# Is $SL(n,\mathbb{R})/SL(n, \mathbb{Z})$ a Hausdorff space?

The special linear group $$SL(n, \mathbb{R})$$ 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 $$SL(n, \mathbb{Z})$$ the group of $$n \times n$$ matrices with integer entries and determinant equals 1. Note that $$SL(n, \mathbb{Z})$$ is a discrete subgroup of $$SL(n, \mathbb{R})$$.

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:

1. $$SL(n, \mathbb{Z})$$ is a lattice in $$SL(n,\mathbb{R})$$. Moreover, the quotiont $$SL(n,\mathbb{R})/SL(n, \mathbb{Z})$$ is not compact.

2. (Mahler Criterion) For a sequnece $$(g_{m})_{m\in \mathbb{N}}$$ of $$SL(n,\mathbb{R})$$, the sequence $$(\pi (g_{m}))_{m\in \mathbb{N}}$$ of $$SL(n,\mathbb{R})/SL(n, \mathbb{Z})$$ 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 $$\pi :SL(n,\mathbb{R}) \to SL(n,\mathbb{R})/SL(n, \mathbb{Z})$$ is the natural projection.

The second result suggest to me that $$SL(n,\mathbb{R})/SL(n, \mathbb{Z})$$ is a Hausdorff space, but I can't find any reference. So, I do not know if it is true.

Thanks