Total ergodicity

A measure-preserving transformation {T} of a (finite) measure space {(X,\Sigma,\mu)} is called ergodic if whenever {A\subset X} satisfies {T^{-1}A=A} we have {\mu(A)=0} or {\mu(A^c)=0}. To aid intuition, it is useful to keep in mind the ergodic theorem, which roughly states that ergodicity equals equidistribution.

Following a discussion today, I’m concerned here with the ergodicity of {T^n} for {n\in\mathbf{N}}. This is an entertaining topic, and I thank Rudi Mrazovic for bringing it to my attention. Here’s an amusing example: While {T} being ergodic does not generally imply that {T^2} is ergodic, {T^2} being ergodic does imply that {T^4} is ergodic.

Let’s start with a quite simple observation.

Lemma 1 If {n\in\mathbf{N}} and {T^n} is ergodic then {T} is ergodic.

Proof: Suppose {T^{-1}A=A}. Then {T^{-n}A=A}, so {\mu(A)=0} or {\mu(A^c)=0}. \Box

What is the picture of {T} being ergodic and {T^n} not being ergodic? Certainly one picture to imagine is a partition {X = E\cup T^{-1}E\cup\cdots\cup T^{-(d-1)}E} with {1<d\mid n} and {T^{-d}E=E}, or in other words a factor {X\rightarrow \mathbf{Z}/d\mathbf{Z}}. Possibly surprisingly, this is the whole story.

Theorem 2 If {T} is ergodic and {T^n} is not ergodic then for some divisor {d>1} of {n} there exists a partition of {X} as {E\cup T^{-1}E\cup\cdots\cup T^{-(d-1)}E \cup N} where {T^{-d}E=E} and {\mu(N)=0}.

Proof: Since {T^n} is not ergodic there exists a set {A} with {T^{-n}A=A} and {\mu(A)>0} and {\mu(A^c)>0}. For any subset {S\subset \mathbf{Z}/n\mathbf{Z}} let {X_S} consist of those {x\in X} such that

\displaystyle \{j\in\mathbf{Z}/n\mathbf{Z} : T^j x \in A\} = t + S

for some {t\in\mathbf{Z}/n\mathbf{Z}}. (Note that because {x\in A} iff {T^n x\in A}, the sentence “{T^j x \in A}” makes sense for {j\in \mathbf{Z}/n\mathbf{Z}}.) Note that {T^{-1}X_S = X_S}, so {\mu(X_S)=0} or {\mu(X_S^c)=0}. Since {X = \bigcup_S X_S}, we must have {\mu(X_S^c)=0} for some {S}.

Let {d} be the smallest positive period of {S}. Note that {d=1} iff {S=\emptyset}, contradicting {\mu(A)>0}, or {S=\mathbf{Z}/n\mathbf{Z}}, contradicting {\mu(A^c)>0}, so {d>1}. Let

\displaystyle E = \bigcap_{j\in S} T^{-j} A.

(Again note that {T^{-j}A} makes sense for {j\in\mathbf{Z}/n\mathbf{Z}}.) Then {\mu(T^{-i}E \cap T^{-j} E) = 0} unless {-i+S = -j+S}, i.e., iff {i-j} is divisible by {d}, in which case {T^{-i}E=T^{-j}E}. Thus we have the desired partition

\displaystyle X_S = E \cup T^{-1}E \cup \cdots \cup T^{-(d-1)}E.

This finishes the proof. \Box

Note as a corollary that “only the primes matter” insofar as which {n\in N} make {T^n} ergodic: if {n>1} and {T^n} is not ergodic then for some prime {p|n}, {T^p} is not ergodic. Combining this with the initial lemma, which shows that if {T^n} is ergodic then {T^p} is ergodic for every {p|n}, we have thus proved one half of the following theorem.

Theorem 3 Given a set {S\subset\mathbf{N}}, there exists a measure-preserving system {(X,\Sigma,\mu,T)} such that {S = \{n\in\mathbf{N} : T^n \text{ is ergodic}\}} if and only if {S} is empty or the set of {n\in\mathbf{N}} not divisible by any {p\in\mathcal{P}}, for some set of primes {\mathcal{P}}.

It remains only to construct a measure-preserving system for a given set {\mathcal{P}} of primes. For {\mathcal{P}} finite this can be achieved even with finite spaces. In general it suffices to look at {\prod_{p\in\mathcal{P}} \mathbf{Z}/p\mathbf{Z}}.

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