12.1. Kazhdan's property (T) 349
Set So = ~ 2-k7r(sk)S7r(sk)* E Co for an enumeration r = { sk}~ 1 and take
a sequence ((n) of unit vectors such that llc(n - So(nll ---* 0. Since
II L 2-k7r(sk)S7r(sk)*ll = c = L 2-kll7r(sk)S7r(sk)*ll,
we have by convexity that
Vs Er, lim n lic(n - 7r(s)S7r(s)*(nll = 0.
We claim that the sequence ((n) is almost 7r(r)-invariant. It suffices to
show (modulo passing to a subsequence) that the vector (w = ((n)n-+w E Hw
is 7rw (r)-invariant. Here w is a fixed free ultrafilter on N and 7rw is the
ultrapower unitary representation of r on the ultrapower Hilbert space Hw.
This means that every vector ~w in Hw is represented as a bounded sequence
(~n) in 1t in such a way that (rJw, ~w) = limw('T/n, ~n) and that 7rw(s)~w =
(7r(s)~n)n-+w· (See Appendix A for details.) Similarly, we consider Sw E
IIB(Hw) given by the constant sequence S. Now the equations (*) and (**)
read
Vh E HiUH2, llSw-1rw(h)Swll s Cl and Vs Er, Sw1rw(s)(w = C7rw(s)(w.
Let JC, C Hw be the 7rw(Hi)-invariant Hilbert subspace spanned by 7rw(r)(w.
It follows from the above equations that Sw IK: = cl JC and
cl cl
Vh E Hi U H2, lllJC - 7rw(h)IJCll s - s --=: c2.
c 1-co
Since we have chosen c so that c2 < 0, this implies that 7rw(h)IJC = 1/C for
all h E Hi by Lemma 12.1.5 (and the proposition following it). Since r is
generated by H 1 and H2, this proves the claim.
Since ( (n) is almost 7r(r)-invariant, we have limn 11 (n -Pi(n II = 0 for each
i, by relative property (T). This implies that limn II (n -T(n II = 0. However,
since
n-+w lim ll(n - Q(nll S 2co + n-+w lim llc(n - S(nll = 2co <^1
by the choice of c, we have
(Recall that Q = X[l-(c:/i,;)2, 1 _ 0 1(T).) This is a contradiction.
Finally, that (S, 10-^3 ) is a Kazhdan pair for SL(3, Z) follows from The-
orem 12.1.10, Lemma 12.1.12 and .the proof above. D
A spectral condition which ensures property (T). Let r be a discrete
group which is generated by a finite symmetric set S with e fj. S. The link
of (r, S) is a graph L(r, S) whose vertex set is S and whose edge set is
E = {(s, t) c S^2 : s-^1 t E S}. Since S is symmetric, the link is a simple
undirected graph. We denote by v(s) = l{t ES: (s, t) E E}I the degree of