# Geodesics on a Hyperbolic Paraboloid

Let $S \subset\mathbb{R}^3$ be a hyperbolic paraboloid and let $a,b \in S$.

I'm wondering how the Gauss-Bonnet Formula tells us that there is at most one geodesic $\gamma: [0,1] \rightarrow S$ such that $\gamma(0)=a$ and $\gamma(1)=b$.

$$\int_{S}KdA + \int_\gamma K_gds = 2 \pi$$

Any help would be much appreciated!

• The gamma in the formula you're citing corresponds to the boundary of $S$, which I'm pretty sure is empty in this case, so it probably won't help you. – Robin Goodfellow Oct 14 '16 at 14:25

Choose a smooth normal vector field of unit length on the hyperbolic paraboloid and fix positive orientation on it, so that when you look from the tip of the vector down on the surface the positive orientation is counterclockwise. Assume you have two different geodesics $\gamma_1$ and $\gamma_2$ connecting points $a$ and $b$. Then the two geodesics cut out a topological disc $S$, a geodesic bi-gon on the surface. Then, orient these geodesics as follow: $\gamma_1$ from $a$ to $b$ and $\gamma_2$ from $b$ to $a$. Let angle $\alpha > 0$ be the counterclockwise angle at $a$ from $\gamma_1$ to $\gamma_2$ and Let angle $\beta > 0$ be the counterclockwise angle at $b$ from $\gamma_2$ to $\gamma_1$. Then clearly $$\alpha + \beta > 0$$ The geodesic curvature of the closed piecewise geodesic loop $\gamma_1 \cup \gamma_2 = \partial \, S$ is zero everywhere except at the vertices $a$ and $b$ where it is $\pi - \alpha$ at $a$ and $\pi - \beta$ at $b$. Recall that the Gaussian curvature $K$ of the hyperbolic paraboloid is negative! Then by Gauss-Bonnet $$\int_{S} \,K \, dA + \int_{\gamma_1 \cup \gamma_2} k_g = 2 \, \pi \, \chi(S) = 2 \pi$$ so $$\int_{S} \, K \, dA + 2 \pi - \alpha - \beta = 2 \pi$$ But then $$0 > \int_S \, K \, dA = \alpha + \beta$$ which is in a contradiction with the fact that $\alpha+\beta > 0$.

Hyperbolic paraboloid $S$ is given by $$f(x,y)=(x,y,xy)$$

Consider $$(x,0,0),\ (0,y,0)$$ which are geodesics in $\mathbb{R}$ Hence they are geodesics on $S$

In further $$f_x=(1,0,y),\ f_y=(0,1,x),\ E=1+y^2,\ G=1+x^2,\ F=xy$$

\begin{align*} dA&=\sqrt{1+r^2}dxdy\\ \int_S K&= \int \frac{-1}{(1+r^2)^2} \sqrt{1+r^2} dxdy \\&=\int \frac{-1}{(1+r^2)^\frac{3}{2}}\ rdrd\theta =2\pi (1+r^2)^\frac{-1}{2}|_0^\infty=-2\pi\ \ast\end{align*}

Consider a geodesic triangle $\Delta$ : $(t,0,0),\ (0,t,0),\ (0,0,0),\ t>0$

Then $\int_\Delta K +\sum_i \theta_i=2\pi$ so that $$\frac{\pi}{2}<\int_\Delta K+ \pi=\sum_i(\pi-\theta_i)$$

Here first inequality is followed since we have $\int_\Delta K> \frac{1}{4}\int_S K$ by observing integrand in $\ast$ And last term is sum of internal angles of $\Delta$ So we can not have a contradiction

• The hyperbolic paraboloid does not have constant negative curvature. – Ted Shifrin Oct 14 '16 at 21:59
• I see : $z=xy,\ K= \frac{-1}{(1+x^2+y^2)^2}$ Thank you – HK Lee Oct 15 '16 at 0:39