Let $K$ be a compact abelian group with its probability Haar measure, and let $S:K\to K$ be an ergodic translation automorphism. Suppose that $T:K\to K$ is a measure-preserving map that commutes with $S$. Is $T$ itself necessarily a translation (almost everywhere)?

This was an optional problem from a class exam, but we never got a solution set and the professor didn't reply to an email query, so we still can't figure it out. We've tried a few things that seemed somewhat promising (namely considering the map $T-\text{Id}$, or using character theory on $L^2$) but didn't end up getting anywhere.


2 Answers 2


Below are two arguments. The first one combines the two pieces you mentioned, and the second one ("Added") assumes that the group is metrizable. The second argument actually does not use the hypothesis that the group is abelian (see ($\dagger$) and ($\dagger\dagger$) below), although the existence of an ergodic translation implies that the group is abelian by one of the arguments I presented at Left Translation Action is Ergodic with respect to Haar.

Claim: Let $G$ be a compact abelian (not necessarily metrizable) group, $\eta_G$ be its Haar probability measure, $R_\bullet:G\curvearrowleft G$ be right translation action: $R_g(x)=x+g$. Let $f:G\to G$ be a $\eta_G$-almost everywhere defined measurable and $\eta_G$-preserving self-map of $G$. If $g\in G$ is such that $R_g$ is ergodic w/r/t $\eta_G$, and $\eta_G$-almost everywhere $f\circ R_g=R_g\circ f$, then there is an $h\in G$ such that $\eta_G$-almost everywhere $f=R_h.$

Denote by $G_0$ the full $\eta_G$-measure subset of $G$ on which both $f$ is defined and the commutation relation $f\circ R_g = R_g\circ f$ holds.

Proof: Let us also denote by $\widehat{G}=\text{Hom}_{\text{LCA}}(G;S^1)$ the Pontryagin dual of $G$. Note that the elements of $\widehat{G}$ (i.e. continuous characters $\chi$ of $G$) separate points of $G$, that is, given any $x,y\in G$, if $x\neq y$, then there is a character $\chi\in\widehat{G}$ such that $\chi(x)\neq\chi(y)$ (see e.g. Rudin's Fourier Analysis on Groups, p.24).

For any $\chi\in\widehat{G}$, put $f_\chi = \chi\circ(f-\text{id}_G):G\to \mathbb{C}$, $x\mapsto \dfrac{\chi\circ f(x)}{\chi(x)}=\chi\circ f(x)\, \overline{\chi(x)}$. Since $f$ is $\eta_G$ preserving we have that $f_\chi\in L^2(G,\eta_G;\mathbb{C})$. Further note that for any $x\in G_0$

$$f_\chi\circ R_g(x)=\chi(f(x+g)-(x+g)) = \chi(f(x)+g-g-x)=\chi(f(x)-x)=f_\chi(x),$$

where in the second equality we used the fact that $f$ commutes with $R_g$ on $G_0$, so that $f_\chi$ is a square integrable function $\eta_G$-almost everywhere invariant under the ergodic $R_g$; consequently it is constant (which constant possibly depends on the character $\chi$) (see e.g. Understanding the proof how an invariant function w.r.t. an ergodic transformation is constant a.e.).

For any $h\in G$, define the level set $S_h = (f-\text{id}_G)^{-1}(h)=\{x\in G_0 \,|\, f(x)-x=h\}$. Then we have that each $S_h$ is a Borel measurable subset and $G_0=\biguplus_{h\in G}S_h$. Note that the level set $S_h$ is precisely the set on which $f$ acts as translation by $h$ (a priori there may be different regions of $G$ on which $f$ acts as translations by different elements; our aim is to show that $f$ acts as a translation by an element that works globally $\eta_G$-almost everywhere). Some level sets are possibly negligible; but since they are disjoint there is an $h_1\in G$ such that $\eta_G(S_{h_1})>0$.

Suppose $f$ is not a translation $\eta_G$-almost everywhere. Then $\eta_G(S_{h_1})<1$ and consequently there is an $h_2\in G$ such that $h_2\neq h_1$ and $\eta_G(S_{h_2})>0$. Continuous characters of $G$ separate points of $G$; thus there is a $\chi=\chi_{h_1,h_2}\in \widehat{G}$ such that $\chi(h_1)\neq \chi(h_2)$. But then $f_{\chi}$ takes different values on the disjoint measurable subsets $S_{h_1}$ and $S_{h_2}$ of positive $\eta_G$-measure, a contradiction to $f_\chi$ being constant $\eta_G$-almost everywhere.

Added: Under the extra assumption that $G$ is metrizable, one can bypass the uncountable union used in the above proof. For this we'll use a version of Lusin's Theorem (see e.g. Lusin Theorem conditions). Specifically consider the version of Lusin's Theorem given in Kechris' Classical Descriptive Set Theory, p.108:

Theorem (Lusin): Let $X$ be a metrizable topological space and $Y$ be a second countable topological space. Then for any Borel probability measure $\mu$ on $X$ and any measurable function $\varphi:X\to Y$ the following holds: for any $\epsilon\in\mathbb{R}_{>0}$, there is a closed subset $F_\epsilon\subseteq X$ such that

  • $1-\epsilon < \mu(F_\epsilon)$,
  • $\left.\varphi\right|_{_{F_\epsilon}}:F_\epsilon\to Y$ is continuous.

If $X$ is Polish (i.e. completely metrizable and separable topological space), then $F_\epsilon$ can be taken to be compact.

Second Proof (assuming $G$ is in addition metrizable): Fix a compatible distance function $d$ on $G$. We may apply Lusin's Theorem (see Every compact metrizable space is second countable, Compact metric spaces is second countable and axiom of countable choice, A metric space is separable iff it is second countable) to get a compact subset $K\subseteq G$ such that

  • $0<\eta_G(K)$,
  • $\left.(f-\text{id}_G)\right|_{_K}:K\to G$ is continuous.

Denote for any $h\in G$ and for any $\delta\in\mathbb{R}_{>0}$ by $G[h|<\delta]=(G,d)[h|<\delta]$ the open ball centered at $h$ with radius $\delta$ (w/r/t the distance function $d$); thus

$$G[h|<\delta]=(G,d)[h|<\delta] = \{x\in G\,|\, d(x,h)<\delta\}.$$

By the continuity of $\left.(f-\text{id}_G)\right|_{_K}$, for any $h$ and $\delta$, the set $\left(\left.(f-\text{id}_G)\right|_{_K}\right)^{-1}(G[h|<\delta])$ is open in $K$ and further for any fixed $\delta$ we have an open cover

$$K=\bigcup_{h\in G}\left(\left.(f-\text{id}_G)\right|_{_K}\right)^{-1}(G[h|<\delta]).$$

By the compactness of $K$, there is a finite set $F_\delta\subseteq G$ such that

$$K=\bigcup_{h\in F_\delta}\left(\left.(f-\text{id}_G)\right|_{_K}\right)^{-1}(G[h|<\delta]).$$

As $K$ has positive $\eta_G$-measure, there is some element $h_\delta\in F_\delta$ such that

$$0<\eta_G\left(\left(\left.(f-\text{id}_G)\right|_{_K}\right)^{-1}(G[h_\delta|<\delta])\right).\quad\quad (\star)$$

Note that for any $h$ and $\delta$ we have that

\begin{align*} \left(\left.(f-\text{id}_G)\right|_{_K}\right)^{-1}(G[h|<\delta]) &=\{x\in K \,|\, d(f(x)-x,h)<\delta\}\\ &=K\cap \left(f-\text{id}_G)\right)^{-1}(G[h|<\delta])\\ &\subseteq \left(f-\text{id}_G)\right)^{-1}(G[h|<\delta]) \end{align*}

Thus by ($\star$) we have that $0<\eta_G\left(\left(f-\text{id}_G)\right)^{-1}(G[h_\delta|<\delta])\right)$. Further, the measurable set $(\left(f-\text{id}_G)\right)^{-1}(G[h_\delta|<\delta])$ is invariant under the ergodic translation $R_g$ (since $f$ commutes with $R_g$; ($\dagger$) one can show this without using the abelian nature of $G$: $f(xg)(xg)^{-1}=f(x)gg^{-1}x^{-1}=f(x)x^{-1}$), whence we have:

$$\forall \delta\in\mathbb{R}_{>0}, \exists h_\delta\in G: \eta_G\left(\left(f-\text{id}_G)\right)^{-1}(G[h_\delta|<\delta])\right)=1.$$

In particular, discretizing $\delta$, we have:

$$\forall k\in\mathbb{Z}_{\geq1}, \exists h_k\in G: \eta_G\left(\{x\in G \,|\, d(f(x)-x,h_k)<1/k\}\right)=1.$$

$G$ is compact, thus there is a subsequence $k_\ell\subseteq k$ and an element $h^\ast\in G$ such that $\lim_{\ell\to\infty} d(h_{k_\ell},h^\ast)=0$. Put $E=\bigcap_{\ell\in\mathbb{Z}_{\geq1}}\left(f-\text{id}_G)\right)^{-1}(G[h_{k_\ell}|<1/{k_\ell}])$. Thus $E$ is a measurable subset of $G$ with full $\eta_G$ measure.

We claim that on $E$, $f$ coincides with translation by $h^\ast$. Indeed, let $\epsilon\in\mathbb{R}_{>0}$. Then there is an $L\in\mathbb{Z}_{\geq1}$ such that for any $\ell\in\mathbb{Z}_{\geq L}$:

$$\dfrac{1}{k_\ell}<\dfrac{\epsilon}{2},\quad\quad d(h_{k_\ell},h^\ast)<\dfrac{\epsilon}{2}.$$

Let $x\in E$. Then

\begin{align*} d(f(x)-x,h^\ast) \leq d(f(x)-x,h_{k_L}) + d(h_{k_L},h^\ast) <\epsilon. \end{align*}

$\epsilon$ was arbitrary, whence on the full $\eta_G$-measure subset $E$, $f=R_{h^\ast}$. (($\dagger\dagger$) Without using the abelian nature of $G$ the same argument gives that on $E$ $f$ coincides with the left translation by $h^\ast$.)

  • $\begingroup$ I'm confused how you conclude that there exists $\eta_G(S_{h_1})>0$. Couldn't all the level sets be negligible? Also, you're using a fact like "every square integrable function invariant by an ergodic action is constant almost everywhere," right? Why is that true? $\endgroup$
    – xir
    Jan 17 at 20:02
  • 1
    $\begingroup$ This was an impressive read (+1) but I don’t understand all of it. I always associate your name on MSE with dynamical system answers. May I ask, from which sources did you “learn your stuff”? I read through about half of “Operator Theoretic Aspects of Ergodic Theory” a year ago, but in the future I intend to learn about it in more breadth and depth $\endgroup$
    – FShrike
    Jan 19 at 7:53
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    $\begingroup$ @FShrike Given the level of interactions I've seen you have here, I'm sure if there is any confusion it's due to my terse notation/phrasing. Thanks to your questions from the Eisner et al book I became aware of it actually; it's a nice book. $\endgroup$
    – Alp Uzman
    Jan 19 at 8:10
  • 1
    $\begingroup$ I see, thank you for responding. I hope to have a good (PhD?) supervisor one day! $\endgroup$
    – FShrike
    Jan 19 at 9:46
  • 1
    $\begingroup$ You don't need any character theory to conclude that every fiber has to be full or zero measure, since they're all invariant under an ergodic transformation. $\endgroup$
    – xir
    Jan 19 at 17:48

Here is another proof, which is due to Adler (see his paper "Generalized Commuting Properties of Measure-Preserving Transformations", p.3, Prop.1). It seems the statement goes back to Halmos & von Neumann's paper "Operator Methods in Classical Mechanics, II" (p.347, Cor.2). Both of these arguments are for compact separable (hence metrizable) abelian groups; the argument that Halmos and von Neumann give is essentially the first argument I gave in my previous answer; here is said argument in totality:

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Note that the content of the footnote is precisely where separability is used, and this was also a point of contention earlier.

Proof (assuming $G$ is in addition separable): Let $\chi\in \widehat{G}$. Then note that

$$\chi\circ R_g(x)=\chi(x+g)=\chi(x)\chi(g)=\chi(g)\chi(x),$$


$$\chi\circ f \circ R_g(x) = \chi\circ R_g\circ f(x)=\chi(f(x)+g)=\chi\circ f(x)\chi(g)=\chi(g)\chi\circ f(x).$$

Thus both $\chi$ and $\chi\circ f$ are square-integrable eigenfunctions of the Koopman operator of $R_g$, associated to the eigenvalue $\chi(g)$. Since $R_g$ is ergodic, the eigenvalue $\chi(g)$ is simple ($\blacktriangle$), whence there is a function $\Phi:\widehat{G}\to S^1$ (that possibly depends on $f$) such that for any $\chi\in\widehat{G}$ we have $\chi\circ f= \Phi(\chi)\,\chi$ almost everywhere. Since $G$ is compact, $\widehat{G}$ is discrete, whence $\Phi$ is continuous. Further, if $\chi,\rho\in\widehat{G}$, then (writing the group operation in $\widehat{G}$ multiplicatively) we have

$$\Phi(\chi\rho)\, \chi\rho = (\chi\rho)\circ f = (\chi\circ f) (\rho\circ f) = \Phi(\chi)\,\chi\, \Phi(\rho)\,\rho = \Phi(\chi)\Phi(\rho)\,\chi\rho,$$

so that $\Phi\in\widehat{\widehat{G}}$. Thus by Pontryagin duality there is a unique $h\in G$ such that for any $\chi\in\widehat{G}$, $\Phi(\chi)=\chi(h)$.

We claim that $f=R_h$ $\eta_G$-almost everywhere. Let $\chi\in\widehat{G}$. Then $\chi\circ f(x)$ $= \Phi(\chi)\, \chi(x)$ $=\chi(g)\chi(x)$ $=\chi\circ R_g(x)$ for almost every $x\in G$. Since $G$ is separable $\widehat{G}$ is countable, so the set

$$N=\bigcap_{\chi\in\widehat{G}}\{x\in G\,|\, \chi(f(x)-R_h(x))\neq1\}$$

is measurable and $\eta_G$-negligible. Further no character can separate $f(x)$ and $R_h(x)$ for $x\in G\setminus N$, which means that $f=R_h$ on the full measure set $G\setminus N$.

($\blacktriangle$) Taking the modulus of $\phi\circ R_g=_{\eta_G}\chi(g) \phi$ for nonzero $\phi$ gives that $|\phi|$ is invariant almost everywhere, hence by ergodicity is a nonzero constant; if $\psi\circ R_g=_{\eta_G}\chi(g)\psi$ also for nonzero $\psi$, then $\dfrac{\phi}{\psi}$ is invariant almost everywhere, hence constant almost everywhere.

  • $\begingroup$ oh wow, thanks! is it possible, in that case, that the result is not even true if one drops the separability condition? I still haven't seen a convincing proof in that generality. $\endgroup$
    – xir
    Jan 22 at 4:48
  • $\begingroup$ @xir Of course it is possible, but I still believe that the statement is true in full generality, however handling the uncountable union involved will require some sophisticated measure theory. $\endgroup$
    – Alp Uzman
    Jan 22 at 4:58

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