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Hodge structure on K3 surface (the middle line of Hodge diamond is 1 20 1) is similar to the Hodge structure of cubic fourfold (the middle line of Hodge diamond of primitive cohomology is 0 1 20 1 0). But the two Hodge structures have different signature. To make the signatures the same, we may remove a class of polarization from $H^2(S)$ and an extra class from $H^4_{prim}(X)$.

We will consider a cubic fourfold $X$ with a hyperplane class $h\in H^{1,1}(X)$ and a nontrivial class $T$ in $H^{2}_{prim}(X,\mathbb Z)$, in this case we put $d=\mathrm{disc}(h^2, T)$ is the discriminant of lattice generated by $T$ and $h^2$ (it is a sublattice of $H^{4}(X, \mathbb Z)$).

Question. Are there examples (besides two given below) of geometric constructions (for example, a construction of a cycle in $X\times S$) of isometries between $H^4_{prim}(X)^{\bot T}$ and $H^2_{prim}(S)$ for some cubic fourfold $X$ and a K3 surface $S$?

Example $d=6$. Fourfold $X$ contains a double point $p$ and projection from point $p$ defines a birational map from $X$ to $\mathbb P^3$ and a regular map $$ Bl_p(X)\to\mathbb P^4, $$ which contracts a K3 surface $S$.

Example $d=14$ (Beauville-Donagi). Let $P=\mathbb P(\Lambda^2 \mathbb C^6)$ be 14-dimesional projective space of skew two-forms. Plucker embedding gives $$ Gr(2,6)\subset P. $$ In the dual space $P^*$ of dual forms we have a variety of degenerated (i.e., of rang 4 or less) skew 2-forms: $$ Pf(4,6)\subset P^*. $$ Obtaining $X$ as an intersection of $Pf(4,6)$ with a general $\mathbb P^5\subset P^*$, K3 surface will be given by the intersection of $Gr(2,6)$ with $(\mathbb P^5)^\bot\subset P$. It is not difficult to construct a correspondence (and Fourier-Mukai kernel) for $X$ and $S$.

There is a non-example for $d=8$, it has deal with $X$ and a K3 surface with a distinguished Brauer class. Also the following theorem holds.

Theorem (Addington-Thomas). If $X$ is as above with $d$ satisfying some numerical condition ($d=6,14,26,38\ldots$), then there exist a polarized $K3$ surface $S$ of degree $d$ and an algebraic cycle in $A^3(S\times X)_{\mathbb Q}$ which induces a Hodge isometry $H^2_{prim}(S,\mathbb Z)\to(h^2,T)^{\bot}$.

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    $\begingroup$ Change "cubic threefold" to "cubic fourfold" everywhere, and especially in the title of your question! And $P^3$ to $P^4$. $\endgroup$
    – Sasha
    Feb 15, 2015 at 20:47
  • $\begingroup$ I'm not an expert on cubic fourfolds, but it seems to me like you could maybe unwind the proofs in Hassett's paper "Some Rational Cubic Fourfolds" using the explicit descriptions of surfaces contained in cubic fourfolds (e.g. in the examples section of his "Special Cubic Fourfolds") to get an answer to this? $\endgroup$
    – dhy
    Feb 15, 2015 at 21:53
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    $\begingroup$ Unfortunately, no one knows any other explicit examples at the moment. (I tried to post this as an answer below but a moderator deleted it.) $\endgroup$ Feb 20, 2015 at 3:56
  • $\begingroup$ Maybe to elaborate a bit: for the theorem you cite, Richard and I start with Hassett's rational cubics containing a plane (as dhy suggests), but then there are several more steps and the cycle we get in the end can hardly be called explicit. We tried hard to make it more explicit, even just to say that it's ch_3 of a sheaf (rather than a complex of sheaves), but we didn't succeed. $\endgroup$ Feb 20, 2015 at 16:00

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