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s(r,b). Then for any a E M,
0(a) = i(pap) = 40(toPaPto) = ço(t o at o). Q.E.D. Let M be a VN algebra, ço,I,G E M. and ço > t,b > O. If X Theorem 1.10.4. is a complex number with ReX > 1, then there exists h E M with 0 < h < 1 such that 0(a) = Aço(ha) +(p(ah), Va E M. Moreover, if ço is faithful, then h is unique.
Proof.
Since {h EMI0 p,
?
f (a),
0 V L,
then there exists a
V f E f.
Write a = a+ — a_, where a± G M+ and a+ • a_ = 0, and pick a projection P of M such that pa+ = a+ ,pa_ = 0. Then
tga + ) > ji, ? Aço(pa) ± Aio(ap) = 2ReA  ça(a) ? ço(a + ). It is impossible since io > 1,b. Thus V) E L, i.e., there exists h E M with 0 < h < 1 such that Ca) = Aço(ha) + Xv(ah), Va E M. Moreover let ça be faithful, and suppose that h and k satisfy our condition simultaneously. Since (A + A)(h
—
k ) 2 = [P (h — k) + X(h — k)h]
—WO — k) +X(h — k)k],
53
it follows that ço((h
—
10 2 ) = 0 and h= k.
Q.E.D.
Proposition 1.10.5.
Let M be a VN algebra on a Hilbert space H, ço be a a(or weakly, or strongly) continuous linear functional on M. Then ço can on be extended to a a(or weakly, or strongly) continuous linear functional B(H), and 11011 = IIço. Moreover, if io > 0, it can be choosed that .t,b > 0.
Proof. First, we assume 0 < ço E M. Since M. = T(H)/M I , we have t* = t E T(H) such that io(a) = tr(ta), Write t =t +
—
Va E M.
t_, where t ± E T(H) + and t + t_ = 0. Then tr(t + a) > ço(a) > 0,
Va
e M.
By Theorem 1.10.3, there is t o E M with 0 < to < 1 such that
ço(a) = tr(t + t oat o) = tr(t ot + t oa),
Va E M.
Let OH =tr(tot + to.). Then V) is a acontinuous linear functional on B(H). Clearly, .1,b is an extension of ço, and i > 0, and licou = (P( 1 = = 11011For general io E M., let io .R,co be the polar decomposition of v. From the preceding paragraph, there is a acontinuous positive linear functional g on B(H) such that 9IM = w)II911 = 11w II PH. Let 1i = R eg. Then .1,b is acontinuous on B(H), and is an extension ço. Clearly, IlçoM ÇMOH liv1111911 Thus MOH = lkoH. When ço is weakly (or strongly) continuous, according to the above procedure it suffices to show there exists t E F(H) such that ço(a) = tr(ta), Va E M. If ço is weakly continuous, then there exists a weak neighborhood U = 1,1 < i < n) of zero U(0; 67 • • • 7 en; rii) • • • , ?in ; 1) = {a E M I *Lei, ni)I such that ko(a)I < 1,Va E U. Define a seminorm p() on B(H): )
p(b) =
1(ben rii)
Vb E B(H).
i=1
Then ko(a)I < p(a),Va e M. By the HahnBanach theorem, ço can be extended to a linear functional tk on B(H) such that kb(b) I < p(b),Vb E B(H). Clearly, 1,b is weakly continuous on B(H). By Proposition 1.2.7, there is t E F(H) such that ?P (b ) =tr(tb),Vb E B(H). In particular, ço(a) =tr(ta), Va E M. If ço is strongly continuous, replacing above U = U(0, " • en; rib • • ' )rin; 1)
and X.)
Ê 1(.6, 77)1 by U= U(0; ei, • •  , en; 1) = {a E Miiiajii < 1,1
0, then we can choose 6 = ni , vi;
each weakly or strongly continous functional ça on M has the following form: n
ça(a) = Daei,r/i), Va E M, i= 1
and if ça > 0, then we can choose ei = ?ii , 1 < I < n. As an application of Proposition 1.10.5, we have the following. Proposition 1.10.8. Let M be a VN algebra on a Hilbert space H. Then the following statements are equivalent: 1) a(M, M.) — (weak topology 1M); 2) s(M, M.) — (strong topology 1M); 3) ss(M, M.) — (strong * topology 1M); n
4) Each acontinuous linear functional on M has the form E( • 6, ni);
1=1
n
5) Each acontinuous positive linear functional on M has the form
Clearly, 1), 2) and 3) are equivalent; 4) and 5) are equivalent. From 4), each acontinuous functional on M is weakly continuous. Thus (weak topology 1M) — a(M, X). By Proposition 1.10.5, it is clear that 1) implies 4). Q.E.D.
Proof.
The conditions in Proposition 1.10.6 are possible. For example, see Sections 1.13 and 2.11. Remark.
References.[32], [127], [147], [159].
55
1.11.
The equivalence of the topologies
s*
and
T
in a
bounded ball Let M be a VN algebra on a Hilbert space H, (M) 1 = { x E M I 11x11 5 1} be its unit ball. In this section, we shall prove that s*(M,M.) — r(MM) in any bounded ball of M. Lemma 1.11.1. Let {ay,* = an} be in (M), and an —> 0 (strongly). Then for any .6 > 0 there is a sequence {pn } of projections of M such that
gstrongly), and IlanPnil < .5, Vn. „ 1 Let an = f Acie‘An) be the spectral decomposition of a n , and define i pn —>
Proof.
5 Pn = f de (xn) ) 8 Then .6 2,2 > 'A'n —
qn 72

1
Xs)
Vn.
(f 6+f6 1.6 2 d ex(n) — qn 1 Vn. )A — 2
i
>
Since a2n —> 0 (weakly), it follows that qn —> 0 (strongly), pn —* 1 (strongly).
s
,, So the sequence {pn } is just what Moreover, Ilanpnii = ii s AdeP ii < 8,Vn.
f
Q.E.D.
we want to find.
Let v be a faithful normal positive linear functional on
Lemma 1.11.2. M, and define
d(a,b) = ço((a — Os (a — b)) 112 ) Va,b E (M) 1 . Then ((M)1,d) is a complete metric space, and the topology generated by d is equivalent to s(M,M) in (M) 1 .
Proof.
By Propositions 1.2.2 and 1.10.5, there is a sequence
that
E I en11 2 < 00,
ço(a) =
n
Da en, en), Va G
{
en}
C
H such
M.
n
Thus d(a,b) = (E II (a — b) en11 2 ) 1 I 2 is a metric on (M) 1 . n
We claim that [M i en in] is dense in H. In fact, let p be projection from H onto [M'en in]. Then p e M, and pen = en ,Vn. Since (,c)(1 — p) = Eql —
P)en) en) = 0 and ço is faithful, it follows that p = 1, i.e., [Mi en in] is dense in H. .1 ,
From this claim, our lemma is easy.
Q.E.D.
56
Lemma 1.11.3. Let çoic)(Po be in Ms , and (,0k —) ço0 (a(M,„M)). In addition, suppose that {a n } is a sequence of (M) 1 and an —> 0(s*(M, Ms)). Then lirn v k (an) = 0 uniformly for k. n
Obviously, {lIvkll k} is bounded. So we may assume that 11(Pkil 5 1/2,Vk. By Theorem 1.9.8, for any k we can write
I
Proof.
i
(1)
(2), , ., (3)
(4) 1
P k = VP k — P k 1 1 J O° k — P k 1 ) where 0 < çoil) E M,„ 1 < j < 4, and
1 > iko(: ) 1 + livi2) 1 = livi1 — vi2) I I ) )
1
> I P(3) II F
° k V ks ) ) v(k1) — Plc2) = 1 (ç+
Pit° II 1 II P (k3) — P (k4) II ) P13) — P (k4) = 4i (P k — PO • 4
Clearly, Ilhodll _< 1 and 0 < (,,r) E
Mt ,
where [o k ] = E ço,Vk, and ço =
i=1
c.a, 1 2_, g k 0 kl • Let p = s(v). Then çoii) (1 — p) = 0, and co(a) = çoif) (pap),Vk, j and a E M. Further,
(Pk(a) = sok (pap), Vk, and a E M. Clearly, pap —) 0(ss(M,M* )). Considering the problem to pMp, we may assume that p = 1, or ça is faithful. Let d be the metric on (M) 1 as in Lemma 1.11.2. For any fixed E > 0 and m, define
Hm = {a E (M) 1 I kok(a) — co o (a)1 5 e ) Clearly, Hm is a closed subset of follows that
Vk > m}.
((MI) d). Since çok —) Soo(a(Ms,M)), it 00
(M) 1 = U Hm.
m=1
Now by the Baire category theorem and Lemma 1.11.2, there is a ao E (M)1 and a it > 0 and a mo such that {II G (M)1
I
d(ao , b) .5_ p.}
C Hmo .
Since {(a+ a)} and {(an — an* )} n C (M) 1 and they converge to 0 with respect to s(M,Ms ), we may assume that an = an*,Vn. Now pick 6 = li e. By Lemma 1.11.1, there is a sequence {pn} of projections of M such that
pn —> 1(s (M, Ms)),
< 5, IlanPnll —
Vn.
57
Let
i,bk = (sOk
Po ,Vk. Then , k (,PnanPn) + 10k ( ( 1 Pn)anPn) I
10k(an)1
p) a(1
+10k (Pnan( 1 — Pn)) + kbk ( ( 1 38 + 10k((1 — P n)an (1 Pn))1,
5,
Pn)
(1)
I
Vk,n.
Let bn = p naop n + (1 — p n)an (1— pn), Vn. Then b n e (M) i , Vn, and b n —> ao (s(M,M*)). Thus there is n i such that bn E Hm o , Vn > n1. By the definition of Hmo , we have (2) 10k(bn)I < E l Vk > m o , n > ni. Since pnaoPn —) ao(s(M, M.)), it follows that there is n 2 such that PnaoPn E > n2. Thus
1114(Pna0Pn)1 < E Vk > m o , n > n2. Now if n > n i , n2, and
kok (an) —
k > mo)
(3)
by ( 1 )1 ( 2)) ( 3))
(an) I =
(an) I
< 38 +
10k ( ( 1 — Pn)an(1 —
0 there exists a 6 > 0 and a finite subset F of A such that if a E with kol(a*a + ace) < b,Vço E F, then ka(a)1 < ,Vço e A. There [ça] = Sal+ V2 +ça3+ P4 and (Sal — P2)) (V3 — so4) are the orthogonal decompositions of 12(ço (3° 4 ), li(ça ça*) respectively. Proof. Suppose that our assertion is false for some E > 0. Then for 12 and any fixed (polo E A, we have al E (M), and (pi E A such that
1 kod(a *l ai + aian < — 2'
01(421)1 > E.
For and {po , p i ) c A, we also have a2 E (M) 1 and ço2 E A such that
ko1 ](a 2sa2
1 a2 a2*) < — 22'
= 0,1,
. Generally, we have {Pin n = such that
c A,
[o i]( asi ai + ai a; < )
{
1V.i( a5)1 ? E l
1(,02 (a2 )1> E.
If,
fan in =
0 < I < j — 1,
j= 1,2, • • •
1, 2, • • •1 c
(M),
58
Since A is cr(M* ,M)compact, by the Eberlein§mulian theorem there is a subsequence {nk} C {0, 1, • • •} and E A such that (Pnk —> 00
Clearly, A is bounded. Thus 0 < ça =
E2  k[vnk ] c M. Notice that k=1
k° (,a;ai aj a;)1 < E2kikoni(a;ai kz 1
+ E 2'llkonkill k>m
j —
1. Since
j —> oo.
Let p = s(p). By the Schwartz inequality, it is easy to see that vapai p)*(pai p)) —> 0, çoapaip)(pai p)*) —> O. Since ça is faithful on pMp, by Lemma 1.11.2 we have pap —) 0(s* (M, M* )). By Lemma 1.11.3 and p> is(ko n,j) (Vk), it follows that limpn,(pai p) = limço nk (ai ) = 0 .7
uniformly for k. Then we get a contradiction since Ron, (ank )1 > E, Vk.
Q.E.D.
Lemma 1.11.5.
Let A be a cr(M„M)compact subset of M. Then we > 0 with the following property: for any E > 0, there is .5 > 0 have V) E such that if a E (M) 1 and l,b(asa 0 such that if a E (141) 1 and kol(asa+aas) <S,Vça E Fn , then ko(a)I < !,Vço E A. Define 00
Proof.
= E 2 (n+m" ) E n=1
[ç
].
pEFn
where mn = °Fn ,and the definition of [ça] is as in Lemma 1.11.4. For any E > 0, pick no < E, and let .5 = .5„0 12" Fmno . If a E (M)1 and Igasa aas) 0 uniformly for ço E A. For that A and any E > 0, pick 0 < ik E M, and b > 0 as in Lemma 1.11.5. Since al —> O (8* 0/, M* )), it follows that there is an index / 0 such that
Iga;c2.1 + al ai') < 8, Vi > 1 0 . Then by Lemma 1.11.5, Iço(a1 )1 < ,V1 > 10,V E A. This means that ça(ai ) —> 0 Q.E.D. uniformly for ça E A.
Proposition 1.11.7. Let M be a VN algebra, and dimM = oo. Then in whole M, r(M, M s ) ,, s*(M,M,,). Moreover, in (M)1, uniform topology 9/, r(M,M.). In consequence, if a VN algebra M is reflexive as a Banach space, then dimill < oo. Let M be infinite dimensional. First we claim that M contains an Proof. infinite orthogonal sequence {pn } of nonzero projections. In fact, if dimZ = oo, where Z is the center of M, then we can easily find such {pn} c Z. If dimZ < oo, then we may assume that M is a factor. By the theory of factors (see Section 7.1), such {pn } exists. Now using above { pn}, similar to the proof of Proposition 1.2.5 we can see that r(M,M) is not equivalent to s*(M,M* ) in whole M. 00
Again using above {pn}, let p = E pn. By Proposition L2.10 and Theorem n=1
1.11.6,
—› p with respect to r(M,M). On the other hand, we have
lip En=1Prill = 1,
Vm.
Thus, in (M) 1 the uniform topology is not equivalent to r(M,M). If M is reflexive as a Banach space, then r(M, M.) — uniform topology. Q.E.D. Therefore, M must be finite dimensional. If A is a C*algebra, and A is reflexive as a Banach space, then dimA < oo. Indeed, A** is a VNalgebra (see Section 2.11), and is reflexive as a Banach space. Remark.
60
It was S. Sakai who initiated the study of the Mackey topology Notes. of a Von Neumann algebra. He showed that the Mackey topology on any bounded ball of a finite Von Neumann algebra M agrees with s*(M,M * ). Theorem 1.11.6 was due to C. Akemann who gave an affirmative answer for Sakai conjecture. Further in Section 4.5, we shall study the characterizations of a(M,M)compact subsets of M. References.[2], [146].
1.12. Normal * homomorphisms Definition 1.12.1. A * (algebraic) homomorphism from a VN algebra M into a VN algebra N is said to be normal, if for any bounded increasing net {ai } C M+ we have sup (I)(ai ) = (I)(sup ai ).
1
1
Let (I) be a * homomorphism from M to N. First, (I) will preserve the order, i.e., (I)(M+ ) C N+ . This is obvious because any element of M+ has the form a* a (for some a E M). Secondly, OH < 1. In fact, (I)(1 m) = p is a projection of N. For any h = h* e M, since —11h11  l m < h < 11h11 • 1m , it
follows that —11h1lp _< (I)(h) < 11h1lp. Thus 11(I)(h)11 5,_ 11h11. Further, 11(I)(a)11 2 = 11 (1) (a* a)11 Ilar)Va E M. Hence WI < 1. From these facts, if { ai } is a bounded increasing net on M+ , then {(1)(a1 )} is also a bounded increasing net of N. Now by Proposition 1.2.10, sup al and sup (I) (ai ) exist. Proposition 1.12.2. Let (I) be a * homomorphism from a VN algebra M to a VN algebra N. Then the following statements are equivalent: 1) ID is aa continuous; 2) ID is normal; 3) (I) is completely additive, i.e., (I)(E pi ) = E 4)(pi ) for any orthogonal
/
1
family {pi } of projections of M. Moreover, (I)(M) is a aclosed * subalgebra of N if (I) is normal. By Proposition 1.2.10, we can easily see that 1) implies 2). Moreover, it is clear that 2) implies 3). Now let (I) be completely additive. Then for any 0 < ça E N* ,(g) o (I) is a completely additive positive functional on M, and by Theorem 1.8.6, ça o 4) E M. Further, by Corollary 1.9.9, ça o 4) E Ms , V(p e Ns . Therefore 0 is aa continuous. Proof.
61.
Now if (I) is normal and N C B(K), let I = {a E M I 0(a) = 0}, then I is a aclosed twosided ideal of M. Thus there is a central projection z of M such that I = M(1 — z), and (I) is a * isomorphism from Mz into B(K). Similar to the proof of Proposition 1.8.13, the unit ball of (I)(M) is weakly Q.E.D. closed. Therefore (I)(M) is aclosed. Let (I) be a * isomorphism from a VN algebra M onto Proposition 1.12.3. a VN algebra N. Then (I) is normal and isometric. Let {ai} be a bounded increasing net of M+ , and a = sup al . Then i b = sup (Nat ) 5_ (I)(a). Moreover, since (I)' is a * isomorphism from N onto i
Proof.
M, it follows that
'((a1 )) = a < (V I (b). i Thus b = 0(a), i.e., (I) is normal. Another conclusion is clear. sup
Q.E.D.
Let M, N be two VN algebras on Hilbert spaces H, K respectively, (I) be a normal * homomorphism from M onto N. Then Theorem 1.12.4.
(1)
(1) 3 C) (1) 2
0 (1) 1 )
where (1) 1 is an ampliation of M, i.e., there exists a Hilbert space L such that = a 0 1L,Va E M; (1)2 is an induction of M 0 001L, i.e., there is a projection p' of (MoT)Œ1 L ) 1 such that 2 (a ® 1L) = (a 0 1L)p',Va E M; (1) 3 is a spatial * isomorphism from (MOT1L )pi (a VN algebra on )9' (H 0 L)) onto N. Proof.
First we assume that N admits a cyclic vector ri. Define ça(a) = ((a)r, 77),
Va E M.
Obviously, ça is a normal positive functional on M. By Proposition 1.10.5, there is a sequence {en} c H with E Ilen11 2 < oo such that ça(a) = E(ctn,È,n) ) Va E M. Let L=1 2 ,
e = (en) c
n
n
HO L,0 1 (a)= a® 1L , Va E M. Then
ça(a) = (0 1 (a) e ,
e),
Va e M.
Suppose that p' is the projection from H 0 L onto (I) 1 (M). Then PI E (1) 1(M ) ' = (M6501L) ' . Let (I) 2 (a 0 1L) = (a 0 1L)pl , Va E M. Clearly )
p(a)= ((0 2 0
4),)(c)e, 0,
Va E M.
62
Define a linear map u from 0 (M)ri to pi(H 0 L) as follows: u0(a)ri = (0 2 0 01)(a) e = Pqaen) = ((ten)) Va E M. Since (0(a)n) ri) = v(a) = 020 4, 1)(a),e), Nva E M, it follows that u is isometric. Moreover, since 0 (M)n = Nn and (0 2 0 0,) )E = are dense in K and )91 (H 0 L) respectively, u can be extended to a unitary operator from K onto )91 (H ® L). Clearly,
(m
u0(a)u 1 =
02 0 01(a),
Then we can define a spatial * isomorphism
Va E M. 03
by the operator u, and 0 =
(1)3 0 0 2 O 0 1For general case, write
K = E EDK,,
Ki = NnI ,V 1.
i
Let q't be the projection from K onto Ki ; then q't E NI, V/. For each 1,0 1 = q11 0 is a normal * homomorphism from M onto Ni = Nql. By the above argument 0 1 = 441) 00 (21) o0 (11) , V/. Define 0 i = E 0010 , i = 1,2,3. Then 0 = 0 3 0020411 ) 1
and 01 7 02 ) 03 satisfy our conditions.
Q.E.D.
Proposition 1.12.5. Let 0 be a * isomorphism from a VN algebra M onto a VN algebra N. Then there exists a VN algebra V and two projections p', q' E V' with c(p 1 ) = e(q 1) = 1 such that M, N are spatially * isomorphic to Vql, VP' respectively, and 0 corresponds to the * isomorphism: ve —> vp'(Vv E V).
Proof.
Keep the notations in Theorem 1.12.4, and let V = MWV1 L . Since 0, 0 1 , 0 3 are * isomorphisms, it follows that 02 is also a * isomorphism. By Proposition 1.5.10, the central cover of pl in V' is 1. Further from Theorem 1.12.4, N is spatial * isomorphic to Vpi. Let q' = 1 H 0 q, where q is a rank one projection on L. Then q' E V'. Since V 1 q1 (H 0 L) D H 0 B(L)qL, it follows that the central cover of q' in V' is 1. Clearly, M is spatially * isomorphic to V q l , and 0 corresponds to the * isomorphism: vq' —* vpl (Vv E V). Q.E.D. Let Mi , Ni be the VN algebras on the Hilbert spaces Hi , Ki respectively, 0 i be a normal * homomorhism from Mi onto Ni ,i = 1,2. Then there exists unique normal * homomorphism 0 from MOM onto N1 ON2 such that
Theorem 1.12.6.
0 (a i 0 a2 ) = 01(al) 0 0 2(a2),
Va l E MI, a2 E
M2.
63
Moreover, if (D i and 412 are * isomorphisms (or spatial * isomorphisms),then 41 is also a * isomiphism (or spatial * isomorphism). Proof.
By Theorem 1.12.4, we can write (Di (ai ) = ui (ai 0 1i )pli u,7 1 , Va iE Mi ,
where 1i is the identity operator on a Hilbert space L i , p: E (MiTgrl i ) / , ui is a unitary operator from p:(Hi 0 Li ) onto K,2* = 1,2. Let L = L 1 (8) L2. Then
PI = pi 0 p2 E (M1 6:)011) 165(M2 0012) 1 = (Wg5M2 00 1L) 1 and u = u1 0 u2 is a unitary operator from p'i (H1 0 L1) 0 P 12( 112 0 L2) = )9'W ' 0 H2 0 L) onto K1 0 K2. Now define
(I)(a) = u(a 0 1L)p lu', Va E M1 6M2.
Clearly. (1) is a normal * homomorphism from MI 0M2 onto u(MI0M204T 1 L)P1 1 such that u (D (ai 0 a2) — (1)1(al) 0 (1)2(a2),
Val E MI,
a2 E M2 .
Since (I) is normal and MI0M2 is generated by {a i 0a2 I a l E MI , a2 E M2), it follows that u(MC6M2001L)pl u 1 will be generated by {(1)1 (al) 0 (1)2(a2)1al E MI, a2 E M2). Therefore,
N10 N2 = u(MIT§M2 0 Œ1L)p lu 1 , i.e., 4:11 is the normal * homomorphism from WW2 onto N1 ON2 which we want to find. From the normality, obviously (1) is unique. Now suppose that Il i is a * isomorphism. Then c(psi ) = 1 by the proof of Proposition 1.12.5, i = 1,2. Further c(p') = 1. By Proposition 1.5.10, ID is also a * isomorphism. Q.E.D. References. [28], [109], [113], [185].
1.13 Comparison of cyclic projections and spatial * iso
morphic theorem Definition 1.13.1. Let M be a VN algebra on a Hilbert space H. For each E H, we denote by Pe (resp. pie ) the projection from H onto M'e (resp. and call it the cyclic projection of M (resp. M') determined by .
ma
e
64
Theorem 1.13.2. Let M be a VN algebra on a Hilbert space H, e, ri E H. Then the relations p,,, < in (in M) and p'n < p'e (in M') are equivalent. Proof.
Let p',7 < i9' (in M'), i.e. there is a u' E M' such that U
t*
U
t
= p ,7, I
uIte* < pIe .
Clearly, u'u' s = P ite ro Pri = Putri. Replacing ri by ter), we may assume that ri E /t//. Let p(a) = (cm, ),Va E M, and (p. = Rv w be the polar decomposition of ça. Then w = RieV5S0 = RveÇo, and (ti — vv*q) E (MO L . On the other hand, 77 E Itl,(77 — vv*ri) E M Thus
e
ri = vv*ri.
(1)
Since pg = and w = Re v), it follows that w = Lp4 w. In addition w > 0, so w = Rp o.), i.e., (v*rhae) = (inv*ri,ae), Va E M. Thus (v*ri — pe v*ri) E
(Me)'.
But ri G itl,(v*ri — p e vsq) E iti, so we have v*77 E p e H = M'.
(2)
By (2), v*Miri = M'v*ri C M'e; by (1), vv*Msri = M'ri. It follows that v*pn is a partial isometry from Mr/ into Ape. Therefore, pn < pe (in M). Similarly, we can prove the converse. Q.E.D. Definition 1.13.3. Let M be a VN algebra on a Hilbert space H. A vector of H is said to be cyclic for M, if 11/1 = H (see Lemma 1.4.9). A vector ri of H is said to be separating for M, if a E M is such that ari = 0, then a = O. It is easy to see that a vector ri of H is separating for M if and only if 7i is cyclic for M'.
e
e
Proposition 1.13.4. Let M be a VN algebra on a Hilbert space H, be a cyclic vector for M, and ri be a separating vector for M. Then there is a vector g of H such that g is cyclic and separating for M. Proof. Since p'e = 1 > pin , it follows from Theorem 1.13.2 that Ai .< p e (in M). But Ai = 1, so by Proposition 1.5.3 we have in — pro i.e., there exists V E M such that v*v = in, vv* = p,, = 1. Let g = v. Then /1/Pg = v/i/Pe = H, and Mg D Mv*ve = me H, i.e., Mg = H. Therefore, g. is cyclic and separating for M. Q.E.D.
=
65
Theorem 1.13.5. Let ivi, be a VN algebra on a Hilbert space Hi , i = 1,2. Suppose that M1 and M2 admit a cyclic and separating vector. Then every * isomorphism ID from M1 onto M2 is spatial.
Proof.
By Proposition 1.12.3, (I) is normal. Thus
M = {a e (1)(a) 1 a E
MI}
is a VN algebra on H1 e H2. Let pli be the projection from (H 1 e H2 ) onto Hi,i = 1,2. Clearly,
PiDAE M i ,
Mpil
= M1)
Mp12 = (1) (M1) = M2.
Now by Proposition 1.5.2, it suffices to show that p ii , p'2 in M'. Let 6(E Hi) be the cyclicseparating vector for Mi , i = 1,2. Then p'i ( H1 e H2 ) = Hi = Mi 6 = M6,1 = 1,2. From Theorem 1.13.2, it suffices to show that p i — p2 in M, where pi is the projection from (H1 H2 ) onto M' 1 ,1 = 1,2. Now we prove a more stronger result: p i = p2 = 1. In fact, since
{4 e a12 1 di E MI, i = 1,2} C Mi , M,= Hi , i = 1, 2, it follows that pi > p'i , (1 — pi)pli = 0,1= 1,2. From (1 — pi ) E Mi= 1,2, and the definition of M, there are a, b E M1 such that
a e (I)(a) = 1 — p i ,
b e 4i(b) = 1  P2.
Therefore, a = (1)(b) = 0. Further, a = b = 0 and Pi = p 2 = 1.
Q.E.D.
e
Proposition 1.13.6. Let M be a VN algebra on a Hilbert space H, (E H) be a separating vector for M. Then for any normal positive linear functional io on M, there is ri E Me such that ço(a) = (aq, ri), Va E M. In consequence, we have cr(M,M.) — (weak topology 1M), s(M,M,$ ) — (strong topology 1M),
8*(M,M.) — (strong * topology 1M).
Proof.
Let irv,, Hv,} be the * representation of M generated by ço, 7/ = H ED Hr , and ig = {a = a e 7rv (a) 1 a E M}. By Proposition 1.8.13, ftl is a VN algebra on g. Put {
{Ma) = (alv,, IÇO) = (w (a)1 v , 110 ) = ço(a), Va E M, where a = a e rv,(a),1,„ = (0,1 v). Let .' = (e,o) E Ti. Obviously, t' is a separating vector for M, 50 iil''' = 17 D /411r . By Theorem 1.13.2, 71,, ‹ in
.A7'.
Since 74 'f/"." = Mir = irv (M)1 p = Hv„ it follows that p',, = p'v, is the 1, to
66
projection from Ti onto H. Then there exists I, E 14' such that v'v' = and v'v' < p'z. Noticing that pp' = M = (Me, o), we can write
v'Tv, =
(rh o)
=
for some ri e M. Further,
(alvI iv,) = (ci
v v"v 1 19 ) ,
(aviiv„ viiv,) = 05, 77) =
(ari ,
Va E M. Q.E.D.
Corollary
Let M be a VN algebra on a Hilbert space II. If M e H such that admits a separating vector, then for any io E M there are ç(a) = (ae,n),Va E M. Proof.
1.13.7.
It is obvious by the polar decomposition of (p.
Q.E.D.
The main results (1.13.2, 1.13.5 and 1.13.6) in this section are the socalled spatial theory of Von Neumann algebras. It is due to F. Murray and J. Von Neumann. B.J. Vowden once gave another proof. The present proof here is taken from R. Herman and M. Takesaki. Notes.
References.[69], [111], [177], [192].
1.14. 7Finite Von Neumann algebras Definition
1.14.1. A VN algebra M is said to be afinite (or countably decomposable), if every family of nonzero pairwise orthogonal projections of
M must be countable. Let M be a VN algebra on a Hilbert space H. Then the following statements are equivalent: 1) M is afinite; 2) M admits a separating sequence of vectors { „}(c H), i.e. if a E M with aen = 0, Vn, then a = 0; 3) M' admits a cyclic sequence of vectors {n„}(c H), i.e., [(z i n n ia' E M' ,n] is dense in H; 4) There is a faithful normal positive linear functional on M. Proposition 1.14.2.
67
Proof. Since a sequence {en} is separating for M if and only if {en } is cyclic for .0, it follows that 2) and 3) are equivalent. Now let {en } be separating for M, and {Pi}/EA be an orthogonal family of projections of M. Since \.‘ I1Pienii 2 < iien11 2 )Vn) it follows that there is lEA
a countable subset J of A such that pie n = 0,Vn and 1 V J. From {en } is separating for M, then pi = 0, V/ V J. Thus M is afinite. Suppose that M is afinite. Write
H=
EEDH„ Hi = M1 771 = p i H, Vi. i
Clearly, {/Ipi 0 0} is countable. Thus H =
E ED.Wri n ,
and the sequence {ti n }
n
is cyclic for M'. Therefore 1), 2) and 3) are equivalent. Suppose that there is a faithful normal positive linear functional io on M. Let {pi}lEA be any orthogonal family of projections of M. Since > ça(pl) = icA. (p(E pi) < oo, it follows that there is a countable subset J of A such that LEA
ça(p i ) = 0,V1 V J. Then pi = 0, Vi V J, i.e., M is afinite. Finally, if M is afinite, then M admits a separating sequence vectors. We may assume that E II en11 2 < oo, and let
{ }
of
n
ça(a) = Da en , rt
)
1
Va E M.
n
Clearly, ça will be a faithful normal positive linear functional on M.
Q.E.D. Proposition 1.14.3. Let M be a VN algebra on a Hilbert space H. If M. is separable, then M is afinite. In particular, if H is separable, then M is afinite.
Proof. Let {ço n } be a countable dense subset of {i,b E M. i 0 ? 0 )11011 1 }, and ça = E 2n io n . Clearly, 0 < ça E M.. If a E M+ is such that {,o(a) = 0, n
then çon (a) = 0,Vn. Further, 0(a) = 0,V0 < t,b E M.. By Corollary 1.9.9, a = O. Therefore io is faithful, and M is afinite. Now let H be separable. We prove that M. is separable. In fact, for a countable dense subset of H, Let n,,(a) = (aen , em,),Va E M, n,m. If {C4)nm I n,m} is not dense in M., then there exists 0 0 a E M such that G.) = 0, Vn, m. Since { n } is dense in H, it comn (a) = 0, Vn, m, i.e., (a follows that a = O. This is a contradiction. So {co nm, I n, m} is dense in M.. {
en}
Q.E.D.
68
An algebra A is called abelian or commutative, if ab = ba,Va,b E A.
Proposition 1.14.4.
Let M be a afinite abelian VN algebra on a Hilbert space H. Then M admits a separating vector. From the proof of 1.14.2, we can write
Proof.
H=
>
63H, Hn = pH = Mi?i n ,Vn.
n
Then {n r
,}
E 11774 2
O. Define a continuous function f on [ 11h11, IIhII] such tEfl that J(0) = 0, f (A) = A1, VA E L  11h11) 114 1\( 6 )e), and pick a sequence {pn (.)} of polynomials with a zero constant term such that max{ i pn(x) — f(A) 1 1 — 1 1h11 5 A 5. 11 h 111 —> O. Then
IIP(h) h(t)) — h(t)  1 I ( n — h i ii = ntacanx IPn i —> 0. 11h714)_ mil I P" (A) — f (A)
Q.E.D.
Proposition 2.1.6.
Let A be a C*algebra with an identity 1, B be a C*subalgebra of A and 1 E B. Then for any b E B , 693(0
Proof.
By Lemma 2.1.5, aB(b*b) = cf,t(b*b),aB(bb*) = (7,1(bb*). Thus if b is invertible in A, then b*b and bb* are invertible in B. So b has a left and a right inverses in B. Therefore, b is invertible in B. Q.E.D.
74
Proposition 2.1.7. Let A be a Csalgebra with an identity 1, u be a unitary element of A (i.e., usu = uu* = 1). Then ci(u) C {À E 01 PI = 1}. Moreover, A is the linear span of all unitary elements of A. From Theorem 2.1.4 and Proposition 1.3.4, it is easy.
Proof.
Q.E.D.
Proposition 2.1.8. Let A be a Csalgebra with an identity 1, a be a B be the C*subalgebra generated by normal element of A (i.e., 42*a. = C(a(a)), and the function corresponding to a is a(À) = {1, a}. Then B A, VA E a(a). In particular, 11a11 = v(a).
a(t) is injective Proof. C(11). Clearly, the map t By Theorem 2.1.4, B and continuous from II onto c(a). Since I/ and c(a) are compact, it follows Q.E.D. c(a). that f2 Proposition 2.1.9. on A such that
Let A be an abelian Csalgebra, 11'11 I be another norm
Ilab III 5 Hall' 011 1) Then 11x11 Proof.
Va, b E A.
114 1,Vx E A. We may assume that A has an identity. Then A
fl = {t E f2 1 if {x„} C A and 11xn11
CM. Let
0, then x(t)
0}.
We claim that II I is dense in f2. In fact, suppose that there is a nonempty open subset U of f2 such that 1 nU = O. Then we have a e A with a(t) = 1, Vt E and a(s) = 0, Vs E U. Let A 1 the completion of (A, 11' III). If a is not invertible in A1 , then there is a nonzero multiplicative linear functional p on A 1 such that p(a) = O. Let plA = t. Then t E SI ' . Thus a(t) = 1 and p(a) = a(t) = O. This is a contradiction. So a is invertible in A l . On the other hand, there is b E A such that suppb() C U. Then ab = O. It is impossible since a is invertible in Al. Therefore fi 1 is dense in n. Finally, let X be the spectral space of A l . Then for any z E A,
max{IP(x)I I P E X}
= max
lx(t)I = max lx(t)I
11x11. Q.E.D.
Proposition 2.1.10. such that
Ilablli
Let A be a Csalgebra 11 111 be another norm on A
Ilbll 1)
Ila* alll = 11a111)
Va, b E A.
75
Then Ilail = Ilalli,Va E A.
Proof.
By Proposition 2.1.2, we may assume that A has an identity. Let h = h* E A, and B be the C*subalgebra generated by {1,h},B 1 be the completion of (B, ill . Il l ). Suppose that 11, and SZ are the spectral spaces of B1 and B respectively. By the proof of Proposition 2.1.9) {(PIB) I P E Il i } is dense in 0. Thus OH 1 'ITELVIcIP(h)1 ' Vox Ih(t)I ' IIhII. Further, for any a
e A, 'lair
= Ila * all ' Ila * alli = 114) i.e., IA = IlalliQ.E.D.
Notes.
Theorem 2.1.4 is due to I.M. Gelfand. The aim of the theory of Csalgebras is to understand a uniformly closed selfadjoint algebra of operators on a Hilbert space (concrete Csalgebra). A Banach * algebra satisfying the axioms in Definition 2.1.1 is still sometimes called a /3 4 algebra. The name Csalgebra was coined by I.E. Segal. Presumably the C is meant to indicate that a Csalgebra is a noncommucative analogue of C(0), whereas the * recalls the importance of the *operation.
References. [51], [81].
2.2. Positive cones of C*algebras Definition 2.2.1. Let A be a C*algebra. An element a of A is said to be positive, denoted by a > 0, if a* = a, and a(a) c 11T + = [0, oo). We shall denote by A + the subset of all positive elements of A. Proposition 2.2.2. Let A be a C 4 algebra, le = h E A. Then there are unique h+ , h_ G A+ such that
h = h+ — h_, h+ h_ = O. Proof. We may assume that A has an identity 1. Let B be the abelian C 4 subalgebra generated by {1, h}. By Theorem 2.1.4, there are h+ ,h_ E B+ such that h = h+ — h_ and h+ • h_ = O. From Proposition 2.1.6, h+ , h_ E A. Now if there exist other hi+ , h' E A + such that h = h'+ — h' and h'+  h' = 0, then the set h, h'+ , h' } is commutative. Let C be the abelian C 4subalgebra generated by {1, h, h'+ , h'_} . Clearly C D B. Using Theorem 2.1.4 for C, we can see that le+ = h+ , h' = h_. Q.E.D. {
76
Proposition 2.2.3.
Let A be a Cealgebra with an identity 1, and h h* E A with 11h11 < 1. Then h > 0 if and only if Ill — hil < 1 
Proof.
Using Theorem 2.1.4 for the abelian Cesubalgebra generated by {1,0, it is easy to get the conclusion. Q.E.D.
Proposition 2.2.4.
Let A be a Cealgebra. Then A + is a cone, i.e., if a, b E A+ , we have (a + b) E A. Moreover, A + n (—A+ ) = {0}. We may assume that A has an identity 1. By Proposition 2.2.3,
Proof.
a
II < 1 ) —
Ill
—
Then
Ill
a+ b hall +
Ilall Ill II 5
a II +
' Ill
Ilall + Ilbll
Again by Proposition 2.2.3, (a + b) E A. Moreover, if h E A+ n (—A + ), then a(h) h = 0, i e., A+ n (—A + ) = {0}.
11 Ilbll . < 1.
{0}. From Proposition 2.1.3,
,
Q.E.D.
We introduce a partial order ">" in AH = {h E A I h* Remark. i.e., a? b if (a — b) > O.
h},
Proposition 2.2.5. Let A be a Cealgebra, and a E A. Then there exists unique al /2 c A + such that a l /2 • a = a • al /2 and (a 1 /2 ) 2 z= a. Moreover, this a l /2 can be approximated arbitrarily by the polynomials of a with zero constant terms. Proof.
Q.E.D.
It is similar to Proposition 2.2.2.
Lemma 2.2.6.
Let B be an algebra with an identity, and a, b E B. Then ()lab) U {0} = cr(ba) Li fol. Proof.
Let 0
A V ()(ab), and u = (ab (bua — 1) (ba — A) = (ba
it follows that Oa — A) is invertible.
A)'. Since ))  (bua — 1) = A,
Q.E.D.
77
Proposition 2.2.7. Let A be a C'algebra, and a E A. Then a E A+ if and only if there exists b E A such that a = b* b. The necessity is clear by Proposition 2.2.5. Now let a = b* b. Clearly a* = a. By Proposition 2.2.2 and 2.2.5, we can write a = u2 — v 2 , where u,v E A+ and uv = O. Then
Proof.
(bv)*(bv) = vav = —v 4 < 0.
(1)
Let by = h + ik, where h* = h, k* = k. Then by Proposition 2.2.4, (bv)(bv)* = —(bv)* (bv) + 2 ( h2 ± k2) = y 4 + 2 (h 2 + k2) > 0.
(2)
From (1), (2) and Lemma 2.2.6, (bv)* (bv) .v 4 E A+ n( A + ) = {o}. Further Q.E.D. by Proposition 2.2.5, v = 0, and a =. u2 E A. —
Let A be a Csalgebra on a Hilbert space H (i.e. A Proposition 2.2.8. is a uniformly closed * subalgebra of B(H)), and a E A. Then a E A+ if and only if a is a positive operator on H.
Proof.
The necessity is clear from Proposition 2.2.7. Now let a(E A) be a positive operator on H. At least we have a* = a. By Proposition 2.2.2, a = a+ — a_, where a+ , a_ E A+ and a+  a_ = O. For any E H,
e
0 < (aa_
e, a_ ) = Hal e) 5. O.
Therefore, a3 = 0, a_ = 0, and a = a+ E A+ .
Q.E.D.
Proposition 2.2.9. Let A be a C*algebra. 1) If a, b E A+ and a < b, then Hail < Il_h ill, and c* ac < c*bc,Ve E A. 2) A+ is a closed subset of A. 3) If A has an identity, a, b E A + , a < b,and a, b are invertible, then b1 < a1 .
Proof.
1) we may assume that A has an identity 1. Then 0 < a < b < 114 1. Using Theorem 2.1.4 for the abelian C*subalgebra generated by {1, a} , we can get that Mail < libil. From Proposition 2.2.7, it is clear that c* ac < c* bc,Ve E A. 2) Let {an } C A + , and an —> a. We need to prove a E A+ . Clearly, a* = a. Write a = a+ —a_, where a+ , a_ E A+ and a+ a_ = O. Let bn = a_ an a_ E A+ . 0. Then bn —> b = a_aa_ = —a 3 < O. Further 0 < —b < b n — b < Ilb n — NI Thus b = 0, a... = 0, and a = a+ E A+ . 3) Since (a 1)1/2(1, _ a)(a1)1/2 > 0, it follows that (a 1 ) 1 /2 b(a1 ) 1/2 > 1. By the function representation, a i/2bl a i/2 < 1 = ai/2alai/2. Therefore, Q.E.D.
78
Proposition 2.2.10. Let A be a Cealgebra, a, b E A + and a < b. Then for any A E [0 1 1], we have a)* < bx , where aA ,bA are defined as in Proposition
2.1.8. We may assume that A has an identity 1. First, suppose that a, b are invertible. Let E = {A E [0,1] I aA < bx}. Since A + is closed, it follows that E is closed. Clearly, 0,1 E E. So it suffices to show that for any A,ji E E, we have E E. Fix X, G E. Since
Proof.
b  2aq  2
b  4aA b  4 x and z> y.
Proposition 2.2.11.
Proof.
Consider the problem in (Ai0) . Let a= x(1 — x) 4 ,b = y(1 — y) 1 , z = (a + b)(1 +
a+
=1—
Then a, b,z E A + , and z E Int(S) nA+ . The inequality z > z is equivalent to the following inequality
1 1 ( 1 + a+ b) 1 >
or
(1—
X)
> ( —21 + a + br l .
22 By Proposition 2.2.9, the latter is equivalent to the
(1 — x)  1 < ( 1 + 2a + 2b). It is clear that (1+2a) > (1— x) 1 by the definition of a. Thus (1+ 2a+2b) > Q.E.D. (1 — x ) l. So z> z. Similarly, z > y.
79
Proposition 2.2.4 is due to M. Fukamiya, IL. Kelley and R.L. Vaught. Proposition 2,2,7 is due to I. Kaplansky. These results are the key lemmas for the GelfandNaimark conjecture (see Section 2.14). Proposition 2.2.10 is known as the LiiwnerHeinz theorem, and the proof here is due to Notes.
G.K. Pedersen. References. [50], [81], [90], [126], [177].
2.3. States and the GelfandNaimarkSegal construction Let A be a C*algebra, p be a linear functional on A. p is said to be positive, denoted by p > 0, if p(a) > 0,Va E A + .p is called a state if p > 0 and ilPli = 1. The set of states of A is denoted by S(A). Clearly, if p > 0, then we have p(a) , p(a),Va E A; and the Schwartz inequality: IP(b * ( )1 2 n2 , Vn. By Proposition 2.2.9, A+ is closed. Thus x =
E lxr, E n
positive integer N, N1
N
0 for any positive element (a+ A) Proof. of (A210).
81
Since (a + A) > 0 in 040, it follow that a* = a,X = A. Let B be the abelian Cesubalgebra of (A jŒ) generated by {1, a}. Then B'=# C(1I), and A Fa(t) _> 0,Vt G O. Since a is not invertible in (A 4), and o./3 (a) = sotheri o e SI such that a(t o ) = O. Thus A = A + a(t o) > O. 0 If a > 0 or IIPil = 0, then p(a) + Atio > 0 immediately. Now let p and a = a+ — a_ with a_ 5Z 0, where a+ , a_ G A+ and a+  a_ = O. Clearly inf{a(t) I t E fn = — I a_ I . Further, 

0 < A + inf { a(t)
I
t G
n} =
A — IlaII
5. A — IIPII 1 P(a) 5 IIPII i Nzo + p(a) — p(a_)} i.e. p(a) + Ap.0 > O.
Q.E.D.
Corollary 2.3.6. Let p be a state on a C*algebra A. Define # (a + A) = p(a) } A,Va E A, A E C. Then "P is a state on (AiŒ), and (IA) = p.  
Proposition 2.3.7. Let A be a Cealgebra. Then the state space S(A) of A is a convex subset of A*. Proof. For any so10,02 E 8 (A ) and A G (0,1), we need to show that XVI + (1 A)ço 2 E 8(A). For any e > 0, pick a G A+ with Hall < 1 such that çoi (a) > 1— E. Now by Proposition 2.3.4, —
1> 114)1+ (1— A)c 0 211 = sup{Aço i (b) + (1 — 4,02(b) I b
> Aio l (a) + (1— A) sup{ço2(b) > A(1 — E) + (1
—
l
e A+, iibil
b E A + , libil I
1, b _.> a} 1,b ? a}
A).
Since e is arbitrary, it follows that Açoi + (1
—
4,o2 G 8(A).
Q.E.D.
Definition 2.3.8. Let A be a Cealgebra, 8(A) be its state space. Each extreme point of 5(A) is called a pure state on A. The set of pure states is denoted by P(A).
Proposition 2.3.9. Let A be a Cealgebra. 1) For p E P(A), let ga + A) = p(a) + A,Va E A, A E OE. Then p' is a pure state on (A10). 2) If 0" is a pure state on (AkV), p = (p1A), then either p = 0 or p G P(A). 

1) Suppose that there are two states 0 17 .#2 on (A jC) and A E (0,1) Proof. such that 0 "= AP1 +(1A) 152 . Let pi = (ilA),i = 1,2. Since IIPIII 5._ 1) Ilp2 II 5_ 1
82
and 1
= MPH 5 AllPill + ( 1 A)I1P211
11
it follows that pi E S(A),i = 1,2. By p E P(A) and p = Ap i + (1 )t)p2 on A, we have p = Pi = P2. Moreover, # (1) = 0 1( 1) = A  (1)  1. Thus 0" = d 'I = d 2) i.e., ' is a pure state on (100). 0. Let lip,II = p,o < 1. By Proposition 2.3.5, ar 2) Suppose that p is a state on (AŒ), where 6  (a + A) = 11,6 1 p(a) + A,Va E Adt E Œ. Let  o is a (pure) state on (A40). 60(a + A) , A(Va E A, A E 0). Obviously, d Moreover, we have the equality: 
It is a contradiction since P" is pure. Therefore IIPII = 1, i.e. p E S (A). Now let pi, P2 C S(A) and A c (0,1) be such that p = Api+ (1— X)p 2 . p i and p2 can be naturally extended to states # 1 and 02 '. on (A40). Clearly, 'd = VI+ (1— A ) 62. Since P' is pure, it follows that ' = d i ' 0' '2 and p = pi = P2Therefore p E P(A). Q.E.D.
Proposition 2.3.10.
Let A be an abelian C*algebra. Then p E P(A) if and only if p is a nonzero multiplicative linear functional on A. By Proposi,tion 2.3.9, we may assume that A has an identity 1. Let p E P(A). Since A = [A+ ], it suffices to show that p(ab) = p(a)p(b) for any a,b e A. If p(a) = 0, then by the Schwartz inequality, Proof.
i p ( ao r = i p ( a 1/2 . a l/20 1 2 < p (a) p (bab) = 0.
Thus p(ab) = 0 = p(a)p(b). Now we may assume that 0 < a < 1, 0 < p(a) < 1.
Let pi() 7= P(a) 1 P(a . ),p2( . ) = P(1 — a)_ i P((1 a).). Clearly, PIIP2 E S(A), and p == p(a)Pi + ( 1 — P(a))P2. Since p is pure, it follows that p =7 pi  P2—
Therefore p(ab)  p (a) p (b) . Conversely, let p be a nonzero multiplicative linear functional on A. If A ''2,' C ( I), then there is to E 11 such that p (a) = a(to) =
L
a(t)clb to (t), Va E A.
By the Riesz representation theorem, each state on A corresponds to a probability measure on a Therefore p E P(A) since p corresponds to a point Q.E.D. measure.
83
There is another direct proof (without using the Riesz representative theorem) as follows. Suppose that p is a nonzero multiplicative linear functional on A. Clearly p E S(A). Let pi) P2 E S(A) and A E (0,1) such that Remark.
P = API ± (1  X) i9 2 . For any h : h* E A,
API (h2 ) + (1 — A)P2(h 2 ) = p(h 2 ) = p(h) 2 77*
[API (h) + (1 — A)p 2 (h)] 2 .
By the Schwartz inequality, p i (h) 2 < pi (h2 ),i = 1,2. Then 0 , —[Ap i (h) + (1 — A)p2 (h)1 2 + API (h2) + (1 — A) p2 (h2 )
> —EV p i (h) 2 + (1 — X) 2 P2(h) 2 + 2A(1 — A)p i (h)p 2 (h)] + Ap i (h) 2 + (1 — A) p 2 (h) 2 = A(1 — MP1(h) — p2(h)] 2 .
Thus p i (h) , p 2 (h),Vh = h* E A. Further p = Pi =
P2
and p E P(A).
Proposition 2.3.11. Let A be a C*algebra with an identity 1, E be a * linear subspace (i.e., if a E E, then a* E E) of A and 1 E E. Let f is a state on E, i.e., f is a linear functional on E, 1 7 =If 1 f (a*) = f (a), Va E E, f (b) ?.._ 0,Vb E E n A + , and f (1) = 1 f '
Then we have that: 1) Each element of I can be extended to a state on A; 2) Each extreme point of I can be extended to a pure state on A. Proof. 1) Let f El. For any h , h* V E, since —1144 < h < llhll 4,1 G E, and A+ is a cone, we can define f (h) such that
sup{ f (b) lb = b* E E ,b < h} < 1(h) _< inf{f (e) I c , c* G E, c > h} .
Then f is a state on E 1[h]. In fact, for (a+ Ah) E A + , where a E E,A EC, we need to prove f (a + Ah) > 0. Clearly a* = a, A = A. When A = 0, it is clear. If A> 0, then h > —A la. By the definition of f (h), f (h) > —A 1 f (a), i.e., f (a + Ah) > 0. If A 0. By the above procedure and the Zorn lemma, f can be extended to a state on A.
84
2) Let f be an extreme point of I. Put
r = {p E S(A)
I
(plE) = f}.
By 1), .0 O. It is easy to see that f is a weak * compact convex subset of A. From the KreinMilmann theorem, f has an extreme point p at least. Now it suffices to show p E P(A). Let p 1 ,p 2 E S(A) and A E (0,1) be such that p = Api + (1 )t)p 2 . Clearly, L 7 . (p i lE) E 7,1  1,2. Since Ail + (1 — A)f2 = (plE) = f and f is an extreme point of 7, it follows that f  fi  12. Thus PI)P2 E L. Since p is an extreme point of AP 7 Pi 7 P2Therefore p E P(A). Q.E.D. 

Corollary 2.3.12. Let A be a C*algebra, and B be a C*subalgebra of A. Then each state (or pure state) on B can be extended to a state (or pure state) on A. Each state (or pure state) on B can be extended to a state (or pure state) on (/344;) (see 2.3.6 and 2.3.9). Further it can be extended to a state (or pure state) on (A+Œ) (see 2.3.11). By Proposition 2.3.9, its restriction is still a state (or pure state) on A. Q.E.D.
Proof.
Suppose that A has an identity 1, and 1 E B,io E S (B ). By Proposition 2.3.3, each extension of ço preserving the norm is a state on A.
Remark.
Proposition 2.3.13. Let A be a C*algebra, and h , h* E A. If 0 a(h), then there is a pure state p on A such that p(h) = A.
AE
By Proposition 2.3.9, we may assume that A has an identity 1. Let Proof. B2d c(n) be the abelian Csubalgebra generated by {1, h}. Then there is t E I/ such that h(t) = A. Define 1(b) = b(t),Vb E B. By Proposition 2.3.10, f is pure state on B. By Corollary 2.3.12, f can be extended to a pure state p on A. Clearly, p(h) = f (h) = h(t) = A. Q.E.D.
Remark. necessary.
If A itself has an identity, then the condition "A
0" is not
Corollary 2.3.14. Let A be a C*algebra, and h = h* E A. Then there is a pure state p on A such that Ip(h)1= 11h11. Consequently, 11h11 = supflp(h)1 1 p E
P(A)}. Corollary 2.3.15. Let A be a C'algebra, and a E A such that p(a) > 0,Vp E P(A). Then a E A.
85
It suffices to show a* = a. If a2. :.I. 0 0, then there is p E P(A) such that p(a4) 0 O. Thus p(a) V IR, a contradiction. Therefore a* = a. Q.E.D. Proof.
The GNS construction was once discussed in Section 1.8. The same procedure can be carried for the C*algebras. Due to its importance, we shall study it again in detail. Definition 2.3.16. Let A be a C*algebra. {7r, H} is called a * representation of A, if r is a * homomorphism from A into B(H), where H is a Hilbert space, i.e., r(Aa + Ab) = Ar(a) + pir(b), r(ab) = 7r(a)r(b), 7r(a*) = Va, b E A, A, p, E Œ. If there is a vector E H such that ir(A) = H, then is called a cyclic vector for {r, H}, and {7r, HI is said to be cyclic. The * representation {7r, H} of A is said to be faithful, if 7r(a) = 0 implies a = O. Two * representations {r i , H1 } and {7r2 , H2} of A are unitarily equivalent, denoted by {ri , Hi } ''=" {7r2 , H2 }, if there is a unitary operator u from Hi onto H2 such that ur i (a)u 1 = 7r2 (a),Va E A.
e
Let { ir, H} be a * representation of a C'algebra A. Proposition 2.3.17. < 1, and r preserves order, i.e., 7r(A + ) C B(H) + . Moreover, if r is Then 114 — faithful, then r is isometric, and 7r 1 (7r(A) + ) = A+ . Consider (A40), and put 7111) = 1H. Thus we may assume that A has an identity 1, and 7r(1) = 1H. Then a (ir (a)) C a (a),V a E A. Further, Proof.
11 70)11 = sup{IAI I A E (f(r(h))} < sup{lAI I Aea (h)} = PM)
V h = h* E A.
Hence 117r(a)II =11 7 (a * a)11 1/2 0 *(4 1/2 = Ilall,Va E A, i.e.)114 < 1. Clearly, 7r(A+ ) C B(H) + . Now let 7r be faithful. If there is e E A such that 7r(e) = 1H, then e is just the identity of A. If 1H V 7r(A), considering (A4 ) and putting 7r(1) = 1H, then r is still faithful on (Ai0). In other words, we may assume that A has an identity 1, and 7r(1) = 1H. Now our conclusion can be obtained from the proof of Proposition 1.8.13. Q.E.D.
86
Let A be a C*algebra, and ça E S(A). Put 4, = {a E A I (g)(a a) = 0 } . 4, is called the left kernel of (p. By the Schwartz inequality, 4, is closed left
ideal of A. Let (Vac A)
be the quotient map from A onto A14. On A/L9 , define = io(b*a),Va,b E A.
Then (,) is welldefined, and is an inner product on AlL v . Denote by 14, the completion of (A/L 9 , (,)). For any a E A, define a linear map A9 (a) : AlL r —> AlL r as follows: rga)b p = (ab) PI
Vb E A.
Since b* a* ab < liaii2b*bl it follow that lirso(a)bp11 2 = ço(b* a*ab) < ilair • Ilbsoir
Therefore 79 (a) can be uniquely extended to a bounded linear operator on Ilr , still denoted by r9 (a). It is easy to show that {irv„1/9 } is a * representation of A.
V b E A.
Proposition 2.3.18.
Let A be a C*algebra, and io E S(A). 1) If {r9 ,1/9 } is the * representation of A generated by (g) as above, then {719 , H9 } admits a cyclic vector and can be chosen such that
e„,
= a,„, ço(a) = (r v,(a)e„, e9 ),
Va E A.
2) Let (7) be the natural extension of (g) on (AîŒ) (i.e., (Ma + A) = ça(a) + A,Va E A, A E 0), and lirs;,141 be the * representation of (A40) generated by (65. Then there is a unitary operator u from lir on 11‘; such that vir9 (a)u 1 = 7(a) r 5 Va E A. Define ua„ = az„Va E A. Then u can be extended to an isometry from H9 into lica. n 1. From the Schwartz inequality, ça(a) = Man) < (g)(a r,2 ) 1 / 2 < 1, it follows that ça(a) —> 1. Further y6((1 — a n) 2 ) —> 0, i.e., u(an ) 9 —> 1 Ç; in lice. Therefore, u is a unitary operator from I4, onto lis;. Moreover, since ur,p (a)b v, = u(ab) v, = (ab) ir(a)b ‘7, = rz,(a)ub v„Va,b E A, we have u 1 1. That comes to the un9 (a)u 1 = r(a),Va E A. Finally, pick P conclusion. Q.E.D. Proof.
e,, =
87
Let A be a C*algebra, and A be a subset of 5(A) Proposition 2.3.19. such that sup{v(a) I ço E A} = liall5Va E A. Then {76, =
E EDirv,, H6, = E e.r4} 
pEA
pEA
is a faithful * representation of A. Proof.
For any a E A, by Propositions 2.3.17 and 2.3.18 11a11 2 ? 11 7,6(a)11 2 = suP{Il 1r(a * a)11 I ça E A} _.> suPOr(a * a)e„, =
co)
i ço E
A}
sup{ ço(a*a) 1 it) E A} = 11a11 2 .
Q.E.D.
Therefore, 114 = 11 1rA(a)11,Va E A.
By Corollary 2.3.14, P (A) or any a(A*, A)dense subset of S(A) can be chosen as A. Remark.
Theorem 2.3.20. Any C*algebra can be isometrically * isomorphic to a concrete C*algebra on some Hilbert space. Proposition 2.3.21. Let A be a C*algebra, { ir, H} be a * representation of A. 1) If ir admits a cyclic vector e, let p(a) = (r(a)e,e),Va E A, then {irp, Hp} c'=d { 1r) H} 2) There exists A C 5(A) such that { r, 1/) is unitarily equivalent to the direct sum of some zero representation and {rv„14,}(ço E A).
= a,va E A. Then
u can be extended to a unitary operator from H onto Hp ; and it is easy to see that ur(a)u 1 = ro (a),Va E A. 2) By the Zorn lemma, we can write Proof.
1) Let tor(a)e
H=
E @Hi ED 1/0 , LEA
where Hi = r(A)eh and 11611 = 1,V/ E A, and Ho = {E H I Ir(a) = 0, Va E A}. For any 1 E A, let çoi (a) = (r(a)6, ei ),Va E A. Then A = {vi 1 1 E A} satisfies our condition. Q.E.D. The following proposition is a version of the RadonNikodym theorem. Proposition 2.3.22. Let ço, V) be two positive linear functionals on a C*algebra A, and ço < i,b (i.e. ça(a) < i,b(a),Va c A+ ). Then there exists
88
unique t' E 70 (4,0 < t' < 1, such that io(a) = (7 0 (a)t 1 ev„ ev,),Va E A, where { ro , Hip, Eip} is the cyclic * representation of A generated by 0 (as in Proposition 2.3.18). Proof.
On the dense subspace AiLii, of 1/0 , define la„ 41 = Eiro (a)ev„r v,(b)e ipi
= {,o(b* a),
Va, b E A.
Since („c) < i,b, it follows that l[aip,b11 < 11(411 • 11411, Va, b E A. Thus there is unique t' E B(H) such that
p(b* a) = (ti ro (a) „ ir,i, ( b)
),
Va, b E A.
Now by the proof of Lemma 1.10.1, we can get the conclusion.
Q.E.D.
Now we study the orthogonal (Jordan) decomposition of a herrniatian functional. Let A be a C*algebra, and X = { p e A* I p > 0 and IIPII < 1}. Clearly, X is a compact Hausdorff space with respect to a(A*, A). Denote by Cr (X) the set of all real continuous functions on X. For a E AH (the set of all selfadjoint elements of A), define a(p) = p(a),Vp E X; then a(.) E C,.(X). By 2.3.14 and 2.3.15, the map: a —> a(.) is isometric (i.e. OM = sup la(p)I)
pex and preserves order (i.e. a(.) > 0 if a E A + ) from AH into Cr (X), and also its inverse preserves the order (i.e. a E A+ if a(.) > 0). Suppose that f is a hermiatian continuous linear functional on A, i.e., f E A* and f* = f, where f* is defined by f*(a) = f(a*),Va E A. Then Ilfhl= II.f1 111/11. Put F(a(.)) = f (a), Va E AH. Clearly, F can be extended to a continuous linear functional on C,.(X), still denoted by F, with the same norm 11111. By the Riesz representation theorem, we can write F = F+ — F)
11111  11F11  11F+11 +
IIF11)
where F+ and F__ are positive on Cr (X). Restricting F+ ,11, to fa(.) I a E AH}, we get positive functionals 1+ , f_ on AH. Let f± (a + a) = f±(a) + if± (b),Va,b E AH. Then 1+ and f_ are positive on A, and f = f+ — f_. Moreover, since Ilf 11 — 11F11 = 11F+11 + 11F11 1 11F±11 ? Ilf±11) and 11111 = 111+ — 111 _._ 111+11 + Ilf11 5 it follows that 11111 = 111+11 + 11111. The above decomposition is called the orthogonal decomposition (or Jordan decomposition) of the herniation functional f. When A is commutative, it is the ordinary Jordan decomposition of a signed measure exactly. Now we prove that the above decomposition is unique. By Proposition 2.3.18, for each p E X, there is a cyclic * representation {rp , Hp , G} of A such that
p(a) = (r p (a)ep ,ep ), Va E A.
89
Let
=
eirp ,
=
pcx
pEx Then {7,H} is a faithful * representation of A. Let M = r(A)". Then M is a VN algebra on H. We may assume 11111 < 1; so f± E X. Write e± = ef±; then f± (a) = (r(a)e ± , ± ),Va E A. Identifying A with r(A), f and f± can be naturally extended as follows:
(be,
±), 1(b) = f+(b) f(b), Vb E M. Denote by 11111m) Ilf±Ilm the norms of f, f± as the functionals on M respectively. By the Kaplansky density theorem, we have II f± Il m = 111±11. Since it follows that 11111 S 11111m S II f+Ilm + 11f11M = 11 1+11 + IIIII = I I Ilm = f+ + IlfIlm. Now by Theorem 1.9.8, we obtain the following theorem.
=
Theorem 2.3.23. Let A be a C*algebra, f be a hermition continuous linear functional on A, i.e., f E A* and f (a*) = f (a),Va E A. Then there exist unique positive linear functionals 1+ and f_ on A such that
f = 1+ — Corollary 2.3.24.
and
11111 = 111+11 + III II.
Let A be a C*algebra. Then A* is the linear span of
S(A). Notes. The GNS construction was studied first by I.M. Gelfand and M.A. Naimark. Then I.E. Segal gave its perfected form.
References. [521, [1551.
2.4. Approximate identities and quotient C*algebras Proposition 2.4.1.
Let A be a C*algebra, and L be a left ideal of A. Then there is a net {d i } C L with di E A + ,11di 11 < 1, V1, and di < dit,V1 < 1',
such that lixdi
—÷ 0,Vx EL.
Let A be the set of all finite subsets of L. And A is directed by the Proof. inclusion relation. For any 1 = {x 1 , • 5 xn } G A, put
hi =
E x: xi , i=1
di = nhi (1
90
Clearly, hi, d1 E L n A + , and 1. P = {x Let 1 ,.. • , T,4,• ,xm} = L,1 y, it follows that
I Ix + All _?. II(x + A ) C1111 ?. 11(X + A ) C11Y11 ?. II X + A l  E for/ enough late. Therefore, Ilx+ All = lin II xdi + Adi ll and Ilx+All = llx * +All = lim Ilx*di i
+Xdill = limild i s + AcliiI)Vs E i 
A, A E C.
Q.E.D.
Let {r, II} be a * representation of a Cealgebra A. The closed linear span of {ir(a) I a E A, E H} is called the essential subspace of fir, H}. The * representation {r, HI of A is said to be nondegenerate, if its essential subspace is H. Clearly, the orthogonal complement of the essential subspace is the null subspace, Le.
Definition 2.4.5.
e
fir(a)e la E A,
E H} 1 = {n E HI Ir(a)n = 0,Va E A}. 
Therefore, the null subspace of a nondegenerate * representation is trivial. In this case, the weak closure of r(A) is a VN algebra on H (see Theorem 1.3.9).
Proposition 2.4.6. Let A be a Cealgebra, and {d i } be an approximate identity for A, {ir, H} be a * representation of A. Then r(di ) —> p (strongly), p is the projection from H onto the essential subspace [r(A)H] ofwher fir, H}. In particular, if {7, H} is nondegenerate, then r(d i ) > 1 (strongly). —
Proof.
By Proposition 1.2.10, we have r(d i )
=
> p=
—
sup
i
r(d i )
(strongly). Let
o, vi, and pri = 0. On the other MAW]. Then for any ri E K  L. ,r(d i )n hand, for any a E A, E H, since 117r(dia)e —, 0, it Ir(a)11 < lidia — all ' il follows that pr(a)e r(a)e. Therefore, pH = K. K=
e =
—
il
Q.E.D.
If fir, HI is a nondegenerate * representation of A, by 2.4.4 and 2.4.6, then for any E H with 1, E S(A). Remark.
e
Proposition 2.4.7. Then I* = fa* la E I}
'lei' = (.e, e)
Let I be a closed twosided ideal of a Cealgebra A.
= I.
92
Proof. By Proposition 2.4.1, there is a net d1 } c I such that adi —> a, Va E I. Then for any a E I, {
Ildia * — all = 11 (a di — a) * II = Hach — all —) o. Since di a* E ',V/ and I is closed, it follows that a* E /,Va E I.
Q.E.D.
Now let A be a C*algebra, and I be a closed twosided ideal of A. By Proposition 2.4.7, Ai/ is a Banach * algebra with respect to the quotient norm. Let {d i } be an approximate identity for I,anda—>a=a+Ibe the cononical map from A onto AR. We claim that
Ilall = lifnlladi— all, va
E A.
In fact, fix a E A. For any b E I, since bd i —> b, it follows that limilad i — all = limlladi — a + bd i — bil i i = limil(a + 0(1— c11)11 5 Ila + bll. i Thus limlladi — all < inf{Ila + bll l b E I} =
i adj E I, we have
Ilall.
lifnlladi — all ? liTML__ Hach — all
On the other hand, since
Ilall.
all) Va E A. Now for any a E A, b E /, since bd i —> b, we have
Therefore,
Ilall = lip iiadi —
adi r (a — ad i )II 11a11 2 = lim i Mach — a11 2 = li m i II(a —
= lim — doeao.— i 11( 1
din
b) (1 — di)II = lim i II ( 1 — di) (a*a +
Ma * a +b11.
Hence
1Ia112 < inf{llea + bll I
b E I} = lic7all
Hall • Pik
Va E A.
Furthermore, lia * all = 10112)Va E A. Therefore we have the following.
Theorem 2.4.8. Let A be a C*algebra, and I be a closed twosided ideal of A. Then Ai/ is a C*algebra in a natural way.
Proposition 2.4.9.
Let (1) be a * homomorphism from a C*algebra A into another C*algebra B. Then (I)(A) is a C*subalgebra of B. In particular, if
93
{r,1/ } is a * representation of a C*algebra A, then r(A) is a C*algebra on H.
By Theorem 2.3.20, it suffices to consider the case of {ir, H} and A. Proof. Let I = {a E A I r(a) = 0 } . Then I is a closed twosided ideal of A. Define Ff(a) = ir(a), Va E AI I and a E
a.
Clearly, {Ff, H} is a faithful * representation of the quotient C*algebra A/./. By Proposition 2.3.17, r(A) = Fr(A//) is a C*algebra on H. Q.E.D. Let A be a C*algebra, I be a closed twosided ideal Proposition 2.4.10. of A and B be a C*subalgebra of A. Then (B +1) = {(b c) I be B,c G I} is a C*subalgebra of A, and the C*algebras (B 1)/I and BAB n I) are canonically isomorphic. Let a —> a = a + / be the canonical map from A onto AR. Clearly, it is also a * homomorphism. By Proposition 2.4.9, B = {1" b E B} is a C*subalgebra of Ai/. It suffices to show that (B I) is closed. Let {x y,} C (B + I) and xr, —> x. Since E = B TiI is a C*subalgebra of A//, it follows that Then In + E h, i.e., s E (B /). Now it is easy to see that Proof.
'
b (13 n > b 4 I is a * isomorphicm from BAB n I) onto (B
(Vb E B) 1)/I.
Q.E.D.
Proposition 2.4.11. Let A be a C*algebra, and I be a closed twosided ideal of A. If p is a state (or pure state) on A and p(I) = {0}, let 0 (a) = p(a)(va E A//, a E a), then P' is a state (or pure state) on Ali. Conversely, if P' is a state (or pure state) on A//, then there is unique state (or pure state) p on A such that p(I) = {0} and p(a) = P(a),Va E A.
op)
Proof. Let p be a state on A with p(I) = {0}. Then we can define = p(a),Va E A//, a E a. Clearly, ' is a positive linear functional on A//, and 'In< 1. On the other hand,
1= suP{IP(a)I = suP{115(a)1
I
a E A , IIaII
1}
aE
= 1 and ' is a state on AII. Thus Conversely, let P" be a state on Ai/. Define p(a) = P'(a),V a E A. Clearly, p is a positive linear functional on A, and p(I) = {0}. From the preceding paragraph, we can see that IIPli s° IIPII = 1 and p is a state on A.
94
Now let p be a pure state on A with p(I) = {o}. By the preceding paragraph, P" is a state on AII. Suppose that there are states Ai, .#2 on Ai/ and A E (0,1) such that p= VI + (1 W2 . Define p i (a) = P"i (a),Va E A,i = 1,2. Then p i and p2 are two states on A with MI) = P2(I) = {0}. Clearly, P = AP' + (1 A)p2. Since p is pure, it follows that p = Pi = P2. Further P'^ = 0' '1 = P. Therefore ' is pure on A/./. Finally, let P" be a pure state on AR. Then there is unique state p on A such that p(I) = {0} and p(a) = 0 (a),Va E A. Suppose that there two states )t)p2. For any a E /r1 A+ , PI, P2 On A and A E (0 5 1) such that p = Ap i + (1 from p(a) = 0 we have pi(a) = p 2 (a) = 0. Further, MI) = p2(1) = { 0}. = pi (a),Va E Aii, a E Thus ' = A 0 ' + (1 — A ) 02 , where P"i is defined by a,i = 1,2. Since 0' is pure, it follows that P" = 0 1 = '02. Further p = pi = P2. Q.E.D. Therefore p is pure on A. —
—
ma)
I.E. Segal showed the existence of an approximate identity in a C* algebra.
Notes.
References. [251, [521, [811, [155], [156].
2.5. Extreme points of the unit ball and the existence of an identity Let A be a C*algebra, and S = {a E A I Mail < 1} Theorem 2.5.1. be its unit ball, x E S. Then x is an extreme point of S if and only if (1 — x* x) A(1 — xx* ) , {o}. Moreover, if x is an extreme point of S, then x is a partial isometry, i.e. x*x and xx* are projections. Proof. Let x be an extreme point of S. First we prove that x*x is a projection. In fact, let B be the abelian C*subalgebra generated by x*x. Then B r=d Cr(n). If there is to E n such that x*x(t o ) E (0,1). By the continuity, we can find an open neighborhood u(c n) of to and E E (0,1)
such that 0 < X * X(t) < 1 — E ) Vt E U.
Pick d E B such that 0 < d(t) < 1,Vt e
1,d(t0)=1,d(1\u) =
{0},
and ri E (0,1) such that 2r/ ± 77 2 < e. Then 0 < (1 ±rid(t)) 2 x*x(t) =
I x*x(t)(< 1),
Vt V U,
1 < (1 + 0 2 (1 — e)(< 1), Vt E U.
95
Since d and x*x commute, it follows that
11x ± rixd11 2 .= II(x(1± rIC* . (x(1 ± rid))11 = 11( 1 + 7/ d) 2 ' x*xli < 1. Now from x = 1(x + rixd) + (x — r 1 xd), we get xd = 0. Further ex. d = 0. It is impossible since (fx • d)(t o) = x*x(t o ) > 0. Therefore x*x(t) = 0 or 1, Vt E fI, i.e. x*x is a projection. Similarly, xx* is a projection since x* is still an extreme point of S. < 1, then py = O. Let p = e x, q = xx* . If y E (1 — p)A(1 q) with WI — Further, 0 = y*py = (xy)* • (xy), and xy = 0. By Theorem 2.3.20, —
11x± Y* 11 2 =
Ilf± YI1 2 = II(x*± Y)* • (x* ±Y)II
= Ilxx* + Y*Y11=11qxfq + ( 1  q)Y* Y( 1  011 = max{11xx* I1,11Y* Y11}
1.
Since x is an extreme point of S and x = 1(x + y*) + 1(x— y*), it follows that y* = 0. Therefore, (1 — p)A(1 — q) = {0}. Conversely, suppose that (1 ex)A(1 xe) = {0}. Then —
0 = x(1 — xe)x(1 — x* x) = x* x  (1 — e x) 2 . Thus a(x*x) c {0,1 } , i.e., x*x is a projection. Similarly, xx* is also a projection. Let p = e x, q = xx* . Since (xp — x)*(xp — x) = px*xp — px*x — x*xp ± x* x = 0, it follows that xp = x, pz *= e. (1)
Suppose that there are a, b E S and A E (0,1) such that x = Aa + (1
A)b. Then p = x* xp = Ax*ap+ (1— A) x* bp. By (1), p• x* ap = e ap •p. Thus the set {p, x* ap, x* bp} is commutative, and it can generate an abelian C*subalgebra with an identity p. By the Gelfand transformation, we can see that p = x* ap = x* bp.
—
(2)
By (1), (2) and q = xx* , we have
x = qap = qbp.
( 3)
From (2), (3), pa*qap = pa* x = (x* ap)* = p, hence
1 ? lipa*apll = 11pa*qap+ pa*(1
—
q)apil
= llp + pa*(1— q)apil.
But pa*(1 — q)ap is a positive element of the C*subalgebra pAp, and pAp has an identity p, so pa*(1 — q)ap = 0,(1 q)ap = 0, ap = qap. By (3), we obtain —
x = ap.
(4)
96
Since x = Aa + (1 — A)b, it follows that y = Ac + (1 — A)d, where y = x*,e = a * , d = b*. Replacing {x, a, b, p, q} by {y,e,d,q,p} in above procedure (1)—(4), we obtain y = eq since yay = q and yy* = p. Thus
x = qa.
(5)
Since (1— q)a(1—p) E (1 q)A(1—p) = {0}, it follows that a = apdqa —qap. Now by (3), (4), (5), x = a. Further, x = a = b. Therefore, x is an extreme point of S. Q.E.D. —
Corollary 2.5.2. If a Caalgebra has an identity, then the identity is an extreme point of its unit ball. Theorem 2.5.3. Let A be a Caalgebra, and S = {a E A I Hall < 1} be its unit ball. Then A has an identity if and only if S has an extreme point at least.
The necessity is clear from Corollary 2.5.2. Now suppose that S has an extreme point x. Let p = xax,q = xx* , and {d i } be an approximate identity for A. By Theorem 2.5.1, (1 — q)d 1 (1 — p) = 0, V/. Thus Proof.
di > p + q — qp. Clearly, e = p + q — qp is an identity of A.
Q.E.D.
Proposition 2.5.4. Let A be a C*algebra, and S(A) be its state space. Then A has an identity if and only if S(A) is compact with respect to the watopology a(A* , A).
The necessity is clear. Now suppose that A has no identity. We prove that S(A),a(A*, A)) is not compact. It suffice to show that 0 E S(A) 7 , where S is the o (A*, A)closure of 5(A) in A*. Let U = U(0; al, • • • ) an; E) = {f E A * I If (ai)I < E ) 1 < i < n} be any a(A*, A)neighborhood of 0, we need to prove that U n 51 (A) 0 O. Since A = [A1 ] , we may assume that a; E A + ,1< i < n. Let a = a l + • • • + an . It suffices to show U(0; a; E) n 5(A) 0 O. By Theorem 2.3.20, we may assume that A c B(H) (some Hilbert space H), and A is nondegenerate on H. Since A has no identity, it follows that a is not invertible in B(H). Thus there is e E H with Heil = 1, such that (at, e) < E. Let p(.) = (e,). By Proposition 2.4.6, p E S(A) n U(0; a; E). Therefore, 5(A) n U(0; a; E) 0 O. Q.E.D. Proof.
Let E be a linear space, K be a convex subset of E, and F be a subset of E. We shall denote by ExK and CoF the sets of all extreme points of K and convex hull of F respectively.
97
Now let A be a C*algebra, S(A) and P(A) be its state space and pure state space respectively. For any E c A 4 , the a (A* , A)closure of E in A* is denoted by E° . Clearly, { p E A* I P > 13511P11 < 1} is a a(A*,A)compact convex subset of A*. And also it is easily verified that Ex{p E A* I p > 0 )11 Pli < 1} = {0, P(A)}. By the KreimMilmann theorem, we have
{P E A* I P ? 0 ,11PII 1. Then i
(E 41) )_ 1 E AM ) —> ço (a (A* , A)). Thus S(A) C CoP (A) 0 , and we have the following.
Proposition 2.5.5.
Let A be a C 4 algebra. Then
Ex{p E A* I p _?:. 0,11p11 O, II pll < 1} = C o {0 , P (A)} . Moreover, if A has no identity, then we have
{P E A *
I
P ? 0 ,11PII 0, Vp E S(A). By Corollary 2.3.15, a e + E > a. Since > 0 is arbitrary and a, G L 1 n A + it follows from Lemma 2.7.4 that a E L I . Thus we have L2 n AcL 1 n A. By Proposition 2.4.1, there is a net d1} C L2 n A+ such that adi —> a Va E L2. But de E L2 n A + C L1 n A + ,V1, so it must be L2 C LI. Further L1 = L2. {
,
Q.E.D.
106
Q.E.D. Theorem 2.7.6.
Let L be a closed left ideal of a C'algebra A. Then L is the intersection of all regular maximal left ideals containing L. Proof.
Let
n=
{p E A* I P ? 0 3 IIPII 5 1 3 and p(L) = {0}}. Clearly, n{L p I p E n} D L,
where Lp is the left kernel of p,Vp E n. By Lemma 2.7.5, we have L = nfLp Ip E n). Since n is a cr(A*, A)compact convex subset of A*, it follows from the KreinMilmann theorem that Co(Exil) is r(A*, A)dense in n, where Exil is the set of all extreme points of n,Co(.. ) is the convex hull of (• • •). If a E A such that **a) = 0,Vp E Exil, then p(a* a) = 0,Vp E Co(Exn), and further p(as a) = 0,Vp E 11 . Therefore, we have
n{Lp I p E Esn} = n{L p I p E n} = L.
(1)
Noticing that 1/ 0 {0} (otherwise, by Lemma 2.7.5, we have L = A, a contradiction), from Theorem 2.7.3 it suffices to show that p is a pure state on A for each p E Exil and p 0 0. Now let ça E Exil and ça 0 0. Clearly, rp is a state on A. Suppose that there are two states ço i , ço2 on A and A E (0,1) such that (pa = Ap i + (1— A)v2 . For any a E L, from (1) it must be a E 4, i.e., (p(a* a) = 0. Further, çoi(a*a) = 0,1 = 1, 2, and by the Schwartz inequality, ça(a) = 0,1 = 1,2. Hence ça(L) = {0} and (pi E n o: = 1,2. Since ça E Exil, it follows that Q.E.D. ça = PI = (,02. Therefore, io is a pure state on A. Theorem 2.7.7.
Let L be a maximal left ideal of a C'algebra A. Then L is regular if and only if L is closed. The sufficiency is clear from Theorem 2.7.6. Now let L be a regular maximal left ideal of A. So there is xo E A such that (bx o — b) E L,Vb E A. Let L = L40(i  xo ). We claim that L is a maximal left ideal of (A+0). In fact, suppose that J is aleft ideal of (AŒ) and J D L. If y=a+AE J, where a E A, A E 0, then Axo + a = y — A(1 — xo) EJnA since (1 — xo ) E L C J. Since L is a maximal left ideal of A, it follows that J n A = L. Thus y E Lj4J (1  x0) = L, and J = I.. Now L is closed since (AiŒ) has an identity. Therefore, L=LnA is Q.E.D. closed. Proof.
Theorem 2.7.8.
Let L be a regular maximal left ideal of a C'algebra A. Then there is unique state p on A such that N(p) = {a E A I p(a) = 0} D L. Moreover, this p is pure, and its left kernel is L, and N(p) = L + L*.
107
By Theorem 2.7.7 and the proof of Theorem 2.7.6, there is a pure state p on A such that Lp D L and p(L) = {0}. Since L is maximal, it follows that L = Lp. Further, by Theorem 2.7.3, N(p) = L + L* . Now suppose that ço is another state on A with (p(L) = {0}. Then (09(N (p)) = {0}. If 20 (E A) is a modular unit for L and {c11 } is an approximate identity for A, then we have Proof.
p(x0 ) = lina p(d i xo ) = lina p(d i )
and ço(x0 ) = limio(dix o ) = lirnço(cli). i i By Proposition 2.4.4, we get p(x 0 ) = ço(x0 ) .= 1. Moreover, it is clear that A = N(p)iŒx o . Therefore, ço = p. Q.E.D.
Corollary 2.7.9. For any C*algebra A, there is an one to one correspondence between the set of all pure states on A and the set of all regular maximal left ideals of A. Theorem 2.7.10. Let A be a C*algebra, p be a state on A, and L be the left kernel of p. If L is regular, then the following statements are equivalent: 1) p is a pure state on A; 2) L is a maximal left ideal of A; 3) N(p) = {a EA I p(a) =0} = By Theorems 2.7.3 and 2.7.8, it is clear that 1) and 2) are equivalent and 1) implies 3). Now let N(p) = L+ L* and xo (E A) be a modular unit for L. Suppose that there are two states p i , p2 on A and A E (0,1) such that p = Api + (1 — À)P2. If x E L, then we have x * x E N(p), p i (x* x) = 0, and pi (x) = 0,1 = 1,2. Further, pi (N(p)) = {0},i = 1,2. By the proof of Theorem 2.7.8, we can get p(x 0) = p 1 (x0) = p 2 (x0) = 1. Moreover, it is clear that A = N(p)4Œx o . Therefore, p = p i = p 2 . So p is a pure state on A. Q.E.D. Proof.
References. [791, [81], [155], [156].
2.8. Ideals and quotient C*algebras Let A be a C*algebra, and P (A) be the set of pure states on A (see Definition 2.3.8). Denote by Â the set of all unitarily equivalent Definition 2.8.1.
108
classes of irreducible * representations of A, and let Prim(A) be the set of all primitive ideals of A. J is called a primitive ideal of A, if there is an irreducible * representation ir of A such that kerr = J. Clearly, a primitive ideal is a closed twosided * ideal. Since two unitarily equivalent representations have the same kernel, so we have a natural map from Â onto Prim(A). Moreover, for each p E P (A ), the * representation generated by p is irreducible from Theorem 2.7.1. Conversely, if fr,H) is an irreducible * representation of A, then by Proposition 2.4.6, p(.) = (71He, is a state on A for any H with 1. And also from Proposition 2.3.21, the * representation {rp , Hp } generated by p is unitarily equivalent to {ir, H}, and is irreducible. Furthermore, p is pure by Theorem 2.7.1. Therefore, we have a map from P(A) onto Â by the GNS construction.
eE
Proposition 2.8.2. Then J=
e)
Il ell =
Let J be a closed twosided ideal of a C*algebra A.
n{/
E Prim(A) I I D .1)
= n{kerirp I p e P(A),
and (61) = 01.
Since J is also a closed left ideal of A, it follows from the proof of Theorem 2.7.6 that Proof.
n{L, I p e P(A), and (pIJ) =
J=
0)
n{ker rp I p E P(A), and (pIJ) = 0).
D
Now let a e J. For any p E P(A) and (0) = 0, since J is twosided, it follows that Ilirp(a)bp112 = p(b*a* ab) = 0,Vb e A. Hence, i  (a) = 0, and a e kerrp . Therefore, we obtain J=
nfkerri, I p e P(A),
= n{i e Prim(A) I I
and (pIJ) = 01
D J}.
Q.E.D. Corollary 2.8.3. Let fi be a compact Hausdorff space, and J be a closed ideal of C(fl). Then there is a closed subset 0 0 of n such that
.1 = {f EC(n) Proof.
I
f(t)=
0,Vt E no)•
It is clear from Propositions 2.8.2 and 2.3.10.
Q.E.D.
109
Definition 2.8.4. of A. Define
Let A be a C*algebra, and J be a closed twosided ideal
Pj (A) = {p E P(A) I (pIJ) = 0} , P J (A) = {p E P(A)
I
(pIJ) 0 0} = P(A)\PJ(A),
A.7 = fr E A I ker 7 D J1, .ii`i = 22iVii j , Primj (A) = {I E Prim (A) I I D J}, Prim(A) = Prim (A) \Primj (A).
Theorem 2.8.5. Let J be a closed twosided ideal of a C*algebra A. 1) For any ir E :4j, let rr(a) = r(a),Va E A/J, a E a, where a —> a = a + J is the canonical map from A onto A/J. Then r —> i f is a bijection from Â onto (A/J)^. 2) r —> (rIJ) is a bijection from Ay onto Î. Proof. 1) It is obvious. 2) Let {r, H} be an irreducible * representation of A, and rIJ 0 O. Since J is a twosided ideal, it follows that the linear span K of {ir(a) a E J, E HI is a nonzero invariant subspace for r(A). Now since 7 is irreducible, it must be that K = H, and {71J) H} is a nondegenerate * representation of J. If {di}(C J) is an approximate identity for J, by Proposition 2.4.6, we have ir(cli ) —) 1 (strongly). Then r(adj ) —) r(a) (strongly), Va E A, i.e., r(J) is strongly dense in r(A). So {r 1113 H} is an irreducible * representation of J by Proposition 2.6.2. Conversely, let {r, H} be an irreducible * representation of J. By Theorem 2.6.6, H = [ir(b) I b E J)e E H ]  For any bi, • • • ,b„ E J and 63 . • • , en E H, E (70*., • bi )ei , e.i ) 1 IIJ is a bijection from Primj (A) onto Prim(A/J). 2) / —> In J is a bijection from PrimJ (A) onto Prim(J). Proof. 1) Let I E Primj (A), and {7r,1/} be an irreducible * representation of A such that kerr = I. For any a E 11/J, since J C I, we can define = ir(a), here a E a = a+J. Then frr, H} is an irreducible * representation of AIJ, and kerrr = II.I. Thus IIJ EPrim(A/J). Moreover, let frr,1/1 be an irreducible * representation of AIJ. Define r(a) = (a), Va E A, here a —> a is the canonical map from A onto AIJ. Then fir, H} is an irreducible * representation of A and kerr = ID J, kerrr = IIJ. Thus, I —> IIJ is a map from Primj(A) onto Prim(A/J). Now let I1 IJ = 12 IJ, where 11,12 EPrim j (A). For any a E Ii , then we have b E 12 such that (a — b) E J. Since J C 12, it follows that a E 12 . Thus II c 12 . Similarly, 12 C Ii . Thus II = 12 . Therefore, I —> IIJ is a bijection from Primj (A) onto Prim(A/J). 2) Let I E Prim J (A), and fir, H} be an irreducible * representation of A such that kerr = I. Since kerr = I 75 J, it follows by Theorem 2.8.5 that {(1r1J))/1} is an irreducible * representation of J, and ker(ir I J) = I n J EPrim(J). Now if {cr,./1} is an irreducible * representation of J, then by the proof of Theorem 2.8.5 {a, H} can be uniquely extended to an irreducible * representation of A. So / —> In J is a map from PrimJ (A) onto Prim(J). Now let II n J =12 n J, where 11,12 EPrimi (A). Then 12 D II n J D 11 J. But 12 75 J, by Proposition 2.8.8 we have 12 D I. Similarly, II D 12 . Thus I = 12 . Therefore, / —> In J is a bijection from Prim(A) onto Prim(J). Q.E.D. Theorem 2.8.10. Let J be a closed twosided ideal of a C*algebra A. 1) p —> P' is a bijection from P(A) onto P(A/J), where xa) = p (a) ,va E A/J, a E a. 2) p —> (pIJ) is a bijection from P(A) onto P (J ). Proof.
1) It is Proposition 2.4.11 exactly.
112
}
2) Let p E P(A), and {7, H, e be the irreducible cyclic * representation generated by p. If J Ckerr, then p(a) = (ir(a)e, ) = 0, Va E J. This is a contradiction since p(J) 0 {0}. Thus J Okerr. Then by Theorem 2.8.5, {rIJ, H} is an irreducible * representation of J. Further by Proposition 2.3.21, {rIJ, H} is unitarily equivalent to the * representation generated by the state (pIJ)(.). Thus (pIJ) E P(J). Moreover, if {di}(c J) is an approximate identity for J, by Proposition 2.4.6 we have r (di ) —> 1 (strongly). Then p(a) = (7(a) lim(r(adi) Va E A. This means that the behaviour of p on A is determined by (pIJ).
=
e, e) =
e, e),
Hence, if pi, P2 E Pi (A) and (PIP) = (P214 then PI = P2. Finally, let a E P(J), and {r,„, H, be the irreducible cyclic * representation of J generated by a. By Theorem 2.8.5, {r,„, H} can be uniquely extended to an irreducible * representation {ir, H} of A. Then p(.) = (r() e, e (V. E A) is an extension of a, and the * representation of A generated by p is unitarily equivalent to {r, H} (see Proposition 2.3.11). Thus p E P J (A) and (pIJ) = a. Therefore, p —> (pIJ) is a bijection from P(A) onto P (J ). Q.E.D.
e}
)
From Theorem 2.8.5, 2.8.9 and 2.8.10, we have the following diagram: P(Al J) i (AIJ) ^ I Prim(A/J)
4
P(A)
A
4—
—>
Prim(A)
4—
J 1 Prim(J)

It is easily checked that this diagram is commutative. About the converse of Proposition 2.8.8, J. Dixmier gave the following result. Let A be a separable or Type I (GCR) C*algebra, and J be a prime closed twosided ideal of A, then J is also primitive. But the question for the general case is still open. Notes.
References. [231, [28], [33], [1561.
2.9. Hereditary C*subalgebras Let A be a C*algebra, a, z, y E A and a > 0. Suppose that Lemma 2.9.1. there are number A, ii, > 0 with (A + A) > 1 such that x*x < a,
Let u Ilull < lia
yy* < aA. — x( 1 + a) ly,Vn. Then there is u E A, such that Ilun — ull —> 0 and A±Zil
2
II'
113
Proof.
Let dnm = ( 7 1 + a)  —
+
, Vn, rn. Then
Ilun um11 2 = Ilx4mY11 2 = IlY * dnmexdnmYll 11Y * dnmaA 4mYll
lialdnmyir
= II a IdnmWdnma II < IlaNnntednma II = IldnIna P11 2 . We may assume that A has an identity 1. Let B be the abelian C*subalgebra generated by fa, Then B C(f)). For each t E n,((,14 + )(t) a 2 (t). By the Dini theorem, this convergence is uniform for t E n. Thus II — O and Ilun — utnii ) 0, and there is u E A such that Ilun —ull —> O. Similarly, we can prove that Muni' < ar4aPII,Vn. Therefore, hull < Q.E.D. Ila 2 II' —
Proposition 2.9.2. Let A be a C*algebra, x, a E A and a > 0, x*x < a. Then for any A E (0, f), there is u E A such that x = ua A , and iluli< Proof.
Let un = x( 1 + a)la
lla2
II.
Vn. By Lemma 2.9.1, there is u E A such
that
Ilun ull 13> and hull < Moreover, since x*x < a, it follows ,tohat Ilx Therefore, z
unaA ll Ç 11a 1 [ 1—
= uaA
Q.E.D.
Corollary 2.9.3. Let A be a there is u G A such that
= tqx yv/2, Proof.
+ar l a 4 ]11 —*().
It is clear by Proposition
C*algebra, z E A, and a E (0,1). Then and huh ç 11(x* x) 11 11.
2.9.2 and picking a = f x, A = a/2. Q.E.D.
Remark.
A factorization with a = 1 (polar decomposition) is not possible
in a general C*algebra.
114
Definition 2.9.4. Let A be a C*algebra. A cone M(C A + ) is said to be hereditary, if a E A + with a < b for some b E M implies a E M.
For a hereditary cone M, we define
L(M) = {x E A I x * x G M}. It is easy to see that L(M) is a left ideal of A. A C*algebra B of A is said to be hereditary, if B+ is hereditary. Theorem 2.9.5. Let A be a C*algebra. 1) B —> B+ is a bijection from the set of all hereditary C*subalgebras
of A onto the set of all closed hereditary cones of A+ . Its inverse is M —> L(M) n L(M)*. 2) M > L(M) is a bijection from the set of all closed hereditary cones of A + onto the set of all closed left ideals of A, and M = L(M) + . Its inverse is L > L + . 3)L—)LnL* is a bijection from the set of all closed left ideals of A onto the set of all hereditary C*subalgebras of A, and L + = (Ln L*) + . Its inverse is B —> L(B+ ). Proof. Let B be a hereditary C*subalgebra of A. By the definition, B+ is a closed hereditary cone of A + . Since B = [B+ 1, so the map B —> B+ is
inject ive. Let M be a closed hereditary cone of A. Clearly, L(M) is a closed left ideal of A. Assume that z E A is such that ex G .L(M). By Corollary 2.9.3, we can write z = u(x*x) 1/4 for some u G A. Thus x E L(M), and L(M) + = {x*x I z E LUIS». By the definition of L(M), we have M = L(M) + . Thus the map M —> L(M) is injective. Let L be a closed left ideal of A. Clearly, L n L* is a C*subalgebra of A. Since L + C L n L* C L, it follows that L+ = (L n L*) + . So in order to prove that L n L* is hereditary, it suffices to show that L + is hereditary. Suppose that a E A+ and a < b for some b G L. By Proposition 2.9.2, we have a factorization al = vb 1 /3 , where y E A. Thus al E L + , and a E L + , i.e., L + is hereditary. Suppose that L is a closed left ideal of A. From the preceding paragraph, M = L + is a closed hereditary cone of A. Let z E A be such that x*x G M. By Corollary 2.9.3, we can write z = u(x*x) 1 /3 . Thus x G L. Further, by the definition we have L = L(M). So 2) is proved. Suppose that M is a closed hereditary cone of A + . From preceding paragraphs, L(M) is a closed left ideal of A; L(M) n gm)* is a hereditary C*subalgebra of A; and M = L(M) + = (L (M ) n L(M)*) + . So 1) is proved. Finally, let B be a hereditary C*subalgebra of A. Thus B+ is a closed hereditary cone of A + ; L (B+ ) is a closed left ideal of A; L (B+ ) n L(B + )* is a
115
hereditary C*subalgebra of A. Since (L(B + ) n L(B+ )*) + = L(B +) + = B+ , it follows that B = L(B +)n L(B +)*. So the map L —> L n L* is surjective. Moreover, if L is a closed left ideal of A such that B =LnL*, then B + = (Ln L*) + = L+ , and L = L(B + ). So the map L —> L n L* is also injective; Q.E.D. and its inverse is B —* L(B + ). Therefore, 3) is proved.
Lemma 2.9.6.
Let 4) be a * homomorphism from a C*algebra A onto a C*algebra B, a E A +l b E B with b*b < (I)(a). Then there is x E A such that b = (I)(x) and x*x < a.
Proof.
Pick y E A such that (I)(y) = b. Write y*y — a = h— k,
where h l k E A+ and hk = O. Since (1)(y)*(1)(y) = b*b < (I)(a), it follows that 0 < 4)(h) 0, Va E A. Thus 1,1i(a) = limi,b(d i adi),Va E A. From the preceding paragraph di Ad i C B, so we have i
i,b(a) = linaço(d i adi ),
Therefore, the extension Notes.
is unique.
Va E A.
Q.E.D.
Theorem 2.9.5 is due to E.G. Effros.
References. [33], [1271.
2.10. Comparison, disjunction and quasiequivalence of * representations Definition 2.10.1. Let A be a C*algebra, and fir, HI be a * representation of A. If K is a closed linear subspace of H, and K is invariant for 7 (i.e., 7r(a) E K,Va E A, E K), then {7, K} is also a * representation of A, and {7,K} is called a * subrepresentation of {7, H}. Suppose that {7r i , H1 } and { r2 , H2 } are two * representation of A. The symbol "7 1 ‹ 7 2" means that {7 1 ,1/1 } is unitarily equivalent to a * subrepresentation of { i12 , H2 } . Proposition 2.10.2. Let {r i , HI } and {72 , H2 } be two * representations of a C*algebra A. 1) Let 7 = 7 1 ED 72 ,H = H1 ED 112 ) and let p be the projection from H onto Hi (clearly, p E 7r(A)'), i=1, 2. Then ri ‹ 72 if and only if p < p'2 in 7(A) 1 . 2) If 7 1 < 72 and 72 ‹ ri , then we have {ri, H1} {r23112}3 i.e.) firi, and {7 2 , H2 } are unitarily equivalent. Proof.
1) It is clear. Moreover, from 1) and Proposition 1.5.3, we can get
2) immediately.
Q.E.D.
Let fr i ,Hi l and {7 2 , H2 } be two * representations of Definition 2.10.3. a C*algebra A. 7 1 and 72 are said to be disjoint, denoted by 7 1 172 , if any nonzero * subrepresentat ion of 71 is not unitarily equivalent to any nonzero * subrepresentation of r2 . Proposition 2.10.4. Let {7 1 , H1 } and {72 , H2 } be two * representations of a C*algebra A. Let 7 = 7 1 0 72 , H = H1 ED H2; and let p be the projection
117
from H on Hi (clearly, p E irply), i = 1,2. Then the following statements are equivalent: 1) r 1 _i_7 2 ; 2) c(pli )  c(p12 ) = 0, where c(p) is the central cover of A in r(A) 1 , I = 1,2; 3) psi is a central projection of ir(A) 1 , I = 1,2.
Proof. Since p'1 ED 1912 = 1, it follows that the statements 2) and 3) are equivalent. Clearly, r 1 1r2 if and only if there are no projections qs1 and q'2 of 71 (A) 1 such that 0 0 q'i < p'0 2: = 1,2, and q q12 in ir(A) 1 . Then by Proposition 1.5.9, the statements 1) and 2) are equivalent. Q.E.D.
Definition 2.10.5. A nondegenerate * representation {ir, H} of a C*algebra A is said to be factorial, if the VN algebra on H generated by r(A) (i.e. r(A)") is a factor. Proposition 2.10.6. Let {ri , Hi } and {72, 112 } be two factorial * representations of a C*algebra A. Then one of the relations ir1 in 2 , 7 1 < r2 , 1r2 holds. Proof. Let r , r 1 ED 72 , H , Hl ED H2 1 M = r(A) 11 and p'i be the projection from H onto Hi , i = 1,2. Then pii ,p12 E M'. By the assumption, Mp'i is a factor on Hi , i = 1,2. If c(p) is the central cover of pii in M', by Proposition 1.5.10 Mp'i and Me(p) are * isomorphic, i = 1,2. Thus M c(p i) is also a factor on He(g); and e(g) is a minimal central projection of M (i.e., if z is a central projection of M and z < c(psi ), then either z = 0 or z = c(pli )),i = 1,2). Therefore, we have either c(p'1 ) • c(p) = 0 or c(pii ) = 02 ). When e(A) • e(p12 ) = 0, by Proposition 2.10.4 r1 lir2 holds. Now let c(pli ) = c(p) = z. Since Mz is a factor, by Proposition 1.3.8 A/1 z is also a factor. From Theorem 1.5.4, we have either p'i . A or p12 .< pi, in Miz (also in M'). Further by Proposition 2.10.2, either r1 < 72 or 72 < 71 holds. Q.E.D. Let {71 , H 1 } and {1 2 , H2} be two irreducible * repProposition 2.10.7. resentations of a C*algebra A. Then r1 ln2 if and only if in and 72 are not unitarily equivalent.
Proof. Since any nonzero * subrepresentation of in must be 7ri itself, I = 1,2, the conclusion is obvious from Definition 2.10.3. Q.E.D. Proposition 2.10.8.
If 7r_Dr1 ,V/, then ri
E ED7ri . i
118
Let H, Hi be the action spaces of 7r, 7r1 respectively, V/. Let a = ir ED E Or' , K = H ED E OA; and let p', p; be the projections from K onto 1 1 H, Hi respectively, V/. Then p', A E a(A)', V/. By Proposition 2.10.4 we have = 0,V/. From Proposition 1.5.8,
Proof.
c(p') 1 sup c(p) = c(sup p;) = c(E pa .
Again by Proposition 2.10.4, ir_i_ EEDiri .
Q.E.D.
i
Definition 2.10.9. Let {7r i , Hi } and {A2 , H2 } be two nondegenerate * representations of a C*algebra A, and Mi = ri (A)", 1 = 1, 2. r i and 7r2 are said to be quasiequivalent, denoted by r i r,.:7r2 , if there is a * isomorphism 4) from Mi onto M2 1 such that 4)(7r i (a)) = 72 (a) , Va E A. Proposition 2.10.10. Let {r i , HI } and {r2 , H2 } be two nondegenerate * representations of a C*algebra A. Then the following statements are equivalent 1) ri r..'s, r2;
2) No nonzero * subrepresentation of ri is disjoint from ri , 1 < i j < 2; 3) Let r = ri ED 71 2 , H = H1 0 H2 ) and pli be the projection from H onto fli(E 7r(A) 1 ), i = 1,2. Then c(A) 4) There exists an ampliation 7r of ri (i.e., there is a Hilbert space K such that r(a) = ri (a) 0 1 K , Va E A) and a projection il of 7r(A) I with central cover 1(= 1H 1 vr) such that irpt 5) There exist ampliations of r i and 72 which are equivalent.
1). 4). It is clear Theorem 1.12.4 and Proposition 1.12.5. 4) = 1). Define 413 ) 43.2 1 41. 1 as in Theorem 1.12.4; and let 4) = (113 0 41 2 0 cliki. Then it is immediate that ir i :•zss 7r2 . 4) = 5). From the condition 4), 7r2 is unitarily equivalent to a * subrepresentation of in (8) 1 K , where K is some Hilbert space. Since 1) and 4) are equivalent, it follows that in is also unitarily equivalent to a * subrepresentation of 7r2 0 1 L , where L is some Hilbert space. Let R be an infinite dimensional Hilbert space, and dimR >dim K, dimL. Then
Proof.
A2 0 1R < ri 0 lic 0 1R ri 0 1R < A2 0 1L 0 1R 1r2 0 1R• By Proposition 2.10.2, we have ri 0 1R Sf.7r2 0 1 R . That comes to 5). 5) == 2). Suppose that ri 0 1 R . 7r2 0 1 R , where R is a Hilbert space. If ai is a nonzero * subrepresentation of ri, then we can regard ai as a nonzero * subrepresentation of in 0 1R . By Definition 2.10.3, oi is not disjoint from 7r1 0 1R . Further, by Proposition 2.10.8 ai is not disjoint from ri.
119
c(A), we may assume that e(A)
e(A). Then z = c(p 12 ) e(A) • c(p) is a nonzero central projection of 71 (A)', and z < c(A) and z I c(A). By Proposition 1.5.8, zy12 0. Clearly, c(zpi2 ) 1 e(A). By Proposition 2.10.4, {7r2,zH2}1{711,H1 } . That contradicts the condition 2). Thus e(A) 3) == 1). Let z = c(A) = e(A), and M = 70)". By Proposition 1.5.10, Mpii and Mz are * isomorphic, i = 1,2. Thus we have a * isomophism (11. from MA such that (1)(bps1 ) = bp'2 ,Vb E M. In particular, for = MA onto M2 any a E A, since ri (a) = = 1,2, it follows that (1)(711 (a)) = 7r(a)1912 = r 2. Q.E.D. r2 (a). Therefore, r i 2)
3).
If c(A)
Proposition 2.10.11.
Let {ri , Hi } and {712 , H2 } be two nondegenerate * representations of a C*algebra A. r 2 1 then iri Pss 7r2 . 1) If ir i 712 . 2) If ri and 7r2 are irreducible, and r i Ps° 1 2 , then r i
1) It is obvious. Now we prove 2). By Proposition 2.10.10, 1r1 is not Proof. disjoint from 712 . Since 7r1 and 7r2 are irreducible, it follows from Definition Q.E.D. 2.10.3 that ri is unitarily equivalent to 7r2 . Proposition 2.10.12. Let {ri, HI } and {712 , H2 } be two factorial * representations of a Cealgebra A. Then we have either A 1 1712 or ri Pes 712 . Proof. By Proposition 2.10.6, we may assume that 71 1 712 . Then there is a projection p' E i12 (A)' such that A1 Ss= 712 p'. But 7r2 (A)' is a factor, so the central cover of p'2 in 7r2(A) 1 is 1. By Proposition 1.5.10, 7r2 Psd7r2p1 . Therefore Q.E.D. 71 Ps, 72.
References. [28], [104], [105].
2.11
The enveloping Von Neumann algebra
Definition 2.11.1. Let A be a Cealgebra, and (A) be its state space. For each cp E 8(A), we have a cyclic * representation {7r,, Hv,, v } of A (see Proposition 2.3.18). Then the faithful * representation
=E 'ES(A)
oirça,
=E
efiv,
vES(A)
is called the universal * representation of A. And rts (A)" is called the enveloping VN algebra of the Cealgebra A, denoted by A =
120
Suppose that 0 is a normal state on A. Since A and ru (A) are * isomorphic, it follows that there is a state cp on A such that ça(a) , 0(ru (a)),Va E A. By the GNS construction, we have Er E Hr C Hu such that
Cru(a)) = ço(a) = (rso(a)‘) ‘) = (Irts(a)‘, E r ),Va E A. Further, t,b(b) = (b‘, ‘),Vb E A = ru (A)". Therefore, every normal state on the VN algebra M  ru (A)" is a vector state. From Proposition 1.10.6, we have a(M,M* ) — (weak operator top. 1M), s(M,M * ) ,, (strong operator top. 1M) and s*(M,M* ) ,, (strong * operator top. 1M) . Now we study the relation between the enveloping VN algebra A and the second conjugate space A**. By Proposition 1.3.3, A is the conjugate space of the Banach space A * = T(H)/A±, where
711 = ft ET(Hu) I tr(tb) =
0,Vb E
711.
Through the following way, the Banach space A * and the conjugate space A* of A are isometrically isomorphic. For any f E A * , let
F(a) = ru (a)(f), Va E A. Then F E A* and 11F11 = 11 111. Conversely, any element of A* must be of above form. In fact, if f E AI*, by density theorem 1.6.1 11 111 = suP{Irts(a)(f)I I a E A, Hall 1 } = IIF11. Now let ça E S(A). Then p(a) = (ru(a)‘,4,),Va E A. Let pr be the one rank projection of Hu onto [Er ], and f be the canonical image of pr in 71*  T(Hu )/71± . Then
Iru(a) (f) = tr(ru(a)Pr)
(Irts(a)‘)
4,) '
P(a), Va E A.
Since A* is the linear span of S(A), it follows that for each F E A* there is unique f E71* such that
F(a) = ru (a)(f), Va E A. Denote the above isomorphism from A * onto A* by re, i.e.,
r* (f)(a)  r u (a)(f), Vf E 71 * , a E A. Then (r * )* is an isometrical isomorphism from A** onto A, and it is olA**, Ala(A,A * ) continuous. Moreover, since for any aEAC A**
(7 *) * (a)(f) = irs(f)(a) = ru(a)(f),v f E 71*, it follows that (r*) * IA = iis) i.e., (r e )* is an extension of the * isomorphism ru from A onto ru (A). So we can write (r* )* = ru simply. Now we have the following theorem.
121
Theorem 2.11.2.
Let A be a C*algebra. Then the second conjugate space A** of A is isometrically isomorphic to the enveloping VN algebra A of A. So we can introduce a multiplication and a * operation on A** such that A** becomes a C*algebra, and A becomes a C*subalgebra of A**. Moreover, if A has an identity, then this identity is also an identity of A**.
In the above discussion, the multiplication and the * operation on A** are defined through A. But we have another way. It depends on A and A* directly.
Theorem 2.11.3.
Let A be a C*algebra, and define a * operation and a multiplication (Arens multiplication) on A** as follows: X* (F) = X(F*), F* (a) = F(a),
XY (F ) = X(EY, Fll,[Y, F](a) = Y (L a F), (L a n (b) = F(ab), Va,b E A, F E A* , X, Y E A** . Then this * operation and multiplication on A** are the same as in Theorem 2.11.2. Keep the above notations: 71, At , 711 , r* : 71* —) A*, and (ir*)* = ru : Proof. A** —) A. For any X E A**, pick a net {xi} C A such that xi › X(o. (A** , Al). Since (X xi)* (F) = (X xi )(F*),VF E A* , it follows that xt ) X* (a(A** , Al). But ru = (7 * )* is a(A**,A*)cr(A,A*) continuous, thus we have —
—
—
—
ru(X1 = r u (X)*. For any t E T(Hu) and a E A, denote the canonical images of t and tru (a)(E T (Hu )) in 71* = T(Hu)/71± by f and La i respectively. Since for any b EA (Lar*(f))(b) = r*(i)(ab) = ru(ab)(f)
= tr(tr i,(ab)) , ru(b)(La f) ) it follows that rt (L a f) = La r* (f)For Y E A**, let g be the canonical image of ru (Y)t(E T(H u )) in A * .=T(Hu)1 A I . From
[Y, ir*Cn](a) =
Y(L a r*(f)) = Y(r * (La f)) ru (Y) (La f) =
tr (rts (Y)tru (a))
= r u (a) (g) = ir * (g) (a) , Va E A, we have [Y,71 * (f)] = r * (g).
122
Now for any X, Y E A** by
ru (XI') (f) = (XI') (7r* (f)) = X([ 37) 7r*(f)]) = X(r(g)) = r u (X)(g) = tr(r u (X)7(Y)t) (irts (X)ru (Y))(f),
VI EA,
Q.E.D.
we have ru (XY) = ru(X)ru(Y)•
Proposition 2.11.4. Let A be a C*algebra, and B be a Cesubalgebra of A. Then the Cealgebra B** is * isomorphic to the a(A** , Alclosure 746 of B in A**. For any X E B**, let (D(X)(F) = X(FIB),VF E A*. Then (I) is an Proof. isometric linear isomorphism from B** onto . By Theorem 2.11.3, (1) also keeps the * operation and multiplication. Q.E.D. Notes. The second conjugate space of a Cealgebra is very important since it is a W*algebra (see Chapter 4) , The Theorem 2.11.2 (A" A) is due to S. Sherman and Z. Takeda. Moreover, let B be a Banach algebra. We can introduce two kinds of Arens multiplication on B**. The first Arens multiplication is as in Theorem 2.11.3, i.e. )
(XY) (F)
XCY, Fll,
[Y, 1](a) Y (L a F) ,
(L a F) (b) F (ab),
Va,b E B,F E B* ,X,Y E B**. The second Arens multiplication is as follows: (X Y)(F) Y (PC, FY),
EX, Ma) = X(R a F), (Ra F)(b) = F(ba),
Va, b E B,FE B* , X, Y E B* . A natural question is when we have XY = X. Y,V X, Y E B** . Definition (P. Civin and B. Yood). A Banach algebra B is said to be regular, if XY = X. Y, VX, Y E B** . Let Z(B) {X E B** I XY = X. Y,VY E B ** }.
Z(B**) is called the topological center of B**. Clearly, B is regular 444 the map • —> X. is continuous in (B**,a(B**, Z(B**) B**; X E Z(B) B*)); and B C Z(B ** ) C B** . By Sakai theorem (see Section 4.2), any C*algebra A is regular, i.e., Z ( A") = A*. Therefore, for any C*algebra A, two kinds of Arens multiplication are the same on A**.

123
References. [16], [64], [161], [168].
2.12.
The multiplier algebra
Definition 2.12.1. Let A be a C*algebra, and see the second conjugate space A** of A as the enveloping VN algebra. Let
M(A) = {a E A** I aA U Aa C A}. Then M(A) is called the multiplier algebra of A. Clearly, M(A) is a C*subalgebra of A**; A C M(A) and A = M(A) A has an identity; A is a closed twosided ideal of M(A). The C*algebra Q(A) = M(A)/A is called the out multiplier algebra of A. Let A be a nondegenerate C*algebra on a Hilbert
Definition 2.12.2. space H. Let
L H (A) = {x E B(H) I xA c A}, RH(A) = {x E B(H) I Ax c A}, and
MH (A)
=
L H (A)
n R H (A ).
L 13 (A) is called the set of left multipliers of A on H; R H (A) is called the set of right multipliers of A on H; MH(A) is called the multiplier algebra of A on H. Suppose that {d i } is an approximate identity for A. Since A is nondegen1(= 1H) (strongly). Thus erate on H, it follows that di L H (A), R H (A ), MH(A) Clearly, A
c MH (A),
and A = MH (A)
1
c74 8 = e
A".
A.
Let A be a C*algebra. A linear map p : A —> A is Definition 2.12.3. called a left (or right) centralizer, if p(xy) = p(x)y
(or = xp(y)), V x, y E A.
Proposition 2.12.4.
Let p be a left (or right) centralizer of a C*algebra A. Then p is continuous (bounded).
Proof.
Suppose that there is a sequence {xn } C A such that
Ilxn11 < —1)
and lip(x)II > n, Vn.
124
Let
E a
Since (4.4 )*4, write
< a (p
=
, x nx n ,
left) or xn*xn
xn
=
if p is left, if p is right.
< a (p
right), by Proposition
2.9.2
we can
a 1/3 n u (p left)
or
xn where
Ilunll < 11a11'611,Vn. n < 11P(xn)11
= un a1/3
( p right),
Thus when p is left, we have
Ilga l/3 )11 • Ilunll
Ila 1/6 11 • IIP(a 1/3)11
Vn;
when p is right, we have
liga l/3)11
0 ) Va E A, it follows that xi —> x(s(M(A), A)). Therefore, (M(A), s) is complete. Q.E.D. Furthermore, M(A) is the completion of (A, slA). Propositions 2.12.9 and 2.12.12 are due to R. Busby. About further developments, see references. Notes.
References. [4], [5], [14].
129
2.13. Finite dimensional C*algebras Lemma 2.13.1. Let M be a finite dimensional factor on a Hilbert space H. Then M is spatial * isomorphic to B(H)01K, where Hn is a ndimensional Hilbert space, n 2 =dim M, and K is some Hilbert space. A nonzero projection p of M is said to be minimal, if a projection Proof. q E M with q < p implies either q= 0 or q = p. Since M is finite dimensional, it follows that there is an orthogonal family {19 1 , • • • , NI of minimal projections of M such that E7_, pi 1. For each i E {1, • • • ,n},Mi = piMpi is a factor on NH, and any projection of Mi is either zero or pi (= 1 H,) since pi is minimal. By Proposition 1.3.4, Mi = ON.
For any 1, j E {1,• • • , n}, since M is a factor, it follows from Theorem 1.5.4 that we have either pi ‹ pi or pi ‹ pi. But pi and pi are minimal, thus pi is equivalent to pi (relative to M). Let p = pl and K = p i n'. Since Mp = 01K, it follows by Theorem 1.5.6 that M is spatial * isomorphic to B(H)01K. Q.E.D.
Corollary
Let M be a finite dimensional factor. Then M is * isomorphic to the algebra of n x nmatrices, where n2 =dimM. 2.13.2.
Theorem 2.13.3. If A is a finite dimensional C*algebra, then A is * isomorphic to a direct sum E7.1 1 EDMni , where Mn, is the algebra of ni x matrices, 1 < I < m. Moreover, the sequence fni, • • • , n 7,1 is a complete invariant for the algebraic structure of A, i.e., if B is another finite dimensional C*algebra with the associated sequence {ril l • • • , ni mi}, then A and B are * isomorphic if and only if {n l , • • • ,n} = {W I , • • • , WT4.
Proof.
Let {A, H} be a nondegenerate faithful * representation of A. Then r(A) is nondegenerate on H and dimr(A) . • • > zn (.) > • • •. We claim that z n (.) E
c(n),vn,
limz n (P) = 0, Vp E n
n.
In fact, for any p E 12\{0}, let {rp , H, p } be the cyclic * representation of A generated by p. Then Z(p) z
—
(rp(xi dnx 1 ) GI G) •
By Lemma 2.14.7, irp (a) Hp is dense in H. Moreover, rp (dn )rp (a)77 = rp (a 1 + tt) 7.7 —) rp (a)77,Vri E H. Thus r(d) —) 1 (strongly), and Zn (p) —) 0,Vp E ft. Now by the Dini theorem, we get
niax{Izn(P)1 ho E n} —> 0 Further, lizn il —> 0 by Corollary 2.3.14, i.e. xi drix —) x. Therefore,
11( 1 — dn) xii 2 < 4 11x11  11( 1 — dn) 1/2x1/2112 = 4 11x11 . 11 xi/2 (1 — dn )xV2 11 —> O.
11xdn — x11 2 =
Q.E.D. Theorem 2.14.9. Let A be a separable C*algebra. Then A has a strictly positive element at least.
135
n
Let {x7, } be a countable dense subset of A + S and a = En 2 nxn) < 1} is the closed unit ball of A. For any state p where S = {h E A i ilbii ___ on A, since p(x n) > 0 for some n, it follows that p(a) > O. Thus a is strictly Q.E.D. positive.
Proof.
Proposition 2.14.10.
If A has a strictly positive element, then the set of strictly positive elements is dense in A.
Proof.
Let a be strictly positive. For any b E A + , (b + ti a) is also strictly positive, and (1) + 71.4 a) —> b. Therefore, the set of strictly positive elements is Q.E.D. dense in A.
Banach * algebras
Definition 2.14.11.
A is called a Banach * algebra if A is a complex Banach algebra and admits a map: x —> x* (E A) with the following properties: (Ax + ity)* = Ax* + rzy*, (xy)* = y*x*, (xl* = z,
Vx, y E A, X, p, E Œ. The * operation on A or A itself is said to be hermitian, if for any x* = X E A, its spectrum o(x) c 1R. z E A is said to be positive, denoted by x > 0, if f = x and a(x) c 1R = [0, oo). Moreover, a> b if (a — b) > O.
Lemma 2.14.2.
Let A be a Banach * algebra with an identity, and B be a maximal abelian * subalgebra of A. Then B is closed, and o(b) = crA(b),Vb E
B. It is easily verified that B is closed. Now suppose that b E B, A E Œ Proof. and (b — A)  ' exists in A. Since {(b — A) 1 , (b* — X)', B} is commutative and B is maximal abelian, it follows that (b — A ) ' E B. Therefore, for any b E B we have a B(b) = cr A (h) . Q.E.D.
Lemma 2.14.13.
Let A be an abelian semisimple Banach * algebra with an identity. Then the * operation is continuous automatically.
Proof. Suppose that [1 is the spectral space of A. For any p E 12, define T(a) = p(a*),Va E A. It is easy to see that # E n. Now let {x,4 } C A and x, y E A be such that
Ilxn — xll —) o l and
114, —
yll —*
O.
136
Then for any p E
n, 1P(x
— Y11
— x)1 + Igsn — Y*) 1 = 1P(xn — x)1 + 170(en — Y)1 Ilxn — x11 + 114 — Yll —> 0) 1P(xn
i.e., p(x — y*) = 0. Since A is semisimple, it follows that z = y*. Thus the * operation is a closed linear operator on the real Banach space A. Further, Q.E.D. the * operation is continuous. Theorem 2.14.14. Let A be a Banach * algebra with an identity, a E A, a > 0, and a be invertible in A. Then there is u E A such that: 1) u > 0 and u is invertible in A; 2) u2 = a; 3) if B is any maximal abelian * subalgebra of A, and a E B, then u E B too.
We may assume that 11a11 < 1. Thus v(1 — a) < 1, and there is E G (0,1) and a positive integer no such that 11(1 — a)nIllin < 1 — E l Vn > no. Since the complex function
Proof.
oo
(1 + z)h/2 =
E AnZn n=0
is analytic in 1z1 < 1, it follows that the sequence k {ak = E An(a — 1) n
I
k = 0,i,•
n=0
.1
is convergent. Suppose that
a k —* u + iv, where u* = u, v* = v. Then we have
(u + iV) 2
=
a.
(1)
Since a* = a, it follows that (2)
Now let B be a maximal abelian * subalgebra of A and a E B. Clearly, a k E B,Vk. From Lemma 2.14.12, we have (u + iv) E B. So it is obvious that u l v E B,
and uv L vu.
(3)
Summing up (1), (2), (3) , we obtain a = u2 — v 2 1
Ul V E
B, and uv  0.
(4
)
137
Let R be the radical of B. Clearly, R* = R. Hence B/R is an abelian semisimple Banach * algebra. By Lemma 2.14.13, the * operation is continuous on B/R. Suppose that b —) b z. bF R is the canonical map from B onto B/R. Then ..... * , (ak — u) = (ak — u) (i—y) := (i— y)* . Thus il  r), and v E R. If 0 E a(u), then by Lemma 2.14.12 there is p E fI(B) such that p(u) = 0, where 11(B) is the spectral space of B. Since y c R, so p(y) = 0. Then p(a) = p(u 2 — y 2 ) = 0, and 0 E aB(a). This is a contradiction since a is invertible in A and aB(a) = GrA(a). Therefore, u is invertible in A, and B. So from (4) we can see that y  u l uy 0, and a = u2 . u Finally, for any p E 12(B), by Lemma 2.14.12 we have that A = p(a) E (0, 1) . Then k
P(ak) = E An(À — 1)n —> (1 + (A — 1))1/2 = A112 > 0, Ts= 1
and p(u) = limk p(ak) > 0. Again by Lemma 2.14.12, a(u) C JR+ , i.e. u > 0. Q.E.D. Theorem 2.14.15. Let A be a hermitian Banach * algebra. Then A i_ ,fa EA I a > 01 is a cone, i.e., if a,b E A + , then (a + b) E A. Proof. We may assume that A has an identity. First step. To show the following inequality:
v(x) < v(x*x) 1 /2 ,
Vx E A.
In fact, fix x E A and E > 0, and let y = (v(x*x) + er 1 /2 x. Then v(y*y) 0, and (1— y*y) is invertible in A. By Theorem 2.14.14, we have an invertible element w of A such that w > 0 and w 2 = 1 — y*y. Notice the equality: (1 + y*)(1 — y) = w[1 + w 1 (y* — y)w 1 ]w. Since a(iw 1 (y* — y)w 1 ) C IR, it follows that the right side of above quality is invertible. Further, (1 — y) has a left inverse. Suppose that v(y) > 1. Pick A E a(y) such that lAi = v(y). Since v(y*y) < 1, it follows that (1 — IXI'y*y) is positive and invertible. Similar to the preceding paragraph, (1 — A  ly) has a left inverse. Let z be the left inverse of (y — A). Since A is a boundary point of a(y), we can pick a sequence {A n } of regular points of y such that A ri —> A. Then il( Y — Xn) i ii —* oo, and 1 — 11 4Y— x)(Y— xn)  '11.11(Y— Anr ill  i
= I lz + (An — x)z(y — Àn) 1 11 . 11(Y — xn) 1 11 1 5 IIzII . II(Y — An) 1 1 1 + lAn —
AI
.114 —) ci.
138
This is impossible. Thus v(y) < 1, and v(x) < (v(x*x) + e)V 2 . Since arbitrary, we have v(x) < v(x*x) 1 /2 ,Vx E A. Second step. To prove that
E
is
v(hk) < v(h)v(k), Vh* = h,k* = k E A. In fact, from the first step we have v(hk) 2
0, similarly we have (1) 1% all + 11): ), i.e., —A V ° ( a + b). Moreover, ° ( a + b) C .11? since A is hermitian. Therefore, (a + b) > O. Q.E.D.
Theorem 2.14.16. Let A be a Banach * algebra. Then A is hermitian if and only if a* a > 0,Va E A. Suppose that a*a > 1R + . a(h) .E, then o(h2) Thus A is hermitian. Conversely, suppose that A identity. Suppose that there is
Proof.
= h E A such that This is a contradiction since h2 = h*h > O.
0,Va E A. If there is h*
is hermitian. We may assume that A has an x E A such that
6  inf{A I A E a(x* x)} < O. Replacing x by /Ix (some p> 0), we may assume that 6 E (1,P. Put y , 2x(1 + ex)  i
Then 1 — ysy = (1 — x*x) 2 (1 + x*x) 2 ?__ 0, and o(y*y) C (—oo, 1]. Write y 7 h + ik, where h* = h, k* = k. By Theorem 2.14.15,
1+ yy* = 2(h2 + k2 ) + (1 — y*y) > 0,
139
and o(yy*) c [1, oo). Since a(y*y)\{0} : cr(yy*)\{0} (see Lemma 2.2.6), it
follows that o(y*y) c [1,1].
From 6 E
ale z)
and y*y = 4x*x(1 + x*x) 2 , we have 46/(1 + .6) 2 E o(y*y).
< i.e., 6161 < 1+6 2 . Further VI< 1/3 since 1 +6 2 < 2. (1 + .6) 2 I — 1, Then we obtain a contradiction since 6 E (1,1/3). Therefore, we have ex > 0,Vx E A. Q.E.D.
Thus 1
4. 6
C*equivalent algebras
Definition 2.14.17. Let A be a Banach * algebra with an identity. A linear functional p on A is called a state, if p(1) = 1,
and p(a) > 0, Va E A.
If A is also hermitian, then for any h* = h E A, p(h) E 11? since Ilhil+h > O. Further p(al = p(a),Va E A. Moreover, by Theorem 2.14.16 we have also the Schwartz inequality:
Ip(b* a)I 2 < p(a* a)p(b*b), Va,b E A.
Lemma 2.14.18.
Let A be a hermitian Banach * algebra with an identity, and h* = h E A. Then for each A E [Ai, X 2 ], where A1 = minfil I IL E alit» ) A2 = max{kt I A E cr(h)}, there is a state p on A such that p(h) = A. Proof.
On the linear subspace [1,h] of A, define p(a + f3h) = a + (A,
Va, 0 E Œ.
Suppose that a+ f3h > 0 for some a,3 E C. In particular, a, (3 E IR. Then the real number (a + #A) is between (a + PI ) and (a + f3A 2 ). Since (a + Pi) E a(a + Oh), it follows that (a + flAi) > 0, j = 1,2. Thus (a + fiA) > O. This means that p is a state on [1,4 Now by Theorem 2.14.15 and the fact that the * operation is hermitian, and by a similar proof of Proposition 2.3.11, p can be extended to a state on A.
Q.E.D. Lemma 2.14.19. Let A be a compact subset of 0, and 0 E A. Then for each A E 0, we have 1 max{IÀ + til I il E A} ? i(max{lizi I kt E Al + IA1).
140
Proof.
Since IA + AI > Ip.1 — IAI, it follows that
max{IÀ + Al I IL E A} ? max{111 1 1 'IL E A} — lAl. In addition, by 0 E A we have 2 max{IÀ + Al I 11 E A} > 21AI. Therefore, Q.E.D. max{IÀ +Al I A E Al ? .1(max{1A1 I ti E A} + 1A1). Lemma 2.14.20. Let A be a positive constant K such that
Banach * algebra. Suppose that there is a
KIlh11 < v(h),
Vh* = h E A.
Then the * operation on A is continuous.
Proof. Let H= {a E A I a* = a}. It suffices to show that H is closed. For any h E H, there is a sequence fhn l c H such that IIhn — hll —* O. Then for each E > 0, we have that v(h) + E > v(h) > KIIhn ll if n is sufficiently large. Hence we get (1) v(h) > KINII,Vh E H. Now let {hr,} C H,hn —> k, and k* = —k. Since (14n + hn) 2 E H and (h,n + hn ) 2 —> (k + h n) 2 as m  oo, it follows from (1) that
< v((k + h n) 2)  v(k + h n) 2 KIRk + h) 2 11 — n)1 2 :: v(h n — k) =v(k+h 2 iihn — kii 2 —' 13 )
as
n ' —
Thus
11 k2 +11!II = ill(k + hn) 2 + k — hn) 2 1 (
i ( I1(k + h ) 2 11 + Il k — hnI 1 2 ) —>
O.
Further, 211k2 I1 < Ilk 2 + qt11 + Ilk2 — hIl —> 0, so Ilk2 11 = O. Again by (1) we have 0 = Ilk 2 11 > v(k 2) = v(k) 2 ? K 2 I1k11 2)
and k = O. Therefore, H is closed.
Q.E.D.
A Banach* algebra (A,1111) is said to be C*equivalent Definition 2.14.21. , if there is a new norm 11 ' 111 on A such that . 111 '''s 11' 11 OA ' 111) is a
C*algebra.
Theorem 2.14.22.
Let A be a hermitian Banach * algebra. If there exists
a positive constant K such that
KIIhIl 5._ v(h),
 h E A, Vh* =
141
then A is C 5equivalent. If A has no identity, then we consider the Banach * algebra A44. Clearly, (A4) is still hermitian. Suppose that (h+ A) is a selfadjoin element of (A4 4 ). Obviously, h* = h(E A), and A = X(E IR). Since A has no identity, it follows that 0 E a(h). By Lemma 2.14.19 and K < 1, we have
Proof.
v(h + A)
=
max{ IA + AI
I
11, E a(h)}
?_ i(v(h) + I AI) ?1 (11h11+ IAD ' 4 (11h+ All). Thus, we may assume that A has an identity. By Lemma 2.14.20, there is a positive constant M such that
Va E A.
Ila * II 5 M211a11)
Let p be a state on A, and L p = {a E A I p(a* a) = 0 } . By the Schwartz ap = a + Lp is the inequality, Lp is a left ideal of A. Suppose that a canonical map from A onto AlL p , and define an inner product on
(a p ,b p ) = p(b* a),
Va,b E A.
Denote the completion of (A/Lp ,(,)) by H. For any a E A, define a linear map r(a) on AlL p :
r p (a)b p = (ab) p ,
Vb E A.
For any E > 0, by Theorem 2.14.14, there is u* = u E A such that
IIa*aII +
E 
a* a = u2 .
Then by Theorem 2.14.16, h* (iia*ail + E  a*a)b = (ub)*(ub) > 0,Vb Hence, Ila*allp(b*b) + ep(b*b) > p(b*a*ab). Let E > 0+, then we get
E A.
Ilir p (a)bp 1 2 = p(b* a* ab) 5
iia * ail • iibpir
5 1 f 2 11a11 2 . Ilbp11 2 )
Vb E A.
So irp (a) can be uniquely extended to a bounded linear operator on Hp , still denoted by ir(a). Clearly,
Illrp (a) II 5 Milaii) P(a) :__ (r(a)1,1), and {rp , Hp } is a * representation of A.
Va G A,
Let S(A) be the state space of A. Construct the universal * representation of A: Ir 7
E
pES(A)
orp)
H=
E pE S (A)
EDH„,
142
and let Ila Ili = 117(011)Va E A. Suppose that there is a E A such that r(a) 0. Write a = a l + ia 2 , where a; = al , a; = a 2 . Then ir(ai) = 7r(a 2) = 0. In particular, p(a i ) = P(a2) = 0, Vp E S(A). By Lemma 2.14.18, v(a i ) = v(a 2) : 0. But v(ai) > KIlaillli = 1,2, so a l  , a 2 = 0. Therefore, II • III is a norm on A. Clearly Ila*alli = Ilall?)Va E A. Now it suffices to show II • II ' II • Ili on A. Obviously, II • Mil • II. Moreover, let {a n } C A and Ilanill + 0. We may assume that a n* = a n since 1141,111 = Ila n ili,Vn. By Lemma 2.14.18,
Ilan Ili
= suP{11 7 p(an)11 IPES (A)}
?__ suP{IP(an)1 IPES (A)} = v (an)
Ma rdi,
Q.E.D.
Further, Ma n i' + 0 too. Therefore, II • II " II • III on A.
Theorem
2.14.23.
Vn.
Let A be a Banach * algebra. If there is a positive
constant K such that
KlIa * all ? Ilal ' Ilall for any normal element a of A (i.e. a* a = aa*), then A is C*equivalent Proof. For any h* = h G A, by the assumption Ifilh2 II _ > 11h112. Generally, we have K2n —liii 2n 11 4 II ? 1111112n) Vn. Thus Kv(h) _ > 11h11)vhs = h E A. Now by Theorem 2.14.22, it suffices to prove that the * operation on A is hermitian. Let h* = h E A. By Lemma 2.14.20, the * operation is continuous on A. (ity is a normal element of A, and f (t h)* = Thus f (th) = e ith _ 1 = Encii f (—th),Vt E E. Then from preceding paragraph and the assumption, we have K v (2 _ e ith _
e —ithN _= )
K v(f (th)* f (th)) ii f (th)* f (th)II
> K  Illf(th)"ll • Ilf(th)11 vt E ?._ Ic 'v(f(th)) 2 ,
E.
Let fi = max{ IImA I I A Eo7 (h)}. Since ci(h) = a(h), it follows that there is a E 17 such that (a + 1 f3) E cr(h). Then for t > 0, 2(1
+ e at ) > v(2 — eith — cith) > K 2 1./(f (th)) 2
> K 2 11
— eit( aifl ) 1 2
= K 2 (1 + e2/3t — 2e 9t cos at).
143
This is impossible if # > 0, so it must be that # = 0, i.e., * operation on A is hermitian.
a(h) C
IR, and the
Q.E.D.
The axioms for C*algebras
Let A be a Banach * algebra with an identity. If Theorem 2.14.24. there is a positive constant K such that Mel < K,Vh* = h E A, then A is C*equivalent. Moreover, if K = 1, then A itself is a C*algebra. Proof. First step. To show that A is hermitian. Suppose that h* = h E A, and (a + ifi) E a(h), where a, (3 E IR. By a(h) = c(h), we may assume that /9 0,
K > Ile ith il > le"("W) i
e fit
Thus # 0, and a(h) C Second step. To prove that h E A, 11h11
h*
inf{11h2 II
1} = E > 0.
In fact, let h* = h and 11h11 = 1. Put 11h2 11 = ii. Clearly 0
0.
Third step. We claim that
v(h) ? EMI')
Vh* = h E A,
where e is as in second step. In fact, from second step we have 11h2 II _ > 6 114 2 > II2n. h E A. Generally) 11h2 Therefore, v(h) > for any h* hil 6
IIhII)Vh* = h E
A.
Now by first, third steps and Theorem 2.14.22, A is C*equivalent.
144
If K = 1, consider identity map I: (A )11 ' MO —> (A, ii ' II)) where'll' '11'11 on A, and (A,11'111) is a Cealgebra. By Theorem 2.14.5, we have that 11 1 11 < 1, i.e.) Ila II < IlaIIIIVa E A. Suppose that there is a () E A such that MaoIli > Ilao11. By Proposition 2.1.8,
* * * v(ao* ao) 5 Had '110011 < Ilao* Ili • 00111 = Ilaoaoll = v(aoa0)• This is a contradiction. Therefore,
Ila II = lialli)Va E
A.
Q.E.D.
Let A be a Banach * algebra with an identity, and Ila *all = Ila s 11 ' Ila II for any normal element a E A. Then A is a C*algebra.
Lemma 2.14.25.
Proof.
Let h* = h
E
A, and an(h)
=
normal, and an (h)* = an (—h). Then
k ,Vn. En 1—(ih) k!
Clearly, o(h) is
k=0
Ilan(h) . an(  01 = Ilan(h)11 •11 (112( 41) Vn. Let n —> co, we get 11 eihll •Il eihll :". 1. By Theorem 2.14.23 and the assumption, A is C*equivalent. In particular, the * operation is hermitian. Thus a(h) C E. Further, we have Hell —> 1 ) Ile  ii 1 ilII 1. Therefore, Me h l' = 1, Vh* = h E A. Now by Theorem 2.14.24, A is a C*algebra. Q.E.D. Let A be a Banach * algebra. If Ila* all =1101'114 for any normal element a E A, then A is a C*algebra. Theorem 2.14.26.
By Lemma 2.14.25, we may assume that A has no identity. By Theorem 2.14.23, A is C*equivalent. Suppose that 11 • II' is a norm on A such that 11'11 '11'11' and (A, II 'Ill is a C*algebra. By the assumption, we have 11h11= v(h) = lihil% Vh* = h E A.
Proof.
In particular,
Ildi ll = II dill' ) V1) where {c11 } is
an approximate identity for (A,
11'11').
We say that for any a E A,
Ilall = suP{Ilabil 1 b E 21 )114
1} .
Indeed, since 11.11 1 and II ' II are equivalent, it follows that lladi — all —* O. Then
suP{Ilabll l b E Adibil Thus Mall = sukilabil i b E Adibil < 11. On A10, define liall
1}
Madill —*• OM.
lia + All = sukilab+ Abll l b E A) libil
11)
145
Va G A, A E C. Suppose that for some a G A and A G 0 we have ab + Ab = 0,Vb G A. Since A has no identity and (A, ll'in is a C*algebra, it follows from Proposition 2.1.2 that a = 0 and A = O. Thus (AiO, 1111) is a Banach * algebra with an identity 1, and (A, 11 *II) is a Banach * subalgebra of (A4 0,11 • ID. We need to prove that (A40, il ' II) is a C*algebra. By Theorem 2.14.24, it suffices to show that
Vh* = h G A.
lieihii 5_ 1,
Fix h* = h E A. By the definition of the norm on (A40 there is a sequence {bn} c A with Ilbnii 5_ 1,Vn, such that Mehl' :. li4n iieth bnii. Let B be the closed * subalgebra of A generated by {h, bn.b:i I n } . Then (BIli ' Ir) is a separable Cesubalgebra of (A, II ill. If B has an identity p, by Lemma 2.14.25 then (B, fi * fi) is a Cealgebra. Thus
ileih ii = limiie sh bnil = limli(P + n n
i=1
' (ihY
E )bnll il
1!
Now suppose that B has no identity. On B40, define
lib + 41 = sup{libe + Xell I c E 13 )114 5 lb Vb G B, A E OE. Similarly, (B4,11 'Hi) is a Banach * algebra with an identity 1, and 011 = libil i ,Vb E B. By Theorem 2.14.9, (B, II ' In has a strictly positive element a. And by the proof of Theorem 2.14.8, (B1 il ' In. Since lib+ Ali?
{4 = (aillall i ) ,;} n is
an approximate identity for
11 0 + Wn11 ? I I(b + x) cin ell —> 1 1 (b + A ) ell
Vc E B and Hell < 1, it follows that
+ A)dnii) lib+ Alii = limii(b n
Vb E B, A E C. In particular, iie ia b ' Pia Ill =
lirn n Il eiadnil
l eiadnil
11111 I ( eia dn) * ( ei(dn) I n lin n l Il d2n11 = linl n 11411 1 = 1 '
Moreover, since or( a ) C 117 and Ile ±ia lli > v(e) = 1, it follows that Ile±ia Iii = 1. By Theorem 2.14.9 and Proposition 2.14.10, the set of strictly positive
146 elements is dense in (B,
11 ' 111)+• Noticing that 'L,3
e"2 Ill and II • II
"s 11 '
b2 E B
7 1 1 11 0i — "•
we have
Ile±" Ili = 1,
Vb E
II
Now we come back to consider h. Write h = h+ h_ where h± E (B,II•11') + , and h+ • h_ = O. Since 1
= Palk
it follows that
II
eih
ea_
1le  lliL= 1
< Ileih ili
)
Ileih Ili = 1. Further,
11 ethIl
11m Ileihbnll
= lleull i =
suP{ 11 eih bli 1 b E B 1 lIbll
0, noticing that (XY* iix)(AY + iix * — (AY* — Ax)(XY )
2Xigyee + xy),
iix * )
147
and
(Ay* + ttx)(Ay ittx*) — (Xy* — ip,x)(Xy
ittx*)
intr,(—y*x* xy), and by the Cscondition, we have 4
14+ JLX 11 2 + 14 — ihx* 11 2
44*i
+14 — iits* 11 2 + 14+ ix* 11 2
5 4 (40 +1.1 11x* 11) 2 . Let À =
= 11x* I1  1 , then we get
11xYll S 4 11x*1111Y11,
Vx, y E A.
2) We prove that 11x* I1 = 11xII In fact, if h* have
and
h E A, then
vx, y E A.
11xyll S 411x11 IIYII)
11h2 11 = 11h11 2 by the C.condition. Generally, we Vn.
1 11 11 = '
Thus for any a G A, by 1)
Ila* all 2n = 11(ea) 2f 11 = lia s (ace) 2nl all < 42 011 2 ' 11("1 2n1 11)Vn. Noticing Ilh 2k 11 = 11h11 2k (Vk have that
and h*
=

h E A) and the claim 1), we
11(ae) 2i ll = 11(aal l+2++2n1 11 • 11 (aa * ) 2 4 n1 II aa* I111 2n1 ,Vn.
• • II (aa * ) 2n II
Thus
Vn. HearS 41"1 1 1a11 2 .11aa*112" 1, Further, 11a*all < Ilaa"loVa E A; and Heal' = Ilaa*II,Va E A. By the C*condition, 11x112 = 11x * x11 11xx* I1 Ilx* 11 2 i.e.) 11x*11 11x11,Vx E A. Again by 1 ), 11xYll 5 4 11x1111Y11,vx , Y E A. Now define a new norm II '= 411 II on A. Then by 1), 2), (A II • ) is a Banach * algebra, and 11x*11 1 = 11x1111Vx E A. Moreover, by the C*condition
Hear = 4 11a*all = 411a11 2
•
Hall%
Va
E A.
148
From Theorem 2.14.23, (A, Il 'Ill is C*equivalent, i.e., II
II ' II' II' =
•
on A, where lai = v(a*a) 112 (Va E A) is the C*norm on A. Furthermore, by the C*condition of II • II
lal 2 = v(a * a) = lip
Ra ) '1 2 n
= liP11(a * a) 2n II 2 n = Ila * all = 011 2, Va E A. Therefore, (A, II • II) is a C*algebra.
Q.E.D.
Notes.
In 1943, I.M. Gelfand and M. Naimark proved the following theorem: If A is a Banach * algebra with an identity, and satisfies: 1) Masai' = Mall • Ilall)Va E A; 2 ) Ilell = Ilall,Va G A; 3) (1 + a*a) is invertible, Va G A, then A is isometrically * isomorphic to a uniformly closed * operator algebra on some Hilbert space. This is a fundamental theorem for the theory of C*algebras. Also, they conjectured that the conditions 2) and 3) are not necessary. I. Kaplansky pointed out that the condition 3) can be canceled easily. Then we have the usual definition of C*algebras (2.1.1) and the Theorem 2,3.20. Proposition 2.14.1 is due to J. Glimm and R.V.Kadison. Using this proposition, they answered affirmatively the GelfandNaimark conjecture for unital case. Theorem 2.14.14 is due to J.W.M. Ford, Theorem 2.14.15 is due to V.Pt6k; and Theorem 2.14.16 is due to S.Shirali and J.W.M. Ford. These results also have their own interest in the theory of Banach algebras. Theorem 2.14.22 and 2.14.23 are due to R.Arens. Theorem 2.14.24 is due to B.W. Glickfeld. And Theorem 2.14.26 is due to G.A. Elliott. The question on the submultiplication of a linear C*norm was presented by R.S. Doran. Theorem 2.14.28 is due to H. Araki and G.A. Elliott.
References. [6], [7], [30], [41], [48], [50], [52], [53], [56], [66], [87], [90], [131], [162], [191], [201].
2.15 Real C*algebras Definition 2.15.1.
Let A be a real Banach * algebra ( i.e., a Banach * algebra over the real field .11?) .A is called a real C*algebra, if A, = A4iA can be normed to become a ( complex ) Csalgebra such that the original norm on A remains unchanged.
149
A real Banach * algebra A is a real Ce—algebra if and only if A can be isometrically * isomorphic to a unifomly closed * algebra of operators on a real Hilbert space.
Proposition 2.15.2.
Proof. Let ir be an isometrical * isomorphism from A into B(H), where H is some real Hilbert space. Consider the complex Hilbert space He = and define Va,b E A lia + ibii iir(a) + It is easily verified that A, = A4A is a ( complex ) Ce—algebra by this norm. Moreover, it is obvious that hall = 11 7r(a)11 = 117(a)111.1c)Va E A. Therefore, A is a real Ce—algebra. Conversely, let A be a real Ce—algebra. Then A, = /14iA is a ( complex ) Ce—algebra. We may assume that A, C B(H) , where He is a complex Hilbert space. Consider He H as a real linear space, and define () 71)r = Re () 7/), V,T7
E H.
Then H is a real Hilbert space, and A is a uniformly closed * algebra of Q.E.D. operators on H. Let A be an abelian real Ce—algebra. Then A is isometrically * isomorphic to
Proposition 2.15.3.
{f
e cr(n)if(i) = f(t),Vt E nl,
where f1 is a locally compact Hausdorff space, and bar "—" is a homeomorphism of such that I = t,Vt E !)
Proof.
Assume that A, = A4iA Cr(n), and for each t E!) define
t(a ib) = (40 ib(t),
Va,b E A,
where z —> x(.) is the Gelfand transformation from A onto Cr (12). Clearly, t 1 is a homeomorphism of 10 t,Vt E 11; and {aHla E A} C {f E Cr(11)1f(i) = f(t),Vt E fn. Now let f E cr(n) with f(t) = f(t),Vt E Suppose that f(.) = a(.) ib(), where a,b e A. Then a (t)
ib(t) = (a + ib)(t) = f(t) f
a(1)
ib(t) = a(t)
Hence, b(t) = 0, Vt E 11, b = 0, and f = Now let A be a real Ce—algebra, and define J(a ib) = a — ib, Va, b E A.
T,Vt E 12. Q.E.D.
150
Then J is a conjugate linear isometric algebra A, = AiiA, i.e.,
J(Xx + Ay)
*
isomorphism of the
(
complex
) C"
= XJxFlay,Jx* = (Jx)*,Jxy= Jx • Jy,
J2 = id) 11Jx11 = 114 V x, y E A„ A, it E Œ. In fact, it suffices to show that J is isometric. We may assum that A where H is a real Hilbert space. Since for any a,b G A, ,ri E H,
C B(H),
 11(1 — bnl1 2 + Ilb+ anl1 2 — II(a — ib)( — in)I1 2 ) g + in11 2 = IK —in11 2 = R11 2 + 11n112) we have 11Jx11 = 11x11)Vx E 11(a + ib)( + n)II
and A. Conversely, let J be a conjugate linear isometric * isomorphism of a ( complex ) C"algebra A„ and A =. {x E A c kix = x}. Then it is easy to see that A is a real C*algebra, and A, = AiiA. Therefore, we have the following Proposition 2.15.4. There is a bijection between the collection of all real C"algebras and the collection {(A„ J) 1 Ac is a complex C"algebra, J is a conjugate linear isometric * isomorphism of Ae } .
Proposition 2.15.5.
Let A be a real C"algebra, and define
f (a + ib) = f (a) + i f (b), Va,b E A, f E A". Then A* can be embedded isometrically into A: such that A: where A, = Ai I A.
= A"IA",
Proof. We may assume that A C B(H), where H is a real Hilbert space. Since f (E Al can be extended to a linear functional on B(H) with the same norm, we may also assume that A = B(H) and 11111 = 1. Then there exists a net {0 c B(1/). = T(H) such that Pill 1
< 1,Vi; and fi (b) = tr(t i b) —) f(b),Vb E B(H).
Clearly, we have fi 4 f in topology 413(He )",B(He )), where He = HiiH and t i as an element of T(H) maintains its trace norm, so fi as an element of B(H,)* maintains also its norm, i.e., < 1. Hence, f as an element of B(H,)* has the norm one. Therefore, A* can be embedded isometrically into A.
Q.E.D. Proposition 2.15.6. Let A be a real C"algebra. p E A* is called a state , if p(a*a) _.?_ 0, p(a*) = p(a),Va E A, and 1111 P = 1. Denote the state space of A by S(A). Then 8(A) = Re S(A) = {Re p e lp, is a state on Ae } , where
151
A, = A4iA is a (complex) ) C*algebra. Moreover, for each p E S (A ), by the GNS construction there is a cyclic * representation {r,, Hp , p l of A such that p(a) = Or p (a) , ,,), Va E A, where
Hp
is a real Hilbert space; and {ir = E3pEs(A) 71p)
H 7 pEs(A)Hp}
is a faithful * representation of A. The proof is similar to Section 2.3.
Definition 2.15.7. Let A be a real Banach * algebra. For each x E A, let ci(x) be the spectrum of x in the (complex ) algebra A, = A kA, and v(x) = max{lAll A E a(x)} be the spectral radius of x. An element a of A is said to be positive , denoted by a > 0, if a* = a and a(a) C [0 7 °4 A is said to be hermitian, if for each h* : h E A we have cr(h) C .1R; A is said to be skewhermitian, if for each k* = k E A we have o(k) C OR; A is said to be symmetric, if for each a E A we have a* a > O. Theorem 2.15.8. Let A be a real Banach * algebra. Then A is symmetric if and only if A is hermitian and skewhermitian. Moreover, if A is symmetric, then A +  {a E Ala > 0} is a cone, i.e., if a,b E A + , then we have (a + b) E A. The proos is similar to Theorem 2.14.5 and 2.14.6. Let A be a real symmetric Banach * algebra with an Lemma 2.15.9. identity 1. A linear functional p on A is called a state, if p(1) = 1, p(x* x) _>: 0, p(f) =. p(x),Vx E A.
Then for each h* = h E A there is a state p on A such that p(h) = v(h). Proof. Let AH = {h E Alh* = h} and AK :7 {k E Alk* = k}. Then A  AH: AK. Now for h E AH, define a linear functional p on [1, h] = {a + Ala, 10 E JR} :
p(a + Ph) = a + fiv(h),
Vot, f3 E IR.
Clearly, p(a) > 0,V 0 < a E [1,4 and p(h) , v(h). Let I
E = {(E,pE)
I I
E is a linear subspace of Ali ) and1, h E E; PE is linear on E, andp(a) > 0,V0 0, by Theorem 2.14.14 ( it also holds for real case) there is b E A such that ilair + e — a* a.
Hence, for any state p on A, by the GNS construction there is cyclic * representation {rp , Hp , ep } of A such that p (a)  (rp (a) G )
el)) ) Va E A.
Now by Lemma 2.15.9, A admits a faithful * representation {A , H}, where H is a real Hilbert space. Define
11x111 = 117r(x)11) Vx E
A.
Then 11' Ili is a norm on A, and Ilexili = 11x111,Vx E A. For any h* = h E A, clearly we have a(r(h)) C a(h). Hence, 11h111 ; v(r(h)) < v(h) = 11h11,Vh E A. On the other hand, for any state p on A we have
11 10)11 ? 1(rp(h)ep)ep)1 7 1P(h)1. By Lemma 2.15.9, 11h111 = 11 7 (h)11 v(h) = 11h11,Vh* ; h E A. So 1141 = 11h11)Vh * = h E A. Further,
I X11 1 = 11X *X11 1 12 = IIX*X 1 1/2 = 114) VX Therefore, A is a real C*algebra.
E A.
Q.E.D.
In above theorem, the hermitian condition is necessary. Indeed, consider Œ with norm 11A11 = Al and * operation A* . A(VA E 0), then the real Banach * algebra 0 is not hermitian. So Œ is not a real C*algebra. Remark.
153
Definitioln 2.15.11. A real Banach * algebra (A, 11 11) is said to be real C* equivalent, if there is a new norm 11 ' 11 on A such that 11 ' 11 11 11 i) and is a real C*algebra. (A, 11 11 i) Let A be a real Banach * algebra. Suppose that A satisfies one of the following conditions: 1) A is symmetric, and there is a positive constatnt K such that ifv(x) > of A 11x11 for each z* = z or z* 2) A is hermitian, and there is a positive constant K such that K11x*x11 11x* 11 11x11 for each normal z of A (i.e., x*x = xx*); 3) there is a positive constant K such that ifIlx*x Y * YII Ilx * II '114 for any normal z, y G A and xy Then A is real C'equivalent. Theorem 2.15.12.
1) By Lemma 2.14.19, we may assume that A has an identity 1. By Lemma 2.14.20, AH = {h E Alh* = h} is closed in A. Now let {k„} C AK = {k G Alk * = — IC} be such that k r, E A H For any k E AK let zn = kkn + kn k,Vn. Then zn E AH and z n z = kh hk. Since AH is closed, it follows that z E AH n AK = {0}. Hence, Proof.
,
.
• kh hk 0,Vk E AK 
In particular, knh hk = 0,Vn, and h 2 = 0. But 0 = v (h2) = v (h)2 K2 11h112, so h = 0, i.e., AK is also closed. Further, the * operation is continuous on A. For each state p on A, by the GNS construction there is a cyclic * representation {r,, Hp , p } of A such that P(x) =
( 1r 1, (X ) e G) , vx
E A.
where H,, is a real Hilbert space. Further, let { 7r =
(DpEs(A) 7p, H = (DpEs(A)Hp}
where 5"(A) is the state space of A. Since * is continuous, it follows that
1 7 (x) = 117(x*x)111/2 = v(r(x*x))1/2 5_ v(x*x) 112
< I ex11
Ift I I xll
11x111
= 11r(x)11,Vx E A. Vz E A, where IC is some positive constant. Define We say that h + k, where is a norm on A. In fact, let = 0 and x h* = h, le = k. Then for each state p on A we have
11x111
11 111
IP
I = I P( h ) I  I (7rp (x)
)I
11 7 (x)
I
 0.
154
By Lemma 2.15.9, we can see that h = O. Further, from 0 = r(x*x) = r(10 2 we have k = 0, and x :. O. Now it suffices to show that 11'11 is continuous with respect to 11' Ili Let 'Iznik  O. Since 11411 1 = uII xn 11 1 ,Vn, it follows that IIhnIll  0 and 11k„, Ili  0 ) where h hn i ky't = kn,xn = hn + kn ,Vn. By 11x111 ? v(x) .?_ K 1 11x11, ye . +x E A, we can see that 11hnll ' 0 ) II kn II  0, and 11x4  O. Therefore, 11'11 is continuous with respect to 1111 i , and A is real C*equivalent. 2) By 1), it suffices to show that A is skewhermitian. Similar to the 1), * is continuous. For k* = k E A, consider f(tk) = e tk _ 1,Vt E E. Then f(a) is normal and f (tk)* = f (tk),Vt E IR. Now by the proof of Theorem 2.14.23, we can see that a(k) C ilR. 3) For h* = h E A, let x = cos h  1,y = sin h. Then x,y are normal and xy .T yx. Hence II cos h  1112 _ 1 in A,.
155
Proof. If 11x0+ibll _< 1, then by Proposition 2.15.2 we can see that Ilxo —2:4 = 11x0 + ibll 5 1. This contradicts the fact xci E ExS e . Therefore, 11x0 + ibll > 1.
Q.E.D.
Lemma 2.15.15. Let H be a complex Hilbert space, and A(C B(H)) be a real uniformly closed * algebra. If A n iA = {0}, then A, = AkA is a( complex ) C*—algebra on H, and lia + ibll ? max(11a11,11b11),Va,b E A.
Moreover, if M is a real weakly closed * operator algebra on H, and M im , {0}, then Mc = MitiM is also weakly closed.
n
Since (H, Re(,)) is a real Hilbert space, it follows from Proposition 2.15.2 that A is a real C*—algebra. Then there is a norm 11'111 on A, such that (At, 11' Ili) is a (complex ) C*—algebra, and 11x11 1  11x11,Vx e A. Clearly, the operator norm is also a C*—norm on A. By Proposition 2.1.10, II ' Ili  = II ' II on A„ i.e., A, is a C*—algebra on H. Now suppose that {r, K} is a faithful * representation of A, where K is a real Hiblert space. Then {A , K,} is a faithful * representation of Ac , where If, = K1iK. Hence, for every a,b E A, Proof.
Ila + ibll
' 11 1r(a) + ir(b)II
meil 5 il ? max{ sup 11r(a)11, sup 11r(b)ell} . max(11a11,11b11).
? suP{11 7 (a)e + ir(b)ell le E IC)
11t11.1 11E11 1 Finally, it suffices to show that the closed unit ball of Me is weakly closed. Suppose that (al + tib i )  z E B(H) weakly, where a1 , b1 E M, Vi, and Hai + ib i ll < 1,V1. Since max(Ilaill) Pill) __< 1,V/, and the closed unit ball of M is weakly compact, we may assume that al —+ a and bi —+ b weakly, where Q.E.D. a,b E M. Therefore, z = a + ib E A. Let H be a complex Hilbert space, M be a real weakly closed * operator algebra on H, and 1 = 1 H E M. Then Co {ultt G M is unitary } is weakly dense in S : {x E Millx11 < 1}.
Lemma 2.15.16.
Proof. Since S is weakly compact, so by the Krein—Milmann theorem it suffices
to show that x 0 E Co{u E Mitt is unitary
}W
for any z, E ExS, where "w" means the weak closure. Now let x o E ExS. By Proposition 2.15.13, z o is a partial isometry of M, and (1 — x*,,x0 )/t/(1 — xo xo ) = {0}. For any projection p of M, denote by c(p)
156
the minimal central projection of M containing p, i.e., c(p) is the projection from H onto [MpH]. By Proposition 1.5.9, we have
41 — xo*xo) • 41 — xoxo*) = O. Let z = 1— 41 — xi;x0). Then x8x0 > z,x 0x8 > 1 — z. Replacing {x0 ,M, H} by {xo z,Mz,zH} and {x(1 — z),M(1 — z),(1— z)H} respectively, we may assume that x8x0 = 1, and x0x40` :. p < 1. Now it suffices to show that for any 6, •  • , rii,   • ,n,,,, E H there is a unitary element u of M such that
1((x0—u)eimi)1 n+1
1( ( x 0 — u)eonal = I(Dx0 — u)gkeio1j)1 k>n
_,._ lixo — ull . I E gked. him n
1 < i < m.
Q.E.D.
Proposition 2.15.17. Let A be a real C'—algebra with an identity, and xo be a normal element of A with lixo ll < 1. Then xo belongs to the closure of Co{cos b • eala,b E A,a* = —a,b* = b,a — b — so } where "x ,, y" means that xy := yx. Proof. We may assume that A is abelian. Then by Theorem 2.14.2, x o belongs
to the closure of Co{e k lk E A c ,k* = —k}
i.e., there are A(in) > 0 )
EA
1 and k i( n) *
'7 — k in) (
E A, such that
i
I E A(in) eXp(14n) —
so
il —4 O.
157
Write k(n) = a.(in) + ib2(n) , where a(in) * = a(n) , b i( n) * = b i( n) E A, Vn, j. Since A is abelian: and A, = AiiA, it follows that 3 11E A (in) COS b i( n) • eXP(a i( n ) — X0 11 —4 O. i
Q.E.D. Corollary 2.15.18. Let H be a complex Hilbert space, M be a real weakly closed * operator algebra on H, 1 :. 1 H E M, and S ; { x E Milix11
of Ai , 1 < ï < n1
a(u),
Vu E 07_ 1 Ai. a( • ) is also a C*norm on 07_ 1 Ai . In fact, if a(u) = 0 for some u E 07L 1 Ai, then 0 1 pi(u) = 0, Vpi E Ai , 1 < 1 < n. By the definitions of L,.  • , An , we have u = O. For any pi E Ai ,1 0, Vpi E Ai , 1 < 1 < n,u, v E 07_ 1 A1 . By the definitions of A i ,.   , A n it is clear ,
that 07.60i(vIa u*u) (
—
teu)Y) ? 0,
Vçoi E Si , 1 < ï < n, and u, y G Ø 1 A. Now from Theorem 3.2.6, a(u) > Q.E.D. ao(u),Vu E 0 1 111 . Therefore, ao (•)  a(•) on Ø 1 A.
Let A be a C*algebra on a Hilbert space H, and A = Lemma 3.2.9. fw (. ) = (e) ) I e E H, 11 = 1}. Then the cr(A*, A)closure of C o A contains the state space S(A) of A. Proof.
Let p G S(A). By Corollary 2.3.12, ço can be extended to a state on B(H), still denoted by ço. Since the unit ball of T(H) is w*dense in the unit ball of B(H)*, it follows that there is a net {t1 } C T(H) with p i ll ' < 1,V/, such that tr(t i b)  ço(b), Vb E B(H). By io* := (p, we may assume that tt = t 1 ,V1. Write t i = tiF — ti  , where tt G T(H) + and tiE • ti .= 0, V/. Clearly, Will < litilli —‹ 1,V1. Since the unit ball of B(H) * is w*compact, so we may assume that the nets {tiF} and {tï} are w*convergent in B(H)*. Let ltd . tt+tr,vi. Then there is 1,b E B(H)* such that
170
0(b),Vb E B(H). Obviously, V) is positive on B(H) ÷ . Moreover, tr(It i lb) since ço(1) 1 = lipatr(ti ) < linatr(It i l) = i,b(i) and Illt11111 Ilti Ili 5. 1,V/, it follows that
is a state on B(H) and lipatr(til = 0. Further, for any
_< a E B(H) + , 0 < 0. Thus, tr(ttb) and we may assume that ti > 0, V/. Now write
=E
tr (tib)
where Il
)t.!) (
be),
ço(b),Vb E B(H),
Vb E B(H),
1, M,.9 > 0, Vn, /, and let
(E a(: )) ' An.d ) , e n, n= 1 k=1 N N
C4) NJ()
Clearly, coN it E CoA, and coN,i(a)
VN,1.
ço(a),Va E A.
Q.E.D.
Let {ir1 , Hi } be a faithful * representation of a C*algebra Theorem 3.2.10. Ai ,1 < < n. Then
a0(u)
= II 0:3.1 ri(u)II)
Vu E
where ao(•) is defined by Theorem 3.2.5.
Proof.
By Proposition 3.2.8,
cro(u) 2
sup
__p
II 07=1 ri(u)II ri (uv)
2
07=1
eill2
II O irl=i Ir(v) 07=1 (V* U*UV) ®71Pi or; ipi(v*v)
= sup{ ®= rp,(u)11 2 I
y
e
n, 07=1 Ai and 07=1 7i(v) 07=1 ei
E
Hi)
1
Pi E Ai, y
pi
E 07_ 1 A1
e
I&II  1,1 < j
0}
A i ,1 < I < n},
I
1), and re, is the Vu E V:=1 Ai, where L = {4(1r(.) ) eil1 , 6) i E Hl * representation of A generated by pi, Vpi E Ai , 1 < I < n. Further, from Proposition 3.2.8 and Lemma 3.2.9 it follows that sukil
rp1(u)11 I Pi E Ail 1 Ç j < n} = c o (u),
Vu E 07_ 1 Ai . That comes to the conclusion.
Q.E.D.
1 Ai is From this theorem, the geometric sense of the C*norm ao () on given. Therefore, acl () is called the spatial Cenorm on 07_ 1 A 1 . Later, we
171
shall see that a0 (•) is the minimal C*norm on 07_ 1 ,41 indeed. So sometimes we also denote ao  07_ 1 A1 by min 07_ 1 Ai . Proposition 3.2.11.
01'.... 1 A: is w*dense in (a00:L 1 Ai)*.
From Corollary 3.2.7, 07_ 1 A: C (a007_ 1 Ai )*. We may assume that Ai is a Cealgebra on a Hilbert space Hi ,1 < i < n. By Theorem 3.2.10, (Iv 0 1 Ai is the uniform closure of 07_ 1 Ai in B (ViL I Hi ). Further, from Lemma 3.2.9, the weclosure of Co{07_ 1 (.6, 6) I ei E Hi , 1 < i < n} contains the state space of aoV l ili. Therefore, 07_ 1 A: is w*dense in (a007_ 1 Ai )*. Q.E.D. Proof.
Notes. The tensor products of Ce algebras was studied first by T. Turumaru. c 0 Ø 1 A i is also called the injective tensor product of A 1 ,   • , A n . Proposition 3.2.2 seems very simple, but we got it quite later. B.J. Vowden
gave first proof. The present proof here is taken from C. Lance. Theorem 3.2.10 is due to A. Wolfsohn. References. [95], [171], [1851, [193], [1981.
3.3.
The maximal C * norm
Let A 1 ,  • • , An be C*algebras. From Theorem 3.2.5, there is a C*norm (Il • ) on 07_ 1 A 1 . By Proposition 3.2.2, we have a() < 1() for any Cenorm a() on 07_ 1 Ai . Thus we can define the maximal Cenorm & I N on 07_ 1 ,4i : a l (u) = sup{a(u) I a is a C*norm on 07_, Ai },
Vu E Ø 1 A. And sometimes we also denote a 1 07_ 1 Ai by max07_ 1 A1 . Clearly, ai () < /() on As in Definition 2.4.5, a * representation {7r, H} of 07_ 1 Ai is said to be nondegenerate, if [7(u)e I H,uE 07_ 1 A1l is dense in H.
eE
Lemma 3.3.1. Let {r, H} be a nondegenerate * representation of 0 1 Ai . Then there is unique nondegenerate * representation {ri , H} of Ai , 1 < i < n, such that ' ri(a.i)ri(ai) f ri(adri(a.i) 1 ( r(Ø
a) = ri (ai )    r(a)
Va iE Ai , 1 the left side. Now let ct(.) be a C*seminorm on 07_ 1 Ai . Then I = {v E ® 1 A I a(v) = 0) is a twosided * ideal of Ø 1 A. Suppose that Proof.
v —+ = v
is the canonical map from 07_ 1 ,41 onto 0,7L I A 1 i/, and let a(v), Vi E OriL l Ai //, V E
Then ii(.) is a C*norm on Ott l Ai i/. Pick a faithful * representation 6  (ViL 1 Ai //), and define
of
r(v) = Fr(V), Vv E
Then A is a * representation of ViL l Ai . Now from Proposition 3.3.2, al(u)
illqu)11 =
Vu E ViL l Ai . That comes to the conclusion.
Proposition 3.3.4.
=
a(u))
Q.E.D.
There is a bijection from the set {a(*) I a(.) is a C*  norm on Ai } onto the set {/ I I is a closed twosided ideal of al and I n Ø 1 Ai 0) as follows: Let ot(.) be a C*norm on 07_ 1 Ai . Then there exists unique closed twosided ideal 1,, of ai ® 1 A i such that the map u + = u /a (Vu E ø1A)
174
can be extended to a * isomorphism from a07_ 1 A1 onto ai 07_ 1 Ai iia . In addition, this lc, satisfies the condition: I, n 0A1 = {0}. Conversely, for each closed twosided ideal I of a 1 07_ 1 /41 with I n 07_ 1 Ai = { o}, there is unique C*norm a(.) on 07_ 1 A1 such that I = L. Consequently, for any C*norm a(.) on 07_ 1 ,4i , oc(u) : inf{oc i (u + v) IV E
L},
Vu E 07_ 1 .441.
Let a(.) be a C*norm on 07_ 1 A1 . Clearly, ot(•) < ot 1 (•). Thus the identity map id on 07Ai can be uniquely extended to a * homomorphism from ai 07_ 1 /1 1 to a ® 1 A. But id(o407_ 1 /11 ) is a C*subalgebra of containing O 1 A1 , so id (a 1 07_ 1 111 ) = a07_ 1 A1 . Denote the kernel 1 A1 containing Ø of this * homomorphism by L. Clearly, L is a closed twosided ideal of o1Ø 1 A1. Moreover, since id(u) = u,Vu E 07_ 1 A1, it follows that Ict n 07_ 1 111 ={0}. Then we get a natural * isomorphism from ot07_ 1 /4i onto a 1 07_ 1 A1 /L such that u ÷ a = u + L, Vu E 0:L I & Now let I be a closed twosided ideal of a1 07_ 1 11 1 , and W be a * isomorphism from a Ø 1 Ai onto oc i 07_ 1 A1 ii such that W(u) = a = u + /,Vu E O l Ai. Define W(a) = W 1 (a), Va E ai  07_ 1 Ai , Proof.
where a = a + I is the canonical image of a(E a 107_ 1 A1 ) in a i 07_ 1 Ai ii. Then xli is a * homomorphism from cx 1 07_ 1 Ai onto cr ® 1 A, and its kernel is I, and W(u) . u,Vu E 07_ 1 ./41 . Thus W is the extension of the identity map on 0:L I & and I =kerW = L. Conversely, let I be a closed twosided ideal of a 1 Ø 1 A, and /n o 1 A, = {0}. Then O 1 111 can be embedded into oc i 07_ 1 A1 ii, and the norm on a 1 07_ 1 Ai i/ determines a C*norm ce(•) on Ø 1 A. Then a07_ 1 A1 and a1 are * isomorphic, and this * isomorphism maps u to ii = u + I, Vu E Q.E.D. 07_ 1 111 . Then from preceding paragraph, we obtain that I = L. Remark.
In Section 3.9, we shall discuss the maximal C*norm a i (.)
furthermore. Lemma 3.3.1 is due to M. Takesaki. a i 07_ 1 /11 is also called the projective tensor product of A 1 ,• • • , An ; it was considered first by A. Guichardet. The maximal C*norm a i (.) is important and is more natural than a0 (.). But very little is known about a i (.). The theory of nuclear C* algebras will be important in this aspect (see Section 3.9 ). Notes.
References. [651, [951, [171].
175
3.4. States on algebraic tensor product Let A 1 , • • • ,A„ be C"algebras, and 07_ 1 Ai be their algebraic tensor product.
If ço is a positive linear functional on 07_ 1 A1 , then
Proposition 3.4.1.
there is a positive constant K such that Kai ll ••• Ilan'',
Vai E Ai , 1 < i < n.
Fix ai E (Ai) + , 2 < i < n. Then ço(• 0 07_ 2 a1 ) is a positive linear Proof. functional on A 1 , so it is continuous (Proposition 2.3.2). Similarly, p(Ø 1 a) is continuous for each variable. Now the result follows from the principle of uniform boundedness. Q.E.D.
Proposition 3.4.2.
Let AS' ) = Aii1J1i, 1 < i < n, and ço be a positive linear functional on 0,t 1 Ai. Then ça can be extended to a positive linear functional (,25 on 07_ 1 4 ) such that for any subset ff of {1, • • • , n} and ai E Ai , 1 ff, (i) ro, Awn (125 (Oic i l i 0i01 ai) = ri m (P(Oi E l ,4 Li q i woo ww i Offai), lEff
where {d i( li) } is an approximate identify for Ai , 1 < I < n. Proof. For any ai E Ai, 2 • • • > 07_ 1 ai ,
Thus (,25(07_ 1 1i ) > v(ViL i ai). 1,V/i, 1 < I < n, it follows that
in 07_ 1 4 ) .
Since cilf) E (Ai ) + , 1141
0(1 1 o 07_ 2b1) >
(b3 a3)
07_4bi n
>
07_ 2 b,) + 6.
Further,
1
sup{(,3(1 1 Ø O 2 b,)1 bi E (A1 ), b1 > a, 11b111 sup{v(h) I ço E E}.

184
Clearly, p(h)
max{AI A E o(h)}. However, by the assumption sup{(g)(h) I (g) E E} = max{A I A E a(h)}. This is a contradiction. Therefore, E = S (A).
Q.E.D. Lemma 3.5.8.
Let A 1 , • • . A n be C*algebras, and a(.) be a C*norm on 011_ 1 21i . Then a(.) > a0 (.) on V 1 /11 if and only if for each pi E Si (the state space of Ai), 1 < I < n, 07_ 1 (pi is continuous with respect to cx(.).
The necessity is clear from Corollary 3.2.7. Conversely, if ViL i vi is Proof. continuous with respect to ot(.),V(p i E Si, 1 c o (.) on
Lemma 3.5.11.
Let (I) be a * homomorphism from a C*algebra A onto a C*algebra B. Then V' is an isometric map from B* to A*.
Proof. Let I = {a E A I (1)(a) = 0}. Then I is a closed twosided ideal of A, and Ai/ is * isomorphic to B. Thus for any b E B, IbM =inf {Hall I a E A, (1)(a) = b}. In consequence,
{b E BI111911
cr*(.) > cr*I (.), it suffices to show that c14(.) = a o ( .) on Or_ 1 24:. Clearly, there is a * homomorphism (I) from cy 1 0,t 1 Ai onto cx00 1 Ai such that (1)(u) = u,Vu E O 1 A1 . For any co E Oti/. ._ 1 A; 1 it is easy to see that c4(w) is the norm of co as an element of (a00:L I Ai)*. Now by Lemma 3.5.11,
cti(w) =
sup{ Ico(u)1 lu E 07=1Ai, a l (u) _
E A. Thus E 0, a.e.A. So we can find a Borel subset no of n such that
Va
,
= 0,
and EXi Ai fii (t) > 0,
190
Vt V no and any complex rational numbers A1 , • • • , A n . But any complex number can be approximated arbitrarily by complex rational numbers, hence we have
A,, • • • , A
E Toy fii (t) > o, vt
E Œ.
Therefore,
ei) =
•i fn (a:ai )(t)fii (t)dp,(t)
= fn Mid ai (t)ai(t)fii (t))dp,(t)
O. Q.E.D.
Theorem 3.6.7. Let A be a C*algebra, K be a Hilbert space, and (I) be a completely positive linear map from A to B(K). Then there exists a * representation {7r, HI of A , a normal * homomorphism W from the VN algebra B = (1)(A )" b(h) ,and that
(I)(a) = v*n(a)v,Va E A, W(h)y = vb,Vb E B, and W (id) C 7r(A) 1 ,H = [7r(A)y1C],11y11 = 011 112 . Moreover, if A has an identity 1, and (I)(1) = 1K, then y can be isometric.
Proof. Let A OK be the algebraic tensor product of the Banach spaces A and K. Define (Ea i 0 6, E bi ?J3 ) =
e
K. Since (I) is completely positive, if follows that (,) is a nonnegative inner product on A 0 K. Let N={xEAOKI(x,x)= 0}, N be the canonical map from A 0 K onto (A 0 10/N. and let Then we get an inner product on (A 0 K)/N,
Vai , bi E A, ei,ni
"g) = (x, y), ba,
e A®
x E z y E 5. ,
Denote the completion of (A 0 KIN,(,)) by H, and let
7r(a)Ei ai
® 6 = Ei aai
W(b)E i ai 0 =
Ei ai
bei ,
Va E A,b E B = 01)(A) 1 (c B(K)), and ai E A, ei E K, Vi. Since W is completely
191 positive, it follows that
i n
11 7r(a) Ei (7i6 0 2
E (41) (a*i a* aai
,
)
6,
ei )
id.1
,
0)* (a a)
0 .
0) a)
• •• • ••
[
ai
•••
0
• ••
• ••
• ••
0
• ••
an ) 0 • "
0 j
J
aon )) (
ei ) ( ei
0.
en.
' • •
'
• ••
*
..
) '
et:3.
Thus 7r(a) can be uniquely extended to a bounded linear operator on H, still denoted by 7r(a) . Clearly, {7r, H} is a * representation of A. From B = (I) (AY, we have n
.•
W(b)Eai 0 6
2
n
= 1j:=1 i,j
i=1
But (4:1)(a i*a3 )) 1 1. For m and if c {1, • • • ,n  1}, we can write u = E u, 0 xj , where {xilj} is linearly independent in A ri , and lc.; E 07:11 Ai, V/. Since u = 0, it follows that uj = 0 in Oinf:Ai,Vj. By induction assumption, ul is also the zeroelement of 0{Ai li //,1 < I < n  1} 0 B,Vj, and hence,
À ± n ,
it follows that (1 + n 2 ) 1 /2 ? IIPa( 1 — P) + nPII II P (pa(1 — p) ± np)II = lia' ± npli > A ± n. But this is impossible when ml is sufficiently large. Therefore, 1(palp + pep) = 0. Replacing n by in, we can prove that 1 (palp — pep) = 0 similarly. So we obtain that pa ip = 0.
(2)
Since e = P ((1— p) a* p), it follows from a similar discussion that (1 —p)e (1 — p) = O. Further we get that (1 — p)a 1 (1 — p) = 0.
(3)
Suppose that (1 — p)aip 0 0. By (2), (3), we have lia' + n(1 — p)a'pll == IIPa l ( 1 — P) + (n + 1 )( 1 — P)411
== max{11Pa l ( 1 — P)II)(n + 1 )11( 1 — P)411} = (n + 1 )11( 1 — P)411 if n is sufficiently large. However, since (1 — p)alp E B, it follows that
Ilai + n( 1 — P)a113 11 = 11 P(pa(1 — p) + n(1 — P)a 1 P)11
II Pa ( 1 — 13) +n( 1 — P) &PH = n11 ( 1 — P)&1311 if n is sufficiently large. That is a contradiction. Therefore, we have .*.
(1 — p)a l p = 0.
(4)
By (2), (3), (4), a' = pa'(1 — p), i.e., P(pa(1 — p)) . pP(pa(1 — p))(1 — p).
(5)
218
Replacing p by (1 p), similarly we have p)ap) = (1 p)P((1 — p)ap)p.
(6)
By Pa
P(pap) P(pa(1 — p)) P((1 — p)ap) P((1 — p)a(1 — p)) and (1), (1'), (5), (6), we can see that p • Pa • (1 —
= P(pa(1 — p))
and
p • Pa p= P (pap).
Therefore p • Pa = P(pa). That comes to the conclusion 2). For any n, b 1 , ••• , b B, al, • • • , ay, E A, by the conclusion 2) and P(A+ ) C B+ , b,* P(a:ai )bi = P(I),* c4ai bi ) j,i id = PaE ai bi r • (E (LA)) > o.
E
E
Therefore, P is completely positive. Finally, for any x E A,
P(x* x) — (P x)* P(x) = P (x* x) P (P x* • — P (x* • P x) P(P x* • 1 • P = P((x — Px)* • (x Px)) i.e. (Px)* • (Px)
0,
P(x* x).
Q.E.D.
Proposition 4.1.6. Let M, N be VN algebras on Hilbert spaces H,K respectively. Then there is a aa continuous projection (I) of norm one from MON onto 1H® N N. Proof.
Fix a normal state ço on M. By Lemma 3.8.5, we can define
(I)(x)(f) = x(ço f), Then (I) satisfies the conditions.
vx e MTON, f E N. Q.E.D.
Notes. In general, a linear map from a C*algebra A onto its C*subalgebra B satisfying the conditions 1), 2), 3) of Theorem 4.1.5, is called a conditional expectation, and was studied first by H. Umegaki. A conditional expectation is clearly a projection of norm one from A onto B. Conversely, J.Tomiyama proved that a projection of norm one from A onto B is automatically a conditional expectation.
219
References. [183], [184], [187].
4.2. W*algebras and their * representations Definition 4.2.1. A Cmalgebra M is called a W* algebra, if there is a Banach space M* such that (Mi ) * = M. For a Wealgebra M,M* is called the predual of M (see Section 1.1.) From Proposition 1.3.3, every VN algebra is a Wealgebra. By Theorem 2.11.2, if A is a C'algebra, then A" is a Wealgebra. Lemma 4.2.2.
Let M be a W'algebra. Then M has an identity.
Proof. Since the closed unit ball S of M is a(M,M*)compact and convex, it follows from the KreinMilmann theorem that S admits an extreme point at least. Now from Theorem 2.5.3, M has an identity. Q.E.D. Now let M be a W'algebra, 1 be its identity. By Theorem 2.11.2, M" is also a Wealgebra; 1 is also an identity of M"; and M is a Cesubalgebra of M". Let M* be the predual of M. See M* as a closed linear subspace of Me, and let P : M" 4 M as follows
P(X) = XIM* , VX E M". Clearly, P is a projection of norm one from M" onto M, and is a(M**,M*)  a(M,M.) continuous. Let
/ = {X E M" I P X = o}. Clearly, I is just the orthogonal complement M*1 of M., as a closed linear subspace of M*. Thus, I is a(M**,M*)closed. By Theorem 4.1.5, P (aXb) = a • P X • b, V X E M** , a, b E M.
By Theorem 2.11.2, M" _r÷f M (the enveloping VN algebra of the Cealgebra M). Thus, the multiplication on M" is a(M**,M*)continuous for each variable. Moreover, M is a a(M**,M*)dense subset of M". Therefore, I is a a(M**,M1closed * twosided ideal of M". By Proposition 1.7.1, there is a unique central projection z of M" such that
M.,1 = 1= M**(1 — z). Since P = P 2 and I is an ideal, it follows that
(PX  X) E /,
and
(PX — X)17 E /
220
VX, Y G M**. By Theorem 4.1.5, we can see that P(XY) = P X • PY, VX,Y
E M**.
Thus, P is a * isomorphism from M**z onto M. Let QG M ÷ M** z) be the inverse of (PIM**z). Since P(xz) = x,Vx E M, it follows that Q(x) = xz, Vx E M. For any X G M**, we can write X = PX + (X  PX), where PX E M, (X — PX) E / = M;L = M**(1 — z). IfxeMni,thenx=Px= O.
Therefore, we get M** = Mi Me'.
For any F G M*, RF and R( l _ z )F (see Section 1.9) E M* since the multiplication on M** is cr(M**, M*)continuous for each variable. Hence, M* = Rz M*4R(l z)M*. Now we claim that M* = Rz M* . In fact, since M* is a closed linear subspace of M*, it follows that M„ = (M*1 )1 = (M** ( 1  z))1 • Thus, M, D R z M*. Conversely, if f E M* , then by the definition of z, f (X(1  z)) = 0,1(X) = f (X z) = (R z f)(X),VX E M**. Thus f = Rz f E Rz M* , and M* = RzM * • We say that the * isomorphism Q(: M ÷ M** z) is also cr(M,M * ) cr(M**,M*) continuous. In fact, let {xi } be a net of M, and xi ÷ 0(cr(M, M* )). Then for each F E M*, F(Q(xi )) = F(xi z) = f (xi ) ÷ 0
since f = RF E M. From the above discussion, we obtain the following. Proposition 4.2.3. Let M be a W*algebra, and M„ be its predual. Embedding M, M* canonically into M**, M* respectively, then we can find a central projection z of M** and a projection P of norm one from M** onto M
such that: 1) P is also a * homomorphism from M** onto M, and is cr(M**, M*) cr(M,M* ) continuous; 2) P is a * isomorphism from M**z onto M. If Q(: M + M** z) is the inverse of (PIM**z),then Q(x) = xz,Vx E M, and Q is also a(M,M * ) cr(M**,M*) continuous; 3) AI = M**(1  z), M, = Rz M* , and M** = M4M, M* = M*4R(1z
221
Definition 4.2.4. Let M be a Wealgebra, and M,, be its predual. {7r,1/} is called a W*representation of M, if 7r is a * homomorphism from M to B(H), and ir is a(M,M.)a(13(H),T(H)) continuous. If { ir, H} is a W*representation of M, then 7r(M) is a weakly closed * subalgebra of B(H) by the proof of Proposition 1.8.13. Moreover, if 7r is nondegenerate, then 7r(1 m ) = 1 H , and 7r(M) is a VN algebra on H. If 7r is faithful, by Proposition 1.2.6 and a(M, M.)compactness of the closed unit ball of M, then the * isomorphism 7r 1 from 7r(M) onto M is also a(B(H),T(H))a(M,M 4,) continuous. Theorem 4.2.5. Let M be a Wealgebra. Then M admits a faithful nondegenerate Werepresentation. In consequence, M is * isomorphic to a VN algebra on some Hilbert space, and this * isomorphism is aa continuous.
Let { ir, H} be the universal * representation of M as a Cealgebra. By the discussion of Section 2.11, {7r, 1/} can be extended to a faithful nondegenerate Werepresentation of the W*algebra M**, which is denoted by {7r, H} still. By Proposition 4.2.3, there is a aa continuous * isomorphism Q from M onto M**z, where z is a central projection of M**. Then {7roQ,7r(z)H} is a faithful nondegenerate Werepresentation of M. Q.E.D. Proof.
By Theorem 4.2.5, we can regard a Wealgebra as a VN algebra. In particular, we have Proposition 4.2.6. Let M be a Wealgebra, and M., be its predual. Then the * operation on M is a(M,M* )continuous; the multiplication is a(M, M,,)continuous for each variable; M* is the linear span of normal positive linear functional on M, in consequence, M4, is unique; for any normal positive linear functional ço on M, by the GNS construction there is a cyclic Werepresentation ir,, H,i} of M. {
Proposition 4.2.6.
Denote the normal state space on M by Sr,(M). Then the normal universal * representation / Ir =
E tp ES „ (M)
ED7r
H=
E
@Hp
}
40 E4 (M)
is a faithful nondegenerate W*representation of M. Now we discuss some properties of W*representations. Theorem 4.2.7. Let A be a Cealgebra, and r, H} be a * representation of A. Then there exists a unique Werepresentation {Fr, H} of Ae* such that 'if is an extension of 7r, and rr(A**) is the weak closure of 7r(A). Consequently, {
222
there is a bijection between the set of * representations of A and the set of W 5 representations of Ass.
A ÷ B(H) and 7r* : B(H) 5 —÷ A*. Regard T(H) as a closed linear subspace of B(H) 5 (since T(H) 5 = B(H)), and let 7r 5 = 7r*IT(H). We claim that Ff = (7r * )*(A** —+ T(H) 5 = B(H) ) satisfies the conditions. In fact, since 7r * : T(H) + A* and if = (7r* )*, it follows that 51 is a(A**, A*)a(B(H), T(H)) continuous. Notice that Proof.
Notice that
7r :
if (a)(t) = a(ir * (t)) = a(r* (t)) =
A. Thus Fr is an extension of 7r. Further, if is a W 5 of A55 since A is a(A**, A)dense in A55 and is a  a continrep sntaio uous. Clearly, Fr is the unique a  a continuous extension of 7r, and if(A 55 ) is the weak closure of r(A). Q.E.D. Vt E T(H),a E
Proposition 4.2.8. Let {ri , Hi } and {7r2 , H2 } be two nondegenerate W 5 representations of a IValgebra M, and kerri = {a e mliri(a) = = 1 1 2 . If kern cker7r2 , then {7r 2 , H2 } is unitarily equivalent to an induction of some amplication of {n, HI }, i.e. there is a Hilbert space K and a projection p' of (n(M) 501K) 1 such that {7r 2 , H2 } pi(H1 IC)), where 7r(a) E M. (ri (a) 0 Let Mi = ri (M), i = 1,2. Then M, is a VN algebra on Hi, i = 1,2. Since kern C ker 7r2, there is a normal * homomorphism ID from M1 onto M2 such that (I) o in = 7r2. Now by Theorem 1.12.4, we can get the conclusion. Q.E.D. Proof.
Proposition 4.2.9. Let {71 , HI ) and {7r2 , H2 } be two W*representations of a W*algebra M, and ker ir = {a E M I ri (a) = 0), i = 1,2. If ri (M) admits a cyclicseparating vector in Hi , j = 1,2, and ker n = ker 7r2 , then {71, Hi} `.1 fir2, H2). Let Mi = ri (M), then Mi is a VN algebra on Hi , i = 1, 2. Since kern =kern, thus there is a * isomorphism (I) from M1 onto M2 such that (I) o n = r2. Now by Theorem 1.13.5, the conclusion can be obtained.Q.E.D. Proof.
Notes. We have a definition of abstract C 5 algebras (see Chapter 2). A natural question is how to define abstract VN algebras. This question received considerable attention during 1950's. Theorem 4.2.5 is due to S. Sakai, and it gives an answer for the above question. However, the proof of Theorem 4.2.5 presented here is due to J. Tomiyama based on his result, Theorem 4.1.5. The uniqueness of the predual of a W 5 algebra, due to J. Dixmier, answered
223 completely the question concerning to what extent the algebraic structure of a IValgebra determines its topological structure.
References. [19], [143], [183], [184].
4.3. Tensor products of W*aigebras Let M, N be two IValgebras. We want to define the tensor product MN of M and N such that MN is still a W'algebra. If we regard M, N as VN algebras, using the tensor product of VN algebras and by Theorem 1.12.6, we can define MN.  O But in this section, we shall define MN from M and N themselves. Let M, N be two W'algebras, and M., N. be their preduals respectively. As C'algebras, there is a spatial C'norm a0 (.) on the algebraic tensor product M 0 N. Then we get a C'algebra ao  (M 0 N). Let a ( .) be the dual norm of a(.) on M* 0 N'. Then
(a0(M 0 N))* D ot',; (r 0 N*) D ai;(M„ 0 N.), where ai;(M* 0 N*) is the completion of (M* 0 N* , a(.)); and cei; (M. O N,) is the completion of (M, 0 N„ a(.)), and is equal to the closure of M. 0 N. in ceor(M* 0 N*). Let 1= (c(M. 0 N.)) ± (C (a0(M 0 N))"), i.e., I is the orthogonal complement of t*(M, 0 N,) which is regarded as a closed linear subspace of (a0(M 0 N))'. Suppose that Y E / 3 X E (cx0(M 0 N))". Pick a net {x i } CMON such that zi —* X with respect to the w'topology in (a0(M 0 N))". For any f E M, 0 N., since L z , f and Ra E M., 0 N„ it follows that
f (xi Y) = (L z: f)(Y) = 0,
f (Y xi ) = (R z , MY) = 0,
VI. Taking the limits, we get XY and YX E I. So I is a cfclosed twosided ideal of the 1Valgebra (a0(M 0 N))". Therefore, (a0(M 0 N))** / I is a IValgebra, and its predual is ceo`(M, O N,). Definition 4.3.1. The IValgebra (c o(M 0 N))** / I is called the tensor product of W'algebras M and N, which is denoted by MN. (From preceding paragraph, MON = (a'i; (M, 0 N,)', and (MN), = c4(M. 0 N.).
224
Lemma 4.3.2. M* 0 N. is w*dense in (a0(M 0 N))*. By Proposition 3.2.10, M* 0 N* is w*dense in (a0(M 0 N))*. Notice that the unit balls of M„ N* are w*dense in the unit balls of M* , N* respectively, and a ( .) is a crossnorm on M* ON*. Then it is easy to see that M* ON* is dense in M* ON* with respect to the w*topology in (a0(MON))*. Q.E.D. Therefore, M* 0 N* is w*dense in (o 0(M 0 N))*. Proof.
Proposition 4.3.3. ao(M 0 N) n I = { o}, in consequence, c o(M 0 N) can be embedded in M(g)N. Moreover, c o(M 0 N) is w*dense in MIN. Proof.
Let x E ao(M 0 N) n I. Then
f 0 g(x) = 0, Vf E M*, g E N.. By Lemma 4.3.2, we have x = 0. Now if )7 E MN = (a0 (M 0 N))**//, and X E )7, then there is a net {x i } C ao(M 0 N) such that xi 3 X with respect to the w*topology in ao(MON))**. Further, for any F E (MN), = a(M* ON.) C (a0 (MON))*, 
—
1 (li
—
50 (F)I = 1 (Xi
—
X)(F)I + O.
Therefore, ao(M 0 N) is w*dense in MN.
Q.E.D.
Let {ri , Hi } be a nondegenerate W*representation of a W*algebra A, i = 1,2. Then there exists a unique W*representation {ir, H} of MI g M2, where H = H1 0 H2 such that Theorem 4.3.4.
r(a l 0a2) = r i (a l ) 0 r2 (a2 ), Va iE Mi, i = 1, 2,
and ir(MO N) = r(m)0 71(N) ( the tensor product of VN algebras r(M) and r(N)). Moreover, if ri is faithful, i. = 1,2, then 71" is also faithful. Proof. By Proposition 3.2.7, there is a unique * representation { ro , H} of ao(Mi 0 M2 ) such that ro(al 0 a2) = ri (al) 0 r2 (a2 ), Val E MI, a2 E M2.
By Theorem 4.2.7, { ro , H} can be uniquely extended to a W*representation {Fro, II} of (co(Mi 0 M2))* * . For any 61ni c Hi, let L(.) = (ri(16,771) c (Mi),„i = 1,2. Then fi 0 h E (M1)* 0 (M2)* C (a0(Mi 0 M2))*. By the definition of I (see 4.3.1), we have fi 012(I) = {0}.
Since H1 0 H2 is dense in H1 0 H2, it follows that î 0 (I) = { o}. Thus fiT0 ,H} induces a W*representation fir, HI of M1 M2 = (ao(MI 0M2))**//. Clearly)
225
{r, H} satisfies the conditions, Moreover, the uniqueness of such {7,H) is
also obvious. Now suppose that ri is faithful, j = 1, 2. For any fi E (Mi )„ since Mi is * isomorphic to ri (Mi ), there are two sequences {C)} and {n)}(C Hi ) with oo such that En(IIC ) 11 2 +
117e1 2)
E lv,({1})1 — 17 0. lEA
IEF,
Since v(l) —÷ 0, V/ E A, there exists n2 such that E Ivn2({/})1 E lv,({/})1 — IEF2 icA
TO .
• • . Generally, we can find a sequence {Fk } of finite subsets of A and a subsequence {n k } such that E Ivn, ({/}) ieFk
I
> E Ivnk({/})1 — lEA
and Fk n F.; = 0, Vk j. Fix m such that m> — le° suPn II vn II . Let E1 z F1 ,/2 1 = vn ,. If 00
v (tt i ) (
U Fm i + p) > — 1Eo'
1 < p < m,
1=1
then 11A111 ?. v(p.1)(_1 LIT1 Fini+p) TYL
E
= Ev(P1)(171iFtni+P) a ?.
P=1
m
11/2' II. This is a contradiction. Thus, there is an integer p i with 1 < Pi < m such that >
sUPrt 11 111211
E
v(iii)( 1171Fmi+pi) < —
10 . Let E2 = F _ ml_p i , /22 = Vn m+pi , and F.; = Frni +pi , Vi. Similarly, there is an integer p2 with 1 < p 2 < m such that v (A2) ( 117_3=IF'mj+p2)
E = v(A2)( 11 711Fm(tni+P2)+P1) < To'  • •. Generally, we have {Ps I S = 1, 2, . • •} such that
1 < p, < m,
231
E 1113({/}) > E ip,,({/})i — 10' lEA
lEEs
(2)
v(A s )(Up s Ei ) < —
10
(noticing (1)), Vs, where vn b ,, Es = Fb., and b 1 = 1, b2 Pi, • • .,bs = mbs1 +Ps,' • •• (A) as follows: Now define a f E f0,
f ( 1) =
mbi +
if / Lfr_ i Ei ,
argiz 1 ({1}),
if 1 E Ei,Vj.
By 1 E5 < oo(VA and (2), for any s we have
E liza({/})11
IEL
+
Ef
iCs Ei
L
. >sEi
L.
lizs({ 1})II 1EE,
f
Ii({/}) f<siEEi
I + v(tis)(ui>sEi)
0, since
0 ((P  Pi) * (P 
pi) + (p — pi )(p — pi)') = 20 (p  pi ) 4
ol
there is /0 such that 1,1)(p  pi) < 6 = 6(e),V1 >10. By 2), we have Iço(P  POI < c, Vço e A,/ > le . That is lip Ap i ) = ça(p) uniformly for (pc) E A. Moreover, by the same proof of 2) == 3), A is bounded. 7) 4). Let {pn } be a decreasing sequence of projections of M, and inf pn = O. Then {(1  pn)} is an increasing sequence of projections of M n and supn (1  pn) = 1. By 7), lim (i)(1  p n) = (p(1) uniformly for ça E A, i.e., n
(P(Pn —÷ 0) uniformly for ça E A. 8) == 7). It suffices to notice that: if {pi } is an increasing net of projections of M, then {q i = pi + (1  p)} is an increasing net of projections of M, and supi qi = 1, where p = sup i pi. 2) == 8) . Let {pi } be an increasing net of projections of M, and supi pi = 1. Pick 1,b as in the condition 2). For any E > 0, there is 1(e) such that .0(1  p i ) < 1S(E), V/ > 1 (E) . Then for any a E M with hail < 1 and 1 > 1(e), we have
1/4(1 p1 )a*(1  pi)a(1 pi ) + (1 — pi)a(1  pi )a*(1  pi)) < 20(1 p i ) < 6(e) By 2), Iço(1  pi )a(1  pi )I < E) \ 7%0 E A and 1 _?.. 40,114 < 1. Therefore, 'ILO1)R(10PM + 0 uniformly for ça E A. Moreover, A is bounded by the same proof of 2) 3). Q.E.D.
Proposition 4.5.2.
Let M be a W'algebra, M. be its predual, and A C (M.) + . Suppose that the a(M.,M)closure of A is o(M.,M)compact. Then
236
the a(M ., M)closure of E = {Raço I a E M, Hail _< 1, ço E A} is also a(M, M)compact.
Proof.
Clearly, E is bounded. Let {pn} be a decreasing sequence of projections of M and inf n pn = O. By Theorem 4.5.1) V(p) + 0 uniformly for ço E A. From the Schwartz inequality, iRaSo(Pn) I
V(a * a) 112 S0 (Pn) 112 < 11(P11 1/2 V(Pn) 1/2
Va E M and Mail < 1. Thus p(p) —÷ 0 uniformly for p E E. Now again by Theorem 4.5.1, the a(M,M)closure of E is a(M,M)compact. Q.E.D.
Theorem 4.5.1 is a combination of results due to several mathematiNotes. cians: A. Grothendieck, S.Sakai, M.Takesaki, H. Umegaki and finally, C.A. Akemann.
References. [2], [62], [146], [169], [188].
Chapter 5 Abelian Operator Algebras
5.1. Measure theory on locally compact Hausdorff spaces Let SI be a localy compact Hausdorff' space, and B be the collection of all Borel subsets of ft (i.e. the uBoot ring generated by compact subsets of SO.
Define = {ECSII EnKEB ,V K compact c [O. Bi„ is a uBool algebra. Each subset in Bi „ is called a locally Borel subset. Clearly, E E [31„, if and only ifEnFEB,VFEB. A complex function f on 11 is said to be measurable, if it is 8measurable. f is said to be locally measurable, if it is Bloc measurable. Clearly, A measurable function is locally measurable. And a locally measurable function f is measurable if and only if {t E fl 1 f(t) o E B. Let v be a regular Borel measure on SI. F(C 0) is called v zero, if F E B and v(F) = 0; E(C [0 is called locally v zero, if E E B i„ and v(EnK) = 0, VK compact C SI. A Proposition about P(t) on il holds almost everywhere with respect to y (a.e.v), if {t E SI I P(t)does not hold} is a subset of some vzero set; P(t) on SI holds locally almost everywhere with respect to v (1.a.e.v), if {t E n i P(t) does not hold} is a subset of some locally vzero set. Let v be a regular Borel measure on CI. Then }


Vo = U{V C II 1 V is open and locally vzero) is the maximal locally vzero open subset. Let suppv = (SAVO. It is called the support of v, and clearly it has the following property. Let U(C SO be a Borel open subset. Then v(U) = 0 if and only if u nsuppv = 0.
238
Lemma 5.1.1.
Let v be a nonzero regular Borel measure on rt. Then there is a nonempty compact subset K(C 12) such that u(K nu) > 0 for any open subset U of 12 with U n K 0.
Since supp y is a nonempty closed subset of fl, we can find an open subset V such that V compact and K = V n suppv 0. Then K is what we want to find. In fact, suppose that there is an open subset U with unic 00 such that v(K n U) = O. Then v(u n V n suppv) = 0 and v(u n 17) = where E = n VAsuppv. But E is open,and E n suppv = 0, so we have Proof.
v(u n v) = v(E) = 0. From the definition of supp y U n V n supp u = O. On the other hand, pick teUnK. Since U is an open neighborhood of t and tEK=Vn suppv, it follows that Un V nsuppv 0 0. ,
We get a contradiction. Therefore, K is what we want to find.
Q.E.D.
Let v be a nonzero regular Borel measure on fl. Proposition 5.1.2. Then there is a disjoint family {Ki}i cA of nonempty compact subsets of fl such that N = fl\il lEA K i is a locally vzero subset, and the family {Kd iEA has the locally countable property, i.e., for any compact subset K of n the index set {/ E AIK1 n K 0} is countable.
By Lemma 5.1.1 and the Zorn lemma, there is a maximal disjoint > 0 family {Ki} lEA of nonempty compact subsets of fl such that v(K, n for any open subset U of SI with U n K1 o Suppose that V is an open subset of 12 and V is compact. Then Proof.
E v(Ki nv) v(v) 01 is countable. However if some 1 E A is such that v(Ki (IV)  0, then Ki n V = 0 by the property of Ki. Thus {1 E A IK1nV 0} is countable. From this discussion, it is easily verified that the family {Ki}l EA has the locally countable property. In consequence, Lli caft E Bloc and N = 11\ [' LEA Kt E Bioc• Now we prove that N is locally vzero. Suppose that there is a compact subset H C N such that v(H) > O. Applying Lemma 5.1.1 to H and (OH), we can find a nonempty compact subset K C H such that v(uH n K) > 0 for each open subset UH of H with UH n K 5L 0. Thus for any open subset U of 12 with UnK 0 0 we have also v(U n K) = v((UnH)nlf)> O. Clearly, K n K1 = O,V1 G A. This is a contradiction since the family {ift}tEA Q.E.D. is maximal. Therefore, N is locally vzero.
239
Let f be a locally measurable function on SI, and V be a regular Borel measure on IL f is said to be locally essentially bounded with respect to u, if there is a constant C such that 1.a.e.v.
C,
if WI
The minimum of such C is called locally essentially supremum of f, denoted by 11f1100. Let L'''' (SI, v) = If
Clearly, (Lœ(ft, 0,
f is locally measurable on and is locally essentially bounded
II. 000) is an abelian Cealgebra.
f'
Indeed, it is a Wealgebra.
Theorem 5.1.3. L 1 (11,0* = VIII, v). Proof.
Suppose that f E L'(11,v). Define F(g) ...,
ft)
Vg E L l (n, v).
f(t)g(t)dv(t),
Clearly, F E L l (fl,v)* and VII = Ilf1100. Now let F E L 1 (1/ 1 0*. For any compact subset K of SI, since v(K) f (t)} is open and G C E. Since 12 is a Baire space and E is first category, it follows that G = 0, i.e., 1(t) > g(t),Vt E 12. Suppose that h E Cr (1Z) and h > AV' E A. Then h(t) > g(t),Vt E 11, and h(t) > f(t),Vt V E. In addition, (12\E) is dense in fi. Thus h(t) > f(t),Vt En, and f is the least upper bound of A. 3) == 2). It is obvious. 2) == 1). Let U be an open subset of 12. Let A = ft G Cr (ft) I 0 _. f (t) — e. Further limg(t i) > 1(t) — e, v.,' — and limg(e) > f (t) since c is arbitrary. Therefore
f (t) = prntg (e) , Vt E il. Q.E.D. Let II be a Stonean space, and ,u be a regular Borel Definition 5.2.4. measure on SI (i.e. a positive linear functional on C(I)). kt is said to be normal, if p(f) = supi k2 (f1 ) for any bounded increasing net ffi l of nonnegative functions of Cr (n), where f is the least upper bound of fi in Cr (1Z).
{ }
Proposition 5.2.5. Let si be a Stonean space and A be a normal regular Borel measure on St Then kt(F) = ,u(E) = 0 for any rare closed subset F and first category Borel subset E. Let F be a rare closed subset of il. Then (11\F) is open and dense
Proof. in
11,
and
n\F = U{suPPf I f E Cffi),0 5 f 5_ 1, suppf C (n\F)}, where suppf = ft E fl I f(t) 0},Vf E COI). Since II is a Stonean space, it follows that n\F = ll{G C fl\ F I G is open and closed}. By the inclusion relation with respect to G, fx a I G C SI\F, and G is open and closed } is a bounded increasing net in Cr (G). Clearly, the least upper bound of {xG} in Cr (G) is 1. Thus we have u(11) = sup{/2(G) I G is as above} since A is normal. Further, A(F) = O. Now suppose that E is a first category Borel subset of FL We can write E = U Fn , where each Fn is rare. Then Fn is closed and rare, Vn. Therefore n
tI(E) = 0 from the preceding paragraph.
Q.E.D.
247
Proposition 5.2.6. Let S 1 be a Stonean space, and it be a normal regular Borel measure on n. Then supp ti is an open and closed subset of n. Proof. Let F = suppit. Then F is a closed subset, and F\ Int(F) is a rare closed subset. By Proposition 5.2.5, 11(F) , ,u(Int(F)). Let E be the closure of Int(F). Then E is open and closed, and Int(F) C E C F. Thus ,u(E) = ,u(F). By the definition of suppv, we have E = F = suppv. Q.E.D.
Proposition 5.2.7. Let 2 be a Stonean space, and h be a bounded measurable function on FL Then there is f G C(1.1) such that f(t) = h(t)„a.e.A
for any normal regular Borel measure it on 2. We may assume that h is real valued. Then g(t) = limh(ti) is a ti—it bounded real valued lower semicontinuous function on FL By Theorem 5.2.3, there is a f c Cr (11) and a first category Borel subset E of 2 such that
Proof.
Vt V E.
f(t) = g(t),
For any normal regular Borel measure it on II, by the Lusin theorem there is a disjoint sequence {K n } of compact subsets of II such that h is continuous
on Kn ,Vn, and A(2 \ U Kn) =0.n Then h(t) = g(t),
Vt E
11 int(Kn). n
Since (K„\ Int (Ku )) is rare and closed, it follows from Proposition 5.2.5 that ,u(K\Int(Kn )) = 0,Vn. Thus ,u(EU(2\ U Int(Kn)) = 0 and n
f (t) = h(t),
Vt E (11Int(K n ))
n
(n\E),
n
i.e., f(t) = h(t), a.e.p.
Q.E.D.
Definition 5.2.8. 11 is called a hyperstonean space, if it is a stonean space,and for any 0 < f G C(S1) and f 0 there is a normal regular Borel measure A on ft such that p(f) > 0.
Let n be a hyperstonean space. Then there is a family Proposition 5.2.9. {Ad of normal regular Borel measure on 2 such that suppkti nsuppktv = ø,V1 0 1', and ilsupp,ut is dense in ft. t
248
Proof.
Let {A } be a maximal family of normal regular Borel measures on
CI such that
suppiti
Put
r = U suppgi .
n suppto = 0,
V/ 0 1'.
By Proposition 5.2.6, T is on open subset of 11. Then
i
nv
r
is open and closed. If E = 0, then 0 < xE E C(I1) and xE 0 0. From such that Definition 5.2.8, there is a normal regular Borel measure iti on AU) > 0. Let
,(A) = ,21(6,\r),
n
v Borel subset A.
Clearly, it is a normal regular Borel measure on SI, and
0 0 supptc C E =
n\r.
This is a contradiction since the family {j i } is maximal. Therefore,
r=
FL Q.E.D.
Notes.
The concept of Stonean spaces was introduced by M. Stone. The presentation here follows a treatise due to J. Dixmier. References. [20 ] , [164], [177].
5.3. Abelian W*algebras Theorem 5.3.1. Let Z be a afinite abelian W*algebra, and fl be its spectral space. Then is a hyperstonean space, and there is a normal regular Borel measure v on II such that
n
suppv = II,
and
Z "=*" C01) = Lœ 01, v).
Proof.
Suppose that Z C B(H), here H is some Hilbert space. By Proposition 1.14.5, Z admits a separating vector eo (E H). Let f 4 m t be the * isomorphism from COI) onto Z. From Theorem 5.2.3 and Proposition 1.2.10, CI is a Stonean space. Clearly, there is a regular Borel measure v on fl such that
(rnf ea, eo) = fn f (t) du ( t) ,
Vf G C(1.1).
By Proposition 1.2.10, v is normal. Suppose that there is a nonempty open Borel subset U of 11 such that v(U) = 0. Pick f G COI), f 0, f 0 0,and suppf C U. Then (rnfeoleo) = 0.
249
Since eo is separating for Z, it follows that f = 0, a contradiction. Thus suppv = SI. In consequence, I/ is a hyperstonean space, and C(1Z) can be embedded into VIII, v). Let {fi } be a net of C(S1 ),11fill — < 1,V/ and fi 4 f (E L' (fl, v)) with respect to w* topology in L' (fl, v). Put mi = m.t.,(E Z),V/. Then limill < 1 . Replacing {m1 } by its subset if necessary, we may assume that mi 4 mg g E C(12). Then for any h E c(n), weakly,hr
I /cf.,
—
ohdvl =
I
((till
—
mOrnheo, ',3)1 4 0.
Since C(12) is dense in L 1 (11, v), it follows that fi 4 g with respect to w*topology in L' (SI, v). Hence f (t) = g(t) , a.e.v. From above discussion, c(n) is w*closed in Loo(n, 0. Clearly, CO is w*dense in L' (11, v) . Therefore, Q.E.D. C (1 ) = L'(I, v). Proposition 5.3.2. Let SI be a compact Hausdorff' space, and v be a regular Borel measure on SI. Then L°°( 2,v) is a ufinite abelian W*algebra.
Proof.
By Theorem 5.1.3, L' (11, v) is an abelian W*algebra. Let co(f) =
4
f (t)dv(t),
V f c L' (11, v).
Since 1 E L i (11, v) , it follows that (.=.4.) is a faithful ucontinuous positive functional on L'IS1, v). From Proposition 1.14.2, L' (11, v) is ufinite. Q.E.D. Theorem 5.3.3. Let 11 be a hyperstonean space. Then C( 12) is an abelian W*algebra. Moreover, if there is a normal regular Borel measure v on 11 with suppv = IL then CO = L'I12,v) is ufinite. First suppose that there is a normal regular Borel measure v on SI Proof. such that suppv = SI. Then C(ft) can be embedded into L'(11, v). Moreover, for any h E L°° (1, u), by Proposition 5.2.7 there is f E CO such that f (t) = h(t) , a.e.v. Thus C(I2) = L' (fl, v) . Further, C ( I) is a ufinite abelian W*algebra from Proposition 5.3.2. Generally, by Proposition 5.2.9 there is a family {v1 } of normal regular Borel measures on SI such that suppvi n supp vi , = 0,V/ 0 /',and r = usupp
i ii, is dense in [2. By Proposition 5.2.6, supp vi is open and closed, V/. Then r is a locally compact Hausdorff' space. Let ti = E elm. Then v is a regular Borel i
measure on r and supp v = F. Consequently, f 4 f ir is an injective map from C(1/) to VII', v). Moreover, for any h E L°° (r, v), let h(t) = 0,Vt E n \r. Then by Proposition 5.2.7 there is f E C(I2) such that f (t) = h(t) , a.e.v .
250
Thus 1(t) = h(t), 1.a.e.t/ on F. Further C(11) is * isomorphic to VII', v), and is a W*algebra. Q.E.D.
Theorem 5.3.4.
Let Z be an abelian W*algebra, and SI be its spectral space. Then 11 is a hyperstonean space, and there is a locally compact Hausdorff space r and a regular Borel measure on r with suppv = r such that Z is * isomorphic to L"°(r, Proof. Let Z C B(H), and f n2f be the * isomorphicm from C ( I) onto Z. Then for any e E H, there is a regular Borel measure tie such that (
mfe,e) =L f(t)dti e (t),
VI E C(n).
From Theorem 5.2.3 and Proposition 1.2.10, SI is a Stonean space, and tJ is normal, ye E H. If f is a nonzero positive element of C(I1), then there is e E H such that (rrif e, e) > 0, i.e., v(f) > 0. Therefore, 11 is hyperstonean. The rest conclusion is contained in the proof of Theorem 5.3.3 indeed. Q.E.D.
Definition 5.3.5. Let M be a W*algebra. E(C M) is called a generated subset for M, if M is the smallest W*subalgebra containing E. Moreover, if M admits a countable generated subset, then M is called countably generated. A generated subset for a C*algebra is understood similarly. Let n be a compact Hausdorff space. If the C*algebra is generated by a sequence {py,} of projections, then C(S/) can be generated by an invertible positive element.
Lemma 5.3.6.
Proof.
Let
h=
E 4(2p
+
1
n=1 3
Then h is an invertible positive element of COIL For any t1,t 2 t 1 t 2 , there is a minimal positive integer k such that
En
and
Pk(t2)
Pk(ti)
since fpn l is a generated subset for COI). Thus 1h(ti)
00
h(t2)I
(p 72 (t i ) 
21 E
P72(t2))I
n=k
2 k
3
001
1
2 n=k+1
Now by the Stone—Weierstrss theorem and Lemma 2.1.5, C(1/) is generated by {h}. Q.E.D.
251
Let Z be a countably generated abelian Wealgebra. Theorem 5.3.7. Then Z can be generated by an invertibel positive element. In particular, every abelian VN algebra on a separable Hilbert space is generated by a single operator. * ), we Proof. Let {a n } be a generated subset for Z. Replacing a, by (aFarn may assume that a n* = a n , Vn. From the spectral decomposition of {an }, Z can be generated by a sequence {N } of projections. Let A be the C*subalgebra of Ce algebra Z generated by {N } . By Lemma 5.3.6, A is generated by an invertible positive element a. Clearly, A is also a generated subset for Z. Thus, Z is generated by a. Moreover, each VN algebra on a separable Hilbert space is countably generated. That comes to the rest conclusion. Q.E.D.
Theorem 5.3.8.
Suppose that Z is an abelian VN algebra on a separable Hilbert space H, and Z contains no minimal projection (a projection p of Z is said to be minimal, if p 0 O and any projection q of Z with q < p implies either q = O or q = p ) . Then Z is * isomorphic to L'([0, 1 ] ), where measure on [0, 1] is Lebesgue measure.
Proof. Let SI be the spectral space of Z. By Theorem 5.3.1, SI is a hyperstonean space, and there is a normal regular Borel measure v on SI with suppv = SI such that Z!"` =' 01) = L' (SI, ti) . From Theorem 5.3.7, Z is generated by a positive element a. We may assume that O < a < 1. Put I = [O, 1 ] , and let z 4 z(.) be the Gelfand transformation from Z to C(SI). Then a(.) is a continuous map from SI to I. Define a Borel measure A on I and a * homomorphism (I) from Lœ ) (I , A) to L' (11, v) as follows:
A(E) = v(a 1 (E)),
V
Borel subset
E C I,
G SI, f E L' (I , IL).
(I:0M=fat),V
Clearly, 1:1:0(p) = p(a) for any polynomial p(.) on I. Thus (I)(L'(/ 40) is w*dense in L°°(SI,v) since Z is generated by {a}. We claim that (1:0(L'(/,,i)) is dense in Ll (SI, v). In fact, suppose that there is some g G L °° (n l v) such that
I,
g (0(I:0 ( f)(t)clv(t) = 0, V f E L' (/,A).
Since (1:0(L'(./,p)) is w*dense in L'(SI, v), there is a net that
(GIL° ° (11, v) 1 L l (11 v))).
{A}
C L' (I , A) such
252
Clearly, g
G Ll (SI, ti)
0
too. Thus
= fn
g(t)(1:0(f1 )(t)dv(t) 4
fa ig(t)1 2 dv(t)
and g = O. Therefore, (1:0(L°T,A)) is dense in Ll (11, v).
Now we say that 40 is aa continuous. It suffices to show that (DM 4 0 (w*topology) for any net {fi} c L' (I „ u) and Ilfill 1,V1 , and f i —4 0 (w*topology). For any g E L l (11, v) and E > 0, from preceding paragraph we can pick f E L°° (/ 1 p) such that
fo 10) — (I)(f)(t) I dv(t) Then I
O. Put E = a 1 (01). Then v(E) > 0. So xE is a nonzero projection of L'In,v). Since (I)(x{x}) = XE and xol is a minimal projection of L°°(/„u), it follows that xE is a minimal projection of Lc°(11,v)('="' Z). This contradicts the assumption. Let f(A) = A([O, Aj),VA E I. Then f is a continuous increasing function with f(0) = 0 and f(i) = 1 (we may assume that v(S1) = 1). Further, let
g(A) = min{A' GII f (A i) = A}, VA G
I.
Then g is a left continuous strictly increasing function on I, and has countable jump points at most. Suppose that {A 1 < A2 < • • • < A n < • • •} is the set of jump points of g. then there is a sequence {Ain } with A1 < All < A2 < Al2 < • • • n means that jt e > . A rl . Moreover, e E H is said to be maximal, if e > rhvii E H. 
258
Lemma 5.4.3.
If n E He = 7r(A)e, then ri < C.
Proof. By Theorem 5.4.1, there is a unitary operator u from He onto L2 ( 144 e ) such that
u(n(a)IHe )u 1 = (1:0 0 (a),
Va E A.
Let f = uri (E L 2 (11 , 1 I e )) . Then
(1 (a)n,n) = f a (t) I f WI' di.1 e (t) , Thus, A R =
Va E A.
1112. /./ e and An < 1 z e , n < C.
Lemma 5.4.4.
If H =
Q.E.D.
E alik , where Hk = r(A)ek and Ilea < 1, Vk, then k
e , E 2 ki2 ek k
is maximal. Proof.
Clearly, ,ue =
E 2 k ,u, .
Fix ri G H,and write ri =
k ilk E
E rik, where k
Hk,Vk. Then An = E,u,7„ and for each k,,u,v, ..< ;L ei by Lemma 5.4.3. k
If E is a Borel subset of SI such that j(E) = 0, then kt e,(E) = 0, Vk. Thus A nk (E) = 0,Vk, and u,.1 (E) = O. Therefore, A n ‹ Ae, i.e., 77 < Q.E.D.
C.
Lemma 5.4.5. in H.
If r(A)' is afinite, then the set of maximal vectors is dense
Proof. By Lemma 5.4.4, there is a maximal vector e(E H) at least. Define He = n(A)e, and fix ri E H. We can write ri = ni + 772, where r i E He, n2 E Ht . Let Hi = r(A)ri l c He , and write
e = Ci + e2,
where
Ci E Hi,
e2 E Hi'.
Suppose that Ai = ize,,i = 1,2. Then f.t e = Al + a2. For any Arn+ce2 — An , + E 2 /12. By Lemma 5.4.3, AI ‹ Aric Thus
E
>
0, clearly
Arri+ce2 >. Ai + E 2 A2 '"" Ai + /2 2 = lie and (ri +ce2) is also maximal since e is maximal. Notice that G He and ri + ce2 E He. Then SO 6 = e —
el
12 t71 1 ce2+ 172 = 11 t71+ce2
4 011172
el G I/1 C He.
259
and n
+ e2 =
711 4 Ee2 +172 is maximal. Clearly,
II(n + € e. 2)
— till 5
eileii.
Therefore, the set of maximal vectors is dense in H.
Q.E.D.
If r(A) 1 is afinite, then we have a decomposition
Lemma 5.4.6.
00
H =Earl,
H k = 77 (A)
k=1
and
ek,
e1> e 2 › • • • › ek > • • ..
Proof.
Let {„} be a cyclic sequence of vectors for r(A), and {TA
I k = 1,2,•..}
{S411 S41) S421 S41) S42 1 S431 S411 g.2) S'3) S441 ' **)•
=
By Lemma 5.4.5, pick a maximal vector
ii el
—
el (E H)
such that
nill < 1 .
Denote the projection from H onto H1 = r(A) Ci by p l . Similarly, there is a maximal vector e2 in HiL = (1 — p 1 )1/ such that
11C2 — (1— P1)11211 < 1 / 2 . Again let p2 be the projection from H onto H2 = 7r(A)e2. . • • . Generally, suppose that we have • • , C i, and pi is the projection from H onto Hi =
el ,.
If(a)ei l 1
k1
p)1/ such that k 1
Il ek — (1 — E pond'
< 1/k.
i= 1
Further, let Pk be the projection from H onto lik = r(A) sequence {G} satisfies:
el
>
e2
>•••>
ek > • • • ,
Now it suffices to show that H =
H,±H, ,
Ek efik.
di 0 j.
G.
Clearly, the
260
For fixed k, by the definition of {rim } there is a subsequence {k n } of
{1,2, • • .} such that nk „ = çk, 'dn. Then
11(eic„ + E j=1
=
II ek„ ( 1
pik)
k„ —1
E E etHi,Vk. Since
1
E i =1
00
Thus s'k
 kII
[7(a)0,
I
< kn 4
0.
a E A, k] is dense in H, it follows that
i=1 H=>
Q.E.D.
Hk.
Lemma 5.4.7. Let ,u, v be two regular Borel measures on 11, and v be a bounded linear operator from L 2 (11,A) to L 2 (11,0 such that
Va E A.
v(a) =
Then vf = af,V f E L2 (1Z,ht), where a = v1 E L 2 (fl,v). Proof.
Since va =
0 (01 = 1:0,(a)a =ca, it follows that
f ia(t)a(t)12dv(t) 5 114 2 f I,a(t)I 2 dp(t), Va E C (f1). Then Ilv11 2 ,u > lar • v, and af E L2 (fl,v) for any f E L2 (1l,,u). Further, from va = aa(Va E C(12)) and the density of C(II) in L2 (II,u) we get vf = Q.E.D. af,V f E L2 (1l,,u).
Lemma 5.4.8. on rt, and
Let {AO, {vic } be two sequences of regular Borel measures H = E elL2 (1t, AO, K =
iaL,2 (1, Vic ).
Suppose that there is an isometry u from H to K such that 1:0H(a) =
K (a)u,
Va G A,
where (1:0 1/ (a)(fi , • • • , fk , • • .) = (a f 1 , • • • , a fk , • • .) for any a E A and (fi , ..• , fk , • • .) G H (i.e., fk E L2 (1i,14),VIc), and (1:0K(a) is defined similarly. Moreover, if we assume that v2
then
v3
•••
• ••,
261
Proof. Let Pk be the projection from H onto Hk = L 2 (1/, tt k ),and qi be the projection from K onto K1 = L2 (SI, vi ), and uik = qi upk ,Vj, k. It is easy to see that uik (D A ,(a) = (k i (a)uik, W I k, a C A. Put aik = tlikge L 2 (11, Vi) = Ki). Then by Lemma 5.4.7, u(0, • • • , fk, 0 • • .) =
Vfk C
(ctik.fic7. •  1 ctikfk,
Hk•
Since u is isometric, it follows that
f ifk(ordtik(t) = Ei f ictik(t)fk(t)1 2 dvi(t)
(1)
Vfk E Hk. Now let E be a Borel subset of n such that v2 (E) = O. Clearly, vi (E) = 0,Vj > 2. Then for any a E A, '
aancxE
[aancXE) 0
)
u: (kth)
=
,
(2) • • • • •
actjkxE k
.
Further, by (2)
f aii Nan (t)a(t)xE(t)dvi (t) aaliXE) = 2) generally.
Lemma 5.4.9.
E
Let u be an isometry from H =
031,2 (n, /2k ) to
K
(ft, vk ),and
u(D H (a) = 'K(a)u, If /1, k2 2 >• • •, and v) vk › 14 ) Vk > j.
v3+1
• • •, where
Va E A.
j is an integer with j
> 2, then
Proof. When j = 2, it is exactly the Lemma 5.4.8. Now we assume that the Lemma holds for (j  1), (j > 2).
263
E E13L 2 (11, vo k>j1
is invariant for the * representation (1:0 K . By Lemma 5.4.6
and Theorem 5.4.1, there is a sequence {yk 1 k > j — 1} of regular Borel measures on 11 with i >. yi >. • • • such that the * representations {K', K'} and {L,41:0 1,} are unitarily equivalent, where
Ki = E
;Tv 01, vic), L=
k>j1
E
eL201,10.
k>j1
From Lemma 5.4.8, vk › ryk ,Vk > j. Now for H and j2
E 631,2(n,14) e L k=1
we have Fyk >. ,uk(Vk > j — 1) by induction. Therefore, vk ›
Let u be a unitary operator from H = E
Lemma 5.4.10.
klic ,Vk
> j. Q.E.D.
ei,201, ilk ) onto
k
K=E
031,2(n, vo,and
k
u40 H (a)u = (1)K(a), Va E A. If A I › ,u2 ›  • • and vi >. v2 › • • •, then vk — ,uk,Vk > 1.
Proof.
Let e
= u(1,0,• • •).
f a(t)d
Then for any a E A,
i (t) = (4) H (a)(1,0,•• •),(1,0,• • •))
e, e) = f a(t)clu(t), where t/e is the measure determined by e and 40 K . Thus A i = ve . = (4•K (a)
Clearly, ( 4°K(a)r7k)17k) = f a(t)dvk(t),Va E A, where rik = (0, • • • ,1,0• • .) ( E
10. Put
?I
 E(1117421)lr1k. Then n is maximal in K by Lemma 5.4.4. k
So v,.7 >. ve = A I . Since vri = E(Ilnk1121)ivk and vi › vic (Vk > 2), it follows
vi . Thus vi >. u1. By Lemma 5.4.9, vk › ,uk,Vk > 2. So we get that vri vk >. ,uk,Vk > 1. Similarly, ilk > Vic (Vk > 1) since u is unitary. Therefore, ,uk  vk,Vk > 1. Q.E.D. —
Theorem 5.4.11. Let A be an abelian C*algebra with an identity, r/ be its spectral space, and {7, I/} be a * representation of A such that r(A) 1 is
264
afinite. Then there is a sequence {P k } of regular Borel measures on SI with A i › IL2 › • • • such that {r, II} L"
{4),E eL2(11, A0), k
where (I)(a)(fi , • • • , fk , • • .) = (afi , • • • , afk , • • .) and (afk )(t) = a(t)fk (t),Vk, Va E A and (f1, • • • , fic, • • .) E H. And the sequence {il k } is unique in the sense of equivalence. Moreover, for each k > 1 the measure ILk is equivalent to k1
min {,u 7
ve,,
ri is a maximal vector in (E er(a)ei ) ± j= 1 • • • , ek_i c H such that n(A)e117(A)ei,vi o
i•
}
Where "mm" is taken according to the absolute continuity of measures. Proof. The result follows immediately from Lemma 5.4.10, 5.4.6., 5.4.9 and Theorem 5.4.1.
Q.E.D. Remark. The determination of { /L k } is very similar to the Courant principle. If a is a completely continuous nonnegative operator on a Hilbert space H, and { A 1 > A2 > • • .} is the sequence of eigenvalues of a, then for any k, Ak =
min max C1,—,Ck_IEHoxr7e[ei1...,eki1l ( 771 77) .
Proposition 5.4.12. With the assumptions and notations of Theorem 5.4.11, the * representation {ir, H} is faithful if and only if Supp,u i = II. Proof. Suppose that suppiz i = O. If a E A is such that 4:0(a) = 0, then a f = 0,Vf E L2 (1l,,u1). Picking f = i, we can see that a(t) = 0, a.e.u i . Put U = ft E II I a(t) 0 01. Then U is open and p i (U) = 0. But supp,u i = n, so U = 0, i.e., a = 0. Thus 1 is faithful. Conversely, if there is a nonempty open subset U such that Al(U) = 0. Then /1k (U) = 0, Vk > 1. Pick a G A such that suppa(.) C U. Then 40(a) = 0. Thus r is not faithful. Q.E.D.
Definition 5.4.13. A function n(.) on I/ is called a multiplicity function, if n(.) is measurable, and n(t) E {1,2,• • • , oo},Vt E n. For any given regular Borel measure kt and multiplicity function n(.) on fl, define a * representation { (1) 4011 Ho1n } of A('24 as follows:
coo)
HA 1 ri =
Eetrik,
k Ak = XEk ' AI
Ilk = L2 (1,pk),
Ek = {t E
n
I
n(t) > 14,Vk > 1,
265
and 4) 14 ,n( a )(fi , '' 1 ficl•• .) = (afi,• • • ,afk, • • .),
V(.1.1, • • • 1 fk•  • ) E H and a E A.
Lemma 5.4.14. Let A be a regular Borel measure on S1,0 G p E L 1 (1,,u), V = p . A, and E = ft E SI I p(t) > 01. Then u • XE • kt• Let F be a Borel subset of O such that v(F) = O. Since v(F) = p(t)clA(t) and p > 0, it follows that p(t) = 0,a.e.A,t E F. Thus there is a
Proof.
fBForel subset F1 C F such that ,u(Fi ) = 0 and p(t) = 0,Vt (XE • ti)(F) = ,a(E
E F\Fi . Then
n F) = IL(E n (F\F1)).
But p(t) > 0,Vt E E, and p(t) = 0,Vt E F\FI, so E n (F\FI ) (x E . u)(F) = it(E n (F\FI )) = O. Thus xE • A .< v. Conversely, let F be a Borel subset such that (x E • /2)(F) = A(E Clearly, p(t) = 0,Vt E F\E. Then from p E L i (fl,A),
v(F) = f p(t)du(t) = f F
F riE
= 0,
and
n F) =
O.
p(t)clA(t) = O.
Hence, u < XE • it. Therefore u • XE • II.
Q.E.D.
Let A be an abelian C*algebra with an identity, fl be Theorem 5.4.15. its spectral space, and {7r, H } be a nondegenerate * representation of A such that 7r(A)' is afinite. Then there is unique (in the sense of equivalence ) regular Borel measure u on SI, and unique (in the sense of a.e.A) multiplicity function n(.) on n such that {7,
H}
1.
{ 41)
4,n,
H4,n} •
Moreover, 7r is faithful if and only if supp,u = 12. Pick the sequence {Ak } as in Theorem 5.4.11. By Theorem 5.1.4, there is O < Pk c Li(n,A) such that ilk = Pk • otilVk > 1, where It = 11 1 and PI = 1. Let Ek = ft E n I Pk(t) > 01, Vic. Proof.
Since ,uk › 14+1, we may assume that SI = E l 9 E2 Now let n(t) = { i(c):),
D • • • D
Ek
D • • • .
if t E Ek\Ek+i, if t E nkEk.
266
Clearly, n(.) is a multiplicity function S2, and Ek = ft E n I n(t) > Icl,Vk. By Lemma 5.4.14, ilk XEk . it, Vk. Thus, {ir, H} From the uniqueness of {Pk }, it is easily verified that A and n(.) are unique. Finally, by Proposition 5.4.12, 7r is faithful if and only if n = supp,u i = suppA. Q.E.D. —
Notes. Using the theory of type (I) VN algebras, we can also obtain the main results in this section. The presentation here follows a treatment due to A.A. Kirillov. References. [10], [28], [91].
Chapter 6
The Classification of Von Neumann Algebras
6.1. The classification of Von Neumann algebras Definite 6.1.1.
Let M be a VN algebra. A projection p of M is said to be finite, if any projection q of M with q < p and q  p implies q = p . p is said to be infinite, if it is not finite, i.e., there exists a projection q of M such that q < p,q .., p and q 0 p. p is said to be purely infinite, if p contains no nonzero finite projection, i.e., if q is a projection of M with q < p and q 0 0 , then q is infinite. Moreover, M is said to be finite, infinite, purely infinite, if its identity is a finite, infinite, purely infinite projection respectively. Proposition 6.1.2. projection z 1 .
In a VN algebra M, there is a maximal finite central
Proof.
Let z1 = sup{z I z is a finite central projection of M}. It suffices to show that z 1 is finite. Suppose that p is a projection of M with p < z 1 and p  z 1 . If z is any finite central projection of M, then z =. zz ! .., zp < z. Thus Q.E.D. zp = z, i.e. p > z. Further, p = zl and z1 is finite.
Proposition 6.1.3. Let p,q be two projections of a VN algebra M, q < p and p be finite. Then q is also finite.
q and vv* = q1 < q. Define u = y + (p  q). Then u*u = p,uu* = (p  q) + q i < p. Since p is finite, it follows that (p  q) + q 1 = p, i.e., q1 = q . Therefore, q is also finite. Q.E.D. Proof.
Let y be a partial isometry of M such that vv
,.
268 Proposition 6.1.4. In a VN algebra M, there is a maximal purely infinite central projection z3 . Let z3 = sup{zIz is a purely infinite central projection of Al}. It suffices to show that z3 is purely infinite. Suppose that p is a finite projection of M with p < z 3 . If z is a purely infinite central projection of M, then pz is finite by Proposition 6.1.3. Since pz < z and z is purely infinite, it follows that pz = 0. Further, p = pz 3 = 0. Therefore, z3 is purely infinite. Q.E.D.
Proof.
Definition 6.1.5. A VN algebra M is said to be semifinite, if z3 = 0 , where z3 is defined by Proposition 6.1.4, i.e. any central projection of M is not purely infinite. M is said to be properly infinite, if z 1 = 0, where z1 is defined by Proposition 6.1.2, i.e, any nonzero central projection of M is infinite. Moreover, a projection p of M is said to be semifinite, or properly infinite , if the VN algebra Mp is semifinite, or properly infinite. Theorem 6.1.6.
Let M be a VN algebra. Then there is a unique decom
position: M = M1 ED M2 ED Ms,
where M1 = Mz i is finite, M3 = M Z3 is purely infinite, M2 = M Z2 is semifinite and properly infinite, and z1 + z2 + z3 = 1. From Propositions 6.1.2 and 6.1.4, such decomposition exists. Now suppose that M = Mp i ED Mp2 ED Mp 3 is another such decomposition. Clearly, Pi < z1 , p3 < z3 , and the central projection (z 1 p i )pi is finite , i = 2,3. Then we have (z i  p i )pi = 0(1 = 2,3) since Mp2 and Mp 3 are properly infinite. So z1 = p i . Moreover, if the central projection (z3  p3)pi is not zero, then it is purely infinite, i = 1 or 2 . But Mp i and Mp2 contain no purely infinite central projection, so (z3 p3)pi must be zero, i = 1, 2, and z3 = p3 . Therefore, zi = pi, i = 1,2,3. Q .E.D . Proof.
Definition 6.1.7. A VN algebra M is said to be discrete, if for any nonzero central projection z , there is a nonzero abelian projection q ( see Definition 5.3.18) such that q < z . M is said to be continuous, if M contains no nonzero abelian projection. Moreover, a discrete VN algebra is also said to be type (I) ; a purely infinite VN algebra is also said to be type (III) ; a semifinite and continuous VN algebra is said to be type (II) ; A finite type (II) VN algebra is also said to be type (HO, and a properly infinite type (II) VN algebra is also said to be type (Moo ) . Clearly, each abelian projection is finite. Thus a type (I) VN algebra is
semifinite.
269
Theorem 6.1.8.
Let M be a VN algebra. Then there is a unique decom
position :
Al = All ®M2 ED M3 , where mi = Mz i , i = 1,2,3 are type (I), (II), (III) VN algebras respectively, and z1 4 z2 + z3 = 1.
Proof. By Proposition 6.1.4, there is a maximal purely infinite central projection z3 in M. Then M3 = M Z3 is type (III). Let z1 = sup{zIz is a central projection of M such that Mz is type (I)}. We claim that Mz i is also type (I). In fact , suppose that p is a nonzero central projection of M with p < z 1 . Then there is a central projection z of M such that pz 0 0 and Mz is type (I) . So pz is a nonzero central projection of type (I) VN algebra Mz . By Definition 6.1.7, there is a nonzero abelian projection q of Mz such that q < pz . Clearly, q is also a nonzero abelian projection of Mz i . Therefore, Mz i is type (I) . Since each abelian projection is finite, it follows that z1 z3 = O. Now let z2 , 1  z1  z3 . Clearly , M2 = MZ2 is semifinite. If p is a nonzero abelian projection of M2, then c(p) < z 2 . We say that Mc(p) is type (I) . In fact , suppose that z is a nonzero central projection of Mc(p). By Proposition 1.5.8, zp 0 O. Since (pM23)z , zp(Mc(p))zp , z contains a nonzero abelian projection zp . Thus , Mc(p) is type (I). By the definition, we have p < z 1 . This contradicts the fact that c(p) < z2 . Therefore , M2 contains no nonzero abelian projection, i.e., M2 is type (II). Moreover, since z1 and z3 are maximal, this decomposition is unique.
Q.E.D. Theorem 6.1.9.
Let M be a VN algebra. Then there is a unique decom
position:
M = M11 ED M12 ED M21 ED M22 ED M3 , where M11 is finite type (I) , M12 is properly infinite type (I) ( and semifinite also ), M21 is type ( 111) ) M22 is type (II) ( and semifinite also ) , M3 is type (III) ( purely infinite). Consequently, there are only five classes of factors.
References.
[21 ] , [28], [82], [111].
6.2. An ergodic type theorem for Von Neumann algebras
270
Let H be a Hilbert space, h* = h E B(H), and p be a projection on H with hp = ph. Let
M(h) = sup{(he, e)leE pH )11 ell = 1 1) m(h) = inf{(he, 01 e E pH, II ell = 11, co p (h) = M(h)  m p (h). Clearly, Mp (h),m p (h) are the maximal, minimal spectral points of (hIpH) respectively. If p = 1 , we denote WO, m i (h), w i (h) by M(h),m(h),w(h) respectively. If I is a family of projections on H with ph = hp, Vp E I, then we define wi(h) = suP{wp(h) IP E I}. Lemma 6.2.1. Let M be a VN algebra on H, Z = MnMI, and h* = h E M. Then there is a projection z E Z and a selfadjoint unitary operator u E M such that 1
1
m(h)f, + n(h)ez + n(h)(f z  h) )
271
it follows that
1(h + uhu1 )z
l(rn(h) + n(h))(fi + ez) + n(h)( f z  fl)
> 1(m(h) + n(h))z. Noticing that
M (h)  lw(h) = .41. M (h) + lm(h)
= 1. (rn(h) + n(h)), we have
M (h)z > 1(h + uhu1 )z > (M (h) 
i.e.,
1 3 coz ( i (h + uhul) < i w(h). Similarly, from
hz' < n(h)ez' ± M (h) f z' = n(h)e i + M (h) f z' + n(h)(ez i  ei) and
(uhu 1 )z i < n(h)fz i +M(h)e i + n(h)(ez i  el), we have
m(h)z' < '.(h + uhu1 )i < 1(n(h) + M (h))(f z' + e i ) + n(h)(ezi  e i ) < 1.(n(h) + M (h)) zi = (m(h) + !co(h)) zi . Thus
1 wiz( (h + uhu 1 )) < 2 4
Q.E.D. Lemma 6.2.2. Let M be a VN algebra, Z = M n M', h* = h E M, and z = 1 . Then 7 be a finite orthogonal family of projections of Z with
E
zE7
there is a finite orthogonal family l' of projections of Z with a selfadjoint unitary element u of M such that
cor(il (h+ uhul) < 43 w7 (h).
E
ziEli
z' = 1 and
272
Let J = {z 1 , • • • , z r,} . For each i E {1, • • • , n } , by Lemma 6.2.1 there Proof. is a central projection cil of Mi = M z i and a selfadjoint unitary element ui of Mi such that
1
( (h 1 2
3 ui ui 1 )) <  cozdh), j = 1,2, 4
where hi = hz i , c12 = z  cil . Define u =
ui . Then u is a selfadjoint
unitary element of M and
(h uhu 1 ))
0. Then there is some f E Q and some zEZ=Mn M.' such that
Ilf • h  zil < Proof. By Lemma 6.2.1, there is a central projection p of M and fl G Q such that
w31(fi • h) < 34 w(h), where 11 = fp,1  pl. Now we assume that for some positive integer j there is a finite orthogonal family of projections of Z with p = 1 and h G Q
E
pE
such that li (fi
• h) < (!4 )' w(h).
273
Using Lemma 6.2.2 for fi  h and fi , then there is a finite orthogonal family p = 1 and g E Q such that iiii of projections of Z with
E
pE7i+ 1
W rj+I(g
• (fi . h))
i CkI rj (fi . h)
(4) 2+1 w(h). Therefore, for any positive integer k there is a finite orthogonal family lk of projections of Z with p = 1 and fk E Q such that
E
pElk
(734 ) k w(h).
wh(fk • h)
Pick k such that (1)kw(h) < E. For any c E ik, let Ac = ii (fk  h) IcHil (H is the action space of M). Then II(fk • h)c — A c cil coc(fk  h). Now let f = fk and z = Ac. Then we have
E
C Elk
lif • h — zii
= max{I1(f • h)c — A c cil I c e ik }
< wyi (fk  h) < (Dkw(h) < E. Q.E.D. Lemma 6.2.5. Let {a l ,  •  ,a ri } c M and E. >0. Then there is a f E Q and {z1 ,  • • , zn } CZ=Mnill' such that Ilf  ak — zk il < E, 1 < k < n. Proof. We may assume that 4 = ak ,Vk. When n = 1 , this is just Lemma 6.2.4. Now we assume that the conclusion holds for n. For a l ,  • • , an+ i E M and E > 0 , first pick z 1 ,•   ,z, E Z and f G Q such that
Ilf • ak — zkil
(p(fi ),VI.
Then ci — fi and p(e i ) > (p(fi ). Consequently,
I I Since p p. ci , fi and M is finite, it follows by Proposition f' < and fi 0 6.3.2 that (p— c i ) , (p fi). Thus, we get ci < p and ci 0 p. By the maximum of {ei} and {fi }, we have (p(e) < p(f) for any projections e and f with e — f, and e < p — ci, f < p fi . Let —
—
pt > 0, and (p(e) 5_ ,up(f) for any projections 1 kto = inf {A e and f with e  f , and e < p el, f < P — fi J —
Clearly, i.to < 1. We say that 0 < p(p — ci) < yo. In fact, if p(P — el) > ',to, then there is kt G [ktchP(P ci)) such that (p(e) < p(f) for any projections e and f with e — f, and e < p el, f < P fi. In particular, P(P — ci) < 1, a AP (P — fi) < (p(p — e l )(p(p — fi ) and P(/) — fi) > 1. But 40 (7) — fi) contradiction. Thus , 0 < (p(p — ci) < itto Now pick € > 0 such that 0 < (to — E)_ l ,u0 < 1 + ;!i . By the definition of fi kto, there are projections e2 and f2 with e2  12, and e 2 < p el, f2 < P —
—
—
—
—
278
such that p(e2 ) > (y o 4p(f2 ). Clearly e2 and f2 are not zero. We claim that there are nonzero projections e3 and h with e3 r..., h , and e3 < e2 , h < h, such that p(e) > (kto  4,o(f) for any projections e and f with e  f, and e < e3, f < h . In fact , if such e3 and h don't exist, then e2 and f2 are not such e3 and h . Thus there are projections e and f with e , f, and e 5., e2 , f __G_ f2 , such that p(e) < (itto  e)p(f). Further, (e 2  e) and (f2  f) are not such e3 and h , we have also . By the Zorn lemma, we can write e2 = eet and f2 = efi such that t t
E
E
ei  fi ,
and
p(e 1 ) < (ti o 
or(fi ),
VI.
Since cp is normal , it follows that cp(e 2 ) < (/Lo  E)(P(h). This contradicts that (P(e2) > ('uo  E)(P (f2) . Thus e3 and h exist. Let y E MI, be such that vv = e, vv* =. h , and define
0(x) = p(v*xv),Vx E f 3M f3 . Clearly , 0(1.3) = p(e3), and p(e3 ) > 0 since cp is faithful on MI, . If r and q are projections of f3 Mf3 with r , q, by (v*q)*(v*q) = q we have ✓ , q  v*qv in MI, and v* qv < e3 . By the property of (e3 , f3) and the definition of ,u0 , we get (izo  Op(r) < (p(v*qv) < ptop(r). In particular, p(v*rv) ttoP(r). Then ( to  E)p(r)
OM = cp (v* qv)
< tioço(r) < i17:;p(v*rv) < (1+ DOH.
Finally, let pa = f3 (_< p) and
Po(x) = 0(h) 1 0(x),
Vx E Mo = Mpo .
Clearly, po is a normal state on Mo . If z G Mo is such that p o (x* x) = 0, then vx = 0 since p is faithful on MI, . Further, x = x f3 = xvv* = 0, i.e., po is faithful on Mo. From the preceding paragraph, we have
1 )(po(r) 400(0 Ç (1 + n for any projections r and q of Mo with r , q. By Lemma 6.3.4., we obtain Q.E.D. Po(a*a) Ç (1 + ili )(po(aa*),Va E Mo.
Lemma 6.3.7.
Let M be a finite VN algebra. Then for any positive integer n there is a normal state Or, on M such that
( x * x) G (1+ n1 )7,1),(xx*),Vx G M.
279
Proof. By Lemma 6.3.7, for fixed n there is a non—zero projection pa of M and a faithful nonmal state p o on Mp,„ such that
po (a*a) G (1 + — n1 )(p 0 (aa*),Va E Mpo . Let {p i , • • • , pm } be a maximal orthogonal family of projections of M such that pi  pa,' < à < tn ( notice that m is finite since M is finite). By Theorem 1.5.4, there is a central projection z of M such that
(1 — E pi)(1 — z).
(1 — Epj )z po z, po (1— z)
Since {pi } is maximal , it follows that p o z 0 O. Let v:v i = po z, vv : = pz, 1 < i < rn, and
vm+ ,vm+ * , .., (1— Epi )z,
vm * +I vm+1 < Paz,
i and define
m+1
pn(z) , E ro(v:xvi), vz E
M.
i=1
Then for any z G M, ip n (X * X)
m+1
tn+1
i=1 < (1 + 7
i,/= 1
=
f* * E po(v:x*xv i ) . E pom x vi v i* xv i ',)i
1 ) Epo(v;xvi v:ev i )
iii = ( 1 + rt, ) E P o ( v; xx * vi) = (1 +
71, )ço n (xx * ).
j
Moreover, ço n (1) > trup o (po z), and po (po z) > 0 since po z 0 0 and (Po is faithful on Mpo . Therefore , On(.) = Pn(l) i pn() is what we want to find. Q.E.D. Theorem 6.3.8. Let M be a finite VN algebra. Then for any a G M,f K(a) = 1, where K(a) is defined as in Theorem 6.2.7. Proof. By Proposition 6.2.8 and Theorem 6.2.7, we may assume that a > 0 and Oil < 1/2. Suppose that there are c 1 , c 2 E K(a) and cl 0 c2. Clearly, cl, c 2 > 0 and
i licl —c211 < 1. Let c l — c2 = f 1ydz i, be the spectral decomposition of (c i — c 2 ) , where zi, is a central projection of M, Vu. Since c l 0 c2 , there is A > 0 such 0 or (1 — z),) 0 O. Let z z_ x for the case of z 0, or that either z z = 1 — zA otherwise. Then we have c 2 z > c i z + Az
Or
CIZ > C2Z + Az.
280
c2 z and By the symmetry we may assume that c i z > c2 z + Az. Since c i z c 1 z,c 2 z G if(az), and replacing M by Mz , we may also assume that z = 1. Pick {On } as in Theorem 6.3.7. Then for any unitary element u of M, 7,141(u s au) = i,b((ctiti)*(alu)) :5_ (1+
n1 )7,bn (u*au). 1,bn(a) = 1,G n ((alu)(a 4 u)*) < (1+ —
Thus for any f, g E Q ( see Definition 6.2.3),
On(f • a)
(1+ — n1 )0n(a)
Let {fk},{g k } c Q be such that fk  a On(ci)
( 1 + 71) 2 0n(9 • a)c l , gk  a
c 2 . Then
(1+ — n1 ) 2 0n(c2).
Further , from c l > c2 + A it follows that n
(c 2 ) + A < On (ci)
(1+ — n1 ) 2 0n(c2)
When n is sufficiently large , we get a contradiction since A > O. Therefore, If(a) contains only one element, Va G M.
Q.E.D.
Remark. In the end of section 6.4, we shall prove that : if K(a) =1, Va E M, then M is finite. Now we start to characterize finite VN algebras by normal tracial states.
Lemma 6.3.9. Let M be a finite VN algebra. Then there is a normal tracial state on M at least. Proof.
By Theorem 6.3.8, we can define a map T() from M to Z = MnM i
such that = {T(a)}, Va G M.
From Proposition 6.2.8 and the definition of If(),T is linear, and T(z) = z, Vz Z, T(M + ) c Z+ .
Since K(u*xu) = K(x) for any x E M and any unitary element u of M , it follows that T(u*xu) =T(x). Further , T(xy) = T(yx), Vx, y E M. Let V) be the same as the 0 1 in Lemma 6.3.7, and define cp(a) = 0(7' (a)),
Va G M.
281
From the preceding paragraph, ço is a tracial state on M. Now it suffices to show that p is normal. Let {b1 } be a bounded increasing net of M+ , and b = supi bi . Put al = b — bi , V/. Then al > 0(a(M, M.)). We need to prove that p(a t ) ) O. For any e> 0 , there is / 0 such that
0 < I,b(ai ) < E ) Vi > / 0 since V) is normal. Pick fi G Q such that 11ft  al —T(ai)II < E. , V/. By Lemma 6.3.7, for / > / 0 we have
0 < cp(ai ) = 71)(7'(al )) O.
Q.E.D.
Theorem 6.3.10 A VN algebra M is finite if and only if there is a faithful family of normal tracial states on M , i.e., for any non—zero a G M+ there is a normal tracial state yo on M such that p(a) > O. Proof. Suppose that M is finite. Then there is a normal tracial state ço on M by Lemma 6.3.9. Let z = s(p). Then z is a non—zero central projection of M, and cp is faithful on Mz . Again we continue this process for finite VN algebra M(1 — z),• •  , and so on. By the Zorn lemma, there is a family {pi} of normal tracial states on M such that
s(pi ) • s(pp) = 0, V/
P,
and
E s(pi ) = 1. i
It
is easily verified that the family {(p t } is faithful. Conversely, let I be a faithful family of normal tracial states on M. If w E M is such that w*w = 1, then p(1— p) . cp(iew)—(p(ww*) = 0,Vp E 7, where p = ww*. Since 7 is faithful, it follows that p , 1. Therefore, M is finite.
Q.E.D. Another characterization of finite VN algebras is as follows.
Lemma 6.3.11.
Let p be a projection of a VN algebra M, and let v G M
be such that v*v = p,
Put qn = vny*n,
n = 1, 2, •  • , q0 = p,
en = qn — qn+ h
n = 0,1,2,— • .
282
Then {e n } is an orthogonal sequence of non—zero projections of M with en etn ,Vn,m, and en 0 (strongly).
■••••.•
Proof. Since q < p, it follows that pv = v and v*nvn = p,Vn > 1. Thus qn is a projection for each n. From qnqn+1 = q n+i we have
p
qo > qi >
Further , en em = 0, Vn m, and e n Let u n = vqn , n = 0, 1, 2, . Then
q2 > • • • •
0 (strongly).
(u n — un+t) * (un — un+t) = en, (u n —
Un+1)(Un Un+1) * = en+17
i.e., e n  en+i)Vn > O. Moreover , by eo = p
—
vv* y 0 we get en 4 0, Vn > O. Q.E.D.
Theorem 6.3.12.
A VN algebra M is finite if and only if the * operation
is strongly continuous in any bounded ball of M.
Let M be finite. By Theorem 6.3.10, there is a faithful family Y" of Proof. normal tracial states on M. Suppose that {x i } is a net of M with Ilx i ll < 1,V/, and xi 0 (strongly). Then for any a G M+ and p E
ILa p(x i x) I = ko(xi'ax1)1 'llaiko(xi xi)
O.
If [La ça E M, (,o E 3] is dense in M* , then xixi' 0 (weakly) ) . Further, 0 (strongly) . So the * operation is strongly continuous in any bounded ball of M. Now we need to prove that [Lapla G M,p E .7 ] is dense in M* . Let b E M be such that Lap(b) = 0, Va G M,cp E J. Then cp(bb*) = 0,Vp E J. Since J is faithful, it follows that b = 0. So the above assertion holds. Conversely, suppose that the * operation is strongly continuous in any bounded ball of M. If M is not finite, then there is v G M such taht vv = 1,vv* 0 1. By Lemma 6.3.11, there is a sequence {e n } of non—zero projections e rn , Vn m, and en 0 (strongly). Let tvn G M of M with . en, = 0, e be such that wn*tvn = en, wnw;', = el, Vn. Clearly, wn 0 (strongly ), I wn II < 1,Vn. By the assumption, wn* 0 (strongly). Thus , e i = wn wn* 0 (weakly) , a contradiction. Therefore, M is finite. Q.E.D. In the proof of Lemma 6.3.9, we introduce a map T ( ) from a finite VN algebra to its center. Now we discuss the properties of that map in detail. Let M be a finite VN algebra. The map T from M to Z=M n is defined by {T (a)} = K (a) (V a G M) and is called the centra/ valued trace on M, where KO is defined as in Theorem 6.2.7.
Definition 6.3.13.
283
Proposition 6.3.14.
Let M be a finite VN algebra, and T : M > Z = M n M' be the central valued trace. Then: 1) T is a projection of norm one from M onto Z, and is oo continuous. Consequently, T(a) > 0,Va E M+ ;T(za) = zT(a),Va E M,z G Z ; T (a)* T (a) < T (a* a) , Va E M; 2) T(ab) = T(ba, \fa, b E M; 3) T(a*a) = 0 if and only if a = 0; 4) {p(T(•))1p is a normal state on M} is a faithful family of normal tracial states on M; 5) p :5_ q if and only if T(p) < T(q), where p and q are two projections of M.
Proof. 1) From the proof of Lemma 6.3.9, it suffices to show that T is a a continuous. Since T is positive, this is equivalent to prove that cdo(TH) is normal for any normal state p on M, i.e., to prove
p(T(a)) = sup i ço(T(a i )) for any bounded increasing net {ad of M+ , where a = sup i a i . But {T(a i )} is also a bounded increasing net of Z+ , by the normality of yo we need only to show that T(a) = sup i T(a i ). Clearly, T(a) > sup1 T(a1 ). If they are not equal, then there is a non—zero central projection z and a positive number A such that
zT(a) > zsup i T(al ) + Az. Let 7 be a faithful family of normal tracial states on M. For any e> 0 and 1, pick fi E Q such that II fi • (a — adz — T((a — a i )z)II < 6. Then for any 1,1) E F,
liP(T((a — adzni < 17,14fi  (a — ad z)I + e
< E fi (u)kb(u(a — = 1; ((a —
ai )zu*)I + e
ai ) z)I + E.
Since V) is normal, it follows that
0(T(a — a i )z) > 0,
VO G F,
i.e., 0 (T (a) z) = lifn 0(T(ai)z) = 0(zsup I T (al )) , VO E F. However, 1,1(T (a) z) > tgzsuPiT(at)) + 4(z),V0 E J. Thus, .1,1)(z) = 0,V0 E .T, and z = O. This is a contradiction. Therefore, T(a) = sup1 T(a 1 ). 2) It is contained in the proof of Lemma 6.3.9.
284 3) Let / = {a E MIT(a* a) = 0 } . Clearly , I is a s(M,M,)closed twosided * ideal of M. By Proposition 1.7.1, there is a central projection z of M such that I = M z . In particular, z E I, i.e.,
z = T (z) = T (z* z) = 0. Therefore , I = 0, This is just our conclusion. 4) If a E M+ is such that p(T(a)) = 0 for any normal state on M , then T (a) = 0. By 3) , we have a = 0 . Therefore, {yo o Tip is a normal state on M } is faithful.
5) Suppose that p  q i G q. By 2) , we have T (p) = T (q 1 ) ..< T (q). Conversely, let T(p) < T (q). By Theorem 1.5.4, there is a central projection z such that pz 3, qz, q(1 — z) L‹ p(1 — z). Thus, T (q) (1 — z) < ( p) (1 — z) 5, T (q) (1 — z) , i.e., T (q)(1 — z) = T (p) (1 — z). Let q(1 — z) , 23 1 < p(1 — z). Then T(p i ) = T (q) (1 — z) = T(p)(1 — z), and T(p(1 — z) — p 1 ) = O. By 3), p i = (1 —z)p, i.e., q(1 — z) ,, p(1 —z). Therefore, Q.E.D. P " q. 
Now consider some properties of o finite and finite VN algebras. Proposition 6.3.15. Let M be a VN algebra. Then the following statements are equivalent: 1) M is a—finite and finite; 2) M is finite, and Z = M n M' is a—finite; 3) There is a faithful normal tracial state on M. Proof.
3) = 1).
It is immediate from Theorem 6.3.10 and Proposition
1.14.2. 1) = 2). It is obvious. 2) = 3) . By Proposition 1.14.2, there is a faithful normal state 0 on Z.
Let
p(a) = 04T (a)) , Va E M. By Propsition 6.3.14, p is a faithful normal tracial state on M.
Q.E.D.
Proposition 6.3.16. Let M be a finite VN algebra. Then M is a direct sum of {MI}, where Mi is a—finite and finite, V/. Consequently, each finite factor is a—finite. From the proof of Theorem 6.3.10, there a family {p i } of normal tracial states on M such that s(r i ) • s(p ii) = 0,V/ l' and s(p i ) = 1. Then Proof.
E t
m = E em, , m, = Ms(pi),
Vi.
t
285
Since p i is a faithful normal tracial state on MI , it follows from Proposition 6.3.15 that Mi is afinite and finite, di. Q.E.D.
Notes.
Using a fixed point theorem , F. J. Yeadon gave another proof of the existence of a trace. Theorem 6.3.12 is due to S.Sakai.
References.
[18 ] , [78 ] , [144 ] , [150 ] , [199].
6.4. Properly infinite Von Neumann algebras Proposition 6.4.1.
Let M =
only if Mi is properly infinite, V/.
E EDMI . Then M is properly infinite if and i
The necessity is obvious. Now let Mi = Mz i be properly infinite, V/, and z be some finite central projection of M. Then zz i is a finite central projection of Mi , V/. Thus zz i = 0 since Mi is properly infinite, V/, and z = 0 . So M is properly infinite. Q.E.D.
Proof.
Proposition 6.4.2. A VN algebra M is properly infinite if and only if there is no normal tracial state on M. Let p be a normal tracial state on M. Then s(p) is a nonzero central projection, and Ms(p) is finite by Proposition 6.3.15. Thus M is not properly infinite. Conversely, if M is not properly infinite, then there is a nonzero central projection z of M such that Mz is finite. There is a normal tracial state IP on Mz at least. Let p(.) = 0(.z). Then p is a normal tracial state on M.
Proof.
Q.E.D. Proposition 6.4.3. If M is a properly infinite VN algebra, then the * operation is not strongly continuous in unit ball of M.
Proof.
It is immediate from Theorem 6.3.12.
Q.E.D.
Theorem 6.4.4. Let M be a VN algebra. Then the following statements are equivalent: 1) M is properly infinite; 2) There is an orthogonal infinite sequence {p,,} of projections of M such that p ri = 1, Pn '''''' 1, Vn.,
En
286
3) There is a projection p of M such that p
1.
(1 — p)
00 Proof. 2) = 3) . Pick p =
P2n+ 1 •
Then we have
n=0
(1 — p)  1.
p
3) = 1) . Let z be any non—zero central projection of M. Then pz (1 p)z z. Thus z is not finite, and M is properly infinite. 1) 2). Since the identity 1 is infinite, it follows that there is v E M such that vv = 1 and vv* < 1, and vv* 0 1. Let —
qn = v n v *n ) en = qn — qn+i) n = 0, 1, 2, • • • .
By Lemma 6.3.11, en 0 0, e n em = 0, en  em , Vn 0 m. By the Zorn lemma, there is a maximal orthogonal family { el e G A} of non—zero projections of M such that {el k E A} D {e n }, and ei el), Vi, P G A. Define p = 1— el. By eEA
Theorem 1.5.4, there is a central projection z of M such that
pz
eo z,
e0 (1
—
z)
p(1
—
z).
Clearly , z 0 by the maximum of the family {e l k E A}. Since #A is infinite , then we can write 00
A=
where Ai n Ai = 0, Vi
Ai,
j, and #Ai = #A,Vi. Let
r i = pz +
Eez
,
r; =
tEn i
E
ei z, Vj
2.
/EA '
00
Clearly, ri ri = 0, Vi
j, r i
r2
z, and z =
E ri .
Since M(1 — z) is
still properly infinite, we can make the same process, • . • . Then by the Zorn lemma, there is an orhtogonal family fzi l of non—zero central projections of M with E z1 = 1 such that for any 1 there exists an orthogonal infinite sequence {rtnin = 1 ) 2 2
} of projections of M satisfying r12 ••• • • • •• zt,
Now let p r.,
E rin , n = 1, 2, . .. PnPm = 0, Vn
m,
and
E rin =
Then
E pn = 1,
and
pn 1,Vn.
Q.E.D.
287
Using Theorem 6.4.4, we can get a property of finite projections. Proposition 6.4.5. Let p, q be two finite projections of a VN algebra M. Then sup{p, q} is also finite.
Proof.
We may assume that sup{p, q} = 1. By Proposition 1.5.2, (1  p)  (q  inf{p,q}) < q.
Thus (1  p) is also finite. Suppose that z is a nonzero central projection of M such that Mz is properly infinite. Noticing that pz and qz are finite and sup{pz,qz} = z,
we may also assume that M is properly infinite By Theorem 6.4.4, we can write 1 = r + (1  r), where r  (1  r)  1. From Proposition 1.5.5, there is a central projection z such that
(1  p)(1  z).
rz < pz, (1  r)(1  z)
Since pz and (1 p)(1 z) are finite, it follows that z( rz) and (1 z)((1  r)(1  z)) are finite. This contradicts that M is properly infinite. Therefore, sup{p, q} is finite. Q.E.D. 


In the end of this section, we prove the conclusion of the Remark after Theorem 6.3.8. Proposition 6.4.6.
Let M be a VN algebra. If 4 K(a) = 1,Va E M, then
M is finite. Proof. Suppose that z is a nonzero central projection of M such that Mz is properly infinite. By Theorem 6.4.4, there is a projection p of M such that p < z and p  (z  p)  z. Let u, y E Mz be such that tett = V * V = z ) 1.1,U * =
p,
vv* = z  p.
Define ito : M —± Z = M n M .' such that K(a) = {4to(a)},Va E M. By the definition of K(.) , it is clear that '1(ab) = (I)(ba), Va, b E M. Then
z = c1(z) = (I)(p) = (I)(z  p), and 2z = 240(z) = 40(p) + 41:0(z  p) = 40(z) = z. This contradicts that z O. Therefore, M is finite. Q.E.D. References. [18], [82].
288
6.5. Semifinite Von Neumann algebras Definition 6.5.1. Let M be a VN algebra. A trace on M+ is a function p on M+ , taking nonnegative, possibly infinite, real values, possessing the following properties: 1) cp(a + b) = cp(a) + p(b), Va,b E M+ ; 2) cp(Act) = Ap(a),Va E M+ , A > 0 ( with the convention that 0 • +oo = 0); 3) p(x* x) = p(xx*),Vx E M. A trace p on M+ is said to be faithful , if a E M+ is such that p(a) = 0, then a = 0. A trace p on M+ is said to be semifinite, if for any 0 a E M+ , there is 0 y4 bEM_F and b O. By Proposition 6.5.4, we can find a projection p of M with p < z and 0 < 0(p) < oc. Thus there exists a normal tracial state on M. From Propositon 6.4.2, Mp is not properly infinite. This is a contradiction since z is purely infinite and 0 0 p < z . Therefore, M must be semifinite.
Q.E.D. Theorem 6.5.8.
A VN algebra M is semifinite if and only if there is a faithful semifinite normal trace on M+. Proof. The sufficiency is clear from Proposition 6.5.7. Now suppose that M is semifinite. By the Zorn lemma, there is a family {pi } of semifinite normal traces on M+ such that s(pi ) • s(p ie) = 0,V1 0 1', and s(p 1 ) = 1. t (pi is faithful semifinite normal trace on M. Then cp = Q.E.D.
E
E i
Remark.
If p is a faithful semifinite normal trace on M+ , then we can get a faithful W*representation of M by the GNS construction. In fact, let M, .M be as in Proposition 6.5.2. Define an inner product on )1/41:
(z,y) = p(y*x) = yo(xy*),Vx,y E M. Denote the completion of (M,(,))  by Hv, . For any a E M, define 1,p (a)x ia = (ax), Vx G M, where x > x,p (Vx E M) is the embedding of M in H. It is easy to see that 7r,p (a) can be uniquely extended to a bounded linear operator on H,p , still denoted by 7r,p (a). Then we obtain a * representation {7,p , Ii",p } of M. If a G M is such that 7,p (a) = 0, then ax = 0,Vx E M. Since M(C )1/41) is udense in M, it follows that a = O. Thus A9 is faithful. Moreover, if a net {at} C Al") M ai II < 1,Vi, and al (L 0, then for any x,y E M, —
p(y*a i x) = p(xy*ai )  0 by Proposition 6.5.3. Therefore, nv, is also a W*representation of M.
Proposition 6.5.9.
Let M be a VN algebra.
293
1) If M is semi—finite, p and p' are the projections of M. and M' respectively, then MI, and My , are also semi—finite. 2) If M = E em, then M is semi—finite if and only if Mi is semi—finite t
for each 1.
Proof. 1) Let r be a faithful semi—finite normal trace on M. Clearly Ao is still a faithful semi—finite normal trace on (4,) + . Thus , My, is semi—finite from Theorem 5.3.8. Moreover, MI,, is a * isomorphic to Mc(pi), where 41) is the central cover of p' in M'. Since Mc(p 1) is semi—finite, it follows that Mi,, is also semi—finite. 2) It is obvious. Q.E.D.
Theorem 6.5.10.
Let M be a VN algebra. Then the following statements
are equivalent: 1) M is semi—finite; 2) There is an orthogonal family {p i } of finite projections of M such that
E pi = 1; t
3) There is an increasing net {0} of finite projections of M such that (strong)lim qi = 1; 4) There is a finite projection p of M such that c(p) = 1, where c(p) is the central cover of p in M.
Proof. 1) = 2) . By the Zorn lemma, there is a maximal orthogonal family {pi } of finite projections of M. If p = 1 — E pi 5 0, then there is a non—zero t finite projection q of My since MI, is still semi—finite. Clearly , qp i = 0,V/. This is a contradiction to the maximum of {pd. Therefore, P1 = 1. 1 2) = 3) . It is immediate from Proposition 6.4.5. 3) = 1). Let z be any non—zero central projection of M. Then there is an O. Since zq i is finite, it follows that z is not purely index 1 such that zo infinite. Therefore, M is semi—finite. 1) = 4). By the Zorn lemma, there is a maximal family {Pi} of finite projections of M such that c(pi ) • c(p is) = 0,V/ l'. Put p = E pi. We claim t that p is also finite. In fact, if r is a projection of M with r < p and r ,, p, then we have that
>
rc(PO  pc(pi) = Pt)
rc(Pt)
5_ pc(pi) = pi, vi.
Since pi is finite , it follows that rc(pi) = pi, V/. Further,
p=
EI n = rEt c(pi) _, rp Et c(pi ) = rp = r.
294
Thus p is finite. Now it suffices to show c(p) = 1. If 1 — c(p) 0 0, then there is a non—zero finite projection q < (1 — c(p)) since M(1 — c(p)) is semi—finite. Clearly c(q) • c(p i ) = 0,V1. This is a contradiction since the family {pi } is maximal. 4) .1.). Let z be a non—zero central projection of M. By Proposition 1.5.8, zp 0 0, where p is a finite projection with c(p) = 1. Thus z contains a non—zero finite projection zp, and z is not purely infinite. Therefore, M is semi—finite.
Q.E.D. Lemma 6.5.11.
Let N be a VN algebra on Hilbert space K, and e be a cyclic and separating vector for N with Hell = 1. If cp(•) = (•, e is a tracial state on N, then there exists a conjugate linear isometry j with .7.2 = 1 such that a * jaj(da E N) is a conjugate linear * algebraic isomorphism from N onto N'. e )
,
Proof. Define jae = a*e,Va E N. Then j can be uniquely extended to a conjugate linear isometry on K, still denoted by j, since p is a tracial state and e is cyclic for N. Clearly, 3.2 = 1. For any a, b,c E N, we have jajbce = bca*e= bjajce. Thus jNj C N'. Further, noticing that ((jaj)*be,ce) = p(ac*b) = cp(c*ba) = (ja*jbe,ce), Va,b,c E N, we get (jaj)* = ja*j,Va E N. It is clear that jaj = 0 implies a = 0. Now it suffices to prove jNj = N'. Let ate N' with 0 < a' < 1. Define 1,1)(a) = (aa' e,e),
Va E N.
Then 0 < Tib < p. By Theorem 1.10.3, there is t o E N with 0 < t o < 1 such that T,b(a) = p(toato),Va E N. Thus, at e = tge = itgie. Since e is also a separating vector for N', it follows that a' = jaj. Therefore, jNj = N'.
Q.E.D. Lemma 6.5.12.
Let M be a VN algebra on a Hilbert space H, e E H, and p, p' be the projections from H onto M.' e, Me repectively. Then p is finite if and only if 23' is finite.
Proof. Clearly, p E M,p' E M' ( see Definition 1.13.1) . Suppose that p is a finine projection of M. Consider the VN algebra L= pp'Mp'p on pp'H. Then e(E pp1 H) is cyclic and separating for L. By Proposition 6.3.1, L is finite. Clearly, L is also a—finite. Hence, there is faithful normal tracial state
295
io on L from Proposition 6.3.15. Let {ir r , H 0 , e,p } be the faithful cyclic W*representation of L generated by p. Put N , 719,(L). Then N is * isomorphic to L, and is finite. Further, by Lemma 6.5.11, N' is also finite. Since N, L admit cyclic and separating vectors respectively, it follows from Theorem 1.13.5 that N is spatially * isomorphic to L. Therefore, L' is also finite. Now if 2 E MI is such that pp' x'ilp = 0, then
e92, e
0 = yp'xipipe = ypi x'pi e = p'x'p'ye, Vy E M. But Me = 731H, so plx 1131 = 0. Thus , pi x'pl —p ppl xi pi p(Vx' E M') is a * isomorphism from p'M'p' onto pp'Mipip. Further, plIMIp' is finite since L' = ppi Mipip, i.e., pi is a finite projection of M. Similarly, p is finite if /I is finite. Q.E.D.
Propositon 6.5.13. Let M be a semifinite VN algebra on a Hilbert space H. Then M' is also semifinite. Suppose that there is a nonzero central projection z of M n M' such that Miz is purely infinite. Clearly , Mz is still semifinite. Therefore, we may assume that M is semifinite , and also M' is purely infinite. By Theorem 6.5.10, there is a finite projection p of M with c(p) = 1. Then M' is * isomorphic to M;. So we may further assume that M is finite, and also M' purely infinite. Pick a nonzero vector H, and let p,71 be cyclic projections of M,M 1 determined by respectively. Clearly, p is finite. Then by Lemma 6.5.12, p' is also nonzero finite. This contradicts that M' is purely infinite. Therefore, M' is semifinite. Q.E.D. Proof.
e
eE
Proposition 6.5.14. A VN algebra M is semifinite if and only if M is * isomorphic to some VN algebra N so that N' is finite. The sufficiency is obvious from Proposition 6.5.13. Now let M be semifinite. Then M' is also semifinite. By Theorem 6.5.10, there is a finite projection p' of M' with c(pi) = 1. Pick N = Mr ,. Then M is * isomorphic to N, and N' = M;) is finite. Q.E.D. Proof.
Proposition 6.5.15. Let M be a semifinite VN algebra, and p be a projection of M. Then p is finite if and only if there is a faithful family 7 of semifinite normal traces on M+ such that p(p) < oo, Vcp E I. Proof.
Sufficiency. Let q be a projection of M with q < p and q , p. Then
p(q) = p(p) 0 ( strongly), and Ilxi II < 1, x ip = xi , V/. For any ça G 7 and 0 < a E .M,p , where .M,,, is the definition ideal of p, by Propositions 6.5.2 and 6.5.3, we have
IlLaP(xixn1 = Ip(ax i x;11 = Icp(xi' ax1 )1 5 IlallP(4 zi) = Ilall(lipP)(xixi) 'O.
Hence, it suffices to show that the set [La c,* E 1, a E (M 92 ) + ] is dense in M. Let x E M be such that
p(ax) = 0,
Vcp G F,aE
Then p(x* ax) = 0 since x* a G ./vi ci, for any a E .tvi v , and cp G F. Since (p(E ,7) is semi—finite and normal. it follows from Propositions 6.5.4 and 1.7.2 that p(x*x) = 0,Vp G F. But ,7 is faithful, so x = O. That comes to the conclusion.
Q.E.D. As preliminaries of next section, we study the semi—finite and properly infinite VN algebras on a separable Hilbert space.
Lemma 6.5.17.
Let H be a separable Hilbert space, and M be a semi— finite and properly infinite VN algebra on H. Then there is an orthogonal infinite sequence {pri } of finite projections of M with /Jr, , p,Vn, m, such that pr, = 1.
E Ti
297
Proof. Let q be any non—zero finite projection of M, and {q/ } 1EA be a maximal orthogonal family of projections of M with qi , q, Vi G A. Put p= c(q) —>q 1 .
tEn By Theorem 1.5.4, there is a central projection z such that pz .L‹ qz, q(1 — z) 
p(1
—
z).
Clearly, qz 0 0 since the family {qi } is maximal . Further, z 1 = c(q)z 0 0,
and z1
. E qi z i + pz i , pz i ‹ qz i . tEA
If #A is finite, then by Proposition 6.4.5 z 1 is a non—zero finite central projection. It is impossible since M is properly infinite. Thus #A is infinite. Indeed q i z i , i.e., there is #A is countably infinite since H is separable. Then z1 ,,
E
IEA y
G M such that
vv *
= z,,
v* v
= E qi z i . lEA
For each 1 G A, let pi = vqiz i v*. Then p i pis = 0,Y1 0 l', and
p i ,, (voz i )*  (vq i z i ) = q 1 z 1 ,V1 E A. Clearly, {pi}iEn is an orthogonal infinite sequence of finite projections of M, pi ,, pir 7 V1,P, and p1 = z 1 0 O. Further, by the Zorn lemma we can get the
>
IEA
Q.E.D.
conclusion.
Proposition 6.5.18. Let M, N be two VN algebras on a separable Hilbert space H, and M', N' be semi—finite and properly infinite. If (1) is a * isomorphism from M onto N, then 40 is also spatial. Proof. By Proposition 1.12.5, we may assume that there is a VN algebra L on H, and projections p', q' of L' with c(p 1 ) = c(q') = 1 such that M = Lp s, N = Lq s , and (1)(api) = aqi, Va E L. By Lemma 6.5.17, there are two orthogonal infinite sequences {p}, {q} of finite projections of L' such that pt
= E p i,7
q
,_E c
i
p:  pli ,
i
q:  q},
Write {1,2,.—} = UAn such that A n countably infinite.
Vi,j.
n Am = 0,Vn
m, and each A n is
298
Fix n. Then there is a central projection z of M such that
E pii ,
zg in ..< z
(1
—
E
z)
iEA rg
pi'
iEA n
E
A n \{S}. Clearly ,
Fix s E An, and write A
(1— z)g ni. pli ,
iEA n
(1 — z)
E
g  (1— z)
icA n
E
pi'. Then
iEA:,
E 7): icAl„
Pi. p.
iEnu
Since n is arbitrary, it follows that g' ..< p'. Similarly, p' ‹ g'. Therefore, p' — g', and (I) is spatial. Q.E.D. 
References.
[28 ] , [144], [150].
6.6. Purely infinite Von Neumann algebras Proposition 6.6.1. Let M be a VN algebra. 1) If M = E EDMI , then M is purely infinite if and only if Mi is purely i infinite, V/. 2) If M is purely infinite, then so is M'. 3) Let M be purely infinite, and p, p' be projections of M, M' respectively. Then MI, and MI,' are also purely infinite.
Proof. 1) The necessity is obvious. Now let Mi = Mz i be purely infinite, V/. If p is a finite projection of M, then pzi = 0,V/, and p = 0. Therefore, M is also purely infinite. 2) It is immediate from Proposition 6.5.13. 3) By 2) , it suffices to show that Mps is purely infinite. But Mp, is * isomorphic to Mc(p 1), and Mc(pl) is purely infinite obviously, so Mps is purely infinite. Q.E.D.
299
Proposition 6.6.2. A VN algebra M is purely infinite if and only if there is no non—zero semi—finite normal trace on M+ . If there is a non—zero semi—finite normal trace (i) on M+ , then its support s(p) is a non—zero central projection of M. By Theorem 6.5.8, Ms(io) is semi—finite, i.e. M is not purely infinite. Conversely, if there is a non—zero central projection z of M such that Mz is semi—finite, then we have a non—zero semi—finite normal trace on (Mz) + . Therefore, there is a non—zero semi—finite normal trace on M+ . Q.E.D. Proof.
Proposition 6.6.3. A VN algebra M is purely infinte if and only if the * operation is not strongly continuous in any bounded ball of Mp for any non—zero projection p of M. Proof. Let M be purely infinite, and p be a non—zero projection of M. Since p is infinite, there is y E M such that vv = p, vv < p, and vv* 0 p. By Lemma 6.3.11, we can find an orthogonal sequence {e n } of non—zero projections of M such that e n  em ,Vn,m, en < p, Vn, and en > 0 ( strongly ) . Suppose that wwn = en , wn wn* = e i ,Vn. Clearly , wn * 0 ( strongly) , < 1, wnp = wn ,Vn. However , {w} does not converge to 0 strongly, so ilwnll — the * operation is not strongly continuous in any bounded ball of Mp. Conversely, if M is not purely infinite , then M contains a non—zero finite projection p. By Proposition 6.5.16, the * operation is strongly continuous in any bounded ball of Mp. Q.E.D.
Proposition 6.6.4, Let M be a afinite and purely infinite VN algebra, and p, q be two projections of M with c(p) = c(q). Then p  q. Proof. From the Zorn lemma, we can take a maximal orthogonal family fzi l lEA of central projections of M such that qz i ..< pz i ,V1 G A. Let
z =
E zi , p' = p(1 — z), q' = q(1 —
z).
t EA By Theorem 1.5.4 and the maximum of the family {zt}ten)vve can see that p' < q' ( relative to M(1 — z)). Suppose that p' 0 O. Pick a maximal orthogonal family {q's } 3E7 such that §", , 21, q's < q',Vs E I. For Theorem 1.5.4, there is a central projection z' of M with z' < (1— z) such that (q' —
E q:)zi .
q, we have 
p = pz =
E pzi _>:: E p i = q. lEA
1EA
Similarly, we can prove that p ..< q. Therefore, p , q.
Q.E.D.
Let M be a afinite and purely infinite VN algebra, Proposition 6.6.5. and a be a nonzero element of M. Then K (a) 0 {0}, where K(a) is defined in Theorem 6.2.7. Proof. By Proposition 6.2.8. and K(a*) = K(a)*, we may assume that a* = a. Also we may assume that Ilall <  1 and a + 0 0 ( otherwise, replace a by a ) . Then we can find a nonzero projection p of M and a positive integer n such that
q>
1 
n
p

(1

p).
 1. Further , If p contains a nonzero central projection z, then a for any b E K(a), we have b_> 7 le z  1. Thus , K(a) 0 {0}n. Now suppose that p contains no nonzero central projection. Replacing M by Mc(p), we may assume that c(p) = 1. Since p >. 1  c(1  p), it follows that c( 1  p) = 1. By Proposition 6.6.4, we have
p ,. (1

p) ,. 1.
301
From Theorem 6.4.4, there are pairwise orthogonal projections {e1, • • • )en+1} such that n+1
P=
ei)
ei p, Vi.
and
Pick vi E M such that
=
Vi Vi* =
ei 5
ej+i ,
0
N40(z) is spatial by Theorem 1.13.5. Moreover, (I) : M(1 — z) * N(1 (1)(z)) is also spatial from Proposition 6.5.18. Therefore, the * isomorphism 4:1:0 from M onto N is spatial. Q.E.D. —
References.
[18 ] , [28 ] , [144].
6.7. Discrete
(
type (I )) Von Neumann algebras
Theorem 6.7.1.
Let M be a VN algebra on a Hilbert space H. Then the following statements are equivalent: 1) M is discrete ( type (I)); 2) M' is discrete ( tyep (I)); 3) M is * isomorphic to some VN algebra N such that N' is abelian; 4) there is an abelian projection p of M such that c(p) =1; 5) any nonzero projection of M contains a nonzero abelian projection of
M. Proof. 1) = 4). Pick a maximal family {p i } of nonzero abelian projections of M such that c(pi ) • c(pi) = 0,V1 y 1', and put p = p i. If c(p) 1, then
S t
(1 — c(p)) contains a nonzero abelian projection since M is discrete. This contradicts the maximum of the family {p i }. Thus c(p) :=. 1. Moreover, by Proposition 1.5.9, pi Mp ii = {0},V1 1'. Therefore, p is also abelian.
303
2) = 3). Suppose that M' is discrete. From the preceding paragraph, M' admits an abelian projection p' with c(pi) := 1. Now let N = Mpi. Then M is * isomorphic to N, and N' M;, is abelian. 3) = 5). Suppose that (I) is a * isomorphism from M onto N, where N is a VN algebra on a Hilbert space K, and N' is abelian. Let p be any nonzero projection of N, e be a nonzero vector of pK, and q be the cyclic projection of N determined by C. Since the VN algebra NIg on qK = Nte is abelian, and admits a cyclic vector e, it follows from Proposition 5.3.15 that Nq' = (Ng;)' = Ng , and q is abelian. Clearly, 0 q < p. Therefore, any nonzero projection of M contains a nonzero abelian projecton of M. 5) = 1) . It is obvious by Definition 6.1.7. 4) = 2). Suppose that p is an abelian projection of M with c(p) = 1. Let L = M, . Then M' is * isomorphic to L, and L' = M,, is abelian. Now by 3) 5) = 1), M' is discrete. Q.E.D. Remark. For any VN algebra M, the VN algebra N generated by M U M' is discrete. In fact, N' = M n M' is abelian.
Proposition 6.7.2. Let M be a VN algebra. 1) If M = EDM1 , then M is discrete if and only if Mi is discrete, V/.
E i
2) Let M be discrete, and p, p' be projections of M, M' respectively. Then Mp ,Mpi are also discrete.
1) It is abvious from Definition 6.1.7. Proof. 2) Since My) ) is * isomorphic to Mz, where z = c(p'), it follows that My,/ is discrete. Moreover, Mp is also discrete by (Mp = MI9 • Q.E.D. ) '
Proposition 6.7.3. Let M be a discrete ( type (I) ) factor. Then M is * isomorphic to B(K), where K is some Hilbert space. Proof. From Theorem 6.7.1, M can be * isomorphic to a VN algebra N on some Hilbert space K such that N' is abelian. Clearly, N is a factor, and N' is an abelian facor. Therefore, N' =0, and N .= B(K). Q.E.D. Remark. From this proposition, we can get the results on finite dimensional C* algebras ( see Section 2.13), too.
Lemma 6.7.4. Let p,q be two projections of a VN algebra M, and p be abelian, p < c(q) . Then p ‹ q.
304 Proof.
From Theorem 1.5.4, there is a central projection z such that
qz :3 pz, p(1

z) :_
,
E
LEA
rEE
Proof.
Clearly, c(Pt) = c(q,.) = 1, V/, r. So by Lemma 6.7.4, we have Pi qr ,A G A, r E ff. If #A < oc, then M is finite by Proposition 6.4.5, and #ff must also be finite. Thus #A and if are finite or infinite simultaneously. Consider the case that #A and # ff are finite. We may assume that #A < 4 /. Then 1 :=, Pi ^..' qr 5.. qr  1,
E
E
E
LEA
rEi s
LEE
where II' c if and #.//' = # A. Since M is finite, it must be II' = if, and #A = #ff. Now let both A and II be infinite index sets. Fix p E {p i }. The abelian VN algebra Mp is finite obviously. So by Propositon 6.3.16 and 1.3.8, there is a nonzero central projection z of M such that Mpz is a finite. Considering Mz, {Az}, {q,z}, we may assume that z , 1. For any 1 G A, MN is a afinite VN algebra on pi H. By Proposition 1.14.2, there is a countable subset .M 1 of pi H such that [MI .Mi] is dense in pi H. Let ffi , {r E ffig,..Mi y4 {O}}. Since {gr it E II} is pairwise orthogonal, II must be a countable subset of if, Vi E A. Moreover, if there exists r E ff\ ULEA ffi, then gr .M1 := {0}, V/ E A. Further, qt.M' M i = {0}, i.e. qr pi = 0,V1 E A. This is a contradiction since pi = 1 I # 11 . So < and gr #ff < # A. Similarly, O. Therefore , if := UtEnit, and #A #A , #ff. Q.E.D.
>
Definition 6.7.6. A VN algebra M is called type ( In ) or nhomogeneous , where n is a finite or infinite cardinal number, if there is an orthogonal family {pill G Al of abelian projections of M such that p i  p 1 1,V1,1' E A and pi = 1, and #A = n.
E
LEA
By Lemma 6.7.5, the definition of type (in) is independent of the choice of the family {p i ll G A}.
305
Proposition 6.7.7. Let M be a VN algebra. 1) If M is type (in ) , then M is type (I). 2) M is type (in ) if and only if M is spatially * isomorphic to NB(H) ) where N is an abelian VN algebra, and kin is a ndimensional Hilbert space. Consequently, A factor of type (/n ) must be * isomorphic to B(B n)• 1) Let {p i } be as in Definition 6.7.6. Then c(p i ) = 1,A. By Theorem 6.7.1, M is discrete. 2) The necessity is immediate from Definition 6.7.6 and Theorem 1.5.6. Conversely, suppose that {e l l/ E A) is a normalized orthogonal basis of Hn , where #A = n. Let p i be the projection from Hn onto [el]. Then {pi I/ E A) is an orthogonal family of abelian projections of B(H) with pi  pi i,V1,1' and pi = 1. Therefore, NB(H) is type (/n ) since N is abelian. Q.E.D. t Lemma 6.7.8. Let M be a VN algebra, and {z1 } be an orthogonal family of nhomogeneous central projections of M where n is a cardinal number. Then z = Ezi is also nhomogeneous. Proof.
E
i Let { pna E I} be an orthogonal family of abelian projections of Mz i withp(j)  p a(1),Va, a', and Ep(j) = z1, Vi, where 4 1 := n. Put pa = Ep(j). a i Then {pa la E I} is an orthogonal family of abelian projections of M such that par ^# Pal, Vol, al , and = z. Therefore, Mz is type (In). Proof.
EN a
Q.E.D. Lemma 6.7.9. Let M be a VN algebra, and zi be a n ihomogeneous central projection of M,i = 1,2. If n 1 n 2 , then z 1 z2 = O. Let {P i) 1/ E Ai} be an orthogonal family of abelian projections of Mz i such that p p,V1,1 1 , and E = zi , where #A i = ni , i = 1, 2. Then tEni {W )z2 11 E A 1 ) and {4 2) zi1 1 E A21 are also two orthogonal family of pairwise Proof.
40
equivalent abelian projections of Mz1 z2 with
E lEA,
23 1) z2 =
E ezi 
ziz2.
iEn2 If z 1 z2 0 , then #A i = #A 2 by Lemma 6.7.5. This contradicts n1 y n2. Therefore, z 1 z2 =, O. Q.E.D.
Theorem 6.7.10. Let M be a type (I) VN algebra. Then there is unique decomposition M = E ITIM„, where E is some set of different candinal numriE
bers, and Mn is type (/n ),Vn E E.
306
Proof. Pick a nonzero abelian projection p of M, and let {p i } be a maximal orthogonal family of projections of M such that pi  p,V1. By Theorem 1.5.4, there is a central projection z such that
(1  q)z < pz, p(1  z) _ (1  q)(1  z), where q =
E pi. By the maximum of {p i }, z is not zero.
If (1  q)z := 0, then
t z = Ep i z is homogeneous. Now let (1  q)z 0 O. Then z 1 = zc(1  q) is a 1 nonzero central projection. Suppose that (1  q)z  q i < pz. Since pz is abelian , it follows from Proposition 1.5.8 that q 1 = c(q 1 )pz =c((1  q)z)pz = pz i . Clearly, (1  q)z i = (1 q)z. Thus (1  q)z i  q i = pz i ,, pi z i , VI, and z 1 =
E pi z i + (1 q)z i is homogeneous.
I From above discussion, we get a nonzero homogeneous central projection of M. Further, by the Zorn lemma there is an orthogonal family {Z r } of homogeneous central projections of M such that 1. Finally, by Lemma Zr
E =
6.7.8 and 6.7.9, the conclusion can be obtained.
References.
r
Q.E.D.
[82 ] , [86 ] , [88 ] .
6.8. Continuous Von Neumann algebras and type (II) Von Neumann algebras Proposition 6.8.1. A VN algebra M is continuous if and only if there is no nonzero central projection z such that Mz is discrete. In consequence, a purely infinite VN algebra must be continuous. Proof. If M is not continuous, then M contains a nonzero abelian projection p. Let z = c(p). By Theorem 6.7.1, Mz is discrete. Conversely, suppose that z is a nonzero central projection such that Mz is discrete. Then Mz contains a nonzero abelian projection p. Clearly, p is also an abelian projection Q.E.D. of M. Thus, M is not continuous.
Proposition 6.8.2. Let M be a VN algebra. 1) If M =EDM1 , then M is continuous ( or type (II)) if and only if Mi
E
is continuous ( or type (II) ) , V/.
1
307
2) Suppose that M is continuous ( or type (II) ), then so is M'. 3) Let M be continuous ( or type (II) ) and p, p' be projections of M, M' respectively. Then My, and Mr, are continuous ( or type (II) ).
1) It is obvious by Definition 6.1.7. 2) Let M be continuous. If there is a non—zero central projection z of M' such that Miz is discrete. Then by Theorem 6.7.1, (Mlz)' = Mz is also discrete, a contradiction. Therefore, M' is also continuous. Moreover, by Proposition 6.5,13, M' is type (II) if M is type (II). 3) From (Ma = M, the conclusion 2), it suffices to show that Mp, is continuous. Since Mp, is * isomorphic to Mc(p 1 ), the conclusion is obvious. Q.E.D.
Proof.
Theorem 6.8.3. A VN algebra M is continuous if and only if every projection p of M can be writen as p = p i + p2 , where p i , p2 are projections of M with p 1 p2 = 0, and p i p 2 •
Proof.
Sufficiency. If p is an abelian projection of M, then from the assumption we can write p p i + p 2 , where p i ,p2 are projections of M with p2. Clearly , c(P1) = c(p)• Since p i < p and p is abelian, it P1P2 = 0,Pi follows from Proposition 1.5.8 that p i = c(p i )p = p. Therefore, p = 0, and M is continuous. Now let M be continuous, and p be any non—zero projection of M. Then Mp is not abelian, and there is a non—zero projection q of My, such that q My n Mpl. By Theorem 1.5.4, we can find a central projection z of M such that
qz
(p— q)z, (p q)(1— z)
q(1 — z).
If qz = (p q)(1 — z) = 0, then q = p(1 z) G Mr, n go a contradiction. Thus we have either qz 0 or (p q)(1 z) O. If qz 0 0, let qz = r i r2 < (p — q)z, then r i r2 = 0, and (r i r2 ) < p. If (p q)(1 z) 0 0, let (p— q)(1 — z) ri r 2 < q(1 — z), then r i r2 = 0, and r i r 2 < p. Therefore, p contains two non—zero projections r i , r 2 such that r i r 2 = 0 and r i r 2 . Continue this process for (p— (r i r 2 )). • • • . Then by the Zorn lemma, we can get a decomposition p = p i + p2 with p i p2 = 0 and p i — p2 . Q.E.D.
Theorem 6.8.4. A VN algebra M is type (II) if and only if there is a decreasing sequence {A i } of projections of M such that p i is a finite projection with c(p i ) = 1, and (Pn r+1) Pn+i)Vn.
Proof.
Let M be type (II) . From Theorem 6.5.10, there is a finite projection
308
Pi of M with c(p i ) = 1. By Theorem 6.8.3., we can write Pi = P2 +
172,
where
Pn = Pn+ 1 + qn+ 1)
P2q2 = 0,
where
and
P2 '''' q2)
n CI .D n+lin+i = )
and Pn+i  qni1)
Then {pn } satisfies the conditions. Conversely, suppose that {pn } is a decreasing sequence of projections of M, where p i is a finite projection with c(p i ) = 1, and (Pn  Pn11) ,, pn .“ ,Vn. From Theorem 6.5.10, M is semifinite. If Afpi is continuous, then (Mpi )' = Mi , is also continuous. Further, M' is continuous ( since M' is * isomorphic to and M is type (II) . So it suffices to show that Mr , is continuous, and we) may assume that p i = 1, i.e., M is finite. By Propositions 6.3.16 and 6.8.2, we may also assume that M is afinite. Then there exists a faithful normal tracial state yo on M. If p is an abelian projection of M, then by Theorem 1.5.4 there is a sequence {zn } of central projections of M such that qn < pZn) 13(1 z n ) .< p n (1  z n ),Vn•
Since p is abelian, it follows that qn = c(q)pz n = c(p)pz. Noticing pn 0(Pn i — pa ), we have c(p) ._?_ Pai. Thus , c(pn ) = 1, and qn = pz n i.e., pn zn  pz n , and p .< pn ,Vn. On the other hand, from Pn — Pn+ 1 + (Pn — Pn+ 1) ) Pn+ 1 ' (Pn — Pfl+1) )
it follows that p(Pn)'= IP(Pn+i). Noticing that p(pi) = (10(1) = 1, we get that p(p) = 2  n+ 1 ,Vn. In addition, cp(p) _G_ p(p)  2  n+ 1 ,Vn, thus p(p) .=' 0. Further p = 0 since p is faithful. Therefore, M contains no nonzero abelian projection, and M is continuous.
Q.E.D. References.
[18], [82].
6.9. The types of tensor products of Von Neumann al
gebras Let Mi be a VN algebra on a Hilbert space Hi ,i = 1,2. Their tensor product Ali012 is a VN algebra on Hi 0 I/2. In this section, we consider the relations between the types of Mh M2 and the type of MITOM2.
309
Proposition 6.9.1. The VN algebra MI OM2 is finite if and only if both VN algebras MI and M2 are finite. Proof.
Let MI OM2 be finite. Since MI is * isomorphism to Mi 0 1 2 , it must
be that MI is finite. Similarly, M2 is finite too. Now suppose that both MI and M2 are finite. By Proposition 6.3.16, we may also assume that both Mi and M2 are ufinite. Then there are faithful normal tracial states p i , p2 on MI , M2 respectively. We can write =
where { di)} c Hi and
En (. eS:) eS:)), v. E A, ,
< 00,1 ,_, Eller n
1,2. Consider pi 0 p2(.) n,m
Clearly, ço i 0 cp 2 is a normal tracial state on M1 0M2 . Since pi is faithful on A, fen is a cyclic sequence of vectors for A/I:,i = 1,2. Thus {d1 ) 0 is cyclic for (MI OM2 )', and p i 0 yo 2 is also faithful on Q.E.D. M1 042. Therefore, MIOM2 is finite. /.4" ( 1 ) M Aq2 ) Aqi)t21eS,V). 1
`.4n.
' Ap. Since M is semifinite, there is a nonzero finite projection q of M with q < p. Then, a > Aq, and 0 G r(Aq) < oc by 1). Therefore cp is semifinite. 4) If M is purely infinite, then any nonzero projection of M is infinite. Therefore, p(a) = doo,Va E M+ \{0}, i.e., ça is uniquely determined. Now suppose that M is semifinite. By 3) and Theorem 6.5.8, such p is existential. Let p i , cp2 be two faithful semifinite normal traces on M+ . We need to prove that p i = Ap2 for some positive constant A. First, let M be finite, and put ça = p i + p2 . By 1) and Proposition 6.5.2, 1°) P1 (P2 can be extended to faithful normal traces on M. From Theorem 1.10.3, there is t E M with 0 < t < 1 such that (
Pi (a) = p(ta),
Va E M.
Then ça (tab) = pi (ab)
p i (ba) = p(tba)
p(atb),
i.e., p((ta at)b) = 0,Va, b E M. Since cp is faithful, it follows that t E M n M' =O. Further, we get p i = Ap 2 for some positive constant A. Secondly, suppose that M is semifinite and properly infinite. From Theorem 6.5.10, there is an increasing net {q i } of finite projections of M with supi qi = 1. From preceding paragraph, for each index 1 there is a positive constant A i such that Pi(a)
A 1 402(a),
Va E (Mq ,) + .
Since {q i } is increasing, it follows that A i is independent of the index 1. Put A = Ai, V/. Then PI (magi) = Ap2(qtaqi),
Va E M+,
V/.
Moreover, by Proposition 6.5.2, pi (qi ctql ) pi (aimai),Va G M+ , i = 1,2. Further, from the normality of p i and p2 , we have pi (a) = Ap2(a),Va E M+. Q.E.D.
315
Remark.
For type (I) factor B(H), the unique (up to multiplication by a positive constant) faithful semifinite normal trace on B(H) + is as follows tr(.)
E(.6,
e,),
v.
G B(H) + ,
f ed
is a normalized orthogonal basis of H. For a finite factor, there is unique faithful normal tracial state on it. For any semifinite factor, there is unique (up to multiplication by a positive constant) faithful semifinite normal trace on its positive part. For purely infinite factors, the case is trivial. where
Proposition 7.1.3. Let M be a factor, P =Proj(M) be the set of all projections of M, ça be as in Proposition 7.1.2, and D {cp(p)lp E P}. Then multiplying cp by a proper positive constant, we can get the following: 1) D = {o, 1 , • • • , n}, when M is type (In ) (n finite or infinite). In particular, if M = B(H r,), where dimHn = n, then P(P) = dim pH, Vp E P;
2) D [O, 1 ] , when M is type (IL); 3) D := [0, +00], when M is type (II„) ); 4) D = {0, +oo}, when M is type (III). Proof. 1) and 4) are obvious. 2) Let p be the unique faithful normal tracial state on a type(II 1 ) factor M. By Theorem 6.8.3,
{2k11 < k < 2',n = 0,1, • • •} C V. For any A E [0,1i, pick pn E P such that ça(p) = An / A. From Proposition 7.1.2, p n Pn+i,Vn. Let q 1 = p i , and qi (1 — q). q < p2 . By Proposition 6.3.2, (1 — q i ) Thus, (p2 — q) (1 q i ). Let (p 2 — q) r + r. (1 — q i ). Then p2 Put q2 = q 1 + r. Then q2 > q l , pi ,i = 1,2. Generally, we can get { q} PnyVn. Let q = sup,., qn . Then p(q) = sup,., ça(p) = A. with qn qn+hqn —
Therefore, V = [0 ) 11. 3) Suppose that {p/ } is an increasing net of finite projections of M with sup = 1. Clearly, p(p1 ) p(1) = +oo. By 2) [0, p(pt)] C V , V1. Therefore, D = [0, +a]. Q.E.D. Lemma 7.1.4. Let M be a finite factor, and P =Proj(M) be the set of all projections of M. Suppose that D P [0, +oo) satisfies: 1) if pi, P2 G P and pip2 =0, then D(pi p2) = D(p i ) + D(p 2 ); 2) for any unitary element u E M and p E P,D(upus) = D(p);
316
3) D(1) > O. Then D = cpIP, where cp is as in Proposition 7.1.2.
Proof. If M = B(H), where dimH,., = n < oo, then by 2), there is some value A such that D (p) = A for each minimal projection p of M. By 1), D(1) = nA. By 3), A > O. We may assume that A = 1. Since each projection of M is an orthogonal sum of several minimal projections of M, it follows that D(P) = {0, 1,• • •, n}. By Proposition 7.1.3, D = (PIP. Now let M be a type (II I ) factor. We may assume D(1) = 1. Suppose that cp is the unique faithful normal tracial state on M. We need to prove D = pIP. Notice the following fact: if q ‹ p, then D(q) < D (p), Indeed, let q ,,,, p i < p. From Proposition 6.3.2, (1— q) — (1—p i ). Thus, there is a unitary element u E M such that Pi = uqu*. Further, D(q) = D (p 1 ) < D(p). Fix p E P. Since D(1) = ço(1) = 1, from Theorem 6.8.3 there is a subset {Pn,kin = 0, 1, • • • , 0 < k G 2n } C P such that D(Pn,k) = (P(Pn,k) "= 2  nk, Vfl, k. Thus, we can find {p m} c P such that D(p m )= ço(pm ) / p(p). By Proposition 7.1.2, p m ._. p, Vm. From preceding paragraph, D(p) .? D(p m ) = p(p m ) ço(p). Hence, D(p) __?_ p(p). Similarly, D(1 — p) _?_ (p(1 — p). Therefore, Q.E.D. D(p) .=' P(P)) and D = PIP.
Definition 7.1.5. Let M be factor, and P be the set of all projections of M. A function D : P  [0, dool is called a dimension function, if: 1) D(p) = 0 4>. p = 0; 2) for any unitary element u E M and p E P, D(upus) = D(p); 3) if p,q E P and pq = 0, then D(p + q) = D (p) + D(q); 4) if M contains a nonzero finite projection, then there is a nonzero projection po of M such that D(p o) < 00. Theorem 7.1.6. Let M be a factor, P be the set of all projections of M, and D(.) be a dimension function on P. Then D = (pIP, where ço is as in Proposition 7.1.2.
Proof. First, we claim that: if p(E P) is infinite, then D(p) = +00. In fact, p is properly infinite. By Theorem 6.4.4, we can write p = En Al) where PnPnt
—
n,mPrz)
6
Pn ^.."
p,
Vfl,
trt.
Clearly, we have a unitary element u, of M such that unmpn u,s = p m , Vn, m. Thus D(p) = D(p,,,,),Vn,m. Of course, D(p) > O,Vn. Therefore, D(p) =
+ 00
From the preceding paragraph, we get D = (pIP if M is purely infinite. Now let M be semifinite, and pa be as in Definition 7.1.5. Clearly, p a is finite (otherwise, D(po ) = doo, a contradiction). Pick ço as in Proposition
317
7.1.2 such that p(p0) = D(p0 ). Now we prove that D = cpIP. It suffices to show that D(p) = cp(p) for any finite projection p of M. Let p(E P) be finite, and q =suP{P)Po}• By Proposition 6.4.5, q is also finite. From Lemma 7.1.4, we have
 pl(P n Mq ),
DI(P nMq ) and
D(p) , cp(p)
consequently.
Q.E.D.
Corollary 7.1.7. The dimension function is uniquely determined up to multiplication by a positive constant. References.
[28 ] , [111].
type (HO
7.2. Hyperfinite
factors
Let M be a type (II I ) factor. Then there is unique faithful normal state ço on M. Define
Vx G M.
11x112 = p(x*x)i, Then II
' 112
tracial
is a norm on M, and
11xY112 5_ min{11x11 ' 11Y112) 11x112 ' 'MI}, Vx, y G Al. topology generated by 11.11 2 is equivalent to the strong
11x112 = 11x * 112 _. 11x11)
From Lemma 1.11.2, the (operator) topology in the unit ball (M), of M. Lemma 7.2.1.
Let
p
be a projection of M, and a*
is a spectral projection q of a such that a > 0, then
11 a 1 — P11 2 Proof.
Let E
G (0,1/2),a = 1 _1 1 Ade A ,
11 q — P112
Then there = a E (M)1, i 911a — pill. Moreover, if
13 11 a — P11 211 • and
q = 1— e l _ e , q i = ee — e, q2 = 1 — q — q l . When A
V [—E) El
U (1  E ) 1], we have
IA2 —
E  E 2 > E/2. Then
lEllq2112 5._ 11 (a 2 — a)q2112 5 IIa2  a 11 2 1/2, then we have immediately
11q — P112 5_ IIq112 + lia — PII2 + 11a112 .5. 2 + (lia 112 + IIP112) 1/2 11a — P11212
 li a — PM V 2 (4 + ( 11 a 11 2 + 0112) 112 ) 5 9 11a — pli V 2
5
Now let a > 0, and keep above notations. Then
11aq — q11 5_ E)
1
IIa 3 qi II < E.
From Ilq2112 5 611 E.Il a — P112 (see preceding paragraph), we have that
I1a / — qII2 5_ Pig — qII2 + Ilaigi ll2 + Ilalq21 1 2 < E + E l/2 + Ilq2112 < E + E1/2 + 211a —pp112. If
Ila — pliV2
0, there is a type (12n) subfactor N of M (n sufficiently large) and b 1 , • • ,b,,E N such that Mai — bill2
E)
1 < < rn.
First for 6/2, there is a finite dimensional * subalgebra A of M, and ci, • • • cn, E A such that Proof.
Ilai
cil12
E/2,
i
m.
We may assume 1 E A, where 1 is the identity of M. By Section 2.13, there is an orthogonal finite set {zi } of central projections of A with > = 1 such that Ai = Az i is a finite dimensional factor, Vi. Suppose that {p P ( } i is an orthogonal set of minimal projections of Ai with E i p i( i) = zi ,Vi. Clearly, p
(relative to Ai), ri "' 1 = P
Vj,k. Thus, we have {tv} C Ai such that (i)*
v./ • (i), tv i(i)W 1(i)*p i(i), y).
Wi W.  p i
tv P is a matrix unit of Ai ,Vi. Further, {c 1( 2 ( tvit). 1i, j,k) Then {w 1 Wk is a basis of A. Now it suffices to show that: for enough small 6 > 0(6 depends on E and c l , • • • , c m), there exists a type (12n) subfactor N of M(n sufficiently large) and {q) } C N such that 114 ) — v i) 112 < 6, Vi, j. Pick sufficiently large n, such that 2 —n < 6 2 and 2 —n < p(p),Vi. From Proposition 7.1.3, for each i we can find an orthogonal set q} of projections of M such that < Pp, ço(q (i) ) = 2, Vk, {
and
p(p
E qii) ) < 2.
Now let N be a type (12 .1) subfactor of M such that {4 ) q(ki) li,j, k} C N, and define q) = tiT ) Ek q(k i) ,Vi,j. Then (4) 4))*( tv (p q))
p)
E
322
and
iitv"1)1)E=r(pçoEqii))< 2n < 5 2 , di,j. k
Q.E.D. Lemma 7.2.6. For any a l , • • • , am E M, any projection p E M and E > 0, if (p(p) = 2n 5 then there exists a type (I2r) subfactor N of M (where r > n), b 1 , • • • ,bm, E N, and a projection q E N such that e)
Il ai — bill2
1 _,_ i < tri)
(P(q) = 2n .
IIP — q112 . e)
By Lemma 7.2.5, we can find a type ((I2 r) subfactor N of M and b i , • • • ,bm+I E N such that Proof.
r _?_ n,
Ilai  bilk < 45,
1 < i < m,
lip  bm+1112 5_ 6 )
* +1 = bm.o . Let where +5(> 0) will be determined later, and bm b = 2b m+1 (1± bm 2 +XI.
Clearly, b E N, 11b11 < 1. Since p = 2p(1 + p ) l , it follows that b i +  (p  bm+i)P. 4
1  (1)  p) = (1 + b 2m+1 ) 1 (bm.0  P)(1 + p) 2
Thus, Ilb  p11 2 < such that
36.
By Lemma 7.2.1, there is a spectral projection q i of b i
II @  P I I2 5_ 9 1 1 b  p111 5_ 156. Then, Ir(qi) — 2  n1  Ir(P — qi)1
5 IIP  g1112 __
156t.
Now pick a projection q of N such that (lo(q) =_ 2n and either q > q1 or q < ql . .4 and Then Ilq  gifl = Go(qi) — 2n 1 < 156.., il q  P112
5_ Mg  qil12 + Ilqi PII2 __ 1515i +
fEbi.
If take 6 > 0 is such that 6 < E and 1564 + VT56i < E ) then we can get the Q.E.D. conclusion.
Lemma 7.2.7. Let a l ,  • am E M and p be a projection of M with cp(p) = 2n 5 and pa i= aip = a 1 ,1 < I < m. Then for any E > 0 there is a type (I2 r) N such that subfactor N of M with r > n, and b 1 , • • • ,b p E N,pbi = bip = bi) liai  bib
0) will be determined later. Then p , q. By Lemma 7.2.4, we have a unitary element u of M such that
73 = u * qu ,
Ilu — 1 112
36
1IP — q11V8 < 366 1/8 .
Let N = u* Au,b 1 ., pu*c iup,1 < i < m. Then N is also a type ( 12 r) subfactor of M, b i ,    ,b,n ,p E N,pb i = b ip = b i ,1 < i < m and
Ila= — b=112 55
Ilu * ciu — a=112 5 11d= — uaiu*I12 11c= — a=112 II + II aiu — uaill2 11a= — c=112 + 2 11a=11 '11u — 1 112
< 8 + 72 iiaiii 81I8 It is enough to pick 6(> 0) such that
8 + 728118 max Had 5 e • ICI:Cm
Now let p
L, where L is a type (I2 n) subfactor of M. Suppose that {pi p, P2, )232n} is an orthogonal set of minimal projections of L. By Theorem 1.5.6, M is spatially * isomorphic to MpTOB(K), where dimK = 2 2 . This spatial * isomorphism also maps L to LAB(K) = ŒlorToB(K), where H is the action space of M. From the preceding paragraph, there is a type (12 r)(r > n) subfactor A of M with p E A, and b 1 , • • • ,b,n E A such that Ila= — bilk < E3Pb1 = bip = b,1 < I < m. Clearly, P3 bi 3 • • • ,L E Ar,. Since io(p) = 2n , Ap should be * isomorphic to a matrix algebra of order 2rn . Let N = (10 1 (40B(K)), where (I) is the above spatial * isomorphism from M onto MAB(K). Clearly, L C N, and p,b 1 ,• • • ,b,„ E N, and N is type ( 12 r). E

• •
Q.E.D. Lemma 7.2.8. Let L be a type ( 12 n) subfactor of M, al, • • • ,a m E M, and e > O. Then there is a type ( 12r) subfactor N of M, and b 1 , • • • , bm E N such that
r > n,
L C N, Hai — bil12 < E) 1 < i < rn
324
Suppose that {pi I 1 < i < 2 2 } is an orthogonal set of minimal Proof. projections of L, and { wi } c L such that
tvi =Pi)
w;wi =. Ph
wiw; = Pi ,
vi.
Let p ,= P13 aijk = w:aktui. Then paiik = aiikP=. aiyk,V1 < T. , 3. < 2 2 ,1 < k < m. From Lemma 7.2.7, there is a type (/27.) subfactor N of M with r > n, and biik E N such that
L
C
N, pbilk = biik p = biik ,
liaiik — biikii2
8)
Vi,j,k, where 8 > 0 and 2 2 n 8 < E. Put
E* wit,„k wi , 1 < k < m.
bk =
1 0, then we have ,u(V n E) > 0 for some n, where V = V. Further, pick a Borel subset F of f/ with F c VnE and 0 < u(F) < oo. From mfiXF = m h u g x F , we get
I
.fi (t)XF (t) = f2(t)r g (t) 1 /2 xF(g i t), a.e.i.t.
333
From FngF = 0 and F C E, the above equality does not hold at each t G F. This is a contradiction since /4/1) > O. Therefore, it(E) = 0, i.e., fi = 0, and M n mug = {0}. Now from Lemma 7.3.7, r (M. ) is maximal commutation in M x c, G. Let f E L'ISI,A) be such that eg m f ug = mf,Vg E G. Then f (gt) = f(t), a.e.,u,Vg G G. We may assume that f is real. If f is not a constant function, then there are real number r1,r2 with r 1 < r 2 such that A(E) > 0 and /2(11\E) > 0, where E . ft E n I ri < f(t) < r2 } . On the other hand, since f (t) = f (gt), a.e.A,dg G G, and G is countable, it follows that ,u((E U g ENE
n g E)) =
0,
dg E G.
Then we get either ,u(E) = 0 or ,u(SI\E) = 0 since (C, fi, 11) is ergodic, a contradiction. Therefore, f is a constant function, i.e., {a E M I a9 (a) = a,Vg E G} , 01 11 .
Q.E.D.
Finarrly, by Lemma 7.3.8, M x a G is a factor.
Lemma 7.3.12.
Let (C, 1l,/1) be a free and ergodic group measure space, and y be a Ginvariant afinite measure on all Borel subsets of n with v and v({t}) = 0,Vt G G. 1) If v(II) < oo, then Mx a G is a type (III ) factor. 2) If v(n) =.7 +00, then M x G is a type (IIoo ) factor.
Proof.
Define cp(m f ) =
fn f (t)dv(t), V f
E L'
(n, it) +.
Then cp is faithful on M+ since y  A. Let fm0 be a bounded increasing net of M+ , and rnf = sup/ m h . By Theorem 5.3.13, fi  f with respect to wstopology in L'IrZ,A) or L'in,v). Since v is afinite, we can write 11 = Un E,,, where E} is an increasing sequence of Borel subsets of 11, and v(E) < oo,Vn. Thus xE„ E L I M, v) and {
f fiXE„dv  f f
XEn dli,
Vn.
Further, sup i f fi dv = f sup fidv,
1
i.e., io is normal. The semifiniteness of cp is obvious from the afiniteness of v. Moreover, since v is Ginvariant, it follows that p(u*g m fu g ) = f f (gt)dv(t) = f f (t)dv(t) =
334
Vrn f G M+, g G G, i.e., yo is also Ginvariant. Now by Lemma 7.3.6 and 7.3.11, M x a G is a semifinite factor. If 0.1) < oc, then cp is finite, and M x a G is also a finite factor from Lemma 7.3.5. If v(11) = +oo, then p is not finite, and M x a G is an infinite factor from Lemma 7.3.5 and Proposition 7.1.2. Now it suffices to show that M x a G is ciontinuous. Let p be any nonzero projection of M with yo(p) < oc. Then by Lemma 7,3.5, = çoo(1) is a faithful semifinite normal trace on (Mx a G).i_ and cp = ;bor. Thus T,b(n(p)) < oc, and 1 (p) is a nonzero finite projection of Mx a G by Proposition 7.1.2. If Mx a G is not continuous, we may assume that Mx a G B(K), where K is some Hilbert space. Then dim7r(p)K < oc, and M contains a nonzero minimal projection Therefore, (< p). This contradicts the assumption: v({t}) = 0,Vt E
M x a G is continuous.
Q.E.D.
Lemma 7.3.13. Let (G,1l,,u) be free and ergodic. measurable, then M x a G is a type (III) factor.
If (G,n,p) is non
Proof. If M x a G is semifinite, then by Lemmas 7.3.11 and 7.3.6, there is a Ginvariant faithful semifinite normal trace ça on M+ . For any Borel subset E of ri, define v(E) = p(mx ).
Then u is a measure on all Borel subsets of fl. Since io is faithful, it follows that u A. From the Ginvariance of ça, u is also Ginvariant. By the Zorn lemma and the semifiniteness of (p, there is an orthogonal family {p i } lEA of projections of M such that EleA pi = 1 and yo(p i ) < oo, V/. Since H = 1,2 (11,11) is separable, A is countable. Suppose that pi = mxEi , where Ei is a Borel subset of ft,V/. Then u(E1 ) = yo(pi) < oo,V1, and pi) = 0.
u(n\ lli EA Ei) = p(1 — lEA
Thus, If is ufinite. From Definition 7.3.10, (G, CI, it) is measurable, a contradiction. Therefore, M x a G is not semifinite, and is a type (III) factor. Q.E.D. Lemma 7.3.14.
Let (G, [Z, A) be a group measure space, and
Go = {g G Glrg (t) = 1, a.e.A} Then Go is a subgroup of G. (G, Si, A) is nonmeasurable.
If (Go , SI, p) is ergodic, and Go 0 G, then
Clearly, Go is a subgroup of G. Now let (Go , SI, A) is ergodic, and Proof. Go 0 G. Suppose that u is a ufinite measure on all Borel subsets of 11 with
335
v  it, and y is Ginvariant. For any g G Go, since A g ,that dz (t) dv (t) = d/1(t) = d,u(gl t)
=
A  v = vg ,
it follows
0, _ i (g l t)dv(g l t) = 3T (g t)dv(t).
d
Thus ) (18 (g  lt),Vg G Go . Now (Go , f"/„u) is ergodic, so t(t) = dv (t) = 48 dv constant (a.e.p) by a similar discussion of Lemma 7.3.11. Further, A is also Ginvariant, i.e., G = Go , a contradiction. Therefore, (G,11, it) is nonQ.E.D. measurable. From above discussions, we have the following.
Theorem 7.3.15. and
Let (G, 11, A) be a free and ergodic group measure space,
H = L2(n, A),
M = {m f I f
E Looffl, ko }
(tt,f)(t) =___ r9 (0112 f (g i r) Vf E H, g E G, IN ,
ag (m f ) — u g m f u*g , V f E
Loo(n, /./), g
G G3
where r9 (.) = (c/A g idA)(.) and ditg (.) = dp(gi .),Vg E G. 1) If there is a afinite Ginvariant measure on all Borel subsets of 11 with If  A and v({t}) , 0,Vt E si, then M x c, G is a type (II I ) factor when 0 < v(11) < 00, and M x c, G is a type (IIoo ) factor when v(f./) = doo. 2) Let Go .7 {t E G I rg (t)  1, a.e.p} (a subgroup of G). If (Go, 12, /2) is ergodic and Go G, then M x a G is a type (III) factor.
Example 1. Let f/ be one dimensional circle group (compact group), i.e., n = {z E 0 I 1z1 = 1},A be the Haar measure on I/ with it(f./) = 1,G be a countable infinite subgroup on 12 , and the action a of G to fl be the multiplication of numbers. Clearly, (G,11,p) is free, p, is Ginvariant, and p({z}) = 0,Vz G f/. Suppose that E is a Borel subset of si such that ,u((E LI gE)\(E n gE)) = 0, Vg E G. Write xE(z) = E Anzn, n
where {zn I n E
El
is a normalized orthogonal basis of L2 (1l,A). Then
E Anzn = XE(z) =. XE(gz) = E An g nzn, n n
Vg G G. Thus, An = 0,Vn (G,SI,A) is ergodic.
a.e.p,
0, i.e., either ,u(E) = 0 or p(fAE)  0,and
338
Now by Theorem 7.3.15, M x a G is a type (III ) factor.
Example 2.
Let f/ = (a locally compact abelian group), be the Haar measure on 12, G be a countable infinite dense subgroup of f/ (for example, G = {r E /R I r is rational}), and the action a of G on 12 be the addition of numbers. Clearly, (G,S1, A) is free, A is Ginvariant, and AO = oo,,u({n})= O, V,1 G
ft Suppose that E is a Borel subset of SI such that A((E U (E
ONE n (E
Vri e G,
n )))
i.e., ti ii*mx url = mx „ V77 G G, where ri 4 ti n is the regular representation of 12 on V ( 1, A). Since G is dense in fl, it follows that mxz un = u ri mx ,Vri Thus, we have either A(E) = 0 or /2(12\ E) = 0, i.e., (G,f1, A) is ergodic. Now by Theorem 7.3.15, M x a G is a type (Iioo ) factor.
Example 3.
Let (12, A) be as in Example 2, G = {(p, ci) I p > 0, p,u rational},
and
a(p,0)ri = pi) o, V(p , u) EG,rj E 1Z. Clearly, (C,12,) is free, and A is quasiinvariant under G. Let Go = {(1,a) I u rational}. By Example 2, (Go) fl,) is ergodic. Clearly,
G.
Go
Now by Theorem 7.3.15, M x a G is a type (III) factor. Theorem 7.3.16. On a separable Hilbert space, there exist five classes of factors: type (In ), (I„) ) (III ) (ILO ) (III) factors. Type (II„) ) factors can be indeed constructed through type (III ) factors. Proposition 7.3.17. A factor M is type (lla,) if and only if M = NOB(H a„), where N is a type (III) factor, and H,„ is a infinite dimensional Hilbert space.
Proof.
The sufficiency is obvious from Theorem 6.9.12. Now suppose that M is a type (lloa ) factor. Pick a nonzero finite projection p of M, and let {pi } lEA be a maximal orthogonal family of projections of M such that pi p, V/. Then q=1— p by Proposition 6.4.5, IA = oc. Thus, lEtt.
=
pi + q
pi .
iEn iEn Further, there exists an orthogonal family {q1 } 1EA of projections of M such that E = 1, qi p,V1. tEA
337
Now by Theorem 1.5.6, M = MAB(B 00 ), where Mr is a type (III ) factor, Q.E.D. and dimI/00 = oo. We have another method to construct type (Ii i ) factors,
Let G be a discrete group, and g 4 )t g , p9 be the left, right regular representations of G on 1 2 (G) respectively, i.e., (Air f)(') = f(g 1 .))
(Pg(f))(.)= f('9))
V f G 1 2 (G), g G G. Let R(G) {)tg I g E G}".
Lemma 7.3.18.
R(G) is a ufinite and finite VN algebra on 1 2 (G).
For each g G G, let E9 (k) =15g .k. Clearly Proof. Let e be the unit of G, and define p(a)
(ace , ce ), Va E
Eg
is a unit vector of 1 2 (G).
MG).
Then p is a normal state on R(G). If a E R(G) satisfies ace = 0, then O p g laE, = ap g ice = acg ,
But bg
Vg E G.
I
g E G] is dense in / 2 (G), so a = 0, i.e., p is faithful. Moreover, since p(Ag Ah) = cp(AhA g ),Vg, h E G, it follows that p(ab) p(ba),Va,b E R(G). Thus, p is also a trace. Now by Proposition 6.3.15, R(G) is ufinite and finite. Q.E.D. Definition 7.3.19. An infinite countable discrete group G is said to be of infinite conjugacy class, if for any e0gE G, the conjugacy class {hglt1 I h E G} of g is infinite. We often abbreviate such a group as an /CCgroup. For example, the group of all finite permutations of IN = {1,2, • • .} is an ICCgroup, and the free group of two or more generators is also an ICCgroup,and etc.
Proposition 7.3.20. on 1 2 (G). Proof.
If G is an ICCgroup, then R(G) is a type (III ) factor
Let a E R(G) n R(G) 1 . Then for any g E G, ace = A g a A g 1 ce
A g ap g c, = p g A g ace ,
i.e., (ace )(.) = (ac e )(g 1 g),Vg G G. Since (aE,) E 1 2 (G) and G is ICC, it follows that (ace )(h) = 0, Vh e, i.e., ace = AEe for some A E T. By the proof of Lemma 7.3.18, a = A. Thus, R(G) is a factor, Moreover, R(G) is infinite dimensional. Now by Lemma 7.3.18, R,(G) is type (II1)• Q.E.D.
338
Let G be an ICCgroup,and {G n } be an increasing Proposition 7.3.21, sequence of finite subgroups of G with G = Un Gn . Then R(G) is a hyperfinite type (III ) factor on 1 2 (G). Clearly, for each n, [)tg I g E G rd is a finite dimensional * subalgebra of R(G), and Lln [Ag I g E G n] is o (M,M.,)dense in M. Now by Theorem Q.E.D. 7.2.12 and Proposition 7.3.20, we get the conclusion. Proof.
Remark. Let G be the group of all finite permutations of IN = {1, 2, . • • , 1, and Gn be the finite subgroup of all permutations of {1, .. • , n},Vn. Then G = UnGn.
the construction of factors in the section is standard. It is called the Notes. group measure space construction (of MurryVon Neumann). References, 1113], 11191, [132].
7.4 The existences of nonhyperfinite type MO factors and nonnuclear C*algebras Consider a discrete group G. Let cg (h) = bg,h,Vg, h E G. Then {cg ig E G) is an orthogonal normalized basis of 12 (G) . Suppose that g > A g ,p g are the left, right regular representations of G on 1 2 (G) respectively, and R(G) = {) g ig E Gr. By Lemma 7.3.18, R(G) is a a—finite and finite VN algebra on  (•E„ Ee) is a faithful normal tracial state on R(G), where e is the 1 2 (G); i0(*) unit of G; and Ee is a cyclic—separating vector for R(G). Proposition 7.4.1. Define jxc e = ec„Vx E R(G), Then j can be uniquely extended to a conjugate linear isometry on 1 2 (G), still denoted by j, and
.7.2 =
I;
i X I Ee =
(.7. e) in) = (n) X i* E„V Xi G
e))ve) 77 G 12 (G);
R(G)'; jA gj = p g ,Vg G G.
Moreover, jR(G)j , R(G)' = {p g lg E G}". Since io() is tracial and ce is cyclic for R(G), j can be uniquely Proof. extended to a conjugate linear isometry on 1 2 (G). Clearly, 52 = I, and jA gi =p g ,Vg G G. By (jzE„jyE e) , (yE„xE,),Vx,y E R(G), we have (je,jri) =
339
(77,
) ,ve,r, E 1 2 (G). For any 2 E R(C)', x E R(C), (j2exe e )  (jx'E„jx*c e ) = (x* E„ 2 Ee) = (2* E„xe.e ).
Hence, jx 1E, , x"Ee ,Vx / E R(G)'. Further, by j2jy'z'E, , jxlzi*yi*E, = y'zix"c e = 1Jij2 jzic„ V2 , y i , z' E R(G), we have jR(G)Ij C R(G). On the other hand , jR(G)j = fpg lg E GI" C R(G)'. Therefore, we obtain that jR(G)j = R(G) I = {p 9 1g E Q.E.D. G}".
For any
e, 77 E 1 2 (G), let (e * 0 (g) z E
(h)ri (141 g ),
e* (g)=• e (g  9, Vg
E G.
hEG
Clearly,
I(e * n)(01 5 Ilell . 11q11) Ile* II —
Ho.
Proposition 7.4.2. Let
B , {to E 1 2 (G)
Then: (i)
there is a positive constant K = K (b) such that (b * c) E 1 2 (G)andllb*cll G KlIcII,Vc E 12 (G).
B = fb E 1 2 (G)I b* c E / 2 (G),Vc E / 2 (G)I
,____ { b E 12(G\ there is A(b) E B(1 2 (G))such that 1 . A(b)Eg = b* Eg = p*g b,Vg E G f ' ) (ii) B is a * algebra, and b  A(b) is a faithful * representation of B on 12 (G), where A (b)c z b* c,Vb EB,cE 1 2 (G); (iii) R(G) = A(B) = {À(b)lb E B). Consequently, if we define Ilbll =11A(b)11,Vb E B, then B is a afinite and finite W*algebra, and p(b) = (A(b)e e , e e ) (V b E B) is a faithful normal tracial state on B; (iv) jA(b)j = p(b) , where p(b)c = c * b* ,Vb E B,c E 1 2 (G) . And R(G)' =p(B) = {p(b)lb E B } . Proof. (i ) Let b E 1 2 (G), and b * c E 1 2 (G), Vc E 1 2 (G). Then we can define a linear operator A(b) on 1 2 (G) : A (b)c = b * c, V c E 1 2 (G).
340 We claim that A(b) is continuous. It suffices to show that MO is a closed is sequence of 1 2 (G) such that operator. Suppose that
{ e„}
en
,
0, and
in 1 2 (G), where ri E 1 2 (G). We need to prove n = O. For any (Theg)
, lim(b* n , 
iim n
E
g
G,
en Eg ) ,
E b(h)en (hi g ) = iim(en , c) = o, n
hEG
where c(.) 4 b(g. 1 ) E 1 2 (G). Hence, ri  0, and MO is continuous. Further, we have B = {b E 1 2 (G)Ib* c E 1 2 (G),Vc E1 2 (G)}. Now let b E 1 2 (G), and suppose that there is A(b) E B(1 2 (G)) such that A(b)E 9 = b* Eg ,Vg E G. Clearly, such À(b) is unique, and A(b)c = b* c,Vc E [eg lg E G]. For any c E 1 2 (G) and any finite subset F of G, let CF =
E (c,
eg)eg•
gEG
Then
C I?
 c in 1 2 (G), and A(b)c = limA(b)c F = lim(b* (A(b)c)(g).= (A(b)c, E g) .=. liin(b * C F )
E g) =
C I?).
In particular,
(b* c)(g)
since 1(b* cF )(g) — (b * 0(01 5_ PH • licF — C11 4 0,Vg E G. Hence, b * c A(b)c E 1 2 (C), VC E 1 2 (G), and b E B. (iii) For any b E B and x E R(G), we have xA(b)c,  xp:b = p:xb,
Vs E G.
By (0 , we get xb E B and A(xb)  xA(b),Vx E R(G),b E B. In particular, X
= xA(Ee) = A(xce) E A(B), Vz E R(C).
On the other hand, for any b E B, A (b)p g ch = A (b) chg 1 = f49 _1 b  p g A(b)e h , Vg,h E G. Hence, MO E {p g lg E G .= R(G),Vb E B. (ii) For any a, b E B, }'
À(a)A(b)E 9 = A(a)pg*b= p(a)b, Vg E
G.
Then by (i) we have A(a)b = a*bEB, and A(a* b) , A (a)A(b). Moreover,
WO* Eg , Eh) = (EgIb*
Eh)
= (I)* * Eg )Eh),
=.7
341
Vg,h E G,b E B. Hence , A(b)*Eg = b* egl and by (i) we get b* E B and A(b)* = A(V),Vb E B. Therefore, B is a * algebra, and b A(b) is a * representation of B on 1 2 (G). Moreover, if A(B) = 0 for some b E B, then b* Ee = b. 0= A(b)E, (iv) It is obvious. Q.E.D.
Let M be a finite factor, and cp(.) be the ( unique ) Definition 7.4.3. faithful normal tracial state on M. We say that M has the property (r) , if for any x1 ,xm E M, and e > 0, there is a unitary element u of M such that P(u) = 0, and Ilu*x i u  xi ll 2 < e, 1 < < m, where 114 = cp(x*x),Vx E M. Proposition 7.4.4. property (r).
If M is a hyperfinite type (III ) factor, then M has the
Proof. By Proposition 7.2,10, there is an increasing sequence {Min > 0 } of subfactors of M such that Un M„ is weakly dense in M, where Mn is type (h.), Vn. Now for any x l , • , xrn E M, we can find n and yi)  )ym E Mn such that Ilxi Yi112 < < < m. Mn+1 is * isomorphic to the tensor product of Mn and a type ( 12 ) subfactor. Hence, there is a unitary element u of Mn+1 such that p(u) = 0 and uyi = yu, 1 < i < m. Further, Iluixiu
YiI12
YOull2 + Ilxi
1 < I < m. Therefore, M has the property
e
(r).
Q.E.D.
Proposition 7.4.5. Let G be an ICC group. If there is a nonempty subset F of G and elements g1 ,g2 ,g3 of G with following properties: (i) F U U {e} = G, (ii) the subsets F,g2 Fgji and g3 FgV are disjoint, then R(G) has no the property (r). Therefore, by Proposition 7.3.20 and 7.4.4 R(G) is a nonhyperfinite type (III ) factor.
Proof. Suppose that R(G) has the property (r). Then for E > 0, by Proposition 7.4.2 there is b E B such that: A(b) = u is a unitary element of R(G);p(u) = (uE e ,E,) = 0; and Ilxi u*xi u11 2 < E.) where xi = A(e9 ) =Ag„ 1 < < 3. Noticing that (x
u*xi u) = (u*
x:u*xi )(u
we have
Ilxi
I
U Xi U 2
*
= 11U —
u 11 2 = 11A ( b
= Ilb e g*,* b*E g,.II< e,
e*gi * b* 691)112 1 < < 3.
342
Since 0 = p(u) = b(e) and libil = Hulk = p(u*u) 1/2 = 1, it follows from the properties of F and gi ,g2 ,g3 that
E lb(g)1 2 + E lb(g)1 2 gEgiFgi1 E 1(e*gi * b* 6g1)(g)i 2 = gEF E lb(g)1 2 + gEF
1Ç
gEF
and
(E + E + E )1b(g)1 2 gEF gEg2F61 gEg3Fg;1 E 1(Eg*2 * b * Eg2)(g)1 2 + gEF = gEF E lb(g)1 2 + gEF EI
1 ?,
*
(
b * Eg3)(g)I
2
•
Let bi = E; 1 * b* Eg„1 < i < 3. Then by Ilb  bill < E,1 < 1 < 3, we have
Ibi(g)1 2 = E 1(bi(g) —14)) E gEF gEF
+ b(02
= E ib( g ) — b(9)1 2 + E lb(g)1 2 + 2e E (bi(g) — gEF
gEF
+ E lb(g)2 + 2( bi(g) — b(02 gEF gEF < E2 + 2E + E lb(g)1 2 1
,u(E U gi1 Eg i ) = ,u(G) = 1 . Hence, Since
U
max{/1(E),A(gil Eg i )} __?_ 1/2.
( 5)
Now by (4) and (3) , ,u(E) < lg + 2t2; and by (5) and (3) , it(E) > } — 2t i . Hence , } —2t i < li + 2t2, i.e., (t1 + t2) > Therefore, for j = 1 or 2 we have
A.
t1
=
E le(g) — e(g1gg 2: ')1 2 ?
gEG
1
25 == 6.
3) a(1 — p) < (4 — 6 2 )(1 — p), where 6 = 1/25, and a is the same as in 1). In fact, by ap = pa it suffices to show that
< (4 — E2), V E ( 1 P)II and Heil = 1. Fix e E (1 — p)H and II ell , 1. Clearly, e(e) ,(e, Ee) = (pe, ce) , O. since (a,) = 2Re[(A; i p;,e, 0 + (A;2 p;2 e, 01 —
and
E le(g) — e(g1gg i ') 1 2 — II E

gEG
= 'lee + wg1 p;1 e112 — = 2[1 — ReV; i p;i e, 01,
A g. i p;,e11 2
0
j = 1,2,
it follows from the conclusion 2) that 2
(ae, ) = 4 —
E EgEGle(g) — e(g1ggii)12 _. 4 — 62
5=1
Now we prove that p E C. Let b  (4 + a)/8. Then by ap = pa = 4p, a* = a < 4 and the conclusion 3) we have
b E C , bp .= pb :=. p,0 < b(1 — p) 5_ 45(1 — p), where 45  (8 — c 2)/8, and 6 = 1/25. Hence, for any positive integer n,
bnp = p, and 0 Ç Further, Ilbn limb" E C. n
—
bn (i —
p)
0 such that [3r7,3n] C W, and 11  eist 1 < 45,Vs E E and It1 _< 3q. Let K = V = [rim], and k 1 ghl2n, where g,h E L 2 (/R) with 79 = Xxir = XH2r 7,217 1) 7h = xv = x[1 respectively. Clearly, K +V  V = [ 3?), 3q] c W. Then by Lemma 9.1.1, we have 
0 < k < 1,
k a
1
on [ ii, n ] ; and supp
kcw.
Moreover,
IlkIll 5 11g112 • 11h112/ 2 ti = 11 7g112 11711112/2n
< 2.
371
Since
J.
f (s)ds .
f(0) = 0, it follows that
( f * k)(t) . fR f (s)[k(t — s) — k(Olds, Vt E E.
Then
IV * kill 5._ Ligswilka—kilids (L+fd1f(s)1.Ilka—kbds) where E' = 1R\E , Ic s (.) = k(. — s), Vs. Clearly, — kil i ds < 445, Li If ()I ' ilk!
and
111111 BEE .suP Ilka — klli.
Lif(s)1 ' lika — kilids
Then by the definition of 45, it suffices to show that s ucEp Ilka — kill < 445.
Since k = ghl2q, it follows that 2r/ (k. — k) = g (h, — h) + (g, — g)ha .
Notice that for any s E E,
Or?
11g8  g 11; == 11 1g3  1g11; = j 217 I l  ei st l 2 dt < 47745 2 , and similarly ) lihs — hfl < 20 2 . Therefore,
ilk — kill _. 4{111g112 • 11h. h112 + 11g3 g112• 111h3112}
< 4 45 ,
Vs
E E.
Q.E.D. In the following, we put KI (E) = { f E V (11?) I suppf is compact}
Lemma 9.1.3.
10 (1R) is dense in
V (1 ?).
Denote all continuous functions with a compat support on 11? by K (1R). Clearly, K (1R) is'dense in L2 (li?). Then {f E L 2 (R) I f E K(R)} is dense in L2(E). Proof.
372
For any 0 < f E V(l?) and e > 0, pick g E L 2 (li?) such that ilg — f 1/2 112 < 6 amd
? E K(IR).
Then we have 11 g2 — f ill 5 11g • i ll2 — f Ili + llg • fi/2 — g 2 11i
5 1111/2 112 • 11g — 1 1/2 112 + 11g112 • 11g — f1/2 112 < e(2 11P/2 112 + E). Moreover, since * # E K(R) and the inverse Fourier transform of ? * ? is Q.E.D. g2 E L' (JR), it follows that (g 2 )^ = * ? E K(IR). 
For a closed ideal I of Li (AT), let
I
IL = ft E IR = n{d14/(1)
Pt) = 0,V f E /1
I f E /},
where ),I( j) is the zero point set of f. Clearly IL is a closed subset of R. Conversely, for any closed subset E of li?, let I(E) = {f E L' (la) I (TIE)= 01. Then /(E) is a closed ideal of V (li?). Lemma 9.1.4. 1) If E is a closed subset of IR, then I(E) L = E. 2) If E is a compact subset of li?, then I(E) is a regular closed ideal of L i (IR) , i.e., I(E) is a closed ideal of L i (14 and I(E) admits a modular unit / (i.e., (f — f *1) E I(E),V f E L i (IR)). 1) Clearly, E C 1(E)'. Conversely, if t V E, then by Lemma 9.1.1 Proof. _, O. Thus f E I(E) and there exists f E OR) such that At)  1 and (f E) — t V 1(E)'.. Furhter, I(E) L C E,and I(E) L  E. 2) Since E is compact, by Lemma 9.1.1 there exists / E OE) such that (11E) EE 1. Then (f — f 1)AIE 1.= 0, i.e., (f — f *1) E I(E),Vf E L l (E). Q.E.D. Lemma 9.1.5. Let U1 , U2 be two open subsets of IR, K be a compact subset of li? with K C U2, f E L i (E), I be a closed ideal of L 1 (1R), and 11)12 E I such that fi zz f on Ui , i = 1,2. Then there exists g E I such that # = f
on
Ul U K.
373
Pick e E L i (E) such that êlK —II 1 and supp C U2. Let 2 = Proof. = J on  * e + f2 * e. Then 2 E I, and = f2 * e * ( 1 e) K(C U2); = fi = J on Ui\U2; # = f2F — fi e = fF f fF = J on n U2. Notice that U1 = (U1 \U2) u (Ui n U2 ). Therefore, we have "4= f on il K. Q.E.D. Lemma 9.1.6. Let I be a closed ideal of L 1 (11?), E If f vanishes on a neighbourhood of El then f E I.
L l and f E
(li?).
1) Suppose that to V E. We say that there is h E I such that Tz 1 on some neighborhood of to. In fact, pick a compact neighborhood K of to with E n K 0. Let J I(K). Then by Lemma 9.1.4 J is a closed ideal of L l (R), and J admits a 1 on K. Clearly, (I J) is still a regular ideal of modular unit / with f V (l ?), then there is a maximal L i (E),and 1 is its modular unit. If (1+ J) regular ideal L of OE) such that (I J) C L. Since L = I({s}) {g E L'(11?) 1 #(s) — 01 for some s E ll?, it follows from Lemma 9.1.4 that s E EnL. But EnK = 0,a contradiction. Thus, I J = L i (1R). Consequently, we can write / = ii + i.r) where l E /,/j E J. Now let h = II E I. Then Proof.
TzIK = (Of) — (61K) =illf =_HE 1.
2) If to E, then there is g E / such that
= f on some neighborhood of
to .
In fact, pick h as in 1), and let g = f * h. Then g satisfies the condition. 3) Suppose that K = sup/4' is compact. Since f EE 0 on some neighborhood of El it follows that E n K = 0. By 2) for each t E K, there is a gt E I and an open neighborhood Ut of t such that on
U.
Further, for each t E K, pick an open neighborhood Vt of t such that t E Vt C Vt C Ut and Vt is compact. Now by the compactness of K, there are t 1 , • • • , tr, E K,g t = gt , E I,V = Vti ,U; = Uto l < i < n, such that K,
and
1  = f on Ui , 1 < < n.
Further, let 0 = gn+i (E I). Clearly, #,L+1 = J on /R\K = By Lemma 9.1.5, there is gq E I such that =
on
Ui Li V2 DV1U V2.
374
Aganin by Lemma 9.1.5, there is g 2i E I such that
912 = .f)
on
Ui U V2 Li 173
D VI Ll V2 U 173 .
• • •. So we can get g' E I such that on
VI U • • • u VT, D K.
Now by Lemma 9.1.5, there is g E I such that on
Un+1 Li
K = E.
From the uniqueness of Fourier transform, we obtain f = g E I. 4) General case. Since L I (I ?) admits an approximate identity, for any e > 0 we can pick z E V(E) such that 11f — f * z11 1 < e/2. From Lemma 9.1.2, there is u E K l (li?) such that Ilu — zili < Ei211 fill. Then
Ilf — f *uiii 5 Ill — f *di + lit ill ' Ilu — zili < e• Clearly, supp (f. * u)A (Csuppii.) is compact, and (f * u)^(= fû) vanishes on a neighborhood of E. By 3), f * u E I. Now since e(> 0) is arbitrary and I is closed, it follows that f E I. Q.E.D. References. [136].
9.2. The Arveson spectrum
For our purpose, we just consider a W*system (M, IR, a), where M is a W*algebra; for each t E IR, at is a * automorphism of M; and t 4 p(at (x)) ,, (cr t (x), p) is continuous on li?, Vx E M, p E M.. Denote the collection of all bounded Radon measures on IR by M(E), i.e., M(R) = Cr(R)*. By the convolution ((it * v)(f) = ff f(s +t)cliz(s)dv(t),Vit, v E M(E), f E Cr ( li?)), M(JR) is an abelian Banach algebra with an identity So (f) = f (0),V f E Cr ( li?). Moreover, 1, 1 (1R) is a closed ideal of M ( R).
Proposition 9.2.1.
For each A E M(R), there exists a(A) E B0 (M) such
that (a(u) (x), p)  fi, p ( at ( x ) ) clit ( t )
375
Vx E M, p E M„ and II II II II where B0 (M) is the set of all cr(M,M.)a(M,M) continuous linear operators on M. In particular (cr(i)(x)) P) = fR P(at(x)).1.
11a(1)11
11111 1)
VI E L l (E),x E M, p E M. Proof. Clearly, E M (J ?). Now it suffices to show that for any normal positive functional p on M, the positive functional (a(,u)(.),p) on M is normal, where ht E M(IR) + . Let {xi } be a bounded increasing net of M+ , and x = sup xi . Then for any
t E IR,(at(st), p) f (at(x), p) • By the Dini theorem , (at(xt),P) f (crt(x),P) uniformly for t E K, where K is any compact subset of li?. Now by the regularity of ,u and the boundedness of {zI}, we can see that (a(u)(x),p) = f p(a t (x))d,u
lip f p(a t (z i ))d,u= sup f p(cr t (si nd,u sup(a(12)(xi), p).
Q.E.D. Definition 9.2.2. Let (M, /R, a) be a W*system. 1) Define the Arveson spectrum of a by spa = {f E L1(JR) I a(f) =O}
{t E IR I if f E L i (R) with alp = 0, then 1(0 = 0}
nogh I f E L 1 (IR)
and a(f) = 0}
where )4/(h is the zero point set of 1, and clearly ffE a closed ideal of Li (IR ). 2) If xE M, let
(E) I cr(f) = 01 is
sp,(x) = {f E 1, 1 (li?) I a(f)(x) = 0} ± ft E li? I if f E V (J ?) with cr(f)(x) = 0, then :f(t) = 01
= no/(i) I
f E L 1 (IR) and cr(f)(x) = 01.
3) If E is a closed subset of IR, define the associated "spectral subspace" by
M(cr,E) ={x
MI sp0 (x) c
E}.
376
Clearly, by the definition spa is a closed subset of li? and 0 E spa (since a(f)(1) = f (0),V f E L 1 (1R)); sp,(x) is closed and sp,(x) C spa, Vx E M; spa (0) = 0 (from Lemma 8.4.1); and 0 E M(a, E) for any closed E C E. Proposition 9.2.3.
Let (M, 1R,a) be a W*system. Then:
(a) spa = U zEmsp c,(x); (b) spo (x*) = —spc (x), spc (at (x)) = sp0 (x),Vx E M, t E JR; (c) sp,(x) = 0 ,;=>. x= 0; (d) sp, (a(f)(x)) c spc (x) n suppf, Vx E M, f E L i (IR); (e) at (M(o, E)) = M (a, E), Vt E h? and closed E C E; (f) for any closed E C 11?, x E M (a , E) ,; a (f)(x)= 0, VI E L' (JR) and fF __ 0 on some neighborhood of E. Consequently, M(o, E) is a a(M,M.)closed linear subspace of M; (g) if x G M and p e M(E) satisfy it' ;_ 0 on some neighborhood of spc (x), then o(tt)(x) = 0, where P.M = fR ei•tcliz(s). Moreover, if f E L' (JR) satisfies either f —= 0 or f_ __ 1 on some neighborhood of sp,(x), then either a(f)(x) = 0 or o(f)(x) = x. Proof.
(b) Since a(f)(x)* = a(7)(x*), it follows that spc (x*) =
n{),/(2) I a(f)(x*) = 0}
=—
n {.Al (1) I
cr(f)(x)
= 0} = —sp,(x),
Vx E M. Moreover, by a(f)(at (x)) = o(ft )(x), ft (s) = Al (h , where ft (s) = f (s — t), we have sP0(crt(x)) =
tf( s ) and
.W(ft)
=_
nfAi(h I cr(f)(at(x)) = 0}
= n01(:4) I a(h)(x) = 0} = sPo (x), Vt E li?,x E M.
(e) It is immediate from (b). (f) Let x G M(a,E), and f e L' (JR) with f —= 0 on some neighborhood of E. Let I = fg G L i (IR) I a(g)(x)= 01. Then spa (x) = IL C E. Now by Lemma 9.1.6 we have f E I, i.e., a (f)(x) = O. Conversely, let x E M and a(f)(x) = 0 for any f e V (J ?) with f E 0 on some neighborhood of E. If there is s E (spa (x)\E), then by Lemma 9.1.1 we can find k E L I (J ?) with k(s) = 1 and supp k C F, where F is a closed neighborhood of s and F n E = O. Then a(k)(x) = 0 since 'ic E 0 on the open
377
neighborhood ( E\F) of E. But s E sp, (x), so k(s) must be 0, a contradiction. Therefore, spa (x) C E and x G M(a, E). Consequently, M(cr, E) is a linear subspace of M. Aganin by Proposition 9.2.1, M(a,E) is a(M, M ) closed. (c) Clearly, sp,(0) = O. Now let x G M and spa (x) = 0. By (0, we have a(f)(x) , 0,V f E L i (11?). Thus,
JR f (t) p(cr t (x))dt =
0, VI E
L i ( E),
p G Ms
.
Since t 4 p(a t (x)) is a bounded continuous function on .11?, it follows that p(at (x)) = 0 ,Vt E IR, p E M,, . Therefore, x = O. (a) Clearly, E = UzemsP a (z) C spa. Now if s V E, then by Lemma 9.1.1 we can find k E V (l ?) such that k(s) = 1, and IC L2. 0 on some neighborhood of E. By (f), we have a(k)(x) = 0,Vx E M, i.e., a(k) = O. Since k(s) = 1, it follows from Definition 9.2.2 that s V spa. Thus, spa c E, and E = spa. (d) If a(g)(x) = 0, then a(g)(a(f)(x)) = a(g * f)(x) = a(f * g)(x) = a(f)(a(g)(x)) = O. Thus, 4! E L 1 (11?)
I a(g)(x) = 01
C fi z G L l ( E)
I a(h)(a(f)(x))
and sp,(x) D sp0 (a(f)(x)). Moreover, if s V suppf, then bi Lemma 9.1.1 we can pick k E L l (E) such that k(s) = 1 and supp kn suppf = O. Thus ,kf=0andk*f=f*k= O. Further a(k)(a(f)(x)) = 0, and spo (a(f)(x)) c )41(). But rc(s) = 1, so s V spa (a ( f)(x)) . Therefore, sp, (a ( f) (x)) C suppf.. (g) Since f * A E V (l ?) and f * A = f A _,: 0 on some neighborhood of spa (x), it follows from (f) that a ( f)(o (A) (x)) = ° ( f * A) (x)= 0,V f E L i (E). Thus by the proof of (c), we have a(A)(x) ,.  O. Now let f E OE) and f E_ 1 on some neighborhood of sp,(x). Pick 0 on It = 60 . Clearly, cr(p)(x) = x, and j2(s) = 1, Vs G R. Thus (f — some neighborhood of spa (x). By the preceding paragraph, we get
0 = a(f — A)(x) = a(f)(x) — x. Q.E.D. Proposition 9.2.4. Let (M, li?,a) be a W*system, E 1 and E2 be two closed subsets of .E,E = E1 + E2, zi E M(a, Ed), 1 = 1, 2, and x = x1x2. Then x E M(cr,E). Consequently. sp0 (xix2) C spa (zi) + sP0 (x2),
\ix', x2 E M.
378 Proof. 1) First, we assume that sp,(xi ) is compact, i = 1,2. Replacing Ei by spc (xi ), we may assume that Eli is compact, i = 1,2. Then E = Ei E2 is also compact. By Proposition 9.2.3 (f), it suffices to show that
o(f)(x) = 0 for any f E (11?) with f 0 on some neighborhood of E. Fix f E 1, 1 (1R) with f 0 on (E + V + V), where V is a compact neighborhood of O. Pick fi E 1, 1 (1R) such that fi _= 1 on some neighborhood of and suppri C Vl i= 1,2. From Proposition 9.2.3 (g), we have
a(fi )(xi ) = xi , 1=1,2. Then for any p E M
(o (f)(x),
=
,
by the Fubini theorem we get
fR p(cr s (x)) f (s)ds • cr (f2) (x2)), p)ds
=
frf
dt 1d t2 ( ), • a a+t 2 ( ( „c r a+t, ,xi f(,8) f (,t „)f 2 ,t2 ,x2) p, _s__
fff f (s) fi (s i — s) f2 (8 2 s i — s) (a 8,(zila , 81 82 (z2), p)dsds i ds 2 . Let k(si,s2) = f f(s)fi(si — .9)12(32 + 81— s)ds = (f * fi. f2,32)(s1)
Fix s2 , and take Fourier taansform for s i . Then
, s 2 ) = (t)Cf; * (t) , where g( • ) =
= f2(• + 8 2). Noticing that
supp ( *
C supp
supp4
= suppj; supp:f; C EdV + V
and f E 0 on E V + V, we have 0,82) = 0,Vt. Thus for any .52 , we get k(s i ,s2) = 0, a.e. for s i . Further, by the Fubini theorem we can see that (o (f)(x), = 0 ,V p E M., i.e., a( f)(x) = 0. 2) Let { zn } be the approximate identity for 1, 1 (E) as the beginning of this section. Then it is easy to see that cr(z„)y y(a(M, M.)),Vy E M. Thus y E {a(f)(y) I f E L l (R))11f111 iY7 5 Vy E M. Now by Lemma 9.1.3 and Proposition 9.2.1, we have
E fa(f)(Y)
I
f E 10(11?)
and 11f III
379
Vy E M. Further, from Proposition 1.2.8 and 1.2.1 we have x E fa(f)(xi) • a (g)(x2)
I
f,g
E K l ( E) )
111111 and
119111
1Y.
Now by 1) and Proposition 9.2.3 (d), sp,(a(f)(x i )  cr(9)(x2)) C E, Vf,g E K i ( E))111 1 1 and 119111 it follows that x E M(a,E).
1. Finally, since M(o, E) is o (M, M)closed,
Q.E.D.
Let t E 1R, K be a compact subset of E, and there exists a compact neighborhood V of t such that
Lemma 9.2.5.
E>
0. Then
ilas (x) — eist xil < 6 114)
Vs E K, and z E M(a, V). Proof. Pick a compact neighborhood WI of t and f E K l ( E) such that ..7 ,, 1 on WI . For each s e K , let
t
f3 (r) . f (r — s) — eist f (r).
Then fa(t) = 0,Vs E K. By Lemma 9.1.2, there is le E L i ( E) and some neighborhood W. of t such that ic8 E.. 1 on W. and Il f 8 * le III < e. Since K is compact and s —4 fa is continuous from IR to OE), there is a compact neighborhood W2 of t such that for each s E K, we can find k E L i ( E) with kE.. 1 on W2 and Ilfs*klli < e • Now let W  WI n W2 ) and V be a compact neighborhood of t with V C the interior of W. For each s E K and x E M(cr, V), pick k E L l ( E) such that iz, 1 on W and lif e *kill < e. Clearly, f E 1 on W, and f * k a 1 on W. By Proposition 9.2.3 (g), we have a( f * k) (x) = x. Then we obtain .
licra(x) — e ist x11 = lia8(cr(f * k)(s)) — eist a (f * k)(x)II = Ilc r(f 3 * k)(x)II
Ilfs *41.114 < EllxII. Q.E.D.
Let (M, IR, a) be a Wesystem. Then the following statements are equivalent: Theroem 9.2.6. 1) t Espa;
Oh 2) For any closed neighborhood V of t, M(a, V) 3) There exists a net {xi} C M with Ilxill = 1,V/, such that Has (xl) — eistzlii
380
uniformaly for s E K, where K is any compact subset of 117; 4) IRO I < 11 6 (f)ii,Vf E L 1 (11?).
Let t E spa. If there is a closed neighborhood V of t such that M (a, V) = {0 } , then pick f E 1. 1 (11?) with 1(0 = 1 and supp f C V. By Proposition 9.2.3 (d), spo (a (f)(x)) C Vl VX E M, i.e., a ( f) (x) E M(a,V),Vx E M. Since M(a, V) = {0 } , it follows that a (f) = O. By the definition of Spa, t(t) must be 0, a contradiction. Therefore, M (a ,V) {0 } for each closed neighborhood V of t. 2) 3). For each closed neighborhood V of t, we can pick xv E M (a , V) with Ilxv I = 1 by the condition 2). From Lemma 9.2.5, the net {sv I V} satisfies the condition 3). 3). 4). Let {xi} be as in 3), and f E V (IR). Then Proof.
1) == 2).
11 6 (f) II ? Ila(f)(s1)11 = ?
I fR cra (x0f(s)dsll
I L e'" f (s)dsx i ll — fR ik' 8 (x l ) — e's t xi ll • I f (s)Ids.
Now by the condition 3) and f E V (11?), we can see that IRO I 4) == 1). It is immediate from the definition of spa.
116(.011.
Q.E.D.
Let (M, la, a) be a W*system, A be the abelian Banach subalgebra of B (M) generated by {a (f) I f E L 1 (./R)}, and n(A) be the spectral space of A. Then spa .. n(A). Theorem 9.2.7.
Clearly, a is a cotinuous homomorphism from OE) to A, and the image of a is dense in A. For each p E 11( A), (p, a(.)) is a nonzero multiplicative linear functional on V (11?). Thus, there is unique t E JR such that Proof.
(p, o (f)) = t(t),
VIE V
(E).
Clearly, t(t) = 0 if a (f) = O. So t Espa. Put t = a* (p). Then a* is a map from 14A) to spa: a*
(P)(f) = (P)cr(f)) = 7 f(t),
Vf E L I (JR ).
We say that a* is injective. In fact, if o* (pi ) = a* (p 2 ) for some Pi, p2 E r/(A), then V f E OE). p2, u(f))  0, (pl —
But a (L 1 (IR)) is dense in A, so p l = p 2 . Now if t E spa, define (p, a (f)) = t(t),
VI E L i ( II?),
381
then p is a nonzero multiplicative linear functional on a(L 1 (IR)). From Theorem 9.2.6 and t E spa, we have l(P,a(M1 = it(t)1
iia(f)II)
VI
E L 1 (11?).
Thus, p can be uniquely extended to a nonzero multiplicative linear functional on A, i.e., the map a* is also surjective. If al1 (m) , ti 4 t , a* (p) in spa, then (pi,cr(f)) = 1( t1) 4 t(t) = (p,a(f)),V f E L i (E). Since a(L 1 (AT)) is dense in A and 11P11 ' 11PI 11 = 1,V 1 , it follows that p i 4 p in f/(A). Conversely, if pi 4 p in 11(A), then 1 (t1) 4 t(t),Vf E Ll (E), where ti = a* (pi),V1, and t = a* (p). Further, by Lemma 9.1.1 we can see that t i 4 t in spa. Therefore, a* is a homeomorphism from Q.E.D. 11(A) onto spa. Theorem 9.2.8. Let (M, IR,a) be a Wesystem. Then t 4 at is uniformly continuous if and only if spa is compact. Let spa be compact, and pick f E 10(//?) such that I E.. 1 on some open neighbourhood of spa. By Proposition 9.2.3 (g), we have a(f)(x) =x,Vx E M. Further Proof.
Ilat(x) — x11 = Ila(St * ns — a (f)(x)11 lift —
fill 11x11 4 0 (as t 4 0),
uniformly for x E M with 11x11 < 1, i.e., 11 04 — idll 4 0 as t 4 O. Conversely, let t 4 at be uniformly continuous, and { zn} be an approximate identity for 1, 1 (1R). Then 11a(zn)x — x11 < f mat(x) —
0 is such that suppf C [r + e, oo),Vy E M. Clearly, sp,(z) is compact, where z = x* a ( f)(y). Now pick g E Ki (E ) with supp? c ( 1 — To) co) and '4 E 1 on a neighborhood of spo (z). Then a(g)(z) = z. By 1), we have er _ ro a(g)(z) = 0, i.e er _ ro x*a(f)(y) = 0,
383 Vy E M, f E 10 (1R) with supp, c (r, 0o). Further, from 1) we get er _ ro x* (1— er ) = 0, i.e.,
(1— er )xe r ..." = 0, IR and z E M with spo(x) C (700, To].  c (0, co) respec5) Pick f, g E 10 (IR) with suppf c (—co,0), supp 4 tively. Let rob E 1R, and M.) = f(.)e  iqr 1 o),g 1 (.) = gHe  i'r. Then fi,, 91 E 10(1R) and suppfi C (00,7 — r0), SUPP41 C (r, oc) . By 3), we have ei 7.0 f fi (s)ua ds = f fi (s)ua ds and (1— er ) f g i (t)u t dt = f g i Mu t dt. Now let M C B(H),and x E M with spo (r) C (00, ro ]. Then by 4) we get
VT ) To E
(x f fi(s)usdse) f 91(t)utdtri) a
✓, T)
0,
E H. Notice that
(x f fi (s)ua cise, f g i (t)u t dtn) = ff (14 xutua t
e, 77) h(s)gi(t)dsds
= ff (f3t(z)use,ri) f i (s — t)g i (—t)cisdt = f h(s)e's(r  ro)ds
=
iz(r o — 7),
where 13t (x) = ut xu, and h(s) = f (f3t (x)u 8 e, 77) f (s — t)g(—t)e  t tri ' dt. Since kro — r) =' 0, VT E IR, it follows that h(s) = 0(a.e.). continuous, so h(s) = 0, Vs E IR. In particular, 0 = h(0) = f (f3t (x)e,,)kmdt
vem E
But h(.) is
„
H, where k(t) = f(—t)g(—t)e it". Thus, 0(k)(x) = 0, and spo(x) C
)4/(i). Notice that k(s) =
Li 4(r) f(ro — s + r)dr
r
'4 (r) f( To — s
+ r)dr,
where E > 0 is such that supp4 c (e, oc). Since suppf c (00,0), if follows that i(s) = 0 if s < To + e. Further, since spo (x) is closed and e, f,g are arbitrary, we can see that spo (x) c (— oc, TobVx E M with spo (x) C (00, rob i.e., M(a , (oo , Top c M (f3 , (—oo, Top , Vro E IR.
384
Now by Proposition 9.2.3 (b), we have also M(cr, [70 , oc)) C
M(0,
[To, 00 ))) Vro E R.
6) Let r E /R,and f,g E 10(/R) with supp f c (—co,0), supp4 c (0, oc) respectively. Let fi (.) = f(.)e ir . ,g1 (.) = g(.)e ir'. Then f1 , g1 E K i (R) and suppfi c (—oo,r), supp'4 1 C (r,00). For any z E M, since a(fi )(x) E M(o,(—oo,r]) c M( 13, ( co, 71), it follows that 41 s_. 0 on a neighborhood of spp(cr(f i )(x)). Thus, 13(M (cr(fi )(x))  0. Notice that —
0(g1)(cr(L)(z))
= if f3t(as(x))gi(t)fi(s)dsdt = if tt(crs + t(x))gi(t)h(s)dsdt = ff tt(a8(x))
As
— t)g(t)e isrdsdt
= f h(s)e  iards =
where (I) t = A 0 a t,Vt 
E
R, and
h(s) = f 4:1)t (a 3 (x)) f (s
—
t)g(t)dt.
Now since h(1), 0 ) VT E R, and h(.) is continuous , it follows that h(s) 0,Vs E E. In particular,
=
0 = h(0) = f do t (x)k(t)dt = 40(10(4
where k(t) = f( t)g(t). Thus, spo (x) C .W(i). Notice that —
k(s) = f 4(0 ffr — s)dr = 
L e° 4(r)f(r
— s)dr,
where e > 0 with supp4 = (e, oc). Since suppt C (00,0), it follows that k(s) = 0 if s < E. Furhter, since sp.(x) is closed, and e, f,g are arbitrary we can see that spo (s) c ( oc, 0], Vs E M. Similarly, from [3(fi )(a(g 1 )(z)) = 0, we have spo (x) c [0, oo),Vx E M. Thus, sp.(x) c {0}, Vx E M. Finally, by Lemma 9.2.5 we get do t = id, i.e., at = A, or at(z) = utxu, Vt E IR, and z E M. —
Q.E.D. Notes. Spectral subspaces were introduced by R. Godement. It may be viewed as an attempt to extend the Stone theorem. A systematic study of spectral subspaces and their applications to dynamical systems was presented
385
by W.B. Arveson. Theorem 9.2.6 is due to A. Connes. And Theorem 9.2.8 and 9.2.9 are due to D. Oleson.
References. [9], [17], [57 ] , [122 ] , [123].
9.3. The Connes spectrum Let (M, E, a) be a W*system, and denote by M° the fixed point algebra:
M° = {x E M I at(x) = z,Vt E E } .
Clearly, M° is a W*subalgebra of M. For a projection e E MY , a induces an action a on Me such that o(eze) = eat (x)e,Vt E IR, x E M. Then we obtain a W*system (Me = eMe, JR, cf = alMe) ) and denote its Arveson spectrum by spa ' .
Definition 9.3.1. by
The Connes spectrum of W*system (M, li?, a) is defined
1' (a) = n{spue I 0 e E Proj (M/ } , where Proj(Mg) is the collection of all projections of the fixed point algebra M" . Clearly, r(a) is a closed subset of JR and 0 E Nu).
Lemma 9.3.2. JR, we have
For any e EProj(Mi with e
Me (cre , E) = M (a , E)
0, and a closed subset E of
n M.
where We , E) = {z E Me I spoe(x) c E}. Proof. If z E M„ then a(z) = at (x),Vt E JR and cre(f)(x) = a(f)(x),V f E Li (J ?). Thus by Definition 9.2.2, we have spo (x) = spOe (x),
Vx E M.
That comes to the conclusion.
Q.E.D.
Proposition 9.3.3. Let (M, IR,a) be a Wesystem. Then r(a)dspa = spa.
Proof.
First, since 0 E
r(o),
it follows that
spa c spa + I' (a).
386
Now let A 1 E r(a), A2 E spa. We need to prove A = A 1 + A2 E spa. From Theorem 9.2.6, it suffices to show that M(cr, V) 0 fol for every compact neighborhood V of A. Fix a compact neighborhood V of A, and pick compact neighborhood Vi of = 1,2, such that Vi + V2 C V. Since A2 E V2 ) it follows from Theorem 9.2.6 that M(a, V2) 0 {0 } . Let E M(a, V2) with z2 0 O. Then ot(4) O,Vt E E. Let at (4) = vt ht be the polar decomposition of at(4),et = vtv;,Vt E ./R, and e = supfet I t E Clearly, et 0,Vt E E, and e O. We say that e EProj (Mc). In fact, if M C B(H), then eH = [ot(x;)H I t E II?]. Thus o 8 (e)H = eH , i.e., ,(e) = e, Vs E E. Now A i E r(o) = n{spaP IO0pE Proj (M/ } . In particular, A i E spas. From Theorem 9.2.6 and Lemma 9.3.2, we have
M(cf,v,) n Me = Me (cre,v,) o fol. Then there is z i E M(o ) vi) n Me with zi 0 O. By the definition of e and exi = x l 0, there exists t E JR such that et xi O. So we can find T./ E H such that 0 (ets1, 71) (xle,ett7)) where M c B(H). But et ri E O. Thus at (x2 ) zi O. then there is ç E H such that (x i Put z = at (x2) z i (0 0). By Propositions 9.2.3 (b) and 9.2.4, we get
e,
o
spa (x C sp0 (x2) + sp,(xi) C V2 + V1 C V.
Therefore, z E
Af(a,
Proposition 9.3.4. subgroup of IR. Proof. r(ce)
Q.E.D.
V), and M(a, V) 0 {o}.
Let (M,IR,o) be a W*system. Then r(a) is a closed
0, from Proposition 9.3.3 we have For any e EProj(M/ with e + spa' = spa'. Clearly, r(a) c r(ag), and r(a) c spa'. Then
+
c F(a) + spa'
= spat,
VO y4 e EProj (Ml. Further from definition 9.3.1, we obtain that r(a) + F(a) c F(a). Moreover, from Proposition 9.2.3 (a) and (b) we have spae(x) = spa' =LI sPae(x*) = —sPae) Li xEM. zekf
VO e EProj/W. Thus, I'M = nfspae I 0 e E Proj = —r(a). Finally, since r(a) is closed and 0 E F(a), r(a) is a closed subgroup of IR. Q.E.D. Lemma 9.3.5. such that
If H is a proper closed subgroup of E, then there is A > 0 H = EA.
387
Proof. We may assume that H 0 {0}. Then we claim that there is A > 0
such that H n (0, A) = 0, and A E H. In fact, if such A does not exist, then there is a sequence { An} C H with An > 0,Vn, and An 4 0. For any A E IR and n, we can find N(n) such that N(n)A„ < p < (N(n) + 1) An. Thus dist (A, H) < A
0.
Since H is closed, it follows that A E H,and H = IR, a contradiction. Hence such A exists. Clearly, H D EA. If A E (H\EA) with A > 0, then there is n such that rtA < p < (n + 1)A.
Now 0 < (p— nA) < A and Ca — nA) E H. This contradicts H n (0, A) = O. Therefore, H = ZA. Q.E.D. Remark. From Proposition 9.3.4 and Lemma 9.3.5, the Connes spectrun F(o) of a Wesystem (M, 1R, a) is one of the following forms:
IR,
{0 } ,
and
Ell
(some A > 0).
So the subgroup er (Q ) of the multiplicative group (0, oc) is one of the following forms: (0, oo),{1}, and {A n I n E Z} (some AE (0,1)). Lemma 9.3.6. Let {Vi I j E A} be an open cover of JR, and z E M with z 0 0. Then there exists a f E L i (E) and some j E A such that supp
fc
17i,
and
a(f)(x) 0 0.
Let /0 = {f E K l (11?) I supp f C Vi , for some j E A}, and .1 = lo. Clearly, 10 is an ideal of L i (IR). For any t E E, there is j E A such that t E Vi. By Lemma 9.1.1, we can find f E 10(//?) such that f(t) = 1 and suppf C V. Then f E 10 C I, and t V I'. Thus I' = (A, and I = L i (E) by Lemma 9.1.6. Moreover, by the proof of Lemma 9.2.5 a(z„)(s) 4 z 0 0, so there is te with a(z n)(x) 0 0. Now since 10 is dense in L i (/R), thus we can find f E /0 such that cr(f)(x) 0 0. Q.E.D. Proof.
Lemma 9.3.7. Let el, e2 E Proj (Mc),and e l 0 0,e2 0 0. If ei — e2 (relative to M), then r(crel) = r(cre 2 ).
388
Proof. Let A E r (Gel). We must prove A E r(cre2), i.e., A E spoh,V0 0 12 E Proj (M1 and h e2. By Theorem 9.2.6, for 0 0 12 E Proj (M1 with 12 < e2 and a compact neighborhood V of A, we need to show that M(cr, V) n M12 = M12 (ah,V)
{0 } .
Let U,W be compact neighborhoods of A,0 respectively such that U+ W c V. Pick an open cover {Vi I j E Al such that Vi — V' C W, Vj. Since e l — e 2 , there 0, we have is u E M such that u*u = el, uu* = e. BY f2uu * = f2e2 = 12 that f2u O. From Lemma 9.3.6, there exists g E V(/R) such that supg C Vi for some j, and a(g)(f2 u) 4 O. Since f2,e1 E Al' and f2u = f2 • f2u • el , it follows that x = f2 xe l , where x = a(g)(f2 u). Clearly, spa (x) C suPP 4 ) and Vi) c W. Let M c B(H),and (sp,(x) — spo (x)) c (supg supp6) C (V.; fl be the projective from H onto [ci t (e)Hlt E /RI. Then h E Proj (M1 and O. Moreover, by e l at (e) = at(eix s ) = ot(e),Vt E E, we have fl < e i . Now A E U and A E r(cr e l) C spoil, then from Theorem 9.2.6 and Lemma 9.3.2 we have o {0 } . m(a, n mf, mf, (ah , Pick 0 y E M(a,U) n Mt.,. By the proof of Prosition 9.3.3, we can also see that crt i (x)yh = at, (x)y 0) for some t l E E. O. Again by the definition of f, , there is t2 E IR such that ati (x)ya t2 (e) Put z = ati(s)Yat2(x * ). Since f2 x = x, x* f2 = e,and at (12 ) = f2 ,Vt, it follows that z = f2 zf2 E Mh . Now by Propositions 9.2.4 and 9.2.3 we have spc,(z) c sp c,(x)
spa (y) — spc,(x)
C U±Wc V.
Therefore, M(cr, V) n mf2
o {0 } .
Q.E.D.
Two actions a,r of IR on M are said to be outer equivDefinition 9.3.8. of alent, if there is an oneparameter strongly continuous group fut I t E unitary elements of M such that ut+, = ut u(u8 ),
Tt(X) = ut at (x)ut*,
Vs,t E IR, zEM.
Lemma 9.3.9. Suppose that a and r are outer equivalent. Then there exists an action y of IR on 14 such that
f Pit (x Prt(x
e ll ) = ot (x) 0 e n , e22) = Tt(X) 0e22,
389
Vt G 1R,x G M, where Ai = M 0 m2(Œ)
{ eii I 1 < i, j O. Then p = mrE is a nonzero projection of M. By Lemma 9.5.4, f). = 71(p) is a nonzero projection P.1i, is a faithful normal state on M, and the modular of M. Clearly, ‘,25=i Suppose that automorphism group of Mpd corresponding to 6i; is ' = Ar, is the modular operator corresponding to (6i. Let x E /4 and b. = m f. be the function (: G M) corresponding to z. Then it is easy to see that xE
•:;=> suppfk C E
n kE,
E G.
Thus, similarly we have the following.
Lemma 9.5.6.
az is spatially isomorphic to the operator hE =
ot,
1:13techt,E
on etEGL 2 (nt)vt), where nt = E n tE,v t = t and ht,E is the operator on L2(11t ,vt ) corresponding to multiplication by (rt1(")I n),vt E G.
Lemma 9.5.7. A E S(M) if and only if for each Borel subset E of f/ with IL(E) > 0 and e> 0, there exists a nonzero projection q of M and t E G such that sup{ q, a t (q)} < p and sp(htiqlqH) c (A
—
e,
e) ,
399
where p = mxE , and ht i is as in Lemma 9.5.5. Proof.
By Proposition 9.5.3 and Lemma 9.5.4, we have S(M) = n{spair I f5= 7r(p), 0 p E Proj(M)}.
Hence, from Lemma 9.5.6 we obtain that AE
SA
.4=>
A E spar) , Vi5= 71 (p), p = mrE , and ,u(E) >0 A E sph E , VA(E) > 0
.4=>
,;
A E U sph t ,E , Vii(E) > 0 tEG Va(E) > 0
and e> 0, there exists t E G
such that (A — e, A + e) n sp ht 1 1 E 0 0. 
If A E S(14), then for each Borel subset E of 11 with A(E) > 0 and e > 0, there exists t E G such that (A — e, A + e) n sp ht 1,E 0 0. Let F = {s E E n t'E I re (s) E (A projection of M such that sup{q, 04(0} < p,
and
—
e, A + e)}. Then q = m r, is a non zero 
sp (htigigH) C (A — el A + e),
where p = mxE . Conversely suppose that A has the following property: for any Borel subset E of n with 12(E) > 0 and e > 0, there exists 0 0 q EProj(M) and t E G such that sup{q, at(q)}
P,
and sp (ht iqlqH) 
C
(A
—
e, A + e).
We may assume that q = mx,. and F c E. Since sup{q, ext (q)} G p, it follows that FCEn C'E. Moreover, sp(ht igIgH) = (A e,A + e) nsp h t 1,E 0. Therefore, A E S(M). Q.E.D. 
—

Definition 9.5.8. Let (G, (1,A) be a group measure space. Define the ratio set r(G)(C [0, cc)) as follows. A(> 0) E r(G) if and only if for any Borel subset E of n with it(E) > 0 and e > 0, there exists a Borel subset F of n and t E G such that A(F) > 0, F Li tF C E, and d,a o t < e, dA (s) — A
Noticing that rt(s) =
_pd ot1 (s) , dp
Vs E F.
from Lemma 9.5.7 we have the following.
400
Proposition 9.5.9. Let (G,f1„u) be a free and ergodic group measure space. Further, suppose that fl is compact, A is a probability measure on Il, and for each t E G there are two positive constants Et and ij such that 0 < et < V(s) < ri t < oo,Vs E G. Then there is a factor M such that S(ii) = r(G).
Let fln (n = 1,2, • • .) be the additive groups of integers, reduced mod 2, i.e., (In is a compact (discrete) group composed of two elements fol 1} as follows: 0+0=0,0+1=1+0=1,and1+1=0. Let fl = x:t i nn be the direct product of { iln I n , 1,2,• • .}. Thus n is a compact Hausdorff space satisfying the second countability axiom, and fl is a compact group. Let G be the set of those a = (as) E f/ for which an 0 0 occurs for a finite number of n only. Then G is a countable group. For b E G, define a homeomorphism of [1: a 4 b(a) = a + b(Va E n). Let An be a probability measure on nn with = Pn,
An ») ({
= qn,
where pn E (0, 1) , and pn + qn = 1, Vn. Let A = x nw_ ban be the infinite product measure of {An } on Q. Then, A is a probability measure on Q. If a is a permutation of 0,11, i.e., a(0) = 1, a(1) = 0, then it is easy to see that 2s1
du n o a (s) . (E' , s = 0,1, d ,u qn the element of G such that the nth component of cn is 1 and Vn. Let c other components of en are 0, Vn. Then we have )
A 0 en = and
X 14 ntlig X
(An 0 a)
) 2an —1
c/A o en (a) = ( Li , Va = (ak)k E n, du qn Vn. For any b E G and b 0 0, there is unique finite sequence fil < • •  < 1} of positive integers such that b = cii ± • • • + cik . Hence, we have dp 0 b
diz
0,0
(pn) (2an l)bn (a) = 11 , n=1 qn
Va = (an) E n, b = (IQ E G; and (G, [1,A) is a group measure space. Let b E G and b 0 0. Clearly, {a E n I b(a) = a} = O. Thus, (G, [LA) is
free. For any Borel subset E of subset of
n, and
n, let
F=
11 b(E).
Clearly, F is also a Borel
bEG
b(F) = F,Vb E G. Then for any a E F we have
a
E F
401
whenever a is obtained by changing any finitely many components of a. Thus, for any positive integer n, F has the following form: F=
xLi nk x Fn,
where Fn is a Borel subset of X (r n+lnk. Now if C Cn x xcktn+i nk is a cylinder subset of [1, where Cn is a (Borel) subset of xL i nk, then it is obvious that n C) = ,u(Cn x Fn) ti(C),u(F). If K is a compact subset of [1, then we can see that K = n(7rn(K)x xft where Irn is the projection from Q onto x'L i flk. Thus, A(K n F) = lim,u(nn (K) x xr=n+l nk)11(F) =
a(K)I(F)
Further, by the regularity of i we have /1(C n F)
for any Borel subset C of Q. Now if E satisfies the following: ,u(bEAE) = 0,
Vb E G,
then we have ,u(E) = ,u(F) = ,u(F n F) = A(F) 2 = ,u(E) 2 ,
and either A(E) = 1 or ,u(E) = O. Therefore, (G, n, A) is ergodic. Now let A E (0, 1), and pn = A(1 + A) 1 ,Vn. Then from Definition 9.5.8 we have r(G) = {0, An I n E Z}. Thus by Propositions 9.4.6 and 9.5.9 I't;/ =Mx„Gis a type (III) factor. Proposition 9.5.10.
Type (MA) factors (0 < A < 1) do exist.
Remark. By Proposition 9.5.9 and above construction, we can also obtain the examples of type (III) and (III) factors (see [165 ] ). Now keep the above notations: [In = {0, 1}, Vn; Q = x ncci_,nn; G; An ({ l») = Pn E (0 , 1 ) An({ 1 }) = qn, Pn qn = Vn; = xItin; and the element en of G, Vn. Let H = L 2 (f u), 11= H 1 2 (G), M =
f E Lœ (n , A )} ,
402
and { (71(x) ') (b) = ab1 (x)Êd (b) ab(mf) = ubmftei — mfb , fb(') = (A(c)e) (b) = È.(b + c),
f ( ' + b) ,
Vb,c E G, x E M, f E Ltx)(n, it), and eA" E ff. Clearlyt the crossed product M =Mx a G= {71 ( M) , A(G)}" admits a cyclic vector 6: f 1, if b = 0, 1 0, otherwise,
where "1" is the constant function 1 on n(E L2 (12, /1) = H). For any ai E {0,1},1 < i < n, let p(a i , ... , an) be the operator on H = L 2 (12, /1) corresponding to multiplication by xgai ,..., an ), where E (al , . • • , an) = (al , ... , an) x xr_n+I nk(c n), and put i5(al, • • • , an) = 7r(P(a1, • • • , an)),
An =
Vn. Then it is easy to check that
A/4'11(4, • • • , an) Ak = gal, — , ak + 1, — , an), Va,i E {0, 1}, 1 < i < n, 1 < k < n. Thus
{gal).* • , an), Alc
I
ai E {0,1},1 < i < n, 1 < k < 71,1
generates a type (I2n) subfactor of gi,Vn. Now define
= Ak )
Vk,
1)
00
Alf (e a l
® ... 0
ean
® 0 1 ) = /5 (ai,..
. ,a n )i )
n+ 1
where e0 =
(
(0 0 \ 1 0\ ,Va i E {0, 1}, 1 < i < n, Vn. Then All can be 1:) OP el = 0 1 ) 00
uniquely extended to a * isomorphism from min0 M2(Œ) into rt= 1
It is easily verified that: (i5(ai, • • • , an) É .0) 'È'CI) = A (E (a i , . .. , an)), (Aki5(ai, • ' • , an) 'È'o, 6' )) = 0 ,
(Ak É' do, È' do) = 0,
Xi
= it4 x a G.
403
Va, E {0, 1 , 1 }
(
0, i.e., x E U and b (xn , x)  0. n SO U as a subspace of E is also a Polish space. Now let F = nn Un, where {Un } is a sequence of open subsets of E. Suppose that bn is a metric on Un as in the preceding paragraph, Vn, and define dF (x, y) =
(x, y) E ,41 y1 1 +bn bn(xl y) 5 VX, y E F. n
Since fintop. and dtop. are equivalent in F, Vn, it follows that dF top. is equivalent to the relative top. in F. If {xn }(c F) is a Cauchy sequence with respect to dF, then for each k there is yk E Uk such that 45k(x)Yk)  0.
407
Clearly, we have also d(x n ,y k ) 0,Vk. Hence, there is z E F such that 0. Therefore, F as a subspace of E is Polish. z,Vk, and dF (x„,x) Yk Conversely, let F(c E) be Polish as a subspace of E, and 4, be a proper metric on F. For each n , put f x Er there is an open neighborhood Uof 1/n such that Dd,.(U n F)
Clearly , F c nnFn. Conversely, if z E Un Fn , then for each n there is an open neighborhood Un of z such that Ddi.(Un n F) < 1/n. We may assume that U1 D U2 D • • • and Dd(Un) 0. Pick zn E Un n F,Vn• Then {zn } is Cauchy in (F,dF ). Thus , there is y E F such that dF(xn,y) 0. Clearly, 0. Hence , z y E F, i.e., d(xn ,y) 0,n nUn = {z}, and d(x„,$) F = rInFn• If z E Fn I then there is an open neighborhood U of x such that Ddp.(U n F) < 11n . By the definition of Fn , it is obvious that U n F C F. Thus , Fn is an open subset of P', i.e., there is an open subset Gn of E such that Fn = n Gn ,Vn. Put Um = {x E Eld(x,P) < 1/m},Vm.
Clearly, Um is open , Vm, and F = n mUrn . Therefore, n n Fn nn (Gn n
F
P)
n m ,n (Gn n Urn
is a as—subset of E.
)
Q.E.D.
Proposition 10.1.4. Any Polish space must be homeomorphic to a subset of [0,1]' ( the countale infinite product of [0,1] ).
as
Proof. Let E be a Polish space , d be a proper metric on E, and {an } be a countable dense subset of E. Then X
d(a n,x) d(an ,z) n
(Vx E E)
is a homeomorphism from E into [0,1 1 00 , and also by Proposition 10.1.3, its Q.E.D. image must be a as—subset of [0,1 ] 0 . Proposition 10.1.5. Let CI be a locally compact Hausdorff space. Then is a Polish space if and only if n satisfies the second countability axiom.
n
Proof. The necessity is clear. Now let fi satisfy the second countability axiom, and fl oc, = flu{oo} be the compactification of II. Clearly, no° is a Polish space. Q.E.D. Now fl is an open subset of noo , so fl is also a Polish space.
408
Definition 10.1.6.
The Polish space /V' is the set
fn = (nk)Ink nonnegative integer, k = 1,2, • • •} with the topology generated by the metric d(n,m)
=E1
In k — Mk1
k 2k 1 + Ink — enk I '
Vn = (nk),m , (mk) E IN' . Clearly , fn = (nk ) the number
I
of nonzero components of n is finite a countable dense subset of /Vœ. Moreover, for any n z (nk) E 1Noe , INZ,...,n, = fm =
(me) E
}
is
Nœ imi = ni,1 < i < k},k = 1,2,•••
is a neighborhood basis of n. Proposition 10.1.7. Let E be a Polish space. Then there exists a continuous map from /V' onto E. Proof. Let d be a proper metric on E, and Dd(E) < 1. For n 1 = 0,1,• • • , let F(ni ) = E. For each n 1 , pick a countable closed cover {F(ni, n2) In2 = 0, 1, • • .} of F(ni) such that Dd(F(ni, n2)) _._ 1/2, Vn2. Further, for each (ni, n2), pick a countable closed cover {F(ni,n2, n3)Ins = 0,1,...} of F(n i , n2) such that Dd(F(ni, n2, n3)) < 1/2 2 ,Vn 3 , • • • , Generally, we have a family {F(n i , • • • , np) IN = 0,1, • • • , 1 < i < p,p = 1,2, • • .} of closed subsets of E such that F(n i ) = E,F(ni ,• • • ,np ) = Uckt.o F(n i , • • • ,np ,k), and Dd(F(n i ,••• ,np )) < 2  ( 91 ),Vn i ,• • ,np ,p = 1,2,•. Since (E, d) is complete, it follows that 4 {n cikt i F(n i ,•
• • ,nk )} = 1, Vn = (n k ) E /V".
Let {f (n)} = nr_ i ll(n i , • • • ,nk),Vn = (nk) E INoe . Clearly, f is a map from IN' onto E. Suppose that n(k)  n in /V'. For any 6 > 0, pick p such that 2  (P  ') < e. Then we have n?) = n1 ,1 < i < p, if k sufficiently large. Hence, 1 (n) and f (n(k)) G F(ni , • • • , np ) and d(f (0)), f (n)) < 2  (P1 ) < e if k sufficiently large. Therefore, f is also continuous. Q.E.D. Lemma 10.1.8. Let E be a Polish space with no isolated point, and d be a proper metric on E. Then for any e > 0 there is an infinite sequence {E n } of nonempty Gysubsets with no isolated point of E such that Dd(En) < e, Vn; En n Ent = 0, Vn 0 m; and UnEn = E. Proof. We may assume that E < Dd(E). Pick a countable open cover {Kt} of E such that Vn 0 0, Dd(Vn) < 6,Vn. Let E1 = VI. Clearly , El is a
with no isolated point. By induction, define En = rfn \Fn , where subetFn wih =
409
Uit=.1Ek,Vn > 1. Since Fn = Ult_TIVk is closed, it follows that En is a G5 — subset. Moreover, from (Vn\Fn ) C En C VT —,\F'n, En has no isolated point. If 4 {niEn 0 0} = oo, then {En} satisfies our conditions. Otherwise, notice that 0 0 El 0 E ( since Dd (Ei ) < e < D(E)), then the same process can be carried to Ei ( for some el with 0 < e' < Dd (E1 )). In this way, we can complete the proof. Q.E.D. Let E be a Polish space with no isolated point. Then for any nonnegative integers n 1 , • • • , nk , there is a nonempty Qssubset E?1) with no isolated point such that 1) if (n 1 , •   , nk) 0 (m i ,    , mk ), then Lemma 10.1.9.
E (ki) 2 ) E $tic1 )
."' tnk
k+ E ( 1)
 uoe p=0
•
•n bnk
n E(k)
",mk
Vn i , • • • , nk;
is a proper metric on E(k) then the diameter of E+1) ni,•••,nk ,nk+1 ni 3 — ink i with respect to (d + d(n1i) + • • • + d(k) •••,nk ) is less than (k + 1) 1 ,Vn 1 , • • • ,nk+i , where d is a proper metric on E. 3) if d(k)
Proof. Using Lemma 10.1.8 to (E, d) and e = 1, we get {E) In i = 0, 1, • • .}. Again using Lemma 10.1.8. to (EZ) , d + dZ)) and e = 1/2, we get {E ) n2 In2 . 0, 1,   },Vn i . Continuing this process, we can get the conclusion. Q.E.D.
Let E be a nonempty Polish space. Then there exists an injective continuous map from N" onto E, if and only if, E has no isolated point.
Proposition 10.1.10.
Proof. Since IN
has no isolated point, the necessity is obvious. Now suppose that E has no isolated point. Pick fE ) .. ,n,} as in Lemma 10.1.9. Since is homeomorphic to IN' , it follows from Proposition 10.1.7 that from Nnoei,.. ,nk onto E' ), ,.. ,nk , Vn i , • • , nk. there is a continuous map f4ki ) Fix k. Since f...Z N 3 •••,nk I n ') • • • , nk } is a closed and open cover of N00 , we can define a continuous map f( k) from N' onto E, such that f(k) I Eit)...,n, ' .f(ki ),...,nk •
For any n = (nk) E Noe and integers p, q with p < q, noticing that f (q ) (n) G E(q) ni,•••,nq C E (P)
so by Lemma 10.1.9 we have d(f(P)(n), f(q)(n)) < 1) 1 . Thus , {f (k) (n)} k is a Cauchy sequence of (E, d), and there is f(n) E E such that d(f( k )(n), 1(n)) 0, uniformaly for n E Noe . Further, f is continuous. By Lemma 10.1.9, {f ( v ) (n)} p>k is a Cauchy sequence of (Eltki),... ink ,4ki)„. ink ) 1 Vk. Hence , 1(n) E rl iTL I .K.ki), ... ,nk , Vn = (nk ) E Noe. But Dd (E tki).,, ,nk )
1} of Borel subsets of E has the following properties: 1) IL I , ... ,nk
2) A,
n Aml ,...
= 0,V(n i ,
nk)
(m i , • , m k );
C An11 ... ,nkl Vni, • • • ,
3) j(117Z ,... ,nk ) C In fact , we can prove f
C
) 1 \in ' 9
f C
• • •
nk •
by induction, and the rest facts
are obvious. Since f is injective, it follows that f (n) = = nckt1 f( IVnœl, the = (nk) E . By proof of Theorem 10.2.11, we have also {f(n)} = nckt., f (INnœi, ...). Further, from above property 3) we get { f (n)} = Vn = (nk) E 1N . Now we prove that f (INoe) = n 143 _ . In fact, since {f (n)} = Vn = (nk) E INoe , it follows that f(Îi) C Coversely, let z E . For k = 1 , there is m 1 such that :1
418
X E Ami . From above properties 1) and 2) , we have x E fl1 U nip  lnk ( A ni,. i nk n Ami)
=
A m,
n (n172 11 n2,,nk
A nti,n2,,nk)•
Repeating this process, there is m = (m k ) E IN" such that x E n kcb3_ 1 Ami ,...,ntk { f (m)} . Hence, x = 1(m) E f(1PT) ), and f (DP°) = nckc:=1 Ani,...,nk• Therefore, f (IN') is Borel. Q.E.D. Let E, F be two Polish spaces, and f be an injective Lemma 10.3.11. Borel map from E to F. Then f(E) is a Borel subset of F, and f is a Borel isomorphism from E onto 1(E).
Proof. It suffices to show that for any Borel subset B of E, 1(B) is a Borel subset of F. Fix a Borel subset B of E. Let G = {(x, f (x)) E Bl, d be a proper metric on F, and {an } be a countable dense subset of F. Put U = {y E Fld(y,ak)
1.
For each equivalent class X of E, by 1) f'(X) is a nonempty closed subset of /V'. From Lemma 10.4.1, r i (X) has a minimal element (pk) = p = p(X).
Then we claim that: a) A(p i , • • • , pk)
n X = B(p i , • • • ,pk) n X 0 0, Vk;
b) if In fact
(ni, • • • , nk) 0 (P1)• • • , pk), then A(n 1 , • • • , nk) n X = 0, Vk. , since 1(p) E X and {f (p)} = nr_ i fl(p i , • • • , pt), it follows that B(pi, • • • 1 pk) n X 0 0, Vk. By the definition, A(p i , • • • , p k ) n X c B(p i , • • • , P/c) n X, Vk. Now let x E B(p i ) n X. Suppose that there is m 1 < pi such that z E fj(m i ). Pick y E B(m i ) with y — z. Clearly, there is m = (m k ) E IN' such that y = f(m). Then m E f' (X). But p is the minimal element of ri (X), so P! < ml . This contradicts m1 < pi . Thus , z E B(23 1 )\U„„ i p. By (n i , • • • , nk) 0 (pl l • • • ,pk), there exists j(< k) such that ni = pi l l < i < j and pi < ni . Then A(Pil• • • ,Piil ni)
C
n 13 (pil • • • IP;)
[E\ up 0,
430 2) The subsets of W(X*) with the following form
{E* E W(X* )111x+ EIJI < A}, where Pi = {y E XI f(y) = 0,Vf E Es },Vx E X, A > 0, generate a standard Borel structure of W(X*). Proof. 1) It is obvious from Propositions 10.3.15, 11.1.6 and 11.1.5. 2) Notice that E*  El (V E* E W(X*)) is a bijection from W(X*) onto Q.E.D. C(X). Then by 1) we can get the conclusion.
Let H be a separable Hilbert space, and W(H) be Proposition 11.1.8. the collection of all closed linear subspaces of H. Then the subsets of W(H) with the following form
{E E W(H)Ille +
Ell < A},ve E H, A >0
generate a standard Borel structure of W(H), and E  E L (V E a Borel isomorphism on W(H).
E W(H)) is
Proof. From Theorem 11.1.7 and Proposition 10.3.2, it suffices to show that for any e E H, A > 0, {E E W(H)iii + Elm < Al is a Borel subset of W (H). If A > lick then we have {E E W(H)iii + ELii < Al = W(H) obviously. Thus, we may assume A < Hell. For E E W(H), let p be the projection from H onto E. Then we have
lie + Eiii = iipeii, Let IL =
(Heir _ A2)1/2.
IIE + Ell =11(1 p)eii.
Then
I He+ Elm 0 {E E W(H)
= W(H)\ nn {E E W(H)
I He + Eli < Til + 0.
Therefore, it is a Borel subset of W(H).
Q.E.D.
References. [34], [177 ] .
11.2. Sequences of Borel choice functions First, we study the process of the HahnBanach theorem. Let X be real Banach space, E be a linear subspace of X, f be a linear functional on E with
431 norm < 1, and z E X\E. We want to extend f from E onto (E4[x]) still with norm < 1, i.e.,
if (x + w)i
iix +will
V w E _E.
So we need to pick the value of f(x) satisfying Hz+ uii
— 1(u)
vii — 1(v),
f(x)
Vu l v E E.
•
Then the value of f (s) must satisfy the following inequality: sup{( 11x+ull — f(u))1u E EI
f(x)
inf{(11x vil — 1(v)) Iv E El.
Conversely, if the value of f (s) satisfy the above inequality, then f is a linear functional on E4[xj still with norm < 1.
Definition 11.2.1. Let X be a real Banach space, E be a linear subspace of X, and z E X ( maybe z E E) . For any linear functional f on E with norm < 1 , define
L(f) = sup{(—Iix+ ull f(u))1u E E}. and
M(f) Since
11111 < 1,
= inf{(11x+ vil — f(v))1v E
El.
it follows that 4) (f) < Mk ) (f).
Lemma 11.2.2. Let X, E, x and f be as in Definiton 11.2.1. 1) If z E E, then 4) (f) = f(s) = MP(f). 2) f can be extended to a linear functional on E [z] still with norm < 1 ( ) (f) < f(z) < if and only if the value of f(x) must satisfy the inequality: L ;
(z) ME (f).
Proof.
1) Suppose that x E E. Then
If(x+w)1
lix+wli, Vw E E.
Further, we have
f(u) Thus ,L(f) that
f(x)
lix + vil +PO/
< 1(x) < Mk ) (f). On the
4) (f) — f (—x) = 1(x),
and
Vu, v E E.
other hand, since s E E, it follows
MP(f)
< —f(—x) = 1(x).
432
Therefore L(f) = f (x) = Mk ) (f). 2) It is obvious from 1) and the discussion of HahnBanch theorem. Q.E.D. Lemma 11.2.3. Let X be a real Banach space, E be a linear subspace of X, and x E X, and S=
{fif is a linear functional
on E,andlifIl < 1}.
Write 4) (.) = L(.) and Mr(.) = MO simply. Then L(.) is a convex function on S, and L(.) = M() is continuous in the interior of S. 
Proof.
Let A E [0, 1], and f, g E S. For any u E E, we have 
11x+ ull  (A1 + (1 A)g)(u)
= A(  11s+ ull  1(u)) + ( 1  A)(  11x+ ull  g(u)) < AL( f) + (1 A).L(g). Thus, L(Af + (1 A)g) < AL( f) ± (1 A)L(g), i.e., L(.) is convex on S. Now let fo E S and 11./011 < 1n for some n E (0, 1). On V = ff E S ill fll < n}, define F(f). L( f + fo) L(f0 ), V f E V. We need to show that F(f) is continuous at f = 0. Clearly, F(0) = 0, M.) is convex on V / and 
F(f)
5._ m(f
+ fo)

WO 5._

11x11 — L( fo), V f
Put a = 11x11  L(f0 ). For any e E (0, 1) and f E S with f, ±e  ' f E VI it follows from the convexity of M.) that F(f)
= F((1

E V.
11111 < n e,
since
e) • 0 + e • e 1 f)
< eF(e1 f) Ç ea and
0 = F((1+ e) i f + e(1 + E)  '  (  E1 n)
Ç (1 + E ) ' F(f) + E(1 + W i ll C 1 A From the second inequality, we get F(f) > eF( e1 f) > 1F(1)1 ca,Vf E S with 11111 < ne, i.e., M.) is continuous at 0. 


ea. Thus, Q.E.D.
Theorem 11.2.4. Let X be a separable Banach space, and W(X*) be as in Theorem 11.1.7 ( a standard Borel space). Then there is a sequence { f'} of Borel maps from W(X*) to (X*, a(X*, X)) such that : for any E* E W(X) and n, fn (E*) E (Et), ( i.e., fn(E t ) E Es and 11./n(E * )11 1); and { fn (E*)1n} is tedense in (El i , VE* E W(X*).
433
Proof. First, let X be real. Suppose that { xn In = 1,2, • • .} is a dense subset of X, and fix E* E W(X*). Then {2;', = xn + Elm = 1, 2, • • • is dense in WI' , where El = {x E XI f (s) = 0,V f E El. Put Bo = {0}, Bn = rii) • • • 1 j, Vn. These are finite dimensional linear subspace of XLEI. Moreover, since (X/EI)* E* , we shall identify them in the following. For each t = (t1, • • • tn • • *) where tn E [0,1], Vn, we say that there is a linear functional fr on X/E'l such that 11ftEa 11 < 1 (i.e., itEa E (E ) i) and (1)
(zz,) = tn+iLn(ftE.) + (1 — tn+i)mn(ftE . ),
where _Lin () = a:+1) (') and Mn (.) = Mgn +1) (•),Vn > 0. We prove this by induction. Assume that such fr exists on B. Put = tn+iLn(ftEs ) + ( 1 — tn+i)Mn(itE* ). Since L n ftE • ) A < Mn(ftE* ), it follows from Lemma 11.2.2 that fp* can be extended to a linear functional on Bn+i with norm < 1 still and LE' (in+1) = A. Therefore, there exists a linear functional Jr on X/EI with norm < 1 and satisfying (1). Define Q = fr = (ri, • • • , rn, • • .)Irn is rational and E [0, 1],Vn; and 4 {nirn 0 } < °el, and fix f E E* with 11f 11 0, since = (1 271)111111 and I f(1)1 < there is a rational number rr) E [0,1] such that l(f,.E* — f)("i i )I < e,Vr = (rn) E Q with ri = rr. For n = 2 and any e > 0, by Lemma 11.2.3 there is ri > 0 with following property : for any g E B; with l(g — f)(1 1 ) I < j ( thus II g — (f IB 011 is very small, and 11g11 < 1), we have ILI f) (

(2)
L1(01 < 6 1 IM1(f) — M1(01 < 6.
( From the preceding paragraph, there is a rational number r 0) that ( o) Vr = (rn) E Q and ri = ri • l(frE *  f)("±1)1
0
438 will generate a standard Borel structure of 7. Since M  (M, M')  M n Ail is a Borel map from A to A, 7= {M E Alm n M' =Œ1H} is a Borel subset of A. Q.E.D.
Proof.
Notes. Effros.
The Borel spaces of Von Neumann algebras was introduced by E.G.
References.
[34], [35], [177].
11.4. Borel subsets of factorial Borel space Let H be a separable Hilbert space, A be the standard Borel space of all VN algebras on H , and F be the standard Borel space of all factors on H. Lemma 11.4.1. Let G be the set of all unitary operators on H. Then G is a Polish topological group with respect to strong ( operator ) topology. Proof. Clearly, G is a topological group with respect to strong topology. Let S be the unit ball of B(H). Then S is a Polish space with respect to
strong topology. If ek } is a dense subset of and only if IluGli = ilu* Gil = 1, Vk. Thus {
fe E Hill
11 = 11, then u E G if
G = nk{u E Silluekil = 1 }n
nk,,, um fu E
s I Rek ,uem)1> 1 —ni 1
is a G5 —subset of (S, s—top ) . That comes to the conclusion.
Q.E.D.
Proposition 11.4.2. For any M E A, let s(M) = fuMuslu E Gl, where G is as in Lemma 10.4.1. Then s(M) is a Borel subset of A, VM E A. Proof. Fix M E A , and put Go = {u E GluMu* = M } . Define an equivalent
relation — in G : u — y if y E uGo . By Lemma 11.4.1 and Theorem 10.4.2, there is a Borel subset E of G such that #(E n uGo) = 1, Vu E G. Then , s(M) = fuMu* lu E El. We say that u  uMu* is a Borel map from G to A. In fact, if {an } is a countable dense subset of ((M) 1 , a(M, Me )), then { an (u) In} generates uMu*, Vu E G, where an (u) = uanu*(Vu E G) is a continuous map from G to (B(H), a), Vn. Now by Proposition 11.3.4, u  uMu* is a Borel map from G to A.
439
In particular, u  uMu* is an injective Borel map from E to A. Therefore, Q.E.D. by Theorem 10.3.12 s(M) is a Borel subset of A. Proposition 11.4.3. Let M E A. Then a(M) = {N E AIN is * isomorphic to M } is a Borel subset of A.
Proof. A = A(H) is the collection of all VN algebras on H. We shall denote the collection of all VN algebras on H 0H by A(H 0H). Define a map 4) : A(H)  A(H 0 H) as follows (NM) = AfOr1H, VM E A(H).
Let M,N E A(H). Then M and N are * isomorphic if and only if 4)(M) and 4)(N) are * isomrorphic. Notice that 4)(M)' = M rOB(H) and 4)(N)' = NrroB(H) are properly infinite ( here dim H = oo; if dim H < oo, then a(M) = s(M) is a Borel subset of A ) . By Proposition 6.6.7, 4' (M) and 4)(N) are spatially * isomorphic if 4)(M) and (10(N) are * isomorphic. Thus, a(M) = V' (s(4)(M))). From Proposition 11.3.4, it suffices to show that 4) is a Borel map. By Proposition 11.3.3. there is a sequence { an (.)} of Borel maps from A to (S, a), where S is the unit ball of B(H) , such that {a n (N)In} generates N,VN E A. Then {a n (.) 01H In} is a sequence of Borel maps from A to (B (H 0 H),a) , and {an (N) 0 1H1n} generates 4)(N), VN E A. Now by Proposition 11.3.4, 4) is a Borel map. Q.E.D. Proposition 11.4.4. Denote the collection of all type (Is) factors on H by Fin , Vn. Then .1;n is a Borel subset of 1, n = co, 1, 2, ... .
Proof. Noticing that all type (Is) factors are * isomorhpic, the conclusion Q.E.D. is obvious from Proposition 11.4.3.
Lemma 11.4.5. Denote the projection from H onto [MipH] by em (p). Then (M,p)  e m (p) is a Borel map from A x P to P, where P is the collection of all projections on H, and with strong (operator)) topology P is a Polish space. Clearly, from Theorem 11.1.7 the standard Borel spaces P and W(H) are Borel isomorphic. So we need to show that (M, p)  [MipH] is a Borel map from A x P to W(H) . By Theorem 11.2.5, it suffices to find a sequence {q,.(., .)} of Borel maps from A x P to (B., w), where "w" means the weak topology in H, such that {nn (M,p)In} is a dense subset of [MipH],VM E A, p E P. By Proposition 11.3.3, there is a sequence { an ()} of Borel maps from A to (S,a) such that {a(M)In} is a r(M,M.,) —dense subset of (M) i , VM E A. Let
Proof.
440
{G be a countable dense subset of H, and define }
g'n ,k(M , p) = a n (Mipe k , V
n, k. Clearly, S'n,k(., ) is Borel from A x P to (H, w), 'In, k, and kn,k(M , p) In, kJ = [Mipa],VM E A, p E P. A x P  (H,w),Vn, such that Therefore, we can find Borel maps i Q.E.D. {il n (M,p)In} is dense in [MspH1,VM E A, p E P. Lemma 11.4.6. Denote the collection of all infinite factors on H by Fif . Then 7if is a Sousline subset of F.
A factor M is infinite if and only if there is v E M such that vv = 1 and vv* 1. By Proposition 11.3.3, there are Borel maps an (.) : A (S,a)(n = 1,2, • • .) such that {an (N)In} is r—dense in (N) I ,VN E A. Notice Proof.
that 1M E 1 v E S• v*v =1,vv* 1;1 ' = van (M 1), Vn f ' (M 1)v E = { (M ' v) 1 and a„ = (Ix {v E Slv*v = 1, vv* 0 1})n
nn {(M, v) lan(M 1) v = van(All} =
(I x {v E Slv*v = 1, vv* 0 1})n
nn ,ii {(M, v) I ((an (A41v — va n (M I))6,ei ) = ol is a Borel subset of 7 x (S, r), where feil is a countable dense subset of H. If 7r is the projection from 7 x S onto 7, then 7if = 7rE. Therefore, Fif is a Sousline subset of J. Q.E.D.
Proposition 11.4.7. Denote the collection of all type (III ) factors on H by Then In , is a Borel subset of I. Proof. Denote the collection of all finite factors on H by if. From Proposition 11.4.4 and Lemma 11.4.6, it suffices to prove that ii• is a Sousline subset of I. Since H is separable, a factor M on H is finite if and only if there exists a faithful normal tracial state on M, i.e. there is a sequence {ek} of H such that 0, Va, b E M, and [aieklas E M', k] is < oo, Eqab — ba)ek ,
Ek Ilea
ek) =
k
441
dense in H. Then by Lemma 11.4.5,
E= {(M, (k))
M E 7, (G) E Hoo; [(L'ek la' E MI, k] = H; 1 and Ek ((ab ba) 6,, ek) = 0, Va, bEM f —
M E 7, (G) E Hoe , and em (p) = 1, = { (, m, (t )) , jg " where p is the projection from H onto [Gild
}
n
M Gi, (ek) E H00 , and nn,„ {(M, (60) Eqa„(M)a nt (M) — a nt (M)a n (M)) ek , ek) = 0 k
}
00
is a Borel subset of 7 x Hoe , where Hoe =
E e)H; a(.) : A  (S, a) is Borel n= 1
, Vn, and {a(M)In} is r—dense in (M) i , VM E A( see Proposition 11.3.3 ) . Let 7r be the projection from 7" x Hoo onto J. Then 7f = 7rE is a Sousline
Q.E.D.
subset of J.
Lemma 11.4.8. Denote the collection of all semifinite factors on H by J. Then Isf is a Sousline subset of J. A factor M on H is semi—finite if and only if there exists a finite projection p of M with c(p) = 1, i.e., there is a projection p of M with [MpH] = H, and there exists a sequence {G} of pH with I ek112 < cc such
Proof.
Ek
that
>(k, k)
is a faithful trace on (pMp).
k
Consider a subset E of 7 x P x Ho.. (M, p, (G)) E E, if pG = = an (Mi)p,Vn; [ate', lk, a' E M1= pH;[MpH]= H, and E((pa n (M)pa,„,,(M)p — pa, n (M)pan (M)p)G,G)
elc 1 Vk; pan (M')
= o,
k
Vn,m where an () : A  (S, a) is Borel , Vn, and {an (M)In} is r—dense in (M) I ,VM E A ( see Proposition 11.3.3 )• By Lemma 11.4.5, the map (M,p)  [MpH] is Borel. Thus , E is Borel . Let Ir be the projection from IxPxn0 onto I. Then 73f = TrE is a Sousline subset of F. Q.E.D. ,3
Lemma 11.4.9. Let M be a factor on H, and 4:1) be a * automorphism of M. If there is a non—zero element a of M such that 41)(b)a = ab,Vb E M, then 4:1) is inner, i.e., there exists a unitary element u of M such that (10(b) = ubu*, Vb E M. Proof.
From (1)(b)a = ab, we have a*(1)(b*) = b*a* ,Vb E M. In particular, if
442
b is unitary , then we get b* (a* a)b = a* to(b*) • 41)(b)a = a* a, 4:0(b)aa*(1)(b*) = ab • b* a* = aa*. Thus, ea and aa* E M n Aft = Œ1H. Now let u = liar l a. Then u is a unitary element of M, and CO = ubu*,Vb E M. Q.E.D.
Let G be the group of all unitary operators on H. Then 1 M E 1,u E G;uMu* = M, 1 E 1 but •  u • u*is not inner for M f ' = { (Mu) is a Borel subset of 3" x G.
Lemma 11.4.10.
By Proposition 11.3.3, we have Borel maps an (.) : A  (S, a), n = 1,2, • • • , such that {an (M)In} is rdense in (M) I ,VM E A. Since (M, u) uan (M)u* is a Borel map from ,.1" x G to (S,a),Vn, it follows that Proof.
E = {(M,u)1M E .7",uE G, and uMus = MI
= nn,m {W., u)
ME1,uE G, and uan (M)u* • a(M 1) = am (M 1) • uan (M)e}
is a Borel subset of Y" x G. Let d be a proper metric on (S,a) ( see Definition 10.1.1 ) , and consider a subset E(j,k,m, n) of .7 x G. (M, u) E E(j,k,m,n), if uMu* = M and satisfies one of following conditions: 1) d(a5 (M),0) < n 1 ; 2) d(uak(Ml),0) 2n ', for any m we can choose j,k such that
d(ai (M), y) < (2mn)  1 , d(uak(M 1 ), y) < (2mn) 1 . Thus (M, u) V E(j,k,m,n). Further, (M, u) V nn U n„ nijg E ( ?' , k , m, n) . Conversely, let (M,u) E F xG,uMu* = M, and (M, u) v nr, Li m ni,k E(j,k, m, n). Then there is n such that for any m, we have j(m),k(m) and (M, u) V E(j(m), k(m), m, n). Thus, d(ai (m)(M), 0) > n1 , d(ua k (7,,)(M 1),0) _>_ n 1 , and d(ai(7,,,,)(M),ua k (7„)(M 1)) < m' ,Vm. Since the unit balls of M and M' are acompact, there exist acluster points a, a' of fai(m)(M)Iml, fuak( m)(M 1)1m} respectively. Hence, we get
a = ua' , d(a , 0) > n1 , d(uas , 0) > n1.
443
Now for any b E M, we have ubusa = uba' = uasb = ab. Thus , by Lemma 11.4.9 • > u • u* is an inner * automorphism of M, i.e., (M,u) V E. Q.E.D. Therefore, E = nn um ni,kE(j,k,m,n). Lemma 11.4.11. Let G be the group of all unitary operators on H ( G is a Polish topological group with respect to strong operator topology, see Lemma 11.4.1 ) , and Go = {u G Gil is not an eigenvalue of u } . Then Go = nnik um n i jo and k > 2 , then
ell
lie — eili,v3..
< 1/4. Then Ilfm(u)e.foli _?_ 3/4,Vm. Hence, if
u V {v E G10'7 ,1(06°11 < k 1 },Vm. Conversely, let u E Go , and e(.) be the spectral measure of u on {z E Œ 1  1z1= 1 } . Put Pin
= e({z I Izi = 1,11 — zl _Ç 2 m}), Vm.
Then pm+i < pm ,0 < fm+i (u) < pm < fm (u),Vm, and pm  0 ( strongly ). For any n,k, choose m sufficiently large such that sup{liPme.i111 1 i  n} < k'. Then we have
Ilfm+i(u )&il l = I l fm +I (u )Pm ei l l . 1 1 Pmeill < k 1 , 1
_< j < n. Q.E.D.
Lemma 11.4.12. Let X,Z be two Polish spaces, Y be a Borel space, and f be a map from X x Y to Z such that: 1) for each y E Y, f( . , y) is continuous from X to Z ; 2) for each z E X, f(x,.) is Borel from Y to Z. Then f is Bord.
444 Let d, 45 be proper metrics on X, Z respectively, and fxk l be a Proof. countable dense subset of X. If F is a closed subset of Z, then
f (F)
{(x,y) E X x Ylf (x, y) E d(x, xk ) < n1 , and 1 nn uk {(x y) 45(f (xlc,Y),F) < n_ 1 f ' = nn Uk ((Xk, n x Y) n (X x
Yk,n))
is a Borel subset of X x Y, where Xk m = E Xld(x,x0 < n  '1, and Yk,n = f (xk ,.) 1 ({z G ZI45(z, F) < n ' } ). Now by Proposition 10.3.2, f is Borel. Q.E.D.
Lemma 11.4.13. Let G be the group of all unitary operators on H, G o = {t./ G Gll is not an eigenvalue of u}, and f (t,z) = exp(—t(z + 1)(z — 1) 1 )),Vt E R, z G Œ with Izi = 1 and z 1. Then f is a Borel map from JR x G o to G , where the Borel structures of G and Go are generated by strong operator topology.
f (t,u) is a continuous map from JR to G. Proof. Fix u E Go . Clearly , t Now fix t E R, and let r {z E CI lz1 = 1}. Then g(z) = it(z + 1)(z — 1) 1 is a real valued continuous function on (r\{1}). Pick a sequence {g n} of real valued continuous functions on r such that gn (z) g(z),Vz G Then ,exp(ign (u)) exp(ig(u)) = f (t ,u) ( strongly ) , Vu G Go . Let b be a proper metric on the Polish space G, and F be a closed subset of G. Since b(exp(ig n (u)), f (t,u)) 0,Vu E G o , it follows that
f (t,
(F) = fu E Golf (t, = nk
Urt>4 fu
G
G
11 1
G0 145(exp(ign (u)),
< k 1 1.
But exp(ign (•)) is continuous on G, Vn, and Go is a Borel subset of G ( Lemma 11.4.11) , thus f (t, V(F) is a Borel subset of G o , and f (t,.) is a Borel map from Go to G. Now by Lemma 11.4.12, f is a Borel map from R x Go to G. Q.E.D. Lemma 11.4.14. Let E be a Sousline subset of F. Then s(E) = fuMulM G E,u G Gl is also a Sousline subset of .7, where G is the goup of all unitary operators on H. Proof. By Proposition 11.3.3, we have Borel maps an (.) : 7 (S, u), n = 1, 2, • • • , such that {a n (M)In} is r dense in (M)I,VM G F. Let bn (M,u) ,uan (M)u*,Vn. Then {bn (., •)} is a sequence of Borel maps from 7 x G to
445
(S,a) , and {b r,(M,u)In} generates uMu*,V(M,u) G F x G. By Proposition 11.3.4, tb : (M, u)  uMu* is a Borel map from 7 x G to F. Therefore, by Proposition 10.3.5, s(E) = ti)(E x G) is a Sousline subset of F. Q.E.D.
Now let D = {z E CIO < Imz f is complex valued, bounded and continuous on D, } A(D) = ff t and f is analytic in the interior of D CND) be the collection of all complex valued continuous functions on D vanishing at oc. By maximal modulus, C(D) is a Banach space. Clearly, f(z) , exp(1Rezpf(z) is an injective map from A(D) into Cr (D); and this map transforms ff E A (D) I If (z) I < r, Vz E DI to a closed subset of Cr (D); further, A(D) can be regarded as a Borel subset of Cr (D). Thus, A(D) admits a standard Borel structure induced by the Borel structure of Cr (D). —
—
Proposition 11.4.15. Denote the collection of all type (III) factors on H by Z„. Then 7„, is a Borel subset of F. Proof. From Lemma 11.4.8, it suffices to show that Fin is a Sousline subset of F. Pick 6 E H with Roil 1, and consider a subset E of .7" x Go x R. (M, u, s) E E, if : 1)Meo ...z APeo :: H; 2) for any t E IR, f (t,u) 6; 3) f(t,u)Mf( t,u) = M,Vt G JR; 4) x p f(s,u)xf ( s, u) is not inner for M, where Go and f are as in Lemma 11.4.3. By Proposition 11.3.3 and 11.2.3, the condition 1) determines a Borel subset of F. For any rational number r, from the proof of Lemma 11.4.13 f (r, .) is a Borel map from Go to G. Thus , the condition 2) determines a Borel subset of Go . By the proof of Lemma 11.4.10, { (M., v) IvMv* :: MI is a Borel subset of 7 x G. Since it is enough that the condition 3) holds for all rational numbers, hence the condition 3) determines a Borel subset of 7 x Go . Moreover, the condition 4) determines a Borel subset of 7 x Go x IR from Lemmas 11.4.10 and 11.4.13. Therefore, E is a Borel subset of 1 x Go x R. From Proposition 11.3.3, we have Borel maps an () : I  (S, a), n =1,2, • • • , such that {a(M)In} is r—dense in (M) I ,VM G F. For any positive integers j, k, consider a subset E(j,k) of F x Go x R x A(D). (M,u,s,g) E E(j, k), if : 5) (M, u, s) E E; 6) g(t) = cp(f (t,u)ay(M)f (—t, u)ak(M)),Vt G JR; .
eo ...,
—
—
—
7) g(t + i) = p(ak(M) f(t,u)a i (M)f (—t,u)),Vt E JR, where p(.) = (6, to).
Since it is enough that the condition 6) holds for all rational numbers , hence the condition 6) determines a Borel subset of F x Go x A(D). The case of the condition 7) is similar. Thus, E(j,k) is a Borel subset of 7 x Go x it x A(D). Let rri be the projection from 7 x Go x JR x A(D) onto 7 x Go x JR, and 7r2 be the projection from F x Go x IR onto F. Then E0 =, 1r2(n1,oriE (i, k))
448 is a Sousline subset of I. By Lemma 11.4.14, s(Eo ) is also a Sousline subset of F. If M E .I;„, then from Proposition 6.6.6 there is a cyclicseparating vector e ce Hdi eii = 1) for M. Then we have u E G such that 6 is cyclic separating for uMu*. Now it suffices to show that E0
= fm G F.H16
is cyclicseparating for MI.
Let M G 7,„, and 6 be cyclic separating for M. Then p(s) = (.6, 6) is a faithful normal state on M. Suppose that {at it E R} is the modular automorphism group of M corresponding to p. Since M is type (III), from Theorem 8.3.6 there is s G R such that cr. is not inner. By the invariance of p for { at }, for each t E R we can define a unitary operator ut on H such that
ut aeo = at (a) 6,
Va E M.
Clearly,
ut eo = 6,
utaut = crt (a),Vt E R, a E M,
and t f u t is strongly continuous. If u is the Caley transformation of the generator of {Ut} I then u G Go and ut = f (t,u),Vt G R. Let gp, be the KMS function corresponding to (cp, {at }, ay (M), ak(M)). Then (M, u, s,gik) E E(j,k),Vj,k. Thus , M G E0 Conversely, let M E Eo , then there is a u E Go and asER such that —
(M,u,$) E n y,kir i E(j,k). Hence , M E F, 6 is cyclicseparating for M, {at (.) = f (t,u) • f ( t,u)} is an oneparameter strongly continuous * automorphism group of M, and cr,(•) is not inner. For any a, b E (M) 1 , we can find 
a,
ak (n)(M)
b.
For each n, since (M, u, s) E 711 E(j(n),k(n)), there is gn E A(D) such that g(t) = 40(at(ai(n)(Mnagn)(M)), gn (t + 1)
Vt E R. Noticing that f(t,u)e o = maximum modulus theorem we have ign(z)  gm (z) i —p 0,
6 and
f(t,u) E G,Vt E R, from the
uniformly for
z E D.
Thus, there is g G A(D) such that g(z) ,— g(z),Vz G D. Consequently,
g(t) = p(a t (a)b), g(t + i) = p(bat(a)),
447
Vt E R. By Theorem 8.2.10, fat (•):z. f (t, u) • f( — t, u)It E /RI is the modular automorphism group of M corresponding to p(.) = (• eo, to). Since a, is not Q.E.D. inner, it follows from Theorem 8.3.6 that M G .I;„.
From above discussions, we obtain the following.
Theorem 11.4.16.
Tin (n.
= 1,2,• • .), Y.,. )I:allF..1
and 1„, are Borel subsets
of F. Notes.
Proposition 11.4.15 is due to O. Nielsen. It is a key result for Theorem
11.4.16.
References. [120],
[154].
Chapter 12 Reduction Theory
12.1. Measurable fields of Hilbert spaces Let (E, B) be a Borel space. A complex valued function f on E is said to be measurable, if it is B—measurable. H(.) is called a field of Hilbert spaces over E , if for each t E E,H(t) is a Hilbert space. is called a field of vectors ( relative to H(.)) over E, if
e(t) G H(t),Vt E
e
E.
Definition 12.1.1. A field HO of Hilbert spaces over a Borel space (E, B) is said to be measurable, if there is a sequence } of fields of vectors over E such that : 1) the function (t))t is measurable on E,Vn,m, where (,) t is the inner product in H(t),Vt E E; 2) for any t G E,{en (t) In} is a total subset of H(t) ( in particular, each H(t) is separable). In this case, a field of vectors over E is said to be measurable ( with respect to the measurable field H(.) of Hilbert spaces ) , if the function (t is measurable on E, Vn. Denote the collection of all measurable fields of vectors by 0.
(en(t) ern,
{e„(.)
e(.)
( ) en(o) t
Proposition 12.1.2. Let HO be a measurable field of Hilbert spaces over a Borel space (E, B). Then: 1) from any n = oc, 0, 1,...,
En = ft E EldimH(t) = n1 is a Borel subset of E; 2) there exists a sequence fri n (.)1 C 0 with following properties: (a) if t E E with dim H(t) = oc, then fri n (t)In1 is an orthogonal normalized basis of H(t); (b) if t E E with dim H(t) = n < oc, then fn, MI. • • )nn(t)} is an
449
orthogonal normalized basis of H(t), and 77k(t) = 0,Vk > n; (c) e() E 0 if and only if the function ((t), q n (t)) t is measurable on E , Vn.
Proof.
Suppose that we have fni(),• •  ,qn(•)} c 0 with following properties: 1) if t E E with dim H(t) > n , then (77i(t),77i(t)) t = 45,V1 < i,j Ç_ n; 2) if t E E with dim H(t) = k < n, then fri i (t), • • • ,r/k(t)1 is an orthogonal normalized basis of H(t), and ni(t) = 0,Vk < i < n; 3) [(t)I1 < à _ n; 4) if G 0, then the function (e(t),ni(t)) t is measurable on E, V1 < i < n. For each t E E, let p(t) be the projection of H(t) onto [qi(t)I1 < t: < n ] . Clearly, if 0, then
e(.)
e(.) G
n t ' Prt(t)
e(t) i=1
is still a measurable field of vectors over For j > 1 , let
E.
n (t))e n+i (t) 0 0, and 1 It G El (1— p Fi = (1 — p n (t))G(t) =< n + 3. f 0,1
and
Foe := ft G El(1— p n (t))ei(t) = Ol Vil = ft E EldimH(t) .___ n1.
Since 11(1—Pn(t))6(t)Ilt is a measurable function on E, it follows that {Foe) F1 , F2 ) • • •} is a Borel partition of E. Let
0, tin Fi(t) = { ((1' 1— 11Pn (t) en+i (t) 11 '
if t E
Foe ,

if t E Fi '
i=
Clearly, {771(.),• • • ,qn + 1(•)} also satisfies the conditions (1)—(4). Repeating this process, we get a sequence fqn (.)In = 1,2,  • .1 of 0 . If e is a field of vectors over E such taht is measurable on E, Vi, then
(.)
(0 q t))t ),
(
i
(
(t )) en (o)t = D (t ), ni(o)t • (rii(t), en (0)t t
is also measurable on E,Vn, and further, e(.) E 0. Thus , the conclusion 2) is proved. Notice that II qi(t)li t is a measurable function on E,Vi. Therefore,
En = ft E Elrgt) 00,1 < is a Borel subset of E,Vn.
n; 74(0 = 0,j > n1 Q.E.D.
450
Definition 12.1.3. The rin (.)} is Propostion 12.1.2 is called an orthogonal normalized basis of the measurable field HO. Moreover, a sequence {S`n(.)} of measurable fields of vectors is said to be fundamental , if {sn (t) In} is a total subset of H(t),Vt E E. {
Proposition 12.1.4. Let lg.) be a measurable field of Hilbert spaces over a Borel space (E, B). 1) a field e(.) of vectors over E is measurable if and only if is measurable on E,Vn, where {sn} is a fundamental sequence of measurable fields of vectors over E. 2) if is a measurable field of vectors over E, then II (t) j, is measurable on E. 3) if () are two measurable fields of vectors over E, then ((t), ii (t)) is measurable on E. 4) let {sm (.)} c 0, and suppose that for each t G E, there is ç(t) E H(t) such that (WO —f G H(t). Then 0.) is also a measurable field of vectors over E.
(e(t), sn(t)),
e• e(),
?I
ot), ) t
Proof. 3) Let {71,1 G 0, then
(')
0,v e
} be an orthogonal normalized basis of H(.). If
e(.) ,r/ (.)
( ect) 1 n(t))t 7 E( e (t) 1 Tin(t))t • t
is measurable on E. 2) It is obvious from the conclusion 3). 4) Let { en (.)} be as in Definition 12.1.1. Then WO) en(t))t = li111 (57n(t), en (t))t is measurable on E, Vn. Thus sq•) E O. 1) The necessity is obvious from the conclusion 3). Now let ((t), (t)) be measurable on E,Vn, and {en (.)} and 0 be as is in Definition 12.1.1. Let 0' != {77 () I ( 77 (t) , s'n (t)) t is measurable on E, Vnl.
Then e (.) G 0', and {en (.)} c 0'. Applying the conclusion 3) to 0', ( (t ), en (t)), is measurable on E,Vn. Therefore, e(.) E 0. Q .E.D . Example 1. The constant measurable field of Hilbert spaces. Let (E, B) be a Bord space, Ho be a separable Hilbert space, and {en } be
a total subset of H0 . Define 11(t) = Ho ,
en (t) = en ,
Vt G E,
451
and 0 = { .)1(0) , en (o) t = ((t), )o is measurable on E,Vn1. This measurable field of Hilbert spaces is called the constant field corresponding to Ho . Clearly) e(.) G 0 if and only if ((t), ri) 0 is measurable on E,Vn E Ho. Consequently , 0 is independent of the choice of the total subset { en}. (
Example 2. Let A be a separable C*–algebra, and S(A) be its state space. Consider (S(A),a(A*,A)) as a Borel space. For each p E S(A), through the GNS construction we get a Hilbert space H. If fan 1 is a countalbe dense * an) is a subset of A, then {(an)„ } n is dense in Hp , and ((an) p ,(a) 0) = p(a,,,,, continuous function of p on S(A),Vn,m. Let H(p)=110 ,Vp E S(A), and 0 = feol(e(p), (a„)„),, is measurable on S(A),Vn1. Then we get a measurable field HO of Hilbert spaces. Clearly, e(.) G 0 if and only if (e(p),a p ),, is measurable on S (A), Va G A. Consequently, 0 is independent of the choice of the countable dense subset fanl.
Proposition 12.1.5. Let H(.) be a measurable field of Hilbert spaces over a Borel space (E, B). For n = oo,O, 1, • • ., define En = ft E EldimH(t) = n}. Suppose that Hn is a fixed n–dimensional Hilbert space, n = oo, 0, 1, • • • . Then there exists u(.) satisfying: 1) for any t E En ,u(t) is a unitary operator from H (t) onto Hn ,Vn; 2) e(.) E 0 if and only if for any n and ri E Hn , (0)e(t),on is measurable on En , where (,) n. is the inner prodeuct in H. Proof. Let {TA N} be an orhogonal normalized basis of the field H(.), and friin) 11 < k < n1 be an orthogonal normalized basis of Hn ,Vn. Define : u(t)rik(t) = qi n) ,Vt G En ,1 < k < n,Vn. Then u() satisfied the condition 1). Moreover, notice that e (.) E 0 if and only if (e(t),77k(t)) t is measurable on E,Vk. Thus , u(.) satifies the condition 2) also . Q.E.D.
Proposition 12.1.6. Let Ho be a countably infinite dimensional Hilbert space, and H(.) be a measurable field of Hilbert spaces over a Borel space (E, B). Then there exists u() such that for each t G E,u(t) is an isometry H(t) into Ho ; and t —b u(t)H(t)  (t) is a Borel map from (E, B) to W(Ho) ( see Proposition 10.1.8 ) . Moreover, e(.) G 0 if and only if (u(t) (t), 00 is measurable on E,Vri E Ho. Conversely , if HO is a field of Hilbert spaces on E, and for each t E E there is an isometry u(t) from H(t) into Ho such that t —p u(t)H(t) is a Borel map from (E ,B) to W(H 0) , then (H0,0) is measurable, where 0 = { e(.) 1(0) e(t),q)0 is measurable on E,Vri G Ho}.
Let H(.) be a measurable field over (E, B), fnn ()1 be an orthogonal normalized basis of H(.), and {n n } be an orhtogonal normalized basis of Ho. Proof.
452
For any t G E, define
nn
u(t)n n (t)
u(t)nn (t) = 0,
if
n < dimH(t),
if n> dimH(t).
Then u(t) is an isometry from H(t) into Ho ,Vt E E. If pn is the projection from Ho onto [ni, • • • ,nnj , then for any 77 E Ho we have 1l77 +0)11 (0110
11(1— pn)77110, vt
Ent
where En = ft E EldimH(t) = n}. Thus , 117/ + u(t)H(t)lb o is measurable on E. By Proposition 11.1.8, t u(t)H(t) is a Borel map from E to W(Ho) • Moreover, from
(u(00),o0 = E((t),qn(0), (uconn(0,00 and
JO,
n> dimH(t), (0)) nn(t))t = (u(t) (t) , ?in) ° , if n dimH(t), we can see that eo E 0 if and only if (u(t) (t) , 0 0 is measurable on E for any n of Ho. Conversely, let u(t) be an isometry from H(t) into Ho,Vt E E, such that t u(t)H(t) is Borel from E to W(Ho) . Denote the projection from Ho onto u(t)H(t) by p(t),Vt G E. Then for each e E Ho, the function + u(t)B (0110 = 11(1 p(t))0o is measurable on E. Further, (p(t)elti)o is measurable on E, ve,n E Ho. Suppose that {en} is a countable dense subset of Ho . Let en (t) = u(t)p(t)e n ,tit G E and n. Since {en (t)In} is dense in H(t),Vt E E, and ((t), em,(0), (p(t) en , ent) o is measurable on E,Vn,m, we can get a measurable field H(.) over E with Notice that (0) , en(t))t =__ (u(t) (t) , en) 0 for any field e(.) of vectors over E. Now by the density of { en} in Ho, we can see that e = fe(.)1(u(t) e(t), 77)0 is measurable on E,Vn G Hol. Q.E.D.
{ en(.)}.
Definition 12.1.7. Let (E, 8) be a Borel space, ii be a measure on 8, and H(.) be a measurable field of Hilbert space over E. Let
H(t)dv(t)
= fe(.)
G
el fE lie(t)gdv(t)
0 such that un = hn •
v.
465
H(t) and a sequence {„ (t) In = 1, 2, • • •} of vectors of H(t) such that ç(t) = 0 if h(t) = 0, and {ç(t)In with h(t) > 01 is an orthogonal normalized basis of H(t). Since (s`n (t), s,,,,(0) t :,finmX..pphn (t) is measurable on [1, Vn, m, H(.) becomes a measurable field of Hilbert spaces over rl with a fundamental sequence {sn(') 1m = 1, 2, • • •} of vector fields. We say that H(t) 0 {0 } , a.e.v. In fact, if there is a Borel subset E of û with v(E) > 0 such that H(t) = {0}, Vt E E, then hn (t = 0, Vn > 1 and t G E. Futher, v(E) = (hn • v)(E) .. 0, Vn > 1 and v(E) = >2 2  n vn (E) = 0, n>1 a contradiction. Thus, H(t) 0 {0 } , a.e.v. Let ff =l e H(t)dv(t), and define u : H  11 as follows For each t E fl, we construct a Hilbert space
n
umfqn = f Ohn(.) 112 s.n(.), Clearly, u is an isometry. If g ( .) E
tin > 1 and f G C(û).
ii satisfies
(f(•)hn (•) 112 sn (.),sq.)) = 0, Vn _.> 1 and f G C(û), then for each n > 1, we have h,(02 (s`n (t), s`(t)) t = 0, a.e.v.. By the definition of { gn ()} , we get ç(t) = 0, a.e.v . Thus , u is unitary. Moreover, it is easy to see that umfu* = fit' i , where r'nf is the diagonal operator on ri corresponding to f,Vf G C(û). By the proof of Theorem 5.3.1, we have also supp v = û and C(û) = L(û, v). Therefore, we obtain the following.
Theorem 12.4.1. Let Z be an abelian VN algebra on a Hilbert space H, Z' be a—finite, and û be the spectral space of Z. Then there is a regular Borel measure v on û with supp v = n, a measurable field H(.) of Hilbert spaces over
sl, and a
unitary operator u from
H(t) 0 {0}, a.e.v;
H onto
17 ,
je n
H(t)dv(t),
such that
c(n)
umfu* =_ r'nf , VI E L°°( 1,v), where f  mi is the Gelfand trandformation from c(n) onto Z, m 1 is the diagonal operator on if' correRonding to f. Consequently, uZu* = 2 , where 2 is the diagonal algebra on H. Theorem 12.4.2. Let H be a separable Hilbert space, M,Z be VN algebras and Z c A/ n Ad.'. Then there is a finite Borel measure on R, a measurable field H(.) of Hilbert spaces over R , a measurable field M(.) of
on
H,
e VN algebra on H(.) ,and a unitary operator u from H onto Hs = f H(t)du(t), R
such that
uzu* = Z, umu* = f e M(t)dv(t), R
466
where 2 is the diagonal algebra on ii. Moreover, if Z , M n Af' , then M(t) is factorial, a.e.v .
Proof.
From Theorem 5.3.7, Z can be generated by a self—adjoint operator a. Let A be the C*—algbera generated by {1, a} . Then A is weakly dense in Z, and the spectral space of A is a(a). By the proof of Theorem 12.4.1, we have also v, HO and u: H  h's" = such that umfu* = J rri f ,Vf G C(11), here I/ = a(a),f  m f is the * isomorphism from C(11) onto A. From Lemma 5.4.4 and the proof of Theorem 12.4.1, we can see that v >ve , \g` G H, where ve is defined by (m f e, f(t)dv e (t),Vf G c(n). Since supp v  [1, c(n) can be embedded into Lœ?n, v). Also by v › ve, Ve E H, f —+ m f (: C(il) —+ A) is a(L',V)weakly continuous. Thus, the map f —+ m f can be extended to a * isomorphism from Lcc(n, v) onto Z. Further, we have uZu* = 2. From Z CMn Mi, ZCMC Zi, we have 2 C uMu* C 2'. Now since
e) . f
ii = uH is separable, by Proposition 12.3.10 uMu* , f ED M(t)dv(t) is decomn posable. Moreover, Li, H() and MO can be trivially extended from (1= a(a) onto R. That comes to the conclusion. Finally, if Z = MnAl s, then 2,_ (uMu*) n (uMus) 1 . By Proposition 12.3.9, M(t) is factorial, a.e.v. . Q.E.D. Theorem 12.4.3. Suppose that il is a locally compact Hausdorff space, and Q is a countable union of its compact subsets. Let va , v2 be two regular Borel measures on fl, and Ha (.), H2 (•) be two non—zero measurable fields of Hilbert a) spaces over Q. If there is a unitary operator u from Ha = f Ha (t)dv i (t) onto n 9 112 = LI H2 (t)dv 2 (t) such that um(Pu* = m(2) f ' Vf G Cr (SI) , where mf(i) is the diagonal operator on Hi corresponding to f, t: = 1, 2, then va — v2 , and there exists a measurable field v(.) of operators from Hi(.) to H2 ( • ) such that v(t) is a) unitary from Ha (t) onto H2 ( t ) 1 a.e.va , and u = wv, where v = fv(t)dv a (t) n 9 , and w is the canonical isomorphism from fH2 (t)dv 1 (t) onto H2, i.e., if n v2 = p • v a , then
we(') = (1) 1 ) ('),
Proof.
\IV')
E
foe 112(t)dv1(t).
Let K be a compact subset with u1 (K) = 0. Since H2 ( • ) is non—zero, we can pick a measurable field ri(.) of vectors of H2 ( • ) with 117/ Mil t = 1, Vt E Q.
467
Suppose that U is an open neighborhood of K and the closure U of U is compact. Then (Xun)() E H2. Put e(•) = u*(xuq(.))(E HI ). For any e > 0, since vi (K) = 0, we can pick an open subset V such that
KcVCU,
and
f
Now let f E C43° (II) with 0 < f < 1; 1(t) = 1, Vt E K; 1(t) 0, Vt V V. Then we have v2 (K) < f f 2 (t) di/2 (t) = 11(fn)(t)11 dv2(t)
f
IIu* T42) (xun)( .)11 H2 1
1174) e( .)11 2H1 < e
*
vi . Similarly, Since 6 is arbitrary, it follows that v2 (K) = 0. Thus, 1/2 Vi ‹ V. Hence u1 V2. Now let H2 = 2 f H (t)dv i (t). Then Fra f = ym(Py*, where y = w*u, and
is the diagonal operator on 712 corresponding to f ,V f E c(gr (n). Since a) , by Theorem 12.2.10 y qx) (n) is we—dense in Lc. ° (1, j y(t)dv i (t) is f
n
decomposable. Moreover, y is unitary from HI onto 712 . Thus, y(t) is unitary Q.E.D. a.e.vi . from Hi (t) onto H2
(t)
Lemma 12.4.4. Let (Ed , B) be a standard Borel space, and u be a a—finite 1,2. If there exists a * isomorphism ir from Loe (El , Bi , v1 ) measure on onto L°° (E 2 , 82 //2) then there is a Borel subset Fi of Ei , I = 1,2, and a Borel isomorphism t from (E2\F2) onto (E, \F,) such that ,
vi (F1 ) = v2 (F2) = 0, u1 u2 o and for any g G
L°° (E i ,Bi , A i ), 71 ( g)(t) = g(4)(t)),Vt E (E2V2), a.e.v2.
From Theorem 10.3.16, we may assume that E I = E2 = [ 0 1 BI and 82 are the collectionof all Borel subsets of [0,1 ] ,and u, and v2 are two probability measures on [0,1]. Let fi(t) = t(E D1E1, v1)) and 12 = Lc° (E2, v2)). Clearly, we may assume that 0 < f2(t) < 1,Vt G E2 . Then f2(t) is a Borel map from E2 to El. The function 1 ( E L i (E2, v2)) determines a faithful normal state w2 on L(E 2 , v2 ), i.e.,
Proof.
w2 (h)
= fo h(t)dv 2 (t), Vh G L°° (E2,112).
Then col = (4)2 o ir is also a faithful normal state on L°° (E i ,
Thus there is
(Ei , vi) such that co i (g) = f f (t)g(t)dv i (t),Vg Also we may assume that 1(t) > 0,Vt E El .°
E L°°(Ei,vi).
unique
f E
468 If p( • ) is a polynomial , then we have 74(fi))(t)
= P(f2)(t) = P( 4) (t)), Vt G E2.
For any g E C[0, 1],pick a sequence fpnl of polynomials such that p(t) + g(t), uniformly for t E [0, 1 ] . Then in L' (E2, u2) , we have ir(pn (f,)) > ir(g). Thus ir(g)(t) = g(4)(t)), a.e.v2 . Further, 
fo' g(t) f (t)dv i (t) =
= w2 (7r (g))
=
,
o g( 4) (t)
)
dv2 (t),
f
Vg E C[0,1]. So 40(112) = f • vl , where 4)(v2) = v2 o For g E Loe (E l , v1 ), pick gn E C[0,1 such that gn 1> g. Then ]
L
1
iir (9n) (t)  Ir (9) (t) I 2 dv2(t) = w2( 7r ((9  9n) * (9  9n)))
= (4)1((gn 
g)
1
* (g  g)) =
fo f (t)i gn (o _ g (01 2 dv i (t) + O.
Since 1(t) > 0,Vt E E, we can find a subsequence {gn,} such that gnk (t)

+ g(t),
a.e.v i ,
and
lr(gnk)(t) > 7r(9)(t) , a.e.v2.
From 71 (g,,,,)(t) = gnk (4) (t)) ) a.e.v2, Vk, it follows that gn,(4)(t)) + 7r(g) (t), a.e. L2. On the other hand, gn, (t) + g (t), a.e.v i , 4) (u2) = f • vl and gn,(4)(t)) + g(4)(t)), a.e.v2, so we get 
7r(g)(t) = g(4)(t)), a.e.v21
Vg E
r ° (E 1 ) vi) •
Replacing 71 by 71 1 (: L'(E 2 , v2 ) + Loe(E i ,v1 )), there is a Borel map * from E I to E2 such that *(v1 ) = vl o xlv ' ‹ v2 , and (h)(t) = h(W (t)) , a .e.v i , Vh E r ° (E2 ) v2) •
Thus we have (go 4)) (11, (t))
(h 0 Alf)(4)(t))
== g(t), a.e.v i ,Vg E Loe (El , vi ); :z= h(t) , a.e.v2,Vh E Lœ (E2, v2).
In particular, from t E lic° (El, vi)
4) o W(t) = t,
n L'(E2, v2) we get
a.e.vi ; * o 40 (0  t,
a.e.v2.
So there is .17‘ E 8 2 with u2 (F) = 0 such that *o 4' (t) = t,dt G (E2 \F). Let FI = * 1 (F) . Then vi (F) = 0 since vl 0 41 1 ‹ v2 . It is easy to see that W(E i \FI) = E2\n
4:1)(E2\F) C
EMI,
(0
469
and (I) is injective on (E2\F2'). Further, pick Fr E Bi with Fr c (E i \FD and vi (Fn = 0, such that 4:1:0
°W (t) = t, Vt E Ei \fli ,
(2)
where F1 = FIU Fr . Clearly , vi (F0 = 0. Let Ff = (1)"(n) n (E2 \F). Then v2 (4) = 0 since 1/2 0 40 1 ‹ vi . Further, v2(F2) = 0, where F2 = .qU F41 =114 U (1:0  '(Fn. Then, (I) maps (E2\F2) into (E 1 \F1 ) injectively. Now let t E (E i \FO. By (1) , we have T(t) E (E2\F). If T(t) E fl C 4:1) 1 (Fr), then (I) o T(t) E 1111 . But by (2) , (I) o T(t) = t V Fi , a contradiction. Thus W(t) V Ff , and T(t) E (E2\F2). Further, by (2) we get (1)(E2V2) =(Ei \FO.
From Theorem 10.3.12 and Proposition 10.3.15, (I) is a Borel isomorphism from (E2\F2) onto (E i \FO, and its inverse is T. Now by v2 0 (1) 1 ‹ vil vi 0 Q.E.D. T'‹ 1/2, we can see that vi — v2 o (1) 1 on (Ei\FI)• Let (Ei , Bi) be a standard Borel space, vi be a afinite measure on Bi , Hi (.) be a nonzero measurable field of Hilbert spaces over Ei , Mi (.) be a measurable field of VN algebras on Hi(.), and Zi be the diagonal e algebra on Hi = fHi (t)dvi(t), I = 1,2. If there is a unitary operator u from Theorem 12.4.5.

E,
Hi onto H2 such that UMiu s
= 1.12, tai te = Z2
5
e
where Mi = f Mi(t)dvi(t),i = 1, 2, then there is a Borel subset Fi of Ei with Ei Vi(Fi) = 0,1 z 1,2, a Borel isomorphism (1:0 from (E2 \F2 ) onto (EI\FO, and a measurable field uH of unitary operators from H1 (.) to H2(411 (*)) 5 such that 1) u(t)M i (t)u(t) = M2(4)"(t)),Vt E (EIVI); 2) (I)(v2) = v2 0 (1)" — vi; ED ( dt (v2) (0 )1/2 u (t) dui (t). 3) u =Lv i i k dv i
By Proposition 12.2.8 and uZi u* = Z2 5 there is a * isomorphism 71Proof. from L(E I ,Bi , vi ) onto L' (E 2 , 8 2 5 112). From Lemma 12.4.4, we have Fi G Bi with v1(F) = 0, 1 = 1,2, and a Borel isomorphism (I) from (E2\F2) onto (E 1 \110 , such that (I)(v2 ) — u1 , and 7r(g)(t) = g(t(t)),tit E (E2 V2), a.e. v2 , and g E L'I.E 1 ,81 ,v1 ). Replacing E1 5 E2 5 v1 by (E i \Fi ), (E2\F2), ts(v2) respectively, with the Borel isomorphism we can identify (E1 , 8 1 , v1 ) with (E2 , B2 5 1/2). Now our case becomes the following. Let (E, B) be a standard Borel space, y be a afinite measure on B, Hi (.) be a measurable field of Hilbert e spaces over E,i = 1,2, and u be a unitary operator from H I , f H1 (t)dv(t) E
470
onto 112 f H2 (t)dv(t), such that tiMite M 2 2 UMV ) us
where Mi =
f Mi(t)dv(t),
= mV) VI G
(E B v),
4) is the diagonal operator on
Hi
corresponding
to f, s = 1, 2. By Theorem 12.2.10, u = f u(t)dv(t), a.e.v. From Proposition
12.3.7, there is a sequence {an = f an (t)dv(t)In} of decomposable operators on H1 such that M1 is generated by Zi and {ain}, and M1 (t) is generated by {a n (t)In}, a.e.v. Then M2 is generated by Z2 and fuan u*Inl, and M2(t) is generated by fu(t)a n (t)u(t)* In}, a.e.v. Therefore, M2 (t) = U(t)Mi (t) U(t) * 1 a.e.v. Q.E.D. References.
[28], [120], [158].
12.5. The relations between a decomposable Von Neumann algebra and its components Let (E, B) be a standard Borel space, v be a a—finite measure on B, HO be a measruable field of Hilbert spaces over E H = fEED H (t)dv(t), and M =
M (O&M be a decomposable VN algebra on H. In this section, we shall discuss the relations between M and M(t)'s. Proposition 12.5.1. Let p
of M, M' respectively. Then a)
ED
f p(t)dv(t),
=f
(t)dv(t) be projections
E
Mp = f M(t) pmciv(t),
a)
Mr), = f M (t) p r(t)d1/(t)
and e(p) = f: c(p(t))dv(t).
Proof.
From Proposition 12.3.7, we can get the expressions of Mr, and Mp t. Now suppose that M is generated by the diagonal algebra Z and a see quence {an n (t)dv(t) In} of decomposable operators, and M(t) is genEa erated by {an (t) In}, Vt E E. Through a suitable treatment, we may assume that {a n (t)In} is strongly dense in M(t),Vt E E. Let {em(')Im} be a fundamental sequence of measurable fields of vectors. Then fan (opmem (oln, rn}
471
is a total subset of c(p(t))H(t),Vt E E. By the method in Proposition 12.1.2, we can construct an orhtogonal normalized basis {rik()} of c(p(.))H() from Clearly, q k (.) is measurable, Vk. Then
(e (p(t)) en (t) , em (0), =
Den(t), nk(t)), • (77 k (t) , em (t)) t k
is measurable on E,Vn,m. Thus, the field c(p(.)) of operators is measurable.
e Put z = f c(p(t))dv(t). Clearly, E
zank(') = ang(•)) Vn and
e(.) E H.
0 Now write e(p) , f q(t)clv(t), where q(t) is a central projection of M(t),Vt G E. Since e(p)a nPE = anP,Vn, it follows that q(t)a n (t)p(t) = an (t)p(t), a.e.v,Vn, i.e., q(t) ? c(p(t)),a.e.v. Therefore, c(p) ? z, and c(p) =
So z >
LED
Q.E.D.
c(p(t))dv(t).
If M is discrete , then M(t) is also discrete , a.e.v.
Proposition 12.5.2.
By Theorem 6.7.1, there is an abelian projection p of M with c(p) = Proof. 1H . Further from Proposition 12.5.1, p(t) is also an abelian projection of M(t) Q .E.D. with c(p(t)) =1 H (t ), a.e.v. Therefore, M(t) is discrete, a.e.v.
If M is properly infinite, then M(t) is also properly
Proposition 12.5.3. infinite, a.e.v. Proof.
It is immediate from Theorem 6.4.4.
If M(t) is finite, a.e.v, then M is also finite.
Proposition 12.5.4.
Proof.
Q.E.D.
It is immediate from the definition of finite VN algebras.
Q.E.D.
If M is continuous, then M(t) is also continuous, a.e.
Proposition 12.5.5. V.
Put E k =, ft G EldimH(t) :7 kl, and let z k be the diagonal operator corresponding to xEk , Vk. Clearly, z k is a central projection of M,Mz k is conProof.
e
tinuous, and Mzk = fM(t)dv(t), Vk. Thus , we may assume that H(.) = HO Ek is a constant field over E. Suppose that M, M' are generated by the diagonal algebra Z and {an = f: an (t)dv(t) In}, fain ..., f: a'n (t)dv(t) In} respectively, and M(t),M(t)' are
472
generated by {a n (t) In}, {a in (t)In} respectively, Vt E E. We may assume that Ilan11)11411111an(t)11 and Mani(t)II .5._ 1, Vn and t E E, and fan(t)inl * = {a(t)In} , I n}* ., {a(t) I n}, Vt E E. {4(0 Let S be the unit ball of B(B0). Clearly, S is a Polish space with respect to the strong operator topology. Consider a subset G of S x E. (a,t) E G, if : 1) aa(t) ., a'n (t)a,Vn; 2) a is a non—zero projection; 3) aan (t)aa,,,,(t)a = aant (t)aa n (t)a,Vn, m. Noticing Proposition 10.3.14, G is a Borel subset of S X E. Let 1 be the projection from S X E onto E. Then from Theorem 10.4.5 there is a Borel subset F C 7rG, and a Borel map p(.) from F to S such that (p(t),t) G G,Vt E F, and (AG\F) C some v—zero subset. e Let p(t) = 0,Vt E E\F, and p = f p(t)dv(t). Then p is an abelian But M is continuous, so p = 0 and v(F) = O. Thus , ir(G) = ft E EIM(t) is not continouos } is contained in some y —zero subset, i.e., M(t) is continuous, a.e.v. Q .E.D .
projection of M.
Proposition 12.5.6.
If M is purely infinite, then M(t) is also purely
infinite, a.e.v.
Proof. With the same reason as in Proposition 12.5.5, we may assume that H(.) = Ho is a constant field. Keep the notations of fan , an (t), ain , 4(0 In} ) S and etc. in Proposition 12.5.5. Further, let (B0). = E 03H. Consider a n
E. (a, (n 0 , t) E G, if : 1) aa'n (t) = ain (t)a, Vn; 2) a is subset G of S x (H0) a non—zero projection; 3) arik = nk , Vk; 4) E Ilnk11 2 = 1; 5) for any finite sets k
Ai, A2
of positive integers,
E((arincA i anmallnien,a,„.(t)a — allmEn ,a,n (t)all nEA ,an (t)a)n,,no
= 0.
k
Clearly, G is Borel subset. Let 7r be the projection from S x (H0),x, x E onto E. From Theorem 10.4.5, there is a Borel subset F C AG, and Borel maps pH, (TAN) from F to S, (B0 ), such that (A G\F) C some v—zero subset , and for any t G F,(p(t),(rik(t)),t) G G. Define p(t) = Om k (t) = 0, Vk and t V F. Then p = LED p(t)dv(t) projection of M. We may assume that Y(E) L2 (R,v()) is a measurable field of Hilbert spaces over //i/ —, and (t —> 74) is a measurable field of VN algebras on (1 —> L 2 (R,fri)), where Zr is the multiplicative algebra on
L 2 (IR,vr)),V1 IRI . Since L2 (R, frr) is separable, there exists unique positive integer n(i) such is * isomorphic to Zna),Vt E RI , where pril is defined as in that Lemma 12.7.8. We say that (t —> n(0) is a Borel map from RI to IN . By Proposition 12.1.2, {ildimL2 (R, = k} is a Borel subset of RI Vk. Thus , we may assume that dimL2 (R, fr() is a constant, VF. Then (t —> L 2 (R,11)) is unitarily isomorphic to a constant field K over RI , and (t—> Zi..) is spatially * isomorphic to (t —> N(i)), where N(.) is a Borel map from RI to A(K). For fixed No E A (K) a(No) is a Borel subset of A (K) ( see Proposition 11.4.3 ). Then N1 (a(No)) is a Borel subset of RI . Further (t —> n(i)) is Borel. —
—
—
—
,
We can also construct a measurable field WU of * isomorphisms such that W(i) is a * isomorphism from Zit onto Zna),Vt E RI . In fact, from the preceding paragraph, (i > Zt) is spatially * isomorphic to (t N()), where N(.) is a Borel map from RI to A(K). Since the function 1(E L 2 (R,11)) is cyclic and separating for Zit, N(i) admits also a cyclicseparating vector in —
485
K, \WE Ifil — . By Corollary 5.3.9, we may assume that all N(t)(ii E RI —) are * isomorphic. Fix a ido E RI —. Then by Theorem 1.13.5, N(i) is spatially * isomorphic to N (t0 ) , VFE IR I — . Further, from Theorem 12.6.5 we can get the field TH. By Lemma 12.7.10, 4) is a Borel isomorphism from RI — onto M(JR) . Let k(A) = n o 4) 1 (4 VA E Wiri), and k(A) = 1,VA E (A\M(R)). Then k(.) is a Borel map from A to IN . Let 0 = i; o (1) 1 = u o M 1 . Then i. is a finite measure on A, and supp O c M(1?). In the following , the symbol " '" means the * isomorphism between VN algebras. Since 4) is a Borel isomorphism, by Proposition 12.3.12 and Lemma 12.7.10 we have
fAe (Aoz, (A ) dp(A ) =_f
=
e
MR ()
f ED
4' o 4' 1 ( A ) 0 Zk o 0 . 0 1( A ) dri o
i M(R)
, fe
R/,
(1110Zk(A))dr/(11)
e
4)(EYOZ TAdr/(0 ' fR/, (M(cr(kOZOdt)(0.
From Lemma 12.7.11, supp 11c 11 ({F}), a.e.iï. Clearly, M(t) = M(cr(0),Vt E 711 ({î}). Thus, e
L
M (s)clii(s) =
e
L. io) M (a(i))47(s)
= M (a (0)  0 7 47 ')
a.e.V.
e
Now by Lemma 12.7.12, M is * isomorphic to f (A OZk(A))cW(A). Noticing that supp P c M(R) c 7,„, we get A
fiie
(A OZk( A))(1P(A)
e
= E ED f e ( AOZ; ) di) ( A) = E. ED fA (24:ozi )dP(A), A .nli n i '
1
1
where Ai = {A E AI/C(A) = 3 } VI'. Further, pick A EED .M such that A = D on (A1 n 7„,) x {j}, Vj. Then , M is * isomorphic to f (A OZ n)c/A(A, n). 7,„xiv Again by Lemma 12.7.8, M is * isomorphic to B(A). Q.E.D. Lemma 12.7.14. Let E be a Sousline subset of A. Then a(E) = {N E AIN is * isomorphic to some element of El is also a Sousline subset of A. Proof. Define 4)(N) = N01 1H . Then 4' is a Borel isomorphism from A = A(H) into A(H 0 H). From the proof of Proposition 11.4.3, we have a(E) = 40 1 (s(4'(E))) = 4:0 1 (s(40(E)) n
4)(A)).
486
From Lemma 11.4.14, s(4)(E)) is a Sousline subset of A(H 0 H) . Then there is a Polish space P and a Borel map f from P to A(H 0 H) such that f(P) = s(t(E)) n CA). Now (to' o f is a Borel map from P to A. Thus , by Q.E.D. Proposition 10.3.5 a(E) = 40 o f(P) is a Sousline subset of A. Proposition 12.7.15.
A„, is a Borel subset of A.
By Lemma 12.7.5, it suffices to show that A„, is Sousline . From Proof. Lemma 12.7.8, B(.1/4) is a Sousline subset of A. By Theorem 12.5.10, we have B(.M) c A m . Again by Theorem 12.7.13, A m = a(B(.14)). Now from Lemma Q .E.D. 12.7.14, A„, is a Sousline subset of A. Theorem 12.7.16. A 1 ( the collection of all finite VN algebras on H) ,A 31 the collection of all semi—finite VN algebras on H) , Api ( the collection of all properly infinite VN algebras on H) ,A,, ( the collection of all type (4) VN algebras on H), k = oc, 1, 2, • • • , A, ( the collction of all type (I) VN algebras on H) , A, ( the collection of all continuous VN algebras on H), A„, ( the collection of all type (III ) VN algebras on H), A„. ( the collection of all type (II co ) VN algebras on H) , A„ (the collection of all type (II) VN algebras on H ) , and A„, ( the collection of all type (III) VN algebras on H) are Borel (
subsets of A. Proof. From Propositions 12.7.1, 12.7.2, 12.7.3 and 12.7.15, A ik , A„ Af and A„, are Borel subsets of A. For the case A31 , by Lemma 12.7.4 it suffices to prove that (A\A,f) is a Sousline subset of A. Notice that M E (A \ Asf ) if and only if M = MI ED M2
on and M2 is type (III), i.e., M is * isomorphic to a VN algebra (MI M2) H e H , where MI, M2 E A and M2 is type (III) . From Proposition 11.3.3 ) is a Borel map from A x A to A(H et H), and 11.3.4, (MI, M2) —> (Mi M2
and this map is injective obviously. Then by Proposition 12.7.15, E = {(
M1 M2)
M1 ) M2 E A, and M2 is type (III) }
is a Borel subset of A(HeH) . Let y be a unitary operator form H ED H onto H. Similarly, from Proposition 11.3.3 and 11.3.4 y • y* is a Borel isomorphism from A (H ED H) to A. Then by Lemma 12.7.14, (A\A,/ = a(vEvs) is a Sousline subset of A. Notice that M V A pi if and only if M = MI ED M2 and M2 is finite. Then by the same method as in the preceding paragraph, we can see that (A\Api) is Sousline. Moreover, M E Api if and only if there is a projcetion p of M such that 1 — p — (1 — p). Let S be th unit ball of B(H), and consider a
487
subset E of A xSx S. (M, v 1 , u2) E E if : 1) an (Miv i = vi an (M1), Vn, and j = 1,2; 2) vi v; = 1, j = 1,2; 3) * 1 • *2 = 0; 4) * 1 = 1, where { an (.) : A Sin} is as in Proposition 11.3.3. Clearly, E is a Borel subset of AxSxS. Then Api is Sousline. Further, Api is a Borel subset of A. By Theorem 6.8.4 and the same method, we can see that A„ is Sousline. Moreover, M A„ if and only if M = M1 ED M2 and M2 is either type (I) or type (III) . Then we can prove that (A\A„) is Sousline. Thus , A u is a Borel subset of A. Further, ;L i = A„ n A f and A„. = A„ n AO are Borel subsets of A. Finally, since A, = A„ U A 11, U {M E AIM = M2 M3 and M2 , M3 are type (II) , (III) respectively , A, is Sousline. Moreover, M V A, if and only if M = M 1 ED M2 and M1 is discrete. Thus, (A\A e ) is Sousline , and A, is a Borel subset of A. Q.E.D. Notes. Lemma 12.7.5 was proved by J.T.Schwartz, and the hard result, Propositon 12.7.15, is due to 0.Nielsen.
References.
[36], [120], [121], [154], [177].
12.8. Borel subsets of the state space of a separable C* —algebra Let A be a separable Ce—algebra with an identity, and S(A) be its state space. By a(A* , A), S(A) is a compact Polish space. In fact, let {an} be a countable dense subset of the unit ball of A, define d(P) (an) I,VP, =
E
ti) E S(A). Then d is a proper metric on (S(A),a(A*, A)).
For any n = oo, 1, 2, • • • , let Hn be n—dimensional Hilbert space, and Rn be the collection of all nondegenerate * representations of A on II. In Rn , the topology is defined as follows: 7ri ir if 11( 70) —7 (a)) eii —> Va E A, E Hn . By Lemma 12.6.1, Rn is a Polish space, Vn.
Lemma 12.8.1. 4': ir(A)" is a Borel map from Rn to A n, where is the collection of all VN algebras on Hn ,Vn.
An
Proof. Let {a k } be a countable dense subset of the unit ball of A, and define ak(.) : Rn > (B(Hn),a), ak(ir) = 1 (ak),V7r E R n and k. For any e, ij E Hn , (a k(.) , q) is continuous on R n obviously. Thus, {ak(•)} is a sequence of Borel maps from Rn to (B(Hn),a). Moreover, for each 7r E Rn , 7r(A)" is generated by fak(7010. Now by Proposition 11.3.4, (I) is a Borel map from Rn to A. Q.E.D. —
488
Proposition 12.8.2. R) = {7r E Rn I7r(A)" is a type (t) VN algebra on HnI is a Borel subset of R,„ Vn, where (t) is factorial, finite, semi—finite, properly infinite, (Ik)(k = 00 ,1,2,...) , ( I) , (II 1 ), ( II) , (II) , (III) , (c) ( continuous ) , or (irr) ( irreducible). Except the case of (t) = (kr), the conclusions are obvious from Theorem 11.3.6, Theorem 12.7.16 and Lemma 12.8.1. Now let (t) = (kr). Notice that 7r E R$!) if and only if 7r(A)" = B(Bn), i.e., the unit ball of 7r(A) is strongly dense in S, where S is the unit ball of B(Hn). Let 45 be a proper metric on the Polish space (S, strong top.) , {b k } be a countable dense subset of (S, 5), and {ak } be a countable dense subset of the unit ball of A. Then Proof.
R$!)
= n icip
um
E Rnifi(7r(arn.),bk) cto > fo (r). So z E Uf octo and y V Ffoao . Therefore, {z} =
n{Ffa l(f, a)
E L.},
Vx E P.
Now let z E P, and we prove that {Fia l(f ) a) E L z } is a basis for the neighborhood system of z in K. Indeed, let U be an open subset of K, and z E U. Since {z} = r1{Ff a l(f, a) E L.} c U, it follows that (K\U) C il{(K\F fa )1(f, a) E Lz l. By the compactness of (K\U), there exist (fi , a i ), • • • , (fa, an) E L z such that (KW) C LI7_ 1 (K\Fi), where Fi = Ffi ,„; , 1 < i < n, i.e., U D n7_ 1 Fi . Let Ui = Uh,„;,Ki = K\Ui,1 < i < n. Then F a Co iltit, I Ki is a compact subset
494
of K. Clearly, x E Ui and x V Ki ,\V1 < i < n. So by x E P, we have x V F. Then we can find f E E* and a E JR such that f(z) a} is convex (VI E E*,a E R) , and the sequence {(K\Ffn ,an )In} is increasing, we can see that (KV) is also convex. Since F nP cUnn n Un nP=Vnn n Vn , it suffices to show that F n P L O. Indeed, let x be an extreme point of F. If x is extremal in K, we are done. If not , let 5 be a line passing through x and such that x is an interior point of the segment K n 5, we then can show that one of the endpoints of F n 45 is Q.E.D. an extreme point of K since F and (KV) are convex. Lemma 13.1.9. Let X be a Baire space, Y be a topological space, and T be a continuous open map from X onto Y. Then Y is also a Baire space.
Let {Vn be a sequence of open dense subsets of Y, and V be any open subset of Y. We need to show that nnvn n V O. Let U = 714 (V),Un = T 1 (Vn),Vn. If W is any open subset of X. Then TW is also open. Hence, TW n Vn 0, dn. Let x E W and Tx E Vn . Then x E Proof.
}
495
W nT 1 (Va) = W nUn . So W n Un 0, and Un is an open dense subset of X, Vn. Now U n nnun 0 since X is Baire. Therefore, V n nn Vn = T (U n nn un) O. Q.E.D.
Lemma 13.1.10. Let A be a C*algebra, {{7ri,11/ }11 E Al be a family of nondegenerate * representations of A, I = n{ker7r1 11 E A}, and p be a state on A with plI = O. Then p belongs to the a(A* , A) closure of following subset of A* : coOri 06, 6)16 E Hi and 1161H 1,1 E AI Let 7r = e1E,011,1/ = elE AHL. Then fr, HI is a faithful * representation of Ai/. By plI = 0, p can be regarded as a state on A//. Now from Lemma 16.3.6 (it is easy and elementary), p is a a (A* , A) l imit of states which belong to the following subset: Proof.
Co{( 7r(*), .)1 . E II,
Ha = 11. Q.E.D.
That comes to the conclusion.
Let A be a C*algebra, P(A) be the pure state space of A, and p i , p 2 E P(A).p i and /32 are said to be unitarily equivalent, and denoted by pl /22 , if there exists a unitary element u E (A4FŒ) such that pi(a) = p2 (u*au),Va E A. —
Proposition 13.1.11. Let A be a C*algebra, p i , p2 E P(A), and {7r,, I/1 , e,}, {72) H2, W be the irreducible cyclic * representations of A generated by p,, P2 respectively. Then we have that PI — p2 4—>. {7ri , Hi} '=s fr21 H21 .,1=. there is n G 1/1 with Ilnll = 1 such that p2 (a) = (7ri (a)n, 0,
Va E A.
Proof. Let pi(a) = P2(u*au),Va E A, where u is a unitary element of (AŒ).
Define Uri (a) ei = 7r2 (au)e2,
Va E A.
Then U can be uniquely extended to a unitary operator from HI onto H2 1 and Ur i (a)U* = 7r2 (a), Va E A. Therefore, {7ri, HI} "=' {72 ) 112}. Conversely, let U be a unitary operator from 1/1 onto H2 such that thr i (a) U* = 7r2 (a),Va E A. Then we have that p2 (a) = (r i (a)17,17),Va E A , where 77 = U* 6 E HI, and Ilnll = 11611 = 1. Further, we can find a unitary operator
496
V on I/1 such that V 1 = n By Theorem 2.6.5, there exists a unitary element u of (A4Œ) such that 711 (u)6 = 77. Therefore, we get .
P2(a) = (7ri(u * au)6, ei) = p,(u*au), va E A.
Q.E.D. Theorem 13.1.12. Let A be a C*algebra. Then (P (A), a (A* , A)), Â and Prim(A) are Baire spaces.
O111111 subset of (A* , a (A* , A)). By Lemma 13.18, (E rlf, a (A* , A)) is a Baire space. From Proposition 2.5.5, we have that ExIC = P(A) U {0 } . Now it is easy to see that (P (A) , a (A* , A)) is a Baire space. By the GNS construction, p r p is a surjective map from P(A) to Â. We claim that the map p ri, is continuous. Indeed, by Proposition 13.1.5 any open subset of Â has the form of Â', where I is some closed twosided ideal of A. Then inverse image of Al under that map is as follows: Proof.
U={pE
P(A)Iirp E AI 4= rid/
0}
Let p E U. If for any a (A* , A)—neighborhood U(p,F, 1) = fço E P (A)11p (x) — cp(x)1 < 1,Vx E Fl of p in P (A), where F is a finite subset of A, there is PF E U (p, F,1)\U . Then we have that pp + p in a (A* , A) and r pp lI = 0, VF. Consequently, pp(axb) = 0, Vx E /, a, b E A, and F. Thus we obtain that p(azb) = 0,Vx E /, a, b E A, and rid/ = 0, a contradiction. Therefore, U is open, and the map p + r p is continuous. Now let U be an open subset of (P(A),a(A*, A)), and V = {r E AI there is p E U such that 71"'.=' 7r I) . For each subset E of P(A), put }
E/ = {cp E P(A) I there is p E E such that q' p}. .. such that Clearly , V = {7r E AI there is p E 0 element u of (AiŒ), let
7r '''' 711,}.
For each unitary
u(U) = fp(u* • u)lp E Ul.
Clearly, u(U) is open. By Proposition 13.1.11, '0 = IA u(U)lu is a unitary element of (A10)}
is also open. Let F = P(A)\U. Then F is closed and F = i. Let I = nfkerri, lp E Fl. If cp E P(24) and ker r 3 ./(41. cp1/ = 0), then by Lemma 13.1.10 and F = P we have that cp E C  Y'r . Since F = .141` and cp is also an
497 extreme point of Cac'7 , it follows from the KreinMilmann theorem ( see [89, Theorem 15.2] ) that ço E F. This means that
{7r E AI there is p G F such that 7r''='' 7r p } is a closed subset of A. Therefore, V is an open subset of Â, and p + 7r p is
an open map. Now by Lemma 13.1.9, ii. is a Baire space. It is wellknown that ker : Â. + Prim(A) is a continuous and open surjection. So Prim (A) is also a Baire space from Lemma 13.1.9. Q.E.D.
Proposition 13.1,13. Let A be a C*algebra, and x E A. Then 7T + 117(x)11 is a lower semicontinuous function on Â. For any k > 0, we need to show that
Proof.
E = Ir E A111 7r(z)11 5 k} = { 7r E :4 111 7r(ex)11
k2 }
= {7r E 2.12917r(x*x)) c [k 2 ,k 2 ]} is a closed subset of A. Let a = e x, L = [k 2 , k 2 ], and I = rl7r1 17r1 E El. Suppose that 7r E .E, and there is some A E JR with A G ol7r(a)AL. Pick a continuous function f on JR such that f IL = 0 and f(A) O. Then we have that 7e(f(a)) = f (7r 1 (a)) = 0,V7e E E, i.e., f (a) E I; and 7r( f (a)) = f (7r (a)) O. Let T = {.I E Prim(A)IJ D /}. Then T is a closed subset of Prim(A), and ker1 (T) is a closed subset of Â. Clearly, E C ker1 (T). Since 7r E E, it follows that 7r E ker 1 (T) and ker7r D I. Hence , we get f (a) E I C kerr and 7r(f (a)) = 0, a contradiction. Therefore, E must be closed. Q.E.D.
Lemma 13.1.14.
Let H be a Hilbert space, 6, • •  , .rs G H, and 6 > O. Then there exists 45 > 0 with the following property: if th,  • • , nn E H such
that 1(ni,tii) — (nimi)
I
a.
Therefore , for each a> 0, {7r E Altr7r(x)> a} is an open subset of Â , i.e., 7r + tr7r(x) is lower semicontinouos on A. Q.E.D. References. [27], [46], [73].
13.2. Elementary C*algebras and CCR ( liminary ) al
gebras A C*algebra A is said to be elementary isomorphic to C(H), where H is some Hilbert space.
, if A is *
Definition 13.2.1.
Proposition 13.2.2. Let A = C(H) be an elementary C*algebra. 1) Each positive linear functional on A has the following form : E Ai (•ei , i
, where {&} is an orthogonal normalized sequence of H; Ai > 0, Vi; and
ei)
E Ai < I
00.
2) Each pure state on A must be the following form : and 110 = 1 
(e, e),
where
EH
3) 4 :4.' = 1.
4) A is simple , i.e., if I is a closed twosided ideal of A , then I must be either {0 } or A. 5) If B is a C*subalgebra of A , and B is irreducible on H, then B = A.
502 6) If B = C(111) is * isomorphic to A, then there exists a unitary operator
U from H onto H' such that U*BU = A. Proof. 1) It is immediate from C(H)* = T(1/). 2) The conclusion is obvious by 1). 3) Let p(.) = (.e, e), a(.) = be two pure states on A, where e, t, E H and Hell = Hid = 1. Since A is irreducible on H, from Theorem 2.6.5 there is a unitary operator u E (A4OE) such that tt." = n Then we have a(.) = p(u*  u), i.e., p — a . By Proposition 13.1.11, { ro , Hp } and {7r„, H,} are unitarily equivalent. Therefore , #.2i =1. 4) If I is a proper closed twosided ideal of A, then we have I = np E Prim(A)IJ D 11. By 3) , Prim(A) = {0 . Therefore, I = {0 } . 6) Let 40 be a * isomorphism from A = C(H) onto B = C(II') . Then {id, H} and {40, H'} are two irreducible * representations of A. By 3) , 40 must be spatial. 5) Let B be a C*subalgebra of A = C(H), and B be irreducible on H. Then B is weakly dense in B(H). If t E T(H) = C(H)* is such that tr(tb) = 0,Vb E B, then we also have tr(tb) = 0,Vb E B(1/). Since T(H)* = B(H) , it follows that t = O. Therefore, B = A. Q.E.D.
oh n
)
.
}
Now we consider C*algebras of compact operators, and the C*subalgebras of an elementary C*algebra. Proposition 13.2.3.
Let A be a C*subalgebra of C(11), where H is some
Hilbert space. 1) If p is a nonzero projection of A, then the rank of p is finite , i.e., dim pH < oo; and p is an orthogonal sum of minimal projections of A. Moreover, p is minimal in A if and only if pAp =Op. 2) A = [Proj(A)], where Proj(A) is the set of all projections of A. 3) Let p be a minimal projection of A, e E pH with R11 = 1, and K = [A t]. Then A is irreducible on K, and AK = C(K). Proof.
Since each slefadjoint element of A has the form of
E )pi , where
{pi } is an orhtogonal sequence of projections on H; Ai j; and Ai :Vi A, + 0, it follows that A = [Proj(A)]. Now if p E Proj(A), then we also have pAp = [Proj(pAp)]. Hence , p is minimal in A if and only if pAp = Op. Now let p be a minimal projection of A, E pH with Heil = 1, and K = [At]. For any b E (AIK)' and s,t E A, we have (WIte)
e) = A(be, e) = (se, te)  (be, e), = (PebsPe, 0 = (Pt * sPbel
503
where pt* sp = 4 and A E 0 ( since p is minimal) . Thus b = (be, ) 1K , i.e., A is irreducible on K. Further, from Proposition 13.2.2 we have Alif = C(K). Q.E.D. Let {Ai 11 e 11} be a family of C*—algebras. lima/ = 0, where al E 241 , VI E I, means that for any e > 0 there is a finite subset Fe of I such that liai II cx i — i 2 el
and V3 { 7r E
AI tr7r(z) > a2 — —63 }
are open subsets containing 7r0 . Then V = V 1 n V2 n V3 is a neighborhood of 7r0 in A. If 7r E V is such that tr7r(y) > ai ± E, then
tr7r(x) = tr7r(y) ± tr7r(z) >
al + a2 + 6/3 = a + 6/3,
a contradiction. Thus, we have that 2 2 cx1 + —3 e > tr7r(y) > al — —3 6,
Mir E V.
Since e is arbitrary, tr7r(y) is continuous at wo. 2) We may assume that 7ro(z)11 = 1. Since the operator 7r0 (x) is trace class, 1 is an eigenvalue of 7r0 (x). Further, there is an one rank projection p on Ho such that 7r0(x)p = p. From Lemma 13.2.6, we have C(H0) C ro (A). So we can find z 1 E A+ such that 7r0 (z i ) = p. Let / t, f(t) = 0, 1,
if t E [0,1], if t < 0, if t > 1,
511
and z2 = f (z i ). Then z2 E A+, 11z211 5 1, and 7r0 (z2 ) = f (p) = p. Further, let z3 = x 1 '2 z 2 x 1 ' 2 . Then 0 < z3 < z, and 7ro (z3) = Iro(r) 1/2P7r0(x) 1/2 = From the preceding paragraph, tr7r(z3 ) is continuous at 7r0 . So there is a neighborhood V1 of 7r0 in A such that tr7r(z 3 ) < 5/4, V7r E Iii. By Proposition 13.1.13, we also have a neighborhood V2 of 7r0 in Â such that 117r(z3 )II > 3/4, V7r E V2. Let V= Vi n V2. Then Pr(z3)II > 4/3)
and
tr7r(z3 ) < 5/4, V7r E V.
Since 7r(z 3 ) is nonnegative and trace class, 7r(z3 ) has only one eigenvalue A, with multiplicity 1 and A, > 3/4, and other eigenvalues of 7r(z3 ) belong to [0,1/2), V7r E V. Now let z = g(z 3 ), where 0 , if t < 1/2 00 = f .— ' 1, if t > 3/4, and g is continuous on IR. Then 0 < z < z3 , and 7r(z) = g(7(z 3 )) is the spectral projection of one rank corresponding to the eigenvalue X, of 7r(z3 ),V7r E V. Q.E.D. Proposition 13.3.11. Let A be a C*algebra which is not NGCR. Then there exists 0 0 z E A+ such that the rank of 7r(z) is either 0 or 1 , Vir E Â.
Since A contains a nonzero CCR closed twosided ideal, by Lemma 13.3.9 we can find 0 0 y E A+ such that the rank of 7r(y) is finite, \/7r E A. Now tr7r(y) is nonnegative finite lower semicontinuous function on Â. By y 0 0,{7r E :4'Itr7r(y) > 0} = U is a nonempty open subset of Â. From Proposition 13.1.5 and Theorem 13.1.12, U is also a Baire space. From Lemma 13.3.8, tr7r(y) will be continuous at some 7r0 E U. Now by Lemma 13.3.10, there is some z E A+ and some neighborhood Al of 7r0 , where I is a closed twosided ideal of A, such that the rank of 7r(z) is one, V7r E AI . We claim that /z 0 fol . Otherwise, let J = [AzA]. Then J n I = [I.I] = fol. From Proposition 13.1.5, it follows that :4' . n A/ = O. Thus, we get 7r(z) = 0,V7r E .A./ , a contradiction. Now let wz 0 0 for some w E /, and x = z*w*wz. Then 0 0 x E A+ n /, and clearly the rank of 7r(z) is either 0 or 1 , V7r E .ii/ . Further, for any 7r E A the rank of 7r(z) is also either 0 or 1 . Q.E.D. Proof.
Proposition 13.3.12. Let A be a C*algebra. Then A is NGCR if and only if A satiesfies the Glimm condition, i.e., for any 0 0 z E A+ , there is an irreducible * representation {7r, H of A such that dim 7 (x)H > 2.
}
The sufficiency is obvious from Proposition 13.3.11. Now let A be NGCR. If there exists 0 0 z E A+ such that the rank of 7r(z) is either 0 or 1 Proof.
512
, V7r E A, let J = [AzA ] , then J is a non—zero CCR closed two—sided ideal of A. This is impossible since A is NGCR. Therefore, A must satisfy the Glimm condition.
Q.E.D.
Reference. [23], [47], [55], [84].
13.4. The existence of type (III) factorial * representations of a NGCR algebra Proposition 13.4.1. Let A be a NGCR algebra. 1) For any 0 0 h = h* E A, there is an irreducible * representation {7r, HI of A such that dim 7r(h)H > 2. 2) For any 0 0 a E A and a*a = aa*, there is an irreducible * representation {7r, H} of A such that dim 7r(a)H > 2. 3) If A has no identity, (A4) is also NGCR. 1) Let h = h+ h_, where h+ and h_ E A+ , and h+ • h_ = O. We may assume that h+ 0 O. Then by Proposition 13.3.12, there is an irreducible * representation {7r, Ill such that dim 7r(h + )H > 2. Pick n11/72161 e2 E H such Proof.
that
(ci, ci) = bii,
and
7r(h÷ )tii =
ei , 1 < i,i
dim[7r(h)thli = 1,2] = 2. 2) Write a = h l + 1 h2 , where h: = hi E A,i = 1,2, and h1h2 = h2h 1 . We may assume that hl 0 O. By 1), there is an irreducible * representation {7r, HI of A swirl that dim 7r(h i )1/ > 2. Pick n1,n2, 6, e2 E H such that and
7r(hOti1 =
ei , 1 < i,j
dim[r(a)ni li = 1,2] = 2. 3) Let I be a non—zero CCR closed twosided ideal of (A40). Clearly, it must be that A n / , {0}. Thus , we have / = [e — 1] for some e E A. Since AI = IA = {0 } , e is an identity of A, a contradiction. Therefore, (A2F4J ) is Q.E.D. also NGCR.
513
Lemma 13.4.2. Let A be a NGCR algebra with an identity 1, d E A + with = 1, and t E (0, 1]. Then we can find w, , E A such that 1) MwM = Mug = = 1,w > 0, d' > 0, and w"w = 0; 2) ft (d)w = w, ft(d)w' = Iv% 3) w 2dr = di , t e w fdr = wher
0, ft (r) = / affine, 1,
if r < 1 t, if 1 – t < r < 1 – 1, (0 < t < 1) if r > 1 –
Proof.. Let s = 03, pick u,c E A with Mull < 1,0 < c < 1, and put do = h8(d)cf28(d), di = Lis(d) – do . Clearly, 0 < do < 1,1 < di < 1. Since f4sf2s = f2., if follows that fo(d)do = do = hs(d). Then {1A, i48(d)} can generate an abelian C*—subalgebra B of A. Let B .=' c (n) . By i48(d)d0 – do, we have
P( f48(d)) = 1, P(c11) = 1 – V p E 11 and p(do ) 0. Hence, if g : 1R —+ 1R is continuous and g vanishes on [0, }], then by 0 Ç MO :5_ 1 we have p(g(c1,0)g(d 1 )) = g(p(d o))0p(d i )) = 0,Vp E n. It follows that g(d i )g(do) = 0, in particular, f2 (d 1 )f2 (d0 ) = 0. Let y = fs (d i )uf3 (4). Then we have
IIM
= 1 ) vv = is(do)u * fs(di) 2 01,(do).
and f2s(do )v* v = tf` v
=
vf23(4), v* f2 .(di)
= v* ,
v* (v*v) = v* f2, (di ) f2, (do )v*y = 0. Furthermore, fa,,(d)do = do and f83 (d)d 1 = d1 , hence f83(d)p(do) = P(d0),
f83 (d)p(d i ) =
if p is a polynomial with no constant term, hence also if p is a continuous function vanishing at 0. In particular f83 (d)y = y, and f88 (d)y* = v*. Finally, put d' = f . 4(v* v) , w = f 12 (v*0 1/2 , w' = vk(v* 0 1 where k : JR —+ JR is the function which is equal to (fi (t)t') 1 / 2 if t 0, and to 0 if t = 0. Clearly, 0 < d' < 1,0 <w < 1. Since v*(y* 2y) = 0 and fl/2(0) = 0, it follows that wi*w = 0. Since f83 (d)v = v, f83 (d)v* = v*, we have f83 (d)v*v = vv, f8 8 (d)w = w,
f83 (d)w' =
That comes to 2) . Further, = /1/2(V V) ) ID
f*
7.0 =k 2 (V * V)V * V = f112(V*V)
514
and w 2 d = hi2(v 4i v)fi/4(v * v) = d' and similarly w"w' = d'. That comes to 3) . If we can choose u, c such that VII > 1, then it must be 1141 = 1. Since w 2 > d' and 1 > w"w` > d', it follows that 11w11 = Ilte11 = 1. That comes to 1) , and the proof will be completed. Now it suffices to find u,c such that 11 d' 11 > 1. By the Glimm condition, there is an irreducible * representation {7r, HI of A such that dim r(f,(CH > 2. Pick e, ri E 7r(f3 (d))11 with Ileil = 11n11 = 1 and (e,t7 ) = O. By Theorem 2.6.5, there exists h* = h E A such that 7r(h)e
=
e,
7,(h),
O.
Let g : JR + JR be the function which is equal to 0 if r < 0, to 1 if r > 1, and to affine on [0, 1]. Pick c = g(h). Then 0 < c < 1 and 7r(cg = e,r(c)ri = 0. By Theorem 2.6.5, there is also a unitary element u of A such that
Since f2313 = Is and e,t7 E 71 (f „(d))H, it follows that 711/23(d)g
= e, 7r(f23(d))/7 =
n.
Similarly, 7r (f 4a (C17 = ri. Hence, we have r(do) = 7r(f28(d))7(c)74/28(d)g = e, 7,(4), = 0,
7
.(dot, = 7r(f43(d))77 — 7r(07
=17,
7r(v*oe = 7r(f3(do))7r(urr(fmo)27(07(f8(do))e 71(d)e = 71(t1 / 4(viov))e = f 1 / 4(1)e = e.
=
Therefore, 1141 ? 11 74 (011 > 1.
e,
Q.E.D.
Let A be a NGCR algebra with an identity 1. Then there exist nonzero elements v(ai , • • • , an) and b(n) in the unit ball of A, where al, • • • , a. E {0, 1 },n = 0,1,2, • • , with the following properties: 1) if j < k and (al , • • • , ai) 0 (/3I , • • • ,fli ), then Proposition 13.4.3.
v(ai ,  • • , ai)*v(f3 1 , • • ,fik ) = 0;
2) if k > 1, then v(a l ,• • • , ak) = *ell • • • )ak1)v(Ok1lak); 3) if j < k , then v(a l , • • • ,ai rv(a i , • • • ,ai )v(O k _ i ,ak ) = v(O k _ i , ak );
4) v(0) = 1, v(0k) > 0;
515
5) v(oc i , • • , an )* v (a l , •  • , an )b(n) = b(n), b(n) > O 0, 1, 2, • .
,
and lb(n)11 = 1,n =
For n = 0 , put v(0) = b(0) = 1. Now suppose that nonzero elements v(a 1 , •  , ai) in the unit ball of A and b(j) of norm 1 in A + have been constructed for j < n and they satisfy these properties. Using Lemma 13.4.2 to d = b(n), we get w,w 1 and d', then let Proof.
v(O
1 ) = w, v (On ,
and
= w`,
b(n + 1) = .
By Lemma 13.4.2, we have v(a i , • • , an )* v(a i , • • , an)v(On , an+1 ) =
*E l , • • • , anrv(ai, • •  , an) ft(b(n))v(On, an+i)
= ft (b(n))v (O n , an+ 1 ) = Since b(n) = v (O n )* v(O n )b(n) = v(O n ) 2 b(n), it follows that v(On ) 2 b(n) = b(n) v(o n) 2 , b (n) v (O n ) = v(O n )b (n) , and b(n) 2 = (v(O n )b (n)) 2 . Hence , b (n) =
v(O n)b(n) and v(On)v(On , an+i ) = v(O n )ft (b(n))v(O n , an+i ) =
Then for j n, and Dn = B(n)p(n ± 1) is * isomorphic to Bn. }
519
Further, let A(n) be the C* subalgebra of A generated by {B(i)Ii < n } . From (1) , we have A(n)P(n +1) = B(n)P(n +1) = D.
Let I(n) = fx E A(n)Ixp(n + 1) = 01. Then I(n) is a closed twosided ideal of A(n), and A(n) I I(n) is * isomorphic to Bn ( the 2' X 2' matrix algebra) . Clearly, A(n) c A(n+ 1), and I(n) C I (n + 1). But A(n) I I (n) is simple , so that A(n) n I (n + 1) = 1(n). Now let B be the C*sebalgebra of A generated by iln A(n), and I be the closure of iln/(n). Then I is a closed twosided ideal of B. Moreover, since A(n) I I(n) is simple, A(n) n I = I(n),Vn. Consider the quotient algebra Bg. By A(n) I I (n) = A(n) I (A(n) n I) (A (n) FI)/I,BII =il(A(n)+I )/ Iisan(UHF) algebra of type (2°, 2 1 , • • • , 2n ,  • ). Therefore, we have the following. Let A be a NGCR algeba with an identity 1. Then there is a separable C*subalgebra B of A with 1 E B and a closed twosided ideal I of B such that /3// is an (UHF) algebra of type {2n}. Proposition 13.4.4.
Lemma 13.4.5. Let X, Y* be two Banach spaces, Y = (Y*)*, and B(X, Y) be the Banach space of all bounded linear operators from X to Y. Then through the following way of(x 0 f) = (Tx, f)
Vz E X, f E 17.,T E B(X,Y),g E (ey  (X 0 Y* ))*, where 1(.) is the largest cross norm on X 0 Y* ( see Proposition 3.1.2 ) , and y (X 0 Y* ) is the tensor product of X and y with respect to y(.), B(X, Y) is isometrically isomorphic to (y  (X 0 Y* ))*. ,
Proof.
For any T E
MX,
Y), define g(u) = i
E xi 0 fi E X0 Y,..
Clearly) 1001 _< MTH E 1141 . Ilfili. Further, we 2 i have 1001 < y (0 IIT 11 , V u E X 0 Y,,. So g can be uniquely extended to a linear functional on y  (X 0 Y), and Ha IITH. For any 6 > 0, pick z E X I f E Y* with 11x11 = 11/11 = 1 such that (Tx, f)  11711 < c. Since 1(x 0 f) = 1, it follows that Ma T  . Thus , we have 11911 = II TI. 19(x® f)I = 1(Tx1f)1 Conversely, let g E (I  (X 0 y)r. Since Ig(x 0 ni Ma • 11z11 1 fillVz E X, f E Y* , there is T E MX, Y) such that g(x 0 f) = (Tx, f). Q.E.D. where u =
Lemma 13.4.6. Let M be a hyperfinite VN algebra on a Hilbert space H, i.e., M = ( ilp Mp )" , where 1 E M1 C • • • C Mp C • • • C M, and for each p, MI,
520
is a matrix algebra. Then there exists a projection of norm one from B(H) onto M'. For any x E B(II), denote by C(x) the weak closure of Co{u*xulu E M and is unitary }. We say that C(x) n M' O. In fact, for any p, U, = fu E Mplu is unitary 1 is a compact group. So there is an invariant Haar measure A on Up with p(Up ) = 1. Let zp = fu u*zudp,(u). Since v*xp v = xp ,Vv E Up , it follows that xp E C(x) n M. Thus , 6 (x) n It/4 0, Vp. Now {C(x)n/t/41,73} is a decreasing sequence of nonempty weakly compact subsets of B(1/). Therefore Proof.
c(x) n M. Vz G
'
= c(x) n (np n = np (C(x)
n M;) 0,
WI).
Denote by B(B(H)) the Banach space of all bounded linear operators on B(H). By Lemma 13.4.5, B(B(H)) = (7  (B(H) 0 T(H))*. Thus , any bounded ball of B(B(H)) is w*compact. Let U be the set of all unitary elements of M, and for any u E U define Tu E B(B(H)) as follows: Tux = u*zu, Vx E B(H).
Clearly, J = Cogults G Ur)* is a w*compact convex subset of B(B(H)). Introduce a partial order " To. But To is maximal, so we have T1 = To . On the other hand, T1 is a w*cluster point of {Ta}. It follows that a' = Ti x E Mi . Therefore, C(To x) = C(Tix) = {d} = {Tox} C C(x), Vx E B(H). Now we have a linear map E(x) = To z(Vx E B(H)) from B(H) to Mi . Clearly, 11E11 < 1, and E(a1) = a', Va' E MI . Therefore, E is a projection of norm one from B(H) onto Mi. Q.E.D. Remark. A VN algebra M on H has the property ( P ) , if c(x) n M' 0 0,Vx E B(1/). From the proof of Lemma 13.4.6, we can see that: there is a projection of norm one from B(H) onto M1 if M has the property (P); and any hyperfinite VN algebra has the property (P).
Let A be a C*algebra with an identity 1, B be a CeLemma 13.4.7. subalgebra of A with 1 E B, and M be a type (III) factor on a separable Hilbert space H. If there is a linear map P from A to M satisfying : i) P(a) > 0, Va E A + ; ii) P(b i ab 2) = POOP(a)P(b 2 ), Vb1,b2 E B,a E A; iii) P (B) is weakly dense in M,
then A admits a type (III) factorial * representation. 1) Denote by SI the set of all linear maps Q from A to M satisfying : Q(a) > 0,Va E A + ;Q(b i ab 2 ) = Q(N)Q(a)02),Vbi,b2 E B,a E A; and Q(b) = P(b),Vb E B. We claim that n is a compact convex subset of (B(A,M),a(B(A,M),7  (A ®M„))) ( see Lemma 13.4.5 ) , and Q(x*x) > Q(z)*Q(z),Vx E A. In fact, if Q E n, then we have  1111111H < Q(h) < 11h11 1H,Vh * = h E A. Hence, f/ is a bounded subset of B(A,M). By Q = P on B,VQ E 12, 12 is also convex. Morevoer, it is easily verified that n is w*closed. Thus, f/ is a w*compact convex subset of B(A,M). For any Q G f/, and z E A, by the Kaplansky density theorem there is a net {el} c P(B) such that cl —+ Q (z) ( * strongly) , and Ilcill II Q (x)117v 1 . For any (p E (M t ) + , since Q*(p > 0, it follows from the Schwartz inequality that Proof.
(Q (x)* Q (z), (00)
= lim(Q(x)* Q (b i ), (p) t =lim(x*b i ,Q*(p) t < Q*(p(x*x) 1/21ifri Q*(p(brb 1 ) 1 /2
= (Q(x*x),(p) 1/2 1irn(cici, ;0) 1 12 = (Q(ex)74 0 ) 112 ' (Q (x)* Q (x) 1 sor / 2 ,
where bi G B and P(b 1 ) = cl, Vi. Hence , Q(ex)> Q(z)*Q(z),VQ E fl,z E A.
522
2) Since H is separable, there is a faithful normal state (p on M. For any Q E SI, let (p Q (a) = (p(Q(a)), Va E A.
Then (p c) is a state on A since Q(1) = P(1) = 1 H • Define e = {PQ IQ E n}.
Then by 1) 6 is a lecompact convex subset of the state space on A. Now fix Qo E n such that (po = io (4 0 is an extreme point of 6. Let {7ro , Ho, 6 } be the cyclic * representation of A generated by (po , and N = 7r0 (A)"( a VN algebra on 1/0 ). If 710 (x) = 0 for some z E A , then we have ço(Qo(x*x)) = (p o (x* x) = O. Since Q 0 (ex) > 0 and (p is faithful, it follows that Q 0 (ex) = O. By 1) , Q 0 (x)*Q 0 (x) < Q 0 (ex), hence Q 0 (x) = O. Now for any fixed f E M* , we can define a linear functional F on 710 (A) as follows: F(ro (x)) = f (Q 0 (x)),
Vz E A.
We say that F is strongly continuous on the unit ball of 7ro(A). In face, let {ad be a net of A such that 117r0 (ai )11 < 1,V1, and 7ro (a1 ) + 0 ( strongly ). Then p o (atai ) = p(Q 0 (atai )) > (,o(Q 0 (ai )*Q 0 (az )) + O. We may assume that Mad < 2, Vi. Then {11Q0 (a1 )1111} is bounded. Further, since ça is faithful and normal, so we have Q 0 (a1 ) * 0 with respect to stop. of M. Hence F(7ro(at)) = f ((Mal)) + O. For any a E N = 7ro(A)", we can find a net fai l C 7r0(A) such that al + a( 0 ( strongly ) , and from the preceding ar) strongly ) . Then (al paragraph {F(a1 )} is a Cauchy net of numbers. Hence, we can define —

F(a) = lifnF(ai ),
and this definition is independent of the choice of fai l obviously. In such way, F is extended to a linear functional on N, still denoted by F. We claim that this extension F is strongly continuous on the unit ball of N. In fact, for any e > 0 by the strong continuity of F on the unit ball of wo(A) we can find
el ,
6,, 6.) = {a E . o(A)iilali < 1 dlaedi < 5 ,Vi} 6, E Ho. Let such that IF(a)1 < e,Va E V, where V = V (0,
• • • ,
ei, • • • ,
U = U(0, ei , • • • , G ) 45) = {a E Nillali < Liked' 0 and a non—zero projection p of M such that t* (e) > 4. )"
526
Oi l of B such that Q 0 (b1)
a ( strongly 1111 r 1 (a) ( strongly) , and (10*(4' 1 1 (a)) ( weakly) . On the other hand, by the T definition of r, 4/, and the properties of 41)*, we have For any a E M, pick a net . Then 111 ' o ri(Q000)
4:10* OF 1 0 ri (Q0 00))
=
4:10*(7ro OOP) = (10 (7r0 (bi ))410*(11)
= Qo 00 4)* (P) + asIP* (FI)
(strongly).
Thus, we obtain that
= a4TP* (Fi ), 4:10* (A11 1 0 F _1 (a*)) = 4:1)* (Ina* , Va E M. Now let { az } be a net of Mp with Mad < 1, Vi, and al + 0( strongly 0( strongly . Since e is finite, it follows . Clearly, IF' o r 1 (a1 )e from Propsition 6.5.16 that eq/ 1 o F' (a) + 0( strongly . By (1)*(a*a) > 4:1)*(a*)(1)*(a),Va E N, and the a  a continuity of 40*, we also have 4*(e4'  i Fi(a ) + 0( strongly ) . Since xli1 0 ri(m) c 7ro (B)"F C 7ro(B)", e < F' and fai l C Mp, it follows that
n
a;
= {P(1)* (e)P + ( 1  741 1 {/34)* (e)P + ( 1  Pilai = {PV(e)P + ( 1  73)10 * (e)ai = {p(I)*(e)p (1 13)114* (eF1)4
= OA)* (e)P + ( 1  1410 * (e) 4)* (r) ai = {p(1)*(e)p ± (1  p)}p4)* (Or 0111 =
/AP* (e)p
n
(1  p )} pt*M1 1 o Fi(a ) + 0(strongly).
This means that the * operation is strongly continuous on the unit ball of Mp. That contradicts the facts: p 0, and M is type ( see Proposition 6.6.3). Therefore, N is type (III) . Q.E.D. Proposition 13.4.8. Let A be a NGCR algebra. Then A admits a type (III) factorial * representation.
Proof.
If A has no identity, and {7r, 1/} is a type (III) factoral * representation of (A 214"), then by Theorem 1.3.9 there is a projection po E 7r (A )' n 7r(A)" such that 7 = 7r(A)"po . Hence, {74.)po ,p0 1/} is a type (III) factorical * representation of A. Moreover, (A1(V) is also NGCR. Thus, we may assume that A has an identity 1. From Proposition 13.4.4, there is a separable C*subalgebra B of A with 1 E B and a closed twosided ideal I of B such that Bi/ is a (U H F) algebra of type {2' } .
527
Write Bi/ = ao  0,,,MP ) , where MP ) = M2 is the 2 x 2 matrix algebra, dn. Fix A E (0, 1), pick a state ça on M2 as follows: So
((a7
 Acx+ (1 A)b, 155)) 
d(7 a fib
.)
E
M2,
and let V) = Onpn, where çon = ço,Vn. By Theorem 9.5.11, The cyclic * representation {7r, 1/, } of Bi/ generated by ti) is a type (III) factional * representation. From Proposition 3.8.7, we have
7r = 0,17rn,
H = (8) ! Hn 1
e — on en,
where {rn, ILI eft} is the cyclic * representation of Mln) generated by ço n ,dn. Clearly, 7r(M) Ls M2, B(H) L' M4, and 7rn (MP) )' Ls M2 ) Vn. Further, by Proposition 3.8.6 7r(BII)' is generated by {(7rn(Mln) )' 0 0,74n1,n)Inl.
Hence , 7r(B I I)' is hyperfinite. Now from Lemma 13.4.6, there is a projection E of norm one from B(H) onto 7r(B I I)". Summing the above discussion, we may assume the following: there is a state (p on B, and a projection E of norm one from B(H) onto the type (III) factor M = 70)", where {7r, 1/} is the * representation of B generated ça, and H is separable. Pick a state ç7 c) on A such that OB = ça, and let {7T,H} be the * representation of A generated by ;O. Then we have H C H,Tr(b)H C H, and fr(b)11/ = 7r(b),Vb E B. Let p be the projection from H onto H, and define P:A+Mas follows: P(a) = E(pTr(a)p), Va E A. By Theorem 4.1.5, {A, B, P, M = 7r(B)" , H} satisfies the conditions of Lemma 13.4.7. Therefore, A admits a type (III) factorial * representation. Q.E.D. Let H be a separable infinite dimensional Hilbert space. Clearly, B(H) is not CCR and C(1/) is a CCR closed towsided ideal of B(H). By Proposition 1.1.2 and 13.3.6, the Calkin algebra A = B(H)IC(H) must be NGCR. Thus by Proposition 13.4.8, the Calkin algebra A admits a type (III) factorial * representation. Remark.
Notes. Quasimatrix systems ( Proposition 13.4.3) were introduced by J. Glimm. Proposition 13.4.8 is due to S.Sakai. The property (P) (see the Remark under Lemma 13.4.6) was introduced by J.T.Schwartz.
References. [55], [150], [153].
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13.5. Type I C*—algebras Definition 13.5.1. A C*algebra A is said to be type I , if for any nondegenerate * representation {7r, HI of A,7r(A)" is a type (I) VN algebra on H. Proposition 13.5.2.
If A is a GCR algebra, then A is type I.
Let {7r, HI be a nondegenerate * representation of A, and {Pi} be a maximal orthogonal family of nonzero central projections of 7r(A)' such that 71 (A)"p i is a type (I) VN algebra on pi ll, V/. By Proposition 6.7.2, it suffices to show that >pi = 1. Suppose that Proof.
t p = 1  En is not zero. Then {7rp,Hp} = PrIPH)PHI is a nonzero * t representation of A. Since 7r(A) is * isomorphic to A/ ker7rp and A is GCR, 7r(A) is GCR. By Proposition 13.3.11, there is 0 L a E rp (A) ÷ such that dim 7e(a)1/1 < 1 for any irreducible * representation 7r', Ir } of rp (A). Thus {
, azp (A)a is commutative. Further, aMa is also commutative, where M = 00 7r (A)" ( on Hp = pH). Let a = f AdeA be the spectral decomposition of a,
o 00 and ye = f AI deA . If e ( > 0) is small enough, then f = ay. = 1  e. is a e
nonzero projection of M. Clearly, faif • fad = a(yeaiye)a • a(Y e a2ye )a = a(yea2Ye)a • a(yea2ye)a = fa2 f • fa1f,Va1,a2 E M. So fMf is commutative. Let z be the central cover of f in M. Then Miz and Mif are * isomorphic. But (M}Y = fM f is commutative, from Theorem 6.7.1 Mif is type (I) . Then M; and (M)' = Mz are also type (I). Clearly, Mz = 7rp (A)"z = 7r(A)"pz = 7r(A)"z,z 0, and zp i = O, Vi. This is impossible since the family {p i } is maximal . Therefore, = 1. Q.E.D. 1 Definition 13.5.3. A C*algebra A is said to be smooth , if for any nonzero irreducible * representation {i r , H} of A, we have 7r(A) n c(H) {0 } . By Lemma 13.2.6, this condition is equivalent to C(H) C 7r(A).
En
Proposition 13.5.4.
If A is a smooth C*algebra, then A is GCR.
Proof. Clearly, (A4Œ) is also smooth. Moreover, by Proposition 13.3.15, A is GCR if (Aj(C) is GCR. Thus , we may assume that A has identity 1.
529
If A is not GCR, then by Proposition 13.3.6 there is a closed twosided ideal I of A such that Ai/ is NGCR. Further, from Proposition 13.4.4 there is a C*subalgebra B of A with 1 E B and a closed twosided ideal J of B such that BP is a (UHF) algebra of type {2' . By the Remark under this Proposition, B/J is simple ( i.e., BP contains no nonzero proper closed twosided ideal ) , so J is the largest proper closed twosided ideal of B. Fix a pure state w on B and wIJ = {0}. Let {7r„,H„} be the irreducible * representation of B generated by w. Then ker7r,„ = J, and {7r,„14} can be regarded as a faithful * representation of B/./. Since B/J is infinite dimensional, dim I1,, = oo. Let E = {p G P (A)1p1B = 4 For any p G 6, let fro Hp } be the irreducible * representation of A. Clearly , 1I,, C Hp 7 so dim H i, = oo, and the identity operator 7rp (1) = lp on Hp is not contained in C(H). Since A is smooth, it follows that 7rp (A) D C(H). Let I(p) = ir; 1 (C(Hp )). Then I(p) is a closed twosided ideal of A, and 1 V I(p). Thus I(p) B, and B n I(p) C J. Introduce a partial order " O. Since 7r(A)' is a factor, the central cover of 17113 in 7r(A)' is 1H . Thus, a  ap'Et is a * isomorphism from 7r(A)" onto 7r(A)l/. In particular, we have 11 7441 =1A7r(z)11,
Vz E A.
Fix 0 0 x E A. By Proposition 12.2.2, t + 1170(41 is a measurable function on SI Thus for any n
Bzn = {t E 1l1110)(z)11 is a Borel subset of XB x „ • Then
n.
11 7 (41
—
— n i
}
Denote by pizn the diagonal operator corresponding to
117Yzn 7r(x)11 = 11 f: n 7r(t)(x)dv(t)11 < 11 7141 — —1n . From the preceding paragraph, it must be that v(B) = 0, Vx E A, n. Thus, we obtain that 117(t)(x)11 = 117r(z)112Vt V B z , where Bz = UnBz n, and v(B) = 0,x G A. Suppose that D is a countable dense subset of A. Then we have 110)(41 =117r(z)111
Vx E A, t V B
where B = U zE DBx , and v(B) = O. Consequently, ker 7r(t) = ker r, Vt V B. Since {r(t), 1/(t)} is irreducible, Vt E f/, by the condition 2) we may assume that {7r(t),H(t)} ' ="' {7ro, ,ffo},Vt G fl, where { ro , 110 } is some irreducible * representation of A. Now by Proposition 13.6.4, the * representation {7r, H} is unitarily equivalent to {1 0 wo ,L2 (n,v) ® Ho}. Clearly, 7r0 (A)" = B(1/0 ) is type (I) factor. Therefore, 7r(A)" is type (I) . Q.E.D. Notes. Theorem 13.6.9 is due to J. Glimm But we still don't know whether this theorem holds in the absence of separability.
References.
[55].
Chapter 14 Decomposition Theory
14.1. Choguet theory of boundary integrals on compact convex sets Let E be a ( real ) locally convex ( Hausdorff topological linear ) space, X be a nonempty compact convex subset of E, and ExX be the set of all extreme points of X. A probability measure on X means a non—negative regular Borel measure and its total mass is 1. Definition 14.1.1. Let A be a probability measure on X. A point z of X is said to be represented by p,, if f(z) f dp.,V f E E*.
Clearly, such z is one at most. We also say that p. is a representing measure of z , and z is the barycenter or resultant of p. , denoted by r(p.) = z. Proposition 14.1.2.
The barycenter of a probability measure p, on X is
existential. For any f E E* , let Hf = fy E Elf(Y) = closed. We need to show that Proof.
n{Hf n
)(If
E
fjc f 44. Clearly, Hf is
0.
Since X is compact, it suffices to prove that for any finite set • • • , fn of < 1 0, consider the following subsets of
Proof. E x IR :
J1 = {(x,r)lx E A1 r = g(x)},
.12 =
{(x, r)lx E X, r = g(x) + el.
Clearly, J1 and J2 are two disjoint nonempty compact convex subsets of E x IR. By the separation theorem, there is some L E (E x JR)* and some A E IR such that supfLOI. E J11 < A < sup{g.)1. E J2}. If (x,r) E Ji, then (x, r ± E) E J2 . Hence L(x, r) < L(x,r + e) = 1,(x, r)
+ EL(0,1),
and L(0,1) > O. Define 1(z) .= L(x,0),Vx E E. Then f E E* , and m(.) = L(0,1)  '( A  f(.)) E M(X). For any x E X, we have L(x,g(x)) = L(x,0) + g(x)1(0,1)
1, Va > al . Now for any 1 G A, pick al as in (ii) . Then AG, = p,d( a) > Az ( C.M. ) , Va > al . Hence, we have A > p.1( C.M. ) ,V1 E A, i.e., p, is a upper bound of {p,ill G A} in Z. By the Zorn lemma, Z admits a maximal element p. at least, and this p. is what we want to find. Q.E.D. Theorem 14.1.8. Let p, be a maximal (C.M. ) probability measure on X. Then p. is pseudoconcentrated on ExX in the sense that p.(B) = 0 for each Baire subset B of X disjoint from ExX. In particular, supp p. C ExX, and if ExX is a Baire subset of X ( for example, X is metrizable), then p, is concentrated on ExX,i.e., p(ExX)= 1. Moreover, for any x E X, there is a probability measure p. on X such that : p. is a representing measure of z; and p, is pseudoconcentrated on ExX in the above sence. Proof.
See [128].
Q.E.D.
Remark. Generally, we can't require that p,(B) = 0 for each Borel subset B of X disjoint from ExX.
Definition 14.1.9. Assume that X is contained in a closed hyperplane of E which misses the origin ( There is no generality lost in making this assumption,
542
since we may embed E as the hyperplane E x {1} of E x R; the image X x {1} of X is affinely homeomorphic with X), and let P = {aria > 0,x E X}. X is called a simplex (in the sense of Choquet), if (P— P) with the positive cone P is a vector lattice ( i.e., if each pair x,y in (P — P) has a least upper bound sup{x, y } in (P — P)) . Such definition of a simplex coincides with usual one in case X is finite dimensional.
Theorem 14.1.10. The following statements are equivalent: (i) X is a simplex ( in the sense of Choquet); (ii) For each z E X, there is unique maximal (C.M.) probability measure p. on X such that p, is a representing measure of x. Proof. Notes.
Q.E.D.
See [128].
Theorem 14.1.8 is due to G. Choquet, P.A. Meyer, E.Bishop and K.
de Leeuw, and its metrizable case is due to G.Choquet. Theorem 14.1.10 is due to G.Choquet and P.A.Meyer. Proposition 14.1.3 is due to H.Bauer.
References. [128].
14.2. The C—measure and C—isomorphism of a state In this section, let A be a C*—algebra with an identity 1 , S = 5(A) be the state space of A, ça be a fixed state on A, and {7r,, I19„,1 9„} be the cyclic * representation of A generated by so. Clearly, S is a compact convex subset of (A*, a(A* , A)). For any a E A, define a(p) = p(a),Vp E S. It is obvious that a + ao is a positive linear map from A into C(S). By Corollary 2.3.14, E = {a(.)la E A} is a closed * linear subspace of C(S) containing the constant function 1 . Now ça can be regarded as a state on E. Then by Propostion 2.3.11 cp can be extended to a state on C(S). Hence, there is a probability measure y on S such that ço(a) = is a(p)dv(p), Va G A. Definition 14.2.1.
(1)
Let
Ç(ça) =
Iv u
is a probability measure on S } • such that (1) holds.
Clearly, for each u E 1(p), the barycenter ( or resultant) of v is r(v) = so; and v is a representing measure of the point io(E S).
so,
i.e.,
543
Moreover, by the above discussion, f*,o) O. Hence, f/((p) is a non—empty compact convex subset of (C(S)*,a(C(S)*, C(S))). Proposition 14.2.2. 1) If C is an abelian VN algebra on Hp and C C rip (A) 1 ,p is the projection from Hr onto C1 r , then : c + cp is a * isomorphism from C onto Cp; and Cp is a maximal abelian a—finite VN algebra on pH,, i.e., Cp = (Cp)' = pCgp; the central cover of p in C' is 1 = 1Hoi.e., c(p) = 1; and p is a maximal abelian projection of C';
pz p (a)rr p (b)p = rr p (b)p7r p (a)p, Va, b E A,
i.e., prgA)p is abelian; and prv,(A)p C pCip = Cp; and C = {r ,(A), /T . 2) Suppose that p is a projection on Ilp such that pl p = 1 p and pr p (A)p is abelian. Then pflp = C1 r , C is abelian, and C C 7r9,(A)', where C = { 7192(A) ) P }' .
3) There is a bijection between the collection f c, IC is an abelian VN algebra on lip , } and C C r p (A)'
and the collection
1p
p is a projection on li v,such that pry,(A)p is abelian
pi, = 1
}
such that plip = C1 r and C = frv,(A),pli. Proof. 1) By c(p)11 9„ = [Cp1 19„] = [C'Cl ip ] D r(A)1, = Hp , we can see that c(p) = 1. Since Cp is abelian and admits a cyclic vector Iv on Flip , thus Cp is 
maximal abelian and a—finite from Proposition 5.3.15. If q is an abelian projection of C' and q > p, then by Proposition 1.5.8 we have p = c(p)q = q, i.e., p is maximal abelian in C. Since pzv,(A)p C pC` p = Cp, it follows that prga)p7r9,(b)p = pz p (b)p7r p (a) p, Va,b E A. Now pr p (A)p is abelian and admits a cyclic vector 1 v, on pH,, so by Proposition 5.3.15 we have
(Pirso(A)P)" = (P 7rço(A)P)' = (PNP)' = MP,
where N = {7tw (A),p}". Moreover, from Cp = (Cp)' = pC'p D pz r (A)p, we get (P792(A)P)' D (CP) ' = CP D (P7rso(A)P)".
544
Hence, Cp = Nip. Noticing that [Np14] D irv (A)1 y, = 1190 the central cover of p in N is also 1 . Thus, c + cp(Vc E C) and xi + E N') are * isomorphisms from C onto Cp and from Ni onto Nip respectively. Now by C C Ni, we must have C = Ni = frv,(A),p1 1 . 2) Let N = {7r 9,(A),p}". Then the central cover of p in N is 1 , and —+ zip is a * isomorphism from Ni onto Nip. Since pNp = (px v,(A)p)" is abelian and admits a cyclic vector 1 v, on it follows that pNp = (pNp)i = N'p. Hence, Nip is abelian. Further, C = Ni is abelian, and C c 7rv (A) i , and C1 Nip1 v, = pNp1 v, = p1192 . 3) It is obvious from 1) and 2). Q.E.D. Now fix an abelian VN algebra C c w v,(A) 1 , and let p be the projection from 14, onto C1 p . Lemma 14.2.3. that
There exists unique * homomorphism A(a)p = pr(a)p,
A
: C(S) + C such
Va E A,
(2)
i.e., the following diagram is commutative: C(S)
C
Cp
1 id A
1"rso(A)P
Moreover, AC(S) is strongly dense in C, and A(a, • • an )p = prgai )p p7rv (an)p, Vai , • , an E A.
Proof. Let B be the abelian C*algebra generated by prgA)p. Then B C Cp, and B P=s C(T), where T is the spectral space of B. For any t E T, define 0(t)(a) = (pw w (a)p)(t),Va E A. Clearly, 00 is a continuous map from T to S. Further, for any al ,• • • , an E A and any polynomial P of nvariables, define T\P(all...
I an)
= PO:nrip(ai)p)...
2 P7V2 ( an)P) •
Since IIPCIErso(ai)P , • • • )Inr(an)P)II
sup Inprgai)p)(t),... ,(px p (an )p)(t))1 tET
= sup I P (t)), • • • , an (0 (0 )) tET < sup 1P(a1(a),• • •,an(p))I = 11P(ai,• • • ,an)II, pEs
it follows from the StoneWeierstrass theorem that A can be extended to a * homomorphism from C(S) to B c Cp. By C Cp, we can get a *
545
homomorphism A : C(S) + C such that A(a)p = pz v (a)p,Va G A.
Again by the StoneWeierstrass theorem, such A must be unique. Clearly, for any al , • • • , an E A we have A(ai , • • • , an)p = A(FLOP • • • A (art)P = Furgai)p  pr v,(A n »
since p E C'. Finally, since Cp = pC ip = p{w v (A), p}"p = (rr p (A)p)", AC (S) = B is strongly dense in Cp. Therefore, AC(S) is strongly dense in C. Q.E.D. Now (A()1 92 , 1 v,) is a state on C(S). Hence, there is unique probability measure A on S such that (A(f)1 92 ,1 v ) =
is
f (p)dp.(p), Vf G C ( S).
In particular, by Lemma 14.2.3 we have
4
ai (p) • • • an(p)dgp) = (p7rgai)p  • • mrgan )p1 w , 1 v,),
( 3)
Va l , • •  , an E A, and p. E n(p) ( see Definition 13.2.1). Clearly, the measure p, satisfying (3) is unique from the StoneWeierstrass theorem. Definition 14.2.4. The unique p.(E 00)) determined by (3) is called the C measure of the state ça.
Now we point out that A can be extended to a * isomorphism from Loe(S, p,) onto C. First, for f E C(S) we say that A(f) = 0 ..#, f = 0 a.e.A.
This is immediate from the following equality: (A ( f)mr v (a l )p • • • pxv (an )p1 9, 2 1(p) =
(AU) A (a i ) • • • A (a) i, 1 9,)
:= f f ( P)a 1 (p)
•••
an(p)dgp),
Val , • • • , an E A, and by the StoneWeierstrass theorem.
546
Secondly, if {fi} is a net of C(S) with 0(0(173 , L 1 )), then we have
Ilfi II < 1, Vi, such that
h +
(A(fi)p7r  (a)4, rço (b)1 v,) = (A(h) • P790 (a)P1 so 1 Mrço(b)P 1 90) = f it(P)a(P)RP)dii(P) + 0 ) Va, b E A. Hence, A(fi )p and A (f1) —+0 ( weakly). Then, A can be extended to a normal * homomorphism from L°° (S, p,) to C, still denoted by A, such that ( A ( f ) 1p,1 )
=f f (p)dp.(p), V f
E L oe (S, ii).
On the other hand, by (pr p (A)p)" = Cp (see that proof of Proposition 14.2.2 ) and A(C(S))p D pr v,(A)p, we have A(.1,m(S,A))p = Cp and A (Lc* (S, p.)) = C. Further, if f E L oe (S, p,) is such that A(f) = 0, then
f f (p)i(p) • • • â(p)d(p) =
(A(f) A (ai , • • • , an.)1,0 , ip)
=0,
Val , • • • , an E A. Hence, f = 0, a.e.p., and A is a * isomorphism from Lc°(S, it) onto C. Denote by r the inverse map of A. Then r is a * isomorphism from C onto Lc° (S, p,), and ,x7rp ( a) 1 v,,1, = L( rx)( p) a(p) dp. (p) ,
(4)
Va E A,x E C. Lemma 14.2.5. Let u E II((p), and r i be a * isomorphism from C onto Lc° (S, v) such that ( .7rp ( a) 1p,ip ) = L (r,.)( p) a(p) dv (p) ,
Va E A, z E C. Then v = il is the Cmeasure, and r, = r. Proof.
For any a E A there is z E C such that r i x =
a.
Then
(xpy1 p , zi p) = (xey1 v,,19,) = f (f i x? y)(p)dv(p) = f (r l e y)(p)a(p)dv(p) = (z*y7r p (a)4,1,p ) = (p7rp (a)py1 9„ zip),
547
Vy, z E C. Since C1 v, = pH,, it follows that xl) = P7rço(a)P•
Then by mrp (a)p = A(a)p we have x = A(a), and rx = r,(x, • • • Xn ) =
a = r i x.
Further,
a, • • • an = r(x„, • • • , z,),
where ai E A, xi E C, and r i xi = ai , < I < n. Hence,
fs a, (p) • • a„(p)dv(p) =
r i (x, • • • xn )(p)dv(p)
(xi • • • z1 1, 1) — ((z i p) • • • (z np)4,1 p )
(pri,(ai )p • • rrgan )p1 p ,1 v,), Va i , • • • an E A. By Definition 14.2.4, y = p. is the Cmeasure. r i (x, • • • zn ) = ai • • • an = r( • • • xn ) we have r
(al
• • • an), a.) = r  1(ai
Also by
Val , • • • ,a1 E A.
Since C(S) is wtdense in L'3 (S,p.) , it follows from the StoneWeierstrass theorem that r = T. Q.E.D. The unique * isomorphism r from C onto Lc*(S,A) Definition 14.2.6. determined by (4) is called the C isomorphism of the state ça , where p. is the Cmeasure of ça. From the above discussion, we have the following.
Theorem 14.2.7. Let A be a C1 algebra with an identity 1, S = (A) be the state space of A (a compact convex subset of (A* ,cf(A* , A)))0,o E S, {rp , H 0 , i} be the cyclic * representation of A generated by ça, and C be an abelian VN algebra on Hp with C c r p (A)'. Then there is a unique probability measure p, on S (the Cmeasure of (p) and a unique * isomorphism r from C onto L'(S,p,) ( the Cisomorphism of ça) such that p. E n(w),
L a,
• • • andA = (P7ro(al)P • • • Prço(an)P 1 p) 110)
and (x7r v,(a)1 v„
=
Va, ai , • • • , an E A, where II(p) is as in Definition 14.2.1, and p is the projection from Hp onto Cl p •
548
Lemma 14.2.8. Let n be a compact convex subset of a locally convex topological linear space E, and p, be a probability measure on A. Then
p. =
L
where 0 < hi E
J
if ai = f
gdp, = lim f gdo p } , Vg G CO), / {hi} n
Loe (n, p.),vi, Et/J (0 = 1, a.e.p. ; Apo = E a5 ] f_ 1 , for
each
hidp" ti Eli such that
ai f (ti) =
.4,
f (t)hi(t)dp(t), VI E E*
(notice that if ai > 0, then cxi l hip. is a probability measure on SL Hence, from
Proposition 14.1.2, there is unique ti E II such that '(t 1 ) = ai' f f (0 hi (t)d p,(t),V f E Et); {h lk }7_, < {hi } =1 if there is a partition {II ,. • • ,./ } of {1, • • • , n} such that hl = E hi ,1 < k < m ( Clearly for any {14 } and {hi }, {hlh i } > jElk {h 1} and {h1 }. So that {{hi } l() < hi E .ric p.),v i, and E hi = 1, a.e.p.} is
in,
J
a directed set with respect to " O. Further, i f gdp,  f gdpo ill = I f gdp, 
E f hi dp,g(ti)I i
< E f 100  g (t i )lhi (0440
{hi }717 then there is a partition {II, • • • , L } of {1, such that hi = E /4,1 < 1 < n. Thus , for any j and 1 E Ii we have Lei, 14(t) = 0,
Further, si E Vi if aii > 0, V/ G
a.e.p, on SZ\Int(Vi ).
4,
where a; = f h4dp,, and a; f (si)
f f h;dp,,V f E E*. Then 1 f gdp.  f
>
,
f g (t)  0801/4(044m
= EE f 100  030114(04M < i Icti v.i
E.
550
Therefore, p.(t) = lim p{h i}(g), Vg E
co. Q.E.D.
Let A,S = S(A),(,0 ES , fir y„ 119 1 9,1 , C C rgAY Proposition 14.2.9. and plip = C1 v, be the same as in Theorem 14.2.7. Suppose that p, is the C —measure of yo. Then = a (C (S) * , C (S))lim A{B;} {Bi }
where Bi E CF , Vi, S such that
1
L
gdp = lim f gdp.{B i }, Vg E C (S),
j
j
{B} s
E 131 = 1; p.{13 1 } = E aibpj , for each j, ai = ( 131 1 p ) 1 99) ) Pi E ai pi (a) = (r v,(a)B1 1 v„4),
Va E A;
{B1 }17, 1 < { B1 7=1 if there is a partition {II •  • , I77,} of {1, • . • , n} such that .51 = Bi ,1< k < m. Morevoer, if {Bik } < {/31 } , then p.{131) ‹ p { ki } .< IL ( i Eik C .M. ) } .
,
E
Proof.
Let
r be the C—isomorphism of (,o from C
onto Loe (S, p,) . Then
(xxga)1 9„,1 9,) = f a(p)(rx)(p)dp,(p), 90 Va E A, z E C. For any {Bi }, let h i = rBi , Vj. Now using Lemma 14.2.8 to E = (A* , a(A* , A)) , S and the C—measure p. of (be), we can get the
n=
conclusion. Q.E.D. Theorem 14.2.10. Let A be a C*—algebra with an identity 1, S = S (A) be the state space of A.0,o E S, {7rço , 14,1 w } be the cyclic * representation of A generated by ço. Suppose that Ci is an abelian VN algebra on 14, with Ci c ir9,(A)', and tii is the Ci —measure of cp on S, I = 1,2. Then the following conditions are equivalent: (i ) Ai ‹ P.2 ( C.M.), i.e., for any convex function g E cr ( s) , 
fs gdp, 1 5._ fs a(p) 2 dp. i (p) <J a(p) 2 (1122(P), Va* = a. E A; s
551
(iii) Cl C C2.
Consequently, if p, i = 122 , then C1 = C2. (0 = (ii) • It is obvious since ao2 is convex on S ,Va* = a E A. (ii) == (iii) . Suppose that the inequality (ii) holds. Let pi be the projection from 1/90 onto ci iv„ r i be the C • isomorphism of (p from Ci onto Lc* (S, Ai ), and Ai =i = 1,2. Notice that Ai (a)pi = parr (a)pi,Va E A,I = 1,2. For any a = b ± ic,b* = b,c* = c E A, we have Proof.
iv,
(porr (a) 1 9„, 7rv (a)1 v,)
= (Pirço(a1POrp(a)Pi 1 s07 1so) = (AI (a*a) 110, 1‘0) =
is r(p)2
is rb(p)2 ±(,9)14.(p)
±  (,9)142(p) = (A2(a*a)1, iço)
= (P27=90 (a) 1 90) 7rso (a) lso) • Hence we get p i < p2 since 1 r is cyclic for xr (A), which means that Ca r C C2 1,p . Let U be the isometry of L 2 (S, /2 2 ) onto C2 1, given by U f = A2 (f)i, VI E Loe(S,A 2 ). For any z E C1 , by Car C C2 4, there exists f E L2 (S,122) such that U f = z1 p . For any n = 1,2, , • , let En = p E sllf(P)I < n} and en = A2 (XEn). Then we have {
enx1 r = A2(xEn )Uf = A2 (XEn ) linrin U(xE„, f ) = A2 (XEn ) linrin A2 (XEm f)i ip = A2 (xEn f)1 95.
since xEm f + f in L2 (S,122 ). But 1 r is separating for 7r ,(A)', so we get enz = A2(XEn i) E C2 ,Vn. Moreover, since xEn + 1 in a(Lcc,1)) and A2 is normal, it follows that en —+ 1 ( weakly). Hence, z E C2 1 and CI C C2. (iii) == (i). Suppose that CI C C2. By Proposition 14.2.9, we have Al = lim p.{ Ei} VBi E
(CO + C
(C2 ) ÷ , and
Corollary 14.2.11.
If
and
p,{ Ei } ‹
E B., = 1. Therefore, A I ‹ 12 2
( C.M.).
Q.E.D.
is abelian, let C = rr (A)', then the Cmeasure A of ça is maximal (C.M.) on S = 5(A). Consequently, /2 is pseudoconcentrated on the pure state space P(A) in the sense that p,(E) = 0 for every Baire subset E of 5(A) disjoint from P(A). Moreover, if A is separable, then 12 is concentrated on P(A). 7r ,(A)'
552
Let v be a probability measure on S, and v > p. ( C.M. ). By Lemma 14.2.8, it suffices to show that p. ›. v{hi} (C.M. ) , VO < hi G L'(S,v), and E hi = 1, a.e.v. Proof.
i
Let v{ hi } =
E aibi,i ,
where ai = f hidv, pi G S I and ai pi (a) = f ahidv, V
i a E A. Since v > p,( C.M.), v > v{h i}( C.M. ) and p. G n(o), it follows that V,
v { i} E Il (so). Hence, f
adp. = ço (a) = E ctip, (a), Va
E A. Consequently,
i io > ai pi > 0, Vj. Hence, for each j, there is Bi E 7r9,(A )!E = C+ such that ai pi(a)= (7 92 (a)Bil y„1 9,),Va E A. Clearly, that
E B5 = 1, and
v{hi } =
A{B i). Now by
Proposition 14.2.9, we obtain
i IL >" A{B i } = Q.E.D.
Remark. Applying Choquet theory ( Section 14.1) to 5(A), and by n(p) 0( see Definition 14.2.1), there is a maximal ( C.M. ) probability measure p. on 5(A) such that
yo(a) = f a(p)dp.(p),
Va E A.
5 (A)
If 7r ,,(A)' is abelian, then by Corollary 14.2.11 we can pick that p. is the C— measure of so, where C = p, is pseudoconcentrated on the pure state space P(A). So (if) seems to be an "integral" of pure states. In particular, if A is separable, then we have p(a) = f
P (A)
Va G A.
a(p)dp.(p),
Proposition 14.2.12. Let p, E n(p) ( see Definition 14.2.1 ) . Then p, is the C—measure of ço for some abelian VN algebra C C xv,(A)' if and only if for each Borel subset E of S there is a projection PE of 7r ,,(A)' such that
is
XE
(P)a(p)d/./.(p) = (p Erço (a) i p , 1 90, Va E A.
The necessity is obvious from Theorem 14.2.7. Conversely, for any f E Loe (S, p.) with 0 < f 0, we have *1 (z) = O. So we can define a * homomorphism ri from Z onto Loe(S,p,') as follows: ilir y,w (z)) = Ar(z), Vz E Z. 
If 'Iii (z) = 0 for some z E Z**, then by Definition 14.3.3 we have (71;1 (z)ir(a)1 v ,1 v ) = (p(za) = f V(z)(p)a(p)diZ(p) =
0,
Va E A. Since 1 9, is cyclic for 7rv,(A), it follows that 71(z) = O. Hence r is a * isomorphism. Moreover, by (x7rv,(a)1 v,, 4) = (ir v (z)7r(a)1 92 , 4,) = (p(za) = (z)(p)a(p)diL l (p) = f (rx)(p)a(p)diAp),Vx G Z, a G A, where z G Z** fand‘111 71(z) = z, and by Lemma 14.2.5, pi is the Z measure p, of and rf is —
the Z—isomorphism r of (p. Further, xle = P.
yo
Q.E.D.
Now we consdier a geometrical characterization of the central measure in n(P). Let A be a C*—algebra with an identity 1 , and S = 5(A) be the state space of A. Let 7r be a W* representation of A**. The support of 7r, denoted by s(ir), is the central projection of A** such that ker 7r = A**(1 — s(r)). Let (pi , (p2 be two positive linear functionals on A. (pi and (p2 are said to be disjoint, if s(741 • 8(741 = 0, where 7rr is the W*—representation of A** generated by çoi ,i = 1, 2. Definition 14.3.5.
558
Now let v be a probability measure on S. For each Borel subset E of S, we can define a positive functional LIE on A as follows: LIE(a) = a(p) dv (p) , JE v is said to be semicentral
Va E A.
, if for any Borel subset E of S, LIE and vs \E are
disjoint. Fix ça E S. Let n e (p) = {v E (1(ç0)1v is semi central }, 
where r/((p) is the same as in Definition 14.2.1. Theorem 14.3.6. Let Z = ir v,(A)" n 7rp (A)'. Then the Zmeasure ( central measure ) of ça is semicentral, and there is a bijection between ne (ç ) and the collection of all abelian VN subalgebras of Z, i.e., for each v E (1,(ç) there is ( unique ) abelian VN subalgebra C of Z such that v is the Cmeasure of ça; conversely, if C is an abelian VN subalgebra of Z, then Cmeasure of (60 is semicentral. Proof. Let C be a VN subalgebra of Z,v and I' be the Cmeasure and Cisomorphism of ça respectively, and E be a Borel subset of S. Let pi = r  lixE,p2 — r 'xs\E. Then pi, p2 are projections of C I pp2 0, pi +p2 = 1, and by Theorem 14.2.6, 
LIE (a) = f xE (p)a(p)dv(p)  = f (rp i )(p)a(p)dv(p) ... (P i rga) 1 V, 3 1 92 ) 1
vs\E(a) = (p 2 719,(a)4,1 9,) ,
Va E A.
C, so we can find projections zi, z 2 of Z** such that z1 z2 = 0, z1 ± z2 = 1, and e(z) = pi , i = 1,2. Then
Since 71(Z**) = Z
D
LI E (a) = (,o(z i a), v s\E (a) = ço(z 2 a), Va E A.
Hence, LI E (Z2 ) = v s \E (z 1 ) = O. Let rE ,7rs\E be the W*representations of A** generated by LIE, vs\E respectively. Since zi, z2 G Z**, it follows that z1 E ker r E , z2 E ker irs \E . Further, zji < (1  sfrE)), z 2 < (1  s(7,\E)), and sfrE) • s(rs\E) = O. Therefore, v is semicentral. Now let v E ne (yo). By the proof of Proposition 14.2.12, there is a bounded positive linear map f + xf from L'(S, v) to rp (A)' such that
f fadv = (x171 v,(a)1 p ,1 p ), V f E I,' (S, v), a E A.
559
f = xE .
By Proposition 14.2.12, it suffices Let E be a Borel subset of S, and to show that xf is a projection of Z. Let g = xs \E . Then
L
i
adz/ = v1 (a) = (xf 71 (a)1 0 ,4) ,
s\E
adv = v g (a) = (xg r v,(a)1 9„ 1 v,), Va E A.
Further, let rf, Hf , 1 f } and irg , Hg , 1g } be the cyclic W*—representations of A** generated by vf and vg respectively. Since y E ne (cp), it follows that s(71 4. s(ir g) = 0. Write z = s(lr f), z' = s(71 g). Then z, 2 is the central projections of A**, and z + z' < 1. Since r f (z) = If ( the identity operator on Hf) and 0 < xf < 1, for any a E A+ we have {
{
(719,(za)4„1 9,) ? (71 (za)x1 1 92 ,1 92 ) = v f (z a) = (A f (z a)1 f 1 14 = (A f (a)1 f 1
if)
= v f (a) = (xf r(a)1 9„, 4).
Hence , ((ir y,w(z) — x 47 ga)4, 4) > 0,Va E A. Further, AZ(z) ? xf . Similarly, ev,(i) > xg . Since z ± z' < 1, and xf ± xg = xf+g = 1, it follows that 1— A9,w(z) > irv,w(d) > xg  1 — xf , and xf ? 71;(z). Hence, xf = irv,w(z) is a projection of Z since AZ(Z**) = Z. Q.E.D. Corollary 14.3.7. Let A E n(p). Then A is the central measure of (p if and only if A E 1L( ça) and 1.t is the largest (C.M.) measure of SI c (yo) . Proof.
It is immediate from Theorems 14.2.10 and 14.3.6.
Q.E.D.
Notes. The formulation of the central decomposition of a state was first given by S.Sakai. Theorem 14.3.2 in non separable cases is due to W.Wils.
References.
[148], [197].
14.4 Ergodic decomposition and tracial decomposition Let (A, G, a) be a dynamical system, where A is a C*—algebra with an identity 1 , G is a group, and a is a homomorphism from G to the * automorphism group Aut(A) of A. Further, let S = S (A) be the state space of A, and
Sa = fit, E Slyo(a,(a)) = yo(a),V s E G,a E A},
560 i.e., SG is the set of all G—invariant states. Clearly, SG is a closed ( compact ) convex subset of (S,a(A*, A)). Fix (p E SG, let {w v,,14,1 v,} be the cyclic * representation of A generated by ça , and define
ti v,(8)7rp (a)1 w, = Va E A, s E G. Then up (s) can be uniquely extended to a unitary operator on I19„ denoted by u p (s) still, Vs E G. Clearly, s + ugs) is a unitary representation of G on I I ço , and frv„ uv ,11yo l is a covariant representation of (A, G , a), i.e., ugs)rga)ugs)* = Vs EG,aE A. Moreover, ugs)1 y0 = 1 w,,Vs E G. Let Ev ,= Vs E G}, and pv, be the projection from I I v E lisoluso(s)e = onto Ev . Clearly, Ev, is a closed linear subspace of 1190 , and 1 90 E Ew .
e,
{e
Proposition 14.4.1.
With the above notations, we have that: (0 p ip ti v,(s) = ti v (s)pp = py,,Vs E G, and pv, E Cou v (G) ° , where " — 8" means the strong closure; (ii) { rp (A) , p v,} 1 = { 719,, (A) , u v,(G)Y; (iii) Let Mço = { irço (A), p,p }" = {7 90 (4 ugG)}" . Then the central cover of pw, in M90 is 1 , and x' + 2 p ço is a * isomorphism from Itf; onto it/pi pv ; (iv) Let Nyo = (p v,wv,(A)pX ( a VN algebra on p p liv, = 44 . Then bliw, =
(Pçorço(A)Pv)' (0 We need to show that for any nil ... 7 tin G H90 and e> 0, u(p v ,tii, 0, where U(ppl nil • • • 7 tin, 6 ) = f x E B ( IIV) 111 (PP — • • • , tin, 6) n cougG) X )ni 11 < e, 1 Api (> 0) on A, Vi.
i
Then for each j, there is Bj E irp (A ) + such that Ap i (a) = (r p (a)B1 1 p ,1 p ), Va E A.
By pi E SG and irp (A)1 v, = 14, we can see that 
B1 4 = u(s)B1 1,,
Vs E G, and j.
On the other hand, rp (a)ti v (s)Bi ugs)* = u p (s)ry,(a9 (a))Bi u v,(s)* = u v (8)Bir so (a9 (a))u v,(s)* = u p (s)Bi u v,(8)*r(a),
Va E A.
Hence, u p (s)Biugsr E rv (A)'. Further ugs)Bi ti p (s)*1 9, = u(s)B1 1 = B1 1 v .
But 1 v, is separating for rp (A)', so we get tiv (s)Bi ugsr = Bi,Vs E G, i.e., Bj E u v,(G)' . Further, by Proposition 14.4.1, Bj E 79„(A)' n ugGY = 0,EB1 = 1, and v{h i } .= 11{13;}• {71 gA),ugG)}1 = C,Vi, Clearly, Bj ?
Therefore, A
i
> IL{B i } = V{h i } ( C.M. ) ,
V{hi }, and p.› v( C.M.) on SG. Q.E.D.
Definition 14.4.5. io E Sc is said to be ergodic , if ça is an extreme point of SG, i.e., ço E ExSG. Proposition 14.4.6. Let yo E SG. Then ça is ergodic if and only if pry,(A),pa =0l14. Moreover, if dim .Ev, , 1, then ça is ergodic. 
565
Let p be ergodic, and h E {71p (A),pp }i = {719,(A),u p (G)} 1 with 0 < h < 1. Then (p h is Ginvariant, and 0 < p h < ça, where p h (a) = (71gaPi1 v,,1 v,),Va E A. Since cp E EzSG, it follows that (p h = Ap for some A E [0,1]. Now by ((h A)A9,(a)4,1 v ) = 0,Va E A, we get h = A. Therefore, Proof.
{rv,(A), p v }' =Œ1H,. Now let {irv,(A),pv,} 1 = {719,(A),u w (G)}1 = Œ1H. If (p is not ergodic, then there is some p E SG and some A E (0,1) such that ça ? Ap and ça P. Further, we can find h E (719,(A)'VV1 H,) + such that p(a) = (rp (a)h1 p , 4), Va E A.
By p E
SG 2 ti y,(8)hi v =
h1 9,,Vs E G. On the other hand,
ugs)hu w,(srirga) = ti yo (s)hirga„(a))u y,( 5)* = r v,(a)u v,(s)hugs)*, Va E A, s E G, i.e., ti v,(s)hugs)* E rv,(A)'. Further, since uw (s)hu v,(s)*1 v, = ti v,(s)h1 v, = h y,,Vs E G , and 1v, is separating for i1 ,(A)', it follows that h E u w (G)'. Hence, we get h E yry,(A) I flu y,(G)' = {r y,(A), tt v,(G) y .= Œfi„,, a contradiction. Therefore, (p is ergodic. Finally, let dim Ey, = 1. Then by Proposition 14.4.1, {Ap ( A),papp = (pv,719„(A)N,Y = Cpp . Moreover, 2  2p v, is a * isomorphism from {71v,(A),p 0 }1 onto fir v (A),N,Yp v, ( see Proposition 14.4.1). Therefore, {71v (4139,}' = Q.E.D. 01 14 , and p is ergoduic.
Definition 14.4.7. The system (A, G, a) is said to be G abeli an, if for any (p E SG, p ço r y9 (11)1390 is commutative.
Proposition 14.4.8. Let (A,G,a) be Gabelian, and (ho E Sc. (0 (p p ir v,(24)p p )" = (p p rv,(24.)p p ) 1 = {irp (A),p v,} 1N, is a maximal abelian VN algebra on pv,liv, = Ev ; (ii) {rv (A),pv,}'= {r v,(A),u v,(G)} 1 is abelian; (iii) (ho is ergodic 4=;, { rw (A), pa = ŒlH,, 4_;,dimEp = 1. (0 Since (pv,zv,(A)pv )" is an abelian VN algebra on p y„Il and it admits a cyclic vector 4, hence (p y,r y,(A)p v,)" = (p w r y,(A)p v,) 1 is maximal abelian on p v,Hp . (ii) By Proposition 14.4.1, (p,,71 ,,(A)N,)' = {719„(A),N,}'pv, is abelian, and 2 —+ 2p y„ is a * isomorphism from {nv,(A),N,}1 onto {A v (4 pap p . Thus, {ir v (A),N,}' = fir w (A), u v (G) Y is abelian. (iii) Let ça be ergodic. By Proposition 14.4.6, we have firv,(A),N,Y = 0 1H• Proof.
Then from (1) , (Pprço(A)Psor = (P92 7rço(A)Pp)' = (rPço.
566
Therefore, dim Ev, = 1. Now by Proposition 14.4.6, the conclusion is obvious. Q.E.D. Theorem 14.4.9. ça E Sa,a
The system (A, G, a) is G—abelian if and only if for any
a E A,i = 1,2,
inf a ll ECo{a.(ai)IsEG}
140([4,a2])1= O.
Th necessity is obvious from Theorem 14.4.2. Conversely, it suffices to show that
Proof.
7145v (a2 )Ao rgai))e) = 0,
(e,(7 v (ai)pv irço(a2)
< 1,1 = 1,2; and e E p , H, Ev with Vço E SG; a = a E A with I— = 1. For any e> 0, by Proposition 14.4.1 there are Ai > 0, s E G and L Ai = 1 such that
IRE Aiuso(si) Pço)rço(ai)ell < 6/2. Let a'IL E Ai cita i(a i ), then by u(t)p, iippirp(aoe — up(8). v,(4)eii
pp (it E G) we have
uppirp(ai) — rço(aimeii
< e/2,
> 0) ti E G, and Ep.i
Vs E G. By the sufficient condition, there are such that
ICLEILjatiM),a21)1 < e,
where .00 = ((.),)(E Se ). Then 1(e, (71v(a1)pvrga2) — Irga2)Pvrv(a1))
0) is arbitrary, it follows that (e, (7rp(al)p çorp(a2) — 7rp(a2)pprp(ai))e)
,
O.
Q.E.D.
Clearly, if A is abelian, then (A, G, a) is G—abelian. Moreover, if for any a: = ai E A, I, = 1,2 and so E SGI inf{l(P(Eas(a1), a2D1Is E GI  0, then by Theorem 14.4.9 (A, G, a) is G—abelian. Remark.
Let (A, G, a) be a G—abelian system. Then SG is a simplex ( in the sense of Choquet). Therefore, for any so E SG there is a unique probability measure p. on SG such that Theorem 14.4.10. ( Ergodic decomposition )
'sa
a i (p) • • • an(p)diL(P) = (Pprp(ai)Pp • • •Pçorso(an)N.41 110) ,
Val, • •  , an E A, and u ‹ A ( C.M. ) for each probability measure v on SG with f a(p)du(p) = (p(a),Va E A. Consequently, p. is pseudoconcentrated on sa ExSG in the sense that a(E) = 0 for each Baire subset E of SG disjoint from ExSG. Moreover, if A is separable, then p,(EsSG ) = 1. For any so E SG, let p. be the C—measure of so, where C = Orv,(A),p p l'. By Theorem 14.4.4, p. is the unique largest ( C.M. ) probability measure on SG such that p. ›. v ( C.M. ) for each v as above. Now by Theorem 14.4.10, SG is a simplex, and p, is pseudoconcentrated on ExSG. Moreover, since p. is the C—measure of sc, and supp p. C SG, it follows from Theorem 14.2.6 that Proof.

La a() • • • an (p)dgp) = (ppir,(ai)p ço • • • p(a)p1, 1 w ), Val , • • • , an E A
Q.E.D.
Let A be a C*—algebra with an identity 1 , G be the group of all unitary elements of A, and S = 5(A) be the state space of A. For each y E G, define a(a) = vav* ,Va E A. Then (A, G, a) is a dynamical system. Clearly, SG is the tracial state space T = T (A) of A, i.e., Definition 14.4.11.
T = T (A) = {so E So(ab) = ço(ba),Va,b E Al. Proposition 14.4.12.
Let (A, G, a) be the same as in Definition 14.4.11.
Then it is G—abelian. Proof.
It is immediate from Theorem 14.4.9.
Q.E.D.
568
Theorem 14.4.13. ( Tracial decomposition ) Let A be a C*algebra with an identity , and T = T (A) be its tracial state space ( a compact convex subset of (A*, o(A*, A ))) . Then T is a simplex ( in the sense of Choquet). Therefore, for any so G T there is unique probability measure li on T such that
f
a1 () • • • an (p)dp.(p) = (pp 7rgai)pp • • • p(a)p,1, 1 v ),
Vai , •  • , an E A, and p, > v( C.M.) for each pprobability measure v on T with = ço(a),Va E A. Moreover, p. is indeed the central measure of 'p, fanda(p)d(p) p. is pseudoconcentrated on ExT = Tn.7 in the sense that /L(E) = 0 for
any Baire subset E of T disjoint from ET , where 7= 7(A) is the factorial state space of A. In particular, if A is separable, then p.(T n 7) = 1. By Proposition 14.4.12 and Theorem 14.4.10, it suffices to show that ExT = T n 7, and {rp (A),N,} 1 = rv (A)" n 719,,(A)' for each so E T. Let so E T, Z = 719,(A)" n rp (A)'. For any x E Z, pick a net fai l C A such that rv,(ai ) + x( strongly ). Then for any unitary element y of A we have Proof.
u p (v)x1 v, = 11 mu y,(Orgai)u p (v)*4 = limwv,(av i(a/))1 v, = 11 mwv,(v*ai v)1 v, = r p (v)*xr p (v)1 v,
Hence, Z1 9,
Conversely, for any a G A,
C Ev,.
pp rv,(a)1 p = lirnE AV u v,(4))7rv (a)up(q))*1 9„
.= limE t
.
1
1
iv,
avi
v,
where A ) > 0,E xy) = 1,V1, yli) E G ( the gropu of unitary elements of A) i
tt) such that p p = slim E Açou 1 V or.1 ) (see Proposition 14.4.1 ) . We may assume t
that
.
1
w E Aloirv (vV* avn ___+ x E
71p(A)u
.i
( replacing {E 4) 71gyravnli by a subnet if necessary). Then j (x1 0 ,71 p (b)1 p ) = (pprp(a)19„,nv(b)1p)
569
Vb E A, and py,xv,(a)1 v, = x1 v,. Pick {a i } c A such that r(a i ) + s ( strongly). Then by py,ry,(a)l y, E Ev, we have x1 v, =
ti v (v)x1 v,
= 11 mu v (v)7rvjai)u v (v)*1 v,
= limxv (v*ai v)1 9, =
Vv E G. Since ça E T, it is easy to see that 1 v, is also separating for Av (A)". Hence, z
Vv E G, and x E
Z. Therefore, pço rga)1 v, E Z1 v ,Va E A, and Ev, = Z1 v . By
Proposition 14.2.2, we have
nv (A) 11 n wv (A)' = {7rv (A), pv }%
\Air, E
T.
Now let (p E ExT . By Proposition 14.4.8, dim Ev, =1. From the preceding paragraph, Œp,, = Ev = Z1 v,, and z1 At 1 v,, where A, E Œ,Vz E Z. Further, znv (a)1 9, = rço (a)z1 v = Azir v (a)1 9,,,Va E A,
i.e. z = Az 1H,„Vz E Z. Therefore, rv (A)" is a factor, and ça E T n T. Conversely, let peTn Y. Then Ev = Z1 v, = 01 v , and dim Ev = 1. From Proposition 14.4.8, ça E ExT . Therefore, ExT = T n Y. Q.E.D. Theorem 14.4.10 is due to O. Lanford and D. Ruelle. Theorem 14.4.9 is due to D.Ruelle. Proposition 14.4.8 is due to G.G. Emch. Notes.
References. [44], [98], [139].
Chapter 15
(AF)Algebras
15.1. The definition of (Analgebras A C*algebra A is said to be approximately finite Definition 15.1.1. dimensional, or (AF) simply, if there is an increasing sequence {A„} of finite dimensional * subalgebras of A such that UA „ is dense in A, i.e., UA „ , A. Let A = UA „ be an (AF)algebra. Then A has an Proposition 15.1.2. identity 1 if and only if there exists n o such that 1„ = 1, Vn > no , where 1„ is the identity of A„,Vn. Proof. Suppose that A has an identity 1. If there is a subsequence {n k } with 1,4 0 1, then i n 0 1,Vn, since {A„} is increasing. On the other hand by LI„A„ = A, we have no and x E A„„ such that Ilx — 111 11 1 — x11
Ile so = He — x1„„ ell — Ril — —
12
a contradiction. Therefore, 1„ must be equal to 1 for all enough large n.
Q.E.D. Lemma 15.1.3.
For any 6 E (0,1), there exists y = y(e) > 0 with the following property: if A is a C*algebra on a Hilbert space H, p is a projection on H, and a E A with lia — Pll < I, then we have a projection q of A with IIP — q11 < E. We may assume that a* = a. Let 45 E (0, I), and m(> 0) be the minimal value of the function IA' — AI on the following set: Proof.
[  2,2 ]\[( — S, 5) U (1 — 5, 1 ± 5)].
571
Now pick 7 = ry(e) > 0 such that y2 + 37 < min{le,
7}.
Notice that
max{1A2 – AI I A E a(a)} ' 11a 2 – all 11a2 – aP – Pa + P II + 1113 (a – P )II + II(a – P)PII + II P – all I Raand
P) 2 11 + 3 11a – P II
AI 1A2 — Al 1A2 —
>
1
>
m
2,
if if
A E [2,2]\[(5,0 LI (1 – 5,1 + 5)].
Thus, a(a) c (45,5) li (1 – 5,1 ± 5) . Pick a continuous function f on IR such that 1 (A) = 0 if A G ( 5,0 and 1 (A) = 1 if A E (1 — 5,1 ± 5). Then q = 1(a) Q.E.D. is a projection of A, and I Iv – q11 c IIP – all + Ila – q11 IV ,d(n,m)= dim (Az))'/ 2 (where {zin) , • • • , 4n(2) } is the set of all minimal central projections of An ),V(n,m) E D;U = • • tn, • • •I (where 4 n is the embedding matrix from An into An+ VT?) •
Example 1. The Bratteli diagram of a (UHF)algebra of type {N } . (m ni l where m n = AVI Pn+i,dn, In this case , r(n) = 1, d(r1,1) = v,n , n i.e., 7711
P2
P1
Inn _4
in2 + • 4 • • •
+
P3
Pn
• • •
Pn+1
Let H be a separable infinite dimensional Hilbert space, Example 2. K = C(II) be the set of all compact linear operators on H, and A = ICI01 H .
If {n} is an orthogonal normalized basis of H, and pn , qn , rn are the projections
Lei,".
from H onto A = Un An , where
en] 1 [ell "
e11111[en]
respectively, dn, then we can write
A n = pnB(11)pn.101H = p nB(II)p n ED Cq n ,Vn. Clearly, any minimal projection of pnB(H)p n is still a minimal projection of Pn+1B(H)Pn+i)qn qn+i +rn + i, and rn+1 is a minimal projection of p n+1 B(H) 'Pn+i,Vn. So D(A,{A n}) is as follows
r(n) = 2, ci(n,1) = n,d(ri,2) =1, 4:b n
589
Vn.
The GICAR (gauge invariant canonical anticommutation Example S. relation ) algebra. It has a diagram with r(n) = n ± 1,d(n,m) = C n 1 , and /1 11 4•,, .= 1
0 1 (n + 2) x (n + 1),
Vn = 0,1,2 • • • ,m = 0,1, • • • , n, i.e., the Pascal's triangle.
Proposition 15.3.4. Let A = UnA n ,B = UnBn be two (Analgebras, and D(A, {Ar = D(B, {B„}). Then A and B are * isomorphic. ,})
Proof. Let do n , Wn be the * isomorphisms from An into An+1 , B r, into BnFi respectively. By the assumption and Lemma 15.3.2, for each n we can construct a * isomorphism en from An onto Br, such that the following diagram is commutative: 4, 1 ___4 .2 •n , ___4 + • • • 4 A1 A3 A2 4 4 4
0 11
B1
*1
821 B2
*2 +
831
____,
B3
• ••
'n
+
.....i. • • • .
Thus we get a * isomorphism 0 from Un An onto U nBn such that el An = en , Vn. Therefore, A and B are * isomorphic. Q.E.D. Clearly, any (Analgebra has a Bratteli diagram at least. Conversely, let D = {D,d,u} be a diagram, and each (I)„(E 1/) satisfy (1), (2), i.e., r(n)
E Od(n,j), 1 < i
r(n + 1),
j=1
and
r(n1 1)
E i=1
4) > 0, 1 6/ 23
a be the canonical map from A onto A/J.
Since
can be naturally embedded into A/J, it follows that 11 1n11 = inf{lIzn — YIllY G
%In}
_>: 6/2, Vn ? no ,
and 11111 _> 6/2,x V J. Therefore, we have J = Un.in
Q.E.D.
Definition 15.4.2. Let D = {D,d,11} be a diagram of an (Analgebra, D = UnDn 7 Dn = {(n,m)11 < m < r(n)},Vn311 = { c1 n = (4)10. A point (n +1, 0 is called a descendant of a point (n, j), if 4) >0. In general, a point y G Dm is called a descendant of a point z E D n , which is denoted by z y, if m> n, and there exist points xk E Dig , n < k < m, such that xn = x7 xnt = 7f 7 and xk+i is a descendant of xk , n < k < m 1. Let z = (n,j), y = (m,1). Clearly, z y if any only if the (i, j) element of the matrix (40,n_i • • • sl.n) is not zero. 
Definition 15.4.3. Let 1) = {D,d,11} be a diagram of an (Analgebra. A subset E of D is called an ideal, if : 1) any descendant of x belongs to E, Vx E E; 2) suppose that x G D n and {y G DnFIly is a descendant of z} c E, then x E E.
591
Let A = Un An be an (AF)algebra, and NA, {An } ) = D = {D, d, II} be the corresponding diagram. If J is a twosided ideal of U nAn , then there exists an ideal subset E of D such that
Lemma 15.4.4.
J = U n ED {An,k1(n,k) E E},
(1)
where each A ni k is a matrix algebra, and An = ED rk (fl A ni k l Vii,. Conversely, if E is an ideal subset of D, then Un ED {An ,k1(n, k) E El determines a two sided ideal J of iln An , and J n A n = ED{A n Al(n, k) E E } , Vn. Let J be a twosided ideal of U n An . Since J = iln (J n An), there is a subset E of D such that (1) holds. Now we must prove that E is an ideal subset of D. (n + 1,1). Clearly, if p is a minimal projection Let (n,k) E E, and (n,k) of A nik ) then pz 0 0, where z is the minimal central projection of An+i with An+i z = AnFl i t. Since pz E J and pz E Anz C An+I pt , it follows that J n An+1,i 0 {0}. But An+ i,t is a matrix algebra, hence An+1 11 C J, i.e., (n +1,1) E Proof.
E.
Now let (n,k) E D, and {(n + 1, j)(n, lc) A n,k
c e{An+i,j 1(n, k)
(n +1, j)} C E. Then (n + 1 1.7.)}
C ED{An+1,j1(n ± 1 1 j) E El c J and (n, k) E E. Therefore, E is an ideal subset. Conversely, let E be an ideal subset of D, and J = U n ED { A rt, ic I (fl, k) E E}. Put Jn = ED{An,k1(n,k) E E},Vn. If (n,k) E E, then An,k CED{An+i i 1(n, k) (n + i, j)} C Jn_F i. So Jn C Jn+i,Vn, and J = Un .in is a twosided ideal of U n An . Moreover, if A ni k C J , then there is m(> n) such that AO C Jtn . (m, r)} C E. Since E is an ideal subset , it follows that Thus {(m, 01(n, k) Q.E.D. (n, k) E E. Therefore, J n An = Jn ,Vn.
Theorem 15.4.5. Let A = UAn be an (Analgebra, and D(A, {An}) = {D, d,11} be the corresponding diagram. Then there are bijections between the following collections: 1) the collection of all closed twosided ideals of A; 2) the collection of all twosided ideals of UnAn; 3) the collection of all ideal subsets of D. Proof. and 3).
By Lemma 15.4.4, there is a bijection between the collections of 2)
From Lemma 15.4.1, J 71 is a map from the collection 2) onto the collection 1). Now let J1 , J2 be two different twosided ideals of Un A n . We
592
must prove that .7 1 0 .72 . By Lemma 15.4.4, we may assume that there is A n,k C J1 , but A ni k n J2 = {0}. Thus z V J2 1 where z is the minimal central projection of An such that Az = A nik. Further, for any m> n, z ± (J2 n Am ) is a nonzero projection of A m /(J2 n Am ), i.e., infaz — YIllY E Am n J21 = 1, Vm > n.
Since J2 =
ilni > n (J2
n Am ), it follows that inf{ II z — YillY E J2}
= 1,
and z V 7/2. Therefore, .71 72.
Q.E.D.
Remark. Let A = UA nand B = il„B„ be two (Analgebras, and An = ./3„,Vn. But they can pick different Bratteli diagrams such that they
have different sets of twosided ideals. Hence, the structure of an (Analgebra A = UAn depends on not only each An but also each embedding way from An into An+1 . Let A = UAn be an (AF)algebra, and D = {D, d,11 } be the corresponding diagram. If J is a closed twosided ideal of A, then J and A/./ are also (Analgebras, and they have diagrams:
Proposotion 15.4.6.
{E, OE ,U1E } , {D\E, di(D\E),111(D\E)}
respectively, where E is the ideal subset of D corresponding to J. Moreover, if ti = {Un = (4) )1tn e {Aprly
(1g, r)}.
By the condition 2), we have J, n .72 {0 . Further, J1 n J2 = Ji n .72 n (UA) o {0 } from Lemma 15.4.1. Thus there exists (p, r) E D such that Apr C ./1 n J2 . By the definition of J1 , we can find (p i , ri) E D with z (p i , ri ) and (p, r) (p i , r i ). Thus, Apo., C J1 n J2. Again by the definition of .12/ there is (p2,r2) E D with y (p2,r2) and (pi, ri) (p2,r2). Let z = (p2, r2). Then z z and y z. 3) == 1). Since any finite subset of D has a common descendant, we can find a subsequence {nk} and a function j(.) such that for each k,j(k) E }
{1, • • • ,r(nk + i)}
and (nk , i)  (nk+1) . i (0),
1 < i < r(n k ).
Now we may assume that
(n, i)  (n + 1,1), V1 < 2: < r(n) and n. Thus , there is a minimal pojection pn of An , Vn, such that pn > pn+1 ,Vn. If write
PrzaPn — Pn(a)Pn, Va E A n , then pn is a pure state on An ,Vn. Since Pn+ 1 ( a)Pn+ 1 = Pn+ 1 aPn+ i = Pni1Pn aPnPn+ 1 = Pn (a)Pn+1/
nn+i IA Va E An , it follows that ,,  n = p, Vn. Hence there is a state p on A such that plAn = pn ,Vn. Moreover, it is easily verified that p is pure. Now it suffices to show that ker 7r1, = {0 } , where 7tp is the irreducible * representation of A generated by p. From Lemma 15.4.1, it is equivalent to prove that ker 7r, n A n = {0},Vn. Let z be a minimal central projection, and Az = Ank• Since (n,k) (n + 1,1), it follows that pn+ 1 ‹ p relative to An+i, where p is
594
a minimal projection of link. Thus , there is y E An+i such that vv = and vv*
0 if a E P; and a > b if (a — b) E P. Definition 15.5.1.
(—
Definition 15.5.2. An element u of an ordered group (G, P) is called an order unit , if any a E P, there exists n E /V such that a < nu. Since P is unperforated, it follows that u G P. An homomorphism p from an ordered group (C, P) to JR is called a state
relative to an order unit u, if p is positive ( denoted by p > 0), i.e., p(a) > 0, Va E P; and p(u) = 1. We shall denote by S(C) the set of all states relative to u. Clearly, IRG = x G IR with product topology is a locally convex Hausdorff topologyical linear space. With the embedding: p (P(a)) ace , Su (G) is a closed convex subset of /RG obviously. We claim that Su (G) is also a compact subset of IRG . In fact, for any a E G there is na E IV such that —na u 0
(1)
for any h E H and m E Z with (h + ma) E P. If h + na > 0,h' — n'a > 0, where h, h' E H , n, ni E IN, then nit' > nn` a > —n! h
and —p(h)In < p(h')In` . Since u E H, we can pick
A E [sup
{ — PO) h E H,n E IV ,}
(h + na) > 0 '
inf f p(hY I h' E h,n1 E _NJ' 1 re n I Of — n' a) > 0 f J•
Clearly, such A satisfies the condition (1).
Q.E.D.
Let (G, P) be an ordered group. A subgroup J of G is called an order ideal , if J = 4 — 4, where 4 = J n P, and for any a, b E P with a < b and b E 4 we have also a E 4. An order ideal J is said to be prime , if J1 and J2 are two order ideals with J = J1 n J2 7 then either J = J1 or J = J2. Definition 15.5.4.
Definition 15.5.5.
A group G with the following form G
is called a dimension group, where 40,,, =
kV)
is a r(n + 1) x r(n) matrix
r(n11)
of nonnegative integers with
E s!rit) > 0,1 < i < r(n),Vn. In detail, every t=i element of G has the following form: 41) noo (in) — ( 0 1 ' ' • ) 0 1 in, inI11 • • •) + / 1
where t, E Zr( 8) and t 8+1 = (I),(t i ), Vs > n, and 1 = f(t i ,•••,t,n ,0,••,0,••)1m > 1,4 E Ern.
596
Proposition 15.5.6.
Let G = lim{Zr (n) , t n } be a dimension group, and
P = ilrA rtoo (Z). Then (G,P) is a countable ordered group, and (G,P) has also the Riesz interpolation property, i.e., if a, b, c, d E G with a, b < c, d, then there exists e E G such that a, b < e < c, d. Notice that (Io n keeps the order, and (Z r(n ) arF(n) ) has the Riesz interpolation property, Vn. Thus the conclusions are obvious. Q.E.D. Proof.
Let G = lim{Zr (n) , t n } be a dimension group, Dn = {(n,m)11 < m < r(n)},D = U nDn and 11 = ft n in}. Then {D, 11 } is called a
Definition 15.5.7. diagram of G.
Clearly, any (AF)algebra admits a dimension gruop. Conversely, if G is a dimension group with a diagram (D,11), the we can construct an (AF)algebra A such that A admits a diagram {D, d,11} . Indeed, it suffices to pick { d(n, such that r(n) d(n + 1, 1) > E .sriod(n, j), 1 < I < r(n 1), i=1
where (Io n = ( 8 in) ) Vn.
Proposition 15.5.8. Let G = lim{Zr (n) , IQ be a dimension group with a diagram Then there is a bijection between the collection of all order ideals of G and the collection of all ideal subsets of D ( see Definition1.4.3). Moreover, if J is an order ideal of G and E is the ideal subset of D corresponding to J, then J is a dimension group with a diagram {E, 111E}. Let J be an order ideal of G, and E = {(n, E J} , where ein) < k < r(n)} is the canonical basis of Zr(n). If (n,k) E E and (m, p), by Definition 15.4.2 we have (n,k) Proof.
t rloo (gin) ) = Cnoo (Cn_ i 0 • • • 0 > Cnoo (4171) ) >
O.
Since J is an order ideal, it follows that (m, p) E E. Now let z = (n, k) E Dn , and {y E Dn+11Z y} c E. Noticing that 4 + J+ c 4 and 4)nœ (e) =
E s çn) sin +i, (e sçrtF1) N)1 sk
(n11,i)EE
we have 4),.00 (e) E
4, i.e.,
(n,k) E E. Therefore, E is an ideal subset of D.
597
Further, let J(E) be the subgroup of G generated by {(ein) )1(n,k) E El. Clearly, J(E) C J, and J(E) is a dimension group with a diagram {E, I E}. Let a G J. By G + = Un(1)„„,„(7Z) there are nonnegative inS ). ince J is an order ideal, it tegers A i ,. • • , Ar(n) such that a = t n.(E Akein) follows that (Ionœ (ein) ) c 4 if Ak > 0, i.e., (n, k) E E if Ak > O. Therefore, a E J(E), and J = J(E). Conversely, let E be an ideal subset of D, and define J = J(E) as above. We claim that J is an order ideal of G. In fact, since E is an ideal subset , it follows that 4) noo
({ E Akein)lAk
EZI) C (Dn+1,00 ({ E Aie!n+ 0 1A;
E Z1)
(n+1,i)EE
(n,k)EE
Vn. Thus , we have
J=
UndP noo ({
E Z 1)
e(n) 1À k
kk (nok)€E
J+
U n 4b noo {
\‘ A ke(n) k IÀ
E
})
(n,k)EE
and J = 4  J+ . If a, b E G + with a < b and b E J+ , from the expression of 4 we can see that a G J+ • Thus J is an order ideal of G. Moreover, if (n,k) E D with nœ(€) E J+ , then we can write (1) noo
(er) =
(m) (ei ),
E (m,i)EE
where m> n and Ai E MF ,Vj. Thus, there is p with p> m,n such that tylp(e in)
=
E Aier ) )
4P inP
(m.,i)EE
E (p i ) E E
(P)1
!Ai
E
This means that every descendant of (n, k) in Di, belongs to E. Since E is an ideal subset of D, it follows that (n, k) E E. Q.E.D. Proposition 15.5.9. Let G = lim{Zr (n) ,(1) n} be a dimension group with a diagram {D, lib and J = J(E) be an order ideal of G, where E is an
598
ideal subset of D. Then G/J is also a dimension group with a diagram {D\E, 1.11(D\E)}. Proof.
For any n, let zr(n) = zp(n)4_7'q(n) 2
[e(: ) I(n, k) E E ] , (n) = [4:1) 1(n,k) V E], and { 4n) 11 < k < r(n)} is the canonical basis of Zr(n). By this decomposition, we have projecZP(n) and Qn = (1 — Pn) : Z r(n) Eq(n) , Vtl. Further, let tons Pi,. : Zr(n) qin = QnFi ( 4) n I Zq ( n) ) 2 Vn. Then the dimension group lim{Zq (n) , AF„} admits a diagram {D\E,U1(D\E)}. We have that:
where
ZP(n) =
tin — tin+i 0 Ilin G I J and 'i(t) = 4:Drioo(tn) ± J,Vtn E where rin : Zq(n) Indeed, since E is an ideal subset of D, it follows that
nn (in)
= (1) n+ 1,00 Qn+ 1 (I) n(tn) + (Dr1+1,00 Pn+14)n(tn) +
'q(n), and n.
J
= t n+1,co x1sn(in) ± .1 = nn+1 Olin (in)) 2 Vi n E Zq (n) and n.
as follows:
Hence, we can define a map ri : iim{z(n), w n } —4 G/J 4'
n ( 411100 (tn) ) — 4) n00 (tn) + li, Vtn
E Zq(n) and n.
If t n E Zq(n) With tnoo (tn) E J, then t„ = 0 since E is an ideal subset ,i.e., /7 is injective. From G I J = Unebnoo(Z r(n) ) + .1)
= il n eDnco (Zq (n) ) + J), 77 is also surjective. Moreover, since 77 and 77' keep the order, G/J is order Q.E.D. isomoprphic to lim{Zq (n) , Wnl.
Proposition 15.5.10. Let G be a dimension group with a diagram {D, U}, and J = J(E) be an order ideal of G, where E is an ideal subset of D. Then J is prime if and only if E is prime (see Definition 15.4.7.) Proof. By Proposition 15.5.9 and replacing G by G/J, we may assume that J = {0} and E = O. Let the order ideal {0} be prime. For any xi E D, put Ei = {z E DIzi z, i.e., z is a descendant of z}, J1 = J(Ei ),I = 1,2. Suppose that Fi is the ideal subset of D generated by Ei , I = 1,2. Then Ji = J WO 1 i = 1,2. By the assumption, we have J1 n J2 0 {0}. Thus, Fi n F2 0 O. Pick y E F1 n F2. By z1
599
and y E F1 , there is z1 E D such that x 1 z1 and y z1 . From y E F2 we z have also z1 E F2. Further, by z1 and x2 E F2 there is z E D such that z1 and x2 z. Therfore, x 1 z and x2 z, i.e., 0 is a prime subset of D. Conversely, let 0 be prime. Suppose that Ji = J(E1) is a nonzero order ideal of G, where Ei is an ideal subset of D,i = 1,2. Since E1 n E2 0 1 it follows that J1 n J2 0 {0}. Therefore, {0} is a prime order ideal of G. Q.E.D. Example 1. The CAR (canonical anticommutation relation) algebra i.e., the (UHF)algebra of type {2n }.
It has a diagram as follows: 2
•
2
2
•
•••
•
2
••
2'
23
22
2n+'
Thus we need to consider the dimension group: G = lim{Zr (n), 41) n },
where r(n) = 1,
= [2],Vn. Define a map: (0 ) • • • 01
in , tn+i, • • •)
tn 2n '
/
where t n E r(n) , tn+r = 2 r tn , Vr. Then we can see that G is order isomorphic to the dyadic rationals E Z,n = 1,2, • • •I = Z[1/21 ( relative ordering in JR). Example 2. Let H be a separable infinite dimensional Hilbert space ilnd K = C(H). From Section 1.3, K has a diagram as follows: •
1
1
• ••
• • •
1 •
So we have a dimension group G = lim.{Zr (n) ,t n l, with r(n) = 1 and (Dn [1],Vn. Clearly, G =s ( usual ordering). . The dimension group of the GICAR algebra. From Example 3 of Section 1.3, its dimension group G will be the inductive limit of the following system: Example 3.
1
7Z —EB7Z
—4
ZEDZeZ•••,
i.e. G =lim{Zn+ 1 (I) n n >
= Un>otn(Zn+1 ) )
600
where
=
(n + 2) x (n + 1),Vn > O.
Lemma 15.5.11. Let a, b E JR with 4b > a2 . Then there exists a positive integer N such that all coefficients of the polynomial (x + 1) 1i (x2 — ax +b) are
nonnegative. Proof.
Clearly, b > 0. So we may assume that a> 0. Write N
(x + 1) N (x2 — ax + b) =
E i=2 (a
N!
Cs+2x1+2.
. ±
2)!(N
—
i)!
Then for 0 < i < N — 2 we have aCk i + bCk 2 }
Ci÷2 =
= (7: + 1)(i + 2) — a(i + 2)(N — i) + b(N — OW — i — 1) = (1 + a + 00: — (b + 1)(1 + a + 01 N) 2 2
+0 — 1)(1 + a + 0 1 N2 — (2a + OW — i) + (3i + 2) 2
__?:
(b — 1)(1 + a ± b) 1 N2 — (2a 4 bp V.
Moreover,
Co= (N;12)! b,
e (N
CNFi = (N
C1 = (N +11)! A*, (Nb — a),
a),
CN+2 =
(N+2)1 NI
•
Therefore, if N is large enough, any coefficient of the polynomial (z+1) N (x2 — ax + b) is nonnegative. Q.E.D. The dimension group G = lim{Zn+ 1 ,41)„In > 0} of the GICAR algebra is order isomorphic to Theorem 15.5.12.
(Pz ([ o,11), n([o, 1])), where P([0, 1]) is the additive group of all polynomials on [0,1] with integer coefficients , and
Pi' ([o,11) = ff E Pz Ool iDlf(t) > ol Vt E (0,1)}
Li {0}.
601
Proof. Let u E G. Then there is n(> 0) and an element (ao , • • • , an ) of 4' n+ 1 such that u = 4) noo((ao, • • • I an)).
From (ao, • • • , an ), we have unique (b0 , • • • , bn ) E ao xn + • + an = bo (x + i)
Define a homomorphism 4) : G
l + • . •
Zn+i
+ b,
such that Vx > 0.
Pz ([0,1]) as follows:
4)(u) = p(t) = bo + bi t — • • • + bn tn
where (b 0, • , bn ) E 7Zn+1 is determined by u as above. First, we must show that 4) is welldefined. If (a,13 , • • ,dn+i ) = 4) n((ao, • • • an )), then a = a0 ;
= ai_ i + ai , 1 < < n; dn+1 = an .
Thus do xn+ 1 + • • • + dn+i = (x + 1)(ao xn + • • • + an ), Vx. Let (g,•••,bni +1 ), in Zn+ 2 , satisfy bio (x + 1) n+1 ± • • • + tin± i = Ceo X n+1 ± • • • ± ain±i VX. Then we have g(x+1)n +1 +• • •+tin+1 = (x+1)(ao xn+• • +an) = (x+1)(bo(r Fir+•••+bn),
O < j < n, and b4 1 = 0. Thus, the definition of 40 is independent of the choice of n. Moreover, if 40 nœ ((ao , • • • , an )) = 0, then there is m(> n) such that (1)„,„((ao , • • , an )) = 0, where Czni = 4)m o • • • o Since each 4) k is injective, it follows that 03 = •• = an = 0. Therefore, 4) is
VX. So , b = b ,
,
 1
welldefined. Clearly, 40 is an isomorphism from G onto Pz ([0,1]). So it suffices to show that 40 is also an order isomorphism. Let u E G+ \{0}. Then there is (ao , • • • , an ) E Zr'\{0} such that u = tnoo nao, • , an )). Thus bo (x +
1) n ± • • • +
bn = aox n ± • • • ± an > 0 , Vx > 0.
Therefore, we have p(t)
=
+ 1) n + • • • ± bn (x + i))
> 0, Vt E (0.1). t= xI1 1
Conversely, let p(t) = h o + •  • + b ntn E PI ([ O, 11)\{0}. Then f (x) = b o (x ±
1) n ± • • • ±
bn = ao ± • • • +
We need to prove that tnœ((ao, • • /an)) E GF m(> n) such that 4:1) nin((ao, • •
> 0, Vx > O.
or to show that there exists
an )) E Z+171+1 .
Clearly, it is equivalent to prove that there exists a positive integer N such that all coefficients of the polynomial (x ± 1) N f (x) are nonnegative.
602
Since f(z) > 0,Vx > 0, we can write
f (x) = c Ilix + Ai) Ilix –
ai)(x –
where C > 0, Ai > 0, and ai E ŒVR,Vi,j. Now applying Lemma 15.5.11 to each (z – ai )(x– taci ), we can find a positive integer N such that all coefficients Q.E.D. of the polynomial (z ± 1)'f(z) are nonnegative. Notes. Proposition 15.5.6 is indeed a characterization of dimension groups. We have the following EffrosHandelmanShen theorem: if G is a countable ordered group, and G satisfies the Riesz interpolation propperty, then G is a dimension group. Let A be an (AF)algebra with a diagram {D, d, U} . If D = Li n Dn ; Dn = {(n, m)11 <m < r (n)}; U = {41:P n In} , then the dimension group G = lim{ Zr(n) , 4} is indeed the K0group of A.
References. [39], [40], [43], [59], [103], [133], [160].
Scaled dimension groups and stablly isomorphic theorem
15.6.
Definition 15.6.1. Let G be a dimension group. A subset r of G + is called a scale for G, if: 1) G + is generatred by r, i.e., G + = r + r ± • • ; 2) for any a, b E G+ with a < b and b E r, we have also a E T. In this case, we say that (C, C + = P, r) is a scaled dimension group. For example, if G has an order unit u, then r = [0,u] = {v E GIO < y < ti is a scale for G. In a scale r, we can define a partial addition , i.e., a,b E r is said to be additive , if (a + b) E r. }
Lemma 15.6.2. Let ai ,f3i E E+ ,1 < i < r,1 < j < s, and al + • • • + a, = Pi + — + fie . Then there is a subset flii 11 < i < r, 1 < j < s} of E+ such that s r ai =
E Pia 1 13i = E rlici )
k=1
1 < i < r,1 < j < s.
k=1
If PI > al , let /11 = cxi ,•y ii = 0,2 < j < s, then we need to find Proof. {yii l 2 < i < r,1 < j < s}(c Z+ ) such that r
a
ai =
2 < 1 < r, k=1
yki = th  al) E k=2
603
and
r
Pi = E ,iki)
2 < j < s.
k=2
If al > Pk , let In = Pi oil = 0, 2
a. Pick m(> n) with 2m1 > N. Then nm = 2' 1 8 1 • • s,n_i > Nsn • • sm_i. Since skik+i,Vk, it follows that n,n t,n > n„,1„, >
o
o (1) n (N1 n ).
Then from 0 < a < ifo nc,o (N1 n ) < Cn œ (nintm ) c r, we have a E r. Therefore,
r = G(A) + .
Q.E.D.
606
Let A = il„A„,B = il„B„ be two (Al)algebras, and G(A),G(B) be their dimension groups respectively. Then G(A) and G(B) are order isomorphic if and only if oto(A 0 K) and ao(B 0 K) are * isomorphic, where K = C(H) and H is a separable infinite dimensional Hilbert space. Theorem 15.43.8.
The sufficiency is obvious from Corollary 15.6.6 and G(A) = G(a 0(A0 K)),G(B) = Now let G(A) and G(B) be order isomorphic. Pick fn i < n2 < • • •I and {mi < m2 <  • .} such that the scales of the dimension groups corresponding to UkAk 0 Kn k ,Uk(Bk 0 Kmk ) are G(A) + ,G(B) + respectively. Thus, the scaled dimension groups of U k (Ak 0 Knk ) and Uk(Bk 0 Knik ) are scaled isomorphic obviously. By Theorem 15.6.5, ao(A 0 K) and ac  (B (8) K) are * isomorphic. Proof.
Q.E.D. References.
[11], [38], [40], [103], [160].
15.7. The tracial state space on an (AF)algebra Let A = UA nbe an (AF)algebra with an identity e and a diagram as follows: e E Ai 'Li*
where Ar, = M(t (n) ) = Ep rk(:1Mt (n), t (n) = (t (in) , • • , t (:4) E /On ) , Vii , and Mk is k the (k x k) matrix algebra, Vk. Then we have the scaled dimension group: (G,G + =
where
G
= iim{ Er(n) , 4, n).)
P = Un (Dno0(41(n) ))
and r = [0, ti] = {v E GIO _< v
til
= Un t n00 (10 3 tn) 1) )
where u = 4:1)„,,(t 0n))(Vn) is an order unit of G ( notice that the definition of u is independent of n since e E Ai and t (n+1) = Let 7 be a tracial state on A, i.e., T be a state on A with ? ( ab) = r(ba),Va,b G A. For any y G r we can find some (si, • •  , sr(n)) E
607
with sk < dn) (1 < k < r(n)) such that y = 4)„00((si, • ,s,( n))). Let q be a projection of A n such that q = q1 ED • • • ED qn , where qk is a rank s k projection of k Ç r(n). In equivalent sense, q is uniquely determined by v. Thus, 1 we can define p(v) = r(q).
Then p is an additive function from r to [0,1]. From the proof of Proposiiton 1.6.3 , p can be extended to a state on the ordered group (G u). Conversely, let p E S(C), and A crt) = p(4) noo ((0, • • • 0 1 1
0,Vk, and A(I n) ± • • • ± A!,r(2) = p(t noo (t (n) )) = p(u) = 1.
For any a E A n , we can uniquely write a = al ± • • • ± ar(n),
where ak E M(0,1 < k < r(n). Then define r(n) rn (a) =
E A :( ) tr(a k ),
k=1
where tr() is the unique cononical tracial state on Mk (Vk). Clearly, rn is a tracial state on A n , Vn. We claim that Tn+lIAT, Vii.
Fix n and k E {1, • • • ,r(n)}. It suffices to show that Tn+ 1
(P) = Tn (P) =
Ain) itin) ,
where p is a minimal projection of Mtkoo. In fact, let do n = (4))1 0, if p(a* a) > 0,Va E L i (G)( or M(G)).{7r, HI is a * representation of Li(G) ( or M(G)), if H is a Hilbert space, and r is a * homomorphism from the Banach * algebra L i (G) ( or M (G )) to B(H). We have the following. 1) Let p be a positive linear functional on Li (G) ( or M(G)). Since Li (G) admits an approximate identity, it follows from the Cohn factorization theorem (see F.F. Bonsall and J. Duncan, Complete normed algebras, Berlin, Springer, 1973.) that p is bounded and hermitian. Moreover, we have 11P11= liumP(zV * zu)
and 1P(b * 01 5_ P(a * a) 1I2 • p(b * 0 112 1
1/3 (01 5 ilPilv(asa)1/25 
633
Va, b E L l (G) ( or M (G )), where v(.) is the function of spectral radius. 2) If 7r is * representation of 1,1 (G) ( or M (G)), then 114 1 . 3) If {71, HI is a nondegenerate * representation of V(C), then it can be uniquely extended to a * representation of M(G). It suffices to define that 7r(v)7r(f) = 7r(v *
fg
Or
7r(v) = lirr(v* zug )
V f E L l (G), e E H. 4) A positive linear functional p on 1)(G) is called a state , if liP11 = 1. For each state p on Ll(G), by the GNS construction there is a cyclic * reperentation {7r„, H, GI of L' (G) such that p(a) = (irp (a) p , '),Va E L I (G ).
Then p can be extended to a state on M (G ). Clearly, each nondegenerate * representation of L' (C) is a direct sum of a family of cyclic * representations, and each cyclic * representation of L' (G) is unitarily equivalent to the * representation generated by a state. 5) Let p be a stete on Li (G). Then p is a pure state ( an extreme point of the state space on Li (G)) if and only if the * representation {7r,„ Hp } generated by p is toplogically irreducible. 6) For each nonzero a E LW), there exists a toplogically irreducible * representation 7r of Ll(G) such that r(a) 0 O. (7) The left regular representation {A, L2 (G)} of I) (G) is faithful, where A(f)g = f * g ,V f E L i (G), g E L 2 (G). Indeed, let A(f) = 0 for some f G L i (G). Since zu E L' (C) n L2 (c) for any compact neighborhood U of e and A(f)zu = f * zu = 0, it follows that f = Il • 111  lipif
*zu =
O.
Unitary representations of G and nondegenerate * representations of Li(G)
Let G be a locally compact group. {u., HI is called a unitary representation of G, if u, is a unitary oerator on H for each s E G,ti st = ui ut , u, = 1, and s + (u,,n) is a continuous function on E H. In this case, s > u, is also continuous from G to (B(H),T(B(H),T(H))). Moreover, we have the following facts. 1) Let {u., H} be a unitary representation of G. If for any zi G M(G) define
G,ve, n
7r(v) = f u s civ(s), G
—
634
17)dv(s),v, n E H, then {7r, H} is a * representation
i.e., (7r(v) , t» = f of M(G), and
G
= u„
Vs E G.
Moreover, f + r(f) = f f (s)u sds is a nondegenerate * representation of G
L' (G).
Conversely, let {7r, HI be a nondegenerate * representation of L' (G). Then it can be uniquely extended to a * representation of M(G), still denoted by {7r, 1/}. Further { u. = 7r(5.), HI is a unitary representation of G. Therefore, there is a bijection between the collection of all unitary representations of G and the collection of all nondegenerate * representations of L i (G). 2) Let fu., HI be a unitary representation of G, and {7r, H} be the nondegenerate * representation of M(G) correspoinding to { u., H} , i.e.,
r CO = f f (s)u i ds,
r(v) = f u s dv(s),
V f E 1,1 (G), v E M (G ). Then we hvae . = 7r(1, 1 (G))" = {u s ls E GI". 3) For each s E G with s e, there exists a topologically irreducible unitary
representation fu., HI such that u, 1H . 4) Let {A., L2 (0} be the left regular representation of G, i.e., (A. f)(t) = f (s i t), Vs EG,f E L 2 (G). Then A(f) = f f (s)A s ds, V f E 1.1 (G)
is exactly the left regular representation of I) (G). Positive linear functionals and continuous positivedefinite functions n
A function (p on G is said to be positivedefinite
, if E xiAk(45,— isk) > k,1=1
O l Vs i , • • ., sn E G, A l , • • • , An E Œ. We have the following facts. 1) Let (p be a positivedefinite function on G. Then we have
yo(e) __?__ 0, io(8 1 ) = (p(s),ko(s)1 < (p(e), Vs E G. 2) Let fu., HI be a unitary representation of G, and is a continuous positivedefinite function on G.
(u.e, e)
e E H. Then (p(.) =
635 Conversely, if ça is a continuous positivedefinite function on C, then there exists a cyclic unitary representation {u., H, } of G such that V() =
3) Let p be a positive linear functional on L i (G) , {7r „, H p ,G} be the cyclic * representation of L i (G) generated by p, and {u!"), H} be the unitary representation of G corresponding to {7, Hp } . Then ça() = (ti!P) C„ ep) is a continuous positivedefinite function on C, and 13( f) = f f (s)cp(s)ds,V f E L i (G); MPH = IIS°1100 = Ae). Conversely, if cp is a continuous positive  definite function on C, then p(f) =
f E L i (G)) is a positive linear functional on Li (G). J fIn(s)(p(s)ds(V particular, there is a bijection between {pp is a state on L i (G)}
and
{VIS0 is continuous and positivedefinite on C, and cp(e) = 1 } .
4) Let ço i o,02 be two continuous positivedefinite functions on G. Then piso2 is still positivedefinite. 5) Let {p i } be a net of continuous positivedefinite functions with çMe) = 1,V1, and p i be the state on L i (G) corresponding to (pi, V/. Then {p i } converges to a state p in w*toplogy, i.e., there is a continuous positivedefinite function
ça on G
with ça (e) = 1 such that f f (s)(p i (s)ds + f f (Op (s)ds ,V f E L i (G) 2
if and only if, cho z (s) * ça(s) uniformly for s E K, where K is any compact subset of C. 6) ( R.Godement's theorem ) Let yo be a continuous positivedefinite function on C, and ça E L 2 (G). Then there exists 0 E L 2 (G) such that ic)( • ) = (A.0) 10 . The enveloping Cs algebra of a Banach * algebra 
Let A be a Banach * algebra, and suppose that A admits a bounded approximate identity {ai } , and Ma*I1 = Mall,Va E A. A positive linear functional p on A is continuous automatically, and MPH = 1} = lira p(a t ) = lip p(ch* al ). sup{ p(a* a) la E 24)011
For any * representation {7r, HI of A , we have also 11 7r11 < 1 . Let p be a state on A ( i.e., p > 0 and MPH = 1). By the GNS construction, there is a cyclic * representation {7rp , Hp , e„} such that p(a) = (7r p (a)ep ,) ,\ 91a E A. Moroever, p is pure if and only if 7r p is topologically irreductible. For any a E A, we define Halle = suP{117r(a)1117r is a * representation of A}.
636
Then we can prove that
Ila Ile
, sup{117r(a) 1117 is topologically irreducible } = sup{p(a*a) h /2 1p is a state on Al = sup{p(aea) 1/2 1p is a pure state on Al = sup{cx(ae a) 1 12 1cx is a C*seminorm on AI .5_
Va G A. In other wored, II •
Ile
Mali,
is the largest Ceseminorm on A. Let
N = {a E Alliall c = 0 . }
Clearly, N is a closed twosided ideal of A, and 11 • 11, can become a Cenorm on A/N. Then completion of (A/N, 11 • Il e ) is called the enveloping C* algebra of A, and denoted by Ce (A). Now let A admit a faithful * representation. Then N = {o}, II • 11 e is the largest C*norm on A, and CIA) is the completion of (A, II • IL). Moreover, since Mall e .5_ mI all, Va G A, {a1 } is still an approximate identity for Ce (A). If p is a state on A, then by Ip(a) 1 = 1(7ri,(a)G, ep)I 117p(a)11 < Hall e , Va E A, and p(al ) —> 1, p can be uniquely extended to a state on C* (A). Conversely, if p is a state on C(A) , by p(a 1) —> 1 then (p1A) is a state on A. Therefore, the state spaces of A and C* (A) are the same. Group C*algebras and reduced group C . algebras Definition 16.3.1. Let G be a locally compact group, and II • 114 be the largest C*norm on 1)(G) (notice that the left regular representation of L' (G) is faithful). Then the enveloping C*algebra of L l (G), i.e., the completion of (L i (G), II '114)1 is called the Ce algebra of the group G, and denoted by C*(G). From the preceding paragraph, {zu} is still an approximate identity for Ce(G); and the state space of Ce (C ) is equal to the state space of Ll(G). Proposition 16.3.2. Let G be abelian. Then Ce(G) is * isomrophic to Cr(d), where d is the dual of G. Since Cs (G) is abelian, so the spetral space of Cs (G) is the pure state space on CIO. But pure state spaces of C* (G) and Li (G) are the same. Therefore, the spactral space of C*(G) is d, the spectral space of V (C ). Q.E.D. Proof.
Definition 16.3.3. Let G be a locally compact group, and {A, L2 (G)} be the left regular representation of V(C). Then Ilf Ilr = IIA(1)11(Vf E V(G)) is
637
a C* norm on Ll(G). The completion of (Li (G ), • Ir) is called the reduced Ce algebra of the group G, and denoted by C(G). Clearly, Ulleldf E L l (G). So the identity map on Li (G) induces a * homomorhpism from C* (G) onto C:(G). Therefore, C:(G) is * isomorphic to a quotient C*algebra of C* (G). Moreover, the VN algebra R(G) = {A, I s E G}" on L 2 (G) is called the VN algebra of the group G.
Amenable groups Definition 16.3.4. Let G be a locally compact group. G is said to be amenable, if there exists a left invariant mean m on L°°(G), i.e., m is a state on L°°(G) ( see L°°(G) as a C*algebra ) and m(8f)=
where ,f(.) = f(s 1 .),Vs G G, f G L'(G). If G is amenable, then we can prove that there exists also a right invariant mean and a twosided invariant mean on L'(G). Remark. If G is discrete, then G is amenable if and only if R(G) has the property (P) (see the Remark under Lemma 13.4.6 and [153]).
If G is a compact group, then G is amenable. Indeed, we have an invariant Haar measure p, on G with p,(G) = 1. Define Example 1.
m( f) = f (s) dp.(s) ,
Vf E L'(G).
Clearly, m is an invariant mean on Lc° (G). If G is abelian, then G is amenable. In fact, let M be the mean ( state ) space on L'(G). Clearly, M is a compact convex subset of (Lœ(G)*, w* top.). For any s E G, define Example 2.
(713 m)(f) = m( s f),
Vm E M, f E Lcc(G).
Then T, is an affine continuous map from M to M, Vs E G. Since G is abelian, it follows that 7171 = TtT8 3 \149 3 t E G. Now by the MarkovKakutani fixed point theorem, there exists mo G M such that Ts mo = mo, Vs E G. Clearly, mo is an invariant mean on LNG). Let F2 be the free group of two generators u, y with discrete topology. We say that F2 is not amenable. Example S.
638 In fact, if m is a left invariant mean on 1 00 (F2 ), let Ez be the set of elements in F2 beginning with z,Vx E {u, v, tr i , v 1 }, then 1= m(G) = m({e}) ± m(Eu) ± rrt(E ui) ± m(E) + m(E„1).
On the other hand, by the left invariance of m we have 1 = m(G) = m(Eu) ± m(uEui) = m(Eu) ± m(E ui)
= m(E) ± m(vE v i) = m(E) ± m(E,i).
This is a contradiction. Therefore, F2 is not amenable. For amenability, there are many classical descriptions. But for our purpose, it suffices to point out the following Godement' condition: G is amenable if and only if there is a net {tk i } C L 2 (G) such that Ohl Ai) + 1
uniformly for t E K,
where K is any compact subset of G.
Main theorem of this section
Definition 16.3.5. Let A be a C*algebra, and {7r, HI be a * representation of A. A state ( or positive linear functional ) p on A is said to be associated with 7r, if there exists e E H such that p(a) = (7r(a)e,),Va E A. Now let 7r 1 , 7r2 be two * representations of A. We say that In is weakly contained in 7r2 ( or 7r2 weakly contains 7r1 ), if ker 7r2 c ker 7ri .
Lemma 16.3.6. Let H be a Hilbert space, and p be a state on the C*algebra B(H). Then p belongs to the a(B(H)* , B (H)) closure of Co{(•e,e)ie E H,
li eu = 1
}
.
If the conclusion is not true, then by the separation theorem there is a E B(H) such that
Proof.
Rep(a) > sup{Re(ae,)1 E 1111111 = 1 . }
Let h = il (a ± al, then we have P(h)
> supf(he, .)1. E Ill 1101 = 1} = max{ AA E a(h)}
639
On the other hand, it is obvious that p(h)
max{AIÀ E a(h)},
a contradiction. Therefore, the conclusion holds.
Q.E.D.
Proposition 16.3.7. Let A be a C*algebra, and {711 , HI }, {72 , HI be two * representations of A. Then the following statementas are equivalent: 1) 7r1 is weakly contained in 7r2 ; 2) Each positive functional on A associated with 7r1 is a welimit of sums of positive functionals associated with 7r2 ; 3) Each state .on A associated with 7r1 is a w*limit of states which are sums of positive functionals associated with 7r2. Proof.
1) == 3). Let p be a state on A associated with 7r1 . Since ker 7r2 c
ker 711 , p can become a state on A/ ker 7r2 ( Proposition 2.4.11). Clearly, we may assume that Al ker 7r2 C B(H2). Since p can be extended to a state on B(H 2), then by Lemma 16.3.6 we have the statement 3). 3) == 2). It is obvious. 2) == 1). For any a E ker 7r2 , and E H, by the condition 2) we have (711 (a)e, e ) = O. Therefore, a E ker 7r 1 , and ker 7r2 c ker 7r1 . Q.E.D.
Theorem 16.3.8. Let G be a locally compact group. Then the following statements are equivalent: 1) G is amenable; 2) Any * representation of C(G) is weakly contained in its left regular representation, where the left regular representation of C*(G) is the unique extension of the left regular representation of Li(G); 3) The left regular representation of C*(G) is faithful; 4) C* (G) = C(G).
Clearly, the statements 2), 3) and 4) are equivalent. 1) == 2). Let G be amenable. By Godement's condition, there is a net C L 2 (G) such that (Ai)013 Or) + 1 uniformly on any compact subset of G. Since K(G) is dense in L 2 (G), we may assume that E MG), Vi. Clearly, (A.01,01) E K(G),V/. If p is any positive functional on C* (G), then there exists unique continuous positivedefinite function (p on G such that p(g) = f(0,o(s)ds,df E L l (G). Clearly, (p(t)(Atikilth) ço(t) uniformly on any compact subset of G, and for each 1, io(.)(A.0 1 ,01 ) E L 2 (G) and ioN(A.iki , IA) is continuous positivedefinite. By the Godement's theorem, we can write Proof.
i0( (A.11)1311)1) = .)
(Pi),
640
where
{11A11211}
E L 2 (G),\11. Since a(A* , A), where A = C*(G) and (pi
Pl(f) =
ff
is bounded, it follows that p i —+ p in
( 8 )(AiSoil Sol)ds = (A(A 011401),
VI E V (C ) and
1. Therefore, any positive functional on C* (G) is a welimit of positive functionals associated with the left regular representation. By Proposition 16.3.7 , we have the statement 2). 2) == 1). By Proposition 16.3.7, for any continuous positivedefinite function io on G with ço(e) = 1 there are cp (11) ,  • • , io,ç!) E L 2 (G) such that
Pt(t) = E(Atp (1) ,A1) )  p(t) i
uniformly on any compact subset of G and yo l (e) = 1, V1. Since IC(G) is dense in L2 (G), we may assume that A l) E K(G),V1, i. By (A.A1) , çoSI) ) E L2 (G),W, and the Godement's theorem, we can write çoi(t) = (Atikhlki),
where 01 E L 2 (G),V1. Picking ço amenable.
Vt E G,
1 and by Godement's condition, G is
Q.E.D.
References. [27], [58], [61], [125], [127].
16.4. C*crossed products Definition 16.4.1. (Ay G y a) is called a C*dynamical system y if A is a Cealgebra, G is a locally compact group, a is a homomorphism from G into Aut (A), where Aut (A) is the group of all * isomorphisms of A, and t + at (a) is continuous from G to A, Va E A. Definition 16.4.2.
Let (A, G, a) be a C*system. Define
L l (G, A, a) =
If
f is measurable from G to A, and
By the norm
Milk =
f 1 f (8)11 G
„Ads
< Co
L mgsHAds,
the multiplication
(f * g)(t) = L
f (s) a, (g(s10)ds,
641
and the * operation f*(t) = A(t)_ i at (f(F 1 ))*, Vf,g E L' (G, A, a), VP, A, a) becomes a Banach * algebra, and Ilf * Ill = 11/.1113 Vf E L l ( G , A, a). Clearly, V (G ) 0 A is dense in Li (G, A, a); and Li (G,Œ , id) = L' (G).
Proposition 16.4.3. Let (A, G, a) be a C*system , {z u } be an approximate identity for V (G ) as in section 1.3, and fai l be an approximate identity for A. Then fzu (t) at (a1 )1(U ,1)} is an approximate identity for I) (G, A, a). Proof.
It suffices to show that
Ilzu()a•(at) *ga — gab + 0, and m g a * zu() 0i.( al ) — g all i  0, Vg E L l ( G), a E A. Notice that
1 g a * zu (•)a.(al ) — g a I li = =
j do Lg (s ) aas( zu (s l t)a s i t )(a1))ds — g Wall
L di ll L o =f
s ) zu (sl o aat(ao ds _
dtl(g * zu)(t) — g(t)
gw all
I • Mall
± f dtl(g * z u )(01 • Maat (ai )
—
all
and * zu)(01 • Ilaat (ai ) — all
f dtKg
0 uniformly for b E B, where B is any compact subset of A, it follows that {
Moreover, by
1 g a * zu (•)a.(a i ) — gab + f zu (s)ds = 1 we have
0,Vg E L l ( G), a E A.
IlzuNa.(az) * ga — gall'
=
L do L
.e.ti (s)g (s  1 t) as (ala)ds — Z zu(s)g(t)dsll
IS f dt f z u (s) • lg (5 1 t) — g(t)Ids • Ila i a — all G G
+11g11 1 ju zu(s)Mas(a) — allds.
642
Thus, Ilzu(•)a.(ai) * ga — gall' + 0, Vg E Remark.
Q.E.D.
(C), a E A.
A, a) and e > 0, we can find gi E L' (G),a1 E A
For any f E
such that
E giai
< 61
where j(t) = at i(f(t)),Vt E G. Thus [g(.) a.(a)Ig E L l (G), a E A] is also dense in I) , A, a). Then we can prove that zu ai (U, 1)} is also an approximate identity for LI (G , A, a). {
(G, A, a) admits a faithful * representation.
Lemma 16.4.4.
Proof. We may assume that A c B(H) for some Hilbert space. Define a * representation {7r, 1,2 (G, H)} of L' (G, A, a) as follows:
(r(f)e)(t) = L ati(f (s))e(s  q)ds, VI E V (G , A, a), e E 1,2 (G,H). Now let f E 1)(G, A, a) be such that r(f) = O. Then for any g, h E IC(G) and e, ti E H, we have 0 = (r(f)g =
h
ri)
(at 1( f (8)) e, n)g(s  't)h,(t)dsdt.
Since h E K (G) is arbitrary, it follows that (at i(f (s))e,008 l ods = 0, ;Le., ve,
E H.
Notice that f (at i (f (8)) e, n)0.5iods —
< f mat i
t
(ar i (f(s))e,
008Iodsi
osit) ids
+ 1(ar1 (f(s))e,17)1 • 108  10 — 08'04s
ai.(,.f) is continuous from G to L l (G, A, a),V f E K(G, A). Thus, fu, HI is a strongly continuous unitary representation of G. Since ur7r(x)4p(pe
=
p(ar(x)f)
=
= urp(xari (7.1 fil e
r(ar(z))p(f),
Vx E A, r E G, f E L l (G, A, a), e E H, {7r ) U 3 HI is a covariant representation of (A, G, a). Finally , for any f,g,h E L i (G, A, a), e, r, E H, we have f (r(f (t))u t p(g)e, p(h)r,)dt
=
f (p(f(t)a t (t g)), p(h)17)dt
= (P(f * g)e) P(h)n) = (P(f)P(9)e) P(h) 17).
Therefore, p = 7r x u. Moreover, since p is nondegenerate and p(f ) = f ut7r(ati(f (t))dt, V f 71"
E L i (G, A, a),
is also nondegenerate. Remark.
Q.E.D.
It is easily verified that (7r x u)(A x cr G)" = { 7 (4 u slx E A, s E C}".
Definition 18.4.8. Let (A, G, a) be a C*system. A map 40 : G + A* is said to be positivedefinite, if n
E 4•(.5isj )(as,_,(a:ai )) ? o ii.i
646
Vn, s i , • • • ,sn E G, and a l , • • • ,a„ E A. 4): G + A* is said to be continuous positivedefinite, if 40 is positivedefinite, and t + t(t)(x) is continuous on G, Vx E A. Proposition 16.4.9. Let (A, G, a) be a C*system, and (1) : G + A* be positivedefinite. 1) t —+ 410(t)(x*a t (x)) is a positivedefinite function on G,Vx E A. Consequently, (1)(e) > 0 on A. 2) 11 4) (01 2 11 4) (0112Vt E G. 3) If (p(.) is a positivedefinite function on G, then cp(•)(1)(.) : G  A* is also positivedefinite. 4) For a covariant representation {7r, u, H} of (A, G, a), and E H, let
e
t(t)(x) = ( 7 (z)ute, ),Vx E A, t E G.
Then 40 : G + A* is continuous positivedefinite. Proof.
1) For any s i , • • • ,sn E GI AI,
•,
An ET',
we have
E (DK'si)(x*ces iii,(zilxiAi id
= E 4)(szis i )(asii(a„(kx)* • a,i (Ai x))) ? O. ij
So t + 4)(t)(x*a t (x)) is positivedefinite on C,Vx E A. 2) Define [y, zi t = 4:10(t)(xsa t (y)),Vx,y E A. By 1) and Section 1.3, we have liz, z1 t 1 _< [z, z] e = 4)(e)(z* z),Vz E A.
Then from polarization, we can see that 1 4) (t)(eat(Y))1 = 1[Ylxlil 4) (e)(Y * Y + ex)
Il 4) (e)11(114 2 +110 2 ).
Let {ai } be an approximate identity for A. Then it(t)(e)I = li[11 0(t)(eat(a1))1 11 4) (e)11(11x11 2 + 1).
Further, lit(t)li = sup I4)(t)(x*)1 11z111
2 10(e)II1Vt E G.
3) For s i , •  ,sn E G, al , • • • , an E A, write Eio(s'si)(1)(s'si)(a871(a:ai)) = E kistii) id
id
647 where Aii = ço(sï l si ),p,ij = 4(sTis1)(cx81(a;a5 )),Vi, j. Since NO and (p,ii ) are two n x n positive matrices, it follows that = E Aifikeyki
E
ijk
= E{E kfikirki) 0, k ij
where (pii) = (flii)* • (14). Thus ( hot is also positivedefinite. 4) It is obvious.
Q.E.D.
Theorem 18.4.10. Let (A, G, a) be a Ce system. Then there is a bijection between the collection of all positive linear functionals on LI (G, A, a) and the collection of all continuous positivedefinite maps from G to A*. In detail, let 40 G —+ A* be continuous positivedefinite, and let
F(f) = t(t)(f(t))dt, V f E L l (G, A, a).
Then F is positive on Li (G, A, a) . Conversely, let F be a positive hear functional on 1)(G, A, a), fp, H, e l be the cyclic * representation of L' (G, A, a) generated by F, and fir, u, HI be the covariant representation of (A, G such that p = 7r x u. Define Ct)(x) = (7r(z)u t e, ) l dt E G,x E A. Then 4) : G + A* is continuous positivedefinite, and F(f) = (1)(t)(f (t))dt, V f E L l (G, A, a).
Moreover, we have that 11 4) (011 dence.
10(e)11 = ilF11
G G) in above correspon
Let F be a positive linear functional on Li (G, A, a), and fp, H, e l, {7r, u, HI and 4 be as above. Then Proof.
F(f) =
=
(P(.0e3) = ((x" x
(74 f (t))u t , e )dt =
e) f (t))dt,
V f E L l (G, A, a). If ffi l is a bounded approximate identity for Li (G, A, a), then
11Fii =
= iirri(p(foe, e) = ei12.
648 On the other hand, since follows that
{7r,
HI is a nondegenerate * representation of A, it
= sup{140(e)(x)11x E A, 11x11 < 1} = sup{1(7r(x)e,
0 Hz E A l lIzil < 11
= lip I Mai )e, e)I = iier = iiFil _>
ow II, vt E G,
where {ai is an approximate identity for A. Now let 4) : G + A* be continuous positvedefinite. Since Ll(G) 0 A is dense in Ll (G , A, a), t + 41)(t)(f (t)) is measurable on G, Vf G L i (G , A, a). By Proposition 16.4.9, we have 140(t)(f (MI < 2 11 4) (01'11f (t)111Vf E 1,1 (G, A, a). Thus , we can define a linear functional }
F(f) = f t(t)(f (t))dt, V f E 1, 1 (G, A, a).
For any
913 • • •
3
9k E K(G),a1, • •  , ak E A, si,  • • , si E G, notice that
E4)(8,,18,„)(a:a iw i sm (aiNi(sri gi (s,n ) iinm
=
E 4.(sin18,,n)(a87 1(a:naini )) ? 0 In
(i,n),(f,m)
where
sin = SnI ain =
g(s)a, n (ai) ,Vi, n. It follows that
E f f 4)(8  '0 (a:a.It (ai))gi(s)gi(t)dsdt > 0 id
i.e., E f 4)(t) (a: at (aj ))(g: * gj)(t)dt > 0, Vg i , •  • , gk E MG), a i , •  • , ak G A. Therefore, F is positive on L l (G, A, a).
Q.E.D. Corollary 16.4.11. and
There is a bijection between the state space of A x a G
40 : G + A* is continuous positivedefinite, 1 and 10( ) I 1 = 1, i.e., 40(e) is a state on A f ' {4)
Let (A, G, a) be a Cesystem, and {7r, HI be a * reperesentation of A. Define
f (Tr(x)e)(t) = wiceti(x))e(t),
t (A(8 ))(t) = Vz G A, s G G," G L 2 (G , M. Then {Tr, A,L2 (G, TI)} is a covariant representation of (A, G, a), i.e., A(s)r(x)A(s) * = Tr(a s (x)),
Vx E A, s E G
649
( it is similar to Proposition 16.1.5). Further, we have a * representation {Tr x A, V(G,H)} of Ax a G: (Tr x A)(f) =
G
Tr( f (0) A(t)dt ,
f E L i (G , A, a).
The * representation {Tr x A, V(G, _MI is called the regular representation of A x a G induced by the * representation {7r, HI of A, and denoted by {Ind7r,L 2 (G, H)}, i.e., Definition 16.4.12.
Ind7r(f) = LTr ( f (8 )) A (s ) ds Or
(Ind(f))(t) =
Vf E
(G, A, a),
7r o at i(f(s))e(s i t)ds,
e E L 2 (G,H).
Now we make the following discussions. 1) Let {7r, HI be a * representation of A, E Al(C H) be a cyclic set of vectors for 7(A), and ffi lj E 7\1(C L2 (G)) be a cyclic set of vectors for A(Li(G)), where {A,L2 (G)} is the left regular representation of Ll(G). Then fi J. } is cyclic for Indr(A x a G). In fact, let e E {Indr(A x a G)(fj 0 6)1i, j}1 . Then for any f E IC(G) and E A , we have 0 = (Indr( f x)(fi 6 ), e)
e(t))dsdt = f (7r o at1(x)6, e(0)(f. fmodt. =f
(7r(ati(x))f(s)fi(s i t)Ei,
Since { f * f = A(f) fil f E K(G),j} is total in L 2 (G), it follows that (7r o ati (x) (0) = 0, a.e., Vx E A, i. For any compact subset E of G with 1E1 > 0, and e > 0, by the Lusin theorem there is a compact subset F of G with F C E and IEVI < e such that "(.) :F+His continuous, where 1 BI = xB (t)dt for any Borel subset B of G. From Proposition 5.1.2 and 0 < IFI O. Now it is easily verified that (7r0 at i(x) 1 e(t)) = 0, vt E Un Kn ,x E A, i. .
660
Since {7r (x)ei lx E A, i} is total in H , it follows that e(t) = 0,dt E IJ„K„. But e ( > 0) is arbitrary, so e (0 = 0, a.e., on E for any compact subset E of G, i.e., e = 0 in L2 (G, H). 2) Let ço be a positive linear functional on A, and fn f2 G K(G). Define
Ofif2(x) = f f fi(8 1 0Mtlp 0 at i(x(s))dsdt,
(1)
Vx E V(C, A, a) . If {7rv„ 14, ev } is the cyclic * representation of A generated by (p , then (Indxp (x)fi
=
® e,, f2 0 4,)
f (f 7rp 0 ati(x(8))fi(s i t)ev,ds, f2 (t)e p )dt
(2)
= fjpo ati(x(s)) ii(s l t) .f ij dsdt = QS fif2 (x),
Vx E L i (G I A, cx). Now notice the following fact: ■
{Ind7rgy)f
® epif E K(G),y E K(G, A)}
is dense in L 2 (G,Iiv ), where y C K(G, A) means that y(.) : G + A is continuous and supp y is compact. Indeed, for any g E IC(G) and a E A, we have (Ind7rgg 0 a)zu 0 ep)(t) = (g * zu)(t)rço 0 at 1 (agv„
where {z u } is an approximate identity for V (G) as in Section 16.3. Moreover, li(g* zu(.)7r. 0 a.1(a)4, — gOr p o a,1(ag v Ili iii,2(G,H,) =
f 117r,0 (3 ati(a)Ir . 19 * zu(t) — g(t)I 2dt 114 2 ' lkoll '
Ji
* zu(g) — g(t)1 2dt
Mall 2 'IlSoll* (110100 + 1191100 • s2 IA (s)I 1 )119* zu
—
9111
Thus , the closure of {Indz gEgi 0 ai)zu 0 evlai E A, g 1 E K(G),U} i
in L2 (G, H) contains the following subset L;
r = {Egi(*) 7rço ° a. i(a)la, E Al gi E K(G)}. i
).
O.
651
.e
Similar to 1), is dense in L 2 (G,H 9,). Therefore, {Ind 7rgy)f K(G), y E IC(G, A)} is dense in L 2 (G,Hv,). Now from (1) , (2) and above fact, we get sup f
I 1 Ind7rv, (x) 11 =
Ilind7rv,(xy)f 0 ev,ji I f
1 IlIncl7rv,(y)f (8)
eSo I I I
® e,i f
E
E K(G), y E K(G, A), and } 11 111(17r40 (V) f ® eSo II >0
Off (V z* xy) 1/2 y E MG, A), f E K(G), I = sup { Off (y* y) 112 and Off (y*y) > 0 f. 3) For any y E
MG, A) and f
1(y, f) =
t0
E K(G), define
cp is a positive functional on A, } and Off (y*y) > 0
And for any * representation 7r of A, let A4(7r) =
t0
Clearly, there is a subset
1 ça _> 0 on A, and 7rw is weakly contained in 7r, i.e., ker7r C ker7rv, f •
A of .M (7r)
such that
7r = EDpen7rp ED 0. Then Ind7r = EN,EA Ind7r9, @ 0, and by (3), (4) we have
IlInd7r(x)11 = sup{Ilincl7rgx)111S0 E Al
I I Indri (x) I 1 , V/ , in particular, IlIncl7r(z)11 > I I Ind7rço (x) II 3 Vz E L I (C , A, a). n
(ii) Let (p =
E vi, where vi () = (7r(.), "i ), and ei E HT ,1 < I < n. We
claim that
IlInd7r9,(x)11  mflx I I indri (x) II 1
Vx E L l (G, A, a),
where 7ri = 7r9„.,1 < à < n. From this inequality and (i ) , we shall get ‹ Ilind7r(z)11,Vx E V(C, A, a). Ilindwp(x)11 
652
In fact, for any x E Ll (G, A, cx),y E K (C , A), f E IC(G) and
Off (y*y) > 0,
let ai = (ef(Y s y)ligi = (4)1(Y s z * xY), where ‘,3!Pf = (pi) ff ,1 < i < n. Since (6!Pf is positive on 1,1 (G, A, a), it follows that A = 0 if ai = O. Then by (3),
(,25 11(VerY) _ Pi + ' • • ± i6n < max {gili with ai > 0} ai l  a l + • • • + an — Off (Y * Y)
< max IlInd7ri(x) 11 2 . I
Further, by (3) we have that IlInd7rv,(x)11 ‹ max IlIndiri (x) II, Vx E L i ( G, A, a).
(iii) Let (p E .M(7r) with lkoll = 1. From Proposition 16.3.7, there is a net {(p i } of states on A such that each (p i is a sum of positive functionals associated with 7r, and (p i (a) + (p(a) ,V a E A. Clearly, for any a E A, (pl(a t (a)) —+ p(a t (a)) uniformly on any compact subset of G. Thus (8) a) —+ ç'ôff (g 0 a) , Vf,g E K(G), a E A, where OPf = (pi) ff ,Vi. By (1) , Ik3(if311
11f1100•11f111,vi.
It follows that
0.(»f (z) + Off(z),Vz E 1, 1 (G, A, a). Now from (3) and (ii) , we get 11inchso (x) li <sup II Ind7rçoi (x) II < IlInd7r(z)11,
Vz E LI (G, A, a). Therefore, we have that
Il 1nd 7r(z)11 =supfilInd7rgx)Ilko E Al(701
=
O 11 (y*x*xy) 1 /2 1 y E K(G, A), f Elf (G), sup { , * 12 SO E M(7r) n n(v)f) SOff(Y Y) /
(6) }
,
V x E Li (G , A, a).
From the proof of Lemma 16.4.4, if {7r,H} is the universal * representation of A, then {Ind7r„, L 2 (G, H)} is a faithful * representatio of L' (G, A, a) • Thus, 11x1Ir = sup{ I 1 Ind7r(z)111 7r is a* representation of AI will be a C 5norm on 1,1 (G , A, a). Definition 16.4.13. Let (A, G, a) be a C5system. The completion of (V (G , A, a), 11 ' Ilr) is called the reduced crossed product of A by the action a
653
of G, and denoted by A x ar G, where 11x11r = suPfilindr(z)111 7r is a * representation of Al, Vz E V (C , A, a). By (6), we have that 11x11r = sup { Vz E
) f f(Y * z e xy)112 y E off (y * y) 1/2
A), f E
E
MG),
}
(7)
n(y1f)
(G, A, a).
Lemma 16.4.14. Let {7r, HI be a * representation of A, and t E G. Then the covariant representations {rr, A, L2 (G,11)} and fir o at , A, L 2 (G, H)} of (A, G, a) are unitary equivalent, and the * representations {Ind7r,L2 (G, H)} and {Inch 0 at , L 2 (G, H)} of A x a G are unitary equivalent. In detail, if define (Ute )( • ) = A(01/2e•),ve G L 2 (G, H), then we have Ut*Tr(a)Ut = r o at (a), U(t)A(s) = Ut*Incl7r(x)Ut = Incl(7r o Va E A, 8,t E G l x E A x a G.
For any
Proof.
E L 2 (G, H), a E A, r, s, tEG,xE
(Ut*Tr(a)Ut e)(s)
(G, A, a), we have
= (Tr(a)Ut e)(st 1 ) A (t)  0 = ir o a 8 1(a)(ut )(st  ') A (t)  V2 = (7r o at ) o asi(a)(s) = (7r o
and (Ut* Ind7r(z)Ute)(r) = A(t) 1 /2 (Indr(z)Ute)(rt 1 )
f 7r o at,.1(.(s))(ute)(s i rtlds
f
A
(7r o at ) o ar _1(x(s))(s l r)ds = (ind(7r o
that comes to the conclusion.
Q.E.D.
Theorem 16.4.15. Let (A, G, a) be a Cesystem, and {7r, HI be a * representation of A. Then the following statements are equivalent: 1)E ED7r o at is a faithful * representation of A; tEG
(G, A, a). 2) Ilind7r(x)11 = 11x11,.,Vx E Consequently, if {7r, H} is a faithful * representation of A, then {Ind7r,L2 (G, H)} can be uniquely extended to a faithful * representation of A x ar G.
654
Proof. Suppose that IlInd7r(z)II = IlzIlr)Vz E V (C ) A, a). If a G A satisfies ir o at (a) = 0,Vt e G, then
(Ind7r(g 0 a)e)(t) = f g(s)7r o a t 1(a)(8 1 t)cl8 = 0, Vg E K (G),
e E L2 (G, H).
Thus we have
Mg ® 4 = IlIncl7r(g 0 a)Il = 0,Vg
E K(G),
E ED7r o at is a faithful * representation of A. Conversely, let E (Dr . at be a faithful * representation of A.
and a = O. So
tEG
By Proposi
tEG
tion 16.3.7, any state on A is a welimit of states which are sums of positive functionals associated with {7r o at It E G}. By Lemma 16.4.14, we have IlInd7r(z)11= 11 111* ° at )(x)II,
Vt E G, x E L l (G, A, a).
Then from the above discussion 3) (ii) and (iii), and the formulas (6) , (7), we get
Ilind7r(z)11
=
SUp {
Off ( Y * z* xy) 1 /2 (7) f f (y s 0 1/2
i6ff (Y*X*XY)1/2
Off (y $ 0112
= SUp {
=• 11xlirl
Off (y * X S ZY) 1/2 ç (y * 0112
aiff
y e IC(G, A), f E K(G), and 1 yo E n(y,f) n (utEG Al (7 oat)) I
MG, A), f E K(G),(,o E 1/(y, f), P = Sol + • • • + Son1Soi E .M (7r o a ti ),Vi y E
y E
MG, A), f So
E
E K(G),
}
and }
n(v, f)
Vz E V(G, A, a).
Q.E.D.
Let a = id, and 7r be a faithful * representation of A. Then for any g E L i (G) and a E A we have Example.
Ind7r(g 0 a) = A(g) 0 7r(a)
on 1,2 (G, H) = L 2 (G) 0 H. Therefore, we get A x ar G = min(A 0 C,*. (0), where II • Ilmin(= act() in Chapter 3) is the spatial Cenorm.
655
Proposition 16.4.16. Let (A, G, a) be a Cesystem, and B be a C*subalgebra of A with at (B) = B ,Vt E G. Then we have that B x ar G c* A x ar G. Proof.
It suffices to show that E L i (G, B, a).
11x11BxarG =
Fix a z E (G, B, a). Clearly, 11x1lAxarG Conversely, let {7r, H} be a * representation of B. Since each state on swEA7r,o , where A is a subset B can be extended to a state on A, and 7r of the state space on B, there is a * representation {p, K} of A such that H C IC, p(b)H C H, and p(b)IH = 7r (b),Vb E B. Then we have (Indp(x))(t) = Lp o = L7ro at 1(x(s))e(s l t)d8 = (Ind7r(x))(t), \9/ . E L 2 (G,H) C L2 (G, K). Thus, kUAXaTG
Further, we get IzIAx G ?y
?_ IlindP(x)11? Ilind7r(z)11. BxarGI
andlizlisxarG = lixIlAxafa•
Q.E.D.
Theorem 16.4.17. Let (A, G, a) be a Cesystem, and G be amenable. Then A x a G = A x o„. G.
It suffices to show that Mir M,Vx E (G, A, a). For any state it) on L i (G, A, cx), by the GNS construction there is a cyclic * representation {7rv,, 14 e„,} of A x a G with = 1. By Theorem 16.4.7 , we have unique covariant representation p, u, Ii v,} of (A, G, a) such that
Proof.
,,
{
7rv, = p X u.
Since G is amenable, by Godement's condition there is a net {0 C L2 (G) such that (Atgt, gt) —+ 1
uniformly on any compact subset of G. We may assume that 1191112 = 1,v1 . Let iot = (Indp(y)ei , ei) ,V y E Li (G, A, a), where e,(s) = g1 (.$) vs E
656 G,1. Clearly,
6
E L 2 (G,Hy,) and 11611 = 1, Vl. Notice that
Sot(Y) =
(79(y(t))A(t)6, eodt
= f f (p o a8i(0))6(ri 8), 6 (x))dsdt gi(t s)g i (s)(u 8 (p o a 8 _1(y(t)))u:u t 4,ep )dsdt
=
=
(kg:3M) • (p(0))ute,,
f (p(0))u t e,, ev,)dt
e,)dt
= ço(y),
Vy E L l (G, A, a). Further,
(p(x* = lipacpz (zer) = lipa(Indp(zer)i,
6) _< I I x
V x E L' (G, A, a). Therefore, we get
11x1 12
= sup{p(x* x)lio is a state on V (G , A, a)} 1141
Vx G V (G , A, a). Q.E.D.
Corollary 16.4.18. Let (A, G, a) be a Cesystem, G be amenable, and B A x a G. be a Cesubalgebra of A with at (B) = B,Vt E G. Then Bx.G Now we discuss the embedding of A and G into M(A x a G) for a Cesystem (A, G, a), where M(A x a G) is the multiplier algebra of A x a G (see Section 2.1.2).
Let (A, G, a) be a Cesystem, and fp, HI be a nondegenerate faithful * representation of Ax a G. Clearly, { p, H} is still nondegenerate and faithful for (G, A, a). By Theorem 16.4.7, there is a covariant representation {7r, u, HI of (A, G, a) such that p = 7r x u. Since p(f) = L7r (f( s)) usds = Lusw (asi ( f (s))) ds, V f E Li (G, A, a) and p(g 0 a) = L7r (g (s) a) usds = 7r(a) Lg (s ) usds,
Vg E L l (G) and a E A, it follows that {7r, H} is also faithful and nondegenerate
for A.
657
For any a E A and f E L i (G , A, a), define (L a f)(t) = a f (0, (R a f)(t) = f (t)a t (a), Vt E G.
Clearly, La and Ra are bounded on Li (G, A, a) with norm < Mall. Further we have 'O a f) = 7r(a)p(f), and p(R a f) = p(f)ir(a).
)
Hence, 1lLaf IlAxaG =11P(Laf)II < Mall • Ilf IlAxac , and 11Ra f 1lAx ac Mail .11IMAxaG. Then, La and Ra can be extended to bounded linear operators on A x a G with 114, 11711RJ —‹ 114) and L ax = p 1 (7r (a)p(x)), R ax = 131 (p(x)7r (a)) , Va E A,z E A X a G. Of course , these expressions are independent of the choice of fp, HI. Moreover, it is easily verififed that
L a (xy) = (L a x)y, , R a (xy) = x(R a (y)),
(L ay) = (R a x)y,,
Vx, y E A X a G. So that (La , Ra) is a double centralizer of A x a G, Va E A (see Definition 2.12.5). Then we get a * isomorphism a —> (L a , Ra) from A into M(A x a G). By Section 2.12, we can write that a = (a or s)lim(L a fi) = (a or s) lim(Rafi),
Va E A, where fi } is an approximate identity for Li (G, A, a) (then for A x a G), and a or stopology is in (A x o, G)** . Now for any s E G and f E L i (G , A, a), define {
(L , f)(t) = a, (f (8 1 0), (R, f)(t) = A(8) 1 f (t 8 1 ) ,
Vt E G.
Clearly, L,, R, are bounded on L' (G, A, a). Further, we have P(I, 8 f) = usp(f), p(R8f)= p(f)u,.
Hence, L, and R, can be extended to bounded linear operators on A x a G, and L s x = p 1 (u,p(x)), R s x = p 1 (p(x)u,), Vs E G,x E A >< G, G. It is easily verified that (L 8 , R3) is a double centralizer of A x a G, and (L 3 , R,) is a unitary element of M(A x a G). Then we get a faithful representation s —> (.1.3 , R 3 ) of G into the group of all unitary elements of M(A x a G). By Section 2.12, we can write that
t = (a or s) lim(Lt fi ) = (a or s)lim(Rtft) Vt E G.
7
658
We say that s —> (L., R3) is continuous with respect to the strict topology in M(A x a G) (see Definition 2.12.11). In fact, it suffices to show that
11L8f — f I —> 0,
and
11R8 f — f 11 —' 0
inAx a Gass—>einG,VfEK(G, A). But it is obvious since iiLif — fi < fo lif(s10—AoliAdtFLII(as — 1)(f (0)11 A dt , IIR8f — flli
LIA(s1)1.11f(ts1)— f (t)I1Adt + L1 , (8)111.11 1(t) II Adt ,
ilglii,dg E L i (G I A, a). Finally, regarding A and G as subset of M(A x a G), and regarding {p = 7r X u, HI as a faithful * representation of M(A x a G) (see Proposition 2.12.9),
and 011AxaG
we have that
p(a) = % ( a),
p(s) = u„
and sas 1 =
Va E A, s E G. In fact, the equalities p(a) = 7r(a) and p(s) = u, are obvious. Further, by
p(sas 1 ) = u 3 7r(a)u: = 7r(a,(a)) = p(a i (a)) we have s as 1 = as (a) in M (A )( a G). Summing up the above discussion, we have the following.
Proposition 16.4.19. Let (A, G, a) be a Cesystem, and M(A x„ G) be the multiplier algebra of A x C. Then A and G can be embedded into M(A x a G) such that
se = s* s = 1, sas 1 =
(s f)(t) = (L, f)(t) = (f 8)(0 =
(Rs Mt) = A(8) — 'fits — i), (af)(t) = (L. f )(t ) = a f (0 ,
(f a)(t) = (R a f)(t) = f (t)a t (a),
in the sense of M(A X ct G),Vs, t E G, a E A, and f E L i (C , A, a). Moreover, if {7r, u, H} is a covariant representation of (A, G, a) such that f7r x u, HI is faithful and nondegenerate for A x a C, then we have p(a) = 7r  (a) , Xs) = u„ Va E A l s E G I where {p, H} is the faithful * extension of fr x u, HI on M(A x a G).
659
Crossed products of Cealgebras with discrete group were introNotes. duced by T.Turumaru. Later G.ZellerMeyer carried out a penetrating analysis. General crossed products were defined by S.Doplicker, D.Kastler and D.W.Robinson, and Theorem 16.4.7 is also due to them. The reduced crossed products were defined by G.Zellermeyer for discrete groups, and generalized by H.Takai. References.
[29], [127], [166], [186], [202].
16.5. Takai's duality theorem Two Cesystems (A, G, a) and (B, G, 0) are said to be isomophic, denoted by (A, G, a) ''=' (B, G, 0), if there is a * isomorphism 40 from A onto B such that 4) o at o 4) 1 = fit , Vt E G. Definition 16.5.1.
Proposition 16.5.2. * isomorphic.
Proof.
If (A, G, a) .=' (B , G, 0), then Ax a G and Bx fi G are
Let 4) be a * isomorphism from A onto B with 4)o at = fit 0 4) ,Vt E G.
Define W (f)(t) = 4)(f (t)),V f E L' (G, A, a). Clearly, * is a * isomorphism from V (C , Aa) onto L i (G , B, 0). Moreover, if 7r is a * representation of V(G,B 4 O), then 7r 0 * is a * representation of L' (G, A, a) obviously. Conversely, for any * represetation p of L' (G, A, a) , there is a * represntation 7r of Ll(G,B, 0) such that p = ir o W. Therefore, we have 11*(f)11 = suglw ° xli (f)111 7r is a * represntation of Li (G , B, (3)}
= sup{11P(f)111P is a * representation of Li (G, A, a)}
= 11f112
V f E L l (G, A, a),
and * can be uniquely extended to a * isomorphism from A x a G onto B Xfi G. Q.E.D. Proposition 16.5.3. Let (A, G, a) and (B, G, (3) be two Cesystems. Then there is a ( tensor product ) Cesystem (min(A 0 13), C, a 0 0) such that
(a 0 M t (a 0 b) = a t (a) 0 A (0 , Vt E G, a E A, b E B, where min(A 0 B) means the injective tensor product, and 11 • Ilmin( = a0(*) in Chapter 3) is the spatial Cienorm on A 0 B.
660
Proof.
It suffices to show that n
1 t at(ai) vt
0 Pt(bi)Ilmin = II Eai 0 bill,
E G, ai E A, bi E B,1 p t be the right regular representation of Con L 2 (G), i.e., (pt )(8) = A (t) 1 /2 (st),V E L 2 (G) . Then (C(L 2 (G)),G,adp) is a C* system, where adp t (a) = p t ap;,Vt E G, a E Lemma 16.5.4.
C (L2 (0). Proof.
It suffices to show that IIPtaP; — all = IIPta — apt il + 0
as
t—>einG,
Va E C(L 2 (G)). Since the subset of all finite rank operators is dense in C(L 2 (G)), we may assume that a is an onerank operator e 0 th where e, n E L2 (G). For any ç E L2 (G) with lk 1 1 2 5._ 1, we have II(Pta — aPOS 112 = il(S )n)Pt — (Ptçl /7)02 II(S )n)(Pt
— . )112+
11(çoi) — (s,ptin)02
11n112 . IIPte — 02 + lie112  11/7 —
Pt0711
* 0
as t + e in G. That comes to the conclusion.
Q.E.D.
Now let (A, G, a) be a Cesystem, and Cr (G, A) = min (C°(C) ®A)
=
ff:G—>Alfis continuous, and Ilf 011 E Cr (G)}.
By Proposition 16.5.3, (Cr (G, A), G, y) is a Ciesystem, where y = A® a, and (At g)(s) = g(t 1 8),Vg E Cr (G) (notice that g is uniformly continuous on G). Lemma 16.5.5.
Let (A, G, a) be a Cesystem. Then
(Cr (G, A) x i G,G, p)
is also a Cesystem, where (P(t)f)(s) = Ptf (s),Vt E G, f E
Ll (G, Cr (GI) A))
'y),
661
and (AO (3) = g(st),Vt E G,g E C(G, A). Proof. Since Po/ = '78Pt on Cr(G, A) and p(s)p(t) = p(st),V8,t E G, p t is a * isomrophism of V (G , Cr (G, A), 7), Vt E G. If {7r, u} is a covariant representation of (q(G, A), G) '7), then {r O Pt, u} is also a covariant representation of (Cr (G, A), G, ry),Vt E G. Thus, we have
IIP(t)f II {7r, u} is a covariant = sup {ii 7r o pt (f(s))u s dall representation of (Cr (G, A), G, y) J = suP{
I
r(f(s)) 8ds li 1{7r, u}is as above} = 11111m
(G, A), ry), and p(t) can be uniquely extended to a * automorVf E L' (G, phism of Cr (G, A) x 7 G,Vt E G. If f = g h, where g E V (G ), h E K(G, A), then
liP(t)f – 111 c 11P(Of – fill f() f (8)11c r ,A)ds
=
Ig(s)cis • sup lh(rt) – h(r)IIiiA —+ 0
=
rEG
as t e in G. Therefore, (Cr(G, A) x 7 G, G, p) is a C* system.
Q.E.D.
Theorem 1 8.5.13. Let (A, G, a) be a C*system. Then the C*systems min(A (8) C(L 2 (G)), G, a adp) and (C1°,°(G, A) x G,G, p) are isomorphic. We may assume that A c B(H). Define a faithful * representation {7r, L2(G, FM of Cr (G, A) as follows: Proof.
(7r( f))(t) = f (t)
0),
(G, A),
VI E
It generates a faithful * representation {Inch = Cr(G, A) x 7 G, i.e., 
(Ind7r(z)e)(s,t) = = (fir o
(z(r,
e E 1,2 (G,11). x A, L 2 (G x G, H)} of
(TrIz(r,.))A(r)e)(s, .)dr)(t)
.)dr)(t)
cx, (z(r, st))e(r  i s, t)dr,
Vz E K(G,q, 3 (G, A)),
E L 2 (G x G, H). Define a unitary operator w on
662
L 2 (G X G, H) as follows:
f (we) (x, t) = A(0 2 e(st, t), t (w* , = A(01/2 e(st1,
ve E L2(G X G , H).
Then we have (w*Ind7r(z)tve)(s, t) = (Indr(z)we)(st 1 , t) • A(t) 112 a t8 i(z(r,$))(tve)(r 1 st  , t)dr A (t) 112 ats i (z(r,
Vz G K (G , C (G, A)), s) g (s), where a E
s)) e V13,0dr,
e G L2 (G x G, . In particular, if A, f , g E K(G), then
(w*Ind7r(z)tve)(s,
E L2(G
X
= g(s) at (a)
f (r  1 s) (7 1 s , t) dr
J = g(s)at (a) J1 (r)
G, H) . Further, define a faithful
z(r,$) = ot,(a) f
(1) (r,t)dr
I A (0,
* representation { if, L2 (G , H)}
of A as follows :
(t) = at (a)e(t), V E L2 (G, Fi), a E A.
(f(a)
Then we have w*Ind7r(z)w = f(a)
E f(A)
(2)
C(L 2 (G)),
where z(r,$) = as (a) f (r  s)g(s), f g G IC(G), fi(r) = f (r) / A (0, and v ftg is an onerank operator on L2 (G) : vfig n = ( n ,
Vt7 E L2 (G).
Now we claim that
If (r  1 s)g (s)a,(a) is dense in Cr(G, as follows:
E A, f,g E K(G)]
xt G. In fact, define 40 : IC(G x G,
( 4)y)(r)s) =
K (G X G,
A)
( 4)1 y)(r)s) = Y(sr i /s),
Vif E K(G x G, A). Clearly, t is a bijection, and keeps the maximal norm. For any y E K(G x G, A) and c > 0, there is a compact subset F of G such that suppy C
F X F.
663
We may assume that e E F and F = F '. Pick o = c /1F2 1, and fi ,gi E K (F), ai E A such that ,T611Y(r5s) —
>
f(r)g(s)a 3 (a1)IA < 0.
Then we have max taEG 11( 4FY) ( 1.1 8) — Z(r ) 8 )11A
1 uniformly on any compact subset of G as p —> i' in Ô. Therefore, we obtain a new C*system (A x a G, 6,&). Definition 16.5.7. Let (A, G, a) be a C*system, and G be abelian. The above Cesystem (A x a G, 6,6 is called the dual system of (A, G, a), and ôt )
is called the dual action of a. Proposition 16.5.8.
If (A, G, a) L' (B, G 06), and G is abelian, then we
have (A x a G,6,&) _ , (B x 0 G,6,,à). Let 40 be a * isomorphism from A onto B with (Do a t r igto4P,vt E G. Then xli is a * isomorphism from A x a G into B x i3 G by Proposition 16.5.2. Now it suffices to show that 'If 0 exp 0 W1 = Sp ,Vp E G. For any p E d and f E Lii (G, B, [3), let g = 4i1 CO (E 1,1 (G , A, a)). Then Proof.
IF
0 Ôtp 0 XII — 1 ( f ) (i) = XII 0 & ID
(OW
= (1) ( 6i p(g)(t)) = (t, 0 (9 (t)) =
(At) f (t) =
That comes to the conclusion. Theorem 16.5.9.
Q.E.D.
Let (A, G, a) be a Ce system, and G be abelian. Then
we have that ((A x a G) x,,,,, 6,G,&) 2f (min(A 0 C (.1. 2 (G ))) , G, a® adp).
665
1) Let {7, HI be a faithful * representatio of A. By Theorem 16.4.15, {Ind 7r = rr X A, L 2 (G, H)} is a faithful * representation of A x a G. Further, {fr, L 2 (6 x G, H)} is a faithful * representation of (A x. G) x t, 6, where = Ind(Indir) = Indir x A, and {Ind7r, A, L 2 (6 x G, H)} is a covariant representation of (A x a G, 6, &). Now let f E IC(6 x G, A), e G L2 (d X G, H), and consider Fr( f) e. For any p E 6, t G 6, we have Proof.
(fr(f)e)(p, t) = “Indir x A) ( f) e)(p, t)
= (f Ind7r(f (q, .))Ag edq)(p, t) = (f Ind7r o "cip i(f (q, .))e( qlp, .)dq)(t) .,dq f dsx  o at i(6C1,1(f (q, .)) (.5) e(q  1 p, s l t) = 1G G =
ï dq f ds(s, p)r 0 a t i(f (q, sile(q 1 p, 8 1 0 G G
Thus , we can write that fr(f) =
f faxG f (q, s)u(s)v(q)dsdq,
Vf E IC(6 X G, A), where
(ae)(p,t) : 7r 0 (u (8) ) (p, t) — (8, p) e(p, 8  '0, {
(v(q)
) (p, t) —
Va E A, s, t E G5p,q E 6, e E L2 (Ô x G, H). 2) Define a unitary operator on L2 (d x G, H) as follows:
(./ ) (7), t)  (t, p)e(p, t),
(J* ) (p, t) = (t, p) e(p, 0 ,
vt E G,p E d, e E L2 (6 x G, _M. By 1), we have (./(f)./ * e)(p, 0
= (t,p) .[TGx0 dqds(s, p»r 0 = f LxG
dqds(s,q)(t, q)x  o at i(f(q, s)) e (q 1 p, s  1 t).
Thus, we can write that .1 i (f)J* = f faxG f (q, 3W WTI (q)dsdq,
666
V f E K(d x G, A), where
{
(ae)(P) t) = 7r ° ati(a)e(P)t)) (u1 (3) ) (p, t) = ID__ ;1, ( vr (q ) 0 (p, t) = (t, q) e(q lp, t),
d, e E L2(6 x G,H). 3) Consider the Ce system (A, 6, id). Clearly,
Va E A, s, t E G, p, q E
A xid 6
min(A 0 Cr (G)) = Cr (G, A).
Define a homomorphism /3: G —* Aut(L 1 (Ô,A,id)) as follows: fit(h(P) = (t1P)at(f(P)) , Vt E G,p E 6,:t E L 1 (Ô, A,id). If {L, v} is a covariant representation of (A, 6, id), then L(a)Vp = VpL(a),Va E A,p E G. Thus {L o at , (t, .)v } is also a covariant representation of (A, d, id),Vt E G, and
IIA (t) II = sup fil f
{L, v} is a covariant
L([3t(i)(P)) V p dPII
}
representation of (A, 6, id)
= sup{ 11 f L 0 at(f(P))(t1P) Vp dPIII{LIV } is as above} = Ilf II) Vf E L1 (6, A,id),t E G
Further, fit can be uniquely extended to a * automorphism of A xid d, vt E G. Moreover, notice that
ilf3t(h
Al 1
ma 1111
= Lli(t3P)at(AP)) — 1(P)11 (1P
5_
fa I (t, p) — i 11 1(p)IIdP + fa llatv(p )) Rp)IldP)
VIE L 1 (6,A,id). As t —> ein G, since (t,p) —> 1 uniformly on any compact
subset of 6, we can see that the first term
L 1 (t,p) _11.11.f(oidp __„ o; if
pick Ap) = g(p)a, where g E L 1 (6) and a E A, then we also have
L
II at (1(p)) — 1(r) I1dP + o
as t
e in G.
Therefore, (A x id 6, G, fi) is a C*system. 4) Let {7r, HI be a faithful * representation of A. Then {(7r 0 id) x A, L2 (d, TI)} is a faithful * representation of A x id Ô. Further, {if, L 2 (6 x
667
G, H)} will be a faithful * representation of (A x id 6) Xfi G, where if = Ind((x 0 id) x A). For any f E K(6' X G, A) and e E L2 (6 X G, H), compute
ff(f)e
as follows: (
r(f) e)(p3
= (fG (7r 0 id)
x A)(f (., s))A s eds)(p,t)
=
x A) 0 fit i (f(., s))(.,s l t)ds)(p)
=

dqds(t,q)7r o a t i(f (q,$))e(q  p, 3'0 . J LxG.
Thus, we can write that
faxa f (q, s)v i (q)u.' (s)dqds,
(f) = VI E K(6 X G, A), where
e) (PI t) = CI (u' (s) (p, t) = t(p,s lt), (11 (q) e) t) = (t, q)
{
Va E A, s, t E G, p, q E Ô it follows from 2) that
,
e E L2 (6 X G, H).
Since vi (q)ui (8) = (s, q)ti' (s)t1 (q),
= f faxGg(q,$)12 (s)t1 (q)dqds = ft (Or , where g (q, s) = (s q) f (q,$),V f E KO X G, A). 5) Notice that K(Ô x G, A) is dense in (A xid d) So if define (1)(f)(p,t) =
(t, p)f (p, t), Vf
E
X
G and (A x. G) x a Ô.
K(6 X
then by 2), 4) 4:1) can be uniquely extended to a * isomorphism from (A Xp G onto (A x. G) x a 6. 6) It is wellknown that the Fourier transform E L1 (6) + f(s) =
k (s, p) Ap)dp E
can be uniquely extended to a * isomorphism from
f
a —+
f
a,
(G)
C* (Ô) onto Cr (G).
vf E L 1 (6), a E A
Y —id
Then
668
can be uniquely extended to a * isomorphism * from min(C*(6) 0 A) A xid 6 onto min(Cr (G) (s) A) = cr(G, A). Morevoer, by 3) we have
=
IF 0 f3t 0 * 1 (f 0 a) = * 0 ,6t(f 0 a) = 4I (4, 40 at (a)) = f (')P)(t)P) .7(P) dp 0 a t (a) = f (t 1 .) 0 at(a), 'If E Cr (G), a E A. Thus * is an isomorphism from (A xid 6, G, 0) onto (Cr (G, A), G, y), where 7 = A® a. By Proposition 16.5.2, the Fourier trans
formation
7 E K(6 x G, A) As, t) = h (s,p)A p,t) dp can be uniquely extended to a * isomorphism from (A xid 6) Cf(G, A) x 7 G. 7) By 5) and 6), the map
Xp G
onto
3 : z(s, t) = a f (s)g(t) —+ (s, p)a f (s)0p), Va E A, f E K(G),"# G L' (Ô) and g(t) = ki (t, p)d(p)dp E Cf (G), can be
uniquely extended to a * isomorphism from Cr (G, A) x 7 G onto (A x,,, G) xa 6. Further, for any z(s,t) = a f (s)g(0, and r E G, since S(P(r)z)( 8 )P) = $(af Ogr0)( 8 1P) = (s, p) (r, p) a f (4(p) = (r. , p)3(z)(s, p)
= Va E A, f E K(G),"0 E L' (â) and 0.) = fa (.,p)Op)dp, it follows that $ o p(r) = ko$,Vr E G. Therefore, (Cr (G,A)x 7 G,G,p) '''' ((Ax.G)x cA,G,a) Finally, from Theorem 16.5.6, we obtain that
((A x
xa 6, G, '.a). — ''''
(min (A 0 C (1, 2 (G))),G, a 0 adp).
Q.E.D. In the study of the structure of Von Neumann algebras of type (III), M. Takesaki obtained a duality theorem for crossed products of Von Neumann algebras ( Theorem 16.2.10). At the same time, he also conjectured about its C*algebra version. Then H. Takai showed that conjecture is affirmative. Notes.
669
References.
[127], [166], [176].
16.6. Some examples of crossed products 1) The relation between C* and Wecrossed products. Let (M ,G, a) be a Wesystem. By Definition 16.1.19, the W*crossed prod
uct is M x a G= {a E Meg B(L 2 (G))10 t (a) = a, Vt E GI,
where et = at 0 adpt,Vt E G. If pick a faithful Werepresentation {7r, HI of M, and let (rr(x)e)(s) = a8  i(z)e(s), (A(t) )(s) = (t 1 .5 )
ye )
E L 2 (G,H),s,t E G. Then M x a G is * isomorphic to the VN algebra tir(M), A(G)1" on H 0 L 2 (G) = L 2 (G , ,H). Hence, M x. G is the We subalgebra of MTOB(1, 2 (G)) generated by {(7r 0 :W)  4:141 0 Ai lz E M, s E (.
GI. Now let A be a Cesubalgebra of M, A be a(M, M)dense in M, a t (A) C A,Vt E G, and (A, G, a) be a Cesystem. Further, we assume that G is amenable. If {7r, H} is a faithful Werepresentation of M, then by Theorem 16.4.15 {Ind7r = Tr x A, L 2 (G, H)} is a faithful * representation of A x. G. From the norm on MB (L2 (G)), the largest Cenorm on L l (G, A, a) is as follows: ii(7, ® idy l Lir ( , (8)) A (.0 8 11 , Vf E LI (G, A, a). Moreover, (Tr 0 A) (A x. G)" = {Tr(M), A(G)I". Hence, A x. G is the norm closure of {(7r 0 id) i f Tr( f (s))A(s)dslf E L' (G, A, a)} in Mr0B(L 2 (G)), and the aclosure of A x. G in itfrOB(L 2 (G)) is M x a G. 2) Let G be an amenable group, M = Loe (G), A = CS° (G), and at (f)(s) = f (t 1 8),V f E L c*(G),t,s E G. Then A is adense in M, (M, G , a) is a W*system, and (A, G, a) is a Ce system . By the discussion 1), A x. G is adense in M x a G. Put H = L 2 (G) , and
7 r(f)g = fg, V f E 173 (G), g E L2 (G).
Then M x a G is * isomorphic to the VN algebra tif(M), A(G)}" on L2 (G x G) where J (Tr(f)e)(5, t) = f (stMs, t), 1 (AM ) (s, 0 = efr is, 0,
670
E L 2 (G x G),s,t,r E G. Define a unitary operator W on L 2 (G x G) as follows: {
Re) (8, = A(01/2e(st,t), ovs
(8 ) t) = A(t) 1 /2 e(st i ,
,
E L 2 (G x G). Then it is easy to check that
W*YrCOW = 7r(f) 0 1, W*A(r)W = A,. 0 1 on L2 (G) L 2 (G) = L2 (G x G),Vf E /7)(G),r E G. Since 7r(M) is maximal commutative in B (1,2 (G)), it follows that Mx. G is * isomorphic to B(L 2 (G)). Moreover, if gr, s) = f (r l s)g(s)(E Ll (G, A, a)), where f,g E K(G), then
we have Wt ( ïr X A)(z)W = vfl g (8) 1,
where v fig is one rank operator on L 2 (G), i.e., v ftg ri = ( n , 75g, Vn E L 2 (G) ( see the proof of Theorem 16.5.6). Therefore, A x. G is * isomorphic to C (L2 (0). 3) Every (UHF) algebra can be represented
as a C*crossed product.
(i) Let m(.) be a function from IN to IN. For each n E N, Put Gn = Zni(n) • Clearly, Gn is a finite abelian group, and ân = Gn , i.e.,
(j, k)
=
e2riPcim(n),
Vi E dn, k G
Vn. Now consider the discrete abelian group
G = ED
i Gn = ts = (s i , • • • , sn , 0, • • • , 0, • • •)n = 1,2, • • • , E Gi, 1 < < n.
Then G = fi Gn = fa n>1
=
(ak)k>ilak E Gic) VIC}
is a compact abelian group ( product topology (0.
, =
11 ( ak s k) k>1
), and
Va E 6, s E G.
(ii) For each n, denote by Cn = C(Gi x • • x Gn ) the set of all complex functions on G1 x • • • x G. Clearly, C11 = C(G1) 0 • • • C (Gn) C*tensor product). If identify Cn with Cn O ln÷i, then we have Cn Cn+ 1, i.e., f(51,.
3 Sn3 Sn11) = f (81).
..,
s,.) ,
V f E CnI si E Gi,1 < < n +1. Further, Cn can be embedded into C(6), i.e., f(s) = f (si, • • • , s n),V f E Cn ,s = (sk) E G.
671
We claim that C(6) = U n Cn = act 0„°°_ 1 C(G) ( infinite Cetensor product). In fact, for any f E C (â ) and 6 > 0, since f is uniformly continuous on G, there is a neighborhood W = Wi x • • • x Wn X Gn+1 X Gn+2 X • • • of 0 (here O. E Wi C Gi, 1 < 1< n) such that 
I f (3)
—
f WI < e,Vs,t E d and (s — t) G W.
If define fn(si, • • • ,sn) = f (si, •   , s n , 0, • • .),Vn,si E Gi, 1 < I < n, then fn E Cn and Ilfn — f1103 < 6 * NO For each n, let Hn = 1 2 (Gn). Then Hn is m(n) dimensional Hiblert space. Put d n) (8 n) = 458 0 I sn E G, where On is the zero element of Gn ,Vn, and 6 = oneijn ) . Then we have 1 2 (G) O!')Hn ( see Section 3.8). Now the (UHF) algebra A00 = act On B(H) = air On
=
Mm (n )
is a C*algebra on H = 12 (G). Define (ua )(8 ) =
ve
ug is a strongly continuE 1 2 (G) = H,a E d, s E G. Clearly, a ous unitary representation of G on H, and ug = On ug„,Va = (an ) E d, where (u,g)(sn ) = (on , sn)g(s n ),Vg E Iin = 1 2 (Gn), sn , an E Gn ,Vn. Hence , (A„., d, 6i) is a Cesystem, where
Mt)
= ug tu:, Vt E A r,o , a E Ô.
For each f E C (G ), define (m f e)(8) = f (s) e(s),
ve E H = 12 (G), s E G.
( notice that G can be regraded as a dense subset of 6), and let A = {mf lf E
C(d)}. We claim that A is the fixed point algebra of the system (A œ , d, ei). Clearly, &OE (mf ) = ug mf u: = rri f ,Va G d, f E C(Ô). Conversely, suppose that z G A„,, and ti g (x) = z, Va E Ô. For e > 0, pick y E 0L 1 B(Hk) 0 03."1i such that My — x11 < 6. Define z = (m(1) • • • m(n)) 1
E
aec i x...xan
a, (y).
Then ecc,(z) = z,Va G dl z E 011=1 B(Hk ), and Hz — z11 < e. Since zu, = ug z,Va E Gi x • • x Gn , it follows that z = mf for some f E Cn C C(6). 6(> 0) is arbitrary. Hence, z E A.
672
(iv) Let s —> As be the left regular represectation of G on 1 2 (G), and define cx,(x) = A, xA*8 , Vz E .1 3(1 2 (G)).
Then (A, C, a) is a C*system, and as (m.f ) = m8/1 where (8 f)(t) = f (t — 8),Vs E G,t E 6, f E C (â ). We say that the (UHF) algebra A„,:, is * isomorphic to the crossed product A x a G. Define (7r(mi)e)(s , 0 = (ots1 (mf) e (.5)) (t)
= f (8 + t)e(s, t), NO 0(8, 0 = e(s — VI E C(â), G 12 (C, H) = 12 (G x G), r, s, t E G. Since G is abelian, {7r x A,1 2 (G,H)} is a faithful * representation of A x. G. In particular
((7r x A) (z) 0(8, t)
= (E a3 1(z(r, •)) Mr) 0 (8)) (t) rEG
= E zfr, s ± tMs
— r, t),
rEG
Vz E K(G,C (6)), as follows:
E 1 2 (G x G). Define a unitary operator W on 1 2 (G x G)
f (IV e)(s,t) = 1 (IV* )(8 ,t) = e (.5
ve E 1 2 (G x C). Then we have W*(7r x A)(z)W = (Em gr k) 0 1, rEG
where gr ( )= z(r,•) EC(6),VzE K(G, C(6)). Hence, Ax a G is * isomorphic to the C*algebra on 1 2 (G) generated by {mf , Ar lf E C(6), r E G}. On the other hand, 07=1/3(Hk) is generated by {A,.,mf lf E C,, r E G1 x • • • x G,} obviously, Vn. Since C(6) = L.1C„ and G = ED.G„, A o. and A x a G are * isomorphic. 4) Semidirect product.
(i ) Let H, K be two groups, and p be a homomorphism from K to Aut(H), where Aut(H) is the automorphism group of H. Define (h, k)(11,1 , le) = (h p (0 (hi , kle),
673
Vh, h! E H, k, le E K. By this multiplication, H x K becomes a group. This group is called the semidirect product of (H, K, p) , and denoted by G = H x K. Now if H, K are two locally compact groups, and the map (k, h) —> p(k)h is continuous from K x H to H, then we have the following facts: (1) G = H x p K is also locally compact with respect to product topology; (2) If dh is the left invariant Haar measure on H, then for any k G K there is a positive constant 45(k) such that dp(k)(h) = 45(k)dh. Moreover, 4.) is continuous on K , b ( ei() , 1 and b(ic1lc2 ) = 45(Ici)t5(k 2 ),Vki,k2 E K; (3) If dk is the left invariant Haar measure on K, then d(h,k) = 00 1 dhdk is the left invariant Haar measure on G = H X,, K; (4) If All ) AK, AG are the modular functions on H, K, G respectively, then we have AG (h , k) = ö(k) 1 AH (0 AK (0 ) Vh E H, k E K. Moreover , AH is pinvariant, i.e.,
AH(P(k)(h)) = AH(h),
Vh E H, k E K.
The proof of these facts can be found in [70]. (ii) Let G = H x,, K. If define
(akf)(h) = Vk E K,hE H, f E On then we can obtain a Cesystem (C*(H), K, a). Moreover, let (1)(f)(k,.) = b(k) i f(.,k), Vf E Ll (G), k E K,. E H. Then (I) is a * isomorphism from V (G ) onto L i (K , On a). The proof is easy.
(iii) Let G = H x,, K. Then we have
C* (G) '''' C s (H) x. K. In fact, by Ll(G) '=' Ll(K, Ll (H), a) C OK, C*(H), a) C C*(H) x c, K, it suffices to show that every * representation of L 1 (K, L i (H), a) can be extended to a * representaton of Ll (K, C* (H), a). Put B = On A = C* (H). Let {gi } be a bounded approximate identity for Ll(G). Then {fi = ( ( D)} is a bounded approximate identity for L l (K,B, a). Since L i (K, B, a) is dense in Ll(K, A, a) , {L} is also a bounded approximate identity for Ll(K, A, a). Now let {p,H} be a nondegenerate representation of Ll (K, B, a), and define 7r(b) = slim p(b f i), uk = s lim p(a k (fi (k 1 .))) , 1
674
M.)
( notice that E B,V. E K),Vb E B, k E K. Then {71, u, HI is a covariant representastion of (B, K, a), and p = 7r X u. Clearly, {7r, HI can be uniquely extended to a * representation of A, and fir, u, HI is also a covariant representation of (A, K, a). Hence {p = 7r X u, F} can be extended to a * representation of Ll (K, A, cx). Example. as follows:
Let G be a locally compact group, and define p : G
> Aut(G)
—
p(s)(t) = sts 1 , Vs,t E G.
Then (cx, f)(t) = A(s) f (s  1 ts) , V f E L l (G) ,s,t E G,
and (C* (G), G, a) is a C*system. By above discussion, we have C* (G) x. G L'' Cs (G X i, G). 5) The periodic action and mapping torus.
Let (A, Z, a) be a Cesystem, and an = id, where n is a fixed positive integer, and a is a * automorphism of A. (i) Let A be the closure of l nit (Z, A, a) = {f E l' (Z, A, a)I f (k) = 0 ,Vk 0 0(mod n)} in A x,„ Z. Then A is a Cesubalgebra of A x a Z, and
J E An 1 1 (71,A,a)
+ JE 1 1 (71,A,id)
can be uniquely extended to a * isomorphism from A onto A xid Z''=s C(T, A), where j(k) = f (kn),V k E Ms , and T is the group of unit circle, i.e., T = {z G Œl lzl= 1}. Proof.
Since ( f g)(k) = E 1(m) atm (g(k — m)), f* (k) = a k (f (—k)*),V f, g G mEZ
0( 71 , A, a) and k E Z, A is a Cesubalgebra of A x a Z, and g homomorphism from 0( 71 , A, id) to /(Z, A, a), where to _f0, I 1 g(m),
Then g we have
A
A
is a *
if k 0 0(mod n) if k = mn. '
can be extended to a * homomorphism from A xid Z to A, and
11 6\
Ilitxaz
1101Axi4z,Vg
E 1 1 (71, A, id).
675
Now suppose that A C B(H) for some Hilbert space H. By Theorem 16.4.15, there is a faithful * representation { p, 1 2 (Z, H)} of A x. Z ( noticing that Z is amenable):
(9(f)e)(k) V f E 1 1 (Z, A, a),
e E 1 2 (Z, H),
jEZ
and k E E. Let
Br) = qt(Z, Br) 6 12x (Z) H) be an orthogonal decomposition of 12(, H), where 12 (Z)

1(7Z, H) = .U" E 12 (E, HMO = 0,Vk $ 0(mod n)}
and
12x (7Z,H) = U E 1 2 (7Z, H)le(kn) = 0,Vk E 711. Clearly, if g E 1 1 (Z, A, id), then 1!(Z, H) and 1!, (Z, H) are invariant for p (6). Hence , we get
11 liAxaz = lipki ? iip() lc
(awl
H)11.
Also by Theorem 16.4.15, there is a faithful * representation {( 7,1 2 (Z , H)} of A x id Z : (a(g)e)(k) =
iEz
e
Vg E 1 1 (Z, A, id), E 1 2 ( 71 , H), and k E Z. Define a unitary operator U from 12 ( 71 , H) onto 171, H) as follows: (Ue)(k)
if k = nj for some j , = { (i), otherwise,
ve E 1 2 (Z , H). Clearly, we have U a(g)U* = p(g)11!(71, H), Vg E 11 (Z, A, id).
Thus,
11911Axi4z = licr(9)11 = IIPMI 1 (ZInr)11
11 9 IlAx.z.
Further, we obtain that
11911Ax14z = 11 g 1lAxaz, and A is * isomorphic to A
Xid
Z.
Vg El l (Z,A,i(1),
Q.E.D.
(ii) If define ak (a + A) = ak (a) + A, Vk E Z, a E A, A E (V, then we have C*system (AiŒ, Z, a). Since 1 1 (71, A, a) is a * twosided ideal of 1 1 (Z, AiŒ, a),
676
and E is amenable, it follows from Proposition 16.4.16 that A x a E is a closed twosided ideal of (A4Œ) x a Z. Put A E 1 11 ( 71, A 24, a) with A(k) = 0 if k
0, and A(1) = 1.
Then A is an invertible element of (A 44) x. Z, and Ai(k) = bi,k)
(fAi )(k) = f (k
—
j),
Vj, k E 71 f E 1 1 (E, A, a). Hence AV is the closure of { f E 1 1 (Z, A, a)1 f = 0, Vk j(mod n)} in A x a 7Z Vj E Z. We claim that ,
,
A x a Z= .11AAA1••4AAn1 . In fact, for any f E 1 1 (Z, A, a) and j E {0, • • • ,n — 1 } , let f f(k), fi (k = l ø,
Then fi E A Ai , and f = b
+ • • • +
if k j (mod n), otherwise. Hence
1 1 (E, A, a) C A + AA+ • • • + AA'.
Notice the following fact. If f E 171, A, a), g E 1 1 (Z, A, a) and g(kn) = 0,Vk E Z, then ilf +911Ax a z Indeed, along the notations : p, 1 2 (71, = 1 (71 , H) q,(Z, H) in (i ) , we have p(g)1(2Z, H) C l (Z, H). Hence,
+00 2 = IIP(M11 2 +11P(g)e11 2 By the proof of (0 we obtain
MP(/' E
(Z, H).
+911Axaz =11P(f
liP(f)1 1!( 7 ' ) ,H)11= IlfliAxaz•
Now let ai E A, 0 < < n 1, be such that ao aiA +... an _iAn 1 = 0.
For each j E {0,• • • , n — 1 } , pick a sequence {fli) } of / ( Z, A, a) such that ai in A x a E. Then n1
f(i) V
0 as k
oo.
By the fact of the preceding paragraph, we have fr) —> 0 in A x a E. Hence, act 0. Similarly, from (al ± a2A + • • • ± aniAn2 )À = 0 we have al = 0. Generally, ai = 0, VO < j < n— 1. So A + A + • • • + On1 = Aj AAA • • • jAA711.
677 Finally, for any a E A x. E, pick sequences {ink }( () < Ini (E, A, cx) such that
i
< n  1) of
n1
fkoxi + a as E i....o
k + oo.
n1
Then E(fIci) must be
* 0 as k,1 + co. From the preceding paragraph, it
(fli)  f (i) ) + 0
as
k,1
oo,0 < j < n  1.
So for each j E {0, • • • ,n  1} there is ai E A such that fi) > al . Therefore, a = ao FaiA ± • • • ± a n_lAn 1 , and A x a E = A4AA4 • • • 4 A An11 . 
(iii) By (0 and Proposition 16.3.2, the Fourier transformation: f E 1(Z, A, a) > F(z) =
E f(nk)z k E C(T, A) kEZ
can be uniquely extended to a * isomorphism from A onto C(T, A). Denote this * isomorphism by t. Now on C(T, A) x • • • x C(T, A)(n times ) define multiplication, * operation and norm as follows: ()0
H21 172, I/21;
= 1,2, • • • , is a sequence of faithful nondegenerate Werepresentations of M, then (ii) If {7r,
E dimm (Bi) ;
dimm( EDHi) =
(iii) The natural trace Tr L' 2 (m) is exactly the restriction of the unique faithful normal tracial state T 1 of A(M)' = p(M) on A(M) 1+ , and dimm(L2 (M)) = 1; (iv) If {7r, H} is a faithful nondegenerate Werepresentation of M, then
7r(M)' is finite
< > Trill is finite on 7r(MY+
1. Then by Proposition 17.1.9 M' O IL adimits a separating vector E H 0 L. Thus , s is cyclic for (M' 1L)' = M B(L). Now it = (SI,. • • is easy to see that {SI,. • • 01 is a cyclic subset for M. Pick and let Clearly, = ( 1 74)çi) 2 < < 1, where At is the projection from H onto
Proof.
ei = [med.
694
we have (174)H such that
H=
= [Ms112 1. Thus M 0 1K admits a separating vector in H O K. By Proposition 1.13.6, for any cp E (M) + there exists /7 = (t 1 3 • • 3thn) EHOK such that
p(x) = ((x
Vz
1 .0 17) /7) =
E
M. Q.E.D.
Corollary 17.2.3. Let Mi be a finite factor on a Hilbert space Hi , and MI be finite too, i = 1, 2. If 40 is a * isomorphism from M1 onto M2 3 then we can write
=
4)3 0 4'2 o 4' 1
where (Mx) = 1K,Vx E MI , and K is a finite dimensional Hilbert space; NO = 11, V. E M1 0 1 K, and )7' is a nonzero projection of (M1 0 1K )'; and ■IP 3 is a spatial * isomorphism from (M1 0 1 K)/' onto M2 . (
Proof.
It is immediate from Lemma 17.2.2 and the proof of Theorem 1.12.4.
Q.E.D. Proposition 17.2.4. Let M be a finite factor, N be a subfactor of M, and {7r, H} be a faithful nondegenerate Werepresentation of M. If dimm(H) 1. We consider the inductive limit of {MIck))4)kik}. By Section 3.7, let
x kMick) = { (ak)».0 1 ak E A = {(ak ) E x k ite
and
I = {(a k ) E
A
Ate ) , Vk
>0 , }
there exists /co such that } ,
a8+1 = (I),(a,),Vs > ko
there exists ko such that } a, = 0, Vs > ko
Then x k ite ) is a * algebra in a natural way, A is a * subalgebra of x km1k), I is a * two—sided ideal of A , and
lim{M1k) , tk ik} = AII.
704
Let Mk =
exists (0, • • • 7 0 7 ak, akFi l ' • `) E a, } fa E Air 1 therewhere a8+1 = (Ma.), Vs > k
)
Vk > O. Then we have the following commutative diagram: 1E
Mitcl)
to
Iwo 1 E Mo C
MP) 1 111 1 MI
'Ii i
7. 11/1
(2)
0••
C M2
where %Ilk is a * isomorphism from Ml ic) onto Mk as follows: for any ak E let 41 k (ak) = a, then (0, • • • , 0, ak , 40k (ak) , • • •) E ai, Vk > O. Define ek = W kFl(Pk),Vk > 1. Since Mlic÷+1 1) = (4:1:0 k (1t41 k),Pk ), it follows that MkF1 = (Mk, ek), i.e., MkFi is generated by Mk and a projection ek ,Vk > 1. Clearly, Mk is a finite factor, and
[Mk+1 : Mk ] = [M: NI,
Vk > O.
Definition 17.3.6. The above chain 1 E Mo C • • • C Mk+1 = (Mkt ek) C • • • is called the tower of finite factors induced by a pair {N C MI of finite factors with [M : N] < oo. Theorem 17.3.7. Let M be a finite factor, N be a subfactor of M with be the [M : N] < oo, and 1 E Mo C MI C • •  C Mk+i = (Mk, 4) C
tower of finite factors induced by {N C M}. Then we have the following: (0 the pair Mo C MI is * isomorphisc to N C M; (ii)Mk is a finite factor, and [MkF i : Mk] = [M: Nl I Vk > 0; (iii) If TA is the unique faithful normal tracial state on MA I then TA has the Markov property of modulus of 13 = [M: N], i.e., Tk+11Mk := TA, and firk+! (xek ) = Tk(z),
Vz G Mk,
Vk > 1. In particular, r(e k ) = (3 1 ,Vk > 1, where T is the tracial state on ii kMk such that T 1Mk = Tic , Vk; (iv) Mk+i = (Mk, el , • • • ,ek ), i.e., Mk+i is generated by Mk and {el ,. • • , ek l,Vk > 1; (v) the sequence fek k > 11 of projections satisfies the following relations:
I
ek ek+i ek = r i ek , ek + iekek+i = 0 l ek+i,Vk > 1;
and ek ei = eiek, if lk — il> 2. Proof.
It suffices to prove (sr) .
705
Let (k
—
j) > 2 and j > 1 . Notice that m(k I) l'_
m (k) k
k1
I 111 k1 Mk1 C
sk
I *lc Mk C
M(k+ 1 ) k+1 I 1 kF1 Mk+1
and ei E Mk1,4 E Mk+1. So the relation ekei = ejek is equivalent to that I'k 0 (1) k1 0 W ici!i(ei) and xliN(ek) commute. Define N = 4:141(M1 k1 1) ) 7 M =M1 ic) ,T3 = Pic , and M1 = ile++1 1) . Then P is the projection from L 2 (M) onto L2 (N), and M1 = (A(M),T)' = {A(M),T3}". Since (1) k o 40 k_i 0 41 1(e5 ) = (1) 1c ° 41. 1c1(P.i) E A(N), and Wilde k) = P, it follows from Proposition 17.3.2 (iii) that (1) k 0 tic _1(1i) and P commute. Therefore, we have that ek e' eiek ,V1k — A > 2. For k > 1, let N = M1k)1 ,M = Mil k) , 4:10 = (1) k (= A); M i = le++1 1) ,* . 41)k÷1 (= A); and M2 = MLFk+2 2) . Then we have
(17 c )m 4 À (4) (A7), A )
=M1 '•: (*( m i), P2 ) = r121
where P1 is the projection from L2 (M) onto L 2 (N),M 1 is a finite factor on L2 (M) generated by (1)(M) := A(.14) and P1 ; P2 is the projection from L2 (M1 ) onto L2 (4)(M)), and M2 is a finite factor on L 2 (M 1 ) generated by *(M 1 ) = A(M i ) and P2. Denote by E, F the conditional expectation from M onto N,M 1 onto CM) respectively, i.e., E = (NM), Ft= (P2 1M1 ). Then the relations: e k ek+i ek =r 1 4, and ek+i ek ek+i = 131 4+1 are equivalent to P2 4/(P1 )P2 = r I P2 ,
and
‘11 (PI )P2 *(P1 ) = 0 1 *(P1 ).
First we show that F(P1 ) = P1 1. Indeed, since F(Pi) = P2(P1), it follows that F(P2 ) = r'l .== (PI  0 4) I 1,2 (4(M))
in L2 (M1 )
e,, ((Pi — 13 1 1), Mx1) = 0, Vz E M. If tr is the canonical tracial state on MI , then the equality F(Pi ) = 0 '1 is equivalent to the following:
fltr(A(x)P i ) = tr(A(x)) := TM,
Vz G M I
where r is the canonical tracial state on M. This is exactly the Markov property of modulus P of T ( see Proposition 17.3.5). Hence, F(Pi ) Now for any z E M 1 (C L2 (M 1 )), we have P2 *(P1 )P2x = FM • F(x)) = F(Pi )F(x)
:= r i F(x) = (31P2x.
706
Hence, P2*(P1)P2 = /3 1 /32. By Proposition 17.3.4 and Theorem 1.6.1, the set {E (1) (xi)PICYi)lxi7Yi E H} i
is dense in L 2 (M 1 ). Now it suffices to show that W(A)P2*(Pi)( 4) (x)Pit(Y)) = 0 141 (P1)( 4) (x)PICY)),
Vx,y E M. Fix z, y E M. Since Pol)(x)Polo(y)z := E(xE(yz)) = EME(yz) = 4(E(x))P01)(y)z, Vz E M(c L 2 (M)), it follows that Pi (z)Pi (y) = CE(x))/31 4)(y). Then 41(A)P2 41(Pi)( 4)(x)PI(Y))
= Pi F(P010(x)Pi t(y)) = P i F(41)(E(x))131 41)(0) '= Pi(EN)F( 131)(y) Noticing that
Pi t(E(x))z = E(E(x)z) = E(z)E(z) = E(xE(z)) = Pi (x)Pi z, Vz E M(C L 2 (M)), we have tli(Pi)P2C/31)(4)(x)Pi4(y)) = r i PiCx)PICY) =0141 (A)( 4) (x)PICY)), Vx,y E M. Therefore, W(A)P2 *(P1 ) = 131 41 (Pi ).
Q.E.D.
Remark. Let M be a type (1,7,) factor, and N be a type (in) subfactor of M. Then it must be Trim. From Chapter 15, there is a Bratteli diagram as follows: P •
n
).
'
m1
where m = np. Clearly, L2 (M) = M, L 2 (N) = N, and P = E : L 2 (M) L2 (M. By Proposition 17.3.4, we have (M, P) = {A(M), P } " = {E A(z i )P A i (Yi)lx17Y1 G MI. Further, (M, P) = End (M)
, It
m :MI
t is t(ry) ,,=__
and E M ,y E N. 1.
707
In fact, it is easy to see that Enc17;%r (M.) = WO'. Since P = E is the conditional expectation from M onto N, it follows that P p(y) = p(y)P,Vy E N. Hence (M 7 P)
1= {E A(xi)P A(N)Ixi7Yi E M} i
C p(NY = EndrN (M).
Conversely, if t G (M, P , then )'
tx = a(x)1 = A(x)t1 = p(t1)x,
Vx E M,
and t = p(t1). Moreover, since P(t1) = t P1 = t1, it follows that (t1) E N. Hence, t E p(N), i.e., (M, P)' C p(N). Therefore, we obtain that (M, P) = p(N)' = EndrN (M). By [(M, P) : M] = [M: N] = t722 /n2 , (M, P) must be a type (Ii) factor, where 1 = m2 /n.
Let 1 E Mo C MI C M2 = (Wei) C
C M i = (Mk,ek) C
be the tower of finite factors induced by {N c M}. Then Mk is a type (inp k) factor, Vk > 0. Thus, we have a Bratteli diagram as follows: •
P
P np
n
P np 2
P
P
np k
The technique of towers of algebra and Theorem 17.3.7 are due to V.F.R.Jones. About Proposiiton 17.3.4(ii), M.Pimsner and S.Popa proved that (M, P) = {E A(xi)PA(yi ) lxi , yi G MI indeed. Notes.
i
References. [60], [75], [129].
17.4. The values of index for subfactors Define a sequence {Pk (A)lk = 0,1,2, • • •} of polynomials as follows: Po = 1,P1 = 1, Pk+1(A) ' Pk(A)  APk1(A) I V k > 1. In particular, P2 (A) = 1 — A, PI N = 1— 3A + A 2 , P6 (À) = 1 — 5A + 6A 2 — A3 ,
P3 (A) = 1  2A,
P5 (A) = 1— 4A + 3A2 , P7 (A) = 1 — 6A + 10A 2 — 4A3 .
708
Lemma 17.4.1. Consider an integer k > 0, and set m = W. Then (i) The polynomial Pk is of degree m. Its leading coefficient is (1)m if k = 2m is even and (1)m(m + 1) if k = 2m + 1 is odd. Consequently, lim Pk (A) = +00. A' 0° 1 • (ii) Pk has m distinct roots which are given by for j = 4 cos2 ( 37 ) k+1
1,2,•••,m.
(iii) Assume k> 1. Let A be a real number with 1 4 cos2(
k+2
1
< A
0, P2(A) > 0, • • • ,Pk(A) > 0, and Pk+1 (A) < 0. Proof. Claim (i ) is easily checked by induction. For (ii), we consider A > 1/4 and let
1 p.i =  (1+ i(4A  1) 1 / 2 ), 2
1 A2 ',=  (1  i(4A  1) 1 /2 ). 2
We say that Pk(A) = (p, 1  p, 2) 1 (p,t+ 1  pt") for each k > 0. In fact, it holds for k = 0,1. Now assume the conclusion holds for each j < k. Then by A := p. i p.2 we have that Pk+1(A) = Pk(A) — APk1(A) = Gil — /42) 1 [4+'

A l26+1 — 11 1A2(At — 1 4129]
= Gi l — A 2 ) 1 (14+2 — 4+2 ).
Hence, we have Pk(A) = (11 1  /1 2)  '(14÷ '  /4"),VA > 1/4 and k > 0. Consider now a real number 6 with 8 E (0, ir ) and set A = 1 2 (> 1/4), so 4 cos 6 e Then ' 0 and p. 2 that pq = 2 cos 2cos0' Pk (A) =
sin((k + 1 )6 )
2k cosk (0) sin 6
7 with j = 1, • • • , m. which vanishes when 6 = ki+1 Claim (iii) is obvious for k = 1, and we may assume k > 2. For 1 G 1 Ir and Pi (A) > 0 for A < {2,• • • ,k}, the smallest root of P1 is 4 cos2 )' (1 + 1 1 7 1 7r . As A < we have Pi (A) >0. The < 1 7r 4 cos2 ( T_71)
4cos 2 ( k +1 )  4cos 2 ( 1 +1 ) ,
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two smallest roots of Pk+1 are 1 4cos2 ( k
+ 2)
1 A2
1
= 4cos
and Pk+i < 0 on (A I , A2). As A 1 < A
e,,1 <j < m, then we have p(145,,„÷I ) = (1 — 5,n+i )p = (1— 5,n+i ) since (1— 5,n÷i ) is a linear combination of monomials in 4 1 , • • • , cm }, i.e., (1 — 45,n÷i ) < p. By (iv) it is obvious that (1 — b,,n+i ) > 6 j ,1 < j < m. Therefore, we have (1 — bm+i ) = sup{e i , • • • ,em l. Q.E.D. Lemma 17.4.3. Assume that k> 3. (i) If bk_ i = bk, then ek1 < 1 — t5k2; (ii) If k > 4 and 5k_ i = 45k , then 5k_ 36i ci = O, Vi, j > k — 2 and Ii — A > 2 .
Proof. (0 Let p = ck1(5k2 — 5k1). By Lemma 17.4.2 (iv) and bk 1 = 5k, we have ek_ibki = Ekibk = O. Since 51,2 is a linear combination of monomials
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in {1, e l , • • • 1 61C317 it follows that ek_ibk2 = bk2 6 k1. Hence, p a projection. Further,
_
is
G k 1 kvk1 — 5k2)( 15k1 — 5k2)&k1 i Pk3 g = e klk 44242 45k2) 2 6 k1 k2
2 :=_ P k 3(0 Li \ ck1 442kc k2vk2)kc k2vk2 Jo k1 
Pk2V ) 2 Pk_ 3 t k1bk2 6 k2 15k2 6k1
Since
Pk_i(t) 0
(ii) Put q =
0, it follows that p = 0, i.e., ek_ 1 bk_ 2 := 0, and eic 1 < 1 — bk  2 . 43 6 k2 6 k•
Clearly, q is a projection. Since
6 k 5k2 = 13 441 5k2 6 k = 0 7
it follows that q = co ic_ 2 (4 3 — 4 2). Then q = qq *
= 6 kfek20 5k3 — bk2) 2 6 k21 6' k
=
Pk _ 4 (t) 2 r
2 ekiCk2 ( 5k343) 05k3 6k30k3e k2}ek
Pk3(t) Pk4
6 {6k245k36k345k36k2}6k i
kbk3 6 k2 45k3ek = (I
Pk2
q.
As Pk _ 2 (t) 0 0, we get q = 0, i.e., 0k342 6 k := O.
iilc
If j > k, let vi = f3
ek ek+i • • • ei , then
v;e k vi = ei ,
and
vi(bk36k2) = (45k34_2)vi.
Hence, bk_3426'5 = v;0k3 6 k2ekv.i = O.
i—k 2
+2
Finally, if k  2 < I < j2, let u i , 0 6 k2 • • • ei I then u:ek_ 2 u1 =and ui (bk_ 3 65) = (bk_ 3 e5)u1. Hence , bk_3eiei = u:45k_ 3 6k_26jui := 0. Q.E.D. Lemma 17.4.4.
Assume that 4 cos 2
k> 7r
k—1
Then P5 (t) 0 0,V j < k — 1 , and
4 and < 13
< 4 cos 2
7r —
k.
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By Lemma 17.4.1 (iii) we can see that Pi (t) 0 0, Vj < k  1, and Pk2(t) > 0,Pk_ i (t) sup{ Re f (x) Ix E IC}. Consequently, a convex subset of X is ( norm  ) closed if and only if it is a (X, Xlc losed.
(ii) Let K* be a a(X*, X)closed convex subset of X*, and fo E X*\K* . Then there is z E X such that Refo (x) > supfRe f (x)If E K*1. Theorem A.3. ( Bipolar theorem ) (i ) Let A be a subset of X. Then
A° = {f E X*11f(x)1