So fix an algebraic isomorphism f : [E] [F ]. It follows from 4.7 that
if T [E], 1 = T, T
= 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
) we get f
))) ⊥(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
−1U2 −1
T1V1T2V2V1−1T1 −1V2−1T2 −1
and, since Ti,Vj commute, this is equal to
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
on X/T, so if 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
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 )).
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