I would like to know why the space of Lattices in the euclidean plane
$$SL(2, \mathbb{Z}) \backslash SL(2, \mathbb{R})$$
is not compact. I am told to consider a height function and observe it can tend to infinity, and to set the area of the rhombus to = 1 (by re-scaling)
$$ \bigg( \frac{\min_{x\in L \backslash \{ 0\}} ||x||}{\text{Vol}(L)}\bigg)^{-1} \tag{$\ast$}$$
To me this space being non-compact is not a bad thing. It's just the same as when $\mathbb{R}$ is not compact, and there are several directions in which things can tend to infinity. This is an aspect of our number system. And we just deal with it.
Compact subsets of the Euclidean space $\mathbb{R}^n$ are
- closed
- bounded
so either the space of lattices is not closed or the space is not bounded. And ($\ast$) shows it's the second one.
Some of these identifications are a bit confusing. I have also found the space of lattices written:
$$ PSL(2, \mathbb{Z}) \backslash PSL(2, \mathbb{R}) = T^1 \Big( PSL(2, \mathbb{Z}) \backslash PSL(2, \mathbb{R}) \Big) $$
I have not drawn any pictures. Is there a more visual way of understanding these lattices are not compact?
There is a nice article on lattices and geodesics by Caroline Series which I just leave here.
Here's a specific continuous function $f$ on the space of lattices which takes values in $[0,\infty)$. Given a lattice $L \subset \mathbb{Z}^2$, let $v_1 \in L$ be a nonzero element whose norm $\sigma_1 = |v_1|$ is minimal, and let $v_2 \in L$ be an element that has smallest norm $\sigma_2 = |v_2| \ge \sigma_1$ amongst all elements of $L$ that are linearly independent of $v_1$. Let $$f(L) = \sigma_2 / \sigma_1 \in [1,\infty) $$ This continuous function is surjective onto $[1,\infty)$: for each real number $r \ge 1$ I can take $L$ to be the lattice spanned by $(1,0)$ and $(0,r)$, in which case $v_1 = (1,0)$ and $\sigma_1 = 1$, and $v_2 = (0,r)$ and $\sigma_2 = r$, so $f(L)=r$.
Since the image of $f$ of noncompact, the domain is also noncompact.