# 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.

• Algebraically, $d^2=0$ is usually viewed as a complex, so that the image of one $d$ lies in the kernel of the next $d$. Commented Mar 14, 2017 at 14:34
• Commented Mar 14, 2017 at 14:39
• Commented Mar 14, 2017 at 14:40

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.

$$\frac{\partial^2f}{\partial x^i\partial x^j} = \frac{\partial^2f}{\partial x^j\partial x^i}.$$

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 $\int_A d^2\omega = \int_{\partial A} d\omega = \int_{\partial\partial A} \omega = \int_{\emptyset}\omega=0$