Constant Curvature Metric and Biholomorphic Equivalence This is probably a dumb question, but let's try it anyway.
I know two versions of the uniformization theorem, and I don't understand their equivalence. The first says that every Riemann surface has a metric under which it has constant Gauss curvature, and the second says that every Riemann surface is biholomorphically equivalent to the plane, the sphere or the disk (or the upper half plane).
Why are these things the same? I have seen several proofs of uniformization, but they use PDE or variational methods in order to prescribe curvature on a surface. I'm not sure I see how this implies anything complex analytic at all though..
It is a known fact that if two surfaces have the same curvature at a point, they are locally isometric. Since the uniform metric gives it constant curvature, it is tempting to say they are globally isometric. I'm not quite sure that's right though. I know that for compact surfaces, I can construct fundamental domains in the disk which after taking quotients by some isometry gives the surface, but this is just another statement of uniformization, if I'm understand correctly.
Can someone help clear up my confusion? How do I get from knowing the curvature is constant to a biholomorhic equivalence between the two?
 A: I want to start by clarifying your statement a little. The metric statement of the uniformization theorem is, to be precise, "Every conformal class of metrics contains a complete metric of constant scalar curvature." The holomorphic statement is "Every simply connected Riemann surface is biholomorphic to the unit disc, plane, or Riemann sphere."
Now for why these are equivalent.
1) Riemann surfaces are the same thing as smooth, oriented surfaces with a given conformal structure. (Use the multiplication-by-$i$ map on the tangent spaces to define an orthogonal basis and write the conformal structure; conversely the conformal structure and orientation let you define a multiplication-by-$i$ map, which comes from a complex structure by the existence of isothermal coordinates). Holomorphic maps with nonzero derivative are the same thing as conformal maps (maps that preserve the conformal structure), and hence biholomorphisms are the same thing as conformal isometries.
2) Let $\Sigma$ be a Riemann surface, with conformal class $[g]$. Suppose $g' \in [g]$ has constant scalar curvature and is complete. Pass to the universal cover $\tilde \Sigma$; $g'$ pulls back to a complete metric on this, again with constant scalar curvature. Now invoke the following famous theorem.
Theorem (Killing-Hopf): a simply connected complete surface with constant scalar curvature is isometric to either the 2-sphere with its round metric; the plane with its Euclidean metric; or the disc with its hyperbolic metric.
Thus $\tilde \Sigma$ is conformally isometric to one of the three examples; hence is biholomorphic to it.
3) Conversely, suppose we have the holomorphic statement. Going to the metric statement is a little more subtle, since the holomorphic statement only talks about simply connected things. But the proof goes as follows. Pass to $\tilde \Sigma$; we know it's either the sphere, plane, or disc. We know that $\Sigma$ is a quotient of this by a group of oriented conformal isometries, but we want to quotient by an honest group of isometries to get a constant scalar curvature metric in the quotient. a) The sphere's oriented conformal isometries all fix some point, hence do not act freely. So if $\tilde \Sigma = S^2$, $\Sigma = S^2$. b) Conformal isometries of the plane are of the form $f(z)=az+b$. It is trivial to see that these have fixed points iff $a \neq 1$; so we see that the conformal isometries without fixed points are actual, honest isometries, and we get $\Sigma$ as an isometric quotient of the plane. c) Oriented isometries of the disc (actually, better to think of it as the upper half plane here) form the group $PSL_2 \Bbb R$; the oriented conformal isometry group is larger, the group $GL_2(\Bbb R)/\pm I$. It is a true result, that I forget how to prove and forget a reference for, that any group $G \subset GL_2(\Bbb R)/\pm I$ that acts without fixed points is conjugate to a subgroup of $PSL_2 \Bbb R$, hence proving the theorem.

I tend to think that part of uniformization's power and beauty is the way it links these two different statements: one is about curvature, and the other is about complex geometry! So I think a good understanding of what's going on here is key to appreciating the theorem.
