8
$\begingroup$

Let's recall that a group $G$ is amenable if for any finite subset $E\subset G$ and any $\epsilon>0$ there is a finite subset $F\subset G$ such that $$\max_{s\in E} |s F \mathbin{\triangle} F| \le \epsilon |F|.$$ The group $G$ is said to be uniformly amenable if one can find such $F$ with $|F|\le \alpha(|E|,\epsilon)$ for some $\alpha(m,t)$, i.e., the size of $F$ can be bound in terms of $|E|$ and $\epsilon$. See Bozejko's paper and Wysoczanski's paper for details. It is proved by Wysoczanski that a group $G$ is uniformly amenable if and only if its ultrapower is amenable. In particular solvable groups are uniformly amenable, while the locally finite group of finite permutations on a countable set is not.

Now, let's recall that a group $G$ is amenable at infinity (a.k.a. exact) if for any finite subset $E\subset G$ and any $\epsilon>0$ there are a finite subset $F\subset G$ and a function $\eta\colon G\to\mathrm{Prob}(F)$ such that $$\sup_{s\in E,\, x\in G} \|s\eta_x-\eta_{sx}\|_1 \le\epsilon.$$ Here $\mathrm{Prob}(F)$ is the space of probability measure on the finite set $F$, viewed as a subset of $\ell_1(G)$. For example, when $G$ is amenable, one can take $\eta_x=|F|^{-1}1_{F}$ for all $x\in G$. As in the previous case, the group $G$ is said to be uniformly amenable at infinity (or uniformly exact) if one can find such $F$ and $\eta$ with $|F|\le \alpha(|E|,\epsilon)$.

Are there interesting examples of groups that are uniformly amenable at infinity?

In particular, are linear groups uniformly amenable at infinity?

It is proved by Guentner–Higson–Weinberger (MathSciNet link) that linear groups are amenable at infinity and so are their ultrapower. However, since the $\sup$ is the above definition is taken over an infinite set, this doesn't imply (at least in a obvious way) that linear groups are uniformly amenable at infinity. Still, I inclined to conjecture that linear groups are uniformly amenable at infinity. Even the simplest case of the free groups looks very challenging.

$\endgroup$
3
  • $\begingroup$ Have you checked whether the ultraproduct characterization also works ($G$ uniformly exact iff $^\eta G$ is exact for some/every nonprincipal integral ultrafilter)? The result would indeed follow. In the case of amenability, the verification is pretty straightforward. $\endgroup$
    – YCor
    Sep 14, 2021 at 12:02
  • $\begingroup$ @Ycor Isn't your question answered in the last paragraph of the question? $\endgroup$ Sep 14, 2021 at 12:09
  • $\begingroup$ @MikaeldelaSalle ah, thanks, indeed it seems to suggest it doesn't follow easily... And actually, if $S_{(\omega)}$ is the group of finitary permutations of $\omega$, its ultraproduct contains all countable residually finite groups if I'm correct. And there exists non-exact such groups, as proved by Osajda ArXiv link. $\endgroup$
    – YCor
    Sep 14, 2021 at 12:46

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.