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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 $S$?

3 Answers 3

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No.

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.2010-10-24
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    @Nate Eldredge Yes, as the nice two page paper that Max linked to shows. I was clearly too quick to blow the question off.2010-10-24
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No.

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 Max2010-10-24
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    @hao: What is an example of a complete, connected, noncompact Riemannian manifold of finite area? I'm having trouble thinking of one.2010-10-24
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    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](http://en.wikipedia.org/wiki/Pseudosphere), which has finite area. This is the image invoked by the short paper that Max linked to.2010-10-24
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    @yasmar: Thanks. I was trying to think of something like that, but I kept coming up with http://en.wikipedia.org/wiki/Gabriel%27s_Horn instead.2010-10-25
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Sort of.

http://www.ams.org/proc/1982-086-01/S0002-9939-1982-0663893-8/S0002-9939-1982-0663893-8.pdf