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Sep 18, 2023 at 3:08 comment added Nicholas Kuhn @TimCampion For this splitting problem, Jie Wu's memoir that you found sends one to [Selick, Paul and Wu, Jie, On functorial decompositions of self-smash products, Manuscripta Math. 111 (2003), 435–457]. It is nice to see that I gave the same answer as they found! Their techniques are the same as those used by Jeff Smith, and Lionel Schwartz (see his unstable A-modules book), and me with Dave Carlisle in the 1980s, or with Chris Lloyd just recently. In all these cases, there is no difference between stable splittings and splittings after just one suspension.
Sep 17, 2023 at 20:07 comment added Tim Campion I found a reference for this: Jie Wu, homotopy theory of suspensions of the projective plane (working unstably, in fact). Is there a good reference for the stable decomposition of $X^{\otimes n}$ under the action of $\Sigma_n$? This stuff was also important in Devinatz-Hopkins-Smith's proof of the nilpotence theorem, but it seems that Jeff Smith's work never appeared in print...
Sep 12, 2023 at 22:09 comment added Nicholas Kuhn @DaveBenson Yes, idempotents lift. And at the prime 2 only, I know which primitive idempotents have nonzero image on the n-fold tensor power of a vector space (char 2) of a fixed dimension d. Answer: those which correspond to 2-regular partitions of n with at most d columns. (The proof of this is something I figured out awhile ago, and is too long to put into a MathOverflow comment, or answer.)
Sep 12, 2023 at 21:33 comment added Dave Benson Hi Nick, I have some confusions which you may be able to clear up for me. Does $\mathbb{Z}_2$ for you mean the $2$-adics or the integers mod two? If the latter, since $2$ does not act as the zero endomorphism of $S/2$, how does the mod two group ring of $\Sigma_n$ act on the $n$-fold smash product of $S/2$? Fortunately, idempotents in the mod two group ring lift to the $2$-adic group ring. Is this what you intend? Finally, you seem to know which idempotents act as zero on the tensor powers. Can you explain this? Sorry for all the questions - Cheers, Dave
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