1550075568-C-Algebras_and_Finite-Dimensional_Approximations__Brown_

368 12. Approximation Properties for Groups

Proof. It is not hard to see that D = {a~ f : llall < 1, llfll < 1} is a

convex subset of the unit ball of Ai§E* and that ll<plfcb =sup l(<p, a~f)I for

every <p E CB(A, E). It follows that the duality pairing.from CB(A, E) into

(Ai§E*)* is a surjective isometry and n is dense in the unit ball of Ai§E*, by

the Hahn-Banach separation theorem. To prove the second part, we observe

that the map x is continuous since each Ei is finite-dimensional. Therefore,

it suffices to show llu1 x u2ll < 1 for Ui = ai ~Ji, where ai E MnJA)

and fi E MnJEi)* with llaillllfill < 1. In this case, u1 x u2 =a~ f for

a= ai © az E Mn 1 n 2 (A1 © Az) and f =Ji© fz E Mn 1 n 2 (E1 © E2)* (modulo

shuffling of indices) and llu1 x u2ll::; llallllfll = lla1lllla2llllfillll!zll < 1. D

We will use repeatedly the following lemma, known as the small pertur-

bation argument.

Lemma 12.3.15. Let F C A be a finite-dimensional subspace with basis

(xk)k=l C F, and let (xk)k=l C F* be a dual basis: xj(xk) = Oj,k· Then,

for any (ak)k=l c A, there is a map e on A such that e(xk) = ak for every

k and

Proof. By the Hahn-Banach Theorem, we may assume that x'k E A*. The

map e defined by e(a) = a+ l:x'k(a)(ak - Xk) has the desired property.

(Note that llx'kllcb = llx'kll-) D

It follows that Acb (A) < C if and only if for any finite-dimensional

subspace F CA there is a finite-rank map <p on A such that <plF = idF and

ll<pllcb < C.

Lemma 12.3.16. Let F C E be finite-dimensional subspaces of A and let

C > 0. Then, there is a map <p E CB(A, E) such that <plF = idF and

ll<pllcb::; C if and only if I tr(u)I ::; CllullA@E* for every u = l:k ak © fk E

F 8 E*, where tr(u) = l:k fk(ak)·

Proof. If <p E CB(A, E) is such that <plF = idF and ll<pllcb ::; C, then

ltr(u)I = l(<p,u)I::; ll<pllcbllullA@E*::; CllullA@E*

for every u E F 8 E. Conversely, if II tr IF('.)E II ::; C, then by the Hahn-
Banach Theorem we can find <p E (Ai§ E) = CB(A, E) which is a norm-
preserving extension of tr IF0E. Then <pip= idp and ll<pllcb = II tr IF0E II ::;
C. D

Proof of Theorem 12.3.13. We only prove that

Acb(A1 © Az) = Acb(A1)Acb(A2)