Timeline for Fourier-Mukai kernels for Fano threefolds
Current License: CC BY-SA 4.0
7 events
| when toggle format | what | by | license | comment | |
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| Oct 5, 2022 at 18:42 | comment | added | Sasha | 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. | |
| Oct 5, 2022 at 18:09 | comment | added | mathphys | 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)? | |
| Oct 5, 2022 at 17:59 | comment | added | mathphys | 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 $P$? | |
| Oct 4, 2022 at 16:12 | comment | added | Libli | 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? | |
| Oct 4, 2022 at 12:59 | history | edited | mathphys | CC BY-SA 4.0 |
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| Oct 4, 2022 at 12:50 | history | edited | mathphys | CC BY-SA 4.0 |
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| Oct 4, 2022 at 12:41 | history | asked | mathphys | CC BY-SA 4.0 |