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  "body": "\n## Statement\n\nLet $\\Delta = \\set{z \\in \\C^n : \\abs{z_j} \u003c r_j \\text{ for } j = 1, \\dots, n}$ be a [[polydisc|DJBWP7D3M8]] with radii $0 \u003c r_j \\le \\infty$.\nLet $p \\ge 0$ and $q \\ge 1$ be integers, and let $\\alpha$ be a smooth [[(p,q)-form|DC81TCR98Y]] on $\\Delta$ with $\\bar\\partial \\alpha = 0$, where $\\bar\\partial$ is the [[dbar operator|M1VKCW3RFW]].\nThen there is a smooth $(p,q-1)$-form $\\beta$ on $\\Delta$ with $\\bar\\partial \\beta = \\alpha$.\n\n## Remarks\n\nThe key case is $n = 1$, $q = 1$: for a smooth function $g$ with compact support in $\\C$, the function\n$$u(z) = \\frac{1}{2\\pi i} \\int_\\C \\frac{g(w)}{w - z}\\, dw \\wedge d\\bar w$$\nis smooth and satisfies $\\partial u / \\partial \\bar z = g$.\nIn general one first proves the result on a slightly smaller polydisc, by induction on the largest index $k$ such that some $d\\bar z_k$ occurs in $\\alpha$.\nThe coefficients involving $d\\bar z_k$ are holomorphic in $z_{k+1}, \\dots, z_n$, and solving the one-variable equation in $z_k$ with a cutoff removes $d\\bar z_k$ without destroying this holomorphy.\nAn exhaustion of $\\Delta$ by smaller polydiscs, together with Runge approximation to correct the successive solutions, gives the result on $\\Delta$ itself.\nExample: on $\\C$, the $(0,1)$-form $\\bar z \\, d\\bar z$ is $\\bar\\partial$ of $\\bar z^2/2$.\nThe condition $q \\ge 1$ is necessary to make the statement meaningful; for $q = 0$ a $\\bar\\partial$-closed $(p,0)$-form is a holomorphic $p$-form and need not vanish.\nThe conclusion fails on general open sets: on $\\C^2 \\setminus \\set{0}$ the Dolbeault group $H^{0,1}$ is not zero.\n\n## Explanations\n\n### L1 human\n\nThe lemma says that on a polydisc every form with $\\bar\\partial \\alpha = 0$ is of the form $\\bar\\partial \\beta$.\nIt is the complex analogue of the Poincaré lemma, which says that closed forms on a ball are exact.\nIn particular, on a small polydisc around any point of a complex manifold, the equation $\\bar\\partial \\beta = \\alpha$ can always be solved.\n\n### L2 human\n\nThe lemma is the local input to the Dolbeault isomorphism.\nIt says that the sequence of sheaves $0 \\to \\Omega^p \\to \\mathcal{A}^{p,0} \\to \\mathcal{A}^{p,1} \\to \\cdots$ is exact, where $\\Omega^p$ is the sheaf of holomorphic $p$-forms and $\\mathcal{A}^{p,q}$ is the sheaf of smooth $(p,q)$-forms.\nSo every Dolbeault cohomology class is locally trivial, and the global groups measure only the topology and the complex structure of the manifold as a whole.\n",
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  "name": "Dolbeault–Grothendieck lemma",
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