Timeline for Can one define fields in stable homotopy theory via invertibility?
Current License: CC BY-SA 4.0
13 events
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| Aug 23, 2021 at 21:41 | comment | added | Emily | @TimCampion Thanks, Tim! This sounds like a really good idea! | |
| Aug 23, 2021 at 21:40 | history | edited | Emily | CC BY-SA 4.0 |
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| Aug 23, 2021 at 16:18 | comment | added | Tim Campion | In the cited result (saying that any field spectrum is a $K(p,n)$-module for some $p,n$), one must allow $n = \infty$ and not just $n \in \mathbb N$ to include the case of $H \mathbb F_p$. Regarding the question, I would probably start first by considering the question in $D(\mathbb Z)$ -- the derived category of abelian groups -- rather than in Spectra. Already there, it seems unclear to me how to characterize "differential graded fields" via "invertibility conditions", but if you work out such a characterization, you could then try to generalize it to spectra. | |
| Aug 21, 2021 at 2:02 | history | edited | Emily |
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| Aug 21, 2021 at 1:56 | history | edited | Emily | CC BY-SA 4.0 |
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| Aug 13, 2021 at 4:25 | comment | added | Emily | @AchimKrause This is a wonderful point! Thanks, Achim! | |
| Aug 12, 2021 at 22:47 | comment | added | Achim Krause | if you fix it - the underlying $E_\infty$ space only depends on the connective cover. Connective ring spectra are only field spectra if they are ordinary fields, so whatever you define in terms of underlying $E_\infty$ spaces won't be related to field spectra. | |
| Aug 11, 2021 at 22:40 | comment | added | Emily | @MaximeRamzi Ah, that's right! The naive definition I proposed definitely doesn't work, at least as it currently is. | |
| Aug 11, 2021 at 22:24 | comment | added | Maxime Ramzi | I don't get what you mean by that, but let me say it in a different way : if the Péroux-Shipley result is indeed true for $S_*$, then all $E_\infty$-comonoids are of the form $X_+$ - this is never the case for $E_\infty$-monoids of the form $\Omega^\infty E$ - in particular your "to some $E_\infty$-monoid, which is then a Hopf algebra" doesn't make sense, because there is no such "then" | |
| Aug 11, 2021 at 22:16 | history | edited | Emily | CC BY-SA 4.0 |
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| Aug 11, 2021 at 22:16 | comment | added | Emily | @MaximeRamzi Thanks! I corrected the two points you mentioned. Also, I wonder if the problem with $\Omega^{\infty}E$ would go away if we relaxed the requirement for it to be an $\mathbb{E}_\infty$-Hopf algebra to it being isomorphic (in some appropriate weak sense) to some $\mathbb{E}_\infty$-monoid in $\mathcal{S}_*$, which then is an $\mathbb{E}_{\infty}$-Hopf algebra there? | |
| Aug 11, 2021 at 22:08 | comment | added | Maxime Ramzi | It's a bit odd to cite "Maximilien-Shipley" : one of them is a first name and the other one a last name. Also note that $\mathbb Z$ (and friends) is not the free ring on one element, but the free ring on $0$ element (i.e. the initial ring). I don't know about your questions - note that unlike in the Set-context, not every pointed space is of the form $X_+$, so not every pointed space admits a canonical $E_\infty$-comonoid structure - in particular $\Omega^\infty E$ is never of this form, except for discrete $E$ (because all components are equivalent) | |
| Aug 11, 2021 at 21:58 | history | asked | Emily | CC BY-SA 4.0 |