Curvature of a non-compact complete surface

Question

Assume $\Sigma$ is a non compact, complete surface. Assume the integral $$\int_{\Sigma}K$$ is convergent, where K is the Gauss curvature of $\Sigma$. Is it always true that $$\frac{1}{2\pi}\int_{\Sigma}K$$ is an integer?

Answer 1

No. Start with a one-nappe circular cone. Cut off the tip and replace with a subset of a sphere, resulting in what is now a $C^1$ infinite surface. Smooth near the circle where they join, so the result is at least $C^2.$ The result has zero curvature except near the spherical part. Since we did not specify the central angle of the cone, we get whatever we like in the way of total Gauss curvature, up to an effective maximum given by a hemisphere and an infinite cylinder.

Answer 2

Better luck on this for complete minimal surfaces. In that case, $K$ can't change sign and complex variable theory tells you the Gauss map can't miss many values.

Answer 3

I believe that Alexander horned sphere is a good example for as a visual representation for as an answer.

Answer 4

Sure, I am going to have to go with the contrapositive on this one: $\int{\Sigma}$div$\gamma$$\nu$ = $\int{\partial\Sigma}< \nu, N> \sigma$ #. In other words Stokes theorem.