The statement that for a 0 -form
\[\omega^0 , \: d (d \omega^0) =0\]
is equivalent to the statement that for a function \[f, \: \mathbf{\nabla} \times (\mathbf{\nabla} f)=0\]
where \[\omega^0 \: f\]
are both twice differentiable.Proof
Let
\[\omega^0 =f(x_1,x_2,x_3)\]
Then
\[\begin{equation} \begin{aligned} d(d \omega^) &= (\frac{\partial^2 d}{\partial x_2 \partial x_2} - \frac{\partial^2 d}{\partial x_1 \partial x_1}) dx_1 \wedge dx_2 + (\frac{\partial^2 d}{\partial x_3 \partial x_1} - \frac{\partial^2 d}{\partial x_1 \partial x_3}) dx_3 \wedge dx_1 \\ &+ (\frac{\partial^2 d}{\partial x_2 \partial x_3} - \frac{\partial^2 d}{\partial x_3 \partial x_2}) dx_2 \wedge dx_3 \\ &=0 \end{aligned} \end{equation}\]
On the other hand,
\[\begin{equation} \begin{aligned} \mathbf{\nabla} \times (\mathbf{\nabla} f) &= (\frac{\partial}{\partial x_1}, (\frac{\partial}{\partial x_2},(\frac{\partial}{\partial x_3}) \times (\frac{\partial f}{\partial x_1}, (\frac{\partial f}{\partial x_2 \partial x_3}) \\ &=(\frac{\partial^2 f}{\partial x_2 \partial x_3}- \frac{\partial^2 f}{\partial x_3 \partial x_2} , \frac{\partial^2 f}{\partial x_3 \partial x_1}- \frac{\partial^2 f}{\partial x_1 \partial x_3}, \frac{\partial^2 f}{\partial x_1 \partial x_2}- \frac{\partial^2 f}{\partial x_2 \partial x_1} )\\ &=0 \end{aligned} \end{equation}\]
The components of
\[\mathbf{\nabla} \times (\mathbf{\nabla} f)\]
are the component functions of \[d ( d \omega^0 )\]
. The Theorem is proved.