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Vrije Universiteit Brussel Department of Mathematics

C-algebras

Docent :

K. De Commer

Master Wiskunde

20132014

Contents

1 Elementary theory of C-algebras 1

1.1 Banach algebras . . . . . . . . . . . . . . . . . . . . . . . . . . 11.1.1 Denition and examples . . . . . . . . . . . . . . . . . 11.1.2 Spectrum of an element in a Banach algebra . . . . . . 51.1.3 Spectrum of a commutative Banach algebra and the

Gelfand representation . . . . . . . . . . . . . . . . . . 111.1.4 Banach -algebras . . . . . . . . . . . . . . . . . . . . . 18

1.2 C-algebras . . . . . . . . . . . . . . . . . . . . . . . . . . . . 211.2.1 Denition and examples . . . . . . . . . . . . . . . . . 211.2.2 The commutative Gelfand-Neumark theorem . . . . . . 221.2.3 Positive elements in a C-algebra . . . . . . . . . . . . 281.2.4 Positive functionals on a C-algebra . . . . . . . . . . . 321.2.5 The non-commutative Gelfand-Neumark theorem . . . 351.2.6 Non-unital C-algebras . . . . . . . . . . . . . . . . . . 401.2.7 Approximate units and quotient C-algebras . . . . . . 49

2 Examples of Banach and C-algebras 55

2.1 The Calkin algebra and the concrete Toeplitz C-algebra . . . 552.1.1 Compact operators . . . . . . . . . . . . . . . . . . . . 552.1.2 The Calkin algebra . . . . . . . . . . . . . . . . . . . . 572.1.3 The Fredholm alternative . . . . . . . . . . . . . . . . 592.1.4 Compact operators on a Hilbert space . . . . . . . . . 642.1.5 The Fredholm index . . . . . . . . . . . . . . . . . . . 662.1.6 The concrete Toeplitz C-algebra . . . . . . . . . . . . 74

2.2 The abstract Toeplitz algebra . . . . . . . . . . . . . . . . . . 782.2.1 Enveloping C-algebra . . . . . . . . . . . . . . . . . . 782.2.2 The abstract Toeplitz C-algebra . . . . . . . . . . . . 80

i

ii CONTENTS

2.2.3 Quantum discs and elds of C-algebras . . . . . . . . 822.2.4 More on representations of C-algebras . . . . . . . . . 91

2.3 von Neumann algebras . . . . . . . . . . . . . . . . . . . . . . 101

Chapter 1

Elementary theory of C-algebras

1.1 Banach algebras

1.1.1 Denition and examples

In the following, all vector spaces will be taken over the eld of complexnumbers C.

Denition 1.1.1. An algebra consists of a vector space A together with abilinear associative map

m : A AÑ A, px, yq ÞÑ xy,

called the multiplication map. The algebra A is called unital if there existsan element 1A P A (necessarily unique) such that 1Aa a a1A for alla P A.Denition 1.1.2. A norm on a vector space A is a map

: AÑ R

such that

x y ¤ x y for all x, y P A, λx |λ|x for all x P A and λ P C,

1

2 Chapter 1. Elementary theory of C-algebras

x 0 implies x 0.

If only the rst two conditions are satised, is called a seminorm.

A Banach space consists of a vector space A together with a norm forwhich A is complete under the metric dpx, yq x y.Denition 1.1.3. A Banach algebra consists of a triple pA,m, q such thatpA,mq is an algebra, pA, q is a Banach space, and the following submulti-plicativity condition is satised:

xy ¤ xy.

In the following, we will most of the time abbreviate pA,m, q by just thesymbol A.

Examples 1.1.4. 1. Let V be a Banach space. The algebra BpV q ofbounded linear maps V Ñ V forms a Banach algebra for the norm

X suptXv | v P V, v ¤ 1u.

2. Let A be a Banach algebra. If B A is a norm-closed linear subspacewhich is also a subalgebra, then B is a Banach algebra. It is said to bea Banach subalgebra of A.

3. Let A be a Banach algebra and B A a subset of A. Let B be thenorm-closure of the linear span of (non-empty) products of elements inB: with

Bn tb1b2 . . . bn1bn | bi P Bu, n P Nzt0u,

we dene

B norm-closure tn

i1

λixi | n P N, λi P C, xi P8¤k1

Bku.

Then B is a Banach algebra. (Check that B is closed under multipli-cation.) One calls B the Banach algebra generated by B (inside A).Verify that B is the smallest Banach subalgebra of A containing B.

1.1 Banach algebras 3

4. Let B be an algebra, and let be a submultiplicative norm on B.Let pB, q pB, q be the Banach space completion of B. Thenthere exists a unique Banach algebra structure pB,m, q on B suchthat m restricts to the multiplication map on B.

5. If A is a Banach algebra and J A is a Banach subalgebra which isalso a (two-sided) ideal1, then the quotient space B AJ is a Banachalgebra for the norm

a J : infta x | x P Juand the quotient algebra structure.

6. Let A be a Banach algebra. Dene a new algebra Aop with underlyingset consisting of elements tao | a P Au, such that the map

o : AÑ Aop, a ÞÑ ao

is linear, and such that aobo pbaqo. Endow Aop with the Banach spacestructure ao a. Then Aop is again a Banach algebra, called theopposite of A.

7. Let X be a discrete set. The Banach space cbpXq of bounded functionsf : X Ñ C for the supremum norm

f : supt|fpxq| | x P Xuforms a Banach algebra for pointwise multiplication.

8. Let X be a topological space. The Banach space CbpXq of boundedcontinuous functions f : X Ñ C for the supremum norm

f : supt|fpxq| | x P Xuforms a Banach algebra for the pointwise multiplication.

9. Let X be a topological space. Let C0pXq CbpXq be the space offunctions vanishing at innity,

C0pXq tf P CbpXq | @ε ¡ 0, DK X compact , @x R K : |fpxq| εu.Then C0pXq is a Banach subalgebra of CbpXq.

1Recall that a left (resp. right) ideal in an algebra A is a subset J such that for alla P A and all x P J , one has ax P J (resp. xa P J). An ideal is called two-sided if it isboth a left and right ideal. An ideal J is called proper if J A.


10. Let Γ be a semi-group.2 The Banach algebra l1pΓq of absolutely summablefunctions f : Γ Ñ C on Γ with the norm

f :¸xPΓ

|fpxq|

forms a Banach algebra under the convolution product

pf gqpxq :¸y,zyzx

fpyqgpzq.

11. Let A and B be two Banach algebras. Let A ` B be their direct sumalgebra, that is A`B is the vector space sum with algebra structure

pa, bqpa1, b1q paa1, bb1q.

Endow A ` B with the Banach space norm pa, bq maxta, bu.Then A ` B is a Banach algebra, called the 8-sum of A and B. It issometimes also denoted A`8 B.

12. Let A and B be two Banach algebras. Let A ` B be their direct sumalgebra. Endow A`B with the Banach space norm pa, bq ab.Then A ` B is a Banach algebra, called the 1-sum of A and B. It issometimes also denoted A`1 B.

Remark 1.1.5. Recall that a Fréchet space V is a vector space with a countable familyof semi-norms which separate points of V (in the sense that 0 is the only element on whicheach semi-norm is zero) and for which V is complete (in the sense that a sequence whichis Cauchy for each semi-norm admits a limit). A Fréchet algebra is a Fréchet space withalgebra structure for which the above seminorms can be chosen to be submultiplicative.Although Fréchet algebras form a valiant generalization of Banach algebras, the theory ismuch less advanced.

Denition 1.1.6. Let A be a Banach algebra. We say A is a unital Banachalgebra, or a Banach algebra with unit, if the underlying algebra admits aunit element 1A with 1A 1.

In this case we identify C A by λ ÞÑ λ1A. If B A is a subalgebra, wecall it a unital subalgebra if the unit of A belongs to B.

2A semigroup is a set Γ with an associative multiplication map ΓΓ ÞÑ Γ, for exampleZ, N and Nzt0u are semigroups under addition.


Denition 1.1.7. Let A be a Banach algebra. The unitization of A is thevector space AI A` C with the Banach norm pa, λq a |λ| and thealgebra structure

pa, λqpb, µq pab λb µa, λµq.

Note that the map A Ñ AI sending a to pa, 0q is isometric, and that p0, 1qis a unit for AI making AI a unital Banach algebra. In the following, wewill identify A AI , and again just write 1 1AI for the unit of AI . Notehowever that if A already has a unit, the unitization AI is distinct from A,and the unit 1A of A is no longer a unit for AI !

1.1.2 Spectrum of an element in a Banach algebra

Let A be a unital algebra. An element x P A is called invertible if thereexists a (two-sided!) inverse, i.e. an element x1 (necessarily unique) suchthat xx1 1A x1x.

Denition 1.1.8. Let A be a unital Banach algebra. If x P A, the spectrumSpApxq C of x is dened by

SpApxq tλ P C | λ x is not invertible in Au.

The complement ρApxq C of SpApxq is called the resolvent of x.

If A is an arbitrary Banach algebra, we write Sp1Apxq SpAI pxq (the quasi-spectrum of x) and ρ1Apxq ρAI pxq (the quasi-resolvent).

Whenever A is clear from the context, we write Sppxq SpApxq, and similarlyfor the other denitions. Note that 0 P Sp1pxq for all x P A. Also note thatthe spectrum of an element only depends on the algebra structure of A, noton its Banach space structure.

Remark 1.1.9. It is obvious that if A is a unital Banach algebra, and B aunital Banach subalgebra, then SpApxq SpBpxq for all x P B. However, ingeneral this inclusion can be strict.

Examples 1.1.10. 1. Let A M2pCq BpC2q, the algebra of oper-

ators on the two-dimensional Hilbert space C2. If x λ1 00 λ2


then Sppxq tλ1, λ2u. More generally, if

λ1 λ3

0 λ2

, one has Sppxq

tλ1, λ2u.2. For n P N0, let A MnpCq. Then for x P A, the set Sppxq is precisely

the set of eigenvalues of x.

3. Let A CbpXq with X a topological space. Then for f P CbpXq, wehave Sppfq ClpfpXqq, the closure of fpXq.

4. Let X be a topological space, and let A C0pXq. Then for f P C0pXq,we have Sp1pfq ClpfpXqq Y t0u.

Lemma 1.1.11. Let A be a unital Banach algebra. Let x P A with x 1.Then 1 x is invertible with inverse

°8n0 x

n, and

p1 xq1 ¤ 1

1 x .

Proof. Pick k ¤ l. Thenl

n0

xn k

n0

xn

l

nk1

xn

¤

l

nk1

xn

¤l

nk1

xn.

As x 1, it follows that°l

n0 xnlis a Cauchy sequence in A, hence

converges to an element°8n0 x

n. Clearly,

p1 xq

l

n0

xn

l

n0

xn

p1 xq 1 xl1.

As x 1, limlÑ8 xl1 ¤ limlÑ8 xl1 0. Hence

p1 xq

8

n0

xn

8

n0

xn

p1 xq 1.


The estimate p1 xq1 ¤ 11x

follows immediately from the triangle in-equality.

Lemma 1.1.12. Let A be a unital Banach algebra, and let G A be thesubset of invertible elements. Then G is open (for the topology induced bythe norm), and the operation

inv : GÑ G, x ÞÑ x1

is continuous.

Proof. Assume that x is invertible, and take y P A with x y x11.We can write

y x px yq xp1A x1px yqq.As x1px yq ¤ x1x y 1, we have by Lemma 1.1.11 that theelement 1A x1px yq admits an inverse. As y is a product of invertibleelements, it is itself invertible. Hence G is open.

Let now xn Ñ x in G. Then

x1 x1n x1

n pxn xqx1 ¤ x1n x1xn x.

By the calculation of the previous paragraph and the norm estimate inLemma 1.1.11, we have that for n large

x1n ¤ x1

1 x1px xnq ¤ 2x1.

It follows that x ÞÑ x1 is continuous on G.

Proposition 1.1.13. Let A be a unital Banach algebra. Then Sppxq is aclosed subset of Dp0, xq, the closed disc of radius x around the origin.

Proof. Pick x P A.If |λ| ¡ x, then λ1x 1. Hence, by Lemma 1.1.11, the element 1λ1xis invertible, and so also λ x is invertible. Hence Sppxq is contained inDp0, xq.By Lemma 1.1.12, it follows immediately that the resolvent ρpxq is open, andhence Sppxq is closed.


Example 1.1.14. Let A Bpl2pNqq and S P Bpl2pNqq the shift operatorpSaqn an1.

We use Proposition 1.1.13 to determine the spectrum of S.

Clearly S ¤ 1. It follows that SppSq Dp0, 1q.On the other hand, if |λ| 1, we can form the element a P l2pNq withan λn, and then Sa λa. Hence S λ can not be invertible, and so theopen unit disc Dp0, 1q belongs to SppSq. As SppSq is closed by Proposition1.1.13, it follows that in fact SppSq Dp0, 1q.Example 1.1.15. Let A Bpl2pZqq and consider again the shift operatorS P Bpl2pZqq,

pSaqn an1.

Again, S ¤ 1 implies that SppSq Dp0, 1q. However, in this case S isinvertible with

pS1aqn an1.

As also S1 ¤ 1, it follows that for 0 |λ| 1, the operator S λ λSpλ1 S1q is invertible. Hence SppSq S1 tz P C | |z| 1u.Let us show the equality SppSq S1. Take λ P S1. Suppose that S λ wereinvertible. Then there would exist a P l2pNq with

pS λqa e0,

where e0 is the sequence which is zero everywhere except at position 0, whereit takes the value 1. By a simple induction argument, we nd that thennecessarily "

an1 λnp1 λa0q for n ¥ 0,an λna0 for n 0.

Clearly, as |λ| 1, such an a can not be in l2pNq for any value of a0, leading toa contradiction. It follows that Sλ is not invertible, and hence SppSq S1.

The boundedness of the spectrum can also be used to establish a powerseries calculus for elements in a Banach algebra. We write Dp0, 1q for theopen unit disc, and HolpDp0, 1qq for the algebra of holomorphic functions onDp0, 1q. When f P HolpDp0, 1qq, we write fpzq °8

n0 fnzn for its power

series expansion.


Recall that a homomorphism φ : A Ñ B between algebras is a linear mapsuch that φpxyq φpxqφpyq for all x, y P A.Proposition 1.1.16. Let A be a unital Banach algebra, and a P A witha 1. Then the map

φ : HolpDp0, 1qq Ñ A, f ÞÑ8

n0

fnan

is a well-dened algebra homomorphism.

Proof. Let f P HolpDp0, 1qq. Then °8n0 |fn|rn is convergent for all r 1. It

follows that°8n0 fna

n is a convergent series in A. Hence φ is a well-denedlinear map.

As pfgqn °nk0 fkgnk for f, g P HolpDp0, 1qq, it follows from absolute

convergence that one can interchange the order of summation to obtainφpfgq φpfqφpgq.

The above calculus can be extended to the holomorphic calculus, associatingto any holomorphic function f on an open neighborhood of Sppxq an elementfpxq in the Banach algebra. We will not introduce this holomorphic calculushere, but will use the theory of holomorphic functions in another way toestablish a fundamental property of the spectrum, namely the fact that it isnever empty.

Denition 1.1.17. Let Ω C be an open set. Let A be a Banach algebra.A function f : Ω Ñ A is called holomorphic if for each z P Ω, there existsf 1pzq P A such that

limhÑ0

1

hpfpz hq fpzq hf 1pzqq 0.

Example 1.1.18. Let A be a unital Banach algebra, and x P A. Then themap

f : z ÞÑ pz xq1

is analytic on the resolvent ρpxq, with

f 1pzq pz xq2.


Proposition 1.1.19. Let A be a unital algebra. Then Sppxq H for eachx P A.

Proof. Suppose by contradiction that x P A is such that Sppxq H. Let ωbe a continuous functional on A. Then the map

λ ÞÑ ωppλ xq1q

is analytic on C and tends to zero when |λ| Ñ 8 (since pλ xq1 ¤p|λ|xq1 for λ large). By Liouville's theorem, we obtain ωppλxq1q 0for all ω. It follows that pλ xq1 0 for all λ, and in particular x1 0, acontradiction.

Corollary 1.1.20 (Gelfand-Mazur). Let A be a unital Banach algebra suchthat A is a eld. Then A C.

Proof. Let x P A. As Sppxq H, we can choose λ P C with x λ notinvertible. As A is a eld, x λ.

Denition 1.1.21. Let A be a unital Banach algebra, and x P A. Thespectral radius of x is the number

rpxq : maxt|λ| | λ P Sppxqu.

If A is non-unital, we replace Sp by Sp1 in the above denition.

As we have seen, rpxq ¤ x, and in general this inequality is strict. Never-theless, the norm (i.e. the metric structure) can still be used to determinethe spectral radius (i.e. the algebraic structure).

Proposition 1.1.22. Let A be a Banach algebra. Then for each x P A onehas

limnÑ8

xn1n rpxq.

Proof. We may assume that A is unital.

If λ P Sppxq, then λn P Sppxnq, as xn λn px λqy ypx λq for somey P A. Hence |λ|n ¤ xn, and rpxq ¤ inf xn1n.

Let now ω be a continuous functional on A. For r ¡ 0, let γr be the standardanti-clockwise parametrization of the circle of radius r in C. Take r ¡ x.


Then for each n ¡ 0, we have

1

2πi

»γr

znωppz xq1qdz 1

2πi

»γr

8

k0

znk1ωpxkqdz

8

k0

ωpxkq

1

2πi

»γr

znk1dz

ωpxnq,

where we could interchange the integral and series by absolute convergence(since r ¡ x). However, by contour deformation we can conclude that

ωpxnq 1

2πi

»γr

znωppz xq1qdz, @r ¡ rpxq.

Hence

|ωpxnq| ¤ rn1ωmaxtpz xq1 | |z| ru, @r ¡ rpxq.Taking the maximum over all ω with ω ¤ 1, taking n-th roots and lettingn tend to innity, we arrive at

lim supn

xn1n ¤ r, @r ¡ rpxq.

Hence lim supn xn1n ¤ rpxq, nishing the proof.

1.1.3 Spectrum of a commutative Banach algebra and

the Gelfand representation

We now concentrate on some algebraic properties of Banach algebras.

Proposition 1.1.23. Let A be a unital Banach algebra. Then the followinghold.

1. Any maximal3 (left, right or two-sided) ideal J in A is closed.

2. The closure of any proper (left or right) ideal is still a proper (left orright) ideal.

3Recall that an ideal is called maximal if it is maximal amongst all proper ideals


Proof. Let J be a proper ideal. It is clear that its closure is still an ideal.Suppose its closure was equal to A. Then there exists a sequence of elementsxn P J converging to 1A. As the set of invertibles is open, it follows thatxn is invertible for n large. This means 1A is already in J , a contradiction.Hence the closure of J is still proper.

If J is a maximal ideal, then its closure is a proper ideal containing J . HenceJ equals its closure.

Denition 1.1.24. Let A be a Banach algebra. A character on A is a non-zero algebra homomorphism AÑ C.

Note that if A is unital, then χp1Aq 1 for any character χ on A. If A isnon-unital, then any character χ on A extends uniquely to a character χ onthe unitization AI by dening χpx λ1AI q χpxq λ.

Clearly the kernel of any character forms a maximal two-sided ideal. Recallnow that a linear functional on a Banach space is continuous if and only if itskernel is closed. It then follows by Proposition 1.1.23.1. that any characteron a Banach algebra is continuous. The next lemma gives a direct proof ofthis result.

Lemma 1.1.25. Let A be a Banach algebra. Then for any x P A and anycharacter χ on A, one has χpxq P Sp1pxq, and χpxq P Sppxq if A is unital.In particular, any character χ : A Ñ C is bounded with χ ¤ 1, equalityholding if A is unital.

Proof. If A is not unital, then we can extend χ to the unitization AI bydening χpa λ1q χpaq λ for a P A and λ P C. Hence we may assume Ato be unital with χp1Aq 1.

Let x P A. By applying χ, we see that x χpxq can not be invertible in A.Hence χpxq P Sppxq. It follows that |χpxq| ¤ x for all x, so χ ¤ 1. Asχp1Aq 1, we have in fact χ 1.

Remark 1.1.26. The automatic boundedness of characters on Fréchet algebras is still anopen problem, known as Michael's conjecture.

Denition 1.1.27. Let A be a commutative Banach algebra, so xy yx forall x, y P A. The set of characters on A is called the spectrum SppAq of A.Proposition 1.1.28. Let A be a commutative Banach algebra. Then there


is a bijection between the set of characters on A and the set of maximal idealsof A by the map χ ÞÑ kerpχq.

Proof. Let χ be a character on A. As the kernel has codimension 1, it formsa maximal ideal of A. Hence the above map is well-dened.

Assume χ and ρ are two characters with the same kernel. As for any xand y one has χpxy χpxqyq 0, it follows that ρpxy χpxqyq 0, soρpxqρpyq χpxqρpyq. Choosing y with ρpyq 0, we nd ρpxq χpxq. Hencethe above map is injective.

Finally, if J is a maximal ideal of A, then we know by Proposition 1.1.23that J is closed. Hence AJ is a Banach algebra. Since J is maximal (andA is commutative), AJ is a eld. By the Banach-Mazur theorem, we haveAJ C. Hence the composition A Ñ AJ C is a character, whoseassociated kernel is clearly J . Hence the above map is surjective.

Proposition 1.1.29. Let A be a commutative Banach algebra. Then SppAqbecomes a locally compact Hausdor space with respect to the weak-topologyinherited from the dual Banach space A.

Proof. Suppose rst that A is unital. By the Banach-Alaoglu theorem, theunit ball of A is compact in the weak-topology. As we know that SppAqis contained in the unit ball of A, it is thus sucient to show that SppAqis closed in A. But let χα be a net in SppAq converging to χ P A. Bycontinuity, χp1Aq 1 and χpxyq χpxqχpyq. Hence χ is a character.

If A is not unital, consider its unitization AI . Then AI has the distinguishedcharacter ε for which εp1AI q 1 and εpxq 0 for all x P A. As this is clearlythe unique character on AI which vanishes on A, it easily follows that we havea natural homeomorphism SppAIqztεu SppAq by restricting characters toA. Hence SppAq is locally compact, being the complement of the closed settεu in SppAIq.Example 1.1.30. Let X be a compact Hausdor space. For x P X andf P CbpXq, denote χzpfq fpzq. Let us show that the map

φ : X Ñ SppCbpXqq, x ÞÑ χx

is a homeomorphism.


Proof. Clearly, φ is a continuous map. Hence it suces to show that φ isbijective.

As X is compact and Hausdor, it satises Urysohn's lemma. Hence we cannd for any x, y P X with x y a function f P CbpXq with fpxq 1 andfpyq 0. It follows that φ is injective.

Let us show that φ is surjective. Let χ P SppCbpXqq. Let us say a compactsubset K of X is χ-adapted if there exists a function f P CbpXq with supportin K such that χpfq 0. For example, X is χ-adapted since χp1CbpXqqq 0.Moreover, if we have a nite set of χ-adapted compact subsets tKi | 1 ¤i ¤ nu, then Xn

i1Ki is non-empty since χpf1 . . . fnq 0 for functions fiwith support on Ki and χpfiq 0. As X is compact, it follows that theintersection C of all χ-adapted compact subsets of X is not empty.

Take x P C. Assume that f P CbpXq has compact support disjoint fromx. Then the support of f can not be χ-adapted, so necessarily χpfq 0.Suppose now more generally that f P CbpXq vanishes in x. Take ε ¡ 0, andlet U be an open neighborhood of x such that |fpyq| ε on U . Let K Ube a compact neighborhood of x, and choose a partition of unity with respectto the covering tU,Kcu of X, say g1, g2 P CbpXq with g1 0 on U c, g2 0on K, |gipyq| ¤ 1 for all y P X and g1pyq g2pyq 1 for all y P X. Then

|χpfq| χpg1f g2fq χpg1fq χpg2qχpfq χpg1fq¤ g1f¤ ε,

where we used the continuity of χ. As ε was arbitrary, we obtain χpfq 0.

Take now f P CbpXq arbitrary. Then the function f fpxq vanishes in x.Hence χpf fpxqq 0, and so χpfq fpxq. It follows that χ χx, so φ issurjective.

It follows from the above example that X can be reconstructed from itsalgebra of continuous functions CbpXq. Later on, we will give an abstractcharacterization of those Banach algebras which arise as some CbpXq, seeTheorem 1.2.7.


The following lemma calculates the spectrum of a unital Banach algebragenerated by a single element.

Lemma 1.1.31. Let A be a unital commutative Banach algebra. Supposex P A is such that the Banach algebra generated by t1A, xu equals A. Thenthe natural map

Φ : SppAq Ñ SpApxq, χ ÞÑ χpxq

is a homeomorphism.

Proof. By lemma 1.1.25, we have χpxq P SpApxq for all characters χ : AÑ C.Hence Φ is well-dened.

The map Φ is clearly continuous. To show that Φ is a homeomorphism,it suces to show that Φ is bijective, as both domain and co-domain arecompact.

If χ1 and χ2 are two characters onA such that χ1pxq χ2pxq, then χ1pP pxqq χ2pP pxqq for any polynomial P in x. By continuity, it follows that χ1 χ2.Hence Φ is injective.

If λ P SpApxq, then by denition xλ is not invertible in A. By Proposition1.1.23.2, the closure J of the ideal4 generated by x λ in A is proper. Asany polynomial in x becomes a scalar multiple of the unit in AJ , it followsthat AJ C. Hence we obtain a character χλ : A Ñ AJ C such thatχλpxq λ. It follows that Φ is surjective.

Theorem 1.1.32. Let A be a unital commutative Banach algebra. Then themap

φ : AÑ CbpSppAqq, x ÞÑ pφx : SppAq Ñ C, χ ÞÑ χpxqqis a well-dened algebra homomorphism of norm 1 and

SpApxq SpCbpSppAqqpφpxqq, @x P A.

The map φ is called the Gelfand representation.

4As A is commutative, any left ideal is automatically a two-sided ideal. Why is thisimportant for the argument?


Proof. It is straightforward to verify that φ is an algebra homomorphism. Asφpxq supχPSppAq |χpxq| ¤ x and φp1Aq 1CbpSppAqq, it follows that φ hasnorm 1.

Let us show that for x P A, we have SpApxq SpCbpSppAqqpφpxqq. By Example1.1.10.3 (and the fact that SppAq is compact), this is equivalent with

SpApxq tχpxq | χ P SppAqu.

The inclusion has already been shown before (see Lemma 1.1.25). For theinclusion , we modify the end of the proof of Lemma 1.1.31. Let λ P SpApxq.Then the two-sided ideal generated by x λ in A is proper (note that weuse here the commutativity of A). Hence we can nd a maximal ideal J of Awhich contains x λ. As J is necessarily closed by Proposition 1.1.23.1, wehave AJ C by the Gelfand-Mazur theorem. Hence we obtain a characterχλ : AÑ AJ C such that χλpxq λ. This proves the inclusion .

The Gelfand representation is in general not bijective.

Examples 1.1.33. 1. Take

A tλ1 λ2

0 λ1

| λ1, λ2 P Cu M2pCq.

Then A is a 2-dimensional unital commutative Banach algebra withSppAq consisting of a single point, so CbpSppAqq C. Hence theGelfand representation is not injective in this case.

2. Take A l1pZq with the convolution product. Let λk be the sequencewhose only non-zero entry is the value 1 at position k, so pλkqn δk,n.Then λkλl λkl for all k, l P Z. As in particular λn1 λn, it followsthat the set t1A, λ1, λ1u generates l1pZq.Since λ1 is the inverse of λ1, any algebra homomorphism χ : A Ñ Cis determined by its image on λ1. By the equality λ1 λ1, sucha χ also satises |χpλ1q| 1. Conversely, if |z| 1, the map

χz : a ÞÑ¸nPZ

anzn

denes a character of l1pZq.


The assignment χ ÞÑ χpλ1q then identies SppAq homeomorphicallywith the subset S1 : tz P C | |z| 1u C. Hence the Gelfandrepresentation may be seen as a map l1pZq Ñ CbpS1q, assigning toa the function z ÞÑ °

nPZ anzn on S1. However, there are continuous

functions on S1 whose Fourier coecients are not absolutely summable.Hence the Gelfand representation is not surjective in this case.

Corollary 1.1.34 (Wiener's Lemma). Let a P l1pZq and put

fpzq ¸nPZ

anzn, z P S1.

If fpzq 0 for all z P S1, then 1fpzq

°nPZ bnzn for some b P l1pZq.

Proof. Consider l1pZq with the convolution product. Consider the Gelfandrepresentation

φ : l1pZq Ñ CbpS1q, x ÞÑz ÞÑ

¸nPZ

xnzn

as above. To prove the corollary, it is sucient to show that a has an inversein l1pZq, i.e. that 0 R Spl1pZqpaq. But Spl1pZqpaq SpCbpS1qpφpaqq fpS1q,which is bounded away from zero by assumption.


1.1.4 Banach -algebras

Denition 1.1.35. A Banach -algebra is a Banach algebra A together withan anti-linear, anti-multiplicative, isometric involution , that is, a map

: AÑ A

such that

1. pa λbq a λb for all a, b P A and λ P C (anti-linear),

2. pabq ba for all a, b P A (anti-multiplicative),

3. a a for all a P A (isometric),

4. paq a for all a P A (involutive).

An element a P A is called self-adjoint if a a.

An element a P A is called normal if aa aa.

An element a P A is called unitary if aa aa 1.

An element a P A is called an isometry (resp. a co-isometry) if aa 1(resp. aa 1).

Examples 1.1.36. 1. If H is a Hilbert space, then BpH q is a Banach-algebra upon taking to be the adjoint of an operator.

2. If A is a Banach -algebra, then the examples 1.1.4 2.,3. and 5. giveexamples of Banach -algebras if one assumes the respective sets B,Band J are closed under the operation .

3. Let A be a unital Banach -algebra. Then Aop is again a Banach -algebra by putting paoq paqo.

4. Let X be a topological space. Then CbpXq forms a Banach -algebrafor the -operation

fpxq : fpxq.

5. Let Γ be a group. The Banach algebra l1pΓq forms a Banach -algebrafor the -operation

fpxq fpx1q.


6. Let A and B be two Banach -algebras. Then the8-direct sum Banachalgebra A`B is again a Banach -algebra by the -operation

pa, bq pa, bq.

A unital Banach -algebra will be a unital Banach algebra which is at thesame time a Banach -algebra. Obviously, 1A 1A in this case. If A is aBanach -algebra, its unitization AI becomes a Banach -algebra by puttingpa λ1AI q a λ1AI .

Denition 1.1.37. Let A be a Banach -algebra. A -representation of Aon a Hilbert space H is a -preserving algebra homomorphism

π : AÑ BpHq.A -representation π is called non-degenerate if the vector space

πpAqH tn

i1

aiξi | n P N, ai P A, ξi P Hu

is dense in H. It is called faithful if π is injective.

If A is unital, a -representation π is non-degenerate if and only if πp1Aq idH. Such

-representations will then also be called unital -representations.

One can add two representations together by means of the direct sum. Namely,if one has a Banach -algebra A and pH1, π1q and two -representationspH2, π2q , then one can form the direct sum -representation

π1 ` π2 : AÑ BpH1 `H2q, a ÞÑ π1paq ` π2paq.Proposition 1.1.38. Let A be a Banach -algebra. Let π be a -representationof A on a Hilbert space H. Then π is continuous, with π ¤ 1.

Proof. If A is not unital, then we can extend π to its unitization by puttingπpa λ1q πpaq λidH. Hence we may suppose that A is unital.

One obviously has Sppaq Sppπpaqq for any a P A. Hence, with rpqdenoting the spectral radius, we have for any x P A that

rpπpxqπpxqq rpπpxxqq¤ rpxxq¤ xx¤ x2.


It then suces to prove that rpπpxqπpxqq πpxqπpxq πpxq2. Theseequalities hold for any operator in BpHq, as we prove in the next lemma.

Lemma 1.1.39. Let H be a Hilbert space and x P BpHq. Thenxx x2 (1.1)

andxx rpxxq.

Proof. For any ξ P H, we have

xξ2 xxξ, xξy xxxξ, ξy¤ xxξξ¤ xxξ2.

Hence

x2 supξ¤1

xξ2

¤ xx.On the other hand, xx ¤ xx ¤ x2. Hence xx x2.

Write now y y0 xx, and inductively dene yn1 y2n. Then yn y2n .

As a a and pabq ba for any a, b P BpHq, we obtain by induction thatyn yn. Moreover, by (1.1) we have yn1 yn2. Hence by inductiony2n y2n . Again by (1.1), we have

x2 y y2n2n .

But by Proposition 1.1.22, this value converges to rpyq as n tends to innity.Hence x2 rpxxq.

The property (1.1) does not hold in an arbitrary Banach -algebra.

Example 1.1.40. Consider l1pZq with the convolution product and the -operation as in Example 1.1.36.5. Then

pf fqn ¸mPZ

fmfnm,

1.2 C-algebras 21

hence

f f ¸nPZ

¸mPZ

fmfnm

.However, f2 °nPZ

°mPZ |fmfnm|. Convince yourself that these expres-

sions are distinct in general.

As a corollary of Proposition 1.1.38, it follows that one can take a direct sumof an arbitrary collection of -representations.

1.2 C-algebras

1.2.1 Denition and examples

Those Banach -algebras in which (1.1) holds for all elements turn out tohave a very rich theory that motivates their study for their own sake.

Denition 1.2.1. A C-algebra is a Banach -algebra A for which

xx x2, @x P A.

The main power of C-algebras is the fact that they form a bridge betweenthe abstract and the concrete. On the one hand, their axiomatic denitionallows one to create examples by universal constructions. On the other hand,as we shall see later, every C-algebra is in fact -isomorphic to a closed sub--algebra of BpHq for some Hilbert space H.

Examples 1.2.2. 1. Let X be a topological space. Then CbpXq is a C-algebra.

2. Let H be a Hilbert space. Then BpHq is a C-algebra (by Lemma1.1.39).

3. Let A be a C-algebra. If S A, the Banach subalgebra generated byS Y S inside A is a C-subalgebra. We denote it as CpSq.

4. Let A be a C-algebra. Then Aop is again a C-algebra, called theopposite of A.


5. Let A and B be two C-algebras. Then the 8-direct sum Banach -algebra A ` B is again a C-algebra. (This is no longer true for the1-sum.)

One of the strong properties of a C-algebra is that the norm is uniquelydetermined by the algebraic structure. Recall the notion of spectral radiusfrom Denition 1.1.21.

Proposition 1.2.3. Let A be a unital C-algebra. Then for all x P A, onehas

x rpxxq12.

Proof. We can copy the proof of Lemma 1.1.38.

Corollary 1.2.4. Let A and B be two unital C-algebras. Let π : AÑ B bea unital -algebra homomorphism. Then π is contractive, i.e. π ¤ 1.

Proof. Let x P A. As Sppπpxxqq Sppxxq, we have

rpπpxqπpxqq rpπpxxqq ¤ rpxxq.

By Proposition 1.2.3, we conclude πpxq ¤ x, hence π is contractive.

1.2.2 The commutative Gelfand-Neumark theorem

Lemma 1.2.5. Let A be a unital commutative C-algebra and χ P SppAq.Then χpxq χpxq for all a P A.

Proof. By Lemma 1.1.25, we have χ ¤ 1.

Take rst x P A with x x. Then for t P R, we compute

|χpx itq|2 ¤ x it2

px itqpx itq px itqpx itq x2 t2¤ x2 t2.

1.2 C-algebras 23

But also

|χpx itq|2 |χpxq it|2 pRepχpxqqq2 pImpχpxqq tq2.

Hence|χpxq|2 2tImpχpxqq ¤ x2, @t P R.

It follows that Impχpxqq 0, hence χpxq P R.

Take now x P A general. Then we can write x y iz with y 12px xq

and z 12ipx xq. Both y and z are self-adjoint elements. It follows that

χpxq χpyq iχpzq χpyq iχpzq χpxq.

Corollary 1.2.6. Let A be a unital C-algebra. If x P A is self-adjoint, thenSpApxq R.

Proof. Let B Cp1A, xq be the unital C-algebra generated by x. As xis self-adjoint, B is a commutative C-algebra. By Lemma 1.1.31, we havea natural homeomorphism SpBpxq SppBq. As any element in SppBq is-preserving by Lemma 1.2.5, we deduce by the explicit formula of the abovehomeomorphism that SpBpxq R. As B A, we hence have SpApxq SpBpxq R.

The next theorem (which we will call the commutative Gelfand-Neumarktheorem) shows that unital commutative C-algebras are in one-to-one cor-respondence with compact topological Hausdor spaces.

Theorem 1.2.7 (Gelfand-Neumark). Let A be a unital commutative C-algebra. Then the Gelfand representation φ : A Ñ CbpSppAqq (cf. Theorem1.1.32) is a -isomorphism.

Let us recall rst the Stone-Weierstrass theorem5

5Cf. for example Theorem 7.33 in W. Rudin, Principles of mathematical analysis,International Series in Pure and Applied Mathematics, McGraw-Hill Book Co., New York,3rd edition (1976), x+342 pp.


Theorem 1.2.8. Let X be a compact Hausdor space. Let A CbpXq be aunital sub--algebra which separates points of X: for each pair of elementsx, y in X with x y, we can nd f P A with fpxq fpyq.Then the closure of A equals CbpXq.

Proof (of Theorem 1.2.7). By Lemma 1.2.5, any element χ P SppAq is -preserving. It follows that φ is a -preserving map between C-algebras.

Let us show that it is injective. It is sucient to show that φ is isometric(i.e. φpxq x for all x P A). Take x P A. Then by Proposition 1.2.3 andTheorem 1.1.32,

x2 rpxxq rpφpxxqq rpφpxqφpxqq φpxq2.

To prove that φ is surjective, we use the Stone-Weierstrass theorem. As φ isisometric, the image B of φ will be a Banach sub--algebra of A. It containsthe constant functions (which are the images of scalar multiples of the unit ofA), and for any distinct characters χ1 χ2 in SppAq we can nd x P A withχ1pxq χ2pxq. Hence B separates points in SppAq. The Stone-Weierstrasstheorem lets us conclude that B A.

The Gelfand representation gives a contravariant functorial correspondencebetween unital commutative C-algebras and compact spaces. Namely, ifπ : A Ñ B is a unital -homomorphism between C-algebras, we obtain acontinuous map

π : SppBq Ñ SppAq, χ ÞÑ χ π.Because source and range are interchanged, the correspondence is called con-travariant. It is called contravariant functorial because one has

pπ θq θ π.

In particular, one sees that π is a -isomorphism if (and in fact only if) π isa homeomorphism. The following lemmas elucidate what injectiveness andsurjectiveness correspond to.

1.2 C-algebras 25

Lemma 1.2.9. Let A and B be two unital C-algebras. Let π : AÑ B be aunital -homomorphism.

Then π is injective if and only if π : SppBq Ñ SppAq is surjective.

Proof. By the Gelfand-Neumark theorem, we may assume that A CbpXqand B CbpY q for compact Hausdor spaces X and Y .

Assume π is not surjective. Then πpY q is a proper closed subset of X.By Urysohn's Lemma, there exists a non-zero function f P CbpXq such thatfpπpY qq 0. Hence πpfq 0, and π is not injective.

Conversely, if π is not injective, choose a non-zero f P CbpXq with πpfq 0.Then fpπpxqq πpfqpxq 0 for all x P Y . It follows that there exists a non-zero function on X which vanishes on πpY q, hence π is not surjective.Lemma 1.2.10. Let A and B be two unital C-algebras. Let π : A Ñ B bea unital -homomorphism.

Then π is surjective if and only if π : SppBq Ñ SppAq is injective.

Proof. We may again assume that A CbpXq and B CbpY q for compactHausdor spaces X and Y .

Assume π is not injective. Then there exist two distinct elements x, y P Ywith πpxq πpyq. As x is distinct from y, there exists f P CbpY q withfpxq fpyq. It follows that f is not in the image of π, i.e. π is not surjective.

Conversely, assume that π is injective, and take f P CbpY q. By the Tietzeextension theorem, we can nd a function g P CbpXq such that gpπpxqq fpxq for all x P Y . Hence πpgq f , and π is surjective.

The following corollary should be compared with Corollary 1.2.4.

Corollary 1.2.11. Let A and B be two unital C-algebras. Let π : A Ñ Bbe an injective unital -algebra homomorphism. Then π is isometric.

Proof. Let x P A. We are to show that x πpxq. As y yy12 forall y, we may suppose that x is self-adjoint. Moreover, by restricting π toCp1A, xq, we may also assume that A and B are abelian. Hence A CbpXqand B CbpY q for compact Hausdor spaces X and Y by Theorem 1.2.7.


By Lemma 1.2.9, we obtain that the associated map π : Y Ñ X is surjective.Hence for any f P A, we have

πpfq supyPY

|fpπpyqq| sup

xPX|fpxq|

f,

hence π is isometric.

As an application, we have the following independency of realization for thespectrum of an element in a C-algebra.

Proposition 1.2.12. Let A and B be two unital C-algebras. Let π : AÑ Bbe an injective -algebra homomorphism. Then for x P A one has SpApxq SpBpπpxqq.

Proof. The inclusion SpBpπpxqq SpApxq is immediate. For the inverseinclusion it is sucient to show that if x P A and πpxq invertible in B, thenx invertible in A.

Assume rst that x is self-adjoint. Then SpApxq P R by Corollary 1.2.6.Hence x iε is invertible in A for any ε ¡ 0, say with inverse yε. Then πpyεqis an inverse for πpx iεq. As πpxq is invertible, it follows by continuity ofthe operation a ÞÑ a1 on the set of invertible elements that πpyεq convergesto πpxq1. As the range of π is closed by Corollary 1.2.11, we conclude thatπpxq1 πpzq for some z P A. But then πpxzq 1 πpzxq. As π isinjective, z is an inverse of x.

Assume now that x P A is arbitrary. Then as πpxq is also invertible, wehave that πpxxq is invertible. By the rst part of the proof, we can nd anelement z P A with pzxqx zpxxq 1. Similarly, we have that πpxxq isinvertible, so we can nd y P A with xpxyq pxxqy 1. Hence x admitsa left and a right inverse, so x is invertible.

As a corollary, we arrive at the following powerful tool of continuous func-tional calculus for self-adjoint elements in a C-algebra.

1.2 C-algebras 27

Corollary 1.2.13. Let A be a unital C-algebra, and x P A a self-adjointelement. Then there exists a unique isometric unital -isomorphism

CbpSpApxqq Ñ A, f ÞÑ fpxqsuch that the identity function z ÞÑ z is sent to x.

Proof. By the Stone-Weierstrass theorem, Theorem 1.2.8, polynomials in theidentity function and its adjoint are dense in CbpSpApxqq. This proves theuniqueness of the above map. Let us show existence.

Let B Cp1A, xq be the unital C-algebra generated by x inside A. As x isself-adjoint, this is as well the unital Banach algebra generated by x. HenceLemma 1.1.31 provides us with a homeomorphism

SppBq Ñ SpBpxq, χ ÞÑ χpxq.In turn, this induces a -isomorphism CpSpBpxqq Ñ CpSppBqq such that

f ÞÑ pχ ÞÑ fpχpxqqq .On the other hand, by the Gelfand-Neumark theorem we have a -isomorphism

B Ñ CpSppBqq, y ÞÑ pχ ÞÑ χpyqq .As SpApxq SpBpxq by Proposition 1.2.12, we hence obtain

CbpSpApxqq CbpSpBpxqq CbpSppBqq B ãÑ A,

in such a way that the identity function on SpApxq is sent to x.

The continuous functional calculus also works for normal elements in a C-algebra, i.e. elements x with xx xx. For this, we need the followingresult with which we can then repeat the previous proof in exactly the samemanner.

Proposition 1.2.14. Let A be a unital commutative C-algebra. Assumethat A Cp1A, xq where x is a normal element. Then the natural map

φ : SppAq Ñ SpApxq, χ ÞÑ χpxqis a homeomorphism.


Note that we can not immediately apply Lemma 1.1.31 because the C-algebra generated by x is not the same as the Banach algebra generated byx!

Proof. Clearly φ is continuous.

Let us show that φ is injective. Take χ1, χ2 P SppAq and assume χ1pxq χ2pxq. By Lemma 1.2.5, we also have χ1pxq χ2pxq. Using that χ is acharacter, we have χ1pppx, xqq χ2pppx, xqq for any polynomial p in x andx. By continuity of χ and the fact that A Cp1A, xq, we deduce thatχ1 χ2.

The surjectivity of φ follows as in the proof of Theorem 1.1.32.

Also the equality of norm and spectral radius generalizes to normal elements.

Proposition 1.2.15. Let A be a unital C-algebra, and x P A a normalelement. Then x rApxq, with rA the spectral radius.

Proof. By Corollary 1.2.13 and Proposition 1.2.12, the spectrum and norm ofx coincide with respectively the spectrum and norm of the identity functionon SpApxq. But for any f P CbpXq for X a compact topological space, it isclear that f rCbpXqpfq supt|fpxq| | x P Xu.

1.2.3 Positive elements in a C-algebra

A key notion in the theory of C-algebras is that of positivity. We will rstexamine the notion of positive element in a C-algebra, and then pass to thenotion of positive functional in its dual.

Denition 1.2.16. Let A be a C-algebra. An element x P A is calledpositive if x is self-adjoint and Sp1pxq R. If x, y P A are self-adjoint, wewrite x ¤ y if y x is positive.

Remark 1.2.17. Let A B be a unit-preserving inclusion of unital C-algebras. By Propoposition 1.2.12, an element a P A is positive in B if andonly if it is positive in A, i.e. the notion of positivity does not depend on thesurrounding C-algebra.

1.2 C-algebras 29

Lemma 1.2.18. Let A be a unital C-algebra. Let a be a self-adjoint elementof A. Then there exist positive elements a and a in Cpaq such that

a a a

andaa aa 0.

Proof. We know that Sppaq R. Let f be the function z ÞÑ maxt0, zu onSppaq, and f the function z ÞÑ maxt0,zu. Applying the Stone-Weierstrasstheorem to CbpSppaqq, we see that these functions can be approximated bypolynomials in the function z ÞÑ z. Clearly, we may assume that the ap-proximation polynomials have zero constant term. By continuous functionalcalculus, a fpaq and a fpaq are obtained as limits of polynomialsin a with constant zero term. Again by continuous functional calculus, aand a satisfy the conditions of the lemma.

Lemma 1.2.19. Let A be a unital C-algebra. Then a P A is positive if andonly if a is self-adjoint and t a ¤ t for all (or any) t P R with t ¥ a.

Proof. If a is self-adjoint, then for any t P R, we have Sppt aq t Sppaqand ta rAptaq (where we recall that rA denotes the spectral radius).

Hence if a is positive and t ¥ a, then Sppaq r0, ts, so Sppt aq r0, tsand t a ¤ t.

Conversely, if t P R is such that t a ¤ t, then Sppt aq rt, ts, henceSppaq r0, ts and a is positive.

Lemma 1.2.20. Let A be a unital C-algebra. If a, b P A are positive, thenalso a λb is positive for any λ ¥ 0.

Proof. It suces to consider λ 1. Write t a b. We nd

t a b pa aq pb bq¤ a a b b¤ a b t,

where we used Lemma 1.2.19 in the last step. By that same lemma, a b ispositive.


It is somehow remarkable that one does not need to suppose that a and bcommute to arrive at this conclusion.

Proposition 1.2.21. Let A be a unital C-algebra. Then x P A is positiveif and only if there exists y P A with x yy.

Proof. Assume x is positive. Let

f : Sppxq Ñ Sppxq, z ÞÑ ?z.

Then by continuous functional calculus x12 : fpxq satises px12qx12 x.

Conversely, let y P A and put x yy. Then x is self-adjoint, so by Lemma1.2.18 we can nd positive elements u and v in Cpxq with x u v anduv vu 0. We are to prove that v 0.

Write w yv12. Then ww v12xv12 xv v2, so Sppwwq P R.

On the other hand, we can write w a ib with a 12pw wq and

b 12ipw wq self-adjoint. Then

v2 ww

pa ibqpa ibq a2 b2 ipab baq.

and

ww pa ibqpa ibq a2 b2 ipab baq.

It follows that

ww 2a2 2b2 v2.

By Lemma 1.2.20, we conclude that ww is positive.

Take now λ 0. As Sppwwq R, we can form the element

z pλ wwq1.

A straightforward computation then shows however that λww is invertiblewith inverse λ1p1 wzwq. It follows that Sppwwq R.

1.2 C-algebras 31

As we had already concluded that Sppwwq R, we nd Sppwwq t0u,hence ww 0. But then

v2 ww 0,

and so v 0.

Remark 1.2.22. In the paper of Gelfand and Neumark where C-algebraswere introduced6, an extra axiom was added demanding that 1xx is invert-ible for all x. As Proposition 1.2.21 shows, this axiom is in fact automaticallysatised. The rst proof of this was given by Fukamiya7 and Kaplansky. Infact, in his paper M. Fukamiya proved that the sum of positive elements isagain positive. A little earlier, I. Kaplansky showed independantly that, ifone knew that the sum of two positive elements is again positive, then nec-essarily the spectrum of an element of the form xx must be positive. Thecombined result never appeared in article form as such, but was included(on demand of Kaplansky) in a review article by J. Schatz8 on Fukamiya'sarticle.

Corollary 1.2.23. Let A be a unital C-algebra. Let a, b P A. If a is positive,also bab is positive.

Proof. As a is positive, we can nd c P A with a cc. Then bab pcbqpcbq, which is positive by Proposition 1.2.21.

Corollary 1.2.24. Let A be a unital C-algebra. Let x P A. Then xx PSppxxq for all x P A.

Proof. By Proposition 1.2.3, we have xx rpxxq. As xx is positiveby Proposition 1.2.21, rpxxq must be the supremum of Sppxxq R. AsSppxxq is closed, it follows that rpxxq P Sppxxq.

6I. Gelfand and M. Neumark, On the imbedding of normed rings into the ring ofoperators in Hilbert space, Mat. Sb. 12 (1943), 197213

7M. Fukamiya, On a theorem of Gelfand and Neumark and the B-algebra, Kumamoto

J. Sci. Ser. A 1 (1952), 17228J. Schatz, Math. Rev. 14 (1953), 884


1.2.4 Positive functionals on a C-algebra

Denition 1.2.25. Let A be a unital C-algebra. A linear functional

ϕ : AÑ C

is called positive if ϕpaq ¥ 0 for all positive a P A. A state on A is a positivefunctional such that ϕp1q 1. We dene

SpAq : set of states on A.

Examples 1.2.26. 1. LetX be a compact Hausdor space, and µ a Borelprobability measure on X. Then we obtain a state ϕµ on CbpXq by

ϕµpfq »X

fpxqdµpxq.

Conversely, if ϕ is a state on CbpXq, the Riesz representation theoremsays that there exists a unique Radon probability measure9 µ such thatϕ ϕµ.

2. If A is a unital C-algebra, ϕ a state on A and x P A a normal element,we obtain in particular that the restriction of ϕ to Cp1A, xq determinesuniquely a measure µϕ on Sppxq. This measure is called the spectralmeasure on Sppxq associated to ϕ.

Lemma 1.2.27. Let A be a unital C-algebra. Let ϕ be a positive state onA. Then ϕpxq ϕpxq for all x P A.Remark 1.2.28. An alternative, more direct proof of this lemma is con-tained in the footnote to Lemma 1.2.29.

Proof. Suppose rst that x is self-adjoint. Then by Lemma 1.2.18, we canwrite x u v with u and v positive. As ϕ is positive, it follows thatϕpxq ϕpuq ϕpvq is real.

9A probability measure µ on a compact Hausdor space X is a Radon measure ifµpBq suptµpKq | K compact and K Bu for any Borel set B. If X is metrisable, anyBorel probability measure is automatically Radon.

1.2 C-algebras 33

For general x we can write x a ib with a 12px xq and b 1

2ipx xq

selfadjoint. Hence as ϕpaq and ϕpbq are real,

ϕpxq ϕpa ibq ϕpaq iϕpbq ϕpaq iϕpbq ϕpxq.

Lemma 1.2.29. Let A be a unital C-algebra, and let ϕ be a state on A.Then the sesquilinear form

xa, by : ϕpabq

is conjugate symmetric and positive.

Proof. By Lemma 1.2.27

xb, ay ϕpbaq ϕpabq xa, by.

Moreover10, xa, ay ϕpaaq ¥ 0 for all a P A by Proposition 1.2.21. Hencex , y is a Hermitian (possibly degenerate) positive form.

Proposition 1.2.30. Let A be a unital C-algebra. Then any state ϕ on Ais continuous, with ϕ 1.

Proof. Consider the form x , y from Lemma 1.2.29. Then by the Cauchy-Schwarz inequality, we have

|ϕpaq| |x1A, ay|¤ x1A, 1Ay12xa, ay12

ϕp1q12ϕpaaq12 ϕpaaq12.

10In fact, positivity already implies conjugate symmetry by the polarisation identity

4xξ, ηy °3n0 i

nxη inξ, η inξy.


However, as aaaa is a positive element in A, we have ϕpaaaaq ¥ 0,and hence ϕpaaq12 ¤ aa12 a. It follows that ϕ ¤ 1. As ϕp1Aq 1,we have ϕ 1.

Remark 1.2.31. More generally, if ϕ is a positive functional on A, one hasthat ϕ is continuous with ϕ ϕp1Aq.The following lemma oers a converse of Proposition 1.2.30. It states that abounded functional is positive if its norm is attained in the unit element.

Lemma 1.2.32. Let A be a unital C-algebra, and ϕ a bounded functionalon A. If ϕ 1 ϕp1q, then ϕ is a state.

Proof. Assume by contradiction that there exists a positive a P A with ϕpaq Rr0,8q. Then as Sppaq r0,8q, we can nd λ P C such that Sppaq Dpλ, |λ|q, the closed disc with radius |λ| around λ, but ϕpaq R Dpλ, |λ|q.Put b aλ. As b is normal, we know by Proposition 1.2.15 that b rpbq,the spectral radius of b. But as Sppbq Sppaq λ, we have by constructionthat rpbq ¤ |λ|. It follows by ϕ ¤ 1 and ϕp1Aq 1 that |ϕpaq λ| |ϕpbq| ¤ b ¤ |λ|, which is in contradiction with ϕpaq R Dpλ, |λ|q.Corollary 1.2.33. Let B A be a unital C-subalgebra. Let Res : A Ñ B

be the restriction map. Then Res maps SpAq surjectively onto SpBq.In other words, any state on B extends to a state on A.

Proof. Let ϕ be a state on B. Then by the Hahn-Banach theorem, ϕ canbe extended to a continuous functional ϕ on A with ϕ ϕ 1. Asnecessarily ϕp1Aq 1, it follows from Lemma 1.2.32 that ϕ is a state.

Lemma 1.2.34. Let A be a unital C-algebra, and a P A a normal element.Then

a maxt|ϕpaq| | ϕ P SpAqu.

Proof. As ϕ 1 for ϕ a state, by Proposition 1.2.30, we have |ϕpaq| ¤ a.Conversely, from Proposition 1.2.15 we deduce that there exists z P Sppaqwith |z| a. Let χ be the character on Cp1A, aq CbpSppaqq sendinga to z. Obviously, any character on Cp1A, aq is a state on Cp1A, aq. ByCorollary 1.2.33, it can be extended to a state ϕ on A. Hence ϕ is a state onA with a |ϕpaq|.

1.2 C-algebras 35

Corollary 1.2.35. Let A be a unital C-algebra. Let a, b P A be positiveelements with a ¤ b. Then a ¤ b.

Proof. By Lemma 1.2.34, there exists a state ϕ on A with a ϕpaq. Butϕpaq ¤ ϕpbq ¤ b.

1.2.5 The non-commutative Gelfand-Neumark theorem

Our next peak is to prove that any unital C-algebra can be realized as aC-subalgebra of BpHq for some Hilbert space H. As before, one exploitsthe notion of positivity present in a C-algebra. Let us revisit in more detailthe auxiliary structure which appeared in Lemma 1.2.29.

Denition 1.2.36. Let A be a unital C-algebra and ϕ P SpAq. The leftkernel of ϕ is the set

Nϕ ta P A | ϕpaaq 0u A.

Lemma 1.2.37. For any ϕ P SpAq, one has

Nϕ tn P A | ϕpanq 0 for all a P Au.

Proof. If n satises ϕpanq 0 for all a P A, then in particular ϕpnnq 0,hence n P Nϕ. Conversely, if n P Nϕ and a P A, then using the Hermitianform x , y from Lemma 1.2.29, we see by the Cauchy-Schwarz inequalitythat, for all a P A,

|ϕpanq| |xa, ny|¤ xa, ay12xn, ny12

aϕpaaqϕpnnq

0.

Hence ϕpanq 0 for all a P A.Lemma 1.2.38. Let A be a unital C-algebra and ϕ P SpAq. The left kernelNϕ of ϕ is a left ideal.


Proof. Let m,n P Nϕ and a P A. Then by Lemma 1.2.37, we have for allb P A that

ϕpbpm anqq ϕpbmq ϕppbaqnq 0.

Hence b an P Nϕ.

Lemma 1.2.39. Let A be a unital C-algebra and ϕ P SpAq. Then

xaNϕ, bNϕy ϕpabq

denes a (non-degenerate) Hermitian inner product on the quotient vectorspace ANϕ.

Proof. By Lemma 1.2.37, the inner product from Lemma 1.2.29 descends toANϕ. Clearly the resulting form on ANϕ is non-degenerate by denitionof Nϕ.

Denition 1.2.40. Let A be a unital C-algebra, and ϕ P SpAq. The GNS-space11 associated to pA,ϕq is the Hilbert space completion Hϕ of ANϕ withrespect to the inner product

xaNϕ, bNϕy ϕpabq.

The distinguished element 1A Nϕ P ANϕ Hϕ will be denoted ξϕ.

As Nϕ is a left ideal, ANϕ becomes a left A-module. As Aξϕ ANϕ, wehave that Aξϕ is a dense subset of Hϕ.

Lemma 1.2.41. The left A-module structure on ANϕ extends to a unital-representation πϕ of A on Hϕ.

Proof. Take a, b P A. Then aa ¤ aa, and hence baab ¤ aabb byCorollary 1.2.23. By positivity of ϕ, one has

abξϕ2 ϕpbaabq¤ aaϕpbbq a2bξϕ2.

It follows that left multiplication with a on ANϕ is bounded for the Hilbertnorm. Hence left multiplication with a extends uniquely to an operator on

11GNS stands for Gelfand-Neumark-Segal

1.2 C-algebras 37

Hϕ, which we will denote by πϕpaq. It follows immediately by the uniquenessof these extensions that πϕpabq πϕpaqπϕpbq. Moreover, for a, b, c P A,

xcξϕ, πϕpaqbξϕy xcξϕ, abξϕy ϕpcabq ϕppacqbq xacξϕ, bξϕy xπϕpaqcξϕ, bξϕy.

Hence πϕpaq πϕpaq by density of Aξϕ in Hϕ.

Denition 1.2.42. Let A be a unital C-algebra and ϕ P SpAq. The triplepHϕ, ξϕ, πϕq constructed above is called the GNS-construction with respect toϕ. The representation πϕ is called the GNS-representation.

The following notion will allow one to characterize abstractly those represen-tations which can be realized as GNS-representations with respect to somepositive state.

Denition 1.2.43. Let A be a unital C-algebra, H a Hilbert space andπ : A Ñ BpHq a unital -representation. A vector ξ P H is called cyclic ifthe set πpAqξ is dense in H.

It is clear that if ϕ is a positive state on A, and pHϕ, ξϕ, πϕq the associatedGNS-construction, then ξϕ is a cyclic vector in Hϕ.

Proposition 1.2.44. Let A be a unital C-algebra. Let π be a unital -representation of A on a Hilbert space H, and let ξ P H be a cyclic unitvector. Then

ϕξpaq xξ, πpaqξydenes a positive state on A, and there exists a unique unitary

u : Hϕξ Ñ H

such that uξϕξ ξ and

uπϕξpxq πpxqu, @x P A.


Proof. Clearly ϕξ is a linear functional. As

ϕξpaaq xξ, πpaaqξy xξ, πpaqπpaqξy xπpaqξ, πpaqξy πpaqξ2,

Proposition 1.2.21 gives that ϕξ is positive. As ξ is a unit vector, we havethat ϕξ is a state.

Let now pHϕξ , ξϕξ , πϕξq be the GNS-construction associated to ϕξ. Then forany a, b P A, we have

xaξϕξ , bξϕξy ϕξpabq xξ, πpabqξy xπpaqξ, πpbqξy.

It follows that we can dene an isometric linear map

Aξϕξ Ñ Aξ, aξϕξ ÞÑ πpaqξ.

Let u P BpHϕξ ,Hq be its closure. Then as ξ is cyclic, we see that u is in facta bijective isometry, hence unitary. Since

uπϕξpbqaξϕξ ubaξϕξ πpbaquξϕξ πpbquaξϕξ ,

a density argument lets us conclude that u satises the properties of theproposition. By cyclicity of ξϕξ , it is clear that u is uniquely determined.

Example 1.2.45. Let X be a compact Hausdor space and µ a Radonprobability measure on X. Let CbpXq act by left multiplication on L 2pX,µq,

Lf : L 2pX,µq Ñ L 2pX,µq, g ÞÑ fg.

Then the constant function 1Xpxq 1 is a cyclic vector in L 2pX,µq, and

x1X , Lf1Xy »X

fpxqdµpxq ϕµpfq,

1.2 C-algebras 39

where ϕµ is the state associated to µ. We hence obtain a unitary

u : L 2pX,µq Ñ Hϕµ

such that 1X is sent to ξϕµ and such that

uLfu πϕµpfq, @f P CbpXq.

Proposition 1.2.46. Let A be a unital C-algebra, H a Hilbert space andπ : AÑ BpHq a unital -representation. Then π `απα where πα are unital-representations admitting a cyclic vector.

Proof. Let ξ be any non-zero vector inH. WriteHξ for the closure of πpAqHξ.Then clearly π restricts to a unital -representation on Hξ. Using that πpreserves the adjoint, it then follows that also HK

ξ is invariant under π. Wecan then nd a decomposition as above by induction (or by Zorn's lemma).

The following theorem will be called the non-commutative Gelfand-Neumarktheorem.

Theorem 1.2.47 (Gelfand-Neumark). Let A be a unital C-algebra. Thenthere exists a Hilbert space H and a faithful unital -representation

π : AÑ BpHq.

If A is separable, then H can be chosen separable.

By Corollary 1.2.11, this means that any abstract unital C-algebra has aconcrete representation as a C-subalgebra of some BpHq.

Proof. By Lemma 1.2.34, we can nd for each non-zero a P A a state ϕa PSpAq with ϕapaaq aa. Denote12 H `aPAHϕa together with the -representation13 π `aPAπϕa . Then if a 0, there exists a vector in H on

12Recall that one can take a direct sum of an arbitrary family of Hilbert spaces byrst taking their algebraic sum, dening on this a direct sum inner product by bilinearity,and then completing the resulting space with respect to the norm associated to this innerproduct.

13Recall that this is possible by Lemma 1.1.38 and the fact that the norm of a directsum of two operators is equal to the maximum of their norms.


which a does not vanish, namely the vector which is ξϕa at component a andzero everywhere else. It follows that π is faithful.

If A is separable, we replace the index set of the direct sum by A , where Ais any dense countable subset of A. As Hϕa is separable for any a P A , itfollows that the countable direct sum àPA Hϕa is also separable.

We claim that the representation àPA πϕa is still faithful. Indeed, supposethis were not the case. Then we can nd x P A with πϕapxq 0 for all a P A .In particular, ϕapxxq 0 for all a P A . However, we can nd a P A with

xx aa ¤ xx3

. It follows that

0 ϕapxxq ϕapxx aaq ϕapaaq¥ aa xx aa¥ 2

3xx 1

3xx

1

3xx.

Hence we obtain x 0, a contradiction. It follows that àPA πϕa is faithful.

1.2.6 Non-unital C-algebras

We have a closer look at non-unital C-algebras.

Non-unital C-algebras arise even in situations where one is only interestedin the unital case, for example when one studies ideals in a unital C-algebra.We will show that non-unital C-algebras are `nearly as good' as unital ones:they allow a bounded approximate identity. Before we discuss this notion,we rst treat the procedure of adding a unit to a non-unital C-algebra. Inthe classical world of topological spaces, this corresponds to the procedureof compactication.

We rst consider the analogue of the one-point (or Alexandrov) compacti-cation, which is the minimal compactication.

1.2 C-algebras 41

Proposition 1.2.48. Let A be a C-algebra. Then there exists a uniqueC-norm on the unital -algebra AI . The resulting embedding A Ñ AI isisometric.

Note that the Banach space structure which we dened earlier on AI will notbe appropriate.

Proof. By Lemma 1.2.3, the C-norm is determined by the -algebra struc-ture. Hence the C-norm on AI , if it exists, must necessarily be unique.

Let us explicitly construct a C-norm on AI . For x P A, denote Lx P BpAqfor the bounded operator

Lx : AÑ A, a ÞÑ xa.

Dene x λ1 : Lx λidA for x P A.

Let us rst check that the map

L : AÑ BpAq, x ÞÑ Lx

is isometric: for x 0, we compute

Lx supa¤1

xa

¤ x Lxx

x¤ Lxx

x

Lx.

Hence Lx x.

Let us now verify that with the above introduced norm, AI becomes a unital


Banach -algebra. As L is isometric, we have for x P A that

px λ1q supa¤1

xa λa

¤ supa¤1

ax λa

supa,b¤1

axb λab

¤ supb¤1

xb λb

x λ1.

By symmetry, we have px λ1q x λ1.Now we verify the C-identity. We have

x λ12 supa¤1

xa λa2

supa¤1

axxa λaxa λaxa |λ|2aa

¤ supa¤1

xxa λxa λxa |λ|2a

px λ1qpx λ1q.

As px λ1qpx λ1q ¤ px λ1q2, this shows that the C-identity issatised.

Example 1.2.49. Let X be a locally compact Hausdor space, and let

X X \ t8u

be the one-point compactication. Then clearly C0pXqI CbpX q as unital-algebras. By uniqueness of the C-norm, this is an isometric -isomorphism.

Remark 1.2.50. 1. By adding a unit as above, we see that Proposition1.2.21 still holds for non-unital C-algebras. Indeed, let A be a non-unital C-algebra, and x P A a positive element. Then by Proposition1.2.21, we can nd y P AI with x yy. However, write y z c1with z P A. Then we have x pzz cz czq |c|21. Hence c 0and y z P A.

1.2 C-algebras 43

2. Let A be a non-unital C-algebra, and let a P A be a normal element.Then we can consider Cpa, 1q AI and the associated Gelfan'd iso-morphism

Cpa, 1q Sp1paq.Under this correspondence, the C-algebra Cpaq generated by a is sentto the C-algebra of functions f on Sp1paq with fp0q 0. Hence forany continuous function f on Sp1paq with fp0q 0, we can constructby functional calculus an element fpaq P A.

We now treat the analogue of the Stone-Cech compactication, which is themaximal compactication.

The following lemma introduces the terminology `multiplier'.

Denition 1.2.51. Let A be a Banach algebra. An element Tl P BpAq iscalled a left multiplier if Tlpabq Tlpaqb for all a, b P A.Example 1.2.52. Let A be a Banach algebra and let c P A. Then theelement Lc P BpAq dened by Lcpaq ca for all a P A is a left multiplier.

Remark 1.2.53. Assume A is a unital algebra. Then any left multiplier Tlis of the form Lc, where c Tlp1Aq.Denition 1.2.54. Let A be a C-algebra. The multiplier algebra of A isthe set MpAq of elements Tl P BpAq for which there exists Tr P BpAq suchthat

aTlpbq Trpaqb, @a, b P A.The elements of MpAq are called multipliers of A.

It is easily seen that the multiplier algebra is indeed a subalgebra of BpAq,with the identity operator on A as its unit.

Lemma 1.2.55. Let A be a C-algebra. If Tl is a multiplier, then Tl is a leftmultiplier.

Proof. For all c P A, we have

cTlpabq Trpcqab pTrpcqaqb cTlpaqb.


Hence cpTlpabq Tlpaqbq 0 for all c P A. Taking c pTlpabq Tlpaqbq, wend that Tlpabq Tlpaqb 0, hence Tlpabq Tlpaqb.Lemma 1.2.56. Let Tl P BpAq be a multiplier. Then the associated elementTr is uniquely determined.

Proof. Assume S is another element such that Spaqb aTlpbq for all a, b P A.Then pTrpaq Spaqqb 0 for all a, b P A. Hence S Tr.

In the following, we will call Tr the mate of Tl.

Example 1.2.57. Let A be a C-algebra. For c P A, dene again byLc P BpAq the operator of left multiplication with c, and by Rc P BpAqthe operation of right multiplication with c. Then Lc is a multiplier, and Rc

is the mate of Lc.

Lemma 1.2.58. Let Tl P BpAq be a multiplier. Dene

T l paq pTrpaqq , @a P A.

Then T l P BpAq is a multiplier, and pT

l q Tl.

Proof. Clearly T l is a bounded linear map, hence T

l P BpAq. For a, b P A,we compute

aT l pbq apTrpbqq

pTrpbqaq pbTlpaqq pTlpaqqb.

Hence T l is a multiplier, and pT

l q Tl.

Proposition 1.2.59. Let A be a C-algebra. Then the multiplier algebrabecomes a unital C-algebra for the norm inherited from BpAq and the -operation Tl ÞÑ T

l .

Proof. It is clear that if Sl, Tl are multipliers, and λ P C, then Sl λTl andSlTl are multipliers with respectively Sr λTr and TrSr as their mates, andSl λT

l and T l S

l as their adjoints. Moreover, the identity map idA is clear

a self-adjoint multiplier. It follows that the map Tl ÞÑ T l is an anti-linear

anti-multiplicative involution, and so MpAq becomes a unital -algebra.

1.2 C-algebras 45

Let us show that Tl ÞÑ T l is isometric on MpAq. We showed in the proof of

Proposition 1.2.48 that the map x ÞÑ Lx is isometric. Similarly, the map

Rx : A ÞÑ BpAq, a ÞÑ ax

has norm x for all x P A. Hence for Tl PMpAq, we have

T l sup

a¤1

Tlpaq

supa,b¤1

bTlpaq

supa,b¤1

Trpbqa

supb¤1

Trpbq

supb¤1

Trpbq

T l .

By symmetry, T l Tl.

Let us show that Tl ÞÑ T l satises the C-identity. For Tl PMpAq, we have

Tl2 supa¤1

Tlpaq2

supa¤1

TlpaqTlpaq

supa¤1

TrpTlpaqqa

¤ supa¤1

TrpTlpaqq

supa¤1

T l pTlpaqq

supa¤1

pT l Tlqpaq

T l Tl.

It follows that T l Tl ¤ Tl2 ¤ T

l Tl, and the C-identity is satised.

Let us nally show that MpAq is closed in BpAq. Let Tn be a sequenceof multipliers converging to T P BpAq. As the map S ÞÑ S on MpAq is


isometric, it follows that T n is a Cauchy sequence in BpAq. Let S be its

limit. Then for a, b P A, we haveaT pbq lim

naTnpbq

limpT n paqqb

Spaqb.Hence the map a ÞÑ Spaq is a mate for T , and T is a multiplier.

We still need to embed A into MpAq.Lemma 1.2.60. Let

L : AÑ BpAqbe the map appearing in Proposition 1.2.48. Then the image of L lies inMpAq, and L is an isometric -algebra homomorphism into MpAq.

Proof. Take x P A. Then for a, b P A, we haveaLxpbq axb

pxaqb.Hence Lx is a multiplier with Lx Lx . The fact that L is isometric wasalready proven in Proposition 1.2.48. The lemma follows.

Example 1.2.61. Let X be a locally compact space, and let βX be theStone-Cech compactication. Then

MpC0pXqq CbpβXq CbpXq.Indeed, if f P CbpXq, then fg gf P C0pXq for g P C0pXq. This denes aunital -algebra homomorphism CbpβXq Ñ MpC0pXqq. Moreover, as X isdense in βX, an element f P CbpXq such that fg 0 for all g P C0pXq is thezero element. Hence the map CbpβXq Ñ MpC0pXqq is injective, and thusisometric.

To show that the map is surjective, consider f P MpC0pXqq. Take x P Xand g P C0pXq with gpxq 1. Dene fpxq pfgqpxq. This is independentof the chosen g. Indeed, if also hpxq 1, then

pfgqpxq ppfgqhqpxq ppfhqgqpxq pfhqpxq.

1.2 C-algebras 47

Thus we have associated with f a function x ÞÑ fpxq.Let us show that it is continuous. Take x P X. Then there exists a compactneighborhood K of x and a function g P C0pXq such that gpyq 1 for ally P K. Hence fpyq pfgqpyq on K, and so f is continuous in x.

Let us show that it is bounded. Take x P X. Then we can nd g P C0pXqwith gpxq 1 and g ¤ 1. Thus |fpxq| |pfgqpxq| ¤ fg ¤ f.Any multiplier f P MpC0pXqq has thus been identied with a continuousbounded function x ÞÑ fpxq. It is easy to check that the multiplier associatedto this function is precisely f . Hence the map CbpβpXqq Ñ MpC0pXqq issurjective.

Although we dened MpAq as a quite abstract C-algebra, one can give aconcrete presentation of it once A has been concretely represented.

Proposition 1.2.62. Let A be a C-algebra. Let H be a Hilbert space, andπ : A Ñ BpHq a faithful non-degenerate -representation. Then π extendsto a -isomorphism MpAq MπpAq where

MπpAq : tT P BpHq | tTπpaq, πpaqT u πpAq, @a P Au.

Proof. If A has a unit, there is nothing to prove.

If A is not unital, let AI be the C-algebra dened in Proposition 1.2.48.By Proposition 1.1.38, we can extend π to a unital -homomorphism AI ÑBpHq. As π is injective and A is not unital, this extension is still faithful.By Corollary 1.2.11, the map π is isometric. Hence any x P Mπpaq denes aleft multiplier Lx of A by the formula

Lxa π1pxπpaqq.It clearly is a multiplier, with Rxa π1pπpaqxq as its mate. We obtain inthis way a -homomorphism

ρ : MπpAq ÑMpAq, x ÞÑ Lx,

which by the non-degeneracy of π is injective. The map ρ is the identity onA. It remains to show that ρ admits an inverse.

Let x P MpAq. We claim that there exists a unique πpxq P BpHq such thatπpxqπpaqξ πpxaqξ for all a P A and ξ P H.


Namely, put πpAqH t°ni1 πpaiqξi | n P N, ai P A, ξi P Hu. First dene

πpxq as a linear map πpAqH Ñ πpAqH by

η ¸i

πpaiqξi ÞÑ πpxqη ¸i

πpxaiqξi.

We have to show that this is independent of the presentation of η as sucha nite linear combination. But suppose that

°i πpaiqξi 0. Then for any

b P A and ζ P H, we compute that

xπpbqζ,¸i

πpxaiqξiy xζ,¸i

πpbxaiqξiy

xπpxbqζ,¸i

πpaiqξiy

0.

As πpAqH is dense in H, it follows that also°i πpxaiqξi 0.

Let us show now that πpxq : πpAqH Ñ πpAqH extends to a bounded operator.Let η °i πpaiqξi P πpAqH. Then on MpAq we have the linear functional

ωη : y ÞÑ xη, πpyqηy.Now for any y PMpAq we have

ωηpyyq ¸i,j

xπpaiqξi, πpyyqπpajqξjy

¸i,j

xξi, πpai yyajqξjy

¸i,j

xπpyaiqξi, πpyajqξjy

¸i

πpyaiqξi2.

It follows that ωη is a positive functional onMpAq. Hence, by Remark 1.2.31,we nd

πpxqη2 ωηpxxq¤ xxωηp1Aq x2η2.

1.2 C-algebras 49

So πpxq is bounded by x on πpAqH, and extends to an operator πpxq PBpHq by continuity.

This proves our claim concerning the existence of πpxq. It is then easilychecked that π extends to a unital -algebra homomorphism

π : MpAq ÑMπpAq, x ÞÑ πpxq,and that ρpπpxqq x for all x PMpAq. This proves that π : MpAq ÑMπpAqis a -isomorphism.

Example 1.2.63. Let H be a Hilbert space. A compact operator is anoperator T P BpHq such that14

tTξ | ξ ¤ 1u is pre-compact in the norm-topology of H.

Evidently, any nite rank operator15 is compact. As we shall later see, anycompact operator is the norm-limit of nite rank operators. The set of allcompact operators forms a C-subalgebra B0pHq of BpHq. It is non-unitaland distinct from BpHq if and only if H is innite-dimensional.

We show that MpB0pHqq BpHq by a -automorphism xing B0pHq. Infact, B0pHqH H, so the natural representation of B0pHq on H is non-degenerate. As it is faithful, we know that

MpB0pHqq tT P BpHq | tTx, xT u B0pHqu.But B0pHq is an ideal in BpHq. Hence MpB0pHqq BpHq.

1.2.7 Approximate units and quotient C-algebras

Denition 1.2.64. Let A be a Banach algebra. An approximate identityfor A is a net of elements peαqαPI such that

limαPI

x eαx limαPI

x xeα 0, @x P A.

It is called a bounded approximate identity if lim supαPI eα 8, and con-tractive if lim supαPI eα ¤ 1.

14A subset of a topological space is pre-compact if its closure is compact.15An operators T is nite rank if TH is nite-dimensional


If A is a C-algebra, an approximate identity teαu will be called positive if

0 ¤ eα ¤ eβ, @α ¤ β.

We aim to show that any C-algebra admits a contractive positive approxi-mate identity. The main part consists in proving the following proposition.

Proposition 1.2.65. Let A be a C-algebra. Let

DA tx P A | x 1u, and A tx P A | x positiveu.Then DA X A is a directed set for the order relation ¤ between self-adjointelements.

Proof. Take a, b P DA X A. Inside AI , the element 1 a is positive andinvertible. Hence we can form the positive element a1 ap1 aq1 insideAI . As A is an ideal in AI , we have in fact a1 P A. Moreover, we can thenwrite a a1p1 a1q1. Similarly, b b1p1 b1q1 for b1 bp1 bq1. Writec c1p1 c1q1 where c1 a1 b1. We claim that c P DA X A and a, b ¤ c.

By Lemma 1.2.20, c1 is positive. Hence c c11c1

, so c 1.

Let us show that a ¤ c, as the inequality b ¤ c follows by symmetry. Bydenition, we have a1 ¤ c1. Hence p1 a1q ¤ p1 c1q. Multiplying both sideswith p1 c1q12 on the left and on the right, this implies

p1 c1q12p1 a1qp1 c1q12 ¤ 1,

and thusp1 c1q12p1 a1q1p1 c1q12 ¥ 1.

Multiplying both sides with p1 c1q12 on the left and right once again, wend p1 a1q1 ¥ p1 c1q1. Hence

a a1p1 a1q1

1 p1 a1q1

¤ 1 p1 c1q1

c1p1 c1q1

c.

1.2 C-algebras 51

With this theorem in hand, we can now easily prove the existence of a con-tractive positive approximate identity.

Theorem 1.2.66. Let A be a C-algebra. Then A admits a contractivepositive approximate identity.

Proof. The set I A X DA forms a net. It is sucient to show then thatlimaPI x xa 0 for all x P A. As any element is a linear combinationof hermitian elements, we may assume x hermitian. By Lemma 1.2.18 (or,rather, its non-unital generalization, cf. Remark 1.2.50.2), we may moreoverassume that x is positive.

Fix then x P A and 0 ε 1. Let"fεptq 1ε

εt, 0 ¤ t ¤ ε,

fεptq 1 ε, ε t.

Using the continuous functional calculus, we can dene the positive elementfεpxq P A X DA. Then as |t tfεptq| εp1 tq for all t ¥ 0, we have

x xfεpxq ¤ εp1 xq.Moreover, if a P A with a ¤ 1 and fεpxq ¤ a, then inside AI , we have byCorollary 1.2.35 that

x xa2 xp1 aqp1 aqx xp1 aq2x¤ 1 axp1 aqx¤ 2xp1 fεpxqqx¤ 2xx xfεpxq¤ 2xp1 xqε.

It follows that limaPI x xa 0.

As a corollary, let us show that quotients of C-algebras are again C-algebras.

Corollary 1.2.67. Let A be a unital C-algebra. Let J A be a closed 2-sided ideal. Then J is closed under the involution , and AJ with its Banachquotient norm is again a C-algebra.


Proof. Let x P J . Then xx P J , hence the C-algebra Cpxxq generated byxx is included in J . Let yα be a contractive positive approximate identityfor Cpxxq. Then

pxxq12p1 yαq Ñ 0,

hence alsoxp1 yαq Ñ 0

by the C-identity. Applying the -operation, we see

p1 yαqx x yαx Ñ 0.

But by construction, yα P Cpxxq J . As J is an ideal, yαx P J . By

closedness of J , we nd x P J .Consider now B AJ . Then B is a Banach -algebra. Let us verify thatthe C-identity holds in B.

As J is a C-algebra, we can nd a contractive positive approximate identityuα of J . We claim that x J limα x xuα for x P A. Indeed,x J ¤ lim infα x xuα is immediate. On the other hand, if y P J , wehave

x y ¥ px yqp1 uαq x xuα y yuα¥ x xuα y yuα.

As limα y yuα 0, we nd x y ¥ lim supα x xuα. This proves theclaim.

Hence

x J2 limαx xuα2

limαp1 uαqxxp1 uαq

¤ limαxxp1 uαq

xx J.

For further reference, let us record the following identity derived in the proofof the previous corollary.

1.2 C-algebras 53

Lemma 1.2.68. Let A be a unital C-algebra, and J A a closed 2-sidedideal in A. Let uα be a positive contractive approximate identity for J . Then

x J limαx xuα for all x P A,

where the left hand side denotes the quotient norm of AJ .Corollary 1.2.69. Let A,B be unital C-algebras, and π : A Ñ B a unital-homomorphism. Then πpAq is closed in B.

Proof. By Corollary 1.2.67, C AKerpπq is a unital C-algebra. Then πinduces an injective -homomorphism π : C Ñ B, which is isometric byCorollary 1.2.11. As πpCq πpAq, the corollary follows.

Chapter 2

Examples of Banach and

C-algebras

2.1 The Calkin algebra and the concrete Toeplitz

C-algebra

2.1.1 Compact operators

Denition 2.1.1. Let V be a Banach space. An operator x P BpV q is calledcompact if the image of the unit ball of V under x is pre-compact in thenorm-topology.1 We denote

B0pV q tx P BpV q | x compactu.

Equivalently, x P BpV q is compact if and only if the image under x of eachbounded set is pre-compact.

Example 2.1.2. Let V be a Banach space. Then a nite rank operator2

x P BpV q is compact. Indeed, then the image ImpV q is a nite-dimensionalBanach space, hence all bounded sets in ImpV q are pre-compact.

1A subset of a topological space is called pre-compact if its closure is compact.2An operator is called nite rank if its image is nite-dimensional, or equivalently, if

its kernel is of nite co-dimension.

55

56 Chapter 2. Examples of Banach and C-algebras

Let V be a Banach space. We want to show that the set B0pV q forms aclosed 2-sided ideal.

We rst recall the following general fact.

Lemma 2.1.3. Let pX, dq be a complete metric space. Let Y X be aclosed subset. Then Y is compact if and only if Y is totally bounded: foreach ε ¡ 0, there exists in X a nite number of balls with radius ε whoseunion contains Y .

Proof. The proof is a variation on the classical Heine-Borel Theorem.

Proposition 2.1.4. Let V be a Banach space. Then B0pV q BpV q is aclosed 2-sided ideal.

Proof. For x P BpV q, write Wx for the closure txv | v ¤ 1u.If A,B V are compact subsets, then clearly AB ta b | a P A, b P Buis again compact. As Wxy WxWy, it follows that B0pV q is closed underaddition.

Let now x P B0pV q and y P BpV q. Since Wxy yWx, it follows that xy iscompact. As y is continuous and Wyx yWx, it follows that yx is compact.Hence B0pV q is a 2-sided ideal.

It remains to show that B0pV q is closed. Let xn be a sequence of compactoperators which converge to x P BpV q. By Lemma 2.1.3, it is sucientto show that Wx is totally bounded. So, take ε ¡ 0. Take n ¡ 0 withx xn ε3. As xn is compact, we can choose a nite number of yi P Vsuch that the open balls Bpyi, ε3q of radius ε3 around yi coverWxn . It thenfollows easily that the Bpyi, εq cover Wx. As ε was arbitrary, we see that Wx

is totally bounded.

Remark 2.1.5. The closure B00pV q of the space of all nite rank operatorson V is also a closed 2-sided ideal. As we shall see later, B00pV q B0pV qfor V a Hilbert space, but this is not true in general.3

3A Banach space V is said to have the approximation property (AP) if B0pV q B00pV q.Per Eno constructed a separable Banach space without the approximation property in"P. Eno, A counterexample to the approximation property in Banach spaces. ActaMath. 130, 309317 (1973)." For the construction of this example, Per Eno earneda live goose from Stanislaw Mazur! For a picture, see http://www-history.mcs.st-and.ac.uk/Biographies/Mazur.html.

2.1 The Calkin algebra and the concrete Toeplitz C-algebra 57

Example 2.1.6. Let x pxnq be a bounded sequence of complex numbers.Then the operator

Mx : l2pNq Ñ l2pNq, panq ÞÑ pxnanqis compact if and only if limnÑ8 xn 0.

Indeed, we clearly have that Mx ¤ supn |xn| (in fact, this is an equality).Hence if limnÑ8 xn 0, then Mx is the normlimit of operators MxN where

pxNqn "xn if n ¤ N,0 if n ¡ N.

As such MxN are nite rank, they are compact. Hence Mx is compact as anorm-limit of compact operators.

Conversely, suppose that Mx is compact. Suppose on the contrary thatlimn xn 0. Then we can choose ε ¡ 0 and a subsequence xnk with |xnk | ¥ εfor all k. Let en be the natural orthogonal basis of l2pNq consisting of se-quences which are zero everywhere except for the value 1 at the n-th posi-tion. Then xen xem

a|xn|2 |xm|2 for n m, hence in particularxenk xenl ¥

?2ε for k l. But by compactness of x, the sequence pxenkq

contains a Cauchy subsequence, giving a contradiction.

2.1.2 The Calkin algebra

Let V be a Banach space. As B0pV q is a closed 2-sided ideal in BpV q, wecan form the quotient Banach algebra QpV q BpV qB0pV q with quotientmap

πQ : BpV q Ñ QpV q.In some ways, the operators in B0pV q can be seen as innitesimal operators.If we then want to consider an operator modulo innitesimal changes, that is,`up to compact perturbations', it makes sense to consider rather the imageof the operator inside QpV q. For example, one can modify the notion ofspectrum in this way.

Denition 2.1.7. Let V be a Banach space. Let x P BpV q. The essentialspectrum of x is

Spepxq SpQpV qpπQpxqq.


Now suppose that V H, a separable innite-dimensional Hilbert space(which is unique up to isomorphism). Then as B0pHq BpHq is a norm-closed 2-sided ideal, we know from Corollary 1.2.67 that B0pHq is automati-cally closed under the involution . Moreover, by considering an orthonormalbasis, it is immediate that the unit idH P BpHq is not compact. Hence B0pHqis a proper 2-sided ideal.

Denition 2.1.8. Let H be a separable innite-dimensional Hilbert space.The Calkin algebra is dened as the C-algebra

QpHq BpHqB0pHq.

The Calkin algebra is very `large' C-algebra. For example, it is not separa-ble, and even has no representation on a separable Hilbert space.

Example 2.1.9. Let S be the shift operator on l2pNq, so4

S : pa0, a1, a2, a3, . . .q ÞÑ p0, a0, a1, a2, a3, . . .q.

Then

S : pa0, a1, a2, a3, . . .q ÞÑ pa1, a2, a3, a4, . . .q.

We see that SS idl2pNq while SS idl2pNq p, where p is the projec-

tion onto the one-dimensional subspace Ce0 spanned by the vector e0 p1, 0, 0, 0, . . .q. Hence πQpSq is an inverse for πQpSq in Qpl2pNqq.Let us show that SpepSq S1 tz P C | |z| 1u. For z P S1, consider

Mz : l2pNq Ñ l2pNq, panq ÞÑ pλnanq.

Then Mz is a unitary with inverse Mz Mz, and an easy calculation shows

that

MzSMz zS.

It follows that the spectrum of πQpSq is invariant under rotation. But asπQpSq is a unitary, its spectrum is a non-empty subset of S1. Hence SpepSqmust be the whole of S1.

4We take the shift in the other direction than in the rst chapter.


2.1.3 The Fredholm alternative

The main result we want to prove is the following.

Theorem 2.1.10. Let V be a Banach space. For each x P B0pV q, one hasSppxq Y t0u PSppxq Y t0u, where PSppxq is the point spectrum of x,

PSppxq tλ P C | Dv P V zt0u, xv λvu.

Moreover, this set is at most countable.

The two parts of the theorem will be proven in Proposition 2.1.13 and Corol-lary 2.1.18.

We rst prove that PSppxq Y t0u is at most countable.

Lemma 2.1.11 (Riesz' Inequality). Let V be a Banach space. For W Va closed subspace and 0 ε, there exists v P V with v 1 and

dpv,W q inftv w | w P W u ¥ 1 ε.

Proof. Take u P V zW , and put r dpu,W q. Then r 0 as W is closed.

Take δ ¡ 0. Then we can take w P V with dpu,wq r δ. Write v pr δq1pu wq. Then v 1, and for any w1 P W we have

v w1 pr δq1pu wq pr δqw1 pr δq1u pw pr δqw1q¥ r

r δ.

Hence for δ small enough, we have dpv,W q ¥ 1 ε.

Lemma 2.1.12. Let V be a Banach space and x P B0pV q. Then for λ PCzt0u, the space Kerpλ xq is nite-dimensional.

Proof. Assume that this were not the case. Then by the Riesz' inequality,we can construct inductively a set of vn P Kerpλ xq with vn 1 and

vn vm ¥ 1

2for n m. (2.1)


As x is compact, we may assume that xvn is convergent. However, as vn PKerpλxq, we have xvn λvn. Since λ 0, this implies that vn is convergent.This is in contradiction with (2.1).

Proposition 2.1.13. For x P B0pV q, the set PSppxqY t0u is at most count-able, with only 0 as a possible limit point.

Proof. The proof is similar to the proof of Lemma 2.1.12.

Namely, let λ be a limit point of PSppxq. It is sucient to show that λ 0.

Assume on the contrary that λ 0. Then we can take λn P PSppxqzt0u withλn λm for n m and with limnÑ8 λn λ. Write Vn for the space spannedby all Kerpλm xq with m ¤ n. As Vn is the direct sum of its subspacesKerpλm xq, and as the λn are all distinct, it follows by Lemma 2.1.12 thatthe Vn are closed subspaces and Vn Vn1.

Hence, by the Riesz lemma, we can inductively construct unit vectors vn P Vnwith vm vn ¥ 1

2for n m, and we may then as well assume that

vn P Kerpλn xq since Vn is a direct sum. As x is compact, we may assumethat xvn converges, say to a vector w P V . Then as xvn λnvn, it followsthat vn converges to λ1w. This is in contradiction with the choice of thevn.

We now prove that in fact Sppxq Y t0u PSppxq Y t0u.

Proposition 2.1.14. Let V be a Banach space, and let x P B0pV q. LetW V be a closed subspace. Then for any λ 0,

pλ xqW tpλ xqw | w P W u

is a closed subspace of V .

Proof. As Kerpλ xq is a nite-dimensional Banach space, say of dimensionn, we can nd a continuous linear isomorphism

π : Kerpλ xq Ñ Cn, v ÞÑ pω1pvq, . . . , ωnpvqq.

Let ρ be the inverse of π. Extend the ωi to continuous linear functionals onW , and consider the corresponding extension π : W Ñ Cn. Let U be the


kernel of π. Then U is a closed subspace of W and U X Kerpλ xq t0u.Moreover, w pw ρpπpwqqq ρpπpwqq for each w P W , so we see thatW U Kerpλ xq, hence pλ xqU pλ xqW . It follows that we mayas well assume U W .

Assume now that wn P U is a sequence such that λwn xwn converges to anelement v P V .We want to show rst that the wn are uniformly bounded. Assume onthe contrary that this is not the case. Then we may suppose that wn 0for all n and limn wn 8. Put w1

n 1wn

wn. Then it follows that

λw1n xw1

n converges to zero. However, by the compactness of x, we mayas well suppose that xw1

n converges. This implies that also λw1n converges.

Since λ 0, it follows that w1n converges to a vector w1 P U . But then

λw1 xw1 limnÑ8 λw1n xw1

n 0. So w1 P Kerpλ xq. As λ x isinjective on U , we conclude w1 0. But also w1 limnÑ8 w1

n 1, givinga contradiction.

Hence wn is a bounded sequence. But then again by compactness of x, wemay as well assume that xwn converges. This in turn implies that λwn, andhence also wn converges, say to an element w. Hence λwn xwn convergesto λw xw v. It follows that the image of λ x is closed.

For the next lemma, we will need a result from classical analysis.

Theorem 2.1.15 (Arzelà-Ascoli theorem). Let X be a compact Hausdorspace. Then a set of functions F in CbpXq is pre-compact (in the uniformnorm-topology) if and only if F is equicontinuous and pointwise bounded.

Here a family F of functions is called equicontinuous if

@ε ¡ 0, @x P X, D open Ux Q x, @f P F , @y P Ux, |fpxq fpyq| ε.

A family F of functions is called pointwise bounded if

@x P X, supfPF

|fpxq| 8

Lemma 2.1.16. Let x P B0pV q. Let V be the Banach space dual of V , andlet xt be the dual operator

xt : V Ñ V , ω ÞÑ ω x.Then xt is also compact.


Proof. LetWx be the closure of the set txv | v P V, v ¤ 1u. By assumption,Wx is a compact Hausdor space. Let

F tω|Wx | ω P V , ω ¤ 1u CbpWxq.Then f ¤ x for each f P F , and |fpwqfpw1q| ¤ ww1 for each f P Fand w,w1 P Wx. Hence we can apply the Arzelà-Ascoli theorem to concludethat F is a pre-compact set for the norm topology.

However, consider the map

ρ : V Ñ CbpWxq, ω ÞÑ ω|Wx .

Then ρ is a linear map with ρpωq xtω for any ω P V . Moreover,ρpωq P F when ω ¤ 1. It follows that the image of the unit ball of V

under xt is pre-compact.

Theorem 2.1.17 (The Fredholm alternative, preliminary version). Let x PB0pV q. Then idV x is injective if and only if idV x is surjective.

This property is evident for linear maps between nite-dimensional vectorspaces, but is in general not true for arbitrary operators on Banach spaces!

Proof. For n ¥ 1, put Wn pidV xqnV . By induction, we conclude byProposition 2.1.14 that eachWn is closed. Moreover, as xw wpidV xqwfor each w P V , we have that xWn Wn for each n.

Assume now that idVx is injective. Suppose that idVx were not surjective.Then the injectivity of idV x implies that Wn1 Wn for all n. By theRiesz inequality, we can then construct a sequence of unit vectors wn P Wn

with dpwn,Wn1q ¥ 12for all n. Now for k ¡ 0, we have

xwn xwnk wn ppidV xqwn xwnkq .As both pidV xqwn and xwnk lie in Wn1 pidV xqWn, it follows thatxwn xwnk ¥ 1

2. This clearly contradicts the fact that the sequence

pxwnqn has a convergent subsequence.

Assume now on the other hand that idV x is surjective. Then the dualoperator idV xt on V is injective. As idV xt is compact by Lemma2.1.16, it follows by the rst part of the proof that idV xt is surjective. Bythe Hahn-Banach theorem, this implies idV x injective.


Corollary 2.1.18. If x P B0pV q, then Sppxq Y t0u PSppxq Y t0u.

Proof. Assume that λ P Sppxqzt0u. Then idV λ1x is not invertible. Asa bijective map automatically has a bounded inverse, we conclude by theFredholm alternative that idVλ1x has non-zero kernel, i.e. λ P PSppxq.Remark 2.1.19. If V is innite dimensional, we have in fact that Sppxq PSppxq Y t0u for any compact operator x. Indeed, if 0 R Sppxq, this wouldimply that x is an invertible compact operator, hence that the unit of BpV qis compact. However, by Riesz' inquality this can only happen for V nitedimensional.

In the following, we will need a slightly stronger assertion that the one ofTheorem 2.1.17.

Theorem 2.1.20 (The Fredholm alternative, full version). Let V be a Ba-nach space, and let x P B0pV q be a compact operator. Then KerpidV xq andCoKerpidV xq are nite-dimensional, and moreover5

dimpKerpidV xqq dimpCoKerpidV xqq.

Proof. The fact that KerpidV xq is nite-dimensional is Lemma 2.1.12. AsImpidV xq is closed by Proposition 2.1.14, we have that CoKerpidV xq KerpidV xtq. But by Lemma 2.1.16 and Lemma 2.1.12, this vector space isalso nite-dimensional.

Write now V0 KerpidV xq. As in the proof of Proposition 2.1.14, we canchoose a closed subspace V1 V such that V0XV1 t0u and V0V1 V . Inparticular, any element v P V can be written in a unique way as v0 v1 withvi P Vi. It follows that the quotient map V1 Ñ V V0 is a continuous bijectivemap. By the bounded inverse theorem, its inverse is also continuous. Itfollows that the projection map

p1 : V Ñ V, v ÞÑ v1

is continuous. The projection map p0 idV p1 onto V0 is moreover niterank.

5Recall that if V,W are vector spaces and φ : V ÑW a linear map, then CoKerpφq V Impφq, the quotient of V by the image of φ


Let us further write W1 ImpidV xq, and take again a nite-dimensionalsubspace W0 V such that the restriction of the natural projection mapV Ñ CoKerpidV xq to W0 is a linear isomorphism. Then we also have thatany element v of V can be written uniquely as v10 v11 with v1i P Wi. ByLemma 2.1.14, we obtain a continuous projection map

q1 : V Ñ V, v ÞÑ v11.

Again, the projection map q0 idV q1 onto W0 is nite rank.

Let then y0 : V0 Ñ W0 be a linear map such that y0 is either an injective orsurjective map. Then we obtain a continuous nite rank map

y0 : V Ñ V, v ÞÑ y0p0.

AsV1 Ñ W1, u ÞÑ pidV xqu

is bijective, it follows that also

y : V Ñ V, v ÞÑ pidV xqv y0v

is either injective or surjective. As y idV px y0q with x y0 compact,it follows from Theorem 2.1.17 that y is in fact bijective. This forces y0 tobe bijective, hence dimpKerpidV xqq dimpCoKerpidV xqq.

2.1.4 Compact operators on a Hilbert space

If V is a Hilbert space, it is especially easy to write down the general formof a compact operator. Let us rst introduce the following general polardecomposition of an operator on a Hilbert space.

Proposition 2.1.21. Let H be a Hilbert space, and let x P BpHq. Let p bethe orthogonal projection onto KerpxqK, and let q be the orthogonal projectiononto the closure of Impxq. Write |x| pxxq12. Then there exists a uniqueoperator v P BpHq with x v|x|, vv p and vv q.

Remark 2.1.22. A partial isometry in a unital C-algebra A is an elementv such that p vv and q vv are projections. One then calls p the sourceprojection and q the range projection of v.


Proof. Let ξ P H. Then

|x|ξ2 x|x|ξ, |x|ξy xξ, |x|2ξy xξ, xxξy xxξ, xξy xξ2.

In particular, Kerpxq Kerp|x|q. Since ImpyqK Kerpyq for any y P BpHq,and since |x| is self-adjoint, we have that pH is the closure of the image of|x|. It follows that we can dene an isometric operator v : pH Ñ H suchthat |x|ξ ÞÑ xξ.

Dene v vp P BpHq. Then clearly x v|x| and vv p. Moreover,vv vv is the projection onto the range of v, which is precisely qH. Hencevv q.

It follows that if x P BpHq is a compact operator, we can write x v|x| withv a partial isometry, and |x| pxxq12 again a compact operator as B0pHqis a closed 2-sided ideal. Hence it suces to describe the structure of positivecompact operators.

Proposition 2.1.23. Let H be a Hilbert space, and let x P B0pHq be apositive compact operator. Then there exist strictly positive numbers λ1 ¡λ2 ¡ . . . and a Hilbert space decomposition

H H0 `H1 `H2 ` . . . ,

such that

1. Each Hn with n ¡ 0 is nite-dimensional.

2. For each ξ P H0, one has xξ 0.

3. For each ξ P Hn with n ¡ 0, one has xξ λnξ.

If H is innite-dimensional, then limnÑ8 λn 0, while the λn are nite innumber if H is nite-dimensional.

Proof. By Proposition 2.1.13 and the positivity of x, the set PSppxqzt0u isat most a countable subset of the positive reals. Hence we can arrange its


elements in a strictly decreasing sequence λ1 ¡ λ2 ¡ . . . Again by Proposition2.1.13 and Remark 2.1.19, limnÑ8 λn 0 in case H is innite-dimensional. IfH is nite-dimensional, then one has of course that the number of eigenvaluesof x is nite (as their eigenspaces are orthogonal subspaces).

For n ¥ 1, putHn tξ P H | xξ λnξu.

By Lemma 2.1.12, the spaces Hn are nite-dimensional.

Let H0 be the orthogonal complement of `n¥1Hn H, and suppose thatH0 is not empty. As x is self-adjoint, x restricts to a compact operator x0

on H0. However, the point spectrum of x0 is clearly either empty or t0u.By Corollary 2.1.18, we deduce that Sppx0q t0u. Hence x0 0, as x0 isself-adjoint. It follows that H0 Kerpxq.Corollary 2.1.24. Let H be a Hilbert space. Then B0pHq is the closure ofthe algebra of nite rank operators.

Proof. Let x be a compact operator on H. Then x u|x| with u a partialisometry and |x| a positive compact operator. As the algebra of nite rankoperators is also an ideal, it thus suces to prove that any positive compactoperator can be approximated in norm by nite rank operators.

So, take x a positive compact operator. As in Proposition 2.1.23, choose aHilbert space decomposition H H0 ` H1 ` H2 ` . . . with H0 Kerpxqand, for n ¡ 0, Hn an eigenspace at eigenvalue λn ¡ 0 where λ1 ¡ λ2 ¡. . . and λn Ñ 0. For n ¡ 0, let xn be the operator which equals x onH1 `H2 ` . . . `Hn, and which is zero on the orthogonal complement. Asthe Hn for n ¥ 1 are nite-dimensional, xn is a nite rank operator. AslimnÑ8 λn 0, it follows that x xn Ñ 0.

Remark 2.1.25. The proofs of subsection 2.1.3 become a little easier inthe Hilbert space case, as one can avoid the Riesz' inequality arguments bytaking orthogonal complements of subspaces in the Hilbert space.

2.1.5 The Fredholm index

Recall that if V is a linear space and x a linear operator V Ñ V , the cokernelCoKerpxq is the quotient space V xV .


Denition 2.1.26. Let V be a Banach space. An element x P BpV q is calledFredholm if both Kerpxq and CoKerpxq are nite dimensional.

Example 2.1.27. Let x be a compact operator on V . Then idV x isFredholm by Theorem 2.1.20.

The following lemma hence extends Proposition 2.1.14 to arbitrary Fredholmoperators.

Lemma 2.1.28. Let V be a Banach space, and let x P BpV q be Fredholm.Then the image xV of x is a closed subspace of V .

Proof. As CoKerpxq is nite-dimensional, we can take a nite-dimensionalsubspace W0 V such that the restriction of the natural projection mapp0 : V Ñ CoKerpxq to W0 is a linear isomorphism. Consider a Banach spacedirect sum V Kerpxq `W0. Then we have a bijective continuous map

V Kerpxq `W0 Ñ V, pv Kerpxq, wq ÞÑ xv w.

By the open mapping theorem, it follows that the image of V Kerpxq ` 0 isclosed under this map. However, this image is exactly the image of x.

Remark 2.1.29. Let H be a Hilbert space, and let x P BpHq. As the closureof Impxq is the orthogonal complement of Kerpxq, one sees from the abovethat an operator x P H is Fredholm if and only if the image of x is closedand both the kernel of x and x are nite-dimensional. In this formulationit is essential to include that the image of x is closed!

Lemma 2.1.30. Let V be a Banach space. An operator x P BpV q isFredholm if and only if there exists y P BpV q and nite rank projectionsp, q : V Ñ V such that

xy idV p, yx idV q.

Proof. Write V0 Kerpxq. As in the proof of Theorem 2.1.20, we can choosea closed subspace V1 V such that any element v P V can be written in aunique way as v0 v1 with vi P Vi, and then the projection map

p1 : V Ñ V, v ÞÑ v1

is continuous.


Let us further write W1 Impxq, and take again a nite-dimensional sub-space W0 V such that the restriction of the natural projection mapV Ñ CoKerpxq to W0 is a linear isomorphism. Then we also have thatany element v of V can be written uniquely as v10 v11 with v1i P Wi. ByLemma 2.1.28, we obtain a continuous projection map

q1 : V Ñ V, v ÞÑ v11.

It follows now that we have a continuous bijective map

x : V1 Ñ W1, v ÞÑ xv

between Banach spaces. Let y be its (continuous) inverse. Then we candene an element y P BpV q by

y : V Ñ V, v ÞÑ yv11.

We see thatxy q1, yx p1.

However, q q0 idV q1 is the projection onto the nite-dimensionalsubspaceW0, hence nite rank. Similarly, p p0 idV p1 is nite rank.

Recall that for V a Banach space, we write πQ : BpV q Ñ QpV q BpV qB0pV qfor the canonical quotient map.

Proposition 2.1.31 (Atkinson's theorem). Let V be a Banach space. Anoperator x P BpHq is Fredholm if and only if πQpxq P QpV q is invertible.

Proof. Assume rst that x is Fredholm. By Lemma 2.1.30, it follows thatwe can nd y P BpV q and nite rank projections p, q P BpV q such thatxy idV p and yx idV q. It follows that πQpxq is invertible in QpV qwith inverse πQpyq.Conversely, suppose that πQpxq is invertible in QpV q. Then there exists anelement y P BpV q and compact operators K,L P BpV q such that

xy idV K, yx idV L.

As Kerpxq Kerpyxq KerpidV Lq, it follows from Lemma 2.1.12 thatKerpxq is nite-dimensional.


Similarly, from Theorem 2.1.20 we deduce CoKerpidV Kq CoKerpxyq isnite-dimensional. As we have a surjective map CoKerpxyq Ñ CoKerpxq, itfollows that CoKerpxq is nite-dimensional.

Remark 2.1.32. In the theory of dierential operators, a Fredholm operatory such that πQpyq is an inverse to πQpxq for x Fredholm is often called aparametrix for x.

We can reformulate the previous proposition in terms of the essential spec-trum of an operator.

Corollary 2.1.33. An element x P BpV q is a Fredholm operator if and onlyif 0 R Spepxq.Corollary 2.1.34. Let V be a Banach space, and let FpV q be the set ofFredholm operators. Then FpV q is open (for the norm topology).

Proof. This follows since FpV q is the inverse image of the open set of invert-ibles in QpV q under the continuous map BpV q Ñ QpV q.

If x is a Fredholm operator, we can associate to x two natural numbers,namely dimpKerpxqq and dimpCoKerpxqq. However, these numbers are notstable, in the sense that a slight perturbation of x may drastically changethem.

Example 2.1.35. Let ε P R, and let p be a projection onto a nite-dimensionalsubspace of some Hilbert space H. The idHεp are Fredholm for all ε. How-ever, KerpidH εpq is t0u for ε 1, while KerpidH pq pH.

It turns out that a much stabler invariant is obtained if one considers insteadthe dierence of the above two numbers.

Denition 2.1.36. Let V be a Banach space, and let x be a Fredholm oper-ator. The (Fredholm) index of x is dened as

Indpxq dimpKerpxqq dimpCoKerpxqq.Examples 2.1.37. 1. Let x P B0pV q be a compact operator. Then by

Theorem 2.1.20, we see that idV x is Fredholm with IndpidV xq 0.

2. Let S be the shift operator as in Example 2.1.9. Then we deduce fromAtkinson's theorem that S is Fredholm. Clearly KerpSq t0u andCoKerpSq l2pNql2pN0q. It follows that IndpSq 1.


The key property to show the stability of the index is its logarithmic be-haviour with respect to the multiplication of Fredholm operators.

Theorem 2.1.38. Let V be a Banach space. Then the set FpV q of Fredholmoperators is closed under multiplication. Moreover, if x, y P FpV q, then

Indpxyq Indpxq Indpyq.

In fact, this theorem has no bearing on the Banach space structure, andholds as well for arbitrary vector spaces, a Fredholm operator now just beinga linear map with nite-dimensional kernel and cokernel.

Proof. As Kerpxyq Kerpyq and CoKerpxyq CoKerpxq, it follows thatFpV q is multiplicatively closed.

Take now x, y P FpV q.Consider the map

Impyq XKerpxq Ñ KerpxyqKerpyq, yw ÞÑ w Kerpyq.

It is easily seen that this is a well-dened injective map. As the codomainis nite-dimensional, it follows that this map is in fact bijective. By nite-dimensionality of Kerpxq and Kerpxyq, we nd that

dimpKerpxqqdimpKerpxqpImpyqXKerpxqqq dimpKerpxyqqdimpKerpyqq.

On the other hand, let π : CoKerpxyq Ñ CoKerpxq be the natural projectionmap. Then we have a well-dened surjective map

V pImpyq Kerpxqq Ñ Kerpπq, v Impyq Kerpxq ÞÑ xv CoKerpxyq.

Again by nite-dimensionality of both sides, this map is in fact bijective. As

V pImpyq Kerpxqq CoKerpyq pImpyq KerpyqImpyqq ,

we nd that

dimpCoKerpyqq dimppImpyq KerpxqqImpyqq dimpCoKerpxyqq dimpCoKerpxqq.


It follows that

Indpxyq Indpxq Indpyq dimppImpyq KerpxqqImpyqq dimpKerpxqpImpyq XKerpxqqq.

But the right hand side is zero by the second isomorphism theorem for vectorspaces.

This logarithmic behaviour of the index is the key to prove the many stabilityproperties of the index.

Corollary 2.1.39 (Stability under compact perturbations). Let V be a Ba-nach space, and let x P BpV q be a Fredholm operator. Let K P B0pV q be acompact operator. Then xK is still a Fredholm operator, and IndpxKq Indpxq.

Proof. It follows that xK is Fredholm by Atkinson's theorem.

Again by Atkinson's theorem, we can choose a Fredholm operator y P BpV qsuch that πQpyq is an inverse for πQpxq in QpV q. Then there exists a compactoperator L P B0pV q with xy idV L. By Example 2.1.37.1 and Theorem2.1.38, we nd that Indpxq Indpyq 0.

On the other hand, we also have px Kqy idV pL Kyq with L Kystill compact. So IndpxKq Indpyq 0.

It follows that IndpxKq Indpxq.

The next theorem shows that the notion of index is stable under small norm-perturbations.

Theorem 2.1.40. Let V be a Banach space. Let FpV q be the set of Fredholmoperators. Then the map

Ind : FpV q Ñ Z, x ÞÑ Indpxq

is locally constant.

Proof. Let x be a Fredholm operator. By Atkinson's theorem, we can nd aFredholm operator y P BpV q and a compact operator K P B0pV q such thatxy idV K.


Assume now that z is a Fredholm operator in BpHq with z x y1.Then pzxqy 1, hence idV pzxqy is invertible. In particular, we haveIndpidV pz xqyq 0. It then follows by Theorem 2.1.38 and Corollary2.1.39 that

Indpzq Indpyq Indpzyq Indpxy pz xqyq IndpidV pz xqy Kq IndpidV pz xqyq 0.

But we have as well that Indpxq Indpyq 0. Hence Indpzq Indpxq,proving the theorem.

When V is a separable Hilbert space, one can say something more. We rstprove the following lemma.

Lemma 2.1.41. Let H be a separable Hilbert space, and write G BpHqfor the set of invertible operators. Then G is pathconnected.

Proof. Suppose rst that H L 2pX,µq for some probability Borel measureµ on a compact subset X of the unit circle S1 tz P C | |z| 1u. Then wecan represent the -algebra of essentially bounded Borel functions L 8pS1qon L 2pX,µq by left multiplication,

Lf : L 2pX,µq Ñ L 2pX,µq, g ÞÑ fg.

In particular, let z represent the identity function z ÞÑ z on X, and let h bethe real-valued Borel function

hpeiθq θ, θ P pπ, πs.

Then for the continuous functional calculus, we have

eiLh Leih Lz.

Hence we can make a normcontinuous path from idH to z through unitariesby putting

ψ : r0, 1s Ñ BpHq, t ÞÑ eitLh ,


where the norm-continuity follows from the estimate

eitLh eit1Lh ¤ eith eit

1h 1 eipt

1tqh¤

?2π|t t1|.

In fact, we see that we can take ψ Lipschitz with constant?

2π.

Let now H be a general Hilbert space and u P BpHq a unitary.

Suppose rst that there exists ξ P H such that tukξ | k P Zu is dense in H.Then CpidH, uq CbpSppuqq has ξ as a cyclic vector. By Proposition 1.2.44and Example 1.2.45, it follows that up to unitary conjugation, we are againin the situation as above. Hence there exists a continuous path betweenidH and u, which we may assume to be Lipschitz with respect the universalconstant M ?

2π.

In general, by Proposition 1.2.46, H splits into a direct sum `Hα of GNS-representations for CpidH, uq. As the norm of a direct sum of operators isthe maximum of the respective norms, it follows that if ψα is anM -Lipschitzpath linking idHα to u|Hα , then ψptq `αψαptq is a continuous path linkingu to idH.

Finally, assume x P G is general. By the polar decomposition, Proposition2.1.21, we have x u|x| with |x| invertible and u a unitary.

Now

φ : r0, 1s Ñ G, t ÞÑ p1 tqx t|x|is a well-dened path from the scalar x to |x|: since |x| is positive andinvertible, all p1 tqx t|x| are invertible for 0 ¤ t ¤ 1. Let further ψ bea continuous path linking idH to u. Then

θ : r0, 1s Ñ G, t ÞÑ ψptqφptq

is a continuous path linking the constant operator x to x. It follows thatG is pathconnected.

Proposition 2.1.42. Let H be a separable innite-dimensional Hilbert space,and let FpHq be the set of Fredholm operators. Then x and y belong to thesame connected component of FpHq if and only if Indpxq Indpyq.


Proof. By continuity of the index, elements with dierent indices lie in dif-ferent connected components.

Suppose now that x has index zero. LetH0 be the kernel of x, G1 the image ofx, and H1 and G0 their respective orthogonal complements. By assumption,we can nd a unitary u : H0 Ñ G0. Let u be the extension of u to H bydening it to be zero on H1. Consider

φ : r0, 1s Ñ BpHq, t ÞÑ x tu.

As x tu is invertible for each t ¡ 0, it follows that φt is a continuous pathwithin FpHq from x to an invertible element y x u. By Lemma 2.1.41,we can then nd a continuous path in FpHq joining x with idH.

It follows that Ind1p0q is path-connected.Suppose now that x is a general Fredholm operator. By taking the adjoint,we may suppose that Indpxq n ¥ 0. We may also assume that H l2pNq. Let S be the shift operator on l2pNq. Then by Example 2.1.37.2 andTheorem 2.1.38 (or a direct computation), we nd that, for n ¥ 0, one hasIndpSnq n. Then again by Theorem 2.1.38, IndpSnxq 0. It followsby the rst part of the proof that Snx is pathconnected to idH inside FpHq.But then x pSqnSnx is pathconnected to pSqn inside FpHq. This provesthat Ind1pnq is pathconnected.

2.1.6 The concrete Toeplitz C-algebra

Denition 2.1.43. The concrete Toeplitz C-algebra is the C-algebra CpSq Bpl2pNqq generated by the shift

S : pa0, a1, a2, . . .q ÞÑ p0, a0, a1, a2, . . .q.

Lemma 2.1.44. One has B0pl2pNqq CpSq.

Proof. Let n ¥ 0. Then it is easily seen that idl2pNqSnpSqn is the projectionon the space of all sequences with only the rst n entries possibly non-zero.We deduce by induction that CpSq contains the rank one projections pnonto Cen, with en the standard unit vector of l2pNq.


Now for k ¥ 0, one has Skpn enk,n, where em,n is the rank one operatorsending en to em and all other unit vectors ek to zero. By taking adjoints,we deduce that CpSq contains all operators en,k.By linearity and continuity, it then follows that CpSq contains all nite rankoperators, and hence also B0pHq by Corollary 2.1.24.

Recall now that πQ denoted the quotient map

πQ : Bpl2pNqq Ñ Qpl2pNqq Bpl2pNqqB0pl2pNqq.By Example 2.1.9, πQpSq is unitary with spectrum S1, so the C-algebragenerated by πQpSq is isomorphic to CbpS1q. So, we have a short exactsequence

0 Ñ B0pl2pNqq ãÑ CpSq πQ CbpS1q Ñ 0

of C-algebras.6

We would like to have a linear splitting of the above map, that is, a linear(but not multiplicative!) map s : CbpS1q Ñ CpSq such that πQ s idCbpS1q.

For this, let us rst realize our Hilbert space in a dierent way. Let L 2pS1qbe the space of square integrable functions on S1 with respect to Lebesguemeasure. Then L 2pS1q has an orthonormal basis consisting of the functions

en : S1 Ñ C, z ÞÑ zn, n P Z.

Let H H2pDp0, 1qq L 2pS1q be the Hardy space of the unit disc,

H closed linear span of ten | n ¥ 0u.We can identify H with the space of analytic functions on Dp0, 1q whosepower series coecients are square integrable. We then have a unitary

l2pNq Ñ H, panq ÞÑ8

n0

anen.

Let now pH be the orthogonal projection map from L 2pS1q onto H,

pH : L 2pS1q Ñ H,¸nPZ

anen ÞÑ¸nPN

anen.

6A sequence of homomorphisms 0 Ñ AπÑ B

ρÑ C Ñ 0 between algebras A,B,C is

called exact if π is injective, ρ is surjective and the kernel of ρ equals the image of π.


For f P CbpS1q, write Mf for the multiplication operator

Mf : L 2pS1q Ñ L 2pS1q, g ÞÑ fg,

and writeTf pHMfpH P BpHq

for the compression ofMf toH. Note that under the identicationH l2pNq,we have Tz ÞÑz S and Tz ÞÑz1 S with S the shift operator.

Lemma 2.1.45. For f P CbpS1q, one has T f Tf and Tf ¤ f.

Proof. We have

T f ppHMfpHq

pHMf pH

pHMfpH

Tf .

Furthermore,

Tf pHMfpH¤ pH2Mf¤ f.

Remark 2.1.46. In fact, one can prove that Tf f πQpTf q.The map M is a representation, i.e. for f, g P CbpS1q one has MfMg Mfg.This is no longer true for the map T . Nevertheless, T is a representation upto compacts, as follows from the corollarly to the next proposition.

Theorem 2.1.47. One has

CpSq tTf K | f P CbpS1q, K P B0pHqu,

and the maps : CbpS1q Ñ CpSq, f ÞÑ Tf

is a continuous section of the projection CpSq πQÑ CbpS1q.


Proof. By the estimate Tf ¤ f, the map s is continuous. Now πQpTz ÞÑznq πQpSq is the map z ÞÑ zn for n ¥ 0, and πQpTz ÞÑznq πQppSqnq πQpSnqcorresponds to the map z ÞÑ zn zn for n ¥ 0. As πQ is a -homomorphism,it is continuous, so we can conclude by the Stone-Weierstrass theorem thatπQpTf q f for all f P CbpS1q.

Hence s is a section of πQ restricted to CpSq. As the kernel of πQ is B0pHq,it thus follows that any x P CpSq can be written x TπQpxq px TπQpxqqwith x TπQpxq compact.

Corollary 2.1.48. Let f, g P CbpS1q. Then TfTgTfg is a compact operator.

One can give a topological interpretation for Fredholm operators in the con-crete Toeplitz algebra CpSq.Theorem 2.1.49. Let f P CbpS1q. Then Tf is a Fredholm operator if andonly if 0 R Impfq. In that case, IndpTf q wpfq, where wpfq is the windingnumber of f : S1 Ñ C around zero.

Proof. As πQpTf q f P CbpS1q, it follows from Lemma 2.1.31 that Tf isFredholm if and only if 0 R Impfq.Suppose now that f, g are functions on S1 which don't vanish. Suppose thatf, g are homotopic as maps S1 Ñ Czt0u, and let

φ : r0, 1s S1 Ñ S1, pt, zq ÞÑ φpt, zq

be a homotopy between f and g with φp0, zq fpzq and φp1, zq gpzq.Let CbpS1q CbpS1q be the subset of invertible elements. Then by uniformcontinuity, we obtain a normcontinuous map

φ : r0, 1s Ñ CbpS1q, t ÞÑ pφt : z ÞÑ φpt, zqq

with φp0q f and φp1q g. As h ÞÑ Th ÞÑ IndpThq is a continuous map byTheorem 2.1.40 and Theorem 2.1.47, we have IndpTf q IndpTgq.Now as wpfq wpgq if and only f and g homotopic, and as π1pCzt0uq Z,it suces to prove the theorem for fpzq zn for some n P Z. But then forn ¥ 0 we have IndpTf q IndpSnq nIndpSq n wpfq, and similarlyfor n ¤ 0.


2.2 The abstract Toeplitz algebra

2.2.1 Enveloping C-algebra

Let A be a -algebra, that is, an algebra equipped with an anti-linear, anti-multiplicative involution a ÞÑ a. In general, the resulting notion of positivityfor elements in A can be pathological, for example by the existence of rela-tions of the form

°nk1 a

kak 0. We will however be mostly interested in

-algebras which admit a faithful representation on a Hilbert space.

Examples 2.2.1. 1. Let CrXs be the polynomial algebra in one variable,equipped with the unique -algebra structure such thatX X. LetHbe a separable Hilbert space, and x P BpHq a self-adjoint element withinnite spectrum. Then we have an injective unital -homomorphism

πx : CrXs Ñ BpHq, ppXq ÞÑ ppxq.

The closure of πxpCrXsq is the C-algebra generated by idH and x, andhence isomorphic to CbpSppxqq.

2. Let Γ be a discrete group, and CrΓs the group algebra of Γ. Denote thenatural basis elements of CrΓs by λg. Then CrΓs is a -algebra withthe -structure uniquely determined by λg λg1 . We then have aninjective unital -homomorphism, the regular representation,

πreg : CrΓs Ñ Bpl2pΓqq, λg ÞÑ ug,

where pugfqphq fpg1hq for f P l2pΓq. The closure of πregpCrΓsq, thatis, the C-algebra generated by the ug, is called the reduced C-algebraC

redpΓq of Γ.

Another question one can pose for -algebras is the following: given a -algebra A, when does there exist a C-algebra B and a -homomorphismπu : AÑ B which is a solution to the universal problem of -representing Ain C-algebras? That is, pB, πuq should be such that for any C-algebra Cand any -homomorphism π : AÑ C, there exists a unique -homomorphismΠ : B Ñ C such that Π πu π.

2.2 The abstract Toeplitz algebra 79

Bπ //

πu

C

AΠ

?? (2.2)

In this case, we say A admits a universal C-envelope, namely, the C-algebraB. It is then unique up to -isomorphism by abstract nonsense. Note thatwe do not require the associated homomorphism πu : AÑ B to be injective.

Lemma 2.2.2. Let A be a -algebra with generating set S (so any elementin A is a polynomials expression in the elements of S YS). Then A admitsa C-envelope if and only if for every x P S, there exists Cx ¥ 0 such thatfor any -representation π : AÑ BpHq, one has πpxq ¤ Cx.

Proof. Assume rst that A admits a universal C-envelope pB, πuq. Thenany -homomorphism π : A Ñ BpHq extends to a -representation Π : B ÑBpHq. As a -representation of a C-algebra is contractive, it follows thatπpxq ¤ πupxq for any x P A.Conversely, assume for each x P S the existence of Cx ¥ 0 such that πpxq ¤Cx for all -representations π. Then clearly such Cx's also exist for linearcombinations of products in the elements of S and S. As S is generating,we deduce that such Cx's exist for all x P A.We may assume that the Cx are minimal with respect to this condition. Thenin fact for each x we can nd a -representation πx on a Hilbert spaceHx withπxpxq Cx. Indeed, we can take πx `nπn where πn is a

-representationwith πnpxq ¥ Cx 1

n.

Dene then πu `xPAπx on the Hilbert space H `xPAHx. Let B be theclosure of πupAq. Then πupxq Cx.

We claim that pB, πuq is the universal C-envelope of A. Indeed, if C is aC-algebra, we may assume that C BpGq for some Hilbert space G. Ifthen π : AÑ C is a -homomorphism, it follows that πpxq ¤ Cx πupxqfor every x P A. This implies that there exists a unique -homomorphismΠ : B Ñ C such that Πpπupxqq πpxq for all x P A.Examples 2.2.3. 1. Let A CrXs with X X. Then A does not

admit a universal C-algebra envelope. Indeed, for each self-adjoint


element x P BpHq, there exists a -representation πx of A on H withπxpXq x. Hence πx ÞÑ πxpXq is unbounded.

2. Let A be a Banach -algebra. By Proposition 1.1.38, it follows that Aadmits a universal C-envelope.

3. Let Γ be a discrete group. Then l1pΓq is a Banach -algebra, cf. Ex-ample 1.1.36.5. It follows that l1pΓq admits a universal C-envelopeC

upΓq, called also the universal group C-algebra of Γ. Note that wehave a factorisation

πred : Cu pΓq Ñ C

redpΓq

to the reduced C-algebra from Example 2.2.1.2. In general however,this map is not injective. We will come back to this in a later chapter.

Remark 2.2.4. One can also form the universal C-algebra of a -algebrawith given generators and a priori norm estimates. Namely, in Lemma 2.2.2one then only considers those representations in which the norm estimate issatised. For example, the universal C-envelope of CrXs with the norm-condition X ¤ 1 is the C-algebra Cbpr1, 1sq. Similarly, one can add theassumption of a priori positivity for elements.

2.2.2 The abstract Toeplitz C-algebra

Denition 2.2.5. The abstract Toeplitz -algebra T is the unital -algebragenerated7 by the single element s and the single relation ss 1.

Lemma 2.2.6. The abstract Toeplitz -algebra T admits a universal C-envelope.

Proof. Let π : T Ñ BpHq be a -representation. Then p πp1q is a self-adjoint projection, and moreover πpsqπpsq p. It follows from Lemma 2.2.2that T admits a universal C-envelope.

Denition 2.2.7. The abstract Toeplitz C-algebra T is dened as the uni-versal C-envelope of T .

7We recall that a unital -algebra A is generated by a set B A if any element of Acan be written as a polynomial expression in elements of t1u YB YB.


Clearly, there exists a unital -homorphism

T Ñ CpSq Bpl2pNqq, s ÞÑ S

into the concrete Toeplitz algebra. It then follows that it extends to a sur-jective -homomorphism

T Ñ CpSq.Our aim is to prove that this is in fact a -isomorphism. This is equivalent toshowing that CpSq is a universal solution to the C-homomorphism problemfor T . This, in turn, is implied by the following.

Theorem 2.2.8 (Coburn's theorem). Let H be a Hilbert space, and U P Han isometry. Then there exists a unique unital -representation

π : CpSq Ñ BpHq, S ÞÑ U.

Moreover, if UU idH, then π is injective.

Proof. Of course unicity is clear, so it suces to show existence.

Let X be a set, and consider the Hilbert space HX l2pX Nq. Let SX bethe isometry

SX : HX Ñ HX , f ÞÑ pSXfq :

" pSXfqpx, 0q 0pSXfqpx, nq fpx, n 1q, n ¥ 1.

Then under the natural identication

HX `xPX l2pNq,

the operator SX corresponds to`xPXS. Hence we have a unital-isomorphism

CpSq CpSXq sending S to SX .

On the other hand, let H be a Hilbert space and u a unitary. Then bythe spectral theorem, the inclusion Sppuq S1 gives a -homomorphismCbpS1q Ñ Cpuq sending the identity function to u. Composed with thenatural map CpSq Ñ CbpS1q, we obtain a -representation CpSq Ñ BpHqsending S to u.

Let now U be a general isometry. To prove the theorem, it suces to nd aset X, a Hilbert space G, a unitary u P BpGq and a unitary

v : H Ñ l2pX Nq ` G


such thatvUv SX ` u.

Such a decomposition is known as a Wold decomposition.

Let X be a set of cardinality dimpKerpUqq (which by the above we mayassume ¡ 0, else U is a unitary and we're done). Fix an orthonormal basistex | x P Xu of KerpUq. If n ¡ m and ξ, η P KerpUq, then Unξ K Umη,since

xUnξ, Umηy xξ, pUqnUmηy xξ, pUqnmηy 0.

Hence we can dene an isometry

v1 : l2pX Nq Ñ H, f ÞÑ¸

xPX,nPN

fpx, nqUnex.

Let G be the orthogonal complement of the range of v1. Then

v : l2pX Nq ` G Ñ H, ξ ` η ÞÑ v1ξ η

clearly is a unitary.

We claim that G is invariant under U and U, and that moreover the restric-tion u of U to G is a unitary. Indeed, we have η P G if and only if η K Unξfor any n ¥ 0 and any ξ P KerpUq. Hence G is invariant under U and U.By construction, KerpUq X G 0. It follows that also u is an isometry,hence u a unitary.

It is now clear that vUv SX ` u.

2.2.3 Quantum discs and elds of C-algebras

The Toeplitz C-algebra T is generated by a single element S with SppSq D, the closed unit disc. One might say that T is a non-commutative algebraof functions on the disc, or even more provocatively, an algebra of functionson a quantum disc. In the following, we are going to give some more meaningto this statement. First of all, let us show that one can interpolate betweenthe `quantum disc' and the real disc.


Denition 2.2.9. Let 0 ¤ q 1. The quantum disc algebra at parameterq is the unital -algebra Tq generated by the single element sq with the singlerelation

sqsq qsqsq 1 q.

We see that T0 T .

Lemma 2.2.10. The quantum disc algebra Tq admits a universal C-envelopeTq, and then sq 1.

Proof. Let π be a (unital) -representation of Tq, and put y πpsqq. Thenyy p1 qq qyy. It follows that (with rpq the spectral radius)

y2 yy rpyyq rpp1 qq qyyq p1 qq qrpyyq p1 qq qyy p1 qq qy2.

Hence p1 qqy2 p1 qq, and y 1. As Tq is generated by sq, thelemma follows.

The denition of Tq would of course also make sense for q 1, which givesthe universal (commutative) -algebra generated by a single normal elements1. However, this clearly does not admit a universal C-algebra. But if weconsider the universal problem with respect to -representations sending s1

to a contractive operator, then clearly the universal C-algebra solution tothis problem is precisely the C-algebra CbpDq.We aim to show now that in fact Tq T for any 0 ¤ q 1. This is not truefor the underlying Tq.

Let us rst nd a concrete realization of Tq.

Denition 2.2.11. Let 0 ¤ q 1, and let ten | n P Nu denote the standardorthonormal basis of l2pNq. Dene a weighted shift on l2pNq by

Sq : l2pNq Ñ l2pNq, en ÞÑa

1 qn1en1.

We dene the concrete quantum disc C-algebra at parameter q to be CpSqq.


It is clear that Sq e0 0 and

Sq en a

1 qnen1, n ¡ 0,

and that Sq satises the quantum disc algebra relation

Sq Sq qSqSq p1 qqidl2pNq.

Proposition 2.2.12. One has CpSqq CpSq, and πQpSqq πQpSq in theCalkin algebra.

Proof. Let Tq 1 Sq Sq. Then Tqen qn1en, hence Tq P B0pl2pNqq. SinceSq Sp1 Tqq12, it follows that πQpSq πQpSqq.Now B0pl2pNqq CpSq, hence Sq P CpSq and so CpSqq CpSq. Con-versely, since Tq P CpSqq and p1 Tqq12 invertible, we deduce S P CpSqq.We conclude CpSq CpSqq.Theorem 2.2.13. All Tq with 0 ¤ q 1 are -isomorphic C-algebras.

Proof. Let us show that the canonical map πq : Tq Ñ CpSqq sending sq toSq is a

-isomorphism. The theorem will then follow by Proposition 2.2.12.

By Coburn's theorem, we may suppose that q ¡ 0. By the non-commutativeGelfand-Neumark theorem, we may also assume that Tq BpHq for someHilbert space H. Write tq 1 sqsq. We then claim that Spptqq r0, qs.Indeed, let us rst show that Sppsqsq q t0, 1 q, 1 q2, . . .uY t1u. Supposethis were not the case. Then we can nd λ P Sppsqsq q with λ 1 andλ 1 qn for all n P N. But since Sppxyq Y t0u Sppyxq Y t0u for anycouple of operators x, y, we deduce from the dening relation for sq that alsoq1pλ p1 qqq is a non-zero element in Sppsqsq q. By induction, we obtainthat qnpλ 1q 1 P Sppsqsq q for all n P N. But qnpλ 1q 1 0 for nlarge enough, giving a contradiction with Sppsqsq q r0,8q.Hence also Sppsqsqq t0, 1 q, 1 q2, . . .u Y t1u. Now if 0 P Sppsqsqq, itfollows from the dening relation of sq that p1q1q P Sppsqsq q, contradictingthe positivity of the operator sqs

q . Hence Sppsqsqq t1q, 1q2, . . .uYt1u.

We deduce that indeed Spptqq r0, qs, hence 1 tq is invertible inside Tq.Dene then s sqp1tqq12. By its denition, s is an isometry. By Coburn's


theorem, we nd that we have a -homomorphism

CpSq Ñ Tq, S ÞÑ s.

As clearly πqpsq S (under the isomorphism of Proposition 2.2.12), it followsthat πq is bijective.

Although the Tq are all C-isomorphic, one would like to say Tq is morecommutative than Tq1 if q ¡ q1. To substantiate this, one would like to havea more natural way of comparing dierent Tq when the parameter q varies.The notion of eld of C-algebras is ideally suited for this.

Denition 2.2.14. Let X be a compact Hausdor space. A CbpXq-algebrais a unital C-algebra A together with a unital inclusion

π : CbpXq ZpAq,

where ZpAq is the center of A.8

For x P X, the ber C-algebra Ax at x is dened as the quotient Ax AJxwhere Jx is the closed 2-sided ideal in A generated by the functions in CbpXqwhich vanish at x P X. We call the canonical map πx : AÑ Ax the evaluationmap at x.

One can thus interpret a CbpXq-algebra A as consisting of anX-parametrizedcollection of C-algebras Ax. A map x ÞÑ ax P Ax is then considered a`continuously varying' family of elements in the Ax if and only if there existsa P A with πxpaq ax for all x P X. We say that the assignment x ÞÑ Axforms a eld of C-algebras.

Lemma 2.2.15. Let X be a compact space, and A a CbpXq-algebra. Thenfor all a P A, the map

x ÞÑ πxpaqis upper semi-continuous.

Proof. Let a P A. We are to prove that for any net xα Ñ x, one haslim supα πxαpaq ¤ πxpaq.

8We recall that the center of an algebra A is the set of z P A with zx xz for allx P A.


Fix x P X. Take ε ¡ 0. By denition of the quotient norm, we can ndnitely many fi P CbpXq and ai P A such that fipxq 0 for all i and suchthat, with t °i fiai, one has

πxpaq ¥ a t ε.

Now for each δ ¡ 0, we can nd an open neigborhood U of x and a functiong P CbpXq such that

a) g ¤ 1,

b) gpyq 1 for y P U ,c) gfi δ for all i.

It then follows that gt ¤ δ°i ai. Hence by choosing δ small enough, we

can assume gt ¤ ε. But then

πxpaq ¥ a t ε

¥aq

gpa tq ε

¥gt¤ε

ga 2ε

a p1 gqa 2ε.

But since gpyq 1 for all y P U , we have p1 gqpyq 0. So we conclude bydenition of πy that

πxpaq ¥ πypaq 2ε, @y P U.

This proves the upper semi-continuity.

In general however, there is no reason why the map x ÞÑ πxpaq should becontinuous.

Example 2.2.16. Consider X r1, 1s and Y r1,8q. Let

A CbpX Y q.

Then A is a CbpXq-algebra by the map CbpXq Ñ CbpX Y q associated tothe projection XY Ñ X. Let φ be a continuous bounded positive functionsuch that φpx, yq 0 for x 0 and φpx, yq 1 for x ¡ 0 and y ¡ 1

x(such


a function clearly exists). Then in the ber Ax with x ¡ 0, we have thatπxpφq is a positive element with 1 inside its spectrum, hence πxpφq ¥ 1. Byupper semicontinuity, we conclude also π0pφq ¥ 1. But clearly πxpφq 0for x 0. Hence x ÞÑ πxpφq is not lower semicontinuous in 0.

Hence, to avoid such pathological situations, we often impose that the abovenormfunction should be continuous.

Denition 2.2.17. Let X be a compact Hausdor space, and A a CbpXq-algebra. We say that the assignment x ÞÑ Ax forms a continuous eld ofC-algebras if and only if for each a P A, the map

x ÞÑ πxpaq

is continuous.

A general way to produce continuous CbpXq-algebras is the following.Proposition 2.2.18. Let X be a compact Hausdor space. Let A be aCbpXq-algebra, and suppose there exist states ϕx on Ax such that

a) For all x, the GNS-representation πϕx of Ax is injective.

b) For all a P A, the function x ÞÑ ϕxpaxq is continuous.Then x ÞÑ Ax is a continuous eld of C-algebras.

Proof. As the πϕx are injective, it follows that they are isometric. As theGNS-vectors ξϕx are cyclic for Ax (and as the maps AÑ Ax are surjective),we have for all a P A and x P X that

ax supt|xbxξϕx , πϕxpaxqcxξϕxy | b, c P A, bxξϕx ¤ 1, cxξϕx ¤ 1u supt|ϕxpbxaxcxq | b, c P A,ϕxpbxbxq ¤ 1, ϕxpcxcxq ¤ 1u.

Pick now x P X, a P A and 0 ε. Choose b, c P A with ϕxpbxbxq ¤1, ϕxpcxcxq ¤ 1 and

ax ¤ |ϕxpbxaxcxq| ε.

By continuity of x ÞÑ ϕxpfxq for all f P A, we can nd an open neighborhoodU of x such that for all y P U , ϕypbybyq ¤ 1 ε, ϕypcycyq ¤ 1 ε and

ax ¤ |ϕypbyaycyq| 2ε.


Hence

ax ¤ ayp1 εq 2ε

¤ ay εpa 2q.

It follows that x ÞÑ ax is lower semi-continuous, hence continuous byLemma 2.2.15.

Let us now show that the Toepliz algebras Tq can be collected into a eld ofC-algebras for 0 ¤ q ¤ 1, where we dene T1 CbpDq (with D the closedunit disc). In T1, we dene s1 as the identity function z ÞÑ z, and we deneT1 as the -algebra generated by s1.

Proposition 2.2.19. Let I r0, 1s. Let A be the unital -algebra generatedby elements Q, T with Q Q, QT TQ and

T T QTT 1Q.

Then A admits a universal C-envelope A with respect to the extra conditionsT ¤ 1 and 0 ¤ Q ¤ 1.

Moreover, the assigment p ÞÑ ppQq on polynomials admits an extension to animbedding CbpIq Ñ A, and the associated CbpIq-algebra A satises Aq Tq.

Proof. Since we impose norm restrictions on generators of A , it is clear thatthe universal C-envelope A as above exists. Moreover, from the univeralrelations of A , we conclude that there exists a unital -homomorphism

πq : AÑ Tq, Q ÞÑ q, T ÞÑ sq.

As 0 ¤ Q ¤ 1 in A, we also have a -homomorphism

CbpIq Ñ A, f ÞÑ fpQq.

Since πqpfq fpqq, we conclude that f ÞÑ fpQq is injective. Hence A is aCbpIq-algebra.Let now Iq Kerpπqq. Then clearly fpQq P Iq if fpqq 0. It follows thatJq Iq, with Jq A the ideal generated by functions in CbpXq vanishing atq. On the other hand, since Q q P Jq, it follows that AJq is generated by


a single element Tq T Jq satisfying T q Tq qTqT

q 1 q. Hence we havea map

Tq Ñ AJq, sq ÞÑ Tq.

As this is clearly an inverse to the projection map πq : Aq Ñ Tq, we concludethat Iq Jq.

We want to show that the eld q ÞÑ Tq obtained from the CbpIq-algebra A iscontinuous. We will need some preparations.

Denition 2.2.20. For 0 ¤ q 1, dene ϕq as the state

ϕq : Tq Ñ C, x ÞÑ p1 qq8

n0

qnxen, xeny.

For q 1, let ϕ1 be the state associated to the normalized Lebesgue measureon D.

Remark that ϕq is obviously well-dened since qn is absolutely summable. Itis then easily seen to be a state.

Lemma 2.2.21. Let 0 ¤ q ¤ 1 and x P Tq. Then

ϕqpsqxq qϕqpxsqq.

Proof. For q 1 this is obvious. For 0 ¤ q 1, we compute

ϕqpsqxq p1 qq8

n0

qnxen, sqxeny

p1 qq8

n0

qnxsqen, xeny

p1 qq8

n1

p1 qnq12qnxen1, xeny

p1 qq8

n0

p1 qn1q12qn1xen, xen1y

qp1 qq8

n0

qnxen, xsqeny

qϕqpxsqq.


Lemma 2.2.22. For 0 ¤ q 1 and m,n P N, we have

ϕqpsmq psq qnq δm,nqn 1 q

1 qn1.

We follow here the `convention' that 00 1.

Proof. First remark that from the dening relation sqsq qsqsq 1 q, it

follows by induction that

psq qnsq p1 qnqpsq qn1 qnsqpsq qn. (2.3)

Now from the denition of ϕq, it is clear that ϕqpsmq psq qnq 0 when m n.Write then an ϕqpsnq psq qnq. From Lemma 2.2.21 and Equation (2.3), wededuce that

an qϕqpsn1q psq qnsqq

qp1 qnqan1 qn1an

Hence an q 1qn

1qn1an1. Since a0 1, we deduce an qn 1q1qn1 .

Corollary 2.2.23. For a P A, the map q ÞÑ ϕqpaqq is continuous on r0, 1s.

Proof. By norm-density of A in A, and since ϕq πq ¤ 1, it suces toprove the statement for a P A . Now from the commutation relations in A ,it follows easily that each element in A is a linear combination of productsQkT lpT qm for k, l,m P N. It follows that it is sucient to show that thefunctions

q ÞÑ ϕqpslqpsq qmqare continuous for l,m P N.

Now from Lemma 2.2.22, we see that ϕqpslqpsq qmq δm,lqm 1q

1qm1 for 0 ¤q 1. Clearly, this is a continuous function on r0, 1q. It remains to checkcontinuity in 1. But

ϕ1psl1ps1qmq 1

π

»Dzlzmdz δm,n

1

π

»D|z|2mdz 1

m 1.

Since limqÑ1 qm 1q

1qm1 1m1

, the corollary follows.


Theorem 2.2.24. The eld of C-algebras q ÞÑ Tq induced from the CbpIq-algebra A is continuous.

Proof. By combining Corollary 2.2.23 with Proposition 2.2.18, it is sucientto see that the GNS-representations πϕq are injective on Tq for all 0 ¤ q ¤ 1.

Now for 0 q 1 and x P Tq, we have

xξϕq , πϕqpxxqξϕqy p1 qq¸nPN

qnxen2,

hence πϕqpxq 0 implies x 0.

A similar reasoning works for q 1.

Finally, for q 0, we have for all m,n ¥ 0, that

xsmq ξϕ0 , πϕ0pxqsnq ξϕ0y ϕ0pps0qmxsn0 q xsm0 e0, xs

n0e0y

xem, xeny.We conclude that also πϕ0 is injective.

2.2.4 More on representations of C-algebras

Let A be a unital C-algebra. We have seen that A admits an injective -representation AÑ BpHq on a Hilbert space H. Let us consider the generalrepresentation theory of a C-algebra in more detail.

Let us rst present some general terminology.

Denition 2.2.25. Let A be a unital C-algebra, H1 and H2 two Hilbertspaces, and π1 and π2 unital -representations on respectively H1 and H2.

An intertwiner between π1 and π2 is an operator x P BpH1,H2q such thatπ2paqx xπ1paq for all a P A. We denote the set of all intertwiners by

Morpπ1, π2q tx P BpH1,H2q | x intertwiner between π1 and π2u.

We say π1 is a subrepresentation of π2 if there exists an isometric intertwinerbetween π1 and π2.


We say π1 and π2 are unitarily equivalent if there exists a unitary intertwinerbetween π1 and π2.

Note that if x is an intertwiner from π1 to π2, then x is an intertwiner from

π2 to π1. It is also clear that if x P Morpπ1, π2q and y P Morpπ2, π3q, thenyx P Morpπ1, π3q.Let us say that a closed subspace H1 H is invariant under a representationπ of a C-algebra A if πpAqH1 H1.

Denition 2.2.26. Let A be a unital C-algebra. A unital -representationπ of A on a Hilbert space H is called irreducible if H is non-zero and theonly invariant Hilbert spaces in H are t0u and H itself.

Remarks 2.2.27. 1. As π is a -representation, one easily sees that H1 isinvariant if and only if its orthogonal complement is invariant. It followsthat a -representation is irreducible if and only if it is atomic, i.e. cannot be written as a direct sum of two (non-trivial) -representations.

2. If π is irreducible, any non-zero vector ξ P H is necessarily cyclic.

3. As we shall see later, every irreducible representation is even alge-braically irreducible: if π is irreducible and t0u V H is an invariantlinear subspace (not necessarily closed), then either V t0u or V H.

Lemma 2.2.28 (Schur's lemma for C-algebra representations). Let A be aunital C-algebra.

1. A unital -representation π of A on a Hilbert space H is irreducible ifand only if Morpπ, πq CidH.

2. If π1 and π2 are irreducible unital -representations of A on respec-tive Hilbert spaces H1 and H2, then either Morpπ1, π2q t0u or elseMorpπ1, π2q Cu with u a unitary intertwiner.

Proof. Assume π is irreducible, and let x be an intertwiner from π to itself.We want to show that x is a scalar multiple of the identity operator.

Let us suppose rst that we could nd a normal intertwiner x which is nota scalar multiple of the identity operator. Then, as x is also an intertwinerfrom π to itself, it follows by continuity that each element in CpidH, xq is anintertwiner of π. Now as x is not a scalar, we can nd, by functional calculus,two real nonzero functions f, g P CbpSppxqq such that fg 0. As fpxq 0,


the Hilbert space closure of fpxqH is a non-zero Hilbert space H1. Similarly,the Hilbert space closure of gpxqH is a non-zero Hilbert space H2. But since

xfpxqξ, gpxqηy xξ, fpxqgpxqηy xξ, pfgqpxqηy 0,

H1 is orthogonal to H2.

It follows that t0u H1 H. But since fpxq P CpidH, xq is an intertwiner,we have πpaqfpxqξ fpxqπpaqξ for all a P A and ξ P H. We deduce bycontinuity that H1 is invariant. This is in contradiction with the assumptionthat π is irreducible.

Hence any normal intertwiner is scalar. Let us take now an arbitrary inter-twiner x. Then x, xx and xx are also intertwiners. From the rst partof the proof, it follows that xx and xx are scalars. As xx and xx arepositive operators, these scalars are positive. As xx and xx have the samenorm, we see that in fact these scalars are equal. It follows that x is normal(and in fact a positive multiple of a unitary). Then by the rst part of theproof we conclude that x must necessarily be a scalar multiple of the identity.

This proves the rst part of the proposition. To prove the second part,suppose x is a non-zero intertwiner between π1 and π2. By the rst part ofthe lemma, it follows that xx xx λ for some λ ¡ 0. Hence u λ12xis a unitary intertwiner between π1 and π2.

If then y is another intertwiner from π1 to π2, the rst part of the lemmagives that uy c for some c P C. Hence y cu.

Assume now that ϕ is a state on a unital C-algebra A. When is the asso-ciated GNS-representation πϕ irreducible? It turns out that this property isdetermined by the geometrical position of ϕ inside the state space.

Denition 2.2.29. Let A be a unital C-algebra, and SpAq its state space.We call ϕ P SpAq a pure state if ϕ is an extreme point of SpAq.Indeed, SpAq is clearly a compact convex set inside the linear space of contin-uous functionals on A (with the weak- topology). Then, by denition, ϕ isan extreme point of SpAq if and only if it can not be written as tϕ1p1tqϕ2for two states ϕ1, ϕ2 with 0 t 1 and ϕ1 ϕ2.


Theorem 2.2.30. Let ϕ be a state on a unital C-algebra A. Then theGNS-representation πϕ is irreducible if and only if ϕ is extremal.

Proof. Assume that πϕ is not irreducible. Then H H1 `H2 with H1 andH2 non-trivial orthogonal invariant subspaces. Let ξϕ ξ1ϕ ξ2ϕ be theorthogonal decomposition of ξϕ with respect to this direct sum. Then as ξϕis cyclic, it follows that neither ξ1ϕ or ξ2ϕ are zero. Writing t ξ1ϕ2, we haveξ2ϕ2 1 t since ξϕ is a unit vector. In particular, 0 t 1. Put

ϕ1paq 1

txξ1ϕ, πϕpaqξ1ϕy, ϕ2paq 1

1 txξ2ϕ, πϕpaqξ2ϕy.

Then ϕ1 and ϕ2 are states, and ϕ tϕ1 p1 tqϕ2. If we can show thatϕ1 ϕ2, then we can conclude that ϕ is not pure. But if ϕ1 ϕ2, thennecessarily ϕ1 ϕ. This would mean that for all a P A,

xξϕ, πϕpaqξϕy 1

txξ1ϕ, πϕpaqξ1ϕy

1

txξ1ϕ, πϕpaqξϕy.

By cyclicity of ξϕ, we deduce that ξϕ 1tξ1ϕ, which gives a contradiction.

Conversely, suppose that πϕ is irreducible. Suppose that ϕ were not pure,say ϕ tϕ1 p1 tqϕ2 for 0 t 1 and states ϕ1, ϕ2 with ϕ1 ϕ2. Then

ϕ1paaq 1

ttϕ1paaq ¤ 1

tptϕ1paaq p1 tqϕ2paaqq ¤ 1

tϕpaaq.

It follows that we can dene a surjective bounded operator x P BpHϕ,Hϕ1qsuch that

x paξϕq aξϕ1 , @a P A.As ξϕ is cyclic, it follows from this formula that x is an intertwiner. But asxx is an intertwiner for πϕ, the irreducibility of πϕ gives that xx is a scalarλ ¡ 0. Hence for all a P A,

ϕ1paq xξϕ1 , aξϕ1y xxξϕ, x paξϕqy xξϕ, xx paξϕqy λxξϕ, aξϕy λϕpaq.


As ϕ and ϕ1 are both states, this forces λ 1 and hence ϕ ϕ1. Bysymmetry, also ϕ ϕ2, contradicting ϕ1 ϕ2.

Let us show now that any C-algebra has enough irreducible representations,giving a strengthening of the non-commutative Gelfand-Neumark theorem.

Theorem 2.2.31 (Gelfand-Raikov theorem9). Let A be a unital C-algebraand a P A. Then there exists an irreducible unital -representation π of Awith πpaq 0.

Proof. Let χ be the character fpaaq ÞÑ fpaaq on B Cp1A, aaq. Asχ is a one-dimensional representation, it is clearly irreducible. Hence it is apure state on B. Now by Corollary 1.2.33, the restriction map SpAq Ñ SpBqis surjective. Hence the inverse image of χ forms a non-empty extremal facein the compact convex set SpAq. By the Krein-Milman theorem, there existsan extremal point ϕ of SpAq in the inverse image of χ. Then ϕ is a purestate which by construction satises ϕpaaq aa. It follows that the GNS-representation of ϕ is an irreducible representation. Moreover, πϕpaqξϕ2 ϕpaaq 0. Hence πϕpaq 0.

For commutative C-algebra, irreducible representations are easily classied.

Proposition 2.2.32. Let A be a unital commutative C-algebra. Then aunital -representation π is irreducible if and only if π is one-dimensional.

In other words, all irreducible representations are characters.

Proof. Let π be an irreducible representation of A. Then for each a P A,the operator πpaq is itself an intertwiner of π. By Schur's lemma, πpaq isa scalar. As this holds for all a P A, it follows by irreducibility that π isone-dimensional.

On the other hand, each one-dimensional representation is necessarily irre-ducible.

9The actual Gelfand-Raikov theorem refers to the representation theory of locally com-pact Hausdor groups, but as we will see later, it is a direct corollary of its C-algebraversion.


We also have the following corollary concerning elds of C-algebras, sayingin a sense that elements in the global C-algebra are determined by theirvalues in the bers.

Corollary 2.2.33. Let X be a compact Hausdor space, and A a CbpXq-algebra. Let a P A, and assume πxpaq 0 for all x P X. Then a 0.

Proof. Let a P A be non-zero. By the Gelfand-Raikov theorem, we can ndan irreducible representation π of A with πpaq 0. As π is irreducible,it follows that the self-intertwiner space of π is CidHπ . Hence π|ZpAq givesa character on ZpAq. In particular, there exists x P X such that πpfq fpxqidHπ for all f P CbpXq. We conclude that π factors through Ax. Henceπpaq ¤ πxpaq, and πxpaq 0.

From Proposition 2.2.32, it follows that for A a commutative unital C-algebra, the set SppAq can be identied with the set of all irreducible repre-sentations up to unitary equivalence. There is no obstruction to extend thisdenition to the non-commutative case.

Denition 2.2.34. Let A be a unital C-algebra. We dene the set-theoreticspectrum of A as

SppAq IrrpAq ,the set of all irreducible representations of A modulo unitary equivalence.

Unlike for commutative C-algebras however, the spectrum SppAq can beill-behaved and retain very little of A itself. Nevertheless, SppAq can still beequipped with a compact (but not necessarily Hausdor!) topology. Thistopology is the one inherited from a cruder kind of spectrum.

Denition 2.2.35. Let A be a unital C-algebra. A 2-sided ideal J of Ais called primitive if J Kerpπq for some (non-zero) irreducible unital -representation π. The primitive spectrum PrimpAq of A is the set of allprimitive ideals of A.

Remark 2.2.36. The primitive spectrum can be very small. Indeed, thereexist many simple C-algebras, whose only 2-sided ideals are t0u and thewhole C-algebra. For example, the Calkin algebra Q is simple.


Proposition 2.2.37. Let A be a unital C-algebra. For I a closed 2-sidedideal of A, dene

VI tJ P PrimpAq | I Ju.Then

V tVI | I a 2-sided closed idealuis the set of all closed subsets of a topology on PrimpAq.The above topology on PrimpAq is referred to as the Jacobson topology.

Proof. Cleary Vt0u PrimpAq and VA H. For Iα a collection of closed2-sided ideals, write I for the 2-sided ideal generated by all Iα. Then clearly

VI XαVIα .

It remains to show that V is closed under nite unions.

Let I, J be two closed 2-sided ideals. Then clearly VI Y VJ VIXJ .

Conversely, take P P VIXJ . Let π be an irreducible representation withKerpπq P . Then the closure of πpJqHπ is clearly an invariant subspace ofHπ, since J is a two-sided ideal. By irreducibility, one has

either πpJqHπ 0 and hence πpJq 0, or

πpJqHπ is dense in Hπ.

In the rst case, we obtain J P , hence P P VJ . In the second case,we deduce from I X J P that πpxqπpyq 0 for all x P I, y P J , henceπpIqπpJqHπ t0u. By density of πpJqHπ in Hπ, this entails πpIq 0, henceI P and P P VI . In both cases, we obtain P P VI Y VJ .

Hence VI Y VJ VIXJ .

Denition 2.2.38. The topological spectrum of A is the set SppAq equippedwith the topology inherited from the Jacobson topology from PrimpAq underthe map

Ker : SppAq Ñ PrimpAq, rπs ÞÑ Kerpπq.

In other words, a subset of SppAq is open if and only if it is the inverse underKer of an open in PrimpAq.


Remark 2.2.39. In general, the map SppAq Ñ PrimpAq is not injective.Indeed, one can show that for any innite-dimensional separable simple C-algebra the set SppAq is not a singleton.

Proposition 2.2.40. The topological spaces SppAq and PrimpAq are com-pact.

Proof. As Ker : SppAq Ñ PrimpAq is surjective, it is open by denition ofthe topology on SppAq. It hence suces to prove that PrimpAq is compact.

We use the following criterion for compactness: if Fα is any collection ofclosed subsets for which the intersection of any nite subfamily is neverempty, then the intersection of all Fα is not empty.

For our situation, choose closed 2-sided ideals Iα and suppose that XαVIα H. Then the closure of the C-algebra generated by all Iα contains A, andin particular contains the unit of A. Now the C-algebra generated by theIα is just the closure of the linear span of all Iα. Hence we can nd nitelymany αi and xi P Iαi such that 1 °i xi 1

2. In particular, we deduce

that°i xi is invertible. But this means that the 2-sided ideal generated by

the Iαi is already A, hence XiVIαi H.

When A is a commutative unital C-algebra, the space SppAq already carriedthe topology of weak-convergence. Let us show that it coincides with thetopology inherited from PrimpAq. Note that for A commutative, one of coursehas that the map SppAq Ñ PrimpAq is bijective.Proposition 2.2.41. Let A be a commutative unital C-algebra. Then a netof characters χα on A converges to a character χ for the weak-topology ifand only if Kerpχαq converges to Kerpχq for the Jacobson topology.

Proof. Suppose rst that χα Ñ χ for the weak-topology. Then by denitionwe have Kerpχαq Ñ Kerpχq if and only if for any 2-sided closed ideal I ofA with I Kerpχq, one has eventually I Kerpχαq. But suppose that thiswere not the case. Then we can nd a subnet χβ such that I Kerpχβqfor all β. But then for each x P I, we have χpxq limβ χβpxq 0. HenceI Kerpχq, a contradiction.

Hence the map Ker : SppAq Ñ PrimpAq is a continuous bijection. To show


that it is a homeomorphism, it thus suces to prove that PrimpAq is Haus-dor.

Now by the Gelfand-Neumark theorem, we may identify A CbpSppAqq(where SppAq has the weak-topology). Take χ, χ1 P SppAq, and let U andU 1 be open neighborhoods of respectively χ and χ1 with U X U 1 H. LetI be the closed 2-sided ideal of functions which vanish on U c, and similarlyI 1 the closed 2-sided ideal of functions which vanish on U 1c. Then VI PrimpAq consists of all Kerpχ2q such that fpχ2q 0 for all f P I. Hence,by the Urysohn lemma, VcI is an open neighborhood of Kerpχq, and similarlyVcI 1 is an open neighborhood of Kerpχ1q. Clearly this description also givesVcI X VcI 1 H, as U X U 1 H. Hence PrimpAq is Hausdor.

As an example, let us examine the topological spectrum of the Toeplitz C-algebra.

Lemma 2.2.42. Up to unitary equivalence, any irreducible -representationof the Toeplitz C-algebra T is either the standard representation of T onl2pNq, or else a character

χz : T Ñ C, S ÞÑ z

for some z P S1. These representations are all mutually inequivalent.

Proof. The fact that the above representations are mutually inequivalent isclear. It is also clear that all characters χz are irreducible.

Let us show that the standard representation T Bpl2pNqq is irreducible.In fact if ξ, η in l2pNq, the rank one operator

Tξ,η : l2pNq Ñ l2pNq, ζ ÞÑ xξ, ζyηlies in B0pl2pNqq T . It follows that any non-zero vector in l2pNq is cyclicfor T , hence the representation of T on l2pNq is irreducible.It remains to show that the representations in the statement of the lemmaexhaust all irreducible representations up to unitary equivalence.

But let π be an irreducible representation of T . From the Wold decomposi-tion appearing in Theorem 2.2.8, we may assume H l2pX Nq ` G withthen πpSq SX ` u with u unitary. From the irreducibility of π, we con-clude that either G t0u or X H. In the rst case, irreducibility implies


that X must be a singleton (otherwise we have a direct sum decompositionl2pX Nq l2pX1 Nq ` l2pX2 Nq as representations for any decompo-sition X X1 \ X2), in which case this is just the ordinary representationon l2pNq. In the second case, π quotients to an irreducible representation ofCbpS1q, which is then necessarily a character. We deduce that in this caseπ χz for some z P S1.

Corollary 2.2.43. The map

Ker : SppT q Ñ PrimpT q

is bijective.

Proof. If π is the standard representation, then Kerpπq t0u. Hence Kerpπq Kerpχzq for any character z. Since clearly Kerpχzq Kerpχwq for z w, thecorollary follows.

From the above, we can write SppT q S1 Y t u, with representing thestandard representation on l2pNq. It remains to determine the topology ofSppT q.Proposition 2.2.44. A subset X SppT q is closed if and only if X SppT qor X S1 is closed (for the ordinary topology on S1).

Proof. Under the identication SppT q PrimpT q, the point correspondsto the zero ideal t0u. But clearly t0u P VI for a closed 2-sided ideal I if andonly if I t0u. It follows that SppAq is the only closed set containing .It now suces to show that χzα Ñ χz in SppT q if and only if zα Ñ z P S1.This follows along similar lines as in Proposition 2.2.41.

One may hence visualize the spectrum of T as being the closed unit disc withall points in the interior identied into a `fat open point' whose closure is thewhole unit disc.

2.3 von Neumann algebras 101

2.3 von Neumann algebras

Motivated by the commutative Gelfand-Neumark theorem, one could call C-algebra theory `non-commutative topology'. In this section, we will see thatoperator algebras can also be used to develop a `non-commutative measuretheory'. To avoid certain technical complications, we will assume in thissection that all Hilbert spaces are separable (and hence have an at mostcountable orthonormal basis), although all statements are valid in the non-separable case.

Denition 2.3.1. Let H be a Hilbert space, and let S BpHq. The com-mutant S 1 of S is dened as the set

S 1 tx P BpHq | @y P S, xy yxu.

Note that S 1 is automatically a closed unital subalgebra of BpHq. If S isinvariant under , then also S 1 is invariant under , and hence a C-algebra.

Denition 2.3.2. Let M be a unital C-algebra with a concrete realizationM BpHq on a Hilbert space H.10 Then M is called a von Neumannalgebra if the bicommutant M2 (that is, the commutant of the commutant ofM) equals M :

M2 M.

A C-algebra M is called a W-algebra if M is -isomorphic with a vonNeumann algebra.

Just as the basic abstract theory of C-algebras was developed almost single-handedly by Gelfand and Neumark, the theory of von Neumann algebras wasinitiated and carried to quite an advanced state by John von Neumann andhis occasional collaborator Francis Murray, in a series of articles called `OnRings of Operators'. It is currently still a very active research area, with quitea distinct avor from ordinary C-algebra theory. In these notes, we will haveto restrict ourselves to the bare minimum of von Neumann algebra theory,and will treat them mainly as a tool in understanding the representationtheory of C-algebras.

10We will always assume that the unit of M is the identity operator on H.


The rst important theorem concerning von Neumann algebras is von Neu-mann's double commutant theorem, which gives a topological characteriza-tion of von Neumann algebras. To formulate this theorem, we will need tointroduce some concepts rst.

Denition 2.3.3. Let H be a Hilbert space. There are the following seventopologies on BpHq.

1. The norm topology.

2. The weak operator topology (WOT), which is the locally convex topol-ogy induced from the seminorms

pξ,η : BpHq Ñ R, x ÞÑ |xξ, xηy|, ξ, η P H.

3. The strong operator topology (SOT), which is the locally convex topol-ogy induced from the seminorms

pξ : BpHq Ñ R, x ÞÑ xξ, ξ P H.

4. The strong operator topology (SOT), which is the locally convextopology induced from the seminorms

pξ : BpHq Ñ R, x ÞÑ xξ xξ, ξ P H.

5. The σ-weak operator topology (σWOT), which is the locally convextopology induced from the seminorms

ptξn,ηnu : BpHq Ñ R, x ÞÑ¸n

|xξn, xηny|,¸n

ξn2,¸n

ηn2 8.

6. The σ-strong operator topology (σSOT), which is the locally convextopology induced from the seminorms

ptξnu : BpHq Ñ R, x ÞÑ¸

n

xξn2

12

,¸n

ξn2 8.


7. The σ-strong operator topology (σSOT), which is the locally convextopology induced from the seminorms

ptξnu : BpHq Ñ R, x ÞÑ¸

n

xξn2

12

¸

n

xξn2

12

where°n ξn2 8.

In other words, we have convergence of a net xα Ñ x

in the WOT if and only if xξ, xαηy Ñ xξ, xηy for all ξ, η, in the SOT if and only if xαξ Ñ xξ for all ξ if and only if pxxαqpxxαqconverges to zero in the WOT.11,

in the SOT if and only if xαξ Ñ xξ and xαξ Ñ xξ for all ξ if andonly if px xαqpx xαq and px xαqpx xαq converge to zero in theWOT.

On the other hand, for n P N Y t8u and H a Hilbert space, let us writeHpnq `n

k1H for the direct sum of n copies of H with itself, which weview as (possibly innite) column vectors. Then for any x P BpHq, we canconsider the diagonal operator xpnq n

k1x,

xpnq

ξ1

ξ2...

xξ1

xξ2...

.

We easily see that the xα converge in the σWOT to x if and only if thediagonal operators x

p8qα converge in the WOT to xp8q, and similarly for the

σSOT and σSOT.

Since |xξ, xηy| ¤ ξxη, we also see immediately that the SOT is indeedstronger than the WOT (in that it is more dicult for a net to converge), andin fact the WOT is weaker than any of the other topologies. Similarly, theσWOT is stronger than the WOT, but bears no direct relation to the SOT.Of course, the norm topology is stronger than any of the other topologies,while the σSOT is stronger than any of the other topologies except for thenorm topology.

11Easily seen by the polarisation identity xξ, xηy °3k0 i

kxξ ikη, xpξ ikηqy.


Example 2.3.4. The adjoint operation x ÞÑ x is continuous map for theWOT, since

xξ, xαηy Ñ xξ, xηy ô xξ, xαηy Ñ xξ, xηyô xxαη, ξy Ñ xxη, ξyô xη, xαξy Ñ xη, xξy

for all ξ, η P H.

However, this map is not continuous for the SOT if H is innite dimensional.Indeed, let S be the (isometric) shift operator on l2pNq. Then for any nitelinear combination ξ of basis vectors en, we have that pSqkξ 0 for k bigenough. As pSqk is uniformly bounded, we deduce that limkÑ8 pSqkξ 0 for all ξ P l2pNq, i.e. pSqk converges in the SOT to the zero operator.However, the sequence of adjoints Sk does not converge in the SOT to thezero operator, since each Sk is isometric.

This is the reason why the SOT is introduced, in which the adjoint is forcedto be continuous.

Similarly, one can use this same example to show that the multiplication mapis not jointly continuous for the WOT, i.e. if xα and yα are nets convergingto respectively x and y in the WOT, then the net xαyα does not necessarilyconverge to xy. On the other hand, for y xed, one does have that if xαconverges to x in the WOT, then xαy converges to xy in the WOT, andsimilarly for multiplication on the other side or for the other topologies.

Remark 2.3.5. For the SOT, one obtains an open neighborhood systemaround zero of the form UpFq, where F is a nite set of vectors ξ1, . . . ξn and

UpFq tx P BpHq |n

k1

xξk2 1u.

It is important here to take F an arbitrary nite set, not just a singleton!

For the σSOT, one allows innite sets F with the condition that the sum ofthe squares of their norms is nite.

As a last remark concerning elementary properties, let us show in the follow-ing lemma that the σWOT equals the WOT on bounded sets (and similarlyfor the SOT and SOT-topologies).


Lemma 2.3.6. Restricted to the unit ball of BpHq, the WOT and σWOTcoincide.

Proof. Let xα be a net with xα ¤ 1 for all α, and suppose xα Ñ x in theWOT. Assume ξn, ηn are sequences of vectors with

° ξn2,° ηn2 8.

Then¸n

|xξn, px xαqηny|

¤n0

n0

xξn, px xαqηny| x xα8

nn01

ξnηn

¤n0

n0

xξn, px xαqηny| 2

8

nn01

ξn2

12 8

nn01

ηn2

12

,

which is smaller than a given ε for n0 and α large enough. Hence also xα Ñ xin the σWOT.

The rst important non-trivial property to note is that, as far as functionalsare concerned, weak, strong or strong behave the same. We will prove thisfor the σ-topologies.12

Denition 2.3.7. Let H be a Hilbert space. We denote by BpHq BpHqthe subspace of all functionals which are continuous for the σWOT.

Proposition 2.3.8. Let H be a Hilbert space. Then ω P BpHq if and onlyif ω is continuous with respect to the σSOT.

To prove this proposition, we need some preparations.

Lemma 2.3.9. The unit ball of the compact operators B0pHq is dense in theunit ball of BpHq for the σSOT.

Proof. By Lemma 2.3.6, we can replace the σSOT with the SOT. Choosex P BpHq with x ¤ 1, and let e1, e2, . . . be an orthonormal basis of H. Letpn be the orthogonal projection onto the linear span of

te1, e2, . . . , en, xe1, . . . , xen, xe1, . . . , x

enu.12A similar statement can be proven for the non-σ-topologies. On the other hand, the

set of functionals continuous for the WOT is strictly smaller than the set of functionalscontinuous for the σWOT when H is innite dimensional.


Then xn pnxpn is a compact (and in fact nite rank) operator. For ξ P H,write ξpnq xen, ξy, so that ξ °n ξpnqen. Then

pxn xqξ 8

k1

ξpkqpxn xqek

8

kn1

ξpkqpxn xqek

¤ 2x

8

kn1

|ξpkq|212

,

which is smaller than 1 for n large enough. Hence xn ÑSOT

x. Similarly

xn ÑSOT

x, hence xn ÑSOT

x.

Proposition 2.3.10. Let ω P B0pHq. Then there exist sequences ξn, ηn P Hwith ξm K ξn and ηm K ηn for m n,

°8n1pξn2 ηn2q 8, and

ωpxq 8

n1

xξn, xηny, @x P B0pHq.

Proof. For convenience, let us suppose H to be a separable innite dimen-sional Hilbert space, which we may identify with l2pNq. Write Eη,ξ for therank one operator

Eη,ξ : H Ñ H, ζ ÞÑ xξ, ζyη.Let ten | n P Nu be an orthonormal basis of H, and write Emn Eem,en forthe associated matrix units. If ξ P H, we write ξ °

nPN ξpnqen. Finally,write em,n for the natural orthonormal basis of l2pN Nq.If now ξ P l2pN Nq, the sum °

m,n ξpm,nqEmn is normconvergent since forpartial nite sums we havexη,

¸m,n

ξpm,nqEmnζy

¸m,n

ξpm,nqζpnqηpmq

¤¸m,n

|ξpm,nq|212¸

m,n

|ζpnq|2|ηpmq|212

¤¸m,n

|ξpm,nq|212

ηζ.


In particular, we see that we have a contractive embedding

T : l2pN Nq Ñ B0pHq, ξ ÞÑ Tξ ¸m,n

ξpm,nqEmn.

It follows that ω T gives a continuous functional on l2pNNq. By the Rieszrepresentation theorem, we can nd ξ P l2pN Nq such that

ωpTηq ¸m,n

ξpn,mqηpm,nq, @η P l2pN Nq.

In particular, since T°

m,n ηpmqζpnqem,n Eη,ζ , we have

ωpEη,ζq ¸m,n

ξpn,mqηpmqζpnq.

Now since Tξ is compact, we can nd, by Proposition 2.1.23 and the remarkabove it, orthonormal bases tηku and tζku of l2pNq, as well as a sequencea P c0pN0q (the Banach space of sequences converging to zero13), such that

Tξηk akζk.

From the denition of Tξ, we deduce that, for each m,¸n

ξpm,nqηkpnq akζkpmq.

Let now b be a function of nite support on N0, say bk 0 for k ¥ N . Then

ω

¸k

bkEηk,ζk

¸k

bk¸m,n

ξpn,mqηkpmqζkpnq

¸k,n

bkak|ζkpnq|2

¸k

bkak.

But since p°k bkEηk,ζkq p°k bkEηk,ζkq

°k |bk|2Eζk,ζk , we deduce

¸k

bkak

¤ ωb,

13We denote by N0 the set of strictly positive integers.


where b denotes the supremum norm. By choosing bk with bk|ak| ak fork ¤ N , we nd that

N

k0

|ak| ¤ ω.

We deduce that in fact a P l1pNq.Put then

ω1pxq ¸k

akxηk, xζky, x P B0pHq.

It is easily seen that ω1 is a well-dened continuous functional. Since a similarcomputation as above shows that ωpEηk,ζlq δk,l ak ω1pEηk,ζlq, and sincethe linear span of the Eηk,ζl is dense in B0pHq, it follows that ω ω1.

Finally, choose ck with c2k ak, and write ζ 1k ckζk and η1k ckηk. Then°

k ζ 1k2 8,°k η1k2 8 and

ωpxq ω1pxq ¸k

xη1k, xζ 1ky, @x P B0pHq.

This proves the proposition.

We can now prove Proposition 2.3.8.

Proof (of Proposition 2.3.8). Let ω be a functional on BpHq which is contin-uous for the σSOT. Then certainly ω is normcontinuous, and hence as wellits restriction to B0pHq. Then by Proposition 2.3.10, we can nd sequencestξku and tηku of vectors in H with

°k ξk2 8,

°k ηk2 8 and

ωpxq ¸k

xξk, xηky, @x P B0pHq.

Clearly the expression on the right denes a functional on BpHq which iscontinuous for the σWOT, hence a fortiori for the σSOT. As B0pHq isdense in BpHq for the σSOT by Lemma 2.3.9, Proposition 2.3.8 follows.

Corollary 2.3.11. We have an isometric linear isomorphism

BpHq Ñ B0pHq

by the restriction map.


In particular, it follows from this corollary that BpHq is a Banach subspaceof BpHq.

Proof. By Lemma 2.3.9 and Proposition 2.3.10, it follows that the restrictionmap is a linear isomorphism.

If now ω P BpHq, Proposition 2.3.8 tells us that we can nd orthogonalbases tξnu and tηnu of H, as well as a sequence panq P l1pN0q, such that

ωpxq 8

n1

anxξn, xηny, @x P BpHq.

It follows that ω ¤ °8n1 |an|. On the other hand, choose bn of modulus 1

with an bn|an|, and consider for k ¥ 1 the compact operators

Tkζ k

n1

bnxηn, ζyξn.

Then clearly Tk ¤ 1, and

ωpTkq k

n1

bnan k

n1

|an|.

It follows that the norm of ω is°n |an|, both as a functional on BpHq and

B0pHq.

In fact, the following proposition shows that we have the full analogue of theclassical dualities c0pNq l1pNq l8pNq.Lemma 2.3.12. The natural map BpHq Ñ pBpHqq is an isometric iso-morphism of Banach spaces.

Proof. As BpHq contains all functionals of the form x ÞÑ xξ, xηy, it is clearthat the above map is injective and isometric. To show that this map issurjective, choose X P pBpHqq. For η P H, denote ωη P H for the linearfunctional ξ ÞÑ xη, ξy. Denote also ωξ,η for the functional x ÞÑ xξ, xηy. Thenusing the natural isometric isomorphism H H, we obtain a linear map

x : H Ñ H H : ξ ÞÑ pωη ÞÑ Xpωη,ξqq .


As

|Xpωη,ξq| ¤ Xωη,ξ ¤ Xηξ,we nd that x P BpHq. By construction, we have Xpωη,ξq ωη,ξpxq for allξ, η. As any element in BpHq is a normlimit of linear combinations of suchωη,ξ by Corollary 2.3.11 and Proposition 2.3.10, we nd that Xpωq ωpxqfor all ω P BpHq, establishing surjectivity.

It follows that the σWOT topology on BpHq is nothing but the σ-weaktopology of BpHq as the dual space of BpHq.Another corollary of Proposition 2.3.8 is the following.

Corollary 2.3.13. Let S be a convex set. Then S is closed in the σWOT ifand only if S is closed in the σSOT.

Proof. This follows immediately from the fact that for any locally convexspace, a closed convex set coincides precisely with the intersection of all thehalfspaces (determined by continuous functionals) containing it.

We are now ready to present von Neumann's topological characterization ofvon Neumann algebras. The following easy lemma will be needed.

Lemma 2.3.14. Let S BpHq. Then the commutant S 1 is closed in theWOT.

Proof. Assume that xα P BpHq is a net converging in the WOT to x P BpHq.Let a P BpHq. Then axα Ñ

WOTax and xαa Ñ

WOTxa. Hence if a P S and all

xα P S 1, it follows that xa limα xαa limα axα ax and x P S 1.

Theorem 2.3.15 (von Neumann's bicommutant theorem). Let A BpHqbe a unital C-algebra. Then the closure of A in either the WOT or theσSOT equals A2.

Proof. By Lemma 2.3.14, we have that A2 is closed in the WOT. Since onetautologically has A A2, we have to prove that A2 is the closure of A inthe σSOT. In fact, by Lemma 2.3.14, it is sucient to prove that A2 is theclosure of A in the σSOT.


Fix x P A2. By denition of the σSOT, it suces to nd for each at mostcountable set F of ξk P H with square summable norms an a P A with°k px aqξk2 1.

Assume rst that F tξu. Let p be the orthogonal projection onto theclosure of Aξ. We claim that p P A1. Indeed, since aAξ Aξ for all a P A,and since Aξ is dense in pH by construction, we have that ap pap for alla P A. Applying the -operation, we see pa pap for all a P A. Henceap pa for all a, and p P A1. As obviously pξ ξ, and as x P A2, it followsthat xξ xpξ pxξ P pH. Hence there exists a P A with px aqξ 1.

Assume now more generally that F tξku with #F n P N Y t8u. Wewill reduce this case to the previous one by an important trick known asamplication. Namely, consider the direct sum Hilbert space Hpnq `n

k1H.Then for each y P BpHq, we have the diagonal operator ypnq `n

k1y onHpnq, where we recall that

ypnq

η1

η2...

yη1

yη2...

.

Write Apnq for the collection of all apnq with a P A. Then we have a -isomorphism AÑ Apnq sending a to apnq.

We claim that pA2qpnq pApnqq2. Indeed, any operator y on BpHpnqq can bedecomposed as a matrix of operators on H,

y

y11 y12 . . .y21 y22 . . ....

.... . .

.

We then compute directly that y commutes with some zpnq for z P BpHq ifand only if each yij commutes with z. It follows from this that y P pApnqq1 ifand only if yij P A1 for all i, j. In turn, this implies that pA2qpnq pApnqq2.Conversely, take z P pApnqq2. Then in particular z commutes with the matrixunits Eij, where

E11

1 0 . . .

0 0 . . ....

.... . .

, E12

0 1 . . .

0 0 . . ....

.... . .

, . . .


This implies immediately that zij 0 if i j, and zii zjj for all i, j.Hence z wpnq for some w P BpHq. But since in particular wpnq commuteswith all upnq for u P A1, we conclude that w P A2. This shows that indeedpApnqq2 pA2qpnq.We can now nish the proof. Indeed, by applying the rst part of the proof

to Apnq, we can nd for ξ

ξ1

ξ2...

an a P A such that pxpnq apnqqξ 1.

This is precisely the statement that°k pxaqξk2 1 for all 1 ¤ k ¤ n.

Corollary 2.3.16. Let H be a Hilbert space, and let S BpHq be a setwhich is invariant under the adjoint operation . Then the commutant S 1 isa von Neumann algebra.

Proof. Let A be the unital C-algebra generated by S. We obtain immedi-ately that S 1 A1, which is in particular a C-algebra. The corollary thenfollows from the combination of Lemma 2.3.14 and Theorem 2.3.15.

Corollary 2.3.17. Let H be a Hilbert space and let M BpHq be a unitalC-algebra. Then M is a von Neumann algebra if and only if M is closed inthe σWOT.

Proof. As M M2, we deduce from Lemma 2.3.14 that M is closed in theσWOT. Conversely, if M is closed in the σWOT, we deduce from Theorem2.3.15 that M is a von Neumann algebra.

Another easy consequence of the bicommutant theorem is that any nite-dimensional C-algebra is a direct sum of nite-dimensional matrix alge-bras.14 We leave the proof of this corollary as an exercise.

A technical result which is quite useful is Kaplansky's density theorem. Itsays that when approximating in the σSOT topology, one may always workwith uniformly bounded nets.

Theorem 2.3.18 (Kaplansky's density theorem). Let H be a Hilbert spaceand A BpHq a unital C-subalgebra. Then the unit ball of A is dense in

14Of course, for this particular situation we can simplify the topological content in theproof of the bicommutant theorem.


the unit ball of A2 for the σSOT. Moreover, a self-adjoint element in thelatter can by approximated in the σSOT by a net of self-adjoint elements inthe former.

To prove Theorem 2.3.18, we will need the notion of strongly continuousfunction.

Denition 2.3.19. Let H be a Hilbert space, and denote BpHqsa for thespace of self-adjoint elements. A function f : R Ñ R is called stronglycontinuous if xα Ñ

SOTx in BpHqsa implies fpxαq Ñ

SOTfpxq.

Of course, a strongly continuous function is automatically continuous. Forthe proof of the following lemma, we refer to Lemma 1.7.2 of K. Davidson,C-algebras by Example, Fields Institute Monographs (1996).

Lemma 2.3.20. Let f : RÑ R be a continuous function such that

lim sup|t|Ñ8

|fptq||t| 8.

Then f is strongly continuous.

Proof (of Theorem 2.3.18). Let x P A2 with x ¤ 1. Suppose rst thatx x. Then as x is in the σSOT-closure of A, it follows that we can nda net xα of self-adjoint elements in A s.t. xα Ñ x in the SOT.

Let f be any real continuous function with f ¤ 1 and fptq t on r1, 1s.Then by Lemma 2.3.20, f is strongly continuous, and fpxαq Ñ fpxq x inthe SOT. Now it is easy to see that on the unit ball of BpHq, the SOT andσSOT topologies coincide. As the fpxαq are selfadjoint and as fpxαq ¤f ¤ 1, it follows that fpxαq is a net in the unit ball of A which convergesin the σSOT to x.

Let us now take x P A2 general with x ¤ 1. Consider Hp2q H ` H,which we represent as column vectors. Then we can identify BpHp2qq withM2pBpHqq, the C-algebra of 2 by 2 matrices over BpHq (with the ordinarymatrix product and pXqij pXjiq),

X X11 X12

X21 X22

B

HH

.


In particular, we can consider the C-algebraM2pAq of 2 by 2 matrices over A.As in the proof of Theorem 2.3.15, one easily checks that M2pAq2 M2pA2q.In particular, the element

X

0 xx 0

is a self-adjoint element in M2pAq2 with X ¤ 1. By the rst part of theproof, we can nd a net of self-adjoint elements Xα P M2pAq with Xα ¤ 1and Xα Ñ X in the σSOT. Now by self-adjointness, Xα is of the form

Xα yα xαxα zα

.

It then follows in particular that xα Ñ x and xα Ñ x in the σSOT, i.e.xα Ñ x in the σSOT.

Corollary 2.3.21. Let X be a compact Hausdor space, and let µ be aregular Borel probability measure on X. For f P L 8pX, dµq, consider

mf : L 2pX, dµq Ñ L 2pX, dµq, g ÞÑ fg.

Then M tmf | f P L 8pX, dµqu is a von Neumann algebra, and the map

L 8pX, dµq Ñ BpL 2pX, dµqq, f ÞÑ mf

is a weak to σWOT continuous isomorphism from L 8pX, dµq to M .

Here the weak-topology on L 8pX, dµq is induced by the isomorphism

L 8pX, dµq Ñ L 1pX, dµq.

Proof. It is obvious that mf is a well-dened operator, and that m : f ÞÑ mf

is a unital -homomorphism. Moreover, since

xξ,mfηy »X

ξpxqηpxqfpxqdµpxq

for ξ, η square integrable functions, it is also clear that m is injective and acontinuous map into M from the weak-topology to the σWOT.

Let us show that M is a von Neumann algebra. Now L 8pX, dµq is inparticular a C-algebra for the uniform norm. As m is injective, we deduce


that mf f. Assume now that x is in M2. By Theorem 2.3.18, wecan nd a bounded net gα P L 8pX, dµq such that mgα Ñ x in the σWOT.As the unit ball of L 8pX, dµq is compact for the weak-topology, we canassume that there is a g P L 8pX, dµq such that gα Ñ g weak. But thenx mg.

It is now also clear that m is in fact a homeomorphism.

Conversely, one can show that for any commutative von Neumann algebraM BpHq on a separable Hilbert space, there exists a compact Haus-dor (metric) space X and a probability measure µ on X such that M L 8pX, dµq by a σWOT to weak-continuous isomorphism. This statementcan be shown to be equivalent with the following theorem, which is at rstsight weaker.

Theorem 2.3.22 (Borel functional calculus). Let H be a separable Hilbertspace, and x P BpHq a normal operator. Let W pxq txu2. Then W pxqis a von Neumann algebra, and there exists a Borel probability measure µ onSppxq and a unique isomorphism

L 8pSppxq, dµq Ñ W pxq, f ÞÑ fpxq

which is continuous from the weak-topology to the σWOT and such that theidentity map is sent to x.15

Hence our continuous calculus for C-algebras can be extended to a measur-able calculus when the target algebra is a von Neumann algebra.

To prove the theorem, we will make use of the notion of separating vector,which is in a sense dual to the notion of cyclic vector.

Denition 2.3.23. Let H be a Hilbert space, and A BpHq a unital C-algebra. A vector ξ P H is called separating if aξ 0 for some a P A impliesa 0.

Lemma 2.3.24. Let M BpHq be a von Neumann algebra. Then ξ P H iscyclic (resp. separating) for M if and only if ξ is separating (resp. cyclic) forthe commutant M 1.

15Be aware again that this theorem denes fpxq, in that there is no other `obvious' wayto interpret the element fpxq!


Proof. By the von Neumann bicommutant theorem, it is enough to showthat ξ is cyclic for M if and only if ξ is separating for M 1.

Assume rst that ξ is cyclic for M . Then if x P M 1 and xξ 0, it followsthat xyξ yxξ 0 for all y PM . As ξ is cyclic for M , it follows that x 0.

Conversely, suppose that ξ is separating forM 1. Let p be the projection ontothe closure of Mξ. Since the closure of Mξ is invariant under M , we havepmp mp for all m P M . Taking adjoints, we deduce that pm mp for allm PM , so p PM 1. Since p1pqξ 0 by construction, we deduce p 1 sinceξ separating for M 1. Hence ξ is cyclic for M .

Lemma 2.3.25. Let H be a separable Hilbert space, and M BpHq anabelian von Neumann algebra. Then M admits a separating vector.

Proof. By induction, one can write H `nHn where Hn is the closure ofM 1ξn for certain unit vectors ξn. Write ξ °

n 2nξn. We claim that ξ iscyclic for M 1. It then follows from Lemma 2.3.24 that ξ is separating for M .

To prove the claim, let pn be the projection onto Hn. Then as before, wehave that pn P M2 M . But since M is abelian, we have M M 1. Itfollows that M 1ξ M 1pnξ M 1ξn. Hence the closure of M 1ξ contains allHn, therefore ξ is cyclic for M

1.

Proof (of Theorem 2.3.22). We rst invoke Fuglede's theorem, whose proofis left as an exercise: if x is a normal operator, then y commutes with x ifand only if y commutes with x. It then follows that W pxq Cp1, xq2.Now since Cp1, xq is commutative by normality of x, we deduce from theWOT-density of Cp1, xq in W pxq that also W pxq is commutative. Thenby Lemma 2.3.25, there exists a separating unit vector ξ for M W pxq.Let K be the closure of Mξ, and let π be the restriction map

π : M Ñ BpKq, m ÞÑ pπpmq : ξ ÞÑ mξq .

We claim that π is a -isomorphism and a homeomorphism for the σ-weaktopology.

Clearly, π is a -isomorphism since K isM -invariant and since ξ is separating.It is also clear that π is continuous for the σWOT topology. It remains toshow that π1 is continuous for the σWOT. But since the σWOT topology


on BpKq coincides with the weak-topology on B0pKq, the unit ball of BpKqis compact. Hence since π is isometric, we deduce that π1 is continuous onthe unit ball for the σWOT topology. If then f P BpHq, an argument asin Proposition 2.3.8 shows that f π1 is σWOT-continuous. Hence π1 isσWOT continuous.

We may hence suppose in the following that in fact H K. Let ω be thestate ωpyq xξ, yξy on Cp1, xq. By the Riesz representation theorem, weknow that there exists a unique probability measure µ on Sppxq such thatωpfpxqq ³

Sppxqfpzqdµpzq for all f P CbpSppxqq. It is then easily seen that

we have a unitary transformation

U : L 2pSppxq, dµq Ñ H, f ÞÑ fpxqξ, f P CbpSppxqq.

Now since we have shown already that L 8pSppxq, dµq is the von Neumannalgebra obtained as the bicommutant of CbpSppxqq, represented on L 2pSppxqqby left multiplication operators, we deduce that UL 8pSppxq, dµqU W pxq.By Corollary 2.3.21, the identication L 8pSppxq, dµq Ñ W pxq is a homeo-morphism from the weak to the σWOT topology.

As a particular consequence, this means that we can form projections pEpxqassociated to Borel subsets E Sppxq. The assignment E ÞÑ pEpxq is knownas the spectral measure associated to x. As the pEpxq can be obtained asσWOT-limits of elements in Cp1, xq, it follows that x P M for some vonNeumann algebra M implies pEpxq PM for all Borel sets E.

We can now also substantiate a remark we made concerning irreducible rep-resentations of C-algebras.

Theorem 2.3.26. Let A be a unital C-algebra. Let π be an irreducibleunital -representation of A on a Hilbert space H. Then if V H is a linearsubspace (not necessarily closed) such that AV V , then either V t0u orV H.

Proof. Let x P BpHq. Let K H be a nite-dimensional subspace, andε ¡ 0. We claim that there exists a P A with πpaq ¤ xε and πpaqξ xξfor all ξ P K. This of course implies the theorem, since it means that anynon-zero vector can be sent to any other vector by an element of πpAq (takingx a suitable rank one operator).


The above claim (which is much stronger than what we actually need for thecase at hand) is known as Kadison's transitivity theorem. To prove it, noterst that πpAq2 BpHq, since πpAq1 Morpπ, πq CidH by Schur's lemma.Let p be the orthogonal projection onto K.

Let us rst construct a0 P A with a0 ¤ x 12ε and px πpa0qqp ε

8.

Let te1, e2, . . . , enu be an orthonormal basis of K. By Theorem 2.3.18, wecan nd x0 P πpAq with x0 ¤ x and

n

k1

px x0qek2

12

ε

8.

By Cauchy-Schwarz, this implies

px x0qξ ε

8ξ, @ξ P K,

hence px x0qp ε8. As AKerpπq Ñ BpHq is isometric, we may take

x0 πpa0q with a0 ¤ x ε2.

Assume now that we have constructed, by induction, an element an P A withan ¤ x p1 2n1qε and px πpanqqp 2n3ε. Then by applyingthe argument in the previous paragraph with x replaced by pxπpanqqp andε by 2n3ε, we can nd bn1 P A with bn1 ¤ 2n2ε and

px pπpanq πpbn1qqqp 2n4ε.

We then put an1 an bn1 to complete the induction step.

It is clear then that the normlimit a limn an exists. By construction, itsatises πpaqξ xξ for all ξ P K, and a ¤ x ε.

Remark 2.3.27. In fact, if π : A Ñ B is a unital surjective -homorphismbetween C-algebras, one can show that the unit ball of A is sent surjectivelyonto the unit ball of B. This allows one to simplify some of the estimates inthe above argument.

Corollary 2.3.28. Let A be a unital C-algebra, and let ϕ be a pure stateon A. Let Nϕ be the left kernel of ϕ. Then ANϕ is a Hilbert space for theinner product xaξϕ, bξϕy ϕpabq.


Proof. The only thing we have to show is that ANϕ is complete under theHilbert space norm. But by Theorem 2.2.30, πϕ is irreducible. As Aξϕ isclearly an A-invariant linear subspace, we deduce from Theorem 2.3.26 thatAξϕ Hϕ, proving the corollary.

For the next corollary, recall that a unital ring R over C is called semi-simpleif the simple modules of R separate points, i.e. for any element of R one cannd a simple (=irreducible) module such that the element does not act asthe zero endomorphism.

Corollary 2.3.29. Let A be a unital C-algebra. Then A is semi-simple.

Proof. By Theorem 2.2.31, the (topologically) irreducible representations ofA separate points of A. By Theorem 2.3.26, these are in fact simple modules.

We end this section with stating the following important theorems, for whoseproofs we refer to the literature. Recall that a W-algebra is a C-algebrawhich is -isomorphic to a von Neumann algebra.

Theorem 2.3.30 (Sakai). A C-algebra M is a W-algebra if and only ifthere exists a Banach space M such that pMq M by an isometric Ba-nach space isomorphism. Moreover, the Banach space M is then uniquelydetermined up to isometric isomorphism, and is called the space of normalfunctionals on M .

If M BpHq as a von Neumann algebra, then one can dene M as theBanach space

M tω PM | Dω1 P BpHq, @x PM, ω1pxq ωpxqu M.

Thus we see that the duality L 1pX, dµq L 8pX, dµq for measure spacespX, dµq has a general operator algebraic analogue. It also shows that theσWOT on a von Neumann algebra M is intrinsically associated to the ab-stract -algebra M .

The second theorem concerns the notion of traciality (and lack thereof) fora general normal state on a von Neumann algebra. A normal trace on a vonNeumann algebra M is a normal state τ such that

τpxyq τpyxq, x, y PM.


If M is commutative one has of course that any normal state is a trace, butthis is no longer true in the non-commutative setting. Surprisingly however,the lack of traciality is compensated in a dynamical way by the existence ofa one-parameter group of automorphisms of M , that is, a family of automor-phisms pσtqtPR of M which are σ-weakly continuous and such that

σstpxq σspσtpxqq, @s, t P R and x PM.

The precise theorem is the following.

Theorem 2.3.31. Let M be a von Neumann algebra, and let ω be a state onM . Then there exists a canonical one-parameter group σωt of automorphismson M , the modular automorphism group of ω, such that for any x, y P Mthere exists a continuous function

Fx,y : tz P C | 0 ¤ Impzq ¤ 1u Ñ C

which is continuous, analytic in its interior and such that for all t P R, onehas

Fx,yptq ωpσtpxqyq,Fx,ypt iq ωpyσtpxqq.

The content of this theorem can be elaborated upon, and is known underthe name `Tomita-Takesaki theory'. Note that putting formally t 0 inthe rst formula and t i in the second formula, one obtains the `identity'ωpxyq ωpyσipxqq, which can be made precise if x and y are nicely behaved.This then clearly shows how σ compensates for the lack of traciality of ω.

For a general unital C-algebra A, one says that a state ω satises the KMS-condition with respect to a one-parameter group of automorphisms σωt if andonly if for any x, y P A, one can nd a function Fx,y as above. Here KMSstands for the physicists Kubo, Martin and Schwinger. Indeed, the KMS-condition has a physical signicance, and appears in thermodynamics whereit is used to characterize equilibrium states for systems with innitely manyparticles.

Bibliography

[1] [Bro08] N. P. Brown and N. Ozawa, C-algebras and nite-dimensionalapproximations, Graduate Studies in Mathematics 88, American Math-ematical Society, Providence, RI (2008).

[2] [Dav96] K. Davidson, C-algebras by Example, Fields Institute Mono-graphs (1996).

[3] [Tak79] M. Takesaki, Theory of operator algebras I (2nd printing of the1979 ed.) Encyclopaedia of Mathematical Sciences. Operator Algebrasand Non-Commutative Geometry 124 (5) Berlin: Springer.

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