# Riemannian metric for surface of negative Euler characteristic

So, I want to equip a surface of negative Euler characteristic with a Riemannian metric of negative curvature.

I know from the uniformization theorem, that a metric of constant curvature exists

Now, if M is compact (for example a sphere with a finite number > 2 of puncture points):

-I know from Gauss-Bonnet, that this metric has to be of negative curvature

If M is not compact:

-the curvature can't be $> 0$ because the universal covering of M is not the sphere (because the sphere is compact)

Is there a way to rule out the possibility that the metric is of constant curvature = $0$ (flat)

• Something's fishy here. Any meaning I can construe of "not compact" will not lead to a universal covering possibly being the sphere. So, what precisely do you mean by a (non-compact) surface of negative Euler characteristic? – Ted Shifrin May 29 '13 at 20:18
• Ah, yeah, I meant that we can rule out the possibilty that M has positive curvature, because its universal covering IS NOT the sphere. I edited the question. I hope it's a clearer now :) – Richard2264 May 29 '13 at 20:25
• It's still not right. If you have a punctured sphere, it's no longer compact. Handles, ok ... I think you're misinterpreting the uniformization theorem; would you please give me a precise, complete statement? And am I correct that you're thinking only of manifolds without boundary? – Ted Shifrin May 29 '13 at 20:37
• Ok, let me try it again: I want to show, that there is a metric of constant negative curvature on, let's say the sphere with 4 points removed. As I see it, the uniformization theorem implies, that there exists a complete metric of constant curvature on this surface. This metric could in principle be of positive curvature or flat and I want to eliminate these 2 possibilities, ending up with a metric of negative curvature. – Richard2264 May 29 '13 at 20:53
• Now, if our surface had a metric of constant positive curvature, its universal cover would be the sphere. This can't be true for obvious reasons. Hence we are left open with two possibilities for the metric: constant negative curvature or flatness. – Richard2264 May 29 '13 at 20:55

## 1 Answer

The Gauss-Bonnet theorem applies equally well in the case of a complete metric of finite area. Therefore negativity of Euler characteristic implies negative sign of curvature.

• Are you saying that, if you have a complete metric of finite area on such an open surface, then Gauss-Bonnet applies? Is it even clear, that such a metric of finite area exists? – Carl123235 May 31 '13 at 16:46
• Concerning your first question: think of cutting off the cusp at infinity, and applying Gauss-Bonnet to the resulting surface with a bunch of boundary circles. Then note that the boundary contribution tends to zero as the circle recedes to infinity. – Mikhail Katz Jun 2 '13 at 7:20
• As for the second question: I assume the surface has finite topological type to begin with. Then such a metric always exists. For infinite topological type things are more complicated. I can look up some references if you are interested. – Mikhail Katz Jun 2 '13 at 7:21