How does the symplectic form $\omega$ manifests itself in the Euler-Lagrange equation? + Extreme confusion with time

Question

Let $\omega$ be a symplectic manifold on $\mathbb{R}^n$ and the smooth function $H : \mathbb{R}^n \times \mathbb{R}^n \to \mathbb{R}$ be a Hamiltonian. For $p,q \in \mathbb{R}^n$ let us assume that \begin{equation} p \to H(p,q) \text{ and } q \to H(p,q) \text{ are both convex } \end{equation} and \begin{equation} \lim\limits_{p \to \infty} \frac{H(p,q)}{\lvert p \rvert} = \infty \end{equation}

Then, it is well-known that the Lagrangian $L : \mathbb{R}^n \times \mathbb{R}^n \to \mathbb{R}$ defined by \begin{equation} L(v,q):=\sup_{p \in \mathbb{R}^n} \bigl\{ p \cdot v - H(p,q) \bigr \} \end{equation} satisfies the above properties assumed for $H$.

Now, the Hamilton's equation for the Hamiltonian vector field $X_H : \mathbb{R}^n \to \mathbb{R}^n$ may be expressed in the coordinate-free way by \begin{equation} \omega(X_H, \cdot) = -dH \end{equation}

Here, some very frustrating confusion arises...

1. In the above formalism, there was no such thing as the time derivative. But, the Euler-Lagrange equation based on $L(v,q)$ is written in terms of time derivative as \begin{equation} \frac{d}{dt} \frac{\partial L}{\partial v} - \frac{\partial L}{\partial q}=0 \end{equation} However, the Hamilton equation $\omega(X_H, \cdot) = -dH$ is supposed to be "EQUIVALENT" to the Euler-Lagrange equation. What have I missed?

2. Moreover, the Euler-Lagrange equation does NOT contain the symplectic form $\omega$. What kind of trick has happened? Why has the symplectic form $\omega$ just "EVAPORATED" in the middle of nowhere? Is it possible to write the Euler-Lagrange equation with $\omega$ explicitly appearing on?

These two confusions are quite frustrating and never explained explicitly in any standard textbook..... Could anyone please clarify for me? I have spent hours and days trying to make sense of all the stuff but only get more confused in the end....

Answer 1

There is a time derivative implicit in forming the flow of a vector field. Writing out explicitly $X_H(p,q)=(H_q,-H_p)$, using subscripts for partial derivatives, the equations of flow lines of $X_H$ are $\dot{p}=H_q(p,q)$, $\dot{q}=-H_p(p,q)$. (Your signs may differ.) To have a unique $v$ giving our supremum above, we need it to occur at a critical point for $vp-H(p,q)$, in the $p$ variable, i.e. differentiating in $p$ and setting derivative to zero, the supremum occurs where $v=H_p$, i.e. the Legendre transform $(p,q)\mapsto (v,q)=(H_p,q)$ is well defined. By the same reasoning its inverse map has to be the inverse Legendre transform $(v,q)\mapsto (p,q)=(L_v,p)$. Then $H(p,q)+L(v,q)=vp$ at corresponding points. So then $\dot{p}=H_q$ is precisely $$\frac{d}{dt}\frac{\partial L}{v}=L_q$$, and similarly $\dot{q}=-H_p$ is precisely $\dot{q}=-v$, so $v$ is (up to some messing around with signs) the velocity. We can write $\omega=dq\wedge dp=dq\wedge L_{vv} dv$, and this leads us to see the importance of the $L_{vv}$ matrix. Feel free to clean up the signs.