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Let $X_1,X_2\subset\mathbb{P}^{n+1}$ be two smooth complex cubic hypesurfaces, then I know the following Torelli theorems:

[$n=1$] In this case, $X_1\cong X_2$ if and only if there exists a Hodge isometry $H^1(X_1,\mathbb{Z})\cong H^1(X_2,\mathbb{Z})$.

[$n=3$] In this case, $X_1\cong X_2$ if and only if there exists a Hodge isometry $H^3(X_1,\mathbb{Z})\cong H^3(X_2,\mathbb{Z})$.

[$n\geq 3$] In this case, $X_1\cong X_2$ if and only if there exists an isomorphism of their Fano varieties of lines $F(X_1)\cong F(X_2)$. This isomorphism is asked to be polarized when $n=4$.

Do we know a similar result for cubic surfaces? Explicitly, I would like to know if there are some invariant that can determine a smooth complex cubic surface.

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  • $\begingroup$ Your $n\geq 3$ result is of a different character to the first two. The first two reduce an isomorphism of algebraic varieties to an isometry of Hodge structures, while the last one reduces an isomorphism of algebraic varieties to another isomorphism of algebraic varieties. Are you interested ideally in a result of the first type, constructing a Hodge structure from a cubic surface? Or is there something specific you like about the last isomorphism where the Fano varieties are easier to deal with in some sense? Or are you just interested in some interesting criterion, of no particular form? $\endgroup$
    – Will Sawin
    Commented Apr 26, 2023 at 19:03
  • $\begingroup$ @WillSawin Thank you, this is a good point. I have edited my question. $\endgroup$
    – user503580
    Commented Apr 26, 2023 at 19:19

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The Hodge structures of cubic surfaces do not allow to distinguish them. But there are some ways around. For instance, given a cubic surface $X \subset \mathbb{P}^3$, you can associate with it a cyclic covering $Y(X) \to \mathbb{P}^3$ of degree 3 ramified over $X$ (thus, $Y(X)$ is a cubic threefold) and use the Hodge structure of $Y(X)$.

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    $\begingroup$ 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? $\endgroup$
    – Will Sawin
    Commented Apr 26, 2023 at 19:05
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    $\begingroup$ 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. $\endgroup$
    – Sasha
    Commented Apr 27, 2023 at 5:35
  • $\begingroup$ Could you explain more why the Hodge structures of cubic surfaces do not allow to distinguish them? $\endgroup$
    – user503580
    Commented Apr 28, 2023 at 12:59
  • $\begingroup$ Because all cubic surfaces have the same Hodge structure. $\endgroup$
    – Sasha
    Commented Apr 28, 2023 at 13:37

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