# Use of Steenrod's higher cup product and the graded-commutativity

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

• 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. – user51223 Mar 23 '19 at 19:51
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.