Notes of Francois Dahmani’s Cambridge lecture 9-01-2016

The normal closure of a big Dehn twist in the mapping class group

Joint with Vincent Guirardel and Denis Osin.

1. Introduction

We are interested in quotients of the mapping class group. Influenced by the theory of lattices in Lie groups, Farb and Ivanov asked what is the normal closure of a pseudo-Anosov element. The idea is that pseudo-Anosov elements reflect the hyperbolic character of MCG, whereas reducible elements reflect the higher rank behaviour.

Theorem 1 (Dahmani-Guirardel-Osin) There exists ${n=n(\Sigma)}$ such that the normal closure of the ${n}$-th power a pseudo-Anosov element is free and purely pseudo-Anosov.

This is ping-pong.

Fix a simple closed curve ${\alpha}$ on a closed surface ${\Sigma}$. Let ${\tau}$ be the corresponding Dehn twist. What happens with the nroaml closure of a high power ${\tau^n}$ ?

Koberda: given a finite family ${\alpha_1,\ldots,\alpha_k}$ of distinct simple closed curves, the group generated by ${<\tau_{\alpha_1}^{n},\ldots,\tau_{\alpha_k}^{n}>}$ is a right-angled Artin group in its natural presentation (Dehn twists around disjoint curves commute) for ${n}$large enough.

Is ${\ll \tau^n \gg}$ a right-angled Artin group ? No. Artin groups have too many automorphisms. For instance, one which maps a genrator to its inverse, without changing other generators. This cannot arise from a surface homeomorphism.

Brendle-Margalit: ${Aut(\ll \tau^n \gg)}$ is the extended MCG acting by conjugation.

2. Result

Theorem 2 There exists ${N_0}$ such that, for ${n>N_0}$, ${\ll \tau^n \gg}$ has a partially commutative presentation over some conjugates of ${\tau^n}$. I.e., there is a suitable infinite set of generators which are conjugates of ${\tau^n}$, and relators of the form ${[s,gs'g^{-1}]}$, where ${g}$ are words in ${S}$.

In fact, a relator ${[s,gs'g^{-1}]}$ arises iff the corresponding curves are disjoint.

3. Proof

3.1. The pseudo-Anosov case

We use rotation families of subgroups. Let ${G}$ be a group acting on a hyperbolic space ${X}$. Let ${\gamma\in G}$ be loxodromic in ${X}$, with axis ${A_\gamma}$. Assume hat the collection of conjugates ${g\gamma g^{-1}}$ of ${\gamma}$ has the following property: axes are

• either equal,
• or do not overlap on a length ${>20\Delta}$.

Let us glue a hyperbolic cone of large radius, with infinite angle at the tip, to each axis. Do this equivariantly. In the resulting space, loxodromic translations become rotations (whence the term rotation family). Then, for ${n}$ large enough, the group ${\ll \gamma^n \gg}$ is free.

Indeed, let a ball centered at one cone apex grow, adding progressively its images under rotations around the other apices it encounters. The picture obtained is a tree, since branches are local quasi-geodesics.

This argument applies to the MCG action on the curve complex

3.2. The Dehn twist case

Bestvina-Bromberg-Fujiwara: there exists a normal finite index subgroup ${G_0}$ in MCG and a ${G_0}$-invariant finite coloring of the set of simple close curves of ${\Sigma}$ such that two curves with the same color intersect.

Theorem 3 (Bestvina-Bromberg-Fujiwara) There exists a finite collection of quasi-trees on which ${G_0}$ acts and for which every Dehn twist is loxodromic on one tree and elliptic on all others. Furthermore, axes satifies the rotation family condition.

If ${\alpha}$ is loxodromic on the first quasi-tree ${QT_1}$, we get that ${\ll \tau_\alpha^n \gg_{G_0}}$ is free. We want the full normal subgroup, so we need to perform the rotation family argument simultaneously on all quasi-trees. There is a difficulty: one needs a control on ${\alpha}$ on the secondary quasi-trees, a bound on its behaviour on cones. This uses more “projection” theory.