Another question about amenability and Følner sequences - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T20:53:57Z http://mathoverflow.net/feeds/question/108255 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/108255/another-question-about-amenability-and-folner-sequences Another question about amenability and Følner sequences Simone Virili 2012-09-27T15:07:01Z 2012-09-27T21:06:59Z <p>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). </p> <p>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:</p> <p>(1) $F_{n}\subseteq F_{n+1}$, for all $n\in \mathbb N$;</p> <p>(2) $\bigcup_{n\in\mathbb N}F_n=G$.</p> <p>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.</p> http://mathoverflow.net/questions/108255/another-question-about-amenability-and-folner-sequences/108258#108258 Answer by Marcin Kotowski for Another question about amenability and Følner sequences Marcin Kotowski 2012-09-27T15:59:27Z 2012-09-27T15:59:27Z <p>Yes - see Lemma 5.1 of Gabor Pete's book on probability and groups: <a href="http://www.math.bme.hu/~gabor/PGG.html" rel="nofollow">http://www.math.bme.hu/~gabor/PGG.html</a> (by the way, these are excellent lecture notes)</p> http://mathoverflow.net/questions/108255/another-question-about-amenability-and-folner-sequences/108259#108259 Answer by R W for Another question about amenability and Følner sequences R W 2012-09-27T15:59:46Z 2012-09-27T21:06:59Z <p>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 $A$.</p> <p>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 <strong>connectivity</strong> condition on a Følner exhaustion which is imposed there, which completely changes the situation.</p> <p>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.</p> <p>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 $|\partial\Phi_{n+1}|/|\Phi_{n+1}|&lt;\epsilon_{n+1}$.</p> <p>This argument is totally trivial (and I would say redundant) from geometrical point of view. </p>