# What properties does the flow of a Hamiltonian vector field have compared to the flow of a symplectic vector field?

Let $$(P,\omega)$$ be some real $$2n$$-dim symplectic manifold. A symplectic vector field, $$Z\in\mathfrak{X}_{sp}(P)$$, is one for which the Lie derivative satisfies $$\mathcal{L}_Z \omega =0$$ or, equivalently, for which the exterior derivative satisfies $$\mathbf{d}(Z\cdot \omega)=0$$ (I'm writting $$Z\cdot \omega:=\omega(Z,\cdot)$$ for the interior product). If $$\psi_t$$ is the flow of $$Z\in\mathfrak{X}_{sp}(P)$$, then $$\omega=\psi_t^*\omega$$ and so $$\psi_t$$ is a (global) symplectomorphism (right? Or am I already mistaken at this stage?)

Now, a Hamiltonian vector field, $$X^H\in\mathfrak{X}_{hm}(P)\subset\mathfrak{X}_{sp}(P)$$, is a symplectic vector field for which there exists some function $$H\in\mathcal{F}(P)$$, such that $$X^H=\omega^{-1}(\mathbf{d}H,\cdot)\in\mathfrak{X}_{hm}(P)$$. Equivalently, $$\mathbf{d}H=\omega(X^H,\cdot)$$. In symplectic/canonical/Darboux coordinates, $$z^i$$, integral curves of $$X^H$$ satisfy Hamilton's canonical equations $$\dot{z}^i=J\frac{\partial H}{\partial z^i}$$.

question 1: If $$\phi_t$$ is the flow of $$X^H\in\mathfrak{X}_{hm}(P)$$, and $$\psi_t$$ is the flow of $$Z\in\mathfrak{X}_{sp}(P)$$, what properties/relations does $$\phi_t$$ have/satisfy which $$\psi_t$$ does not?

question 2: Are there particular properties that arise when $$P=T^*Q$$ is the cotangent bundle of smooth manifold $$Q$$? The only property that I am aware of in this case is that $$\mathcal{L}_{X^H}\theta = \mathbf{d}S-\mathbf{d}H$$, where $$\theta\in\Omega(P)$$ is the canonical 1-form ($$\omega=-\mathbf{d}\theta$$) and $$S=\theta\cdot X^H\in\mathcal{F}(P)$$ is the action.

I have seen sources imply that "Hamiltonian diffeomorphisms" are a subset of symplectic diffeomorphisms (i.e., symplectomorphisms). What is the distinguishing properties of the former? Perhaps something to do with cotangent lifts?

One may show that any symplectic vector field is given by $$\omega^{-1}(\alpha,\cdot)$$ as follows: Assume that $$X$$ is a symplectic vectorfield, i.e. the local flow of $$X$$ has $$\phi^*_t\omega=\omega$$. This in turn means that $$\mathcal{L}_{X}=d\iota_{X}\omega+\iota_Xd\omega=d\iota_X\omega=0$$. This means the form $$\iota_X\omega$$ is closed.
The Poincare lemma tells us that locally we may write $$\alpha=d f$$ for a smooth function $$f: U\to \mathbb{R}$$. This means that all symplectic vector fields are locally Hamiltonian vector fields. As such, symplectic vector fields may only be distinguished from Hamiltonian vector fields globally, e.g. by their global dynamical/Homological behavior.
Since symplectic vector fields are obtained from closed forms, many natural question about them can be answered using the first De Rham Cohomology group. For example, if $$H^1_{DR}(P)=0$$ then all closed forms are exact and there are no non-Hamiltonian symplectomorphisms which are isotopic to the identity through symplectomorphisms.
• In your first paragraph do you mean to say $\phi_t^* \omega= \omega$? And I was aware that, locally, symplectic vector fields can treated as Hamiltonian fields. But what distinguishes a Hamiltonian flow from a symplectic flow? I admit I know nothing about cohomology and little of topology in general. Commented Aug 8, 2023 at 4:09