Principal G-covers with G finite abelian

Question

Let $X$ be a smooth, complex projective variety, and $G$ a finite abelian group. We want to study $G$-principal bundles over $X$, or, in other words, étale $G$-covers over $X$.

Topologically, these objects are classified (up to isomorphism of $G$-covers over $X$) by $H^1(X, G)$, or equivalently, by homomorphisms $\phi:H_1(X, \mathbb{Z}) \to G$.

In algebraic geometry these objects are classified (up to isomorphism) by morphisms $L: G^{\vee} \to Pic^0(X)$. Here I write $G^{\vee}$ for the group of characters of $G$, or in other words the homomorphisms from $G$ to $\mathbb{C}^*$.

(For example, think of the well-known correspondence between $2$-torsion line bundles over $X$ and étale double covers on $X$).

There is then an association $L \to \phi$ given by forgetting the structure of algebraic variety and keeping only the topological structure. This is an homomorphism $$ \lambda: Hom(G^{\vee}, Pic^0(X)) \to Hom(H_1(X, \mathbb{Z}), G). $$

The question is: can one make sense of the latter homomorphism in a purely algebraic way? In other words, can you define $\lambda$ in a simple way that does not use geometry (as I did)?

(I apologize if this is too simple, I just cannot see it)

Answer 1

Note that there is a morphism of etale sheaves

$G \to \mathcal{H}om(G^\vee , \mathcal O_X^\times)$

given by $g \mapsto (\chi \mapsto (\chi(g)))$, where we consider $\chi(g)\in \mathbb C^\times$ as a constant function.

I claim that this is in fact an isomorphism of sheaves. For cyclic groups, this follows from the Kummer exact sequence idea: any invertable function is locally an n^th power (see Will's comment). Then any finite abelian group is a product of cyclic groups.

Hence there is an isomorphism

$H^1(X,G) \to H^1 (X, \mathcal{H}om(G^\vee , \mathcal O_X^\times))$.

Now by universal coefficients , $H^1(X,G) = Hom(H_1(X,\mathbb Z),G)$. The right hand side is not what you claim it is (see my comment - EDIT: see edit).

For each choice of character $\chi : G \to \mathbb G_m$, we can consider line bundles with a reduction of structure group to $G$ via $\chi$. The thing above is what manages all of these at once.

EDIT: Sorry, didn't spot that $X$ should be projective. In that case, I guess that $H^1(X,\mathcal{H}om(G^\vee , \mathcal O_X^\times))$ will be isomorphic to $Hom(G^\vee , H^1(\mathcal O_X^{\times})$. This can probably be seen from the spectral sequence in terms of sheaf Exts... Sorry for the error, and feel free to edit.

Answer 2

This is a proof of the last claim in Sam's answer. It would be hard to read as a comment.

Since both groups in question naturally split into direct sums when $G$ is a direct sum, you can reduce to $G$ cyclic. There is an exact sequence:

$0 \to \mathcal Hom(G^{\vee},\mathcal O_X^{\times}) \to \mathcal O_X^\times \to \mathcal O_X^\times \to 0$

with the map from $\mathcal O_X^\times$ to itself being the $n$th power map with $n$ the order of $G$. Taking cohomology gives, after taking cokernels and kernels:

$0\to H^0(X,\mathcal O_X^\times)/H_0(X,\mathcal O_X^\times)^n \to H^1(\mathcal Hom(G^{\vee},\mathcal O_X^\times))\to Hom(G^\vee,H^1(\mathcal O_x^\times))\to 0$

$H^0(X,\mathcal O_X)^\times= \mathbb C^\times$, and $\mathbb C^\times$ is a divisible group, so the $n$th power subgroup is the whole group. This gives the isomorphism.