Behaviour of (principal) polarizations of (singular) surfaces under birational maps

Question

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.

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