Does your argument imply “if a loop is virtually geometric then it’s geometric”? That is, if the “lifts” of a loop in a finite cover are homotopic to an embedded multicurve in the boundary then the original loop was homotopic to an embedded curve in the boundary?

If so, that’s false. There are loops that are virtually geometric but not geometric. The easiest example is probably the Baumslag–Solitar relator in . It’s not geometric, as the corresponding Baumslag–Solitar group is not a 3-manifold group. But there’s a degree-6 cyclic cover in which its “lift” is geometric. Cameron and I construct an explicit Heegaard diagram on which its “lift” is embedded.

H

the answer to the stronger question is surely no. Somehow, if the original curve kills all splittings, then so should the pre-image, shouldn’t it?? If this is the case, the it should follow that every homotopy equivalence of the cover mapping pre-image to pre-image is homotopic to a unique homeo. This should mean that the homotopy to the boundary descends and hence the original thing was homotopic into the boundary… well, all this may be bogus, but the answer is surely no.

Juan

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?

We actually ask whether something stronger is true – rather than asking whether just one lift of g can be made simple, we want the total collection of lifts of g to be embedded.

To be precise, think of g as a continuous map from the circle to H. If H’ is a finite-sheeted covering space of H then we can think of the collection of all lifts of g to H’ as a map g’ from a finite disjoint union of circles to H’. We say that g is virtually geometric if there is a choice of handlebody H and finite-sheeted covering space H’ such that the resulting g’ is homotopic to an embedding. Our question is now “Is every element of a free group virtually geometric?”

More generally, we could focus on the one relator group N = Group(F, g = 1). If this is residually finite, then I think you can do a similar thing where you take a cover of H where g becomes a basis element of corresponding subgroup. Just make sure that all proper cyclic subwords of g are non-trivial in your chosen finite quotient… Also, an old conjecture of Baumslag is that N = Group(F, g^n) is residually finite for all large n (or maybe all n > 1).

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?

Asked this narrowly, isn’t the answer always yes? Given g in F an elementary covering space construction gives finite-index subgroup G of F where g part of free basis for G. Then g can clearly be geometrically represented in the cover of H corresponding to G.

More generally, we could focus on the one relator group N = . If this is residually finite, then I think you can do a similar thing where you take a cover of H where g becomes a basis element of corresponding subgroup. Just make sure that all proper cyclic subwords of g are non-trivial in your chosen finite quotient…

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.

The curve will usually project to a power of g. The basic example is the word in the free group F generated by a and b. This doesn’t have a simple representative on the surface of a handlebody, as 3-manifold groups never contain Baumslag-Solitar subgroups (such as the one-relator group with relator w). But if you look at a certain index-two subgroup F’ of F, and consider w^2 in F’, then it turns out that w^2 has a simple representative on the boundary of a handlebody for F’. In our terminology, this means that w is not geometric but is virtually geometric.

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.

Not quite…. We’ve proved that if the 1-relator group has a finite-index subgroup with positive second Betti number then so does the double. A very nice theorem of Danny Calegari (which you can think of as Dehn’s Lemma for graphs of free groups, if you like) then enables us to deduce that the double has a surface subgroup. But there are lots of interesting examples of one-relator groups in which we can find (virtual) second homology, but without Calegari’s theorem it seems hard to deduce the existence of surface subgroups. And, indeed, there are certainly examples in which the double contains a surface subgroup but the one-relator group doesn’t (like the Baumslag-Solitar word I mentioned above).

I’d be very happy to write a little more about this at some point, though I should consult with Cameron about the status of our preprint first.