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 =
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 =
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
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