If this has been answered already, please let me know and I'll delete the question.
ADDED: I'd prefer to assume a smooth structure, rather than a triangulation.
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asked Apr 14, 2010 at 14:26
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5$\begingroup$ I gave a list of various proofs in my answer to this question : mathoverflow.net/questions/20438/… $\endgroup$– Andy PutmanCommented Apr 14, 2010 at 14:34
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3$\begingroup$ If you want to start with a smooth structure, then use Morse theory. It's easy -- all you have to check is that if you start with a (possibly disconnected) cpt orientable surface with bdry that's covered by the classification and attach a k-handle for k=0,1,2, then you get a surface covered by the classification (a Morse fcn gives a handle decomp of your mnfld). Adding a 0-handle is taking the disjoint union with a disc and adding a 2-handle is gluing a disc to a bdry cpt, so these cases are easy. For a 1-handle, once you draw the picture you'll see that there are 3 special cases to check. $\endgroup$– Andy PutmanCommented Apr 14, 2010 at 15:21
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$\begingroup$ @Andy: Thanks! The proof in Hirsch's book, Differential Topology, looks like what I want. $\endgroup$– Deane YangCommented Apr 14, 2010 at 15:22
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$\begingroup$ I like the good old fashioned combinatorial method of Rado.Yes,it's tedious,but it has the advantage of being constructive and highly visual. $\endgroup$– The MathemagicianCommented Apr 14, 2010 at 17:02
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$\begingroup$ @Andrew L - All of these methods are constructive and visual; it just depends on how they are presented. In any case, if you like the combinatorial approaches, then I recommend you check out Conway's "ZIP proof", which is combinatorial but even more constructive and visual than the old-fashioned one. $\endgroup$– Andy PutmanCommented Apr 14, 2010 at 17:06
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4 Answers
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A very simple and low tech solution to this problem is the very first proof, given by Möbius in 1863. He assumes that the surface is smoothly embedded in $\mathbb{R}^3$, and slices it by a family of parallel planes. Assuming that the orientation of the planes is general and that they are sufficiently close together, this cuts the surface into simple pieces -- either disks, annuli, or pairs of pants.
It is then quite easy to show that the result of assembling such pieces is always a sphere with handles.
answered Apr 15, 2010 at 10:41
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Count the number of disjoint non-separating embedded circles.
answered Apr 14, 2010 at 14:57
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$\begingroup$ Oh, man. That sounds good to me. But I'm an aging geometric analyst short on both time and ability to think geometrically (I prefer formulas and PDE's). Any chance you could add a few more sentences? Or provide a reference? $\endgroup$ Commented Apr 14, 2010 at 15:11
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$\begingroup$ A smooth closed orientable (connected) 2-manifold S is determined, up to diffeomorphism, by its genus: the number of disjoint embedded curves needed to cut S into a planar surface. The two-sphere is planar, by convention. <p> So the sphere has genus zero. the torus has genus one, and so on... One way to think of this: the genus is the number of "handles" the surface has. A suitcase with g handles has genus g. $\endgroup$– Sam NeadCommented Apr 14, 2010 at 15:38
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$\begingroup$ I think the OP wanted a PROOF of the classification, not merely an invariant distinguishing the different surfaces. And I also think you want your embedded circles to be non-homotopic. $\endgroup$ Commented Apr 14, 2010 at 17:09
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1$\begingroup$ Thanks, Andy, for saying out loud what I didn't have the nerve to ask about. So does this approach lead to a proof or not? $\endgroup$ Commented Apr 14, 2010 at 17:32
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$\begingroup$ If you go back to my list of proofs above, the main point of the proof in Armstrong's "Basic Topology" is to show that if a compact surface has no nonseparating embedded simple closed curves on it, then it is a sphere. However, this takes a fair amount of work. $\endgroup$ Commented Apr 14, 2010 at 17:35
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Consider harmonic functions $f$ with exactly 2 log-singularities of weight $\pm 1.$ (locally $f(z)=a log\parallel z\parallel+g,$ $g$ being smooth at $z=0$, $a$ being the weight) on your compact surface equipped with a Riemann metric. They exists by standard elliptic theory ( the two weights $a_1$ and $a_2$ have two add to zero). Consider $\partial f,$ the complex linear part of the differential. This is a meromorphic section of your canonical bundle. Then $deg\partial f=-\frac{1}{2\pi i}\int KdA,$ as the Levi-Civita connection defines a complex linear connection on the canonical bundle.
This shows that if the total curvature is large enough $\geq 4\pi,$ $f$ will not have critical points (only two singularities). Moreover $e^f$ is the real part of a holomorphic bijection onto $CP^1.$
If $f$ has a critical point, then you can easily construct a non-sepreating loop, as in Morse theoretic proofs. You cut your surface, and add two disc (with the right orientation). One, can easily see, that this must increase the total curvature by $4\pi,$ and you end up with the two-sphere after a finite number of steps.
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Check out this nice paper by Thomassen for a short self-contained proof.
answered Apr 14, 2010 at 15:15