Geometric Structure in the Space of 1-Jets

This is based upon my reading of the first chapter of Arnold’s PDE.

Key takeaway: in the space of 1-jets, we can view first order PDE’s as algebraic equations, i.e. a first order PDE describes a hypersurface living in this space.

By definition, the space of k-jets of functions of $(x_1, ..., x_n)$ is the space of Taylor polynomials of degree k.

For example, a point in the space of 1-jets is given by $(x_1, ..., x_n, y, p_1, ..., p_n)$ . The first n components gives us the point $(x_1, ..., x_n)$ in the domain $\mathbb{R}^n$ . The (n+1)-th component is the value of the function. The last n components $(p_1, ..., p_n)$ gives us the derivative of the function in the direction of $x_i$ , where i = 1, 2, …, n. Each point in the space of 1-jets gives us the first degree Taylor expansion of a function.

In the space of 1-jets, we have an invariantly defined field of 2n-dimensional hyperplanes given by, $dy = p \cdot dx$ . (Note: since this field is a geometric object, it is invariant under change of coordinates, but why does Arnold also say that under a change of coordinates $(x, y, p) \mapsto (\tilde{x}, \tilde{y}, \tilde{p})$ , this field of planes is still given by $d\tilde{y} = \tilde{p} \cdot d\tilde{x}$ ?)

Important geometric property of this field of planes in $J^1$ : All tangent planes of all 1-graphs at a given point in $J^1$ lie in the same hyperplane, $dy = p \cdot dx$ .

Proof: At the point $(x_1, ..., x_n, y, p_1, ..., p_n)$ , any tangent vector of the 1-graph of a function $u$ must satisfy $dy = \Sigma \frac{\partial u}{\partial x_i} dx_i = \Sigma p_i dx_i$ (*).

(*) can be explained in the following way:

Let $v$ be a tangent vector of the 1-graph and $\gamma(t) = (x(t), y(t), p(t))$ be a curve on the 1-graph such that $v = \gamma'(0)$ . Since $\gamma$ is a curve on the graph, $y(t) = u(x(t))$ .

Using the chain rule, we have: $\frac{dy}{dt} = \Sigma \frac{\partial u}{\partial x_i} dx_i$ . By definition of the space of 1-jets, $\frac{\partial u}{\partial x_i} = p_i$ . $\square$

Think about: In $J^1(\mathbb{R}^n, \mathbb{R})$ , what are the n-planes in the hyperplanes (dim. 2n) of the contact structure that cannot be tangent planes of a 1-graph?

For n = 1, at the point $(x_0, y_0, p_0)$ , this would be the line given by $x = x_0$ and $y = y_0$ . Notice the projection of all other lines into the (x, y)-plane gives us the line $y = p_0 x$ (i.e. the tangent line of all functions that satisfy $f(x_0) = y_0, f'(x_0) = p_0$ ). The projection of the line given by $x = x_O$ and $y = y_0$ into the (x, y)-plane gives us the point $(x_0, y_0)$ . If a 1-graph were to be tangent to this line at $(x_0, y_0, p_0)$ , then the corresponding function would have a singularity at $x_0$ .

What about when n = 2 or greater?

Think about: What are the dimensions of the integral manifolds (manifolds such that the tangent plane at each of its points is a subspace of the contact plane) for the field of contact planes?

Answer (we take for granted for now): the dimension cannot be greater than half the dimension of the contact plane.

Definition (Legendre manifold): integral sub manifold of a field of contact planes whose dimension is maximal (from above, we know that this would be half the dimension of the contact plane).

Example of Legendre manifold: 1-graphs

Converse is not true: not all Legendre manifolds are 1-graphs.

Example of a Legendre manifold that is not a 1-graph: consider the smooth curve parameterized by $x = t^2, y = \frac{2}{3}t^3, p = t$ . It is easy to check that it satisfies $dy = pdx$ , i.e. this curve is a Legendre manifold. The projection of this curve into the (x,y)-plane is not smooth; we have a cusp singularity at the origin. Hence, this curve is not a 1-graph.

First order PDE’s in $J^1(\mathbb{R}^n, \mathbb{R})$

Such an equation describes a 2n-dimensional submanifold $\Gamma^{2n}$ in $J^1$ .

Our goal: at each point in $\Gamma^{2n}$ , we want to distinguish a characteristic direction determined by the tangent space of $\Gamma^{2n}$ and the contact structure.

At a point $p \in \Gamma^{2n}$ , we consider the intersection of the tangent space (dim. 2n) and the contact plane (dim. 2n). Since $J^1$ has dimension 2n+1, it follows that these two spaces either coincide ( $p$ is singular) or we have a (2n-1)-dimension intersection ($latex p) is regular).

Claim: for a surface $\Gamma$ in general position, the singular points are isolated.

Proof: In a small neighbourhood, we can describe $\Gamma$ as the zero-set of a function, i.e. $\{ x \in J^1: f(x) = 0 \}$ . Let $p$ be a point in this neighbourhood. Every vector tangent to $\Gamma$ at $p$ is orthogonal to the gradient direction of $f$ , i.e. $T_p \Gamma = \text{Ker}(df_p)$ (or $df_p(v) = 0$ for any $v \in T_p \Gamma$ ).

If $T_p \Gamma$ coincides with the contact plane, then $\alpha = dy - p \cdot dx$ vanishes on $T_p \Gamma$ . Hence, we have $df_p = \lambda \alpha_p$ for some $\lambda \in \mathbb{R}*$ .