Left Translation Action is Ergodic with respect to Haar

Question

I am trying to solve exercises on ergodic group actions, from the A. Ioana's lecture notes "Orbit Equivalence of Ergodic Group Actions". The following exercise (p.3, Exr.1.14) has two parts, and I am stuck on the second.

Let $G$ be a compact metrizable group with a dense countable subgroup $\Gamma \le G$. Assume $(G, \mu)$ is the standard probability space defined by the Haar measure. Show that the left translation action of $\Gamma$ on $G$ is free and ergodic.

• The action is free iff every non-neutral $g \in \Gamma$ satisfies $\mu(\{x \in G \mid gx = x\}) = 0$.
• The action is ergodic iff any $\Gamma$-invariant Borel subset $B \subseteq G$ satisfies $\mu(B) \in \{0,1\}$.

The fact the action is free seems fairly trivial, since the left translation has no fixed points for any non-neutral $g \in \Gamma$. I'm having a hard time proving it is ergodic. Starting from some $\Gamma$-invariant Borel subset $B \subseteq G$, we can assume that $\mu(B) > 0$. To conclude the proof, we'd then need to show that $\mu(B) = 1$. I'm not sure which properties I'm supposed to leverage, but the density of $\Gamma$ seems key somehow. One of my attempts involved the set $\bigcup_{k \in \mathbb{N}} \Gamma^kB \subseteq B$ and attempting to use it to provide a lower bound to the measure, but I didn't get much farther with that.

Answer 1

1. Recall that a compact group is unimodular, i.e., if $\mu$ is a right-Haar probab. measure and $\nu$ is a left-Haar probab. measure, then $\mu=\nu$. Here is Kakutani's remarkably short proof of this, which also implies uniqueness of Haar measure:

For $f \in C(G)$, $$\int f(x) \, d\mu=\int f(xy) \, d\mu(x) \, d\nu(y) $$ $$=\int f(xy) \, d\nu(y) \, d\mu(x) = \int f(y) \, d\mu(x)\, d\nu(y) = \int f(y) \, d\nu(y)\,. $$

2. For $y \in G$ and $f \in L^2(G,\mu)$, define $\Psi_f(y)=\int_G f(yx) f(x) \,d\mu(x)$.

3. For $h \in C(G)$ (i.e., $h$ continuous), $\Psi_h$ is continuous by bounded convergence.

4. Given $f \in L^2(G,\mu)$, there exists $h_n \in C(G)$ such that $h_n \to f$ in $L^2(G,\mu)$ and $\|h_n\|_2 \le 2\|f\|_2$. Then by the triangle inequality and Cauchy-Schwarz, $$|\Psi_{h_n}(y)-\Psi_f(y)| $$ $$\le \int_G |h_n(yx)-f(yx)| \cdot |h_n(x)| \,d\mu(x)+ \int_G |f(yx)| \cdot |h_n(x)-f(x)| \,d\mu(x) $$ $$ \le \|h_n-f\|_2 \cdot \|h_n\|_2 +\|f\|_2 \cdot \|h_n-f\|_2 \le 3\|h_n-f\|_2 \cdot \|f\|_2 \to 0 $$ $\qquad \qquad$ as $n \to \infty $. Thus $\Psi_{h_n} \to \Psi_f$ uniformly, so $\Psi_f \in C(G)$.

5. For $f \in L^2(G,\mu)$, $$\int_G \Psi_f(y) \, d\mu(y)= \int_G \int_G f(yx) \, d\mu(y)f(x) \,d\mu(x)$$ $$= \int_G \int_G f(y) \, d\mu(y) f(x) \,d\mu(x)= (\int_G f d\mu)^2 \,.$$ Added remark: In the second equality, we employed the invariance of Haar measure: $\int_G f(yx) \, d\mu(y) = \int_G f(y) \, d\mu(y)$ holds for all $f$ of the form $1_B$ by definition, for simple functions by linearity, and for all $f \in L^1 (G,\mu)$ by approximation. In fact, we just need the case $f=1_B$.

6. Given $B$ which is $\Gamma$ invariant, define $f=1_B$ and note that $$ \forall \gamma \in \Gamma, \quad \Psi_f(\gamma)=\int_G 1_B(\gamma x) 1_B(x) \, d\mu(x)=\mu(B)\,.$$

7. So if $\Gamma$ is dense in $G$ and $B$ is $\Gamma$ invariant, then continuity of $\Psi_f$ for $f=1_B$ implies that $\Psi_f(z)=\mu(B)$ for all $z \in G$.

8. By 5 (for $f=1_B$) and 7, we have $\mu(B)=\mu(B)^2$, so $\mu(B)\in \{0,1\}$.

Edit: Motivational remarks.

A. The basic idea is that if $X$ and $Z$ are chosen independently according to Haar measure $\mu$, then $X$ and $ZX$ are also i.i.d. (Think of circle rotations) , i.e. $$P(X \in B \; \text{and} \; ZX \in B)= P(X \in B)P(ZX \in B)= P(X \in B)^2.$$ Formalizing this leads to the definition of $\Psi_f$ and the calculation in step 5.

B. $\Psi_f$ is the convolution of $f$ with its reflection $x \mapsto f(x^{-1})$ and the fact that convolving two functions in $L^2$ yields a continuous function is standard. (Proved in the same way as step 4, see The convolution of two $L^2(\mathbb R)$ functions is continuous). Step 5 is a special case of https://en.wikipedia.org/wiki/Convolution#Integration

Answer 2

Below I present a more topological argument. The underlying theme is that the density of $\Gamma$ really is a property in the realm of topological dynamics (see Topological Invariance of Unique Ergodicity).

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Claim: Let $G$ be a compact metrizable topological group, $\Gamma\leq G$ be a subgroup. Consider the statements

1. $\Gamma$ is dense.
2. The left translation action $l^\Gamma_\bullet:\Gamma\curvearrowright G$ of $\Gamma$ on $G$ is uniquely ergodic (with the Haar probability measure $\eta_G$ the unique $l^\Gamma$ invariant probability measure).
3. $l^\Gamma$ is ergodic w/r/t $\eta_G$ on $G$.

Then $1\iff 2\implies 3$, and if additionally $G$ is connected, then $3\implies 1$ as well.

Proof: Since $G$ is a compact metrizable topological group, there is a distance function $d:G\times G\to\mathbb{R}_{\geq0}$ compatible with its topology that is invariant under both left translations and right translations (see e.g. Einsiedler & Ward's Ergodic Theory, with a view towards Number Theory, Lemma C.2 on p.430). Note that $\eta_G$ is $l^\Gamma$-invariant.

($1\implies 2$) Suppose $\Gamma$ is dense, and let $\mu$ be an arbitrary $l^\Gamma$-invariant Borel probability measure on $G$. We claim that $\mu=\eta_G$. Let $f:G\to\mathbb{R}$ be continuous and $g\in G$. It suffices to show that

$$\int_G f(gx)\, d\mu(x) = \int_G f(x)\, d\mu(x).$$

Since $G$ is compact, $f$ is uniformly continuous, and together with the density of $\Gamma$ we have:

$$\forall \epsilon\in\mathbb{R}_{>0},\exists \delta\in\mathbb{R}_{>0}, \exists \gamma=\gamma_{g,\delta}\in \Gamma: d(\gamma,g)<\delta \,\,\text{ and }\,\,\sup_{x\in G} |f(gx)-f(\gamma x)|<\epsilon.$$

Then for $\epsilon\in\mathbb{R}_{>0}$ arbitrary we have

\begin{align*} &\left|\int_G f(gx)\, d\mu(x) - \int_G f(x)\, d\mu(x)\right|\\ &= \left|\int_G f(gx)\, d\mu(x) - \int_G f(\gamma x)\, d\mu(x)\right|\\ &\leq \int_G |f(gx)-f(\gamma x)|\, d\mu(x) <\epsilon. \end{align*}

Thus any $l^\Gamma$-invariant measure is also $l^G$-invariant, hence is $\eta_G$.

($2\implies 1$) Suppose $\Gamma$ is not dense and put $\overline{\Gamma}=H\lneq G$ and $U=G\setminus H\neq\emptyset$ be the (set-theoretical) complement of $H$. Then $H$ is a compact metrizable group that is $l^\Gamma$-invariant, denote by $\lambda^\Gamma:\Gamma\curvearrowright H$ the restriction of $l^\Gamma$. The Haar probability measure $\eta_H$ of $H$ is $\lambda^\Gamma$-invariant. Define a new Borel probability measure on $G$ by

$$\nu: \mathcal{B}(G)\to [0,1], B\mapsto \eta_H(B\cap H).$$

Then $\nu$ is $l^\Gamma$-invariant. $\eta_G$ is also $l^\Gamma$-invariant and due to the regularity of the Haar measure (see measure of open set with measure Haar or Haar measure on compact group completely positive) we have $\eta_G(U)>0=\nu(U)$, so that $\eta_G$ and $\nu$ are distinct $l^\Gamma$-invariant measures. Consequently $l^\Gamma$ can't be uniquely ergodic.

($2\implies 3$) This follows from the standard argument that the ergodic Borel probability measures of an action is exactly the extremal points of the Choquet simplex of all invariant Borel probability measures of the action (unique ergodicity means that this simplex is a singleton).

($3\implies 1$) Suppose now additionally that $G$ is connected. Suppose $\Gamma$ is not dense; put $\overline{\Gamma}=H\lneq G$, $U=G\setminus H\neq\emptyset$ the complement of $H$, and define

$$\phi=d(\bullet,H): G\to\mathbb{R}_{\geq0}, x\mapsto \inf_{y\in H} d(x,y).$$

Then $\phi$ is continuous and for any $x\in G$ and for any $h_1,h_2\in H$, $\phi(h_1xh_2)=\phi(x)$; in particular $\phi$ is $l^\Gamma$-invariant. If $\eta_G$ were ergodic w/r/t $l^\Gamma$, then $\phi$ would need to be constant $\eta_G$-ae. In particular since $\eta_G$ is regular we have $\eta_G(U)=1$, and consequently there is a unique $c\in\mathbb{R}_{>0}$ ($c=\int_G \phi(x)\, d\eta_G(x)$) such that $K=\{x\in G\,|\, \phi(x)=c\}\subseteq U$ has full $\eta_G$-measure. $K$ is closed, hence $U\setminus K$ is open. Since $\eta_G(U\setminus K)=0$, we have that $U=K$. Thus $G=H\uplus K$ disconnects $G$, a contradiction.

Answer 3

Here is a third argument, coming from Kechris & Miller's Topics in Orbit Equivalence, pp.4-5, Ex.3.2:

Claim: Let $G$ be a compact metrizable topological group, $\Gamma\leq G$ be a subgroup. If $\Gamma$ is dense, then the left translation action $l^\Gamma_\bullet:\Gamma\curvearrowright G$ is ergodic with respect to the Haar probability measure $\eta_G$ on $G$.

Proof: Denote by $U_\bullet:G\curvearrowright \mathcal{F}$ the induced Koopman action on any function space $\mathcal{F}\subseteq L^0(G,\eta_G;\mathbb{R})$; thus $U_g:\phi\mapsto[x\mapsto \phi(g^{-1}x)]$. First note that $U_\bullet:G\curvearrowright L^1(G,\eta_G;\mathbb{R})$ is continuous (with respect to the $L^1$ norm; i.e. in strong topology), that is,

$$g_n\to g \text{ and } |\phi_n-\phi|_{L^1}\to 0 \implies |U_{g_n}(\phi_n)-U_g(\phi)|_{L^1}\to 0.$$

(See e.g. Continuity of $L^1$ functions with respect to translation ; note that the same is true for $L^p$ spaces for $p\in[1,\infty[$ but not $p=\infty$; see e.g. $L_{p}$ distance between a function and its translation). By duality we have that $U_\bullet:G\curvearrowright L^\infty(G,\eta_G;\mathbb{R})$ is continuous with respect to the weakstar topology, that is,

\begin{align*} &g_n\to\ g\text{ and } \forall \phi\in L^1(G,\eta_G;\mathbb{R}): \langle \phi \,,\, \psi_n\rangle\to \langle \phi \,,\, \psi\rangle \\ &\implies \forall \phi\in L^1(G,\eta_G;\mathbb{R}): \langle \phi \,,\, U_{g_n}(\psi_n)\rangle\to \langle \phi \,,\, U_g(\psi)\rangle, \end{align*}

where $\langle\bullet\,,\, \bullet\rangle: L^1(G,\eta_G;\mathbb{R})\times L^\infty(G,\eta_G;\mathbb{R})\to \mathbb{R}, (\phi,\psi)\mapsto \int_G \psi(x)\phi(x)\,d\eta_G(x)$ is the standard pairing.

Let $B\subseteq G$ be measurable with $\eta_G(B)>0$ and suppose $B$ is $l^\Gamma$-invariant, so that its characteristic/indicator function $\chi_B\in L^\infty(G,\eta_G;\mathbb{R})$ is $U^\Gamma$-invariant, where $U^\Gamma_\bullet$ is the Koopman action of $\Gamma$. Since $\Gamma\leq G$ is dense, $\chi_B$ is $U$-invariant. Then by Fubini

\begin{align*} &\forall g\in G, \forall x\in_{\eta_G} G : U_g(\chi_B)(x)=\chi_B(x)\\ &\implies 0 = \int_G \int_G |U_g(\chi_B)(x) - \chi_B(x)|\, d\eta_G(x)\, d\eta_G(g)\\ &\phantom{\implies 0 }= \int_G \int_G |U_g(\chi_B)(x) - \chi_B(x)|\, d\eta_G(g) \, d\eta_G(x)\\ &\implies \forall x\in_{\eta_G} G, \forall g\in_{\eta_G} G : U_g(\chi_B)(x)=\chi_B(x)\\ &\implies \forall x\in_{\eta_G} G, \forall g\in_{\eta_G} G : g^{-1}x\in B \iff x\in B. \end{align*}

(Here "$y\in_{\eta_G}G$" means "for $\eta_G$-almost all $y$ in $G$".)

Since $\eta_G(B)>0$, in particular $B$ is not empty, thus $x_0\in B$ for some $x_0\in G$. Then there is a full measure set $G_0\subseteq G$ such that $G_0x_0\subseteq B$, whence $\eta_G(B)=1$.