Cameron Gordon asked an interesting question in his talk yesterday morning (at 8 AM!) here at the JMM. Given a non-primitive element g of a free group F, which we identify with the fundamental group of a handlebody H, you can ask whether there is a simple loop in the boundary of H representing the element g in the fundamental group. There is always an infinite family of curves representing g, but in general, these curves may all have self intersections. If you do happen to find a simple representative then you can glue a 2-handle to H along this curve. The fundamental group of the resulting 3-manifold with boundary M is the 1-relator group that you get by quotienting F by the relation g. If you compress the boundary of M into the manifold as much as possible, the result is an incompressible surface, implying that the 1-relator group contains a surface subgroup (i.e. a subgroup isomorphic to the fundamental group of a surface.) This property is important to geometric group theorists for various reasons. (Perhaps I can convince Henry, who has signed up to write occasional posts, to explain why…)
Of course, most elements of F don’t have simple representatives so here’s something else to try: If we could find a subgroup of the 1-relator group that contains a surface subgroup then we’d be done. A finite index subgroup of the 1-relator group is determined by a finite cover of H in which there is a closed curve whose projection into H represents g. If we can find such a finite cover of H in which there is a simple closed curve that projects to g then the fundamental group of the manifold it defines (which is a subgroup of the original 1-relator group) contains a surface subgroup. Cameron asks whether this always happens. In other words, is every non-primitive element of a free group F represented by a simple closed curve in a finite cover of a handlebody with fundamental group F?
Incidentally, the motivation for this question comes from studying certain delta hyperbolic groups in which one takes the free product of two copies of F amalgamated along the subgroup generated by g. He and Henry have shown that the amalgamated free product will contain a surface subgroup if and only if the corresponding 1-relator group does. (At least I think that’s what he said…) He also described another method for finding surface subgroups that I didn’t understand as well, though it sounded like this other method has been more effective.