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Two K3 surfaces show up when talking about cubic fourfolds containing a plane. Let $P\subset X\subset \mathbb{P}^5$ be the plane inside the cubic. Since $P$ is cut out by 3 linear equations then $X$ has equation $L_1Q_1+L_2Q_2+L_3Q_3=0$, with $L_i$ and $Q_i$ respectively of degree 1 and 2. Hence $X$ contains the octic K3 complete intersection of the 3 quadrics.

On the other hand by projecting off $P$ one gets a quadric fibration $\pi: Bl_P(X) \to \mathbb{P}^2$, and the relative Hilbert scheme of lines in the fibers is a Brauer Severi variety over a degree 2 K3.

Is there any known relation between these two K3s? Are they FM partners or have similar cohomology?

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    $\begingroup$ Is the octic K3 uniquely determined up to isomorphism? I mean we can choose the L_i differently, which changes the Q_i, which conceivably -- to me -- changes the K3. $\endgroup$
    – R.P.
    Commented Sep 21, 2020 at 14:04
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    $\begingroup$ @RP: good point, I nedd to think about it but probably you are right. But still the question makes (some) sense - I guess. $\endgroup$
    – IMeasy
    Commented Sep 21, 2020 at 15:33
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    $\begingroup$ Indeed what is well-defined is a net of 2-dimensional quadrics in $X$ (Weil divisors). This does not define a K3 in $\Bbb{P}^5$. $\endgroup$
    – abx
    Commented Sep 21, 2020 at 16:25
  • $\begingroup$ FM partners of a given K3 form a discrete set in the moduli space, so a general member of a non-trivial family of K3 (like intersections of three quadrics from your question) cannot be a partner of the K3 associated with the quadric fibration. $\endgroup$
    – Sasha
    Commented Sep 21, 2020 at 17:19
  • $\begingroup$ Still that degree 8 (and sectional genus 5) K3 must have something to do with the cubic fourfold. In the $\mathcal{C}_d$ cases where there is an associated K3, the surface has degree $d$ and genus $2d-2$. This is why I suspect that when such a cubic 4fold has an associated K3 (for example if it is in $\mathcal{C}_8\cap \mathcal{C}_{14}$ or 26), then this K3 may be related to the octic. $\endgroup$
    – IMeasy
    Commented Sep 22, 2020 at 19:44

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