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Let us say that I have a complex abelian variety $A$, an ample line bundle on $A$, $L$, and an effective divisor $E\in|L|$. It is well known that it exists an isogeny $\varphi:A\rightarrow B$ and a principal polarization $M$ on $B$ such that $L\simeq \varphi^*M$. My question is: can we say something about the divisor $\varphi(E)$? For example it is possible that the map $\varphi$ restricted to $E$ has degree 1? I do not think this may happen, but I do not know how to see it...\ is there any chance that $\varphi(E)$ is a $\Theta$-divisor in $B$?

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up vote 6 down vote accepted

There are different possible behaviours.

Here's an example: let $B$ be a ppav, $\Theta$ an effective theta divisor, let $f\colon A\to B$ be a connected e'tale double cover given by an element $\eta\in Pic^0(B)[2]$ and set $E=f^*\Theta$. Of course $E\to \Theta$ is 2-to-1 by construction.

By the projection formula for double covers, we have $H^0(A, E)=H^0(B, \Theta)\oplus H^0(B,\Theta+\eta)$: if we consider on $L:={\mathcal O}_A(E)$ the pullback linearization, the first summand of this decomposition is the $+1$-eigenspace and the second one is the $-1$-eigenspace. So $|E|$ is a $1$-dimensional linear system, $E$ and $E':=f^*(\Theta+\eta)$ are mapped 2-to-1 onto theta divisors and the remaining elements of $|E|$ are mapped birationally onto their images in $B$.

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Thank you very much! That was really helpful! – Rurik Oct 4 '11 at 7:46
You're welcome. – rita Oct 4 '11 at 8:28

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