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The original paper on Steenrod squares, Steenrod's "Products of cocycles and extensions of mappings", 1947, 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, due originally to Adem, "The iteration of the Steenrod squares in algebraic topology", 1952 and beautifully explained by Bullett and Macdonald, "On the Adem relations", 1982, 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. 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 — I really do want a combinatorial description that fits into the framework of Steenrod's original paper.

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    $\begingroup$ I should remark that, as Bullet and Macdonald explain, the Squares have to do with the symmetric group $S_2$ action on $X \to X \times X$, and the Adem relation has to do with a homotopy between maps $X \to X^{\times 4}$ having to do with the fact that different embeddings of the dihedral group $D_8$ into $S_4$ are conjugate. Surely this geometric picture can be used to produce the primitives I am looking for. But I find I am not sufficient combinatorially adept to do it, so I am hoping someone else has. $\endgroup$ – Theo Johnson-Freyd Feb 6 '17 at 2:07
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    $\begingroup$ The earliest study of cochain level operations that I am aware of is by Kristensen and Madsen. Our work with Baues might be also relevant, it is in his book on the secondary Steenrod algebra. This is however not so directly based on the cochain approach. $\endgroup$ – მამუკა ჯიბლაძე Feb 6 '17 at 7:40
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    $\begingroup$ @მამუკაჯიბლაძე Thanks! Kritensen+Madsen's work is very helpful for me. $\endgroup$ – Theo Johnson-Freyd Feb 6 '17 at 21:28
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    $\begingroup$ I think that the sequence operad from McClure-Smith can be used to produce explicit reasons why the Adem relations hold in cohomology. I did this for a simple Cartan-relation in small degrees by hand (and afterwards I thought it would be nice if this was written up somewhere and stopped thinking about it). mathoverflow.net/questions/268031/… seems related $\endgroup$ – HenrikRüping Nov 30 '17 at 14:08
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The following article might be relevant. This is the review I wrote for it a number of years ago.

MR1622630 (99h:55027) Reviewed Real, Pedro(E-SEVLIS-AM1) On the computability of the Steenrod squares. (English, Italian summary) Ann. Univ. Ferrara Sez. VII (N.S.) 42 (1996), 57–63 (1998). 55S05 (55S10)

Let $C_∗(X)$ denote the normalized chains on a simplicial set $X$. The Eilenberg-Zilber theorem can be proved using very explicit operators $AW:C_∗(X×Y) \rightarrow C_∗(X)\otimes C_∗(Y)$, $EM:C_∗(X)\otimes C_∗(Y) \rightarrow C_∗(X×Y)$, and $SHI:C_∗(X×Y)→C_{∗+1}(X×Y)$. The contracting homotopy SHI is least familiar, but like EM it involves an explicit summation over shuffles. The author gives an explicit and clean cochain level formula for the mod 2 Steenrod operation $Sq^i$ in terms of AW and SHI. Like SHI, it is exponential in terms of computational complexity. {Reviewer's remark: The author systematically uses the word "idempotent'' where he seems to mean "involution''.}

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