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In topology, one learns how to classify the compact surfaces up to homeomorphism. And in fact, since "homeomorphic" and "diffeomorphic" coincide in dimension 2, we can classify the compact (smooth) surfaces up to diffeomorphism.

This makes me wonder about classifying compact Riemannian 2-manifolds up to isometry. In particular:

Is there a classification of all Riemannian 2-manifolds that are diffeomorphic to the 2-sphere?

I imagine this to be a very difficult question. As such, I have two follow-up questions:

  • If this is a tractable question, how much progress has been made in this direction? What is known and what isn't?

  • If the question is considered too difficult to have a real answer (is this the case?), then I imagine there to be simpler, related questions. What are some examples of these?

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If you replace isomorphic by pointwise-conformal equivalent, then the answer is given by the uniformization theorem of Riemannian surfaces. – Sam Feb 5 '13 at 9:51
Oh, I see... is this why everyone always talks about metrics "within a conformal class"? Because we already have a classification up to pointwise-conformal equivalence? Mmm. – Jesse Madnick Feb 5 '13 at 9:56
The point about conformal equivalence is the following: Any Riemannian metric $g$ induces a complex structure $J$ on $S^2$, by $\omega(\cdot,J\cdot) = g(\cdot,\cdot)$. Here $\omega$ is the volume form of $g$ and $J \in End(TS^2)$ a tensor which induces maps $J_p : T_pS^2 \rightarrow T_pS^2$. It is easy to check that $J^2 = - id$. However it is non-trivial so see that there exists a complex atlas (with holomorphic charts) und which $J$ corresponds to multiplication by $i$. Now, two metrics induce the same complex structures if and only if they are conformal equivalent. – Sam Feb 5 '13 at 14:23

There are a couple of directions to go with your question, but the general feeling I get are that these kinds of questions are considered hopeless. I mean, pick any Riemannian metric on $\mathbb{R}^2$. There is an infinite dimensional space of such metrics. Via a partition of unity, one can view this as the metric chosen on the northern hemisphere of $S^2$.

On the other hand, here are two results which may be to your liking.

First, due to Kazdan and Warner (see the linked lecture notes near the bottom of Kazdan's website) states the following:

Suppose $\kappa: S^2\rightarrow\mathbb{R}$ is any smooth function satisfying $\int_{S^2} \kappa dA = 4\pi$, then there is a Riemannian metric $g$ on $S^2$ with curvature given by $\kappa$.

(By the Gauss-Bonnet theorem, the integral condition is necessary. Kazdan and Warner prove sufficiency.)

Second, due to Weinstein (see here for the original paper):

Every smooth manifold, other than $S^2$ has a metric $g$ and for which, in the tangent space at some point $p$, the cut locus and conjugate locus do not intersect. On the other hand, on $S^2$, for any metric and any point, the cut locus and conjugate locus must intersect.

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