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  "body": "\n## Statement\n\nLet $n \\ge 2$.\nLet $\\mathbb{B}^n = \\set{z \\in \\C^n : \\abs{z_1}^2 + \\cdots + \\abs{z_n}^2 \u003c 1}$ be the [[unit ball|PDQ9KW9FAP]] and $\\Delta^n = \\set{z \\in \\C^n : \\abs{z_j} \u003c 1 \\text{ for all } j}$ the unit [[polydisc|DJBWP7D3M8]].\nThen there is no [[biholomorphism|FX92PP3HZR]] from $\\mathbb{B}^n$ onto $\\Delta^n$.\n\n## Remarks\n\nOne proof compares [[automorphism groups|0K1SCPBNKJ]].\nThe holomorphic automorphisms of $\\mathbb{B}^n$ act transitively, so a biholomorphism $F \\colon \\Delta^n \\to \\mathbb{B}^n$ may be assumed to satisfy $F(0) = 0$.\nThen conjugation by $F$ maps the automorphisms of $\\Delta^n$ fixing $0$ isomorphically onto the automorphisms of $\\mathbb{B}^n$ fixing $0$.\nBy [[Cartan's uniqueness theorem|EQHPSKC5SY]] both groups consist of linear maps.\nFor $\\mathbb{B}^n$ it is the unitary group $U(n)$, which is connected and not abelian for $n \\ge 2$.\nFor $\\Delta^n$ it consists of the maps $z \\mapsto (e^{i\\theta_1} z_{\\sigma(1)}, \\dots, e^{i\\theta_n} z_{\\sigma(n)})$ with $\\sigma$ a permutation, and its identity component is abelian.\nSo the two groups are not isomorphic as topological groups, a contradiction.\nFor $n = 1$ both sets are the unit disc, so the hypothesis $n \\ge 2$ is needed.\n\n## Explanations\n\n### L1 human\n\nIn one complex variable, the Riemann mapping theorem says that all simply connected proper domains of the plane look the same to complex analysis.\nIn two or more variables this is false already for the two simplest domains: the round ball and the product of discs.\nBoth are bounded, convex and contractible, so the difference is not topological; it is a rigidity of holomorphic maps.\n\n### L2 human\n\nThe theorem shows that there is no Riemann mapping theorem in several complex variables.\nIt opened the study of biholomorphic invariants of domains, such as automorphism groups, invariant metrics and the geometry of the boundary.\nAnother proof uses the boundary.\nThe boundary of $\\Delta^n$ contains nonconstant holomorphic discs, while the sphere bounding $\\mathbb{B}^n$ contains none.\nA limiting argument applied to a would-be biholomorphism turns this difference into a contradiction.\nPoincaré proved the result for $n = 2$ by computing automorphism groups.\n",
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