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Let $\mathbb{H}$ be an upper half plane (this is a Riemann surface), then $PSL(2,\mathbb{Z})$ acts on $\mathbb{H}$ and it is well-know that $$ \mathbb{H}/PSL(2,\mathbb{Z})\cong \mathbb{C} $$ is again a Riemann surface. There are however three fixed points on $\mathbb{H}$, namely $$ e^{\frac{2\pi i}{6}}, \ \ \ i, \ \ \ e^{\frac{2\pi i}{3}}. $$ I hence think the image of these points should be considered as orbifold point of $\mathbb{H}/PSL(2,\mathbb{Z})$. I know there is a map $z\mapsto z^n$ which gives a new local parameter at each orbifold point, but is it really natural? It seems to me that this new chart is not really natural as it is not conformal at the orbifold point (it maps an angle $\theta$ to $n\theta$, right?)

More generally if a finite group $G$ acts on a Riemann surface $S$, we can ask a similar question about the quotient $S/G$. Should one think of $S/G$ as a smooth Riemann surface or an orbifold?

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Yes, I'd agree with the conclusion of your observation. The traditional local re-parametrization to avoid discussion of orbifolds is made possible by the fact that the isotropy group of point on a Riemann surface is a subgroup of a circle-group, so a discrete subgroup of that isotropy group is a finite cyclic group, thus admitting "unwinding" by the $z\rightarrow z^n$ maps the question mentions. In higher dimensions, orthogonal groups $SO(n)$ (with $n>2$) have non-abelian finite subgroups, so this dodge is not reliably available.

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Thank you for the answer. I am not yet totally convinced that the re-parametrization is really reasonable. I need to think a bit more. As to higher dimensional case, I think as long as the fixed locus is of codimension 1, the quotient can be endowed with a natural smooth holomorphic structure (the situation is pretty much the same as Riemann surface case). –  M. K. Sep 18 '12 at 0:12
    
Yes, if the fixed locus is codimension one, then the relevant isotropy group is again cyclic... But, yes, it's not so much that the re-parametrization is reasonable, but that it is possible. One might easily argue that it is mildly perverse. –  paul garrett Sep 18 '12 at 1:28
    
I see. Anyways I am glad to know that I am not the only one who think this argument is a bit perverse. Thanks, paul! –  M. K. Sep 18 '12 at 5:02
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