While reading the paper *Seifert Fibred Homology 3-Spheres and the Yang-Mills Equations on Riemann Surfaces with Marked points* by M. Furuta and B. Steer, I stumbled upon the following statement:



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> Any compact orbifold Riemann surface, with $n\geq 3$ singular points or $n=2$ and $\alpha\_1=\alpha\_2$ if the genus $g$ is zero, has the form $N/D$, where $N$ is a compact Riemann surface and $D$ is a finite group of diffeomorphisms.
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However, there is no comment why this should be true. The paper cites the article of P. Scott on the geometrization conjecture, but on that paper the only thing that is assured is that if this conditions are satisfied, then the orbifold is a global quotient, by some unspecified manifold. However,



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> How can we deduce that this manifold must be a Riemann surface?
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This also leads me to ask,



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> If $N$ is a compact Riemann surface and $D$ is a finite group of diffeomorphisms, is the quotient $N/D$ a compact Riemann orbifold surface?
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Thanks in advance for your answers.



When is a compact orbifold Riemann surface a global quotient of a Riemann surface
Let $M$ be a your compact orbifold Riemann surface and let $S\subset M$ be the finite set of orbifold singularities.


**Theorem:**
The following are equivalent:


* $M$ is the quotient of a closed Riemann surface by a finite group
* $M$ admits a conformal metric of constant curvature (with the correct cone angles at the singularities)


*Proof:*
If $M = N/D$, $N$ admits a metric of constant curvature by Poincaré--Koebe'uniformization, which is essentially unique and thus invariant by $D$ (one has to be a bit careful in the positive curvature case).


Conversely, a metric of constant curvature on $M$ gives rise to a holonomy representation $\rho$ of the fundamental group $\pi\_1(M\backslash S)$ into $\mathrm{Isom}(X)$, where $X$ is the sphere/the euclidean plane/the hyperbolic plane depending on the sign of the curvature. The image of $\rho$ is a finitely generated linear group which admits a torsion-free subgroup of finite index $\Gamma$ (by Selberg's lemma). The preimage of $\Gamma$ is a normal finite index subgroup of $\pi\_1(M\backslash S)$ which thus defines a finite ramified normal cover $N$ of $M$. One verifies that the ramifications have the correct degree at the singularities, so that $N$ is a smooth Riemann surface covering $M$. QED


Now, by a theorem of Troyanov, every closed orbifold of non-positive Euler characteristic admits a conformal metric of constant non-positive curvature. This shows that every orbifold is covered by a smooth Riemann surface except for the few ones that have positive Euler characteristic, namely, spheres with 1 or 2 singularities, or 3 singularities of angles $\pi, 2\pi/3, 2\pi/3$ or $\pi, \pi, 2\pi/3$.


These cases are not too hard to treat. For instance, the sphere with 3 singularities of angle $\pi, 2\pi/3, 2\pi/3$ is the double of a spherical triangle with angles $\pi/2, \pi/3, \pi/3$.


