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