I'm trying to understand some aspects of the geodesics of surfaces without conjugate points and the following question came out: consider the hyperbolic space and two geodesics wich are asymptotic to each other in the "future". It is true that the distance between these two geodesic increases monotonically in the "past" (see the figure).
I'm trying to see if this is true for universal coverings of complete surfaces without conjugate points. To be more precise, consider two geodesics wich are asymptotic to each other on the "future". Is it true that the distance between them does not decrease on the past? I was thinking the following: if it was false, something like the following should happen:
Perhaps it is possible to use some "shortcut" argument to prove this is impossible (if the result I want is true...) but I'm not able to construct "the smart shortcut", perhaps using the fact that the "upper" geodesic approaches a little bit to the "bottom" geodesic before the local maximum as we see in the figure. My idea lies on the fact that the "no conjugate points" conditions implies that on the universal covering every geodesic is minimizing.
Thanks a lot on advance!
I think this is false. The condition of having no conjugate points allows for a moderate amount of positive curvature, which is enough to destroy the monotonicity of distance between geodesics. Here is an idea for a counterexample: let $S$ be the part of the surface of revolution $x^2+y^2=1/(z^2+1)$ lying in the sector $x>0,y>0$. It is simply-connected and has no conjugate points. The curves obtained by intersecting $S$ with planes through the $z$-axis (for example, $y=x$ and $y=2x$) are geodesics. In both directions (up and down) these geodesics are asymptotic to each other. Their mutual distance is maximal when they cross the plane $z=0$. Of course, this surface is not complete. To fix this, multiply the metric by a function like $\phi=\max(y/x, x/y, 3)^2$. This makes the surface complete and preserves the two geodesics mentioned above. One still has to check that multiplication by $\phi$ created no conjugate points, but intuitively this is clear, because $\phi$ introduces a lot more negative curvature toward the edges.
Alternatively, you can modify the Poincaré metric $ds^2=dx^2/(1-|x|^2)^2$ on the unit disk by making it spherical near the center $x=0$. That is, cut out the part $|x|<r$ and replace it by a cap of a sphere of radius slightly greater than $r$. Since the spherical cap is less than half-sphere, it has no conjugate points. Two nearly parallel geodesics entering the spherical part will travel along arcs of great circles, which will create a local maximum of distance near the middle of the cap. It remains to smoothen the edge between spherical cap and the hyperbolic disk, which will not seriously affect the geodesics.