# 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!

• Do you mean Liouville's theorem (the Hamiltonian vector field preserves the symplectic form) or Liouville integrability (maximal set of first integrals in involution)?
– Ari
Commented Apr 7, 2011 at 22:29
• I mean the former. I think that the thing I'm looking for is (at least in local coordinates) along the lines an almost-symplectic analogue of: From the Darboux Theorem, $\exists$ local symplectic coordinates $(x^i,y^i)$ Now if I write the canonical volume form as $x^1 \wedge x2 \wedge \cdots \wedge x^n \wedge y^1 \cdots \wedge y^n$ then this volume form is invariant under the flow of the Hamiltonian Vector Field. I want the same thing but for an almost symplectic structure. Commented Apr 8, 2011 at 0:45
• The lack of the Darboux Theorem is the biggest issue here Commented Apr 8, 2011 at 0:46
• The paper you link to imposes a condition on the Hamiltonian, not on the manifold. If $\omega$ is the non-degenerate two form, they define the Hamiltonian vector field the same way as usual, by $i_X \omega = - dH$. They then define strongly Hamiltonian as requiring additionally $i_X d\omega = 0$. If $X$ is a strongly Hamiltonian vector field, by Cartan's formula, $L_X \omega = 0$, which then implies $L_X \omega^n = 0$ (Liouville's theorem). What kind of generalization are you looking for? Commented Apr 8, 2011 at 23:47
• I suppose I was wondering if the strongly Hamiltonian case is the weakest situation in which the volume form is invariant under the action of the Hamiltonian flow. In coordinates, I want the weakest case for something like, $$\varphi^∗ dV_g=dV_g$$ where $\varphi : M \rightarrow M$ is the flow of the Hamiltonian vector field and $dV_g$ is the volume form associated to a metric $g$. Commented Apr 9, 2011 at 19:43

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'.