In the paper Products of Cocycles and Extensions of Mappings, Steenrod introduced the cup-i product (and Steenrod square). I would like to ask if Steenrod's cup-i product associative or not? The paper did not discuss this property (it seems).
1 Answer
The cup-i products are not associative for $i > 0$.
For example, Steenrod's cup-1 product has the following description (taken mod 2 for expedience). For cocycles $f$ of degree $p$ and $g$ of degree $q$, and a simplex of degree $p+q-1$ with vertices $[v_0, v_1, \dots, v_{p+q-1}]$, the cup-1 product is defined by $$ (f \cup_1 g) [v_0 \dots v_{p+q-1}] = \sum_{i=0}^{p-1} f[v_0, \dots, v_i, v_{q+i}, v_{p+q-1}] \cdot g[v_i, \dots, v_{q+i}]. $$ In other words, you sum up over all ways to apply $g$ to a "middle" portion of the simplex.
If $f$ and $g$ have degree 2 and $h$ has degree 1, we find $$ \begin{align*} ((f \cup_1 g) \cup_1 h) [v_0, v_1, v_2, v_3] =& (f[v_0, v_2, v_3] g[v_0, v_1, v_2] + f[v_0, v_1, v_3] g[v_1, v_2, v_3]) \\&\cdot (h[v_0,v_1]+ h[v_1, v_2] + h[v_2, v_3]) \\ (f \cup_1 (g \cup_1 h)) [v_0, v_1, v_2, v_3] =& f[v_0, v_2, v_3] g[v_0, v_1, v_2] \cdot (h[v_0, v_1] + h[v_1, v_2]) \\&+ f[v_0, v_1, v_3] g[v_1, v_2, v_3] \cdot (h[v_1, v_2] + h[v_2, v_3]) \end{align*} $$ and the difference between the two is $$ f[v_0, v_2, v_3] g[v_0, v_1, v_2] h[v_2, v_3] + f[v_0, v_1, v_3] g[v_1, v_2, v_3] h[v_0, v_1]. $$
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$\begingroup$ Thank you very much. I wonder are there any theories to discuss those associators. $\endgroup$ Apr 26, 2017 at 22:41
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4$\begingroup$ @Xiao-GangWen Yes! See here: arxiv.org/abs/math/0106024. In section 2 they discuss a collection of operations that make up the "sequence operad". The cup-0 product is $\langle 12\rangle$, cup-1 is $\langle 121\rangle$, cup-2 is $\langle 1212\rangle$, etc. They give rules for how these interact with the boundary operators and how to compose them. $\endgroup$ Apr 26, 2017 at 23:15
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$\begingroup$ For instance, in their notation: $(f \cup_1 g) \cup_1 h = \langle 121\rangle (\langle 121\rangle(f,g), \langle 1 \rangle(h))$, which expands to $\langle 12131\rangle(f,g,h) + \langle 12321\rangle(f,g,h) + \langle 13121\rangle(f,g,h)$. On the other hand, $f \cup_1 (g \cup_1 h) = \langle 121\rangle (\langle 1 \rangle(f), \langle 121\rangle(g,h))$, which expands to $\langle 12321\rangle(f,g,h)$. $\endgroup$ Apr 26, 2017 at 23:15