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Reading through McDuff, Salamon, I came across the following extract:

ing geometrically. Because the functions $F_i$ are constants of the motion, the level sets $T_c=\{z\in \Omega|F_j(z)=c_j\}$ for $n$-dimensional invariant submanifolds of the Hamiltonian flow of $H$. These submanifolds are also invariant

We are supposing a Hamiltonian system has $n$ independent integrals $F_1,\dotsc,F_n$, that is there are $n$ functions $F_i$ whose Poisson brackets with the Hamiltonian $H$ are 0 and whose differentials are linearly independent at every point, and actually they Poisson commute, i.e. for any $j,k$ we have $\{F_j,F_k\}=0$. My question is: how does one prove those level sets are manifolds? It is evident that they are invariant under the flow because $\{F,H\}=0$ is equivalent to $F$ being a constant of motion, and the flow has to move along the integral curves of the field and the fact that $F_i$ are constants of motion means the integral curves are contained inside the level sets of those functions. But how do I show they are manifolds? If they were one, linearly independent differential would equate to everywhere nonzero differential, which in turn would equate to everywhere surjective differential, and we're done. This could serve as a basis for induction. I have to show that $(F_1,\dotsc,F_n)$ has surjective differential, because then by the regular value theorem the level sets are indeed manifolds. The differential of that multi-valued function is $(dF_1,\dotsc,dF_n)$. I tried doing the step from $n-1$ (induction hypothesis) to $n$, but all I could get is the induction hypothesis, i.e. there are $n-1$ linearly independent images of that differential for any choice of $n-1$ of those differentials, but I cannot show how adding the $n$th differential gives me another image and renders the "greater" differential also surjective. Any ideas?

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  • $\begingroup$ If you read the question carefully, you will see that was precisely what I tried and got stuck on. This confirms I was on the right track, but I don't quite see how "it follows that the differential is everywhere surjective". $\endgroup$ – MickG Aug 14 '15 at 17:46
  • $\begingroup$ Indeed: row rank and column rank both coincide with that. And that matrix represents the differential of $g$. Rank $n$ mens surjective. I guess you should post that as an answer. $\endgroup$ – MickG Aug 14 '15 at 18:24
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Take coordinates on a neighborhood. Then the differential of every function can be thought of as a row vector. Set$$g:\Omega\to\mathbb{R}^n,\quad p\mapsto(F_1(p),\ldots,f_n(p)).$$The differential of $g$ is the matrix formed by the differentials of the $F_i$'s. Since they are linearly independent, the rank of this matrix is equal to $n$, which means it is surjective.

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