MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

subj: etale covers of line bundles on an abelian variety

Is there an explicit decryption of finite etale covers of a line bundle $L$ on an abelian variety and its associated C*-bundles $L^o = L \setminus A\times {0}$ (i.e. the C*-bundle $L^o$ is $L$ without the zero section) ?

Pull-back along multiplication by $n$ map $n:A\rightarrow A$ gives a pull-back $n^*_A L' \rightarrow L$. A tensor power map $L'\mapsto L'^{\otimes n}$ gives rise to an etale map of C*-bundles
$L^o \rightarrow L^{o\otimes n}$, and thus if $L$ happens to be a tensor power, to an etale cover of $L^o$.

Can we obtain all etale covers of $L^o$ this way ?

share|cite|improve this question
You need to consider pullback along isogenies of $A$ followed by pulbacks along tensor power maps. After that, I think it's everything, but I don't have a decent proof yet. – Will Sawin May 11 '12 at 20:31
up vote 1 down vote accepted

For clarity, the best way to work with this is complex-analytically. I am sure there is a, probably more involved, algebraic proof.

Lemma: Let $M$ be a complex manifold and let $X$ be a $\mathbb C^\times$-bundle on $X$. Let $Y$ be a finite etale cover of $X$. Then $Y$ is a $\mathbb C^\times$ bundle on an etale cover of $M$, with that bundle being an $n$th tensor root of the pullback of $X$.

Since etale covers of abelian varieties are just isogenies, that gives you the explicit description.

Proof of the lemma: Consider the inverse image in $Y$ of a fiber of $X$ over $M$. This is a union of connected components. The components, being etale covers of $\mathbb C^\times$, are copies of $\mathbb C^\times$ that map to it along an $n$th power map. Let $N$ be $Y$ with each connected component contracted. That is, it is the quotient by the equivalence relation that two points are equivalent if they are in the same connected component of a fiber over $M$. Then $Y$ is a $\mathbb C^\times$-bundle on $N$.

$N$ has a map to $M$. We prove that it is etale. This is local on $M$, so consider an open ball on which $X$ is trivial. Then $X$ is just $\mathbb C^\times$ cross an open ball. The fundamental group is $\mathbb Z$, so all etale covers are just the obvious $n$th power maps, and in all of these the map $N\to M$ is etale.

Furthermore these obvious $n$th power maps are locally $n$th power maps, and $n$ is the same in the entire open ball, therefore locally constant, therefore constant. So the map from the $Y$ bundle to the pullback of $X$ is an $n$th power map, so the pullback of $X$ is the $n$th tensor power.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.