44 I. MEASURE PRESERVING AUTOMORPHISMS
7. The outer automorphism group
(A) The group [E] is normal in N[E] and the quotient
Out(E) = N[E]/[E]
is called the outer automorphism group of E. If E is aperiodic and N[E]
has the Polish topology τN[E], then Out(E) with the quotient topology will
be a Polish group iff [E] is closed in N[E].
The group [E] may or may not be closed in N[E] (we will say more
about this below) but it is always a Polishable subgroup of N[E] with the
corresponding Polish topology being equal to the uniform topology on [E].
This is because τN[E]|[E] u|[E]. Moreover we have that [E] will be closed
in N[E] exactly when τN[E]|[E] = u|[E], i.e., for Tn [E], TnSTn
−1
S
uniformly, ∀S [E], implies Tn 1 uniformly. If E =
X
, this is also
equivalent to the assertion that Tn 1 weakly and ∀γ Γ(TnγTn
−1
γ
uniformly) implies Tn 1 uniformly.
Proposition 7.1. If E is ergodic and {gn} is dense in ([E],u), then
[E] = N[E]
n
{T : δu(T, gn) 1}.
In particular, [E] is open in (N[E]),u) and an Polishable subgroup of
(N[E],τN[E]).
Proof. If T N[E] and δu(T, gn) 1, then T(x) = gn(x) and thus
T(x)Ex on a positive measure set. Then, by the ergodicity of E, T(x)Ex,
a.e., i.e., T [E].
Since every open ball in u is in w and w|N[E] τN[E], it follows that
[E] is in (N[E],τN[E]).
The preceding result should be contrasted with the fact that, for ergodic
E, [E] is Π3-complete
0
in (Aut(X, µ),w) (see the paragraph after 3.1).
(B) We next study the complexity of Out(E), when E is ergodic, hy-
perfinite.
Theorem 7.2 (Hamachi-Osikawa [HO]). If E is ergodic, hyperfinite,
then [E] is dense in N[E].
Proof. Take E = E0 on
2N.
Consider the basic open sets {Ns}s∈2n and
let Gn be the permutation group of
2n.
We view each g Gn as a member
of [E0] (a dyadic permutation) by letting g(sˆx) = g(s)ˆx. So G1 G2 . . .
and G∞ =
n
Gn is dense in ([E0],u). Finally note that if for s
2n,gs
is
the transposition gs =
(0n
s) that switches
0n,s
(with g0n = id), then Gn is
generated by {gs}s∈2n (as (a b) = (x a)(x b)(x a)).
If T N[E0],n 1, we will find Tn [E0] such that
Tn(Ns) = T(Ns),TngsTn
−1
= TgsT
−1,
for any s
2n.
Then TngTn
−1
= TgT
−1
for any g Gn, so TngTn
−1
TgT
−1
uniformly for any g G∞, and Tn T weakly, thus Tn T in N[E].
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