Let $\Gamma \leq PSL_2(\mathbb{R})$ be a Fuchsian group. What is the relation between $N(\Gamma) = \{ \alpha \in PSL_2(\mathbb{R}) \mid \alpha \Gamma \alpha^{-1} = \Gamma \}$ and $Aut(\Gamma \backslash \mathcal{H})$, the group of automorphisms of the Riemann surface $\Gamma \backslash \mathcal{H}$?

Here $\mathcal{H}$ is the upper half-plane.

  • $\begingroup$ If $\Gamma$ is torsion free, then they are the same. Any automorphism of the Riemann surface, by the lifting criterion, lifts to an automorphism of the upper half plane (i.e. an element of $SL(2,{\mathbb R})$ . This element normalises $\Gamma$ almost by construction. $\endgroup$ Feb 27, 2013 at 20:20
  • $\begingroup$ Yes, but what about the case where $\Gamma$ is not torsion free? $\endgroup$
    – expmat
    Feb 27, 2013 at 20:29
  • $\begingroup$ It can happen, I think, that $\Gamma \backslash {\mathcal H}$ may even be the projective line, if $\Gamma$ has torsion; in that case, the automorphisms of the Riemann surface is $PGL(2, {\mathbb C})$, much larger than $N(\Gamma)$. $\endgroup$ Feb 27, 2013 at 20:34
  • 2
    $\begingroup$ In general, the relation (for nonelementary Fuchsian groups $\Gamma$) is: $N(\Gamma)< Aut(\Gamma)$ (since such $\Gamma$ has trivial centralizer in $PSL(2,R)$) and $N(\Gamma)/\Gamma< Aut(H^2/\Gamma)$. If you treat the quotient $H^2/\Gamma$ as an orbifold (and its best if you do), then the latter inclusion is the equality. $\endgroup$
    – Misha
    Feb 27, 2013 at 21:42
  • 1
    $\begingroup$ An obvious case is $\Gamma=\operatorname{PSL}_2(\mathbf{Z})$ : in this case $N(\Gamma)/\Gamma$ is trivial while the quotient $\Gamma \backslash \mathcal{H}$ is isomorphic to $\mathbf{C}$, whose automorphism group is much larger. $\endgroup$ Feb 27, 2013 at 22:48

1 Answer 1


Unfortunately, as far as I know, nobody really explains such things as they are considered to be "too elementary". The most basic, briefest and down-to-earth reference I know for the needed background in Fuchsian groups is S.Katok's book "Fuchsian groups." Here are some proofs:

A Fuchsian group is called elementary if its limit set consists of at most 2 points; a group is called nonelementary otherwise. (E.g., every Fuchsian group $\Gamma$ with $Area(H^2/\Gamma)<\infty$ is nonelementary.) You also need to know that a nonelementary group contains at least two hyperbolic elements $\gamma_1, \gamma_2$ with disjoint fixed-point sets. (All this should be in Katok's book.)

Suppose now that $\Gamma$ is nonelementary Fuchsian and $g\in G:=PSL(2,R)$ belongs to the centralizer of $\Gamma$. Then $g$ would have to fix the fixed points of $\gamma_1$ and of $\gamma_2$, so it fixes four distinct points on the unit circle. Thus, $g=1$. Suppose now that $g\in N(\Gamma)$, the normalizer of $\Gamma$ in $G$. Then, unless $g=1$, the automorphism of $\Gamma$ induced by conjugation via $g$, is nontrivial. Thus, we obtain an injection $N(\Gamma)\to Aut(\Gamma)$. In particular, $g\in N(\Gamma)$ induces an inner automorphism of $\Gamma$ if and only if $g\in \Gamma$. Now, if $g\in G$ satisfies $g\Gamma g^{-1}\subset \Gamma$, then $g$ projects to an endomorphism $[g]$ of $S=H^2/\Gamma$ (just check that the above inclusion forces $g$ to preserve $\Gamma$-equivalence relation on $H^2$). If $g\in N(\Gamma)$ then the same argument shows that $[g]$ is an automorphism since $[g^{-1}]$ is its inverse. Clearly, $N(\Gamma)\to Aut(S), g\to [g]$ is a homomorphism whose kernel contains $\Gamma$. It is also immediate that $\Gamma$ is the kernel of this homomorphism. Thus, you get an embedding $N(\Gamma)/\Gamma \to Aut(S)$.

There are many sources for orbifolds (start with wikipedia article), but the key is that in the context you are interested in, orbifolds provide the "correct" generalization of the covering theory to the case of Fuchsian groups which do not act freely on $H^2$. For instance, you probably know that if $p: X\to Y$ is the universal cover (in the standard sense), then every homeomorphism $f: Y\to Y$ lifts to a homeomorphism $X\to X$ which normalizes the deck-group. The same works for orbifolds. Thus, if you treat $S$ as an orbifold $O$, then every (say, conformal) automorphism $f$ of $O$ lifts to a conformal automorphism $\tilde{f}$ of $H^2$ (the universal cover of $O$) which normalizes $\Gamma$. Hence, $\tilde{f}\in PSL(2,R)$ and, thus, belongs to $N(\Gamma)$. This establishes that the homomorphism $$ N(\Gamma)/\Gamma \to Aut(O) $$ is also surjective. (The fact that for $g\in N(\Gamma)$, $[g]$ is an automorphism of $O$, is immediate once you understand the orbifold definitions.)

  • $\begingroup$ Where are you using the fact that $\Gamma$ is nonelementary in the proof that $N(\Gamma) / \Gamma \leq Aut(S)$ ? $\endgroup$
    – expmat
    Feb 28, 2013 at 15:52
  • $\begingroup$ Also, is it clear that the map $[g]$ respects the complex structure of the Riemann surface $\Gamma \backslash \mathcal{H}$? $\endgroup$
    – expmat
    Feb 28, 2013 at 16:24
  • $\begingroup$ 1. Yes, conformality is immediate since it is a local property: First check it away from the projections of fixed points and then, if you like, use Riemann extension theorem. 2. If a group is elementary then (with one exception) it will have centralized of real dimension 1 or 3. Every element of centralizer projects trivially to the quotient surface. $\endgroup$
    – Misha
    Feb 28, 2013 at 18:28

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