The reason for symplectomorphism to conserve the canonical form of the Hamilton equations.

If I have $$(M,\omega)$$ with Hamiltonian a symplectic manifold, let $$(q_1,p_1,...,q_n,p_n)$$ be the Darboux coordinates. With these coordinates, the integral curves of the Hamiltonian vector field satisfy the Hamiltonian differential equations (canonical form) :
$$\dot{q}(t)=\frac{\partial H}{\partial p}$$
$$\dot{p}(t)=-\frac{\partial H}{\partial q}$$
Le $$f$$ be a symplectomorphism, i.e. $$f : M \to M$$, such that $$f^*\omega=\omega$$. Then apparently $$f$$ should be a canonical transformation, i.e. a change of variable for which the Hamiltonian equations are also in the canonical form in the new variables $$\hat{q}$$, $$\hat{p}$$. So for me it seems sufficient to show that in the new coordinates, $$\omega = \sum d\hat{q}_i\wedge d\hat{p}_i$$ as well, is it?.
But $$\omega = f^*(\omega) = f^*(\sum dq_i\wedge dp_i)=\sum d(q_i \circ f)\wedge d(p_i \circ f) =\sum d\hat{q}_i\wedge d\hat{p}_i$$. Is this the proof that symplectomorphism conserve the canonical form of the Hamilton equations?

• as far as I remember, given the symplectic transformation $f$, you just need to make sure that the formula for $\omega$ remains valid in new coordinates. that's basically what you saying I guess
– Jane
May 13 '19 at 23:57

• Darboux coordinates satisfy, by definition, $$\omega = \sum_{i=1}^n {\rm d}q^i\wedge {\rm d}p_i$$.
• For any set of Darboux coordinates, the Hamiltonian vector field of $$H\colon M \to \Bbb R$$ is expressed by $$X_H = \sum_{i=1}^n \left(\frac{\partial H}{\partial p_i}\frac{\partial}{\partial q^i} - \frac{\partial H}{\partial q^i}\frac{\partial}{\partial p_i}\right)$$
Thus $$\sum_{i=1}^n \left(\frac{\partial H}{\partial p_i}\frac{\partial}{\partial q^i} - \frac{\partial H}{\partial q^i}\frac{\partial}{\partial p_i}\right)=X_H=\sum_{i=1}^n \left(\frac{\partial H}{\partial \hat{p}_i}\frac{\partial}{\partial \hat{q}^i} - \frac{\partial H}{\partial \hat{q}^i}\frac{\partial}{\partial \hat{p}_i}\right).$$
• It is an if and only if right? Darboux coordinates are sent to Darboux coofdinates if and only if $f^*\omega = \omega$ right? Because of $\omega = f^*(\omega) = f^*(\sum dq_i\wedge dp_i)=\sum d(q_i \circ f)\wedge d(p_i \circ f) =\sum d\hat{q}_i\wedge d\hat{p}_i$ May 14 '19 at 9:56
• Yes, $f$ will be at least a local symplectomorphism. May 14 '19 at 16:07