Reducing degrees of freedom with first integrals

Given is a Hamiltonian function with $k$ first integrals. Suppose these $k$ first integrals are closed under the Poisson bracket, is it then possible to reduce the number of independent variables by $2k$?

In the case that the integrals are involutive it is of course possible. But I guess the above must follow in some way from the Frobenius theorem.

Any help is welcome, thanks in advance.

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I imagine you mean: you have a function $H$ (the Hamiltonian) on a symplectic manifold $M$, functions $L_1,\dots,L_k$ (the 1st integrals) such that $\{H,L_i\}=0$, and you suppose there are constants $c_{pq}^r$ such that $\{L_p,L_q\}=\sum_r c_{pq}^r L_r$. In this case $c_{pq}^r$ are structure constants of a Lie algebra $\mathfrak{g}$ (the Lie algebra spanned by $L_i$'s); this Lie algebra acts on $M$ and $(L_1,\dots,L_k)$ can be seen as a map $\mu:M\to\mathfrak{g}^*$, so-called moment map. If you choose particular values for $L_i$'s, i.e. a point $P\in\mathfrak{g}^*$, you can reduce the number of variables by $k+m$ where $m$ is the dimension of the stabilizer of $P$ under the coadjoint action. More precisely, if $P$ is a regular value of $\mu$ then $\mu^{-1}(P)/stabilizer$ is a symplectic manifold. Notice that $m$ depends on $P$; if $P=0$ then $m=k$, but in general it is smaller. Google "symplectic reduction" for details.
Thanks for your reply. I am still not sure how this moment map is defined though. $\mu$ is just $(L_1, \ldots, L_k)$? How does this map into $\mathfrak{g}^*$? – Novo Jul 7 '12 at 13:26