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.

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About metric2011

metric2011 is a program of Centre Emile Borel, an activity of Institut Henri Poincaré, 11 rue Pierre et Marie Curie, 75005 Paris, France. See http://www.math.ens.fr/metric2011/
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