## Equivalence of Second Differential of a 1 Form and div curl of a Vector

Theorem
Consider a vector field
$\mathbf{f}=(f_1,f_2,f_3)$

The statement that for a 1 -form
$\omega^1=f_1 dx_1 +f_2 dx_2 +f_3 dx_3 , \: d (d \omega^1) =0$
is equivalent to the statement
$\mathbf{\nabla} \cdot (\mathbf{\nabla} \times \mathbf{f})=0$
where
$\omega^0 \: f$
are both twice differentiable.
Proof
\begin{aligned} d(d \omega^1) &= \sum^3_{i,j,k=1} \frac{\partial f_i}{\partial x_j \partial x_k} \\ &= (\frac{\partial^2 f_3}{ \partial x_1 \partial x_2}- \frac{\partial^2 f_2}{\partial x_1 \partial x_3}+ \frac{\partial^2 f_1}{\partial x_2 \partial x_3}- \frac{\partial^2 f_3}{\partial x_2 \partial x_1}+ \frac{\partial^2 f_2}{\partial x_3 \partial x_1}- \frac{\partial^2 f_1}{\partial x_3 \partial x_2})dx_1 \wedge dx_2 \wedge dx_3 \\ &=0 \end{aligned}

On the other hand
$\mathbf{\nabla} \cdot (\mathbf{\nabla} \times \mathbf{f})= \frac{f_3}{\partial x_1 \partial x_2}- \frac{f_3}{\partial x_2 \partial x_1}+ \frac{f_1}{\partial x_2 \partial x_3}- \frac{f_1}{\partial x_3 \partial x_2}+ \frac{f_2}{\partial x_3 \partial x_1}- \frac{f_2}{\partial x_1 \partial x_3}=0$