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Mar 4, 2013 at 18:56 vote accept expmat
Feb 27, 2013 at 23:45 answer added Misha timeline score: 8
Feb 27, 2013 at 22:48 comment added François Brunault 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.
Feb 27, 2013 at 22:45 comment added expmat @Misha: can you give me a reference for those things?
Feb 27, 2013 at 21:42 comment added Misha 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.
Feb 27, 2013 at 20:50 comment added expmat Or some other (more complicated) relation that holds in general for any $\Gamma$?
Feb 27, 2013 at 20:42 comment added expmat Right, but can we at least say that $N(\Gamma)$ is a subgroup of $Aut(\Gamma)$?
Feb 27, 2013 at 20:34 comment added Venkataramana 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 $N(\Gamma)$.
Feb 27, 2013 at 20:30 history edited expmat CC BY-SA 3.0
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Feb 27, 2013 at 20:29 comment added expmat Yes, but what about the case where $\Gamma$ is not torsion free?
Feb 27, 2013 at 20:20 comment added Venkataramana 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.
Feb 27, 2013 at 20:15 history asked expmat CC BY-SA 3.0