Let's assume we are in $\mathbb{R}^3$ and we have a $f(x,y) \in C^\infty$. Then a regular surface is given by $$M:=\{(x,y,z,)\in\mathbb{R}^3 | f(x,y)=z\},$$ and clearly we have a local parametrisation given by $$X(u,v) := (u,v,f(u,v)).$$

I now want to calculate the integral over the Gaussian curvature using the Gauss-Bonnet theorem.

My idea is that $$F(u,v,w):=X(u,v)+we_3$$ is a diffeomorphism between the $xy-plane$ and $M$ so the Euler characteristic of $M$ should be $$\chi(M)=2.$$

Using now the global Gauss-Bonnet theorem we get that for all $M$ induced by such an $f$ the integral over the Gaussian curvature satisfies $$\int_M KdA = 2\pi\chi(M)=4\pi.$$

But this seems strange to me because if we take for example the cone induced for $$f(x,y)=\sqrt{x^2+y^2},~~~and~~~z>0$$ $M$ should be flat and therefore the integral should vanish.

I am thankful for every helping hand.


1 Answer 1


There are at least two issues with your cone example:

  1. $M$ is only a regular submanifold if zero is a regular value of the level set function $g(x,y,z) = f(x,y)-z$. This is not the case for the cone since $g$ is not differentiable at the cone point $(0,0,0)$. This is not a fatal issue, since one might imagine "rounding off" the cone's tip; however

  2. Gauss-Bonnet holds only for compact surfaces, which the cone is not. You could restrict yourself to a graph over some compact region of the plane, in which case Gauss-Bonnet will hold, though you will need to include the geodesic curvature terms that arise at the boundary of the domain.

  • 1
    $\begingroup$ +1. Of course, even if you round off the cone point, there will be some Gaussian curvature there that you need to account for. $\endgroup$ Oct 23, 2017 at 3:05
  • $\begingroup$ okay, lets forget about the cone example. What about the rest of the proof? This would still be an interessting result for me. Could there be a Problem with the $F$? My idea to show that it is a diffeomorphism was with the invers function theorem. But this only gives us a local diffeomorphism. Is the Euler characteristic still preserved? $\endgroup$
    – Bara
    Oct 23, 2017 at 7:36
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    $\begingroup$ @Bara No, there is a fundamental problem here: any graph over the entire plane will not be compact. As I mention above, probably the simplest correction is to restrict $f$ to a bounded region of the plane, but then you will still need to account for the boundary curvature. $\endgroup$
    – user7530
    Oct 23, 2017 at 8:04
  • $\begingroup$ oh, I see what you mean. So if we restrict $f$, in a way that $M$ is bounded. Why is there boundary curvature popping up? Is $M$ in this case not compact since it is closed and bounded? $\endgroup$
    – Bara
    Oct 23, 2017 at 8:10

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