onumak essay 1

7/29/2019 OnumaK Essay 1

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Linear Mappings Between Banach

Algebras That Preserve Spectral

Properties

by

Kenneth Emeka Onuma

A project

presented to the University of Waterloo

in fulfillment of the

research project requirement for the degree of

Masters of Mathematics

in

Pure Mathematics

Waterloo, Ontario, Canada 2011

c Kenneth Emeka Onuma 2011


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Authors Declaration

I hereby declare that I am the sole author of this research project. This is

a true copy of the project, including any required final revisions, as accepted

by my examiners.

I understand that my project may be made electronically available to the

public.

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Abstract

In order to build an algebraic model of a given physical system, mathemati-

cians often begin by considering the geometric properties of the given system.

A fundamental question in physics asks to what extent the algebraic model is

predetermined by the geometric data obtained from the system. In a quan-

tum mechanical system, the geometric data are the measurements, that is, the

spectral values of the observables, which are represented by self-adjoint opera-

tors on some Hilbert space. Following the approach by Jordan, von Neumann

and Wigner, the mathematical model is one of a Jordan algebra. Therefore,

the question reads in this case: if all the spectral values of the observables

are known, will the algebraic model be uniquely determined up to Jordan

isomorphism?

In 1970, Kaplansky reformulated the above question mathematically as fol-lows. If : A B is a unital, surjective invertibility preserving linear mapbetween semisimple unital complex Banach algebras A and B, is necessarily

a Jordan homomorphism? This question remains open but there are a number

of partial results in the literature, especially in the case where is spectrum

preserving. In this essay, we discuss three of these results. The first one shows

that a surjective spectrum preserving linear map between von Neumann alge-

bras must be a Jordan homomorphism. The second result gives an affirmativeanswer to Kaplanskys question in the case where A = B(X) and B = B(Y),

with X and Y Banach spaces. The third result shows that a surjective spectral

isometry between finite-dimensional semisimple Banach algebras is a Jordan

homomorphism. None of them answers the question completely, but they are

all motivating results on which further work is being done and which hopefully

may lead to the eventual solution of the problem.

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Acknowledgements

I am profoundly grateful to my project supervisor, Professor L.W Marcoux,

for his inexhaustible patience and encouragement to me by way of advice,

suggestions and guidance which ultimately made the completion of this project

possible. I am also indebted to all the staffand students of the Department Of

Pure Mathematics, University of Waterloo. I owe the successful completion of

this program to you all!

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Dedication

To Osita Onuma,

for all your support to me

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Chapter 1

Definitions and Examples.

In this first chapter, we give a number of definitions of terms which we shall

use throughout this essay. There are also examples as well as brief discussions

illustrating some of the definitions. We begin with the following.

Definition 1.1 By a Banach algebra A, we shall mean a pair(A, ) con-sisting of a C-algebra A and a norm on the vector space A which is com-plete, such that

xy x y

for all x, y A. If A is unital (in the sense that it has a multiplicativeidentity, say e), we also require that

e

= 1. The centre of A is the set

Z(A) = {x A : xy = yx for all y A}.

If A = Z(A), then A is a commutative algebra.

We point out at this juncture that in this essay all vector spaces are over

the complex field C, and that all rings and Banach algebras will be assumed

unital.

1


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CHAPTER 1. DEFINITIONS AND EXAMPLES. 2

Example 1.2 Let X be a compact Hausdorff space and denote by C(X) the

set of all continuous complex-valued functions on X. Forf, g C(X) andk C, define f + g, kf, f g C(X) by

(i) (f + g)(x) = f(x) + g(x);

(ii) (kf)(x) = kf(x); and

(iii) (f g)(x) = f(x)g(x),

forx X. With these operationsC(X) becomes a commutative unital algebra.If we define : C(X) R by

f := sup {|f(x)| : x X} ,

thenf g f g for all f, g C(X). Thus the pair (C(X), ) isa Banach algebra.

The next example will feature prominently in this essay.

Example 1.3 Let X be a Banach space and let B(X) denote the set of all

bounded linear operators on X. Multiplication on B(X) is defined by compo-

sition. If B(X) is equipped with the operator norm:

T = sup {T(x) : x X, x 1}

for T B(X), we see that B(X) is a Banach algebra. It is unital, with unite = I, the identity operator on X. Recall that B(X) is not commutative if

dim X 2.

Definition 1.4 If A and B are algebras (resp. rings), then we will call a

linear (resp. additive) map

: A B


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a homomorphism if

(xy) = (x)(y)

for every x, y A. We call an anti-homomorphismif

(xy) = (y)(x)

for every x, y A. Finally, is called a Jordan homomorphism if

(xy + yx) = (x)(y) + (y)(x)

for every x, y A.

Observe that the definition of a Jordan homomorphism given above is

equivalent to the statement that (x2) = (x)2 for every x A.

Definition 1.5 Let A be a Banach algebra. An involution on A is a map : A A satisfying the following properties:

(i) (a) = a for all a A;

(ii) (a + b) = a + b for all a, b A and all, C;

(iii) (ab) = ba for all a, b A.

The pair (A, ) is called an involutive Banach algebra. If L A has theproperty that x L whenever x L, then we say that L is self-adjoint. IfA and B are involutive Banach algebras and : A B is a homomorphismsatisfying

(x) = (x)

for all x A, we say that is a -homomorphism.


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If A and B are Banach algebras with identities 1A and 1B respectively, we

can form the direct sum A B of A and B, which is a unital Banach algebrawith product

(a1, b1) (a2, b2) = (a1a2, b1b2)

and norm

(a, b) = max{aA , bB}.

The identity ofA B is 1 = (1A, 1B). It follows that for any a A and b B,

AB ((a, b)) = A(a) B(b),

and hence

rAB ((a, b)) = max{rA(a), rB(b)}.

Definition 1.10 If A is a unital ring, the radical (or more exactly, the Ja-

cobson radical) rad(A) of A is defined as the intersection of all the maximal

left ideals of A. If rad(A) = {0}, we say that A is semisimple.

For the following basic facts concerning the Jacobson radical, we refer to

[Rot02, p. 544].

Proposition 1.11 Let A be a unital ring. Then:

(a) rad(A) is the intersection of all the maximal right ideals of A. As such,

rad(A) is a two-sided ideal of A.

(b) A/rad(A) is semisimple. Moreover, for x A, if x A/rad(A) de-notes the image of x under the quotient map A A/rad(A), then x isinvertible in A if and only if x is invertible in A/rad(A).

(c) For x A, the following are equivalent:


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(i) x rad(A).

(ii) 1 ax A1 for all a A.

(iii) 1 xa A1 for all a A.

Corollary 1.12 LetA be a unital Banach algebra. Then

rad(A) = {x

A : r(ax) = 0 for all a

A}

= {x A : r(xa) = 0 for all a A}.

Moreover, let A = A/rad(A), and suppose that x A has image x A underthe quotient map. ThenA(x) = A(x) and rA(x) = rA(x).

Proof. We begin with the proof of the equality concerning rad(A). By sym-

metry, it is enough to verify the first equality. If x rad(A), a A and

0 = C, then since 1 1ax A1 by Proposition 1.11 (c), it followsthat / (ax). Hence (ax) = {0} for all a A. Conversely, if x has theproperty that r(ax) = 0 for all a A, then in particular, 1 / (ax) and so1 ax A1 for all a A. By Proposition 1.11 again, x rad(A).

As for the last claim, let C. Then by Proposition 1.11 (b), 1A x isinvertible if and only if1A x is invertible. In other words, A(x) = A(x),which in turn leads to rA(x) = rA(x).

Lemma 1.13 LetA be a Banach algebra. Then

rad(Z(A)) = Z(A) rad(A).

In particular, if A is semisimple, then so is Z(A).


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Proof. First, let x Z(A) rad(A). Then by Proposition 1.11, 1 ax A1

for all a A. In particular, if a Z(A), then 1 ax Z(A) and so(1 ax)1 Z(A). By Proposition 1.11 again, x rad(Z(A)). Hence,Z(A) rad(A) rad(Z(A)).

Conversely, let x rad(Z(A)). Then x Z(A), and r(x) = 0 by Corollary1.12. For any a A, since xa = ax, we have (ax)n = anxn for all n. By the

spectral radius formula,

r(ax) = l imn

(ax)n 1n = limn

anxn 1n limn

an 1n xn 1n = r(a)r(x).

This implies that r(ax) = 0 for all a A. By Corollary 1.12, x rad(A). Butx Z(A) as well. Therefore, we have rad (Z(A)) Z(A) rad(A), whichfinishes the proof.

Important examples of semisimple Banach algebras include all C-algebras,

and in particular, all von Neumann algebras. In the next example, we demon-

strate that the Banach algebra B(X) is semisimple.

Example 1.14 LetX be a Banach space. By a standard operator algebra

on X, we mean any closed unital subalgebra A of B(X) that contains the ideal

F(X) of finite-rank operators.

LetA be a standard operator algebra on X. We show thatA is semisimple.

Indeed, for any 0 = x X, set Ix = {T A : T x = 0}. Then Ix is a properleft ideal of A. We show that it is a maximal left ideal. Now suppose thatL

is any left ideal of A containing Ix, with L = Ix. Then there exists A Lsuch thatAx = 0. Take a linear functionaly X such thaty(Ax) = 1, andconsider the rank-one operator B = x y A. Then E = BA L satisfiesEx = x. For an arbitrary T A, we have T E L and T T E Ix, so


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T L, and hence L = A. Accordingly, rad(A) x=0Ix = {0}. Thus A issemisimple.

Semisimple Banach algebras will feature regularly throughout this essay.

We briefly review Gelfands theory of commutative Banach algebras. A

multiplicative linear functional (or character) of a unital Banach algebra

A is a non-zero algebra homomorphism : A C. Such must be unital,i.e. (1) = 1. Moreover, is continuous with = 1. We record its proof,as the argument will be useful later (specifically in the proof of the Gleason-

Kahane-Zelazko Theorem 2.2). Since is a unital homomorphism, it preserves

invertible elements in the sense of Definition 1.21 below. Now let x A, andtake any C \ (x). Then 1 x A1. Therefore, (x) = . Hence(x) (x), from which we deduce |(x)| r(x) x. This proves that 1, and since (1) = 1, we conclude that = 1.

Suppose that A is a unital commutative Banach algebra. The spectrum

(or maximal ideal space) of A is the set A of all multiplicative linear func-

tionals on A. It is a weak-closed subset of the unit ball of the dual space A.

It follows from the Banach-Alaoglu theorem that A is compact. Each x Ainduces a continuous function x C(A), given by

x() = (x) ( A).

The function x has range

ran(x) = (x),

and so

x = r(x).


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The assignment

x A x C(A)

defines the unital homomorphism

A : A C(A)

known as the Gelfand transform. It follows from this fact and the equality

r(x) =

x

that

r(x + y) r(x) + r(y), r(xy) r(x)r(y)

for all x, y A. Also, we have ker(A) = rad(A). Indeed, the equality

ker(A) = {x A : x = 0} =

A

ker()

shows that ker(A) is the intersection of all maximal ideals of A, and the

latter is by definition rad(A). We conclude that A induces an injective unital

homomorphism

A/rad(A) C(A).

In particular, when A is semisimple, we see that for every non-zero x A,there exists A such that (x) = 0. (In this case, we also say that Aseparates the points of A.)

Definition 1.15 Let A be a Banach algebra. An element x

A is called

nilpotent if xm = 0 for some natural number m 1. If x satisfies r(x) = 0(and hence(x) = {0}), we declare x quasinilpotent.

Suppose x A is nilpotent, with xn = 0 for some n 1. If (x), thenn (xn) = (0) = {0} by the spectral mapping theorem, so that = 0.This shows that every nilpotent element is quasinilpotent. The converse is

false in general, as we see in the next example.


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Example 1.16 Consider the following unilateral forward weighted shift on

H = 2(N). Let w = (wn)n=1 (N) and define W B(H) byW en = wnen+1 for all n N.

If we set

wk :=

1 if k is odd,

2m if k = 2m is even,

then

r(W) = limsupn

(wkwk+1 wk+n1)1/n = 0.

We see therefore thatW is quasinilpotent. On the other hand, for every n 1,

Wne1 = (w1 wn)en+1 = 0.

Hence W is not nilpotent of any order.

There is a useful spectral characterisation of the radical of a Banach alge-

bra, due to J. Zemanek. The proof is contained in [Aup91] and [Pta79].

Theorem 1.17 LetA be a Banach algebra and let x A. The following areequivalent:

(a) x rad(A);

(b) (x + y) = (y) for all y A;

(c) there exists a neighbourhoodU of0 so thatr(x+y) = 0 for al l quasinilpo-

tent elements y U;

(d) there exist a neighbourhoodU of x and K > 0 so that r(y) Ky xfor all y U.


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Definition 1.18 Let A be a ring. If e A satisfies e2 = e, we say that eis an idempotent. If, in addition e Z(A), we say that e is a centralidempotent. Furthermore, by an orthogonal family of idempotents, we

mean a family {ei}iI of idempotents of A such that eiej = 0 whenever i = j.

Definition 1.19 LetA and B be algebras. A linear map : A B is said topreserve idempotents if(e) is idempotent whenever e

A is idempotent.

Lemma 1.20 Let A and B be algebras. If : A B is a linear map whichpreserves idempotents, then transforms any orthogonal family of idempotents

of A into an orthogonal family of idempotents of B.

Proof. It is enough to prove the claim for a pair of orthogonal idempotents.

Let e, f A be such a pair, so that ef = f e = 0. Then e +f is an idempotent,

and hence (e), (f) and (e + f) are all idempotents ofB. Then the equality

((e) + (f))2 = (e) + (f)

reduces to

(e)(f) + (f)(e) = 0.

Multiplying to this relation first on the left and then on the right by (f), we

get

(e)(f) = (f)(e)(f) = (f)(e).

Hence (e)(f) = (f)(e) = 0.

We have collected just enough of the basic definitions from the theory of

Banach algebras. Next, we turn to the introduction of several classes of linear

maps between Banach algebras. Their study will be the theme of our essay.


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Definition 1.21 A linear map from a unital Banach algebra A to a unital

Banach algebra B is said to preserve invertibility if (x) B1 wheneverx A1.

We mention that when studying invertibility preserving linear maps between

unital Banach algebras, there is no loss of generality in assuming that the map

is unital. Indeed, if : A B preserves invertibility, then (1) is invertiblein B and we can instead work with the linear map : A B, defined by(x) = (1)1(x) for all x A. This map clearly preserves invertibility andis unital.

Definition 1.22 LetA and B be Banach algebras. A linear map : A Bis said to be

(a) spectrum compressing if for all x A,

((x)) (x);

(b) spectrum preserving if for all x A,

((x)) = (x).

(c) a spectral isometry (or preserves spectral radius) if for all

x A,r((x)) = r(x).

It is clear from the definition that a spectrum preserving linear map is a

spectral isometry.

Note. It is understood that when we refer to an invertibility preserving map

(resp. a spectrum compressing map, a spectrum preserving map or a spectral

isometry), we are assuming that the given map is linear.


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At the end of this chapter, we collect some properties of the maps defined

above for later use.

Lemma 1.23 LetA and B be Banach algebras, and let : A B be a linearmap.

(a) If is spectrum compressing, then preserves invertibility.

(b) If is surjective and spectrum preserving, then(A1

) = B1

.

Proof. (a) Suppose that is spectrum compressing. If a A1, then0 C \ (a) C \ ((a)), so that (a) B1.

(b) Suppose that is surjective and spectrum preserving. Let b B1.Then b = (a) for some a A. Since 0 C \ (b) = C \ (a), it follows thata A1. Hence b (A1).

Proposition 1.24 LetA andBbe Banach algebras. If : A B is a unital,invertibility preserving map, then is spectrum compressing. Moreover, if

is bijective with(A1) = B1, then is spectrum preserving.

Proof. If x A and C, then (.1 x) = .1 (x). Since preservesinvertibility, / ((x)) whenever / (x). In other words, ((x)) (x).Furthermore, if is bijective with (A1) = B1, then 1 is also unital and

invertibility preserving. Hence (1(y)) (y) for y B. Putting y = (x)gives (x) ((x)), and we are done.

Lemma 1.25 Let A and B be Banach algebras, and let : A B be aspectral isometry. Then ker() rad(A). It follows that if A is semisimple,then must be injective.


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Proof. Suppose that x ker(). Then for any y A,

r(y) = r((y)) = r ((x + y)) = r(x + y).

In particular, if y A is quasinilpotent, then

0 = r(y) = r(x + y).

It follows from Theorem 1.17 that x rad(A). Thus, ker() rad(A).

Lemma 1.26 Let A and B be Banach algebras, with B semisimple. If

: A B is a surjective, spectrum preserving map, then (1) = 1.

Proof. To begin, note that

((1)) = (1) = {1}.

Then (1) = 1 + q, where q B is quasinilpotent. For any x A, we have:1 + (q+ (x)) = (1 + q+ (x))

= ((1) + (x))

= ((1 + x))

= (1 + x) = 1 + (x).

Hence (q+ (x)) = (x) = ((x)). Since is assumed surjective, it fol-

lows from Theorem 1.17 that q rad(B) = {0}. Thus, (1) = 1.

Let A be a Banach algebra, and let : A A/rad(A) be the quotientmap. The centre modulo the radical of A is defined by

Z(A) := 1 [Z(A/rad(A))]

= {x A : xa ax rad(A) for all a A}.


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Of course, if A is semisimple, then Z(A) = Z(A). There is a characterisation

of Z(A) in terms of the spectral radius, due to V. Ptak [Pta79]. Below is but

a short list extracted from the ten equivalent properties of that paper.

Theorem 1.27 Let A be a unital Banach algebra and x A. The followingare equivalent:

(a) x Z(A).

(b) r(x + a) r(x) + r(a) for all a A.

(c) r(xa) r(x)r(a) for all a A.

(d) sup{r (x eaxea) : a A} < .

It is not hard to see that (a) implies (b), (c) and (d), by passing to A/rad(A).

Ptak used elementary methods to show that each of (b) and (c) implies (d).

The most difficult part, that (d) implies (a), involves derivations on A. It

relies on the fact (also proved in [Pta79]) that, if D is a bounded derivation

on A and x A, and if the function z C r (x exp(zD)x) is bounded,then that same function is identically zero.

Lemma 1.28 LetA and B be Banach algebras and : A B be a surjectivespectral isometry. Then

[Z(A)] = Z(B).

If both A and B are semisimple, then(Z(A)) = Z(B).

Proof. The proof of the first claim is a repeated application of the equivalence

between (a) and (b) in Ptaks Theorem 1.27. Let x Z(A), and let y B


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be arbitrary. Since is surjective, (a) = y for some a A. Since is aspectral isometry, we obtain

r((x) + y) = r ((x + a))

= r(x + a)

r(x) + r(a)= r((x)) + r((a))

= r((x)) + r(y),

so that (x) Z(B). This proves that [Z(A)] Z(B). For the reverseinclusion, suppose that y Z(B). Then y = (x) for some x A. For anya A,

r(x + a) = r((x + a))

= r(y + (a))

r(y) + r((a))= r(x) + r(a).

Hence x Z(A), and so y [Z(A)].

Now the second claim follows, since Z(A) = Z(A) when A is semisimple.


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Chapter 2

Kaplanskys question.

In this chapter, we introduce Kaplanskys question, and briefly address some of

the landmark results regarding the question and its close analogues. We shall

prove one of the pioneering results, the Gleason-Kahane-Zelasko Theorem,

which states that any unital, invertibility preserving linear functional on a

Banach algebra is multiplicative.

We begin with a general observation concerning Jordan homomorphisms.

Recall that when X and Y are rings, a Jordan homomorphism : X Y isan additive map that satisfies (s t) = (s) (t) where s t := st + ts fors, t X.

Proposition 2.1 LetX andY be rings with identities1X and1Y respectively,

with Y having the property that

2y = 0 y = 0 y Y. (2.1)

Let : X Y be a Jordan homomorphism such that 1Y ran(). Thenpreserves invertibility.

17


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CHAPTER 2. KAPLANSKYS QUESTION 18

Proof. Write (a) = 1Y for some a X. Now we have

2a = 1X a.

Consequently,

2.1Y = 2.(a) = (2a)

= (1X

a)

= (1X) (a)= 2.(1X).

By (2.1), we get 1Y = (1X) which shows that is unital. Next, a straight-

forward calculation shows that for any a, b X,

2(b a) a b (a a) = 4 aba.

Since is a Jordan homomorphism, the above and (2.1) imply that

(a)(b)(a) = (aba) a, b X.

In particular, given x X1, we obtain

(x) = (xx1x) = (x)(x1)(x). (2.2)

Set s = (x)(x1

) and t = (x1

)(x). We shall be done if we can showthat s = t = 1Y. First observe that s

2 = s and t2 = t, by (2.2). Next,

2.1Y = 2.(1X) = (x x1) = (x) (x1) = s + t. (2.3)

Accordingly,

(2.1Y s)2 = t2 = t = 2.1Y s,


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and this implies that 2 (s 1Y) = 0. Thanks to (2.1), it follows that s = 1Y.Finally, this leads to t = 1Y as seen from (2.3). Thus,

1Y = s = (x)(x1) = (x1)(x) = t = 1Y,

from which we conclude that (x1) = (x)1.

A natural question at this point (and this question turns out to be theessence of this essay) is the following: is the converse of the above proposition

true? In other words, if : X Y is additive and preserves invertibility,must be a Jordan homomorphism?

It was in the nineteenth century that Frobenius [Fro97] proved that a linear

map : Mn(C) Mn(C) satisfies det ((T)) = det(T) for all T Mn(C) ifand only if is of the form

(A) = P AQ or (A) = P AtQ

for some fixed invertible matrices P, Q Mn(C) such that det(P Q) = 1 .J. Dieudonne [Die49] proved that if : Mn(F) Mn(F) is an invertible-semilinear map, where is any automorphism of the field F, and if pre-

serves the set of singular matrices in Mn(F), then must be of the form

(A) = P AQ or (A) = P(A)tQ,

where P and Q are invertible matrices. Observe that preserves singular

matrices precisely when 1 preserves invertibility.

For the infinite-dimensional case, A. Gleason [Gle67], and J.-P. Kahane

and W. Zelazko [KZ68] showed the following:


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Theorem 2.2 Let : A B be a unital, invertibility preserving map from aBanach algebra A into a semisimple commutative Banach algebra B. Then

is a homomorphism.

Proof. We begin by proving that is continuous. First of all, we point out that

any unital, invertibility preserving map : A C is continuous and = 1.The proof is identical to the one given in our discussion of multiplicative linear

functionals in Chapter 1. Now take any sequence (xn) in A with xn 0, andwe assume that (xn) y for some y B. If y were non-zero, then sinceB is semisimple, there exists a multiplicative linear functional on B such

that (y) = 0. Since is bounded, ( )(xn) (y). On the other hand, : A C preserves invertibility, and hence it is continuous, so that()(xn) ()(0) = 0. This contradiction shows that y = 0. The closedgraph theorem implies that is continuous.

We now set out to prove that is a homomorphism. If this were false,

then (xy) = (x)(y) for some x, y A. Since the multiplicative lin-ear functionals on B separate points of B, there exists B such that( )(xy) = ( )(x)( )(y). But note that is unital and invert-ibility preserving. Therefore, our problem reduces to proving that any unital,

invertibility preserving map : A C is a homomorphism. From the above,

any such map has norm equal to 1.

We begin by showing that any such is a Jordan homomorphism, i.e.

(x2) = (x)2 for any x A. Let x A be fixed. Consider the function : C C defined by (z) := (exp(zx)), which is analytic since A.Since every value of the exponential function on A is invertible, is entire

without zeros. So, there is an entire function f with (z) = ef(z) for every

z C, such that f(0) = 0 and Re(f(z)) |z| x for all z C. By Schwarzs


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lemma, f(z) = z for some complex constant . Expanding both sides of the

equation (exp(zx)) = ez,

1 + z(x) +1

2!z2(x2) + = 1 + z+

1

2!2z2 + .

By comparing coefficients, we get (x2) = 2 = (x)2, so that is a Jordan

homomorphism.

Finally, in order to prove that (xy) = (x)(y), it is enough to show

that

([x, y] ) = [(x),(y) ] = 0,

where [x, y] := xy yx. We begin by noting that

(xy yx)2 + (xy + yx)2 = 2(x.yxy + yxy.x) = 2.x (yxy),

so that

[x, y]

2

= 2.x (yxy) (x y)2

.

Applying the Jordan homomorphism , we get

([x, y])2 = 4.(x)(yxy) 4.((x)(y))2. (2.4)

There are now two ways of showing that ([x, y]) = 0. First of all, (2.4)

implies that ([x, y] ) = 0 i f x ker(). For general x A, note thatx := x(x)1 ker() and [x, y] = [x, y]. Therefore, ([x, y]) = ([x, y]) = 0

for all x. For the second approach, recall from the proof of Proposition 2.1that the product yxy can be expressed entirely in terms of Jordan products,

leading to (yxy) = (y)(x)(y). Putting this equality into (2.4) gives

([x, y]) = 0.

Motivated by the discussions above, I. Kaplansky in 1970 [Kap70] asked

the following question:


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Let : A B be a unital, invertibility preserving map between unitalBanach algebras A and B. Is a Jordan homomorphism?

Let us quickly state that the above question of Kaplansky is too general

and the answer to it is negative in this generality. The first example below,

given in Sourour [Sou96], shows that if is not surjective, then it may not be

a Jordan homomorphism.

Example 2.3 Let H be a Hilbert space and let : B(H) B(H) be linearsuch that(I) = 0. Consider the map

: B(H) B(H H)

defined by

(X) =

X (X)

0 X

for X B(H). Then is clearly unital and preserves invertibility. However,for X B(H) we have

(X2) (X)2 = 0 (X

2) X(X) (X)X0 0

.

Hence, is a Jordan homomorphism if and only if(X2) = (X)X+ X(X)

for all X B(H), or equivalently,(X Y) = (X) Y + X (Y)

for all X, Y B(H). If satisfies this property, then is a Jordan deriva-tion of B(H).

It is now easy to exhibit a counterexample to Kaplanskys question. Our

shall take the form(X) = f(X)I, where f B(H) satisfies f(I) = 0. Let


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dim H 2, and choose an orthogonal pair of unit vectors h, k H. Definef B(H) by

f(X) = Xh,k.

Take the rank-one operator X = (k h). ThenX2 = 0 and f(X) = 1, sothat

(X)X+ X(X) = 2X = 0 = (X2).

The associated invertibility preserving map is therefore not a Jordan homo-

morphism.

In the next example, Aupetit [Aup79] showed that Kaplanskys question

may not have a positive answer if the Banach algebras A and B are not

semisimple.

Example 2.4 Let

A =

W X

0 Y

: W,X,Y M2(C)

and define : A A by

Z =

W X

0 Y

(Z) =

W X

0 Yt

.

Clearly, is surjective, unital, and invertibility preserving. But then

(Z2) (Z)2 = 0 X(Y Y

t)

0 0

is not always zero. Hence is not a Jordan homomorphism. However, note

that(Z2) (Z)2 rad(A) for all Z A.


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on C-algebras. There they indicated their belief that Kaplanskys conjecture

is true for C-algebras. However, this case is still open.

Recently, Aupetit [Aup00] proved that every surjective, spectrum preserv-

ing map from a von Neumann algebra onto another von Neumann algebra is a

Jordan isomorphism. This is one of the deepest results so far in this direction,

and it is close to a positive solution to Kaplanskys conjecture.

Bresar and Semrl [BS03] proved that every unital, surjective, invertibility

preserving map from a von Neumann algebra onto a standard operator algebra

is a Jordan homomorphism.

It is often said that the best way to approach a difficult problem is to

generalize it. Mathieu [Mat94] introduced the notion of spectral isometry.

Over the past decade, Mathieu has been looking at the following spectral

radius analogue of Kaplanskys conjecture:

Let A and B be unital semisimple Banach algebras. Let : A B be aunital surjective spectral isometry. Must be a Jordan isomorphism?

Mathieu and Sourour [MS04] gave a positive result for this conjecture in the

case where A and B are finite-dimensional semisimple Banach algebras.

To sum up, many researchers have been trying their hands at Kaplanskys

question. The aim of this essay is to present a sample of positive resultsobtained so far. As mentioned before, none has answered the question com-

pletely but they are all motivating results on which further work may lead to

the eventual solution of the problem.

In Chapter 3, we sketch Aupetits work on spectrum preserving maps be-

tween von Neumann algebras. Then in Chapter 4, we present Sourours solu-

tion in the framework of invertibility preserving maps between Banach algebras


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of the form B(X). We shall end our discussions in Chapter 5, where we shall

give the exciting result obtained by Mathieu and Sourour in the more general

setting of spectral isometries.


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Chapter 3

Spectrum preserving maps

between von Neumann algebras

Recall that Kaplanskys question essentially asks for a characterisation of linear

maps which preserve invertible elements between semisimple Banach algebras.

This falls into the class of the so-called Linear Preserver Problems (LPP)

[Jaf09]. It turns out that a large number of solutions to the Linear Preserver

Problems involve the consideration of maps preserving finite-rank operators.

This is the approach used by Sourour, and which is going to occupy us in

Chapter 4.

Another method that has been used by researchers in trying to solve Ka-planskys question hinges on the fact that, for von Neumann algebras A and

B, a continuous linear map : A B that preserves idempotents must bea Jordan homomorphism (Theorem 3.6). Based on this fact, Aupetit solved

Kaplanskys question affirmatively in the case of spectrum preserving maps

between von Neumann algebras. In this chapter, we take a closer look at this

result.

27


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CHAPTER 3. SPECTRUM PRESERVING MAPS 28

Theorem 3.1 ([Aup00]) Let A and B be von Neumann algebras, and let

: A B be a surjective, spectrum preserving linear map. Then is aJordan isomorphism.

We begin with the following observation regarding the bicontinuity of a

surjective spectrum preserving map. Note that such a map is a spectral isom-

etry.

Lemma 3.2 Let A and B be semisimple Banach algebras, and let

: A B be a surjective spectral isometry. Then : A B is a Banachspace isomorphism.

Proof. (Sketch) This is a consequence of [Aup82, Theorem 1], which states that

for Bsemisimple, any surjective linear map T : A Bsuch that r(T x) r(x)for all x

Amust be continuous. By Lemma 1.25, is bijective. Both and

1 are spectral isometries. The aforementioned fact now implies that both

and 1 are continuous.

The next ingredient is an observation about idempotents.

Lemma 3.3 LetA be a unital Banach algebra and lete A be an idempotent.For every a

A we have

(a) D (0;(e + 1 e) a e) D (1;(e + 1 e) a e) .

Proof. If e = 0 or e = 1 the result is clear. Suppose that e is a non-trivial

idempotent, in which case (e) = {0, 1}. Let C \ {0, 1}. Then, since e = ( 1)e + (1 e),

( e)1 = ( 1)1e + 1(1 e)


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which implies that (( e)1) = {1, ( 1)1}. Hence,

r

( e)1 = max{||1, | 1|1} = 1dist(, (e))

.

Thus we have( e)1 | 1|1 e + ||1 1 e

(e + 1 e) r ( e)1 .(3.1)

Suppose that the result is false. Then there exists a

A and

(a)

such that we have

dist (, (e)) > (e + 1 e) a e ,

and in particular = 0, 1. Consequently, by (3.1) above, we conclude that

dist (, (e)) =1

r [( e)1] e + 1 e( e)1 .

Hence

a

e

0, the set

K+ := {z C : dist(z, K) }

is a neighbourhood of K. (A more familiar notation is K.)


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Theorem 3.4 ([Aup00]) An elemente of a semisimple Banach algebraA is

idempotent if and only if(e) {0, 1} and there exist C, r > 0 so that

(x) (e) + Cx e

for every x A with x e < r.

The proof of the above theorem is not trivial. The easy direction follows from

Lemma 3.3 above (just choose C = e + 1 e and r > 0 arbitrary). Forthe converse (which is the harder part), as well as more results on idempotents

of a semisimple Banach algebra, we refer to [Aup00].

We now prove the two key steps involved in the proof of Theorem 3.1.

Theorem 3.5 ([Aup00]) Let A and B be semisimple Banach algebras, and

let : A B be a surjective spectrum preserving map. Then preservesidempotents.

Proof. Let e A be an idempotent. Since (1) = 1 by Lemma 1.26, we mayassume that e = 0, 1. Then (e) = {0, 1}, which implies that ((e)) = {0, 1}.By Theorem 3.4, there exist C, r > 0 such that

(x) {0, 1} + Cx e

for all x A with x e < r. By Lemma 3.2, we can find > 0 such that x (x) for all x A. Consequently,

((x)) {0, 1} + C

(x) (e)

for all x A such that (x) (e) < r. Since is surjective, we deducethat

(y) {0, 1} + C

y (e)


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for all y B such that y (e) < r. Therefore, by Theorem 3.4 again,(e) is idempotent.

Theorem 3.6 Let A be a von Neumann algebra and B any Banach algebra.

Let : A B be a continuous linear map that preserves idempotents. Then is a Jordan homomorphism.

Proof. Let Asa denote the set of self-adjoint elements of A. Take any h Asawith finite spectrum. Then

h =n

k=1

tkpk,

where {tk}nk=1 are real numbers and {pk}nk=1 is an orthogonal family of pro-

jections (i.e. self-adjoint idempotents). By Lemma 1.20, {(pk)}nk=1 is an

orthogonal family of idempotents of B. It follows that

(h2) = (h)2.

But the set of all self-adjoint elements with finite spectrum is norm-dense in

Asa. Thus, by the continuity of, we get that (h)2 = (h)2 for every h Asa.Now, replacing h by h1 + h2 for arbitrary h1, h2 Asa, we get

(h1h2 + h2h1) = (h1)(h2) + (h2)(h1).

Finally, an arbitrary element a A can be written as a = h1 + ih2 withh1, h2 Asa. Then

a2 = (h21 h22) + i(h1h2 + h2h1).

It follows readily that (a2) = (a)2. Hence is a Jordan homomorphism.


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It is worth noting that the only property of von Neumann algebras which

was used in this argument is the fact that every self-adjoint element of a

von Neumann algebra can be approximated in norm by a self-adjoint element

with finite spectrum. As such, Aupetits argument may be extended mutatis

mutandis to the setting of C-algebras of real-rank zero. (These are precisely

the C-algebras where every self-adjoint element can be approximated by self-

adjoint elements with finite spectrum.)

Proof of Theorem 3.1. We can now complete the proof of Theorem 3.1. Sup-

pose that : A B is a surjective, spectrum preserving map between vonNeumann algebras A and B. By Lemma 3.2, is continuous. By Theorem

3.5, we know that preserves idempotents, which by Theorem 3.6 shows that

is a Jordan homomorphism.


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Chapter 4

Invertibility preserving bijective

maps from B(X) to B(Y)

In this chapter we present Sourours answer to Kaplanskys question in the

setting where the Banach algebras consist of the sets of bounded operators on

given Banach spaces. We have shown in Chapter 2 that such Banach algebras

are semisimple. The main result in this chapter is the following:

Theorem 4.1 (Sourour [Sou96]) Let X and Y be Banach spaces. Then a

bijective linear map : B(X) B(Y) preserves invertibility if and only if takes either one of the following forms:

(a) (T) = AT B, for all T B(X); or

(b) (T) = CTD, for all T B(X),

where A B(X, Y), B B(Y, X), C B(X, Y) and D B(Y, X) areinvertible operators.

33


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CHAPTER 4. INVERTIBILITY PRESERVING BIJECTIONS 34

In order to prove this result, we shall need a characterisation of rank-one

operators in terms of the spectrum (Theorems 4.4 and 4.5 below), which is of

independent interest.

4.1 Spectral characterisation of rank-one op-

eratorsFor preparation, we gather here some frequently used notation as well as recall

some basic definitions. Let W be a locally convex topological linear space. We

shall abbreviate this to a locally convex space. Denote by B(W) the algebra

of all continuous linear operators on W. The dual space of W is denoted by

W. Denote by , the duality between W and its dual W, so that

x, y = y(x)

for all x W and y W. If A B(W), then A B(W) stands for theadjoint (or transpose) of A, which is defined by the equation

x, Ay = Ax,y

for x W and y W. The rank of A is rank(A) := dimran(A). Recall

that A is said to be finite-rank if rank(A) < . Throughout this essay, weshall denote the set of finite-rank operators on W by F(W). Also, for each

n N, by Fn(W) we denote the set of operators of rank at most n. Note thatF(W) = span F1(W).

For 0 = w W and 0 = g W, the symbol w g will stand for therank-one operator on W given by (w g)y = g(y)w, for every y W. Everyrank-one operator on W can be written in this form.


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Lemma 4.2 LetV be a vector space, v V, g a linear functional on V, andA L(V) an invertible operator. Then the operator A (v g) is invertibleif and only if g(A1v) = 1. If g(A1v) = 1, then A (v g) is invertiblein every unital subalgebra of L(V) that contains {A, A1, A (v g)}.

Proof. If g(s) = 1, then s g is a rank-one idempotent. Then I (s g)is a proper idempotent and is not invertible. Conversely, if g(s)

= 1, then by

direct computation, we have that

(I (s g))1 = I + (1 g(s))1(s g),

which gives an inverse of I (s g) in every unital algebra containingI (s g). We have just shown that I (s g) is invertible if and onlyif g(s) = 1. But we know that A (v g) is invertible if and only ifA1 (A

(v

g)) = I

((A1v)

g) is. The result then follows.

Proposition 4.3 Let W be a locally convex space. Let A B(W), w W,w W and z C \ (A). Then z (A + (w w)) if and only if(z A)1w, w = 1.

Proof. This is a direct consequence of Lemma 4.2. Indeed, since zI A

is invertible, it follows that (zI A) (w w) is invertible exactly when(zI A)1w, w = 1.


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Theorem 4.4 ([JS86]) LetW be a locally convex space and A B(W). Thefollowing are equivalent:

(a) rank(A) 1.

(b) (B + kA) (B + sA) (B) for all B B(W) and k = s C.

(c) (B + kA) (B + sA) (B) for all B F2(W) and k = s C.

Proof. Since (b) and (c) hold trivially when A = 0, we assume that A = 0from now on.

[(a) (b)] First suppose that rank(A) = 1. We can find w W andw W so that A = w w. Let B B(W) and let z / (B). Then byProposition 4.3 above, z (B + kA) if and only if k(z B)1w, w = 1.Therefore z cannot belong to (B + kA) for two distinct values of k. But this

amounts to proving (b).

[(b) (c)] This is trivial.

[(c) (a)] We assume (c) holds and assume also that rank(A) 2.

Suppose A = I for some = 0. Let B F2(W) with (B) {0,}.Then (B +A) (B +2A) = {2}, which contradicts our assumption. HenceA is not a non-zero scalar multiple of the identity.

We now consider two cases.

Case 1: There is a vector w W such that {w,Aw,A2w} is linearlyindependent. SetU = span {w,Aw,A2w} and choose any (closed) complement


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V of U in W. Define N B(W) by

N(w) = w Aw,N(Aw) = Aw 2A2w,

N(A2w) = 12

w +3

2Aw 2A2w,

and N(v) = 0 for every v V. Then one obtains that N3 = 0,

(N+ A)w = w and (N+ 2A)Aw = Aw, so that N F2(W), (N) = {0} and1 (N + A) (N + 2A). Thus we get a contradiction in this setting. Theclaim of the theorem holds in this case.

Case 2: {y,Ay,A2y} is linearly dependent for every y W. By a re-sult of Kaplansky (see [Aup91, p. 84] for a proof), A satisfies a quadratic

polynomial equation p(A) = 0. By considering four subcases according as

p(t) = (t

)(t

), p(t) = (t

)2, p(t) = t(t

) or p(t) = t2,

= 0

=

= ,

and by the standard decomposition of algebraic operators, we see that A has

a finite-dimensional invariant subspace M such that the A|M has one of the

following four matrix representations for a suitable basis of M.

(a)

0

0

;

(b) 1

0

;

(c)

0 0

0 0

0 0 0

; or


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Theorem 4.5 ([Sou96]) LetW be a locally convex space. The following are

equivalent for given A B(W):

(a) rank(A) 1;

(b) For all B F2(W), there exists a compact subset KB C so that

(B + kA) (B + sA) KB

for all k = s C.

Proof. [(a) (b)] This follows from [(a) (c)] of Theorem 4.4 and the factthat finite-rank operators have finite spectrum.

[(b) (a)] We assume (b) and show that rank(A) 1. We begin byshowing that (A) does not contain more than one non-zero complex num-

ber. Suppose to the contrary that ,

(A) with = and both are

non-zero. Putting B = 0 into the assumption of (b), we get a compact set

K C such that (kA) (sA) K for all scalars k = s. But note that1 (1A) (1A), so that

z (z1A) (z1A) K

for all z C. But then C K, which is absurd. So we can fix a non-zero

scalar, say k, such that (A) {0, k}.Take any non-zero w W and w W. Set B = w w, and define the

analytic function : C \ {k1} C by

(z) := (I zA)1w, w.

Note that the only possible singularities of are k1 and . We shall provethat they are at worst poles.


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First, take any M > 0 such that KB D(0; M). We show that for any C with || > M, the equation (z) = cannot have more than onesolution. To verify this claim, we first show that (z) = if and only if

(B + zA). Indeed, thanks to Proposition 4.3:

= (z) if and only if = (I zA)1w, wif and only if 1 = (I zA)11w, wif and only if 1 (zA + 1w w)if and only if (zA + B).

As a consequence, if(z1) = (z2) = for some z1 = z2, then

(B + z1A) (B + z2A) KB,

contradicting the choice of .

Recall Picards Big Theorem [Tit39, p. 283]: If an analytic function f

has an essential singularity at a point , then on any deleted neighbourhood

of, f takes on all possible complex values (with at most a single exception)

infinitely often.

It follows that the function has poles (or removable singularities) at k1

and . Consequently, is a rational function, and in fact either = 0 or(z) = P(z)/Q(z), where P and Q are polynomials with P(k1) = 0 andQ(z) = (kz 1)n for some integer n 0. We will show that deg P 1 anddeg Q 1. For all with || > M, we know that the equation

P(z) Q(z) = 0 (4.1)

has at most one solution. If the degree of P Q is greater than 1, then anyz satisfying (4.1) above, being a multiple root, must also satisfy the equation

P(z) Q(z) = 0. (4.2)


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So, one of the subspaces must be all of W, implying that either

(A2 kA)w = 0 or A2w = 0. Then W in turn is a union of the two sub-spaces

{w W : (A2 kA)w = 0}

and

{w W : A2w = 0}.

By the same reasoning, we conclude that either A2 kA = 0 or A2 = 0.Finally, we argue that the assumption rank(A) > 1 must lead to a contra-

diction. There are two cases to consider.

Case 1: Assume that A2 = kA. Let x, y ran(A) be linearly independentvectors. Then of course, Ax = kx and Ay = ky. Let g W be such thatg(x) = 0 and g(y) = 1, and set

B = y g.

Take any z C, and observe that

B + k1zA

x = 0 + zx = zx,

B + k1(z 1)A y = y + (z 1)y = zy.

This implies that

z B + k1zA B + k1(z 1)A KB.

But then C = KB, violating compactness of KB.

Case 2: Assume that A2 = 0. Let x, y ran(A) be linearly independentvectors. If x = Au and y = Av for some u, v W, then {x,y,u,v} is linearlyindependent. Set

M := span {x,y,u,v} .


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Then M is a subspace ofW which is clearly invariant for A. Furthermore, the

matrix representation of A|M is

0 0 0 0

1 0 0 0

0 0 0 0

0 0 1 0

relative to the ordered basis {u,x,v,y}. Now choose g, h W such that

g(x) = 1, g(u) = g(v) = g(y) = 0

while

h(y) = 1, h(u) = h(v) = h(x) = 0.

Then set

B := (u g

) + 4(v h

),

a rank-two operator for which M is also invariant. Note that B|M has matrix

representation

0 1 0 0

0 0 0 0

0 0 0 4

0 0 0 0

relative to the basis {u,x,v,y}. We see that every complex number z satisfies

z (B + z2A) (B + z24

A). As in Case 1, this leads to an absurdity. Thus,

in both cases, we must have rank(A) 1.


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4.2 The rank-reducing property of some

invertibility-preserving maps

Let X and Y be locally convex spaces, let be a subspace of B(X) that

contains F(X), and let : B(Y) be a linear map. We call rank-reducing if rank((T)) rank(T) for every T F(X). Since for each

n N and T B(X), T Fn(X) if and only if T is a sum of no morethan n rank-one operators, it is clear that is rank-reducing precisely when

[F1(X)] F1(Y).

Lemma 4.6 Let X be a Banach space and Y a locally convex space. Let

be a unital subspace of B(X) containing all finite-rank operators on X. If

: B(Y) preserves invertibility and if ran() contains all finite-rank

operators on Y, then

is rank-reducing.

Proof. From the discussion preceding this lemma, it is enough to show that

[F1(X)] F1(Y).

Define : B(Y) by (T) = (I)1(T). Then (I) = I. Moreover, preserves invertibility, and ran() F(Y). Note that the set F1(Y) is closed

under multiplication by elements of B(Y). Hence, our goal is equivalent toshowing that

[F1(X)] F1(Y).

If A F1(X), then for every B and k = s C,

(B + kA) (B + sA) (B).


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Since is unital and preserves invertibility, it is spectrum compressing (Propo-

sition 1.24). We get

((B) + k(A)) ((B) + s(A)) (B)

for scalars k = s. Since (B) is compact, and since ran() contains F2(Y), weapply Theorem 4.5 to conclude that rank ((A)) 1.

We develop here some elementary facts about rank-reducing linear maps

for later use. See [Hou89] for a more detailed study.

Let X be a locally convex space. For every x X and x X, define thesets

Lx := {x w : w X}, Rx := {w x : w X }.

These are subspaces of B(X) contained in F1(X). Clearly,

Lx Rx = C(x x).

The following simple observation shows that ifx = 0 and x = 0, then Lx andRx are maximal among all subspaces of B(X) that are contained in F1(X).

Lemma 4.7 LetX be a locally convex space. Let S be a non-trivial subspace

of B(X) consisting of rank-one operators. Let s = x

x

S be non-zero.

(a) Ifdim S = 1, then S = Lx Rx.

(b) Ifdim S > 1, then either S Lx or S Rx, but not both.

Proof. Part (a) is clear. To prove (b), take any t = y y S. Since

s + t = (x x) + (y y)


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belongs to S, it is of rank one, and hence either y Cx or y Cx. Accord-ingly, either t Lx or t Rx. This means that

S Lx Rx.

Since S, Lx and Rx are subspaces, it follows that either S Lx or S Rx.As Lx Rx has dimension 1, we cannot have both S Lx and S Rx.

Lemma 4.8 LetX and Y be locally convex spaces. If : F(X) B(Y) is arank-reducing linear map, then at least one of the following is true:

(i) for every x X, (Lx) Lyx for some yx Y;

(ii) for every x X, (Lx) Ryx for some yx Y.

Proof. For each x

X, (Lx) is a subspace of B(Y) consisting of rank-one

operators. Hence by Lemma 4.7, either (Lx) Lyx for some yx Y, or(Lx) Ryx for some yx Y.

If dim(Lx) 1 for every x X, then both (i) and (ii) are true. Sowe assume that there exists x0 X with dim (Lx0) 2. Then either(Lx0) Ly0 for some y0 Y, or (Lx0) Ry0 for some y0 Y.

We first suppose that (Lx0) Ly0, and fix x X. We are done if

(Lx) Lyx for some yx Y. Otherwise, suppose that (Lx) Ryx for someyx Y. There exist linear transformations A : X Y and B : X Ysuch that

(x0 w) = y0 Aw, (x w) = Bw yx

for all w X. Their sum

(y0 Aw) + (Bw yx) = ((x0 + x) w)


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is a rank-one operator. Hence for each w X, either {Aw, yx} or {y0, Bw}is a linearly dependent set. In other words,

X = {w X : Aw Cyx} {w X : Bw Cy0},

expressing X as a union of two subspaces. One of these subspaces must be

all of X. But rank(A) = dim (Lx1) 2, and so the latter case prevails.

Therefore, (Lx) C(y0 y

x) Ly0.Summarizing, if dim(Lx1) 2 and (Lx1) Ly1, then for each x X,

(Lx) Lyx for some yx Y. Similarly, if dim (Lx1) 2 and (Lx1) Ry1 ,then for each x X there is yx Y such that (Lx) Ryx.

Lemma 4.9 Let V and W be vector spaces, and let A, B : V W be linear

transformations. Suppose that for every x V, there is a scalarx C suchthat

Ax = xBx.

Then A = B for some C.

Proof. The equation Ax = xBx implies that ker(B) ker(A). LetU = ran(B), a subspace of W. Define T : U U by

T(Bx) := Ax.

Note that T is well-defined. Indeed, if x, y V satisfy Bx = By, thenx y ker(B) ker(A) and so Ax = Ay. Observe that T is linear andthat, for each u U, there is u C such that T u = uu. Now it is easy toshow that T = I for some C.


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Lemma 4.10 LetX andY be locally convex spaces, and let : F(X) B(Y)be a linear map.

(a) Suppose that for each x X, there exists yx Y such that(Lx) Lyx, and that there exist x1, x2 X such that yx1 and yx2 arelinearly independent. Then there exist linear transformations A : X Yand B : X Y such that (x x) = Ax Bx for all x X andx X.

(b) Suppose that for each x X, there exists yx Y such that(Lx) Ryx, and that there exist x1, x2 X such that yx1 and yx2 arelinearly independent. Then there exist linear transformations

C : X Y and D : X Y such that (x x) = Cx Dx forall x X and x X.

Proof. We prove (a) only, the proof of (b) being similar. Thus, we assume the

following:

(i) (Lx) Lyx for all x X.

(ii) There exist x1, x2 X such that yx1 and yx2 are linear independent.

In (i), we insist that yx = 0 if and only if(Lx) = 0. For each x

X, there is

a linear transformation Bx : X Y such that

(x x) = yx Bxx

for all x X. When yx = 0, we choose Bx = 0. On the other hand, ifyx = 0, then necessarily Bx = 0. Let us write

B = Bx1+x2.


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Note that for all x X,

yx1+x2 Bx = ((x1 + x2) x)= (yx1 Bx1x) + (yx2 Bx2x) .

As yx1 and yx2 are linearly independent, it follows that

{Bx1x, Bx2x

}

CBx

for all x X. So, by Lemma 4.9, there are non-zero scalars , C suchthat

Bx1 = B, Bx2 = B.

In particular, Bx1 and Bx2 are linearly dependent.

Now let x X. If yx = 0, then Bx = 0B. If yx = 0, then yx must be

linearly independent of one of yx1 and yx2. The above argument shows thatBx is a scalar multiple of B. We conclude that for each x X, Bx = xB forsome x C. Then

(x x) = yx Bxx = xyx Bx

for every x X and x X. Define A : X Y by Ax = xyx. Then

(x

x) = Ax

Bx

for all x X and x X. Finally, as B = 0, it follows that A is linear.

The following characterisation of rank-reducing maps will be used in the

proof of Theorem 4.13 in the next section.


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Case I.a: There exist x1, x2 X such that yx1 and yx2 are linearly indepen-dent.

Case I.b: There exists 0 = y0 Y such that (Lx) Ly0 for all x X.In Case I.a, Lemma 4.10 shows that takes the form (i) listed above. In

Case I.b, there is a function

: X

X

Y

such that (x x) = y0 (x, x) for all x X and x X. Since y0 = 0and (x, x) (x x) is bilinear, it is clear that is also bilinear. Hencethis subcase leads to the form (iv).

It remains to consider Case II. Once again, there are two subcases:

Case II.a: There exist x1, x2 X such that yx1 and yx2 are linearly indepen-dent.

Case II.b: There exists 0 = y0 Y such that (Lx) Ry0 for all x X.As above, Case II.a implies that takes the form (ii), whereas Case II.b leads

to the form (iii).

4.3 The main theorem

Recall from Proposition 2.1 that if : A Bis unital Jordan homomorphism,then preserves invertibility. The next theorem gives us more in the case

when A = B(X) and B = B(Y). In fact, the main theorem of this chapter is

equivalent to the following result.


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Theorem 4.12 Let X and Y be Banach spaces and suppose that

: B(X) B(Y) is a unital bijective linear map. The following are equiva-lent:

(a) preserves invertibility;

(b) is a Jordan isomorphism;

(c) is either an isomorphism or an anti-isomorphism;

(d) either

(i) Y is isomorphic to X and (R) = A1RA for all R B(X), withA B(Y, X) an isomorphism; or

(ii) Y is isomorphic to X and(R) = B1RB for allR B(X), withB

B(Y, X) an isomorphism.

That (b) implies (a) is precisely Proposition 2.1. That (d) implies (c) and that

(c) implies (b) are clear. So it suffices to show that (a) implies (d). We give

this in Theorem 4.13 which comes next.

Theorem 4.13 ([Sou96]) Let X be a Banach space, and let Y be a locally

convex space. Let be a standard operator algebra on X and let : B(Y)

be a unital, invertibility preserving map whose range contains F(Y). Theneither|F(X) = 0 or is injective. If is injective, then either

(i) there is a weak-weak continuous bijective linear transformation

P : Y X such that(T) = P1T P for all T ; or

(ii) there is a weak-weak continuous bijective linear transformation

P : Y X such that(T) = P1TP for all T .


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Remark. When Y is a Banach space, the weak-weak continuity ofP in Case (i)

of the conclusion and the closed graph theorem imply that P B(Y, X). Theopen mapping theorem implies that P1 is bounded. Likewise, the weak-weak

continuity of P in Case (ii) implies that P B(Y, X) and has a boundedinverse.

Proof. We know from Lemma 4.6 that is rank-reducing, so that maps

rank-one operators to rank-one operators. For any x X and x X, wemay write

(x x) = yx,x yx,x,

where yx,x Y and yx,x Y.

Fix an operator 0 = E , and let C. Observe that if || < E1,then I E is invertible in . Since preserves invertibility, I (E) is

invertible in B(Y).

For x X and x X, consider the analytic functions

x,x, x,x : D(0; E1) C

defined by

x,x(z) := (I zE)1x, x = Tr

(I zE)1(x x)

and

x,x(z) := (I z(E))1 yx,x, yx,x= Tr

(I z(E))1 (x x) .

Note that

(I E (x x)) = I (E) (yx,x yx,x).


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Ifx,x() = 1, then by Lemma 4.2, I E (x x) is invertible in andso I(E) (yx,x yx,x) is invertible in B(Y), implying that x,x() = 1.Equivalently, ifx,x() = 1, we get that x,x() = 1. Since x,x and x,x

are linear in x and x, we see that if x,x() = = 0, then x,x() = also. Since both x,x and x,x are analytic, we see that ifx,x 0, thenx,x x,x.

We claim that unless x,x 0 for all x X and x

X

, we must havex,x x,x for all x X and x X. Indeed, ifv,v is not identically zerofor some v X and v X, then we can write X = J K as a union of twosubspaces where

J = {x X : v,x 0}

and

K = {x X : v,x v,x} .

So one of the two subspaces must be the whole of X. Since v,v 0 byassumption, we see that K = X. In other words,

v,x v,x

for all x X. A similar argument applied to v shows that

x,v x,v

for all x X.Fix w X and w X. If w,w 0, then from the above, we have

w,w w,w. Suppose that w,w 0. For any k C, consider

w+kv,w+kv k(v,w + w,v) + k2v,v.

Since v,v 0, we must have w+kv,w+kv 0 for some k = 0. Thus,

w+kv,w+kv w+kv,w+kv.


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Finally, expanding both sides by linearity and noting that v,w v,w,w,v w,v and v,v v,v, we get

w,w w,w.

So, we have shown that for all 0 = E , either x,x x,x for all x X,x X, or x,x 0 for all x X, x X.

Taking derivatives at z = 0, we get that for all E , either

Ex,x = Tr[(E)(x x)] (4.4)

for all x X and x X, or

Tr [(E)(x x)] = 0 (4.5)

for all x

X and x

X. Thus itself is a union of two subspaces cor-

responding to the equations (4.4) and (4.5). Therefore, either (4.4) or (4.5)

holds for every E .

First, suppose that (4.5) holds. Given u Y and u Y, since ran()contains F(Y), there is Eu,u such that (Eu,u) = u u. Then we findthat

(x x)u, u = Tr[(u u)(x x) ] = 0

for all x X, x X, u Y and u Y. This implies that (x x) = 0for all x X and x X. Therefore, (E) = 0 for all E F(X).

On the other hand, if (4.4) holds, then it is clear that is injective. By

Lemma 4.11, there are four possibilities for .

Case 1: There are linear transformations A : X Y and B : X Y suchthat (x x) = Ax Bx for all x X and x X.


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Case 2: There are linear transformations C : X Y and D : X Y suchthat (x x) = Cx Dx for all x X and x X.Case 3: There exist y0 Y and a bilinear map : X X Y such that(x x) = (x, x) y0 for all x X and x X.Case 4: There exist y0 Y and a bilinear map : X X Y such that(x x) = y0 (x, x) for all x X and x X.

First of all, if dim Y = 1, then B(Y) = CI. Since : CI is unital andinjective, it follows that dim F(X) = dim = 1, forcing dim X = 1. Thus,

(T) = P1T P for all T , where P : Y X is any isomorphism.

From now on, we assume that dim Y > 1. We argue that Cases 3 and 4

cannot occur. Suppose that Case 3 holds. Then (4.4) becomes

Ex,x = (E)(x, x), y0.

Since y0 = 0 and dim Y 2, there exist y1 = y2 Y such thaty1, y0 = y2, y0 = 1. Take any 0 = y Y. Choose S1, S2 suchthat (Si) = yi y for i = 1, 2. Then S1 = S2. However, for all x X,x X and i = 1, 2,

Six, x = (Si)(x, x), y0=

yi, y

0

(x, x), y

= (x, x), y.

This implies that S1 = S2, which is absurd. A similar argument shows that

Case 4 does not hold.

Next, assume that Case 1 holds. Then (4.4) becomes

Ex,x = (E)Ax,Bx. (4.6)


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In particular, A = 0. Pick x0 X and y0 Y such that Ax0, y0 = 1. Foreach y Y, there is a unique Ey such that (Ey) = y y0. Now (4.6)implies that

Eyx0, x = (Ey)Ax0, Bx = y,Bx.Define P : Y X by P y = Eyx0. Clearly, P is linear, and

P y , x = y,Bx. (4.7)

Note that the weak-weak continuity of P follows from (4.7). Combining (4.6)

and (4.7), we see that for all x X and x X,

Ex,x = P(E)Ax,x,

or equivalently,

E = P(E)A. (4.8)

Putting E = I gives I = P A, so that P is surjective. To show that P is

injective, take any 0 = y Y. Fix 0 = y Y, and take E such that(E) = y y. Then (4.8) gives

0 = E = P(y y)A = (P y y)A,

proving that P y = 0. Now that P is bijective, we also have A = P1, and(4.8) is equivalent to

(E) = P1EP.

Finally, suppose that Case 2 holds. Then (4.4) becomes

Ex,x = (E)Cx, Dx. (4.9)

In particular, C = 0. Pick x1 X and y1 Y such that Cx1, y1 = 0. Giveny Y, there is a unique Ey such that (Ey) = y y1. Consequently,

x, Eyx1 = Eyx, x1 = (Ey)Cx1, Dx= y,Dx.


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Define P : Y X by P y = Eyx1, which is linear. Then

x , Py = y,Dx. (4.10)

Now (4.10) shows that P is weak-weak continuous, and that (4.9) is equivalent

to the statement

Ex,x = x, P(E)Cx,

or

E = P(E)C. (4.11)

We can now proceed as in Case 1 to prove that P is bijective, C = P1 and

(E) = P1EP.


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Chapter 5

Spectral isometries between

finite-dimensional semisimple

Banach algebras

In Chapter 4, we looked at Kaplanskys question in terms of invertibility pre-

serving maps between Banach algebras of the form B(X) for some Banach

space X. Recall from Proposition 1.24 that a unital, invertibility preserving

map : A B between Banach algebras A and B satisfies ((x)) (x)for all x A; moreover, equality holds if is bijective and (A1) = B1. Ofcourse, a spectrum preserving map is a spectral isometry. In this final chap-

ter, we shall study the analogue of Kaplanskys question regarding spectral

isometries. In this context, Kaplanskys question is reformulated as follows:

Let A and B be unital semi-simple Banach algebras. Let : A B be aunital surjective spectral isometry. Must be a Jordan isomorphism?

59


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CHAPTER 5. SPECTRAL ISOMETRIES 60

One of the most satisfactory results in this direction so far is that obtained

by Mathieu and Sourour. They showed that every unital, surjective spectral

isometry between finite-dimensional semisimple Banach algebras is a Jordan

isomorphism. This result is the theme of the present chapter.

5.1 Nagasawas theorem

We start with a result due to M. Nagasawa (Theorem 5.3 below), which states

that a unital, surjective spectral isometry between commutative semisimple

Banach algebras is a homomorphism. We point out that this result answers

Kaplanskys question for the case where the Banach algebras are commutative.

Our proof of Nagasawas theorem follows [Aup91]. To prepare for the proof,

we need a definition.

Definition 5.1 Let A be a commutative Banach algebra. A spectral state

on A is a linear functional f : A C such that |f(x)| r(x) for all x Aand f(1) = 1. The set of spectral states on A is denoted byA.

If f is a spectral state on A, the above definition shows that f = 1 andrad(A)

ker(f). Moreover, it is readily verified that A is a weak

-closed

(hence compact) convex subset of the unit ball of A. The extreme points of

A are naturally called the extreme spectral states on A. It is clear that a

multiplicative linear functional is a spectral state, since (x) (x) for allx A. So, co(A) A.

We quote from [Aup91] two basic results regarding spectral states. If S is

a subset of a vector space V, then co(S) denotes the convex hull ofS. The set


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of extreme points of a convex set C is denoted by ex(C).

Lemma 5.2 Let A be a commutative Banach algebra, and let f be a linear

functional on A.

(a) The functional f is a spectral state on A if and only if f(x) co((x))for all x A.

(b) Iff is an extreme spectral state, then f is multiplicative.

Part (b) of the lemma says that ex(A) A. It follows from the Krein-Milman theorem that A is equal to the weak

-closure of co(A).

Theorem 5.3 (Nagasawa) LetA andBbe commutative semisimple Banach

algebras. If : A B is a unital spectral isometry from A onto B, then is

an isomorphism.

Proof. First, note that since is a spectral isometry and A is semisimple,

we know from Lemma 1.25 that is injective. Thus, 1 is also a spectral

isometry.

Let f ex(B) and set g = f . As is a spectral isometry, clearlyg A. It is easy to see that g is extreme. Indeed, assume that g1, g2 Asuch that g = 12(g1 + g2). We shall verify that g = g1 = g2. Indeed, as

f = g 1, we have

f =1

2

g1 1 + g2 1

.

Since both g1 1, g2 1 B and f is extreme, it follows that

f = g1 1 = g2 1.


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Accordingly, g = g1 = g2.

Since f and g = f are extreme spectral states on B and A respectively,by Lemma 5.2, they are multiplicative. This implies that

f((xy)) = f((x))f((y)) = f((x)(y))

for all x, y A and f ex(B). By the Krein-Milman theorem,

f((xy)) = f((x)(y))

for all x, y A and f B. In particular, the above holds for all B. SinceB separates points ofB, we deduce that (xy) = (x)(y) for all x, y A.

Corollary 5.4 LetX and Y be semisimple Banach algebras and suppose that

: X Y is a unital surjective spectral isometry. Then restricts to an

algebra isomorphism from Z(X) onto Z(Y).

Proof. By Lemmas 1.25 and 1.28, we saw that maps Z(X) onto Z(Y).

Recall from Lemma 1.13 that both Z(X) and Z(Y) are semisimple. Now,

|Z(X) : Z(X) Z(Y) satisfies by inheritance all the properties of . ButZ(X) and Z(Y) are commutative. So Nagasawas Theorem 5.3 applies and

the proof is complete.

5.2 Spectral isometries and central idempo-

tents

Let X and Y be semisimple Banach algebras. Suppose that a Z(X) is anidempotent. We know from Corollary 5.4 that if : X Y is a unital surjec-


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tive spectral isometry, then (a) Z(Y) is also an idempotent. Therefore, maps central idempotents to central idempotents.

Let X and Y be semisimple Banach algebras, and let : X Y be a unitalsurjective spectral isometry. Now let a Z(X) be a central idempotent of Xand suppose b = (a). By the preceding remarks, b is a central idempotent

of Y. Take any x X and consider (1 b)(ax) Y. Multiplying x by a

complex number of modulus 1 if necessary, we can arrange that

r ((1 b)(ax)) ((1 b)(ax)) .

Suppose r((1 b)(ax)) > 0. Then for any > 0, we have

r((1 b)((ax) + 1)) > 1.

Furthermore, for > 0 small enough, we will get that r(ax + 1 a) = 1. Butthen,

1 = r(ax + 1 a) = r((ax) + 1 (a)) r((1 b)((ax) + 1) > 1,

a contradiction. We have shown that r((1 b)(ax)) = 0. Hence (1 b)(ax)is quasinilpotent. Therefore, (1 b)(aX) is a subspace of Y consisting ofquasinilpotent elements.

The next lemma states that, in fact, (1 b)(aX) = {0}.

Lemma 5.5 Let X and Y be semisimple Banach algebras, and suppose that : X Y is a unital surjective spectral isometry. Let a Z(X) be a centralidempotent and let b = (a). Then

(1 b)(aX) = {0}.

Furthermore, the map a : aX bY defined by

a(ax) := b(ax)


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is a unital surjective spectral isometry.

Proof. (Sketch) While we must refer the proof of the first claim to [MS04],

it is easy to derive the rest. The proof of the first claim depends upon the

subharmonicity of the spectrum, a theory we have not developed in this essay.

To see that a is a spectral isometry, take any x

X. Recall from the

preceding remark that (1 b)(ax) is quasinilpotent. Therefore,

r(ax) = r ((ax)) = r (b(ax) + (1 b)(ax))= max {r (b(ax)) , r ((1 b)(ax))}= max {r (b(ax)) , 0} = r (b(ax)) .

That a is unital follows from the fact that a and b are idempotents. It

remains to show that a(aX) = bY. The first claim, when applied to thecentral idempotent 1 a, implies that b ((1 a)X) = {0}. Let y Y. Since is surjective, y = (x) for some x X. It follows that

by = b(x) = b (ax + (1 a)x) = b(ax),

or, by = a(ax).

Proposition 5.6 Let X and Y be semisimple Banach algebras, and suppose

that : X Y is a unital surjective spectral isometry. Let a X be a centralidempotent and letb = (a). Then|aX is a unital surjective spectral isometry

fromaX onto bY.

Proof. In fact, we shall prove that |aX coincides with the map a defined in

Lemma 5.5, from which our claim follows. For any x X, by Lemma 5.5 we


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have (1 b)(ax) = 0. Hence,

(ax) = b(ax) + (1 b)(ax) = b(ax) = a(ax),

as required.

Corollary 5.7 Let Mi (i = 1, 2) be semisimple Banach algebras with the

property that every unital spectral isometry fromMi onto a semisimple Banach

algebra is a Jordan isomorphism. Then M1 M2 has the same property.

Proof. Consider the semisimple Banach algebra X = M1 M2, and let : X Y be a unital surjective spectral isometry. Let a = (1, 0) X, so thataX = M1 and (1 a)X = M2. Clearly, both a and 1 a = (0, 1) are centralidempotents of X. By Proposition 5.6, the maps |M1 : M1

(a)Y and

|M2 : M2 (1 (a))Y are unital spectral isometries which, by hypothesis,are both Jordan isomorphisms. Simple algebra shows that must be a Jordan

isomorphism as well. Indeed, for any x X, write x = (m1, m2) wheremi Mi for i = 1, 2. Then, since (m1) (a)Y and (m2) (1 (a))Y,

(x2) =

(m21, m22)

= (m21) + (m22) = (m1)

2 + (m2)2 = (x)2,

as desired.

5.3 The theorem of Mathieu and Sourour

The following theorem is classical (see [Rot02] or [Her68] for instance).


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Theorem 5.8 (Artin-Wedderburn) LetA be a finite-dimensional semisim-

ple algebra over C. Then there exist integers {ni, : 1 i k} N so that

A = Mn1(C) Mnk(C).

We shall quote one more result. Let sln be the space of n n complexmatrices with trace zero. Note that Mn(C) = CI sln. We remark that sln

is the linear span of the set Nn of nilpotent elements of Mn(C). Finally, letGLn(C) denote the group of invertible elements of Mn(C).

Theorem 5.9 ([BPW83]) Fix n 1. Let T : sln sln be an invertiblelinear transformation such that T(Nn) Nn. Then there exist C andA GLn(C) such that either

T(X) = AXA1 (X sln),

or

T(X) = AXtA1 (X sln).

Lemma 5.10 Forn 1, if : Mn(C) Mn(C) is a unital surjective spectralisometry, then is a Jordan automorphism of Mn(C).

Proof. First of all, in any finite-dimensional Banach algebra A, any quasinilpo-

tent element is nilpotent. Indeed, suppose that x A is quasinilpotent. SinceA is finite-dimensional, the powers of x must be linearly dependent; that is,

x is algebraic over C. Let p(t) be the minimal polynomial of x, which is the

unique monic polynomial of lowest degree such that p(x) = 0. By the spectral

mapping theorem, the roots of p(t) belong to (x) = {0}. But then p(t) = tn

for some n 1, which means that xn = 0.


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Specializing to Mn(C), it follows that the spectral isometry must pre-

serve the nilpotent elements of Mn(C). By Theorem 5.9, there exist C

and A GLn(C) such that either (X) = AXA1 for all X sln or(X) = AXtA1 for all X sln. Suppose that the former case occurs.Since is a spectral isometry, = 1. Therefore, for any C and X sln,

(I + X) = I+ AXA1 = A(I+ X)A1.

Hence (Y) = AY A1 for all Y Mn(C). Similarly, the latter case impliesthat (Y) = AYtA1 for all Y Mn(C). In either case, is a Jordan auto-morphism of Mn(C).

We are now ready to state and prove the main result of this chapter.

Theorem 5.11 ([MS04]) LetX andY be finite-dimensional semisimple Ba-nach algebras. If : X Y is a unital surjective spectral isometry, then isa Jordan isomorphism.

Proof. By the Artin-Wedderburn Theorem 5.8, we can write

X = Mn1(C) Mnk(C),Y = Mm1(C)

Mmp(C).

Let {a1, , ak} and {b1, , bp} be the families of minimal central idempo-

tents in X and Y respectively. Note that ajX = Mnj(C) and bjY = Mmj(C)for each index j, and also

Z(X) = Ca1 Cak,Z(Y) = Cb1 Cbp.


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