Take $G := \{Id,h,\ldots,h^{q-1}\}$ the group of isometries of $S^3$ seen as subset of $\mathbb{C}^2$, where each $$h(z_1,z_2) := (\exp({\frac{2\pi i}{q})z_1},\exp({\frac{2\pi ir}{q})z_2}),$$ $(z_1,z_2) \in S^3, \gcd{(q,r) = 1}, q,r\in \mathbb{Z}.$
We can show that $S^3$ is the covering space of $S^3\big/G$ and with the covering metric, these spaces are locally isometric. In particular, the geodesics of $S^3\big/G$ are closed since they are the projections of the great circles in $S^3.$ However, it seems that the geodesics on the quotient does not have the same length. Can anyone show me that this is true?
Here we are considering $S^3$ with the induced metric.
Thanks.
Fix two pt $\overline{p},\ \overline{q}\in X$ where $$\pi: S^3\rightarrow X:= S^3/G,\ \pi (p)=\overline{p}$$
If $\overline{p},\ \overline{q}$ are close then there is unique geodesic $\overline{c}$ joining them in $X$ If $c$ is a geodesic which is a lifting of $ \overline{c}$ then it is also $unique$ In further $c$ can be uniquely extended into a great circle $C$
Since $\pi$ is locally isometric so $\pi (C)$ is a closed smooth geodesic That is any geodesic in $X$ is a closed smooth geodesic If $G=\mathbb{Z}_3$, and $$ g\cdot (Z_1,Z_2)=(gZ_1,gZ_2),\ g\in G$$ then $\{(e^{it},0) | t\in [0,2\pi] \}$ is projected into a closed smooth geodesic of length $\frac{2\pi}{3}$
Consider great circle which is union of $$ (\cos\ t,\sin\ t),\ (-\sin\ t,\cos\ t),\ (-\cos\ t,-\sin\ t),\ (\sin\ t,-\cos\ t),\ t\in [0,\frac{\pi}{2}]$$ This its projection is a closed smooth geodesic whose length $2\pi$ in $X$