I just want to clarify a point that I haven't seen made explicitly made: there is no globally defined tautological one-form or symplectic potential on a closed symplectic manifold (i.e. a compact manifold without boundary). If there were such a one-form $\theta$ (in local coordinates, expressable as $p_i dq^i$) such that $\omega = -d\theta$, then on a $2n$ dimensional closed manifold $M$,
$$ V = \int_M \omega^n = - \int_M d\theta \wedge \omega^{n-1} = -\int_M d \left( \theta \wedge \omega^{n-1} \right) = -\int_{\partial M} \theta \wedge \omega^{n-1} = 0$$
since $\partial M = \emptyset$ (and $d \omega = 0$). So we can't use the symplectic form $\omega$ as a volume form.
EDIT: I.e. $\omega$ being exact would imply that the volume of the compact manifold $M$ vanishes. This suggests (but does not constructively prove) that $\omega$ is not exact, and that there is no globally defined tautological one-form or symplectic potential whose exterior derivative yields $\omega$ on a compact manifold. Is there a more direct proof of the statement in italics, or does it suffice to assert that $\omega$ give rise to a volume form on $M$?
I'd take the 2-torus $T^2$ as an example of such an $M$.
I'm trying to square this with Darboux's theorem – presumably on $T^2$, taking $\omega = d\psi \wedge d\phi = -d(\phi d \psi)$ where $\psi, \phi$ are angular coordinates for the two cycles, is only true locally (see e.g. this post), so there's no contradiction.
EDIT: To clarify, my understanding is that Darboux's theorem allows for $\omega = -d(\phi d \psi)$ on a given coordinate patch of $T^2$, but these local definitions cannot be made to agree with each other. Hence $\omega$ is not globally exact, and will yield a positive volume for $T^2$.
Is the above reasoning correct?