Questions tagged [steenrod-algebra]

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Triviality of Steenrod operation on $\Sigma^{2k}H(r,s)$

I want to prove that any Steenrod operation for $k>0$ $$ \phi: H^{2^j}(\Sigma^{2k}H(r,s)) \rightarrow H^{2^l}(\Sigma^{2k}H(r,s))$$ is zero. Recall that $H(r,s)$ is the Milnor manifold which can be ...
Devendra Singh Rana's user avatar
4 votes
2 answers
359 views

Triviality of Steenrod operation on $\Sigma^{2k}\mathbb{CP}^n$

I was going through this paper by Tanaka. I am actually stuck at Lemma 5.2, page 365, given below also The argument he gives above works, in particular for $\operatorname{Sq}^{2^r-2^j}$ but I am not ...
Devendra Singh Rana's user avatar
10 votes
1 answer
364 views

Are all degree-1 cohomology operations Bocksteins?

I'm interested in cohomology operations (in ordinary cohomology) $$H^i(-, G)\rightarrow H^{i+1}(-, H)\;,$$ that is, elements of $$H^{i+1}(K(G, i), H)\;.$$ I know that $K(G, 1)=BG$, so for $i=1$, those ...
Andi Bauer's user avatar
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Is there a local simplicial formula for the Steenrod squares which commutes with the derivative on cochain level?

There is a well-known formula for the cup product of an $i$-cochain $A$ and $j$-cochain $B$ in simplicial homology given by $$(A\cup B)(0\ldots i+j) = A(0\ldots i) B(0\ldots j)\;.$$ This formula ...
Andi Bauer's user avatar
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8 votes
1 answer
329 views

Analogue of Bockstein for crossed module extensions and higher Steenrod square

It is well known that in $\mathbb{Z}_2$-valued simplicial cohomology (and other cohomologies) $$ Sq^1 = \beta\;,$$ where $Sq^1$ is the first Steenrod square and $\beta$ is the Bockstein homomorphism ...
Andi Bauer's user avatar
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9 votes
2 answers
284 views

What is an unstable dual-Steenrod comodule?

$\newcommand\Sq{\mathit{Sq}}$Recall that a (graded) module $V^\ast$ over the Steenrod algebra $\mathcal A^\ast$ is said to be unstable if $\Sq^i v = 0$ for $i > |v|$. The motivating example, of ...
Tim Campion's user avatar
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9 votes
2 answers
327 views

Over which (graded) rings are all modules decomposable into indecomposables?

A module is decomposable if it is the direct sum of two modules. The process of splitting summands off of a decomposable module does not need to terminate, so infinitely generated modules do not ...
Tilman's user avatar
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4 votes
0 answers
214 views

Eilenberg-Maclane spectrum and $E_{\infty}$-algebra structure on singular cochain complex

I'm trying to understand how the $E_{\infty}$-algebra structure on the singular cochain complex $C^{\bullet}(X)$ of a topological space $X$, in at least somewhat down-to-earth terms. (I'm coming at ...
Aaron Wild's user avatar
12 votes
2 answers
506 views

Modules over the integral dual Steenrod algebra as linear functors

Let $\text{Latt}$ denote the category of lattices, i.e., finitely generated free abelian groups. In the appendix to Lecture 4 of Condensed.pdf, Scholze considers functors $F \colon \text{Latt} \to \...
jmc's user avatar
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7 votes
1 answer
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Why does $\iota_4^2 \in H^8(K(\mathbb Z/2,4);\mathbb Z/2)$ not come from $H^8(K(\mathbb Z/2,4);\mathbb Z)$?

In Hatcher's Chapter 5 (https://pi.math.cornell.edu/~hatcher/AT/ATch5.pdf) on page 574 (page 57 in the pdf), he states that $\iota_4^2 \in H^8(K(\mathbb Z/2,4);\mathbb Z/2)$ is not in the image of $H^...
Dolly Wu's user avatar
4 votes
1 answer
144 views

Wall's presentation of the Steenrod algebra

In the paper "Generators and Relations for the Steenrod Algebra" (C. T. C. Wall, Annals of Mathematics, Second Series, Vol. 72, No. 3 (Nov., 1960), pp. 429-444) Wall shows that there is a ...
Pedro's user avatar
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5 votes
1 answer
383 views

Examples of non-zero negative Steenrod operations

In JP May's paper A general algebraic approach to Steenrod operations, Steenrod operations are constructed in wide generality. In this context, it is not necessarily true that negative Steenrod ...
V. Pofek's user avatar
9 votes
0 answers
205 views

Two $\mathbb Z$-algebra structures on $\mathbb Z\otimes_{\mathbb S} R$

$\newcommand{\Sph}{\mathbb S} \newcommand{\Z}{\mathbb Z} \newcommand{\F}{\mathbb F}$ In this question I will abuse notation by writing $A$ for the (generalized) Eilenberg-MacLane spectrum associated ...
Maxime Ramzi's user avatar
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1 vote
1 answer
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Steenrod operations from the delooping viewpoint

Let $F$ be a finite field, $Sq^i$ be the $i$-th Steenrod operation $$ H^*(-;F) \to H^{*+i}(-;F).$$ By Yoneda lemma, such operation is a map $\phi_i: B^{*}F \to B^{*+i} F$, where $B$ denotes the ...
Student's user avatar
  • 4,560
6 votes
1 answer
417 views

Action of Steenrod algebra on Chern classes

This is question about result from Brown and Peterson $H^*(MO)$ as an algebra over the Steenrod algebra. Unfortunately, the paper is not available on the Internet, so I can't find the proof. One of ...
user avatar
6 votes
0 answers
206 views

Generators for unitary bordism ring $\pi_*(MU)$

I’m reading Pengelley’s paper “The mod 2 homology of $MSO$ and $MSU$ as $\mathfrak A^*$ comodule algebras, and the cobordism ring”. He has chosen very special generators $z_n \in H_n(MO; \mathbb F_2)$...
user avatar
7 votes
2 answers
412 views

How does the Steenrod algebra act on $\mathrm{H}^\bullet(p^{1+2}_+, \mathbb{F}_p)$?

Let $p$ be an odd prime. The $\mathbb F_p$ cohomology of the cyclic group of order $p$ is well-known: $\mathrm{H}^\bullet(C_p, \mathbb F_p) = \mathbb F_p[\xi,x]$ where $\xi$ has degree 1, $x$ has ...
Theo Johnson-Freyd's user avatar
3 votes
0 answers
310 views

mod $p$ homology of Thom spectra MSU

Using pairing in Atiyah-Hirzebruch spectral sequence one can show that homology of $BU(n)$ is a free abelian group with basis $\alpha_{k_1}\cdots\alpha_{k_t}$, $k\leqslant n$, where $\alpha_{i} = \big(...
user avatar
5 votes
1 answer
342 views

Why does the Steenrod algebra act faithfully on $H^\ast(BC_p)$?

Define the Steenrod algebra $A^\ast$ to be the algebra of all stable mod $p$ cohomology operations. Without actually computing $A^\ast$, is it possible to see that $A^\ast$ acts faithfully on $H^\ast(...
Tim Campion's user avatar
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2 votes
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Computation of mod p homology of $MSU$

I am trying to proof Novikov theorem \begin{equation} MSU_*\otimes \mathbb Z[\frac 1 2] \cong \mathbb Z[\frac 1 2][y_2, y_4, \ldots],\quad \deg y_i = 2i. \end{equation} This can be proved by using ...
user avatar
2 votes
0 answers
111 views

Stable homology operations

Let $x\in (H\mathbb F_2)_n(X)=[S^n,H\mathbb F_2 \wedge X]$ be a homology class for a space $X$. Is there a description of $$[S^n\overset x\to H\mathbb F_2 \wedge X\overset{Sq^r\wedge id}\to \Sigma^r H\...
syzyg's user avatar
  • 21
4 votes
1 answer
188 views

specific modules over the Steenrod algebra with one generator

I'd be happy to clarify the following. Consider the module which is a quotient of the Steenrod algebra mod $2$ by the left ideal generated by $\operatorname{Sq}^1, \operatorname{Sq}^2, \operatorname{...
Dr.Martens's user avatar
5 votes
1 answer
308 views

Ádem relations for the Steenrod and the Dyer–Lashof algebra

In this paper by Nondas Kechagias, the Steenrod algebra and the Dyer–Lashof algebra are compared. The rough difference ist: The Steenrod algebra arises by dividing out the “cohomological” Ádem ...
FKranhold's user avatar
  • 1,613
4 votes
0 answers
140 views

Dual Steenrod squares

Fix the ground ring $\mathbb{F}_2$ and let $X$ be a space with finite homology. Then we have an isomorphism $\Phi^i_X:H_i(X)\to H^i(X)^*,a\mapsto \langle-,a\rangle$ which allows us to define the dual ...
FKranhold's user avatar
  • 1,613
4 votes
1 answer
160 views

Steenrod algebra: Ádem relations from Milnor product formula

The question is how to deduce the Ádem relations from the Milnor product formula. Straightforward approach leads to certain relation on binomial coefficients mod p. Could anyone tell me if there is a ...
Dr.Martens's user avatar
5 votes
0 answers
205 views

Generalizing the formula between Wu class and the Steenrod square

I know that on the tangent bundle of $M^d$, the corresponding Wu class and the Steenrod square satisfy $$ Sq^{d-j}(x_j)=u_{d-j} x_j, \text{ for any } x_j \in H^j(M^d;\mathbb Z_2) . \tag{eq.1}$$ ...
wonderich's user avatar
  • 10.1k
11 votes
1 answer
677 views

Power operations from a Tate construction

In an action-packed three pages of Lurie's DAG-XIII: Rational and p-adic Homotopy Theory, section 2.2: Power Operations on $\mathbb{E}_{\infty}$-algebras, one finds a construction of the power ...
pupshaw's user avatar
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7 votes
1 answer
293 views

Why is $\pi_{-*}F(H\mathbb{F}_p, H\mathbb{F}_p)$ the mod $p$ Steenrod algebra?

Why is $\pi_{-*}F(H\mathbb{F}_p, H\mathbb{F}_p)$ the mod $p$ Steenrod algebra? (This is quite a common statement, seen, for instance, in EKMM.) To be more precise, stable mod $p$ cohomology ...
user09127's user avatar
  • 755
5 votes
1 answer
208 views

Pontryagin square, Postnikov square and their consistency formulas

$\mathcal{P}_2$ is Pontryagin square $$H^{2i}(M,\mathbb Z_{2^k})\to H^{4i}(M,\mathbb{Z}_{2^{k+1}}).$$ $\mathfrak{P}$ is the Postnikov square $$H^2(M,\mathbb Z_3)\to H^5(M,\mathbb Z_9).$$ question (i)...
wonderich's user avatar
  • 10.1k
3 votes
1 answer
593 views

Bockstein homomorphism and Square Operations: Their consistency formulas

Here are various ways to define "Bockstein homomorphism:" Let $\beta_p:H^*(-,\mathbb{Z}_p) \to H^{*+1}(-,\mathbb{Z}_p)$ be the Bockstein homomorphism associated to the extension $$\mathbb{Z}...
wonderich's user avatar
  • 10.1k
20 votes
1 answer
829 views

Odd primary dual Steenrod algebra

My question is related to this, this, and this older questions. Let $\mathcal A_*$ be the dual Steenrod algebra. This is a super-commutative Hopf algebra, and so its $Spec$ is an algebraic super-group....
André Henriques's user avatar
11 votes
0 answers
195 views

What are examples of spectra whose mod 2 cohomology contain A//A(n)?

Let $//$ denote the Hopf algebra quotient. We know that: $$HF_{2}^*(ko) \simeq A//A(1)$$ $$HF_2^*(tmf) \simeq A//A(2)$$ By Hopf invariant one, we know there is no $X$ such that $HF_2^*(X) \simeq A//...
Catherine Ray's user avatar
18 votes
1 answer
835 views

Steenrod squares as power operations vs. as cohomomology operations

There are already several excellent questions and answers on MO regarding Steenrod squares, understanding them in various ways and relating them to power operations and I think I get this. Still, I am ...
KotelKanim's user avatar
  • 2,240
8 votes
1 answer
323 views

Adams spectral sequence and short exact sequences. Some clarifications

as the title suggests I'm looking for some clarifications in the computations of the ext charts of some $A(1)$-modules arising as extensions of other modules. In particular, I've the following example ...
Luigi M's user avatar
  • 503
12 votes
2 answers
552 views

Steenrod powers of Pontryagin classes

It is well known that the Stiefel–Whitney classes $w_i$ of a smooth manifold are generated, over the Steenrod algebra, by those of the form $w_{2^{i}}$. I wonder if it the same statement is known/true ...
CuriousUser's user avatar
  • 1,374
8 votes
2 answers
646 views

Adem relations of Steenrod square without modding out the coboundaries

In the paper Products of Cocycles and Extensions of Mappings, Steenrod introduced the cup-$i$ product and Steenrod square $Sq^k$: $$ Sq^k(x_n) \equiv x_n \smile_{n-k} x_n,\ \ \ x_n \in C^n(M^d;\...
Xiao-Gang Wen's user avatar
7 votes
1 answer
798 views

Associativity of Steenrod's cup-i product

In the paper Products of Cocycles and Extensions of Mappings, Steenrod introduced the cup-i product (and Steenrod square). I would like to ask if Steenrod's cup-i product associative or not? The paper ...
Xiao-Gang Wen's user avatar
3 votes
1 answer
172 views

Cartan Formula for Steenrod square on cocycles

Let $x_n,y_n,\cdots$ be cocycles in $Z^n(X,\mathbb{Z}_2)$ (not cohomology classes in $H^n(X,\mathbb{Z}_2)$). Let $Sq^k(x)\equiv x_n \cup_{n-k} x_n$ be the Steenrod square (This definition is valid for ...
Xiao-Gang Wen's user avatar
18 votes
2 answers
573 views

A cochain-level Adem relation?

The original paper on Steenrod squares, Steenrod's "Products of cocycles and extensions of mappings", 1947, uses an explicit combinatorial formula for the squares in terms of simplicial cochains: ...
Theo Johnson-Freyd's user avatar
9 votes
0 answers
217 views

The Steenrod Algebra of the Dihedral Group $D_{2n}$, $n=0 \pmod{4}$

As the tile suggests, I'm interested in computing the action of the Steenrod Algebra on $H^*(D_{2n};\mathbb{Z}_2)$, for $n=0 \pmod{4}$. Let us start with some definitions/facts: $$D_{2n} = \langle x,y ...
Riccardo's user avatar
  • 1,878
6 votes
2 answers
498 views

An integral cohomology operation related to Steenrod square?

Let $\beta: H^n(X, \mathbb{Z}_2)\to H^{n+1}(X, \mathbb{Z})$ be the Bockstein homomorphism. Is it possible to define a cohomology operation $f: H^{n+1}(X, \mathbb{Z})\to H^{n+k+1}(X, \mathbb{Z})$ such ...
No_way's user avatar
  • 373
17 votes
2 answers
719 views

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 ...
Bogdan's user avatar
  • 335
25 votes
2 answers
2k 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 $$H^i(X,...
user84144's user avatar
  • 2,709
7 votes
0 answers
365 views

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 ...
Shiquan Ren's user avatar
  • 1,950
13 votes
0 answers
505 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 ...
Piotr Achinger's user avatar
6 votes
0 answers
380 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 ...
Daniel Grady's user avatar
6 votes
1 answer
503 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 ...
DavidT's user avatar
  • 61
3 votes
1 answer
169 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 ...
QSR's user avatar
  • 2,203
10 votes
2 answers
1k 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 ...
QSR's user avatar
  • 2,203
9 votes
1 answer
446 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 ...
QSR's user avatar
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