Why are abelian groups amenable?

Question

A (discrete) group is amenable if it admits a finitely additive probability measure (on all its subsets), invariant under left translation. It is a basic fact that every abelian group is amenable. But the proof I know is surprisingly convoluted. I'd like to know if there's a more direct proof.

The proof I know runs as follows.

1. Every finite group is amenable (in a unique way). This is trivial.

2. $\mathbb{Z}$ is amenable. This is not trivial as far as I know; the proof I know involves choosing a non-principal ultrafilter on $\mathbb{N}$. This means that $\mathbb{Z}$ is amenable in many different ways, i.e. there are many measures on it, but apparently you can't write down any measure 'explicitly' (without using the Axiom of Choice).

3. The direct product of two amenable groups is amenable. This isn't exactly trivial, but the measure on the product is at least constructed canonically from the two given measures.

4. Every finitely generated abelian group is amenable. This follows from 1--3 and the classification theorem.

5. The class of amenable groups is closed under direct limits (=colimits over a directed poset). This is like step 2: it seems that there's no canonical way of constructing a measure on the direct limit, given measures on each of the groups that you start with; and the proof involves choosing a non-principal ultrafilter on the poset.

6. Every abelian group is amenable. This follows from 4 and 5, since every abelian group is the direct limit of its finitely generated subgroups.

Is there a more direct proof? Is there even a one-step proof?

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Update Yemon Choi suggests an immediate simplification: replace 1 and 4 by

1'. Every quotient of an amenable group is amenable. This is simple: just push the measure forward.

4'. Every f.g. abelian group is amenable, by 1', 2 and 3.

This avoids using the classification theorem for f.g. abelian groups.

Tom Church mentions the possibility of skipping steps 1--3 and going straight to 4. If I understand correctly, this doesn't use the classification theorem either. The argument is similar to the one for $\mathbb{Z}$: one still has to choose an ultrafilter on $\mathbb{N}$. (One also constructs a Følner sequence on the group, a part of the argument which I didn't mention previously but was there all along).

Yemon, Tom and Mariano Suárez-Alvarez all suggest using one or other alternative formulations of amenability. I'm definitely interested in answers like that, but it also reminds me of the old joke:

Tourist: Excuse me, how do I get to Edinburgh Castle from here?

Local: I wouldn't start from here if I were you.

In other words, if a proof of the amenability of abelian groups uses a different definition of amenability than the one I gave, then I want to take the proof of equivalence into account when assessing the simplicity of the overall proof.

Jim Borger points out that if, as seems to be the case, even the proof that $\mathbb{Z}$ is amenable makes essential use of the Axiom of Choice, then life is bound to be hard. I take his point. However, one simplification to the 6-step proof that I'd like to see is a merging of steps 2 and 5. These are the two really substantial steps, but they're intriguingly similar. None of the answers so far seem to make this economy. That is, every proof suggested seems to involve two separate Følner-type arguments.

Answer 1

Here is a simpler argument, combining 1--6 into one step.

Let $G$ be a countable abelian group generated by $x_1,x_2,\ldots$. Then a Følner sequence is given by taking $S_n$ to be the pyramid consisting of elements which can be written as

$a_1x_2+a_2x_2+\cdots+a_nx_n$ with $\lvert a_1\rvert\leq n,\lvert a_2\rvert\leq n-1,\ldots,\lvert a_n\rvert\leq 1$.

The invariant probability measure is then defined by $\mu(A)=\underset{\omega}{\lim}\lvert A\cap S_n\rvert / \lvert S_n\rvert$ as usual.

A more natural way to phrase this argument is:

1. The countable group $\mathbb{Z}^\infty$ is amenable.
2. All countable abelian groups are amenable, because amenability descends to quotients.

But I would like to emphasize that there is really only one step here, because the proof for $\mathbb{Z}^\infty$ automatically applies to any countable abelian group. This two-step approach is easier to remember, though. (The ideas here are the same as in my other answer, but I think this formulation is much cleaner.)

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2016 Edit: Here is an argument to see that $S_n$ is a Følner sequence. It is quite pleasant to think about precisely where commutativity comes into play.

Fix $g\in G$ and any finite subset $S\subset G$. We first analyze the size of the symmetric difference $gS\bigtriangleup S$. Consider the equivalence relation on $S$ generated by the relation $x\sim y$ if $y=x+g$ (which is itself neither symmetric, reflexive, or transitive). We will call an equivalence class under this relation a "$g$-string". Every $g$-string consists of elements $x_1,\ldots,x_k\in S$ with $x_{j+1}=x_j+g$.

The first key observation is that $\lvert gS\bigtriangleup S\rvert$ is at most twice the number of $g$-strings. Indeed, if $z\in S$ belongs to $gS\bigtriangleup S$, then $z$ must be the "leftmost endpoint" of a $g$-string; if $z\notin S$ belongs to $gS\bigtriangleup S$, then $z-g$ must be the "rightmost endpoint" of a $g$-string; and each $g$-string has at most 2 such endpoints (it could have 1 if the endpoints coincide, or 0 if $g$ has finite order).

Our goal is to prove for all $g\in G$ that $\frac{\lvert gS_n \bigtriangleup S_n\rvert}{\lvert S_n\rvert}\to 0$ as $n\to \infty$. Since $\lvert abS\bigtriangleup S\rvert\leq\lvert abS\bigtriangleup bS\rvert+\lvert bS\bigtriangleup S\rvert= \lvert aS\bigtriangleup S\rvert+\lvert bS\bigtriangleup S\rvert$, it suffices to prove this for all $g_i$ in a generating set.

By the observation above, to prove that $\frac{\lvert g_iS_n \bigtriangleup S_n\rvert}{\lvert S_n\rvert}\to 0$, it suffices to prove that $\frac{\text{# of $g_i$-strings in $S_n$}}{\lvert S_n\rvert}\to 0$. Equivalently, we must prove that the reciprocal $\frac{\lvert S_n\rvert}{\#\text{ of $g_i$-strings in $S_n$}}$ diverges, or in other words that the average size of a $g_i$-string in $S_n$ diverges.

We now use the specific form of our sets $S_n=\{a_1g_1+\cdots+a_ng_n\,|\, \lvert a_i\rvert\leq n-i\}$. For any $i$ and any $n$, set $k=n-i$ (so that $\lvert a_i\rvert\leq k$ in $S_n$). The second key observation is that every $g_i$-string in $S_n$ has cardinality at least $2k+1$ unless $g_i$ has finite order. Indeed given $x\in S_n$, write it as $x=a_1g_1+\cdots+a_ig_i+\cdots+a_ng_n$; then the elements $a_1g_1+\cdots+bg_i+\cdots+a_ng_n\in S_n$ for $b=-k,\ldots,-1,0,1,\ldots,k$ belong to a single $g_i$-string containing $x$. If $g_i$ does not have finite order, these $2k+1$ elements must be distinct. This shows that the minimum size of a $g_i$-string in $S_n$ is $2n-2i+1$, so for fixed $g_i$ the average size diverges as $n\to \infty$.

When $g_i$ has finite order $N$ this argument does not work (a $g_i$-string has maximum size $N$, so the average size cannot diverge). However once $N<2k+1$, the subset containing the $2k+1$ elements above is closed under multiplication by $g_i$. In other words, once $n\geq i+N/2$ the set $S_n$ is $g_i$-invariant, so $\lvert g_iS_n\bigtriangleup S_n\rvert=0$.

I'm grateful to David Ullrich for pointing out that this claim is not obvious, since the quotient of a Følner sequence need not be a Følner sequence (Yves Cornulier gives an example here).

Answer 2

I believe that the neatest proof for amenability of abelian topological groups, is via the Markov-Kakutani fixed point theorem: see the half-page proof of this result as Theorem G.2.1 in: MR2415834 (2009i:22001) Bekka, Bachir; de la Harpe, Pierre; Valette, Alain Kazhdan's property (T). New Mathematical Monographs, 11. Cambridge University Press, Cambridge, 2008. xiv+472 pp. ISBN: 978-0-521-88720-5

Answer 3

There is a slightly quicker approach than that outlined by Mariano, using the Markov-Kakutani fixed point theorem. I first learned of this from Rudin's Functional Analysis -- more accurately, I remember skimming over that part as a struggling undergraduate, and then years later, once I'd heard of amenability, realizing the connection. (As far as I know, the word "amenable" is never mentioned in the book.)

The proof can be found via Theorems 5.23, 5.24 and 5.25 of the aforementioned book (2nd ed. if that makes any difference) and I don't think I can improve on the exposition there.

(Regarding the approach Tom outlines: it might be worth observing that amenability passes to quotient groups (this is particularly obvious using the invariant mean, but probably isn't too hard in most of the other formulations). Therefore, going from 3 to 4 doesn't need the classification theorem.)

A vague remark on possible alternative routes. From one point of view, the "invariant mean" formulation of amenability is just a convenient way to avoid epsilon-delta arguments with Folner nets. Now, Folner sequences can be easily constructed in ${\mathbb Z}^n$ for any finite $n$ -- just take cubes centred at zero of ever increasing width -- but at present I can't quite nail down how to transfer the construction over to arbitrary abelian groups. (Of course, things would be easier if we could pass to the formulation with an invariant mean; but then one might as well work with the invariant mean throughout.)

Answer 4

The simplest argument for me to keep in my head is (none of this is original, even to this thread):

1. $\mathbb{Z}^n$ is amenable (because the cube of radius $k$ gives a Følner sequence).
2. Thus finitely generated abelian groups are amenable (because amenability descends to quotients).
3. Thus (discrete) abelian groups are amenable (because locally amenable groups are amenable).

Each of these properties is most easily proved using a different characterization of amenability, so this might not be pedagogically optimal, but since we already know the characterizations are equivalent we don't have to worry about that. Each of the parentheticals above is a property that everyone should have in mind when dealing with amenable groups, and so this argument also serves to remind me what's true in case I forget. Of course 1 and 2 could be combined (the ball of radius $k$ gives a Følner sequence in any f.g. abelian group, no need to be free) or deduced differently (you might prefer to justify 2 instead by "virtually amenable groups are amenable" or "amenable-by-amenable groups are amenable"), but I prefer this approach because it doesn't depend on the classification of f.g. abelian groups.

[Edit: upon rereading, I see that more of this than I realized was already present in Yemon's answer and the comments, so I'm making this community wiki.]

Second edit: in comments, Tom Leinster asked about seeing directly that f.g. abelian groups are amenable. Here are my thoughts: if you assume the classification and write $G=\mathbb{Z}^n\times T$ for $T$ finite, this is easy. Geometrically the most natural Følner sequence is the cube-of-radius-$k\times T$, but anything you try will work. If you don't want to use the structure theorem, I would note that if $G$ is abelian and has rank $n$, then it has polynomial growth of rank at most $n$; it follows that the ball of radius $k$ gives a Følner sequence. (Above I implicitly criticized this approach; that was because to bound the growth rate without knowing the structure of the group, you dominate it by the growth of the free abelian group. Thus this seemed to just be a hidden appeal to my step 2 above. It's perfectly valid though.)

Both approaches give a Følner sequence (because that's the definition of amenable I understand best). If you prefer the invariant mean definition, that's fine, but there is still value in sticking with Følner sequences as long as you can. For the simplest possible proof without relying on any equivalences, I would do the following:

1. Finitely generated abelian groups admit Følner sequences (e.g. the ball of radius $k$).
2. If $G$ is countable and every f.g. subgroup of $G$ admits a Følner sequence, then $G$ is amenable (admits an invariant f.a. probability measure).

To see 2, choose an increasing sequence of f.g. subgroups $G_i < G$ which exhausts $G$, and let $S_i^n$ be a Følner sequence for $G_i$; we can consider the $S_i^n$ as subsets of $G$. Then the measure we get is the "asymptotic density"

$\mu(A) = \underset{i\to \omega}{\lim}\underset{n\to\omega}{\lim}\ \ \vert A\cap S_i^n\vert / \vert S_i^n\vert$

where $\omega$ is a non-principal ultrafilter so that the limits exist. This is clearly a finitely additive probability measure, and to see that it's invariant note that every $\gamma\in G$ lies in all $G_N$ for $N$ sufficiently large.

Answer 5

The way I've seen it was also with the Markov-Kakutani fixed point theorem. The steps are these:

First define $$ K= \lbrace \Phi\in l^{\infty}(G)^\ast \mid \Phi(1) = 1 ,\Phi(F)\ge0\text{ whenever }F(g)\ge 0\text{ for all }g\in G\rbrace. $$

1. Equip $l^{\infty}(G)^\ast $ with the weak$^\ast$-topology. Prove that $K$ is weak$^\ast$ compact and convex.

2. Define for every $g\in G$ the map $$ T_g:K\to K:(T_g(\Phi)(F)=\Phi(F\cdot g). $$ Prove that every $T_g$, $g\in G$, is weak$^\ast$ continuous and affine.

3. Since $G$ is a commutative group, we can apply the Markov-Kakutani fixed point theorem to the family $\lbrace T_g \mid g\in G\rbrace$ of affine maps from $K$ to $K$. So, we get $\Phi\in K$ such that $T_g\Phi=\Phi$ for all $g\in G$. Define $m(A)=\Phi(\chi_A)$ whenever $A\subset G$ and check that $m$ is an invariant mean on $G$.