7.4. Two more examples 253Proof. We first recall a basic operator theory fact: If T E IIB('h'.) is a con-
traction, then
[T Jl -TT*]
U = Jl -T*T -T*is a unitary operator on 'h'. EB 'h'..^10
Since C(IB' 00 ) C C(IB'2), it suffices to prove the theorem when n = 2. So,
fix a faithful representation C*(IB'2) c IIB('h'.) on a separable Hilbert space and
let Pk :S Pk+l be finite-rank projections converging strongly to the identity.
If u 1 , u2 are the canonical free generators of IF 2, then for each k we can dilate
the contractions Pku1Pk, Pku2Pk, as above, to get unitaries vfk\ v~k) acting
on the finite-dimensional Hilbert space Pk 'h'. EB Pk 'h'.. Note that for each j,
(k) [Uj 0 ]
VJ· --+ Q -uj 'ask--+ oo, where convergence is in the strong operator topology. It follows
that for every noncommutative -polynomial p of two variables, we have
p ( Vl (k) 'V2 (k)) --+ [p(u1, Q u2) p ( -ul, ^0 -U2 ) ] '
where convergence is in the strong operator topology, as k--+ oo.
By universality there exist -homomorphisms Trk : C (IB' 2) --+ JIB( Pk 'h'. EB
Pk'h'.) such that 7rk(uj) = v}k). The previous paragraph implies that for
every element x E C[IB'2] in the group algebra (i.e., x = p(u1, u2) for some p)
we have the inequality
llxllo*(JF 2 ) :S II [xo p .( -u*O *)]II :S liminf llrrk(x)ll·
1 , -u 2 k-+ooIt follows that
EB 'Trk: C (IB'2) --+ II IIB(Pk 'h'. EB pk 'h'.)
k k
is isometric on a dense subspace and hence must be injective on the whole
C-algebra. D
Remark 7.4.2. At first glance it may seem that RFD algebras can't be too
exotic due to the abundance of finite-dimensional representations. However,
nothing could be further from the truth, as C(IB'n) is, in many ways, as
exotic as they come. Every single separable unital C -algebra arises as a
quotient of this beast (if n = oo) and hence there is no hope of understanding
it at a very deep level.
lOHint: Observe that if p is any polynomial, then T*p(TT*) = p(T*T)T* and hence the
same relation passes to continuous functional calculus.