---
tag: 0X308R2DBE
kind: statement
role: theorem
name: The exterior derivative squares to zero
aliases: [d-squared-is-zero]
msc: [58A10]
uses: [004FRCXH6M, 0S166D363M, 0NNQY8WCW6, 0M2JRW1J0N, 0F56CMMC5P, 0F1AVGX26N]
sources:
  - url: https://en.wikipedia.org/wiki/Exterior_derivative
    locator: In terms of axioms
  - url: https://en.wikipedia.org/wiki/Exterior_derivative
    locator: Closed and exact forms
  - url: https://en.wikipedia.org/wiki/Vector_calculus_identities
    locator: Curl of gradient is zero
---

## Statement

Let $U \subseteq \mathbb{R}^n$ be open, and let $\omega$ be a differential $k$-form on $U$ with coefficients of class $C^2$.
Then
$$d(d\omega) = 0.$$

## Remarks

Proof: for $\omega = f \, dx_I$, the [[exterior derivative|004FRCXH6M]] gives $d(d\omega) = \sum_{j, l} \partial_l \partial_j f \, dx_l \wedge dx_j \wedge dx_I$.
By [[Schwarz's theorem|0NNQY8WCW6]], $\partial_l \partial_j f = \partial_j \partial_l f$, while $dx_l \wedge dx_j = -dx_j \wedge dx_l$. So the terms cancel in pairs, and the terms with $j = l$ are zero.
In $\mathbb{R}^3$, the cases $k = 0$ and $k = 1$ are the identities [[curl of a gradient is zero|0M2JRW1J0N]] and [[divergence of a curl is zero|0F56CMMC5P]].
It follows that each [[exact form|0F1AVGX26N]] with $C^2$ coefficients is closed.

## Explanations

### L1 human

Taking the exterior derivative twice always gives zero, because mixed second partial derivatives are equal while the wedge product changes sign.
This one rule contains the facts that gradients have no curl and curls have no divergence.