pith. sign in

arxiv: 1705.00198 · v3 · pith:CSMOS2MWnew · submitted 2017-04-29 · 🧮 math.GR · math.OA

Computational explorations of the Thompson group T for the amenability problem of F

classification 🧮 math.GR math.OA
keywords groupfrac1thompsonsqrtsqrt2amenablemathbbproblem
0
0 comments X
read the original abstract

It is a long standing open problem whether the Thompson group $F$ is an amenable group. In this paper we show that if $A$, $B$, $C$ denote the standard generators of Thompson group $T$ and $D:=C B A^{-1}$ then $$\sqrt2+\sqrt3\,<\,\frac1{\sqrt{12}}||(I+C+C^2)(I+D+D^2+D^3)||\,\le\, 2+\sqrt2.$$ Moreover, the upper bound is attained if the Thompson group $F$ is amenable. Here, the norm of an element in the group ring $\mathbb{C} T$ is computed in $B(\ell^2(T))$ via the regular representation of $T$. Using the "cyclic reduced" numbers $\tau(((C+C^2)(D+D^2+D^3))^n)$, $n\in\mathbb{N}$, and some methods from our previous paper [arXiv:1409.1486] we can obtain precise lower bounds as well as good estimates of the spectral distributions of $\frac1{12}((I+C+C^2)(I+D+D^2+D^3))^*(I+C+C^2)(I+D+D^2+D^3),$ where $\tau$ is the tracial state on the group von Neumann algebra $L(T)$. Our extensive numerical computations suggest that $$\frac1{\sqrt{12}}||(I+C+C^2)(I+D+D^2+D^3)||\approx 3.28,$$ and thus that $F$ might be non-amenable. However, we can in no way rule out that $\frac1{\sqrt{12}}||(I+C+C^2)(I+D+D^2+D^3)||=\, 2+\sqrt2$.

This paper has not been read by Pith yet.

discussion (0)

Sign in with ORCID, Apple, or X to comment. Anyone can read and Pith papers without signing in.