$Sq^1$ cohomology of spaces - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T22:32:46Z http://mathoverflow.net/feeds/question/73889 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/73889/sq1-cohomology-of-spaces $Sq^1$ cohomology of spaces Mark Grant 2011-08-28T09:16:13Z 2011-08-29T17:49:36Z <p>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). </p> <p>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 $\mathbb{Z}_2$.</p> <p>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:</p> <blockquote> <ol> <li>What is the precise relationship between the $Sq^1$-cohomology of a space $X$ and $2$-torsion of higher order in $H^\ast(X;\mathbb{Z})$?</li> <li>Is there a reference with specific calculations of the $Sq^1$-cohomology of the Eilenberg-Mac Lane spaces $K(\mathbb{Z}_2,n)$?</li> <li>Are there any canonical references I should know about (besides McCleary and Mosher-Tangora)?</li> </ol> </blockquote> http://mathoverflow.net/questions/73889/sq1-cohomology-of-spaces/73890#73890 Answer by Neil Strickland for $Sq^1$ cohomology of spaces Neil Strickland 2011-08-28T09:41:39Z 2011-08-28T09:41:39Z <p>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 <a href="http://neil-strickland.staff.shef.ac.uk/research/thesis.pdf" rel="nofollow">http://neil-strickland.staff.shef.ac.uk/research/thesis.pdf</a> and the relevant material is in Section 5.1.</p> http://mathoverflow.net/questions/73889/sq1-cohomology-of-spaces/73891#73891 Answer by Torsten Ekedahl for $Sq^1$ cohomology of spaces Torsten Ekedahl 2011-08-28T09:43:43Z 2011-08-29T17:49:36Z <p>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 $\mathbb Z/2^{n-1}$.</p> <p>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 $H^{n+1}(X,\mathbb Z)$.</p> <p>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.</p> http://mathoverflow.net/questions/73889/sq1-cohomology-of-spaces/73915#73915 Answer by Shaun Ault for $Sq^1$ cohomology of spaces Shaun Ault 2011-08-28T21:31:16Z 2011-08-28T21:31:16Z <p>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.</p> <p>The following references may be useful (both available on arXiv):</p> <p>Ault, Singer. On the Homology of Elementary Abelian Groups as Modules over the Steenrod Algebra.</p> <p>Ault. Relations among the kernels and images of Steenrod squares acting on right $\mathcal{A}$-modules.</p> http://mathoverflow.net/questions/73889/sq1-cohomology-of-spaces/73930#73930 Answer by John Palmieri for $Sq^1$ cohomology of spaces John Palmieri 2011-08-29T00:31:15Z 2011-08-29T00:31:15Z <p>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</p> <ul> <li>Adams and Margolis, "Modules over the Steenrod algebra", Topology 10 (1971)</li> <li>Anderson and Davis, "A vanishing theorem in homological algebra", Comment. Math. Helv. 48 (1973)</li> <li>Margolis, <em>Spectra and the Steenrod algebra</em> (1983)</li> </ul> <p>I also wonder if there is anything helpful in </p> <ul> <li>Adams and Priddy, "Uniqueness of BSO".</li> </ul> <p>You might also search for the phrase "Bockstein acyclic", since $\textrm{Sq}^1$ is the mod 2 Bockstein.</p>