Surface subgrops of graphs of free groups with cyclic edge groups
1. Gromov’s questions
Tits’ Ping-pong Lemma. Let be a non-elementary hyperbolic group. If do not commute, then some high powers of and generate a free subgroup.
Hence there exist heaps of free subgroups in hyperbolic groups. What about one-ended subgroups? Known: they willcome out in at most finitely many conjugacy classes. Need to rule out free groups, of course.
Question (Gromov). Does every one-ended hyperbolic group contain a surface subgroup ?
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.
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 ?
2. Free groups and relative questions
Relative means relative to a given set of relators. I.e. study group pairs where is e finitely generated free group and a finite collection of words. A surface pair is where is a surface with boundary and represents its boundary components.
Note that a pair can be doubled into
Bestvina-Feighn: is hyperbolic iff is not a proper power.
Shentzer: is one-ended iff is not contained in a proper free factor. In this case, we say that the group pair is irreducible.
Theorem 1 (Wilton) Every irreducible group pair admits an essential admissible map of a surface with boundary .
Essential is a bit more restrictive than injective. Admissible means that along the boundary, the gluing map has the same degree on all components.
3.1. Graphs of free groups
Corollary 2 For any one-ended graph of free groups with cyclic edge groups ,
- either containes a Baumslag-Solitar subgroup,
- or is hyperbolic and contains a surface subgroup.
There were earlier results with C. Gordon and Kim. In 2008, Calegari proved this under condition that . He reduced the problem to linear equations, and the homology assumptions provided a solution. In 2010, Kim and Oun solved the case of doubles of rank 2 free groups, where Calegari’s equation could be scrutinized.
3.2. Residually free groups
A group is fully residually free if every finite subset is mapped injectively by some homomorphism to some free group.
Corollary 3 If is a finitely generated non-free fully residually free group, then contains a surface subgroup and for any free group .
This follows from a structure theorem for such groups. This answers Remeslennikov’s question that arised in Alan Reid’s course.
3.3. Rigid groups
Say that a group is rigid if it does not split over 1 or .
Theorem 4 (Louder-Touikan’s Strong Accessibility) If is a one-ended hyperbolic group, then cntains a quasi-convex subgroup such that
- either is rigid,
- or s a graph of free groups with cyclic edge-groups.
Hence Gromov’s question is reduced to the rigid case.
4. Ideas in the proof
- Study all essential admissible maps with irreducible. There is at least one, itslef.
- Look at
Argue that there is a map that maximizes .
- Show that this is a surface pair.
One can maybe implement the algorithm, but it does not seem to be efficient.
Arzhantseva: one cannot fix the genus of surfaces (1997).