Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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.

share|improve this question
    
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
show 1 more comment

3 Answers

up vote 6 down vote accepted

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.

share|improve this answer
1  
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
add comment

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 $G$.

share|improve this answer
    
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
1  
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
show 2 more comments

Edit: Misread the question.

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.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.