Assume we have two p.p. simple abelian surfaces $(A_i,D_i)$, i=1,2, over $\mathbb{C}$ with the following commutative diagram:

$\require{AMScd} \begin{CD} A_1 @>{birational}>> A_2\\ @V{2:1}VV @VV{2:1}V \\ A_1/(-1) @>{birational}>> A_2/(-1)\\ @V{\iota_1}VV @VV{\iota_2}V\\ \mathbb{P}^3 @>{birational}>> \mathbb{P}^3 \end{CD}$

Here the horizontal arrows, are birational maps (not morphisms) and the composition of $(2:1)$ and $\iota_i$ is actually the morphism induced by the linear system $|2D_i|$ for $i=1,2$. That is we have $|2D_i|: A_i \rightarrow \mathbb{P}^3$.

The birational map on top must be an isomorphism, since $A_1$ and $A_2$ are abelian. So we have $A_1\cong A_2$ as unpolarized abelian surfaces.

$\textbf{Question:}$ Do we actually get $(A_,D_1)\cong (A_2,D_2)$ respectively under which conditions can we conclude that we have an isomorphism of principally polarized abelian surfaces using this diagram? Or are there pairs $(A_i,D_i)$ where such an diagram cannot exist?

We may assume $NS(A_i)\geq 2$ because for $NS(A_i)=1$ we must have $(A_1,D_1)\cong(A_2,D_2)$.

I thought one could start with the polarizations $\mathcal{L}_i$ on $A_i/(-1)$ coming from $\mathcal{O}_{\mathbb{P}^3}(1)$. These pullback to $\mathcal{O}_{A_i}(2D_i)$. It would be nice if one could now show that $A_1 \xrightarrow{\cong}A_2$ pulls $D_2$ back to $D_1$. This isomorphism preserves amplenes and self-intersection. So the pullback of $D_2$ should be a principal polarization on $A_1$ but is it $D_1$? But for this we need to know what happens with $\mathcal{L}_2$ under $A_1/(-1) ---> A_2(-1)$? But I have no idea how to get any information about this.

  • $\begingroup$ Every rational transformation from a smooth variety to an Abelian variety is everywhere regular by the Weil extension theorem. Thus, all of your horizontal arrows are isomorphisms. $\endgroup$ Oct 20, 2017 at 13:56
  • $\begingroup$ @JasonStarr: There is no abelian variety in the bottom horizontal arrow so how do you apply Weil's theorem? $\endgroup$
    – naf
    Oct 21, 2017 at 4:34
  • $\begingroup$ @ulrich Hello again. If the top horizontal arrow is an isomorphism, then the middle arrow is also an isomorphism, since it is just the associated map of Kummer varieties. Of course an Abelian variety can have more than one principle polarization, but my interpretation of the following cryptic sentence is that the OP was focused on regularity of the top morphism, "Do we actually get $(A,D_1)\cong (A_2,D_2)$ under which conditions can we conclude that we have an isomorphism of principally polarized abelian surfaces using this diagram?" $\endgroup$ Oct 21, 2017 at 9:54
  • $\begingroup$ @JasonStarr: I think the OP is actually just trying to ask his earlier question [mathoverflow.net/questions/281677/… in a slightly different way. $\endgroup$
    – naf
    Oct 22, 2017 at 6:03

1 Answer 1


@ulrich points out that the OP might be asking a question different than the one I answered, and I completely agree. So let me explain what the comment above actually proves.

Let $(A_1,\mathcal{L}_1)$ and $(A_2,\mathcal{L}_2)$ be Abelian varieties of dimension $g$ together with a specified ample invertible sheaf. Let $U_1\subset A_1$ and $U_2\subset A_2$ be dense open subsets. Let $f_U:U_1\to U_2$ be an isomorphism of $k$-schemes. Let $\phi_U:\mathcal{L}_1|_{U_1}\to f_U^*\mathcal{L}_2$ be an isomorphism of invertible sheaves on $U_1$.

Proposition. There is a unique $k$-isomorphism $f:A_1\to A_2$ whose restriction to $U$ equals $f_U$, and this extends to a $k$-isomorphism of the associated Kummer varieties. If the complement of $U_1$ in $A_1$ has codimension $\geq 2$ everywhere, then there exists a unique isomorphism $\phi:\mathcal{L}_1\to f^*\mathcal{L}_2$ that restricts on $U$ to $\phi_U$. There are counterexamples if the complement has codimension $1.$

Proof. The first result follows from what is often called "Weil extension": every $k$-rational transformation from a smooth, projective $k$-variety to an Abelian variety extends uniquely to a regular $k$-morphism on the entire smooth $k$-scheme. If the complement of $U_1$ in $A_1$ has codimension $\geq 2,$ then by $S2$ extension, there exists a unique extension $\phi$ of $\phi_U$. By Krull's Hauptidealsatz, if $\text{Coker}(\phi)$ is nonzero, then the support has pure codimension $1.$ Since $\phi_U$ is an isomorphism on $U_1,$ then the cokernel is zero. Thus, $\phi$ is surjective. A surjective morphism of invertible sheaves is an isomorphism.

Every $g$-dimensional principally polarized Abelian variety $A$ has many ample invertible sheaves $\mathcal{L}$ with $c_1(\mathcal{L})^{g}$ equal to $g!,$ for example the pullbacks by translations of $A$. Given two effective Cartier divisors $D_1$ and $D_2$ whose invertible sheaves $\mathcal{L}_i=\mathcal{O}_A(D_i)$ are not isomorphic yet have $c_1(\mathcal{L})^g$ equal to $g!,$ on the open $U_1=U_2=U$ equal to $A\setminus (D_1\cup D_2),$ there is an isomorphism $\phi_U$ of $\mathcal{L}_1$ and $\mathcal{L}_2$ on $U.$ Yet $\phi_U$ does not extend to an isomorphism $\phi.$ QED


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