§ Symplectic version of classical mechanics


§ Basics, symplectic mechanics as inverting ω\omega:


I've never seen this kind of "inverting ω\omega" perspective written down anywhere. Most of them start by using the inteior product iXωi_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 ω:TpM×TpM→R\omega: T_pM \times T_pM \rightarrow \mathbb R.
Now, given a hamiltonian H:M→RH: M \rightarrow \mathbb R, we can construct a vector field XH:M→TMX_H: M \rightarrow TM under the definition:
partially apply ω to see ω as a mapping from Tp to Tp∗Mω2:TpM→Tp∗M≡λ(v:TpM).λ(w:TpM).ω(v,w)ω2−1:Tp∗M→TpM;dH:M→Tp∗MXH≡λ(p:M)→ω2−1(dH(p))(p:M)→dHdH(p):Tp∗M→ω2−1ω2−1(dH(p)):TpMXH=ω2−1∘dH \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→RH: M \rightarrow \mathbb R, we can construct an associated vector field XHX_H, in a pretty natural way.
We can also go the other way. Given the XX, we can build the dHdH under the equivalence:
ω2−1∘dH=XHdH=ω2(XH)∫dH=∫ω2(XH)H=∫ω2(XH) \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 dHdH being integrable. But this works, and gives us a bijection between XHX_H and HH as we wanted.
We can also analyze the definition we got from the previous manipulation:
ω2(XH)=dHλ(w:TpM)ω(XH,w)=dHω(XH,⋅)=dH \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 XHX_H and dHdH. Exploiting this, we can notice that dH(XH)=0dH(X_H) = 0. That is, moving along XHX_H does not modify dHdH:
ω2(XH)=dHλ(w:TpM)ω(XH,w)=dHdH(XH)=ω(XH,XH)=0 ω is anti-symmetric \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 ω\omega


We wish to show that XH∗(ω)=ωX_H^*(\omega) = \omega. That is, pushing forward ω\omega along the vector field XHX_H preserves ω\omega.
TODO.

§ Moment Map


Now that we have a method of going from a vector field XHX_H to a Hamiltonian HH, 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 g∈G\mathfrak{g} \in \mathfrak{G} to a vector field XgX_{\mathfrak g}, performing:
t:R↦etg:Gt:R↦ϕ(etg):MXg≡ddt(ϕ(etg))∣t=0:TM \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 HgH_{\mathfrak g} from XgX_{\mathfrak g}. If we get a hamiltonian from this process, then it will have the right symmetries.