Timeline for A question about spectral sequences
Current License: CC BY-SA 4.0
17 events
| when toggle format | what | by | license | comment | |
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| Sep 22 at 15:50 | history | edited | Mehmet Onat | CC BY-SA 4.0 |
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| Sep 22 at 15:27 | history | edited | Mehmet Onat | CC BY-SA 4.0 |
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| Sep 22 at 15:16 | history | edited | Mehmet Onat | CC BY-SA 4.0 |
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| Sep 22 at 15:09 | history | edited | Mehmet Onat | CC BY-SA 4.0 |
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| Jun 21 at 17:03 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
| May 22 at 16:20 | history | edited | Mehmet Onat | CC BY-SA 4.0 |
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| May 22 at 14:32 | comment | added | John Palmieri | Please put the citation information in the question itself, so that people don’t have to search the comments for it. The goal is to give appropriate credit to others for their work. | |
| May 22 at 7:45 | comment | added | Mehmet Onat | @DaveBenson Can't $H^i(M_G,F_G;\mathbb{Q})$ and $H^i(M_G-F_G;\mathbb{Q})$ be identified here? | |
| May 22 at 7:31 | comment | added | Mehmet Onat | @JohnPalmieri The pontrjagin numbers of an orbit map and generalized G-signature theorem by Hsu-Tung Ku & Mei-Chin Ku. | |
| May 20 at 21:13 | comment | added | John Palmieri | When quoting someone else's work, please provide a proper citation, including names of authors and the title of the publication. | |
| May 20 at 20:34 | comment | added | Dave Benson | In the long exact sequence of a pair, shouldn't we identify the subspace to a point rather than remove it altogether? Otherwise, there isn't a map, or if there is, it goes the wrong way. | |
| May 20 at 19:28 | answer | added | Nicholas Kuhn | timeline score: 3 | |
| May 17 at 10:32 | comment | added | Mehmet Onat | @AchimKrause it is the long exact cohomology sequence of pair $(M_G,F_G)$ | |
| May 17 at 8:46 | comment | added | Mehmet Onat | @AchimKrause This theorem is mentioned in this article. link.springer.com/chapter/10.1007/BFb0085610 | |
| May 17 at 8:28 | comment | added | Mehmet Onat | @AchimKrause Yes you are right I didn't notice that. | |
| May 17 at 8:22 | comment | added | Achim Krause | I'm confused way earlier, doesn't the inclusion $j$ go $j: M-F\to M$, hence on cohomology, $j^*: H^s(M)\to H^s(M-F)$? | |
| May 17 at 8:02 | history | asked | Mehmet Onat | CC BY-SA 4.0 |