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I found the following result on some lecture notes, but there was no proof... I am not even sure whether it is correct or not... I tried to prove it, but I'm having some troubles:

Let $f$ be a hyperbolic transformation with fixed points $x$ and $y$, and let $g$ be a Moebius transformation mapping $x$ to $y$. Then the following hold:

1) $fgf^{-1}g^{-1}$ is hyperbolic

2) $fgfg^{-1}$ is parabolic

Here is what I tried:

Firstly, I assumed $x=0$ and $y=\infty$ (indeed, the thesis is invantiant by conjugation, since the trace is). Hence, I wrote $f$ and $g$ as $f(z)=\lambda z$ and $g(z)=\frac{\gamma}{z}+\delta$. Now, an explicit computation gives me:

1) $(fgf^{-1}g^{-1})(z)=...=\lambda^2 z+\lambda\delta(1-\lambda)$

Now I should compute the trace to show that $fgf^{-1}g^{-1}$ is hyperbolic when $f$ is. Is that correct? And how can I do this computation?

2) $(fgfg^{-1})(z)=...=z+\delta(\lambda-1)$

Now, if $\delta(\lambda-1)\neq 0$, clearly the transformation is parabolic, with only fixed point $\infty$. We need to check that $\delta(\lambda-1)= 0$ cannot hold.

If $\gamma=1$ we have $f(z)=z$, which contradicts the hypothesis that f is hyperbolic.

If $\delta=1$, we have $g(z)=\frac{\gamma}{z}$, which is a mobius transformation sending $0$ to $\infty$, and I cannot find an absurd in this. It seems to me that $fgfg^{-1}$ can also be the identity (hence, not parabolic). Does it make sense? Any help?

dpskrz
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1 Answers1

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Your counterexample to 2) is correct.

Regarding 1), I would not do this using trace calculations, instead I would use the dynamics of the action on $\mathbb{R} \cup \infty$. One of the points $x,y$ is an attracting fixed point for the action of $f$ on $\mathbb{R} \cup \infty$, the other is a repelling fixed point. Let me assume that $x$ is the repelling fixed point for $f$. It follows that $x$ is the attracting fixed point for $f^{-1}$. Since $g(x)=y$ it follows that $y$ is the attracting fixed point for $gf^{-1}g^{-1}$. By composition, $y$ is an attracting fixed point for $fgf^{-1}g^{-1}$. Now complete the proof by using that every Möbius transformation with an attracting fixed point on $\mathbb{R} \cup \infty$ is hyperbolic.

Lee Mosher
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