Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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 $Sq^1(a\cup b) = Sq^1(a)\cup b + a\cup Sq^1(b)$ (there are no signs since we are working in characteristic two).

Hence we may form the $Sq^1$-cohomology of the space, $$H\left(H^\ast(X;\mathbb{Z}_2),Sq^1\right)$$ which will be a graded algebra over $\mathbb{Z}_2$.

I am looking for references on this object. From McCleary's "User's guide to spectral sequences", I know that this is related to the Bockstein spectral sequence. More specifically, I would like to know:

  1. What is the precise relationship between the $Sq^1$-cohomology of a space $X$ and $2$-torsion of higher order in $H^\ast(X;\mathbb{Z})$?
  2. Is there a reference with specific calculations of the $Sq^1$-cohomology of the Eilenberg-Mac Lane spaces $K(\mathbb{Z}_2,n)$?
  3. Are there any canonical references I should know about (besides McCleary and Mosher-Tangora)?
share|improve this question
2  
i think this is also related to Margolis homology. you might want to look at that. The issues there are more related to classification of modules over subalgebras of the steenrod algebra though. –  Sean Tilson Aug 28 '11 at 21:31

4 Answers 4

up vote 7 down vote accepted

I think the easiest way to understand the Bockstein spectral sequence is through the exact couple coming from the long exact sequence of cohomology associated to $0\to\mathbb Z\to\mathbb Z\to \mathbb Z/2\to0$. This shows first that indeed the first differential is $Sq^1$ and tells you that the next page is the direct sum of the cokernel and kernel (shifted one step) of multiplication by $2$ on $2H^\ast(X,\mathbb Z)$. Hence it is like what you would get from applying the universal coefficient formula to $2H^\ast(X,\mathbb Z)$ (instead of $H^\ast(X,\mathbb Z)$). When each cohomology group $H^\ast(X,\mathbb Z)$ is finitely generated this means concretely that you "keep" each $\mathbb Z$-factor (as well as odd torsion) and downgrade each $\mathbb Z/2^n$ to $\mathbb Z/2^{n-1}$.

In particular the difference between the dimension of $H^n(X,\mathbb Z/2)$ and that of the $Sq^1$-cohomology is equal to the number of $\mathbb Z/2$-factors in $H^n(X,\mathbb Z)$ and $H^{n+1}(X,\mathbb Z)$.

I found a reference to Q2. In Madsen, Milgram: The classifying spaces for surgery and cobordism of manifolds, Ann of Math Studies 92 where they refer to Browder: Torsion in H-spaces, Ann of Math 74 for the Bockstein s.s. of $K(\mathbb Z_{(2)},n)$ and $K(\mathbb Z/2,n)$. The Madsen-Milgram book also contains other examples of computations with the Bss.

share|improve this answer
    
Hi Torsten. This was helpful, thanks. There are some words missing from the last sentence of your first paragraph, perhaps "finitely generated"? –  Mark Grant Aug 29 '11 at 8:13
    
Thanks again. The Browder paper looks like the reference I was seeking. –  Mark Grant Aug 29 '11 at 15:34
    
This really does answer all 3 of my questions! For anyone interested, the Bockstein SS for $K(Z_2,n)$ is given a fairly explicit description in Theorem 5.5 of the Browder paper. –  Mark Grant Aug 30 '11 at 9:24

I remember that when I wrote my thesis I was unable to find references for some quite basic facts about this that everyone knew. It is quite possible that there were references that I did not succeed in finding (there was no Google then!) or that someone has written a good exposition in the intervening time. For what it's worth, my thesis is at http://neil-strickland.staff.shef.ac.uk/research/thesis.pdf and the relevant material is in Section 5.1.

share|improve this answer

Bill Singer and I have looked at the "dual" question... Given a right $\mathcal{A}$-module $M$ (such as the homology of a space), $Sq^1$ acts (on the right) as a differential. For any $s \geq 1$ and $M = \tilde{H}_*((\mathbb{R}P^{\infty})^{\wedge s}, \mathbb{Z}/2)$, $Sq^1$-homology is trivial, indicating that the kernel of $Sq^1$ is the same as the image of $Sq^1$ there. There are interesting things to say even for higher squares, though they won't be differentials in general.

The following references may be useful (both available on arXiv):

Ault, Singer. On the Homology of Elementary Abelian Groups as Modules over the Steenrod Algebra.

Ault. Relations among the kernels and images of Steenrod squares acting on right $\mathcal{A}$-modules.

share|improve this answer

Several people have addressed question 1 (Torsten Ekedahl and Neil Strickland). Question 2 is interesting, but I don't have a good answer for it. For question 3, as Sean Tilson points out, this is a special case of "Margolis homology", a.k.a. $P^s_t$-homology. Try

  • Adams and Margolis, "Modules over the Steenrod algebra", Topology 10 (1971)
  • Anderson and Davis, "A vanishing theorem in homological algebra", Comment. Math. Helv. 48 (1973)
  • Margolis, Spectra and the Steenrod algebra (1983)

I also wonder if there is anything helpful in

  • Adams and Priddy, "Uniqueness of BSO".

You might also search for the phrase "Bockstein acyclic", since $\textrm{Sq}^1$ is the mod 2 Bockstein.

share|improve this answer
    
Thanks for the references! –  Mark Grant Aug 29 '11 at 8:57

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.