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 $\mathcal{F}$)



About the geometry of  completely integrable systems
### 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.


