Edit: here is a potential proof

Let $X,Y$ be two smooth cubic threefolds over an algebraically closed field. Suppose that they are birationally equivalent. Then are they isomorphic? This seemingly important question has no answer in the literature, to my knowledge, although it seems to be approachable with current tools.

The only known non-vanishing birational invariant is the intermediate Jacobian. $X$ and $Y$ being birationally equivalent means that their intermediate Jacobians are equivalent modulo Jacobians of curves. Smooth cubic threefolds are non-rational since their intermediate Jacobians are not Jacobians of curves.

There is a unique factorization of p.p.a.v.'s into irreducible ones. If we consider the factorizations of the intermediate Jacobians of $X,Y$ this means that the non-Jacobian components must be the same. However, the intermediate Jacobian of a smooth cubic threefold is irreducible. Then it would follow from the Torelli theorem that a smooth cubic threefold is unique is its birational equivalence class.

Does this proof follow?


1 Answer 1


Since this is more or less an answer to the question, I hope people do not mind it not being a comment.

The result you claim is indeed in the literature (I’m surprised it is not in one of Huybrecht’s books). Kuznetsov’s paper Derived Categories of Cubics and $V_{14}$ threefolds states your claim on page 8 and the “proof” they claim is more or less what you have said (is isn’t explicitly written down) — it is due to the Torelli Theorem.


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