Consider a cubic threefold $Y$ and its associated degree $14$ prime Fano threefold $X$, we have the equivalences of non-trivial components of $D^b(Y)$ and $D^b(X)$, i.e, $\mathcal{A}_X\cong\mathcal{B}_Y$ and this equivalence is given by a Fourier-Mukai functor $\Phi:=\Phi_{I_Z(H_Y)}:D^b(X)\rightarrow D^b(Y)$, where $Z\subset X\times Y$ is an irreducible four dimensional subvariety. Consider the projection map $\pi:Z\rightarrow Y$, my question is how to describe the fiber of this map $\pi^{-1}(y)$ for every point $y\in Y$, which is $Z\cap X_y$.

Since $X_y\cong X$ is a prime Fano threefold, its Picard number is 1, so it can not contain any divisor which is not multiple of the degree 14 $K3$ surface, by playing some game with matrices, it seems to me that $\pi^{-1}(y)$ is degree 2, so it looks like it is of dimension 1 and degree $2$, which looks like a conic on $X$, but I have difficulty to show this fiber is connected to exclude the possibility that two disjoint lines.


1 Answer 1


This is a rational quartic curve.

Indeed, the FM kernel is induced by the HPD kernel, which is, essentially, the locus $\mathbf{Z}$ of pairs $(U,y)$, where $U$ is a 2-dimensional subspace in the fixed 6-dimensional space $V_6$ and $y$ is a degenerate skew form on $V_6$ such that $$ U \cap \operatorname{Ker}(y) \ne 0. $$ The fiber $\mathbf{Z}_y$ of this locus over a general degenerate $y$ is $$ \mathbf{Z}_y \cong \operatorname{Cone}(\mathbb{P}^1 \times \mathbb{P}^3). $$ Finally, $Z_y$ is a liner section of $\mathbf{Z}_y$ of codimension 4, and it is easy to see that it is a rational quartic curve.


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