Operator theory and exotic Banach spaces

(Banach spaces with small spaces of operators)Bernard Maurey

This set of Notes is a largely expanded version of the mini-course “Banach spaceswith small spaces of operators”, given at the Summer School in Spetses, August 1994.The lectures were based on a forthcoming paper [GM2] with the same title by Tim Gowersand the speaker. A similar series of lectures “Operator theory and exotic Banach spaces”was given at Paris 6 during the spring of ’95 as a part of a program of three mini-coursesorganized by the “Equipes d’Analyse” of the Universities of Marne la Vallee and Paris 6.

We present in section 10, 11 and 12 several examples of Banach spaces which we call“exotic”. The first class is the class of Hereditarily Indecomposable Banach spaces (inshort H.I. spaces), introduced in [GM1]: a Banach space X is called H.I. if no subspace ofX is the topological direct sum of two infinite dimensional closed subspaces. One of themain properties of a H.I. Banach space X is the following: every bounded linear operatorT from X to itself is of the form λIX + S, where λ ∈ C, IX is the identity operator onX and S is strictly singular. It is well known that this implies that the spectrum of T iscountable, and it follows easily that a H.I. space is not isomorphic to any proper subspace.More generally, we present in section 11 a class of examples of Banach spaces having “few”operators. The general principle is the following: given a relatively small semi-group ofoperators on the space of scalar sequences (for example, the semi-group generated by theright and left shifts), we construct a Banach space such that every bounded linear operatoron this space is (or is almost) a strictly singular perturbation of an element of the algebragenerated by the given semi-group. We obtain in this way in section 12 a new prime space,a space isomorphic to its subspaces with finite even codimension but not isomorphic to itshyperplanes, and a space isomorphic to its cube but not to its square.

We have chosen to present a fairly detailed account of all the tools of general interestthat are necessary to the analysis, although these appear already in many classical books(but probably not in the same book); we develop elementary Banach algebra theory insection 2, Fredholm theory in section 4, strictly singular operators and strictly singularperturbations of Fredholm operators in section 6, an incursion into the K-theory for Banachalgebras in section 9. Ultraproducts and Krivine’s theorem about the finite representabilityof `p are also presented in sections 7 and 8, with some emphasis on the operator approachto these questions. The actual construction of our class of examples appears in section 11,and the applications to some specific examples in the last section 12.

1. Notation

We denote by X, Y, Z infinite dimensional Banach spaces, real or complex, and byE, F finite dimensional normed spaces, usually subspaces of the preceding. Subspaces areclosed vector subspaces. We write X = Y ⊕ Z when X is the topological direct sum oftwo closed subspaces Y and Z. The unit ball of X is denoted by BX . We denote by K thefield of scalars for the space in question (K = R or K = C).

We denote by L(X, Y ) the space of bounded linear operators between two (real orcomplex) Banach spaces X and Y . When Y = X, we simply write L(X). We denote

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by S, T, U, V bounded linear operators. Usually, S will be a “small” operator; it could besmall in operator norm, or compact, or finite rank or strictly singular. . . By IX we denotethe identity operator from X to X. An into isomorphism from X to Y is a bounded linearoperator T from X to Y which is an isomorphism between X and the image TX; this isequivalent to saying that there exists c > 0 such that ‖Tx‖ ≥ c‖x‖ for every x ∈ X. LetK(X, Y ) denote the closed vector subspace of L(X, Y ) consisting of compact operators (wewrite K(X) if Y = X).

A normalized sequence in a Banach space X is a sequence (xn)n≥1 of norm one vectors.The closed linear span of a sequence (xn)n≥1 is noted [xn]n≥1. A basic sequence is aSchauder basis for its closed linear span [xn]n≥1. This is equivalent to saying that thereexists a constant C such that for all integers m ≤ n and all scalars (ak)n

k=1 we have

∥∥m∑

k=1

akxk

∥∥ ≤ C∥∥

n∑

k=1

akxk

∥∥.

The smallest possible constant C is called the basis constant of (xn)n≥1.An unconditional basic sequence is an (infinite) sequence (xn)n≥1 in a Banach space

for which there exists a constant C such that for every integer n ≥ 1, all scalars (ak)nk=1

and all signs (ηk)nk=1, ηk = ±1, we have

∥∥n∑

k=1

ηkakxk

∥∥ ≤ C∥∥

n∑

k=1

akxk

∥∥.

The smallest possible constant C is called the unconditional basis constant of (xn)n≥1. Aquestion that remained open until ’91 motivated much of the research contained in theseNotes: does every Banach space contain an unconditional basic sequence (in short: UBS).The answer turned out to be negative and lead to the introduction of H.I. spaces.

2. Basic Banach Algebra theory

(For this paragraph, see for example Bourbaki, Theories spectrales, or [DS], or manyothers.) A Banach algebra A is a Banach space (real or complex) which is also an algebrawhere the product (a, b) → ab is norm continuous from A× A to A. This means that theproduct and the norm are related in the following way: there exists a constant C such thatfor all a, b ∈ A, we have

‖ab‖ ≤ C‖a‖ ‖b‖.It is then possible to define an equivalent norm on A satisfying the sharper inequality

∀a, b ∈ A, ‖ab‖ ≤ ‖a‖‖b‖.

We shall call a norm satisfying this second property a Banach algebra norm. In order todefine an equivalent Banach algebra norm from a norm satisfying the first property witha constant C, we may for example consider

|||a||| = sup‖ab + λa‖ : ‖b‖ ≤ 1, |λ| ≤ 1.

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We say that A is unital if there exists an element e ∈ A such that ea = ae = a for everya ∈ A; we write usually 1A for this element e. If A is unital, we get an equivalent Banachalgebra norm on A using the formula

|||a||| = sup‖ab‖ : b ∈ A, ‖b‖ ≤ 1and for this norm |||1A||| = 1. A Banach algebra norm with this additional property will becalled unital Banach algebra norm.

A C∗-algebra is a complex Banach algebra A with a Banach algebra norm and withan anti-linear involution a → a∗ (i.e. a∗∗ = a, (a+ b)∗ = a∗+ b∗, (λa)∗ = λa, (ab)∗ = b∗a∗

for every a, b ∈ A and λ ∈ C) and such that

∀a ∈ A, ‖a∗a‖ = ‖a‖2.If A is unital, then 1∗A = 1A and ‖1A‖ = 1. An element x ∈ A is Hermitian (or self-adjoint) if x∗ = x. Every a ∈ A can be written as a = x+ iy, where x and y are Hermitian(x = (a + a∗)/2, y = i(a∗ − a)/2). At some point we will need the self-explanatory notionof a C∗-norm on a not complete algebra with involution.

The above definition is not satisfactory in the real case. Indeed, if A is any real unitalsubalgebra of the complex algebra C(K) of continuous functions on a compact topologicalspace K, and if we define on A the trivial involution f∗ = f for every f ∈ A, then allproperties of the preceding definition hold (because λ is now restricted to R), but A isnot necessarily what we want to call a real C∗-algebra. In order to obtain a reasonabledefinition for the real case, we need to add an axiom which is consequence of the others inthe complex case, but not in the real case, for example

∀a, b ∈ A, ‖a∗a‖ ≤ ‖a∗a + b∗b‖.

Adding 1When A has no unit it is possible to embed A in a larger unital Banach algebra, by

considering on A+ = A⊕K the product (a, λ)(b, µ) = (ab+λb+µa, λµ). Then 1A+ = (0, 1)is the unit of A+ and A is a closed two-sided ideal in A+. When A is a C∗-algebra, it ispossible to define on A+ a C∗-norm.

An important example of unital Banach algebra is L(X) where X is a (real or complex)Banach space. The operator norm is a unital Banach algebra norm on L(X). The subspaceK(X) is a non unital closed subalgebra of L(X), actually a closed two-sided ideal of L(X).The algebra (K(X))+ is isomorphic to the subalgebra of L(X) consisting of all operatorsof the form T = λIX + K, K compact (recall that X is infinite dimensional). The Calkinalgebra C(X) = L(X)/K(X) is another important example. It will play a role for thenotion of essential spectrum later in this section.

When H is a Hilbert space, L(H) is a C∗-algebra. It is a fundamental example sinceevery C∗-algebra can be ∗-embedded in some L(H). The quotient of a C∗-algebra by aclosed two-sided ideal is a C∗-algebra for the quotient norm (it is true but not obviousthat for any such ideal I, we have x ∈ I ⇒ x∗ ∈ I); in particular the Calkin algebra C(H)of a Hilbert space H is a C∗-algebra.

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Complexification

Most of spectral theory is done when the field of scalars is C. It is therefore importantto be able to pass from the real case to the complex case. If X is a real Banach space, thecomplexified space is XC = X ⊕X with the rule

i(x, y) = (−y, x).

In this way we have (0, y) = i(y, 0) and identifying (x, 0) ∈ XC with x ∈ X, we see thatevery z ∈ XC can be written as z = x + iy with x, y ∈ X.

In order to define a complex norm on XC it is useful to think about XC as beingC⊗X; it is clear that any tensor norm on C⊗X will give in particular a complex normon XC (of course we choose the modulus as norm on the 2 dimensional real space C). Wecan consider for example the injective tensor norm C⊗ε X,

‖x + iy‖ = sup|x∗(x) + ix∗(y)| : x∗ ∈ X∗, ‖x∗‖ ≤ 1.

With this norm C⊗X is isometric to (the real space) L(X∗,C). A (real) linear functionalx∗ on X is extended to a complex linear functional on XC by setting simply x∗(x + iy) =x∗(x) + ix∗(y), and it is easy to see that every complex linear functional on XC can beobtained as x∗ + iy∗, where x∗, y∗ ∈ X∗. Given T ∈ L(X, Y ), one gets a complex linearoperator TC ∈ L(XC, YC) by setting TC = Id ⊗ T in the tensor product language. In theprevious language we can write TC(x + iy) = Tx + iTy for all x, y ∈ X.

When A is a real Banach algebra, it can be checked that if we norm AC by AC = C⊗πA,we get a complex Banach algebra norm (if A had one); if A = L(X), X real, then ACidentifies with L(XC) and any complex norm on XC gives a Banach algebra norm on AC;given V = T + iU in AC, T, U ∈ A = L(X), we associate the operator on XC = X ⊕ Xgiven by the matrix

V =(

T −UU T

).

Invertible elements

Let A be a unital Banach algebra over K. We say that a ∈ A is invertible (in A) ifthere exists x ∈ A such that ax = xa = 1A.Lemma 2.1. Let A be a unital Banach algebra (with a Banach algebra norm). If b ∈ Aand ‖b‖ < 1 then 1− b is invertible in A.

Proof. (1− b)−1 =∑∞

n=0 bn.Corollary 2.1. Let A be a unital Banach algebra. The set of invertible elements is openin A. Furthermore, if K = C, when a ∈ A is invertible, the function f(z) = (a − zb)−1 isanalytic in a neighborhood of 0 in C for every b ∈ A.Proof. Write a− b = a(1− a−1b) and use Lemma 2.1. If ‖b‖ < ‖a−1‖−1, then ‖a−1b‖ < 1,and we obtain the following formula for the inverse

(a− b)−1 = u + ubu + ububu + · · · ,

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where u = a−1. Applying to zb instead of b clearly gives an analytic function of z in aneighborhood of 0 in C,

(a− zb)−1 = u + z ubu + z2 ububu + · · · .

Remark 2.1. The same proof works when a is replaced by T ∈ L(X, Y ) invertible, andb is replaced by a small operator S ∈ L(X, Y ); the above formulas hold with u = T−1 ∈L(Y, X), provided ‖S‖ < ‖T−1‖−1.

Spectrum of an element, resolvent set

Let A be unital over C and let a ∈ A. The resolvent set ρ(a) is the set of all λ ∈ Csuch that λ1A−a is invertible in A. The spectrum σ(a) is the complementary set C \ ρ(a).This set ρ(a) is open by Corollary 2.1 and is clearly a neighborhood of infinity, hence σ(a)is a closed and bounded subset of C.Exercise 2.1. Let a, b ∈ A. Show that σ(ab)\0 = σ(ba)\0 (hint: if z(λ1A−ab) = 1A,find u ∈ A such that u(λ1A − ba) = λ1A). Why did we exclude 0?

Real spectrum

If A is a real unital Banach algebra and if a ∈ A we may define a real spectrum by

σR(a) = λ ∈ R : (a− λ1A) not invertible in A.

The problem is that this spectrum may be empty (contrary to what happens in the complexcase: we shall recall that the spectrum is non empty in the complex case). If we extend thescalars and consider the same a ∈ A as an element of AC, we can work with the complexspectrum σAC(a).

Exercise. Let A be a real unital Banach algebra and let a ∈ A.1. Show that σAC(a) = σAC(a)2. Show that σR(a) = R ∩ σAC(a).

In what follows we shall denote by σK(a) the real spectrum when A is real and thecomplex spectrum if A is complex.

Changing the ambient algebra

Assume that A is a closed subalgebra of a unital Banach algebra B, with the inducednorm and containing 1B . If x ∈ A is invertible in A, it is obviously also invertible inB, but it is possible for x ∈ A to be invertible in B but not in A. It is necessary inthis case to distinguish the spectrum of x relative to B or relative to A; we denote thespectrum by σKA(x) or σKB(x), and similarly for the resolvent sets. If x ∈ A, it is clear thatσKA(x) ⊃ σKB(x) (equivalently, ρKA(x) ⊂ ρKB(x)). We also have ∂σKA(x) ⊂ ∂σKB(x) from thenext Lemma.

Lemma 2.2. Let B be a unital Banach algebra, and let A be a closed subalgebra with1B ∈ A. For every x ∈ A, the resolvent set ρKA(x) is open and closed in ρKB(x). Hence, forevery connected component ω of ρKB(x), either ω ⊂ ρKA(x) or ω∩ρKA(x) = ∅. It follows that∂ρKA(x) = ∂σKA(x) ⊂ ∂σKB(x) = ∂ρKB(x). If ρKB(x) is connected, then σKA(x) = σKB(x).

5

Proof. Let x be fixed in A and write simply ρA, ρB for ρKA(x) and ρKB(x). We know that ρA isopen. Let (λn) ⊂ ρA, λn → λ ∈ ρB . There exists b ∈ B such that (x−λ)b = b(x−λ) = 1B .On the other hand, for every n there exists an ∈ A such that (x−λn)an = 1B . Multiplyingby b we get

b = (bx− λnb)an = (1B + (λ− λn)b)an;

when n →∞ we see that (1B + (λ− λn)b)−1 tends to 1B , thus an → b, hence b ∈ A andλ ∈ ρA.

Let λ ∈ ∂ρA. Then λ /∈ ρA since ρA is open in K, and therefore λ /∈ ρB since ρA isclosed in ρB ; this implies that λ ∈ ∂ρB .

Remark. When σA(x) 6= σB(x), we see that the interior int(σA(x)) must be non empty.

Examples 2.1.1. It will be shown later (section 6, Proposition 6.1) that the spectrum (in B = L(X))

of a strictly singular operator S on a complex Banach space X is a countable compactsubset of C. This implies that the resolvent set ρB(S) is connected, hence the inverse ofλIX − S, when it exists, belongs to the closed subalgebra A of L(X) generated by IX andthe operator S.

2. Suppose that A is a complex unital C∗-subalgebra of L(H) and let u ∈ A behermitian. Since the spectrum of u in B = L(H) is contained in R and bounded, it is clearthat the resolvent set ρB(u) is connected, hence σA(u) = σB(u). This remark implies easilythat σA(x) = σB(x) for every x ∈ A (it is enough to show that when x ∈ A is invertiblein B its inverse belongs to A; if x is invertible in B, then the hermitian operators x∗x andxx∗ are invertible in B, hence in A, therefore x is invertible in A: there exist y, z ∈ A suchthat y(x∗x) = (xx∗)z = 1A, and yx∗ = x∗z is the inverse of x in A). More generally, ifan isomorphism from H ⊕H onto H is given by a matrix (a, b) with a, b ∈ A, the inverseoperator from H to H ⊕H is given by a matrix with two entries in A.

3. We can always embed a Banach algebra A in the space of bounded linear operatorson some Banach space. Suppose that A is a unital Banach algebra with a unital Banachalgebra norm. We simply embed A into L(A) by mapping each a ∈ A to the operator Ma

of left multiplication by a, defined by Ma(x) = ax for every x ∈ A. It is clear that thisgives an isometric embedding from A into L(A). We show now that the spectrum of a ∈ Ais the same relative to A or to the larger algebra B = L(A). We only need to show that ais invertible in A iff Ma is invertible in B. One direction is obvious. In the other direction,assume that Ma is invertible in B. Then Ma is onto and there exists u ∈ A such that1A = Ma(u) = au. We see that MaMu = 1B . Since Ma is invertible in B, this impliesMuMa = 1B and ua = au = 1A shows that a is invertible in A.

4. Suppose that A is a real unital Banach algebra; if a ∈ A is invertible in AC, thena is invertible in A.

5. Consider the embedding of L(X) into L(X∗) given by the adjoint (this is notexactly an algebra morphism since (UT )∗ = T ∗U∗); the spectrum of T ∗ in L(X∗) is thesame as the spectrum of T ∈ L(X) (of course this is totally obvious when X is reflexive).

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Spectral radius

Let a ∈ A. The spectral radius of a is defined by

r(a) = limn‖an‖1/n.

Exercise. Show that the above limit exists.Example 2.2. If u is hermitian in a C∗-algebra, we get ‖u2n‖ = ‖u‖2n for every integern ≥ 1, thus r(u) = ‖u‖.

Notice that the spectral radius is not changed if the norm on A is replaced by anequivalent norm. Also, r(a) ≤ ‖a‖ if the norm is a Banach algebra norm.

Proposition 2.1. (K = C and 1A 6= 0) The spectrum σ(a) is contained in the closed disc∆(0, r(a)) in C centered at 0 and of radius r(a), and it intersects the circle of radius r(a).In other words,

r(a) = max|λ| : λ ∈ σ(a).In particular, σ(a) is non empty.

Proof. The function g(z) = (1A − za)−1 is clearly defined and analytic when |z| < r(a)−1,hence σ(a) ⊂ ∆(0, r(a)). Since 1A 6= 0 we may assume ‖1A‖ = 1 for some equivalent unitalBanach algebra norm. If a is invertible, it is then easy to see that r(a)r(a−1) ≥ 1, hencer(a) > 0. If r(a) = 0, we know therefore that a is not invertible, thus 0 ∈ σ(a) and finallyσ(a) = 0 in this case.

Assume now r(a) > 0. If the inverse (1A − za)−1 exists for every z on the circle|z| = r(a)−1, we can show that the function g is analytic in a neighborhood of a closed discof radius R > r(a)−1. It follows then from Cauchy’s inequalities that for some constantM , we have

∀n ≥ 0, ‖an‖ ≤ M

Rn

yielding r(a) ≤ 1/R and contradicting the choice of R.

Resolvent equation. Spectral projections

Let A be a unital Banach algebra over C and let a ∈ A. The resolvent operator of ais defined for z ∈ ρ(a) by R(z) = (z1A − a)−1. Note that R(z) and R(z′) commute andcommute with a. We have

R(z′)−R(z)z′ − z

= −R(z′)R(z).

Let λ ∈ C be isolated in σ(a). Let ∆r be the closed disc ∆(λ, r) with radius r centeredat λ; let r0 > 0 be such that λ is the unique point of σ(a) contained in ∆r0 ; let γr bethe boundary of ∆r, oriented in the counterclockwise direction. Let f be holomorphic ina neighborhood V of λ; for r > 0 such that ∆r ⊂ V and r ≤ r0 we know that γr ⊂ ρ(a)and we set

Φ(f) =1

2iπ

∫

γr

f(z)R(z) dz ∈ A.

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Since f is holomorphic, the result does not depend upon the particular value r ∈ (0, r0]such that ∆r ⊂ V . Also Φ(f) commutes with a and with every R(µ). The mappingf → Φ(f) is clearly linear with respect to f . What is more interesting is that

Φ(fg) = Φ(f) Φ(g).

For the proof let 0 < r < s ≤ r0 be such that f and g are holomorphic in a neighborhoodof ∆s and write

Φ(f) Φ(g) =1

2iπ

∫

z∈γr

12iπ

∫

z′∈γs

f(z)g(z′)R(z)R(z′)dzdz′;

use then the resolvent equation and Cauchy’s formula. It follows that for every r with0 < r ≤ r0

p = Φ(1) =1

2iπ

∫

γr

(z1A − a)−1dz

is an idempotent (p2 = p), commuting with a and with every R(µ); furthermore for everyelement b = Φ(g) ∈ A we have

pb = bp = b = pbp

since pΦ(g) = Φ(1) Φ(g) = Φ(g).To every idempotent p in A we may associate the Banach algebra Ap = pAp of all

elements of the form pap, a ∈ A. As a Banach algebra, the norm and the product in Ap arethose of A, but the unit of Ap is p. For example, let p be a bounded projection defined onsome Banach space X, and let Y = p(X) be the range of p. We see that L(X)p identifieswith L(Y ) (with an equivalent norm).

Let us come back to our isolated λ ∈ σ(a) and let again p = Φ(1). The above remarkshows that Φ(f) ∈ Ap for every f holomorphic in a neighborhood of λ. Also notice that

Φ(z)− ap =1

2iπ

∫

γr

(z1A − a)R(z) dz = 0,

so that ap = Φ(z). Suppose that f(λ) 6= 0; then g = 1/f is holomorphic in a neighborhoodof λ, and Φ(f)Φ(1/f) = Φ(1) = p. It follows that Φ(f) is invertible in Ap when f(λ) 6= 0.This applies in particular to f(z) = z − µ when µ 6= λ to show that Φ(z − µ) = ap− µp isinvertible in Ap. This shows that the spectrum of pa = pap in Ap reduces to λ. Lettingq = 1A − p, the spectrum of qa in Aq does not contain λ. This follows from

(λ1A − a)( 1

2iπ

∫

γr

R(z)z − λ

dz)

= q.

More generally, when the spectrum σ(a) can be decomposed into two subsets σ1 andσ2 open and closed in σ(a), we can construct similarly a spectral projection p by replacingthe above circle γr by a curve γ around σ1, such that σ2 is exterior to γ. We still getpa = ap and σ(pa) = σ1 (in Ap) and similarly in Aq we have that σ(qa) = σ2.

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Exercise. Suppose that ‖a2 − a‖ < ε < 1/4; show that there exists an idempotent p suchthat ap = pa and ‖p − a‖ < f(ε, ‖a‖) (prove that σ(a) is contained in the union of theinteriors of two circles γ1 and γ0 with radius 1/2 and centered at 1 and 0, for example byconsidering (a−t)(a−(1−t)) for |t(1−t)| ≥ 1/4; give an upper bound for ‖(z−a)−1‖ whenz belongs to γ0 or γ1; let Φ0 and Φ1 be the operators as above associated with the twocircles, and consider p = Φ1(1); use b = Φ1(1/z) for proving that (p − ap) = (1A − a)abpis small, and similarly for (1A − p)a).

Essential spectrum

Let X be an infinite dimensional Banach space. The ideal K(X) is then proper andthe Calkin algebra C(X) = L(X)/K(X) is not 0. For every T ∈ L(X), the essentialspectrum σK(T ) is the spectrum of the image T of T in C(X). We also consider thecorresponding resolvent set ρK(T ). A scalar λ ∈ K belongs to this essential resolvent setiff T − λIX is invertible modulo compact operators. We shall recall in section 6 that thishappens iff T − λIX is a Fredholm operator on X.

Commutative Banach algebras

A (complex) Banach algebra A which is a (skew-)field is isomorphic to C. Indeed, leta ∈ A and λ ∈ σ(a). Since a − λ1A is not invertible, we must have a − λ1A = 0. Henceevery element of A is of the form λ1A for some λ ∈ C.

Let A be a unital commutative Banach algebra over C. A maximal ideal I is closedand is an hyperplane: first, I is closed because the set of invertible elements is open, hencethe closure of I is still a proper ideal, equal to I since I is maximal; second, A/I is a Banachfield, therefore isomorphic to C, and I is thus an hyperplane. The linear functional χ suchthat kerχ = I, normalized by the condition χ(1A) = 1, is called a character. A characterχ is a non zero bounded linear functional on A which is also multiplicative. Indeed, ifa ∈ A and χ(a) 6= 0, let g(x) = χ(ax)/χ(a). Then g vanishes on I, so g is proportional toχ, but χ(1A) = g(1A) = 1, thus χ = g and χ(ax) = χ(a)χ(x) for every x ∈ A.

The set of characters on A is called spectrum of A, and denoted by Sp(A); supposethat A is equipped with a Banach algebra norm; then Sp(A) is a subset of the unit sphereof the dual space A∗; to see this, observe that the sequence (an)n≥0 is bounded when‖a‖ ≤ 1, so that (χ(an)) = (χ(a)n) is bounded, therefore |χ(a)| ≤ 1, and ‖χ‖ = 1 sinceχ(1A) = 1.

An element a ∈ A is invertible in A iff χ(a) 6= 0 for every character χ on A. Indeed,it is clear that χ(a) 6= 0 when a is invertible; if a is not invertible, aA is a proper idealin A, thus contained in a maximal ideal I. If χ is the character such that I = ker χ thenχ(a) = 0. It follows that for every a ∈ A,

σ(a) = χ(a) : χ ∈ Sp(A);

since χ(a − χ(a)1A) = 0, we see that χ(a) ∈ σ(a) for every χ ∈ Sp(A). Conversely, ifa− λ1A is not invertible, there exists a character χ such that χ(a− λ1A) = 0.

Furthermore Sp(A) is w∗-closed in the unit sphere; this allows to map A to the spaceC(Sp(A)) of continuous functions on the compact space Sp(A); for every a ∈ A, let j(a)

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be the continuous function on Sp(A) defined by j(a)(χ) = χ(a); this map j need not beinjective in general.

When A is a unital (complex) commutative C∗-algebra, this embedding is isometric;we first observe that when u is hermitian, a = eiu is unitary, i.e. a∗a = aa∗ = 1A; thena∗ = a−1 and ‖a‖ = ‖a−1‖ = 1; this implies that the sequence (an)n∈Z is bounded, thus(χ(an))n∈Z is bounded and therefore 1 = |χ(a)| = | eiχ(u) |, hence χ(u) is real for everyhermitian u; this implies

∀a ∈ A, χ(a∗) = χ(a).

If u is hermitian, we have r(u) = ‖u‖ by Example 2.2, hence there exists χ ∈ Sp(A) suchthat |χ(u)| = ‖u‖. This implies that for every a ∈ A, there exists χ such that

|χ(a)|2 = χ(a∗a) = ‖a∗a‖ = ‖a‖2,

showing that the mapping j : A → C(Sp(A)) is isometric in this case. The image of A intoC(Sp(A)) is now a subalgebra closed under complex conjugation and obviously separatingpoints of Sp(A), therefore our embedding is onto by Stone-Weierstrass’ theorem.

Suppose that ϕ is a (unital) ∗-homomorphism between two unital commutative C∗-algebras, then ‖ϕ‖ ≤ 1; by the preceding paragraph we may think that A = C(K) and B =C(L), where K and L are two compact topological spaces. Since ϕ is a ∗-homomorphism,it sends every function f ≥ 0 on K to ϕ(f) ≥ 0 (introduce the hermitian element g =

√f).

If ‖f‖ ≤ 1, then f∗f ≤ 1, so that ϕ(f∗f) ≤ 1 and ‖ϕ(f)‖ ≤ 1; furthermore if ϕ isinjective then ϕ is isometric; this is because the adjoint map ϕ∗ : (C(L))∗ → (C(K))∗

sends Sp(B) ' L into Sp(A) ' K, and must be onto by the preceding result (otherwisewe may find a continuous function f 6= 0 supported on K \ϕ∗(Sp(B)), and then ϕ(f) = 0,contradicting the injectivity). Observe that we don’t need B to be commutative in theabove argument, because the range ϕ(A) is commutative, so that its closure in B is acommutative unital C∗-algebra.

Proposition 2.2. Let B be a non necessarily commutative unital C∗-algebra. For everyunital C∗-algebra C, every unital ∗-homomorphism ϕ : B → C satisfies ‖ϕ‖ ≤ 1. If ϕ isinjective, then ϕ is isometric.

Proof. Given an hermitian element u ∈ B, we may consider the unital subalgebra A of Bgenerated by u. This is a commutative C∗-algebra. We have seen that the spectrum of uin A is real, hence σA(u) = σB(u) by Lemma 2.2. Suppose that ϕ is a ∗-homomorphismfrom B to some C∗-algebra; restricting ϕ to A, it follows from the preceding remark that‖ϕ(u)‖ ≤ ‖u‖, and this is true for every hermitian u ∈ B. For a general b ∈ B, we write

‖ϕ(b)‖2 = ‖(ϕ(b))∗ϕ(b)‖ = ‖ϕ(b∗b)‖ ≤ ‖b∗b‖ = ‖b‖2.

Suppose further that ϕ is injective; then ϕ is isometric; indeed, the preceding remarksshow that ‖ϕ(u)‖ = ‖u‖ when u is hermitian; for a general b ∈ B, we write

‖ϕ(b)‖2 = ‖(ϕ(b))∗ϕ(b)‖ = ‖ϕ(b∗b)‖ = ‖b∗b‖ = ‖b‖2.

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Wiener’s algebra

The Wiener algebra W is the algebra of continuous (complex) functions on T withabsolutely summable Fourier coefficients; the product is the pointwise product and thenorm is the `1-norm of Fourier coefficients. This algebra W is clearly isometric to `1(Z)with its usual norm and the convolution as product.

A function f in W is invertible in W iff f does not vanish on T. This amountsto showing that the characters on W reduce to evaluation at points of T. Let χ be acharacter on W and let λ = χ(eiθ). Since the family (einθ)n∈Z is bounded in W , itfollows that the sequence (λn = χ(einθ))n∈Z is bounded, hence |λ| = 1. For every functionf(θ) =

∑n∈Z an einθ in W ,

χ(f) =∑

n∈Zanλn = f(λ).

If f does not vanish on T, we see that χ(f) 6= 0 for every character χ, hence f is invertiblein W .

C∗-algebra summary

(see first pages of Pedersen’s book [Pd])Let A be a unital (complex) C∗-algebra. For every hermitian element b ∈ A, we may

consider the unital subalgebra B generated by b; it is a commutative C∗-algebra. Weknow that jB : B → C(Sp(B)) is an onto isomorphism. For example, when b = b∗ andσA(b) ⊂ [0,∞), then σB(b) = σA(b) by Lemma 2.2; the function jB(b) is a non-negativecontinuous function on Sp(B), therefore there exists c ∈ B which is the “square root” ofb, i.e. c = c∗, b = c2 and σ(c) ⊂ [0,∞). Consider

C = a ∈ A : a = a∗ and σ(a) ⊂ [0,∞).

If b = b∗ and a = b2, then a ∈ C; this follows from the commutative theory, but alsosimply from the fact that σ(b) ⊂ R which implies that for every t > 0, b2 + t1A =(b + i

√t1A)(b− i

√t1A) is invertible.

If a = a∗ and ‖t1A − a‖ ≤ t for some t ≥ 0, then a ∈ C. This is clear when t = 0;when t > 0, we have ‖1A − a/t‖ ≤ 1 and we may define b =

√1A − (1A − a/t) = b∗

from its Taylor series. Then a = (√

t b)2 ∈ C. Conversely, if a ∈ C and t = ‖a‖, then‖t1A− a‖ ≤ t, so that we got a characterization of C; this last fact is because we know forevery hermitian element a that r(a) = ‖a‖ = t, and

‖t1A − a‖ = r(t1A − a) = max|t− λ| : λ ∈ σ(a) ⊂ [0, t] ≤ t.

It follows immediately from this characterization that a1, a2 ∈ C implies a1 + a2 ∈ C,so that C is a closed convex cone in A, with C ∩ (−C) = 0.

The next essential step is to prove that a∗a ∈ C for every a ∈ A. This is easy once weknow that we may embed A as a ∗-subalgebra of some L(H), but it is not obvious fromthe abstract definition, and it is a main ingredient for the proof of the representation ofan abstract C∗-algebra as a subalgebra of some L(H).

11

Observe first that for every b = r + is ∈ A, r, s hermitian, we have that b∗b + bb∗ =2(r2 +s2) ∈ C. Let a ∈ A. The commutative theory implies that we can write a∗a = u−v,with u, v ∈ C and uv = vu = 0. Let b = a

√v; then b∗b = −v2 ∈ −C, and bb∗ ∈ C since

b∗b + bb∗ ∈ C; by Exercise 2.1, this implies that σ(b∗b) = 0, hence r(b∗b) = ‖b∗b‖ =‖b‖2 = 0, thus v2 = 0, and finally v = 0, a∗a = u ∈ C.

For every u 6= 0, we know that −u∗u /∈ C. We may therefore find by Hahn-Banacha linear functional ξ on A such that −ξ(u∗u) ≤ inf ξ(C); this yields that ξ ≥ 0 on C. wedefine a scalar product on A by

〈a, b〉 = ξ(a∗b).

To every a ∈ A we associate the operator Ta(b) = ab. Let a ∈ A with ‖a‖ ≤ 1. We have

〈Ta(b), Ta(b)〉 = ξ(b∗a∗ab);

Since ‖a‖ ≤ 1, we know that 1A − a∗a = c2 ∈ C and b∗b − b∗a∗ab = b∗c2b ∈ C, thereforeξ(b∗a∗ab) ≤ ξ(b∗b) = 〈b, b〉. This shows that

∀a ∈ A, ‖Ta‖L(H) ≤ ‖a‖,

where H is the Hilbert space obtained from the above scalar product. It is easy to checkthat (Ta)∗ = Ta∗ , so that we have a ∗-homomorphism from A to L(H). We may improvethe argument and obtain ‖Tu‖ = ‖u‖ for the given u and then find an isometric embeddingof A into some L(H) using a direct sum of such embeddings (if ‖u‖ = 1, observe that theclosed convex cone u∗u + C is disjoint from the open unit ball, and use the separationtheorem as before to obtain ξ such that ξ(u∗u) = ‖ξ‖ = 1).

Suppose now that I is a closed two-sided ideal in A, and let a ∈ I. Let |a| =√

a∗a;then |a| ∈ I; indeed, there exists a sequence (Pn) of polynomials with real coefficients suchthat Pn(0) = 0 and such that (Pn(t)) converges to

√t uniformly on any compact interval

[0, T ], so that for every x ∈ C ∩ I, we have√

x ∈ I. For every ε > 0, ε21A + a∗a isinvertible, thus there exists uε ∈ A such that

a =√

ε21A + a∗a uε.

We obtain from the commutative theory

‖uε‖2 = ‖(ε21A + a∗a)−1/2a∗a(ε21A + a∗a)−1/2‖ ≤ 1.

When ε → 0 we obtain that a = limε→0 |a|uε hence

a∗ = limε

u∗ε|a| ∈ I.

Finally, let us say some words about the real case. Let A be a real Banach algebrawith involution and with a Banach algebra norm such that

∀a, b ∈ A, ‖a‖2 ≤ ‖a∗a + b∗b‖.

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This implies as in the complex case that every hermitian element x ∈ A as a real spectrumin AC (we don’t claim so far that AC can be equipped with a C∗-norm). Indeed, a = cos txand b = sin tx are hermitian for every real t, and a2 + b2 = 1A, therefore by our hypothesis‖ cos tx‖ ≤ 1 and ‖ sin tx‖ ≤ 1 for every real t, which implies that eitx is bounded in AC,from which it follows that the spectrum of x is real. With this information, it is possibleto reproduce the arguments from the beginning of this paragraph and to ∗-embed A inL(H), for some real Hilbert space H.Exercise. Complete the details. Show first that if B is the (real) subalgebra generated byan hermitian element a ∈ A, and if x, y ∈ B,

‖x + iy‖ = ‖x2 + y2‖1/2A

is a C∗-norm on BC. If v ∈ A is anti-hermitian, i.e. v∗ = −v, observe that ev∗ ev = 1A

and ‖ ev ‖ = 1. It follows that σ(v) ⊂ iR, and v2 − t21A is invertible for t real, t 6= 0, thusσ(−v2) ⊂ [0, +∞).

3. Some operator theory: finitely singular operators

See for example [LT1], section 2.c.Lemma 3.1. Let X and Y be real or complex Banach spaces.

1. Trivial principle: if T is an isomorphism from X into Y , a small norm perturbationT + S of T is still an into isomorphism.

2. Fundamental principle: if T is an isomorphism from X onto Y , a small normperturbation T + S is still an isomorphism from X onto Y (this is Remark 2.1).

Lemma 3.2. Let T ∈ L(X,Y ) be an into isomorphism and k ≥ 0 an integer.1. Suppose that codim TX ≥ k. There exists c > 0 such that codim(T + S)X ≥ k

whenever ‖S‖ < c.2. Suppose that codim TX = k. There exists d > 0 such that codim(T + S)X = k

whenever ‖S‖ < d.

Proof. If codim TX ≥ k we can find a subspace F ⊂ Y such that dimF = k and TX∩F =0. Let πF denote the quotient map Y → Y/F . Then πF T is an isomorphism from Xinto Y/F . If c > 0 is small enough and ‖S‖ < c, πF (T + S) is also an into isomorphismby Lemma 3.1. This implies that F ∩ (T + S)X = 0, hence codim(T + S)X ≥ k.

In the second case the proof is similar, but we can now select a subspace F such thatY = TX ⊕ F , dim F = k. Then πF T is an onto isomorphism, hence there is d > 0 (wemay choose d < c), such that πF (T + S) is an onto isomorphism when ‖S‖ < d. By theabove argument we already know that F ∩ (T + S)X = 0; furthermore, for every y ∈ Y ,there exists x ∈ X such that πF (y) = πF ((T + S)x), so y − (T + S)x ∈ F , showing thatY = F + (T + S)X. Finally Y = F ⊕ (T + S)X and codim(T + S)X = k.

Proposition 3.1. Let T ∈ L(X, Y ) be an into isomorphism. Then T + S is an intoisomorphism and codim(T+S)X = codim TX (finite or +∞) for every S in a neighborhoodof 0 in L(X, Y ).Proof. Let c > 0 be such that T + S is an into isomorphism whenever

S ∈ Bc = U ∈ L(X,Y ) : ‖U‖ < c.

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Observe that Dk = U ∈ Bc : codim(T + U)X = k is open and closed in Bc for everyinteger k ≥ 0. Indeed, the set W ∈ Bc : codim(T + W )X ≥ k + 1 is open by Lemma3.2, part 1, while each Dj , j = 0, 1, . . . , k is open by part 2 of the same Lemma. Since Bc

is connected, each Dk is empty or equal to Bc. The result follows.

Boundary of spectrum lemma

Lemma 3.3. If U ∈ L(X) is an into isomorphism but is not invertible in L(X), then 0belongs to the interior int(σK(U)) (relative to K) of σK(U).

Proof. Since U is an into isomorphism but not invertible, U is not onto and codim UX ≥ 1.This remains true under small perturbation by Lemma 3.2, part 1: there exists ε0 > 0such that U − εIX is not onto, therefore not invertible, for every ε ∈ K such that |ε| < ε0.This shows that B(0, ε0) ∩K ⊂ σK(U).

Corollary 3.1. Let T ∈ L(X). If λ ∈ ∂σK(T ) (of course this boundary is relative to K),there exists a normalized sequence (xn) in X such that (T − λIX)xn → 0 (this sequenceis possibly constant).

Proof. What we want to prove is equivalent to saying that U = T − λIX is not an intoisomorphism. By our assumption, we have that U is not invertible, but 0 is not interior toσK(U), hence U is not an into isomorphism by Lemma 3.3.

Exercise. If X is real, T ∈ L(X) and if λ = r(cos θ + i sin θ), r sin θ 6= 0, belongs tothe boundary of σ(TC), then there exist two sequences (xn) and (yn) in X such that‖xn‖+ ‖yn‖ = 1, Txn − r(cos θ xn − sin θ yn) → 0 and Tyn − r(sin θ xn + cos θ yn) → 0.

Remark 3.1. Let A be a unital Banach algebra. If λ belongs to the boundary of σK(a),there exists a normalized sequence (bn) in A such that abn − λbn tends to 0 in A. Thisfollows from section 2, example 2.1,3 (we could also get a normalized sequence (cn) suchthat cna− λcn goes to 0).

Definition 3.1. Let T ∈ L(X, Y ); we say that T is finitely singular if there exists c > 0and a (closed) finite codimensional subspace X0 ⊂ X such that

‖Tx‖ ≥ c‖x‖

for every x ∈ X0. In other words the restriction of T to X0 is an into isomorphism.

Let c(T ) denote the supremum of all c > 0 for which the above property holds, andset c(T ) = 0 if T is not finitely singular.

Our terminology is not classical and perhaps a little strange, since we call a nonsingular operator, for instance an onto isomorphism, “finitely singular”; it would be betterto say “at most finitely singular”, but this is definitely too long.

Remark 3.2. Suppose that T ∈ L(X, Y ) is finitely singular. It is clear that a small normperturbation of T is still finitely singular (precisely, T +S is finitely singular if ‖S‖ < c(T );actually it is enough that ‖S|X1‖ < c(T ) for some finite codimensional subspace X1 of X).It is also clear that the restriction of T to any infinite dimensional subspace Z of X isfinitely singular.

14

If T ∈ L(X,Y ) and if UT is finitely singular for some U ∈ L(Y, Z), then T is finitelysingular.Exercise 3.1. Suppose that T ∈ L(X,Y ) is finitely singular. Show that

1. kerT is finite dimensional.2. For every (closed) subspace Z of X, T (Z) is a closed subspace of Y .3. If (xn) is a bounded sequence in X and if (Txn) converges in Y , then there exists

a norm-converging subsequence (xnk); in other words, the restriction of T to any bounded

subset of X is a proper map.4. One can choose X0 in Definition 3.1 in such a way that X = kerT ⊕X0. Hence, if

kerT = 0, then T is an isomorphism from X into Y .5. Let T ∈ L(X, Y ). Show that T is finitely singular if and only if TX is closed and

dimker T < ∞.6. If T is finitely singular from X to Y and K = R, the complexified operator TC is

finitely singular from XC to YC.

Lemma 3.4. Let T ∈ L(X,Y ). Then T ∗ is finitely singular from Y ∗ to X∗ if and only ifTX is closed and finite codimensional in Y .

Proof. Suppose that T ∗ is finitely singular. Since kerT ∗ = (TX)⊥ is finite dimensional,we know that TX is finite codimensional. It is enough to show that for some c > 0 andfor every y ∈ TX, there exists x ∈ X with ‖y − Tx‖ ≤ ‖y‖/2 and ‖x‖ ≤ ‖y‖/c (the endof the proof is by iteration: one constructs a convergent series x′ =

∑xn in X such that

y = Tx′). If the preceding claim is not true, we can find for every integer n ≥ 1 a vectoryn ∈ TX such that ‖yn‖ = 1 and

yn /∈ nT (BX) +12BY .

By Hahn-Banach there exists y∗n ∈ Y ∗ such that y∗n(yn) = 1 and ‖y∗n‖ ≤ 2, ‖T ∗(y∗n)‖ ≤ 1/n.Since T ∗ is finitely singular, we know from Exercise 3.1,3 that there exists a subsequence(y∗nk

) converging to some y∗; it follows then that T ∗y∗ = 0, thus y∗ ∈ kerT ∗, which impliesthat y∗(yn) = 0 for every n, contradicting y∗n(yn) = 1 and y∗nk

→ y∗.Conversely, assume TX closed and finite codimensional in Y . By the open mapping

theorem, T induces an isomorphism from X/ kerT onto TX. Let Y0 = TX and let c > 0be such that for every y0 ∈ Y0, there exists x ∈ X such that y0 = Tx and ‖y0‖ ≥ c‖x‖. LetY = Y0 ⊕ F and let Q be the projection from Y0 ⊕ F onto Y0. Given y∗ ∈ F⊥, ‖y∗‖ = 1,there exists y = y0 + f ∈ Y0 ⊕ F such that ‖y0 + f‖ ≤ 1 and y∗(y0 + f) = y∗(y0) > 1/2.Then, since ‖y0‖ = ‖Qy‖ ≤ ‖Q‖ there exists x ∈ X such that y0 = Tx and ‖x‖ ≤ ‖Q‖/c,hence ‖Q‖

c‖T ∗y∗‖ ≥ T ∗(y∗)(x) = y∗(y0) > 1/2,

showing that ‖T ∗y∗‖ ≥ c‖y∗‖/(2‖Q‖) for every y∗ in the finite codimensional subspaceF⊥ of Y ∗, hence T ∗ is finitely singular.

We say that T is infinitely singular if it is not finitely singular.Proposition 3.2. Let T ∈ L(X, Y ). Then T is infinitely singular if and only if forevery ε > 0, there exists an infinite dimensional subspace Z ⊂ X such that ‖T|Z‖ < ε.

15

Furthermore, we may assume that this subspace Z has a Schauder basis (zn)n≥1 and thatthe norm of the restriction of T to [zn, zn+1, . . .] tends to 0 when n →∞; in particular wemay assume that T|Z is compact.

Proof. Suppose that T is infinitely singular. We construct a normalized basic sequence(zn) in X such that ‖Tzn‖ < ε′2−n for every n ≥ 1, where 0 < ε′ < ε/4. If z1, . . . , zn arealready selected, let An be a finite subset of BX∗ which is 1/2-norming for the linear span[z1, . . . , zn], that is

∀x ∈ [z1, . . . , zn], ‖x‖ ≤ 2 maxx∗∈An

|x∗(x)|.

We may assume that An ⊃ An−1. Let X0 =⋂

x∗∈Ankerx∗; since X0 is finite codimensional

and T infinitely singular, we may find zn+1 ∈ X0 such that ‖zn+1‖ = 1 and ‖Tzn+1‖ <ε′2−n−1. We let Z be the closed linear span of the sequence (zn)∞n=1; it is easy to showthat this sequence is a Schauder basis with constant 2 for Z; indeed, if m < n, sincezm+1, . . . , zn were chosen in the kernel of all x∗ ∈ Am, we obtain for all scalars (ak)n

k=1

‖m∑

k=1

akzk‖ ≤ 2 maxx∗∈Am

∣∣x∗(m∑

k=1

akzk)∣∣ = 2 max

x∗∈Am

∣∣x∗(n∑

k=1

akzk)∣∣ ≤ 2‖

n∑

k=1

akzk‖.

For z =∑

k≥n akzk this implies that |ak| ≤ 4‖z‖ for every k, hence

‖Tz‖ ≤ 4‖z‖∑

k≥n

ε′2−k ≤ 8 2−nε′‖z‖ < ε‖z‖.

We obtain that T|Z is compact and that ‖T|Z‖ < ε. The other direction is clear.

Exercise. T is finitely singular iff the image of every closed subspace Z ⊂ X is closed.Remark. What we have showed is that when c(T ) = 0, there exists a subspace Z ⊂ X suchthat ‖T|Z‖ is small; it does not seem possible to obtain in general a quantitative resultof the form ‖T|Z‖ ≤ Mc(T ) + ε for some universal constant M when c(T ) > 0. This ishowever true with M = 1 in a Hilbert space or in `p, 1 ≤ p < ∞ (or in c0).

4. Basic Fredholm theory

Definition 4.1. We say that T ∈ L(X, Y ) is a Fredholm operator from X to Y if thereexists a (closed) finite codimensional subspace X0 of X such that T|X0 is an isomorphismfrom X0 onto some finite codimensional subspace Y0 = TX0 of Y .

In particular T is finitely singular. We know by Exercise 3.1, part 4, that one canchoose X0 such that X = ker T ⊕X0. In this case Y0 = TX0 = TX.

Exercise 4.1.1. Let T be a Fredholm operator from X to Y , and let X0, Y0 be as above. Prove that

codimX X0 − codimY Y0 = dimkerT − codimY TX.

This integer is called index of T , and denoted ind(T ).

16

Hint. Write TX = Y0 ⊕ F , X = X0 ⊕ T−1F , T−1F = ker T ⊕ E and count dimensions.

2. Show that T ∈ L(X, Y ) is Fredholm if and only if kerT is finite dimensional, andTX closed and finite codimensional in Y (this is the usual definition).

3. If T is Fredholm on a real Banach space X, then TC is Fredholm on XC, with thesame index (counting of course complex dimensions for TC).

4. Direct sums; if T1, T2 are Fredholm from X1 to Y1 and from X2 to Y2, then T1⊕T2

is Fredholm from X1 ⊕X2 to Y1 ⊕ Y2. Check that ind(T1 ⊕ T2) = ind(T1) + ind(T2).5. When U1T and TU2 are Fredholm, then T is Fredholm.6. If Q : X → Y is a quotient map with finite dimensional kernel E, then Q is

Fredholm and ind(Q) = dimE. If T : Z → X is such that QT is Fredholm, show that Tis Fredholm.

Perturbation by a small norm operator or a finite rank operator

Proposition 4.1. Let T ∈ L(X, Y ) be Fredholm. There exists d > 0 such that ‖S‖ < dimplies that T + S is Fredholm and ind(T + S) = ind(T ).

Proof. Let X0, Y0 be as in Definition 4.1. The result follows immediately from Lemma3.2, part 2, applied to the operator T0 = T|X0 , considered as operator from X0 to Y .This operator is an into isomorphism, hence for d > 0 small and ‖S‖ < d we know that(T + S)|X0 is an into isomorphism and that codim(T + S)X0 = codim TX0.

Lemma 4.1. If T ∈ L(X, Y ) is Fredholm and if S has finite rank, then T +S is Fredholmand ind(T + S) = ind(T ).

Proof. This is because we may choose X0 contained in the finite codimensional subspacekerS of X. Then T + S and T coincide on X0.

Exercise 4.2.1. If T ∈ L(X,Y ) is Fredholm, there exists U ∈ L(Y,X) such that UT − IX and

TU − IY have finite rank.2. If U1T − IX and TU2 − IY have finite rank, then T is Fredholm. Hence T is

Fredholm iff it is invertible modulo finite rank operators.

Proposition 4.2. Composition formula. If T : X → Y and U : Y → Z are Fredholm,then UT is Fredholm from X to Z and

ind(UT ) = ind(T ) + ind(U).

Proof. We can find X0, Y0, Y1, Z1 finite codimensional such that T defines an isomorphismfrom X0 onto Y0 and U an isomorphism from Y1 onto Z1. We simply replace these finitecodimensional subspaces by smaller finite codimensional subspaces given by Y2 = Y0 ∩ Y1,X2 = X0 ∩ T−1Y2 and Z2 = UY2; now T|X2 is an isomorphism from X2 onto Y2 and U|Y2

an isomorphism from Y2 onto Z2 and we compute

ind(T ) + ind(U) = (codim X2 − codim Y2) + (codim Y2 − codim Z2) = ind(UT ).

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Duality

Lemma 4.2. An operator T ∈ L(X, Y ) is Fredholm if and only if T and T ∗ are finitelysingular.

Proof. If T is Fredholm, we know that T is finitely singular, and that TX is closed andfinite codimensional, so that T ∗ is finitely singular by Lemma 3.4. Conversely, when Tand T ∗ are finitely singular, we have dim kerT < ∞ by Exercise 3.1,1 and TX closed andfinite codimensional by Lemma 3.4.

Lemma 4.3. Let T ∈ L(X, Y ). Then T is Fredholm iff T ∗ is Fredholm, and in this casewe have ind(T ∗) = − ind(T ).Proof. If T is Fredholm, T ∗ is finitely singular by the preceding Lemma, hence dim kerT ∗ <∞ and T ∗Y ∗ is closed. Furthermore, since kerT = (T ∗Y ∗)⊥, it follows that T ∗Y ∗ is finitecodimensional and T ∗ is Fredholm. Conversely, if T ∗ is Fredholm, then T ∗ is finitelysingular, hence TX is closed and finite codimensional by Lemma 3.4, and dim kerT < ∞for the same reason as before.

Let X0 and Y0 be finite codimensional subspaces of X and Y such that T0 = T|X0 isan isomorphism from X0 onto Y0. Let i : X0 → X and j : Y0 → Y be the natural inclusionmaps. Now j∗ : Y ∗ → Y ∗

0 and i∗ : X∗ → X∗0 are the natural projections. They are clearly

Fredholm (quotient maps with finite dimensional kernel; ker i∗ = X⊥0 , ind(i∗) = codim X0).

By the composition formula

ind(T ∗0 ) + ind(j∗) = ind(i∗) + ind(T ∗)

so thatind(T ∗) = ind(j∗)− ind(i∗) = codimY0 − codim X0 = − ind(T ).

When T is Fredholm, it follows that T ∗∗ is Fredholm and ind(T ∗∗) = ind(T ). We candeduce it from the duality statement but it is actually clear directly.

Lemma 4.4. The set of T ∈ L(X,Y ) which are finitely singular and not Fredholm isopen in L(X,Y ), as well as the set of T ∈ L(X,Y ) such that T ∗ is finitely singular andnot Fredholm from Y ∗ to X∗.Proof. Let T ∈ L(X, Y ) be finitely singular and not Fredholm. We can find a finitecodimensional subspace X0 of X such that T is an isomorphism from X0 into Y . If T isnot Fredholm, it implies that codim TX0 = ∞. By Proposition 3.1 this remains true ina neighborhood of T , hence codim(T + S)X0 = +∞ and codim(T + S)X = +∞ also forT + S in this neighborhood. The second part is similar.

Semi-Fredholm operators

If T is finitely singular from X to Y , there exists a finite codimensional subspace X0

of X such that T is an isomorphism from X0 into Y . Either codim TX0 < ∞ and T isFredholm, or codim TX0 = ∞ (and thus codim TX = +∞) and we define the generalizedindex by ind(T ) = −∞.

More generally, an operator T : X → Y is called semi-Fredholm if TX is closed and ifthe kernel or the cokernel is finite dimensional; the generalized index is defined as before,

18

with the possible values +∞ and −∞. An operator T is semi-Fredholm iff T or T ∗ isfinitely singular by Lemma 3.4 and Exercise 3.1, 5. Observe that a Fredholm operator isprecisely a semi-Fredholm operator with finite index.

Lemma 4.5. The set of semi-Fredholm operators is open in L(X,Y ), and the generalizedindex is locally constant.

Proof. By Lemma 4.4 and Proposition 4.1.

Lemma 4.6. If Tt is a continuous path in L(X, Y ), t ∈ [0, 1], such that Tt is semi-Fredholmfor every t ∈ [0, 1], then ind(T1) = ind(T0). In particular, under the same assumptions, ifT0 is Fredholm then T1 is Fredholm with same index.

Proof. We know that the generalized index is locally constant.

Corollary 4.1. If T is finitely singular and ‖S‖ < c(T ), then T + S is finitely singularand the generalized index satisfies ind(T + S) = ind(T ). In particular, if T is Fredholmand ‖S‖ < c(T ), then T + S is Fredholm and the index satisfies ind(T + S) = ind(T ).

Proof. When ‖S‖ < c(T ), we see that ‖tS‖ < c(T ) for every t ∈ [0, 1], therefore T + tSis a continuous path consisting of finitely singular operators by Remark 3.2. The resultfollows from Lemma 4.6.

Corollary 4.2. If T is Fredholm and K compact from X to Y , then T + K is Fredholmand ind(T + K) = ind(T ).

Proof. It is enough to show that T + K is finitely singular for every compact operatorK (because then T + tK will be finitely singular for every t ∈ [0, 1]); this will be aconsequence of Lemma 6.1, but we give here a different proof: for every ε > 0, there existsa finite codimensional subspace X0 such that ‖K|X0‖ < ε; indeed, there exists a finite setx∗1, . . . , x

∗n in X∗ such that

K∗(BY ∗) ⊂n⋃

j=1

(x∗j + εBX).

This implies that ‖Kx‖ ≤ ε‖x‖ when x ∈ ⋂nj=1 kerx∗j . If we choose ε < c(T ), we see that

T + K is finitely singular (Remark 3.2).

Corollary 4.3. Let T ∈ L(X,Y ). Then T is Fredholm if and only if T is invertible modulocompact operators, i.e. iff there exists U ∈ L(Y, X) such that TU − IY and UT − IX arecompact.

Proof. This condition is necessary by Exercise 4.2,1. Conversely, if TU − IY is compact,we know that TU is Fredholm by Corollary 4.2. In the same way, UT is Fredholm and weknow then that T is Fredholm by Exercise 4.2,2.

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5. More on operators

See [LT1], section 2.c; Kato, [Kt1], [Kt2].We prove now another version of Lemma 3.2. Suppose that T ∈ L(X, Y ) has closed

range. Then T induces an isomorphism between X/ kerT and TX, and there exists aconstant c > 0 such that for every y ∈ TX, we can find x ∈ X such that y = Tx and‖y‖ ≥ c‖x‖.Lemma 5.1. Let T1, T2 ∈ L(X,Y ). Suppose that for every y ∈ T1X, there existsx ∈ X such that y = T1x and ‖y‖ ≥ c‖x‖; if ‖T1 − T2‖ < c and if T2X is closed, thencodim T2X ≤ codim T1X, finite or infinite.

Remark. We can recover from the above Lemma the fact proved in Proposition 3.1 thatcodim T1X remains constant in a neighborhood of an into isomorphism T1. Indeed, if T1 isan into isomorphism from X into Y , with ‖T1x‖ ≥ c‖x‖ for every x ∈ X, and if we assume‖T2 − T1‖ < c/2, we have ‖T2x‖ ≥ (c/2)‖x‖ for every x ∈ X; this shows that T2 is an intoisomorphism, therefore T2X is closed; we obtain codim T2X ≤ codim T1X by the aboveLemma; since ‖T2 − T1‖ < c/2, we can exchange the roles of T1 and T2 and conclude thatcodim T2X = codim T1X, finite or infinite.Proof of the Lemma. If codim T2X > codim T1X, there exists by the next sublemma,applied to Z = T1X, Y = T2X and ε = 1 − ‖T1 − T2‖/c > 0 a vector z ∈ T1X, ‖z‖ = 1such that dist(z, T2X) > ‖T1 − T2‖/c. Then there exists x ∈ X such that z = T1x and‖x‖ ≤ 1/c; now ‖z − T2x‖ ≤ ‖T1 − T2‖ ‖x‖ gives a contradiction.Sublemma. Let Y and Z be two finite codimensional subspaces of X. If codim Y >codim Z, there exists for every ε > 0 a vector z ∈ Z such that ‖z‖ = 1 and dist(z, Y ) > 1−ε.This is also valid if Y is closed, codimY = ∞ and codim Z < ∞.Proof. Let ρ be a continuous lifting (not necessarily linear!) from the unit sphere S(X/Y )of X/Y to the ball of radius (1− ε)−1 in X. We may assume that ρ(−x) = −ρ(x). ThenπZ ρ is an odd mapping from the sphere of X/Y to X/Z and dim X/Y > dim X/Z. ByBorsuk’s antipodal mapping theorem, there exists x ∈ S(X/Y ) such that πZ(ρ(x)) = 0,i.e. ρ(x) ∈ Z. Take z′ = ρ(x) and z = z′/‖z′‖. When codim Y = ∞, it is enough to selectF ⊂ X/Y such that dim F > dim X/Z and to apply the same reasoning.

Corollary 5.1. Let T1, T2 ∈ L(X, Y ). If ‖T1x‖ ≥ c‖x‖ for every x ∈ X and if ‖T1−T2‖ <c, then codim T2X = codim T1X, finite or infinite.Proof. It follows from the hypothesis that every T on the segment [T1, T2] is an intoisomorphism, and we know that codim TX is locally constant by Proposition 3.1 or theRemark following the above Lemma.

Lemma 5.2. Let T ∈ L(X); if T or T ∗ is finitely singular (in other words if T is semi-Fredholm) and if 0 ∈ ∂σK(T ), then T is Fredholm with index 0.

Proof. Since 0 ∈ ∂σK(T ), we can find invertible operators, in particular Fredholm operatorswith index 0, arbitrarily close to T , hence T is Fredholm with index 0 by the continuity ofthe index of semi-Fredholm operators (Lemma 4.5).

Remark. We obtain a slightly different proof for the “Boundary of spectrum lemma”: ifT ∈ L(X) and λ ∈ ∂σK(T ), then there exists a (possibly constant) normalized sequence

20

(xn) in X such that (T −λIX)xn → 0. Indeed, let U = T −λIX . If U is infinitely singular,we know the result by Proposition 3.2. If U is finitely singular and 0 ∈ ∂σK(U), then U isFredholm with index 0 by Lemma 5.2. Since 0 ∈ σK(U), it follows that U is not invertible,thus kerU 6= 0.Lemma 5.3. If T ∈ L(X) is finitely singular, then Y =

⋂∞n=0 TnX is closed and TY = Y .

Furthermore T|Y : Y → Y is Fredholm and the constant value of ind(T|Y −λIY ) for λ in aneighborhood of 0 (in K) is the (constant and finite) dimension of the kernel of T − λIX

for small λ 6= 0. Precisely, there exists ε > 0 such that

∀λ ∈ K, 0 < |λ| < ε ⇒ dimker(T − λIX) = ind(T|Y ).

Proof. We prove that TnX is closed by induction using Exercise 3.1, part 2, hence Y isclosed. It is clear that TY ⊂ Y . We also know that N1 = kerT is finite dimensional. Itfollows that there exists an integer k such that, setting Yj = T jX, we have N1∩Yk = N1∩Y .Let y ∈ Y ; for every integer j, there exists a vector zj ∈ Yj such that y = Tzj . For j > kwe get zj − zk ∈ N1∩Yk = N1∩Y . In particular zj − zk ∈ Y ⊂ Yj and thus zk ∈ Yj . Sincethis is true for every j > k, we obtain zk ∈ Y , and finally y = Tzk ∈ TY . We have thatkerT|Y ⊂ ker T is finite dimensional; since TY = Y , it follows that T|Y is Fredholm fromY to Y .

Suppose that λ 6= 0 and (T − λIX)x = 0. We have x = Tnx/λn for every integern ≥ 1, yielding x ∈ Y . Hence the kernel of T −λIX coincides with the kernel of T|Y −λIY .Furthermore, if we choose Y0 finite codimensional in Y such that Y = kerT|Y ⊕ Y0, we seethat T|Y0 is an isomorphism from Y0 onto Y . This remains true for T − λIX for small λ,say |λ| < ε, hence T|Y − λIY remains onto and for 0 < |λ| < ε we obtain

dimker(T − λIX) = dim ker(T|Y − λIY ) = ind(T|Y − λIY ) = ind(T|Y ).

Remark. Suppose that kerT = 0 and ker(T − λnIX) 6= 0 for a sequence (λn) tendingto 0. Then T is infinitely singular. Indeed, there exists a normalized sequence (xn) suchthat Txn = λnxn tends to 0; if T is finitely singular, there exists by Exercise 3.1,3 asubsequence (xnk

) converging to some x; then x 6= 0 and Tx = 0, a contradiction. Moregenerally, we see that when T is finitely singular, dim ker(T − λIX) ≤ dimker T when λ issmall (we may apply Lemma 5.1 to T ∗ and T ∗ − λIX∗).

Proposition 5.1. Let T ∈ L(X); if T or T ∗ is finitely singular and if 0 ∈ ∂σK(T ), thereexists an integer k ≥ 1 such that kerT k = ker T k+1 and T kX = T k+1X. The space X isthen the direct sum of two invariant subspaces for T , Y = T kX and the finite dimensionalsubspace N = Nk = kerT k 6= 0. The operator T|Y is an isomorphism from Y onto Y .

Furthermore 0 is isolated in σK(T ).

Proof. If T or T ∗ is finitely singular and 0 ∈ ∂σK(T ) then T is Fredholm with index 0 byLemma 5.2; also ker T 6= 0 since 0 ∈ σK(T ). Furthermore by Lemma 5.3, there existsε0 > 0 such that dim ker(T − λIX) is constant for all λ ∈ K such that 0 < |λ| < ε0. Since0 ∈ ∂σK(T ), this constant dimension of ker(T − λIX) must be 0; we may assume that ε0

21

is so small that T −λIX is still Fredholm with index 0 when 0 < |λ| < ε0. These two factsimply that (T − λIX)X = X for such λ, hence T − λIX is invertible when 0 < |λ| < ε0.Therefore 0 is isolated in the spectrum of T .

We also know by Lemma 5.3 that ind(T|Y ) = dim ker(T −λIX) = 0 when 0 < |λ| < ε0

and since TY = Y , it yields that T|Y is an isomorphism and so kerT ∩ Y = 0; thereexists therefore an integer k such that 0 = kerT ∩ Y = ker T ∩ T kX, and this yieldsthat kerT k = kerT k+1 (if T k+1x = 0, then T kx ∈ kerT ∩ T kX, hence T kx = 0); itfollows that T kX = T k+1X = Y , because T k and T k+1 are Fredholm with index 0 byProposition 4.2. We obtain a decomposition of the space into two invariant subspaces, Yand Nk = kerT k (this one is finite dimensional). Indeed, let V ∈ L(Y ) be the inverse ofT|Y and set Q = V kT k, considered as a map from X to X. Then Q is a projection fromX onto Y and kerQ = kerT k = N .Remark. In the complex case, the restriction of T to N decomposes into a finite numberof Jordan cells with 0 on the diagonal. The spectral projection defined in section 2 has Yas kernel and N for range.

Exercise 5.1. If K ∈ L(X) is compact, we know that T = IX −K is finitely singular bythe proof of Corollary 4.2. Find a direct proof that Nk = kerT k stabilizes. If 0 ∈ σ(T ),show that 0 is isolated in σ(T ).Hint. If Nk does not stabilize, let xk ∈ Nk be such that 1 = ‖xk‖ = dist(xk, Nk−1). Ifyk ∈ Nk−1 is arbitrary, observe that ‖(xl − yl) − (xk − yk)‖ ≥ 1 when l 6= k. Apply toyk = Txk to obtain a contradiction to the compactness of K = IX − T .

Boundary of essential spectrum lemma

Recall that the Calkin algebra of a real or complex infinite dimensional Banach spaceX was defined by C(X) = L(X)/K(X). Let T ∈ L(X). It follows from Corollary 4.3 thatλ ∈ ρK(T ) iff T − λIX is Fredholm. Let

σK∞(T ) = λ ∈ K : T − λIX infinitely singular,σ∗K∞ (T ) = λ ∈ K : T ∗ − λIX∗ infinitely singular.

We know that T − λIX is semi-Fredholm if and only if λ /∈ σK∞(T ) ∩ σ∗K∞ (T ); the twosets σK∞(T ) and σ∗K∞ (T ) are compact, and non-empty in the complex case (see Lemma 5.4below); we know that T − λIX is Fredholm iff λ /∈ σK(T ), and by Lemma 3.4 T − λIX isFredholm iff T − λIX and (T − λIX)∗ are finitely singular, therefore

σK(T ) = σK∞ ∪ σ∗K∞ .

Example. Let R be the right shift on `2(N) (complex case). The spectrum of R is theclosed unit disc; σ∞(R) = σ∗∞(R) = T.Exercise. Suppose that K = C and let T be an isometry from X into X. Show thatσ∞(T ) ∩ σ∗∞(T ) ⊂ T and that they are equal if T is not onto.

Lemma 5.4. Let λ ∈ ∂σK(T ). Then T − λIX and (T − λIX)∗ are infinitely singular,

∂σK(T ) ⊂ σK∞(T ) ∩ σ∗K∞ (T ).

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Proof. We know that T − λIX is not Fredholm since λ ∈ σK(T ) but T − λIX is arbitrarilyclose to Fredholm operators since λ ∈ ∂σK(T ). It follows then from Lemma 4.5 thatT − λIX is not semi-Fredholm.

Corollary 5.2. If K = C and if dim X = ∞, then for every T ∈ L(X) there exists λ ∈ Csuch that T − λIX is infinitely singular (and also (T − λIX)∗).Proof. Since dim X = ∞ the algebra L(X)/K(X) is not 0 hence the spectrum of theimage T in C(X) is not empty, and we simply have to pick any boundary point λ of thisspectrum.Exercise. If X is a real Banach space and T ∈ L(X), then either there exists λ ∈ R suchthat T − λIX is infinitely singular, or there exists p, q ∈ R with p2 − 4q < 0 such thatT 2 + pT + qIX is infinitely singular.

Lemma 5.5. Let T ∈ L(X) and let K be a compact subset of K such that σK∞(T ) ∩σ∗K∞ (T ) ⊂ K. If Ω is a connected component of intσK(T )\K, the boundary ∂Ω is containedin K.

Proof. Suppose that λ ∈ ∂Ω but λ /∈ K; then λ ∈ ∂σK(T ). Let U = T −λIX ; since λ /∈ K,U or U∗ is finitely singular, and 0 ∈ ∂σK(U). By Proposition 5.1, 0 is isolated in σK(U),a contradiction to the fact that λ ∈ Ω.

Corollary 5.3. Every λ ∈ σK(T ) belonging to the unbounded connected component ofρK(T ) is an isolated eigenvalue of T with finite multiplicity (by this we mean that Xsplits as X = E ⊕ Y , where E and Y are invariant under T , E is finite dimensional andσ(T|E) = λ, and T|Y is an isomorphism from Y onto Y ).

Proof. Let K be the complement in K of the unbounded connected component ω of ρK(T );then K is compact and contains σK∞(T )∩ σ∗K∞ (T ). We want to prove that intσK(T ) \K isempty. The open subset intσK(T ) \K of ω is bounded, hence different from ω; if it is nonempty it is not closed in ω; there must therefore exist some µ ∈ ∂σK(T )∩ω, contradictingLemma 5.5). Assume that λ ∈ σK(T )∩ ω. By the preceding remark, λ does not belong tothe interior of σK(T ); also U = T −λIX is finitely singular, and 0 belongs to the boundaryof the spectrum of U . The result follows by Proposition 5.1.

Remark. We know by Proposition 5.1 that every non isolated point in ∂σ(T ) belongs toσ∞(T ).

Corollary 5.4. (K = C) If σ∞(T )∩ σ∗∞(T ) cannot contain the boundary of any boundednon empty open subset of C, every λ ∈ σ(T ) \ (σ∞(T ) ∩ σ∗∞(T )) is isolated and is aneigenvalue with finite multiplicity.Proof. By Lemma 5.5 the spectrum has an empty interior. If λ ∈ σ(T ) but λ /∈ σ∞(T ) ∩σ∗∞(T ), the operator U = T − λIX is semi-Fredholm and 0 belongs to the boundary ofσ(U). The result follows by Proposition 5.1.

This Corollary applies for example when σ(T ) is countable; also if A is hermitian inL(H) (complex case) and if K is a compact operator, then the essential spectrum of A+Kis real, hence does not contain the boundary of any bounded non empty open subset of

23

the complex plane; every non real λ in the spectrum of A + K is an isolated eigenvaluewith finite multiplicity.

If σ(T ) does not contain the boundary of any bounded non empty open subset of Cand if S is strictly singular, then. . .

6. Strictly singular operators. More on Fredholm operators

See Kato [Kt1], PeÃlczynski [Pe2]; also [LT1], section 2.c. for a concise presentation.Definition 6.1. We say that S ∈ L(X,Y ) is strictly singular if for every (infinite dimen-sional) subspace Z of X and every ε > 0, there exists z ∈ Z such that ‖Sz‖ < ε‖z‖.

Let S(X, Y ) denote the set of strictly singular operators from X to Y . When Y = Xwe simply write S(X).

Exercise 6.1.1. Let S ∈ S(X,Y ). For every (infinite dimensional) subspace Z ⊂ X and every ε > 0,

there exists (an infinite dimensional) subspace Z ′ ⊂ Z such that ‖S|Z′‖ < ε (compare toProposition 3.2).

2. If S1, S2 ∈ S(X,Y ), then S1 +S2 is strictly singular. Show that S(X, Y ) is a closedvector subspace of L(X,Y ).

3. Let S ∈ S(X, Y ). For every T ∈ L(W,X) and U ∈ L(Y,Z), show that ST and USare strictly singular. When Y = X, S(X) is a closed two-sided ideal of L(X).

4. Show that K(X,Y ) ⊂ S(X, Y ), and that they coincide when X = Y = H is aHilbert space, or when X = Y = `p. Give an example of S ∈ S non compact. Give anexample where S ∈ S(X, Y ) but the adjoint S∗ is not strictly singular.

5. Matrix of strictly singular operators. Let T ∈ L(Xn, Y m). Then T can be repre-sented by a m×n matrix (Ti,j) of operators from X to Y . Show that T is strictly singulariff each Ti,j is strictly singular.

6. Complexification of a strictly singular operator. If S is a strictly singular operatorbetween two real spaces X and Y , then SC is strictly singular from XC to YC.

7. Show that L(`1, `2) = S(`1, `2).

Remark. The essentially dual notion of strictly cosingular operators was defined by PeÃl-czynski [Pe2] in the following way: an operator T : X → Y is strictly cosingular if, forevery linear q from Y onto some Banach space Z, the map q T is not onto.

Lemma 6.1. Assume that T ∈ L(X,Y ) is finitely singular and that S ∈ S(X,Y ). ThenT + S is finitely singular.

Proof. There exists a finite codimensional subspace X0 ⊂ X and c > 0 such that ‖Tx‖ ≥c‖x‖ for every x ∈ X0. Assuming T +S infinitely singular, there would exist by Proposition3.2 a subspace Z ⊂ X (infinite dimensional) such that ‖(T+S)|Z‖ < c/2. Then Z ′ = Z∩X0

is infinite dimensional, hence there exists z ∈ Z ′ ⊂ X0 such that ‖Sz‖ < (c/2)‖z‖. Butthis implies ‖Tz‖ < c‖z‖, contradicting the choice of X0 and c.Remark 6.1. Let U ∈ L(X, Y ). In order that the above proof works for T + U , it isenough that

s(U) = supZ⊂X

dim Z=∞

inf‖Uz‖ : z ∈ Z, ‖z‖ = 1

24

is strictly less than c(T ). Note that U is strictly singular iff s(U) = 0.

Lemma 6.2. Let K = C. For every U ∈ L(X),

r(U) ≤ s(U).

Proof. There exists λ ∈ ∂σ(U) such that |λ| = r(U) and U − λIX infinitely singular byLemma 5.4, hence there exists an infinite dimensional subspace Z on which U ∼ λIX byProposition 3.2, so that s(U) ≥ |λ|.Corollary 6.1. Let T, U ∈ L(X, Y ). If T is Fredholm and if s(U) < c(T ), then T + U isFredholm, and ind(T + U) = indT .

Proof. For every t ∈ [0, 1], we have s(tU) < c(T ), hence T + tU is finitely singular forevery t ∈ [0, 1] by Remark 6.1. The result follows by Lemma 4.6.

Remark. The corresponding result holds also if T is finitely singular and not Fredholm.

Corollary 6.2. Let S, T ∈ L(X, Y ). If T is Fredholm and S strictly singular, then T + Sis Fredholm, and ind(T + S) = indT .

Corollary 6.3. T is Fredholm iff it is invertible modulo strictly singular operators.

Corollary 6.4. T is Fredholm iff it is invertible modulo compact operators.

Proposition 6.1. Let S ∈ L(X) be strictly singular, X complex. Then every λ 6= 0 in thespectrum of S is isolated in σ(S) and is an eigenvalue with finite multiplicity. It followsthat the spectrum of S is finite or consists of a sequence converging to 0.

(This is a particular case of a Riesz operator; for the proof below it is enough to knowthat σ(S) = 0.) If X is real and S ∈ S(X) we obtain considering SC a complex spectruminvariant under complex conjugation and consisting of a sequence converging to 0; the realspectrum is at most a sequence converging to 0.

Proof. If S is strictly singular it is clear that σ(S) = 0 by Lemma 6.2. It follows thenfrom Corollary 5.3 that every λ 6= 0 in σ(S) is isolated and is an eigenvalue with finitemultiplicity.

Exercise. Let T ∈ L(X), S ∈ S(X). If λ belongs to the unbounded component of ρK(T )and if λ ∈ σK(T + S), then λ is isolated in σK(T + S) and is an eigenvalue with finitemultiplicity for T + S.

7. Ultraproducts

Ultraproducts appeared in model theory (ÃLos) and as models for non-standard analy-sis. The notion of “restricted ultraproduct”, also called ultraproduct of Banach spaces, ismore suitable to classical analysis and was developed among others by Dacunha-Castelleand Krivine in [DK] (see also [Ja], approximately at the same period).

Let I be an infinite index set and let U be a non-trivial ultrafilter on I. Let (Xi)i∈I bea family of Banach spaces indexed by I. Let L = `∞(I, (Xi)) be the space of all bounded

25

families x = (xi)i∈I such that xi ∈ Xi for every i ∈ I and supi ‖xi‖Xi< ∞. The norm of

x = (xi)i∈I ∈ L is given by ‖x‖ = supi∈I ‖xi‖Xi . We define a seminorm on L by

p(x) = limU‖xi‖Xi

(≤ ‖x‖)

and we let N = x : p(x) = 0. Then N is a closed subspace of L and we define theultraproduct X =

∏Xi/U to be the quotient Banach space L/N . When all spaces Xi are

equal to the same space X, we call X an ultrapower of X. We can embed X isometricallyin the ultrapower X by mapping each x ∈ X to the constant sequence (xi)i where xi = xfor every i ∈ I.

By a slight abuse, we shall consider (xi)i as representing an element in X, insteadof the correct formulation which refers to the equivalence class modulo U (same traditionin measure theory when speaking about a “function” in L1, instead of a class modulonegligible functions).

Exercise.1. Show that X = X when Xi = X is finite dimensional, and that X 6= X when X is

infinite dimensional. In this case X is non separable.2. Show that X is finite dimensional iff d = limU dim Xi is finite. In this case d is the

dimension of X.

The index set is usually the set N of integers, but more general sets are useful whendealing for example with ultrapowers X of spaces X with non-separable dual; in this casea natural index set is the set of finite subsets of the dual space X∗ (or finite subsets ofa dense subset in X∗: this is why the case of separable dual reduces to the index set N);equivalently we can work with the set I of (closed) finite codimensional subspaces of X.

The weakly null part X0 of the ultrapower X of X consists of elements x that have arepresentative (xi)i∈I such that w- limU xi = 0 and plays a role in several questions.

Exercise. If X is reflexive, thenX = X ⊕ X0.

When H is a Hilbert space, the ultrapower H is also a Hilbert space: we may define ascalar product on H that extends the scalar product of H, and such that the correspondingnorm is the norm of H. Indeed, let for x = (xi) and y = (yi)

〈x, y〉 = limi,U〈xi, yi〉.

Then‖x‖2 = 〈x, x〉.

It is also true, but more complicated, that for any given p ∈ [1, +∞), the class of Lp

spaces is stable under ultraproduct (Dacunha-Castelle and Krivine [DK]).

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Ultrapower of an operator

We fix an index set I and an ultrafilter U throughout this paragraph; all ultrapowersof possibly different spaces are taken with respect to I and U . Given T ∈ L(X,Y ) we get inthe obvious way a bounded linear operator T from the ultrapower X to the correspondingultrapower Y : if x = (xi)i we simply let T x = (Txi)i. It is clear that ‖T‖ = ‖T‖. IfT ∈ L(X,Y ) and U ∈ L(Y, Z), then UT = U T . Also IX is the identity of X. It followsthat T is invertible when T is invertible. In the case of a Hilbert space, it is easy to seethat the ultrapower of the hilbertian adjoint T ∗ of T is the adjoint of T .

When Y = X, we see that T → T is a unital Banach algebra morphism from L(X)to L(X). Furthermore, it is a ∗-morphism when X = H is a Hilbert space.

Let X0 be the weakly null part of the ultrapower that was defined before. Then, forevery T ∈ L(X, Y ), T X0 ⊂ Y0. For a compact operator T , T X0 = 0. Conversely, whenthe index set is rich enough for coding every weakly null net, T|X0

= 0 implies that T iscompact.

Exercise. Fredholm and ultrapowers.1. If T is finitely singular, then so is T .2. When T is Fredholm from X to Y , show that T is Fredholm from X to Y , with

the same index.3. When λ ∈ ∂σK(T ), there exists an eigenspace for T corresponding to λ. What is

the spectrum of T?

Finite representability

Definition 7.1. We say that Y is finitely representable into X if for every finite dimensionalsubspace F of Y and every ε > 0, there exists a linear map A : F → X such that

∀y ∈ F, (1− ε)‖y‖ ≤ ‖Ay‖ ≤ (1 + ε)‖y‖.

Proposition 7.1. If every Xi is finitely representable in X then the ultraproduct X ofthe family (Xi) is finitely representable in X.

Proof. Let F ⊂ X be finite dimensional and let (y(α))α be an algebraic basis for F ; let(y(α))i∈I be a representative of y(α). By assumption there exists for every i ∈ I a linearmap Ai : Fi = [y(α)

i ]α → X such that

∀y ∈ Fi, (1− ε)‖y‖ ≤ ‖Aiy‖ ≤ (1 + ε)‖y‖.

We obtain a linear map A : F → X by setting Ay(α) = (Aiy(α)i )i. If x =

∑α aαy(α) ∈ F ,

then xi =∑

α aαy(α)i ∈ Fi and

(1− ε)‖x‖ = (1− ε) limU‖xi‖ ≤ lim

U‖Aixi‖ = ‖Ax‖ ≤ (1 + ε)‖x‖.

27

Spreading models ([BS], [BL])

Let U be a non trivial ultrafilter on N; consider the successive ultrapowers of X definedin the following way. Let X1 be the usual ultrapower with index set N. Let X2 be thevector space of classes of double sequences x = (xn1,n2) with the norm

‖x‖ = limn1,U

limn2,U

‖xn1,n2‖.

Similarly X3 is defined from triple sequences, and so on. . . There is a natural isometricembedding from Xn into Xn+1, which allows to consider Xn as a subspace of Xn+1 andthen to define the completion X∞ of the union

⋃n Xn (this type of construction is very

similar to the notion of ultralimit in model theory). Let us describe the embedding ik fromXk into Xk+1: to x = (xn1,...,nk

) ∈ Xk we associate ik(x) = (yn1,...,nk+1) ∈ Xk+1 definedby yn1,...,nk,nk+1 = xn1,...,nk

for every (n1, . . . , nk+1). For every operator T ∈ L(X, Y ),there exists an operator T∞ : X∞ → Y∞ defined in the obvious way. This space X∞ isfinitely representable into X. We can define on X∞ an isometric shift D in the followingmanner: if x = (xn1,...,nk

) belongs to Xk, let

Dx = (yn1,...,nk+1) ∈ Xk+1, where yn1,n2,...,nk+1 = xn2,...,nk+1 .

The spreading model operation corresponds then to an iterated action of this shift operatoron an element x ∈ X1. This point of view was popularized by Krivine (the original approachof Brunel and Sucheston to spreading models uses a precise extraction of subsequence,with the help of Ramsey’s theorem). The ultrapower point of view has the advantage ofbeing “functorial”: if Y is a second Banach space, and if we construct the correspondingspace Y∞, there exists a similar shift D′ on Y∞, and for every T ∈ L(X,Y ) we haveD′T∞ = T∞D.

Let (xn)n≥1 be a sequence in X with no Cauchy subsequence. We consider in thesuccessive ultrapowers the vectors e1 = (xn), e2 = De1 (note that e2 is not the image ofe1 under the canonical embedding of X1 into X2), and generally ek = Dk−1e1 for everyk ≥ 1. The norm in X∞ of a linear combination of the vectors e1, . . . , ek is given by

‖k∑

i=1

aiei‖ = limn1,U

. . . limnk,U

‖a1xn1 + a2xn2 + · · ·+ akxnk‖.

This norm on [en]n≥1 is invariant under spreading, which means that

‖k∑

i=1

aiei‖ = ‖k∑

i=1

aiemi‖

for every k, all scalars (ai)ki=1 and all m1 < m2 < . . . < mk. Note that ‖e1 − e2‖ > 0,

otherwise (xn) would have a Cauchy subsequence.

28

Let f1 = e1− e2, f2 = e3− e4, . . . Then ‖fn‖ = ‖f1‖ > 0 and (fn) is a monotone basicsequence (see below). The space generated by this sequence is contained in X∞ and hencefinitely representable in X. The norm is invariant under spreading. We call this spacegenerated by (fn) a monotone spreading model of X, generated by the sequence (xn).

We prove that (fn)n is a monotone basic sequence. By the spreading invarianceproperty, we obtain

‖k∑

i=1

aifi‖ = ‖k−1∑

i=1

aifi + ak(el − el+1)‖

for every l ≥ 2k − 1. Taking averages from l = 2k − 1 to l = 2k + n − 2 we obtain‖∑k

i=1 aifi‖ ≥ ‖∑k−1i=1 aifi + ak(e2k−1 − e2k+n−1)/n‖, hence letting n →∞

‖k∑

i=1

aifi‖ ≥ ‖k−1∑

i=1

aifi‖.

Block finite representability

Let (xn) be a sequence in a Banach space X, with no Cauchy subsequence. We saythat a space Y with a basis (fn) is block finitely representable in the span of (xn) if forevery finite sequence y1, . . . , yk of (successive) blocks in Y and every ε > 0 there exists alinear map A : [y1, . . . , yk] → X such that Ay1, . . . , Ayk are successive linear combinationsof the (xn), and

∀y ∈ [y1, . . . , yk], (1− ε)‖y‖ ≤ ‖Ay‖ ≤ (1 + ε)‖y‖.

A (monotone) spreading model generated by a sequence (xn) is block finitely repre-sentable into this sequence.

Ultrapowers of commuting or almost commuting operators

When U, T ∈ L(X) commute then U and T commute on X. In some situations U

and T do not exactly commute but the restrictions of T and U to the weakly null partX0 commute. For example, suppose that TU − UT is compact. Then T U − U T vanisheson the subspace X0 of X. We shall give an easy application to the existence of commonapproximate eigenvectors.Lemma 7.1. (Complex scalars) Let T, U ∈ L(X) with TU = UT . If T is not an intoisomorphism, there exists λ ∈ C and a normalized sequence (xn) (possibly constant) suchthat

Txn → 0, (U − λIX)xn → 0.

Proof. We only have to find λ ∈ C such that for every ε > 0, there exists a norm onevector x ∈ X with ‖Tx‖ < ε and ‖Ux− λx‖ < ε. Since T is not an into isomorphism, wecan find a normalized sequence (yn) ⊂ X such that Tyn → 0 (this sequence (yn) may beconstant). If we consider y = (yn) in the ultrapower X, we get T y = 0. This shows thatZ = ker T 6= 0. Since U and T commute, we know that UZ ⊂ Z. Let V = U|Z ∈ L(Z)

29

and let λ ∈ ∂σ(V ). There exists z such that z ∈ Z and V z ∼ λz. Pulling back z to Xgives the desired vector x.

Exercise. If X is a real Banach space and T,U are as in Lemma 7.1 we can find r > 0,θ ∈ R and two sequences (xn), (yn) in X with ‖xn‖ + ‖yn‖ = 1, Txn → 0, Tyn → 0,Uxn − r(cos θ xn − sin θ yn) → 0 and Uyn − r(sin θ xn + cos θ yn) → 0.Corollary 7.1. (K = C) Let T1, . . . , Tn ∈ L(X) commute. There exist λ1, . . . , λn ∈ Csuch that for every ε > 0, there exists x ∈ X, ‖x‖ = 1 and

∀i = 1, . . . , n, ‖Tix− λix‖ < ε.

Proof. The proof is by induction, using the argument of Lemma 7.1. When n = 2, wechoose λ1 ∈ ∂σ(T1); then T = T1 − λ1IX is not an into isomorphism and we can find byLemma 7.1 some λ2 for which there exist common approximate vectors. Passing to anultrapower we see that the eigenspaces for T1 and λ1, and for T2 and λ2 intersect, andtheir intersection is stable under T3 since the operators commute. It is therefore possibleas before to find an approximate eigenvector for T3 in this intersection.

There is a variant of Lemma 7.1 where one assumes that T is infinitely singular andthen the normalized sequence (xn) in the result can be chosen basic. In order to constructthis basic sequence, we only need to show that the vector x in the above proof can be chosenin any given finite codimensional subspace of X. Let I be the set of finite codimensionalsubspaces of X and let U be a ultrafilter on I containing the set Z ∈ I : Z ⊂ Y for everyY ∈ I. Since T is infinitely singular, we may choose for every i = Y ∈ I a norm one vectoryi ∈ i such that ‖Tyi‖ < (codim i)−1. The corresponding net (yi)i belongs to the weaklynull part X0 of the ultrapower, and X0 is stable under T and U . It is easy then to adaptthe above argument to the present case. The final z is now a weakly null net, so it can bepulled back in any finite codimensional subspace.Exercise. Assume that T1, . . . , Tk commute and that there exists a normalized (basic)sequence (xn) such that Tixn → 0 for every i = 1, . . . , k. Let U commute with each Ti.Then there exists λ ∈ C and a normalized (basic) sequence (yn) such that Tiyn → 0 foreach i = 1, . . . , k and (U − λ)yn → 0.

The above variant of Lemma 7.1 remains true if T and U weakly commute.

Lemma 7.2. (Complex scalars) Let T,U ∈ L(X). Assume that T is infinitely singularand that TU − UT is compact. We can then find λ ∈ C and a normalized basic sequence(xn) in X such that Txn → 0 and (U − λIX)xn → 0.

Wiener’s algebra again

Let f =∑

n∈Z an einθ ∈ W and T =∑

n∈Z anRn where R is the right shift on `1(Z).Then

‖f‖W = ‖T‖L(`1(Z)).

Note that T commutes with R. Conversely, to T =∑

n∈Z anRn with∑Z |an| < ∞ we

associate f = ϕT =∑

n∈Z an einθ ∈ W . If f does not vanish on T, we show first that Tis an into isomorphism. Otherwise there exists by Lemma 7.1 a λ ∈ T and a normalized

30

sequence (xn) such that Txn → 0 and Rxn− λxn → 0; on the other hand, Txn ∼ f(λ)xn,hence f(λ) = 0, contradiction. It follows that ∂σ(T ) ⊂ f(T); indeed, if λ /∈ f(T), we seethat ϕT−λId = ϕT − λ does not vanish on T, hence T − λId is an into isomorphism andtherefore λ /∈ ∂σ(T ) by the boundary of spectrum Lemma.

Assuming that ϕT does not vanish on T, we can find a trigonometric polynomial gsuch that |gϕT − 1| < 1/2 on T. Let U be the finite linear combination of powers of Rsuch that ϕU = g. Then, since ϕTU = ϕT ϕU = ϕT g, the boundary of the spectrum of TUis contained in the disc ∆(1, 1/2), hence the spectrum itself is contained in the same discand it follows that TU is invertible; similarly UT is invertible, therefore T is invertible.The inverse must commute to R, therefore T−1 =

∑n∈Z bnRn, where the sequence (bn)

is defined by T−1e0 =∑

n∈Z bnen; finally∑

n∈Z |bn| < ∞ since T−1e0 ∈ `1. If we seth =

∑n∈Z bn einθ, we see that hf = 1 on T. Summing up, we found an alternate proof for

the fact that a non-vanishing function in W is invertible in W (end of section 2).

Cuntz algebras

See Cuntz [C1].Let P be the unital complex algebra generated by six elements (uj) and (u∗j ), j =

0, 1, 2, satisfying the relations

u∗i uj = δi,j1P ;2∑

j=0

uju∗j = 1P .

Notice that pj = uju∗j is an idempotent, and pipj = 0 when i 6= j. We shall also consider

the unital complex algebra Q generated by six elements (vj) and (v∗j ), j = 0, 1, 2, satisfyingthe relations

v∗i vj = δi,j1Q.

Again qj = vjv∗j is an idempotent, qiqj = 0 when i 6= j, but q = 1Q −

∑2j=0 qj is a non

zero idempotent in Q.Exercise. Let (q) denote the ideal generated by q in Q. Show that P ' Q/(q).

Let T be the ternary tree⋃∞

n=00, 1, 2n. The root is the empty word, denoted ∅. Ifs, t ∈ T , let (s, t) stand for the concatenation of s and t. For t ∈ T , we define ut and u∗t inP (and in a similar way vt and v∗t in Q) inductively by u(t,i) = uiut and (u(t,i))∗ = u∗t u

∗i .

(We also let u∅ = u∗∅ = 1P .) Then because u∗i uj = 0 when i 6= j, every w ∈ P has adecomposition

w =N∑

l=1

cl uαlu∗βl

,

where αl and βl are words in T and cl ∈ C. Notice that u∗αuα = 1P and that uαu∗α is anidempotent pα for every word α ∈ T . Similarly every w ∈ Q has a decomposition

w =N∑

l=1

cl vαlv∗βl

,

31

where αl and βl are words in T and cl ∈ C.We shall describe a model for Q and two models for P. Let Y00 be the vector space

of finitely supported scalar sequences indexed by T . Denote the canonical basis for Y00

by (et)t∈T and denote the length of a word t ∈ T by |t|. We shall now describe someoperators on Y00. Let Id denote the identity operator on the space of sequences. Let Vi,for i = 0, 1, 2 be defined by their action on the basis as follows:

Viet = e(t,i).

Thus Vi can be thought of as the map taking each vertex of T to the ith vertex immediatelybelow it. The adjoints V ∗

i act in the following way: V ∗i et = es if t is of the form t = (s, i),

and V ∗i et = 0 otherwise. The following facts are easy to check: V ∗

i Vj = δi,jId; if Q denotesthe natural rank one projection on the line Ce∅, then

∑2i=0 ViV

∗i = Id−Q. This is a model

for Q: we have a representation ρ : Q → L(Y00) defined by ρ(vi) = Vi, ρ(v∗i ) = V ∗i for

i = 0, 1, 2. We shall see below that ρ is injective.

For constructing our first model for P, consider the subset T0 of T consisting of allwords t ∈ T that do not start with 0 (including the empty sequence). Let L0 be the vectorsubspace of Y00 generated by the (et)t∈T0 . In order to define U0 on L0, we modify thedefinition of V0 slightly, by letting U0e∅ equal e∅ instead of e0. Operators U1 and U2 aredefined exactly as V1 and V2 were. We still have that the UiUi

∗ are projections and thatUi∗Uj = δi,jId, but this time

∑2i=0 UiUi

∗ = Id. We have a model for P, that we call P0.The mapping ρ0 from P to L(L0) which takes ui to Ui and u∗i to U∗

i is a representationof P into L(L0). It is injective.... There is a simpler way to present this model: define onc00(N) the operators

U ′ien = e3n+i−2, i = 0, 1, 2.

This is the same model, up to isomorphism; indeed, to each s = (i1, . . . , in) ∈ T0 wecan associate the integer ns = 3n−1i1 + · · · + 3in−1 + in + 1, (with n∅ = 1), and thisdefines a bijection between T0 and N such that U ′

i = ϕUiϕ−1 for i = 0, 1, 2, where ϕ is the

isomorphism from L(L0) to L(c00) deduced from that bijection.Our second model for P uses the index set T∞ = Z×T . We consider the vector space

L∞ of finitely supported complex functions on T∞, with its natural basis en,t, n ∈ Z,t ∈ T . Now define

U0(en,∅) = en+1,∅; U0(en,t) = en,(t,0) if t 6= ∅,and for j = 1, 2

Uj(en,t) = en,(t,j).

Then we set U∗0 (en,∅) = en−1,∅ and for j = 1, 2, U∗

j (en,∅) = 0; when t = (s, k) is not theempty word (k = 0, 1, 2) we let U∗

j (en,t) = 0 when t does not end with j, that is j 6= k,and U∗

j removes that last j otherwise: U∗j (en,(s,j)) = en,s. We write en instead of en,∅;

one can think of en as a vector et with an infinite t, having infinitely many 0s at the leftof the nth place. We have a second model P∞ for P, with a representation ρ∞ from P inL(L∞). This vector space L∞ admits a natural graduation by

L∞,n = spanem,t : m + |t| = n, n ∈ Z.

32

It should be observed that each Uj sends L∞,n to L∞,n+1 while each U∗j sends L∞,n to

L∞,n−1.

Consider in P the subset Bk generated by products uαu∗β where |α| = |β| ≤ k. It iseasy to see that Bk is a subalgebra of P. Furthermore, every element in Bk is a linearcombination of words uαu∗β with |α| = |β| = k. This follows from the relation

uαu∗β =2∑

j=0

uαuju∗ju∗β

which shows that products of length r can be expressed as sum of products of length r+1.Let fα,β = uαu∗β , for |α| = |β| = k. It is easy to check that they generate an algebraisomorphic to the matrix algebra M3k , thus Bk is isomorphic to M3k .

Let B denote the subalgebra of P obtained as union of the increasing sequence (Bk).

Lemma 7.3. Let L 6= 0 be a complex vector space and let (Ui), (U∗i ), i = 0, 1, 2 be

operators on L such that

U∗i Uj = δi,j1L(L);

2∑

j=0

UjU∗j = 1L(L).

Then ρ(ui) = Ui, ρ(u∗i ) = U∗i for i = 0, 1, 2 defines an injective representation of P into

L(L). Similarly suppose (Vi), (V ∗i ), i = 0, 1, 2 are operators on L such that

V ∗i Vj = δi,j1L(L).

Then ρ(ui) = Vi, ρ(u∗i ) = V ∗i defines an injective representation of Q into L(L).

Proof. It is clear that ρ exists. Let Li = Ui(L); it is easy to check that L is the direct sumof L0, L1 and L2; for every t ∈ T let Lt = Ut(L); it is also clear that for every k, L is thedirect sum of the Lt, |t| = k. This implies that the restriction of ρ to Bk is injective. ...

Using B, it is possible to give a useful representation for every w ∈ P. For everyn ≥ 0, let 0n denote the word consisting of n zeros and let rn be the idempotent rn =u0nu∗0n = un

0 (u∗0)n.

We shall also define the weight q(w) of an element w ∈ P by letting q(10 = 0, q(ui) = 1and q(u∗i ) = −1, and defining the weigth of w as the sums of weights of the factors (notethat the fundamental relation between the generators is compatible with this definition).

Proposition 7.2. Every w ∈ P has a unique representation

w =∑n<0

(u∗0)−nbn +

∑

n≥0

bnun0 ,

where bn ∈ B satisfy

bn = bnrn for n ≥ 0, bn = r−nbn, n < 0.

33

Proof. It is possible to express w as w =∑

n∈Z wn, where each wn is of the form

wn =k∑

l=1

cl uαlu∗βl

,

with |αl| − |βl| = n for each l = 1, . . . , k. Then for n ≥ 0 we have

wn = (wn(u∗0)nun

0 (u∗0)n)un

0 = bnun0

and for n < 0 we may write wn = (u∗0)−nbn; it is easy to check that bn ∈ B and bn = bnrn

for n ≥ 0, bn = r−nbn for n < 0. This shows the existence. For the uniqueness, assumethat for some N ≥ 0, we have

0 =∑n<0

(u∗0)−nbn +

∑

n≤N

bnun0 = w

and that this is a representation with the above properties; we want to show that bN = 0(the case where the largest index N in the summation is < 0 can be treated in a similarway and will be left to the reader). For proving bN = 0 we use the model P∞. Observe thatfor v ∈ B, V = ρ∞(v) leaves each subspace L∞,n invariant. We have, letting Bn = ρ∞(bn)

0 =∑n<0

(U∗0 )−nBn +

∑

n≤N

BnUn0 = W = ρ∞(w).

Let y ∈ L∞,0. Then PNWy = BNUN0 y, where PN denotes the projection of L∞ onto

L∞,N ; since bN ∈ B, it belongs to some BM and since BN = BNRN we get, settingFt,t′ = ρ∞(ft,t′)

BN =∑

t,t′′at,t′′Ft,(t′′,0N ),

where |t| = M , |t′′| = M −N . Choosing y = eN−M,t′′ we get BNUN0 y =

∑t at,t′′eN−M,t =

0 hence all at,t′′ are 0.Remark. We may obtain analogous results for Q.....

It follows that the projection πn in P on the set of elements of weight n is well defined.It also follows that for every λ ∈ T the transform defined by

ϕλ(uj) = λuj ; ϕλ(u∗j ) = λu∗j

extends to a morphism of P. Indeed, if

0 =∑

l

cl uαlu∗βl

= w

then πn(w) = 0, but ϕλ multiplies by λn the set of elements of weight n hence

0 =∑

l

clλ|αl|−|βl|uαl

u∗βl

and ϕλ is well defined on P. In the same way the ∗-transform is well defined on P.

34

Hilbertian representations of P and QSuppose that H is a Hilbert space. A Hilbertian representation of P is a representation

ρ : P → L(H) such that U∗i = ρ(u∗i ) is the Hilbertian adjoint of Ui = ρ(ui). This implies

immediately that Ui is an (into) isometry on H. We obtain an orthogonal decompositionof H into three subspaces Hi, with Hi = UiH, i = 0, 1, 2, and each of these subspacesagain decomposes into a sum of three. . . For every Hilbertian model of P, we obtain aC∗-algebra norm on P. We shall prove the result of Cuntz [C1] that the norm of ρ(w) doesnot depend from the representation, or in other words that there exists a unique C∗-normon P. We first observe that this is the case for the subalgebra B: for every k, it is easy tocheck that ρ is injective on Bk; the image ρ(Bk) is a finite dimensional C∗-algebra, hencehas a unique C∗-norm (namely, the norm of operators on `3

k

2 ).Suppose that ρ is a ∗-representation of P in some L(H), H a Hilbert space. Then,

considering an ultrapower H of H we may define

∀w ∈ P, ρ(w) = ρ(w);

this is a ∗-representation of P in L(H). We define a Hilbertian model from P∞ in the obvi-ous way: we associate to the vector space L∞ a Hilbert space H∞ admitting (en,t)n∈Z,t∈Tfor hilbertian basis. Every Ui extends clearly to an isometry of H∞ and U∗

i is the Hilber-tian adjoint of Ui. We want to show that P∞ with this `2-norm is minimal among thehilbertian models of P.Lemma 7.4. For every ∗-representation ρ of P into some L(H), there exists an invariantsubspace M ⊂ H for ρ and an onto isometry j : H∞ → M such that

∀w ∈ P, ρM (w) = jρ∞(w)j−1,

where ρM denotes the restriction of ρ to M . Hence

∀w ∈ P, ‖ρ∞(w)‖ = ‖ρM (w)‖ ≤ ‖ρ(w)‖ = ‖ρ(w)‖.Proof. Let Uj = ρ(uj), j = 0, 1, 2. Let y ⊥ U0(H), ‖y‖ = 1; it is easy to check that thesequence (Un

0 y)n≥0 is orthonormal. We define in H the vectors

fn,t = (UtUk+n0 y)k≥0 ∈ H.

It is not hard to show that these vectors are normalized and pairwise orthogonal, so thatj(en,t) = fn,t defines an isometry from H∞ into H. We set M = j(H∞) and the conclusionfollows easily.

In a similar way one can show that the model P0 is also minimal. Let

zn =1√n

(n−1∑

k=0

Uk0 y

),

where y is chosen as before; then zn is almost fixed under U0, and z = (zn) is fixed underU0; let

j(et) = Utz, t ∈ T0.

35

Proposition 7.3. Suppose that ρ is a representation of P in a Banach algebra A such thatUj = ρ(uj) and U∗

j = ρ(u∗j ) have norm one for j = 0, 1, 2 and such that ϕλ is isometric, i.e.‖ρ(ϕλ(w))‖ = ‖ρ(w)‖ for every w ∈ P. Let Φλ(W ) = ρ(ϕλ(w)) where W = ρ(w), w ∈ P.Then λ → ϕλ(W ) is well defined and continuous on T for every W in the closure of ρ(P).We can write

Φλ(W ) ∼+∞∑−∞

λnWn,

where Wn is uniquely defined by

Wn =∫

Tλ−nΦλ(W ) dµ(λ),

(µ is the invariant probability on T); we have for n ≥ 0

Wn = BnUn0 , Bn ∈ ρ(B), Bn = BnRn, Rn = ρ(rn),

and for n < 0Wn = (U∗

0 )−nBn, Bn ∈ ρ(B), Bn = R−nBn.

Proof. It is clear that λ → ρ(ϕλ(w)) is continuous from T to A for every w ∈ P, andsince ϕλ is isometric λ → Φλ(W ) is well defined and is the uniform limit of continuousfunctions on T for every W in the closure of ρ(P). Let w =

∑n wn ∈ P, where each wn

has weight n ∈ Z; then Φλ(ρ(w)) =∑

n λnρ(wn), and the integral formula above impliesthat ‖ρ(wn)‖ ≤ ‖ρ(w)‖. Observe that, writing wn = bnun

0 as in Proposition 7.2???, weget ‖ρ(bn)‖ ≤ ‖ρ(wn)‖. When ρ(w) converges in A to some W belonging to the closureof ρ(P), we see thus that the corresponding ρ(wn) converges in A to some Wn, and thatρ(bn) converges to some Bn ∈ ρ(B). The equations for Wn and Bn follow by continuity.

Proposition 7.4. Suppose that Y∞ is equipped with a norm such that (en,t) is 1-uncondi-tional (this is in particular true for the Hilbert space H∞). Then Dλen,t = λn+|t|en,t definesan isometry on Y∞ and

∀w ∈ P, Dλρ∞(w)Dλ−1 = ρ∞(ϕλ(w)).

Proof. Immediate.

It follows that ‖ρ∞(ϕλ(w))‖L(H∞) = ‖ρ∞(w)‖L(H∞) for every w ∈ P and every λ ∈ T.In particular, the Hilbertian model P∞,2 satisfies the hypothesis of Proposition 7.3.

Theorem 7.1. There exists a unique C∗-norm on P.

Proof. We already know a C∗-norm on P, namely the norm given by ρ∞ : P → L(H∞).We also know by Lemma 7.4 that this norm is smaller than any other C∗-norm on P.Conversely, let ρ : P → L(H) be a Hilbertian representation of P. We already know that‖ρ(w)‖ ≥ ‖ρ∞(w)‖. Define

|w| = supλ∈T

‖ρ(ϕλ(w))‖ ≥ ‖ρ(w)‖.

36

This norm corresponds to the direct sum of the family of representations (ρ ϕλ)λ∈T andis therefore a C∗-norm on P. Let A denote the completion of P under this norm. It is aC∗-algebra and we may define a ∗-representation ψ from A to the closure A∞ of ρ∞(P)in L(H∞). All we have to show (by Proposition 2.2) is that ψ is injective. It will followthat for every w ∈ P, we have |w| = ‖ρ∞(w)‖, so that finally since

‖ρ∞(w)‖ ≤ ‖ρ(w)‖ ≤ |w|,

the three norms coincide.

We first observe that ψ is isometric on B, since B has a unique C∗-algebra norm.Next, if n ≥ 0 and Wn = BnUn

0 , with Bn ∈ B and Bn = BnRn, we get ‖ψ(Bn)‖ = ‖Bn‖,‖Rn‖ ≤ 1 since U0 is an isometry, hence ‖Wn‖ ≤ ‖Bn‖ and ‖Bn‖ = ‖Wn(U∗

0 )n‖ ≤ ‖Wn‖.Finally

‖Wn‖ ≤ ‖Bn‖ = ‖ψ(Bn)‖ = ‖ψ(BnRn)‖ ≤ ‖ψ(Wn)‖ ≤ ‖Wn‖for every such Wn. A similar computation gives the case where Wn = (U∗

0 )−nBn, forn < 0. Suppose now that W ∈ A and that ψ(W ) = 0. We have that ϕλ is isometric inA by construction and also isometric in A∞ (because the hypothesis of Proposition ?? issatisfied), so that we may apply Proposition 7.3 and obtain

Φλ(W ) ∼∑

n∈ZλnWn;

taking the image under ψ gives

Φλ(ψ(W )) ∼∑

n∈Zλnψ(Wn) = 0,

hence ψ(Wn) = 0 thus Wn = 0 for every n and W = 0.

The unique C∗-algebra constructed from P is called O3. The above result says thatany unital C∗-algebra generated by three elements Uj , j = 0, 1, 2 such that U∗

j Uj = 1 and∑UjU

∗j = 1 is isometric to O3 (in the C∗-case, it is easy to see that the property U∗

j Ui = 0when i 6= j follows from the two properties above).

Simplicity: let I be a proper two-sided closed ideal in O3; the quotient algebra O3/I isa C∗-algebra generated by three elements U ′

j = π(Uj), such that U ′∗j U ′

i = δi,j1,∑

j U ′jU

′∗j =

1, therefore this quotient map is isometric and I = 0. Thus O3 is simple.

Extensions by compacts. Let E be a Hilbertian model for Q. We know ??? that wehave a map from E to O3: it is a quotient map.

We have described O3; all the proofs generalize easily to the C∗-algebra On generatedby n partial isometries such that

∑n−1j=0 = 1.

Embedding Q in O4;

37

8. Krivine’s theorem

Theorem 8.1. ([K], see also [Le], [MiS]) Let X be a Banach space and let (xn) be asequence in X with no Cauchy subsequence. There exists p ∈ [1,∞] such that `p (or c0 ifp = ∞) is block finitely representable in the span of the given sequence. In other words,there exists p ∈ [1,∞] such that for every k and ε > 0, we can find successive blocksy1, . . . , yk of the sequence (xn) such that for all scalars (ai)k

i=1

(1− ε)(k∑

i=1

|ai|p)1/p ≤ ‖k∑

i=1

aiyi‖ ≤ (1 + ε)(k∑

i=1

|ai|p)1/p.

We proceed by successive reductions of the problem, each time constructing a spacewith basis, block finitely representable in the preceding, thus block finitely representable inthe given sequence. We assume that the scalars are real. The first reduction is to a spacewith a monotone basis and a norm invariant by spreading. This is given by any monotonespreading model generated by the sequence (xn), as explained before in section 7.

Building unconditionality

We shall use an operator trick with the right shift defined on our spreading invariantspace. After the first reduction we have a Banach space Y with a monotone basis (fn)∞n=1

and a norm invariant under spreading. In particular the right shift R on Y defined byRfn = fn+1 is an isometry on Y . It follows that the real spectrum of R is contained in[−1, +1]. Also, it is easy to check that R + IY is not onto (check that f1 is not in therange) hence −1 belongs to the boundary of the real spectrum of R. It follows by Lemma3.3 that one can find for every ε > 0 a vector y ∈ Y such that ‖y‖ = 1 and ‖y + Ry‖ < ε.One can assume that y has finite support, y =

∑N−1i=1 aifi. Consider y0 = y, y1 = RNy,

y2 = R2Ny, . . . , yk = RkNy, and so on. . . It is easy to check that this sequence (yk) isinvariant under spreading and that changing one sign in a linear combination gives

‖∑

i 6=i0

aiyi − ai0yi0‖ ≤ ‖∑

i 6=i0

aiyi + ai0Ryi0‖+ ε|ai0 | = ‖∑

i

aiyi‖+ ε|ai0 |.

For every given integer n > 0 and ε = 1/n we can find such a vector y(n) with‖y(n) + Ry(n)‖ < 1/n and we form the sequence y

(n)1 , . . . , y

(n)k , . . . as above. This sequence

is spreading invariant. Then in the ultrapower Y we obtain for every k ≥ 1 a vectorek = (y(n)

k )n with the property that ‖R ek + ek‖ = 0. This sequence (ek) is invariant underspreading and also 1-unconditional because we obtain in the limit

‖∑

i 6=i0

aiei − ai0 ei0‖ ≤ ‖∑

i

aiei‖

for every i0 and all scalars (ai)i.

Exercise. Construct the unconditionality in the case of complex scalars.

38

Lemberg’s method: a space on Q+

At this point we have a space with a 1-unconditional basis (yn)∞n=1 invariant underspreading. We may define in a further ultrapower the vectors

fq = (y1+[nq])n,

for every non-negative rational q. Let Ξ be the closed subspace generated by (fq)q≥0 andlet Ξ0 be the closed subspace of Ξ generated by (fq)0≤q<1. The family (fq) is still invariantunder spreading in the following sense: if q1 < . . . < qn and r1 < . . . < rn, then

‖∑

i

aifqi‖ = ‖

∑

i

aifri‖

for all scalars (ai)ni=1 (and it is equal to ‖∑

i |ai| fqi‖ because the family is 1-unconditio-nal). We can consider elements of Ξ as (real) functions defined on Q+ (for example, fq isthe function equal to 0 at every s ∈ Q+, except fq(q) = 1). We define operators Dn on Ξ0

by∀t ∈ Q+, (Dnf)(t) = f(nt mod 1).

The operators Dn commute. It is easy here to complexify the space Ξ by simply definingΞC to be the space of complex functions f on Q+ such that |f | belongs to Ξ, and with thenorm ‖f‖ = ‖ |f | ‖.Common approximate eigenvectors

Since the operators (Dn) commute, it is possible by Corollary 7.1?? to find, for everyinteger N ≥ 2, scalars λ2, . . . , λN and a common approximate eigenvector g ∈ Ξ0 suchthat ‖g‖ = 1 and Dig ∼ λig for all Di, i = 2, . . . , N . We can replace each λi by |λi| and gby |g| because |Dig| = Di|g|; we assume therefore that λi ≥ 0, i = 2, . . . , N , and g ≥ 0 inwhat follows. One shows that ln λi/ ln i is constant; this is not totally obvious: let R bethe right shift by 1 on Ξ (defined by (Rf)(t) = f(t− 1)); if 2m < 3n, we see that

2m−1∑

j=0

Rjg ≤3n−1∑

j=0

Rjg

in the Banach lattice ΞC, thus

‖(D2)mg‖ =∥∥

2n−1∑

j=0

Rjg∥∥ ≤ ∥∥

3n−1∑

j=0

Rjg∥∥ = ‖(D3)ng‖,

therefore λm2 ≤ λn

3 , and this yields that lnλ2/ ln λ3 ≤ ln 2/ ln 3; the argument can bereversed to get ln λ2/ ln 2 = ln λ3/ ln 3, hence there exists p ∈ [1,+∞] (1/p = ln λi/ ln i)such that λi = i1/p for i = 1, . . . , N .

39

Construction of `mp

Assume p < ∞. We choose N so large that the set of all vectors in Rm with coordinatesof the form (k/N)1/p, k integer with 0 ≤ k ≤ N , gives a good approximation for the positivepart of the unit ball of `m

p (in more precise terms: an ε-net for some small ε > 0). Lety1, . . . , ym be defined by yi = Riy for i = 1, . . . , m, where y ∈ Ξ0 satisfies ‖y‖ = 1,Diy ∼ i1/py for i = 1, . . . , Nm. We obtain

‖(k1/N)1/py1 + · · ·+ (km/N)1/pym‖ '

N−1/p‖Dk1y + RDk2y + · · ·+ Rm−1Dkmy‖ =

= N−1/p‖Dk1+k2+···+kmy‖ ' (k1 + k2 + · · ·+ km

N

)1/p.

If p = ∞, we have D2y ∼ y and then D2ny ∼ y. If yi = Riy, i = 1, . . . , 2n, we getthat

‖∑

i

±yi‖ ' ‖y‖ = 1

hence (y1, . . . , y2n) is well equivalent to the usual vector basis for `2n

∞ .

9. K-theory of Banach algebras

See [Bl], [Ta], [WO], [C2], [Sk]. We work in this section with complex Banach spacesand complex Banach algebras. Let A be a unital Banach algebra over C. We denote byMn(A) the unital algebra of n × n matrices with entries in A. It can be identified withMn⊗A, where Mn = Mn(C). It is easy to see that Mn(A) is a Banach algebra, but we willnot insist on defining a Banach algebra norm on it. We denote by 1n and 0n respectivelythe unit matrix and the zero matrix in Mn, and by 1n,A = 1n⊗1A and 0n,A = 0n⊗0A theunit matrix and the zero matrix in Mn(A). Given a ∈ Mn(A) and b ∈ Mp(A), we denoteby a⊕ b the matrix in Mn+p(A) equal to

(a 00 b

).

When X is a Banach space and A = L(X), the algebra Mn(A) is naturally identifiedto L(Xn). Let GLn = GLn(C) denote the group of complex n × n invertible matrices,and GLn(A) the (topological) group of n × n invertible matrices with entries in A. ByGL(A) = GL1(A) we denote the group of invertible elements in A, and by GL(0)(A) theconnected component of the identity 1A in this group.

Exercise 9.1. Show that GLn(C) is connected (let M be an invertible matrix; since σ(M)is finite and does not contain 0, one can find µ 6= 0 such that the half-line R+µ does notintersect σ(M)∪1; consider then Mt = (1− tµ)−1(M − tµId), t varying from 0 to +∞).

The following fact will be very useful to the discussion.Proposition 9.1. For every unital Banach algebra B, the set of all finite products

eb1 eb2 . . . ebn ,

40

for bi ∈ B and n ∈ N, is equal to the connected component GL(0)(B) of 1B in GL(B).

Proof. Our first remark is that there is an obvious continuous path from eb to 1B , namelyt → etb, t varying from 1 to 0, hence all finite products of exponentials belong to GL(0)(B).For proving the converse, we may assume that the norm on B is a Banach algebra norm.We observe that the Taylor series of ln(1B + x) converges when ‖x‖ < 1. It follows thateach a ∈ B such that ‖1B − a‖ < 1 is an exponential a = eb, where b = ln(1B + (a− 1B)).If u is invertible and if ‖v−u‖ < ‖u−1‖−1, this implies that v = eb u for some b ∈ B. Withthese remarks it is easy to see that the set of finite products of exponentials is open andclosed in GL(B) and since it contains 1B , it is equal to GL(0)(B).

Corollary 9.1. Let B be a unital Banach algebra and let J be a closed two-sided idealin B. Then every invertible element in GL(0)(B/J) can be lifted to an invertible elementin GL(0)(B).

Proof. Simply write our invertible element in GL(0)(B/J) as product of exponentialseb1 . . . ebn , with bi ∈ B/J , and lift each bi arbitrarily in B.

Similarity, equivalence and homotopy

Let p and q be two idempotents in A, i.e. p2 = p, q2 = q. We say that p and q areequivalent in A if there exist x, y ∈ A such that p = yx, q = xy. We say that p and q aresimilar in A if there exists an invertible u in A such that q = upu−1. It is clear that similarimplies equivalent. We say that p and q are homotopic in A if there exists a continuouspath t → pt from [0, 1] into A such that p0 = p, p1 = q and p2

t = pt for all t ∈ [0, 1].We shall also say that two invertible elements a, b ∈ GL(A) are homotopic in GL(A)

if there exists a continuous path in GL(A) joining a and b.

Let B be a unital Banach algebra. Consider the following path in GL2(B):

rθ,B =(

(cos θ)1B −(sin θ)1B

(sin θ)1B (cos θ)1B

).

Using (rθ,B) for θ varying from θ = 0 to θ = π/2 we build a continuous path rθ,B (a ⊕b) r−θ,B in M2(B) between a ⊕ b and b ⊕ a, for any a, b ∈ B. Observe that when a andb are invertible in B this path is contained in GL2(B). When a, b are idempotents, it isa path of idempotents in M2(B). We get in this way an homotopy in GL2(B) betweena⊕ 1B and 1B ⊕ a. Multiplying it by 1B ⊕ a−1, we get an homotopy between a⊕ a−1 and12,B . In particular a⊕a−1 ∈ GL

(0)2 (B). This is an important fact that will be used several

times later. If we apply this to B = Mn(A), and if a ∈ GLn(A), we get an homotopy inGL2n(A) between a⊕ 1n,A and 1n,A⊕ a, and an homotopy between a⊕ a−1 and 12n,A. Inparticular a⊕ a−1 ∈ GL

(0)2n (A). We have thus obtained

Lemma 9.1. If p and q are idempotents in Mn(A), then p⊕ q and q⊕ p are homotopic inM2n(A). If a, b ∈ GLn(A), then a ⊕ b and b ⊕ a are homotopic in GL2n(A), and a ⊕ a−1

belongs to GL(0)2n (A).

41

Exercises and examples 9.2.1. Show that equivalence is an equivalence relation (if q = vu and r = uv, consider

uqx and yqv).2. In Mn(C), two idempotents are equivalent iff they have the same rank. They are

then similar and homotopic (because GLn(C) is arcwise connected).3. Let X be a Banach space and let p and q be two projections in L(X). Show

that p and q are equivalent in L(X) iff the ranges pX and qX are isomorphic Banachspaces. Consequently, any rank one projections p and q are equivalent (and also similarand homotopic) in L(X).

For example, assume that X is a Banach space such that X ' X2, one has thatIX ⊕ IX and IX ⊕ 0 are equivalent in L(X2). Indeed, let U : X → X ⊕ X be an ontoisomorphism and let V : X ⊕X → X be its inverse. In M2(L(X)) we have the equations

e =(

IX 00 IX

)= ( U 0 )

(V0

); f =

(IX 00 0

)=

(V0

)( U 0 ) .

This means that the two idempotents e and f in L(X2) = M2(L(X)) are equivalent.

4. If p is an idempotent in B, then there is an homotopy in M2(B) between p⊕(1B−p)and 1B ⊕ 0B : consider the path of idempotents in M2(B)

(p + (1B − p) sin2 θ (1B − p) sin θ cos θ(1B − p) sin θ cos θ (1B − p) cos2 θ

),

where θ varies from 0 to π/2. More generally, if p and q are two idempotents in Bsuch that pq = qp = 0, then there exists an homotopy in M2(B) between p⊕ q and(p + q) ⊕ 0B . Applying this to B = Mn(A), if p and q are two idempotents in Mn(A)such that pq = qp = 0, then there exists an homotopy in M2n(A) between p⊕ q and(p + q)⊕ 0n,A.

We shall now investigate the relations between the three notions of equivalence, simi-larity and homotopy.

Proposition 9.2. Let B be a unital complex Banach algebra. If two idempotents p and qare similar in B, they are equivalent in B; if p and q are homotopic in B, they are similarin B.

Proof. The first assertion is obvious; for the second we need the following lemmaLemma 9.2. For two idempotents p and q in a unital Banach algebra B (with a Banachalgebra norm), ‖p− q‖ < (‖p‖+ ‖q‖)−1 implies that p and q are similar.

Proof. Let u = qp + (1B − q)(1B − p). Then qu = qp = up, and u = 1B − p − q + 2qp =1B − q(q − p) + (q − p)p is invertible when ‖q − p‖(‖p‖+ ‖q‖) < 1.

Suppose that pt is a continuous path of idempotent elements in B, t ∈ [0, 1]. Then‖pt‖ is bounded by some M , and the condition ‖pt − ps‖ < 1/(2M) implies that pt

and ps are similar by Lemma 9.2. By uniform continuity we can find ε > 0 such that‖pt − ps‖ < 1/(2M) whenever |t − s| < ε. We can then pass from p0 to p1 by a finitenumber of similarities (actually, we may find a continuous path of invertible elements (ut)such that pt = ut p0 u−1

t for every t ∈ [0, 1]).

42

Proposition 9.3. Let B be a unital complex Banach algebra. If two idempotents p andq are equivalent in B, then p⊕ 0 and q ⊕ 0 are similar in M2(B); if p and q are similar inB, they are homotopic in M2(B).Proof. Suppose first that p and q are equivalent, with p = xy, q = yx. We may assumethat x = pxq and y = qyp. Indeed, we have p = xyxyxyxyxy = (xy)x(yx) (yx)y(xy),which gives that p = (pxq)(qyp) and similarly for q. We can write

(0 00 p

)=

(1− q y

x 1− p

)(q 00 0

)(1− q y

x 1− p

),

and we know that p⊕ 0 and 0⊕ p are similar (actually homotopic).Suppose now that p and q are similar in B; we can write q = upu−1, with u invertible

in B. Let vt denote a path in GL2(B) from 1B ⊕ 1B to u⊕ u−1 (Lemma 9.1). We get anhomotopy between p⊕ 0B and q ⊕ 0B in M2(B) given by rt = vt (p⊕ 0B) v−1

t .

Examples 9.1.1. Triangular matrices. Let T2(A) denote the algebra of 2×2 upper triangular matrices

with entries in A. Every triangular idempotent is homotopic to its diagonal, using the path(

λ−1 00 1

)(a b0 d

)(λ 00 1

)=

(a b/λ0 d

),

λ varying from 1 to +∞. A similar reasoning applies to the algebra T (Y,X) of operatorson a Banach space X that leave a complemented subspace Y ⊂ X invariant; let Z be thekernel of a projection from X onto Y . Then every idempotent in this algebra T (Y, X) ishomotopic to a projection sending Y to Y and Z to Z.

2. Vector bundles. Let K be a compact topological space. Consider the Banachalgebra C(K) of continuous complex functions on K. An idempotent in Mn(A) identifieswith a continuous function p from K to Mn, such that p(x) is a projection for every x ∈ K.This data defines a (complex) vector bundle over K; for every x ∈ K, the fiber at x is thecomplex vector space Fx = p(x)(Cn) ⊂ Cn; the dimension of the fiber at x ∈ K is therank of p(x). Assume that K = [0, 1]. Setting ps(t) = p(st), we get an homotopy betweenp and the constant function p(0) ∈ Mn. The same reasoning applies to any contractiblecompact space.

Example of the Hopf fibration. The space P1(C) admits a canonical (complex) linebundle. Recall that P1(C) is the set of complex lines in C2. To every line t ∈ P1(C)we associate the one-dimensional vector space t ⊂ C2. Every line t can be parametrizedby a non zero point ζ = (z1, z2) ∈ t. To t ∈ P1(C) containing (z1, z2) we associate theorthogonal projection on the line t,

p(t) =1|ζ|2

(z1z1 z1z2

z2z1 z2z2

);

if we represent t by (z, 1) (this is possible except for the line passing trough (1, 0)) we geta continuous idempotent defined for z = ρ eiθ ∈ C by

p(z) =1

ρ2 + 1

(ρ2 ρ eiθ

ρ e−iθ 1

),

43

and this converges when z tends to infinity to

p(∞) =(

1 00 0

)

that corresponds to the line t passing trough (1, 0). It is clear that P1(C) is homeomorphicto the one-point compactification of C, or also to the sphere S2. Identifying S2 with theclosed unit disc D in C, with all points on the unit circle identified to a single point, weget from the preceding p, setting r = tan(πρ/2), for z = r eiθ ∈ D

q(z) =12

(1− cos πr eiθ sin πre−iθ sinπr 1 + cos πr

)

(note that q(eiθ) = 1 ⊕ 0 for every θ, so that q is constant on the unit circle T and cantherefore be considered as a function on S2). This idempotent q is not equivalent to aconstant function on S2 (more generally, for every integer k ≥ 0, q ⊕ 0k is not equivalentin Mk+2(C(S2)) to a constant idempotent; this is a relatively difficult exercise, where thedecisive argument is Brouwer’s fixed point theorem or the notion of degree).

3. Cuntz algebras. For every n ≥ 2 let On be the C∗-algebra generated in L(`2)by n into isometries U1, . . . , Un (so U∗

i Ui = I for every i) such that∑n

i=1 UiU∗i = I (it

follows that U∗j Ui = 0 when i 6= j). It is proved in [C1] that On does not depend upon the

particular choice of (Ui) (see section 7 for O3). For every i = 1, . . . , n, we have a projectionpi = UiU

∗i equivalent to I since pi = UiU

∗i and I = U∗

i Ui.The algebra En is generated in L(`2) by n into isometries V1, . . . , Vn such that

n∑

i=1

ViV∗i < I

(it also follows that V ∗j Vi = 0 when i 6= j). The projections qi = ViV

∗i , i = 1, . . . , n are

again equivalent to I.

The semi-group of classes of idempotents

We introduce now the algebra M∞(A) equal to the union of the Mn(A), the embeddingof Mn(A) into Mn+p(A) being given by a → a⊕0p,A; we say that two idempotents a and bare equivalent, similar or homotopic in M∞(A) iff there is some n such that a, b ∈ Mn(A)and a and b are equivalent, similar or homotopic in Mn(A); in M∞(A) the three notionsof comparison for idempotents coincide. Let Pr(A) denote the set of equivalence classesof idempotents in M∞(A). Let p denote the equivalence class of an idempotent p ofM∞(A) in Pr(A).Additive structure on Pr(A)

If p and q are two classes in Pr(A), we define their sum by p+ q = p⊕ q.Exercise. Show that this operation is well defined, associative and that 0 is a neutralelement. In other words, Pr(A) is a monoid.

This addition is commutative; indeed, we know by Lemma 9.1 that p ⊕ q and q ⊕ pare homotopic.

44

Examples 9.2.1. In Mn(C), we know that two idempotents are equivalent iff they have same rank.

This shows that Pr(C) can be identified to Z+. Furthermore, the addition correspondsto the addition of ranks. This shows that Pr(C) is simply Z+ (as monoid), where 1 ∈ Zis identified to the class of rank one projections. If we consider Pr(Mn), it is clear thatM∞(Mn) ' M∞(C) and hence Pr(Mn) ' Z+, where again 1 ∈ Z+ corresponds to theclass of rank one projections.

2. 1n,A = n 1A.3. If p is an idempotent in Mn(A), then p+ 1n,A− p = 1n,A. More generally, if

p and q are two idempotents in Mn(A) such that pq = qp = 0, then p+ q = p + q.This follows directly from Example 9.2, 4 above.

4. Let A,B be two unital Banach algebras. It is clear that Pr(A × B) ' Pr(A) ×Pr(B). Let T2(A) be the algebra of 2 × 2 upper triangular matrices with entries in A. Amatrix in Mn(T2(A)) can be considered as an element of T2(Mn(A)), after some reindexing.Applying the deformation from Example 9.1, 1 with A replaced by Mn(A), we see thatPr(T2(A)) ' (Pr(A))2. For the algebra T (Y, X) of operators on a Banach space X thatleave a complemented subspace Y ⊂ X invariant, we see that Pr(T (Y, X)) ' Pr(L(Y ))×Pr(L(Z)), where Z is the kernel of a projection from X onto Y .

5. Vector bundles. We said that an idempotent in Mn(A), A = C(K) identifies to acontinuous map from K to the space of idempotents in Mn. If K splits into two closedand open subsets K1 and K2, then Pr(C(K)) is isomorphic to Pr(C(K1)) × Pr(C(K2)).If K is connected, the rank of p(t) is constant when t varies in K, but this rank is notenough to characterize the class of p in Pr(C(K)). This rank only gives the dimension (orrank) of the associated vector bundle on K. When K is contractible, every idempotentp ∈ M∞(C(K)) is equivalent to a constant function p(x0) ∈ Mn, hence Pr(C(K)) ' Z+ inthis case. In any case Z+ is always a submonoid of Pr(C(K)), corresponding to the classesof constant functions p (or of trivial bundles).

The example of the Hopf fibration yields an idempotent p ∈ M2(C(S2)) which is notequivalent to a constant function on S2; the monoid Pr(C(S2)) contains at least Z2

+.6. When A = L(X), the class IX in the monoid Pr(L(X)) is the class of comple-

mented subspaces of Xn isomorphic to X. If X is a Banach space such that X ' X2, wehave seen in Example 9.2, 3 that IX ⊕ IX = IX in M∞(L(X)). One has therefore2IX = IX+ IX = IX. Indeed, we know that IX ⊕ IX and IX ⊕ 0 are equivalent,and this implies by definition of the addition that IX+ IX = IX.

Several examples show that the monoid Pr(A) may fail the cancellation property (butit is true in Pr(C) ' Z+): for A = L(L1) for example, we have

`1 ⊕ L1 ' L1 ⊕ L1 ' L1 ' 0⊕ L1,

hence identifying projections and ranges (see Exercise 9.2,3),

`1+ L1 = L1+ L1 = L1.When the Banach space X is isomorphic to its hyperplanes, we have in A = L(X),

IX = IX − p when p is a rank one projector, hence

IX+ p = IX.

45

7. Cuntz algebras. In On we have n projections pi = UiU∗i , each equivalent to I, and

pipj = 0 when i 6= j. Then the relation∑n

i=1 UiU∗i = I implies that n1On = 1On,

using case 3 above.In En, Q = I−∑n

i=1 ViV∗i is a non zero projection and we get n1En

+ Q = 1En.

Group associated to an additive monoid M . Additive group K0(A)

On the set of couples (m,n) ∈ M2, we define the equivalence relation (m, n) ∼ (m′, n′)if there exists r ∈ M such that m+n′+r = m′+n+r. The quotient of M by this relationis an additive group G. If φ denotes the map from M to G that sends m ∈ M to the class[(m, 0)] of (m, 0) in G, there exists for every monoid morphism f from M to a group Ha unique group morphism f : G → H such that f = f φ. Every element α ∈ G can bewritten α = φ(m)− φ(n) for some m,n ∈ M .

Definition 9.1. If A is a unital Banach algebra, we denote by K0(A) the group associatedto the monoid Pr(A). We denote by [p] the image φ(p) by φ : Pr(A) → K0(A) of theclass p in Pr(A) of an idempotent p ∈ M∞(A). Every element of K0(A) can be written[p]− [q] for some idempotents p, q in M∞(A).

Every α = [p]−[q] ∈ K0(A) can be written [p′]−[1n,A] for some n and some idempotentp′ in M2n(A). This follows from Example 9.3, 3 below.

We have [p] = [q] if and only if there exists r such that p+ r = q+ r. Sincer is an idempotent in some Mn(A), there exists s ∈ Mn(A) so that r + s = 1n,A(we may simply choose s = 1n,A− r), and finally [p] = [q] if and only if there exists n suchthat p+ 1n,A = q+ 1n,A.Examples 9.3.

1. For C, we obtain of course K0(C) ' Z. Recall that 1 ∈ Z corresponds to the classof rank one projections. We also have K0(Mn) ' Z for every integer n ≥ 1, where again1 ∈ Z corresponds to the class of rank one projections.

2. [1n,A] = n [1A] (since we had 1n,A = n 1A).3. If p is an idempotent in Mn(A), [p] + [1n,A − p] = [1n,A]. This is because p +

1n,A − p = 1n,A. More generally, if p and q are two idempotents in Mn(A) such thatpq = qp = 0, then [p] + [q] = [p + q].

4. Let p be a finite rank projection in some Banach space X isomorphic to its hyper-planes. Then [p] = 0. Indeed, we saw in Example 9.2, 6 that IX+ q = IX if q hasrank one, thus [q] = 0 and when p has rank n, we get [p] = n[q] = 0.

5. Suppose A = L(X), where X is a Banach space such that X ' X2, we have[IX ] + [IX ] = [12,A] = [IX ], thus [IX ] = 0. Since IX is the class in Pr(L(X)) ofcomplemented subspaces of Xn isomorphic to X, the class of [IX ] is 0 in K0(L(X)) if andonly if there exists an integer n such that Xn ' Xn+1. It is relatively usual for manyclassical spaces that X ' X2. I know of no example where X 6' X2 but X2 ' X3.

Suppose that X ' `p(X) for some 1 ≤ p < ∞. Then every complemented subspaceY of X containing a complemented copy of X is isomorphic to X. This is PeÃlczynski’sdecomposition method. If X = Y ⊕ Z and Y = X ⊕ U we have

X ⊕ Y ' `p(X)⊕X ⊕ U ' `p(X)⊕ U ' Y

46

andX ⊕ Y ' `p(Y ⊕ Z)⊕ Y ' `p(Y ⊕ Z) ' X.

This implies that X is isomorphic to its square and isomorphic to its hyperplanes.6. K0(L(X)) for a primary Banach space isomorphic to its square. The space X

is said primary when for every decomposition X = Y ⊕ Z, one has Y ' X or Z ' X.We know for example after Enflo [E] that Lp[0, 1] is primary when 1 ≤ p < ∞ ([AEO],[M1]). The space `p is more than primary, it is prime; recall that a Banach space is said tobe prime if it is isomorphic to every infinite-dimensional complemented subspace of itself.The only known examples before [GM2] were c0 and `p (1 ≤ p ≤ ∞). These were shownto be prime by PeÃlczynski [P], apart from `∞ which is due to Lindenstrauss [L1]. If weassume that X2 is primary, it follows that X ' X2, and Xn ' X is primary for everyn. Let A = L(X). For every idempotent p ∈ Mn(L(X)), either [p] = [1n,A] = [IX ] or[1n,A − p] = [IX ]. We also know that [IX ] = 0 because X ' X2. In either case [p] = 0,hence K0(L(X)) = 0. In particular for a Hilbert space H we obtain K0(L(H)) = 0.This shows that in most classical cases, the K0-theory of L(X) is trivial (and thus mostlyuseless). It will not be so for the exotic spaces of sections 10, 11and 12.

7. It is clear that K0(A × B) ' K0(A) × K0(B). Since Pr(T2(A)) ' (Pr(A))2, itfollows that K0(T2(A)) ' K0(A)2. For the algebra T (Y, X) introduced previously, we seethat K0(T (Y,X)) ' K0(L(Y ))×K0(L(Z)), where Z is the kernel of a projection from Xonto Y . On the contrary, for M2(A) as well as for Mn(A), we obtain K0(Mn(A)) ' K0(A).

8. Let A = C(K), where K is compact and connected. We see that K0(C(K))contains a canonical copy of Z, corresponding to idempotents in Mn(C(K)) given byconstant functions from K to Mn. When K is contractible, we know that every idempotentin Mn(C(K)) is homotopic to a constant function, hence we get K0(C(K)) ' Z.

9. Simpler presentation of the case of a Banach space X such that X⊕X is isomorphicto a complemented subspace of X. In this case we can avoid the use of M∞(L(X)) andthe symmetrisation from Pr to K0 in the following way: let X ' X ⊕ X ⊕ Y . To everyT ∈ L(X) we associate, in block notation with respect to the decomposition X = X⊕X⊕Y

i1(T ) =

T 0 00 0 00 0 0

; i2(T ) =

0 0 00 T 00 0 0

.

Then i1(T ) and i2(T ) are equivalent to T ; when T and U are equivalent, then i1(T ) andi2(U) are homotopic in L(X) and given T1, . . . , Tn we can find using products of i1 and i2operators U1, . . . , Un such that Ui ∼ Ti and such that the Ui appear as disjoint diagonalblocks in a larger decomposition of X. We may then define the sum T1+ · · ·+ Tn asU1 + · · ·+ Un.

10. Cuntz algebras. In On the relation n1On = 1On implies that (n−1)[1On ] = 0.Cuntz proved in [C2] that K0(On) = Z/(n− 1)Z. In En we obtain (n− 1)[1En ] + [Q] = 0.

K0 functor

Let ϕ be a morphism of unital Banach algebras from A to B. Letting ϕ act oneach entry of a matrix in Mn(A), we get a unital algebra morphism ϕ(n) from Mn(A) to

47

Mn(B) for every n ≥ 1, sending idempotents in Mn(A) to idempotents in Mn(B). Clearly,homotopic idempotents have homotopic images. This gives a morphism of monoids fromPr(A) to Pr(B), then a group morphism ϕ∗ from K0(A) to K0(B). One can check thatA → K0(A), ϕ → ϕ∗ defines a functor from the category of unital Banach algebras to thecategory of additive groups.

An example. For every n ≥ 0 let ϕn be the algebra morphism from M3n to M3n+1 definedby ϕn(a) = a⊕a⊕a. For every n, we have that K0(M3n) = Z, and the map ϕn sends rankone projections to rank three projections, hence ϕn,∗(1) = 3. If we consider the Banachalgebra A obtained as inductive limit of the sequence (M3n) with the (ϕn) as successiveembeddings, we obtain a chain of maps

Z→ · · · → Z→ Z→ · · ·

where each arrow is the map ϕ∗(k) = 3k; this implies (with some work) that

K0(A) ' k

3n∈ Q : k ∈ Z, n ≥ 0.

We may associate to 1 ∈ Q the class [1A]. This algebra A is closely related to the algebraB appearing in section 7 with the discussion of P; actually A is the completion of B underits (unique) C∗-norm.

K0 for non unital algebras

Let A be a Banach algebra without unit. We consider the unital algebra A+ =A⊕ C from section 2. Then A is a closed two-sided ideal of A+ and A+/A is canonicallyisomorphic to C. Let π be the projection from A+ onto C, and let i : C→ A+ be given byi(λ) = λ1+. We have πi = IdC. By the functorial character we get π∗ : K0(A+) → K0(C),i∗ : K0(C) → K0(A+), and π∗ i∗ = Id. Hence K0(C) ' Z appears as factor in K0(A+);we define K0(A) as the kernel of π∗:

K0(A) = kerπ∗ ⊂ K0(A+).

Let α ∈ K0(A). We know that we can write α = [p] − [1n,A+ ] ∈ K0(A+), where p isan idempotent in M2n(A+); by definition of K0(A) we have π∗(α) = 0. This means thatπ(p) and 1n ⊕ 0n are equivalent idempotents in M2n(C). Hence there exists u ∈ GL2n(C)such that uπ(p)u−1 = 1n ⊕ 0n. Then, setting u = u ⊗ 1A+ , we see that r = u p u−1 is anidempotent in M2n(A+), equivalent to p, and r = 1n,A+ ⊕ 0n,A + a, with a ∈ M2n(A).Finally, replacing p by r we obtain:

Every α in K0(A) can be expressed as α = [r]− [1n,A+ ], where r is an idempotent inM2n(A+) of the form r = 1n,A+ ⊕ 0n,A + a, with a ∈ M2n(A).

Remark 9.1. The above construction is certainly necessary for Banach algebras withoutunit that have no idempotent, except 0 (for example, the algebra A = C0(K) of continuousfunctions on a compact connected space K, vanishing at some point x0 ∈ K). Somealgebras without unit, like K(X), already have idempotents (finite rank projections). If p

48

is an idempotent in A, its class in the above construction is given by [1A+ ]− [1A+ − p] or[1A+ ⊕ p] − [1A+ ]. One can check that when A is already unital, the above constructiondefines the same group K0(A). Indeed we have the exact sequence A

i−→A+ π−→C, giving

K0(A) i∗−→K0(A+) π∗−→K0(C),

where K0(A) here is defined according to our first definition for unital Banach algebras.By functoriality it is clear that i∗(K0(A)) is contained in the kernel of π∗. Conversely, letα ∈ K0(A+) belong to kerπ∗. We write α = [p]− [1n,A+ ], where p = 1n,A+ ⊕0n,A +a is anidempotent in M2n(A+), with a ∈ M2n(A). We see that q = 1A+ − 1A is an idempotent inA+ such that qA = Aq = 0. We may write p as p = p′ + q′, where p′ = 1n,A ⊕ 0n,A + a isan idempotent in M2n(A), and q′ is an idempotent in M2n(A+) (q′ is the direct sum of ncopies of q) such that q′p′ = p′q′ = 0, hence [p] = [p′] + [q′]; similarly [1n,A+ ] = [1n,A] + [q′]and finally [p]− [1n,A+ ] = [p′]− [1n,A] belongs to the image of K0(A) in K0(A+).

The K0 functor extends now to the category of Banach algebras, not necessary unital.

Examples.1. Let K be a compact connected topological space. Let C0(K) = Cx0(K) denote

the closed ideal of C(K) consisting of continuous functions on K vanishing at some pointx0 ∈ K; we see that C(K) ' C0(K)+, hence

K0(C(K)) ' K0(C0(K))⊕ Z.

When A = C([0, 1]), we have seen that K0(C([0, 1])) ' Z (true for any contractible space).For a contractible compact space K we get K0(C0(K)) = 0. For C0(T), we also haveK0(C(T)) = Z, thus K0(C0(T)) = 0.

2. We have seen that P1(C) ' S2 admits a canonical (complex) line bundle (Hopffibration). This bundle gives a non zero class in K0(C0(S2)): if we identify C0(S2) tothe space of continuous functions on C vanishing at infinity, the idempotent described inExample 9.1, 2 gives a non zero element [p]− [1⊕ 0] in K0(C0(S2)).

Example of K0(S)Here X is a Banach space, and we study the ideal S(X) of strictly singular operators.

Lemma. Let X = U ⊕ V , dimU = dim V = +∞, and let πU be the projection of U ⊕ Vonto U . Let A denote the unital subalgebra of L(U ⊕ V ) generated by πU and S(U ⊕ V ).Let p be a projection in L(X), with the form p = πU + S, where S ∈ S(U ⊕ V ). Thenp is equivalent (in A), either to a projection on Y , finite codimensional subspace of U , orto a projection on U ⊕ E, where E is a finite dimensional subspace of V . In other words,p is equivalent to a projection (πU − r1)⊕ r2, where r1 and r2 are finite rank projectionsin L(U) and L(V ) respectively. The quantity rank(r2) − rank(r1) is an invariant of thesimilarity class of p in A.Proof. Let q = πU p πU = IU +S′, S′ ∈ S(U), considered as operator on U . By Proposition6.1, we know that q has a finite codimensional invariant subspace Z on which q gives anisomorphism a ∈ L(Z). Considering a new decomposition Z ⊕W of U ⊕ V (notice that

49

finite rank projections belong to A, so the projections on Z and W belong to A), we getin block notation in Z ⊕W

p =(

a bc d

)= p2,

with a invertible, b, c, d strictly singular; then p is similar in A to(

a−1 0−ca−1 IW

) (a bc d

)(a 0c IW

)=

(IZ b′

0 d′

),

(we know that a−1 ∈ A by Lemma 2.2 and Proposition 6.1) with b′, d′ strictly singular. Itfollows from the fact that the above matrix is an idempotent that b′d′ = 0, d′2 = d′; next

(IZ b′

0 IW

) (IZ b′

0 d′

)(IZ −b′

0 IW

)=

(IZ 00 d′

)

is an idempotent equivalent in A to an idempotent of the desired form; indeed, d′ isa strictly singular projection, thus has finite rank. It only remains to move the finitedimensional image F of d′ in the correct position; if k denotes the codimension of Z in Uand if dim F ≤ k, we can move F to G inside U , in such a way that Z ⊕G is a direct sum;if dim F > k, we will move a k-dimensional subspace of F to some G inside U in such away that U = Z ⊕G, and the remaining part of F into V .

Let us give now a partial proof for the claim in the last line of the Lemma. Let rbe a finite rank non zero projection in L(V ), and let F = rV 6= 0; we will show thatq = πU ⊕ r is not similar to πU in A. If it was similar, we could find an invertible elementu in A such that πUu = uq. Then quq gives an isomorphism from U ⊕ F onto U , thus aFredholm operator in L(U ⊕ F ) with non zero index. But quq has a block decompositionin U ⊕ F (

λIU + s brrc rdr

),

with s strictly singular and λ 6= 0 (because u is invertible). This operator is a strictlysingular perturbation of λπU , and should therefore have zero index, a contradiction.

We arrive to the computation of K0(S(X)). Here A = S(X) and we assume of coursethat dimX = +∞; we can then identify A+ to the subalgebra of B = L(X) consistingof all operators λId + S, S ∈ A and λ ∈ C. Let now α ∈ K0(S(X)). There exists aprojection p on X2n of the form 1n,B +S, with S ∈ S(X2n), such that α = [p]− [1n,B ]. Bythe Lemma, we know that p is equivalent either to a projection on Y , finite codimensionalsubspace of Xn, or to a projection on Xn ⊕F , dim F < ∞. Furthermore we said that theinteger k, equal to − codim Y or to dim F characterizes α, hence K0(S(X)) is isomorphicto Z. We may choose for generator of K0(S(X)) the class of rank one projections, (seeremark 9.1).

The above proof also applies to K(X). We get that K0(K(X)) ' Z. In the case ofthe Hilbert space, a more natural approach uses the fact that K(`2) is the inductive limitof the algebras (Mn).

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Short exact sequence in K-theory

Suppose given a short exact sequence

0 → Ii−→A

π−→A/I → 0,

where I is a closed two-sided ideal in A, i the inclusion map and π the canonical quotientmap. We get

K0(I) i∗−→K0(A) π∗−→K0(A/I)

exact at K0(A) (we just remark that π∗ i∗ = 0 by functoriality; the proof that kerπ∗ isprecisely equal to the range of i∗ uses Corollary 9.1 and Lemma 9.1).

Exercise. Prove the above: let α = [p] − [1n,A], with p idempotent in M2n(A), such thatπ∗(α) = 0. We know that for some m, the idempotent π(p)⊕1m,A/I is similar to 1n+m,A/I

in Mn+m(A/I); apply now the last part of Lemma 9.1 to M2n+2m(A/I) and next applyCorollary 9.1.

Suspension and the group K1(A)

Let A be a Banach algebra. Denote by SA the (non unital) algebra of continuousmaps f : T → A such that f(1) = 0A, the product being the pointwise product in A.We will set K1(A) = K0(SA), but some comments are necessary. For defining K0(SA)we consider idempotents in Mn((SA)+). Such an idempotent p is a n × n matrix withentries in (SA)+. An element of (SA)+ identifies to a continuous function f from T to A+

such that f(1) = λ1+. The matrix p identifies to a continuous function p(t) with valuesin idempotents of Mn(A+) such that p(1) is a “scalar” matrix i.e. a matrix q ⊗ 1n,A+ ,q ∈ Mn(C). Up to equivalence we can assume that p(1) = 1k,A+ ⊕ 0n−k,A+ . By definitionwe get an homotopy from p(1) to p(1) obtained by travelling around the circle. This givesas explained after Lemma 9.2 a similarity up(1) = p(1)u. But the special form of p(1)implies that u ∈ GLn(A+) leaves both factors (A+)k and (A+)n−k invariant, so u = v⊕w,with v ∈ GLk(A+) and w ∈ GLn−k(A+). We have therefore associated to any idempotentp in M∞((SA)+) an invertible v ∈ GLk(A+), for some k.

Conversely, let v ∈ GLk(A+); we may associate to v some w ∈ GLn−k(A+) suchthat v ⊕ w ∈ GL

(0)n (A+) (for example n = 2k, w = v−1). We can find a continuous

path ut in GLn(A+) such that u0 = v ⊕ w and u1 = 1n,A, and the path of idempotentsp(t) = ut (1k,A+⊕0n−k,A+)u−1

t ; we check that p(1) = p(0) (because u(0) and u(1) commuteto p(0)) and p(0) is scalar), thus p identifies to an idempotent in Mn((SA)+). If v′ belongsto the same connected component of GLk(A+), we may find a continuous path (vt) fromv to v′ in GLk(A+), that gives us an invertible v in GLk(C([0, 1], A+)). Applying theabove construction to the algebra C([0, 1], A+), we associate to v an idempotent p ofMn(C([0, 1], A+)), i.e. a continuous function from [0, 1] to the idempotents of Mn(A+).The values at 0 and 1 of that function p are the idempotents p et p′ associated to v et v′

by the above reasoning, so p and p′ are homotopic. Finally, p and p′ are homotopic whenv and v′ belong to the same component. This explains why K0(SA) is equivalent to thestudy of connected components of GLn(A+) (for varying n).

51

The usual definition of K1(A) uses the family of groups GLn(A). Let A be a unitalBanach algebra. Let GL∞(A) denote the group equal to the union of the GLn(A), n ∈ N,where the injection from GLn(A) into GLn+p(A) is now of course given by u → u ⊕1p,A. The group K1(A) is the set of connected components of GL∞(A), that is the setof equivalence classes of u ∈ GL∞(A) for the homotopy relation in GL∞(A): we saythat u, v ∈ GL∞(A) are homotopic in GL∞(A) if there exists an integer n such thatu, v ∈ GLn(A) and u, v are homotopic in GLn(A). The product of [u] and [v] is thecomponent of the product uv. This product is commutative because [uv] = [uv ⊕ 1] =[u ⊕ v] = [v ⊕ u] = [vu] by Lemma 9.1. We shall actually use the additive notation inK1(A).

For an algebra without unit we set K1(A) = K1(A+). We obtain a second functorfrom the category of Banach algebras to the category of groups, the K1 functor. Indeed,every homomorphism from A to B induces a map from GL∞(A) to GL∞(B) which givesa group homomorphism from K1(A) to K1(B).

Examples 9.4.

1. Since GLn(C) is connected we get K1(C) = 0. Compare to K0(C0(T)) = K0(SC).

2. Products and triangular matrices; it is clear that K1(A×B) = K1(A)×K1(B). Itis also clear that a triangular matrix is invertible iff its diagonal elements are invertible. Wealready explained that an element in Mn(T2(A)) can be seen as an element of T2(Mn(A));if it is invertible we may deform it inside GLn(T2(A)) to the diagonal form as explainedbefore, but also trivially by letting the non diagonal entries go to 0. It follows thatK1(T2(A)) ' (K1(A))2. For T (Y, X), we see that K1(T (Y, X)) ' K1(L(Y ))×K1(L(Z)),where Z denotes the kernel of the projection from X to Y .

3. Show that K1(S(X)) = 0, K1(K(X)) = 0.Hint. Use Proposition 6.1 and a reasoning similar to the one used for proving the connect-edness of GLn(C) in Exercise 9.1.

4. It is known that the linear group of a Hilbert space is connected [CL], henceK1(L(H)) = 0. This proof uses the functional calculus for isometries on a Hilbert space.We can give a more Banach space theoretic proof (which is essentially Kuiper’s lemma from[Ku]), that also works for `p, when 1 ≤ p ≤ ∞ (and for c0). Let T be an invertible operatorin L(`p). We shall prove that T ⊕ I`p is homotopic to I`p ⊕ I`p in L(`p⊕ `p) = M2(L(`p)).Let us represent the second factor `p in the sum as X = `p(`p ⊕ `p). In each component`p⊕ `p of X, we may find an homotopy from I`p ⊕ I`p to T ⊕T−1. This gives an homotopyfrom IX , written symbolically as `p(I`p ⊕ I`p) to `p(T ⊕ T−1). Then T ⊕ IX is homotopicto T ⊕ `p(T ⊕ T−1); using a different grouping of the T and T−1 we may deform this lastoperator back to I`p⊕IX . Finally K1(L(`p)) = 0. The same proof works for any Banachspace X such that X ' `p(X).

Actually, more is true: Neubauer proved that GL(L(`p)) is contractible; see alsoMityagin [Mt] for more examples. To the contrary, the linear group of `p ⊕ `q, 1 ≤ p <q < ∞, is not connected (Douady [Do]). We shall compute later K1(L(`p, `q)).

5. Let α ∈ K1(C([0, 1])); it corresponds to an invertible v ∈ GLn(C([0, 1])), that isto say a continuous map from [0, 1] to GLn. Letting vs(t) = v(st) we define an homotopy

52

between v and the constant function v(0) ∈ GLn(C). Since GLn is connected we getK1(C[0, 1])) = 0. This argument generalizes to any contractible compact space K.

The situation is different for C(S1). In this case the determinant of v(t) can make anon trivial loop around 0 in C, therefore K1(C(S1)) 6= 0; we get actually K1(C(S1)) ' Z.This will correspond to a special case of Bott’s theorem.

The index map

Let I be a closed two sided proper ideal of A. We are going to define a map ∂ fromK1(A/I) to K0(I) that plays the role of the connecting map in homological theories. Recallthat for every unital Banach algebra B, every invertible element in GL(0)(B/J) can belifted to an invertible element in GL(0)(B).

In order to simplify the discussion of the index map let us assume that A is unital;let I be a proper closed two-sided ideal in A, let i : I → A be the inclusion map andπ : A → A/I the quotient map; let us identify I+ with the subalgebra of A consistingof all elements λ1A + x, λ ∈ C and x ∈ I. Let u be invertible in GLn(A/I). Noticethat Mn(A/I) ' Mn(A)/Mn(I), so that we can apply the above remarks to B = Mn(A)and to the ideal J = Mn(I) in B. We have seen that u ⊕ u−1 belongs to GL

(0)2n (A/I).

We know that every element in GL(0)2n (A/I) can be lifted to GL

(0)2n (A). Let v ∈ GL2n(A)

be any lifting of u ⊕ u−1. Consider p = v (1n,A ⊕ 0n,A) v−1. This is an idempotent andπ(p) = 1n,A/I ⊕ 0n,A/I . The matrix p is thus in M2n(I+). If we set

∂[u] = [p]I+ − [1n,A]I+

we get an element of K0(I+) such that q∗(∂[u]) = 0, where q : I+ → I+/I ' C is thecanonical quotient map; therefore we get an element of K0(I) (of course one has to showthat this class only depends upon [u]; the notation [p]I+ means that this is a class computedin K0(I+); notice that as idempotent in M2n(A), p is similar to 1n,A ⊕ 0n,A, so that

[p]A − [1n,A]A = 0 ∈ K0(A);

in other words i∗∂[u] = 0). It is easy to check that ∂ is a group morphism.

Example. The right shift on `2(N).Let R be the right shift on `2(N), and let L denote the left shift; we see that LR =

I = I`2 and RL = I − e1 ⊗ e1. The image R of R in the Calkin algebra C = L(`2)/K(`2)is thus invertible, and the inverse is the image of L. Apply the preceding discussion tou = R ∈ C. Let P be the rank one projection P = e1 ⊗ e1. We obtain an explicit lifting ofu⊕ u−1 given by

v =(

R P0 L

), and v−1 =

(L 0P R

),

(Notice that LP = PS = 0). The idempotent p from the general discussion is now

p = v (1⊕ 0) v−1 =(

RL 00 0

)=

(I − P 0

0 0

).

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According to the discussion about K0(S), the element ∂R = [p] − [I] is identified to theopposite of the codimension of the image of p in the first factor `2, which gives here∂R = −1. This value is equal to the index of R. This is not an accident:Exercise. Generalize the above discussion to the quotient algebra L(X)/S(X) and to anarbitrary Fredholm operator T on any Banach space X.

Exactness

The construction of the index map extends to the non unital case. Suppose given ashort exact sequence

0 → Ii−→A

π−→A/I → 0.

We deduce a sequence

K1(A) π∗−→K1(A/I) ∂−→K0(I) i∗−→K0(A) π∗−→K0(A/I)

exact at K1(A/I) and at K0(I). We shall only explain that i∗ ∂ and ∂ π∗ vanish. Wealready observed that i∗∂ = 0; if w ∈ GLn(A) and u = π(w), we may choose as liftingv ∈ GL2n(A) of u⊕u−1 simply v = w⊕w−1, and with this choice it is clear that ∂[u] = 0;this reasoning shows that ∂ π∗ = 0.

Triviality of ∂; when for each integer n, every invertible u ∈ GLn(A/I) can be lifted toan invertible v ∈ GLn(A), we see by definition of ∂[u] that ∂[u] = 0. This is in particulartrue if the quotient map π : A → A/I is splitted by an algebra morphism σ such thatπ σ = Id. In this case ∂ = 0.

The above exact sequence is related to the discussion about K1(A) ' K0(SA). Sup-pose that A is unital, and represent SA as the algebra of continuous functions from [0, 1]to A such that f(0) = f(1) = 0A. Let SA be the algebra of continuous functions from[0, 1] to A such that f(0) = 0A. It is clear that SA/SA ' A in a canonical way. The aboveexact sequence gives

K1(SA) → K1(A) ∂−→K0(SA) → K0(SA);

using the homotopy ft(s) = f(ts) which moves every element in Mn(SA) to the zeromatrix when t decreases from 1 to 0, it is easy to see that K1 and K0 vanish for SA, sothat ∂ gives our isomorphism between K1(A) and K0(SA).

K2(A) and Bott’s periodicity theorem

We set now K2(A) = K1(SA); then Bott’s periodicity theorem states that K2(A) 'K0(A). We define K2(A) = K1(SA) = K1((SA)+); assume that A is unital for simplicity.We describe again SA as the algebra of continuous functions from the circle T to A suchthat f(1) = 0A, and (SA)+ as the algebra of continuous functions from T to A such thatf(1) = λ1A for some λ ∈ C. We say that a ∈ Mn(A) is “scalar” if a = Λ ⊗ 1A, whereΛ ∈ Mn. The group K1((SA)+) is defined using invertible elements in Mn((SA)+). Butgiving such an invertible is the same as giving a continuous path u from T into GLn(A), such

54

that u(1) is a “scalar” matrix. We can find an equivalent element v such that v(1) = 1n,A

in the following way: let u(1) = Λ ⊗ 1A and let (µt) be a continuous path in GLn from1n to Λ−1; then ut = (µt ⊗ 1A)u is a path from u to v in GLn((SA)+), and v satisfiesv(1) = 1n,A.

We define a map j : K0(A) → K2(A) which is easy to describe; the difficult partwill be to show that j is onto; to each idempotent p in Mn(A) we associate the “loop”f(z) = zp + (1n,A − p) in GLn(A), z ∈ T. Then f(1) = 1n,A, f belongs to GLn((SA)+),and we set j(p) = [f ] in K2(A). We can see easily that the image of p ⊕ q in K2(A) isthe sum of images, hence our map j is a monoid morphism from Pr(A) to K2(A), hencegives a group morphism still denoted j from K0(A) to K2(A). As we said, the difficultpart in the proof of Bott’s theorem is the fact that j is onto from K0(A) to K2(A); thefact that j is injective from K0(A) into K2(A) is not obvious but will follow from the factthat j is onto for every Banach algebra. Let us indicate the main steps of the proof thatj is onto. We first observe that the path zk1n,A of invertible defines the same class asz1kn,A in K2(A). Multiplying a path of invertible by zk1n,A amounts thus to add the loopj(1kn,A) associated to the idempotent 1kn,A.

Let α ∈ K2(A). We can find a continuous map w from T to GLn(A) such that [w] = α,and we may assume that w(1) = 1n,A. We want to find an element β in K0(A) such thatj(β) = [w] = α.

1. We begin by approximating our continuous map w from T to GLn(A) by a mapz → w′(z) from T to GLn(A) which is a trigonometric polynomial w′(z) =

∑Nk=−N vkzk,

with each vk ∈ Mn(A), and such that w′(1) = 1n,A; the standard way is to use convolutionwith a Fejer kernel; we have [w′] = [w] if the approximation is good enough; multiplyingby zN we get a polynomial u(z) = zNw′(z) in z ∈ T with coefficients in Mn(A). Withthis multiplication we have added to [w] the class of the loop associated to 1nN,A, that weshould substract at the end. We are going to find an idempotent p in M∞(A) such thatj(p) = [u]. It will follow that j([p]− [1nN,A]) = [w] = α, thus proving that j is onto.

2. Passing to a larger dimension K we may find an equivalent path u1 that is linearin z, namely u1(z) = a + bz, a, b ∈ MK(A) and u1(1) = 1K,A. We may describe this (ina way that is not the cheapest on dimensions) in the following way: for every polynomialP (z) with coefficients in a unital Banach algebra B and such that deg P ≤ m, we maywrite

P (z) = (z − 1)2Q(z) + R(z),

where deg R ≤ 1 and deg Q ≤ m− 2. Let for every λ ∈ [0, 1]

ϕλ(z) =(

1B λ(z − 1)1B

0 1B

)(P (z) 0

0 1B

) (1B 0

−λQ(z) 1B

)=

(P (z)− λ2(z − 1)2Q(z) λ(z − 1)1B

−λ(z − 1)Q(z) 1B

).

For every λ, we get that z → ϕλ(z) is a path of invertible and ϕλ(1) = 12,B , ϕ0 = P (z)⊕1B .When λ = 1, the result is a path of invertible elements with degree ≤ m− 1 in z,

ϕ1z) =(

R(z) (z − 1)1B

−(z − 1)Q(z) 1B

).

55

When λ varies from 0 to 1 we get an homotopy in GL2((SB)+) between P (z) ⊕ 1B andthis path ϕ1(z) of degree ≤ m− 1 in z.

3. Using spectral theory one finally shows that a linear path of invertible elements isequivalent to a loop z → zp + (1− p) where p is an idempotent. We shall prove a crucialLemma. We recall the notion of spectral projection. Let B be a unital Banach algebra.Assume that the spectrum of b ∈ B does not meet the imaginary axis in C. It is thenpossible to find a circle γ centered at some real point M , with M > 0 large, and withradius M − ε, ε > 0 small, such that every λ ∈ σ(b) with Re λ > 0 will be contained in theinterior of γ. Let

p =1

2πi

∫

γ

(z − b)−1 dz.

Then p is a spectral projection commuting with b.

Lemma 9.3. Let B be a unital Banach algebra. Assume that b ∈ B is such that b− it1B

is invertible for every t ∈ R, and let p be the above spectral projection corresponding tothe half complex plane C+ = z ∈ C : Re z > 0. Then b + sp is invertible for every realnumber s ≥ 0. In other words: if σ(b) does not meet the imaginary axis iR, the same istrue for σ(b + sp) for every real s ≥ 0.

Proof. Let Bp = pBp. This is a Banach algebra with unit p. We know that bp = pb = pbpand the spectrum of bp = pbp in Bp is contained in C+. It follows that bp + sp is invertiblein Bp for every real s ≥ 0, hence there exists u ∈ B such that

(pbp + sp)pup = pup(pbp + sp) = p.

Let q = 1B − p. The spectrum of qbq in Bq is contained in C−, hence bq is invertible in Bq

and there exists v ∈ B such that

(qbq)qvq = qvq(qbq) = q.

Now(b + sp)(pup + qvq) = (b + sp)(p + q)(pup + qvq) =

(bp + sp)(pup) + bq(qvq) = (pbp + sp)(pup) + (qbq)(qvq) = p + q = 1B

and similarly for the other direction, (pup + qvq)(b + sp) = 1B .

For the sake of completeness, let us give a sketch of a proof that the spectrum of bp

is contained in C+. First of all, it is clear that the spectrum of bp in Bp is contained inthe spectrum of b in B (if (b − λ1B)u = u(b − λ1B) = 1B , then u commutes with p and(pbp − λp)(pup) = (pup)(pbp − λp) = p). If the spectrum of bp contains elements λ ∈ Cwith Re λ ≤ 0, then the boundary of the spectrum will also contain such λ. Then thereexists by Remark ?? a norm one d = pcp ∈ Bp such that (bp − λp)d ∼ 0. But in Bp,

p = p2 =1

2πi

∫

γ

(zp− bp)−1 dz.

56

We have bpd ∼ λd, thus (zp− bp)d ∼ (z − λ)d and (zp− bp)−1d ∼ (z − λ)−1d, therefore

d = pd ∼ 12πi

(∫

γ

(z − λ)−1 dz)d = 0.

This contradicts ‖d‖ = 1.

Corollary. If b ∈ B is as in the Lemma, there exists an homotopy (bs) in GL(B) from bto p such that for every s, bs + it1B is invertible for every t ∈ R.Proof. For s real varying from 1 to +∞ we set bs = (1 + s)−1(b + sp).

Let us come back to the proof of the third step. We have a linear path u1(z) = a+ zbin GLK(A), with u1(1) = 1K,A, thus a+b = 1K,A. We see that 1K,A +(z−1)b is invertiblefor every z ∈ T; this yields that the spectrum of b in MK(A) does not meet the lineRe λ = 1/2. Applying the preceding Corollary with B = MK(A) to b− (1/2)1K,A, we getan homotopy (bs) from b to an idempotent p, such that bs + λ1K,A is invertible for everyλ ∈ C with Re λ = 1/2. It follows that for every s, the path 1K,A + (z − 1)bs, for z ∈ T, isa path of invertible. We have thus found a deformation in GLK((SA)+) from the originalpath 1K,A + (z− 1)b to a loop 1K,A + (z− 1)p associated to an idempotent p, as was to beproved.

This proof of the third step is nicer with the full strengh of the functional calculus.Let us find two circles γ1 and γ2 containing the spectrum of b, where γ1 is contained inRe ζ < 1/2 and γ2 in Re ζ > 1/2. We have

b =1

2iπ

∫

γ1

z

z − bdz +

12iπ

∫

γ2

z

z − bdz.

What we do is to find a deformation ϕ1(z, t), t ∈ [0, 1], of the identity function z → z toz → 0 at the left of the line Re ζ = 1/2 and a deformation ϕ2(z, t) from z → z to z → 1at the right of the same line in such a way that ϕ1(z, t) and ϕ2(z, t) never meet the lineRe ζ = 1/2. There is an easy choice: ϕ1(z, t) = (1− t)z if z ∈ γ1 and ϕ2(z, t) = (1− t)z + tfor z ∈ γ2. We obtain a continuous path

bt =1

2iπ

∫

γ1

ϕ1(z, t)z − b

dz +1

2iπ

∫

γ2

ϕ2(z, t)z − b

dz

of elements such that 1K,A +(z− 1)bt is invertible for every z ∈ T, with b0 = b and b1 = p.Finally, let us explain why j : K0(A) → K2(A) is injective; suppose that α = [p]− [q]

and j(α) = 0; this means that the two loops zp + (1 − p) and zq + (1 − q), z ∈ T, arehomotopic in GL∞((SA)+). We can therefore construct a continuous map ϕ(t, z), t ∈ [0, 1],such that ϕ(0, z) = zp + (1 − p) and ϕ9!, z) = zq + (1 − q); we may consider this as amap from T to the space of continuous maps from [0, 1] to GL(A). More precisely, thevalues at 0 and 1 belong respectively to the unital subalgebras Cp and Cq generated by pand q (thus Cp is the subalgebra of elements λ1A + µp, and similarly for Cq). Let B bethe algebra of continuous functions from [0, 1] to A such that f(0) ∈ Cp and f(1) ∈ Cq;assume for simplicity that p and q are different from 1A; the two preceding subalgebras

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are then isomorphic to C2. Since the map j is onto for the algebra B, we may deform ϕto a loop zP + (1− P ), where P is an idempotent in M∞(B), i.e. a continuous map from[0, 1] to M∞(A) such that zP (0)+(1−P (0)) is homotopic to zp+(1−p) and similarly fort = 1. Since the subalgebra is isomorphic to C2, this implies that the ranges of P (0) andp have equal dimensions (use determinant??), and are therefore equivalent as projectors;similarly for P (1) and q; finally, P (t) is a continuous path of idempotents from [p] to [q],and therefore α = [p]− [q] = 0.

Examples 9.5.1. K1(C(S1)) corresponds to K2(C). Indeed, K2(C) is K1((SC)+), and (SC)+ iden-

tifies with C(S1). It follows from Bott’s theorem that K1(C(S1)) = K2(C) ' K0(C) ' Z.We get a generator of K1(C(S1)) by considering a continuous map v(t) from T to GLn

such that the determinant of v(t) makes a loop around 0 in C with index 1.2. C(S2). The suspension SC identifies to the space of continuous functions on [0, 1],

vanishing at 0 and 1. Then SSC identifies to continuous functions on the square [0, 1]2,vanishing on the boundary. But this algebra can also be identified to continuous functionson S2, vanishing at a given point (here, the point obtained by identifying all points onthe boundary to a single point). Therefore, K0(C0(S2)) ' K2(C) ' K0(C) ' Z. This isa fundamental example, and it is possible to deduce the general case from it by a tensorproduct technique.

With this Bott’s isomorphism we get a new connecting map ∂0 from K0(A/I) 'K2(A/I) = K1(S(A/I)) to K1(I) = K0(SI), (observe that S(A/I) ' SA/SI) and a newexact sequence. Call now ∂1 the connecting map that was defined earlier. Suppose givena short exact sequence

0 → Ii−→A

π−→A/I → 0.

We obtain a cyclic exact sequence with period 6

K1(A/I) ∂1−→K0(I) i∗−→K0(A) π∗−→K0(A/I) ∂0−→K1(I) i∗−→K1(A) π∗−→K1(A/I) ∂1−→ . . . .

Some examples with the cyclic exact sequence

1. Let A = L(`p) and I = K(`p), A/I = Cp = C(`p). We know that K1(I) = 0,K0(I) ' Z, K1(A) = 0 (see Example 9.4, 4) et K0(A) = 0 because `p is prime. Weget

0 π∗−→K1(A/I) ∂1−→Z i∗−→ 0 π∗−→K0(A/I) ∂0−→ 0,

thus K1(Cp) ' Z, K0(Cp) = 0.When X ' `p(X), we have K1(L(X)) = 0 by Example 9.4, 4, and we know that

i∗(1) = 0 because X is isomorphic to its hyperplanes (Example 9.2, 6); this gives

K0(L(X)) ' K0(C(X)); K1(L(X)) ' Z.

2. Let now A = L(`p ⊕ `q). Assume 1 ≤ p < q < ∞. Then every operator from `q to`p is compact (Pitt’s Theorem, see [LT1], Proposition 2.c.3). This implies that the Calkin

58

algebra Cp,q = C(`p ⊕ `q) is triangular, hence K0(Cp,q) ' K0(Cp) × K0(Cq) = 0. Also,K1(Cp,q) ' K1(Cp)×K1(Cq) = Z× Z. We get using Z ' K0(Kp,q)

0 i∗−→K1(A) π∗−→K1(Cp,q)∂1−→Z i∗−→K0(A) π∗−→K0(Cp,q),

this gives0 i∗−→K1(A) π∗−→Z× Z ∂1−→Z i∗−→K0(A) π∗−→ 0.

The map ∂1 from Z × Z to Z is simply the sum ∂1(n, m) = n + m. Finally K1(A) isisomorphic to the kernel of this sum map, thus isomorphic to Z. Since this ∂1 is onto, weget that i∗ = 0, but i∗ is also onto and thus K0(A) = 0 (for this last statement onecan use directly Edelstein-Wojtaszczyk [EW] who say that every projection on `p ⊕ `q isequivalent to a direct sum of projections in `p and `q).

3. Cuntz algebras. We want to show that [1On ] 6= 0 for the Cuntz algebra On whenn ≥ 3. To this end Cuntz introduces the auxiliary algebra En generated by n into isometriesV1, . . . , Vn such that

∑ni=1 ViV

∗i < 1. The projection Q = 1 −∑

ViV∗i generates an ideal

I isomorphic to K, with Q playing the role of a rank one projection in K, and On ' En/Iby the uniqueness of On; we obtain

K1(On) ∂1−→K0(I) ' Z i∗−→K0(En) π∗−→K0(On) ∂0−→K1(I) = 0.

One can show that ∂1 = 0 (for example by the reasoning of Corollary 12.1). Since i∗(1) isthe class of rank one projections in K we have i∗(1) = [Q], and since ∂1 = 0 we know thati∗ is injective, thus [Q] generates a subgroup of K0(En) isomorphic to Z. If [1On ] = 0, wemust have [1En ] = m[Q] for some m ∈ Z by exactness. On the other hand (Example 9.3,10),

n[1En ] + [Q] = [1En ],

hence (m(n− 1) + 1)[Q] = 0, which is impossible since n− 1 ≥ 2 and since [Q] generatesa group isomorphic to Z. Cuntz proved in [C2] that K0(On) = Z/(n − 1)Z. This issignificantly more difficult than the above remark (see also Pimsner-Voiculescu).

10. Exotic Banach spaces

H.I. spaces

Definition 10.1. Let X be an infinite dimensional Banach space, real or complex. We saythat X is Indecomposable if X cannot be written as the topological direct sum of two infinitedimensional closed subspaces Y1 and Y2. We say that X is Hereditarily Indecomposable(in short, H.I.) if every closed infinite dimensional subspace Y of X is indecomposable,that is if no subspace Y of X can be written as the topological direct sum of two infinitedimensional closed subspaces Y1 and Y2 of X.

Obviously, if X is H.I. then every (infinite dimensional) subspace Y ⊂ X is H.I.Exercise.

1. A Banach space X is H.I. iff for all infinite dimensional subspaces Y and Z of X,we have

inf‖y − z‖ : y ∈ Y, z ∈ Z, ‖y‖ = ‖z‖ = 1 = 0.

59

In other words, X is H.I. when the “angle” between any two (infinite dimensional) sub-spaces of X is equal to 0.

2. X is H.I. iff for every (infinite dimensional) subspace Y ⊂ X, the quotient mapπY : X → X/Y is strictly singular.

Fact [GM1]. There exist H.I. Banach spaces.

Actually the first example in [GM1] of a H.I. space was also the first example of anindecomposable space. The existence of an indecomposable space answered a question ofLindenstrauss [L2]. This example in [GM1] is a special case of the construction presentedin section 11.

When X is a H.I. Banach space, then X contains no (infinite) Unconditional BasicSequence (UBS in short). This is clear, because a space with unconditional basis is easilydecomposable, for example into the two subspaces generated by basis vectors with oddindices and even indices. Tim Gowers and the author of these Notes were actually lookingfor an example of a space without UBS. The example turned out to have the strongerproperty of being H.I. (this was observed by W.B. Johnson). This was not totally accidental(although not deliberate), as Gowers’ dichotomy theorem will explain (see Theorem 10.2below).

Since every Banach space contains a subspace with basis, it is formal from the existenceof any H.I. space that there exist H.I. spaces with monotone basis; actually the examplein [GM1] is also reflexive; much more difficult is another example due to Gowers [G2] of aH.I. space without any reflexive subspace. Being H.I. this last example does not containc0 or `1, solving another longstanding conjecture in Banach space theory (see [LT, ???]).

Ferenczi [F1] has constructed an example of uniformly convex H.I. space. The proofadds to the ideas of the construction of [GM1] the notion of complex interpolation forfamilies of Banach spaces developed by Coifman, Cwikel, Rochberg, Sagher and Weiss in[CW]. Argyros-Delyanni [AD] constructed asymptotically `1 H.I. spaces, using a techniquecloser to the original Tsirelson example [T] instead of working with a modification ofSchlumprecht’s example as it is done in [GM1] or [F1].

Kalton [K] has constructed an example of a quasi-Banach space X with the verystrange property that there is a vector x 6= 0 such that every closed infinite dimensionalsubspace of X contains x. It follows that this quasi-Banach space does not contain anyinfinite basic sequence. This example is related to an example of Gowers [G1] of a spacewith unconditional basis not isomorphic to its hyperplanes; Kalton’s construction uses thetechnique of twisted sums together with the properties of the space in [G1].

“Germs” of H.I. spaces; in some sense all subspaces of a H.I. space intersect; we maydefine a net of subspaces that captures a good part of the structure of a H.I. space; theorder of this net is not the inclusion, as it is not true that any two infinite dimensionalsubspaces have an infinite dimensional intersection, but almost. . . . We say that Y ≤ Zif there exists a compact operator K : Y → X such that iY,X + K)(Y ) ⊂ Z. GivenZ1, Z2 ⊂ X there exists Y such that Y ≤ Z1 and Y ≤ Z2. We could call “germ” of H.I.space an equivalence class of such nets, in a way to be made precise. An interesting classof examples is the family of spaces containing no UBS but having only a finite set of germs

60

of H.I. spaces. The example of Gowers [G2] of a non reflexive H.I. space is mainly anexample of non reflexive germ. See also Remark 10.1 below.

Spectral theory and consequences

Theorem 10.1. Let X be a complex H.I. space. Then every T ∈ L(X) can be writtenas T = λIX + S, where λ ∈ C and S is strictly singular. Thus every T ∈ L(X) is eitherstrictly singular or Fredholm with index 0. Furthermore, the spectrum of T is either finite,or consists of a sequence (λn) converging to λ. In this second case, each λn 6= λ is aneigenvalue of T with finite multiplicity.

Proof. Let T ∈ L(X). We know by Lemma 5.4 or Corollary 5.2 that there exists λ ∈ Csuch that T −λIX is infinitely singular. Let U = T −λIX . For every ε > 0 there exists byProposition 3.2 an infinite dimensional subspace Yε ⊂ X such that ‖U|Yε

‖ < ε. Now letZ be any infinite dimensional subspace of X. Since X is H.I., we can find a vector z ∈ Zsuch that ‖z‖ = 1 and dist(z, SYε) < ε. It follows that ‖Uz‖ < (1 + ‖U‖)ε, showing thatU = T − λIX is strictly singular. The rest is given by Proposition 6.1.

Remark 10.1. Ferenczi [F2] has shown that, given a complex H.I. space X and a boundedlinear operator T from a subspace Y of X to X, one can write T = λiY,X +S, where λ ∈ C,S is strictly singular and iY,X denotes the inclusion map from Y to X. This property wasshown to be true for the specific H.I. example in [GM1]. Conversely, it is easy to see thatany Banach space X with the property that for every subspace Y , every T ∈ L(Y, X)can be written as λiY,X + S is a H.I. space, so that the above result is a characterizationof complex H.I. spaces. Ferenczi’s proof consists essentially to show that the space of“germs” of operators on X is a Banach field, hence isomorphic to C. In the case of a realH.I. Banach space X, his proof shows that the quotient L(X)/S(X) is a division ring,hence isomorphic to R,C or H. A germ of operator is an equivalence class for the relationwhere T1 ∈ L(Z1, X) and T2 ∈ L(Z2, X) are equivalent if there exists Y ≤ Z1, Z2 suchthat T1 (iY,X + K1) − T2 (iY,X + K2) is compact on Y , and K1,K2 are the compactoperators from the definition of the order.

Exercise 10.1.1. Operators on real H.I. spaces. Let X be a real H.I. space and let T ∈ L(X). Either

there exists λ ∈ R such that T − λIX is strictly singular, or there exists λ ∈ C \ R suchthat T 2 − 2Re λ T + λλIX is strictly singular. Check that T is either strictly singular orFredholm with index 0. The spectrum of the complexified operator TC is invariant undercomplex conjugation, and the part of the spectrum contained in the upper half plane isfinite or consists of a convergent sequence with its limit.

2. Operators on Xn. If X is a complex H.I. space and if T ∈ L(Xn), there exists amatrix Λ ∈ Mn(C) such that T = Λ⊗ IX + S, with S strictly singular on Xn.

3. If X is H.I. then Xn 6' Xm when m 6= n.

4. If X is a complex H.I. space, then K1(L(X)) = 0. Also, K0(L(X)) 6= 0.The hyperplane problem

Corollary 10.1. Let X be a H.I. space, real or complex. Then X is not isomorphic toany proper subspace. In particular, X is not isomorphic to its hyperplanes.

61

The first example of a space not isomorphic to its hyperplanes appeared in [G1].

Proof. Let T be an isomorphism from X into itself; then T is not strictly singular, henceit must be Fredholm with index 0 by Theorem 10.1 or Exercise 10.1 and thus TX = X.

Exercise.

1. If X is H.I. and Z ⊂ Y ⊂ X, Z 6= Y , then Z and Y are not isomorphic.2. Show that an H.I. space X is not isomorphic to any quotient X/Y .

Hint. Use the fact that the spectrum of every operator on X is countable. Let T : X/Y →X be an isomorphism, consider T πY , then its adjoint and the spectrum of the adjoint.

3. Homotopy of subspaces. Two subspaces Y and Z of a complex H.I. space X areisomorphic if and only if there exists an homotopy in L(Y,X) from iY,X to T consistingof into isomorphisms from Y to X, where iY,X denotes the injection from Y to X and Tis an into isomorphism from Y to X such that TY = Z.

Ferenczi showed that the dual of the example in [GM1] is also H.I. and even that everyquotient of this space is still H.I. This question is not at all clarified in general. What isclear is that the dual of a reflexive indecomposable space (not necessarily hereditarilyindecomposable) is indecomposable; therefore, if every quotient of a reflexive space is H.I.,then every subspace of a quotient is indecomposable and this property clearly passes tothe dual. However Ferenczi gave an example of a H.I. space such that the dual is not H.I.

Gowers’ dichotomy theorem and homogeneous Banach spaces

Recall that a Banach space X is said homogeneous if X is isomorphic to all its infinitedimensional closed subspaces. What can we say about a homogeneous Banach space? Sinceevery Banach space contains a subspace with a basis, X must have a basis. Furthermore,every subspace will also have a basis. It follows from the work of Enflo on the approximationproperty, extended by Szankowski, that every Banach space with this property must havetype 2 − ε and cotype 2 + ε for every ε > 0 ([LT2], Theorem 1.g.6). These results havebeen obtained in the ’70s; more recently, Komorowski and Tomczak proved in [KT] thefollowing result:

Theorem. Let X be a Banach space with finite cotype and not containing any subspaceisomorphic to `2. Then there exists a subspace Y of X without unconditional basis.

The proof of [KT] is rather difficult and complicated, and uses techniques differentfrom those of these Notes.

Corollary. [KT] If X is a homogeneous Banach space not isomorphic to `2, then X doesnot contain any UBS.

Proof. We said that X must have finite cotype if it is homogeneous. If X is not isomorphicto `2, we know by the Theorem above that X contains a subspace Y without unconditionalbasis. Since X is homogeneous, it follows that X has no subspace with unconditional basis.

It was then very tempting to try to relate the fact that a space does not containany UBS to the H.I. property. This was done by T. Gowers in a beautiful “dichotomyTheorem”; Gowers obtains more general combinatorial statements (in [G4] and [G5]) thatwe shall not give here, and which are somewhat analogous to the infinite versions of

62

Ramsey’s Theorem; he then deduce the result about UBS and H.I. from them. We shallonly prove the particular case which is needed here.

Theorem 10.2. (Gowers’ dichotomy theorem, [G4], [G5]). Let X be an arbitrary (infinitedimensional) Banach space. Either X contains an infinite unconditional basic sequence,or X contains a H.I. subspace.

This result gives a very good reason for introducing this notion of H.I. spaces. If one isinterested to know whether a general Banach space contains a subspace with unconditionalbasis, one has to encounter H.I. spaces some day.

Proof of Theorem 10.2. For the proof we need a more quantitative result; let ε > 0; weshall say that a Banach space X is HI(ε) when for all subspaces Y and Z of X, there existtwo vectors y ∈ Y , z ∈ Z such that

‖y − z‖ < ε‖y + z‖.

Setting λ = ‖(y + z)/2‖ > 0 and y′ = y/λ, z′ = z/λ we see that 1− ε < ‖y′‖, ‖z′‖ < 1 + εand ‖y′ − z′‖ < 2ε. It is thus clear that X is H.I. if and only if it is HI(ε) for every ε > 0.

Lemma 10.1. Let X be a Banach space. For every ε > 0, either X contains an UBS withconstant 2/ε, or X contains an HI(2ε) subspace Z.

Proof (from [M2]). The approach is combinatorial. We shall need to discretize the problemto make the situation countable, and even finite later on. We shall restrict now to the realcase. Let us choose in X a normalized basic sequence (xn)n≥1 with constant 2 (say)and denote by X0 the Q-vector subspace generated by this sequence. This space X0 iscountable and infinite dimensional (over Q); furthermore, for every infinite dimensionalQ-vector subspace Y of X0, the closure Y in X is an infinite dimensional Banach spaceover R or C.

From now on, in the proof of the Lemma, the notation Y , Z, or U , V, W will be usedfor infinite dimensional Q-vector subspaces of X0. Let us consider the set

A = (x, y) ∈ X0 ×X0; ‖x− y‖ < ε‖x + y‖.

This set A is countable and symmetric. We introduce a convenient terminology, inspired by[GP]. Let (x, y) be a couple of vectors in X0 and let Z be an infinite dimensional subspaceof X0. We say that (x, y) accepts Z if for all subspaces U, V of Z there exists (u, v) ∈ U×Vsuch that (x+u, y +v) ∈ A. Since A is symmetric, acceptation is also symmetric. Observealso that if (x, y) ∈ A, then (x, y) accepts every subspace Z: just take u = 0 and v = 0.

We say that (x, y) rejects Z if no subspace Z ′ ⊂ Z is accepted by (x, y). Rejectionis also symmetric, and saying that (x, y) rejects some subspace Z implies that (x, y) /∈ A.Observe that when (x, y) accepts or rejects a subspace Z, it remains true for every subspaceZ ′ of Z, and it is also true for “supspaces” of Z of the form Z + F , when F is finitedimensional; combining these two observations, we see that when (x, y) accepts or rejectsZ, the same is true for every Z ′ such that Z ′ ⊂ Z + F , when F is finite dimensional. Thissimple remark is the basis for our first step:

63

Claim: there exists an infinite dimensional subspace Z0 of X0 such that for every couple(x, y) ∈ X0 ×X0, either (x, y) accepts Z0 or (x, y) rejects Z0.

We use for this a very usual diagonal argument; since X0 is countable, we may form thelist (xn, yn)n≥1 of all elements of X0×X0. We then construct a decreasing sequence (Xn)n

of subspaces in the following way: if (xn+1, yn+1) rejects Xn, we simply let Xn+1 = Xn.Otherwise, there exists a subspace Xn+1 of Xn such that (xn+1, yn+1) accepts Xn+1. Weconsider then a diagonal subspace Z0 built by taking one vector in each Xn, in such a waythat Z0 is infinite dimensional.

From now on the whole construction will be performed inside our “stabilizing” sub-space Z0. For every couple (x, y) in Z0 × Z0, (x, y) accepts ou rejects, and we don’t needanymore to specify “accepts or rejects a subspace Z ′ ⊂ Z0”.

There are two possibilities: either the couple (0, 0) accepts, or it rejects. If (0, 0)accepts, we see that the Banach space Z = Z0 is HI(ε + ε′) for every ε′ > 0, in particularHI(2ε). Indeed, if U and V are two subspaces of Z, we may approximate them by twoQ-subspaces U ′ and V ′ of Z0. Since (0, 0) accepts, there exist u′ and v′ in U ′ and V ′ suchthat (u′, v′) ∈ A which gives by approximation two vectors u ∈ U and v ∈ V such that‖u− v‖ < (ε + ε′)‖u + v‖.

Suppose now that (0, 0) rejects; we will find in Z0 an unconditional sequence (ek)k≥1

with constant 2/ε, namely such that

‖∑

k

bkek‖ ≤ 2ε‖

∑

k

ηkbkek‖

for all scalars (bk)k and every choice of signs (ηk)k (signs appear usually on the other sideof the inequality, but it is clearly equivalent to put them on the right). In order to dealwith this in a combinatorial manner, we discretize our problem as follows: it is easy to seethat we only need to make sure that 1 < ‖ek‖ < 2 for every integer k ≥ 1 and that

(∗) ‖K∑

k=1

akek‖ ≤ 1ε‖

K∑

k=1

ηkakek‖

for every integer K ≥ 1, all choices of signs (ηk)Kk=1 and all scalars (ak)K

k=1 taking thevalues ak = j/(N2k), j = −N2k, . . . , N2k, where N is an integer larger than 16/ε. Wecall reasonable such a choice of scalars (ak)K

k=1, and reasonable combination (of length K)a linear combination of the form

∑Kk=1 akek. Relation (∗) means that ‖x− y‖ ≥ ε‖x + y‖

whenever x =∑

k∈I akek and y =∑

k∈J akek, where the coefficients are reasonable and(I, J) is the partition of 1, . . . , K corresponding to the signs (ηk)K

k=1. We call such acouple (x, y) a partition of a reasonable combination. In other words, we want to makesure that (x, y) /∈ A whenever (x, y) is a partition of a reasonable combination

∑Kk=1 akek

with arbitrary length K. As we observed, it is enough to know that every partition of areasonable combination rejects.

64

Sublemma: if (x, y) rejects, then for every infinite dimensional subspace W of Z0 thereexists a further subspace W ′ ⊂ W such that for every w′ ∈ W ′, the couple (x + w′, y)rejects.

(Otherwise, for every subspace U ⊂ W , there would exist u0 ∈ U such that (x + u0, y)accepts; then, for every subspace V ⊂ W we could find a couple (u1, v) in U × V suchthat (x + u0 + u1, y + v) ∈ A, which implies that (x, y) accepts, contrary to the initialhypothesis.)

Let us finish the proof of Lemma 10.1. Assuming that (0, 0) rejects, we build byinduction a sequence (ek)∞k=1, such that for every integer n ≥ 1, every partition (x, y)of a reasonable combination with length n rejects. If e1, . . . , en are already constructed,consider the finite list of all partitions (xi, yi) of reasonable combinations of length n.By our induction hypothesis every such couple (xi, yi) rejects; applying successively thesublemma to each (xi, yi) from the list, we obtain a subspace W such that for everyw ∈ W and every i, (xi + w, yi) rejects; observe that (yi, xi) also belongs to the list, hence(yi + w, xi) rejects, and since A is symmetric (xi, yi + w) also rejects. We choose now avector en+1 in W , such that 1 < ‖en+1‖ < 2. We check that the induction hypothesis isverified for n + 1. Indeed, every partition (x′, y′) of a reasonable combination of length(n + 1) is either of the form (x + aen+1, y) or of the form (x, y + aen+1), where (x, y) isa partition of a reasonable combination of length n. It follows from the choice of W anden+1 that (x′, y′) rejects.

We can now finish the proof of Theorem 10.2. Assume that X does not contain anyUBS. Let Y be a Banach subspace of X. Since Y does not contain an UBS, it follows fromLemma 10.1 that for every ε > 0, Y contains a subspace Z which is HI(ε).

Taking successively ε = 2−n, we can construct a decreasing sequence (Zn) of subspacescorresponding to εn = 2−n. Let Z be a subspace obtained by a diagonal procedure fromthe sequence (Zn), which means that for every n this space Z is contained in Zn up tofinitely many dimensions. Let ε > 0, and let U and V be infinite dimensional subspacesof Z. Let n be such that 2−n < ε. We can find infinite dimensional subspaces U ′ of Uand V ′ of V such that U ′ ⊂ Zn, V ′ ⊂ Zn. By the construction of Zn there exists a couple(u, v) such that u ∈ U ′, v ∈ V ′ and ‖u − v‖ < ε‖u + v‖. Therefore Z is HI(ε) for everyε > 0, so Z is H.I.

Exercise. Finite field.

Theorem. Every homogeneous Banach space is isomorphic to `2.

Proof. Let X be a homogeneous Banach space, not isomorphic to `2. We know that Xdoes not contain any UBS by the Corollary of [KT]. By Gowers’ result, X must contain aH.I. subspace, hence X itself is H.I. But an H.I. space is obviously not homogeneous, andin a very strong way, since we have seen that it is not isomorphic to any proper subspaceby Corollary 10.1.

65

11. A class of examples of exotic spaces

The contents of this section come from [GM2].Let c00 be the vector space of all scalar sequences of finite support. Let (en)∞n=1 be

the standard basis of c00. Given a vector a =∑∞

n=1 anen its support, denoted supp(a),is the set of n such that an 6= 0. Given two subsets E, F ⊂ N, we say that E < F ifmax E < min F . If x, y ∈ c00, we say that x < y if supp(x) < supp(y). We also writen < x when n ∈ N and n < min supp(x). If x1 < . . . < xn, then we say that the vectorsx1, . . . , xn are successive. An infinite sequence of successive non-zero vectors is also calleda block basis and a subspace generated by a block basis is a block subspace. Given a subsetE ⊂ N and a vector a as above, we write Ea for the vector

∑n∈E anen. An interval of

integers is a set of the form n, n + 1, . . . , m and the range of a vector x, written ran(x),is the smallest interval containing supp(x).

Let X stand for the set of Banach spaces obtained as the completion of c00 for a norm‖.‖ such that the sequence (en)∞n=1 is a normalized bimonotone basis. A first extremelyimportant example in this class is the space T constructed by Tsirelson [T] (see also [FJ]).Let B∗

T be the smallest convex subset of B(c0) ∩ c00 containing ±en for each n ≥ 1 andsuch that

(x∗1 + · · ·+ x∗n) ∈ 2B∗T

whenever x∗1, . . . , x∗n are successive in B∗

T and n < x∗1. The norm is then defined on c00 by

‖x‖T = sup|x∗(x)| : x∗ ∈ B∗T .

A second extremely important example is the space S constructed by Schlumprecht[S1], [S2], which is a very useful variation of the construction of T . Let f(t) = log2(t + 1)for t ≥ 0. The relevant properties of this function will be listed below. Let B∗

S be thesmallest convex subset of B(c0) ∩ c00 containing ±en for each n and such that

(x∗1 + · · ·+ x∗n)/f(n) ∈ B∗S

whenever x∗1, . . . , x∗n are successive in B∗

S and n ≥ 2. The norm is then defined on c00 by

‖x‖S = sup|x∗(x)| : x∗ ∈ B∗S.

The basic idea for the construction of our class of examples uses the technology oflower f -estimate introduced by Schlumprecht in [S1], [S2]. Given X ∈ X , we shall say thatX satisfies a lower f -estimate if, given any vector x ∈ X and any sequence of intervalsE1 < . . . < En, we have ‖x‖ ≥ f(n)−1

∑ni=1 ‖Eix‖. In the dual formulation, this property

means that whenever x∗1, . . . , x∗n are successive functionals with norm ≤ 1 in X∗, we have

‖(x∗1 + · · ·+ x∗n)/f(n)‖X∗ ≤ 1.

The norm of S appears then as the smallest norm for a space in X satisfying a lower-festimate.

66

Schlumprecht introduced the important notion of Rapidly Increasing Sequences, inshort RIS. Let us mention first that every subspace Y of S generated by a block basiscontains for every n ≥ 1 a sequence y1 < . . . < yn of normalized vectors which is almostisometrically equivalent to the unit vector basis of `n

1 (see Lemma 11.1 below). Roughlyspeaking, a RIS is a normalized sequence x1 < . . . < xk where each xi is the average of an`ni1 sequence, with n1 < n2 < . . . < nk growing extremely rapidly (a precise definition will

be given later). Schlumprecht proved that the norm of the sum of RIS sequences has analmost minimal behaviour; indeed, Schlumprecht’s space satisfies a lower f -estimate, hence‖∑n

i=1 xi‖ ≥ n/f(n) for every sequence of successive norm one vectors. Schlumprecht’sLemma states that for an RIS, we almost get an equality, ‖∑n

i=1 xi‖ ≤ (1+ ε)n/f(n). Weobtain in this way one of the most important features of Schlumprecht’s example: on onehand, we can find `n

1 in every subspace; on the other hand, we can always combine verydifferent `ni

1 in a RIS and get a behaviour arbitrarily far from the `1 behaviour. But thesenew vectors can again be combined to give further `n

1 , and so on. . .

For our construction, we need to work with more than one function f . To this end weintroduce the family F of functions g : [1,∞) → [1,∞) satisfying the following conditions:

(i) g(1) = 1 and g(t) < t for every t > 1;(ii) g is strictly increasing and tends to infinity;(iii) limt→∞ t−qg(t) = 0 for every q > 0;(iv) the function t/g(t) is concave and non-decreasing;(v) g(st) ≤ g(s)g(t) for every s, t ≥ 1;It is easy to check that f(t) = log2(t+1) satisfies these conditions, as does the function√

f(t). Note also that some of the conditions above are redundant. In particular, it followsfrom the other conditions that g(x) and x/g(x) are strictly increasing.

Let X ∈ X and y ∈ X. For every n ≥ 1, let

‖y‖(n) = supn∑

i=1

‖Eiy‖

where the supremum is extended to all families E1 < . . . < En of successive intervals. Thisquantity is clearly increasing with n, and ‖y‖ = ‖y‖(1) since (en) is a bimonotone basis forthe space X. Observe that the basis (ei) satisfies ‖ei‖(n) = 1 for every n ≥ 1.Lemma 11.1. Let X ∈ X satisfy a lower f -estimate. Given n ≥ 1 and ε > 0, there existsan integer N(n, ε) such that for every sequence x1, . . . , xN of successive norm one vectorswith N ≥ N(n, ε), we may find x of the form x = λ

∑i∈A xi, where A is some subinterval

of 1, . . . , N such that ‖x‖ = 1 and ‖x‖(n) ≤ 1 + ε.

The proof of this Lemma uses a variant of a well known blocking procedure for con-structing `n

1 , originating in James [J3]; let us also mention Giesy [Gi], Pisier [P1]; andmuch more elaborated results of Elton (E], (Pajor [Pa] in the complex case).

Corollary 11.1. Let X ∈ X , satisfying a lower f -estimate. Then for every n ∈ N andε > 0, every subspace Y of X contains a vector y such that ‖y‖ = 1 and ‖y‖(n) ≤ 1 + ε.

Proof. By the standard gliding hump procedure, we may find for every N a normalizedsequence y1, . . . , yN of vectors in Y and successive vectors x1 < . . . < xN in X such

67

that ‖yi − xi‖ < ε/nN . The result follows from Lemma 11.1 and an easy approximationargument.

Given a subspace Y ⊂ X, we will be interested in a seminorm |||.||| defined on L(Y, X)as follows. We say that a sequence (xn) is a sequence of almost successive vectors if thereexists a sequence (yn) of successive vectors such that limn ‖xn − yn‖ = 0. Let MY be theset of sequences (xn)∞n=1 of almost successive vectors in Y such that lim supn ‖xn‖(n) ≤ 1.Now, given T ∈ L(Y, X) let

|||T ||| = supx∈M(Y )

lim supn‖Txn‖ .

Suppose that X is reflexive and that (xn) is a weakly null sequence in a subspace Y suchthat lim supn ‖xn‖(m) ≤ 1 for every integer m. Let T ∈ L(Y, X) and t = lim sup ‖Txn‖.We can find a subsequence (x′n) such that t = limn ‖Tx′n‖. Since (x′n) is weakly null, wecan extract a further subsequence (x′′n) which is almost successive, and we may also arrangethat ‖x′′n‖(n) ≤ 1+2−n. Now (x′′n) belongs to MY , therefore t = lim ‖Tx′′n‖ ≤ |||T |||. Let ussay the same thing in a slightly different way: let Pm denote the projection on the interval1, . . . , m; for every T ∈ L(Y, X) and for every ε > 0, there exists integers m,n ≥ 1 suchthat, for every y ∈ Y , the condition ‖Pny‖ ≤ 1/n implies that ‖Ty‖ ≤ (|||T |||+ ε)‖y‖(m).

Lemma 11.2. Suppose that X ∈ X satisfies a lower f -estimate; then for every subspaceY of X and T ∈ L(Y, X)

|||T ||| = 0 ⇒ T is strictly singular;if X is reflexive and T compact, then |||T ||| = 0;if for every z in some infinite dimensional subspace Z of Y , we have ‖Tz‖ ≥ ‖z‖, then

|||T ||| ≥ 1.

Proof. By the preceding Lemma, every subspace Y contains for every ε > 0 normalizedsequences in MY . Hence every (infinite dimensional) subspace of Y contains a norm onevector x such that ‖Tx‖ ≤ (1 + ε) |||T |||. As a consequence, for every U ∈ L(Y, X), we seethat s(U) ≤ |||U ||| (where s(U) was defined in section 6); in particular, if |||T ||| = 0, then Tis strictly singular. Suppose that X is reflexive; then every normalized sequence of almostsuccessive vectors is weakly null, therefore lim ‖Txn‖ = 0 if T is compact, hence |||T ||| = 0.Lastly, suppose that ‖Tz‖ ≥ ‖z‖ for every z in some infinite dimensional subspace Z ofY . We know from Corollary 11.1 that Z contains a normalized sequence (zn) of almostsuccessive vectors with lim ‖zn‖(n) = 1. By definition,

|||T ||| ≥ limn‖Tzn‖ ≥ 1.

Remark. if U ∈ L(X), then r(U) ≤ s(U) ≤ |||U |||. If T − λIX is infinitely singular, we seethat |λ| ≤ |||T |||.

The second ingredient inspired by Schlumprecht is that of Rapidly Increasing Se-quences, in short RIS. Let X ∈ X . For 0 < ε ≤ 1, we say that a sequence x1, . . . , xN ofsuccessive vectors in X satisfies the RIS(ε) condition if there is a sequence

2N2/ε2< n1 < · · · < nN

68

of integers such that ‖xi‖(ni) ≤ 1 for each i = 1, . . . , N and

ε√

f(ni) > | ran(i−1∑

j=1

xj)|

for every i = 2 . . . , N .Given g ∈ F , M ∈ N and X ∈ X , an (M, g)-form on X is defined to be a functional

x∗ of norm at most one which can be written as∑M

j=1 x∗j for a sequence x∗1 < . . . < x∗Mof successive functionals all of which have norm at most g(M)−1. Observe that if x∗ is an(M, g)-form then ‖x∗‖∞ ≤ 1/g(M) and |x∗(x)| ≤ g(M)−1‖x‖(M) for any x.Lemma 11.3. Let X ∈ X . Suppose that (x1, . . . , xN ) satisfies RIS(ε) in X. If g ∈ F ,√

f ≤ g and if x∗ is a (k, g)-form on X, we have

|x∗(x1 + · · ·+ xN )| ≤ maxj=1,...,N

‖xj‖+ ε +N

g(k).

In particular, |x∗(x1 + · · ·+ xN )| ≤ max ‖xj‖+ 2ε when k ≥ 2N2/ε2.

Proof. Let n1 < n2 < . . . < nN be the sequence of integers associated to the RIS property.Let i ∈ 1, . . . , N be such that ni < k ≤ ni+1. Observe that the RIS condition implies‖xj‖∞ ≤ 1 for each j = 1, . . . , N . The result follows from three easy inequalities,

∣∣x∗(i−1∑

j=1

xj)∣∣ ≤ ‖x∗‖∞

∣∣ran(i−1∑

j=1

xj)∣∣ ≤ 1

g(k)ε√

f(ni) ≤ ε,

|x∗(xi)| ≤ ‖xi‖ ≤ maxj=1,...,N

‖xj‖,

and for j ≥ ni+1,

|x∗(xj)| ≤ 1g(k)

‖xj‖(k) ≤1

g(k)‖xj‖(nj) ≤

1g(k)

.

When k ≥ 2N2/ε2, we get g(k) ≥

√f(k) ≥ N/ε.

The next Lemma is a variation of a main Lemma due to Schlumprecht. We alreadymentioned that in the case of the space constructed by Schlumprecht, this Lemma says thatthe norm of the sum of RIS sequences has an almost minimal behaviour. Our situationis technically more complicated; the space we want to construct will satisfy a lower f -estimate, but in some parts of our space the behaviour of RIS will be larger than n/f(n),namely it could sometimes be as big as n/

√f(n). We need a more general statement that

allows to play between the two possibilities. To this end we introduced the family F offunctions. The next Lemma is similar to Lemma 3 from [GM2] or Lemma 7 from [GM1].

Lemma 11.4. Let X ∈ X , g ∈ F ,√

f ≤ g, and let x1 < . . . < xn in X satisfy ‖xi‖(pn) ≤ 1for every i = 1, . . . , n and some integer p ≥ 2. Let x =

∑ni=1 xi and suppose that

‖Ex‖ ≤ 1 ∨ sup|x∗(Ex)| : x∗ is a (k, g)-form, 2 ≤ k ≤ p

69

for every interval E. Then‖x‖ ≤ ng(n)−1.

Proof. Let G(t) = t/g(t) when t ≥ 1 and G(t) = t when 0 ≤ t ≤ 1. This function G isconcave and increasing on [0,+∞). For every interval E and every integer l ≥ 0, let

σl(E) =n∑

i=1

‖Exi‖(pl).

We shall prove by induction on l, 1 ≤ l ≤ n that

(∗) ‖Ex‖ ≤ G(σκ(E)(E)),

where κ(E) is the number of i ∈ 1, . . . , n such that Exi 6= 0 (if κ(E) = 0, then Ex = 0and this case is obvious). Once this is done, we obtain the result for l = n, E = ran(x),

‖x‖ ≤ G(σn(ran(x))) = G( n∑

i=1

‖xi‖(pn)

)≤ G(n) =

n

g(n).

Observe first that when ‖Ex‖ ≤ 1, we have ‖Ex‖ = G(‖Ex‖) ≤ G(∑n

i=1 ‖Exi‖) ≤G(σl(E)). This shows that (∗) is true when κ(E) = 1. Assume (∗) true when κ(E) ≤ l < n,and suppose there exists an interval E such that κ(E) = l + 1 and ‖Ex‖ > G(σl+1(E));since (∗) is not true for E we know that l ≥ 1 and ‖Ex‖ > 1. From our assumption thereexists a (k, g)-form x∗ = (

∑kj=1 Ajx

∗j )/g(k), 2 ≤ k ≤ p, ‖x∗j‖ ≤ 1 and A1 < . . . < Ak, such

thatG(σl+1(E)) < |x∗(Ex)|.

Assume first that κ(AjE) ≤ l for every j = 1, . . . , k. We have ‖AjEx‖ ≤ G(σl(AjE)) bythe induction hypothesis, and using the concavity of G we obtain

|x∗(x)| ≤ k

g(k)

k∑

j=1

1k

G(σl(AjE)) ≤ k

g(k)G

(1k

k∑

j=1

σl(AjE))

=k

g(k)G

(1k

n∑

i=1

k∑

j=1

‖AjEx‖(pl)

)≤ k

g(k)G

(1k

n∑

i=1

‖Exi‖(pl+1)

)=

k

g(k)G

(σl+1(E)k

).

If σl+1(E) ≤ k, this last expression is σl+1(E)/g(k) ≤ σl+1(E)/g(σl+1(E)) = G(σl+1(E)),otherwise it is equal to

σl+1(E)g(k)g(σl+1(E)/k)

≤ σl+1(E)g(σl+1(E))

= G(σl+1(E)),

so that we have reached a contradiction.

70

In the remaining case there exists j0 ∈ 1, . . . , k such that Aj0Exi 6= 0 for everyi such that Exi 6= 0. Assume for example j0 < k (otherwise 1 < j0 deserves a similartreatment). Let m be the last integer i such that Exi 6= 0. Let Bj0 = Aj0 \ ran(Exm),B′

j0+1 = Aj0 ∩ ran(Exm), B′′j0+1 = Aj0+1, Bj0+1 = B′

j0+1 ∪B′′j0+1 and Bj = Aj otherwise.

We see that

‖Aj0Ex‖+ ‖Aj0+1Ex‖ ≤ ‖Bj0Ex‖+ ‖B′j0+1Exm‖+ ‖B′′

j0+1Exm‖ ≤

≤ ‖Bj0Ex‖+ ‖Bj0+1Exm‖(2).

Every Bj satisfies κ(BjE) ≤ l, so that the induction hypothesis applies and since pl ≥ 2we obtain

k∑

j=1

‖AjEx‖ ≤∑

j 6=j0

‖BjEx‖+ ‖Bj0+1Exm‖(2) ≤∑

j 6=j0

G(σl(BjE)) + G(σl(Bj0+1E)),

and the conclusion follows as before.

The next simple Lemma is useful in conjunction with the preceding.

Lemma 11.5. Let X ∈ X , and let x1 < . . . < xl in X be such that

‖xi‖ ≤ 1; ‖∑

i∈A

xi‖ ≤ |A|f(|A|)

for every interval A ⊂ 1, . . . , l such that m ≤ |A| ≤ l. Then for every integer n ≥ 1

f(l)l

∥∥l∑

i=1

xi

∥∥(n)

≤ f(l)f(m)

+2nmf(l)

l.

Proof. Let x =∑l

i=1 xi and let (Ej) be a sequence of n successive intervals. By adding atmost n cuts, we may assume that we have a family (Ej) of at most 2n intervals and thatfor every j = 1, . . . , 2n there exists an interval Aj ⊂ 1, . . . , l such that Ejx =

∑i∈Aj

xi.Let

J = j : |Aj | ≥ m.We get

2n∑

j=1

‖Ejx‖ =∑

j /∈J

‖Ejx‖+∑

j∈J

‖Ejx‖ ≤ 2nm +∑

j∈J

|Aj |f(|Aj |) ,

and the result follows.

71

Proper semi-groups of spreads.

Given two infinite subsets A = a1, a2, . . . and B = b1, b2, . . . of N, define the spreadfrom A to B to be the map SA,B on c00 that sends en to zero if n /∈ A, and sends eak

toebk

for every k ∈ N. SA,A is just the projection on to A. Note that SB,CSA,B = SA,C .Note also that SB,A is (formally) the adjoint of SA,B . Given any set S of spreads, we shallsay that it is a proper set if it is closed under composition (note that this applies to allcompositions and not just those of the form SB,CSA,B) and taking adjoints; we also makein [GM2] a technical assumption which means roughly that our semi-group is rather small:For every (i, j) 6= (k, l), there are only finitely many spreads U ∈ S for which e∗i (Uej) 6= 0and e∗k(Uel) 6= 0. Note that a proper semi-group is countable.

A good example of such a set is the collection of all spreads SA,B where A = m,m+1,m + 2, . . . and B = n, n + 1, n + 2, . . . for some m,n ∈ N. This is the proper setgenerated by the shift operator. The next theorem is the main result of [GM2].

Theorem 11.1. Given a proper set S of spreads, there exists a Banach space X = X(S) ∈X such that:

1. The space X is reflexive, satisfies a lower-f estimate (and the natural basis (en) ofc00 is a bimonotone basis for X by definition of the class X ).

2. ‖U‖ ≤ 1 for every U ∈ S. It follows that ‖SA,Bx‖ = ‖x‖ when SA,B ∈ S andsupp x ⊂ A.

3. |||TU ||| ≤ |||T ||| |||U ||| for all T,U ∈ L(X).

4. Let A be the algebra generated by S. For every subspace Y of X, every ε > 0 andevery T ∈ L(Y, X), there exists U ∈ A such that |||T − U iY,X ||| < ε, where iY,X denotesthe injection from Y into X.

Recall some facts from the general discussion: property 1 implies that every subspaceY ⊂ X contains normalized sequences in M. Since X is reflexive with a basis, normalizedsequences of (almost) successive elements are weakly null, in particular sequences in M areweakly null. It follows that |||K||| = 0 when K is compact from Y ⊂ X to X. If T ∈ L(Y, X)and |||T ||| = 0, then T is strictly singular. We have seen that r(U) ≤ s(U) ≤ |||U |||;Corollary 11.2. All spaces X = X(S) from Theorem 11.1 have the property that theydo not contain any UBS.

Proof. This is because property 4 immediately implies that for every subspace Y of X,L(Y ) is separable for |||.|||; on the other hand, if X contains an infinite dimensional subspaceY with unconditional basis (yn), then we can find an uncountable set P of projections inL(Y ) such that |||P −Q||| ≥ 1 when P 6= Q in P; for every infinite set L ⊂ N, let YL denotethe span [yn]n∈L of the corresponding subsequence, and let PL denote the correspondingprojection from Y onto YL. If L and M are two subsets of N with infinite difference D,we obtain that ‖(PL − PM )z‖ ≥ ‖z‖ for every z in YD, hence |||PL − PM ||| ≥ 1 by Lemma11.2. Finally, it is well known that we may find uncountably many infinite subsets of Nwith pairwise infinite differences.

These examples are not necessarily H.I. spaces however. They give a good illustrationfor the dichotomy result of Tim Gowers (Theorem 10.2).

72

It follows from Gowers’ dichotomy theorem that every Banach space X not containingan UBS has the property that every subspace Y of X contains an H.I. subspace. By theCorollary above this is the case for all the examples given by Theorem 11.1. It is possibleactually to see this directly by a reasoning close to the proof of the Corollary. Supposethat X = X(S) for some proper set of spreads, and that Y ⊂ X contains no H.I. subspace.We can build by a standard ordinal construction a family (Yα)α<ω1 of subspaces of Y inthe following way: since Yα is not H.I. we can find a direct sum Uα ⊕ Vα in Yα. Let Tα

be Id on Uα and 0 on Vα. We choose then Yα+1 = Uα; for a limit ordinal, we constructYβ by a diagonal procedure in such a way that Yβ is almost contained in the precedingspaces, up to finite dimension. If α < β, Tβ − Tα = Id on a finite codimensional space ofVβ , hence |||Tα − Tβ ||| ≥ 1 by ???. On the other hand there should exist by Theorem 11.1for every ordinal α < ω1 an Aα ∈ A such that |||Uα − Tα||| ≤ 1/4, and this contradicts theseparability of A.

We start now the construction of the spaces X(S) and the proof of Theorem 11.1. Weintroduce a lacunary subset J of N, which we split into two disjoint parts K and L. LetJ ⊂ N be a set such that, if m < n and m,n ∈ J , then log log log n ≥ 2m. Let us writeJ in increasing order as j1, j2, . . .. We also need f(j1) > 256. Now let K, L ⊂ J be thesets j1, j3, j5, . . . and j2, j4, j6, . . ..

We mentioned before the statement of Theorem 11.1 two important ingredients of theconstruction. The third and last important ingredient is the notion of special sequence.Let us recall the definition from [GM1] of the special functionals on a space X ∈ X . LetQ ⊂ c00 be the (countable) set of sequences with rational coordinates and maximum atmost 1 in modulus. Let σ be an injection from the collection of finite sequences of successiveelements of Q to the set L introduced above. Given X ∈ X such that X satisfies a lowerf -estimate and given an integer m ∈ N, let A∗m(X) be the set of (m, f)-forms on X, i.e.the set of all functionals x∗ of the form x∗ = f(m)−1

∑mi=1 x∗i , where x∗1 < . . . < x∗m and

‖x∗i ‖ ≤ 1 for each i = 1, . . . , m. Note that these functionals have norm at most 1 by thelower f -estimate. If k ∈ N, let ΓX

k be the set of sequences y∗1 < . . . < y∗k such that y∗i ∈ Qfor each i, y∗1 ∈ A∗j2k

(X) and y∗i+1 ∈ A∗σ(y∗1 ,...,y∗i)(X) for each 1 ≤ i ≤ k − 1. We call these

special sequences. Let B∗k(X) be the set of all functionals y∗ of the form

y∗ =1√f(k)

k∑

j=1

gj

such that (g1, . . . , gk) ∈ ΓXk is a special sequence. These, when k ∈ K, are the special

functionals (on X of size k). Note that if g ∈ F and g(k) = f(k)1/2, then a specialfunctional y∗ of size k is also a (k, g)-form, and the same is true for every EUy∗, for everyinterval E and every U ∈ S. The idea behind this notion of special fonctionals is that theirnormalization is very different from the usual normalization of functionals obtained by the“Schlumprecht operation” (x∗1 + · · ·+x∗n)/f(n), so they produce “spikes” in the unit ball ofX∗, but they are extremely rare and easily identified: a relatively weak information abouta part of a special functional f(k)−1/2

∑kj=1 gj , namely knowing simply the integer l ∈ L

such that gj ∈ A∗l , allows us to trace back all the past of its construction, since there is atmost one sequence (g1, . . . , gj−1) such that l = σ(g1, . . . , gj−1).

73

Now, given a proper set S of spreads, we define the space X(S), inductively. It is thecompletion of c00 in the smallest norm satisfying the following equation.

‖x‖ = ‖x‖c0∨ sup

f(n)−1

n∑

i=1

‖Eix‖ : 2 ≤ n ∈ N, E1 < . . . < En intervals

∨ sup|x∗(Ex)| : k ∈ K, x∗ ∈ B∗

k(X), E ⊂ N an interval

∨ sup‖Ux‖ : U ∈ S

In the case S = Idc00 the fourth term drops out and the definition reduces to that of thespace constructed in [GM1]. The fourth term is there to force X(S) to have property (2)claimed in Theorem 11.1. The second term ensures that X satisfies a lower f -estimate. Itis also not hard to show that X(S) is reflexive. (A proof can be found in [GM1], end ofsection 3, which works in this more general context.) It is also useful to understand theconstruction of X in a way similar to what we said about T and S: we construct a densesubset of the unit ball BX∗ in a sequence of steps, producing an increasing sequence (Bn)of convex subsets of Bc0 . We start with B0 = B`1 ∩ c00. If Bn is defined, we add to it

— all (m, f)-forms x∗ = f(m)−1∑m

j=1 x∗j using elements x∗j from Bn, for any integerm ≥ 2

— all functionals λEUx∗ where |λ| = 1, E is an interval, U ∈ S and x∗ is any specialfunctional x∗ = (

∑kj=1 gj)/

√f(k) with gj ∈ Bn for j = 1, . . . , k;

we let finally Bn+1 be the convex hull of the union of Bn and of the set of all these newfunctionals. We let B =

⋃n Bn and we can see that the above defined norm is equal to

‖x‖ = sup|x∗(x)| : x∗ ∈ B.

If we want to compute the norm of x ∈ X, either ‖x‖ = ‖x‖c0 or, given ε > 0 such that‖x‖c0 < ‖x‖ − ε, there exist a first n ≥ 0 such that |x∗(x)| > ‖x‖ − ε for some x∗ thatwas adjoined to Bn in the construction of Bn+1, namely either an (m, f)-form or someEUy∗, with y∗ special functional and U ∈ S. It should be observed that if g ∈ F is suchthat g =

√f on K, then all these functionals of the form EUy∗ are (k, g)-forms for some

k ≥ 2 (observe that the images of successive functionals by a spread are still successive).The next technical Lemma is taken from [GM1]. It is just a painful exercise using onlyelementary calculus.Lemma 11.6. Let K0 ⊂ K. There exists a function g ∈ F such that f ≥ g ≥ √

f ,g(k) =

√f(k) whenever k ∈ K0 and g(x) = f(x) whenever N ∈ J \ K0 and x is in the

interval [log N, expN ].

Lemma 11.7. Let 0 < ε ≤ 1, M ∈ L and let N be such that N ∈ [log M, exp M ]. Assumethat x = x1 + · · ·+ xN satisfies the RIS(ε) condition and let x = x1 + · · ·+ xN . Then

‖(f(N)/N)x‖ ≤ 1 + 2ε.

Assume further that 0 ≤ δ < 1, and let n be such that N/n ∈ [log M, expM ] and f(N) ≤( 1 + δ )f(N/n). Then ‖(f(N)/N)x‖(n) ≤ (1 + δ)(1 + 3ε).

74

Proof. Let g be the function given by Lemma 11.6 in the case K0 = K. It is clear thatevery vector Ex such that ‖Ex‖ > 1 is normed by a (k, g)-form for some k ≥ 2; furthermoreif k ≥ 2N2/ε2

, then |x∗(x)| ≤ 1 + 2ε by Lemma 11.3, so the conditions of Lemma 11.4 aresatisfied for x′i = (1 + 2ε)−1xi, and thus ‖∑N

i=1 x′i‖ ≤ N/g(N). Since g(N) = f(N) weobtain the first estimate. The second follows by Lemma 11.5. ?????

We start now the proof of assertion 3 in Theorem 11.1. Let X0 be the weakly nullpart of X, consisting of all classes x of weakly null sequences (xn) in X. Similarly weshall consider Y0 for every subspace Y of X. Let Ξ0(Y ) be the space of finitely supportedsequences of elements of Y0. We shall use the following notation,

(y1, . . . , yk, 0, 0, . . .) =k∑

j=1

yj ⊗ fj ∈ Ξ0(X),

where (fn) is the canonical basis for the space of sequences (to avoid confusion, we chose anotation different from the previous (en)). We define a norm on Y ⊕ Ξ0(Y ) by inductionon k

‖y +k∑

j=1

yj ⊗ fj‖ = limn,U

‖y + y1,n +k∑

j=2

yj ⊗ fj‖,

where y1 = (y1,n). We can write this directly with an iterated limit,

‖y +k∑

j=1

yj ⊗ fj‖ = limn1,U

. . . limnk,U

‖y + y1,n1 + · · ·+ yk,nk‖.

We may also observe that for every integer k ≥ 1, there exists an ultrafilter U⊗k on Nk

defined byA ∈ U⊗k ⇔ lim

n1,U. . . lim

nk,U1A(n1, . . . , nk) = 1.

For every integer m ≥ 1 we extend the norm ‖.‖(m) to Y ⊕ Ξ0(Y ) in the following way

‖y +k∑

j=1

yj ⊗ fj‖(m) = limn1,U

. . . limnk,U

‖y + y1,n1 + · · ·+ yk,nk‖(m).

Letting n = (n1, . . . , nk) and yn = y +∑k

j=1 yj,nj we may write this iterated limit as

limn,U⊗k

‖yn‖(m).

Observe that since (yj,n)n is weakly null for each j, the vectors y1,n1 , . . . , yk,nkare almost

successive when n1 < . . . < nk is lacunary enough, which always happens when we considerthe iterated limit. It follows from the lower-f estimate that

‖k∑

j=1

yj ⊗ fj‖ ≥ (k∑

j=1

‖yj‖)/f(k).

75

For ξ =∑

j xj ⊗ fj ∈ Ξ0, we define its support by as j : xj 6= 0. It is possible togeneralize the above in the following way:Lemma. If ξ1, . . . , ξk are successive in Ξ0, then

‖ξ1 + · · ·+ ξk‖ ≥ (k∑

j=1

‖ξj‖)/f(k).

Let ΞY be the completion of Y ⊕ Ξ0(Y ). We do the same for X, writing simply Ξ.We will work now with the triple norm. Let T ∈ L(Y, X). Recall that for every ε > 0,there exists m ≥ 1 such that

(∗) ∀y ∈ Y, (‖Pmy‖ ≤ 1/m) ⇒ ‖Ty‖ ≤ (|||T |||+ ε)‖y‖(m).

For ξ ∈ Ξ we define |||ξ||| to be the (increasing) limit of ‖ξ‖(m), when m tends to+∞; this limit may be +∞. For every bounded operator T from Y to X, we know thatT (Y0) ⊂ X0, and we define an operator TΞ : ΞY → Ξ by the formula

TΞ(y +k∑

j=1

yj ⊗ fj) = Ty +k∑

j=1

T (yj)⊗ fj ∈ Ξ.

It is clear from the iterated limit formula that ‖TΞ‖ = ‖T‖.Lemma. Let T ∈ L(Y, X). For every ε > 0, there exists an integer m ≥ 1 such that

∀ξ ∈ Ξ0(Y ), ‖TΞ(ξ)‖ ≤ (|||T |||+ ε)‖ξ‖(m).

It follows that ‖TΞ(ξ)‖ ≤ |||T ||| |||ξ|||.Proof. Given ε > 0, we find m ≥ 1 such that (∗) holds. Let ξ =

∑kj=1 yj ⊗ fj and let

β > ‖ξ‖(m). Let ξn =∑k

j=1 yj,nj . Since each yj is weakly null, we get

limn,U⊗k

‖Pm(ξn)‖ = 0.

It is therefore possible to find A ∈ U⊗k such that ‖Pm(ξn)‖ ≤ β/m for every n ∈ A.There exists B ∈ U⊗k such that ‖ξn‖(m) ≤ β for every n ∈ B. It follows by (∗) that‖T (ξn)‖ ≤ (|||T |||+ ε)β for n ∈ A ∩B, hence ‖TΞ(ξ)‖ ≤ (|||T |||+ ε)‖ξ‖(m).

Let V (L) =⋃

N∈L[log N, exp N ]. Let NY be the part of Y0 of all x such that

f(l)l

( l∑

j=1

x⊗ fj)

is bounded when l ∈ V (L). For every x ∈ N we consider the family

m(x) =(f(l)

l(

l∑

j=1

x⊗ fj))

l∈L

76

as representing a new vector in a further ultrapower Yω of ΞY , where the index set is L. Bythe lower-f estimate, we have ‖m(x)‖ ≥ ‖x‖. If x ∈ M, then ‖m(x)‖ ≤ 1 by ???, so that1 ≥ ‖m(x)‖ ≥ ‖x‖. Given T ∈ L(Y,X), we can extend TΞ to an operator Tω ∈ L(Yω, Xω)in the usual way. Then Tω(m(x)) = m(T x).Lemma. Let T ∈ L(Y, X). Then

∀y ∈ Y0,∣∣∣∣∣∣∣∣∣m(T y)

∣∣∣∣∣∣∣∣∣ ≤ |||T ||| |||y||| .

Proof. Let ε > 0 and m ≥ 1 satisfying (∗). For every l ∈ V (L) let

ξl = y ⊗(f(l)

l

k∑

j=1

fj).

Suppose |||y||| ≤ 1. Let N ∈ L and let M1 > 2N2/ε2. We can find A1 ∈ U such that

‖Pmyn1‖ < 1/m and ‖yn1‖(M1) ≤ 1 for every n1 ∈ A1. For every n1 ∈ A1, let M2(n1) besuch that ε

√f(N2(n1)) > | ran(yn1)|; we can find A2(n1) ∈ U such that ‖PM2(n1)yn1‖ <

ε/N and ‖xn2‖(M2(n1)) ≤ 1 for every n2 ∈ A2(n1); continuing in this way we construct

A = (n1, . . . , nN ) : nj ∈ Aj(n1, . . . , nj−1), j = 2, . . . , N ∈ U⊗N

such that yn1 , . . . , ynN is a small perturbation of a RIS(ε) whenever (n1, . . . , nN ) ∈ A.This implies that

∥∥f(l)l

l∑

j=1

ynj

∥∥(m)

≤ 1 + ε

when log N ≤ l ≤ N and m ????? by Lemma 11.7, therefore by (∗) we get for everyl ∈ [log N, N ]

∥∥f(l)l

l∑

j=1

Tynj

∥∥ ≤ |||T |||+ ε.

We obtain by Lemma 11.5 that

∥∥f(l)l

l∑

j=1

Tynj

∥∥(p)≤ |||T |||+ ε

when p = ????? Since this holds for every n ∈ A we obtain ‖TΞξl‖(p) ≤ |||T ||| + ε, so thatfinally ‖Tωm(x)‖(p) ≤ |||T ||| |||x||| for every p ≥ 1.

With these elements it is easy to prove property 3 of Theorem 11.1. Suppose that wepick y such that |||y||| = 1 and

‖ST y‖ = |||ST ||| .

77

We obtain

‖ST y‖ ≤ ‖m(ST y)‖ = ‖Sω(m(T y))‖ ≤ |||S|||∣∣∣∣∣∣∣∣∣m(T y)

∣∣∣∣∣∣∣∣∣ ≤ |||S||| |||T ||| |||y||| .

The next Lemma is the main part of the analysis in [GM2], where the properties ofspecial functionals are fully used, as well as the structure of S. Note first that a proper set Sof spreads must be countable, and if we write it as U1, U2, . . . and set Sm = U1, . . . , Umfor every m, then for any x ∈ X(S), x∗ ∈ X(S)∗, we have limm sup|x∗(Ux)| : U ∈S \ Sm ≤ ‖x‖∞ ‖x∗‖∞.Lemma 11.8. Let S be a proper set of spreads, let X = X(S), let Y ⊂ X be aninfinite-dimensional subspace and let T be a continuous linear operator from Y to X. LetS =

⋃∞m=1 Sm be a decomposition of S satisfying the condition just mentioned. Then

for every ε > 0 there exists m such that, for every x ∈ Y such that ‖x‖(m) ≤ 1 and‖Pmx‖ ≤ 1/m,

d(Tx, m convλUx : U ∈ Sm, |λ| = 1) ≤ ε .

Proof. We may also assume that ‖T‖ ≤ 1. Suppose that the result is false. Then, for someε > 0, we can find a sequence (yn)∞n=1 with yn ∈ Y , ‖yn‖(n) ≤ 1 and ‖Pn(yn)‖ ≤ 1/n suchthat, setting Cn = n convλUyn : U ∈ Sn, |λ| = 1, we have d(Tyn, Cn) > ε, and we canalso find a sequence (En) of successive intervals such that if zn is any one of yn, Tyn orUyn for some U ∈ Sn and zn+1 is any one of yn+1, Tyn+1 or V yn+1 for some V ∈ Sn+1,then ‖(N \ En)zn‖ ≤ ε2−n and ‖Enzn+1‖ ≤ ε2−n.

By the Hahn-Banach theorem, for every n there is a norm-one functional y∗n such that

sup|y∗n(x)| : x ∈ Cn + εB(X) < y∗n(Tyn) .

It follows that y∗n(Tyn) > ε and sup |y∗n(Cn)| ≤ 1. Therefore |y∗n(Uyn)| ≤ n−1 for everyU ∈ Sn. We may also assume that the support of y∗n is contained in En (up to 1/n) ???.(The case of complex scalars requires a standard modification.)

Given N ∈ L define an N -pair to be a pair (x, x∗) constructed as follows. Letyn1 , yn2 , . . . , ynN

be a subsequence of (yn)∞n=1 satisfying the RIS(1) condition, which impliesthat n1 > N2. Let x = N−1f(N)(yn1 + · · ·+ ynN

) and let x∗ = f(N)−1(y∗n1+ · · ·+ y∗nN

),where the y∗ni

are as above. Lemma 11.7 implies that ‖x‖ ≤ 4 and ‖x‖(√N) ≤ 8.If (x, x∗) is such an N -pair, then x∗ ∈ A∗N (X) and, by our earlier assumptions about

supports,

x∗(Tx) = N−1N∑

i=1

y∗ni(Tyni) >

ε

2.

Similarly, |x∗(Ux)| ≤ N−2 for every U ∈ SN .Let k ∈ K be such that (ε/24)f(k)1/2 > 1. We now construct sequences x1, . . . , xk

and x∗1, . . . , x∗k as follows. Let N1 = j2k and let (x1, x

∗1) be an N1-pair. Let M2 be such

that |x∗1(Ux1)| ≤ ‖x1‖∞ ‖x∗1‖∞ if U ∈ S \ SM2 . The functional x∗1 can be perturbed sothat it is in Q and so that σ(x∗1) > maxM2, f

−1(4), while (x1, x∗1) is still an N1-pair.

78

In general, after x1, . . . , xi−1 and x∗1, . . . , x∗i−1 have been constructed, let (xi, x

∗i ) be an

Ni-pair such that all of xi, Txi and x∗i are supported (up to ???) after all of xi−1, Txi−1

and x∗i−1, and then perturb x∗i in such a way that, setting Ni+1 = σ(x∗1, . . . , x∗i ), we have

|x∗i (Uxi)| ≤ ‖xi‖∞ ‖x∗i ‖∞ whenever U ∈ S \ SNi+1 and we also have f(Ni+1) > 2i+1 and√f(Ni+1) > 2| ran(

∑ij=1 xj)|.

Now let x = (x1 + · · · + xk) and let x∗ = f(k)−1/2(x∗1 + · · · + x∗k). Our constructionguarantees that x∗ is a special functional, and therefore of norm at most 1. We thereforehave

‖Tx‖ ≥ x∗(Tx) > εkf(k)−1/2 .

Our aim is now to get an upper bound for ‖x‖ and to deduce an arbitrarily large lowerbound for ‖T‖. For this purpose we use Lemma 11.4.

Let g be the function given by Lemma 11.6 in the case K0 = K \ k. It is clearthat all vectors Ex are either normed by (M, g)-forms or by spreads of special functionalsof length k, or they have norm at most 1. In order to apply Lemma 11.4 with this g, itis therefore enough to show that |U∗z∗(Ex)| = |z∗(UEx)| ≤ 1 for any special functionalz∗ of length k and U ∈ S. Let z∗ = f(k)−1/2(z∗1 + . . . + z∗k) be such a functional withz∗j ∈ A∗mj

. Suppose that U ∈ SM+1 \SM , and let j be such that Nj ≤ M < Nj+1. Let t bethe largest integer such that mt = Nt. Then z∗i = x∗i for all i < t, because σ is injective.For such an i, |z∗i (Uxi)| = |x∗i (Uxi)| < N−2

i , if M < Ni. If M ≥ Ni+1, then U /∈ SNi+1 , so|x∗i (Uxi)| ≤ ‖xi‖∞ ‖x∗i ‖∞ ≤ 2−i. If Ni ≤ M < Ni+1, the only remaining case, then i = jand at least we know that |z∗i (Uxi)| ≤ ‖xi‖ ≤ 4.

If l 6= i or l = i > t, then z∗l (Uxi) = U∗z∗l (xi), and we have U∗z∗l ∈ A∗mlfor some

ml. Moreover, because σ is injective and by definition of t, in both cases ml 6= Ni. Ifml < Ni, then, as we remarked above, ‖xi‖(√Ni)

≤ 8, so the lower bound of j2k for m1

tells us that |U∗z∗l (xi)| ≤ k−2. If ml > Ni, the same conclusion follows from Lemma ?.There are at most two pairs (i, l) for which 0 6= z∗l (UExi) 6= z∗l (Uxi) and for such a pair|z∗l (UExi)| ≤ 1.

Putting all these facts together, we get that |z∗(UEx)| ≤ 1, as desired. We alsoknow that (1/8)(x1, . . . , xk) satisfies the RIS(1) condition. Hence, by Lemma 11.4, ‖x‖ ≤24kg(k)−1 = 24kf(k)−1. It follows that ‖T‖ ≥ (ε/24)f(k)1/2 > 1, a contradiction.

We can now explain how the main assertion 4 of Theorem 11.1 follows by a fixedpoint argument. Suppose that m(x) = m1(x), m2(x),. . . are successive copies of m(x).For example,

m2(x) =(f(l)

l

l∑

j=1

x⊗ fl+j

)l∈L

.

For every l, the vector

ξ1,l + · · ·+ ξk,l =k∑

i=1

f(l)l

l∑

j=1

x⊗ f(i−1)l+j

79

is the sum of kl successive vectors x⊗ fi in Ξ0, therefore

‖k∑

j=1

ξj,l‖ ≥ kl

f(kl)f(l)

l‖x‖

and f(kl)/f(l) tends to 1 for fixed k when l →∞, so that

‖m1(x) + · · ·+ mk(x)‖ ≥ k‖x‖.

More generally, if x1, . . . , xk ∈ N , we get

‖m1(x1) + · · ·+ mk(xk)‖ ≥k∑

j=1

‖xj‖.

Indeed,. . .Lemma. Suppose that T ∈ L(Y, X) and that m and ε are as in the above Lemma. Let

Am = convλU : U ∈ Sm, |λ| = 1.

Then there exists U ∈ Am such that |||T − U iY,X ||| < 8ε.

Proof. If the Lemma is false, then for every U ∈ Am there is a sequence xU ∈ Y0 suchthat |||xU ||| ≤ 1 and ‖ ˜(T − U)xU‖ > 17ε. Our first aim is to show that these xU can bechosen continuously in U . Let (Uj)k

j=1 be a covering of Am by open sets of diameter lessthan ε in the operator norm. For every j = 1, . . . , k, let Uj ∈ Uj and let xj be a sequencewith the above property with U = Uj . By the condition on the diameter of Uj , we have‖ ˜(T − U)xj‖ > 16ε for every U ∈ Uj . Let (φj)k

j=1 be a partition of unity on Am with φj

supported inside Uj for each j.Now let us consider in Yω the vector y(U) =

∑kj=1 φj(U)mj(xj). We shall show that

y(U) is a “bad” vector for U , by showing that ‖(Tω − Uω)y(U)‖ > 8ε. To do this, letU ∈ Am be fixed and let J = j : φj(U) > 0. Note that

∥∥∥( ˜T − U)xj

∥∥∥ > 16ε for everyj ∈ J , hence

‖(Tω − Uω)y(U)‖ = ‖k∑

j=1

φj(U)mj(( ˜T − U)xj)‖ ≥k∑

j=1

φj(U)‖( ˜T − U)xj‖ > 16ε.

The function U 7→ y(U) is clearly continuous. We now apply a fixed-point theorem. Forevery U ∈ Am, let Γ(U) be the set of V ∈ Am such that

∥∥∥( ˜T − V )y(U)∥∥∥ ≤ 8ε. Clearly

Γ(U) is a compact convex subset of Am. By the previous lemma, Γ(U) is non-empty forevery U . The continuity of U 7→ y(U) gives that Γ is upper semi-continuous, so thereexists a point U ∈ Am such that U ∈ Γ(U). But this is a contradiction.

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12. Applications to some specific examples

In this section we present some specific examples which are special cases of Theorem11.1.

Construction of a H.I. space

Let S = Id, let X = X(S), let Y be any subspace of X and let iY,X be the inclusionmap from Y to X. Then given any operator T from Y to X, there exists by Theorem11.1, for every ε > 0, some λ such that |||T − λiY,X ||| < ε. Since |λ| ≤ |||T ||| + ε, an easycompactness argument then shows that there exists λ such that |||T − λiY,X ||| = 0 and thusthat T − λiY,X is strictly singular, which is one of the main results of [GM1]. It implieseasily that X is hereditarily indecomposable. Recall (Exercise 10.1) that Xn is isomorphicto Xm if and only if m = n.

Shift space Xs

Let S be the proper set generated by the right shift R on c00. This set S consists ofall maps of the form SA,B where A = [m,∞) and B = [n,∞). Let Xs = X(S). We willwrite L for the left shift, which is (formally) the adjoint of R, and Id for the identity onc00. Then LR = Id and every operator in S is of the form RmLn. Since RL − Id is ofrank one, every operator V in A is a finite-rank perturbation of an operator of the form

U =−1∑

n=−N

anL−n +N∑

n=0

anRn,

so the difference V − U is of |||.|||-norm zero, hence strictly singular.For every such U ∈ A we define a fonction ϕU on the unit circle T by

ϕU (λ) =N∑

i=−N

aiλi.

For every λ ∈ T it is easy to find a normalized sequence (x(λ)n ) in M such that Rx

(λ)n −

λx(λ)n → 0 (such vectors are simply obtained by Lemma 11.1 by normalizing a sum like∑k+Nj=k λ−jej , where N is much larger than k). This implies for any such sequence (x(λ)

n )

that Ux(λ)n ' ϕU (λ)x(λ)

n , and we know that lim supn ‖Ux(λ)n ‖ ≤ |||U |||. It follows that

‖ϕU‖∞ ≤ |||U ||| (the uniform norm is taken on the unit circle T).For a general V ∈ A we notice that |||V − U ||| = 0 implies that limn V (x(λ)

n ) −ϕU (λ)x(λ)

n = 0. We may thus define ϕV (λ) to be the only scalar µ such that limn V (x(λ)n )−

µx(λ)n = 0 for any sequence (x(λ)

n ) ∈M such that lim(R−λIX)(x(λ)n ) = 0; then |||V − U ||| = 0

implies that ϕV = ϕU ; it follows easily that ϕV1ϕV2 = ϕV1V2 . By properties 4 and 5 wecan therefore extend the map ϕ to an algebra homomorphism ϕ : U → ϕU from L(X) toC(T).

Proposition 12.1. Let T ∈ L(Xs); the operator T is finitely singular iff ϕT does notvanish on T.

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Proof. Suppose that λ ∈ T and ϕT (λ) = 0, choose U ∈ A such that |||T − U ||| ' 0.Then ϕU (λ) ' 0 which implies that U(x(λ)

n ) ' 0 therefore T (x(λ)n ) ' 0 and T is infinitely

singular.In the other direction, assume T infinitely singular. We can find a block subspace Y

such that ‖T|Y ‖ < ε by Proposition 3.2, hence a vector x ∈ Y such that 1 = ‖x‖ ≤ ‖x‖n ≤1 + ε (by Lemma 11.1) and ‖Tx‖ < ε. Next we get a normalized weakly null sequencex = (xn) in M(Y ) such that Txn → 0. We have ‖x‖ = 1, x ∈ X0 (the weakly null partof the ultrapower X) and T x = 0. Let T0 denote the restriction of T to X0. Let U ∈ Abe such that |||T − U ||| < ε. Then RU − U R = 0 on X0 because RU − UR has finite rank.Now ‖U x‖ ≤ ε because x ∈ M. Since R is an isometry preserving successiveness, we seethat Rx ∈ M, and RU x = U Rx, so ‖T Rx‖ ≤ 2ε. We get T Rx = 0, and similarly forevery k ≥ 1 we have T Rkx = 0. It follows that we can find an invariant space for R onwhich T0 = 0. We can then find an approximate eigenvector y for R such that T y = 0,and the eigenvalue must be some λ ∈ T. Then ϕT (λ) = 0.

Projections in Xs

Suppose that P is a projection on Xs. Since ϕ is an algebra homomorphism, it followsfrom P 2 = P that ϕ2

P = ϕP , hence 0 and 1 are the only possible values for ϕP (λ). Bycontinuity we get either ϕP = 0 or ϕP = 1. In the second case P is finitely singular byProposition 12.1, hence has finite dimensional kernel. In the other case we get a finitedimensional range. We see that Xs is indecomposable. However, Xs is not H.I. For everyλ ∈ T, we can find an H.I. subspace Xλ of Xs by considering a subspace generated bya normalized basic subsequence (x′n) of the sequence (x(λ)

n ) such that Rx′n − λx′n → 0rapidly. If λ 6= µ, it is easy to see that Y = Xλ + Xµ is closed, which implies that Y isdecomposable and Xs is not H.I.Remark-Exercise. These spaces Xλ are pairwise non-isomorphic for λ ∈ T. We have anuncountable family of different germs indexed by the unit circle T.

This space Xs is a new prime space. The only known examples before [GM2] werec0 and `p (1 ≤ p ≤ ∞). The space Xs is prime by virtue of having no non-trivial com-plemented subspaces and being isomorphic to its subspaces of finite codimension. Indeed,we know that every projection P on Xs is of finite rank or corank. Thus, if PXs isinfinite-dimensional, then it has finite codimension. Since the shift on Xs is an isometry, itfollows that Xs and PXs are isomorphic, which proves the following theorem (using nextExercise).Theorem 12.1. The space Xs is prime.

Exercise. The hyperplanes of a given Banach space X are mutually isomorphic. Moregenerally, all subspaces of X of a fixed finite codimension are isomorphic.

We note here that the argument in the above proof can be generalized to show thatif m and n are integers with m > n, then Xn

s does not contain a family P1, . . . , Pm ofinfinite-rank projections satisfying PiPj = 0 whenever i 6= j. Indeed, given any projectionP ∈ L(Xn), we can regard it as an element of Mn(L(X)). Acting on each entry with ϕ, weget a function h ∈ Mn(C(T)). The map taking P to h is an algebra homomorphism so h is

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an idempotent. Regarding h as a continuous function from T to Mn(C), we have that h(t)is an idempotent in Mn(C) for every t ∈ T. By the continuity of rank for idempotents, wehave that if h(t) = 0 for some t, then h is identically zero. But then P is strictly singularand hence of finite rank. Applying this reasoning to the family P1, . . . , Pm above, we obtainh1, . . . , hm such that, for every t ∈ T, h1(t), . . . , hm(t) is a set of non-zero idempotents inMn(C) with hi(t)hj(t) = 0 when i 6= j. But this is impossible if m > n. It follows thatXn

s and Xms are isomorphic if and only if n = m.

Another simple consequence of properties 4 and 5 is that, up to strictly singularperturbations, any two operators on Xs commute. Indeed, U1U2 − U2U1 has finite rankwhen U1, U2 ∈ A, hence |||U1U2 − U2U1||| = 0. By approximation we get |||T1T2 − T2T1||| = 0for every pair of operators on Xs.

One can actually get better estimates relating Xs to the Wiener algebra.

Lemma. (Lemma 11 of [GM2].) Let U =∑N

n=0 λnRn +∑N

n=1 λ−nLn. Then

‖U‖ = |||U ||| =∑

n∈Z|λn|.

It follows that the homomorphism ϕ takes values in the Wiener algebra W and itis possible to improve in this case property 5 by saying that for every T ∈ L(Xs), thereexists U =

∑∞n=1 a−nLn +

∑∞n=0 anRn such that

∑n∈Z |an| < ∞ and |||T − U ||| = 0. Using

this and the properties of invertible elements in W we get in [GM2] an easier approach toProposition 12.1.

Exercise. Compute K0(L(Xs)) and K1(L(Xs)).

The result in this section can be compared to those of Mankiewicz [Mz]; we have hereanother example of a Banach space such that there exists an algebra homomorphism fromL(X) into a commutative Banach algebra. It follows that X is not isomorphic to anypower Y n, for n ≥ 2. Indeed, if ϕ is a non zero multiplicative functional from L(X) toC, and if X = Y n, there is a natural homomorphism i from Mn to L(X). But then ϕ iwould be a non zero multiplicative functional on Mn, which is not possible for n ≥ 2.

Double shift space Xd

This example Xd is isomorphic to its codimension 2 subspaces but not to its hyper-planes. Let S be the proper set generated by the double shift R2. That is, S is as in theprevious example but m and n are required to be even. We show that every Fredholmoperator T on Xd = X(S) has even index. By property 4, and by the fact that everyoperator in S differs by a finite-rank operator from some even shift, we can find, for anyε > 0, some linear combination U of even shifts such that |||T − U ||| < ε. Then s(T−U) < εand we know that ind(T ) = ind(U) when ε is small by Corollary 6.1. Hence it is enoughto show that every U ∈ A has even index.

Lemma. Let V be a Fredholm isometry on a Banach space X with a left inverse W , andlet T : X → X be a Fredholm operator which can be written in the form P (V ) + Q(W )for polynomials P and Q. Then the index of T is a multiple of the index of V .

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Proof. Suppose first that the scalars are complex. It is clear since V is isometric thatV − λIX is an into isomorphism when |λ| 6= 1 (and it is onto when |λ| > 1). By Lemma4.6, we get ind(V − λIX) = ind(V ) when |λ| < 1 (because we can connect V − λIX to Vby a path of semi-Fredholm operators) and ind(V − λIX) = 0 when |λ| > 1. If V − λIX isfinitely singular for some λ ∈ T, then ind(V − λIX) = 0 for the same reason. In all cases,the only possible values for the index are 0 and ind(V ).

Now suppose that T is as in the statement of the lemma. For sufficiently large N ,TV N can be written F (V ) for some polynomial F and is still Fredholm. Writing F (V ) =c∏

i(V − λiIX), we must have V − λiIX finitely singular for TV N to be Fredholm, soind(V − λiIX) is either 0 or ind(V ). It follows from the composition formula, Proposition4.2, that the index of F (V ), and hence that of T , is a multiple of the index of V as stated.

When the scalars are real we may complexify V to an isometry VC of XC, for exampleusing the injective norm C⊗ε X on XC.

Remark. When K = C and V is not invertible, the spectrum of V contains T.

Putting these facts together, we find that no continuous operator on Xd can be Fred-holm with odd index. We therefore have the following result.Theorem 12.2. The space Xd is isomorphic to its subspaces of even codimension whilenot being isomorphic to those of odd codimension. In particular, it is isomorphic to itssubspaces of codimension two but not to its hyperplanes.

Remark. The proof of Theorem 12.1 gives for this space also that every complementedsubspace has finite dimension or codimension. Combining this observation with Theorem12.2, we see that the space Xd has exactly two infinite-dimensional complemented sub-spaces, up to isomorphism. It is true for this space as well that it is isomorphic to nosubspace of infinite codimension. Note that the methods of this section generalize easilyto proper sets generated by larger powers of the shift.

Exercise. Compute K0(L(Xd)).

Ternary space

This application is more complicated than the previous ones. The aim is to constructa space Xt which is isomorphic to Xt ⊕ Xt ⊕ Xt but not to Xt ⊕ Xt. This question isrelated to the Schroder-Bernstein problem for Banach spaces, first solved (by the negative)by T. Gowers [G3]. We have seen that in some cases, we can deduce that X ' Y from thefact that X and Y embed complementably in each other. The Schroder-Bernstein problemfor Banach spaces is the question whether this is true in general. Constructing a Banachspace X such that X ' X3 but X 6' X2 gives a strong negative question to the problem,because X and X2 are in this case complementably embeddable in each other.

There is a very natural choice of S in this case, strongly related to the algebra Pfrom section 7 (or to the Cuntz’ algebra O3). For i = 0, 1, 2 let Ai be the set of positiveintegers equal to i + 1 (mod 3), let U ′

i be the spread from N to Ai and let S′ be thesemigroup generated by U ′

0, U ′1 and U ′

2 and their adjoints. It is shown in [GM2] that thisis a proper set. The space Xt = X(S′) is easily seen to be isomorphic to its cube, andthis isomorphism is achieved in a “minimal” way. (The primes in this paragraph are to

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avoid confusion later.) This is one of the models introduced in section 7 for the algebra P.We equipped it here with the norm of L(X), where X is an exotic Banach space given byTheorem 11.1.

We shall indeed consider the space X(S′) defined above. However, we define it slightlyless directly, which helps with the proof later that it is not isomorphic to its square. Thealgebra A′ arising from the above definition is, if completed in the `2-norm, isometric tothe Cuntz algebra O3 ([C1], see section 7). Our proof is inspired by his paper [C2]. Recallsome notation from section 7: T is the ternary tree

⋃∞n=00, 1, 2n, Y00 is the vector space

of finitely supported scalar sequences indexed by T and the canonical basis for Y00 isdenoted by (et)t∈T ; we write e for e∅, denote the length of a word t ∈ T by |t| and (s, t)stands for the concatenation of s, t ∈ T . Let Id denote the identity operator on the spaceof sequences. Let Vi and Ti, for i = 0, 1, 2 be defined by their action on the basis as follows:

Viet = e(t,i), Tiet = e(i,t).

Thus Ti takes the whole tree T on to the ith branch. The adjoints V ∗i and T ∗i act in the

following way: V ∗i et = es if t is of the form t = (s, i), and V ∗

i et = 0 otherwise, whileT ∗i et = es if t = (i, s), and T ∗i et = 0 otherwise. The following facts are easy to check:ViTj = TjVi, V ∗

i Vj = T ∗i Tj = δi,jId; ViV∗i and TiT

∗i are projections; if Q denotes the

natural rank one projection on the line Ce, then∑2

i=0 ViV∗i =

∑2i=0 TiT

∗i = Id−Q. Let S

and A be respectively the proper set generated by V0, V1 and V2, and the algebra generatedby this proper set. (Strictly speaking, S is not a proper set, but it is easy to embed T intoN so that the maps V0, V1 and V2 become spreads as defined earlier. Note that S is thesemigroup generated by the Vi and the V ∗

i , that it contains Id and that A contains Q, aswe have just shown.

In order to obtain the space Xt, consider the subset T0 of T consisting of all wordst ∈ T that do not start with 0 (including the empty sequence). We modify the definitionof V0 slightly when defining U0, by letting U0e equal e instead of e0. Operators U1 and U2

are defined exactly as V1 and V2 were. We still have that the UiUi∗ are projections and

that Ui∗Uj = δi,jId, but this time

∑2i=0 UiUi

∗ = Id. We noticed in section 7 that we canassociate the integer ns = 3n−1i1 + · · ·+ 3in−1 + in + 1, (with n∅ = 1), and this defines abijection between T0 and N. The operators U0, U1 and U2 then coincide with the spreadson c00 defined earlier, so we can define S′ to be the proper set they generate and obtainthe space Xt = X(S′). Let A′ be the algebra generated by S′.

For t ∈ T , we defined Vt inductively by V(t,i) = ViVt. Let V ∗t be the adjoint of Vt. We

now from section 7 that every W ∈ A has a decomposition

W =N∑

l=1

cl VαlV ∗

βl,

where αl and βl are words in T . Define β(W ) to be the smallest value of maxl |βl| over allsuch representations of W . We make the obvious modifications to the above definitions forA′. The remarks are still valid, except that the actions of Vt and Ut on e will be different if

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the word t begins with 0. The next lemma is similar to Lemma 11 in [GM2]. The notation‖x‖1 is for the norm in `1.Lemma 12.1. (Lemma 20 of [GM2].) Let U ∈ A′. Then for |t| > β(U), we have theinequality ‖Uet‖1 ≤ |||U |||.Proof. Let |t| > β(U) and suppose that Uet =

∑Mk=1 ckesk

, where the sks are distinct.Since |t| > β(U), we have Ueu,t =

∑Mk=1 ckeu,sk

for every u ∈ T . Pick a sequence (uj)∞j=1

lacunary enough to guarantee that the sequences (euj ,t)∞j=1 and (Ueuj ,t)∞j=1 are successive.Then by the construction of Xt, we obtain the inequality

∥∥U(N∑

j=1

euj ,t)∥∥ ≥ N

f(MN)

M∑

k=1

|ck|.

By Lemma 11.7 we know that for some infinite subset L ⊂ N and for every N ∈ L

‖N∑

j=1

euj ,t‖ ≤ N

f(N).

Letting N → ∞, this gives∑M

k=1 |ck| ≤ ‖U‖. For the inequality for |||U |||, see [GM2] orLemma 11.7. ?????

We now consider the algebra A. Let Y = `1(T ) be the completion of Y00 equippedwith the `1 norm and let E denote the norm closure of A in L(Y ). Note that every Vi orTi is an isometry on Y , and ‖V ∗

i ‖ ≤ 1, ‖T ∗i ‖ ≤ 1.Lemma 12.2. Every Fredholm operator in E has index 0. More generally, every Fredholmoperator T : Y n → Y n given by a matrix in Mn(E) has index 0.

Proof. Since the Fredholm index is stable under small perturbations, it is enough toconsider operators in A (as operators on Y ). For any such operator W we associate theoperator

W# =2∑

i=0

TiWT ∗i .

We claim that W# is a finite rank perturbation of W . It is enough to show that V #i is a

rank-one perturbation of Vi (and to observe that (W ∗)# = (W#)∗). But

V #j =

2∑

i=0

TiVjT∗i = Vj

2∑

i=0

TiT∗i = Vj(I −Q) = Vj − VjQ ;

(instead of using approximation of elements in E by elements in A, we could observedirectly that for every V ∈ E , V − V # is compact). Consider the projections Qi = TiT

∗i .

Then QiQj = 0 for i 6= j and

Y = Ce⊕Q0Y ⊕Q1Y ⊕Q2Y.

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Each TiWT ∗i represents an operator on QiY , equivalent (in the obvious sense) to W onY , so that ind(TiWT ∗i ) = ind(W ) (the first operator is obtained from W by compositionwith onto isomorphisms) and W# is 0 on the component Ce. It follows that ind(W#) =3 ind(W ). On the other hand ind(W#) = ind(W ) since it is a finite rank perturbation ofW . It follows that ind(W ) = 0.

The proof is essentially the same for the more general statement. Given T ∈ L(Y n),represented by a matrix A ∈ Mn(E), use the #-operation on each entry. The resultingmatrix is equivalent to three copies of A plus the zero matrix in Mn. This zero matrixcontributes n to the dimension of the kernel and n to the codimension of the image, fromwhich we obtain the equation

ind(T ) = ind(T#) = 3 ind(T ) + n− n.

Remark. In the rectangular case, if T : Y m → Y n is Fredholm, then 2 ind(T )+m−n = 0.This shows that there is no Fredholm operator from Y m to Y n when m− n is odd.

Let I be the closed two-sided ideal in E generated by Q. This ideal contains allrank-one operators of the form e∗s ⊗ et with s, t ∈ T . Hence, every finite rank operatoron Y n which is w∗-continuous (considering Y n as the dual of (c0)n) belongs to Mn(I).Indeed, the matrix of such an operator consists of entries which are finite sums of the form∑

k yk ⊗ xk, with yk ∈ c0. We can approximate yk and xk by finitely supported sequencesy′k and x′k, and

∑k y′k ⊗ x′k certainly belongs to I. (In fact, I consists exactly of the

compact w∗-continuous operators on `1.)Lemma. If V ∈ Mn(E) is Fredholm then there exists W ∈ Mn(I) such that V + W isinvertible in Mn(E).Proof. By Lemma 12.2 the index of V is zero. Let x1, . . . , xN and z1, . . . , zN be bases forthe kernel and cokernel. We can construct a w∗-continuous projection

∑Nk=1 yk ⊗ xk on

the kernel. Then W =∑N

k=1 yk ⊗ zk will do.

Let O denote the quotient algebra E/I. Since I consists of compact operators, weknow that every operator on Y (or on Y n) which is invertible modulo I (or mod Mn(I))is Fredholm on Y or on Y n by Corollary 4.3. Hence any lifting in Mn(E) of an invertibleelement in Mn(O) is Fredholm on Y n. As an immediate consequence of the precedingdiscussion we have the following statement.Corollary 12.1. Every invertible element of Mn(O) can be lifted to an invertible elementof Mn(E).

It follows easily from Lemma 12.1 that |||.||| is actually a norm on A′. Let G be theBanach algebra |||.|||-completion of A′. Recall that by properties 3 and 4 of Theorem 11.1there is a unital algebra homomorphism ϕ : L(X) → G.Lemma. There is a norm-one algebra homomorphism θ from G to O.Proof. Define a map θ0 : A′ → O as follows. Given U ∈ A′, write U =

∑Nl=1 cl Uαl

Uβl

∗

in some way, consider the corresponding sum∑N

l=1 clVαlV ∗

βlas an element of E and let

θ0(U) be the image of this operator under the quotient map from E to O. To see that

87

this map is well defined, observe that for any pair of words α and β we have the equationUαUβ

∗ =∑2

i=0 U(i,α)U(i,β)∗. If n is sufficiently large, we can therefore write U as above in

such a way that all the αl are words of length n. Let Wn be the set of all words of lengthn. Then what we have said implies that U can be written as a sum

∑α∈Wn

U ′αT ∗α, where

each T ∗α is some linear combination of distinct operators of the form Uβ∗. It is easy to see

now that U = 0 if and only if Tα = 0 for every α ∈ Wn, and moreover that distinct Uβ∗ are

linearly independent. Therefore any U ∈ A′ has at most one representation in the aboveform. In A we know that for any pair of words α and β the images in O of the operatorsVαV ∗

β and∑2

i=0 V(i,α)V∗(i,β) are the same. It follows that θ0 is well defined. Similarly, one

can show that it is a unital algebra homomorphism.We may want to argue in this way: E/I is an algebra with six elements which are

the classes wi of Ui and w∗i of U∗i , and these elements satisfy the defining properties of

our algebra P from section ???, because U∗i Ui = Id and Id−∑

UiU∗i ∈ I, therefore there

exists an algebra homomorphism ρ from P to E/I such that ρ(ui) = wi and ρ(u∗i ) = w∗i .....

Let Pn denote the projection on to the first n levels of the tree T , so that Pn ∈ I forevery n. If U ∈ A′, then Lemma 12.1 implies that

limn‖U(I − Pn)‖L(Y ) ≤ |||U ||| .

It follows that we may extend θ0 to a norm-one homomorphism θ : G → O, as claimed.

We work with complex scalars for the rest of this section. The proof given in [GM2]works also in the real case, but we want to apply here directly K-theoretic results that areproved in the complex case.Theorem 12.3. The spaces Xt and Xt ⊕Xt are not isomorphic.

Proof. If Xt and X2t are isomorphic we know that [IXt ] = 0 in K0(L(Xt)) by ?????. Taking

the image under θ φ : L(Xt) → O, this yields [1O] = 0 in K0(O). All we have to shownow is that [1O] 6= 0. For this we follow the proof given by Cuntz for Theorem 3.7 of [C2].By the definition of equivalence for idempotents, 1E = V ∗

i Vi and ViV∗i are equivalent. The

relation Id−Q =∑2

i=0 ViV∗i implies in K0(E) that

[1E ]− [Q] = 3[1E ],

and therefore that [Q] = −2[1E ]. Now consider the short exact sequence

0 → I j−→E π−→O → 0

and the corresponding exact sequence in K-theory

K1(O) ∂1−→K0(I)j∗−→K0(E) π∗−→K0(O) ∂0−→K1(I).

It is easy to see that K1(I) = 0 and K0(I) ' Z as they are for the ideal of compactoperators. Corollary 12.1 and the definition of ∂1 (see section 9) immediately imply that∂1 = 0, so we get an exact sequence

0 → K0(I)j∗−→K0(E) π∗−→K0(O) → 0.

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Now, we know that r = [Q] generates j∗(K0(I)) = kerπ∗ ' Z. If 0 = [1O] = π∗([1E ]), itfollows by exactness that [1E ] = nr for some integer n ∈ Z. But we know that r = −2[1E ],so (2n + 1)r = 0, contradicting the fact that r generates a group isomorphic to Z.

The proof of Theorem 12.3 generalizes in a straightforward way to give, for everyk ∈ N, an example of a space X such that Xn is isomorphic to Xm if and only if m = n(mod k). It is likely that every Fredholm operator on the space X of this section has zeroindex, so that X is not isomorphic to its hyperplanes. Working with a dyadic tree maythen give an example of a space X isomorphic to X2 but not isomorphic to its hyperplanes.

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