58 I. MEASURE PRESERVING AUTOMORPHISMS
measure), ii) the shift action of Γ on
2I
admits non-trivial almost invariant
sets (equivalently: admits a non-trivial asymptotically invariant sequence),
iii) the Koopman representation associated to the shift action, restricted to
the orthogonal of the constant functions, admits non-0 almost invariant vec-
tors (see Section 10, (C) for an explanation of these notions). In particular,
if these conditions hold, the shift action of Γ on
2I
is not orbit equivalent to
the shift action of Γ on
2Γ,
provided Γ is not amenable.
We finally note the following (almost) characterization of weakly com-
mutative groups.
Proposition 9.10. Let Γ be a countable group.
i) If Γ is a non-discrete (i.e., not closed) normal subgroup of a Polish
group, then Γ is weakly commutative.
ii) If Γ is weakly commutative with trivial center, then Γ is a non-discrete
normal subgroup of a Polish group.
Proof. i) Assume that Γ is a non-discrete normal subgroup of the Polish
group G. Consider the map
π : G Aut(Γ),
π(g) =
gγg−1).
This is a Borel homomorphism, so it is continuous, where Aut(Γ) has the
pointwise convergence topology. Since Γ is not discrete in G, there is a
sequence {γn} Γ \ {1} with γn 1 (in G). So
∀δ∀∞n(γnδγn −1
= δ), i.e.,
∀δ∀∞n(γnδ
= δγn), where
∀∞
means “for all but finitely many”. Thus Γ is
weakly commutative.
ii) Consider the map
ρ : Γ Aut(Γ) = G,
ρ(γ) =
γδγ−1).
Since Γ has trivial center, ρ is an isomorphism, so we can identify Γ with
ρ(Γ), which is a normal subgroup of G. Let {γn} Γ\{1} witness the weak
commutativity of Γ. Then γn = 1,γn 1 in G, so Γ is not discrete in G.
Thus a centerless group is weakly commutative iff it is (up to algebraic
isomorphism) a non-discrete normal subgroup of a Polish group (or equiv-
alently a dense normal subgroup of a non-discrete Polish group). In the
case of groups with non-trivial center (which are automatically weakly com-
mutative) it is not clear if there is a simple characterization of when they
are non-discrete normal subgroups of Polish groups. Tsankov pointed out
that if Γ has non-trivial finite center and countable automorphism group
Aut(Γ), then Γ cannot be a non-discrete normal subgroup of a Polish group
G. Otherwise if π : G Aut(Γ) is defined as in the proof of 9.10, i)
and γn are distinct in Γ with γn 1, then π(γn) 1 in Aut(Γ), which
is discrete, so π(γn) = 1 for all large enough n and thus the center of Γ
is infinite. For example, one can take Γ = × Z2, where is a simple
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