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

Question

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?

Answer 1

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