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 $0$, 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.

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    $\begingroup$ 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. $\endgroup$
    – Dan Rust
    Feb 5, 2013 at 12:22
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    $\begingroup$ I would go on by fixing a loop, considering the disk it bounds and also considering its complement.. $\endgroup$
    – Berci
    Feb 5, 2013 at 12:45
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    $\begingroup$ 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. $\endgroup$ Feb 5, 2013 at 13:51
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    $\begingroup$ 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. $\endgroup$
    – Sam
    Feb 14, 2013 at 21:20
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    $\begingroup$ @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 $\endgroup$ May 12, 2013 at 11:43

3 Answers 3


The following proof is a bit over the top, but here it is:

  1. Prove Gauss' isothermal coordinates theorem (every 2-dimensional Riemannian manifold is conformally flat). As a corollary, conclude that every oriented 2-dimensional Riemannian manifold $(M,g)$ has a natural structure of a Riemann surface (the holomorphic atlas is given by charts $\phi: U\subset {\mathbb C} \to V\subset M$ such that $\phi^*(g)= \varphi(z)|dz|^2$, $\varphi(z)>0$).

  2. Prove the Riemann-Roch theorem for compact Riemann surfaces (Theorem 16.9 in [1]). Here the genus of the Riemann surface $X$ is defined as the dimension of $H^1(X, {\mathcal O})$, where ${\mathcal O}$ is the sheaf of holomorphic functions on $X$. Then prove that the genus equals half of 1st Betti number (Theorem 19.14 in [1].) This step one can, in principle, shorten by proving that if $X$ is a simply-connected Riemann surface, $H^1(X, {\mathcal O})=0$: Use the Dolbeault isomorphism and the fact that every holomorphic 1-form $\omega$ on a simply-connected surface $X$ is exact, $\omega= df$, $f=\int_{z_0}^z \omega$. (See [1], Theorem 15.14.)

  3. Use the Riemann-Roch theorem to prove that if $X$ is a Riemann surface of has genus $0$ then $X$ is biholomorphic to $CP^1$ as it admits a degree 1 holomorphic map to it. (Corollary 16.13 in [1].)

  4. Lastly, if $M$ is a smooth compact simply-connected surface, it hs to be oriented; picking an arbitrary Riemannian metric on $M$ converts it to a Riemann surface $X$ as above. By Hurewitz theorem, $b_1(M)=0$. Since $X$ is biholomorphic to $S^2$, $M$ is diffeomorphic to $S^2$.

Needless to say, this proof is much more complicated than the standard one (still, simpler than the one via the Ricci flow, although both are analytical in nature).

[1] O.Forster, "Lectures on Riemann surfaces." Springer Verlag, 1981.


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! :)

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    $\begingroup$ 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 $2$. $\endgroup$
    – Sam
    Feb 7, 2013 at 10:28
  • $\begingroup$ 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? $\endgroup$
    – Lor
    Feb 7, 2013 at 12:19
  • $\begingroup$ 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. $\endgroup$ Feb 7, 2013 at 15:19
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    $\begingroup$ 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. $\endgroup$ May 10, 2013 at 13:38

As suggested by Dylan, you can do Morse theory.

Lemma. On a connected surface, there is a Morse function with exactly one local maximum and one local minimum.

For a proof, see e.g. this post. In fact, we only need half of the lemma, but the other half simplifies the proof slightly.

Now, take such a Morse function and obtain a handlebody decomposition of the surface $S$. Start with a disk coming from the only point of index $0$, and start attaching $1$-handles. Homotopically, you start with a point and attach circles, to get a bouquet of circles. Thus, after attaching $k$ $1$-handles you get something with first homology group equal to $\mathbb{Z}^k$. Attaching the final $2$-handle, which is a disk, may ''kill'' a subgroup of first homology generated by one element, but if $k>1$, we will not get trivial homology at the end. But $\pi_1 = 0$ implies $H_1 = 0$, contradiction.

The remaining cases are $k=0$ and $k=1$. For $k=0$, you get a sphere. For $k=1$, you have two ways to attach the $1$-handle. Either you get an unorientable surface (and then $\pi_1 \neq 0$, as you observed), or you get two boundary components which can't be capped off with one disk.

The sad thing about this answer is that the classification of surfaces does not ask for much more than the proof above. But the same things is deeply satisfying: while classification of surfaces may sound like a large and difficult theorem, it is in fact rather easy (I could risk saying that it is not harder than the problem you stated).


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