# Conjugate Groups of (quasi) Fuchsian Groups

I apologize in advance if this question is so trivial or too low level.

Let $\Gamma$ be a Fuchsian group. Let $\mathcal{F}$ be the set of pairs $(\mu,f)$, where $\mu \in L^\infty(\mathbb{C})$ such that $\mu(\overline{z})=\overline{\mu(z)}$, $||\mu|| < 1$, and $f$ is a quasiconformal mapping of the plane satisfying beltrami differential equation with beltrami coefficient $\mu$:

$$\mu f_z = f_{\overline{z}}$$

The solution exists up to a Mobius transformation.

Let $\mathcal{F}(\Gamma)$ pairs $(\mu,f) \in \mathcal{F}$ be such that $\mu \circ \gamma \frac{\overline{\gamma'}}{\gamma'} = \mu$ for all $\gamma \in \Gamma$. In this case, $\Gamma_f = f\circ \Gamma \circ f^{-1}$ is a Fuchsian group.

I'm wondering when $\Gamma = f\circ \Gamma \circ f^{-1}$ would hold. (Here, by "=" I mean equal as subsets of $PSL(2,\mathbb{R})$.

What I do know are some special (trivial) cases.

1) $\Gamma = 1$. No conditions on $f$ needs to be imposed.

2) $\Gamma$ is generated by a single parabolic element $g$. Then, the following conditions are sufficient: a) $f$ fixes the fixed point of $g$. b) $f$ fixes $b$ and $g(b)$ for some other point $b$.

3) $\Gamma$ is generated by a single hyperbolic element $g$. Then, the following conditions are sufficient: a) $f$ fixes the fixed points of $g$. b) $f$ has some other fixed point.

But outside of these cases, I don't have the slightest clue.

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When $\Gamma$ is nonelementary, I think the condition is simply that the map is the identity on the limit set of $\Gamma$.
Aren't "Teichmuller trivial" differentials those solutions whose restriction to a real line is that of an element in $PSL(2,\mathbb{R})$? – BrainDead Dec 3 '10 at 13:34
Yes, but the normalized solutions for these are the identity on $\mathbb{R}$. Maybe I don't have the right terminology for the differentials in the infinite covolume case, but that should be the correct condition. – Richard Kent Dec 3 '10 at 13:39
Sorry, perhaps my question was a bit round-about. What I'm really interested in getting out of this question is whether there are appropriate normalizations on the solutions so that $T(\Gamma)$ becomes an honest group from composition of the solutions. I imagine this is not going to be possible in general. – BrainDead Dec 3 '10 at 14:21
Sorry about all of my silly comments this morning, you can ignore all my previous comments (I'm a little doped up on cold medicine). You were right about the definition of Teichmuller trivial. I do think the answer that I have up now should be the correct one, and you should be able to derive it from the cases you mentioned. I do know that $\mathrm{Teich}(1)$ is a group as you say, but you have to be careful about the order of composition, or else the group law isn't continuous. I'm not sure about the more general situation. I can try and dig up a reference if you like. – Richard Kent Dec 3 '10 at 14:44