Timeline for What's the relationship between the different versions of the BBD decomposition theorem?
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
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| Oct 19 at 12:59 | comment | added | Toan | Thanks @DanPetersen, I have asked a new question in mathoverflow.net/q/480919/89665! Hopefully it is clearer now. | |
| Oct 19 at 12:25 | comment | added | Dan Petersen | @Toan at a glance I don’t think I understand what you’re asking. Maybe you should ask this as a new question. | |
| Oct 19 at 11:23 | comment | added | Toan | @DanPetersen Sorry for digging this up, but I'm also confused about the same thing, spend few days on it. I know we can choose a stratification as in 1.a. so that $f_*IC_X$ constructible wrt such stratification. How exactly do you show that only $IC(\overline{S_{\lambda}},\chi)$ appear? I can only show that $IC(\overline{\sqcup S_{\lambda}},\chi)$ appear from constructibility of $f_*IC_X$. | |
| Jul 30, 2015 at 10:58 | comment | added | Dan Petersen | Regarding whether you can deduce Version 1.a from Version 1, I don't believe that there's a direct argument. What you need to know is that $f_\ast \mathrm{IC}_X \in D^b_c(Y)$ is actually constructible wrt the stratification $\{S_\lambda\}$. | |
| Jul 30, 2015 at 10:54 | comment | added | Dan Petersen | I agree with Ben Webster - what's your definition of geometric origin? I would say that $K$ is of geometric origin if it can be obtained from the trivial perverse sheaf over a point by applying the standard six functors and forming subquotients; in particular, every summand of $f_\ast K$ in Version 2 ought to be of geometric origin by definition, and the only nontrivial part of the statement is semisimplicity. | |
| Jul 30, 2015 at 4:33 | comment | added | Will Sawin | I don't think $\chi$ must have finite monodromy. Do you mean that the local monodromy is quasi-finite? | |
| Jul 30, 2015 at 2:08 | history | edited | Balerion_the_black | CC BY-SA 3.0 |
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| Jul 30, 2015 at 2:06 | comment | added | Balerion_the_black | @BenWebster That's right, $IC_X$ is of geometric origin in Version 1. I was rather wondering about the $IC(Z,\chi)$'s occuring in the decomposition. Version 2 states they should be of geometric origin, Version 1 does not. So my question was: is there some (hidden) assumption in Version 1 which guarantees that any simple perverse sheaf is necessarily of geometric origin? It seems to me now that this is not the case - Version 1 is probably indeed weaker, and fails to mention that only $IC(Z,\chi)$'s of geometric origin (i.e., where the monodromy of $\chi$ is finite) can occur. | |
| Jul 30, 2015 at 1:29 | comment | added | Ben Webster♦ | I'm confused by your confusion. Version 1 doesn't mention that the summands are of geometric origin, but I believe they are obviously are by the definition of the geometric origin, so there's no need to mention it. Even if it is a stronger statement, that would just mean that Version doesn't state the theorem in its strongest form, which we've already established. | |
| Jul 30, 2015 at 0:56 | history | asked | Balerion_the_black | CC BY-SA 3.0 |