Margulis superrigidity theorem

Remarks

The proof builds a $\Gamma$-equivariant measurable map from the boundary $G(\R)/P$, with $P$ a minimal parabolic subgroup, to a projective variety of $H$. This uses the amenability of $P$ and the ergodicity of the action of $\Gamma$ on $G(\R)/P$. The higher rank is then used to show that this map is rational, and a rational boundary map gives the virtual extension.

Example: take $\Gamma = \mathrm{SL}_3(\Z)$ in $G = \mathrm{SL}_3$. Every 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 $\mathrm{SL}_3(\R)$.

The rank condition is necessary. The group $\mathrm{SL}_2(\R)$ has real rank $1$, and $\mathrm{SL}_2(\Z)$ contains a free group $F_2$ of finite index, which is a lattice in $\mathrm{SL}_2(\R)$. A subgroup of index $2$ in $F_2$ is a free lattice $F_3$, and it maps onto $F_2$ by killing one free generator $x$. Composing 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. Suppose a continuous $\varphi \colon \mathrm{SL}_2(\R) \to \mathrm{PSL}_2(\R)$ agreed with $\rho$ on a subgroup $\Lambda$ of finite index in $F_3$. Then $\varphi$ would be nontrivial, so its kernel, a proper normal subgroup of $\mathrm{SL}_2(\R)$, would lie in $\set{\pm I}$. But 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.

Going deeper

For a student

Superrigidity 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. It strengthens Mostow rigidity, which concerns isomorphisms between lattices, to arbitrary homomorphisms. The rank one groups $\mathrm{Sp}(n, 1)$ and $F_{4(-20)}$ also satisfy superrigidity, by work of Corlette and of Gromov and Schoen.