7. THE OUTER AUTOMORPHISM GROUP 45

Definition of Tn: Let R ∈ [E0] be such that R(N0n ) = T(N0n ). Then

define Tn on Ns,s ∈

2n,

as follows:

Tn(x) = TgsT

−1Rgs(x).

(*)

As gs,TgsT

−1,R

∈ [E0], clearly Tn ∈ [E0]. Also Tn|N0n = R|N0n , Tn(Ns) =

T(Ns). We now check that TngsTn

−1(y)

= TgsT

−1(y),

∀s ∈

2n,

∀y. If y ∈

T(Ns) ∪ T(N0n ),TgsT

−1(y)

= TT

−1(y)

= y and TngsTn

−1(y)

= TnTn

−1(y)

=

y. If y ∈ T(Ns) ∪ T(N0n ) this follows from

(∗).

✷

Since clearly [E0] = N[E0] (e.g., any infinite A ⊆ N induces an element

of N[E0] \ [E0] via x → x + χA(mod 2)), it follows that for any ergodic,

hyperfinite E, Out(E) is not Polish. In fact one can say quite a bit more.

Call Out(E) turbulent if the action of ([E],u) by (right) translation on

N[E] is turbulent. In that case, the equivalence relation induced by the

cosets of [E] in N[E] admits no classification by countable structures (i.e.,

Out(E) cannot be Borel embedded in the isomorphism types of countable

structures).

Theorem 7.3 (Kechris). If E is not E0-ergodic, then every element of [E]

(closure of [E] in N[E]) is turbulent for the translation action of ([E],u) on

N[E]. In particular, if also [E] is not closed in N[E], the translation action

of ([E],u) on [E] is turbulent and the associated coset equivalence relation

admits no classification by countable structures.

Corollary 7.4. If E is ergodic, hyperfinite, then Out(E) is turbulent.

Proof. Let G = [E]. Since the periodic g ∈ [E] are dense in ([E],u), fix

a countable set {gj} of periodic elements in [E] which is dense in ([E],u).

Then a basic nbhd of g ∈ G has the form

U =

n

j=1

{S ∈ G :

δu(SgjS−1,ggjg−1)

}.

Also fix a basic nbhd V of 1 in ([E],u).

We will show that O(g, U, V ) is dense in U. Now {gT

−1

: T ∈ [E]} is

dense in G, so {gT

−1

∈ U : T ∈ [E]} is dense in U. Therefore {ggj

−1

∈ U :

j = 0, 1,... } is dense in U, thus it is enough, for each ρ 0, and h ∈ {gj}

such that

gh−1

∈ U, to show that O(g, U, V ) ∩ {S : δu(S,

gh−1)

ρ} = ∅.

Note that for S ∈ G:

gS ∈ U ⇔ ∀j ≤

n[δu(gSgjS−1g−1,ggjg−1)

]

⇔ ∀j ≤

n[δu(SgjS−1,gj)

].

It is thus enough, for each ρ 0, to find a continuous path λ → hλ in

([E],u) with h0 = 1 such that δu(hλ

−1gjhλ,gj)

, ∀j ≤ n, and δu(h, h1) ρ.

Let α be such that

δu(h−1gjh,

gj) α, ∀j ≤ n. Using the notation of 5.2,

we can find

¯

h, ¯j,j g ≤ n, which are in

[π−1(EN

) ∩ E] for some large enough

N and δu(h,

¯)

h δ0,δu(gj, ¯j) g δ0, ∀j ≤ n, where 0 δ0 ρ, α + 6δ0 .

Then

δu((¯)−1¯j¯

h g h, ¯j) g 3δ0 + α + δ0 = α + 4δ0. Then, as in the proof of