Clarifying the action of $\mathrm{PSL}(2,\mathbb{R})$ on $\mathbf{H}^2$.

35 Views Asked by At

I’m working my way up towards understanding the proof of the fact that the geodesic flow on $\mathbf{H}^2$ (the hyperbolic space) is ergodic, and, in doing so, I’ve come across the group $\mathrm{PSL}(2,\mathbb{R}) = \mathrm{SL}(2,\mathbb{R})/\lbrace\pm I_2\rbrace$, where $I_2$ denotes the $2\times2$ identity matrix, repeatedly, and I’d just like a little bit of help clarifying the action of this group on $\mathbf{H}^2$. In particular, in Bekka’s book, $\mathrm{PSL}(2,\mathbb{R})$ is said to act on $\mathbf{H}^2$ by $$ gz = \frac{a + bz}{c + dz},\quad g\in\mathbf{H}^2,\quad g=\left(\begin{array}{c c}a & b \\ c & d\end{array}\right)\in\mathrm{SL}(2,\mathbb{R}). $$ I was basically wondering the following: since $\mathrm{PSL}(2,\mathbb{R})$ is a quotient group, its elements are given, using set comprehension, by $$ \mathrm{PSL}(2,\mathbb{R})=\lbrace\lbrace g,-g\rbrace:g\in\mathrm{SL}(2,\mathbb{R})\rbrace, $$ and yet the action of this group on $\mathbf{H}^2$ seems to ignore this completely, and just makes use of the elements of $\mathrm{SL}(2,\mathbb{R})$, directly. What, therefore, is the significance of taking the quotient of $\mathrm{SL}(2,\mathbb{R})$, and defining its action on $\mathbf{H}^2$, instead of just using $\mathrm{SL}(2,\mathbb{R})$, directly? Moreover, what even is the point of taking this quotient? Intuitively, I imagine we take this quotient because the orbits of the point $z\in\mathbf{H}^2$ under $g\in\mathrm{SL}(2,\mathbb{R})$ and $-g$ are identical, but is this the sole reason for working with this quotient in the context of hyperbolic geometry? As an aside, my group theory is very (very) minimal, so it’s probably in your best interest to explain this to me like I’m a 5-year-old, sorry! Any help in this regard is kindly appreciated.