Question

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?

Answer 1

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.

Answer 2

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?

Answer 3

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.