The image of divisor under isogeny - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T15:55:36Z http://mathoverflow.net/feeds/question/77040 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/77040/the-image-of-divisor-under-isogeny The image of divisor under isogeny Rurik 2011-10-03T13:39:09Z 2011-10-03T16:12:02Z <p>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 $B$?</p> http://mathoverflow.net/questions/77040/the-image-of-divisor-under-isogeny/77053#77053 Answer by rita for The image of divisor under isogeny rita 2011-10-03T16:12:02Z 2011-10-03T16:12:02Z <p>There are different possible behaviours.</p> <p>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.</p> <p>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 $B$. </p>