# multiplicity free chain of Lie subgroups gives a completely integrable system on symplectic manifold

Assume we have a symplectic manifold $N$ and a Hamiltonian action of a Lie group $G$, such that the algebra of all $G$-invariant functions on $N$ is commutative under Poisson bracket. Furthermore, assume we have a ascending chain $\{1\} =G_l\subset G_{l-1} \subset \dots \subset G_1 = G$ of Lie subgroups, such that the coadjoint-actions of $G_{i+1}$ on the coadjoint orbits of $G_i$ acting on $\mathfrak{g}_i^*$ is multiplicity-free. That means, that the algebra of all $G_{i+1}$-invariant functions on the $G_i$-orbits of $\mathfrak{g}_i^*$is commutative under the induced Poisson bracket.

Then we get that the Moment map of the $G_{i+1}$ action is just the projection of $\mathfrak{g_i^*}$ to $\mathfrak{g_{i+1}^*}$ restricted to the $G_i$-orbit on $\mathfrak{g_i^*}$.

Guillemin and Sternberg prove in their paper "On collective complete integrability according to the method of Thimm" that under these assumptions, $N$ is completely integrable. But I don't understand, how this assumptions already imply, that we find enough first Integrals, that are functionally independent on an open and dense subset on $N$.

I checked their book "symplectic techniques in physics" and their paper mentioned before, but couldn't find any answer to this problem.

They are just saying, if we take the trivial Lagrangian foliation on ${0}=\mathfrak{g}_l^*$ and pullback this foliation step by step to a foliation on $\mathfrak{g^*}$ we are done.

But then: How does this follow, or why are we done?

I hope someone can help me to understand this problem.

• EDIT: I think I'm missing one thing to prove this. I'm always assuming we have orbits of maximal dimensions. Let $x\in \mathfrak{g}_i^*$ and $\alpha = G_i \cdot x$ be its $G_i$ orbit through $\alpha$. For each $y \in \alpha$ we find locally $\text{codim}(G_{i+1} \cdot y)$ many $G_{i+1}$-invariant, functionally independent functions on $\alpha$. But now I have to get these functions to be $G_{i+1}$-invariant and functionally independent almost everywhere on $\mathfrak{g}_i^*$. Is this possible? Even if we don't have proper actions? – Olorin Jul 14 '15 at 6:48