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 $S$.

  • $\begingroup$ So, the pull-back of the metric has the same curvature as the original metric. I thought this was not the case in general for a conformal map, that was my concern... I'm probably just a bit confused. Is it true? $\endgroup$ – Second Apr 21 '13 at 21:14
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    $\begingroup$ @Second: See my edit. $\endgroup$ – Christian Blatter Apr 22 '13 at 8:38
  • $\begingroup$ @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. $\endgroup$ – Second Apr 22 '13 at 12:15
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    $\begingroup$ @Second: Yes. Any Riemann surface can be equipped with a constant curvature metric. $\endgroup$ – Christian Blatter Apr 22 '13 at 16:55

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