Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

From the Classification Theorem for closed (i.e. compact and boundaryless) surfaces, it follows that $S^2$ is the only closed surface with trivial $\pi _1$. That's easy because the fundamental group classifies closed surfaces.

I'd like to get the same conclusion (i.e. that a simply-connected closed surface is isomorphic to $S^2$, in your favorite category) without that theorem. Is it possible? Any discussion about (generalized) Poincaré conjecture I could find starts saying that in dimension $1$ and $2$ it's true because of the classification theorems (available in those dimensions), dimension $3$ was settled by Perelman and then switches to high-dimensional wild cases. No insights on a direct proof in dimension $2$.

My attempt: let $S$ be a closed surface with $\pi_1(S)=1$; trivial $\pi_1$ implies that $S$ is orientable, that $S$ is covered only by itself, that every embedded loop bounds a disk (so $S$ has genus 1, since cutting along every embedded loop disconnects $S$); compactness rules out $\mathbb{R}^2$ and $\partial S = \emptyset$ rules out the unit disk. But here I don't know how to go on. Maybe one should try to find an explicit isomorphism to $S^2$, but I can't see how.

share|improve this question
4  
I kind of feel like any attempt at an alternative proof would just naturally lead to the beginning of the classification theorem. Defining the genus and orientability of a surface just seems so natural. –  Daniel Rust Feb 5 '13 at 12:22
4  
I would go on by fixing a loop, considering the disk it bounds and also considering its complement.. –  Berci Feb 5 '13 at 12:45
4  
I believe I once heard that Ricci flow (which is what Perelman used to settle the dimension 3 case) can also be used to handle the dimension 2 case. Perhaps someone more well versed in Ricci flow can confirm or deny. –  Jason DeVito Feb 5 '13 at 13:51
3  
I think, you could probably solve it "Yamabe-style" by first fixing an arbitrary metric $g_0$ on $M$ and then finding a conformally equivalent metric $g = e^{2u} g_0$ with constant curvature $R$. Since $\chi(M) = 2$, it follows from Gauss-Bonnet that $R>0$. A simply-connected manifold of positive constant curvature is diffeomorphic to a sphere (space form). I hope someone more knowledgeable in these matters can fill in some more detail. I'm unsure about the solvability of the "Yamabe-problem" for $n=2$... The resulting PDE is quite non-linear. –  Sam Feb 14 '13 at 21:20
3  
@Jason DeVito: yes, take any Riemann metric on a compact 2-manifold and Ricci flow will deform it to a constant curvature metric. So the result then follows from the classification of constant-curvature geometries, which is a standard theorem in many intro differential geometry courses. See for example: arxiv.org/abs/1103.4669 –  Ryan Budney May 12 '13 at 11:43

1 Answer 1

EDIT: Of course (as suspected), this argument is wrong. The problem is that I may well have lots more critical points, as long as they are canceled out by critical points of index 1. Some massaging is necessary, but then the proof is less fun. Anyway, I'll leave this here in case anyone can find a Morse theoretic approach. (And make it community wiki so they can add it.)

This might go against the spirit of the question, but it's fun anyway. (Assuming it's correct; I have a tendency to be careless at this hour...)

If you're willing to assume a little Morse theory and that the surfaces are smooth than this is immediate:

Pick a Morse function. The indices of the critical points can be either 0, 1, or 2. But you know by the Morse inequalities that there is at least one critical value of index 0 and one of index 2 (since you know the Betti numbers). Moreover, the Euler characteristic of the surface is 2, so by the Morse inequalities again you actually know that there are precisely two critical points- one of index 0 and another of index 2.

Morse theory (Reeb's theorem) gives us a construction showing that manifolds with Morse functions having exactly two critical points must be homeomorphic to a sphere.

It's plausible that with enough unwraveling the above argument could be made elementary... after all- everything is in 1 and 2 dimensions so one can draw pictures! :)

share|improve this answer
2  
There are Morse functions on the sphere, which have any number of critical points of index $0$ and $2$. For example, take the height function of a sphere deformed to a peanut. This has $2$ critical points of index $2$. –  Sam Feb 7 '13 at 10:28
    
The Morse approach is really nice! So I have at least one $p$ of index $0$ and one $q$ of index $2$, by homological constraints. I can choose two disjoint balls $U_p$ and $U_q$. Let $\gamma$ and $\delta$ be the boundaries of these two balls, so they are embedded loops in $S$. By simply-connectedness of $S$ they have to be (freely-)homotopic. I think we can choose this homotopy to stay outside of $U_p$ and $U_q$. As $\gamma$ is homotoped to $\delta$ it traces out a cylinder, so I get a cylinder+balls=sphere inside $S$; but by connectedness $S$ is exactly this sphere. Am I missing something? –  Lor Feb 7 '13 at 12:19
    
You need some more- the homotopy might not be injective as a map to the sphere, cuz I may have made a bad choice. Maybe you could solve this by taking a flow from an index 0 to an index 2 guy and working from there... Not sure. –  Dylan Wilson Feb 7 '13 at 15:19
1  
Dylan, you can make this kind of argument work but all you'd be doing is giving the standard proof Lor explicitly stated he wants to avoid. –  Ryan Budney May 10 '13 at 13:38

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.