The power of exterior differential calculus in CM

Question

Classical Mechanics has a beautiful interpretation in terms of symplectic geometry and exterior differential calculus. For example, the Hamilton equations of motion can be written in a coordinate free way as $$ dH = \omega(\cdot,\Delta) $$ where $H$ is the Hamiltonian, $\omega$ is the canonical two-form, and $\Delta$ is the Hamiltonian vector field. But students naturally wonder what the practical usefulness of all of this is. Are there any examples where we can solve this equation (say, obtain the periods of some dynamical systems) without going to coordinates and integrate the equation (or solve the Hamilton-Jacobi equation, .. )?

Or, at the very least, are there any proofs (say Arnold-Liouville theorem or something like that) that are significantly simplifed by the use of exterior differential calculus, Lie groups, etc?

Answer 1

Or, at the very least, are there any proofs (say Arnold-Liouville theorem or something like that) that are significantly simplifed by the use of exterior differential calculus, Lie groups, etc?

Liouville's theorem (assuming by that you mean the preservation of phase volumes) is one such example actually.

Suppose that $(M,\omega, H)$ is a Hamiltonian system with symplectic form $\omega$, and $\dim M=2n$. Then $$ \Omega=\omega^{\wedge n}\ \text{(n-fold wedge product)} $$ is the Liouville volume form.

Suppose then that $\mathcal D$ is a compact domain of integration in $M$. The volume is $$ \mathrm{vol}\ \mathcal D=\int_\mathcal D\Omega. $$

Now, let's fix some conventions which may be opposite of what everyone is using, since I don't want to check it now (the reasoning's the same nontheless), if $X\in\mathfrak X(M)$ is a vector field, then $\imath_X\omega$ is its dual where the interior product inserts into the first argument. The dual of the symplectic form is $\omega^\ast\in\mho^2(M)$. If $H$ is the Hamiltonian (then following OPs convention), $$ \mathrm dH=-\imath_{X_H}\omega. $$

Let $\phi_t$ denote the flow of the Hamiltonian vector field $X_H$. The change of volume of the region $\mathcal D$ after evolving it for time $t$ is $$\mathrm{vol}\ \mathcal D(t)=\int_{\phi_t(\mathcal D)}\Omega=\int_\mathcal D\phi_t^\ast\Omega. $$

The time-derivative of this at $t=0$ is $$ \frac{\mathrm d}{\mathrm dt}\mathrm{vol}\ \mathcal D(0)=\frac{\mathrm d}{\mathrm dt}\int_\mathcal D\phi^\ast_t\Omega|_{t=0}=\int_\mathcal D\mathscr L_{X_H}\Omega=\int_\mathcal D \mathrm d\imath_{X_H}\Omega. $$

Here we have used the Cartan formula ($\mathscr L_X=\mathrm d\imath_X+\imath_X\mathrm d$) and the interior product in the last term can be expanded as $$ \imath_{X_H}\Omega=\imath_{X_H}\omega\wedge...\wedge\omega+\omega\wedge\imath_{X_H}\omega\wedge...\wedge\omega=n\imath_{X_H}\omega\wedge...\wedge\omega=-n\mathrm dH\wedge...\wedge\omega. $$

All factors in this product are closed, so $\mathrm d\imath_{X_H}\Omega=0$, hence $$ \frac{\mathrm d}{\mathrm dt}\mathrm{vol}\ \mathcal D(0)=0. $$