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2
votes
0answers
132 views

Twisting of the power functor

Let $k$ be a field of characteristic $p$ and $D^b(k)$ be the infinity (equivalently, DG) category of perfect complexes over $k$. Let $C_p(=\mathbb{Z}/p)$ be the cyclic group on $p$ elements. For a ...
6
votes
3answers
499 views

Steenrod operations in algebraic geometry

What are some applications of Steenrod operations (or similar constructions) in algebraic geometry? I am dimly aware of the the use of these Voevodsky's work on motivic cohomology, and would be ...
3
votes
1answer
517 views

Do people still use Massey Products for computations in the Adams Spectral Sequence

Hey everyone, It seems to me like in the literature of the Adams Spectral Sequence, older publications (Toda, May, Tengora+Mahowald) make heavy and explicit use of Massey Products for computations. ...
13
votes
0answers
419 views

Steenrod algebra at a prime power

Let $n=p^k$ be a prime power. When $k=1$, the algebra of stable operations in mod $p$ cohomology is the Steenrod algebra $\mathcal{A}_p$. It has a nice description in terms of generators and ...
4
votes
2answers
703 views

Why are cup-i products and Steenrod Squares often (always?) unary?

One way to define the Steenrod Operations is to use the cup-i product, as in Mosher and Tangora's book. It basically says, given the chain complex from mod-2 homology $C_\ast$, define $D_0 : ...
8
votes
4answers
584 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 ...
3
votes
2answers
801 views

Why does one consider the dual of the Steenrod algebra?

Why does one consider the dual of the Steenrod algebra?
4
votes
2answers
379 views

Adem-Wu relations from Bullett-Macdonald identities

Question. Let $p$ be a prime. Let $q$ be a power of $p$. Let $P^0$, $P^1$, $P^2$, ... be elements of some associative $\mathbb F_q$-algebra $A$. (Here, $P^i$ does not mean $P$ to the $i$-th power; ...
31
votes
8answers
4k 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 ...