Exercise 7, § 3

Let $G$ be the unimodular group $\mathrm{SL}(n, \mathbf{R})$, $\Gamma$ the subgroup of $G$ formed by the matrices of determinant 1 with integer entries. Show, by induction on $n$, that every invariant measure $\mu$ on the homogeneous space $G/\Gamma$ is bounded and that, for $f \in \mathscr{K}(\mathbf{R}^n)$, one has, with $\mu$ suitably chosen,

$(\dot{X} = X \Gamma)$. (Let $G'$ be the subgroup of $G$ leaving invariant the basis vector $(1, 0, 0, \ldots, 0)$, so that $G/G'$ may be identified with $\mathbf{R}^n$. Let $G'' = G' \cap \Gamma$. Let $H$ be the subgroup of matrices $\begin{pmatrix} 1 & w \\ 0 & Y \end{pmatrix}$ of $G'$ for which $Y \in \mathrm{SL}(n-1, \mathbf{R})$ has integer entries. Then $H/G''$ is compact and, by the induction hypothesis, $G'/H$ has a finite invariant measure, therefore by Prop. 12, §2, No. 8, $G'/G''$ has a finite invariant measure. Applying Exer. 6 of §2, and Exer. 20 b) of A, VII, §4, show that, for $f \in \mathscr{K}(\mathbf{R}^n)$,

where the summation is over the set of vectors $m = (m_1, \ldots, m_n)$ whose coordinates are setwise coprime integers (A, VII, §1, No. 2, Th. 1). Applying this to the functions $f(2x)$, $f(3x)$, ... and summing, prove (1). Then, replacing $f$ by the function $x \mapsto \varepsilon^n f(\varepsilon x)$ with $f \geq 0$, and letting $\varepsilon$ tend to 0, show that every compact subset of $G/\Gamma$ has measure $\leq 1$.