# A question on the uniformization theorem

Wikipedia reads, on the uniformization theorem:

In mathematics, the uniformization theorem says that every simply connected Riemann surface is conformally equivalent to one of the three domains: the open unit disk, the complex plane, or the Riemann sphere. In particular it admits a Riemannian metric of constant curvature.

My question is, does it follow so easily? Let's consider a conformal map $F$ between two Riemannian 2-manifolds $(A,g)$ and $(B,g_1)$, where the notation means (manifold, metric). If we consider a conformally related metric to $g_1$, say $g_2 := e^{2u}g_1$, where $u \in C^\infty(B)$, the map $F$ remains conformal, but the curvature of $g_2$ is changed! I would put it like that: curvature is not a conformal invariant. I think you need a little bit of effort to achieve the statement in bold. What am I missing?

A "naked" Riemann surface $S$ carries no metric, and therefore doesn't have a curvature either. It is just a two-dimensional manifold provided with a so-called conformal structure. This structure is encoded in the local charts $z_\alpha:\ U_\alpha\to{\mathbb C}$ which are related by conformal maps among each other.
But given any Riemann surface $S$ with local coordinate patches $(U_\alpha,z_\alpha)_{\alpha\in I}$ you can define on $S$ various Riemannian metrics $g$ compatible with the given conformal structure. In terms of the local coordinates $z_\alpha$ these metrics appear in the form $ds^2=g_\alpha(z_\alpha)|dz_\alpha|^2$. The statement in bold says that if $S$ is simply connected you can choose the $(g_\alpha)_{\alpha\in I}$ in such a way that the resulting Riemannian manifold $(S,g)$ has constant curvature $-1$, $0$, or $1$. Just transport the well known constant curvature metrics on $D$, ${\mathbb C}$, or $S^2$ via the map guaranteed by the uniformization theorem to your surface $S$.
• @CristianBlatter: so, if I understand it correctly, the uniformization theorem only provides the compatibility of the pull-back metric with the conformal structure already present on $S$. I guess the same reasoning would prove that, given an oriented Riemannian 2-manifold $(M,g)$, there is always a conformally related metric to $g$, say $g'$, of constant curvature. This is because $g$ induces a complex structure (not trivially, though!) and, for Riemannian metrics, being compatible with the same complex structure means being conformally related. – Second Apr 22 '13 at 12:15