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

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

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

3. $$\mathfrak g_\mu\subset\mathfrak p$$

4. $$(\mathfrak p\oplus\bar{\mathfrak p})\cap \mathfrak g$$ is a Lie subalgebra of $$\mathfrak g$$.

If we take $$\mathfrak d=\mathfrak g\cap \mathfrak p$$ then in complex polarization $$\mathfrak d=\mathfrak g_\mu$$

• Invariant polarizations determined by parabolic Lie sub algebras $\mathfrak{g_{\mu}}\subset \mathfrak {p}$ (where $\mu \in \mathfrak{g}^*$)so why $D$ must be zero?
– user21574
Mar 5 '14 at 15:50
• I still don't undrestand. What do you mean the former is $g_{\mu}$ and the latter is $\{0\}$. You know that for metaplectic correction for inner product we take integral on $M/D$ where $D=P\cap\bar P\cap TM$. So, by your comment, $D=0$ on coadjoint orbit
– user21574
Mar 5 '14 at 16:02
• Your question (what is the leaf space $O_\mu/D$?) is independent of metaplectic considerations. Its answer is that $O_\mu/D$ is $O_\mu$ itself. Mar 6 '14 at 14:00
• Why $O_μ/D$ is $O_μ$ where $D=P\cap\bar P\cap TO_μ$ and $P$ is polarization of coadjoint orbit.
– user21574
Mar 6 '14 at 14:06
• You're welcome. I'll make this an answer so that we can erase this unwieldy discussion if you wish. Mar 6 '14 at 17:00

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 $D=\{0\}$.