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0
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0answers
62 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 ...
32
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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 ...
2
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0answers
135 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
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3answers
501 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 ...
4
votes
2answers
705 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 : ...
3
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1answer
525 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
425 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 ...
8
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4answers
593 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
806 views
4
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2answers
382 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; ...