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  "body": "\n## Statement\n\nLet $G$ be a connected semisimple algebraic group defined over $\\R$ whose real rank is at least $2$.\nAssume that the identity component of $G(\\R)$ has no compact factors.\nLet $\\Gamma$ be an irreducible lattice in $G(\\R)$.\nLet $H$ be a connected simple algebraic group over $\\R$ with trivial center.\nLet $\\rho \\colon \\Gamma \\to H(\\R)$ be a [[group homomorphism|0JYTG5K9Y4]] such that $\\rho(\\Gamma)$ is Zariski dense in $H$ and its closure in $H(\\R)$ is not compact.\nThen there is a continuous homomorphism $\\varphi \\colon G(\\R)^\\circ \\to H(\\R)$, where $G(\\R)^\\circ$ is the identity component.\nThe map $\\varphi$ agrees with $\\rho$ on a subgroup of finite index in $\\Gamma$.\nIn other words, $\\rho$ virtually extends to $G(\\R)^\\circ$.\n\n## Remarks\n\nThe proof builds a $\\Gamma$-equivariant measurable map from the boundary $G(\\R)/P$, with $P$ a minimal parabolic subgroup, to a projective variety of $H$.\nThis uses the amenability of $P$ and the ergodicity of the action of $\\Gamma$ on $G(\\R)/P$.\nThe higher rank is then used to show that this map is rational, and a rational boundary map gives the virtual extension.\n\nExample: take $\\Gamma = \\mathrm{SL}_3(\\Z)$ in $G = \\mathrm{SL}_3$.\nEvery homomorphism of $\\Gamma$ to such an $H(\\R)$ with Zariski dense and unbounded image agrees, on a subgroup of finite index, with a continuous homomorphism of $\\mathrm{SL}_3(\\R)$.\n\nThe rank condition is necessary.\nThe group $\\mathrm{SL}_2(\\R)$ has real rank $1$, and $\\mathrm{SL}_2(\\Z)$ contains a [[free group|0KTW26C80J]] $F_2$ of finite index, which is a lattice in $\\mathrm{SL}_2(\\R)$.\nA subgroup of index $2$ in $F_2$ is a free lattice $F_3$, and it maps onto $F_2$ by killing one free generator $x$.\nComposing with the inclusion of $F_2$ into $\\mathrm{SL}_2(\\R)$ and then the projection to $\\mathrm{PSL}_2(\\R)$ gives a homomorphism $\\rho$ with Zariski dense, unbounded image.\nSuppose a continuous $\\varphi \\colon \\mathrm{SL}_2(\\R) \\to \\mathrm{PSL}_2(\\R)$ agreed with $\\rho$ on a subgroup $\\Lambda$ of finite index in $F_3$.\nThen $\\varphi$ would be nontrivial, so its kernel, a proper normal subgroup of $\\mathrm{SL}_2(\\R)$, would lie in $\\set{\\pm I}$.\nBut some power $x^k$ with $k \\ge 1$ lies in $\\Lambda$ and in the kernel of $\\rho$, and $x^k \\ne \\pm I$ because a free group has no torsion, so no such $\\varphi$ exists.\n\n## Explanations\n\n### L1 human\n\nSuperrigidity says that certain discrete groups of matrices can only be represented in the ways the surrounding continuous group allows.\nAny reasonable homomorphism of such a discrete group into another matrix group agrees, after passing to a subgroup of finite index, with a continuous homomorphism of the big group.\nThis holds for lattices in groups like $\\mathrm{SL}_n(\\R)$ with $n \\ge 3$, but not for $\\mathrm{SL}_2(\\R)$.\n\n### L2 human\n\nSuperrigidity is the main step in the proof of the Margulis arithmeticity theorem: every irreducible lattice in a group of real rank at least $2$ is an [[arithmetic group|4XWG6S7R4Z]].\nIt strengthens Mostow rigidity, which concerns isomorphisms between lattices, to arbitrary homomorphisms.\nThe rank one groups $\\mathrm{Sp}(n, 1)$ and $F_{4(-20)}$ also satisfy superrigidity, by work of Corlette and of Gromov and Schoen.\n",
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