Intuitive meaning of $d^2=0$ for differential forms / de Rham theory I am just starting to learn de Rham cohomology, and I see this equation $d^2=0$ often.
I am curious what is the intuitive meaning of $d^2=0$ in this context?
If $d$ represents derivative, $d^2=0$ seems to imply that differentiating every function twice becomes zero, which does not seem to make sense.
Thanks for any explanation.
 A: The intuitive meaning of $d^2=0$ which I took away from mathematical physics and working with differential forms is the following:
"The boundary of a boundary is an empty set" (or alternatively: "A boundary has no boundary")
Which I take from $\int_A d^2\omega = \int_{\partial A} d\omega = \int_{\partial\partial A} \omega = \int_{\emptyset}\omega=0$
A: Let $f$ be a smooth function, then 
$$df = \sum_{j=1}^n\frac{\partial f}{\partial x^j}dx^j$$ 
and 
\begin{align*}
d^2f = d(df) &= \sum_{i=1}^n\sum_{j=1}^n\frac{\partial^2f}{\partial x^i\partial x^j}dx^i\wedge dx^j\\ 
&= \sum_{i<j}\frac{\partial^2f}{\partial x^i\partial x^j}dx^i\wedge dx^j + \sum_{j<i}\frac{\partial^2f}{\partial x^i\partial x^j}dx^i\wedge dx^j\\
&= \sum_{i<j}\frac{\partial^2f}{\partial x^i\partial x^j}dx^i\wedge dx^j - \sum_{j<i}\frac{\partial^2f}{\partial x^i\partial x^j}dx^j\wedge dx^i\\
&= \sum_{i<j}\frac{\partial^2f}{\partial x^i\partial x^j}dx^i\wedge dx^j - \sum_{i<j}\frac{\partial^2f}{\partial x^j\partial x^i}dx^i\wedge dx^j\\
&= \sum_{i<j}2\left(\frac{\partial^2f}{\partial x^i\partial x^j} - \frac{\partial^2f}{\partial x^j\partial x^i}\right)dx^i\wedge dx^j.
\end{align*}
The fact that $d^2 = 0$ follows from the fact that partial derivatives commute for a smooth function, i.e.
$$\frac{\partial^2f}{\partial x^i\partial x^j} = \frac{\partial^2f}{\partial x^j\partial x^i}.$$
