Let $Q$ be a manifold Let $q^\alpha$ for $\alpha \in 1 .. n$ be a chart of $Q$.
The tangent bundle $TQ$ has a chart $q^\alpha, \dot{q}^\alpha$. It has an almost-tangent structure:
$$ J = dq^\alpha \otimes \frac{\partial}{\partial \dot{q}^\alpha} \text{ is a (1,1) tensor on $TQ$} \\ \Delta = \dot{q}^\alpha \frac{\partial}{\partial \dot{q}^\alpha} \text{ is a vector field on $TQ$ } $$ A vector field $X$ on $TQ$ is "second order" if $JX = \Delta$, ie if $X(q^\alpha) = \dot{q}^\alpha$. A second order vector field on $TQ$ is the same thing as a second order ODE on $Q$.
Let $L : TQ \to \mathbb{R}$ be a Lagrangian. The Euler-Lagrange equation is
$$ JX = \Delta \\ \mathscr{L}_X (dL \cdot J) = dL $$
The cotangent bundle $T^*Q$ has a chart $q^\alpha, p_\alpha$, and a symplectic structure $\omega = dq^\alpha \wedge dp_\alpha$.
The Legendre transform gives a map $\tau : TQ \to T^*Q$ given by $p_\alpha = \frac{\partial L}{\partial \dot{q}^\alpha}$. This is a local diffeomorphism if and only if $\frac{\partial^2L}{\partial \dot{q}^\alpha \partial \dot{q}^\beta}$ is not singular.
Define $\Omega = -d(dL \cdot J) $ This 2-form is the pullback of the symplectic form on the cotangent bundle. $\Omega = \tau^*\omega$. It will give a symplectic structure to $TQ$ if $\frac{\partial^2L}{\partial \dot{q}^\alpha \partial \dot{q}^\beta}$ is not singular.
Define $H = \Delta(L) - L$. The Euler-Lagrange equations can be rewritten as Hamilton's equations:
$$ JX = \Delta \\ \Omega(X, Y) = Y(H) \text{ for all $Y$}$$
Assume $\frac{\partial^2L}{\partial \dot{q}^\alpha \partial \dot{q}^\beta}$ is not singular.
What I'm confused by is that this pair of equations is overdetermined. $\Omega(X, Y) = Y(H)$ is enough to fully determine $X$, so it should be possible to prove $JX = \Delta$. But I can't seem to figure out how. Are additional assumptions necessary to prove this?
