Tangent Planes of 1-Graphs in Contact Structure

Basis of a hyperplane in the contact structure

For simplicity, we consider the case n = 2. Let the coordinate for $J^1(\mathbb{R}^2, \mathbb{R})$ be $(x_1, x_2, y, p_1, p_2)$ .

We consider the contact plane at the point $(x_1, x_2, y, p_1, p_2)$ . Let $X = a_1 \frac{\partial}{\partial x_1} + a_2 \frac{\partial}{\partial x_2} + b \frac{\partial}{\partial y} + c_1 \frac{\partial}{\partial p_1} + c_2 \frac{\partial}{\partial p_2}$ be a vector in the contact plane. Since $X$ vanishes on the 1-form $dy - p \cdot dx$ , it follows that:

$b - a_1 p_1 - a_2 p_2 = 0$ ,

or equivalently $b = a_1 c_1 + a_2 c_2$

Thus, we can rewrite the tangent vector as $X = a_1(\frac{\partial}{\partial x_1} + p_1 \frac{\partial}{\partial y}) + a_2(\frac{\partial}{\partial x_2} + p_2 \frac{\partial}{\partial y}) + c_1 \frac{\partial}{\partial p_1} + c_2 \frac{\partial}{\partial p_2}$ .

This immediately gives us a basis for the contact hyperplane: $\{ E_1 = \frac{\partial}{\partial x_1} + p_1 \frac{\partial}{\partial y}, E_2 = \frac{\partial}{\partial x_2} + p_2 \frac{\partial}{\partial y}, F_1 = \frac{\partial}{\partial p_1}, F_2 = \frac{\partial}{\partial p_2}$ .

Tangent planes of 1-graphs (when n = 2)

Let us consider the 1-graph of the function $u = u(x_1, x_2)$ . Then each point on the 1-graph is of the form $(x_1, x_2, u(x_1, x_2), u_{x_1}, u_{x_2})$ . The tangent plane at a point on this 1-graph is spanned by:

$X_1 = (1, 0, u_{x_1}, u_{x_1, x_1}, u_{x_1, x_2})$ and

$X_2 = (0, 1, u_{x_2}, u_{x_2, x_1}, u_{x_2, x_2})$ .

Thus, a vector in the contact plane $X = a_1(\frac{\partial}{\partial x_1} + p_1 \frac{\partial}{\partial y}) + a_2(\frac{\partial}{\partial x_2} + p_2 \frac{\partial}{\partial y}) + c_1 \frac{\partial}{\partial p_1} + c_2 \frac{\partial}{\partial p_2}$ that is also a vector of this tangent plane must satisfy:

$c_1 = u_{x_1, x_1} a_1 + u_{x_2, x_1} a_2$

$c_2 = u_{x_1, x_2} a_1 + u_{x_2, x_2} a_2$ .

Hence, the map $(a_1, a_2) \mapsto (c_1, c_2)$ is given by the Hessian $H = (u_{x_i, x_j})$ , which is symmetric.

Thus, every tangent plane of a 1-graph in the contact plane is the graph of a symmetric linear map,

$H: \text{span} \{E_1, E_2 \} \mapsto \text{span} \{F_1, F_2 \}$ .

Equivalently, a tangent plane of a 1-graph is the space given by $\{(v, H(v)) : v \in \text{span} \{E_1, E_2 \}$ (where H is a symmetric linear map from the space spanned by $\{E_1, E_2 \}$ to the space spanned by $\{F_1, F_2 \}$ ).

Example of a 2-plane in the contact hyperplane that cannot be a tangent plane of a 1-graph:

Consider the plane spanned by the two vectors:

$X_1 = \frac{\partial}{\partial x_1} + p_1 \frac{\partial}{\partial y} + \frac{\partial}{\partial p_2} = E_1 + F_2$

$X_2 = \frac{\partial}{\partial x_2} + p_2 \frac{\partial}{\partial y} = E_2$

Then, the corresponding map $H: \text{span} \{E_1, E_2 \} \mapsto \text{span} \{F_1, F_2 \}$ satisfies:

$H(E_1) = F_2$

$H(E_2) = 0$ .

Hence, in matrix form, $H = \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}$ , which is not symmetric.

Since the space of 1-jets only determine functions up to their first derivatives, any 2-planes in the contact planes satisfying the above condition (existence of such a symmetric linear map) is a tangent plane of a 1-graph.

This can easily be generalize to all $n > 2$ .

Think about: Closure of the union of all the tangent planes of 1-graphs is the contact plane.

Proof:

Important clarification – this is NOT saying that the symmetric invertible matrices are dense in the space of invertible matrices. We are trying to prove:

$\overline{\cup_{H = H^{T}} (v, H(v))}$ (where $v \in \text{span} \{E_1, E_2 \}$ and $H(v) \in \text{span} \{F_1, F_2 \}$ ) is the contact plane.

What vectors of the contact plane belong to a tangent plane of a 1-graph and what vectors do not?

Note: even though $X_1 = \frac{\partial}{\partial x_1} + p_1 \frac{\partial}{\partial y} + \frac{\partial}{\partial p_2} = E_1 + F_2$ and $X_2 = \frac{\partial}{\partial x_2} + p_2 \frac{\partial}{\partial y} = E_2$ do not span a tangent plane of a 1-graph, $X_1$ belongs to a tangent plane of a 1-graph.

In fact, for every non-zero vector $(v, q)$ where $v \neq 0$ , we can find a symmetric linear map $H$ such that $H(v) = q$ .

Thus, the only vectors of the contact plane that do not belong to tangent planes of 1-graphs are vectors of the form $(0, q)$ with $q \neq 0$ . It is easy to see that the closure of the union of all such vectors is the contact plane.

Tangent Planes of 1-Graphs in Contact Structure

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