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Sep 20, 2023 at 13:54 answer added Tyler Lawson timeline score: 8
Sep 18, 2023 at 17:27 comment added Tim Campion @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.
Sep 18, 2023 at 3:36 comment added Nicholas Kuhn I don't understand your sentence "Certainly it restores ..." How does this work when h = 1 and n = 2?
Sep 17, 2023 at 21:21 comment added Maxime Ramzi 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
Sep 17, 2023 at 20:57 history edited Tim Campion CC BY-SA 4.0
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Sep 17, 2023 at 19:53 history asked Tim Campion CC BY-SA 4.0