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Question: Do there exist amenable Thompson-like groups?

I realise that my question is vague, but defining and studying groups which look like Thompson's groups $F$, $T$ and $V$ seems to be an independent field of research in the litterature, so my question should make sense.

Among the examples of such groups I know, typically either they contain a non-abelian free group or they contain Thompson's group $F$. In the former case, of course the group is not amenable; and in the latter case, the amenability of the entire group would imply the amenability of $F$, so a proof of the amenability should not be available.

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    $\begingroup$ It's indeed very vague, especially given the number of inequivalent ways to define Thompson's groups. At least some full-topological groups including ones occurring as subgroups of IET have been proved (by Juschenko-Monod + generalizations) to be amenable (while they're not elementary amenable). $\endgroup$ – YCor Jul 8 '18 at 13:41
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    $\begingroup$ Also some piecewise projective f.g. groups of self-homeomorphisms of the segments, very similar to Thompson's $F$ have been proved by Monod to be non-amenable. $\endgroup$ – YCor Jul 8 '18 at 13:47
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There are two types of groups which are sometimes considered similar to the R. Thompson group $F$.

  1. Some "fractal groups" acting on rooted trees. One of the "closest" to $F$ groups in this class is probably the Basilica group discovered by Grigorchuk and Zuk whose amenability was proved by Bartholdi and Virag. You can try finding lectures by Grigorchuk where he advertised the amenability problem for the Basilica group and its similarity to $F$. The reason for similarity is that both $F$ and the Basilica group act nicely on the Cantor set.

  2. Topological full groups of minimal subshifts. Their amenability was proved by Juschenko and Monod. The reason for similarity is that $F$ can also be viewed as topological full group of some amenable (but not cyclic) group (an affine group of homeomorphisms of $\mathbb{Z}[1/2]$) acting on a (non-compact) metric space.

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This is not an answer but rather a personal thought related to Matt Zaremsky's answer.

Thompson's groups $F$, $T$ and $V$ can be interpreted as groups of partial isometries of the $3$-regular tree $A_3$. A partial isometry is the data of two finite subtrees $R,S \subset A_3$ and an isometry $f : A_3 \backslash R \to A_3 \backslash S$. Two partial isometries are considered as equal if they are well-defined and agree on the complement of a finite subtree. Similarly, the Higman-Thompson groups $F_{n,r}$, $T_{n,r}$, $V_{n,r}$ are subgroups in $\mathrm{PIsom}(A_{n+1,r})$ where $A_{n+1,r}$ is the forest having $r$ trees which are all $(n+1)$-regular.

Naturally, a partial isometry of $A_3$ extends to a quasi-automorphism (aka almost isometry) of $A_3$, i.e. a bijection of the vertices of $A_3$ that preserves adjacency and non-adjacency for all but finitely many pairs of vertices. This leads to extensions $QF$, $QT$, $QV$ of $F$, $T$, $V$, which can be thought of as symmetrisations of Thompson's groups because of the exact sequences $$1 \to S_\infty \to QF \to F \to 1$$ and similarly for $T$ and $V$, where $S_\infty$ denotes the group of finitely supported permutations of a countably infinite set.

Generalisation. Now, consider an arbitrary tree $A$ (but infinite, otherwise there is nothing interesting to say). You still have an exact sequence $$1 \to S_\infty \to \mathrm{QAut}(A) \to \mathrm{PIsom}(A),$$ but the third arrow may not be surjective in general (see Houghton's groups below). So $\mathrm{QAut}(A)$ can be thought of as a symmetrisation of its image in $\mathrm{PIsom}(A)$.

Houghton's groups. As a particular case, instead of a regular tree, consider the graph $R_n$ which is a union of $n$ infinite rays with a common origin. Then the group of partial isometries $\mathrm{PIsom}(R_n)$ is isomorphic to $\mathbb{Z}^n \rtimes S_n$ and the group of quasi-automorphisms $\mathrm{QAut}(R_n)$ is isomorphic to $H_n \rtimes S_n$ where $H_n$ denotes the $n$th Houghton group. From the exact sequence above, $H_n$ can be thought of as a symmetrisation of $\mathbb{Z}^{n-1}$.

Of course, $H_n$ is quite different from Thompson's groups, but I think that the point of view described here explains the similarity in the techniques which can be used to study both Houghton's and Thompson's groups.

Lamplighter groups. Another example I find interesting is when $A$ is a bi-infinite horizontal line $\mathbb{R}$ with an infinite vertical descending ray attached at each integer. Then $\mathrm{PIsom}(A)$ is isomorphic to $\left( \bigoplus\limits_{n \in \mathbb{N}} \mathbb{Z} \right) \rtimes S_\infty$. In the same way that $T$ lies in $\mathrm{PIsom}(A_3)$, the lamplighter group $\mathbb{Z} \wr \mathbb{Z}$ lies in $\mathrm{PIsom}(A)$. By looking at the pre-image under $\mathrm{QAut}(A) \to \mathrm{PIsom}(A)$ of a subgroup isomorphic to $\mathbb{Z} \wr \mathbb{Z}$, it is possible to introduce a symmetrisation of $\mathbb{Z}\wr \mathbb{Z}$. Up to my knowledge, this example does not appear in the literature, but this is probably because such a group is not finitely presented.

A symmetrisation of $\mathbb{Z} \wr \mathbb{Z}$ should be compared to $H_2$, and a problem I am interested about is: similarly, what would be a good analogue of $H_n$? It should be a (free abelian)-by-$\mathbb{Z}^{n-1}$ group which is of type $F_{n-1}$ but not of type $F_n$.

Anyway, if $H_n$ is considered as an amenable Thompson-like group, so should be the lampligher group $\mathbb{Z} \wr \mathbb{Z}$.

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  • $\begingroup$ What you call "quasi-automorphism" was initially called "almost automorphism" by Truss in the 80s. $\endgroup$ – YCor Sep 5 '20 at 9:51
  • $\begingroup$ Good point, $\mathbb{Z} \wr \mathbb{Z}$ is certainly Thompson-like in many ways. Still it would be nice to find an amenable-not-EA example (besides, you know, possibly $F$). $\endgroup$ – Matt Zaremsky Sep 5 '20 at 16:27
  • $\begingroup$ @YCor: The terminology "almost automorphism" seems to be more common indeed. I added a remark in my answer, but I kept the terminology "quasi automorphism" for coherence with the notation $QF$ (which is not mine). $\endgroup$ – AGenevois Sep 6 '20 at 16:20
  • $\begingroup$ Note that actually some authors (ignoring Truss' pioneer work) have, notably in the context of Neretin group, subsequently named "almost automorphism groups" (and I say "near automorphism group") something different: namely the set of isomorphisms between cofinite subgraphs (in particular it kills finitely supported permutations). $\endgroup$ – YCor Sep 6 '20 at 17:06
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I think the Houghton groups $H_n$ deserve to be called "Thompson-like", and they are amenable (although they are even elementary amenable, so this is maybe not what you were aiming for). Recall that $H_n$ is the group of permutations of $\mathbb{N}\times\{1,...,n\}$ that are eventually translations in the first component (and constant in the second).

Here are some reasons I think they should count as Thompson-like:

  1. One can view $H_n$ as arising from an "edge replacement rule" on graphs, as in Belk-Forrest: https://arxiv.org/pdf/1510.03133.pdf. As they show, Thompson's group $V$ arises by taking a single edge as the base graph and using the edge replacement rule that an edge gets replaced by a pair of disjoint edges (see Table 21 on page 23). One can discover that $H_n$ arises by taking $n$ disjoint edges as the base graph and using the edge replacement rule that an edge gets replaced by an edge and a disjoint vertex. Since "one edge becomes two edges" and "one edge becomes an edge and a vertex" are such similar-sounding rules, I declare $V$ and $H_n$ are similar.

  2. The proof that $V$ is of type $F_\infty$ is remarkably similar to the proof that $H_n$ is of type $F_{n-1}$. Brown proved these facts in the same paper (Finiteness properties of groups, JPAA 1987), and before proving the Houghton result he comments, "The proof is an imitation of what we did for the Thompson-Higman groups." Also, in both cases easier proofs were later found by putting a CAT(0) cubical structure on the respective complexes.

There are also some more surface-level similarities between $V$ and $H_n$, like arising as groups of bijections of an infinite set subject to certain finiteness constraints. But anyway, this is all just to advertise the Houghton groups and say that I think they should count as "Thompson-like" in some sense.

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  • $\begingroup$ Anyway amenability (unlike finiteness properties) of Houghton groups is trivial. $\endgroup$ – YCor Sep 4 '20 at 17:52
  • $\begingroup$ Yeah, true, and I suppose the spirit of this question was asking more for non-elementary amenable examples, but I still find the various Thompson-Houghton connections interesting. $\endgroup$ – Matt Zaremsky Sep 5 '20 at 0:03

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