§ Symplectic version of classical mechanics

§ Basics, symplectic mechanics as inverting $\omega$:

I've never seen this kind of "inverting $\omega$" perspective written down anywhere. Most of them start by using the inteior product $i_X \omega$ without ever showing where the thing came from. This is my personal interpretation of how the symplectic version of classical mecanics comes to be. If we have a non-degenerate, closed two-form $\omega: T_pM \times T_pM \rightarrow \mathbb R$. Now, given a hamiltonian $H: M \rightarrow \mathbb R$, we can construct a 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.