etale covers of line bundles on an abelian variety - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T18:23:02Z http://mathoverflow.net/feeds/question/96680 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/96680/etale-covers-of-line-bundles-on-an-abelian-variety etale covers of line bundles on an abelian variety o a 2012-05-11T15:53:59Z 2012-05-12T04:57:47Z <p>subj: etale covers of line bundles on an abelian variety</p> <p>Is there an explicit decryption of finite <em>etale covers</em> 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) ?</p> <p>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<br> $L^o \rightarrow L^{o\otimes n}$, and thus if $L$ happens to be a tensor power, to an etale cover of $L^o$. </p> <p>Can we obtain all etale covers of $L^o$ this way ? </p> http://mathoverflow.net/questions/96680/etale-covers-of-line-bundles-on-an-abelian-variety/96737#96737 Answer by Will Sawin for etale covers of line bundles on an abelian variety Will Sawin 2012-05-12T04:57:47Z 2012-05-12T04:57:47Z <p>For clarity, the best way to work with this is complex-analytically. I am sure there is a, probably more involved, algebraic proof.</p> <p>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 $X$.</p> <p>Since etale covers of abelian varieties are just isogenies, that gives you the explicit description.</p> <p>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 $N$.</p> <p>$N$ 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.</p> <p>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.</p>