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76 votes
9 answers
15k views

understanding Steenrod squares

There is a function on $\mathbb{Z}/2\mathbb{Z}$-cohomology called Steenrod squaring: $Sq^i:H^k(X,\mathbb{Z}/2\mathbb{Z}) \to H^{k+i}(X,\mathbb{Z}/2\mathbb{Z})$. (Coefficient group suppressed from ...
Aaron Mazel-Gee's user avatar
18 votes
2 answers
617 views

A cochain-level Adem relation?

The original paper on Steenrod squares, Steenrod's "Products of cocycles and extensions of mappings", 1947, uses an explicit combinatorial formula for the squares in terms of simplicial cochains: ...
Theo Johnson-Freyd's user avatar
12 votes
4 answers
1k views

$Sq^1$ 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
  • 35.9k
8 votes
2 answers
723 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
3 votes
1 answer
691 views

Bockstein homomorphism and Square Operations: Their consistency formulas

Here are various ways to define "Bockstein homomorphism:" Let $\beta_p:H^*(-,\mathbb{Z}_p) \to H^{*+1}(-,\mathbb{Z}_p)$ be the Bockstein homomorphism associated to the extension $$\mathbb{Z}...
wonderich's user avatar
  • 10.5k