Left Translation Action is Ergodic with respect to Haar

dynamical systemsergodic-theorygroup-actionsmeasure-theorytopological-dynamics

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.

Best Answer

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

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

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

  3. 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 $\Psi_f \in C(G)$.

  4. 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 $f=1_B$.

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

  6. 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 $z \in G$.

  7. 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

Related Question