# 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.

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 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 ${\prod_{p\in\mathcal{P}} \mathbf{Z}/p\mathbf{Z}}$.

## One thought on “Total ergodicity”

1. alameda says:

Sean, again nicely stated – why can't more mathematicians write like you. I get parts but not the whole but I am old.

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