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?

Thanks for your help!

  • $\begingroup$ @Landscape, actually, in general, $|\theta_j|\le\pi$. Here we have $\theta_j<\pi$ because these are geodesic segments. Sam, your argument is good. $\endgroup$ May 15 '13 at 15:34
  • $\begingroup$ @TedShifrin: My mistake. Thank you. $\endgroup$
    – 23rd
    May 15 '13 at 16:05
  • $\begingroup$ @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. :) $\endgroup$
    – Sam
    May 15 '13 at 17:35

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.