230 6. Amenable Traces
Proof. We can find a representation u: C*(I') --+ Iffi(H) and a finite-rank
projection P E IIB(H) such that <p may be identified with x 1--+ Pu(x)P. It
follows that
JJu(g )Pu(g )* - PJJ2 = ( 2 _ 2 Tr(Pu(g )Pu(g )*))
1
1
2
JJPJJ2 Tr(P)
= (2tr(l - <p(g)<p(g*)))^1 l^2
"5. c/'i,
for all g E r. Hence the Hilbert-Schmidt unit vector lll~ll 2 P is almost in-
variant under the conjugation action T 1--+ u(g)Tu(g). By the quantitative
statement above, we can thus find a Hilbert-Schmidt operator R such that
JJRll2 = 1, u(g)Ru(g) = R for all g Er, and
1
IJR-IJPll2 PJl2 < 2c.
In other words, RE u(C(r))' and it is close to 11 )- 112 P. Observe that
I Tr( u(x )RR) - tr( <p(x)) J
= I Tr(R* ( )R) - Tr(Pu(x)P) I
u x Tr(P)
= ITr(Ru(x)(R-IJ:iJ)) +Tr((R-JJ:il2)u(x)ll:il2)I
"5. 2c;JJu(x*)RJl2 + 2cJJu(x) ll:il
2
JJ2
"5. 4cJJxJJ,
for all x E C (r). Since RR commutes with u( C (r)), so do all its spectral
projections and hence we can find a finite-rank positive operator h which
commutes with u(C(I')), has rational eigenvalues, Tr(h) = 1 and JJRR -
hJJ1 < c. For such R we have
J Tr(u(x)RR) -Tr(u(x)h)J "5. JJu(x)JIJJRR* - hJJi < cJJxJJ
and thus
J Tr(u(x)h) -tr(<p(x))J < 5cJJxJJ,
for all x E C(r). By Lemma 6.2.5, there is a u.c.p. map 11"^1 : IIB(H)--+ Mm(C)
such that
tr(7r^1 (u(x))) = Tr(u(x)h) Rj tr(<p(x))
and
tr (1-7r^1 (u(g))7r^1 (u(g))) = 0
for all g E I' (since h E u(C(I'))'). Since tr is faithful, it follows that
11"^1 ( u(g)) is a unitary element for every g and hence all of u( C (r)) falls in the
multiplicative domain of 11"^1 • Defining 7r = 11"^1 o O", the proof is complete. D