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 = EΓ

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 Fσ 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 Fσ in w and w|N[E] ⊆ τN[E], it follows that

[E] is Fσ 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].