# Universal covering is holomorphically isomorphic to the complex plane

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.

• Your question is very confusing. As stated this is not true. Consider the unit disc with Poincare metric. But then you have written something about genus one only which is the case when your statement is valid. So could you rephrase the question. – tessellation Apr 21 '17 at 2:39

## 1 Answer

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.

• Sorry, I meant multiply connected. I suppose the question is then, given a metric $g,$ how would one show that the Riemann surface $X$ is of genus 1? @Igor Rivin – Sergio Charles Apr 21 '17 at 2:53
• Given that you know your surface is compact (I assume it is), you find the integral of Gauss curvature. If the integral is zero, you are cooking with gas. – Igor Rivin Apr 21 '17 at 2:55