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Let $G$ be a Lie group acting Hamiltonian on some real analytic symplectic manifold $(M, \omega)$, with an $G$-equivariant momentum map $\Phi \colon M \to \mathfrak{g}^*$.

Assuming I can find polynomials $f_1, \dots, f_n \colon \mathfrak{g}^* \to \mathbb{R}$, such that $f_1 \circ \Phi, \dots, f_n \circ \Phi \colon M \to \mathbb{R}$ are functionally independent in at least one point in $M$. Is it now possible to conclude, that $\Phi^*f_1, \dots \Phi^*f_n$ are functionally independent on some open and dense subset in $M$? Are the functions $\Phi^*f_j$ for instance real analytic?

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If group, action, manifold, and symplectic form $\omega$ are real analytic then also the momentum map is real analytic. The reason for this is that the Poincar\'e lemma also works in the real analytic setting. More precisely, let $\xi\in\mathfrak g$. Then (at least locally) there is a real analytic function $f_\xi$ on $M$ with $df_\xi=\omega(\xi x,\ast)$. The component function $\Phi_\xi(x):=\langle\Phi(x),\xi\rangle$ has the same property, so it differs from $f_\xi$ by a constant.

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  • $\begingroup$ Thank you again for your quick response. So if my symplectic manifold $(M, \omega)$ and my Lie group $G$ are smooth and paracompact and $G$ acts properly and hamiltonian, with a $G$-equiv. momentum map $\Phi$, on $M$, then using a theorem of Illman we find a real analytic structure on $G$ and a smooth $G$-equivariant symplectomorphism $f \colon (M, \omega) \to (M^a, \omega^a)$, such that $(M^a, \omega^a)$ is a real analytic $G$-space. Then $G$ acts also hamiltonian on $M^a$ with a $G$-equivariant momentum map $\Phi^a = \Phi \circ f^{-1}$. $\endgroup$
    – Olorin
    May 5, 2016 at 14:44
  • $\begingroup$ Now pulling the functions $f_j \colon \mathfrak{g}^* \to \mathbb{R}$ back on $M^a$, we can conclude that the functions $f_j\circ \Phi^a$ are functionally independent on an open and dense subset of $M^a$. Since $f$ was a diffeomorphism we now can conclude that the functions $f_j \circ \Phi$ are functionally independent on an open and dense subset of $M$. Is this right or is there a mistake? $\endgroup$
    – Olorin
    May 5, 2016 at 14:45
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    $\begingroup$ This appears correct to me. $\endgroup$
    – Friedrich Knop
    May 5, 2016 at 16:03

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