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 EΓ

X

(on (X, µ)) is isomorphic to EΓ

X

×

E0 (on (X, µ) ×

(2N,ν),

where ν is the usual product measure on

2N).

In