4
$\begingroup$

Is it known whether the Brin-Thompson 2V contains a distortion element? By this I mean an element $f$ such that the word norm $|f^n|$ grows sublinearly, and $f$ is of infinite order. If such an element exists, what is known about distortion functions (how slowly can the word norm grow)? It is easy to construct subgroups of Brin-Thompson 2V that are distorted in the big group, I am only interested in cyclic ones.

It would suffice to embed a group having distortion elements, for example the discrete Heisenberg group or the Baumslag-Solitar group $\mathrm{BS}(1, 2)$. I am not aware of such an embedding and did not find one on a quick search, but this means very little.

I give one definition of the Brin-Thompson 2V for completeness, this is not the most standard definition (and wouldn't be the first time I manage to mangle it), see any reference for other definitions. The Brin-Thompson 2V is the group acting on $\{0,1\}^{\mathbb{Z}}$ by word-rewritings near the origin. So an element of 2V is a homeomorphism $f : \{0,1\}^{\mathbb{Z}} \to \{0,1\}^{\mathbb{Z}}$ such that you have an in-radius $r$ and a function $\{0,1\}^{2r+1} \to \{0,1\}^* \times {\mathbb{Z}}$, such that if $|u| = r$, $|v| = r+1$ and $f(uv) = (w, n)$, then $f(xu.vy) = \sigma^n(x.wy)$ for any tails $x, y$. Here $\sigma$ is the shift map, $.$ indicates the origin, and juxtaposition is concatenation.

$\endgroup$
11
  • 3
    $\begingroup$ Up to my knowledge, this is not known. For many Thompson-like groups, proper actions on CAT(0) cube complexes can be constructed, which implies that infinite-order elements are undistorted. But the usual construction fails in higher dimensions. It is something I would like to know: does 2V acts properly on some CAT(0) cube complex? $\endgroup$
    – AGenevois
    Commented May 12, 2022 at 7:22
  • $\begingroup$ By the way, some elements are proved to be undistorted in Burillo and Cleary's article Metric properties in higher dimensional Thompson's group. $\endgroup$
    – AGenevois
    Commented May 12, 2022 at 7:23
  • $\begingroup$ Thanks for your "no" vote. It's hard to find negative information in papers... I think I did notice the result of Burillo and Cleary, and it's what made me suspect that this is not known. $\endgroup$
    – Ville Salo
    Commented May 12, 2022 at 7:28
  • 1
    $\begingroup$ Yes, this is a result due to Haglund. I think the question is interesting because: (1) if there exists such a proper action, then this implies interesting properties for 2V automatically; and (2) if there is no such action, then this constrasts with other Thompson-like groups and asks the question of which geometry would be relevant and how it is different from the one-dimensional case. $\endgroup$
    – AGenevois
    Commented May 12, 2022 at 7:38
  • $\begingroup$ Sorry I removed my comment asking about connections between CAT(0) cube complexes and distortion, and whether "not acting on any CAT(0) complex" is interesting. (I removed it because after reading your first comment a few times I realized you already answered the technical question.) Thank you for the answer though. $\endgroup$
    – Ville Salo
    Commented May 12, 2022 at 7:42

0

You must log in to answer this question.

Browse other questions tagged .