1550075568-C-Algebras_and_Finite-Dimensional_Approximations__Brown_

10.3. Quasidiagonal C* -algebras 307

Proof. Let p: A -+ M(I) be the natural extension of the inclusion I ~

M(I) ([198, 2.2.14]). Then the map pE97r: A-+ M(I) E9B is injective. The

previous proposition (plus the trivial fact that direct sums of QD C* -algebras

are QD) now implies quasidiagonality. D

The proof shows that M(I) is QD whenever I has a separating family

of unital QD quotients. Hence the proposition above remains true for ideals

of the form I= J@ B where J is RFD and Bis unital and QD. We still

lack a description of those algebras whose multiplier algebras are QD.

Another instance where quasidiagonality passes to extensions is when

the extension is quasidiagonal. Confusing, yes, but not a tautology.

Definition 10.3.3. Let 0 -+ I -+ A ~ B -+ 0 be a short exact sequence.

We call this a quasidiagonal extension if I has a quasicentral approximate

unit consisting of projections.

It is important to note that in general an extension being quasidiagonal

has nothing to do with whether or not the middle algebra A is QD - it is

simply a statement about projections.

Proposition 10.3.4 (Exercise 7.1.6). Let 0 -+ I -+ A ~ B -+ 0 be a

quasidiagonal extension where both I and B are QD. Then A is QD.

Proof. Let {Pn} CI be an approximate unit of projections which is quasi-

central in A. Consider the c.c.p. maps <fJn: A-+ IE9B, <fJn(x) = PnXPnE91f(x).

Evidently these maps are asymptotically multiplicative, so we must check

that they are asymptotically isometric.

Let P EI** CA** be the (weak) limit of the Pn's. Then P is central in

A** and yields a decomposition A** = I** E9 B**. Hence for each x E A we

have llxll = max{llPxPll, 11(1-P)x(l-P)ll}. But 117r(x)ll = 11(1-P)x(l-P)ll

and llPxPll :::; liminfn llPnXPnll since PnXPn -+ PxP in the strong operator

topology. The other inequality is obvious. D

When all algebras in a sequence are nuclear and satisfy the UCT, K-

theory can help determine quasidiagonality. See [31] for some partial results.

Quotients. Since C*(JF 00 ) is QD (even RFD - Theorem 7.4.1), it is clear
that quasidiagonality need not pass to quotients. There is a simple sufficient
condition, however.
Proposition 10.3.5 (Exercise 7.1.5). Let 0-+ I-+ A~ A/ I-+ 0 be locally
split and short exact. If A is QD and the extension is quasidiagonal, then
A/I is QD.