Noncompactness of $SL_n(\mathbb R)/SL_n(\mathbb Z)$

Question

For $n\ge 2$, I saw from a paper that $SL_n(\mathbb R)/SL_n(\mathbb Z)$ is noncompact. I wonder how to prove this fact. If the proof for the general case $n\ge 2$ is too complicated, I am also glad to learn the proof just for the case $n=2$.

Answer 1

Not only is $SL_n(\mathbb R)/SL_n(\mathbb Z)$ noncompact, there is a well known criterion (called Mahler's compactness criterion) which gives a necessary and sufficient condition of when a subset $A\subset SL_n(\mathbb R)/SL_n(\mathbb Z)$ is compact.

You can read a nice proof in the book 'An introduction to the geometry of numbers' by J.W.S Cassels, which also helps understand why this set it noncompact.

Answer 2

Here's a sketch of a proof: $SL_n(\mathbb{R}) / SL_n(\mathbb{Z})$ parametrizes lattices in $\mathbb{R}^n$ with fundamental domain of volume $1$. To see this is non-compact just consider the lattices $L(r)$ generated by $\frac{1}{r}e_1, \frac{1}{r}e_2, \dots, \frac{1}{r}e_{n-1}$, and $r^{n-1}e_n$ where $e_i$ are the standard basis vectors. As $r \to \infty$ this sequence of lattices does not approach any sort of limit (or have any convergent subsequence) and therefore the space of lattices must be non-compact.