Let $G$ be a locally compact Hausdorff group. Assume that $\theta:G\to U_d$ is a group homomorphism where $U_d$ is a finite dimensional unitary group. Consider a action of $G$ on $U_d$ by $g.u:=\theta(g)u,$ $u\in U_d.$ Consider $U=\overline{\theta(G)}.$ Is it true that $G$ acts ergodically on $U$ where $U$ is equipped with the invariant probability measure of $U_d$?
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asked May 29, 2020 at 6:28
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Since $U_d$ is connected, every subgroup proper of it had infinite index, so unless $U=U_d$ it must be a set of measure zero there, and the literal answer is "no".
On the other hand, the action of $G$ on $U$ is ergodic for the Haar measure of $U$. Indeed, let $f\in L^1(U)$ be $G$-invariant (e.g. the characteristic function of a $G$-invariant set). Then by the continuity of the action of $U$ on $L^1(U)$, $f$ is $U$-invariant, hence constant a.e.
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$\begingroup$ ($U_d$ being connected, $U$ of finite index means $U=U_d$.) $\endgroup$– YCorCommented May 29, 2020 at 8:49
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$\begingroup$ Yes, you're right. I was thinking "compact Lie group". $\endgroup$ Commented May 29, 2020 at 14:16
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$\begingroup$ Not clear what you mean by saying that the literal answer is "no". Look at the action of $\mathbb Z$ on the unit circle (aka $SO(2)$) determined by an irrational rotation. $\endgroup$– R WCommented May 29, 2020 at 15:27
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1$\begingroup$ @A beginner mathmatician Please forgive me I think my use of the English language caused some confusion I meant it was a dumb question on my part and that is why I used the verb feel, I was just trying to confirm my understanding of the problem with the answerer, I also try to learn here so we are in the same boat, my apologies to you I have deleted the comment and maybe I should rephrased it once again here as "Could you explain why this doesn't contradict the Weyl's ergodic theorem on $R/ Z\cong SO(2)$? $\endgroup$– DabedCommented May 30, 2020 at 14:24
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1$\begingroup$ Consider the case of a rational circle rotation: the map $x\mapsto x+\alpha$ on $\mathbb{R}/\mathbb{Z}$. When $\alpha$ is irrational, it generates a dense subgroup and hence acts ergodically wrt the invariant measure on the circle. But when $\ alpha=p/q$ is rational, it generates the finite subgroup $U=\frac1q\mathbb{Z}/\mathbb{Z}$. This subgroup has measure zero in the circle, so the action on $U$ is not ergodic for Lebesgue measure on the circle. However, the action is ergodically for the discrete measure on the finite group $U$. $\endgroup$ Commented May 30, 2020 at 16:05