Definition of $SL(2,\mathbb{Z})\backslash \mathbb{H}$

Question

I've found the notation $SL(2,\mathbb{Z})\backslash \mathbb{H}$ in some notes, where $$SL(2,\mathbb{Z})= \left\{ \begin{pmatrix} a & b \\ c & d \end{pmatrix} \in GL(2,\mathbb{Z})\ \bigg \rvert \ ad-bc=1 \right\} $$

$$ \mathbb{H}= \left\{ z \in \mathbb{C} \ \big\rvert \ \text{Im}(z)>0 \right\} $$

But I'm not sure what the author meant by it. Does it mean: $\left\{SL(2,\mathbb{Z})\cdot z \ \big \rvert \ z \in \mathbb{H} \right\}$? It doesn't quite makes sense to me since $\ \forall z \in \mathbb{H}, \ SL(2,\mathbb{Z})\cdot z=\mathbb{H}. $

Answer 1

It is not true that $\ \forall z \in \mathbb{H}, \ SL(2,\mathbb{Z})\cdot z=\mathbb{H}. $. $SL(2,\mathbb{Z})$ is discrete so for any point $z \in \mathbb{H}$ the orbit is that countable collection of points inside $\mathbb{H}$. As $z$ varies, these collections vary but some points give the same orbits. These become the same point in the quotient.

You might be thinking of $SL(2,\mathbb{R})$ in which case the action is transitive.

There is a convenient fundamental domain for this. See this image.

Answer 2

This is a "left quotient". The idea is that one constructs cosets by allowing the set of matrices to act on the points of the upper half-plane by left-multiplication.

I'm used to seeing the quotient by $\mathrm{PSL}(2,\Bbb{Z})$, which partitions the half-plane as shown here. The difference in the two matrix groups is an orientation reversal, so, for example the orientation reversal $(x,y) \mapsto (-x,y)$ folds the fundamental domain in gray in that image in half -- and similarly all the triangles are folded in half when you quotient by $\mathrm{SL}(2,\Bbb{Z})$ instead of $\mathrm{PSL}(2,\Bbb{Z})$.