TITLE: Flow of a Hamiltonian vector field
QUESTION [5 upvotes]: Smooth vector fields are in a one-to-one relationship with flows $\Phi: D \subseteq M\times \mathbb{R} \rightarrow M$, 
$$X_m = {\frac{d}{d t}}_{t=0} \Phi(m, t),$$
and by the symplectic form also with 1-forms
$$X \longleftrightarrow \; - \iota_X \omega. $$
The following is well known (e. g. Abraham Marsden, Proposition 3.3.6):
$X$ is local Hamiltonian iff $\iota_X \omega$ is closed iff the flow $\Phi$ of $X$ is symplectic.
$X$ is Hamiltonian iff $\iota_X \omega$ is exact iff ??.
Question: What is the corresponding property for flows of (global) Hamiltonian vector fields?   
I am a bit confused about this, as the existence of flows can be guaranteed only for small times and only in neighborhoods of each point (so it is a strongly local object), but the difference between closed and exact forms is determined by global/topological characteristics.

A short proof of the cited proposition about local Hamiltonian vector fields:
The flow $\Phi$ leaves the symplectic form invariant iff $\Phi_t^* \omega = \omega$ iff $0 = L_X \omega = d(\iota_X \omega)$ iff $\iota_X \omega$ is closed.

REPLY [4 votes]: Given a symplectic vetor field, you may try to build a hamoltonian function the following way. Fix some point P. For any point Q, choose some curve from P to Q, and define H(Q) to be the flux of the vector field across the chosen curve (infinitesimal symplectic area swept by the curve when you push it by the flow during an infinitesimal amount of time). If you choose a second curve which is homotopic to the first one, you get the same value for H(Q), because your vector filed is symplectic. But this may not be the case if you choose an other curve which is not homotopic to the first one (think of a translation of the 2-torus). Thus the field is hamiltonian if and only if the flux across every closed curve is zero. (In dimension two, the flux is called the mean rotation vector).