geodesics in Lens spaces

Question

For which Lens spaces, all the simple closed geodesics have the same length?

Is it just $S^n$ and $S^n/\mathbb{Z}_2$ or there are more?

Answer 1

this is true for all spherical spaceforms. given two non-antipodal points on the sphere, the geodesic circle passing through them is unique. therefore given any point $p\in\mathbb S^n$ with $n>1$ at most finitely many geodesic circles through $p$ can project to closed geodesics of length less than $2\pi$ (or $\pi$ if the group contains $-Id$). On the other hand there will obviously be some shorter closed geodesics in the quotient too.

Answer 2

Consider for a moment the $n=3$ case and the $(p,q)$ action of $\mathbb Z_p$ on $S^3$. A geodesic of $S^3$ is a great circle, meaning the intersection of a linear $2$-dimensional subspace of $\mathbb R^4$ with $S^3$. An isometry preserves a great circle gives you a complex eigenspace of your isometry (this is Hopf's argument, it can be found in Thurston's "Three-Dimensional Geometry and Topology" -- its the standard argument used in into ODEs courses to go between complex eigenvalues and real periodic solutions).

But if we fix the action ahead of time so that the eigenspaces are $\mathbb R^2 \times \{0\}^2$ and $\{0\}^2 \times \mathbb R^2$ respectively that means the only great circles fixed by the action are those two -- unless of course we're in the situation where the eigenvalues agree -- but that's the case of $S^3$ and $\mathbb RP^3$.

I think this argument works in all dimensions. Of course there's the degenerate case of the 1-dimensional lens spaces $S^1 / \mathbb Z_n$.