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.

topmorphism, "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?" $\endgroup$ – Jason Starr Oct 21 '17 at 9:54