# Low Dimensional Topology

## February 14, 2011

### The good, the bad and the one-sided

Filed under: 3-manifolds,Heegaard splittings — Jesse Johnson @ 11:44 am

If you manage to embed a non-orientable surface into an orientable 3-manifold, the embedding will be what’s called one-sided:  If you choose a normal vector somewhere on the surface, then drag it by its endpoint around a non-orientable loop in the surface, it will come back on the other side.  An equivalent definition is that the boundary of a regular neighborhood of the surface will have a single boundary component.  (In many popular accounts of topology, one-sidedness seems to be the property of Mobius bands and Klein bottles that is described, since non-orientable is harder to define.)  These surfaces are notoriously misbehaved.  For example, unlike two-sided surfaces, a one-sided surface that is geometrically incompressible (no embedded disk has essential boundary in the surface) will often have a non-injective fundamental group.  However, Loretta Bartolini has found [1] that in some cases they can be understood better than their two-sided counterparts.

A one-sided surface in a 3-manifold $M$ whose complement is an open handlebody is called a one-sided Heegaard surface.  The fact that such things even exist takes a little getting used to, but here’s how you can think about it:  A regular neighborhod $N$ of a one-sided surface $S$ is a twisted interval bundle whose boundary is a two-sided separating surface.  If the complement of $N$ is a handlebody, then you can extend it into the open set $N \setminus S$.  As you extend the handlebody, it will approach each point in $S$ from two different directions, and the result is a handlebody whose boundary double covers $S$.

Because the handlebody is on both sides of $S$, the boundary of a compressing disk $D$ for the handlebody will be immersed in $S$, but may not be embedded.  Thus the inclusion map between the fundamental groups of $S$ and $M$ will have a very large kernel, even though the surface may be geometrically incompressible.  In particular, every non-Haken 3-manifold with non-trial $\mathbf{Z}_2$ second homology will contain a geometrically incompressible one-sided Heegaard surface.  Being geometrically incompressible corresponds roughly to the strongly irreducible condition on two-sided Heegaard splittings.

Because of this, the study of one-sided surfaces requires adapting a mixture of techniques from both 2-sided incompressible surfaces and 2-sided Heegaard surfaces.  Bartolini’s early work, such as [2], adapted classical techniques from Heegaard splittings (particularly, Waldhausens proof of the uniqueness of Heegaard splittings of the 3-sphere) and found the limitations of these methods in the one-sided setting.  However, her more recent papers have exploited the incompressible nature of these surfaces to prove results whose analogues in the two-sided setting are not known.

For example, in [1] she classifies the geometrically incompressible one-sided Heegaard surfaces in all Dehn fillings of the figure-eight knot complement.  The equivalent problem of classifying strongly irreducible Heegaard splittings in Dehn fillings of the figure eight knot complement is at best only partially understood.   (Two-sided Heegaard splittings are unique for a large class of Dehn fillings on the figure eight, and there is (in theory) an algorithm to figure out which fillings these are.  But that’s the best that one can say given current technology.)

1. Maybe if an irreducible 3-manifold contains infinitely many non-isotopic 1-sided
Heegaard surfaces, then it is Haken?

Comment by Ian Agol — February 14, 2011 @ 6:38 pm

2. That’s a good question. The equivalent theorem for 2-sided splittings comes from normal surface theory (via branched surfaces). With respect to normal surface theory, geometrically incompressible surfaces act like 2-sided incompressible surfaces, so they behave quite nicely. (In particular, they can always be made normal.) If you take a repeated Haken sum with a one-sided surface, you are effectively Haken summing with the two-sided boundary of its regular neighborhood, so if the result stays geometrically incompressible, it seems reasonable to expect that the two-sided surface is incompressible.

Comment by Jesse Johnson — February 15, 2011 @ 8:46 am

3. This is interesting… are presentations of the MCG of 1-sided surfaces fairly well understood? Also, could you recover the 3-manifold from a one-sided Heegaard surface and a mapping class?

Comment by dmoskovich — February 15, 2011 @ 11:37 am

4. As far as I know, the mapping class groups of these surfaces (i.e. the symmetries of the manifold/surface pair) have not been studied. A one-sided Heegaard splitting is defined by a fixed-point-free, order-two, orientation-reversing automorphism of the surface and knowing the handlebody and the automorphism is enough to construct the 3-manifold.

Comment by Jesse Johnson — February 15, 2011 @ 12:20 pm

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