Let $X$ be an abelian variety of dimension $n$, and let $L$ be a polarization, that is, an ample line bundle on $X$, with $\chi(L)=3$. In my specific case, I have that $L=\mathcal{O}_X(\Theta + D)$, where $\Theta$ is an ample divisor with $\chi(\Theta)=1$ and $D$ is an effective Cartier divisor.

I want to show that $(D^2)=0$ (self-intersection of $D$), or equivalently that $(\Theta^{n-2}.D^2)=0$.

For $n=2$, $X$ is a surface, and using Riemann-Roch I have that $2\chi(L)=6=(\Theta^2)+2(\Theta.D)+(D^2)$, where the first two intersection numbers are strictly positive because of ampleness of $\Theta$ (in particular $(\Theta^2)=2$). If I suppose $(D^2)\ne 0$, then $(D^2)=2$ and so $(\Theta.D)$ must be 1. But this is impossible by the index theorem, because we have $4=(\Theta^2)(D^2)\le (\Theta.D)^2$.

But for dimension $n>2$, I don't know how to procede, because in Riemann-Roch formula $n!$ increases too fast, so it seems impossible to make the same argument.

Thanks for help!

Note: I have already posted this question on Math StackExchange, but maybe it is better to post it here.

  • $\begingroup$ Why do you expect to get 3 in general? $\endgroup$ – Angelo May 10 '20 at 5:20
  • $\begingroup$ @Angelo, what do you mean? 3 is simply the degree of the polarization. $\endgroup$ – TartagliaTriangle May 10 '20 at 7:38

For any $P\in Pic^0(A)$ consider the map $|\Theta +P|\times |D-P|\to |\Theta +D|\cong \mathbb P ^2$. Since $D$ is effective, there is an abelian subvariety $T\subset Pic ^0(A)$ such that $|D-P'|\ne \emptyset$ for any $P'\in P+T$ and if $t=\dim T$ the $D^t\ne 0$ but $D^{t+1}=0$. If $t\geq 3$, then a general element in the image of the above map may be written as $\Theta _{P'}+D_{P'}$ in for infinitely many $P'\in P+T\subset Pic ^0(A)$. Thus $\Theta _{P'}+D_{P'}=\Theta _{P''}+D_{P''}$, but then $D_{P''}\geq \Theta _{P'}$ (as $\Theta _{P'}\in |\Theta +P'|$ is unique and different from $\Theta _{P''}$) and hence $\chi (\Theta +D)\geq \chi (2\Theta )>3$. Finally, if $t=2$, the above argument shows that any element $G\in |\Theta +D|$ can be written as a sum of elements $\Theta _{P'}\in |\Theta +P'|$ and $D_{P'}\in |D-P'|$ and $\dim |D-P'|=0$. Thus, the corresponding rational map $T\to \mathbb P ^2$ is generically finite, and of degree $>1$ (as $T$ is not rational). But then, for general $G\in |\Theta +D|$, we have $G=\Theta _{P'}+D_{P'}=\Theta _{P''}+D_{P''}$ which implies $\Theta _{P'}=D_{P''}$ and hence $\chi (L)=4$.

NB I originally misread the question so I have edited the answer appropriately

  • $\begingroup$ Thanks for the answer! A couple of things I don't understand: why there exists such an abelian subvariety T? And why its dimension determines self-intersections of D? $\endgroup$ – TartagliaTriangle May 12 '20 at 7:52
  • 1
    $\begingroup$ Let $D$ be an effective divisor on an abelian variety $A$. If $D$ is not ample, there exists a quotient abelian variety $A\to B$ such that $D$ is the pull-back of an ample divisor on $B$. Thus $D^t\ne 0$ and $D^{t+1}=0$ where $t=\dim B$. Also $|D+P|\ne 0$ for any $P\in Pic^0(B)\subset Pic ^0(A)$. If $D$ is nef, then there will be some $P\in Pic^0(A)$ such that $|D+P|\ne \emptyset$. Since you are actually assuming $D\geq 0$ then we can assume $P=0$. $\endgroup$ – Hacon May 12 '20 at 14:17
  • $\begingroup$ Another question: why for $t \ge 3$ there are infinitely many ways for writing an element in the image, while for $t=2$ there is only one? I was thinking about $V^0(D)$, but I don't know how to see its dimension. Thanks for your help $\endgroup$ – TartagliaTriangle May 16 '20 at 8:28
  • 1
    $\begingroup$ If $t\geq 3$, then there is a $t+d-1\geq 3$-dimensional family of divisors in $|D-P|$ where $d=h^0(D)$ and $P\in Pic ^0(B)\subset Pic ^0(A)$; in fact these divisors are parametrized by a $\mathbb P ^{d-1}$ bundle over $T$ and they are all distinct. The point is that $D$ is the pullback of an ample divisor say $\bar D$ on $B$ and then $h^0(\bar D-P)=h^0(\bar D)\geq 1$ for any $P\in Pic^0(B)$. If $t=2$ and $h^0(D)=1$, then we only have a 2 dimensional family of divisors $D_P\in |D-P|$. But in this case the family is parametrized by the abelian surface $Pic^0(B)$. $\endgroup$ – Hacon May 18 '20 at 2:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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