1550075568-C-Algebras_and_Finite-Dimensional_Approximations__Brown_

106 3. Tensor Products

Then a perturbation argument (Lemma 3.9.7 plus Arveson's Extension The-

orem) would imply that we can find u.c.p. maps 'I/Jn: Ms(n) (C) -----+ JB(Ji) such

that

ll'l/Jn('Pn(x)) - <p~^1 1cpn(E)('Pn(x))ll = 11'1/Jn('Pn(x)) - xii-----+ 0

for all x E E. If this held for all finite-dimensional operator systems E c A,

then we could evidently conclude that A is exact.

The miraculous fact (Proposition 3.9.6) is that if A is @-exact, then the

completely bounded norms of <p~^1 1'Pn(E) really do tend to 1 for every finite-

dimensional operator system E c A. The bulk of the proof of Theorem

3.9.1 goes into showing this.

Let's first show that @-exactness is a local property. That is, a C*-

algebra A is @-exact if and only if all of its finite-dimensional operator

subsystems are @-exact. By definition, an operator system E is @-exact if

we have an isometric identification

E@ B ~ E @ (BI J)

E@J

for all C* -algebras B and ideals J <JB. Note that there is always a contractive

map

E@B

E @ J _____, E @ (BI J)'

since the kernel of the contraction E @ B -----+ E @ ( B / J) contains E @ J.

Lemma 3.9.2. If EC A is an operator system and J <J B is an ideal, then

there is an isometric inclusion

E@B A@B

E@l c A@J"

Proof. We must show that if x E E @ B, then its norm down in 1:~ is

equal to

inf{llx + yJI : y EE@ J}.

This, however, is easily seen since the norm in 1:~ is equal to

lim llx(l@ (1-ei)) IJ,

where { ei} is an approximate unit for J. D

Proposition 3.9.3. A C* -algebra A is @-exact if and only if all its finite-

dimensional operator subsystems are @-exact.

Proof. The "if' direction is not too hard, since the union of the subspaces

1i::~, where E C A is a finite-dimensional operator subsystem, is dense in

AOB

AOJ"