Low Dimensional Topology

December 20, 2007

Lower bounds on stabilizing Heegaard splittings

Filed under: 3-manifolds,Heegaard splittings,Hyperbolic geometry — Jesse Johnson @ 2:32 pm

A lot of excitement at last week’s AIM workshop was caused by two announcements of lower bounds on the number of stabilizations needed to flip over certain Heegaard surfaces. Recall that a Heegaard splitting is a decomposition of a 3-manifold into two handlebodies with a common surface called a Heegaard surface. In certain cases, there is an isotopy of the 3-manifold that takes the Heegaard surface off of itself, then back onto itself in a way that interchanges the two handlebodies. (We will call this “flipping” the Heegaard splitting.) One can create a new Heegaard splitting called a stabilization of the given splitting by by adding one or more trivial handles to each handlebody. Whether or not a Heegaard splitting can be flipped, it has a stabilization that can be flipped. The flip genus of a Heegaard splitting is the genus of its smallest stabilization that can be flipped.

A simple construct (left to the reader) shows that the flip genus of a Heegaard splitting is at most twice the original genus. Last week, Abby Thompson announced joint work with Joel Hass and Bill Thurston showing that Heegaard splittings with high Hempel distance have flip genus equal to precisely twice their genus. (In otherwords, they find a sharp lower bound.) Dave Backman announced a similar, independent result. This theorem is not necessarily surprising, but it is important because it is the first non-trivial lower bound on stabilization for Heegaard splittings. I will give a brief overview of the Hass-Thompson-Thurston proof below. Perhaps I can discuss Bachman’s proof (which is considerably more complicated, but has some interesting ideas) in a future post.

As one might expect from a proof with Bill Thurston’s name attached to it, the Hass-Thompson-Thurston approach is based on hyperbolic geometry. The idea is that a 3-manifold with a high distance Heegaard splitting can be given a geometric structure consisting of a long, thin, negatively curved region that looks like a piece from the cyclic cover of a surface bundle (See the project by Namazi, Souto, Brock and Minsky.) capped off by two handlebodies with pinched negative curvature. The long thin part is homeomorphic to the Heegaard surface cross an interval and one of these surfaces cuts the region in half. Given a sweep-out for a Heegaard splitting in this 3-manifold, there is a leaf of the surface that cuts the volume of the 3-manifold exactly in half. One then asks if this leaf cuts each half of the surface-cross-interval region in half. If we can flip the Heegaard surface by an isotopy then at some point in the isotopy, a surface must cut the whole 3-manifold in half and cut each half of the surface-cross-interval region in half. The key is to show that such a surface has high genus.

There’s a lot of hyperbolic geometry involved in this last step and I’m going to gloss over most of it (in particular, a part about isotoping the sweep-outs towards harmonic maps in order to get area bounds) but here’s the topological idea: If the Heegaard surface intersects the leaves of the surface cross interval roughly in half then one gets a path in the curve complex from the boundary of a disk in one handlebody to the boundary of a disk in the other handlebody (essentially the situation in Scharlemann and Tomova’s bound on the genus of non-isotopic Heegaard splitting) so the genus of the leaf of the sweep-out must be at east half the distance of the original Heegaard splitting. If some of the intersections of the leaf of the sweep-out with the leaves of the long thin region are trivial then the fact that it cuts the volume in half means that there must be a loop bounding a small region on one side and another loop bounding a small region on the other side. In between, these loops, the leaf of the sweep-out must have at least as much genus surface that’s crossed with an interval. Since each half of the long thin region is divided exactly in half, we can apply this argument twice, to see that the leaf of the sweep-out must have genus at least twice the genus of a leaf of the long thin region (i.e. twice the genus of the original Heegaard splitting).


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