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12 votes
4 answers
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$Sq^1$ cohomology of spaces

For any space $X$, the first Steenrod square cohomology operation $$Sq^1\colon H^\ast(X;\mathbb{Z}_2)\to H^{\ast +1}(X;\mathbb{Z}_2)$$ is a derivation, meaning that $Sq^1\circ Sq^1 = 0$ and $Sq^1(a\...
Mark Grant's user avatar
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12 votes
2 answers
558 views

Modules over the integral dual Steenrod algebra as linear functors

Let $\text{Latt}$ denote the category of lattices, i.e., finitely generated free abelian groups. In the appendix to Lecture 4 of Condensed.pdf, Scholze considers functors $F \colon \text{Latt} \to \...
jmc's user avatar
  • 5,504
8 votes
2 answers
724 views

Adem relations of Steenrod square without modding out the coboundaries

In the paper Products of Cocycles and Extensions of Mappings, Steenrod introduced the cup-$i$ product and Steenrod square $Sq^k$: $$ Sq^k(x_n) \equiv x_n \smile_{n-k} x_n,\ \ \ x_n \in C^n(M^d;\...
Xiao-Gang Wen's user avatar
5 votes
1 answer
263 views

Pontryagin square, Postnikov square and their consistency formulas

$\mathcal{P}_2$ is Pontryagin square $$H^{2i}(M,\mathbb Z_{2^k})\to H^{4i}(M,\mathbb{Z}_{2^{k+1}}).$$ $\mathfrak{P}$ is the Postnikov square $$H^2(M,\mathbb Z_3)\to H^5(M,\mathbb Z_9).$$ question (i)...
wonderich's user avatar
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