The action of $PSL_2(\mathbb{R})$ on $\mathbb{H}$ is proper

Question

Consider the upper half-plane $\mathbb{H}:=\{z\in \mathbb{C}:\Im (z)>0\}$ with the hyperbolic metric, and the group $PSL_2(\mathbb{R})=SL_2(\mathbb{R})/\{\pm I_2\}$, where $SL_2(\mathbb{R})$ is the set of $2\times 2$ real matrices with determinant equal to $1$, which acts on $\mathbb{H}$ by Möbius transformations. I'd like to show that this action is proper, i.e.

For any compact set $P\subset \mathbb{H}$ there exists a compact set $L\subset PSL_2(\mathbb{R})$ such that $z,g(z)\in P$ for some $g\in PSL_2(\mathbb{R})$ implies $g\in L$.

My reasoning was that, if such an $L$ where to exists, then surely

$$ \bigcup_{z\in P} \{g\in PSL_2(\mathbb{R}): g(z)\in P\}\subset L$$

and thus the closure of the LHS would be compact. Hence, a first candidate for $L$ would be

$$L:=\text{cl}\left( \bigcup_{z\in P} \{g\in PSL_2(\mathbb{R}): g(z)\in P\}\right ).$$

I'm stuck in showing that this is indeed compact. I tried to show sequential compactness, taking $g_n\in L$, which gives $z_n\in P$ such that $g_n(z_n)\in P$ from which we can extract a subsequences $z_{n_k}$ and a $z\in P$ such that $z_{n_k}\to z$, but I don't see how to fabricate a $g$ to act as a cluster point for the original sequences. Any hints? For reference, this is a problem in Ergodic theory with a view towards number theory by M. Ensiedler and T. Ward.

Answer 1

• Given $z_0$ and $P$.

  Find finitely many balls $B_\epsilon(\tau_k)=\{\tau,|\tau-\tau_k|\le\epsilon_k\}$ such that $P \subset \bigcup_{k=1}^K B_\epsilon(\tau_k)$

  take $u,v_k \in PSL_2$ such that $u(z_0) = i$, $v_k(i) = \tau_k$

  thus $$Lu^{-1} = \{ gu^{-1} \in PSL_2, g(z_0) \in P\}=\{ g \in PSL_2, g(i) \in P\} \\ \subset\quad \bigcup_{k=1}^K \{ g \in PSL_2, |g(i)-v_k(i)| \le \epsilon_k\}$$

• Since $g(i) = \frac{i+(ac+bd)}{c^2+d^2}$ and $ad-bc=1$ you'll find that $|g(i)-i| \le \epsilon$ implies $f = \sqrt{c^2+d^2} \in (1-2\epsilon,1+2\epsilon)$ and $|(\frac{ci+d}{f}) \cdot (fai+fb)| \le \epsilon, | (\frac{ci+d}{if}) \cdot (fai+fb)| = 1$ which implies $\sqrt{a^2+b^2} \in (1-4\epsilon ,1+4\epsilon)$.

  (where $\cdot$ is the inner product in $\mathbb{R}^2$)

• Thus $\{ g \in PSL_2, |g(i)-i| \le \epsilon\}$ is bounded, as well as $\{ g \in PSL_2, |g(i)-v_k(i)| \le \epsilon_k\}$ and $Lu^{-1}$ and $L$. And since $L$ is clearly closed it means it is compact.

Of course all this works iff we stay carefully in some compact included in the upper half-plane. If $P \cap \mathbb{R} \ne \emptyset$ it doesn't work. The same holds for the balls $\{ g \in PSL_2, |g(i)-i| \le \epsilon\}$.

Answer 2

Another definition of a map $f$ being proper: the preimage of every compact set is compact.

The group you describe, $PSL(2,\mathbb R)$, is the group of isometries of the hyperbolic plane $\mathbb H^2$. To see this, first we note that it's the group of biholomorphic maps from $\mathbb H^2$ to itself (see the groupprops wiki), and that in a space of constant, non-zero Gaussian curvature, any map that preserves angles will also preserve area.

Isometries and their inverses map compact sets to compact sets, and so it follows that the group of actions you describe are proper.