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

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

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.