9
$\begingroup$

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!

$\endgroup$
4
  • $\begingroup$ See also mathoverflow.net/q/268181/27004 Associativity of Steenrod's cup-i product $\endgroup$
    – wonderich
    Mar 23, 2019 at 16:39
  • $\begingroup$ mathoverflow.net/q/270302/27004 Adem relations of Steenrod square without modding out the coboundaries $\endgroup$
    – wonderich
    Mar 23, 2019 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 $\endgroup$ Mar 23, 2019 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, 2019 at 19:51

1 Answer 1

2
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ thank you +1, appreciate it. $\endgroup$ Apr 3, 2019 at 18:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.