Timeline for Commutativity of pairs of reflective localizations
Current License: CC BY-SA 4.0
16 events
| when toggle format | what | by | license | comment | |
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| S Nov 26 at 2:08 | history | bounty ended | CommunityBot | ||
| S Nov 26 at 2:08 | history | notice removed | CommunityBot | ||
| S Nov 18 at 0:08 | history | bounty started | user39598 | ||
| S Nov 18 at 0:08 | history | notice added | user39598 | Canonical answer required | |
| Nov 6 at 1:34 | answer | added | user39598 | timeline score: 0 | |
| Nov 6 at 0:39 | comment | added | user39598 | @TimCampion I posted a possible approach and was wondering if it is related to some of the ideas you suggested. | |
| Oct 30 at 23:26 | comment | added | Zhen Lin | Reflective localisations are idempotent monads. For monads, there is a notion of distributive law. As far as I can tell it does not degenerate in the idempotent case to commutativity of endofunctors. | |
| Oct 30 at 18:47 | comment | added | user39598 | I think 7) is right. My question seems to be equivalent to whether the Beck-Chevalley condition holds for the commutative diagram of right adjoints (the inclusions of local objects) | |
| Oct 30 at 17:29 | comment | added | Tim Campion | (7.) It feels like something something Beck-Chevalley condition could be relevant... (8.) My (4.) is incorrect, since the composite of two localization endofunctors is generally not a localization endofunctor (and that's precisely the obstruction to commutativity!) | |
| Oct 30 at 17:25 | history | edited | LSpice | CC BY-SA 4.0 |
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| Oct 30 at 17:23 | comment | added | Tim Campion | ... localizations form a (4.) band (i.e. an idempotent semigroup) under composition. Commutative bands = semilattices. Varieties of bands are classified... (not sure what to do with that) (5.) example of (2) : if $R_1$ has a further adjoint (i.e. $L_1$ is smashing), then $L_1$ commutes with all other localizations (er -- in some contexts?). It's in the center of the band. (6.) Stuff on the Bousfield lattice or left-exact localizations (which form a distributive lattice (does that mean they commute?) -- see Borceux-Kelly) might be relevant. | |
| Oct 30 at 17:20 | history | edited | user39598 | CC BY-SA 4.0 |
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| Oct 30 at 17:20 | comment | added | Tim Campion | A few thoughts -- (1.) The case of a recollement is somehow the "opposite extreme" where composing one way gives zero but composing the other way records enough info to recover the ambient category. Maybe one could try build up a general pair of localizations in terms of recollements and commuting pairs? (2.) If you impose conditions like "$L_i$ exact" or "$R_i$ has a further adjoint" things might simplify. (3.) Favorite example: Nisnevich / $\mathbb A^1$ localization don't commute... | |
| Oct 30 at 17:17 | comment | added | user39598 | Thanks. I mentioned commutativity in the title, but will make a more explicit reference in the post. | |
| Oct 30 at 16:57 | comment | added | Paul Taylor | Rather than generalise the problem I suggest you start by considering the simpler one of two idempotents (retracts) on a set. The intersection of the two fixed subsets need not be fixed unless there is some commutativity condition. However, a weaker condition such as $e\cdot f\cdot e=f\cdot e$ may suffice, but giving two isomorphic subsets. | |
| Oct 30 at 15:16 | history | asked | user39598 | CC BY-SA 4.0 |