The original paper on Steenrod squares, [Steenrod's "Products of cocycles and extensions of mappings", 1947](https://www.jstor.org/stable/1969172), uses an explicit combinatorial formula for the squares in terms of simplicial cochains: given a simplicial cochain $\alpha$ on some simplicial set, Steenrod defines a cochain $\mathrm{Sq}^i\alpha$; if $\alpha$ is a cocycle, so is $\mathrm{Sq}^i\alpha$, and the cohomology class of $\mathrm{Sq}^i\alpha$ only depends on that of $\alpha$, and so $\mathrm{Sq}^i$ descends to a (linear!) map on cohomology. (In general, $\mathrm{Sq}^i$ is not linear on cochains and does not commute with $\mathrm d$, so this is a nontrivial statement. But it is not too hard to find an operation that agrees with $\mathrm{Sq}^i$ on cocycles and that does commute with $\mathrm d$, and its failure to be linear, it is not hard to show, is a total derivative.) (Steenrod makes his definition with $\mathbb Z$-coefficients, but I will only care about $\mathbb Z/2$ coefficients, and so assume that throughout.)


The well-known [Adem relations](https://en.wikipedia.org/wiki/Steenrod_algebra#Adem_relations), due originally to [Adem, "The iteration of the Steenrod squares in algebraic topology", 1952](https://dx.doi.org/10.1073/pnas.38.8.720) and beautifully explained by [Bullett and Macdonald, "On the Adem relations", 1982](http://dx.doi.org/10.1016/0040-9383(82)90015-5), are relations of the form $0 = \sum\_{\text{certain pairs }(i,j)} \mathrm{Sq}^i \mathrm{Sq}^j$ that hold in cohomology. The first few are
 $\mathrm{Sq}^1 \mathrm{Sq}^1 = 0$, $\mathrm{Sq}^1\mathrm{Sq}^2 = \mathrm{Sq}^3 \mathrm{Sq}^0$, $\mathrm{Sq}^1 \mathrm{Sq}^3 = 0$, $\mathrm{Sq}^3 \mathrm{Sq}^1 = \mathrm{Sq}^2 \mathrm{Sq}^2$, ...


At the cochain level, these relations do not hold on the nose. The fact that they hold on cohomology just says that, if $\alpha$ is a cocycle, then $\sum\_{\text{certain pairs }(i,j)} \mathrm{Sq}^i (\mathrm{Sq}^j(\alpha))$ is a coboundary, which of course depends on $\alpha$ (and on the particular relation in question).


Is there a combinatorial formula (like the one Steenrod gives for $\mathrm{Sq}^i$) that takes in $\alpha$ a cocycle and produces a primitive for $\sum\_{\text{certain pairs }(i,j)} \mathrm{Sq}^i (\mathrm{Sq}^j(\alpha))$? Where can I find it? I care only about the early Adem relations up to $\mathrm{Sq}^3\mathrm{Sq}^1 = \mathrm{Sq}^2\mathrm{Sq}^2$.


A related question was asked by [Kapustin](https://mathoverflow.net/questions/252018/). Note that I am not after more categorical descriptions of the Adem relations and the Steenrod squares like those available in the many excellent answers to [this old MO question](https://mathoverflow.net/questions/461/) — I really do want a combinatorial description that fits into the framework of Steenrod's original paper.



A cochain-level Adem relation?
Explicit homotopies for a cochain-level Adem relation were first worked out in:


[Greg Brumfiel, Anibal M. Medina-Mardones, John Morgan. A Cochain Level Proof of Adem Relations in the Mod 2 Steenrod Algebra. 2020](https://arxiv.org/abs/2006.09354)


The odd-prime case is still open.


