# Stokes' theorem and symplectic geometry

Let $V = \mathbb{R}^2,$ as a vector space then the Poincaré invariant is an integral

$\int_{\gamma} \theta$ where $\theta = p dx$ is the symplectic 1-form and $\gamma$ a closed curve. Now, it is interesting that in the 2d-case, that by Stokes' theorem $\gamma$ natually defines an area $A$ with $\partial A = \gamma$ and $\int_{\gamma} \theta = \int_{A} d \theta = \int_{A} dp \wedge dx.$ So what this invariant actually defines is the phase space volume.

The situation gets more difficult when we are dealing with higher-dimensional spaces

Let $V = \mathbb{R}^{2n}$ with $n>1$, then we can still define

$\int_{\gamma} \sum_{i} p_i dx_i.$ Now, I started wondering: Can we still identify this with some area or interpret what this quantity actually tells us?

A natural thing to do would be to assume

$\int_{\gamma} \sum_{i} p_i dx_i = \int_{A} \sum_{i} dp_i \wedge dx_i.$

This basically means that we are summing up the area of the projection down to each $x_i,p_i$ plane, but now I don't know whether $\gamma$ naturally defines such an $A$ so that we can apply Stokes' theorem?

But even if you cannot make any sense out of my last equation, it would be interesting for me to know whether we can interpret what $\int_{\gamma} \sum_i p_i dx_i$ actually tells us?

If anything is unclear, please let me know.

First, a couple of terminological corrections.

$\int_{\gamma} \theta$ where $\theta = p dx$ is the symplectic 1-form and $\gamma$ a closed curve.

"Symplectic 1-form" is a misnomer. A symplectic form is by definition a 2-form. The form $p\, dx$ is called a symplectic potential, or in the case of a phase space, the tautological 1-form or canonical 1-form.

Now, it is interesting that in the 2d-case, that by Stokes' theorem $\gamma$ natually defines an area $A$ ... .

"Area" is the wrong word here. An area is a nonnegative real number. A region in $\mathbb R^2$ or a surface can have an area, but cannot be an area. What you want to say is that $\gamma$ defines a region in the plane.

With those corrections in mind, here's what can be said in the higher-dimensional case. Given a smooth (or piecewise-smooth) closed curve $\gamma$ in $\mathbb R^{2n}$, there are typically many smooth compact surfaces that have $\gamma$ as a boundary. For any such surface $A$, Stokes's theorem yields $$\int_{\gamma} \tau = \int_{A} \omega,$$ where $\tau = \sum_{i} p_i dx_i$ is the tautological 1-form and $\omega=d\tau = \sum_i dp_i \wedge dx_i$ is the symplectic 2-form. This is not in general equal to the area of $A$, but because $\omega$ is a calibration, it is always less than or equal to the area. It is equal exactly when $A$ is a holomorphic curve or antiholomorphic curve, meaning that it can be locally parametrized by $n$ holomorphic or antiholomorphic functions (thinking of $(x_1 + ip_1,\dots,x_n+ip_n)$ as complex variables).

• thank you for the corrections, you say that $\int_{A} \omega$ is always equal of less than the area, but is it legitimate to say that it is exactly the area we get from summing up the projected area on all the $x_i,p_i$ planes? Commented Jul 18, 2015 at 20:25