---
tag: T3DR2V0J21
kind: proof
name: Proof that the divergence of a curl is zero
msc: ["26B12"]
proves: 0F56CMMC5P
uses: [02NSMQVQWK, 0GDNHS9J2X, 0NNQY8WCW6, 0TCBZMEYGN, 02RSXVM1ZN, 0M0TKFRX2A]
sources:
  - url: https://en.wikipedia.org/wiki/Vector_calculus_identities
    locator: Divergence of curl is zero
---

## Proof

We compute in coordinates and use the symmetry of second derivatives.

1. Let $A = (A_1, A_2, A_3)$ be of [[class $C^2$|0TCBZMEYGN]] on the open set $U \subseteq \R^3$. For $u, v \in \set{x, y, z}$, write $\partial_u \partial_v A_k$ for the [[second partial derivative|02RSXVM1ZN]] of $A_k$, first in $v$ and then in $u$.
2. By the definition of the [[curl|0GDNHS9J2X]], $\nabla \times A = (\partial_y A_3 - \partial_z A_2, \ \partial_z A_1 - \partial_x A_3, \ \partial_x A_2 - \partial_y A_1)$. Its components are $C^1$, since $A$ is $C^2$.
3. By the definition of the [[divergence|02NSMQVQWK]] and [[linearity of differentiation|0M0TKFRX2A]], $\nabla \cdot (\nabla \times A)$ equals
$$(\partial_x \partial_y A_3 - \partial_y \partial_x A_3) + (\partial_y \partial_z A_1 - \partial_z \partial_y A_1) + (\partial_z \partial_x A_2 - \partial_x \partial_z A_2).$$
4. By [[Schwarz's theorem|0NNQY8WCW6]] for the $C^2$ functions $A_1$, $A_2$ and $A_3$, each bracket in step 3 is $0$ at each point of $U$.
5. Hence $\nabla \cdot (\nabla \times A) = 0$ on $U$.

## Idea

Expanding the divergence of the curl gives six second derivatives. They pair up as the same mixed derivative of one component taken in two orders, with opposite signs. For $C^2$ fields the two orders agree, so everything cancels.