Math-2002 · mathematics · Group theory

Math-2002

Questions in Geometric Group Theory — Q 12.12

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Statement

Definitions and mathematical statement

Let $S$ be a connected orientable surface of finite type with negative Euler characteristic, and let $\operatorname{Mod}(S)$ be its orientation-preserving mapping class group. A mapping class is pseudo-Anosov if it has a representative preserving a pair of transverse measured singular foliations and multiplying their transverse measures by reciprocal factors $\lambda$ and $\lambda^{-1}$, with $\lambda>1$.

Teichmüller space $\mathcal T(S)$ consists of marked conformal structures, with the Teichmüller metric. A subgroup $H\leq\operatorname{Mod}(S)$ is convex cocompact in this metric sense if an orbit $H\cdot x$ is quasiconvex: there is $R$ such that each geodesic segment between two orbit points lies in the $R$-neighborhood of the orbit. A Schottky subgroup here is a finitely generated free subgroup which is purely pseudo-Anosov and convex cocompact.

Does there exist such a surface and a finitely generated nonabelian free subgroup $H$ for which \[ h\ne1\Longrightarrow h\text{ is pseudo-Anosov}, \qquad H\text{ is not convex cocompact}? \] Purely pseudo-Anosov is a requirement on every nonidentity word in the subgroup, not only on a chosen free generating set. Non-convex-cocompactness must concern the orbit geometry in Teichmüller space. Infinite-rank free subgroups and groups containing a reducible or periodic mapping class do not answer this finitely generated question.

Short English statement

Can a finitely generated free group consist entirely of pseudo-Anosov mapping classes yet fail to be Schottky?

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