Let $Y\xrightarrow{2:1}\mathbb{P}^3$ be the double cover, branched over a quartic K3 surface $S$, known as quartic double solid. Assume $S$ is generic, we know that there is a Torelli theorem for $Y$ in terms of intermediate Jacobian. I was wondering if there is an argument that goes with Torelli theorem for $Y$ breaks down to Torelli theorem for the K3 surface $S$.How is the Hodge structure of $Y$ related to Hodge structure of $S$?
Now, we look at a more complicated example, let $X\xrightarrow{2:1} Y$ be a double cover of a cubic threefold $Y$, branched over the smooth anti-canonical divisor, which is a degree $6$ K3 surface $S$, I also assume $S$ is generic, then we get $X$ being an index one degree 6 prime Fano threefold, which is an intersection of a cubic and a quadric in $\mathbb{P}^5$. Note that all the index one degree 6 prime Fano threefold constructed in this way is a sub-locus of the whole moduli space.
Now, I would like to prove a Torelli theorem for $X$ in terms of intermediate Jacobian, the argument I would like to see is something like Hodge structure of $X$ is related to Hodge structure of $Y$ and that of $S$, and the Torelli theorem for $X$ breaks down to that of $Y$ and $S$. It looks to me that this is quite "classical", but I did not find anything in the literature.
For the first example, we know that by [Kuznetsov-Perry 19'], the equivariant category $\mathcal{K}u_{\mathbb{Z}_2}(Y)$ under the action of the involution of double cover is exactly derived category $D^b(S)$, then one can show categorical Torelli theorem for quartic double solid like this: $$\mathcal{K}u(Y)\simeq\mathcal{K}u(Y')\Longrightarrow\mathcal{K}u_{\mathbb{Z}_2}(Y)\simeq\mathcal{K}u_{\mathbb{Z}_2}(Y')\Longrightarrow D^b(S)\simeq D^b(S')\Longrightarrow S\cong S',$$ where the first implication is because the Serre functor is just $\iota\circ[2]$, so the equivalence of Kuznetsov components would induce the equivalence between equivariant category. Since I assume that $S,S'$ are very general, then by some results regarding the number of Fourier-Mukai partner of Picard number one K3 surfaces, we get $S\cong S'$. So from this proof, one see that categorical Torelli theorem for $Y$ is determined by "Torelli theorem" for S since the ambient space $\mathbb{P}^3$ is rigid and this is why I expect the classical Torelli theorem also has similar fashion.
For the second example, I am now in progress to show that $$\mathcal{K}u_{\mathbb{Z}_2}(X)=\langle\mathcal{K}u(Y), D^b(S)\rangle.$$ Then I believe that using the recent results in [Kuznetsov-Perry 23'], one can also have similar logic chain to prove the categorical Torelli theorem for $X$: $$\mathcal{K}u(X)\simeq\mathcal{K}u(X')\Longrightarrow\mathcal{K}u_{\mathbb{Z}_2}(X)\simeq\mathcal{K}u_{\mathbb{Z}_2}(X')\Longrightarrow\langle\mathcal{K}u(Y),D^b(S)\rangle\simeq\langle\mathcal{K}u(Y'),D^b(S')\rangle.$$ Then I can apply Alex Perry's intermediate Jacobian "functor" to get $\mathrm{J}(\mathcal{K}u(Y))\cong\mathrm{J}(\mathcal{K}u(Y'))\Longrightarrow J(Y)\cong J(Y')\Longrightarrow Y\cong Y'$(There is no intermediate Jacobian for K3 surface, then $\mathrm{J}(D^b(S))=0$). Finally, I think can probably use the topological K-theory of category developed by Blanc and the techniques in Dell-Jacovskis-Rota to show a Hodge isometry $H^2(S,\mathbb{Z})\cong H^2(S',\mathbb{Z})$, then by Torelli theorem for K3 surface, we show $S\cong S'$, then we show $X\cong X'$. This is why I believe that the classical Torelli statement for such $X$ would have to be determined by the Torelli statement for $Y$ and for $S$.
