Why has $\mathrm{GL}(n, \mathbb{R})/\mathrm{GL}(n, \mathbb{Z})$ infinite co-volume?

Question

The space $X=\mathrm{SL}(n, \mathbb{R})/\mathrm{SL}(n, \mathbb{Z})$ can be identified as the space of unimodular lattices in $\mathbb{R}^n$ and it is well-known that if we take the Haar measure on $\mathrm{SL}(n, \mathbb{R})$ then $X$ has finite co-volume. I have read a few times that this is not the case for the space $Y = \mathrm{GL}(n, \mathbb{R})/\mathrm{GL}(n, \mathbb{Z})$. Why not? While these quotient groups seem to be important in the theory of Lie groups I'm unable to find an argument or a source explaining that $Y$ has infinite co-volume.

Answer 1

The problem of $GL(n,\Bbb Z)$ is that it only has matrices with determinant $\pm 1$, while $GL(n,\Bbb R)$ can have matrices with high determinant. The matrices having determinant $\pm 1$ form a subgroup $H \subset GL(n,\Bbb R)$. We have $GL(n,\Bbb Z) \subset H \subset GL(n,\Bbb R)$ and $GL(n,\Bbb R)/H \approx \Bbb R^{+*} $ has dimension $1$, so really, $GL(2,\Bbb Z)$ is small inside $GL(2,\Bbb R)$.