Timeline for Are cofibrant commutative S-algebras flat?
Current License: CC BY-SA 3.0
15 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 28, 2022 at 5:23 | comment | added | Bruno Stonek | @DavidWhite I was a bit hasty in my previous comment. This question is not about flatness of cofibrant S-modules (for which I give a reference in the original post), but about flatness of cofibrant commutative S-algebras, of which the reference you give seems to make no mention of. | |
| Jun 27, 2022 at 19:40 | comment | added | David White | Yep, that's what I was referring to! Not a proof but a place one could cite. I think the result is true | |
| Jun 27, 2022 at 18:28 | comment | added | Bruno Stonek | @DavidWhite You're thinking of the sentence "Although the we do not know of a general principle that would imply the second property, it holds in all presently known monoidal model categories of spectra [3,4,9]" (where [3] is EKMM)? It's encouraging that the authors would claim so, but do you know where's a proof? | |
| Jun 26, 2022 at 12:11 | comment | added | David White | I came back to this question today when asking my own about S-modules, and now I think a reference for what you wanted in your question is just after (ii) on page 25 of "Modules in monoidal model categories" by Lewis and Mandell. They don't even need R to be cofibrant. core.ac.uk/download/pdf/82044183.pdf | |
| S Dec 5, 2017 at 13:16 | history | bounty ended | Bruno Stonek | ||
| S Dec 5, 2017 at 13:16 | history | notice removed | Bruno Stonek | ||
| S Nov 28, 2017 at 2:54 | history | bounty started | Bruno Stonek | ||
| S Nov 28, 2017 at 2:54 | history | notice added | Bruno Stonek | Draw attention | |
| Nov 26, 2017 at 14:10 | comment | added | Bruno Stonek | The second is a quote of the paper "Topological Hochschild Homology" by Schwänzl-Vogt-Waldhausen: "We have to distinguish between the associative and commutative case, because the forgetful functor $RCAlg\to RAlg$ does not preserve q-cofibrant objects. This is a well-known phenomenon: in ordinary algebra free associative resolutions use tensor algebras, while free associative and commutative resolutions use symmetric algebras". (Also, thank you for the interest.) | |
| Nov 26, 2017 at 14:10 | comment | added | Bruno Stonek | @DavidWhite Two remarks: the first one is that the unit $R\to A$ of a cofibrant $R$-algebra or commutative $R$-algebra is a cofibration of underlying $R$-modules (EKMM, right after VII.4.14). | |
| Nov 26, 2017 at 12:00 | comment | added | David White | Thanks for the response. What I meant to ask was: can it be cofibrant as a commutative R-algebra but not as an R-algebra? When you forget all the way to just R-modules, you can lose cofibrancy just as you point out, but what's usually true is that a cofibration with cofibrant source forgets to a cofibration. So you have a kind of relative cofibrancy, and that's enough for the flatness property you asked about. | |
| Nov 26, 2017 at 7:14 | comment | added | Bruno Stonek | @DavidWhite $\mathbb S$ is a cofibrant commutative $\mathbb S$-algebra (it is cell), but $\mathbb S$ is not a cofibrant $\mathbb S$-module. | |
| Nov 25, 2017 at 21:15 | comment | added | David White | You wrote that "the underlying R-modules of cofibrant commutative R-algebras are not generally cofibrant" - do you have an example in mind? This is why in my answer below I suggested Shipley's approach of working with positive cofibrations (which is also what I did in my thesis), but I thought that for EKMM S-modules one could often avoid the shift to positive cofibrancy, since the unit is already not cofibrant (so Lewis's obstruction doesn't apply) | |
| Nov 25, 2017 at 21:12 | answer | added | David White | timeline score: 3 | |
| Nov 24, 2017 at 7:17 | history | asked | Bruno Stonek | CC BY-SA 3.0 |