15
$\begingroup$

I am interested in the amenability properties of infinite products of solvable groups. The following facts are well-known:

  1. Any solvable group is amenable.
  2. The class of solvable groups is closed under finite products and quotients (but not infinite products).
  3. Infinite products of solvable groups of a fixed derived length are solvable.
  4. Finite products and quotients of amenable groups are amenable (but not infinite products).

My specific question is whether an infinite product of solvable groups is still amenable?

The motivation arises from uniformity in multiple recurrence within ergodic theory. My coauthor and I wish to prove that the largeness of certain sets of multiple return times is independent of the underlying measure-preserving system and the acting group. Currently, we have achieved this for solvable groups of a fixed derived length, and more generally, for a class of uniformly amenable groups, where we use the device of ultraproducts as a key tool in our approach.

$\endgroup$
1
  • $\begingroup$ I'm pretty sure this is a duplicate of the same question, here or at MathSE. Of course the usual counterexamples work. $\endgroup$
    – YCor
    Commented Aug 11, 2023 at 16:48

1 Answer 1

21
$\begingroup$

The free group $F_2$ is residually nilpotent, meaning that the intersection of its lower central series is trivial, because the length of an element in the $k$th term of the lower central series is bounded below by $k$. It follows that $F_2$ embeds in a direct product of nilpotent groups. So there is an infinite product of nilpotent groups that is not amenable.

$\endgroup$
4
  • 10
    $\begingroup$ Indeed free groups are residually (finite $p$-groups), so there is an infinite product of finite nilpotent groups which is not amenable. $\endgroup$ Commented Aug 7, 2023 at 19:34
  • 1
    $\begingroup$ @Carl-FredrikNybergBrodda Yes, even better... $\endgroup$ Commented Aug 7, 2023 at 19:37
  • 1
    $\begingroup$ @Carl-FredrikNybergBrodda For a very concrete example, there is a natural embedding of the automorphism group $\mathrm{Aut}(T_2)$ of an infinite rooted binary tree into the product of the autuomorphism groups of finite complete rooted binary trees of depth $n$, and nonabelian free subgroups of $\mathrm{Aut}(T_2)$ have been described explicitly as automata groups. $\endgroup$
    – Jim Belk
    Commented Aug 9, 2023 at 20:22
  • 1
    $\begingroup$ @JimBelk Very nice! (And, for completeness, elementary arguments show that the depth-$n$ automorphism group is a $2$-group for all $n$). $\endgroup$ Commented Aug 9, 2023 at 21:32

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .