Angle between geodesics at different fixed points of a Riemannian isometry

Question

Suppose I have an isometry $f$ of a Riemannian manifold $(M,g)$. Suppose further that $p$ and $q$ are fixed points of $f$. If $\gamma$ is a geodesic segment from $p$ to $q$, then so is $f(\gamma)$.

Let $\theta_p$ be the angle between $\gamma$ and $f(\gamma)$ at $p$ and $\theta_q$ be the angle between $\gamma$ and $f(\gamma)$ at $q$. Must we always get $\theta_p=\pm\theta_q$?

I’ve been trying to come up with a simple counterexample, but my inability to find one is starting to convince me that it might actually be true.

Answer 1

No. There is in fact a very tangible model that gives a counterexample.

1. Take a piece of paper. Draw a straight line parallel to one of the edges.
2. Roll the paper up into a cone, with the vertex on the edge that you chose.
3. The straight line that you drew earlier is now a geodesic $\gamma$ on your Riemannian manifold (the cone).
4. $\gamma$ can be observed to intersect itself once, call the point $p$. On one side $\gamma$ forms a loop. On the other side of $p$ the two branches of $\gamma$ fly off to infinity. Forget about those two branches.
5. Start from $p$, travel along the loop part of $\gamma$. Call the half-way point $q$.
6. Your cone has a reflection symmetry that preserves the straight lines that connect $p$ to the vertex, and from vertex to $q$ that swaps the two halves. Call this isometry $f$.
7. $\gamma$ and $f(\gamma)$ are parallel at $q$. They are not at $p$ (with angle related to the defect angle of your cone).

If you don't have a piece of paper handy, here's what the result looks like:

N.B.: You can make the manifold smooth and complete by cutting off the vertex and replace by a small smooth cap. You can make it compact by putting a cap on the other end. In both cases you can make it so that the new pieces also respect the reflection symmetry.