Low Dimensional Topology

June 21, 2013

Lots and lots of Heegaard splittings

Filed under: 3-manifolds,Heegaard splittings,Knot theory — Jesse Johnson @ 12:28 pm

The main problem that I’ve been thinking about since graduate school (so around a decade now) is the following: How does the topology of a three-dimensional manifold determine its isotopy classes of Heegaard splittings? Up until about a year ago, I would have predicted that most three-manifolds probably don’t have many distinct Heegaard splittings, maybe even just a single minimal genus Heegaard splitting and then all of its stabilizations. Sure, plenty of examples have been constructed of three-manifolds with multiple distinct (unstabilized) splittings, but these all seemed a bit contrived, like they should be the exceptions rather than the rule. I even wrote a blog post a couple years back stating what I called the generalized Scharlamenn-Tomova conjecture, which would imply that a “generic” three-manifold has only one unstabilized splitting. However, since writing this post, my view has changed. Partially, this was the result of discovering a class of examples that disprove this conjecture. (I’m hoping to post a preprint about this on the arXiv in the near future.) But it turns out there is an even simpler class of examples in which there appear to be lots and lots of distinct Heegaard splitting. I can’t quite prove that they’re distinct, so in this post I’m going to replace my generalized Scharlemann-Tomova conjecture with a conjecture in quite the opposite direction, which I will describe below.

Recall that given a Heegaard surface $S \subset S^3$ and a knot $K \subset S^3$, we say that $S$ is a bridge surface for $K$ if $K$ intersects each of the two handlebodies bounded by $S$ in a collection of boundary parallel arcs. A few months ago, I wrote a post about Alex Zupan’s work on the bridge spectrum, and in particular how the notion of a meridional stabilization can be use to link bridge surfaces of different genera. A meridional stabilization consists of attaching a tube to $S$ along one of the arcs of $K \setminus S$. In other words, we remove two disks from $S$ that are regular neighborhoods of consecutive points of intersection in $K$, then attach an annulus to $S$ along the boundaries of these disks, such that the annulus follows the arc. It is a relatively straightforward exercise to show that if $S$ is a bridge surface for $K$, then the resulting surface $S'$ will also be a bridge surface, but with genus one greater than that of $S$. But there’s one exception: If $K$ is one-bridge with respect to $S$ (i.e. there is one arc of $K$ on either side of $S)$ then the resulting surface $S'$ will be disjoint from $K$, and will in fact be a Heegaard surface for the complement of $K$.

We can use this to construct Heegaard splittings for knots from any bridge surface. For example, if $S$ is a sphere and $K$ is $n$-bridge with respect to $S$ then we can choose all the arcs on one side of $S$ and meridionally stabilize along them. This gives us a tunnel system for the knot in which the tunnels are horizontal and connect all the local maxima (if we stabilized along the arcs below $S$) or connect all the local minima (if we stabilized above $S$). That’s two potentially different Heegaard surfaces, and in the case of two-bridge knots Morimoto and Sakuma  showed that if the knot is defined by a sufficiently complicated braid then these two Heegaard surfaces are not isotopic. It’s conceivable that they should also be distinct for knots with higher bridge number and “sufficiently complicated” braids, but why stop there?

Instead, lets do the meridional stabilizations one at a time and see if there are other choices. Given a knot $K$ with an $n$-bridge sphere $S$, we’ll start by picking an arc $\alpha$ below the bridge sphere and we’ll meridionally stabilized to get an $(n-1)$-bridge torus $S'$ with respect to $K$, as in the middle of the Figure below. Most of the bridge arcs of $S'$ are the same as the bridge arcs of $S$, but there’s one exception: One of the bridge arcs of $S'$ is actually the union of our original arc $\alpha$ with the two upper bridge arcs that were adjacent to it. With respect to $S'$, this union of three arcs is no different from the original bridge arcs with respect to $S$, so why not meridionally stabilize along it? This meridional stabilization, shown on the right below, adds a tube to $S'$ that runs through the tube that we originally added along $\alpha$, so it looks funny. But, as noted above, meridional stabilization always produces a new bridge surface (in particular, the new surface will still be a Heegaard surface for $S^3$) so this is a perfectly reasonable construction. Looking at it a different way, for our initial bridge surface $S$, we have $2n$ choices for which bridge arc to meridionally stabilize along. In the resulting surface $S'$, we have $2(n-1)$ choices and so on. However, we have to be careful about double counting. For example, if we choose all the original top arcs, we’ll get the same surface no matter what order we pick them in. By my calculations (which I won’t describe here, but it’s a good combinatorics problem) there should be ${2n}\choose{n}$ possibilities for the final surface.

Are these surfaces distinct (up to isotopy)? Well, they clearly won’t be for some knots, such as if $K$ is the unknot. But if $K$ is a two-bridge knot then the ${{4}\choose{2}} = 6$ possibilities determine the six known unknotting tunnels for any two-bridge knot. Morimoto-Sakuma  show that for a sufficiently complicated braid, these unknotting tunnels will be distinct up to isotopy. What about for higher bridge number? Here’s my conjecture:

Conjecture: If $K$ has an $n$-bridge surface $S$ such that the distance (with respect to the curve complex) is greater than $4n$ then:

1. The ${2n}\choose{n}$ Heegaard surfaces of genus $n$ defined by repeated meridional stabilization of $S$ as above are the only (up to isotopy) minimal genus Heegaard surfaces for the complement of $K$,
2. No two of these suraces are isotopic to each other and
3. The stable genus of any two of these Heegaard surfaces is $2n-1$.

I should add that to get the first two, we probably only need a distance greater than $2n$. If part 3 is true, the conjectural stable genus is higher (relative to the original genus) than the stable genus of any currently known examples, so that would be an exciting (at least to me) result. One reason I’m willing to make this conjecture is that it seems like it might be possible to prove it by generalizing the spanning/splitting machinery that I introduced in . This would involve comparing the sweep-out of the knot complement defined by the bridge surface to the sweep-outs defined by the Heegaard surfaces. I don’t know exactly how to do it, though, so I’ll leave it as an open conjecture.

 Morimoto, Kanji; Sakuma, Makoto, On unknotting tunnels for knots. Math. Ann. 289 (1991), no. 1, 143–167.

1. Hi Jesse,

This is a nice post and an interesting conjecture. It’s quite surprising to think that there could be 3-manifolds with many distinct minimal genus Heegaard splittings (this somehow seems very complicated), all of which are obtained by generic operations performed on a single high distance bridge surface (this somehow seems very uncomplicated).

With regards to part 1 of your conjecture, I think Maggy Tomova’s paper “Multiple bridge surfaces restrict knot distance” (http://arxiv.org/abs/math/0511139) might be helpful. She shows that if S and Q are two distinct irreducible bridge surfaces for a knot K, then the distance d(S) bounds a function of the Euler characteristic of Q from below. Although I haven’t read the paper closely, it appears that the result also holds when Q is a Heegaard surface. Thus, if S is a bridge surface, d(S) is above some threshold (2n for S an n-bridge sphere), and Q is a low genus Heegaard surface for the exterior of K, it should follow from Maggy’s main theorem that Q is one of the your candidate surfaces.

Comment by Alex Zupan — June 22, 2013 @ 2:07 pm

• That’s a good point, and I agree that part 1 probably follows from Maggy’s paper. The two-bridge case, which was proved by Tsuyoshi Kobayashi, uses a double sweep-out argument similar to the argument in Maggy’s paper, but it takes advantage of the fact that the two-bridge surface (a four-punctured sphere) has a Farey graph rather than a curve complex.

Comment by Jesse Johnson — June 25, 2013 @ 9:19 am

2. Should anyone be interested, I’ve proved a slightly altered form of this conjecture here: http://arxiv.org/abs/1507.07231

Comment by George Mossessian — December 28, 2015 @ 6:16 pm

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