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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.

joshuaheckroodt
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    If you think of it as an action of $SL_2(\mathbb{R})$ then it's not faithful. Its kernel is exactly the center, and quotienting by this kernel gives you $PSL_2(\mathbb{R})$. So it's $PSL_2(\mathbb{R})$ that acts faithfully. – Qiaochu Yuan Feb 24 '23 at 02:40
  • @QiaochuYuan I just want to clarify a few things about your comment: firstly, by "the center" of $\mathrm{SL}(2,\mathbb{R})$, you mean $\lbrace\pm I_2\rbrace$. Since $\pm I_2z = z$, for all $z\in\mathbf{H}^2$, the action of $\mathrm{SL}(2,\mathbb{R})$ is not a faithful action, as you said. On the other hand, by taking the quotient of $\mathrm{SL}(2,\mathbb{R})$, we identify $I_2$ with $-I_2$, so $I_2$ and $-I_2$, as elements of $\mathrm{PSL}(2,\mathbb{R})$, produce the same action on $\mathbf{H}^2$. Am I understanding this correctly? – joshuaheckroodt Feb 24 '23 at 12:09
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    You mean "as elements of $SL_2(\mathbb{R})$," but other than that, yes. – Qiaochu Yuan Feb 24 '23 at 18:50
  • Besides making actions faithful, using $PSL$ instead of $SL$ also means the action of $PSL_2\Bbb R$ on the tangent bundle of the upper half-plane is regular (or sharply transitive), and similarly the action of $PSL_2\Bbb C$ on the Riemann sphere is sharply $3$-transitive. – anon Mar 26 '23 at 21:29

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