All Questions
Tagged with at.algebraic-topology steenrod-algebra
49 questions
6
votes
1
answer
348
views
Detailed exposition of construction of Steenrod squares from Haynes Miller's book
$\DeclareMathOperator\Sq{Sq}$Chapter 75 of Haynes Miller's book on algebraic topology contains a beautiful construction of the Steenrod squares $\Sq^i$.
Roughly speaking, it goes as follows. All ...
- 63
8
votes
2
answers
602
views
Derivations in the Steenrod algebra
Let $\mathcal A^\ast$ be the (mod 2) Steenrod algebra.
Question 1:
Is there a classification of homogeneous elements $D \in \mathcal A^n$ such that $D^2 = 0$?
Question 2: Is there a classification of ...
- 64k
7
votes
1
answer
179
views
Generalisation of Hirsch formula for the associativity of Steenrod's higher $\cup_2$ product with $\cup_1$ and cup products
For $f$, $g$ and $h$ cochains, the Hirsch formula is given as
$$ (f\cup g)\cup_1 h=f\cup (g\cup_1 h)+(-1)^{q(r-1)}(f\cup_1 h)\cup g.$$
Is there a more general formula that relates the associativity of ...
- 71
3
votes
1
answer
260
views
Cohomology ring $H^*(\operatorname{SL}(3,\mathbb{Z}),\mathbb{Z}_2)_{(2)}$
$\DeclareMathOperator\SL{SL}$In Soulé's paper "The cohomology of $\SL_3(\mathbb{Z})$" the cohomology ring $H^*(\SL(3,\mathbb{Z}),\mathbb{Z})_{(2)}$ is determined in Theorem 4.iv. I'm wanting ...
- 545
4
votes
2
answers
409
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 ...
10
votes
1
answer
418
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 ...
- 3,001
4
votes
0
answers
120
views
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 ...
- 3,001
8
votes
1
answer
374
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 ...
- 3,001
10
votes
2
answers
350
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 ...
- 64k
4
votes
0
answers
356
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 ...
- 241
12
votes
2
answers
558
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 \...
- 5,504
8
votes
1
answer
195
views
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^...
- 83
5
votes
1
answer
439
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 ...
- 53
1
vote
1
answer
230
views
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 ...
- 5,230
6
votes
0
answers
210
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 $z_n \in H_n(MO; \mathbb F_2)$...
user131113
7
votes
2
answers
494
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 ...
- 54.6k
3
votes
0
answers
345
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(...
user131113
5
votes
1
answer
386
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(...
- 64k
5
votes
1
answer
387
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 ...
- 1,623
5
votes
0
answers
185
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 ...
- 1,623
4
votes
1
answer
167
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 ...
- 171
5
votes
0
answers
225
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
$$
Sq^{d-j}(x_j)=u_{d-j} x_j, \text{ for any } x_j \in H^j(M^d;\mathbb Z_2) .
\tag{eq.1}$$
...
- 10.5k
12
votes
1
answer
782
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 ...
- 858
5
votes
1
answer
263
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)...
- 10.5k
3
votes
1
answer
691
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 $$\mathbb{Z}...
- 10.5k
20
votes
1
answer
874
views
Odd primary dual Steenrod algebra
My question is related to
this, this, and this older questions.
Let $\mathcal A_*$ be the dual Steenrod algebra.
This is a super-commutative Hopf algebra, and so its $Spec$ is an algebraic super-group....
- 43.2k
19
votes
1
answer
1k
views
Steenrod squares as power operations vs. as cohomology operations
There are already several excellent questions and answers on MO regarding Steenrod squares, understanding them in various ways and relating them to power operations and I think I get this. Still, I am ...
- 2,320
8
votes
1
answer
363
views
Adams spectral sequence and short exact sequences. Some clarifications
as the title suggests I'm looking for some clarifications in the computations of the ext charts of some $A(1)$-modules arising as extensions of other modules. In particular, I've the following example ...
- 503
12
votes
2
answers
597
views
Steenrod powers of Pontryagin classes
It is well known that the Stiefel–Whitney classes $w_i$ of a smooth manifold are generated, over the Steenrod algebra, by those of the form $w_{2^{i}}$. I wonder if it the same statement is known/true ...
- 1,452
8
votes
2
answers
724
views
Adem relations of Steenrod square without modding out the coboundaries
In the paper Products of Cocycles and Extensions of Mappings,
Steenrod introduced the cup-$i$ product and Steenrod square $Sq^k$:
$$
Sq^k(x_n) \equiv x_n \smile_{n-k} x_n,\ \ \ x_n \in C^n(M^d;\...
- 4,826
18
votes
2
answers
617
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: ...
- 54.6k
9
votes
0
answers
239
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:
$$D_{2n} = \langle x,y ...
- 2,018
6
votes
2
answers
570
views
An integral cohomology operation related to Steenrod square?
Let $\beta: H^n(X, \mathbb{Z}_2)\to H^{n+1}(X, \mathbb{Z})$ be the Bockstein homomorphism. Is it possible to define a cohomology operation $f: H^{n+1}(X, \mathbb{Z})\to H^{n+k+1}(X, \mathbb{Z})$ such ...
- 383
17
votes
2
answers
771
views
Massey products in the Steenrod algebra
When building $kU/2$ via its Postnikov tower, there are some interesting Massey products that show up in the Steenrod algebra, and I'd like to understand them. I bet these appear somewhere in the ...
- 335
25
votes
2
answers
2k
views
Steenrod operations in etale cohomology?
For $X$ a topological space, from the short exact sequence
$$ 0 \rightarrow \mathbb{Z}/2 \rightarrow \mathbb{Z}/4 \rightarrow \mathbb{Z}/2 \rightarrow 0 $$
we get a Bockstein homomorphism
$$H^i(X,...
- 2,809
7
votes
0
answers
424
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 ...
- 1,990
13
votes
0
answers
561
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 ...
- 16.2k
7
votes
0
answers
430
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 ...
- 928
6
votes
1
answer
559
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 ...
- 61
3
votes
1
answer
180
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 ...
- 2,223
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 ...
- 2,223
9
votes
1
answer
487
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 ...
- 2,223
3
votes
0
answers
180
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 $...
- 7,972
7
votes
1
answer
1k
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.
...
- 1,147
18
votes
0
answers
760
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 ...
- 35.9k
5
votes
2
answers
2k
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 : C_\ast\...
- 1,147
12
votes
4
answers
1k
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 $Sq^1(a\...
- 35.9k
11
votes
3
answers
2k
views
Why does one consider the dual of the Steenrod algebra?
Why does one consider the dual of the Steenrod algebra?
- 417
76
votes
9
answers
15k
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 ...
- 6,069