---
tag: 3B451RD8AJ
kind: statement
role: theorem
name: Borel–Harish-Chandra theorem
msc: [20G30, "22E40", 11F06]
uses: [GA81VVQHEP, 4XWG6S7R4Z, NS2057MFWP]
about: [4XWG6S7R4Z, NS2057MFWP]
sources:
  - url: https://en.wikipedia.org/wiki/Arithmetic_group
---

## Statement

Let $G \subseteq \mathrm{GL}_n$ be a [[linear algebraic group|GA81VVQHEP]] defined over $\Q$, and let $G^\circ$ be its identity component.
A *$\Q$-character* of $G^\circ$ is a homomorphism of algebraic groups $G^\circ \to \mathrm{GL}_1$ defined over $\Q$.
Let $G(\Z) = G(\R) \cap \mathrm{GL}_n(\Z)$, an [[arithmetic group|4XWG6S7R4Z]].
Then $G(\Z)$ is a [[lattice|NS2057MFWP]] in $G(\R)$ if and only if $G^\circ$ has no nontrivial $\Q$-character.

In particular, if $G$ is semisimple, then $G(\Z)$ is a lattice in $G(\R)$.

## 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.

## Explanations

### L1 human

Take a group of matrices defined by polynomial equations with rational coefficients, like the matrices of determinant $1$.
Its integer points form a discrete subgroup of its real points.
The theorem says when this discrete subgroup is large enough that the quotient has finite volume.
The answer is: unless the group has a homomorphism to the multiplicative group, defined over $\Q$, that would stretch the volume to infinity.

### L2 human

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|FEJK846N5X]] 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.