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May 26, 2019 at 13:21 comment added Tyler Lawson @TimCampion Right. The reason it works here is that the d2 differential does vanish for all spaces, because the K-groups of a point are concentrated on even degrees.
May 25, 2019 at 21:01 comment added Tim Campion @TylerLawson I’m trying to parse this... I can understand how this shows the d2 differential must be a cohomology operation, but even if the d2 differential vanishes at X, it seems to me that because it doesn’t vanish for all spaces, it results that the E3 page is not a cohomology theory and so the d3 differential is not a natural transformation of cohomology theories and so I don’t see how to conclude that it’s a cohomology operation
Aug 18, 2017 at 23:07 comment added Tyler Lawson @54321user I believe that I learned that from Adams. To be a stable cohomology operation it merely needs to (a) be a natural transformation (which is true because the Atiyah-Hirzebruch spectral sequence is natural), and (b) be compatible with the suspension isomorphism (which one can show straightforwardly when the (based) space is a point and then progressively bootstrapped up to the result for a general space).
Aug 15, 2017 at 3:51 comment added 54321user Why do the differentials for the AHSS have to be a stable cohomology operation? Do you have a reference?
Apr 24, 2011 at 5:14 vote accept Sam Nariman
Apr 24, 2011 at 5:11 comment added Sam Nariman But computing the K group of $\mathbb{R}P^k$ needs to apply AHSS. Actually at the appendix of the paper by Atiyah, titled Analytic cycle on complex manifold, He computed $d_{2p−1}$ in terms of Steenrod p-power by attaching a cell to some$\Sigma^k\matbb{C}P^n$ and using some relation of differentials and chern character which I don't quite understand. But I expect for d3 there should be some straight forward example
Apr 22, 2011 at 17:58 history answered Tyler Lawson CC BY-SA 3.0