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
\[ \begin{equation} \begin{aligned} d(d \omega^1) &= \sum^3_{i,j,k=1} \frac{\partial f_i}{\partial x_j \partial x_k} \\ &=
(\frac{\partial^2 f_3}{ \partial x_1 \partial x_2}- \frac{\partial^2 f_2}{\partial x_1 \partial x_3}+ \frac{\partial^2 f_1}{\partial x_2 \partial x_3}- \frac{\partial^2 f_3}{\partial x_2 \partial x_1}+ \frac{\partial^2 f_2}{\partial x_3 \partial x_1}- \frac{\partial^2 f_1}{\partial x_3 \partial x_2})dx_1 \wedge dx_2 \wedge dx_3 \\ &=0 \end{aligned} \end{equation} \]
On the other hand
\[\mathbf{\nabla} \cdot (\mathbf{\nabla} \times \mathbf{f})= \frac{f_3}{\partial x_1 \partial x_2}- \frac{f_3}{\partial x_2 \partial x_1}+ \frac{f_1}{\partial x_2 \partial x_3}- \frac{f_1}{\partial x_3 \partial x_2}+ \frac{f_2}{\partial x_3 \partial x_1}- \frac{f_2}{\partial x_1 \partial x_3}=0 \]