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There is a well-known formula for the cup product of an $i$-cochain $A$ and $j$-cochain $B$ in simplicial homology given by $$(A\cup B)(0\ldots i+j) = A(0\ldots i) B(0\ldots j)\;.$$ This formula commutes with the derivative in the sense that $$d(A\cup B)=dA\cup B+(-1)^j A\cup dB$$ holds exactly on the level of cochains. This property ensures that if we apply the cup product to two cocycles, the result will again be a cocycle.

Another operation which can be defined on the cochain level is the $x$th Steenrod square of a $i$-cocycle $A$, $$Sq^x(A)=A\cup_{i-x} A$$ via higher order cup products, see, e.g., appendix B in this paper. Again, the formula applied to cocycles yields cocycles. However, the map does not commute with the derivative on the level of cochains. Specifically, using the first formula in the mentioned appendix, we get $$d Sq^x(A) = Sq^x(dA)+d(A\cup_{i-x+1} dA)$$ up to some minus signs I'm potentially missing (I'm mostly thinking about $\mathbb{Z}_2$-valued cohomology anyways).

Question: Is there a local combinatorial formula for the Steenrod squares which commutes with the derivative on the level of cochains, such as the one for the cup product? Or is there some obstruction to this?

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  • $\begingroup$ I'm not sure whether they achieve your precise conditions, but some of Anibal Medina and collaborators' work may be relevant, including this and this. $\endgroup$
    – Tim Campion
    Commented Jan 20, 2022 at 23:23
  • $\begingroup$ @TimCampion Thanks for the references! The second one explictly says they are using the original definitions via higher cup products. The first one sounds a bit like they are not considering different formulas for $p=2$ but only extending them to $p\neq 2$, but I'll have a closer look. $\endgroup$
    – Andi Bauer
    Commented Jan 21, 2022 at 9:12
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    $\begingroup$ I am skeptical such a thing can exist. It is known that $\operatorname{Sq}^i$ is not a map of $H\mathbb{Z}$-modules for $i>1$, and this precludes it being given by a map of chain complexes. I guess there could be a non-additive map that commutes with the differential, but it feels unnatural... $\endgroup$
    – Denis Nardin
    Commented Jan 21, 2022 at 12:37
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    $\begingroup$ @AndiBauer Every stable cohomology operation $H^*(-;\mathbb{Z}/2)\to H^{*+k}(-;\mathbb{Z}/2)$ is induced by a map of spectra $H\mathbb{Z}/2\to \Sigma^k H\mathbb{Z}/2$. If the cohomology operation comes from a natural transformation of chain complexes this map is $H\mathbb{Z}$-linear for the standard $H\mathbb{Z}$-module structure on the spectrum $H\mathbb{Z}/2$. However, it is known that $\operatorname{Sq}^i$ is not $H\mathbb{Z}$-linear for $i>1$. $\endgroup$
    – Denis Nardin
    Commented Jan 26, 2022 at 16:19
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    $\begingroup$ @AndiBauer Yes, the (∞-)category of spectra is symmetric monoidal and there is a lax symmetric monoidal functor $H$ from abelian groups to spectra sending $A$ to $HA$ (the spectrum representing ordinary homology with coefficients in $A$). But giving the whole story of stable homotopy theory is going to be hard in the comment boxes :). $\endgroup$
    – Denis Nardin
    Commented Jan 26, 2022 at 17:29

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