I heard a theorem in differential geometry course.
State of the theorem is
"There is no closed (regular) surface having only negative Gaussian curvature."
I tried to prove the theorem using Gauss-Bonnet theorem, but coudn't have any progress.
How can I get proof?
+) I guess that above theorem is also true when the word 'negative Gaussian curvature' is replaced with 'nonpositive Gaussian curvature'. Is this right?
Actually there are such surfaces in every genus $g=2,3,4,\ldots$ and in fact loads of them.
The hypothesis that seems to be missing is that the surface is embedded in Euclidean $3$-space. Then indeed you can't have nonpositive curvature. The simplest argument to prove this is to use compactness to find a point on the surface at maximal distance from the origin. Then you can show that Gaussian curvature must be positive at such a point.
Here the Gauss-Bonnet theorem you mentioned does not help because as I mentioned such surfaces do exist if you allow higher-dimensional Euclidean spaces.
