Timeline for Torsion in Deligne cohomology
Current License: CC BY-SA 3.0
19 events
| when toggle format | what | by | license | comment | |
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| May 3, 2018 at 22:28 | history | notice removed | user95222 | ||
| May 2, 2018 at 22:25 | history | notice added | user95222 | Reward existing answer | |
| S Apr 23, 2018 at 16:06 | history | bounty ended | CommunityBot | ||
| S Apr 23, 2018 at 16:06 | history | notice removed | user119470 | ||
| S Apr 22, 2018 at 16:06 | history | bounty started | CommunityBot | ||
| S Apr 22, 2018 at 16:06 | history | notice added | user119470 | Reward existing answer | |
| Feb 8, 2018 at 5:38 | vote | accept | CommunityBot | ||
| Feb 8, 2018 at 4:14 | answer | added | user87684 | timeline score: 4 | |
| Feb 8, 2018 at 3:10 | comment | added | user92332 | As you note, $H^i_{\mathcal{D}}(X,\mathbf{Z}(p))$ contains a copy of a quotient of a $\mathbf{Q}$-vector space, hence lots of divisible elements. If you ask yourself the question whether $J^{i,p}(X/\mathbf{C})$ contains torsion divisible elements, then the answer is yes as soon as $H^i(X,\mathbf{Z}(p))$ is not contained in $F^pH^i_{\rm dR}(X/\mathbf{C})$. If so, then $J^{i,p}(X/\mathbf{C})$ is a $\mathbf{Q}$-vector space, hence torsion free. It typically never happens that $H^i(X,\mathbf{Z}(p))\subset F^pH^i_{\rm dR}(X/\mathbf{C})$, so usually $J^{i,p}(X/\mathbf{C})_{\rm tor}$ has div elements | |
| Feb 8, 2018 at 2:52 | history | edited | user113393 | CC BY-SA 3.0 |
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| Feb 8, 2018 at 2:51 | history | undeleted | user113393 | ||
| Feb 8, 2018 at 1:48 | history | deleted | user113393 | via Vote | |
| Feb 8, 2018 at 1:25 | history | edited | user113393 | CC BY-SA 3.0 |
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| Feb 7, 2018 at 20:40 | history | edited | user113393 | CC BY-SA 3.0 |
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| Feb 7, 2018 at 19:40 | comment | added | K.K. | After the edit, $j = p$ in my comment above. | |
| Feb 7, 2018 at 18:33 | history | edited | user113393 |
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| Feb 7, 2018 at 18:21 | history | edited | user113393 | CC BY-SA 3.0 |
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| Feb 7, 2018 at 18:18 | comment | added | K.K. | The Deligne cohomology group with $i = 2$ and $j=1$ is isomorphic to the Picard group of $X$, so, in this case, the answer to your second question also depends on $X$. | |
| Feb 7, 2018 at 17:58 | history | asked | user113393 | CC BY-SA 3.0 |