Weakest condition for an integrable, almost-symplectic manifold? I was recently speaking with someone who works in Computational Chemistry and they mentioned that in a lot of the computational simulations they do, they have systems that are not symplectic but still integrable. I believe that this means symplectic in the formal sense, but for the record the computation was described to me in terms of Jacobians associated to Hamiltonian flows (Something along the line of the treatment in the Wikipedia article on Symplectic Integrators). The system still has a well-defined non-degenerate bilinear form, but it may not be closed (i.e. an Almost Symplectic Structure). From his examples, I believe that Liouville's Theorem holds for a slightly bigger subcategory of the category of manifolds than the category of symplectic manifolds. I was just wondering what the most general conditions are for having a $2n$-dimensional smooth manifold with global, non-degenerate $2$-form $\omega$ satisfy Liouville's Theorem. I found this paper and I was just wondering if these are the most general conditions for such an integrable system.
Thanks!
 A: Actually, there is a much weaker condition than 'strongly Hamiltonian' if all you want to do is preserve the associated volume form:  For a given 'Hamiltonian' $H$ and a non-degenerate $2$-form $\omega$, if you define $X$ so that $\iota_X\omega = -\mathrm{d}H$, then the condition that $L_X(\omega^n) = 0$ simply becomes
$$
0 = \mathrm{d}\bigl(\iota_X(\omega^n)\bigr) = \mathrm{d}\bigl(-n\,\mathrm{d}H\wedge\omega^{n-1}\bigr) = n\,\mathrm{d}H\wedge \mathrm{d}(\omega^{n-1}),\tag1
$$
and this is a single linear first-order partial differential equation for $H$.  In fact, if one (uniquely) defines the vector field $Y$ by the relation
$$
n\,\mathrm{d}(\omega^{n-1}) = \iota_Y\bigl(\omega^n\bigr),\tag2
$$
then the partial differential equation is just $\mathrm{d}(H)(Y) = 0$, i.e., $H$ must be constant along the flow lines of $Y$.  To see this, just note that, for dimension reasons, we have $\mathrm{d}H\wedge\omega^n = 0$, so 
$$
0 = \iota_Y(\mathrm{d}H\wedge\omega^n) 
= \mathrm{d}H(Y)\,\omega^n - \mathrm{d}H\wedge\iota_Y\bigl(\omega^n\bigr)
= \mathrm{d}H(Y)\,\omega^n - \mathrm{d}H\wedge (n\,\mathrm{d}(\omega^{n-1})),
$$
which, since $\omega^n$ is non-vanishing, shows that (1) is equivalent to 
$\mathrm{d}(H)(Y) = 0$.
Of course, if $Y$ vanishes identically (which happens in the symplectic case, of course, but much more generally), then this PDE is trivial and all 'Hamiltonian' flows preserve the 'symplectic' volume.  On the other hand, if the flow of $Y$ is ergodic, then only the trivial 'Hamiltonian' flow preserves the volume, and, in particular, only the constants are 'strongly Hamiltonian'.
