls) = x∗ +N r

s

where x ∈ H ls. It is easily shown that j is a conjugate-linear isomorphism of

Hilbert spaces. Let xs ∈ B such that σ(xs)(s) = 1 and v = (Qrs⊗Ql

s)(ρ(xs)).

Define a linear functional γ : Hrs ⊗H l

s → C by

γ = tr (id⊗ j)

where id is the identity on Hrs and tr : Hr

s ⊗Hrs → C is the trace map

defined by tr(x, y) = 〈x, y〉. For any representation ρ(xs) =∑∞

i=1 ai ⊗ bi

where ai ∈ EA and bi ∈ EAσ(B) we have

γ(v) = tr (id⊗ j) (Qrs ⊗Ql

s)(ρ(xs))

= tr (Qrs ⊗ (j Ql

s))

(∞∑i=1

ai ⊗ bi

)

=∞∑i=1

tr(πrs(ai)(η

rs)⊗ j(πls(bi)(η

ls))).

Note that πls(bi)(ηls) = bi + N l

s and so j(πls(bi)(ηls) = b∗i + N r

s = πrs(bi)(ηrs).

78


Also πrs(b∗i )(η

rs) = πrs(bi)

∗(ηrs) so

γ(v) =∞∑i=1

〈πrs(ai)(ηrs), πrs(b∗i )(ηrs)〉

=∞∑i=1

〈πrs(bi)πrs(ai)(ηrs), ηrs〉 =∞∑i=1

〈πrs(aibi)(ηrs), ηrs〉

=∞∑i=1

fs(aibi) = fs

(∞∑i=1

aibi

)= fs π ρ(xs)

= fs(σ(xs)) = σ(xs)(s) = 1.

Since γ(0) = 0 we see that v defined above is non-zero. Hence for any

s ∈ EBσ(B),

Φ(s, s) = ‖ϕs,s(v0)‖2 6= 0.

Lemma 5.21. Let p, p′ ∈ B such that p 6= p′. Suppose s ∈ Θ−1(p) and

t ∈ Θ−1(p′). Then Φ(s, t) = 0.

Proof. Let x ∈ B be such that x(p) = 1 and x(p′) = 0. (This can be

constructed using the Urysohn lemma.) Then we have

σ(x)(s) = Θ(s)(x) = p(x) = 1

and similarly σ(x)(t) = p′(x) = 0. Let xs = x2. Then since ρ is a morphism

in A-mod-B, ρ(xs) = ρ(x) · σ(x). Observe that, for any a ∈ EA,

Qlt(aσ(x)) = πlt(a)π

lt(σ(x))

(ηlt)

= πlt(a)σ(x)(t)ηlt = 0.

If we write ρ(x) =∑∞

i=1 ai ⊗ bi where each ai ∈ EA and bi ∈ EAσ(B). Then

(Qrs ⊗Ql

t)(ρ(xs)) =∞∑i=1

Qrs(ai)⊗Ql

t(biσ(x)) = 0.

Hence Φ(s, t) = 0 as claimed.

We will now show that the spectrum of B is discrete. Recall that Θ :

EBσ(B) → B as defined earlier is a perfect map.

79


Claim. Let p0 ∈ B. Then

W :=⋃

p∈B:p6=p0

Θ−1(p)

is closed in EBσ(B).

Proof. Suppose it is not true. Then W does not contain all its cluster points

so there exists s0 ∈ Θ−1(p0) of which every neighbourhood contains a point

s ∈ W . We know, by the above, that Φ(s0, s) = 0 for any s ∈ W but

Φ(s0, s0) 6= 0. This contradicts the continuity of Φ and proves the claim.

Since Θ is closed, this implies that Θ(W ) = B \ p0 is closed in B. Hence

p0 is open, and since p0 was arbitrary, it follows that B is discrete.

80

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