math lessons - Følner sequence

In mathematics, a Følner sequence for a group is a sequence of sets satisfying a particular condition. If a group has a Følner sequence with respect to its action on itself, the group is amenable.

Definition

Given a group $G$ that acts on a set $X$ , a Følner sequence for the action is a sequence of finite subsets $F_1, F_2, \dots$ of $X$ that "don't move too much" when acted on by any group element. Precisely,

$\lim_{i\to\infty}\frac{|gF_i\,\triangle\,F_i|}{|F_i|} = 0$ for all group elements

g

G

Explanation of the notation used above:

$g F i$ is the set $F i$ acted on the left by $g$ . It consists of elements of the form $g f$ for all $f$ in $F i$ .
$\triangle$ is the symmetric difference operator.
$| A |$ is the cardinality of a set $A$ .

Thus, what this definition says is that for any group element $g$ , the percent of elements of $F i$ that are moved away by $g$ goes to 0 as $i$ gets large.

Examples

Any finite group $G$ trivially has a Følner sequence $F i = G$ for each $i$ .

Consider the group of integers, acting on itself by addition. Let $F i$ consist of the integers between $- i$ and $i$ . Then $g F i$ consists of integers between $g - i$ and $g + i$ . The symmetric difference has size $2 g$ , while $F i$ has size $2 i + 1$ , so the ratio is $2 g / (2 i + 1)$ , which goes to 0 as $i$ gets large.

Proof of amenability

We have a group $G$ and a Følner sequence $F i$ , and we need to define a measure $μ$ on $G$ , which philosophically speaking says how much of $G$ any subset $A$ takes up. The natural definition that uses our Følner sequence would be

$\mu(A)=\lim_{i\to\infty}{|A\cap F_i|\over|F_i|}$ .

Of course, this limit doesn't necessarily exist. To overcome this technicality, we take an ultrafilter $U$ on the natural numbers that contains intervals $[n,\infty)$ . Then we use an ultralimit instead of the regular limit:

$\mu(A)=U{\textrm-}\lim{|A\cap F_i|\over|F_i|}$ .

It turns out ultralimits have all the properties we need. Namely,

$μ$ is a probability measure. That is, $\mu(G)=U\textrm{-}\lim1=1$ , since the ultralimit coincides with the regular limit when it exists.
$μ$ is finitely additive. This is since ultralimits commute with addition just as regular limits do.
$μ$ is left invariant. This is since
$\left|{|gA\cap F_i|\over|F_i|}-{|A\cap F_i|\over|F_i|}\right| = \left|{|A\cap g^{-1}F_i|\over|F_i|}-{|A\cap F_i|\over|F_i|}\right|$
$\leq{|A\cap(g^{-1}F_i\,\triangle\,F_i)|\over|F_i|}\to0$