Timeline for $D$-equivalence between schemes and dimension
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
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| Mar 16, 2023 at 23:50 | comment | added | Dave Benson | Thanks! Interesting. | |
| Mar 16, 2023 at 23:49 | comment | added | AT0 | @DaveBenson If you relax the condition on the divisor just a little and require big instead of ample then it already doesnt work. You only get a birational equivalence ( this is due to Kawamata ) and so the tensors are different by Balmer's construction. | |
| Mar 16, 2023 at 23:39 | comment | added | Dave Benson | I've edited my answer to reflect this conversation. | |
| Mar 16, 2023 at 23:38 | history | edited | Dave Benson | CC BY-SA 4.0 |
added 32 characters in body
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| Mar 16, 2023 at 23:37 | comment | added | Dave Benson | Yes it does. So I guess I should have said that you can recover $X$ from $D^{\mathrm{perf}}$ as a tensor triangulated category. For smooth algebraic varieties with ample canonical or anticanonical sheaf, Bondal and Orlov prove this without using the tensor structure. So do you know smooth examples where the tensor structures are different? | |
| Mar 16, 2023 at 23:33 | comment | added | R. van Dobben de Bruyn | Doesn't the Balmer spectrum require not just the triangulated structure, but also the monoidal structure on it? | |
| Mar 16, 2023 at 23:28 | comment | added | Dave Benson | Are they topologically Noetherian? How does this fit with Theorem 6.3 of the second paper of Balmer quoted above? | |
| Mar 16, 2023 at 23:25 | comment | added | Jason Starr | There are non-isomorphic smooth schemes whose derived categories are equivalent as triangulated categories. | |
| Mar 16, 2023 at 22:44 | history | answered | Dave Benson | CC BY-SA 4.0 |