15.2. On bi-exactness 413Lemma 15.2.6. Let r be an exact group, Y c r be an amenable subgroup
and g be a family of subgroups of r. If there is a map
(: r ----t gl (r /Y)
such that
lim ll((sxt) - s.((x)ll =
0
x->oo/g [[((x)[[
for every s, t E I', then I' is bi-exact relative to Q.Proof. We defineμ: r ----t Prob(r/Y) by μ(x) = ll((x)ll-^1 l((x)I. Then,
[[μ(sxt) - s μ(x)ll < 11-ll((sxt)ll I+ ll((sxt) - s.((x)ll
· - ll((x)ll [[((x)ll
< 2ll((sxt)-s.((x)ll ----t 0 as x ----t oo/Q- ll((x)ll
for every s, t E r. Let μ: g^00 (r /Y) ----t goo(r) be the u.c.p. map defined by
μ(J)(x) = (μ(x), f). It is not hard to see that μ(!) E C(f'g) and com-
posed with the quotient map, it gives rise to a I'-equivariant u.c.p. map from
g^00 (r /Y) into C(.6.gr). We view g^00 (r /Y) as the C-subalgebra of right Y-
invariant functions in goo (r). Since Y is amenable, by taking an "average"
over the right Y-action, one can find a (left) r-equivariant conditional ex-
pectation from goo (r) onto goo (r /Y). Combining these two r-equivariant
u.c.p. maps, we obtain a r-equivariant u.c.p. map from g^00 (r) into C(.6.gr).
Amenability of .6.gr now follows from Exercise 15.2.2. D
Proposition 15.2.7. Let r be a group. For families g and Q' of subgroups
of r, define
g /\y^1 = {AnsA's-^1 : A E Q, A' E Q', s Er}.
If r is bi-exact relative to g and to Q', then r is bi-exact relative to g /\ Q'.
Proof. For notational simplicity, set f = rxr, A= g^00 (r) and I= co(r; Q),
I'= c 0 (r; Q'). Then, the natural short exact sequence
o ____,,. (I/ (In I')) ~ f ____,,. (A/ (In I')) ~ f ____,,. (A/ I) ~ f ____,,. o
is exact. By assumption, (I/(I n I'))~ f ~((I+ I')/I') ~ f <J (A/I')~ f
and (A/I)~ fare nuclear. Hence the middle algebra (A/(InI')) ~ f is also
nuclear. Therefore, it suffices to show that In I' = co(r; g /\ Q'). We may
assume that g is saturated in the sense that sAs-^1 E g for any A E g and
s E I', and likewise for Q'. It is not hard to see that I n I' is generated by
a function whose support is contained in At n A't'. Pick any x E At n A't'
(unless it is empty) and observe that At n A't' = (An A')x. This completes
the proof. D