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 a cyclic
dyadic permutation of rank n. In that case 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) +
Previous Page Next Page