# Polarization of an abelian variety made by the sum of two divisors

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.

• Why do you expect to get 3 in general? – Angelo May 10 '20 at 5:20
• @Angelo, what do you mean? 3 is simply the degree of the polarization. – 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 $$\chi (L)=4$$.
• 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 $P=0$. – Hacon May 12 '20 at 14:17
• 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 – TartagliaTriangle May 16 '20 at 8:28
• 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 $Pic^0(B)$. – Hacon May 18 '20 at 2:53