Exterior Derivative and $d^2=0$

The exterior derivative sends a $k$-form to a $(k+1)$-form. On a scalar function it is the gradient one-form,

$$df=\frac{\partial f}{\partial x^i},dx^i,$$

and on all differential forms it satisfies the fundamental identity $d^2=0$.

A coordinate-free implementation pattern

Egison's tensor derivative can map the partial-derivative operator across the coordinate vector. The polymorphic definition below works for a scalar or a tensor-valued expression.

declare symbol x, y, z : MathValue

def params : Vector MathValue := [| x, y, z |]

def d {a} (X : a) : DiffForm a := !(flip ∂/∂) params X

def f : MathValue := x ^ 2 + y ^ 2 + z ^ 2

For $f=x^2+y^2+z^2$, the first exterior derivative is the radial gradient one-form $2x\,dx+2y\,dy+2z\,dz$.

d f
$\begin{pmatrix} 2 x \\ 2 y \\ 2 z\\ \end{pmatrix}$

Applying the tensor derivative a second time first produces the raw Hessian. This intermediate object has not yet been projected onto its alternating differential-form part.

d (d f)
$\begin{pmatrix} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \\ \end{pmatrix}$

The Hessian of a smooth scalar is symmetric, while a two-form is antisymmetric. Their alternating projection therefore vanishes.

dfNormalize (d (d f))
$\begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \\ \end{pmatrix}$

This computation is the coordinate expression of $d^2f=0$: mixed partial derivatives cancel pairwise after antisymmetrization. Showing both the Hessian and its normalized form separates ordinary tensor differentiation from the geometric exterior derivative.

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