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There is a question that arises, while I'm trying to understand Guillemin & Sternbergs paper "On collective complete integrability according to the method of Thimm".

Assume $M$ is a symplectic manifold of dimension $2n$. Assume $G$ is a Liegroup, $\mathfrak{g}$ be the Liealgebra and $\mathfrak{g^*}$ the corresponding dual. Endow the $\operatorname{Ad}^*_G$ orbits of $G$ with the Kirillov-Kostant-Souriau symplectic form. By $\mathcal{O}$ we will denote a maximal dimensional $G$-orbit in $\mathfrak{g^*}$ under the $\operatorname{Ad}_G^*$ action. Let $\dim \mathcal{O} =2k$.

Assume $G$ acts on $M$ in a hamiltonian fashion, with $\operatorname{Ad}^*_G$-equivariant momentum map $\Phi \colon M \to \mathfrak{g}^*$ and such that $\Phi(M) =W$ is a submanifold.

Assume we have functions $f_1, \dots, f_k \colon W \to \mathbb{R}$ such that $\{f_i,f_j\}=0$ and such that they are functionally independent on the maximal dimensional $G$-orbits.

(So $f_1|_\mathcal{O}, \dots, f_k|_\mathcal{O} \colon \mathcal{O} \to \mathbb{R}$ are functionally independet, so $d(f_1|_\mathcal{O}) \wedge \dots \wedge d(f_k|_\mathcal{O}) \neq 0$)

Now we know, that the levelsets of $f_1|_\mathcal{O}, \dots, f_k|_\mathcal{O} \colon \mathcal{O} \to \mathbb{R}$ give us a foliation on $\mathcal{O}$, which is Lagrangian on $\mathcal{O}$ and so it is coisotropic. Since the levelsets are coisotropic submanifolds of $\mathcal{O}$ and $\Phi$ is a Poisson-map, the levelsets of $\Phi^*f_1|_\mathcal{O}, \dots, \Phi^*f_k|_\mathcal{O}$ give us a coisotropic foliation of $\Phi^{-1}(\mathcal{O})$.

If we now assume, that the $G$-orbits on $M$ are $\textbf{coisotropic}$ submanifolds, we can conclude that $\ker d_x\Phi$ is isotropic for all $x \in M$ and that we get a foliation of $M$, such that the maximal dimensional leaves are lagrangian submanifolds of $M$.

Now the authors say, that if we find $n-k$ many functions $h_1, \dots h_{n-k} \colon W \to \mathbb{R}$ whose common level surfaces are the $\operatorname{Ad}^*_G$-orbits in $W$, then the $n$ functions $\Phi^*f_1, \dots,\Phi^*f_k, \Phi^*h_1, \dots \Phi^*h_{n-k} \colon M \to \mathbb{R}$ give us a completely integrable system. That means, that the functions Poisson-commute and that they are functionally independent on $M$.

As the momentum map is a Poisson-map, it is clear that the functions Poisson-commute. But why should they be functionally independent?

Am I missing some crucial point of the momentum map for this?

Edit: Made a mistake in one assumption. We are assuming that the $G$-orbits are coisotropic and not isotropic submanifolds.

Edit2: If we write $F := (f_1, \dots, f_k)$ and assume that $\Phi$ has clean intersection with the levelsets ${F|_\mathcal{O}}^{-1}(c)$, i.e. $(d_x\Phi)^{-1}\left(T_\alpha ({F|_\mathcal{O}}^{-1}(c))\right) =T_x\left(\Phi^{-1}({F|_\mathcal{O}}^{-1}(c))\right)$, then I'm able to prove that $\Phi^*f_1, \dots,\Phi^*f_k, \Phi^*h_1, \dots \Phi^*h_{n-k} \colon M \to \mathbb{R}$ give us a completely integrable system.

But do I really have to assume that I have a clean intersection or is this somehow given by the momentum map`?

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  • $\begingroup$ I think that, if you first consider the case in which $G$ is abelian, i.e., $k=0$, you will see what is going on. The rest is linear algebra. $\endgroup$ Feb 15, 2016 at 16:07
  • $\begingroup$ @RobertBryant: Thank you again for your reply. I will sit down tomorrow and think about it, first in the abelian case. But shouldn't I need some more assumptions, for instance that $G$ is compact or the momentum map is proper? $\endgroup$
    – Olorin
    Feb 15, 2016 at 18:47
  • $\begingroup$ Let's assume $G$ is abelian. The corresponding Lie algebra can be identified with $\mathbb{R}^l$ and $l\geq n$. Now I choose a basis $\xi_1, \dots \xi_l$ of $\mathcal{g}$ and I'm assuming that the flows of the induced hamiltonian vectorfields $(\xi_j)_M(x) = \frac{d}{dt} e^{t\xi_j}.x$ are complete. If $f_j$ is the corresponding hamiltonian function to $(\xi_j)_M$, then $f_1, \dots, f_l \colon M \to \mathbb{R}$ is my corresponding momentum map. In that case I am done. But I don't see how it helps me to see why my pulled-back functions on $M$ are functionally indep. if $G$ isn't abelian. $\endgroup$
    – Olorin
    Feb 18, 2016 at 12:57

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