I posted my questions in a previous post MO, but it seems that a more refined version for question on the Euler-Lagrange equation is needed. So, I post my question again.

In standard symplectic geometry, we assume the "canonical coordinates" $(p_i,q_i)$ at least locally, under with the symplectic form $\omega$ is written as \begin{equation} \omega=\sum_i dp_i \wedge dq_i \ . \end{equation}

Then the Hamilton's equations are $\dot{p_i}= -\frac{\partial H}{\partial q_i}$ and $\dot{q_i}= \frac{\partial H}{\partial p_i}$ while the Euler-Lagrange equation is \begin{equation} \frac{d}{dt}\frac{\partial L}{\partial \dot{q}_i}=\frac{\partial L}{\partial q_i} \end{equation}

However, I am curious about how to write the Hamilton's equations and Euler-Lagrange equation in "arbitrary" coordinates $(p'_i,q'_i)$ where $\omega \neq \sum_i dp'_i \wedge dq'_i$.

In "Classical Dynamics : A Contemporary Approach" (1998) by Jose Saletan, the Euler Lagrange equation is expressed in a coordinate-free way by \begin{equation} \mathcal{L}_\Delta \theta_L = d L \end{equation} where $\Delta$ is the vector field generating time translation and $\theta_L$ is the canonical $1$-form corresponding to the Lagrangian $L$. However, he only defines these quantities in terms of canonical (local) coordinates as \begin{equation} \theta_L : =\sum_i \frac{\partial L}{\partial \dot{q}_i} dq_i \text{ while } \Delta:= \sum_i \dot{q_i} \frac{\partial}{\partial q_i} + \ddot{q_i} \frac{\partial}{\partial \dot{q_i}} \end{equation}

So, I don't think it "truly" coordinate-free...

Could anyone please provide the general formula for the Euler-Lagrange equation either

- In a "truly" coordinate-independent way (which is more desired for me)
- In "arbitrary" local coordinates $(p'_i,q'_i)$ in which $\omega \neq \sum_i dp'_i \wedge dq'_i$

I would deeply appreciate any help...