TITLE: Explicit free subgroup in Thompson's group $V$
QUESTION [9 upvotes]: R. Thompson introduced three groups $F\subset T\subset V$. The question concerning amenability of $F$ is still unanswered and has attracted much attention. I have read that Thompson group $V$ contains a copy of the free group $F_2$ (with two generators), in particular it is not amenable. Does anyone know an explicit embedding or can give me a reference for it? Thank you very much for the help.

REPLY [2 votes]: As already mentioned, a standard strategy to construct (non-abelian) free subgroups in $T$ or $V$ is to play ping-pong on the Cantor space. Just for fun, I would like to describe another point of view.
A topological interpretation of $T$ I really like is the following. Let $T_3$ denote the $3$-regular tree $T_3$, thought of as drawned on the plane (such that its vertex-set is discrete), and let $\mathscr{S}$ denote the ribbon surface obtained by thickening it. For each edge of $T_3$, fix a transverse arc in $\mathscr{S}$ with its endpoints on the boundary of $\mathscr{S}$. This collection of arcs decomposes $\mathscr{S}$ as a union of hexagons. We refer to a homeomorphism of $\mathscr{S}$ which sends all but finitely many hexagons to hexagons as an asymptotically rigid homeomorphism.

Theorem. The asymptotically rigid mapping class group $$\{ \text{orientation-preserving asymptotically rigid homeomorphisms of $\mathscr{S}$} \}/ \text{isotopy}$$ is isomorphic to the Ptolemy-Thompson group $T$.
See for instance Section 1.3 in Kapoudjian and Funar's article The braided Ptolemy-Thompson group is finitely presented (and references therein).
Some isometries of $T_3$ induce asymptotically rigid homeomorphisms of $\mathscr{S}$. For instance, the rotation $\alpha$ of order $3$ around a vertex (say $v$) and the rotation $\beta$ of order $2$ around the middle point of an edge (having $v$ as one of its endpoints). Clearly, the subgroup $\langle \alpha, \beta \rangle$ acts on the subdivision of $T_3$ with a single orbit of edges and with trivial edge-stabilisers. It follows that
$$\langle \alpha, \beta \rangle = \langle \alpha \rangle \ast \langle \beta \rangle \simeq \mathbb{Z}/2 \mathbb{Z} \ast \mathbb{Z} /3 \mathbb{Z}.$$
Consequently, $T$ contains a non-elementary free product (and a fortiori a non-abelian free subgroup).