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Assume we work over $\mathbb{C}$.

Let $S\subset \mathbb{P}^3$ be a quartic surfaces with 16 nodes (ordinary double points). Then there is a simple principally polarized abelian surface $(A,\theta)$ such that $S\cong A/\{\pm 1\}$ given by the linear system $|2\theta|: A \rightarrow \mathbb{P}^3$ and the singularities are exactly the images of the 2-torsion points. The Torelli theorem states that there is a unique smooth projective curve $C$ of genus $g=2$ such that $(A,\theta)=(Jac(C),C)$.

Now, let $S_C$ and $S_{C'}$ be two quartic surfaces with 16 nodes in $\mathbb{P}^3$ with corresponding curves $C$ and $C'$ of genus 2. Assume $S_C$ and $S_{C'}$ are birational. This implies that their minimal resolutions (blowing up the 16 nodes) $X_C$ and $X_{C'}$ are also birational. But $X_C$ and $X_{C'}$ are $K3$-surfaces, which means they are already isomorphic.

$\textbf{Question:}$ How much information about a curve $C$ is encoded in the $K3$-surface $X_C$? For example in the setting described above:

Does $X_C\cong X_{C'}$ imply $C\cong C'$?

If $C\cong C'$ is not true not in general, can we relate $C$ and $C'$ in any other way in this setting?

My thoughts on this: assume the Picard numbers of $S_C$ and $S_{C'}$ are one, that is the associated abelian surfaces also have Picard number one. Using $X_C\cong X_{C'}$ we get $A\cong A'$ as abelian surfaces by a result of Inose, because they both have Picard number one and a principal polarization. But the pullback of this isomorphism must map $\theta'$ to $\theta$ so that it is in fact an isomorphism of principally polarized abelian surfaces $(A,\theta) \cong (A',\theta')$ which implies $C\cong C'$.

$\textbf{Question 2:}$ What happens for $\rho(A)>1$? The case $\rho(A)=1$ is the very general case in the moduli space of principally polarized abelian surfaces. Can we get the other curves by some limit or deformation argument maybe?

$\textbf{Question 3:}$ If we assume $X_C\cong X_{C'}$ and $J(C)\cong J(C')$ (as abelian surfaces) is this enough to see that $C\cong C'$?

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    $\begingroup$ For K3 surfaces $S$ and $S'$, for dense open subsets $U\subset S$ and $U'\subset S'$, for an isomorphism $\phi_U:U\xrightarrow{\cong} U'$, there exists a unique isomorphism $\phi:S\xrightarrow{\cong} S'$ whose restriction to $U$ equals $\phi_U$. So instead of working in the birational category, I suggest that you work in the biregular category. $\endgroup$ Jun 8, 2017 at 11:26
  • $\begingroup$ @Jason Starr : Okay. I phrased the question in terms of isomorphims and the K3 surfaces. My observation for Picard number one still remains. Can something be done for higher Picard numbers? $\endgroup$
    – Bernie
    Jun 8, 2017 at 14:58
  • $\begingroup$ You can reconstruct the genus 2 curve by taking the double cover of the Theta divisor (which is a conic in the singular Kummer - namely the intersection of the Kummer with one of the tropes) ramified along the 6 nodes contained therein. $\endgroup$
    – IMeasy
    Jun 9, 2017 at 8:54
  • $\begingroup$ @Imeasy : Thank you. I knew that. But I don't know the behaviour of the 16 nodes, the tropes etc. with respect to the birational map of the singular Kummers $S_C$ and $S_{C'}$. So I can recover the curves, but what is their relation ship now? Must they be isomorphic? $\endgroup$
    – Bernie
    Jun 9, 2017 at 9:43
  • $\begingroup$ One possibility to find an answer is by using global Torelli theorem for K3 and analyzing possible embeddings of the lattice $16A_1$ into the second cohomology of a K3 ($3U + 2E_8$). Probably much information in this direction can be extracted from works of Nikulin. $\endgroup$
    – Sasha
    Jun 14, 2017 at 6:33

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