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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.

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  • $\begingroup$ 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
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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

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  • $\begingroup$ 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
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    $\begingroup$ 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
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    $\begingroup$ 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

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