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  "body": "\n## Proof\n\nWe compute in coordinates and use the symmetry of second derivatives.\n\n1. 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 $u$.\n2. 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 $C^2$.\n3. By the definition of the [[divergence|02NSMQVQWK]] and [[linearity of differentiation|0M0TKFRX2A]], $\\nabla \\cdot (\\nabla \\times A)$ equals\n$$(\\partial_x \\partial_y A_3 - \\partial_y \\partial_x A_3) + (\\partial_y \\partial_z A_1 - \\partial_z \\partial_y A_1) + (\\partial_z \\partial_x A_2 - \\partial_x \\partial_z A_2).$$\n4. 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 $U$.\n5. Hence $\\nabla \\cdot (\\nabla \\times A) = 0$ on $U$.\n\n## Idea\n\nExpanding 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.\n",
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