Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $X,Y$ be two hyperbolic Riemann surfaces (i.e. they have universal cover the upper half plane $\mathbb{H}$). Let $\pi_X:\mathbb{H}\to X, \pi_Y:\mathbb{H}\to Y $ be the corresponding covering maps. Let $f:X\to Y$ be a homeomorphism. Then $\forall x\in X,f_*: \pi_1(X,x)\to\pi_1(Y,f(x))$ is an isomorphism of their fundamental groups. Now view the fundamental groups of $X,Y$ as the deck transformation groups $G_1,G_2$ respectively, which are Fuchsian groups. I am interested in knowing :

1) What is $f_*:G_1 \to G_2$, viewed as a group isomomorphism of the Fuchsian groups (deck transformation groups)?. After studying the (base point $x$ and its chosen lift $\tilde{x}\in \mathbb{H}$-dependent isomorphism between $ \pi_1(X,x)\to G_1 $, and similar base point dependent isomorphism : $\pi_1(Y,y=f(x))\to G_2$, I realize that $f_*$ becomes conjugation by $\tilde{f}$, where $\tilde{f}$ is the unique lift of $f$ sending $\tilde{x}$ to the chosen lift $\tilde{y}$ of $f(x)$. Am I correct ? The reason I am asking this question 1) is to double check whether it is exactly conjugation by $\tilde{f}$ or something different ?

N.B. : the assumption of hyperbolicity is redundant in my question 1), I just put them in order to keep parity with the literature I have been asking this question from.

2) In this question,I will require hyperbolicity. If the answer to my question 1) is "yes, it is nothing but conjugation by $\tilde{f}$", then in this case : $\tilde{f}\circ G_1 \circ \tilde{f}^{-1}$ is another Fuchsian group, being isomorphic to the fundamental group of $Y$. I want to make sure that $ h \circ G_1 \circ h^{-1}$ is $\textbf{not}$ in general a Fuchsian group for any arbitrary quasiconformal homeomorphism $h : \mathbb{H} \to \mathbb{H} $, where $h$ is NOT the lift of any homeomorphism $X \to Y$. Then $ h \circ G_1 \circ h^{-1}$ is $\textbf{not}$ a Fuchsian group, right ? Please give me your detailed opinion on it.

share|improve this question
    
I think you're wrong about 1)... wouldn't that imply that every automorphism of a surface group induced from a homeomorphism was an inner automorphism? That would contradict the Dehn-Nielson theorem (which states that the group of isotopy classes of homeomorphisms is isomorphic to the outer automorphism group). –  MartianInvader Sep 12 '13 at 22:03
add comment

1 Answer 1

I think you are right on both counts. The conjugation by a general quasiconformal homeomorphism of $\mathbb H$ does not preserve Fuchsian groups. Indeed, let $U\subset \mathbb H$ be a fundamental domain for the group. Define $h$ to be some nonsmooth perturbation in a part of $U$, and identity elsewhere. If $g$ is any group element other than identity, then $h^{-1}gh$ is not smooth in $U$, and in particular is not a Möbius map.

The conjugation of a Fuchsian group $G$ by a quasiconformal homeomorphism $h$ gives another Fuchsian group if and only if the Beltrami coefficient of $h$ is invariant under $G$. The lift considered in 1) has this property.

share|improve this answer
add 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.