Take the tour ×

Følner's characterization of Amenability says that a group $G$ is amenable if there exists a directed set $(I,\leq)$ and a net {$F_i:i\in I$} of finite subsets of $G$ such that for every $γ ∈ G$, $$\lim_{i\in I}\frac{|γF_i\Delta F_i|}{|F_i|}\ \ \ \rightarrow 0\ ,$$ where $\Delta$ is the symmetric difference of two sets. It is also known that, if $G$ is countable, the word "net" can be substituted by "sequence" (that is $I=\mathbb N$ with the usual order).

Is it true that for countable (or at least finitely generated) groups we can always find a Følner sequence as above, which satisfies the following conditions:

(1) $F_{n}\subseteq F_{n+1}$, for all $n\in \mathbb N$;

(2) $\bigcup_{n\in\mathbb N}F_n=G$.

The motivation for my question comes from the paper "The Abramov-Rokhlin Entropy Addition Formula for Amenable Group Actions" by Ward and Zhang (Mh. Math, 1992). In fact, their Theorem 2.6 (that they attribute to Ornstein and Weiss) is proved for a Følner sequence as the one above but it is applied to actions of arbitrary countable Amenable groups. So... it seems that such sequences always exist.

share|improve this question
add comment

2 Answers

up vote 3 down vote accepted

Yes - see Lemma 5.1 of Gabor Pete's book on probability and groups: http://www.math.bme.hu/~gabor/PGG.html (by the way, these are excellent lecture notes)

share|improve this answer
Thanks! It seems that these lecture notes are exactly what I need! –  Simone Virili Sep 27 '12 at 16:06
add comment

Yes, it follows from the following simple lemma: if $(F_n)$ is a Følner sequence, then $(A\cup F_n)$ is also a Følner sequence for any finite set $A$.

EDIT. It is a quite common misconception to formulate the Følner condition in terms of exhausting sequences only. The bottom line is that existence of almost invariant functions or measures (be it for a group/groupoid action or the Riemannian/combinatorial Laplacian) is equivalent to existence of sets with small boundary (Følner sets). There is nothing in this equivalence which would require these sets to be nested or to exhaust the whole state space. In what concerns the lecture notes you mention you miss the connectivity condition on a Følner exhaustion which is imposed there, which completely changes the situation.

Let me give a more detailed argument showing how the lemma above implies existence of a nested exhausting sequence of Følner sets on any amenable group or graph. Actually, for the sake of simplicity let me do it just for graphs with uniformly bounded vertex degrees - which also takes care of finitely generated groups; for infinitely generated groups one has to consider symmetric differences between individual translates instead of dealing just with boundaries in the Cayley graph.

By the classical theorem of Gerl a graph $X$ is amenable if and only if there exist finite subsets $F$ with arbitrarily small ratio $|\partial F|/|F|$. I claim that one can always choose a nested sequence $\Phi_n$ which exhausts $X$ with $|\partial \Phi_n|/|\Phi_n|\to 0$. Indeed, take a sequence $F_n\subset X$ with $|\partial F_n|/|F_n|\to 0$, an exhausting nested sequence $A_n\subset X$, and a sequence $\epsilon_n\to 0$. Put $\Phi_1=F_1$, and then inductively (by using the above lemma) $\Phi_{n+1}=\Phi_n \cup A_n \cup F_N$, where $N=N(n)$ is chosen to satisfy the inequality $|\partial\Phi_{n+1}|/|\Phi_{n+1}|<\epsilon_{n+1}$.

This argument is totally trivial (and I would say redundant) from geometrical point of view.

share|improve this answer
I did know that fact but how can you conclude using just this? –  Simone Virili Sep 27 '12 at 16:06
Take a Følner sequence $F_n$ and an auxiliary sequence $A_n$ exhausting the group, then by using the lemma build a new sequence of Følner sets $\Phi_n$ such that $\Phi_{n+1}\supset \Phi_n\cup A_n$. –  R W Sep 27 '12 at 16:45
I still do not see this proof, the lemma you mention holds for a single set $A$. So, even applying it many times I do not see what you can obtain. Looking at the proof suggested by Marcin Kotowski it seems that the right argument is not much more complicated but still you need something more than what you are suggesting. In particular, in that lecture notes, a Følner sequence satisfying (1) and (2) is called Følner exhaustion. It turns out that not every graph (which is not a Cayley graph) with a Følner sequence has a Følner exhastion, still the lemma you suggest to use holds on graphs. –  Simone Virili Sep 27 '12 at 17:02
See the edit of my answer. –  R W Sep 27 '12 at 20:16
Thanks, now I see what you wanted to say! –  Simone Virili Sep 27 '12 at 21:12
add comment

Your Answer


