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During a conversation I heard an assertion that I found at least dubious for the lack of adeguate hypothesis, but I am not able to imagine a counterexample, even if it is probably obvious to some of you.
My question: is there someone who can point out to me a counterexample for the following implication?

This is the setting:
Let $(M,\omega)$ be a $2n$-dimensional symplectic manifold, and $f_1,\ldots,f_n$ independent functions on $M$ mutually Poisson-commuting, and such that the hamiltonian vector fields $X_{f_1},\ldots,X_{f_n}$ are complete.
Let $\mathcal{F}$ denote the lagrangian foliation of $M$ determined by the integrable distribution $D$ generated by $X_{f_1},\ldots,X_{f_n}$.

This is the conclusion that I find not well justified:
For any $x$ in $M$ there exists a local manifold $\Sigma_x$ which is lagrangian, transversal to $D$, and doesn't intersect any leaf of $M$ at two distinct points.
(I know that this condition is necessary and sufficient for the existence of the manifold of the leaves of $\mathcal{F}$)

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preliminary remark

I assume that being independant for functions here means that their differentials at any point are linearly indenpendant (and not at almost any point, like it is sometimes assumed... in which case a counter-example is easy to find for $n=2$: take $f_1$ to be the square of the norm on $\mathbb{R}^2$, and look at the origin).

the statement

I claim that one can prove the following weak version of the action-angle coordinate Theorem, under the hypothesis of the question:

For any $x\in M$ there exists Darboux coordinates $(p_1,\dots,p_n,q_1,\dots, q_n)$
around $x$ such that the leaves of the Lagrangian foliation are $(q_1,\dots,q_n)=cst$.

Then the local manifold $\{p_1=\dots=p_n=0\}$ satisfies your requirement.

To prove the statement, consider $q_i=f_i-f_i(x)$ and extend them to a Darboux chart $(q_1,\dots,q_n,p_1,\dots,p_n)$ around $x$.

source of confusion

  1. foliations may be very wilde... but here we actually have a submersion.

  2. in usual action-angle coordinates Theorem on needs some properness assumption. But the usual Theorem tell us about properties of semi-global coordinates. Here we were dealing with a purely local statement and we don't need any properness assumption.

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  • $\begingroup$ Dear DamienC, thanks a lot for your attention. About the question: my doubt arose just because I was unable to conclude the reported statement, or to construct a counterexample, within the hypothesis in the initial setting. Thanks also for the linked first chapter, surely it will be useful. $\endgroup$
    – agt
    Commented May 4, 2011 at 5:46
  • $\begingroup$ Dear Giuseppe. The proof given in the limked Lecture Notes extend to any complete lagrangian leaf $\Lambda$ (without properness assumption) as soon as it does not contain critical points for $F$. So to make the conclusion work one could either say "for almost all $x\in M$". $\endgroup$
    – DamienC
    Commented May 4, 2011 at 7:16
  • $\begingroup$ Dear DamienC, By the theorem of Carathedory-Jacobi-Lie I am able to extend $f_1,\ldots, f_n$ to a local symplectic chart $f_1,\ldots,f_n,g_1,\ldots,g_n$ around an arbitrary point $x$. So taking level sets of $g_1,\ldots,g_n$, I get a Lagrangian submanifold $\Sigma$ passing through $x$ and transversal to $\mathcal{F}$. But I am unable to exclude that the intersection of $Sigma$ with some leaf of $mathcal{F}$ has $x$ as accumulation point. $\endgroup$
    – agt
    Commented May 4, 2011 at 12:26
  • $\begingroup$ But the leaves of $\mathcal F$ are not arbitrary: they are precisely level sets of $f_1,\dots,f_n$. $\endgroup$
    – DamienC
    Commented May 4, 2011 at 13:19
  • $\begingroup$ Dear DamienC: Thank you very much, "perhaps" I have realized what I was missing. This dialogue with you was very useful to me. $\endgroup$
    – agt
    Commented May 4, 2011 at 13:29

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