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?



multiplicity-free action on $SO(n+1)/SO(n-1)$
Let $G\_\mathbb{C}$ be the complexification of a compact group $G$. Let $H\subseteq G$ be a closed subgroup. Then there is the following criterion: $M=T^\*(G/H)$ is multiplicity-free (as a Hamiltinian manifold) if and only if $X=G\_\mathbb{C}/H\_\mathbb{C}$ is spherical (as a $G\_\mathbb{C}$-variety, i.e. a Borel subgroup $B$ has an open dense orbit). This is a consequence of Brion's criterion.


In your case, $G\_\mathbb{C}=SO(n+1,\mathbb{C})\times SO(2,\mathbb{C})$ and $H\_\mathbb{C}=SO(n-1,\mathbb{C})\times SO(2,\mathbb{C})$ where the second factor is embedded diagonally. Now we take the quotient of $X$ by the central $SO(2,\mathbb{C})$ and obtain a morphism of $X$ to $\overline X:=SO(n+1,\mathbb{C})/H\_\mathbb{C}$. So $B$ has a dense orbit on $X$ if and only if it has one on $\overline X$. But $\overline X$ is spherical since it is a symmetric variety.


