Two geodesics cannot form a simple region

Question

Suppose S is an orientable surface with nonpositive Gaussian curvature. How can I prove that two geodesics that start from the same point $p\in S$ cannot meet again at another point $q\in S$ such that their traces form the boundary of a simple region in $S$?

Answer 1

Think of the universal cover $\mathbb{H}^2$ of $S$. Then the two geodesics will be lifted on geodesics in $\mathbb{H}^2$, and we can fit the starting point of the two lifting. Then we can see that as the two geodesics in $S$ are homotopic, the end points of the two lifted geodesics coincide. But in $\mathbb{H}^2$ this cannot happen, as Hadamard's theorem says exponential map is diffeomorphism.

Answer 2

Have a look at Gauss-Bonnet:

$$ \int_G K \mathrm{dvol}_G + \int_{\partial G} \kappa_G (t) dt + \sum_i \alpha_i = 2 \pi \chi (G) \quad (1)$$ where $K$ is the Gaussian curvature, $\kappa_G$ the geodesical curvature and $\alpha_i$ the angle at the $i$th corner of the polygone in question.

For your problem, let us suppose suche a simple region $G$ exists. The Euler charactersitic of $G$ is $\chi (G) = 1$. The geodesic curvature $\kappa_G$ vanishes, as the bondary of $G$ is composed of geodesics and you have two corners, what means the sum of all angles in the corners is less than $2\pi$: $\sum_i \alpha_i < 2 \pi$. Finally as your surface has non-positive curvature, the integral over the Gaussian curvature is less or equal zero: $\int_G K \mathrm{dvol}_G \leq 0$. Summing everything up you see that the LHS of (1) $< 2 \pi$ whereas the RHS of (1) $= 2 \pi$ which is clearly a contradiction.