Classifying nonorientable surfaces

One of the strange side-affects of my number theory –> representation theory –> quantum topology journey during graduate school is that I often find myself severely lacking in a basic topology education. In cleaning up an argument from my recent surfaces paper I learned quite a bit about the topology of surfaces. This is all stuff which I’d probably know if I’d had a more thorough topology background, but hopefully it’ll still be interesting to some of you.

Consider a nonorientable surface S. It is not difficult to see that there is a finite collection of nonintersecting circles in S which seperates S into orientable parts. (At least its easy to see this in the piecewise linear category, which is where quantum topologists usually work. In the smooth category this is probably much harder.) Ben pointed out to me yesterday that it is enough to only have one such closed curve.

Let’s think about what properties this closed curve X would have. Consider some other closed curve Y in S, if the orientation changes as you go around that curve, then we’d better have X intersect Y nontrivially. Let’s rephrase this condition in the language of algebraic topology.

  • closed curve where the orientation changes when you go around it –> element of \pi_1(S) which is not an element of \pi_1(S') where S’ is the orientation cover of S.
  • curves which intersect –> elements of H_1(S, \mathbb{Z}_2) whose intersection pairing is nontrivial. (Actualy, the latter is much stronger.)

Well, let’s look at the image of \pi_1(S') in H_1(S, \mathbb{Z}_2). It is easy to see that this image is codimension 1. By Poincare duality in characteristic 2 we have that the intersection pairing is nondegenerate, hence we can find an element which is in the perpendicular complement to the image of \pi_1(S'). Take a representative of this element of H_1 in pi_1 and that has the required properties of X!

This result has some fun consequences.

Let’s look at a neighborhood of this element X. Topologically it must either be an annulus or a Mobius strip (this requires a little bit of work). In the former case, removing the neighborhood of X and gluing in two caps yields an orientable surface whose Euler characteristic is 2 larger than that of S. In the latter case, removing the neighborhood of X and gluing in one cap yields an orientable surface whose Euler characteristic is 1 larger than that of S. Since all orientable surfaces have even Euler characteristic (Poincare duality again), we see that the neighborhood of X is a Mobius band when the Euler characteristic of S is odd, and an annulus when the Euler characterstic of S is even. (This is essentially the fact I wanted for my paper.)

Let’s pause for a moment to think about these two cases topologically. In RP^2 you can take X to be the equator. Then when you remove X you get the disc. To recover RP^2 from the disc you glue in a Mobius band. In the Klein bottle, on the other hand, you can take X to be the longitude of the cylinder. Removing X gives a cylinder. To recover the Klein bottle glue the cylinder to an anulus but identifying the two boundary circles in different ways. We can understand all cases in terms of these two simple models. When the neighborhood of X is a mobius strip then S is the connect sum of a closed orientable surface with RP^1, and when the neighborhood of X is an annulus then S is the connect sum of a closed orientable surface with the Klein bottle. (You should object at this point that I haven’t showed that said orientable surfaces are connected, but it’s easy to see that if X disconected the surface than it would be a boundary and thus trivial in H_1.)

Let’s get a fun corollary from this. Consider S = RP^2 # RP^2. This is a nonorientable surface of Euler characteristic 0. In particular, it must be the connect sum of a orientable surface with Euler characteristic 2 with the Klein bottle! Thus, if we believe the 2-dimensional Poincare conjecture (that any homology 2-sphere is the 2-sphere) we conclude that RP^2 # RP^2 is the Klein bottle. Similarly, if we know the classification of orientable surfaces of genus 0, then we can prove that RP^2 # RP^2 # RP^2 = T^2 # RP^2. Thus, we’ve reduce the classification of all surfaces to the classification of orientable surfaces.

(This is a little silly as one can prove that RP^2 # RP^2 is the Klein bottle directly without using algebraic topology, and in particular without the 2d Poincare conjecture.)

Algebraic topology methods as above can almost classify orientable surfaces as well. Suppose that S is an orientable surface with Euler characteristic other than 2. Take a curve X that represents a nontrivial class in H_1 (here’s where we use the assumption on Euler characteristic). Remove a neighborhood of X and cap off the two boundary components. This operation increases Euler characteristic by 2, and keeps things connected. Undoing this operation is just taking connect sum with a torus. By induction we see that any orientable surface is the connect sum of a homology 2-sphere and a bunch of tori. Thus we’ve reduced the entire classification of surfaces (both orientable and nonorientable) to the 2-dimensional Poincare conjecture. Anyone know an algebraic topology proof of this?

3 thoughts on “Classifying nonorientable surfaces

  1. Hey Noah,

    You’re going to have a hard time if you want to stay completely in the realm of algebraic topology. Here are a couple sketches of how one might approach this…

    The easiest proof I know uses morse theory. You pick a morse function and this decomposes your surface into m 0-handles, n 1-handles and p 2-handles. Since you know the euler characteristic is 2, we have m + p = 2 + n. The idea is then to simplify the handle decomposition using “handle slides” and “birth/death” cancellation. You can immediately cancel (m-1) of the 1-handles using (m-1) of the 0-handles. Then, since you know your manifold is orientable, it is not to hard to show you can use (p-1) 2-handles to cancel the rest of 1-handles. What you have left is a surface made by attaching a single 0-handle to a single 2-handle, i.e. it is made by gluing two disks together.

    Careful! we’re not done yet! Now we have to show that any way you glue two disks together you get the regular old 2-sphere. This is actually a serious issue because in higher dimensions if you just arbitrarily glue disks together you’ll get an exotic sphere! Here however we’re in luck. When you glue two manifolds together along a diffeomorphism of their boundaries the result only depends on the isotopy class of the diffeomorphism, i.e. the connected component of Diff(B). Here we’re gluing two disks together along thier boundary, the circle. You can show without too much trouble that Diff(circle) has exactly two components: orientation preserving and orientation reversing. Using either of these will give you a 2-sphere. See also wikipedia on exotic spheres.

    You can probably get a similar result using just algebraic topology. Since our surface is oriented (and connected) and the euler characteristic is 2, we know the dimensions of the homology groups: 1, 0, 1 in dimensions 0,1,2. Thus if we remove a disk we’d like to say what we’re left with is a disk. Then we’re back to gluing two disks together. As I see it, this involves showing two things: (1) after removing a disk you’re left with a conected contractible manifold with boundary (2) such a manifold is the disk. You might laugh, but these are actually serious problems. Let’s look at (2) for example. In dimensions larger than 2, it is FALSE. Look up Whitehead manifold on wikipedia. In fact J. H. C. Whitehead ran into the same problem when he was trying to solve the Poincare conjecture! In dimension 2, however, we’re saved. With effort we can use the Reimann mapping theorem or the Jordan–Schönflies theorem. However we all know the proofs of these theorems are non-trivial.

    The first step (1) is also non-trivial. We’re done if we can show that the fundamental group is trivial. We know the first homology is trivial, so all we seem to know is that the abelianization of the fundamental group is trivial. It could be that we are unlucky and we’ve found a so-called homology sphere, a manifold which has the homology of a sphere but whose fundamental group can be a perfect group. These show up starting in dimension 3.

    I think you can still use algebraic topology to argue in dimension 2 that this can’t happen unless the fundamental group is trivial. What I have in mind would go something like: choose a CW structure (with only one 0-cell). For such CW-complexs with only 1-cells and 2-cells there is a nice way to relate this to generators and relations of the fundamental group. The fact that you have a surface gives you constraints on what these can be. For example every edge must be contained in exactly two faces. Now you try to use these constraints and go from there.

    ok. That’s probably more then anyone wants to know about this. If anyone thinks of a purely algebraic topology proof, I’d be interested to hear it.

  2. I am really sorry for posting such an irrelevant and perhaps annoying question here but can you recommend me a comprehensive source about differential geometry on non-orientable surfaces.

  3. It doesn’t quite make sense to expect an algebraic topology proof of the 2D Poincare conjecture, because it’s not an algebraic topology fact! Even the higher-genus case has a geometric step: You have to know that non-trivial cycles can be cleaned up to non-separating circles.

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