vector field $X_H: M \rightarrow TM$ under the definition:
$\begin{aligned}
&\text{partially apply $\omega$ to see $\omega$ as a mapping from $T_p$ to $T_p^*M$} \\
&\omega2: T_p M \rightarrow T_p*M \equiv \lambda (v: T_p M). \lambda (w: T_p M) . \omega(v, w) \\
&\omega2^{-1}: T_p^*M \rightarrow T_p M; dH: M \rightarrow T_p* M \\
&X_H \equiv \lambda (p: M) \rightarrow \omega2^{-1} (dH(p)) \\
&(p: M) \xrightarrow{dH} dH(p) : T_p*M \xrightarrow{\omega2^{-1}} \omega2^{-1}(dH(p)): T_pM \\
&X_H = \omega2^{-1} \circ dH
\end{aligned}$

This way, given a hamiltonian $H: M \rightarrow \mathbb R$, we can construct
an associated vector field $X_H$, in a pretty natural way.
We can also go the other way. Given the $X$, we can build the $dH$
under the equivalence:
$\begin{aligned}
&\omega2^{-1} \circ dH = X_H\\
&dH = \omega2(X_H) \\
&\int dH = \int \omega2(X_H) \\
&H = \int \omega2(X_H)
\end{aligned}$

This needs some demands, like the one-form $dH$ being integrable. But this
works, and gives us a bijection between $X_H$ and $H$ as we wanted.
We can also analyze the definition we got from the previous manipulation:
$\begin{aligned}
&\omega2(X_H) = dH \\
&\lambda (w: T_p M) \omega(X_H, w) = dH \\
&\omega(X_H, \cdot) = dH \\
\end{aligned}$

We can take this as a *relationship* between $X_H$ and $dH$. Exploiting
this, we can notice that $dH(X_H) = 0$. That is, moving along $X_H$ does
not modify $dH$:
$\begin{aligned}
&\omega2(X_H) = dH \\
&\lambda (w: T_p M) \omega(X_H, w) = dH \\
&dH(X_H) = \omega(X_H, X_H) = 0 ~ \text{$\omega$ is anti-symmetric}
\end{aligned}$

#### § Preservation of $\omega$

We wish to show that $X_H^*(\omega) = \omega$. That is, pushing forward
$\omega$ along the vector field $X_H$ preserves $\omega$.
TODO.
#### § Moment Map

Now that we have a method of going from a vector field $X_H$ to a Hamiltonian
$H$, we can go crazier with this. We can *generate vector fields* using
Lie group actions on the manifold, and then look for hamiltonians corresponding
to this lie group. This lets us perform "inverse Noether", where for a given
choice of symmetry, we can find the Hamiltonian that possesses this symmetry.
We can create a map from the Lie algebra $\mathfrak{g} \in \mathfrak{G}$ to
a vector field $X_{\mathfrak g}$, performing:
$\begin{aligned}
&t : \mathbb R \mapsto e^{t\mathfrak g} : G \\
&t : \mathbb R \mapsto \phi(e^{t\mathfrak g}) : M \\
&X_{\mathfrak g} \equiv \frac{d}{dt}(\phi(e^{t\mathfrak g}))|_{t = 0}: TM
\end{aligned}$

We can then attempt to recover a hamiltonian $H_{\mathfrak g}$ from
$X_{\mathfrak g}$. If we get a hamiltonian from this process, then it
will have the right symmetries.