Timeline for Chromatic representation theory of the symmetric groups?
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
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| Sep 21, 2023 at 21:23 | comment | added | Tyler Lawson | Ach, yes, please interpret central as central in pi_*. It's effectively showing that the Ore localization exists. | |
| Sep 21, 2023 at 20:53 | comment | added | Maxime Ramzi | I find the claim hard to believe - I guess I can prove it for an idempotent in pi_0(R) which is (graded) central in pi_*(R) but without this assumption it seems unlikely. I guess this is still not a serious obstruction in the given situation though, so you are probably right | |
| Sep 21, 2023 at 16:40 | comment | added | Tyler Lawson | @MaximeRamzi good point! (I believe someone explained to me once that central idempotents in pi_0(R) automatically lift to THC(R), which would make this not too serious) | |
| Sep 21, 2023 at 15:36 | comment | added | Maxime Ramzi | I think classically, natural splittings of the identity correspond to idempotents in $\pi_0THC(R)= \pi_0 Map_{R\otimes R^op}(R,R)$ (idempotents in $\pi_0(R)$ correspond to splittings of the forgetful functor). Because the target in our case is $K(h)$-local, where you compute the hom does not matter so much. I agree that you will not get the same as over $\mathbb Q$, though | |
| Sep 21, 2023 at 12:11 | comment | added | Tyler Lawson | If you wanted to do something similar with your result, you'd have to use something like the internal hom in $K(h)$-local spectra - and that's fine, I think you can probably identify natural splittings with idempotents as well. But even though Tate vanishing is true, it doesn't give you an idempotent in the same way that the transfer does in characteristic zero (fixed-point objects are definitely not natural summands). I guess in my parenthetical I was expressing a feeling that this difference is related to changing the ground category from $Sp$ to $Sp_{K(h)}$, which changes $X_{hG}$. | |
| Sep 21, 2023 at 12:02 | comment | added | Tyler Lawson | How I interpreted the desired result was as follows. To classify natural splittings of a module category $Mod_R$, you describe the identity functor as $Hom_{Mod_R}(R,-)$, whose natural endomorphisms are identified with the ring $\pi_0 R$. Natural splittings are then idempotents in this ring. For example, $\Bbb Q[G]$ has an idempotent that splits off the trivial representation $\Bbb Q$, and so the fixed-point object $X^{hG}$ is always a split summand of $X$ in characteristic zero. | |
| Sep 21, 2023 at 11:58 | comment | added | Tyler Lawson | @MaximeRamzi Yes, that description is true. When I said that there is not a Schwede-Shipley result, I meant in the original sense that it's not a module category over the spherical group algebra in $Sp$. Here you're using the module category in $K(h)$-local spectra. | |
| Sep 21, 2023 at 8:13 | comment | added | Maxime Ramzi | I'm confused by your claim about Schwede-Shipley. Surely there is an equivalence $(Sp_{K(h)})^{B\Sigma_n} \simeq Mod_{\mathbb S_{K(h)}[\Sigma_n]}(Sp_{K(h)})$, and $\mathbb S_{K(h)}[\Sigma_n]$, being a finite sum, is unambiguous. Also, I think one might want something a bit more than idempotents in $\pi_0$ to get relevant decompositions of the representation $\infty$-category | |
| Sep 20, 2023 at 13:54 | history | answered | Tyler Lawson | CC BY-SA 4.0 |