Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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.

share|improve this question
add comment

2 Answers

up vote 2 down vote accepted

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.

share|improve this answer
    
Dear Richard Montgomery, thanks for your answer. I have only a rough knowledge of dual pairs, but I am interested to learn more about them, (by the way I posted a previous question requesting for reference). So, if it is possible, could you give me some suggestion for further reading about dual pairs with a focus on concrete examples of interest for mechanics? Anyway thanks again. –  Giuseppe Tortorella Jul 6 '11 at 18:11
add comment

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 $(x,y)\in T^{\times}S^n$.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.