# Using Gauss-Bonnet to prove that geodesics have at most one point of intersection

Given an oriented Riemannian manifold $(M,g)$ of dimension $2$, such that $M$ has negative Gaussian curvature everywhere and $M$ is diffeomorphic to $\mathbb R^2$, I'm looking for a way to show that two geodesics are either disjoint, coincide or intersect in precisely one point, using the Gauss-Bonnet theorem.

Lacking any insight, I'm trying to find a way by contradiction: If there are two distinct geodesics intersecting in two points $p$, $q$, then we obtain a simple curve in $M$, consisting of two geodesic segments, by first following along one geodesic from $p$ to $q$ and then following along the other geodesic back from $q$ to $p$. Since $M\cong \mathbb R^2$, this simple closed curve bounds a simply connected domain $\Omega \subset M$ ($\Omega$ is diffeomorphic to a disk).

Let $\theta_1, \theta_2$ be the exterior angles between the two geodesic segments at the end-points $p$ and $q$. The Gauss-Bonnet theorem applied to this situation yields $$\int_\Omega K + (\theta_1 + \theta_2) = 2\pi \chi(\Omega) = 2\pi.$$ (where the geodesic curvature parts drops out, because we have two geodesic segments). Since $K<0$, this implies that $\theta_1 + \theta_2 > 2\pi$.

Can anyone see how to finish the argument from here? Or maybe someone has a better idea?

• @Landscape, actually, in general, $|\theta_j|\le\pi$. Here we have $\theta_j<\pi$ because these are geodesic segments. Sam, your argument is good. May 15 '13 at 15:34
• @TedShifrin: Ah, of course, we have $\theta \in [-\pi, \pi]$ and not $\theta \in [0,2\pi)$... How silly of me. Alright, thanks a lot for reading through my argument and commenting on it! If you could just put your comment as an answer, then I can accept and this question can be ticked off. :)
In general, at a corner, we have $|\theta_j|\le \pi$. Here we have $|\theta_j|<\pi$ because these are geodesic segments, so $\theta_1+\theta_2<2\pi$. Sam, your argument is good.