# Ergodic action of a subgroup

Are there any examples of $H < G$ such that for any pmp ergodic action of the group $G$ on a standard proba space $(X,\mu)$ there exists a set $A$ of $\mu(A)>C$ such that the action of the subgroup $H$ on $A$ is ergodic? It seems that this can happen only in finite index case...

Conclusion: The combination of the answers of Alex and Clinton give the complete answer to my question.

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Thanks for editing, I've tried several times to post correctly but it did not show up in a right form... –  Kate Juschenko Oct 4 '12 at 21:44
About finite index. What if, say, $H$ is a finite index normal subgroup of $G$ and we consider the action $G$ on $G/H$, where $G/H$ is equipped with the uniform measure? –  Mark Sapir Oct 4 '12 at 21:44
I was keeping in mind standard Bernoulli action on $G/H$, where the action of H is never ergodic, since it fixes itself. But now I am asking completely opposite question... What I mean is that there exists a set A of positive measure in X such that H is ergodic on it. I will correct the post. –  Kate Juschenko Oct 4 '12 at 22:03
Kate, if you're interested in the discrete case (as indicated below), then unless I'm totally misunderstanding something there will be no examples with $H$ infinite index below $G$. The action of $G$ by shifts on $\{0,1\}^{G/H}$ with product measure is ergodic on a nonatomic probability space, and the induced action by $H$ is trivial. (Isn't this what you just wrote in the previous comment?) –  Clinton Conley Oct 4 '12 at 23:06
Yes, this is what I meant in my previous comment. I corrected the post accordingly. The constant C is universal over the actions. –  Kate Juschenko Oct 4 '12 at 23:24

We seem to be talking past each other in the limited space provided by the comments, so maybe I can express myself better in the room provided by the answer box. You indicated that you were focused on countable discrete groups. For countable discrete $G$ and $H < G$ the following are equivalent.

1. There exists a constant $c > 0$ such that for any measure-preserving ergodic action of $G$ on a standard probability space $(X,\mu)$ there is a measurable $H$-invariant subset $A \subseteq X$ on which $H$ acts ergodically and with $\mu(A) \geq c$.

2. $H$ has finite index in $G$.

(not 2) $\Rightarrow$ (not 1). If $H$ has infinite index, consider the ergodic action of $G$ by shifts on $[0,1]^{G/H}$ with the usual product measure. $H$ doesn't act ergodically on any set of positive measure, as the components of its ergodic decomposition are null. More concretely, if $A \subseteq [0,1]^{G/H}$ is an $H$-invariant set of positive measure, there are disjoint sets $B,C \subseteq [0,1]$ such that a positive measure of elements of $A$ send the coset $H$ to something in $B$, and the same for $C$. Since the $H$ action doesn't shift this coset, this shows $H$ doesn't act ergodically on any set of positive measure. (A better way of writing this argument is simply that $f \mapsto f(H)$ is a null-to-one $H$-invariant Borel function from $[0,1]^{G/H}$ to $[0,1]$, which is enough to preclude ergodicity of the $H$ action on any non-null set.) Thanks to Robin for fixing the error in the original answer.

2 $\Rightarrow$ 1. Say $H$ has index $n$ in $G$, and fix coset representatives $g_1, \ldots, g_n$. Fix a measure-preserving ergodic action of $G$ on $(X, \mu)$. Suppose that $B \subseteq X$ has positive measure and is $H$-invariant. Then $\bigcup_{i \leq n} (g_i \cdot B)$ is $G$-invariant and $\mu$-positive, so by ergodicity has measure $1$. This implies that $\mu(B) \geq 1/n$. So picking $A$ to be an $H$-invariant set of smallest positive measure, $H$ acts ergodically on $A$ and $\mu(A) \geq 1/n$. Thus $c = 1/n$ works for all actions.

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Clinton it looks like $\{ 0,1 \}$ needs to be replaced by a non-atomic base space in general if $H$ is malnormal for example. If the orbits of $H$ on the coset space $G/H - \{ H \}$ are infinite then the set of functions in $\{ 0, 1 \} ^{G/H}$ that evaluate to $1$ at the coset $H$ is $H$-invariant and ergodic. This is fixed with a non-atomic base space since then the ergodic decomposition of the H action will be non-atomic –  Robin Tucker-Drob Oct 5 '12 at 2:06
the proble here is that H does not seem to fix everything. It fixes say sequences that have 0 at point H, I.e., the half of the Bernoulli space... –  Kate Juschenko Oct 5 '12 at 2:36
Oh, Robin's right. Thanks for catching that error! –  Clinton Conley Oct 5 '12 at 4:00
I am tending to accept this as the answer, since this is what I was looking for... The case of $\{0,1\}^{G/H}$ made me blind. Thanks Clinton and Robin for clarifying this to me. –  Kate Juschenko Oct 5 '12 at 16:14

You can take $G$ to be any simple Lie group with finite center, and $H$ to be any non-compact Lie subgroup. Then, if the action of $G$ is ergodic, so is the action of $H$. This statement is called the "Moore ergodicity theorem".

In fact, both the actions of $G$ and $H$ will be automatically mixing. This follows from the theory of unitary representations of $G$.

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To clarify, $H$ should be a (closed) Lie subgroup? –  Ian Agol Oct 4 '12 at 22:00
@Agol, thanks, fixed. –  Alex Eskin Oct 4 '12 at 22:03
@Alex: You need simple'' rather than semisimple, to avoid actions factoring through a direct factor... –  Alain Valette Oct 4 '12 at 22:09
The property alluded to by Alex, is the Howe-Moore property: every unitary rep of $G$, without non-zero fixed vector, has coefficients vanishing at infinity. Two results from arxiv.org/pdf/1003.1484.pdf 1) A (2nd countable) locally compact group $G$ has the Howe-Moore property if and only if every pmp ergodic $G$-action is mixing. 2) Assume $G$ is a closed subgroup of $GL_n(K)$, $K$ a local field; the group $G$ has the Howe-Moore property iff it is isomorphic to the subgroup generated by unipotent elements in a simple algebraic $K$-group. –  Alain Valette Oct 4 '12 at 22:25
@Alain: thanks. I am being very careless today :( –  Alex Eskin Oct 4 '12 at 22:37

Let $X = \lbrace 0,1 \rbrace^\mathbb{Z}$ with the $(\tfrac{1}{2},\tfrac{1}{2})$ Bernoulli measure and let $T$ be the shift map $(T\omega)(n) = \omega(n+1)$. Put $G = \mathbb{Z}^2$ and define a $G$-action $S$ on $X$ by $S^{(n,m)} = T^{n + m}$. Take $A = X$ and $H = \mathbb{Z} \times \lbrace 0 \rbrace$. Since $T$ is ergodic both $S$ and $S$ restricted to $H$ are ergodic.