Let $X=\{0,1\}^{\mathbb{N}}$ and $\theta$ be the partition of $X$ induced by the equivalence relation $x \sim x'$ when $x$ and $x'$ differ only at a finite number of coordinates (see this related question).

Given a Bernoulli measure $m$ on $X$, let ${\cal H}$ be the group of transformations $S$ of $X$ satisfying $\theta(x)=\theta(S(x))$ for almost all $x$ and let ${\cal G}$ be the subgroup of ${\cal H}$ consisting of measure-preserving transformations.

Is it possible to explicitely describe ${\cal G}$ ? Under which condition on $m$ the group ${\cal G}$ is ergodic ?

EDIT: I am also interested in the case when $m$ is a stationnary Markov probability on $X$.


1 Answer 1


It's easy to see that the action is always ergodic since $\mathcal G$ contains the group of finite permutations on the indices, which acts ergodically. In fact, the group $\mathcal G$ (which equals $\mathcal H$ in the case $m = \mu^{\mathbb N}$ with $\mu(0) = 1/2$.) that you are describing is the full group of the ergodic hyperfinite measurable equivalence relation. It, and other full groups are discussed in Sections I.3 and I.4 in the book by Alexander Kechris: Global aspects of ergodic group actions, Mathematical Surveys and Monographs, 160, American Mathematical Society, 2010.

  • $\begingroup$ Are you sure you're right when $m=\mu^{\mathbb{N}}$ with $\mu(0) \in ]0, \frac{1}{2}[$ ? (I had been trying to understand an ergodic theoretic paper and that would contradict my understanding) $\endgroup$ May 5, 2012 at 17:01
  • $\begingroup$ Oops sorry, in the case I am interested in, $m$ is Markov. I'm going to add this question to my initial question. $\endgroup$ May 5, 2012 at 17:14
  • $\begingroup$ Yes, you are right. In the case when $m = \mu^{\mathbb N}$ with $\mu(0) \in (0, 1/2)$ you don't have $\mathcal G = \mathcal H$. But $\mathcal G$ still acts ergodically and gives the full group of the ergodic hyperfinite measurable equivalence relation. $\endgroup$ May 5, 2012 at 18:18
  • $\begingroup$ Thank you Jesse. This is more clear for me now. In a stationnary Markov non-Bernoulli case, I think there are only two transformations: the idendity and $x \mapsto -x$ (with $\\{-1,1\\}$ instead of $\\{0,1\\}$), hence the full group is not ergodic in this case. $\endgroup$ May 6, 2012 at 7:16

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