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Jan 12, 2012 at 20:32 comment added Alexander Chervov @Peter Thank You very much ! Is it possible to see the result in this toy model ?
Jan 12, 2012 at 16:37 comment added Peter Dalakov @Alexander: I should've been more explicit about the section, sorry. Use the hermitian metric on $\mathbb{C}^2$: given a point $p\in\mathbb{P}^1$ take a unit vector $v_1\in p$. Then take an orthogonal unit vector, say $v_2$. Let $P$ be the matrix with columns $v_i$. The (smooth) section is $p\mapsto PDP^{-1}$, where $D=\textrm{diag}(\sqrt{\lambda},-\sqrt{\lambda})$. Here $\lambda\in \mathbb{C}^\times$ is the value of the determinant, which is fixed on the orbit.
Jan 12, 2012 at 15:37 comment added Peter Dalakov @Alexander: You're right.The orbit is an affine bundle over $G/P$, not a vector bundle. In the toy example you map the orbit to $\mathbb{P}^1$ by sending a (regular) matrix to the flag determined by an eigenvector. The affine bundle doesn't have a holomorphic section, but has a smooth one.
Jan 12, 2012 at 12:54 comment added Alexander Chervov Interesting fact. Can we understand it somehow ? Toy model for such considerations - take sl_2 = C^3, regular orbit - is qaudric defined by Casimir = Const. Let me point out that Orbit - affine complex manifold and TP - is NOT, since contains a projective P. So they cannot be isomorphic as complex manifolds. But according to what I heard Orbit - is deformation of TP.
Jan 12, 2012 at 11:21 comment added Yuji Tachikawa Thank you; this paper springerlink.com/content/kfvevf4a13g0t3pp seems to explicitly construct involutive Hamiltonians on $T^*(G/P)$ (I'm confused if I'm talking about hol. sympl. spaces or just sympl. spaces here...) Do you have any comments on nilpotent orbits?
Jan 12, 2012 at 9:47 history answered Peter Dalakov CC BY-SA 3.0