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In Steenrod's ``Products of Cocycles and Extensions of Mappings (1947),'' which derives [Theorem 5.1]

$$ \delta(u\cup_{i} v)=(-1)^{p+q-i}u\cup_{i-1}v+(-1)^{pq+p+q}v\cup_{i-1}u+\delta u\cup_{i}v+(-1)^pu\cup_{i}\delta v $$ where $u\in C^p$, $v\in C^q$ are cochains.

Take $u\in C^2$, $v\in C^3$.

Suppose $u=\delta w\in B^2 \subset C^2$ is a coboundary and $v\in C^3$ is a cochain. We have the following result the cup-1 result:

$$ \delta(\delta w\cup_{1} v)=+(\delta w)\cup v+(-1)v\cup (\delta w)+\delta (\delta w)\cup_{1}v+(\delta w)\cup_{1}\delta v $$ This means that $$ \delta(\delta w\cup_{1} v)=+(\delta w)\cup v+(-1)v\cup (\delta w)+ (\delta w)\cup_{1}\delta v $$

In other words, whether the commutativity of $(\delta w)\cup v-v\cup (\delta w)$ boils down to $$ (\delta w)\cup v-v\cup (\delta w)=- (\delta w)\cup_{1}\delta v +\delta(\delta w\cup_{1} v) \tag{1} $$

My questions:

(I) Whether a 2-coboundary $u=(\delta w)$ and a 3-cohain $v$ are commutative in the normal cup product up to a cobounary term $\delta(...)$? What is the property of graded-commutativity (up to (-1) power of combinations of dimensions) of $(\delta w)\cup v-v\cup (\delta w)$?

(II) Namely, is the right hand side in eq (1), this term $(\delta w)\cup_{1}\delta v$ is also a coboundary? If so $$ (\delta w)\cup_{1}\delta v =\delta \alpha? $$ What is $\alpha$?

See a related issue:

"Higher cup-1 product of coboundaries is also a coboundary?" https://math.stackexchange.com/q/3159473/141334

Thank you!!! <3 Please help/comment/advice/give Refs!

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  • $\begingroup$ See also mathoverflow.net/q/268181/27004 Associativity of Steenrod's cup-i product $\endgroup$ – wonderich Mar 23 at 16:39
  • $\begingroup$ mathoverflow.net/q/270302/27004 Adem relations of Steenrod square without modding out the coboundaries $\endgroup$ – wonderich Mar 23 at 16:40
  • $\begingroup$ thank you! I also ask a related question: "Higher cup-1 product of coboundaries is also a coboundary?" math.stackexchange.com/q/3159473/141334 $\endgroup$ – annie heart Mar 23 at 16:45
  • $\begingroup$ The construction of operas has its origin in a 1967 paper of Peter May where without any mention of the phrase “operad” the machinery is defined using a categorification of Steenrod operations! So, if I were to find an answer then I would look into operads bearing in mind that the square i operation $Sq^i$ is defined using cup $i$ operations on the cochain level. $\endgroup$ – user51223 Mar 23 at 19:51
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For $(\delta w) \cup v - v \cup (\delta w)$ to be a coboundary, it would need to also be a cocycle: so we would have to have $$ 0 = \delta((\delta w) \cup v - v \cup (\delta w)) = (\delta w) \cup (\delta v) - (\delta v) \cup (\delta w) $$

Let $X$ be the standard 6-simplex $[v_0,v_1,\dots,v_6]$. Define the following cochains on $X$: $$ \begin{align*} v(\sigma) &= \begin{cases} 1 &\text{if }\sigma = [v_3,v_4,v_5,v_6]\\ 0 &\text{otherwise} \end{cases} \\ w(\sigma) &= \begin{cases} 1 &\text{if }\sigma = [v_0,v_1]\\ 0 &\text{otherwise} \end{cases} \end{align*} $$ Then we have $$ \begin{align*} [(\delta w) \cup (\delta v) - (\delta v) \cup (\delta w)]([v_0,v_1,\dots,v_6]) =\ & (\delta w)([v_0,v_1,v_2]) \cdot (\delta v)([v_2,v_3,v_4,v_5,v_6]) \\&- (\delta v)([v_0,v_1,v_2,v_3,v_4]) \cdot (\delta w)([v_4,v_5,v_6]) \\=\ & w(\partial[v_0,v_1,v_2]) \cdot v(-\partial[v_2,v_3,v_4,v_5,v_6]) \\&- v(-\partial[v_0,v_1,v_2,v_3,v_4]) \cdot w(\partial[v_4,v_5,v_6]) \\=\ & \dots \\=\ & -1 \end{align*} $$ Therefore, $(\delta w) \cup v - v \cup (\delta w)$ is not a cocycle.

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  • $\begingroup$ thank you +1, appreciate it. $\endgroup$ – annie heart Apr 3 at 18:11

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