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In the paper Products of Cocycles and Extensions of Mappings, Steenrod introduced the cup-$i$ product and Steenrod square $Sq^k$: $$ Sq^k(x_n) \equiv x_n \smile_{n-k} x_n,\ \ \ x_n \in C^n(M^d;\mathbb{Z}_2) $$ Note that the above definition works for any $\mathbb{Z}_2$-valued cochain $x_n$.

My question is that if the Adem relations $$ Sq^i Sq^j = \sum_{k=0}^{[i/2]} {j-k-1 \choose i-2k} Sq^{i+j-k} Sq^k $$ are still valid when acting on cochains (or cocycles) without modding out the coboundaries?

(All the references that I can find state Adem relations up to the coboundaries.)

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    $\begingroup$ I don't know, but I'd find it extremely doubtful. The Adem relations have their origin in the cohomology of symmetric groups, why should they hold at the chain level? $\endgroup$ – Denis Nardin May 20 '17 at 22:20
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    $\begingroup$ It feels like, if they did, there wouldn't be higher order cohomology operations? I don't know how to turn that intuition into a proof. $\endgroup$ – Dylan Wilson May 20 '17 at 22:39
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    $\begingroup$ This question seems relevant: mathoverflow.net/questions/261467/a-cochain-level-adem-relation $\endgroup$ – Denis Nardin May 21 '17 at 0:12
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    $\begingroup$ I dared to add that link to the Mathematics Genealogy Project because somebody out there suggested that the OP had not meant Adem but Adams... $\endgroup$ – José Hdz. Stgo. May 21 '17 at 2:48
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Theorem 3.1 in Peter May's "A General Algebraic Approach to Steenrod Operations" gives a better behaved definition for $Sq^k(x)$ for (co)chains $x$ which are not assumed to be cocycles. It probably makes sense to think about these questions (Adem relations or Cartan formula on the cocycle level) using them rather than just the cup-(n-k) formula. I find it highly implausible that either of these (Adem relations or Cartan formula) should hold 'on the nose', no matter what definition extending the usual one on (co)cycles one gives.

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I will make use of the surjection operad to produce an explicit example of a (functorial) cochain whose boundary is the Adem-Relation. Let us show that $Sq^1Sq^1([x])=0$ for a cohomology class $x$ in degree 1. Thus we have on cochains $$(x\cup x)\cup_1(x\cup x)=\langle 121\rangle (\langle 12\rangle(x^{\otimes 2}))=(\langle 13412\rangle +\langle 12342\rangle)(x^{\otimes 4}).$$ The boundary of $\langle 123412\rangle\in Hom ((C^*)^{\otimes 4},C^*)$ is $$\langle 23412 \rangle +\langle 13412 \rangle+\langle 12342 \rangle+\langle 12341 \rangle.$$ If we plug in the cycle $x^{\otimes 4}$, we can use that relabeling the numbers $1,..,4$ corresponds to swapping the indices and thus does not affect the result. Thus we have $$\langle 12341\rangle(x^{\otimes 4})=\langle 23412\rangle(x^{\otimes 4}).$$ So if we plug in $x^{\otimes 4}$, two summands cancel out and the boundary is just the Adem-relation on cochain level.

Now it remains to find a simplicial set $X$ and $1$-cocycle $c$ such that $\langle 123412\rangle(c^{\otimes 4})\neq 0$, i.e. there is a two simplex on which the cochain $\langle 123412\rangle(c^{\otimes 4})$ does not vanish.

This happens for the $1$-cochain on the $2$-simplex $012$ given by $c(01)=1,c(12)=1,c(02)=0$.

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