# Uniformization theorem and metrics on Riemann surfaces

I know next to nothing about Riemannian geometry, and have some trouble interpreting some things I read in relation to the uniformization theorem for Riemann surfaces.

To me the uniformization theorem says that the universal cover of any Riemann surfaces is biholomorphic to either $\mathbb{C}P^1$, $\mathbb{C}$ or $D$, the open unit disk with its complex structure given by that induced from $\mathbb{C}$(?).

In particular, for $S$ of genus $g>1$ we have a universal cover $D$, so $$\pi:D\to S$$ Now I do not see what this has to do with a metric of constant negative curvature? $D$ is just an open subset of $\mathbb{C}$, and so it seems to me that it carries a flat metric? I just don't see where in this story a metric of negative curvature shows up.

• Its well known that the disc $D$ is a model of non-euclidean geometry, "lines" are circles orthogonal to the boundary. This this is probably what is behind whatever you are talking about. – Rene Schipperus Mar 10 '17 at 12:27
• This is the discovery that Poincare was very proud of: Conformal automorphisms of the unit disk $D$ are precisely the orientation-preserving isometries of the hyperbolic metric on $D$. Can you now conclude that every Riemannian metric on $S$ is conformal to a hyperbolic metric? – Moishe Kohan Mar 10 '17 at 14:05
• @MoisheCohen Using that theorem I can proof that the metric on $D$ descents to $S$, so $S$ carries a constant $-1$ curvature metric. How do I show that this metric is in the same conformal class defined by the complex structure though? – user2520938 Mar 10 '17 at 19:33
• @MoisheCohen Essentially I do not see how to relate the hyperbolic metric on $D$ to the conformal structure on $S$? – user2520938 Mar 10 '17 at 20:34
• Do you know how to show that the hyperbolic metric on the unit disk is conformal to the Euclidean metric on this disk? – Moishe Kohan Mar 10 '17 at 20:44