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