Definition of a derivative of differential form

Question

While reading a paper I encountered the following:

Let $(\mathbf{q,p}) \in \mathbb{R}^{2n}$ be canonical coordinates and let $H: \mathbb{R}^{2n} \to \mathbb{R}$ be a smooth function.

The continuous-time Hamiltonian system

$\mathbf{q}_t = + \nabla_p H(q, p)$

$\mathbf{p}_t = - \nabla_q H(q, p)$

preserves exactly the symplectic form $\mathbf{\omega = dp \wedge dq}$, that is $(d/dt)\omega = 0$

Here $\mathbf{p} , \mathbf{q} \in \mathbb{R}^n$

How is $(d/dt)\omega $ even defined? I have seen exterior derivatives of differential forms, but this is evidently something else.

Also, how can we prove that (for the given Hamiltonian system) $(d/dt)\omega = 0$?

Any help will be greatly appreciated.

Answer 1

$\def\L{\mathcal L}$From what you've written there is no dependence of the Hamiltonian on time (indeed a time variable is not introduced at all), so the interpretation in the comments doesn't feel right. Without more context, I would assume that by $(d/dt)\omega$ the author means the Lie derivative $\L_X \omega$ where

$$X = \sum_j \left(\nabla_{p_j} H \frac{\partial}{\partial q^j}-\nabla_{q^j} H \frac{\partial}{\partial p_j}\right)$$ is the Hamiltonian vector field, whose flow gives the time evolution of the system (thus the notation $d/dt$).

Answer 2

As mentioned in the comments, you may simply differentiate componentwise.

Alternatively, given any time-dependent $k$-form $\omega_t$ on a manifold $M$, note that at each point $p \in M$, $\omega_t(p)$ may be viewed as a curve $\mathbb{R} \to \bigwedge^k(T_p^*M)$ in a finite-dimensional vector space, defined by $t \mapsto \omega_t(p)$. There is a canonical smooth structure on any finite-dimensional vector space, and so it makes sense to differentiate this curve. Verify that this agrees with the previous definition!