12.1. Kazhdan.'s property (T) 353For a II1-factor M, we let Aut(M) denote the group of *-automorphisms
of M, Int(M) c Aut(M) denote the normal subgroup of inner automor-
phisms, and we let Out(M) = Aut(M)/Int(M) be the quotient group. It
can be shown that the hyperfinite type II1 -factor and free group factors have
uncountable outer-automorphism groups (see Exercises 12.1.2 and 12.1.3).
In stark contrast, the following result of Connes was the first application of
property (T) to operator algebras.
Theorem 12.1.19. Let M be a II1-factor with property (T). Then Out(M)
is countable.
Proof. Take a finite subset J C M and 8 > 0 as in Definition 12.1.16, for
E = 1/2. We claim that"( E Aut(M), satisfying ll"f(x) - xll2 < 8 for every
x E J, is inner. Indeed, relative property (T) implies that ll1(a)-all2 < cllall
for all a E M. Let z be the circumcenter (see Exercise D.l) of {"f(u)u :
u E M unitary} C L^2 ( M, T). It follows that z is a nonzero element in M
(why?) and z = 1(u)zu for every unitary element u EM. Let z =viz! be
the polar decomposition of z. Since I z I belongs to the center of M, it is a
positive scalar and"(= Ad(v).
Suppose now that Out(M) is uncountable and choose a lift a E Aut(M)
for each a E Out(M). Since Mis separable in 2-norm (by property (T)), a
simple cardinality argument shows that there exist two distinct lifts a and
/3 such that lla(x) - /3(x)ll2 < 8 for every x E J. Thus, "( = jj-^1 & is not
inner, but it satisfies ll"f(x) -xll2 < 8 for every x E J: contradiction. D
A similar argument can be used to show that the fundamental group
J(M) of a II 1 -factor with property (T) is countable. (Recall that J(M) is
the set oft> 0 such that Mt~ M, where Mt= Pt(lE(H) ® M)Pt for some
projection Pt such that (Tr@T)(Pt) = t.)
Exercises
Exercise 12.1.1. Let r be a group and e be an automorphism of r. Define
e E Aut(L(r)) by B(,(s)) = ,(e(s)). Prove that if e is inner, then there
exist t EI' and a finite index normal subgroup I'o CI' such that e(s) = tsr^1
for every s E I'o.
Exercise 12.1.2. Let 600 be the group of all finite permutations of N.
For any permutation cp of N which does not belong to 600 , we define an
automorphism e of 600 by e(n) = cpncp-^1. Prove that 0 E Aut(L(6 00 )) is
not inner.