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)

  • $\begingroup$ I think what you have written is not quite right as stated. For example etale double covers of $X$ are in correspondence with line bundles $L$ with a specified isomorphism $\mathcal O_X \equiv L^{\otimes 2}$. This is not the same as 2-torsion elements of $Pic^0$. For example, on $\mathbb A^1 - {0}$ there are no non-trivial line bundles, but there is a double cover. $\endgroup$ Jul 5, 2012 at 12:50
  • $\begingroup$ The symbol $\equiv$ was meant to be $\cong$... $\endgroup$ Jul 5, 2012 at 12:51
  • $\begingroup$ @Sam: Franz does specify that $X$ is projective, whereas your example is not projective. For an 'etale double cover $f:Y\to X$ of a projective variety $X$, the invertible sheaf $\text{Ker}(\text{Tr}_f:f_*\mathcal{O}_Y \to \mathcal{O}_X)$ is trivial if and only if $h^0(Y,\mathcal{O}_Y)$ equals $2$ if and only if $Y$ is disconnected if and only if $Y$ equals $X \sqcup X$ as a covering of $X$. $\endgroup$ Jul 5, 2012 at 13:09
  • $\begingroup$ @Jason: Ah, thanks I didn't spot that $X$ should be projective. I'll edit my answer. $\endgroup$ Jul 5, 2012 at 13:16

2 Answers 2


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.

  • $\begingroup$ I don't understand how your first claim is related to the Kummer exact sequence at all. The isomorphism is an isomorphism of presheaves as well as sheaves because it is defined on every open set, since field extensions of $\mathbb C$ don't give you any new characters of a group. $\endgroup$
    – Will Sawin
    Jul 5, 2012 at 15:46
  • $\begingroup$ Remark 1.3.1, p. 84, of "Enriques Surfaces" by Cossec and Dolgachev gives this result also. They fram this as part of "Cartier duality". $\endgroup$ Jul 5, 2012 at 17:11
  • $\begingroup$ By the way, I also think a spectral sequence proof does exist. $\endgroup$
    – Will Sawin
    Jul 5, 2012 at 20:41
  • $\begingroup$ As for the spectral sequence. It would be enough (using the exact sequence in low degrees) to prove that 1) $Ext^1(G^{\vee},\mathcal{O}_X^*)$ equals $Hom(G^{\vee},H^1(\mathcal{O}_X^*)$ 2)Vanishing of $H^0(\mathcal{E}xt^1(G^{\vee}, \mathcal{O}_X^*))$. Are these true? $\endgroup$
    – calc
    Jul 6, 2012 at 7:22
  • $\begingroup$ Most important thing: Thank you and Will very much for your answers! $\endgroup$
    – calc
    Jul 6, 2012 at 7:34

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.

  • $\begingroup$ Do you mean $H^0$ when you write $H_0$? $\endgroup$
    – calc
    Jul 5, 2012 at 18:01
  • $\begingroup$ Yes. I'll fix that. $\endgroup$
    – Will Sawin
    Jul 5, 2012 at 20:37
  • $\begingroup$ Combining Sam Gunningham's answer and yours I see how the (iso)morphism that I want. I am still a bit confused about one thing. I was hoping to see clearly from the proof of the isomorphism $Hom(H_1(X, \mathbb{Z}), G) \to Hom(G^{\vee}, Pic^0(X))$ that one really needs to dualize $G$ into $G^{\vee}$. Is this clear? $\endgroup$
    – calc
    Jul 6, 2012 at 9:07
  • $\begingroup$ @Franz -- "... Is this clear?" Since $G$ and $G^\vee$ are isomorphic as finite Abelian groups, I agree that it might be unclear that one "must" dualize $G$. This becomes more clear when one works in char $p$, with finite flat Abelian group schemes. In this case, when $G$ equals $\mathbb{Z}/p\mathbb{Z}$, then $G^\vee$ is the non-reduced group scheme $\mathbf{\mu}_p$. These group schemes are not isomorphic. Yet the isomorphism above extends to this case only for $G^\vee = \mathbf{\mu}_p$, not for $G^\vee = \mathbb{Z}/p\mathbb{Z}$, cf. Cossec-Dolgachev (or others). $\endgroup$ Jul 6, 2012 at 13:29
  • $\begingroup$ A different way to say it is this: It's because $G$ switches places in the $Hom$. $Hom (X,G)=Hom (G^\vee,X^\vee)$. $\endgroup$
    – Will Sawin
    Jul 6, 2012 at 15:06

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