If we consider in a symplectic N-dimensional manifold $(M,\omega)$ the infinitely differentiable functions at all points of $M$ denoted by $C^\infty(M,\mathbb{R})$, and a Hamiltonian vector field defined from a $C^\infty(M,\mathbb{R})$ function $H$ by $\omega( V, \xi_H) := dH(V)\,\,, \forall \, V \in \mathfrak{X}(M)$, the set of smooth vector fields, then when considering a Hamiltonian vector flow $\Phi: \mathbb{R} \times M \rightarrow \mathbb{R} \times M$ given by:
$$\hspace{3.2cm}\frac{d}{dt}\bigg\rvert_{\Phi_t(X)} = \xi_H\big\rvert_{\Phi_t(X)} \,\,\,\,\,, \, \Phi((t,X))\equiv \Phi_t(X)\, \text{ for } (t,X) \in \mathbb{R} \times M$$
Where $\xi_H = \sum\limits_{i=1}^N \alpha^i\frac{\partial}{\partial x^i}$ in the basis $\{x^1,...,x^N\}$, and therefore the equality can be understood as:
$$\frac{df}{dt}\bigg\rvert_{\Phi_t(X)} = \xi_Hf\big\rvert_{\Phi_t(X)}= \sum\limits_{i=1}^N \alpha^i\frac{\partial f}{\partial x^i}\bigg\rvert_{\Phi_t(X)} \,\,\,\,\,\,\, \forall f \in C^\infty(M,\mathbb{R}) $$
If $[\omega^{ij}]$ is the inverse matrix of $[\omega_{ij}]$ the matrix of $\omega$ in the basis $\{x^1,...,x^N\}$, then the components of the vector field $\xi_H$ can be written as:
$$\xi_H^i = \sum\limits_{i=1}^N \omega^{ij}\frac{\partial H}{\partial x^i}$$
What are the sufficient and necessary conditions for $\xi_H$ to be a complete vector field? That is, for the flow equation to be solvable on any point of $M$ and the curve $\Phi_t$ to be arbitrarily extendable over the on manifold $M$ ?
Arnol'd mentions that $H$ having all its level sets compact is sufficient. Abraham and Marsden have some other sufficient conditions such as:
Proposition: Suppose $X$ is a $C^k$ vector field on $M$, $k \geq 1$ and $f:M \rightarrow \mathbb{R}$ is a $C^1$ proper map ( this means that the inverse image of compact sets is compact). Suppose $X$ admits the estimate: $$|X(f)(m)|\leq A |f(m)|+B \,\,\,\,\,\,, \,\text{for}\, A,B \geq 0,\, m \in M$$ Then the flow of $X$ is complete.
My questios is then, is there a sufficient and necessary condition for the specific vector field given by $\xi_H$ to be a complete vector field? In case there isn't (probable), the restrictions of considering $M$ a smooth manifold or $H$ in $C^\infty(M, \mathbb{R})$ or only considering functions to be evaluated in the flow equation to be of $C^\infty(M, \mathbb{R})$ give sufficient conditions for $\xi_H$ to be a complete vector field? Or are some other properties that can be required of $M, H$ or the evaluated functions $f$ on the flow equation such that no condition has to be directly imposed in $\xi_H$ for it to be complete?
Part of the reason because I am asking this is because in a book I am reading it is said that Poisson Manifolds are actually a generalization of symplectic manifolds and that there are Poisson Manifolds that aren't symplectic manifolds, and in those there exist at least singular symplectic foliations of M that separate Hamiltonian flows, so that these can be said to have classical superselection rules. But Abraham and Marsden put as an exercise (possibly originally solved by Wolfgang Pauli and R. Jost) that:
Let $\{f,g\}$ be an $\mathbb{R}$-bilinear bracket defined on $\mathscr{F}(M) \times \mathscr{F}(M) \rightarrow \mathscr{F}(M)$ that makes $\mathscr{F}(M)$ into a Lie algebra. Suppose $\{\,,\,\}$ is a derivation in each factor, and $\{f,g\}=0$ for all $g$ implies $f$ is constant. Show that $\Lambda(df_x,dg_x) = \{f,g\}(x)$ defines a two tensor on $M$. Then show $\Lambda$ is nondegenerate and that the corresponding two-form $\omega$ is a symplectic structure.
Which seems to imply that all Poisson Manifolds are exactly symplectic Manifolds. On what am I wrong?
Can u define and integrate $X(f(m))$ and by some kind of fundamental theorem of gradient get the result of integration as $f(m)$ and hence can bound $|X(f(m))|$ as a linear function of $|f(m)|$ ?
If $M$ is embedded in $\mathbb{R}^n$ then,
$$\int_{0}^1 <\nabla f(\gamma(t)),\gamma'(t)> = f(\gamma(1)) - f(\gamma(0)) = \int_{0}^T \omega(df(\Phi_t(a)),X(\Phi_t(a))) dt$$
Check if last equality can be made true for a given symplectic form $\omega$.
Fix $$\gamma(0) = a = m - \epsilon \in M \ and \ \gamma(1) = m \ and \ \Phi_{tT}(a) = \gamma(t)$$ $$\int_{0}^T \omega(df(\Phi_t(m)),X(\Phi_t(a))) dt = \int_{0}^1 <\nabla f(\gamma(t)),\gamma'(t)> = f(\gamma(1)) - f(\gamma(0)) $$ $$\inf_{t \in [0,1]} |\omega(df(\Phi_t(a)),X(\Phi_t(a)))| = \inf_{t \in [0,T]} |<\nabla f(\gamma(t)),\gamma'(t)>| \leq |f(\gamma(1)) + f(\gamma(0))| \leq 3 |f(m)|$$.
Now make $\epsilon \rightarrow 0$.
I mentioned the integration w.r.t vector field just for analogy of $M$ with $\mathbb{R}^n$. You have to define and check.