Recently, I have been studying the concept of amenability on groups. This concept can be defined in a “combinatorial” or “finitary” fashion, using Følner sequences, and also in a more “functional-analytic” or “infinitary”‘ fashion, using invariant means. I wanted to get some practice passing back and forth between these two definitions, so I wrote down some notes on how to do this, and also how to take some facts about amenability that are usually proven in one setting, and prove them instead in the other. These notes are thus mostly for my own benefit, but I thought I might post them here also, in case anyone else is interested.
— 1. Equivalent definitions of amenability —
For simplicity I will restrict attention to countable groups . Given any and , I define the left-translation by the formula . Given as well, I define the inner product whenever the right-hand side is convergent.
All spaces are real-valued. The cardinality of a finite set is denoted . The symmetric difference of two sets is denoted .
A finite mean is a non-negative, finitely supported function such that . A mean is a non-negative linear functional such that . Note that every finite mean can be viewed as a mean by the formula .
The following equivalences were established by Følner:
Theorem 1 Let be a countable group. Then the following are equivalent:
- (i) There exists a left-invariant mean , i.e. mean such that for all and .
- (ii) For every finite set and every , there exists a finite mean such that for all .
- (iii) For every finite set and every , there exists a non-empty finite set such that for all .
- (iv) There exists a sequence of non-empty finite sets such that as for each . (Such a sequence is called a Følner sequence.)
Proof: We shall use an argument of Namioka.
(i) implies (ii): Suppose for contradiction that (ii) failed, then there exists such that for all means . The set is then a convex set of that is bounded away from zero. Applying the Hahn-Banach separation theorem, there thus exists a linear functional such that for all means . Since , there thus exist for such that for all means , thus . Specialising to the Kronecker means we see that pointwise. Applying the mean , we conclude that . But this contradicts the left-invariance of .
(ii) implies (iii): Fix (which we can take to be non-empty), and let be a small quantity to be chosen later. By (ii) we can find a finite mean such that
for all .
Using the layer-cake decomposition, we can write for some nested non-empty sets and some positive constants . As is a mean, we have . On the other hand, observe that is at least on . We conclude that
for all , and thus
By the pigeonhole principle, there thus exists such that
and the claim follows.
(iii) implies (iv): Write the countable group as the increasing union of finite sets and apply (iii) with and to create the set .
(iv) implies (i): Use the Hahn-Banach theorem to select an infinite mean , and define . (Alternatively, one can define to be an ultralimit of the .)
Any countable group obeying any (and hence all) of (i)-(iv) is called amenable.
Remark 1 The above equivalences are proven in a non-constructive manner, due to the use of the Hahn-Banach theorem (as well as the contradiction argument). Thus, for instance, it is not immediately obvious how to convert an invariant mean into a Følner sequence, despite the above equivalences.
— 2. Examples of amenable groups —
We give some model examples of amenable and non-amenable groups:
Proof: Trivial (either using invariant means or Følner sequences).
Proof: One can take the sets as the Følner sequence, or an ultralimit as an invariant mean.
Proposition 4 The free group on two generators is not amenable.
Proof: We first argue using invariant means. Suppose for contradiction that one had an invariant mean . Let be the set of all words beginning with , , respectively. Observe that , thus . By invariance we conclude that ; similarly for , , . Since the identity element clearly must have mean zero, we conclude that the mean is identically zero, which is absurd.
Now we argue using Følner sequences. If were amenable, then for any we could find a finite non-empty set such that differs from by at most points for . The set is contained in and in , and so
Similarly for permutations. Summing up over all four permutations, we obtain
leading to a contradiction for small enough (any will do).
Now we generate some more amenable groups.
Proposition 5 Let be a short exact sequence of countable groups (thus can be identified with a normal subgroup of , and can be identified with ). If and are amenable, then is amenable also.
Proof: Using invariant means, there is a very short proof: given invariant means , for , we can build an invariant mean for by the formula
for any , where is the function defined as for all cosets (note that the left-invariance of shows that the exact choice of coset representative is irrelevant). (One can view as sort of a “product measure” of the and .)
Now we argue using Følner sequences instead. Let , be Følner sequences for respectively. Let be a finite subset of , and let . We would like to find a finite non-empty subset such that for all ; this will demonstrate amenability. (Note that by taking to be symmetric, we can replace with without difficulty.)
By taking large enough, we can find such that differs from by at most elements for all , where is the projection map. Now, let be a preimage of in . Let be the set of all group elements such that intersects . Observe that is finite. Thus, by taking large enough, we can find such that differs from by at most points for all .
Now set . Observe that the sets for lie in disjoint cosets of and so . Now take , and consider an element of . This element must take the form for some and . The coset of that lies in is given by . Suppose first that lies outside of . By construction, this occurs for at most choices of , leading to at most elements in .
Now suppose instead that lies in . Then we have for some and , by construction of , and so . But as lies outside of , must lie outside of . But by construction of , there are at most possible choices of that do this for each fixed , leading to at most . We thus have as required.
Proof: We first use invariant means. An invariant mean on induces a mean on which is invariant with respect to translations by . Taking an ultralimit of these means, we obtain the claim.
Now we use Følner sequences. Given any finite set and , we have for some . As is amenable, we can find such that for all , and the claim follows.
Proof: Every virtually solvable group contains a solvable group of finite index, and thus contains a normal solvable subgroup of finite index. (Note that every subgroup of of index contains a normal subgroup of index at most , namely the stabiliser of the action on .) By Proposition 5 and Proposition 2, we may thus reduce to the case when is solvable. By inducting on the derived length of this solvable group using Proposition 5 again, it suffices to verify this when the group is abelian. By Proposition 6, it suffices to verify this when the group is abelian and finitely generated. By Proposition 5 again, it suffices to verify this when the group is cyclic. But this follows from Proposition 2 and Proposition 3.