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

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

1. the equivalence relation generated by $\rho$ is ergodic and amenable,
2. the action $\rho$ is faithful,
3. 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.

• This question is discussed in: Alexander S. Kechris "Global Aspects of Ergodic Group Actions and Equivalence Relations", pp. 31-34. It is available on line here: citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.67.3195 – Jesse Peterson Mar 27 '11 at 7:18
• You can use the construction known as Mackey range: take a cocycle of an amenable equivalence relation with values in a non-amenable group G. Then G induces an amenable action on the space of skew product orbits. See Zimmer's paper for details. – SIB Mar 27 '11 at 12:21
• @Jesse Peterson - Could you be a bit more specific? – R W Mar 27 '11 at 12:31
• The discussion is in Chapter 1, section 4, subsection (E). In the on line version I've linked to this actually begins on page 27 (page 37 of the pdf file). Thanks RW. – Jesse Peterson Mar 27 '11 at 13:30
• @SIB - Yes, but the problem is still to find such a cocycle for which the resulting action is faithful - which is more or less equivalent to the original question. – R W Mar 27 '11 at 13:40

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 $\mathbb Z$.

• I may miss something, but I don't see how it becomes an action. Take the sake of example the following labelling: $a$ on the edge $(0,1)$, and $b$ on all other edges $(n,n+1)$. Then, according to what you say, $a$ maps 0 to 1 and preserves all other vertices, whereas $a^{-1}$ maps 1 to 0 and preserves all other ones (I presume you meant that if you reverse orientation of an edge, then the label changes to its inverse). But then their composition is not the identity map. – R W Mar 27 '11 at 13:37
• I agree that now there is an action of the free product on $\mathbb Z$. However, I still don't see how you define an action of this free product on your space of symbol sequences (your description "just by the method described above, done orbit by orbit" is not clear to me at all), and why this hypothetical action should be orbit equivalent to the shift. – R W Mar 27 '11 at 22:27
• It is really easy; I added more details above. – Andreas Thom Mar 28 '11 at 0:15
• Simple and elegant! Thanks for sharing it! – Kate Juschenko Mar 28 '11 at 13:22

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"
• Let me also mention that by taking generic'' (think Baire category) generators one can show that the free groups have such an action. Also, by a result of Kirchberg (ams.org/mathscinet-getitem?mr=1282231), if $\Gamma$ has property (T) and also has such an action then it must be residually finite. – Jesse Peterson Mar 24 '11 at 12:53

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.