Borel–Harish-Chandra theorem

Remarks

The proof uses reduction theory. One builds an explicit Siegel set $\mathfrak{S}$ in $G(\R)$, of finite Haar measure, such that $G(\R) = \mathfrak{S} \, F \, G(\Z)$ for a finite set $F$. For $\mathrm{GL}_n$ and $\mathrm{SL}_n$, Siegel sets come from Minkowski's reduction of quadratic forms; the general case reduces to these.

Example: $G = \mathrm{SL}_n$ has no nontrivial characters, so $\mathrm{SL}_n(\Z)$ is a lattice in $\mathrm{SL}_n(\R)$.

The condition on characters is necessary. For $G = \mathrm{GL}_1$, the identity map is a $\Q$-character. Here $G(\Z) = \set{\pm 1}$, and $\R^\times / \set{\pm 1}$ has infinite Haar measure $\int_0^\infty dx / x$. So $G(\Z)$ is not a lattice in $G(\R) = \R^\times$.

A companion result, proved by Borel and Harish-Chandra and by Mostow and Tamagawa, says when $G(\R)/G(\Z)$ is compact. For reductive $G$, this happens exactly when $G$ has no nontrivial unipotent elements in $G(\Q)$ and no nontrivial $\Q$-characters.

Going deeper

For a student

This theorem is the source of most lattices in semisimple Lie groups. Groups like $\mathrm{SL}_n(\Z)$, $\mathrm{Sp}_{2n}(\Z)$ and integral orthogonal groups of indefinite forms are lattices because of it. The Margulis arithmeticity theorem gives a converse in higher rank: every irreducible lattice arises from this construction. For $\mathrm{SL}_2$ it recovers the fact that the modular surface $\mathrm{SL}_2(\Z) \backslash \mathbb{H}$ has finite area.