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In a paper I saw the following statement:

Let $M$ be a connected symplectic manifold and $G$ be a compact Liegroup acting symplectically and hamiltonian on $M$. Let $\Phi \colon M \to \mathfrak{g^*}$ be the corresponding $G$-equivariant momentum map w.r.t. the coadjoint-action on $\mathfrak{g^*}$. Let $s$ be the codimension of a maximal dimensional orbit in $\mathfrak{g^*}$ contained in $\Phi(M)$.

Then there exists $s$ $G$-invariant functions $f_1, \dots f_s \colon \Phi(M)\to \mathbb{R}$, such that for almost all $x\in M$ we have $df_1 \wedge\dots\wedge df_s(\Phi(x))\neq 0 $.

It would be great, if someone could provide some ideas how to prove this statement.

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  • $\begingroup$ This is not correct. If $M$ is a co-adjoint orbit, it is an homogenous space, and every invariant function is constant. It is however true if $s$ denotes the dimension of the polytope $\Phi(M)\cap C,$ if $C$ is a Weyl chamber. $\endgroup$
    – Thomas
    Commented Dec 18, 2015 at 14:33
  • $\begingroup$ @Thomas Thank you very much for your counter-example. But if $M$ is a co-adjoint orbit, it wouldn't be a problem. My setting is, that we have $k$ functions $h_1, \dots, h_k \colon \Phi(M) \to \mathbb{R}$, that make the generic co-adjoint orbits Liouville integrable. If $G \in \{ SO(n), U(n)\}$ and the orbits in $M$ are now co-isotropic, then $M$ should also be Liouville integrable. So I'm assuming that $\dim M = 2n$ and we find $n-k$ functions as mentioned in my first question. And now I'm trying to find the reason, why we find such functions $f_1, \dots, f_{n-k}$. $\endgroup$
    – Olorin
    Commented Dec 18, 2015 at 15:43

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