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Oct 25, 2023 at 18:45 vote accept Aditya De Saha
Oct 24, 2023 at 11:34 answer added Tom Goodwillie timeline score: 4
Oct 24, 2023 at 3:43 comment added Aditya De Saha @TomGoodwillie thanks for your response. I was (am) stuck at figuring out what these differential operators look like. Would you please expand on why they are multiplication by $m_1m_2...m_k$?
Oct 23, 2023 at 23:06 comment added Tom Goodwillie Did you look at the spectral sequence with $E_2^{i,j}=H^i(BS^1;H^j(S^n))$? It has only two non-zero rows, the nontrivial groups $E_2^{2p,0}$ and $E_2^{2p,n}$ being infinite cyclic. The computation all comes down to knowing the differential $E_2^{0,n}\to E_2^{n+1,0}$, which takes generator to $m_1\dots m_k$ times generator.
Oct 23, 2023 at 19:21 history asked Aditya De Saha CC BY-SA 4.0