Let $(M,\omega)$ be a sympletic manifold and $\{ \cdot, \cdot \}$ the corresponding Poisson-bracket. Assuming $M$ is completely integrable w.r.t $f=f\_1$, so we find $n = \frac{1}{2}\dim M$ functions $f\_1, \dots , f\_n \colon M \to \mathbb{R}$ such that they are functionally independent and mutually Poisson-commute on an open and everywhere dense subset of $M$. We write $F := (f\_1, \dots, f\_n) \colon M \to \mathbb{R}^n$.


For my question I'll assume that the functions $f\_1, \dots, f\_n$ are functionally independent everywhere on $M$. So $F$ is a submersion and the levelsets of $F$ are lagrangian submanifolds (if I'm not mistaken). So the levelsets of $F$ define a Lagrangian Foliation of $M$.


If we now consider some additional functions $g\_1, \dots, g\_k$ on $M$ such that
$\{f\_i,g\_j\}=\{g\_i,g\_j\}=0$, what do we know about the levelsets of $(F,g\_1, \dots, g\_k) \colon M \to \mathbb{R}^{n+k}$.
Do we know, that the levelsets give us the same foliation as given by the levelsets of $F$?



Lagrangian foliation
Yes since functions which Poisson commute are constant on one another's Hamiltonian flows.


