# Gauss Bonnet Theorem for half a cone

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!

• the surface is not smooth. There is some curvature hidden in the origin of the cone. Commented May 4, 2021 at 21:02
• 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. Commented May 4, 2021 at 22:40

## 1 Answer

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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