Is there a way to classify incompressible surfaces in $\Sigma \times [0,1]$ ?

Question

Is there a way to classify incompressible surfaces in $\Sigma \times [0,1]$ where $\Sigma$ is any closed surface? I know of the Hatcher-Thurston classification of incompressible surfaces in 2-bridge knot exteriors, but I wonder if a classification of incompressible surfaces can be carried out in this other simple setting.

Answer 1

An incompressible and boundary incompressible connected surface is isotopic to either (a) a vertical annulus or (b) a horizontal surface. A vertical annulus is of the form $\alpha \times I$ where $\alpha$ is an essential simple closed curve. A horizontal surface is of the form $\Sigma \times \{t\}$.

Here is a sketch of the proof. Let $F$ be the given incompressible, boundary incompressible, connected surface. Suppose that $\beta \subset \Sigma$ is essential simple closed curve. Let $B = \beta \times I$ be the corresponding vertical annulus. An innermost disk/outermost bigon argument simplifies the intersection between $F$ and $B$ until it is a disjoint union of either vertical arcs or horizontal curves (ie, copies of $\beta \times \{t\}$).

Now cut $\Sigma \times I$ along $B$ to get a handlebody with a product structure. Repeat the above argument, replacing the vertical annulus with a sequence of vertical rectangles.

I believe that you can find all of the tools you need for this kind of thing in Gordon's lecture notes on normal surfaces.

There is also a proof of a similar fact by Scharlemann and Thompson in their paper "Heegaard splittings of (surface)×I are standard."

Allowing boundary compressible surfaces makes the classification more annoying. I haven't thought that through.

Answer 2

It is a difficult problem and as far as I know there is no general answer. Topologists usually look for incompressible and $\partial$-incompressible surfaces, and in that case there are only horizontal $\Sigma$'s and vertical annuli as Sam said.

For irreducible manifolds with toric boundary, every incompressible connected surface is either boundary-parallel or $\partial$-incompressible, hence $\partial$-incompressibility is not an important hypothesis. But in general it is an important hypothesis, even in a manifold as simple as $\Sigma \times I$.

You can construct plenty of incompressible surfaces in $\Sigma \times I$ as follows. Take two homologous oriented multicurves $\mu_0$ and $\mu_1$ in $\Sigma \times 0$ and $\Sigma \times 1$: a multicurve is a collection of disjoint homotopically non-trivial circles, and it is oriented if every component is oriented. With such an orientation a multicurve determines an element in $H_1(\Sigma, \mathbb Z)$ and we require that $\mu_0$ and $\mu_1$ determine the same object.

Since $\mu_0$ and $\mu_1$ are homologous they cobound some 2-cycle which can be transformed to be an orientable surface in $\Sigma \times I$. If you suppose that this surface has least genus among all possible such 2-cycles, it is certainly incompressible. You can construct a simple example by picking $\mu_0$ equal to two oriented curves, and $\mu_1$ equal to one curve obtained by "summing" the two. Then one pair-of-pants bounds both and is certainly incompressible. It is natural to extend this construction and hence define a graph where the vertices are the oriented multicurves and the edges are such pair-of-pants: this graph has been defined and investigated in Ingrid Irmer's thesis.

Two homologous multicurves define at least one incompressible surface. But what is its genus? How many incompressible surfaces do we get? Can we get infinitely many of them? The same questions can be asked in the non-orientable case, see this other question on MO

Answer 3

Since you are asking for incompressible but not $\partial$-incompressible, the classification is more complicated. As pointed out in Sam Nead's answer, the classification of incompressible and boundary-incompressible surfaces has a straightforward description. Any incompressible surface has a sequence of boundary compressions which takes it to a boundary-incompressible surface, and one may reverse this process by adding a tunnel ("arc-extension"). The subtlety in reversing this process is to tell when adding an arc extension keeps the surface incompressible. In principle there is an algorithm since there is Haken's algorithm to tell if a surface is incompressible (so one adds the tube and applies the algorithm). Thus the set of such surfaces is recursively enumerable, but this is not a very satisfying description.

One result says that (apart from a few exceptional cases) one may always take an incompressible surface and arc extend it to another incompressible surface. One also knows from Hugh Howards' thesis that incompressible surfaces can be created with arbitrarily many non-parallel components. On the other hand if the euler characteristics of the components are bounded then the number of disjointly embedded non-parallel components are bounded by a result of Freedman-Freedman. These are the only structural results that I am aware of, so it seems like there hasn't been much work on this topic since the 90s.

Off the top, I'm not even sure how to answer the following question: given a subgroup of $\pi_1(\Sigma)$, can it be realized by an incompressible surface in $\Sigma\times I$?

Answer 4

Please check this paper--http://comp.uark.edu/~yoav/kobayashi-qiu-wang.pdf. In Proposition 1, they have proved that there are infinitely many incompressible surface based on a simple curve. So I guess it is hard to find a simple classification of all incompressible surfaces in a surface I bundle