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 $F$?


1 Answer 1


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

  • $\begingroup$ You can locally build Darboux coordinates out of these $f_i$ and conjugate variables. $\endgroup$
    – Ben McKay
    Mar 30, 2016 at 18:11
  • $\begingroup$ I was thinking quite some time about that and when you are mentioning it, it is just so obvious! Thank you very much for your answer! $\endgroup$
    – Olorin
    Mar 30, 2016 at 18:14

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