11.3. Inner quasidiagonality 325Proof. First we must argue that A;* is equal to pA ** p + ( 1 - p) A** ( 1 - p).
The inclusion A;* c pA**p + (1-p)A**(l - p) is immediate.
Since p is the central cover of Ap -+ pAp, 1 - p is open in A** - i.e.,
there is an increasing net of positive elements ai E A such that ai -+ 1-p in
the strong operator topology. Hence (1-p)A(l -p) n A is weakly dense in
(1-p)A**(l-p). Since (1-p)A(l-p)nA c Ap, we have (1-p)A**(l-p) c
A;*. Proving that pA**p c A;* is virtually identical to the proof of the
previous lemma. Hence A;*= pA**p + (1-p)A**(l - p), as desired.
Having identified the weak closure of Ap, the remainder of the proof
is not hard. First, note that d(a, Ap) = d(a, A;*) (by convexity and the
Hahn-Banach Theorem). Now define
x =pap+ (1 - p)a(l - p) E pA**p + (1 - p)A**(l ~ p) =A;*.
A simple calculation shows II~ - xii= ll[a,p]ll and thus d(a,Ap) ~ ll[a,p]ll·
On the other hand, if y E Ap, thenlla-yll 2:: max{ll(l-p)(a-y)pll, llp(a-y)(l-p)ll} = ll[a-y,p]ll = ll[a,p]ll,
which completes the proof. 0
Corollary 11.3. 7. The C* -algebra A is inner QD if and only if there is a
sequence of c.c.p. maps i.pn: A-+ Mk(n)(<C) such that llall = lim ll<pn(a)ll and
d( a, Arpn) -+ 0 for all a E A, where Arpn is the multiplicative domain of cpn.
Proof. The "only if' direction follows from the previous result and Remark
11.3.2.
For the other implication we fix a finite set -6 c A and c; > 0. Let
cp: A-+ Mn(<C) be a c.c.p. map such that d(a,Arp) < c; and ll<p(a)ll > llall-c;
for all a E -6. Define B = <p(Arp), decompose B = JB(H1) E9 · · · E91B(Hk), for
finite-dimensional Hilbert spaces Hi, and consider the corresponding decom-
position cplAcp = 7r1 E9 · · · E91fk· Eliminating repetitive maps, if necessary, we
may assume that the 1ri's are pairwise disjoint representations of Arp (that
is, the corresponding projections in Z(A~*) are pairwise orthogonal),
Regarding each 1ri as an irreducible representation, we can find irre-
ducible representations ifi: A-+ $(Hi), 1 ~ i ~ k, such that Hi c Hi and
if Pi: Hi-+ Hi is the orthogonal projection, then Pdri(x)Pi = 1ri(x) for all
x E Arp. Sadly the ii"/s need not be pairwise disjoint, so we assume that
ii\, ... , ii"j (j ~ k) are pairwise disjoint and each ii"i with i > j is equivalent
to one of these. Hence, for each i > j there is a unitary conjugating the
representation ii"i: A--'-+ $(Hi) over to ii"m: A-+ $(Hm), for some m ~ j,
and this allows us to regard the projection Pi as acting on Hm· Note that
under this identification, Pi is orthogonal to Pm since the representations 1ri
and 1fm are disjoint. ·