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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