The steenrod-algebra tag has no usage guidance.

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### Massey products in the Steenrod algebra

When building $kU/2$ via its Postnikov tower, there are some interesting Massey products that show up in the Steenrod algebra, and I'd like to understand them. I bet these appear somewhere in the ...

**23**

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790 views

### Steenrod operations in etale cohomology?

For $X$ a topological space, from the short exact sequence
$$ 0 \rightarrow \mathbb{Z}/2 \rightarrow \mathbb{Z}/4 \rightarrow \mathbb{Z}/2 \rightarrow 0 $$
we get a Bockstein homomorphism
...

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### kernel of the mod $2$ Bockstein on the first cohomology group

Let $M$ be a path-connected finite $CW$-complex. Suppose the first integral homology group is $H_1(M;\mathbb{Z})= \mathbb{Z}_2^{\oplus r}\oplus A$ where $r\geq 1$ and $A$ is a finite abelian group of ...

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201 views

### When does an $E_\infty$ algebra come from a commutative differential graded algebra?

Suppose that $K$ is an $E_\infty$-algebra on a space $X$ (more generally, any ringed topos; also, feel free to assume that $X$ is a point). That is, $K$ is a cochain complex of sheaves on $X$, endowed ...

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### algebraic structure of Integral Steenrod squares

It is well known that the classical Steenrod squares $Sq^a$ satisfy the Adem relations
$$Sq^aSq^b= \sum_c \binom{b-c-1}{a-2c}Sq^{a+b-c}Sq^c\;.$$
In the case where $a$ is odd, one can define an ...

**3**

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**1**answer

201 views

### 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 ...

**3**

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**1**answer

131 views

### 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 ...

**6**

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**2**answers

321 views

### 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 ...

**7**

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**1**answer

208 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 ...

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162 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 ...

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586 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 ...

**5**

votes

**1**answer

622 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.
...

**15**

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484 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 ...

**5**

votes

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840 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 : ...

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votes

**4**answers

684 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 ...

**6**

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1k views

### Why does one consider the dual of the Steenrod algebra?

Why does one consider the dual of the Steenrod algebra?

**4**

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405 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; ...

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5k 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 ...