## Notes of Henry Wilton’s Cambridge lecture 21-06-2017

Surface groups in graphs of groups

1. Gromov’s questions

Today, all infinite groups are torsion-free.

Question (Gromov). Does every one-ended hyperbolic group contain a surface subgroup ?

Ping-pong produces heaps of free subgroups. However, one-ended subgroups arise in only finitely many sorts up to conjugacy, so it ought to be much harder to construct surface subgroups. Let us relax the question a bit: look for infinite index one-ended subgroups. One needs rule out surface group.

Question (Gromov). Does every one-ended hyperbolic group which is not a surface group, contain a finitely generated one-ended subgroup of infinite index?

Question. Does every one-ended finitely presented group contain either a surface group or a Baumslag-Solitar subgroup ${BS(1,m)}$?

The most important development is Kahn-Markovic’s positive result for 3-manifold groups. In 2012, Calegari-Walker gave a positive answer for random groups.

2. Free groups and relative questions

Here is a much easier class of examples.

Relative means relative to a given set of relators. I.e. study group pairs ${(F,\omega)}$ where ${F}$ is e finitely generated free group and ${\omega}$ a finite collection of words. A surface pair is ${(\pi_1(\Sigma),\partial \Sigma)}$ where ${\Sigma}$ is a surface with boundary and ${\partial\Sigma=(c_1,\ldots,c_b)}$ represents its boundary components.

Note that a pair can be doubled into

$\displaystyle \begin{array}{rcl} D(\omega)=F\star_{\langle \omega\rangle} F. \end{array}$

Bestvina-Feighn: ${D(\omega)}$ is hyperbolic iff ${\omega}$ is not a proper power.

Shentzer: ${D(\omega)}$ is one-ended iff ${\omega}$ is not contained in a proper free factor. In this case, we say that the group pair is irreducible.

2.1. History

In 2008, Calegari proved this under condition that ${H_2(\Gamma,{\mathbb Q})\not=0}$. He reduced the problem to linear equations, and the homology assumptions provided a solution.

There were earlier results with C. Gordon and Kim, we could treat infinite families of examples with ${H_2=0}$.

In 2010, Kim and Oum solved the case of doubles of rank 2 free groups, where Calegari’s equation could be scrutinized. This solves question 1 in this special case.

I gave a complete affirmative answer to Question 1 for doubles in 2010.

2.2. Results

Theorem 1 (Wilton) If ${\Gamma=D(w)}$ and ${\Gamma}$ is hyperbolic, then it contains a surface subgroup.

Theorem 2 (Delzant-Potyagailo, Louder-Touikan) If ${\Gamma}$ is a one-ended hyperbolic group, then${\Gamma}$ contains a quasi-convex subgroup ${H}$ such that

1. either ${\partial H}$ is connected without local cutpoints (rigid case).
2. or ${H}$ satisfies the assumptions of the main theorem.

Hence for Gromov’s questions, one can reduce to the rigid case.

Corollary 3 Finite covers and linear algebra suffice to recognize primitive elements in free groups.

Corollary 4 Il ${L}$ is a finitely generated fully residually free group, and if ${\hat L=\hat F}$ for some free group ${F}$, then ${L=F}$. Finite covers and linear algebra suffice to recognize primitive elements in free groups.

3. Ideas in the proof

1. Study all essential admissible maps ${(H,u)\rightarrow (F,\omega)}$ with ${(H,u)}$ irreducible. There is at least one, ${(F,\omega)}$ itself.
2. Define a cone ${\mathcal{C}\subset {\mathbb R}_+^n}$ defined by integral equations, whose integer points correspond to pairs ${(H,u)}$ of a surface with boundary and a boundary map.
3. Look at

$\displaystyle \begin{array}{rcl} \rho(H,u)=\frac{-\chi(H)}{n(u)}. \end{array}$

Argue that there is a map that maximizes ${\rho}$.

4. Show that this ${(H,u)}$ is a surface pair.

One can maybe implement the algorithm, but it does not seem to be efficient.