In Steenrod's ``[Products of Cocycles and Extensions of Mappings (1947)](https://www.jstor.org/stable/1969172?seq=1#metadata_info_tab_contents),'' 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!


* Ref: [N. E. Steenrod, Products of cocycles and extensions of mappings, Annals of Mathematics 48 290–320 (1947)](https://www.jstor.org/stable/1969172?seq=1#metadata_info_tab_contents)



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


