5.2, there is a continuous path λ in ([E],u) such that h0 = 1,h1 =
and δu(hλ
g ¯j) g α + 4δ0, ∀j n. Then δu(h, h1) = δu(h,
h δ0
ρ, δu(hλ
2δu(gj, ¯j) g + α + 4δ0 α + 6δ0 .
Remark (Tsankov). As we noted earlier, N[E] is closed in the space
(Aut(X, µ),u) and so δu is a complete metric on N[E]. Also as we saw in the
proof of 7.1 δu(S, T) = 1 if S [E],T N[E] \ [E], so for any two distinct
cosets [E]S = [E]T of [E] in N[E],δu(S , T ) = 1, ∀S [E]S, ∀T [E]T.
It follows that (N[E],u) is separable (i.e., Polish) iff Out(E) is countable iff
τN[E] = u|N[E].
Remark. Consider the shift action of a countable group Γ on
given by γ · p(δ) =
with corresponding equivalence relation E. Any
S Aut(X, µ) induces

given by
= S(p(γ)).
N[E], in fact
∀γ Γ. It is easy to check that
= 1, for every S = 1,T [E]. It follows that S
is a
Borel, therefore, continuous embedding of (Aut(X, µ),w) into N[E] with
image having trivial intersection with [E]. In fact, it is a homeomorphism of
(Aut(X, µ),w) with (a necessarily closed) subgroup of N[E] since, in view
of 6.2, Sn

in N[E] iff Sn

weakly and this is easily equivalent to
Sn S weakly. If π : N[E] Out(E) is the projection map and G =
S Aut(X, µ)}, then it is easy to check that π|G is a homeomorphism,
so (Aut(X, µ),w) embeds topologically as a subgroup of Out(E). So, if
moreover Out(E) is Polish, this copy of (Aut(X, µ),w) in Out(E) is closed,
i.e., [E]G is closed in N[E]. Finally note that for any distinct T1,T2
Aut(X, µ),δu(T1
∗,T2 ∗)
= 1, i.e., G is discrete in u. In particular, taking
Γ = Z, we see that if E is ergodic, hyperfinite, then N[E] contains a closed
subgroup G isomorphic to (Aut(X, µ),w) with G [E] = {1} and moreover
G is discrete in u.
Remark. Assume now that E is ergodic, hyperfinite. Then Connes
and Krieger [CK] have shown that Out(E) has the following very strong
property: Every two elements of Out(E) which have the same order (finite
or infinite) are conjugate. Since, by the preceding remark, Out(E) contains
a copy of Aut(X, µ), which is simple and has elements of any order, it
follows that Out(E) is simple. Also since in Aut(X, µ) every element is a
commutator and a product of three involutions (see Section 2, (D)) the same
is true for Out(E). Going up to N[E] we see that [E] is the unique non-trivial
normal subgroup of N[E]. Moreover, using again the preceding remark and
the result of Connes-Krieger, we see that there is a copy G of Aut(X, µ) in
N[E] so that for every T N[E] some conjugate of T is of the form S1S2,
where S1 [E] and S2 G. It follows (using also 2.10 and the comments in
Section 4) that every element of N[E] is the product of 2 commutators and
6 involutions. I do not know if every (algebraic) automorphism of Out(E)
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