I want to prove the following statement:
$\phi: (M,\omega_1) \rightarrow (M_2,\omega_2)$ is a symplectomorphism if and only if $\phi^* X_H=X_{H\circ \phi}$ for any $H: U \subset M_2 \rightarrow \mathbb{R}$, $U$ open set in $M_2$.
I have proven the necessary condition. But I am stuck proving that $\phi$ is a symplectomorphism supposed $\phi^* X_H=X_{H\circ \phi}$.
In the book Foundations of Mechanics by Ralph Abraham and Marsden, page $194-195$, Theorem$3.3.19$ (Jacobi Theorem), there is a proof for this I don't understand:
$d(H\circ \phi)=\phi^* dH=\phi^*i_{X_H}\omega=i_{\phi^* X_H}\phi^* \omega=i_{X_{H\circ \phi}}\phi^* \omega$ and like every vector in $T_x M_1$ has the form $X_{H\circ \phi}(x)$ for some $H$, then $\omega_1=\phi^* \omega_2$ and $\phi$ is a symplectomorphism.
I don't understand the last sentence, why is it true?
There are two separate parts to the bolded statement, and so I'll address both.
Firstly, sticking to one symplectic manifold $(M,\omega)$: since $\omega_x$ is non-degenerate for each $x\in M$, the map $^\flat:T_xM\to T^*_x M$ given by $(X_x)^\flat = i_{X_x}\omega_x$ is an isomorphism, with inverse $^\sharp:T^*_x M \to T_xM$. In particular, since $T_x^*M$ is spanned by elements $d_xf$ for $f\in C^\infty(M)$, $T_xM$ is spanned by elements $(X_f)_x = (d_xf)^\sharp$. Applying this in the case that $(M,\omega) = (M_1,\omega_1)$, the map $C^\infty(M_1) \to C^\infty(M_2)$ given by $f\mapsto f\circ \phi^{-1}$ is a bijection, with inverse $H\mapsto H\circ\phi$. Hence $T_x^*M_1$ is spanned by elements of the form $(X_{H\circ\phi})_x$ for $H\in C^\infty(M_2)$.
Secondly, we have $i_{X_{H\circ\phi}}\omega_2 = d(H\circ\phi) = i_{X_{H\circ\phi}}\phi^*\omega_1$ for all $H\in C^\infty(M_2)$. Since the vector fields $X_{H\circ\phi}$ span $T^*M_1$ at each point $x\in M_1$, this implies that $\omega_2 = \phi^*\omega_1$.