# Two hyperbolic surfaces corresponding to conjugate Fuchsian groups are isometric

I have a basic question :

a) Suppose $\gamma$ and $\gamma'$ be conjugate Fuchsian groups acting freely and properly discontinuously on the upper half-plane H to produce two Riemann surfaces $S$ and $S'$. How can we prove that $S$ and $S$ are isometric ? A detailed answer would be appreciated .Do I need to have $S$ and $S'$ closed for this question ?

I was trying to send the equivalence class $[\gamma(z) ]$ in $S$ to $[\delta\gamma\delta^-1]$ . Would that work ?

b) Also, how can we prove that two hyperbolic surfaces ( again, do we need closed surfaces ) are isometric by $f$ implies the lift $\tilde{f}$ of $f$ is an isometry ( it is obviously a local isometry, but why is it bijective,i.e. have an inverse ) of the upper half plane $H$ ? Is the result true for arbitrary quotients of differential manifolds and diffeomorphism of quotients rather than isometry ? Thanks !

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You can also mention a reference where I can find the details . –  Mathmath Feb 11 '11 at 17:07
Your question would be better if you gave us some of the background. For example, is this homework from a class? What level is the class? What what is your background knowledge? What are we allowed to assume you know? –  Sam Nead Feb 11 '11 at 20:17
You can assume that I know the basic covering space theory and some hyperbolic geometry, including the hyperbolic structure of closed hyperbolic surfaces. –  Mathmath Feb 12 '11 at 5:15

As $\gamma$ is generally used for curves, i will be using $\Gamma$ and $\Gamma'$ for the fuschian groups and $p$ and $p'$ the corresponding covering maps.

a) Let the two groups be conjugate by a function $h$ i.e $h\Gamma h^{-1}= \Gamma'$. Define $f(x)=h(y)$ for some $y\in p^{-1}(x)$. We will check that this is well defined. That is we need to check $p(y)=p(y')$ then $p'(h(y))=p'(h(y'))$. $p(y)=p(y')$ implies that $y'=g(y)$ for some $g\in \Gamma$. $h\circ g\circ h^{-1} = g'\in \Gamma'$ which implies that $h\circ g = g^{-1}\circ h$. $h(y')=h(g(y))=g'(h(y))$ thus $p'(h(y))=p'(h(y'))$.

This map is clearly a local isometry. An assumption that $f$ is nor injective(surjective) will imply a similar statement for $h$, a contradiction. Thus $f$ is bijective.

To prove part b) i will use the following standard results from covering space theory(c.f Hatcher: Algebraic topology)

1) For a fixed point $z_0\in p^{-1}(x_0)$ ($p$ is the covering map) the set $\Gamma$ corresponds one-one to $p^{-1}(x_0)$ via $g\in\Gamma\leftrightarrow g(z_0)\in p^{-1}(x_0)$.

2) For a fixed point $z_0\in p^{-1}(x_0)$ the group of deck transformations $\Gamma$ and $\pi_1(\mathbb{H}/\Gamma,x_0)$ are isomorphic. The isomorphism is: an element $\gamma\in \pi_1(\mathbb{H}/\Gamma,x_0)$ is sent to the deck transformation such that $z_0\mapsto \tilde{\gamma}(1)$.($\tilde{\gamma}$ denotes the lift of $\gamma$).

b)If $S$ and $S'$ are isometric, the isometry being $f$. Fix points $x_0\in S$, $y_0\in p^{-1}(x_0)$, $x_1\in S'$ and $y_1\in p'^{-1}(x_1)$. This induces a map $\tilde{f}$ from $\mathbb{H}$ to itself as follows: Consider any point $y\in \mathbb{H}$ and take a curve $\gamma\in \pi_1(S,x_0)$ such that the lift starting at $y_0$ ends at $y$. Define $\tilde{f}(y):= \tilde{\alpha}(1))$ where $\alpha = f\circ \gamma$. This clearly a local isometry as you said. I will prove that it is bijective also.

First i will prove that it is injective. Suppose $\tilde{f}(z)=\tilde{f}(z')=y$. Let $p(z)=x$ and $\tilde{\gamma}$ be any curve from $z$ to $z'$ and $\gamma:= \tilde{\gamma}$. Now, $f(\gamma)\in \pi_1(S',f(x))$ and $\tilde{f(\gamma)}$ starts and ends at $y$ hence $f{(\gamma)}$ is the identity in $\pi_1(S',f(x))$. But a homeomorphism between two spaces induces an isomorphism between their fundamental groups. Hence $\gamma$ has to be identity, which implies that $x=x'$.

Surjectivity:Consider a $y\in \mathbb{H}$. If $x=p'(y)$ then there exists atleast one $y'\in p^{-1}(x)$ such that $y'=\tilde{f}(z)$ for some $z$, as otherwise $x$ will not be in the image of $f$. Fix this $y'$. Take an element $\gamma'\in\pi_1(S',x)$ such that $\tilde{\gamma'}(1)=y$. If $\gamma$ is defined as $f^{-1}(\gamma')$ then $\tilde{f}(\tilde{\gamma}(1))=y$.

Pardon me if i have made some silly error.

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Thanks very much Divakaran, it was really helpful and I sincerely appreciate it ! –  Mathmath Feb 15 '11 at 15:41

Here are pointers to get you started.

1) If the two groups $\Gamma$ and $\Gamma'$ are conjugate by an isometry of the upper half plane, then you can use that given isometry to define an isometry of the quotients.

2) On the other hand, using some covering space theory, an isometry of the quotients lifts to a conjugacy of the deck groups. For example, you can use the lifting property for covering maps (see this page on covering spaces).

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