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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.

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2 Answers 2

up vote 9 down vote accepted

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. http://homepages.warwick.ac.uk/~masgar/Articles/Gordon/normal.pdf

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.

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Thanks for the answer and reference! My only concern is that i am not sure the surfaces (surfaces constructed using Culler-Shalen theory) i want to deal with can be assumed boundary incompressible. –  Renaud Detcherry May 30 '13 at 7:30

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

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