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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?

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  • $\begingroup$ 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 $\endgroup$
    – Jane
    May 13 '19 at 23:57
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Yes. What you have is correct. It is just a matter of piecing together the following facts:

  • 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 $$X_H = \sum_{i=1}^n \left(\frac{\partial H}{\partial p_i}\frac{\partial}{\partial q^i} - \frac{\partial H}{\partial q^i}\frac{\partial}{\partial p_i}\right)$$

  • Symplectomorphisms take Darboux coordinates to Darboux coordinates.

Thus $$\sum_{i=1}^n \left(\frac{\partial H}{\partial p_i}\frac{\partial}{\partial q^i} - \frac{\partial H}{\partial q^i}\frac{\partial}{\partial p_i}\right)=X_H=\sum_{i=1}^n \left(\frac{\partial H}{\partial \hat{p}_i}\frac{\partial}{\partial \hat{q}^i} - \frac{\partial H}{\partial \hat{q}^i}\frac{\partial}{\partial \hat{p}_i}\right).$$

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  • $\begingroup$ 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 $\omega = f^*(\omega) = f^*(\sum dq_i\wedge dp_i)=\sum d(q_i \circ f)\wedge d(p_i \circ f) =\sum d\hat{q}_i\wedge d\hat{p}_i$ $\endgroup$
    – roi_saumon
    May 14 '19 at 9:56
  • $\begingroup$ Yes, $f$ will be at least a local symplectomorphism. $\endgroup$
    – Ivo Terek
    May 14 '19 at 16:07

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