This problem arose when trying to understand the stack of twisted stable maps into a stack (specifically BG), as introduced by Dan Abramovich, Angelo Vistoli and several co-authors (Olsson, Graber, Corti,...). However, my specific question can be formulated without mentioning such beasts and I will do so. If anyone wants background I can give some.

Suppose that we have the following set-up. One is given:

  • A smooth projective curve C
  • and a finite abelian group G acting on C,
  • such that the quotient map $\pi : C \to C/G$ is an admissible cover.

Suppose moreover that $C/G \cong \mathbf{P}^1$. Over the complex numbers, one can make the following computation. For each ramification point of the covering, consider a small loop centered around that point, with orientation induced by the complex structure. Choose a lifting of that loop to a path on C. The path will start and end in the same fiber. Since the restriction of $\pi$ away from the ramification locus is a G-torsor, the difference between start- and endpoint will give us a well-defined element of G. (This uses that G is abelian -- in general, one would only get an element up to conjugation, since there are several choices of liftings of the loop.) Now consider the product of all these elements over all ramification points. By considering the fundamental group of $\mathbf{P}^1$ minus the ramification locus, it is clear that this product is the identity in G.

I would like to express this computation algebraically, i.e. without recourse to any monodromy or the classical fundamental group, working over an arbitrary base where the order of G is invertible. Unless I'm mistaken, one can still define an evaluation map associating an element of G to each ramification point, since the following (to me rather mysterious) construction should work. Consider the stack quotient $[C/G]$. This is a twisted curve in the sense of Abramovich et al, which basically means that the ramification points on C/G have been cut out and replaced by cyclotomic gerbes ("stacky points"). Now this twisted curve has an actual G-torsor over it, so we get a map $[C/G] \to BG$. Restricting it to one of the stacky points, we obtain a cyclotomic gerbe over the base scheme with a G-torsor over it. But this is by definition an object of the rigidified inertia stack of BG, and the points of the (rigidified) inertia stack correspond to the elements of G.

Question 1: Is there a way of formulating this without using the language of twisted curves, i.e. associating an element of G to each ramification point only in terms of the admissible cover? There should be, since the moduli space of maps from twisted stable curves to BG is isomorphic to the moduli space of stable curves with an admissible cover which is a G-torsor away from the branch locus, but I don't see any sensible way of expressing such a construction algebraically.

Question 2: Can one show that in the algebraic setting, the product of the elements over all ramification points is the identity?


(written hurriedly, hope it is semi-helpful and -correct)

Let's say you're over some field k. If it's not separably closed, basechange until it is. The strictly completed local ring of a ramification point in your base P^1 is going to look like k[[t]]. Now remove that point to give yourself Spec k((t)). The restriction of your cover to this punctured formal neighborhood is going to be a G-cover of Spec k((t)), which is to say a map from the absolute Galois group of k((t)) to G. But since you've wisely demanded that the order of G is prime to char k, this map kills wild inertia, and factors through the tame inertia group, which is cyclic. So the image of this map at least gives you a well-defined cyclic subgroup of G. You have to be a bit more careful to get an element of G, and indeed there isn't really a canonical choice locally; there is a choice of generator of tame inertia involved. Via class field theory you can make these choices "compatibly" at different ramification points (e.g. this will make the product vanish, as you say) but I don't think there's a "correct" such compatible choice.


What naturally associates to each ramification point (or more generally, irreducible divisor in the ramification locus) is a pair $(H,\psi)$ where $H$ is a cyclic subgroup of $G$ (with $\ge 2$ elements) and $\psi$ is an injective character. If the ground field is the complex numbers, you can have a bijection between such pairs $(H,\psi)$ and the nonidentity elements of $G$ by mapping an element $g$ of order $n$ to the subgroup generated by $g$ and the character $\psi$ which has value $e^{2\pi i/n}$, as explained by Arend.

This is written in very classical, nonstacky terms in Pardini's Abelian Covers paper. On the other hand, it works in arbitrary dimensions, so the paper is a bit more complicated than it would be if only curves where involved.


On question 1: The short answer is: Given a point on C fixed by a subgroup of $G$, the stabilizer will be cyclic, say $\mathbb Z/n$. There will be a unique generator of this group that acts on the tangent space by multiplication with $e^{2\pi i/n}$. This is the group element you are looking for.

Longer answer: To get a group element from an object of the cyclotomic inertia stack, you need to fix a generator of $\mu_r$. So assume we have done that for all relevant $r$ compatibly, say by embedding your field, which contains all relevant $r$-roots of unity, into $\mathbb C$.

Given a map $\mathcal S \to BG$ from a gerbe $\mathcal G$ with cyclic stabilizer groups over your base $S$ only gives you an object of the inertia stack if you can identify all stabilizer groups with $\mu_r$. In our situation, this identification is given by the action of the stabilizer group of the marked point on the tangent space of the twisted curve. This is the same as the action of the stabilizer of the corresponding point on $C$ on the tangent space of $C$.

  • $\begingroup$ As for question 2: I don't have a nice explanation but one can argue for $G = \mu_r$: - the twisted curve is obtained from P^1 via $r$-th root constructions - a morphism to $B\mu_n$ is given by an $n$-torsion line bundle - this $n$-torsion line bundle is given by tensoring some $O(k)$ on $P^1$ with tensor powers of the ``tautological line bundles'' (coming from the $r$-th root constructions) - the coefficient of each tautological line bundle determines the group element - the $n$-th power of this line bundle can only be trivial if the product of the corresponding group elements is trivial. $\endgroup$ Jul 23 '10 at 19:45

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