Can two elliptic isometries generate a free group?

Question

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?

Answer 1

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 $PSL(2, {\mathbb R})$.

Answer 2

Here’s my fairly extended example:

If you’re dealing with two centers of rotation, you might as well put them both on the imaginary axis (in Upper Half Plane model), and one of them at $i$, the other at $Ki$ with $K>1$. In case you needed to know, the distance then would be $\log K$. I made another decision, to do the contrary of what you suggested, and take both rotations to be of $90^\circ$ CCW. To make things easier, I’ll prefer the matrix representation: $$ \text{about $i$:}f(z)=\frac{z+1}{-z+1}\>, \text{i.e.} \begin{pmatrix}1&1\\-1&1\end{pmatrix}\>;\qquad\text{about $Ki$:}g=\begin{pmatrix}K&K^2\\-1&K\end{pmatrix}\,. $$ Then we get $$ f\circ g=\begin{pmatrix}K-1&K^2+K\\ -K-1&K-K^2\end{pmatrix}\,,\quad g\circ f=\begin{pmatrix}K-K^2&K^2+K\\-K-1&K-1\end{pmatrix}\,. $$ In both cases, the condition to be hyperbolic is $K^2-6K+1>0$; the roots of the quadratic are $K=3\pm\sqrt8$, so for an explicit example, I chose $K=6$. Here, the condition for $(f\circ g)(z)=z$ is $0=7(z^2+5z+6)=7(z+2)(z+3)$ so that the fixed points of the hyperbolic $f\circ g$ are $-2,-3$, while the fixed points of $g\circ f$ are $2,3$. Thus if what you say about hyperbolic with different fixed-point sets is correct, there’s a rank-two free group sitting in $\langle f,g\rangle$