# Compact surface with Gaussian curvature is positive, negative, and zero

I have a question regarding an exercise of do Carmo, Differential geometry, p. 282:

Let $S$ be a regular, compact, orientable surface which is not homeomorphic to a sphere. Prove that there are points on $S$ where the Gaussian curvature is positive, negative, and zero.

I think a torus could be an example, but that is, of course, no proof. Any ideas? Thanks!

(Strictly speaking we need $S$ to be connected as well, otherwise the disjoint union of two spheres provides a counterexample.)
"Homeomorphic" is a tipoff that we have to look for a connection between topology and geometry. Let's start with the Gauss-Bonnet theorem: the integral of the total curvature is equal to $2\pi$ times the Euler characteristic.
• thank you very much for the answer! you mean the integral $\int_S K dA$, right? can you explain what it means? if i get your answer right it is something like a sum of all curvatures, that would be why we need points with gaussian curvature, when we have points with positive gaussian curvature... – JadonD87 Sep 16 '17 at 14:20
• @JadonD87 indeed, I mean $\int_S KdA$. The Gaussian curvature is a function $K:S\to\Bbb{R}$ and so it can be integrated (i.e. summed) over the surface $S$. Your intuition is correct. – Neal Sep 16 '17 at 14:41