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 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
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