Universal covering is holomorphically isomorphic to the complex plane

Question

I apologize for the vagueness of this question:

How would one show that a Riemann surface $(X,g),$ endowed with a Riemannian metric $g,$ is conformally equivalent to the complex plane? (i.e. is there a standard program to prove such a result?)

Heuristically, I assume you would need to show it is a Riemann surface of genus one. The Uniformization Theorem would then imply that the universal cover of the Riemann surface is the complex plane.

Answer 1

A simply connected Riemann surface is not of genus 1, and is generally not conformally equivalent to the complex plane. As you note, this is only sure to be true if the surface is of genus 1.