I'm trying to show that the Lie group $G=SO(n+1) \times SO(2)$ acts multiplicity-free on the cotangentbundle $T^* (SO(n+1)/SO(n-1))$.
That means:
1) There exists an $\operatorname{Ad}^*_G$-equivariant momentum map $$\Phi \colon T^* (SO(n+1)/SO(n-1)) \to \mathfrak{g}^*$$ 2) For $\alpha \in \mathfrak{g}^*$ the isotropygroup $G_\alpha=K$ acts transitively on the connected components of $\Phi^{-1}(\alpha)$.
If $\mathfrak{k}$ is the Lie algebra for $K$, $\mathfrak{k}^0$ the annihilator of $\mathfrak{k}$ in $\mathfrak{g}^*$ and $\mathcal{O}$ a $\operatorname{Ad}^*_G$-orbit in $\mathfrak{g}^*$, then this can be rephrased as:
$\mathcal{O} \cap \mathfrak{k}^0$ is a finite union of $\operatorname{Ad}^*_K$-orbits.
Let $SO(n-1)$ be identified as the subgroup of $SO(n+1)$, by the injection $A \mapsto \begin{pmatrix} E_2 & 0 \\ 0 & A \end{pmatrix}$ and $SO(2)$ as a subgroup through $R \mapsto \begin{pmatrix} R & 0 \\ 0 & E_{n-1} \end{pmatrix}$.
If $SO(2)$ induces a left action on $SO(n+1)$ by right-multiplication and $SO(n+1)$ by left-multiplication, then these actions commute and so $G =SO(n+1) \times SO(2)$ acts on $M=SO(n+1)/SO(n-1)$.
Let $q_0 = \begin{pmatrix} E_2 & 0\\ 0 & SO(n-1) \end{pmatrix} \in G$ and denote by $K=G_{q_0}$ the isotropy group for $q_0$. Now I computed the corresponding Lie algebra $$\mathfrak{k} = \left\{ \middle( \begin{pmatrix} X & 0 \\ 0 & Y \end{pmatrix}, \begin{pmatrix} X & 0 \\ 0 & 0 \end{pmatrix}\middle) \middle| X \in so(2), \ Y \in so(n-1) \right\}$$ and the annihilator $\mathfrak{k}^0 \subset \mathfrak{g}^*$ using the identification of $so(n+1)$ with $so(n+1)^*$ by $${so(n+1) \ni Z \mapsto \alpha \in so(n+1)^*}$$ for $\alpha(X) = \operatorname{tr}(ZA), \ \forall A \in so(n+1)$.
So the corresponding set for $\mathfrak{k}^0$ is $$\tilde{\mathfrak{k}} = \left\{\left( \begin{pmatrix} Z_1 & Z_2 \\ -Z_2^T & 0 \end{pmatrix}, -Z_1 \right) \middle| Z_1 \in SO(2), \ Z_2 \in \mathbb{R}^{2 \times (n-1)} \right\} .$$
Since our identification of $so(n+1)$ with $so(n+1)^*$ is $Ad_G$-invariant, we identified the adjoint-action of $G$ on $\mathfrak{g}$ with the coadjoint-action of $G$ on $\mathfrak{g}^*$. With all this in mind we now have to prove that for $\operatorname{Ad}_G$-orbits $\mathcal{O}' \subset \mathfrak{g}$ we have that $\mathcal{O}' \cap \tilde{\mathfrak{k}}$ is the finite union of $K$-orbits.
Now take an element $Z = \left(
\begin{pmatrix}
Z_1 & Z_2 \\
-Z_2^T & 0
\end{pmatrix},
-Z_1 \right) \in \tilde{\mathfrak{k}}$ and let $\mathcal{O}'$ be the $\operatorname{Ad}_G$ orbit through $Z$.
Since $SO(2)$ is an abelian Lie group, we have that each element $X \in \mathcal{O}' \cap \tilde{\mathfrak{k}}$ is of the form
$X =
\left(
\begin{pmatrix}
Z_1 & X_2 \\
-X_2^T & 0
\end{pmatrix}
,
-Z_1 \right)$.
But from here on, I don't know what to do. One idea was to calculate the Eigenvalues of $\begin{pmatrix} Z_1 & X_2 \\ -X_2^T & 0 \end{pmatrix}$ so that I get conditions on the entries of $X_2$, then using this conditions to show that $K$ acts locally transitive on $\mathcal{O}' \cap \tilde{\mathfrak{k}}$ and finally showing that $\mathcal{O}' \cap \tilde{\mathfrak{k}}$ has only finitely many connected components. Would that be the right way, or is there some easier one?
