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 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
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