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Let $Y_1$ and $Y_1'$ be index two degree one Fano threefolds. Suppose we have a Fourier-Mukai equivalence $\Phi_P : \mathrm{D}^b(Y_1) \to \mathrm{D}^b(Y_1')$. Can anything be said about the kernel $P$, i.e. is there some kind of a classification in this Fano case?

Thank you.

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    $\begingroup$ If $Y_1$ and $Y_1'$ are Fano, doesn't Bondal and Orlov Theorem implies that they are isomorphic and that, up to a shift and a twist by a line bundle, $P$ is the graph of the isomorphism? $\endgroup$
    – Libli
    Commented Oct 4, 2022 at 16:12
  • $\begingroup$ It's indeed true that $\mathrm{Auteq}(\mathrm{D}^b(Y_1)) = \mathrm{Aut}(Y_1) \ltimes (\mathrm{Pic}(Y_1) \oplus \mathbb{Z})$, but how do you mean that $P$ should be the graph of the isomorphism $\phi : Y_1 \cong Y_1'$? I see how by Bondal-Orlov we essentially have that $\Phi_P \in \mathrm{Auteq}(\mathrm{D}^b(Y_1))$, but I'm not sure how to translate this to saying something about the object $P$? $\endgroup$
    – mathphys
    Commented Oct 5, 2022 at 17:59
  • $\begingroup$ I guess $\Phi_P(-) = p'_*(p^*(-) \otimes P) = p_*(p^*(-) \otimes P)$ should be a combination of automorphisms, twists and shifts. Maybe we can consider the $p$'s as automorphisms of $Y_1$, so $P$ has to be some kind of line bundle (up to shift)? $\endgroup$
    – mathphys
    Commented Oct 5, 2022 at 18:09
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    $\begingroup$ Bondal and Orlov proved that two varieties with ample or antiample canonical class (in particular, two Fano varieties) are derived equivalent if and only if they are isomorphic, so you may assume $Y'_1 = Y_1$ and consider an autoequivalence of $Y_1$. They also proved that under these assumptions every autoequivalence is standard, i.e., a composition of an automorphism, a line bundle twist, and a shift. This makes a description of the FM kernel straightforward. $\endgroup$
    – Sasha
    Commented Oct 5, 2022 at 18:42

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