Past exam of the geometry of surfaces course of the University of Oxford / 2023 / 1 / d

The uniformization theorem says that every simply connected Riemann surface is biholomorphic to the Riemann sphere, the complex plane, or the unit disc. Its consequence for a compact Riemann surface of genus at least two is that the universal cover is the unit disc, equivalently the hyperbolic plane.

To explain the restriction, a spherical universal cover would make the fundamental group finite. Since a finite covering map multiplies Euler characteristic, $2=m(2-2g)$ is impossible for $g\ge 2$. If the universal cover were the complex plane, its deck transformations would be biholomorphisms $z\mapsto az+b$. A nonidentity such map with $a\ne 1$ has a fixed point, impossible for a deck transformation. Thus the deck transformation group would consist of translations. A discrete subgroup of translations with compact quotient space is a rank-two Euclidean lattice, whose quotient space is a torus, again incompatible with $g\ge 2$.

On the unit disc, the Poincare metric on a Riemann surfacehas Gaussian curvature $-1$ and is invariant under every biholomorphism of the unit disc. It therefore descends through the deck transformation group to a hyperbolic metric on $X$. Its local expression is a positive multiple of $|dz{|}^2$ in every holomorphic coordinate chart, so it is compatible with the given complex structure. In fact, this hyperbolic metric is unique within that conformal class.