What does the topology on $SL(n, \mathbb{R}) / SL(n, \mathbb{Z})$ intuitively look like? I have come across Mahler's compactness criterion, and am having trouble wrapping my head around the topology of the moduli space of unit volume lattices.
Is there an intuitive way to think about it, rather than being a quotient topology of the special linear group?
 A: For $n=2$ (but much can be generalized a bit less elegant) we can map $A=\begin{pmatrix}a&b\\c&d\end{pmatrix}$ to the Möbius transform $\phi_A\colon z\mapsto \frac{az+b}{cz+d}$ of $\overline{ \mathbb C}$. This is a group homomorphism with kernel all matrices with $b=c=0$ and $a=d$, which is just $\pm$ the identity. As this kernel is also contained in $SL(2,\mathbb Z)$ we can ignore it.
Note that we can only obtain Möbius transforms with real coefficients, i.e., those that let $\mathbb R\cup\{\infty\}$ invariant.
Note that $\phi_A$ has  one real or two conjugate fixed points: $\frac{az+b}{cz+d}=z\iff cz^2+(d-a)z-b=0 $. We identify $A$ with the fixedpoint of $\phi_A$ in the closed upper half plane $\overline{\Bbb H}$. We lose a dimnesion here, i.e., every point of $\overline {\Bbb H}$ actually represents a onedimensional set of matrices, but let us postpone this.
With the Möbius transforms belonging to  matrices of the form $\begin{pmatrix}1&*\\0&1\end{pmatrix}\in SL(2,\mathbb Z)$ we can transport an arbitrary point of $\overline {\Bbb H}$ to the strip $-\frac12\le \Re z\le \frac12$. Similarly, $\begin{pmatrix}0&1\\-1&0\end{pmatrix}\in SL(2,\mathbb Z)$ takes reciprocals, so interchanges inside and outside of the unit circle $S^1$. Thus all points have representatives modulo $SL(2,\mathbb Z)$ in the strip $-\frac12\le \Re z\le \frac12$ and outside (or on the boundary of) $S^1$. Moreover, we should wrap around and glue the vertical boundary lines together, as well as the two halves of the arc at the bottom of this shape. We obtain a mostly smooth object, but it has a pointy cusp at $\infty$ and vertices at $i$ and at $\frac{1+i\sqrt 3}{2}$. To complete the image, you need to contemplate how the additional dimension(the one we postponed) must be added to this, with special care at the three special points.
