## orbifold covering

Given two compact surfaces $S_1$ and $S_2$ of genus at least $2,$ it is easy to tell when $S_1$ covers $S_2:$ whenever $\chi(S_2)$ divides $\chi(S_1).$ Now, suppose I have two orbifolds of negative Euler characteristic. There is still the divisibility obstruction, but there must be others. Or must there?

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 Stupid question: if $S_2$ is actually a manifold but $S_1$ isn't, can $S_1$ cover $S_2$? Apologies in advance; I know nothing about orbifolds really. – Kevin Buzzard Apr 28 2011 at 18:49 [I think what I'm really asking is what "cover" means. In algebraic geometry the analogue of this question is a question about algebraic stacks, and there are morphisms between these guys that aren't representable; if you are allowing such gadgets in your setting then an orbifold can map to a manifold---but then you don't have to have the divisibility of Euler characteristics, I don't think, because the degree of the map may not be a positive integer.] – Kevin Buzzard Apr 28 2011 at 18:51 Mind if I ask for the definition of $\chi$(orbifold)? Or first, the definition of an orbifold? Is it just a (proper) DM-stack (let me limit myself to algebraic stacks rather than analytic ones, though I don't know if any compact 2-dim orbifold admits a global "covering" by a compact smooth surface so that one can use GAGA) over $\mathbb C$ (whose coarse moduli has dimension 1, in your setting)? Or one requires in addition that there are only finitely many stacky points? Then I guess one defines $\chi$ using a triangulation that involves all stacky points (the inertia stack ... – shenghao Apr 28 2011 at 23:26 is generically flat, so there is always an alg. stratification, which refines to a triangulation, such that...). And as Kevin pointed out, by a "covering" we probably want "representable finite etale maps", so that $\chi$ divides. – shenghao Apr 28 2011 at 23:34 @shenghao: definition of $\chi(orbifold)$ is the one in @hungrygrad's not-quite-answer below. @Kevin's question is not stupid, and the answer is: no. A subgroup of a torsion-free subgroup is torsion-free also (see @Agol's comment below). – Igor Rivin Apr 29 2011 at 0:11

Since $\chi(S_1)$ and $\chi(S_2)$ are non-zero, you know what the degree a possible covering should be: it is $$d=\chi(S_1)/\chi(S_2).$$

Let $S_2$ have $k$ cone points of order $$d_1, \ldots, d_k \geqslant 2.$$ If a covering $S_1 \to S_2$ exists, then the preimage of the cone point of order $d_i$ is a collection of $k_i \leqslant d$ cone points of order $d_{i1}, \ldots, d_{i_{k_i}}\geqslant 1$ (order 1 = smooth point). Every $d_{ij}$ divides $d_i$ and we have

$$d = d_i/d_{i1} + \ldots + d_i/d_{i_{k_i}}.$$

A necessary condition for having a covering $S_1 \to S_2$ is therefore the following:

By adding some auxiliary 1's to the set of all cone orders of $S_1$, we must get a set of natural numbers which can be partitioned into subsets ${d_{i1}, \ldots, d_{i_{k_i}}}$ such that every $d_{ij}$ divides $d_i$ and by summing the natural numbers $d_i/d_{ij}$ along $j$ we get $d$, for every $i$.

The non-trivial problem here is: is this numerical condition enough to guarantee the existence of a covering? The same problem can be rephrased in therms of branched coverings of surfaces, and is called the Hurewitz existence problem. The Hurewticz problem has a positive solution when $S_2$ is not a sphere, i.e. when it is a surface with genus (and cone points), as proved by Husemoller in 1962. I think that this implies that an orbifold covering exists when $S_2$ has positive genus.

When $S_2$ is a sphere there are some cases where the Hurewitcz problem has no solution, i.e. the necessary conditions above do not guarantee the existence of a covering. The general case when tha base hyperbolic orbifold $S_2$ is a sphere with some cone points is open, see some recent papers of Pascali, Pervova and Petronio.

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I may be misinterpreting your question, but if you calculate the Euler characteristic of an orbifold using the Riemann-Hurwitz Formula, then it is multiplicative in covering.

Here, $\chi(S)=2g-2-m+\sum_{i=1}^m\frac{1}{p_i}$ with $m$ cone points $x_i$ with respective order $p_i$.

I learned about it in the "Primer on Mapping Class Groups" by Farb and Margalit, available on both their websites.

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 Yes, you are correct, it is multiplicative, which is what I say in my question: if the ratio of Euler characteristic is not an integer, there is no cover. But if it IS an integer, it is still not clear that there is. – Igor Rivin Apr 29 2011 at 0:06

Kevin's comment gives easy counterexamples -- take an orbifold that's topologically a sphere, but has lots of Z_2 points -- we could give it any negative orbifold characteristic that we want, and yet this would obvious never cover a two holed torus with no orbifold structure, say.

You might rule that specific example out with also asking something about the euler characteristic of the coarse moduli spaces, but I don't think it would help much.

A little more broadly, you have the usual condition of covering spaces, that the fundamental group of the cover is a subgroup of the fundamental group downstairs. It would be interesting to know whether this was sufficient: if $\pi_1(Y)$ is a subgroup of $\pi_1(X)$, does $Y$ cover $X$?

I haven't thought about this hard, though. I suspect this is probably known, somewhere in the literature on Fuchsian groups. Again, not being careful, but you might hope this reformulated question would be related to something along the lines of whether if G was abstractly a subgroup of H as groups, and now we consider them as subgroups of the automorphisms of the hyperbolic plane, can we conjugate G into H?

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Yes, the fundamental group of a closed orientable 2-orbifold determines it uniquely. An orientable 2-orbifold has finitely many cone points. Each cone point is determined uniquely by a conjugacy class of maximal finite cyclic subgroups. The genus is then determined by the Euler characteristic and the types of cone points. – Agol Apr 28 2011 at 20:22
Ah, this is much easier than I could see at the beginning. I didn't process that essentially I was asking if the fundamental group determined the orbifold surface, and it gets much cleaner when you think of it that way. Thanks! – Paul Johnson Apr 28 2011 at 21:16