How to express the Euler-Lagrange equation in arbitrary coordinates where $\omega \neq \sum dp_i \wedge dq_i$

Question

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

1. In a "truly" coordinate-independent way (which is more desired for me)
2. 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...

Answer 1

The answer is in this short paper by Erik Curiel. The Geometry of the Euler-Lagrange Equation

The key idea is to express the Euler-Lagrange equation in the tangent bundle, in terms of a structure tensor that the tangent bundle automatically comes equipped with.

Let $Q$ be the coordinate manifold. Pick coordinate chart $q_i$ of $Q$. Then $q_i, \dot{q}_i$ is a chart of the tangent bundle $TQ$.

Define a $(1,1)$ tensor $J$ on $TQ$ by

$$ J \begin{bmatrix} \vec{\dot{q}} \cr \vec{\ddot{q}} \end{bmatrix} = \begin{bmatrix} 0 \cr \vec{\dot{q}} \end{bmatrix} $$

You can check that this is well defined and doesn't depend on the coordinate chart.

Let $\xi$ be a vector field in $TQ$, which is to be the solution of the Euler-Lagrange equation. Then the equation can be expressed as:

$$ \mathscr{L}_\xi ({dL} \cdot J) = dL $$