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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 ?

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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
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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.

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