Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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 $\mu$:

\begin{equation} \mu f_z = f_{\overline{z}} \end{equation}

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.

share|improve this question
add comment

1 Answer

When $\Gamma$ is nonelementary, I think the condition is simply that the map is the identity on the limit set of $\Gamma$.

This should be in Gardiner's "Teichmuller Theory and Quadratic Differentials" somewhere in the chapter on the Teichmuller space of a fuchsian group.

share|improve this answer
    
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, by "these" I meant the maps whose boundary values agree with a Mobius transformation. –  Richard Kent Dec 3 '10 at 13:41
    
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
show 1 more comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.