2. SOME BASIC FACTS ABOUT Aut(X, µ) 5

2. Some basic facts about Aut(X, µ)

(A) An element T ∈ Aut(X, µ) is periodic if all its orbits are finite a.e.

It is periodic of period n if for almost all x, T

0(x)

= x, T

1(x),...,T n−1(x)

are distinct but T

n(x)

= x. It is aperiodic if T

n(x)

= x for all n = 0, a.e.

Consider now X =

2N,µ

= the usual product measure. A basic nbhd of

rank n is a set of the form

Ns = {x ∈

2N

: s ⊆ x},

where s ∈

2n.

A dyadic permutation of rank n is determined by a permuta-

tion π of

2n

via

Tπ(sˆx) = π(s)ˆx.

It is clearly in Aut(X, µ). If π is cyclic, we call the corresponding Tπ a cyclic

dyadic permutation of rank n. In that case Tπ is periodic of period

2n.

We now have the following approximation theorem.

Theorem 2.1 (Weak Approximation Theorem, Halmos [Ha]). The

cyclic dyadic permutations are dense in (Aut(X, µ),w). In fact, every open

set contains cyclic dyadic permutations of any suﬃciently high rank.

Remark. The following nice argument for the special case of 2.1 that

states that a measure preserving homeomorphism of X can be weakly ap-

proximated by a dyadic permutation comes from [LS], where it is attributed

to Steger.

Fix T ∈ Aut(X, µ),T a homeomorphism. It is enough to show that for

any given 0 and f1,...,fm ∈ C(X), there is n such that for any N ≥ n,

there is π of rank N such that |f (Tπ

−1(x))

− f (T

−1(x))|

, ∀x ∈ X, =

1,...,m. Choose n large enough so that the oscillation of every f , f ◦ T

−1

on each basic nbhd of rank n is /2. Fix now N ≥ n and enumerate the

basic nbhds of rank N as D1,...,D2N . Next define a bipartite graph G on

two disjoint copies of {1,...,

2N

} by:

(i, j) ∈ G ⇔ Di ∩ T(Dj) = ∅.

Claim. This satisfies the criterion for the application of the marriage

theorem.

Granting this, by the marriage theorem, there is a bijection

ϕ : {1,...,

2N

} → {1,...,

2N

}

with Di ∩ T(Dϕ(i)) = ∅. Consider now x ∈ X. Say x ∈ Di, so that

Tϕ(x) ∈ Dϕ(i). Also T

−1(Di)

∩ Dϕ(i) = ∅, so find z ∈ Di with T

−1(z)

= y ∈

Dϕ(i) Then |f (Tϕ(x)) − f (y)| /2 and |f (y) − f (T

−1(x))|

/2, thus

|f (Tϕ(x)) − f (T

−1(x))|

. Finally, put π =

ϕ−1.

Proof of the claim. Fix A ⊆ {1,...,

2N

} of cardinality M. We will

check that there are ≥ M points adjacent to elements of A. If not, then there

is B ⊆ {1,...,

2N

} of cardinality

2N

− M such that Di ∩ T(Dj) = ∅ for

all i ∈ A, j ∈ B. Then (

i∈A

Di) ∩ (

j∈B

T(Dj)) = ∅, so 1 ≥

∑

i∈A

µ(Di) +