Contact Manifolds (Arnold’s Geometrical Methods of ODE)

Definition: a contact manifold is a manifold equipped with a field of hyperplanes in its tangent spaces satisfying the condition of “maximal non-integrability”.

Example:

Recall the space of 1-jets of functions that we have seen when working with quasi-linear PDEs. Suppose the coordinates in this space to be $(x_1, ..., x_n, u, p_1, ..., p_n)$ . The field of hyperplanes is given by:

$du - pdx = 0$ .

Goal: In this post, we will understand what it means for a contact manifold to be non-integrable (and what it means for it to be maximal non-integrable) and we will see why the space of 1-jets with the field of planes is indeed a contact manifold.

Step 1: non-integrability

Unlike a field of one-dimensional directions, a field of planes may not have integral surfaces (i.e. surface whose tangent space is a subspace of the plane at each point) whose dimension is equal to the dimension of the planes.

How can we tell if a field of hyperplanes is not integrable?

Let $O$ be a point on the manifold and $\mathcal{P}$ be the plane of the field at $O$ . We choose coordinates $(x_1, ..., x_n)$ so that $\mathcal{P}$ is given by the hyperplane spanned by $x_1, ..., x_{n-1}$ and $O$ is the origin. Let $x_1, ..., x_{n-1}$ be the horizontal coordinates and $x_n$ be the vertical coordinate.

Let $\xi, \eta$ be two vectors starting at $O$ on $\mathcal{P}$ . Consider the parallelogram $\Pi$ spanned by $\xi, \eta$ on $\mathcal{P}$ ; there are two paths along the boundary of $\Pi$ from $O$ to the opposite vertex. Construct the vertical parallelepiped with $\Pi$ as its base and its lateral surface parallel to the vertical direction.

The intersections of the field of planes with the lateral surfaces of the parallelepiped gives a direction field on the lateral surfaces. We can now lift the two paths so that they are integral curves of this direction field. Note: the projection of these two integral curves onto $\mathcal{P}$ gives us the original two paths.

The end points of these two integral curves are two points above the opposite vertex which are different in general. This difference is what prevents us from constructing integral surfaces.

The principal bilinear part of this difference, with respect to $\xi, \eta$ , gives us the degree of non-integrability of the field. This is the geometric equivalence of the exterior derivative $d\alpha$ (where $\alpha$ is the 1-form that gives us the field of planes).

The difference being non-zero is equivalent to saying the exterior derivative $d\alpha$ of $\alpha$ is a non-degenerate 2-form on every plane $\alpha = 0$ .

In fact, a 1-form $\alpha$ which is nowhere equal to the zero form on $M$ is called a contact form if $d\alpha$ is non-degenerate.

If $d\alpha$ happens to be degenerate at one point or a discrete set of points, then is the corresponding contact manifold integrable?

In the example of the space of 1-jets,

$d\alpha = \sum_{k=1}^n dx_k \wedge dp_k$

The corresponding matrix has the form:

$\begin{bmatrix} 0 & -I \\ I & 0 \end{bmatrix}$ .

The determinant of this matrix is equal to 1. Hence, $d\alpha$ is non degenerate.

Note: non-degenerate skew-symmetric bilinear forms can only be defined on even dimensional spaces. Why?

The corresponding matrix for the bilinear form is a skew-symmetric matrix, i.e. $\mathcal{A}^T = -\mathcal{A}$ . The bilinear form being non-degenerate is equivalent to $\text{det}(\mathcal{A}) \neq 0$ .

$\text{det}(\mathcal{A}^T) = (-1)^n \text{det}(\mathcal{A}^T)$

$\Rightarrow \text{det}(\mathcal{A}) = (-1)^n \text{det}(\mathcal{A})$

This implies n has to be even.

Thus, contact structures exist only on odd-dimensional manifolds.

Claim: the dimension of an integral manifold of a field of contact planes on a contact manifold of dimension 2n+1 does not exceed n.

Suppose $Y \subset M$ is an integral manifold. Let $\alpha$ be the 1-form defining the contact structure and $I: Y \mapsto M$ be the embedding.

Since the tangent plane of $Y$ belongs to the contact plane at every point, the pullback of $\alpha$ on the tangent space of $Y$ vanishes, i.e. $i^{*} \alpha = 0$ .

Thus, $i^{*}d\alpha = di^{*}\alpha = 0$ (the exterior derivative commutes with the pullback, why?).

Hence, every two vectors in the tangent space of $Y$ are skew-orthogonal, i.e. the tangent space of $Y$ is isotropic.

The dimension of an isotropic subspace cannot exceed $n$ . Why? Suppose $m$ be the dimension of this isotropic subspace. Then, $2n+1 - m$ is the dimension of its skew-orthogonal complement. By definition, the isotropic subspace contains its skew-orthogonal complement. Hence, $2n+1 -m \leq m$ . This implies $m \leq n$ .