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  "body": "\n## Proof\n\nWe expand $d(d\\omega)$ in coordinates. The mixed second derivatives are symmetric, while the wedge symbols are antisymmetric, so the terms cancel in pairs.\n\n1. Write $\\partial_j = \\partial / \\partial x_j$. For a finite sequence $J = (j_1, \\dots, j_m)$ of indices in $\\set{1, \\dots, n}$, let $e_J = dx_{j_1} \\wedge \\dots \\wedge dx_{j_m}$, as in the definition of the [[wedge product|0S166D363M]]. It is $0$ if two indices are equal. Otherwise it is $\\operatorname{sgn}(\\sigma) \\, dx_K$, where $K$ is the increasing rearrangement and $\\sigma$ the sorting permutation.\n2. Fact A: let $J$ have no repeated index, with $e_J = \\varepsilon \\, dx_K$. Then $e_{(l, J)} = \\varepsilon \\, e_{(l, K)}$ for each index $l$.\n3. Proof of fact A. If $l$ occurs in $J$, then it occurs in $K$, and both sides are $0$. Otherwise, sort $(l, J)$ in two stages: first sort the last $m$ places, turning $J$ into $K$, then sort $(l, K)$.\n4. The first stage has [[sign|0SK90238BH]] $\\varepsilon$. The sign is multiplicative, because [[the sign is a homomorphism|0JEJJCB0EX]]. So the sign of sorting $(l, J)$ is $\\varepsilon$ times the sign of sorting $(l, K)$, which is fact A.\n5. Fact B: for indices $l, j$ and a sequence $I$, $e_{(l, j) I} = -e_{(j, l) I}$, where $(l, j) I$ means $(l, j, i_1, \\dots, i_k)$. If there is a repeated index, both sides are $0$.\n6. Proof of fact B without repeats. The two sequences differ by the [[transposition|02DTRYKGE8]] of the first two places. So their sorting permutations differ by a transposition, which has sign $-1$, and the homomorphism property gives the sign change.\n7. Let $\\omega = \\sum_I f_I \\, dx_I$ be a [[$k$-form|0QHTX7WQEY]] whose coefficients $f_I$ are of [[class $C^2$|0TCBZMEYGN]]. By the definition of the [[exterior derivative|004FRCXH6M]], and since $dx_j \\wedge dx_I = e_{(j) I}$, we have $d\\omega = \\sum_I \\sum_j \\partial_j f_I \\, e_{(j) I}$.\n8. For $j \\notin I$, write $e_{(j) I} = \\varepsilon(j, I) \\, dx_{K(j, I)}$ with $\\varepsilon(j, I) = \\pm 1$. For $j \\in I$, $e_{(j) I} = 0$. Collecting terms gives $d\\omega = \\sum_K g_K \\, dx_K$, where $g_K$ is the sum of $\\varepsilon(j, I) \\, \\partial_j f_I$ over the pairs with $j \\notin I$ and $K(j, I) = K$.\n9. Each $g_K$ is a finite sum of first partial derivatives of $C^2$ functions, so it is $C^1$. So $d(d\\omega)$ is defined, and $d(d\\omega) = \\sum_K \\sum_l \\partial_l g_K \\, e_{(l) K}$.\n10. Expanding $g_K$ by step 8 and regrouping the finite sum, $d(d\\omega) = \\sum_I \\sum_{j \\notin I} \\sum_l \\partial_l \\partial_j f_I \\, \\varepsilon(j, I) \\, e_{(l) K(j, I)}$.\n11. By fact A, $\\varepsilon(j, I) \\, e_{(l) K(j, I)} = e_{(l, j) I}$. For $j \\in I$ the sequence $(l, j) I$ has a repeated index, so $e_{(l, j) I} = 0$. Hence $d(d\\omega) = \\sum_I \\sum_{j, l} \\partial_l \\partial_j f_I \\, e_{(l, j) I}$.\n12. Fix $I$. The terms with $j = l$ vanish, because $(l, l) I$ has a repeated index.\n13. For $j \\ne l$, take the two terms for $(j, l)$ and $(l, j)$. By fact B their sum is $(\\partial_l \\partial_j f_I - \\partial_j \\partial_l f_I) \\, e_{(l, j) I}$.\n14. By [[Schwarz's theorem|0NNQY8WCW6]] for the $C^2$ function $f_I$, $\\partial_l \\partial_j f_I = \\partial_j \\partial_l f_I$. So each such pair contributes $0$.\n15. By steps 11 to 14, every term of $d(d\\omega)$ cancels, and $d(d\\omega) = 0$.\n\n## Idea\n\nApplying $d$ twice produces each second derivative $\\partial_l \\partial_j f$ together with $dx_l \\wedge dx_j$. Swapping $l$ and $j$ leaves the derivative the same for $C^2$ functions but changes the sign of the wedge. So the terms cancel in pairs, and the diagonal terms vanish because $dx_l \\wedge dx_l = 0$.\n",
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