# Riemann surface with punctures corresponds to a hyperbolic surface with cusps

I am reading a paper on Riemann surfaces and the author used the fact that

$\{$Riemann surfaces with genus $g$ and $n$ punctures$\}$ is in one-to-one correspondence with $\{$ hyperbolic surfaces with genus $g$ and $n$ cusps $\}$. The paper says to equip a Riemann surface with a complete hyperbolic metric, we need to move punctures to infinity. How do we do that?

Can someone help me understanding why this is true, please?

• Hey, do you remember which paper it was? – draks ... Feb 28 '16 at 15:04
• I don't remember but if I find it, I'll let you know. – yaa09d Feb 29 '16 at 9:45
• I am not able to comment since reputation is low. How about infinitely many punctures? Also, given a compact Riemann surface with boundary, we take punctures both at the boundary and in the interior, can we do the uniformization so that the geodesic curvature of the boundary of the resulting surface is zero? – Xiaoxiang Chai Oct 30 '17 at 21:19
• P.S. The name of the paper is Moduli spaces of hyperbolic surfaces and their Weil–Petersson volumes by Norman Do – yaa09d May 1 '19 at 22:04

First suppose that $n=0$ on a genus $g \geq 2$ surface. Then the surface can be endowed with a hyperbolic metric (this is essentially the uniformization theorem). Note that $g \geq 2$ is necessary, e.g. because of Gauss-Bonnet theorem.