Timeline for Are cofibrant commutative S-algebras flat?
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Dec 5, 2017 at 13:16 | history | bounty ended | Bruno Stonek | ||
| Dec 4, 2017 at 13:11 | comment | added | David White | @BrunoStonek I'm back from the hike. Sorry to see this question hasn't gotten more attention. If you have more questions let's just talk by email. I've been meaning to think about S-modules, and this is a good opportunity. However, I'm getting on a ferry tomorrow and will not have email till December 9. If you send me something in the next 20 hours I promise to think about it while on the ferry. | |
| Nov 29, 2017 at 2:43 | comment | added | David White | @BrunoStonek at the moment I'm in Patagonia, about to start the W trek. So, I don't think I'll have any internet for the next 5 days. Towards Dylan's point, because I never verified the axiom for S-modules, I sketched how to get around it using Shipley's work. Basically, cofibrant commutative monoids don't forget to honestly cofibrant objects, but should forget to a class of objects contained in the flat objects. I agree with your comment answering Dylan. Good luck! | |
| Nov 28, 2017 at 8:51 | comment | added | Bruno Stonek | (Above, I meant: if X is an R-module which is positively flat. Couldn't edit it on time). An analogous result is true in EKMM (III.5.1): if R is a cofibrant commutative S-algebra and $X$ is a cell $R$-module, then $\pi$ is a homotopy equivalence of spectra. | |
| Nov 28, 2017 at 8:49 | comment | added | Bruno Stonek | @DylanWilson: using different terminology, Proposition 4.2 says: if $f$ is a map in $CAlg(R)$ which is a projective cofibration in $CAlg(R)$, then it is a positive flat cofibration in $CAlg(R)$. This implies that that it is a positive flat cofibration in $Mod(R)$ (so in particular, a flat cofibration in $Mod(R)$, but not necessarily a projective cofibration in $Mod(R)$). As for your last statement: Shipley's Proposition 3.3 proves that an $R$-module which is positive flat, then the standard map $\pi:X^{\wedge_R n}_{h\Sigma_n}\to X^{\wedge_R n}_{\Sigma_n}$ is a stable equivalence. | |
| Nov 28, 2017 at 3:59 | comment | added | Dylan Wilson | If I understand correctly (which I probably don't), Shipley's result says that if you have some flavor of positively cofibrant commutative algebra, then the underlying module is cofibrant (but not necessarily positively cofibrant). But the OP is precisely not asking that the algebra A be some type of positively cofibrant. Also, I think I'd be surprised if S-modules satisfies the 'strong commutative monoid axiom' since it looks to me like $S^{\wedge 2}_{\Sigma_2}$ is just $S$, which isn't homotopically correct. (But again, I'm likely wrong since I'm not used to these things.) | |
| Nov 28, 2017 at 2:56 | comment | added | Bruno Stonek | I'm starting a bounty on the question -- more details on your answer would be very welcome, thanks. | |
| Nov 25, 2017 at 21:12 | history | answered | David White | CC BY-SA 3.0 |