Minimal geodesics in $S^{n+1}$

Question

Let $\Omega^d(M)$ the space of minimal geodesics on a smooth manifold $M$. How can I prove that if $M= S^{n+1}$, $\Omega^d(S^{n+1}) \simeq S^n$?

Answer 1

Arguing by symmetry, we can classify all geodesics of $M = S^{n+1}$.

Claim: Every geodesic of $M=S^{n+1}$ is a constant velocity parametrization of $P\cap M$, where $P\subseteq\mathbb{R}^{n+2}$ is a $2$-plane passing from the origin.

Proof: Notice that $M$ is invariant under the action of $O(n+2)$ on $\mathbb{R}^{n+2}$. Let $x\in M$ be any point, $\gamma:\mathbb{R}\rightarrow M$ a (maximally extended) geodesic with $\gamma(0) = x$. Let $$P=span\left\{\mathbf{x},\frac{d\gamma}{d t}(0)\right\}$$ where $\mathbf{x}$ denotes the vector going from the origin to $x\in M$. Assume that the trace of $\gamma$ is not contained in $P$. Let $S\in O(n+2)$ be the reflection on $P$. Then $\gamma'=P\circ\gamma$ is again a geodesic with $\frac{d\gamma'}{d t}(0) = \frac{d\gamma}{d t}(0)$ and $\gamma'\neq \gamma$, which is a contradiction to the uniqueness of geodesics with given boundary conditions.

$\square$

I hope this helps. If you would define what a "minimal" geodesic is, then I might be able to help you with the rest of your problem.

Answer 2

If the letter "d" in your notation $\Omega^d(M)$ is referring to a basepoint $d$ (thought of as the north pole), then the space of closed geodesics (for example with the compact-open topology of maps from the circle to the sphere) starting from $d$ is clearly identifiable with a sphere of one dimension lower, because a directed geodesic is uniquely determined by a point on the equatorial hypersphere.