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 $\mathfrak d=\mathfrak g_\mu$

  • $\begingroup$ 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? $\endgroup$ – user21574 Mar 5 '14 at 15:50
  • $\begingroup$ 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 $\endgroup$ – user21574 Mar 5 '14 at 16:02
  • $\begingroup$ 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. $\endgroup$ – Francois Ziegler Mar 6 '14 at 14:00
  • $\begingroup$ Why $O_μ/D$ is $O_μ$ where $D=P\cap\bar P\cap TO_μ$ and $P$ is polarization of coadjoint orbit. $\endgroup$ – user21574 Mar 6 '14 at 14:06
  • 1
    $\begingroup$ You're welcome. I'll make this an answer so that we can erase this unwieldy discussion if you wish. $\endgroup$ – Francois Ziegler 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 $D=\{0\}$.

  • $\begingroup$ It is very nice answer :). I am studying some parts of your thesis :) $\endgroup$ – user21574 Mar 6 '14 at 17:02

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