The special linear group $\operatorname{SL}(n,\mathbb{R})$ of degree $n$ over $\mathbb{R}$ is the set of $n \times n$ matrices with determinant $1$, with the group operations of ordinary matrix multiplication and matrix inversion. We denote by $\operatorname{SL}(n,\mathbb{Z})$ the group of $n \times n$ matrices with integer entries and determinant equals 1. Note that $\operatorname{SL}(n,\mathbb{Z})$ is a discrete subgroup of $\operatorname{SL}(n,\mathbb{R})$.
Let $G$ to be a topological group and $H$ to be a subgroup of $G$. We will say that a regular Borel measure $\mu$ on the quotient $G/H$ is a left invariant Haar measure if for all Borel sets $E \subseteq G/H$ and all $g \in G$ we have $\mu(gE) = \mu(E)$.
If $G$ is a locally compact Hausdorff group and $\Gamma$ is a discrete subgroup such that $G/H$ carries a finite left G-invariant Harr measure, then we say that $\Gamma$ is a $\textbf{lattice}$ in $G$.
We have the following results:
$\operatorname{SL}(n,\mathbb{Z})$ is a lattice in $\operatorname{SL}(n,\mathbb{R})$. Moreover, the quotiont $\operatorname{SL}(n,\mathbb{R})/\operatorname{SL}(n,\mathbb{Z})$ is not compact.
(Mahler Criterion) For a sequnece $(g_{m})_{m\in \mathbb{N}}$ of $\operatorname{SL}(n,\mathbb{R})$, the sequence $(\pi (g_{m}))_{m\in \mathbb{N}}$ of $\operatorname{SL}(n,\mathbb{R})/\operatorname{SL}(n, \mathbb{Z})$ does not have a convergent subsequence if, and only if, there exists a sequence $v_{m} \in \mathbb{Z}^{n}$ with $v_{m} \neq 0$ such that $g_{m}(v_{m})$ tends to $0$.
where $\pi\colon\operatorname{SL}(n,\mathbb{R}) \to \operatorname{SL}(n,\mathbb{R})/\operatorname{SL}(n,\mathbb{Z})$ is the natural projection.
The second result suggest to me that $\operatorname{SL}(n,\mathbb{R})/\operatorname{SL}(n, \mathbb{Z})$ is a Hausdorff space, but I can't find any reference. So, I do not know if it is true.
Thanks
