I would like to find all regular values for a map on a symplectic manifold such that the level sets are lagarangians. Precisely, the following example :

The $4$-dimensional symplectic manifold $(\mathbb{R}^4,\omega=dx_1\wedge dx_3+dx_2\wedge dx_4)$ and the map $F : \mathbb{R}^4\to\mathbb{R}^2$ defined by $F(x_1,x_2,x_3,x_4)=(x_1^2+x_2^2+x_3^2+x_4^2,x_1x_2x_3x_4)$. For which regular value $c\in\mathbb{R}^2$, the fiber $F^{-1}(c)$ is a lagrangian submanifold ?

I only succeed to find the set of of regular values, as follows: $DF=\begin{pmatrix}2x_1&2x_2&2x_3&2x_4\\ x_2x_3x_4&x_1x_3x_4&x_1x_2x_4&x_1x_2x_3\end{pmatrix}$ and $\mathrm{rank}~ DF<2$ iff $$x_1x_2(x_3^2-x_4^2)=x_1x_3(x_2^2-x_4^2)=x_1x_4(x_2^2-x_3^2)=x_2x_3(x_1^2-x_4^2)=x_2x_4(x_1^2-x_3^2)=0$$ The set of singular values is $\{(X,0),\ X\geq0\}\cup\{(4x_1^2,\pm x_1^4),\ x_1\in\mathbb{R}\}$.

The image of $F$ is $\{(x,y),\ |y|\leq\frac{x^2}{16}\}$, since $\frac{x_1^2+x_2^2+x_3^2+x_4^2}{4}\geq(x_1^2x_2^2x_3^2x_4^2)^{\frac{1}{4}}$.

So the set of regular values of $F$ (for which $F^{-1}(c)$ is not empty) is: $$R=\{(x,y),\ |y|<\frac{x^2}{16}, y\neq0\}.$$

The two functions $f=x_1^2+x_2^2+x_3^2+x_4^2$ and $g=x_1x_2x_3x_4$ are not in involution ($\{f,g\}\neq0$), we can't deduce (as in the Arnold-Liouville theorem) that the submanifold $L=F^{-1}(c)$ (for $c\in R$) is lagrangian since the hamiltonian vector fields $X_f$ and $X_g$ are not tangent to $L$: $$\omega(X_f,X_g)=\{f,g\}\neq0.$$

So the question remains:

For which $c\in R$, we have $DF(X)=DF(Y)=0\Longrightarrow\omega(X,Y)=0$?

Thank you for any help!


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