17.2. Property (T) and Kazhdan projections 437Definition 17.2.3 (Kazhdan projections). If r has property (T) and the
representation 1f: C* (r) ---> Mn ( C) is irreducible, then the central cover c( n)
is also known as the Kazhdan projection associated with n.
Theorem 17.2.4 (Structure theorem for property (T) groups). Let r be a
discrete group with property (T). For each finite-dimensional irreducible rep-
resentation n: C*(r)---> Mn(C), the Kazhdan projection c(n) is an element
of C*(r).Proof. Let O": C (r) ---> IIB(H) be an essential representation such that (1)
nEBO": C(r) ---> IIB(.e; EBH) is faithful and (2) O" contains no subrepresentation
which ·is unitarily equivalent to 1f.^6 ·
It suffices to show that O" can't be faithful. Indeed, if O" has a kernel
0 -/= J <1 C (r), then n I J must be faithful (since n EBO" is faithful and O" I J = 0).
Hence J is finite-dimensional and, as such, has a unit p'lr which is a central
projection in C(r). Note that n(p'lr) = 1, since n is irreducible,· and thus
we may identify 1f with the -homomorphism A---> p'ITA, a 1-+ p'lra (i.e., 1f is
an isomorphism on p'ITA). _Since p'lr is also a central projection in C(r),
the representation C* (r)* ---> p'ITC (r) is quasi-equivalent to n and hence
has the same central cover as n - i.e., p'lr = c( n).
So, why isn't O" faithful? Well, if it were, then Voiculescu's Theorem
would imply that O" is approximately unitarily equivalent to O" EB 1f. In other
words, O" weakly contains 1f and thus, by Proposition 17.2.2, it actually
contains n. This contradicts our assumption (2) above and the proof is
complete. D
Here are two nice applications.
Corollary 17.2.5. If r has ·property (T), then it has at most countably
many nonequivalent finite-dimensional representations.Proof. A separable C-algebra (e.g., C(r)) can have at most countably
many orthogonal projections (since it can be represented on a separable
Corollary 17.2.6. Assumer has property (T) and let J1 <1 C*(r) be the
ideal generated by all of r's Kazhdan projections. Then
C*(r)/ J1
has no amenable traces.
6There are many ways to do this. Perhaps the simplest is to take a direct sum of all the
irreducible representations except 7r and then inflate to force the essential part. However, one
wants to stay on a separable Hilbert space so one should restrict to a countable set of such
irreducible representations while preserving condition (1).