# Another exercise in differential geometry using Gauss Bonnet

Consider the surface $$M=\{(x,y,z) \mid x^2+y^2-z^2=1 \,\text{and} -1 and calculate $$\int_{M}KdA$$ where $$K$$ is the Gaussian curvature.

Solution; Geometrically this is the part of the one-sheeted hyperboloid between $$-1$$ and $$1$$. We simply apply Gauss-Bonnet for compact surfaces and get that $$\int_{M}KdA=0$$ due to the Euler characterisitc being $$F-E+V=2-2+0$$

Is this correct?

• Your intuition should tell you that on this surface $K<0$ everywhere, so could the result be correct? Aug 12 '19 at 16:45
• The surface is not compact. Aug 19 '19 at 15:46

No, this is not correct. At each point of the hyperboloid the gaussian curvature is negative so the integral over the curvature must be negative aswell.

The problem is that $$M$$ is not compact so you cannot use Gauss-Bonnet directly. If you want to use Gauss-Bonnet you could consider $$\bar M=\{(x,y,z) \mid x^2+y^2-z^2=1 \,\text{and} -1\leq z\leq 1\}$$. The integral of the curvature is the same since the difference Between $$\bar M$$ and $$M$$ is a measure zero set.

Now $$\bar M$$ is a compact two-dimensional Riemannian manifold with boundary so by Gauss-Bonnet

$$\int_{ M} K\;dA=\int_{\bar M} K\;dA=2\pi\chi(\bar M)-\int_{\partial \bar M}k_g\;ds=0-\int_{\partial \bar M}k_g$$

so what's left to do is to calculate the geodesic curvature of the boundary.

$$\textbf{Edit:}$$

Let $$S\subseteq \mathbb R^n$$ be a compact two-dimensional Riemannian manifold with boundary. The question is how to compute $$\int_{\partial S}k_g\ ds$$. This can be done as follows:

Let $$B_i$$ be the connected components of the boundary. Then each $$B_i$$ is diffeomorphic to a circle. For fixed $$i$$ choose a regular paramatrization $$\gamma: [a,b]\to B_i$$ (not necessarily unit-speed but atleast $$C^2$$) and set $$T=\gamma'/|\gamma'|$$. Let $$\eta(t)$$ bet the unit tangent vector at $$\gamma(t)$$ which is orthogonal to $$\gamma'(t)$$ and points inwards into the manifold. Then $$\kappa_g\circ\gamma=\frac{1}{|\gamma'|}\langle T',\eta\rangle$$ and $$\int_{ B_i}k_g\ ds=\int_a^b\langle T',\eta\rangle \ dt$$. Finally $$\int_{\partial S}k_g\ ds=\sum_i\int_{ B_i}k_g\ ds$$.

One remark: If $$M\subseteq \mathbb R^3$$ is orientable and $$N$$ is the gauss map (for some chosen orientation) then $$\eta=\pm\ T\times (N\circ\gamma)$$. How to get the sign right? Well let $$x(u,v):W\to M$$ be a regular parametrization arround $$\gamma(t)$$ where $$W\subseteq\mathbb R_{\geq 0}\times\mathbb R$$ is open. Then we have to choose the sign such that $$\langle x_u(\gamma(t)),\eta(t)\rangle>0$$.

Now let's do this for $$\bar M=\{(x,y,z) \mid x^2+y^2-z^2=1, \,-1\leq z\leq 1\}$$. The boundary has two connected components: $$B_1=\{(x,y,z) \mid x^2+y^2-z^2=1, \ z=1\}$$ and $$B_2=\{(x,y,z) \mid x^2+y^2-z^2=1, \ z=-1\}$$.

We can parametrize $$B_1$$ by $$\gamma(t)=(\sqrt 2\cos t,\sqrt 2\sin t,1)$$, $$t\in [0,2\pi]$$. Then $$T=(-\sin t,\cos t,0)$$ and $$T'=-(\cos t, \sin t,0)$$. Now what is $$\eta(t)$$? For fixed $$t$$ consider the curve $$\alpha_t(s)=(s\cos t,s\sin t,\sqrt{s^2-1})$$. Then it geometrically clear that $$\alpha_t$$ is a curve on $$M$$ which runs towards $$B_1$$ and intersects $$B_1$$ orthogonally at $$s=\sqrt2$$ so we should have $$\eta(t)=\frac{-1}{\sqrt3}(\cos t,\sin t,\sqrt 2)$$ (which also can be checked more rigorously e.g. by using polar coordinates near $$B_1$$). Hence

$$\int_{ B_1}k_g\ ds=\int_0^{2\pi}\langle T',\eta\rangle \ dt=\int_0^{2\pi}\frac 1{\sqrt 3} \ dt=\frac{2\pi}{\sqrt 3}$$

In a similar way or using symmetry $$\int_{ B_2}k_g\ ds=\frac{2\pi}{\sqrt 3}$$ so we have $$\int_{\partial \bar M}k_g\ ds=\frac{4\pi}{\sqrt 3}$$ and hence

$$\int_{ M} K\;dA=-\frac{4\pi}{\sqrt 3}$$

• Typo? Next.. how are tangent rotations found? Aug 13 '19 at 0:14
• Id like to see them as well. Also how does one account for two curves which are disjoint enclosing region? Aug 15 '19 at 9:57
• @user1 : I edited my answer to adress the question of how the geodesic curvature of the boundary can be computed. Aug 20 '19 at 13:52