1550075568-C-Algebras_and_Finite-Dimensional_Approximations__Brown_

242 7. Quasidiagonal C* -Algebras

we will see counterexamples after describing the other obstruction to qua-

sidiagonality. In some cases (e.g., crossed products of either abelian or AF

algebras by Z) it is true that stable finiteness implies QD (though this is

not easy to prove). In some cases there are questions which may turn out to

have positive or negative answers and it is not clear which way things should

go. For example, a question of Blackadar and Kirchberg asks whether every

stably finite nuclear C* -algebra is QD. There is some evidence suggesting a

positive answer, but we feel it is rather thin and we would not be surprised

by counterexamples. Even the case of stably finite type I algebras is not

presently known, as pointed out to us by David Kerr.

Obstruction 2: Every unital QD C*-algebra has an amenable

trace.

Since amenable traces are characterized by a finite-dimensional approx-

imation property (Theorem 6.2.7), this is a very simple observation indeed.

Proposition 7.1.16. Every unital QD C*-algebra has an amenable trace.^5

Proof. Let cpn: A -+ Mk(n) (C) be a sequence of u.c.p. maps which are

asymptotically multiplicative. If tr is the unique tracial state on Mk(n) (C),

then the functionals tr ocpn are states on A, any cluster point of which is

easily seen to be an amenable trace. D

Combining with Proposition 6.3.3, we get the following result.

Corollary 7· .1.17. If r is a discrete group and C~ (r) is QD, then r is

amenable.

Jonathan Rosenberg first observed the previous corollary and conjectures
the converse. The problem appears to be very hard, however.
Here are some nonquasidiagonal, yet stably finite, examples.
Corollary 7.1.18. If r is any discrete, nonamenable group, then C~(r) is
not QD. In particular, C~(lFn) is stably finite (since it has a faithful trace)
and exact, but not QD.
Exercises
Exercise 7 .1.1. Show that A is QD if and only if all its separable sub alge-
bras are QD.
Exercise 7.1.2. If A is nonunital and QD, then the unitization A is also
QD. (Use Proposition 2.2.1 and check that the unitized maps are still asymp-
totically multiplicative and isometric.)

(^5) The assumption of a unit is necessary. The compact operators, being AF, are QD but have
no tracial states.