Exterior derivative

There is a unique family of $\mathbb{R}$-linear maps

such that:

1. (On functions) If $f\in\Omega^0(M)=C^\infty(M)$, then $df$ is the usual differential (a 1-form) given by $df_p(v)=v(f)$.
2. (Graded Leibniz rule) For $\alpha\in\Omega^k(M)$ and $\beta\in\Omega^\ell(M)$,
   $$d(\alpha\wedge\beta)=d\alpha\wedge\beta + (-1)^k\,\alpha\wedge d\beta.$$
3. (Nilpotence) $d\circ d = 0$.

A key naturality property is that for any smooth map $F:M\to N$, the pullback of differential forms satisfies

for every $\omega\in\Omega^k(N)$.

The notions of closed and exact forms are defined using $d$, and their quotient defines the de Rham cohomology groups.

In a coordinate chart $(x^1,\dots,x^n)$, write a $k$-form as

where $I=(i_1<\cdots<i_k)$ and $a_I$ are smooth functions. Then

1. A function on $\mathbb{R}^2$. For $f(x,y)$, the exterior derivative is
   $$df = \frac{\partial f}{\partial x}\,dx + \frac{\partial f}{\partial y}\,dy.$$

1. A 1-form on $\mathbb{R}^2$. Let $\alpha = P(x,y)\,dx + Q(x,y)\,dy$. Then
   $$d\alpha = \left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right)\,dx\wedge dy.$$

1. A 2-form on $\mathbb{R}^3$. If $\omega = f(x,y,z)\,dx\wedge dy$, then
   $$d\omega = df\wedge dx\wedge dy = \frac{\partial f}{\partial z}\,dz\wedge dx\wedge dy,$$
   since the terms involving $dx\wedge dx$ and $dy\wedge dy$ vanish by alternation.