Timeline for Isomorphic Jacobian Varieties Just Like Abelian Varieties — Torelli's Theorem
Current License: CC BY-SA 4.0
6 events
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| Dec 25, 2019 at 1:19 | comment | added | ssx | @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. | |
| Dec 24, 2019 at 22:21 | comment | added | Manoel | 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 $C$? | |
| Dec 24, 2019 at 22:21 | comment | added | Manoel | 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. | |
| Dec 24, 2019 at 20:59 | vote | accept | Manoel | ||
| Dec 24, 2019 at 12:06 | history | edited | ssx | CC BY-SA 4.0 |
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| Dec 24, 2019 at 2:38 | history | answered | ssx | CC BY-SA 4.0 |