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12
votes
Why does one consider the dual of the Steenrod algebra?
Hal Sadofsky's answer is great, and you should accept it.
If you want a more obscure point of view, one which is purely algebraic, then (at least at the prime 2), the dual of the Steenrod algebra rep …
4
votes
Accepted
Wall's presentation of the Steenrod algebra
Question 1: "Closed forms" for the relations are known, due to Grant Walker in what seems to be unpublished work. The relations are described in Wood's paper "Problems in the Steenrod algebra," (PDF) …
9
votes
$Sq^1$ cohomology of spaces
Several people have addressed question 1 (Torsten Ekedahl and Neil Strickland). Question 2 is interesting, but I don't have a good answer for it. For question 3, as Sean Tilson points out, this is a …
7
votes
Over which (graded) rings are all modules decomposable into indecomposables?
In the book Spectra and the Steenrod Algebra, Margolis proves the following (Theorem 21 in chapter 11): if $A$ is a graded connected algebra over a finite field and if $M$ is an $A$-module which is fi …
4
votes
Derivations in the Steenrod algebra
I have a guess for question 1. Fix $n \geq 0$ and let $E(n)$ be the Hopf subalgebra dual to $\mathbb{F}_2 [\xi_{n+1}, \xi_{n+2}, \dots] / (\xi_i^{2^{n+1}})$. Every $x\in E(n)$ satisfies $x^2=0$, and m …
4
votes
Accepted
Adams spectral sequence and short exact sequences. Some clarifications
The red dot in (3,0) comes from a map $\Sigma^3 D \to \mathbb{F}_2$, and this map is the image of a map $\mathbb{R}P^\infty \to \mathbb{F}_2$, so it goes to zero under the coboundary map. This agrees …