# Uniformization Theorem

I understand this statement of Uniformization Theorem:

Every simply connected Riemann surface is conformally equivalent to the unit disk, the complex plane, or the Rieman $n$ sphere.

However, the following statement totally throws me out:

Every connected 2-manifold is diffeomorphic to a quotient of one of the three constant curvature model surfaces listed above by a discrete group of isometries acting freely and properly discontinuously. Therefore, every connected 2-manifold has a complete Riemannian metric with constant Gaussian curvature.

From the context, "the three constant curvature model surfaces listed above" are zero curvature, $1/R^2$ and $-1/R^2$.

Could anyone explain how these two are equivalent? I am particularly lost at the terms "a quotient of" and "a discrete group of isometries acting freely and properly discontinuously".

Thanks!

## 1 Answer

Riemann surfacehttp://en.wikipedia.org/wiki/Riemann_surface is different from Riemannian manifoldhttp://en.wikipedia.org/wiki/Riemannian_manifold.

Riemann surface is 1-(complex)dimensional complex manifold. Riemannian 2-manifold is 2-(real)dimensional smooth manifold with a Riemannian metric.