# the local structure of an immersed incompressible surface

Assume that $M$ is a closed, irreducible, orientable 3-manifold. Suppose that we have a closed, immersed, incompressible surface $F$ of genus at least 1. Since we only required $F$ to be immersed in $M$, there may be some self-intersections along arcs or curves. Of course, we can assume that all these intersections are efficient, i.e. transverse. So my first question is:

Are these self-intersecting components some graphs without valence-1 vertices?

My second question is:

If we remove all these self-intersection components from $F$, what does the complement look like?

## 1 Answer

For the first question, if the map of the surface in the manifold is made generic, then the singular set will consist of just double and triple points, and will be a graph with vertices of valence six (and maybe closed loops). Near the triple points, it will look like:

## As for the second question, the complement of the self-intersection locus will just be subsurfaces, and can change quite a bit under homotopy. If the surface is made minimal area (in its homotopy class), then some information can be gleaned from a paper of Freedman-Hass-Scott. Minimal area maps of $\pi_1-$injective surfaces are immersions, and may factor through a finite covering. They show that the minimal immersions intersect minimally in a certain sense. From their results, one can show that the complementary regions of the singular locus will be $\pi_1$-injective (of course, this is meaningless if all of the complementary regions are disks). I think one could perturb this immersion to be generic while retaining this property. Not sure what exactly you're fishing for though.

• thank you, this is a great answer, specially for the figure, May I ask that if we require that $F$ is least area, is there any disk in the complement of the self intersection locus? – yanqing Nov 16 '16 at 6:46
• @yanqing: If the surface is homotopic to an embedding (so the manifold is Haken), then the minimal area representative will be embedded, hence the singular locus is trivial, and the complement will not have any disk. Similarly if the singular locus has only closed curves of intersection. On the opposite extreme, the surfaces constructed by Kahn and Markovic will likely have all of the complementary regions of the singular locus being disks. – Ian Agol Nov 16 '16 at 18:09