Not long ago, Jesse was kind enough to blog about my recent work with Cameron Gordon. Now that the preprint has finally appeared on the arXiv, I thought I’d say a few more words about what we do, and why.
The context is Gromov’s famous question about surface subgroups.
“Does every one-ended word-hyperbolic group contain a subgroup isomorphic to the fundamental group of a closed surface?”
[The hypothesis of one-endedness excludes stupid counterexamples like virtually free groups. And of course the fundamental groups of the sphere and the projective plain don’t count as surface subgroups.]
A (word-)hyperbolic group is a group that satisfies a certain coarse form of negative curvature. The standard examples are free groups and the fundamental groups of closed hyperbolic manifolds, but in fact word-hyperbolic groups are much more general than that. A randomly chosen finitely presented group is almost surely word-hyperbolic, in a suitable sense of almost surely.
In the case of the fundamental groups of hyperbolic 3-manifolds, this question is the famous Surface Subgroup Conjecture. So Gromov’s question is a natural generalization. Perhaps hyperbolic 3-manifolds have surface subgroups because all word-hyperbolic groups do?
There are lots of open questions about the nature of word-hyperbolic groups, but this one is striking for how little we know, even about some very basic examples. For example, let w be any element of a finitely generated free group F that’s not a proper power. Then the double
is word-hyperbolic, and one-ended as long as w isn’t in a proper free factor of F. A couple of years ago, though, no one had any idea whether any such doubles contained surface groups (except in a few obvious cases – any closed surface of even genus is obviously a double, for instance!). At conferences, you’d hear speculation that such doubles should include a counterexample to Gromov’s question. I engaged in such speculation myself. After all, why should a double along a randomly chosen word contain a surface subgroup?
This all changed when Danny Calegari used his marvellous stable commutator length machinery to prove the following, which he advertises as a substitute for Dehn’s Lemma in the world of graphs of free groups.
Theorem (Calegari): Suppose that G is the fundamental group of a graph of free groups with cyclic edge groups, G is word-hyperbolic and G has positive second Betti number. Then G contains a surface subgroup.
In the case of doubles, the positive second Betti number condition translates into the hypothesis that w is in the commutator subgroup of F. When this appeared, my first thought was something like “Well, OK. You might expect to find surfaces in that case. But what about for other w?” After thinking a little harder, though, I realized that Calegari’s theorem also helps with the general case.
Corollary: The double D(w) has a surface subgroup if and only if it has a finite-index subgroup D’ with positive second Betti number.
One direction is immediate; the other follows from the fact that every finitely generated subgroup of a double is a virtual retract.
This shifts the focus to virtual homology of doubles. Unfortunately, a general approach to computing the virtual homology is still elusive. In our paper, though, Cameron and I provide several very large classes of examples of doubles with surface subgroups. This is really addressed at the doubters (like my former self) who couldn’t see why doubles should contain surface subgroups, other than for trivial reasons.
Our main trick is to consider the one-relator group
Theorem: If G(w) has positive virtual second Betti number then so does D(w).
Now, one-relator groups can be fairly badly behaved – indeed, doubles are particularly nice examples of one-relator groups – so this may not seem like much of an improvement. But on the other hand one-relator groups are much-studied, and furthermore in the case when the free group F is of rank 2, a simple calculation with Euler characteristic gives the following.
Corollary: If F is of rank 2 and G(w) has virtual first Betti number greater than one then D(w) contains a surface subgroup.
This enables us to produce lots of examples. Briefly, we get some from the Alexander polynomial, some from the mod p Alexander polynomial using work of Jim Howie, and some by adapting Dani Wise’s work on positive small-cancellation one-relator groups.
Unfortunately we can’t deal with all doubles this way, even in the case when F is of rank 2. Edjvet and Pride proved that Baumslag-Solitar groups, which have presentation
have virtual first Betti number equal to one whenever p and q are relatively prime and not both equal to plus or minus one.
Nevertheless, there is some evidence that a generic two-generator one-relator group might have positive second Betti number, in some suitable sense of ‘generic’. Jack Button has used a computer to attack this sort of problem for a huge number of examples.
Finally, we came up with an intriguing and entirely different idea for dealing with the Baumslag-Solitar examples. This idea brings us back to the world of 3-manifolds, and is what Jesse posted about.
If w is the relator of a Baumslag-Solitar group then D(w) is not a 3-manifold group; if it were, then G(w) would be too, but it isn’t. Nevertheless, we do have the following.
Theorem: If w is a Baumslag-Solitar relator with p,q coprime then D(w) has a finite-index subgroup that is a 3-manifold group. Therefore, D(w) has a surface subgroup, coming from the boundary.
If w has this property then we say it is virtually geometric. We have been unable to prove that any words are not virtually geometric. Indeed, we suggest a tricky word in the paper, but already I’ve had an email which seems to show that it is virtually geometric. Perhaps I’ll comment about that later, when I’ve had time to think it over. So we’re left with the following question, a positive answer to which would imply a positive answer to Gromov’s question in the case of doubles.
Question: Is every double D(w) virtually a 3-manifold group?
Opinion seems to be divided about what the correct answer should be. Personally, I don’t know what to think.