show/hide this revision's text 4 fixed typo

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 counterimage 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.

show/hide this revision's text 3 rephrasing

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 counterimage 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

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 above equality holds natural numbers $d_i/d_{ij}$ along $j$ we get $d$, for each 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.

show/hide this revision's text 2 rephrasing

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 above every the counterimage of the cone point of order $d_i\geqslant 2$ d_i$ is a collection of $S_2$ there must be some k_i \leqslant d$ cone points of order $d_{i1}, \ldots, d_{i_{k_i}}\geqslant 1$ (order 1 = smooth point is allowed) such that $d_i$ is divided by each 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 the above equality holds for each $i$.

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.

show/hide this revision's text 1