The steenrod-algebra tag has no usage guidance.

**2**

votes

**0**answers

176 views

### 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**

votes

**1**answer

193 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**

votes

**1**answer

128 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 ...

**5**

votes

**2**answers

282 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 ...

**6**

votes

**1**answer

180 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 ...

**3**

votes

**0**answers

161 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

**3**answers

568 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

594 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

**0**answers

471 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

**2**answers

808 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

**4**answers

661 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

**2**answers

906 views

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

Why does one consider the dual of the Steenrod algebra?

**4**

votes

**2**answers

402 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; ...

**41**

votes

**9**answers

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 ...