# Quotient Riemann surfaces

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?

-
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.