The fibers of the momentum map for the $SO(n+1)$ symmetry of the geodesic flow on $S^n$

Question

My question is: Are the orbits of the geodesic flow on $S^n$ determined as the fibers of the momentum map for its $SO(n+1)$ symmetry?

I started by considering the analog problem for the orbits of the hamiltonian flow of the n-dimensional harmonic oscillator and the fibers of the momentum map for its $U(n)$-symmetry. If I am not wrong in such a case the answer is yes.

Below I give some other details, hoping to be sufficiently clear.
Any kind of correction and/or suggestion is really welcome.

The n-dimensional harmonic oscillator Let us consider the n-dimensional isotropic harmonic oscillator as the hamiltonian system on $\mathbb{C}^n$ with the symplectic form $\omega=\sum_{k=1}^n d\overline{z}^k\wedge dz^k$ and the Hamilton function $H(z)=1/2|z|^2$.
The natural action of $U(n)$ leaves $H$ invariant and has an equivariant momentum map given by $\langle J,A\rangle(z)=1/2\sqrt{-1}\langle z,Az\rangle$ for $z\in\mathbb{C}^n,\ A\in\mathfrak{u}(n)$.
The fibers of $J$ are exactly the orbits of the hamiltonian flow, infact $J^{-1}(J(z))=S^1.z\equiv\{e^{i\phi}z|\phi\in\mathbb{R}\}$.

The geodesic flow on $S^n$ We can consider the geodesic flow on $S^n$ as the hamiltonian system on $TS^n\equiv\{(x,y)\in T\mathbb{R}^{n+1}\equiv\mathbb{R}^{n+1}\times\mathbb{R}^{n+1}||x|=1,\langle x,y\rangle=0\}$ with the symplectic form induced on it by the canonical symplectic form $\sum_{k=1}^{n+1} dy^k\wedge dx^k$ on $T\mathbb{R}^{n+1}\equiv\mathbb{R}^{n+1}\times\mathbb{R}^{n+1}$, and the Hamilton function $H(x,y)=\frac{1}{2}|y|^2$.
$TS^n$ is invariant under the lifting to $T\mathbb{R}^{n+1}$ of the natural action of $SO(n+1)$ on $\mathbb{R}^{n+1}$.
This action leaves $H$ invariant and has an equivariant momentum map given by $\langle J,A\rangle(x,y)=\frac{1}{2}(\langle x,Ay\rangle-\langle y,Ax\rangle)$ for $(x,y)\in TS^n,\ A\in\mathfrak{so}(n+1)$.

On $T^{\times}S^n\equiv(TS^n)\setminus S^n$, the punctured tangent space to $S^n$, the momentum map has constant rank $2n-1$. So, for any $(x,y)\in T^{\times}S^n$, we have that $J^{-1}(J(x,y))$ is a $1$-dimensional submanifold and includes $\{(e^{t.J(x,y)}x,e^{t.J(x,y)}y)|t\in\mathbb{R}\}$, the orbit of the hamiltonian flow through $(x,y)$, but I don't know if this is its only component.

My question amounts to: Is $J^{-1}(J(x,y))$ exactly equal to $\{(e^{t.J(x,y)}x,e^{t.J(x,y)}y)|t\in\mathbb{R}\}$? or not? Where $J:T^{\times}S^n\to\mathbb{so}(n+1)^\ast\cong\mathfrak{so}(n+1)$ is given by $J(x,y)=\frac{1}{2}(y^Tx-x^Ty)$, (with the linear isomorphism $\mathbb{so}(n+1)^\ast\cong\mathfrak{so}(n+1)$ realized through the scalar product trace).

Edit Now I have realized that there is an easy positive answer to my question and I have posted it below. Again any kind of comment is welcome.

Answer 1

You may want to look into the notion of a dual pair in symplectic and Poisson geometry to place your examples in a more general context. Your first example is one of the canonical examples of a dual pair. Weinstein defined dual pairs in symplectic/Poisson geometry as a symplectic analogue of Howe's dual pairs important in representation theory. (Look up `dual pair' in wiki.) The special case I recall is a pair of Hamiltonian group actions on a fixed symplectic manifold with the property that the reduced spaces for one group are co-adjoint orbits for the OTHER group's lie algebra. In symplectic terms the components of one group's momentum map generate the invariants for the other group's action.

Answer 2

I have found that the answer to my question is yes. Excuse me, having found it, now it is easy, but I did not realized it earlier.

Let be given an element $(x,y)$ of $T^{\times}S^n$, i.e. $(x,y)\in T\mathbb{R}^{n+1}$ such that $|x|=1$, $\langle x,y\rangle=0$, $|y|\neq 0$.

We have that $2J(x,y)=(y^Tx-x^Ty)\in\mathfrak{so}(n+1)$ annihilates the ortogonal complement of $\mathrm{span}\{x,y\}$ and maps $x$ in $y$, and $y$ in $-|y|^2x$.
So $e^{2t.J(x,y)}\in SO(n+1)$, fixes pointwise $(\mathrm{span}\{x,y\})^\perp$, and maps $x$ in $\cos(t|y|)x+\sin(t|y|)|y|^{-1}y$, and $y$ in $-\sin(t|y|)x+cos(t|y|)|y|^{-1}y$.

Therefore (*) $e^{2t.J(x,y)}$, periodic function of $t\in\mathbb{R}$ with period $2\pi|y|^{-1}$, constitues the subgroup of $SO(n+1)$ which stabilizes pointwise $(\mathrm{span}\{x,y\})^\perp$ in $\mathbb{R}^{n+1}.

Now let be given $(x,y),(x',y')\in T^{\times}S^n$ such that $J(x,y)=J(x',y')$.
By (*) we have that $\mathrm{span}\{x,y\}=\mathrm{span}\{x',y'\}$ and $|y|=|y'|$.
This implies that $\{x,|y|^{-1}y\}$ and $\{x',|y|^{-1}y'\}$ are two ordered orthonormal basis of the same plane in $\mathbb{R}^{n+1}$ and have the same orientation, so there exists $t\in\mathbb{R}$ such that $x'=e^{tJ(x,y)}x$, $y'=e^{t.J(x,y)}y$.

This proves that $J^{-1}(J(x,y))=\{e^{tJ(x,y)}.(x,y)|\ t\in\mathbb{R}\}$, for any $(x,y)\in T^{\times}S^n$.