Timeline for Understanding (the wiki page on) Verdier duality
Current License: CC BY-SA 3.0
16 events
| when toggle format | what | by | license | comment | |
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| Jan 11, 2012 at 8:02 | comment | added | DamienC | @euklid345: nobody's complaining (IMO spotting a typo and/or a mistake is not complaining). | |
| Aug 29, 2011 at 23:47 | comment | added | euklid345 | Hey guys: instead of complaining about mistakes on the wiki page: fix the wiki page already! if you think it's hard to understand: make it easier! (You might also have this whole discussion on the talk page of the wikipedia article.) | |
| Jul 7, 2011 at 9:16 | history | edited | DamienC | CC BY-SA 3.0 |
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| Jun 24, 2011 at 13:39 | history | edited | DamienC | CC BY-SA 3.0 |
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| Jun 24, 2011 at 6:40 | history | edited | DamienC | CC BY-SA 3.0 |
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| Jun 24, 2011 at 6:27 | history | edited | DamienC | CC BY-SA 3.0 |
$*$ replaced by $pt$ to avoid ambiguity
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| Jun 23, 2011 at 22:22 | comment | added | DamienC | Sorry. The point is that there is another mistake in the wiki: Poincaré duality is - either between $H^*$ and $H^{BM}_*$ (Borel-Moore homology) - either between $H_c^*$ and $H_*$ - not between $H^*$ and $H_*$. Then one has $H^k(X,F)=[F[-k],F]$, $H_k(X,F)=[F[-k],D_X]$, and $H^k_c(X,F)$ is given by the formula in you comment. | |
| Jun 23, 2011 at 21:33 | comment | added | James D. Taylor | Okay, okay, okay. So, for example, for Poincare duality we have $F$'s everywhere (where $F$ is say, $\mathbb{Q}$ or maybe $\mathbb{Z}$), but for Serre duality we have $O_X$. Got it. Perfecto. Things fall into place. This was extremely helpful. | |
| Jun 23, 2011 at 21:28 | comment | added | James D. Taylor | Right, so in this case replace $O_X$ by $F$. | |
| Jun 23, 2011 at 21:26 | comment | added | DamienC | In principle there is no $\mathcal O_X$ here, as we are dealing with sheaves of $F$-modules. | |
| Jun 23, 2011 at 21:21 | comment | added | James D. Taylor | Sorry, that should be $Hom(O_X,C[-n])$ is dual to $H^n(X,$ dual of $C)$. | |
| Jun 23, 2011 at 21:19 | comment | added | James D. Taylor | Aha! Here's what's missing. For some reason: $Hom(O_X,C)\cong$ to the dual of $H^(X,$ dual of $C)$. If I can only figure out why that's true, I'll be satisfied. | |
| Jun 23, 2011 at 21:17 | comment | added | James D. Taylor | Oh, I guess you're saying that $Hom(F,F[a])\neq H^a(X,F)$, but rather $H^a(X,F)=Hom(O_X,F[a])$. But still, how does one get from $Hom(F,F[k-n])$ to the desired $H^{n-k}(X,F)$ dual? I guess $Hom(F,F[k-n])=Hom(O_X,F[k-n]$ dual $)$. Now what? | |
| Jun 23, 2011 at 21:11 | comment | added | James D. Taylor | I'm still a little confused. Under some conditions, in Poincare duality we have $D_X=F[-n]$, right? So $H_k(X,F)=Hom(F[-k],F[-n])=Hom(F,F[k-n])=H^{k-n}(X,F)=0$. I'm doing something wrong. | |
| Jun 23, 2011 at 21:09 | history | edited | DamienC | CC BY-SA 3.0 |
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| Jun 23, 2011 at 20:59 | history | answered | DamienC | CC BY-SA 3.0 |