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5
votes
0answers
87 views

Steenrod operations on cohomology of grassmannians

Let $G_k(\mathbb{R}^n)$, $n\geq k$ and $G_k(\mathbb{R}^\infty)$ be the finite-dimensional and infinite-dimensional grassmannians respectively. Their cohomology rings are expressed in terms of ...
0
votes
0answers
68 views

Dyer-Lashof algebra structures over graded modules

In Lecture Notes in Mathematics, Vol. 533, The homology of iterated loop spaces, Chapter 3, The homology of $C_{n+1}$-spaces, F. Cohen, Section 2, page 222, line 4, 5, 6: for an arbitrary graded ...
3
votes
0answers
149 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
531 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
559 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
446 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
738 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
607 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
845 views

Why does one consider the dual of the Steenrod algebra?

Why does one consider the dual of the Steenrod algebra?
4
votes
2answers
389 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; ...
35
votes
9answers
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 ...