Prevalence of ergodic rotations on a compact group

Question

So I know that any compact group $G$ carries a Haar probability measure $\mu$. For any $g \in G$, let $T_g : G \to G$ be the rotation $T_g : x \mapsto gx$. In the cases where $G = \mathbb{R}^n / \mathbb{Z}^n$ and $g = (t_1, \ldots, t_n)$, the rotation $T_g$ is ergodic iff $\{ t_1, \ldots, t_n, 1 \}$ is linearly independent over $\mathbb{Q}$.

In the case of toral rotations, I notice that almost every rotation is ergodic, i.e. $T_g$ is ergodic for almost all $g$. Is this typical behavior for group rotations, i.e. that almost every element of the compact group gives rise to an ergodic rotation?

Answer 1

Let $G = \{0,1\}^\mathbb{N}$ with component-wise mod $2$ addition. Then for all $g$, the rotation $x \mapsto gx$ is not ergodic, since $g^2 = 0$.

Answer 2

This also fails for nonabelian compact connected nontrivial Lie groups (e.g. $G=SO(3)$): Every $g\in G$ is contained in a compact abelian subgroup (a compact torus) $T$ and the latter does not act ergodically on $G$ because $G/T$ is a manifold which is not a singleton.

Edit. I'll say that a compact Hausdorff topological group $G$ satisfies Property E if almost every (for a Haar measure on $G$) $g\in G$ acts ergodically on $G$ via left multiplication. The class of groups satisfying Property E is very restrictive:

Theorem. $G$ satisfies Property E if and only if $G$ is the inverse limit of a family of compact tori $T^{n_\beta}$, where $n_\beta\in {\mathbb N}\cup \{0\}$ for each $\beta$.

Besides tori, examples of groups satisfying E include (but are not limited to) solenoids and infinite products of circles.

Proof of the theorem (one direction). Suppose that $G$ satisfies Property E and $g\in G$ acts ergodically on $G$. Let $H$ denote the closure in $G$ of the cyclic subgroup $\langle g\rangle$ generated by $g$. Then $H$ is abelian (since it is the closure of an abelian subgroup). If $G$ is nonabelian, $G\ne H$. Let $Hx\subset G$ be a coset different from $H$. Since $G$ is Hausdorff, there exists $H$-invariant disjoint neighborhoods $U, V$ of $H$ and $Hx$ in $G$. Thus, $H$ does not act ergodically on $G$ and neither does $g$.

We, thus, conclude that $G$ is abelian.

I next claim that $G$ is connected. Let $G_0< G$ be the identity component ; it is a closed subgroup of $G$. The quotient $Q=G/G_0$ is a totally disconnected Hausdorff compact topological group. Then $1_Q$ has a basis of clopen neighborhoods $V_\alpha$ consisting of subgroups of $Q$. If $Q$ is nontrivial, there exists $V_\alpha\ne Q$. The preimage $U_\alpha$ of $V_\alpha$ in $G$ has positive measure (since it is open). Taking any element $g\in U_\alpha$, we see that $g^nU_\alpha=U_\alpha$ for all $n\in {\mathbb Z}$. At the same time, if $yV_\alpha\subset Q$ is a coset different from $V_\alpha$ and $x\in G$ is an element projecting to $y$, then $xU_\alpha$ has positive measure and $g^n x U_\alpha\cap U_\alpha=\emptyset$ for all $n$. Thus, $g$ does not act ergodically on $G$, which is a contradiction.

We conclude that $G$ has to be connected. Now, it follows from the classification of (locally) compact connected Hausdorff topological groups that $G$ is isomorphic (as a topological group) to the inverse limit of a sequence of quotient groups $G_\beta$ of $G$ (obtained by dividing $G$ by "small" subgroups of $G$) such that each $G_\beta$ is a Lie group. In our case, each $G_\beta$ is compact, connected and abelian, hence a torus. Thus, $G$ is isomorphic to an inverse limit of a family of tori. If $G$ were also 1st countable (equivalently, metrizable) we would could say that this family of tori is a sequence.

The following reference contains much more than you need regarding (locally) compact groups:

Montgomery, Deane; Zippin, Leo, Topological transformation groups, Mineola, NY: Dover Publications (ISBN 978-0-486-82449-9). xi, 289 p. (2018). ZBL1418.57024.

I will add a proof for the converse later on...