Standard applications of the ping-pong lemma can be used to show that two hyperbolic isometries or two parabolic isometries of $\mathbb{H}^2$ generate a free group. (Assuming they have disjoint fixed points and after passing to high enough powers, of course.)
I'm wondering if it's ever possible for two elliptic isometries to generate a free group (rank $2$). Clearly they would have to be irrational rotations about different fixed points. The action of each rotation on the boundary of $\mathbb{H}^2$ has dense orbits, so the standard ping-pong argument doesn't go through.
If they don't generate a free group, is there a (preferably geometric) way to see where a relation would come from?
Best Answer
Here is an indirect proof. Let $G$ denote $SL(2, {\mathbb R})$. Then given any nonempty reduced word $w$ in the alphabet $\{x_1^{\pm 1}, x_2^{\pm 1}\}$, we have a map $$ P_w: G\times G\to G $$ sending $(g_1, g_2)$ to $w(g_1, g_2)$ (you substitute $g_i$ for $x_i$, $i=1, 2$). The map $P_w$ is polynomial in terms of the matrix coefficients of $g_1, g_2$. Since $P_w^{-1}(\{\pm 1\})\ne G\times G$ (as $G$ contains rank 2 free subgroups), it follows that each $X_w=P_w^{-1}(\{\pm 1\})$ is a closed subset with empty interior in $G\times G$. Thus, by Baire's theorem, the union $$ X:= \bigcup_{w} X_w $$ has empty interior in $G\times G$. (The union is taken over all nonempty reduced words $w$.) The subset $E\subset G$ consisting of matrices projecting to nontrivial elliptic elements of $PSL(2, {\mathbb R})$ is open (and, of course, nonempty). It follows that $E\times E$ has nonempty intersection with $$ G\times G \setminus X. $$ Every $(g_1, g_2)\in E\times E \setminus X$ will project to a pair of elliptic isometries of ${\mathbb H}^2$ generating a free subgroup of rank 2 in $PSL(2, {\mathbb R})$.