amenable equivalence relation generated by an action of a non-amenable group - MathOverflow

amenable equivalence relation generated by an action of a non-amenable group

2011-03-23T16:30:16Z

Question. Give a (possibly elementary) example of a probability measure preserving action $\rho\colon G \curvearrowright X$ of a finitely-generated discrete group $G$ on a standard borel space $X$ with a probability measure, such that

the equivalence relation generated by $\rho$ is ergodic and amenable,

the action $\rho$ is faithful,

the group $G$ is non-amenable.

A friend of mine asked me this question couple of days ago, which led us to another question, but perhaps there is an easier way to provide an example.

Answer by Kate Juschenko for amenable equivalence relation generated by an action of a non-amenable group

2011-03-23T17:32:24Z

I believe in this paper there is an example of such group:

Rostyslav Grigorchuk, Volodia Nekrashevych, "Amenable actions of nonamenable groups" (which can be downloaded from Volodia's webpage)

also, this is related (but not quite clear if one can built examples that you ask from it):

Yair Glasner, Nicolas Monod, "Amenable actions, free products and a fixed point property"

Answer by Nicolas Monod for amenable equivalence relation generated by an action of a non-amenable group

2011-03-26T13:08:40Z

EDIT: I misread "faithful" for "free" (thanks RW). I removed my answer...

Answer by R W for amenable equivalence relation generated by an action of a non-amenable group

2011-03-27T06:00:30Z

Let me join the discussion. Nicolas rightly says that amenability of an action is equivalent to amenability of the orbit equivalence relation and of a.e. stabilizer. It is also true that for a finite invariant measure amenability of the action is equivalent to amenability of the acting group. However, there is no contradiction here as (at least a priori) it might be possible that the orbit equivalence relation of an action of a non-amenable group is still amenable - due to the presence of huge stabilizers (which in this case must necessarily be non-amenable).

I think that the paper by Grigorchuk and Nekrashevych quoted by Kate does provide a requested example. Indeed, they construct (Sections 3 and 4) a non-amenable group which has a faithful self-similar action on a homogeneous rooted tree. This action preserves the uniform measure on the boundary of the tree. Moreover, the orbit equivalence relation is a subrelation (mod 0) of the tail (or co-final in authors' terminology) equivalence relation. Since the latter one is hyperfinite ($\equiv$ amenable), the orbit equivalence relation is also amenable.

Answer by Andreas Thom for amenable equivalence relation generated by an action of a non-amenable group

2011-03-27T13:17:28Z

The answer is yes, such an action exists.

What is needed for the construction is the following very nice example of an action of a non-amenable group on $\mathbb Z$, which I just learned from Gabor Elek.

Consider a graph with vertices given by $\mathbb Z$ and unoriented edges between $n$ and $n+1$.

Pick a random labelling of the edges by the letters $a,b$ and $c$ with no $a$, $b$ or $c$ adjacent to the same letter. This defines an action of the group $G=\mathbb Z/2 \mathbb Z \ast \mathbb Z/2 \mathbb Z \ast \mathbb Z/2 \mathbb Z$. Indeed, just act according to existing labels or fix the element.

This action has the nice feature that it keeps invariant all counting measures on $\mathbb Z$, i.e. all $\mathbb Z$-Folner sequences sets are also Folner sequences for the $G$-action.

Now, the space of labellings (as above) of the graph is itself a probability measure space (a Bernoulli space), which carries an ergodic p.m.p. $\mathbb Z$-action by shifting. It is easy to see that $G$ acts on this space by measure preserving transformations (just by the method described above, done orbit by orbit) and induces an action as required. Indeed, the orbits are just the $\mathbb Z$-orbits, so its ergodic and amenable. Faithfulness follows the fact that you considered all labellings, so that with positive probability (on the space of labellings), an element will act non-trivially. Note also that $G$ is not amenable.

EDIT: As requested, more details on the action. The elements of the shift space are maps $f: \mathbb Z \to \lbrace a,b,c \rbrace$ with $f(n) \neq f(n+1)$. A letter shifts $f$ to the right if $f(1)$ equals that letter, it shifts to the left, if $f(0)$ is equal to the letter; otherwise you fix $f$. It is obvious that the orbits are just the orbits of the shift-action of $\mathbb Z$. Hence, the induced equivalence relation is just the one induced by the action of $\mathbb Z$.