TITLE: What is Symplectic Area?
QUESTION [6 upvotes]: In classical Mechanics, momentum and position can be paired together to form a symplectic manifold.  If you have the simple harmonic oscillator with energy $H = (k/2)x^2 + (m/2)\dot{x}^2$.  In this case, the orbits are ellipses.  How is the vector field determined by the (symplectic) gradient, then?  
Also, does anyone know an interpretation for the area inside a closed curve in phase space?

REPLY [2 votes]: You can view $\mathbb{R}^{2n}$ as a quotient of the real Heisenberg group $\mathcal{H}^{2n+1}$ modulo its center. For a closed loop $\alpha$ in $\mathbb{R}^{2n}$ and a point in  $\mathcal{H}^{2n+1}$ over $\alpha(0)$ there's unique lift $\tilde{\alpha}$ of $\alpha$ to $\mathcal{H}^{2n+1}$ going through this point. The symplectic area enclosed by $\alpha$ expresses the signed distance from $\tilde{\alpha}(0)$ to $\tilde{\alpha}(1)$ with respect to a left invariant Riemannian metric on $\mathcal{H}^{2n+1}$.