Timeline for Is there a Dold-Kan theorem for circle actions?
Current License: CC BY-SA 4.0
13 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 15, 2022 at 17:55 | vote | accept | Tim Campion | ||
| May 15, 2022 at 17:55 | comment | added | Tim Campion | @MaximeRamzi Oh, duh --yes, of course, thanks. I must have been thinking about the other way around. | |
| May 11, 2022 at 7:31 | comment | added | Maxime Ramzi | (Because $S^1\times S^1$ is $2$-dimensional, you can also view this as being in $[S^1\times S^1, K(\mathbb Z,2)]$, i.e. the nonzero cohomology class(es) of the torus in degree $2₩ | |
| May 11, 2022 at 7:30 | comment | added | Maxime Ramzi | Tim : there are - the map that collapses the $1$-skeleton has degree $1$ and is therefore non null | |
| May 11, 2022 at 0:36 | comment | added | Tim Campion | @JohnRognes Thanks! I was definitely misled by the fact that there are no nonconstant maps $S^1 \times S^1 \to S^2$. | |
| May 7, 2022 at 20:45 | comment | added | John Rognes | For a proof that s^2 = \eta s (where s is the degree 1 operation induced by the S^1-action) you might look at Proposition 3.3 of arxiv.org/abs/2008.09095 , where there are also references to earlier discussions. Here \eta arises as the Hopf construction for the multiplication S^1 x S^1 --> S^1. | |
| May 5, 2022 at 19:35 | comment | added | Maxime Ramzi | Mhm I'm not sure how easy to see it is. Let me try to see if I can come up with a quick explanation (or if someone else can :D ), but I'm not sure (funnily enough, I wasn't aware of this and learned about it while learning about the analogue of that in motivic homotopy theory) | |
| May 5, 2022 at 19:34 | comment | added | Maxime Ramzi | where $\mathbb S^1\otimes X\to \mathbb S[S^1]\otimes X$ is the inclusion of the top cell, and $\mathbb S[S^1]\otimes X\to X$ is the action. In particular, this is in general nonzero. For instance it is sometimes nonzero when $X$ is $THH$ of some ring spectra - this features in some recent/in progress work (not of mine !) on a spherical HKR | |
| May 5, 2022 at 19:34 | comment | added | Tim Campion | Okay, I think I understand your statement. Is this easy to see? And isn't it dependent on the choice of splitting? | |
| May 5, 2022 at 19:32 | comment | added | Maxime Ramzi | In particular, if you have an $S^1$-action on $X$, $\mathbb S[S^1]\otimes \mathbb S[S^1]\otimes X\to \mathbb S [S^1] \otimes X\to X$ is going to be, "on the top cell", not $0$ but the action of $\eta$ on $X$. I should be more precise here, I specifically mean the $\eta$ coming from $S^1$, not coming from $\mathbb S$. So, precisely, $\Sigma^2X\to \Sigma X \to X$ is going to be $\mathbb S^2\otimes X \xrightarrow{\Sigma\eta \otimes X} \mathbb S^1\otimes X\to \mathbb S[S^1]\otimes X\to X$ | |
| May 5, 2022 at 19:29 | comment | added | Maxime Ramzi | It doesn't rule it out, indeed :) As for my comment, the point is that as you indicated, the top cell of $S^1\times S^1$ splits off. Actually it splits off after one suspension, and so the multiplication $S^1\times S^1\to S^1$, after one suspension, looks like some map $S^2\vee S^2 \vee S^3\to S^2$. The $S^2\to S^2$ maps are just identities, and the $S^3\to S^2$ map is $\eta$ | |
| May 5, 2022 at 19:26 | comment | added | Tim Campion | Thanks, this is a great obstruction! I'm not fully satisfied, because it doesn't rule out that the incoherent construction I indicated above might underly a functor which might be fully faithful or something.... Also, could you explain your comment above? I agree it seems reasonable that $\eta$ should appear somehow in what it means to have a $S^1$-action, but I'm having trouble seeing how exactly. | |
| May 5, 2022 at 19:16 | history | answered | Maxime Ramzi | CC BY-SA 4.0 |