the free group Fm and En by a similar action of the free group Fn (1
m, n ∞), then if m = n, Em

= En. Can one detect this by looking
at their full groups [Em], [En]? One can look for example at topological
generators. Recall that given a topological group G a subset of G is a
topological generator if it generates a dense subgroup of G. Denote by t(G)
the smallest cardinality of a set of topological generators of G. (Thus t(G) =
1 iff G is monothetic.) Clearly 1 t(G) ℵ0, if G is separable.
In the notation above, and keeping in mind that if Γ is dense in ([E],u),
then E = X , Gaboriau’s [Ga1] theory of costs immediately implies that
t([En]) n. In fact, Ben Miller pointed out that one actually has t([En])
n + 1. To see this, assume, towards a contradiction, that T1,...,Tn are
topological generators for [En] and let Γ be the group they generate. Since
En =
, and En has cost n, it follows from Gaboriau [Ga1], I.11, that
Γ acts freely. Let then S1,S2 be two distinct elements of Γ and (as in the
proof of 3.10) find disjoint Borel sets A1,A2 of positive measure such that
S1(A1),S2(A2) are also disjoint. Then there is S [En] with S|Ai = Si|Ai,
for i = 1, 2. Clearly, S cannot be in the uniform closure of Γ.
In particular, t([E∞]) = ℵ0. Note that, by 3.9, for any ergodic E we have
t([E]) 2. In an earlier version of this work, I have raised the question of
whether t([En]) (even in the case n = 1). This has now been answered
by the following result.
Theorem 4.12 (Kittrell-Tsankov [KiT]). An ergodic equivalence rela-
tion E is generated by an action of a finitely generated group iff t([E])
(where [E] is equipped with the uniform topology). Moreover if En is gen-
erated by a free, measure preserving, ergodic action of Fn, then t([En])
3(n + 1).
Kittrell and Tsankov [KiT] also proved that if n = 1, i.e., for ergodic
hyperfinite E, we have t([E]) 3. It is unknown whether in this case the
value of t([E]) is 2. For any n ∞, we have n + 1 t([En]) 3(n + 1).
It is unknown if t([En]) is independent of the action and, in case this has a
positive answer, what is the exact value of t([En]). In any case, the preceding
shows that t([En]) t([Em]), provided m+1 3(n+1). This appears to be
the first result providing a topological group distinction between [Em], [En],
provided m, n are sufficiently far apart.
We remark that it is known that t(Aut(X, µ),w) = 2 and in fact the set
of pairs (g, h) Aut(X,
that generate a dense subgroup is dense in
(see Grzaslewicz [Gr], Prasad [Pr], and Kechris-Rosendal [KR]
for a different approach). Also it is not hard to see that if U(H) is the
unitary group of an infinite dimensional Hilbert space, then t(U(H)) = 2,
and again, in fact, the set of pairs (g, h)
that generate a dense
subgroup of U(H) is dense in
This is because if U(n) is the
unitary group of the finite-dimensional Hilbert space
then (with some
canonical identifications) U(1) U(2) · · · U(H) and
U(n) = U(H).
Then since each U(n) is compact and connected, it follows, e.g., by a result
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