Let $S \subset \mathbb{R}^3$ be given by $$ S = \{ (x,y,z) \in \mathbb{R}^3 \, : \, x^2+y^2=z^2 , \quad y \ge0, 0\le z \le 1 \}. $$ Verify the Gauss Bonnet Theorem, by computing $\int_S K\, dA$ and $\int_{\partial S} \kappa_g \, ds $.

Computation of $\int_S K \, dA$: The surface (minus the pointy end) is a local isometry of a plane so the Gauss curvature $K$ is everywhere $0$, so this integral contributes nothing. I think this can also be found easily via $K = (LN-M^2)/(EG-F^2) = 0 $ via parametrizing the half-cone.

Computation of $\int_{\partial S} \kappa_g \, ds $: The meridians of the cone are geodesics and have zero geodesic curvature, so the meridians don't contribute. The other boundary of the half-cone is the semi-circle parametrized by $\alpha(t) = (\cos(t), \sin(t), 1)$ for $ t \in (0, \pi)$. To find the geodesic curvature $\kappa_g$ on the cone, one computes the Gauss map $N \circ \alpha(t) = (\cos(t), \sin(t), -1)/\sqrt{2}$ and then use the formula $$ \kappa_g = \frac{1}{|| \alpha'||^3} (\alpha' \times \alpha'') \cdot (N \circ(\alpha)) = \frac{-1}{\sqrt{2}} $$ Note: I am aware more careful analysis of tangent direction is required to determine the sign - but I'm quite sure it's $1/\sqrt{2}$ up to $\pm$. The length of the semi-circle is $\pi$, making this integral $\pm \pi/\sqrt{2}$.

And then finally the Euler Characteristic of the half-cone is $1$. (Homeomorphic to a disc).

But then it doesn't add up with the Gauss Bonnet theorem: $$ \int_S K \, dA + \int_{\partial S} \kappa_g \, ds + \sum_{i=1}^n \theta_i = 2 \pi \chi(S) $$ $$ \implies 0 \pm \frac{\pi}{\sqrt{2}} + \sum_{i=1}^n \theta_i = 2 \pi$$

And I can't see any way the exterior angles (given by $\sum_{i=1}^n \theta_i$) can make this work. By my computation, we have three right-angles, giving this sum to be $3 \pi /2$ which can't cancel the $1/\sqrt{2}$.

There must be an error somewhere. Any insight or help is greatly appreciated!

  • 3
    $\begingroup$ the surface is not smooth. There is some curvature hidden in the origin of the cone. $\endgroup$
    – user8268
    May 4, 2021 at 21:02
  • 3
    $\begingroup$ If you're interested in exploring this phenomenon more deeply, do exercise 13 on pp. 90-91 of my undergraduate differential geometry text, freely linked in my profile. $\endgroup$ May 4, 2021 at 22:40

1 Answer 1


The calculations $K=0$, $\kappa_g = \pi/\sqrt{2}$ and Euler Characteristic $1$ are correct. The subtlety comes down to the fact that the surface is not smooth at the vertex. Therefore, the three angles are not all $\pi/2$. When 'unfolded', the cone actually has two angles of $\pi/2$ but a third angle of $\pi - \pi/\sqrt{2}$ coming from the vertex.

Now, when you use the Gauss Bonnet Theorem with these results, everything works.


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