I’ve been thinking about thin position a lot lately. It seems that thin position is more of a philosophy than anything else, but I think it can be made into a fairly formulaic way of approaching a wide variety of problems. (I’ve been thinking of writing an article about it for the soon-to-be Tricki site, but it may be too far down my to-do list to ever happen.) In this post and the next few posts, I want to describe exactly how the thin position formula works. In this post I’ll describe a cell complex I call the complex of surfaces, which leads to a nice description of both Scharlemann-Thompson’s thin position for 3-manifolds  and Dave Bachman’s topological index . In this post, I’ll explain how it relates to topological index, then in a future post I’ll describe what it has to do with thin position.
Let M be a 3-manifold. The complex of surfaces S(M) will be a cell complex whose vertices are isotopy classes of closed, separating, transversely oriented, (not necessarily connected) surfaces embedded in M. In fact, I want the surfaces to not only be separating, but what I call strongly separating: It must be possible to label the components of the complement + and – so that each component of the surface bounds both a positive component and a negative component. A transverse orientation for the surface will consist of a choice of labelings for the complement.
Some of the surfaces in this complex will be compressible, i.e. will contain an essential loop bounding a disk whose interior is disjoint from the surface. Compressing along such a disk produces a new strongly separating surface. For each compressing disk in a surface, we will include an edge in S(M) connecting this surface to the that results from the compression. We will also add edges corresponding to adding or removing sphere components. Note that some pairs of surface will be connected by multiple edges.
Given two disjoint compressing disks in a surface, producing two disjoint compressed surfaces, we can compress along one and then the other, but we have a choice of which one we compress first. These two disks thus define a square of edges in S(M). We will add in a 2-cell bounded by this square for every such pair of disjoint compressing disks. In fact, any set of disjoint compressing disks, whose boundaries do not cut a planar surface out of the surface they sit in, defines an n-dimensional cube of edges, and we will add in an n-cell bounded by these edges.
We will define the complexity of a connected surface to be its genus plus one. The complexity of a vertex in S(M) will be the n-tuple of complexities of its components, in non-increasing order. We will define a partial ordering on vertices in S(M) by applying lexicographic or dictionary ordering to these n-tuples. In other words, given two complexities, we’ll compare the first entry of each. If they’re the same, we’ll try the second entries and so on until we find a pair of entries that are different. The vertex that has the larger entry in this place is larger than the other vertex. This ordering is a bit complicated to define but it has the nice property that it strictly decreases when one compresses a surface. (This is a simple exercise for the reader.)
In a previous post I mentioned Dave Bachman’s definition of topological index for a surface: He looks at the disk complex for a surface, in which vertices are compressing disks and simplices are bounded by sets of pairwise disjoint compressing disks. The topological index is one plus the dimension of the first non-trivial homotopy group. As mentioned above, each compressing disk defines an edge in S(M) along which the complexity decreases. A set of disjoint compressing disks defines a cube, and we can think of the simplex in the disk complex as defining a corner of the corresponding cube. In other words, the disk complex for a surface is isomorphic to the part of the link of the corresponding vertex along which the complexity decreases. I’ll call this part of the link the descending link.
In Morse theory, when we have a critical point, the set of unit vertices of the tangent space along which the function is decreasing is homotopy equivalent to a sphere. The dimension of this sphere is precisely one less than the index of the critical point. In particular, the first non-trivial homotopy group of this set is one less than its index. Thus it seems reasonable to define the index of a vertex in S(M) to be one plus the dimension of the first non-trivial homotopy group of its descending link. This is precisely Dave’s definition of topological index, so this seems like a nice explanation for why topological index deserves to be called an index. Next time I’ll write about how the complex of surfaces S(M) relates to thin position.