In this post I already ask a similar question.

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.

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 and $d_x\Phi(T_xM)=T_{\Phi(x)}W$.

By $\mathcal{O}$ denote a maximal dimensional $G$-orbit in $\Phi(M) \subset \mathfrak{g^*}$ under the $\operatorname{Ad}_G^*$ action. Let $\dim \mathcal{O} =2k$.

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 in $\Phi(M)$.

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 $M$.

In this post I assumed that we find $n-k$ functions $h_1, \dots h_{n-k} \colon W \to \mathbb{R}$ such that the common level surfaces are the $\operatorname{Ad}^*_G$ orbits in $W$. If the functions $h_1, \dots, h_{n-k}$ are functionally independent on $W$ then I understand now, that in this case, we have that the 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 integrabel system. (They Poisson-commute and are functionally independent a.e.) But the crucial part was, that I already found $n=\frac{1}{2}\dim M$ functionally independent functions on $W$. But with this method we are able to construct at most $\frac{1}{2}\dim G.\Phi(x) + (\dim W - \dim G.\Phi(x))=\dim W - \frac{1}{2}\dim G.\Phi(x)$ many functionally independent functions. Since $\ker d_x\Phi = (\mathfrak{g}_M(x))^\omega$ we have $\dim W = \dim G.x$. So we find at most $\dim G.x - \frac{1}{2}\dim G.\Phi(x)$ many functions on $W$ which are functionally independent

But in Guillemin & Sternbergs paper "On collective complete integrability according to the method of Thimm" the authors say, that in the case, that $G = U(n)$ or $O(n)$ we find automatically enough Poisson-commuting and functionally independent functions, provided that $G$ acts hamiltonian and such that the orbits $G.x$ are coisotropic in $M$. For that they use, that each coadjoint orbit in $\mathfrak{g}^*$ is completely integrable.

Shouldn't we still need the assumption, that $\dim M \leq 2\dim G.x - \dim G.\Phi(x)$? Since the $G$-orbits are coisotropic we only know, that $\dim M \leq 2\dim G.x$.


1 Answer 1


Maybe, I misunderstand your problem. After doing the math I get even an equality $$ \dim M=2\dim G.x-\dim G.\Phi(x) $$ for any coisotropic action. More precisely, the multiplicity freeness of $M$ is equivalent to $$ M/G\to W/G $$ being bijective (i.e., all symplectic reductions are points). This translates into $$ \dim M-\dim G.x=\dim W-\dim G.\Phi(x). $$ Now you get the result from $\dim W=\dim G.x$.

  • $\begingroup$ My problem was, that I didn't consider the fact, that multiplicity-freeness not only implies that the $G$-orbits are coisotropic, but also that the Marsden-Weinstein-spaces are points. Considering this, I also get the equalitiy. Thank you very much for your help. $\endgroup$
    – Olorin
    May 15, 2016 at 13:26

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .