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7
votes
A cochain-level Adem relation?
The following article might be relevant. This is the review I wrote for it a number of years ago.
MR1622630 (99h:55027) Reviewed
Real, Pedro(E-SEVLIS-AM1)
On the computability of the Steenrod square …
6
votes
Accepted
Detailed exposition of construction of Steenrod squares from Haynes Miller's book
Hatcher's book also proceeds by first showing that there is the power operation $P(x)$ as you say, and he proves all the properties.
Hatcher makes quite a bit of use of the fact that cohomology is rep …
4
votes
What is an unstable dual-Steenrod comodule?
Your question seems to be equivalent to asking for a description of the unstable condition in terms of the Milnor basis. This is easy to do: Given $r = (r_1,\dots, r_s)$, let $P(r)$ be dual to $\xi_1 …
8
votes
Triviality of Steenrod operation on $\Sigma^{2k}\mathbb{CP}^n$
Recall that $H^*(\mathbb CP^{\infty};\mathbb Z/2)= \mathbb Z/2[y]$ with $|y|=2$. Let $F_k \subset \mathbb Z/2[y]$ be the span of $y^m$ such that $\alpha(m)\leq k$. It is well known, and easy to chec …
9
votes
Accepted
How does the Steenrod algebra act on $\mathrm{H}^\bullet(p^{1+2}_+, \mathbb{F}_p)$?
If $P$ is the group of order $p^3$ and exponent $p$, its mod $p$ cohomology ring is known to have its depth = its Krull dimension = rank of a maximal elementary abelian subgroup = 2. A theorem of Jon …