How do you calculate an exterior derivative on forms in $\mathbb{R}^3$? If we have a form, say, $\omega = f(x,y,z) \, dx + g(x,y,z) \, dy + h(x,y,z) \, dz$, what is the formula for the exterior derivative $d \omega$?
 A: Roughly speaking, it is the anti-symmetric part of the full derivative due to the "odd permutation = sign change" trait of the differential forms, for a 1-form in $\mathbb{R}^3$ you gave if we write it as $\displaystyle\omega = \sum^{3}_{j=1} f_i dx_i$, it is :
$$
\begin{aligned}
d\omega &= \sum^{3}_{i=1} \sum^{3}_{j=1} \frac{\partial f_i}{\partial x_j}dx_j\wedge dx_i
\\
&= \sum^{3}_{j=1} \frac{\partial f_1}{\partial x_j}dx_j\wedge dx_1 + \sum^{3}_{j=1} \frac{\partial f_2}{\partial x_j}dx_j\wedge dx_2 + \sum^{3}_{j=1} \frac{\partial f_3}{\partial x_j}dx_j\wedge dx_3
\\
&= \frac{\partial f_1}{\partial x_2}dx_2\wedge dx_1 + \frac{\partial f_1}{\partial x_3}dx_3\wedge dx_1
\\
 &\quad+  \frac{\partial f_2}{\partial x_1}dx_1\wedge dx_2 + \frac{\partial f_2}{\partial x_3}dx_3\wedge dx_2
\\
&\qquad + \frac{\partial f_3}{\partial x_1}dx_1\wedge dx_3 + \frac{\partial f_3}{\partial x_2}dx_2\wedge dx_3
\\
&=  (\frac{\partial f_2}{\partial x_1} - \frac{\partial f_1}{\partial x_2})dx_1\wedge dx_2 + (\frac{\partial f_3}{\partial x_2}-\frac{\partial f_2}{\partial x_3})dx_2\wedge dx_3 +  (\frac{\partial f_1}{\partial x_3} - \frac{\partial f_3}{\partial x_1})dx_3\wedge dx_1
\\
&= (\frac{\partial g}{\partial x} - \frac{\partial f}{\partial y})dx\wedge dy + (\frac{\partial h}{\partial y}-\frac{\partial g}{\partial z})dy\wedge dz +  (\frac{\partial f}{\partial z} - \frac{\partial h}{\partial x})dz\wedge dx
\end{aligned}
$$
We could see it behaves the $\nabla \!\!\times$ operator and you get a 2-form, since we are talking in $\mathbb{R}^3$, by using the Hodge dual operator $\star$, you could get a formula looking more like the old curl formula we learned in vector calculus in the following way:
$$
\nabla \times\omega = \star d\omega =
\begin{pmatrix}
dx & dy & dz
\end{pmatrix} 
\begin{pmatrix}
0 & -\partial_z & \partial_y
\\
\partial_z & 0 & -\partial_x
\\
-\partial_y & \partial_x & 0
\end{pmatrix}
\begin{pmatrix}
f\\g\\h
\end{pmatrix}
$$
here we could see pretty clear that the anti-symmetricity comes from the anti-commutativity of the wedge product of the exterior algebra.
