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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!

Dover87
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1 Answers1

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(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.

Neal
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  • Since the surface is compact, there is at least one umbilic point: Is this a well-known theorem? – anderstood Sep 16 '17 at 14:18
  • 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... – Dover87 Sep 16 '17 at 14:20
  • @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. – Neal Sep 16 '17 at 14:40
  • @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
  • so we look - so to speak - at every point of S? thats ok with my intuition, but i thought we just look at the boundary according to stokes' theorem, that was what i thought in class. but you can not use this intuition with the boundary of $S$ right? because in this case we wouldn't know if there are positive or negative gaussian curvatures of points on the surface – Dover87 Sep 16 '17 at 14:58
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    Of course, you mean the Intermediate Value Theorem, as opposed to the Mean Value Theorem. – Ted Shifrin Sep 17 '17 at 05:14
  • @TedShifrin these aren't the droids you're looking for – Neal Sep 17 '17 at 14:08
  • @Neal, I understood your argument, but I have a doubt when you said "Gauss curvature takes on negative values". The Gauss Curvature can be $K \leq 0$. Thus, we could have just parabolic, planar and elliptic point (we have this last kind of point by the fact $S$ is compact), but how you can ensure that $K$ have hyperbolic points. I can see that if we have hyperbolic and elliptic points, then we have points where $K = 0$ by the continuity of $K$, but how you ensure that $K$ have hyperbolic points? – George Nov 23 '17 at 09:51
  • @George Think of it this way. Gauss-Bonnet guarantees that $\int K \leq 0$. If $\int K < 0$ then surely there exist points where $K < 0$. The other case is $\int K = 0$ Because $K$ is continuous, if there exists $p$ such that $K(p) > 0$, then there exists some small neighborhood $U$ of $p$ such that $K > 0$ on $U$. So compute $0 = \int_S K = \int_U K + \int_{S - U} K$ and because $\int_U K > 0$ we must have that there are hyperbolic points in $S - U$. – Neal Nov 25 '17 at 14:33
  • @Neal, I understood, thanks a lot! – George Nov 26 '17 at 21:07