Invitation to C*-algebras-Exercises

Problems by Karen R. Strung

Solutions by

Micha l Banacki

14.10.2016

Contents

1 Banach algebras and spectral theory 2

2 Gelfand representation 10

3 C*-algebra basics 12

4 Positive elements 18

5 Positive linear functionals and representations of C*-algebras 24

6 Further examples of C*-algebras 28

7 A very short introduction to classification for simple nuclear C*-algebras 38

1

Introduction

The following lecture notes were prepared as a supplementary material connected to the course An

invitation to C*-algebras by Karen R. Strung which was organized in Warsaw (19.09-14.10.2016)

as a part of the Simons Semester Noncommutative geometry the next generation. The aim of

this notes is to present solutions of selected problems which were discussed by the author during

his contact hours with participants of Simons Semester at IMPAN Warsaw. All exercises cov-

ered here are contained in the notes by Karen R. Strung (available at http://bcc.impan.pl/

16Simons-III/index.php/programme).

The author of this notes wish to thank Karen R. Strung for a series of inspiring lectures and

clarifying remarks. He also would like to thank Paul F. Baum, Alan Carey, Piotr M. Hajac

and Tomasz Maszczyk for organizing the Simons Semester Noncommutative geometry the next

generation.

1 Banach algebras and spectral theory

Exercise 1.1: We are going to show that for any Banach space X, the space

LpXq “ tT : X Ñ X| T linear, boundedu

with a norm

T “ sup tTx |x P X, x ď 1u

and natural operations is a Banach algebra (the proof that the other algebras described in example

are in fact Banach algebras is analogous). Indeed, ¨ defines a norm since

Tx ě 0, Tx “ 0 ô Tx “ 0,

αTx ď |α| Tx ,

pT1 ` T2qx ď T1x ` T2x ď pT1 ` T2q x .

Note that by the above discussion and the following expression

T1T2x ď T1 T2x ď T1 T2 x

LpXq is a norm algebra. To show completeness with respect to the given norm consider a Cauchy

sequence pTnq Ă LpXq. For any ε ą 0 there exists n0 P N such that for any n,m ě n0 the following

inequality holds

Tn ´ Tm ă ε.

For any (chosen) x P X we have

Tnx´ Tmx ď Tn ´ Tm x ă ε x . (1.1)

2

Therefore, pTnxq Ă X is a Cauchy sequence and by completeness of X there is a linear map

T : X Ñ X such that limnÑ8 Tnx “ Tx. Taking mÑ8 in (1.1) we get

Tnx´ Tx ď ε x .

Since this is true for all x P X we obtain

Tn ´ T ď ε.

Observe that T “ pT ´ Tnq ` Tn and therefore T P LpXq which ends the proof.

Exercise 1.2: Consider the following sequence panq Ă Cpzq given by

an “nÿ

k“0

zk

k!

Observe that the function ez is a limit of panq with respect to the given norm. Therefore, in

particular, panq is a Cauchy sequence, but its limit does not belong to the considered algebra.

From that Cpzq can not be a Banach algebra with respect to the given norm.

Exercise 1.3: Continuity of multiplication in Banach algebra simply follows for triangle inequality

and submultiplicativity of a norm

a1b1 ´ a2b2 ď a1b1 ` a1b2 ´ a1b2 ´ a2b2 ď a1 b1 ´ b2 ` b2 a1 ´ a2 .

Exercise 1.5: Let a P InvpAq. Then a´1 P InvpAq with inverse given by a. Obviously 1 P

InvpAq and if a, b P InvpAq, then ab is invertible with inverse defined by b´1a´1, so InvpAq is

a multiplicative group (associativity of multiplication follows automatically from associativity of

multiplication in A).

Exercise 1.6: When A “MnpCq, then a´λ1 P A is not invertible if and only if it is not invertible

as a matrix. Therefore spectrum of a PMnpCq is precisely the set of eigenvalues of a.

If A “ CpXq, then f is invertible if and only if fpxq ‰ 0 for all x P X, so f´λ1 is not invertible

if and only if there exist x P X such that fpxq “ λ. Therefore we obtain

sppfq “ tfpxq| x P Xu .

3

Exercise 1.7:

a) Form the definition of bilateral shift operator S we know that for any pλnq P H “ `2pZq

Spλnq “ pµnq

where µn “ λn´1. If so, then

Spλnq “

¨

˝

8ÿ

|n|ě0

|µn|2

˛

‚

12

“

¨

˝

8ÿ

|n|ě0

|λn|2

˛

‚

12

“ pλnq

and S “ 1 (S is an isometry), so in particular S P BpHq.

b) By the definition of inner product (here we take the convention that inner product is linear in

the first variable) in H “ `2pZq we have (for any pλnq, pµnq P H)

xSpλnq, pµnqy “8ÿ

|n|ě0

λn´1µn “8ÿ

|n|ě0

λnµn`1 “ xpλnq, S˚pµnqy .

Therefore, S˚ is given by

S˚pλnq “ pµnq

where µn “ λn`1. From that this is obvious that S˚S “ 1 “ SS˚, so S˚ “ S´1 (S is a unitary

operator).

c) Suppose that there exists a nonzero vector pξnq P H “ `2pZq such that

Spξnq “ λpξnq

for some λ P C. Then for any n P Z we have

Spξnq “ λpξnq (1.2)

Since pξnq is nonzero, there is some i P Z such that ξi “ c ‰ 0. Let |λ| “ 1, then by (1.2)

8ÿ

|n|ě0

|ξn|2 “

8ÿ

|n|ě0

|c|2 “ |c|8ÿ

|n|ě0

1

which can not be finite (so pξnq would not be in H). Similarly, when |λ| ‰ 1 we have that

8ÿ

|n|ě0

|ξn|2 “

8ÿ

|n|ě0

|λ|n´i|c|2

is not convergent, so pξnq R H. Therefore, S has no eigenvalues.

4

d) Since S is unitary we may conclude that sppSq Ă T. Consider some λ P T. We will show that

S´λ1 is not invertible. Indeed, suppose that S´λ1 is invertible in BpHq. Let pφnq Ă H “ `2pZq

be a sequence of norm one elements. Then for any n P N we have

1 “ φn “›

›pS ´ λ1q´1pS ´ λ1qφn›

› ď›

›pS ´ λ1q´1›

› pS ´ λ1qφn . (1.3)

For each n P N define φn P H by

φpmqn “1?n

$

’

&

’

%

λm

for m P r1, ns

0 for m R r1, ns

Obviously φn “ 1. Observe that

ppS ´ λ1qφnqpmq “

1?n

$

’

’

’

’

&

’

’

’

’

%

´1 for m “ 1

λn

for m “ n` 1

0 for m R t1, n` 1u

,

hence for any (thus arbitrary large) n we get

pS ´ λ1qφn “

c

2

n.

But from the inequality (1.3) we obtain a contradiction (as›

›pS ´ λ1q´1›

› should be bounded by

assumption), hence S ´ λ1 can not be invertible in BpHq, so sppSq “ T.

Exercise 1.8: To calculate the spectral radius of Volterra operator T we will use Gelfand-Beurling

formula. We stat with the proof (by induction) that

Tnpfqptq “

ż t

0

fpxqpt´ xqn´1

pn´ 1q!dx. (1.4)

Indeed, if n “ 1 then above formula coincide with definition of T . Suppose that (1.4) holds for

some n. Observe that

Tn`1pfqptq “

ż t

0

Tnpfqpsq ds (1.5)

“

ż t

0

ż s

0

fpxqps´ xqn´1

pn´ 1q!dx ds

“

ż 1

0

ż 1

0

χr0,tspsqχr0,sspxqfpxqps´ xqn´1

pn´ 1q!dx ds,

where χra,bspxq denotes characteristic function of ra, bs, i.e.

χra,bspxq “

$

’

&

’

%

1 for x P ra, bs

0 for x R ra, bs

.

5

One can check that χr0,tspsqχr0,sspxq “ χrx,tspsqχr0,tspxq and (1.5) become

Tn`1pfqptq “

ż 1

0

ż 1

0

χr0,tspxqfpxqχrx,tspsqps´ xqn´1

pn´ 1q!dx ds “

ż t

x

ż t

0

fpxqps´ xqn´1

pn´ 1q!dx ds.

By Fubini theorem we obtain

Tn`1pfqptq “

ż t

0

ż t

x

fpxqps´ xqn´1

pn´ 1q!ds dx “

ż t

0

fpxqps´ xqn

n!dx

which ends the proof.

To estimate operator norm of T consider the following inequalities

Tnpfq2“

ż 1

0

ˇ

ˇ

ˇ

ˇ

ż s

0

fpxqps´ xqn´1

pn´ 1q!dx

ˇ

ˇ

ˇ

ˇ

2

ds

ď

ż 1

0

ˆż s

0

|fpxq|

ˇ

ˇ

ˇ

ˇ

ps´ xqn´1

pn´ 1q!

ˇ

ˇ

ˇ

ˇ

dx

˙2

ds

ď

ż 1

0

ˆż s

0

|fpxq|1

pn´ 1q!dx

˙2

ds

ď

ż 1

0

ˆż 1

0

|fpxq|1

pn´ 1q!dx

˙2

ds

“1

ppn´ 1q!q2

ż 1

0

ds

ˆż 1

0

|fpxq| dx

˙2

.

Because of the Holder inequality

ˆż 1

0

|fpxq| dx

˙2

ď

ˆż 1

0

|fpxq|2 dx

˙ˆż 1

0

dx

˙

we get

Tnpfq ď1

pn´ 1q!f ,

which leads to Tn ď 1pn´1q! . By Gelfand-Beurling formula and Stirling inequality n! ą

?2πnnne´n

we arrive at

rpT q “ limnÑ8

Tn1n ď lim

nÑ8

ˆ

1

pn´ 1q!

˙1n

“ 0.

From that sppT q “ t0u. Volterra operator T is an example of so called quasinilpotent operator -

rpT q “ 0 despite the fact that for any n P N we have Tn ‰ 0.

Exercise 1.9: Let X be a compact space and A denotes a Banach algebra with norm ¨A.

Clearly, CpX,Aq “ tf : X Ñ A| f continuousu with natural pointwise operations is an algebra.

Define a norm on CpX,Aq by

f “ supxPX

fpxqA .

Since ¨A is a submultiplicative norm, it follows that CpX,Aq equipped with ¨A is a normed

algebra. What remain is to show that this normed algebra is in fact complete. Consider a Cauchy

sequence pfnq Ă CpX,Aq For any ε ą 0 there exists n0 P N such that for all n,m ě n0 and for

6

any x P X we have

fmpxq ´ fnpxqA ď fm ´ fn ă ε. (1.6)

so pfnpxqq Ă A (for any x P X) is a Cauchy sequence. From completeness of A there exists a limit

fpxq of this sequence (with respect to the norm in A). Taking mÑ8 in (1.6) we obtain

fpxq ´ fnpxqA ď ε.

which is valid for any x P X. Therefore f is a limit of a sequence pfnq and by uniform limit

theorem f is continuous (f P CpX,Aq) so the proof is completed. Note that in the case when

A “MnpCq we have CpX,MnpCqq –MnpCpXqq.

Exercise 1.10: Consider the algebra of polynomials in one variable Crzs. Starting form Crzs we

construct the fields of fractions Cpzq defined as a quotient of CrzsˆCrzsz t0u „ by the equivalence

relation given by

pa, bq „ pc, dq ô ad “ bc.

Addition an multiplication are define naturally by

rpa, bqs ` rpc, dqs “ rpad` bc, bdqs

and

rpa, bqsrpc, dqs “ rpac, bdqs.

The inverse of rpa, bqs is given by rpb, aqs (if a ‰ 0), the unit and the zero element are defined by

rp1, 1qs and rp0, 1qs respectively. In the considered example the field of fractions can be seen as a

field of rational functions. Since all nonzero elements of Cpzq are invertible, to show that spppq

can be empty consider for example

p “z3

z3 ` z ´ 2.

Indeed, for any λ P C

p´ λ1 “z3

z3 ` z ´ 2´ λ1 “

p1´ λqz3 ´ λz ` 2λ

z3 ` z ´ 2‰ 0,

so spppq “ H.

Exercise 1.11: Let |λ| ą a, then›

›λ´1a›

› ă 1 and 1´ λ´1a is invertible with inversion

p1´ λ´1aq´1 “

8ÿ

n“0

pλ´1aqn.

From that we have

pλ1´ aq´1 “ λ´1p1´ λ´1aq´1,

7

so λ R sppaq. Therefore, sppaq Ă Bp0, aq.

Observe that sppaq is in fact closed. Indeed, define the following function φ : CÑ A given by

φ : λ ÞÑ a´ λ

for some chosen a P A. It is obvious that φ is continuous and Czsppaq “ φ´1pInvpAqq. Since

InvpAq is open, sppaq must be closed.

Exercise 1.12: Consider the flowing function ψ : CÑ InvpAq defined by

ψ : λ ÞÑ pa´ λq´1.

Observe that ψ “ ϕ ˝ φ, where φ : CÑ A and ϕ : InvpAq Ñ InvpAq are given by

φ : λ ÞÑ a´ λ

and

ϕ : a ÞÑ a´1.

Since we know that φ, ϕ are differentiable (hence continuous), its composition must be a continuous

function. Because of that ψ is bounded on a disc of radius 2 a as it is compact (being closed and

bounded subset of C).

Exercise 1.13: It is obvious that A “ A‘ C with multiplication given by

pa, λqpb, µq “ pab` λb` µa, λµq

is an algebra. Simple calculations show that pa, λq “ a ` |λ| is indeed a norm:

• αpa, λq “ αa ` |αλ| “ |α| pa, λq for any α P C,

• pa, λq ě 0 and pa, λq “ 0 ô a “ 0^ |λ| “ 0 ô pa, λq “ p0, 0q,

• pa, λq ` pb, µq “ a` b ` |λ` µ| ď a ` |λ| ` b ` |µ| “ pa, λq ` pb, µq.

Considered norm is also submultiplicative

pa, λqpb, µq “ pab` λb` µa, λµq “ ab` λb` µa ` |λµ|

ď ab ` λb ` µa ` |λµ|

ď a b ` |λ| b ` |µ| a ` |λ||µ|

ď a pb ` |µ|q ` |λ|pb ` |µ|q

“ pa, λq pb, µq .

8

To see that A is complete with respect to that norm consider a Cauchy sequence ppan, λnqq Ă A.

For any ε ą 0 there exists n0 P N such that for all n,m ě n0

pam, λmq ´ pan, λnq ă ε.

This implies that

am ´ an ă ε

and

|λm ´ λn| ă ε,

so sequences panq Ă A and pλnq Ă C are Cauchy sequences (hence convergent sequences since

A and C are complete). Let a “ lim an and λ “ limλn. It is obvious that pa, λq is a limit of

ppan, λnqq - for each ε we can find such n0 P N that for all n ě n0

pa, λq ´ pan, λnq “ a´ an ` |λ´ λn| ă ε.

Exercise 1.14: Recall that for a nonunital Banach algebra A spectrum of an element a P A is

defined by (we identify a with pa, 0q)

sppaq “!

λ P C| λ1A ´ a R InvpAq)

.

We will show that 0 P sppaq. Indeed, pa, 0q is not invertible because

pa, 0qpb, µq “ pab` µa, 0q ‰ p0, 1q

for all b P A and µ P C.

Exercise 1.15:

a) Let B be a subalgebra of a unital algebra A such that the unit in A belongs to B. Consider

InvpBq Ă BXInvpAq. Since InvpAq is an open subset, then InvpBq is an open subset in BXInvpAq.

To see that InvpBq is also closed consider a sequence panq P InvpBq such that

limnÑ8

an “ a P B X InvpAq.

If so, then the sequence pa´1n q Ă InvpBq converges to a´1 P A and from that a´1 P InvpBq (since

a´1 P B).

b) If b P B is invertible in B, then b is also invertible in A. From that it is clear that

spApbq Ă spBpbq.

9

To see that

BspBpbq Ă BspApbq,

observe that there exists a sequence xn P CzspBpbq which converges to some x P C. Since

b´xn P InvpBq and b´x P CzspBpbq so by above considerations b´x R InvpAq, but x P CzspApbq

and x P BspApbq, thus we are done. Finally, suppose that CzspApbq is connected. Because CzspBpbq

is a clopen subset, then by the previous discussion we have CzspApbq “ CzspBpbq or equivalently

spApbq “ spBpbq.

2 Gelfand representation

Exercise 2.1: It easy to observe that ker φ is a two-sided ideal

φpabq “ φpaqφpbq “ 0,

φpbaq “ φpbqφpaq “ 0

for any a P A and any b P ker φ. By definition φ is continuous map from A to B and therefore,

for any a P A such that a is a limit of some sequence panq Ă ker φ we get

φpaq “ limnÑ8

φpanq “ 0,

so ker φ is also closed.

Exercise 2.2: Let I be a maximal ideal in a unital Banach algebra A. Observe that for any a P I

we have 1´ a ě 1, because if not then

8ÿ

n“0

p1´ aqn “ p1´ p1´ aqq´1 “ a´1

and a´1a “ 1 P I, so I “ A which contradicts that I is maximal (hence proper). Let us consider

any b P I. By triangle inequality we get

1 ď 1´ a ď 1´ b ` b´ a

for all a P I and since b´ a can be make arbitrary small (as b P I) we get that 1 ‰ b for all

b P I. Therefore I ‰ A and I Ă I so by maximality of I we have I “ I.

Exercise 2.3:

a) To show that every proper modular ideal is contained in a maximal modular one we shall use

Kuratowski-Zorn lemma. Let I Ă A be a proper modular ideal. Therefore there exists u P A such

that for all a P A, a ´ ua P I and a ´ au P I. It is clear that u R I (if not, then for any a P A

10

we have a P I which is contradiction with I being proper). Let us denote by S family of all ideals

Is such that u R Is and I Ă Is. Let S be partially ordered by inclusion. Clearly S is not empty

since I P S. Observe that all Is P S are modular and proper. Consider any totally ordered subset

S1 Ă S and IS1 defined as a union of all ideals Is contained in S1. Obviously u R IS1 and IS1 is an

ideal such that I Ă IS1 . Therefore, IS1 is an upper bound for S1. By the above discussion, S has

a maximal element (Kuratowski-Zorn lemma) IM . If there is any other ideal IM Ă I, then either

IM “ I or u P I (if it would not be the case then IM could not be a maximal element in S), so

IM must be maximal (modular) ideal in A.

b) Let I Ă A be a modular maximal ideal in a commutative Banach algebra A. We take convention

in which equivalence classes in AI will be denoted by a ` I where a P A is some representative

of a given equivalence class. By purely algebraic considerations one can show that AI is a unital

algebra if and only if I is modular. If so, then there is a unit in AI given by i ` I, i P A. Let

I 1 denotes an ideal in AI. Then q´1pI 1q is an ideal in A and I Ă q´1pI 1q (q stands for canonical

projection q : A Ñ AI). Therefore, either q´1pI 1q “ A or q´1pI 1q “ I. From this follows the

fact that there are only two ideals in AI: t0u and AI itself. For any a` I P AI we can define

Ia “ pa ` IqAI which is an ideal in AI. Either Ia “ t0u (which implies that a ` I “ I, so it

is a zero element) or Ia “ AI (in this there exists b ` I such that pa ` Iqpb ` Iq “ pi ` Iq). All

nonzero elements in AI are therefore invertible which ends the proof.

Exercise 2.4: As always we identify A with pA, 0q Ă A. It is obvious that A is a closed ideal

since

pa, 0qpb, µq “ pab` µa, 0q P A,

pb, µqpa, 0q “ pba` µa, 0q P A

and A is itself a Banach space. Observe that A is proper (p0, 1q R A). To show that A is maximal

consider an ideal I 1 generated by A and some pb, µq R A. Then p´b, 0q ` pb, µq “ p0, µq P I 1 and

p0, 1q “ p0, µ´1qp0, , µq P I 1 so I 1 “ A which ends the proof.

Exercise 2.5: It is known that for any unital Banach algebra A and any a P A

sppaq “ tτpaq|τ P ΩpAqu .

Le us consider some nonunital Banach algebra A. The space of characters of its unitization ΩpAq

is given by

ΩpAq “ tτ |τ P ΩpAqu Y tτ0u ,

11

where τ denotes unique extension of a character τ : AÑ C provided by

τppa, λqq “ τpaq ` λ

and τ0 is defined as τ0ppa, λqq “ λ. Form the definition of spectrum for elements of nonunital

Banach algebra and by previous remark on spectrum in the unital case we can write

sppaq “ spppa, 0qq “!

τppa, 0qq|τ P Ω´

A¯)

“ tτpaq|τ P ΩpAquYtτ0ppa, 0qqu “ tτpaq|τ P ΩpAquYt0u ,

so we get the thesis.

3 C*-algebra basics

Exercise 3.1: Suppose that λ R sppaq. Then pa ´ λ1qpa ´ λ1q´1 “ 1. Taking adjoint of this

formula we obtain (since 1 “ 1˚) ppa´ λ1q´1q˚pa˚ ´ λ1q “ 1 and therefore λ R sppa˚q. Similarly

λ R sppa˚q implies λ R sppaq. From that

sppa˚q “

λ P C| λ P sppaq(

.

Exercise 3.2: If p P A is a nonzero projection, then by C*-condition we obtain p “ 1, so

spppq Ă Bp0, 1q. What is more, since p is selfadjoint, spppq Ă r0, 1s. Let γ : Cpspppqq Ñ A be a

˚-isomorphism defying continuous functional calculus. Consider f P Cpspppqq such that fpxq “ x.

We have γpfq “ p “ pp˚ “ γpffq, so for any x P spppq : x “ |x|2. Therefore, spppq Ă t0, 1u.

In fact we have spp1q “ t0u and if 0 R spppq, then p “ 1. Indeed, in that case p is invertible so

p “ p2p´1 “ pp´1 “ 1. If 1 R spppq, then 1 ´ p is invertible and 0 R spp1 ´ pq, so 1 ´ p “ 1 and

p “ 0. To sum up, if p is nontrivial projection (not equal to 0 or 1), then spppq “ t0, 1u.

Exercise 3.3: Since A is a Banach algebra, for any a P A and for all t P R

fptq “ eta “8ÿ

n“0

ptaqn

n!

is well define by›

›

›

›

›

8ÿ

n“0

ptaqn

n!

›

›

›

›

›

ď

8ÿ

n“0

›

›

›

›

ptaqn

n!

›

›

›

›

ď

8ÿ

n“0

tan

n!.

Obviously f 1ptq “ afptq and fp0q “ 1. Suppose that there is another function g : RÑ A with the

same properties. Then one can define the following function h : RÑ A given by

hptq “ fptqgp´tq.

This function is differentiable as a product of differentiable function and by Leibniz rule of

12

differentiation we get that h1ptq “ afptqgp´tq ´ afptqgp´tq “ 0. Form that hptq “ 1, since

hp0q “ fp0qgp0q “ 1. This shows that gp´tq “ e´ta (by uniqueness of inverse element), so

gptq “ fptq and f is indeed the unique function form R to A such that f 1ptq “ afptq and fp0q “ 1.

Exercise 3.4: The algebra of compact operators KpHq can be define in several equivalent ways.

For the purpose of this exercise we will assume that KpHq “ F pHq Ă BpHq, where F pHq denotes

the algebra of finite rank operators and the closure is taken with respect to the operator norm

(uniform topology). We want tho show that KpHq is an example of a simple C*-algebra. We

require that all considered ideal are closed and two-sided.

We will start by recalling a well know fact that F pHq is linearly spanned by rank one projec-

tions. Our first aim is to show that for any nonzero closed ideal I Ă BpHq, F pHq Ă I. Indeed,

let a P I be nonzero, then there exists x P H such that apxq ‰ 0. For fixed a, x as before and for

any fixed y P H of norm one define b P BpHq by (here we take the convention that inner product

is linear in the second variable)

bpzq “xapxq, zy

apxqy.

Clearly bapxq “ y and from that

py “ bapxa˚b˚,

where py denotes rank one projection on the subspace spanned by y, i.e. pypxq “ xy, xy y for any

x P H. Since a P I, py P I for any norm one element y P H, so F pHq Ă I.

Consider any C*-algebra A (e.g. BpHq). Let I Ă A be an ideal, then any ideal J Ă I (ideal

with respect to I) is automatically an ideal in the whole algebra A. Recall that any ideal in a

C*-algebra is a C*-algebra itself (the only nontrivial part of this statement is selfadjointness of an

ideal I - to see that it is true, one should consider C*-algebra I X I˚ and its approximate unit).

Since any element a P J is a linear combination of some positive elements in J we can restrict our

considerations to positive elements. Let a P J`, then a12 P J` Ă I (since the smallest C*-algebra

which contains a must be contained in J). For any b P A we have

ba “ ba12 a

12 P J,

because ba12 P I and J is an ideal in I.

Obverse that if a P F pHq and b P BpHq then ab, ba P F pHq. From this it is clear that KpHq

is an ideal in BpHq. Consider any nonzero ideal I Ă KpHq. Then by the previous discussion I is

an ideal in BpHq and F pHq Ă I. Since I is closed, we may conclude that I “ KpHq. Therefore,

KpHq is simple C*-algebra.

Exercise 3.5: Let CpXq denotes an algebra of continuous functions on compact Hausdorff space

13

X. Define

IF “ tf P CpXq| f |F “ 0u .

For any g P CpXq and any f P IF we have gf P TF . If pfnq Ă IF such that

limnÑ8

f ´ fn “ 0

then in particular

limnÑ8

fnpxq “ fpxq

for all x P F and IF is therefore a closed ideal. Consider any other closed ideal I Ă CpXq and

define

E “ tx P X| @fPI fpxq “ 0u .

Since E is closed as an intersection of closed sets (E “Ş

fPI f´1p0q), I Ă IE . We shall show that

I “ IE . Consider any f P IE . For any ε ą 0 define

Eε “ tx P X| |fpxq| ă εu

and

Dε “ CzEε.

Obviously Dε is closed, hence compact (as X is compact). For any x P Dε there exists a function

h P I such that hpyq ‰ 0 for all y P Ux, where Uxis appropriately small neighborhood of a given

x (this holds because of the definition of E and the fact that Dε X E “ H). Taking the union

of all Ux for x P Dε, we obtain the open cover of Dε. Since Dε is compact we can chose a finite

subcover of that cover. In other words there exists a finite set of points x1, x2, . . . xn P Dε such

that Uxi form an open cover of Dε. As it was stated above, for any i we can find hi P I such that

hipyq ‰ 0 for all y P Uxi .

h “nÿ

i“1

|hi|2.

Note that h P I. For any n ě 1 construct

gnpxq “nhpxq

1` nhpxq.

It easy to see that gn is well defined positive function such that |gnpxq| ă 1 fora any x P X.

Moreover, observe that the sequence pgnq Ă I uniformly converge to the unit 1 P Cpxq. Define

pfnq “ fgn P I. For all x P Eε and all n ě 1 we have (since |fpxq| ă ε)

|fpxq ´ fnpxq| “ |fpxq| ` |fpxq||gnpxq| ă 2ε

and by the fact that pgnq uniformly converge to 1, there exists n0 P N such that for any n ě 1 and

any x P Dε

|fpxq ´ fnpxq| ď |fpxq||1´ gnpxq| ď f |1´ gnpxq| ă 2ε

14

The above expressions are true for any given ε, therefore one can construct a sequence (net)

pknq Ă I such that

limnÑ8

f ´ kn “ 0.

If so, then f P I, since I is closed, and we may conclude that I “ IE .

Exercise 3.6: Firstly, observe that in fact the proof given in the previous exercise (exercise

3.5) can be apply also if we consider the algebra C0pXq of continuous functions (on some locally

compact Hausdorff space X) vanishing at infinity. By Gelfand-Naimark theorem all commutative

C*-algebras are of that form so classification of simple commutative C*-algebras turns out to be

equivalent to classification of closed ideals in C0pXq. Because every closed ideal in C0pXq is of

the form

IF “ tf P CpXq| f |F “ 0u

and the only such ideals in a simple C*-algebra A are A itself and t0u, it follows that if A is

commutative, simple and nonzero C*-algebra then A – C – C0ptxuq.

Exercise 3.7: Let A be a concrete C*-algebra, i.e. A Ă BpHq is a closed ˚-subalgebra of BpHq.

Consider an algebra MnpAq consisting of n ˆ n matrices with entries belonging to A. One can

think of MnpAq as a ˚-subalgebra in BpKq, where K is a direct sum of Hilbert spaces

K “

nà

i“1

Hi

with all Hi “ H and a norm defined by

x2K “

nÿ

i“1

xi2,

where x “ px1, x2, . . . , xnq P K. If so then MnpAq can be naturally equipped with a norm induced

from BpKq

B2Mn

“ supxď1

nÿ

i“1

›

›

›

›

›

ÿ

j

bijxj

›

›

›

›

›

2

,

where B P MnpAq, pBqij “ bij and x “ px1, x2, . . . , xnq P K. It is clear that MnpAq equipped

with such a norm is a norm algebra and that ¨Mnfulfill C*-condition (all this fact follows form

the property of operator norm of BpKq). The remaining part of our construction is to show that

MnpAq is complete with respect to the discussed norm. As usually, consider a Cauchy sequence

pBnq ĂMnpAq. Then for any ε ą 0 there exists n0 P N such that for any n,m ě n0

Bm ´Bn2Mn

“ supxK“1

nÿ

i“1

›

›

›

›

›

ÿ

j

pbpmqij ´ b

pnqij qxj

›

›

›

›

›

2

ă ε2,

15

Therefore for any i and any x P K such that xK “ 1 we get

›

›

›

›

›

ÿ

j

pbpmqij ´ b

pnqij qxj

›

›

›

›

›

2

ă ε2.

Taking x “ p0, 0, . . . xk, . . . 0q with any xk P H such that xk “ 1 we get

›

›

›pbpmqik ´ b

pnqik qxk

›

›

›

2

ă ε2.

From that we obtain

›

›

›pbpmqik ´ b

pnqik q

›

›

›

2

“ supxk“1

›

›

›pbpmqik ´ b

pnqik qxk

›

›

›

2

ď ε2

and by the fact that A Ă BpHq is complete, for any i, k there exists bik P A such that pbpnqik q Ă A

converge to it (with respect to the norm in A). If we define B P MnpAq such that pBqij “ bij ,

then by previous considerations B is a limit of pBnq (with respect to ¨Mn). Therefore, MnpAq is

a C*-algebra.

Exercise 3.8: By the C*-condition

a2“ a˚a “

›

›a2›

› .

From that we obtain

a2n“

›

›

›a2n

›

›

›.

Gelfand-Beurlig formula give us

rpaq “ limnÑ8

an1n “ lim

nÑ8

›

›

›a2n

›

›

›

12n

“ limnÑ8

a “ a .

Note that by more or less the same reasoning one can prove that rpaq “ a is true also for normal

elements.

Now, assume that for a given ˚-algebra A there are two norms ¨1 , ¨2 such that A is a

C*-algebra with respect to both of them. Observe that for any a P A

a1 “ pa˚a1q

12 “ rpa˚aq “ pa˚a2q

12 “ a2 ,

so the C*-norm is unique.

Exercise 3.9: Let a, b P A be normal and let u P A be a unitary element such that b “ uau˚.

Suppose λ R sppaq, then λ R sppbq because of

upa´ λ1qu˚ “ b´ λ1.

16

Conversely, by the same reasoning, λ R sppbq implies λ R sppaq. Therefore, sppaq “ sppbq and by

continuous functional calculus

C˚pa, 1q – Cpsppaqq “ Cpsppbqq – C˚pb, 1q

we obtain the desired isomorphism.

Exercise 3.10: Let a P Asa with sppaq “ r0, εs Y r1 ´ ε, 1s for ε ă 12 . If f P Cpsppaqq such that

fpxq “ x then γpfq “ a (by continuous functional calculus). Consider g P Cpsppaqq given by

gpxq “

$

’

&

’

%

0 for x P r0, εs

1 for x R r1´ ε, 1s

.

Because g “ g “ g2, γpgq “ p is a projection in A. Since γ is a ˚-isomorphism we get

p´ a “ γpg ´ fq “ supxPsppaq

|gpxq ´ fpxq| ď ε.

Exercise 3.11: By theorem 2.32 we know that for any unital C*-algebra B and nay normal

element b P B there is an isometric (hence injective) ˚-homomorphism

γ : Cpsppbqq Ñ B

such that

γpfq “ b

for fpxq “ x and γpCpsppbqqq “ C˚pa, 1q. Consider a nonunital C*-algebra A and its unitization

A. We will show that if a P A is a normal element, then there exists an isometric ˚-homomorphism

γp0q : Cp0qpsppaqq Ñ A

such that

γpfq “ a

for fpxq “ x and γp0qpCp0qpsppaqqq “ C˚ppa, 0q, p0, 1qq “ C˚paq (where Cp0qpsppaqq denotes the

space of continuous functions on sppaq vanishing at 0). Indeed, from previous discussion of the

unital case we get a ˚- isomorphism

γ : Cpsppaqq Ñ C˚ppa, 0q, p0, 1qq Ă A

(where clearly sppaq “ spppa, 0qq). Observe that γ´1pC˚ppa, 0qqq “ Cp0qpsppaqq and restricting γ

to Cp0qpsppaqq we get the desired isomorphism γp0q.

17

Exercise 3.12: Consider a continuous inclusion i : p0, 1q ãÑ R and induced map i˚ such that

i˚pfq “ f ˝ i for any f P C0pRq. Take g P C0pRq such that g|p0,1q “ 1. Then i˚pgqpxq “ 1 for all

x P p0, 1q, but if so then i˚pgq R C0pp0, 1qq. Therefore, i˚ does not define the ˚-homomorphism

form C0pRq to C0pp0, 1qq.

Exercise 3.13: Consider C˚pa, 1q with a P A being normal element. Observe that

sppfpaqq “ tτpfpaqq| τ P ΩpC˚pa, 1qqu

and

fpsppaqq “ tfpτpaqq| τ P ΩpC˚pa, 1qqu .

It easy to see that for polynomial function p we have ppτpaqq “ τpppaqq. For any f P C0psppaqq

and polynomials pn we get

|τpfpaqq´fpτpaqq| “ |τpfpaqq`pnpτpaqq´τppnpaqq´fpτpaqq| ď τ fpaq ´ pnpaq`|pnpτpaqq´fpτpaqq|

and form that (taking ppnq such that limnÑ8 pn “ f), fpτpaqq “ τpfpaqq, so

sppfpaqq “ tτpfpaqq| τ P ΩpC˚pa, 1qqu “ tfpτpaqq| τ P ΩpC˚pa, 1qqu “ fpsppaqq.

To solve the second part of the exercise, observe that C˚pfpaq, 1q Ă C˚pa, 1q. Therefore any

τ P ΩpC˚pa, 1qq has a unique restriction to τ P ΩpC˚pfpaq, 1qq. For all τ P ΩpC˚pa, 1qq we get

τpg ˝ fpaqq “ gpfpτpaqqq “ gpτ fpaqq “ τpgpfpaqqq,

so g ˝ fpaq “ gpfpaqq.

4 Positive elements

Exercise 4.1:

a) Obviously vv˚ and v˚v are selfadjoint. To see that they are projections observe that

pvv˚q2 “ vv˚vv˚ “ vpv˚vv˚q “ vv˚

and

pv˚vq2 “ v˚vv˚v “ v˚pvv˚vq “ v˚v,

where we use v “ vv˚v and v˚ “ v˚vv˚.

b) By the discussion bellow (exercise 4.2 c) and the fact that trace is invariant under cyclic

permutations, condition trppq ď trpqq is equivalent to r “ rk p ď rk q “ k. Without the loss of

18

generality we can chose such a basis that q is a diagonal matrix with respect to it (with only first k

entries different than 0:q “ Ik). We may find unitary element u PMnpCq such that p “ uIru˚. If

so, then Iru˚uI˚r “ IrI

˚r “ Ir. Clearly Ik ´ Ir “ Ik´r is a projection, thus in particular a positive

element.

c) We will show that the presented definition give a rise to the equivalence relation.Let p, p, r P A

denote projections. Obviously p “ p˚p “ pp˚ “ p, so relation is reflexive. When p “ v˚v and

q “ vv˚ then clearly p “ uu˚ and q “ u˚u with u “ v˚ so relation is symmetric. Consider

p “ v˚v, q “ vv˚ and q “ u˚u, r “ uu˚. Define w “ uv, then

r “ uu˚ “ uu˚uu˚ “ uqu˚ “ uvv˚u˚ “ ww˚

and

p “ v˚v “ v˚vv˚v “ v˚qv “ v˚u˚uv “ w˚w,

so relation is transitive, hence it is equivalence.

Let A “ MnpCq. Since for any v P MnpCq, v˚v and vv˚ has the same rank, it follows that

equivalent projections p and q also have the equal rank. Now suppose that p, q are projections

with the same rank rk p “ rk q “ r. Recall that any rank r projection can be diagonalized by

some unitary element (as a selfadjoint element), i.e. there exist unitaries u, v PMnpCq such that

p “ uIru˚

and

q “ vIrv˚

where by Ir we denote nˆn matrix with all entries equal to zero except for first r diagonal entries.

Observe that

I˚r Ir “ Ir “ IrI˚r

and from that

p “ uI˚r Iru˚.

This is equivalent to

Iru˚uIr “ I˚r Ir “ Irv

˚vIr

and that is equivalent to

vIrIrv˚ “ vIrv

˚ “ q.

Therefore, projections in MnpCq are Murray-von Nenumann equivalent if and only if they have

the same rank (or equivalently if they agree on normalized trace).

19

Exercise 4.2:

a) We will show that „ define an equivalence relation. Let a P A, where A is separable C*-algebra.

In that case there exists a countable approximate unit punq Ă A. Observe that putting vn “ un

we get

vnav˚n ´ a “ unau

˚n ´ aun ` aun ´ a “ un una´ a ` aun ´ a ,

therefore

limnÑ8

vnav˚n ´ a “ 0

and the Cuntz relation „ is reflexive. By the definition of „ (a „ b if and only if and ) there

is obvious that „ must be symmetric. To conclude we shall see that the Cuntz relation is also

transitive. It is enough to show that a À b and b À c implies a À c. Supose that there exist

sequences punq, pvnq Ă A such that

limnÑ8

vnbv˚n ´ a “ 0

and

limnÑ8

uncu˚n ´ b “ 0

One can write

vnuncu˚nv˚n ´ a ď vnuncu

˚nv˚n ´ vnbv

˚n ` vnbv

˚n ´ a ď ` vn

2uncu

˚n ´ b ` vnbv

˚n ´ a .

From that we have

limnÑ8

pvnunqcpvnunq˚ ´ a “ 0

and the proof is done.

b) Let p, q P A denote Murray-von Neumann equivalent projections, i.e. p “ v˚v and q “ vv˚ for

some v P A. Observe that

vpvvn ˚ ´q “ vv˚vv˚ ´ vv˚ “ vv˚ ´ vv˚ “ 0,

so taking vn “ v for any n P N we obtain the thesis.

d) To show that a À an observe that by the continuous functional calculus we have γpfq “ a and

γpfnq “ an, where f P C0psppaqq such that fpxq “ x. From that, the problem is reduced to the

previous one because

emfnem ´ f “ γpemf

nem ´ fq “ γpemqγpfnqγpemq ´ γpfq “ γpemqa

nγpemq ´ a .

Therefore, there exist a sequence pvmq “ pγpemqq Ă A such that

limmÑ8

vmanv˚m ´ a “ 0.

20

In particular aa˚ À paa˚q2 “ aa˚aa˚ and there exists a sequence pvnq Ă A such that

limnÑ8

vnaa˚aa˚v˚n ´ aa

˚ “ limnÑ8

pvnaqa˚apvnaq

˚ ´ aa˚ “ 0,

so aa˚ À a˚a. Similarly, starting form a˚a À pa˚aq2 “ a˚aa˚a, we get a˚a À aa˚, therefore aa˚

and a˚a are Cuntz equivalent.

One can also observe that an À a, where a P A`. Indeed, define the sequence pvmq Ă A by

vm “ uman´12 ,

where pumq P A denotes an approximate unit. Then we have

limmÑ8

vmav˚m ´ a

n “ limmÑ8

umanu˚m ´ a

n “ 0.

From that a „ an for any positive element.

Exercise 4.3:

a) Let φ : A Ñ B be a ˚-homomorphism. Any C P MnpAq` is of the form C “ D˚D for some

D P MnpAq (i.e. cik “ř

j d˚jidjk, where pCqik “ cik and pDqik “ dik). Consider φN : MnpAq Ñ

MnpBq. If C PMnpAq`, then the matrix entries of φnpCq are of the form

pφnpCqqik “ÿ

j

φpd˚jiqφpdjkq “ÿ

j

φpdjiq˚φpdjkq “

ÿ

j

e˚jiejk “ pE˚Eqik

for some E PMnpBq, which shows that φ is completely positive map.

b) Consider A “ B “M2pCq and a transpose map τ : M2pCq ÑM2pCq. For any positive element

C PM2pCq we have

pτpCqqik “ cki “ÿ

j

djkdji “ÿ

j

djidjk “ pDTDqik “ pE

˚Eqik,

so τ is positive. Note that if A PMnpCqsa is positive if and only if all eigenvalues of A are nonneg-

ative. To see that τ is not completely positive consider the following element F P M2pM2pCqq –

M4pCq

F “

¨

˚

˚

˚

˚

˚

˚

˝

1 0 0 1

0 0 0 0

0 0 0 0

1 0 0 1

˛

‹

‹

‹

‹

‹

‹

‚

.

21

Define also G “ τ2pF q

G “

¨

˚

˚

˚

˚

˚

˚

˝

1 0 0 0

0 0 1 0

0 1 0 0

0 0 0 1

˛

‹

‹

‹

‹

‹

‹

‚

.

Observe that the characteristic polynomial of F is given by

fpλq “ ´λ3pλ´ 2q.

Therefore, F has the following eigenvalues: λ1 “ 2, λ2 “ λ3 “ λ4 “ 0, but detpGq “ ´1, so G has

at least one negative eigenvalue. If so, then τ is not completely positive (it is even not 2-positive).

c) Consider the linear map ψ : AÑ B given by ψpaq “ v˚φpaqv for some ˚-homomorphism φ and

some v P B. By argument similar to the given in point a), for C PMnpAq` we get

pψnpCqqik “ v˚φpcikqv “ v˚

˜

ÿ

j

φpd˚jiqφpdjkq

¸

v “ÿ

j

pφpdjiqvq˚pφpdjkqvq “

ÿ

j

e˚jiejk “ pE˚Eqik,

which ends the proof of complete positivity of ψ.

Exercise 4.4: Le X be a compact metric space. Fix a finite subset F Ă CpXq and ε ą 0. There

exists an open cover U “ tUα|α P Au (A denotes just some indexing set) of X such that all f P F

are ε-constant on each Uα, i.e. for any Uα P U , any f P F and any x, x1 P X the following is true

|fpxq ´ fpx1q| ă ε.

Since X is a compact space we can chose a finite subcover of this cover, so without the loss of

generality we can assume that the starting cover already consists only of finite number of open

subsets U “ tUi|i “ 1, 2 . . . , nu. Finally, take a partition of unity on X

1 “nÿ

i“1

hi

subordinate to the considered cover (i.e. each positive function hi is supported on Ui). Obviously

hipxq ď 1 for all x P X and all i “ 1, 2, . . . , n.

Consider a finite dimensional C*-algebra of continuous functions on a finite set of n points

CpXnq. One can think of this C*-algebra via the isomorphism CpXnq – Cn, where for any

y “ py1, y2, . . . , ynq, z “ pz1, z2, . . . , znq P Cn

y ` z “ py1 ` z1, y2 ` z2, . . . , yn ` znq,

yz “ py1z1, y2z2, . . . , ynznq,

22

involution is given by complex conjugation on each yi and the norm is defined by

y “ maxi“1,2,...,n

|yi|.

Let us describe ψ : CpXq Ñ Cn by

ψpfq “ pfpx1q, fpx2q, . . . , fpxnqq,

where x1, x2, . . . , xn P X are some fixed (once and for all) points such that xi P Ui for any

i “ 1, 2, . . . , n. Obviously ψ is a ˚-homomorphism, hence a completely positive map (by exercise

4.4 point a)). Moreover, for any f P CpXq

ψpfq “ maxi“1,2,...,n

|fpxiq| ď supxPX

|fpxq| “ f ,

so ψ is contractive. Now define a ˚-homomorphism (hence completely positive map) ϕ : Cn Ñ

CpXq by

ϕpyq “nÿ

i“1

yihi.

ϕpyq “ supxPX

ˇ

ˇ

ˇ

ˇ

ˇ

nÿ

i“1

yihipxq

ˇ

ˇ

ˇ

ˇ

ˇ

ď maxi“1,2,...,n

|yi| “ y ,

so ψ is also a contractive map (we use the fact that hi give a resolution of identity). To conclude

we shall show that

ϕψpfq ´ f ă ε. (4.1)

Indeed, for any x P X we have

ˇ

ˇ

ˇ

ˇ

ˇ

fpxq ´nÿ

i“1

fpxiqhipxq

ˇ

ˇ

ˇ

ˇ

ˇ

“

ˇ

ˇ

ˇ

ˇ

ˇ

nÿ

i“1

pfpxq ´ fpxiqqhipxq

ˇ

ˇ

ˇ

ˇ

ˇ

ď

nÿ

i“1

hipxq |fpxq ´ fpxiq| ă εnÿ

i“1

hipxq “ ε.

Form that follows (4.1) and we are done.

Exercise 4.6: Let A be a separable C*-algebra with an approximate unit puλq Ă A. Then, there

exists a sequence pxnq Ă A such that its elements form a dense subset of A. For any m P N we

can define vm “ uλ (for some λ P Λ) such that

xkvm ´ xk ă1

m

and

vmxk ´ xk ă1

m

for any 1 ď k ď m. Obverse that for any a P A we have

avm ´ a “ avm ´ xkvm ` xkvm ´ xk ` xk ´ a ď vm a´ xk ` xkvm ´ xk ` xk ´ a

23

and

vma´ a “ vma´ vmxk ` vmxk ´ xk ` xk ´ a ď vm a´ xk ` vmxk ´ xk ` xk ´ a .

Therefore, the sequence pvkq Ă A is a countable approximate unit for A (since pxnq is dens in A).

Exercise 4.7: Let A be a unital C*-algebra (if not consider its unitization) and let a P A be a

positive element. It is obvious that aAa is a ˚-algebra, so taking aAa we obtain a C*-subalgebra

in A. We will show that it is a hereditary subalgebra. Indeed, let b P aAa`, c P A and 0 ď c ď b.

Moreover, let puλq Ă aAa denotes an approximate unit of aAa. Observe that c 12ď b

12 and

p1´ uλqcp1´ uλq ď p1´ uλqbp1´ uλq ď p1´ uλq p1´ uλqb ď 2 p1´ uλqb ,

because p1´ uλqcp1´ uλq ď p1´ uλqcp1´ uλq. Therefore, we obtain

limΛp1´ uλqcp1´ uλq “ lim

Λ

›

›

›c

12 p1´ uλq

›

›

›

2

“ 0.

From this it is clear that

limΛuλcuλ “ c

and since puλcuλq Ă aAa, then c P aAa as well.

Exercise 4.8: Suppose that A is a unital C*-algebra. Let a P InvpAq an p P A be a projection

such that ra, ps “ 0. In the trivial situation p “ 0, 1 we have pAp “ 0 and pAp “ A, so we omit

this cases. Observe that the unit element in pAp is p and apa´1p “ paa´1p “ p, so a is indeed

invertible in the corner.

Now, suppose that a P pAp is invertible in pAp. We will show that this not implies a P InvpAq.

To see this take a “ p, clearly a “ ppp P pAp and aa “ p, so a is invertible in pAp, but p is not

invertible in A (as long as p ‰ 1).

5 Positive linear functionals and representations of C*-algebras

Exercise 5.2: Let n “ m, then the faithful C*-algebra representation

π : MnpCq ÑMmpCq

can be simply define as an identity map. When m ą n then we may define desired representation

by

MnpCq Q a ÞÑ

¨

˝

a 0

0 0

˛

‚PMmpCq

24

Consider the case when m ă n. By Eij P MnpCq we will denote matrix units (i.e. matrices with

all entries equal to zero except for 1 in the i-th row and j-th column). Set of all different matrix

units is linearly independent and consists of n2 elements. Suppose that there exists a faithful

representation π of MnpCq in MmpCq. Because m ă n, elements of the form πpEijq have to be

linearly dependent. Therefore, there exist n2 complex constants αij such that at least one of them

is different that zero and the following expression holds

0 “ÿ

i,j

αijπpEijq.

Form that we get

0 “ πpÿ

i,j

αijEijq,

but Eij are linearly independent thus

ÿ

i,j

αijEij ‰ 0

which is a contradiction with faithfulness of π.

Now consider any two faithful representations

π1 : MnpCq ÑMmpCq,

π2 : MnpCq ÑMmpCq.

In geenral, it is not true that they are unitarily equivalent, i.e. there is no unitary element

u PMmpCq such that for any a PMnpCq we have

π1paq “ uπ2paqu˚.

Suppose that m “ 2n and consider faithful representations π1, π2 given by

π1 : MnpCq Q a ÞÑ

¨

˝

a 0

0 0

˛

‚PMmpCq

and

π2 : MnpCq Q a ÞÑ

¨

˝

a 0

0 a

˛

‚PMmpCq.

If π1 and π2 are unitarily equivalent then in particular π1p1q “ uπ2p1qu˚ so

¨

˝

1 0

0 0

˛

‚“ u

¨

˝

1 0

0 1

˛

‚u˚

which leads to¨

˝

1 0

0 0

˛

‚“

¨

˝

1 0

0 1

˛

‚

25

and we obtain a contradiction. However, unitary equivalence of such representations is true if we

restrict our attention to the case of unital representations.

Exercise 5.3 To show that the universal representation of a C*-algebra A is in fact faithful,

assume thatà

φPSpAq

πφpaq “ 0

for some a P A. From the construction of universal representation (direct sum of GNS representa-

tions with norm being supremum over norms of all representations indexed by φ - direct sumands

in the universal representation) we have that πφpaq “ 0 for any φ P SpAq. If so then for any b P A

and any state

πφpaqpb`Nφq “ Nφ,

so in particular aa˚ P Nφ. Then for all states φppaa˚q˚aa˚q “ φppaa˚q2q “ 0. Recall that for

any normal element a P A there exists a state φ such that ϕpaq “ a and by previous discussion

0 “›

›paa˚q2›

› “ aa˚2“ a

4. Therefore,

À

φPSpAq πφpaq “ 0 implies a “ 0 thus

à

φPSpAq

πφ : AÑ B

˜

à

φPSpAq

Hφ

¸

is an injective map.

Exercise 5.5 Let the net paλq Ă BpHq be convergent to a P BpXq in the operator norm topology,

i.e.

limΛaλ ´ a “ lim

Λ

˜

supξď1

paλ ´ aqξ

¸

“ 0.

This immediately implies that for any ξ P H (if ξ has norm different that one, we shall simply

rescale it by some complex number) we have

limΛpaλ ´ aqx “ 0,

so paλq converges (with respect to strong operator topology) to a. Since weak operator topology

is weaker that the strong operator topology we also have a convergence in the weak sense. That

conclusion could be seen directly, namely if there is a SOT-convergent net paλq, i.e. for all ξ P H

limΛpaλ ´ aqξ “ 0.

By the Cauchy-Schwarz inequality we get

| xpaλ ´ aqξ, ηy | ď pan ´ aqξ η

26

for any ξ, η P H, so

limΛ| xpaλ ´ aqξ, ηy | ď lim

Λpan ´ aqξ η “ 0

which ends the proof.

Exercise 5.6 Consider some subset S Ă BpHq and its commutant

S1 “ ta P BpHq| as “ sa,@s P Su .

Let paλq Ă S1 be a net convergent to a P BpHq with respect to weak operator topology (WOT-

convergent net), i.e.

limΛxaλξ, µy “ xaξ, µy

for any ξ, µ P H. By the definition of S1 we obtain (for any ξ, µ P H, s P S)

limΛxaλsξ, µy “ xasξ, µy

and

limΛxsaλξ, µy “ lim

Λxaλξ, s

˚µy “ xaξ, s˚µy “ xsaξ, µy .

If so then for all ξ, µ P H

xasξ, µy “ xsaξ, µy .

Therefore a P S1 and commutant is indeed closed subset (with respect to weak operator topology).

Now assume that S “ S˚. It is clear that S1 is an algebra since if a, b P S1, then a` b, ab, αa P S1

for any α P C. Suppose that

as “ sa

for any s P S, then also

s˚a˚ “ pasq˚ “ psaq˚ “ a˚s˚

and we may conclude that S1 is a ˚-algebra.

Exercise 5.7 Let Π : AÑ BpHq denotes the universal representation of a C*-algebra A. Define

A “ ΠpAqSOT

Ă BpHq.

It is obvious from the definition of A that

ASOT“ A.

It is clear that A is an algebra. Therefore to prove that A is a von Neumann algebra it is enough to

show that A acts on H in a nondegenerated way, i.e. if ξ P H and aξ “ 0 for all a P A then ξ “ 0.

This follows from the fact that H “À

φPSpAqHφ and each GNS representation πφ : A Ñ BpHφq

27

acts in a nondegenerated way on Hφ.

Exercise 5.7 Let A Ă BpHq be a concrete C*-algebra. Consider an approximate unit puλq Ă A

of A. Let a P A, obviously

Exercise 5.9 Let paλq P A be a WOT-convergent net, i.e.

limΛxaλξ, ηy “ xaξ, ηy

for some a P A and any ξ, η P H. Observe that form the Riesz representation theorem, each φ P H˚

is of the form (here we take the convention that inner product is linear in the first variable)

φηpξq “ xξ, ηy

for some η P H. From this we can conclude that the net pφpaλξqq (for any ξ P H,φ P H˚) is

bounded. Therefore, paλξq must we bounded as well. To see this consider ˜aλξ P H˚˚, defined

by

˜aλξpφq “ φpaλξq

for any φ P H˚. Using the Riesz theorem once more we may identify η P H with φη P H˚ (and

their norms), so we obtain the following equality

›

›

›

˜aλξ›

›

›“ supφηď1

| ˜aλξpφq| “ supφηď1

|φηpaλξq| “ supηď1

| xaλξ, ηy | “ aλξ .

From it follows that the net paλξq must be bounded (for any ξ P H), since´›

›

›

˜aλξ›

›

›

¯

is bounded

(for any ξ P H) by the uniform boundedness principle and the fact that p ˜aλξpφqq “ pφpaλξqq is

bounded (for any ξ P H,φ P H˚). Once more, by the uniform boundedness principle we obtain

that the net paλq is bounded which ends the proof.

6 Further examples of C*-algebras

Exercise 6.2 Let A denotes the inductive limit of the inductive sequence pAn, φnq and B be a C*-

algebra such that there exist ˚-homomorphism ψ : AÑ B and ˚-homomorphisms ψpnq : An Ñ B

defined for any n P N such that the following diagram is commutative

An A

B

φpnq

ψpnqψ

28

Firstly, assume that ψ is injective. Let a P kerpψpnqq Ă An for some n P N. By the above

diagram we get

ψ ˝ φpnqpaq “ ψpnqpaq “ 0

so by injectivity of ψ, φpnqpaq “ 0. Therefore, kerpψpnqq Ă kerpφpnqq for any n P N. Conversely,

suppose that kerpψpnqq Ă kerpφpnqq for all n P N. Then the following relation

ψ ˝ φpnqpaq “ ψpnqpaq “ 0

implies a P kerpψpnqq which leads to φpnqpaq “ 0. From that ψ must be injective onŤ

n φpnqpAnq,

which is a dense subset of A. Observe that ψ restricted to each φpnqpAnq (which is a C*-algebra

as an image of a C*-algebra by ˚-homomorphism) is an injective ˚-homomorphism to ψpnqpAnq,

so form general theory of ˚-homomorphisms it must be isometric. Therefore ψ is isometric on the

dense subsetŤ

n φpnqpAnq and it must be isometric on the whole A, hence ψ : A Ñ B is indeed

an injective map.

To see that the second part of the claim is also true, consider the case when B “Ť

n ψpnqpAnq.

If so, then any b P B is of the form

b “ limiÑ8

ψpniqpaiq,

where ai P Ai. Observe than by the above diagram and the fact that ψ is a ˚-homomorphism we

have

b “ limiÑ8

ψ ˝ φpniqpaiq “ ψp limiÑ8

φpniqpaiqq.

Therefore, there exist a P A such that b “ ψpaq, so ψ is surjective. Conversely, assume that

B “ ψpAq, then by reversing the previous formula we may conclude that B “Ť

n ψpnqpAnq.

Exercise 6.3 Consider an inductive sequence pAn, φnq and te following inductive limits

A “ limÑpAn, φnq

B “ limÑpAnk , φnk,nk`1

q.

For any n,m P N, one can construct the following commutative diagram

An Am

Ankn Ankm

B

φn,m

φn,nknφm,nkm

φnkn ,nkm

φpnkn

qψpnkm

q

where nkn is the smallest natural number such that n ď nkn and φn, n “ idAn . The above diagram

29

can give a rise to another commutative diagram

An Am

B

φn,m

ψpnqψpmq

with ψpnq “ φpnkn q˝φn,nkn . From that we may conclude that there exists a unique ˚-homomorphism

ψ : AÑ B such that

An A

B

φpnq

ψpnqψ

is commutative for any n P N. Therefore, we may use the previous exercise (exercise 6.2) to end

the proof. Indeed, observe that form the definition fo B

B “ď

k

φpnkqpAnkq “ď

k

ψpnkqpAnkq “ď

n

ψpnqpAnq,

so ψ is surjective. Moreover, we can deduce that ker ψpnq Ă ker φpnq for any n P N. Therefore, ψ

is also injective and we obtain the desired isomorphism

A “ limÑpAn, φnq – lim

ÑpAnm , φnk,nk`1

q “ B.

Exercise 6.4 Consider two UHF algebras A and B defined by the UHF sequences pknqnPN and

plmqmPN respectively, i.e.

A “ limÑpAn, φnq “ lim

ÑpMknpCq, φnq

B “ limÑpBm, ϕmq “ lim

ÑpMlmpCq, ϕmq

Let pknqnPN and plmqmPN be UHF decompositions of the same supernatural number p. If so, then

one can chose a subsequences pkniqiPN and plmiqiPN such that for any i P N

kni |lmi |kni`1.

By the previous exercise (exercise 6.3) we have the following isomorphisms

limÑpAn, φnq – lim

ÑpAni , φni,ni`1

q

and

limÑpBm, ϕmq – lim

ÑpBmi , ϕmi,mi`1

q

Therefore, we can work on the level of subsequences defined above. To simplify the notation,

30

without the loss of generality, we may assume that mi “ ni “ i (since it is just a renumbering).

Because ki ď li, in particular we have a unital ˚-homomorphism ρ1 : A1 “ Mk1pCq Ñ Ml1pCq “

B1. Similarly we have a ˚-homom.orphism κ1 : B1 “ Ml1pCq Ñ Mk2pCq “ A2. Consider the

following digram

A1 A2

B1

φ1

ρ1κ1

By the discussion from exercise 5.2, we know that So κ1 can adjusted by unitary element in order

to make the above diagram commutative. We may continue this procedure by defying ρi, κi and

adjusting all maps ρi, κi by appropriate unitaries (on each step of this construction) in order to

obtain the following commutative (commutative in each triangle) diagram

A1 A2 A3 A4 . . .

B1 B2 B3 . . .

φ1

ρ1

φ2

ρ2

φ3

ρ3

φ4

ρ4

ϕ1

κ1

ϕ2

κ2 κ3

ϕ3

From that we derived at maps

ρ “ď

ně1

ρn : AÑ B

and

κ “ď

ně1

κn : B Ñ A.

Note that ρ and κ are inverse to each other (by the above construction), hence we get the desired

isomorphism between A and B. This proves that UHF algebra described by some supernatural

number p is define in a unique way (it is not dependent - up to isomorphism - of the particular

choice of UHF sequence).

Exercise 6.5 Let A be a UHF algebra, i.e.

A “ limÑpBn, φnq

with all An being full matrix algebras (in particular finite dimensional C*-algebras). Observe that

A “ď

n

φpnqpBnq “ď

n

An,

where φpnqpBnq “ An and for any n,m P N there is k P N such that An, Am Ă Ak. Therefore,

there is an upwards-directed set pAnqnPN such thatŤ

nAn is dense in A and any An is a nuclear

C*-algebra (see exercise 5.4). We will show that from this follows nuclearity of A. Indeed, consider

two C*-norms ¨α , ¨β on algebraic tensor product AbB (where B is any C*-algebra) and two

31

completions Abα B and Abβ B with respect to this norms. We may define

A “ď

n

An bB.

SinceŤ

nAn is dense in A, A is dense in both Abα B and Abβ B. By nuclearity of each An we

may conclude that for any a P A

aα “ aβ

If so, then we can extend identity map on A to the ˚-homomorphism

π : Abα B Ñ Abβ B.

Consider some b P B and a P A with panq PŤ

nAn such that

a “ limnÑ8

an.

Then we get (both in Abα B and Abβ B)

ab b “ limnÑ8

an b b.

From that we obtain

πpab bq “ limnÑ8

πpan b bq “ limnÑ8

an b b “ ab b,

so π acts as identity on the whole AbB. Therefore two C*-norms ¨α and ¨β coincide on AbB

and Abα B – Abβ B, thus A is nuclear (since B was arbitrary). Note that practically the same

arguments work for any AF algebra (since all finite dimensional C*-algebras are nuclear).

Exercise 6.7 Consider the UFH algebra U28 and the C*-algebra CpX,U28q of continuous U28-

valued functions on some compact Hausdorff space X. Since all C*-algebras of our interest are

nuclear (see exercise 6.5) we will omit any subscript of (unique in this situation) tensor product.

Since

CpX,U28q – CpXq b U28

we get

CpX,U28q b U28 – CpXq b U28 b U28 .

But by the previous exercise (exercise 6.6) we know that in particular U28 b U28 – U28 , so we

obtain

CpX,U28q b U28 – CpX,U28q

and the proof is completed.

32

Exercise 6.9 Since AF algebra A can be nonunital (maps φn in the definition of incutive sequence

may not be unital), consider such a case. We have already proved (exercise 2.4) that A can be

seen as a two-sided closed ideal in int unitization A. Observe that in a natural way A is also an

AF algebra, therefore not all AF algebras are simple C*-algebras.

Exercise 6.12 By tr we denote the normalized trace on MnpCq, i.e.

trpaq “1

n

nÿ

i“0

aii,

where a P MnpCq. Suppose that there exists another tracial state φ on MnpCq. Since φ is linear,

it is completely determined by its values on matrix units Eij P MnpCq (see exercise 5.2), i.e. for

any a PMnpCq

φpaq “nÿ

i,j“1

aijφpEijq.

For any i ‰ j we get

φpEijq “ φpEikEkjq “ φpEkjEikq “ φp0q “ 0.

Since Eii “ E˚ii and Eii “ E2ii for all i we know that φpEiiq P R`. For all i, j we can find a matrix

v PMnpCq such that

Eii “ vEjjv´1.

Form that we obtain

φpEiiq “ φpvEjjv´1q “ φpv´1vEjjq “ φpEjjq

Because

1 “ φp1q “nÿ

i“1

φpEiiq

finally we get

φpEiiq “1

n.

Therefore, φ “ tr. Existence and uniqueness of a tracial state on UHF algebra follow then form

its construction as an inductive limit of inductive sequence consisting of full matrix algebras.

Exercise 6.13 If G is discrete then L1pGq “ CrGs. Consider a, b, c P CrGs given by

a “ÿ

gPG

agδg,

b “ÿ

hPG

bhδh,

c “ÿ

iPG

ciδi,

33

where ag, bh, ci P C (in each case only finite number of those coefficients is not equal to 0) and

δgphq “

$

’

&

’

%

0 for g ‰ h

1 for g “ h

Observe that for any t P G we obtain

δg ˚ δhptq “ÿ

sPG

δgpsqδhps´1tq “ δgpgqδhpg

´1tq “ δghptq.

Since ˚ is obviously a linear operation we have

a ˚ pb ˚ cq “ a ˚

˜˜

ÿ

hPG

bhδh

¸

˚

˜

ÿ

iPG

ciδi

¸¸

“ a ˚

˜

ÿ

h,iPG

bhciδh ˚ δi

¸

“

˜

ÿ

gPG

agδg

¸

˚

˜

ÿ

h,iPG

bhciδhi

¸

“ÿ

g,h,iPG

agbhciδg ˚ δhi

“ÿ

g,h,iPG

agbhciδghi

and

pa ˚ bq ˚ c “

˜˜

ÿ

gPG

agδg

¸

˚

˜

ÿ

hPG

bhδh

¸¸

˚ c “

˜

ÿ

g,hPG

agbhδg ˚ δh

¸

˚ c

“

˜

ÿ

g,hPG

agbhδgh

¸

˚

˜

ÿ

iPG

ciδi

¸

“ÿ

g,h,iPG

agbhciδgh ˚ δi

“ÿ

g,h,iPG

agbhciδghi,

so we are done.

Now suppose that G is no longer discrete. Let f, g, h P LpGq. Then we have

ppf ˚ gq ˚ hq ptq “

ż

G

pf ˚ gqpsqhps´1tq dµpsq “

ż

G

ˆż

G

fprqgpr´1sq dµprq

˙

hps´1tq dµpsq

“

ż

G

ż

G

fprqgpr´1sqhps´1tq dµprq dµpsq

and

pf ˚ pg ˚ hqq ptq “

ż

G

fpsqpg ˚ hqps´1tq dµpsq “

ż

G

fpsq

ˆż

G

gprqhpr´1s´1tq dµprq

˙

dµpsq

“

ż

G

fpsq

ˆż

G

gps´1pqhpp´1tq dµps´1pq

˙

dµpsq

“

ż

G

fpsq

ˆż

G

gps´1pqhpp´1tq dµppq

˙

dµpsq

“

ż

G

ż

G

fpsqgps´1pqhpp´1tq dµpsq dµppq,

where we use substitution p “ sr. This ends the proof.

34

Exercise 6.14 Suppose that there is a given nondegenerated representation π : L1pGq Ñ BpHq

(for some locally compact group G). Since π is nondegenerated, the following set

D “ span

πpfqξ| f P L1pGq, ξ P H(

is dense in the Hilbert space H. For any s P G consider the operator upsq : D Ñ H defined by

upsqπpfqξ “ πpfsqξ,

where

fsptq “ fps´1tq

for any s, t P G and f P L1pGq. It is easy to see that

u´1psqπpfqξ “ πpfs´1qξ,

give an operator inverse to upsq. Moreover, since upsq ď 1 and›

›u´1psq›

› ď 1, we get sppupsqq Ă T

and because upsq is normal, by the continuous functional calculus, we may conclude that upsq :

H Ñ H is indeed a unitary operator (for any s P G).

Exercise 6.15 Recall that the left-regular representation λ : GÑ BpL2pGqq of a locally compact

group G is given by

λpsqfptq “ fps´1tq,

where s P G and f P L2pGq. We will show that in fact λ : G Ñ UpL2pGqq. Indeed, consider

g, f P L2pGq and any s P G. Then we have (here we take the convention that inner product is

linear in the second variable)

xg, λpsqfy “

ż

G

gptqλpsqfptq dµptq “

ż

G

gptqfps´1tq dµptq “

ż

G

gpst1qfpt1q dµpst1q

“

ż

G

gpst1qfpt1q dµpt1q “

ż

G

λ˚psqgpt1qfpt1q dµpt1q

“ xλpsq˚g, fy .

Therefore

λ˚psqfptq “ fpstq

for any s P G and f P L2pGq. Obviously

λ˚psqrλpsqf sptq “ λpsqfpstq “ fps´1stq “ fptq

and

λpsqrλ˚psqf sptq “ λ˚psqfps´1tq “ fpss´1tq “ fptq,

so λpsq P UpL2pGqq for all s P G.

35

Exercise 6.16 Recall that the Fourier-Plancheral transform ˆ: L1pGq Ñ L1pGq is a map defined

by

ˆpfqpγq “

ż

G

γptqfptq dµptq

for any γ P G and f P L1pGq. Clearly it is a linear map because of

ˆpf ` gqpγq “

ż

G

γptqpfptq ` gptqq dµptq “

ż

G

γptqfptq dµptq `

ż

G

γptqgptq dµptq “ fpγq ` gpγq

and

αfpγq “

ż

G

γptqαfptq dµptq “ α

ż

G

γptqfptq dµptq “ αfpγq

where f, g P L1pGq and α P C. To see that the Fourier-Plancheral transform is also multiplicative

consider

ˆf ˚ gpγq “

ż

G

γptqpf ˚ gqptq dµptq “

ż

G

γptq

ˆż

G

fpsqgps´1tq dµpsq

˙

dµptq

“

ż

G

ˆż

G

γptqfpsqgps´1tq dµpsq

˙

dµptq “

ż

G

ˆż

G

γptqfpsqgps´1tq dµptq

˙

dµpsq

“

ż

G

ˆż

G

γpst1qfpsqgpt1q dµpst1q

˙

dµpsq “

ż

G

ˆż

G

γpst1qfpsqgpt1q dµpt1q

˙

dµpsq

“

ż

G

ˆż

G

γpsqγpt1qfpsqgpt1q dµpst1q

˙

dµpsq “

ż

G

γpsqfpsq

ˆż

G

γpt1qgpt1q dµpt1q

˙

dµpsq

“

ż

G

γpsqfpsq pgpγqq dµpsq “ fpγqgpγq.

Finally, to show that the considered map is a ˚-homomorphism, it remains to prove that it preserves

involution. Indeed, observe that

f˚pγq “

ż

G

γptqf˚ptq dµptq “

ż

G

γpt´1qfpt´1q dµptq “ fpγq

which ends the proof.

Exercise 6.17 Consider a one point space txu and C*-algebra of continuous functions Cptxuq – C.

Suppose that a locally compact group G acts trivially (via action α) on Cptxuq (or equivalently

on txu), i.e. for any z P C and any g P G

αgpzq “ z.

Observe that in this case

L1pGq “ L1pG,Cq – L1pG,Cptxuqq.

Therefore, representations of L1pGq coincide with representations of L1pG,Cptxuqq and we may

conclude that by the construction of reduced and universal versions of group C*-algebras and

cross products we obtain

C¸r G – C˚r pGq,

36

C¸f G – C˚f pGq.

Exercise 6.22 Let s1, s2, . . . , sn P A be isometries (i.e. s˚i si “ 1). It is clear that for any i we get

psis˚i q˚ “ sis

˚i

and

psis˚i q

2 “ sis˚i sis

˚i “ sis

˚i ,

so sis˚i are projections. Now suppose that

nÿ

i“1

sis˚i “

nÿ

i“1

pi “ 1.

We will show that all pi are orthogonal. Since any A is faithfully represented on some BpHq, we

may treat each pi as a projection which belongs to BpHq. Observe that for any given j

ÿ

i‰j

pipj ` p2j “ pj .

From that we obtainÿ

i‰j

pipj “ 0.

For any ξ, η P H

0 “

C

nÿ

i‰j

pipjξ, η

G

“

nÿ

i‰j

xpipjξ, ηy “nÿ

i‰j

xpjξ, piηy

In particular for any ξ such that ξ P im pj we obtain

0 “nÿ

i‰j

xξ, piξy “nÿ

i‰j

xpiξ, piξy “nÿ

i‰j

piξ2.

Therefore, for any i ‰ j and all ξ P im pj we get piξ “ 0. Since we can chose arbitrary j we may

conclude that

pipj “ sis˚i sjs

˚j “ 0

for all i ‰ j.

Let us consider the sequence psiq Ă A of partial isometries such that for n P N

nÿ

i“1

sis˚i “

nÿ

i“1

pi ď 1.

Suppose that there are two indexes i, j P N such that i ‰ j and pipj ‰ 0. Using once more

representation of A in some BpHq we can treat each pi as a projection which belongs to BpHq.

There exists such m P N such that i, j ď m. Since pi and pj are not orthogonal we can find a

37

nonzero ξ P im pj such piξ ‰ 0. If so, then

C

ξ, 1´mÿ

i

piξ

G

“ xξ, ξy ´mÿ

i

xpiξ, piξy “ ´mÿ

i‰j

piξ2.

The above expression gives a rise to negative number, which is a contradiction with positivity of

1´řni pi. Therefore, pipj “ 0 for all j ‰ i and we are done.

7 A very short introduction to classification for simple nu-

clear C*-algebras

Exercise 7.1 Consider an inductive sequence pGn, φnq consisting of abelian groups Gn and group

homomorphisms φn

φn : Gn Ñ Gn`1.

Observe that the infinite product

G “ź

n

Gn

with pointwise defined operation is itself a group. Consider the following subgroup of G

G1 “!

g “ pg1, g2, . . . q P G| DN @j ě N gj`1 “ φjpgjq)

.

To conclude define the following normal subgroup N in G1

N “

g “ pg1, g2, . . . q P G1| DN @j ě N gj`1 “ ej`1

(

,

where en denotes neutral element in Gn. The inductive limit G of a given inductive sequence

pGn, φnq is defined by

G “ limÑpGn, φnq “ G1N.

Exercise 7.3 Let A be a unital and simple C*-algebra with T pAq ‰ H. Consider any τ P T pAq

and suppose that there exists x P A such that τpx˚xq “ 0 but x ‰ 0. Define

I “ tx P A| τpx˚xq “ 0u .

Obviously I is a linear subspace. By the Cauchy-Schwarz inequality we get (for any x P I, a P A

τppaxq˚paxqq “ τpx˚a˚axqq ď τpx˚xqτppa˚axq˚pa˚axqq “ 0.

Moreover, we have

τppxaq˚pxaqq “ τpa˚x˚xaq “ τpx˚xaa˚q ď τpx˚xqτppxaa˚q˚pxaa˚qq “ 0,

38

where we use the tracial property of τ . From that I is a two-sided ideal. Suppose that

a “ limnÑ8

an

with panq Ă I. Clearly

τpa˚aq “ limnÑ8

τpa˚nanq “ 0,

since any positive map is automatically norm continuous. Therefore, I is also closed. Because

x P I and 1 R I (τp1q “ 1) we obtain a contradiction with simplicity of A. If so, then there is no

nonzero x P A such that τpx˚xq “ 0 (τ is faithful).

Now, suppose that there exists a nonzero projection p P A such that p “ v˚v and 1 “ vv˚ for

some v P A. Observe that

1 “ τp1q “ τpvv˚q “ τpv˚vq “ τppq “ 1

and what follows

τp1´ pq “ 0.

We have

τpp1´ pq˚p1´ pqq “ τp1´ pq “ 0,

so by faithfulness of τ we get p “ 1. Therefore, A is finite.

Exercise 7.6 Suppose that there are two projections p, q P MnpAq such that pq “ qp “ 0. This

implies that p ` q is a projection. We will show that rps ` rqs “ rp ` qs. By the definition of

addition in P pAq „ we get

rps ` rqs “ rp‘ qs.

Observe that for any p PMnpAq (in our case also for p` q since it is a projection) and any m P N

we have

p‘ 0m „ p,

where 0m PMmpAq is a zero matrix. Indeed, define v PMn,pn`mqpAq by

v “´

p 0n,m

¯

,

then

vv˚ “´

p 0n,m

¯

¨

˝

p

0m,n

˛

‚“ p

and

v˚v “

¨

˝

p

0m,n

˛

‚

´

p 0n,m

¯

“

¨

˝

p 0n,m

0m,n 0m

˛

‚.

39

Now define u PM2npAq by

u “

¨

˝

p q

0n 0n

˛

‚

Observe that

uu˚ “

¨

˝

p q

0n 0n

˛

‚

¨

˝

p 0n

q 0n

˛

‚“

¨

˝

p` q 0n

0n 0n

˛

‚

and

u˚u “

¨

˝

p 0n

q 0n

˛

‚

¨

˝

p q

0n 0n

˛

‚“

¨

˝

p pq

qp q

˛

‚“

¨

˝

p 0n

0n q

˛

‚,

so we get the desired equivalence p‘ q „ pp` qq ‘ 0n „ p` q.

40