Commutativity of the exponential spectrum

(1)

Commutativity of the Exponential Spectrum

Aram Gevorgyan

Maîtrise en mathématiques

(2)

(3)

Résumé

Pour l’algèbre de Banach complexe A ayant l’élément unité, on dénote par G(A) l’ensemble des éléments inversibles de A, et par G1(A) on dénote la composante qui contient l’unité. Le

spectre de a ∈ A est l’ensemble de tous les nombres complexes λ tels que λ1 − a /∈ G(A), et le spectre exponentiel de a est l’ensemble de tous les nombres complexes tels que λ1−a /∈ G₁(A). Évidemment, pour chaque élément de l’algèbre, son spectre exponentiel contient le spectre habituel. Il est bien connu que le spectre habituel a une propriété que l’on nommera «propriété de commutativité». Cela signifie que, pour chaque choix des deux éléments a, b ∈ A, nous avons Sp(ab) \ {0} = Sp(ba) \ {0}, où Sp est le spectre. Avons-nous la même propriété pour les spectres exponentiels? Cette question n’est toujours pas résolue. L’objectif de ce mémoire est d’étudier le spectre exponentiel, et plus particulièrement sa propriété de commutativité.

Dans le premier chapitre, nous donnerons les définitions d’algèbre de Banach complexe, spectre et spectre exponentiel de ses éléments, et leurs propriétés de base. Aussi nous établirons des relations topologiques entre les spectres exponentiel et habituel.

Dans le deuxième chapitre, nous définirons les fonctions holomorphes sur une algèbre de Banach, et discuterons du problème de la propriété de commutativité de spectre exponentiel, en établissant des résultats positifs connus.

Dans le troisième et dernier chapitre, nous examinerons quelques exemples d’algèbres de Ba-nach, décrivant les ensembles G(A) et G₁(A), et discuterons de la propriété de commutativité pour ces algèbres.

(4)

(5)

Abstract

For a complex Banach algebra A with unit element, we denote by G(A) the set of invertible elements of A, and by G1(A) we denote the component of G(A) which contains the unit. The

spectrum of a ∈ A is the set of all complex numbers λ such that λ1−a /∈ G(A), and the exponential spectrum of a is the set of all complex numbers λ such that λ1−a /∈ G₁(A). Of course for each element of the algebra its exponential spectrum contains the usual spectrum. It is well known that the usual spectrum has the so-called commutativity property. This means that, for any two elements a and b of A, we have Sp(ab) \ {0} = Sp(ba) \ {0}, where Sp denotes the spectrum. Does this property hold for exponential spectra? This is still an open question. The purpose of this memoir is to study the exponential spectrum, and particularly its commutativity property.

In chapter one, we will give definitions of a complex Banach algebra, the spectrum and expo-nential spectrum of its elements, and their basic properties. Also we will establish topological relations between exponential and usual spectra.

In chapter two, we will define holomorphic functions on a Banach algebra, and also discuss the commutativity property problem for the exponential spectrum, establishing some known positive results.

In the last chapter, we will consider some examples of Banach algebras, describing the sets G(A) and G1(A), and discuss the commutativity property for these algebras.

(6)

(7)

Acknowledgments

Several people have been supporting me during my studies and my integration into the social life of Quebec City, since the very first day of my arrival in this town. They were always there to help me with any advice and services I asked for. Without them, my education in Laval university simply couldn’t have progressed to the final stage.

First of all I would like to thank my supervisor Professor Thomas Joseph Ransford. Professor Ransford helped me with the very first steps, personally introducing me the Laval University and its personnel who I needed to know during my education. The next and the most valuable help of him was the professional support, scientific discussions, advice he provided through per-sonal meetings. I’m deeply grateful to Professor Ransford, his valuable support was essential at every stage of my education.

I also would like to thank all the staff and professors of the Department of Mathematics and Statistics with whom I collaborated or asked for help. They were always kind and ready to help me with any service I needed. Specially I would like to mention Sylvie Drolet and Robert Guénette, whose help and support were very significant for me.

I am grateful to my friends in Quebec City, for their advice and support in daily life that was especially valuable at the starting period of my stay in Quebec. Specially I want to thank my friend Anush Stepanyan who advised me to start my education in Laval University and helped me very much with everything I needed.

Finally, and most of all, I’m deeply grateful to my family and my friends (in Armenia) who encouraged me and provided all the necessary help to make the decision to move to Canada and to start my education in Laval University.

(14)

(15)

Chapter 1

Banach Algebras, Spectrum and

Exponential Spectrum

1.1 Banach Algebras, Definitions and Examples

The material of this chapter is mainly from the book of B. Aupetit Aupetit[1991].

Definition 1.1.1. An algebra is a vector space A over the field C, with a multiplication satisfying the following properties:

x(yz) = (xy)z, (x + y)z = xz + yz, x(y + z) = xy + xz,

λ(xy) = (λx)y = x(λy) for all x, y, z ∈ A, and λ ∈ C.

If moreover A is a Banach space for a norm k · k and satisfies the norm inequality kxyk ≤ kxkkyk,

for all x, y ∈ A, we say that A is a complex Banach algebra.

In the definition, C could be replaced by any field K, in particular by R, but in this memoir we will deal mainly with complex Banach algebras.

A Banach algebra A is called commutative, if for all a, b ∈ A we have ab = ba, and with unit, if there is an element 1 ∈ A with k1k = 1, such that for all a ∈ A we have 1a = a1 = a. Obviously if the algebra has a unit, then the unit is unique. Indeed, if there were another e ∈ A, such that ea = ae = a for all a ∈ A, then e = e1 = 1e = 1. Let us give some examples of Banach algebras.

(16)

Remark 1.1.2. If A is a Banach algebra without unit, it is always possible to embed it isomet-rically in the Banach algebra with unit A₊ _{= A × C, where the operations and the norm are} defined by

(x, α) + (y, β) = (x + y, α + β), λ(x, α) = (λx, λα),

(x, α)(y, β) = (xy + βx + αy, αβ), k(x, α)k = kxk + |α|

.

Here e = (0, 1) will be the unit element of A₊, and the function x 7→ (x, 0) is an isomorphism from A into A₊.

Example 1.1.3. Let K be a compact set. Then C(K), the space of all complex continuous functions on K, with the supremum norm, is a Banach algebra with unit. If K has only n elements then C(K) = Cn with coordinatewise multiplication, and of course C is the simplest commutative Banach algebra. If K is a locally compact space, then the Banach space of all complex continuous functions on K which go to zero at infinity, that is, continuous functions f such that {x : x ∈ K, |f (x)| ≥ } is compact for all > 0, with the supremum norm, is a Banach algebra, which has no unit if K is not compact.

Example 1.1.4. If K is compact, every closed subalgebra of C(K) is a Banach algebra. For instance if K is a compact subset of C, the following examples of Banach algebras are interesting: The algebra P (K), of continuous functions on K which are uniform limits of polynomials on K; the algebra R(K), of continuous functions on K which are uniform limits of rational functions with poles outside of K; and the algebra A(K), of continuous functions on K which are holomorphic on the interior of K. We have the inclusions P (K) ⊂ R(K) ⊂ A(K) ⊂ C(K), and there are examples of K for which all these inclusions are strict.

Example 1.1.5. Let G be a locally compact commutative group and let µ be its Haar measure. Then L1(G) is a Banach algebra if we define multiplication by convolution

(f ∗ g)(x) = Z

G

f (xy−1)g(y)dµ(y) and the norm by the L1-norm kf k1=

R

G|f (x)|dµ(x). For instance if we consider the additive

group L1(Z) = `1(Z), then we have

(f ∗ g)(n) =X k∈Z f (n − k)g(k) kf k₁ =X k∈Z |f (k)|.

In fact, this algebra `1_{(Z) can be identified with the Wiener algebra W of continuous functions} f (t) =P

n∈Zaneint on [0, 2π] having Fourier coefficients an such that

P

n∈Z|an| < +∞, with

the operations of pointwise addition and multiplication and the norm kf k =P

n∈Z|an|.

(17)

Example 1.1.6. The Banach space of integrable functions on [0, 1] with multiplication defined by convolution (f ∗ g)(x) = Rx 0 f (x − t)g(t)dt and L 1_{-norm kf k} 1 = R1 0 |f (t)|dt is a Banach algebra.

The following examples of Banach algebras are non-commutative.

Example 1.1.7. Let X be a Banach space. Then L(X), the algebra of all bounded linear operators on X, is a Banach algebra with unit for the usual operator norm. If X is finite dimensional, with dim X = n, then L(X) can be identified with M_n_{(C). If dim X > 1, then} L(X) is not commutative. Except for finite-dimensional X, the simplest non-commutative Banach algebra is L(H) where H is an infinite-dimensional Hilbert space. Every closed sub-algebra of L(X) is also a Banach sub-algebra. For instance, the sub-algebra of compact operators LC(X) is a Banach algebra which has no unit if X is infinite-dimensional. In fact, every Banach algebra can be isomorphically represented as a closed subalgebra of L(X) for some Banach space X, but in practice this does not help very much.

Starting with some Banach algebras, how can new ones be obtained?

For instance, given A a Banach algebra, it is possible to consider L(A), the algebra of bounded linear operators on A, and all its closed subalgebras. Given a family of (Ai)i∈I of Banach

algebras we can define the Banach algebra product Q

i∈AAi as the set of all families (xi)i∈I

such that xi ∈ Ai and supi∈Ikxik < +∞ with the norm kxk = supi∈Ikxik. It is easy to

verify that Q

i∈IAi is a Banach algebra which contains an isometric copy of each Ai0 (by the

mapping x → (x_i)i∈I where xi = x if i0 = i and xi = 0 otherwise). If I = 1, . . . , n, obviously

this product coincides with A1× · · · × An.

Given a Banach algebra A it is also possible to consider the n × n matrix algebra Mn(A)

with the standard operations

(aij) + (bij) = (aij+ bij), (aijbij) = ( n X k=1 aikbkj)

and the norm ka_ijk = sup_i=1,...,n(kai1k + · · · + kaink).

Mn(A) is not commutative for n > 1 and it contains an isometric copy of A (associating to

x the diagonal matrix having only x on the diagonal).

A much more useful tool is the following. Let I be a closed two-sided ideal of A. Then A/I is a Banach space for the norm ||| ˙x||| = infu∈I||x + u||, where ˙x denotes the coset x + I of x.

(18)

of A by the two-sided idealI. This comes from

||| ˙x + ˙y||| ≤ ||x + u + y + v|| ≤ ||x + u|| + ||y + v||, ||| ˙x ˙y||| ≤ ||xy + xv + uy + uv|| ≤ ||x + u||||y + v||, for u, v ∈ I, so

||| ˙x + ˙y||| ≤ ||| ˙x||| + ||| ˙y|||, ||| ˙x · ˙y||| ≤ ||| ˙x||| · ||| ˙y|||.

This implies in particular that ||| ˙1||| = ||| ˙12_{||| ≤ ||| ˙1||| · ||| ˙1||| so 1 ≤ ||| ˙1||| ≤ ||1|| = 1, and}

consequently ||| ˙1||| = 1 if A has a unit.

Example 1.1.8. Let X be a Banach space. Then F and LC(X) are closed two-sided ideals of L(X), (where F is the linear subspace of finite-rank operators, and LC(X) is the linear subspace of compact operators), so we can consider L(X)/F and L(X)/LC(X). The latter is called theCalkin algebra of X. If X is a Hilbert space then they coincide, and we then have a Banach algebra with involution because LC(X) is stable by involution.

1.2 Invertible Elements and Spectrum

In this section we give some basic properties of invertible elements and spectrum in Banach algebras.

Definition 1.2.1. Let A be a Banach algebra. We denote by G(A) the set of invertible elements of A. It is obviously a group containing the unit.

We now prove that G(A) is an open subset of A.

Theorem 1.2.2. Suppose that A is a Banach algebra, x ∈ A and kxk < 1. Then 1 − x is invertible and (1 − x)−1 = ∞ X k=0 xk.

Proof. Let kxk = r < 1 and let sn=Pn_k=0xk. Obviously, for n < m, we have

ksn− smk ≤ m

X

k=n+1

kxkk ≤ rn+1/(1 − r),

so (sn) is a Cauchy sequence converging to some element a =P∞k=0xk. Because xsn= snx =

sn+1− 1 we conclude that a(1 − x) = (1 − x)a = 1.

Theorem 1.2.3. Suppose that A is a Banach algebra and that a is invertible. If kx − ak < 1/ka−1k, then x is invertible. Moreover the mapping x 7→ x−1is a homeomorphism from G(A) onto G(A).

(19)

Proof. We have x = a + x − a = a(1 + a−1_{(x − a)). Now ka}−1_{(x − a)k ≤ kx − ak · ka}−1_{k < 1}

so, by the previous theorem, 1 + a−1(x − a) is invertible, and consequently x is invertible. Moreover x−1 = (1 + a−1(x − a))−1a−1=P∞ k=0(a −1_{(a − x))}k_a−1_{. Consequently, we have} kx−1− a−1k = ∞ X k=1 (a−1(a − x))ka−1 ≤ ka−1k2kx − ak ∞ X k=0 (ka−1k · kx − ak)k. So x 7→ x−1 is continuous, and since it is its own inverse, it is a homeomorphism.

Definition 1.2.4. As with operators we can define the spectrum of x, denoted by Sp x, as the set of λ ∈ C such that λ1 − x is not invertible in A. Also we define the spectral radius by ρ(x) = sup{|λ| : λ ∈ Sp x}.

We have λ1 − x = λ(1 − x/λ), for λ 6= 0. So Theorem 1.2.2implies that λ1 − x is invertible for kxk < |λ|, and consequently Sp x is a bounded subset of C. In other words, it says that ρ(x) ≤ kxk.

Theorem 1.2.5. Let A be a Banach algebra and let x, y ∈ A. Suppose that xy = 1 and yx 6= 1. Then Sp x and Sp y contain a neighbourhood of 0.

Proof. By hypothesis x is not invertible. Let p = yx 6= 1. Then p2 _{= y(xy)x = p. Moreover}

(x − λ1)y = xy − λy = 1 − λy and y(x − λ1) = p − λy 6= 1 − λy. If 1 − λy is invertible, then (x − λ1)y(1 − λy)−1= 1.

Because y and (1 − λy)−1 commute we have

y(1 − λy)−1(x − λ1) = (1 − λy)−1y(x − λ1) = (1 − λy)−1(p − λy) 6= 1.

and consequently x − λ1 is not invertible, that is λ ∈ Sp x. So we have B(0, 1/ρ(y)) ⊂ Sp x. The argument for Sp y is similar.

Theorem 1.2.6. Let A be a Banach algebra and let x ∈ A. Then for every non-constant polynomial p with complex coefficients we have Sp p(x) = p(Sp(x)).

Proof. Let q(z) = C(z − z1)...(z − zn) be a given arbitrary polynomial. Then q(x) = C(x −

z1)...(x − zn). So q(x) is invertible if and only if each of the x − zi1 is invertible. Applying

this to q(z) = p(z) − λ, we get the result.

The next theorem is very important and is due to I.M. Gelfand.

Theorem 1.2.7. ([Aupetit, 1991, Theorem 3.2.8]). Let A be a Banach algebra and x ∈ A. Then

(20)

1. λ 7→ (λ1 − x)−1 _{is analytic on C \ Sp x and goes to 0 at infinity,}

2. Sp x is compact and non-empty, 3. ρ(x) = lim

n→∞kx n_k1/n_.

Proof. We know that ρ(x) ≤ kxk. Moreover Sp x is closed by Theorem1.2.3, because λ0 ∈ Sp x/

and |λ − λ0| < 1/k(λ01 − x)−1k implies λ /∈ Sp x. So Sp x is compact. Again by Theorem

1.2.3, λ → R(λ) = (λ1 − x)−1 _{is continuous on the open set C \ Sp x. Let λ, µ /}∈ Sp x. We have (λ1 − x) − (µ1 − x) = (λ − µ)1, so multiplying by R(λ)R(µ) we get R(µ) − R(λ) = (λ − µ)R(λ)R(µ). Consequently lim

µ→λ

R(µ)−R(λ)

µ−λ = −R(λ)2 by continuity of R. So R is analytic

on C \ Sp x. Moreover (λ1 − x)−1 = _λ1(1 − x/λ)−1 for λ /∈ Sp x. So, for |λ| ≥ 2kxk, we have by Theorem1.2.2 k(λ1 − x)−1k ≤ 2/|λ|. Hence it goes to zero at infinity.

We now prove that Sp x is non-empty. Suppose this is false and let f be a bounded linear functional on A. Then by part1, the function λ 7→ f((λ1 − x)−1) is entire and goes to zero at infinity so, by Liouville’s theorem, it is identically zero. This being true for all f ∈ A0, by one of corollaries of the Hahn-Banach theorem, we have (λ1 − x)−1 = 0 for all λ /∈ Sp x, but this is absurd.

We now prove 3. We know that ρ(x) ≤ kxk, for all x ∈ A. So, by Theorem 1.2.6applied to zn, we conclude that ρ(x)n≤ kxn_{k. Let f ∈ A}0_{. Then λ 7→ f ((λ1 − x)}−1_{) is holomorphic on}

C \ Sp x by part 1, so in particular in |λ| > ρ(x). For |λ| > kxk, by Theorem1.2.2, we have f ((λ1 − x)−1) = 1 λ(f (1) + f (x) λ + · · · + f (xn) λn + · · · ).

Consequently this is also true for |λ| > ρ(x) by the Identity Principle. Let λ be fixed such that |λ| > ρ(x). Then for every f ∈ A0 we have sup_n|f (xn_)/λn_{| < +∞. By the Banach-Steinhaus}

Theorem applied to A0 and to the sequence of T_n : A0 → C, defined by Tn(f ) = f (xn)/λn,

we conclude that there exists a constant C, depending on λ, such that kxnk ≤ C|λ|n _{for all}

n ≥ 1. Then

lim sup

n→∞

kxnk1/n ≤ |λ| for all |λ| ≥ ρ(x). So finally, using the last inequality with ρ(x)n≤ kxn_{k, we get:}

ρ(x) ≤ lim inf

n→∞ kx

n_k1/n _{≤ lim sup} n→∞

kxnk1/n≤ ρ(x), and the theorem is proved.

The formula given by part 3 is called the Beurling-Gelfand formula.

Corollary 1.2.8. (Gelfand-Mazur)([Aupetit,1991, Corollary 3.2.9]). If A is a Banach algebra in which every non-zero element is invertible then A is isometrically isomorphic to C.

(21)

Proof. Let x ∈ A. By part 2 of Theorem1.2.7, Sp x is non-empty. Let λ ∈ Sp x. Then x − λ1 is not invertible, consequently x = λ1. This implies that Sp x contains only one point, which we call α(x). The formula x = α(x)1 implies that α is an isomorphism from A onto C. It is an isometry because we have kxk = |α(x)|k1k = |α(x)|.

Corollary 1.2.9. Let A be a Banach algebra. Suppose that x, y ∈ A satisfy xy = yx. Then ρ(x + y) ≤ ρ(x) + ρ(y) and ρ(xy) ≤ ρ(x)ρ(y).

Proof. Because (xy)n _{= x}n_yn _{for every integer n ≥ 1, using part} _{3 of Theorem} _1.2.7 _we

conclude that ρ(xy) = lim n→∞k(xy) n_k1/n_{≤ lim} n→∞kx n_k1/n _lim n→∞ky n_k1/n_{= ρ(x)ρ(y).}

Let α > ρ(x), β > ρ(y) and a = x/α, b = y/β. Then ρ(a) < 1 and ρ(b) < 1. So there exists some integer N such that n ≥ N implies max(ka2nk, kb2n

k) < 1. Defining γn= max 0≤k≤2nka k_{k · kb}2n_−k k, we have k(x + y)2nk1/2n _{= k} 2n X k=0 (2_kn)xky2n−kk1/2n _{≤ [} 2n X k=0 (2_kn)αkβ2n−kkak_kkb2n−k_k]1/2n _{≤ (α + β)γ}1/2n n .

But it is easy to see that the sequence (γ_n) is decreasing for n ≥ N , because max 0≤k≤2n+1ka k_kkb2n+1−k_{k = max( max} 0≤k≤2nka k_kkb2n+1−k_k, _max 2n_≤k≤2n+1kb 2n+1−k_kkak_k) ≤ γ_nmax(ka2nk, kb2nk) So we have ρ(x + y) = lim n→∞k(x + y) 2n k1/2n ≤ (α + β) lim sup n→∞ γ1/2_n n ≤ (α + β) lim sup n→∞ γ_N1/2n = α + β, for arbitrary α > ρ(x), β > ρ(y). Hence the theorem is proved.

Suppose that A is a Banach algebra and that B is a closed subalgebra containing the same unit 1. If x ∈ B, what is the relation between the spectrum of x relative to B, denoted by Sp_Bx, and the spectrum of x relative to A, denoted by Sp_Ax? Of course, we have Sp_Ax ⊂ Sp_Bx, but is it possible to say more?

Theorem 1.2.10. Let (x_n) be a sequence of invertible elements of A converging to a non-invertible element. Then lim

n→∞kx −1

(22)

Proof. Suppose the result is false. Then there exist C > 0 and a subsequence of (xn), which

we denote in the same way, such that kx−1_n k ≤ C. Let x be the limit of the sequence (x_n). We have x = xn(1 + x−1n (x − xn)) and kx−1n (x − xn)k ≤ Ckx − xnk < 1 for n large enough

so, by Theorem 1.2.2, 1 + x−1_n (x − xn) is invertible, and hence x is invertible. So we get a

contradiction.

Definition 1.2.11. Let A be a Banach algebra. We say that a ∈ A istopological divisor of zero if there is a sequence (an) ∈ A such that kank = 1, and limn→∞aan= limn→∞ana = 0.

Corollary 1.2.12.Let x ∈ A and let α be in the boundary of Sp x. Then x−α1 is a topological divisor of zero.

Proof. Because α ∈ ∂ Sp x, there exists a sequence (αn) of complex numbers converging to α

with α_n∈ Sp x. We take x/ _n= (x − αn1)−1/k(x − αn1)−1k. Then we have

(x − α1)xn= (x − αn1)xn− (α − αn)xn= 1/k(x − αn1)−1k − (α − αn)xn. Consequently k(x − α1)x_nk ≤ 1/k(x − α_n1)−1k + |α − α_n|. By Theorem1.2.10, lim n→∞k(x − αn1) −1_{k = +∞, so lim}

n→∞k(x − α1)xnk = 0. The other result

is proved similarly.

Lemma 1.2.13. Let A be a Banach algebra with unit. Then a topological zero divisor is not invertible.

Proof. If x is a topological zero divisor, then there exists a sequence {xn} ⊆ A such that kxnk

and xxn → 0. If x is invertible, i.e. x−1x = xx−1 = 1, then we have xn = 1xn = x−1xxn =

x−1(xxn) → 0 therefore xn→ 0, a contradiction.

Definition 1.2.14. Let A be a Banach algebra, and a ∈ A. We define thesingular spectrum of a by τ (a) = {α ∈ C : a − α1 is a topological zero-divisor in A}

So, in fact with Lemma 1.2.13and Corollary1.2.12we can state the next theorem:

Theorem 1.2.15. (Harte [1976]) Let A be a Banach algebra with unit and let x ∈ A. Then ∂ Sp(x) ⊆ τ (x) ⊆ Sp(x).

Theorem 1.2.16. Let A be a Banach algebra and let B be a closed subalgebra containing the unit 1. We have

1. G(B) is the union of some components of B ∩ G(A), and furthermore the set ∂G(B) ∩ B ∩ G(A) is empty,

(23)

2. if x ∈ B, then SpBx is the union of SpAx and a (possibly empty) collection of bounded

components of C \ SpAx. In particular ∂ SpBx ⊂ ∂ SpAx.

Proof. 1 It is obvious that G(B) ⊂ G(A), so G(B) ⊂ B ∩ G(A). These last two sets are open in B. We prove that B ∩ G(A) contains no boundary point x of G(B). If such an x = lim xn,

with xn ∈ G(B), exists and is in G(A), by continuity of x 7→ x−1 on G(A) we conclude

that x−1 = lim x−1_n . In particular (kx−1_n k) is bounded so, by Theorem1.2.10, x is not in the boundary of G(B), a contradiction. Let Ω be a component of B ∩ G(A) that intersects G(B), and let U be the complement ofG(B). Since B ∩ G(A) contains no boundary points of G(B), Ω is the union of the two open sets Ω ∩ G(B) and Ω ∩ U . But Ω is connected, so Ω ∩ U is empty. Hence Ω ⊂ G(B). So part1 is proved.

2 Let ΩA = C \ SpAx and ΩB = C \ SpBx. Obviously ΩB ⊂ ΩA. A similar argument

shows that ∂Ω_B∩ Ω_A = ∅, so that ΩB is the union of some components of ΩA. Hence 2. is

proved.

Corollary 1.2.17. With the same hypotheses, suppose that Sp_Ax does not separate the com-plex plane. Then SpBx = SpAx.

1.3 Holomorphic Functional Calculus

If x is in a Banach algebra A and if p(λ) = α0 + α1λ + · · · + αnλn is a polynomial with

complex coefficients, we can define without any problem p(x) = α₀1 + α1x + · · · + αnxn. Now

if r(λ) = p(λ)/q(λ) (where p and q are relatively prime) is a rational function with poles outside Sp x then, by Theorem 1.2.6, q(x) is invertible, so we can define r(x) by p(x)q(x)−1. Is it possible to extend such definition to a larger class of functions?

Suppose that f is holomorphic on the disk |λ| < R and ρ(x) < R, so that f (λ) =P∞

k=0αkλk

for |λ| < R. This implies in particular thatP∞

k=0αkkxkk converges, so that f (x) =

P∞

k=0αkxk

converges in A to an element we call f (x). This argument can be applied in particular to define ex, the exponential of x.

But if f is only holomorphic on a neighbourhood of Sp x, is it possible to define f (x)? Of course this function cannot be defined by series because there is a problem in glueing the local pieces. But we now show that it can be done using the Cauchy formula for contours.

Let K be a compact subset of C supporting a measure µ and let f be a continuous function from K into a Banach algebra A. Then, exactly as is done in the scalar situation, it is possible to define R_Kf (λ)dµ(λ), and of course we have φ(R_Kf (λ)dµ(λ)) = R_Kφ(f (λ))dµ(λ) for all bounded linear functionals φ on A (see for instance Rudin[1987], pages 73-78).

(24)

Let x ∈ A be fixed and let Ω be an arbitrary open set containing Sp x. Let Γ be a smooth oriented contour contained in Ω \ Sp x, surrounding Sp x and not surrounding any point of C \ Ω. The set Sp x may be very complicated, but we may suppose that Ω is a finite union of polygonal closed curves, without multiple points, and such that Sp x is contained in the union of the bounded components they limit.

For f ∈ H(Ω), the algebra of holomorphic functions on Ω, the integral 1

2πi Z

Γ

f (λ)(λ1 − x)−1dλ

is well-defined because λ 7→ (λ1 − x)−1 is defined and continuous on Γ. Moreover this integral is independent of the contour Γ surrounding Sp x for the following reasons.

Suppose that there exist two contours Γ₁, Γ2⊂ Ω \ Sp x, surrounding Sp x and not

surround-ing any point of C \ Ω such that x1= 1 2πi Z Γ1 f (λ)(λ1 − x)−1dλ 6= 1 2πi Z Γ2 f (λ)(λ1 − x)−1dλ = x2.

By the Hahn-Banach theorem there exists φ ∈ A0 such that φ(x₁) 6= φ(x2). We have

φ(x1) = 1 2πi Z Γ1 h(λ)dλ, φ(x2) = 1 2πi Z Γ2 h(λ)dλ

where h(λ) = f (λ)φ((λ1 − x)−1). But by Theorem 1.2.7 (1 ), the function h is holomorphic on Ω \ Sp x, so by Cauchy’s theorem φ(x1) = φ(x2) and we get a contradiction. So we can set

f (x) = 1 2πi

Z

Γ

f (λ)(λ1 − x)−1dλ

for an arbitrary contour Γ having the previous properties. But before doing that we must at least verify that this definition coincides with the standard one given for rational functions. Lemma 1.3.1. Let x ∈ A, let Γ be a smooth contour surrounding Sp x and let r(λ) be a rational function having no poles surrounded by Γ. Then r(x) = 1

2πi

R

Γr(λ)(λ1 − x) −1_dλ_.

Proof. We first prove this for r(λ) = (α − λ)n_{, with n ∈ Z, and α not surrounded by Γ. Let}

rn= 1 2πi Z Γ r(λ)(λ1 − x)−1dλ. When λ /∈ Sp x we have the relation

(λ1 − x)−1 = (α1 − x)−1+ (α − λ)(α1 − x)−1(λ1 − x)−1. So we have rn= 1 2πi(α1 − x) −1Z Γ (α − λ)ndλ + (α1 − x)−1rn+1. 10

(25)

But the first integral is zero because λ(α − λ)n _{is holomorphic, so we have}

rn+1= (α1 − x)rn.

Consequently the formula will be proved for (α − λ)n_{, n ∈ Z, if we succeed in proving it for} n = 0. We have 1 2πi Z Γ (λ1 − x)−1dλ = 1 2πi Z ΓR (λ1 − x)−1dλ = 1 2πi Z |λ|=R (1 λ+ x λ2 + . . . )dλ for R > kxk. So 1 2πi Z Γ (λ1 − x)−1dλ = 1.

Thus the first part is proved. Now if r is an arbitrary rational function it can be written as r(λ) = p(λ) + a1,1 α1− λ + · · · + a1,k1 (α1− λ)k1 + · · · + an,1 αn− λ + · · · + an,kn (αn− λ)kn

where p is a polynomial in λ and where the α1, . . . , αn are the poles with their respective

multiplicities k1, . . . , kn(decomposition in simple elements). A formal calculation implies that

we also have

r(x) = p(x) + a1,1(α11 − x)−1+ · · · + a1,k1(α11 − x)

−k1_{+ · · · + a}

n,1(αn1 − x)−1+ . . .

· · · + a_n,k_n(αn1 − x)−kn.

So the first part implies the result.

Theorem 1.3.2. (Cayley-Hamilton theorem)([Aupetit, 1991, Corollary 3.3.2]). Let a be a n × n complex matrix and let p(λ) = det(a − λ1) be its characteristic polynomial. Then p(a) = 0.

Proof. Let α1, . . . , αkbe the different eigenvalues of a and let Γ be the union of k small disjoint

circles with centres respectively at α1, . . . , αk. By Lemma1.3.1 we have

p(a) = 1 2πi

Z

Γ

p(λ)(λ1 − a)−1dλ

But (a − λ1)−1 = _det(a−λ1)1 b(λ), where b(λ) is an n × n matrix depending analytically on λ since its (j, i) entry is a cofactor of a − λ1, and so is a polynomial in λ of degree ≤ n − 1. We then have p(a) = − 1 2πi Z Γ b(λ)dλ = 0 by Cauchy’s theorem.

Theorem 1.3.3. (Holomorphic Functional Calculus)([Aupetit, 1991, Theorem 3.3.3]). Let A be a Banach algebra and let x ∈ A. Suppose that Ω is an open set containing Sp x and that Γ

(26)

is an arbitrary smooth contour included in Ω and surrounding Sp x and not surrounding any point of C \ Ω. Then the mapping

f 7→ f (x) = 1 2πi

Z

Γ

f (λ)(λ1 − x)−1dλ from H(Ω) into A has the following properties:

(i) (f1+ f2)(x) = f1(x) + f2(x),

(ii) (f1f2)(x) = f1(x)f2(x) = f2(x)f1(x),

(iii) 1(x) = 1 and I(x) = x (where I(λ) = λ),

(iv) if (fn) converges to f uniformly on compact subsets of Ω, then f(x) = lim fn(x),

(v) Sp f(x) = f(Sp x).

Proof. (i) is obvious by definition. (iii) follows immediately from Lemma 1.3.1. (iv) is also easy because (f_n) converges uniformly to f on Γ and k(λ1 − x)−1k is bounded on this set. We now prove (ii). By Runge’s theorem there exist two sequences of rational functions (r1_k) and (r2

k), with poles not surrounded by Γ, converging uniformly respectively to f1an f2. By Lemma

1.3.1 we have (r_k1r2_k)(x) = r1_k(x)r2_k(x). So by (iv), property (ii) is true. We finish by proving (v). If f has no zero on Sp x then g = 1/f is holomorphic on a neighbourhood Ω1 of Sp x. If

necessary we can replace Γ by a contour Γ1⊂ Ω1 surrounding Sp x. We have f (λ)g(λ) = 1 on

Γ1 and consequently, applying (ii) and (iii) for Γ1 and Ω1, we get f (x)g(x) = 1. Thus f (x) is

invertible in A. On the other hand, if f (α) = 0 for some α ∈ Sp x, then there exists h ∈ H(Ω) such that f (λ) = (α − λ)h(λ) on Ω, and consequently f (x) = (α1 − x)h(x). But α1 − x is not invertible, so f (x) is not invertible, so f (x) is not invertible. Therefore β1 − f (x) is not invertible if and only if f takes the value β on Sp x.

We now give several elementary applications of this extremely important theorem.

Theorem 1.3.4.Let A be a Banach algebra. Suppose that x ∈ A has a disconnected spectrum. Let U0, U1 be two disjoint open sets such that Sp x ⊂ U0∪ U1, Sp x∩U0 6= ∅and Sp x∩U1 6= ∅.

Then there exists a non-trivial projection p commuting with x, such that Sp(px) = (Sp x ∩ U1) ∪ {0}, Sp(x − px) = (Sp x ∩ U0) ∪ {0}.

Proof. We consider the holomorphic function f defined on U = U0∪ U1 by f (λ) = 0 on U0

and f (λ) = 1 on U1. We set p = f (x). Because f (λ)2 = f (λ), we get p2= p. By construction

p commutes with x, by Theorem 1.3.3 (ii) and (iii). By part (v) of Theorem 1.3.3 we have Sp p = {0, 1}, so p is non-trivial. Moreover, we have px = f (x)I(x) = (f I)(x). But (f I)(λ) =

(27)

λ on U1 and (f I)(λ) = 0 on U0, so by Theorem1.3.3 (v) we have Sp(px) = (Sp x ∩ U1) ∪ {0}.

For Sp(x − px) the result is obtained similarly.

Theorem 1.3.5. Let A be a Banach algebra. Suppose that x ∈ A and that α /∈ Sp x. Then we have

dist(α, Sp x) = 1

ρ((α1 − x)−1).

Proof. Let Ω be an open set containing Sp x, but not α. Then f(λ) = 1/(α−λ) is holomorphic on Ω. So by Theorem1.3.3 (v) we have

Sp(α1 − x)−1 = {1/(α − λ) : λ ∈ Sp x}. So in particular,

ρ((α1 − x)−1) = sup{1/|α − λ| : λ ∈ Sp x}

= 1/ inf{|α − λ| : λ ∈ Sp x} = 1/ dist(α, Sp x).

Theorem 1.3.6. Let A be a Banach algebra. Suppose that x ∈ A has a spectrum which does not separate 0 from infinity. Then there exists y ∈ A such that x = ey_{. Hence, for every}

integer n ≥ 1 there exists z ∈ A such that zn_{= x}_.

Proof. By hypothesis 0 is in the unbounded component of C \ Sp x. Hence there exists a simply connected open set Ω containing Sp x and f ∈ H(Ω) such that ef (λ)= λ on Ω. Then we apply Theorem 1.3.3 and take y = f (x). For each n ≥ 1 we take z = ey/n.

1.4 Exponential Spectrum

Definition 1.4.1. Let A be a Banach algebra. We denote by exp(A) the set of all products of exponentials ex1_{. . . e}xn_{, where x}

1, . . . , xn∈ A

Definition 1.4.2. The connected component of G(A) which contains the unit, is called the principal component of G(A). We denote it by G1(A).

It is obvious that exp(A) ⊂ G(A). But t 7→ etx1_{. . . e}txn _{is a continuous function from [0, 1]}

into G(A) which connects 1 and ex1_{. . . e}xn_{. So in fact exp(A) is included in G}

1(A). We now

prove that the inverse is also true.

(28)

Proof. First we prove that exp(A) is open in G(A). Let a ∈ exp(A) and suppose that kx−ak < 1/ka−1k. Then k1 − a−1xk = ka−1(a − x)k < 1 and so ρ(1 − a−1x) < 1. Consequently Sp(a−1x) is included in the open disk of center 1 and radius 1. But for <λ > 0 there exists a holomorphic function f such that ef (λ)= λ. So by Theorem 1.3.3 there exists y = f (a−1x) such that ey = a−1x. Then x = aey ∈ exp(A). We now prove that exp(A) is closed in G(A). If an∈ exp(A) and limn→∞an= a ∈ G(A) then (a−1n a) converges to 1 and so k1 − a−1n ak < 1

for n large. As before, we conclude that a−1_n a = ezn _{for some z}

n ∈ A, so a ∈ exp(A).

Because exp(A) is closed and open in G(A) and is contained in G1(A), we conclude that

exp(A) = G1(A).

Definition 1.4.4. For x ∈ A we define the exponential spectrum of x, denoted by (x), by the set of λ ∈ C such that λ1 − x /∈ exp(A).

Obviously we have Sp x ⊂ (x). Let \(Sp x) be the polynomially convex hull of Sp x, that is the union of Sp x with the bounded components of C \ Sp x, and let λ0 ∈ \/ (Sp x). Then by

Theorem 1.3.6 there exists y such that λ01 − x = ey, and consequently λ0 ∈ (x). In other/

words, Sp x ⊂ (x) ⊂ \(Sp x). This implies in particular that (x) is a non-empty compact subset of C because G1(A) is open in A. Actually we have the following inclusions:

Theorem 1.4.5. (Harte[1976]) Let A be a Banach algebra with unit and let x ∈ A. Then ∂(x) ⊆ τ (x) ⊆ Sp x ⊆ (x) ⊆ \(Sp x)

.

Proof. All the inclusions already has been proved, except the first one: ∂(x) ⊆ τ(x). Suppose that s ∈ (x) is the limit of a sequence (sn) of points outside (x). The elements x − sn1 will,

in particular, be invertible, and we claim that

|s_n− s|k(x − s_n1)−1_{k ≥ 1, ∀n ∈ N.} If not, then for some n the element

(x − s1)(x − sn1)−1= 1 − (s − sn)(x − sn1)−1

will be in exp(A), hence also the element x − s1, contradicting s ∈ (x). Define bn= (x − sn1)−1/k(x − sn1)−1k.

We have kb_nk = 1 and

k(x−s1)b_nk = k1−(s−s_n)(x−sn1)−1k/k(x−sn1)−1k ≤ 1/k(x−sn1)−1k+|s−sn| ≤ 2|s−sn| → 0,

giving s ∈ τ (x).

(29)

So, in particular, we have ∂(x) ⊆ Sp x ⊆ (x), which means that (x) is therefore obtained from Sp x by filling in some of the holes of Sp x.

We now prove a simple and nice result characterizing the spectrum of ˙x, for ˙x ∈ A/I, where I is a closed two-sided ideal of A.

Theorem 1.4.6. Let T be a continuous morphism from a Banach algebra A onto a Banach algebra B. Then T (exp(A)) = exp(B). So we have

(T x) = \

y∈ker T

(x + y), Sp T x ⊂ \

y∈ker T

Sp(x + y) ⊂ \(Sp T x). In particular, if I is a closed two-sided ideal of A then

Sp ˙x ⊂ \

y∈I

Sp(x + y) ⊂ \(Sp ˙x).

Proof. It is clear that T (ex1_{. . . e}xn_{) = e}T x1_{. . . e}T xn_{, so the first part is obvious. If λ /}_{∈ (T x),}

then T x−λ1 = T u for some u ∈ exp(A). So, taking y = u−x+λ1, we have x+y−λ1 ∈ exp(A) and T y = 0, and consequently

\

y∈ker T

(x + y) ⊂ (T x). The other inclusion is obvious. Moreover

\

y∈ker T

Sp(x + y) ⊂ \

y∈ker T

(x + y) = (T x) ⊂ \(Sp T x)

(30)

(31)

Chapter 2

Commutativity of Exponential

Spectrum: Positive Results

2.1 Special Case, When at Least One of the Two Elements is a

Limit of Invertible Elements

In this chapter we discuss the commutativity of exponential spectrum and bring some positive results.

Theorem 2.1.1. Let A be an algebra with unit 1 and let x, y ∈ A, λ ∈ C, with λ 6= 0. Then λ1 − xy is invertible in A if and only if λ1 − yx is invertible in A.

Proof. Suppose that λ1−xy has an inverse u in A. Then we have (λ1−xy)u = u(λ1−xy) = 1. So we have:

(λ1 − yx)(yux + 1) = λyux + λ1 − y(xy)ux − yx

= λyux + λ1 − y(λu − 1)x − yx = λ1, (yux + 1)(λ1 − yx) = λyux + λ1 − yu(xy)x − yx

= λyux + λ1 − y(λu − 1)x − yx = λ1. Consequently λ1 − yx is invertible in A.

Theorem 2.1.1 can be reformulated as saying that Sp(xy)\{0} = Sp(yx)\{0} and conse-quently ρ(xy) = ρ(yx), for all x, y ∈ A. It is not proved yet that the corresponding result is true for the exponential spectrum in general. But now we show that in some special cases the result is true.

(32)

Proof. If b ∈ exp(A) ⇔ ∃ continuous map t 7→ [0, 1] → exp(A) such that b0 = 1 and b1 = b.

Hence ab₀a−1 = a1a−1= 1 and ab1a−1 = aba−1. Therefore aba−1 ∈ exp(A).

Theorem 2.1.3. (Murphy [1992]) Let A be a Banach algebra with unit and suppose that a, b ∈ A. Then

(ab) \ {0} = (ba) \ {0} (2.1) provided that either a or b is a limit of invertible elements of A.

Remark 2.1.4. Note that (2.1) is equivalent to saying that

1 − ab ∈ exp(A) ⇐⇒ 1 − ba ∈ exp(A). (2.2) Proof. We may suppose that a = limn→∞an, where the elements anbelong to G(A). To prove

the result it clearly suffices to show that 1 − ab ∈ exp(A) if and only if 1 − ba ∈ exp(A). We show the forward implication only- the reverse implication is proved similarly. Since exp(A) is an open set in A, it follows that 1 − ab ∈ exp(A) implies that 1 − anb ∈ exp(A) for all

sufficiently large n. Since 1−ba_n= a−1_n (1−anb)an, by Lemma2.1.2, we have 1−ban∈ exp(A)

for such values of n. Because 1 − ba = limn(1 − ban), and exp(A) is closed in G(A), therefore

1 − ba ∈ exp(A).

We do not define thetopological stable rank of Banach algebra A in general, (for the definition and more about the stable rank see Corach [1986] and Rieffel [1983]), but when topological stable rank is one, the definition is equivalent to saying that G(A) is dense in A. So we can conclude the next corollary:

Corollary 2.1.5. If A is a unital Banach algebra of topological stable rank one, then for all a, b ∈ A,

(ab) \ {0} = (ba) \ {0}.

Definition 2.1.6. Thecentre Z(A) of a Banach algebra A is the set Z(A) = {a ∈ A|xa = ax, ∀x ∈ A}.

Corollary 2.1.7. Suppose that A is a unital Banach algebra and suppose that a, b ∈ A. Then 1 − ab ∈ exp(A) ⇐⇒ 1 − ba ∈ exp(A). (2.3) provided that a (or b) ∈ Z(A)G(A).

Proof. Case 1. Suppose that a ∈ Z(A)G(A). Then a = zx, where z ∈ Z(A) and x ∈ G(A). 1 − ab = 1 − zxb = 1 − xbz = x(1 − bzx)x−1= x(1 − ba)x−1

(33)

The map y 7→ xyx−1 sends exp(A) onto exp(A). So 1 − ab ∈ exp(A) ⇐⇒ 1 − ba ∈ exp(A) in this case.

Case 2. Now suppose that a ∈ Z(A)G(A). There exists an → a with an ∈ Z(A)G(A).

Suppose that 1−ab ∈ exp(A). Then since an→ a, hence 1−anb → 1−ab, so 1−anb ∈ exp(A)

for n large enough. Hence 1 − ba_n ∈ exp(A), by case 1. Hence 1 − ba ∈ exp(A) ∩ G(A) = exp(A).

2.2 Other Special Cases

For further results we need to define the radical of Banach algebras and give some properties. Lemma 2.2.1. Let A be a ring with a unit 1. Every left (respectively right) ideal of A is contained in a maximal left (respectively right) ideal.

Proof. Let L0 be such a left ideal and let F be the family of all left ideals containing L0. This

family is partially ordered by inclusion. This order is inductive because if C is a chain in F then it is easy to see that either S

I∈CI is a left ideal or

S

I∈CI = A. But this last case is

impossible, because 1 /∈ I for all I ∈ C. By Zorn’s lemma there exists a maximal left ideal in F.

Theorem 2.2.2. Let A be a ring with unit 1. Then the following sets are identical: (i) the intersection of all maximal left ideals in A,

(ii) the intersection of all maximal right ideals in A,

(iii) the set of x such that 1 − zx is invertible in A, for all z ∈ A, (iv) the set of x such that 1 − xz is invertible in A, for all z ∈ A.

Proof. By Theorem 2.1.1,(iii) and (iv) are equivalent. We only prove the equivalence of (i) and (iii), as the equivalence of (ii) and (iv) can be proved by a similar argument. Let x be in the intersection of all maximal left ideals of A. If a = 1 − zx is not invertible then Aa is a left ideal of A so, by Lemma 2.2.1, it is contained in some maximal left ideal L0. Then zx ∈ L0

and 1 − zx ∈ L₀, so 1 ∈ L₀, that is L₀ = A which is a contradiction. Conversely, suppose that 1 − zx is invertible for all z ∈ A. If x is not in the intersection of all maximal left ideals, it means that there exists a maximal left ideal L0 such that x /∈ L0. Then L0+ Ax = A

and consequently 1 − zx ∈ L₀, but this gives a contradiction because it implies that we have 1 ∈ L0.

(34)

Definition 2.2.3. This two-sided ideal of A having properties(i)-(iv) is called the radical of A and denoted by Rad A.

Definition 2.2.4. If Rad A = {0} we say that A issemi-simple.

There are many examples of semi-simple Banach algebras, for instance Examples 1.1.3and

1.1.4.

Theorem 2.2.5. Let A be a ring with unit 1. Then A/ Rad A is semi-simple. Moreover, if x ∈ Athe coset ˙x is invertible in A/ Rad A if and only if x is invertible in A.

Proof. Let L0 _{be a maximal left ideal of A/ Rad A. Then L = {x : x ∈ A, ˙}_{x ∈ L}0_{} is a left}

ideal of A. It is maximal because if J is a left ideal containing L, then L0 ⊂ ˙J = { ˙x : x ∈ J }, so that L0 = ˙J and thus L = J . Conversely, if L is a maximal left ideal of A, then ˙L is a maximal left ideal of A/ Rad A for similar reasons. By property (i) of Theorem 2.2.2 this implies that {x : ˙x ∈ Rad(A/ Rad A)} is in the intersection of all maximal left ideals of A, that is the radical of A. Hence A/ Rad A is semi-simple. It is obvious that if x is invertible in A then ˙x is invertible in A/ Rad A. So suppose the latter property is true. There exists y ∈ A and u, v ∈ Rad A such that xy − 1 = u, yx − 1 = v. By property(iii) of Theorem2.2.2, 1 + u and 1 + v are invertible in A, so we have xy(1 + u)−1 = 1 and (1 + v)−1yx = 1. Hence x is invertible.

Theorem 2.2.5 implies that Sp x = Sp ˙x, for the coset ˙x of x in A/ Rad A.

Theorem 2.2.6. Every left (respectively right, respectively two-sided) ideal of A is disjoint from the open unit ball with centre at 1. Consequently every maximal left (respectively right, respectively two-sided) ideal of A is closed. In particular Rad A is closed, so A/ Rad A is a semi-simple Banach algebra.

Proof. Let L be a maximal left ideal of A. Then L contains no invertible elements, so the intersection of L with the open unit ball centred at 1 is empty by Theorem1.2.2. Consequently L ∩ B(1, I) = ∅ and also L 6= A. This implies that L = L if L is maximal. By Theorem2.2.2

(i), Rad A is closed.

Remark 2.2.7. Also Theorem 2.2.2 is equivalent to saying that the radical of A is the set of x ∈ A such that ρ(xz) = 0 for all z ∈ A. In particular it implies that the radical is included in the set ofquasinilpotent elements, that is the set of elements a such that ρ(a) = 0. The set of all these elements will be denoted by QN(A).

From now on, by an ideal in A we mean a two-sided ideal in A.

(35)

Definition 2.2.8. An ideal I in A is calledinessential whenever for every a ∈ I the spectrum of a in A is either a finite set or a sequence converging to zero.

The set of inessential elements relative to I we denote by kh(A, I). We can describe this set as kh(A, I) = {a ∈ A|a + I ∈ Rad(A/I)}.

Definition 2.2.9. If I is a closed ideal in A the set of Riesz elements relative to I is the set R(A, I) = {a ∈ A|a + I ∈ QN(A/I)}

It is clear that

I ⊂ kh(A, I) ⊂R(A, I). (2.4) Definition 2.2.10. For a, b ∈ A, we denote by [a, b] the commutator ab − ba.

The next three theorems have an auxiliary character. Therefore we introduce them without proofs.

Theorem 2.2.11. (Pták [1979]) Let A be a Banach algebra with unit, and let a ∈ A. Then the following conditions are equivalent:

• [b, a] ∈ Rad(A), for all b ∈ A, • [b, a] ∈ QN(A), for all b ∈ A.

Theorem 2.2.12. ([Aupetit,1991, Corollary 5.7.5]) Let I be an inessential ideal of a Banach algebra A. Let x ∈ A and suppose that ρ( ˙x) = 0, where ˙x denotes the coset of x in A/I. Then the spectrum of x has at most 0 as a limit point and, for every non-zero spectral value of x, the associated projection is in I.

Theorem 2.2.13. ([Aupetit,1991, Theorem 5.1.3]) Let A be a Banach algebra and let a, b ∈ A. Suppose that a(ab − ba) = (ab − ba)a. Then ρ(ab − ba) = 0.

Theorem 2.2.14. (Lindeboom and Raubenheimer[1999]) If A is a Banach algebra, then exp(A) + Rad(A) ⊂ exp(A). (2.5) Proof. If a ∈ exp(A) and r ∈ Rad(A), then a + r = a(1 + a−1_{r). Since the spectrum of}

1 + a−1r consists of the point 1 only, it does not separate 0 from ∞. Hence by [Rudin,1973, Theorem 10.30] 1 + a−1r has a logarithm, which means that 1 + a−1r ∈ exp(A). Since exp(A) is closed under multiplication, a + r ∈ exp(A).

Corollary 2.2.15. Let I be a closed inessential ideal of a Banach algebra A. If either a or b is Riesz in A relative to I then (ab) \ {0} = (ba) \ {0}.

(36)

Proof. If a is Riesz relative to I then, by Theorem2.2.12, Sp(a) is either finite or a sequence converging to zero. Hence a is the limit of invertible elements in A and our conclusion follows from Theorem2.1.3.

(37)

Chapter 3

Examples of Exponential Spectra

In this chapter we will consider some examples of Banach algebras, describe their group of invertible elements and the component of the group containing the unit element, and discuss the commutativity property of the exponential spectrum.

3.1 The Wiener Algebra

Let A be a commutative Banach algebra with unit. Consider the set hom(A, C) of all homo-morphisms ω : A → C. An element ω ∈ hom(A, C) is a complex linear functional satisfying ω(xy) = ω(x)ω(y) for all x, y ∈ A; notice that we do not assume that ω is continuous, or that kωk ≤ 1, but it is not difficult to prove it (we omit the proof). The Gelfand spectrum of A is defined as the set

SPg(A) = {ω ∈ hom(A, C) : ω 6= 0} of all nontrivial complex homomorphisms of A.

Remark 3.1.1. Every element ω ∈ SPg(A) satisfies ω(1) = 1. Indeed, for fixed ω the complex number λ = ω(1) satisfies λω(x) = ω(1x) = ω(x) for every x ∈ A. Since the set of complex numbers ω(A) must contain something other than 0, it follows that λ = 1.

The Gelfand map. Every element x ∈ A gives rise to a function ˆx : SPg(A) → C by way of ˆx(ω) = ω(x), ω ∈ SPg(A); ˆx is called theGelfand transform of x, and x → ˆx is called the Gelfand map. The functions ˆx are continuous by definition of the weak∗_{-topology on SPg(A).}

For x, y ∈ A we have

ˆ

x(ω)ˆy(ω) = ω(x)ω(y) = ω(xy) =xy(ω)._c

Moreover, since every element ω of SPg(A) satisfies ω(1) = 1, it follows that ˆ1 is the constant function 1 in C(SPg(A)). It follows that the Gelfand map is a homomorphism of A onto a unital subalgebra of C(SPg(A)) that separates points of SPg(A).

(38)

The Gelfand map exhibits spectral information about elements of A in an explicit way. Theorem 3.1.2. Let A be a commutative Banach algebra with unit. For every element x ∈ A, we have

Sp(x) = {ˆx(p) : p ∈ SPg(A)}.

Proof. Since for any x ∈ A and λ ∈ C, \x − λ = ˆx − λ and Sp(x − λ) = Sp(x) − λ, it suffices to establish the following assertion: An element x ∈ A is invertible iff ˆx never vanishes.

Indeed, if x is invertible, then there is a y ∈ A such that xy = 1; hence ˆx(ω)ˆy(ω) =xy(ω) =_c 1, for ω ∈ SPg(A), so that ˆx has no zeros.

Conversely, suppose that x is a noninvertible element of A. We must show that there is an element ω ∈ SPg(A) such that ω(x) = 0. For that, consider the set xA = {xa : a ∈ A} ⊆ A. This set is an ideal that does not contain 1. As proper ideal of A it is contained in some maximal ideal M ⊆ A, necessarily closed. We will show that there is an element ω ∈ SPg(A) such that M = ker ω. Indeed, A/M is a simple Banach algebra with unit; therefore it has no nontrivial ideals at all. Since A/M is also commutative, this implies that A/M is a field (for any nonzero element ζ ∈ A/M , ζA/M is a nonzero ideal, which must therefore contain the unit of A/M ). Therefore A/M is isomorphic to C. Choosing an isomorphism ˙ω : A/M → C, we obtain a complex homomorphism ω : A → C by way of ω(x) = ˙ω(x + M ). It is clear that ker ω = M , and finally ˆx vanishes at ω because x ∈ xA ⊆ M .

As we defined in chapter one, the Wiener algebra W is the algebra of all continuous functions on the unit circle whose Fourier series converges absolutely, that is, all functions f : T → C whose Fourier series have the form

f (eiθ) ∼ ∞ X n=−∞ aneinθ, whereP

n|an| < ∞. This algebra is a subalgebra of C(T), and obviously contains the constant

functions. And as we mentioned in chapter one, this algebra can be identified with `1_{(Z) =} L1(Z).

Define B(W ) = {f ∈ W : f (eiθ) 6= 0, ∀eiθ ∈ T}. It is clear that G(W ) ⊂ B(W ). But the inverse is also true due to Wiener’s theorem:

Theorem 3.1.3. ([Arveson,2002, Theorem 1.10.6]) If f ∈ W and f has no zeros on T, then the reciprocal 1/f belongs to W .

Proof. Consider the Banach algebra A = `1_{(Z), with multiplication defined by convolution ∗.}

The unit of A is the sequence 1 = (en), where e0 = 1 and en = 0 for n 6= 0. We show first

that SPg(A) can be identified with the unit circle T.

(39)

Indeed, for every λ ∈ T we can define a bounded linear functional ωλ on A by ωλ = ∞ X n=−∞ anλn, a = (an) ∈ `1(Z).

Obviously, ωλ(1) = 1, and one verifies directly that ωλ(a ∗ b) = ωλ(a)ωλ(b). Hence ωλ ∈

SPg(A).

We claim that every ω ∈ SPg(A) has the form ω_λ _{for a unique point λ ∈ T. To see} that, fix ω ∈ SPg(A) and define a complex number λ by λ = ω(ζ), where ζ = (ζn) is the

sequence ζn = 1 if n = 1, and ζn = 0 otherwise. Then ζ has unit norm in A, and hence

|λ| = |ω(ζ)| ≤ kζk = 1. Another direct computation shows that ζ is invertible in A, and its inverse is the sequence η = (ηn), where ηn = 1 for n = −1, and ηn = 0 otherwise. Since

kηk = 1 and |1/λ| = |1/ω(ζ)| = |ω(η)| ≤ kηk = 1, we find that |λ| = 1. Notice that ω = ωλ.

Indeed, we must have ω(ζn) = λn= ωλ(ζn) for every n ∈ Z, ζn being the unit sequence with

a single nonzero component in the nth position. Since the set {ζn _{: n ∈ Z} obviously has} `1_{(Z) as its closed linear span, it follows that ω = ω}

λ. Then λ = ω(ζ) is obviously uniquely

determined by ω.

These remarks show that the map λ 7→ ω_λ _{is a bijection of T on SPg(A). The inverse of} this map (given by ω ∈ SPg(A) 7→ ω(ζ) ∈ T) is obviously continuous, so by compactness of SP g(A) it must be a homeomorphism. Thus we have identified SP g(A) with the unit circle T and the Gelfand map with the Fourier transform, which carries a sequence a ∈ `1(Z) to the function ˆ_{a ∈ C(T) given by} ˆ a(eiθ) = ∞ X n=−∞ aneinθ.

Having computed SP g(A) and the Gelfand map in concrete terms, we observe that the range of the Gelfand map {ˆa : a ∈ A} is exactly the Wiener algebra W . The proof can now proceed as follows. Let f be a function in W having no zeros on T and let a be the element of A = `1_{(Z) having Gelfand transform f . By Theorem} 3.1.2, there is an element b ∈ A such that a ∗ b = 1; hence ˆa(λ)ˆb(λ) = 1, for λ ∈ T. It follows that 1/f = ˆb ∈ W , as asserted.

Therefore G(W ) = {f ∈ W : f (eiθ) 6= 0, ∀eiθ ∈ T}. Now we will try to describe the set exp(W ).

The next theorem is very familiar in complex analysis and we state it here without proof. Theorem 3.1.4. ([Rudin,1987, Theorem 10.10]) Let γ be a closed path, let Ω be the comple-ment of the image of γ (relative to the plane), and define

Wind(γ, z) = 1 2πi Z γ dζ ζ − z (z ∈ Ω).

(40)

Then Wind(γ) is an integer-valued function on Ω which is constant in each component of Ω and which is 0 in the unbounded component of Ω.

We call Wind(γ, z) the winding number of z with respect to γ. Note that the image of γ is compact, hence it lies in a bounded disc D whose complement Dc is connected; thus Dc lies in some component of Ω. This shows that Ω has precisely one unbounded component. The winding number of a closed curve in the plane around a given point is an integer representing the total number of times that curve travels counterclockwise around the point. The winding number depends on the orientation of the curve, and is negative if the curve travels around the point clockwise.

Let s be a continuous, nowhere zero function on the unit circle T. The winding number Wind(s, 0) of a curve s(θ), 0 ≤ θ ≤ 2π, about 0 can be defined geometrically as the increase in the argument of s(θ) as θ goes from 0 to 2π, divided by 2π. Since

log s(θ)|t₀= Z t

0

s0(θ) s(θ)dθ

for s continuously differentiable, for such functions this can be expressed analytically as Wind(s, 0) = 1 2πi Z 2π 0 s0(θ) s(θ)dθ = 1 2π= Z 2π 0 s0(θ) s(θ)dθ.

From this geometric definition it is clear that Wind(s, 0) depends continuously on s (where s ∈ G(W )). And because Wind(s, 0) takes on only integer values, therefore Wind(s, 0) is invariant under continuous deformation (within the class of continuous nonvanishing functions). Theorem 3.1.5. ([Lax and Zalcman, 2012, Lemma 3.2 (iv)]) Let s be a continuous complex-valued function on the unit circle T that does not vanish. Then Wind(s, 0) = 0 if and only if s has a single valued logarithm, i.e., there exists a continuous function l on T such that s(θ) = el(θ), 0 ≤ θ ≤ 2π.

Proof. It suffices to prove this for s continuously differentiable since such functions are dense in the continuous functions on T. If s = el, then

Wind(s, 0) = Z 2π 0 s0(θ) s(θ)dθ = Z 2π 0 l0(θ)dθ = l(2θ) − l(0) = 0, so s has winding number 0. On the other hand, if Wind(s, 0), we can set

l(t) = log s(0) + Z t

0

s0(θ) s(θ)dθ.

Clearly l is continuous; and since Wind(s, 0) = 0, l(2π) = l(0), i.e., l is a continuous function on T. Finally, a simple calculation shows that the derivative of the function s(t)e−l(t) is 0 on [0, 2π]. Hence s(t)e−l(t) is constant on [0, 2π]. But s(0)e−l(0)= 1, therefore

s(t)e−l(t) = 1 for 0 ≤ t ≤ 2π.

(41)

Thus s = el _{on T, as required.}

From this theorem we conclude that

exp(W ) = {f ∈ G(W ) : Wind(f, 0) = 0}.

Since f ∈ exp(W ) if and only if there is a continuous map t 7→ f_t : [0, 1] → G(W ), such that f0 ≡ 1 f1≡ f , therefore the last conclusion about exp(W ), could be done from the proof of

the next useful theorem.

Theorem 3.1.6. Within the class of continuous, complex-valued, nonvanishing functions on T, two functions can be continuously deformed into one another if and only if they have the same winding number.

Proof. We have already talked about the invariance of the winding number under continuous deformation, so we need to prove only the opposite direction of the theorem. For that, consider first the case in which the winding number of s is zero. Such a function has a single-valued logarithm log s(θ). Deform this function to zero as t log s(θ). Exponentiation yields

s(θ, t) = et log s(θ) 0 ≤ t ≤ 1, a deformation of s(θ) into the constant function 1.

Given s of winding number N , we write it as

s(θ) = eiN θ(e−iN θs(θ)).

The second factor has winding number zero and therefore can be deformed into the constant function 1. So s(θ) can be deformed into eiN θ, N = Wind(s, 0).

Since the Wiener algebra is commutative, then the commutativity property ((ab) \ {0} = (ba) \ {0} ∀a, b ∈ W ) obviously holds.

3.2 The Algebra of Bounded Linear Operators on

Infinite-Dimensional Hilbert Space

Let H be an infinite-dimensional Hilbert space.

Definition 3.2.1. We denote by L(H) the Banach algebra of all bounded linear operators T on a Hilbert space H 6= {0}, normed by

(42)

For T ∈ L(H) we denote by T∗ the adjoint operator of T . Definition 3.2.2. An operator T ∈ L(H) is said to be (a) normal if T T∗ = T∗T ,

(b) self-adjoint (or hermitian) if T = T∗,

(c) unitary if T∗T = I = T T∗, where I is the identity operator on H, (d) aprojection if T2 = T .

It is clear that self-adjoint operators and unitary operators are normal.

The next three theorems are well known in operator theory and we give them without proofs (for the proofs see for exampleRudin[1973], theorems 12.26, 12.32 and 12.33 respectively). Theorem 3.2.3. A normal operator T ∈ L(H) is

(a) self-adjoint if and only if Sp(T ) lies in the real axis, (b) unitary if and only if Sp(T ) lies on the unit circle. Theorem 3.2.4. Suppose T ∈ L(H). Then

(a) (T x, x) ≥ 0 for every x ∈ H if and only if (b) T = T∗ _{and Sp(T ) ⊂ [0, ∞).}

Definition 3.2.5. If T ∈ L(H) satisfies (a), we call T apositive operator and write T ≥ 0. Theorem 3.2.6. Every positive T ∈ L(H) has a unique positive square root S ∈ L(H). If T is invertible, so is S.

Definition 3.2.7. For T ∈ L(H), if there exist U ∈ L(H) unitary and P ∈ L(H) positive, such that T = U P , then we call U P apolar decomposition of T .

Theorem 3.2.8. If T ∈ L(H) is invertible, then T has a unique polar decomposition T = UP . Proof. If T is invertible, so are T∗ _{and T}∗_{T , and Theorem}_3.2.6_{shows that the positive square}

root P of T∗T is also invertible. Put U = T P−1. Then U is invertible, and U∗U = P−1T∗T P−1 = P−1P2P−1 = I,

so that U is unitary. Conversely, if T = U P , then T∗T = P U∗U P = P2, so necessarily P is the positive square root of T∗T , and since P is invertible, it is obvious that T P−1 is the only possible choice for U .

(43)

Theorem 3.2.9. ([Rudin,1973, Theorem 12.37]) The group G(L(H)) is connected, and every T ∈ G(L(H)) is the product of two exponentials.

Here an exponential is, of course, any operator of the form exp(S) with S ∈ L(H).

Proof. Let T = UP be the polar decomposition of some T ∈ G(L(H)). Recall that U is unitary and that P is positive and invertible. Since Sp(P ) ⊂ (0, ∞), then log is a continuous real function on a neighbourhood Sp(P ). It follows from the symbolic calculus that there is a self-adjoint S ∈ L(H) such that P = exp(S). Since U is unitary, Sp(U ) lies on the unit circle, so that there is a real bounded Borel function f on Sp(U ) that satisfies

exp{if (λ)} = λ [λ ∈ Sp(U )].

Put Q = f (U ) (Q is well defined, for the justification see [Aupetit, 1991, Theorem 6.3.4]). Then Q ∈ L(H) is self-adjoint, and U = exp(iQ). Thus

T = U P = exp(iQ) exp(S).

From this it follows easily that G(L(H)) is connected, for if T_r is defined, for 0 ≤ r ≤ 1, by Tr= exp(irQ) exp(rS)

then r → Tr is a continuous mapping of the unit interval [0, 1] into G(L(H)), T0 = I and

T1 = T . This completes the proof.

Remark 3.2.10. As G(L(H)) is connected, hence exp(L(H)) = G(L(H)), and hence the com-mutativity property for exponential spectrum is satisfied.

3.3 The Calkin Algebra

In this section we will omit the proofs of some results, because these results are well known in the theory of bounded linear operators on Hilbert space, and also because their proofs are very long. The proofs can be found for example inDouglas [1998],Lax [2002] orArveson [2002].

In this section H will be a separable infinite-dimensional Hilbert space.

Definition 3.3.1. An operator T ∈ L(H) is a finite rank operator if the dimension of the range of T is finite and a compact operator if the image of the unit ball of H under T is a relatively compact subset of H. Let LF (H), respectively, LC(H) denote the set of finite rank, respectively, compact operators.

In the definition of compact operator it is often assumed only that T (B(0, 1)) has a compact closure (B(0, 1) is the unit ball of H, with centre 0). The two conditions are equivalent.

(44)

Lemma 3.3.2. If H is an infinite-dimensional Hilbert space and T is a compact operator, then the range of T contains no closed infinite-dimensional subspace.

Theorem 3.3.3. If H is an infinite-dimensional Hilbert space, then LC(H) is the norm closure of LF(H).

Definition 3.3.4. If H is a Hilbert space, then the quotient algebra L(H)/LC(H) is a Banach algebra called theCalkin algebra. The natural homomorphism from L(H) onto L(H)/LC(H) is denoted by π.

Definition 3.3.5. If H is a Hilbert space, then T in L(H) is a Fredholm operator if π(T ) is an invertible element of L(H)/LC(H). The collection of Fredholm operators on H is denoted by F (H).

Definition 3.3.6. An operator T on the Hilbert space H is bounded below if there exists > 0 such that kT f k ≥ kf k for f in H

Remark 3.3.7. If T is bounded below, and if {T fn}∞n=1 is a Cauchy sequence in the range of

T , then the inequality

kf_n− f_mk ≤ 1

kT fn− T fmk implies {f_n}∞

n=1 is also a Cauchy sequence. Hence if f = limn→∞fn, then T f = limn→∞T fn

is in the range of T . Thus the range of T is a closed subspace of H.

Lemma 3.3.8. If H is a Hilbert space, M is a closed subspace of H, and N is a finite-dimensional subspace of H, then the linear span of M + N is a closed subspace of H.

Theorem 3.3.9. (Atkinson’s theorem). If H is a Hilbert space, then T ∈ L(H) is a Fredholm operator if and only if the range of T is closed, dim ker T is finite, and dim ker T∗ _{is finite.}

Proof. If T is a Fredholm operator, then there exist an operator A ∈ L(H) and a compact operator K such that AT = I + K. If f is a vector in the kernel of I + K, then (I + K)f = 0 implies that Kf = −f , and hence f is in the range of K. Thus,

ker T ⊂ ker AT = ker(I + K) ⊂ ran K

and therefore by Lemma3.3.2, the dimension of ker T is finite. By symmetry, as the adjoint of compact operator is compact, the dimension of ker T∗ is also finite. Moreover, by Theorem

3.3.3, there exists a finite rank operator F such that kK − F k < 1/2. Hence for f in ker F , we have

kAkkT f k ≥ kAT f k = kf + Kf k = kf + F f + Kf − F f k ≥ kf k − kKf − F f k ≥ kf k/2.

(45)

Therefore, T is bounded below on ker F , which implies that T (ker F ) is a closed subspace of H (by Remark 3.3.7). Since (ker F )⊥ is finite dimensional, it follows from Lemma 3.3.8 that ran T = T (ker F ) + T [(ker F )⊥] is a closed subspace of H.

Conversely, assume that the range of T is closed, dim ker T is finite, and dim ker T∗ is finite. The operator T₀ defined T₀f = T f from (ker T )⊥to ran T is one-to-one and onto and hence is invertible. If we define the operator S on H such that Sf = T₀−1f for f in ran T and Sf = 0 for f orthogonal to ran T , then S is bounded, ST = I − P1, and T S = I − P2, where P1 is

the projection onto ker T and P₂ is the projection onto (ran T )⊥ = ker T∗. Therefore, π(S) is the inverse of π(T ) in L(H)/LC(H), and hence T is a Fredholm operator which completes the proof.

Atkinson’s theorem describes the group of the invertible element of the Calkin algebra: G(L(H)/LC(H)) = { ˙T : T ∈ F (H)}.

Definition 3.3.10. If H is a Hilbert space, then theclassical index ind is the function defined from F (H) to Z by ind(T ) := dim ker T − dim ker T∗. For n ∈ Z, set Gn = {T ∈ F (H) :

ind(T ) = n}.

Theorem 3.3.11. ([Lax,2002, Chapter 27, Theorem 3]) Suppose that T is an operator on a Hilbert space H with finite rank, and K is a compact operator on H. Then

ind(T + K) = ind(T ).

Theorem 3.3.12. ([Douglas, 1998, Theorem 5.36]) If H is a Hilbert space, then the compo-nents of F(H) (G(L(H)/LC(H))) are precisely the sets {Gn: n ∈ Z}. Moreover, the classical

index ind is a continuous homomorphism from F(H) onto Z which is invariant under compact perturbation.

The last theorem allow us to describe the set exp(L(H)/LC(H)). As for an operator T its index is defined by ind(T ) = dim ker T − dim ker T∗, then ind(I) = 0, because I = I∗. Therefore

exp(L(H)/LC(H)) = { ˙T : T ∈ F (H), ind(T ) = 0}. Now we will discuss the commutativity property for the Calkin algebra. Theorem 3.3.13. Every compact operator T can be factored as

T = U A, (3.1)

(46)

The operator A is called theabsolute value of T , and (3.1) is called the polar decomposition of T .

When T is compact, so is its absolute value A. The nonzero eigenvalues of A, denoted as {sj}, are positive numbers that tend to zero; we index them in decreasing order. The numbers

sj are called thesingular values of the operator T , and denoted as sj(T ).

Definition 3.3.14. A compact operator T of a Hilbert space H into H is intrace class when

∞

X

1

sj(T ) < ∞.

This sum is called thetrace norm of T : kT ktr=

X sj(T ).

A bounded operator T in Hilbert space can be represented as an infinite-dimensional matrix with respect to any orthonormal basis {fn}. The (m, n)th element of this matrix is (T fn, fm).

Therefore the trace of this matrix is

X

n

(T fn, fn), (3.2)

provided that this series converges.

Theorem 3.3.15. ([Lax,2002, Chapter 30, Theorem 3]) For every trace class operator T the series (3.2) converges absolutely to a limit that is independent of the orthonormal basis chosen. It is called the trace of T , and is denoted by tr T .

Theorem 3.3.16. Let A, B be bounded operators on infinite-dimensional Hilbert space H, and suppose that I − AB is Fredholm. Then I − BA is Fredholm, and

ind(I − AB) = ind(I − BA).

In the course of the proof, we shall need the following result, which characterizes Fredholm operators in terms of partial inverses, and the index in terms of trace. The details can be found for example inMurphy [1994].

Theorem 3.3.17. Let T be a bounded operator on H. Then T is Fredholm if and only if there exists an operator S such that I − ST and I − T S are both of finite rank, and then

ind(T ) = tr(T S − ST ).

Furthermore, if such an S exists, then it may be chosen so that T ST = T .

(47)

Now we begin the proof of the Theorem 3.3.16.

Proof. By the Theorem3.3.17, since I − AB is Fredholm, there exists an operator V such that F := I − (I − AB)V and G := I − V (I − AB) are of finite rank and

(I − AB)V (I − AB) = I − AB. (3.3)

Furthermore

ind(I − AB) = tr((I − AB)V − V (I − AB)) = tr(G − F ). Define W := I + BV A. Then

I − (I − BA)W = I − (I − BA)(I + BV A) = BA + BABV A − BV A = BF A, I − W (I − BA) = I − (I + BV A)(I − BA) = BA + BV ABA − BV A = BGA, both operators of finite rank. By the Theorem3.3.17, it follows that I − BA is Fredholm, and

ind(I − BA) = tr((I − BA)W − W (I − BA)) = tr(BGA − BF A). Now, using (3.3), we have F (I − AB) = 0 and (I − AB)G = 0. Hence

G − F − BGA + BF A = (I − AB)G − F (I − AB) + [A, BG] − [F A, B] = [A, BG] − [F A, B]. As BG and F A are of finite rank, tr[A, BG] = 0 and tr[F A, B] = 0. Hence, finally, we obtain

ind(I − AB) − ind(I − BA) = tr(G − F − BGA + BF A) = tr([A, BG] − [F A, B]) = 0.

From the theorems 3.3.16 and 3.3.11, we conclude that the commutativity property holds for the Calkin algebra.

Remark 3.3.18. The last theorem was proved by Thomas Ransford and Malik Younsi but was not published. Later, independently it was proved and published inGrobler and Raubenheimer

[2008].

3.4 Banach Algebra, With a Disconnected Group of Invertible

Elements

In this section we describe another example of non-commutative Banach algebra with discon-nected group of invertible elements. The idea is to replace the Hilbert space in the example of Section 3.2 by an appropriately chosen Banach space.

Commutativity of the exponential spectrum