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4 questions
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\...
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 $F \colon \text{Latt} \to \...
8
votes
2
answers
724
views
Adem relations of Steenrod square without modding out the coboundaries
In the paper Products of Cocycles and Extensions of Mappings,
Steenrod introduced the cup-$i$ product and Steenrod square $Sq^k$:
$$
Sq^k(x_n) \equiv x_n \smile_{n-k} x_n,\ \ \ x_n \in C^n(M^d;\...
5
votes
1
answer
263
views
Pontryagin square, Postnikov square and their consistency formulas
$\mathcal{P}_2$ is Pontryagin square
$$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)...