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If $S$ is a surface which is the complement of finitely many points in a compact surface, and the metric in $S$ is complete, then is Gauss-Bonnet theorem still valid for $S$?

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Perfect, thanks! – hao Oct 24 '10 at 15:48


Take the sphere $S^2$ and remove one point, giving you a surface $S$ diffeomorphic to $\mathbb{R}^2$. So we can equip $S$ with a metric making it isometric to $\mathbb{R}^2$, which is certainly complete. $\mathbb{R}^2$ has zero curvature but its Euler characteristic is $1$, so $$ 0 = \int_{\mathbb{R}^2} K dVol \ne \chi(\mathbb{R}^2) = 1.$$

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Yes, I think I should add one more condition, $S$ has finite area, then it will be true, as shown by Max – hao Oct 24 '10 at 15:47
@hao: What is an example of a complete, connected, noncompact Riemannian manifold of finite area? I'm having trouble thinking of one. – Nate Eldredge Oct 24 '10 at 22:38
If a small disk in the sphere is replaced by a 'spike' extending indefinitely we can get such a surface. I.e., something like half of the Tractricoid, which has finite area. This is the image invoked by the short paper that Max linked to. – yasmar Oct 24 '10 at 23:24
@yasmar: Thanks. I was trying to think of something like that, but I kept coming up with instead. – Nate Eldredge Oct 25 '10 at 16:48


If you remove a finite set of points from a compact surface, it will no longer be complete. If you change the metric so that you do have a complete surface, you cannot apply the Gauss Bonnet theorem anyway, because the surface isn't compact.

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You can't apply the Gauss-Bonnet theorem, but its conclusion could still hold. – Nate Eldredge Oct 24 '10 at 22:35
@Nate Eldredge Yes, as the nice two page paper that Max linked to shows. I was clearly too quick to blow the question off. – yasmar Oct 24 '10 at 23:15

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