Timeline for Derived category of coherent sheaves with a codimension $\geq$ 1 support
Current License: CC BY-SA 4.0
14 events
| when toggle format | what | by | license | comment | |
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| Sep 6, 2019 at 2:40 | comment | added | Pavel Čoupek | Maybe I am pointing out the obvious, but note that the coh. sheaves supported in codim $\geq1$ form a Serre subcategory (even better, torsion class) in the category of coh. sheaves. The question in this setting is somewhat considered in stacks project: stacks.math.columbia.edu/tag/06UP but the criterion for equivalence there does not seem to fit this setting. | |
| Sep 4, 2019 at 18:33 | comment | added | GTA | @Arkadij Bojko Sorry, I realized I misunderstood your point before you replied. | |
| Sep 4, 2019 at 18:20 | comment | added | Arkadij | @WilleLiou Huybrecht in his proof of Proposition 3.5 in Fourier–Mukai transforms in algebraic geometry is satisfied with showing the existence of a quasi-isomorphic complex of the abelian subcategory, so I think it should still work. I think one can use the composition rule for the roofs to replace the objects at the top of the roof. | |
| Sep 4, 2019 at 18:13 | comment | added | Wille Liu | @Arkadij Bojko But one category contains more objects which are quasi-isomorphic to the given complex than the other. Remember that the definition of hom in the derived category takes all roofs. | |
| Sep 4, 2019 at 18:13 | comment | added | Arkadij | @GTA yes, that is how I can see why it is true for 1., but why should a complex with vanishing cohomologies at the generic point be quasi-isomorphic to one with each term being trivial at the generic point? | |
| Sep 4, 2019 at 18:08 | comment | added | Arkadij | @WilleLiou I think once one shows that there is a quasi-isomorphic complex of that form, then by looking at the morphisms in terms of roofs, it should follow that the Homs are the same. | |
| Sep 4, 2019 at 18:04 | comment | added | Arkadij | @GTA I see that the condition for the complex restricted to a generic point being acyclic is equivalent to it lying in 1., but it seems that it only restates the question about why this should be equivalent to it being in 2. | |
| Sep 4, 2019 at 17:57 | comment | added | Wille Liu | However, even if we are able to construct such a complex, it doesn't mean that the hom must be the same in both categories. | |
| Sep 4, 2019 at 17:53 | comment | added | Wille Liu | I think one can construct the complex by taking a large enough subscheme of codimension $\ge 1$, for example the subscheme given by $\ker(\mathcal{O}_X\to \mathcal{End}(\mathcal{H}^*F))$. Then $\mathcal{H}^*F$ will be acyclic with respect to the pullback onto this subscheme | |
| Sep 4, 2019 at 17:49 | comment | added | GTA | What I'm saying is that being supported in codimension >=1 for coherent sheaf is the same as having zero stalk for each generic point, and taking stalk at a generic point is exact so it is actually a functor of derived categories. Say there is one generic point $i:\eta\rightarrow X$ then $Li^{*}=i^{*}:D^{b}Coh(X)\rightarrow D^{b}Coh(\eta)$ and both conditions are equivalent to complexes mapped to zero complex via $i^{*}$ because a complex with zero cohomology is quasi-isomorphic to zero complex. | |
| Sep 4, 2019 at 17:03 | comment | added | Arkadij | I am not sure I understand what you mean. Are you saying that I can construct a quasi-isomorphic complex by taking stalks at the generic points? | |
| Sep 4, 2019 at 16:41 | comment | added | GTA | Isn't this just checkable at the generic points? Taking stalks is exact so I don't think there is a problem. | |
| Sep 4, 2019 at 16:34 | history | edited | Arkadij | CC BY-SA 4.0 |
had an additional remark/idea
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| Sep 4, 2019 at 11:19 | history | asked | Arkadij | CC BY-SA 4.0 |