Let $S^{2n+1}$ be the $m$-dimensional sphere in $\mathbb{C}^{n+1}$. Endow $S^{2n+1}$ with the standard metric. Let $S^1$ act by multiplication on $S^{2n+1}$. Then $S^1$ and the canonical action of $SU(n+1)$ are by isometries. Since the actions of $S^1$ and $S^{2n+1}$ commute, $G = SU(n+1) \times S^1$ acts by isometries on $S^{2n+1}$. Lifting the action to $TS^{2n+1}$ we get, that this action is hamiltonian. 


In their paper "New examples of manifolds with completely integrable geodesic flows", Paternain and Spatzier say, that $G$ acts on $TS^{2n+1}$ multiplicity-free/coisotropic.


That means: 


1)
Their exists an $\operatorname{Ad}^\*\_G$-equivariant momentum map 
$$\Phi \colon TS^{2n+1} \to \mathfrak{g}^\*$$
2)
For $\alpha \in \mathfrak{g}^\*$ the isotropygroup $G\_\alpha$ acts transitively on the connected components of $\Phi^{-1}(\alpha)$.


I'm not sure how to show now, that this action is really multiplicity-free/coisotropic, because I don't know how to calculate the momentum map and the preimage effectively.



coisotropic action on $TS^{2n+1}$
1) Since $S^{2n+1}$ is a Riemannian manifold one can identify $TS^{2n+1}$ with the cotangent bundle $T^\*S^{2n+1}$. The latter is well-known to carry a canonical symplectic structure and a momentum map.


2) This condition can be rephrased as $\Phi/G:T^\*S^{2n+1}/G\to\mathfrak{g}^\*/G$ being finite to one. Let $G\_0\cong SU(n)\times S^1$ be the isotropy group of a point $x\in S^{2n+1}$. Then $T^\*S^{2n+1}/G=T^\*\_xS^{2n+1}/G\_0=V/G\_0$ where $V=\mathbb{C}^n\oplus \mathbb{R}$. Hence $V/G^0=\mathbb{R}\_{\ge0}\times \mathbb{R}$. Now the first factor is mapped to the quotient of $SU(n+1)$ while the second to $(Lie S^1)^\*$. This shows that $\Phi/G$ is finite to one.


Remark: More generally, the cotangent bundle $T^\*X$ of a homogeneous space $X=G/H$ is the fiber product $G\times^H\mathfrak{h}^\perp$ where $\mathfrak{h}^\perp\subseteq\mathfrak{g}^\*$ is the annihilator and the moment map is $[g,\xi]\mapsto Ad^\*(g)\xi$. So multiplicity-freeness means that $\mathfrak{h}^\perp/H\to\mathfrak{g}^\*/G$ is finite to one.


