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
    Commented 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. $\endgroup$ Commented 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.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .