Timeline for Local cohomology groups and linearity
Current License: CC BY-SA 3.0
18 events
| when toggle format | what | by | license | comment | |
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| Sep 13, 2015 at 12:33 | comment | added | Jason Starr | So, "Vlad Dezamagit" now? Does this refer to Vlad during his imprisonment by Corvinus? | |
| Sep 13, 2015 at 12:18 | vote | accept | user45397 | ||
| Sep 13, 2015 at 12:18 | comment | added | user45397 | @Vlad: Thank you very much for the very detailed and helpful answer. I could not find any reference where they mention this linearlity property. In all literatures, they work with the category of abelian groups as soon as they apply the cohomology functor/right derived functor and not in the category of modules. I thought the boundary map must be linear but was puzzled by the tensor product conclusion, since I was not expecting it. Thanks once again. | |
| Sep 13, 2015 at 12:13 | comment | added | darx | @user45397 My confusion about your question was: how can you know about cohomology with supports but not know that the boundary map is linear? But I did not express myself very well. Apologies. | |
| Sep 13, 2015 at 12:09 | comment | added | user45397 | @Vlad: Thank you very much. So, the answer to both the questions is yes, am I right? I thought you were giving a counterexample earlier, so I was confused. | |
| Sep 13, 2015 at 12:03 | history | edited | darx | CC BY-SA 3.0 |
Was asked by the moderator to say more.
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| Sep 13, 2015 at 11:57 | comment | added | user45397 | @WillSawin I guess the counterexample above is to disprove that "$\delta$ is $\mathcal{O}_U(U)$-linear". Could you please elaborate on how I can show this. | |
| Sep 13, 2015 at 6:11 | vote | accept | user45397 | ||
| Sep 13, 2015 at 6:11 | |||||
| Sep 12, 2015 at 21:07 | comment | added | Todd Trimble | @user45397 Stefan and Will might not have seen your comments because you need to write 'at StefanKohl' or 'at WillSawin' for messages to appear in their inboxes. Anyway, I think Stefan was addressing the answerer Vlad who ends his answer with a 'question'. It's a Socratic question, obviously. I would encourage Vlad to expand this into more of a detailed explanation, lest certain reviewers in the review queue have his answer deleted via Community vote. | |
| Sep 12, 2015 at 20:12 | comment | added | user45397 | @Kohl: I am confused. The previous question is not a new question but rather a follow up of the answer above. There is nothing new in it. Am I missing something? | |
| Sep 12, 2015 at 19:06 | comment | added | user45397 | @Sawin: I am sorry, I still do not understand how to show that $\delta$ is not $\mathcal{O}_U(U)$-linear. Could you please elaborate a bit. | |
| Sep 12, 2015 at 18:59 | review | Low quality posts | |||
| Sep 12, 2015 at 20:22 | |||||
| Sep 12, 2015 at 18:57 | comment | added | Will Sawin | @user45397 In this case it's easy to figure out $H^0(U, \mathcal O_U)$ and $H^0(U-Z, \mathcal O_U)$, without knowing anything about local cohomology. So you just need to find the map between them. An injective resolution is not needed. | |
| Sep 12, 2015 at 18:55 | history | rollback | Will Sawin |
Rollback to Revision 1
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| Sep 12, 2015 at 18:54 | history | edited | Will Sawin | CC BY-SA 3.0 |
added 2 characters in body
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| Sep 12, 2015 at 18:34 | review | Low quality posts | |||
| Sep 12, 2015 at 18:54 | |||||
| Sep 12, 2015 at 18:31 | comment | added | user45397 | Thank you. I am not very familiar with computations in local cohomology. Could you please suggest a hint/reference how to compute $H^1_Z(\mathcal{O}_U)$ in this case. In particular, I do not know of any particular injective resolution of $\mathcal{O}_U$, which could help in the computation. | |
| Sep 12, 2015 at 18:17 | history | answered | darx | CC BY-SA 3.0 |