28

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?)

Eric Wofsey
  • 342,377
Qiaochu Yuan
  • 468,795

1 Answers1

20

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 :-)

Mariano Suárez-Álvarez
  • 139,300
  • $\alpha\beta=-\beta\alpha$, so your operation seems to be zero, no? – Grigory M Jun 17 '13 at 06:22
  • That equality holds in cohomology, not on the level of (singular) cocycles. Notice that $\beta$ is not even a cocycle. – Mariano Suárez-Álvarez Jun 17 '13 at 06:25
  • 1
    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. – Mariano Suárez-Álvarez Jun 17 '13 at 06:38
  • 1
    Oh, you're right. (What anti-commutativity actually implies is that the definition doesn't depend on choice of $\beta$.) – Grigory M Jun 17 '13 at 06:54
  • (I'm upvoting this, but the question why this operation is nontrivial still remains...) – Grigory M Jun 17 '13 at 07:16
  • @mariano how does this definition occur to you? – user17786 Jun 17 '13 at 07:18
  • 4
    @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) – Mariano Suárez-Álvarez Jun 17 '13 at 07:28
  • 1
    Interesting! Does this Massey product line of reasoning suggest a reason why this operation ought to land in $3$-torsion? – Qiaochu Yuan Jun 17 '13 at 07:32
  • If this is non-zero, the example should be on $K(Z,3)$, mapping the identity mapping in $H^3(K(Z,3))$ to a non-zero element of $H^8(K(Z,3))$. This should work already on sufficiently good approximations to the E-M space, like $(S^3)^n/\Sigma_n$ (with $n=3$, probably)... – Mariano Suárez-Álvarez Jun 17 '13 at 08:24
  • 10
    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 $0\to \mathbb{Z}\to \mathbb{Z}\to \mathbb{Z}/3\to 0$. – Charles Rezk Jun 17 '13 at 13:50
  • 1
    @QiaochuYuan I suspect it's something like Jacobi identity. (Notice that the operation is $P(\alpha)=[\alpha,\beta]$ and $d\beta=[\alpha,\alpha]$, where $[-,-]$ is the supercommutator...) – Grigory M Jun 17 '13 at 15:46
  • @GrigoryM, triple products are indeed closely related to the Jacobi identity; this is explained in McCleary's book spectral sequences, iirc. – Mariano Suárez-Álvarez Jun 17 '13 at 16:05
  • @CharlesRezk, that's the sort of thing Kraines gets for Massey powers $\langle a,\dots,a\rangle$, in fact. – Mariano Suárez-Álvarez Jun 17 '13 at 16:09