A couple of months ago I wrote about advanced positions of knots in Heegaard surfaces: Embeddings of a given knot into a Heegaard surface (usually in the 3-sphere) such that the knot is not primitive in either of the two handlebodies that make up the Heegaard splitting. (Recall that a knot in the boundary of a handlebody is *primitive* if it intersects the boundary of a properly embedded disk in a single point.) Today, I want to mention two updates on the subject.

First, Alice Stevens posted to the archive a result about positions of knots in Heegaard splittings [1]. For a knot/Heegaard surface pair *(S, K)*, she defines K-stabilization of *S* to be a stabilization of *S* such that *K* is in the new surface and one of the meridian disks for the new handle is disjoint from *K*. She shows that any two positions of a knot in a Heegaard surface, with the same slope, have a common K-stabilization. (K-stabilization does not change the slope of the embedding.) Alice had told me about this result before I posted the first installment of “advanced knots” and it was one of the things that started me thinking about them.

Second, I was talking to Mike Williams a couple of weeks ago and we worked out a characterization of all positions of torus knots. Mike pointed out that if the Heegaard surface *S* is incompressible in the complement of *K* then after a Dehn surgery on *K* along the slope defined by *S*, the image of *S* is a separating incompressible surface that comes from *S* by removing a neighborhood of *K* and gluing in two disks. (This follows from Jaco’s 2-handle addition lemma.) If *K* is a torus knot then every Dehn surgery is either a Seifert fibered space or a connect sum of lens spaces. The only separating incompressible surfaces in these 3-manifolds are spheres, so if *K* is a torus knot sitting in a surface *S* so that *S* is incompressible in the complement of *K*, then *S* is a torus.

If we start with a torus knot *K* sitting in any Heegaard surface *S* in the 3-sphere, we can compress *S* in the complement of *K* until we get an incompressible surface *S’ *containing *K*. By the above argument, *S’* is a torus. There are only two tori in the 3-sphere that contain a given torus knot: The unknotted torus that defines it as a torus knot and the boundary of a solid torus parallel to the knot. In the second of these cases, the knot is primitive with respect to the solid torus. With a careful argument, one can show that if *K* is primitive in *S’* then it was also primitive in *S.*

On the other hand, if *S’* is the unknotted torus, then *S’* is a Heegaard splitting for the 3-sphere. There’s an argument in my paper on flipping Heegaard splittings [2] that shows that if *S* and *S’* are Heegaard surfaces such that *S’* is the result of compressing *S* some number of times then *S* is a stabilization of *S’*. By modifying this argument slightly, one can show that if *K* is contained in both *S* and *S’* under these conditions then *S* is a K-stabilization of *S’*. Thus is *K* is a torus knot sitting in a Heegaard surface *S* then either *K* is primitive in *S* or *S* is a K-stabilization of the unknotted torus containing *S*.

Of course, there are still a few things to work out about positions of torus knots in Heegaard surfaces. Every primitive position comes from a Heegaard splitting for the knot complement by pushing the knot into the Heegaard surface, and Heegaard splittings of torus knot complements are classified. However, there seem to be infinitely many ways push the knot into the surface, most likely with most of them inequivalent. Even the different K-stabilizations of the one genus one position need to be classified, since K-stabilizations are not unique. So there are still lots of interesting questions.

I am curious about some parts of the statement of Alice Stevens’s result — the surface slope.

In 3-sphere, for a knot K in Heegaard surface S, the surface slope has a corresponding integer value, which can be calculated as the linking number of K and boundary curve of S-K.

In 3-manifold M, given two distinct pairs (S,K) and (S’,K), how do we compare their surface slopes? The boundary curves of S-K and S’-K would be called longitude (or surface slope) for K and the slope is

0(=0/1). So I am curious that the two surface slopes are same, from the beginning before any

K-stabilizations.

Comment by JungHoon Lee — March 18, 2009 @ 12:08 am |

For a knot in an arbitrary 3-manifold, the slope isn’t defined as an integer, but as an isotopy class of loops in the boundary of a regular neighborhood of the knot that intersects a meridian disk in a single point. Such isotopy classes are still in one-to-one correspondence with the integers, but unless the original 3-manifold is a homology sphere, the correspondence is not canonical.

Comment by Jesse Johnson — March 27, 2009 @ 3:29 pm |

Thanks. So we just isotope K to second K and compare the accompanying slopes. I was confusing something.

Comment by JungHoon Lee — March 28, 2009 @ 1:55 am |