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.



A Fourier-Mukai equivalence between non trivial component of cubic threefold and degree 14 prime Fano threefold
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.


