1550075568-C-Algebras_and_Finite-Dimensional_Approximations__Brown_

438 17. K-Homology

Proof. If C'*(I')/J1 had an amenable tracer, then it would also have a

finite-dimensional quotient (Remark 6.4.11). Hence r would have a finite-

dimensional irreducible representation which factored through C'*(r)/ J1 and

this is impossible. D

17.3. Ext need not be a group

With the structure theorem for C'*(r) in hand, there is little more to do. We

start with an infinite residually finite discrete group r which has property

(T) (say SL(3, Z)). Then r necessarily has an infinite (but countable) num-

ber of nonequivalent finite-dimensional irreducible representations which we

denote by 1f'k: C'*(I') -----+ .IB\(1ik), k EN. Let

1[' = EB7rk: c*(r)-----+ .IB\(E91ik)·

kEN kEN

Here is the key observation/ exercise: If J1 <JC'* (r) is the ideal generated by

all of the Kazhdan projections, then

7r(J1) = EB.IB\(1ik) c oc(ffi1ik) n II.IB\(1ik) c .IB\(E91ik)·

kEN kEN kEN kEN

Hence, passing to the Calkin algebra, we have an induced *-homomorphism

<p: C*(r)/J 1 -----+ Q(E91ik)·

kEN

Theorem 17.3.1. Let C' = <p(C'*(I')/J1) c Q(EBkEN1ik)· Then Ext(C') is

not a group.

Proof. We will show that if Ext( C') were a group, then C' would have an
amenable trace, which contradicts Corollary 17.2.6 (since C' is a quotient of
C'*(I')/J1)·
If there existed a u.c.p. lifting to the natural inclusion C' C Q(EBkEN 1ik),
then we could compose with a conditional expectation to force it to take
values in rrkEN .IB(1ik) c .Ill( EBkEN 1ik). Evidently this implies C' has an
amenable trace, so we are finished. D

For some time, experts wondered whether a quasidiagonal set of op-
erators would always descend to a QD C* -algebra in the Calkin algebra.
Wassermann's work also answers that question, negatively, since the pull-
back of C' is obviously a quasidiagonal set of operators.
Corollary 17.3.2. Since C' has no amenable trace, it is not QD.

Remark 17.3.3. Since C' is MF, by definition, it follows that inductive
limits of RFD algebras need not be QD (Proposition 11.1.8).