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Let $G$ be a connected reductive group over $\mathbb{C}$, let $Gr$ be the affine grassmannian of $G$. On $Gr$, we know that there is a canonical line bundle $L$ (the generator of $Pic(Gr)$). Now $G(\mathbb{C}((t)))$ acts by left multiplication on $Gr$ and the obstruction to lift this action to an action on $L$ is given by a class in $H^{2}(G(F),\mathbb{G}_{m})$, which will give rise to the central extension of $G$

Is there a way, without writing cocycles to show that this class is non-trivial?

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  • $\begingroup$ 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. $\endgroup$
    – Allen Knutson
    Commented Oct 22, 2015 at 12:35
  • $\begingroup$ Why the fact that $\Omega K$ is not a coadjoint orbit of $LK$ implies that the extension is non-trivial. $\endgroup$
    – prochet
    Commented Oct 22, 2015 at 16:04
  • $\begingroup$ 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". $\endgroup$
    – Allen Knutson
    Commented Oct 23, 2015 at 0:51

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