4. THE RECONSTRUCTION THEOREM 23

So fix an algebraic isomorphism f : [E] → [F ]. It follows from 4.7 that

if T ∈ [E], 1 = T, T

2

= 1, then f([T]) = [f(T)].

Let for T ∈ [E] (and similarly for [F ]),

⊥(T) = {S ∈ [E] : supp(S) ∩ supp(T) = ∅}.

Lemma 4.8. If T = 1 is an involution, then f(⊥(T)) = ⊥(f(T)).

Proof. Note that ⊥(T) is really the same as [E|(X \supp(T))], so every

element of ⊥(T) is the product of commutators in ⊥(T), by 4.2. Thus the

same holds for f(⊥(T)). We will show that f(⊥(T)) ⊆ ⊥(f(T)). Applying

this to f

−1,f(T

) we get f

−1(⊥(f(T

))) ⊆ ⊥(T), so we have equality.

Claim. If S ∈ C[T ], where C[T

]

is the centralizer of [T] in [E], then

S = S1S2, where S1 ∈ [T],S2 ∈ ⊥(T).

Granting this claim, we complete the proof of 4.8. Take any commutator

in f(⊥(T)), say [U1,U2], with U1,U2 ∈ f(⊥(T)). Then, as ⊥(T) ⊆ C[T ],

Ui ∈ f(C[T ]) = C[f(T

)]

(this is because [T] and thus C[T

]

is preserved under

f, by 4.7). So, by the claim, applied to f(T), Ui = TiVi with Ti ∈ [f(T)],Vi ∈

⊥(f(T)). Thus

U1U2U1

−1U2 −1

=

T1V1T2V2V1−1T1 −1V2−1T2 −1

and, since Ti,Vj commute, this is equal to

V1V2V1−1V2−1

∈ ⊥(f(T)).

Proof of the claim. If S ∈ C[T ], then S ∈ CT , so S(supp(T)) =

supp(T), thus S = S1S2, where S1 = S|supp(T) ∪ id|(X \ supp(T)),S2 =

S|(X \ supp(T)) ∪ id|supp(T). Clearly S2 ∈ ⊥(T), so it is enough to show

that S1 ∈ [T]. By restricting everything to the support of T, if necessary, it

is enough to show that if supp(T) = X and S ∈ C[T

]

, then S ∈ [T]. Now

S descends to

S∗

on X/T, so if S ∈

[T],S∗

= 1 on X/T, thus there is a

T-invariant set A ⊆ X of positive measure with S(A) ∩ A = ∅. Then S(A)

is also T-invariant. Define now U ∈ [T] to be the identity on X \ S(A) and

switch any two elements of S(A) in the same T-orbit. Then US = SU, a

contradiction. ✷

Now take any A ∈ MALGµ. Then there is an involution T in [E] with

supp(T) = A (split A into two sets of equal measure and let T map one onto

the other and be the identity off A). Put

Φ(A) = supp(f(T)) ∈ MALGµ.

Claim. This is well-defined.

Proof of the claim. Indeed, if T is another involution with supp(T ) =

A, we have ⊥(T) = ⊥(T ), so ⊥(f(T)) = ⊥(f(T )), by 4.8. Note now that

for any U, V ∈ [F ], supp(U) ⊆ supp(V ) ⇔ ⊥(V ) ⊆ ⊥(U), so supp(U) =

supp(V ) ⇔ ⊥(V ) = ⊥(U), thus supp(f(T)) = supp(f(T )). ✷