Jun 08

Dealing In Triangles Part Deux: Revenge Of The Sugarman

When last we left my humble post, I’d attempted to do the student community a favor by making more widely known the short proof by Moran and Doyle of the existence of a triangulation for a compact 2 manifold. Feeling proud of the service I performed, I went back to my studies and pathetic excuse for a life until my friend JS (his name is being withheld to protect the innocent from harm-namely ME) brought up during a discussion about a possible student conference we’re discussing for Queens College that the matter was quite a bit murkier then it initially looked. Quote from the math society message board:
That's interesting. I was confused because I heard that all surfaces are compact and I believed it without looking at any proofs. Also when I looked at it on wikipedia, they were unclear as to whether or not the surfaces had to be compact.I imagine that since every 2-manifold has a differentiable structure, and every manifold that has a differentiable structure has a piecewise linear triangulation you should be able to triangulate every surface regardless of whether or not it is compact.I could be wrong about a lot of things though, since I don't know the proofs for any of these things and I haven't even glanced at the papers. It would be awesome if you could clarify all this stuff for me and even more awesome if you could talk about it in August.

Leave it to JS to point out the gaping hole in the ceiling. He was right,though-I’d seen Rado’s proof stated for noncompact and compact surfaces, something clearly wasn’t clear here. Does an arbitrary surface admit a triangulation or not? It turns out it depends on what you call a surface-and mathematicians are not of a single mind on this.
This is the definition of a manifold I was always taught: An n dimensional manifold is a Hausdorff topological space that is locally homeomorphic to an open subset of R n . That definition clearly does NOT require n dimensional manifolds to be compact: consider the very simple example of a hyperplane in R n . This is clearly a manifold of dimension n-1, but it’s not compact. It turns out the necessary condition for 2-manifolds to be triangulable is that they must be 2nd countable (i.e. have a countable basis). This in turn implies that the spaces are separable (i.e. has a countable dense subset). If the surface is 2nd countable, then every surface is triangulable by the aforementioned VERY lengthy proof by Rado. ( Yes, to make sure, I DID read Rado’s original proof. There is an excellent statement and discussion of the proof in Zieschang, Vogt and Coldewey’s Surfaces And Planar Discontinuous Groups, Springer-Verlag, 1980). The proof hinges on a decomposition of the surface into disjoint closed regions which are then separated using carefully selected arcs and the Jordan-Shoenflies theorem. It turns out without second countability, there are counterexamples to this construction, namely the so-called Prufer surface, which is a seperable but NOT second countable and therefore non-triangulable complex surface. (Note that for a manifold, second countability implies separability but not vice versa.) A terrific discussion of this counterexample can be found in the paper linked HERE:
(It turns out this surface is also nonmetrizable ,which has deep consequences for both the general theory of Riemann surfaces and homotopy group of an associated CW complex, namely it doesn’t HAVE one. )
So all we have to do is assume all manifolds are second countable and end of problem, right? WRONG. It turns out differential geometers always define manifolds to be second countable partly to avoid this problem and algebraic topologist and algebraic geometers usually drop the condition since the counterexamples such as a Prufer surface is useful to them. (I’m too damn tired at this point to answer why, just trust me, they like the wacko manifolds……….) So as usual in mathematics-context, context ,context. This is why we need to DEFINE everything carefully people. Since most of us are a lot more interested in the geometry of manifolds then their topology per se-I’m inclined to give them all countable bases and be done with it.
By the way, the geometers have found there’s an even better reason to assume all manifolds are 2nd countable spaces: Since manifolds that are not 2nd countable are not necessarily seperable, the result is that the smooth structure on said manifold MAY NOT BE UNIQUE. That kind of ruins your day if you plan to do differential geometry on them, doesn’t it………….?
Off to bed. Discuss this among yourselves and no need to thank me………………..

Leave a Reply