I'm trying to get a better understanding of multiplicities in geometric quantization, and so I've been concentrating on a specific simple case: let $\mathcal{O}\subset\mathfrak{g}^*$ be an integral coadjoint orbit of a compact semisimple Lie group $G$ with KKS symplectic form $\omega$, defined by $\omega_\mu(-\mathrm{ad}_\xi^*\mu,-\mathrm{ad}_\zeta^\*\mu)=\mu([\xi,\zeta])$, and compatible line bundle-connection pair $(L,\nabla)$. Let $(L^\*,\nabla^*)$ be the dual bundle, with base $(\mathcal{O}^-,-\omega)$. Put a totally complex polarization $F$ on $\mathcal{O}$, with corresponding complex structure $\mathcal{J}_F$, and the complex conjugate polarization $F^\*$ on $\mathcal{O}^-$, with corresponding complex structure $\mathcal{J} _{F^*} = -\mathcal{J}_F$. The momentum map corresponding to the diagonal coadjoint $G$-action on $\mathcal{O}\times\mathcal{O}^-$ is $\mathrm{J}(\mu,\nu)=\mu-\nu$. The covariantly constant sections of $L$ form an irrep, and those of $L^\*$ the dual irrep, so the $G$-invariant sections of $L\otimes L^*$ should then be one-dimensional, and I'm trying to verify this explicitly. Following some ideas in Guillemin and Sternberg's 1982 paper "Geometric quantization and multiplicities of group representations", I using the complex structure to to extend the $G$-action on $\mathcal{O}\times\mathcal{O}^-$ to a $G^\mathbb{C}$-action, by defining the infinitesimal generator corresponding to $i\xi\in i\mathfrak{g}$ to be $(i\xi)_{\mathcal{O}\times\mathcal{O^-}}=\mathcal{J}\xi\_{\mathcal{O}\times\mathcal{O^-}}$, (where $\mathcal{J}$ is the combined complex structure) and exponentiating.

I want to show that the so-defined $G^\mathbb{C}$-action is transitive on the whole of $\mathcal{O}\times\mathcal{O}^-$, but I'm having trouble doing so (and maybe it's not?). The problem is that the action isn't free everywhere, and I don't really understand how it spans the tangent space at a general point $(\mu,\nu)$. I think it would help if I understood what $\ker T_{(\mu,\nu)}\mathrm{J}$ was everywhere. For example, if I can show that $\ker T_{(\mu,\nu)}\mathrm{J}\subset \mathfrak{g}\cdot(\mu,\nu)$, then since $\left(\mathfrak{g}\cdot(\mu,\nu)\right)^\omega = \ker T_{(\mu,\nu)}\mathrm{J}$, $\;\mathfrak{g}\cdot(\mu,\nu)$ will be coisotropic, and so $(i\mathfrak{g})\cdot(\mu,\nu)$ will be transversal to it (by general properties of complex structures). Then its just comes down to counting dimensions. But maybe this is the wrong track.

Any help appreciated. Thanks.

  • 1
    $\begingroup$ The $G^{\mathbb C}$ action isn't transitive, it just has a dense orbit, which I think is enough for your argument. You can identify each coadjoint orbit with $G^{\mathbb C}/B$, and then the orbits of $G^{\mathbb C}$ on that are $B\backslash G^{\mathbb C}/B \cong W$. Finitely many orbits, hence one of them is open dense. One $T$-fixed point in this open dense orbit is $(1, w_0)$, where you should be able calculate. (Here $w_0$ is the long element in $W$.) $\endgroup$ – Allen Knutson Sep 21 '12 at 12:20
  • $\begingroup$ Thanks a lot for the clarification. I guess I was pretty naive in my thinking on this, but your comment makes it clear what I need to read next. $\endgroup$ – user26687 Sep 22 '12 at 6:51
  • $\begingroup$ Also, anyone with sufficiently high reputation, feel free to close this question. $\endgroup$ – user26687 Sep 22 '12 at 6:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.