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Oct 23, 2015 at 0:51 comment added Allen Knutson On a symplectic manifold $(M,\omega)$, the map $Fun(M) \to Vec(M)$ taking $f\mapsto \omega^{-1}(df)$ is a Lie algebra homorphism and central extension of its image. Pull back that central extension to $L\mathfrak k$. If it were trivial, then $L\mathfrak k$ would factor through $Fun(\Omega K)$, giving dually an $LK$-equivariant map $\Omega K\to (L\mathfrak k)^*$. But none exists. This general idea (not this example) is in Guillemin-Sternberg "Symplectic Techniques in Physics".
Oct 22, 2015 at 16:04 comment added prochet Why the fact that $\Omega K$ is not a coadjoint orbit of $LK$ implies that the extension is non-trivial.
Oct 22, 2015 at 12:35 comment added Allen Knutson In the more topological version, with $K$ a maximal compact group of $G$, $Gr$ sits densely inside the symplectic manifold $\Omega K$. The group $LK$ acts symplectically and transitively on it, but $\Omega K$ is not a coadjoint orbit of $LK$, only of this central extension -- hence it's nontrivial. Lots of this is in Pressley and Segal's Loop Groups.
Oct 22, 2015 at 2:13 history asked prochet CC BY-SA 3.0