Symplectic Structure & Characteristic Vector Field

Sep 25—2025

This is post a continuation of posts I am making based upon my reading of the first lecture of Arnold’s PDE.

Recall our goal: we are trying to find a characteristic field for the general first-order PDE in the $J^1(\mathbb{R}^n, \mathbb{R})$ space. The field should arise from the field of contact planes, $dy = pdx$ and the surface $\Gamma^{2n}$ given by the PDE. The problem we have encountered is that the intersection of the surface and the contact planes (which are hyperplanes of dimension 2n) at regular points is a (2n-1)-dimensional space. How are we suppose to determine a vector in each one of these (2n-1)-dimensional space to obtain a smooth vector field that uniquely corresponds to our PDE?

Key idea: We endow the space of tangent vectors lying on each contact plane (this is a 2n-dimensional vector space) with the so-called ‘symplectic structure’. Unlike in the Euclidean space, the orthogonal complement to a hyperplane would be a 1-dimensional subspace ‘lying outside’ of the hyperplane, in this case, the orthogonal complement would be a 1-dimensional subspace lying inside the hyperplane. Thus, the characteristic vector field we want is obtained by taking the orthogonal complement (2n-1)-dimensional intersection space in terms of the symplectic structure.

Step 1: Defining the symplectic structure on the contact planes.

Recall, in the last post, we defined the 2-form $\omega^2 = d\alpha$ . This is the linear part of integrating $\alpha = dy - pdx$ around the parallelogram with sides $\xi, \eta$ and vertex at 0 ( $\xi, \eta$ are the 2 inputs for our 2-form). We have shown that $\omega|_{\alpha = 0}$ is defined up to a non-zero multiple at each point, i.e. the 2-forms obtained from $\alpha$ and $f \alpha$ (where $f$ is a function of x, y, p) by exterior differentiation are the same when restricted to the contact planes.

Last time, we found the expression for the 2-form to be: $d\alpha = - \sum dp_i \wedge dx_i$ .

Given two arbitrary vectors $\xi = (x_1, ..., x_n, y, p_1, ..., p_n)$ and $\eta = (\tilde{x_1}, ..., \tilde{x_n}, \tilde{y}, \tilde{p_1}, ..., \tilde{p_n})$ on the same contact plane:

$d\alpha(\xi, \eta) = \sum(-x_k \tilde{p_k} + p_k \tilde{x_k})$ .

A new view: Just like how we can interpret a 1-form as the map $\xi \mapsto \vec{v} \cdot \xi$ for some fixed $\vec{v}$ , we can also interpret the 2-form as the following map:

$(\xi, \eta) \mapsto (\eta)^{T} \bf{A} \xi$ for some fixed square matrix $\bf{A}$ .

Let’s find the corresponding matrix for $d\alpha$ .

In $\mathbb{R}^3$ , the action of the 2-form is:

$\bf{A}_1 := (\begin{bmatrix} x \\ y \\ p \end{bmatrix}, \begin{bmatrix} \tilde{x} \\ \tilde{y} \\ \tilde{p} \end{bmatrix}) \mapsto -x \tilde{p} + p \tilde{x}$

It is easy to see that the corresponding matrix is:

$\begin{bmatrix} 0 & 0 & 1 \\ 0 & 0 & 0 \\ -1 & 0 & 0 \end{bmatrix}$

in the coordinates $x, y, p$

Generalizing this to $\mathbb{R}^{2n+1}$ , we obtain:

$\bf{A}_n := \begin{bmatrix} 0_{n \times n} & 0_{n \times 1} & I_{n \times n} \\ 0_{1 \times n} & 0 & 0_{1 \times n} \\ -I_{n \times n} & 0_{n \times 1} & 0_{n \times n} \end{bmatrix}$

Notice that this matrix is invertible. Hence, for any non-zero $(x_1, ..., x_n, y, p_1, ..., p_n)$ , there exists $(\tilde{x_1}, ..., \tilde{x_n}, \tilde{y}, \tilde{p_1}, ..., \tilde{p_n})$ such that: