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.

  • $\begingroup$ 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). $\endgroup$ Sep 12, 2013 at 22:03

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.


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