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.

I’m glad to hear that Cameron’s talk made our work sound intriguing! Thanks for writing about it, Jesse. I have a couple of comments.

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

powerof 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.

Comment by Henry W — January 8, 2009 @ 2:14 pm |

Jesse, thanks for summarizing Cameron’s very interesting talk.

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

ginFan elementary covering space construction gives finite-index subgroupGofFwheregpart of free basis forG. Thengcan clearly be geometrically represented in the cover ofHcorresponding toG.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…Comment by Nathan Dunfield — January 8, 2009 @ 10:34 pm |

Grr, that should be:

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).

Comment by Nathan Dunfield — January 8, 2009 @ 10:37 pm |

You’re quite right, Nathan – you can always arrange for g to be part of a basis of a finite-index subgroup of F.

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 geometricif 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?”Comment by Henry W — January 9, 2009 @ 10:57 am |

Oh, I forgot to mention that we insist that the image of g’ be contained in the boundary of H’.

Comment by Henry W — January 9, 2009 @ 10:59 am |

Henry, do you mean that g’ is homotopic to an embedding on each component?

Comment by Richard Kent — January 9, 2009 @ 3:10 pm |

Richard, no. It needs to embed the disjoint union.

Comment by Henry W — January 9, 2009 @ 3:19 pm |

hola,

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

Comment by Juan Souto — February 18, 2009 @ 11:12 pm |

Juan,

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

Comment by Henry Wilton — February 19, 2009 @ 1:11 pm |