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We know that in characteristic 0, the group ring of the symmetric group $\Sigma_n$ splits via one idempotent for each partition of $n$.

In characteristic $p$, I believe the analogous statement is that there is one idempotent for each partition of $p$ which doesn't have any $\geq(p-1)$-fold repetitions of numbers. This has to do with certain denominators becoming noninvertible mod $p$.

But chromatically, it seems like ambidexterity might restore some of these characteristic zero idempotents? Certainly it restores the idempotent projecting onto the trivial representation for $\Sigma_p$.

Question: What is a complete set of idempotents for the $E_1$ ring spectrum $K(h)[\Sigma_n]$? How about for $\mathbb S_{T(h)}[\Sigma_n]$ or $\mathbb S_{K(h)}[\Sigma_n]$?

Any partial answers would be interesting.

It seems like the character theory of Hopkins-Kuhn-Ravenel ought to be relevant, but I'm not sure how.

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  • $\begingroup$ For representation theory purposes, you probably want central idempotents (in a strong homotopy coherent sense). For both $K(h)$ and $\mathbb S_{K(h)}$ it seems relevant to start by computing the ones for $E_h$ (in fact for $K(h)$ the answer might be the same), and there the answer will look like a computation of the idempotents in $\pi_0(E_n^{LB\Sigma_n})$ with some interesting ring structure (here $L = $ free loop space). I don't know if HKR character theory contains enough information to do this because it works away from the torsion $\endgroup$
    – Maxime Ramzi
    Commented Sep 17, 2023 at 21:21
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    $\begingroup$ I don't understand your sentence "Certainly it restores ..." How does this work when h = 1 and n = 2? $\endgroup$
    – Nicholas Kuhn
    Commented Sep 18, 2023 at 3:36
  • $\begingroup$ @NicholasKuhn Maybe I'm misunderstanding something, but the thought is that these idempotents correspond naturally to functorial splittings of spectra with $\Sigma_n$ action. Tate vanishing tells us that there is a functorial splitting of the fixed points = orbits, so this should correspond to an idempotent in $\mathbb S_{T(h)}[\Sigma_n]$ which doesn't exist in $\mathbb Z_{(p)}[\Sigma_n]$ -- the one onto fixed points. $\endgroup$
    – Tim Campion
    Commented Sep 18, 2023 at 17:27

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The coefficient ring of $K(h)[\Sigma_n]$ is, in degree zero, $\Bbb F_p[\Sigma_n]$, the group algebra on $\Sigma_n$ over $\Bbb F_p$. As a result, the list of idempotents in this ring is the same as for the ordinary group algebra over $\Bbb F_p$. I believe that Hensel's lemma will also put these in bijection with the idempotents over $E_n$ and over the spherical group algebra over the $K(h)$-local sphere; in both cases $\pi_0$ is a group algebra over a complete local ring with residue field $\Bbb F_p$.

(This might be a little bit unsatisfying but perhaps one complicating factor is the usual, that the unit in $K(h)$-local homotopy theory is not compact. As a result, there is not a Schwede-Shipley type equivalence between the category of $K(h)$-local spectra with $\Sigma_n$-action and a module category over the $K(h)$-local spherical group algebra. The idempotents you're looking at are most relevant to the latter, but ambidexterity-type results are more relevant to the former.)

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  • $\begingroup$ 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 $\endgroup$
    – Maxime Ramzi
    Commented Sep 21, 2023 at 8:13
  • $\begingroup$ @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. $\endgroup$
    – Tyler Lawson
    Commented Sep 21, 2023 at 11:58
  • $\begingroup$ 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. $\endgroup$
    – Tyler Lawson
    Commented Sep 21, 2023 at 12:02
  • $\begingroup$ 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 $X_{hG}$. $\endgroup$
    – Tyler Lawson
    Commented Sep 21, 2023 at 12:11
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    $\begingroup$ Ach, yes, please interpret central as central in pi_*. It's effectively showing that the Ore localization exists. $\endgroup$
    – Tyler Lawson
    Commented Sep 21, 2023 at 21:23

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