The set of leaves of the distribution $D$ on coadjoint orbit $O_{\mu}$ Let $G$ be a compact connected Lie group and $O_{\mu}$ be a coadjoint orbit where $\mu\in  \mathfrak{g}^*$ and  $\mathfrak{g}^*$ is the dual of the Lie algebra of $\mathfrak{g}=\mathrm{Lie}(G)$.  Let $P$ be an invariant polarization of the coadjoint orbit $O_{\mu}$ and $D=P\cap\bar P\cap TO_μ$ (here $\bar P$ i.e. complex conjugate of $P$  ) so how can we find the set of leaves of the distribution $D$ in  $O_{\mu}$  .i.e. How can we find $O_{\mu}/D$?
ps: I need this for metaplectic correction. This question come from metaplectic correction on coadjoint orbits
Note that a polarization of the coadjoint orbit $G/G_\mu$, is given by the left invariant extension of complex Lie subalgebra $\mathfrak p\subset \mathfrak g^{\mathbb C}$ with the properties

*

*$\mu^{\mathbb C}([\mathfrak p,\mathfrak p])=0$


*$\dim \mathfrak g/\mathfrak g_\mu=\dim \mathfrak g^{\mathbb C}/\mathfrak p$.


*$\mathfrak g_\mu\subset\mathfrak p$


*$(\mathfrak p\oplus\bar{\mathfrak p})\cap \mathfrak g$ is a Lie subalgebra of $\mathfrak g$.
If we take $\mathfrak d=\mathfrak g\cap \mathfrak p$ then in complex polarization $\mathfrak d=\mathfrak g_\mu$
 A: The leaves of $D$ are points and the leaf space $O_\mu/D$ is $O_\mu$ itself. 
Indeed, a $G$-invariant polarization $P$ (involutive lagrangian subbundle of $(TO_\mu){}^\mathbf C$) is determined by the preimage $\mathfrak p\subset\mathfrak g^\mathbf C$ of its value $P_\mu\subset (T_\mu O_\mu){}^\mathbf C=(\mathfrak g/\mathfrak g_\mu){}^\mathbf C$. Likewise your $D$ is the $G$-invariant distribution whose value at $\mu$ is the subspace $D_\mu=\mathfrak d/\mathfrak g_\mu$ of $T_\mu O_\mu=\mathfrak g/\mathfrak g_\mu$, where $\mathfrak d=\mathfrak p\cap\bar{\mathfrak p}\cap\mathfrak g$.
Now in your case ($G$ compact) the possible $\mathfrak p$ are known (parabolics containing $\mathfrak g_\mu^\mathbf C$) and all complex, i.e. $\mathfrak p\cap\bar{\mathfrak p}\cap\mathfrak g=\mathfrak g_\mu$. (E.g., if $G=\mathrm{SU}(2)$ and $O_\mu=S^2$ then $\mathfrak p$ and $\bar{\mathfrak p}$ are opposite Borels (upper and lower triangulars in $\mathfrak{sl}(2,\mathbf C)$) intersecting in the diagonals.) So we have $\mathfrak d/\mathfrak g_\mu=\{0\}$ and hence $D=\{0\}$.
