46 I. MEASURE PRESERVING AUTOMORPHISMS

5.2, there is a continuous path λ → hλ in ([E],u) such that h0 = 1,h1 =

¯

h

and δu(hλ

−1¯jhλ,

g ¯j) g α + 4δ0, ∀j ≤ n. Then δu(h, h1) = δu(h,

¯)

h δ0

ρ, δu(hλ

−1gjhλ,gj)

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

(XΓ,µΓ),

given by γ · p(δ) =

p(γ−1δ),

with corresponding equivalence relation E. Any

S ∈ Aut(X, µ) induces

S∗

∈

Aut(XΓ,µΓ)

given by

S∗(p)

= (γ → S(p(γ)).

Clearly

S∗

∈ N[E], in fact

S∗γ

=

γS∗,

∀γ ∈ Γ. It is easy to check that

δu(S∗,T)

= 1, for every S = 1,T ∈ [E]. It follows that S →

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

∗

→

S∗

in N[E] iff Sn

∗

→

S∗

weakly and this is easily equivalent to

Sn → S weakly. If π : N[E] → Out(E) is the projection map and G =

{S∗

:

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)