Commutativity of the Exponential Spectrum

Aram Gevorgyan

Maîtrise en mathématiques

Québec, Canada

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.

iii

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 exponential 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.

Contents

Résumé iii

Abstract v

Contents vii

Acknowledgments xiii

1 Banach Algebras, Spectrum and Exponential Spectrum1 1.1 Banach Algebras, Deﬁnitions and Examples...... 1 1.2 Invertible Elements and Spectrum ...... 4 1.3 Holomorphic Functional Calculus...... 9 1.4 Exponential Spectrum...... 13

2 Commutativity of Exponential Spectrum: Positive Results 17 2.1 Special Case, When at Least One of the Two Elements is a Limit of Invertible Elements ...... 17 2.2 Other Special Cases ...... 19

3 Examples of Exponential Spectra 23 3.1 The Wiener Algebra ...... 23 3.2 The Algebra of Bounded Linear Operators on Inﬁnite-Dimensional Hilbert Space 27 3.3 The Calkin Algebra...... 29 3.4 Banach Algebra, With a Disconnected Group of Invertible Elements ...... 33

Conclusion 37

Bibliography 39

vii

To my family

God used beautiful mathematics in creating the world.

Paul Dirac (1902-1984)

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.

Aram Gevorgyan

xiii

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.

1 Remark 1.1.2. If A is a Banach algebra without unit, it is always possible to embed it isometrically 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 n elements then C(K) = C 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 Z (f ∗ g)(x) = f(xy−1)g(y)dµ(y) G 1 R and the norm by the L -norm kfk1 = G |f(x)|dµ(x). For instance if we consider the additive 1 1 group L (Z) = ` (Z), then we have X (f ∗ g)(n) = f(n − k)g(k) k∈Z X kfk1 = |f(k)|. k∈Z

1 In fact, this algebra ` (Z) can be identiﬁed with the Wiener algebra W of continuous functions P int P f(t) = n∈Z ane on [0, 2π] having Fourier coeﬃcients an such that n∈Z |an| < +∞, with P the operations of pointwise addition and multiplication and the norm kfk = n∈Z |an|.

2 Example 1.1.6. The Banach space of integrable functions on [0, 1] with multiplication deﬁned R x 1 R 1 by convolution (f ∗ g)(x) = 0 f(x − t)g(t)dt and L -norm kfk1 = 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 Mn(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 subalgebra of L(X) is also a Banach algebra. For instance, the 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 Q algebras we can deﬁne the Banach algebra product i∈A Ai as the set of all families (xi)i∈I such that xi ∈ Ai and supi∈I kxik < +∞ with the norm kxk = supi∈I kxik. It is easy to Q verify that i∈I Ai is a Banach algebra which contains an isometric copy of each Ai0 (by the mapping x → (xi)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), n X (aijbij) = ( aikbkj) k=1 and the norm kaijk = supi=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. With this norm it is easy to verify that A/I is a Banach algebra, called the quotient algebra

3 of A by the two-sided ideal I. 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˙||| = |||1˙2||| ≤ |||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 the Calkin 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 ∞ X (1 − x)−1 = xk. k=0

Pn k Proof. Let kxk = r < 1 and let sn = k=0 x . Obviously, for n < m, we have m X k n+1 ksn − smk ≤ kxk ≤ r /(1 − r), k=n+1

P∞ k so (sn) is a Cauchy sequence converging to some element a = k=0 x . 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−1 is a homeomorphism from G(A) onto G(A).

4 Proof. We have x = a + x − a = a(1 + a−1(x − a)). Now ka−1(x − a)k ≤ kx − ak · ka−1k < 1 so, by the previous theorem, 1 + a−1(x − a) is invertible, and consequently x is invertible. −1 −1 −1 −1 P∞ −1 k −1 Moreover x = (1 + a (x − a)) a = k=0(a (a − x)) a . Consequently, we have

∞ ∞ X X kx−1 − a−1k = (a−1(a − x))ka−1 ≤ ka−1k2kx − ak (ka−1k · kx − ak)k. k=1 k=0

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 1.2.2 implies 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

5 −1 1. λ 7→ (λ1 − x) is analytic on C \ Sp x and goes to 0 at inﬁnity,

2. Sp x is compact and non-empty,

3. ρ(x) = lim kxnk1/n. n→∞

Proof. We know that ρ(x) ≤ kxk. Moreover Sp x is closed by Theorem 1.2.3, because λ0 ∈/ Sp x −1 and |λ − λ0| < 1/k(λ01 − x) k implies λ∈ / Sp x. So Sp x is compact. Again by Theorem −1 1.2.3, λ → R(λ) = (λ1 − x) 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 R(µ)−R(λ) = −R(λ)2 by continuity of R. So R is analytic µ→λ µ−λ −1 1 −1 on C \ Sp x. Moreover (λ1 − x) = λ (1 − x/λ) for λ∈ / Sp x. So, for |λ| ≥ 2kxk, we have by Theorem 1.2.2 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 part 1, 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 1.2.6 applied to zn, we conclude that ρ(x)n ≤ kxnk. Let f ∈ A0. 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 Theorem 1.2.2, we have 1 f(x) f(xn) f((λ1 − x)−1) = (f(1) + + ··· + + ··· ). λ λ λn Consequently this is also true for |λ| > ρ(x) by the Identity Principle. Let λ be fixed such that 0 n n |λ| > ρ(x). Then for every f ∈ A we have supn |f(x )/λ | < +∞. By the Banach-Steinhaus 0 0 n n Theorem applied to A and to the sequence of Tn : A → C, defined by Tn(f) = f(x )/λ , we conclude that there exists a constant C, depending on λ, such that kxnk ≤ C|λ|n for all n ≥ 1. Then lim sup kxnk1/n ≤ |λ| for all |λ| ≥ ρ(x). n→∞ So finally, using the last inequality with ρ(x)n ≤ kxnk, we get:

ρ(x) ≤ lim inf kxnk1/n ≤ lim sup kxnk1/n ≤ ρ(x), n→∞ n→∞ 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.

6 Proof. Let x ∈ A. By part 2 of Theorem 1.2.7, 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 = xnyn for every integer n ≥ 1, using part 3 of Theorem 1.2.7 we conclude that

ρ(xy) = lim k(xy)nk1/n ≤ lim kxnk1/n lim kynk1/n = ρ(x)ρ(y). n→∞ n→∞ n→∞

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(ka2n k, kb2n k) < 1. Deﬁning

k 2n−k γn = max ka k · kb k, 0≤k≤2n we have

2n 2n 2n 1/2n X 2n k 2n−k 1/2n X 2n k 2n−k k 2n−k 1/2n 1/2n k(x + y) k = k (k )x y k ≤ [ (k )α β ka kkb k] ≤ (α + β)γn . k=0 k=0

But it is easy to see that the sequence (γn) is decreasing for n ≥ N, because

n+1 n+1 n+1 max kakkkb2 −kk = max( max kakkkb2 −kk, max kb2 −kkkakk) 0≤k≤2n+1 0≤k≤2n 2n≤k≤2n+1 2n 2n ≤ γn max(ka k, kb k)

So we have

2n 1/2n 1/2n 1/2n ρ(x + y) = lim k(x + y) k ≤ (α + β) lim sup γn ≤ (α + β) lim sup γN = α + β, n→∞ n→∞ n→∞ 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

SpB x, and the spectrum of x relative to A, denoted by SpA x? Of course, we have SpA x ⊂

SpB x, but is it possible to say more?

Theorem 1.2.10. Let (xn) be a sequence of invertible elements of A converging to a noninvertible element. Then lim kx−1k = +∞. n→∞ n

7 Proof. Suppose the result is false. Then there exist C > 0 and a subsequence of (xn), which −1 we denote in the same way, such that kxn k ≤ C. Let x be the limit of the sequence (xn). −1 −1 We have x = xn(1 + xn (x − xn)) and kxn (x − xn)k ≤ Ckx − xnk < 1 for n large enough −1 so, by Theorem 1.2.2, 1 + xn (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 is topological 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 −1 −1 with αn ∈/ Sp x. We take xn = (x − αn1) /k(x − αn1) k. Then we have

−1 (x − α1)xn = (x − αn1)xn − (α − αn)xn = 1/k(x − αn1) k − (α − αn)xn.

Consequently −1 k(x − α1)xnk ≤ 1/k(x − αn1) k + |α − αn|.

−1 By Theorem 1.2.10, lim k(x − αn1) k = +∞, so lim k(x − α1)xnk = 0. The other result n→∞ n→∞ 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 −1 −1 −1 and xxn → 0. If x is invertible, i.e. x x = xx = 1, then we have xn = 1xn = x xxn = −1 x (xxn) → 0 therefore xn → 0, a contradiction.

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

So, in fact with Lemma 1.2.13 and Corollary 1.2.12 we 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,

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

components of C \ SpA x. In particular ∂ SpB x ⊂ ∂ SpA x.

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, −1 with xn ∈ G(B), exists and is in G(A), by continuity of x 7→ x on G(A) we conclude −1 −1 −1 that x = lim xn . In particular (kxn k) is bounded so, by Theorem 1.2.10, 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 of G(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 part 1 is proved.

2 Let ΩA = C \ SpA x and ΩB = C \ SpB x. Obviously ΩB ⊂ ΩA. A similar argument shows that ∂ΩB ∩ ΩA = ∅, 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 SpA x does not separate the complex plane. Then SpB x = SpA x.

1.3 Holomorphic Functional Calculus

n If x is in a Banach algebra A and if p(λ) = α0 + α1λ + ··· + αnλ is a polynomial with n complex coefficients, we can define without any problem p(x) = α01 + α1x + ··· + αnx . 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?

P∞ k Suppose that f is holomorphic on the disk |λ| < R and ρ(x) < R, so that f(λ) = k=0 αkλ P∞ k P∞ k for |λ| < R. This implies in particular that k=0 αkkx k converges, so that f(x) = k=0 αkx 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 R R R to deﬁne K f(λ)dµ(λ), and of course we have φ( K f(λ)dµ(λ)) = K φ(f(λ))dµ(λ) for all bounded linear functionals φ on A (see for instance Rudin[1987], pages 73-78).

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

0 By the Hahn-Banach theorem there exists φ ∈ A such that φ(x1) 6= φ(x2). We have 1 Z 1 Z φ(x1) = h(λ)dλ, φ(x2) = h(λ)dλ 2πi Γ1 2πi Γ2 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 1 Z f(x) = f(λ)(λ1 − x)−1dλ 2πi Γ 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 1 R −1 rational function having no poles surrounded by Γ. Then r(x) = 2πi Γ r(λ)(λ1 − x) dλ.

n Proof. We ﬁrst prove this for r(λ) = (α − λ) , with n ∈ Z, and α not surrounded by Γ. Let Z 1 −1 rn = r(λ)(λ1 − x) dλ. 2πi Γ 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 Z 1 −1 n −1 rn = (α1 − x) (α − λ) dλ + (α1 − x) rn+1. 2πi Γ

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

rn+1 = (α1 − x)rn.

n Consequently the formula will be proved for (α − λ) , n ∈ Z, if we succeed in proving it for n = 0. We have Z Z Z 1 −1 1 −1 1 1 x (λ1 − x) dλ = (λ1 − x) dλ = ( + 2 + ... )dλ 2πi Γ 2πi ΓR 2πi |λ|=R λ λ for R > kxk. So 1 Z (λ1 − x)−1dλ = 1. 2πi Γ Thus the ﬁrst part is proved. Now if r is an arbitrary rational function it can be written as

a1,1 a1,k1 an,1 an,kn r(λ) = p(λ) + + ··· + k + ··· + + ··· + k α1 − λ (α1 − λ) 1 αn − λ (αn − λ) n 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

−1 −k1 −1 r(x) = p(x) + a1,1(α11 − x) + ··· + a1,k1 (α11 − x) + ··· + an,1(αn1 − x) + ...

−kn ··· + an,kn (αn1 − x) .

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, . . . , αk be the diﬀerent eigenvalues of a and let Γ be the union of k small disjoint circles with centres respectively at α1, . . . , αk. By Lemma 1.3.1 we have 1 Z p(a) = p(λ)(λ1 − a)−1dλ 2πi Γ

−1 1 But (a − λ1) = det(a−λ1) 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 1 Z p(a) = − b(λ)dλ = 0 2πi Γ 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

11 is an arbitrary smooth contour included in Ω and surrounding Sp x and not surrounding any point of C \ Ω. Then the mapping 1 Z f 7→ f(x) = f(λ)(λ1 − x)−1dλ 2πi Γ 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 −1 easy because (fn) converges uniformly to f on Γ and k(λ1 − x) k is bounded on this set. We 1 now prove (ii). By Runge’s theorem there exist two sequences of rational functions (rk) and 2 (rk), with poles not surrounded by Γ, converging uniformly respectively to f1 an f2. By Lemma 1 2 1 2 1.3.1 we have (rkrk)(x) = rk(x)rk(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 2 2 and f(λ) = 1 on U1. We set p = f(x). Because f(λ) = f(λ), we get p = 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) = (fI)(x). But (fI)(λ) =

12 λ on U1 and (fI)(λ) = 0 on U0, so by Theorem 1.3.3 (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 1 dist(α, Sp x) = . 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 Theorem 1.3.3 (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

x xn of exponentials e 1 . . . e , where x1, . . . , 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 . . . etxn is a continuous function from [0, 1]

x xn into G(A) which connects 1 and e 1 . . . e . So in fact exp(A) is included in G1(A). We now prove that the inverse is also true.

Theorem 1.4.3. If A is a Banach algebra we have exp(A) = G1(A).

13 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). −1 −1 If an ∈ exp(A) and limn→∞ an = a ∈ G(A) then (an a) converges to 1 and so k1 − an ak < 1 −1 zn for n large. As before, we conclude that an a = e for some zn ∈ 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 y Theorem 1.3.6 there exists y such that λ01 − x = e , 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

−1 |sn − s|k(x − sn1) k ≥ 1, ∀n ∈ N.

If not, then for some n the element

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

−1 −1 bn = (x − sn1) /k(x − sn1) k.

We have kbnk = 1 and

−1 −1 −1 k(x−s1)bnk = k1−(s−sn)(x−sn1) k/k(x−sn1) k ≤ 1/k(x−sn1) k+|s−sn| ≤ 2|s−sn| → 0, giving s ∈ τ(x).

14 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) = (x + y), Sp T x ⊂ Sp(x + y) ⊂ (Sp\T x). y∈ker T y∈ker T

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

Proof. It is clear that T (ex1 . . . exn ) = eT x1 . . . eT 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 \ (x + y) ⊂ (T x). y∈ker T

The other inclusion is obvious. Moreover \ \ Sp(x + y) ⊂ (x + y) = (T x) ⊂ (Sp\T x) y∈ker T y∈ker T by the previous remark. The last part follows immediately, taking B = A/I and T x =x ˙.

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 consequently 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.

Lemma 2.1.2. If a ∈ G(A) and b ∈ exp(A) then aba−1 ∈ exp(A).

17 Proof. If b ∈ exp(A) ⇔ ∃ continuous map t 7→ [0, 1] → exp(A) such that b0 = 1 and b1 = b. −1 −1 −1 −1 −1 Hence ab0a = a1a = 1 and ab1a = aba . Therefore aba ∈ 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 an belong 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 −1 suﬃciently large n. Since 1−ban = an (1−anb)an, by Lemma 2.1.2, 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 the topological 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. The centre 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

18 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 − ban ∈ 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 S S then it is easy to see that either I∈C I is a left ideal or I∈C I = 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.

19 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 is semi-simple.

There are many examples of semi-simple Banach algebras, for instance Examples 1.1.3 and 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 ∈ A the 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˙ ∈ L0} 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 Theorem 2.2.2, 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 = Spx ˙, 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 Theorem 1.2.2. Consequently L ∩ B(1,I) = ∅ and also L 6= A. This implies that L = L if L is maximal. By Theorem 2.2.2 (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 of quasinilpotent 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.

20 Deﬁnition 2.2.8. An ideal I in A is called inessential 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−1r). 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}.

21 Proof. If a is Riesz relative to I then, by Theorem 2.2.12, 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 Theorem 2.1.3.

22 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 homomorphisms 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 xˆ : SPg(A) → C by way of xˆ(ω) = ω(x), ω ∈ SPg(A); xˆ is called the Gelfand transform of x, and x → xˆ is called the Gelfand map. The functions xˆ are continuous by deﬁnition of the weak∗-topology on SPg(A). For x, y ∈ A we have

xˆ(ω)ˆy(ω) = ω(x)ω(y) = ω(xy) = xyc(ω). Moreover, since every element ω of SPg(A) satisﬁes ω(1) = 1, it follows that 1ˆ 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).

23 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) = {xˆ(p): p ∈ SPg(A)}.

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

Indeed, if x is invertible, then there is a y ∈ A such that xy = 1; hence xˆ(ω)ˆy(ω) = xyc(ω) = 1, for ω ∈ SPg(A), so that xˆ 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 xˆ 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 ∞ iθ X inθ f(e ) ∼ ane , n=−∞ P where n |an| < ∞. This algebra is a subalgebra of C(T), and obviously contains the constant 1 functions. And as we mentioned in chapter one, this algebra can be identiﬁed with ` (Z) = 1 L (Z).

iθ iθ Deﬁne B(W ) = {f ∈ W : f(e ) 6= 0, ∀e ∈ 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 .

1 Proof. Consider the Banach algebra A = ` (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.

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

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 ωλ 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. n n n n Indeed, we must have ω(ζ ) = λ = ωλ(ζ ) for every n ∈ Z, ζ being the unit sequence with n a single nonzero component in the nth position. Since the set {ζ : n ∈ Z} obviously has 1 ` (Z) as its closed linear span, it follows that ω = ωλ. Then λ = ω(ζ) is obviously uniquely determined by ω.

These remarks show that the map λ 7→ ωλ 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 1 T and the Gelfand map with the Fourier transform, which carries a sequence a ∈ ` (Z) to the function aˆ ∈ C(T) given by ∞ iθ X inθ aˆ(e ) = ane . n=−∞

Having computed SP g(A) and the Gelfand map in concrete terms, we observe that the range of the Gelfand map {aˆ : 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 1 of A = ` (Z) having Gelfand transform f. By Theorem 3.1.2, there is an element b ∈ A such that a ∗ b = 1; hence aˆ(λ)ˆb(λ) = 1, for λ ∈ T. It follows that 1/f = ˆb ∈ W , as asserted.

iθ iθ Therefore G(W ) = {f ∈ W : f(e ) 6= 0, ∀e ∈ 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 complement of the image of γ (relative to the plane), and deﬁne 1 Z dζ Wind(γ, z) = (z ∈ Ω). 2πi γ ζ − z

25 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 Z t 0 t s (θ) log s(θ)|0 = dθ 0 s(θ) for s continuously differentiable, for such functions this can be expressed analytically as 1 Z 2π s0(θ) 1 Z 2π s0(θ) Wind(s, 0) = dθ = = dθ. 2πi 0 s(θ) 2π 0 s(θ) 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 l in the continuous functions on T. If s = e , then Z 2π s0(θ) Z 2π Wind(s, 0) = dθ = l0(θ)dθ = l(2θ) − l(0) = 0, 0 s(θ) 0 so s has winding number 0. On the other hand, if Wind(s, 0), we can set Z t s0(θ) l(t) = log s(0) + dθ. 0 s(θ) Clearly l is continuous; and since Wind(s, 0) = 0, l(2π) = l(0), i.e., l is a continuous function −l(t) on T. Finally, a simple calculation shows that the derivative of the function s(t)e 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.

26 l Thus s = e 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→ ft : [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) = eiNθ(e−iNθs(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 eiNθ, 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

kT k = sup{kT xk : x ∈ H, kxk ≤ 1}.

27 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 TT ∗ = T ∗T ,

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

(d)a projection if T 2 = 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 example Rudin[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 a positive 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 = UP , then we call UP a polar 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 = TP −1. Then U is invertible, and

U ∗U = P −1T ∗TP −1 = P −1P 2P −1 = I, so that U is unitary. Conversely, if T = UP , then T ∗T = PU ∗UP = P 2, so necessarily P is the positive square root of T ∗T , and since P is invertible, it is obvious that TP −1 is the only possible choice for U.

28 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 = UP = exp(iQ) exp(S).

From this it follows easily that G(L(H)) is connected, for if Tr 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 commutativity 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 in Douglas[1998], Lax[2002] or Arveson[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.

29 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 the Calkin 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 fk ≥ kfk 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 1 kf − f k ≤ kT f − T f k n m n m ∞ implies {fn}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 Lemma 3.3.2, 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 fk ≥ kAT fk = kf + Kfk = kf + F f + Kf − F fk ≥ kfk − kKf − F fk ≥ kfk/2.

30 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 −1 invertible. If we define the operator S on H such that Sf = T0 f for f in ran T and Sf = 0 for f orthogonal to ran T , then S is bounded, ST = I − P1, and TS = 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 the classical 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 components 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 = UA, (3.1)

where A is a positive symmetric operator, and U ∗U = I on the range of A.

31 The operator A is called the absolute 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 the singular 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 in trace class when

∞ X sj(T ) < ∞. 1

This sum is called the trace norm of T : X kT ktr = 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 (T fn, fn), (3.2) n 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 in Murphy[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 − TS are both of ﬁnite rank, and then

ind(T ) = tr(TS − ST ).

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

32 Now we begin the proof of the Theorem 3.3.16.

Proof. By the Theorem 3.3.17, 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 + BVA. Then

I − (I − BA)W = I − (I − BA)(I + BVA) = BA + BABV A − BVA = BF A,

I − W (I − BA) = I − (I + BVA)(I − BA) = BA + BV ABA − BVA = BGA, both operators of ﬁnite rank. By the Theorem 3.3.17, it follows that I − BA is Fredholm, and

ind(I − BA) = tr((I − BA)W − W (I − BA)) = tr(BGA − BFA).

Now, using (3.3), we have F (I − AB) = 0 and (I − AB)G = 0. Hence

G − F − BGA + BFA = (I − AB)G − F (I − AB) + [A, BG] − [F A, B] = [A, BG] − [F A, B].

As BG and FA 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 + BFA) = 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 in Grobler 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 disconnected 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.

33 Theorem 3.4.1. (Pitt’s theorem). Let 1 ≤ q < p < ∞. Then every bounded linear operator from `p into `q is compact.

The following proof is taken from Delpech[2009].

Proof. Let T : `p → `q be a norm-one operator. As 1 < p, the dual of `p is separable. Hence every bounded sequence in `p has a weakly Cauchy subsequence. Thus, for proving the compactness of T , it is enough to show that T is weak-to-norm continuous. So, consider a p weakly null sequence (hn) in ` . We have to show that limn→∞ kT (hn)k = 0. We claim that r r for every x ∈ ` , 1 ≤ r < ∞, and for every weakly null sequence (ωn) in ` ,

r r r lim sup kx + ωnk = kxk + lim sup kωnk . (3.4) n→∞ n→∞

Indeed this is obvious when x is ﬁnitely supported, because the coordinates of (ωn) along the support of x tend to 0 in norm. The general case is true by the density of ﬁnitely supported elements in `r and since the norm is a Lipschitzian function.

p Fix 0 < < 1. By deﬁnition of the norm of T , there exists x ∈ ` such that kxk = 1 and 1 − ≤ kT (x)k ≤ 1. Moreover, for all n ∈ N and for all t > 0

kT (x) + T (thn)k ≤ kx + thnk. (3.5)

q In the left side of (3.5), we apply (3.4) in ` , with x = T (x) and the weakly null sequence

(T (thn)).

We apply (3.4) to the right side of (3.5) with r = p, x = x and the weakly null sequence

(thn) to obtain

1/q 1/p n q q qo n p p po kT (x)k + t lim sup kT (hn)k ≤ kxk + t lim sup khnk . n→∞ n→∞

Recall that kxk = 1, 1 − ≤ kT (x)k ≤ 1 and that (hn) is weakly convergent, thus bounded by some M > 0. This gives

q 1 n p p q/p qo lim sup kT (hn)k ≤ q (1 + t M ) − (1 − ) . n→∞ t Taking t = 1/p here, we get

q 1 n q p o lim sup kT (hn)k ≤ 1 + M − (1 − q) + o() . n→∞ q/p p q Now, letting → 0 here, we get that lim supn→∞ kT (hn)k ≤ 0, and therefore the sequence (T (hn)) norm-converges to 0.

34 For ﬁxed 1 ≤ q < p < ∞ denote E := `p ⊕ `q. Now we construct an example of a bounded linear invertible operator T : E → E, such that T/∈ exp(L(E)). This example is due to Douady[1965]. ! AB Deﬁne T : E → E to be T = , where the operators A, B and D are as follows: 0 D

p p A : ` → ` is the right shift operator: A :(x1, x2,... ) 7→ (0, x1, x2,... ),

q p B : ` → ` is the projection onto ﬁrst coordinate: B :(x1, x2,... ) 7→ (x1, 0, 0,... ), and

q q D : ` → ` is the left shift operator: D :(x1, x2, x3,... ) 7→ (x2, x3,... ).

Proposition 3.4.2. The operator T is invertible in L(E).

Proof. We will prove the invertibility of T , by proving that T is bijective.

p q p 1.T is surjective. Let (x, y) ∈ E = ` ⊕` , where x = (x1, x2,... ) ∈ ` and y = (y1, y2,... ) ∈ q p q ` . Set u := (x2, x3,... ) ∈ ` and v := (x1, y1, y2,... ) ∈ ` . Calculate ! ! ! ! ! AB u Au + Bv (0, x , x ,... ) + (x , 0, 0,... ) x = = 2 3 1 = . 0 D v Dv (y1, y2,... ) y

Hence T is surjective.

p 1 q 2. T is injective. Let (x, y) ∈ E, where x = (x1, x2,... ) ∈ ` , and y = (y1, y2,... ) ∈ ` . ! x Assume that T = 0. y

! ! ! ! ! ! x AB x Ax + By (y , x , x ,... ) 0 T = = = 1 1 2 = . y 0 D y Dy (y2, y3,... ) 0

Therefore,  ! ! y1 = 0, x1 = 0, x2 = 0,... x 0 ⇒ = . y2 = 0, y3 = 0,... y 0 Hence, T is injective, consequently bijective and consequently invertible.

Proposition 3.4.3. T/∈ exp(L(E)).

Proof. Suppose that T ∈ exp(L(E)). Then there is continuous path t 7→ Tt : [0, 1] → G(L(E)) such that T0 = I and T1 = T . Write " # At Bt Tt = , Ct Dt

35 p p q p p q q q where t 7→ At,Bt,Ct,Dt are continuous maps from [0, 1] into L(` , ` ), L(` , ` ), L(` , ` ), L(` , ` ) respectively. Consider the Calkin algebra L(E)/LC(E). The coset of Tt in the Calkin algebra has the form " # A˙t B˙t T˙t = , C˙t D˙ t

p p p p q p q p where A˙t, B˙t, C˙t, and D˙ t lie in the quotient spaces L(` , ` )/LC(` , ` ), L(` , ` )/LC(` , ` ), L(`p, `q)/LC(`p, `q), and in L(`q, `q)/LC(`q, `q) respectively. As p > q, by Pitt’s theorem p q p q p q p q L(` , ` ) = LC(` , ` ). So the quotient L(` , ` )/LC(` , ` ) = {0}. So C˙t = 0 for all 0 ≤ t ≤ 1.

Since Tt is invertible in L(E), there exists an operators W, X, Y, Z such that " #" # " # " #" # WX A B I 0 A B WX t t = = t t . YZ Ct Dt 0 I Ct Dt YZ

Further, since Y ∈ L(`p, `q), it is compact. Hence in the Calkin algebra, we have Y˙ = 0, and so " #−1 " # A˙ B˙ W˙ X˙ t t = . 0 D˙ t 0 Z˙ From where, particularly we have  A˙tW˙ = I˙ . W˙ A˙t = I˙

p p p Hence A˙t is invertible in L(` )/LC(` ). Therefore At is Fredholm in L(` ), for all t ∈ [0, 1]. As t 7→ ind(At) is a continuous integer-valued function on [0, 1], it is constant. Hence ind A1 = ind A0. But ind A1 = ind A = −1 and ind A0 = ind I = 0, which is contradiction.

36 Conclusion

We began this mémoire by deﬁning a Banach algebra over C, its group of invertible elements, the spectrum and the exponential spectrum for the elements of the algebra. Then we described some examples of Banach algebras.

In the second chapter, we explored the commutativity property for the exponential spectrum of a Banach algebra and established some positive results.

In the last, third chapter we studied some examples of Banach algebras, describing their group of invertible elements, the component of the group which contains the unit element, and we discussed the commutativity property for the exponential spectrum. In the ﬁrst example we demonstrated a commutative Banach algebra with not connected group of invertible elements, and since the algebra is commutative, hence the commutativity property for the exponential spectrum holds. In the second example, although the algebra is non-commutative its group of invertible elements is connected, so the commutativity property for the exponential spectrum holds because the usual and exponential spectra coincide. In the third example the algebra is non-commutative and its group of invertible elements is not connected, but we proved that the commutativity property holds for the exponential spectrum. In the last example also we exhibited a non-commutative Banach algebra with disconnected group of invertible elements. This algebra is a candidate as a counterexample for the commutativity property of the exponential spectra. However, this is still an open question.

Remark 3.4.4. After the dépôt initial of this memoir, a counterexample to the commutativity of exponential spectra was found by H. Klaja and T. Ransford (preprint in preparation).

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