Timeline for Torelli theorem for smooth complex cubic surfaces?
Current License: CC BY-SA 4.0
6 events
when toggle format | what | by | license | comment | |
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Apr 28, 2023 at 13:37 | comment | added | Sasha | Because all cubic surfaces have the same Hodge structure. | |
Apr 28, 2023 at 12:59 | comment | added | user503580 | Could you explain more why the Hodge structures of cubic surfaces do not allow to distinguish them? | |
Apr 28, 2023 at 12:55 | vote | accept | CommunityBot | ||
Apr 27, 2023 at 5:35 | comment | added | Sasha | I don't know if this is true (although I remember I have seen this approach somewhere), but if you want you can consider pairs $(Y,\tau)$, where $\tau$ is an automorphism of order $3$ (whose fixed points is $X$), then the answer to your question is positive, and on the other hand, $\tau$ acts on the intermediate Jacobian of $Y$ (and can be reconstructed from this action), so eventually you can look at the Hodge structure of $Y$ endowed with an automorphism of order 3. | |
Apr 26, 2023 at 19:05 | comment | added | Will Sawin | The claim that two cubic threefold constructed this way are isomorphic only if the original cubic surfaces are isomorphic is plausible, but is it obvious? | |
Apr 26, 2023 at 18:50 | history | answered | Sasha | CC BY-SA 4.0 |