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.

Because our surface is not a sphere, its Euler characteristic is nonpositive. Since the surface is compact, there is at least one point with all positive principal curvatures, so its Gauss curvature is positive in at least a small open set. It cannot be positive everywhere because otherwise its integral would be positive, in contradiction to Gauss-Bonnet. Thus the Gauss curvature takes on negative values. By the intermediate value theorem, there must exist a set where the Gauss curvature is equal to zero.

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  • $\begingroup$ Since the surface is compact, there is at least one umbilic point: Is this a well-known theorem? $\endgroup$ – anderstood Sep 16 '17 at 14:18
  • $\begingroup$ 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... $\endgroup$ – JadonD87 Sep 16 '17 at 14:20
  • $\begingroup$ @anderstood I think I may have misused "umbilic point" to mean "point with all positive principal curvatures," but Wikipedia informs me my memory of the definition is incorrect. I'll change that. $\endgroup$ – Neal Sep 16 '17 at 14:40
  • $\begingroup$ @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. $\endgroup$ – Neal Sep 16 '17 at 14:41
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    $\begingroup$ Of course, you mean the Intermediate Value Theorem, as opposed to the Mean Value Theorem. $\endgroup$ – Ted Shifrin Sep 17 '17 at 5:14

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