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Suppose that we have an orbifold, for example $S^2(3,5,7)$. There are many references that this is a good orbifold and so finitely covered by a surface. By Riemann-Hurwitz, the surface would have Euler number as multiple of -34.

My question is, does the surface of Euler number exactly -34 cover the orbifold?

More generally, if $k\chi(O)=2m$, is there a surface of Euler number $2m$ covering $O$?

B. Peet
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1 Answers1

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I suspect that it was known earlier, but you can apply a theorem of Edmonds, Ewing and Kulkarni to determine the existence of torsion-free subgroups of the given index in a Fuchsian group, such as $F=\pi_1(S^2(3, 5, 7))$. It follows from their work that $F$ contains a torsion-free subgroup $F'$ of index $k$ if and only if $k$ is divisible by $3\times 5\times 7$. This is equivalent to the existence of a connected surface $S$ which is a $k$-fold cover of your orbifold. Since $$ \chi(S)=k \chi(S^2(3, 5, 7)) = k (\frac{1}{3} + \frac{1}{5} + \frac{1}{7} -1) $$ the problem reduces to simple arithmetic which I will leave to you to work out.

[1] Allan L. Edmonds, John H. Ewing, and Ravi S. Kulkarni, Torsion free subgroups of Fuchsian groups and tessellations of surfaces, Bull. Amer. Math. Soc. (N.S.) Volume 6, Number 3 (1982), 456-458.

If I remember it correctly, a full proof of their theorem appeared in Inventiones Mathematicae in 1984 or so.

Related.

Moishe Kohan
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  • Thanks, this is exactly what I was looking for. What I want to show is that if there is a $G$ action on a surface of order $105p$ for $p$ prime, with quotient $S^2(3,5,7)$, then there is an element of order $p$ that acts freely. Does this follow? – B. Peet Sep 03 '17 at 11:22
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    @B.Peet: No, it does not follow from the stated result; if I had to guess, it is just false. Anyway, this is a separate question. – Moishe Kohan Sep 03 '17 at 23:10