Notes on Operator Algebras

John Roe

Fall 2000

Abstract

These are the lecture notes for the Penn State course Math 520 held inFall 2000. They will be revised and extended as the course progresses.

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1 C∗-Algebra Basics

The key property that relates the norm and the involution onB(H) is theC∗-identity:

‖T ∗T‖ = ‖T‖2.

The proof follows from Cauchy-Schwarz: if‖v‖ = 1, then

‖T‖‖T ∗‖ > ‖T ∗T‖ > |〈T ∗Tv, v〉| = ‖Tv‖2

and so by taking the supremum over allv we find

‖T‖‖T ∗‖ > ‖T ∗T‖ > ‖T‖2

whence we obtain both‖T‖ = ‖T ∗‖ and theC∗-identity.

(1.1) DEFINITION: A C∗-algebraA is a complex involutive Banach algebrasatisfying theC∗-identity: ‖a∗a‖ = ‖a‖2 for all a ∈ A.

One easily deduces that the involution is isometric. We will show eventuallythat everyC∗-algebra is faithfully represented as an algebra of operators on Hilbertspace, but it is useful to begin with the abstract definition. Compare ‘permutationgroup theory’ versus ‘abstract group theory’.

(1.2) REMARK : about units. We shan’t necessarily assume that aC∗-algebraAhas a unit. Indeed, non-unitalC∗-algebras are extremely important — the firstfundamental example is the algebraK(H) of compactoperators on a Hilbert spaceH. However it is the case that every non-unitalC∗-algebraA can be embedded ina unitalC∗-algebraA as a closed maximal ideal. To figure out the construction, ithelps to consider the case thatA is a subalgebra ofB(H). ThenA should be thealgebra generated byA and 1 insideB(H). We thus arrive at the following idea:A is the algebra of pairs(a, λ), with a ∈ A andλ ∈ C. Multiplication is definedin the obvious way if we think of(a, λ) asa+ λ1, namely

(a, λ)(b, µ) = (ab+ λb+ µa, λµ).

One shows this is a Banach algebra, and indeed aC∗-algebra with the involution(a, λ) 7→ (a∗, λ), by representing it as an algebra of left multiplication operatorsonA, and using the operator norm. The algebraA is called theunitalizationof A.

An important example of aC∗-algebra isC(X), the algebra of continuouscomplex-valued functions on a compact Hausdorff spaceX. If X is just locallycompact thenC0(X) denotes the algebra of continuous complex-valued functionswhich vanish at infinity. It is a non-unitalC∗-algebra.

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An important example of an involutive Banach algebra which isnot a C∗-algebra is thedisk algebra: this is the algebra of continuous complex-valuedfunctions on the closed disk inC which are holomorphic in the interior of thedisk, with the involution

f∗(z) = f(z).

1.1 Banach Algebra Basics

We review some basic information about Banach algebras. LetA be a unitalBanach algebra.

(1.3) DEFINITION: Thespectrumsp(a) of a ∈ A is the set of complex numbersλsuch thata − λ1 doesnot have an inverse inA. The complement of the spectrumis called theresolvent set.

We’ll use the notationspA(a) if it is necessary to specify the algebraA —although, as we shall see later, this will seldom be necessary forC∗-algebras. IfA is non-unital, we define the spectrum by passing to the unitalization. For therecord, ‘invertible’ means ‘having a two-sided inverse in the algebraA’. Note thatthis is a purely algebraic notion — the norm ofA is not involved.

(1.4)LEMMA : The resolvent set ofa ∈ A is open (and so the spectrum is closed).

PROOF: It suffices to show that every element sufficiently close to1 ∈ A isinvertible, which follows from the absolutely convergent power series expansion

(1− x)−1 = 1 + x+ x2 + · · · ,

which is valid for‖x‖ < 1 in any Banach algebra.

(1.5) COROLLARY: Maximal ideals in a unital Banach algebraA are closed.Every algebra homomorphismα : A→ C is continuous, of norm6 1.

PROOF: Let m be a maximal ideal inA. Thenm does not meet the open set ofinvertible elements inA. The closurem then does not meet the set of invertibleseither, so it is a proper ideal, and by maximalitym = m.

Since the kernel ofα is a maximal ideal, it is closed. Thusα is continuous, bythe usual continuity criterion for linear functionals. The norm estimate is obtainedby refining the argument: if‖α‖ > 1 then there isa ∈ A with ‖a‖ < 1 andα(a) = 1. But then1 − a is invertible, so1 − α(a) is invertible too, and this is acontradiction.

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(1.6) PROPOSITION: LetA be a unital Banach algebra, and leta ∈ A. Then thespectrumsp(a) is a nonempty compact subset ofC. Moreover, thespectral radius

spr(a) = infr ∈ R+ : sp(a) ⊆ D(0; r)

is given by the formulaspr(a) = lim

n→∞‖an‖1/n.

PROOF: The power series expansion

(λ− a)−1 = λ−1 + λ−2a+ λ−3a2 + · · · (1.7)

converges for|λ| > ‖a‖ and shows that the spectrum ofa is contained withinD(0; ‖a‖) (note in particular thatspr(a) 6 ‖a‖). We already showed that thespectrum is closed, so it is compact. It is nonempty since otherwise the function

λ 7→ (λ− a)−1

would be a bounded, entire function (with values inA), contradicting Liouville’sTheorem from complex analysis.1 Finally let us prove the spectral radius formula.Since by hypothesis the resolvent functionλ 7→ (λ − a)−1 is holomorphic for|λ‖ > spr(a), elementary complex analysis (Cauchy’sn’th root test) yields theestimate

lim sup ‖an‖1/n 6 spr(a)

for the coefficients of the Laurent expansion 1.7 above. On the other hand, it iseasy to see that for any polynomialp, the spectrum ofp(a) is p(sp(a)). Therefore

spr(a)n = spr(an) 6 ‖an‖

and sospr(a) 6 inf ‖an‖1/n.

Putting together our two estimates for the spectral radius ofa, we complete theproof.

(1.8) LEMMA : (Gelfand-Mazur theorem) The only Banach algebra which is alsoa field isC.

PROOF: Let A be such an algebra,a ∈ A. Thena has nonempty spectrum,so there is someλ ∈ C such thata − λ1 is not invertible. But in a field the onlynon-invertible element is zero; soa = λ1 is scalar.

1This often seems a surprising argument. In finite dimensions the existence of eigenvalues followsfrom the Fundamental Theorem of Algebra. But one way to prove the Fundamental Theorem is byway of Liouville’s Theorem — ifp is a polynomial without root then1/p is a bounded entire function— so the Banach algebra argument is not really so different.

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(1.9) PROPOSITION: LetA be a unitalC∗-algebra. Ifa ∈ A is normal (that is,commutes witha∗) then‖a‖ = spr(a). For everya ∈ A, ‖a‖ = spr(a∗a)1/2. Thusthe norm onA is completely determined by the algebraic structure!

PROOF: If a is normal then

‖a‖2 = ‖a∗a‖ = ‖a∗aa∗a‖1/2 = ‖a2a∗2‖1/2 = ‖a2‖

using theC∗-identity several times. Thus the spectral radius formula gives‖a‖ =spr(a). Whatevera is, the elementb = a∗a is normal (indeed selfadjoint), so thesecond part follows from the first applied tob, together with theC∗-identity again.

Gelfand used the results above to give a systematic analysis of the structure ofcommutativeunital Banach algebras. IfA is such an algebra then itsGelfand dualA is the space of maximal ideals ofA; equivalently, by 1.8 and 1.5 above, it is thespace of algebra-homomorphismsA→ C. Every such homomorphism is of norm6 1, as we saw, soA may be regarded as a subset of the unit ball of the Banachspace dualA∗ of A (namely, it is the subset comprising all those linear functionalsα which are alsomultiplicative in the sense thatα(xy) = α(x)α(y)). This isa weak-star closed subset of the unit ball ofA∗, and so by the Banach-AlaogluTheorem it is a compact Hausdorff space in the weak-star topology. Ifa ∈ A thenwe may define a continuous functiona on A by the usual double dualization:

a(α) = α(a).

In this way we obtain an algebra-homomorphismG : a 7→ a, called theGelfandtransform, from A to C(A), the algebra of continuous functions on the Gelfanddual.

(1.10) THEOREM: LetA be a commutative unital Banach algebra. An elementa ∈ A is invertible if and only if its Gelfand transforma = Ga is invertible.Consequently, the Gelfand transform preserves spectrum: the spectrum ofGa isthe same as the spectrum ofa.

PROOF: If a is invertible thenGa is invertible, sinceG is a homomorphism.On the other hand, ifa is not invertible then it is contained in some maximal idealm, which corresponds to a point of the Gelfand dual on whichGa vanishes. SoGaisn’t invertible either.

The algebraC(A) has an involution, given by pointwise complex conjugation.One might thus ask: ifA is a Banach∗-algebra, mustG be a∗-homomorphism, thatis, preserve the involutions? To see that the answer is no, look at the disk algebraagain. The Gelfand dual of the disk algebra is the closed disk (exercise!) and the

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Gelfand transform is the identity map, but the involution on the disk algebra is ofcourse not pointwise complex conjugation.

It turns out that this is related to the existence in the disk algebra of “selfad-joint” elements whose spectrum is not real (one says that the disk algebra is notsymmetric). For instance the functionz is “selfadjoint” in the disk algebra, but itsspectrum is the whole disk. One has a simple

(1.11) LEMMA : Let A be a commutative unital Banach∗-algebra. Then theGelfand transform is a∗-homomorphism if and only if the spectrum of every“selfadjoint” element ofA is real.

PROOF: In view of the Theorem above, the hypothesis implies that a “selfad-joint” element ofA has real-valued Gelfand transform. Now split a generala ∈ Ainto real and imaginary parts:

a =a+ a∗

2+a− a∗

2and apply the Gelfand transform separately to each.

(1.12) PROPOSITION: A selfadjoint element of aC∗-algebra has real spec-trum. Consequently, the Gelfand transform for a commutativeC∗-algebra is a∗-homomorphism.

PROOF: Let a ∈ A with a = a∗ whereA is a unitalC∗-algebra. Letλ ∈ R.Then from theC∗-identity

‖a± λi‖2 = ‖a2 + λ21‖ 6 ‖a‖2 + λ2.

Consequently the spectrum ofa lies within the lozenge-shaped region shown in thediagram??. As λ → ∞ the intersection of all these lozenge-shaped regions is theinterval[−‖a‖, ‖a‖] of the real axis.

(1.13)COROLLARY: The Gelfand transform for a commutativeC∗-algebra is anisometric∗-homomorphism (in particular, it is injective).

PROOF: This follows from the facts thatG preserves the spectrum and theinvolution, and that the norm on aC∗-algebra is determined by the spectral radius.

What about therangeof the Gelfand transform?

(1.14)LEMMA : The Gelfand transform for a commutativeC∗-algebra is surjec-tive.

PROOF: The range ofG is a∗-subalgebra ofC(A) which separates points ofA (for tautological reasons). By the Stone-Weierstrass theorem, then, the range ofG is dense. But the range is alsocomplete, since it is isometrically isomorphic (viaG) to a Banach algebra; hence it is closed, and thus is the whole ofC(A).

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Putting it all together.

(1.15)THEOREM: If A is a commutative unitalC∗-algebra, thenA is isometri-cally isomorphic toC(X), for some (uniquely determined up to homeomorphism)compact Hausdorff spaceX. Any unital∗-homomorphism of commutative unitalC∗-algebras is induced (contravariantly) by a continuous map of the correspond-ing compact Hausdorff spaces. Consequently, the Gelfand transform gives riseto an equivalence of categories between the category of commutative unitalC∗-algebras and the opposite of the category of compact Hausdorff spaces.

This is theGelfand-Naimark Theorem. There is a non-unital version as well,which is proved by unitalization: every commutativeC∗-algebra is of the formC0(X) for some locally compact Hausdorff spaceX.

Let A be a unitalC∗-algebra, not necessarily commutative, and leta ∈ Abe a normal element. Then the unitalC∗-subalgebraC∗(a) ⊆ A generated bya is commutative, so according to Gelfand-Naimark it is of the formC(X) forsome compact Hausdorff spaceX. We ask: What isX? Notice that everyhomomorphismα : C∗(a) → C is determined by the complex numberα(a). Thus

a defines a continuous injectionC∗(a) → C.

(1.16)PROPOSITION: Leta be a normal element of a unitalC∗-algebraA. Thenthe Gelfand transform identifiesC∗(a) withC(sp(a)), the continuous functions onthe spectrum (inA) of a. Under this identification, the operatora corresponds tothe identity functionz 7→ z.

Notice that it is a consequence of this proposition that, ifa ∈ A ⊆ B, whereA andB areC∗-algebras, thenspA(a) = spB(a). It is for this reason that wedo not usually need to specify theC∗-algebra when discussing the spectrum of anelement.

PROOF: What is the image of the injectiona : C∗(a) → C defined above? It issimply the spectrum of the functiona considered as an element of the commutative

Banach algebraC(C∗(a)). By Gelfand’s theorem above, this is the same thing asthe spectrum ofa, considered as an element of the commutative Banach algebra

C∗(a). Thusα is a homeomorphism2 from C∗(a) to spC∗(a)(a).It remains to prove thatspC∗(a)(a) = spA(a). Clearly, if x ∈ C∗(a) is

invertible in C∗(a), then it is invertible inA. On the other hand suppose thatx ∈ C∗(a) fails to be invertible inC∗(a). Thenx is a continuous function on a

compact Hausdorff space which takes the value zero somewhere, say atα ∈ C∗(a).Using bump functions supported in neighborhoods ofα one can produce, for eachε > 0, an elementyε ∈ C∗(a) such that‖yε‖ = 1 and‖xyε‖ < ε. But if x

2A continuous bijective map of compact Hausdorff spaces is a homeomorphism.

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were invertible inA, this would be impossible for sufficiently smallε (less than‖x−1‖−1).

A slight reformulation of this yields the so-calledfunctional calculusfor C∗-algebras. Namely, givena a normal element of aC∗-algebraA, and given acontinuous functionf on sp(a), there is a unique element ofA (in fact ofC∗(a))which corresponds tof under the Gelfand transform forC∗(a). What is usuallycalled the functional calculus is simply the following:

DECISION: We denote the element described above byf(a).

This procedure has all the properties which the notation would lead you toexpect (so long as you confine yourself to thinking about only a single normalelement). For instance,(f + g)(a) = f(a) + g(a), (f · g)(a) = f(a)g(a),(f g)(a) = f(g(a)) and so on. The proofs all follow from the fact thatthe Gelfand transform is an isomorphism, except the last which uses polynomialapproximation.

(1.17)REMARK : Finally we make some comments about how all this goes in thenon-unital case. LetA be a non-unitalC∗-algebra and leta ∈ A be normal. Wedefine thespectrumof a in this case to be its spectrum inA. It always containszero, and it is easy to see that

C∗0 (a) ∼= C0(sp(a) \ 0),

whereC∗0 (a) denotes the non-unitalC∗-subalgebra ofA generated bya. So, inthis case we have a functional calculus for continuous functions on the spectrumthat vanish at zero.

1.2 Positive Elements and Order

Let T be a selfadjoint operator on a Hilbert spaceH. One says thatT is positive3

if the quadratic form defined byT is positive semidefinite, that is, if

〈Tv, v〉 > 0 for all v ∈ H.

The positive operators form acone— the sum of two positive operators is positive,as is any positive real multiple of a positive operator. It is easy to prove that thefollowing are equivalent for a selfadjointT ∈ B(H):

(a) The spectrum ofT is contained in the positive reals.

3This means ‘non-negative’ in this context, not ‘strictly positive’.

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(b) T is the square of a selfadjoint operator.

(c) T = S∗S for some operatorS.

(d) T is positive.

Indeed, (a) implies (b) by using the functional calculus to construct a square rootof T , (b) obviously implies (c) which implies (d) since〈S∗Sv, v〉 = ‖Sv‖2, andfinally if (d) holds let us define

T+ = f(T ), T− = g(T ) (1.18)

wheref(x) = max(x, 0) andg(x) = −min(x, 0). We haveT = T+ − T−, andfor anyw ∈ H,

〈TT−w, T−w〉 = −〈T 2−w, T−w〉 = −‖T 3/2

− w‖2 6 0.

So positivity implies thatT− = 0, which yields (a).A key challenge in setting up the theory of abstractC∗-algebras is to prove

these properties without using quadratic forms on Hilbert space as a crutch. Thiswas done by Kelley and Vaught some years afterC∗-algebras were invented! (Thistime lapse is responsible for a now-obsolete terminological difference in the olderliterature: there was a distinction between ‘B∗-algebras’ and ‘C∗-algebras’, andKelley and Vaught’s theorem showed that this distinction was otiose.)

(1.19)PROPOSITION: LetA be a unitalC∗-algebra. The following conditions ona selfadjointa ∈ A are equivalent:

(a) The spectrum ofa is contained in the positive reals.

(b) a is the square of a selfadjoint element ofA.

(c) ‖t1− a‖ 6 t for all t > ‖a‖.

(d) ‖t1− a‖ 6 t for somet > ‖a‖.

PROOF: Conditions (a) and (b) are equivalent by a functional calculus argu-ment. Suppose that (a) holds. Thensp(a) is contained in the interval[0, ‖a‖] ofthe real axis and so the supremum of the functiont1− λ, for λ ∈ sp(a), is at mostt; this proves (c) which implies (d). Finally, assume (d). Then|t − λ| < t for allλ ∈ sp(a); soλ is positive onsp(a), yielding (a).

We will call a ∈ A positiveif it is selfadjoint and satisfies one of these fourconditions. Part (d) of this characterization implies that the positive elements forma cone:

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(1.20)PROPOSITION: The sum of two positive elements ofA is positive.

PROOF: Let a′, a′′ be positive elements ofA. Let t′, t′′ be as in (d). Leta = a′ + a′′ and t = t′ + t′′. Then by the triangle inequalityt > ‖a‖ and‖t1− a‖ 6 ‖t′1− a′‖+ ‖t′′1− a′′‖ 6 t′ + t′′ = t. Soa is positive.

We can therefore define a (partial) order onAsa by settinga 6 b if a − bis positive. This order relation is very useful, especially in connection with vonNeumann algebras, but it has to be handled with care — not every ‘obvious’property of the order is actually true. For instance, it is not true in general thatif 0 6 a 6 b, thena2 6 b2. In Proposition 1.22 we will give some true propertiesof the order relation. These depend on the following tricky result.

(1.21)THEOREM: (KELLEY–VAUGHT) An elementa of a unitalC∗-algebraAis positive if and only ifa = b∗b for someb ∈ A.

PROOF: Any positive element can be written in this form withb = a1/2. Toprove the converse we will argue first thatb∗b cannot be anegativeoperator (exceptzero). Indeed, putb = x+ iy with x andy selfadjoint; then

b∗b+ bb∗ = 2(x2 + y2) > 0.

Thus if b∗b is negative,bb∗ = 2(x2 + y2) − b∗b is positive. But the spectra ofb∗bandbb∗ are the same (apart possibly from zero), by a well-known algebraic result;we deduce thatsp(b∗b) = 0, and henceb∗b = 0.

Now for the general case: suppose thata = b∗b; certainlya is selfadjoint, sowe may writea = a+ − a− where the positive elementsa+ anda− are defined asin Equation 1.18. We want to show thata− = 0. Let c = ba

1/2− . Then we have

c∗c = a1/2− aa

1/2− = −a2

−

soc∗c is negative. Hence it is zero, by the special case above; soa− = 0 anda ispositive.

(1.22)PROPOSITION: Letx andy be positive elements of a unitalC∗-algebraA,with 0 6 x 6 y. Then

(a) We havex 6 ‖x‖1 (and similarly fory); moreover‖x‖ 6 ‖y‖.

(b) For anya ∈ A, a∗xa 6 a∗ya.

(c) If x andy are invertible, we have0 6 y−1 6 x−1.

(d) For all α ∈ [0, 1] we havexα 6 yα.

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PROOF: The first part of (a) is obvious from the functional calculus, and itimplies the second part by arguing that

x 6 y 6 ‖y‖1

so x 6 ‖y‖1 whence‖x‖ 6 ‖y‖. Item (b) becomes obvious if we use Theo-rem 1.21; ifx 6 y theny − x = c∗c, soa∗ya − a∗xa = (ca)∗ca is positive.For (c), if x 6 y put z = x1/2y−1/2; thenz∗z = y−1/2xy−1/2 6 1 (by (b)). Butz∗z 6 1 if and only if zz∗ 6 1, which is to say thatx1/2y−1x1/2 6 1; and applying(b) again we gety−1 6 x−1. Finally (d) follows from (c) once we observe that thefunctionλ 7→ λα, defined onR+, is a norm limit of convex combinations of thefunctions

gξ(λ) =1ξ

(1− 1

1 + ξλ

);

this follows from the integral formula

λα = cα

∫gξ(λ)ξα dξ

which is valid for0 6 α 6 1.

1.3 Moore–Smith Convergence

The strong topology and other related matters forC∗ and von Neumann algebrasare frequently discussed in terms ofnets. Here is a brief reminder about theirtheory. The proofs will be left to the reader.

(1.23)DEFINITION: A directed setΛ is a partially ordered set in which any twoelements have an upper bound. Anet in a topological spaceX is a function froma directed set toX. Usually we denote a net like this:xλλ∈Λ.

For example, a sequence is just a net based on the directed setN. The followingexample of a directed set will be important to us:

(1.24)LEMMA : LetA be aC∗-algebra and letD ⊆ A be the set of allu ∈ Asawith 0 6 u < 1. ThenD is directed by the order relation.

PROOF: We must show that any two elements ofD have an upper bound. Let

f(t) =t

1− t, g(t) =

t

1 + t= 1− 1

1 + t.

The functionf is a homeomorphism of[0, 1) on [0,∞), andg is its inverse. Givenx′, x′′ ∈ D let x = g(f(x′) + f(x′′)), defined by the functional calculus. Thenx is in D also. Clearlyf(x′) 6 f(x′) + f(x′′), and since the functiong isorder-preserving by Proposition 1.22(c) we getx′ 6 x. Similarly x′′ 6 x, sox will do as the required upper bound.

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Note that since the functionsf andg send 0 to 0, this argument is valid even inanon-unitalC∗-algebra.

(1.25)DEFINITION: LetU ⊆ X. A netxλλ∈Λ in X is eventually inU if thereis someλ0 ∈ Λ such thatxλ ∈ U for all λ > λ0. If there is some pointx ∈ X suchthatxλ is eventually in every neighborhood ofx, we say thatxλ convergestoX.

As an exercise in manipulating this definition, the reader may prove that a spaceX is Hausdorff if and only if every net converges to at most one point. Similarlya functionf : X → Y is continuous if and only if, wheneverxλ is a net inXconverging tox, then netf(xλ) converges inY to f(x). The basic idea here isto use the system of open neighborhoods of a point, directed by (reverse) inclusion,as the parameter space of a net.

Warning: a convergent net in a metric space need not be bounded.A subnetof a net is defined as follows. Letxλλ∈Λ be a net based on the

directed setΛ. Let Ξ be another directed set. A functionλ : Ξ → Λ is called afinal function if, for everyλ0 ∈ Λ, there isξ0 ∈ Ξ such thatλ(ξ) > λ0 wheneverξ > ξ0. For such a function the netxλ(ξ)ξ∈Ξ is a net based on the directed setΞ;it is called asubnetof the originally given one. Notice that this notion of subnetis more relaxed than the usual one of subsequence; we allow some repetition andbacktracking.

(1.26)PROPOSITION: A topological spaceX is compact if and only if every netin X has a convergent subnet.

For completeness we mention the notion ofuniversal net. A net isuniversalif, for every subsetA of X, the net is either eventually inA or eventually in thecomplement ofA. Using the axiom of choice it can be shown that every net has auniversal subnet. This makes possible a rather short proof of Tychonoff’s theorem.

(1.27)THEOREM: A product of compact topological spaces is compact.

PROOF: Let X =∏Xα be such a product,πα the coordinate projections.

Pick a universal netxλ in X; then a one-line proof shows that each the netsπα(xλ) is universal inXα, and hence is convergent (a universal net with aconvergent subnet must itself be convergent), say toxα. But then by definitionof the Tychonoff topology the netxλ converges to the pointx whose coordinatesarexα. Thus every universal net inX converges, and since every net has a universalsubnet, this proves thatX is compact.

1.4 Approximate Units, Ideals, and Quotients

(1.28)DEFINITION: LetA be aC∗-algebra, probably without unit. Anapproxi-

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mate unitfor A is a netuλ in A which is increasing, has0 6 uλ 6 1 for all λ,and such that moreover for eacha ∈ A, limuλa = lim auλ = a.

For example, letA = C0(R). Then the sequence of functions

un(x) = e−x2/n ∈ A

tends to 1 uniformly on compact sets, and so constitutes an approximate unit.A very similar argument produces an approximate unit for anycommutativeC∗-algebraC0(X) (if A is separable, so thatX is second countable, our approximateunit will be a sequence; otherwise we must use a net).

(1.29)THEOREM: EveryC∗-algebra has an approximate unit.

PROOF: For our approximate unit we just take the setD of all elementsu with0 6 u < 1), ordered by the usual order relation. It is directed (Lemma 1.24).

To prove the approximate unit property we first note that for anypositivea ∈ Athere is asequenceun in D with ‖(1− un)x‖ and‖x(1− un)‖ tending to zero.Indeed,C∗0 (a) is a commutative, separableC∗-algebra and we may takeun tobe an approximate unit for it, as in the example above. Ifu, u′ ∈ D with u 6 u′,then4

‖(1− u′)x‖2 = ‖x(1− u′)2x‖ 6 ‖x(1− u′)x‖6 ‖x(1− u)x‖ 6 ‖x‖‖(1− u)x‖.

SinceD is directed this allows us to infer that the existence of the sequenceunin D for which limn ‖(1 − un)x‖ = 0 implies that the limitlimu∈D ‖(1 − u)x‖exists and is zero also. This shows thatD acts as an approximate identity at leaston positive elements, and sincex∗x is positive for anyx, another simple argumentshows thatD acts as an approximate identity on all elements.

By anideal in aC∗-algebra we mean a norm-closed, two-sided ideal. We couldalso require∗-closure but this fact is automatic:

(1.30)PROPOSITION: Any idealJ in aC∗-algebraA is closed under the adjointoperation (that isJ = J∗).

PROOF: Let x ∈ J and letuλ be an approximate unit for theC∗-algebraJ ∩ J∗. Using theC∗-identity

‖x∗(1− uλ)‖2 = ‖xx∗(1− uλ)− uλxx∗(1− uλ)‖

which tends to zero. It follows thatx∗uλ → x∗, whencex∗ ∈ J .

4It is convenient here and elsewhere to write the approximate unit condition in the formlim ‖x(1 − u)‖ = 0. Formally speaking this is an identity in theunitalizationof A. However,the elementsx(1 − u) = x − xu in fact belong toA itself.

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It is an important fact that the quotient of aC∗-algebra by an ideal (which iscertainly a Banach algebra in the induced norm) is in fact aC∗-algebra. We needapproximate units to check this.

(1.31)THEOREM: LetA be aC∗-algebra andJ an ideal inA. ThenA/J is aC∗-algebra with the induced norm and involution.

PROOF: A/J is a Banach algebra with involution (note that we needJ = J∗

here). Recall that the quotient norm is defined as follows:

‖[a]‖ = infx∈J

‖a+ x‖.

Here[a] denotes the coset (inA/J) of the elementa ∈ A. I claim that

‖[a]‖ = limu‖a(1− u)‖

whereu runs over an approximate identity forJ . Granted this it is plane sailing totheC∗-identity, since we get

‖[a∗][a]‖ = limu‖a∗a(1− u)‖ > lim

u‖(1− u)a∗a(1− u)‖

= limu‖a(1− u)‖2 = ‖[u]‖2.

To establish the claim notice that certainly‖a(1 − u)‖ > ‖[a]‖ for everya. Onthe other hand, givenε > 0 there existsx ∈ J with ‖a − x‖ 6 ‖[a]‖ + ε. It isfrequently true that‖x(1− u)‖ 6 ε also and then

‖a(1− u)‖ 6 ‖[a]‖+ 2ε

by the triangle inequality. This establishes the claim.

We can use this result to set up the fundamental ‘isomorphism theorems’ in thecategory ofC∗-algebras.

(1.32)LEMMA : Any injective homomorphism ofC∗-algebras is an isometry.

PROOF: Let α : A → B be such a homomorphism and leta ∈ A. Assumewithout loss of generality thatA, B are unital. It suffices to show that‖a∗a‖ =‖α(a∗a)‖, by theC∗-identity. But now we may restrict attention to theC∗-subalgebras generated bya∗a and its image, and thus we may assume without lossof generality thatA andB are commutative. In this caseA = C(X), B = C(Y ),andα is induced by a surjective mapY → X; this makes the result obvious.

14

(1.33)COROLLARY: Letα : A→ B be any homomorphism ofC∗-algebras. Thenα is continuous with norm6 1, and it can be factored as

A A/ ker(α) B

where the first map is a quotient map and the second is an isometric injection whoserange is aC∗-subalgebra ofB.

PROOF: Continuity ofα follows from the spectral radius formula, and the restof the proof is standard from algebra.

We can also prove the ‘diamond isomorphism theorem’.

(1.34)PROPOSITION: LetA be aC∗-algebra. LetJ be an ideal inA and letBbe aC∗-subalgebra ofA. ThenB + J is aC∗-subalgebra ofA and there is anisomorphismB/(B ∩ J) = (B + J)/J .

PROOF: The ∗-homomorphismB → A → A/J has range aC∗-subalgebraof A/J , isomorphic toB/B ∩ J by the previous theorem. Thus the inverse imagein A of thisC∗-subalgebra (namelyB + J) is closed, and so is aC∗-algebra also.The result now follows by the usual algebra argument.

WhenB is another idealJ there is a useful addendum to this result: for twoidealsI andJ in A, their intersection is equal to their product. To see this, observe

(I ∩ J)2 ⊆ IJ ⊆ I ∩ J.

But for anyC∗-algebraB, we haveB = B2, by an easy functional calculusargument.

1.5 Topologies onB(H)

Let H be a Hilbert space. TheC∗-algebraB(H) carries a number of importanttopologies in addition to the familiarnorm topology. These areweak topologies, inother words they are defined to have the minimal number of open sets required tomake certain maps continuous.

Theweak operator topologyis the weakest topology making the mapsB(H) →C given byT 7→ 〈Tu, v〉 continuous, for all fixedu, v ∈ H. Similarly, thestrongoperator topologyis the weakest topology making the mapsB(H) → H given byT 7→ Tu continuous. It is clear that the weak operator topology is weaker (feweropen sets) than the strong operator topology, which in turn is weaker than the normtopology. These relations are strict.

15

(1.35)REMARK : A setW ⊆ B(H) is a strong neighborhood ofT0 ∈ B(H) ifthere existu1, . . . , un ∈ H such that

n∑i=1

‖(T − T0)ui‖2 < 1 =⇒ T ∈W.

Similarly,W is a weak neighborhood ofT0 if there existu1, . . . , un andv1, . . . , vn

such thatn∑

i=j=1

|〈(T − T0)ui, vi〉| < 1 =⇒ T ∈W.

Here are some easy properties of these topologies. Addition is jointly continu-ous in both topologies; multiplication is not jointly continuous in either topology.However, multiplication is jointly strongly continuous when restricted to the unitball. Multiplication by afixedoperator (on the left or the right) is both weakly andstrongly continuous. The unit ball ofB(H) is weakly compact, but not stronglycompact. A weakly (or strongly) convergentsequenceis bounded (use the uniformboundedness principle).

It is easy to see that the adjoint operationT 7→ T ∗ is weakly continuous.However it isnot strongly continuous: ifV denotes the unilateral shift operator on`2, defined byV (x1, x2, x3, . . .) = (0, x1, x2, . . .), then the sequence(V ∗)n tendsstrongly to 0 while the sequenceV n does not. It is sometimes worth noting thatthe adjoint operationis strongly continuous on the setN of normal operators. Theidentity

‖Tu‖2 = 〈T ∗Tu, u〉 = 〈TT ∗u, u〉 = ‖T ∗u‖2

which is valid forT ∈ N, shows that if a netTλ of normal operators convergesstrongly to zero, then the netT ∗λ converges strongly to zero also. Unfortunatelythe general case does not follow from this, sinceN is not closed under addition.Instead, use the identity (valid forS, T ∈ N)

‖(S − T )∗u‖2 = ‖Su‖2 − ‖Tu‖2 + 2<〈(T − S)T ∗u, u〉

to obtain the inequality

‖(S − T )∗u‖2 6 ‖(S − T )u‖(‖Su‖+ ‖Tu‖) + 2‖(T − S)T ∗u‖‖u‖.

From this it follows that ifSλ is a net of normal operators converging strongly toa normal operatorT , thenS∗λ converges strongly toT ∗. Warning: It is importantthatT be normal too.

16

(1.36)PROPOSITION: The only strongly continuous linear functionals onB(H)are of the form

T 7→n∑

i=1

〈Tui, vi〉

for certain vectorsu1, . . . , un andv1, . . . , vn in H. (Notice that these functionalsare also weakly continuous.)

PROOF: Let ϕ : B(H) → C be a strongly continuous linear functional. Thenby definition of the strong operator topology there exist vectorsu1, . . . , un suchthat

|ϕ(T )| 6

(n∑

i=1

‖Tui‖2

) 12

.

Let L be the linear subspace ofHn comprised of those vectors of the form(Tu1, . . . , Tun) for T ∈ B(H). We may define a linear functional of norm6 1onL by

(Tu1, . . . , Tun) 7→ ϕ(T ).

Extend this functional to the whole ofHn by the Hahn-Banach theorem, and thenrepresent it by an inner product with a vector(v1, . . . , vn) in Hn. Restricting backdown toL we find that

ϕ(T ) =n∑

i=1

〈Tui, vi〉

as required.

Since the weak and strong duals ofB(H) are the same we find

(1.37) COROLLARY: The weak and strong operator topologies have the sameclosed convex sets.

There is an intimate relation between positivity and the strong (and weak)operator topologies:

(1.38)LEMMA : If a bounded net of positive operators converges weakly to zero,then it converges strongly to zero.

PROOF: Let Tλ be such a net inB(H). Letv ∈ H. By the Cauchy-Schwarzinequality

‖Tλv‖4 = 〈T 1/2λ v, T

3/2λ v〉2 6 〈Tλv, v〉〈T 3

λv, v〉 6 ‖Tλ‖3‖v‖2〈Tλv, v〉

and the right-hand side tends to zero by weak convergence.

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(1.39)PROPOSITION: Let Tλ be an increasing net of selfadjoint operators inB(H), and suppose thatTλ is bounded above, byS say. ThenTλ is stronglyconvergent to an operatorT with T 6 S.

PROOF: For each fixedv ∈ H the limit of the increasing net〈Tλv, v〉 of realnumbers exists and is bounded by〈Sv, v〉. The functionv 7→ limλ〈Tλv, v〉 is aquadratic form, and using the polarization identity one may define a linear operatorT such that

〈Tv, v〉 = limλ〈Tλv, v〉.

By construction, the netTλ converges toT in the weak operator topology. SoT − Tλ is a bounded net of positive operators converging weakly to zero; byLemma 1.38 it converges strongly to zero.

1.6 Continuity of the Functional Calculus

In this section we’ll look at the following question: letf be a function onR, andconsider the map

T 7→ f(T )

from the selfadjoint part ofB(H) to B(H). When is this a continuous map,relative to the various topologies that we have been discussing onB(H)?

(1.40) PROPOSITION: Let f : R → C be any continuous function. ThenT 7→f(T ) is continuous relative to the norm topologies onB(H)saandB(H).

PROOF: Since neighborhoods in the norm topology are bounded, and thespectral radius is bounded by the norm, only the behavior off on a compactsubinterval ofR is significant for continuity. Thus we may and shall assume thatf ∈ C0(R).

If f(λ) = (λ + z)−1, for z ∈ C \ R, thenT 7→ f(T ) is continuous; for wehave

f(S)− f(T ) = f(S)(T − S)f(T ) (1.41)

and‖f(S)‖ and‖f(T )‖ are bounded by|=z|−1.Let A denote the collection of all functionsf ∈ C0(R) for whichT 7→ f(T )

is norm continuous. It is easy to see thatA is aC∗-subalgebra. Since it containsthe functionsλ 7→ (z + λ)−1 it separates points ofR. ThusA = C0(R) by theStone-Weierstrass theorem.

Now we will look at continuity in thestrong topology. Note that strongneighborhoods need not be bounded, so that the behavior of the functionf atinfinity will become important. We say that a functionf : R → C is of lineargrowth if |f(t)| 6 A|t|+B for some constantsA andB.

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(1.42)PROPOSITION: Let f : R → C be a continuous function of linear growth.ThenT 7→ f(T ) is continuous relative to the strong topologies onB(H)sa andB(H).

PROOF: Let V denote the vector space of all continuous functionsf : R → Cfor which T 7→ f(T ) is strongly continuous, and letV b denote the boundedelements ofV . Note that iff belongs toV (or V b), then so doesf ; this followsfrom the strong continuity of the adjoint on the setN of normal operators. Weclaim first thatV bV ⊆ V ; this follows from the identity

fg(S)− fg(T ) = f(S)(g(S)− g(T )) + (f(S)− f(T ))g(T )

together with the following properties of the strong topology: ifTλ → T strongly,then(Tλ − T )R → 0 strongly for any fixed operatorR, whileRλ(Tλ − T ) → 0strongly for any bounded netRλ. In particular,V b is closed under multiplication,so it is an algebra. One sees easily thatV b is a C∗-algebra. Arguing withEquation 1.41 exactly as above, and using the properties of multiplication in thestrong topology that we just mentioned, we see thatV b ⊇ C0(R).

Now letf be a function of linear growth. Then the functiont 7→ (1+t2)−1f(t)belongs toC0(R), hence toV b. The identity function of course belongs toV , henceso does the functiong : t 7→ t(1 + t2)−1f(t), as it is the product of a member ofV and a member ofV b. The functiong is bounded, so belongs toV b, hence thefunctiont 7→ tg(t) = t2(1 + t2)−1f(t) belongs toV . Finally, then, the function

t 7→ (1 + t2)−1f(t) + t2(1 + t2)−1f(t) = f(t)

belongs toV , as required.

1.7 Von Neumann Algebras

LetH be a Hilbert space.

(1.43)DEFINITION: A unital∗-subalgebra ofB(H) which is closed in the strongoperator topology5 is called avon Neumann algebraof operators onH.

If M is any selfadjoint6 set of operators onH its commutantM ′ is the set

M ′ = T ∈ B(H) : ST = TS ∀ S ∈M.

It is a von Neumann algebra. For tautological reasons,M ′′ (thedouble commutant)containsM . Thedouble commutant theoremis

5By Corollary 1.37 it makes no difference to replace ‘strong’ by ‘weak’ here.6This means thatT ∈ M impliesT ∗ ∈ M .

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(1.44)THEOREM: A unitalC∗-subalgebra ofB(H) is a von Neumann algebraif and only if it is equal to its own double commutant. In general, the doublecommutant ofS ⊆ B(H) is the smallest von Neumann algebra of operators onHthat containsS.

PROOF: The thing we have to prove is this: ifA is a unitalC∗-subalgebra ofB(H) andT ∈ A′′, thenT is the strong limit of a net of operators inA. In otherwords, we have to show that for any finite listv1, . . . , vn of vectors inH one canfind an operatorR ∈ A such that

n∑i=1

‖(T −R)vi‖2 < 1.

Let us first address the casen = 1, andv1 = v. LetP be the orthogonal projectionoperator onto theA-invariant closed subspace generated byv (that isAv). ThenPbelongs to the commutantA′ of A, because ifR ∈ A thenR preserves the rangeof P andR∗ preserves the kernel ofP . SinceT ∈ A′′, T commutes withP , andin particularPTv = TPv = Tv, soTv belongs to the range ofP . That is,Tv isa limit of vectors of the formRv,R ∈ A, which is what was required.

Now consider the general case. LetK = H ⊕ · · · ⊕ H (n times) and letv ∈ K be the vector(v1, . . . , vn). Let ρ : B(H) → B(K) be the ‘inflation’∗-homomorphism which is defined by

T 7→

T...

T

(in a more sophisticated language this isT 7→ T ⊗ In). A routine calculationshows thatρ(A)′ consists ofn × n matrices all of whose elements belong toA′,and therefore thatρ(A)′′ = ρ(A′′). Now apply the previous case to theC∗-algebraρ(A), the operatorρ(T ), and the vectorv. We find that there isρ(R) ∈ ρ(A) suchthat‖(ρ(R) − ρ(T ))v‖ < 1, and when written out componentwise this is exactlywhat we want.

It follows from the double commutant theorem that a von Neumann algebrais weakly closed. This can also be proved directly; in fact, each strongly closedconvex set inB(H) is weakly closed.

(1.45)REMARK : Our hypothesis thatA must be unital is stronger than necessary.Suppose simply thatA is non-degenerate, meaning thatAH is dense inH, orequivalently thatRu = 0 for all R ∈ A impliesu = 0. Then the conclusion of thedouble commutant theorem holds. To see this, we need only check that the vector

20

v still belongs toAv — the rest of the proof will go through as before. Indeed, onecan writev = Pv + (1 − P )v, whereP is the projection ontoAv as above, andsinceP ∈ A′ we have, for everyR ∈ A,

R(1− P )v = (1− P )Rv = Rv −Rv = 0;

so by non-degeneracy(1− P )v = 0 andv = Pv ∈ Av.

TheKaplansky density theoremstrengthens the double commutant theorem byshowing that every element ofA′′ can be strongly approximated by aboundednetfrom A (note that strongly convergent nets need not be bounded). The main stepin the proof has already been taken: it is the strong continuity of the functionalcalculus discussed in the preceding section.

(1.46)THEOREM: LetA be a unitalC∗-subalgebra ofB(H) and letM = A′′

be the von Neumann algebra that it generates. The the unit ball ofA is stronglydense in the unit ball ofM , the unit ball ofAsa is strongly dense in the unit ballofMsa, and the unit chunk ofA+ is strongly dense in the unit ball ofM+. Finally,the unitary group ofA is strongly dense in the unitary group ofM .

PROOF: Let’s start with the case ofMsa. Let T ∈ Msa with ‖T‖ 6 1.Using the Double Commutant Theorem, we see thatT is the strong limit of anetTλ of selfadjoint elements ofA.7 The norms of the operatorsTλ need not bebounded; but if we replaceTλ by f(Tλ), wheref(x) = min(1,max(−1, x)), thenf(Tλ) → f(T ) = T by Proposition 1.39. The same argument works forM+, andfor the unitary group if we use the fact (a consequence of the spectral theorem)that in any von Neumann algebraM , any unitaryU can be writtene2πiT with Tselfadjoint and of norm6 1. The last remaining conclusion is that the whole unitball ofA is dense in the unit ball ofM ; this can be proved by a2× 2 matrix trick,or as a consequence of the Russo–Dye Theorem below.

One pleasant feature of von Neumann algebras is the existence in them of apolar decomposition theory, analogous to the familiar representationz = reiθ of acomplex number as a product of a positive real number and a complex number ofmodulus one.

(1.47) DEFINITION: Let H be a Hilbert space. Apartial isometryon H isan operatorV ∈ B(H) which satisfies one of the following four (equivalent)conditions:

7There is a technical point here: The double commutant theorem might produce a netTλ whichis not selfadjoint. We can replaceTλ by 1

2(Tλ + T ∗

λ ) but now we get a net which only convergesweakly to T . However, by taking convex combinations, we can replace weak convergence bystrong convergence (Corollary 1.37), and of course a convex combination of selfadjoint operatorsis selfadjoint.

21

(a) V ∗V is a projection (the ‘initial projection’),

(b) V V ∗ is a projection (the ‘final projection’),

(c) V = V V ∗V ,

(d) V ∗ = V ∗V V ∗.

It is easy to see that these conditions are equivalent, since (c) and (d) are adjointare one another, (c) implies (a) (multiply on the left byV ∗), and if (a) is true thenI − V ∗V is a projection with rangeKer(V ), so thatV (I − V ∗V ) = 0 which is(c). One should think ofV as giving an isomorphism from the range of the initialprojection to the range of the final projection.

(1.48) DEFINITION: Let T ∈ B(H). A polar decompositionfor T is a factor-izationT = V P , whereP is a positive operator andV is a partial isometry withinitial spaceran(P ) and final spaceran(T ).

Here we are using the notationran(T ) for the closure of the range of theoperatorT . Note thatV and P need not commute, so that there are reallytwo notions of polar decomposition (left and right handed); we have made theconventional choice.

(1.49)PROPOSITION: Every operatorT ∈ B(H) has a unique polar decomposi-

tion T = V P . The operatorP = |T | = (T ∗T )12 belongs toC∗(T ). The partial

isometryV belongs toW ∗(T ), the von Neumann algebra generated byT . If Thappens to be invertible, thenV is unitary; it belongs toC∗(T ) in this case.

PROOF: Let T = V P be a polar decomposition. ThenT ∗T = P ∗V ∗V P =P 2, soP is the unique positive square root ofT ∗T , which we denote by|T |. FixingthisP let us now try to construct a polar decomposition. Forv ∈ H,

‖Pv‖2 = 〈P ∗Pv, v〉 = 〈T ∗Tv, v〉 = ‖Tv‖2

so an isometryran(P ) → ran(T ) is defined byPv 7→ Tv. Extending by zeroon the orthogonal complement we get a partial isometryV with the property thatT = V P ; and it is uniquely determined since the definition of polar decompositionrequires thatV (Pv) = Tv and thatV = 0 on the orthogonal complement ofran(P ).

Let us check thatV ∈W ∗(T ); using the bicommutant theorem, it is enough toshow thatV commutes with every operatorS that commutes withT andT ∗. WehaveH = Ker(P ) ⊕ ran(P ) so it suffices to check thatV Sv = SV v separatelyfor v ∈ Ker(P ) = Ker(T ) and forv = Py. In the first case,STv = TSv = 0 soSv ∈ Ker(T ), and thenV v = 0, V Sv = 0. In the second case

V Sv = V SPy = V PSy = TSy = STy = SV Py = SV v

22

using the fact thatP belongs to theC∗-algebra generated byT .Finally, if T is invertible then so isT ∗T , and we haveV = T (T ∗T )−

12 ; this

shows thatV ∈ C∗(T ) in this case, and unitarity is a simple check.

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2 Representation Theory

We will discuss representations ofC∗ and von Neumann algebras on Hilbert space,and prove various forms of the spectral theorem.

2.1 Representations and States

(2.1) DEFINITION: Let A be a C∗-algebra. A representationof A is a ∗-homomorphismρ : A → B(H) for some Hilbert spaceH. The representationis faithful if ρ is injective, andnon-degenerateif ρ(A)H = H. The representationis irreducibleif there are no proper closed subspaces ofH that are invariant underρ(A).

(2.2) REMARK : Suppose thatρ : A → B(H) is a representation and thatK isa closed subspace ofH. Then the orthogonal projectionP ontoK belongs tothe commutantρ[A]′ if and only if K is invariant under the action ofA. It is aconsequence of the spectral theorem that every von Neumann algebra larger thanC1 contains a proper projection, and so we see thatρ is irreducible if and only ifthe commutantρ[A]′ is equal toC1, or equivalently if and only if the bicommutantρ[A]′′ is equal toB(H).

From the bicommutant theorem,ρ[A]′′ is the strong closure ofρ[A]. It followsthen that, ifρ is irreducible, then given any two vectorsu, v ∈ H with u 6= 0,and givenε > 0, there isa ∈ A such that‖ρ(a)u − v‖ < ε. By an iterativeargument, Kadison improved this result: he showed that in fact there isa ∈ A withρ(a)u = v. This is calledKadison’s transitivity theorem, and it if course impliesthat an irreducible representation of aC∗-algebra hasno invariant subspaces,closed or not. We will (perhaps) prove this theorem later on.

Representations are studied by way ofstates.

(2.3) DEFINITION: A linear functionalϕ : A → C on aC∗-algebraA is calledpositiveif ϕ(a∗a) > 0 for all a ∈ A. A stateis a positive linear functional of normone.

Any positive linear functional is bounded; for, ifϕ were positive and un-bounded, we could find elementsak ∈ A+ of norm one withϕ(ak) > 4k, andthena =

∑2−kak is an element ofA+ satisfyingϕ(a) > 2−kϕ(ak) > 2k for

all k, contradiction. Thus there is little loss of generality in restricting attentionto states. A state onC(X) is a Radon probability measure onX, by the Rieszrepresentation theorem from measure theory. Ifρ : A→ B(H) is a representationandv ∈ H is a unit vector, a stateσv : A→ C (called avector state) is defined byσv(a) = 〈ρ(a)v, v〉. This is the link between states and representations. Note thatthe state defined by the Radon probability measureµ is a vector state ofC(X);

24

takeH = L2(X,µ), ρ the representation by multiplication operators, andv theconstant function 1.

If σ is a state (or just a positive linear functional) onA then we may define a(semidefinite) ‘inner product’ onA by

〈a, b〉σ = σ(b∗a).

The Cauchy–Schwarz inequality continues to hold (with the usual proof): we have

|〈a, b〉σ|2 6 〈a, a〉σ 〈b, b〉σ.

Note that〈a, a〉σ is different from theC∗-norm‖a‖2.

(2.4) PROPOSITION: A bounded linear functionalϕ : A → C is positive if andonly if limϕ(uλ) = ‖ϕ‖ for some approximate unituλ for A; and in particular, ifand only ifϕ(1) = ‖ϕ‖ in caseA happens to be unital.

PROOF: There is no loss of generality in supposing that‖ϕ‖ = 1.Suppose thatϕ is positive. Thenϕ(uλ) is an increasing net of real numbers,

bounded above by 1, and therefore convergent to some realr 6 1. For eacha inthe unit ball ofA, we have by Cauchy–Schwarz

|ϕ(auλ)|2 6 ϕ(a∗a)ϕ(u2λ) 6 ϕ(uλ) 6 r.

Thus, sinceuλ is an approximate unit,|ϕ(a)|2 6 r, whencer > 1 on taking thesupremum overa in the unit ball ofA. Since we already knowr 6 1, this givesr = 1 as required.

Conversely suppose thatlimϕ(uλ) = 1. First we will prove thatϕ takes realvalues on selfadjoint elements. Take a selfadjoint elementa ∈ A, of norm one, andwriteϕ(a) = x+ iy ∈ C; we want to prove thaty = 0; assume for a contradictionthaty > 0. For eachn one can extract aun from the approximate unit which hasϕ(un) > 1− 1/n3 and commutes well enough witha that we have the estimate

‖nun + ia‖2 6 n2 + 2

(of course if it were the case thatuna = aun exactly then the bound would ben2 + 1). Therefore

x2 +(n(1− 1/n3)2 + y

)2

6 |ϕ(nun + ia)|2 6 n2 + 2

as this yields a contradiction asn→∞.We have now shown thatϕ is a selfadjoint linear functional; to show positivity,

let x ∈ A with 0 6 x 6 1. Then for eachλ, uλ − x is in the unit ball and soϕ(uλ−x) 6 1. Letλ→∞ to obtain1−ϕ(x) 6 1, whenceϕ(x) > 0 as required.

25

(2.5)COROLLARY: Every state on a non-unitalC∗-algebra extends uniquely to astate on its unitalization.

PROOF: Let σ : A → C be a state, and defineσ on the unitalizationA in theonly possible way,

σ(a+ µ1) = σ(a) + µ.

We need only to know thatσ is positive (and hence continuous), and this followsif we write

σ(a+ µ1) = limσ(a+ µuλ)

using the previous proposition.

2.2 The Gelfand–Naimark–Segal construction

(2.6) DEFINITION: LetA be aC∗-algebra. A representationρ : A → B(H) iscyclic if there is a vectorv ∈ H (called acyclic vector) such thatρ[A]v is densein H.

(2.7) PROPOSITION: Every non-degenerate representation of aC∗-algebra is a(possibly infinite) direct sum of cyclic representations.

PROOF: Let ρ : A → B(H) be a representation. For each unitv ∈ H, letHv be the cyclic subspace ofH generated byv, that is the closure ofρ[A]v. Bynon-degeneracy,v ∈ Hv (see Remark 1.45), and it clearly follows thatHv is acyclic subrepresentation ofH with cyclic vectorv. Define a partially orderedsetS as follows: a member ofS is a setS of unit vectors ofH such that thecorresponding cyclic subspaces are pairwise orthogonal, andS is partially orderedby setwise inclusion. Zorn’s Lemma clearly applies and produces a maximalS ∈ S. LetK =

⊕v∈S Hv 6 H; I claim thatK = H. If not, there is a vectoru

orthogonal toK, and thenHu is orthogonal toK sinceK is ρ[A]-invariant. Thusumay be added toS, contradicting maximality. The claim, and thus the Proposition,follows.

(2.8)THEOREM: LetA be aC∗-algebra and letσ be a state onA. Then there exista representationρ of A on a Hilbert spaceHσ, and a unit cyclic vectoru ∈ Hσ,such that

σ(a) = 〈ρ(a)u, u〉Hσ ;

that is,σ is the vector state corresponding tou.

PROOF: The Hilbert spaceHσ is built out ofAwith the inner product〈a, b〉σ =σ(b∗a) defined above, and the representation is the left regular representation. Thatis the idea: here are the details.

26

Let N ⊆ A be defined byN = a ∈ A : 〈a, a〉σ = 0. By virtue of theCauchy–Schwarz inequality,N is a closed subspace: in fact

N = a ∈ A : 〈a, b〉 = 0∀b ∈ A.

Since〈ca, b〉σ = 〈a, c∗b〉σ,N is actually closed under multiplication on the left bymembers ofA; it is a left ideal.

The quotient spaceA/N inherits a well-defined positive definite inner productfrom the inner product onA (this uses the fact thatN is a left ideal). We completethis quotient space to obtain a Hilbert spaceHσ. SinceN is a left ideal, the leftmultiplication representation ofA onA descends to a representationρ (clearly a∗-representation) ofA onA/N . This representation is norm-decreasing because

‖ρ(a)‖2 = supσ(b∗b)61

σ(b∗a∗ab) 6 ‖a∗a‖ = ‖a‖2,

where for the inequality we used the relation

b∗a∗ab 6 ‖a∗a‖b∗b

between positive operators. Thusρ extends by continuity to a representation ofAonHσ.

Now we must produce the cyclic vectoru in Hσ. Let uλ be an approximateunit forA. The inequality

‖[uλ]− [uλ′ ]‖2 = σ((uλ − uλ′)2) 6 σ(uλ − uλ′)

shows that the classes[uλ] form a Cauchy net inHσ, convergent say to a vectoru.For eacha ∈ A we haveρ(a)u = lim[auλ] = [x], soρ(A)u is dense inHσ anduis cyclic. Finally, for any positive elementa∗a of A,

〈ρ(a∗a)u, u〉σ = 〈[a], [a]〉σ = σ(a∗a).

Hence by linearity〈ρ(b)u, u〉σ = σ(b) for all b ∈ A.

(2.9) REMARK : There is a uniqueness theorem to complement the existencetheorem provided by the GNS construction. Namely, suppose thatρ : A→ B(H)is a cyclic representation with cyclic vectoru. Then a state ofA is defined byσ(a) = 〈ρ(a)u, u〉, and from this state one can build according to the GNSconstruction a cyclic representationρσ : A → B(Hσ) with cyclic vectoruσ. Thisnewly-built representation is unitarily equivalent to the original one: in other wordsthere is a unitary isomorphismU : H → Hσ, intertwiningρ andρσ, and mappingu to uσ. It is easy to see this: from the GNS construction the map

ρ(a)u 7→ [a]

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is a well-defined isometric map from the dense subspaceρ[A]u of H to thedense subspaceA/N of Hσ; so it extends by continuity to the required unitaryisomorphism.

We are now going to show that everyC∗-algebra has plenty of states, enoughin fact to produce a faithful Hilbert space representation by the GNS construction.

(2.10)LEMMA : LetA be aC∗-algebra. For any positivea ∈ A, there is a stateσofA for whichσ(a) = ‖a‖.

PROOF: In view of Corollary 2.5 we may assume thatA is unital. The commu-tativeC∗-algebraC∗(a) = C(sp(a)) admits a stateϕ (in fact, a∗-homomorphismto C) corresponding to a point of(a) where the spectral radius is attained; forthis ϕ, one hasϕ(a) = ‖a‖, and of courseϕ(1) = 1, ‖ϕ‖ = 1. Using theHahn–Banach theorem we may extendϕ to a linear functionalσ on A of normone. Sinceσ(1) = 1, Proposition 2.4 shows thatσ is a state.

(2.11)REMARK : By the same argument we may show that ifa ∈ A is self-adjoint,there is a stateσ such that|σ(a)| = ‖a‖.

This allows us to complete the proof of the Gelfand–Naimark RepresentationTheorem.

(2.12)THEOREM: EveryC∗-algebra is∗-isomorphic to aC∗-subalgebra of someB(H).

PROOF: Let A be aC∗-algebra. For eacha ∈ A, use the previous lemma tomanufacture a stateσa havingσa(a∗a) = ‖a‖2, and then use the GNS constructionto build a representationρa : A → B(Ha), with cyclic vectorua, from the stateσa. We have‖ρa(a)ua‖2 = 〈ρ(a∗a)ua, ua〉 = σa(a∗a) = ‖a‖2. It follows thatρ =

⊕a∈A ρa is a faithful representation, as required.

WhenA is separable the enormously large Hilbert space we used above can bereplaced by a separable one; this is a standard argument.

(2.13) REMARK : We have applied the GNS construction toC∗-algebras. Theconstruction, however, has a wider scope. LetA be a unital complex∗-algebra(without topology). A linear functionalϕ : A → C is said to bepositive ifϕ(a∗a) > 0 for all a ∈ A. It is representableif, in addition, for eachx ∈ Athere exists a constantcx > 0 such that

ϕ(y∗x∗xy) 6 cxϕ(y∗y)

for all y ∈ A.By following through the GNS construction exactly as above, one can show

that to any representable linear functionalϕ there corresponds a Hilbert spacerepresentationρ : A→ B(H) and a cyclic vectoru such thatϕ(a) = 〈ρ(a)u, u〉.

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Representability is automatic for positive linear functionals on (unital)Banachalgebras:

(2.14)LEMMA : Any positive linear functionalϕ on a unital Banach∗-algebra8

A is continuous and representable.

PROOF: Supposex = x∗ ∈ A with ‖x‖ < 1. Using Newton’s binomial serieswe may construct a selfadjoint

y = (1− x)12 = 1− 1

2x+

38x2 − · · ·

with y2 = 1 − x. We conclude thatϕ(1 − x) > 0 by positivity, soϕ(x) 6 ϕ(1).Now for anya ∈ A we have

|ϕ(a)|2 6 ϕ(1)ϕ(a∗a) 6 ϕ(1)2‖a∗a‖ 6 ϕ(1)2‖a‖2

(the first inequality following from Cauchy-Schwarz and the second from the pre-ceding discussion), soϕ is continuous and‖ϕ‖ 6 ϕ(1). To show representability,we apply the preceding discussion to the positive linear functionala 7→ ϕ(y∗ay)to get the desired conclusion withcx = ‖x∗x‖.

It was shown by Varopoulos that this theorem extends to non-unital Banachalgebras having bounded approximate units.

Remark on universal algebras defined by generators and relations.

2.3 Abelian von Neumann Algebras and the Spectral Theorem

In this section we discuss the GNS construction in the abelian case, and what canbe obtained from it. Let us begin by constructing a standard example of an abelianvon Neumann algebra.

Let (X,µ) be a finite measure space. For each essentially bounded functionf ∈ L∞(X,µ), the corresponding multiplication operatorMf on the Hilbert spaceH = L2(X,µ) is bounded, with operator norm equal to theL∞-norm off ; and inthis way one identifiesL∞(X,µ) with a∗-subalgebra ofB(H).

(2.15) PROPOSITION: The algebraL∞(X,µ) is a von Neumann algebra ofoperators onL2(X,µ).

PROOF: We will show thatM = L∞(X,µ) is equal to its own commutant.Suppose thatT ∈ B(H) commutes with eachMf , and letg = T · 1, where

8By definition, we require that the involution on a Banach∗-algebra should be isometric.

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1 denotes the constant function 1, considered as an element ofH. Then for allf ∈ L∞(X,µ) ⊆ L2(X,µ) we have

Tf = TMf1 = MfT1 = Mfg = fg.

SinceL∞ is dense inL2 it follows thatT = Mg, and standard estimates show that‖g‖∞ 6 ‖T‖, so thatg ∈ L∞. ThusM ′ ⊆ M ; butM ⊆ M ′ sinceM is abelian,and consequentlyM = M ′.

The proposition still holds for aσ-finite measure space(X,µ), with essentiallythe same proof.

The content of Spectral Theorem is that this is essentially theonly example ofan abelian von Neumann algebra. However, to explain what is meant here we needto distinguish between two notions of isomorphism for von Neumann algebras (orother algebras of operators on Hilbert space). LetM ⊆ B(H) andN ⊆ B(K)be von Neumann algebras. We will say that they areabstractly isomorphicif thereis a∗-isomorphism (an isomorphism ofC∗-algebras) fromM to N ; and we willsay that they arespatially isomorphicif there is a unitaryU : H → K such thatUMU∗ = N . Clearly spatial isomorphism implies abstract isomorphism but theconverse is not true: the 1-dimensional von Neumann algebras generated by theidentity operators on two Hilbert spaces of different finite dimension are abstractlyisomorphic, but they are not spatially isomorphic.

(2.16) PROPOSITION: Let X be a compact metrizable space. Every cyclicrepresentation ofC(X) is unitarily equivalent to a representation ofC(X) bymultiplication operators onL2(X,µ), whereµ is some regular Borel probabilitymeasure onX.

PROOF: Let ρ be a cyclic representation ofA = C(X) with cyclic vectoru.Thenσ(f) = 〈ρ(f)u, u〉 defines a state onA. By the uniqueness of the GNSconstruction (Remark 2.9),ρ is unitarily equivalent to the GNS representationconstructed from the stateσ. Now we appeal to the Riesz Representation Theoremfrom measure theory (see Rudin,Real and complex analysis, Theorem 2.14, forexample): each stateσ of C(X) is the functional of integration with respect tosome regular Borel probability measureµ. The GNS construction then producesthe spaceL2(X,µ) and the multiplication representation.

Let M be an abelian von Neumann algebra inB(H). We say thatM ismaximal abelianif it is contained in no larger abelian von Neumann algebra. It iseasy to see that this condition is equivalent toM ′ = M . In particular, the proof ofProposition 2.15 shows thatL∞(X,µ) is a maximal abelian von Neumann algebraof operators onL2(X,µ).

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(2.17)THEOREM: LetM be an abelian von Neumann algebra inB(H), whereH is separable. Then the following are equivalent:

(a) M has a cyclic vector;

(b) M is maximal abelian;

(c) M is spatially equivalent toL∞(X,µ) acting onL2(X,µ), for some finitemeasure spaceX.

PROOF: Suppose thatM has a cyclic vector. We begin by remarking thatMhas aseparableC∗-subalgebraA such thatA′′ = M . (Proof: The unit ball ofB(H) is separable and metrizable in the strong topology, hence second countable.Thus the unit ball ofM is second countable, henceM itself is separable, inthe strong topology. LetA be theC∗-algebra generated by a countable stronglydense subset ofM .) The representation ofA on B(H) is cyclic (sinceA isstrongly dense inM ) so by Proposition 2.16 it is unitarily equivalent to therepresentation ofC(X) onL2(X,µ) by multiplication operators. NowL∞(X,µ)is a von Neumann algebra which containsC(X) as a strongly dense subset (bythe Dominated Convergence Theorem) so the unitary equivalence must identifyL∞(X,µ) with M . This proves that (a) implies (c). We have already remarkedthat (c) implies (b). Finally, suppose thatM is anyabelian von Neumann algebraof operators onH. DecomposeH as a countable direct sum of cyclic subspacesfor the commutantM ′, say with unit cyclic vectorsun; and letu =

∑2−nun.

The projectionPn : H → Hn belongs toM ′′ = M ⊆M ′, using the bicommutanttheorem and the fact thatM is abelian. Hence eachun = 2nPnu belongs toM ′x,sou is a cyclic vector forM ′. We have shown that the commutant of any abelianvon Neumann algebra on a separable Hilbert space has a cyclic vector; in particularif M is maximal abelian, thenM = M ′ andM has a cyclic vector.

We can use this to get the structure of a general abelian von Neumann algebra,up to abstract (not spatial!) isomorphism:

(2.18) THEOREM: Each commutative von Neumann algebra on a separableHilbert space is abstractly isomorphic to someL∞(X,µ).

PROOF: LetM be an abelian von Neumann algebra onB(H). Recall from thelast part of the preceding proof that the commutantM ′ has a cyclic vectoru; letP denote the orthogonal projection ontoMu. ThenP is a projection inM ′. Themapα : T 7→ TP is a∗-homomorphism fromM onto the von Neumann algebraMP , and, by construction,u is a cyclic vector forMP (thought of as acting onthe range ofP ). ThusMP is isomorphic to an algebraL∞(X,µ) by the previoustheorem. I claim thatKer(α) = 0; this will show thatM andMP are isomorphic

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and so will complete the proof. IfT ∈ Ker(α) thenTu = 0. For anyR ∈ M”,then,TRu = RTu = 0; but sinceu is cyclic forM ′, the vectorsRu are dense inH, and it follows thatT = 0.

(2.19)REMARK : (about spectral multiplicity theory)

Let us specialize these constructions to the case of a single operator.

(2.20) COROLLARY: Let T be a normal operator on a separable Hilbert spaceH. Then there exist a measure space(Y, µ) and a unitary equivalenceU : H →L2(Y, µ), such thatUTU∗ is the multiplication operator by someL∞ function onY .

PROOF: Let X be the spectrum ofT , so that the functional calculus gives arepresentationρ ofC(X) onH. By Proposition 2.7, we can breakρ up into a directsum of cyclic representations; there are only countably many summands sinceHis separable. According to Proposition 2.16, each such cyclic representationρn

is unitarily equivalent to the representation ofC(X) by multiplication operatorson L2(X,µn) for some regular Borel probability measureµn on X. Put now(Y, µ) =

⊔n(X,µn) to get the result.

As a corollary we obtain theBorel functional calculus. Given any boundedBorel functionf on the spectrum ofT , we may definef(T ) as follows: ifT isunitarily equivalent to multiplication by someL∞ function g on L2(Y, µ) (notethat the closure of the essential range ofg is the spectrum ofT ), thenf(T ) is thebounded operator that is unitarily equivalent to multiplication byf g. Observethat if fn is a uniformly bounded sequence of Borel functions that tends pointwiseto f , thenfn(T ) tends strongly tof(T ), since∫

Y|(fn − f) g(y)|2|u(y)|2 dµ(y) → 0

asn→∞, for all u ∈ L2(Y, µ), by Lebesgue’s dominated convergence theorem.

(2.21) REMARK : An alternative, classic formulation of these results makes useof the notion ofspectral measure(also calledresolution of the identity). Let Hbe a Hilbert space and let(X,M) be a measurable space, that is a space equippedwith a σ-algebra of subsets. Aspectral measureassociated to the above data is amapE from M to the collectionP (H) of selfadjoint projections inH, having thefollowing properties:

(a) E(∅) = 0, E(X) = 1;

(b) E(S ∩ T ) = E(S)E(T );

(c) if S ∩ T = ∅, thenE(S ∪ T ) = E(S) + E(T );

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(d) E is countably additive relative to the strong operator topology; that is,if Sn is a sequence of mutually disjoint subsets ofX with union S, thenE(S) =

∑E(Sn) (with strong operator convergence).

Given such a spectral measure it is straightforward to define an ‘integral’∫Xf(x)dE(x) ∈ B(H)

for every functionf on X which is bounded andM-measurable; one simplymimics the usual processes of measure theory. For instance iff is the charac-teristic function of someS ∈ M, then we define

∫X f(x)dE(x) = E(S). By

linearity we extend this definition to simple functions (finite linear combinationsof characteristic functions), and then by a limit argument we extend further to allmeasurable bounded functions.

We may then formulate the spectral theorem as follows: for every normaloperatorT on a separable Hilbert space, there is a resolution of the identityEon theσ-algebra of Borel subsets ofsp(T ), such that

T =∫

sp(T )λdE(λ).

For the proof, one need only verify that the definitionE(S) = χS(T ) (using theBorel functional calculus described above) describes a resolution of the identity.To show thatT has an integral decomposition as above one approximates theintegral by Riemann sums and uses the fact that, when restricted to the range ofthe projectionE((k/n, (k + 1)/n]), the operatorT lies within1/n in norm of theoperator of multiplication byk/n.

2.4 Pure States and Irreducible Representations

In this section we will begin a more detailed study of the representation theory ofaC∗-algebraA. By application to groupC∗-algebras, this theory will include thetheory of unitary representations of groups. We’ll discuss this connection.

Recall that a representationρ : A → B(H) is irreducible if there are noclosedA-invariant subspaces ofH. In Remark 2.2 we drew attention to Kadison’stransitivity theorem, which states that our topological notion of irreducibility isequivalent to the algebraic notion according to which there are no invariant sub-spaces at all, closed or not. We will now prove this theorem.

(2.22)KADISON TRANSITIVITY THEOREM Letρ be a (topologically) irreduciblerepresentation of aC∗-algebraA on a Hilbert spaceH. Then for any two vectors

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u, v ∈ H, with u nonzero, there existsa ∈ A with ρ(a)u = v. In particular, thereare noA-invariant subspaces ofH, closed or not.

PROOF: We begin by establishing a quantitative, approximate version of thetheorem. Suppose thatu andv are unit vectors inH. We assert that there existsa ∈ A with ‖a‖ < 2 and‖ρ(a)u − v‖ < 1

2 . Indeed, irreducibility implies thatρ[A]′′ = B(H) (see Remark 2.2), so by the Kaplansky Density Theorem the unitball of ρ[A] is strongly dense in the unit ball ofB(H). In particular there is anelementρ(a) of this unit ball such that‖ρ(a)u − v‖ < 1

2 ; and we may take‖a‖ < 2 sinceρ is the composite of a quotient map and an isometric injection(Corollary 1.33).

Now we apply this argument, and an obvious rescaling, iteratively as follows:givenu andv = v0, produce a sequenceaj in A and a sequencevj of vectorsof H with

vj+1 = vj − ρ(aj)u, ‖vj‖ < 2−j , ‖aj‖ < 4 · 2−j .

The seriesa =∑aj then converges inA to an element of norm6 4 having

ρ(a)u = v. (With more care we can reduce the norm estimate from4 to 1 + ε. Wecan also moven linearly independentu’s simultaneously onto arbitraryv’s, ratherthan just one. )

Let A be aC∗-algebra (unital for simplicity). Recall that a linear functionalσ : A→ C is a state if it has norm one and moreoverσ(1) = 1. Thus the collectionof states onA is a weak-star compact9, convex subset ofA∗. It is called thestatespaceof A, and denoted byS(A).

The general yoga of functional analysis encourages us to consider the extremepoints ofS(A), called thepure states. By the Krein–Milman Theorem,S(A) isthe closed convex hull of the pure states.

(2.23) LEMMA : Let σ be a state of aC∗-algebraA, and letρ : A → B(H)be the associated GNS representation, with unit cyclic vectoru. For any positivefunctionalϕ 6 σ onA there is an operatorT ∈ ρ[A]′ such that

ϕ(a) = 〈ρ(a)Tu, u〉

and0 6 T 6 1.

This is a kind of ‘Radon-Nikodym Theorem’ for states.PROOF: Recall thatH is obtained by completing the inner product spaceA/N ,

whereN is the seta ∈ A : σ(a∗a) = 0. Define a sesquilinear form onA/N by

9Warning: In the non-unital case the state space is not closed in the unit ball ofA∗, and hence isnot compact; because the conditionlim σ(uλ) = 1 is not preserved by weak-star convergence. Thiscauses some technical gyrations, see Pedersen’s book for details.

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(a, b) 7→ ϕ(b∗a). This form is well-defined, positive semidefinite, and bounded by1, so there is an operatorT onH such that

ϕ(b∗a) = 〈T [a], [b]〉 = 〈Tρ(a)u, ρ(b)u〉.

The computation

〈ρ(a)T [b], [c]〉 = 〈T [b], [a∗c]〉 = ϕ(c∗ab) = 〈Tρ(a)[b], [c]〉

shows thatT commutes withρ[A].

(2.24)PROPOSITION: A state is pure if and only if the associated GNS represen-tation is irreducible.

PROOF: Suppose first that the representationρ : A → B(H) is reducible, sayH = H1⊕H2 whereH1 andH2 are closed,A-invariant subspaces. Write a cyclicvectorv = v1 + v2 wherevi ∈ Hi, i = 1, 2. We cannot havev1 = 0, otherwiseρ[A]v ⊆ H2; and for similar reasons we cannot havev2 = 0. Now we have

σ(a) = 〈ρ(a)v, v〉 = 〈ρ(a)v1, v1〉+ 〈ρ(a)v2, v2〉

which is a nontrivial convex combination of the states corresponding to the unitvectorsv1/‖v1‖ andv2/‖v2‖. Thusσ is impure.

Conversely, suppose thatσ = tσ1 + (1− t)σ2 whereσ1, σ2 are states. By theLemma above there is an operatorT in the commutantρ[A]′ such thattσ1(a) =〈ρ(a)Tu, u〉. If ρ is irreducible thenρ[A] has trivial commutant and soT = cIfor some constantc and thusσ1 = cσ. Normalization showsσ1 = σ. Similarlyσ2 = σ and thusσ is pure.

(2.25)REMARK : It is important to know that there are enoughirreducibleunitaryrepresentations of aC∗-algebraA to separate points ofA, that is to sayC∗-algebrasaresemi-simple. To show this we need to refine Lemma 2.10 to show that, for anypositivea ∈ A, there exists apure stateσ with σ(a) = ‖a‖. Here is how to dothis: Lemma 2.10 shows that states having this property exist. The collectionΣ ofall states having this property is a closed convex subset of the state space, so (bythe Krein–Milman Theorem) it has extreme points — indeed it is the closed convexhull of these extreme points. Letσ be an extreme point ofΣ. If σ = tσ1+(1−t)σ2,0 < t < 1, then the inequalities

σ(a) = ‖a‖, σ1(a) 6 ‖a‖, σ2(a) 6 ‖a‖

imply thatσ1, σ2 ∈ Σ. Henceσ1 = σ2 = σ sinceσ is extreme inΣ. Thusσ isextreme inS and we are done.

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(2.26)DEFINITION: LetA be aC∗-algebra. ThespectrumofA, denotedA, is thecollection of unitary (= spatial) equivalence classes of irreducible representationsofA. We topologize it as a quotient of the pure state space: that is, we giveA thetopology which makes the surjective mapP (A) → A, which associates to eachpure state its GNS representation, open and continuous. This is called theFelltopology.

(2.27)REMARK : LetA = C(X) be commutative. The states ofA are measuresonX; the pure states are the Diracδ-measures at the points ofX. On the otherhand, it is shown in the exercises that each irreducible representation ofA is1-dimensional, that is, is a homomorphismA → C. These homomorphisms areagain classified by the points ofX. In this case therefore we haveP (A) = A = X.Needless to say the general case is not so simple.

Let ρ : A→ B(H) be a representation. Then the kernel ofρ is an ideal ofA.

(2.28)DEFINITION: An ideal in aC∗-algebraA is calledprimitive if it arises asthe kernel of an irreducible representation. The space of primitive ideals is denotedPrim(A).

(2.29)LEMMA : Every primitive ideal of aC∗-algebra is prime.

PROOF: We begin by proving a general fact. Suppose thatρ : A → B(H)is a representation andJ is an ideal inA. Then the orthogonal projection ontoρ[J ]H10 lies in the center of the bicommutantρ[A]′′. To see this, note that thesubspaceρ[J ]H is bothρ[A]-invariant (becauseJ is an ideal) andρ[J ]′-invariant.ThereforeP lies inρ[A]′ ∩ ρ[J ]′′. But this is contained inρ[A]′ ∩ ρ[A]′′, which issimply the center of the bicommutant.

If we apply this whenρ is irreducible, the bicommutant ofρ[A] is B(H)which has trivial center; soρ[J ]H is either zero orH, andJ ⊆ Ker(ρ) preciselywhenρ[J ]H = 0. It is now easily seen that ifJ1 andJ2 are ideals such thatJ1J2 ⊆ Ker(ρ), then at least one of them is itself contained inKer(ρ). That is,Ker(ρ) is prime.

The converseof this result is also true, at least for separableC∗-algebras.However, to prove this will require something of a detour.

LetR be any ring. The collection of prime ideals ofR always has a topology,called theZariski topology, according to which the closed sets of prime ideals arejust those sets of the form

hull(X) = p : X ⊆ p10This subspace is closed by the Cohen Factorization Theorem.

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asX ranges over the set of subsets ofR. (To prove that this defines a topology youneed to use the defining property of a prime ideal, namely that if an intersectionj1 ∩ j2 ⊆ p, then eitherj1 or j2 already lies inp. The closure of any setS of prime

ideals is the sethull(⋂

p∈S p)

. The intersection appearing here is sometimes

called the ‘kernel’ ofS, a terminology that I find confusing.) Since primitiveideals in aC∗-algebraA are prime, the Zariski topology gives rise to a topology onPrim(A), which in this context is called theJacobson topologyor thehull-kerneltopology.

(2.30)REMARK : The Zariski topology is rather bad from the point of view of theseparation axioms. A single point ofPrim(A) is closed iff it represents a maximalideal. Consequently, unless all prime ideals are maximal (a rare occurrence) thetopology is not evenT1, let alone Hausdorff. However, it isT0 — given any twopoints, at least one of them is not in the closure of the other one.

(2.31) THEOREM: Let A be a separableC∗-algebra. The canonical mapA → Prim(A) is a topological pullback: that is, the open sets inA are preciselythe inverse images of the open sets inPrim(A).

We will give the proof in a moment.

(2.32)COROLLARY: The canonical mapP (A) → Prim(A) is a quotient map.

Here is an important deduction from this:

(2.33) PROPOSITION: (DIXMIER ) In a separableC∗-algebraA every closedprime ideal is primitive.

PROOF: (Recall our standing assumption thatA is unital.) The spaceP (A) is aclosed convex subset of the unit ball ofA∗, and hence is compact and metrizable; inparticular it is second countable and it is a Baire space. This latter assertion meansthat the Baire category theorem holds: each countable intersection of dense opensets is dense. Both properties pass to quotients, and so we conclude thatPrim(A)is also a second countable Baire space.

Now suppose that 0 is a prime ideal ofA (the general case can be reducedto this one by passing to quotients), in other words that any two nontrivial idealsof A have nontrivial intersection. We want to prove that 0 is a primitive ideal,that is,A has a faithful irreducible representation. Since nontrivial ideals intersectnon-trivially, it follows that we can’t writePrim(A) as the union of two properclosed subsets; which is to say that any two nonempty open subsets ofPrim(A)intersect; so every nonempty open subset is dense. Applying the Baire categorytheorem to a countable base for the topology we find that there is a dense point inPrim(A). This point corresponds to a primitive ideal which is contained ineveryprimitive ideal, and since the irreducible representations separate points ofA, the

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only possibility for such an ideal is(0) which is therefore primitive, as required.

Now we will work towards a proof of Theorem 2.31.

(2.34)LEMMA : LetA be aC∗-algebra, letE be a collection of states ofA, andsuppose that for everya > 0 in A we have

supσ(a) : σ ∈ E = ‖a‖.

Then the weak-∗ closure ofE contains every pure state.

PROOF: Once again, we stick with unitalA. We’ll show in fact that the weak-∗closed convex hullC of E is the entire state spaceS of A. This will suffice, sincethe extreme points of the closed convex hull ofE must belong to the closure ofE .

Think ofAsaas a real Banach space, and think ofC andS as compact convexsubsets of the LCTVSE which is the dual ofAsa equipped with the weak-∗topology. The dual ofE is thenAsaonce again. Suppose for a contradiction thatC 6= S, so that there existsϕ ∈ S \ C. By the Hahn-Banach theorem applied tothe spaceE, there exista ∈ Asaandc ∈ R such thatϕ(a) > c whereasσ(a) < cfor all σ ∈ C. We may assumea is positive (add a large multiple of 1). Now wegetϕ(a) > supσ(a) : σ ∈ E = ‖a‖ which is a contradiction.

PROOF OFTHEOREM 2.31: LetΠ ⊆ A be a set of irreducible representationsof A, and letρ be another such representation. Thenρ belongs to the closure ofΠfor the quotient topology fromP (A) if and only if

(a) some state associated toρ is a weak-∗ limit of states associated to represen-tationsπ ∈ Π.

On the other hand,ρ belongs to the closure ofΠ for the pull-back of the Jacobsontopology fromPrim(A) if and only if

(b) the kernelKer(ρ) contains⋂Ker(π) : π ∈ Π.

We need therefore to prove that conditions (a) and (b) are equivalent.Suppose (b). By passing to the quotient, we may assume without loss of

generality that⋂Ker(π) : π ∈ Π = 0. The direct sum of all the representations

π is then faithful, and so for everya ∈ A

supσ(a) : σ ∈ E = ‖a‖,

whereE denotes the collection of vector states associated to the representationsbelonging toΠ. By the lemma, every pure state ofA (and every state associated toρ in particular) belongs to the weak-∗ closure ofE . This proves (a).

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Conversely suppose (a). LetJ denote the ideal⋂Ker(π) : π ∈ Π.

Suppose that the stateσ(a) = 〈ρ(a)v, v〉 is a weak-∗ limit of states associatedto representationsπ ∈ Π. All such states vanish on everya ∈ J , soσ(a) = 0 forall a ∈ J . But then for ally, z ∈ A,

〈ρ(a)ρ(y)v, ρ(z)v〉 = σ(z∗ay) = 0

sinceJ is an ideal. Becausev is a cyclic vector forρ this implies thatρ(a) = 0, soKer(ρ) ⊇ J , which is (b).

(2.35)REMARK : Using Voiculescu’s theorem one can show that two points ofAmap to the same point ofPrim(A) if and only if the corresponding representationsare approximately unitarily equivalent.

2.5 Representations of compact operators

To begin our study of representation theory, we need a few lemmas about represen-tations of algebras of compact operators.

(2.36)LEMMA : LetA be aC∗-algebra, letJ be an ideal inA, and letρJ : J →B(H) be a non-degenerate representation. There is a unique extension ofρJ to arepresentationρA : A → B(H). Moreover,ρA is irreducible if and only ifρJ isirreducible.

PROOF: Since we went to the trouble of proving it in an exercise, let’s use theCohen Factorization Theorem here (we could avoid this by a marginal complicationof the argument). By Cohen, everyv ∈ H is of the formv = ρJ(j)w for somej ∈ J , w ∈ H. Define then

ρA(a)v = ρJ(aj)w.

To see that this is well-defined, take an approximate unituλ for J and write

ρJ(aj)w = lim ρJ(auλj)w = lim ρJ(auλ)v,

which also shows thatρA(a) is the strong limit of the operatorsρJ(auλ) and there-fore has norm bounded by‖a‖. It is routine to verify thatρA is a∗-representationof A.

It is obvious that ifρA is reducible, so isρJ . On the other hand, ifρJ has aproper closed invariant subspaceH ′ say, the construction by way of an approximateunit shows thatρA will mapH ′ toH ′. SoρA is reducible too.

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(2.37)LEMMA : LetH be a Hilbert space and letA ⊆ K(H) be aC∗-algebra ofcompact operators onH that is irreducibly represented onH. ThenA = K(H).

PROOF: SinceA consists of compact operators, it contains finite-rank pro-jections by the functional calculus. LetP be such a projection, having minimalrank. I claim thatP is of rank one. To see this, note first thatPTP = λTPfor all T ∈ A (one may assumeT to be selfadjoint, and if the assertion werefalse then a nontrivial spectral projection ofPTP would have smaller rank thanP ,contradiction). Now if there are orthogonal vectorsu, v in the range ofP , then forall T ∈ A,

〈Tu, v〉 = 〈PTPu, v〉 = λT 〈u, v〉 = 0.

Thus Au is a nontrivial invariant subspace, contradicting irreducibility. Thisestablishes the claim thatP has rank one.

Any rank-one projectionP has the form

P (w) = 〈w, u〉u

for some unit vectoru. Letv′, v′′ be two more unit vectors. According to Kadison’stransitivity theorem there areS′, S′′ ∈ A such thatS′u = v′ andS′′v′′ = u. ThusA contains the operatorS′PS′′, which is given by

w 7→ 〈w, v′′〉v′.

The operators of this sort span all the operators of finite rank, hence they generateK(H) as aC∗-algebra.

(2.38)COROLLARY: The algebraK(H) is simple (it has no non-trivial ideals).

PROOF: Let J ⊆ K(H) be a non-zero ideal. The subspaceJ · H of H isinvariant underK(H), so it is all ofH. ThusJ is non-degenerately representedonH. By Lemma 2.36, this representation ofJ is irreducible. By Lemma 2.37,J = K(H).

(2.39) COROLLARY: If A is an irreducibleC∗-subalgebra ofB(H) whichcontains a single non-zero compact operator, thenA contains all ofK(H).

PROOF: Consider the idealJ = A∩K(H) inA. The spaceJ ·H isA-invariant,andA is irreducible, soJH = H andJ is non-degenerate. Now we argue asbefore: by Lemma 2.36J is irreducible, and then by Lemma 2.37,J = K(H).

(2.40)REMARK : Of course it follows thatPrim(K(H)) consists of a single point(corresponding to the zero ideal). Let us also go on to prove thatK(H) has only

one irreducible representation, so thatK(H) also consists just of a single point.

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Let ρ : K(H) → B(H ′) be an irreducible representation. LetP be a rank-oneprojection inK(H). SinceK(H) is simple,ρ is faithful; soρ(P ) is a nonzeroprojection onH ′. Now let v′ ∈ H ′ be a unit vector in the range of the projectionρ(P ), and letσ be the associated state, that is

σ(T ) = 〈ρ(T )v′, v′〉.

Sinceρ is irreducible,v′ is a cyclic vector. By the GNS uniqueness theorem,ρ isunitarily equivalent to the GNS representation associated to the stateσ. But

σ(T ) = 〈ρ(PTP )v′, v′〉 = λT = Trace(PTP )

and the GNS representation associated to this state is the standard one.

2.6 The Toeplitz algebra and its representations

Let H = `2 be the usual sequence space. Theunilateral shift operatorV is theisometryH → H given by

V (x0, x1, x2, . . .) = (0, x0, x1, . . .).

Notice thatV ∗V = 1, whereasV V ∗ = 1− P , whereP is the rank one projectiononto the span of the basis vectore0 = (1, 0, 0, . . .). ThusV is a Fredholm operator,of index−1.

(2.41) DEFINITION: TheToeplitzC∗-algebraT is theC∗-subalgebra ofB(H)generated by the unilateral shift operator.

(2.42)LEMMA : The algebraT is irreducible and contains the compact operators.

PROOF: Suppose thatK ⊆ H is a closed invariant subspace forT. If K 6= 0,thenK contains a nonzero vector, and by applying a suitable power ofV ∗ ifnecessary we may assume thatK contains a vectorv = (x0, x1, x2, . . .) withx0 6= 0. SinceP ∈ T we find thatPv = v0e0 ∈ K and soe0 ∈ K. Nowapplying powers ofV we find that eachen = V ne0 ∈ K, soK = H. ThusT isirreducible. Since it contains the compact operatorP , the second statement followsfrom Corollary 2.39.

The unilateral shift operator is unitary modulo the compacts, and so itsessen-tial spectrum, that is the spectrum of its image in the Calkin algebraQ(H) =B(H)/K(H), is a subset of the unit circleS1.

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(2.43)LEMMA : The essential spectrum ofV is the whole unit circle.

PROOF: This is a consequence of Fredholm index theory. Suppose the con-trary: then there is a continuous path of complex numbersλ, running fromλ = 0to λ = 2, that avoids the essential spectrum ofV . But thenλ 7→ (V − λ) is acontinuous path of Fredholm operators, andIndexV = −1 as we observed above,whereasIndex(V − 2) = 0 sinceV − 2 is invertible. This is a contradiction.

We see therefore that theC∗-algebraT/K is commutative, generated by asingle unitary element whose spectrum is the whole unit circle. Consequently thereis a short exact sequence ofC∗-algebras

0 → K → T → C(S1) → 0

called theToeplitz extension.IT is helpful to make the link with complex function theory.

(2.44)DEFINITION: LetS1 denote the unit circle inC. TheHardy spaceH2(S1)is the closed subspace ofL2(S1) spanned by the functionszn, forn > 0. AToeplitzoperatoronH2(S1) is a bounded operatorTg of the form

Tg(f) = P (gf) (f ∈ H2(S1)),

whereg ∈ L∞(S1) andP is the orthogonal projection fromL2(S1) ontoH2(S1).The functiong is called thesymbolof Tg.

(2.45)LEMMA : TheC∗-algebra generated by the Toeplitz operators is isomorphicto T, and the mapg 7→ Tg is a linear splitting for the quotient mapT → C(S1)that appears in the Toeplitz extension.

PROOF: We haveV = Tz (in the obvious basis) so certainly theC∗-algebragenerated by the Toeplitz operators containsT. In the other direction, supposethatp(z) = a−nz

n + · · · + a0 + · · · + anzn is a trigonometric polynomial. Then

Tp = a−n(V ∗)n + · · · + a01 + · · · + anVn belongs toT. Since‖Tg‖ 6 ‖g‖, we

find using the Stone-Weierstrass theorem thatTg belongs toT for each continuousfunctiong, and therefore that theC∗-algebra generated by theTg is contained inT.

Toeplitz operators serve as a prototype for the connection betweenC∗-algebras,index theory, andK-homology. It is not the purpose of this course to focus onthese connections, but we should at least mention the fundamental index theoremfor Toeplitz operators. Recall that thewinding numberWinding(g) ∈ Z ofa continuous functiong : S1 → C \ 0 is its class in the fundamental groupπ1(C \ 0), which we identify withZ in such a way that the winding number ofthe functiong(z) = z is +1.

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(2.46)TOEPLITZ INDEX THEOREM If Tg is a Toeplitz operator onH2(S1) whosesymbol is a continuous and nowhere vanishing function onS1 thenTg is a Fred-holm operator and

Index(Tg) = −Winding(g).

PROOF: Denote byMg the operator of pointwise multiplication byg onL2(S1). The set of all continuousg for which the commutatorPMg −MgP iscompact is aC∗-subalgebra ofC(S1). A direct calculation shows thatg(z) = zis in thisC∗-subalgebra, for in this case the commutator is a rank-one operator.Since the functiong(z) = z generatesC(S1) as aC∗-algebra (by the Stone–Weierstrass Theorem) it follows thatPMg−MgP is compact for every continuousg. IdentifyingTg with PMg, we see that

Tg1Tg2 = PMg1PMg2 = PPMg1Mg2 + compact operator

= PMg1g2 + compact operator

= Tg1g2 + compact operator,

for all continuousg1 andg2. In particular, ifg is continuous and nowhere zero thenTg−1 is inverse toTg, modulo compact operators, and so by Atkinson’s Theorem,Tg is Fredholm. Now the continuity of the Fredholm index shows thatIndex(Tg)only depends on the homotopy class of the continuous functiong : S1 → C \ 0.So it suffices to check the theorem on a representative of each homotopy class; saythe functionsg(z) = zn, for n ∈ Z. This is an easy computation, working in theorthogonal basis zn | n > 0 for H2(S1).

(2.47) REMARK : on C∗-algebras defined by generators and relations. Wold–Coburn theorem:T is the universalC∗-algebra generated by an isometry.

Now let us classify the irreducible representations of the Toeplitz algebraT.Let ρ : T → B(H) be such a representation. We consider two cases:

(a) ρ annihilates the idealK ⊆ T;

(b) ρ does not annihilate the idealK ⊆ T.

In case (b),ρ(K) · H is a T-invariant subspace, hence is equal toH; so therestriction ofρ to K(H) is a non-degenerate representation. Hence it is irreducible(Lemma 2.36) and thus it is unitarily equivalent to the identity representation. Itfollows (Lemma 2.36 again) that the representationρ of T is unitarily equivalentto the identity representation ofT onH2(S1).

In case (a),ρ passes to an irreducible representation ofT/K = C(S1). Such arepresentation is 1-dimensional, determined by a point ofS1; conversely all such1-dimensional representations ofT are (of course!) irreducible.

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ThusT = Prim(T) consists of the union ofS1 and a disjoint point. What is thetopology? It is easy to see that the circleS1 gets its usual topology. On the otherhand, the disjoint point corresponds to the zero ideal, which is of course containedin every other ideal; so that the only neighborhood of that point is the whole space.Thus the ‘extra point’ cannot be separated from any other representation. Wesummarize:

(2.48)PROPOSITION: The spaceT of irreducible representations of the ToeplitzalgebraT is of the formS1 t p0, whereS1 has its usual topology and the onlyopen set containingp0 is the whole space.

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3 Unitary Group Representations

3.1 Basics on GroupC∗-Algebras

Let G be a group. Usually, later on in the course, we will assume thatG is acountable, discretegroup; but for the moment we will allow continuous groups aswell, so let us assume thatG is a topological group, with aHausdorff and secondcountabletopology. A unitary representationof G is a (group) homomorphismπ : G → U(H), whereU(H) is the unitary group of a Hilbert spaceH, andπis required to be continuous relative to thestrong topology onU(H). That is, ifgλ → g, thenπ(gλ)v → π(g)v, for each fixed vectorv ∈ H.

(3.1)REMARK : It would not be a good idea to require thatπ benormcontinuous.To see why, recall that a locally compact topological groupG carries an (essentiallyunique)Haar measureµ: a regular Borel measure which is invariant under lefttranslation. One can therefore form the Hilbert spaceH = L2(G,µ) and theunitary representation ofG onH by left translation

π(g)f(h) = f(g−1h)

which is called theregular representation. It is a simple consequence of theDominated Convergence Theorem that this representation is strongly continuous;but it is not norm continuous for instance whenG = R.

The Haar measureµ onG, which is invariant underleft translation, is usuallynot invariant underright translation. However, for each fixedg ∈ G, the mapE 7→ µ(Eg) is anotherleft-translation-invariant measure onG, and hence it differsfrom µ only by a scalar multiple which we denote∆(g): thus

µ(Eg) = ∆(g)µ(E).

It is easily checked that∆: G → R+ is a topological group homomorphism.∆is called themodular functionof G; and groups such as discrete groups or abeliangroups, for which∆ ≡ 1, are calledunimodular. An example of a non-unimodulargroup is the “(ax+ b) group” — the Lie group of matrices(

a b0 a

)with a > 0. (insert discussion of the modular function and Lie groups here.)

We’ll use the notationdg instead ofdµ(g) for integration with respect to theHaar measure. Notice the identities (for fixedh)

d(hg) = dg, d(gh) = ∆(h)dg, d(g−1) = ∆(g−1)dg.

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To prove the last identity note that both sides defineright Haar measures, so theyagree up to a scalar multiple; consider integrating the characteristic function of asmall symmetric neighborhood of the identity (where∆ ≈ 1) to show that thescalar is1.

(3.2)DEFINITION: We makeL1(G,µ) into a Banach∗-algebra by defining

f1 ∗ f2(g) =∫

Gf1(h)f2(h−1g)dh, f∗(g) = ∆(g)−1f(g−1).

Exercise: Check that the involution is isometric and that(f1 ∗ f2)∗ = f∗2 ∗ f∗1 .

(3.3) THEOREM: There is a one-to-one correspondence between strongly con-tinuous unitary representations ofG and non-degenerate∗-representations of theBanach algebraL1(G).

PROOF: To keep matters simple let us assume first of all thatG is discrete. (Inthis case the Haar measure onG is just counting measure, and we write`1(G) forL1(G).) Letπ : G→ U(H) be a unitary representation. We define a representationρ : L1(G) → U(H) by

ρ(∑

ag[g]) =∑

agπ(g);

the series on the right converges in norm. This representation hasρ([e]) = 1,so it is non-degenerate. Conversely, given a representationρ as above, the mapπ : G→ U(H) defined by

π(g) = ρ([g])

is a unitary representation ofG.In the general case we must replace summation by integration, handle the

complications caused by the modular function, and use an approximate unit todeal with the fact that the ‘Dirac masses’ at points ofG are no longer elements ofL1(G). This is routine analysis, the details are on page 183 of Davidson’s book.

The L1-norm onL1(G) is not a C∗-norm in general. To manufacture aC∗-algebra related to the representation theory ofG, we should form acompletionof L1(G) in an appropriateC∗-norm. There are several different ways to form sucha completion (as will become apparent, this kind of issue turns up many times inC∗-algebra theory.)

(3.4) DEFINITION: ThemaximalC∗-algebra ofG is the envelopingC∗-algebraof the Banach∗-algebraL1(G).

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Here is the explanation of the term ‘envelopingC∗-algebra’. As we sawin Remark 2.13, the GNS construction can be applied in any Banach∗-algebra(having a norm-bounded approximate unit) and it shows that each state of such analgebra (continuous positive linear functional of norm one) gives rise to a cyclicHilbert space representation. In particular there exists auniversalrepresentationρA : A → B(HA) for each Banach∗-algebraA, obtained as the direct sumof the GNS representations coming from all the states; and theC∗-subalgebraof B(HA) generated byρA[A] is called theenvelopingC∗-algebra ofA. Ithas the universal property that any∗-homomorphism fromA to aC∗-algebraBfactors uniquely through the envelopingC∗-algebra. In particular, then, there isa 1:1 correspondence between unitary representations ofG and non-degenerate

representations ofC∗(G). We usually denoteC∗(G) by G, regarding it as thespace of irreducible unitary representations ofG — the ‘unitary dual’.

There is a canonical∗-homomorphism from any Banach∗-algebraA to itsC∗-envelope. In general this∗-homomorphism need not be injective (think aboutthe disk algebra again!); it will be injective precisely whenA has a faithful Hilbertspace representation. To see that this is so in the case ofA = L1(G) we consideronce again the left regular representationλ of L1(G) onL2(G):

(λ(f1)f2)(g) =∫

Gf1(h)f2(h−1g)dh.

It is an easy measure-theoretic fact that this representation is faithful. Thus the∗-homomorphismL1(G) → C∗(G) is injective.

The special role of the regular representation in the above argument promptsus to define another ‘groupC∗-algebra’. ThereducedC∗-algebraC∗r (G) of Gis theC∗-subalgebra ofB(L2(G)) generated by the image of the left regularrepresentation. Thus we have a surjective∗-homomorphism

C∗(G) → C∗r (G)

coming fromλ via the universal property. Dually, the spectrum ofC∗r (G) isidentified with a closed subspace of the spectrumG of C∗(G); this is thereduceddual Gr of G.

Of course we want to know what all these weird spaces are and whether thereis any difference between them. Let us first consider the case ofabeliangroups.

(3.5) LEMMA : LetG be a locally compact abelian group. Then its unitary dualG is equal to itsPontrjagin dualHom(G,T), the space ofcharactersof G, thatis continuous homomorphisms ofG into the circle group. The topology onG isthe restriction of the weak-∗ topology onL∞(G); that is, a netαλ of characters

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converges to the characterα if and only if∫Gαλ(g)f(g)dg →

∫Gα(g)f(g)dg

for all f ∈ L1(G). If G is discrete, therefore, the topology onG is the topology ofpointwise convergence.

(if G isn’t discreteG isn’t compact — here my insistence on considering onlyrepresentations of unitalC∗-algebras is coming back to haunt me.)

PROOF: SinceG is commutative, so isC∗(G). Thus by the theory of commuta-tiveC∗-algebras,C∗(G) = C0(G), andG is the space of∗-homomorphisms fromC∗(G) → C, that is one-dimensional representations ofC∗(G). By the discussionabove, these are the same thing as one-dimensional unitary representations ofG,that is characters. The topologies match up since (by construction) states onC∗(G)are all obtained by extending states onL1(G), which themselves are given by(positive)L∞ functions onG.

(3.6) LEMMA : Let G be a locally compact abelian group. ThenC∗(G) =C∗r (G) = C0(G).

PROOF: All we need to do is to show thatC∗(G) = C∗r (G), and for this it willbe enough to show that every characterα of G extends to a continuous linear mapC∗r (G) → C.

Let us first establish the following: pointwise multiplication by a character ofG preserves the norm on the regular representation. In other words, iff ∈ L1(G)andα is a character, then

‖λ(αf)‖ = ‖λ(f)‖.

To see this writeλ(αf) = Uαλ(f)U∗α

whereUα : L2(G) → L2(G) is the unitary operator of pointwise multiplication byα.

It follows that if α is any character ofG, andβ is a character which extendscontinuously toC∗r (G) (that is,β ∈ Gr), then the pointwise productαβ belongsto Gr also. SinceG is a group under pointwise product, andGr is nonempty (afterall, C∗r (G) must havesomecharacters), we deduce thatG = Gr.

(3.7) LEMMA : LetG be a discrete (or a locally compact) group. The commutantof the left regular representation ofG on L2(G) is the von Neumann algebragenerated by the right regular representation (and vice versa of course).

PROOF: Obviously the left and right regular representations commute. ThecommutantR = λ[G]′ of the left regular representation consists of those operators

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onL2(G) that are invariant under the left-translation action ofG onL2(G), and thecommutantL = ρ[G]′ of the right regular representation consists of those operatorsthat are invariant under the right-translation action. It will be enough to show thatL andR commute, since thenρ[G] ⊆ R ⊆ (L)′ = ρ[G]′′.

So letS ∈ L andT ∈ R. For simplicity, assume thatG is discrete. Then(by translation invariance)S is completely determined byS[e] =

∑sx[x], and

similarly T is completely determined byT [e] =∑ty[y].

For group elementsg andh we have

〈TS[g], [h]〉 = 〈S[g], T ∗[h]〉 =⟨∑

sx[gx],∑

ty−1 [yh]⟩

=∑

sxthx−1g−1 .

By a similar calculation

〈ST [g], [h]〉 =∑

tysg−1y−1h.

The substitutionx = g−1y−1h shows that these sums agree, and thusTS = ST .

3.2 Commutative Harmonic Analysis

We will use our results to discuss the bare bones of the Pontrjagin duality theoryfor abelian groups.

For starters letG be a discrete abelian group. As we saw above,G is acompact abelian also, andC∗(G) = C∗r (G) = C(G). The implied isomorphismC∗r (G) → C(G) is called theFourier transform, and denoteF . For an elementf ∈ L1(G) it can be defined by the formula

Ff(α) =∫

Gf(g)α(g)dg

whereα is a character ofG. (Of course this integral is really a sum in the discretecase, but we use notation which is appropriate to the continuous case as well.)Notice that it is immediate that the Fourier transform converts convolution intomultiplication.

(3.8) PLANCHEREL THEOREM The Fourier transform extends to a unitary iso-morphism ofL2(G) withL2(G) (for an appropriate choice of Haar measure).

The phrase ‘for an appropriate choice of Haar measure’ conceals all the variouspowers of2π which show up in the classical theory of Fourier analysis.

PROOF: The regular representation ofC∗r (G) is a cyclic representation withcyclic vector[e]. The associated vector state

τ(f) = 〈f ∗ [e], [e]〉

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corresponds to a measure onG (by the Riesz representation theorem). Translationby a characterα onC(G) corresponds to pointwise multiplication byα onC∗r (G);and sinceα(e) = 1 we see that the measure corresponding toτ is translationinvariant, so it is (up to a multiple) the Haar measureν on G. By the uniqueness ofthe GNS construction, the regular representation ofC∗r (G) is spatially equivalentto the multiplication representation ofC(G) onL2(G, ν), with cyclic vector equalto the constant function 1. In particular there is a unitaryU : L2(G) → L2(G)such that

U(f ∗ h) = (Ff) · (Uh)for f ∈ L1(G), h ∈ L2(G). In particular, takingh = [e] (which is the cyclicvector, so thatUh = 1), we see thatUf = Ff for f ∈ L1. Thus the unitaryU isan extension of the Fourier transform, as asserted.

The Plancherel theorem extends to non-discrete locally compact groups too.Now the representing measureν on G is not finite, and we cannot appeal directlyto the GNS theory. Instead we build the representing measure by hand, in thefollowing way. For eachx ∈ L1 ∩ L2(G) we define a measureνx by the GNSconstruction, so that ∫

bGFfdνx = 〈f ∗ x, x〉L2(G).

If both x andy belong toL1 ∩L2(G) then a simple argument using commutativityshows that

|Fx|2νy = |Fy|2νx

as measures onG. For any compact subsetK of G one can find anx ∈ L1∩L2(G)such thatFx is positive onK. Now we may define an unbounded measureν onG via the Riesz representation theorem as a positive linear functional onCc(G): iff ∈ Cc(G) is supported withinK we put∫

fdν =∫

f

|Fx|2dνx,

for anyx with Fx > 0 onK; and this is independent of the choice ofx. Thus weobtain a measureν on G.

Let uλ be an approximate unit forC∗r (G); then the functionsFuλ increasepointwise to the constant function 1. Forg ∈ L1 ∩ L2(G) we have

‖g‖2 = lim〈uλ ∗ g, g〉 = lim∫Fuλdνg =

∫dνg =

∫|Fg|2dν.

Consequently, the Fourier transform extends to a unitary equivalence betweenL2(G) andL2(G). A similar argument with approximate units shows thatν isa Haar measure onG.

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The next step in the duality theory is to show thatG = G. We won’t do thishere.

3.3 Positive Definite Functions and States

From now on we will restrict attention to discrete, but not necessarily abelian,groups. LetG be such a group. A functionp on such a group is calledpositivedefinite if for every positive integern and everyn-tuple (x1, . . . , xn) of groupelements, then× n matrix whose entries arep(x−1

j xi) is positive as a matrix; thatis, for allα1, . . . , αn ∈ C we have

n∑i,j=1

αiαjp(x−1j xi) > 0.

The collection of positive definite functions onG is denoted byB+(G). Letus observe that every positive definite function is bounded: since for every suchfunction the matrix [

p(e) p(x)p(x−1) p(e)

]is positive, which implies that|p(x)| 6 p(1) for all x.

(3.9)LEMMA : Letϕ be a positive linear functional onC∗(G). Then the mappingp : G→ C defined byp(x) = ϕ([x]) is positive definite; and every positive definitefunction onG arises in this way from a unique positive linear functional. The statesofC∗(G) correspond to the positive definite functions withp(e) = 1.

PROOF: If ϕ is positive linear and we definep as suggested above, then

n∑i,j=1

αiαjp(x−1j xi) = ϕ(f∗ ∗ f) > 0

wheref =∑n

i=1 αi[xi]. Thusp is positive definite. Conversely ifp is positivedefinite and we defineϕ to be the linear extension ofp to `1[G], then the samecomputation shows thatϕ is a positive linear functional on the Banach algebra`1[G]. Every such functional is automatically continuous for theC∗-norm and soextends to a positive linear functional onC∗[G].

We letB+(G) denote the space of positive definite functions onG, and welet B(G) denote the vector space of complex-valued functions onG spanned byB+(G), that is, the setp1 − p2 + i(p3 − p4) wherep1, . . . , p4 are positivedefinite functions. It is not hard to show (Exercise 6.22) that every self-adjointlinear functional on aC∗-algebra is the difference of two positive linear functionals

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and thus we find thatB(G) is isomorphic, as a vector space, to the dual ofC∗(G).We makeB(G) a Banach space by giving it the norm it acquires fromC∗(G)∗ viathis isomorphism. (Notice that the weak-∗ topology onC∗(G)∗ corresponds to thetopology of pointwise convergence onB(G).)

(3.10)PROPOSITION: The pointwise product of two positive definite functions isagain positive definite. Consequently,B(G) is a Banach algebra under pointwisemultiplication.

PROOF: To prove that the product of positive definite functions is again positivedefinite, it is necessary to prove that the pointwise product (Schur product) ofpositive matrices is positive. This is a standard result of matrix theory. Wereproduce the proof. LetA = (aij) andB = (bij) be positive matrices. TheirSchur productC = (aijbij). Now becauseB is positive,B = X∗X for somematrixX; that is,bij =

∑k xikxjk. It follows that for any vectorξ,

ξ∗Cξ =∑

k

∑i,j

aij(xjkξj)(xikξi)

and for eachk the inner sum is positive. HenceC is a positive matrix, as required.To finish the proof it is necessary to show that‖pq‖ 6 ‖p‖‖q‖ for the norm

of B(G). This is obvious ifp, q are positive definite because positive definitefunctions attain their norm at the identity element ofG. The general case followsfrom this same fact if we decomposep and q as differences of positive definitefunctions.

Can every positive definite function onG be approximated (in some sense) bycompactly supportedpositive definite functions? This turns out to be a delicatequestion, and its answer leads us to consider the relationship between the maximalC∗-algebraC∗(G) and the reducedC∗-algebraC∗r (G) which is associated tothe regular representation. To see why, suppose thatp(g) = 〈λ(g)v, v〉 is thepositive definite function associated to a vector state of the regular representation,v ∈ `2(G). Then we have

‖p‖B(G) 6 ‖v‖2;

and therefore, since vectorsv of compact (= finite) support are dense in2(G), wefind that every positive definite function associated to a vector state of the regularrepresentation ofG belongs to the closure (inB(G)) of the space of functions offinite support. Conversely we also have:

(3.11) LEMMA : Any positive definite function of finite support is associated to avector state of the regular representation.

52

PROOF: Let p be such a function. The operatorT of right convolution bypis bounded on2(G), by finiteness of the support; and the fact thatp is a positivedefinite function shows thatT is a positive operator. Moreover,T commutes withthe left regular representationλ[G]. Now putv = T

12 [e] (notice thatv need not

have finite support). Then forg ∈ G,

〈λ(g)v, v〉 = 〈λ(g)T12 [e], T

12 [e]〉 = 〈[g], T [e]〉 = 〈[g], p〉 = p(g)

as required.

Denote byA+(G) the space of positive definite functions onG of the form

p(g) =∑

i

〈λ(g)vi, vi〉

wherevi is a sequence in2(G) having∑‖vi‖2 < ∞ (notice that this implies

that the above sequence converges uniformly to a positive definite function.) Morecanonically we may write

p(g) = Trace(λ(g)T )

where T is a positive trace-class operator on`2(G). We have‖p‖B(G) =∑‖vi‖2 = ‖T‖1, the trace norm ofT , and it follows from the fact that the space

of trace-class operators is a Banach space under the trace norm thatA+(G) is aclosed cone inB+(G). In fact the arguments above prove

(3.12) PROPOSITION: A+(G) is the closure ofC[G] ∩ B+(G) (the compactlysupported functions onG) in B+(G).

Indeed, Lemma 3.11 shows that every compactly supported element ofB+(G)belongs toA+(G), and the remarks preceding the lemma show that every elementof A+(G) can be approximated by compactly supported elements ofB+(G).

We can also introduce the notationA(G) for the subalgebra ofB(G) generatedbyA+(G). It is the closure ofC[G] in B(G). Elements ofA(G) are functions ofthe form

p(g) =∑

〈λ(g)vi, wi〉 = Trace(λ(g)T )

where∑‖vi‖2 <∞ and

∑‖wi‖2 <∞ (the trace-class operatorT need no longer

be positive). In other words, the corresponding elements ofC∗(G)∗ are preciselythose that arise, via the regular representation, from the ultraweakly continuouslinear functionals onB(`2(G)).

53

(3.13) REMARK : The algebraA(G) is called theFourier algebraof G; B(G)is the Fourier–Stieltjesalgebra. To explain this terminology, consider the caseG = Z. The positive definite functions onZ are the Fourier transforms of positivemeasures on the circle, and thus the algebraB(G) is just the algebra of Fouriertransforms of measures on the circle, classically known as the Stieltjes algebra. Onthe other hand the algebraA(G) is generated by sequences of the form

n 7→∫ 2π

0eintu(t)v(t) dt

whereu, v ∈ L2(T). This is the sequence of Fourier coefficients of theL1 functionuv, and the collection of products of this sort has dense span inL1, so we see thatA(Z) is just the algebra of Fourier transforms ofL1 functions on the circle.

(3.14)LEMMA : LetG be a discrete group. The weak-∗ topology on a boundedsubset ofC∗(G)∗ is equivalent to the topology of pointwise convergence on thecorresponding bounded subset ofB(G).

PROOF: It is apparent that weak-∗ convergence in the dual ofC∗(G) impliespointwise convergence inB(G). Conversely suppose thatpλ is a bounded net inB(G) converging pointwise top. Then the corresponding functionals

ϕλ

(∑ag[g]

)=∑

agpλ(g)

converge toϕ(∑

ag[g])

=∑

agp(g)

for all a =∑ag[g] belonging to 1(G). Since theϕλ are uniformly bounded and

`1(G) is norm dense inC∗(G), it follows thatϕλ(a) → ϕ(a) for all a ∈ C∗(G).

(3.15)THEOREM: LetG be a countable discrete group. Then the following areequivalent:

(a) C∗r (G) = C∗(G),

(b) The Banach algebraA(G) possesses a bounded approximate unit,

(c) Every function inB(G) can be pointwise approximated by a norm boundedsequence of compactly supported functions.

54

PROOF: Suppose (c). Then (by lemma 3.14) each stateσ of C∗(G) is a weak-∗limit of vector statesx 7→ 〈λ(x)v, v〉, wherev is a unit vector in the regularrepresentation. It follows that for each such stateσ,

‖σ(a)‖ 6 ‖λ(a)‖

for all a ∈ C∗(G); and consequently‖a‖C∗(G) 6 ‖λ(a)‖, whenceC∗(G) =C∗r (G), which is (a).

Now suppose (a), thatC∗(G) = C∗r (G). Let a ∈ C∗(G) haveϕ(a) = 0 foreach ultraweakly continuous linear functionalϕ onB(`2(G)). Thenλ(a) = 0 andthusa = 0 sinceλ is faithful onC∗(G). It follows that the space of ultraweaklycontinuous functionals on the regular representation is weak-∗ dense inC∗(G)∗.Using lemma 3.14, this implies thatA+(G) is dense inB+(G) in the topologyof pointwise convergence. Now the constant function 1 belongs toB+(G) (itcorresponds to the trivial representation ofG) and thus there is a net inA+(G)converging pointwise to 1. SinceA(G) is generated by compactly supportedfunctions, such a net is necessarily an approximate unit forA(G), and it is boundedbecause the norm inA(G) is given by evaluation ate. This proves (b).

Finally, suppose (b). Letfλ be an approximate unit forA(G). We may assumewithout loss of generality thatfλ consists of compactly supported functions. Letf ∈ B(G); thenffλ is a net inA(G) and it converges pointwise tof .

To connect this discussion with the geometry ofG we introduce the notion ofamenability.

3.4 Amenability

LetG be a discrete group. Aninvariant meanonG is a statem on the commutativeC∗-algebra ∞(G) which is invariant under left translation byG: that ism(Lgf) =m(f), whereLgf(x) = f(g−1x).

(3.16) DEFINITION: A groupG that possesses an invariant mean is calledamenable.

EXAMPLE : The groupZ is amenable. Indeed, consider the states of`∞(Z)defined by

σn(f) =1

2n+ 1

n∑j=−n

f(j),

and letσ be a weak-∗ limit point of these states (which exists because of thecompactness of the state space). Since for eachk ∈ Z, and eachf ,

|σn(Lkf − f)| 6 2k2n+ 1

‖f‖

55

tends to zero asn→∞, we deduce thatσ is an invariant mean.

EXAMPLE : Generalizing the construction above, suppose thatG is a group andthat Gn is a sequence of finite subsets having the property that for each fixedh ∈ G,

#Gn 4 hGn

#Gn→ 0

asn→∞. (Such a sequence is called a Følner sequence.) Then any weak-∗ limitpoint of the sequence of states

f 7→ 1#Gn

∑g∈Gn

f(g)

is an invariant mean onG. It can be proved that every amenable group admits aFølner sequence, but we will not do this here.

EXAMPLE : The free group on two or more generators isnotamenable. To see this,denote the generators of the free groupG byx andy, and define a bounded functionfx on the groupG as follows:fx(g) is equal to 1 if the reduced word forg beginswith the letterx, and otherwise it is zero. Notice then thatfx(xg)−fx(g) > 0, andthatfx(xg)− fx(g) = 1 if the reduced word forg begins withy or y−1. Definefy

similarly; then we have

[fx(xg)− fx(g)] + [fy(yg)− fy(g)] > 1

for all g ∈ G. But this contradicts the hypothesis of amenability: an invariant meanm would assign zero to both terms on the left-hand side, yet would assign 1 to theterm on the right.

There are numerous equivalent characterizations of amenability. For our pur-poses the most relevant one is the following.

(3.17) PROPOSITION: A countable discrete groupG is amenable if and only ifthere is a sequencevn of unit vectors in 2(G) such that the associated positivedefinite functions

g 7→ 〈λ(g)vn, vn〉

tend pointwise to 1.

PROOF: Suppose that we can find such a sequencevn. Sincevn andλ(g)vn

are unit vectors we may write

‖vn − λ(g)vn‖2 = 2− 2<〈λ(g)vn, vn〉 → 0

56

so that, for each fixedg, λ(g)vn − vn → 0 asn → ∞. Now define statesσn on`∞(G) by

σn(f)〈Mfvn, vn〉

whereMf is the multiplication operator on2 associated tof , and letσ be a weak-∗limit point of the statesσn. Since

|σn(Lgf − f)| 6 ‖f‖‖vn − λ(g)vn‖

we see thatσ(Lgf − f) = 0, that is,σ is an invariant mean.Conversely, suppose thatG is amenable and letm be an invariant mean. Then

m is an element of the unit ball of∞(G)∗ = `1(G)∗∗. A theorem of functionalanalysis (Goldstine’s Theorem — a consequence of the Hahn-Banach theorem)states that for any Banach spaceE, the unit ball ofE is weak-∗ dense in the unitball of E∗∗. Thus we can find a netϕλ in the unit ball of`1(G) that convergesweak-∗ to the invariant meanm. We have∑

|ϕλ(g)| = 1,∑

ϕλ(g) 6 1;

and these facts together show that we may assume without loss of generality thateachϕλ is a positive function of norm one. For eachg ∈ G we haveLgϕλ−ϕλ →0 weakly in `1(G) as λ → ∞. Now we use the fact (a consequence of theHahn-Banach Theorem) that the weak and norm topologies on`1 have the sameclosed convex sets to produce a sequenceψ1, ψ2, . . . of convex combinations oftheϕλ such that for eachg ∈ G, Lgψn − ψn → 0 in the norm topology 1(G).Eachψn is a positive function of1-norm equal to one. Takevn to be the pointwisesquare root ofψn.

Now we can finally prove

(3.18)THEOREM: A discrete groupG is amenable if and only ifC∗r (G) = C∗(G).PROOF: Suppose thatG is amenable. Then by Proposition 3.17 there is

a sequencepn of positive definite functions onG of norm one , associatedto vector states of the regular representation, that tend pointwise to 1. Thesefunctionspn constitute a bounded approximate unit forA(G), so by Theorem 3.15,C∗(G) = C∗r (G).

Conversely suppose thatC∗(G) = C∗r (G). ThenA(G) has a bounded ap-proximate unit, which we may assume without loss of generality is made up ofcompactly supported positive definite functions of norm one. By Lemma 3.11,such functions are associated to vector states of the regular representation. ByProposition 3.17,G is amenable.

57

3.5 Example: A crystallographic group

In the next few sections, following Davidson, we shall look in detail at theC∗-algebras of some discrete groups. Our first example is acrystallographic group—the groupG of isometries of the plane generated by the translationτ0,1 : (x, y) 7→(x, y+ 1) together with the glide reflectionσ : (x, y) 7→ (x+ 1

2 ,−y). One verifieseasily thatσ2 is the translationτ1,0 and that the translations

τm,n : (x, y) 7→ (x+m, y + n)

form a free abelian subgroup ofG of index two. ThusG fits into a short exactsequence

0 → Z2 → G→ Z/2 → 0.

A discussion analogous to the present one can be given for any extension of a freeabelian group by a finite group. Notice thatG is amenable soC∗(G) = C∗r (G).

We partitionG as the union of two cosets ofA = Z2: G = A t Aσ. Thus`2(G) = `2(A) ⊕ `2(A) = L2(T2) ⊕ L2(T2) by Fourier analysis. The elementsof A act diagonally with respect to this decomposition so that the subalgebragenerated byA in C∗r (G) is equal toC∗r (Zn) = C(T2) acting diagonally: itconsists of matrices( f 0

0 f ) wheref ∈ C(T2).The relationsτm,nσ = στm,−n, σ2 = τ1,0 show that the action ofσ in this

representation is by the matrix( 0 zJJ 0 ), whereJ is induced by the automorphism

(z, w) 7→ (z, w) of the torusT2. It is convenient to adjust this representationslightly by applyingJ to the secondL2(T2) summand, that is by conjugating by thematrix ( 1 0

0 J ). Then the actions of the group generators are given in the followingtable:

σ 7→(

0 z1 0

), τ1,0 7→

(z 00 z

), τ0,1 7→

(w 00 w

).

It follows thatC∗(G) = C∗r (G) may be identified with theC∗-algebra of two bytwo matrices of the form (

f(z, w) zg(z, w)g(z, w) f(z, w)

)(3.19)

wheref andg are continuous functions onT2.A function of this kind is completely determined by its values on the cylinder

(half-torus) consisting of those points(z, w) where=w > 0. We may thereforewrite w = eiπt with 0 6 t 6 1 and identifyC∗(G) with the algebra of functionsT× [0, 1] →M2(C) generated by

σ 7→(

0 z1 0

), τ1,0 7→

(z 00 z

), τ0,1 7→

(eiπt 00 e−iπt

).

58

By the nature of the construction, we obtain a two-dimensional representationVz,t

of C∗(G) by fixing z ∈ T andt ∈ [0, 1].

(3.20) PROPOSITION: For t ∈ (0, 1) the representationVz,t is irreducible. Fort = 0 and t = 1, Vz,t splits as a direct sumWt,u ⊕ Wt,−u, whereWt,u is theone-dimensional representation on whichτ0,1 acts aseiπt ∈ ±1, andσ acts asu ∈ T with u2 = z.

PROOF: We look at the commutant. Ift ∈ (0, 1) thenτ0,1 acts as a diagonalmatrix with distinct eigenvalues; the only things that commute with such a ma-trix are other diagonal operators. And one easily checks that the only diagonaloperators that commute with the matrix( 0 z

1 0 ) representingσ are multiples of theidentity. Thus the representation is irreducible in this case. Ift = 0 then τ0,1

acts as the identity. The projection12( 1 uu 1 ) commutes with the action ofσ and

thus with the whole representation; it decomposes it into the two one-dimensionalrepresentations mentioned above. Similarly fort = 1.

(3.21) PROPOSITION: The only irreducible representations ofC∗(G) are thosementioned above.

PROOF: Left to the reader. Here is the start of the proof. Sinceσ2 = τ1,0

is in the center ofG, it must act as a scalarzI in any irreducible representation,with z ∈ T by unitarity. Sinceσ itself acts unitarily it is not hard to deduce that itmust act either as a scalaruI, u2 = z, or as( 0 z

1 0 ) with respect to some orthogonaldirect sum decomposition. In the second case, use the relationσ−1τ0,1στ0,1 = e todeduce thatτ0,1 acts as a diagonal matrix( w 0

0 w ). Continue from here. . .

Thecharacterof a finite-dimensional representationπ of a groupG is the classfunctiong 7→ trπ(g) onG; it is invariant under isomorphism of representations.The characterχz,t of Vz,t sendsτ1,0 to 2z andτ0,1 to 2 cosπt. Thus the represen-tationsVz,t are mutually non-isomorphic.

(3.22) COROLLARY: The spectrum of the algebraA = C∗r (G) is equal to itsprimitive ideal space. It is a non-Hausdorff space consists of an open cylinderT × (0, 1) together with two copies ofT attached two-to-one onto the ends of thecylinder.

The famousKadison–Kaplansky Conjectureis this: ifG is a torsion-free groupthenC∗r (G) contains no non-trivial idempotent. We can verify this explicitly forour crystallographic groupG. Indeed, a non-trivial idempotent inC∗r (G) mustbe a function of the form 3.19 whose value is a rank-one projection. Look atthe boundary circle given byt = 0, w = 1 in this parameterization. The onlyprojections of this sort are

12

(1 uu−1 1

)59

whereu2 = z; and simple topology shows that there is nocontinuousdetermina-tion of such a projection around the whole circleT = z : |z| = 1.

3.6 The reducedC∗-algebra of the free group

Let G be a discrete group. Thecanonical traceon C∗r (G)) is the vector stateτassociated with[e] ∈ `2(G); that is,

τ(a) = 〈λ(a)[e], [e]〉.

The associated positive definite function is equal to 1 at the identity and 0 at everyother group element. Notice that we may also write

τ(a) = 〈λ(a)[g], [g]〉

for any fixedg ∈ G (the right-hand side agrees withτ on group elements, and it iscontinuous, so it agrees withτ everywhere. Otherwise stated,τ([g]∗a[g]) = τ(a),soτ(a[g]) = τ([g]a) and by linear extension we deduce

(3.23) PROPOSITION: The stateτ is a trace, that is,τ(aa′) = τ(a′a) for everya, a′ ∈ C∗r (G).

The traceτ is faithful, meaning that ifa ∈ C∗r (G) is positive andτ(a) = 0,thena = 0. Indeed, this is a consequence of the fact that the regular representation(a faithful representation) is the GNS representation associated toτ . To prove itfrom first principles, use the Cauchy–Schwarz inequality to estimate the matrixcoefficients ofλ(a):

|〈λ(a)[g], [g′]〉| 6 〈λ(a)g, g〉12 〈λ(a)g′, g′〉

12 = τ(a).

We are going to use the faithful traceτ as a key tool in analyzing the reducedgroupC∗-algebraC∗r (F2). Notice that since the free group is non-amenable, thisis different from the fullC∗-algebraC∗(F2). We are going to distinguish these byC∗-algebraic properties. Specifically, we will prove thatC∗r (F2) is simple, whichmeans that it has no non-trivial ideals. This is very, very different fromC∗(F2): letG be any finite group with two generators, then according to the universal propertyof C∗(F2) there is a surjective∗-homomorphismC∗(F2) → C∗(G), and of coursethe kernel of this∗-homomorphism is a non-trivial ideal.

Here is the key lemma. Denote byu, v the unitaries inC∗r(F2) correspondingto the generatorsx, y of the group.

60

(3.24)LEMMA : Leta ∈ C∗r (F2). Then the limit

limm,n→∞

1mn

m∑i=1

n∑j=1

uivjav−ju−i

exists in the norm ofC∗r (F2) and equalsτ(a)1.

Before proving this lemma let’s see how we can apply it.

(3.25)THEOREM: TheC∗-algebraC∗r (F2) has no non-trivial ideals (it is simple).

(3.26) REMARK : As we earlier pointed out (corollary 1.5), maximal ideals inC∗-algebras are closed. Thus it makes no difference whether we saytopologicallysimple(no nontrivial closed ideals) oralgebraically simple(no nontrivial ideals,closed or not).

PROOF: Let J be a nonzero ideal and leta ∈ J be positive and nonzero. Thenthe norm limit

limm,n→∞

1mn

m∑i=1

n∑j=1

uivjav−ju−i

belongs toJ . But this limit equalsτ(a)1 by Lemma 3.24, andτ(a) 6= 0 by thefaithfulness of the trace. Consequently1 ∈ J soJ is the wholeC∗-algebra.

(3.27)THEOREM: The trace onC∗r (F2) is unique.

PROOF: If τ ′ is another trace, apply it to the left side of the formula inLemma 3.24 to find thatτ ′(a) = τ(a)τ ′(1) = τ(a).

Let us now begin the proof of Lemma 3.24. The basic observation that isneeded is this: if Hilbert space operatorsA andB have orthogonal ranges, then‖A+B‖2 6 ‖A‖2 + ‖B‖2. We are going to argue then that ifa = [g], g 6= e, thenthe ranges of the terms in the sum

m∑i=1

n∑j=1

uivjav−ju−i

are ‘essentially’ orthogonal, so that the norm of the sum is of orderO((mn)12 ), and

thus becomes zero in the limit after we divide bymn.Here is a more precise statement of the idea in the form that is needed.

(3.28)LEMMA : LetH = H1⊕H2 be a Hilbert space equipped with an orthogonaldirect sum decomposition. LetT ∈ B(H) be an operator mappingH2 into H1,

61

and letU1, . . . , Un ∈ B(H) be unitary operators such that all the compositionsU∗j Ui, j 6= i, mapH1 intoH2. Then

∥∥ n∑i=1

UiTU∗i

∥∥ 6 2√n‖T‖.

PROOF: Start with a special case: assume thatT mapsH (and not justH2)intoH1. Write

∥∥ n∑i=1

UiTU∗i

∥∥2 =∥∥T +

n∑i=2

U∗1UiTU∗i U1

∥∥2.

The first termT in the sum above has range contained inH1, whereas the remainingsum has range contained inH2 (sinceU∗1Ui mapsH1 intoH2. Thus the ranges ofthe two terms are orthogonal and we get

∥∥ n∑i=1

UiTU∗i

∥∥ 6 ‖T‖2 +∥∥ n∑

i=2

UiTU∗i

∥∥2.

By induction we get ∥∥ n∑i=1

UiTU∗i

∥∥26 n‖T‖2

giving the result (without the factor 2) in the special case. The general case followssince any operatorT that mapsH2 into H1 can be written asT = T ′ + (T ′′)∗,where‖T ′‖ 6 ‖T‖, ‖T ′′‖ 6 ‖T‖, and bothT ′ andT ′′ mapH intoH1 (to see this,putT ′ = TP2 andT ′′ = P1T

∗, whereP1 andP2 are the orthogonal projections ofH ontoH1 andH2 respectively.

Now we will use this to prove Lemma 3.24. It will be enough to prove that ifa = [g], then the average1

mn

∑mi=1

∑nj=1 u

ivjav−ju−i converges in norm to 1 ifg = e and to 0 otherwise. For simplicity work with one sum at a time, so considerthe expression

Sn(a) =1n

n∑j=1

vjav−j .

Clearly,Sn([g]) = [g] if g is a power ofy. Otherwise, I claim,‖Sn([g])‖ 6 2n−12 .

To see this decompose the Hilbert spaceH = `2(G) asH1 ⊕H2, whereH1 is thesubspace of2 spanned by those free group elements represented by reduced wordsthat begin with a non-zero power ofx (positive or negative), and similarlyH2 isspanned by those free group elements represented by reduced words that begin with

62

a non-zero power ofy (positive or negative), together with the identity. Supposethatg is not a power ofy. Then the reduced word forg contains at least onex andby multiplying on the left and right by suitable powers ofy (an operation whichcommutes withSn and does not change the norm) we may assume thatg beginsand ends with a power ofx. Now we apply Lemma 3.28 withT the operation of leftmultiplication byg andUi the operation of left multiplication byvi. Once we havechecked that these elements indeed mapH2 andH1 in the way that Lemma 3.28applies, the result follows immediately.

To complete the proof of Lemma 3.24 all we need to do is to reiterate the aboveargument to show that

‖ 1mn

m∑i=1

n∑j=1

uivjav−ju−i‖ 64

min√m,√n

whenevera = [g] andg is not the identity.

3.7 The Kadison–Kaplansky conjecture forF2

In this section we are going to prove that the reducedC∗-algebraC∗r (G) for G =F2 has no non-trivial projections. ThusC∗r (G) is an example of a simple, unitalC∗-algebra with no non-trivial projections. Such examples had long been soughtin C∗-algebra theory. This is not the first such example (that was constructed byBlackadar in the seventies using the theory of AF algebras), but it is the ‘mostnatural’.

We begin with some abstract ideas which lie at the root ofK-homology theoryfor C∗-algebras. They are developments of a basic notion of Atiyah (1970) toprovide a ‘function-analytic’ setting for the theory of elliptic pseudodifferentialoperators.

(3.29) DEFINITION: Let A be a C∗-algebra. AFredholm moduleM overA consists of the following data: two representationsρ0, ρ1 of A on HilbertspacesH0 andH1, together with a unitary operatorU : H0 → H1 that ‘almostintertwines’ the representations, in the sense thatUρ0(a) − ρ1(a)U is a compactoperator fromH0 toH1 for all a ∈ A.

It is often convenient to summarize the information contained in a Fredholmmodule in a ‘supersymmetric’ format: we formH = H0⊕H1, which we consideras a ‘super’ Hilbert space (that is, the direct sum decomposition, which may berepresented by the matrixγ = ( 1 0

0 −1 ), is part of the data), we letρ = ρ0 ⊕ ρ1, andwe put

F =(

0 U∗

U 0

),

63

which is a self-adjoint operator, odd with respect to the grading (that is, anticom-muting with γ), havingF 2 = 1, and commuting modulo compacts withρ(a)for eacha ∈ A. Thus we may say that a Fredholm module is given by a tripleM = (ρ,H, F ).

(3.30)LEMMA : Suppose that a Fredholm module over theC∗-algebraA is given.Then for each projectione ∈ A the operatorUe = ρ1(e)Uρ0(e) is Fredholm,when considered as an operator from the Hilbert spaceρ0(e)H0 to the Hilbertspaceρ1(e)H1.

PROOF: We have

UeU∗e = ρ1(e)Uρ0(e)U∗ρ1(e) ∼ ρ1(e)UU∗ρ1(e) = ρ1(e)

where we have used∼ to denote equality modulo the compact operators. SimilarlyU∗eUe ∼ ρ0(e). ThusUe is unitary modulo the compacts.

It follows that each Fredholm moduleM onA gives rise to an integer-valuedinvariant

IndexM : Proj(A) → Z

defined on the space of projections inA.LetM = (ρ,H, F ) be a Fredholm module over aC∗-algebraA. Its domain of

summabilityis the subset

(A) = a ∈ A : [F, ρ(a)] is of trace class

of A (here[x, y] denotes the commutatorxy − yx). One sees easily thatA is a∗-subalgebra ofA. In fact, it is a Banach algebra under the norm

‖a‖1 = ‖a‖+ Trace |ρ(a)|

which is in general stronger than the norm ofA itself.

(3.31)LEMMA : LetM be a Fredholm module overA, with domain of summabilityA. The linear functional

τM (a) = 12 Trace(γF [F, a])

is a trace11 onA.11That is to sayτ(ab) = τ(ba) for a, b ∈ A.

64

PROOF: We write (suppressing for simplicity explicit mention of the represen-tationρ)

τM (ab) = 12 Trace(γFa[F, b] + γF [F, a]b)

= 12 Trace(γFa[F, b]− γ[F, a]Fb)

= 12 Trace(γFa[F, b]− Fbγ[F, a])

= 12 Trace(γFa[F, b] + γFb[F, a])

and the last expression is now manifestly symmetric ina and b. (We used theidentityF [F, a] + [F, a]F = [F 2, a] = 0, the symmetry property of the trace, andthe fact thatγ anticommutes withF and commutes withb.)

We have now used our Fredholm moduleM to construct an integer-valuedindex invariant of projections inA, and a complex-valued trace invariant of generalelements ofA. These invariants agree where both are defined:

(3.32) PROPOSITION: LetM be a Fredholm module over aC∗-algebraA andlet e be a projection in the domain of summabilityA of M . ThenτM (e) =− IndexM (e). In particular, τM (e) is an integer.

PROOF: We will need to make use of a standard index formula for Fredholmoperators, namely the following: suppose thatP is a Fredholm operator and thatQis a ‘parametrix’ or approximate index forP which has the property that1 − QPand1− PQ are trace-class operators. Then

Index(P ) = Trace(1−QP )− Trace(1− PQ).

One can prove this formula by first showing that any two choices for the parametrixQ differ by a trace-class operator, next using the symmetry property of the traceto show that the right-hand side does not depend on the choice ofQ, and finallymanufacturing an explicitQ for which 1 − QP is the orthogonal projection ontothe kernel ofP and1− PQ is the orthogonal projection onto the kernel ofP ∗.

Let us grant this index formula. Use the computation in the proof of thepreceding lemma to write

τM (e) = τM (e2) = Trace(γFρ(e)[F, ρ(e)]).

Putting

ρ(e) =(e0 00 e1

), F =

(0 U∗

U 0

)we get

[F, ρ(e)]F = Fρ(e)F − ρ(e) =(U∗e1U − e0 0

0 Ue0U∗ − e1

).

65

Thus

τM (e) = Trace(γρ(e)[F, ρ(e)]Fρ(e)) = Trace(U∗eUe − e0 0

0 e1 − UeU∗e

).

SinceU∗e is a parametrix forUe the index formula now gives

τM (e) = − Index(Ue) = − IndexM (e)

as asserted.

(3.33)DEFINITION: We will say that a Fredholm module over a unitalC∗-algebraA is summableif its domain of summability is a dense, unital subalgebra ofA.

(3.34) THEOREM: Let A be a unitalC∗-algebra. Suppose that there exists asummable Fredholm moduleM overA, for which the associated traceτM on thedense subalgebraA ⊆ A is positive, faithful, and hasτM (1) = 1. ThenA has nonon-trivial projections.

PROOF: Suppose first thate is a projection inA. Since 0 6 e 6 1,0 6 τ(e) 6 1; sincee is a projection,τ(e) is an integer by Proposition 3.32.Thusτ(e) equals zero or one. Sinceτ is faithful, if τ(e) = 0, thene = 0 as anoperator; ifτ(e) = 1 thenτ(1− e) = 0 so1− e = 0 ande = 1.

To complete the proof we need only show thatProjA is dense inProjA. Thisis a consequence of two facts. The first is thatA is inverse closed: if a ∈ A isinvertible inA, then its inverse belongs toA. This follows from the well-knownidentity

[F, a−1] = −a−1[F, a]a−1.

Thus an elementa ∈ A has the same spectrum when considered as an element ofA as it does when considered as an element ofA.

The second key fact, or technique, is theholomorphic functional calculusforBanach algebras. This states that ifa is a member of a Banach algebraA, andif f : Ω → C is a holomorphic function, for some openΩ ⊆ C that containssp(a), then one can definef(a) ∈ A in such a way that the usual laws of thefunctional calculus hold:f 7→ f(a) is a homomorphism and the definition off(a)by holomorphic functional calculus agrees with the usual one iff is a rationalfunction with poles off the spectrum. Moreover, thecontinuous functional calculusfor self-adjoint (or normal) elements of aC∗-algebra, which we have been usingthroughout the course, agrees with this holomorphic functional calculus in theinstances in which both are defined.

Let thene ∈ A be a projection, letε be given with0 < ε < 1, and leta ∈ A bea selfadjoint element with‖a−e‖ < ε/2. The spectrum ofa (inA and therefore in

66

A) is contained in the subset(−12ε,

12ε)∪(1− 1

2ε, 1+ 12ε) of R. Hence the function

f defined by settingf(z) = 0 if <z < 12 andf(z) = 1 if <z > 1

2 is holomorphicon a neighborhood of the spectrum ofa. Sof(a) ∈ A and, by the properties of thefunctional calculus,f(a) is a projection. Finally,

‖a− f(a)‖ 6 ε/2

using the norm properties of thecontinuousfunctional calculus, since|f(λ)−λ| <ε/2 for all λ ∈ sp(a). It follows that‖e− f(a)‖ < ε, as required.

So much for abstraction. Following an idea of Cuntz, we will nowconstructa Fredholm module overA = C∗r (F2) that meets the criteria of Theorem 3.34.This Fredholm module makes use of the geometry of thetreeassociated to the freegroupG = F2, which is shown in the figure.

The tree is an infinite graph (1-dimensional simplicial complex) which isequipped with a natural action ofG by translation. LetV denote the set of verticesof the tree,E the set of edges. Considered as a representation ofG, the Hilbertspace 2(V ) is a copy of the regular representation; the Hilbert space`2(E) is thedirect sum oftwo copies of the regular representation, one copy being made up bythe edges in the ‘x-direction’ and the other by the edges in the ‘y-direction’.

We construct a Fredholm module as follows. First we describe the Hilbertspaces:

H0 = `2(V ), H1 = `2(E)⊕ C.

The representations ofC∗r (G) onH0 andH1 are the regular ones, and we use thezero representation on the additional copy ofC. We define the unitaryU : H0 →H1 by first picking a ‘root’ vertex of the tree (labelede in the figure). For eachvertexv 6= e we define

U [v] = ([lv], 0),

where lv is the unique edge of the tree originating fromv and pointing in thedirection ofe. For the vertexe itself we define

U [e] = (0, 1),

thus mapping to the basis vector of the additional copy ofC. It is clear thatUis unitary. Moreover, ifg is a group element, thenλ(g)Uλ(g−1) differs fromUonly in that the special role of the vertexe is now played byge. This affectsthe definition of`(v) only for those finitely many verticesv that lie along thegeodesic segment frome to ge. Thusλ(g)Uλ(g−1)−U is of finite rank. It follows

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thatU commutes modulo finite rank operators with elements ofC[G], and henceby an approximation argument thatU commutes modulo compact operators withelements ofC∗r (G). We have shown that our data define a summable Fredholmmodule.

To complete the proof it is necessary to identify the traceτM on the domainof summability. We need only computeτM ([g]) for group elementsg. If g = eis the identity, we easily calculateτM ([e]) = 1. (For instance this follows fromProposition 3.32 applied to the projection[e] = 1. If g is not the identity, thento computeτM ([g]) = 1

2 Trace(γF [F, [g]]), we must compute traces of operatorssuch as

λ(g)− U∗λ(g)U

onH0. Represent this operator by an infinite matrix relative to the standard basis ofH0. The diagonal entries of the matrix are all zeros, so the trace is zero. It followsthat

τm(∑

ag[g]) = ae

and thusτM agrees with the canonical trace onC∗r (F2), which we know to havethe properties listed in Theorem 3.34.

This completes the proof of

(3.35)THEOREM: TheC∗-algebraC∗r (F2) has no non-trivial projections.

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4 More Examples ofC∗-Algebras

4.1 The Cuntz Algebras

Let us recollect that anisometryin a unitalC∗-algebraA is an elementw such thatw∗w = 1. It then follows thatww∗ is a projection, which we sometimes call therange projectionof the isometry. Any isometry has norm6 1. Any isometry on afinite-dimensional Hilbert space is a unitary, but on infinite-dimensional Hilbertspaces there exist isometries that are not unitaries, the most famous examplebeing the unilateral shift which we have discussed earlier in our work on Toeplitzoperators.

(4.1) DEFINITION: Let n = 2, 3, 4, . . .. TheCuntz algebraOn is the universalC∗-algebra generated byn isometriess1, . . . , sn subject to the single relation∑n

i=1 sis∗i = 1.

For the definition of the universalC∗-algebra defined by generators and rela-tions we refer back to Remark 2.13. Since the projectionssis

∗i sum to 1, they must

be mutually orthogonal.12 Thus we have

s∗i sj = δij1. (4.2)

(4.3) REMARK : One can produce an explicit Hilbert space representation of theCuntz algebra, by taking for generators the isometriesS1, . . . , Sn on the HilbertspaceH = `2(N) defined by

Sk(ej) = enj+k−1

wheree0, e1, . . . are the standard basis vectors forH. We will see later that thealgebra generated by these isometries actually is isomorphic toOn.

Let us use the termword for a finite product ofs’s and their adjoints, somethinglike s1s

22s3s

∗5. Because of the relations that we just noted, every word can be

written in areduced formin which all the terms with adjoints appear to the rightof all those without. Ifµ is a list whose members come from1, 2, . . . , n, weuse the shorthand notationsµ to denote the wordsµ1sµ2 · · · sµk

. Thus any word

12Proof — suppose thatp1 + · · · + pn = 1 where thepi are self-adjoint projections. Multiplyingon the left and on the right byp1 we get

p1p2p1 + p1p3p1 + · · · + p1pnp1 = 0.

The operators here are all positive and sum to zero, so they are all zero. In particularp1p2p1 =(p2p1)

∗(p2p1) = 0, and sop2p1 = 0. Similarly all the other productspipj are zero fori 6= j.

69

can be written in reduced form assµs∗ν . We use the notation|µ| for the number

of members of the listµ, and we call the number|µ| − |ν| theweightof the wordsµsnu

∗. The linear span of the words is a dense subalgebra ofOn.The product of a word of weightl and a word of weightm is a word of weight

l +m, or else zero. Thus the words of weight zero span a subalgebra of the Cuntzalgebra. We are going to identify this subalgebra.

(4.4) LEMMA : LetAkn be the subspace ofOn spanned by the wordssµs

∗ν where

|µ| = |ν| = k. ThenAkn is a subalgebra, and it is isomorphic to the full matrix

algebraMnk(C).PROOF: How does one identify a matrix algebra? Am2-dimensional∗-algebra

over C is a matrix algebra if and only if it is spanned by asystem of matrixunits, which is a basiseIJ , whereI, J = 1, . . . ,m, with respect to which themultiplication law of the algebra is

eIJeKL = δJKeIL

and the adjunction law ise∗IJ = eJI .

(Of course,eIJ corresponds to the matrix with 1 in the(I, J)-position and zeroeselsewhere.) Now, inAk

n, then2k wordssµs∗ν where|µ| = |ν| = k are linearly

independent (for instance because their images in the explicit representation ofRemark 4.3 are linearly independent). Moreover by repeated application of Equa-tion 4.2 we find that

sµs∗νsµ′s

∗ν′ = δνµ′sµs

∗ν′

so that thesµs∗ν comprise a system of matrix units.

The fundamental relation

1 = s1s∗1 + · · ·+ sns

∗n

expresses the identity operator (which we regard as the generator ofA0n = C1) as

a linear combination of generators ofA1n. Generalizing this observation we have

inclusionsAk−1n ⊆ Ak

n of algebras. In terms of the representations ofAk−1n and

Akn as matrix algebras these inclusions amount to then-fold inflation maps

T 7→

T...

T

.

70

It follows that theC∗-subalgebraAn ofOn defined as the closed linear span ofthe words of weight zero is the direct limit of matrix algebras

C →Mn(C) →Mn2(C) →Mn3(C) → · · ·

which is called theuniformly hyperfinite(UHF) algebra oftype n∞. We willdiscuss UHF algebras, and more general inductive limits of matrix algebras, laterin the course; we won’t need any of their theory in our discussion of the Cuntzalgebra for now.

(4.5) DEFINITION: Let A be aC∗-algebra andB a C∗-subalgebra ofA. Aconditional expectationof A ontoB is a positive linear mapΦ: A → B whichis the identity onB. (Positivemeans thatΦ carries positive elements to positiveelements.) The expectationΦ is faithful if a > 0 impliesΦ(a) > 0.

Our analysis of the Cuntz algebra will be based on a certain faithful conditionalexpectation ofOn ontoAn. Rather as we did with the free groupC∗-algebra andits canonical trace, we will represent this expectation in a certain sense as a limitof inner endomorphisms, and we will use this representation to show thatOn issimple.

Let z ∈ T be a complex number of modulus one. The elementszs1, . . . , zsn

of On are isometries which satisfy the defining relation for the Cuntz algebra, soby the universal property there is an automorphism

θz : On → On

which sendssi to zsi for eachi. One hasθz(sµs∗ν) = z|µ|−|ν|sµs

∗ν , so thatAn

is precisely the fixed-point algebra for the one-parameter group of automorphismsθz. It is easy to see thatz 7→ θz(a) is continuous for eacha ∈ On.

(4.6)LEMMA : With notation as above, the mapΦ defined by

Φ(a) =12π

∫ 2π

0θeit(a) dt

is a faithful conditional expectation ofOn ontoAn.

The proof is easy. We will now show how to approximateΦ by inner endomor-phisms:

(4.7) LEMMA : For each positivek one can find an isometrywk ∈ On with theproperty that

Φ(a) = w∗kawk

for everya in the span of those wordssµs∗ν where|µ|, |ν| < k.

71

PROOF: Let us begin with the following observation. Suppose thatm > k,thatx = sm

1 s2, and thaty = sµs∗ν is a word where|µ|, |ν| 6 k. Thenx∗yx = 0 in

all cases except wheny is of the forms`1s∗`1 , whenx∗yx = 1. Indeed, in order that

x∗sµ not equal zero it is necessary thatµ be of the form(1, 1, 1, . . .) and in orderthats∗νx not equal zero it is necessary thatν be of the form(1, 1, 1, . . .). Thus weneed only considery = s`

1s∗`′1 and direct computation yields the stated result.

Now putx = s2k1 s2 and letw =

∑sγxs

∗γ , where the sum runs over all words

sγ of lengthk. Notice thatw is an isometry. Indeed, we have

w∗w =∑γ,δ

sδx∗s∗δsγxs

∗γ =

∑γ

sγx∗xs∗γ =

∑γ

sγs∗γ = 1.

Let us consider what isw∗yw for wordsy = sµs∗ν with |µ|, |ν| 6 k. If |µ| 6= |ν|

then our previous observation shows that each of the terms in the double sum thatmakes upw∗yw is equal to zero, so the whole sum is zero. Suppose now that|µ| = |ν| = k.13 Theny is a matrix unitsµs

∗ν . We have

wy =∑

γ

sγxs∗γy = sµxs

∗ν ,

since only the term withγ = µ contributes, and similarly

yw = sµxs∗ν .

Thusw commutes withy, sow∗yw = y. We have shown that (on the set ofwords with|µ|, |ν| 6 k), the mapy 7→ w∗yw fixes those words of weight zero andannihilates those words of nonzero weight. That is to say,w∗yw = Φ(y) on thosewords.

(4.8)THEOREM: Leta ∈ On be nonzero. Then there are elementsx, y ∈ On suchthatxay = 1.

As we shall see, this property is expressed by saying thatOn is apurely infiniteC∗-algebra.

(4.9) COROLLARY: The Cuntz algebraOn is simple. In particular, therefore, theconcreteC∗-algebra of operators on2 described in Remark 4.3 is isomorphic toOn.

PROOF: To begin the proof, notice that each worda = sµs∗ν has the stated

property; just takex = s∗µ andy = sν . The proof is an approximation argument to

13Remember that the terms with|µ| = |ν| < k are included in the linear span of those with‖mu| = |ν| = k, so we do not have to consider them separately.

72

use this property of the words in the generators. We need to be careful because itis not at all obvious that ifa′, a′′ have the desired property thena = a′ + a′′ willhave it too.

Sincea 6= 0, the faithfulness of the expectationΦ givesΦ(a∗a) > 0, and by asuitable normalization we may assume thatΦ(a∗a) is of norm one. Approximatea∗a by a finite linear combination of wordsb well enough that‖a∗a− b‖ < 1

4 . Inparticular we have‖Φ(b)‖ = λ > 3

4 .Now Φ(b) is a positive element of some matrix algebra overC, and has norm

λ > 34 . The norm of a positive element of a matrix algebra is just its greatest

eigenvalue. So there must be some one-dimensional eigenprojectione for Φ(b), inthe matrix algebra, for whicheΦ(b) = Φ(b)e = λe > 3

4e. We can find a unitaryuin the matrix algebra which conjugatese to the first matrix unit, that is

u∗eu = sm1 s

∗m1 .

It is now simple to check that if we putz = λ−12 eusm

1 , thenz∗Φ(b)z = 1.By the preceding lemma,Φ(b) = w∗bw for some isometryw ∈ On. Put

v = wz; thenv∗bv = 1. Moreover,‖v‖ > 2√3.

Now we compute

‖1− v∗a∗av‖ 6 ‖v‖2‖b− a∗a‖ < 13 ,

and sov∗a∗av is invertible. Lety = v(v∗a∗av)−12 and letx = y∗a∗. Then

xay = y∗a∗ay = 1

as required.

4.2 Real Rank Zero

(4.10)DEFINITION: A projectionp in aC∗-algebraA is calledinfinite if there isa partial isometryw ∈ A withw∗w = p andww∗ = q with q < p. AC∗-algebraA is calledinfinite if it contains an infinite projection.

Clearly the Cuntz algebras are infinite (the identity is an infinite projection).The same applies for instance to the Toeplitz algebraT. We are going to show,however, that the Cuntz algebras are ‘more infinite’ than the Toeplitz algebra.Indeed,T contains the finiteC∗-subalgebraK, but we will see that the analogousstatement is not true forOn. To find out whatis the analogous statement we needthe notion of hereditary subalgebra.

73

(4.11)DEFINITION: LetA be aC∗-algebra. AC∗-subalgebraB ofA ishereditaryif wheneverb ∈ B is positive, and0 6 a 6 b, thena ∈ B also.

It is a consequence of the Cohen factorization theorem thatevery ideal isa hereditary subalgebra. Other examples of hereditary subalgebras include thesubalgebrapAp wherep is a projection inA, or more generallyxAx wheneverx isa positive element ofA (we will prove this below). By a general theorem of Effros,each hereditary subalgebra is of the forml ∩ l∗, wherel is a closed left ideal.

(4.12) LEMMA : LetA be a unitalC∗-algebra and letx ∈ A be positive. ThenxAx is a hereditaryC∗-subalgebra.

PROOF: Clearly it is aC∗-subalgebra. To prove that it is hereditary, let0 6 b 6 xax. Then by the first part of the Cohen factorization theorem we haveb

12 = c(xax)

14 and so

b = (xax)14 c∗c(xax)

14 .

Using the Stone-Weierstrass theorem, approximate(xax)14 uniformly by polyno-

mials inxax, to find thatb ∈ xAx. A slight additional approximation argumenthandles the general case.

(4.13) DEFINITION: A C∗-algebraA is purely infinite if each hereditaryC∗-subalgebra ofA is infinite.

Clearly the Toeplitz algebra is not purely infinite. However, we do have

(4.14)THEOREM: The Cuntz algebrasOn are purely infinite.

PROOF: The proof will depend only on the conclusion of Theorem 4.8, and infact this conclusion isequivalentto the property of being purely infinite, at least forsimpleC∗-algebras. However, we will not prove the whole of this latter assertion,but only the easier direction.

Let B be a proper hereditary subalgebra ofA = On and letb ∈ B be anon-zero positive element. Remark that it is impossible thatb is invertible inA,for then1 6 ‖b−1‖b ∈ B and the hereditary property now tells us thatB = A. Iclaim now that we can findz ∈ A with z∗bz = 1. Indeed, we can findx, y withxb

12 y = 1, hence

1 6 y∗b12x∗xb

12 y 6 ‖x‖2y∗by.

So y∗by is invertible, and we can takez = y(y∗by)−12 . Having foundz put

s = b12 z. Thens is a proper isometry inA, with range projection

p = ss∗ = b12 z∗zb

12 6 ‖z‖2b

belonging to the hereditary subalgebraB. I now claim thatp is an infiniteprojection in B. To see this notice first thatsp = psp belongs toB. Now

74

p = (sp)∗(sp), whereasq = (sp)(sp)∗ = sps∗ is a proper sub-projection ofp.14

We will see that the Cuntz algebras are ‘universal models’ for the behavior ofinfinite simpleC∗-algebras, in a certain rather weak sense.

(4.15)LEMMA : LetA be a simple unitalC∗-algebra, and leta ∈ A be positiveand non-zero. Then there exist finitely many elementsx1, . . . , xn ∈ A such that∑n

i=1 x∗i axi.

PROOF: Because 1 belongs to the ideal algebraically generated bya, there existy1, . . . , yk andz1, . . . , zk in A such that

y1az1 + · · ·+ ykazk = 1.

Note the inequalityyaz + z∗ay∗ 6 yay∗ + z∗az

which follows from the positivity of(y + z∗)a(y∗ + z). We obtain from this

b =k∑

i=1

ykay∗k + z∗kazk > 2.

With n = 2k putxi = y∗i b12 for 1 6 i 6 k andxi = zi−nb

12 for k + 1 6 i 6 n.

(4.16) PROPOSITION: LetA be an infinite simple (unital)C∗-algebra. ThenAcontains isometriesti, i = 1, 2, . . ., such that thetit

∗i are pairwise orthogonal

projections and∑n

i=1 tit∗i < 1 for all n.

PROOF: The projection 1 is infinite so there is an isometrys ∈ A withp = ss∗ a proper projection. By Lemma 4.15 there arex1, . . . , xk ∈ A suchthat

∑ki=1 x

∗i (1− p)xi = 1. Put

t1 = (1− p)x1 + s(1− p)x2 + · · ·+ sk−1(1− p)xk.

Since the projectionssi(1− p) are mutually orthogonal we get

t∗1t1 =∑

x∗i (1− p)xi = 1.

Thust1t∗1 is a projection and, since we easily compute thatt∗1s

k = 0, we have

t1t∗1 6 1− sks∗k.

14Let s be a proper isometry, that is,p = ss∗ < 1. Thensn(s∗)n is a proper sub-projection ofsn−1(s∗)n−1 for eachn. This follows from the fact that the projectionsn(1 − p)s∗n is non-zero; itis equal toxx∗, wherex = sn(1 − p) and we note thats∗nx = 1 − p 6= 0.

75

We can now easily construct the highert’s by induction asti = sk(i−1)t1. Then

tit∗i 6 sk(i−1)(s∗)k(i−1) − ski(s∗)ki

and so we obtain the desired property.

One says that a unitalC∗-algebraA is properly infiniteif it contains isometriest1, t2 such thatt1t

∗1+t2t

∗2 6 1. Thus we have shown that a simple unitalC∗-algebra

is properly infinite. In fact we have shown more:

(4.17) COROLLARY: Every simple infinite unitalC∗-algebraA hasOn as asubquotient (that is, as a quotient of a subalgebra).

PROOF: Let t1, . . . , tn be isometries as in the preceding proposition, letB bethe unital subalgebra ofA generated byt1, . . . , tn, and letJ be the ideal inBgenerated byq = 1 −

∑tit∗i . Notice that sinceq is non-invertible and is in the

center ofB, the idealJ is proper. In the quotient algebraB/J the images of theisometriesti satisfy the relations for the Cuntz algebra, henceB/J ∼= On sinceOn is simple.

Here is a slight generalization.

(4.18) LEMMA : Let A be a simple unitalC∗-algebra and letp be an infiniteprojection inA. Then any other projectionq in A is equivalent to a subprojectionof p (that is, there is a partial isometryt with tt∗ = q andt∗t a subprojection ofp).

PROOF: Use Lemma 4.15 to findz1, . . . , zn such thatq =∑n

i=1 z∗i pzi. Apply

Proposition 4.16 in the hereditary subalgebrapAp to find partial isometriesti witht∗i ti = p and

∑tit∗i < p. Put t =

∑z∗i pt

∗i . Thentt∗ = q, and sot is a partial

isometry; moreover,t∗t = pt∗tp is a subprojection ofp.

(4.19)DEFINITION: A unitalC∗-algebraA hasreal rank zeroif every selfadjointelement ofA is a limit of invertibleselfadjoint elements.

This is a disconnectedness condition. For instance,C(X) has real rank zero ifX is the Cantor set, but not ifX = [0, 1].

(4.20)THEOREM: Each purely infinite simpleC∗-algebraA has real rank zero.

PROOF: Let a be a self-adjoint element ofA and letε > 0 be given. Weare going to find an invertible within2ε of a. Let f(λ) andg(λ) be the functionswhose graphs are shown in the figure:f(λ) = max(λ − ε,min(0, λ + ε)) andg(λ) = max(ε− |λ|, 0). SinceA is purely infinite, there is an infinite projectionpin the hereditary subalgebrag(a)Ag(a). By lemma 4.18, there is a partial isometrys in that subalgebra such thatss∗ = p ands∗s = 1−p. The projectionsp0 = 1−p,

76

p1 = q, andp2 = q − p are mutually orthogonal and sum to 1, and with respect tothe decomposition provided by these projections we have

f(a) =

f(a) 0 00 0 00 0 0

, s =

0 0 01 0 00 0 0

,where we have denoted by1 the isomorphism fromp1 to p0 provided bys. Itfollows that

f(a) + ε(s+ s∗ + p2) =

f(a) ε 0ε 0 00 0 ε

is invertible, and it lies within2ε of a.

4.3 UHF Algebras

In this section we shall studyC∗-algebras (and associated von Neumann algebras)that are constructed as inductive limits of matrix algebras. An important example isthe algebraAn ⊆ On which was used in our earlier discussion of Cuntz algebras.

(4.21)LEMMA : LetMm(C) andMn(C) be matrix algebras. There exists a unital∗-homomorphismMm(C) →Mn(C) if and only ifm|n; and, when this conditionis satisfied, all such∗-homomorphisms are unitarily equivalent to that given by

T 7→

T...

T

with T repeatedn/m times down the diagonal and zeroes elsewhere.

PROOF: Letα : Mm(C) →Mn(C) be a∗-homomorphism of the given sort. Itis injective becauseMm(C) is simple. Leteij be a set of matrix units inMm(C).Thenα(e11) is a projection inMn(C); let v1, . . . , vd be an orthonormal basis forits range. For everyT ∈Mm(C), e11Te11 = λT e11 whereλT ∈ C. It follows thatfor r 6= s,

〈α(T )vr, vs〉 = λT 〈vr, vs〉 = 0,

so that theα(Mm(C))-cyclic subspaces generated byv1, . . . , vd are orthogonal toone another. (We already used this argument — see the proof of lemma 2.37.) Eachsuch cyclic subspace is spanned as a vector space by them vectors

vr = α(e11)vr, α(e12)vr, . . . , α(e1n)vr

77

which one checks are orthonormal. Moreover,Mm(C) acts on thesem vectorsin the standard way. We have shown therefore thatCn has an orthonormal basiscomprisingd sets ofm vectors, on each of which setsMm(C) acts in the standardway. This is what was required.

Suppose now thatk1|k2|k3| · · · is an increasing sequence of natural numbers,each of which divides the next. By the lemma, there is associated to this sequencea unique (up to unitary equivalence) sequence of matrix algebras and unital∗-homomorphisms

Mk1(C)α1

Mk2(C)α2

Mk3(C) · · · .

Let A denote the (algebraic) inductive limit of this sequence. (Reminder aboutwhat this means: Think of unions. More formally, the elements ofA are equiv-alence classes of sequencesTi, Ti ∈ Mki

(C), which are required to satisfyαi(Ti) = Ti+1 for all but finitely manyi, and where two sequences are consideredto be equivalent if they differ only in finitely many places. These may be added,multiplied, normed (remember that theαi are isometric inclusions!), and so on,by pointwise operations.) The algebraA is a normed algebra which satisfies alltheC∗-axioms except that it need not be complete. Its completion is aC∗-algebrawhich is called theuniformly hyperfinite(UHF) algebra(or Glimm algebra) deter-mined by the inductive system of matrix algebras.

(4.22)PROPOSITION: All UHF algebras are simple, separable, unital, and havea unique trace.

PROOF: Obviously they are separable and unital. LetA be a UHF algebra andlet α : A → B be a unital∗-homomorphism to anotherC∗-algebra. Since eachmatrix subalgebraMki

(C) of A is simple, the restriction ofα to it is an injective∗-homomorphism and is therefore isometric. Thusα is isometric on the densesubalgebraA of A, whence by continuityα is isometric on the whole ofA and inparticular has no kernel. Thus,A is simple.

For each matrix algebraMki(C) let τki

: Mki(C) → C denote the normalized

trace, that is

τki(T ) =

1ki

Trace(T ).

The τkithen are consistent with the embeddingsαi so they pass to the inductive

limit to give a traceτ on the dense subalgebraA of A. Moreover,τ is normcontinuous onA (since theτki

are all of norm one on the matrix algebrasMki(C))

and so extends by continuity to a trace, indeed a tracial state, onA. The uniqueness

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of the trace on matrix algebras15 shows that every other trace must agree, up tonormalization of course, with this one.

Suppose thatk1|k2| · · · is the sequence of orders of matrix algebras making upa UHF algebra. For each primep there is a natural number or infinitynp definedto be the supremum of the number of timesp divideski, asi → ∞. The formalproduct

∏p

nprr is called thesupernatural numberassociated to the UHF algebra.

EXAMPLE : The UHF algebra associated to the sequence of embeddings

M2(C) M4(C) M8(C) · · ·

has supernatural number2∞. This example is particularly important; it is calledtheCAR algebra(we will see why in a moment) or theFermion algebra.

(4.23)LEMMA : Let p andq be projections in aC∗-algebraA with ‖p − q‖ < 1.Thenτ(p) = τ(q) for any traceτ onA. In fact,p andq are unitarily equivalent.

PROOF: Definex = qp + (1 − q)(1 − p). Thenxp = qx, and a simplecalculation shows that

x− 1 = 2qp− q − p = (2q − 1)(p− q).

Thus‖x− 1‖ < 1 and sox is invertible. We may define a unitaryu = x(x∗x)−12

by the functional calculus. Nowx∗x commutes withp, and so

up = xp(x∗x)−12 = qx(x∗x)−

12 = qu

as required.

(4.24)THEOREM: Two UHF algebras are isomorphic if and only if they have thesame supernatural number.

PROOF: First we will show that two isomorphic Glimm algebras have the samesupernatural number. Let the two algebrasA andB be obtained from generatingsequencesk1|k2|k3| · · · and l1|l2|l3| · · · respectively. Letα : A → B be anisomorphism. Notice thatα takes the unique traceτA on A to the unique traceτB onB.

We will show that for eachi there is aj such thatki divides lj . Indeed,let p denote the matrix unite11 in Mki

(C) ⊆ A. Thenα(p) is a projection inB. By density there isx = x∗ ∈ Mlj (C) ⊆ B such that‖x − α(p)‖ < 1

20 .Simple estimates give‖x2 − x‖ < 3

16 . Thus the spectrum ofx is contained in

15And how do you provethat? Compute with matrix units to show that every trace-zero finitematrix lies in the span of commutators.

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[−14 ,

14 ] ∪ [34 ,

54 ] and the spectral projectionq of x corresponding toλ > 1

2 has‖q − x‖ < 1

4 . Notice thatq ∈Mlj (C) and that, in particular,‖q − α(p)‖ < 1. Bythe previous lemma and the uniqueness of the trace

τB(q) = τB(α(p)) = τA(p) =1ki.

Since the normalized trace of every projection inMlj (C) is a multiple of1/lj , weconclude thatki divideslj , as required.

Conversely, ifA andB have the same supernatural number, then by passingto subsequences (which doesn’t change the direct limit) we can arrange so thatki|li|ki+1. Then construct inductively embeddings of matrix algebras as in thediagram below

Mk1(C) Mk2(C) Mk3(C) · · ·

Ml1(C) Ml2(C) Ml3(C) · · ·

twisting by appropriate unitary equivalences to make the diagram commute exactly.We obtain a∗-isomorphism between dense subalgebras ofA andB, which extendsautomatically to a∗-isomorphism fromA toB.

Notice as a corollary that there are uncountably many isomorphism classes ofUHF algebras.

EXAMPLE : The Fermion algebra (CAR algebra) arises in quantum field theory.The second quantization of a fermion field leads to the following problem: given aHilbert spaceH, find aC∗-algebraA generated by operatorscv, v ∈ H, such thatv 7→ cv is linear and isometric and

cvcw + cwcv = 0, c∗vcw + cwc∗v = 〈w, v〉1.

These are called thecanonical anticommutation relations. Compare the definitionof a Clifford algebra. We will show that theC∗-algebraA arising in this way is theunique UHF algebra with supernatural number2∞.

Let v be a unit vector onH. Thencv has square zero, andc∗vcv + cvc∗v = 1.

Multiplying through byc∗vcv we find thatc∗vcv is a projection, call itev, and thatcvis a partial isometry with range projectionev and domain projection1−ev. We cantherefore identify theC∗-subalgebra ofA generated bycv: it is a copy ofM2(C),generated by matrix unitse21 = cv, e11 = ev, e22 = 1− ev, e12 = c∗v.

The idea now is to apply the process inductively to the membersv1, v2, . . . ofan orthonormal basis for the Hilbert spaceH. LetA1, A2, . . . be the subalgebras

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of A (all isomorphic toM2(C)) generated bycv1 , cv2 , . . .. If the algebrasAi, Aj

all commuted with one another fori 6= j then we would be done, because thesubalgebra spanned byA1, . . . , An would then be the tensor productM2(C) ⊗· · · ⊗ M2(C) = M2n(C) and we would have representedA as an inductivelimit of matrix algebras. Unfortunately theAi’s do not commute; but they do‘supercommute’ when they are considered as superalgebras in an appropriate way.Rather than developing a large machinery for discussing superalgebras, we willinductively modify the algebrasAj to algebrasBj which are each individuallyisomorphic to2 × 2 matrix algebras, which do commute, and which have theproperty that the subalgebra ofA spanned byA1, . . . , An is equal to the subalgebraof A spanned byB1, . . . , Bn. This, then, will complete the proof.

Let γi be thegrading operatorfor Ai, which is to say the operator correspond-ing to the2 × 2 matrix ( 1 0

0 −1 ). Notice thatγi is self-adjoint, has square 1 andcommutes withcvj for j 6= i, while it anticommutes withcvi ; in particularγi

commutes withγj . Let

B1 = A1 = C∗(cv1), B2 = C∗(cv2γ1), . . . Bn = C∗(cvnγ1 · · · γn−1).

By the same argument as for theAi above, eachBi is isomorphic toM2(C).Moreover the generators of the algebrasBi all commute. Finally, it is clear thatthe subalgebra spanned byB1, . . . , Bn is the same as the subalgebra spanned byA1, . . . , An. The proof is therefore complete.

4.4 UHF Algebras and von Neumann Factors

We are going to use UHF algebras to construct interesting examples of von Neu-mann algebras. (This is something of a reversal of history: Glimm’s UHF-algebraconstruction is explicitly patterned after a construction of Murray and von Neu-mann, that of the so-called hyperfiniteII1 factor.) LetH be a Hilbert space.Recall that avon Neumann algebraof operators onH is a unital, weakly closed∗-subalgebraM ⊆ B(H). Its commutant

M ′ = T ∈ B(H) : TS = ST ∀ S ∈M

is also a von Neumann algebra.

(4.25)DEFINITION: If the center

Z(M) = M ∩M ′

consists only of scalar multiples of the identity, then the von Neumann alegbraMis called afactor.

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There is a sophisticated version of the spectral theorem which expresses everyvon Neumann algebra as a ‘direct integral’ of factors.

We are going to manufacture examples of factors by the following procedure:take aC∗-algebraA (in our examples, a UHF algebra) and a stateσ of A. By wayof the GNS construction, form a representationρσ : A → B(Hσ) and letMσ bethe von Neumann algebra generated by the representation, that isMσ = ρσ[A]′′.If Mσ is a factor, then we callσ a factorial state. As we shall see, many differentexamples of factors can be generated from the factorial states on one and the sameC∗-algebra.

By way of a warm-up let us see that UHFC∗-algebras have trivial center.

(4.26)LEMMA : LetA be a UHF algebra; thenZ(A) = C1.

PROOF: The proof applies to any unitalC∗-algebra with a unique faithful trace,τ . Let z ∈ Z(A). Then the linear functionala 7→ τ(az) is a trace onA, so byuniqueness it is equal to a scalar multiple ofτ , sayλτ . But then

τ(a(z − λ1)) = 0

for all a ∈ A, in particular fora = (z − λ1)∗, so by faithfulnessz − λ1 = 0 andzis a multiple of the identity.

Of course this doesnot prove that various von Neumann algebra completionsof A must have trivial center. We need more work to see that! We will discuss aspecial class of states on UHF algebras, theproduct states.

Recall that a UHF algebraA is a completed inductive limit of matrix algebras

Mk1(C) ⊆Mk2(C) ⊆Mk3(C) ⊆ · · · .

Puttingd1 = k1, di+1 = ki+1/ki for i > 1, we can write the inductive sequenceequivalently as

Md1(C) ⊆Md1(C)⊗Md2(C) ⊆Md1(C)⊗Md2(C)⊗Md3(C) ⊆ · · · .

Notice thatMdi+1(C) is the commutant ofMki

(C) insideMki+1(C). For eachi,

let σi be a state ofMdi(C). (We will discuss the explicit form of these states in a

moment.) Then one can easily define a stateσ on the algebraic inductive limitAby

σ(a1 ⊗ a2 ⊗ · · · ) = σ1(a1)σ2(a2) · · · .

Note that all but finitely many of theai are equal to one. When each of theσi

are the normalized trace, this is exactly the construction we performed in the lastlecture to obtainτ . The functionalσ has norm one, so it extends by continuity to astate on the UHF algebraA. Such a state is called aproduct state.

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(4.27)LEMMA : LetA be a UHF algebra and letσ be a product state on it. LetM = Mki

(C) be one of the matrix algebras in an inductive sequence defininingA. If x ∈ M and y ∈ M ′ (by which I mean the commutant ofM in A), thenσ(xy) = σ(x)σ(y).

PROOF: The result is apparent ify ∈ Mkj(C) for somej > i. For then

the commutation condition allows us to writex = X ⊗ 1, y = 1 ⊗ Y relativeto the tensor product decomposition of the matrix algebra, and the result followsfrom the definition of a product state. To complete the proof we must thereforeshow that anyy ∈ A that commutes withMki

(C) can be approximated in normby elementsyj ∈ Mkj

(C) that commute withMki(C). To do this, first find a

sequencey′j ∈Mkj(C) with y′j → y in norm. Then defineyj by averagingy′j over

the unitary group ofMki(C):

yj =∫

U(ki)u∗y′ju du

wheredu denotes Haar measure on the compact groupU(ki). Thenyj ∈ Mkj(C)

and commutes withU(ki) and hence with the whole ofMki(C) (every matrix is a

linear combination of unitaries). Moreover, since

y =∫

U(ki)u∗yu du

we see that‖y − yj‖ 6 ‖y − y′j‖ → 0. This completes the proof.

Recall that a stateσ is pure if it is an extreme point of the space of states. Itis equivalent to say that ifϕ is a positive linear functional withϕ 6 σ, thenϕ is ascalar multiple ofσ. LetM = Mr(C) be a finite matrix algebra; then the states

σq(T ) = Tqq, q = 1, . . . , r

are pure states ofM . To see this we can argue as follows, takingq = 1 forsimplicity of notation: ifϕ 6 σq andϕ is positive, thenϕ(eii) = 0 for i > 2;thereforeϕe[ij

∗eij) = 0 for i > 2; therefore by Cauchy-Schwarzϕ(eij) = 0 fori > 2, and symmetrically forj > 2. Thusϕ = ϕ(1)σ1.

(4.28) LEMMA : LetA = Md1(C) ⊗Md2(C) ⊗ · · · be a UHF algebra and letσ = σq1 ⊗ σq2 ⊗ · · · be a product state in which each of theσqi is a pure state ofMdi

(C) of the sort described above. Thenσ is a pure state ofA.

PROOF: Letϕ 6 σ and prove by induction oni thatϕ = ϕ(1)σ when restrictedtoMki

(C) ⊆ A, using the argument sketched above which proves that theσq arepure states of the corresponding matrix algebras.

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Let us call the state described above thepure product stateσq associated tothe sequence of indicesq = (q1, q2, . . .). We are going to investigate when theirreducible representations associated to two pure product states are equivalent.We need the following lemma:

(4.29)LEMMA : LetA be a unitalC∗-algebra and letσ, σ′ be pure states ofA.Then the irreducible representations associated toσ andσ′ are unitarily equivalentif and only if there is a unitaryu ∈ A fow whichσ(a) = σ′(uau∗).

In other words, unitarily equivalent irreducible representations are ‘internally’unitarily equivalent. The proof follows from a version of Kadison’s transitivitytheorem; it will be set as an exercise.

(4.30)PROPOSITION: Two pure product statesσq andσq′ define unitarily equiv-alent irreducible representations if and only if the sequencesq andq′ agree exceptfor finitely many terms.

(4.31) COROLLARY: Let A be a UHF algebra. Then the unitary dualA isuncountable, but has the indiscrete topology.

Indeed, the Proposition shows that a UHF algebra has uncountably manyinequivalent unitary representations; but since it is simple the primitive ideal spaceis trivial, and henceA has the indiscrete topology by Theorem 2.31.

PROOF: If the sequencesq andq′ agree for subscriptsi > n, say, then theassociated pure states define two different irreducible representations ofMkn(C).Any two irreducible representations of a full matrix algebra are equivalent, so thereis a unitaryu ∈ Mkn(C) such thatσq(a) = σq′(u∗au) for all a ∈ Mkn(C).This equality extends to alla ∈ A by construction, and then it extends toAby continuity. So the associated irreducible representations ofA are unitarilyequivalent.

Conversely, suppose thatσq(a) = σq′(u∗au) for some unitaryu. Find anelementx ∈ Mkn(C) (n sufficiently large) with‖u − x‖ < 1

2 . Suppose for acontradiction thatqj = 1, q′j = 2 for somek > n, and lete ∈ Mkj

(C) be ofthe form1 ⊗ e11, wheree11 is the matrix unit inMdj

(C). Now σ′(e) = 0, soσ′(x∗ex) = σ′(x∗xe) = σ′(x∗x)σ′(e) = 0 using the multiplicative property ofproduct states and the fact thate commutes withx. Now write

1 = σ(e) = σ′(u∗eu) 6 |σ′((u∗ − x∗)eu)|+ |σ′(x∗e(u− x))|+ |σ′(x∗ex)| < 1

to obtain a contradiction.

Irreducible representations are (of course!) factorial. However UHF algebrasalso possess factorial representations that are not irreducible, In fact we have

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(4.32)PROPOSITION: Each product state of a UHF algebra is factorial.

PROOF: Let σ be a product state of the UHF algebraA and let(ρσ,Hσ, uσ) bea cyclic representation associated withσ via the GNS construction. Suppose thatz ∈ Z(ρσ[A]′′). By the Kaplansky Density Theorem 1.46, there is a sequenceyj

in A with yj → z weakly and‖yj‖ 6 ‖z‖ for all j. Fix n, and letzj be obtainedfrom yj be averaging over the unitary group ofMkn(C) ⊆ A, as in the proof ofLemma 4.27. For each fixed such unitaryu we have

ρ(uyju∗) → ρ(u)z(u∗) = z

in the weak topology, boundedly in norm; so by applying the Dominated Conver-gence Theorem we see thatρ(zj) → z weakly. Sincezj commutes withMkn(C)we have forx, y ∈Mkn(C),

〈z[x], [y]〉 = lim〈ρ(zj)[x], [y]〉 = limσ(y∗zjx) = σ(y∗x)σ(z) = 〈[x], [y]〉σ(z)

using Lemma 4.27. Since this holds for alln, it follows thatz = σ(z)1 is a scalarmultiple of the identity. ThusρσA

′′ is a factor.

4.5 The Classification of Factors

We now know how to build a great variety of factors from product states of UHFalgebras. But how are these factors to be distinguished? In this section we willbriefly outline the classification of factors that was developed by Murray and vonNeumann.

(4.33) DEFINITION: Let M be a von Neumann algebra. Aweight on M isa functionϕ : M+ → [0,∞] (notice that the value∞ is permitted) which ispositive-linear in the sense that

ϕ(α1x1 + α2x2) = α1ϕ(x1) + α2ϕ(x2)

for α1, α2 ∈ R+ andx1, x2 ∈M+.

The canonical example is the usual trace onB(H). We say thatϕ is semi-finiteif the domain

Mϕ+ = x ∈M+ : ϕ(x) <∞

is weakly dense inM+, and thatϕ is normal if

limϕ(xλ) = ϕ(limxλ)

wheneverxλ is a monotone increasing net inM that is bounded above (and henceweakly convergent). Ifϕ(u∗xu) = ϕ(x) for all x ∈ M+ and unitaryu we say

85

thatϕ is a tracial weight, or sometimes just atrace (so long as we remember thatit does not have to be bounded).

bbbb

Finite faithfulnormal trace

Semifinite faith-ful normal trace

No faithful nor-mal trace

Minimal projec-tion

TypeIn TypeI∞ Impossible

No minimalprojection

TypeII1 TypeII∞ Type III

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5 Hilbert Modules

This section represents a discussion which was not delivered in the course. I wantto give a brief introduction to the theory ofHilbert modulesoverC∗-algebras, andto some of their applications.

5.1 Basic Definitions

LetA be aC∗-algebra. AHilbert moduleoverA is a rightA-moduleM equippedwith anA-valuedC-sesquilinear ‘inner product’

〈·, ·〉M ×M → A

satisfying the following axioms analogous to the usual ones for a Hilbert space:

(i) 〈x, ya+ y′a′〉 = 〈x, y〉a+ 〈x, y′〉a′, for all x, y, y′ ∈M anda, a′ ∈ A;

(ii) 〈x, y〉 = 〈y, x〉∗;

(iii) 〈x, x〉 > 0 (the inequality in terms of the order onAsa);

(iv) If 〈x, x〉 = 0 thenx = 0;

(v) M is complete in the norm‖x‖M = ‖〈x, x〉‖12A (we will prove in a moment

that, given (i) through (iv), this really is a norm).

Obviously,A is a Hilbert module over itself, with inner product〈x, y〉 = x∗y.The basic commutative examples are sections of bundles. Indeed, letA = C(X)and letV be a Hermitian vector bundle over the compact spaceX. Let M bethe space of continuous sections ofV , and let〈〉 : M ×M → A be the fiberwiseinner product. ThenM is a HilbertA-module. Later, we will see that allfinitelygeneratedHilbertA-modules are of this form.

(5.1) REMARK : General Hilbert modules overC(X) can be thought of as madeup of sections of ‘generalized bundles’ overX. Indeed, suppose thatM is a HilbertC(X)-module. Then, for each fixedp ∈ X, 〈x, y〉p = 〈x, y〉(p) gives a positivesemidefinite sesquilinear form onM (a ‘pre-inner-product’) and completing weobtain a Hilbert spaceMp. Thusp 7→Mp assigns to each point ofX a Hilbert spaceandM provides a space of ‘continuous sections’ ofp 7→ Mp, whose pointwiseinner products are continuous and whose restrictions to each fiber are dense in thatfiber. This data is said to describe acontinuous field of Hilbert spaces. Conversely,the sections of a continuous field of Hilbert spaces form a HilbertC(X)-module.For an example of such a continuous field that is not a vector bundle, consider the

87

spaceM = C0(0, 1] as a Hilbert module overA = C[0, 1]. This space is the spaceof sections of a field of Hilbert spaces with fiber 0 over 0 andC over the otherpoints of the interval. Note thatM is not finitely generated overA.

Return to the proof that the norm‖x‖M defined above really is a norm. As inthe classical case this will follow from a Cauchy–Schwarz inequality.

(5.2) LEMMA : LetM be a rightA-module satisfying (i)–(iii) above and supposethatx, y ∈M . Then we have the inequality (inA)

〈x, y〉∗〈x, y〉 6 ‖〈x, x〉‖〈y, y〉.

Consequently, with the definition of‖ · ‖M in (v) above we have

‖〈x, y〉‖A 6 ‖x‖M‖y‖M ,

and it follows easily that‖ · ‖M is indeed a norm.

PROOF: Imitate the usual proof, as follows: consider the inequalities inAsa

0 6 〈xa+ y, xa+ y〉= a∗〈x, x〉a+ a∗〈x, y〉+ 〈y, x〉a+ 〈y, y〉6 ‖〈x, x〉‖a∗a+ a∗〈x, y〉+ 〈y, x〉a+ 〈y, y〉

where we have used Proposition 1.22. Puttinga = λ〈x, y〉, λ ∈ C, we get(λ2‖〈x, x〉‖+ 2λ

)〈x, y〉∗〈x, y〉+ 〈y, y〉 > 0.

If 〈x, x〉 = 0, then takingλ large and negative in this inequality we see that〈x, y〉 = 0 also. Otherwise, putλ = −‖〈x, x〉‖−1 to get Cauchy-Schwarz.

Note that the proof of Cauchy-Schwarz only requires (i)–(iii). It follows thatgiven anA-moduleM satisfying (i)–(iii), the setN of vectors of norm zero formsa submodule, and dividing by thatN we obtain a moduleM/N satisfying (i)–(iv).Then by completing we may obtain a Hilbert module. This is the analog for Hilbertmodules of a construction used in the GNS process.

EXAMPLE : An important example of a HilbertA-module is the ‘universal’moduleHA which is comprised of sequencesan of elements ofA such that∑a∗nan converges inA. Using the Cauchy-Schwarz inequality it is not hard

to show that this is a Hilbert module. Note the convergence condition carefullyhowever: it is not equivalent to say that

∑‖an‖2 converges (a series of positive

elements of aC∗-algebra can converge in norm without converging absolutely, forinstance the series

∑ 1nen, whereen is the orthogonal projection onto then’th

basis vector, converges in norm but not absolutely inK(`2).

88

The above example is a special case of a general notion of infinite direct sumof Hilbert modules.

(5.3) LEMMA : LetM be a HilbertA-module. ThenMA is dense inM . In factM〈M,M〉 is dense inM , where〈M,M〉 denotes the subspace ofA spanned byinner products.

Notice that〈M,M〉 need not be dense inA (obvious examples). We say thatM is full if this is so.

PROOF: The closure of〈M,M〉 is aC∗-idealJ inA. Letuα be an approximateunit for J . Then forx ∈M

〈x− xα, x− xα〉 = 〈x, x〉 − uα〈x, x〉 − 〈x, x〉uα + uα〈x, x〉uα → 0

and thusxα → x, giving the result.

(5.4) REMARK : As well as the ordinary norm on a HilbertA-moduleM , wemay define an ‘A-valued norm’|x| = 〈x, x〉

12 (the square root being implemented

by functional calculus inA). From Cauchy–Schwarz one easily deduces theinequalities

|〈x, y〉| 6 ‖x‖|y|, |xa| 6 ‖x‖|a|.However theA-valued norm shares the odd properties of the absolute value on aC∗-algebra, e.g. it isn’t true always that|x+ y| 6 |x|+ |y|.

In the theory of Hilbert space a critical role is played by the existence oforthogonal complements to closed subspaces. IfM is a Hilbert space (or Hilbertmodule) andN a subspace, theorthogonalsubspace is

N⊥ = y ∈M : 〈x, y〉 = 0 ∀x ∈ N.

TheProjection Theoremstates that ifM is a HilbertspacethenN⊥ is complemen-tary toN ; eachx ∈ M can be decomposed uniquely into a component inN anda component inN⊥. The Projection Theorem however fails for Hilbert modules.For instance ifA = C[0, 1], M = A, andN = C0(0, 1] considered as a HilbertA-module, thenN⊥ = 0 and we have no direct sum decomposition ofM intoN ⊕N⊥. This is the fundamental difficulty of Hilbert module theory.

(5.5) DEFINITION: Let M andN be HilbertA-modules. Anadjointable mapfrom M to N is a linear mapT : M → N for which there exists anadjointT ∗ : N →M , necessarily unique, satisfying

〈Tx, y〉 = 〈x, T ∗y〉, x ∈M, y ∈ N.

The set of adjointable maps will be denotedB(M,N , or B(M) if M = N .

Because of the failure of the Projection Theorem, it need not be the casethat a norm bounded linear mapM → N is adjointable. On the other hand, anadjointable map must be bounded (a Uniform Boundedness Principle argument).

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(5.6)PROPOSITION: LetM be a HilbertA-module. ThenB(M) is aC∗-algebra.

PROOF: B(M) is clearly a normed∗-algebra. TheC∗-identity can be provedby mimicking the Hilbert space proof: forx ∈M , ‖x‖ = 1,

‖Tx‖2 = 〈Tx, Tx〉 = 〈x, T ∗Tx〉 6 ‖T ∗T‖.

The supremum of the left side over all possiblex is ‖T‖2, giving the result. Itfollows as usual that‖T‖ = ‖T ∗‖ for all T ∈ B(M), and so we see thatB(M) isa closed subspace of the Banach algebra of all boundedC-linear mapsM → M ,and therefore is complete.

It is just a restatement of the definition to say that ifT ∈ B(M,N) andx ∈ M then ‖Tx‖ 6 ‖T‖‖x‖. In fact, however, there is a similar inequalityfor theA-valued norm (think ofT as operating ‘fiberwise’ on sections of a vectorbundle.) To prove it, we use states ofA to reduce to the scalar case — a devicewhich could also have been used in some earlier arguments.

(5.7) PROPOSITION: LetM andN be HilbertA-modules, and letT : M → Nbe an adjointable operator. Then

|Tx| 6 ‖T‖|x|

for all x ∈M .

PROOF: Let σ be a state ofA. We haveσ(|Tx|2) = σ(〈T ∗Tx, x〉). A simpleinduction using the Cauchy-Schwarz inequality for the positive sesquilinear formσ(〈·, ·〉) gives

σ(〈T ∗Tx, x〉) 6 σ(〈(T ∗T )2nx, x〉)2−n

σ(|x|2)1−2−n.

As n→∞ we obtain

σ(〈T ∗Tx, x〉) 6 ‖T‖2σ(|x|2)

and since this holds for all statesσ we get|Tx|2 6 ‖T‖2|x|2, whence the result.

Let E andF be HilbertA-modules. Arank one operatorfrom E to F is alinear mapE → F of the form

θx,y(z) = x〈y, z〉, x ∈ F, y ∈ E.

It is adjointable, with adjointθy,x. The closed linear span of rank one operatorsis denotedK(E,F ) and called the space ofcompactoperators fromE to F .(Warning: They need not be compact in the sense of Banach space theory!) IfE = F , the subspaceK(E) of compact operators inB(E) is aC∗-ideal.

90

EXAMPLE : If A is a unitalC∗-algebra, and we considerA as a Hilbert moduleover itself, the it is easy to see thatB(A) = K(A) = A. If A is non-unital thenB(A) is much bigger thanA (in fact it is the so-calledmultiplier algebraof A, seelater), but it is still the case thatK(A) = A. To see this, choose an approximate unituλ for A and define an isomorphismA → KA by sendinga ∈ A to lim θa,uλ

,having first checked of course that the net appearing on the right is Cauchy.

Note thatB(E) is not a von Neumann algebra in general.

(5.8) DEFINITION: Thestrict topologyon B(E) is the topology induced by theseminormsT 7→ ‖Tx‖, T 7→ ‖T ∗y‖ for all x, y ∈ E. (Essentially the strong-∗topology.)

Let E be a HilbertA-module and letuλ be an approximate unit forK(E). Ifx ∈ E〈E,E〉 thenx can be written as a linear combination of images of rank oneoperators and so it is simple to see thatuλx → x. From Lemma 5.3 we concludethatuλx → x for all x ∈ E. ThereforeTuλ → T strictly for all T ∈ B(E) andwe conclude:

(5.9) PROPOSITION: Let E be a HilbertA-module. The unit ball ofK(E) isstrictly dense in the unit ball ofB(E).

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6 Exercises

6.1 Exercises Sep 8, due Sep 15

EXERCISE 6.1: Let A be thedisk algebra, that is the Banach algebra (withpointwise operations and supremum norm) of continuous functions on the closedunit diskD in the complex plane which are holomorphic in the interior ofD. Showthat the operation

f∗(z) = f(z)

defines an involution onA which makes it into a commutative Banach∗-algebra inwhich not every selfadjoint element has real spectrum.

EXERCISE 6.2: Show that ifλ is an isolated point in the spectrum of a normaloperatorT thenλ is an eigenvalue ofT . Is this true ifT is not normal?

EXERCISE6.3: LetX be a locally compact Hausdorff space, and letCb(X) denotethe algebra of bounded, continuous, complex-valued functions onX. Show thatCb(X) is a commutativeC∗-algebra. Deduce thatCb(X) = C(Z), whereZ is acompact Hausdorff space containingX as a dense open subset (acompactificationof X). Prove moreover thatZ is the universal compactification, in the sense that ifZ ′ is another compactification ofX then there is a commutative diagram

X Z

Z ′

where the vertical arrow is a uniquely determined surjection. (The spaceZ is calledtheStone–Cech compactificationof X.)

EXERCISE 6.4: LetA = C(X) be a commutative and unitalC∗-algebra. IfZ isa closed subset ofX then the set of those functions inA whose restriction toZ iszero constitutes an ideal inA. Prove that every ideal arises in this way.

EXERCISE6.5: LetA be aC∗-algebra.

(a) Letx, a ∈ A with a > xx∗ > 0. Show that the elements[1j + a

]− 12a

14x, j = 1, 2, . . .

form a Cauchy sequence inA which converges to an elementb ∈ A withx = a

14 b and‖a‖

14 > ‖b‖.

Now letπ : A→ B(H) be a representation.

92

(b) Show that for anyv ∈ π[A]H and anyε > 0, one can findx in the unit ballof A such that‖v − π(x)v‖ < ε. (Use an approximate unit.)

(c) Givenv ∈ π[A]H, use part (b) above to define by induction a sequencexnof elements in the unit ball ofA and a sequencevn of elements ofπ[A]Hsuch thatv1 = v and

vn = vn−1 − π(xn−1)vn−1, ‖vn‖ 6 4−n.

for n > 1.

(d) Show that∑π(xn)vn converges tov.

(e) Leta =∑

4−nxnx∗n. Using part (a) above write2−nxn = a

14 bn with ‖bn‖

bounded. Show that

v = π(a14 )∑

π(bn) (2nvn)

where the series converges in norm, and deduce thatv ∈ π[A]H.

Deduce thatπ[A]H is always closed inH.

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6.2 Exercises Sep 22, due Sep 29

EXERCISE 6.6: LetA be aC∗-algebra. Show that the two by two matrix algebraM2(A) can be made into aC∗-algebra also. (Hint: Use the Gelfand–NaimarkRepresentation Theorem.)

EXERCISE 6.7: Show that every irreducible representation of a commutativeC∗-algebra is one-dimensional.

EXERCISE6.8: LetT be a normal operator on a Hilbert spaceH and letE be theassociated spectral measure. Show that the projectionE(D) corresponding to theopen unit discD has range equal to the subspace

v ∈ H : limn→∞

‖Tnv‖ = 0.

Deduce that if a bounded operatorS commutes withT , then it commutes withE(D). Argue further thatS commutes withE(U) for every open diskU ⊂ C.Deduce finally thatS commutes withT ∗ (Fuglede’s Theorem).

EXERCISE 6.9: LetF be the vector space of finite-rank operators fromH to H ′,whereH andH ′ are Hilbert spaces. Show that an inner product may be defined onF by

〈S, T 〉HS = Trace(S∗T )

where the trace of a finite-rank operator is the sum of its diagonal matrix entriesrelative to an orthonormal basis. This inner product is called theHilbert–Schmidtinner product.

Let ‖ · ‖HS denote the corresponding norm. Prove that

‖T‖ 6 ‖T‖HS

for any T ∈ F. Deduce that the completion ofF with respect to the Hilbert–Schmidt inner product may be identified with a certain space of bounded operatorsfromH toH ′. These are called theHilbert–Schmidt operators.

Prove that every Hilbert–Schmidt operator is compact, and give an exampleof a compact operator which is not Hilbert–Schmidt. If eitherH or H ′ is offinite dimensionn, prove that every bounded operator is Hilbert–Schmidt and that‖T‖HS 6 n

12 ‖T‖.

EXERCISE 6.10: An operatorT ∈ B(H) is said to be oftrace classif it can bewritten T = T1T2, whereT1 andT2 are Hilbert–Schmidt operators. Show that ifthis is so, then the number

Trace(T ) = 〈T ∗1 , T2〉HS

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depends only onT , and that

Trace(ST ) = Trace(TS)

for all bounded operatorsS.Show thatT 7→ Trace |T | defines a norm on the space of trace-class operators

which makes it into a Banach space, and that the dual of this Banach space may beidentified withB(H). Deduce thateveryvon Neumann algebra is isomorphic as aBanach space to the dual of some Banach space.

Here is some discussion of the last exercise, which could be put into the textof the notes later. We take up the story at the beginning of the second paragraph ofthe last exercise above.

(6.11) LEMMA : Suppose thatS ∈ B(H) and thatT is of trace class. Then|Trace(ST )| 6 ‖S‖Trace |T |.

PROOF: LetT = V |T | be the polar decomposition. Think ofTrace(ST ) as theHilbert-Schmidt inner product ofSV |T |

12 and|T |

12 , and use the Cauchy-Schwarz

inequality to get

|Trace(ST )|2 6 Trace(|T |12V ∗S∗SV |T |

12 ) Trace(|T |) 6 ‖S‖2 Trace(|T |)2

where in the second step we used the inequality of positive operators|T |12V ∗S∗SV |T |

12 6

‖S∗S‖|T |12V ∗V |T |

12 = ‖S∗S‖|T |.

(6.12)LEMMA : If S andT are trace-class thenTrace(|S + T |) 6 Trace |S| +Trace |T |.

PROOF: Let S + T = V |S + T | be a polar decomposition. Then

Trace |S + T | = Trace(V ∗S) + Trace(V ∗T ) 6 Trace |S|+ Trace |T |

where we used the previous lemma for the second step.

This shows that the trace-norm is a norm. Since by spectral theoryTrace |T |is the sum of the eigenvalues of|T |, whereas‖T‖ is the supremum of theseeigenvalues, it is clear thatTrace |T | > ‖T‖. It now follows that the spaceL1(H)of trace-class operators onH is a Banach space of compact operators, just as in theHilbert-Schmidt case above (by the way, the space of Hilbert-Schmidt operators issometimes denotedL2(H).)

A mapΦ from B(H) to L1(H)∗ is defined by assigning to a bounded operatorS the linear functional

Φ(S) : T 7→ Trace(ST )

95

on L1(H). It is easy to check that this is a continuous, isometric, linear map. Weneed to see that it is also surjective, which is to say that every linear functional onL1(H) is of this sort. Thus letϕ ∈ L1(H)∗. For every finite-rank projectionPthere is a trace-class (indeed finite-rank) operatorSP = SPP with ‖SP ‖ 6 ‖ϕ‖such that

ϕ(PT ) = Trace(SPT )

(for instance one can use the Riesz representation theorem in the Hilbert spaceL2(H) to see this). TheSP form a net parameterized by the space of finite rankprojections. Since closed balls inB(H) are weakly compact, there is a subnet oftheSP that converges weakly to a bounded operatorS, with ‖S‖ 6 ‖phi‖. Now itis easy to verify that for any trace-class operatorT the functionalX 7→ Trace(XT )is weakly continuous on bounded subsets ofB(H). Consequently,

Trace(ST ) = lim Trace(SPT ) = limϕ(PT ) = ϕ(T )

sincePT → T in trace norm.Theultraweak topologyon B(H) is the weak-∗ topology that it inherits from

its representation as the dual ofL1(H). Confusingly, the ultraweak topology isstronger than the weak topology. We have already noted above that the weakand ultraweak topologies agree on bounded subsets ofB(H). Since according toKaplansky’s density theorem aC∗-subalgebra ofB(H) is a von Neumann algebraiff its unit ball is weakly closed, we see:

(6.13)PROPOSITION: A C∗-subalgebraM ⊆ B(H) is a von Neumann algebraiff it is ultraweakly closed.

(6.14) COROLLARY: Every von Neumann algebra is isomorphic, as a Banachspace, to the dual of a Banach space.

PROOF: LetM ⊆ B(H) be a von Neumann algebra. LetN 6 L1(H) be itspreannihilator, that is,N = T ∈ L1(H) : Trace(ST ) = 0 ∀S ∈ M. N isa closed subspace in a Banach space, and is aM is weak-star closed subspace ofthe dual. By duality theory in Banach spaces,M is the dual of the quotient spaceL1(H)/N .

96

6.3 Exercises Oct 6, due Oct 20

EXERCISE 6.15: Show that every finite-dimensionalC∗-algebra has a faithfulrepresentation on a finite-dimensional Hilbert space.

Solution: By the nature of the GNS construction, the GNS representationassociated to any state on a finite-dimensionalC∗-algebraA is itself a finite-dimensional representation. The intersection of the kernels of all the GNS rep-resentations ofA is zero; hence (by dimension counting) the intersection offinitely many of these kernels is zero. It follows that there are finitely many GNSrepresentations (each finite-dimensional) whose direct sum is faithful.

EXERCISE 6.16: Letρ1, ρ2 be irreducible representations of aC∗-algebraA onHilbert spacesH1,H2. Show that(ρ1 ⊕ ρ2)[A]′ = ρ1[A]′ ⊕ ρ2[A]′ if and only ifρ1 andρ2 arenot unitarily equivalent.

Solution: In general(ρ1 ⊕ ρ2)[A]′ consists of matrices[W XY Z

]whereW ∈ ρ1[A]′, Z ∈ ρ2[A]′, andX andY are intertwining operators forρ1

andρ2, which is to say that

ρ1(a)X = Xρ2(a), ρ2(a)Y = Y ρ1(a).

Sinceρ1 andρ2 are irreducible,W andZ are multiples of the identity. Moreover,X∗X andXX∗ are multiples of the identity onH2 andH1 respectively, sayX∗X = λI2, XX∗ = λI1. If λ 6= 0, thenλ > 0 by positivity andλ−

12X is

a unitary equivalence betweenρ1 andρ2.

EXERCISE 6.17: Show thatT is the universalC∗-algebra generated by an isome-try.

Solution: Suppose thatW is an isometry on a Hilbert spaceH, so thatW ∗W = I. Let K 6 H be the closed subspaceKerW ∗. For n > m > 0the subspacesWnK andWmK are orthogonal, since foru, v ∈ K

〈Wnu,Wmv〉 = 〈Wn−mu, v〉 = 〈Wn−m−1u,W ∗v〉 = 0.

Thus we have an orthogonal decomposition

H = L⊕ L⊥, L = K ⊕WK ⊕W 2K ⊕ · · · .

The operatorsW andW ∗ mapL to itself and thus also mapL⊥ to itself. Therestriction ofW toL is unitarily equivalent to the direct sum ofp = dimK copies

97

of the unilateral shift, and the restriction ofW to L ⊥ is an isometry with zerokernel, that is a unitary. We have provedWold’s theorem: any isometry on aHilbert space is unitarily equivalent to a direct sum of copies of the unilateral shift,plus a unitary. We writeW ≡ V p ⊕ U .

Now we must define a∗-homomorphism from the Toeplitz algebraT to theC∗-algebra generated byW , which sendsV toW . From the Wold decompositionof W above we see

C∗(W ) = Tp ⊕ C(spU).

wherespU ⊆ S1 is the spectrum of the unitary partU ofW . Our∗-homomorphismis the direct sum ofp copies of the identity mapT → T together with the map

T → C(S1) → C(spU)

where the first arrow of the display is the symbol map of the Toeplitz exact se-quence and the second is the restriction map. It is clear that this∗-homomorphismhas the desired propertyV 7→ W , and since the Toeplitz algebra is generated byV , the∗-homomorphism is uniquely determined by this property.

EXERCISE 6.18: Use the spectral theorem to proveStone’s theorem:if π is astrongly continuous unitary representation of the groupR, then there is a spectralmeasureE on the Borel subsets ofR such that

π(t) =∫

ˆReiξt dE(ξ).

(HereR just denotes another copy ofR, one should think of it as the dual group inthe sense of Pontrjagin’s theory of locally compact abelian groups.)

Solution: By the general theory that we have discussed, a representationπ ofthe groupR is the same thing as a non-degenerate representation of theC∗-algebraC∗(R), which in turn can be identified with the commutativeC∗-algebraC0(R).According to the Spectral Theorem, given any non-degenerate representationρ of acommutativeC∗-algebraC0(X) on a Hilbert spaceH, there is a spectral measureE onX (with values in the projections ofH) such that

ρ(f) =∫

Xf(x) dE(x)

and substituting in the explicit form of the duality betweenR and R we obtainStone’s theorem as given above.

98

EXERCISE 6.19: LetA denote the maximalC∗-algebra of the free groupF2,and letρ : A → B(H) be the universal representation. LetB be the algebraof continuous operator-valued functionsT : [0, 1] → B(H) such thatT (0) is amultiple of the identity. Show that there exists a∗-homomorphism

α : A→ B

whose composite with evaluation att = 1 is the representationρ. Deduce thatAcontains no projections other than 0 and 1.

Solution: Letu, v ∈ A be the elements corresponding to the generators of thefree group and letU = ρ(u), V = ρ(v) be the corresponding operators onH. Bythe Borel functional calculus there exist selfadjoint operatorsX andY onH suchthat

U = exp(2πiX), V = exp(2πiY ).

Let U ′ and V ′ be the unitary elements ofB that are given by the continuousoperator-valued functions

U ′(t) = exp(2πitX), V ′(t) = exp(2πitY ).

Notice thatU ′(0) = I, U ′(1) = U , and similarly forV ′. By the universal propertyof the free groupC∗-algebra there exists a∗-homomorphismα : A → B takingutoU ′ andv to V ′, and att = 1, α is the representationρ. Now suppose thatp ∈ Ais a projection. Thenα(p) = P ′ ∈ B is a projection-valued function. In particularthe scalarP ′(0) must be either 0 or 1. But the point 0 is isolated in the space ofprojections (because every nonzero projection has norm 1) and, dually, the point1 is isolated in the space of projections also. ThusP ′ is a constant function withvalue 0 or 1 and, evaluating att = 1, we find thatρ(p) equals 0 or 1, as required.

99

6.4 Exercises Oct 27, due Nov 3

EXERCISE6.20: LetA be a unitalC∗-algebra, and letS = S(A) be its state space.Show that the convex hull ofS ∪ (−S) is the space of all real linear functionalson A of norm 6 1. (Hint: Use the Hahn-Banach Theorem as in the proof ofLemma 2.34.)

Solution — The space of real linear functionals onA is the dual of the realBanach spaceAsa. The convex hullK of S ∪ (−S) is a weak-∗ closed convexset inA∗sa. Suppose, if possible, thatψ is a real linear functional of norm 1 notbelonging toK. Then by the Hahn–Banach theorem there existsa ∈ Asa andα ∈ R with ψ(a) > α andϕ(a) 6 α for all ϕ ∈ K. The latter condition impliesin particular that|σ(a)| 6 α for all σ ∈ S. But ‖a‖ = sup|σ(a)| : σ ∈ S byLemma 2.10 and the remark following, so we getψ(a) > ‖a‖, a contradiction.

EXERCISE 6.21: Show that the following conditions are equivalent for positivelinear functionalsϕ andψ on a (unital)C∗-algebraA:

(a) ‖ϕ− ψ‖ = ‖ϕ‖+ ‖ψ‖;

(b) For eachε > 0 there is a positivea in the unit ball ofA with ϕ(1 − a) < εandψ(a) < ε.

(One then says thatϕ andψ areorthogonal. What does this condition say whenA = C(X)?)

Solution — (WhenA = C(X) this says that the measures defined byϕ andψhave essentially disjoint supports.)

Suppose (a). Then givenε > 0 there is a selfadjointa of norm at most 1 suchthat

ϕ(a)− ψ(a) + ε > ‖ϕ− ψ‖ = ϕ(1) + ψ(1).

Thereforeϕ(1− a) + ψ(1 + a) < ε.

Choosez = 12(1 + a).

Suppose (b). We reverse the above argument. We have

‖ϕ−ψ‖ 6 ‖ϕ‖+‖ψ‖ = ϕ(1)+ψ(1) 6 ϕ(2z−1)+ψ(1−2z)+4ε 6 ‖ϕ−ψ‖+4ε

using the fact that‖2z − 1‖ 6 1. Let ε→ 0 to obtain the desired conclusion.

EXERCISE6.22: Show that every real linear functionalϕ on a unitalC∗-algebraAcan be decomposed asϕ+ − ϕ− whereϕ± are mutually orthogonal positive linearfunctionals.

Can you show the uniqueness of this decomposition? (This is true, but is harderthan existence.)

100

Solution — Suppose thatϕ has norm one. According to the first exercise wemay writeϕ as a convex combinationλσ1− (1− λ)σ2 whereσ1, σ2 are states andλ ∈ [0, 1]. Putϕ+ = λσ1 andϕ− = −(1− λ)σ2. We have

‖ϕ+‖+ ‖ϕ−‖ = λ+ (1− λ) = 1 = ‖ϕ‖ = ‖ϕ+ − ϕ−‖,

so the decomposition is orthogonal. For the uniqueness see p.45 of Pedersen’sbook (this is a theorem of Grothendieck).

EXERCISE 6.23: Let G be a discrete group. Suppose that the trivial one-dimensional representation ofG extends to a∗-homomorphismϕ : C∗r (G) → C.Show that thenG is amenable. (Hint: Using the Hahn-Banach theorem, extendϕ toa state ofB(`2(G)), and then restrict to a state of`∞(G), embedded inB(`2(G))as multiplication operators. Show that this process produces an invariant mean onG.)

Solution — Letσ be the state ofB(`2(G)) produced by the process describedin the hint. We want to prove thatσ(MLgf ) = σ(Mf ), whereMf denotes themultiplication operator by the∞ functionf . Observing that

MLgf = λ(g)Mfλ(g)∗,

we see that it will suffice to show thatσ(T ) = σ(Tλ(g)) for all g ∈ G.This is a general fact: ifA is a unitalC∗-algebra with a stateσ, andu ∈ A is

unitary such thatσ(u) = 1, thenσ(au) = σ(a) for all a ∈ A. To see this note that

σ((1− u∗)(1− u)) = 2− σ(u)− σ(u∗) = 0

so that by Cauchy-Schwarz

|σ(a(1− u))|2 6 σ(a∗a)σ((1− u)∗(1− u)) = 0

as required.

EXERCISE6.24: Let1 → K → G→ H → 1

be a short exact sequence of groups. Show that ifH andK are amenable, thenGis amenable.

Solution — LetmK andmH be invariant means forK andH respectively.Define a unital positive linear mapΦ: `∞(G) → `∞(H) by

Φ(f)(h) = mK(k 7→ f(ks(h))

101

wheres : H → G is a set-theoretic section of the quotient map. Because of theinvariance of the mean onK, Φ is well-defined independent of the choice ofs.Now define an invariant mean onG by

f 7→ mH(Φ(f)).

102

6.5 Exercises Nov 15, due Nov 27

EXERCISE 6.25: LetG be the free group on two generators and leta ∈ C[G] bean element of the algebraic group ring. LetAa be the subalgebra ofB(`2(G))generated (algebraically) byC∗r (G) together with the orthogonal projectionPa

onto the kernel of the operator of (left) convolution bya. Show that the Fredholmmodule overC∗r (G) that we defined in class extends to a summable Fredholmmodule over the larger algebraAa. Deduce that ifa 6= 0 then the operation of leftconvolution bya is injective on`2(G) and has dense range. (This is the essentialidea of Linnell’s proof of the Atiyah conjecture forG.)

Solution (outline). The Fredholm moduleM overC∗r (G) that we have definedhas the following strong commutation property: ifa ∈ C[G] then the commutator[a, F ] is an operator offinite rank. We will show that if[a, F ] has finite rank thenso does[Pa, F ]. By replacinga by a∗a we see that it is enough to assume thata > 0.

In this case we may writePa = limε→0 ε(a+ ε)−1, with the limit in the strongoperator topology (this follows from the spectral theorem). Consequently,

[Pa, F ] = limε→0

−ε(a+ ε)−1[a, F ](a+ ε)−1

with the limit in the strong operator topology. If[a, F ] has rankk, the operatorsappearing in this limit all have rank6 k; and since the collection of operatorsof rank 6 k is strongly closed, it follows that[Pa, F ] has rank6 k. Thisthen establishes thatF defines a Fredholm module over the larger algebraAa.Moreover, the projectionPa lies in the domain of summability of this module.

From our earlier results it now follows thatτ(Pa) is an integer. Notice thatτextends to a faithful trace onAa, indeed it extends to a faithful trace on the groupvon Neumann algebra ofA. We haveτ(Pa) = 0 or = 1; in the former casePa = 0by faithfulness, soa has no kernel as an operator on`2 and is therefore injective;in the latter casePa = 1, so the kernel ofa is the whole of 2 and thusa = 0.

EXERCISE6.26: Prove that the Toeplitz algebra is not properly infinite.

Solution: If P is a projection inT, then its symbolσ(P ) is a projection inC(S1), and is therefore equal either to 0 or 1. Ifσ(P ) = 0 thenP is compact,and if σ(P ) = 1 then1 − P is compact (call such projections ‘cocompact’). Nocompact projection is equivalent to 1, and two cocompact projections cannot beorthogonal, so it is impossible to write 1 as the sum of two orthogonal projectionsboth of which are equivalent to 1.

EXERCISE6.27: Show that ifA is a simpleC∗-algebra, anda ∈ A is positive andnon-zero, then there existx1, . . . , xn ∈ A such thatx∗1ax1 + · · ·+ x∗naxn = 1.

103

Solution: Given in the text.

EXERCISE 6.28: Show thatO3 can be embedded as a subalgebra ofO2 (considerthe subalgebra generated bys1, s2s1, s22). Generalizing this, show that every Cuntzalgebra appears as a subalgebra ofO2.

Solution: Lett1 = s1, t2 = s2s1, t3 = s22 in O2. These are isometries. Wecompute

3∑i=1

tit∗i = s1s

∗1 + s2(s1s

∗1 + s2s

∗2)s

∗2 = 1,

using (twice) the identitys1s∗1 + s2s

∗2 = 1. Thust1, t2, t3 generate a copy ofO3

within O2.A straighforward generalization of the argument shows that

t1 = s1, t2 = s2s1, . . . tn−1 = sn−22 s1, tn = sn−1

2

generate a copy ofOn insideO2.

EXERCISE 6.29: Prove thatM2(O2) ∼= O2. Prove thatM3(O2) ∼= O2. General-ize, if you dare.

Solution: Here is the solution in the case ofM2(O2). Let

t1 =(s1 s20 0

), t2 =

(0 0s1 s2

).

Then we have by calculationt∗1t1 = t∗2t2 = I and tit∗j = eij whereeij are the

matrix units. Sincet1, t2 are isometries andt1t∗1 + t2t

∗2 = 1, it follows thatt1, t2

generate a copy ofO2 insideM2(O2). Since theC∗-algebra generated byt1 andt2 contains all the matrix units, as well as matrices withs1 ands2 as entries, itcontains every matrix whose entries are eithers1, s2, 0, or 1, and therefore it is thewhole ofM2(O2). ThusM2(O2) is isomorphic toO2.

ForM3(O2) we similarly consider the3× 3 matrices

t1 =

s1 s2s1 s220 0 00 0 0

, t2 =

0 0 01 0 00 s1 s2

.

These are isometries such thatt1t∗1+t2t

∗2 = 1, so they generate a copy ofO2 inside

M3(O2). Arguing as above, in order to show that theC∗-algebra that they generateis the whole ofM3(O2) it will suffice to show that it contains every matrix unit; infact it will suffice to show that it containse21 ande31, since these two matrix unitsgenerateM3(C). But in fact we have

e21 = t2t1t∗1, e31 = t22t

∗1

and so we are done.

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6.6 Exercises Dec 1, due Dec 8

EXERCISE6.30: Extend the proof of the Kadison transitivity theorem to show thatif ρ is an irreducible representation of aC∗-algebraA on a Hilbert spaceH, andif K is a finite-dimensional subspace ofH, then for each self-adjoint operatorTon K there exists a self-adjointa ∈ A such thatρ(a)|K = T . Prove the samestatement also with the word ‘unitary’ in place of ‘self-adjoint’.

Solution: For the self-adjoint case, the ordinary version of Kadison transitivityproducessomea ∈ A agreeing withT onK; to make it self-adjoint, replacea by(a + a∗)/2. For the unitary case, writeT = exp(iS) with S self-adjoint. Use thefirst part to produce a self-adjointb agreeing withS onK, then definea = exp(ib).

EXERCISE6.31: Using the previous exercise, prove that ifσ1, σ2 are pure states ofa unitalC∗-algebraA, and if the corresponding representations ofA are equivalent,then there exists a unitaryu ∈ A. such thatσ1(a) = σ2(u∗au).

Solution: Suppose that the GNS representationsρ1, ρ2 are unitarily equivalent.Using the unitary, identify the Hilbert spaces of the two representations so thatwe are in the following situation: we are given an irreducible representationρ : A→ B(H) and two vector states

σi(a) = 〈ρ(a)vi, vi〉

for unit cyclic vectorsv1, v2. On the finite-dimensional subspace spanned byv1andv2 there exists a unitary operatorU such thatU(v2) = v1. By the previousresult, there exists a unitaryu ∈ A such thatρ(u) agrees withU on the subspacespanned byv1, v2. Then

σ2(u∗au) = 〈ρ(u)∗ρ(a)ρ(u)v2, v2〉 = 〈ρ(a)ρ(u)v2, ρ(u)v2〉 = 〈ρ(a)v1, v1〉 = σ1(a).

105