So i've found this lemma in some notes on symplectic manifolds and I think i'm lacking some basic knowledge on flows as I the hint for the proof makes no sense to me. It goes as follows.
Lemma 18.11: If the hamiltonian vector fields $X_{H_1}, . . . , X_{H_n}$ on a 2n-dimensional symplectic manifold satisfy $X_{H_i}H_j=0$ and $f=(H_1, . . . , H_n) : M → R^n$ is transverse to 0, then the connected, compact components of $f^{−1}(0)$ are isomorphic to a torus i.e., are of the form $R^{n}/A$ where A is the lattice structure of $R^n$.
Proof. Exercise (just follow the flows to obtain coordinates).
Im not seeing how the flow comes into the proof? Any help would be great!
The condition $X_{H_i}H_j=0$ implies that the level sets of the mapping $f$ are invariant by the flows of the hamiltonian vector fields $X_{H_i}$.
It is also the case that these vector fields are tangent to the level sets ---here is where Riemannian and Symplectic Geometry behave quite differently, as the gradient of $H_i$ is orthogonal to its level sets.
The condition of $f$ being transverse to zero guarantees that these hamiltonian vector fields generate the tangent space at each point of the level set $f^{-1}(0)$. Now it is readily to see that you can reach any point of the level set $f^{-1}(0)$ by (piecewise) moving along the integral curves of the hamiltonian vector fields $X_{H_i}$.
If one denotes the flow at time $t_i$ of $X_{H_i}$ by $\exp(t_iX_i)$, one can define a $\mathbb{R}^n$-action on $M$ by $\mathbb{R}^n\ni (t_1,\dots,t_n)\mapsto \exp(t_1X_1)\circ\cdots\circ\exp(t_nX_n)\in\mathrm{Diff}(M)$. The level set $f^{-1}(0)$ is, then, just an orbit of that action; and this is why it is of the form $\mathbb{R}^n/A$, $A$ is homomorphic to a stabiliser subgroup of that action on a point where $f$ vanishes.