Exterior Differentiation (Part 2)

The first part of the post is a continuation of the previous post Arnold’s Mathematical Methods of Classical Mechanics: Exterior Differentiation. The second part is based upon my reading of Lecture 1 of Arnold’s PDE.

We left off with the computation of the principal linear part of the integral of the differential 1-form $\omega = a(x_1, x_2)dx_1$ along the boundary of the parallelogram, $\partial \Pi$ , with edges $\xi, \eta$ and vertex at 0 on the $(x_1, x_2)$ -plane.

After some parameterization, we arrived at the step:

$\int_{\partial \Pi} \omega = \int_0^1 [a(\xi t) - a(\xi t + \eta)]dx_1(\xi) - [a(\eta t) - a(\eta t + \xi)]dx_1(\eta)$

Claim: $a(\xi t + \eta) - a(\xi t) = \frac{\partial{a}}{\partial x_1} dx_1(\eta) + \frac{\partial{a}}{\partial x_2}dx_2(\eta) + O(\xi^2, \eta^2)$

Why is this true? This is just the first-order Taylor expansion of $a(x_1, x_2)$ at (0, 0).

Notice: the linear part $\frac{\partial a}{\partial x_1} dx_1(\eta) + \frac{\partial a}{\partial x_2}dx_2(\eta) = \nabla(a) \cdot \eta$ . The $\nabla(a)$ gives us the direction of greatest ascent for $a(x_1, x_2)$ at each point $(x_1, x_2)$ . So, geometrically, this means that when our parallelogram is small enough, the increment of $a(x_1, x_2)$ from $\xi t$ to $\xi t + \eta$ is given by the ‘amount of ascent’ in the $\eta$ direction.

Algebraically, we have:

$a(\xi t + \eta) - a(\xi t) = [a(\xi t) + \nabla(a) \cdot \eta + o(\xi^2, \eta^2)] - a(\xi t) = \nabla(a) \cdot \eta + o(\xi^2, \eta^2)$

which is what we want.

Return to Arnold’s PDE

Recall the differential 1-form we were interested comes from the contact structure in $J^1(\mathbb{R}^n, \mathbb{R}) \cong \mathbb{R}^{2n+1}$ :

$\alpha = dy - p dx$ .

We now look at the 2-form $\omega^2 = d\alpha$ that arises from the exterior differentiation of $\alpha$ .

Example: in $\mathbb{R}^3$

The exterior differentiation would give us:

$d \alpha = d1 \wedge dy + dz \wedge dx = 0 - dx \wedge dz = - dx \wedge dz$

This 2-form gives the negative of the oriented area of the projection of a parallelogram in $\mathbb{R}^3$ onto the (x, z)-plane.

For general $\mathbb{R}^{2n+1}$ :

Suppose the coordinates on $\mathbb{R}^{2n+1}$ are given by $x_1, ..., x_n, y, z_1, ..., z_n$ .

The exterior differentiation would give us:

$d\alpha = \sum_{k=1}^n -dx_k \wedge dz_k$ .

This 2-form gives the sum of the oriented area of the projection of a parallelogram in $\mathbb{R}^{2n+1}$ onto the $(x_k, z_k)$ -planes.

Thinking about: Given two vectors $\vec{u}, \vec{v}$ lying on a contact plane, what would $d\alpha(\vec{u}, \vec{v}$ be?

Claim: The 2-form $\omega^2$ when restricted to the contact field is defined invariantly up to multiplication by a non-zero number at each point.

Proof: Let $\tilde{\alpha} = f\alpha$ (i.e. $f$ is a function of x, y, z.)

Then, $d\tilde{\alpha} = d(f dy + \sum_{k=1}^n f z_k dx_k$ .

The RHS is equal to:

$[df \wedge dy + f (d1) \wedge dy] + \sum_{k=1}^n (df \wedge z_k dx_k + f dz_k \wedge dx_k)$

$= df \wedge (dy + \sum_{k=1}^n z_k dx_k) + f \sum_{k=1}^n -dx_k \wedge dz_k$

$= df \wedge \alpha + f d \alpha$ .

Hence, when restricted to the field of contact planes, i.e. $\alpha = 0$ , we have:

$d\tilde{\alpha}|_{\alpha = 0} = df \wedge 0 + f d\alpha|_{\alpha = 0}$

which is the same as

$d\tilde{\alpha}|_{\alpha = 0} = f d \alpha|_{\alpha = 0}$ .

What does this mean geometrically?

Exterior Differentiation (Part 2)

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