Questions tagged [steenrod-algebra]
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63 questions
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Geometric interpretation of the conjugation operation in the dual Steenrod algebra
As the dual mod 2 Steenrod algebra, $A$, is a Hopf algebra, it has the conjugation operation, $\chi:A\to A$. Milnor also gives a formula for this.
I wonder if there is any source telling about a ...
- 61
3
votes
1
answer
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cohomology ring of infinite iterated loop space
What is the cohomology ring
$$
H^*(\Omega^\infty \Sigma^\infty (S^m\vee S^n);\mathbb{Z}_2)?
$$
I already write out the graded-vector-space basis using Dyer-Lashof operations, but I do not know how to ...
- 2,223
10
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2
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cup product and Steenrod operations in Serre spectral sequence
Let $F\to E\to B$ be a fibration with $B$ simply-connected. Suppose all differentials in the cohomology Serre spectral sequence (corresponding to the above fibration) are zero maps. Then as a graded ...
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9
votes
1
answer
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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 ...
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3
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0
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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 $...
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6
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3
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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 ...
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7
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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.
...
- 1,147
18
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0
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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 ...
- 35.9k
5
votes
2
answers
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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
$D_0 : C_\ast\...
- 1,147
12
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4
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$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 $Sq^1(a\...
- 35.9k
11
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Why does one consider the dual of the Steenrod algebra?
Why does one consider the dual of the Steenrod algebra?
- 417
4
votes
2
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604
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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; ...
- 33.8k
76
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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 ...
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