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 (T −1 π (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(D ϕ(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) +

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