# Geodesics and spontaneously broken symmetry

I have a spacetime $(M,g)$ and a set of discrete symmetries $S_\alpha: M\to M$. Given two fixed points $p_0,p_1\in M$ with $S_\alpha(p_i)=p_i$, I want to argue that a geodesic connecting the two points consists itself only of fixed points.

Obviously, this is not true in general: For instance, take rotations of the 2-sphere as symmetry that leaves the two poles invariant. The geodesics are the great circles and all of them have the same length and they get mapped into each other. However, if we consider the full 3d-space with rotations, then the shortest distance is actually invariant under rotations, because it is the straight line connecting the two poles.

In some sense, the symmetry is spontenously broken in the first case. Is there a way to distinguish the two cases? In particular, if I find a geodesic that is invariant under the symmetry transformations, is there a general argument that other ones that do not respect this symmetry should be longer (only locally shortest geodesics).

The examples given in the question indeed (and in a strict sense) correspond to spontaneously broken and unbroken rotation symmetry around the $z$ axis as stated in the question and as will be clarified in the following.
In the case of (non-relativistic) geodesic motion (of a massive particle), the Hamiltonian can be taken as: $$H = \frac{m}{2}g_{ij}(x)p^ip^j$$ ($g$ is the metric, $x$ is the position, $p$ is the momentum) This Hamiltonian is invariant under any rotation about the $z$ axis for both a round two-sphere and the embedding three space. (The question mentions a discrete symmetry which can be taken any discrete subgroup of the $U(1)$ group of rotations around the $z$ axis, but the discussion is valid for the whole continuous $U(1)$ group of rotations around the $z$ axis) . To see that, it is sufficient to prove that the Hamiltonian Poisson commutes with the generator of $z$ rotations: $$L_z = p_x y - p_y x$$ (In the sphere case, the components are not independent).
In the sphere case there is no possible initial momentum invariant under the rotation about the $z$ axis, since the momentum is tangent to the sphere, therefore parallel to the $x-y$ plane.