6. AUTOMORPHISM GROUPS OF EQUIVALENCE RELATIONS 41

Note that each T ∈ N[E] induces by conjugation an isometry iT of

([E],δu) : iT (S) = TST

−1.

Consider the Polish group Iso([E],δu), with

the pointwise convergence topology and the map i(T) = iT . We first check

that it is an algebraic isomorphism of N[E] with a subgroup of Iso([E],δu),

provided that E is aperiodic. It is clearly a homomorphism. Injectivity

follows from 4.11.

We next verify that the image i(N[E]) of N[E] is closed in Iso([E],δu),

provided E is aperiodic.

To see this let Tn ∈ N[E] and assume iTn → i0 in Iso([E],δu). Then for

every S ∈ [E], {TnSTn

−1}

is δu-Cauchy, i.e.,

lim

m,n→∞

δu((Tm1Tn)S(Tm1Tn)−1,S) − −

→ 0.

Fix A ∈ MALGµ and find S ∈ [E] with supp(S) = A (by 4.10). Since

supp((Tm1Tn)S(Tm1Tn)−1) − −

=

Tm1Tn(A), −

we have

µ(Tm1Tn(A)∆A) −

→ 0,

as m, n → ∞. Similarly, since also (iTn

)−1

= iTn

−1

→ i0

−1

in Iso([E],δu),

we have limm,n→0 µ(TmTn

−1(A)∆A)

= 0. So limm,n→∞[µ(Tn(A)∆Tm(A)) +

µ(Tn

−1(A)∆Tm1(A)] −

→ 0, i.e., {Tn} is a Cauchy sequence in the complete

metric

¯w

δ of Aut(X, µ). Thus there is T ∈ Aut(X, µ) with Tn → T weakly.

Then for S ∈ [E],TnSTn

−1

→ TST

−1

weakly. But also TnSTn

−1

→ i0(S)

uniformly, so TST

−1

= i0(S) ∈ [E]. Thus T ∈ N[E] and clearly iT = i0.

We summarize the preceding discussion as follows.

Theorem 6.1. Let E be aperiodic. The group N[E] is a Polishable subgroup

of (Aut(X, µ),w). The corresponding Polish topology τN[E] is given by the

complete metric

δN[E](T1,T2) =

n

2−n[δu(T1SnT1 −1,T2SnT2 −1)

+ δu(T1

−1SnT1,T2 −1SnT2)],

where {Sn} is dense in ([E],u). We have w|N[E] ⊆ τN[E] ⊆ u|N[E], and

thus if Tn → T in N[E], then Tn → T weakly.

Convention. From now on, when we consider topological properties of

N[E] without explicitly indicating the topology, we will always assume that

we refer to τN[E].

If E = EΓ

X

, where the countable group Γ acts in a Borel way on X,

we also have the following fact, denoting for each γ ∈ Γ also by γ the map

γ(x) = γ · x.

Proposition 6.2. For E = EΓ

X

aperiodic, Tn,T ∈ N[E],

Tn → T ⇔ Tn → T weakly and ∀γ ∈ Γ(TnγTn

−1

→ TγT

−1

uniformly).

Proof. Again it is enough to show that if Tn ∈ N[E],Tn → 1 weakly and

TnγTn

−1

→ γ uniformly, ∀γ ∈ Γ, then TnTTn

−1

→ T uniformly, ∀T ∈ [E].

Fix T ∈ [E]. Then there is a partition X =

i

Ai,Ai Borel, and γi ∈ Γ

with T(x) = γi(x), ∀x ∈ Ai. Fix 0. Choose then M large enough so that