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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?

peter a g
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ThePiper
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  • It’s been long time since I’ve thought of such things. What does the product of two such elliptic transformations look like? – Lubin Oct 07 '20 at 18:20
  • I think they could be elliptic, but that generically maybe they are hyperbolic/loxodromic. My source being the answer here: https://math.stackexchange.com/questions/1751822/product-of-two-elliptic-isometries-with-distincts-centers – ThePiper Oct 08 '20 at 01:12
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    Thanks for the reference. Seems clear what one’s strategy should be to get nonelliptic $g\circ f$. You just take the centers fairly far apart, or the angles fairly large. Very likely $f\circ g$ would get you another nonelliptic, and maybe those two would generate your free group. I’m almost tempted to do some computations… And while we’re at it, why is it clear that you’d need irrational rotations? – Lubin Oct 08 '20 at 01:52
  • @Lubin: with rational rotations the resulting group would have torsion. – Qiaochu Yuan Oct 08 '20 at 02:28
  • Not if the centers are different. But look: I think I have a good example for you, but I need to get a bunch of things taken care of today. I hope to have it written up by (maybe) noon today. – Lubin Oct 08 '20 at 12:55
  • Even with different centers, a rational rotation would have finite order, so a relation in your group. This post contains a very explicit formulation for how to write elliptic transformations with a prescribed center and angle of rotation: https://math.stackexchange.com/questions/2274011/hyperbolic-rotation I guess I will also get to computing! – ThePiper Oct 08 '20 at 14:08

2 Answers2

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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 $\langle f,g\rangle$

Lubin
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  • This shows that $\langle f,g \rangle$ contains a free group of rank two, but it sounds like the original question wanted it to be equal to a free group of rank two. – Ted Oct 12 '20 at 01:44
  • Right, @Ted . I definitely do not have the smarts to say, when the angles are, as OP specified, irrational multiples of $\pi$, what the resulting group would be, much less that it was free. – Lubin Oct 12 '20 at 02:44
  • Yes, this is an interesting and somewhat surprising example. I am mostly interested in when you get a free group on the nose. Thank you for your input! – ThePiper Oct 12 '20 at 19:25
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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 $PSL(2, {\mathbb R})$.

Moishe Kohan
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