Symplectomorphisms preserves Hamiltonian equations

Question

Let $(M_1,\omega_1)$, $(M_2,\omega_2)$ be symplectic manifolds and let $\psi:(M_1,\omega_1)\rightarrow (M_2,\omega_2)$ be a symplectomorphism. Consider a Hamiltonian $H\in\mathcal{C}^\infty(M_2)$. Show that a curve $t\mapsto \gamma(t)\in M_1$ solves Hamilton equation for $\tilde{H}:=H\circ\psi$ $\iff$ the curve $t\mapsto \psi\circ\gamma(t)\in M_2$ solves them for $H$.
It will be great if it is obtained as a corollary of the fact that $X_{H\circ\psi}=\psi^*(X_H)$.
Thank you

Answer 1

My attempt.
Proposition 1
Let $(M_1,\omega_1)$, $(M_2,\omega_2)$ be symplectic manifolds and let $\psi:(M_1,\omega_1)\rightarrow (M_2,\omega_2)$ be a symplectomorphism. Let, moreover $H\in\mathcal{C}^\infty(M_2)$ be a Hamiltonian, with Hamiltonian vector field $X_H\in\Gamma(TM_2)$.
Then $\psi^*(X_H)$ is the Hamiltonian vector field w.r.t. the Hamiltonian $\tilde{H}:=H\circ\psi\in\mathcal{C}^\infty(M_1)$, i.e. \begin{equation} \label{ciao} X_{H\circ\psi}=\psi^*(X_H). \end{equation}

proof \begin{equation*} d(H\circ\psi)=d(\psi^*H)=\psi^*dH=-\psi^*\left(i_{X_H}\omega_2\right)=-i_{\psi^*X_H}\psi^*\omega_2=-i_{\psi^*X_H}\omega_1, \end{equation*} so $\psi^*X_H$ is the unique Hamiltonian vector field w.r.t $H\circ\psi$.

Corollary 1
$X_{H\circ\psi}$ and $X_H$ are $\psi$-related

proof
We can explicit conclusion of Proposition 1, which means that $\forall p\in M_1$, $\forall h\in\mathcal{C}^\infty(M_1)$ \begin{equation} \label{expanded correlation Hamiltonian vector fields} (X_{H\circ\psi})_p(h)=(\psi^*X_H)_p(h)=:(X_H)_{\psi(p)}(h\circ\psi^{-1}), \end{equation} where we have developed the pull-back of vector fields through diffeomorphisms. Now, take any $p\in M_1$ and $g\in\mathcal{C}^\infty(M_2)$, then \begin{equation} \label{first related vector fields} \left[T_p\psi((X_{H\circ\psi})_p)\right](g)=(X_{H\circ\psi})_p(g\circ\psi); \end{equation} apply the first equation with $h:=g\circ \psi$, then we have \begin{equation*} \left[T_p\psi((X_{H\circ\psi})_p)\right](g)=(X_H)_{\psi(p)}(g\circ\psi\circ\psi^{-1})=(X_H)_{\psi(p)}(g). \end{equation*} Since this hold for every $g\in\mathcal{C}^\infty(M_2)$, we conclude that \begin{equation*} \left[T_p\psi((X_{H\circ\psi})_p)\right]=(X_H)_{\psi(p)}(g\circ\psi\circ\psi^{-1})=(X_H)_{\psi(p)}, \end{equation*} which means exactly that $X_{H\circ\psi}$ and $X_H$ are $\psi$-related.

Proposition 2
Let $F:M\rightarrow N$ be a smooth map between manifolds and suppose that $X\in\Gamma(TM)$, $Y\in\Gamma(TN)$ are $F$-related vector fields. Then $F$ takes integral curves of $X$ to integral curves of $Y$.

Proof
Let $\gamma:\mathcal{I}\rightarrow M$ be an integral curve of $X$, we have to show that $\sigma:=F\circ\gamma$ is an integral curve of $Y$: \begin{equation*} \dot{\sigma}(t)=\frac{d}{dt}(F\circ\gamma)(t)=T_{\gamma(t)}F(\dot{\gamma}(t))=T_{\gamma(t)}F(X_{\gamma(t)})=Y_{F(\gamma(t))}=Y_{\sigma(t)}. \end{equation*}

Conclusion
Symplectomorphisms preserve Hamilton's equations.

proof
Let $\psi$ be a symplectomorphism, then, thanks to Corollary 1, we see that Hamiltonian vector fields $X_{H\circ\psi}$ and $X_H$ are related through $\psi$. Moreover by Proposition 2, $\psi$ maps integral curves to integral curves of $\psi$-related generic vector fields. But integral curves of Hamiltonian vector fields are solutions of Hamilton's equations and so $\psi$ preserves Hamilton's equations.

Answer 2

Since you mentioned Abraham-Marsden as a source, here are a few comments which I think you'll find beneficial (the notation is very much identical to how they use it). Here is a more "streamlined approach" (atleast in my opinion) which is at the "mapping level" rather than the "pointwise level".

1. I hope you realize that the conclusion of Proposition 1 can be written as $\psi^*(X_H) = X_{\psi^*H}$, which of course makes it very memorable. Similarly, by replacing $\psi$ by $\psi^{-1}$, and using the fact that $(\psi)_*:= (\psi^{-1})^*$ (i.e push-forward is same as pull-back by inverse (by definition)), we get $\psi_*(X_H) = X_{(\psi_*H)}$ (of course you have to redefine where everything is defined)

2. Recall that if $F:M \to N$ and $X$ and $Y$ are vector field on $M$ and $N$ respectively, then we say $X$ and $Y$ are $F$-related if $TF \circ X = Y \circ F$, and we write $X\sim_F Y$; i.e the following diagram commutes $\require{AMScd}$ \begin{CD} TM @>TF>> TN\\ @A{X}AA @AA{Y}A \\ M @>>F> N \end{CD} Finally, recall the definition of the pull-back of a vector field (this requires $F$ to be a diffeomorphism): $F^*(Y):= TF^{-1}\circ Y \circ F$ (and note that $T(F^{-1}) = (TF)^{-1}$, so simply writing $TF^{-1}$ isn't ambiguous). With this, corollary 1 is simple to prove: \begin{align} T\psi \circ X_{\psi^*H} &= T\psi \circ (\psi^*X_H) \tag{by proposition $1$} \\ &= T\psi \circ (T\psi^{-1}\circ X_H \circ \psi) \tag{by definition} \\ &= X_H \circ \psi \end{align} This says exactly that $X_{\psi^*H} \sim_{\psi}X_H$ that the two vector fields are $\psi$-related.

3. We can rewrite the proof of Proposition $2$ as follows: \begin{align} (F\circ \gamma)' &= T F \circ \gamma' \\ &= TF \circ (X\circ \gamma) \\ &= (Y\circ F) \circ \gamma \tag{since $X\sim_F Y$} \\ &= Y \circ (F\circ \gamma) \end{align} This says exactly that $F\circ \gamma$ is an integral curve of $Y$. Here, I use $\gamma'$ where you use $\dot{\gamma}$; this is a curve in the tangent bundle $I\subset \Bbb{R}\to TM$