It's a fact that if two Fuchsian group are conjugate, the corresponding surfaces are isometric. Is the converse true ?

Take 2 isometric Riemann surfaces $S$ and $S'$(which are covered by the upper half-plane) or equivalently 2 hyperbolic surfaces. You can endow their universal covering with the complex(or hyperbolic) structure, so they can be identified to the hyperbolic plane. Then $S = \mathbb{H} / \Gamma$ and $S'= \mathbb{H} / \Gamma'$ where $\Gamma$ and $\Gamma'$ are two fuchsian groups. Are they conjugate ?

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    $\begingroup$ Yes, this is true: one might regard it as part of the Uniformization Theorem, in fact. Unfortunately I am having trouble nailing down a proof or a reference at the moment. If no one answers this question first, I'll come back and try a little harder (but surely someone else can do better). $\endgroup$ May 21, 2012 at 13:33

1 Answer 1


Let $\varphi\colon S\to S'$ be an isometry. Since $\mathbb{H}$ is simply connected, $\varphi$ lifts to a map $\widetilde{\varphi}\colon \mathbb{H}\to \mathbb{H}$ making the following diagram commute: $$ \begin{array}{ccc} \mathbb{H} & \xrightarrow{\widetilde{\varphi}} & \mathbb{H} \\ \downarrow & & \downarrow \\ S & \xrightarrow{\varphi} & S' \end{array} $$ Then $\widetilde{\varphi}$ is a local isometry. Since $\mathbb{H}$ is simply connected and geodesically complete, it follows that $\widetilde{\varphi}$ is an isometry. We claim that $\widetilde{\varphi}^{-1}\;\Gamma'\,\widetilde{\varphi} = \Gamma$.

Let $\gamma'\in\Gamma'$, and let $p\colon\mathbb{H}\to S$ and $p'\colon\mathbb{H}\to S'$ be the covering maps. We know that $\gamma'$ is a covering transformation for $p'$, i.e. $p'\gamma'=p'$. Since $p'\widetilde{\varphi} = \varphi p$, we have $$ p\widetilde{\varphi}^{-1}\gamma'\widetilde{\varphi} \,=\, \varphi^{-1}p' \gamma'\widetilde{\varphi} \,=\, \varphi^{-1} p' \widetilde{\varphi} = p. $$ Thus $\widetilde{\varphi}^{-1}\gamma'\widetilde{\varphi}$ is a covering transformation for $p$, so $\widetilde{\varphi}^{-1}\gamma'\widetilde{\varphi} \in \Gamma$. This proves that $\widetilde{\varphi}^{-1}\Gamma'\widetilde{\varphi} \leq \Gamma$, and a similar argument shows that $\Gamma \leq \widetilde{\varphi}^{-1}\Gamma'\widetilde{\varphi}$.

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    $\begingroup$ I don't understand how you get to the conclusion. Could you detail a bit more ? $\endgroup$ May 21, 2012 at 15:38
  • $\begingroup$ @SelimGhazouani I'd be happy to. Is there a specific conclusion that you're wondering about? $\endgroup$
    – Jim Belk
    May 21, 2012 at 17:10
  • $\begingroup$ I don't understand why the isometry $\widetilde{\varphi}$ conjugate $\Gamma$ and $\Gamma'$ $\endgroup$ May 22, 2012 at 12:33
  • $\begingroup$ Ok, I've added a more detailed proof of this fact. $\endgroup$
    – Jim Belk
    May 22, 2012 at 14:28

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