56 I. MEASURE PRESERVING AUTOMORPHISMS
Proposition 9.6. Let Γ be an infinite countable group. Then Γ is ICC iff
its conjugacy shift action on
2Γ∗
is ergodic iff its conjugacy shift action on
2Γ∗
is weak mixing.
Proof. View
2Γ∗
as Z2
Γ∗
= the Cantor group. The conjugacy shift action
of Γ on Z2
Γ∗
is an action by automorphisms of Z2
Γ∗
and the corresponding
action on Z2
Γ∗
= Z2
Γ∗
=
Pfin(Γ∗)
(= the set of finite subsets of
Γ∗)
is the
conjugacy action of Γ on
Pfin(Γ∗).
Let Pfin(Γ)

=
Pfin(Γ∗)\{∅}
and denote by
ΓF the stabilizer of F Pfin(Γ)

under this action. Since the trivial character
corresponds to ∅, we have that the conjugacy shift action is ergodic iff it is
weak mixing iff ∀F Pfin(Γ)([Γ : ΓF ] = ∞) (see, e.g., Kechris [Kec4], 4.3).
If γ
Γ∗,
then Γ{γ} has infinite index iff the conjugacy class of γ is
infinite. Also [ΓF : ΓF Γ{γ}] for any γ F , thus ∀F Pfin(Γ)([Γ

:
ΓF ] = ∞) iff ∀γ
Γ∗([Γ
: Γ{γ}] = ∞) iff Γ is ICC.
Also note the following fact.
Proposition 9.7. If Γ is countable ICC, then the conjugacy shift action of
Γ on
2Γ∗
is free (a.e.).
Proof. Fix γ
Γ∗
and let p
2Γ∗
be such that γ · p = p, i.e., ∀δ(p(δ) =
p(γ−1δγ)). Thus if γ acts by conjugation on Γ∗, p is constant on each
orbit. Thus {p : γ ·p = p} is null unless all orbits of this action are finite and
all but finitely many are trivial, i.e., γ commutes with all but finitely many
elements of
Γ∗
in which case its conjugacy class is finite, a contradiction.
Proof of 9.5. Let X =
2Γ∗
, µ = the usual product measure and let Γ
act on
2Γ∗
by conjugacy shift. This action is free, measure preserving and
ergodic. Let {γn}
Γ∗
witness the weak commutativity of Γ, i.e., for each
δ Γ,δγn = γnδ for all large enough n. View each γ Γ as an element of
[EΓ
X
].
(i) γn 1 weakly: It is enough to show for each set A of the form
A = {p
2Γ∗
: p(δi) = ai,i = 1,...,k}, where ai {0, 1},δ1,...,δk
Γ∗,
that µ(γn(A)∆A) 0. Now for all large enough n:
p γn(A) γn
−1
· p A
∀i k((γn
−1
· p)(δi) = ai)
∀i k(p(γnδiγn
−1)
= ai)
∀i k(p(δi) = ai)
p A.
(ii) γnδγn
−1
δ uniformly, ∀δ Γ: This is clear as γnδγn
−1
= δ, if n is
large enough.
(iii) δu(γn, 1) = 1, as Γ acts freely.
(D) Recall from Jones-Schmidt [JS] that a measure preserving action
of Γ on (X, µ) is called stable if
X
(on (X, µ)) is isomorphic to
X
×
E0 (on (X, µ) ×
(2N,ν),
where ν is the usual product measure on
2N).
In
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