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Torelli's theorem states:

Let $R$, $R'$ be compact Riemann surfaces of genus $g$, $J(R)$, $J(R')$ their Jacobian varieties, $\Theta$, $\Theta'$ their respective theta divisors. The Riemann surfaces $R$ and $R'$ are isomorphic if and only if $(J(R), \Theta)$ and $(J(R'), \Theta')$ are isomorphic as principally polarized Abelian varieties.

In this theorem, $J(R)$ and $J(R')$ are required to be isomorphic not only as Abelian varieties but also as principally polarized Abelian varieties. It turns out that the condition for $J(R)$ and $J(R')$ to be isomorphic as Abelian varieties alone need not imply that $R$ and $R'$ are isomorphic.

Where can I find an example that shows that $J(R)$ and $J(R')$ being isomorphic just as Abelian varieties, does not imply that $R$ and $R'$ are isomorphic?

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  • $\begingroup$ The claim that isomorphic surfaces have Jacobians isomorphic as polarized abelian varieties is obvious. So it’s equivalent to ask for an example where they are isomorphic as abelian varieties but not as polarized abelian varieties. $\endgroup$
    – Will Sawin
    Commented Dec 24, 2019 at 2:13
  • $\begingroup$ See also Simplest complex curves with isomorphic Jacobian. $\endgroup$
    – abx
    Commented Dec 24, 2019 at 5:04
  • $\begingroup$ There is Prasad's conceptual approach to isomorphic jacobians via Galois theory of function fields of curves: arxiv.org/pdf/1409.3173.pdf (Theorem 1.1). $\endgroup$
    – Evgeny Shinder
    Commented Dec 24, 2019 at 18:03

1 Answer 1

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Consider the case of curves of genus $2$. If $\mathrm{A}$ is an abelian surface and $\mathrm{C}$ a smooth curve in $\mathrm{A}$ of genus $2$, then $\mathrm{A}\simeq\mathrm{J}(\mathrm{C})$ and $\mathrm{C}$ is the theta divisor of $\mathrm{J}(\mathrm{C})$. The special case $\mathrm{A}=\mathrm{E}\times\mathrm{E}$ (where $\mathrm{E}$ is an elliptic curve) was studied in this paper by Hayashida. It is known that for a given abelian variety $\mathrm{A}$ there are only finitely many curves with Jacobian $\mathrm{A}$ (see this paper by Narasimhan and Nori; for surfaces this was proven much earlier by Hayashida and Nishi). The first paper I mentioned gives in fact formulae (depending on the nature of $\mathrm{End(E)}$) for the number of curves $\mathrm{C}$ with Jacobian $\mathrm{A}=\mathrm{E}\times\mathrm{E}$.

More explicit examples (in the sense that the equations for the curves $\mathrm{C}$ can be written down) were constructed by Howe.

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  • $\begingroup$ Following the reasoning of the first part of your answer, I arrive at the following situation: $(E\times E, C)$ is isomorphic to the Jacobian variety $J(C)$ as principally polarized abelian varieties. On the other hand, for points $a, b \in E$ a divisor $D = a \times E+E \times b$ is effective and $(E \times E, D)$ is also a principally polarized abelian variety. But $(E \times E, C)$ and $(E \times E, D)$ are not isomorphic as principally polarized abelian varieties. $\endgroup$
    – Manoel
    Commented Dec 24, 2019 at 22:21
  • $\begingroup$ That is $(E \times E, D)$ and $(J(C), C)$ are not isomorphic as principally polarized abelian varieties. Why does this imply that I cannot recover the curve $C$? $\endgroup$
    – Manoel
    Commented Dec 24, 2019 at 22:21
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    $\begingroup$ @Manoel: note that $D$ is not a genus $2$ curve or even an irreducible divisor. The point is that one can find non-isomorphic genus $2$ curves $C, C'$ on $E\times E$. In Hayashida's paper one associates a matrix $M(D)$ to every divisor $D$ in $E\times E$ such that $M(D)=M(D')$ if and only if $D$ and $D'$ are algebraically equivalent. He then considers an equivalence relation on the set of all effective divisors on $E\times E$ with self intersection $2$; the relation is such that two genus 2 curves (irreducible effective divisors) are equivalent if and only if they are isomorphic. $\endgroup$
    – ssx
    Commented Dec 25, 2019 at 1:19

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