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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 $x$.

Tell me if I

### sourceofconfusion

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

2. in usual action-angle coordinates Theorem on needs some properness assumption. But the questionusual 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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### preliminary remark

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

### the statement

I clam claim that one can prove the following 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 $x$.

Tell me if I have misunderstood the question.

Isn't it a consequence of the existence of action-angle semi-global coordinates? Namely, one has

### preliminaryremark

I assume that

If $F=(f_1,\dots,f_n)$ is proper, then being independant for functions here means that the their differentials at anylagrangian invariant torus $\Lambda$ there exists point are linearly indenpendant (and not at almost, like it is sometimes assumed... in which case a neighborhood $U$ of $\Lambda$ that counter-example is symplectomorphic easy to
a neighborhood find for $V$ of a fiber of the projection in the cotangent bundle n=2$: take$T^*\mathbb{T}^n$f_1$ the square of the $n$-dimensional torus $\mathbb{T}^n$ (namely, norm on $V=\{(\theta,I), |I-I_0|\leq r\}$).

It seems to me that \mathbb{R}^2$, and look at the zero section $\{\theta=0\}$ satisfies your requirementorigin. EDIT: if you read french then you can find a proof of this ### the statementinLecture Notes by Colin de Verdière(proofofTheorem20,pages50-51).EDIT 2: I think the proof is the same as the clam that one in Colin de Verdiere's notes. Namely, you first can prove it in the dimension$1$situation. Here you don't need properness assumption since the conclusion you are asking for is only a local statement and not a semi-global one (it is about the neighborhood of a point, while action-angle is about the neighborhood of a Lagrangian leaf). The rest following version of the proof is action-angle coordinate Theorem, under the same except that instead of a free action of$\mathbb{T}^n$you have a locally free action hypothesis of$\mathbb{R}^n$(which is the only thing you need for your conclusion). Obvious remark concerning vocabularyquestion: For any$x\in M$there is one more thing to care about. It is what people mean by exists Darboux coordinates$f_1,\dots,f_n$being independant. I personnaly mean (p_1,\dots,p_n,q_1,\dots, q_n)$
around $x$ such that at almost any point their derivatives the leaves of the Lagrangian foliation are linearly indpendant. Therefore in this case $(q_1,\dots,q_n)=cst$.

Then the conclusion is true for almost all points in local manifold $M$. At critical points of \{p_1=\dots=p_n=0\}$ satisfies your requirement. To prove the statement, consider$F$this is very easy q_i=f_i-f_i(x)$ and extend them to construct a counter-example in dimension $2$: take Darboux chart $F$ to be the square of the norm in (q_1,\dots,q_n,p_1,\dots,p_n)$around$\mathbb{R}^2$, and look at what happens at x$.

Tell me if I have misunderstood the originquestion.

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