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

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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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 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.