# Free discrete action vs discrete subgroup

Suppose a group $\Gamma \leq PSL(2,\mathbb{R})$ is acting fixed point freely on the upper half plane $\mathbb{H}$ via Moebius transformations. I am looking for a quick argument showing that the action is properly discontinuous if and only if $\Gamma$ is a discrete subgroup of the Lie group $PSL(2,\mathbb{R})$.

• So presumably you mean this to be the restriction of the usual action of $SL_2$ via Mobius transformations? – Tobias Kildetoft Mar 21 '18 at 10:14
• Correct, I updated the question. – darko Mar 21 '18 at 10:18

First the easy direction: if $\Gamma$ is not discrete, there is a sequence of non-identity elements $\gamma_i$ converging to the identity. For any point $p \in \mathbb{H}$ it follows that the sequence $\gamma_i(p)$ converges to $p$. For any neighborhood $U$ of $p$, there are infinitely many values of $i$ for which $U \cap \gamma_i^{-1}(U)$ contains $p$, hence the action is not properly discontinuous.
Now for the more interesting direction: if the action is not properly discontinuous, then there is a point $p \in \mathbb{H}$, and for every integer $n \ge 1$ there is a non-identity group element $\gamma_n$, such that $\gamma_n^{-1}(B(p,1/n)) \cap B(p,1/n) \ne \emptyset$, and so $\gamma_n(p)$ converges to $p$. Thus $d(p,\gamma_n(p)) < 1/n$. Since the action is not free, $\gamma_n(p) \ne p$. Passing to a subsequence, we may assume that no two elements of the sequence $\gamma_n$ are equal. The subset of $\text{PSL}(2,\mathbb{R})$ consisting of those $\gamma$ such that $d(p,\gamma(p)) \le 1$ is compact. It follows that the sequence $\gamma_n$ has a convergent subsequence. This implies that $\Gamma$ is not discrete.
Note, this proof uses freeness in only a weak fashion. The proof can be made to work without freeness as long as one adopts a weaker (and better) notion of proper discontinuity: for any point $p$ there exists a neighborhood $U$ of $p$ such that the set of group elements $\gamma$ for which $\gamma(U) \cap U \ne \emptyset$ is finite. And this is a good thing, because with that weaker definition there are lots of non-free properly discontinuous subgroups of $\text{PSL}(2,\mathbb{C})$ that we know and love, such as the hyperbolic Von Dyck groups (fundamental groups of hyperbolic orbifolds with underlying space $S^2$ and three cone singularities).