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If you take a $G$ as any group acting on the dual of its lie algebra $g^{*}$ by the coadjoint action it is known that the coadjoint orbit $O_{\alpha}$ is a symplectic manifold with the KKS symplectic form.

Now it is possible to take some copies of these coadjoint orbits and reduce it further then again by KKS the reduced space is a symplectic manifold

The problem is a convenient expression of $\omega_{red}$ in $(O_{\alpha}\times O_{\beta}\times O_{\gamma}//G, \omega_{red})$ does it depend on some choice of a connection form on the principal bundle $\mu^{-1}(0)\rightarrow \mu^{-1}(0)/G$, $\mu$ being the moment map ?

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It's very hard for me to tell what your question is. Do you think you could make be a bit more specific about what you want to know? –  Ben Webster Jan 4 '12 at 3:35
    
Take the $T^{*}G$ of any lie group $G$ with its canonical symplectic form then after lift the left action of $G$ on $G$ to an action of $G$ on $T^{*}G$ , now after performing reduction on $T^{*}G$ by Marsden we know that $(T^{*}G)_\mu= O_{\mu}$ with the reduced symplectic form $\omega_KKS$ on $O_{\mu}$. So if now i start with some copies of coadjoint orbits the product space has a symplectic form (the sum of the pullback of $\omega_{KKS}$ by the projections ), so now we can reduce $\O_{\alpha}\times \O_{\beta}\times \O_{\gammma}//G$, is there any general algorithm to find $\omega_{red}$ now? –  Gourishankar Jan 5 '12 at 1:42
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2 Answers

Firstly, the diagonal action of $G$ on $\mathcal{O}_1\times\mathcal{O}_2\times\mathcal{O}_3$ (I'm altering your notation for later convenience) is not necessarily free: if there are any points $(\alpha_1,\alpha_2,\alpha_3)\in\mathcal{O}_1\times\mathcal{O}_2\times\mathcal{O}_3$ with $\mathfrak{g} _{\alpha_1}\cap\mathfrak{g} _{\alpha_2}\cap\mathfrak{g} _{\alpha_3}\ne\lbrace0\rbrace$, then $G$ will have nontrivial isotropy group there. So it's not clear to me that $(\mathcal{O}_1\times\mathcal{O}_2\times\mathcal{O}_3) / G$ is even a differentiable manifold. Supposing the freeness condition is met (e.g. the $\alpha_i$ are nonzero on different simple blocks of a semisimple Lie algebra), the value of the reduced form is obtained from the lifted expression $$ \sum_{i=1}^3\ \omega_{KKS}(\alpha_i)(-\rm{ad} _{\xi_i}^*\alpha_i, -\rm{ad} _{\zeta_i}^*\alpha_i) = \sum_{i=1}^3\alpha_i([\xi_i,\zeta_i]) $$ where $\sum_{i=1}^3\alpha_i=0$ and $\sum_{i=1}^3(-\rm{ad} _{\xi_i}^*\alpha_i)=\sum_{i=1}^3(-\rm{ad} _{\zeta_i}^*\alpha_i)=0$ (expressing the restriction to $\mu^{-1}(0)$). However, by the general theory of reduction, the $\mathcal{O}_1\times\mathcal{O}_2\times\mathcal{O}_3$ symplectic form, when restricted to $\mu^{-1}(0)$, is degenerate along directions
$$(-\rm{ad} _\eta^*\alpha_1,-\rm{ad} _\eta^*\alpha_2,-\rm{ad} _\eta^*\alpha_3),\qquad\eta\in\mathfrak{g}.$$ So subtract such a vector with $\eta=\xi_2$ from the first slot and $\eta=\zeta_3$ from the second. The expression reduces to $$ \alpha_1([\xi_1-\xi_2,\zeta_1-\zeta_3]). $$

That's the simplest form I can see right now. It would probably help if you provided some context for your question. Why do you want to know this anyway?

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Your question is a bit vague, but it seems like you're asking whether it's possible to perform reduction several times, in sequence, instead of all at once. If that's the case, you may be interested in so-called reduction by stages, cf. http://www.cds.caltech.edu/~marsden/books/Hamiltonian_Reduction_Stage.html

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