4. THE RECONSTRUCTION THEOREM 29

the ergodic components of the action of Γ, with ν non-atomic, then F |Y

is smooth. Consider the cocycle α from Γ × Y into Z : α(γ, x) = (the

unique n such that γ · x = n · x). Then there is Borel f : Y → Z such that

α(γ, x) = f(γ · x) − f(x). Fix a set A of positive ν-measure, thus meeting

every F -class in Y , on which f is constant. Then if x, γ · x ∈ A we have

α(γ, x) = 0, so γ · x = x, i.e., A meets every F -class in exactly one point, so

F |Y is smooth. ✷

There are related questions that came up in a discussion with Sorin Popa.

What countable groups Γ embed (algebraically) into [E], for an ergodic,

hyperfinite E, so that they generate E? More generally, what countable

groups Γ have measure preserving, ergodic actions that generate a hyper-

finite equivalence relation? Every group that has this last property does

not have property (T) and one may wonder if conversely every non-property

(T) group has such an action. This however fails. Fix a countable group

Γ which is simple, not residually finite, and has property (T) (such groups

exist as pointed out by Simon Thomas). Let F be a non-trivial finite group

and let ∆ = Γ ∗ F . Then ∆ does not have property (T). Assume, towards

a contradiction, that ∆ has an ergodic action which generates a hyperfi-

nite equivalence relation E. Then there is a homomorphism π : ∆ → [E].

Clearly π is 1-1 or trivial on Γ. In the second case, π(∆) = π(F ) is a finite

group generating E, which is absurd. In the first case Γ embeds into [E],

which contradicts 4.14.

On the other hand, one can see that the free groups embed into [E], E

ergodic, hyperfinite, so that they generate E. Take, for example, F2. It is

enough to find S, T ∈ [E], which generate a free group that acts ergodically.

But 3.9 shows that in the uniform topology the set of pairs (S, T) in (APER∩

[E])2

that generate a free group is dense Gδ and, using 3.6, it is easy to check

that the set of such pairs that generate a group acting ergodically is also

dense Gδ. Thus the generic pair in (APER ∩

[E])2

works.

In general, it is not clear what countable groups embed into the full

group of a given ergodic equivalence relation, not necessarily hyperfinite. A

result of Ozawa [O] implies that there is no full group [F ] into which every

countable group embeds (algebraically). In particular, there is no full group

[F ] that contains up to conjugacy all full groups or, equivalently, there is no

equivalence relation F such that for any other equivalence relation E there

is E ⊆ F with E

∼

= E . We give an ergodic theory proof of this fact in

Section 14, (C).

Ben Miller has pointed out that in the purely Borel theoretic context

one actually has the opposite answer. Below all spaces are uncountable

standard Borel spaces. If E, F are countable Borel equivalence relations

on X, Y, E

∼

=

B

F means that there is a Borel bijection π : X → Y with

xEy ⇔ π(x)Fπ(y). Also E

B

F means that there is a Borel injection with

xEy ⇔ π(x)Fπ(y). Miller shows that there is a countable Borel equivalence