1550075568-C-Algebras_and_Finite-Dimensional_Approximations__Brown_

6.3. Some motivation and examples 225

Proof. Thanks to the previous result, we only have to observe (1) ::::;:.. (3).
However, if r is amenable, then C~(r) is nuclear (Theorem .2.6.8) and thus
for every trace Ton C~(r), the product map

7r7 x 7r~P: C~(r) 0 C~(r)^0 P---} JB(L^2 (CHr),T))

is continuous with respect to the spatial tensor product norm (Proposition
3.6.12). Thus Theorem 6.2.7 applies and the proof is complete. D

The next result is also an immediate consequence of Proposition 3.6.12
and Theorem 6.2.7.

Proposition 6.3.4. Every trace on a nuclear C* -algebra is amenable.

Let's turn to some easy permanence-type questions.

Proposition 6.3.5. The following assertions hold:

(1) if B c A and T is an amenable trace on A, then TIE is amenable;

(2) if J <J A is an ideal and T is an amenable trace on A/ J, then the

induced trace on A is also amenable;

(3) if TA and TB are amenable traces on A and B, respectively, then

the product trace TA 0 TB : A 0 B --+ C is amenable (hence it is

amenable on A 0max B as well);

( 4) if 0 --+ J ---+ A ---+ A/ J ---+ 0 is a locally split extension and T is

an amenable trace on A such that TJJ = 0, then T drops to an

amenable trace on A/ J.

Proof. The first three statements are trivial and are left to the reader.
There are various ways to prove the last fact. Here is a tensor product argu-
ment. (You may want to give another proof using approximation properties
and Arveson's Extension Theorem.)

By Proposition 3.7.6, the sequence

0-} J 0 A^0 P-} A 0 A^0 P-} A/ J 0 A^0 P-} 0

is exact. But the product map

A 0 A^0 P-} JB(L^2 (A, T))

evidently kills J 0 A^0 P and thus we have a product map

A/J@A^0 P ---+lB(L^2 (A,T)).

Letting A/Jc JB(h'.) be any faithful representation, we can apply The Trick
to the inclusion A/ J 0 A^0 P c JB(h'.) 0 A^0 P and appeal to the last condition
in Theorem 6.2.7. D