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By an integral cohomology operation I mean a natural transformation $H^i(X, \mathbb{Z}) \times H^j(X, \mathbb{Z}) \times ... \to H^k(X, \mathbb{Z})$, where we restrict $X$ to some nice category of topological spaces such that integral cohomology $H^n(-, \mathbb{Z})$ is represented by the Eilenberg-MacLane spaces $K(\mathbb{Z}, n)$. The Yoneda lemma shows that such operations are in natural bijection with elements of $H^k(K(\mathbb{Z}, i) \times K(\mathbb{Z}, j) \times ... , \mathbb{Z})$.

Altogether, these cohomology operations determine a (multisorted) Lawvere theory. There is an obvious subtheory of this theory generated by zero, negation, addition, the cup product, and composition. Based on the table in this paper, it looks like the simplest integral cohomology operation not in this obvious subtheory is a cohomology operation $H^3(X, \mathbb{Z}) \to H^8(X, \mathbb{Z})$ coming from the generator of $H^8(K(\mathbb{Z}, 3), \mathbb{Z}) \cong \mathbb{Z}/3\mathbb{Z}$, but I don't know enough algebraic topology to extract an explicit description of this cohomology operation from the paper.

So: what is this cohomology operation? Where does it come from? What can you do with it? (I know there are some cohomology operations coming from the $\text{Tor}$ terms in the Künneth formula; is this one of them?)

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Suppose $\alpha$ is an integral $3$-cocycle on a space $X$. Then $2\alpha\cup\alpha$ is a coboundary, because the cup product is graded-commutative, and there exists a $5$-cochain $\beta$ such that $\mathrm d\beta=2\alpha\cup\alpha$. One can easily check that $\alpha\cup\beta+\beta\cup\alpha$ is an $8$-cocycle. I think that the class of this cocycle does not depend on the choice of $\beta$, so we get a well defined mapping $H^3\to H^8$. This map is natural because there is a natural proof that the cup product is graded-commutative, so we $\beta$ depends naturally on $\alpha$.

This gives an operation :-)

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    $\begingroup$ This is so obvious that it had to be done ages ago :-) Something very similar was studied by Kraines here He shows that his is zero over the rationals, and non-trivial over odd primes (it is then given by Bocksteins and reduced powers) He does not say anything about integralcoefficients, afaict. $\endgroup$ – Mariano Suárez-Álvarez Jun 17 '13 at 6:38
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    $\begingroup$ Oh, you're right. (What anti-commutativity actually implies is that the definition doesn't depend on choice of $\beta$.) $\endgroup$ – Grigory M Jun 17 '13 at 6:54
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    $\begingroup$ @user17786, if $x$, $y$, $z$ are classes such that $xy=0=yz$, then the product $xyz$ is zero for «two reasons». This observation is used to construct a secondary operation called the Massey triple product (this is explained more or less in Wikipedia) I started with this and tried to make it work better than usual in our case (triple products have an indeterminancy, but it disappears in out case magically) $\endgroup$ – Mariano Suárez-Álvarez Jun 17 '13 at 7:28
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    $\begingroup$ Interesting! Does this Massey product line of reasoning suggest a reason why this operation ought to land in $3$-torsion? $\endgroup$ – Qiaochu Yuan Jun 17 '13 at 7:32
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    $\begingroup$ This operation is presumably the same as $H^3(X,\mathbb{Z}) \to H^3(X,\mathbb{Z}/3) \xrightarrow{P^1} H^7(X,\mathbb{Z}/3) \to H^8(X,\mathbb{Z})$, where $P^1$ is the first Steenrod operation, and the first and last maps are from the coefficient sequence for $0\to \mathbb{Z}\to \mathbb{Z}\to \mathbb{Z}/3\to 0$. $\endgroup$ – Charles Rezk Jun 17 '13 at 13:50

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