6. AUTOMORPHISM GROUPS OF EQUIVALENCE RELATIONS 43

graph(S) ⊆ E, then δu(TST

−1,S)

≤ UT

×T

(f) − f

2.

2

On the other hand

T → UT

×T

is clearly a Borel homomorphism from the Polish group N[E] into

the Polish group

U(L2(E,

M)), thus it is continuous, so if Tn → 1 in N[E],

then UTn×Tn → 1 in

U(L2(X,

M)). So T → UT

×T

is a homeomorphism of

N[E] with a (necessarily closed) subgroup of

U(L2(X,

M)), and we have the

following fact.

Proposition 6.3. For aperiodic E, the map T → UT

×T

from N[E] to

U(L2(E,

M)) is a (topological group) isomorphism of N[E] with a closed

subgroup of

U(L2(E,

M)).

Thus we can identify each T ∈ N[E] with the corresponding unitary

operator UT

×T

on

L2(E,

M).

(C) It follows from 4.1 that for every ergodic E, N[E] can be also iden-

tified with the automorphism group, Aut([E]), of the (abstract) group [E].

More precisely, every S ∈ N[E] gives rise to the automorphism T →

STS−1

of [E] and every automorphism of the (abstract) group [E] is of this form for

a unique S ∈ N[E]. Next notice that [E] is the smallest non-trivial normal

subgroup of N[E]. Indeed, if K ✁ N[E] is another normal subgroup, then,

since [E] is simple (by 4.6), either [E] ≤ K or else [E] ∩ K = {1}. In the

latter case K ⊆ C[E], the centralizer of [E] in Aut(X, µ), which by 4.11 is

trivial, so K = {1}.

Assume now E, F are ergodic and f : N[E] → N[F ] is an (abstract)

group isomorphism. Then f sends [E] onto [F ] so, by 4.1, E

∼

= F and there is

ϕ ∈ Aut(X, µ) such that f(T) = fϕ(T) =

ϕTϕ−1,

∀T ∈ [E]. Thus fϕ maps

[E] onto [F ] and therefore fϕ maps N[E] onto N[F ]. Consider f

−1

◦fϕ. This

is an automorphism of N[E] which is trivial on [E]. Since [E] is simple, non-

abelian, a theorem of Burnside (see Thomas [Tho], 1.2.8) asserts that every

automorphism of N[E] (which can be identified with the automorphism

group of [E]) is inner and thus there is ψ ∈ N[E] such that f

−1

◦ fϕ(T) =

ψTψ−1,

∀T ∈ N[E], and as f

−1

◦ fϕ(T) = T for T ∈

[E],ψTψ−1

= T, ∀T ∈

[E], i.e., ψ ∈ C[E], so ψ = 1. Thus f(T) = fϕ(T), ∀T ∈ N[E].

So we have the following Reconstruction Theorem for N[E].

Theorem 6.4. If E, F are ergodic equivalence relations, the following are

equivalent:

(i) E

∼

= F ,

(ii) N[E],N[F ] are isomorphic as abstract groups.

Moreover, for any algebraic isomorphism f : N[E] → N[F ] there is

unique ϕ ∈ Aut(X, µ) with f(T) =

ϕTϕ−1,

∀T ∈ N[E].

(D) Finally, we note that Danilenko [Da] has shown that (as a topolog-

ical group) N[E] is contractible, when E is ergodic, hyperfinite. We will see

some further properties of such N[E] in the next section.

Comments. For the definition of the topology on N[E], see Hamachi-

Osikawa [HO], Gefter [Ge], Danilenko [Da].