Broken geodesics in the hyperbolic plane and bending angles

Question

Let $\gamma$ be an infinite broken geodesic in the hyperbolic plane, that is a curve formed by consecutive geodesic segments. Assume also that each of these segments is longer than a certain positive constant $C$. For any two consecutive geodesic segments $[p,q]$ and $[q,r]$, we can consider the angle $\theta_q$ between the two segments. I would like to say that if these angles are always close enough to $\pi$ (i.e. the line does not bend too much) than the whole broken geodesic does not go back to itself, as opposite to what would happen in Euclidean geometry. I think this could be proved by some explicit hyperbolic geometry computation, but maybe it's a well known fact and somebody could point me to a good reference. Thank you.

Answer 1

I think that your conjecture is correct. I’ll use the Poincaré half-plane model to describe my attempt at a proof.

Take the line $L$ parallel to the real axis a certain height above, I’ll take it one unit up. In hyperbolic geometry, this is a horocycle, a circle tangent to the horizon; in this case the point of tangency is $i\infty$. Now draw a hyperbolic geodesic tangent to $L$ at the point $i$, and on this “straight line” go a distance $C/2$ in either direction. So you have a segment of length $C$, symmetrically placed. Let its endpoints be at $\pm a+bi$, with $0<b<1$ of course. Now shift the segment over by the transformation $z\mapsto z+2a$, a hyperbolic rigid transformation. You get, on the right, another segment of length $C$, also tangent to $L$, and it makes an angle $\theta_C$ with the original segment. (You see that as $C$ is taken smaller, $\theta_C$ will be closer to $\pi$.) Now I say that if you restrict your angles to be at least as close to $\pi$ as $\theta_C$, your broken geodesic will not come back to itself. That should do it.