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.
$\endgroup$ 32 Answers
$\begingroup$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 $\int_A d^2\omega = \int_{\partial A} d\omega = \int_{\partial\partial A} \omega = \int_{\emptyset}\omega=0$
$\endgroup$ $\begingroup$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.
$$\frac{\partial^2f}{\partial x^i\partial x^j} = \frac{\partial^2f}{\partial x^j\partial x^i}.$$
$\endgroup$
