## Proof That the Second Differential of a Zero Form or Function is Zero

Theorem
If
$\omega^0$
is a twice differentiable zero form in
$\mathbb{R}^n$
ie a function, then
$d(d \omega^0) =0$
.
Proof
We can write
$\omega_{\mathbf{x}} =f(x_1,x_2,...,x_n)$
.
Then
$d \omega^0 =\frac{\partial f}{\partial x_1} dx_1 +...+ \frac{\partial f}{\partial x_n} dx_n$

and
\[ \begin{aligned} d(d \omega^0) &= (\frac{\partial^2 f}{\partial x_1 \partial x_1} dx_1 +...+ \frac{\partial^2 f}{\partial x_n \partial x_1} dx_n) \wedge dx_1 \\ &+ (\frac{\partial^2 f}{\partial x_1 \partial x_2} dx_1 +...+ \frac{\partial^2 f}{\partial x_n \partial x_2} dx_n) \wedge dx_2 +... \\ &+ (\frac{\partial^2 f}{\partial x_1 \partial x_n} dx_1 +...+ \frac{\partial^2 f}{\partial^2 x_n \partial x_n} dx_n) \wedge dx_n \\ &= \sum^n_{i,j=1} \frac{\partial^2 f}{\partial x_i \partial x_j} dx_i ) \wedge dx_j \\&= \sum^n_{i,j=1, \: i j} \frac{\partial^2 f}{\partial x_i \partial x_j} dx_i \wedge dx_j \\ &= \sum^n_{i,j=1, \: i i} \frac{\partial^2 f}{\partial x_i \partial x_j} dx_j \wedge dx_i \\ &= \sum^n_{i,j=1, \: i