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Mar 19, 2016 at 22:47 comment added user84144 I would guess that this can't be derived formally, since it is only true under some assumptions (that the cohomology of the space satisfies a Poincare duality), whereas a formal derivation would presumably work for anything. I'm hoping to be wrong though!
Mar 18, 2016 at 21:50 comment added alpoge (@MatthiasWendt) Having talked with the very anonymous user84144, I think the main issue is that there is a pairing defined in terms of the "classical" Bockstein (i.e. sans twisting) which one wants to write as Sq^1 so as to apply the usual identities satisfied by the Steenrod operations. I think both of us were just confused, though. Thanks for the great answer! By the way, is it possible to prove the equality x*Sq^1(x) = Sq^2Sq^1(x) for x in degree 2 from the usual axioms of the Steenrod operations?
Mar 18, 2016 at 18:39 history edited Matthias Wendt CC BY-SA 3.0
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Mar 18, 2016 at 16:54 comment added Sean Tilson Minor point, but I think Bert spells his last name Guillou.
Mar 18, 2016 at 8:36 comment added Matthias Wendt @user84144: Looking at it, in your motivation/example, you have a cycle class. Probably this should be an element in $H^{2r}(X,\mathbb{Z}/2(r))$ rather than $H^{2r}(X,\mathbb{Z}/2(0))$. Then $Sq^1$ should be the Bockstein associated to the $\mathbb{Z}/4$-sequence (twisted by r), i.e., the cohomology operation appearing in all the abovementioned papers. But that would be trivial if the class lifts to $\mathbb{Z}/4(r)$.
Mar 18, 2016 at 3:59 vote accept user84144
Mar 18, 2016 at 0:44 comment added user84144 In response to your addendum: ah, that's too bad. The equality would have made life much easier, but oh well. (This does beg the question of whether the other boundary map fits into a larger story...) I had noticed that things were okay over field containing $\mu_4$, but was really hoping to get the result for all finite fields.
Mar 17, 2016 at 21:18 history edited Matthias Wendt CC BY-SA 3.0
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Mar 17, 2016 at 21:05 comment added Matthias Wendt @user84144: there is a difference between the preprint and the published version. The published version includes the discussion of étale cohomology.
Mar 17, 2016 at 15:40 comment added user84144 Wow, this is exactly the kind of thing I need! But forgive my ignorance (I'm far from this motivic stuff), I may need more help in seeing why their theorem implies the desired equality. Are you referring to the version people.math.osu.edu/joshua.1/coh-ops.new.pdf? I see a Theorem 1.1 but no part (iii). I guess I should be taking r=0, but I don't quite see how the two things can be identified with my two Bockstein maps.
Mar 17, 2016 at 9:07 history answered Matthias Wendt CC BY-SA 3.0